TITLE: Constructing an injective reduction of equivalence relations
QUESTION [9 upvotes]: [Metastuff: I asked this question in a slightly different way on mathSE last week, and it didn't go anywhere, which is why I am asking here. I added the DST tag because it's basically a problem about Borel equivalence relations stripped of all the Borelness constraints. I do need help, so helpful redirection is appreciated.]
I am trying to give a somewhat constructive definition of a function. It's somewhat constructive because I'll freely assume that I can well-order any set. Aside from that, I want to say what the function looks like.
I have two equivalence relations $E$ and $F$ on spaces $X$ and $Y$, respectively. There are no restrictions on the sizes of anything. I want to define a function $f : X \to Y$ such that
$$ x E y \Leftrightarrow f(x) F f(y)\;\;\;\text{ and }\;\;\;f(x) = f(y) \Rightarrow x = y $$
for all $x,y \in X$. This makes $f$ send all points in an $E$-class to the same $F$-class and also be injective on equivalence classes (i.e., injective as $X/E \to Y/F$) and on the underlying space.
Let $I$ be the class of nonzero cardinals. For every $i \in I$, the number of $F$-classes of size at least $i$ is greater than or equal to the number of $E$-classes of size at least $i$. I want to give a mostly-constructive proof that this is sufficient for there to be a function as described above (from $E$ to $F$), i.e., I want to describe the function.
I have been struggling with this on and off for several weeks. Below are some possible time-savers for you guys. If you already have a solution, you can skip it.

The problem is extremely easy in the slightly nicer situation where, for every $i \in I$, the number of $F$-classes of size exactly $i$ is greater than or equal to the number of $E$-classes of size exactly $i$. Just partition the set of $E$-classes by size and put a well-order on each set in the partition. Do the same for $F$-classes. Then send the $n$th $E$-class of size $i$ to the $n$th $F$-class of size $i$.
The complication for the original case is that you might have to send an $E$-class of size $i$ to an $F$-class of size $j$ with $i < j$. Two problems arise this way.
First, you can't use the larger classes wastefully by sending relatively small classes to them. E.g., if $E$ has solely two classes, one of size $2$ and one of size $5$, and $F$ has solely two classes, one of size $4$ and one of size $6$, you cannot send the class of size $2$ to the class of size $6$. The only way that I can think to avoid this problem is inductively: (i) well-order the classes in some way, (ii) send the least $E$-class to the least $F$-class that is big enough, (iii) remove these, and (iv) repeat from step (ii).
This creates the second problem: how to choose the well-order for step (i). If you try, e.g., to order the classes by increasing size with an arbitrary order among classes of the same size, you run into the following problem (as Brian Scott pointed out to me on mathSE a week ago). Suppose $E$ has $\omega$ many classes of size $1$ and one class of size $2$. Suppose $F$ has one class each of every finite size. Then the above won't work because $F$ has order-type $\omega$, but $E$ has order-type $\omega+1$.
You can fix this case with the same trick that you use to well-order the rationals. Put the $E$-classes of size $i$ into a column and well-order each column. Then move along the diagonals like so. But it's not clear to me what this looks like when you have any number of columns and rows rather than just countably many. 

Edit to explain potential solution: It sounds plausible to me that sending an $E$-class to an $F$-class of the smallest available size that is large enough will avoid fatally wasteful assignments regardless of the order in which you make assignments. E.g., given an $E$-class of size $5$, if $F$-classes of sizes $4,7,$ and $9$ are available, choose one of size $7$.
The problem then is just how to iterate through the $E$-classes. This sounds problematic generally, but my knowledge of ordinals is weak. E.g., is there always some sense in which you can iterate through all the members of an initial ordinal?
Put the $E$-classes into an array like this one so that an $E$-class has an index (column,row). Let the index $(s,p)$ mean that $s$ is the size of the $E$-class and $p$ is its arbitrarily-assigned position in the column. 
Consider the case where you have at most countably many $E$-classes of each size and only countably many possible infinite sizes. That is, $p \in \omega$ and $s \in \omega \times \lbrace 0,1\rbrace$. That is, you have countably many finite sizes (tagged with $0$) and countably many infinite sizes (tagged with $1$). Then you just have two copies of the above array; one for finite sizes and one for infinite sizes. 
Separately snake through each of them in the way depicted in the linked picture. For the array of finite sizes, this hits indices in this order: $((1,0),1), ((1,0),2), ((2,0),1), ((1,0),3), \ldots$, where the $0$ indicates that you're in the "finite" array. For the array of infinite sizes, this hits the indices in this (analogous) order: $((1,1),1), ((1,1),2), ((2,1),1), ((1,1),3), \ldots$. Here, $(1,1)$ denotes some infinite size such as $\omega$; $(2,1)$ might be $2^\omega$, and so on.
Finally, interleave the two orders that you got from snaking through each array. Most simply, you can take one member from each order at a time. This gives:
$$((1,0),1), ((1,1),1), ((1,0),2), ((1,1),2), ((2,0),1), ((2,1),1), ((1,0),3), \ldots$$
I apologize for the somewhat cumbersome notation, but I hope that the pattern becomes clear.
Incidentally, you won't necessarily have an $E$-class for all of the points in the above arrays. E.g., you might not have any $E$-classes of size $2$. I am assuming the fullest possible case for simplicity, as it still defines a well-order when you remove some of the points.

REPLY [4 votes]: This is a very nice problem, which I like very much.
The answer is that yes, indeed, there is such an injective reduction of $E$ to $F$. And one can give a recursive construction. (This answer now incorporates several simplifications to my original construction.)
Specifically, let $\delta_\kappa^E$ be the number of $E$ equivalence classes of
size at least $\kappa$, and similarly $\delta_\kappa^F$ for $F$.
Your assumption is that $\delta_\kappa^E\leq\delta_\kappa^F$ for any cardinal $\kappa$. 
As $\kappa$ increases, this number is non-increasing, and since it can drop only finitely many times, because there is no infinite descending sequence of ordinals, it follows that there are only finitely many values for $\delta_\kappa^E$. 
Let $\kappa_1$ be the least cardinal such that $E$ has no class of size $\kappa_1$ or larger, and so $\delta_{\kappa_1}^E=0$. Below this, there is some minimal $\kappa_0\lt\kappa_1$ where $\delta_\kappa=\delta$ has constant nonzero value $\delta$ for all $\kappa\in[\kappa_0,\kappa_1)$. 
Consider the largest classes of $E$, those of size at least $\kappa_0$. We may enumerate them in a sequence of length $\delta$. We shall map them to corresponding $F$ classes of equal or larger size in a recursive procedure. Specifically, at stage
$\alpha<\delta$, consider the $\alpha^{\rm th}$ class of $E$ of
size at least $\kappa_0$; it has some size $\kappa$; we've used up
only $|\alpha|$ many $F$ classes of size at least $\kappa$; since this is less than $\delta$, we still have $F$ classes of size at least $\kappa$ remaining, and so we may map the $\alpha^{\rm th}$ class to any unused $F$ class of equal or larger size (and there is no need to be optimal or to minimize size here). 
Thus, we are able to map all the $E$ classes of size at least $\kappa_0$ to distinct corresponding $F$ classes of equal or larger size. Consider the $E$ and $F$ classes that remain, a smaller instance of the problem. Notice that this smaller instance of the problem still satisfies the size hypothesis, using the minimality of $\kappa_0$, since for $\kappa<\kappa_0$ we have $\delta^E_\kappa$ is strictly larger than $\delta$, and hence also $\delta^F_\kappa$ is that large. Since we used up exactly $\delta$ many $F$ classes of size at least $\kappa_0$, there are still sufficient $F$ classes of size at least each $\kappa\lt\kappa_0$.  
(To illustrate with the example from your question, where you had infinitely many $E$ classes of size $1$ and only one of size $2$, my algorithm proceeds here by mapping the size $2$ class first, and then realizing that the hypothesis is still true for what remains.)
Furthermore, this smaller instance of the problem now has a strictly smaller version of $\kappa_0$, and it can therefore be handled by induction. So the proof is complete.
To understand the reduction constructively, without induction, one should imagine it working like this. The reduction breaks into finitely many pieces. The first piece consists of the largest classes, on a maximal interval of cardinals $\kappa$ where $\delta_\kappa^E$ is constant value $\delta$, whether finite or infinite. Since there are $\delta$ classes here, we enumerate them in a $\delta$ sequence, and so we never run out of classes on the $F$ side during the course of this $\delta$ process. Then, we move to the preceeding maximal interval of cardinals $\kappa$ on which $\delta_\kappa^E$ has a new strictly larger constant value. Our previous part of the reduction does not interfere with this part, precisely because the new $\delta$ value is now strictly larger than on the first piece, and so there are plenty of $F$ classes of the desired size. And so on for finitely many steps, thereby completing the entire reduction.<|endoftext|>
TITLE: What is the 'non-intuitive' part in sphere eversion (turning inside out)?
QUESTION [9 upvotes]: The question does not mean sphere eversion is intuitive to me! In fact, it is just the opposite and that is the purpose of this question. 
Recently, I was reading about Smale's paradox, the problem of sphere eversion (turning a sphere inside out). The wiki article is quite clear and gave me a good overview of the topic. I happened to see an animation of the eversion process as well. 
The problem of sphere eversion is to construct a homotopy between the inside and outside of a sphere in a three dimensional space. During the continuous deformation self-intersections of the sphere are allowed and creating creases is not allowed. 
Given that we can self-intersect the sphere while the process of eversion what could be a possible obstruction to the eversion? What exactly do we mean by self-intersection? Moreover, I find it difficult to imagine why a similar process cannot be employed in the circle case? Why can't we self intersect a circle with itself to turn it inside out? Is there an easy explanation for this phenomenon? 
This topic is new to me. I hope the question is not too naive.  Thank you in advance.

REPLY [15 votes]: Watch Outside In (something we should all do anyway, to commemorate Bill Thurston's passing).
To understand the mathematics behind sphere eversions, you should first get a good intuition for the concepts of immersion and regular homotopy. I recommend Guillemin and Pollack's "Differential Topology" book for starters.
To see why it is not possible to turn a circle inside out, you should read up on the Whitney-Graustein Theorem. It basically boils down to the calculation $\pi_1(S^1)=\mathbb{Z}$, once you notice that the normalized differential of an immersion $S^1\looparrowright \mathbb{R}^2$ is a map $S^1\to S^1$.
You'll find a few more resources related to sphere eversions on my web page.<|endoftext|>
TITLE: Koszul duality for modular operads
QUESTION [7 upvotes]: Has anyone defined what it means for a modular operad to be Koszul, or what the Koszul dual of a modular operad is? In particular, is it meaningful to say that a modular operad is quadratic? Merkulov, Markl and Shadrin (Wheeled PROPs, graph complexes and the master equation) give a definition for wheeled properads, which are basically modular operads with oriented edges, which are constrained to have at most one output. 
Presumable the role of bar-cobar duality is then played by the Feynman transform and the dual Feynman transform.

REPLY [5 votes]: Here are two answers to this question based on different interpretations of the meaning of ``Koszul.''
One reasonable way to define things is that Koszul duality depends not only on algebraic data (operad, modular operad, algebra, whatever) but also on a presentation of the algebraic data. This is both elegant and a bit of a cheat. For example, in algebras, operads, or properads (let me just say operad for simplicity), take the "maximal" presentation of your operad. That is, take all operations as generators and then all relations of the form [(infinitesimal) composition of two generators] = [first generator] composed with [second generator]. Then the Koszul dual is precisely the bar construction. So you've totally diluted the meaning of Koszul duality by allowing every operad to have a "Koszul" presentation which doesn't tell you anything new. But I think if you're a homotopy theorist, this is a pretty reasonable definition.
On the other hand, maybe you want something more rigid or more amenable to calculation. So let's start over. One motivation for Koszul duality for an operad $P$ is that you have a canonical (large) resolution of $P$, the cobar bar resolution $\Omega B P\to P$ but you'd like something smaller if you can get it without losing the simplicity of the differential. Typically you hope that your resolution will be of the form $\Omega P^!$ for some cooperad $P^{!}$ (these exclamation points should be upside down but I can't figure out how). If $P^!$ has no differential of its own, this will be the unique-up-to-isomorphism minimal model for $P$. So one natural candidate is for $P^!$ to be the homology of $BP$. This inherits cooperadic structure from $BP$ and as such lives in the right category to apply the cobar functor $\Omega$. 
However, there are three problems. The first one, which is serious, is that we don't know that $BP$ is formal, i.e., quasi-isomorphic to its homology not just linearly but as a cooperad. Typically the homology will have to be given higher homotopy cooperations and then there's no hope for $\Omega H_*(BP)$ to be a resolution of $P$. But maybe you get lucky and you can construct this quasi-isomorphism. The second problem is computing this homology. In the classical definition of Koszul duality for operads, you don't take $H_*(BP)$ as a definition, you take one particular linear summand of $H_*(BP)$ which is easy to compute, based on a grading that comes from a quadratic presentation. Then you cross your fingers and hope this summand exhausts the homology. So say it does. The third problem, which is not such a big deal, is that $\Omega$ doesn't preserve all quasi-isomorphisms. In practice you can use the same grading on $BP$ that passes to its homology to ensure that this is one of the good quasi-isomorphisms preserved by $\Omega$.
So let's try to mimic this in modular operads. Define $P^!$ as the homology of the Feynman transform (in this paragraph the exclamation point should not be upside down because of the dual involved in the Feynman transform), and call P Koszul if $P^!$ and the Feynman transform of $P$ are weakly equivalent as modular operads. As you point out, it's not obvious what quadratic means in this context, so let's skip the second problem altogether. Then it should be an exercise to ensure that even though you haven't used the grading of a homogeneous presentation, in most (all?) cases of interest, the Feynman transform preserves quasi-isomorphisms so we get the desired behavior from $P^!$, that it generates the minimal model for $P$.<|endoftext|>
TITLE: Get Largest Inscribed Rectangle of a Concave Polygon
QUESTION [10 upvotes]: I'm looking for an algorithm to find a set of largest inscribed rectangles of a concave polygon where each rectangle must be collinear with one of the edges of the polygon. 
In other words, I want to continually cut my polygon down using rectangles until I am left with regions were the largest-inscribed rectangle is not sharing an edge with the original polygon.
Is there a known algorithm for this? 
An example:

This post is related to another post: Covering a Polygon with Rectangles
I am hoping to split up the questions so I can attract better answers.

REPLY [13 votes]: Let $P$ be your polygon, $e$ an edge of $P$.  I will interpret one aspect
of your question
as seeking all the maximal rectangles with one side flush with each $e$, 
each rectangle $R$ maximal in the sense that 
(a) $R \subset P$ but 
(b) there is no other rectangle $R' \subset P$ with $R' \supset R$.
If you can solve this for one edge $e$, you can iterate over all $n$ edges of $P$.
     
Fix $e$ to be horizontal (by rotating $P$).  Now your problem is to find the maximal
axis-aligned rectangles flush with $e$.
It is clear that $R$ is maximal only if all four sides of $R$ cannot be moved to enlarge
$R$.  The edge $e$ determines one side, so there are three to be blocked by $P$ from
enlargement.  A side of $R$ can be blocked by a vertex of $P$ lying in its interior,
or two sides can be blocked by their common corner resting on an edge of $P$. Various
possibilities are shown above.
A naive $O(n^3)$ algorithm can search through all possible blockers for the three sides,
yielding an $O(n^4)$ algorithm over all edges of $P$.
From among these, you could select, e.g., those with largest area, or largest perimeter, or whatever criterion is relevant to your application.
This naive algorithm can be considerably sped up.  A series of papers led to
an $O(n \log^2 n)$ algorithm, which I think might still be the best available.
Whether it would be worth implementing all the tricks necessary to reach this low time complexity
is another question.
 

K. Daniels, V. Milenkovic, and D. Roth. Finding the largest area axis-parallel
  rectangle in a polygon. Comput. Geom. Theory Appl., 7:125-148, 1997.
  (CiteSeer link)  Fig.1 shown above. 

To address how to "continually cut [your] polygon down" requires some criterion to chose among
all the options.  What I describe above is just a start,
identifying all the maximal rectangles.
I suspect any choice you make will lead to an NP-hard problem, and you'll be forced into
heuristics or approximation algorithms.
One heuristic is the greedy algorithm.  Find, say, the largest-area maximal rectangle $R$ in $P$.
Create several new polygons by subtracting $R$: $P \setminus R$.  Recurse on the pieces.
This would be easy to implement, and equally easy to thwart. :-)

Here is what I meant by thwarting the greedy algorithm:
     
Because no definition of what constitutes an optimal coverage by these rectangles has
been detailed, I cannot prove the greedy algorithm fails to achieve optimality.
But I am convinced that any reasonable definition of optimality will not always
be achieved by the greedy algorithm.<|endoftext|>
TITLE: When is a manifold a tangent bundle?
QUESTION [11 upvotes]: Given a (smooth) manifold $M$, are there any sufficient, intrinsic properties that would tell you there exists a (smooth) manifold $N$ such that $M$ is diffeomorphic to $TN?$  There are some obvious necessary conditions (like $M$ needs to be even-dimensional), but I'm more interested in whether any sufficient conditions are known.

REPLY [8 votes]: Suppose $M$ is an open $2n$-manifold that is homotopy equivalent to a closed smooth $n$-manifold $N$, and suppose $n>2$. Then Haefliger's embedding theorem ensures that the homotopy equivalence $N\to M$ is homotopic to a smooth embedding. Moreover, by Siebenmann's open collar recognition theorem $M$ is diffeomorphic to
the normal bundle to this embedding if and only if $M$ is the interior of a compact manifold
with boundary such that the inclusion of the boundary induces an isomorphism on the fundamental group. Now it remains to check whether the normal bundle and tangent bundle
to the embedding are isomorphic, which of course rarely happens.
A good example is when $N$ is an orientable $3$-manifold and $M=N\times \mathbb R^3$, which is precisely $TN$ because orientable $3$-manifolds are parallelizable. By above arguments, any two homotopy equivalent orientable $3$-manifolds have diffeomorphic tangent bundles.
Specific examples can be found among lens spaces, such as $L(7,1)$ and $L(7,2)$.
The case $n=2$ seems more delicate.

REPLY [4 votes]: One necessary set of conditions:
Step 1: you need $M$ to have the homotopy-type of a submanifold $N$ half the dimension of $M$.  This forces $M$ to be a vector bundle over $N$ with the fibre the same dimension as the base with some reasonable hypothesis (see texts on the h-cobordism theorem, like Kosinski, where these kinds of theorems are proven). 
Step 2: Given a vector bundle, how do you know if it is the tangent bundle of the base space?  For this you need an exponential map, or to show the vector bundle is diffeomorphic to a tubular neighbourhood of $\Delta N$ in $N \times N$.   Or you could compare the classifying map of $TN$ with that of your vector bundle structure on $M$.<|endoftext|>
TITLE: weakening naive comprehension to avoid the paradoxes
QUESTION [7 upvotes]: Weakening the axiom of naive comprehension has not been a popular way of escaping from the set-theoretic paradoxes because no consistent weakenings seem to be particularly well motivated or even to lead to understandable models.  At any rate, that is so of the most famous consistent (well, probably consistent) weakening, New Foundations.  Nonetheless, it could be illuminating to understand the partial order of consistent subtheories of naive set theory.  My question would be: what is known about it?  Unfortunately, however, that question seems ill-posed in that an arbitrary axiom $\psi$ can be coded as comprehension for $(\psi \land x\neq x)\lor (\lnot\psi \land x\notin x)$.  But can anything interesting be said?
Besides NF, I am aware of just one type of set theory that is (almost) naturally thought of as arrived at by admitting only a subset of all possible instances of naive comprehension.  These are the so-called positive set theories.  Unfortunately, I know nothing about them except that they admit a universal set (like NF), and they apparently do require some extra axioms that are not naturally expressed as instances of comprehension.  In particular, according to Wikipedia, the theory known as $GPK^{+}_{\infty}$ requires the axiom of infinity, the empty set axiom (!), and an axiom scheme of "closure" giving, for each formula $\phi$ with one free variable, the intersection of all the sets that contain every $x$ such that $\phi(x)$.  This seems to me arguably in the spirit of restricting naive comprehension because comprehension is still the main set construction principle, and in particular there is no need for powerset or replacement.  Are there other "natural" examples of set theories that can be thought of as arrived at by weakening naive comprehension?  Perhaps ones that don't admit a universal set?
Even non-effective examples (examples where the set of instances of comprehension that is admitted is not computably enumerable) might be interesting.  Also, there might be interesting ways of weakening naive comprehension that are different from simply restricting the allowed instances of the schema.  For instance, maybe some set of disjunctions of instances of naive comprehension is interesting, or maybe it is interesting to consider an axiom that would only guarantee the existence of a set that is in some sense "close" to the class of objects satisfying $\phi$.
The context in which this question came up is that I was trying to explain Russell's paradox to someone, and their reaction was, well you should just throw out the instance of the comprehension axiom that leads to paradox.  Of course, throwing out literally that one instance won't restore consistency.  But pointing out a flaw in any particular proposal someone with this attitude toward the paradoxes might propose wouldn't show that some more sophisticated proposal might succeed.  I was hoping for some sort of general argument that, say, proceeding in this way inevitably leads to a system that is either like NF or like positive set theory (if it is not inconsistent or extremely weak).  (What else could be wrong with $x\notin x$ except that it is unstratified or that it involves negation?)  Or at least an argument that you won't get an extension of ZF by any natural weakening procedure would be nice!  Both NF and positive set theory, if I understand the situation aright, could serve as a foundation for mathematics, but both are less intuitive and convenient than ZFC, and it is sort of an article of faith for set theorists that any alternative to ZFC we might ever find is either somehow worse than ZFC or not deeply different from it, yes?

REPLY [2 votes]: Ackermann set theory seems "pretty close" to having unrestricted comprehension, specifically in the form of the class and set comprehension schemas. Reinhardt proved that it is as strong as ZF (in particular, an additional replacement schema is not necessary). It's not entirely clear whether class comprehension is necessary.
I asked a question here about the strength of an even more minimalist fragment of Ackermann's set theory, which has only 1 or 2 axioms and a single comprehension schema. The form of the comprehension schema is, again, conceptually "quite close" to unrestricted comprehension.<|endoftext|>
TITLE: Expected distance between two points in the plane
QUESTION [8 upvotes]: Let $f(x)$ be a continuous probability distribution in the plane.  It is obvious that if $X$ and $X'$ are two independent random samples from $f$, then $\mathbf{E}(\|X - X'\|) \leq 2 \mathbf{E}(\|X\|)$ by the triangle inequality.  Can this upper bound be made tighter if we assume that $f$ is rotationally symmetric about the origin., i.e. $f(x) = g(\|x\|)$ for some function $g$?

REPLY [5 votes]: Using the pareto distribution $f(x) = \frac{\alpha}{x^{\alpha+1}}$ ($x > 1$) , the ratio $\frac{E(||X-X'||)}{E(||X||)}$ approaches a $2$ as $\alpha$ tends to 1. 
To find such a distribution, consider that all else equal, you want to maximize the difference $||X||-||X'||$ since the angle between the two is independent. This means that you want a distribution that extends to infinity as flatly as possible.

REPLY [4 votes]: The conditional expectation
$$\eqalign{E[ \|X - X'\| | \|X\| = r, \|X'\| = s] &= \frac{1}{2\pi} \int_0^{2\pi} \sqrt{r^2 + s^2 - 2 r s \cos \theta}\ d\theta \cr &=  \frac{2(r+s)}{\pi} EllipticE(2 \sqrt{rs}/(r+s))\cr}$$
where EllipticE is Maple's version of the complete elliptic integral of the second kind.
Note that $0 < 2 \sqrt{rs}/(r+s) \le 1$, with $1$ occurring for $r=s$.  On the interval $[0,1]$, $1 \le EllipticE(x) \le \pi/2$.  Thus 
$$ \frac{4}{\pi} E[\|X\|] = \frac{2}{\pi} E[\|X\|+\|X'\|] \le E[\|X - X'\|] \le E[\|X\|+\|X'\|] = 2 E[\|X\|] $$

REPLY [3 votes]: Under the assumption that $f$ is radial, the angle between $X$ and $X'$ is uniformly distributed in $[0,\pi]$ (since the signed angle that each of $X$ and $X'$ makes with the $x$-axis is uniformly distributed by rotational invariance, hence so is their difference). Thus the distribution of $\|X-X'\|$ is the same as the distribution of $\| (r,0)-s e^{i\alpha}\|$ where $\alpha$ is an angle chosen uniformly at random, and $r, s$ are  two independent samples of the distribution $g$ (on $[0,\infty)$). Its expectation thus equals (after a little calculation)
$$
\pi^{-1}\int_0^\pi \int \int \sqrt{r^2+s^2-2rs\cos(\alpha)} g(r) g(s) dr ds d\alpha,
$$
or
$$
\pi^{-1}\int_0^\pi \int \int \sqrt{(r+s)^2-2rs(1+\cos(\alpha))}  g(r) g(s) dr ds d\alpha.
$$
I suppose you could get some bound in terms of $\mathbb{E}(\|X\|)$ from here which is better than $2\mathbb{E}(\|X\|)$, but there is a much more pleasant expression for the expectation of $\mathbb{E}\|X-X'\|^2$: as before, this is
$$
\pi^{-1}\int_0^\pi \int \int r^2+s^2-2rs\cos(\alpha) g(r) g(s) dr ds d\alpha,
$$
which can be readily evaluated to $2\mathbb{E}(\|X\|^2)-4\mathbb{E}(\|X\|^2)/\pi$.<|endoftext|>
TITLE: Over which fields is the Sylvester law of inertia valid?
QUESTION [9 upvotes]: Short version: 

Over which fields is the (appropriate version of the) "Sylvester law of inertia" valid?

Long version:
Let $V$ be a finite dimensional vector space over the field $\Bbbk$ of characteristic different from $2$ (so quadratic forms are the same as symmetric bilinear forms). Consider the "discriminant" function 
$$\mathrm{discr}: \mathrm{Sym}^2(V^*)\to \Bbbk/\Bbbk^2,$$
where $\Bbbk/\Bbbk^2$ is the quotient of the multiplicative monoid $\Bbbk$ by the squares, which maps a (possibly degenerate) symmetric bilinear form $q$ to the determinant of it's matrix relative to any base, which is well defined up to multiplication by squares. Of course, this is not a complete invariant, as different degenerate quadratic forms all have discriminant $0$.
Given $q\in\mathrm{Sym}^2(V^*)$, and an orthogonal basis $B$ for $V$ in which $q$ has the form $\Sigma_i\lambda_ix_iy_i$, we have a "signature" map 
$$\sigma_q^B:\Bbbk/\Bbbk^2\to\mathbb{N}$$ 
sending a $\lambda$ in the domain to the number of vectors in $B$ for wich the corresponding $\lambda_i$ has image $\lambda$ in $\Bbbk/\Bbbk^2$.

1) Is $\sigma_q^B$ independent from $B$ ? 

-

2) If the answer to 1) is "yes", is $q\mapsto\sigma_q$ a complete invariant for quadratic forms over $\Bbbk$ ? That is, is it true that two quadratic spaces $(V,q)$ and $(V',q')$ are isomorphic if and only if $\sigma_q=\sigma_{q'}$ ?

For $\Bbbk=\mathbb{R}$ or $\mathbb{C}$ the answer to 1) and 2) is "yes": it's is the usual Sylvester law of inertia. Maybe the above general question has a well known or even elementary answer?...

REPLY [7 votes]: OK, I now have a complete answer; I'll delete the other shortly. The answer to question 2 is yes and easily so; I wonder if I am missing something. If one quadratic form is $\sum a_i x_i^2$, and other is $\sum b_i y_i^2$, and $a_i = b_i u_i^2$ for $u_i \neq 0$, then clearly the two quadratic forms are equivalent by the invertible change of variable $x_i = y_i u_i$. 

Now for the interesting part. Let $K$ be a field. The following are equivalent:
(1) The answer to Question 1 is "yes".
(2) For every $a \in K$ we have (Condition 2a) either $a$ or $-a$ is square and (Condition 2b) the sum of two squares is a square.
(3) One of the following two cases occurs: (Case 3a) Every element of $K$ is square or (Case 3b) $K$ is an ordered field in which every positive element is square.
$(1) \implies (2)$: 
For $a \neq 0$, note that
$$\begin{pmatrix} (a+1)/2 & (a-1)/2 \\ (a-1)/2 & (a+1)/2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}  \begin{pmatrix} (a+1)/2 & (a-1)/2 \\ (a-1)/2 & (a+1)/2 \end{pmatrix} = \begin{pmatrix} a & 0 \\ 0 & -a \end{pmatrix}$$
So the quadratic forms $x^2-y^2$ and $ax^2-ay^2$ are equivalent. We deduce that either $a$ or $-a$ is a square.
For any $a$ and $b$, not both zero, note that
$$\begin{pmatrix} a & b \\ -b & a \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}  \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{pmatrix}$$
So the forms $x^2+y^2$ and $(a^2+b^2) x^2 + (a^2+b^2) y^2$ are equivalent, and we deuce that $a^2+b^2$ is a square.
$(2) \implies (3)$: If $-1$ is square, then Condition 2a implies that everything is square. If not, define a total order on $K$ by $x \leq y$ if $y-x$ is square. Condition 2a implies that this is a total order; condition 2b implies that it is transitive and compatible with addition; I leave it as an exercise that it is compatible with multiplication by positive elements. So $K$ is an ordered field where the positive elements are the squares, and we are in condition (3b).
$(3) \implies (1)$. If we are in case 3a, then the statement is trivial because $K/K^2$ only contains one nonzero element. In case 3b, look up your favorite proof that signature is a well-defined invariant of real quadratic forms, and note that it only uses these facts.

I suspect some classical algebraist assigned a name to fields as in 3b. They are more special than formally real field (where $-1$ is not a sum of squares), but less special than real closed fields (where, in addition to the properties in 3b, we require that every odd degree polynomial have a root). 
An example of a field as in 3b is the field of real numbers which have straight-edge and compass constructions.<|endoftext|>
TITLE: How should one understand orbifold fundamental groups? 
QUESTION [23 upvotes]: I am studying orbifold fundamental group (or more generally orbifold homotopy groups). In a nutshell, my questions is: what are they intuitively? In what follows I give definitions and more precise questions. My definition of orbifold fundamental group is via classifying space of groupoid, which is explained in the next paragraph (so you may want to skip it if you know the definition). 
Let $\mathcal{G}$ be a topological groupoid consisting a topological spaces $G_{0}$ of $objects $ and $G_{1}$ of $arrows$ together with usual continuous structure maps. Let $|\mathcal{G}|$ denote the associated topological space $G_{0}/G_{1}$. Let $G_{n}$ be the iterated fibered product $G_{n}=G_{1}\times_{s,t} G_{n-1}$. These $G_{n}$ have the structure of a simplicial topological space, called the $nerve$ of $\mathcal{G}$. Face operads $d_{i}:G_{n}\rightarrow G_{n-1}$ for $i=0,\dots,n$ are given by
$$
d_{i}(g_{1},\dots,g_{n})=(g_{1},\dots,g_{i}g_{i+1},\dots,g_{n})
$$
for $i=1,\dots,n-1$ and 
$$
d_{0}(g_{2},\dots,g_{n})=(g_{2},\dots,g_{n}), \ \ 
d_{n}(g_{1},\dots,g_{n})=(g_{2},\dots,g_{n-1}). 
$$
The classifying space $B\mathcal{G}$ of $\mathcal{G}$ is then defined as 
$$
B\mathcal{G}=\bigsqcup_{n}(G_{n}\times \Delta^{n})((d_{i}(g),x)\sim(g,\delta_{i}(x))),
$$
where $\Delta^{n}$ is the topological $n$-simplex and $\delta_{i}:\Delta^{n-1}\rightarrow \Delta^{n}$ is the standard facemap.\
The $n$-th orbifold homotopy group of $\mathcal{G}$ based at $x\in |\mathcal{G}|$ is defined to be 
$$
\pi_{n}^{orb}(\mathcal{G},x)=\pi_{n}(B\mathcal{G},y),
$$
where $y\in G_{0}$ maps to $x$ under the quotient map $G_{0}\rightarrow |\mathcal{G}|$. 
The following are my questions:

Why is this a reasonable definition? Any manifold $M$ can be thought of topological groupoid via its chart i.e. $G_{0}=\bigsqcup_{i}U_{i}$ and $G_{1}=\bigsqcup_{i,j}U_{i}\times_{M} U_{j}$. It is not clear to me that the definition above reproduce $\pi_{n}(M)$. 
I am aware of an explicit description the orbifold fundamental groups of the orbifold Riemann surface $\Sigma_{g,n,k}$  of genus $g$ and $n$ orbifold points $p_{i}$ of order $k_{i}$: 
$$
\pi_{n}^{orb}(\Sigma_{g,n,k})
=\langle \alpha_{i},\beta_{i},\sigma_{j} \ (1\le i \le g,1\le j \le n)\ | \ 
\sigma_{1}\dots\sigma_{n}\prod_{i=1}^{g}[\alpha_{i},\beta_{i}]=1,,\sigma_{i}^{k_{i}}=1\rangle 
$$
Is it easy to see this explicit presentation by the definition above? 
It seems there are several ways to define the fundamental group of an orbifold, such as covering space etc. How should one understand orbifold fundamental groups? 

Thank you for your assistance.

REPLY [7 votes]: The answer to question 1, which asks whether this definition recovers homotopy groups of a manifold, is yes. As in the question, we'll take $M$ to be manifold, $G_0 = \bigsqcup_i U_i$ the disjoint union of the sets of an open cover of $M$ and $G_1 = G_0 \times_M G_0 = \bigsqcup_{i,j} U_i \cap U_j$. It is easy to see that $G_2 = \bigsqcup_{i,j,k} U_i \cap U_j \cap U_k$, and generally, $G_n$ is the disjoint union of the $(n+1)$-fold intersections of the $U_i$. To prove that $B\mathcal{G}$ has the same homotopy groups as $M$, we'll show in fact that these two spaces are homotopy equivalent.
First of all there is a clear map $B \mathcal{G} \to M$, that sends any point $(x,p) \in G_n \times \Delta^n$ to just $x$. This map is well defined since (1) $G_n$ is just a disjoint union of some open subsets of $M$, so any point of $G_n$ can be thought of as a point of $M$, and (2) all the face and degeneracy maps in $\mathcal{G}$ just perform bookkeeping: they change which multiple intersection of $U_i$'s a point is regarded as being in, but don't actually change the point.
Now, we'll define a section $M \to B \mathcal{G}$. For this, take a locally finite partition of unity $\phi_i$ subordinate to the cover $\{U_i\}_i$. Given $x \in M$, it'll belong to the support of finitely many $\phi_i$, say $\phi_{i_0}, \ldots, \phi_{i_k}$ and we can send $x$ to the equivalence class of $(x, (\phi_{i_0}(x), \ldots, \phi_{i_k}(x))) \in G_k \times \Delta^k$. Here, $x$ is thought of as belonging to the $U_{i_0} \cap \cdots \cap U_{i_k}$ term of $G_k$, and we regard $\Delta^k$ as the set of points $(t_0,\ldots,t_k) \in \mathbb{R}^{k+1}$ such that $t_i \ge 0$ and $t_0 + \cdots + t_k = 1$. It's straightforward to check that the definition above does give a function $M \to B\mathcal{G}$ (for example, adding extra $\phi_{i_j}$ with $\phi_{i_j}(x)=0$, does not change the image of $x$ in $B\mathcal{G}$), and it is clear this map is a section of the map $B\mathcal{G}\to X$ above.
Finally, the composite $B\mathcal{G} \to M \to B\mathcal{G}$ is homotopic to the identity by a straight line homotopy within the $\Delta^k$'s (that is, via homotopy that is the identity on the $G_k$ coordinate and moves in a straight line segment in the $\Delta^k$ coordinate).
This argument has probably been discovered many times and is well-known, I fairly recently found out that it does appear in print at least in Segal's Classifying Spaces and Spectral Sequences, proposition 4.1. It doesn't use that $M$ is a manifold, just that the open cover has a continuous subordinate partition of unity, so for example it works for any cover of a paracompact space $M$. Even without subordinate partitions of unity, for any open cover $U_i$ of any any space $M$ the homotopy groups of $M$ and $B \mathcal{G}$ are the same: these two spaces might not be homotopy equivalent in this general setting, but they are still weakly homotopy equivalent. See Dugger and Isaksen’s Hypercovers in Topology, theorem 2.1. This simplicial space $\mathcal{G}$ is called a Čech cover, and Dugger and Isaksen's paper also deals with the more general notion of a hypercover.<|endoftext|>
TITLE: Can we actually find any fixed points with Brouwer's theorem?
QUESTION [47 upvotes]: Background
At the risk of greatly oversimplifying matters, let me state a heuristic from Granas and Dugundji's beautiful book: fixed point theorems fall into two broad categories. The first class is usually functional analytic and imposes strong conditions on the map $f:X \to X$ whereas the second class is usually algebraic topological and imposes strong conditions on the space $X$ itself.
A typical example of the first class of theorems is the fixed point theorem of Banach. While the spaces it applies to are fairly general (complete metric spaces), the function must have a Lipschitz constant strictly less than $1$. On the other hand, Brouwer's theorem falls into the second class. Any continuous map works, but the domain must be a compact and convex subset of Euclidean space (originally a disk). Of course, both these theorems have been vastly generalized from the versions that I am stating here.
Question
One fundamental advantage of the Banach theorem is that it actually provides a recipe for converging to the fixed point as part of the standard proof: just start at an initial point and iterate. The proofs of the Brouwer theorem that I have seen do no such thing. The best known proof (I think) is the one by contradiction: assuming the domain is a disk, if $f(x)$ and $x$ are always distinct then the ray from $f(x)$ through $x$ to the boundary of said disk provides a deformation-retraction from the disk to its boundary, aha!
Here is my question:

Is there any way to actually find a fixed point when using Brouwer's theorem?

A Possible Idea
One scheme that unfortunately fails is as follows. Consider the sequence of iterates $f^n(x)$ for $n \in \mathbb{N}$ and any initial $x$ in the domain. We have an infinite sequence in a compact set, and hence a convergent subsequence, so the limit point is a candidate. This won't work since a) we haven't used convexity at all, and b) one may just be converging to a periodic orbit of $f$.

REPLY [17 votes]: The paper "Exponential lower bounds for finding Brouwer fixed points"
Addendum by original poster: It was non-trivial to find a copy of this great paper of Hirsch, Papadimitriou and Vavasis. It does answer my general question quite clearly: finding Brouwer fixed points is exponentially hard in the worst case no matter what algorithm you use. Here is a link to this paper for all those who are interested and don't want to run into many, many pay-walls. I will take it down in a few days. -VN<|endoftext|>
TITLE: How to define the equivalence of Maurer-Cartan elements in an $L_{\infty}$-algebra?
QUESTION [8 upvotes]: First let $L^{\bullet}$ be a  pro-nilpotent differential graded Lie algebra (dgla). We have the set of Maurer-Cartan elements in $L^{\bullet}$ ($MC(L^{\bullet})$) which are $\alpha \in L^1$ such that it satisfies the Maurer-Cartan equation
$$
\partial \alpha+ \frac{1}{2}[\alpha,\alpha]=0.
$$
We have a definition of gauge equivalance: $\alpha_0,\alpha_1\in MC(L^{\bullet})$ are called gauge equivalent if and only if there exists a $\xi \in L^0$ such that
$$
e^{\text{ad}\xi}\circ(\partial +\text{ad}\alpha_0)\circ e^{-\text{ad}\xi}=\partial +\text{ad}\alpha_1
$$
or in other words
$$
e^{\text{ad}\xi}\alpha_0-\frac{e^{\text{ad}\xi}-1}{\text{ad}\xi}\partial\xi=\alpha_1.
$$
From the first definition it is easy to see that gauge equivalence is really an equivalent relation. From the second definition we can define a path between $\alpha_0$ and $\alpha_1$. Let
$$
\alpha(t)=e^{t\text{ad}\xi}\alpha_0-\frac{e^{t\text{ad}\xi}-1}{\text{ad}\xi}\partial\xi.
$$
Then $\alpha(t)$ is a power series of $t$ in $L^{\bullet}$, $\alpha(0)=\alpha_0$, $\alpha(1)=\alpha_1$ and we can prove $\partial\alpha(t)+ \frac{1}{2}[\alpha(t),\alpha(t)]=0.$
Now we come to $L_{\infty}$ algebra $L^{\bullet}$ with higher bracket $[\cdot,\ldots,\cdot]_n$ with $n$ auguments. We still have Maurer-Cartan elements in $ L^{\bullet} $ ( $MC(L^{\bullet})$) which are $ \alpha \in L^1$ such that it satisfies the  Maurer-Cartan equation
$$
\partial \alpha+ \sum \frac{1}{k!}[\alpha,\ldots,\alpha]_k=0.
$$
My question is how to define equivalence of Maurer-Cartan elements in $ L^{\bullet} $? 
Of course we can define $\alpha_0,\alpha_1\in MC(L^{\bullet})$ are "equivalent" if and only if there exists a power series $\alpha(t)\in L^{\bullet}$ such that $\alpha(0)=\alpha_0$, $\alpha(1)=\alpha_1$ and $\alpha(t)$ satisfies the $L_{\infty}$ Maurer-Cartan equation. However, it is difficult to show that this is an equivalent relation, for example, how to connect two paths?
It seems that a generalization of gauge equivalence is what we want. But $e^{\text{ad}\xi}$ is not enough since we have higher bracket.

REPLY [9 votes]: To add a bit to what Damien says, addressing your question on how to generalise the gauge approach (which is equivalent to the approach outlined by Damien, as proved by several people):
You can view gauge symmetries in DGLAs via solving the differential equation
 $$
\frac{d\alpha}{dt}=-\partial\xi-[\alpha,\xi],
 $$
where $\xi$ is the given element of degree $0$. This generalises to homotopy Lie algebras as follows: consider the differential equation
 $$
\frac{d\alpha}{dt}=-\partial\xi-[\alpha,\xi]-\frac12[\alpha,\alpha,\xi]-\ldots-\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}\xi]_{p+1}-\ldots,
 $$
where the right hand side is simply the negative of $[\xi]_1^\alpha$, the first structure map of the twisted Lie-infinity structure
 $$
[x_1,\ldots,x_k]_k^\alpha:=\sum_{p\ge0}\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}x_1,\ldots,x_k]_{k+p}.
 $$
From that it is almost obvious that moving along the integral curves of this equation preserves the property of being Maurer--Cartan, since the Maurer--Cartan condition for $\alpha+\beta$, where $\alpha$ is a Maurer--Cartan element, and $\beta$ is infinitesimal becomes
 $$
\partial\beta+[\alpha,\beta]+\frac12[\alpha,\alpha,\beta]+\ldots+\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}\beta]_{p+1}+\ldots,
 $$
that is $[\beta]_1^\alpha=0$, and so $\beta=[\xi]_1^\alpha$ satisfies that, $[\cdot]_1^\alpha$ being a differential of the twisted structure. This circle of ideas is explained in many places, one important reference is ``Lie theory for nilpotent $L\_\infty$-algebras'' by Ezra Getzler (Ann. of Math. (2) 170 (2009), no. 1, 271--301.).<|endoftext|>
TITLE: Advantages of the sequence definition of limits
QUESTION [7 upvotes]: I will be teaching an introductory analysis course in the coming semester. In it the students will learn about limits of real sequences, and then will learn about limits of functions in terms of sequences.
More precisely, we will say that $\lim_{x\to a+}f(x) = L$ if whenever $(x_n)$ converges to $a$ with $x_n>a$ for all $n$, we have $\lim_{n\to \infty} f(x_n) = L$. Likewise, we will say that $\lim_{x\to a-}f(x) = L$ if whenever $(x_n)$ converges to $a$ with $x_n< a$ for all $n$, we have $\lim_{n\to \infty} f(x_n) = L$. Then we say that $\lim_{x\to a} f(x) = L$ if both $\lim_{x\to a+}f(x) = L$ and $\lim_{x\to a-}f(x) = L$.
The students will have already seen the $\varepsilon$-$\delta$ definitions of limits of functions in their calculus course. The question then is, how to properly motivate this second (equivalent) definition of limits of functions?

Are there any arguments which become significantly easier when using the sequence definition of limits of functions in place of the $\varepsilon$-$\delta$ definition? (These should be elementary enough to be understood by first year Mathematics undergraduates.)

For instance, I suppose that once one has the Algebra of Limits for sequences, one gets the Algebra of Limits for functions for free. But I'm not convinced that much is to be gained from doing things this way around.
Edit: Thanks for the answers and comments so far. It seems many people are in favour of teaching the sequence definition of limits alongside the $\varepsilon$-$\delta$ definition. I agree that it should be useful to be aware of both definitions. To be certain of this, however, I would still like to see an example of a proof which is simpler when using the sequence definition.

REPLY [3 votes]: "Are there any arguments which become significantly easier...". There is an obvious choice: Disprove that a certain function has a limit at a certain point. Showing that with epsilon-delta would be a mission impossible for most students. Using the sequence definition makes it much easier and more appealing.<|endoftext|>
TITLE: Is compact flat manifold cusp cross-sections of a complete finite volume hyperbolic manifold?
QUESTION [9 upvotes]: Let $M^{n-1}$ be a closed flat manifold. Is it true that there exists a hyperbolic manifold $N^n$ with finite volume such that $M$ is a cusp cross-section of $N$?
It was proved in "On the geometric boundaries of hyperbolic 4-manifolds" by Long and Reid in Geom. Topol. 4 (2000), 171–178 that there are 3-manifolds are not cusp cross-sections of any complete finite volume one-cusped hyperbolic 4-manifold, due to the obstructions of eta-invariant. They constructed a 3-dim flat manifold with eta invariant = -4/3. if it was the cusp of a hyperbolic manifold then it has to be integer according to Aatiyah-Patodi-Singer's theorem.
My question is can it always a cross-section of several-cusped hyperbolic 4-manifold?

REPLY [11 votes]: Yes.  Every (compact) flat $n$-manifold is diffeomorphic to a cusp cross section of a hyperbolic $(n+1)$-manifold. This is a theorem of McReynolds, Controlling manifold covers of orbifolds, Math. Res. Lett. 16 (2009), 651-662.
In your case, $n=3$, this is a theorem of Nimershiem. B. E. Nimershiem, All flat three-manifolds appear as cusps of hyperbolic four-manifolds, Topology Appl. 90 (1998), 109–133.

REPLY [10 votes]: Since there are countably many finite-volume $3$-manifolds, and uncountably many flat tori, the answer is NO for $n=3,$ at least if your question is geometric, as opposed to topological.<|endoftext|>
TITLE: Example of a form linear in infinitely many variables ?
QUESTION [12 upvotes]: We all know plenty of examples of multilinear forms in finitely many variables (e.g. determinants). However, I am missing an interesting example of a form in infinitely many variables, linear in each. The important point here is the word "interesting" ; the example should occur naturally in some nice situation in algebra, or geometry, or analysis... (One can easily find non-interesting examples.) I thought that maybe regularized determinants could provide examples, but I don't know much about these. Thanks for ideas !

REPLY [2 votes]: Hmmm... I think that the functional determinants (as in http://en.wikipedia.org/wiki/Functional_determinant) of Quantum Mechanics and Quantum Field Theory appear rather naturally. These CAN be defined rigorously, insofar as the defining Feynman path integrals are, in this case, defined rigorously. The most readable introduction to rigorous path integration I've read is "A Modern Approach to Functional Integration" by John R. Klauder.
EDIT: I did some reading. The bosonic path integral expression (as in the wikipedia page I linked to earlier) for the functional determinant may fail to be multilinear (though it is rigorous), whereas the fermionic path integral expression
$\det S = \int\int \exp (\langle \bar{c} | S | c\rangle) \mathcal D c \mathcal D \bar{c}$
should be genuinely multilinear and valid for every "reasonable" S (including nonsymmetric/nonhermitian ones) IF fermionic functional integrals in infinite variables can be consistently defined to yield sensible results (e.g $\int 1 \mathcal D c=0$), which, as far as I know, has not yet been done rigorously.<|endoftext|>
TITLE: Elliptic Curves with CM and Class Field Theory
QUESTION [14 upvotes]: Let $K$ be an imaginary quadratic field with Hilbert class field $H$, and let $E$ be an elliptic curve defined over $H$ with complex multiplication by the ring of integers $O_K$ of $K$. It is known that for an integral ideal $\mathfrak{m}$ of $O_K$, $K(j(E),h(E[\mathfrak{m}]))$ is the ray class field of K modulo $\mathfrak{m}$, where $h$ is the Weber function for $E/H$. (This is stated, for example, on page 135 of Silverman's Advanced Topics in the Arithmetic of Elliptic Curves.)
My question is this: what if $E$ has CM by an arbitrary order? Can any generalization of this statement be made? I've read that if $E$ has CM by an order of conductor $\mathfrak{f}$, then $K(j(E))$ is the ring class field of $K$ with conductor $\mathfrak{f}$, but I'm wondering if anything more can be said.

REPLY [3 votes]: Now as for an intuitive explanation concerning Pete Clark's question:

I would be interested in, at least, a reference for the fact that
  K(j(E),h(E[N])) contains the N-ray class field of K for arbitrary
  orders.

Here is my intuitive point of view on it (I put it in another answer because I won't try to be perfectly rigorous), in terms of moduli.
Let $E_1$ be a an elliptic curve with maximal CM by $O_K$, and $E_2$ be a curve with CM by $O$, an order $O$ of conductor $F$ in $O_K$. $E_2$ is defined over $K(j(E_2)) \supset K(j(E_1))$, so there is a rational isogeny $E_2 \to E_1$ of degree $F$. If we add the field of definition of the points of $m$-torsion of $E_2$ then it is clear that if $F$ is prime to $m$, these points get transported to the points of $m$-torsion of $E_1$, so we already have the $m$-ray class field.
What is more surprising is that it works also when $m$ is not prime to $F$. 
So let's assume that $m \mid F$ and see why we can still transport the $m$-torsion from $E_2$ to $E_1$.
The reason is as follow: when we are in the ring class field of $O$, all isogenies of degree $F$ between $E_1$ and an elliptic curve with endomorphism by $O$ are already rational; this means that the Galois action on $E_1[F]$ is given by a diagonal matrix. (And of course being in the $F$-ray class field means that the Galois action on $E_1[F]$ is the identity.) So in particular the kernel $K_1$ of the isogeny $E_1 \to E_2$ has a rational complement $K_2$.
Now if we have the $m$-torsion on $E_2$, pushing it through the dual isogeny $E_2 \to E_1$, we have that at least the points of $m$-torsion of $K_1$ are all rationals.
But because the points of $m$-torsion are rationals in $E_2$, the $m$-roots of unity are rationals, and so by looking at the Weil pairing we see that the points of $m$-torsion in $K_2$ are rationals. So all points of $m$-torsion in $E_1$ are rationals.<|endoftext|>
TITLE: elliptic curves with and without CM
QUESTION [7 upvotes]: Let $E/\mathbb C$ be an elliptic curve. It is known that if $E$ has CM, then $j(E)$ is an algebraic integer. My first question is: what about the converse? Is there a way to identify the subset of algebraic integers whose elements correspond to [isomorphism classes of] elliptic curves with CM?
The second question is: fixed a number field $K$, does there always exist an elliptic curve over $K$ with CM? And without CM?
Thank you very much!

REPLY [5 votes]: 1)If an elliptic curve has integral $j$-invariant it absolutely DOES NOT NEED to have CM. The class of curves with integral $j$-invariant (let's call that the class of IM Elliptic curves for Integral Modulus) is MUCH MUCH larger than the class of CM Elliptic curves. In fact, one can use Heilbronn's Theorem that class numbers of imaginary quadratic fields tend to infinity to show that over any given number field, there are only finitely many elliptic curves with CM. In particular over $\mathbf{Q}$, there are only 13 CM $j$-invariants. Even for all number fields of a certain degree, there are only a finite number $N(d)$ of elliptic curves with CM over any number field of degree $d$. This gives a way to enumerate all the CM $j$-invariants or "singular moduli," which is about as good as you can hope for in terms of describing the complex numbers which are $j$-invariants of CM elliptic curves. To do this explicitly (say for number fields of degree up to 100) see Mark Watkins' enumeration of imaginary quadratic fields of class number up to 100.
Meanwhile, for any regular integer $n$ (1,2,3, etc) there is at least one elliptic curve over $\mathbf{Q}$ whose $j$ invariant is $n$. Therefore over $\mathbf{Q}$ and therefore over any number field, there are infinitely many non-isomorphic IM elliptic curves.
If you want to say something general about elliptic curves with IM, consider the theorem of Deuring that an elliptic curve has IM if and only if it has potential good reduction. For this and much much more see Serre-Tate's "Good reduction of abelian varieties"
2) Easy proof that the answer is yes, at least as long as you mean "has a CM $j$-invariant" when you say CM: $y^2 = x^3 + 1$ is a CM elliptic curve with $j$-invariant zero defined over any number field. On the other hand, if we take an elliptic curve with $j$-invariant equal to 1/2, say this one: y^2 + x*y = x^3 + 72*x + 13822 , well it doesn't have integral $j$-invariant and therefore can't have CM!<|endoftext|>
TITLE: Finite dimensional "Mountain Pass Lemma"
QUESTION [17 upvotes]: Question Does anyone know of a good reference which I can cite for the finite dimensional version of Mountain Pass Lemma?
Motivation I am writing a paper and found myself using the following result:

Let $f$ be a proper smooth real-valued function on $\mathbf{R}^3$ such that $f(0) = 0$, $f|_{B_1(0)} \geq 0$, $f|_{\partial B_1(0)} \geq 1$ and $\exists p \in {\partial B_2(0)}$ such that $f(p) = 0$. Then $\exists q\in \mathbf{R}^3 \setminus B_1(0)$ such that $f'(q) = 0$ and $f(q) \geq 1$. 

For the time being I referred to Ambrosetti and Rabinowitz's JFA article for the mountain pass lemma, but citing a Banach space version for a finite-dimensional Euclidean space application gives me a funny feeling. (Also, if feels like such a result could in principle be found in not-so-advanced undergraduate textbooks...)

REPLY [4 votes]: I have stumbled across Richard Palais' (co-author Chuu-lian Terng) Critical Point Theory and Submanifold Geometry (Springer Lecture Notes in Math 1353). This is an awesome book!
The "Mountain Pass Lemma" for finite-dimensional manifolds is presented as Theorem 9.2.7 (pg189).<|endoftext|>
TITLE: Reverse map of a homology equivalence.
QUESTION [10 upvotes]: Dear community,
assume that $f\colon X\to Y$ is a homology equivalence, i.e. induces isomorphisms on all integer homology groups. Is there a map $g\colon Y\to X$ which is also a homology equivalence?
Actually, I am only interested in the case if $f\colon X\to X^+$ is obtained by Quillen´s plus construction.
Thanks for any comments in advance.

REPLY [13 votes]: As I was writing this answer, Oscar beat me to the punch. I will keep it posted anyway.
Let $X^3$ be the Poincare homology sphere. Let $\tilde X \to X$ be the universal cover (note: $\tilde X$ is $S^3$).  As in my comment, if there were a map $S^3 \to X$ inducing a homology equivalence, then that map
must necessarily factor through $\tilde X$ (by covering space theory). But this is impossible since $\tilde X \to X$ has degree 120, whereas the composite $S^3 \to \tilde X \to X$ is supposed to have degree one.<|endoftext|>
TITLE: Must the $j$-invariant of an elliptic curve with an isogeny be integral?
QUESTION [6 upvotes]: Let $K$ be a quadratic field, and $E/K$ a non-CM elliptic curve with a $K$-rational $p$-isogeny, for $p$ a prime. I would like to say the following:

For large enough $p$, the $j$-invariant $j(E)$ must be in $O_K$. 

I would also like to know exactly how large $p$ must be. (This lower bound may very well not depend on $K$.)
The above is indeed true when $K$ is replaced with $\mathbb{Q}$; 37 is definitely large enough; (though the true bound may be as small as 17. EDIT: Actually, it's 19. See the comments.)
Corollary 4.3 in Mazur's article [1] gives me hope that the above may be true. Assuming condition A (which I'll describe presently), it says (for my set-up) that the only primes which can divide the denominator of the $j$-invariant are the primes above 2 and 3; (moreover, if 3 doesn't ramify in $K$, then it's only the primes above 2 that need concern us). 
Condition A is that $J_0(p)$, the jacobian of the modular curve $X_0(p)$, possesses an "optimal quotient" whose Mordell-Weil rank over $K$ is zero.
So I guess I'm hoping for two things:

That these small primes can be dealt with (i.e. for $p$ large enough, they don't arise in the denominator of $j$); and 
Condition A can be removed after all of these years (at least for $p$ large enough).

EDIT (after comments from Felipe Voloch and Noam Elkies): Merel's "winding quotient" has rank 0 over $\mathbb{Q}$; but I don't think it will have rank zero over every quadratic field. I also don't know over which quadratic fields the winding quotient has rank zero. 
Noam Elkies is quite right that the desired result is "vacuously true", since it is known (by work of Momose, Theorem B in [2]) that there are only finitely many $p$ for which there is a $K$-rational $p$-isogeny. However, I'd still like a more direct approach to the lower bound question, not using Momose's much stronger result, if there is one…
[1]: Mazur, B. "Rational Isogenies of Prime Degree". Inventiones, 1978.
[2]: Momose, F. "Isogenies of Prime Degree over Number Fields". Compositio, 1995

REPLY [6 votes]: If you are given K, and you want to find a p such that J_0(p) has a rank-0 quotient over K, you might try proceeding as follows.  Let $\chi$ be the quadratic twist whose kernel is the Galois group of K. Then what you really want is to show that there exists a newform f in S_2(p) with the property that both the corresponding quotient A_f and its twist by $\chi$ have rank 0.  This follows if you know that both $L(f,1)$ and $L(f \times \chi, 1)$ are nonzero.  Why should there be any such f?  One way to get at this is by averaging the quantity
$L(f,1) L(f \times \chi, 1)$
over the an orthogonal basis of cuspforms f for S_2(p).  Often there's a nice analytic way to estimate this average as
something that's not zero + something that decays with p
which means that, for sufficiently large p (usually explicit) the average is not 0, which means that one of the guys being averaged is not zero, which means that you have a quotient which does what you want.  See the recent paper of Michel and Ramakrishnan, and the papers it cites, to see how the method works:
http://arxiv.org/abs/0709.4668
BUT NOTE:
Momose's argument actually doesn't go via this route; he uses the same quotient A of J_0(p) that Mazur does.  As you note, this quotient has rank 0 over Q but not over K.  But that's OK; Momose's argument (which goes back to Kamienny) involves mapping the symmetric square of X_0(p), not X_0(p) itself, into A.  And a point on X_0(p) over ANY quadratic field gives a Q-rational point on the symmetric square.<|endoftext|>
TITLE: Most degenerate Weyl tensors in Riemannian and Lorentzian signature
QUESTION [11 upvotes]: Hi all,
This is my first post on Math Overflow!  I've been stuck on the following question and was wondering if anyone might have any insight on it.  Here it is:
Let $n \geq 5$.  Let $G = SO(n)$ or $SO(1,n-1)$, and let $V$ be the irreducible $G$-module corresponding to algebraic Weyl tensors.  What is the "most degenerate" (nonzero) $G$-orbit in $V$ (or the projectivization $\mathbb{P}V$)?  More precisely, I want to find the maximum dimension of the stabilizer of a nonzero Weyl tensor.  More hopefully, I would like to know what a representative element in this minimal orbit (is it unique?) and its stabilizer look like.
The annoying issue here is that we're working over $\mathbb{R}$.  Over $\mathbb{C}$, if $G = SO(n,\mathbb{C})$, then $V$ has highest weight $2\lambda_2$, and the orbit through the highest weight line is the most degenerate orbit.  The stabilizer of a highest weight line is (in the standard representation) the parabolic subgroup which is the stabilizer of a null 2-plane.  These of course don't exist in Riemannian or Lorentzian signature.
Any tips would be greatly appreciated!  Thanks!

REPLY [12 votes]: Re-Amended Answer:
My guess for $\mathrm{SO}(5)$ appears to have been correct in one sense, but not in another. It's true that a nonzero Weyl tensor in this case has to have stabilizer of dimension at most $4$, but there are two distinct candidates with this property.  
First, there is the obvious one, which is the Weyl curvature of $\mathbb{CP}^2\times \mathbb{R}$, which has $\mathrm{U}(2)\subset \mathrm{SO}(4)\subset\mathrm{SO(5)}$ as its stabilizer.  However, it turns out that $H = \mathrm{SO}(2)\times\mathrm{SO(3)}\subset \mathrm{SO}(5)$ (which is also of dimension $4$) fixes a nonzero Weyl curvature as well, and this yields a completely different $6$-dimensional orbit of $\mathrm{SO(5)}$ in the vector space of Weyl curvatures.  Up to multiples, these two are the only $6$-dimensional orbits.
For $n=5$, there are no other nontrivial orbits of dimension $6$ or less.  There is a $7$-dimensional orbit, which is the Weyl curvature of the $5$-dimensional symmetric space $\mathrm{SU}(3)/\mathrm{SO}(3)$.  Its stabilizer is the subgroup of $\mathrm{SO}(5)$ that is the irreducibly-acting subgroup isomorphic to the holonomy of this space, i.e., $\mathrm{SO}(3)$.  Up to multiples, there are no other orbits of dimension $7$ or less than what I have listed.
When you 'bootstrap' the $\mathrm{SO}(2)\times\mathrm{SO(3)}$-stabilizer example to dimensions higher than $5$, you get an element with stabilizer $\mathrm{SO}(2)\times\mathrm{SO(3)}\times\mathrm{SO}(n{-}5)$, which, when $n>5$, has lower dimension than $\mathrm{U}(2)\times\mathrm{SO}(n{-}4)$, so this gives a higher dimensional orbit when $n>5$ than the Weyl curvature of $\mathbb{CP}^2\times\mathbb{R}^{n-4}$.
In low dimensions (but higher than $n=5$) the Weyl curvature of $\mathrm{CP}^2\times\mathbb{R}^{n-4}$ is not optimal.  When $n=2m$, there is the Weyl curvature of $\mathbb{CP}^m$, which has $\mathrm{U}(m)\subset\mathrm{SO}(2m)$ as stabilizer, and this group is bigger than $\mathrm{U}(2)\times\mathrm{SO}(2m{-}4)$ for $2 < m < 7$. However, as soon as $m>7$, the group $\mathrm{U}(m)$ has dimension less than the dimension of $\mathrm{U}(2)\times\mathrm{SO}(2m{-}4)$.  
Last night, I thought that Weyl curvature of $\mathrm{CP}^2\times\mathbb{R}^{n-4}$ wins for large enough $n$, but this morning, I had another idea and realized that there is an even better candidate in the 'stable' range:  I turns out that, for $n\ge 5$, under the subgroup $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)\subset \mathrm{SO}(n)$, the space of Weyl tensors in dimension $n$ has a trivial summand, and hence there is a nonzero Weyl tensor (unique up to multiples) whose stabilizer contains $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$.  Since there is no connected Lie group between this subgroup and $\mathrm{SO}(n)$, it follows that the identity component of the stabilizer of this tensor is $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$.  Thus, this orbit has dimension $2n{-}4$.  I now think that, for sufficiently large $n$ (maybe even $n\ge 9$), this might be the lowest dimensional orbit.  It might not be unique, though.  For example, when $n=8$, there is an orbit of type $\mathrm{SO}(8)/\mathrm{U}(4)$ and one of type $\mathrm{SO}(8)/\bigl(\mathrm{SO}(2)\times\mathrm{SO}(6)\bigr)$.  Both of these orbits have dimension $12$. (NB: Even though these two spaces are isomorphic because of triality, the two Weyl orbits are, of course, quite different.)
There aren't many subgroups of $\mathrm{SO}(n)$ that are larger than $\mathrm{SO}(2)\times\mathrm{SO}(n{-}2)$ when $n$ is sufficiently large.  The one obvious exception, $\mathrm{SO}(n{-}1)$, does not fix a Weyl tensor.  That's why I'm thinking that this one will win for sufficiently large $n$.
I haven't thought seriously about $\mathrm{SO}(1,n{-}1)$ yet, but, probably, there are at least two orbits of minimal dimension, namely $\mathrm{SO}(1,n{-}1)/\bigl(\mathrm{SO}(1,1)\times\mathrm{SO}(n{-}2)\bigr)$ and $\mathrm{SO}(1,n{-}1)/\bigl(\mathrm{SO}(1,n{-}3)\times\mathrm{SO}(2)\bigr)$, and there might be another 'degenerate' one that fills in the gap between these two.<|endoftext|>
TITLE: When do blow-up and quotient commute? 
QUESTION [7 upvotes]: Let a finite group $G\subset SL(n,\mathbb{C})$ act on $\mathbb{C}^{n}$ in a natural way. Assume there is a crepant resolution of $f:X\rightarrow \mathbb{C}^{n}/G$. When is it possible to write $X$ as $Y/G$ for some $Y$ and $G$-action on $Y$? 
This is true for example $\pm id_{\mathbb{C}^{2}} \subset SL(2,\mathbb{C})$ acting on $\mathbb{C}^{2}$. Is it still true for example
$$
\langle diag(e^{\frac{2\pi i}{3}},e^{\frac{2\pi i}{3}},e^{\frac{2\pi i}{3}})\rangle \cong \mathbb{Z}/3\mathbb{Z} \subset SL(n,\mathbb{C})
$$
acting on $\mathbb{C}^{3}$? FYI, this has toric crepant resolution.

REPLY [5 votes]: By definition $C^n/G = Spec(C[x_1,\dots,x_n]^G)$ and $X$ being a blow up of an ideal $I$ on $C^n/G$ can be written as
$$
X = Proj_{Spec(C[x_1,\dots,x_n]^G)}(C[x_1,\dots,x_n]^G \oplus I \oplus I^2 \oplus \dots).
$$
Now assume the ideal $I \subset C[x_1,\dots,x_n]^G$ can be written as 
$$
I = J \cap C[x_1,\dots,x_n]^G = J^G,
$$
where $J \subset C[x_1,\dots,x_n]$ is a $G$-invariant ideal (the simplest thing to do is to take $J = C[x_1,\dots,x_n]\cdot I$). Take
$$
Y = Proj_{C[x_1,\dots,x_n]}(C[x_1,\dots,x_n] \oplus J \oplus J^2 \oplus\dots).
$$
Then $Y/G = X$.<|endoftext|>
TITLE: Intuitionistic consistency of surjection from naturals to reals
QUESTION [6 upvotes]: Is it consistent intuitionistically (in the sense of topos theory) for there to be a surjection from the natural numbers to the (Dedekind, let us say) real numbers? [I've managed to convince myself this happens in the effective topos, but not so convincingly as I'd like; I suspect others will be able to answer more confidently and cleanly...]

REPLY [10 votes]: Theorem (Intuitionistic logic with Dependent Choice): 
For every sequence $a : \mathbb{N} \to \mathbb{R}$ there is $x \in \mathbb{R}$ such that $|x - a_n| > 0$ for all $n \in \mathbb{N}$.
Proof. Define a nested sequence of closed intervals $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ such that the length of $I_n$ is $(1/5)^n$, as follows. Begin with $I_0 = [0, 1]$. Suppose $I_n = [u, v]$ has been defined already. Using dependent choice we choose $I_{n+1}$ to be either $[u, (4 u + v)/ 5]$ or $[(u + 4 v)/5, v]$ so that
$$a_n > (3 u + 2 v)/5 \Rightarrow I_{n+1} = [u, (4 u + v)/ 5]$$
and
$$a_n < (2 u + 3 v)/5 \Rightarrow I_{n+1} = [(u + 4 v)/5, v].$$
(You should draw a picture of $[u,v]$ divided into five equal parts.)
Observe that the choice of $I_{n+1}$ guarantees $|y - a_n| > (1/5)^n$ for every $y \in I_{n+1}$.
There is a unique real number $x$ which is contained in all the intervals. It satisfies $|x - a_n| > (1/5)^n$ for all $n \in \mathbb{N}$. QED.
Remark: With a bit of work we can replace Dependent Choice with Number Choice, as is typical in such situations. I will leave it as exercise. (Hint: for every pair of possible endpoints of the intervals, make the choice before constructing the intervals.) Also, the proof works for Dedekind as well as Cauchy reals.
Corollary: There is no surjection $\mathbb{N} \to \mathbb{R}$ in any realizability topos, such as the Effective topos.
Proof. Dependent Choice is valid in a realizability topos. Thus every sequence of reals misses a real by the above theorem. QED.
Now the really interesting question is whether we can do without choice. I remember thinking about this with someone someplace sometime, but we never arrived at a conclusion.
By the way, it is intuitionistically consistent to assume that there is an injection $\mathbb{R} \to \mathbb{N}$. This is validated in Infinite Time Turing Machine (IITM) realizability. But since IITM realizability is a realizability topos, there is no surjection going the other way in this model.<|endoftext|>
TITLE: When do the $\gamma$-filtration and codimension filtration of K-theory agree?
QUESTION [8 upvotes]: Let $X$ be a smooth quasiprojective algebraic variety over a field $k$. Then the $K$-groups $K_m(X)$ are defined, and there are two standard filtrations on them: the "codimension filtration" given by 
$$\operatorname{Fil}^p_{\mathrm{cod}} K_m(X) :=\bigcup\nolimits_Y \text{Ker}(K_m(X)\to K_m(X-Y))$$
where the limit is taken over all $Y$ of codimension $\ge p$; and the "$\gamma$-filtration" $\operatorname{Fil}^\bullet_\gamma$ defined by Quillen. I gather that $\operatorname{Fil}^p_\gamma \subset \operatorname{Fil}^{p-m}_{\mathrm{cod}}$, and that for $m = 0$, this becomes an isomorphism after tensoring with $\mathbb{Q}$.

Are there other cases (maybe under additional assumptions on $k$ or on $X$) where the two filtrations agree rationally? Grayson's article in "Handbook of $K$-theory" suggests that they should be different when $X = \mathrm{Spec}(k)$ and $k$ has large cohomological dimension, but I confess that I don't understand the reasoning behind this.
The two filtrations both come from spectral sequences: the codimension filtration from the Brown--Gersten--Quillen sequence 
$$E_2^{pq} = H^p(X, \mathscr{K}_{-q}),$$
where $\mathscr{K}_q$ denote the $K$-theory sheaves on $X$; and the $\gamma$-filtration (if I've undstood correctly) comes from the Friedlander--Suslin motivic spectral sequence 
$$E_2^{pq} = \operatorname{CH}^{-q}(X,-p-q) = H^{p-q}_{\mathrm{mot}}(X, \mathbb{Z}(-q)).$$
Both converge to $K_{-p-q}(X)$ (sorry about the convoluted indexing!). Landsburg ("Relative Chow groups", Illinois J Math (35), 1991) has constructed maps $\operatorname{CH}^{-q}(X,-p-q) \to H^p(X, \mathscr{K}_{-q})$ between the $E_2$ terms of the spectral sequences, which he shows are isomorphisms if $p + q = 0$ or $-1$.
Are Landsburg's maps compatible with the differentials in the two spectral sequences, and are there additional hypotheses which would force them to be isomorphisms (maybe up to torsion) in all degrees?

(Apologies if this is a naive question; I'm only just beginning to learn all of this stuff.)

REPLY [3 votes]: Regarding fields, you might already know this, but if you take $p=0$, then $Ch^{-q}(Spec(k),-q)=K_{-q}^M(k)$, whereas $H^0(Spec(k),\mathscr{K_{-q} })=K_{-q}^Q(k)$, where $K^M$ and $K^Q$ are Milnor and Quillen $K$-theory.  These won't agree in general; for example, for finite fields the higher Milnor $K$-groups are all zero.<|endoftext|>
TITLE: Monsky's proof of the finiteness of de Rham cohomology
QUESTION [13 upvotes]: I'd really like to understand the proof that Paul Monsky wrote about the finiteness of the de Rham cohomology of algebraic varieties. I'd like it very much because it seems to explain in concrete terms some of Dwork's deformation theory and it may help me to proceed to concrete computations.
MR301017 — Monsky, P. “Finiteness of de Rham cohomology”. Amer. J. Math. 94 (1972), 237–245.
Unfortunately, I'm stuck quite early in the proof, lemma 2.1 in fact.
If I may, since the paragraph I'm stuck with is short and self-contained, let me show it to you.

(the last words are missing, they say: “the lemma follows”.)
I think I'm OK with every single word of this extract, however, I don't understand why a proof follows...
In fact, I don't even understand why this lemma can conceivably be true...
Indeed, its says that two Koszul homology are isomorphic, but the first one is built with $n$ operators whereas the second is built with $n+1$ operators, so that the $H_{n+1}$ of the first is always zero but the $H_{n+1}$ of the second may be non zero, a priori.
The two questions I'd like to ask you are:

Why am I wrong about the $H_{n+1}$?
Why does “the lemma follow”?

Thank you very much!

REPLY [4 votes]: As for your issue with the $n+1$-st term, 
observe that it is $0$ for both, since $H_{n+1}$ of the second complex is the kernel of $\frac{\partial}{\partial T}+ f$ which is injective (use $f$ is not a zero-divisor).
The "lemma follows" since Monsky has reduced the homology of the first complex to that of the complex $K_\cdot(A[T];E_1+L_{Tf_1},\dots,E_n+L_{Tf_n},\frac{\partial}{\partial T}+ L_f)$ (I assume he uses the notation $L_x$ to mean the operator "multiplication by $x$"). Now, the big exact sequence and the considerations right after it show that the homology of my last complex is the one of the first complex in his statement, by shifting: $H_1$ is the cokernel of the last operator $\frac{\partial}{\partial T}+ L_f$ which is $A_f/A$ and each other $H_i$ is the $\text{ker}/\text{coker}$ of $E_i+L_{Tf_i}$ on $A[T]$ which coincides with $E_i$ on $A_f/A$.<|endoftext|>
TITLE: Cup products and hypercohomology
QUESTION [8 upvotes]: This is a cross-post of the following math.stackexchange question: https://math.stackexchange.com/questions/188760/cup-product-and-hypercohomology
I always found the cup product slightly mysterious. Recently I discovered the following interesting theorem (in Voisin's book Hodge theory and complex algebraic geometry I, chapter 4.3):
For the setup, let $(X, \mathcal{O})$ be a ringed space, $\mathcal{F}$, $\mathcal{G}$ sheaves of $\mathcal{O}$-modules, $\mathcal{F}^\bullet, \mathcal{G}^\bullet$ acyclic resolutions of $\mathcal{F}, \mathcal{G}$, and $\mathcal{H}^\bullet$ an acyclic resolution of $\mathcal{F} \otimes \mathcal{G}$. Suppose given a morphism of complexes $$\phi^\bullet: Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet) \to \mathcal{H}^\bullet,$$ (where $Tot$ denotes the total (simple) complex associated to a double complex). This data naturally yields homomorphisms $$H^p(X, \mathcal{F}) \otimes H^q(X, \mathcal{G}) \to H^{p+q}(X, \mathcal{F} \otimes \mathcal{G}) \quad(*).$$
The theorem is this: if $\phi^\bullet$ is compatible with the resolutions (that is, the evident triangle involving $\mathcal{F}\otimes\mathcal{G}$, $Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet)$ and $\mathcal{H}^\bullet$ is commutative), then the induced morphism $(*)$ on cohomology is the cup product pairing.
The proof says, somewhat mysteriously to me, that the result follows by defining cup products on hypercohomology, and then using commutativity. While I know about hypercohomology, it is unclear what cup products should even mean in this situation. Can you explain what Voisin means, or provide a reference?
Note: the theorem essentially says that all such $\phi^\bullet$ induce the same morphism on cohomology (independent of the resolutions even), so we need not acutally know here what the cup product pairing is.
Thanks in advance.
EDIT: Since this may not have been clear, my question is this: How do you prove the above theorem, potentially by defining a cup product on hypercohomology and exploiting its properties?
It is quite easy to construct, for arbitray left-bounded complexes $\mathcal{F}^\bullet$ and $\mathcal{G}^\bullet$, a canonical product $\mathbb{H}^p(\mathcal{F}^\bullet) \otimes \mathbb{H}^q(\mathcal{G}^\bullet) \to \mathbb{H}^{p+q}(Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet))$ (by using "Godement double-resolutions" and the standard fact that godement resolutions remain resolutions after tensoring). It is not clear to me if this is a sensible construction, since $\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet$ is not stable under replacing $\mathcal{F}^\bullet$, $\mathcal{G}^\bullet$ by quasi-isomorphis complexes.
EDIT 2: the above claim is false; I made a mistake in my computation. As Donu points out below, the natural target for cup product in hypercohomology is the "derived tensor" $\mathcal{F}^\bullet \otimes^L \mathcal{G}\bullet$. This is indeed what the construction I had in mind yields.

REPLY [2 votes]: Any two natural approaches will likely yield the same product up to sign, but this is not much of an answer. If you want a precise reference comparing the Cech-style cup product with the product using resolutions, take a look at Godement's Théorie des Faisceaux. 

After rereading the post-edited question, I realize you were partly wondering about a good
definition of cup product for hypercohomology. Unfortunately, due to lack of time, this will be a bit sketchy, but here goes.
You can identify
$$\mathbb{H}^i(X,\mathcal{F}^\bullet)\cong Ext^i(\mathcal{O}, \mathcal{F}^\bullet)\cong Hom_D(\mathcal{O}[-i], \mathcal{F}^\bullet)$$
where $D$ is the derived category of (bounded  if you like) sheaves of $\mathcal{O}$-modules.
The natural target for the cup product is the hypercohomology of the derived tensor product
$\mathcal{F}^\bullet\otimes^L \mathcal{G}^\bullet\in D$ obtained  by tensoring flat acyclic resolutions (although there are some technical issues that I'm suppressing if want boundedness).
From the above formulas, one gets
$$\mathbb{H}^i(X, \mathcal{F}^\bullet)\otimes \mathbb{H}^j(X, \mathcal{G}^\bullet)\to
\mathbb{H}^{i+j}(X, \mathcal{F}^\bullet\otimes^L \mathcal{G}^\bullet)$$
To prove that this coincides with some other notion of cup product is going to be
a slog.<|endoftext|>
TITLE: Structure of the unitary representation $L^2(N/M)$ when $N$ is a nilpotent Lie group
QUESTION [5 upvotes]: Hi All,
I am new to this (though I seem to be a latecomer); so forgive me if this is not your most favorite question:
I am trying to understand the structure (e.g., decomposition) of the unitary representation $L^2(N/M)$ where $N$ is a nilpotent Lie group acting by left translation on this Hilbert space (coming from the invariant measure on N/M). Surprisingly, I am unable to find any suitable references. Does anyone here know where one should look for an answer in the literature?

REPLY [2 votes]: Is $N$ or $M$ connected? simply connected? Is $N$ commutative? Ronald Lipsman has settled the decomposition of quasiregular representations of Lie groups in many settings. You will find his papers very readable. They are all available on google.<|endoftext|>
TITLE: Nonnegative smooth function as sum of squares of smooth functions
QUESTION [15 upvotes]: There is a famous open problem, whose solution is attributed to Paul Cohen, but no published paper seems to be available: 
There exists $f\in C^\infty(\mathbb R,\mathbb R_+)$ such that $f$ is not a finite sum of squares of $C^\infty$ functions. 
I would be grateful for any hint or reference to that specific question.

REPLY [12 votes]: It seems to be a result of J-M Bony that every nonnegative function in $C^{2m}$ is a sum of squares of two $C^m$ functions, which means that every $C^\infty$ function is the sum of squares of two $C^m$ functions for any $m$ (which, I suppose, does not mean that you can do it with two $C^\infty$ functions -- the counterexamples are attributed to Paul Cohen and D.B.A. Epstein -- see references 1and 4 in the cited paper): the reference is:
Bony, Jean-Michel(F-POLY-CMT)
Sommes de carrés de fonctions dérivables. (French. English, French summary) [Sums of squares of derivable functions] 
Bull. Soc. Math. France 133 (2005), no. 4, 619–639. 
For functions from $\mathbb{R}^k \rightarrow \mathbb{R}_+$ these results are extended in:
Nonnegative functions as squares or sums of squares
Jean-Michel Bonya, Fabrizio Brogliab, Ferruccio Colombinib, Ludovico Pernazzac (J. Func. An, 2006)
EDIT I asked J-M Bony for the scoop, and his response is that no one seems to know what the counterexample actually is, P.J. Cohen was asked about this shortly before his death, but did not remember, Bony himself says he does not even know whether a counterexample exists, and would not know which way to wager. Given that he is THE expert in the field, I would say the problem is open.<|endoftext|>
TITLE: How to get Haar measure on a compact Lie group, given the complexification?
QUESTION [9 upvotes]: This is the first in what may be a series of questions on the theme "a Banach algebraist/Bear Of Little Brain needs help with algebraic geometry".
$\newcommand{\Cplx}{{\mathbb C}}\newcommand{\fg}{{\mathfrak g}}$
Let $\fg$ be a complex semisimple Lie algebra. I have two objects on my mind:
-- the algebraic group over $\Cplx$ obtained by exponentiating $\fg$; and
-- the compact Lie group obtained by exponentiating the compact real form of $\fg$. Wikipedia
To be honest, what I'm really interested in the matrix coefficient ring of the latter, but according to what I've swotted up on, this is naturally isomorphic in a fairly precise sense to the coordinate ring of the former, in the sense of affine algebraic varieties. Let me denote this ring by $R$.
Anyway, if I think of $R$ as an algebra of smooth functions on the compact group, then there is a pairing $R\times R \to \Cplx$ given by $(f,g)\mapsto \int f(p)g(p)\,dp$ where $dp$ denotes a fixed choice of Haar measure on the compact group. My somewhat vague and naive question is:
Q1. What is the right way to "see" this pairing when I think of $R$ as the coordinate ring of an affine algebraic variety over $\Cplx$?
Here, by "see", I mean that I want some way of getting it in principle from the algebro-geometric object $R$, which is not just a concatenation of big existence results.
This may or may not require an answer to another simple-minded question:
Q2. What is the natural/canonical way of "seeing" the compact group, up to suitable isomorphism, if we are just given $R$ + Hopf algebra structure?

REPLY [6 votes]: I'll write $K$ for the compact group. A correction to start with: You write $\int_K f g$, but I assume you mean $\int_K f \bar{g}$. 
Complex conjugation We need to identify the map $\sigma: R \to R$ which has the property that $\sigma(f)|_K = \overline{\sigma(f)}$. Note that this map will NOT be complex conjugation of functions on $G$; all of the elements of $R$ are holomorphic as functions on $G$. For example, if $K=U(n)$ and $G=GL_n$, then $\sigma(f)(g) = \overline{f}(\overline{g}^{-T})$.
We cannot define $\sigma$ completely canonically, because choosing a different $K$ will change $\sigma$. However, if you have a subalgebra $\mathfrak{k}$ of $\mathfrak{g}$ which you want to be the algebra of $K$, then I think that $\sigma$ is the unique $\mathbb{C}$-antilinear Hopf automorphism (commutes with both multiplication and co-multiplication) whose induced action on $\mathfrak{g}$ fixes $\mathfrak{k}$. And, of course, an $\mathbb{R}$-subalgera $L$ of $\mathfrak{g}$ is a possible candidate for $L$ iff $\mathfrak{g} = \mathfrak{k} \otimes \mathbb{C}$ and the Killing form is positive definite  (or maybe its negative definite) on $L$.
Integration Integration is a special case of the Reynolds operator. In general, if $G$ is a reductive group, and $V$ is an algebraic representation of $G$, than $\rho: V \to V$ is the unique $G$-equivariant idempotent whose image is the $G$-invariants. I'm not sure of a really clean way to define it. One fairly algebraic way is the following: Let $v \in V$, and let $U \subset V$ be the sub-rep spanned by the $G$-orbit of $v$. Note that $U$ is finite dimensional. Let $\Omega$ denote the Casimir; let the characteristic polynomial of $\Omega$ acting on $U$ be $x^k g(x)$ with $g(0) \neq 0$; then $\rho|_U = g(\Omega)/g(0)$. Conceptually, $\rho = \lim_{t \to \infty} e^{- t \Omega}$ (the heat equation) but that's not algebraic.
In your case, you want the Reynolds operator for $G$ acting on $R$. Since I think it is easiest to define $\rho$ from the Cassimir, you need to know how to algebraically write down the action of $\mathfrak{g}$ on $R$. I think it is as follows: Let $\epsilon: R \to \mathbb{C}$ be evaluation at the identity (normally called counit). Let $\mathfrak{m}$ be the kernel of $\epsilon$, so $\mathfrak{m}/\mathfrak{m}^2 = \mathfrak{g}^{\vee}$. Then $\mathfrak{g}$ maps $R \to \mathbb{C}$ by $g \cdot f = \langle g, f-\epsilon(f)+\mathfrak{m}^2 \rangle$. This is the standard description of how to see elements of the Zariski tangent space as derivations.
We want to extend from a single vector to an entire left invariant vector field. I think the recipe is $R \stackrel{\Delta}{\longrightarrow} R \otimes R \stackrel{g \cdot \otimes 1}{\longrightarrow} \mathbb{C} \otimes R \cong R$, where $\Delta$ is comultiplication.
Hope that helps!<|endoftext|>
TITLE: Regions of Hyperplane Arrangements
QUESTION [7 upvotes]: Let $\mathcal{A}$ be an arrangement of the hyperplanes $h_1, h_2, \ldots h_n$. $\mathcal{A}$ partitions the underlying space $V$ into connected regions, denoted by $R(\mathcal{A})$. I would like to enumerate the regions using the intersection lattice $L(\mathcal{A})$ of $\mathcal{A}$.
Given a hyperplane $h \in \mathcal{A}$, we can define the following two arrangements:
$\mathcal{A}-h$ is the arrangement obtained by removing $h$.
$\mathcal{A}/h$ is the arrangement obtained by contracting to $h$; that is, the new underlying space is $h$, and the new hyperplanes are the intersections of the old hyperplanes with $h$.
It is not hard to see that $|R(\mathcal{A})| = |R(\mathcal{A}-h)| + |R(\mathcal{A}/h)|$. Indeed, each region in $R(\mathcal{A}/h)$ corresponds to a region in $R(\mathcal{A}-h)$ which $h$ cuts in two.
To review, $L(\mathcal{A})$ is the set of intersections of hyperplanes, ordered by reverse inclusion. It has bottom element $\hat{0} = V$, but only has a top element if all of the hyperplanes intersect at a point. Thus, joins (which are intersections) may fail to exist, while meets do always exist. Each element is the join of the hyperplanes below it. (For a better overview of this material, see www.math.rice.edu/~samans/ZaslavskyTheorem.pdf).
For each $x\neq \hat{0}$, let $f(x)$ be the maximal $i$ such that $h_i \leq x$, and let $h(x) = h_{f(x)}$. Define an increasing chain in $L(\mathcal{A})$ to be a sequence $\hat{0} = x_0 \triangleleft x_1 \triangleleft \cdots \triangleleft x_m$ such that $f(x_i)$ is increasing for $i\geq 1$ ($\triangleleft$ denotes covering in the intersection lattice). Note that $x_i = x_{i-1} \lor h(x_i)$. Let $C(\mathcal{A})$ denote the set of all increasing chains.
It is not too hard to see that $|C(\mathcal{A})| = |C(\mathcal{A} - h_1)| + |C(\mathcal{A}/h_1)|$, given an appropriate ordering of the atoms in $\mathcal{A}/h_1$. It then follows by induction that $|C(\mathcal{A})| = |R(\mathcal{A})|$ and that $|C(\mathcal{A})|$ does not depend on initial order of the hyperplanes.
My question is then: does there exist a "natural" bijection between $R(\mathcal{A})$ and $C(\mathcal{A})$?

REPLY [4 votes]: There is an inductive construction of the bijection you seek in a paper of Ken Jewell and Peter Orlik, in the proceedings of Arrangements in Boston (1998), published in Topology and Its Applications. Your increasing chains are in bijection with nbc (= "no-broken-circuit) sets in the arrangement. (The nbc set associated with $x_0 < \cdots < x_m$ is {f(x_0), ... , f(x_n)}, identifying hyperplanes with their labels.) Jewell and Orlik assign a chamber to each nbc set, bijectively, in Lemma 3.14 of that paper. Their bijection is built up one hyperplane at a time, that is, by deletion-contraction.<|endoftext|>
TITLE: What are the higher morphisms between enriched higher categories?
QUESTION [13 upvotes]: This question is about $n$-categories, or perhaps $(\infty,n)$-categories, or ...  My guess is that the answer will not depend sensitively on the model of higher categories, so rather than have me force you to work in my favorite model, I ask only that you say with some precision what model you are using, and that your model is not too strict.
You may assume that I have, and am comfortable with, some chosen symmetric monoidal $n$-category $(S,\otimes)$.  I am trying to understand the notion of "($n$-)category enriched in $S$".  By this I would like to mean the following.  Roughly,  $A$ is enriched in $S$ if $A$ has a set of $0$-morphisms, and between any pair of $0$-morphisms $X,Y \in A$, there is an $S$-object $A(X,Y)$ of 1-morphisms between them.  The composition should be a morphism $A(X,Y) \otimes A(Y,Z) \to A(X,Z)$ in $S$.
I am fairly satisfied that, at least in my example, I can write all of this down explicitly.  Recall that $S$ comes with a chosen object $1 \in S$.  Then the corepresentable functor $S(1,-) : S \to (n-1)\text{-Cat}$ is symmetric monoidal in an essentially unique way.  The usual thing is to use this functor, and define the de-enrichment $A_\delta$ of $A$ to be the $n$-category with hom-$(n-1)$-categories given by $A_\delta(-,-) = S(1,A(-,-))$.  Certainly in my example I can work out this $(n-1)$-category.  (De-enrichment is often a highly lossy operation, and I am OK with that.)
Suppose that $A$ and $B$ are both $S$-enriched categories.  This is where I start to run into trouble.  I understand what an $S$-enriched functor $A \to B$ is.  But I'm having trouble figuring out what is the correct definition of "natural transformation", and in general of the higher morphisms.

Question: Given a symmetric monoidal $n$-category $S$, what is the $(n+1)$-category of $S$-enriched ($n$-)categories?  In particular, what are the higher morphisms?

I recognize that this "$(n+1)$-category of $S$-enriched categories" is likely itself enriched in $S$-enriched categories.  But I am interested simply in writing down its de-enrichment — I'm looking for it just as an $(n+1)$-category.
A final remark: The notion of enrichment in this question is not the same as in n-categories enriched in an (n+1)-category.  That question concerned $n$-categories in which the collection of $n$-morphisms was an object of the enriching category.  I would like the collection of $1$-morphisms to be an object of the enriching category.

REPLY [3 votes]: Another way to get the definition of an $S$-transformation, if you know what an $S$-functor is and also the tensor product of $S$-categories (so $S$ must be symmetric) and $S$ has an initial object preserved by $\otimes$ in each variable, is as an $S$-functor
$$ C \otimes \mathbf{2} \to D$$
where $\mathbf{2}$ is the $S$-category with two objects $a$ and $b$, hom-objects $1$ from $a$ to $a$, $b$ to $b$, and $a$ to $b$, and the initial object of $S$ from $b$ to $a$.<|endoftext|>
TITLE: What is the minimal number of symmetric generators of the full matrix algebra?
QUESTION [8 upvotes]: Is there any lower bound known for the minimal number of generators needed to generate the full matrix algebra of real $n\times n$ matrices — when using only symmetric matrices for the generators?
Analogous question for complex matrices — when using only Hermitian matrices for the generators.
I am aware that $3$ generators suffice when using only idempotent generators. This is a result of Naum Krupnik (Minimal number of idempotent generators of matrix algebras over arbitrary field, Comm. Algebra 20 (1992), no. 11, 3251–3257). (Tandfonline link, restricted access)
I am not familiar with this type of results, so this might be well known or easy.
Thanks for any tips.

REPLY [3 votes]: Here is a more direct argument, without Burnside's Theorem.  As in M. Karimi's answer, let $A$ be diagonal with distinct entries $\lambda_1,\dotsc,\lambda_n$, and let $B$ be the matrix with all entries equal to $1$.  Note that 
$$ (A^iBA^j)_{st} = \lambda_s^i\lambda_t^j $$
Now let $M$ be an arbitrary $n\times n$ matrix.  By Lagrange interpolation, there is a unique polynomial 
$$ p(x,y) = \sum_{i,j=0}^{n-1} a_{ij}x^iy^j $$ 
with $p(\lambda_s,\lambda_t)=M_{st}$ for all $s$ and $t$.  It is now easy to check that 
$$ M = \sum_{i,j=0}^{n-1} a_{ij}A^iBA^j.  $$

REPLY [2 votes]: As an answer, the lower bound is exactly $2$. To this end take a diagonal matrix of order $n$ with all distinct and non zero entries along diagonal and the matrix of all ones of order $n$. In this way the algebra generated by these two symmetric matrices is simple(i.e. irreducible ), Now  use Burnside's theorem to deduce that it should be $M_n(C)$. See 
Laffey, Thomas J., A structure theorem for some matrix algebras, Linear Algebra Appl. 162-164, 205-215 (1992). ZBL0758.16010.<|endoftext|>
TITLE: Cyclotomic polynomials evaluated at roots of unity
QUESTION [19 upvotes]: Dear MO_World,
I'm working on an ergodic theory question (about a generalization of eigenfunctions for measure-preserving transformations) and have run into a number theory question concerning cyclotomic polynomials that I'm unable to tackle.
The question is this:

Let $p$ be a prime and let $p|n$. When is it the case that $\Phi_n(e^{2\pi i/p})=\pm e^{2\pi ij/p}$ for some $j$?

Here $\Phi_n$ denotes the $n$th cyclotomic polynomial.
I've experimented with Mathematica and have found there are non-trivial cases in which the condition holds, whereas for most cases it does not seem to hold.
Letting $c(n,p)=\Phi_n(e^{2\pi i/p})$, we have $c(105,3)=1$, but $c(105,5)$ and $c(105,7)$ are not on the unit circle. None of $c(15,3)$, $c(21,3)$, $c(15,5)$, $c(35,5)$, $c(21,7)$, $c(35,7)$ are on the unit circle; $c(40,2)=1$, but $c(50,2)=5$...
Not surprisingly it seems to be easiest for the condition to hold for small $p$.
Also, using the relations $\Phi_{p^2n}(x)=\Phi_{pn}(x^p)$; and $\Phi_n(1)=q$ if $n=q^k$ for some prime $q$ and an integer $k$, but $\Phi_n(1)=1$ otherwise, it's not hard to see that the condition holds whenever $p^2|n$, but $n$ is not a power of $p$.
Thanks for any more systematic suggestions...

REPLY [15 votes]: Set $\zeta=e^{2\pi i/p}$. By what Anthony wrote, we may assume $n=pm$ where $p$ does not divide $m$. Note that if $\Phi_n(\zeta)=\pm\zeta^j$, then $\Phi_n(\zeta^k)=\pm\zeta^{jk}$ for each $k$ prime to $p$, because the absolute Galois group of $\mathbb Q$ acts transitively on the $\zeta^k$s (by irreducibility of $\Phi_p$).
Let $r$ be the radical of $m$ (the product of the distinct prime divisors of $m$). From $\Phi_n(\zeta)=\Phi_{pr}(\zeta^{m/r})$, we see that we furthermore may assume that $m$ is squarefree.
We cannot have the case $\Phi_n(\zeta)=-\zeta^j$ unless $p=2$: Suppose that happens. Then $\zeta$ is a root of $\Phi_n(X)+X^j$. On the other hand, $\Phi_p$ is irreducible, so $\Phi_p$ divides $\Phi_n(X)+X^j$. But then $p=\Phi_p(1)$ divides $\Phi_n(1)+1=2$, so $p=2$.
Using the formula $\Phi_n(X)=\prod_{d\mid n}(X^d-1)^{\mu(n/d)}$, we get $\Phi_{pm}(\zeta)=\Phi_m(1)/\Phi_m(\zeta)$. Clearly, if $m$ is a prime, then $\lvert\Phi_m(1)/\Phi_m(\zeta)\rvert>1$. Thus $m$ is not a prime, so $\Phi_m(1)=1$, and $\Phi_m(\zeta)$ is $\pm$ a power of $\zeta$.
Now $\Phi_m(\zeta)$ depends only on the residue class of each prime divisor of $m$ modulo $p$. So given $p$, and the number $t$ of prime factors of $m$, one only has to check which of the $(p-1)^t$ possibilities work. For instance, if each prime divisor of $m$ is $\equiv1\pmod{p}$, then  $\Phi_m(\zeta)=1$. So by Dirichlet, for each prime $p$, there are infinitely many cases.
I don't believe that a more explicit description of the possibilities for $m$ is possible.
Added: Under the assumptions from above ($m$ is squarefree, prime to $p$, and not a prime) $\Phi_m(\zeta)$ (and then also $\Phi_{pm}(\zeta)$) is a power of $\zeta$ whenever $m$ has a prime divisor which is $\equiv1$ or $\equiv-1\pmod{p}$. This follows by induction from the identity $\Phi_m(X)=\Phi_k(X^q)/\Phi_k(X)$ for $m=kq$ and $q$ a prime.
Probably, this sufficient condition is necessary too. I only checked it if $m=uv$ with $u,v$ distinct primes: $\Phi_m(\zeta)=\zeta^j$ gives that $X^p-1$ divides $(X^{uv}-1)(X-1)-X^j(X^u-1)(X^v-1)$, and this easily yields $u\equiv\pm1\pmod{p}$ or $v\equiv\pm1\pmod{p}$.<|endoftext|>
TITLE: Convenient definition of "category of Riemannian manifolds"?
QUESTION [13 upvotes]: Has a notion of "category of Riemannian manifolds" been defined and used in the literature?
For which reasons is it or would it (not) be a useful notion?
I think the objects should be all (perhaps complete) Riemannian manifolds, and two objects should certainly be isomorphic if they are isometric as Riemannian manifolds.
Which "should" be the morphisms of such a category? 
I think some possibilities are:
1) isometries
2) local isometries
3) finite compositions of local isometries and Riemannian submersions
4) conformal maps
5) any of the above localized at local isometries
I tagged it "soft question" because I don't have in mind any specific application of this notion.

REPLY [17 votes]: I'm sure the answer to your question is "it depends on the application".  Here are three categories that come to (my idiosyncratic) mind.
Perhaps the most general category in the direction you're looking is a version of Lawvere's category of metric spaces.  Recall that $\mathbb R_{\geq 0}$ is a category, on account of the fact that it is a poset: there is a unique morphism $x\to y$ whenever $x\geq y$.  It can be given a symmetric monoidal structure by declaring that $\otimes = +$.  Then an $(\mathbb R_{\geq 0},+)$-enriched category is nothing but a (generalized) metric space.  The natural "functors" are the distance-non-increasing maps.
So I would suggest that a good guess, if you must make a guess, for a category of Riemannian manifolds has as its objects all Riemannian manifolds $(M,g_M)$ (of whatever regularity you like) and its morphisms $f : (M,g_M) \to (N,g_N)$ are smooth maps $M \to N$ such that at each $m\in M$, the symmetric bilinear form $g_M - f^* g_N$ on the tangent fiber at $m$ is positive-semidefinite.  This is sort of an "infinitesimal" version of the Lawvere one.  Or just notice that every Riemannian manifold gives a metric space, and use the distance-nonincreasing maps (in infinite dimensions this is more subtle, as there are many important examples in which distinct points are connected by arbitrarily short paths).
But here are two other categories "of Riemannian manifolds" that are important in quantum field theory:

In one approach to understanding $n$-dimensional quantum field theory (in Euclidean rather than Lorentzian signature), one constructs a category whose morphisms are $n$-dimensional compact Riemannian manifolds with boundary, and whose objects are germs of $n$-dimensional Riemannian manifolds around $(n-1)$-dimensional compact manifolds.  For details on this approach, one should start with the work of Stolz and Teichner.
A category I am particularly fond, which provides a "weaker" notion of $n$-dimensional Riemannian quantum field theory, of has as its objects $n$-dimensional open Riemannian manifolds, and its morphisms are isometric embeddings.

And I'm sure that there are other equally interesting categories, especially if you are interested in infinite-dimensional manifolds, which I haven't really even touched.<|endoftext|>
TITLE: In search of a set theory with specific properties
QUESTION [5 upvotes]: I'm in search of a set theory that satisfies the following requirements.

There is a universal set $V$ such that $\forall x(x \in V)$. So for example, $V \in V$.
Sets whose elements are 'large' exist. e.g. I want $\lbrace V,\emptyset\rbrace$ to be a well-defined set with cardinality $2$.
[Edit] Sets form a boolean algebra; in particular, the complement of a set always exists.

REPLY [5 votes]: Here is a CW answer incorporating the answers in the comments.  Both NF (New Foundations) and NFU (New Foundations with urelements, obtained from NF by weakening the axiom of extensionality) satisfy conditions (1) through (3).<|endoftext|>
TITLE: Inter-Kissing Number for Spheres of Different Sizes
QUESTION [5 upvotes]: What is the maximum number of spheres that can be placed in 3D such that all inter-touch?
One can of course place four unit spheres tetrahedrally and then add a smaller sphere in the
middle, so this number must be at least 5.
[By the way, I was trying to extend the "five points in 2D cannot be inter-connected without
a crossing" limitation to 3D with a simple statement, but this was sadly the best I could
do.  If anyone knows a better simple extension, please comment.]

REPLY [12 votes]: In $\mathbb R^n$, the answer is $n+2$.
You can apply an inversion which sends two of the spheres in to two parallel hyperplanes.
The rest of the spheres will have the same radii and their centers lie in a hyperplane.
Hence everything follows.<|endoftext|>
TITLE: What is the importance of the conjectural semi-simplicity of the action of the Frobenius on the etale cohomology of a variety over a finite field ?
QUESTION [7 upvotes]: It is conjectured that the action of the Frobenius acting on the etale cohomology of an algebraic variety over a finite field is semisimple.
A first approximation of my question is :
What is the importance of this statement ?
It is only an approximation because this question is rather unprecise.
More precisely, I would like to know two different things :
1) Is there some conjectural statements which are implied by the semisimplicity conjecture ?
More generally :
2)What the semisimplicity property means ?
I ask for an arithmetic or geometric interpretation of the semisimplicity of the Frobenius.
For example, the eigenvalues of the Frobenius are important because they control the number of points of the variety over finite fields. I have the impress that often people only care of eigenvalues of the Frobenius (often there are remarks of the form : "if the representation is not semisimple, take the semisimplification"). Is there some situation 
where the conjectural semisimplicity would be an important thing ?
Remark : I don't want an answer like "it is important because it is implied by the standard conjectures and so if it would be true it would be a confirmation of a part of a general philosophy ..." I would like some "concrete" example of the use of the semisimplicity property (maybe in the cases where it is actually known : abelian varieties ...)

REPLY [6 votes]: Here is a conjectural statement that follows from  the combination of the semisimplicity conjecture  and the Tate conjecture on algebraic cycles.
Let $X$ and $Y$ be geometrically irreducible smooth projective varieties over a finite field $F$. Suppose that $X$ and $Y$ have the same number of $F^{\prime}$-points for all finite overfields $F^{\prime}$ of $F$. Then there exist a finite overfield $F_0$ of $F$ and a geometrically irreducible closed $F_0$-subvariety $Z\subset X \times Y$ such that 
$\dim(X)=\dim(Y)=\dim(Z)$ and
both projections map $Z \to X$ and $Z \to Y$ are surjective.
The semisimplicity is known to be true for abelian varieties (Weil). Combining the semisimplicity with Tate's theorem on homomorphisms, one may deduce that if $X$ and $Y$ are abelian varieties with the same number of $F^{\prime}$-points (for all $F^{\prime}$) then they are isogenous over $F$.<|endoftext|>
TITLE: Complex structures on a K3 surface as a hyperkähler manifold
QUESTION [6 upvotes]: A hyperkähler manifold is a Riemannian manifold of real dimension $4k$ and holonomy group contained in $Sp(k)$. It is known that every hyperkähler manifold has a $2$-sphere $S^{2}$ of complex structures with respect to which the metric is Kähler.
A K3 surface is a hyperkähler manifold of real dimension $4$. It is classic in algebraic geometry that its complex structure is parametrized by $\mathcal{D}_{K3}/\Gamma$ the period domain mod some arithmetic group (you may want to impose polarization). Note that here we don't think of the K3 surface as a Riemannian manifold. 
My question is, are there any relation between the $2$-sphere $S^{2}$ above and the moduli space $\mathcal{D}_{K3}/\Gamma$? For example, can the moduli space be foliated by such $S^{2}$?

REPLY [10 votes]: These $2$-spheres are called 'twistor lines'. They indeed cover the moduli space (in the non-polarized case) : more precisely, any two points of the moduli space may be linked by a chain of twistor lines.
A reference where this is nicely explained (and used !) is Huybrecht's Bourbaki talk about Verbitsky's Torelli theorem : http://arxiv.org/abs/1106.5573. More precisely, Definition 3.3 gives a lattice-theoretic definition of twistor lines, the link with your description of twistor lines is made in paragraph 4.4, and the result I mentionned above is Proposition 3.7.
In the polarized case, no twistor line is included in the moduli space, as a general member is not projective : see remark 8.1 of http://arxiv.org/abs/1009.0413. This article is particularly interesting in this respect. Indeed, Charles and Markman prove the standard conjectures for some projective hyperkähler varieties (a statement peculiar to projective varieties) using deformations along a twistor line (hence using non-projective varieties).<|endoftext|>
TITLE: Is $\partial X$ a sphere for $X$ a complete CAT$(0)$ space?
QUESTION [16 upvotes]: Let $X$ be a complete CAT$(0)$ metric space, and $\partial X$ its boundary.
One way to define $\partial X$ is as the equivalence class of geodesic rays 
$\gamma(t), \gamma'(t)$
that remain within a constant distance of one another for large $t$.

Under what conditions and for which $n$
  is it known that the boundary of a complete CAT$(0)$ $n$-manifold
  is homeomorphic to the $(n{-}1)$-sphere $\mathbb{S}^{n-1}$ ?

I believe this is known if $X$ is a complete $n$-dimensional Riemannian manifold
of nonpositive sectional curvature, but I have not found clear counterexamples otherwise.
I am especially interested in $n{=}3$.  Pointers would be appreciated, as this area
is relatively new to me.  Thanks! 
Answered. Here is a snippet from the Davis-Januszkiewicz paper Igor cites,
describing an $n{=}5$ example where $\partial X \neq \mathbb{S}^4$:
    
I would still be interested to learn if a similar example is known for $n < 5$.

REPLY [13 votes]: For a piecewise Euclidean or piecewise hyperbolic metric on a PL manifold, the answer is YES. This is proved (p348 in published version) by M. Davis and T. Januszkiewicz in 
M. Davis, T.Januszkiewicz, Hyperbolization of polyhedra. 
J. Differential Geom. 34 (1991), no. 2, 347–388. Link: projecteuclid or (non-paywalled) Davis' page.
For a topological manifold with a piecewise Euclidean or piecewise hyperbolic metric, the answer is NO, as shown in the same paper (Section 5).<|endoftext|>
TITLE: Chern classes of ideal sheaf of an analytic subset
QUESTION [12 upvotes]: Let $X$ be a Kähler manifold of dimension $n$, and $Z \subset X$ an analytic subset of codimension $k$. I have read in a paper the following result, a proof of which I cannot find:
$$c_k(\mathscr{I}_Z) = (-1)^k(k-1)![Z]$$
The form of the expression suggests using GRR, but I cannot figure out how. 

Does anyone know a proof for this (or at least an online reference)?

REPLY [13 votes]: In Fulton's book Intersection Theory, Theorem 15.3 and Example 15.3.1 (p. 297 of my edition) it is proven, by using GRR, that $$c_k(\mathscr{O}_Z)=(-1)^{k-1}(k-1)![Z].$$
By using the short exact sequence
$$0 \to \mathscr{I}_Z \to \mathscr{O}_X \to \mathscr{O}_Z \to 0$$
we obtain $1=c(\mathscr{O}_X)=c(\mathscr{O}_Z)c(\mathscr{I}_Z)$ and this in turn  implies $$c_k(\mathscr{I}_Z)=(-1)^k (k-1)![Z].$$<|endoftext|>
TITLE: Hilbert style axioms for Euclidean and/or hyperbolic geometry without reference to congruence?
QUESTION [8 upvotes]: Hilbert's axioms from Grundlagen der Geometrie involve notions of incidence, between-ness, segment congruence and angle congruence.  
Consider the sub-theories of either Euclidean or hyperbolic geometry involving only the notions of incidence and between-ness.  Of course, a priori, notions of congruence may occur in proofs  (though not statements) of theorems solely about incidence and between-ness.  
As a matter of fact, consider any 2-dimensional compact convex body $B$, with every boundary point extreme, say, but $B$ not affinely equivalent to a round disk.  One can associate to $B$ an incidence geometry after the manner of Beltrami-Klein (with "lines" equal to interiors of chords).  Such a geometry will not have an isomorphism to hyperbolic or Euclidean incidence geometries even though Hilbert's axioms about incidence and between-ness will hold!  Indeed the incidence theory of such a quasi-Beltrami-Klein model actually determines $B$ up to an affine transformation.  (Has this fact been recorded in the literature?)
Main Question:  Can one formulate a finite, or at least an elegant set of incidence and between-ness axioms, extending Hilbert's, so as to capture the theory of Euclidean and/or hyperbolic incidence?
Added later: In light of Will Jagy's comment I'll sketch very briefly why I think the incidence geometry determines $B$ as much as it does, in case I'm obviously wrong.  Given $B$, one can decorate it in many way with three families of evenly spaced parallel chords with triple intersections wherever intersections occur.  The infinite of maximal configurations of this sort constitute approximations sufficient to recover $B$ up to affine.

REPLY [5 votes]: The answer given to this question in the 20th century  is quite complex, and a summary of what was done can be found in  Victor Pambuccian, The axiomatics of ordered geometry
I. Ordered incidence spaces, Expositiones Mathematicae 29 (2011) 24–66.<|endoftext|>
TITLE: Elementary Proof of Basis of Order k
QUESTION [7 upvotes]: Context
According to the FAQ, questions of the form "the sorts of questions you come across when you're writing or reading articles or graduate level books" are acceptable. This falls into the "reading graduate level books."
Problem Statement
Let $N$ be the natural numbers.
$B \subseteq N$ is a basis of order $k$ if $N \setminus kB$ is finite.
I would like to show that there is a basis $B$ of order $k$ s.t.
$|B \cap [1,n]| = O(n^{1/2} \log^{1/k} n)$.
What I've tried
Suppose all we needed was $O(n^{1/2} \log^{1/2} n)$, then I would define $B$ by randomly sampling from $N$ s.t.
$$P(n \in B) = \frac{c\log^{1/2} n}{\sqrt{n}}$$
By the chernoff bound, with high probability we have $|B \cap [1,n]| = O(n^{1/2}\log^{1/2} n)$.
Furthermore, for any $n$, there does not exists $a,b\in B$ s.t. $a+b=N$ with probability at most $(1-\frac{c\log n}{n})^{n/2} \leq 1/n^2$, and we're done.
Unfortunately, however, I need to push this down to $O(n^{1/2}\log^{1/k} n)$.
What I'm stuck on
So far, I've only used $B$ as a order 2 base, rather than an order $k$ base.
Question:
What should I be looking at to go from order 2 to order $k$ and $\log^{1/2} n$ to $\log^{1/k} n$?

REPLY [2 votes]: To elaborate on the above comment, the problem with order $k$ bases is precisely that Chernoff's inequality does not work. The joint independence assumption for Chernoff's inequality is essential; as seen by the following example taken from Tao and Vus' Additive Combinatorics:
Color the elements of $[1, N]$ either black or white independently and with equal probability. For each $A \subset [1, N]$ let $s_A$ denote the parity of black elements of $A$ (so say if $A$ contains 3 black elements then $s_A = 1$). One can check that the $s_A$'s are independent events. Write $X = \displaystyle \sum_{A \subset [1, N]} s_A$. One can check that $\mathbb{E}X = 2^N - 1/2$ and $\textbf{Var} X = 2^{N-2} - 1/4$. Further, $\mathbb{P}(X = 0) = 2^{-N}$. The upper-bound on Chernoff's inequality would be $2\exp(-2^{N-2})$, which is much smaller than $\mathbb{P}(X = 0)$, so the inequality fails.
The reason why a simple argument suffices for additive bases of order 2 is because we have
$$\displaystyle r_{2,B}(n) = \sum_{x < n/2} \mathbb{I}(x \in B) \mathbb{I}(n - x \in B) + E$$
where $E$ is a suitably small error, and $r_{2,B}(n)$ is the number of ways to write $n$ as a sum of two elements in $B$. The key here is that the events $\mathbb{I}(x \in B) \mathbb{I}(n - x \in B)$ are independent for $1 \leq x < n/2$. This is not the case when there are more summands. In the Erdos-Tetali paper cited above, this issue is circumvented via Janson's inequality. In particular, Erdos-Tetali showed that there are additive bases of order $k$ satisfying $| B \cap [1,N]| = \Theta(N^{1/k} \log^{1/k} N)$. 
The main difficulty you have to circumvent is how to deal with the non-independence of the random variables $\mathbb{I}(x_1 \in B) \cdots \mathbb{I}(x_k \in B)$.<|endoftext|>
TITLE: Inter-Kissing Number for Non-Spheres
QUESTION [5 upvotes]: In 3D, the maximum number of spheres which can inter-touch is $5$ (MO question Inter-Kissing Number for Spheres of Different Sizes).  This maximum reduces to $4$ for unit spheres.
Is there a different shape (e.g., an egg, or a pyramid) for which these maximums are not $5$ and $4$?  If so, what shape has the highest maximum?
To avoid "corner touching" (e.g., $8$ cubes could all touch at one corner), please additionally require that every "touch-point" have only one "official connection" (e.g., only $2$ of the $8$ cubes can be declared as touching at the corner).

REPLY [11 votes]: There is no upper bound. There can be arbitrarily many congruent convex solids which pairwise touch face-to-face. See Erickson, J. Kim, S. "Arbitrarily large neighborly families of congruent symmetric convex 3-polytopes," for many references.
    
 (source)<|endoftext|>
TITLE: Maximal subgroups of a certain finite 2-group
QUESTION [10 upvotes]: The following came up in a problem on reconstruction of digraphs.  I determined enough about the answer to satisfy the application completely, but still I am curious to know what the complete solution is.  My group theory is weak, so apologies if this is too simple.
Let $T$ be a full binary tree with depth $k$. Call its levels $L_0,\ldots,L_k$.  Here is the case $k=4$:
      (source)
The number of leaves is $n=2^k$.
Let $A$ be the full automorphism group of $T$ and let $f$ be its (faithful) action on the leaves of the tree, i.e. on $L_k$. Obviously $f(A)$ is an iterated wreath product of $\mathbb{Z}_2$ with itself and has order $2^{n-1}$.  It is, indeed, the Sylow 2-subgroup of $S_n$.
The problem is: what are the subgroups of $f(A)$ of index 2?
Here is what I think the answer is.  Let $X$ be a union of levels of $T$, including at least one level other than $L_0$. Let $P$ be the set of all $f(\gamma)$ such that $\gamma\in A$ and the action of $\gamma$ on $X$ is an even permutation.  Then $P$ is a subgroup of the desired index. I'm guessing there are no others...

REPLY [9 votes]: You can easily count the number of maximal subgroups of $W(k)$, the $k$-fold iterated wreath product of $\mathbb{Z}_{2},$ by calculating the index of the Frattini subgroup. You can inductively prove that the number of generators is $k,$ which is clear for $k =1,2.$
To proceed, note that $W(k) = W(k-1) \wr \mathbb{Z}_{2}.$ Factor out the Frattini subgroup of the base group, and by induction, you are left with $E(k-1) \wr \mathbb{Z}_{2}$, where $E(k-1)$ is elementary Abelian of order $2^{k-1}.$ If $x$ is an element of order $2$ outside the new base group, then $[E(k-1) \times E(k-1),x]$ has order $2^{k-1},$ so that the largest elementary Abelian factor group of the original wreath product does have order $2^{k},$ as claimed.
Hence the group $W(k)$ has $2^{k}-1$ maximal subgroups, since there is a bijection between maximal subgroups of $W(k)$ and maximal subgroups of $W(k)/\Phi(W(k)).$<|endoftext|>
TITLE: Definitive source about Dirichlet finally proving the Unit Theorem in the Sistine Chapel
QUESTION [38 upvotes]: (This question was posted on math.stackexchange a week ago at https://math.stackexchange.com/questions/187315/definitive-source-about-dirichlet-finally-proving-the-unit-theorem-in-the-sistinbut and still has received no answers there, so I hope it's okay to post it here too.) 
In a few books or survey articles (e.g., page 49 of Helmut Koch's "Number Theory: Algebraic Numbers and Algebraic Functions") it is said that Dirichlet figured out a proof of the unit theorem while listening to an Easter concert in the Sistine Chapel. My question is: what is the evidence for this story?
From an internet search I found that Kummer wrote on p. 343 of volume 2 of  Dirichlet's collected works that Dirichlet could work on math in all kinds of situations, and then Kummer says "Als Beispiel hierfür kann ich anführen, dass er die Lösung eines schwierigen Problems der Zahlentheorie, womit er sich längere Zeit vergeblich bemüht hatte, in der Sixtinischen Kapelle in Rom ergründet hat, während des Anhörens der Ostermusik, die in derselben aufgeführt zu werden pflegt" (translation: "As an example I can say that he found the solution to a difficult problem in number theory, which he had worked on for a considerable amount of time without success, in the Sistine Chapel in Rome while he was listening to the Easter music that tends to be played there.") 
Notice Kummer does not say precisely what the "difficult problem" was. Maybe it is just an oral tradition that the problem is the unit theorem, but I would like a more definitive source. 
I don't read German well, but if you do then Kummer's essay on Dirichlet can be read online. It starts on 
http://archive.org/stream/glejeunedirichl00dirigoog#page/n323/mode/1up
and page 343 is 
http://archive.org/stream/glejeunedirichl00dirigoog#page/n355/mode/1up.

REPLY [24 votes]: In a letter to Gauss, on January 3, 1840 (Werke II), Dirichlet stated the unit theorem in the same version as he would later publish it, along with the comment that the proof was very easy and could be given on two or three pages. Indeed the proof he published in 1846 was little more than 2 pages long. In addition, the French version of his proof was contained in a letter to Liouville, and was published already in 1840.  I think this shows convincingly [See the Edit below] that the problem he solved in the Sistine Chapel in 1844 was not related to the unit theorem.
On the other hand Kummer, in his talk to the memory of Dirichlet, said that Dirichlet worked out the class number formulas for the equivalence classes of forms corresponding to fields of the p-th roots of unity while he was in Italy; in fact his articles on units that Dirichlet published before his journey to Italy are directly related to problems showing up in such an investigation. When Dirichlet heard that Kummer was also working out the class number formulas for such fields, he decided not to publish anything on this subject.
Edit. I'd like to add a few quotations from Kummer's papers concerning Dirichlet's results obtained in Italy.

1844, De numeris complexis.

For most numbers $\lambda$, the investigation of all units is very difficult and requires particular principles, which so far we have not yet sufficiently examined. We may skip this question, however, since we have heard that recently Lejeune-Dirichlet in Italy, where his still dwells, has worked out fundamental theorems on these complex units that we eagerly expect him to publish soon.

1846, Zur Theorie der Zahlen

p. 324: This investigation on the real and ideal complex numbers is completly
identical to the classification of certain related forms of degree  $\lambda-1$ in $\lambda-1$ variables, of which Dirichlet has found but not yet published the main results.
p. 325: I have so far not yet studied this area of the theory of complex numbers
in depth; in particular I have not yet worked out the determination of the true number of classes  since I have heard by word of mouth that Dirichlet, using principles similar to those in his famous memoirs on quadratic forms, has already found this number. 

1846, Zerlegung der Wurzeln der Einheit:

The complete determination of those powers of ideal numbers that become real, as well as the determination of the number of inequivalent ideal numbers, requires principles that differ essentially from those given in the present memoir. We do not discuss this important question any further since, as we have already mentioned, the publication of an article by Dirichlet is imminent in which he has completely solved this question for a closely related topic.


Beweis des Fermatschen Satzes


Assume that (A) the class number of $K = {\mathbb Q}(\zeta_p)$ is not divisible 
by $p$, and (B) that each unit in $K$ that is congruent to a rational 
integer modulo $p$ is actually a $p$-th power in $K$. Then the equation 
$x^p + y^p = z^p$ does not have nontrivial solutions in integers.
This article was presented to the Academy of Berlin by Dirichlet, who 
added the following remark. 
The truth of the second assumption of Kummer's astute proof may be 
verified for each given value of $p$ using the general theory of 
complex units, about which I have given a few hints in the report 
from March 1846 and which will be published in one of the next few
volumes of Crelle's journal.
After I had proved that every unit may be represented by $\frac{\lambda-3}2$
fundamental units, which is the result analogous to the general solution
of Fermat's equation $x^2 - Dy^2 = \pm 1$, it was natural to follow
this analogy between quadratic and these higher forms by determining
the number of the latter forms by methods similar to those with which
the same question in the theory of quadratic forms was solved. This
investigation, to which Kummer referred at the beginning of his note,
was luckily completed three years ago with the help of a new principle
that was not necessary for the determination of the number of forms
of the second degree, and has led to a result that is remarkable in its
form and which is as simple as may be expected for a result that 
encompasses forms of all degrees. The expression for the number for forms
of degree $\lambda-1$ that I have found, which, as may be expected by
the analogy with quadratic forms, contains the $\frac{\lambda-3}2$
fundamental units, provides us, as soon as these units are known, with
a method to verify assumption (A) by a rather simple numerical calculation.
Edit (Oct. 2016) I retract my analysis above and claim the opposite: Dirichlet indeed did work out the general unit theorem during his stay in 
Italy. In the preface to vol. I of Dirichlet's Collected Works, Kronecker writes
"Dirichlet read this memoir a year after his return from Italy, but the
fundamental and far-reaching investigations about which he made a few
remarks there he had completed - as I have learned from himself - already
during his stay in Italy."
And in Dirichlet's article "On the theory of units", the last one in vol. I of his Collected Works, he writes
"For the degrees after the second this theorem could be proved without
great problems, and we have presented the result concerning the third
degree in an earlier communication (Monatsbericht October 1841) 
several years ago. The proof of the theorem in its full generality, which
we have found by induction, was obstructed by the greatest difficulties
that could be overcome completely only after many fruitless attempts.
Continued occupation with this problem then allowed me to simplify the
proof to such an extent that its prinicpal moments may be given in a
comprehensible manner in just a few words."
What follows is a description of how to construct the correct number of independent solutions of the unit equation, and it seems that what had eluded him in his early investigations is the role that the regulator is playing in this construction. In his earlier articles Dirichlet announced the general unit theorem, but gave details of the proof only for cubic extensions without stating clearly that he did not have a proof in the general case.<|endoftext|>
TITLE: Elementary computation of direct image sheaves.
QUESTION [8 upvotes]: I am a physicist and would like to understand the section 1 of 
this math paper, which explains how the SYZ conjecture implies topological mirror symmetry. I have some technical problem and would appreciate your help. My question is purely mathematical and nothing scary to most of you, I think. 
The setting is the following; let $f:X\rightarrow B$ be a 3-torus $T^3$-fibration of a Calabi-Yau threefold (which has singular fibers). Since $X$ is a real sixfold, $B$ is a real threefold. Under certain technical assumption, using Leray spectral sequence, the author shows that the dual torus fibration gives another Calabi-Yau threefold with mirrored Hodge diamond.
My questions concern the Leray spectral sequence associated with $f$. First I don't quite understand why $E_2$ of the Leray spectral sequence looks like the diagram in the paper; how can we conclude that
$$
H^{i}(B,R^{j}f_{*}\mathbb{R})=\mathbb{R} \ (i=0,3), \ \ 0 \ (i=1,2)
$$
for $j=0,3$? My understanding is that roughly the direct image sheaves $R^{i}f_{*}\mathbb{R}$ are local system, so on a small open subset we have $R^{i}f_{*}\mathbb{R}\cong H^{i}(T^{3},\mathbb{R})$. How should we understand the computation above? 
Secondly why $\pi_{1}(X)=0$ (or  rather $H^{1}(X,\mathbb{R})=H^{5}(X,\mathbb{R})=0$) implies that 
$$
H^{0}(B,R^{1}f_{*}\mathbb{R})=H^{3}(B,R^{2}f_{*}\mathbb{R})=0?
$$
I know that the problem is caused by my poor understanding of direct image sheaves. I hope to make my understanding clearer with this example. 
I am afraid that my questions are elementary and this may not be a place to ask such a question. Any help, comments and reference suggestions will be appreciated.
Thank you,

REPLY [5 votes]: Edited following algori's comment.
In the context of the paper I think one can make these deductions on rather formal grounds - deep understanding of higher direct images is not needed.
Gross assumes that the base $B$ is a simply connected closed 3-manifold (i.e., it's $S^3$). 
Let $i\colon B_0\to  B$ be the inclusion of the (open and dense) regular locus, and $f_0\colon f^{-1}(B_0)\to B_0$ the restricted fibration. Gross also tells us that $f$ is "simple", meaning that $i_*(R^jf_{0\ast} \mathbb{R})\cong R^j f_\ast \mathbb{R}$ for each $j$. In degree $j=0$, simplicity is automatic: the fibers of $f_0$ are connected, and since disconnectedness of a fiber is an open property on $B$, the fibers of $f$ are also connected. So $f_{0\ast}\mathbb{R}=\mathbb{R}$ and $f_{\ast}\mathbb{R}=\mathbb{R}$.
The fibers of $f_0$ are also oriented. Moreover, after choosing orientations for $X$ and $B$ we have a consistent notion of a "positive" fiber-orientation: one which, after wedging with the pullback of a positive orientation form on $B$, agrees with the orientation of $X$. Hence $R^3f_{0\ast}\mathbb{R}=\mathbb{R}$. By simplicity, $R^3 f_{\ast}\mathbb{R}\cong \mathbb{R}$. We can therefore complete the $0$th and $3$rd rows of $E_2$ - each is the de Rham cohomology of $S^3$:
$$ \mathbb{R}\quad 0\quad 0 \quad \mathbb{R}.$$ 
The spectral sequence converges to $H^\ast(X;\mathbb{R})$, and we know that $E_2^{2,0}=0$. We must have $E_2^{0,1}=0$ because otherwise it would survive to $E_\infty$ and we would have $H^1(X;\mathbb{R})\neq 0$. Similarly, since $E_2^{1,3}=0$ and $H^5(X;\mathbb{R})=0$, we must have $E_2^{3,2}=0$.<|endoftext|>
TITLE: Without choice, can every homomorphism from a profinite group to a finite group be continuous?
QUESTION [20 upvotes]: In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in this question. However, all three constructions given use consequences of the axiom of choice: ultrafilters in the first answer, and "every vector space has a basis", in Milne's notes as referenced in the second answer, and used to compute the number of finite-index subgroups in the third answer.
Is it possible to prove the existence of a discontinuous homomorphism from a profinite group to a finite group without the axiom of choice? Instead is it consistent with ZF that there is none?

REPLY [13 votes]: There are two common definitions of what it means for a topological group to be profinite, which I’ll distinguish as “formally profinite” and “compact profinite”.

Formally profinite: an inverse limit of finite groups, with the inverse limit topology
Compact profinite: a compact Hausdorff totally disconnected topological group

By compact I mean that any open cover has a finite subcover. ZF proves that every compact profinite group is formally profinite: it’s a limit of quotients by normal open subgroups. The other implication is not a theorem of ZF. The compactness is equivalent to the Boolean Prime Ideal Theorem.
ZF proves the existence of a discontinuous homomorphism from a certain formally profinite group to a finite group. I don’t know the answer with compact profinite groups, though there are models ruling out some classes of discontinuous map. $\DeclareMathOperator{colim}{colim}$

Let $k=\mathbb F_2.$

Theorem. (ZF) There is a formally profinite group $V$ and a discontinuous group homomorphism $V\to k.$

If we can find a $k$-vector space $V$ such that the natural map $V\to V^{**}$ is not surjective, then we get a homomorphism $V^*\to k$ that is discontinuous when $V^*$ is given the topology of pointwise convergence. This topology makes $V^*$ into a formally profinite group: it’s the inverse limit of duals of finite subspaces of $V.$
The first thing to try is the space of finitely supported $\omega$-sequences, $V_1=\colim_{n\in\omega} k^n.$ The dual space is the direct product $k^\omega.$ If $V_1\to V_1^{**}$ is not surjective we’re done. So we can assume that every linear $k^\omega\to k$ is a dot product with a finitely-supported vector.
The second thing to try is the space of countably supported $\omega_1$-sequences, $V_2=\colim_{\alpha\in\omega_1} k^\alpha.$ Let $e_\alpha\in V_2$ denote the vector with $e_\alpha(\alpha)=1$ and $e_\alpha(\beta)=0$ for $\beta\neq \alpha.$ For each linear $\phi:V_2\to k$ define $I_\phi=\{\alpha:\phi(e_\alpha)\neq 0\}.$
We’ll show that $I_\phi$ is always finite. Fix $\phi$ and let $\beta$ be the supremum of ordinals $\alpha$ such that $\alpha\cap I_\phi$ is finite. By the assumption that every linear $k^{\omega}\to k$ is a dot product with a finitely-supported vector, the restriction of $\phi$ to $k^{\alpha}$ is a dot product with a finitely-supported vector. But its support is just $\beta\cap I_\phi,$ so $I_\phi$ is finite.
Define a linear functional $F:V_2^*\to k$ by taking $\phi$ to $\sum_{\alpha\in I_\phi} \phi(e_\alpha).$ This is not in the image of $V_2\to V_2^{**}$ because $F(x\mapsto x_\alpha)=1$ for all $\alpha.$ In any case we have found a vector space $V$ such that $V\to V^{**}$ is not surjective. If forced to choose we can take $V=V_1\oplus V_2.$ (Old joke)

Compactness can be a sensible notion without the Boolean Prime Ideal Theorem. Tychonoff’s theorem works fine if all its input is well-ordered. In particular, given an inverse system of finite groups $G_i$ where the union of $G_i$ is well-ordered, the inverse limit is compact.
Any non-principal ultrafilter on $I$ defines a discontinuous function on $k^I,$ whether or not $k^I$ is compact. By a result of Blass that I haven’t seen [1], it is consistent with ZF that every ultrafilter is principal. As BS said in their answer, a model of Shelah deals with the second-countable case. So we have two different models ruling out different classes of discontinuous maps. These two results can be combined:

Theorem. It is consistent with ZF that there is no discontinuous map $G\to H$ where $G$ is a product of second-countable compact profinite groups and $H$ is finite. ($G$ itself is not assumed to be compact.)

Pincus and Solovay [2] built on Blass’s argument to construct a model $N$ without ultrafilters and satisfying the handy axiom Dependent Choice. Working in $N$ for now, suppose we have a finite group $H,$ a family of compact second-countable profinite groups $S_i, i\in I$ with product $G=\prod_{i\in I}S_i,$ and a discontinuous homomorphism $\phi:G\to H.$
For each $J\subseteq I$ let $\phi^J:G\to H$ denote the map $\phi^J(g)=\phi(g^J)$ where $g^J_i=g_i$ for $i\in J$ and $g^J_i=1$ for $i\not\in J.$ Let $\mathcal I$ be the ideal of subsets $J\subseteq I$ such that $\phi^J$ is trivial. Let $B$ denote the Boolean algebra $\mathcal P(I)/\mathcal I.$ We’ll consider two cases, and in each case construct a sequence $g_n\in G$ with $\phi(g_n)\neq 1$ such that for every sequence $x\in \{0,1\}^\omega,$ the product $g_0^{x_0}g_1^{x_1}\dots$ converges. (If $G$ is compact, the last condition is equivalent to $g_n\to 1.$)
If $B$ is infinite, pick a countably infinite antichain, and pick disjoint representatives $J_0,J_1,\dots\subset I$ with $\phi^{J_n}$ non-trivial. Also pick $g_n$ with $g_n=g_n^{J_n}$ (i.e. $(g_n)_i=1$ for $i\not\in J_n$) such that $\phi(g_n)\neq 1.$ Any product $g_0^{x_0}g_1^{x_1}\dots$ converges.
Now consider the case that $B$ is finite. For each atom $b$ pick a set $J_b$ representing $b.$ Since $\phi(g)=\phi(\prod_b g^{J_b})=\prod_b \phi^{J_b}(g)$ (with any choice of ordering), some $\phi^{J_b}$ must be discontinuous. Pick one. The ultrafilter $\mathcal P(J_b)\setminus \mathcal I$ is principal because $N$ has no non-principal ultrafilters. So there is an $i\in J_b$ and a homomorphism $\psi:S_i\to H$ with $\phi^{J_b}(g)=\psi(g_i)$ for all $g.$ By second-countability, there is a descending sequence of open normal subgroups $U_0,U_1,\dots$ of $S_i$ such that $\bigcap_j U_j=\{1\}.$ If $\psi$ is trivial on some $U_j$ then $\psi$ is continuous, a contradiction. Otherwise, we can pick $g_n\in U_{k_n}$ with $\psi(g_n)\neq 1$ and $k_n$ strictly increasing to ensure $g_n\to 1.$ Compactness ensures that each limit $g_0^{x_0}g_1^{x_1}\dots$ exists. Embed these $g_n$ in $G$ using the obvious map $S_i\to G.$
We can assume that $H$ is in the ground model $L.$ Define $\psi: 2^\omega\to H$ by $\psi(x)=\phi(g_0^{x_0}g_1^{x_1}\dots).$ Note $\psi(x)\neq \psi(x’)$ whenever $|\{i:x_i\neq x’_i\}|=1.$ The argument that there are no ultrafilters on $\omega$ from [2, Section 1.2] goes through with a small change to use an inequality instead of an equality. (It would be slightly easier to use Cohen forcing here instead of random reals. I've been conservative and avoided changing the model.)
Fix $h\in H.$ Their event $a$ needs to be modified to
$$a_h=\|s(\dot R|v)\in t_h(\dot R|u)\|$$
where $t_h(\dot R|u)$ is a term defining the preimage $\psi^{-1}(\{h\}).$ Their equality $l(a\cap b)=l(b)/2$ needs to be modified to the inequality $l(a_h\cap b)\leq l(b)/2.$ By the measure preserving transformation argument in [2, Section 1.2], all basic opens $b$ satisfy this inequality. By the monotone class theorem, this inequality is also satisfied by $b=a_h,$ giving $l(a_h\cap a_h)\leq l(a_h)/2.$ Hence $l(a_h)=0.$ (Alternatively, apply the Lebesgue density theorem for $2^\omega.$) This holds for all $h,$ contradicting $\bigvee_{h\in H} a_h=1.$

[1] Blass, Andreas, A model without ultrafilters, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 25, 329-331 (1977). ZBL0365.02054.
[2] Pincus, David; Solovay, Robert M., Definability of measures and ultrafilters, J. Symb. Log. 42, 179-190 (1977). ZBL0384.03030.<|endoftext|>
TITLE: Last term of repeating continued fraction expansion
QUESTION [9 upvotes]: Once again, working with stable vector bundles on $\mathbb{P}^2$ I have run into a question that is really out of my area.  (Thanks to everybody who helped out with my last question!)
Let $D>9$ be a rational number which is not a square, and consider the quadratic irrational 
$$\xi = \frac{-3 + \sqrt{D}}{2},$$
(I'd be willing to force $D$ to be an integer, and even to assume $D \equiv 5 \pmod{8}$, but I don't think it matters).  Let $r$ be the positive integer such that 
$$(2r+1)^2 < D < (2r+3)^2.$$
Numerous examples with Mathematica suggest that the continued fraction expansion of $\xi$ takes the form 
$$\xi = [r-1;\overline{a_1,\ldots,a_k}],$$ where the last term of the repeating part is $a_k = 2r+1$.  
For my particular situtation, I'd be happy enough to know that the number $2r+1$ appears somewhere in the expansion.
As I know almost nothing about continued fractions aside from statements of the basic results, I haven't the slightest idea how to prove something like this.  I also don't have a source which does much more advanced things than show that the expansion of a quadratic irrational always repeats.  Are statements like this well-known?  And where should I look for more advanced theory relevant to this problem?
In case it helps, the original form I came to this number is as follows.  Put  $$ q = \frac{1}{8}(D-5). $$ Then $\xi$ is a solution of the equation $$\frac{1}{2}(x^2+3x+1) = q.$$ Thanks!
EDIT: At request, here is what Mathematica gives for the continued fractions for some $D$:
$D=5: [0;-2,\overline{-1}]$ (but I am requiring $D>9$)
$D=10: [0;\overline{12,3}]$
$D=13: [0;\overline{3}]$
$D=141: [4;\overline{2,3,2,11}]$
(need more examples? Just ask!)

REPLY [13 votes]: It's known that a quadratic irrational has a purely periodic continued fraction expansion if and only if it is greater than $1$ and its conjugate is between $-1$ and $0$.  Your observation amounts to proving that $r+2 + (-3+\sqrt{D})/2$ has this property (note that adding $r + 2$ makes it start with $2r+1$).  This amounts to checking that $-1 < (2r+1 - \sqrt{D})/2 < 0$, which is equivalent to your assumption that $(2r+1)^2  < D <  (2r+3)^2$.<|endoftext|>
TITLE: Determine if a matrix is unimodular
QUESTION [8 upvotes]: Is deciding if an integer square matrix has determinant $\pm 1$ faster that calculating the determinant of the matrix?

REPLY [2 votes]: From Santha and Tan, "Verifying the determinant in parallel":

"for the determinant, there doesn't seem to be a simpler way to verify it than the computation itself."

Other notes: 
Matrix multiplication, inversion, and determinant-finding are all equally complex. I will quote from Kaltofen and Villard, "Computing the sign or the value of the determinant of an integer matrix, a complexity survey":

In algebraic complexity—i.e. when counting the number of operations in
  an abstract domain—we refer to Strassen [52] and Bunch & Hopcroft [13]
  for the reduction of the problem of computing the determinant to
  matrix multiplication. Conversely, Strassen [53] and Bunch & Hopcroft
  [13] reduce matrix multiplication to matrix inversion, and Baur &
  Strassen reduce matrix inversion to computing the determinant [7]

Since unimodularity amounts to invertibility, your question is equivalent to "is determining invertibility easier than actually computing the inverse (if it exists)?"<|endoftext|>
TITLE: Schubert varieties which admit small resolutions of singularities
QUESTION [8 upvotes]: I am looking for an (incomplete) list of partial flag varieties for 
which all Schubert cells admit small resolutions of singularities. 
This is interesting, for many reasons. My motivation is, that a description of a small resolution will give the corresponding IC sheaves very explicitly and hence explicit formulas for the KL-polynomials.
For example I know that Zelevinsky showed that this is the case for all type A Grassmannians. 
What about other $G/P$ for say $P$ Hermitian symmetric? 
I think in this case there are at least explicit formulas for KL polynomials.
What about other partial flag varieties?

REPLY [7 votes]: For the Hermitian symmetric $G/P$, Nicolas Perrin explicitly classified all the Schubert varieties admitting a small resolution.  (More explicitly, he classifies all the minimal models and quotes a theorem that says a small resolution is a smooth minimal model.)  If I remember correctly, except for some very small rank exceptions, type A, and the obvious cases (e.g. projective space), all $G/P$ have a Schubert without a small resolution.  Fortunately, there are few enough minimal models for the Hermitian symmetric cases that even in the non-smooth case, their intersection cohomology have a reasonable description, leading to the (previously known) explicit formulas for KL-polynomials.  (These formulas are summarized in a paper by Boe that proves the final cases and summarizes and refers to the cases done earlier.)
In type A, there is a Schubert without a small resolution for any 2-step flag variety where the two steps are not adjacent.  (The example mentioned in the Zelevinsky paper (which he notes as being due to MacPherson) generalizes to all such cases.)  This also means there is a Schubert without a small resolution for any 3-or-more-step flag variety.
This leaves as the only case (barring low rank exceptions) adjacent 2-step flag varieties.  I have not seen this resolved in the literature.  I suspect there is a Schubert variety in those cases without a small resolution provided the steps are at least two away from both ends.  In fact I have an explicit expected counterexample, but I have never learned the tools necessary to prove it actually is one.<|endoftext|>
TITLE: Trace matrix inequality
QUESTION [7 upvotes]: Hello all,
I come across the following problem.
Is it true that for a positive definite matrix $X^{n\times n}$, the following holds
$\text{trace}(X^{-1})\geq\text{trace}([\text{diag}(X)]^{-1})$,
where $\text{trace}(\;\cdot\;)$ and $\text{diag}(\;\cdot\;)$ are the matrix trace and diagonal operators, respectively.
Remark 1: Note that in the above, there is equality when $X$ is diagonal.
Remark 2: By using the SVD of the matrix $X$, or by using the Hadamard inequality, the above inequality is equivalent to
$\sum\limits_{i=1}^n\frac{1}{\lambda_i}-\sum\limits_{i=1}^n\frac{1}{x_{ii}}\geq0$,
where $\lambda_i$ and $x_{ii}$ are the $i$th eigenvalue of the matrix $X$, and the $i$th diagonal element of the matrix $X$, respectively.

REPLY [19 votes]: It is quite well known that the diagonal of an Hermitian matrix is majorized by sequence of its eiqenvalues (this is known as Schur's theorem). Since all of these numbers are positive, we can use Karamata's inequality for function $x \mapsto \frac{1}{x}$ to conclude that inequality $\sum_{i=1}^{n} \frac{1}{\lambda_{i}} \geqslant \sum_{i=1}^{n} \frac{1}{x_{ii}}$ is true.<|endoftext|>
TITLE: Additivity of Signature
QUESTION [7 upvotes]: The signature of compact oriented $4k$-manifolds has the following additivity property, first observed by S. P. Novikov: If two manifolds are glued by an orientation-preserving diffeomorphism of their boundaries, then the signature of their union is the sum of their signatures.
C.T.C. Wall showed that this additivity property does not hold for the more general situation where one glues two manifolds along a common submanifold of the boundaries, which itself has a boundary. 
My questions is: does it hold when the manifolds has several boundary components and we only glue part of them? (which means the final manifold still has boundary)

REPLY [8 votes]: Yes, the signature is additive in this case.  Wall gave a precise description of the failure of additivity in gluing two $4k$-manifolds along a subset of their boundaries.  (I think the name of the paper is "On the non-additivity of the signature", or something like that, but I don't have it in front of me.)  Let $A$ and $B$ be two $4k$-manifolds, and suppose we are gluing along $S\subset \partial A$ and an identical $(4k{-}1)$-manifold $S\subset \partial B$.  Consider the $(4k{-}2)$-manifold $\partial S$.  It bounds three manifolds: $S$, $\partial A\setminus S$, and $\partial B\setminus S$.  The kernels of these inclusions give rise to three lagrangian subspaces of the symplectic middle-dimensional homology of $\partial S$; call them $L_1, L_2, L_3$.  Then $\sigma(A \cup_S B) = \sigma(A) + \sigma(B) + c(L_1, L_2, L_3)$, where $\sigma$ denotes signature and $c(L_1, L_2, L_3)$ is an integer which, as the notation suggests, depends only on the three lagrangians $L_1$, $L_2$, and $L_3$.
You assume that $\partial S$ is empty.  In this case $c(L_1, L_2, L_3) = 0$ and the signature is additive.

One simple way to define $c(L_1, L_2, L_3)$ (not the way Wall does it) is as follows.  Let $n$ be the dimension of $L_i$ and let $V$ be the symplectic vector space which contains the lagrangians.  If $n=1$ we define $c(L_1, L_2, L_3) = \pm 1$ according to the cyclic order of   $L_1$, $L_2$, and $L_3$, assuming the three lagrangians are distinct.  If they are not distinct we define $c(L_1, L_2, L_3) =0$.  For $n>1$ it is not hard to show that we can find a symplecitc direct sum decomposition $V = \oplus_\alpha V^\alpha$, with $V^\alpha$ 2-real-dimensional, and lagrangians $L_i^\alpha\subset V^\alpha$, such that $L_i = \oplus_\alpha L_i^\alpha$.  We now extend linearly and define $c(L_1, L_2, L_3) = \sum_\alpha c(L_1^\alpha, L_2^\alpha, L_3^\alpha)$.

REPLY [7 votes]: The answer is yes,   and it is an immediate consequence of  the Atiyah-Patodi-Singer  index theorem,  more precisely  Theorem 4.14. of

M.F. Atiyah, V.K. Patodi, I.M.
  Singer, Spectral asymmetry and
  Riemannian geometry I,  Math. Proc.
  Camb. Phil. Soc. vol. 77, 1975, 43-69.

Admittedly the above  argument  feels like mouse-hunting  with a  bazooka. There are more elementary ways    to see this  but they would involve a bit more sweat.
Update.  For  an elementary proof of the this general form of additivity  see Theorem  27.5 in

B.A. Dubrovin, A.T. Fomenko, S.P.
  Novikov, Modern Geometry-Methods and
  Applications. Part III. Introduction
  to Homology Theory, Graduate Texts
  in Math., vol. 124, Springer Verlag, 
  1990.

The above Novikov is the same  Novikov in Novikov additivity, and the proof  in the above reference is Novikov's original argument which   was also sketched in Kevin Walker's answer.<|endoftext|>
TITLE: Is there a "universal" cohomology theory for varieties over p-adic fields?
QUESTION [10 upvotes]: Let $K$ be a $p$-adic field, $X$ a smooth proper algebraic variety over $K$, and $0 \le i \le 2 \dim X$. For a prime $\ell \ne p$ one can consider the $\ell$-adic cohomology $H^i(\overline{X}, \mathbb{Q}_\ell)$, and massage this in the usual way (via Grothendieck's abstract monodromy theorem) to get a Weil–Deligne representation of $K$ with coefficients in $\mathbb{Q}_\ell$. For $p$-adic étale cohomology, there is a more complicated construction starting from $H^i(\overline{X}, \mathbb{Q}_p)$ going via Fontaine's $D_{\text{pst}}$ functor. I gather it is conjectured that all of these Weil–Deligne representations are in fact definable over $\mathbb{Q}$, and they should have the same character and thus be isomorphic up to semisimplification; and this is known in some cases.
Is it expected that there should be a "universal" cohomology theory taking values in the category of Weil–Deligne representations over $\mathbb{Q}$, from which all of the above can be obtained by extending scalars? If so, have there been any attempts to construct such a cohomology theory?

REPLY [15 votes]: I'm seeing this old question just now, and simply wanted to remark that the situation may be slightly better.
Namely, enlarging the category of $\mathbb Q$-vector spaces into the larger semisimple $\mathbb Q$-linear tensor category $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$ defined by Kottwitz (see for example the article Representations of the Kottwitz gerbes of Iakovenko for a definition), one expects that there is a (canonical) cohomology theory valued in Weil–Deligne representations with coefficients in $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$. This should recover the other constructions via natural functors to $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q_\ell})$ for all primes $\ell$. This is a variant of Conjecture 9.5 in my 2018 ICM paper. (Warning: In that paper, I write $\mathrm{Kt}_{\mathbb Q}$ for what I denote $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$ in this answer.)
While the definition of $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$ is as representations of a gerbe $\mathrm{Kt}_{\mathbb Q}$ that is more-or-less explicitly constructed by class field theory — quite similar to the global Weil group —, we still lack a direct linear-algebraic description of $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$, unfortunately.<|endoftext|>
TITLE: Two commuting operad actions
QUESTION [6 upvotes]: If the following is true, it is probably well-known to the experts. Nevertheless, I could not find a reference for it. 
Suppose $P$ and $Q$ are $A_{\infty}$-operads in topological spaces and $X$ is a $P$-algebra as well as a $Q$-algebra. Moreover, assume that the actions of $P$ and $Q$ on $X$ commute, i.e. the following diagram is strictly commutative:
$$
\begin{gather}
Q(n) \times P(m) \times X^{nm} & \to & Q(n) \times X^n \\\\
\downarrow &  & \downarrow \\\\
P(m) \times X^m & \to & X
\end{gather}
$$
where the upper horizontal arrow uses the "diagonal action" of $P$ on $X^n$ and likewise for the left vertical arrow. Now I can deloop with respect to the operad $P$ to obtain $B_PX$, similarly I can obtain $B_QX$, where I either use the delooping machine of Boardman and Vogt or the bar construction - your choice.

Is it true that $B_PX$ carries an action of the operad $Q$ in this case?

Somehow this question boils down to the problem, how well products are respected by the common delooping machines. For example, I know that $B(G_1 \times G_2)$ is homeomorphic to $BG_1 \times BG_2$ for groups (or monoids). Is this still true, if $G_i$ are just $A_{\infty}$-spaces?

REPLY [4 votes]: First of all let me mention that the structure you describe is precisely the Boardman-Vogt tensor product of the operads $P$ and $Q$. I'll denote this by $P\otimes_{BV}Q$. A space is an algebra for the BV tensor product of $P$ and $Q$ if it is a $P$-algebra in the symmetric monoidal category of $Q$-algebras (note that you need the fact that the monoidal structure is the cartesian product to show that the category of $Q$-algebras in spaces is symmetric monoidal).
I don't have an answer to your exact question but here is something close to what you ask. Maybe you are aware of all this. If so I apologize :
The category of $P$-algebras in $Q$-algebras is Quillen equivalent to the category of $Ass$-algebras in $Q$-algebras by standard facts about category of algebras over operads. Therefore you can strictify your $P$-algebra structure. You get a space $X'$ which is homotopy equivalent to $X$ with an action of $Ass\otimes_{BV} Q$. To this you can apply the Bar construction. The important fact is that it doesn't matter if you do it in the category of spaces or in the category of $Q$-algebras because the forgetful functor $Q-Alg\to Spaces$ commutes with geometric realization of simplicial objects.
If you are careful about doing everything homotopically, you are going to end up with a space $Y$ which is weakly equivalent to $B_PX$ and which is equipped with a $Q$-action.
The following paper http://arxiv.org/abs/math.AT/0410367 does something similar to show that $THH$ of an $E_2$-algebra is an $A_{\infty}$-algebra.
Something interesting is that you can only strictify one of the two $A_{\infty}$-structure. Indeed $P\otimes_{BV} Q$ is usually equivalent to the little $2$-disk operad (in fact the derived BV tensor product is equivalent to $E_2$). However $Ass\otimes Ass$ is the commutative operad. This paper http://arxiv.org/abs/1102.1311 talks about the BV tensor product of $A_\infty$ and $E_n$.<|endoftext|>
TITLE: Area of union of random circles in a plane
QUESTION [6 upvotes]: If $n$ circles of radius $r$ are drawn at random in the plane such that their centers lie in some region $S$ (we could take $S$ to be a square), what is the expected area $E$ of their union?
Edit: In light of Noam's comment, it seems like the answer is 
$\displaystyle \int \int 1- (1-f(x,y))^n dx dy$
where $f(x,y) = a(B_r(x,y) \cap S)/a(s)$; the probability a random circle contains $(x,y)$.
If the region $S$ is large compared to $r$ and it has a reasonably "out of the way" boundary, then $f(x,y)$ is approximately constant over all of $S$, and is about $\pi r^2/ a(S)$, and so the expected value $E$ for a fixed $n$ is $n\pi r^2 + O(r^4)$.
However, the more interesting cases occur when $r$ is not small or when $\partial S$ gets "in the way" significantly. 
Suppose $S$ is a square of side length $s$ and the circles have radius $r=s/2$. How does $E$ grow with $n$ asymptotically? What if $S$ is a circle of radius $s$? A regular $k$-gon inscribed in a circle of radius $s$?

REPLY [2 votes]: The question is already interesting for general $r$ when $S=[0,1]$. Then the area of $E$ is $\pi r^2 + 2 r$. Because the area defect to $E$ of two circles being $d$ apart is $\frac{d^3}{12~ r}$, and $d\approx n^{-1}$, for large $n$ only the distance between minimum $a$ and maximum $b$ of the circle centers in $S$ determines the asymptotics.
The independent expectation values for $a$ and $b$ are $1/(n+1)$ and $1-1/(n+1)$ giving an 
asymptotic area defect to $E$ of $4r/(n+1)$.<|endoftext|>
TITLE: a question on TITS' note "Reductive groups over local fields"
QUESTION [6 upvotes]: This note appears in "Proceedings of Symposia in pure mathematics" vol.33 1979 part 1 pp. 26-69.
The question will be about materials on page 31-32. 
Let $G$ be a reductive algebraic group (not necessarily connected) defined over a local field $K$. 
We fix a maximal $K$-split torus $S$ of $G$ and take N(resp. Z) to be normalizer 
(resp. the centralizer) of $S$ in $G$. Let $X_*=Hom_K (Mult, S)$  (Resp. $X^*=Hom_K (S, Mult)$) be the group of cocharacters (resp. characters) of $S$. Let $V=X_*\otimes_{\mathbb Z}\mathbb R$.
We fix a  discrete valuation $\omega: K\to (-\infty, \infty]$. 
Let $\nu: Z(K)\to V$ be the homomorphism defined by 
$$
\chi (\nu (z))=-\omega (\chi (z)) \quad \mbox{for}\  z\in Z(K) \mbox{ and } \chi \in X^*(\mathbb Z).
$$
Let $Z_c$ be the kernel of $\nu$. Then we have a short exact sequence of gorups
$$
0\to Z(K)/Z_c\to N(K)/Z_c\to N(K)/Z(K)\to 0
$$
where $N(K)/Z(K)$ is a finite group.
Then it is claimed that the map $\nu$ induces  a group homomorphism $\phi$ from $N$
to the group of affine transformations of $V$ such that for $z\in Z(K)$ and $x\in V$ one has
$\phi(z)x=x+\nu (z) $. 
I do not understand in which way this function $\phi$ is defined.

REPLY [10 votes]: Of course, grp is right that to justify the whole discussion in Tits about the construction of the apartment you need the full strength of Bruhat-Tits. But for the purpose of Tits' article (who assumes you accept facts 1.4.1 and 1.4.2) there is a rather simple construction to get $\nu$ which doesn't require the Bruhat-Tits books. I think it's what Tits had in mind. I learnt this construction from J.K.Yu, who says "This is one of the questions I got asked most often regarding Tits’ article.". You can find it in his very useful notes on buildings, or (as he notes) in Landvogt's book. Unfortunately JK's notes are not available on the web; they are in this book:
http://books.google.ca/books/about/Ottawa_Lectures_on_Admissible_Representa.html?id=7oBZIxZfPQAC&redir_esc=y
On second thought, I assume it's ok with copyright laws to take just the relevant two pages and put them here: http://www.math.toronto.edu/~herzig/apartment_jkyu04.pdf<|endoftext|>
TITLE: Image of J in the classical Adams Spectral Sequence
QUESTION [7 upvotes]: Hey all,
I know that in some versions of the Adams Spectral Sequence you can easily identify the image of $J$, and I was wondering if there was a way to identify the image of $J$ in the $E_2$ page of the classical mod-2 version, especially for $t-s=3$ (mod 4).  Since the order of this image is known and it is known that the image is a direct summand, it isn't so hard to find it in $E_\infty$.  Of course, if you can identify $Im(J)$ in an earlier page, then you learn a huge amount about the differentials in that column.  This might imply that identifying the image of $J$ is almost as hard as calculating the differentials, so maybe this is too much to hope for, but maybe just maybe there's a trick.  
Thanks

REPLY [7 votes]: The image of $J$ is pretty easy to see in the Adams $E_2$ term: it consists of the elements along the vanishing line, plus, in dimensions 8k-1, of the towers that end near the vanishing line. 
This identification is due to Mahowald, see The order of the image of the J-homomorphism for the announcement ( http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183532412 ) and On bo-resolutions ( http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.pjm/1102736799&page=record ) for details.<|endoftext|>
TITLE: which algebraic integers in a cyclotomic field give you integer absolute value?
QUESTION [8 upvotes]: Does anyone know an answer to this question?
Question: In an cyclotomic field which algebraic integers have integer absolute value?
Revision 1:  -1
I like to add this to the above question, Let's take w to be a primitive n-th root of unity, for which set of exponents A of {0,1,...n-1} we have the absolute value of the sum_{i \in A} w^i is an integer. this might not be any help to make it solvable but at least avoid some repetitions

REPLY [3 votes]: Katie asks further about the case where we have a subset $A$ of $\mathbb{Z}/n$ and asks for $\sum_{k \in A} \zeta^k$ to have integer absolute value. She writes that, for $n$ prime, the only solutions should be the trivial ones $|A| =0$, $1$, $p-1$ or $p$. The point of this note is that she is correct, and that there probably isn't a good description like this for $n$ not prime. This answer is built on Noam Elkies's very helpful comment above.
Let $p$ be prime and let $A \subseteq \mathbb{Z}/p$. Let $a=|A|$. Let $z=\sum_{k \in A} \zeta^k$. Let $b_k = \# \{ (i,j) \in A^2 : i-j \equiv k \bmod p \}$ So
$$z \bar{z} = a + \sum_{k=1}^{p-1} b_k \zeta^k$$
Since the minimal polynomial of $\zeta$ is $1+\zeta+\zeta^2+\cdots+\zeta^{p-1}$, the only way that $z \bar{z}$ can be an integer is if all the $b_k$ are equal to some common value, say $b$. In this case, $z \bar{z} = a-b$. Furthermore, we want $\sqrt{z \bar{z}}$ to be an integer, say $n$. So $a=n^2+b$ for some nonnegative integer $n$.
Now, we must have $a(a-1) = b(p-1)$ since $\sum_{k=1}^{p-1} b_k$ is clearly $a(a-1)$. So $(n^2+b)(n^2+b-1) = b(p-1)$. If $b=0$ then $|A|=0$ or $1$. If not, we can divide by $b$ to write:
$$p=\frac{(n^2+b)(n^2+b-1)}{b}+1 = \frac{(n^2+n+b)(n^2-n+b)}{b}.$$
(That factorization came out of nowhere, as far as I'm concerned.) Since $p$ is prime, that means that at least one of $n^2+n+b$ and $n^2-n+b$ is $\leq b$. But $n^2 \pm n \geq 0$ for nonnegative integer $n$. So $n=0$ or $1$. This gives $p=b$ and $p=b+2$, and then $a=p$ and $a=p-1$ respectively.

Now, if $n$ is not prime, then there are lots of sets $A$ such that all the $b_k$ are equal. The special case $b=1$ is called a perfect difference set, and they are pretty plentiful. For example, $(0,5,6,9,19)$ is a perfect difference set modulo $21$, giving 
$$|1+\zeta_{21}^5+\zeta_{21}^6+\zeta_{21}^9+\zeta_{21}^{19}| = \sqrt{5-1} = 2.$$
According to the mathworld link above, there are perfect difference sets modulo $q^2+q+1$ for every perfect power $q$; taking $q=p^2$ we deduce that we can always find $A$ in $\mathbb{Z}/(p^4+p^2+1)$ with absolute value $p$.
Based on skimming the results of a quick google search, my impression is that there are lots of methods known for constructing perfect difference sets, but no classification.
Probably someone has studied the case $b>1$, but I didn't see it.
But even if there were a classification, that wouldn't be a complete answer. Because $1+\zeta+\cdots + \zeta^{n-1}$ is not the minimal polynomial of $\zeta$ for $n$ composite. And I don't see how to control the other solutions that might come up because of this.<|endoftext|>
TITLE: Alexandrov-Bakelmann-Pucci maximum principle
QUESTION [6 upvotes]: The ABP maximum principle states (roughly) that, if $a^{ij} \partial _i \partial _j u \geq f$, over a domain $\Omega$ in $\mathbb{R}^n$ (where $a^{ij} \geq C Id >0$), then (assuming sufficient regularity of the coefficients), $\sup _{\Omega} u  \leq \sup _{\partial \Omega} u + C (\int _{\Omega} \vert f \vert^n )^{1/n}$. Its proof seems to be quite unconventional (the final inequality is proved by an inequality of the measures of certain sets constructed using the normal mapping). Is there an intuitive explanation of the proof? (Perhaps using some probabilistic ideas or something?) I mean, usually, in order to prove maximum principles, the key idea is to use that at a local max the second derivative is negative-definite. But, this one seems to use some strange ideas..

REPLY [15 votes]: In my view, the key point the ABP estimate is that it ties pointwise information (the PDE) to information in measure (the contact set). This is crucial to regularity theory for non-divergence equations; it enables the proof of a Harnack inequality (due to Krylov and Safonov), and $C^{1,\alpha}$ regularity for viscosity solutions of fully nonlinear elliptic equations. For divergence equations, whose solutions are already defined by integrals, it is a bit easier to obtain regularity.
Note that the technique of looking at a local maximum, we only use the equation at a single point. In the proof of ABP that I know, we use the equation at many points. Here's the theorem I'd like to discuss:
If $a^{ij}(x)u_{ij} \leq 1$ in $B_1$ (uniformly elliptic)and $u|_{\partial B_1} \geq 0$ then 
$|inf_{B_1}u| \leq C |A|^{1/n}$ where $A$ is the set where $u$ agrees with its convex envelope. (The contact set).
Another way to describe $A$ is the collection of points in $B_1$ for which the graph of $u$ can be touched below by a supporting hyperplane. It is reasonable to believe that, given a supersolution, it doesn't have "cone-like" behavior, i.e. few points in $A$, roughly because the Hessian would be too large and contradict the equation at these points. This observation is made rigorous by the Area Formula; all the planes of slope $c|inf_{B_1}u|$ touch $u$ by below somewhere in $B_1$, so the derivative mapping $Du(B_1)$ contains a ball of radius $c|\inf_{B_1}u|$. The Jacobian determinant of this mapping is $detD^2u$, which tells us locally "how much we curved", i.e. the measure of the set of slopes of supporting hyperplanes nearby. The PDE tells us that at each of these points in $A$, $det D^2u$ is not too large, completing the proof.
Prof. Ovidiu Savin uses a similar technique in his paper "Small Perturbation Solutions of Elliptic Equations" to prove an ABP-type estimate; in this version, we slide paraboloids up from below the solution until they touch and show that the set of contact points has measure controlled below by the set of vertices of these paraboloids. This gives information in measure that can be exploited to prove a Harnack inequality and $C^{\alpha}$ regularity in the setting of the paper.
I hope this helps!<|endoftext|>
TITLE: Equivariant homotopy theory: some history questions
QUESTION [12 upvotes]: I have sometimes wondered about the following:
(1) Who was the first articulate that in dealing with $G$-equivariant cohomology theories ($G$ a finite group or a compact Lie group), it is best to work in an $RO(G)$-graded context?
(2) Who was the first to realize that the correct set up for equivariant stable homotopy was to work in a complete universe?
(3) At what time did these ideas first come to the surface? The seventies? The eighties?
How much do they predate the Segal conjecture?
(Maybe I should thank Peter May in advance?)

REPLY [20 votes]: Well if you insist John :) (1) The first explicit formulation I know of is in the nice paper:
Klaus Wirthmuller. Equivariant homology and duality. Manuscripta Math. 11 (1974), 373–390.
He writes: "The ideas developed here partly originate from suggestions made by T. tom Dieck, who introduced me to the subject." They were thinking about equivariant Poincar\'e duality and about equivariant cobordism, which make $RO(G)$ grading inevitable.   (2) That may be tom Dieck and may be me. I'm honestly not sure.  I was
explicitly using universes nonequivariantly in 1972, with good reason.  I think tom Dieck may have at least implicitly used complete universes in his work on cobordism. Certainly I was using complete
$G$-universes by some time around or before 1974-75. The question is confused by the fact that tom Dieck was in Chicago lecturing on equivariant things that year.  See Tammo tom Dieck. The Burnside ring and equivariant stable homotopy. Lecture notes by Michael C. Bix. Department of Mathematics, University of Chicago, Chicago, Ill., 1975.  (3) The Segal conjecture in its simplest form dates from 1970.  However, early work on it was not based on equivariant stable homotopy theory, let alone $RO(G)$ grading. As late as 1983, Frank Adams wrote a paper ``Graeme Segal's Burnside ring conjecture'' in which he barely mentioned equivariant cohomotopy, and then a bit skeptically.  There is a 1982 paper "Classifying $G$-spaces and the Segal conjecture'' by Lewis, McClure, and myself that proves the equivalence of the nonequivariant and equivariant versions of the Segal conjecture,
before Carlsson's proof. 
It is worth emphasizing that, in a sense, $RO(G)$-grading has no philosophical justification. Logically, grading should be on the equivariant Picard group which is considerably larger (at least under space level equivalence) or on the stable equivalence classes of representation spheres (and their negatives), which is considerably smaller.<|endoftext|>
TITLE: Does this Banach manifold admit a Riemannian metric?
QUESTION [19 upvotes]: First, the question; after, the motivation.
Consider 27.6 (pdf pp. 262-263) in The convenient setting of global analysis (AMS, 1997), and, in particular, the example given at the end of it, which concludes with: "Then the same results are valid, but $X$ is now even second countable."

Question: Does this second countable $X$ admit a Riemannian metric?

For the motivation:
This post stems from Jeff Rubin's earlier MO question and a follow-up that I posted. 
The former recalls (but also questions) the following result proved by both Serge Lang (Fundamentals of Differential Geometry, 1999, Springer-Verlag) and Abraham, Marsden, and Ratiu (Manifolds, Tensor Analysis, and Applications, 1988, Springer-Verlag):
Theorem: A connected Hausdorff Banach manifold with a Riemannian metric is a metric space. 

For an earlier incarnation of this question, Wolfgang Loehr gave a short argument (below) indicating that the space $X$ mentioned above is connected. In particular, $X$ is a second-countable, connected Hausdorff Banach manifold, which is separable and not regular, hence non-metrizable by Urysohn's Theorem.
If $X$ admits a Riemannian metric, then it is a counterexample to the "theorem" above. In any case, I am not sure how to prove when a manifold does or does not admit a Riemannian metric, and would appreciate assistance in this direction.

REPLY [5 votes]: Sub-question 2 is easy: $X$ is connected. Assume $A\subseteq X$ is open and closed. $Y\setminus \ker\lambda$ carries the topology inherited from $\ell^2$, hence is connected, hence we may assume w.l.o.g. that it is contained in $A$ (otherwise we take the complement of $A$). Now fix $y$ with $\lambda(y)=1$. For any $x\in\ker\lambda$, $x_n:=x+\frac1n y\in A$ and $x_n\to x$, hence $x\in A$ and $A=X$.<|endoftext|>
TITLE: Rational automorphisms of semisimple algebraic groups
QUESTION [5 upvotes]: Suppose $G$ is a semisimple algebraic group defined over a field $k$.  Let $\mathrm{Aut}(G)$ and $\mathrm{Inn}(G)$ denote the groups of automorphisms and inner automorphisms (respectively) of $G$.  Given a Borel subgroup $B\subset G$ and a maximal torus $T \subset G$, one has an associated system of simple roots $\Delta(B,T)$, and there is a well-known short exact sequence
$$
1 \longrightarrow \mathrm{Inn}(G) \longrightarrow \mathrm{Aut}(G) \longrightarrow \mathrm{Aut}(\Delta(B,T)) \longrightarrow 1.
$$
All of the groups in the sequence are defined over $k$.  (This is true even if $B$ and $T$ are not defined over $k$.)  So are the maps between them.  If we choose a pinning for $(G,B,T)$, then this choice gives rise to a splitting for the sequence, thus realizing $\mathrm{Aut}(G)$ as a semidirect product.
Question:  Can we always choose a pinning so that the splitting is defined over $k$?
Discussion:  If we could, then we could decompose $\mathrm{Aut}_k(G)$ as a semidirect product.  While the standard textbooks all give the short exact sequence above, I have not noticed that any of them have much to say about $\mathrm{Aut}_k(G)$, suggesting that such a decomposition isn't valid.
Can anyone point me to a proof or counterexample?
Partial result:  Josh Lansky and I can show that everything works if $G$ is $k$-quasisplit.  I believe that I can reduce the general case to the case where $G$ is $k$-anisotropic.  If $k$ is $p$-adic, then we know what the $k$-anisotropic groups look like, and everything works.
Motivation:  At one point, Josh and I needed to know that we could lift $k$-automorphisms of $\Delta(B,T)$ to $k$-automorphisms of $G$.  While this ability would make some aspects of our work more explicit, it is no longer essential.  But as Titchmarsh said of the irrationality of $\pi$, ``if we can know, it surely would be intolerable not to know.''

REPLY [5 votes]: I think ${\rm Aut}(\Delta(B,T))$ doesn't always split: there are some very serious obstructions for that.
According to a recent result of Skip Garibaldi ad Philippe Gille on algebraic groups with few subgroups, published in PLMS, there exist groups $G$ of type ${\rm D}_4$ whose all proper connected subgroups defined over the base field are abelian (such groups are necessarily anisotrpic).
If the extension in question splits then $\rm Aut(G)$ contains an involution which acts on the Dynkin graph
by interchanging two nodes and fixing the rest. The fixed point group of such an involution would contain a connected reductive subgroup or semisimple rank $3$ defined over the base field. However, such subgroups cannot exist in the forms of type ${\rm D}_4$ constructed by Garibaldi and Gille.<|endoftext|>
TITLE: Iterating monoid categories
QUESTION [12 upvotes]: Let $(C, \otimes)$ be a symmetric monoidal category (maybe braided is also okay). Then the category $\text{Mon}(C)$ of monoid objects in $C$ is also a symmetric monoidal category with the same monoidal product. The category $\text{Mon}(\text{Mon}(C))$ is, by the Eckmann-Hilton argument, just the category of commutative monoids in $C$; by a second application of Eckmann-Hilton, the process of constructing monoid categories stabilizes in the sense that the corresponding sequence of forgetful functors
$$... \to \text{Mon}(\text{Mon}(C)) \to \text{Mon}(C) \to C$$
stabilizes to a sequence of equivalences by the third functor at worst. However, the same argument shows that if $C$ is itself the category of monoids in some other symmetric monoidal category, then this sequence in fact stabilizes by the second functor, and if $C$ is itself the category of commutative monoids in some other symmetric monoidal category, then this sequence in fact stabilizes by the first functor.
So symmetric monoidal categories are divided up into three classes based on how soon the process of repeatedly constructing monoid categories stabilizes. Are there names for these three classes? Are there theorems characterizing them, for example theorems to the effect that if $C$ stabilizes early then it is nontrivially a category of monoids (not just $\text{Mon}(C)$ in the case that it stabilizes by the first functor)? (Vaguer question: is there some kind of homotopy-theoretic interpretation of these three classes?) 
A somewhat nontrivial example of a category $C$ such that $\text{Mon}(C) \to C$ is already an equivalence is $C = \text{Set}^{op}$ with categorical coproduct (product in $\text{Set}$) as the monoidal product. This is due to the fact that every set has a unique comonoid structure given by the diagonal $X \ni x \mapsto (x, x) \in X \times X$. The corresponding monoid interpretation is that $\text{Set}^{op}$ can be realized as a suitable category of Boolean rings, which are commutative monoid objects in the category of abelian groups (with tensor product).

REPLY [3 votes]: If you take commutative monoids in the first step, there is another natural choice of monoidal product on $\mathrm{CMon}(C)$. Namely, the universal "bilinear" construction, due in this generality to Anders Kock: in case $(C,\otimes) = (\mathrm{Set},\times)$, morphisms $M \otimes N \to K$ correspond to functions $M \times N \to K$ that are monoid morphisms in each argument separately when the other is fixed. In a sense this is more interesting, because no collapse of the kind you describe occurs. In fact, $\mathrm{Mon}(\mathrm{CMon}(C))$ is precisely the category of semirings in $C$. So heightening the "tower" of monoid categories then adds structure that occurs naturally and often in algebra.
See also http://dx.doi.org/10.1016/j.entcs.2008.10.012 and references therein. Incidentally, that paper also has a precise proof of the result Mike Shulman's answers refers to.<|endoftext|>
TITLE: Algorithm for detecting prime powers
QUESTION [6 upvotes]: While reading Peter Shor's paper Polynomial-Time Algorithms for Prime Factorization and Discrete Logarithms on a Quantum Computer, I came across the following quote:

"This scheme will thus work as long as $n$ is odd and not a prime power; finding factors of prime powers can be done efficiently with classical methods."

I have two questions:
(1) How does one efficiently determine if a given number $N$ is a prime power, and
(2) How does one efficiently determine the factorization of a prime power? 
(Note: I have included these as separate questions since I am aware that many of the standard  algorithms for determining a number is composite will not necessarily produce a proper factor.)

REPLY [12 votes]: See Dan Bernstein's paper, ``Detecting perfect powers in essentially linear time.'' Mathematics of Computation 67 (1998), 1253--1283, available at http://cr.yp.to/papers.html#powers.  Here "linear" means "linear in log n".<|endoftext|>
TITLE: Mochizuki's proof and Siegel zeros
QUESTION [35 upvotes]: Granville and Stark (Invent. Math. 139 (2000), 509-523) proved that a uniform version of the abc conjecture for number fields eliminates Siegel zeros for $L$-functions associated with quadratic characters of negative discriminant. Does the recently announced proof
by Mochizuki of the abc conjecture cover this statement?

REPLY [23 votes]: I don't think so. Mochizuki claims to have proved a diophantine result for points of bounded degree, while you need a uniform form of the ABC conjecture for the application that you mention.
EDIT: About your question below, on the version of the ABC conjecture claimed in Mochizuki's work, it is clearly stated in Theorem A of the 4th paper. Anyways, for the benefit of the people that might read this question, I will state in very elementary terms a corollary of Theorem A in the following context: X is the projective line with the usual projective coordinates [x:y], and D is the divisor $[0:1] + [1:0] + [1:1]$ which makes the curve U=X\D hyperbolic (the degree of the canonical divisor $\omega$ of X in this case is -2 and the degree of D is 3, hence the degree of $\omega(D)$ is 1>0). Ok, here is the corollary (the notation is explained below):
Statement: Let $d$ be a positive integer and let $\epsilon>0$. There is a constant $C>0$ depending only on $d$ and $\epsilon$ such that the following is true: If $A,B$ are non-zero algebraic numbers with $A+B=1$, and if the degree over Q of the number field $K=Q(A)$ is at most d, then we have $H(A,B,1) < C(\Delta_K N_K(A,B,1))^{1+\epsilon}.$
Notation: Here I am using the same definition of $\Delta_K$, H(a,b,c) and $N_K(a,b,c)$ as in the paper on Siegel zeros of the question (this notation is explained in the first page of the paper). Well, if you check the reference you'll see that actually there is one difference: the paper uses N(a,b,c), not $N_K(a,b,c)$. However, in the above statement it is crucial that we must compute N(A,B,1) using the number filed K=Q(A), that's why I added this subscript.
I hope that the readers can see the difference between this version and the uniform ABC conjecture for the paper on Siegel zeros: the fact that here the constant C also depends on d, not only $\epsilon$.
A last trivial remark. To get the classical ABC conjecture with coprime integers a+b=c you take A=a/c, B=b/c and hence K=Q which makes $\Delta_K=1$, and N(A,B,1)=rad(abc).<|endoftext|>
TITLE: Curvature of the Cayley projective plane
QUESTION [9 upvotes]: The Cayley projective plane can be realized as the compact homogeneous space $F_4/\mathrm{Spin}(9)$. In this way one can compute the curvature of this symmetric space in terms of a suitable orthonormal basis and the Lie brackets of the basic vectors. But is there any elegant expression for the curvature of $\mathbb{OP}^2$  which is independent of this homogeneous description due to the fact that $\mathbb{OP}^2$ is a rank one symmetric space?

REPLY [16 votes]: I'm not sure what you'll count as 'elegant' or as a 'description', but here's something you may find interesting or useful:  Recall that the space of curvature tensors $\mathcal{K}(V)$ on an inner product space $V$ is essentially the kernel of the natural map $S^2\bigl(\Lambda^2(V^\ast)\bigr)\to \Lambda^4(V^\ast)$.  Now let $V$ be the $16$-dimensional spin representation of $\mathrm{Spin}(9)$.  It is well known (and easy to see via highest weights) that one has
$$
\Lambda^2(V^\ast) = \Lambda^2(W)\oplus\Lambda^3(W)
$$
where $W$ is the standard $9$-dimensional representation of $\mathrm{Spin}(9)$ (which, of course, factors through $\mathrm{SO}(9)$). Each of these summands is irreducible.  It follows that, as $\mathrm{Spin}(9)$-representations 
$$
S^2\bigl(\Lambda^2(V^\ast)\bigr) = S^2\bigl(\Lambda^2(W)\bigr)\oplus \bigl(\Lambda^2(W)\otimes\Lambda^3(W)\bigr) \oplus S^2\bigl(\Lambda^3(W)\bigr),
$$ 
and hence it has two trivial summands, i.e., the space of curvature tensors fixed under $\mathrm{Spin}(9)$ has dimension $2$.  (You don't lose either of these in the map to $\Lambda^4(V^\ast)$ since this latter space has no trivial summands under $\mathrm{Spin}(9)$.)  These represent the Ricci curvature (the Cayley plane is an Einstein manifold) and the Weyl curvature of the Cayley plane.  
To get explicit formulae, you just need to make the above decomposition of $\Lambda^2(V^\ast)$ explicit.  The first summand is obvious, since it is the Lie algebra of $\mathrm{Spin}(9)$, for the second, you need a model of how to see the wedge product of two elements in $V$ as the sum of a $2$-form and a $3$-form on $W$.  Of course, this can be done by using the Clifford algebra, which can be studied using the octonions. I think that's probably as explicit as you can make it.

REPLY [5 votes]: I'm not quite sure what you are looking for, but explicit computation of the curvature appears in arXiv:math/0702631. The authors define $\mathbb{OP}^2$ via an octonionic atlas, write down explicit isometries with which they prove that it is indeed a homogeneous manifold and then they express the curvature tensor at a point using a coordinate frame. The expressions of the components of $R$ involve only octonionic multiplication and inner product.<|endoftext|>
TITLE: Symmetries of probability distributions
QUESTION [9 upvotes]: When talking about a single random variable, knowing only its distribution, the construction of a probability space is quite easy. Namely, let $(X,\mathscr A)$ be a measurable space and let $\mathsf Q$ be some probability measure over this space which we refer to as a distribution of some random variable. The usual definition states that there is some probability space $(\Omega,\mathscr F,\mathsf P)$, the random variable is
$$
  \xi:(\Omega,\mathscr F)\to(X,\mathscr A)
$$
i.e. it is a measurable map, and its distribution is a pushforward measure: 
$$
\mathsf Q:=\xi_*(\mathsf P)
$$
i.e. $\mathsf Q(A) = \mathsf P(\xi^{-1}(A))$ for any $A\in \mathscr A$.
Clearly, given $(X,\mathscr A,\mathsf Q)$ for a single random variable there is no reason to come up with a new sample space and we can take $(\Omega,\mathscr F,\mathsf P) = (X,\mathscr A,\mathsf Q)$ and $\xi:=\mathrm{id}_X$. 
Let us stick to this latter case. It may happen, that there is a map 
$$
  \eta:(X,\mathscr A)\to(X,\mathscr A)
$$
such that $\eta\neq \rm id_X$ but still it holds that $\mathsf Q = \eta_*(\mathsf Q)$. I wonder if the existence of this other maps is studied somewhere.

The brief statement of the problem is thus the following: given a probability space $(X,\mathscr A,\mathsf Q)$ if the identity map $\rm id_X$ is the unique solution of the equation
$$
  \mathsf Q = \xi_*(\mathsf Q) \tag{1}
$$
where the variable $\xi$ is any measurable map from $(X,\mathscr A)$ to itself. As far as I am not mistaken, the space of solutions of $(1)$ is a monoid as it is closed under the composition of maps.
Also, if $\xi$ is a bijection which solves $(1)$ then $\xi^{-1}$ solves it as well:
$$
\xi^{-1}_*(\mathsf Q)(A) = \mathsf Q(\xi(A)) = \mathsf Q(\xi^{-1}(\xi(A))) = \mathsf Q(A).
$$
Hence, bijective solutions of $(1)$ form a group - which may seem to be thought of a group of "symmetries" of $\mathsf Q$. For example, the standard normal distribution over reals $\mathsf Q = \mathscr N(0,1)$ admits at least two representations $\xi(\omega) = \omega$ and $\xi(\omega) = -\omega$. As well as any Haar measure over a group admits representation via $\xi(\omega) = \alpha \omega$ where $\alpha$ is any element of the group.
I've asked this question on MSE, but I have not received any answers.

Edited: To clarify (as requested), my question is exactly as follows: are such groups of symmetries of measures studied somewhere in the literature - may be, providing some interesting results for measures exhibiting such symmetries. I have studied the Lie groups of ODE/PDE symmetries, and I wonder if there is anything similar known for measures.

REPLY [5 votes]: Usually, such symmetries have been either studied in the context of Lebesgue spaces or studied in the context of homogenous measure algebras, where autmorphisms are easy to study. Every automorphism of a probability space gives rise to an automorphism of the corresponding measure algebra.
The easiest case are Lebesgue spaces, or even simpler, studying the uniform distribution on $[0,1]$. This is of course essentially the case of a Haar measure. A nice property is that one can take any automorphism of the measure algebra and find an automorphism of the probability space inducing it. Moreover, two automorphisms of the probability space giving rise to the same automorphism of the measure algebra can differ only on a set of measure zero. Every, homogenous, atomless, separable measure algebra can be represented as a Lebesgue space.
If one starts with a homogenous measure algebra, one may look for probability spaces representing the measure algebra. The two most prominent representations are by the Stone space of the measure algebra or in the form of a product of coinflips $\{0,1\}^\kappa$ with $\kappa$ infinite, which can represent every atomless (normed) homogenous measure algebra by Maharam's theorem. In the case of the Stone space, the automorphisms of the measure algebra correspond essentially to the automorphisms of the representing probability space. In the coin-flipping representation, every automorphism of the measure algebra is induced by an automorphism of the probability space. But very different automorphisms may give rise to the same automorphism of the measure algebra. Actually, there exists an automorphism of $\{0,1\}^\mathfrak{c}$ that induces the identity on the measure algebra but has no fixed point.
The discussion so far is largely adapted from the introduction to Ergodic theory on homogeneous measure algebras. by Choksi and Prasad. The book this has appeared in is likely to be available somewhere on the internet...
One can also study the case of rigid probability spaces, where there is essentially no automorphism but the identity. It is actually possible to find a countably generated and countably separated measurable space in which all automorphisms differ from the identity only on a countable set. This is done in the booklet Borel Spaces by Rao and Rao in Proposition 4. There also is an example of an atomless, countably generated probability space with no autmorphism but the identity (up to a countable set) in Section 48 of Values of non-atomic games by Aumann and Shapley.<|endoftext|>
TITLE: Delooping and unreduced operads
QUESTION [8 upvotes]: Suppose I have an operad $P(n)$ in topological spaces in the sense of the book by Markl, Shnider and Stasheff, i.e. there is no space $P(0)$. In agreement with some of the answers to  this question, I will call such an operad unreduced. 
Let $H$ be an infinite dimensional separable Hilbert space. Then an example for $P$ would be 
$$
P(n) = Iso(H^{\otimes n}, H)
$$
the linear isometric isomorphisms from $H^{\otimes n}$ to $H$. There is no obvious candidate for $P(0)$ (since we insist on isomorphisms, $P(0) = Hom(\mathbb{C},H)$ will not work). 
It is still possible to talk about unreduced $E_{\infty}$-operads in this case. But now it is less clear (at least to me) that I can deloop a space, which is an algebra over an unreduced $E_{\infty}$-operad. Hence my question:

Let $X$ be a space, which is an algebra over an unreduced $E_{\infty}$-operad $P$, such that  $\pi_0(X)$ is a group. Is $X$ an infinite loop space?

If this is not true:

Are there conditions on the space or the operad (other than that it can be reduced) to ensure that $X$ is an infinite loop space?

Using the bar construction by May, I think this would correspond to the question, whether I can drop degeneracies and still get a delooping.

REPLY [15 votes]: The answer to your first question is no: for example, take $X$ to be any connected pointed space, and regard $X$ as a nonunital commutative monoid by saying that the product of
any two points of $X$ is equal to the base point. This endows $X$ with the structure of
an algebra over any operad $\mathcal{O}$ with $\mathcal{O}(0) = \emptyset$, but $X$ need not be an infinite loop space.
However, with a slightly stronger hypothesis, the answer is yes. Let $X$ be a nonunital
$E_{\infty}$-space with the property that translation by any point $x \in X$ is a weak homotopy equivalence from $X$ to itself. Then $X$ is weakly homotopy equivalent to an infinite loop space.
Assuming that your operads are sufficiently nice (so that "algebras up to coherent homotopy" can be rectified), one has the following more precise statement: the homotopy category of $E_{\infty}$-spaces is equivalent to the subcategory of
the homotopy category of "nonunital" $E_{\infty}$-spaces, whose objects are nonunital
$E_{\infty}$-spaces $X$ for which $\pi_0 X$ contains a unit element $e_X$ such that translation by (any representative of) $e_X$ is a weak homotopy equivalence, and whose morphisms are maps of nonunital $E_{\infty}$-spaces $X \rightarrow Y$ which carry
$e_{X}$ to $e_{Y}$.<|endoftext|>
TITLE: Number of conjugacy classes in GL(n,Z)
QUESTION [15 upvotes]: Like every other group also $\mbox{GL}(n,\mathbb{Z})$ acts on the set of all its subgroups, by conjugation: if $\phi \in \mbox{GL}(n,\mathbb{Z})$, then $\phi$ acts by $H \mapsto \phi H \phi^{-1}$, where $H \leq \mbox{GL}(n,\mathbb{Z})$.
A theorem by Jordan (and later Zassenhaus) implies that the number of conjugacy classes of finite subgroups of $\mbox{GL}(n,\mathbb{Z})$ is finite.
About this fact I have a few questions:

Is the actual number of conjugacy classes known for each $n$? Maybe at least asymptotically?
Is it known which of these classes are particularly big for every $n$? (Edit: Doesn't make sense in this context, see comments.)

REPLY [6 votes]: A rough estimate can be obtained by comparison with the number of isomorphism classes of (symmorphic) space groups (see also Agol's comments): The map $Q \mapsto Q \ltimes\mathbb{Z}^n$ induces a bijection between the conjugacy classes $C_n$ of finite subgroups of $GL_n(\mathbb{Z})$ and the isomorphism classes of symmorphic space groups. A proof thereof can be found in my answer  to this question: 
Subgroups of the Euclidean group as semidirect products
In particular, $|C_n| = 73,\; 710,\; 6079,\; 85311$ for $n=3,4,5,6$ respectively: 
http://www.math.ru.nl/~souvi/papers/acta03.pdf
The number of isomorphism classes of (all) space groups has been estimated (cf. Remark 5.5) in 
http://www.unige.ch/math/folks/bucher/Affine/pdfAffine/BuserBieberbach.pdf
and yields the upper bound $|C_n| \le e^{\displaystyle e^{4n^2}}$. 
Actually it is conjectured by Schwarzenberger in a 1974 paper that the number of isomorphism classes of  space groups is asymptotically $O(2^{\displaystyle n^2})$. But I don't know if this has been proved in the meanwhile.<|endoftext|>
TITLE: Can a birational morphism between smooth varieties be dominated by smooth blowup sequences?
QUESTION [5 upvotes]: Suppose $f:X\rightarrow Y$ is a birational morphism between smooth varieties and D is a snc divisor on $Y$. Can we find a smooth blowup sequence on $Y$ which dominates $f$ such that the preimage of D and all other subsequent preimages involved are snc divisors? Thanks for any comments or references.

REPLY [8 votes]: Here's an easy way to do it for morphisms in characteristic zero.
Suppose that $f : X \to Y$ is projective and thus the blowup of some ideal $I$ on $Y$, see Harthsorne Chapter II, Section 7 (if $f$ is not projective, make it projective by compactifying and using Chow's lemma).  
The nice thing about setting things up this way is that now if $Z \to Y$ is any log resolution / principalization of $(Y, I)$, then $Z \to Y$ factors through $X$ by the universal property of blowing up.
Goal: All you need to do is a sequence of blowups $h : Z \to Y$ at smooth centers on $Y$ that principalizes $I$ to become an SNC divisor $E$ and keeps $h^{\star} D \cup E$ SNC (and keeps the pullback of $D$ SNC at each stage) 
Fortunately, modern resolution algorithms do exactly this if I recall correctly.  Indeed, when you are run a modern resolution algorithm, you are running it somewhat recursively on data
$(I, D)$ where $D$ is a SNC divisor obtained by previous blowups (you need to keep track of this data for the resolution algorithm to work at further steps).  
In particular, if you pass your resolution algorithm both $D$ and $I$ (in other words, telling the algorithm that this divisor needs to be handled with care), you should be able to do exactly what you want.  
I would suggest checking out the following sources to see if what I said is actually accurate :-)  (It's been a long time since I thought seriously about this).
Explicitly, in reference 2 below, you should apply Theorem 3.10 to the basic object
$(Y, (I, 1), D)$.

János Kollár's book
A paper by Bravo-Encinas-Villamayor
A paper by Wlodarczyk<|endoftext|>
TITLE: Categorification of coends and ends
QUESTION [13 upvotes]: I describe below a categorified version of the coend construction, "2-coend" for short.  It takes as input a collection of 1-categories $\{W_{xy}\}$ which afford left and right representations of a (weak) 2-category $A$.  The output is a 1-category $C$ which can be thought of as the self-gluing or self-tensor product of $\{W_{xy}\}$ via the left and right $A$ actions.

Questions: Does this construction appear in the literature anywhere (other than here)?  Does the special case where $A$ is replaced by a tensor category (2-category with one object) appear in the literature anywhere?  How about the special case where $\{W_{xy}\}$ is the product of a left $A$ representation and a right $A$ representation?  (So a categorification of tensor product rather than a categorification of the more general coend.)

The Drinfeld double is a special case of the 2-coend, and dually the Drinfeld center is a special case of the 2-end.  (See final remark below.)
The definition I give below seems natural and inevitable to me, so I'm a little surprised I have not (yet) come across it elsewhere.  If this MO question does not turn up any citations, I will tentatively assume that the definition is new.
Final remark before the details: I'm aware of definitions for general $n$ which should, in theory, specialize to this $n=2$ case.  I'm mainly interested in definitions which are specific to $n=2$ are are close is spirit to this one.

Let $A$ be a weak 2-category.  ("Weak" meaning that composition of 1-morphisms is not strictly associative.)  I have in mind the case where the 2-morphisms of $A$ form vector spaces, but I don't think that matters for what follows.
Define a left representation of $A$ to be a (weak) 2-functor from $A$ to the 2-category of (1-categories, functors, natural transformations).  For each object $x$ of $A$ we have a 1-category $Y_x$.  For each 1-morphism $e:x\to y$ of $A$ we have a (notationally overloaded) functor $e: Y_y\to Y_x$.  For each 2-morphism $h:e\to f$ of $A$ we have natural transformation (also denoted $h$) between the functors assigned to $e$ and $f$.  For each pair of composable 1-morphisms $e$ and $f$ there are also an invertible natural transformation $c_{ef}$ which relates the functor for $ef$ to the composition of the $e$ functor and the $f$ functor.  All of this data is required to satisfy some standard relations.
Let $A^{op}$ denote the 2-category where the order of composition of 1-morphisms (but not 2-morphisms) is reversed.  (Reverse horizontally but not vertically.)
Let $\{W_{xy}\}$ be a collection of 1-categories with the structure of a left $A^{op}\times A$ representation.  For each object $(x,y)$ of $A^{op}\times A$ we have a 1-category $W_{xy}$.  For each pair of 1-morphisms $e:x\to u$ and $f:v\to y$ we have a functor $e\times f$ from $W_{xy}$ to $W_{uv}$.  And so on.
Now let's define the 2-coend of $\{W_{xy}\}$ as above.  The data for the 2-coend consist of (1) a 1-category $C$; (2) for each object $x$ of $A$ a functor
$$
 gl_x : W_{xx} \to C ;
$$
and (3) for each 1-morphism $e:x\to y$ of $A$ an invertible natural transformation $s_e$ between the functors $(e\times 1)\bullet gl_y$ and $(1\times e)\bullet gl_x$, as shown in the following diagram.

(source)
(My convention in the above sort of 2-dimensional commutative diagram is to enclose the natural transformations labeling the 2-cells in boxes.)
The above data is required to satisfy conditions 1 and 2 below, and also to be universal in the sense described below.
Condition 1:  For each 2-morphism $h:e\to f$ of $A$, the following 2-sphere-shaped diagram commutes.

(source)
(The natural transformation $s_f$ labels the "2-cell at infinity".  Saying that a diagram like this commutes means that two different natural transformations, built out of two subsets of $s_e$, $s_f$, $1\times h$, $h\times 1$ and some identity natural transformations, are equal.
Condition 2:  For each composable pair of 1-morphisms $e:x\to y$ and $f:y\to z$ of $A$, the following 2-sphere-shaped diagram commutes.

(source)
The data $(C, \{gl_x\}, \{s_e\})$ is required to be universal in the following sense.  For any $(C', \{gl'_x\}, \{s'_e\})$ satisfying conditions 1 and 2 above, there exists a functor $\theta: C\to C'$ and, for all objects $x$ of $A$, a natural transformation $\eta_x: \theta\circ gl_x\to gl'_x$, such that the following diagram commutes for all 1-morphisms $e$ of $A$.

(source)

Some further remarks.

If we reverse all the arrows above, we get a definition of the 2-end of $\{W_{xy}\}$.
The above abstract definitions in terms of a universal property can be converted into more concrete definitions in terms of generators and relations.  In this version the 2-coend $C$ has (by definition) objects the union over all $x$ of the objects of $W_{xx}$.  The morphisms of $C$ are the union of all morphisms of $\{W_{xx}\}$, plus addition invertible morphisms $s_{eu}: e\cdot u \to u\cdot e$, for all $e:x\to y$ and all objects $u$ of $W_{xy}$.  These extra morphisms must satisfy relations corresponding to conditions 1 and 2.  An object of the 2-end is a tuple $(r_x)$, indexed by objects $x$ of A, where each $r_x$ is an object of $W_{xx}$.  In addition, each such tuple is equiped with a collection of invertible morphisms $t_{er}: e\cdot r_y\to r_x\cdot e$ of $W_{xy}$.  A morphism between two such enhanced tuples is a tuple of morphisms which is compatible with the $\{t_{er}\}$ according to conditions 1 and 2.
In the special case where $A$ is a tensor category and $W$ is the regular 2-category bimodule ${}_AA_A$, then the 2-end is the Drinfeld center of A and the 2-coend is what I would call the Drinfeld double of $A$.

REPLY [2 votes]: Fosco Loregian has surveyed the theory of co/lax co/ends in Section 7.1 of his new book, citing Bozapalides's PhD thesis and paper on the subject (see also [MO67083]).
Another reference is Chapter 2 of the Alexander Corner's PhD thesis or [arXiv:1709.01332]. These two references develop the theory of extrapseudonatural transformations, defining bicoends (called "bicodescent objects" there) as universal extrapseudonatural transformations. Corner also cites the following papers:

[Street, 1976] Street, R. (1976). Limits indexed by category-valued 2-functors. J. Pure Appl. Al-gebra, 8(2):149–181.
[Street, 1980] Street, R. (1980). Fibrations in bicategories. Cahiers de Topologie et Géométrie Différentielle Catégoriques, 21(2):111–160
[Lack, 2002] Lack, S. (2002). Codescent objects and coherence. J. Pure Appl. Algebra, 175(1-3):223–241. Special volume celebrating the 70th birthday of Professor Max Kelly.
[Street, 1987] Street, R. (1987). Correction to fibrations in bicategories. Cahiers de Topologie et Géométrie Différentielle Catégoriques, 28(1):53–56<|endoftext|>
TITLE: Examples of manifolds with effective circle actions?
QUESTION [6 upvotes]: I would like to know examples of smooth compact connected manifolds, on which there exists an effective smooth circle action preserving a positive smooth volume, besides the simple example: $[0,1]^d \times \mathbb{S}^1$. Is there a classification of these manifolds?

REPLY [2 votes]: This is more of a comment than an answer.  Any simplectic manifold with an effective circle action preserving the form is an example.  So  any smooth projective toric variety is an example of a manifold with an effective circle action preserving a volume form.
Edit: I forgot to say what the volume is: it's the top power of the symplectic form.<|endoftext|>
TITLE: Separating subspaces in an irreducible representation
QUESTION [7 upvotes]: Suppose $G$ is a semisimple $\mathbb{R}$-algebraic group with finite center, and suppose $G$ acts irreducibly on a vector space $V$. Suppose $U \subset V$ and $W \subset V$ are subspaces. 
$\mathbf{Question 1}.$ 
Is it always possible to find $g \in G$ such that $g U$ and $W$ are in general position, i.e. 
$$\dim(g U \cap W) = \max(0, \dim U + \dim W - \dim V)?$$
I believe the answer to this question is no. I remember thinking about this when I was a graduate student, and there was some simple example, but I no longer remember what the example was. So the real question is: construct a simple counterexample to this assertion. 
$\mathbf{Question 2}.$ It is possible to show using the Weyl unitary trick and orthogonality of characters that there is always $g \in G$ such that
$$\dim(gU \cap W) \leq \frac{(\dim U)(\dim W)}{\dim V}.$$ 
(Actually John Stalker showed this to me when we were graduate students). The question is: can one get a better bound? 
Basically, I am mostly curious if this circle of questions was ever studied, and what keywords do I need to look it up.

REPLY [8 votes]: Let $G=SL_n(\mathbb R) \times SL_m(\mathbb R)$. Let $V$ be the irreducible representation given by taking the defining representation $V_n$ of the first tensor  the defining representation $V_m$ of the second. Let $U$ be a copy of of $V_n$ inside $V_n \otimes V_m$ and let $W$ be a copy of $V_m$. The intersection of $gU$ and $W$ is always $1$-dimensional, showing the bound in question $2$ is tight and constructing a simple counterexample to question $1$ when $n\geq 2$, $m\geq 2$.

REPLY [5 votes]: Counter-example for $(1)$: $G = SO(2,2)$ with $V$ the standard four dimensional rep. To be concrete, $G$ preserves the quadratic form $q(w,x,y,z) = w^2+x^2-y^2-z^2$. $U$ is the two dimensional subspace $\mathrm{Span}((1,0,1,0), (0,1,0,1))$ and $W =  \mathrm{Span}((1,0,1,0), (0,1,0,-1))$.
Geometrically, inside the $3$-sphere $S:=(\mathbb{R}^4 \setminus \{ 0 \})/\mathbb{R}_{+}$, the quadratic $w^2+x^2=y^2+z^2$ cuts out a torus $T$, and $G$ takes this torus to itself. (One can see this by intersecting $w^2+x^2=y^2+z^2$ with $w^2+x^2+y^2+z^2=1$ to get $w^2+x^2=y^2+z^2=1/2$, clearly a torus.) The images of $U$ and $W$ in $S$ are two circles which have intersection number $2$ on the torus, so you can't move them not to intersect.
An algebraic geometer would work in $\mathbb{RP}^3$ rather than $S$, and would say that $T$ is a quadric surface and $U$ and $W$ belong to opposite rulings of the surface.<|endoftext|>
TITLE: Gandhi's quote formalized
QUESTION [5 upvotes]: Hello,
I hope this question is appropriate for Mathoverflow. Gandhi said, "Be the change that you wish to see in the world". I don't understand anything in Game/optimization theory (I don't know exactly what theory), but I am curious whether there are mathematical models, in which if the individual makes his local contribution the best, without thinking about the interaction with the decisions of the other, this makes the whole system act in the best possible way.
Thank you,
Sasha

REPLY [7 votes]: The simplest example is an Edgeworth Box economy, or more generally a Walrasian equilibrium.
The fundamental insight is that when multiple individuals are faced with the same (linear) prices, they choose consumption bundles (or production bundles or something else, depending on the model) where certain level hypersurfaces are tangent to a particular hyperplane. Moreover, equilibrium requires these tangencies to occur at a common point. On the other hand, for one interpretation of "best possible" (i.e. Pareto optimality), the condition for a best possible outcome is that these level hypersurfaces should be tangent to each other.
Because hypersurfaces tangent to the same hyperplane at the same point are tangent to each other, the result follows.  This simple insight has been vastly generalized and underlies all of modern welfare economics.

REPLY [2 votes]: I think in several cases "best self interest" can lead to overall poorer solutions than solutions with interaction. 
More formally, the self-interest maximizing version falls into the domain of traditional "Non-cooperative game theory" while the one with interactions falls  into "Cooperative game theory".<|endoftext|>
TITLE: Finding an axis-aligned ellipsoid of minimal volume which contains a given ellipsoid
QUESTION [5 upvotes]: A friend asked me to post the following question. He's not an MO user and felt it would be better received if asked by someone who was already known to the community. This is not my area, but I'll do my best to answer questions raised in the comments. Also, please feel free to retag if you want.
"Let $\Sigma$ and $\Lambda$ be $n \times n$ positive semi-definite matrices.  We say that $\Lambda$ is "conservative" for $\Sigma$ if, for all $n$-dimensional vectors $x$, we have $x^\top \Sigma^{-1} x \leq x^\top \Lambda^{-1} x$.  Our problem is given $\Sigma$, can we find (and if so how) a conservative diagonal matrix $\Lambda$ that minimizes $\det \Lambda$?
To think about it geometrically, we have an ellipsoid defined by the positive semi-definite matrix $\Sigma$, and we want to bound it with an axis-aligned ellipsoid defined by the diagonal matrix $\Lambda$ that has minimal volume. I'd also be interested in knowing, for example, whether we can bound $\frac{\det \Lambda}{\det \Sigma}$ in terms of the spectrum of $\Sigma$.
One trivial $\Lambda$ that works is $\lambda_n I$, which defines a sphere with radius equal to the largest eigenvalue of $\Sigma$.  It's easy to see that this can be far too large than need, however, for example if $\Sigma$ is already diagonal (say, Diag $[1, \varepsilon,
\varepsilon, \dotsc \varepsilon]$ for small $\varepsilon$)."
EDIT: I finally found time to chat with my friend about the answers here. It turns out that there was a typo, but not the one people (myself included) guessed. The original question should have read $x^\top \Sigma^{-1} x \geq x^\top \Lambda^{-1} x$ in the second paragraph, i.e. find the smallest axis aligned ellipsoid defined by $\Lambda$ containing the one defined by $\Sigma$. I've left the original question in place to avoid confusion. Those answering thought the typo was that the goal is to minimize det $\Lambda^{-1}$ instead of det $\Lambda$, and based on this they solved the related problem of finding the largest axis aligned ellipsoid contained inside the ellipsoid defined by $\Sigma$. Anyway, the good news is that the answers here did help my friend with his problem (which is highly related to the problem the answers solved), so I'm upvoting everyone and accepting the answer which I think helps most. Thanks to everyone for comments and answers!

REPLY [2 votes]: Suppose your "ellipsoid" is a line segment from $(1, 1, \dots, 1)$ to $(-1, -1, \dots, -1).$ Any "axis-aligned ellipsoid" must contain a cube of volume $2^n,$ and pretty obviously (although I haven't checked) must contain a ball or radius $1.$ while the maximal semiaxis of your ellipsoid is $\sqrt{n},$ and all the other semiaxes are of length $0.$ The volume is, of course, also $0.$ This means that your "stupid" algorithm is not so stupid -- the maximal semi axis in some sense controls the size of the smallest axis-aligned ellipsoid, up to a $\sqrt{n}$ multiplicative factor.<|endoftext|>
TITLE: Non-compact Kähler manifolds which admit a positive line bundle
QUESTION [9 upvotes]: A complex manifold which admits a positive line bundle is automatically Kähler. Furthermore, if the manifold is compact, then it is projective by the Kodaira Embedding Theorem. In particular, not every compact Kähler manifold can admit a positive line bundle. What about in the non-compact case? That is:

Are there any restrictions as to which non-compact Kähler manifolds can admit a positive line bundle?

REPLY [5 votes]: Take a K3, or a general $n$-dimensional complex torus $M$, $n>1$, without any integer (1,1)-classes, and remove a point $x$. You will obtain a non-compact Kahler manifold without any non-trivial line bundles (because its second cohomology stays the same). For such a manifold it's not hard to show that no positive bundles exist. Indeed, if there is a positive line bundle, it must be trivial. 
Then you would have a positive form $\eta$ which is exact. By Sibony's lemma
(see e.g. arXiv:0712.4036, Theorem 5.1), $\eta$ is locally integrable around $x$. Then, by 
Skoda-El Mir theorem, the trivial extension of $\eta$ to $M$ is a closed, positive and
hence exact current. This is impossible, because $M$ is compact and Kahler.<|endoftext|>
TITLE: Is there an $E_1$-definition of primality?
QUESTION [7 upvotes]: Here, $E_1$ denotes the set of arithmetic formulas starting with a bounded existential quantifier, followed by a quantifier-free formula. Is there an $E_1$-formula $\phi$ such that $\phi(n)$ holds
iff $n$ is prime? If yes, it is likely to be rather complicated to obtain, as this apparently implies that PRIMES is in P.
Clearly, there is such a $U_1$-formula (i.e. one starting with a bounded universal quantifier instead). And of course there's this famous prime polynomial, but this gives a $\Sigma_1$-statement where the variables correspond to exponentiations and factorials, so certainly there can be no polynomial bounds for them.

REPLY [10 votes]: I don’t see how an $E_1$ formula would give a polynomial-time algorithm. Primality testing has been known to be in NP long before AKS (due to Pratt), and this gives a definition of primes by a $\Sigma^b_1$ formula (a bounded existential quantifier followed by a formula with logarithmically bounded quantifiers); if we extend the language of arithmetic by sufficiently many polynomial-time computable functions, this becomes an $E_1$ formula.
Meanwhile, there are $E_1$ formulas in the basic language of arithmetic expressing some NP-complete predicates, such as $\phi(a,b)=\exists x\le b\,\exists y\le b\,(x^2+ay=b)$ (due to Adleman and Manders). It is even consistent with the current state of knowledge that every NP predicate $R(x)$ is definable by an $E_1$ formula where the bound on the quantifier is allowed to be $2^{(\log x)^c}$ for a constant $c$ (this is known as the “bounded Hilbert’s 10th problem”); there are some conjectures that imply that this is the case, but also some results indicating otherwise, see e.g. Pollett for a relevant discussion. Of course, a positive answer would in particular imply that primality definable in such a way. (Note that $E_1$ itself in the basic language of arithmetic is a strict subset of NP by the nondeterministic time-hierarchy theorem, as every $E_1$ predicate is computable in nondeterministic time $O(n^2)$, or even $O(n\log n)$ using the best known multiplication algorithms.)
Irrespective of general results like that, it is possible that Pratt’s (or another) NP definition of primality can be expressed as an $E_1$ formula, but as far as I am aware, this is unknown.<|endoftext|>
TITLE: Whether Hilbert schemes of 3 points on arbitrary smooth projective varieties are smooth
QUESTION [5 upvotes]: As is well known, the Hilbert scheme of two points on a given smooth projective variety X are blow up along diagonal of product of X and then quotient the Z2 action. It is smooth.
 My question is whether Hilbert schemes of 3 points on arbitrary smooth projective varieties are smooth. If so, why and how to describe the geometry of them?

REPLY [7 votes]: Yes, the Hilbert scheme of 3 points on a smooth variety is smooth.  I don't know of a global description for the resulting Hilbert scheme, but here's the local reason this is true.

Every length 3 scheme is abstractly isomorphic to a subscheme of the plane.
For a zero dimensional subscheme $\text{Spec} A$ of a smooth variety $X$, there is a natural functorial map from embedded deformations of $\text{Spec} A\subseteq X$ to abstract deformations of $\text{Spec} A$, and this map is smooth.

Fact 1 is easy.  Fact 2 takes more work, but it follows from some elementary arguments about deformation theory for affine schemes.  Combining these facts: $\text{Spec} A$ is a smooth point of $\text{Hilb}^3 X$ if and only if the miniversal abstract deformation ring of $\text{Spec} A$ is smooth if and only if, after any reembedding of $\text{Spec} A$ into $\mathbb A^2$ we have that $\text{Spec} A$ is a smooth point of $\text{Hilb}^3 \mathbb A^2$; the last statement is true by Fogarty.
If somebody has a global description of the resulting Hilbert scheme, I would be very curious!<|endoftext|>
TITLE: Solving Lyapunov-like equation
QUESTION [10 upvotes]: The following matrix equation might be a Lyapunov-like equation, but it seems hard for me to develop a simpler way to solve it. From the computation effort, I need some help for solving the special case of the following Lyapunov equation:
Let $X$ be an $n\times n$ symmetric matrix, and $I$ an identity matrix, and $A$ is a matrix whose entries are all between 0 and 1, and $A$ is invertible. I need to solve $X$ in the following equation:
$$AX+XA^T=I$$
Previously, I found some article discussing on using Krylov subspace to solve the following Lyapunov equation:
$$AX+XA^T=b \cdot b^T$$
where $b$ is a vector. Due to $b \cdot b^T$ being a rank-one matrix, Krylov subspace appraoch is highly efficient. Now in my case it is the identity matrix $I$, but $X$ in my case is symmetric. I found that in my equation $AX$ and $XA^T$ are symmetric. So by letting $Y=AX$, my equation can be reduced to :
$$Y+Y^T=I  \quad \textrm{with } \ \ Y=AX$$.
I don't know how to continue this. 
Another common way is to use tensor product to rewrite my equation as:
$$(I \otimes A + A \otimes I) vec(X) = vec(I)$$
but the LHS of the above equation is $n^2 \times n^2$ size, which is too large to solve.
Is there any other efficient way to solve this? Any advices are warmly welcome!

REPLY [5 votes]: I will summarize everything, for future reference.

All Lyapunov equations $AX+XA^T=B$ have a unique, symmetric solution $X=X^T$, unless there is a $\lambda\in\mathbb{C}$ such that $\lambda$ and $-\lambda$ are both eigenvalues of $A$. This holds, for instance, when $A$ is  Hurwitz stable, which is a common case.
As far as I know, reducing to $B=I$ (or to $Y+Y^T=I$ with a nontrivial constraint on $Y$) does not help in its numerical solution.
There are integral representations and reformulations as a system of $n^2$ linear equations in $n^2$ unknowns using Kronecker products, but they are also worthless from the point of view of numerical solution.
Up to $n\approx 1000$, the standard algorithm (also implemented in MATLAB's lyap) is the $O(n^3)$ Bartels-Stewart algorithm. The Hessenberg-Schur algorithm is a variation that helps for the more general Sylvester equation, but not in this case. For a general Sylvester equation $AX+XC=B$, the HS method requires one Schur decomposition of $A$ instead of one of $A$ and one of $C$, but if $C=A^T$ then the second decomposition comes for free.
For values of $n$ larger than $1000$, and in general for the case of $A$ large and sparse, the standard algorithms are either rational Krylov subspace methods (MATLAB code and papers on V. Simoncini's page, http://www.dm.unibo.it/~simoncin/software.html) or the ADI method (Matlab and C code on the page of Peter Benner's group, http://www.mpi-magdeburg.mpg.de/csc). Both can handle problems where $B$ has low rank. 
The case in which $A$ is large and sparse and $B$ has full rank (for instance $B=I$) is more difficult; solving it efficiently is still an active research problem. This preprint seems to go in the right direction http://www.math.cts.nthu.edu.tw/download.php?filename=683_a93bf34d.pdf&dir=publish&title=prep2012-05-005, but the methods is much newer and I have never tried it. EDIT: I should rather recommend this approach by a collaborator of mine https://arxiv.org/abs/1711.05493 ; it should solve exactly this problem for the case of large matrices.<|endoftext|>
TITLE: Comparing layered triangulations of 3-manifolds which fiber over the circle.
QUESTION [14 upvotes]: I am sorry but I am reposting this question because I wasn't logged in when I first asked it.
Ian Agol has produced a method to build an ideal layered triangulation of a hyperbolic 3-manifold which fibers over the circle with pseudo-Anosov monodromy of the fiber through periodic splitting of train tracks. Such triangulation is transverse-taut and veering.
On the converse, he proves that a layered triangulation of a manifold coming from a periodic sequence of Whitehead moves comes from a periodic train track splitting only if it is veering (see arxiv.org/pdf/1008.1606).
Both this construction and the notion of layered triangulation require to specify the fiber and the monodromy.
From Thurston's work, it is known that a hyperbolic 3-manifold $M$ which fibers over the circle fibers in many different ways, and in fact the fibers of the different fibrations are integral points of the cones over certain faces of the unit ball for the thurston norm in $H_2(M)$.
I was wondering how to compare the different layered triangulations of the same manifold that one can construct from the different fiberings. More precisely:

Are two veering layered triangulations built from two different fibrations whose fibers lie in the cone over a common fibering face for the thurston norm combinatorially the same?
If this is true, are their taut structures the same? If not, how can we relate the taut structures?
Is there a way to compare triangulations built from fibers in different fibered faces for the thurston norm?

Thank you very much for your attention.

REPLY [9 votes]: Revision 10/3/12

is true. First of all, by a result of Fried, associated to a fibered face of the Thurston norm on homology, there is a unique (up to isotopy) pseudo-Anosov flow which is transverse to every fibration in (the interior of) the face. Thus, the construction of a veering triangulation associated to each fiber in the face will give a taut ideal triangulation of the same cusped manifold obtained by drilling out the singular fibers. 
We may replace the original manifold with this cusped manifold, and ask if taut ideal
triangulations are the same for every fiber in a fibered face of the cusped manifold?

For any given fibration, some multiple of it will be fully carried by the branched surface associated to the taut ideal veering triangulation (this may be seen by the construction of a veering triangulation as a cycle of Whitehead moves of ideal triangulations of the surface). Since the branched surface is oriented, for each surface carried by the branched surface, one obtains a homologically non-trivial surface (more generally, for non-integral weights, one obtains a measured foliation representing a non-trivial homology class).
One sees that the weight space of the branched surface surjects the relative homology of the cusped manifold (since this is just isomorphic to the simplicial 2-cycles). Moreover, the subspace of cycles which are fully carried with positive weights by the branched surface form an open piecewise linear rational cone in the homology space. One also sees that any such cycle (with positive integral weights) must be fibered. So one obtains an open cover of the rational points in the interior of the fibered face by open rational cones. Since there is always a rational point on the boundary of a rational cone, if one of the boundary points of a cone associated to a veering triangulation lies in the interior of the fibered face, then two of these open sets must intersect, which gives a contradiction unless every boundary point lies in the boundary of the fibered face. Thus, the veering triangulation associated to every fiber in the face will be the same. 
For 2., I see no relation, unless there happens to be a symmetry which permutes the fibered faces. It might be interesting to find ideal triangulations that admit two distinct taut veering structures (although these won't necessarily be associated to fibered faces).<|endoftext|>
TITLE: Is the first order theory of ordered rings without infinitesimals effectively enumerable?
QUESTION [5 upvotes]: We take an ordered ring to be a structure of type $(+ - \times < 0\,\, 1)$ satisfying the usual axioms. If $A$ is an ordered ring then we say that an element $a$ of $A$ is infinitesimal if for all integers $n$ it holds that $-1<na<1$.
Let $T$ be the set of sentences that hold in every ordered ring that has no infinitesimal elements other than 0. 
Question: Is $T$ effectively enumerable? Is the set of universal sentences of $T$ effectively enumerable?

REPLY [7 votes]: The answer to the first question is no. First, let $\chi=\forall x>0\\,\exists y\\,(xy=1)$ and $T'=T+\chi$, so that $T'$ is the first-order theory of archimedean ordered fields. Let $\phi(x)$ be a formula defining $\mathbb Z$ in $\mathbb Q$, and let $\psi$ be the sentence “$\phi(x)$ defines a discretely ordered ring”. Since the only DOR embeddable in an archimedean field are the integers, $\phi(x)$ provides an interpretation of true arithmetic in the theory $T'+\psi$. Thus, $T'+\psi$ is not recursively enumerable, and a fortiori $T$ is not recursively enumerable either. (In fact, $T$ is not even arithmetical.)
Let me spell the argument in more detail. For any sentence $\alpha$, let $\alpha^\phi$ be the sentence obtained by relativizing all quantifiers to $\phi(x)$. Then I claim
$$\mathbb Z\models\alpha\iff T\vdash\chi\land\psi\to\alpha^\phi,$$
which implies that $\mathrm{Th}(\mathbb N)$ is recursively reducible to $T$.
Right to left: $\mathbb Q$ is an ordered ring without infinitesimals, hence $\mathbb Q\models\chi\land\psi\to\alpha^\phi$. Also, $\mathbb Q\models\chi\land\psi$ and $\phi(\mathbb Q)=\mathbb Z$, hence $\mathbb Z\models\alpha$.
Left to right: Let $R$ be an ordered ring without infinitesimals, we have to show $R\models\chi\land\psi\to\alpha^\phi$. Assume $R\models\chi\land\psi$. Then $R$ is a field (by $\chi$) and $S=\phi(R)$ is its discretely ordered subring (by $\psi$). Since $R$ has no infinitesimals, it is archimedean, hence so is $S$, which means $S\simeq\mathbb Z$. Thus, $S\models\alpha$, and $R\models\alpha^\phi$.
As for the second question, this is an interesting problem. It may be related to the notorious open problem whether the universal theory of $\mathbb Q$ is decidable (or equivalently, recursively enumerable).<|endoftext|>
TITLE: Topologies on the field of rationals 
QUESTION [11 upvotes]: Ostrowski's theorem give the answer for valuations, but is there a complete classification of (at least separated) topologies on Q (compatible with the field operations, obviously)?

REPLY [19 votes]: According to this link, there are as many field topologies on $\mathbb{Q}$ as there are subsets of $\mathbb{R}$, so I doubt there is a classification. A reference seems to be Wieslaw's book "topological fields" (it doesn't seem to be on google books, unfortunately).
PS: note that there is only one non-Hausdorff field (or ring) topology on a field (the two open sets one) since the closure of 0 is an ideal, and the others are completely regular, as any Hausdorff group topology. Also there is only continuum many metrizable topologies on $\mathbb{Q}$, so most of the topologies referred to in the above link are quite pathological, non first countable for instance.
PPS: searching "number of field topologies" in MathSciNet returns the following references
Podewski, Klaus-Peter
The number of field topologies on countable fields.
Proc. Amer. Math. Soc. 39 (1973), 33–38. 
Kiltinen, John O.
On the number of field topologies on an infinite field.
Proc. Amer. Math. Soc. 40 (1973), 30–36. 
and they are both freely accessible (thanks AMS!) here and here.
Concerning the proof for a countable field $K$, Podewski manages to define a continuum  $\mathcal{G}$ of (metrizable) field topologies on $K$ such that the suprema of any two distinct susbsets of $\mathcal{G}$ are distinct field topologies.
The details are somewhat complicated, but the idea is quite natural. Suffice it to say
that the metrizable field topologies on $K$ are parameterized --- via fundamental sequences of neighbourhoods of $0$ --- by chains $G$ in a partially ordered set $P$ of "conditions" (is it forcing in disguise?), which specify for a finite number of elements of $K$ wether they belong or not to the $n$-th neighborhood in the sequence.
Strangely, for uncountable fields $K$ the proof is easier, using valuations instead of chains of conditions, and the fact that the transcendence degree of $K$ over the prime subfield is the cardinality of $K$.<|endoftext|>
TITLE: Bitcoin Research
QUESTION [19 upvotes]: I have recently been assigned to advise a student on a senior thesis.  She has taken linear algebra, introductory real analysis, and abstract algebra.  Her interest is in cryptography.  And she has a love of Bitcoin.  The point of a senior thesis is to get a student to teach themselves a subject and learn to find and read mathematical papers.  Original work that could be published would be nice, but is often untenable.  
My question is whether anyone knows of any research that is/has being/been done in cryptography related to Bitcoin.  Thanks.

REPLY [10 votes]: there is a nice preprint server, widely used by cryptographic researchers: http://eprint.iacr.org
If you do a search, there are some papers talking more or less about bitcoin. I guess "Decentralized anonymous credentials" is a good begin for your student. This is indeed a well known area of research, and is the main point of bitcoins (with mining).
You may take a look at Google Scholar http://scholar.google.com/scholar?hl=en&q=bitcoin&btnG=&lr= there are many references dealing with bitcoins, probably some mathematical with a large interest.<|endoftext|>
TITLE: Bruhat decomposition of a quadric hypersurface
QUESTION [5 upvotes]: On page 3 of this paper, the authors give a Bruhat cell decomposition of a quadric hypersurface $Q$ of complex dimension $n$. This may be a stupid question, but it doesn't seem clear to me what exactly the decomposition is. They only explicitly describe the real dimension $2n-2$, $n$, and $n+2$ cells, but say that there is exactly one cell for each even real dimension. Particularly, I would like to know what the complex codimension 2 cells are.

REPLY [10 votes]: Let $Q$ live in $\mathbb{P}^{2n-1}$, so $Q$ defines a symmetric bilinear form on $\mathbb{C}^{2n}$, and the complex dimension of $Q$ is $2n-2$. Take a flag of isotropic subspaces $0 \subset F_1 \subset F_2 \subset \cdots \subset F_n$. So $\mathbb{P}(F_i) \subset Q$. For $1 \leq i \leq n-1$, the class of $\mathbb{P}(F_i)$ spans $H_{2i-2}(Q)$ or, Poincare dually, spans $H^{4n-2i-2}(Q)$.
Let $F_i^{\perp}$ denote the $Q$-orthogonal to $F_i$. So $F_n=F_n^{\perp} \subset F_{n-1}^{\perp} \subset \cdots \subset F_2^{\perp} \subset F_1^{\perp}$. For $1 \leq i \leq n-2$, the intersection $\mathbb{P}(F_i)^{\perp} \cap Q$ is a smooth hypersurface in $\mathbb{P}(F_i^{\perp})$. It is a basis for $H_{4n-2i-2}(Q)$, or for $H^{2i-2}(Q)$.
The remaining case is middle cohomology; $H_{2n-2}(Q)$. The recipe of the first paragraph would suggest taking $\mathbb{P}(F_n)$; the recipe of the second paragraph would suggest taking $\mathbb{P}(F_{n-1}^{\perp}) \cap Q$. In fact, these two classes together form a basis for the two dimensional space $H_{2n-2}(Q)$, but there is a better way to think about it. The quadratic form $Q$, restricted to $F_{n-1}^{\perp}$, has kernel $F_{n-1}$, and hence descends to a nondegenerate pairing on $F_{n-1}^{\perp}/F_{n-1}$. A symmetric nondegenerate bilinear form on a $2$-dimensional vector space has precisely two isotropic subspaces. One of them is $F_n/F_{n-1}$. Call the other one $F'_n/F_{n-1}$, so $F'_n$ is another isotropic plane sitting between $F_{n-1}$ and $F_{n-1}^{\perp}$. Then $(\mathbb{P}(F_n), \mathbb{P}(F'_n))$ form a basis for $H_{2n-2}(Q)$. From the above description, we see that $\mathbb{P}(F_{n-1}^{\perp}) \cap Q = \mathbb{P}(F_n) \cup \mathbb{P}(F'_n)$.
The Bruhat cells are just formed by taking each Schubert variety and removing the smaller Schubert varieties inside it.
In my opinion, it would be best to define a complete isotropic flag to consist of the data $F_1 \subset F_2 \subset \cdots \subset F_{n-2} \subset F_n, F'_n$. (Note that we can recover $F_{n-1}$ as $F_n \cap F'_n$.) If you notice that the containments between these subspaces look like the $D_n$ Dynkin diagram, that's not a coincidence...

$\def\Span{\mathrm{Span}}$
ADDED It might help to write all of this out for $Q$ given by $x_1 x_8 + x_2 x_7 + x_3 x_6 + x_4 x_5$. Let $F_1 = \Span(e_1)$, $F_2 = \Span(e_1, e_2)$, $F_3 = \Span(e_1, e_2, e_3)$ and $F_4 = \Span(e_1,e_2,e_3, e_4)$.  So $F_i^{\perp} = \Span(e_1, e_2, \ldots, e_{8-i})$ and $F'_4 = \Span(e_1, e_2, e_3, e_5)$. 
The Bruhat cells (also known as Schubert cells) are
$$(1:0:0:0:0:0:0:0)$$
$$(t:1:0:0:0:0:0:0)$$
$$(t:u:1:0:0:0:0:0)$$
$$(t:u:v:1:0:0:0:0) \ \mbox{and} \ (t:u:v:0:1:0:0:0)$$
$$(t:u:-wx:w:x:1:0:0)$$
$$(t:-vy-wx:v:w:x:y:1:0)$$
$$(-uz-vy-wx:u:v:w:x:y:z:1)$$<|endoftext|>
TITLE: Trying to find Russian paper from 1947
QUESTION [7 upvotes]: I'm reading a paper by V.S. Vladimirov and I.V. Volovich and they make a particular claim which is supposed to be discussed in:
N.M. Krylov, Dokl. Akad. Nauk SSSR, 60, 687 (1947).
I have not been able to reproduce the claimed result, so, it would be nice to see what precisely is done in the reference above. 
The claim is that hyperderivatives of functions on $\mathcal{A} = \mathbb{R} \oplus j\mathbb{R} \oplus j^2\mathbb{R}$ with $j^3=1$ give three dimensional wave equations. V.S. Vladimirov and I.V. Volovich do show other examples that I've been able to reproduce, but the $j^3=1$ example is elusive. 
Thanks in advance for any insights.
Update:(9/13/2012) I have tried my interlibrary loan service and contacting the author, but neither has been thus far successful. I need to find the title of the article and/or the page numbers for the interlibrary loan service to try to find it. The style of reference I list above is sadly insufficient for them to locate the article.

REPLY [11 votes]: Does this sound like it?
Крылов Н.М. О кватернионах Роана Гамильтона и понятии моногенности// ДАН СССР.- 1947.- т.55, № 9. с.799-800.
In English: N.M.Krylov, On Rowan Hamilton quaternions and the notion of monogeny, Doklady Akademii Nauk SSSR (Proceedings of the Academy of Sciences of the USSR), 1947, vol. 55, no 9, pages 799-800
It disagrees with the data you have in the volume and the issue numbers but it is the closest one I could find as far as the other parameters are concerned. I do not have an access to the article either (I doubt any kind soul was patient enough to scan all old volumes of DAN) but if you confirm that the title sounds reasonable, I can try to get a scan of the article (alas, in Russian). Also, I doubt it'll have a full proof of anything there: most likely it is a short announcement (a la Comptes Rendus). However, it still may give some clues.<|endoftext|>
TITLE: Non trivial vector bundle over non-paracompact contractible space
QUESTION [23 upvotes]: The proof that the set of classes of vector bundles is homotopy invariant relies on the paracompactness and the Hausdorff property of the base space. Are there any known examples of:
Non trivial vector bundles over a paracompact non-Hausdorff contractible space
Non trivial vector bundles over a Hausdorff non-paracompact contractible space
Non trivial vector bundles over a non-Hausdorff non-paracompact contractible space

REPLY [3 votes]: To complement the answer providing a non-Hausdorff example, here is an example on a non-paracompact Hausdorff space as given in Schröer's Pathologies in cohomology of non-paracompact Hausdorff spaces arxiv:1309.2524.
I will slightly simplify the space in question, though.
Let $(X,0)$ be the wedge sum of countably many copies of intervals $([0,1],0)$. We denote the $1$ of the $n$-th copy by $1_n\in X$, $n\geq 1$. Of course, we have to alter the topology a little to get a non-paracompact space. Namely, the open subsets $U\subset X$ are those which are open in the CW-topology, and which either do not contain $0$, or if they do, they have to contain all but finitely many of the half-open intervals $[0,1_n)$.
This space is Hausdorff, as is easily seen, and contractible: each interval $[0,1_n]\subset X$ carries the usual topology and so $\{0\}\subset [0,1_n]$ is a strong deformation retract, which implies that each arm of $X$ can be contracted onto $0$ simultaneously.
Instead of an explicit example of a (complex) line bundle, Schröer shows that $H^1(X,U(1))\not = 0$, as follows.
Using that even for non-paracompact spaces, $H^1$-sheaf cohomology is computed by Čech-cohomology, it is first shown that $H^1(X,C^0(X))\not = 0$; then $H^1(X,U(1))\not = 0$ follows from the real Euler sequence $0\to\underline{\mathbb Z}_X\to C^0(X)\xrightarrow{exp(\cdot 2\pi i)}\underline{U(1)}_X\to0$, and $H^1(X,\underline{\mathbb Z}_X) = 0$.
Instead of giving all the details, let me directly construct a non-trivial real line bundle, but using the same idea.
Consider the open cover of $X$ defined by $U_0 = \bigcup_{n\geq 1}[0,1_n)$ and $U_n = (0,1_n]\subset X$, $n\geq 1$.
Then clearly $U_n\cap U_m=\emptyset$ unless $n = m$ or $n = 0$ or $m=0$.
Hence, we have to specify transition functions $g_n\colon U_0\cap U_n = (0,1_n)\to\mathbb{R}^\times$, $n\geq 1$.
Let $f_n\colon U_n\to\mathbb{R}$, $n\geq 1$, be continuous functions with $f_n(1) = 0$, but $f_n(x)\not=0$ for $x\not=1$, and let $g_n = \tfrac{1}{f_n}$.
(E.g., $g_n(x) = (1-x)^{-1}$.)
Let $L$ be the line bundle defined by those transition functions.
I claim that every section $s\colon X\to L$ is trivial at all but finitely many $1_n$.
In particular, $L$ itself is non-trivial.
Let $s\colon X\to L$ be any section. It defines (and is defined by) continuous functions $s_n\colon U_n\to \mathbb{R}$, $n\geq 0$, satisfying $s_n(x) = f_n(x)s_0(x)$ for all $x\in (0,1_n)$, $n\geq 1$.
From the existence of the limit $\lim_{x\to 1}f_n(x) = f_n(1) = 0$, we conclude that $s_n(1_n) = 0$ as soon as the limit $\lim_{x\to 1_n}s_0(x)$ exists as well.
To see that this happens for all but finitely many $n$, it suffices to observe that $s_0$ is bounded away from finitely many $U_n$, which is a consequence of our choice of topology.<|endoftext|>
TITLE: Examples of q-expansions in a Hida family
QUESTION [16 upvotes]: Let $p$ be a prime number and $N$ a positive integer not divisible by $p$. 
For some easy choices of $p$ and $N$, can anybody provide me with explicit examples of collections $$\{f_k,\quad 2\leq k \leq k_0 \}$$ of q-expansions such that, for each $k$, $f_k$ is the $q$-series of a classical eigenform of weight $k$ and level $Np$, all being members of an ordinary $p$-adic Hida family of tame level $N$? Here $k_0$ is some reasonable upper bound for the weight, and I leave to the reader the meaning of the term "reasonable". Even if $k_0=4$ I will already be happy, and it is also fine if $k$ only runs among some of the integers between $2$ and $k_0$. But of course, the cardinal of $\{ f_k\}$ should be greater than $1$!
I am interested in Hida families which are neither Eisenstein not CM, as explicit examples of those are already well-known.
The simplest example seems to arise when one takes $N=1$ and $p=11$. In this case the associated Hida Hecke algebra is $\mathbf{T}=\Lambda:=\mathbb{Z}_p[[T]]$. I guess that people like Kevin Buzzard, Robert Pollack, William Stein and many others have computed the $q$-expansions $f_k$ in this case, up to some reasonable $k_0$, and I would be happy to have access to these computations.
Yet another interesting example: reading the paper "How can we construct abelian Galois extensions of basic number fields?" by Barry Mazur, he reports on an example computed by Citro and Stein, where $N=1$ and $p=691$. They compute explicitly $\{f_2,f_{12}\}$ where $f_{12}$ is the $p$-stabilization of the discriminant modular form $\Delta$ and $f_2$ turns out to be the only weight 2 newform of level $p$ and character $\omega^{-10}$, where $\omega$ is the Teichmuller character. I would like to know the $q$-expansions $f_k$ in this case for some other integers in between $2$ and $12$. (For integers $k$ larger than $12$ one can possibly compute $f_k$ by multiplying $f_{12}$ by a suitable Eisenstein series of weight $k-12$ and applying Hida's ordinary projector.) 
And of course, any other collection of examples is welcome!

REPLY [16 votes]: In the case when the Hida algebra is simply $\Lambda$, one can use families of overconvergent modular symbols to compute the $q$-expansion of the Hida family where the coefficients are functions of the weight.  
The idea is the following: form a "random" family of overconvergent modular symbols -- that is modular symbols with values in a distribution module tensor power series in the weight.  Then iterate $U_p$ on this family until you are in the ordinary subspace (modulo the accuracy at which you are working).  Since the Hida algebra is just $\Lambda$, the result will be an eigensymbol.  Its Hecke-eigenvalues will then be functions of the weight and these are the coefficients of the formal $q$-expansion I mentioned above.
With this $q$-expansion in hand, just plug in your favorite weight and you'll get the form in the Hida family of that weight (of course, computed modulo your fixed accuracy).  
This approach actually came out of an Arizona Winter School 2011 student project.  Currently three graduate students at University of Madison (Lalit Jain, Marton Hablicsek, and Daniel Ross), together with Rob Harron and myself, are writing up a paper on this. 
Our code is still very much in beta form, but hopefully this example works.  Take $p=3$ and $N=5$.  Then the first few Hecke-eigenvalues (mod $3^3$) of the associated Hida family are:
$$
a_2(w) = 12w^2 + 7w - 1,
$$
$$
a_3(w) = 21w^2 + 11w - 1,
$$
and
$$
a_5(w) = 16w^2 + 17w + 1.
$$
Here $w$ is not exactly the weight variable.  Instead, to specialize these eigenvalues to weight $k$, one simply sets $w = 4^{k-2} - 1$.  (Here is 4 is a topological generator of $1+3{\mathbb Z}_3$.)
Here's the example with $N=1$ and $p=11$ -- below I'm working modulo $11^8$:
$$
a_2(w) = 37w^7 + 880w^6 + 12388w^5 + 151975w^4 + 385840w^3 + 10344442w^2 + 40094463w + 857435522,
$$
$$
a_3(w) = 71w^7 + 681w^6 + 8325w^5 + 94314w^4 + 220797w^3 + 2758794w^2 + 16210985w + 428717761,
$$
and
$$
a_5(w) = 12w^7 + 472w^6 + 2245w^5 + 44445w^4 + 30443w^3 + 14355835w^2 + 180294533w + 1.
$$
As before, setting $w = (1+11)^{k-2}-1$ specializes to weight $k$.  For instance, taking $k=12$ one should recover the first few Fourier coefficients of $\Delta$ modulo $11^8$.<|endoftext|>
TITLE: Sharpening of Lindelöf hypothesis
QUESTION [9 upvotes]: The Lindelöf hypothesis is:
$$
\forall \epsilon >0,\exists C_\epsilon >0,\forall t\ge 1,\quad
\vert\zeta(\frac12+it)\vert\le C_\epsilon t^\epsilon.\qquad \tag{LH}.
$$
It is a weaker statement than the Riemann hypothesis: $(RH)\Longrightarrow (LH)$. The (not-so-easy) texbook result that the estimate above is true for $\epsilon=1/6$ was improved in 1986 by Bombieri and Iwaniec (mathreview#:MR0881101) who found the estimate for $\epsilon=\frac{9}{56}$.
Several works followed, using some variations of their method, but I do not think that the threshold $1/7$ was reached. 
Now my question:
is there a stronger  inequality, e.g. $$
\exists C >0,\forall t\ge 2,\quad\vert\zeta(\frac12+it)\vert\le C(\ln t)^C\tag{LH$^\sharp$}
$$
which would be equivalent to $(RH)$?

REPLY [4 votes]: There are reason to doubt that the size of $\zeta(s)$ alone could be responsible for the truth of the Riemann Hypothesis, however bounds for $\zeta(s)$ could be equivalent to a statement of the form "RH does not fail massively", as I will explain further.
Let me present one argument to convince you that the size of $\zeta(s)$ should be independent of the truth of RH: Suppose that the following (unlikely, but currently not ruled out) configurations of zeros occur in infinitely many intervals $[T; T + 1]$: we have roughly $\asymp \log T / \log\log T$ clusters of $\log\log T$ zeros, then in such interval $\zeta(s)$ should be of size $\exp(c \log T / \log\log T)$ (in particular contradicting the conjecture of Farmer, Gonek and Hughes). And then imagine that there are a few (say $4$) zeros of $\zeta(s)$ lying off the critical line. 
The two behaviors are envisage-able to occur simultaneously, unless of course we prove the falsehood/truth of each statement independently. 
To wit: The size of $\zeta(s)$ is in a sense a local behavior, having $O(\log T)$ badly placed zeros in a $O(1)$ vicinity of a point $1/2 + it$ is enough to produce a very large (if not super large) value of $\zeta(s)$ at that point. Therefore the size of $\zeta(s)$ will not be affected by the truth or a small failure of the Riemann Hypothesis. However good bounds for $\zeta(s)$ can prevent the Riemann Hypothesis from failing badly. For example a result of Turan and Halasz asserts that if the Lindelof Hypothesis is true then there are at most $O(T^{\varepsilon})$ zeros in the half-plane $\sigma > \tfrac 34$.<|endoftext|>
TITLE: p-adic representations of groups
QUESTION [5 upvotes]: Hi, 
is there a possibility to classify irreducible representations of (finite) groups over $ \mathbb{Z_p} $
with $p\neq 2$ a prime? Can one get the number of the irred. representations  here? I tried to solve this for "easy" groups like the cyclic or symmetric groups, but did not really get a result. Maybe there is a connection to representation over $\mathbb{Z}/p\mathbb{Z}$?
Best regards

REPLY [6 votes]: What Alex says is correct, but there is another viewpoint, which is a bit easier in some ways.
You can consider finitely generated $\mathbb{Z}_{p}$-free $\mathbb{Z}_{p}G$-modules which have no proper non-zero pure submodules. These are the modules which are irreducible when the ground ring is extended to $\mathbb{Q}_{p}.$ There are finitely many isomorphism types of such modules. It may be possible to deduce the structure of faithful such modules for the dihedral group using Clifford theory.<|endoftext|>
TITLE: Partitions of $\mathbb{R}^d$ by implicit polynomial equations
QUESTION [17 upvotes]: Given a polynomial
$p(x_1,x_2,\ldots,x_d)$ 
in $d$ variables, with maximum degree $k$,
what is the maximum number of 
components of $\mathbb{R}^d$ minus $p(\ldots)=0$?
In other words, into how many pieces can an implicit 
polynomial equation partition $\mathbb{R}^d$?
For example, the following three equations partition
$\mathbb{R}^2$ or $\mathbb{R}^3$ into $3$, $4$, and $2$ pieces
respectively
(I think!):
$$x^3 y^2+x^3 -3 x^2 y -y^2 +4 x y+x=0$$
$$x^6 y^8+x^3+4 x y-y=0$$
$$x^4+3
   \left(x^2+y^4+z\right)-
\left(x^2+y^2+z^2\right)^2+y^3+z^5 + 2 xy=3$$

   

Of course the answer is $k+1$ in $\mathbb{R}^1$.
I suspect this is well known for $\mathbb{R}^d$; 
if so, I would appreciate a pointer.  Thanks!
Update. Greg Martin's idea (from the comments), using the 5th Chebyshev polynomial of the first kind:

         

As Aaron Meyerowitz points out, here the degree $k=10$, and the plane is partitioned into $28$ pieces. But using Pietro Majer's line-arrangement idea leads to (now corrected:) $56$ pieces for a degree $10$ polynomial.

REPLY [7 votes]: A slightly more general result -- namely on the number of connected components of the non-zeros of a polynomial, restricted to the zeros of another polynomial -- where the degrees of the two polynomials could be different, can be deduced from the main theorem in the paper titled  
"Refined Bounds on the Number of Connected Components of Sign Conditions on a Variety" by Barone and Basu, 
DISCRETE & COMPUTATIONAL GEOMETRY
Volume 47, Number 3 (2012), 577-597, DOI: 10.1007/s00454-011-9391-3
Note that this is a bound only on the number of connected components and not on he higher Betti numbers. I don't know if the added generality is helpful to the original poster.<|endoftext|>
TITLE: The Class Number One Problem for Real Quadratic Fields
QUESTION [11 upvotes]: An approach to the Gauß class number one problem for imaginary quadratic fields is to determine the integral points on the modular curve $Y_{nonsplit}(n)$ for a suitable $n$. Here follows a quick summary, but see Appendix A in Serre's book "Lectures on the Mordell-Weil Theorem" for more details:
Given an imaginary quadratic $K$ of class number 1, one considers an elliptic curve $E$ with CM by $O_K$, which is unique up to $\mathbb{C}$-isomorphism. Its $j$-invariant lies in $\mathbb{Z}$. Given any integer $n$, all of whose prime divisors are inert in $K$, our $E$ yields a unique integral point on $Y_{nonsplit}(n)$. That means, fixing an $n$ and letting $p$ be its largest prime divisor, any imaginary quadratic $K$ of class number 1 whose discriminant is larger than $4p$ in absolute value will furnish a unique integral point on $Y_{nonsplit}(n)$. Hence, determining the integral points on $Y_{nonsplit}(n)$ for one $n$ for which there are finitely many integral points solves the problem. Heegner used $n=24$, as did Stark.

My question is whether a similar approach has been, or can be, used in the real quadratic case;

the goal being to prove that there are infinitely many such fields of class number one. 
For what it's worth, here's what would happen in my pipe dream:

To a real quadratic $K$ of class number 1 one attaches uniquely an abelian surface (or some other kind of object) with real multiplication by $O_K$;
One shows that this object gives rise to a unique rational point on some moduli space, and conversely, all such rational points would come from a $K$; ($Y_{nonsplit}$ plays this role in the imaginary case...)
One shows that the moduli space has infinitely many rational points. 

(Since this is a well-known open problem, I'll follow the advice of the FAQ and make it community-wiki. I am a little embarrassed asking such a speculative question, but I also feel that speculation drives a lot of mathematical research.)

REPLY [8 votes]: I would say there are two issues: first of all, the good thing making the above machinery work is that the moduli problem you look at, represented by $Y_\text{nonsplit}(n)$, works for all quadratic fields at once. In other words, you are interested in looking at all possible curves with CM by some imaginary quadratic field and you end up looking at points always in the same moduli space. This is not the case for Hilbert-Blumenthal surfaces with real multiplication by a real quadratic $F$, who do possess a moduli space (stack, indeed) $\mathfrak{M}_F$ which heavily depends upon $F$.
But suppose this is only a technical issue, either by constructing some monstrous $\mathfrak{M}$ parametrizing abelian surfaces with al multiplication by some quadratic field not specified in the moduli problem; or by hoping, in your "pipe dream" that one proves that there are infinitely many $F$ with at least one element in $\mathfrak{M}_F(\mathbb{Q})$. Then a major (and, to my knowledge and understanding, not a merely technical one) problem is the following: the crucial step in the CM procedure, as you said, is the implication (now $K$ is imaginary quadratic)
$$
h(K)=1\Leftrightarrow j(E_K)\in\mathbb{Q}
$$
(it is indeed integral, but that won't matter, here) where $E_K$ is some/any elliptic curve with CM by $K$. This is false for real multiplication: to convince yourself of the failure of one implication, observe that there are elliptic curves with real multiplication by $\mathbb{Q}$ (what else?) which are not defined over the rationals although $h(\mathbb{Q})=1$. Likewise, the field of definition of the abelian surface tells you almost nohing on the arithmetic of the field of real multiplication, unlike the CM case, preventing you from reducing the class number problem to counting varieties (with bonus structure) defined over $\mathbb{Q}$ - i.e. rational points on some moduli space.<|endoftext|>
TITLE: Philosophy behind Mochizuki's work on the ABC conjecture
QUESTION [290 upvotes]: Mochizuki has recently announced a proof of the ABC conjecture. It is far too early to judge its correctness, but it builds on many years of work by him. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?

REPLY [7 votes]: For the sake of completeness, let me add the references of the published version in Publ. RIMS that appeared earlier in March this year (should be rather a comment, but the references are too long for that):
Mochizuki, Shinichi, Inter-universal Teichmüller theory. I: Construction of Hodge theaters, Publ. Res. Inst. Math. Sci. 57, No. 1-2, 3-207 (2021). ZBL1465.14002.
Mochizuki, Shinichi, Inter-universal Teichmüller theory. II: Hodge-Arakelov-theoretic evaluation, Publ. Res. Inst. Math. Sci. 57, No. 1-2, 209-401 (2021). ZBL1465.14003.
Mochizuki, Shinichi, Inter-universal Teichmüller theory. III: Canonical splittings of the log-theta-lattice, Publ. Res. Inst. Math. Sci. 57, No. 1-2, 403-626 (2021). ZBL1465.14004.
Mochizuki, Shinichi, Inter-universal Teichmüller theory. IV: Log-volume computations and set-theoretic foundations, Publ. Res. Inst. Math. Sci. 57, No. 1-2, 627-723 (2021). ZBL1465.14005.
(Peter Scholze indicates in his review that the versions do not differ with respect to the issues Stix and he raised in 2018).<|endoftext|>
TITLE: A space in which sequences have unique limits but compact sets need not be closed
QUESTION [10 upvotes]: A topological space is KC if every compact subspace is closed.
A topological space is US if every convergent sequences has exactly one limit.
Does someone know an easy example of a US space which is not KC?
Thanks.

REPLY [5 votes]: I refer to COROLLARY 1 of This Article.
In COROLLARY 1 of it, $X^+$ denotes the one point compactification of the topological space $X$:

COROLLARY: Let $X$ be a Hausdorff space.Then:
(a) $X^+$ is always $US$.
(b) $X^+$ is $KC$ if and only if $X$ is a $\kappa$ space.

So it suffices to choose a Hausdorff space $X$, which is not $\kappa$ space. then $X^+$ is $US$ but is not $KC$.
PS: The topological space $X$ is called $\kappa$ space, if A subspace $A$ is closed in $X$ if and only if $A \cap K$ is closed in $K$, for all compact subset $K \subset X$.<|endoftext|>
TITLE: When is an orbit spherical?
QUESTION [8 upvotes]: I asked the following question over at math.stackexchange, but got no answers. Maybe it's less well-known than I thought, but I still wanted to ask here:
Let's assume we have an affine, reductive, algebraic group $G$ acting algebraically on a variety $X$, everything over an algebraically closed field of characteristic zero. Let $x\in X$ be some point with reductive stabilizer $H:=G_x$. Under what conditions on $x$ or $H$ is the orbit $G.x\cong G\newcommand{\qq}{/\hspace{-.8ex}/}\qq H$ a spherical variety? Let me briefly recall that a spherical variety is a homogeneous space $G\qq H$ satisfying one of the following, equivalent properties:

Any Borel subgroup $B\subseteq G$ has an open orbit in $G\qq H$.
Every equivariant completion of $G\qq H$ contains only finitely many orbits.
For every irreducible $G$-module $V$ and any character $\chi$ of $H$,
$$\dim\left\{~v\in V \mid \forall h\in H: h.v = \chi(h)v ~\right\}\le 1.$$

I was hoping that this is well-known, but I cannot find any direct statements of that kind. Searching for the keywords "orbit" and "spherical" is quite fruitless because of property 1. 
Edit: In the cases of interest to me, the orbit $G.x$ is affine.

REPLY [4 votes]: An extended comment:  As Dave points out, Michel Brion has been active in the study of spherical varieties (and some generalizations) in the setting of reductive algebraic groups in characteristic 0.    His work spans by now several decades, up to the present, and includes research papers in both French and English along with numerous surveys and lectures.    His classification result in the affine case (i.e., $H$ affine) was already outlined in his note:
Classification des espaces homogenes spheriques.
C. R. Acad. Sci. Paris S´er. I Math. 301 (1985), no. 18, 813–815.
A number of these papers are published in Comment. Math. Helv., while an influential joint paper with Luna and Vust can be found online at GDZ:
Espaces homogenes spheriques.
Invent. Math. 84 (1986), no. 3, 617–632.
I'm not sure whether this extensive work, or that of Knop et al. linked by Dave, will have all the information you want, but the subject remains active.<|endoftext|>
TITLE: Is every distance-regular graph vertex-transitive?
QUESTION [7 upvotes]: Is every distance-regular graph vertex-transitive?

REPLY [4 votes]: Very small examples of distance-regular graphs that are not vertex-transitive are the Chang graphs on 28 vertices.  They are strongly regular but not vertex-transitive.  They are constructed from the very symmetric graph T(8).  The groups have sizes
http://en.wikipedia.org/wiki/Chang_graphs
http://www.win.tue.nl/~aeb/graphs/Chang.html 
384,360 and 96, which is not even divisibly by 28.
It is not true either that for large diameter, they have to be vertex-transitive.  In 2005, Koolen and van Dam constructed new distance-regular graphs of arbitrary diameter d, by twisting the very symmetric Grassmann graphs (hence the name "twisted Grassmann graph").  They have two orbits on vertices.<|endoftext|>
TITLE: When do blowups ''commute''?
QUESTION [14 upvotes]: Let $M$ be a manifold (variety, scheme, your favorite object) and let $N_1,N_2$ be two submanifolds (subvarieties, closed subschemes, ideal sheafes, etc.) such that $N_1 \cap N_2 \neq \emptyset$. Denote by $\text{BL}_{N_i}M$ the blowup of $M$ along $N_i, i=1,2$ and let $\text{BL}_{N_1}\text{BL}_{N_2}M$ be the blowup of $\text{BL}_{N_1}M$  along the proper transform $N_2^\prime$ of $N_2$. (Define $\text{BL}_{N_2}\text{BL}_{N_1}M$ vice versa). 

Question: Under which conditions does the following hold: $\text{BL}_{N_1}\text{BL}_{N_2}M \cong \text{BL}_{N_2}\text{BL}_{N_2}M$ ?

This should be true if we are in the situation $N_1 \subseteq N_2 \subseteq M$, since ``blowups commute with restriction'' but is it in genreal true? If so, is there some reference available? If not, is there a criterium when blow ups commute (e.g. something like transversal intersection or) or/and can you come up with an easy counterexample?

REPLY [14 votes]: Sándor is exactly right, this generally isn't true.  However, there is a related statement that is always true, instead of blowing up strict transforms of $N_1$ and $N_2$, you should blow up total transforms.

Blowing up total transforms
Suppose that $I_1$ and $I_2$ are the ideals defining $N_1$ and $N_2$ in $M$.  Let $$Y_1 = Bl_{I_1} M = Bl_{N_1} M.$$
Now define the total transform of $I_2$, denoted $\overline{I_{2}}$, to be the ideal formed by extending $I_2$ to $Y_1$, in other words $\overline{I_2} = I_2 \cdot O_{Y_1}$ (note that by the universal property of blowing up $I_1 \cdot O_{Y_1}$ is an invertible sheaf).
$\overline{I_2}$ need not define a manifold, or even a reduced scheme, but we can still blow it up.  Set $$Y_{1,2} = Bl_{\overline{I_2}} Y_1.$$
I claim:  
Theorem: We have $Y_{1,2} = Y_{2,1}$ where the second object is obtained in the same way but blowing up $I_2$ first.
There are at least two ways to see this.  

You can do this from the universal property of blowing up.
Since you already assumed that $M$ is a manifold, it is integral.  Thus the charts of a blowup can be computed as follows.  Suppose $M = \text{Spec } A$ is affine for simplicity.  If $\langle x_1, \dots, x_n\rangle$ generate an ideal $I$, then as $i = 1,...,n$ varies, $Y_{I,i} = \text{Spec } A[x_1/x_i, \dots, x_i/x_i, \dots, x_n/x_i]$ are affine charts covering the blowup.  $Y_I = Bl_I Y$.

EDIT: As Dustin Cartwright points out in a comment below, blowing up $I_1$ followed by blowing up $\overline{I_2}$ is the same as blowing up $I_1 \cdot I_2$.  

A normal crossings example
I'd like to give you an example in $\mathbb{A}^4 = \text{Spec } k[x,y,u,v]$ showing the difference between blowing up strict transforms and total transforms, even when the $N_i$ intersect transversally (are in normal crossings).  Let $N_1 = V(x,y)$ be a plane and let $N_2 = V(y,u,v)$ be a line.  
Suppose we blowup the line $N_2$ first, then the strict transform and the total transform of $N_1$ coincide.  In particular, $Y_{2,1}$ is the object you were considering.  
On the other hand, let's blow up $N_1$ first, in this case the strict transform $\widetilde{N_2}$ of $N_2$ is a line, but the total transform $\overline{N_2}$ of $N_2$ is two lines.  $\overline{N_2}$ contains the strict transform but also one of the new $\mathbb{P}^1$'s lying over the origin of $N_1 \subseteq \mathbb{A}^4$.  The blowups of the strict and total transform of $N_2$ on $Y_1$ are thus manifestly different.  
One can check that $Y_{1,2} = Y_{2,1}$ if you did the total transform blowup, and so blowing up the strict transforms does not commute.
Note: If I recall correctly, you can also do the example when two planes are kissing.  $N_1 = V(x,y)$, $N_2 = V(u,v)$.  The same sort of thing happens, but it's even messier.  I've seen this example in various papers on resolution of singularities.

In the literature
The difference between strict and total transforms plays a key role in modern resolution of singularities algorithms.  Even though conceptually strict transforms are much easier to think about, they are much harder to compute or control.  Thus modern algorithms tend to use total transforms instead in SOME places, and then peel off any copies of the exceptional you can, while possibly leaving embedded components.  
See for example: 

Section 7 of This paper by Bravo-Encinas-Villamayor
Example 2.3 of This paper by Howard Thompson<|endoftext|>
TITLE: New formula for counting semi-standard Young tableaux?
QUESTION [7 upvotes]: I have a proof that given a partition $\lambda=(\lambda_1,\dots,\lambda_l)$
then the number of semi-standard Young tableaux (SSYT) of shape $\lambda$ with entries in $1,2,\dots, n$ is given by
$$\frac{1}{1!2!\cdots (n-1)!} \prod_{1\leq i\lt j\leq n} (\lambda_i-i)-(\lambda_j-j).$$
(We define $\lambda_j:=0$ if $j \gt l.$)
The product is also recognized as a Vandermonde determinant.
There are plenty of product formulas (over boxes in the tableau) 
and determinant formulas (but not in Vandermonde form, as far as I can tell) 
for the number of such SSYTs, but I have not seen a this particular one in the literature or in any article I've come across.
Is this formula known? Is this formula of any interest?

REPLY [4 votes]: I've finally found a place that gives this formula explicitly, without any need for piecing together or rewriting or taking limits using L'Hopital: Martin Aigner, A Course in Enumeration, Springer 2007, Section 8.5, formula (7) and the sentence that follows it.
I am wondering if anyone has ever given a bijective proof, though (after multiplying with the denominator).<|endoftext|>
TITLE: Vector space structure on velocity space of manifold
QUESTION [5 upvotes]: Let $M$ be $C^{\infty}$-manifold and $x\in M$. We define $(k,r)$-velocity space at x as
$(T_k^rM)_x:=J_0^{r}(\mathbb{R}^k,M)_x$.Can we define vector space structure on $(T_k^rM)_x$?

REPLY [2 votes]: Nice answer BS. I was about to post something similar but I didn't have a proof of the non-linearity of the action of $k$-jets of diffeomorphisms.  One additional remark that might help the OP is that you can put  a vector space structure on $J^k_0({\mathbb R}^n, M)_x$ if you have additional structure on $M$.  Basically you need to be able to determine a family of co-ordinates (actually $k$-jets of co-ordinates) that are related by linear transformations to avoid the non-linear action of diffeomorphisms when you change co-ordinates.  Also you want to choose different co-ordinates at each point of $M$. Sufficient would  be to choose at each $x \in M$ the $k$-jet of a diffeomorphism from $M$ to $T_xM$ sending $x$ to $0 \in T_xM$.  For example if $M$ is Riemannian the $k$-jet of the inverse of the exponential map would do or if $M$ is a submanifold orthogonal projection onto the tangent subspace would work. In such a case composition with the chosen $k$-jet of a diffeomorphism defines a bijection 
$$
J^k_0({\mathbb R}^n, M)_x \to J^k_0({\mathbb R}^n, T_xM)_0
$$
and the latter space is a vector space because $T_xM$ is a vector space.<|endoftext|>
TITLE: Can flatness be specified by a natural coherent sheaf?
QUESTION [11 upvotes]: More precisely:

Given a finite-type morphism $f \colon X \to Y$ of nice schemes (say, both of finite type over a field), is there a "natural" coherent sheaf $\mathcal F_f$ on $X$ such that the support of $\mathcal F_f$ is precisely $\{x \in X : f \text{ is not flat at }x\}$?

I'd rather not say what exactly "natural" should mean here, but if you held a gun to my head, I would guess it means that $\mathcal F_f$ is preserved under flat pullbacks; in other words, that if $g\colon Z \to Y$ is a flat finite-type morphism of nice schemes, then $\mathcal F_{f \circ g} = g^* \mathcal F_{f}$.
Example: If $X = \operatorname{Spec} B$, $Y = \operatorname{Spec} A$, and $A \to B$ is a local morphism of local noetherian rings, then $$\ker(B \otimes_A \mathfrak m \to B)$$ is 
a finitely generated $B$-module that is zero iff $B$ is flat over $A$ (by the local criterion for flatness).  [Notation: $\mathfrak m$ is the maximal ideal of $A$.] This construction is respected under pullback by flat local morphisms $B \to C$ of local noetherian rings. Unfortunately, it runs into trouble if you look at flatness anywhere except the closed point of $\operatorname{Spec} B$.
Motivation: There are standard theorems that certain sets (including the one above) are "open," without really specifying a closed subscheme structure on the complement. In some cases, the complement can be described as the support of a coherent sheaf, which gives it a natural scheme structure. For instance, the "indeterminacy locus" of a rational function may be described by the cokernel of the ideal of denominators. I personally find such constructions more satisfying and memorable than more direct proofs that a particular subset is open.

REPLY [4 votes]: Assume the morphism $f$ is projective. Let $L$ be a relatively ample line bundle on $X$. Then there is such $N$ that the nonflatness locus of $f$ coincides with the locus where $F := f_*(L^N)$ is not locally free. If additionally $Y$ is smooth the latter is the union of supports of the sheaves ${\mathcal Ext}^i(F,O_Y)$, $i>0$. 
Of course this is not quite what you want, but still sometimes is useful.<|endoftext|>
TITLE: Solovay's paper from AD+ that all sets are Ramsey
QUESTION [7 upvotes]: For my research I've been trying to locate Solovay's proof that AD+ implies that all sets are Ramsey (or completely Ramsey, depending on the terminology).  I know that it uses Mathias forcing but I can't locate the paper.  Does anyone know where I could find it or another paper that reproduces his proof using that technique?

REPLY [9 votes]: About all you need is in Adrian's paper "Happy families". It is not spelled out as following from ${\sf AD}^+$, of course, since the concept did not exist yet. 
Anyway, if you are familiar with Solovay's arguments in his paper on all sets of reals being measurable (or as discussed in the Feng-Magidor-Woodin paper on universally Baire sets), then all you want to read is section 2 (up to 2.2.3) of my paper with Richard Ketchersid, "A trichotomy theorem in natural models of ${\sf AD}^+$", in Set Theory and Its Applications, Contemporary Mathematics, vol. 533, Amer. Math. Soc., Providence, RI, 2011, pp. 227-258, MR2777751, and also available at my papers page.
Very briefly, what we use of ${\sf AD}^+$ is little more than all sets of reals having $\infty$-Borel codes. If $S$ is an $\infty$-Borel code for a set of reals $A$, then $L[S]$ is a model of choice, so (from determinacy), its reals are countable and in fact $\omega_1^V$ is inaccessible in $L[S]$. One can now run the argument that sets of reals are Ramsey in the Solovay model, noting that there are (in $V$) generics over $L[S]$ for Mathias's forcing, and the proof is concluded from the defining property of $\infty$-Borel sets.
Since the proof uses the $\infty$-Borel machinery so explicitly, it is still open whether ${\sf AD}$ suffices for this result.<|endoftext|>
TITLE: projective representations of a finite group over reals
QUESTION [5 upvotes]: It follows from the theory of Schur multiplier that any $n$-dimensional projective representation $\theta : G\to PGL(n,\mathbb{R})$ of a finite group $G$ is either an ordinary representation of $G$, i.e.
$\theta : G\to GL(n,\mathbb{R})$, or lifts to an ordinary representation $\theta' : 2.G\to GL(n,\mathbb{R})$ of a double cover $2.G$ of $G$.
A direct reference to this fact would be very useful.
Is there a more direct way to see this, preferably suitable for non-algebraist readers? The quickest route I know is to mimick the usual proof that the $|G|$-th power of the cocycle is trivial, as in e.g. Theorem 11.15 in [1]. 
[1]: I.M.Isaacs, Character Theory of Finite Groups, Dover 1994.

REPLY [3 votes]: I think a proof is also in Curtis and Reiner (Wiley, 1962). Not sure whether it counts as a non-algebraic proof, but if you think of the projective representation as a map $\sigma$ from $G$ to ${\rm GL}(n, \mathbb{R})$, defined only up to scalars, and for each $g \in G,$ and you make a particular choice of $g\sigma$ for each $g \in G,$ you can if necessary replace it by a (real) scalar multiple so that $d(g) = {\rm det}(g\sigma) \in \{1,-1\}$ for all $g \in G.$ Then the double cover you need (if you need one at all) is $\hat{G} = \{(g, d(g)):g \in G \}$ with the multiplication of the second component forced by making $\sigma$ a genuine representatin of the new group ${\hat G}.$<|endoftext|>
TITLE: Kronecker's Jugendtraum for real quadratic fields?
QUESTION [10 upvotes]: Kronecker's Jugendtraum (or Hilbert's 12'th problem) is to find abelian extensions of arbitrary number fields by adjoining `special' values of transcendental functions. The Kronecker-Weber theorem was the first realisation of this: i.e. $\mathbb{Q}^{ab}=\mathbb{Q}^{cycl}=\mathbb{Q}(e^{2\pi i \mathbb{Q}})$.
If $K$ is an imaginary quadratic field, then the theory of complex multiplication realises Kronecker's dream: roughly $K^{ab}=K(j(\tau), \wp(\tau,z))$, where $\tau$ is a `special' value of the upper half plane corresponding to an elliptic curve $E\cong\mathbb{C} / \Lambda$ with complex multiplication by the ring of integers of $K$, and $z \in \Lambda \otimes \mathbb{Q}$ (this is like adjoining $j(E)$ and the $x$-coordinates of the torsion points of $E$).
Is there some class of special analytic functions, and some special kind of objects which conjecturally play the role of the $j$-function, the $\wp$-function and CM elliptic curves for real quadratic fields?

REPLY [4 votes]: There is a proposal by Manin to engage noncommutative tori - when the modular parameter of an elliptic curve approaches the real line. It started in his Real Multiplication and noncommutative geometry from 2001 but I don't know about recent developments - what I found is his later update, a thesis from 2006 and a talk by Marcolli from ICM 2010. Maybe somebody knows about current status of this? I am really curious.<|endoftext|>
TITLE: $\infty$-ary tensor product on a category
QUESTION [11 upvotes]: Is there any notion in the literature which captures the idea of an $\infty$-ary tensor product on a category $C$? This should include tensor products of $\alpha$-indexed families of objects in $C$ for every ordinal $\alpha$, as well as some coherence isomorphisms and compatiblity properties. Here are three motivating examples:
1) The tensor product of modules
2) A category with choosen arbitrary (co)products
3) Let $C$ be a cocomplete monoidal category such that every object $x$ is equipped with a natural morphism $1 \to x$. If this data isn't available, just pass to this slice category over $C$. For an ordinal $\alpha$ and a familiy $(x_{\beta})_{\beta < \alpha}$ in $C$ define $\bigotimes_{\beta < \alpha} x_{\beta}$ by recursion on $\alpha$. It's clear what to do for $\alpha=0$ and when $\alpha$ is a successor. When $\alpha$ is a limit ordinal, take the colimit of all the preceding tensor products. The transition maps are induced by the data above.
There are some applications for the construction in 3), for example the largest Hausdorff quotient, the associated sheaf and the coproduct of algebras. In the context of orthogonal classes in presentable categories it is also called "transfinite composition".
If there is no such notion yet, what axioms should we choose?

REPLY [3 votes]: Here's a sketch of a direct way to define a symmetric infinitary-monoidal category.  For a category $C$, let $T C$ be the Grothendieck construction (lax colimit) of the functor $\mathrm{Core}(\mathrm{Set}) \to \mathrm{Cat}$ sending $X$ to $C^X$.  Thus an object of $T C$ is a set-indexed family of objects of $C$, and a morphism is a bijection between indexing sets which indexes a family of morphims in $C$.  There is an obvious functor $C \to T C$ sending an object to the corresponding singleton family.  And there is a functor $T T C \to T C$ induced by coproducts in Set.  This makes $T$ into a pseudomonad on Cat; then a symmetric infinitary-monoidal category is a pseudo $T$-algebra.
If you use the (discrete) category $\mathrm{Core}(\mathrm{Ord})$ instead of the groupoid $\mathrm{Core}(\mathrm{Set})$, then I think you get Zhen's notion of non-symmetric infinitary-monoidal category.  I don't have any suggestions for what a "braided" version would mean, though.<|endoftext|>
TITLE: left- and right- Folner sets
QUESTION [6 upvotes]: Given an amenable group, it is a standard trick to turn a left-invariant mean ( i.e. a continuous positive normalised linear functional $m:\ell_\infty(G) \to \mathbb{R}$ such that $\forall g \in G, m \circ \lambda_g = m$ where $\lambda_g: \ell_\infty(G) \to \ell_\infty(G)$ is the left-regular action of $G$) into a bi-invariant mean (also invariant under pre-composition with the right-regular action of $G$).
From this bi-invariant mean one gets a sequence (or a net, when $G$ is uncountable) of almost invariant probability measures (i.e. $\xi_n \in \ell_1G$ with $\| \lambda_g \xi_n - \xi_n\|_1 \to 0$ and $\| \rho_g \xi_n - \xi_n\|_1 \to 0$). 
$\mathbf{Question}$: Does there exists a bi-invariant Folner sequence? (i.e. a sequence of finite set $F_n$ such that $\xi_n = \chi_{F_n} /|F_n|$ is a sequence of almost invariant probability measure)
In other words, is it obvious that the bi-invariant property'' follow through the same proof asleft-'' and ``right-'' do, or, better, does there exist a simpler argument to show such a sequence exists?

REPLY [6 votes]: Yes, it is obvious. If you have a sequence of probability measures which is approximately invariant on the left, by convolving these measures on the right with the reflected ones you get an approximately invariant sequence of symmetric measures. By applying to this new sequence the standard "slicing" argument (the same as in the one-sided case) one gets symmetric Folner sets.<|endoftext|>
TITLE: Large gaps between P2s
QUESTION [5 upvotes]: Gaps between consecutive primes are $O(n^{\theta+\varepsilon})$ for $\theta=0.525$ and any $\varepsilon>0.$ I was wondering if a better result is known for gaps between numbers with at most two prime factors. (It seems that this would be easier because it removes the parity obstacle.)
I am interested in both unconditional results and those conditional on standard hypotheses.
Related result: Harman (1981) showed that almost all intervals of length $(\log x)^{7+\varepsilon}$ contains a $P_2.$

REPLY [10 votes]: Here is a result, due to Halberstam, Heath-Brown, and Richert that seems to be of the flavor you are looking for (Almost-primes in short intervals. Recent progress in analytic number theory, Vol. 1 (Durham, 1979), pp. 69–101, Academic Press, London-New York, 1981.)
Theorem. For all sufficiently large $x$ the interval $(x−x^{0.455},x]$, contains at least $\frac{1}{121}\frac{x^{0.455}}{\log x}$ integers that are either primes or products of two primes.
Reference chasing on MathSciNet, it seems that Iwaniec & Laborde improved the exponent from $0.455$ to $0.45$ and Wenzhi Luo later improved this slightly. I am not sure if Luo's result is still the "best" exponent.<|endoftext|>
TITLE: Serre duality and Hirzebruch-Riemann-Roch in the non-projective case
QUESTION [8 upvotes]: Serre duality and the Hirzebruch-Riemann-Roch formula are usually stated for $X$ a smooth projective algebraic variety. Do you know of a reference which proves these results for $X$ smooth and proper?
(1) Since Hirzebruch-Riemann-Roch is true for compact complex manifolds, and Serre duality holds for compact Kähler manifolds, one could expect them to hold more generally. 
(2) As both results are traditionally proved by deducing the result from the case of $X=\mathbb{P}^n$ (either by embedding $X$ in $\mathbb{P}^n$ or finding a finite morphism $X\to \mathbb{P}^n$), it should be easy to adapt these proofs to the case when $X$ is $A_2$, that is, embeddable in a toric variety. ($A_2$ is in fact equivalent to the property that every two points admit a common affine open neighbourhood).
(3) In characteristic $0$, we can connect $X$ to a smooth projective $X'$ by blow-ups and blow-downs with smooth centers. So I guess proving relative duality and HRR (that is, Grothendieck duality and Grothendieck-Riemann-Roch) for a blow-up in a smooth center should do the trick. I am more interested in the characteristic $p$ case though.

REPLY [10 votes]: Yes, sure.

See Theorem 15.2 (at least in my edition) of Fulton's Intersection Theory for Grothendieck-Riemann-Roch for a proper map of smooth varieties. Now take the target to be a point to obtain HRR.
This is certainly overkill, but you can find a proof of Grothendieck duality for proper
maps (with finite Tor dimension) between noetherian schemes in chap VII section 3 of Hartshorne's Residues and Duality. Now specialize as above to get Serre duality.
(It just occurred to me that Lipman's Dualizing sheaves, differentials and residues
on algebraic varieties is probably a more reasonable reference for this.)<|endoftext|>
TITLE: Almost Complex Structures: 'Tame' versus 'Compatible'
QUESTION [8 upvotes]: Consider a symplectic manifold $(M,\omega)$ and the space of almost complex structures. These are $J:TM\to TM$ with $J^2=-\text{id}$. A given $J$ is $\omega$-tame when $\omega(v,Jv)>0$, and $J$ is $\omega$-compatible when it is $\omega$-tame and $\omega(J\cdot,J\cdot)=\omega(\cdot,\cdot)$. The set of either such $J$ forms a contractible subspace. Note that in either scenario we can form a Riemannian metric $g_J$ by twisting $\omega$ in some way with $J$.
In practice, I sometimes see $J$ being tame, and more often I see $J$ being compatible, but in either case I am unsure of the restrictions that each imposes.  Is there any context in which it is necessary/helpful to use one condition over the other? This need not be related to pseudo-holomorphic curves in Floer theory. Do certain results fail when I relax "compatible" to "tame"?
A fundamental difference I see (quoted from McDuff-Salamon's big textbook) is that compatible $J$ minimize the energy of a J-holomorphic curve in its homology class, but not necessarily for tame $J$.

REPLY [5 votes]: A high dimensional situation, where one doesn't have the choice to work with a compatible almost complex structure is the case of a contact manifold $(V,\xi)$ weakly filled by a symplectic manifold $(W,\omega)$.  I do not want to give the precise definition of weak fillings here, but the important thing to understand is that to work in a sensible way with $J$-holomorphic curves on a manifold with boundary, we would like to choose a $J$ that is 

(at least!) tamed by the symplectic structure $\omega$ so that we can measure energy and hopefully use Gromov compactness,
furthermore we require $\xi$-convexity along the boundary so that the holomorphic curves cannot move "up" and escape from the symplectic manifold.

The correct definition of weak fillability is a compatibility condition between $\omega$ and $\xi$ that guarantees the existence of such almost complex structures satisfying both conditions.
Unfortunately, it is not always possible to find a $J$ that in this situation is both compatible with $\omega$ and with $d\alpha|_\xi$ (for a contact form $\alpha$), and we have to stick to tamed almost complex structures, see Remark 2.3 in 
Massot, Patrick; Niederkrüger, Klaus; Wendl, Chris, Weak and strong fillability of higher dimensional contact manifolds, Invent. Math. 192, No. 2, 287-373 (2013). ZBL1277.57026..<|endoftext|>
TITLE: Why are $S$-arithmetic groups interesting?
QUESTION [11 upvotes]: Let $K$ be a number field and $S$ a finite set of valuations of $K$, including $\infty$.
Define the $S$-numbers $K_S$ to be the direct product $\prod_{s \in S} K_s$ where $K_s$ denotes the completion of $K$ at the valuation $s$.
Define the $S$-integers $\mathcal{O}_S$ to be the subset of $K$ consisting of the elements $x$ such that $|x|_s \leq 1$ when $s \notin S$.
Example: Let $K = \mathbb{Q}$ and $S = \{\infty, p_1, \ldots, p_n \}$. Then we have
$$K_S = \mathbb{R} \times \mathbb{Q}_{p_1} \times \cdots \times \mathbb{Q}_{p_n}$$
$$\mathcal{O}_S = \mathbb{Z}[p_1^{-1},\ldots,p_n^{-1}]$$
Note that these rings come with topologies induced from the topologies on the completions $K_s$.
Furthermore, we can define algebraic groups over $K_S$ such as, for example
$$\mathbf{GL}_m(K_S) = \prod_{s \in S} \mathbf{GL}_m(K_s)$$
Here are my questions:
Why is it interesting to study groups in the $S$-arithmetic setting such as $\mathbf{GL}_m(\mathcal{O}_S)$ or $\mathbf{GL}_m(K_S)$?
In particular, is there some classical problem that is solved by using $S$-arithmetic groups, or one that served to launch the study of $S$-arithmetic groups? Perhaps some relevant (famous) names would be Borel, Harish-Chandra, Siegel, Weil, Tits, etc.
It is easy to believe that number theorists would be interested in studying a ring such as $\mathbb{Z}[p_1^{-1},\ldots,p_n^{-1}]$, although I don't really know why and I would like to hear more.
I am also aware that $\mathbf{GL}_m(K_S)$ is a natural locally compact group in which one can realize $\mathbf{GL}_m(\mathbb{Z}[p_1^{-1},\ldots,p_n^{-1}])$ as a discrete subgroup. Why one would care about this, I am also not sure. I imagine it has something to do with studying functions on the quotient and things such as Tamagawa numbers. Perhaps some representation theory is involved.

REPLY [11 votes]: Supplementing other comments and @JimHumphreys' answer: Thinking of automorphic forms as living only on quotients of symmetric spaces or of real Lie groups leaves one with an extremely awkward neo-classical version of Hecke operators, and, more pointedly, no action in sight of the corresponding $p$-adic groups, so no way to take advantage of what is known about their representation theory. To "convert" automorphic forms as functions on something like $G(\mathbb Z)\backslash G(\mathbb R)$ to automorphic forms on $G(\mathbb Z[1/p])\backslash G(\mathbb R)\times G(\mathbb Q_p)$, not only makes the $p$-Hecke operators much more tractable, but, in fact, happily, allows the direct application of the representation theory of $G(\mathbb Q_p)$.
(One main virtue of the latter is the wonderful Borel-Casselman-Matsumoto theorem, that shows that not only spherical representations, but admissible repns with Iwahori-fixed vectors, are subrepresentations (and quotients) of unramified principal series. This also does account for the "square-free level" condition of many classical papers on modular forms, since these exactly correspond to Iwahori-fixed vectors...)
For that matter, the physical space $G(\mathbb Z[1/p])\backslash G(\mathbb R)\times G(\mathbb Q_p)$ arises very reasonably, as a sort of non-abelian solenoid, namely, the (projective) limit of $\Gamma(p^n)\backslash G(\mathbb R)$, as $\Gamma(p^n)$ runs over principal $p$-power congruence subgroups. One would find that the limitands in this limit arise inevitably in looking at $p$-power Hecke operators even at level one.<|endoftext|>
TITLE: In which geometries do triangles have an Euler line?
QUESTION [11 upvotes]: In Euclidean geometry, the centroid, orthocenter and circumcenter of a triangle lie on a line.
In which other geometries does this hold?

REPLY [6 votes]: As shown by Richard Baldus in 1929 (Ueber Eulers Dreieckssatz in der absoluten Geometrie, Sitzungsberichte  der  Heidelberger  Akademie  der Wissenschaften. Mathematisch-naturwissenschaftliche  Klasse  11. Abhandlung), the Euler line exists in non-Euclidean geometry only in the case of isosceles triangles. A different proof for that result and a related one for Feuerbach's circle can be found in H. Kuenneth, H.: Der Schwerpunkt, die Eulersche Gerade und der Feuerbachsche Kreis in der absoluten Geometrie.
Jahresber. Deutsch. Math.-Verein. 43 (1933), 65–77.<|endoftext|>
TITLE: Nearly constant curvature implies "nearly isometric" to a space form?
QUESTION [7 upvotes]: It is well known a Riemannian manifold with constant sectional curvature is a quotient of the Euclidean space, hyperbolic space or sphere. In particular we know how their metric looks like locally. 
My question is, is there a quantitative version of the above result? By this, I mean for example, given $\varepsilon>0$, there exists $\delta$ such that if $(M,g)$ satisfies 
 $\left|Rm\right|_g<\delta$ `, does there exist a local diffeomorphism $\phi$ to $\mathbb R^n$ such that $|\phi_*g-g_0|_{g_0}< \varepsilon$, where $g_0$ is the standard metric on $\mathbb R^n$?

REPLY [5 votes]: For the global question, and hyperbolic metrics, in dimension > 3 this is a result of Gromov, stated in his 1978 JDG paper, and in dimension 3 it is an unpublished result of Daryl Cooper, from the late nineties, and Gromov, independently, so while Tian might have a more general result, he is far from the first.<|endoftext|>
TITLE: Quantum mechanics basics
QUESTION [6 upvotes]: Hello. I'm thinking about where does the basic quantum mechanics things comes from. I mean the forms of operators and a Shroedinger equation. The more intuitive explanation is better.
To get forms of operators and Shroedinger eq., we can start from assumption that in our representation square of absolute value of wavefunction is probability density. Then it became clear that coordinate operator is just a multiplying by variable (obvious, looking for example to mean value expression). Next, Shroedinger equation anyway should be of the form $\frac{\partial \phi}{\partial t} = A \phi$ with some A. To see that $A$ is actually Hamiltonian multiplied by something, we can see that  $\frac{\partial \phi}{\partial t}$ is an enegry multiplied by something. I have the only idea to explain this: assume additionally that  in our representation defined momenta states are plane waves. Considering them, $i\frac{\partial \phi}{\partial t} = E$ is visible from De Broglie relation $E=\omega$ for plane waves (as well as $p=\frac{\partial \phi}{\partial x}$ visible from $p=k$). Then, the only thing i want explained deeper is De Broglie relations. Finally, now i also think that i want some explanation of introducing wavefunction as complex-valued. 
Or maybe there is an other way? I guess, it would be more general to start from commutation relations, but i'm afraid it would be hard and abstract. But i appreciate if you try to explain where are they come from.

REPLY [5 votes]: I'll try to help with intuition on why wavefunctions are complex-valued.
First, it's not actually true that they have to be complex. When you quantize electromagnetic waves, the wavefunction is simply the electric and magnetic fields, which are real. The correct statement is that a spin-1/2 particle's wavefunction has to be complex.
As a simple example of why this is, consider the case of two planar sine waves that are moving in antiparallel directions and then merge and superpose.
If the wavefunction is a real scalar, then at a time when the two superposed waves are 180 degrees out of phase, their sum is identically zero. This violates conservation of energy and, perhaps more importantly, conservation of probability.
If the wavefunction is a complex scalar, then one can prove that solutions of the Schrodinger equation always conserve probability. This is easy to check in an example like the superposition of $e^{i(kx+\omega t)}$ with $e^{i(-kx+\omega t)}$.
To see that this argument doesn't prove that wavefunctions are always complex-valued, consider electromagnetic waves in the same situation of superposition of antiparallel plane waves. Because there is a right-handed relationship among $\mathbf{E}$, $\mathbf{B}$, and $\mathbf{k}$, you can't make both $\mathbf{E}$ and $\mathbf{B}$ cancel. For example, you could choose the polarization so that $\mathbf{E}$ would cancel, but then $\mathbf{B}$ wouldn't.
To see that the fundamental issue is conservation of probability, so that this is really something specific to quantum mechanics rather than classical physics, consider the case of sound waves, which can be represented as real scalar functions $f$ measuring the pressure. Rerunning the same argument about superposing antiparallel plane waves, we find that it's possible for $f$ to cancel, but that's OK, because $f$ doesn't have a probability interpretation. We also still have conservation of energy, because the energy depends not just on $f$ (potential) but also on $\partial f/\partial t$ (kinetic), so $f$ can vanish without making the energy vanish.
Complex numbers come up in a lot of places in quantum mechanics, not just in wavefunctions, and it's not always obvious when they're just a notational convenience. For example, Pauli basically reinvented the quaternions in 1924. His spin matrices $\sigma_1$, $\sigma_2$, and $\sigma_3$ are equivalent to the quaternions i, j, and k if you multiply them by i.
Operators can be complex-valued, but expectation values are always supposed to be real, since they correspond to measurable quantities.<|endoftext|>
TITLE: Center and representations of finite group - how are related ? 
QUESTION [10 upvotes]: If finite group G has a center how does it influence the representations of this group  ?
And vice versa - can we see somehow the center (or some of its properties) from representations (from character table, from ring structure,  ... whatever) ?
One has a natural map Z(G)-> G-> G/Z(G),
so we can pull-back representations of G/Z(G) to G, but so what ? How far R(G) is from R(G/Z(G)) ?
Can we claim that 
at least the dimensions of irreps of G are the same or just not bigger, than that of G/Z(G) ? (NO as Xogn Ambandl answer implies). 
(F. Ladisch comment below is some weaker indication that something like this might happen).
In any irrep of G center Z should act by scalars, so it defines some homomorhpism of Z to C^,
is any such homomorphism is realized by some irrep V of G ? Probably not... 
Is it possible to characterize those Z->C^ which occur, depending on the group G ? 
PS
I just learnt from comments by F. Ladisch: 
"It is a general fact that χ(1)^2≤|G:Z(G)| for any irred. character χ of a group G (see Isaacs' book on character theory, Corollary 2.30)." 
PSPS
Another relevant MO-discussion Which finite groups have faithful complex irreducible representations?. Let me quote: "Obvious necessary condition is that the center must be a cyclic group." 
"For finite p-groups, it's a standard fact that having a faithful irreducible representation is equivalent to having a cyclic center. I'm not sure about the general case, but it's been discussed in many books and papers. My impression is that there is no known definitive structural condition for sufficiency. – Jim Humphreys Mar 2 2011 at 16:52"
And further - see answers by Andreas Thom and Rob Harron.

REPLY [3 votes]: There is an easy way to count the number of irreducible characters of $G$ that restrict to $Z = {\bf Z}(G)$ to yield a multiple of a given linear character $\lambda$ of $Z$. By a result of Gallagher, this number is the number of conjugacy classes of "$\lambda$-good" elements of $G/Z$, where an element $Zg$ of $G/Z$ is $\lambda$-good if for every element $x$ of $G$ such that $[g,x] \in Z$, it is true that $\lambda([g,x]) = 1$.
In particular, it follows that the number of irreducible characters of $G$ lying over a given linear character $\lambda$ of $Z$ is at most $|{\rm Irr}(G/Z)|$. It is also true that the sum of the squares of the degrees of the irreducible characters of $G$ lying over $\lambda$ is equal to $|G:Z|$. We conclude from this that the average of the squares of the degrees of these characters is at least the average of the squares of the degrees of all members of ${\rm Irr}(G/Z)$, so one might say that  "on average" the irreducible character degrees of $G/Z$ are no less than the degrees of the irreducible characters lying over each linear character $\lambda$.<|endoftext|>
TITLE: How much do homological knot invariants improve the classification problem of knots?
QUESTION [11 upvotes]: The mutation operation in knots appears to be detected by the Floer homological invariants. See the papers by Ozsvath, Szabo and by Baldwin, Gillam. In addition, the Khovanov homology turns out to be able to detect the unknot. See the papers by Elisenda Grigsby, Wehrli and by Kronheimer, Mrowka.
My question is the following.

How much do homological knot invariants improve the classification problem of knots? Is there something homological invariants cannot distinguish?

REPLY [7 votes]: I don't think that quantum topology or Heegaard Floer homology are useful as knot tabulation tools. Geometric invariants (e.g. hyperbolic volume) and classical invariants (e.g. signature) have been the weapon of choice for knot tabulators. I don't know of any pair of knots which have been separated using quantum invariants or Floer homology which we did not know how to separate using classical or geometric invariants.
There are several invariants which separate knots at least conjecturally, including the knot group (plus peripheral structure), the collection of all twisted Alexander polynomials of a knot (whatever that means), and the knot quandle. Surely HFH separates knots as well, although I don't think that this has been proven, so it detects "everything" in some sense. The issue is computational efficiency. But I think that the strength of HFH is in revealing structure to the space of knots and in analyzing properties of individual knots as opposed to as a tool to distinguish knots.<|endoftext|>
TITLE: $\mathcal{D}$-modules of level m
QUESTION [5 upvotes]: My question is regarding the definition of $\mathcal{D}$-modules of level $m$ given in this paper. As an example, let $X=\mathbb{A}^1$ over $S=\text{Spec }\overline{\mathbb{F}_p}$; I was told that a $\mathcal{D}$-module of level $m$ is a module over $\mathbb{k} \langle x, \partial_x, \frac{{\partial_x}^p}{p!}, \cdots, \frac{{\partial_x}^{p^m}}{(p^m)!} \rangle$; I was wondering how to work this out from first principles, using  Definition $2.5$ given in that paper.
Consider the immersion $X \rightarrow X \times_S X$. Definition $2.1$ from that paper defines what a divided power structure of level $m$ on an immersion is; Definition $2.3$ (and $2.4$) states constructs the divided power envelope of level $m$, $P_{X,m}(Y)$. Subsequently, on pg $5$ they define $\mathcal{P}_{X,m}^n(Y)$; the sheaf of differential operators of level $m$ is defined to be union of the duals of this family of sheaves (as $n$ varies). Most of the details/proofs are done in this other paper.
I'm having trouble properly understanding these definitions and working out what they are in the case of $X=\mathbb{A}^1, S=\text{Spec } \overline{\mathbb{F}_p}$. So, my question is what the above objects look like in this particular example, and how to explicitly calculate everything in this example.

REPLY [10 votes]: I find it helpful to first work through the definition of multiplication on $\mathcal{D}^{(m)}$ when $m = \infty$, in which case it reduces to the "classical" ring of differential operators in the sense of Grothendieck; read sections 16.7 and 16.8 of EGA 4, Quatrième partie.
So let $A$ be a commutative base ring, let $S = Spec(A)$ and let $X = \mathbb{A}^1_S$ so that $B = A[t] = \Gamma(X, \mathcal{O})$. We want to work out $\mathcal{D}^{(\infty)}(X)$. Let $Y = X \times_S X$ and $m = \infty$.
Let $n \geq 0$. First we have to work out the global sections of the sheaf $\mathcal{P}^n_{Xm}(Y)$, considered as a $B$-module. This will turn out to be dual (as a $B$-module) to the $B$-module of all Grothendieck differential operators of order at most $n$.
Now $\mathcal{O}(Y) = B \otimes_A B$ is isomorphic as an $A$-algebra to the polynomial ring $A[t,t']$ where $t \mapsto t \otimes 1$ and $t' \mapsto 1 \otimes t$. The diagonal immersion $X \hookrightarrow Y$ corresponds to the algebra surjection $B \otimes_A B \to B$ which is just the multiplication map. So the ideal of the diagonal, namely the kernel of this map, is generated as an ideal by the element 
$$\tau := t \otimes 1 - 1 \otimes t.$$
Let's view $\mathcal{O}(Y)$ as a $B$-algebra via the map $b \mapsto b \otimes 1$; then $\mathcal{O}(Y) \cong B[\tau]$. By definition (EGA IV, 16.7.1.1), the global sections of $\mathcal{P}^n_{X\infty}(Y)$ are just
$$P^n := \mathcal{O}(Y) / (\tau^{n+1})$$
--- this is the algebra of functions on the $n$-th infinitesimal neighbourhood of the diagonal $\tau = 0$ inside $\mathcal{O}(Y)$ (hence the $n+1$ in the exponent). So in particular it is a free $B$-module of rank $n+1$ with generators (the images of) $\tau^i$ for $0 \leq i \leq n$. By definition, 
$$\mathcal{D}^{(\infty)}_n (X) := Hom_B(P_n, B) =: D_n $$
which is again a free $B$-module of rank $n+1$; let $\{ \partial^{[i]}, i=0, \ldots, n\}$ be the dual basis for this module. 
Now the multiplication map $D_r \times D_s \to D_{r+s}$ is the $B$-module dual of a map $\delta : P^{r+s} \to P^r \otimes P^s$ which is constructed in EGA IV, Lemma 16.8.9.1. Morally $\delta$ sends $a \otimes b$ to $a \otimes 1 \otimes 1 \otimes b$, as Gros/Le Stum/Quirros mention. This turns out to be a $B$-algebra homomorphism, and tts key property is that
$$\delta( \overline{\tau} ) = \overline{ \tau} \otimes 1 + 1 \otimes \overline{\tau}$$
(it is a "primitive element" in an appropriate bialgebra --- see EGA IV.4, 16.8.9.4). Let's now work out how to multiply $\partial^{[i]}$ by $\partial^{[j]}$ (drop the bars for clarity):
$$(\partial^{[i]} \cdot \partial^{[j]})(\tau^k) =  (\partial^{[i]} \otimes \partial^{[j]})(\tau \otimes 1 + 1 \otimes \tau)^k = \sum_{a + b = k} \binom{k}{a} \partial^{[i]}(\tau^a) \partial^{[j]}(\tau^b)$$
which is just $\binom{i+j}{i}\delta_{k,i+j}$. Since $\binom{i+j}{i} \partial^{[i+j]}$ has the same effect on each $\tau^k$, we deduce that
$$ \partial^{[i]} \cdot \partial^{[j]} = \binom{i+j}{i} \partial^{[i+j]}$$
which is hopefully the familiar rule for multiplying divided powers (morally $\partial^{[i]} = \partial^i/i!$). 
The point of the Berthelot construction is that it is possible to vary the divided-power structure on the diagonal, and thereby control just how many divided powers one gets in $\mathcal{D}^{(m)}$. For example, if $m = 0$ then you instead allow all divided powers on the ideal of the diagonal (algebraically this means you consider the divided power algebra of the ideal $(\tau)$ in $B[\tau]$ to get $\oplus_{n=0}^\infty B \tau^{[n]}$), and when you take the $B$-dual, these divided powers in $\tau$ "remove" the divided powers in $\partial$ and you end up with $\mathcal{D}^{(0)}(X) = B[\partial]$, the ring of crystalline differential operators (no divided powers).
Now to answer your question, let the level $m \geq 0$ be fixed. Then as Gros/Le Stum/Quirros explain just before Definition 2.5,
$$\Gamma(Y, \mathcal{P}^n_{Xm}(Y)) = \oplus_{a=0}^n B \tau^{ \{ a \} } $$
where $\tau^{ \{ a \} }$ is a symbol that "behaves like $\tau^a / q_a!$" (where $q_a$ is the integer part of $a / p^m$: thus $a = q_a p^m + r_a$ say).
To understand the multiplication of the dual vectors to these $\tau^{ \{ a \} }$, namely the $\partial^{ \langle a \rangle }$, we need to understand how to comultiply the $\tau^{ \{ a \} }$. So we compute (again dropping bars for convenience)
$$ \delta( \tau^{ \{ a \} }) = \frac{1}{q_a!} \delta(\tau)^a = \sum_{i+j = a} \frac{1}{q_a!} \binom{i+j}{i} \tau^i \otimes \tau^j = \sum_{i+j = a} \frac{q_i! q_j!}{q_{i+j}!}  \binom{i+j}{i} \tau^{ \{ i \} } \otimes \tau^{ \{ j \} }$$ 
and the same computation as above in the case $m=\infty$ shows that 
$$ \partial^{\langle i \rangle} \cdot \partial^{\langle j \rangle} = \frac{q_i! q_j!}{q_{i+j}!}  \binom{i+j}{i} \partial^{\langle i + j \rangle}.$$
The interesting thing is that this structure constant $\frac{q_i! q_j!}{q_{i+j}!}  \binom{i+j}{i}$ is always a $p$-integral rational number (see Lemma 1.1.3(i) in Berthelot's paper), so it makes sense whenever $A$ is an algebra over $\mathbb{Z}_{(p)}$, say, and in particular if $A$ had characteristic $p$. Note that if $A$ was a $\mathbb{Q}$-algebra, then there would be a ring homomorphism from $\mathcal{D}^{(m)}$ to $A[t; \partial]$ which sends 
$$ \partial^{\langle i \rangle} \mapsto \frac{q_i!\partial^i}{i!} $$
since
$$ \left(\frac{q_i! \partial^i}{i!}\right) \cdot \left(\frac{q_j! \partial^j}{j!}\right) = \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \left( \frac{q_{i+j}! \partial^{i+j}}{(i+j)!}\right).$$
Thus morally $\partial^{\langle i \rangle}$ should be thought of as "modified divided powers" $q_i! \partial^i / i!$. 
Finally, you don't need all of the $\partial^{\langle i \rangle}$ to generate $\mathcal{D}^{(m)}$. As is well-known, the full ring of Grothendieck differential operators in characteristic $p$ can be generated by the divided powers $\partial^{[p^a]}$ (for all $a \geq 0$). Since $q_i = 0$ for $i < p^m$ and $q_{p^m} = 1$, the modified divided powers $\partial^{\langle p^i \rangle}$ are equal to the "true" divided powers $\partial^{[p^i]}$ for $0 \leq i \leq m$. If $a > m$ then since $q_{p^a} = p^{a-m}$,
$$\partial^{\langle p^a \rangle} = \frac{ p^{a-m}! }{ p^a! } \partial^{p^a} = \left(\frac{ p^{a-m}! (p^m!)^{p^{a-m}} }{p^a!} \right) (\partial^{\langle p^m \rangle})^{p^{a-m}}$$
shows that $\partial^{\langle p^a \rangle}$ is a $p$-adic unit times a power of $\partial^{\langle p^m \rangle}$ for $a \geq m$ since the $p$-adic valuation of that big fraction is 
$$\frac{p^{a-m}-1}{p-1} + p^{a-m} \frac{p^m-1}{p-1} - \frac{p^a-1}{p-1} = 0.$$
So we see that $\mathcal{D}^{(m)}(\mathbb{A}^1_S)$ in this case is the $A$-algebra generated by $t$ and the divided powers $\partial^{[p^0]}, \partial^{[p^1]}, \ldots, \partial^{[p^m]}$, subject to appropriate natural relations.
Edit: To see what the map $X \to P_{Xm}(Y)$ looks like in the case $X = \mathbb{A}^1_S$,  it is enough to describe the corresponding map $C := \mathcal{O}(P_{Xm}(Y)) \to B = \mathcal{O}(X)$ on functions, because everything in sight is affine. $C$ is a $B$-algebra, generated by symbols $\tau^{ \{a \} }$ for all $a \geq 1$ subject to the relations 
$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \frac{q_{a+b}!}{q_a!q_b!} \tau^{ \{ a + b \} }$$
(note that the structure constant $\frac{q_{a+b}!}{q_a!q_b!}$ is actually an integer, this again follows from Lemma 1.1.3(i) in Berthelot's paper.) The map $C \to B$ sends all of the generators $\tau^{ \{a \} }$ to zero, and the map $C \to P^n$ which corresponds to the closed subscheme $P^n_{Xm}(Y)$ of $P_{Xm}(Y)$ sends all of the $\tau^{ \{ a \} }$ to zero for $a \geq n+1$.
This description makes it easy to see that the algebra of functions $C$ on $P_{Xm}(Y)$ is isomorphic to the polynomial ring $B[\tau]$ when $m = \infty$ (since we can take $q_a$ to be always zero in this case), and to the "free" divided-power algebra $B[\tau^{[n]} : n\geq 1]$ that Gros/Le Stum/Quirros call $\Gamma_\bullet(B \tau)$ when $m = 0$. This is because $q_a = a$ in this case, so the defining relations between the $\tau^{ \{a \}}$ reduce to
$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \binom{a+b}{a} \tau^{ \{ a + b \} }.$$<|endoftext|>
TITLE: Characteristic zero and characteristic $p$ in algebraic geometry
QUESTION [14 upvotes]: Are there  non-trivial (i.e. excluding concepts that can be defined only for $p>0$) statements in algebraic geometry that hold for all fields of characteristic $p$ for all prime $p$ but are known to be false in characteristic zero?

REPLY [2 votes]: The stack $\overline{\mathcal{M}}_{g,n}$ of Deligne-Mumford stable curves and its coarse moduli space $\overline{M}_{g,n}$ are defined over $\mathbb{Z}$. Therefore they are defined over any commutative ring and in any characteristic.
Let $\pi:\mathcal{U}\rightarrow\overline{\mathcal{M}}_{g,n}$ be the universal curve, $\omega_{\pi}$ the relative dualizing sheaf and $\Sigma$ the union of the sections of $\pi$. Then $\mathcal{L}:=\pi_{*}\omega_{\pi}(\Sigma)$ is a line bundle on $\overline{\mathcal{M}}_{g,n}$.
If $p:\overline{\mathcal{M}}_{g,n}\rightarrow\overline{M}_{g,n}$ is the coarse moduli space then $p_{*}\mathcal{L}$ is semi-ample in positive characteristic but this fails in characteristic zero. 
See http://arxiv.org/abs/math/9901149.<|endoftext|>
TITLE: Detecting a hidden convex body with line probes
QUESTION [13 upvotes]: Imagine that, somewhere inside an origin-centered, unit-radius sphere
$S$ in $\mathbb{R}^3$,
sits a convex body $K$ of volume vol$(K)=\alpha (\frac{4}{3} \pi)$,
with $\alpha < 1$ the fraction of the volume of $S$.
$K$ is inside $S$ at an unknown but fixed location and orientation.
My question is: How many line-probes are needed to detect
its presence?
A line-probe is a line $L$ whose intersection
with $K$ includes a point strictly interior to $K$.
One might need many probes to certainly detect the presence of a small-volume $K$.
Let $f(k)$ be the volume fraction $\alpha$ such that
(a) there is some body $K$ that is not detected by any fixed set of
$k$ probes, and (b) every body with vol$(K) > \alpha$ is detectable
by $k$ probes.
I believe $f(1)=\frac{1}{2}$: If $K$ fills a hemisphere, it could 
"hide" in $S$ from any single probe. But any $K$ with more than half
the volume of $S$ necessarily includes the origin, and so a line
through
the origin would detect it.

   

It may be that $f(2)=\frac{1}{3}$ by two orthogonal probes that partition $S$
into two spherical caps and the sandwich between, each of
$\frac{1}{3}$
the volume of $S$.
And perhaps $f(3)=\frac{1}{4}$ via three probes through the origin.
But I am uncertain of these values of $f()$. If anyone can hide bodies of larger volumes
from these probes, please let me know!
This feels like a question that was likely considered before;
if so, a pointer would be appreciated.
Of course, the question generalizes to $\mathbb{R}^d$, with various dimensional probes.
In $\mathbb{R}^1$ with point-probes, $f(k)=\frac{1}{k+1}$.
Edit: Michael Biro suggests in the comments that the $f(2)$ example above could be
generalized to establish that also $f(k)=\frac{1}{k+1}$ in $\mathbb{R}^3$.
Update.
Here is an illustration of Ilya Bogdanov's argument that my 2nd example does not
establish that $f(2)=\frac{1}{3}$:

REPLY [7 votes]: This is visibly closely related to the "maximal empty convex set problem considered in a number of papers, most recently by Dumitrescu et al (see http://arxiv.org/pdf/1112.1124.pdf). That asks for the size of the biggest convex set not containing a fixed point set, and the bound is somewhere between $O(1/n)$ and $O(\log n/n),$ (where $n$ is the number of points) In your question, you are looking at line segments, so this corresponds to the maximal empty convex set problem in the Grassmannian of affine lines, and since the bounds are likely very similar, you get a set of lines of measure somewhere between $O(1/n)$ and $O(\log n/n).$ Going back to $\mathbb{R}^d$ by Crofton, you will be missing a convex set whose surface area is bounded as above, so by the isoperimetric inequality, it's volume is smaller than $O((\log n/n)^{d/(d-1)}).$
This does not answer you question for specific small $n.$<|endoftext|>
TITLE: Do random projections (approximately) preserve convexity?
QUESTION [25 upvotes]: The Johnson-Lindenstrauss lemma implies that any set of $k$ points in $\mathbb{R}^d$ can be randomly projected into $d' \approx \log(k)/\epsilon^2$ dimensions such that the distances between each pair of points are approximately preserved, up to a multiplicative factor of $(1 \pm \epsilon)$. 
My question is whether such a projection will also approximately preserve convexity. Suppose the $k$ points lie on the surface of a convex body in $\mathbb{R}^d$. Does there exist a projection into $d'$ dimensions such that each point lies near the surface of a convex body?

REPLY [9 votes]: Negative result:  
See p. 377 in Chapter 15 of Matousek's book, which can be found 
here. In short, if you want the image of the $k$ points to be between the surface of a convex body $K$ and the surface of $DK$ for some $D>1$, you need the operator to have rank at least $k^{f(D)}$ for some function $f$.
Positive result on a related problem:
In 
Johnson, William B.; Lindenstrauss, Joram; Schechtman, Gideon On Lipschitz embedding of finite metric spaces in low-dimensional normed spaces. Geometrical aspects of functional analysis (1985/86), 177–184, Lecture Notes in Math., 1267, Springer, Berlin, 1987,
it is proved that for some constant $C$, if you have $k$ points on the surface of a symmetric convex body, then you can put the points isometrically into a suitable $\ell_\infty^m$ in such a way that   a random projection of order rank $k^{1/D}$ will place the points between the surface of a symmetric convex body $K$ and the surface of $CDK$; see the paper for a precise statement.  I don't think symmetry places much of a role here. We were interested in the embedding of points into a Banach space and so did not think about general  convex bodies. The embedding theorem we proved was later made obsolete by Matousek when he proved that and metric space with size $k$ embeds into $\ell_\infty^{n}$ with distortion $D$ with $n$ about $Dk^{1/(2D)} \log k$ (see p. 404 at the above given link).<|endoftext|>
TITLE: Restricting representations to lattices
QUESTION [7 upvotes]: Let $V$ be a finite-dimensional irreducible representation of the Lie group $\text{SL}_n(\mathbb{R})$.  Must $V$ remain irreducible when you restrict the action to $\text{SL}_n(\mathbb{Z})$?  More generally, when you restrict it to other lattices in $\text{SL}_n(\mathbb{R})$?

REPLY [8 votes]: The result is essentially the statement of Borel's stability theorem for $\mathrm{SL}_n(\mathbf{R})$, see for example
Theorem 4.39 of the following:
http://people.uleth.ca/~dave.morris/books/IntroArithGroups.pdf
Sometimes Borel's stability theorem is phrased in terms of one of the corollaries, which in this case would be that any lattice in $\mathrm{SL}_n(\mathbf{R})$ is Zariski dense. Then to deduce the result one would also have to note that all finite dimensional representations of $\mathrm{SL}_n(\mathbf{R})$
are algebraic.<|endoftext|>
TITLE: Product of random diagonals on the unit circle
QUESTION [5 upvotes]: Let $P_1, P_2, ..., P_n$ be points randomly placed on a unit circle from a uniform distribution. Consider the product $D$ of all pairwise distances:
$D=\displaystyle \prod_{1\leq i < j \leq n} \overline{P_iP_j}$ 
I wonder...
1) What is the probability density function for $D$? What is the expected value of $D$?
2) When the $P_i$ are equally spaced, we know $D=n^{n/2}$. Is this an absolute maximum for $D$?

REPLY [12 votes]: To 2, the answer is yes. This is easy. Re-enumerate points according to their cyclic order. For every fixed $k$ ($1\le k<n$), the product
$$
 D_k:= \prod_i |P_i P_{i+k}|
$$
(where the indices are taken modulo $n$), equals
$$
 D_k=\prod_i 2\sin (t_i/2) = 2^n\exp \left(\sum_i \log\sin(t_i/2) \right)
$$
where $t_i$ is the (oriented) angle from $P_i$ to $P_{i+k}$. Since $t_i\in(0,2\pi)$, $\sum t_i=2\pi k$, and the function $t\mapsto \log\sin(t/2)$ is concave on $(0,2\pi)$, by Jensen's inequality the maximum is attained when all $t_i$ are equal, i.e., when the points are equally spaced. Since $D=\prod_k D_k^{1/2}$, the result follows.<|endoftext|>
TITLE: Beginning reference for configuration spaces
QUESTION [11 upvotes]: In my mathematical reading and thoughts, I keep running across the notion of configuration spaces, and while I essentially understand the idea behind them, I don't have much intuition for them (not even simple ones like the configuration space of two (indistinguishable) points on a line), and I don't know how computations are done with them. Is there a textbook or paper that would serve as a gentle introduction to configuration spaces, to someone with perhaps a semester of graduate algebraic topology (or maybe a little more?)

REPLY [3 votes]: Let me add one more introductory book.
Hansen, Vagn Lundsgaard. Braids and coverings: selected topics. With appendices by Lars Gæde and Hugh R. Morton. London Mathematical Society Student Texts, 18. Cambridge University Press, Cambridge, 1989. x+191 pp. ISBN: 0-521-38479-6 MR1247697
If I remember correctly, this small book does not require much and explains basic techniques to handle configuration spaces of manifolds, such as the fibration sequence given by selecting points.
Note, however, that most classical books and papers, including this book and the one by Fadell and Husseini, only deal with configuration spaces of manifolds. In order to study configurations spaces of singular spaces, we need completely different techniques.
For example, a graph $\Gamma$, regarded as a 1-dimensional cell complex, is a singular space. We cannot use the basic fibration technique. I recommend this expository article by Abrams and Ghrist for such configuration spaces.<|endoftext|>
TITLE: Coordinate-free derivation of the Einstein's field equation from the Hilbert action.
QUESTION [9 upvotes]: It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional (without matter and cosmological constant, which is irrelevant here):
$$S = \int_M R \mu_g,$$
is given by the Einstein's field equation:
$$Ric -\frac{1}{2}R g = 0, $$
where $\mu_g$ is the canonical volume form given by the metric $g$, $Ric$ is the Ricci curvature and $R$ is the Ricci scalar.
The standard derivation of the above statement seems to be a not so hard but not so pleasant direct calculation, either in coordinates or abstract indices, expanding everything in terms of the Christoffel symbol and eventually in terms of $g$ and then calculus.
My questions is: is there a more geometric and coordinate-free way to derive this?

REPLY [10 votes]: This can be found in Besse "Einstein Manifolds", in chapter 4.
The idea is to use Koszul formula for the Levi-Civitta connection to compute the derivative of the curvature with respect to the metric. Bianchi identities also help.<|endoftext|>
TITLE: Why are Gromov-Witten invariants of K3 surfaces trivial? 
QUESTION [20 upvotes]: Why is GW invariants of K3 surfaces are trivial? My naive guess is that GW invariants are deformation invariant and you can always deform your K3 surface to non-projective one, which has no subcomplex manifold except points. Then GW invariants (or GV invariants) naively count the number of curves, so they must be trivial. 
Are there any more rigorous proof of this fact? Or can we make the argument above rigorous? Since GW invariants are symplectic invariant, I wonder if there is a proof in symplectic geometry too. 
Another question is that, are DT invariants of K3 surfaces also trivial?

REPLY [5 votes]: The most direct answer to the original question is provided by the DMJ paper of Junho Lee. A (2,0) form on a Kahler surface X determines an almost complex structure J on X so that all J-holomorphic curves lie in the zero set of  the (2,0)-form. Since a K3 admits a nonwhere zero (2,0)-form, there are no J-holomorphic curves in K3 for the corresponding almost complex structure J.<|endoftext|>
TITLE: regularity of local ring
QUESTION [10 upvotes]: Let $(R, \mathfrak{m})$ be a Noetherian local ring. It is well know that
$R$ is regular iff $pd(R/\mathfrak{m}) < \infty$ (i.e. $R/\mathfrak{m}$ has finite projective dimension).
Assume that $\dim R > 0$. Is $R$ regular, if $pd(R/\mathfrak{m}^2)< \infty$?

REPLY [7 votes]: If $R$ has positive dimension, then for any $t\ge 1$, the ideal $I=\mathfrak m^t$ is a so called Burch ideal (as defined by Dao,  Kobayashi, Takahashi; Burch ideals and Burch rings )  i.e. $\mathfrak m(I:\mathfrak m)\ne \mathfrak mI$. 
It is a Theorem of Burch that if $I$ is a Burch ideal and $R/I$ has finite projective dimension, then $R$ is regular.<|endoftext|>
TITLE: A generalization of Catalan numbers
QUESTION [7 upvotes]: It is well-known that the $n$th Catalan number is equal to $(n+1)^{-1}\binom{2n}{n}$. A long time ago I had wondered what happens if you look at the sequence generated by $(n+k)^{-1}\binom{pn}{n}$ - for which $p,k$ is it integral?
I found no other integral-producing values except for $(p,k)=(2,1)$ but then I'm probably missing something. (Also, MATLAB, which is my main tool, quickly runs out of precision on this kind of computation).
So - is this known? Trivial, perhaps?

REPLY [7 votes]: You might also be interested in the rational Catalan numbers, defined for $(a,b)=1$ as $\mathrm{Cat}(a,b) := \frac{1}{a+b}\binom{a+b}{a}$.
See these slides by Drew Armstrong for their significance: http://www.math.miami.edu/~armstrong/RCCinDC.pdf
As this paper explains, they are a further generalization of Fuss-Catalan numbers: http://arxiv.org/abs/1305.7286

REPLY [2 votes]: One more related oject is the Catalan's triangle with integer entries 
$$B(n,r)=\frac rn{2n\choose n-r}$$
(see also the free book Catalan Numbers With Applications - T Koshy (Oxford, 2009)<|endoftext|>
TITLE: Measures of full Hausdorff dimension for self-affine sets
QUESTION [6 upvotes]: Consider the iterated function system $T_{1}(x)=(\beta x,\tau y)$, $T_{2}(x,y)=(\beta x+(1-\beta),\tau y+ (1-\tau))$ for $\beta\in(1/2,1)$ and $\tau\in (0,1/2)$ with self affine set $\Lambda_{\beta,\tau}$.

It is known that for almost all $\beta$ and all $\tau$ $\dim_{H}\Lambda_{\beta,\tau}=\dim b=1-\log(2\beta)/\log(\tau)$ where $b=(1/2,1/2)$ is the Bernoulli measure on $\Lambda$. Essential this is due to the fact that the projection of $b$ to the $x$-axis is an infinite Bernoulli convolution which is know to be generic absolutely continues (Solomyak theorem).

On the other hand we have shown that if $\beta^{-1}$ is a Pisot number we have
 $\dim b<\dim_{H}\Lambda_{\beta,\tau}<1-\log(2\beta)/\log(\tau)$ for all Bernoulli measures $b$. The projection of all $b$ in this case is singular with a dimension drop (Erdös theorem). 

Now here comes the question: Is there a (ergodic) measure of full dimension and what is $\dim_{H}\Lambda_{\beta,\tau}$ in the later case?

REPLY [4 votes]: If $\beta^{-1}$ is Pisot, then there is an ergodic measure of maximal dimension. This is a special case of the rather difficult Theorem 2.15 in the paper Dimension Theory of iterated function systems by De-Jun Feng and Huyi Hu. Very roughly speaking, Feng and Hu adapt Ledrappier-Young theory to the IFS setting.
Note that the assumptions that $\beta^{-1}$ is Pisot and $\tau<1/2$ are crucial since they ensure that the weak separation condition holds (this is a key assumption in their theorem). As far as I know it is still not known if an ergodic measure of maximal dimension exists for all $\beta\in (1/2,1)$ (of course, this would follow from the above and the extremely difficult conjecture that the only singular Bernoulli convolutions come from Pisot numbers).
I don't know of any explicit formulas for the Hausdorff dimension of the attractor in the case that $\beta^{-1}$ is Pisot. In my paper Overlapping self-affine sets I gave a fairly explicit upper bound:
$$
\dim_H(\Lambda_{\beta,\tau})\le 1-\frac{\log (2\beta)}{\log\tau} + \tau_\beta(q)-(1-q),
$$
where $\tau_\beta$ is the (negative of the) $L^q$ spectrum of the (uniform) Bernoulli convolution of parameter $\beta$ and $q=\log\beta/\log\tau$ (See Theorem 15 and the remark afterward). It is well known that the multifractality of the BC for Pisot parameters ensures that $\tau_\beta(q)<1-q$.
I used to believe that this upper bound is in fact the Hausdorff dimension, but I don't have a proof.<|endoftext|>
TITLE: Linear system of equations with nonnegative solutions and a recursion rule
QUESTION [10 upvotes]: My question derives from reading a recent preprint (arXiv:1209.0827v1, in particular Section 4.1), but it can be phrased quite independently from that paper. The setup is as follows.
Let $A$ be the tridiagonal $n\times n$ matrix with $a_{ii} = -1$ on the main diagonal and $a_{i,i+1} = a_{i+1,i} = 2$ on the other two diagonals. We take the $n\times 1$ vector 1 (consisting of only 1's) as a right-hand side and look for the solution to the linear system of equations. It is not difficult to show that the determinant of the matrix is always an odd integer, the system has a unique solution for each $n$. 

Question (asked by the authors). Are there infinitely many $n$ such that all entries of the solution vector are positive? 

This would be interesting because each such $n$ gives rise to a solution with a particular property of a toy model for the energy transfer in a nonlinear Schrödinger equation. 

Out of curiosity, I did a numerical search for solutions up to $n \sim 1000$ and got the following list of valid $n$ for which the solution is indeed nonnegative.

2, 3, 4, 8, 13, 18, 23, 42, 61, 80, 142, 204, 347, 490, 633, 776, 919, ...

I noticed the following for the sequence of consecutive differences, which starts with

1, 1, 4, 5, 5, 5, 19, 19, 19, 62, 62, 143, 143, 143, 143, 143...

Observation. This sequence seems to have the property that each entry is either the previous entry or it is the previous entry multiplied by the number of times the previous entry has appeared in the sequence in total (4 has appeared once, 5 has appeared three times) plus the biggest element that is strictly smaller. We do see that indeed $19 = 3\cdot5+4$, $62 = 3\cdot19+5$ and $143 = 2\cdot62+19$. Furthermore, each element bigger than 1 in the sequence of differences seems to be one larger than an element in the sequence of valid $n$.
One can now use the rule to create much larger matrices and check whether they have the property and this seems to work just fine, though I am not sure up to which size Mathematica as used by a numerical layman is trustworthy.

My question. Does the sequence of differences really observe this rule? Assuming it does, does it create all $n$ with the desired property? What decides whether the next term in the sequence of differences is identical to its predecessor or the sum of previous terms?

I am fairly confident that all of this is well-known (there seems to be a sort of construction algorithm behind it) and would be thankful for any references.

REPLY [9 votes]: Yes there are infinitely many such values of $n$ and the sequence satisfies the rule you observed. The proof is straightforward but technical.
Let $x_1,\dots,x_n$ be the solution. Add $x_0=0$ and $x_{n+1}=0$, then $2x_{k-1}-x_k+x_{k+1}=1$ for all $k=1,\dots,n$. Introduce $y_k=3x_k-1$, then $y_k$ satisfies a linear recurrence relation 
$$
2y_{k-1}-y_k+2y_{k+1}=0
$$ and boundary conditions $y_0=y_{n+1}=-1$. And we are looking for solutions satisfying $y_k\ge -1$.
Solving the recurrence by the standard method yields that
$$
 y_k = A\sin(k\alpha)+B\cos(k\alpha)
$$
for some constants $A$ and $B$, where $\alpha=\arccos\frac14$. (This number comes from the fact that the roots of the equation $2x^2-x+2=0$ are $\cos\alpha\pm i\sin\alpha$).
Since $y_0=-1$, we have $B=-1$. Then $y_k=-1$ if and only if 
$$
 A = A_k := \frac{\cos(k\alpha)-1}{\sin(k\alpha)} = -\tan(k\alpha/2)
$$
So for the solution with $y_{n+1}=-1$ we have $A=A_{n+1}$, and the relation $y_k\ge-1$ takes the form
$$
\begin{cases}
 A_k \ge A_{n+1}, & A_{n+1}<0 \\
 A_k \le A_{n+1}, & A_{n+1}>0 
\end{cases}
$$
(note that $\tan(k\alpha/2)$ and $\sin(k\alpha)$ are of the same sign).
So $n$ is included in the sequence iff $A_{n+1}$ is either the minimum of the positive $A_k$'s, or the maximum of the negative $A_k$'s ($k=1,\dots,n+1$). Observe that the order of $-A_k$'s in $\mathbb R$ is the same as of the numbers $k\alpha\bmod 2\pi$ in $(-\pi,\pi)$. So everything  boils down to the study of the sequence $\alpha_k:=k\alpha\bmod 2\pi\in(-\pi,\pi)$. A number $n$ is included in the sequence iff $\alpha_{n+1}$ is the best approximation to 0 among entries of the same sign seen so far. Let's add 1 to all these $n$, so we can consider $\alpha_n$ rather than $\alpha_{n+1}$. The indices of the best approximations are
$$
 1(+), 3(-),4(-),5(+),9(-),14(-),19(-),24(+),43(+),\dots
$$
where the signs indicate whether the approximation is positive or negative. The rule for the best approximations is well-known (and easy to prove): the next index is the sum of the latest "positive" one and the latest "negative" one. For example, $19=5+14$, $24=5+19$, $43=24+19$. In other words, the last "positive" entry keeps adding itself to "negative" entries until a new "positive" one appears. Since $\alpha/\pi$ is irrational, zeroes do not happen.  The sequence is infinite because the set $\{\alpha_k\}$ is dense in $(-\pi,\pi)$.
Which entries are "positive"  depends on number-theoretical properties of the number $\alpha/2\pi$. More precisely, how many "positive"/"negative" ones follow in a row is determined by the expansion of this number into a continued fraction. In our case, this expansion should be aperiodic because the number is probably not a quadratic irrational. (I believe it is transcendental but I am not sure).<|endoftext|>
TITLE: Convex subsets of sumsets
QUESTION [5 upvotes]: There's a famous old problem in additive number theory which asks whether, for any $\epsilon > 0$, if $B$ is a 2-basis for the set $A =$ {$1^2,2^2,...,n^2$}, i.e.: a set for which $B+B \supseteq A$, then $|B| = \Omega(n^{1-\epsilon})$ ? Erdos and Newman showed many years ago that the answer is yes if $\epsilon > 1/3$, but not much beyond that is known, I think. So what is it about the set of squares that makes them hard to "cover by sums" ? When thinking about this, I wondered if convexity might be an important property. A finite set $A =$ {$a_1 < a_2 < \cdots < a_n$} is said to be (strictly) convex if the consecutive differences $a_{i+1}-a_i$ are (strictly) increasing. My question, in its simplest form, is as follows: Let $B$ be a set of $n$ integers and let $A$ be a strictly convex subset of $B+B$. Must $|A| = o(n^2)$ ?

REPLY [9 votes]: The conjecture has been resolved in the negative by I. Ruzsa and myself https://arxiv.org/abs/1708.04901<|endoftext|>
TITLE: Is there any "deep" relation between the localization theorem of equivariant cohomology and the  localization theorem of equivariant K-theory
QUESTION [17 upvotes]: First let's consider equivariant cohomology: if a compact Lie group $G$ acts on a compact manifold $M$. We have the equivariant cohomology $ H_G(M)$ defined as the cohomology of the cochain complex $((\mathbb{C}[\mathfrak{g^*}]\otimes \Omega^{\bullet}{M})^G, d_G)$ (for the definition of equivariant cohomology we can look at chapter 1 and 4 of Guillemin and Sternberg's book "Supersymmetry and Equivariant de Rham Theory"). $K< G$ is a closed subgroup, Let $M^K$ be the points of $M$ which has isotropy groups conjugated to $K$, obviously $M^K$ is a $G$-submanifold of $M$ and let $~i: M^K \rightarrow M$ denote the inclusion map.  we have a version of localization theorem, see Guillemin and Sternberg's book "Supersymmetry and Equivariant de Rham Theory" chapter 11, especially Theorem 11.4.3 in page 178. In more details :
Consider the equivariant cohomology $ H_G(M)$ and $H_G(M^K)$  as $ S( \mathfrak{g^* })^G $ modules. Then the pullback map
$$
i^*: H^ * _G(M)\rightarrow H^ *_G(M^K)
$$
is an isomorphism after localizing at some certain prime ideals of $ S( \mathfrak{g^* })^G $.
On the other hand, we have the equivariant K-theory $K_G(M)$ and we also have the localization theorem in this side, see Segal "Equivariant K-theory" (1967) section 4, proposition  4.1, which also claims that 
Then the pullback map
$$
i^*: K^ * _G(M)\rightarrow K^ *_G(M^K)
$$
is an isomorphism after localizing at some certain prime ideals of $R(G)$, the representation ring of $G$.
We notice the similarity of the above two version of localization theorems. Nevertheless  equivariant cohomology and  equivariant K-theory are different. The first is the cohomology of a differential graded algebra and the latter is the Grothedieck group of modules of the cross product algebra $G \ltimes C(M)$.
My question is: is there any deep relation between them? Are they valid because of the same reason?

REPLY [7 votes]: This is very late and you've no doubt learned this in the last five years, but for completeness, the relation is indeed that they are linked by completion and the Chern character, as suggested in one of the comments. Given a compact $G$-space $X$, one can equip $EG \times X$ with the diagonal $G$-action, and then the projection $EG \times X \to X$ is equivariant. Pulling back an equivariant bundle $V \to X$ to $EG \times X$ gives an $G$-equivariant bundle, which descends to bundle $V_G \to X_G$ over the homotopy orbit space (Borel construction) $X_G = (EG \times X) / G$. Thus there is a natural induced ring map 
$$K^*_G(X) \longrightarrow K^*(X_G).$$
(The theorem of Atiyah and Segal is that that this
can be identified with completion at the augmentation ideal $I(G)K^*_G(X)$.)
The Chern character $K^* \longrightarrow H^*(-;\mathbb Q)$ then gives a map to $H^*_G(X;\mathbb Q)$. The composition can be seen as an equivariant Chern character. Everything in sight is natural, so the inclusion $X^K \hookrightarrow X$ induces a commutative diagam
$$\require{AMScd}\begin{CD}
    K^*_G(X) @>>> H^*_G(X;\mathbb Q)\\
    @V V V @VV V\\
    K^*_G(X^K) @>>> H^*_G(X^K;\mathbb Q).
\end{CD}$$
The claim is that the vertical maps become isomorphisms upon localization at certain ideals of $R(G)$ (resp. $H^*(BG;\mathbb Q)$). One checks now that this map, applied in the case $X = *$, sends the one ideal to the other.
Now, we've constructed natural transformations 
$$K^*_G \to K^*(-_G) \to H_G^*(-;\mathbb Q);$$
the final claim is that this process amounts to tensoring with $\mathbb Q$ and then completing at $I(G) K^*_G \otimes \mathbb Q$. One can extract this from the proof of the completion theorem, which proceeds by looking at compact approximations $X_{n,G}$ of $X_G$ and comparing the inverse limit to the ring completion; since these approximations are compact, the Chern character induces isomorphisms $K^*(X_{n,G}) \otimes \mathbb Q \to H^*(X_{n,G};\mathbb Q)$. 
I have never checked this, but because $R(G)$ is Noetherian, I believe the localization theorem in equivariant cohomology at the level of $\mathbb Z/2$-graded rings then follows from the K-theoretic localization theorem by commutative algebra.<|endoftext|>
TITLE: Distances on generalizations of the symmetric group
QUESTION [5 upvotes]: I'm a computer vision student, and I'm looking for some symmetric group literature guidance. I'm going to provide some context, and finally ask two questions.
The Cayley distance and other distances on permutations ($S_n$) are quite useful in applied machine learning; see the beginning of this paper for some examples of distances on $S_n$. Typically, such distances are used when the absolute intensities of measurements are meaningless, but the relative intensities are meaningful. For example, if I wanted to compare the colors of pixels in a natural scene from an RGB image, I would not directly consider the absolute values of the R, G, and B components, as such values are sensitive to lighting. Instead, I would consider the differences in the ratios of R to G, R to B, etc. In the extreme case, I might only care about the ranking of the intensities of the color channels. The ranking is a permutation, so I could compare colors using a permutation distance.
This approach falls down when there is ambiguity in the ranking, as the ambiguity is a source of noise. For example, for a purely green pixel (R = 0, G = 255, B = 0), the rankings R <= B <= G and B <= R <= G are equally valid. So if I don't choose the rankings consistently, identical pixels may have nonzero distance (breaking a metric axiom). 
Even if I choose rankings consistently (say I choose the unique stable sort), weird behavior can arise; consider the pixels {0, 0, 255} and {255, 0, 0}. In the "spirit" of the Cayley distance, these pixels should have a distance of one: swap the R and B channels. This corresponds to the sorts G <= R <= B and G <= B <= R, which differ by one swap. However, the stable sorts are R <= G <= B and G <= B <= R, which has a Cayley distance of two.
Finally, my questions: 1) What generalization of $S_n$ are there that explicitly account for repeated elements? To give a flavor of what I'm asking for, one possible generalization could replace the single-row representation of a permutation, e.g. [1, 3, 4, 2], with a single row of rows representation, e.g. [[1], [3, 4], [2]]. This latter representation would represent the sorts for [100, 255, 127, 127]. Of course, people smarter than myself will have thought about this and come up with a better solution. 2) For a given generalization, what are some common distances on it?

REPLY [4 votes]: I think the generalization you want is a total preorder, or equivalently a strict weak ordering. On finite sets, these correspond to the orderings of values of a function from the set to $\mathbb{R}$. In a total preorder, if $f(x) = f(y)$ then you say $x \le y$ and $y \le x$. In a strict weak order, if $f(x) = f(y)$ then you say neither $x \lt y$ nor $y \lt x$.
Strict weak orders are in bijection with the faces of a permutohedron whose $1$-skeleton is identified with the Cayley graph of adjacent transpositions (by inverting the permutations). The dimension of a face is the number of equalities. The distance you want might be the distance in a graph $G$ whose vertices are the faces of the permutohedron where each face is adjacent to the faces of one lower dimension it contains and the faces of one higher dimension which contain it. 
This distance on $G$, restricted to the $0$-dimensional faces, is at most twice the Cayley distance, since you can achieve a transposition in two steps, moving from $a \lt b \lt c$ to $a \lt b = c$ to $a \lt c \lt b$. However, the macroscopic properties of this distance are quite different from those of the Cayley graph because you can take big shortcuts through high dimensional faces. To reverse $a \lt b \lt c$ takes $3$ transpositions but only $4$ steps in $G$: $a \lt b \lt c$, $a \lt b = c$, $a=b=c$, $b=c\lt a$, $c \lt b \lt a$.  Every face is of distance at most $2(n-1)$ from every other face since every face is of distance at most $n-1$ from the $n-1$-dimensional face in which all coordinates are equal. 
So, another possible distance on $G$ is to declare that the length of an edge is the number of pairwise inequalities added or subtracted. Adjacent faces are related by merging two equivalence classes or dividing one into two. If an edge merges two equivalence classes of sizes $m_1$ and $m_2$, then let the length of that edge be $m_1\times m_2$. So, the distance between $a\lt b=c$ and $a=b=c$ is $2$, and the path of $4$ edges from $a\lt b\lt c$ to $c \lt b \lt a$ has length $6$. If you restrict this distance to the vertices of the permutohedron, you get twice the Cayley distance.<|endoftext|>
TITLE: Isomorphic general linear groups implies isomorphic fields?
QUESTION [47 upvotes]: Suppose $n > 1$ is a natural number. Suppose $K$ and $L$ are fields such that the general linear groups of degree $n$ over them are isomorphic, i.e., $GL(n,K) \cong GL(n,L)$ as groups. Is it necessarily true that $K \cong L$?
I'm also interested in the corresponding question for the special linear group in place of the general linear group.
NOTE 1: The statement is false for $n = 1$, because $GL(1,K) \cong K^\ast$ and non-isomorphic fields can have isomorphic multiplicative groups. For instance, all countable subfields of $\mathbb{R}$ that are closed under the operation of taking rational powers of positive elements have isomorphic multiplicative groups.
NOTE 2: It's possible to use the examples of NOTE 1 to construct non-isomorphic fields whose additive groups are isomorphic and whose multiplicative groups are isomorphic.

REPLY [40 votes]: The answer is "yes", see below.
Dieudonné in his book "La géométrie des groupes classiques"
considers the abstract group $SL_n(K)$ for a field $K$, not necessarily commutative,
and writes $PSL_n(K)$ for $SL_n(K)$ modulo the center.
In Ch. IV, Section 9, he considers the question whether $PSL_n(K)$ can be isomorphic to $PSL_m(K')$ for $n\ge 2,\ m\ge 2$.
He writes that they can be isomorphic only for $n=m$, except for $PSL_2(\mathbb{F}_7)$ and $PSL_3(\mathbb{F}_2)$.
If $n=m>2$, then the isomorphism is possible only if $K$ and $K'$ are isomorphic or anti-isomorphic.
The same is true for $m=n=2$ if both $K$ and $K'$ are commutative, except for the case $K=\mathbb{F}_4$,
$K'=\mathbb{F}_5$. Dieudonné gives ideas of proof and references to Schreier and van der Waerden (1928), to his paper "On the automorphisms of classical groups"
 in Mem. AMS No. 2 (1951) and to the paper of Hua L.-K. and Wan in J. Chinese Math. Soc. 2 (1953), 1-32.
This answers affirmatively the question for $SL_n$, because if $SL_n(K)\cong SL_n(K')$, then $PSL_n(K)\cong PSL_n(K')$.  In the case $n=2$, $K=\mathbb{F}_4$,
$K'=\mathbb{F}_5$, the orders $|SL_2(\mathbb{F}_4)|=60$ and $|SL_2(\mathbb{F}_5)|=120$ are different, and therefore these groups are not isomorphic.
This also answers affirmatively the question for $GL_n$, because $SL_n(K)$ is the commutator subgroup of $GL_n(K)$, except for $GL_2(\mathbb{F_2})$, see Dieudonné, Ch. II, Section 1. In the case $n=2$, $K=\mathbb{F}_2$, we have $|GL_2(\mathbb{F}_2)|=6$ , which  is less than $|GL_2(\mathbb{F}_q)|=q(q-1)(q^2-1)$ for any  $q=p^r>2$, hence 
$GL_2(\mathbb{F}_2)\not\cong GL_2(\mathbb{F}_q)$ for $q>2$.<|endoftext|>
TITLE: proper quoting of theorems
QUESTION [5 upvotes]: Dear collegues, I am writing an overview paper for an academic journal, where I also need to state theorems proved by other authors. Usually I cite the source and then rephrase the theorem. However in a few cases rephrasing seems counterproductive and the optimal formulation was already given by the author. Is it OK to refer to the original author and then give say a 2 lines long theorem verbatim? Can this be considered as a ``weak" form of plagiarism?  Thank you for your answers.

REPLY [5 votes]: As far as I know, it is not plagiarism to use someone else's exact words as long as you make it clear that they are that other person's words and not your own.  If you do this with a long piece of text, though, it might be copyright infringement, and there I don't think either Barry's suggestions or quotation marks would help.  I think avoiding plagiarism is solely your responsibility, but avoiding copyright infringement is something you and the eventual publisher of your paper should handle together.  (Maybe I should add that I consider plagiarism a serious moral issue; copyright is primarily a legal issue with, admittedly, some connections to morality.)<|endoftext|>
TITLE: Geometric derivation of the Einstein’s field equation from the Hilbert action.
QUESTION [5 upvotes]: It is well-known that the equation for stationary solutions of the Einstein-Hilbert functional is given by the Einstein field equation (for a statement, see previous question). The standard derivation of this is through Koszul's formulae either in coordinates (for example wikipedia), or in abstract index notation (for example, in Wald's General Relativty), or in coordinate-free notation (for example, as pointed out by Thomas Richard in Besse - Einstein manifold). This approach is mainly algebraic by using the definition in terms of Koszul's formulae and then calculus in various notations. Essentially the derivation is a direct calculation without the need to even mention the manifold. 
I am wondering if there is a way to derive/interprete the statement refered to at the beginning using an alternative method which is more geometric, ie. using parallel transport or alike. The criterion for "geometric" being (a) a direct reference to the manifold is necessary; or (b) a picture, at least in principle a mental picture, can be drawn to at least carry the main idea of the derivation (of course, pictures of formulae don't count).

REPLY [2 votes]: Comment: the following is a somewhat convoluted way of deriving the
Euler-Lagrange equation using Clairaut's theorem for the volume functional and
some standard, albeit not simpler, variation formulas (all is $C^{\infty}$ in
the following). Let $g_{0}$ be a Riemannian metric and let $v$ be a symmetric
$2$-tensor. Let $g_{0,s}$ satisfy $\frac{\partial}{\partial s}|_{s=0}
g_{0,s}=v$ and $g_{0,0}=g_{0}$. With $g_{0,s}$ as initial data, let $g_{t,s}$,
$t\in\lbrack0,\varepsilon)$, solve the Ricci-Hamilton-DeTurck flow
$\frac{\partial}{\partial t}g_{t,s}=-2\operatorname{Ric}{}_{g_{t,s}
}+\mathcal{L}_{W_{t,s}}g_{t,s}$, where $W_{t,s}=\operatorname{tr}^{1,2}
{}_{g_{t,s}}(\nabla_{g_{t,s}}-\nabla_{g_{t,0}})$. Note that $\frac{\partial
}{\partial t}g_{t,0}=-2\operatorname{Ric}{}_{g_{t,0}}$. Let $v_{t,s}
=\frac{\partial}{\partial s}g_{t,s}$. We have $\frac{\partial^{2}}{\partial
s\partial t}|_{s=0}g_{t,s}=\Delta_{L}v_{t,0}$, which equals $\frac
{\partial^{2}}{\partial t\partial s}|_{s=0}g_{t,s}=\frac{\partial}{\partial
t}v_{t,0}$ (Lichnerowicz Laplacian heat equation). We compute $\frac{\partial
}{\partial t}\operatorname{Vol}(g_{t,s})=\frac{1}{2}\int\operatorname{tr}
{}_{g_{t,s}}(\frac{\partial}{\partial t}g_{t,s})d\mu_{g_{t,s}}=-\int
R_{g_{t,s}}d\mu_{g_{t,s}}$ since $\int\operatorname{tr}{}_{g_{t,s}
}(\mathcal{L}_{W_{t,s}}g_{t,s})d\mu_{g_{t,s}}=0$ by the divergence theorem.
Now
\begin{align*}
\frac{\partial}{\partial s}|_{s=0}\int R_{g_{t,s}}d\mu_{g_{t,s}}  &
=-\frac{\partial^{2}}{\partial s\partial t}|_{s=0}\operatorname{Vol}
(g_{t,s})=-\frac{\partial^{2}}{\partial t\partial s}|_{s=0}\operatorname{Vol}
(g_{t,s})\\
& =-\frac{1}{2}\frac{\partial}{\partial t}\int\operatorname{tr}{}_{g_{t,0}
}(v_{t,0})d\mu_{g_{t,0}}\\
& =\int\langle-\operatorname{Ric}{}_{g_{t,0}}+\frac{g_{t,0}}{2}R_{g_{t,0}
},v_{t,0}\rangle_{g_{t,0}}d\mu_{g_{g_{t,0}}}
\end{align*}
since $\int\operatorname{tr}{}_{g_{t,0}}(\frac{\partial}{\partial t}
v_{t,0})d\mu_{g_{t,0}}=\int\operatorname{tr}{}_{g_{t,0}}(\Delta_{L}
v_{t,0})d\mu_{g_{t,0}}=\int\Delta_{g_{t,0}}(\operatorname{tr}{}_{g_{t,0}
}(v_{t,0}))d\mu_{g_{t,0}}=0$. Finally, take $t=0$.
December 18, 2013. The notion of volume, in various guises, occurs throughout
the study of Ricci flow, especially in Perelman's work. Now, per unit increase
in scale $t$, the volume form of a metric changes with velocity $\frac
{\partial}{\partial t}d\mu=-Rd\mu$. By Clairaut's theorem, the variation of
$-Rd\mu$ is equal to the change per unit increase in scale of the variation of
the volume form, i.e.,
$$
\frac{\partial}{\partial t}(\frac{\operatorname{tr}_{g}v}{2}d\mu_{g})=\left(
\langle\operatorname{Ric}-\frac{1}{2}Rg,v\rangle+\operatorname{div}
(\frac{\nabla\operatorname{tr}v}{2})\right)  d\mu.
$$
In the $f$-warped or entropy version of this, we have $\frac{\partial
}{\partial t}(fe^{-f}d\mu)=(-R-\Delta f)e^{-f}d\mu$ under $\frac{\partial
}{\partial t}g=-2(\operatorname{Ric}+\nabla^{2}f)$ and $\frac{\partial
f}{\partial t}=-R-\Delta f$. Integrating this yields that $\mathcal{N}
\doteqdot\int_{\mathcal{M}}fe^{-f}d\mu$ satisfies $-\frac{d\mathcal{N}}
{dt}=\mathcal{F}\doteqdot\int(R+|\nabla f|^{2})e^{-f}d\mu$. If $\frac
{\partial}{\partial s}g=v$ and $\frac{\partial f}{\partial s}=\frac
{\operatorname{tr}_{g}v}{2}$, then the variation of the energy integrand is
\begin{align*}
& \frac{\partial}{\partial s}((-R-\Delta f)e^{-f}d\mu)\\
& =\left(  (-L(v,\nabla f)+2\langle\operatorname{Ric}+\nabla^{2}
f,v\rangle)e^{-f}+\operatorname{div}(e^{-f}\{\frac{\nabla\operatorname{tr}
v}{2}-v(\nabla f)\})\right)  d\mu\doteqdot A,
\end{align*}
where $L(v,X)=\operatorname{div}^{2}v+\langle\operatorname{Ric},v\rangle
-2\langle\operatorname{div}v,X\rangle+v(X,X)$ is the linear trace Harnack
quadratic. On the other hand, $\frac{\partial}{\partial s}(fe^{-f}d\mu
)=\frac{\operatorname{tr}_{g}v}{2}e^{-f}d\mu$. So Perelman's version is
$\frac{\partial}{\partial t}(\frac{\operatorname{tr}_{g}v}{2}e^{-f}d\mu
)=\frac{\partial^{2}}{\partial s\partial t}(fe^{-f}d\mu)=A$, using Clairaut's
theorem. Note that integration by parts gives $\int L(v,\nabla f)e^{-f}
d\mu=\int\langle\operatorname{Ric}+\nabla^{2}f,v\rangle e^{-f}d\mu$, from
which one obtains Perelman's energy variation formula.
In Section 6.2 of arXiv:0211159 Perelman argues that the $\mathcal{W}$-entropy
(i.e., $\mathcal{F}$ with scaling) integrand is a warped scalar curvature. So,
without scaling (i.e., $\tau$), we would have the correspondences
$\mathcal{N}\sim\operatorname{Vol}$ and $\mathcal{F}\sim\int Rd\mu$, which is
also clear from taking $f=\operatorname{const}\neq0$ as a special case.
However, in 6.2, Perelman's volume is essentially $\int e^{-f}d\mu$, which is
constant under the above variations.<|endoftext|>
TITLE: At what times were people interested in prime numbers
QUESTION [30 upvotes]: While prime numbers are central objects in mathematics it looks that they were ignored and forgotten for long periods of time. I am interested to get some facts and insights about this matter, in particular:
1) Were prime numbers studied in ancient times only by the ancient Greeks? At what periods were they studied by the ancient Greeks themselves?
2) Is it the case that people largely or even entirely lost their interest in the prime numbers for about fifteen centuries until Fermat? What are the facts of the matter and what are the  reasons that may explain these facts.
(motivated by conversations with Ron Livne.)

REPLY [2 votes]: This is a VERY interesting question to which I do not have an answer, despite some research and much cogitation.  In India and China there was substantial deep mathematical thought well before 500 BCE. I find it hard to believe that people did not think about prime numbers but I have been able to find NO evidence.
It is striking to me that the earliest classical Greek mathematicians lived on the Greek shores of Turkey, not in mainland Greece. Some are said to have visited Egypt and inland Turkey, but the history is so vague as to be probably only conjectures made centuries after these people died.  There is little evidence that previous civilizations within the Greek area of travel, made substantial contributions to the kind of mathematical ideas that Greek mathematicians began to explore.  But why did  Thales of Miletus (ca. 624–548 BCE)  and Pythagoras of Samos (ca. 580–500 BCE) (he moved to Croton later) grow up where they did?<|endoftext|>
TITLE: Beautiful theorems with short proof
QUESTION [6 upvotes]: I like to ask for beautiful mathematical theorems with short proof. A proof is short in my sense if it is at most one page assuming basic notations and very basic results a second year student will know and understand. I do not like to discuss what beauty means...I have now written down a script with 107 such theorems (in my sense) and their proof and I have ideas for another 20. One the one hand I would be very thankful for theorems I have yet not considered. On the other hand I like to see if there is some consensus which theorems with a short proof are beautiful.

REPLY [2 votes]: The proof(via the pigeon-hole principle--continued fractions would need too much preparation) that when D>0 is not a square then the "Pellian equation" xx-Dyy=1 has a non-trivial solution.<|endoftext|>
TITLE: Estimating the volume of a semialgebraic set from above
QUESTION [5 upvotes]: Suppose $S$ is a subset of $\mathbb{R}^n$ of finite volume defined by a system of  finitely many polynomial inequalities with integer coefficients. Can anyone describe an algorithm that, given such a system of inequalities, generates a sequence of rational numbers that converges to the volume of $S$ from above? 
This question expands a comment of Andrej Bauer to a  related question.  
This question was posted on math stackexchange here. There were no responses. 
The case that baffles me is when $S$ is unbounded. The obvious approach would be to find some general way to enlarge $S$ by "a little bit" to a set whose volume is easy to compute. But I don't see how to do this.
Actually, I cannot even describe an algorithm to determine whether or not $S$ has finite volume, and such an algorithm might give a good start to solving the original problem

REPLY [2 votes]: Here is an algorithm: partition $\mathbb{R}^n$ into cubes of side $1/k.$ For each cube $C_i$, use your favorite quantifier elimination algorithm to check whether the set $S$ intersects it. Then, your bound is the number of cubes $S$ intersects divided by $k^n,$ which is obviously rational, and just as obviously converges to the volume of $S$ from above. You may argue that your set $S$ is not known to be bounded, so on $k$-th step make your cubes fill a cube of side $k.$ Granted, the algorithm will be rather slow, but given that even computing the volume of a polytope (that is, a semi-algebraic set where the inequalities are linear) is known to be hard, no really quick algorithm is likely...
UPDATE As the OP points out in the comment, the method as described only works for compact semi-algebraic sets (at least in so far as requiring upper bounds). There is a better (as in, theoretically faster) method that also works for compact sets only, due to Henrion/Lasserre/Savognan.<|endoftext|>
TITLE: Non-probabilistic proof of the Johnson–Lindenstrauss lemma
QUESTION [10 upvotes]: The Johnson–Lindenstrauss lemma states that a small set of points in a high-dimensional space can be embedded into a space of much lower dimension in such a way that distances between the points are nearly preserved (a set of $n$ points in high
dimensional Euclidean space can be mapped into an $O(\log (n)/ \varepsilon^2)$-dimensional Euclidean space such that the distance between any two points changes by only a factor of $1 \overset{+}{-} \varepsilon)$. 
A well-known proof of this exploits the phenomenon of concentration of measure and involves, in a sense, a probabilistic argument (Gaussian random matrices).
Does there exist a non-probabilistic proof of this lemma?

REPLY [11 votes]: You might be interested in something Jelani Nelson wrote me in an email on Oct. 13, 2011:
"Another notion of derandomizing JL is the following: come up with a
distribution over embeddings that can be sampled using as few random
bits as possible so that, for any vector $x$ in $R^d$, a random vector has
its $\ell_2$ norm preserved up to $1+\epsilon$ with probability $1-\delta$ by a
random embedding from that distribution (embedding into
$O(\epsilon^{-2}\log(1/\delta))$ dimensions).  Existentially, it can be shown
that there exists such a distribution which can be sampled from using
$O(log(d/\delta))$ random bits.  An actual explicit such distribution
would imply the two works above, since our deterministic algorithm
could just try all embeddings in the support of the distribution
(there would be poly$(d/\delta)$ of them) and take the best one, with
$\delta = 1/n^2$.
Obtaining a distribution that can be sampled using $O(\log(d/\delta))$
random bits is open.  The best we have to date are:
-- Daniel M. Kane, Raghu Meka, Jelani Nelson: Almost Optimal Explicit
Johnson-Lindenstrauss Families. APPROX-RANDOM 2011: 628-639. (requires
$O(\log d + \log(1/\delta)*\log\log(1/\delta) + \log(1/\delta)\log(1/\epsilon))$
random bits)
-- Zohar Shay Karnin, Yuval Rabani, Amir Shpilka: Explicit Dimension
Reduction and Its Applications. IEEE Conference on Computational
Complexity 2011: 262-272. (requires $(1+o(1))\log d +
O(\log(1/(\epsilon\delta))$ random bits).
In fact, combining the approaches of both works can get $(1+o(1))\log d
+ O(\log(1/\delta)\log\log(1/\delta) + \log(1/\delta)\log(1/\epsilon))$."
I suggest contacting Jelani directly if you want more information.<|endoftext|>
TITLE: When does a planar ternary ring uniquely coordinitise a projective plane?
QUESTION [7 upvotes]: From every projective plane a coordinitisation can be constructed on a planar ternary ring, and conversely from every planar ternary ring a projective plane can be constructed. (For background see Weibel's survey of non-Desarguian planes).
Isomorphic planar ternary rings yield isomorphic projective planes, however there exist projective planes that can be coordinitised by non-isomorphic planar ternary rings.
Which planar ternary rings coordinitise their projective plane uniquely up to isomorphism? 
(In other words there is a surjective function from isomorphism classes of planar ternary rings to isomorpism classes of projective planes; on what domain is it injective?)
As stated in Weibel's survey above two ternary rings are isomorphic if (and only if) the automorphism group of the projective plane maps any quadrilateral into any other quadrilateral. It is not clear to me, however, what this means about the ternary ring itself.
I know this class includes the alternative division rings (see Bruck and Kleinfeld - The Structure of Alternative Division Rings, Theorem B in Section 5).
P.S. I originally stated that this class contains the near-fields, but this is false.

REPLY [7 votes]: Too long for a comment.
Contrary to what wikipedia and the accepted answer say, the following paper proves that there are non-Moufang infinite projective planes where the collineation group is transitive on quadrangles:

Otto H. Kegel, Adolf Schleiermacher, Amalgams and embeddings of projective planes,  Geometriae Dedicata 2 (1973), pp 379-395  http://dx.doi.org/10.1007/BF00181482

See also the last theorem in

Martin Funk, Karl Strambach, On free constructions, manuscripta mathematica 72 (1991), pp 335-374, https://doi.org/10.1007/BF02568284, ResearchGate pdf

On the other hand, the collineation group is sharply transitive on quadrangles iff the plane is pappian:

T. Grundhfer,  Projective planes with collineation groups sharply transitive on quadrangles, Archiv der Mathematik, 1984 https://doi.org/10.1007/BF01190963<|endoftext|>
TITLE: What bounds can we establish on coefficients of Swinnerton-Dyer polynomials?
QUESTION [5 upvotes]: The maximum coefficient of the nth Swinnerton-Dyer polynomial seems to grow very fast with n. These are the maximum absolute values of the first 6 polynomials,

2
10
960
13950764
255690851718529024
1771080720430629161685158978892152599456

What bounds can we establish on the absolute value of coefficients in the nth Swinnerton-Dyer polynomial? A very trivial bound appears to be $B(n) = 2^{2^n} n \sqrt{p_n}$ but this doesn't take into account cancellation of any of the terms. Is it possible to do better?

REPLY [3 votes]: There is a missing exponent in your trivial bound. Letting $u_n = \sum_{k=1}^n \sqrt{p_k}$, we have
$$B_0(n) \equiv \max_i \; \left| [x^i] S_n \right|$$
$$\le B_1(n) \equiv \max_i \; [x^i] (x+u_n)^{2^n} = \max_i {2^n \choose i} u_n^i$$
$$\le B_2(n) \equiv { 2^n \choose 2^{n-1} } u_n^{2^n}$$
$$\le B_3(n) \equiv 2^{2^n} (n \sqrt{p_n})^{2^n}.$$
Of course, in the last step we can get a better bound by estimating $u_n$ less crudely. It should also be possible to find an analytic bound for $B_1(n)$ that is less crude than $B_2(n)$ (regardless, $B_1(n)$ is easy to evaluate numerically).
For reference, I have computed numerical approximations of the actual value $B_0(n)$ up to $n = 20$ (correct up to rounding in the last digit):
0:  1
1:  2
2:  10
3:  960
4:  13950764
5:  2.55690851718529e+17
6:  1.77108072043063e+39
7:  8.57834471403602e+86
8:  4.69693103314689e+187
9:  3.24515842436673e+401
10: 8.31078370973184e+853
11: 4.18601441612784e+1793
12: 1.37441368638541e+3755
13: 7.32398012717744e+7815
14: 4.7364530607185e+16172
15: 8.41442697691365e+33355
16: 6.50154879984207e+68684
17: 2.98829955473397e+141188
18: 7.8161464597922e+289271
19: 2.1050522533847e+591950
20: 2.92232330678161e+1209132

Here is a comparison of the bounds:

(source: fredrikj.net)
It appears that $B_3$, $B_2$ and $B_1$ asymptotically overestimate the number of bits in $B_0$ less than by a respective factor 2.20, 2.02 and 1.76.
As you say, it should possible to do better by taking into account the cancellation that occurs. Perhaps by expressing the coefficients in terms of elementary symmetric polynomials? A lower bound would also be interesting.<|endoftext|>
TITLE: Category of Judgements?
QUESTION [10 upvotes]: I have been able to find a lot of information on the category of contexts -- for example, the page on syntactic categories at the nLab is a good starting point.  However, when I try to find similar information on the category of judgements, I find a whole lot less.  My guess would be that I am simply not looking for the right term.
To be more specific, I am looking for a reference which defines the category of judgements with $\Gamma \vdash t:T$, i.e. term $t$ has type $T$ in context $\Gamma$ as objects, and ???? as arrows (i.e. that is one of the things I am looking for).  I am guessing that the morphisms are likely the same as in the category of contexts, namely the substitutions that respect the underlying type theory.

Edit: on top of Andrej's answer, and Paul's book there is also relevant work by Garner such as the paper Two dimensional models of type theory, and the slides Two dimensional locally cartesian closed categories which are quite relevant.
As far as I understand, Seely's work (see links in Andrej's answer) uses explicit reduction paths (based on explicit generators such as $\beta$ reduction) as 2-cells, while the more recent work uses abstract identity types for the same idea.  If I understand well, these are essentially the same, just that Seely's work gave explicit generators for the 2-cells, while in homotopy type theory one allows generalizations to higher dimensions, and the simplest way to do this is to let the inhabitants be implicit.
Surprisingly, no one mentionned that the category of judgements is mostly easily seen as the slice category of the category of contexts over a single variable -- as explained over at the n-lab.

REPLY [8 votes]: One way to set up a category is to use contexts as objects, and declare that a morphism from the context $\Gamma = x_1 : A_1, \ldots, x_m : A_m$ to the context $\Delta = y_1 : B_1, \ldots, y_n : B_n$ is an $n$-tuple $(t_1, \ldots, t_n)$ where $$\Gamma \vdash t_i : B_i.$$
Composition is given by substitution. Then a judgment of the form $\Gamma \vdash u : B$ is just a special morphism whose codomain is the context $y_1 : B$. This sort of thing can be read about in Paul's book.
If you really insist on having judgments as objects, rather than morphisms, you could impose a further 2-categorical structure. Say that a 2-cell from $\Gamma \vdash u : B$ to $\Gamma \vdash v : B$ is an equality $\Gamma \vdash u = v : B$. What kind of equality you use may depends on what your are doing. This way you will get a groupoid-like structure. You could also use one-way reductions as 2-cells, for example $\beta$-reductions, in which case your 2-category will look like a poset enriched category. See the paper by R. Seely,
"Modelling Computations: A 2-categorical framework", LICS 1987. I think Neil Ghani's work is also relevant. See his PhD thesis, but he will be able to provide better references if you contact him.<|endoftext|>
TITLE: Borel sets preserved under open maps?
QUESTION [11 upvotes]: Given open map f: $R^n$ to $R^n$ such that each open set $U\in R^n$, $f(U)$ is also open. Are Borel sets in $R^n$ preserved under f?
Motivation: Pre-image of Borel sets under continuous map is a Borel set in $R^n$.
The problem of the analogous statement above is when $U$ and $V$ are open, $f(U\cap V) \subset f(U)\cap f(V)$, but they may not be equal. Is it possible to construct a concrete counter-example to the statement?

REPLY [14 votes]: Every analytic set ($\Sigma^1_1$ set) of reals is the projection of a Borel subset of $\mathbb{R}\times\mathbb{R}$, and the projection map $p(x,y)\mapsto x$ is an open map. So the standard examples of non-Borel $\Sigma^1_1$ sets are also examples where Borel sets are not preserved by an open map $\mathbb{R}^2\to\mathbb{R}$.
But you asked for an open map $\mathbb{R}^n\to\mathbb{R}^n$, with the same domain and codomain, and the reasoning above concerned only an open map $\mathbb{R}^2\to\mathbb{R}$. Here is one way to fix the issue and make an open map $\mathbb{R}^3\to\mathbb{R}^3$ having the image of a Borel set being non-Borel. Let $h:\mathbb{R}\to\mathbb{R}^2$ be any function whose restriction to every open interval is onto. One can make such a function by using Cantor's interleaving digits trick, combined with the idea of Conway's base 13 function. This function is an open map, since every nonempty open set maps onto the whole space. Now, define $f(x,y,z)=(x,z_0,z_1)$, where $h(z)=(z_0,z_1)$. It is easy to see that the function $f$ is an open map. Meanwhile, every analytic set $A$ has the form $x\in A\iff \exists y B(x,y)$, where $B\subset\mathbb{R}^2$ is a Borel set. Let $C=B\times\mathbb{R}$, which is Borel. Consider the image set $f[C]$, and note that $(x,0,0)\in f[C]$ if and only if there is some $y$ such that $(x,y)\in B$, since in this case we will find a $z$ with $h(z)=(0,0)$; hence, $(x,0,0)\in f[C]$ if and only if $x\in A$, and so the intersection of $f[C]$ with the $x$-axis is $A$, a non-Borel set. So $f[C]$ cannot be Borel if $A$ is not. So this is a case where we have an open map $f:\mathbb{R}^3\to\mathbb{R}^3$ taking a Borel set to a non-Borel set.

REPLY [3 votes]: No.  Projection from $\mathbb R^2$ onto $\mathbb R$ is open, but the image of a $G_\delta$ set can be any analytic set.  So there are non-Borels among them.  See http://en.wikipedia.org/wiki/Analytic_set 
edit  Not a counterexample from a space to itself as required.  
So a counterexample has to be an open map $\mathbb R^n$ to itself, but not at-most-countable-to-one, since those maps do preserve Borel, as I recall.<|endoftext|>
TITLE: Who named it the Snake Lemma?
QUESTION [23 upvotes]: What is the history behind the colorful name of this result? Cartan-Eilenberg states it without any particular fanfare.

REPLY [2 votes]: If you Google for "diagramme du serpent" it becomes plausible that it was a diagram in Cartan-Eilenberg first of all, before a lemma. Interesting example of how Bourbaki became the standard grad student syllabus.<|endoftext|>
TITLE: Is there an analogue of the Erdős–Gallai theorem for simplicial complexes?
QUESTION [20 upvotes]: The Erdős–Gallai theorem gives a necessary and sufficient condition for a finite sequence of natural numbers to be the degree sequence of a simple graph.
In particular $d_1 \ge d_2 \ge \dots \ge d_n$ is the degree sequence of a graph on $n$ vertices if and only if
(1) $d_1 + d_2 + \dots d_n$ is even, and
(2) $$ \sum_{i=1}^k d_i \le k(k-1) + \sum_{i=k+1}^n\min (d_i, k) $$
holds for $1 \le k \le n$.
Now let $\Delta$ be a $k$-dimensional simplicial complex on $n$ vertices, with a complete $(k-1)$-skeleton.  I.e. $$f_{k-1} (\Delta) = {n \choose k} $$
The degree of a $(k-1)$-dimensional face of $\Delta$ is defined to be the number of $k$-dimensional faces containing it.
What are the possible degree sequences
$$d_1 \ge d_2 \ge \dots \ge d_{n \choose k}?$$
Clearly a necessary condition is that $(k+1)$ divides the sum $d_1 + d_2 + \dots$, but is there something analogous to condition (2) above that makes this into necessary and sufficient conditions?
Is the Kruskal-Katona theorem of any help?

REPLY [6 votes]: Very little is known about the question (and even about the easier case of vertex degrees), and it contains as a special case some notoriously hard questions: For example the case that all $d_i$s are equal to 1 (or to some $\lambda$ is the question on the existence of combinatorial designs of certain parameters. I suppose that even checking if a sequence is a dgree sequence is computationally intractable, for large values of $d$ but I am not sure about that.<|endoftext|>
TITLE: Is $k[X]^G$ integral closed in $k[X]$.
QUESTION [5 upvotes]: May assume field $k=\mathbb{C}$.
Let $X$ be an affine variety and $G$ be a reductive group (may assume connected). 
Is the ring of invariants  $k[X]^G$ integral closed in $k[X]$?
The claim may not true in general, probably we may assume $X$ is normal and the affine quotient $X/G$ also normal.
Any reference?

REPLY [20 votes]: Yes, if $G$ is connected. Let $f$ be a function in $k[X]$ which satisfies a polynomial equation over $k[X]^G$. Everything in the $G$-orbit of $f$ must satisfy that polynomial equation. Thus the $G$-orbit is finite, because it is contained in the set of roots of a polynomial, and is connected because it is the orbit of a connected group action, so it is a single point, so $f\in k[X]^G$.
In fact, for $G$ not connected with $G_0$ the connected component of the identity and $G/G_0$ finite, the integral closure of $k[X]^G$ in $k[X]$ is $k[X]^{G_0}$ because any element in $k[X]^{G_0}$ has a finite orbit and the coefficients of the monic polynomial that vanishes exactly on the orbit are $G$-invariant.<|endoftext|>
TITLE: What are good methods for detecting parabolic components and Siegel disk components in the Fatou set of a rational function?
QUESTION [7 upvotes]: Given a rational function $f$ on the Riemann sphere, I would like to answer the question: Does the Fatou set of the function, $F(f)$, contain any parabolic components or Siegel disk components?  Additionally, if it turns out that $F(f)$ contains parabolic components or Siegel disk components I would like to visualize those components, perhaps by highlighting their boundaries in a plot of the Julia set of $f$.  My question is:

Is there a known computational method for reliably detecting and describing the parabolic components and Siegel disk components of the Fatou set of an arbitrary rational function?

I am aware that a Siegel disk component will contain an indifferent periodic point of $f$ and that a parabolic component will have an indifferent periodic point of $f$ on its boundary, so a preliminary step in the method I seek might be to merely identify the indifferent periodic points of $f$. I have thought of a method for detecting these points, but I'm sure it is unreliable.  The method is to take a large number of sample points, $z_0$, and use any cycle detection algorithm (say tortoise and hare) on the sequence $f^n(z_0)$.  Then if that process detects a cycle of period $m$ at $w$ (equal to one of $z_0$, $f(z_0)$, $f^2(z_0)$, etc.), to numerically measure the magnitude of $(f^m)'(w)$ to see if that magnitude is tolerably close to 1.  My secondary question is:

Is there  a known efficient and reliable computational method for identifying the indifferent periodic points of an arbitrary rational function?

Update: I have made some progress by following Alexandre Eremenko's suggestion to examine the orbits of the critical points.  I would like to share an example application of this technique for identifying (rational) neutral cycles.  
Consider the family of functions $f:z\to z - p(z)/p'(z)$, describing Newton-Raphson iteration on $p$, where $p$ is a cubic polynomial with simple roots.  An element $f$ of this family has three attracting fixed points in $\mathbb{C}$, the roots of $p$, and is conjugate by a Möbius transformation to a function with an attracting fixed point at -1, an attracting fixed point at 1, and an attracting fixed point at $\lambda\in\mathbb{C}$.  The following image is a visualization of types of cycles occurring in $f_\lambda$, for $\lambda$ in the axis aligned square of side length 0.04 with top left corner $-0.834 + 0.861 i$.  The only candidate for a critical point of $f_\lambda$ is $\alpha = \lambda/3$.

In the image a positive green color component at a point $\lambda$ indicates that $f_\lambda^n(\alpha)$ converges to one of the fixed points (1, -1, or $\lambda$).  Brighter green indicates slower convergence.  The green regions are typical of Newton basin renderings.  Points $\lambda$ having no green color component, but positive red and blue color components, indicate that $f_\lambda^n(\alpha)$ converges to a cycle of period greater than 1.  In this case the cycle may be neutral or attracting.  Brighter purple indicates slower convergence, so the bright purple regions indicate values of $\lambda$ such that $f_\lambda$ probably has neutral periodic points and $F(f_\lambda)$ might have parabolic components or Siegel disk components.

REPLY [2 votes]: "Detecting" whether a neutral cycle exists or not will be difficult in general. 
However, when you do know that there is a parabolic cycle (e.g. because you have constructed your map that way), Braverman shows, as quoted by Adam, that you can compute the Julia set in polynomial time. What should be noted here is that the program used (and the time bound) will depend on the parameter in question. 
For irrationally indifferent fixed points, things can get even more tricky. But when the rotation number is nice (e.g. some quadratic irrational such as the 'golden mean'), you will have a critical point on the boundary of the Siegel disk. So you can draw the boundary of the Siegel disk by plotting the orbit of the critical point.
In parameter spaces, rather than trying to 'detect' neutral cycles, you may wish to draw the boundaries of hyperbolic components using Newton's method. That is, take a point in the hyperbolic component that you are interested in (where there is an attracting cycle), and then find a curve along which the modulus of the multiplier tends to one. Then you will have found an indifferent parameter. Now you can similarly change the argument of the multiplier, again using Newton's method, and trace this curve. Some care is required near "cusps". 
For an example of such a picture, in the family of exponential maps, see Figure 1 in my article with Dierk Schleicher, "Bifurcations in the space of exponential maps" (http://arxiv.org/abs/math/0311480).<|endoftext|>
TITLE: Computer-aided homology computations
QUESTION [16 upvotes]: Background
I am currently working on the homology of some moduli space and there exists a much simpler chain complex with the same homology.
It is a quotient of a bisimplicial complex by a subcomplex. One could say that some faces are degenerate, thus are zero.
Unfortunately, it's too hard to compute the homology groups by hand.

Given a free chain complex of finite type over the coefficient ring $\mathbb{Z}$ or $\mathbb{Z}/{p^k}$, with $p$ prime, I cannot compute the homology groups by hand. Some matrices have approximately $18.000.000 \times 15.000.000$ entries.
Question: Is there a C/C++ library or an external program that does the following (in descending order of importance)?

computes the homology groups of the given chain complex
describes the image of one or multiple cycles as a linear combination of homology classes
is efficient (e.g. it is capable of using multiple CPUs)

REPLY [10 votes]: I realize that this is an old question and it is likely that the OP has moved on, but let me summarize the state-of-the-art as it exists now.

The given sizes (18 million by 15 million) are pretty hopeless if you are planning to resort to matrix methods. To put things in perspective, latest version of Linbox has been chugging away at a 12,000 x 11,000 dense integer matrix on my cluster for over three months with no end in sight. So, you would almost certainly need to resort to a reduction method and the most effective ones are based on discrete Morse theory.
Since discrete Morse theory exploits the presence of units in the boundary matrix, it is important to keep in mind that there is a world of difference computationally between computing with $\mathbb{Z}$ coefficients versus computing over a field. Over a field, there is no trouble since everything nonzero is a unit, and you are guaranteed to be able to Morse-reduce very effectively.
If you are working over the integers, then everything depends on whether your matrix has many units (i.e., 1's and -1's) in it. If you are unit-rich, and if the actual homology of your complex is light on torsion, then there is a pretty good chance that the Morse-reductions will be drastic even over the integers; and they might produce much smaller boundary matrices which can then be fed into Linbox.
I've written some C++ software to implement these discrete Morse theoretic reductions before computing homology (over the integers!): see here. Although it doesn't mention this explicitly on the website, you can easily feed the software various matrix representations of boundary operators (rather than simplicial complexes) and produce smaller matrices which fit into a quasi-isomorphic chain complex.<|endoftext|>
TITLE: 1-jet bundle on vector bundle with metric connection
QUESTION [5 upvotes]: Background
I'm working to simplify the Lagrangian formalism of classical field theory for the situation of a vector bundle with a bundle metric and a metric connection. Particularly, I want to specify the Euler-Lagrange equations and the Noether theorem for this case.

Consider a vector bundle $(E,\pi,M,\mathbb R^n)$ with a bundle metric $g$ and a metric connection $\nabla$. Let $J^1E$ be the 1-jet bundle associated to $E$. Is there a canonical way to identify an element $j \in J^1E$ with an element $(\phi,\nabla \phi) \in E\times(E\otimes TM^*)$? I would also be grateful for some bibliography on that subject.
What I know is that there exists a 1:1 correspondence between sections of $J^1E \to E$ and connections on $E$. Furthermore the connection leads uniquely to a splitting $TE = VE \oplus HE$ of the tangent bundle of $E$.

REPLY [8 votes]: Is there a canonical way to identify an Element... ?

Yes: an element $j\in J^1E$ is the same as subspace $R\subset T_{\phi}E$ of dimension $\dim(M)$ transversal to $VE$. Since your metric connection gives a splitting $T_\phi E=V_\phi E\oplus H_\phi E$ and since $V_{\phi}E\cong E_{\pi(\phi)}$ and $H_\phi E\cong T_{\pi(\phi)}M$ canonically, you may interpret $R$ as the graph of a linear map $T_{\pi(\phi)}M\to E_{\pi(\phi)}$, hence as an element in $E\otimes T^*M$.
A reference which might be useful: Symmetries and Conservation Laws for Differential Equations of Mathematical Physics 
Edit: (in response to the comment) I assume your definition of jet is as follows: two sections $\phi,\tilde\phi$ of the bundle have the same 1st jet at $p\in M$ iff their values and their first derivatives coincide at $p$ (one then checks that this is independent of the coordinates). Geometrically this means that the two sections are tangent (picture them as submanifolds in the total space), so the plane $R$ is their tangent space at $p$. In local coordinates $(x_1,\ldots,x_m,\phi_1,\ldots,\phi_n)$ the plane $R$ is spanned by the vectors $\partial_{x_1}+\sum\partial_{x_1}(\phi_j)e_j,\ldots,\partial_{x_m}+\sum\partial_{x_m}(\phi_j)e_j$ and your tensor is $\sum \partial_{x_k}(\phi_j)e_j\otimes dx_j$.<|endoftext|>
TITLE: What are the necessary and sufficient conditions for GL(n,Z/p^lZ) to be isomorphic to GL(n,F_p[t]/t^l)?
QUESTION [7 upvotes]: Let $p$ be a prime number and $n,l$ be natural numbers. I'm interested in the conditions under which the general linear groups of degree $n$ over the following two length $l$ finite discrete valuation rings with residue field $\mathbb{F}_p$ are isomorphic:
$$GL(n,\mathbb{Z}/p^l\mathbb{Z}) \cong GL(n,\mathbb{F}_p[t]/(t^l))$$
I've worked out the following cases:

The case $n = 1$ and $l = 2$: In this case, both groups are isomorphic to $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/(p-1)\mathbb{Z}$.
The case $n = 1$ and $l > 2$: In this case, the only situation that the groups are isomorphic seems to be where $p = 2$ and ($l = 4$ or $l = 5$).
The case $n > 1$ and $l = 2$: The isomorphism seems to be equivalent to the requirement that the quotient map $GL(n,\mathbb{Z}/p^2\mathbb{Z}) \to GL(n,\mathbb{Z}/p\mathbb{Z})$ split. It seems that, for $n = 2$, this happens when $p = 2$ and when $p = 3$, but not when $p = 5$. This in turn seems to have something to do with modular/local representation theory. In particular, it seems that $GL(2,\mathbb{Z}/4\mathbb{Z}) \cong GL(2,\mathbb{F}_2[t]/(t^2))$ and $GL(2,\mathbb{Z}/9\mathbb{Z}) \cong GL(2,\mathbb{F}_3[t]/(t^2))$, but $GL(2,\mathbb{Z}/25\mathbb{Z}) \not \cong GL(2,\mathbb{F}_5[t]/(t^2))$. Is there a general way of figuring out what $p$ work and what don't?
The case $n > 1$ and $l > 2$: Don't know what happens here.

The question can be generalized somewhat to comparing the general linear group over a Galois ring with characteristic $p^l$ with residue field of size $q = p^r$ versus the ring $\mathbb{F}_q[t]/(t^l)$.

REPLY [3 votes]: For $l=2$, the only cases where the extension splits are $n=2$ and $p=2,3$ and $n=3$ and $p=2$. In her PhD thesis, on page 3, Pooja Singla attributes this to [Chih Han Sah. Cohomology of split group extensions. II. J. Algebra, 45(1):17–68, 1977] and [Yuval Ginosar. On splitting of the canonical map: mod(p): GL(n,(p)) → GL(n,p). Comm. Algebra, 29(12):5879– 5881, 2001].<|endoftext|>
TITLE: Relative irreducibility  
QUESTION [6 upvotes]: Let $X$ be a one-dimensional one-step irreducible shift of finite type and let $\pi$ be a one-block factor code from $X$ to a sofic $Y$. Suppose $y$ is a right transitive point of
$Y$ and $\pi(u)=y$ for some $u\in X$. Given $u_0=a$ and a block $B$ of $X$, is there a point $x\in\pi^{-1}(y)$ such that $x_0=a$ and $B$ occurs in $x_{[0,\infty)}$? (Note that the stronger statement is true when $\pi$ is a finite-to-one code: a point is transitive if and only if it's image under $\pi$ is transitive, so $B$ actually occurs infinitely often in the right of $x$.)
Thanks to the helpers.

REPLY [3 votes]: Sorry that this is slightly technical... It's uses some concepts that Mahsa showed me in answering a more simple question that I asked her.
Using results from http://arxiv.org/abs/1001.5323v1, we may assume that there exists a 'magic symbol' $k\in Y$ satsisfying that for any
word $y_m\cdots y_n\in Y$ with $y_m=y_n=k$, we have that any word
$x_m\cdots x_n\in X$ with $\pi(x_m\cdots x_n)=(y_m\cdots y_n)$ can be
extended to a sequence $x\in X$ with $\pi(x)=y$. This is not actually the definition of magic symbol but a theorem about them, see the above paper for a definition.
We assume that the desired word $B$ in $X$ starts and finishes with
elements of $\pi^{-1}(k)$, if not we just extend the word $B$ a
little bit. We let $B=b_1\cdots b_m$ and $C=\pi(B)$.
Let $A=\{a^1,\cdots,a^j\}$ be the set of possible values of $x_1$ for
words $x_1\cdots x_m$ with $\pi(x_1\cdots x_m)=C$. Clearly $b_1\in A$,
we let $a^1=b_1$.
We begin with the following lemma:
\begin{lemma}
There exists a word $W=w_1\cdots w_n$ in
$Y$ with $w_1=w_n=k$ such that for each $a\in A$ there exists
a word $V=v_1\cdots v_n\in X$ with $\pi(V)=W$, $V$ contains $B$ and
$v_1=a$.
\end{lemma}
To prove the lemma we define $W=w_1\cdots w_{i_j}$ in chunks which
deal with the possible values of $v_1\in A$.
We let $w_1\cdots w_{i_1}$ be the word $C$. Then for $v_1=a^1$ we are
done, since $B$ is a preimage of $C$ which starts with $a^1$. Since
$B$ starts and finishes with elements of $\pi^{-1}(k)$ it can be
extended to a sequence $x$ projecting to $y$.
Now we deal with $a^2$. By the definition of $A$, there exists a
preimage of $C$ starting with $a^2$. By the transitivity of $X$, this
preimage, which is a finite word in $X$, can be extended to be another
finite word in $x_1\cdots x_{i_2}\in X$ which finishes with word $B$.
We let $w_1\cdots w_{i_2}=\pi(x_1\cdots x_{i_2})$, noting that
$i_2>i_1$ and $\pi(x_1\cdots x_{i_1})=w_1\cdots w_{i_1}$, so there is
no conflict with the values of $W$ which we already defined.
Now we do the same trick for $a^3$. There is a preimage of $w_1\cdots
w_{i_1}$ $(=c_1\cdots c_m)$ starting with $a^3$, and hence there must
be a preimage of $w_1\cdots w_{i_2}$ starting with $a^3$, because
$w_1, w_{i_1}$ and $w_{i_2}$ are all equal to the magic symbol $k$.
Let $x_1\cdots x_{i_2}$ be this word, and because $X$ is transitive we
can extend it to a finite word $x_1\cdots x_{i_3}$ in $X$ finishing
with block $B$. Let $w_1\cdots w_{i_3}=\pi(x_1\cdots x_{i_3})$.
We continue this process until we have a word $W=w_1\cdots w_{i_j}$,
this word $W$ satisfies the conditions of the lemma, so the lemma is
proved.
Now let $y\in Y$ be a transitive sequence. Then $y$ contains the word
$W$, say $y_l\cdots y_{l+i_j}=W$. For our desired starting symbol $u$
there exists a word $x_1\cdots x_l$ with $x_1=u$ and $\pi(x_1\cdots
x_l)=y_1\cdots y_l$, since there exists a preimage of $y$ starting
with $u$. $x_l$ will be some member of $A$, and then using the lemma
we can extend $x_1\cdots x_l$ to a sequence which contains $B$ and
projects to $y$ as required.<|endoftext|>
TITLE: Frobenius splitting of Fano varieties
QUESTION [13 upvotes]: Dear MO,
Question 1. Do you know of an example of a Fano variety which is not Frobenius split?
Background
(1) A variety $X$ in characteristic $p$ is called Frobenius split if there is a "$p$-th root" map $\sigma: \mathcal{O}_X \to \mathcal{O}_X$, that is, an additive map satisfying $\sigma(f^p g) = f\sigma(g)$ and $\sigma(1) = 1$ (in particular, $\sigma(f^p) = f$, so that $\sigma$ is an $\mathcal{O}_X$-linear splitting of the Frobenius map $F: \mathcal{O}_X \to F_* \mathcal{O}_X$). Such varieties enjoy very nice properties, for example, $H^i(X, L)=0$ for $i>0$ for every ample line bundle $L$ on $X$.
In case $X$ is smooth and projective, $X$ is Frobenius split if and only if the map $F: H^{\dim X}(X, \omega_X) \to H^{\dim X}(X, F^* \omega_X)$ is nonzero. Note that $F^* \omega_X = \omega_X^p$. In particular, by Serre duality, $H^{\dim X}(X, \omega_X^p)^\vee = H^{0}(X, \omega_X^{1-p})$ is nonzero, that is, $(1-p)K_X$ is effective - so Frobenius split varieties are ,,on the Fano side''.
(2) A smooth projective variety $X$ is called Fano if $\omega_X^{-1}$ is ample. One can prove (Brion, Kumar Frobenius splitting methods in geometry and representation theory, Exercise 1.6.E5) that if $X$ is a Fano variety in characteristic $0$, then for $p\gg 0$ the reduction $X_p$ of $X$ mod $p$ is Fano and Frobenius split. This means that counterexamples to Question 1 might be difficult to find.
Further questions
Therefore, I am almost sure that, if counterexamples appear in the answers, they will have $\dim X$ (or other invariants of $X$, for example the degree of $K_X$ or its index) big compared to $p$. So I would like to ask:
Question 2. Can you find an effective bound $M = M(X) = M(\dim X, \ldots)$, depending on the dimension of $X$ and maybe other relatively simple invariants, such as the degree or index, such that whenever $X$ is a Fano variety in characteristic $p>M(X)$ then $X$ is Frobenius split. For example, does $M = 0$ (this is Question 1) or $M = n$ work?
Note. The $M = n$ case reminds me of the requirement in the theorem of Deligne-Illusie about decompositions of the de Rham complex that $p$ has to be $>n$.

REPLY [18 votes]: For question 1, the following example works, and you'll see how to construct many more (in higher characteristic).  
Set $k$ to be a perfect field of characteristic $2$.
$$X = \text{Proj } k[x,y,u,v]/\langle x^3+y^3+u^3+v^3 \rangle \subseteq \mathbb{P}^3_k.$$

Fano check:
I'm certain you already know this but...  Certainly $K_X \sim (K_{\mathbb{P}^3} + X)|_X \sim (-4H + 3H)|_X = -H|_X$ where $H$ is the hyperplane on $\mathbb{P}^3_k$.  Clearly also $X$ is smooth.  Thus $X$ is Fano.

Frobenius splitting check:
This is slightly more involved.  First we need a couple lemmas:
Lemma:  [Probably due to Karen Smith]  A projective variety $X$ is Frobenius split if and only if some/every section ring with respect to an ample divisor
$$S_X = \bigoplus_{n \geq 0} O_X(nA).$$
is Frobenius split.
The proof is pretty easy.  If $X$ is Frobenius split, so is $S_X$ (use the splitting on $S_X$ induced by that on $X$ on degrees divisible by $p$, throw the other degrees out).  For the converse direction, it's not hard to show that if $S_X$ has a Frobenius splitting, it then has an appropriately graded Frobenius splitting (basically, use the fact that $Hom_{S_X}(F_* S_X, S_X)$ is generated by appropriately graded maps).  It then induces maps on the associated sheaves.   Care must be taken since $\widetilde{F_* S_X} \neq F_* \widetilde S_X = F_* O_X$, but the latter is a summand of the former and this is enough.
Now for the next lemma.  If $J = \langle g_1, \dots, g_t \rangle$ is any ideal in a ring of characteristic $p$, we use $J^{[p]}$ to denote the ideal $\langle g_1^p, \dots, g_t^p \rangle$.
Lemma:  [Fedder's Criterion]  Suppose $p = \text{char} k$, then a ring $k[x_1, \dots, x_n]/I$ is Frobenius split at a point $\mathbb{m} \in \text{Spec} k[x_1, dots, x_n]$ if and only if $I^{[p]} : I \nsubseteq \mathbb{m}^{[p]}$.
See Lemma 1.6 in $F$-purity and rational singularity by Richard Fedder (1983, Transactions of the AMS).  It's quite easy, but maybe more than I want to explain.
Corollary: If $p = 2$ and $I = \langle f \rangle$, then $R$ is Frobenius split at $\mathbb{m}$ if and only if $f \notin \mathbb{m}^{[2]}$. (in other characteristics, you have( $f^{p-1} \notin \mathbb{m}^{[p]}$).
Ok, now we verify that $X$ is not Frobenius split.  Consider the section ring given our embedding, $S_X = k[x,y,u,v]/\langle x^3 + y^3 + u^3 + v^3 \rangle$.  It is sufficient to show that $S_X$ is not Frobenius split, and that we can check at the origin since $S_X$ is a graded ring.  Then we simply observe that
$$
x^3+y^3+u^3+v^3 \in \langle x^2, y^2, u^2, v^2 \rangle.
$$  

Generalizations, question 2, and further references
The result you mention about Fanos is actually generalized to log-Fano varieties in this paper:  Globally $F$-regular and log Fano varieties due to Karen Smith and myself.  Log Fano's are varieties where $-K_X$ is not necessarily ample, but where it is close.  Some additional discussion of the lemmas above can also be found here.
Lots of people in commutative algebra have been exploring question of when various rings are $F$-split.  For example, see THIS PAPER of Daniel Hernandez and references.  I know there are people exploring exactly when graded hypersurfaces are $F$-split = $F$-pure but this is not public yet.  
In terms of Question 2: effective bounds, I don't know any great ones off the top of my head.  I'll try to get back to you on this.<|endoftext|>
TITLE: Topologically enriched homotopy colimits commuting with homotopy pullbacks
QUESTION [6 upvotes]: Hi,
I am looking for an enriched analogon of Proposition 4.4 in https://www.google.de/url?q=http://hopf.math.purdue.edu/Rezk-Schwede-Shipley/simplicial.pdf
Concretely, I would like to prove the following statement:
Suppose $K$ is a topologically enriched category, i. e. the morphism sets carry a topology and composition is continuous. For a functor $X: K \to Top$, I can consider an enriched version of the homotopy colimit, namely $hocolim X$ is the realization of the simplicial space $srep X$, whose $n$-th level is given by
$ \coprod_{k_0,\ldots,k_n \in (ob K)^n} K(k_0, k_1) \times \ldots \times K(k_{n-1},k_n) \times X(k_0)$
Then, suppose there is a natural transformation between (enriched) functors $X,Y: K \to Top$, such that the diagram
$$
\begin{array}{ccc}
X(k) & \to & Y(k)
\end{array}$$
$$
\begin{array}{ccc}
 X(l) & \to &Y(l)
\end{array}
$$
is a homotopy pullback for all $k,l \in ob\ K$ and all morphisms $\alpha: k \to l$, which induce the (missing) vertical arrows.
Then the diagram
$$
\begin{array}{ccc}
X(k) & \to & Y(k)
\end{array}$$
$$
\begin{array}{ccc}
 hocolim X & \to &hocolim Y
\end{array}
$$
is a homotopy pullback for all $k \in ob K$.
Has anyone ever seen a statement like this or an idea on how to prove it? If it helps, one may assume that the natural transformations $X \to Y$ is levelwise a Serre fibration of topological spaces, since this is the only case, in which I need the statement to be true.
Thanks in advance,
Alex

REPLY [4 votes]: I haven't thought about this hard (no time) but here are quick observations.
Your homotopy colimit is the bar construction $B(\ast,K,X)$, the geometric
realization of the simplicial space with $n$-simplices $B_n(\ast,K,X)$, as
you state.  The map $X(k) \to B(\ast,K,X)$ you are interested in is the
geometric realization of the map from the constant simplicial space at $X(k)$
to $B_*(\ast, K, X)$ that identifies $X(k)$ with the subspace of $B_n(\ast,K,X)$
that sees only identity maps of the object $k$.
Homotopy pullbacks of diagrams one leg of which is a (Hurewicz) fibration 
are equivalent to actual pullbacks, so one approach might be to try to prove 
that $B(\ast,K,X) \to B(\ast,K,Y)$ is a fibration.   It is standard that 
geometric realization of simplicial spaces preserves pullbacks (takes levelwise
pullbacks to pullbacks).  A variation on the theme of replacing maps by fibrations 
should convince you that geometric realization also preserves homotopy pullbacks
(takes levelwise homotopy pullbacks to homotopy pullbacks).
So you would like your map to be the realization of a levelwise homotopy pullback.
However, your stated hypothesis feels wrong to me, since it does not take the topology 
on the category K into account. Your hypothesis presumably should say that the 
evident square with upper left corner $K(\ell,k)$ and lower right corner 
$Map(X(k),Y(\ell))$ is a homotopy pullback.  Assuming that, you should be able
to prove that your map of simplicial spaces is a levelwise homotopy pullback, 
and then you would be done.
Hope that helps a bit.<|endoftext|>
TITLE: Translation of Kähler's "Über eine bemerkenswerte Hermitesche Metrik"
QUESTION [5 upvotes]: Has anyone translated Erich Kähler's "Über eine bemerkenswerte Hermitesche Metrik" into English or French? (Preferably, but I'll take anything.)

REPLY [7 votes]: If anyone is still interested:
http://www.woflmao.net/publications/Kaehler1932.pdf (Wayback Machine)<|endoftext|>
TITLE: Reals added after Cohen forcing II
QUESTION [7 upvotes]: This is related to my previous question "Reals added after Cohen forcing" Reals added after Cohen forcing
The fact is that what I had in mind by $\lambda-$many Cohen reals was different from what was considered in answers. To be more precise   by $\lambda-$many Cohen reals over $V$ I mean a sequence $(r_{\alpha}: \alpha < \lambda)$ which is $V-$generic for the Cohen forcing $Add(\omega, \lambda)$ for adding $\lambda-$many new Cohen reals (thus it is different from saying that we have $\lambda-$many reals where each of them is $V-$generic for the Cohen forcing $Add(\omega, 1)$ for adding a new Cohen real). Similarly for other forcing notions for adding reals.
Let $V_1$ be a generic extension of $V\models GCH$ obtained by adding $\aleph_{\omega}-$many Cohen reals. Then we have the following:
1- In $V_1$ there are $\aleph_{\omega+1}-$many reals,
2- In $V_1$ there are only $\aleph_{\omega}-$many Cohen reals.
Question. Can we find $\aleph_{\omega+1}-$many new reals which are generic for some forcing notion over $V$?

REPLY [6 votes]: It is a general fact that every intermediate model $M\models\text{ZFC}$ sitting between a model and a forcing extension $V\subset M\subset V[G]$ is itself a forcing extension. Indeed, if $G\subset\mathbb{B}$ is $V$-generic for the complete Boolean algebra $\mathbb{B}$, then there is a complete subalgebra $\mathbb{B}_0\subset\mathbb{B}$ such that $M=V[G\cap \mathbb{B}_0]$.  (This is proved in Jech's book.) 
Thus, if $z$ is any real or indeed any set of ordinals in a forcing extension $V[G]$, then we may form the model $V[z]$, which is a model of ZFC, and by the general fact above it is obtained by forcing with a complete subalgebra of the forcing used with $G$. 
In your case, every real $z\in V_1$ has a hereditarily countable name, and thus $z\in V[g_0]$ for some $V$-generic Cohen real in $V_1$. Since every nontrivial subalgebra of $\text{Add}(\omega,1)$ has a countable dense set, it follows that $V[z]=V[g_0]$ for some Cohen real $g_0\in V_1$. In short, every real in $V_1$ that is not in $V$, and this means all $\aleph_{\omega+1}$ of them, is added by the forcing to add a single Cohen real. (These reals are not themselves Cohen reals, but they have names for the forcing to add a Cohen real which interpret to them by some actual Cohen real in $V_1$.)
But I'm not sure if I have understood your question in the sense that you may have intended it. In particular, if you meant that you wanted a single forcing notion to add an entire $\aleph_{\omega+1}$ sequence of reals, then I would say that you already have it, namely, the forcing $\text{Add}(\omega,\aleph_\omega)$ itself was already adding all those extra reals. 
Edit. Let me answer Goldstern's follow-up question in the
comments, a question I find very interesting.
First, I claim that the $V[G]=V[H]$ situation he mentions is
impossible, where $G$ is $V$-generic for adding $\aleph_\omega$
many Cohen reals and $H$ is $V$-generic for adding
$\aleph_{\omega+1}$ many Cohen reals. The reason is that if these
two extensions were equal, then it would follow that the two
forcing notions $\text{Add}(\omega,\aleph_\omega)$ and
$\text{Add}(\omega,\aleph_{\omega+1})$ are isomorphic below
respective conditions and hence simply isomorphic. But the former
has a dense set of size $\aleph_\omega$ and the latter has dense
sets only of size at least $\aleph_{\omega+1}$, since any smaller
set than this could not mention enough points in the domains of
the conditions to be dense.
Second, a similar argument shows now that we cannot even have that $V[G]$
contains a $V$-generic filter $H$ for
$\text{Add}(\omega,\aleph_{\omega+1})$, for in this case we would
have that $\text{Add}(\omega,\aleph_{\omega+1})$ is isomorphic to
a complete subalgebra of $\text{Add}(\omega,\aleph_\omega)$. But
since this latter Boolean algebra is $\aleph_\omega$-dense, it
follows that every complete subalgebra of it is also
(at most) $\aleph_\omega$-dense. But $\text{Add}(\omega,\aleph_{\omega+1})$
has no dense set of size less than $\aleph_{\omega+1}$, for the reason described in the previous paragraph. This is a
contradiction, and so $V[G]$ contains no $V$-generic filter for
$\text{Add}(\omega,\aleph_{\omega+1})$. 
More generally, essentially the same argument shows that adding $\theta$ many Cohen reals can never add a generic filter for adding $\lambda$ many Cohen reals, if $\theta\lt\lambda$. 
But meanwhile, as my answer at the other question shows, $V[G]$ has a family of $\aleph_{\omega+1}$ many pairwise (and finitely-wise) mutually generic Cohen reals. But these do not rise to the level of mutual genericity necessary to form a generic for $\text{Add}(\omega,\aleph_{\omega+1})$.<|endoftext|>
TITLE: Saavedra's Definition of Tannakian Category
QUESTION [9 upvotes]: I have been reading Deligne-Milne's Tannakian Categories and got to the point at the end of part 3 where they discuss what went wrong with Saavedra's definition. Motivated by their counterexample, they ask the following question:
Let $C$ be a $k$-linear rigid abelian tensor category over a field $k$. Let us further assume that $End (\underline{1})=k$ and there exists a fiber functor with values in a field extension $k'\supseteq k$. Does it follow that $C$ is Tannakian?
What is the current status of this question? A reference to a solution or to a modern discussion would be very useful.

REPLY [11 votes]: The answer is yes; in fact, this is mentioned in a footnote to the definition of Tannakian category in Milne's corrected version of the Deligne-Milne paper, on page 32. This is proved in Deligne's Catégories tannakiennes, Theorem 1.12.<|endoftext|>
TITLE: Can all convex polytopes be realized with vertices on surface of convex body?
QUESTION [18 upvotes]: The following question was asked by me on Mathematics.SE. Unfortunately, no one answered it so I thought I might give it a try one level higher. Below the line you can find the slightly edited question, the original one can be found here.

Every convex polytope $P$ has a combinatorial type, its so-called face lattice. This lattice is just the poset of all faces of $P$ ordered by inclusion. Given one realization of such a combinatorial type, one can easily get many others. Just apply an arbitrary projective map to the given realization and the image will be combinatorially equivalent.
Now it would be very nice if one could realize every combinatorial type with vertices on the sphere. Unfortunately, this is not possible. For example, consider the octahedron with pyramids stacked on each facet -- this cannot have all vertices on the sphere (and still be convex).
So my questions is:
Is there a convex body in $\mathbb{R}^d$ such that every combinatorial type of a d-dimensional convex polytope can be realized with vertices on its surface?

REPLY [24 votes]: Yes, there is such a body. Actually there is one very close to the standard unit ball and containing disjoint representatives of each combinatorial type (but these representatives are very small).
Indeed, every combinatorial type of a $d$-polytope has a realization which looks as follows: there is a "large" $(d-1)$-dimensional facet and the remaining surface is a graph over this facet. To construct such a realization, choose a $(d-1)$-facet, pick a hyperplane parallel to it and very close to it (but not intersectiong the polytope), and apply a projective map which sends this hyperplane to infinity.
We can further "flatten" this realization so that it is very close to its large face.  Choose a very small $\varepsilon>0$, apply a homothety such that the diameter of the polytope becomes less than $\varepsilon$, and place the resulting tiny polytope so that it touches the sphere by a point on its "large" face. Then consider the convex hull of the sphere and the polytope. All vertices will be on the boundary of this convex hull if the polytope is sufficiently "flattened". And the convex hull diverges from the ball only in a neighborhood of size $\sim\sqrt\varepsilon$.
Now pick another combinatorial type of a polytope and repeat the procedure with a much smaller $\varepsilon$ and a location on the sphere chosen so that the neighborhood affected by the second polytope does not interfere with the first one. And so on. Since there are only countably many combinatorial types, they all can be packed into the sphere, provided that $\varepsilon$ goes to 0 sufficiently fast.<|endoftext|>
TITLE: Is the category of vector bundles over a topological space abelian?
QUESTION [29 upvotes]: Consider the trivial bundle $V=\mathbb{R}\times\mathbb{R}$ and the map $f:V\rightarrow V$ given by $(t,x)\mapsto(t,tx)$. This has fibrewise kernels and cokernels, but the ranks jump at 0, so the kernel and cokernel of $f$ (as sheaves) are not vector bundles. This example (or similar) is often given to show that Vect($X$) is not in general abelian (or even preabelian).
But this doesn't strike me as correct. Just because $f$ has a kernel (say) $K$ in Sh($X$) which is not an object of Vect($X$) does not mean that there is not an object of Vect($X$) which is a kernel for $f$ in Vect($X$). In the example above, the zero bundle seems to do the job? Indeed I think you can always fix this rank-jumping behaviour by extending smoothly over the bad points (or am I wrong?).
The situation is further confused by Serge Lang claiming (in Algebra, p 134)
"the category of vector bundles over a topological space is an abelian category."
As a counter-appeal to authority Ravi Vakil has (Foundations of AG notes)
"locally free sheaves (i.e. vector bundles), along with reasonably natural
maps between them (those that arise as maps of $\mathcal{O}_X$-modules), don’t form an abelian
category."
So is the category of vector bundles over a topological space abelian or not?

REPLY [12 votes]: Here is an argument to show that this category doesn't have kernels.  Let's take $X$ to be the interval $[-1,1]$, and consider the self-map of the trivial bundle $X \times \mathbb{R}$ given by
$$
(x,t) \mapsto (x, (x + |x|)t).
$$
Suppose this had a kernel $K$, fitting into a sequence $K \to X \times \mathbb{R} \to X \times \mathbb{R}$.  We get a sequence of modules of global sections
$$
\Gamma(K) \to C(X) \stackrel{x + |x|}{\longrightarrow} C(X)
$$
over the ring $C(X)$ of continuous functions on $X$.  (The Serre-Swan theorem has been mentioned already; this is equivalent data.)
However, the global section functor is another name for $Hom(X \times \mathbb{R}, -)$ in the category of vector bundles, and so the "kernel" property implies that the module $\Gamma(K)$ would actually be the kernel of multiplication by $x + |x|$.
This is impossible.  Every element in $\Gamma(K)$ is annihilated by $x + |x|$ by definition, and $\Gamma(K)$ is nonempty, but every nonzero vector bundle on $X$ is some power of the trivial bundle (by the homotopy invariance property) and has many sections which do not vanish on $(0,1]$.<|endoftext|>
TITLE: a determinantal identity
QUESTION [20 upvotes]: Dusan Pokorny and Jan Rataj have just posted a paper (http://arxiv.org/abs/1209.2305) in which they prove the identity
$$
\det (A-B) = \frac 1{d!} \sum_{k=0}^d (-1)^k \binom dk \det((d-k)A + kB)
$$
where $A,B$ are $d \times d$ matrices (this is the corrected version of the formula, in response to Paseman's initial incredulity).
This is so simple and beautiful that one is tempted to suspect that it is "known". Has anyone seen this before?
I should mention Pokorny-Rataj's application: if $f,g$  are (nonsmooth) convex functions on $\mathbb R^n$ , then there exists a signed measure on $\mathbb R^n$  that stands in for the integral of the determinant of the Hessian of $f -g$. In fact, there exists a closed integral current (in the sense of Federer-Fleming) in $\mathbb R^n \times \mathbb R^n$  that stands in for the graph of $\nabla(f-g)$ . Using the identity, this follows from the classical fact that this is true of any convex function (e.g.
$(d-k)f + kg, \ 0\le k \le d$).

REPLY [8 votes]: See the comments of Elkies and Grinberg: 
Looks like an application of the finite-difference formula to
the polynomial $P(k) := \det(dA - k(A-B))$ which has degree $d$
with leading coefficient $(-1)^d \det(A-B)$. (Noam Elkies) 
More generally: if $A$ is a commutative ring, and $P\in A\left[X\right]$ is a polynomial of degree $\leq d$, then $\dfrac{1}{d!}\sum\limits_{k=0}^{d} \left(-1\right)^k \dbinom{d}{k} P\left(k\right)$ equals $\left(-1\right)^d$ times the $X^d$-coefficient of $P$. (Darij Grinberg)<|endoftext|>
TITLE: What are Penrose Tilings, and how do they relate to Quasicrystals?
QUESTION [8 upvotes]: The question is in the title, but let me elaborate a little.
Background
Penrose Tilings are really pretty and satisfy some remarkable properties. For instance, I believe the following is true: even though a Penrose tiling is aperiodic by definition, given a point and an arbitrarily large neighborhood, there exists another point with precisely the same neighborhood (up to rigid motion). While many such fascinating properties are well-documented (see the linked Wikipedia article and the references therein), it is not so clear to me how the tilings themselves should be defined in the first place.
There are three specific tilings attributed to Penrose: P1 (generated by $5$ tiles), P2 and P3 (generated by two quadrilaterals each). My first question is purely one of terminology,

does the term "Penrose tiling" mean precisely one of these three specific constructions, or is it used as a blanket term for any aperiodic tiling that satisfies some axioms (such as the self similarity property mentioned above)?

Relationship with Quasicrystals
I am slightly tentative about mentioning Quasicrystals here since they may not lie within the strict purview of research-level math. However, it seems that MO has been kind to such questions before: see here and here. I'm not an expert on QCs beyond reading the odd survey article or two, so I am not too attached to any particular "definition". Here's the question:

What, if anything, is the precise mathematical connection between Penrose tilings and Quasicrystals?

In particular, it is just a question of using a mathematical object to model the (growth of the) physical object because of shared properties, or are there some concrete insights into the nature of Quasicrystals that can be attained by the study of Penrose tilings?

REPLY [4 votes]: Usually in the mathematics of long range aperiodic order, the Penrose Tiling refers to any tiling obtained from $P_2$ or $P_3$.
We don't differentiate between those, because they have a property called "mutually locally derivable", which says basically that there is a simple way of deriving one from the other. Basically they are the same tiling [the exact meaning of this is that dynamical systems of those tilings are topologically conjugate, which shows that most of their properties are the same]. 
The tiling $P_1$ is as far as I know not mentioned too often, but this is also "the same" (mutually locally derivable) as $P_2$ and/or $P_3$. This means that $P_1$ is harder to study physically, but it has the same long range properties of $P_2/ P_3$, so it can be studied by studying $P_2$ or $P_3$ instead. 
Relation to quasicrystals
If you mark a point in each of the two (or 5) Penrose tiles, you get a point set which you can think as the positions of atoms in an idealized (infinite) solid.
If you calculate the diffraction measure of this model, you get a diffraction diagram which is  discrete (i.e. consisting of sharp Bragg peaks) and with 10-fold symmetry. This is almost identical to the Diffraction of the first discovered quasicrystal, but as it was mentioned in another answer we don't know if there is any quasicrystal which is produced by the Penrose rules.
The mathematical models we have for quasicrystals (we don't know if this is the right model) are the model sets (or the larger class of Meyer sets if you also allow for a "small" diffuse background in the diffraction, as nothing in nature is perfect), produced by the cut and project schemes. Those models have diffraction properties identical to the discovered quasicrystals.
If by a "mathematical quasicrystal" you understand a model set, than it was shown by Ammann/de Bruijn that the Penrose tiling actually fits in this class. So in this sense there is a connection between the two concepts.<|endoftext|>
TITLE: Does every smoothly embedded surface $\mathbb{R}^3$ inherit a natural complex structure, and if so, which one?
QUESTION [12 upvotes]: Smoothly embed a genus g surface in $\mathbb{R}^3$, and pick a normal vector pointing "out" of the surface at each point.  Then on each tangent plane, I have a map which rotates the tangent plane 90 degrees in the direction given by the right hand rule.  This gives an almost complex structure on the manifold, and every almost complex structure is integrable in real dimension 2, so this defines a complex structure on the surface.
Question 1:  Does the above paragraph make sense?  I am at the point where I think I can make grammatically correct sentences using the words above, but I am still not sure about all of the logical interconnections.
Say the first paragraph is correct.  Then for each smooth embedding of the g-holed torus into $\mathbb{R}^3$ I get a complex structure on that surface.  But I know that there are many different complex structures on the g-holed torus (and that the moduli space of such curves is 3g-3 complex dimensional for g>1).  
Question 2,3,4: How many of these complex structures can I get through different embeddings?  If I give you actual equations for an embedding, can you compute which point in the moduli space I am determining?  In the particular case of the torus, can you tell me which $\tau$ corresponds to a given embedding?

REPLY [3 votes]: Your first paragraph does not only make sense but is a very important observation for studying the differential geometry of surfaces in 3-space: For special surfaces classes one can use the induced Riemann surface structure to reduce the differential geometric problem to an algebraic geometric one. Let me just mention some examples

The most obvious one is about minimal surfaces in $\mathbb R^3.$ The (conformal) immersion is harmonic, and is determined (up to euclidean motions) by a pair of holomorphic spinors.
Further restriction to special behavior at the ends led to an algebraic geometric problem.
Willmore spheres turn out to be (after a moebius transformation) minimal spheres with planar ends. They are all given by special contact curves in $\mathbb CP^3.$ (see the surfaces papers of Bryant).
CMC tori: Their Gauss map is harmonic. This turns out to be an integrable system. There exists an associated family of flat $SL(2,\mathbb C)$-connections, which is completely deterimend by a (special) compact Riemannn surface together with some additional spectral data,
see Hitchin (Harmonic maps froma torus to $S^3$) and Pinkall Sterling (all CMC tori in $\mathbb R^3$).

Q2,3,4: For $g\geq 2$ I am not aware of any explicit embedding, nevertheless there are implicitely constructed surfaces, and for some of them one can compute the complex structure explicitely, see for example Lawson's minimal surfaces.
For tori, it depends (of course) how your formula for your embedding looks like. But there are many examples (even of geometric interesting tori) where you can compute the modulus $\tau$ explicitely. I think the easiest way to obtain all complex structures on a 2-torus through emebddings in $\mathbb R^3$ is given by a work of Pinkall (Hopf tori in S^3): The preimage of a (simple, closed) curve in $S^2$ under the Hopf fibration is a torus, and its modulus can be computeted easily in terms of the enclosed area and of the length of the curve. (After stereographic projection, which is conformal, you get a torus in $\mathbb R^3$.)<|endoftext|>
TITLE: Moduli space of modules over non-commutative rings
QUESTION [5 upvotes]: Let $X=Proj(A)$ be a projective scheme, one can the moduli space of coherent sheaves on $X$ with fixed Hilbert polynomial and stability. Since coherent sheaves on $X$ are all obtained as the sheafification $\widetilde{M}$ of a graded $A$-module $M$, it is reasonable to ask whether we can construct the moduli space of sheaves on non-commutative space, i.e. the moduli space of graded right $B$-modules for some $good$ non-commutative graded ring $B$ with some fixed data. I am pretty sure that some works in this direction have been done (what condition on $B$ and  what data of modules should be fixed, etc). 
Could anyone tell me a reference or paper which discuss the construction? Thank you very much. 
Edit
I have for example a quantum plane in my mind.

REPLY [3 votes]: I think the paper "Abstract Hilbert Schemes" by Artin and Zhang might be useful. This may be one of the other papers Peter had in mind in his answer.
The basic idea as far as I understand it seems to be the following; things work nicely if your ring is strongly noetherian! This condition means that whenever you tensor your ring with a commutative noetherian ring your ring is still noetherian. There are examples of rings which are noetherian but not strongly noetherian (the naive blow-ups of Rogalski), but the quantum plane is certainly strongly noetherian - it is an Ore extension and when you tensor with a commutative ring you can express the result as another Ore extension. Then you can use the version of Hilbert's Theorem for Ore extensions, Theorem 2.6 in Goodearl and Warfield's "Introduction to Noncommutative Noetherian rings".
Back to Artin and Zhang's paper, quoting from the abstract, they show that "For the category of the graded modules over a strongly Noetherian graded ring, the Hilbert functor of graded modules with a fixed Hilbert series is represented by a commutative projective scheme."
There is a recent paper of Nevins and Sierra which I think looks at what can be said for non-strongly noetherian rings, http://arxiv.org/abs/1009.2061. I think that they may work with just point modules though, these being the graded modules with Hilbert series that of a polynomial ring in one variable.
I hope this is the kind of thing you were looking for. Also, I'm sure that someone more expert than myself would be able to explain in more detail what Artin and Zhang do.<|endoftext|>
TITLE: Finding the matroids with a specified set of non-bases
QUESTION [6 upvotes]: I'm a grad student in algebraic geometry, and I've encountered a problem which requires me to produce an algorithm involving matroids. Since this isn't my area of expertise, I'm hoping someone knows an algorithm which is known to do this efficiently. The problem is as follows:
I'm dealing only with matroids of rank exactly $k$ on an $n$-element set. I have a bunch of $k$-element subsets of $\{1,\ldots,n\}$. Call a matroid "good" if all of these $k$-element sets are not bases. I want to find all the matroids $M$ which are minimal among the good ones, in the sense that there is no good matroid whose independent sets are a proper subset of those of $M$.
That's the whole problem; if you care about where it came from, keep reading. I'm looking at subvarieties of the Grassmannian $G(k,n)$ which are given by ideals generated by Plücker variables. The irreducible decomposition of such a subvariety can often (though I don't think always; I'm unclear on this point still) be found by taking the subsets in the above paragraph to be the ones that appear as subscripts of the Plücker variables that generate the ideal. Since primary decomposition in Macaulay2 is slow and checking ideal equality is fast, I'm hoping to handle a lot of cases of this procedure by trying to solve it combinatorially and checking to see if I got the right answer. I'd love to talk more about what I'm doing if anyone cares.

REPLY [4 votes]: What is the number n of points in your matroid?
If n is at most 9, you can simply run by the ~400.000 matroids on at most 9 points. The sage matroid package will enumerate these matroids in half an hour. There is also a database of matroids that goes a bit further here: 
http://www-imai.is.s.u-tokyo.ac.jp/~ymatsu/matroid/index.html
I may not fully understand your original problem, but perhaps this is useful: given the 3-term grassmann-plucker relations and the fact that certain $k$-sets are dependent, many purely multiplicative relations follow. E.g. for $k=2$ and on a 4-set $\{a,b,c,d\}$, there is a g-p relation $[ab][cd]+[ad][bc]+[ac][db]=0$. If $ab$ is not a basis, then $[ab]=0$ and hence $[ad][bc]=-[ac][db]$. The multiplicative group generated by the nonzero brackets [..] and these relations is called the Tutte group in case of a matroid. This group is a lot easier to handle computationally that the ideal generated by the g-p relations. Perhaps creating this group will help you to make your guess for the right answer. If I understand your problem correctly, these multiplicative relations are contained in the ideal you are looking for.
Edit: the 3-way g-p relations can also be used in the converse direction: if $[ad][bc]=-[ac][db]$, then $[ab][cd]=0$ follows, i.e. either $[ab]=0$ or $[cd]=0$. But $[ab]=0$ means that $[ab]$ should be removed as a generator. Branching on these possibilities until $[ad][bc]\not=-[ac][db]$ everywhere may be better even than trying to find that matroid. The non-Pappos matroid may turn up when you just try to find a matroid, but this branching process will not terminate at the non-Pappos matroid.<|endoftext|>
TITLE: Terminology for a notion of "categories parameterized by another (symmetric monoidal) category"
QUESTION [5 upvotes]: Consider the category $Space(G)$ of $G$-spaces. For $H\subset G$, there is a forgetful functor from $Space(G)$ to $Space(H)$. Also, for an object $X$ in $Space(G)$ and another object $X'$ in $Space(G')$, $X\times X'$ is in $Space(G\times G')$.
Consider also the category of $Mod(G)$ of modules over $H_G(pt)$. For $H\subset G$, there is a forgetful functor from $Mod(G)$ to $Mod(H)$. Also for a module $M$ in $Mod(G)$ and another module $M'$ in $Mod(G')$, $M\otimes_{\mathbb{Z}} M'$ is in $Mod(G\times G')$.
My question is whether there is an established terminology for the categories $C(G)$ parameterized by a group $G$ satisfying the axioms abstracting those listed above:

For $H\subset G$ there is a forgetful functor from $C(G)$ to $C(H)$
For $G$, $G'$ there is a "multiplication functor" from $C(G)\times C(G')$ to $C(G\times G')$

satisfying various relations. The equivariant cohomology functor $H_G: Space(G) \to Mod(G)$ is compatible with these two operations. Is there a terminology for such a functor?

Update: Pondering more about it, I think it is better to formulate the concept as "categories $\mathcal{C}$ contravariantly parameterized by another (symmetric) monoidal category $\mathcal{X}$ ", i.e. 

for an object $X$ in $\mathcal{X}$, there is a category $\mathcal{C}(X)$
for a morphism $f:X\to Y$ between objects in $\mathcal{X}$, there is a functor $f:\mathcal{C}(Y)\to\mathcal{C}(X)$
for an object $X$ and $Y$ with its product $X\times Y$ in $\mathcal{X}$, there is a multiplication functor which defines, by $o_1\in \mathcal{C}(X)$ and $o_2\in\mathcal{C}(Y)$, an object $o_1\times o_2 \in \mathcal{C}(X\times Y)$.

REPLY [7 votes]: Zhen is right that you can think of it as a lax monoidal pseudofunctor.  In this paper, I called an equivalent structure a "monoidal fibration".<|endoftext|>
TITLE: First known proof of $\sqrt 2$ is irrational with prime factorization?
QUESTION [6 upvotes]: Do any of you happen to know the history of the standard prime factorization proof of $\sqrt 2$ is irrational?  I know this theorem was known to Aristotle, and that the Fundamental Theorem of Arithmetic, on which the proof rests, is found already in Euclid, but I've not been able to track down the origin of this particular proof.  
These sites I know about:  http://www.cut-the-knot.org/proofs/sq_root.shtml, http://www.math.ufl.edu/~rcrew/texts/pythagoras.html, and of course http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
But any other references, online or in paper form, would be greatly appreciated!

REPLY [3 votes]: Here is a proof based on the well-ordering of the positive integers rather than the FTA:
To begin with, we observe that if $\sqrt{2}$ is rational, then there is some positive integer q such that q × $\sqrt{2}$ is an integer. Since the positive integers are well ordered, we may suppose that q is the smallest such number.
We next observe that since 1 < $\sqrt{2}$ < 2, then $\sqrt{2}$ – 1 < 1, and consequently q × ($\sqrt{2}$ – 1) = (q × $\sqrt{2}$ – q ) is less than q. Let us call this new number r, and observe that it too is a positive integer. But we now have r × $\sqrt{2}$ is also an integer, since r × $\sqrt{2}$ = (q ×$\sqrt{2}$ – q ) × $\sqrt{2}$ = (2q – q × $\sqrt{2}$). In short, r is a positive integer less than q and r × $\sqrt{2}$ is an integer. But we said that q was the smallest positive integer with this property, and so we have a contradiction.<|endoftext|>
TITLE: Pochhammer symbol of a differential, and hypergeometric polynomials
QUESTION [7 upvotes]: I have a minor result which I'm sure has come up somewhere before but I can't seem to find it.
Consider a confluent hypergeometric function of the form 
$$\newcommand{\ff}{{}_1F_1}
\ff(b+k;b;z)\textrm{, for }k\in\mathbb{N}.$$
Numerical tests suggest that this is always a polynomial of degree $k$ multiplied by an exponential. One can prove this in a dull fashion by using $\ff(b;b;z)=e^z$ and then applying recurrence relations, but I found a cleaner way using the series definition,
$$
\ff(b+k;b;z)
=\sum_{n=0}^\infty\frac{(b+k)_n}{(b)_n}\frac{z^n}{n!}
$$
where $(b)_k=b(b+1)\cdots(b+k-1)$ is the Pochhammer symbol. By exploiting the identities
$$
\frac{(b+k)_n}{(b)_n}=\frac{\Gamma(b+k+n)\Gamma(b)}{\Gamma(b+k)\Gamma(b+n)}=\frac{(b+n)_k}{(b)_k} 
\qquad\textrm{and}\qquad
nz^n=z\frac{d}{dz}z^n,
$$
one can easily prove that
$$
\ff(b+k;b;z)=\frac{\left(b+z\frac{d}{dz}\right)_k}{(b)_k}e^z,$$
by being somewhat liberal with the meaning of the Pochhammer symbol. This is clearly the desired polynomial-times-exponential, and provides an explicit expression for the polynomial that looks kind of like a Rodrigues formula.
Even better, if you put this together with Kummer's first transformation, $$\ff\left(a;b;z\right)=e^{z}\ff\left(b-a;b;-z\right),$$ and the expression for Laguerre polynomials in terms of hypergeometric functions, $L^{(\alpha)}_{n}\left(x\right)=\frac{\left(\alpha+1\right)_{n}}{n!}\ff\left(-n,\alpha+1,x\right)$, you get an analogous result for Laguerre polynomials,
$$
L^{(b-1)}_{k}\left(x\right)=\frac{1+k/b}{k!}e^z\left(b+z\frac{d}{dz}\right)_ke^{-z}.
$$
Are these results familiar to anyone? Do they fit inside a larger framework? They are not the best thing since sliced bread but they do have a nice simplicity to them, and particularly I would like to cite the appropriate reference if they have appeared before.

REPLY [7 votes]: Formally using the inverse Mellin transform for x>0:
$$e^x f(x\tfrac{d}{dx})e^{-x}=e^x f(x\tfrac{d}{dx}) \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} \frac{x^{-s}}{(-s)!} ds$$
$$=e^x  \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} f(-s) \frac{x^{-s}}{(-s)!} ds.$$
Let $$f(x)=\binom{x+\alpha+\beta}{\beta},$$
then
$$e^x \binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}e^{-x}=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x)$$
where $L_{\beta}^{\alpha}(x)$ is the generalized Laguerre function and $K(-\beta,\alpha+1,x),$ Kummer's confluent hypergeometric function.
For an elaboration, see the notes The Inverse Mellin Transform, Bell Polynomials, a Generalized Dobinski Relation, and the Confluent Hypergeometric Functions.
See also Rodrigues-like formula (attributed to Poole, 1936) on pg. 59 of Bateman's (et al.) Higher Transcendental Functions Vol. I:
$$(x\tfrac{d}{dx}+\alpha)_{n} h(x) = x^{1-\alpha}D^{n}[x^{n+\alpha-1}h(x)],$$
with $D=\tfrac{d}{dx}$, leading to
$$e^x \binom{x\tfrac{d}{dx}+\alpha+n}{n}e^{-x}=e^x x^{-\alpha}\tfrac{D^{n}}{n!}[x^{n+\alpha}e^{-x}]=L_{n}^{\alpha}(x).$$
This can be generalized by using the fractional integro-derivative representation of $K(a,b,x)$ (see Eqn. 13.2.1 on pg. 505 of Abramowitz and Stegun):
$$K(a,b,x)= e^x \tfrac{(b-1)!}{x^{b-1}}\int_{0}^{x} e^{-t}\tfrac{(x-t)^{a-1}}{(a-1)!} \tfrac{t^{b-a-1}}{(b-a-1)!} dt=e^x \tfrac{(b-1)!}{x^{b-1}}D^{-a}[e^{-x}\tfrac{x^{b-a-1}}{(b-a-1)!}],$$
leading to
$$e^x {x^{-\alpha}}\tfrac{D^{\beta}}{\beta!}[x^{\beta+\alpha}e^{-x}]=L_{\beta}^{\alpha}(x)=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,x).$$
Edit (Sept. 2019):
Abstracting, (as Jeans said) "leaving the operators hungry for something to differentiate":
$$\binom{x\tfrac{d}{dx}+\alpha+\beta}{\beta}={x^{-\alpha}}\frac{:Dx:^{\beta}}{\beta!}x^{\alpha}=L_{\beta}^{\alpha}(-:xD:)$$
$$=\binom{\alpha+\beta}{\beta} K(-\beta,\alpha+1,-:xD:),$$
where, by definition, for any two operators $:AB:^{\beta}=A^{\beta}B^{\beta}$.<|endoftext|>
TITLE: A function that is defined everywhere but has unknown values
QUESTION [7 upvotes]: For pedagogical purposes I am looking for a function $\mathbb{N}\to\mathbb{N}$ that is defined everywhere but has most of its values unknown.  Although such a function cannot be simple by definition, I nevertheless hope that there is such a function that is simple to explain.
I have listed a few examples myself which I am not very satisfied with for various reasons:

The halting function.  This function is one of the uncomputable functions (probably the only uncomputable function) that is most easy to explain.
My problem with this one is that it involves computability theory, and I do not want to draw in computability if that is not necessary.
The busy beaver function.  Also an uncomputable function.  Which is a bit harder to explain.
The function that is $1$ everywhere if Goldbach's conjecture is true, and $0$ everywhere if Goldbach's conjecture is false.  
Problem: this is a constant function.
The function that is $1$ if $n$ is even and every even number is the sum of $n$ primes, or $n$ is odd and every odd number is the sum of $n$ primes.  (And $0$ in all other cases.)  For $n=2$ I believe this is the Goldbach conjecture, and for $n=3$ I believe this is the weak Goldbach conjecture.  So this function is already a headeache for $n=2,3$, let alone for $n>3$.
Personally, this would be my favourite if it wasn't so baroque and over the top.
The Collatz characteristic function. $1$ if the Collatz sequence converges, $0$ if it does not.
Problem: too easy to verify on individual arguments. And it has little illustrative value because the Collatz characteristic function seems to be $1$ everywhere.  (Emphasis on "seems".)
The function $\mathbb{N}\to\mathbb{N}_0:n\mapsto \mbox{number of living people 
aged}(n)$.
Problem: depends on the real world.  Not really a pure Platonic math function. (And function values change constantly throughout time.)

BTW: My question rules out computable functions.  Values of computable functions on their domain of definition are known due to, e.g., dovetailing.  (So strictly speaking, the 3rd item should be ruled out for this reason since constant functions are computable.)
Thank you.

REPLY [2 votes]: Kolmogorov Complexity
This should actually be easy to explain.  Fix a programming language like Java.  Let $K(n)$ be the length of the shortest computer program (in number of ASCII characters for example) that returns $n$ as the only output (in decimal notation for example).  (This is roughly the Kolmogorov complexity of $n$.)
This is not computable (knowable) except for a few small $n$.
The main idea is that the even if we know a computer program that outputs $n$, there may be a shorter one.  It just is taking so long to run, and the program itself uses such complicated math that we can't be sure it will stop and give $n$.  (This can be made formal with the Halting problem or Gödel's incompleteness theorem.)

Note, this is a formalization of Barry's Paradox.

Consider the shortest number not definable in eleven words or less. We just defined it in eleven words. 

We could also just use the paradox itself.  Let $B(n)$ be the shortest number of English words needed to define $n$.  Of course this isn't well-defined and leads to the above paradox.<|endoftext|>
TITLE: Weitzenböck Identities
QUESTION [22 upvotes]: I asked this question at Maths Stack Exchange, but I haven't received any replies yet (I'm not sure how long I should wait before it is acceptable to ask here, assuming there is such a period of time).

The Wikipedia page for Weitzenböck identities is explicitly example based. I am looking for a reference which takes a more rigorous approach (as well as a discussion of the Bochner technique). In particular, I am interested in references which focus on these identities in complex geometry.
I have already consulted Griffiths & Harris which is mentioned in the article, but it only contains one example. Berger's A Panoramic View of Riemannian Geometry doesn't have much more.
My interest in Weitzenböck identities has been motivated by a question arising from the following theorem:

Let $X$ be a Kähler manifold and $E$ a hermitian holomorphic vector bundle with Chern connection $\nabla$. Then for the Laplacians $\Delta_{\bar{\partial}} = \bar{\partial}\bar{\partial}^* + \bar{\partial}^*\bar{\partial}$, $\Delta_{\partial} = \partial\partial^* + \partial^*\partial$, we have $\Delta_{\bar{\partial}} = \Delta_{\partial} + [iF_{\nabla}, \Lambda]$.

Is $\Delta_{\bar{\partial}} = \Delta_{\partial} + [iF_{\nabla}, \Lambda]$ an example of a Weitzenböck identity?

REPLY [8 votes]: The most general version of Weitzenbock identities (with coefficients
in appropriate universal enveloping algebras) is due to Uwe Semmelmann and Gregor Weingart: http://arxiv.org/abs/math/0702031
"The Weitzenböck Machine".<|endoftext|>
TITLE: Homotopy groups and homology groups for the $H\mathbb Z$ module-dg module correspondence.
QUESTION [5 upvotes]: In Shipley's paper http://arxiv.org/abs/math/0209215 she proves a Quillen equivalence between the category of $H\mathbb Z$-modules and dg $\mathbb Z$-modules. So, to a chain complex $C$, she assigns a spectrum $HC$, but is it true in some sense that $\pi_i HC \cong H_i C$? I am aware there are some subtleties involved with homotopy groups of symmetric spectra, but I'm not sure how that plays out in this case.

REPLY [7 votes]: Shipley's result (of which an alternative version can be found in EKMM, IV.2) proves that there is a string of Quillen equivalences between the derived category of chain complexes over $\mathbb{Z}$ and the category of $H\mathbb{Z}$-modules.  This, in particular, gives rise to a chain of natural equivalences between their homotopy categories.  We can then prove the result you're asking for by defining both homotopy and homology in the same terms.
Both homotopy categories have a suspension/shift functor, which is an autoequivalence.  In both cases, this can be described as a homotopy pushout of the diagram $\ast \leftarrow X \rightarrow \ast$, where $\ast$ is the terminal object.  Quillen equivalences preserves homotopy pushouts and so preserves shift (and hence, up to natural isomorphism, its inverse).
In the category of chain complexes, there is an object $\mathbb{Z}$, concentrated in degree zero, and this is explicitly carried to $H\mathbb{Z}$ under the chain of Quillen equivalences.
Finally, we can define both homotopy groups and homology groups using the derived category:
$$
H_i C = [\mathbb{Z}[n], C]
$$
$$
\pi_i HC = [\Sigma^n H\mathbb{Z}, HC]
$$
Since all of the ingredients in this definition (shifting degree, the representing object, maps in the homotopy category) are preserved under Quillen equivalence, the two can be made naturally isomorphic.
(Of course, this really manifests that one of Shipley's string of equivalences is Dold-Kan equivalence between simplicial abelian groups and nonnegatively graded chain complexes, which takes homology groups to homotopy groups.)<|endoftext|>
TITLE: Intuition behind the diagonal intersection
QUESTION [14 upvotes]: Suppose that for all $\alpha<\kappa$ we have that $A_\alpha\subseteq\kappa$. We define the diagonal intersection to be $$\bigtriangleup_{\alpha<\kappa}A_\alpha = \left\lbrace\xi<\kappa\ \middle|\ \xi\in\bigcap_{i<\xi}A_i\right\rbrace$$
One of the most surprising theorems in basic set theory, I think, is that if $A_\alpha$ is closed and unbounded (and $\kappa$ is regular and uncountable) then this diagonal intersection is also a closed and unbounded set.
Looking at it from a measure theoretic point of view now, clubs correspond to sets of measure one. Is there any measure theoretic operation which corresponds to diagonal intersections?
Are there possibly other analogies in mathematics which can be used to describe this construction in a rather simple way that non-set theorists could relate to?
Furthermore, it is quite clear that changing the order of the $A_\alpha$ or taking a subsequence can completely change the resulting set. Is there some invariance? For example, up to order the result is unique modulo a non-stationary set?

REPLY [16 votes]: The diagonal intersection corresponds to the infimum in the boolean algebra $\mathbb{B}_I:=P(\kappa)/I$ (where $I$ is the nonstationary ideal, or more generally where $I$ is any normal ideal on $\kappa$).  More precisely:  if $Z \subset P(\kappa)/I$ and $|Z| = \kappa$, then $Z$ has an infimum in $\mathbb{B}_I$, and this infimum is exactly ``the'' diagonal intersection of representatives from the members of $Z$. (This diagonal intersection does not depend on the particular $\kappa$-enumeration of $Z$ or the choice of representatives from the equivalence classes in $Z$; they'll all yield the same element of $\mathbb{B}_I$).<|endoftext|>
TITLE: Does the Monoid Axiom hold for k-spaces?
QUESTION [7 upvotes]: In “Algebras and Modules in Monoidal Model Categories” Schwede and Shipley introduced the monoid axiom. If a cofibrantly generated monoidal model category $M$ satisfies this axiom and some smallness hypotheses, then the category $Mon(M)$ of monoids inherits a model structure with fibrations and weak equivalences taken from $M$. In that paper and related papers by the same authors, the monoid axiom is shown to hold on sSet, simplicial functors, simplicial abelian groups, $\Gamma$-spaces, symmetric spectra, $S$-modules, orthogonal spectra, Ch(R), and StMod(R). 
The monoid axiom was explored further in Hovey's preprint “Monoidal Model Categories.” Here he showed it to hold on compactly generated spaces and he remarks on page 5 that he doesn't know if it holds on $K$-spaces. The proof he gives for compactly generated spaces fails for $K$-spaces because compact spaces are not finite relative to closed inclusions, but only relative to closed $T_1$ inclusions. This was about 10 years ago, so I have to ask:

Does the monoid axiom hold on $K$-spaces?

I've read in several papers that there is no known model structure on the category of topological monoids. For instance, here is one written by Vogt in 2012 which makes this claim. I don't know if that's because the monoid axiom is known to fail, or if it just isn't known, or if the obstruction to building a model structure has more to do with smallness than with the monoid axiom. This is what I'm trying to get at. 

Is there some known obstruction to $Mon(K$-spaces$)$ being a model category? If not, could experts weigh in on why this is such a hard problem, or on whether or not they think it's true?

My interest in this is for writing the background section of my thesis. I do a lot in my thesis with monoids and commutative monoids, and deal a lot with the monoid axiom. Most of what I do probably won't have an application to categories of spaces, and that was not the motivating application. Still, it would be nice to figure out what my results say in that context (if anything) and that's also what motivated my recent questions on $W$-spaces (see here and here). Incidentally, it was also for this background section that I asked my other question on $K$-spaces and learned that they are not named for Kelley. 
Incidentally, I know that some will say this is the wrong question. Instead of strict monoid structures we should care about $A_\infty$ structures. I'm aware of that argument and of the theory in the $A_\infty$ setting, but I have found there are still interesting things to say about strict monoids and strict commutative monoids. So let's restrict attention to that case and only bring in $A_\infty$ as it helps solve the question about strict monoids (e.g. by rectification). Note that this MO question strikes at the difference between $A_\infty$ and strict monoids for spaces.

REPLY [2 votes]: It seems to me that there is a relatively simple answer to this question but perhaps I am overlooking something.
The category of K-spaces does satisfy the monoid axiom. (If I read the question correctly K-spaces are what is usually called "compactly generated spaces" and compactly generated spaces are what is usually called "weak Hausdorff compactly generated spaces.") We need to check that a (transfinite) sequential colimit of pushouts of maps of the form $X \times i$ where $X$ is an arbitrary K-space and $i$ is an acyclic cofibration is a weak equivalence. First, observe that $X \times i$ is a weak equivalence and while it is not necessarily a Serre cofibration it is a Hurewicz cofibration.
Now, the conclusion will follow if we know the two following facts.

Pushouts of Hurewicz cofibrations which are weak equivalences are again weak equivalences (and of course Hurewicz cofibrations.)
(Transfinite) sequential colimits of Hurewicz cofibrations which are weak equivalences are again weak equivalences.

The first one is proven by Boardman and Vogt in Proposition 4.8 (b) in the appendix of Homotopy Invariant Algebraic Structures on Topological Spaces. (The argument doesn't use any fancy point-set topology, in particular separation axioms play no role, so this holds in K-spaces.)
For the second one let's consider a sequence of Hurewicz cofibrations between spaces $(X_\beta \mid \beta < \alpha)$. We want to show that if they are all weak equivalences then so is $X_0 \to \mathrm{colim}_{\beta < \alpha} X_\beta$. First observe that the canonical map from the telescope $\mathrm{Tel}_{\beta < \alpha} X_\beta \to \mathrm{colim}_{\beta < \alpha} X_\beta$ is a homotopy equivalence so it suffices to show that $X_0 \to \mathrm{Tel}_{\beta < \alpha} X_\beta$ is a weak equivalence. By fattening the stages of the telescope slightly we can write it as a colimit of open subspaces homotopy equivalent to the original ones. The conclusion follows since compact spaces are small with respect to open embeddings.<|endoftext|>
TITLE: Dehn function for undistorted subgroups of a product of free groups
QUESTION [7 upvotes]: Let $G$ be a finitely generated subgroup of a product of two finite rank free groups $F_m \times F_n$. If there is a Lipschitz retraction $F_m \times F_n \to G$ with respect to word metrics, then $G$ is undistorted in $F_m \times F_n$, and the Dehn function of $G$ has a quadratic upper bound.
Suppose now that we require only that $G$ is undistorted in $F_m \times F_n$. Is it still true that the Dehn function of $G$ has a quadratic upper bound?

REPLY [10 votes]: The Bieri-Stallings subgroup of $F_n\times F_n$ is undistorted and finitely generated, but not finitely-presented, so in some sense it has an infinite Dehn function.  It's the kernel of the map $F_n\times F_n\to \mathbb{Z}$ which sends each generator to 1, and it's generated by elements of the form $g_ih_j^{-1}$ where $g_i$ and $h_j$ are generators of the two different factors.

REPLY [9 votes]: The answer is no: actually by Baumslag-Roseblade (1984: [BR]) either $G$ is commensurable to a product of free groups (hence has linear or quadratic Dehn function), or is not finitely presented (so the Dehn function is infinite, or not defined, as you wish). The latter case occurs if $H$ is an infinite word hyperbolic group and $f:F_m\to H$ is a non-bijective surjection and $H$ is the fibre product $$\{(g,h)\in F_m\times F_m:f(g)=f(h)\}.$$ 
Edit 1 (reply to Lee Mosher's comment, copy from a comment dated Sep 14 '12) In general the Dehn function of $H$ is equivalent to the distortion of $G$. This is not completely formal (there's a little Van Kampen diagram cuisine) but in the case of $H=\mathbf{Z}$ it's simple to verify by hand. I saw the general result written somewhere but I can't remember right now.
Edit 2 (Extracted from Robert Young's comment to his reply) There's a proof for [this non-distortion fact] mentioned in Olshanskii, Sapir [0S, Theorem 2].

References:
[BR] G. Baumslag, J. Roseblade. Subgroups of direct products of free groups. J. London Math. Soc. (2) 30 (1984), no. 1, 44-52. Journal link (restricted access). MR link
[OS] A. Ol'shanskii, M. Sapir. Length and area functions on groups and quasi-isometric Higman embeddings. ArXiv Link. Internat. J. Algebra Comput. 11 (2001), no. 2, 137-170. Journal link (restricted access). MR link<|endoftext|>
TITLE: References for the result that $\sqrt{n}$ is equidistributed mod 1
QUESTION [6 upvotes]: It is not difficult to show (even without Weyl criterion) that the sequence $\sqrt{n}$, $n=1,2,\ldots$ is equidistributed mod 1. However, I need a reference to this result. Can you help me? Thanks.

REPLY [4 votes]: My favorite reference on this is G. Polya and G. Szego, Problems and Theorems in Analysis,
vol. 1, second part,  Chap IV, section 4, see for example problem 174.<|endoftext|>
TITLE: What fraction of n x n invertible integer matrices contain at least one unit?
QUESTION [15 upvotes]: The question is simple:

What fraction of matrices in $G_n = \text{GL}_n(\mathbb{Z})$ have at least one unit entry (i.e., either $\lbrace\pm 1 \rbrace$)?

I'm not sure what the correct measure  on $G_n$ would be, so here is a suggestion: for each natural number $m \geq 1$, define $G_n(m)$ to be the set of precisely those matrices in $G_n$ whose entries are bounded in absolute value by $m$ and let $H_n(m)$ be the subset of matrices which have at least one unit entry. These are finite and non-empty sets, so in particular for each $m$ the following ratio is defined: $$r_n(m) = \frac{|H_n(m)|}{|G_n(m)|}$$
Now, one can take limits (or lim-sups?) as $m \to \infty$. Again, this is only a suggested measure and it should not constrain potential answers: all reasonable measures are welcome.
Motivation
I write software that pre-processes large (filtered) cell complexes via discrete Morse theory to produce smaller cell complexes with identical homology groups. Without getting into gory details, the basic idea is to greedily exploit unit incidence among cell-pairs in order to clear out the corresponding row and column from the matrix representation of a boundary operator via obvious row and column operations: once these have been cleared, these paired cells can be removed from the complex altogether.
Recently, I was handed a collection of triangulated homology $4$-spheres with tons of torsion in the fundamental groups. On these complexes, the naive greedy collapsing schemes do not produce a perfect reduced complex (i.e., with one zero-dimensional cell and four dimensional cell). In fact, the boundary matrices of the reduced complexes often contain no units at all, and this is precisely when no more collapses are possible. I would like a quantification of how often should one expect an invertible integer matrix to have exploitable units? in order to judge the performance of discrete Morse theoretic reductions on these spheres.

REPLY [2 votes]: Really a couple of comments on @smokedsalmonsandwich's answer:
for the result $\mod p$ the sledgehammer way of dealing with this is the Lang-Weil bound: restricting some specific entry to 1 defines a proper sub variety of algebraic group $SL(n)$ and so for large $p$ the number of restricted matrices is like $c/p$ times the number of unrestricted matrices. You are in the union of $n^2$ such subvarieties, so you get something very similar to @smokedsalmonsandwich's bound.
Secondly, if the matrices have different restrictions on coefficients, there is relevant work of Ahmadi-Shparlinsky: Distribution of matrices with restricted entries over finite fields
O Ahmadi, IE Shparlinski - Indagationes Mathematicae, 2007
Thirdly, one can get sharp error bounds on the asymptotics restricted to congruence subsets using Nevo/Sarnak and Gorodnik/Nevo.
Fourthly, using all of the above together with sieve machinery (see, e.g., Emmanuel Kowalski's book) should give decent bounds on what the odds are of finding a matrix with a $\pm1$ entry.<|endoftext|>
TITLE: Monomorphisms from natural numbers objects into products.
QUESTION [7 upvotes]: Let $A$ and $B$ be objects in a topos $\mathcal{E}$ with natural numbers object $\mathbb{N}$.  If there is a monomorphism $m: \mathbb{N}\rightarrow A\times B$, is it necessarily the case that there is either a monomorphism from $\mathbb{N}$ to $A$ or a monomorphism from $\mathbb{N}$ to $B$?
This is clearly true when $\mathcal{E}$ is Boolean, although the proof I've seen is more involved than one might expect.  It freely uses the Boolean assumption in several places.

REPLY [10 votes]: Let us call an object $A$ infinite if there is a monomorphism $\mathbb{N} \to A$.  Your question then asks whether $A$ or $B$ must be infinite in order for $A \times B$ to be infinite.
From now on we argue in the internal language of a topos. I am going to show that the Lesser Limited Principle of Omniscience (LLPO), which does not hold in all toposes and is a particular instance of the Law of excluded middle, follows from the statement

For all subobjects $A, B \subseteq \mathbb{N}$, if $A \times B$ is infinite then $A$ or $B$ is infinite.

LLPO can be stated as follows: given two infinite binary sequences such that not both of them contain a 1, then one or the other does not contain a 1.
So assume the statement and suppose $f, g : \mathbb{N} \to \lbrace 0, 1 \rbrace$ are sequences as in the premise of LLPO. Define the sets $A$ and $B$ by
$$A = \lbrace n \in \mathbb{N} \mid \forall k < n . f(k) = 0 \rbrace$$
and
$$B = \lbrace n \in \mathbb{N} \mid \forall k < n . g(k) = 0 \rbrace$$
We claim that $A \times B$ is infinite. Define the sequences $a : \mathbb{N} \to A$ and $b : \mathbb{N} \to B$ by
$$a_n = \begin{cases}
  n, & \forall k < n . f(k) = 0\\\\
  k, & k < n \land f(k) = 1 \land \forall j < k . f(j) = 0
\end{cases}$$
and
$$b_n = \begin{cases}
  n, & \forall k < n . g(k) = 0\\\\
  k, & k < n \land g(k) = 1 \land \forall j < k . g(j) = 0
\end{cases}$$
Notice that $a_m = a_n$ and $m \neq n$ imply that $f(k) = 1$ for some $k \leq \min(m,n)$, and a similar observation holds for the other sequence. The sequence $n \mapsto (a_n, b_n)$ clearly takes values in $A \times B$. It is injective because $(a_m,b_m) = (a_n, b_n)$ and $m \neq n$ imply that both $f$ and $g$ contain a 1, contrary to the premise of LLPO. Therefore, $A \times B$ is infinite and by the statement $A$ or $B$ is infinite. If $A$ is infinite then $f$ does not contain a 1, and if $B$ is infinite then $g$ does not contain a 1.
In summary, the answer to your question is negative because the statement implies LLPO, which is not valid in every topos. In particular, LLPO is violated in the effective topos.
I do not know (yet) whether LLPO implies the statement.
An obvious question to ask is whether we can positively violate the statment, i.e., can we produce two non-infinite objects $A$ and $B$ such that $A \times B$ is infinite. I would search for such objects in the effective topos, where the question will take on a recursion-theoretic flavor.<|endoftext|>
TITLE: Is there a category of topological-like spaces that forms a topos?
QUESTION [5 upvotes]: The category of convergence spaces generalise topological spaces and form a quasi-topos, as topoi are allegedly nicer is there a nicer kind of topological-like space, the category of which forms a topos?

REPLY [2 votes]: Let me point out a couple of other possibilities, they may come handy in certain situations. The realizaibility topos $RT(P(\omega))$ over Scott's graph model $P(\omega)$ contains countably based spaces as a full subcategory. The inclusion preserves countable limits, countable coproducts, and those exponentials which happen to exist in countably based spaces. This can be generalized to spaces of arbitrary weight by using a larger graph model $P(\kappa)$. But as Todd points out, such a topos contains a lot of non-topological junk.
If you do not insist on a topos, and are willing to consider slightly less than a topos, then the exact completion of topological spaces is a good candidate. The objects are topological spaces equipped with a (formal) equivalence relation, so they feel like spaces still. Not too much junk there. For this sort of thing, see

Carboni, A., Rosolini, G. Locally cartesian closed exact completions, J.Pure Appl. Alg., 154, 2000.

A naive way to get spaces inside a topos is to simply consider (pre)sheaves on all spaces, which of course runs against size restrictions. Nevertheless, a useful picture emerges, see

Rosolini, G., Equilogical spaces and filter spaces, Rend. Circ. Mat. Palermo, 64, 2000<|endoftext|>
TITLE: Characterization of schemes whose dualizing complex is perfect
QUESTION [10 upvotes]: I'm wondering if there is a characterization of schemes over a a field $k$ whose dualizing complex is a perfect complex in terms of singularities. E.g. on a proper Cohen-Macauley scheme over a field, the dualizing complex is a sheaf, when is this sheaf have a finite locally free resolution? (Or this never happens unless the scheme is Gorenstein?)
Recall a complex is called a perfect complex if it is locally quasi-isomorphic to a bounded complex of locally free sheaves.

REPLY [14 votes]: As Hailong said in his comment this only happens in the Gorenstein case; here is a sketch of an argument.
Suppose $X$ is a quasi-compact quasi-separated scheme with a dualising complex $D$ and let us assume $D$ is perfect. Recall that $\mathsf{D}^\mathrm{perf}(X)$, the category of perfect complexes over $X$, is a rigid tensor category i.e., it is closed symmetric monoidal and  setting, for E a perfect complex, $E^\vee = hom(E,\mathcal{O}_X)$ we have natural isomorphisms
$$ E^\vee \otimes F \cong hom(E,F)$$
for all perfect complexes F (here I am being notationally lazy - of course the internal hom and tensor are the derived ones).
Now we just notice that there are isomorphisms
$$R \cong hom(D,D) \cong D^\vee \otimes D$$
so $D$ is an invertible object with respect to the tensor product in $\mathsf{D}^\mathrm{perf}(X)$. This implies $D$ is isomorphic to a shift of a line bundle on each connected component of $X$ (see for instance Prop 6.4 of "Gluing techniques in triangular geometry" by Balmer and Favi for a short proof, although, I should mention, the result is pretty well known and older than that paper). Thus $X$ is Gorenstein.<|endoftext|>
TITLE: A discrete random walk that avoids previously visited vertices for an exponentially distributed time interval
QUESTION [8 upvotes]: Imagine a discrete random walk on an infinite one-dimensional lattice where, for every unit interval of time, $(t_1, t_2, ...)$, the walker takes a step with uniform probability to its left or right.  
We add the constraint that each time the walker visits a vertex, the vertex is transiently "blocked", and cannot be revisited by the walker until it is "unblocked", which has a constant probability $p$ of occurring during every unit interval of time $t_i$ prior to the walker taking a step.  In other words, every time the walker visits a vertex, the vertex is assigned an exponentially distributed random variable $X$ (always with the same rate parameter $\lambda = p$) that determines the number of time intervals that need to pass (counting from the interval in which the site is "blocked") before it can be reoccupied by the walker.  If the vertex to the right or left of a walker is blocked during some time interval, it will move to the left or right, respectively.  In the situation where sites to the left and right of the random walker are "blocked", the walker will remain in place for that unit interval of time.  
To provide the extremal examples, if $p = 1$, sites are immediately unblocked prior to the walker taking its next step, and one will have a vanilla one-dimensional random walk with a Gaussian probability distribution.  If, however, we set $p = 0$, sites will never "unblock" and the walk will become fully self-avoiding and, with a direction chosen with uniform probability, continuously move to the right or left without ever revisiting the origin.   
As a function of $p$, what is the probability distribution for this walker?  How does this walk generalize to higher dimensions, specifically $D = 2$?  Is there a "magic term" for this sort of walk, and are there any good literature references?
As a sort of cute extension, provided some $p$ can we predict the mean distance the walker will travel returning to its point of origin?

Note - I would be happy to accept the simplification where, instead of assigning vertices an exponentially distributed random variable, $X$ can adopt an alternative probability distribution.
Note 2 - I suppose one could say that this system is similar to a persistant or correlated random walk with a memory for each step that lasts according to the chosen probability distribution.  My impression is that the literature mostly refers to vertex reinforced random walks as situations where repeated traversal of "blocked" vertices increases or otherwise alters the barrier to revisiting a particular vertex, with no alternative mechanism for a decrease or change in these barriers over time.

REPLY [8 votes]: At the intuitive level, it would seem that the process should behave as if it had finite memory, and be diffusive in the long term: as soon as $p>0$ I would expect a central limit theorem. Of course the variance of the limit should depend on the value of $p$ and be smaller and smaller as $p\to0$. Now proving that is another story, and questions of that kind tend to be very tricky indeed.
One case where everything works out well: assume, in one dimension (but that is not essential here), that the lifetime of a block is always bounded by some fixed number $A$, and that there is a positive probability $\eta$ that it is $0$ (meaning that the backtracking probability is positive). Then you cannot be ballistic forever, because at some point you will jump $A$ times in the same direction while putting traps with no lifetime. The probability of that happening at time $t$ is bounded below by $(\eta/2)^A$. After such a stretch, the world is completely fresh because all traps have expired. This is called a regeneration time, and the existence of such a time implies central limit theorems and convergence to Brownian motion in the scaling limit. It even gives a (ludicrously bad) bound on the variance of the limit, as a function of $\eta$ and $A$, if you really need one.
The other easy case is the one when a blockage has lifetime bounded below, then there is never a way to backtrack and you go to infinity linearly in a randomly chosen direction.
I would think that the regeneration time argument will apply in most cases under similar hypotheses (say, any dimension and positive probability to be zero), as soon as the tail behavior of the trap lifetimes is small enough, but obviously that would need to be checked more carefully.<|endoftext|>
TITLE: The Fukaya category of a simple singularity (reference request)
QUESTION [10 upvotes]: I have heard that for an ADE singularity $f$,
$ D^b\mathrm{Fuk}(f) \simeq D^b(\mathrm{Rep}\ Q)$
where $Q$ is the corresponding Dynkin quiver. (As one would hope, if $\mathrm{Fuk}$ is some kind of categorification of the Milnor fibre...)

Could anyone point me to a reference for this result?

REPLY [7 votes]: This sounds wrong to me.  I think $D^b(Q)$ should be replaced by the derived category of finite length modules over the corresponding preprojective algebra of affine type.
Homological mirror symmetry would say the derived Fukaya category should be equivalent to the bounded derived category of quasi-coherent sheaves over the mirror manifold (in this case, over the exceptional fibre of the resolution of the singularity).  I don't understand this in any detail, and I don't know a reference for it, so I may not be getting this quite right.  
The quasi-coherent sheaves supported on the exceptional fibre form an abelian category equivalent to the finite length modules of the preprojective algebra of the corresponding affine type.  This is discussed in the introduction Bridgeland's Stability conditions and Kleinian singularities.  The reference he gives for the link between the two is Crawley-Boevey and Holland, Noncommutative deformations of Kleinian singularities, Duke Math. J. 92 (1998), no. 3, 605–635.<|endoftext|>
TITLE: Mostow's theorem on algebraic groups
QUESTION [7 upvotes]: In his classical 1956 paper 
Fully reducible subgroups of algebraic groups
Mostow proves the following theorem:

Theorem 7.1.
  Let $G$ be an algebraic group over a field $K$ of characteristic 0, 
  $\mathfrak{N}$ the set of nilpotent elements in the radical of its Lie algebra,
  and $N$ the connected algebraic group with Lie algebra $\mathfrak{N}$.
  Then $G=M\cdot N$ (semi-direct) with $M$ a maximal fully reducible group.
  Furthermore any two maximal fully reducible subgroups of $G$
  are conjugate under an inner automorphism from $\mathfrak{N}$.

This paper was written long before the Three Books on linear algebraic groups were published, 
and I do not fully understand Mostow's terminology.
Is "a fully reducible group" the same as "an algebraic group with reductive identity component" (in characteristic 0)?
Does the following result follow from Mostow's theorem?

Theorem? 
  Let $G$ be a linear algebraic group, not necessarily connected, over a field $K$ of characteristic 0.
  Let $G^0$ denote the identity component of $G$, and let $N$ denote the unipotent radical of $G^0$.
  Then the extension
  $$
1\to N\to G\to G/N\to 1
$$
  splits, i.e. $G=M\cdot N$ (a semidirect product),
  where $M\subset G$ is a $K$-subgroup of $G$ isomorphic to $G/N$.

Are there modern proofs of Mostow's theorem? References will be appreciated.

REPLY [4 votes]: Mostow's theorem actually has a relatively modern textbook treatment, by his early collaborator Gerhard Hochschild: see Theorem 4.3 in VIII.4 of his book Basic Theory of Algebraic Groups and Lie Algebras (GTM 125, Springer, 1981).   Hochschild followed the original attempt by Chevalley (in volume 2 of his Theorie des groupes de Lie) to transfer to affine algebraic groups over arbitrary fields the main ideas of Lie group theory.   Though this approach eventually becomes unsatisfactory in prime characteristic (especially over fields which are not algebraically closed), as Chevalley recognized, it does work fairly well in characteristic 0.   
In particular, Hochschild is able to reformulate in somewhat more modern language some of Mostow's insights into Levi decomposition.   For algebraic groups, the unipotent radical supplants the traditional solvable radical in Lie group theory due to the nicer description of solvable algebraic groups and the good behavior of tori.   It's worth taking a look at Hochschild's proof of Mostow's theorem and related results, since he provides a fairly elementary unified framework that still makes sense to people working today on more sophisticated problems in algebraic geometry.    Any way the proof is approached, it definitely has prerequisites, as the version here by grp illustrates.   It's hard to say what version is "easiest", since that depends heavily on the reader's background.
In any case, Hochschild also points out why the restriction to characteristic 0 is essential.   The notion of "reductive" algebraic group is meaningful in any characteristic, but "fully reducible" (or "linearly reductive" in Hochschild's more current language) applies only to tori among the connected algebraic groups in prime characteristic (Nagata).   Indeed, there are serious problems with the notion of Levi decomposition (and conjugacy of Levi factors) in the latter case, which George McNinch has recently been studying.    Fortunately Mumford's geometric invariant theory survived, due to the proof by Haboush of his conjecture that reductive groups are at least "geometrically reductive" in all characteristics.   But Lie groups don't serve well enough as a model for linear algebraic groups in general.<|endoftext|>
TITLE: A word with maximal number of subwords
QUESTION [6 upvotes]: Let $W$ be a cyclic word of length $n$ in a 2-letter alphabet $\{0,1\}$. It is clear that it has at most $n^2$ different subwords (because the number of subwords of length $i$ is at most $n$ for each $i$) and that the actual number of subwords is less than $n^2$ (because the number of subwords of length $1$ is not $n$, but $2$). What is the maximal possible number of subwords as a function of $n$ and what are words where this upper bound is achieved.

REPLY [10 votes]: If you take a de Bruijn sequence of length $2^k$, then you have every length $k$ 
sequence precisely once. This implies that the number of subwords is maximal, since each subword of length $\geq k$ is determined uniquely by its prefix, and each subword of length $ < k$ occurs (with equal frequency). So this achieves the upper bound when $n=2^k$.<|endoftext|>
TITLE: Finiteness of elliptic curves of a given conductor
QUESTION [10 upvotes]: It follows from the modularity theorem for elliptic curves over $\mathbb{Q}$ that there are finitely many elliptic curves of a given conductor $N$. Moreover, one can algorithmically enumerate them. [Edit: As Emerton comments below, without further argument, this is only true for elliptic curves up to isogeny!]

Was the finiteness and/or algorithmic enumeration of elliptic curves of a given conductor known before the modularity theorem?

Every elliptic curve over $\mathbb{Q}$ can be written in the form  $y^2 = x^3 + ax + b$ where $a, b \in \mathbb{Z}$ with discriminant $\Delta = -16(4a^3 +27b^2) \neq 0$. So the number of elliptic curves of discriminant $D$ is bounded above by number of nontrivial pairs $(a, b) \in \mathbb{Z}^2$ such that $D = -16(4a^3 +27b^2)$.
Let $D \in{\mathbb{Z}}, D \neq 0$ be given. Because $D \neq 0$, the cubic equation  $b^2 = \frac{-D}{16\cdot27} - \frac{4a^3}{27}$ is nonsingular, so by Siegel's theorem there are finitely many solutions. It follows that there are finitely many elliptic curves of a given discriminant. Silverman's book says that Baker even gave an explicit upper bound in this case, which was refined by Stark. 
However, a priori there is no bound on the size of the discriminant of elliptic curves of a given conductor, so it doesn't immediately follow that there are finitely many elliptic curves of a given conductor. Szpiro's conjecture implies that if one fixes the conductor there are only finitely many discriminants that give that conductor. However, this conjecture is open (or not, depending on the status of Mochizuki's work).

Is there a weaker form of Szpiro's conjecture that has been proved giving an upper bound on the discriminant of an elliptic curve of a given conductor? If so, what's the minimum amount of technology needed to get the results? 

Of course there are also issues of effectivity as well, which I also welcome comments on.

REPLY [7 votes]: I just want to mention a couple of things to complement what Joe Silverman and Noam Elkies have already said.
As Brian Conrad pointed out in response to this ill-posed question of mine, you need the Shafarevich conjecture (for abelian varieties of all dimensions as proved by Faltings) to prove modularity of elliptic curves.
I also want to mention this paper of Cremona and Lingham:
Finding all elliptic curves with good reduction outside a given set of primes 
which deals with the algorithmic problems of finding all elliptic curves of a given conductor as discussed in Noam's first comment. They compute the $S$-integral points on $y^2 - x^3 = D$ assuming that a Mordell-Weil basis has been computed. The ideas are implemented in Sage and can be used to find the elliptic curves with everywhere good reduction outside a sufficiently simple finite set of primes $S$. 
For example, if you wanted to find all elliptic curves with everywhere good reduction outside $37$ you would use the following code
sage: egros = EllipticCurves_with_good_reduction_outside_S
sage: time curves = egros([37])
Time: CPU 1.49 s, Wall: 1.50 s
sage: for E in curves:
....:     print E.a_invariants()
....:     
(0, 0, 1, -1, 0)
(0, 1, 1, -23, -50)
(0, 1, 1, -1873, -31833)
(0, 1, 1, -3, 1)
(0, 1, 1, -4563, 116200)
(0, 1, 1, -31943, -2138543)
(0, 1, 1, -2564593, -1581651042)
(1, -1, 0, 3166, -59359)
(1, -1, 0, -2276219, -1321241558)
(1, -1, 1, 2, -2)
(1, -1, 1, -1663, -25680)
(0, 0, 1, -1369, 12663)
(0, 1, 1, -12, -17)
(0, 1, 1, -2602, 50229)
(0, 1, 1, -16884, -647702)
(0, 1, 1, -3562594, 2587011456)<|endoftext|>
TITLE: Examples of excess intersection theory?
QUESTION [12 upvotes]: Let $M$ be a smooth manifold of dimension $m$ and $\pi:E\rightarrow M$ a vector bundle of rank $e$. Given a section $s$ of the bundle $\pi:E\rightarrow M$, we expect that the zero locus $Z(s)$ of $s$ is a submanifold $N\subset M$ of dimension $m-e$. This is true if $s$ can be perturbed into a general position so that $s(M)$ and the zero section intersect transversally. 
Perturbation is not always possible (for example in holomorphic category category). In this case we need "excess intersection theory"; if the section $s$ lies in a subbundle $F\subset E$ and is a transversal section of $F$, the correct $(m-e)$-cycle we should take is the Euler class of the quotient bundle $E/F$, which is homologous to $Z(s)$ if transverse perturbation of $s$ exists. 
My problem is that I don't really know good explicit examples with which I can compute things. Could anyone give me an example or reference, which shows how useful excess intersection theory is?
Edit
My motivation to study excess intersection theory is virtual cycles of moduli spaces, which of course are very good examples of excess intersection theory. But I am looking for some elementary examples on which I can compute things. I want to convince myself that the theory is really reasonable by computing a few simple examples.

REPLY [2 votes]: Surely not the simplest example but certainly one of the reason why excess intersection theory is useful is the theory of virtual fundamental classes. Suppose that A is some moduli space and that you
can find M,E, s as in the question such that A = Z(s). E/F is called the obstruction bundle (it is the obstruction to the transversality of s). Then the pullback to A of the Poincare dual of
the top chern class of the obstruction bundle is an homology class of the "expected " dimension m-e.
In order to simplify suppose that this dimension is 0 then the integral of 1 over the virtual fundamental class has to be interpreted as the number of elements in A for a generic perturbation
of the parameters of which M can depend, even if such a deformation actually does not exist.
In general, there does not exist M,E and s but sometimes it exists locally and one can still use excess intersection theory in order to define a virtual fundamental class.
Most of the  counting theories of curves uses this idea (Gromov-Witten, Donaldaon-Thoms, stables pairs ...)(see for example for a survey "13/2 ways to count curves" of R.Pandharipande and R.P.Thomas).
In this way, one obtains a lot of examples which can be seen as application of excess intersection theory : when A is smooth, the obstruction is really a bundle and you just have to calculate a top chern class (just to cite a real "concrete" simple example : Gromov-Witten invariants for homology class 0 can be written has an integral of caracteristic classes over the moduli space of curves.
There are surely thousands of such examples.)<|endoftext|>
TITLE: Realizable Order Sequences for Finite Groups
QUESTION [14 upvotes]: My post is motivated at least in part by this MO question. 
Has there been any work done on realizable order sequences for finite groups? By an "order sequence" I mean a non-decreasing list of the orders of the elements of the group. By "realizable" I mean there is some finite group that has that particular order sequence.
For example, $(1, 2, 4, 4)$ is a realizable order sequence; it is the order sequence of $\mathbb{Z}/4\mathbb{Z}$.
Is this something that has been studied in any depth? (Perhaps under a different name?) Are there any non-trivial theorems about realizable order sequences? (Examples of trivial theorems: ones that fall immediately out of Lagrange's Theorem; the degree sequence of $\mathbb{Z}/n\mathbb{Z}$)
I know that there are some nice theorems about degree sequences in graph theory (e.g., Erdős-Gallai theorem; Havel-Hakimi theorem), and though I have seen "order sequence" defined in a few Abstract Algebra texts, I have yet to come across any results of much interest.
I also wonder whether results such as this problem solution (Problem 6636, F. Schmidt, Amer. Math. Monthly, Vol. 98, No. 10 (Dec., 1991), pp. 970-972) or this paper (Isaacs et al (2009). Sums of element orders in finite groups. Commun. Alg. 37(9):2978-2980) contain ideas that would be applicable to such a topic.
Finally, is there an easily accessible (and organized) database that lists order sequences for all finite groups up to a certain not-too-small size? 

Edit 1: Is anyone up to computing such a list and posting it somewhere accessible?

Edit 2: Now that Alexander Gruber has kindly posted computations for a fair number of order sequences, I wonder: (a) when does the same order sequence correspond to more than one group? (b) given an order sequence that corresponds to precisely one group, how difficult is it to recover the corresponding group's structure?

Edit 3: Mr. Gruber has most recently pointed me toward a related area of research on "OD-characterizability." One mathematician who has done a fair bit of work in this area is AR Moghaddamfar. See, for example, Recognizing Finite Groups Through Order and Degree Pattern.

REPLY [5 votes]: At least for finite abelian groups the problem has been solved. See
Isomorphism of Finite Abelian Groups,
by Ronald McHaffey, The American Mathematical Monthly, Vol. 72, No. 1 (Jan., 1965), pp. 48-50, Stable URL: http://www.jstor.org/stable/2313001
The author essentially proves that if the sequence of orders of two finite abelian groups are the same then the groups are isomorphic.<|endoftext|>
TITLE: Isogeny classes of elliptic curves
QUESTION [8 upvotes]: Let $E \subset \mathbb{P}_\mathbb{C}^2$ be an elliptic curve. If $E$ has complex multiplication (by anything) then the theory of complex multiplication in particular tells us that if $\sigma \in \textrm{Aut}(\mathbb{C})$ then $E^\sigma$ will be isogenous to $E$. Suppose $\textrm{End}(E) \cong \mathbb{Z}$ do we still have that $E^\sigma \cong E$ or are there some obvious counterexamples? What if we insist that the $j$-invariant, $j(E)$ is an algebraic number?

REPLY [11 votes]: One way to get a counterexample is to take any elliptic curve $E$ over a quadratic field $K$ whose conductor is a prime ideal $\mathfrak{p}$ lying over a split prime $p$ in $\mathbb{Q}$.  This means that $E$ has multiplicative reduction at $\mathfrak{p}$.  If $\sigma$ is the non-trivial automorphism of $K/\mathbb{Q}$, then $E^\sigma$ has conductor $\mathfrak{p}^\sigma \neq \mathfrak{p}$, so $E^\sigma$ has good reduction at $\mathfrak{p}$.  Any isogeny $E \to E^\sigma$ would be defined over a number field, and multiplicative/good reduction are both stable under finite field extension.  Since isogenous curves have the same reduction type, $E$ and $E^\sigma$ cannot be isogenous.
An example: the two curves of minimal norm conductor over $\mathbb{Q}(\sqrt{5})$ have norm conductor 31, but their conductors are the two different ideals of norm 31. Check out William Stein's table:
http://modular.math.washington.edu/Tables/hmf/sqrt5/ellcurve_aplists.txt 
As a double check, you can even tell from the Hecke eigenvalues in the table that the curves are conjugate to one another.<|endoftext|>
TITLE: 'Contactization' and Symplectization
QUESTION [10 upvotes]: Given any manifold $M$, we can get a symplectic manifold by taking the cotangent bundle $T^\ast M$ with symplectic form $\omega=\sum dp_i\wedge dq_i$. Given any manifold $M$, we can get a  contact manifold by taking the projectivization of the cotangent bundle $\mathbb{P}^\ast M=(T^\ast M-\lbrace0\text{-section}\rbrace)/{\sim}$ where the contact form arises from the tautological 1-form on $T^\ast M$.
Given any contact manifold $(N,\lambda)$, we can get a symplectic manifold by symplectization $\mathbb{R}\times N$ with symplectic form $d(e^s\lambda)$. Continuing in the same spirit:
Is there a "contactization" to pass from any given symplectic manifold to a contact one, making use of the symplectic data?
Aside: I came across a paper of Eliashberg-Hofer-Salamon (Lagrangian Intersections in Contact Geometry), and in certain scenarios we do indeed have one. If our symplectic manifold $M$ is exact, i.e. $\omega=d\alpha$, then $(M\times S^1,dz-\alpha)$ is a contact manifold. Now if we don't have exactness, there is at least a way to contactize $M$ when some positive multiple of $\omega$ represents an integral cohomology class in $H^2(M)$, and this is some principal $S^1$-bundle called ''pre-quantization''.  Is ''pre-quantization'' the only way to contactize here?

REPLY [6 votes]: The "pre-quantization" construction of a contact manifold out of symplectic manifold predates prequantization by a couple of decades: 
see Boothby, W. M.; Wang, H. C.
On contact manifolds.
Ann. of Math. (2) 68 1958 721–734. 
The analogue of the theorem for symplectic orbifolds is due to Thomas: Thomas, C. B.
Almost regular contact manifolds.
J. Differential Geometry 11 (1976), no. 4, 521–533.
You may think of the Boothby-Wang construction as constructing a contact fiber bundle over a symplectic manifold with fiber $S^1$.  If we look at the construction this way, it can be generalized.  See my paper  Contact fiber bundles.
J. Geom. Phys. 49 (2004), no. 1, 52–66.<|endoftext|>
TITLE: A 14th and 26th-power Dedekind eta function identity?
QUESTION [28 upvotes]: Given the Dedekind eta function $\eta(\tau)$.  Define $m = (p-1)/2$ and a $24$th root of unity $\zeta = e^{2\pi i/24}$.

Let p be a prime of form $p = 12v+5$. Then for $n = 2,4,8,14$: 
$$\sum_{k=0}^{p-1} \Big(\zeta^{m k}\, \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = -\big(\sqrt{p}\;\eta(p\tau)\big)^n\tag1$$
Let p be a prime of form $p = 12v+11$. Then for $n = 2,6,10,14, and\; 26$:
$$\sum_{k=0}^{p-1} \Big(\zeta^{m k}\, \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = \big(\sqrt{p}\;\eta(p\tau)\big)^n\tag2$$

Note: One can also use the identity $\zeta^N \eta(\tau) = \eta(\tau+N)$ for a further simplification.
It is easily checked with Mathematica that $(1)$ and $(2)$ hold for hundreds of decimal digits, but are they really true? (Kindly also see this related post.  I have already emailed W. Hart, and he replied he hasn't seen such identities yet.) 
----EDIT----
The $n = 26$ for proposed identity $(2)$ was added courtesy of W.Hart's answer below. (I'm face-palming myself for not checking $n = 26$.)

REPLY [14 votes]: (Intended as a comment to Noam Elkies' response.)
My colleague Marco Streng was kind enough to point out that according to "Eta Products and Theta Series Identities", a book of Guenther Koehler MR2766155: "Serre [128] proved that the Fourier series of a modular form f is lacunary if and only if f is of CM-type, i.e., if f is a linear combination of Hecke theta series. In [129] he showed that eta^r for r=2,4,6,8,10,14,26 are the only even powers of eta which are lacunary."
Here, [129] is J.-P.Serre, Sur la lacunarité des puissances de $\eta$, Glasg. Math.
J, (27) 1985, 203--221.
Note that this is only for even powers of eta, not for odd powers or other kinds of eta product, of which there are numerous lacunary expressions. 
Anyway, this suggests that there ought to be an identity for n = 26, (and possibly others for various eta products?).<|endoftext|>
TITLE: A fourier series related to spin Chern numbers almost commuting matrices
QUESTION [6 upvotes]: Let 
$$
f(x)=\sin(x)\sqrt{1+\cos^{2}(x)+\cos^{4}(x)}.
$$
In my study of almost commuting unitary matrices, $U$ and $V$, I
have need for a bound like
$$
\left\Vert \tilde{f}(V)U-U\tilde{f}(V)\right\Vert \leq C\left\Vert VU-UV\right\Vert ,
$$
where $\tilde{f}(e^{\pi ix})=f(x)$ so my functional calculus makes
sense. (Matrix function to any applied listeners). The norm is the 
operator norm.
This is all in
the context spin Chern numbers and the Pfaffian-Bott index, as in
my paper with Hastings, Topological insulators and $C^{*}$- algebras:
Theory and numerical practice'' which came out in 2011. I want precise
bounds on how small a commutator I need to ensure a derived matrix
remains invertible.
One way to estimate $C$ is to writing $f$ and $f^{\prime}$ as Fourier
series and working term by term. Let $g=f^{\prime},$ so
$$
g(x)=\frac{3\cos^{5}(x)}{\sqrt{1+\cos^{2}(x)+\cos^{4}(x)}}.
$$
I need to know, or get a good estimate on $\left\Vert \hat{g}\right\Vert _{1}$,
meaning the $\ell^{1}$-norm of the fourier series of $g(x)$.

What is the $\ell^{1}$-norm of the
  Fourier series of
  $\frac{3\cos^{5}(x)}{\sqrt{1+\cos^{2}(x)+\cos^{4}(x)}}$,
  or what is a good upper bound?

I can prove this is less than $3\pi$ but I want a tighter estimate.
Perhaps someone has seen this before. Is there a stategy to bound
this I should follow? Better still, what is the Fourier
series for $g$?
Any hints on finding $C$ by another route will be welcome.
A little more about $f$: The Bott index and its relatives require three functions with
$f^2 + g^2 + h^2 = 1$ and $gh=0$ and these are to be continuous, real valued and
periodic.  And non-trivial, so $h=0$ is not allowed.  For theoretical work these
are all equally valid, but to use these in index studies of finite systems we
need to pick carefully to control a bunch of commutators.  This $f$ is designed
to be similar to $\sin$ but will big "flat spots."  Notice 
$$
\left(f(x)+\cos^{3}(x)\right)^{2}=1.
$$

REPLY [5 votes]: Let $\|\cdot \|_F $ denote the $\ell^1$ norm of the Fourier series of a function on $[0,2\pi]$.  This is a Banach algebra norm on functions whose Fourier series converge absolutely. 
Note that $$1 + \cos^2(x) + \cos^4(x) = \frac{15}{8} + \frac{1}{2} e^{2ix} + \frac{1}{2} e^{-2ix} + \frac{1}{16} e^{4ix} + \frac{1}{16} e^{-4ix}$$  Write this as $(15/8)(1 + W(x))$, where $\|W(x)\|_F = 3/5$.  So $$g(x) = 3 (15/8)^{-1/2} \cos^5(x) (1+W(x))^{-1/2}= 3 (15/8)^{-1/2} \cos^5(x)\sum_{k=0}^\infty {{-1/2} \choose k}  W(x)^k$$
Thus $$\|g\|_F \le 3 (15/8)^{-1/2} \sum_{k=0}^\infty \left| {{-1/2} \choose k}\right| (3/5)^k = 2 \sqrt{3}$$
Moreover, by explicitly evaluating the norm of a partial sum of the series and bounding the norm of the remainder you can get arbitrarily good approximations to the actual value of $\|g\|_F$.<|endoftext|>
TITLE: symmetry in monoidal categories
QUESTION [5 upvotes]: Suppose that $C,D$ are symmetric monoidal categories and $F:C\rightarrow D$ is a nonsymmetric monoidal functor. 

Does it imply that there exists isomorphic functor $G$ which is a symmetric monoidal functor? 
If not, then maybe additional assumption that $F$ is an equivalence should be imposed? 

In other words what nice properties the 2-functor of inclusion $\mathbf{Symm}(Cat)\rightarrow \mathbf{Mon}(Cat)$ have? 
$\mathbf{Symm}(Cat)$:= 2-category of symmetric monoidal categories and symmetric monoidal functors.
$\mathbf{Mon}(Cat)$:= 2-category of monoidal categories and monoidal functors.

REPLY [10 votes]: The answer to both questions is no, since a monoidal category can have several non-equivalent symmetric structures. For instance, the category of $\mathbb{Z} / 2\mathbb{Z}$-graded vector spaces can be given the "usual" symmetry, 
$$c(x \otimes y) = y \otimes x,$$
or the "super" symmetry,
$$c(x \otimes y) = (-1)^{\lvert x \rvert \lvert y \rvert} y \otimes x.$$
Thus, the identity functor between the category of $\mathbb{Z} / 2\mathbb{Z}$-graded vector spaces with the "usual" symmetry and the category of $\mathbb{Z} / 2\mathbb{Z}$-graded vector spaces with the "super" symmetry is a monoidal equivalence, but not symmetric monoidal.
Edit:
With regard to classifying symmetric structures on monoidal categories, I don't think anything useful can be said in complete generality, but there are some interesting results related to the theory of Tannaka duality. Here's one such statement, which is a special case of a result of Deligne:

Theorem. Let $\mathcal{C}$ be a unitary symmetric fusion category over $\mathbb{C}$. Suppose further that the associated twist $\Theta_X$ is equal to $\operatorname{Id}_X$ for all objects $X \in \mathcal{C}$. Then $\mathcal{C}$ admits a unique fiber functor which identifies $\mathcal{C}$ with the category $\operatorname{Rep} G$ for some finite group $G$.

Thus, given a unitary fusion category $\mathcal{C}$, the set of all unitary symmetric structures with trivial twist is the same as the set of groups $G$ for which $\mathcal{C} \cong \operatorname{Rep} G$ as unitary fusion categories. For a fixed $G_0$, the collection of all $G$ with $\operatorname{Rep} G \cong \operatorname{Rep} G_0$ in this way was classified by Etingof and Gelaki in their paper Isocategorical groups.
There is a generalization, originally due to Doplicher and Roberts, of Deligne's result where we remove the restriction on the twist. Let me describe the analogous result in the setting of unitary fusion categories.
We denote by $\operatorname{SVec}$ the category of finite dimensional super vector spaces, that is, the category I described in my original answer with the "super" symmetry. A finite supergroup is a group $G$ together with an element $x \in Z(G)$ with $x^2 = e$. A representation of a finite supergroup is a representation of the group $G$ on a super vector space $V$ such that $x$ acts by the identity on $V_{\operatorname{even}}$ and by $-\operatorname{Id}$ on $V_{\operatorname{odd}}$. Then we have the following result.

Theorem. Let $\mathcal{C}$ be a unitary symmetric fusion category over $\mathbb{C}$.  Then $\mathcal{C}$ admits a unique fiber functor which identifies $\mathcal{C}$ with the category $\operatorname{Rep} (G, x)$ for some finite supergroup $(G, x)$.

I don't know if anybody has extended the Etingof-Gelaki result to supergroups; I don't imagine this would be difficult.
I would welcome anybody who knows more about this than I do to expound further on the Doplicher-Roberts theorem and friends, or especially to correct anything I wrote here that is wrong!<|endoftext|>
TITLE: Symmetric structures on Vect
QUESTION [9 upvotes]: 1.Is there any classification of symmetric structures on monoidal category $Vect$(with respect to usual tensor product)? 
2.Can one organize them in suitable category?
I hope that this is nice and instructive exercise, which is appropriate for mathoverflow.

REPLY [17 votes]: There's just the one (up to symmetric monoidal equivalence). 
For this answer, I'll need a lemma: if $C$, $D$ are $Vect$-enriched categories and $F, G$ are additive $Vect$-enriched functors $C \to D$, then any natural transformation $F \to G$ is $Vect$-enriched natural. This is a consequence of the fact that the underlying-set functor $\hom(k, -): Vect \to Set$ (where $k$ is the ground field) is faithful; see Kelly's Basic Concepts of Enriched Category Theory, bottom of page 10. 
Let $\otimes_{bil}: Vect \times Vect \to Vect$ be the tensor product. (All tensor products are over the ground field.) The subscript $bil$ indicates that as it stands, the tensor product is not $Vect$-enriched, but at least it is "locally bilinear" in the sense that the structure maps 
$$\hom(V, V') \times \hom(W, W') \to hom(V \otimes W, V' \otimes W')$$ 
are bilinear. Similarly, let $\sigma_{bil}: Vect \times Vect \to Vect \times Vect$ denote the obvious transposition functor, taking a pair $(V, W)$ to $(W, V)$.  A symmetry structure on $Vect$ is, by definition, an invertible natural transformation of the form 
$$\otimes_{bil} \to \otimes_{bil} \circ \sigma_{bil}$$ 
satisfying some extra coherence properties. 
There is an obvious passage from bilinearity to linearity, where we replace $Vect \times Vect$ by $Vect \otimes Vect$ (objects of $Vect \otimes Vect$ are again pairs $(V, W)$, but $\hom((V, V'), (W, W'))$ is the tensor product $\hom(V, V') \otimes \hom(W, W')$ instead of the cartesian product). We have corresponding functors $\otimes: Vect \otimes Vect \to Vect$, $\sigma: Vect \otimes Vect \to Vect \otimes Vect$, and the invertible natural transformations above are in natural bijection with invertible natural transformations of the form 
$$\otimes \to \otimes \circ \sigma$$ 
which are, by the lemma, $Vect$-enriched. 
In short, we are trying to understand $Vect$-enriched natural transformations of the form 
$$V \otimes W \to W \otimes V$$ 
or equivalently, extranatural transformations of the form 
$$V \to \hom(W, W \otimes V) \cong \hom(\hom(k, W), W \otimes V).$$ 
By the enriched Yoneda lemma, these are in natural bijection with enriched natural transformations of the form 
$$V \to k \otimes V \cong V.$$ 
Applying the Yoneda trick a second time, these are in natural correspondence with morphisms $k \to k$, i.e., scalars. Since the transformation $V \otimes W \to W \otimes V$ is to be invertible, the corresponding scalar $\lambda \in k$ is also invertible. 
Additionally, the required coherence condition puts another constraint on the scalar: if you work through the hexagon identity, you find out that $\lambda^2 = \lambda$. Therefore $\lambda = 1$, meaning that the only symmetry structure is the standard one. 
A key to the argument is the faithfulness of $\hom(k, -): Vect \to Set$. This gives a clue that to get more interesting examples where there is a variety of possible symmetric monoidal structures, you might pass to an enriched category context where the functor $hom(I, -): V \to Set$ represented by the monoidal unit $I$ is not faithful, as was the case with $V$ the category of $\mathbb{Z}_2$-graded vector spaces, mentioned in an answer to another question of yours.<|endoftext|>
TITLE: Almost constant bump function
QUESTION [8 upvotes]: I ran into the following situation and it turned out to be more subtle than it looked.
I have a complete Riemannian manifold $M$ and my objective is to construct a sequence of functions $f:M \to [0,1]$ which have compact support, $L^2$ norm bounded away from zero, but all third derivatives converge uniformly to zero.
Trivial case: if the manifold is compact I can take the constant function $1$.
If the manifold is $\mathbb{R}$ I take a smooth bump function $f:\mathbb{R} \to [0,1]$ and then rescale it to $f_\lambda(x) = f(\lambda x)$.  The third derivatives are $f_\lambda'''(x) = \lambda^3 f'''(\lambda x)$ and for the $L^2$ norm you get:
$$|f_\lambda|^2 = \int f(\lambda x)^2 \mathrm{d} x = \frac{1}{\lambda}\int f(y)^2 \mathrm{d} y = \frac{1}{\lambda}|f|^2$$
So that if $\lambda \to 0$ I get what I want.
The same idea works for $\mathbb{R}^k$.
Can this be done on any complete Riemannian manifold $M$?
Idea:  Embed the manifold isometrically in $\mathbb{R}^k$ using Nash's theorem and use the restriction of a sequence of functions constructed for $\mathbb{R}^k$.
Problem:  Maybe the embedding has very large (second and third order) derivatives.  This can be a problem since $M$ is non-compact and the support of the sequence of functions has to grow.  Also this seems too high-tech for the problem at hand.
Any suggestions?

REPLY [13 votes]: Here is a counter-example.
Take a sequence of round 2-dimensional spheres $M_n$ of radii $r_n=n^{-1/2}$, $n=1,2,\dots$. Join them together into a long connected sum, namely connect each sphere to the next one by a tiny handle. Choose the points on the spheres where handles are attached carefully: on $M_n$, the two points (the connections with $M_{n-1}$ and $M_{n+1}$) should be not too close to each other and not too close to being diametrally opposite. (The distance half the diameter is just fine.)
Since $\sum r_n=\infty$, the resulting Riemannian manifold is complete. Since $\sum r_n^2=\infty$, its area is infinite. Now suppose that $f:M\to\mathbb R$ is a function with $|f'''|<\varepsilon$ along every geodesic. Observe that if $\gamma$ is a closed geodesic of length $\lambda$, then the difference between maximum and minimum of $f$ on $\gamma$ is at most $\varepsilon\lambda^3$. Our manifold has lots of short closed geodesics: any two points on neighboring spheres $M_n$ and $M_{n+1}$ can be connected by a chain of, say, at most 5 closed geodesics of length at most $10(r_n+r_{n+1})$. It follows that $|f(x)-f(y)|\le C \varepsilon r_n^3$ for all $x,y\in M_n\cup M_{n+1}$. Since the area is infinite, $f$ must go to 0 at infinity (to keep the $L^2$ norm bounded), therefore for $x\in M_n$ we have
$$
 |f(x)| \le \sum_{k=n}^\infty C\varepsilon  r_k^3 \sim C\varepsilon n^{-1/2} .
$$
Therefore
$$
 \int_M f^2 \le C\varepsilon^2 \sum r_n^2 n^{-1} =C\varepsilon^2 \sum n^{-2}  .
$$
The last series converges, thus $\|f\|_{L^2} \le C\varepsilon$ for some constant $C$.<|endoftext|>
TITLE: About the map $S(\mathfrak{g}^ * )^G\rightarrow S(\mathfrak{h}^ * )^H$ for $H <  G$
QUESTION [7 upvotes]: Let $G$ be a compact connected semisimple Lie group, $\mathfrak{g}$ be its complexified Lie algebra and $\mathfrak{g}^*$ its complex dual space. We can form the symmetric algebra $S(\mathfrak{g}^ * ) $ and its $G$-invariant subalgebra $S(\mathfrak{g}^ * ) ^G$. For a closed connected subgroup $H< G$ we have the same construction and get $S(\mathfrak{h}^ * )^H$.
Since $\mathfrak{h} \hookrightarrow \mathfrak{g}$ we get  $\mathfrak{g}^*\twoheadrightarrow \mathfrak{h}^ * $ hence $S(\mathfrak{g}^ * ) \twoheadrightarrow S(\mathfrak{h}^ * )$. Take invariant subalgebra we get the map 
$$
\phi: S(\mathfrak{g}^ * )^G \rightarrow S(\mathfrak{h}^ * )^H.
$$
Notice that $\phi$ is not always a projection map. For example when $H=T$ is a Cartan subalgebra, then it is well-known that $S(\mathfrak{g}^ * )^G \cong S(\mathfrak{t}^ * )^W$ is the invariant subalgebra under the Weyl group action and $S(\mathfrak{t}^ * )^T=S(\mathfrak{t}^ * )$ since $T$ is abelian. Therefore $\phi: S(\mathfrak{g}^ * )^G \rightarrow S(\mathfrak{t}^ * )^T$ is the embedding (injective but not surjective).
My question is: for which $H$, the map $\phi$ is injective? We know that $H=G$ itself or $H$ is a Cartan subgroup makes $\phi$ an injection. Are these the only cases?

REPLY [11 votes]: Here is a complete answer to this question: the map $\varphi$ is an embedding if and only if the group $H$ contains a maximal torus of $G$. I'm assuming (as in the question) that all groups are complexified but originate from a compact group. 
The map $\varphi$ is injective if and only if no nonzero $G$-invariant function on $\mathfrak g$ vanishes on $X:=G\cdot \mathfrak{h}$. This happens if and only if $X$ is Zariski dense in $\mathfrak g$ (if $X$ is not dense then $f(X)=0$ for some nonzero invariant $f\in S(\mathfrak{g}^*)^G$ by an old result of Nagata). Since the original group was compact the group $H$ is reductive (as $H(\mathbb{R})$ is anisotropic). Let $T_0$ be a maximal torus of $H$ and $\mathfrak{t}_0={\rm Lie}(T_0)$, a Cartan subalgebra of $\mathfrak{h}$. As all maximal tori in $H$ are conjugate and the semisimple elements of $\mathfrak{h}$ are dense in $\mathfrak{h}$, the set $H \cdot \mathfrak{t}_0$ is dense in $\mathfrak{h}$. So $\varphi$ is injective if and only if $G\cdot\mathfrak{t}_0=G\cdot (H\cdot\mathfrak{t}_0)={(G\cdot\mathfrak{h})}_s=X_s$ is dense in $\mathfrak{g}$. Now $T_0$ is contained in a maximal torus of of $G$, say $T$, and it follows from the theorem of the dimension of fibres of a morphism that $\dim X = 
\dim G\cdot \mathfrak{t}_0\le \dim G+\dim \mathfrak{t}_0-\dim T$. So $X$ is dense in $\mathfrak g$ if and only if 
$T_0=T$ as claimed.<|endoftext|>
TITLE: Flipping coins on a budget
QUESTION [60 upvotes]: A coin is flipped $n$ times and you win if it comes up heads at least $k$ times.  The coin is unusual in that you're allowed to pick the probability $p_i$ that it comes up heads on the $i$th flip, subject only to the constraint that $\sum_i p_i \le b$, where $b$ is some predetermined "budget" that you have.  Moreover, you are allowed to wait until you've seen the results of the first $i-1$ flips before choosing the value of $p_i$.  Given $n$, $k$, and $b$, what is your optimal strategy, and what is your probability of winning?
One colorful way to state the problem is that if you're a sports team tasked with winning a best-of-$n$ series and you have limited resources (e.g., a limited bullpen for the World Series of major league baseball), how should you budget them?

Naturally, if $b\ge k$, you can simply pick $p_i=1$ for $k$ of the $n$ flips, and win with probability 1.  So the question is interesting only if $b\lt k$.
I've circulated this problem informally among colleagues, who have obtained miscellaneous partial results but not a full solution.  It would take too much space to summarize all the partial results, but let me mention some of the highlights.

Even the "non-adaptive case," where you're not allowed to see the results of your flips before choosing $p_i$, is not trivial.  The best strategy is to divide the budget evenly over $r$ flips for some $r$, but the exact value of $r$ is more complicated than you might think.  For a given $r$, the probability of $k$ successes is $$\sum_{m=k}^r {r \choose m} \left({b\over r}\right)^m\left(1-{b\over r}\right)^{r-m}.$$ From this it appears that if $b\lt k-1$ then we should choose $r=n$, and if $k-1 \le b \lt k$ then $r\approx (k-1)/3(b-k+1)$, but we have a proof only in special cases.
In the actual stated problem, let's let $d=k-b$, the deficit.  Then, at least in the small-deficit case, the best general strategy we have so far is to make an initial coin flip with probability $1-\lbrace d\rbrace$ (where $\lbrace d\rbrace$ denotes the fractional part of $d$), and then take $p_i=1/2$ until we find ourselves in a situation where we can "clinch" the win by taking the remaining $p_i=1$.  (It's possible to analyze this strategy quantitatively but I'll omit the details here.)  In particular, one can show that adaptive strategies significantly outperform non-adaptive strategies.
If $b$ is small then one can show that the best non-adaptive strategy is within a constant factor of optimum.  For example if $b\le 1$, then one can show that the overall winning probability $p$ satisfies
$${1\over 4}{n\choose k}\left({b\over n}\right)^k \le p \le {n\choose k}\left({b\over n}\right)^k.$$
The upper bound is actually true for all $b$ and the lower bound can be derived from the best non-adaptive strategy.

REPLY [2 votes]: I give here a solution at the limite $b$, $n$, $k$ very large: The optimal strategy is to alway play $\frac{b}{n}$ until you have more budget left than the number of head to get. It will give a probability of win equal to
$$2 \mathcal{N}([\frac{k-b}{\sqrt{b(1-\frac{b}{n})}},\infty])$$
With $\mathcal{N}$ the gaussian measure.
We note $S(i)$ the number of heads up to time $i$ and $b(i)=b-\sum_{j<i}p(j)$ the budget you still have after $i$ flip. We define $X$ as $$X(i)=k-S(i)-b(i) $$
The central observation is the following : whatever the strategy you chose $X$ is a Martingale. Therefore for  $n,b,k $ large and with a correct scaling it will converge to a continuous stochastic process.
We note $t=\frac{i}{n}$ and on $[0,1]$, and we change the notation such that $p(t)=p(i)$, $b(t)=\frac{b(i)}{n}$, $X(t)=\frac{1}{\sqrt{n}}X(i)$. At the limite the system evolves as follow:
$$\begin{cases}dX_t = \sigma(p(t))dB_t \\ db_t=-p(t)dt \end{cases}$$ where $\sigma(p(t))=\sqrt{p(t)(1-p(t))}$. The process $p(t)$ is our strategy and it should be think as $p(t,b_t,X_t)$ where  $0\leq p(t)\leq 1$. With initial condition $X_0 = \frac{1}{\sqrt{n}}(k-b)$ and final condition $b(t=1)=0$.
As soon as $X_t=0$, we are in the situation $b\geq k$ one can set $p(s)=1$ and then $p(s)=0$ to finish. Therefore one have to obtimize $$ \mathbb{P}(\exists t : X_t\leq 0)$$
We have an explicit solution for $X$: $$X_t=X_0+B_{\int_0^t \sigma(p(s))^2 ds} $$
And then $$ \mathbb{P}(\exists t \in [0,1]: X_t\leq 0)= \mathbb{P}(\exists u \in [0,\int_0^1 \sigma(p(s))^2 ds]: B_u\leq -X_0)$$ We recall that $\int_0^1 \sigma(p(s))^2 ds$ is not a constant but a random variable depending on the chosen strategy. However here we remark the probability is monotone in $\int_0^1 \sigma(p(s))^2 ds$ and we can conclude : we should always chose $p$ such that $$\int_0^1 \sigma(p(s))^2 ds = \max_{p'} \int_0^1 \sigma(p'(s))^2 ds=\int_0^1 \frac{b}{n}(1-\frac{b}{n})ds= \frac{b}{n}(1-\frac{b}{n})$$ The last equalities follow from the fact that $x\rightarrow \sigma(x)^2$ is concave. And we have 
$$\max_{p'} \mathbb{P}(\exists t : X_t\leq 0)= \mathbb{P}(\min_{u\in [0,\frac{b}{n}(1-\frac{b}{n})]} B_u\leq -X_0)=2 \mathcal{N}(y \leq \frac{k-b}{\sqrt{b(1-\frac{b}{n})}})$$
With $\mathcal{N}$ the gaussian measure. (the last equality is a well known property of the maximum of the Brownian motion)<|endoftext|>
TITLE: Applications of idempotent ultrafilters
QUESTION [12 upvotes]: Recently Justin Moore has posted a solution to the amenability of Thompson's group F. A key(?) step exploits the existence of idempotent ultrafilters on $\mathbb N$ to construct an idempotent measure on the free non-associative semigroup on one-generator. 
Previously the main applications I knew of idempotent ultrafilters involved Ramsey theory, most specifically Hindman's theorem.  


Question: What are other applications of idempotent ultrafilters?


I have made this a big-list CW question, although a big list would pleasantly surprise me.

REPLY [5 votes]: There are numerous applications of ultrafilters within ergodic theory/combinatorial number theory. Vitaly Bergelson's article gives a very nice review of these. Among the results presented there are the following:

Strengthened Hindman: In a finite partition, one can find a cell that both and additive IP-set and a multiplicative IP-set.
Partition regularity of $a + b = cd $:  In a finite partition, one can find a cell that contains $a,b,c,d$ with $a + b = cd$.
Integer approximation of polynomials: If $f:\ \mathbb{R} \to \mathbb{R}$ is a polynomial with $f(0) = 0$, then for fixed $\varepsilon > 0$, the distance from $f(n)$ to a closest integer is less than $\varepsilon$ for $IP^*$-many $n \in \mathbb{Z}$.
Combinatorial richness of return times: If $f:\ \mathbb{Z} \to \mathbb{Z}$ is a polynomial with $f(0) = 0$, and $(X,T,\mu)$ is an invertible measure preserving system, then for fixed $\varepsilon > 0$, it holds that $\mu(T^{-f(n)}A \cap A) > \mu(A)^2-\varepsilon$ for $IP^*$-many $n \in \mathbb{Z}$. Under Furstenberg correspondence, this turns into the statement that if $E \subset \mathbb{N}$ has positive density $d^*(E) > 0$, then $d^*(E \cap (E − f(n))) > d^∗(E)^2 − \varepsilon$ for $IP^*$-many $n \in \mathbb{Z}$.

In fact, the last two results can be substantially extended, for instance $f$ can be a generalised polynomial, if we take care of some additional assumptions. 
A more sophisticated application can be found in another article by Bergelson and McCutcheon, where a version of Szemeredi's theorem is proved for generalised polynomials. They prove a number of amusing auxiliary results, such as characterisation of weak mixing involving ultrafilters, or some convergence results for ultrafilter convergence in weakly mixing systems. There are also nice results for multiple operators re-proved in an article by Schnell.
I read somewhere a slogan saying approximately that ultrafilters allow one to do ergodic theory without ergodic averages. (So, instead of $\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N T^n $, look at $p\!-\!\lim_{n } T^n $, where $p$ is an ultrafilter (normally: idempotent or even minimal)).
There is a MSc thesis by J.G. Zirnstein, which covers some more applications, including a variety of Ramsey theoretical results (which go far beyond Hindman, but include also van der Waerden and Jin theorems), and is written in a very nice and accessible tone.<|endoftext|>
TITLE: Nonisomorphic number fields with the same Galois group and discriminant
QUESTION [5 upvotes]: It is possible for nonisomorphic fields to have the same discriminant, and even to be arithmetically equivalent, meaning with the same Dedekind zeta function. Previous questions which discussed this are "Number fields with same discriminant and regulator?" and "Are there two non-isomorphic number fields with the same degree, class number and discriminant?". The usual way this comes about involves fields with isomorphic subfields. I would find it useful if the discriminant sufficed to characterize number fields whose splitting field had a simple Galois group, so that there were no subfields. So the question is, can there be two nonisomorphic fields with the same degree, the same discriminant and the same simple Galois group for the splitting field?

REPLY [2 votes]: I just came upon this (old?) thread.  Here is a handy way to produce lots of non-Abelian, non-isomorphic extensions of Q with the same field discriminant:
Let K3/Q be a non-Abelian (resp Abelian) cubic field.  Let h3 be the size of the elementary 2-class group of K3.  Then Heilbronn (On the 2-classgroup of cubic fields, in: Studies in Pure Math}., 1971) showed that there is a (h3-1)  (resp (h3-1)/3) quadruplets of conjugate quartic fields K4 such that disc(K4)=disc(K3) and that the Galois closure of K4 contains K3.
So if you pick a K3 with large elementary 2-rank, then you will get lots of non-isomorphic K4 with Galois group S4 or A4, depending on whether K3 is non-Abelian or otherwise.
There is also a similar (and old) theorem of Hasse for 3-rank of quadratic fields.<|endoftext|>
TITLE: Unbounded complexes, resolutions and computation of derived functors
QUESTION [5 upvotes]: Hey guys, let $F: \mathcal{A} \rightarrow \mathcal{B}$ be a left exact functor between abelian categories with enough injectives, let $K \in Kom(\mathcal{A})$ be an unbounded complex, I've heard that you can construct resolutions and define a derived functor $RF$ for $K$ but I was wondering:
1 - Are the resolutions similar to Cartan-Eilenberg resolutions for bounded complexes, where you find an injective (projective) resolution for each member $K^i$ of $K$ and check it satisfies certain properties?
2 - Can you define filtrations, and a spectral sequence to compute the derived functors?
Any good references on this?

REPLY [3 votes]: All references below (unless otherwise stated) refer to Weibel (We): An Introduction to Hom. Algebra. Futhermore I consider projective resolutions and assume $\mathcal{A}$ has enough projectives and $F$ is right exact since this case is treated in Weibel. The case of injective resolutions can be easily adapted by switching to the opposite category [We, 2.3.4]. 
Let $Ch(\mathcal{A})$ be the category of (unbounded) chain complexes in $\mathcal{A}$. Since $\mathcal{A}$ is abelian, $Ch(\mathcal{A})$ is abelian as well [We, Th. 1.2.3] and has enough projectives [We, 2.2.2]. The functor $F: \mathcal{A} \to \mathcal{B}$ induces a functor 
$Ch(F): Ch(\mathcal{A}) \to Ch(\mathcal{B})$. A morphism $h: C \to D$ of chain complees in $\mathcal{A}$ is epi, iff each $h_i:C_i \to D_i$ is epi [We, Proof of 1.2.3]. Hence, the right exactness of $F$ implies that $Ch(F)$ is also right exact. 
In summary, we have shown: $Ch(F)$ is a right exact functor between abelian categories and $Ch(\mathcal{A})$ has enough projectives. Consequently, $Ch(F)$ has a left derived functor and everything what can be done for $L_\ast F$ (i.e. filtations, spectral sequences, etc.) can also be done for $L_\ast Ch(F)$. 
A discussion of projective resolutions in $Ch(\mathcal{A})$ can be found in my answer to this question: 
On the difference between a projective chain complex and a level-wise projective chain complex
Similar, a chain complex $I$ in $\mathcal{A}$ (which is now supposed to have enough injectives) is an injective object in $Ch(\mathcal{A})$, iff $I$ consists of injective objects $I_i\in \mathcal{A}$ such that all short sequences 
$$ 0 \to \ker(d_i) \to I_i \to \operatorname{im}(d_i) \to 0$$
are exact and do split.<|endoftext|>
TITLE: Matroid representable over $\mathbb{R}$ but not over $\mathbb{Q}$?
QUESTION [6 upvotes]: Does there exist a matroid that is representable over $\mathbb{R}$ but not over $\mathbb{Q}$?
In particular, can one give a positive answer using a nonrational polytope, i.e., a combinatorial polytope that cannot be realized as the convex hull of rational vertices?  (Such things do exist; see, e.g., p.94 of Grünbaum's Convex Polytopes.)  The vertex sets of the faces of a convex polytope certainly form the flats of a matroid, but it's not clear to me why the same matroid could not be realized by affine dependences of a set of points not in convex position.

REPLY [10 votes]: Jeremy, on the very same page 94 you will find a "point and line configuration" called Perles configuration which when viewed as set ov vectors in $\Bbb R^3$ is a matroid that is realizable over $\Bbb Q[\sqrt{5}]$ but not over $\Bbb Q$.  In my book I even prove it (Ex 12.3) - sorry to make a plug, this is the only place with a proof I know.<|endoftext|>
TITLE: Knot security (When to trust your life with a knot)
QUESTION [37 upvotes]: This question is related to a a question about self-tightening knots.  
I am supervising a senior thesis and my student is interested in knots.  My student is also a rock climber and has an interesting idea for a project, but I am not sure how reasonable it is.  She wants to look at the mathematics of knot security.  As a climber, she may tie a single rope to a closed loop or tie two pieces of rope together.  There are some knots you should trust your life with and others you should avoid while hanging from a cliff.  She wants to understand what properties of a knot make it secure.  I like the idea, but I have no idea what kind of model we would use.  The only articles I could find were about knot security for surgeons tying sutures and were based on experimental evidence only.
Does anyone know any mathematical references for studying physical knots and knot security?

REPLY [5 votes]: Louis Kauffman's masterful book "Knots and Physics" has some thoughts on it. Especially the introductory chapter, and then later a small chapter called "The Theory of Hitches". Worth a reading IMO..<|endoftext|>
TITLE: Upper bounds on eigenvalues of PSD matrix?
QUESTION [7 upvotes]: Suppose A is a symmetric positive semidefinite matrix.  Is there a way to upper bound the largest eigenvalue using properties of its row sums or column sums?
For instance, the Perron–Frobenius theorem states that, for positive matrices, the largest eigenvalue can be upper bounded by the largest row sum.  
I'm hoping to find an upper bound that states something like the largest eigenvalue is upper bounded by the largest sum of absolute values of a row.
An example matrix is 
 3    -2     1     0     0     0
-2     3    -2     1     0     0
 1    -2     3    -2     1     0
 0     1    -2     3    -2     1
 0     0     1    -2     3    -2
 0     0     0     1    -2     3

The above matrix has a maximum absolute row sum of 9 and a maximum eigenvalue of about 7.98.
Thanks!

REPLY [8 votes]: Also the fact that $\rho(A) \leq |\!|\!|A|\!|\!|$ can be helpful here (i.e. the maximum modulus of an eigenvalue is bounded by any matrix norm).<|endoftext|>
TITLE: Does every nonempty definable finite set have a definable member?
QUESTION [10 upvotes]: I asked this on MSE yesterday ( https://math.stackexchange.com/q/197873/39378 ) but no one has answered it yet.  I hope it's not too soon to post it here.
Here are a few ways to formalize the question, so you can pick your favorite and answer it. Assume whatever large cardinals you like.
(1) Is it consistent with ZFC that there is an inaccessible cardinal $\delta$ and a nonempty finite set that is first-order definable without parameters over $(V_\delta,\in)$ but has no elements that are first-order definable without parameters over $(V_\delta,\in)$?
(2) Is there any model of ZFC that has a finite nonempty set, first-order definable without parameters over the model, with no element that is first-order definable without parameters over the model?
(3) Is it consistent with ZFC that there is an ordinal-definable finite nonempty set with no ordinal-definable member? (I am aware of the question A question about ordinal definable real numbers, but that question asks about sets of real numbers and I already know the answer to my question for sets of real numbers, or indeed for sets of subsets of any ordinal, because they are definably linearly ordered.)
(4) Any of the above formulations with ZFC replaced by ZF.

REPLY [9 votes]: I believe the answers to these questions are all positive. This kind of problem was discussed by Groszek and Laver in Finite groups of OD-conjugates [Period. Math. Hungar. 18 (1987), 87-97, MR0895774]. Answering a question of Mycielski, they show that there can be two sets of reals $x,y$ such that $\lbrace x,y\rbrace$ is ordinal definable but neither $x$ nor $y$ is ordinal definable. They also prove a lot of other interesting things about OD conjugates.
Here is the brief argument from the intro to that paper. Suppose $u, v$ are two mutually Sacks generic reals over $L$. Both $u$ and $v$ have minimal degree over $L$. Let $x$ and $y$ be the $L$-degrees of $u$ and $v$ respectively. Then $x$ and $y$ satisfy the same formulas with ordinal parameters because Sacks forcing is homogeneous. However, $\lbrace x, y \rbrace$ is definable (without parameters) since these are the only two minimal $L$-degrees in $L[u,v]$.<|endoftext|>
TITLE: Families of ideal sheaves: What's the correct definition?
QUESTION [10 upvotes]: I'm looking at Bridgeland's paper "Flops and Derived categories" and I got confused on what he meant by a family of ideal sheaves.
Let $Y$ be a scheme, and let $S$ be another scheme. A family of sheaves $\mathcal E_Y$ on $Y$ over $S$ is a sheaf on $S\times Y$, flat over $S$. Two such families are equivalent if they differ by tensoring by pullback of a line bundle on $S$.
When we talk about ideal sheaves $\mathcal{I}$ of $Y$, usually an inclusion to the structure sheaf $\mathcal{O}_Y$ is taken as part of the data. Now when we talk about a family of ideal sheaves, what do we really mean by that hidden part of the data? At least a family of sheaves which are flat over $S$, but there should be more. More specifically, when we say "let $\mathcal E_Y$ be a family of ideal sheaves on $Y$ over $S$", is there an inclusion of $\mathcal E_Y$ into $\mathcal O_{S\times Y}$ given part of the data?
(EDIT: according to MartinG's answer, this attempt of definition doesn't seem to be right, some functoriality is missing. Now let me replace my question by the following:


What is the functor $M_I(X)$? What's the definition and why is it functorial? (I'm guessing MartinG's suggestion is right, namely rank 1 sheaves with trivial determinant line bundle, but I'm not 100% sure.)

Why does $M_I(X)$ exist as a scheme under that definition? (In the paper he seems need this fact in a crucial way.)

What's the relation between $M_I(X)$ and $\text{Hilb}(X)$?



End of EDIT)

REPLY [2 votes]: Bridgeland uses another point of view on moduli spaces. His definition of a moduli functor (Def. 3.7) does not uses any assumption of flatness. Instead, he considers DERIVED restriction of the family object to fibers, and the condition which replaces flatness is that these derived restrictions are PURE in the t-structure he considers. If the t-structure is the usual one then this definition is equivalent to the standard definition (flatness = no higher derived pullbacks). 
The moduli space $M_I(X)$ of ideal sheaves is defined by the functor which associates to any scheme $S$ the set of (equivalence classes of) all sheaves on $S\times X$ which after (derived) restriction to fibers over $S$ are ABSTRACTLY isomorphic to sheaves of ideals. On a contrary, the Hilbert scheme functor associates to a scheme $S$ the set of sheaves on $S\times X$ with a FIXED EMBEDDING into $O_{S\times X}$ (again up to an appropriate embedding). Forgetting the embedding we obtain a morphism $Hilb(X) \to M_I(X)$. 
Of course, if $S$ is a closed point, and $I$ is isomorphic to an ideal sheaf, then there is a unique (up to a constant) embedding of $I$ into $O_X$. This shows that the map above is
a bijection on closed points. On the other hand, globally it is not clear why a family of sheaves isomorphic to ideal sheaves has an embedding into $O_{S\times X}$. I think that Bridgeland just does not want to deal with this subtle question, which has nothing to do with his goal, so he just ignores it.<|endoftext|>
TITLE: Manifold whose universal covering is a sphere but which is not a space form? 
QUESTION [12 upvotes]: Let $M^n$ be a smooth manifold whose universal cover is homeomorphic $\mathbb{S}^n$, are there examples where $M^n$ is not homeomorphic to a space form ? 
The answer may vary if you replace homeomorphic by diffeomorphic, and I'm also interested in this question under this restriction.
I came to this question while reading surveys about sphere theorems, where the non simply-connected case is harder to get (while you already know that the universal cover is a sphere, you need more work to show it is actually a space-form).

REPLY [8 votes]: The geometrization theorem implies that 3-dimensional manifolds covered by $S^3$ are diffeomorphic to space forms. 
There are examples of Cappell-Shaneson and Fintushel-Stern
of fake $\mathbb{RP}^4$s (in the Fintushel-Stern case, they actually
show there is a smooth exotic free involution of $S^4$).<|endoftext|>
TITLE: Eisenstein series and 163?
QUESTION [8 upvotes]: Given $q = e^{2\pi i \tau}$ and the Eisenstein series $E_{2k}(\tau)$, i.e.,
$$E_2(\tau) = 1-24\sum_{n=1}^\infty \frac{n q^n}{1-q^n}$$
$$E_4(\tau) = 1+240\sum_{n=1}^\infty \frac{n^3 q^n}{1-q^n}$$
and so on. Define the function,
$$F_{2k}(\tau) = \frac{E_{2k}(\tau)}{\left(E_2(\tau)-\frac{3}{\pi\; \Im(\tau)}\right)^k}$$
for $k \geq 2$, where $\tau = \frac{1+\sqrt{-d}}{2}$, $\Im(\tau)$ is the imaginary part of $\tau$, and $d$ has class number $h(-d) = m$. For example, we have,
$$F_4\left(\tfrac{1+\sqrt{-163}}{2}\right) = \frac{5\cdot23\cdot29\cdot163}{2^2\cdot3\cdot181^2}$$
$$F_6\left(\tfrac{1+\sqrt{-163}}{2}\right) = \frac{7\cdot11\cdot19\cdot127\cdot163^2}{2^9\cdot181^3}$$
$$F_8(\tau) = F_4^2(\tau)$$
and so on.
Question: In general, is it true that for $k \geq 2$ the function $F_{2k}$, like the j-function, is an algebraic number of degree m = h(-d)?  (I've tested it with d with higher class numbers, and it seems to be true.)

REPLY [3 votes]: I inadvertently came across the partial answer to my own question. It turns out Ramanujan had already explored something similar. Let $q = e^{2\pi i \tau}$, $\tau=\tfrac{1+\sqrt{-n}}{2}$, and,
$$P_n = 1-24\sum_{k=1}^\infty \frac{k q^k}{1-q^k}$$
$$Q_n = 1+240\sum_{k=1}^\infty \frac{k^3 q^k}{1-q^k}$$
$$R_n = 1-504\sum_{k=1}^\infty \frac{k^5 q^k}{1-q^k}$$
$$a_n = \frac{b_n}{6} \left(1-\frac{Q_n}{R_n}\left(P_n-\frac{6}{\pi\sqrt{n}}\right)\right)$$
$$b_n = \sqrt{n(1728-j(\tau))}$$
and j-function $j(\tau)$, then,
$$\frac{1}{F_4(\tau)} = \frac{1}{Q_n}\left(P_n-\frac{6}{\pi\sqrt{n}}\right)^2 = \left(1-\frac{6\,a_n}{b_n}\right)^2\left(\frac{-1728+j(\tau)}{j(\tau)}\right)$$
which is Theorem 7.1 (p.13) of Berndt's "Ramanujan's contributions to Eisenstein Series". 
Thus, it remains to establish that the square factor of the RHS is an algebraic number with the same degree as $j(\tau)$. (It seems it suffices to prove it for $a_n$.)
P.S. Note that for $n>3$,
$$\sum_{k=0}^\infty (-1)^k\frac{(6k)!}{(3k)!(k!)^3}\frac{a_n+b_n k}{(-j(\tau))^{k+1/2}} = \frac{1}{\pi}$$
in particular, for $n=163$, we have the Chudnovsky's formula,
$$\sum_{k=0}^\infty (-1)^k\frac{(6k)!}{(3k)!(k!)^3}\frac{163096908+6541681608 k}{(640320^3)^{k+1/2}} = \frac{1}{\pi}$$<|endoftext|>
TITLE: $J$-holomorphic curve as a minimal surface
QUESTION [8 upvotes]: The following is a part of the proof of Gromov nonsqueezing theorem.
The existence of a $J$-holomorphic curve gives an upper bound for the radius of a symplectically embedded ball.
Let $\psi: B(r) \rightarrow M$ be a symplectic embedding of a ball (of dimension equal to dim $M$) of radius $r$ centered at the origin into a symplectic manifold $(M, \omega)$.
An $\omega$-compatible almost complex structure $J$ on $M$ is chosen so that, on the image of $\psi$, it coincides with the pushforward of the standard complex structure by $\psi$.
Let $u: S^2 \rightarrow M$ be a $J$-holomorphic sphere passing through the point $\psi(0)$.
Then the preimage $C$ of this holomorphic sphere by $\psi$ is a minimal surface in $B(r)$ with boundary in $\partial B(r)$.
By the monotonicity formula, the area of $C$ is at least $\pi r^2$, which is the area of the flat $2$-disk.
This gives an upper bound on $r$.
Now I have a question. How do we consider $C$ as a minimal surface in $B(r)$? A minimal surface is a surface with mean curvature zero, and the mean curvature is defined on the image of an immersion. But a $J$-holomorphic curve is not an immersion in general. How do we deal with critical points?
One possible option seems to be taking critial points off. But then the immersion is not proper and the proof of the monotonicity formula seems to use properness. For example, the proof in the book "holomorphic curves in symplectic geometry" uses a compactly supported vector field.
I am not familiar with minimal surfaces, so I must be missing something. Any comment is appreciated.

REPLY [2 votes]: I think what you are missing is that any region of a J-holomorphic curve in a symplectic manifold $(M,\omega)$ is always absolutely area minimizing in its homotopy class relative the boundary of the region, with respect to the compatible metric $g_J$. In particular if it is a closed curve it is minimizing in its homology class.
This is explained for example in McDuff-Salomon's "J-Holomorphic curves in symplectic topology", in one of the first sections. But is actually a fairly simple calculation. 
I don't think we need to assume immersed for this, the singularities will be isolated  so 
that the argument goes through.
This basic phenomenon is generalized by calibration geometry of Lawson in higher dimensions. http://en.wikipedia.org/wiki/Calibrated_geometry<|endoftext|>
TITLE: minimum of two probability densities
QUESTION [5 upvotes]: Consider a smooth probability density $\pi(x)$ on $\mathbb{R}^d$. I am looking for natural for the integral $\iint_{u,v} \ \min\big(\pi(u), \pi(v) \big) \ du \ dv$ to be finite. If $\pi$ is a radially decreasing density, this is equivalent to the condition $\mathbb{E}\big[ \|X\|^{d} \big] < \infty$. Are there smooth densities verifying this moment condition such that $\iint_{u,v} \ \min\big(\pi(u), \pi(v) \big) \ du \ dv = \infty$ ?
No answer given on math.stackexchange. This integral appeared while studying a Metropolis-Hastings Markov chain.

REPLY [5 votes]: If $\mathbb{E}\left[\lVert[ X\rVert^d\right]$ is finite then the integral in the question is necessarily finite. As mentioned, this holds whenever $\pi$ is radially decreasing. However, in the general case, you can swap regions with equal volume in $\mathbb{R}^d$ about in order to move the large probability regions closer to the origin. Doing this has no effect on the integral in question, but can only decrease $\mathbb{E}\left[\lVert[ X\rVert^d\right]$. So, it reduces to the radially decreasing case.
It isn't hard to make this idea more rigorous. If $\pi(x)$ is 'radially decreasing', so that it is a decreasing function of $\lVert x\rVert$, then
$$
\begin{align}
\iint \min\left(\pi(x),\pi(y)\right)\,dxdy&=2\iint_{\lVert y\rVert\le\lVert x\rVert}\pi(x)\,dxdy\cr
&=2K\iint\lVert x\rVert^d\pi(x)\,dx=2K\mathbb{E}\left[\lVert X\rVert^d\right]
\end{align}
$$
where $K$ is the volume of the unit ball in $\mathbb{R}^d$.
In the general case, we have
$$
\mathbb{E}\left[\lVert X\rVert^d\right]=\int_0^\infty\int_{\pi(x)\ge p}\lVert x\rVert^d\,dxdp.
$$
Letting $S_p=\lbrace x\colon\pi(x)\ge p\rbrace$ then, for a given volume $V_p$ for $S_p$, the integral $\int_{S_p}\lVert x\rVert^ddx$ is minimized when $S_p$ is a ball about the origin. In particular, if we define a radially decreasing function $\tilde\pi\colon\mathbb{R}^d\to\mathbb{R}$ by
$$
\tilde\pi(x)=\sup\lbrace p\in\mathbb{R}\colon K\lVert x\rVert^d\le V_p\rbrace
$$
then $\lbrace x\colon \tilde\pi(x)\ge p\rbrace=\lbrace x\colon K\lVert x\rVert^d\le V_p\rbrace$ has volume $V_p$. So,
$
\mathbb{E}\_{\tilde\pi}\left[\lVert X\rVert^d\right]\le\mathbb{E}\_\pi\left[\lVert X\rVert^d\right].
$
Also, the integral $\iint\min(\pi(x),\pi(y))\,dxdy$ is unchanged by passing to $\tilde\pi$. This follows from the fact that $\tilde\pi=\pi\circ f$ (almost everywhere) for some (Lebesgue) measure preserving Borel isomorphism of $\mathbb{R}^d$. Alternatively, the equality can be seen by showing that the integral only depends on $V_p$,
$$
\begin{align}
\iint \min\left(\pi(x),\pi(y)\right)\,dxdy
&=2\int\pi(x)V_{\pi(x)}\,dx\cr
&=-2\int pV_p\,dV_p=\int_0^\infty V_p^2\,dp.
\end{align}
$$
So, better than just finiteness of the integral, we have the inequality
$$
\begin{align}
\iint \min\left(\pi(x),\pi(y)\right)\,dxdy&=\iint \min\left(\tilde\pi(x),\tilde\pi(y)\right)\,dxdy\cr
&=2K\mathbb{E}\_{\tilde\pi}\left[\lVert X\rVert^d\right]\cr
&\le2K\mathbb{E}\_{\pi}\left[\lVert X\rVert^d\right],
\end{align}
$$
which is an equality whenever $\pi$ is radially decreasing.<|endoftext|>
TITLE: Maximal tamely ramified extension of $\mathbf Q_p$
QUESTION [9 upvotes]: Is there an explicit description of the maximal tamely ramified extension of $\mathbf Q_p$?

REPLY [14 votes]: Yes, there is. The maximal unramified extension is obtained by adding all roots of unity of order prime to $p$. The maximal tame extension is obtained by adding on top of that all $n$-th roots of $p$, for $n$ prime to $p$; so its Galois group is isomorphic to $\prod_{\ell \ne p} \mathbf{Z}_\ell$, and conjugation by $\operatorname{Gal}(\overline{\mathbf{Q}}_p^{nr} / \mathbf{Q}_p)\cong \widehat{\mathbf{Z}}$ acts on each $\mathbf{Z}_\ell$ factor via the $\ell$-adic cyclotomic character.
(EDIT: For a reference for this see Pete Clark's notes at http://math.uga.edu/~pete/8410Chapter4.pdf, in particular Theorem 11 and Corollary 12.)<|endoftext|>
TITLE: The discriminant for the plane cubic curve
QUESTION [5 upvotes]: 10 coefficients determine a degree 3 homogenous polynomial in $k[x,y,z]$. I understand that there is a degree 12 polynomial in these coefficients, called the discriminant, with 2040 terms, which vanishes precisely when the curve is singular. I'm working on some generalizations to tropical geometry, and I  would like to understand the classical case. Could someone point me to a reference where that shows how to construct this and check that it has the desired properties? I know that Gelfand, Kapranov, and Zelevinski have an important book about discriminants, but glancing through the table of contents it looks like it has a lot of prerequisits and may not be as concrete as I would like. Google searching has been fruitless so far.

REPLY [5 votes]: Go to http://www.ma.utexas.edu/cnt/cnt-frames.html and click on "jac_cubic" on the right hand panel. It gives a link to download a pari-gp script that computes the Weierstrass equation of the jacobian of a general cubic and, from that, the discriminant, if you want. Artin, Tate and  Villegas have a paper explaining the theory too. It's on Villegas's webpage too.<|endoftext|>
TITLE: Crepant resolutions of cDV singularities?
QUESTION [6 upvotes]: Compound Du Val 3-fold singularities form a good class of singularities in 3-fold singularity theory. I would like to know which singularities admit crepant resolutions. If I remember correctly, $cA_{n}$ admits a crepant resolution. What about $cD_{n}$- and $cE_{n}$-singularities? 
I would really appreciate it if you could give me a reference or explain what is known about crepant resolution of cDV singularities.
Thank you for your help.

REPLY [9 votes]: Background. The threefold compound du Val singularities have been introduced by Miles Reid in the 1980s [R1, R2, R3]. Their geometric description is that a general hyperplane section through the singular point is a du Val singularity. It has been proved by Miles Reid that isolated compound du Val singularities are precisely Gorenstein terminal singularities, and by definition of terminal, every crepant resolution is the same as small. Curiously, a Gorenstein threefold with a small resolution must have compound du Val singularities [R2, (1.2)]. On the other hand, compound du Val singularities always admit so-called $\mathbf{Q}$-factorializations [K, 4.5], which are more natural and less restrictive than small resolutions.
Compound $A_n$. I will consider the case of isolated $cA_n$ singularities, which are the simplest kind. They are characterized by the property that the completion at the singular point is isomorphic to $$k[[x,y,z,w]] / (xy + f(z,w))$$. The condition of being an isolated singularity is equivalent to $f(z,w)$ being reduced, and $n$ in the $cA_n$ stands for the lowest degree of monomial in $f$ minus one. Notation-wise, in the singularities and commutative algebra communities, the emphasis is on $A_n$ singularities $xy + z^2 + w^{n+1}$, which are among $cA_1$ in this definition.
Setting. Let $k$ be algebraically closed field of characteristic zero; in this case $xy$ after changing the basis is same as $x^2 + y^2$ (but $xy$ is the correct expression over non-closed fields too). We work with the local model $X = \mathrm{Spec}(k[x,y,z,w] / (xy + f(z,w))$. Existence of small resolutions is controlled by the existence of sufficiently many Weil divisors which are not Cartier (cf [K, Proof of 4.5]). Note that $\mathrm{Pic}(X) = 0$, because $k[X]$ is a normal graded ring. Thus to say that $X$ is factorial is the same as to say $\mathrm{Cl}(X) = 0$.
Claim 1. A factorial variety does not admit small resolutions, see this answer.
Claim 2. $\mathrm{Cl}(X) \simeq \mathbf{Z}^{r-1}$, where $r$ is the number of irreducible factors of $f(z,w)$. I think it should not be difficult to prove this directly, but I don't know the reference; for a proof using derived categories and Knorrer periodicity, see Lemma 2.22 in this paper. See below for the explicit Weil divisors on $X$.
Putting Claim 1 and Claim 2 together we obtain:
Corollary. If $f(z,w)$ is irreducible and non-smooth (e.g. $f(z,w) = z^2 + w^{2k+1}$), then $X$ admits no small resolutions (cf [R2, Cor. 1.16]).
Now the general result:
Proposition. $X$ admits a small resolution if and only if $f(z,w)$ is a product of distinct smooth curve germs.
Inductive construction for the "if" direction. Assume $f(z,w) = g(z,w) h(z,w)$, where $g(z,w)$ is a smooth curve germ, and $h(z,w)$ is coprime to $g(z,w)$. Consider the Weil divisor $D$ given by $x = g(z,w) = 0$. Let $\pi: \widetilde{X} \to X$ be the blow up of $D$. Writing the local charts of the blow up, we see that $\pi$ contracts a $\mathbf{P}^1$ to a singular point, and $\widetilde{X}$ has a singularity $xy + h(z,w) = 0$.
Obstruction to the "only if" direction. I think it follows from the fact that all $\mathbf{Q}$-factorializations are related by flops (Kawamata), and in particular existence of a small resolution implies that any $\mathbf{Q}$-factorialization is smooth. The inductive construction above will produce $\mathbf{Q}$-factorializations with singularities $xy + h(z,w) = 0$, with $h(z,w)$ irreducible nonsmooth. (One should come up with a more conceptual and direct proof which avoids using $\mathbf{Q}$-factorializations.)
Local vs global. One needs to be aware, that we have only constructed small resolutions in a local model, hence formally, or complex analytically. Indeed, many projective threefolds with $A_1$ singularities do not admit projective small resolutions, and this is also controlled by Weil divisors. For example $1$-nodal cubic hypersurfaces in $\mathbf{P}^4$ are factorial, hence by the Claim 1 do not admit projective small resolutions (cf [K, page 104]).
Classical references.
[R1] Miles Reid, "Canonical threefolds'', 1980
[R2] Miles Reid, "Minimal models of canonical threefolds'', 1983
[R3] Miles Reid, "Young person's guide to canonical singularities'', 1987
[K] Yujiro Kawamata, "Crepant Blowing-Up of 3-Dimensional Canonical Singularities and Its Application to Degenerations of Surfaces", 1988
Fun quote. I do not wish to go at present into the various interesting questions concerned with resolving the cDV points; for many purposes it seems natural to leave them alone! [R1] :-)<|endoftext|>
TITLE: Reference request: a conjecture of Rota on positive functions of a random variable
QUESTION [17 upvotes]: Rota and Shen's On the Combinatorics of Cumulants ends with a conjecture which I'll restate as follows: 

Let $p \in \mathbb{R}[x_1, x_2, ...]$ be a polynomial such that, for any sequence $X_1, X_2, ...$ of i.i.d. random variables on the real line all of whose moments are finite, we have $\mathbb{E}(p(X_1, X_2, ...)) \ge 0$. Then there exists $q \in \mathbb{R}[x_1, x_2, ...]$ such that $\mathbb{E}(p(X_1, X_2, ...)) = \mathbb{E}(q(X_1, X_2, ...))$ for all possible $X_i$ and such that $q$ is a sum of squares.

The idea is that $\mathbb{E}(p)$ is a polynomial function of the moments of $X_1$. The motivating example is when $\mathbb{E}(p)$ is a Hankel determinant $h_n$, which is a positive scalar multiple of
$$\mathbb{E} \left[ \prod_{1 \le i < j \le n} (X_i - X_j)^2 \right].$$
What work has been done on this conjecture? Looking at the papers citing this one didn't turn up anything promising, and neither did various Google searches. 
Edit: There is a familial resemblance to Hilbert's 17th problem, but it's probably worth noting that a counterexample to the strong version of Hilbert's 17th problem is not automatically a counterexample to this problem. For example, it's known that
$$p(x, y, z) = x^6 + y^4 z^2 + y^2 z^4 - 3x^2 y^2 z^2$$
is everywhere non-negative (by AM-GM) so in particular satisfies the hypotheses of the problem, but is not a sum of squares of polynomials. However, $\mathbb{E}(p) = \mathbb{E}(q)$ where
$$q(x, y, z) = (x^3 - xy^2)^2 + \frac{3}{2} (x(y^2 - z^2))^2.$$
Edit #2: I'm now a little concerned that I'm misstating the conjecture because there appears to be a very small counterexample. Taking $p = x_1 x_2$ we have $\mathbb{E}(p) = m_1^2$ (where $m_1 = \mathbb{E}(X_1)$), which is non-negative. However, no sum of squares $q$ exists such that $\mathbb{E}(q) = m_1^2$: any such $q$ must be a sum of squares of linear polynomials, and in that case $\mathbb{E}(q)$ must contain a term $m_2 = \mathbb{E}(X_1^2)$ with positive coefficient. Perhaps one should allow in addition sums of squares of polynomials in the moments...

REPLY [5 votes]: I think your reformulation of the conjecture in the Rota-Shen paper is correct.
Also your counterexample is correct. So I guess the conjecture was not stated properly
in that article. Perhaps one should restrict to translation invariant polynomials only.
By that I mean polynomials in the moments which remain invariant if we change
the basic random variable $X$ to $X+c$ where $c$ is some constant, the first example being
the variance $m_2-m_1^2$. I think one can make the conjecture more precise by
asking the polynomial to be a positive linear combination of evaluations
of squares of products of differences of umbral letters.
It is hard to know what Rota's motivation was. It is not clear to me if for him this
conjecture was a matter of probability theory or of (real) classical invariant theory. The Hankel determinants which determine moment sequences and motivate this conjecture are essentially the catalecticants of binary forms and their umbral expression with squared Vandermondes is the standard representation of these catalecticants using the classical symbolic method.
My feeling is Rota's motivation might be to understand a big problem in classical invariant theory which is how does one know that the evaluation of an umbral polynomial is nonzero (see this very interesting article by Alexandersson and Shapiro).
Also for the proper formulation of the conjecture, I suggest contacting J. Shen.
Her latest article, "Least-squares halftoning via human vision system and Markov gradient descent (LS-MGD): algorithm and analysis", SIAM Rev. 51 (2009), no. 3, 567–589,
has a contact email address for her.
Another person who works in this area and might know what exactly Rota's conjecture was
is Elvira Di Nardo who gave lectures at the SLC 67 on cumulants and umbral calculus.<|endoftext|>
TITLE: Fusion category and Hopf algebra
QUESTION [7 upvotes]: Let $H$ be a semisimple Hopf algebra over an algebraically closed field of characteristic zero. Further, let $K\subseteq H$ be a normal Hopf subalgebra. As we all know, $H$ then can be reconstructed from $K$ by some compatible data [1]. 
I want to know if there exists a similar result on Rep($H$) and Rep($K$), where Rep($H$) is the fusion category of finite-dimensional representations of $H$? If not, what can be said about Rep($H$) and Rep($K$)?
Thank you!
[1]N. Andruskiewitsch, Notes on extensions of Hopf algebras, Canad. J. Math. 48 (1996), 3-42

REPLY [3 votes]: The following article may interest you: C. Pinzari and J. Roberts, A Duality Theorem for Ergodic Actions of Compact Quantum Groups on C *-Algebras, Communications in Mathematical Physics 277 (2008) no 2, 385-421.
However, this article is written in the language of C*-tensor categories and compact quantum groups.
I hope this helps...<|endoftext|>
TITLE: Is $Q_n(x)=\sigma_{n+1}(x)/\sigma_n(x)$ logarithmically convex on $\mathbf{R}$?
QUESTION [10 upvotes]: In 1975 J. van de Lune considered the monotony properties of the canonical Riemann Upper and Lower sums for $\int_0^1 t^xdt$, with  $x>0$.
Writing $\sigma_n(x) := 1^x+2^x+\cdots+n^x$ these sums are
$$U_n:=U_n(x)=\sigma_n(x)/n^{x+1} \quad \text{and}\quad L_n:= L_n(x)=\sigma_{n-1}(x)/n^{x+1}.$$
He proved (by mathematical induction) that $U_n>U_{n+1}$ and $L_n<L_{n+1}$.  
A later proof of $U_n>U_{n+1}$ was obtained by showing  that the function
$$h(x):= h_n(x)=(\sigma_{n+1}/\sigma_n(x))(n/(n+1))^x$$
is strictly decreasing on all of $\mathbf{R}$.
Soon afterwards he came  to realize that the monotonicity of $h(x)$ would be a consequence of the logarithmic convexity of $Q(x):=Q_n(x)=\sigma_{n+1}(x)/\sigma_n(x)$ on all of $\mathbf{R}$.
Various numerical tests on $Q(x)$ were performed, but no proof was found.
Another application: From the logarithmic convexity of $Q(x)$ one may also obtain a simple proof of a conjecture made by H. Alzer and A. A. Jagers:
$$f(x):=f_n(x)=\Bigl(\frac{\frac{1}{n+1}\sigma_{n+1}(x)}{\frac1n \sigma_n(x)}\Bigr)^{1/x}$$
is strictly increasing for $x>0$.

Question: Is for every (fixed) integer $n\ge2$, the function
  $$x\in\mathbf{R}\mapsto Q_n(x):=(1^x+2^x+\cdots+n^x+(n+1)^x)/(1^x+2^x+\cdots+n^x)$$
  logarithmically convex on all of $\mathbf{R}$?

REPLY [3 votes]: OK, I'll post my argument then. It will be a somewhat long story so I'll do it in a few stretches (I don't think I'll have a single continuous time interval to make the full post, so I apologize for bumping it to the front page a few times when editing to add the stuff).
First of all, let us compute the second derivative of the logarithm of $\sigma_n(x)$. It is
$$
\frac{\sigma_n''(x)}{\sigma_n(x)}-\left[\frac{\sigma_n'(x)}{\sigma_n(x)}\right]^2=\frac{\sum_{k=1}^n k^x\log^2 k}{\sum_{k=1}^n k^x}-
\left[\frac{\sum_{k=1}^n k^x\log k}{\sum_{k=1}^n k^x}\right]^2
\\
=\frac 12\frac{\sum_{k,m=1}^n k^xm^x(\log k-\log m)^2}{\sum_{k,m=1}^n k^xm^x}
\\
=\frac 12\frac{\sum_{k,m:1\le k<m\le n} k^xm^x \log^2 \frac km}{\sum_{k,m:1\le k<m\le n} k^xm^x+\frac 12\sum_{m:1\le m\le n} m^xm^x}
\\
=\frac 12\frac{\sum_{m=1}^n m^x\left(\sum_{k:1\le k<m} k^x \log^2 \frac km\right)}{\sum_{m=1}^nm^x\left(\sum_{k:1\le k<m} k^x+\frac 12 m^x\right)}
=
\frac 12\frac{\sum_{m=1}^n w_mE_m}{\sum_{m=1}^n w_m}\,,
$$
where
$$
w_m=m^x\left(\sum_{k:1\le k<m} k^x+\frac 12 m^x\right),\ 
E_m=\frac{\sum_{k:1\le k<m} k^x \log^2 \frac km}{\sum_{k:1\le k<m} k^x+\frac 12 m^x}.
$$
Thus, to show that the second derivative of $\log\sigma_{n+1}(x)$ is at least as large as that of $\log\sigma_{n}(x)$, it is sufficient to show that $E_n$ is a non-decreasing sequence in $n$.
Dividing the numerator and the denominator in the definition of $E_n$ by $n^{x+1}$, we see that for $x>0$, $E_n$ is the ratio of the $n$-th trapezoid sums $T_ng$ and $T_nf$ of the functions $g(z)=z^x\log^2z$ and $f(z)=z^x$ on $[0,1]$. It will be convenient from now on to use $p$ instead of $x$, so that $x$ could be used for something else as needed.
The trapezoid sum of a reasonably smooth on $[0,1]$ function $h$ can be also expressed as
$$
T_n h=\sum_{m:n\mid m}\widehat h(m)\,.
$$
If, in addition, $h$ is real-valued, then we can take the real part of both sides and get
$$
T_n h=\widehat h(0)+2\sum_{m:m>0, n\mid m}\Re\widehat h(m)\,,
$$
so, if the real parts of the Fourier coefficients $\widehat h(n)$ with positive indices have certain monotonicity for $n\ge n_0$, the trapezoid sums $T_nh$ will be monotone in the same way for $n\ge n_0$ (note that the direction of the monotonicity of the real parts of the Fourier coefficients of $h$ determines their signs as well since they tend to $0$ at infinity).
It has been shown in this thread that for $p>1$, the real parts of the Fourier coefficients of $f(z)$ are decreasing all the way from $n=1$ to $n=\infty$, so for $p>1$ the denominators $T_nf$ of $E_n$ form a positive decreasing sequence (for $p\ge 2$ one can also give an elementary real variable proof of this fact). Thus, everything will be fine if we show that the numerators $T_ng$ form an increasing sequence.
That is not always true (and, technically, we do not need that much to show that $E_n$ increase, which seems to be a correct statement for all $p>0$) but the numeric experiments indicate that it is, indeed, so for $p\ge 4$ and we will (try to) prove that it is true for $p\ge p_0$ with some large $p_0>0$.
We shall start with the trivial reason that forces $T_ng$ to increase. Suppose that $g$ is convex on $[0,\frac n{n+1}]$. Then, using the representation
$$
\frac kn=\frac{n-k}{n}\frac k{n+1}+
\frac{k}{n}\frac {k+1}{n+1}
$$
for $k=1,\dots,n-1$, we get
$$
g(\tfrac kn)\le \tfrac{n-k}{n}g(\tfrac k{n+1})+
\tfrac{k}{n}g(\tfrac {k+1}{n+1})\,,
$$
so
$$
T_ng=\frac 1n\sum_{k=1}^{n-1}g(\tfrac kn)\le
\frac 1n\sum_{k=1}^{n-1}[\tfrac{n-k}{n}g(\tfrac k{n+1})+
\tfrac{k}{n}g(\tfrac {k+1}{n+1})]
\\
=\frac{n-1}{n^2}\sum_{k=1}^n g(\tfrac k{n+1})=\frac{n^2-1}{n^2}T_{n+1}g\le T_{n+1}g\,.
$$
Let's now find the range of $z\in[0,1]$ for which $g''(z)>0$. We have $g'(z)=z^{p-1}(p\log^2 z+2\log z)$ and $g''(z)=z^{p-2}[p(p-1)\log^2 z+(4p-2)\log z+2]$. The expression in parentheses is positive at $0$ if $p>1$ and hits $0$ when
$$
\log z=-\frac{(2p-1)+\sqrt{2p^2-2p+1}}{p(p-1)}\approx -\frac{2+\sqrt 2}{p}
$$
for large $p>1$. Thus, for every fixed $\delta>0$, the convexity assumption holds as long as $\frac pn>2+\sqrt 2+\delta$ and $p>p(\delta)$.
We shall show that we have $\Re\widehat g(n+1)\ge \Re\widehat g(n)$ when $\frac pn<4-\delta$ and $p>p(\delta)$.
To this end we shall flip the function and write
$$
\Re\widehat g(n+1)-\Re\widehat g(n)=\int_0^1 e^{2\pi i nz}[e^{2\pi iz}-1](1-z)^p\log^2(1-z)\,dz=\int_0^1 G_{p,n}(z)\,dz\,.
$$
We shall do the contour integration again (like in that thread I mentioned already with an escape to $+i\infty$) but the details will be uglier.
First, let us take care of the return, which will again be done over the right side $1+iy$, $y\ge 0$ of the half-strip. We note that
$$
|G_{p,n}(1+iy)|=\left[\log^2 y+\left(\tfrac \pi 2\right)^2\right](1-e^{-2\pi y})y^p e^{-2\pi ny}\,.
$$
Recall that we are interested in $n>p/4$ and large $p$. Let us consider $|G_{1,1/4}(1+iy)|$. Choose $Y'>0$ so that $y\mapsto ye^{-\frac \pi 2y}$ is decreasing on $[Y',+\infty)$. Now choose $Y>0$ so that
$$
\int_{Y'}^Y|G_{1,1/4}(1+iy)|\,dy\ge \int_{Y}^{+\infty}|G_{1,1/4}(1+iy)|\,dy
$$
Then this inequality will be preserved if we multiply the integrand by any decreasing on $[Y',+\infty)$ function of $y$. In particular, we can multiply it by
$[ye^{-\frac \pi 2y}]^{p-1}e^{-2\pi(n-\frac p4)y}$ to get
$$
\int_{Y'}^Y|G_{p,n}(1+iy)|\,dy\ge \int_{Y}^{+\infty}|G_{p,n}(1+iy)|\,dy
$$
for all $p>1$, $n>\frac p4$. Thus the return integral will be at most
$$
2\int_{0}^Y|G_{p,n}(1+iy)|\,dy
$$
with some fixed $Y$ independent of $p$ and $n$ in our range.
Now we want to escape from $0$ to $+i \infty$ along some curvilinear contour. Note that $\left(\frac{\log(1-z)}z\right)^2$ and $\frac{e^{2\pi iz}-1}z$ are nice analytic functions with bounded argument in some fixed size neighborhood of our half-strip minus the disk of that size centered at $1$. Thus the argument of their product is some bounded (in the same neighborhood) harmonic function $U(z)$ with $U(0)=\frac \pi 2$, which then has bounded derivatives as well. Thus, the integral over the escaping contour is
$$
\int |G_{p,n}(z)|e^{i(-p\alpha+2\pi nx+U(z) +3\arg(z)+\theta)}|dz|=\int |G_{p,n}(z)|e^{i\Phi(z)}|dz|\,
$$
in the same notation as in the thread about monotonicity of Fourier coefficients.
We shall choose below the phase $\Phi$ so that $\cos\Phi\ge c\sin\alpha$ with some absolute $c>0$. Let us show now that this will suffice.
Again, we shall compare $|G_{n,p}(z)|$ with $|G_{n,p}(z')|$ where $z'=1+i\Im z$ when $0\le \Im z\le Y$ and $z$ runs  over our contour. $|e^{2\pi i nz}|=|e^{2\pi i nz'}|$, $|e^{2\pi i z-1}|\ge |e^{2\pi i z'}-1|$, so these parts are fine.
Now $|1-z'|^p=|1-z|^{p}\frac{\Im z}{|1-z|}\sin^{p-1}\alpha$, so we have a huge gain here if $\alpha$ stays separated from $\frac\pi 2$, which will be true if $x$ stays away from $1$ (recall that $\Im z$ is bounded by $Y$ in that comparison). The problem is the $log$ factor.
In $|G_{p,n}(z)|$ we can guarantee just $\alpha^2$ (the square of the imaginary part) while in $|G_{p,n}(z')|$ we have $\log^2\Im z
+\left(\frac\pi 2\right)^2$. Fortunately, this expression times $\frac{\Im z}{|1-z|}$ stays bounded if we keep away from $1$ and below $Y$, so we see that to get $c\sin\alpha |G_{p,n}(z)|>2|G_{p,n}(z')|$, we just need to ensure that
$$
c\sin\alpha \alpha^2\ge C\sin^{p-1}\alpha\,,
$$
which, indeed, holds for large enough $p$ that depend on $c$, $Y$ and on how well we stay separated from $1$ in $x=\Re z$.
It remains to construct the contour.
To be continued...<|endoftext|>
TITLE: An inequality involving sums of powers
QUESTION [5 upvotes]: I asked this question at Stack Exchange but received no answer. The origins of the question are unclear, as I came across it rummaging through old notebooks from highschool, in one of which it was stated without any reference or proof. Let $x, y, z$ and $t$ be positive numbers such that $x+y+z+t=1$. Then the following inequality holds:
$$
\frac{\sqrt[3]{x^4 + y^4 + z^4 + t^4} - (x^2 + y^2 + z^2 + t^2)}{\sqrt[3]{x^4 + y^4 + z^4 + t^4} - \sqrt{x^3 + y^3 + z^3 + t^3}}<4
$$
I tried various approaches, e.g. using some form of power means monotonicity, symmetric reduction, looking up Bullen's "Handbook of Means and Their Inequalities",  even desperate approaches like the y-positivity of Cuttler, Greene & Skandera. It didn't work. 
I doubt this is a research grade question even though numerical experiments show that it can be extended to any number of variables, not only 4. Moreover, I believe that an elementary proof exists otherwise I would not have been able to prove it in highschool.

REPLY [8 votes]: Let $s_k = x^k + y^k + z^k + t^k$. First we check that the denominator is nonnegative. By Holder, we know $s_4^{2/3}s_1^{1/3} \ge s_3$, rearranging that and using $s_1 = 1$ we see that the denominator is indeed at least $0$.
Now we multiply out and rearrange, to see that the given inequality is equivalent to:
$\frac{3\sqrt[3]{s_4s_1^2}+s_2}{4} \ge \sqrt{s_3s_1}$.
By the weighted AM-GM inequality the left hand side is at least $\sqrt[4]{s_4s_2s_1^2}$, so the inequality boils down to showing that $s_4s_2 \ge s_3^2$, which follows from Cauchy-Schwartz.
Note that no step of this proof depends on the number of variables.

REPLY [7 votes]: Let us abbreviate the vector $(x,y,z,t)$ as $\mathbf{x}$. Combining Hölder's inequality and Young's inequality,
$$ |\mathbf{x}|_3^{3/2} \leq |\mathbf{x}|_4 |\mathbf{x}|_2^{1/2} 
\leq \frac{3}{4} |\mathbf{x}|_4^{4/3} + \frac{1}{4}|\mathbf{x}|_2^2. $$
We do not have equality in the first inequality, because the entries of $\mathbf{x}$ are positive. Therefore
$$ |\mathbf{x}|_3^{3/2} < \frac{3}{4} |\mathbf{x}|_4^{4/3} + \frac{1}{4}|\mathbf{x}|_2^2. $$
Rearranging, we obtain the desired inequality. In this last step we use that the denominator is positive, which is another application of Hölder's inequality: 
$$ |\mathbf{x}|_3^{3/2} \leq |\mathbf{x}|_4^{4/3} |\mathbf{x}|_1^{1/6}, $$
where we do not have equality as before, and $|\mathbf{x}|_1=1$ by assumption.<|endoftext|>
TITLE: Multivariable Calculus Lecture Ideas
QUESTION [8 upvotes]: I am teaching a course in multivariable calculus this semester.  We are covering the basics about $\mathbb{R}^n$, including dot products and cross products, curves, and quadric surfaces.  After that we learn differentiation for functions $f:\mathbb{R}\rightarrow \mathbb{R}^n$ and partial differentiation for functions $f:\mathbb{R}^n\rightarrow\mathbb{R}$.  We finish off with Lagrange multipliers.  
My program so far has been to lecture on a chalkboard while they take notes.  I am not really keeping their attention.  I was wondering if anyone had any ideas to involve them more in class.  We are in a computer lab, and have access to Maple.  Your answer need not be computer based, but I have computers available.

REPLY [2 votes]: When introducing the cross product, one thing that always catches my students' attention is a demonstration of precession. We have a bike wheel with a heavy solid rubber tire. (You may be able to borrow one from your physics department.) I get it spinning with its axis in the horizontal plane, and then suspend it by a rope from one side. They are gratifyingly surprised when they see it precess in the horizontal plane rather than flopping. This connects to torque expressed as a vector cross product.
Another example I like to do is to have them imagine a spinning cylindrical space station and lead them through a series of thought experiments leading them to figure out the rules by which the centrifugal and Coriolis forces operate. E.g., a person standing on the deck releases a ball. In the frame of the stars, the ball travels straight. What appears to happen in the rotating frame? Once they see that the centrifugal force operates like a radially directed gravitational field, I do examples that show the existence of a velocity-dependent force. E.g., what happens if the person standing on the deck throws the ball opposite to the direction of rotation, such that he exactly cancels its motion in the frame of the stars? What if a ball is released on the axis? What if the ball is released on the axis with some small radial velocity? Based on this sort of thing, it's fairly easy to get them to conclude that the Coriolis force is an odd function of the ball's velocity, and also an odd function of the angular velocity of the space station. There is only one good way to get a vector-valued function with these properties, so it must be $\mathbf{F}=c\mathbf{\omega}\times\mathbf{v}$. They enjoy imagining the fun and strange things that happen on the space station, and it's a cool example where the uniqueness of certain mathematical operations makes it possible to determine the answers to problems without grotty calculations.<|endoftext|>
TITLE: Computing signature
QUESTION [16 upvotes]: I have a feeling that this might have already been asked, but can't find the question. Anyway, the question is: given a symmetric $n\times n$ matrix, is there a faster way to compute its signature than computing the full eigen-decomposition? (the gt tag is because my application is in computing signatures of four-manifolds, and some others might want to do the same computation...)

REPLY [2 votes]: May I attract your attention on the following formula,
where $B$ is a $n\times n$ non-singular symmetric real-valued matrix
$$
\int_{\mathbb R^n}e^{-2i\pi x\cdot \xi}
e^{i\pi Bx\cdot x}dx=e^{\frac{i\pi}{4}\text{signature } B}\vert\det B\vert^{-1/2}e^{-i\pi B^{-1}\xi\cdot \xi},
$$
so that
$
e^{\frac{i\pi}{4}\text{signature } B}=\vert\det B\vert^{1/2}
\int_{\mathbb R^n}
e^{i\pi Bx\cdot x}dx.
$
Of course the Fresnel-type integral on the rhs of last formula is not absolutely converging but, as an oscillatory integral, will become so after integration by parts.<|endoftext|>
TITLE: Difference between automorphic forms for SL(2) and GL(2)?
QUESTION [8 upvotes]: Hi,
Let $A$ denote the adeles of $Q$.
I know how to decompose $L^2(SL(2,A)/SL(2,Q))$ into irreducible $SL(2,A)$-representations. What is the difference between this decomposition and the corresponding decomposition for $GL(2)$? Can I deduce the $GL(2)$-case from the $SL(2)$-case?
Thanks for answering this basic question.

REPLY [4 votes]: I think GH's link and K.Buzzard summary of Labesse-Langlands explains that there is no easy comparison possible.
But I think at the heart of your question is something else. Are you asking about a decomposition into cuspidal, continuous and residual part? If yes, the decomposition is completly analogous (of course for technical convenience you should rather fix a central character - say trivial - in GL(2)).
You get
1) a direct sum of cuspidal representation (all single multiplicity)
2) a sum over the one-dimensional representations $\chi \circ  \det$ with $\chi$ Hecke character and $\chi^2 =1$ resp. for SL(2) only the trivial rep.
3) a direct integral over parabolic induced representation $\chi_1,\chi_2$ with $\chi_j$
 Hecke quasi character and $\chi_1 \chi_2 =1$
The proof in Gelbart-Jacquet "Analytic ..." translates easily to this situation.<|endoftext|>
TITLE: Non-isomorphic finite simple groups
QUESTION [21 upvotes]: Hello,
The smallest integer $n$ such that there exists two non-isomorphic simple groups of order $n$,  is $n=20160$ (namely for the groups $\mathrm{PSL}_3(\mathbb F _4)$ and $\mathrm{PSL}_4(\mathbb F _2)$).
I read that there are infinitely many integer $n$ such that here exists two non-isomorphic simple groups of order $n$. I have two questions:

Do you have a reference (if possible self contained, but that's probably too much to ask)?
I suspect that it is "rare" to find such an integer. For instance if we denote by $a_k$ the orders of non-cyclic simple groups ($a_1=60$, $a_2=168$, $a_3=360$,....) and $b_k$ the integers such that there exists two non-isomorphic simple groups of order $b_k$, then I guess that $\displaystyle \lim\frac{b_k}{a_k}=+\infty$. Do you know if this is the case?

Thanks

REPLY [18 votes]: Just to summarise the comments: the only nonisomorphic finite simple groups with the same orders are

$A_8 \cong {\rm PSL}_4(2)$ and ${\rm PSL}_3(4)$ of order 20160.
The groups ${\rm P \Omega}_{2n+1}(q)$ and ${\rm PSp}_{2n}(q)$ for all odd prime powers $q$ and $n \ge 3$. These have order

$$(q^{n^2} \Pi_{i=1}^n (q^{2i}-1))/2$$
For references, see Gerry Myerson's comment.<|endoftext|>
TITLE: Status of a conjectural definition of H. Nakajima
QUESTION [5 upvotes]: In his paper '$t$-analogue of $q$-characters of finite dimensional representations of quantum affine algebras' - http://arxiv.org/abs/math/0009231 - H. Nakajima states a conjectural definition of the $t$-analogue of the $q$-character of a standard module $M_{P}$ (with weight $P$) of a quantum affine algebra at level $0$ - this appears as Conjecture 3.1.1 on p. 5 of the arXived preprint above. In summary, Nakajima conjectures that the $q,t$'-character  of a standard module $M_{p}$ can be determined using certain filtrations on individual weight spaces.
I was hoping that someone can let me know of the status of this conjecture - is it true that we can describe the '$q,t$'-character as conjectured?
I imagine that this has been resolved since the paper is a bit more mature now. If this is the case, can someone point me in the direction of a resolution? 
If this conjecture has not been resolved, does anyone know of any progress towards its resolution/any problems that have arisen in resolving this conjecture?
Thanks in advance.

REPLY [6 votes]: See http://arxiv.org/abs/1004.2321.<|endoftext|>
TITLE: Why doesn't this group have a name?
QUESTION [9 upvotes]: $$\{A\in GL_n(\mathbb{C}) : |det(A)|=1\}$$  This seems to me to be a perfectly natural group to study; it is easy to define and contains $U(n), SL_n$, and all the torsion.    Is there any good reason why this group isn't among the usual classical groups that are so well-understood and thoroughly discussed?  I understand that most of those are studied/defined by looking at groups preserving particular inner products, but it still surprises me that I've never heard of any interesting results/properties of this group.  The only guess I currently have is it's not compact.
I could ask the same question with $\mathbb{R}$ but then "morally" the group is just two copies of $SL_n(\mathbb{R})$  so I understand why it's less interesting.  
A perfectly acceptable answer is that I'm totally misinformed and this group is perfectly understood, classical, named, etc., in which case any reference would be appreciated.

REPLY [7 votes]: Denote the group in question by $G$. Then there is a split extension 
$$1 \to SL_n(\mathbb{C}) \to G \xrightarrow{\text{det}} S^1 \to 1$$
where the splitting is given by $S^1 \to G,\; z \mapsto \text{diag}(z,1,...,1)$. Hence, from the group theoretical point of view $G$ is just the semi-direct product 
$$G = SL_n(\mathbb{C}) \ltimes S^1$$
Added: Your guess that $G$ isn't compact (in Euclidean topology) if $n>1$ is correct. For, suppose $G$ is compact. Then, the closed subgroup $SL_n(\mathbb{C}) = \text{det}^{-1}(1)$ is also compact, in contradiction to the fact that it contains the unbounded subset $\lbrace\text{diag}(z,z^{-1},1,...,1) \mid z \neq 0\rbrace$.<|endoftext|>
TITLE: large ccc forcing that preserves CH
QUESTION [9 upvotes]: Can you name a ccc forcing with the following properties?
1) Atomless and separative
2) The least size of a dense set is large, say at least $\aleph_3$, hopefully as big as you like.
3) Existence is consistent with CH.
4) Preserves CH.
If you can think of any, please list as many as you can.

REPLY [6 votes]: Thanks to Michael Blackmon for this idea:
If P has the ccc and preserves CH, then there is a contiuum sized regular suborder R that adds all the reals P will add.  The factor forcing P/R is ccc and $\omega$-distributive, a Suslin algebra.  A theorem (of Jech?) says that any Suslin algebra has size at most $2^{\omega_1}$.  So there is a bound on the size of P.<|endoftext|>
TITLE: Free subgroup of Diff([0,1])?
QUESTION [10 upvotes]: It is well known that the group of diffeomorphisms of the circle contains free non-Abelian subgroups. Is it true (known) that the group of diffeomorphisms of the interval $[0,1]$ contains free subgroups? One approach to get a positive answer can be the following. Consider all functions $f_a=\frac{\exp(ax)-1}{\exp(a)-1}$, $a>1$, which are smooth diffeomorphisms of $[0,1]$. First prove that the group generated by these functions does not satisfy any non-trivial law. Now for any non-trivial word $w$ in two variables consider the function $w(f_a,f_b)$ (for every $a,b$). The set of pairs $(a,b)\in \mathbb{R}^2$ for which this function is the identity function has dimension at most 1. Then by Baire category theorem, since the set of words $w$ is countable, there are $a,b$ such that $f_a,f_b$ freely generate a free subgroup. Does this argument actually work?

REPLY [12 votes]: Much more is true.  The compactly supported diffeomorphism group of any (positive-dimensional, nonempty) manifold contains free subgroups of uncountable rank.  In fact, there are such subgroups that are generated by sets which are arcwise connected!  See the paper
MR0974661 (90b:58031)
Grabowski, Janusz(PL-WASW)
Free subgroups of diffeomorphism groups. 
Fund. Math. 131 (1988), no. 2, 103–121. 
which is available online here.<|endoftext|>
TITLE: Discrete orderings on $\mathbb{Z}[x,y]$ that violate the universal theory of the integers
QUESTION [6 upvotes]: Working in the language of ordered rings, which we take to have type $(+ - \times < 0\, 1)$, can anyone give an example of a discrete ordering on the polynomial ring in two variables $\mathbb{Z}[x,y]$ such that the resulting ordered ring does not satisfy the universal theory of the integers? 
It is not difficult to show that as a ring, i.e. forgetting the less-than symbol, the ring $\mathbb{Z}[x,y]$ does in fact satisfy the universal theory of the ring of integers. The problem is to find a discrete ordering on $\mathbb{Z}[x,y]$ such that some system of inequalities is solvable in $\mathbb{Z}[x,y]$ but not in the integers, or, on the contrary, to prove that there is no such ordering.

REPLY [4 votes]: Here is an example of a discrete ordering on $ \mathbb{Z}[x,y]$ which does not satisfy the universal theory of the ordered ring $ \mathbb{Z}$. 
Let $\alpha$ be a real algebraic number of degree at least 3 over $\mathbb{Q}$. Let $e$ be any irrational between 1 and 2. Let $A$ be the ring 
$\mathbb{Z}[x,\alpha x+x^{-e}]$.  For $f,g\in A$ define $f<g$ to mean that, regarding $f$ and $g$ as functions on the positive reals, $f(r)<g(r)$ for all sufficiently positive $r$. This makes $A$ into an ordered ring. I claim that

$A$ is ring-isomorphic to $\mathbb{Z}[x,y]$. 
$A$ fails to satisfy the universal theory of the integers.
$A$ is discretely ordered.

Proof of 1. We show that if $p(x,y)\in \mathbb{Z}[x,y]$ and $p(x,\alpha x+x^{-e})=0$, then $p$ is the zero-polynomial. 
Write $p$ in the form
$$\sum_n q_n(x)x^{-en},$$
where the $q_n$ are polynomials in $\mathbb{Q}(\alpha)[x]$. It follows from the irrationality of $e$ that no monomial in $q_n(x)x^{-en}$ can appear in $q_m(x)x^{-em}$ if $n\ne m$. Therefore all of the $q_n$ must be zero-polynomials, whence p=0. 
Proof of 2. Fix positive integers $a$ and $b$ such that $1<a/b<\epsilon$. Roth's theorem on diophantine approximation (here) implies that there is a natural number $N$ such that the following holds for all integers $x$ and $y$: 
$$x>N\implies|\alpha x-y|^bx^a>1.$$
Using quantifier elimination for real closed fields and the assumption that $\alpha$ is  algebraic,  it is possible to construct a quantifier-free formula, call it $\phi(x,y)$, that  is equivalent  in any real closed field to the above inequality. (In an arbitrary real closed field $\alpha$ has the meaning of an element satisfying a certain polynomial and lying in a certain interval with rational endpoints.)
Then  $$\mathbb{Z}\models \forall x,y\,\phi(x,y).$$ On the other hand, in the real closure of the ring $A$, with $u=x$ and $v=\alpha x+x^{-e}$, it is easily verified (using $1<a/b<\epsilon$) that 
$$u>N\textrm{ and }|\alpha u-v|^bu^a<1.$$
Since $\phi$ is is quantifier-free, it follows that
$$A\models \neg\phi(u,v).$$
Therefore $A$ fails to satisfy the universal theory of $\mathbb{Z}$.
Proof of 3. We will show that A has no finite elements other than the integers. It is enough to show that if  $H\in \mathbb{Z}[x,y]$ is non-constant then    $H(x,\alpha x+x^{-e})$ has at least one monomial of positive degree. We shall do this first for homogeneous $H$ and then for sums of non-constant homogeneous polynomials of different degrees.
Assume now that $H(x,y)$ is homogeneous of degree $n>0$. Let $$h(y)=H(1,\alpha+y).$$
Writing out the Taylor expansion for $h$ about $y=0$,
$$h(y)=\sum_{k=0}^{n}\dfrac{h^{(k)}(\alpha)}{k!}y^k,$$
where $h^{(k)}$ is the $k$-th derivative of $h$. Substituting $x^{-1-e}$ for $y$, and multiplying both sides by $x^n$, we obtain
\begin{equation}\tag{$*$} H(x,\alpha x+x^{-e})=\sum_{k=0}^{n}\dfrac{h^{(k)}(\alpha)}{k!}x^{n-k(1+e)}.\end{equation}
The exponent $n-k(1+e)$ is positive precisely when $k=0,1,\ldots,\lfloor n/(1+e)\rfloor$. By way of contradiction, suppose that $h^{(k)}(\alpha)$ vanishes for all of these values of $k$. 
Let $f$ be the irreducible polynomial of $\alpha$ over $\mathbb{Q}$, and let $m$ be the degree of $f$. Then $f^{1+\lfloor n/(1+e)\rfloor{}}$ must divide $h$.  Since $h$ has degree at most $n$, we find that $$\tag{$**$}m(1+\lfloor n/(1+e)\rfloor)\le n.$$ But  $$n/(1+e)<1+\lfloor n/(1+e)\rfloor,$$ hence 
$$\tag{$***$}mn/(1+e)<m(1+\lfloor n/(1+e)\rfloor).$$
Combining the inequalities $(**)$ and $(***)$, we have $$mn/(1+e)<n,$$ hence $m<1+e$. But by assumption $e<2$. Therefore $m<3$.  This contradicts the definition of $\alpha$, which is assumed to have degree $m\ge3$. This concludes the homogeneous case.
Now suppose $H\in \mathbb{Z}[x,y]$ has the form $\sum_k H_k$, where $H_k$ is homogeneous of degree $k>0$. It follows from equation ($*$) that the powers of $x$ in $H_k(x,\alpha x+x^{-e})$ all have the form $a+be$, with $a+b=k$. Since $e$ is irrational, no power of $x$ in $H_k(x,\alpha x+x^{-e})$ can appear in $H_m(x,\alpha x+x^{-e})$ if $k\ne m$. But we have shown that  some positive power of $x$ appears in $H_k(x,\alpha x+x^{-e})$ for every $k>0$. This completes the proof.<|endoftext|>
TITLE: Reference for : a Fréchet nuclear space is Montel
QUESTION [6 upvotes]: I'm looking for a reference to cite regarding the property presented in the title: "Closed and bounded sets of a nuclear Fréchet space are compact"
Thank you in advance for the help!

REPLY [2 votes]: Since the garden variety Fréchet nuclear spaces (the Schwartz space, the space of smooth functions on the torus, the space of smooth fuctions with support in a given compact set, etc., Edit: see also Jochen's comment below) are isomorphic as topological vector spaces to $\mathscr{s}$, the space of rapidly decaying sequences, and since the proof of the Montel property for $\mathscr{s}$ is very easy, I might as well explain it here in a self-contained answer.
I will use the convention $\mathbb{N}=\{0,1,2,\ldots\}$ and denote by $\mathscr{s}$ the space of sequences $x=(x_n)_{n\in\mathbb{N}}$ such that
$$
||x||_{\infty,k}:=\sup\limits_{n\ge 0}\ \langle n\rangle^k|x_n|<\infty
$$
for all $k\in\mathbb{N}$. Here $\langle n\rangle:=\sqrt{1+n^2}$. The topology is the one given by the collection of seminorms $||\cdot||_{\infty,k}$, $k\ge 0$. An equivalent system of seminorms defining the same locally convex topology is
$$
||x||_{1,k}:=\sum_{n\ge 0}\langle n\rangle^k|x_n|\ .
$$
I will denote by $\mathscr{s}_{+}$ the subset of $\mathscr{s}$ made of sequences with only nonnegative entries.
Given a subset $A\subset \mathscr{s}$, I will define the envelope of $A$, or ${\rm env}(A)$, as the sequence $z=(z_n)$ given by
$$
z_n:=\sup_{x\in A}|x_n|\ .
$$
This sequence a priori belongs to $[0,\infty]^{\mathbb{N}}$. Using the $||\cdot||_{\infty,k}$ seminorms, the following is trivial.
Proposition: $A$ is a bounded subset of $\mathscr{s}$ if and only if ${\rm env}(A)\in\mathscr{s}_{+}$.
Conversely, given $\omega\in\mathscr{s}_{+}$, let me define the subset ${\rm box}(\omega)$ made of all sequences $x$ such that, for all $n$, $|x_n|\le \omega_n$.
The wanted Montel property immediately follows from the following result.
Proposition: For any $\omega\in\mathscr{s}_{+}$, we have that ${\rm box}(\omega)$ is a compact subset of $\mathscr{s}$.
Proof: (in the real case, the complex case only needs disks instead of intervals) Let $K:=\prod_{n\ge 0}[-\omega_n,\omega_n]$ with the (metrizable) product topology. Of course, by the countable Tychonov Theorem, $K$ is compact. Let $\tau:K\rightarrow \mathscr{s}$ be the obvious tautological inclusion map. Then the compactness of ${\rm box}(\omega)=\tau(K)$ would follow from the continuity of the map $\tau$. Since the spaces are metrizable, it is enough to show sequential continuity. Now switch to the $||\cdot||_{1,k}$ seminorms for $\mathscr{s}$. The needed sequential continuity is immediate from the discrete Dominated Convergence Theorem (DCT). Indeed, let $(x^{(m)})_{m\ge 0}$ be a sequence in $K$ converging to some element $x\in K$. This means that for all $n$, one has the pointwise convergence $\lim_{m\rightarrow\infty}x_{n}^{(m)}=x_n$.
Once mapped inside $\mathscr{s}$, we have, for any $k\ge 0$,
$$
||\tau(x^{(m)})-\tau(x)||_{1,k}=\sum_{n\ge 0}\langle n\rangle^k|x_{n}^{(m)}-x_n|
\longrightarrow 0
$$
by the DCT and the use of the dominating function $n\mapsto 2\langle n\rangle^k \omega_n$.<|endoftext|>
TITLE: Index of derived subgroup in derived group
QUESTION [6 upvotes]: Let $G$ be a group with finite index subgroup $H$. Let $G^\prime = [G,G]$ denote the derived subgroup of $G$.
Is it true that $|G:H|<\infty$ implies that $|G^\prime: H^\prime|<\infty$.
If this is not true in general, is it true for a large class of groups? say, finitely generated.
Thanks to Mark Sapir for providing a simple counter-example to the general statement below.
What about if $G$ is nilpotent for example?

REPLY [3 votes]: This is a old question but I think it is still worthwile to supply a reference to a much more general statement (taken from subsection 2.3.3 of "The Theory of Infinite Soluble Groups" by John C. Lennox and Derek J. S. Robinson). 
Let $\theta(x_1,\dots, x_n)$ be a word in the variables $x_1,\dots, x_n$. 
If $H_1,\dots H_n$ are subgroups of a group $G$, let $\theta(H_1,\dots, H_n)$ be the subgroup generated by all $\theta(h_1,\dots, h_n)$
where $h_i\in H_i$. Then a result of Philip Hall says:
 Theorem Let $G$ be a finitely generated nilpotent group with subgroups $H_i\le K_i$,
$i=1,\dots, n$ such that each index $|H_i:K_i|=m_i$ is finite. Then for every $n$-variable word $\theta$ 
the index $|\theta(H_1,\dots, H_n):\theta(K_1,\dots, K_n)|$ is finite and
it divides some power of $m_1 m_2 · · · m_n$.
In our case $n=1$ and the word $\theta$ is the commutator. Thus if $|G:H|=m$, then $|G^\prime: H^\prime|=m^r$ for some nonnegative interger $r$ (which could be $0$ as happens e.g. when $G$ is abelian).<|endoftext|>
TITLE: closed dual of vector fields
QUESTION [5 upvotes]: Given a closed manifold and a nowhere zero vector field $X$, can one always find a closed 1-form $\alpha$ such that $\alpha(X)=1$ as the constant function? Or under what kind of conditions on $X$ can this be true? Thanks!

REPLY [4 votes]: On the other hand, there are situations where the answer is positive. Actually Schwartzman and Fried gave a very satisfactory criterion for the existence of forms which are positive along the flow (see for example Fried's The geometry of cross sections to flows).
For $X$ a vector field on a compact manifold $M$, denote by $\phi^t$ the associated flow. For $p$ a point in $M$, let $k(p,t)$ be a loop obtained by following $X$ from $p$ to $\phi^t(p)$, and closed by an arbitrary (short) segment. Consider the classes $[k(p,t)]/t$ in $H_1(M, \mathbb R)$, denote by $C_\phi$ the set of such classes, and by $S_\phi$ the set of accumulation points of $C_\phi$. As long as the length of the closing segments is bounded, $S_\phi$ depends on $X$ only. Schwartzman-Fried's theorem states that $X$ admits a closed $1$-form $\alpha$ with $\alpha(X)>0$ everywhere if and only if $S_\phi$ lies in an open half-space in $H_1(M, \mathbb R)$.<|endoftext|>
TITLE: Origin of square-and-multiply algorithm
QUESTION [9 upvotes]: I'm teaching an introductory course in cryptography and explained the square-and-multiply algorithm to the class. 
http://en.wikipedia.org/wiki/Square-and-multiply_algorithm
Someone asked who discovered the algorithm, which I didn't know, so after a short web search that gave no answers, I thought I'd ask on MO. In particular, the above wikipedia article is not helpful, and I didn't see any MO questions that address the issue. This seems like something that Gauss and Euler, or even Fermat, might have known, and ditto for Indian and Chinese mathematicians centuries earlier, but I'm just speculating. Specific references would be appreciated. (Sorry if this isn't really a research level question, although maybe it qualifies as historical research.)

REPLY [12 votes]: This method is indeed over 2000 years old. The history, with references, is discussed by Donald Knuth in Seminumerical Algorithms, volume 2 of The Art of Computer Programming, page 441:

The method is quite ancient; it
  appeared before 200 B.C. in Pingala's
  Hindu classic Chandah-sutra [see B.
  Datta and A.N. Singh, History of Hindu
  Mathematics 1, 1935]; however, there
  seem to be no other references to this
  method outside of India during the
  next 1000 years. A clear discussion of
  how to compute $2^n$ efficiently for
  arbitrary $n$ was given by al-Uqlidisi
  of Damscus in 952 A.D.; see The
  Arithmetic of al-Uglidisi by A.S.
  Saidan (1975), p. 341-342, where the
  general ideas are illustrated for
  $n=51$. See also al-Biruni's
  Chronology of Ancient Nations (1879),
  p. 132-136; this eleventh-century
  Arabic work had great influence.

For a detailed discussion of the earliest history, see A. Kulkarni, Recursion and Combinatorial Mathematics in Chandashaastra. [Chandashaastra = Chandah-sutra]<|endoftext|>
TITLE: motivic t-structure and realisations
QUESTION [7 upvotes]: Let $k$ be a field and $DM_k$ denote the triangulated category of geometric motives with $
\mathbb{Q}$ coeffients over $k$.  Recall that there exists a motive functor $M: Var_k\rightarrow DM_k$, which yields an fully faithful embedding of tensor $
\mathbb{Q}$-categories $CHM_k\rightarrow DM_k$, where $CHM_k$ is athe category of Chow motives with $\mathbb{Q}$-coefficients over $k$.
For $\ell$ prime to the characteristic of $k$ one has the $\ell$-adic realisation functor $r_{\ell}:DM_k\rightarrow D^b(Vec_{\mathbb{Q}\ell})$.  As I understand it, if $X$ is smooth projective variety over $k$ then the cohomology of $r_{\ell}(M(X))$ is just the $\ell$-adic cohomology of $X$.    
Now the conjectural motivic $t$-structure on $DM_k$ has the property that the realisation functor $r_{\ell}$ is $t$-exact.Thus those objects in the heart of this $t$-structure (the conjectural category of mixed motives $MM_k$) have realisation with trivial cohomology outside  of degree zero.  
Here is what is confusing me: for a general smooth projective variety its $\ell$-adic cohomology is not always concentrated in degree zero.  Thus for such $X$, the motive $M(X)$ is not in the category of mixed motives.  This can't be correct as the category of mixed motives should contain the category of pure motives. Where am I going wrong?

REPLY [9 votes]: The image of the inclusion $CHM_k\hookrightarrow DM_k$ is indeed not contained in the heart $M_k$ of the motivic t-structure.
$CHM_k$ does map to the heart, but via a different functor, namely, the projection $CHM_k\to NM_k$ to numerical motives, followed by the inclusion $NM_k\hookrightarrow M_k$ ($NM_k$ should be the subcategory of semi-simple objects in $M_k$). This functor $CHM_k\to M_k$ should be equivalent to the composition of  the inclusion $CHM_k\hookrightarrow DM_k$ followed by the functor $DM_k \to M_k$ which sends $M$ to $\bigoplus_{i\in \mathbb{Z}} H^i(M)$.
I would draw a commutative square here if I knew how, but you can see it in Yves André's book "Une introduction aux motifs", 21.1.5.<|endoftext|>
TITLE: Are negatively pinched manifold locally conformally flat?
QUESTION [12 upvotes]: One knows that hyperbolic manifolds are locally conformally flat.
How about those negatively pinched manifolds, i.e. the sectional curvature $K$ satisfy:
$$
-\Lambda \le K \le -\lambda$$
for $\Lambda>\lambda$.
How about sufficiently pinched, i.e. $\Lambda/\lambda=1+\epsilon$ for $\epsilon$ small?
Are they have vanished Pontryagin classes?

REPLY [6 votes]: Another easy locally homogeneous counterexample to the original question is given by complex hyperbolic manifolds. This includes compact examples. Complex hyperbolic manifolds are Einstein. Curvature tensor of any manifold decomposes into its Weyl part+Ricci part +scalar part. Thus, a conformally flat Einstein manifold must necessarily have scalar curvature operator and hence have constant sectional curvature in dimensions above 2. This is definitely not the case for complex hyperbolic manifolds so they are not locally conformally flat. Also, as Igor mentioned Chern-Weil theory in dimension 4 says that $sig(M^4)=\frac{1}{12\pi^2}\int_M(|W^+|^2-|W^-|^2)$, where $W^\pm$ are self-dual and anti-self-dual parts of $W$.
4-dimensional complex hyperbolic manifolds are conformally semi-flat (i.e they have $W^-=0$) which can be easily derived from the fact that their curvature tensors are $U(2)$ invariant. Thus, for a closed complex hyperbolic 4-manifold its signature is $sig(M^4)=\frac{1}{12\pi^2}\int_M(|W^+|^2\ne 0$. Moreover, the integrant is just a constant (by homogeneity). On the other hand, locally conformally flat closed 4-manifolds have signature 0 by the above formula.<|endoftext|>
TITLE: Pseudo-Anosov map, Heegaard splitting, hyperbolic 3-manfold
QUESTION [8 upvotes]: Hi, I am interested in the relationship between the pseudo-anosov map and volume of the hyperbolic 3-manifold. 
Assume $H_{1}$ and $H_{2}$ are two handlebodies with $\partial H_{1}=\partial H_{2}=S$. 
Question 1:For any pseudo-anosov homeomorphism $\psi: S\rightarrow S$, if the $n\in N$ is large enough, is $M_{\psi^{n}}=H_{1}\cup_{\psi^{n}} H_{2}$ hyperbolic? 
Question 2:Given a pseudo-anosov map $\psi$, suppose $M_{\psi^{n}}=H_{1}\cup_{\psi^{n}} H_{2}$ is hyperbolic, for
any $n\geq k$, where $k\in N$. 
How does the Vol(M) change when $n$ goes to infinity?

REPLY [9 votes]: For Question 1, the answer is no. In fact, there are pseudo-Anosov maps $\psi$ that preserve a handlebody $H_1$ (the existence of such a map follows from work of Masur-Minsky, although probably appears earlier).  $H_1\cup_{\psi^n} H_2 = H_1\cup_S H_2$, so if $H_1\cup_S H_2$ is not hyperbolic, then neither will $M_{\psi^n}$. 
For Question 2, the answer is a bit more complicated. Namazi-Souto prove if $\psi$ is "generic", which means that the stable lamination $\lambda_+$ of $\psi$ is not a limit of meridians of $H_2$, and the unstable lamination $\lambda_-$ of $\psi$ is not a limit of meridians of $H_1$, then $M_{\psi^n}=H_1\cup_{\psi^n} H_2$ is hyperbolic for large $n$. Moreover, for any $\epsilon >0$ and large enough $n$, there are metrics on $M_{\psi^n}$ with curvatures pinched between $-1-\epsilon$ and $-1+\epsilon$, with volume growing linear with $n$. This implies that the hyperbolic volume of $M_{\psi^n}$ grows linearly as well, either by applying volume comparison theorems of Besson-Courtois-Gallot, or by an unpublished preprint of Tian. 
If the map $\psi$ is not generic, (say $\lambda_+$ is a limit of meridians of $H_2$), then Biringer-Johnson-Minsky prove that a power of $\psi$ extends over a compression body inside of $H_2$. This does not necessarily imply that the manifolds $M_{\psi^n}$ is not hyperbolic, but if they are I'm not sure how fast the volume grows; I suspect it would still grow linearly if some power does not extend entirely over $H_1$ or $H_2$ (like in the answer to Question 1).

REPLY [7 votes]: For Question 1:
Souto and Namazi (pdf link) showed that for a generic pseudo-anosov homeomorphism $\psi$ and $\epsilon >0$, there is $n_\epsilon$ such that $M_{\psi^n}$ admits a Riemannian metric with all sectional curvatures between $-1-\epsilon$ and $-1+\epsilon$ for all $n\ge n_\epsilon$.
Namazi (pdf link) used a theorem of Tian to show that for $\epsilon$ small enough these manifolds are actually hyperbolic.
For Question 2:
The manifolds Souto and Namazi construct have injectivity radius bounded below independently of $n$ and $\epsilon$, so the volumes must grow without bound as $n$ goes to infinity.<|endoftext|>
TITLE: canonical bundle on the stack of G bundles on a curve
QUESTION [5 upvotes]: I would like to understand the following statement that I found in some of Sorger's lecture notes (Lectures on moduli of principal G-bundles over algebraic curves).
Let X be a projective and smooth curve and $G$ a reductive group. 
We have the universal $G$-bundle $\mathcal{E}$ on $Bun_G(X)\times X$ and we can form the adjoint vector bundle $\mathcal{E}(\mathfrak{g})$ and then the determinant bundle $D_{\mathcal{E}(\mathfrak{g})}$ on $Bun_G(X)$ (i.e. $\det(Rp_*(\mathcal{E}(\mathfrak{g})))^{-1}$ ). The claim is that this determinant bundle is the canonical bundle on $Bun_G(X)$.
Could someone explain me how to prove and/or see this?

REPLY [7 votes]: The tangent complex is a complex of quasi-coherent sheaves on $Bun_G(X)$, i.e. a compatible family of complexes of quasi-coherent sheaves on every affine scheme $f:U\rightarrow Bun_G(X)$ (i.e. a $G$-bundle $P_0\rightarrow X\times U$) mapping smoothly.
The global sections $\Gamma(U, f^* T_{Bun_G(X)})$ of the tangent complex over $U$ is the complex associated to the Picard groupoid given by the fiber of $Hom(D\times U, Bun_G(X))\rightarrow Hom(\mathrm{pt}\times U, Bun_G(X))$, where $D=\mathrm{Spec}\ \mathbf{C}[\epsilon]/\epsilon^2$. Since this is a Picard groupoid, $\pi_0$ and $\pi_1$ are abelian groups, which are $H^0$ and $H^{-1}$ of the tangent complex.
One can explicitly describe this groupoid as follows: its objects are $G$-bundles $P\rightarrow X\times U\times D$ together with an isomorphism $P|_{X\times U\times\mathrm{pt}}\cong P_0$ on $X\times U$.
$\pi_1$ computed at the trivial bundle $P=P_0\times D$ is the group of $G$-equivariant automorphisms $P\rightarrow P$ which commute with the projection to $X\times U\times D$. This is precisely the space of vertical vector fields $\pi_1=H^0(X\times U, \mathrm{ad}\ P_0)$.
The computation of $\pi_0$ is trickier, let me just write down the answer: $\pi_0=H^1(X\times U, \mathrm{ad}\ P_0)$. The derivation relies on the fact that there are no deformations over an affine scheme, so you just have to pass to an affine cover of $X\times U$ (see Sam Raskin's notes for details).
So, on each affine $U$ mapping to $Bun_G(X)$ the tangent complex has cohomology computed by $\mathbf{R}\pi_*\ \mathrm{ad}\ P_0[1]$, where $\pi: U\times X\rightarrow U$. With some work one can show that the actual tangent complex $f^* T_{Bun_G(X)}$ is quasi-isomorphic to $\mathbf{R}\pi_*\ \mathrm{ad}\ P_0[1]$.
Finally, let me explain why this implies that the tangent complex on $Bun_G(X)$ is $T_{Bun_G(X)}=\mathbf{R} p_*\ \mathrm{ad}\ \mathcal{E}[1]$ for $p:Bun_G(X)\times X\rightarrow Bun_G(X)$. The point is that for nice morphisms of stacks $p$, $\mathbf{R} p_*$ can be defined by the usual pushforward for every base change to an affine scheme. See the proof of proposition 2.1.1 here.
To conclude, the canonical bundle is the inverse of the determinant of the tangent complex, i.e. $K=det(\mathbf{R} p_*\ \mathrm{ad}\ \mathcal{E})$.<|endoftext|>
TITLE: On the existence of a direct summand containing a fixed subgroup
QUESTION [8 upvotes]: Let $G$ be a finite abelian group, and $g_1, \ldots, g_n \in G$ such that the cyclic groups that they generate are in direct sum $\langle g_1 \rangle \oplus \ldots \oplus \langle g_n \rangle$. Is it always possible to find elements $h_1, \ldots, h_n \in G$ and integers $a_1, \ldots, a_n$ such that the following three facts hold
1) $g_i= a_i h_i$, for all $1 \leq i \leq n$,
2) the cyclic subgroups generated by the $h_i$ are in direct sum, $H:=\langle h_1 \rangle \oplus \ldots \langle h_n \rangle$.
3) $H$ is a direct summand of $G$?
(I asked this question on math.stackexchange, see https://math.stackexchange.com/questions/199928/smallest-pure-subgroup-containing-a-fixed-subgroup)

REPLY [10 votes]: I am not convinced that this is true, because the pure subgroup generated by the $g_i$ might
not have the stipulated form as a direct sum of the $h_i$.
Let $G = {\mathbb Z}/16{\mathbb Z} \oplus {\mathbb Z}/4{\mathbb Z}$, $n=1$, and $g_1=(4,2)$.
What could $h_1$ be?
We can prove that there is no such $h_1$ as follows. For any $g \in G$ of order 16, we have $8g = (8,0)$. Since $(8,0) \in \langle h_1 \rangle$, $\langle h_1 \rangle$ cannot be a direct summand of $G$ of order 4. So $\langle h_1 \rangle$ would have to be a direct summand of order 16. But then $4h_1 = (4,0)$ or $(12,0)$ and so $g_1 \not\in \langle h_1 \rangle$.<|endoftext|>
TITLE: Number of 2-dimensional irreducible representations of a finite group ?
QUESTION [25 upvotes]: Question: What is the number of two-dimensional irreducible representations of a finite group ? How it can be expressed in  groups-theoretic terms ? (Number of 1-dimensional irreps is |G/[G,G]| ).
The question is somewhat naive and actually I heard it from our teachers when I was an undergrad many years ago, it was always outspoken with some kind of mysterious flavour - "nobody knows, but may be...". 

Some analogies:
There is a paper by V. Drinfeld 1981, which title is "Number of two-dimensional irreducible representations of the fundamental group of a curve over a finite field". 
The main theorem express the number via the zeta function of the curve over F_q.
(Russian pdf is for free - main formula can be seen from there).
Of course, it is very specific class of the groups, however may be something can be done ?
Another analogy which comes to mind is related to topological quantum field theories,
quantization of Wess-Zumino and Chern-Simons models.
One consider the moduli space of d-dimensional representatation of the fundamental group.
It is natarally symplectic manifold and its VOLUME is somewhat an anologue of the "number"
of irreps for finite group. The volume can be calculated and is related to the famous Verlinde formula.
So, it is of course, both considerations are related to  fundamental (=Galois) groups of CURVES. 
Question: WHY? What makes fundamental (=Galois) groups of curves so specific ? 
Can it be somehow generalized to other classes of curves ?

REPLY [24 votes]: This is not a complete answer in any sense, but I will make a few comments. The irreducible subgroups $G$ of ${\rm GL}(2,\mathbb{C})$ are the primitive ones, which have $G/Z(G)$ isomorphic to $A_{4},S_{4}$ or $A_{5}$, and imprimitive groups, which have an Abelian normal subgroup of index $2.$ On the other hand, any finite group with an Abelian normal subgroup of index $2$ has all its irreducible representations of degree $1$ or $2,$ so the number of $2$-dimensional irreducible representations is easily calculated.
A more careful analysis of the primitive case shows that if $G$ has a faithful $2$-dimensional primitive complex representations, then (as has been known since the late 19th century),
$G = Z(G)E,$ where $ E \cong {\rm SL}(2,3), {\rm GL}(2,3),{\rm SL}(2,5)$ or the binary octahedral group (also, a double cover of order $48$ of $S_{4},$ (as ${\rm GL}(2,3)$ is), but with a generalized quaternion Sylow $2$-subgroup).
Now let $G$ be any finite group, and let $K$ be the intersection of the kernels of the irreducble complex representation of $G$ of degree at most $2$. The above discussion means that the only possible non-Abelian composition factor of $G/K$ is $A_{5}$, though it may be repeated if it appears. The answer to your question only depends on the structure of $G/K,$  so we may reduce to the case that all composition factors of $G$ are cyclic  or $A_{5}.$ Also, by Clifford theory,  we may suppose that the Fitting subgroup $F(G)$ is a direct product of an Abelian group of odd order and a $2$-group, and that all components of $G$ (if there are any) are isomorphic to ${\rm SL}(2,5).$
Further analysis can be carried out, but I believe that the analysis is not entirely straightforward. Perhaps this outline will help others to complete it, so I submit it.<|endoftext|>
TITLE: reference for list of left-regular representations of real associative algebras
QUESTION [5 upvotes]: Suppose $\mathcal{A}$ is a unital associative algebra over $\mathbb{R}$. If we identify $\mathcal{A} = \mathbb{R}^n$ then the $\mathcal{A}$ multiplication corresponds to particular linear maps on $\mathbb{R}^n$. Of course any linear map on $\mathbb{R}^n$ corresponds uniquely to its standard matrix hence we obtain a correspondence between vectors in $\mathcal{A}$ and matrices in $\mathbb{R}^{n \times n}$. These square matrices are known as the left regular representation of the algebra. This is not unique unless we add additional data about the correspondence of the $n$-dimensional algebra and its presentation on $\mathbb{R}^n$.
My favorite examples, $\mathbb{C} = \mathbb{R}^2$ is naturally identified with the subalgebra of $2 \times 2$ matrices of the form:
$$  \left[\begin{array}{cc} a &-b \\ b &a \end{array} \right]  $$
Or the hyperbolic numbers $\mathbb{R}+j\mathbb{R}$ identified with
$$ \left[ {\begin{array}{cc} a & b \\ b & a \end{array}} \right] $$
Or the direct product of $\mathbb{R}$ with $\mathbb{R}$ the element $(a,b)$ is identified with
$$ \left[ \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right] $$
Up to isomorphism the last two examples are actually the same example. I know of one other two-dimensional algebra up to isomorphism.

Question: where can I find a complete tabulation of the low-dimensional left-regular representations of unital algebras? 

I have found many results on google and here, but I can't find one which stands out as a continuation of the list I began at the start of this post. In particular, complete list of complex associative algebra, is great except the base-field is $\mathbb{C}$. If there was a simple theorem that allowed me to extract the list I desire from that list then that would also be a useful answer. But, I'd rather have direct reference for a list of the real associative left regular representations. Ideally this will help me choose a good notation if there already is an agreed notation accepted among those who worked on such classifications.
As always the help of the MO community is greatly appreciated.

REPLY [2 votes]: (Too long for a comment). More (modern and not-so-modern) references, some of them may (partially) contain list(s) you are interested in:

A.A. Albert, Structure of Algebras, AMS, 1939: on page 172 discusses classification of 4-dimensional algebras.
S.C. Althoen, K.D. Hansen, L.D. Kugler, 
A survey of four-dimensional C-associative algebras,
Algebras Groups Geom. 21 (2004), N1, 9-27
R. Ballieu, 
Anneaux finis; systèmes hypercomplexes de rang trois sur un corps commutatif,
Ann. Soc. Sci. Bruxelles Sér. I. 61 (1947), 222-227: presumably contains classification of complex 3-dimensional algebras.
W.A. de Graaf, Classification of nilpotent associative algebras of small dimension,
arXiv:1009.5339.
D. Happel, 
Klassifikationstheorie endlich-dimensionaler Algebren in der Zeit von 1880 bis 1920, 
Enseign. Math. 26 (1980), 91-102 DOI:10.5169/seals-51060 :
A nice historical survey from the modern viewpoint, with a large bibliography.
O.C. Hazlett, 
On the classification and invariantive characterization of nilpotent algebras
Amer. J. Math. 38 (1916), N2, 109-138 http://www.jstor.org/stable/2370262
G. Pickert, Dreidimensionale assoziative nichtkommutative Algebren,
J. Algebra 234 (2000), N2, 280-290 DOI:10.1006/jabr.2000.8550
Scorza, Atti Acad. Sci. Fis. Mat. Napoli 20 (1935), N13 and N14: Classification of 3- and 4-dimensional algebras.
D.A. Suprunenko and R.I. Tyshkevich, Commutative Matrices, 
Acad. Press, 1968 (translation from Russian): On p.61 (of the Russian edition)
there is a discussion of commutative nilpotent algebras of dimension 5.<|endoftext|>
TITLE: Isoperimetric inequality in complex hyperbolic space
QUESTION [9 upvotes]: Let $\mathbb{H}_\mathbb{C}^n$ be n-dimensional complex hyperbolic space. 
This space is a complex analog of hyperbolic space.  It is isometric to the quotient of hyperboloid
$$|z_0|^2-|z_1|^2-\dots-|z_n|^2=1$$
in $\mathbb{C}^{n+1}$
by $S^1$.

Question 1. Is it known that round balls in $\mathbb{H}_\mathbb{C}^n$ minimize the surface area among all bodies of given volume? 

(I am almost sure that the answer is not known.)

Question 2. Was it conjectured somewhere?

REPLY [4 votes]: I am pretty sure it is a somewhat reputed conjecture, but I do not have a clear reference where it is stated. 
Edit: It is stated in Gromov's "Metric structures for Riemannian and non-Riemannian spaces", 6.28 (1/2+) (in fact, it is stated for all rank one symmetric spaces).
It might be evoked in a paper of Hsiang and Hsiang in Inventiones, where they prove that the isoperimetric domains in products of hyperbolic and euclidean spaces are invariant under the group of all isometries fixing the center of gravity. It seems a reasonable conjecture that this is true in all symmetric spaces of non-positive curvature. That conjecture might be stated in the Hsiang and Hsiang paper, and is a broad generalization of the conjecture you are interested in.<|endoftext|>
TITLE: What categories correspond to the typed lambda calculus with parametric types?
QUESTION [5 upvotes]: the unadorned typed lambda calculus correspond to the closed cartesian categories, but if we add in dependent or parametric types how are they then characterised?

REPLY [11 votes]: Dependent type theory, i.e., $\lambda$-calculus with dependent products and dependent sums corresponds to locally cartesian closed categories, which was written up by R. Seely in "Locally cartesian closed categories and type theory", Math. Proc. Phil. Comb. Soc, (1984), 95, 33.
If by "parametric types" you mean the kind of polymorphism that is present in ML, then one possible semantics is that of relationally parametric models. Probably the most influental in the development of this idea was John Reynolds, see for example his "Types, abstraction and parametric polymorphism" from 1983.<|endoftext|>
TITLE: Intuition behind Thom class
QUESTION [48 upvotes]: The Thom class and Thom isomorphism theorem for oriented vector bundles are proven ( at least to my knowledge) by induction on the open covers and some manipulation with Mayer-Vietoris sequences.
What is the "actual reason" behind the existence of Thom class? It seems strange that such an interesting class would exist just because some Mayer-Vietoris sequences routinely produce it.

REPLY [8 votes]: There is a nice formulation and interpretation of the Thom isomorphism in terms of sheaf theory, or "Grothendieck's six functors". The statement of the Thom isomorphism in this setting is that if $f \colon E \to X$ is a rank $n$ orientable vector bundle, then
$$ Rf_! \mathbf Z_E \cong \mathbf Z_X[-n].$$
In this form it is also trivial to prove. Note first that it's clear that $Rf_!\mathbf Z_E$ is concentrated in degree $n$, where it is a rank $1$ local system: the statement is local on $X$, so it is enough to prove this for a trivial bundle. So we should argue that this rank $1$ local system is trivial, i.e. that it has a global section. But a section of this local system is precisely an orientation of the vector bundle; again one argues locally, using that an orientation of the fiber $\mathbf R^n$ is the same thing as a generator of $H^n_c(\mathbf R^n,\mathbf Z)$. 
So why does this global statement on $X$ imply the usual cohomological form of the Thom isomorphism? Let $j \colon E \to \overline E$ be the inclusion into the fiberwise one-point compactification, and $\overline f \colon \overline E \to X$ the projection. For any space $Y$ we let $a^Y$ be the projection from $Y$ to a point. The relative cohomology group $H^\bullet(\overline E, \overline E \setminus E)$ can be computed as $R a^{\overline E}_\ast j_! \mathbf Z_E$ (this is how you compute relative cohomology in general in the six functors language). But
\begin{align*} Ra^{\overline E}_\ast j_! \mathbf Z_E & \simeq Ra^X_\ast R\overline f_\ast j_! \mathbf Z_E && \text{since $a^{\overline E} = a^X \circ \overline f$} \\
 & \simeq Ra^X_\ast R\overline f_! j_!\mathbf Z_E && \text{since $\overline f$ is proper} \\ 
& \simeq Ra^X_\ast Rf_!\mathbf Z_E && \text{since $f = \overline f \circ j$} \\
& \simeq Ra^X_\ast \mathbf Z_X[-n] && \text{by the Thom isomorphism}
\end{align*}
and the final equation manifestly computes $H^{\bullet+n}(X)$.<|endoftext|>
TITLE: Left-Module Structure on the Tensor Product ofTwo Left  Modules
QUESTION [5 upvotes]: Given a noncommutative ring $R$, and two (left) $R$-modules $M$ and $N$, how does one define a left action on the the vector space tensor product $M \otimes N$? Multiplying on just the first factor of the tensor product seems a little unnatrual, but I can't see what else to do.

REPLY [15 votes]: Let $R, S$ be two (unital and associative to be safe) algebras over a commutative ring $k$ and let $M, N$ be respectively a left $R$-module and a left $S$-module. Then we can define the tensor product $M \otimes_k N$ by the usual universal property, and it is naturally a left $R \otimes_k S$-module by functoriality. If $M, N$ are both left $R$-modules, then $M \otimes_k N$ is a left $R \otimes_k R$-module, so defining a left $R$-module structure on the tensor product which is compatible with its $k$-module structure is tantamount to giving a morphism of $k$-algebras
$$\Delta : R \to R \otimes_k R.$$
Note that any such morphism gives a left $R$-module structure to the tensor product of two left $R$-modules, but it won't behave the way you expect it to with respect to multiple tensor products unless $\Delta$ is coassociative and cocommutative. I will be ignoring this.
Now, the ring $R \otimes_k R$, by its universal property, naturally admits two morphisms $R \to R \otimes_k R$, namely the two inclusions $i_1, i_2 : R \to R \otimes_k R$, which correspond to the two $R$-module structures described by Peter Samuelson in the comments. If you don't want to just provide $\Delta$ as extra data, then your question can be reinterpreted as follows: 

What other natural morphisms $R \to R \otimes_k R$ are there?

Suppose that $\Delta_R : R \to R \otimes_k R$ is a family of morphisms which is natural in $R$; that is, it defines a natural transformation. Applying the forgetful functor to $\text{Set}$, we get a natural transformation from the forgetful functor $U : k\text{-Alg} \to \text{Set}$ to the functor
$$U^2 : k\text{-Alg} \ni R \mapsto U(R \otimes_k R) \in \text{Set}.$$
The functor $U$ is representable by the $k$-algebra $k[x]$, so by the Yoneda lemma, natural transformations $U \to U^2$ can be naturally identified with elements of
$$U^2(k[x]) \cong U(k[x, y]).$$
More concretely, any polynomial in two variables $f(x, y) = \sum f_{ij} x^i y^j \in k[x, y]$ over $k$ induces a natural map of sets
$$R \ni r \mapsto \sum f_{ij} r^i \otimes r^j \in R \otimes_k R$$
and these are the only such natural maps. Now the question reduces to the following:

Which of the maps $r \mapsto \sum f_{ij} r^i \otimes r^j$ are always morphisms of $k$-algebras? 

The short answer is that it depends a lot on $k$. For starters, letting $R = k[x], r = x, R \otimes_k R = k[x, y]$ and requiring compatibility with scalar multiplication gives
$$cr \mapsto c f(x, y) = f(cx, cy) \in k[x, y]$$
for all $c \in k$. Comparing coefficients of $x^i y^j$ gives
$$(c - c^{i+j}) f_{ij} = 0$$
for all $i, j$ and for all $c$. This suggests the possibility of other natural morphisms $R \to R \otimes_k R$ in characteristic $p$. In particular, if $k = \mathbb{F}_p$ then we have $c^{p^r} = c$ for all $r \ge 0$, hence $f$ can have components of degree $p^r$ and some of these give rise to genuine natural morphisms such as
$$R \ni r \mapsto r^p \otimes 1 \in R \otimes_k R.$$
Assume for simplicity that $k$ is an integral domain of characteristic $0$ (these hypotheses can probably be considerably weakened). Then the above condition implies that $f$ is homogeneous and linear, so our desired morphism can only have the form
$$R \ni r \mapsto f_{10} r \otimes 1 + f_{01} 1 \otimes r \in R \otimes_k R.$$
Compatibility with multiplication now requires (taking $R = k[x, y], R \otimes_k R = k[x, y, z, w]$)
$$xy \mapsto f_{10} xy + f_{01} zw = (f_{10} x + f_{01} z)(f_{10} y + f_{01} w)$$
and comparing coefficients gives
$$f_{10}^2 = f_{10}, f_{01}^2 = f_{01}, f_{10} f_{01} = 0.$$
Our hypotheses then imply that exactly one of $f_{10}$ or $f_{01}$ can be equal to $1$ and the other must be equal to $0$. (If $k$ is not an integral domain then other things can happen; in general a choice of $f_{01}, f_{10}$ above is equivalent to a direct product decomposition $k = k_1 \times k_2$ with respect to which $f_{10} = (1, 0), f_{01} = (0, 1)$.) In other words, when $k$ is an integral domain of characteristic $0$ then the only natural morphisms $R \to R \otimes_k R$ are
$$R \ni r \mapsto r \otimes 1 \in R \otimes_k R$$
resp. 
$$R \ni r \mapsto 1 \otimes r \in R \otimes_k R$$
and the corresponding natural left $R$-module structures on $M \otimes_k N$ are given by multiplication on the first resp. the second factor. In particular, 

If $k$ is a field of characteristic $0$, then the only natural morphisms $R \to R \otimes_k R$ are the two obvious ones.

Group algebras are highly misleading here: any group algebra $k[G]$ (more generally any monoid algebra $k[M]$) is naturally equipped with a comultiplication given by extending
$$\Delta : k[G] \ni g \mapsto g \otimes g \in k[G] \otimes_k k[G]$$
and $\Delta$ induces the usual tensor product of representations of $G$ over $k$. It is worth mentioning that this map is not defined using any of the structure maps of the group $G$; more generally, for any set $S$ there is a canonical diagonal map $\Delta : S \to S \times S$ given by sending $s$ to $(s, s)$, and this induces a canonical coalgebra structure on the free module $k[S]$ given by extending
$$k[S] \ni s \mapsto s \otimes s \in k[S] \otimes_k k[S].$$
But a generic $k$-algebra is not a free vector space over any distinguished set.<|endoftext|>
TITLE: Surfaces that can be rolled by a ball
QUESTION [7 upvotes]: Let $S$ be a smooth solid body in $\mathbb{R}^3$,
and $B$ a ball of radius $r$.
Say that $B$ is in contact with $S$ if 
(1) they share a point $x$
that is on the surface of each, 
$x \in \partial S$ and $x \in \partial B$,
and (2) neither penetrates the other,
$\operatorname{int} S \cap \operatorname{int} B = \varnothing$.
Say that $S$ can be rolled by $B$ if,
for every pair of points $x,y$ on the surface of $S$, there
is a path $\rho$ connecting $x$ to $y$ on $\partial S$
such that $B$ can be placed in contact at every point $z \in \rho$.
In other words, the ball $B$ can roll between any pair of points of
the surface without ever penetrating $S$.

          


My question is:

Characterize the bodies $S$ that can be rolled by a ball of radius
    $r$.

A necessary condition is that the Gaussian curvature at any point 
on the surface of
$S$ must be $\ge -\frac{1}{r^2}$.
Could that also be sufficient, or are there global obstructions
such that $B$ could be in local contact but penetrate away from
the contact point?
Is the characterization different for surfaces of genus zero than
for surfaces with handles?
Any insights, speculations, or literature pointers would be
appreciated.
Thanks!

Update.  Here is my interpretation of Anton's example:

REPLY [3 votes]: The property you are asking for is called global radius of curvature. The notion combines the condition of lower bound on the principal curvatures as well as the global condition of no interpenetration of a neightborhood. This notion is very well adapted to the calculus of variations.
The heuristic definition from the paper of P. Strzelecki and H. von der Mosel: Global curvature for surfaces and area minimization under a thickness constraint

The main idea can be sketched as follows: Take a continuous parametric surface $X : \mathbb{R}^2\supset \mathbb{B}^2 \to \mathbb{R}^3$ (with possibly infinite area) which possesses a tangent plane on a dense subset $G \subset \mathbb{B}^2$ which may even have zero measure. Consider the radii of all spheres touching the surface  $X(\mathbb{B}^2)$ in one of these points $X(\omega)$,  $\omega \in G$, and containing at least one other point of the surface. We define the infimum of these radii as the global radius of curvature $\Delta[X]$ of the surface $X$ . It turns out that a positive lower bound on $\Delta[X]$ serves as an excluded volume constraint for the surface $X$.<|endoftext|>
TITLE: Extending Jordan loops
QUESTION [5 upvotes]: I encountered this issue recently, but do not know of any general results to deal with it, so I would appreciate any pointers. 

Let $\mathbb T=\{z\in\mathbb C\mid |z|=1\}$, and let $f:\mathbb T\to\mathbb C$ be continuous and injective, so its image $\mathbb T'$ is a Jordan loop. Under what (general) conditions can we ensure that there is a homeomorphism between the unit disc and the interior of $\mathbb T'$ whose extension to the boundary is $f$? 

Moreover, if there are reasonable conditions that ensure this, and $f$ is $C^\infty$, can we further require some nice regularity (perhaps even $C^\infty$) of the extension as well?

REPLY [8 votes]: What you're asking is equivalent to asking whether any homeomorphism $g : S^1 \rightarrow S^1$ can be extended to a homeomorphism of the disc.  This is easy -- write the disc in polar coordinates $(t,\theta)$ with $\theta \in S^1$, and define an extension $G(t,\theta) = (t,g(\theta))$.
The question about whether this can be done smoothly if $g$ is smooth is more subtle.  Observe that the above also works for $S^k$ with $k > 1$.  The smooth version fails in higher dimensions and is responsible for the existence of exotic spheres.  However, for $k=1,2$ there is no problem.  For $k=1$, this is a theorem of Smale; see
Smale, Stephen 
Diffeomorphisms of the 2-sphere. 
Proc. Amer. Math. Soc. 10 1959 621–626.
For $k=2$, it is a much deeper theorem of Hatcher; see
Hatcher, Allen E.
A proof of the Smale conjecture, Diff(S3)≃O(4). 
Ann. of Math. (2) 117 (1983), no. 3, 553–607.<|endoftext|>
TITLE: is there a universal property that characterises the join of two categories?
QUESTION [5 upvotes]: Let A & B be two categories, the join A*B is created by stipulating its class of object is the  disjoint union of the objects of A & B, the morphisms remain the 'same', but we throw in an extra morphism for every object a in A, and b in B.
that is:
A*B[a,a']=A[a,a'] if a,a' are in A
A*B[b,b']=B[b,b'] if b,b' are in B
A*B[a,b]=1 if a is in A, and b in B
A*B[b,a]=0 if b is in B, and a in A
it seems like a pretty ad-hoc construction (its obviously based on a construction coming from algebraic topology), is there a more categorical way of defining this?

REPLY [9 votes]: It's a special case of what's called a collage or cograph construction. Recall that a profunctor or bimodule between categories $B$, $A$ is a functor $R: A^{op} \times B \to Set$. The cograph of $R$ is the category $\bar{R}$ where $Ob(\bar{R}) = Ob(A) \sqcup Ob(B)$, and where $\bar{R}(a, a') = A(a, a')$ if $a, a' \in Ob(A)$, $\bar{R}(b, b') = B(b, b')$ if $b, b' \in Ob(B)$, $\bar{R}(a, b) = R(a, b)$ if $a \in Ob(A), b \in Ob(B)$, and $\bar{R}(b, a) = \emptyset$ if $a \in Ob(A), b \in Ob(B)$. Compositions and identities are the obvious ones. 
A cograph of the terminal object in the category of bimodules from $B$ to $A$ is the join of $A$ and $B$. In turn, the cograph of $R$ is a lax colimit, in the bicategory of small categories, bimodules, and bimodule homomorphisms of the diagram consisting just of $R$ itself. The nLab is a good source of information on this (as it is for many categorical questions).<|endoftext|>
TITLE: Is the following invariant of rooted trees a complete invariant?
QUESTION [13 upvotes]: Recall that rooted trees may be generated by starting with a trivial rooted tree (just a vertex), along with the operations of grafting a number of trees (identify their roots) and adding a new vertex to the tree to be a new minimum element. We will call this second operation "leafing".
Now let us define an invariant of rooted trees. If $T$ is a rooted tree, we will denote $P_T(z)$ to be the associated polynomial.

If the number of edges of $T$ is zero, then $P_T(z)=1$.
If $T'$ is the leafing of $T$, then $P_{T'}(z)=(z+1)P(z)+1$.
If $T$ is the grafting of $T_i, i=1\ldots n$, then $P_T(z)=P_{T_1}(z)P_{T_2}(z)\ldots P_{T_n}(z)$.

This polynomial is an isomorphism invariant of rooted trees. My question is

If $P_T=P_{T'}$, are the rooted trees, $T,T"$ isomorphic? If these trees are not isomorphic, what is the smallest counterexample? Any references to this invariant would be appreciated.

REPLY [3 votes]: The number of different values taken by the polynomial is given by
1, 1, 2, 4, 9, 20, 47, 112, 274, 679, 1717, ...
Comparing with the sequence A000081 given by
1, 1, 2, 4, 9, 20, 48, 115, 286, 719, 1842, ...
one can easily see that this is not at all a complete invariant.<|endoftext|>
TITLE: Coherent MU_*-Modules
QUESTION [7 upvotes]: It is proven by Thom that for a finite cw-complex $X$, its $MU$-homology, which, in honor of the authors I'm currently reading, I'll denote by $\Omega_\ast^U(X)$, is a coherent module over $\Omega_\ast^U$ if $\Omega_\ast^U(X)$ has projective dimension 0 or 1 over $\Omega_\ast^U$.  It is stated that in a series of lecture notes by Larry Smith that this result can probably be extended to other complexes (and spectra...).  However, these lecture notes are from 1970.  Does anyone know if this result has been fully generalized? I guess I mean, is it known precisely how far this result can be extended?
Thanks!

REPLY [7 votes]: The question is also addressed in Lecture 5 of J.F. Adams ``Lectures on generalized cohomology'' in Springer Lecture Notes in Mathematics Vol 99(1969).  Again ancient, but none the worse for that. 
The paper is reprinted in Volume I of "The selected works of J. Frank Adams''.<|endoftext|>
TITLE: Characterization of Stone-Cech compactifications
QUESTION [14 upvotes]: Suppose I have an infinite discrete topological space $X$ of cardinality $\kappa$. Then I know some things about the Stone-Cech compactification, $\beta X$: it is Hausdorff and compact but not sequentially compact, has a basis of clopen sets, etc. My question is the following: is there a "nice" characterization of the spaces $Y$ which are homeomorphic to the Stone-Cech compactification of a discrete space? Certainly, the term "nice" is vague; I have in mind characterizations only using terms from a standard text on point-set topology, but I would consider as an answer to this question really any nontrivial characterization of Stone-Cech compactifications. 
I am particularly interested in nice characterizations that require some set theory, such as "assuming $V=L$, $Y$ is homeomorphic to $\beta X$ for some discrete $X$ iff $Y$ is compact, not sequentially compact, and has a basis of clopen sets" (although I'm certain that statement is extremely false), and I would especially like to know whether there are two incompatible strong set-theoretic assumptions which yield distinct nice characterizations. The only relevant result I know is along these lines: in 1963, Parovicenko showed that assuming CH, the only Parovicenko space (which has a long but elementary definition*) is $\beta\mathbb{N}-\mathbb{N}$; this can be molded into a characterization of $\beta\mathbb{N}$, assuming CH, but says nothing about whether a space is the Stone-Cech compactification of a discrete space of uncountable cardinality. In 1978, van Douwen and van Mill showed that CH was necessary. One more concrete sub-question I have, then, is:
Does Parovicenko's result generalize in some way to characterize Stone-Cech compactifications of larger discrete spaces? If so, how much set theory is needed - is GCH enough?
(One very tempting way to try to rephrase Parovicenko's result is to define "$\kappa$-Parovicenko space" by taking the definition of Parovicenko space and replacing the "weight $c$" condition with "weight $2^\kappa$," and then claiming that - assuming GCH - every $\kappa$-Parovicenko space is homeomorphic to $\beta X-X$ for a discrete space $X$ of cardinality $\kappa$. However, I see absolutely no reason to believe this. A sub-subquestion: is this statement obviously false?)

*For completeness, a Parovicenko space is a topological space which is compact and Hausdorff, has no isolated points, has no nonempty $G_\delta$ set with empty interior, has no two disjoint $F_\sigma$ sets with non-disjoint closures, and has weight $c=2^{\aleph_0}$ - that is, every basis has cardinality $\ge c$, and there is some basis with cardinality $c$.

REPLY [11 votes]: I confirme my comment :
$X$ is the stone-cech compactification of a discrete space if and only if $X$ is compact, haussdorf, extremally disconected, and has a dense set of open points.
here is a sketches of the proof :
If X is a stone-chech compactification of a discret set Y, then it is clear that X is compact, hausdorf, the point of Y form a dense set of open points, and it is well know that X is extremally disconected.
Asume now that $X$ is a topological space satisfying all those hypothesis.
Let $Y$ be the set of open point of $X$.
It's a routine to check to see that the map which to a subset $P$ of $Y$ associate it's closure in $X$, and the map which to a clopen of $X$ associate it's intersection with $Y$, are reciprocal bijection between the parts of $Y$ and the clopen set of $X$.
considere now a point $x \in X$, then {x} is the intersection of clopen set containing $X$, and the set of clopen of $X$ containing $x$ correspond through  the previous bijection to an ultrafilter on $Y$.
After that, consider an ultrafilter $\mathcal{F}$ on $Y$, you can see that $\displaystyle \bigcap_{P \in \mathcal{F}} \overline{P} $ is a singleton (it contains a point because it is an intersection of non-empty compact, and it can't contain two point because of the properties ultrafilter).
those two application will induce an homeomorphism between $X$ and the space of ultrafilter of $Y$.<|endoftext|>
TITLE: Space of solutions of nonlinear Helmholtz equation on a torus
QUESTION [8 upvotes]: On a unit torus $T^n$ (or equivalently, on $\mathbb{R}^n$ with periodic boundary conditions), the linear Helmholtz equation:
$\nabla^2 \phi + k^2 \phi=0$
will have no non-trivial solutions for generic values of $k$, while for special values of $k$ it will have a finite-dimensional vector space of solutions, with a basis of functions:
$\phi_m(x)=\exp(2\pi i m_j x^j)$
for each $n$-tuple of integers $m=(m_1,m_2,...m_n)$ such that:
$m_1^2+m_2^2+...+m_n^2=\left(\frac{k}{2\pi}\right)^2$
Now, consider a non-linear version of this problem, such as:
$\nabla^2 \phi + k^2 \phi + \lambda \phi^3=0$
still on $T^n$.  The solutions will no longer comprise a vector space, but rather a manifold.
My question is:  how can I determine the cardinality of the dimension of that manifold?  Will it again be zero-dimensional generically and finite for some special parameter values, or will the behaviour change?

REPLY [6 votes]: Ignoring boundary conditions, the PDE has solutions $\phi = a\; \text{sn}\left(b x, c\right)$ where $\text{sn}$ is the Jacobi SN function (in Maple's parametrization: note that Mathematics uses a different convention),  ${a}^{2}=2\;{\dfrac {{b}^{2}-{k}^{2}}{\lambda}}$, ${c}^{2}={\dfrac {{k}^
{2}}{{b}^{2}}}-1$.  These are periodic in $x$, with a period depending on $b$ and $c$.  For any given $k$ and $\lambda$ (at least in some region) I believe there should be infinitely many values of $a,b,c$ for which the period divides $2 \pi$. For example, with $k=2$ and $\lambda=1$ I get a period dividing $2 \pi$ for $c = 2.722857918$, $3.502129242$, $4.303773851$, $5.116503598$, etc.  Of course we can replace $bx$ by $\sum_j b_j x^j$ with $\sum_j b_j^2 = b^2$.   
EDIT:  We can think about it this way. Assume $k, \lambda > 0$.
By scaling $y$ and time $t$ we can non-dimensionalize the autonomous differential equation $\ddot{y} + k^2 y + \lambda y^3 = 0$ to $\ddot{y} +  y +  y^3 = 0$.
The phase-plane trajectories of this autonomous differential equation  are the closed curves $\dfrac{v^2}{2} + \dfrac{y^2}{2} + \dfrac{y^4}{4} = C$ for $C > 0$, where $v = dy/dt$.  The period $P$ is of the orbit through $(y=y_0, v=0)$ is, by symmetry, $4$ times the time needed to get from $(y_0,0)$ to $(0,\sqrt{ y_0^2 + y_0^4/2})$, and thus
$$ P = 4 \int_0^{y_0} \dfrac{dy}{\sqrt{ y_0^2 - y^2 +  y_0^4/2  - y^4/2}}$$
Under the change of variables $y = s y_0$ this becomes
$$ P = 4 \sqrt{2} \int_0^1 \dfrac{ds}{\sqrt{2 - 2 s^2 + y_0^2 - y_0^2 s^4}} $$
which goes to $0$ as $y_0 \to \infty$.<|endoftext|>
TITLE: Top cohomology of resolution of singularities
QUESTION [5 upvotes]: Let $X$ be a projective variety over $\mathbb C$ of dimension $n$.  Let $\tilde{X} \to X$ be a resolution of singularities.  Suppose that $H^n(\tilde{X}, \mathcal O_{\tilde{X}}) = 0$.  What can we say about $H^n(X, \mathcal O_X)$?  When is it $0$?

REPLY [2 votes]: A very easy counterexample: let $X$ be a nodal cubic in $\mathbb{P}^3$. Then the resolution of singularities of $X$ is $\bar X = \mathbb{P}^1$, so $H^1(\bar X, \mathcal{O}_{\bar X}) = 0$. On the other hand, the exact sequence of sheaves on $\mathbb{P}^2$
$$ 0\to \mathcal{O}(-3)\to \mathcal{O} \to \mathcal{O}_X \to 0 $$
shows that $H^1(X, \mathcal{O}_X)$ is one-dimensional.<|endoftext|>
TITLE: [Reference Request] The Definition of Adjoint Functors between dg-categories
QUESTION [7 upvotes]: Let $A$ and $B$ be two dg-categories, $F: A \rightarrow B$ and $G: B \rightarrow A$ are two functors. Then what is the definition that $F$ and $G$ form an adjoint pair?
In my mind $F\dashv G$ requires that there is a quasi-isomorphism between the cochain complext $\text{Hom}_B(Fx, y) $ and $\text{Hom}_A(x, Gy)$ for any $x\in \text{Obj}(A)$ and  $y\in \text{Obj}(B)$. Can we make it more precise?

REPLY [3 votes]: A good notion of adjointness in the dg-world should perhaps involve derived categories and quasi-functors.
Definition: A quasi-functor $F: \mathcal A \to \mathcal B$ is a dg-bimodule $F \in \mathcal A \text{-dgm-} \mathcal B$ (written $F=F(B,A)$, covariant in $A$ and contravariant in $B$), such that for all $A \in \mathcal A$, $F(-,A)$ is quasi-isomorphic (that is, isomorphic in the derived category $\mathbf{D}(\mathcal B)$) to a representable right $\mathcal B$-dg-module. As a matter of notation, I will sometimes write $F(A)$ as a shorthand for $F(-,A)$.
Note that any bimodule $F \in \mathcal A \text{-dgm-} \mathcal B$ can be viewed as a dg-functor $F: \mathcal B^{\text{op}} \otimes \mathcal A \to \mathbf{C}_{\text{dg}}(k)$, where $\mathbf{C}_{\text{dg}}(k)$ is the dg-category of cochain complexes over our ground commutative ring $k$. So, $F \in \text{dgm-}(\mathcal B \otimes \mathcal A^{\text{op}})$, a right $\mathcal (\mathcal B \otimes \mathcal A^{\text{op}})$-dg-module.
The category $\mathrm{rep}(\mathcal A,\mathcal B)$ of quasi-functors is defined to be the full subcategory of the derived category $\mathbf D(\mathcal B \otimes \mathcal A^{\text{op}})$ whose objects are the quasi-functors $\mathcal A \to \mathcal B$. Composition of quasi-functors $F: \mathcal A \to \mathcal B$ and $G: \mathcal B \to \mathcal C$ is obtained by the derived tensor product:
\begin{equation}
G \circ F = F \otimes_{\mathcal B}^{\mathbb L} G.
\end{equation}
As expected, given a quasi-functor $F: \mathcal A \to \mathcal B$, one has "composition functors" such as
\begin{equation*}
F^* : \mathrm{rep}(\mathcal B, \mathcal C) \to \mathrm{rep}(\mathcal A,\mathcal C).
\end{equation*}
Now, let $\tau : F \to G$ be a morphism of quasi-functors $\mathcal A \to \mathcal B$. We can speak of its "components" $\tau_A : F(A) \to G(A)$ (morphisms in $\mathbf{D}(\mathcal A)$), for any object $A \in \mathcal A$. In fact, an object $A \in \mathcal A$ can be viewed as an element of $\mathrm{rep}(\mathbf 1 , \mathcal A)$, where $\mathbf 1$ is the dg-category with an object and endomorphism ring $k$: we have the derived Yoneda embedding
\begin{equation*}
h_{-} : H^0(\mathcal A) \hookrightarrow \mathbf{D}(\mathcal A),
\end{equation*}
and we may view $h_A$ as a bimodule, $h_A \in \mathbf{1}\text{-dgm-}\mathcal A$. Next, recall that, given $F: \mathcal A \to \mathcal B$ any quasi-functor (or even bimodule), we have
\begin{equation*}
h_A \otimes_{\mathcal A}^{\mathbb L} F = h_A \otimes_{\mathcal A} F = F(A),
\end{equation*}
without the need of taking the derived tensor product, since $h_A$ is cofibrant as a right $\mathcal A$-module. In other words:
\begin{equation*}
h_A^*(-) = h_A \otimes_{\mathcal A} - : \mathrm{rep}(\mathcal A, \mathcal B) \to \mathrm{rep}(\mathbf 1,\mathcal B),
\end{equation*}
and this functor sends a morphism $\tau: F \to G$ to what we call its component $\tau_A : F(A) \to G(A)$, as needed.
Finally, I may write down my definition of adjoint quasi-functors. If $F: \mathcal A \to \mathcal B$ and $G: \mathcal B \to \mathcal A$ are quasi-functors, then we could say that $F \dashv G$ if there is a morphism of quasi functors $\varepsilon : 1 \to GF$ such that the following composition
\begin{equation}
\mathrm{rep}(\mathbf 1, \mathcal B)(F(A), B) \xrightarrow{G_*} \mathrm{rep}(\mathbf 1, \mathcal A)(GF(A), G(B)) \xrightarrow{\mathrm{Hom}(\varepsilon_A, G(B))} \mathrm{rep}(\mathbf 1, \mathcal A)(A, G(B))
\end{equation}
is an isomorphism, for all $A \in \mathcal A, B \in \mathcal B$.
where $A \in \mathcal A$ and $B \in \mathcal B$ are again shorthands for $\mathcal A(-,A) \in \mathrm{rep}(\mathbf 1, \mathcal A)$ and $\mathcal B(-,B) \in \mathrm{rep}(\mathbf 1, \mathcal B)$.<|endoftext|>
TITLE: When do maximum and expectation commute?
QUESTION [9 upvotes]: Hi, I'm looking for conditions on $G(t,x)$ such that
$$
\sup\limits_{t\in [0,1]}E[G(t,X)]=E[\sup\limits_{t\in [0,1]}G(t,X)]
$$
where $X$ is a random variable (it's easy to see that $\sup\limits_{t\in [0,1]}E[G(t,X)]\leq E[\sup\limits_{t\in [0,1]}G(t,X)]$).
Any suggestion or reference is greatly appreciated!

REPLY [15 votes]: This will require very strong conditions on $G$. The most general result I know of is an "almost upward-filtering" condition, fairly well known in stochastic optimal control theory: Assume $G(t,\cdot)$ is measurable for each $t \in [0,1]$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\text{ess}\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $\epsilon > 0$ and $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $\mathbb{E}[(G(t,X) - G(s,X) \vee G(r,X))^-] \le \epsilon$. I believe this result is originally due to J.A. Yan, in the hard-to-find paper "On the commutability of essential infimum and conditional expectation operators". Note that $\sup_{t \in [0,1]}G(t,X)$ need not be measurable, which is why the essential supremum is used in the aforementioned theorem.
A more transparent condition can be derived from the above if we add a continuity assumption: Assume $G(\cdot,x)$ is continuous for each $x$, $G(t,\cdot)$ is measurable for each $t$, and  $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if there exists $T \in [0,1]$ such that $G(T,X) = \sup_{t \in [0,1]}G(t,X)$ almost surely.
PROOF: The continuity assumption guarantees that $\sup_{t \in [0,1]}G(t,X)$ is indeed measurable (e.g. by Theorem 18.19 of Aliprantis & Border), and thus $\text{ess}\sup_{t \in [0,1]}G(t,X) = \sup_{t \in [0,1]}G(t,X)$. The aforementioned theorem and a simple argument using compactness of $[0,1]$ and Fatou's lemma shows that (under our assumptions) $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $G(t,X) \ge G(r,X) \vee G(s,X)$ a.s.. Since the "if" part is trivial, we now prove the "only if". Consider the set $S := \{G(t,X) : t \in [0,1]\}$ with the partial order given by almost sure inequality. Compactness of $[0,1]$ and continuity of $G(\cdot,x)$ for all $x$ yield the existence of an upper bound in $S$ for any chain of $S$, and thus by Zorn's lemma $S$ contains a maximal element. That is, there exists $T \in [0,1]$ such that there is no $s \in [0,1]$ for which $G(s,X) \ge G(T,X)$ a.s. and $P(G(s,X) > G(T,X)) > 0$. For any $t \in [0,1]$ there exists $r \in [0,1]$ such that $G(r,X) \ge G(T,X) \vee G(t,X) \ge G(T,X)$ a.s., which implies $G(r,X) = G(T,X)$ a.s. and thus $G(T,X) \ge G(t,X)$ a.s.. Hence $G(T,X) \ge G(t,X)$ a.s. for any $t \in [0,1]$.<|endoftext|>
TITLE: Finite Subgroups of $SL_2(R)$
QUESTION [5 upvotes]: Can you show any finite subgroup of $SL_2(R)$ is cyclic without using an invariant form?

REPLY [4 votes]: Here is an answer using character theory.  For an irreducible character $\chi$ of a finite group $G$, define $\nu_2(\chi)$ to be $1$ if $\chi$ is afforded by a real representation, $0$ if $\chi$ is not real valued, and $-1$ otherwise.  Let $t$ be the number of elements of order two in $G$.  As shown in Chapter 4 of Martin Isaacs' ``Character Theory of Finite Groups", the sum of $\nu_2(\chi)\chi(1)$ over all irreducibles of $G$ is $1+t$.
Now assume for contradiction that $G$ is a counterexample of minimal order to the claim that every finite subgroup of $SL_2({\mathbb R})$ is cyclic.  So, every proper subgroup of $G$ is cyclic.  (I believe that such groups were classified in a 1903 paper of Miller and Moreno, but we will not need that.)
As a group of odd order has no irreducible character of even degree and every reducible finite subgroup of $SL_2({\mathbb C})$ is cyclic, $|G|$ is even and and $G$ acts irreducibly on ${\mathbb C}^2$.  As $SL_2({\mathbb R})$ has a unique element of order two, so does $G$.  Therefore, if $G$ is a $2$-group, then $G$ is quaternion of order eight, as follows from an analysis of $2$-groups with a cyclic subgroup of index two.  However, the group $Q_8$ has four linear characters, each of which satisfies $\nu_2(\chi)=1$.  Since $Q_8$ has one element of order two, the unique $2$-dimensional irreducible of $Q_8$ must satisfy $\nu_2(\chi)=-1$, contradicting our assumption.
We assume now that $G$ is not a $2$-group.  A Sylow $2$-subgroup $S$ of $G$ is cyclic by our minimality assumption.  Thus $Aut(S)$ is a $2$-group and $S$ is central in its normalizer $N_G(S)$.  By Burnside's normal $p$-complement theorem, $G$ has a normal subgroup $N$ of index $|S|$.  We know that $N$ is cyclic.  $S$ normalizes every Sylow subgroup of $N$.  Since $G$ is not cyclic, $S$ does not centralize $N$ and therefore there is some Sylow $p$-subgroup $P$ of $N$ that is not centralized by $S$.  By minimality of $|G|$, $G=SP$.  An analysis of $Aut(P)$ shows that $S$ does not centralize the unique subgroup of order $p$ in $P$.  By minimality of $|G|$, $|P|=p$.
Some element $s$ of $S$ acts as an automorphism of order two on $P$ and therefore inverts a generator of $P$.  By minimality of $|G|$, $S=\langle s \rangle$.  Now $s^2$ is central in $G$ and $\langle s^2,P \rangle$ is abelian and has index two in $G$.  Therefore, every irreducible of $G$ has degree at most two.  If $|s|=2$ then $G$ is dihedral and has $p$ elements of order two, giving a contradiction.  If $|s|>4$ then $G$ cannot have a faithful real irreducible character $\chi$ of degree two, as the image of $s^2$ must be a scalar matrix in any irreducible representation of $G$.  Therefore, $|S|=4$.
Since $[G,G]=P$, $G$ has four linear characters and $p-1$ irreducible characters of degree two, as the sum of the squares of the degrees of the irreducible characters is $|G|=4p$. Two of the linear characters satisfy $\nu_2(\chi)=1$ and the other two satisfy $\nu_2(\chi)=0$.
The quotient $G/\langle s^2 \rangle$ is dihedral of order $2p$ and therefore has $(p-1)/2$ characters of degree two, each of which satisfies $\nu_2(\chi)=1$ (either by direct construction or by counting involutions).  Each of these characters lifts to a (nonfaithful) irreducible of $G$.  Since $G$ has a unique element of order two, the remaining $(p-1)/2$ degree two irreducibles of $G$ must satisfy $\nu_2(\chi)=-1$.  Therefore, $G$ has no faithful real irreducible representation and our proof is complete.<|endoftext|>
TITLE: The Odds 3 (or More) Group Elements Commute
QUESTION [8 upvotes]: Some time ago I asked about the odds 2 group elements commute.  I wonder about the odds that 3 group elements commute.  Is there a "closed" formula for the sum
$$ \frac{1}{|G|^3} \sum_{g,h,k} \delta([g,h]=1)\delta([h,k]=1)\delta([k,g]=1)$$
One approach, as mentioned in Kefeng Liu, might be to use the "Heat Kernel" for finite groups.
$$ H(t,x,y) = \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \mathrm{dim}(\lambda) \chi_\lambda(xy^{-1}) e^{-t f(\lambda)}$$
If I'm not mistaken $f(\lambda)$ is the quadratic Casimir, but not sure.  Really, for $t=0$ it reduces to the group theory identity:
$$ \delta(xy^{-1})=   \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \chi_\lambda(x) \overline{\chi_\lambda(y)}  = \frac{1}{|G|} \sum_{\lambda \in \mathrm{Irr}(G)} \chi_\lambda(1) \chi_\lambda(xy^{-1}) $$

REPLY [5 votes]: This question is studied in the article Isoclinism classes and commutativity degrees of finite groups by P. Lescot. The probability that $n+1$ elements of $G$ pairwise commute is called there the $n$-th commutativity degree $d_n(G)$. A kind of recursive formula for $d_n(G)$ in terms of $d_{n-1}$ of centralizers is proved (Lemma 4.1). Lescot also proves that if $G$ is not abelian then $d_n(G) \leq \frac{3 \cdot 2^n - 1}{2^{2n+1}}$ with equality if and only if $G$ is isoclinic to the quaternion group $Q_8$.<|endoftext|>
TITLE: Diophantine Equation with Polynomial Coefficients
QUESTION [8 upvotes]: DISCLAIMER: I'm primarily a graph theorist and am fairly inept when it comes to classical number theory.
Recently I have been looking at the possibility (or impossibility) of embedding various graphs as Euclidean distance graphs in $\mathbb{Q}^3$.  Frequently, constructions I am considering lead to questions of the following form:
Given polynomials $p_1(n), p_2(n), p_3(n)$ with integer coefficients, does there exist $n \in \mathbb{Z}$ such that the equation
$p_1(n)x^2 + p_2(n)y^2 + p_3(n)z^2 = 0$
has a non-trivial rational solution?
I can sometimes show for specific polynomials that the answer is NO by using criteria given by Legendre in the 1700s along with repeated use of the Law of Quadratic Reciprocity.  However, generalized methods for showing the existence of an $n$ leading to non-trivial rational solutions of the above equation are something I am not familiar with.  If someone can point me in the right direction or even just shed a little light on the subject, I would be most appreciative.
If a much more tailored question is to your taste, consider this:
Let $r$ be a square-free positive integer and let $a, b, c \in \mathbb{Z}$ such that $a^2 + b^2 + c^2 = r$.  Characterize the values of $r$ such that for some $n \in \mathbb{Z}^+$ the equation 
$rx^2 + y^2 -(12n^2 - 1)(a^2 + b^2)z^2 = 0$
has a non-trivial rational solution.

REPLY [4 votes]: Let $m^2 = n^3+An^2 + Bn + C$ be any elliptic curve.
Your question includes the question: For which values of $n$ does $y^2 = (n^3+An^2 + Bn + C) x^2$ have a solution or, in other words, "find the integer points on  $m^2 = n^3+An^2 + Bn + C$." There are excellent algorithms for this in practice, but none that are proved to work in all cases. So your question probably doesn't have an answer which is better than this.<|endoftext|>
TITLE: Collisions between rooks taking random flights on an N by M chessboard
QUESTION [7 upvotes]: I randomly place $k$ rooks on an (arbitrarily sized) $N$ by $M$ chessboard.  Until only one rook remains, for each of $P$ time intervals we move the pieces as follows:
(1) We choose one of the $k$ rooks on the board with uniform probability.  
(2) We choose a direction for the rook, $(N, W, E, S)$, with uniform probability. 
(3) We choose a number of squares in which to move the rook along the direction chosen in [2] with uniform probability over the interval consisting of the rook's current position to the edge of the board.
(4) If the rook being moved collides with another piece while being translated in [3], just as in regular chess it will annihilate that piece and remain at the piece's former position.
NOTE - An alternative way of stating [2], [3], and [4] would be to say that the chosen rook samples all possible sets of moves, with uniform probability, and is unable to bypass other rooks without annihilating them and stopping at their former positions.
NOTE 2 - Gerhard Paseman is correct in suggesting that the original formulation for [2] and [3] will bias the rook towards shorter path lengths.  This is in part due to the choice of direction in [2] not being weighted by the resulting possible number of choices in [3], and also the over-counting of positions in [3] due to the lack of consideration that there may be a collision.  There are also problems with [2] near the board's boundaries where a direction can be chosen in which no move can take place.  Instead of [2] and [3], I'll suggest that a better method would be to number all possible position that the chosen rook from [1] can occupy (keeping the collision constraint from [4] in mind), and then use a PRNG to select the next position. 
What does the distribution look like for the number of time intervals, $P$, necessary for only a single rook to remain on the board?

REPLY [3 votes]: It occurred to me it might be of interest to see what happens if you start with a board completely clogged with rooks*, so I decided to pluck the lowest-hanging nontrivial fruit and examine the $2\times2$ case, which features 5 distinct states:  the starting state $S$ with 4 rooks, a trio state $T$, a pair of doublet states $R$ and $D$ with the rooks lined up in a row or along a diagonal, respectively, and the quitting state $Q$ with just one rook.  
In this set-up, state $S$ transitions in one step to $T$ (I'm assuming here that when you pick a rook at random, you actually have to move it).  State $T$ transitions back to itself with probability 1/3, to $R$ with probability 1/3, and to $D$ with probability 1/3.  State $D$ transitions to $R$ with probability 1, while $R$ transitions to $D$ with probability 1/2 and to $Q$ with probability 1/2.  For the expected number of steps to get to $Q$, we thus have
$$E(S) = 1+E(T)$$
$$E(T) = 1 +{1\over3}(E(T)+E(R)+E(D) $$
$$E(D) = 1 + E(R)$$
$$E(R) = 1 + {1\over2}E(D)$$
from which one finds $E(R) = 3$, $E(D) = 4$, $E(T) = 5$, and $E(S) = 6$.
It seems doubtful that the expected values will be integers in general, but it might be worth checking the $2\times3$ and $3\times3$ cases, which ought to be doable.  (The $2\times3$ case, which has 23 essentially different states, might be a good place to experiment with different conventions for the transition probabilities.)  One thing worth noting:  The states $R$ and $D$ are equiprobable when starting from $S$, but not if you create a 2-rook state from scratch by placing rooks at random.  This makes me wonder what Joseph O'Rourke's histogram would look like if you started, say with 16 rooks on a $4\times 4$ board but didn't start counting moves until you were down to the last 4 rooks.
*I wrote all this up before I read Aaron Golden's answer carefully.  His graph shows simulated results for the $8\times8$ case starting with 64 rooks, but he's allowing rooks not to move if they're on an edge of the board.<|endoftext|>
TITLE: Does there exist a non-trivial Ultrafinitist set theory?
QUESTION [8 upvotes]: Does there exist a set theory T-which has not yet been proved to be inconsistent-and in which
one can prove the existence of (1) the empty set (2) sets that are singletons and (3) sets which
have non-empty proper subsets. T has no axiom of infinity but-as with Quine's NF-one can prove in
T the existence of a universal set (i.e a set of all sets). However-unlike Quine's NF-the universal
set of T should be finite. One can think of T as being formalized in the classical first order predicate
calculus, using the same language as ZF.
My motive in seeking a set theory such as T is to find out whether there exist set theories that might
be acceptable to an ultrafinitist (as conforming to the principles of that viewpoint), while still
allowing a certain amount of arithmetic to be carried out in them.

REPLY [5 votes]: I have been studying a theory I call Modular Arithmetic (MA). MA has the same axioms as first order Peano Arithmetic (PA) except Ax( ~S(x)=0 ) is replaced with Ex( S(x)=0 ). MA is consistent because it has arbitrarily large finite models base on modular arithmetic. The upward Löwenheim–Skolem theorem proves MA must have infinite models. MA can be made into an ultra-finite theory by adding an axiom like Ax( x=0 or x=S(0) or ... or x=Sn(0) ) where Sn(0) is some finite number of applications of successor to 0 (a numeral). 
Coming up with a set theory based on MA is problematic. It is simple to show MA is omega inconsistent. The predecessor of 0 must be non-standard (not a numeral) in any infinite model of MA. If the predecessor of 0 is standard we can prove the model is finite using induction. This means Ax( ~S(x)=0 ) is true for all standard natural numbers in any infinite model. A set theory based on MA can't be well ordered or well founded, either. (x =< y) <-> Ez( x+z=y ) is trivially true for any x and y. Using S(x) = x U {x} as a definition of successor is inconsistent with the axioms of MA.
In ZF, even the Axiom of Pairing allows the construction of arbitrarily large sets. Assuming we could come up with a set theory for MA, it would have the properties you ask for. One way to do this would be to encode sets as binary expansions of natural numbers. Element x is a member of the set if the x'th bit of the expansion is true. This would allow us to have sets of size log2(n) where n is the size of the universe. We can equate 0 with the empty set. We can define singleton sets for "small" elements of the universe. We could also define sets with subsets. We could do a reasonable amount of arithmetic by choosing n large enough. We could have sets and do math on sets as large as 100 by having a universe of size 2^100.<|endoftext|>
TITLE: Conceptual explanation of Strassen's trick for matrix multiplication
QUESTION [18 upvotes]: Algorithms for fast multiplication of polynomials and integers have well-known conceptual explanations. A good survey paper is Daniel J. Bernstein's Fast Multidigit Multiplication for Mathematicians.
For example, Karatsuba's trick and FFT-based multiplication both fit into the evaluate-and-interpolate scheme. These rely on the fact that evaluation at a point is a ring homomorphism. Karatsuba's trick cheaply evaluates the product of two linear polynomials at 0, 1 and $\infty$ without multiplying them out explicitly and then interpolates the values to get the polynomial product. Similarly, the n-point FFT efficiently evaluates a polynomial at the nth roots of unity, and the inverse FFT efficiently interpolates a polynomial from its values at those same points.
You can also go beyond evaluation and use any convenient ring homomorphisms.
My question is whether Strassen's algorithm for fast matrix multiplication can be explained in conceptual terms along these lines. The core of the algorithm is Strassen's trick for computing products of 2x2 matrices over noncommutative rings. Looking at Strassen's formula, it's hard to shake the feeling that some related approach should work.

REPLY [6 votes]: To my poor knowledge, fast matrix multiplication is not based on rings homomorphisms, but on the notion of tensorial rank. More precisely, the trilinear form $(A,B,C)\mapsto{\rm Tr}(ABC)$ extends as a linear form $L$ over 
$$M_{n\times m}\otimes M_{m\times p}\otimes M_{p\times n}.$$ 
The tensorial rank of $L$ is the minimal number $r$ in which you can write 
$$L=\sum_{j=1}^re_j\otimes f_j\otimes g_j,$$
with $e,f,g$ linear forms. Then the calculation of $AB$ needs only $r$ multiplication and a certain number of additions. The latter is of little importance, because when you apply the formula recursively to (say square) matrices of size $Nn$, the number of operations grow like $r^N$. This is easily seen in the case $n=2$, where $r=7$ and the formula above involves as many as $16$ additions. For $2^N\times2^N$ matrices, the number of operations is bounded by $C7^N=C(2^N)^\alpha$ with $\alpha=\frac{\log7}{\log2}<3$.
Notice that if you had only two spaces (instead of $3$), then the tensorial rank over $E\otimes F$ is just the rank of linear algebra.<|endoftext|>
TITLE: Automorphisms of a specific type of weighted projective space
QUESTION [8 upvotes]: A question very close to this one was already asked: Automorphisms of a weighted projective space
But the answer given does not satisfy my needs. So avoiding having two questions that are identical, I am interested in a specific type of weighted projective space, namely $\mathbb{P}(1, 1, \cdots, 1, k)$ for some natural number $k$. The case $k = 1$ gives rise the usual projective space. To be more precise, I am considering a real weighted projective space with weight vector $(1, 1, \cdots, 1, k)$. 
So the question is can one characterize, in general, the automorphism group of such a space. If $k = 1$ then we get the projective linear group, namely the group $\textbf{GL}(\mathbb{R}^n)/\mathbb{R}^\ast \cong \textbf{PGL}(\mathbb{R}^n)$. In general, can one characterize the automorphism group of weighted projective spaces of the type $\mathbb{P}(1, 1, \cdots, 1, k)$?

REPLY [8 votes]: As rita said $\mathbb{P}(1,\dots,1,k)$  is naturally isomorphic to the cone in $\mathbb{P}^{N+1}$ over the $k$-th embedding (Take the map which sends $(x_1:\dots:x_n:y)$ onto $((x_1)^k:(x_1)^{k-1}x_2:...:(x_n)^k:y)$ where the $N+1$ first coordinates are the monomials of degree $k$ in $x_1,\dots,x_n$), so there is a natural morphism from the group $G=\mathrm{Aut}(\mathbb{P}(1,\dots,1,k))$ to $\mathrm{PGL}(n,\mathbb{K})$. However, the kernel is not the one which was described in the above answer.
We can in fact give $G$ more explicitly (because there are a priori many extensions given two groups):
We choose $k>1$ (otherwise the description is different and obvious). We identify $\mathbb{K}^{N+1}$ with the set of homogeneous polynomials of degree $k$ in $n$ variables. The group $\mathrm{GL}(n,\mathbb{K})$ naturally acts on $\mathbb{K}^{N+1}$.
Let $H$ be the semi-direct product $\mathbb{K}^{N+1}\rtimes \mathrm{GL}(n,\mathbb{K})$. There is a natural surjective map $H\to G$, that we describe now:
The action of $\mathbb{K}^{N+1}$ on $\mathbb{P}(1,\dots,1,k)$ is given by $(x_1:\dots:x_n:y)\mapsto (x_1:\dots:x_n:y+P(x_1,\dots,x_n))$ where $P\in\mathbb{K}^{N+1}$ is the corresponding polynomial.
The action of $\mathrm{GL}(n,\mathbb{K})$ on $\mathbb{P}(1,\dots,1,k)$ is given by the action on $x_1,\dots,x_n$.
It yields thus a morphism $H\to G$ whose kernel is the subgroup $L$ of $\mathrm{GL}(n,\mathbb{K})$ consisting of diagonal matrices of the form $\{\lambda I| \lambda^k=1\}$.
The group $G=\mathrm{Aut}(\mathbb{P}(1,\dots,1,k))$ is thus equal to the quotient of $\mathbb{K}^{N+1}\rtimes \mathrm{GL}(n,\mathbb{K})$ by the subgroup $L$. 
The surjective morphism $G\to  \mathrm{PGL}(n,\mathbb{K})$ corresponds to the projection on $\mathrm{GL}(n,\mathbb{K})/L$ followed by the quotient by the image of all diagonal matrices (we have first killed only finitely many and then kill all others). The kernel of this map is thus equal to $\mathbb{K}^{N+1}\rtimes\mathbb{K}^{*}/L$.<|endoftext|>
TITLE: A question about Speiser's 1934 result on the Riemann hypothesis
QUESTION [17 upvotes]: A number of sources concerning Speiser's 1934 result state that the Riemann Hypothesis (RH) implies $\zeta'(s)\neq 0$ for all $0<\text{Re}(s)<1/2$. But I have seen some (possibly less reliable) sources without proof suggesting this is an if and only if relationship, i.e. RH$\Longleftrightarrow\zeta'(s)\neq 0$. However, those (perhaps more reliable) sources state only forward implication, i.e. RH$\Longrightarrow\zeta'(s)\neq 0$. My question is this: is Speiser's result an if and only if relationship or not?

REPLY [20 votes]: Yes, Speiser's theorem is an if and only if.
See Theorem 1 and "Corollary to Theorem 1" in Levinson and Montgomery's Zeros of the derivatives of the Riemann Zeta-function. Acta Math. 133 (1974), 49–65.<|endoftext|>
TITLE: Is it possible to have t triangles in some graph on n vertices?
QUESTION [11 upvotes]: Fix $n>4$. Is there a characterization of the set $T_n$ of all natural numbers $t$ such that there is some graph on $n$ vertices with exactly $t$ distinct triangles? For example, it's clear that {$1,2,\ldots,n$} $\subseteq T_n$ and $\binom{n}{3}-1 \notin T_n$; what else can we say?

REPLY [3 votes]: I will argue that the following are true: 

For any $\epsilon \gt 0$ and all large enough $n$, $T_n$ contains all integers less than $\binom{n}{3}-\epsilon n^2.$ 
For any fixed positive integer $c$ there is a finite list of expressions which accounts for all numbers larger than $\binom{n}{3}-cn-n/2$ provided that $n$ is large enough.  

For $n \gt 14$ the largest members of $T_n$ are precisely  

$\binom{n}{3}$ then 
$\binom{n}{3}-n+2$ then 
$\binom{n}{3}-2n+r $ for $r=4,5$ then 
$\binom{n}{3}-3n+r $ for $r=6,7,8,9$ then 
$\binom{n}{3}-4n+r $ for $r=8,9,10,11,12,14$

A simple construction and simple bounds are enough to show that $T_n$ contains all integers less than $\binom{n}{3}-n^2.$ The same simple construction with more sophisticated bounds evidently shows that for any $\epsilon \gt 0$ and all large enough $n$, $T_n$ contains all integers less than $\binom{n}{3}-\epsilon n^2.$ 
First a side result. Every integer can be expressed as a sum of three triangular numbers. I asked elsewhere : What is the smallest positive integer $s=s_m$ which can not be written in the form $$s=\binom{a}{2}+\binom{b}{2}+\binom{c}{2}$$ subject to $\max(a,b,c) \le m?$ For now I will say that obviously, for all $m $, $s_m \ge \binom{m+1}{2} .$
Consider this simple construction: Start with $K_{n-3}$ which already has $\binom{n-3}{3}$ triangles. Adjoin three further vertices $u,v,w$ with no edges between them but allow each  to be connected to all, some or none of the other $n-3$ vertices. This will allow all triangle counts of the form $\binom{n-3}{3}+\binom{a}{2}+\binom{b}{2}+\binom{c}{2}$ subject to $\max(a,b,c) \le n-3$ and hence at least everything from $\binom{n-3}{3}$ up to $\binom{n-3}{3}+\binom{n-2}{2}-1=\binom{n}{3}-(n^2-5n+8).$ We may assume by inductive hypothesis and a few very small examples that using  $n-1$ or less vertices we can get all counts up to $\binom{n-1}{3}-((n-1)^2-5(n-1)+8)=\binom{n-3}{3}+n-5.$ Hence we can get all counts from $0$ up to $\binom{n}{3}-(n^2-5n+8)$ using this construction.

If I understand a rather impressive answer (in a comment) to my question, for any $\epsilon \gt 0$, $s_m \ge (3/2-\epsilon)m^2$ for all large enough $m.$ Hence for large enough $n$ we can get all counts up to $\binom{n-3}{3}+(3/2-\epsilon)(n-3)^2=\binom{n}{3}-(\epsilon n^2+O(n)).$

Consider now the largest counts in $T_n.$ They must result from removing only a few edges from $K_n$. Suppose that $c$ edges are removed and let $H$ be the graph formed by the removed edges. If $H$ is $c$ disjoint edges, this kills $c(n-2)$ triangles. In general the number of killed triangles is $c(n-2)-p-2t$ where $t$ is the number of triangles in H (since these triples get counted three times by $c(n-2)$) and $p$ is the number of paths of length two in $H$ which are not part of a triangle in $H.$ To find the exact spectrum of values requires not an enumeration but simply a classification of which triples $c,p,t$ represent possible values for the edge,path and triangle count of graph. Hence all triangle free graphs with $v$ vertices, $c$ edges and degree sequence $d_1,d_2,\cdots,d_v$ give (when removed) the same count:  $\binom{n}{3}-\sum_1^v\binom{d_i}{2}.$ We need not know how many ways a certain degree sequence can be obtained, only if it can be realized in at least one one.<|endoftext|>
TITLE: Nonstandard analysis in probability theory
QUESTION [36 upvotes]: I am quite new at nonstandard analysis, and recently I became aware of its use in probability theory mainly through the following two books: 

Nelson (1987). Radically Elementary Probability Theory
Geyer (2007). Radically Elementary Probability and Statistics

Although Nelson's book is several decades old, as far as I can see, its approach has not yet caught on. Also, I couldn't find a lot of papers published in the leading probability journals on that topic. I am quite intrigued by that phenomenon. My questions are the following

Why hasn't nonstandard analysis been widely adopted by probabilists?
Were there some success stories in some particular sub-fields of probability theory or statistics?
Does there exist some known fundamental objections in probability theory to the approach in there?

REPLY [2 votes]: Probabilists are no different than the conventional analyst, at least in most places I have been, Eastern Europe (my home), and US. NSA has been seen a lot more as an alternative rather than competitor. Model theorists are more interested in studying different models than actually electing one best and propose a universal transition. The work of Nelson has been revisited. Besides the work of Geyer in your post there is a terrific recent book by Herzberg and articles by Weisshaupt (Journal of Logic and Analysis 2009 and 2011) and Andrade (Positivity, doi 10.1007/s11117-015-0333-9) and also an article by Geyer and Andrade (Journal of Logic and Analysis).
I dont agree with other post saying that Nelson's NSA is more of interest to non mathematicians. Nelson, Herzberg and Weisshaupt are mathematicians, Geyer has degree in Physics but is now statistician as is Andrade (also see work on NS Brownian motion in Physica A, 2015 Volume 429). I say interest is also high in physics and mathematical finance as illustrated by Herzberg's book as in pure maths and stats<|endoftext|>
TITLE: Which 2-coskeletal simplicial sets is the nerve of a category?
QUESTION [7 upvotes]: Let ${\mathrm{tr}}_2$ be the truncation functor that takes a simplicial set and restricts it to dimensions at most 2. Its right adjoint is the 2-coskeleton functor. NLab says that the nerve of a small category is a 2-coskeletal simplicial set.
In a category, composition is defined only if two morphisms abut. Precisely, the composite $g\circ f$ is defined if the target of $f$ is the source of $g$. This translates to saying the nerve of a category is a simplicial set such that every inner horn of a 2-simplex has a filler.
My question is: Does this extension condition characterize those 2-coskeletal simplicial sets that is a nerve of a category? If not, is there a necessary and sufficient condition?
EDIT: Thanks Tyler for your comment. Can we identify a "category" with a "simplicial set truncated in dimension 2 such that every inner horn of a 2-simplex has a unique filler"? How does this account for the associativity of composition?

REPLY [3 votes]: It's not good enough to check inner $2$-horns. Let $K$ be a $2$-coskeletal simplicial set. It follows that for $m>3$ and any $k$, every map $\Lambda_{k}^m\rightarrow K$ admits a unique extension to $\Delta ^{m}$. On the other hand, $K$ is the nerve of a category if and only if for any $m > 1$ and $k$ for which $0<k<m$, every map $\Lambda_{k}^m\rightarrow K$ admits a unique extension to $\Delta ^{m}.$ Since this is automatic for $m>3$ it follows that a $2$-coskeletal $K$ is the nerve of a category if and only if for $m=2,3$ and $0<k<m$, every map $\Lambda_{k}^m\rightarrow K$ admits a unique extension to $\Delta ^{m}$. An example of a $2$-coskeletal simplicial set that fails to be the nerve of a category is $\partial \Delta ^{2}$. For a more interesting example where every map $\Lambda_{1}^2\rightarrow K$ admits a unique extension, but $K$ still fails to be the nerve of a category, consider the following. Let $K$ be the union of $\Lambda_{1}^3$ and $\Delta^2$ with the edges from $0$ to $2$ and $2$ to $3$ on $\Lambda_{1}^3$ glued to the edges from $0$ to $1$ and $1$ to $2$ on $\Delta^2$, respectively. Then $K$ is $2$-coskeletal, and every inner $2$ horn admits a unique extension, but the inclusion $\Lambda_{1}^3\rightarrow K$ does not.<|endoftext|>
TITLE: How to compute the Picard rank of a K3 surface?
QUESTION [13 upvotes]: I'm curious about the following question:

Given a K3 surface, how does one proceed to compute its rank?

Of course the answer may depend on the form of the input, i.e. how the K3 is "given". So

For a given way of writing down a K3 surface, (e.g. quartics in $\mathbb{P}^3$) 
   How does one compute the Picard rank of the K3 surface?

(Aside: What I've seen people sometimes did is avoiding this question by nailing down a K3 surface $X$ with its $NS(X)$ together with the intersection form. Then find an embedding given by the ample class.)

REPLY [4 votes]: In Theorem 6 of the following paper : http://arxiv.org/abs/1111.4117, building on Van Lujik's work, François Charles explains a (theoretical) algorithm that computes the rank of a K3 surface $X$ defined over a number field. This algorithm terminates conjecturally, for instance if $X\times X$ satisfies the Hodge conjecture. 
The main new feature of this article, that allows him to obtain an algorithm, is that the discrepancy between the rank of $X$ and the rank of the reduction of $X$ at a typical prime may be read off the algebra of endomorphisms of the transcendental lattice of $X$.<|endoftext|>
TITLE: Hyperbolic polynomials and group representations
QUESTION [14 upvotes]: Recall that a homogeneous polynomial $P\in{\mathbb R}[X_1,\ldots,X_d]$ of degree $n$ is hyperbolic in the direction of a vector $V\ne0$ if for every vector $W$, the univariate polynomial $t\mapsto P(W+tV)$ has $n$ real roots. Notice that $P(V)$ must be non-zero, and we may assume w.l.o.g. that $P(V)>0$. The notion of hyperbolic polynomials is due to L. Garding, because of hyperbolic PDEs. He proved, among other things, that the connected component of $V$ in $\{P(W)>0\}$ is a convex cone (the cone of future in terms of evolutionary PDEs), and that $P$ is hyperbolic in every direction $V'$ of this cone. This is reminiscent to special relativity.
Examples of hyperbolic polynomials include

$P=X_1\cdots X_d$, where $n=d$ and $V$ can be taken as $(1,\ldots,1)$
$P(S)=\det S$, where $d=\frac{n(n+1)}2$ and ${\mathbb R}^d$ identifies with the space of symmetric matrices with real entries. Then $V=I_n$ and one uses the fact that eigenvalues of symmetric matrices are real. One may replace real symmetric by complex hermitian matrices, with $d=n^2$ instead.

Besides, let me recall a result of I. Schur about (generalized) permanents: let $H$ be a subgroup of ${\frak S}_n$, and $\chi$ be a complex character over $H$. Then, for every $S\in SPD_n$, one has
$$\chi(e)\det S\le d_\chi(S):=\sum_{g\in H}\chi(g)\prod_{i=1}^ns_{ig(i)}.$$
In particular, the connected component of $I_n$ in $d_\chi>0$ contains the cone $SPD_n$.
This suggest that some of the $d_\chi$'s could be hyperbolic in the direction of $I_n$. This is true at least in the following cases:

$H=\frak S_n$ and $\chi=\epsilon$ the signature, where $d_\chi=\det$,
$H=(e)$, where $d_\chi(S)=\prod_i s_{ii}$ (remark that the corresponding Schur's inequality is known has Hadamard inequality),
In between, when $H$ is the natural inclusion of ${\frak S}_m\times\cdots\times{\frak S}_q$ with $m+\cdots+q\le n$, and $\chi$ is the product of the signatures, then $d_\chi$ is the product of hyperbolic polynomials of the types above, hence is itself hyperbolic.

However, the choice $H={\frak S}_n$ and $\chi={\bf 1}$ yields the permanent, which is not hyperbolic in the direction of $I_n$ if $n\ge2$.

My question is whether there are other hyperbolic polynomials among the $d_\chi$'s. If so, how do they classify ? Does hyperbolicity correspond to a classical property of $(H,\chi)$ ?

REPLY [4 votes]: [EDIT]: Recently, I ran into this question again, so decided to undelete this 3 year old answer of mine. 

Although not an answer to the exact question, I thought it worth mentioning an important problem where characterizing certain situations where the permanent leads to a hyperbolic polynomial was an open problem for many years. Details follow below.

(Defn.) Let $A$ be a square real matrix. We say that an $n \times n$ real matrix $A$ is column monotone if it has weakly decreasing entries in each column (when reading downwards). That is, $a_{ij} \ge a_{i+1,j}$ for all $1 \le i \le n-1$ and $1\le j \le n$.

Based on this definition, (MCPC) The Monotone Column Permanent Conjecture of Haglund, Ono, Wagner ("Theorems and conjectures involving rook polynomials with only real zeros", Kluwer, 1999), is as follows.

(Erstwhile conjecture). If $A$ is an $n \times n$ column monotone matrix, then $\text{per}(A+zJ)$ is a polynomial in $z$ that has only real roots, where the direction $J$ is an $n \times n$ matrix of all ones.

This conjecture was proved in its full generality in "Proof of the Montone Column Permanent Conjecture" by Brändén, Haglund, Visontai, and Wagner (2010).
My summary above is taken from the abovecited paper. If I find anything related to immanants, I'll update my answer.<|endoftext|>
TITLE: Reference request: an elementary proof of Brouwer fixed-point theorem.
QUESTION [12 upvotes]: One of the elementary way to prove of the Brouwer fixed-point theorem is,  making it follow from the (smooth) Non-Retraction theorem. The latter is then proven by contradiction by means of a simple computation on the "oriented area" of smooth mappings $g:B\subset \mathbb {R}^n\rightarrow\mathbb {R}^n$
$$\int_B \operatorname{det} D g(x) dx$$
and only involves a differentiation under the sign of integral with respect to the parameter of deformation (I mentioned this proof in this wiki-article) . Due to this fact, I sometimes like to use it in elementary courses as a meaningful application of differential calculus and Lebesgue integration (on the other hand, the geometrical ideas behind remain a bit hidden, but that is an other story). 
However, a slight annoyance to me now is, that I can't remember where I read this proof the first time, several years ago. I would be very glad to learn a reference, and (if it is known) the name of the inventor of this nice proof.

REPLY [2 votes]: There is an interesting essay on Brouwer's Fixed Point theorem, including a contructive proof, at Kevin Brown's MathPages site http://www.mathpages.com/home/kmath262/kmath262.htm
Looking at the home page, I see he has now written a string of books. If his articles are any indication, these books are doubtless excellent and well worth buying.<|endoftext|>
TITLE: Kazhdan-Lusztig C-basis and categorification
QUESTION [5 upvotes]: Does anyone know if there exists some natural way to interpret the Kazhdan-Lusztig C-basis in a categorification of the Hecke algebra ? The category of Soergel bimodules categorifies the C'-basis but it doesn't seem clear to me whether this or some other related structure (Rouquier complexes, sheaves...) could be adapted to categorify the C-basis instead ? I am currently working on some elements of the Hecke algebra which seem to have remarkable positivity properties when written in the C-basis (---> (-1) signs occur when writing them in the C'-basis).
The $C$ and $C'$-basis are defined in Kazhdan-Lusztig 79 :
For each element $w\in W$ there is a unique selfdual $C_w\in H$ such that
$$C_w=\sum_{y\leq w} \epsilon_y \epsilon_w q_w^{1/2} q_y^{-1} \bar{P_{y,w}} T_y$$
with $P_{y, w}\in\mathbb{Z}[q^{1/2}, q^{-1/2}]$ a polynomial in $q$ having degree at most $\frac{1}{2}(\ell(w)-\ell(y)-1)$ for $y\leq w$, $P_{w,w}=1$. The set $(C_w)_{w\in W}$ forms a basis, the $C$-basis of the Hecke algebra. 
For each element $w\in W$, there is a unique selfdual $C'_w\in H$ such that 
$$C'_w=q_w^{-1/2} \sum_{y\leq w} P_{y,w} T_y$$
with $P_{y, w}\in\mathbb{Z}[q^{1/2}, q^{-1/2}]$ a polynomial in $q$ having degree at most $\frac{1}{2}(\ell(w)-\ell(y)-1)$ for $y\leq w$, $P_{w,w}=1$. The set $(C_w')_{w\in W}$ forms a basis, the $C$-basis of the Hecke algebra.
the problem of applying the involution $j$ from 7.9 of Humphreys to my elements is that one gets (-1) signs before some coefficients of $C'$-elements

REPLY [5 votes]: If you identify the Hecke algebra with the Grothendieck principal block of a graded lift of category $\mathcal{O}$ for the corresponding semi-simple Lie algebra (oddly, enough which of the two Langlands dual options you pick doesn't matter) such that $T_y$ is the class of a Verma module with highest weight $-y\rho-\rho$ (so $T_1$ corresponds to the anti-dominant Verma), then the $C'$-basis will match with the classes of tilting modules, and the $C$ basis with the simple modules (I tend to prefer matching the $C'$-basis with projectives, but then the $C$-basis is something horrible).  
I'm not really sure where this is written down properly; philosophically, the point is that everything will be fixed in place once you decide what functor to send the bar involution to.  If you send it the contragredient dual, then simples and tiltings are obvious sets of modules fixed by this duality, and they satisfy the triangularities you've written above.<|endoftext|>
TITLE: Examples of finitely generated elementary amenable groups which are not virtually solvable
QUESTION [7 upvotes]: What are some examples of finitely generated (finitely presented) elementary amenable groups which are not virtually solvable?

REPLY [7 votes]: Houghton groups are (non-split) extensions of the group of the finitely supported permutations of the integers by $\mathbf{Z}^d$ for $d\ge 2$ and are thus elementary amenable and not virtually solvable. They're finitely presented, as shown by K.Brown here (Finiteness properties of groups, JPAA 1987). They were introduced by Houghton as f.g. groups with coset spaces with $2<n<\infty$ ends.<|endoftext|>
TITLE: Are Galois groups of Q with restricted ramification supposed to be finitely generated?
QUESTION [12 upvotes]: Fix a finite set $S$ of places of $\mathbb Q$. Let $G_{\mathbb Q,S}$ be the Galois group of the maximal extension of $\mathbb Q$ unramified outside S$. I believe that it is an open question whether this group is topologically finitely generated, that is, contains a dense, finitely generated subgroup. 

Is there a standard conjecture about whether $G_{\mathbb Q,S}$ should be finitely generated or not?
  Has anyone published an opinion?

JSE mentions a stronger conjecture: $|S|$ generators suffice; further, the existence of a special set of generators labeled by $S$ and inert at the labeling place. Is this refinement standard? (Note that a finite extension would suffice to disprove this hypothesis.)
Weakening the hypothesis, we could restrict the ramification at $S$ to be tame.

REPLY [5 votes]: I think you can look at page 532 of (the first version of) J. Neukirch, A. Schmidt, K. Wingberg, Cohomology of Number Fields, Springer, 1999. They explicitly write that we do not even know precisely what to conjecture for arbitrary number field. They also add "many mathematicians (including the authors) tend to think of $G_S$ as being not finitely generated". In their notation, $G_S$ is the Galois group of the maximal unramified outside of $S$ extension of a number field $K$ which has been fixed once and for all (and $S$ is finite). I do not know if the special case $K=\mathbb{Q}$ is somehow different, but I doubt it (after all, their $G_S$ is a subgroup of your $G_{\mathbb{Q},S'}$ for $S'=S\cap\mathrm{Ram}(K/\mathbb{Q})$...)<|endoftext|>
TITLE: Why additional constraint is need for this two groups to be isomorphic?
QUESTION [6 upvotes]: I'm reading AMS's book Papers on Topology, which collects Poincare's papers on topology.
However, the first paper stops me.
In the paper, he considered the group generated by transformations in $\mathbb{R}^3$, the generators are:
$(x,y,z)\rightarrow (x+1,y,z)$
$(x,y,z)\rightarrow (x,y+1,z)$
$(x,y,z)\rightarrow(\alpha x+\beta y,\gamma x+\delta y, z+1)$
where $\alpha\delta-\beta\gamma=1$.
Obviously, the third transformation is a rotation in $XY$-plane and one move in $Z$.
We denote the group generated above by $(\alpha,\beta,\gamma,\delta)$.
Then Poincare claims that two groups $(\alpha,\beta,\gamma,\delta)$ and $(\alpha',\beta',\gamma',\delta')$ cannot be isomorphism unless the two transformation in $\mathbb{R}^2$:
$(x,y)\rightarrow(\alpha x+\beta y,\gamma x+\delta y)$
$(x,y)\rightarrow(\alpha' x+\beta' y,\gamma' x+\delta' y)$
are conjugate of each other by a linear transformation with integer coefficient.
I'm confused by this claim and can not find why.
I think if we denote the three transformation $\sigma_1,\sigma_2,\sigma_3$ and $\sigma_1',\sigma_2',\sigma_3'$, we can just form an isomorphism: $\phi(\sigma_i)=\sigma_i'$.
Why the additional constraint is needed?

REPLY [3 votes]: Since you are interested in topology, you would do well to read Peter Scott's beautiful paper from 1983, called "The geometries of 3-manifolds". The manifolds you are asking about are Solvmanifolds (at least in the case @Mark Sapir is discussing in his answer), and are discussed in detail starting on page 470 of that paper (including an answer to your question), but you would do well to read the whole thing.<|endoftext|>
TITLE: Definition of torsion sheaf on reducible spaces
QUESTION [9 upvotes]: I need to discuss torsion-free sheaves on reduced, but possibly reducible spaces. Here "torsion" means "element is annihilated by a non-zero-divisor". The standard references (EGA, Hartshorne, ...) restrict themselves to normal varieties or  irreducible spaces and do not seem to cover the definition in this level of generality. 
I saw some papers and survey articles that use torsion-free sheaves on reducibly spaces, but mostly without any discussion of the definition. Some papers even seem to give the "wrong" definition where "torsion" = "annihilated arbitrary non-zero element", which is probably not what the authors had in mind.
Is anyone aware of a reliable reference for the definition and for basic properties of torsion-free sheaves that I could possibly use?
Thank you in advance!

REPLY [5 votes]: Hi Stefan! Torsion-free (coherent) sheaves are pure sheaves of codimension 0; I suggest that you try "The Geometry of Moduli Spaces of Sheaves", by Daniel Huybrechts and Manfred Lehn, where pure sheaves are discussed in detail.<|endoftext|>
TITLE: Double a manifold with boundary
QUESTION [13 upvotes]: Suppose $(M,g)$ is a compact Riemannian manifold with total geodesic boundary. The double of it is given by
$$
D(M) = M\cup_fM
$$
where $f:\partial M\to\partial M$ is an identity map. In general $D(M)$ is not a $C^\infty$ manifold. But I wonder if $D(M)$ is a $C^2$ manifold or not.

REPLY [17 votes]: It is a $C^\infty$ manifold if you define charts properly (e.g. using geodesics normal to the boundary as coordinate lines).
The metric of the double is $C^2$ (but not always $C^3$). Indeed, since the boundary is totally geodesic, the normal derivative of the metric tensor (in the above mentioned coordinates) vanishes, and the second normal derivatives on two copies match due to the symmetry.

REPLY [10 votes]: You can give $D(M)$ a $C^k$ structure, $k=0,\ldots,\infty,\omega$ if $M$ has one. 
Take a $C^k$ function $f$ on $M$ that is $1$ on the boundary with positive normal derivative, and $0\leq f<1$in the interior. 
Then you can take $D(M)=\{(x,t)\in M\times\mathbb{R}: f(x)+t^2=1\}$.
Do you want $D(M)$ to be a riemannian $C^k$ manifold ? With isometric copies of $(M,g)$
this will not be possible in general for $k>2$, but sometimes it will, for instance in the locally symmetric case.<|endoftext|>
TITLE: A product on the square roots of unit matrix
QUESTION [11 upvotes]: There is a strange product that takes two square roots of unit matrix, say $A$ and  $B$, $A^2=I$, $B^2=I$ to a square root again, 
$$ A\star B=(A+B)^{-1}(A-B+2I), \qquad (A\star B)^2=I$$
Could anybody help me with identifying this structure? Where it comes from?
It was obtained from the Caley transform $C(A)=(1-A)^{-1}(1+A)$, $C(C(A))=A$, by expanding $C(C(A),C(B))$ and using $A^2=B^2=I$. Ones we imposed $A^2=B^2=I$ the inverse transform does not exists anymore, so we are at the singular point of the Caley transform. Somehow it looks like adding some points at infinity and extending the action on them.

REPLY [8 votes]: Let me denote $E_\pm(M)$ the eigenspaces of $M$ associated with the eigenvalues $\pm1$. Let me assume that the characteristic of the scalar field $k$ is not $2$. Then your assumption is that 
$$k^n=E_+(A)\oplus E_-(A)=E_+(B)\oplus E_-(B).$$
In addition, the assumption that $A+B$ is non-singular means $E_+(A)\cap E_-(B)=(0)$ and $E_+(B)\cap E_-(A)=(0)$, from which it follows that $\dim E_+(A)=\dim E_+(B)$.
Then one verifies easily
$$E_+(A\star B)=E_+(B),\qquad E_-(A\star B)=E_-(A).$$
Because 
 the dimensions of $E_-(A)$ and  $E_+(B)$ sum up to $n$, one deduces 
$$k^n= E_+(A\star B)\oplus E_-(A\star B),$$
and therefore $(A\star B)^2= I_n$.<|endoftext|>
TITLE: Is it common practice to publish parts of a PhD thesis in advance?
QUESTION [16 upvotes]: I'm interested in publishing parts of my PhD thesis in advance and I'm wondering wether or not this will result in problems later on. One of the problems I'm thinking of is that usually the copyright is transferred to the journal/publisher but at our university you are required to publish your thesis online via the library website, which means you have to give them the right to publicize the thesis (which I'm not sure you still can do, even when a slightly modified version of  the journal paper is only part of the thesis).
So I guess my question is this:
Is it common practice to publish parts of a PhD thesis in advance and if so does this result in legal problems? What are your experiences?
Somehow related
Publishing journals articles without transferring copyright.

REPLY [13 votes]: I think a lot of journals will expressly allow this in their copyright policy.  I just poked around, and I see for example that the AMS allows this:
AMS copyright policies
and even Elsevier allows this in its journals:
Elsevier copyright policies
So you can probably even do this completely legally.  I would just check the web site of the particular journal(s) where you are considering submitting articles.  And journals will even allow you to modify the copyright agreement in many cases, so you could still ask about this if a journal doesn't specifically address the issue already.
Added later: my last comment was inspired by having read the article ``Do Mathematicians Get the Author Rights They Want?'' by Kristine K. Fowler in:
 March 2012 issue of the AMS Notices (appearing in the left column of the issue web page)<|endoftext|>
TITLE: Minimizing the perimeter around an obstacle
QUESTION [5 upvotes]: Let $A\subset \mathbb{R}^n$ be a (measurable) bounded set, and consider the following optimization problem: minimize $P(X)$, the perimeter of a set $X$, where $X$ ranges over all Caccioppoli subsets of $\mathbb{R}^n$ such that $X\supset A$.
If $A$ is convex, then it's "obvious" that the minimizer is $A$ itself (up to a set of measure zero, of course). Is this true and where can I find a proof? Actually I only consider the case where $A$ is a ball.
Note that even in $\mathbb{R}^2$ it's not true that the minimizer is the convex hull of $A$, which can be seen by taking $A$ to be two balls which are placed at a far enough distance.

REPLY [8 votes]: For any sensible definition of perimeter, an adaptation of following argument proves the claim. Consider the nearest-point projection map $p_A:\mathbb R^n\to A$ (that is, for every $x\in\mathbb R^n$, let $p_A(x)$ be the point of $A$ nearest to $x$). It is easy to see that $p_A$ is Lipschitz-1, i.e., $|p_A(x)-p_A(y)|\le |x-y|$ for all $x,y\in\mathbb R^n$. Such a map should decrease perimeters (or your definition of perimeter is flawed), hence $P(X)\ge P(A)$. 
With Caccioppoli definition (which I just learned from Wikipedia), you need to translate this argument to the respective language. The resulting proof is e.g. the following. We may assume that $\partial A$ is smooth. Then consider the following smooth unit vector field $\phi$ in $\mathbb R^n\setminus A$:
$$
 \phi(x) =\operatorname{grad}(\operatorname{dist}(\cdot,A))(x) .
$$
It is easy to see that $\operatorname{div}\phi\ge 0$ on $\mathbb R^n\setminus A$. Extend $\phi$ to $A$ so that it remains smooth and $|\phi(x)|\le 1$ for all $x\in\mathbb R^n$. Then
$$
 P(X) \ge \int_X \operatorname{div}\phi = area(\partial A) + \int_{X\setminus A}\operatorname{div}\phi \ge area(\partial A) 
$$
where the first inequality follows from the definition of the perimeter, the equality from Green's formula, and the second inequality from the fact that $\operatorname{div}\phi\ge 0$ on $\mathbb R^n\setminus A$.<|endoftext|>
TITLE: Sum of three bounded triangular numbers
QUESTION [8 upvotes]: Every nonnegative integer can be written (eventually in many ways) as a sum of three triangular numbers by the Gauss Eureka theorem.

What is the smallest positive integer $n=n_m$ which can not be written in the form $$n=\binom{a}{2}+\binom{b}{2}+\binom{c}{2}.$$ subject to $\max(a,b,c) \le m?$

The answers for $m$ from $1$ to 60 are
$  \begin {array}{cccccccccc} 1&4&8&11&24&29&29&47&68&68
\\95&99&137&141&173&173&245&281&314&314
\\314&407&419&419&470&470&617&617&711&800
\\863&911&911&911&911&1118&1118&1118&1118&1118
\\1383&1433&1433&1679&1679&1679&1868&1868&1868&1868
\\1868&2360&2493&2493&2519&2925&3044&3044&3098&3098
\end {array}
$
The sequence does not seem to be in the OEIS even with a superseeker search. It just seems a curious sequence. Any information would be welcome. What bounds or asymptotics can be established? It would appear that $n_m \lt m^2$ although $m_{105}=11018 \lt 11025=105^2$ leaves some doubt. I'd conjecture (rashly) that $$\limsup \frac{n_m}{m^2}=1$$ but $n_{110}=n_{111}=n_{112}=n_{113}=11625$ and $\frac{11625}{113^3} \approx 0.91$ so the $\liminf$ might be less. I wonder what explains the repeated values and what can be said about them.
UPDATE Noam makes a nice argument that $\lim \frac{n_m}{m^2}=3/2.$ I'd accept it if it was an answer and not a comment. Let me spell out that $8\binom{t}{2}+1=(2t+1)^2$ so $n=\binom{a}{2}+\binom{b}{2}+\binom{c}{2}$ exactly if $8n+3=(2a+1)^2+(2b+1)^2+(2c+1)^2$ and this allows one to pull in results on sums of squares.
I still don't see why things such as $n_{36}=n_{37}=n_{38}=n_{39}=n_{40}$ happen,  but I have not thought about it very deeply. 
For any fixed $k$, $n_m \lt \frac{3m^2}{2}-km$ with finitely many exceptions. so one could wonder about things like $\frac{3m^2}{2}-\sqrt{m^3}.$ However this result was great for the problem I wanted to apply it to. I would have mentioned the connection sooner but the answer came before I ggot to that.

REPLY [6 votes]: I get $n_{m} > m^2$ for the following values of $m$ (up to 322):
$$\eqalign{ &118, 139, 140, 141, 152, 153, 176, 177, 179, 180, 182, 183, 184, 185, 186, 188,\cr     &189, 190, 191, 192, 193, 194, 196, 197, 198, 199, 200, 201, 202, 203, 209, 
210,\cr     &220, 221, 222, 223, 224, 225, 227, 228, 229, 230, 231, 232, 233, 234, 235, 
236,\cr     &242, 243, 244, 249, 250, 251, 252, 253, 254, 259, 260, 261, 262, 263, 264, 
265,\cr     &266, 267, 270, 271, 272, 273, 274, 275, 278, 279, 280, 281, 282, 283, 284, 
285,\cr     &286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 
301,\cr    &302, 303 }$$<|endoftext|>
TITLE: Ideals of affine space curves
QUESTION [5 upvotes]: Given some algebraic closed curve in the affine space $\mathbb{A}^3_\mathbb{C}$, is there a way to decide whether its ideal (polynomials in $\mathbb{C}[X,Y,Z]$ vanishing on the curve) is generated by two elements?
I am mostly interested in the case where the curve is smooth and irreducible.

REPLY [7 votes]: There is a necessary condition: the element in the divisor class group of the affine curve
coming from the cotangent sheaf of the curve must be trivial.  If $I$ is generated by two elements, then $I/I^2$ is a free sheaf of rank $2$ on the curve.  By adjunction, this forces the cotangent sheaf of the curve to be trivial.  This immediately suggests how to make examples where $I$ is not generated by two elements.  Begin with a smooth, projective curve $X$ of genus $g>1$.  Let $x$ be a closed point such that $\omega_X(-(2g-2)\underline{x})$ is nontorsion.  For integers $N\gg 0$, the divisor $N\underline{x}$ is very ample.  For three general sections $f_1,f_2,f_3$ of $\Gamma(X,\mathcal{O}_X(N\underline{x}))$, form the morphism $(f_1,f_2,f_3):X\setminus\{x\} \to \mathbb{A}^3$.  If $f_1,f_2,f_3$ are sufficiently general, this will be a closed immersion.  The image is a smooth curve whose ideal $I$ is not generated by two elements.
$\textbf{Edit.}$  According to my computations, if $X$ is a genus $2$ curve and $x$ is a generic (hence non-hyperelliptic) point, then the linear system $|5\underline{x}|$ is sufficient for this argument.  This will embed $X$ in $\mathbb{P}^3$ as a degree $5$ curve whose "osculating 2-plane" $H$ at $x$ has contact of order $5$.  Thus the complement of $H$ will be an affine space $\mathbb{A}^3$ and $X\setminus (X\cap H)$ equals $X\setminus\{x\}$ is an affine curve whose ideal is not generated by two elements.
$\textbf{Second edit.}$  Using "Serre's Construction" and the Quillen-Suslin theorem (perhaps avoidable), you can show that the necessary condition above is also sufficient.  Let $R$ denote $k[x_1,x_2,x_3]$, and let $I$ denote the defining ideal of the affine curve $C$.  Then $\text{Ext}^1_R(I,R)$ is annihilated by $I$, i.e., equivalent to an $R/I$-module.  As an $R/I$-module, it is isomorphic to the dual of $\bigwedge^2(I/I^2)$, which is isomorphic to $R/I$ by hypothesis.  Choose an element that generates this $R/I$-module, i.e., $0\to R \to F \to I \to 0$.  Since $I$ is locally generated by 2 elements, it is not hard to check that $F$ is a locally free $R$-module of rank 2.  By the Quillen-Suslin theorem (or perhaps something weaker), $F$ is a free $R$-module of rank $2$.
$\textbf{Third edit.}$  I just noticed that there are some nice notes on "Serre's construction" (used above) written by my colleague, Christian Schnell.  Here is the URL:http://www.math.sunysb.edu/~cschnell/pdf/notes/serre.pdf.<|endoftext|>
TITLE: Topics for an Undergraduate Expository Paper in Number Theory
QUESTION [11 upvotes]: I am teaching an undergraduate course in number theory and am looking for topics that students could take on to write an expository paper (~10 pages). No new results are expected of them. Many of the students have an undergraduate course in abstract algebra and a course in real analysis (but few have any complex analysis background).
The only topics that I have come up with are
1.) Elliptic curves
2.) Cryptography
Of course these are related but I think these could be two projects. Other topics (such as the prime number theorem) seem too difficult to me. What other good projects are there? In particular are there good projects based on analysis? References would be greatly appreciated, including references for the two projects above.

REPLY [2 votes]: Arithmetic functions -- there are some nice, very accessible results about arithmetic functions and some questions about these (notably the sum of divisors, $\sigma$) date back to the Greeks (perfect numbers, amicable pairs, etc.)  There is also a nice algebra of arithmetic functions using the operation of convolution.  Many basic number-theoretic ideas (RSA!) have arithmetic functions hiding in the background.  
Example of arithmetic functions include the Euler totient $\phi$-function, $\sigma(n)=$ sum of divisors of $n$, $\tau(n)=$ number of prime divisors of $n$ and the Moebius function $\mu$.  
I once had a small team of undergraduates (who had no higher mathematics in their backgrounds) create their own arithmetic functions and analyze questions about iterates of the functions.<|endoftext|>
TITLE: Noncommutative computational package
QUESTION [5 upvotes]: I am wondering if there is a program which can do simple operations over noncommutative rings, like expand products and substitute one expression for another.
To clarify, consider the following situation.  I have two reductions $ab\mapsto 1$ and $ca\mapsto c-1$.  If I consider the monomial $cab$ I can reduce it in two ways: $cab=c(ab)=c$ or $cab=(ca)b=(c-1)b=cb-b$.  I can combine these computations to arrive at a third reduction $cb\mapsto b+c$.
I'm in a situation where I have upwards of twelve reduction rules, and it gets very complicated doing the reductions.  I find myself making small errors.  Thus, the need for a machine to do these computations for me.
To make this more precise, is there a program where I can first input a number of reductions, and then second have it work on a monomial and spit out a reduced form?

REPLY [3 votes]: There is this non-commutative algebra package for Mathematica that is quite extensive

http://www.math.ucsd.edu/~ncalg/

It can handle the symbolic computations in the question, among many other things.<|endoftext|>
TITLE: Grothendieck ring of "varieties carrying a function"
QUESTION [18 upvotes]: Fix a base ring $R$, and consider pairs $(X,f)$ where $X$ is a scheme
of finite type over $R$ and $f:X\to R$ is an $R$-valued algebraic (not
constructible!) function on $X$.
I want to consider a Grothendieck $R$-algebra of such pairs, where
if $X = Y \coprod Z$, then $[(X,f)] = [(Y,f|_Y)] + [(Z,f|_Z)]$,
but also $[(X,f+g)] = [(X,f)] + [(X,g)]$ and $[(X,rf)] = r[(X,f)]$.

Surely this is a standard extension of the usual notion of the
  Grothendieck ring of varieties (which only has $f=1$, and the first
  sort of relation)? If so, where can I read about it?

Maybe I'm misreading the motivic integration survey literature (by
K. Smith, and E. Looijenga), but it seems like they're insisting 
on constructible functions, not algebraic. 
Ordinarily when a construction like this isn't in the literature, I assume it's
because it has too many relations and is $0$, but if $R = {\mathbb Z}$ 
it seems to me that this ring has many functionals, like
$[(X,f)] \mapsto \sum_{x \in X_p} (f(x) \bmod p) \in {\mathbb Z}/p.$ (I don't
see an analogue of $[X] \mapsto$ the Euler characteristic $\chi(X_{\mathbb C})$.)
EDIT: One problem I see is that $({\mathbb A}^1, f(x)=x)$ is
isomorphic under translation to $({\mathbb A}^1, f(x)=x+1)$. So
$[({\mathbb A^1}, 1)] = [({\mathbb A}^1, (x+1)-x)]
= [({\mathbb A}^1, x+1)] -  [({\mathbb A}^1, x)] = 0$. Of course
this fits with point-counting $\bmod p$.

REPLY [9 votes]: I have just come across this question, and I have no idea if the OP still has any interest in it, but theories built out of pairs $(X, f)$ go under the name of "exponential motives", and there has been quite a lot of recent interest in developing and applying them. Two recent papers that come to mind, emphasising different aspects, are Motivic classes of Nakajima quiver varieties, https://arxiv.org/pdf/1603.03200.pdf, and Exponential motives http://javier.fresan.perso.math.cnrs.fr/expmot.pdf; these papers in turn have several further references.<|endoftext|>
TITLE: Cantor Sets Inside Cantor Sets
QUESTION [7 upvotes]: (Or: "I heard you liked Cantor Sets...")
I'm working on a student project, and the following construction came up very naturally: If $C$ is the usual Cantor Set, build a countable union of copies of this set as follows: start from $C$, and add a one-third scale copy of $C$ in the interval $[1/3,2/3]$, add three one-ninth scale copies of $C$ in the intervals $[1/9, 2/9]$, $[4/9,5/9]$ and $[7/9, 8/9]$, etc.
To Clarify: basically, every time you remove an interval in the classical construction, in this version, you do not remove the whole interval but you replace it with an appropriately scaled copy of the Cantor set. But you also add Cantor sets in the "holes of the holes", so to speak. So that's why in the second stage, there is a copy of $C$ in $[4/9,5/9]$, that fills in the middle third gap that appears in the copy of $C$ that was added at the first stage.
I guess the best way to see it is that at step $n$, you add to the set previously constructed copies of $C$ scaled at $3^{-n}$ in every empty interval where such a copy will fit.
I have my reasons for wanting to look at this construction, but that got me wondering: it looks so natural that it may very well come up in several contexts.
So. Anyone knows where this construction first appeared? Is it especially notable? Does it illustrate any especially interesting property? I would hate to miss something good about it.

REPLY [8 votes]: The set you are considering, obtained  by "filling the gaps" of the Cantor set $C$ by rescaled copies of $C$, is a dense Fσ set of null Lebesgue measure. It may be equivalently described as the invariant superset of $C$ generated by the (non-commuting) mappings $x\mapsto x/3$, $x\mapsto x/3+1/3$, $x\mapsto x/3+2/3$. So, as Andreas says, it can also be described as the set of all points in the unit interval that admit a ternary expansion with finitely many 1's . The complement is a full-measure Gδ  set (thus, of course, uncountable: there are uncountably many ways of having infinitely many 1's in one's ternary expansion). 
Your construction is customary in  the handicraft of examples and counter-examples in Topology and Measure Theory. For instance, a variant of it (with fat Cantor sets) produces a set $S$ that sub-divides any non-empty open set $A$ into two parts of positive measure: $|A\cap S| > 0$ and  $|A\setminus S| > 0$.<|endoftext|>
TITLE: Conjuring phantoms by hand?
QUESTION [25 upvotes]: A map $f:X\to Y$ of CW-complexes is called a phantom if $f$ restricted to the $n$-skeleton of $X$ is contractible for all $n$. The first non-trivial example of such a map, with $X=\Sigma\mathbb{P}^\infty(\mathbb{C})$ and $Y$ an infinite wedge of 4-spheres, was constructed by J. F. Adams and G. Walker in 1964. Subsequently, it was shown that many other examples exist. In particular, in 1966 B. Gray showed that there are continuously many non-homotopy equivalent phantoms $K(\mathbb{Z},2)\to S^3$.
I would like to ask if there is a way to construct a non-trivial phantom map, or at least to prove such maps exist, by hand (say, using the material covered in Hatcher's Algebraic topology).
The motivation is this: a colleague of mine has gone away for a while and has asked me if I could replace him during his topology problem classes. I agreed but the instructions given to me were rather vague ("just show them some cool examples..."). I have enough examples to fill all the sessions (a couple of those found on MO by the way), but still I was wondering if I could construct a phantom by hand, and learn myself how to do this. The students seem to be pretty smart but haven't seen much beyond the basic cohomology and homotopy theory, not yet anyway.

REPLY [12 votes]: For any CW complex $X$ there is a map $\theta_X:X\to \bigvee_{n=1}^\infty \Sigma X_n$, where $X_n$ denotes the $n$-skeleton of $X$. This map can be realized as the cofiber of the folding map $\bigvee_{n=1}^\infty X_n \to X$. By construction, the map $\theta_X$ is phantom, called the Universal Phantom Map out of $X$. 
Gray showed in his thesis that every phantom map $X\to Y$ factors through $\theta_X$ up to homotopy, so essential phantom maps out of $X$ exist if and only if $\theta_X$ is nontrivial. Since we know, for example, that $\mathbb{C}P^\infty$ is the domain of essential phantom maps, it follows that $\theta_{\mathbb{C}P^\infty}:\mathbb{C}P^\infty\to \bigvee \Sigma \mathbb{C}P^n$ is an essential phantom map that is the homotopy cofiber of the map $\bigvee \mathbb{C}P^n \to \mathbb{C}P^\infty$. 
This example is included in McGibbon's article in the Handbook, and is addressed in some detail in Modern Classical Homotopy Theory by Jeff Strom.<|endoftext|>
TITLE: Degeneration of riemannian metrics with curvature bounds
QUESTION [6 upvotes]: In short, I'm curious to know what modes of degeneration of metric might still keep the curvature bounded. More precisely, assume we are keeping the total volume of the manifold fixed and deform the metric, e.g by a conformal factor that preserves the  volume. If we allow the metric to degenerate, namely allow it to become semi-definite, $g_{ij} \geq 0$ on some set, is it possible to keep the Ricci curvature bounded below and yet let the metric degenerate? If not, is it possible to still keep some $L^p$ bound on the sectional, Ricci, or scalar curvature? 
Edit: It seems that the post requires some clarification: I am aware that there is whole field of studying degeneration of metrics, convergence etc. I was mostly curious to see examples of  degeneration of metrics while curvature is bounded. Examples that could, for example, illustrate when you can keep the scalar curvature bounded but Ricci, or any $L^p$ norm of Ricci, might blow up.
More than the conformal deformation I was curious to see an example of the case when metric in a K\"ahler class degenerates as one varies the potential, but one curvature functional, say $\Vert Ric \Vert_p$ remains bounded.

REPLY [6 votes]: John Lott has a number of papers on this sort of thing, many to do with the spectrum of the Laplacian on collapsing manifolds.  Nice thing is they are all available from his home page at Berkeley, http://math.berkeley.edu/~lott/papers.html.  If you look at the most recent preprint, it should point you in all sorts of interesting directions.<|endoftext|>
TITLE: How to tell a paradox from a "paradox"?
QUESTION [6 upvotes]: Russell's paradox showed that naive set theory leads to a contradiction. This was something that was taken seriously and caused a lot of work.
Now, Banach–Tarski paradox is arises from a result that a ball can be decomposed into finite amount of pieces and the pieces can be used to built two identical copies of the decomposed ball. Banach-Tarski paradox is often treated as a "paradox", basicly meaning that, yes, it is counter intuitive but yet there is no problem - mathematics just occasionally is counter intuitive.
To be honest, I have never understood why Banach-Tarski is not a "real" paradox but not being expert of measure theory I chose to accept the common view.
Is there some high level explanation on how to tell a paradox from a "paradox"? What is it that makes a counter intuitive result to a "real mathematical paradox" that we should start worrying about?

REPLY [3 votes]: What you're describing as a "true paradox" is sometimes called an "antimony" and it means an actual logical inconsistency in the underlying theory.  The Burali-Forti paradox is another example and it means there can't be a set of all ordinals (the ordinals are a proper class).  By contrast, the Banach-Tarski theorem is consistent; it's just counterintuitive.  The reason we don't hear much about "true paradoxes" (antimonies) these days is that logicians in the 1920's got the earlier inconsistencies under control, and (in all likelihood) we're not dealing with any actual inconsistent systems today, at least in everyday mathematics.<|endoftext|>
TITLE: Periodic Holomorphic ODE
QUESTION [5 upvotes]: Suppose I have an annulus $U\subset \mathbb{C}$ and a single-valued holomorphic function $V:U\to \mathbb{C}$.
I would like to know if there are (tractable) conditions on $V$ that ensure that the second order linear ODE
$$
\partial_{zz}^2 f+ V f=0
$$
admits a single-valued holomorphic solution $f: U\to \mathbb{C}$ (here $z$ is the usual complex coordinate).
Such a question is easy to answer for the first order linear ODE 
$$
\partial_{z} f+ V f=0
$$
Indeed, in this case separation of variables tells us that a necessary and sufficient condition is that the residue of $V$ is an integer. 
I should note that taking $V=\alpha z^{-2}$ and $U$ the annulus $\lbrace z: 0<|z|<1\rbrace$ allows one to give explicit necessary and sufficient conditions on $\alpha$.
Where are questions of this sort treated?  It seems like a classical problem...

REPLY [4 votes]: When the annulus is $0<|z|<1$, and $V$ has at most a second order pole at $0$, the answer is known,
and is given by Fuchs theory. This is the so-called regular singularity at $0$. The reference is any
book on analytic theory of linear differential equations, for example,
Ince, Ordinary differential equations. In all other cases, the question is much harder,
and there is no simple explicit answer.
Putting $z=\exp(w)$ you obtain a linear differential equation of second order with analytic periodic
coefficients, and you are interested in the existence of a periodic solution with the same period.
This was subject to much research since the 19 century, and of course a lot is known,
but there is no simple answer. The key words are "Hill's equation", "Hill's determinant",
"Lyapunov", etc. You can look into Whittaker Watson Chapter 19 "Mathieu functions" for the
beginning.<|endoftext|>
TITLE: Comparison of the norm of a matrix and its entry-wise absolute value.
QUESTION [15 upvotes]: It is an easy fact that for a matrix $A \in M_n(\mathbb C)$, the matrix $A' = (|A(i,j)|)_{i,j \leq n}$ has a larger operator norm than $A$. By operator norm I mean the norm as an operator on $\ell^2_n$, or equivalently its largest singular value.
My question is: 

What happens when the operator norm is replaced by the Schatten $p$-norm?

Let me recall that for $1 \leq p \leq \infty$ and a matrix $A \in M_n(\mathbb C)$, its
Schatten $p$-norm  $\|A\|_p$ is the $\ell^p$-norm of its singular values, or equivalently $\|A\|_p= Tr((A^*A)^{p/2})^{1/p}$ if $p<\infty$. If $p=\infty$, this is just the operator norm.
Some remarks:

If $p=2$ the equality $\|A\|_2 =\|A'\|_2$ is obvious.
If $p$ is an even integer, the inequality $\|A\|_p \leq \|A'\|_p$ holds (just expand $\|A\|_p^p = Tr( (A^*A)^{p/2})$ and use the triangle inequality to prove that $\|A\|_p^p \leq \|A'\|_p^p$).
When $A$ is a complex Hadamard matrix (a unitary with all entries having absolute value $1/\sqrt n$), $\|A\|_p < \|A'\|_p$ for $p>2$ and $\|A\|_p> \|A'\|_p$ for $p<2$: indeed, $p$-norm of $A$ is $n^{1/p}$ and the $p$-norm of $A'$ is $\sqrt n$. A naive guess would be that these inequalities always hold, BUT: 
When $p<6$ and $p \neq 2,4$, the quantities $\|A\|_p$ and $\|A'\|_p$ are not comparable (with constants independant of $n$). For $p<4$, take $A=\begin{pmatrix} 1 & -1 & 0\\ 0 & 1 & 1 \\ 1 & 0 & 1\end{pmatrix}$, and for$4<p<6$ take $A = \begin{pmatrix} 0 & 1&1 & 0\\ 1&0 & 1 & 0 \\ 1 & 0&0 & -1\\0&1&0&1\end{pmatrix}$.

So the question is really for $p>6$. For example, I could not find any matrix $A$ such that  $\|A\|_7 > \|A'\|_7$.
Update I was informed by Gilles Pisier that this question has already been considered. This question has been first answered by V. Peller, and the answer was refined successively by B. Simon, M. Dechamps-Gondim, F. Lust-Piquard and H. Queffélec and L. Rosen. The counterexamples given in DLQ are the same as the ones I give in my answer, with a slightly different proof.

REPLY [3 votes]: Mikael's explicit construction is extremely nice. Here's an ugly numerical example for those who like matlab based counterexamples.
This problem turned out to be quite hard to defeat numerically by brute force, which is a common characteristic of many singular value inequalities. I tried slightly more structured matrices, and am tempted to believe that I have found computational counterexamples.
Here's the brute-force construction.
a=randn;b=2*randn;c=rand;
m=gallery('tridiag', b*ones(n,1), a*ones(n+1,1), c*ones(n,1));
m=full(m);
m(n,1)=-2;   % this additional entry is important.
lhs = norm(svd(m),p);
rhs = norm(svd(abs(m)),p);

Now, if you loop over the above random "almost" tridiagonal matrices, Matlab quickly reports a counterexample where "lhs" is bigger than "rhs" (I tried $n=9$ and $p=7$).<|endoftext|>
TITLE: Small Ramsey numbers and Brooks' Theorem
QUESTION [5 upvotes]: I'm studying about Graph Ramsey Theory now. Starting this study, I'm reading Chvatal and Harary's series of papers. In the second paper (V.Chvatal, F.Harary, Generalized ramsey theory for graphs,Ⅲ. Small off-diagonal numbers, Pacific Journal of Mathematics 41, No.2, 1972, pp.335-345), I can't understand the proof of $r(C_4,K_4)=10$. Their proof is like following.
Let $G$ is arbitrary simple graph of order 10 with point independence number $<4$. It is sufficient to prove $G$ contains $C_4$. From $G$'s point independence number is $<4$, $G$'s (point) chromatic number is $\ge4$. Hence by Brooks' theorem either $K_4$ (and hence $C_4$) is contained in $G$, or the degree of each point of $G$ is at least four. If the first case occur, we have done. If the second case occur, we also have $C_4$ in $G$ by the following lemma (I omit the proof of this lemma but it's not so difficult). 
Lemma. If a graph $G$ with $p$ points has minimum degree $d$ and $d(d-1)>p-1$, then $G$ contains $C_4$.
I can't understand how to use Brooks' Theorem. I only succeed to derive the maximum degree of $G$ is greater than 3. How to derive that the minimum degree of $G$ is greater than 3 from Brooks' Theorem? Chvatal and Harary's proof is wrong as it is? or not? (If you have other elegant proof of $r(C_4,K_4)=10$, then It also help me.)
supplementation:I got a (awkward?) proof of $r(C_4,K_4)=10$. The proof is like following.
For lower bound, we use Chvatal-Harary theorem.
For upper bound, we think about above graph $G$. Using $r(C_4,K_3)=7$, easily we have there is no vertex with degree $\le2$. By lemma, we have at least one vertex (say $u$) whose degree is 3.
Claim. The subgraph induced by vertices non-adjacent to $u$ contains $2K_3$.
The subgraph induced by vertices non-adjacent to $u$ has 6 vertices. So we have two triangles $T_1,T_2$ in this subgraph. If $T_1,T_2$ has two common vertex, we get $C_4$. If $T_1,T_2$ has only one common point, let $T_j=v_0v_1^jv_2^j$. Let $w$ be the other vertex non-adjacent to $u$. Then $v_2^1w$ isn't an edge by symmetry and avoiding $C_4$, namely $v_0v_1^1wv_2^1$. We also have edge $v_1^2w$ by avoiding 4 independent vertices $v_2^1wv_1^2u$.By symmetry, we have $C_4$, namely $v_0v_1^2wv_2^2$ and it's a contradiction.
Claim. The neighborhood subgraph $N(u)$ of $u$ is $\bar{K_3}$.
If not, the neighborhood subgraph of $u$ is an isolated vertex $v_1$ and an edge $v_2v_3$. $v_2$ and $v_3$ has at least one edge to $T_1\cup T_2$, since their degree $\ge3$. If $v_2$ and $v_3$ has edges to common triangle, we get $C_4$. So if we let $T_j=w_1^jw_2^jw_3^j$, we can assume there are edges $v_2w_1^1, v_3w_1^2$. ($v_1,v_2$ has no other edges to $T_1\cup T_2$.) Then both edges $v_1w_2^1,v_1w_3^1$ cannnot be exist. So we can assume there isn't edge $v_1w_3^1$, then we have edge $w_3^1w_1^2$, since
 otherwise we have 4 independent vertices $v_1v_2w_3^1w_1^2$. By symmetry, we also have edge $w_1^1w_3^2$. So we have $C_4$, $w_1^1w_3^2w_1^2w_3^1$. It's a contradiction.
Now, we have $T_j=w_1^jw_2^jw_3^j$ and 6 edges $v_iw_i^j$. Then we have edge $w_1^1w_1^2$, since otherwise $w_1^1w_1^2v_2v_3$ form 4 independent vertices. By symmetry, we have $C_4$, namely $w_1^1w_1^2w_2^2w_2^1$. It's a contradiction.

REPLY [5 votes]: It seems as if the Chvatal, Harary proof has a logical gap, and your proof seems to be missing some details.  
Here is a proof that is based on Brook's Theorem.  We plagiarize you and start by noting that $r(C_4,K_3)=7$, and so $G$ has minimum degree at least 3.  We then plagiarize Chvatal, Harary and note that $G$ has chromatic number at least 4.  Thus, by Brook's Theorem, $G$ has maximum degree $\Delta(G)$ at least 4.  Let $v$ be a vertex of maximum degree and let $N(v)$ be the neighbours of $v$ and let $S(v)$ be the non-neighbours of $v$.  Since $G$ has no $C_4$, note that each vertex in $S(v)$ has at most one neighbour in $N(v)$.  Thus, the minimum degree of the subgraph induced by $S(v)$ is at least 2.  This rules out $\Delta(G)=9,8$, or $7$.  
If $\Delta(G)=6$, then there are at least three vertices $x,y,z \in N(v)$ which are not adjacent to any vertex in $S(v)$.  Since $x$ has degree at least 3 in $G$, it must be adjacent to at least two other vertices in $N(v)$, which creates a $C_4$.  
If $\Delta(G)=5$, then $G[S(v)]$ is a graph on 4 vertices with minimum degree 2.  Such a graph necessarily contains a $C_4$.   
We now suppose $\Delta(G)=4$.  In this case, $G[S(v)]$ is a graph on 5 vertices with minimum degree 2.  Thus, every cycle of $G[S(v)]$ must be of length 3 or 5.  If $G[S(v)]$ contains a $C_5$, then $G[S(v)]=C_5$, since adding any chord to a $C_5$ produces a $C_4$.  Thus, each vertex in $G[S(v)]$ has exactly one neighbour in $N(v)$.  Hence, $G[N(v)]$ must be a matching $\{ab, cd\}$ of size 2, else $G$ has a vertex of degree 2.  It follows that each vertex in $N(v)$, has at least one neighbour in $S(v)$.  Thus, one vertex (say $a$) has exactly two neighbours in $S(v)$, while $b,c$, and $d$ have exactly one neighbour in $S(v)$. Let $x,y,z$ be the vertices in $S(v)$ which are not adjacent to either $b$ or $c$.  If any of $xy,yz,xz \notin E(G)$, then $G$ has a stable set of size 4.  Thus, $xyz$ is a triangle.  This contradicts that $G[S(v)]=C_5$.   
The only remaining possibility is that $G[S(v)]$ is a bowtie.  In particular, $G$ has two non-adjacent vertices $u$ and $v$ of degree 4 such that $N(u)$ and $N(v)$ are disjoint.  Thus, the subgraph $H$ of $G$ induced by $N(u) \cup N(v)$ has minimum degree at least 2.  In particular $H$ contains a cycle $C$.  It is easy to verify that if $|C|=3,4,5,6,$ or $7$, then $G$ contains a $C_4$ since $H$ cannot contain a vertex with two neighbours in $N(u)$ or two neighbours in $N(v)$.  Thus, $|C|=8$.  But then $H=C_8$ else $H$ contains a cycle of smaller length.  Thus, every other vertex of $H$ is a stable set in $G$, a contradiction.  
Note that this proof avoids the use of the minimum degree lemma.<|endoftext|>
TITLE: Three half circles on the plane may not meet nicely
QUESTION [11 upvotes]: Let $H$ denote the union of the northern hemisphere of the unit circle $S^{1}$ with the interval $[-1,1]$ on the $x$-axis. That is, $H=\{(x,\sqrt{1-x^{2}}):-1\le x\le 1\}\cup\{(x,0):-1\le x\le 1\}$
Let us say that two copies of $H$ meet nicely if they intersect in exactly 6 points e.g. as the two pictures show below:

 or 

Note that the picture on the right shows two half-circles meeting nicely but whose centers do not lie inside their partner´s half disk.
Now, if we have three copies of $H$, it may not seem possible to arrange them so that they meet nicely, i.e. both of the following two conditions hold:
1. Any two meet nicely, and
2  The intersection of the three is empty.
Is this true? Or, if I am mistaken, I would appreciate that someone would show me, or describe, the desired arrangement.
EDIT: I am considering ALWAYS half-circles, i.e. copies of $H$. No need to give answers related to half-disks.
EDIT: It is possible to arrange three copies of $H$ so that both 1. and 2. hold (See the selected answer below).

REPLY [12 votes]: 1. This is the answer under the assumption that the condition 2. means exactly what it says.
Consider a regular triangle with side $2+\varepsilon$ and three diameters in the middles of its sides. If you construct the half-circles towards the triangle on these diameters, you obtain the desired example.


Now about the four copies.
Lemma. Assume that the two copies of $H$ meet nicely. Consider their supporting half-planes determined by the diameters. Then their intersection is an acute angle, and the centers belong to its sides. (Possibly this angle is degenerate; in this case, it should be 0 but not $\pi$, which means that the intersection should be a strip but not a half-plane.)
Proof. If the two diameters do not intersect, then each of three pairs of the form (diameter, half-circle) and (half-circle,half-circle) meet at two points. Now, consider a point $A$ of intersection of the lines supporting the diameters. At least one diameter (say, $d_1$) does not contain $A$. Hence, if the angle in the Lemma statement is not acute, then the half-circle on $d_1$ cannot intersect $d_2$. Next, the center $C_1$ clearly lies on the side of this angle. Finally, the projection of $O_2$ onto $d_1$ should lie on the segment $d_1$, hence $O_2$ is also on the side of the angle (but not on its prolongation).
Finally, assume that the diameters intersect.Then each half-circle can intersect the other diameter in at most one more point, and the total number of the intersection points is less than 6. Lemma is proved.
Now we can prove that the four copies of $H$ cannot pairwise meet nicely. Let $c_{ij}$ be the angle from the lemma for $H_i$ and $H_j$. It is easy to see that $c_{12}$, $c_{13}$, $c_{23}$ should form an acute-angled triangle with the centers $C_1$, $C_2$, $C_3$ on its sides (just try to add the third diameter to $c_{12}$!). But then it is impossible to add the fourth half-plane --- these four half-planes should now form a quadrilateral with four acute angles!
2. Now let us assume that you speak on the half-disks. Then the answer is positive. From the previous paragraph, we see that the three diameters lie inside the three sides of some acute triangle $XYZ$, respectively. 
Now consider the three distances  between the centers. If all three are less than $\sqrt3$, then by Jung's theorem they may be covered by the unit disk, and the center of this disk belongs to all three half-disks.
Otherwise, assume that $C_1C_2\geq \sqrt3$, where $C_1$ and $C_2$ be the centers on the sides $XY$ and $XZ$, respectively. We have $d(C_1,XZ)\leq 1$, otherwise the respective half-circle and segment do not intersect. But then the projection of $C_1$ onto $XZ$ is at least $\sqrt2$ away from $C_2$, hence the first half-circle cannot intersect the second diameter twice. So in this case we also get a contradiction.
I may expand any part of the above sketch.<|endoftext|>
TITLE: Modern proof of Serre's open image theorem?
QUESTION [29 upvotes]: Let $E$ be an elliptic curve defined over a number field $K$ without complex multiplication. Serre's open image theorem (which appears in his book 'Abelian $l$-Adic Representations and Elliptic Curves') says that the image of the representation of $Gal(\bar{K} / K)$ on the $l$-adic Tate module $T_l(E)$ is open in $GL_2(\mathbb{Z}_l)$. 
Is there a modern proof of this written down somewhere using Faltings' Theorem (i.e. the Tate conjecture) or other methods?
Edit: I've just found Ribet's review of Serre's book, which contains fairly detailed sketch of the kind of proof I was after, so I included it below.

REPLY [23 votes]: Here is Ribet's proof (expanding on Ulrich's comment):
Let $G_K:=Gal(\bar{K} / K)$ and $V_l:=T_l(E)\otimes \mathbb{Q}_l$.
The image $\rho_{l,E}(G)$ is a closed subgroup of the $l$-adic Lie group $\text{Aut}(V_l(E)) \cong \text{GL}_{2}(\mathbb{Q}_l)$ and is therefore a Lie subgroup of $\text{Aut}(V_l(E))$. Its Lie algebra $\mathfrak{g}_l$ is a subalgebra of $\mathfrak{gl}_{2}(\mathbb{Q}_l)$. We want to show that $\mathfrak{g}_l=\mathfrak{gl}(V_l)\cong \mathfrak{gl}_2(\mathbb{Q}_l)$ and the result follows. 
(Note that the Lie algebra of the image $\rho_{l,E}(G_K)$ is the tangent space of the identity component of the Zariski closure of $\rho_{l,E}(G_K)$ in $\text{GL}_{2}(\mathbb{Q}_l)$. So $\mathfrak{g}_l$ `measures the representation up to finite extensions of the base field $K$', since a finite index subgroup of an algebraic group has the same identity component). 
Now $V_l$ is irreducible as a $\mathfrak{g}_l$-module (this is a theorem of Shafarevich, and depends on Siegel's theorem on the finiteness of integral points on curves). Secondly, $\mathfrak{g}_l$ can't be contained in the subalgebra $\mathfrak{sl}(V_l)$ of $\mathfrak{gl}(V_l)$ since $\det(\rho_{l,E})=\chi_l$ (where $\chi_l$ is the cyclotomic character giving the action of Galois on $K^{cycl}$).
This leaves two possibilities for $\mathfrak{g}_l$: either $\mathfrak{g}_l$ is $\mathfrak{gl}_2(\mathbb{Q}_l)$ and we're done, or $\mathfrak{g}_l$ is a non-split Cartan subalgebra of $\mathfrak{gl}_2( \mathbb{Q}_l)$ (an abelian semisimple algebra coming from a quadratic field extension of $\mathbb{Q}_l$).
Faltings proved two important facts about represenations $\rho_{l,E}$:

$\rho_{l,E}$ is a semisimple representation of $G_K$ over $\mathbb{Q}_l$
$\text{End}(E)\otimes \mathbb{Q}_l \cong \text{End}_{\mathfrak{g}_l}(V_l)$.

Faltings results then rule out the possibility that $\mathfrak{g}_l$ is a non-split Cartan subalgebra of $\mathfrak{gl}_2( \mathbb{Q}_l)$ and we're done.<|endoftext|>
TITLE: Hyperreal finitely-additive measure on [0,1) assigning $b-a$ to $[a,b)$ or $(a,b]$ and infinitesimals to singletons
QUESTION [9 upvotes]: Is there a hyperreal-valued finitely additive measure on all the subsets of [0,1), or at least the Borel ones, that 

assigns $b-a$ to $[a,b)$ and to $(a,b]$ for all $a\lt b,$ and 
assigns an infinitesimal--ideally, the same one--to each singleton?

It's (1) that's a problem.  The Bernstein-Wattenberg construction yields a finitely-additive measure that gives (1) up to infinitesimals.  But it would be nice to have (1) exactly.

REPLY [7 votes]: Yes, by compactness. 
Let $R$ denote your favorite hyperreal ordered field 
and let $\delta\in R$ be a positive infinitesimal. 
Let $\mathcal{E}$ denote the set of all 
(standard) finite Boolean subalgebras of $\mathcal{P}([0,1))$.
For every $A\in\mathcal{E}$,
let $\lambda_A(I)$ be the (exact) length of $I$ 
for all half-open intervals $I\in A$; 
for all open or closed intervals $I\in A$, 
respectively subtract or add $\delta$ to the length of $I$ 
to define $\lambda_A(I)$;
let $\lambda_A(S)=\delta$ for all singletons $S\in A$.
Extend $\lambda_A$ to a probability measure $\mu_A$ on $A$.
(Specifically, for 
each minimal finite union of intervals $F\in A$, 
let the connected components of $F$ be $[a_0,b_0),\ldots,[a_k,b_k)$
with $p$ elements of $\{a_i,b_i:i\leq k\}$ added and $q$ removed.
Partition $F$ into its atomic subsets $H_0,\ldots,H_n$.
Choose a positive $\mu_A(H_i)\in R$ for each $i$, such that
$\sum_{i\leq n}\mu_A(H_i)=(p-q)\delta+\sum_{i\leq k}(b_i-a_i)$.
Now extend $\mu_A$ from the atoms to all of $A$.)
Let $U$ be a fine ultrafilter on $\mathcal{E}$.
("Fine" means that $\{B\in\mathcal{E}:A\subseteq B\}\in U$
for all $A\in\mathcal{E}$.) 
The ultraproduct measure $\mu_U$ is $R^U$-valued and has 
the two properties you seek.<|endoftext|>
TITLE: Strongly Noetherian property. When is the tensor $A\otimes_{k}B$ Noetherian for Noetherian rings $A$ and $B$?
QUESTION [9 upvotes]: Let $k$ be a field. It is well-known that $A\otimes_{k}B$ is not necessarily Noetherian even if $k$-algebras $A$ and $B$ are Noetherian. For example $\mathbb{R}\otimes_{\mathbb{Q}}\mathbb{R}$. 

When is the tensor $A\otimes_{k}B$ Noetherian for Noetherian "commutative" $k$-algebras $A$ and $B$? 
What if $A$ is noncommutative? Are there any good criteria for $A\otimes_{k}B$ to be Noetherian? If $B$ is a finitely generated $k$-algebra, Hilbert's basis theorem implies that $A\otimes_{k}B$ is again Noetherian. So we need to check this with quite nasty $B$. 

My primary motivation to ask these questions is the second question. Such ring $A$ is called a "strongly Noethrian ring" and has a lot of good properties, but I don't know many examples. Moreover I realized that things are not very clear even in commutative case and I need to understand commutative case first. I would appreciate it if experts on MO could let me know good criteria for this property and provide me with examples. 
Rings I have in my mind are weakly noncommutative in the sense that they are commutative up to scalar multiplication such as quantum planes and their $good$ hypersurfaces.

REPLY [4 votes]: Although the OP is mainly interested in noncommutative results and examples, let me say a few words about the commutative case. 
Let $k\subset K$ be a field extension. N. Bourbaki in Algebre. Chapitre 8,  Modules et anneaux semi-simples (edition 1958), exercise 22, page 99, gives the following criterion: if the extension $k\subset K$ is algebraic, then $K\otimes_kK$ is Noetherian iff $[K:k]<\infty$. 
Later on, in 1978, P. Vamos in the paper On the minimal prime ideals of a tensor product of two fields proves a more general result: for a field extension $k\subset K$, $K\otimes_kK$ is Noetherian iff $k\subset K$ is finitely generated. 
Inspired by Vamos' result, Resco, Small and Wadsworth in the paper Tensor products of division rings and finite generation of subfields prove a (partially) noncommutative result: let $D$ be a division algebra over a field $k$ and $k\subset K$ a commutative subfield of $D$. Then $D\otimes_kL$ is Noetherian iff the extension $k\subset K$ is finitely generated. As a by-product they get that $D\otimes_kD^0$ Noetherian implies $k\subset K$ finitely generated for every commutative subfield $K$ of $D$ containing $k$.<|endoftext|>
TITLE: When is the derived category of representations of a finite poset equivalent to its opposite?
QUESTION [12 upvotes]: If I have a finite partially ordered set $K$, I can look at its derived category of finite dimensional representations $D(K)$. Note that $D(K^{op}) \simeq D(K)^{op}$ by linear duality. 
But when do we have an equivalence $D(K) \simeq D(K^{op})$?
The kind of thing I have in mind is in this paper of Justin Curry: http://www.math.upenn.edu/~jucurry/papers/co_sheaf_dereq.pdf
When the poset comes from a finite cell complex, there is such a duality which interchanges the standard injectives with skyscrapers. 
Does this depend only on the poset, and can this always be done?

REPLY [2 votes]: I don't know a very general answer.  Your duality on cell complexes resembles Verdier duality and has a local nature, but some of these equivalences aren't like that.  E.g. $K = (0 < 1 < 2)$ and $K^{op} = (2 < 1 < 0)$ are isomorphic as posets and so we get $D(K) = D(K^{op})$ from that, but I am pretty sure $K$ doesn't have a Verdier dualizing sheaf.
But here is an answer to a narrower question: when is there an equivalence $D(K) = D(K^{op})$ that is similar to the one on regular cell complexes?  I mean an equivalence that is Verdier-like in the sense that it takes the standard injectives on $K$ to the skyscrapers on $K^{op}$, up to a shift.
For each $x \in K$, define $J_x:K \to \mathrm{Mod}_k$ by
$$
J_x(w) = \begin{array}{cc}
k & \text{if $w \leq x$}\\\
0 & \text{otherwise}
\end{array}
$$
These are the indecomposable injective objects in the abelian category of functors $K \to \mathrm{Mod}_k$.
The $J_x$ have a simple Hom pattern: $\mathrm{Hom}(J_x,J_y) = k$ if $x \geq y$.  All other Homs (and Exts) vanish.  (They form an "exceptional collection" in $D(K)$.)
The simple objects are the skyscrapers $\delta_x$.  You can compute the Homs and Exts between $\delta_x$ and $\delta_y$ by writing down an injective resolution of $\delta_y$ whose $p$th term is
$$
\bigoplus_{y = y_0 \geq y_1 \geq y_2 \cdots \geq y_p} J_{y_p}
$$
The differentials have degree $+1$.  Then $\mathrm{Hom}(\delta_x,\text{that injective resolution})$ is a complex whose $p$th term is
$$
\bigoplus_{y = y_0 \geq y_1 \geq y_2 \cdots \geq y_p = x} k
$$
This is the cochain complex that computes something like the relative cohomology $H^*(N,\partial N)$ where $N$ is the nerve of the interval
$
\{p \in K \mid x \leq p \leq y\}
$
and $\partial N$ is the subcomplex of simplices that don't contain the edge $[x < y]$.  So a necessary condition is for this $\partial N$ to be a homology sphere.  
I think this condition is sufficient too, except you have to worry a little bit about the dimensions of those spheres.  You have to be able to choose integers $d(x)$ for each $x \in K$ so that the dimension of that homology sphere is $d(y) - d(x)$ or something.
Somebody once told me that this condition on the intervals in posets has a standard name, maybe "Gorenstein star posets" but I am not sure I am remembering that right.<|endoftext|>
TITLE: Classification of Hopf algebras (state of the art)
QUESTION [16 upvotes]: I assume that the classification of (certain families of) Hopf algebras is still an open problem, am I right?
My question is the following: What is the current state of the art? What is known about the classification of certain families of Hopf algebras? (To be more precise, say for example that I am interested in finite-dimensional or combinatorial Hopf algebras.)

REPLY [11 votes]: The general problem about the classification of finite-dimensional Hopf
algebras (over $\mathbb{C}$) is widely open. I mention some general results.
Here
Zhu, Yongchang. Hopf algebras of prime dimension. Internat. Math. Res. Notices 1994, no. 1, 53--59. MR1255253 (94j:16072), link
it is proved that a Hopf algebra of prime dimension is isomorphic to a group
algebra. Here
Ng, Siu-Hung. Non-semisimple Hopf algebras of dimension $p^2$. J. Algebra 255 (2002), no. 1, 182--197. MR1935042 (2003h:16067), link
it is proved that the only Hopf algebras of dimension $p^2$ are the group
algebras and the Taft algebras. Hopf algebras of dimension $2p^2$ were also
classified by Hilgemann and Ng, see the following paper:
Hilgemann, Michael; Ng, Siu-Hung. Hopf algebras of dimension $2p^2$. J.
Lond. Math. Soc. (2) 80 (2009), no. 2, 295--310. MR2545253 (2010h:16080), link
Of course, these are not the only general results. For a good accound related
to the classification of Hopf algebras of a given dimension you may want to check
the following papers:
Beattie, Margaret. A survey of Hopf algebras of low dimension. Acta Appl. Math.
108 (2009), no. 1, 19--31. MR2540955 (2010i:16054), link
M. Beattie and G. A. García. Classifying Hopf algebras of a given dimension.
Preprint: arXiv:1206.6529
It is important to mention that it is very interesting to study the classification of certain families of finite-dimensional Hopf algebras. If this is your interest, maybe google can help.<|endoftext|>
TITLE: Left Properness of Simplicial Commutative Algebras
QUESTION [5 upvotes]: A bit of light googling turns up several sources asserting that the model structure on simplicial commutative algebras over a ring is left proper (for example, 2.9 in Charles Rezk's paper Every homotopy theory of simplicial algebras admits a proper model). Does a proof of this fact occur anywhere in the literature?

REPLY [3 votes]: For simplicial commutative rings this is proved in Lemma 3.1.2
of Schwede's “Spectra in model categories and applications 
to the algebraic cotangent complex”,
and the proof there immediately extends to algebras.<|endoftext|>
TITLE: What is the Brauer group of the moduli space of (p.p.) abelian varieties?
QUESTION [11 upvotes]: What is the Brauer group of the moduli space of principally polarized abelian varieties of a given dimension? I am primarily interested in the "open" moduli space, i.e. not a compactification. The question has several levels of generality depending on how general is the ground field (ring even?) but I know nothing, so already the knowing the analytic Brauer group over the complex numbers would be very interesting for me.

REPLY [7 votes]: Edit. The main idea below is incorrect.  I explain the mistake below the original post.
I am posting as an answer instead of a comment, even though I think this might be wrong, because I could not format the weblinks properly in comments.
Since the (orbifold) moduli space is a quotient of the Siegel upper half space by the (orbifold) fundamental group $\textbf{Sp}_{2g}(\mathbb{Z})$, it seems to me that the analytic Brauer group of the moduli space should be $\text{Hom}(H_2,\mathbb{Q}/\mathbb{Z})$, where $H_2 = H_2(\textbf{Sp}_{2g}(\mathbb{Z}))$ is the kernel of the universal central extension of $\textbf{Sp}_{2g}(\mathbb{Z})$.  According to the proof of Proposition 2.3 of 
Finite quotients of symplectic groups vs mapping class groups
by Funar and Pitsch, it is well-known that $H_2$ equals $\mathbb{Z}$ for all $g\geq 3$.  So this implies that the analtyic Brauer group is $\mathbb{Q}/\mathbb{Z}$. 
Correction. Let $X$ be a smooth, complex orbifold with universal cover $f:\widetilde{X}\to X$ and with fundamental group $\Gamma$.  Denote the universal central extension of $\Gamma$ as follows, $$1\to H \to \widetilde{\Gamma} \to \Gamma \to 1.$$  For every homomorphism $\widetilde{\rho}:\widetilde{\Gamma} \to \textbf{GL}_n(\mathbb{C})$ mapping $H$ into the center of $\textbf{GL}_n(\mathbb{C})$, say $$\rho_H:H\to \mathbb{C}^\times,$$ there is an associated group homomorphism, $$\rho: \Gamma \to \textbf{PGL}_n(\mathbb{C}),$$ and vice versa.  There is also an associated $\textbf{PGL}_n(\mathbb{C})$-torsor over $X$ defined as the quotient of the diagonal action of $\Gamma$ on $\textbf{PGL}_n(\mathbb{C})\times \widetilde{X}$.  The class of this $\textbf{PGL}_n(\mathbb{C})$-torsor only depends on $\rho_H$.  In fact, there is a modification of $\widetilde{\rho}$ (not changing $\rho$) so that the image of $\rho_H$ is in the torsion subgroup $(\mathbb{C}^\times)_{\text{tor}} \cong \mathbb{Q}/\mathbb{Z}$.  The Brauer class of $\rho$ comes from the corresponding element in $\text{Hom}(H,\mathbb{Q}/\mathbb{Z})$.
Unfortunately, the kernel of $\rho_H$ contains the kernel of the map of $\widetilde{\Gamma}$ to its profinite completion.  Indeed, there is a finitely generated $\mathbb{Z}$-subalgebra $R$ of $\mathbb{C}$ that contains all the matrix coefficients of $\text{Image}(\widetilde{\rho}).$  The ring $R$ is a Jacobson ring.  Thus, for every element of $H$ that is mapped to a non-identity element, there is a maximal ideal $\mathfrak{m}$ of $R$ such that the image after reduction to $R/\mathfrak{m}$ is also a non-identity element.  Since $\textbf{GL}_n(R/\mathfrak{m})$ is a finite group, it follows that the element of $H$ is not in the kernel of the profinite completion.  Although the group $\Gamma = \textbf{Sp}_{2g}(\mathbb{Z})$ is residually finite, the group $\widetilde{\Gamma}$ is not, and the kernel of the map to the profinite quotient is precisely $2H$ as a subgroup of $H$ (writte in additive notation).  Thus, although $\text{Hom}(H,\mathbb{Q}/\mathbb{Z})$ is $\mathbb{Q}/\mathbb{Z}$, the part of the group that arises from Brauer elements is $(1/2)\mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$.<|endoftext|>
TITLE: Intuition Behind a Decimal Representation with Catalan Numbers
QUESTION [9 upvotes]: From $0 = 0.5 - 0.5 = 0.5 - \sqrt{0.25}$, we can adjust the subtrahend slightly to obtain
$$0.5 - \sqrt{0.249} = 0.001\ 001\ 002\ 005\ 014\ 042\ldots$$
where the decimal representation contains the first few Catalan numbers: $1, 1, 2, 5, 14, 42 
\ldots$
We can see even more Catalan numbers (albeit spaced apart with more $0$s) by using more $9$s. 
(For example, check the decimal representation of $0.5 - \sqrt{0.24999999999}$.)
My question is not how to show this formally; it is a straightforward problem to show how one derives, e.g., the decimal representation given above. (I'll include a derivation below for anyone who doesn't want to think this through her/himself.)
Instead, my question is: why would it make sense, intuitively, for the Catalan numbers to show up in these decimal representations?

Derivation:
Recall that the generating function for the Catalan numbers $c(x)$ satisfies $c(x) = 1 + xc(x)^2$. Rearranging, we find that $c(x) = \frac{2}{1 + \sqrt{1 - 4x}}.$ Then
$$\sum_{n = 0}^{\infty}\frac{C_n}{10^{3n + 3}} = \frac{1}{1000}\sum_{n=0}^{\infty}\frac{C_n}{1000^{n}} = x \sum_{n=0}^{\infty}C_n x^n = xc(x) = \frac{2x}{1 + \sqrt{1 - 4x}},$$
where we have simplified our computations by letting $x = \frac{1}{1000}$.
Evaluating at this value of $x$ yields $0.5 - \sqrt{0.249}$.

REPLY [11 votes]: To make it less surprising is to fade some of the magic! But, ok. 
First let us say that any  real sequence $a_1,a_2,\cdots$ has an ordinary generating function (ogf) $f(x)=\sum a_ix^i$ which may be a formal series with radius of convergence $0$ (and still be useful) BUT if the $a_i$ are positive integers and $f(x)$ converges at $\frac{1}{b^k}$ then $f(\frac{1}{b^k})$ is a number whose base $b$ expansion is the sequence $a_i$ buffered by $0$'s until they start to bump into each other. I'll stick to $b=10$ and I'll use $(10^{-j})f(10^{-k})$ if there is an $a_0$ term I want to shift past the decimal point. 
So the question might be which series have a nice ogf? That the Catalan numbers do is very nice.
From $\frac{1}{(1-x)^k}=\sum \binom{k+i}{k}x^i$ one obtains
$\frac{1}{10\cdot 0.9998}=0.10002000400080016003200640128025605121024$
$\frac{1}{10(0.999)^2}=0.10002000300040005000600070008000900100011$
$\frac{1}{10(0.999)^3}=0.10003000600100015002100280036004500550066$
Since $\binom{i}{2}+\binom{i+1}{2}=i^2$, 
we can use $(\frac{1}{10}+\frac{1}{100000})\frac{1}{(1-x)^3}$ at $x=0.0001$ to get
$\frac{1.0001}{10(0.999)^3}=\frac{100010000000}{999700029999}=0.100040009001600250036004900640081010001210$
But it is more productive to use $\binom{i}{1}+2\binom{i}{2}=i^2$ for 
$\frac{1}{10(1-x)^2}+\frac{1}{5000(1-x)^3}.$ Another function which coincides with the previous one at $x=0.0001.$ So the same rational but obtained another way. This second approach makes it clear how to get any polynomial sequence $a_i=p(i).$
This already seems less fun. So thinking of a dramatic last target, the Fibonacci numbers remind us that anything given by a recurrence relation (linear, with constant coefficients...) has a nice ogf.
At $x=\frac{1}{100}$, 
$\frac{1}{10(1-x-x^2)}=\frac{100000}{998999}=0.1001002003005008013021034055089144$
A cute point is that all my examples (none of which are as nice as yours) lead to a rational number so as the $a_i$ increase in size and overlap they result in an eventually repeating decimal. For example the base $10$ "Fibonacci" rational I gave has period $496620$ while
$\frac{10}{89}=0.\overline{11235955056179775280898876404494382022471910}$<|endoftext|>
TITLE: Numerical Methods for ODEs - History
QUESTION [5 upvotes]: Wikipedia presents a timeline of important developments in Numerical Methods for ODEs, namely:
1768 - Leonhard Euler publishes his method.
1824 - Augustin Louis Cauchy proves convergence of the Euler method. In this proof, Cauchy uses the implicit Euler method.
1855 - First mention of the multistep methods of John Couch Adams in a letter written by F. Bashforth.
1895 - Carl Runge publishes the first Runge–Kutta method.
1905 - Martin Kutta describes the popular fourth-order Runge–Kutta method.
1910 - Lewis Fry Richardson announces his extrapolation method, Richardson extrapolation.
1952 - Charles F. Curtiss and Joseph Oakland Hirschfelder coin the term stiff equations.

but with no links for the original works. 
The question is: are there references with a timeline like this with links pointing to the original works?

REPLY [3 votes]: Here are the sources:
Leonhard Euler: Institutiones calculi integralis (1768)
Augustin Louis Cauchy: Cours d'Analyse: Equations différentielles ordinaires et aux dérivées partielles (1824) 
The Adams-Bashforth multistep-method as well as the Adams–Moulton methods are described here
Carl Runge: Über die numerische Auflösung von Differentialgleichungen, Math. Ann. 46 (1895) 167-178
W. Kutta: Beitrag zur näherungsweisen Integration totaler Differentialgleichungen, Z. Math. Phys. 46 (1901) 435-453. (Remark: The name is Martin Wilhelm Kutta, the correct year is 1901.)
Lewis Fry Richardson: The approximate arithmetical solution by finite differences of physical
problems involving differential equations,with an application to the stresses in a masonry dam Phil. Trans. R. Soc. London Ser. A 210 (1910) 307–57
Of interest in this connection is also E.J. Nyström: Über die numerische Integration von Differentialgleichungen, Acta Soc. Sci. Fennicae 50, 13 (1925) 55
Charles F. Curtiss and Joseph Oakland Hirschfelder: Integration of stiff equations, Proc Natl Acad Sci U S A, 38,3 (1952) 235–243<|endoftext|>
TITLE: Another question about amenability and Følner sequences
QUESTION [6 upvotes]: Følner's characterization of Amenability says that a group $G$ is amenable if there exists a directed set $(I,\leq)$ and a net {$F_i:i\in I$} of finite subsets of $G$ such that for every $γ ∈ G$,
$$\lim_{i\in I}\frac{|γF_i\Delta F_i|}{|F_i|}\ \ \ \rightarrow 0\ ,$$
where $\Delta$ is the symmetric difference of two sets. It is also known that, if $G$ is countable, the word "net" can be substituted by "sequence" (that is $I=\mathbb N$ with the usual order). 
Is it true that for countable (or at least finitely generated) groups we can always find a Følner sequence as above, which satisfies the following conditions:
(1) $F_{n}\subseteq F_{n+1}$, for all $n\in \mathbb N$;
(2) $\bigcup_{n\in\mathbb N}F_n=G$.
The motivation for my question comes from the paper "The Abramov-Rokhlin Entropy Addition Formula for Amenable Group Actions" by Ward and Zhang (Mh. Math, 1992). In fact, their Theorem 2.6 (that they attribute to Ornstein and Weiss) is proved for a Følner sequence as the one above but it is applied to actions of arbitrary countable Amenable groups. So... it seems that such sequences always exist.

REPLY [10 votes]: Yes, it follows from the following simple lemma: if $(F_n)$ is a Følner sequence, then $(A\cup F_n)$ is also a Følner sequence for any finite set $A$.
EDIT. It is a quite common misconception to formulate the Følner condition in terms of exhausting sequences only. The bottom line is that existence of almost invariant functions or measures (be it for a group/groupoid action or the Riemannian/combinatorial Laplacian) is equivalent to existence of sets with small boundary (Følner sets). There is nothing in this equivalence which would require these sets to be nested or to exhaust the whole state space. In what concerns the lecture notes you mention you miss the connectivity condition on a Følner exhaustion which is imposed there, which completely changes the situation.
Let me give a more detailed argument showing how the lemma above implies existence of a nested exhausting sequence of Følner sets on any amenable group or graph. Actually, for the sake of simplicity let me do it just for graphs with uniformly bounded vertex degrees - which also takes care of finitely generated groups; for infinitely generated groups one has to consider symmetric differences between individual translates instead of dealing just with boundaries in the Cayley graph.
By the classical theorem of Gerl a graph $X$ is amenable if and only if there exist finite subsets $F$ with arbitrarily small ratio $|\partial F|/|F|$. I claim that one can always choose a nested sequence $\Phi_n$ which exhausts $X$ with $|\partial \Phi_n|/|\Phi_n|\to 0$. Indeed, take a sequence $F_n\subset X$ with $|\partial F_n|/|F_n|\to 0$, an exhausting nested sequence $A_n\subset X$, and a sequence $\epsilon_n\to 0$. Put $\Phi_1=F_1$, and then inductively (by using the above lemma) $\Phi_{n+1}=\Phi_n \cup A_n \cup F_N$, where $N=N(n)$ is chosen to satisfy the inequality $|\partial\Phi_{n+1}|/|\Phi_{n+1}|<\epsilon_{n+1}$.
This argument is totally trivial (and I would say redundant) from geometrical point of view.

REPLY [5 votes]: Yes - see Lemma 5.1 of Gabor Pete's book on probability and groups: http://www.math.bme.hu/~gabor/PGG.html (by the way, these are excellent lecture notes)<|endoftext|>
TITLE: Honda-Tate in families
QUESTION [7 upvotes]: Let $k$ be a finite field, say with $q=p^a$ elements. Honda-Tate theory states that there is a bijection between isogeny classes of simple abelian varieties over $k$ and $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$-conjugacy classes of algebraic integers in $\mathbb{C}$ all of whose conjugate have absolute value $p^{a/2}$ (these are the so-called "Weil numbers"). For a beautiful survey, see Tate's  Bourbaki seminar .
Given a simple abelian variety $A/k$, the associated Weil number $\pi_A$ (or rather its conjugacy class) corresponds to the Frobenius endomorphism, view as an element of $\mathrm{End}_k(A)$. Let now $C$ be a smooth, projective, irreducible curve over $k$ and let $\mathcal{A}/C$ be an abelian scheme, therefore a family of abelian varieties parametrized by points of $C$. There is a Frobenius operator acting on $\mathcal{A}$ giving the Frobenius operator on each fiber $A_v$ for all $v\in C$, and these fibers are abelian varieties over finite fields for all closed $v$. My question is how Honda-Tate behaves in families, so if we can find a bijection between isogeny types of $\mathcal{A}/C$ and "polynomials" in $\mathcal{O}_C[T]$ (or may be in $\Gamma(C,\mathcal{O}_C)[T]$?) whose specialization at every closed point has roots that are Weil numbers.
EDIT: As Piotr Achinger observes, it seems reasonable that the coefficients be in characteristic $0$. I do not know if hoping the coefficients to live in the Witt vectors of $\mathcal{O}_C$ is enough for giving some sense to "a family parametrized by $C$.
I am tempted to think the answer should be "yes", being so fiber-wise. On the other hand, if we pick a "random" collection of Weil polynomials building an element of $\mathcal{O}_C[T]$, I would be surprised if we can build an abelian scheme over $C$ having the "right" fibers (because, for instance, abelian schemes must have good reduction everywhere, and it seems too strong a condition to be simply controlled by a "nice" collection, if what I write ever makes any sense).
If the answer to my question is "yes" (or if it can be made to be "yes" after some modification of my question...), is this "polynomial" in two variables related to the Hasse-Weil function of the abelian scheme? After all, they are both constructed by looking at Frobenius acting on Tate modules, so I would expect a connection.

REPLY [5 votes]: I think your intuition that Weil numbers generalize to functions is wrong, or rather only partially correct. A Weil "number" is in fact a sheaf on $\operatorname{Spec}\mathbb F_p$: The sheaf corresponding to the Galois representation that sends $Frob_p$ to a matrix whose characteristic polynomial is the minimal polynomial of $\pi_A$. 
Lisse and constructible sheaves, or rather their derived categories / Grothendieck groups, are often the appropriate scheme-theoretic notion of function. One has addition and multiplication of functions, and even integration in the form of higher pushforward maps. Thus scheme-theoretic analogues of convolution, Fourier analysis, etc. often exist.
I will thus discuss the sheaf version of the problem. I would guess that the Witt vector version of the problem can be solved by thinking of the sheaf of Tate modules using crystalline cohomology, but as I don't know anything about crystalline cohomology you should take this with a grain of salt.
One can split the sheaf version of the problem into two parts. First, what are necessary and sufficient conditions on a constructible $l$-adic sheaf on $C$ for it to be the sheaf of Tate modules of an abelian scheme over $C$? Second, if two abelian varieties $A_1$ and $A_2$ have the same sheaf of Tate modules, are they isogenous?
I don't have much to say about the first part, but the second part is a special case of the Tate conjecture. An isomorphism between $R^1 \pi_{1*} \mathbb Q_l$ and $R^1 \pi_{2*} \mathbb Q_l$ produces a Galois-invariant cohomology class in $H^0(C,R^1 \pi_{1*} \mathbb Q_l ^\vee \otimes R^1 \pi_{2*} \mathbb Q_l)$, which is a quotient of $H^2( A_1^\vee \times _C A_2, \mathbb Q_l)$. If the Tate conjecture is true, then that cohomology class is a $\mathbb Q_l$-linear combination of classes induced by algebraic cycles. At least one of those cycles must correspond to a line bundle on $A_1^{\vee} \times_C A_2$ that induces a nontrivial morphism $A_2 \to A_1$. For $A_1$ and $A_2$ simple abelian schemes this must be an isogeny. It is not too hard to see that this means we can also get an isogeny for products of simple abelian schemes.
I have no idea whether this special case of the Tate conjecture has a solution. A similar problem that naively seems about as hard, the Tate conjecture for $H^1 (R^1 \pi_* \mathbb Q_l)$ of an elliptic surface, is the function field BSD conjecture.<|endoftext|>
TITLE: Is the Hasse principle a birational invariant?
QUESTION [10 upvotes]: Is the Hasse principle a birational invariant?

It is probably a very trivial question, but I am a beginner in arithmetics.

REPLY [25 votes]: In this generality, the answer is no. The projective curve $X$ given by $2y^2z^2 = x^4 - 17z^4$ over the rationals satisfies the HP, since it has local points everywhere (the affine part $z \neq 0$ is given by $2y'^2=x'^4-17$, which is the famous Reichardt-Lind equation which is known to be everywhere locally, but not globally, soluble) and it has the unique rational point $(0:1:0)$. However, this point is singular: so now consider the normalization $X'$ of $X$: it has two points above $(0:1:0)$, neither of which is rational. By the parenthetical remark, $X'$ has local points everywhere, but it doesn't have rational points: therefore $X'$ does not satisfy the HP. Also, $X'$ is birational to $X$, being its normalization.
If you restrict to smooth varieties however, the answer is yes: by Lang-Nishimura, if $X$ and $X'$ are smooth varieties over any field $k$ that are birational to each other, then $X$ has a $k$ point iff $X'$ does.<|endoftext|>
TITLE: Spectral radius on 0-1 vectors.
QUESTION [20 upvotes]: Let $A$ be an $n\times n$ symmetric substochastic matrix (i.e. all entries are non-negative and each row adds up to $1$ or less).
Call a vector $v \in \mathbb{R}^n$ an indicator if $v \neq 0$ and each coordinate of $v$ is either $0$ or $1$.  Define the indicator spectral radius of $A$ by:
$$r_I(A) = \max\left\lbrace \frac{1}{|v|^2}\langle Av,v \rangle: v\text{ is an indicator}\right\rbrace$$
where $|v|$ is the standard Euclidean norm and $\langle,\rangle$ the inner product.
If $|A|$ is the operator norm of $A$ then Cauchy-Schwarz and Jensen's inequalities imply that:
$$0 \le r_I(A) \le |A| \le 1$$
Looking at the standard basis one obtains that $r_I(A) = 0$ if and only if $|A| = 0$.
Also, it's not too hard to show that $r_I(A) = 1$ if and only if $|A| = 1$.
Consider the function $f_n:[0,1] \to [0,1]$ defined by:
$$f_n(x) = \max\left\lbrace |A|: A\text{ is }n\times n\text{ symmetric and substochastic with }r_I(A) \le x \right\rbrace$$
And let $f(x) = \lim_{n \to +\infty}f_n(x)$.
My questions are:  Is $f(x) < 1$ whenever $x < 1$? Is $f(x) \le \sqrt{x}$?

Motivation:  In the course of showing his criteria for amenability of a countable group Kesten proved ("Full Banach mean values on countable groups." 1959) that $f(x) \le O(x^{\frac{1}{3}})$ when $x \to 0$ so that in particular $f(x) < 1$ for all $x$ small enough.  He claims the proof actually gives $f(x) \le O(x^{\frac{1}{2}-\epsilon})$.
I've also done some numerical experiments on the indicator norm (analogous definition) which suggests the bound $\sqrt{x}$ might work.

REPLY [2 votes]: With rank one matrices (I mean those of the form $A = \frac {1}{(\max w) \sum w_i}w \otimes w$ for some vector $w$) here is a proof that says $f(r) \leq 2 \sqrt{r}$ if we restrict to that kind of matrices.
Let us assume wlg that $1=w_1\geq w_2 \geq w_3 \geq \dots$ (we can do this by rearranging the indices and dividing by $w_1$). Then we can obtain
$$\|A\| = \frac{\sum w_i^2}{\sum w_i}$$
and
$$r:=r_I(A) = \max_m \frac{\left(\sum_{i\leq m} w_i \right)^2} {m \sum w_i}.$$
Let $k$ be the largest integer so that $w_i \geq \sqrt{r}$, then by definition of $r$ we have that
$$ r \geq \frac{\left(\sum_{i \leq k} w_i \right)^2} {k \sum w_i}.$$
Also
$$\|A\| \leq \frac{\sum_{i \leq k} w_i^2}{\sum w_i} + \frac{\sum_{i > k} w_i^2}{\sum w_i} \leq \frac{\sum_{i \leq k} w_i}{\sum w_i} + \sqrt r \frac{\sum_{i > k} w_i}{\sum w_i}
\leq \frac{\sum_{i \leq k} w_i}{\sum w_i} + \sqrt r $$
And using that $v_i \leq 1$ and $\sum_{i\leq k} w_i \geq k w_k \geq k \sqrt r$, we get
$$\|A\| \leq \frac{\left(\sum_{i \leq k} w_i\right)^2}{k \sqrt r \sum w_i} + \sqrt r \leq 2 \sqrt{r}
$$<|endoftext|>
TITLE: Left invariant metric on ${\rm SL}_n(\mathbb{R})$
QUESTION [10 upvotes]: I am looking for a left invariant metric on $SL_n(\mathbb{R})$. If this is not possible, it would be acceptable to have a metric on $SL_n(\mathbb{R})/SO_n(\mathbb{R})$ or something like that. Is there such a thing?
Note: I am not looking for a Riemannian metric, just an ordinary metric.

REPLY [25 votes]: The OP specifically asked for a(n ordinary) metric, not a Riemannian metric.  While Misha and Paul have given good answers, I think that it's worth pointing out that, if one just takes an arbitrary left-invariant Riemannian metric $ds^2$ on a connected group $G$, there is no guarantee that the associated $dist$ can be computed explicitly.  To do this, one would need to be able to integrate the geodesic equations explicitly enough to be able to construct the distance function; unless the Riemannian metric is quite special, this generally can't be done in any explicit way.  (See for example, what one has to do to describe the free rotations of a rigid body that has 3 distinct moments of inertia, which is essentially computing the geodesics on $\text{SO}(3)$ with respect to a left-invariant metric that is not bi-invariant.)
An $\text{SL}(n,\mathbb{R})$-invariant metric on $\text{SL}(n,\mathbb{R})/\text{SO}(n)$: Since $M=\text{SL}(n,\mathbb{R})/\text{SO}(n)$ is an irreducible Riemannian symmetric space (with nonpositive sectional curvature), it has, up to constant multiples, only one $\text{SL}(n,\mathbb{R})$-invariant Riemannian metric, and the associated $dist$ in this case is not entirely trivial to write down:  We can identify each element $m = A\cdot \text{SO}(n)$ with its associated positive definite, unimodular symmetric matrix $s = \sigma(m) = AA^T$, and the formula for $dist(s_1,s_2)$ is as follows:  Write
$$
\text{det}(ts_1-s_2) = (t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_n)
$$
where each $\lambda_i$ is positive and they satisfy $\lambda_1\lambda_2\cdots\lambda_n=1$.  Then, up to a constant multiple, one has
$$
dist(s_1,s_2) = \left(\sum_{i=1}^n (\log\lambda_i)^2\right)^{1/2}.
$$
Obviously, writing this out as a function of the entries of $s_1$ and $s_2$ would not be easy.  (Of course, this is just one example of the sort of metric that Paul gave; its distinguishing characteristic is that it is the $dist$ of a Riemannian metric, which is not true for Paul's specific example.)
A left-invariant metric on $\text{SL}(n,\mathbb{R})$:
Once an invariant metric on $M$ has been defined, one can use it to define a metric on $\text{SL}(n,\mathbb{R})$ itself:  First, suppose that $n$ is odd, so that $\text{SL}(n,\mathbb{R})$ acts effectively on $M$.  Let $\delta:M\times M\to \mathbb{R}$ be an invariant metric and let $(s_1,\ldots,s_k)\in M\times M\times \cdots \times M$ ($k$ times) 
be a $k$-tuple of symmetric matrices with the property that the simultaneous stabilizer of all of the $s_i$ in $\text{SL}(n,\mathbb{R})$ is the identity matrix.  (I guess $k=3$ suffices to find such a $k$-tuple; $k=2$ does not.)  Set
$$
dist(A,B) = \delta(As_1,Bs_1) + \delta(As_2,Bs_2) + \cdots +  \delta(As_k,Bs_k).
$$
This defines a left-invariant metric on $\text{SL}(n,\mathbb{R})$.  Note, however, that this is not derived from a Riemannian metric.  When $n$ is even, $-I_n$ lies in $\text{SL}(n,\mathbb{R})$ and it acts trivially on $M$, so the above construction won't work.  However, when $n$ is even, just take the metric induced on $\text{SL}(n,\mathbb{R})$ by its natural embedding into $\text{SL}(n{+}1,\mathbb{R})$, and that will do the job.
To get a Riemannian metric on $\text{SL}(n,\mathbb{R})$ whose $dist$ is computable, one should probably take the left-invariant metric $ds^2 = \text{tr}\bigl((g^{-1}dg)^Tg^{-1}dg\bigr)$.  The reason is that this metric is both left-invariant and invariant under right action by $\text{SO}(n)$, and the extra symmetries make the geodesic flow explicitly integrable.  (There is, of course, no bi-invariant Riemannian metric on $\text{SL}(n,\mathbb{R})$.) 
A little calculation shows that the $ds^2$-geodesic starting at $I_n$ with velocity $v\in{\frak{sl}}(n,\mathbb{R})$ is given by the formula
$$
\gamma_v(t) = e^{v^Tt}e^{(v-v^T)t},
$$
where $v^T$ represents the transpose of $v$.  In particular, one has the 'formula', for $A\in\mathrm{SL}(n,\mathbb{R})$,
$$
dist(I_n,A) = \min\bigl\{\bigl(\text{tr}(v^Tv)\bigr)^{1/2}\ |\ e^{v^T}e^{(v-v^T)} = A\bigr\}.
$$
Unfortunately, computing this $dist$ more explicitly is a challenge.  By left-invariance, of course, one has $dist(A,B) = dist(I_n,A^{-1}B)$, so this determines the metric completely.
Computing $dist$ explicitly is nontrivial even in the case $n=2$.  In this case, one has the fortunate circumstance that, unless $\det(v)>0$, there are no conjugate points along the geodesic $\gamma_v$, and, when $\det(v)>0$, the first conjugate point is at $t = \pi/\sqrt{\det(v)}$.  However, describing the exact cut locus of $I_n$ in ${\frak{sl}}(n,\mathbb{R})$ with respect to $ds^2$ does not seem to be trivial.  This does not seem to be a particularly good way to construct a left-invariant metric on $\text{SL}(2,\mathbb{R})$.  Nevertheless, it's not hopeless.  If I have time, I'll add a little note to this describing what one can say.<|endoftext|>
TITLE: Are complex varieties Kahler? - Algebraic, non-projective complex manifolds
QUESTION [8 upvotes]: Let $X/\mathbb{C}$ a nonsingular proper variety and $X_{an}$ it's associated analytic space. Is $X_{an}$ necessarily Kahler? Certainly we know this if $X$ is projective.
A complex torus is algebraic iff it is projective. Are there Kahler manifolds which are algebraic, but not projective?

REPLY [16 votes]: Any abstract algebraic compact complex manifold is Moishezon.  By Moishezon's theorem, any Kähler Moishezon manifold is projective algebraic.  There are non-projective proper complex varieties, so $X_{an}$ is not necessarily Kähler.  This is represented in the diagram at the end of Hartshorne's Algebraic Geometry Appendix B.
In summary, all of your questions have negative answers.<|endoftext|>
TITLE: What are the applications of Dowker's theorem?
QUESTION [12 upvotes]: Let $R \subset X \times Y$ be any relation between sets $X$ and $Y$. CH Dowker constructed two simplicial complexes $K$ and $L$ associated to $R$:

a simplex in $K$ consists of finitely many elements $x \in X$ such that there exists a single $y \in Y$ with $(x,y) \in R$, and
a simplex in $L$ consists of finitely many elements $y \in Y$ such that there exists a single $x \in X$ with $(x,y) \in R$.

Clearly, these are simplicial complexes. The main theorem of Dowker is the construction of a natural isomorphism of homology groups of $K$ and $L$. There is even a proof outline on nlab. In fact there is a homotopy equivalence between the geometric realizations although that requires an ordering on the simplices and is therefore not natural. This may be found in the paper

C. H. Dowker, Homology Groups of Relations, Annals of Math, 56, (1952), 84–95.

The goal of this paper was to prove equivalence of Cech, Vietoris and Alexander (co)homology theories. My question is 

What other applications (if any) have been found for this theorem?

In particular, the fact that the relation itself creates the simplicial complexes seems to make this theorem have very limited uses beyond the consequences proved already by Dowker: I can't seem to find much on the internet at least. I wonder if the experts have some idea...

REPLY [11 votes]: There is a considerable literature on `applications' of Dowker's result to sociology.  This is sometimes doubtful in its depth! The development was started by R. Atkin. 
As an example look at 
http://www.ehu.es/ccwintco/uploads/1/11/Blanca-Cases-Q-analysis.pdf
As an offshoot of this there is fairly recent work in discrete maths (see work by Hélène Barcelo). I will not try to describe this other than saying it looks at an idea of the connectivity of a relation.
Back in the world of algebraic topology, it provides a way of proving that the pro-object in the homotopy category of simplicial sets, that is given by the Cech complex construction is in fact homotopy coherent. This provides a way of linking strong shape theory to the original form of shape theory. (It is not hard to prove this coherence directly although I only know one proof that has been written down in a thesis of one of my ex-students.)
(Edit: I forgot another example.  You start with a group, $G$, and a family of subgroups, and ask to what extent invariants of the family give you invariants of the big group. This was the subject of a paper by Abels and Holz (Higher generation by subgroups , J. Alg, 160, (1993), 311– 341.) The family generates a covering of $G$ by its cosets. The two complexes given by that covering allow proof to be shortened and also in certain circumstances for links to Volodin's alegrbaic K-theory to be given.)<|endoftext|>
TITLE: Number of spin structures
QUESTION [7 upvotes]: I am probably missing something obvious, but still...
Consider an oriented Riemannian $n$-dimensional vector bundle $\pi: E\rightarrow X$ over compact manifold $X$ with $\omega_2(E)=0$ so it has spin structures. Let $P_{SO}(E)$ denote the associated principal $SO_n$ frame bundle. We would like to count the number of different spin structures over $E$.
$1)$ From the fibration $SO_n\xrightarrow{i}P_{SO}(E)\xrightarrow{\pi}X$ we have an exact sequence 
$0\rightarrow H^1(X, \mathbb{Z}_2)\xrightarrow{{\pi}^*}H^1(P_{SO}(E), \mathbb{Z}_2)\xrightarrow{i^*}H^1(SO_n , \mathbb{Z}_2)$ from which we see that the number of different spin structures over $E$ is $H^1(X, \mathbb{Z}_2)$. 
$2)$ On the other hand the short exact sequence $0\rightarrow\mathbb{Z}_2\rightarrow Spin_n\rightarrow SO_n\rightarrow 1$ gives an exact sequence $H^0(X, SO_n)\xrightarrow{\delta} H^1(X, \mathbb{Z}_2)\rightarrow H^1(X, Spin_n)\rightarrow H^1(X, SO_n)$ from which the number of different spin-structures over $E$ is $H^1(X , \mathbb{Z}_2)/\delta(H^0(X, SO_n))$.
So why do we have different answers for the same object?

REPLY [12 votes]: Your first answer is correct, and the second one is almost correct, but the problem is that they count different things:
In the first case, you count all twofold covers of $P = P_{SO} E$ that restrict to a nontrivial cover of each fibre. In other words, you count $Spin (n)$-bundles $Q$, together with an isomorphism $\phi:Q \times_{Spin(n)} SO(n) \cong P$ (up to isomorphism of $(P,\phi)$). Here $Q \times_{Spin(n)} SO(n)$ is just a different way to write $Q/\mathbb{Z}_2$.
This is the usual notion of a spin structure on $E$. 
Your second answer (correctly) counts the number of $Spin (n)$-principal bundles $Q \to X$ such that $Q /\mathbb{Z}_2 $ \emph{admits} an isomorphism with $P$.
The point is that an abstract $Spin (n)$-bundle $Q \to X$ can yield many different spin structures on $P = Q/\mathbb{Z}_2$. Say if $P$ is trivial, then the set of homotopy classes of isomorphisms $Q /\mathbb{Z}_2 \to P$ is in bijection with $[X;SO(n)]$. This accounts for the division by the image $\delta$ in your second answer. 
For nontrivial $P$, your second answer is not quite correct, as your exact sequence is only a sequence of sets.
To see this is an example, let $X=S^1$ and $n \geq 3$. As $SO(n)$ $Spin(n)$ are connected, all principal bundles on $S^1$ are trivial. In this case $H^0 (X;SO(n))=[S^1;SO(n)]=Z/2$; $\delta$ is an isomorpism and the two terms to the left are null. The generator of $[S^1;SO(n)]$ gives an isomorphism of the trivial $SO(n)$-bundle that transforms the two spin structures into each other.<|endoftext|>
TITLE: Description of modules over self-injective algebras of finite representation type
QUESTION [7 upvotes]: Is there any description of indecomposable modules and irreducible morphisms over self-injective algebras of finite representation type? I am interested mainly in such a description for nonstandard algebras of tree type $D_{3n}$. Modulo stable equivalence, such an algebra can be represented by:

a quiver $Q$ whose vertices are $Q_0 = \{0, \ldots, n-1\}$ (considered modulo $n$), and whose arrows are $b\colon 0\to 0$ and $a_i\colon i\to i+1$ for $(0 \leq i \leq n-1)$,
and an ideal $I$ generated by the elements $a_i\cdots a_0 a_{n-1}\cdots a_i$, $b^2 + a_{n-1}\cdots a_0$ and $a_0 a_{n-1} + a_0 b a_{n-1}$.

The characteristic of the field equals $2$.

REPLY [3 votes]: Such algebras were named "penny-farthing algebras" by Gabriel.
Note first that this algebra is socle equivalent to the "standard" penny-farthing algebras, so the representation theory of the standard and non-standard ones is more or less the same. The article https://eudml.org/doc/152308 summarises the representaiton-theory of this algebra, but without giving too much details and its in German.
Chapter 4 of the recent textbook by Skowronski and Yamagata discusse in general how to obtain the Auslander-Reiten quiver of representation-finite selfinjective algebras. It is an exercise there to calculate the Auslander-Reiten quiver for penny farthing algebras with 6 simples modules. Id be interested in a quick solution of this exercise. Full detail solution should take hours to write down with all proofs.<|endoftext|>
TITLE: Does every ellipse inside a tetrahedron inside a ball fit in a triangle inside the ball?
QUESTION [22 upvotes]: In three-dimensional euclidean space, consider the closed unit ball $B$. Let $T$ be a tetrahedron, and $E$ an ellipse, with $E \subset T \subset B$. Does there necessarily exist a triangle $T'$ with $E \subset T' \subset B$?
Clearly if 1 vertex of the tetrahedron is on one side of the plane of of the ellipse, and the other 3 vertices are on the other side, then intersecting the plane with the tetrahedron gives such a triangle. The interesting case is when 2 vertices of the tetrahedron are on each side of the plane.
I work in quantum information theory, and have come up with a conjecture that, remarkably, is true if and only if the answer to the above question is "yes"! Since this is very far from the sort of thing I normally think about, I don't even know where to begin to look in the mathematics literature, so even just a pointer would be a big help.

REPLY [7 votes]: Mihai-Dorian Vidrighin has suggested the following idea. (He's asked me to post it because he can't post images.)
For simplicity, assume that the setup is "tight" in that $E$ meets every face of $T$ and the vertices of $T$ are on the boundary of $B$. The generalisation should be straightforward.
If any vertices of $T$ lie in the plane of $E$ then the cross-section is already a triangle, so assume otherwise. Pick an arbitrary vertex of $T$ and draw a cone staring from there with $E$ (shown in red) as a base. This cone meets the opposite face of $T$ in a new ellipse $E'$ (shown as a dotted black curve):

$E'$ lies inside a triangle (the face of $T$), which itself lies inside a circle (the cross-section of $B$). By our simplifying assumption, $E'$ touches the edges of the triangle and the triangle's vertices are on the circle. Therefore Poncelet's Porism applies. Hence we can find a different triangle around $E'$ with one edge in the plane of $E$.
Define a new tetrahedron (shown in green) using the chosen vertex of $T$ and the new triangle. By construction it still contains the cone identified before, and in particular it still contains $E$. But the cross-section in the plane of $E$ is now a triangle (shown in thick black).<|endoftext|>
TITLE: Graham-Rothschild via Hales-Jewett
QUESTION [5 upvotes]: I am currently reading the recent preprint of Dodos, Kanellopoulos, Tyros, where the ambitiously short proof of Density Hales Jewett theorem is provided. The important ingredient is Graham-Rothschild theorem. The authors say that it follows from the HJ by some standard Ramsey arguments, but I can not find them myself, at least immediately. Is it written anywhere? Original paper of Graham and Rothschild looks too long for being used in "simple self-contained proof" of anything. Polymath's DHJ proof does not use GR at all, on first glance.

REPLY [3 votes]: In Randall - McCutcheon's book "Elemental methods in ergodic Ramsey theory" a stronger version of GR's theorem  is prooved about block subspaces (theorem 2.4.1). The only theorems you need in order to proove it is Hales-Jewett and Folkman's theorem (about finite unions which you can proove using  Hales - Jewett theorem again).
Another proof is given in Graham - Rothschild -Spencer book about Ramsey theory using HJ and linear algebra.<|endoftext|>
TITLE: Higgs mechanism from a deformation quantization point of view
QUESTION [6 upvotes]: Is it possible to describe the Higgs mechanism from a deformation quantization point of view? How would one do it? Are there aspects of the Higgs mechanism and Higgs particle which one may see clearer from such a point of view?

REPLY [6 votes]: The Higgs mechanism in the Standard Model doesn't have much to do with deformation quantization as other people have explained. However there is a version of the Higgs mechanism in string theory which involves stable D-branes arising via the Higgs mechanism from unstable D-branes. This is very hard to study directly in string field theory where one has to resort to approximate numerical techniques, but if one deforms the theory by turning on a B field as discussed in Seiberg and Witten, hep-th/9908142 one obtains a noncommutative version and in the "large B limit" one can obtain exact results. For details see hep-th/0005031 and references therein.<|endoftext|>
TITLE: Representations attached to p-adic modular forms
QUESTION [16 upvotes]: A theorem of Gouvea and Hida (or rather a consequence of it) states that there exist a Galois representation attached to a $p$-adic eigenform $f$ provided the residual representation attached to a classical eigenform $g$ congruent to $f$ is absolutely irreducible. I believe there might be generalizations due to Skinner-Wiles. So my question is: do there always exist a Galois representation attached to an ordinary $p$-adic modular form? What if $g$ has some extra properties like CM but I still want the residual representation to be reducible?

REPLY [17 votes]: I am not sure this answer will satisfy you totally because I am not sure what you mean exactly by $p$-adic modular forms. But at least, for a $p$-adic modular form $f$ in the sense of
Serre, which is an eigenform for  almost all the Hecke operators $T_\ell$ (with eigenvalue $a_\ell)$ there always  exists a semi-simple Galois representation
$r:G_{\mathbb Q} \rightarrow Gl_2( \bar Q_p)$ such that $tr(\rm Frob_\ell)=a_\ell$
for almost all $\ell$. This is quite easy to prove with modern techniques which were not available at the time Serre, Hida, and Gouvêa worked on the subject. Here is how:
Lemma : for every classical normalized form $g$ of weight $k$ and level $\Gamma$ with coefficient in $\mathbb Z_p$ such that $T_\ell g \equiv a_\ell g \pmod{p^n}$ for almost all $\ell$, there exists a unique continuous pseudo-character $T_g : G_{\mathbb Q} \rightarrow \mathbb{Z}/p^n \mathbb{Z}$ such that $T_g(\rm Frob_\ell) = a_\ell$ in $\mathbb Z/p^n \mathbb Z$
Proof: consider the Hecke algebra $A$ generated by the Hecke operators $T_\ell$ acting on
$M_k(\Gamma,\mathbb Z_p)$ (the module of modular forms of weight $k$, level $\Gamma$, coefficients in $\mathbb Z_p$). There is pseudo-character $T : G_Q \rightarrow A$ such that $T(Frob_\ell)=T_\ell$.
To prove this, first prove it with $A$ replaced by $A \otimes_{\mathbb Z_p} K$ where $K$ is is some
sufficiently large extension of $\mathbb Q_p$. Then $A \otimes_{\mathbb Z_p} K = K^d$ where each factor
corresponds to an eigenform in $M_k(\Gamma,K)$. Hence to construct $T : G \rightarrow K^d$, one just takes the sum of the trace of the representations attached to these eigenforms.
Then $T(\rm Frob_\ell)=T_\ell$ by construction, so $T$ takes values in $A$ on a dense subset of $G_{\mathbb Q}$, so $T$ takes values in $A$ everywhere since $A$ is closed in $A \otimes_{Z_p} K$.
To finish the proof of the existence part of the lemma, just compose $T$ with the morphism of ring $A \rightarrow \mathbb Z/p^n \mathbb Z$ which sends $T_\ell$ on $a_\ell$. The uniqueness
is clear by Cebotarev.
Now, back to the main claim, for every $n$, $f$ is by definition congruent to a classical
form $g$ mod $p^n$, hence by the lemma we get a pseudo-character $T_n: G_{\mathbb Q} \rightarrow \mathbb Z/p^n \mathbb Z$ such that $T_n(\rm Frob_\ell)$ $ = a_\ell \pmod{p^n}$. By uniqueness, we can glue those pseudo-characters into
a pseudo-character $T_\infty: G_{\mathbb Q} \rightarrow \mathbb Z_p \rightarrow \mathbb Q_p$ and by a theorem of Taylor, this pseudo-character is the trace of a representation over $\bar Q_p$.<|endoftext|>
TITLE: Hopf algebras and bijective antipodes
QUESTION [11 upvotes]: By a theorem of Larson and Sweedler, the antipode of every finite-dimensional Hopf algebra is bijective.
My question is the following: 
Is it true that in every noetherian Hopf algebra the antipode is bijective?

REPLY [12 votes]: It is conjectured that the antipode is bijective for all noetherian Hopf algebras (Skryabin), but no proof is known. Take a look at this recent short survey, "Noetherian Hopf Algebras", by K.R. Goodearl, where this is listed as conjecture 1.9. Skryabin's original paper is:

S. Skryabin, New results on the bijectivity of antipode of a Hopf algebra, J. Algebra 306
  (2006), 622–633<|endoftext|>
TITLE: Compactness of Sobolev embedding for domains of finite measure
QUESTION [6 upvotes]: Let $\Omega \subset \mathbb{R}^d$ be a domain of finite Lebesgue measure, not assumed to be smooth or bounded. Is it true that the embedding of, say, $W^{1,p}_0(\Omega)$ (Sobolev functions with zero boundary value) into $L^q(\Omega)$ is compact for $1/q > 1/p - 1/d$?

REPLY [3 votes]: You can find a rather general answer in the book of Adams and Fournier, Theorem 6.16. You do have the embedding you wish, under a mild measure theoretical assumption.<|endoftext|>
TITLE: Translation distance in the curve complex
QUESTION [13 upvotes]: Given a map $\psi: S\rightarrow S,$ for $S$ a closed surface, is there any algorithm to compute its translation distance in the curve complex? I should say that I mostly care about checking that the translation distance is/is not very small. That is, if the algorithm can pick among the possibilities: translation distance is 0, 1, 2, 3, many, then I am happy...
I know there are algorithms for computing distances IN the curve complex, but this is not quite the same...

REPLY [7 votes]: Recently, Richard Webb and myself gave a polynomial-time algorithm for computing the asymptotic translation length of a mapping class
$$ \ell(h) = \lim_{n \to \infty} d(x, h^n(x)). $$
This appears as Algorithm 6 of the paper and relies on being able to compute geodesics in the curve complex. Assuming this, the key observation is that a midpoint $c'$ of a curve $c$ and its image under a large enough power of $h$ lies close to the tight axis of $h$. Hence $d(c', h^n(c')) \approx n \ell(h)$. Therefore, since $\ell(h)$ is a rational number with bounded denominator, if we take $n$ large enough then
$$ \ell(h) = \frac{1}{n} d(c', h^n(c')). $$
If you fix a finite generating set $\langle X \rangle = \textrm{Mod}(S)$ then the running time of this algorithm is a polynomial function of the word length of the mapping class $|h|_X$ since this is the running time of our algorithm for computing a geodesic from $c$ to $h(c)$ (Algorithm 4).<|endoftext|>
TITLE: From very many sets of fixed measure in a probability space, can we select many that have a positive intersection?
QUESTION [9 upvotes]: I assume the following Lemma is either well known or, more probably, a
Corollary of a much stronger well known Theorem, and I would be grateful for a
reference:
For all $\delta\in (0,1)$ and all $\ell\in \mathbb N$ there are $M$ and
$\varepsilon>0 $ such that:
Whenever we have a probability space $\Omega$ and a family $(A_i:i\lt M)$ of
sets of measure $\ge \delta$, we can find a subfamily of $\ell$ many
sets whose intersection has measure at least $\varepsilon$.
PS: An (easy) proof is a nice but straightforward application of Ramsey's theorem.
PS2: Only finite additivity of the measure is required/relevant.

REPLY [3 votes]: As Bill Johnson pointed out in his comment, the Lemma that I stated is trivial: simple counting is enough. Therefore, a reference is not required and would not make sense. So Bill Johnson answered my question (thanks!).
I would like to add a remark (unfortunately, a comment does not provide enough space, so I have to abuse an "answer" entry):
Originally we though of the following, slightly stronger variant of the Lemma (for which we used Ramsey's Theorem, but which might have a counting proof as well, who knows): 
For all $\delta\in (0,1)$ and all $k\in \mathbb N$ there is an $\varepsilon(\delta,k)>0 $, and for
all $\ell$ (bigger than $5/\delta$, say) there is a  $M(\ell)$ such that:
Whenever we have a probability space $\Omega$ and a
family $\mathcal A=(A_i:i\lt N)$ of
sets of measure $\ge \delta$, then we can find a subfamily $\mathcal B$ of
size $\ell$ such that every subfamily $\mathcal C$ of $\mathcal B$ of size $k$ has an
intersection of measure at least $\varepsilon$.
(So here $\varepsilon$ does not depend on $\ell$, but on $\delta$ and $k$ only. For $k=\ell$ we get the Lemma of the original question.)<|endoftext|>
TITLE: $\beta\mathbb{N}$  vs $\beta\mathbb{Z}$
QUESTION [5 upvotes]: Just started learning the Stone-Cech compactification of discrete groups this week. My motivation comes from a question on $\beta\mathbb{Z}$. Surprisingly, I realized there are muchhhh more literature devoted to $\beta\mathbb{N}$ than to $\beta\mathbb{Z}$. I wonder why is that? After all, algebraically $\mathbb{N}$ is a semigroup while $\mathbb{Z}$ is a group, and as discrete topological spaces they are homeomorphic. From your experience, how far $\beta\mathbb{N}$ and $\beta\mathbb{Z}$ are different in behaviour? Also, is $\beta\mathbb{N}$ ( or ($\beta\mathbb{N}\setminus\mathbb{N}$) easier to deal with?

REPLY [7 votes]: $\beta\mathbb Z$ consists of just two copies of $\beta\mathbb N$, one at "positive infinity" and one at "negative infinity".  It's generally easier to think about just one copy rather than both, so people tend to write more about $\beta\mathbb N$.  The advantage you mentioned for $\mathbb Z$ over $\mathbb N$, namely that the former is a group while the latter is only a semigroup, doesn't carry over to the Stone-Cech compactifications, both of which are (under the natural extensions of the addition operation) only semigroups.

REPLY [2 votes]: Since $\mathbb N$ and $\mathbb Z$ are homeomorphic, so are $\beta\mathbb N$ and $\beta\mathbb Z$, though of course the semigroup structure will be different.<|endoftext|>
TITLE: Are D_dR and D_st "potentially comparable"?
QUESTION [11 upvotes]: Suppose we have a de Rham Galois representation $G_K\rightarrow GL(V)$ for some $p$-adic field $K$ and some finite dimensional vector space $V$ over $\mathbf{Q}_p$.  Then it is a theorem that there is some finite extension $L/K$ such that $D_{dR}^L(V)\cong L\otimes_{L_0}D_{st}^L(V)$ (this is actually a corollary of the theorem that de Rham implies potentially semi-stable).
My question is whether I can still find such an extension $L/K$ without the assumption that the representation is de Rham.  In other words, is there always some finite extension $L/K$ such that $dim_L D_{dR}^L(V)=dim_{L_0}D_{st}^L(V)$?  I can't imagine that this is true, so what I'm really asking for is a counterexample.

REPLY [15 votes]: Actually, I think that Rebecca is right and that the answer is "no". Here's a sketch of the reason why.
Let $V$ be a $p$-adic representation. If $V$ is Hodge-Tate, then $D_{dR}(V) \neq 0$. So it's enough to find a HT representation such that $D_{st}^L (V) = 0$ for any $L$. Although I can't think of an explicit one, in my paper "Représentations potentiellement triangulines de dimension 2", with Gaëtan Chenevier, we prove the following theorem: if $X$ is the universal deformation space of some mod $p$ representation, and if $X_P$ is the subset of $X$ consisting of representations whose Sen polynomial is $P$, then the subset of $X_P$ consisting of potentially trianguline representations is a "thin subset"  (i.e. most representations in $X_P$ are not potentially trianguline). It remains to observe that if $D_{st}^L (V) \neq 0$, then $V$ is potentially trianguline.
EDIT : I should have said that $V$ is of dimension 2 here.<|endoftext|>
TITLE: Kaplansky's 6th conjecture: dim(Irrep) | dim(algebra) - for semi-simple Hopf algebras 
QUESTION [16 upvotes]: Let $H$ be a semisimple Hopf algebra. One of the Kaplansky's conjectures states that the dimension of any irreducible $H$-module divides the dimension of $H$. 
In which cases the conjecture is known to be true?

REPLY [6 votes]: There is a new survey on Kaplansky's sixth conjecture by L. Dai and J. Dong, available on the arxiv. Among other results, it mentions the following (always assuming $\operatorname{char} k=0$):
Special primes:

If a semisimple Hopf algebras has a simple module of dimension $p$, where $p=2$ or $p=3$, then its dimension is divisible by $p$.

Low dimension:

Semisimple Hopf algebras of dimension less than $60$ satisfy Kaplansky's sixth conjecture.

Particular properties:

A semisimple Hopf algebra $H$ satisfies Kaplansky's sixth conjecture if it satisfies one of the following conditions:

It is quasitriangular.
Its characters are central in $H^*.$
It is semisolvable.


Dividing other invariants:

$H$ a semisimple Hopf algebra. $A$ a transitive $H$-module algebra (e.g. $A=\operatorname{End}_k(V)$ for a simple $H$-module $V$).  Then $\dim A$ divides $(\dim V)^2\dim H$.

For references, see the survey mentioned above.<|endoftext|>
TITLE: Why would dim primitive irrep divide size of some conjugacy class ?  
QUESTION [12 upvotes]: From Isaacs et.al. 2005

Conjecture C. Let χ be a primitive
  irreducible character of an arbitrary
  finite group G. Then χ(1) divides |
  clG(g)| for some element g ∈ G.
Here, of course, we have written
  clG(g) to denote the class of g in G.
  We have checked that Conjecture C
  holds for all irreducible characters
  (primitive or not) of all groups in
  the Atlas 1.

Question 1 What is motivation for this ? Is it possible to describe what 
 conjugacy class(es) should correspond to irreducible representation in this way ?
(at least for some standard groups S_n, A_n, GL_n(F_q),...) What are representative examples?  
Question 2 Is it still open ? 

The authors write:

We now digress to explain our original
  motivation for considering these
  questions. There are numerous
  parallels and analogies between
  theorems concerning the of set
  irreducible character degrees of a
  finite group and theorems concerning
  the set of conjugacy class sizes of
  such groups. This suggests that
  perhaps there are some subtle
  arithmetic connections between these
  two sets of integers associated with a
  given group. One such connection that
  is easy to see is that each prime
  number that divides an irreducible
  character degree of G must also divide
  some class size of G. If G is
  solvable, then S. Dolfi showed that
  more is true. He proved [2] that given
  any two distinct primes p and q such
  that pq divides some irreducible
  character degree of a solvable group
  G, then pq also divides some class
  size of G. One might conjecture that
  the analogous assertion for three or
  more distinct primes is also true, but
  as far as we know, this remains open.


Partial result:
In the following, we use the notation np to denote the p-part of a positive
integer n, where p is a prime number.
Corollary D. Let χ be a primitive irreducible character of a solvable group
G, and let p be a prime divisor of |G|. Then χ(1)p divides (| clG(g)|p)
3 for some element g ∈ G.

Not related results, for complteness:
Denote CV(g) fixed point subspace for g in V.
Our main result is the following.
Theorem A. Let V be a nonzero finite dimensional completely reducible
F G-module, where F is any field and G is any finite group. Assume that
CV (G) = 0 and let p be the smallest prime divisor of |G|. Then there exists
some element g ∈ G such that
$ dim CV (g) ≤ (1/p) ~ dim V $.
The fraction 1/p cannot, in general, be replaced by any smaller quantity.
In particular, this shows that Neumann’s conjecture is valid for odd-order
groups, at least...
Corollary B. Let V be a nonzero finite dimensional completely reducible
F G-module, where F is an arbitrary field and G is any finite group, and
assume that CV (G) = 0. Then
$1/ |G| \sum_{g∈G} dim CV (g) ≤ ((p + 1)/ 2p)~~ dim V$ ,
where p is the smallest prime divisor of |G|.

REPLY [9 votes]: I have not noticed this question before, though it was posted several years ago. As a comment on the question as a whole, and especially Question 1 asked in the text, there are likely to be many such elements $g$ for many groups, and I would not expect there to be any "natural correspondence" between the $g$ associated in this way to a given $\chi.$
The reason I put forward that comment is that a result of J.G. Thompson (which appears in the text by Isaacs on character theory).
The result of Thompson asserts that if $\chi$ is an irreducible character (primitive or not) of  finite group $G$ then $\chi$ takes value $0$ or a root of unity on more than $\frac{|G|}{3}$ elements of $G$.
This does not prove the conjecture, since it could be that $\chi$ vanishes on more than $\frac{|G|}{3}$ elements ( and extraspecial $p$-groups of order $p^{3}$- $p$ a prime - are examples where there is an irreducible character taking root of unity values nowhere, though the irreducible character is imprimitive in that case).
But it does make it seem likely that there will be many groups $G$ which have an irreducible character $\chi$ taking a root of unity value at some $x \in G,$ and often one might expect several such $x$.
For any such $x,$ note that $\frac{[G:C_{G}(x)]}{\chi(1)} = \overline{\chi(x)}\frac{[G:C_{G}(x)]\chi(x)}{\chi(1)} $ is a rational algebraic integer, hence is an integer.
Later edit: A familiar example is the Steinberg character $\chi$ of a finite quasisimple characteristic $p$ Lie type group $G$. For each $p$-regular $g \in G,$ we have $\chi(g) = \pm |C_{G}(g)|_{p},$ and by Brauer's general theory we have $p \not | |C_{G}(g)|$ for some $p$-regular $g \in G$, so that $\chi(1)$ (which is a power of $p$) divides $[G:C_{G}(g)].$
Later edit rewriting badly written earlier addition:
Thompson's argument uses a result of C. Siegel, stating that if $\alpha \neq 1$ is a totally positive (real) algebraic integer with $n$ algebraic conjugates, then the sum of those conjugates is at least $\frac{3n}{2}$ (the bound is attained for $\alpha = \frac{3 \pm \sqrt{5}}{2}),$ as Siegel noted.
We apply this to the orthogonality relations for the columns of the character table (instead of the rows, as Thompson did) to obtain an inequality relevant to this question.
Let $x$ be an element of the finite group $G$ such that $|C_{G}(x)| = c$ such that $s$ irreducible characters of $G$ do not vanish at $x$ and such that $a$ of those characters do not take root of unity values at $x$ either.
Then Siegel's result yields that $(s-a) + \frac{3a}{2} \leq c,$ so that $s \leq c- \frac{a}{2}.$ Note also that $s \geq a + [G:G^{\prime}],$ since every linear character of $G$ takes a root of unity value at $x.$
Hence $c \geq \frac{3a}{2} + [G:G^{\prime}]$ 
and $a \leq \frac{2(c-[G:G^{\prime}])}{3}.$
If there is no non-linear irreducible character of $G$ taking a root of unity value at $x,$ then we have $s = a +[G:G^{\prime}],$ so that 
$s \leq \frac{[G:G^{\prime}]}{3} + \frac{2c}{3}.$ 
We may conclude that if more than $\frac{[G:G^{\prime}]}{3} + \frac{2|C_{G}(x)|}{3}$ irreducible characters of $G$ do not vanish at $x,$ then there is a non-linear irreducible character $\chi$ of $G$ such that $\chi(x)$ is a root of unity (so that $\chi(1)$ divides $[G:C_{G}(x)]).$<|endoftext|>
TITLE: Smith normal form of a Matrix with -1 outside the diagonal
QUESTION [8 upvotes]: I am given a matrix of the following form:
$$M = \begin{pmatrix}
    a_0   & -1     & \cdots & \cdots     & -1 \newline
    -1    &  a_1   & \ddots     &  & \vdots \newline
   \vdots &   \ddots     & \ddots &   \ddots     & \vdots \newline
   \vdots &        &   \ddots     & \ddots & -1 \newline
   -1     & \cdots &    \cdots  & -1 & a_n 
\end{pmatrix}$$
 with $a_i \in \mathbb{Z}$, $a_i > 0$.
Is there an easy way to write down the Smith normal form of this matrix?
greatz Johannes

REPLY [6 votes]: According to http://www-math.mit.edu/~rstan/transparencies/snf.pdf, 
the Smith Normal form for a nonsingular matrix $M$ is 
$$
\textrm{Diag}(e_1,\ldots, e_n),
$$
where $e_1\cdots e_i = \gcd \big(    i\times i\textrm{ minors of }M  \big)$. 
There are infinitely many $a_1,\ldots, a_n$ such that $M$ is nonsingular, so we assume that $M$ is nonsingular for convenience. The singular case may be treated separately.
Case 1 : $M$ is nonsingular
Since the matrix has $-1$ in most of the entries, the $i\times i$ minors are relatively easy to find. 
From this, the $1\times 1$ entry $e_1$ is easily seen to be $1$. 
The $2\times 2$ entry $e_2$ can be obtained from 
$$
e_2=e_1e_2=\gcd\big( \{a_i+1 | i=1,\ldots,n\}, \{a_ia_j-1 | 1\leq i<j\leq n\}\big).
$$
Since $a_ia_j-1=(a_i+1)(a_j+1)-(a_i+1)-(a_j+1)$, we have
$$
e_2=\gcd\big( \{a_i+1 | i=1,\ldots,n\}\big).
$$
The $3\times 3$ entry $e_3$ is obtained from
$$
\begin{align}
e_1e_2e_3&=\gcd\big( \{(a_i+1)(a_j+1) | 1\leq i<j\leq n\}, \{(a_i+1)(a_j+1)(a_k+1) -(a_i+1)(a_j+1) - (a_j+1)(a_i+1) - (a_k+1)(a_i+1)|1\leq i<j<k\leq n\}\big)\\
&=\gcd\big( \{(a_i+1)(a_j+1) | 1\leq i<j\leq n\} \big)
\end{align}
$$
In general, for $2\leq i\leq n-1$, 
$$
e_1\cdots e_i = \gcd\big(   \{\prod_{k=1}^{i-1} (a_{j_k}+1) | 1\leq j_1< \cdots <j_{i-1}\leq n\}\big).
$$
Then $e_n$ is obtained from 
$$
e_n=\frac{\det M}{e_1\cdots e_{n-1}}.$$
Case 2: $M$ is singular
This case, we have 
$$
1 = \sum_{i=1}^n \frac1{a_i+1}.
$$
Since $1\neq \sum_{i=1}^{n-1}\frac1{a_i+1}$, the rank of $M$ is $n-1$ and we can proceed the same method as in Case 1, after putting all zeros on the last column and the last row by row/column operations.<|endoftext|>
TITLE: Herringbone partitions of regions and surfaces
QUESTION [6 upvotes]: Let $R \subset \mathbb{R}^2$ be a region of the plane bounded
by a Jordan curve.  The boundary $\partial R$ could be a polygon,
or a smooth curve—there are variations depending upon boundary assumptions.
I would like to partition $R$ into regions $R_i$ that can be striped by
parallel lines with the property that each stripe meets the boundary
$\partial R_i$ at an angle that excludes the open 
interval $(\frac{1}{4}\pi,\frac{3}{4}\pi)$.
In other words, the stripes meet the boundary of $R_i$ at $45^\circ$ or more sharply; they cannot meet the boundary nearly orthogonally. 
(Here the boundary is the boundary of $R_i$, which might include portions of 
the boundary of $R$.)
And the ultimate goal is to partition $R$ into the minimum number of such regions.
For example, a rectangle can be partitioned into one region,
but it seems a circle may need four regions(?):

          


A number of questions suggest themselves:

Q1. For $R$ a polygon of $n$ vertices, what is the largest number
  of herringbone regions needed for a fixed $n$, and how many regions always suffice for
  a fixed $n$?
Q2. For $R$ a circle in $\mathbb{R}^2$, or the surface of a sphere in
    $\mathbb{R}^3$, what is the optimal (fewest regions) herringbone partition?
    Can the circle be herringbone-partitioned into fewer than four regions?
Q3. What is the optimal herringbone partition of a torus?

I'll stop here, as you can spin off these questions as easily as can I.
My original focus was on compact surfaces in $\mathbb{R}^3$ (the sphere and torus above),
when the pattern is drawn by parallels to a geodesic,
but it already
seems interesting in $\mathbb{R}^2$.
Thanks for any insights and/or pointers to relevant literature!

          
          


Update. Sergei's idea, as articulated by Cristi, leads to a 3-region herringbone partition
of a circle.  Here his Cristi's illustration from his comment:

REPLY [4 votes]: Three regions is the minimum for any convex figure with a smooth boundary. To construct a partition, let $AB$ be the diameter of $R$. The segment $[AB]$ splits $R$ into two pieces $R_1$ and $R_2$. Let $C_i$ be a point at maximal distance from $[AB]$ in $R_i$, $i=1,2$. Draw perpendiculars $C_iD_i$ from $C_i$  to $AB$. The segments $[AB]$, $[C_1D_1]$ and $[C_2D_2]$ split $R$ into 4 regions which can be filled by parallel lines just like the 4 quarters of the circle. Then you can merge two regions by removing the segment $[D_1D_2]$ or using a zig-zag if $D_1=D_2$.
Two regions are impossible. Indeed, parametrize $S^1$ by tangent directions of $\partial R$ (say, oriented counter-clockwise). For each region $R_i$, consider the set $R_i\cap\partial R$. The tangent directions of this set form angles at most $\pi/4$ with a fixed line, hence they belong to the union of two arcs of length $\pi/2$ in the circle of tangent directions. It is possible to divide $S^1$ into 2 pair of arcs like this, but then the two corresponding pairs of arcs in $\partial R$ cannot be connected by disjoint paths (I assume you don't want connected but not path-connected examples).
The same arguments, modulo minor details, works for any $n$-gon with angles sufficiently close to $\pi$. For example, for a regular $n$-gon with $n\ge 10$.<|endoftext|>
TITLE: For which Millennium Problems does undecidable -> true?
QUESTION [36 upvotes]: Three good answers were received — by Alex Gavrilov, Bjørn Kjos-Hanssen, and Terry Tao — and the bounty has been awarded (somewhat arbitrarily) to Alex Gavrilov.  
The answers are summarized below; because they are open-ended and technically subtle, the question has been flagged for conversion to community Wiki.
Thanks are extended to all who contributed.

Summary  Harry Altman cogently commented:

This is essentially asking which of these statements are equivalent to a $\Pi^0_1$ statement, right? 

We embed this insight into a better version of the question:  

Which of the Millennium Prize problems can be stated as a postulate that can be falsified by a $\Pi^0_1$ counterexample?

to which the answers given (as I understand them) amount to:

"The Riemann Hypothesis is true" …a $\Pi^0_1$ counterexample could exist;(per Noam Elkies' comment)
"The Birch and Swinnerton-Dyer Conjecture is true" … conceivably a $\Pi^0_1$ counterexample could be constructed, but not with present knowledge (per Alex Gavrilov's answer);
"$\mathsf{P}\ne\mathsf{NP}$" … no obvious $\Pi^0_1$ counterexample(per Bjørn Kjos-Hanssen's answer);
"Navier–Stokes is globally regular" … no obvious $\Pi^0_1$ counterexample(per Terry Tao's answer);
"Yang–Mills has a mass gap" … no obvious $\Pi^0_1$ counterexample (?);
"The Hodge Conjecture is true" … no obvious $\Pi^0_1$ counterexample (?);

Resource  Wikipedia's article Arithmetical Hierarchy explains the notation of Harry Altman's answer.
What "No Obvious $\Pi^0_1$ Counterexample" Means   As was noted on Dick Lipton and Ken Regan's weblog Gödel's Lost Letter and P=NP, the authority of the Scientific Advisory Board (SAB) of the Clay Mathematics Institute (CMI) extends to:

"The SAB may consider recommending the award of the prize to an individual who has published work that, in the judgement of the SAB, fully resolves the questions raised by one of the Millennium Prize Problems even if it does not exactly meet the wording in the official problem description.” 

In view of the CMI/SAB's supreme executive authority, the logical possibility of amending a Millennium Prize question to accommodate $\Pi^0_1$ counterexamples — via ingenious "burning arrows," to adopt Dick Lipton and Ken Regan's phrase — cannot be formally excluded.

REPLY [19 votes]: Global regularity for Navier-Stokes on the torus is (logically equivalent to) a $\Pi_2^0$ statement; this is essentially an unpublished observation of Bourgain (who made it in the more general context of supercritical equations), which I sketched out in this paper http://arxiv.org/abs/0710.1604 .  Basically, Navier-Stokes is equivalent to the assertion that for all $T, E > 0$, there exists an $M$ such that all initial data with $H^1$ norm at most $E$, there exists a solution up to time $T$ whose $H^1$ norm is always bounded by $M$.  Now if such a claim is the case, it can be verified (using rigorous perturbation theory) by constructing a sufficient number of approximate solutions to a sufficient number of choices of initial data, and verifying that all of these approximate solutions have norm at most $M/2$ (say) up to time $T$, where "sufficient number" is something explicit that depends on $T, E, M$, and the nature of the approximation can be discretised at an explicit scale that also depends on $T,E,M$.  So Navier-Stokes is equivalent to a statement of the form $\forall T,E \exists M: P(T,E,M)$ where $P(T,E,M)$ is something that can be verified in finite time for each $T,E,M$ (which can be taken to be rational numbers).  (In fact, for each $E$ there is an explicit time $T = T(E)$ for which one needs to verify $P(T,E,M)$, and beyond which global regularity is easily obtained from the energy dissipation, so the claim even simplifies a little to $\forall E \exists M: P( T(E), E, M)$.)
It looks very difficult to me, however, to make Navier-Stokes equivalent to a $\Pi^0_1$ statement.  This would basically amount to being able to describe all possible "blowup scenarios" by a countable set, and to be able to determine whether each such blowup scenario can actually happen in finite time.  While some blowup scenarios (particularly "stable", "approximately self-similar" ones) can be described and verified in such a manner, it's not clear whether all of them can.<|endoftext|>
TITLE: How to understand $Ext(\mathcal{O}_{Y},\mathcal{O}_{Z})$ for subvarieties $Y,Z\subset X$? 
QUESTION [10 upvotes]: By standard homological algebra we know that $Ext(A,B)$ of $R$-modules classifies certain equivalence classes of short exact sequences $0\rightarrow B\rightarrow C \rightarrow A \rightarrow 0$ of $R$-modules, where $R$ is a commutative ring. I now would like to understand this fact in geometry. 

Let $X$ be a variety (or a scheme if you want), how should I understand $Ext(O_{Y},O_{Z})$ for subvarieties $Y,Z\subset X$? Of course it classifies extensions of $O_{Z}$ by $O_{Y}$, but are there any geometric or intuitive way to understand $Ext(O_{Y},O_{Z})$? 
More generally are there any geometric way to understand $Ext(\mathcal{E},\mathcal{F})$ for coherent $O_X$-modules $\mathcal{E},\mathcal{F}$? 

I would appreciate any idea about "seeing" these extensions.

REPLY [4 votes]: Here is some special example.
If j : Y --> X is a closed embedding of smooth varieties over a field then 
$\mathcal{Ext}^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) = \wedge^{i} N_{Y|X}$
where$ N_{Y|X}$ is the normal bundle of Y in X.
In particular, il the local global spectral sequence degenerates, for example if all is affine, 
then
$ Ext^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) = H^{0}(\wedge^{i} N_{Y|X})$
The case i = 1 express the intuition that $ Ext^{1}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) $
is a (infinitesimal) measure of the ability to deform Y in X.
By taking X the product of Y by itself and j the diagonal embedding, 
one obtains a interpretation of $ Ext^{i}_{X}(j_{*}O_{Y}, j_{*}O_{Y}) $ in terms
of (dual of) De Rham cohomology.<|endoftext|>
TITLE: When do Hochschild homology and cohomology agree? (Ambidexterity?)
QUESTION [5 upvotes]: Suppose $X$ is a smooth algebraic variety over a field of characteristic $0$. What are the most general conditions under which Hochschild homology and cohomology of $X$ agree?
The existence of a symplectic form will do the job, but is there something more general?
I would be interested in particular into a condition along the lines of Lurie's recent approach to isomorphism of homology and cohomology via the theory of ambidexterous functors (iso between a left and a right adjoint to a given functor, under some specific conditions).

REPLY [5 votes]: If $X$ is Calabi--Yau (that is $K_X = 0$) then Hochschild homology and cohomology agree up to a shift of grading by dimension of $X$.<|endoftext|>
TITLE: Smallest dimension of nontrivial representation of a simple Lie algebra over `$\mathbb{C}$`
QUESTION [11 upvotes]: The question involved here is natural and very classical, but I'm unsure what has been formally stated and proved in the literature.  The only approach I know involves assembling facts that apparently weren't all known until the 1970s.  
Start with a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$, along with some Cartan subalgebra $\mathfrak{h}$ and the resulting weight lattice $X$ and root sublattice $X_r$ inside $\mathfrak{h}^*$ (Bourbaki's $P \supset Q$).  Fix simple roots $\alpha_1, \dots, \alpha_\ell$ and corresponding fundamental dominant weights $\varpi_1, \dots, \varpi_\ell$.   [The symbol \varpi gives an old version of the handwritten letter pi, for "poids".]    Finite dimensional simple modules $L(\lambda)$ are then parametrized by dominant weights $\lambda$ with $L(0)$ the trivial module, while arbitrary finite dimensional modules are direct sums of these.  Everything here just depends up to isomorphism on $\mathfrak{g}$, by standard conjugacy theorems.
Now the natural problem is: Find the smallest nontrivial (hence faithful) finite dimensional modules.   By complete reducibility, it's enough to consider simple modules.  A detailed answer requires the Killing-Cartan classification of simple Lie algebras, but there is a classification-free result:
All weights of $L(\lambda)$ lie in a single coset of $X_r$ in $X$, and these weights consist of Weyl group orbits of various dominant $\mu \leq \lambda$ in the standard ordering (possibly with multiplicity $>1$ if $\mu < \lambda$).  
(1) Fix a coset not containing 0.   Then there is a unique smallest nontrivial
$L(\lambda)$ corresponding to a "minuscule" $\lambda$ (always one of the fundamental weights), with weights consisting of the Weyl group orbit of $\lambda$.
(2) Fix the coset containing 0.    Then there is a unique smallest nontrivial $L(\lambda)$, whose weights consist of 0 together with the Weyl group orbit of $\lambda$.   Here $\lambda^\vee$ is the highest root in the dual root system.

Is this written down and proved somewhere?   

The history might also be interesting (or just messy).   Once this result is in hand, it's not difficult to fill in the details for each simple Lie algebra, but much of that is found only in exercises of books by Bourbaki or me. 
REFERENCES: As I commented to Sasha Premet, I've only seen fragments of the story in earlier textbooks.   There is a short note by H. Freudenthal in Proc. Amer. Math. Soc. 7 (1956), 175-176, concerning occurrence of the 0 weight; but this is superseded by later results.   For general background and details on (1) and (2), I'll refer to [B1] = Bourbaki, Chap. 6 (1968), my book [H] = GTM 9 (1972), [B2] = Bourbaki, Chap. 8 (1975).
[B1]: $\S1$, Exer. 24, and  $\S4$, Exer. 15.
[H]:  Exer. 13.10 (just using root system axioms) and Prop. 21.3 (using basic representation theory),  Exer. 21.3.   Also Exer. 13.13 ("minimal nonzero" = "minuscule").
[B2]: $\S7.2$, $\S7.3$, and $\S7$, Exer. 22.

REPLY [6 votes]: In Bourbaki, Ch. VIII, $\S$ 7, Sect. 2, one can find the notion of an $\mbox{$R$-saturated set}$, and Corollary to Prop. 4 in that section proves that for every $R$-saturated set $\mathcal X$ there is a finite dimensional $\mathfrak g$-module whose set of weights coincides with $\mathcal X$. Prop. 6 in the next sections proves that the smallest $R$-saturated sets have the form $W.\lambda$ where $\lambda$ is minuscule.
This answers Question (1). For Question (2) we take for $\mathcal X$ the union of $0$ and the set of all short roots in $R$. It is easy to see that this is an $R$-saturated set as verifying this reduces to root systems of rank 2 where everything is clear (even in type ${\rm G}_2$). Applying the above-mentioned Corollary we get a $\mathfrak g$-module with desired properties. If $\beta$ is the dominant short root then, by construction, our set will coincide with the set of weigths of $L(\beta)$ as that set cannot be any smaller.<|endoftext|>
TITLE: A question about Quine's set theory NF.
QUESTION [8 upvotes]: This question might not really be considered appropriate for mathoverflow.net but
I'll risk asking it and apologize in advance if I have commited a booboo. It is often
said that in NF one can prove the existence of infinite sets without the help of any
special axiom of infinity. Now Tarski's definition of an infinite set (which includes
more sets than Dedekind's definition when the axiom of choice is not available) states
that a set X is infinite just in case there exists a non-empty set T of subsets of X
such that if we are given any element u of T there is an element v of T which is a
proper subset of u. In order to prove that even the universal set V is infinite one
would need to exhibit a specific non-empty sub-cllection V* of V which is certified
to be a set in NF and is such that for every element y of V* there is an element z
of V* which is a proper subset of y. I have never seen a specific example of such a set
V* and cannot think of how to define one. Note that most infinite sub-collections of V
are not sets in NF because of the stratification requirements. Is there (an example of)
such a V* and if not how can one really say that NF proves the existence of infinite sets?

REPLY [11 votes]: One can give stratified definitions for individual Frege-Russell natural numbers, and then so too for the set $\mathbb{N}$ of all Frege-Russell naturals, so that exists in NF. One can then check that the set $$\mathcal{T} = \{X\subseteq\mathbb{N} : \exists n\in\mathbb{N} (X = \mathbb{N}\setminus\{0,\dots,n\})\}$$ has a stratified definition, and so it too provably exists in NF. This $\mathcal{T}$ is the obvious candidate for witnessing that $\mathbb{N}$ is Tarski-infinite; we just need to know that NF proves it has the desired property.
The only sticking point for that would be showing that $$\mathbb{N}\setminus\{0,\dots,n+1\} \subsetneq \mathbb{N}\setminus\{0,\dots,n\}$$ for all $n$, and specifically that the inclusion really is proper, i.e. that none of our Frege-Russell naturals is empty. But this is where Specker's 1953 heavy lifting comes into play; namely, his proof that the universe $V$ cannot be well-ordered also implies that it is "Frege-infinite," i.e. $\forall n\in\mathbb{N} (V\notin n)$, because NF can prove that all Frege-finite sets are well-ordered. In turn, the fact that $V\notin n$ forall $n$ can be used to show that $n\ne\emptyset$ for all $n$, as required.<|endoftext|>
TITLE: Injective maps on cohomology and Kahler manifolds
QUESTION [7 upvotes]: Compact Kahler manifolds have the property that surjective maps induce injections on cohomology with coefficents in $\mathbb{Q}$ (That is, if $X,Y$ compact Kahler, then a surjective map $\phi: X \rightarrow Y$ induces injections $\phi^*: H^i(Y, \mathbb{Q}) \rightarrow H^i(X, \mathbb{Q})$ for all $i$, [Voisen, Hodge Theory I, p 177]).
Question: I'm wondering if I should think of this as a property of compact Kahler manifolds, or as an instance of something more general. For example, can the Kahler condition be replaced with a more general class of manifolds (not dropping the compactness hypothesis). Perhaps one that includes not just complex manifolds but maybe a few manifolds of odd (topological) dimension? I know that if we require that $\dim X = \dim Y$ then the fact above is true more generally just for compact oriented manifolds, for formal reasons.

REPLY [15 votes]: In order to have this property it is sufficient to require that $\phi^{-1}(y)$ is a non-zero cycle in $H_*(X,\mathbb Q)$, where $y$ is a generic point in $Y$. This holds indeed when $X$ and $Y$ are Kahler.
Let me give an example showing that this does not work when $X$ and $Y$ are just complex. 
Example. Let $X$ be a Hopf surface $X=(\mathbb C^2\setminus 0)/\mathbb Z$ where $\mathbb Z$ is acting on $\mathbb C^2$ by multiplication by (say) $2$. Then there is a fibration $\phi\colon X\to \mathbb CP^1=Y$. This fibration comes  from the standard $\mathbb C^*$-action on $\mathbb C^2$. Now  $\phi^{-1}(y)$ is null-homlogous.

REPLY [2 votes]: Suppose one has compact symplectic manifolds $(X,\omega), (Y,\sigma)$, and a map $f:X\to Y$ such that $f^\ast\sigma=\omega$ (if $f$ is also a diffeomorphism, then this map is a symplectomorphism, but I'm not sure the terminology in this case). Then $f^\ast: H^*(Y;\mathbb{Q})\to H^\ast(X;\mathbb{Q})$ will be injective. If $dim Y=2k$, then $\sigma^k\neq 0\in H^{2k}(Y)$ is a fundamental class, and $dim X=2n \geq 2k$, then $\omega^n\neq 0\in H^{2n}(X)$. So $f^\ast(\sigma^k)= \omega^k \neq 0 \in H^{2k}(X)$. 
Now, consider $\alpha \in H^j(Y)$. By Poincare duality, there exists $\beta\in H^{2k-j}(Y)$ such that $\alpha\cup \beta = \sigma^k$. Then $f^\ast(\alpha\cup \beta)=\omega^k\neq 0$, so $f^\ast \alpha\neq 0$.<|endoftext|>
TITLE: Quasi-linear System of First Order P.D.E.s of "Mixed" type
QUESTION [5 upvotes]: In my research work, I am dealing with a quasi-linear system of first order p.d.e.'s with two independent variables (say $x_1$ and $x_2$) and four dependent variables (say $u_1(x_1,x_2)$, $u_2(x_1,x_2)$, $u_3(x_1,x_2)$ and $u_4(x_1,x_2)$) of the form
\begin{equation}
\mathbf{A}(\mathbf{x},\mathbf{u}(\mathbf{x}))\frac{\partial \mathbf{u}}{\partial x_1}+\mathbf{B}(\mathbf{x},\mathbf{u}(\mathbf{x}))\frac{\partial \mathbf{u}}{\partial x_2}=\mathbf{f}(\mathbf{x}),
\end{equation}
where $\mathbf{x}=[x_1\,\,x_2]^{T}$ and $\mathbf{u}=[u_1\,\,u_2\,u_3\,u_4]^{T}$ are, respectively, the independent and dependent vector variables, constructed out of $x_\alpha$'s and $u_i$'s, $\alpha=1,2$ and $i=1,2,3,4$. $\mathbf{A}$ and $\mathbf{B}$ are two $4\times 4$ real matrices which are functions of $\mathbf{x}$ and $\mathbf{u}$ (dependence on $\mathbf{u}$ makes the system quasi-linear) and $\mathbf{f}$ is a known vector valued function of $\mathbf{x}$.
To determine the type of the system, we consider the eigenvalues of the matrix $\mathbf{A}^{-1}\mathbf{B}$ (assuming $\mathbf{A}$ to be invertible which it is in my case). Now there are 4 possibilities about the nature of these eigenvalues and there are corresponding "type" of the p.d.e. system defined:

All eigenvalues real and distinct, implying hyperbolic system. We have the well-studied system of hyperbolic system of conservation laws falling under this category.
All eigenvalues are complex (two pairs of complex conjugate eigenvalues). We have then an elliptic system. (I don't know about any physical example of this type.)
All eigenvalues are real but with unequal algebraic and geometric multiplicity (i.e. no set of 4 linearly independent eingenvectors). We then have a parabolic system. (Again, no knowledge of physical example.)
Two real and one pair of complex conjugate eigenvalues. This does not fall into the above categories. Probably these are called the quasi-linear system of mixed type.

My first question is about the the above classification scheme which, in fact, imitates the classification scheme of linear system of first order p.d.e.'s. Matrices $\mathbf{A}$ and $\mathbf{B}$ involve unknown function $\mathbf{u}$. So, how to determine the type of the system (from the nature of its eigenvalues) a priori to the determination of the solution?
My second question is about the type 4 systems, because my problem falls into this category. How to deal with this mixed type p.d.e. systems?
My major is mechanical engineering and I am working in theory of elasticity. Any help would be greatly appreciated as I know very little about analytical methods to solve quasi-linear p.d.e. systems of first order.
Just to mention, in a previous question I was concerned about a system of 3 linear first order pdes of mixed type (3 roots of the characteristic polynomial: either all are real, or one real and the others appearing as a complex conjugate pair) which is a special case of this problem. To emphasize, I come up with these systems in a physical problem and I know that there must exist real valued "well-behaved" solutions (bounded). Which I do not know is how to find those solutions which must exist and I am looking for possible methods in this forum.
Thanks in advance for any help.

REPLY [3 votes]: I gave an answer to your earlier question, but although it was technically correct, I see now that since you have only two independent variables my answer wasn't really the right one. Let me try one more time (but I can't guarantee I'm right).
I'm going to address only case 4, since that's the one you care about. Given your assumptions, there exist invertible-matrix-valued functions $P(x,u)$ and $Q(x, u)$ such that the system
$$
P(A\partial_1 u + B\partial_2 u) = Pf
$$
can be written as
$$
\partial_1 v + C(x,v)\partial_2v = g(x,v),
$$
where $v = (v_1, v_2, v_3, v_4) = Q(x,u)u$.
$$
 C(x,v) =
 \begin{bmatrix} 
 a(x,b)& 0 & 0 & 0 \newline 
 0 & b(x,v) & 0 & 0 \newline
 0 & 0 & 0 & -1 \newline
 0 & 0 & 1 & 0
\end{bmatrix}
 $$
If you fix a rectangular domain, say, $D = [0,T]\times[0,S]$, then this system is well-behaved from the point of view of uniqueness and, with the right additional assumptions, existence if you specify initial data consisting of $v_1$ and $v_2$ on $\{0\} \times [0,S]$ and $v_3$ on the boundary of $D$.
This is a coupled hyperbolic-elliptic system that is hyperbolic in $v_1$ and $v_2$ and elliptic in $v_3$ and $v_4$. My guess is that linear systems like this been analyzed and solved before, but unfortunately I don't know any specific reference. You would handle the quasilinear version using the linear theory using the usual techniques (inverse function theorem, fixed point theory, etc.). If you know the theory of linear first order elliptic systems of PDE's and of linear first order hyperbolic systems of PDE's in two independent variables really well, it's reasonably straightforward to adapt it to a coupled system like this.<|endoftext|>
TITLE: Closed formula for heat kernel
QUESTION [5 upvotes]: Is there, similar to the Mehler kernel, a closed formula for the heat kernel of the heat equation associated to the Laplacian
$$ -\sum_j \frac{d^2}{dx_j^2} + 2\sqrt{-1} \sum_j \lambda_j \frac{d}{dx_j} + \sum_{ij} a_{ij}x_ix_j$$
on $\mathbb{R}^n$? Here, the matrix $(a_{ij})$ is supposed to be symmetric and positive definite, while the $\lambda_j$ can be arbitrary.

REPLY [13 votes]: Yes there is.  Here is how you do it. First   find an orthogonal change in variables  
$$ x_j=\sum_{jk} s_{jk}y_k $$
$(s_{jk})$ orthogonal matrix, so that in the new coordinates we have
$$ \sum_{i,j}a_{ij} x_ix_j = \sum_j \mu_j^2 y_j^2, $$
where $\mu_j^2$ are the eigenvalues of the symmetric matrix $(a_{ij})$. 
Note that $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\pa}{\partial}$
$$\frac{\pa}{\pa x_k}=\sum_j\frac{\pa y_j}{\pa x_k}\frac{\pa }{\pa y_j} =\sum_j t_{jk}\frac{\pa}{\pa y_j}, $$
where $(t_{jk})$ is the inverse of the orthogonal matrix $(s_{jk})$ so that $t_{jk}=s_{kj}$.
Then for some real numbers $r_j$ 
$$ -\sum_j\frac{\pa^2}{\pa x_j^2}+2\ii\sum_j\lambda_j\frac{\pa }{\pa x_j}+\sum_{i,j}a_{ij}x_ix_j $$
$$ =-\sum_j\frac{\pa^2}{\pa y_j^2} +2\ii\sum_j r_j\frac{\pa}{\pa y_j} +\sum_j \mu_j^2 y_j^2 $$
$$=\underbrace{\sum_j \left(\ii\frac{\pa}{\pa y_j}+r_j\right)^2 +\sum_j\mu_j^2y_j^2-\sum_j r_j^2}_{=: L}. $$
Next set
$$ R^2 :=\sum_j r_j^2, \;\; w(t,y) :=  R^2t +\sum_j \ii r_j y_j, $$
$$ L_0 :=\sum_j\left(-\frac{\pa^2}{\pa y_j^2} +\mu_j^2y_j^2\right), $$
and observe that
$$\pa_t +L =e^{w}(\pa_t+L_0) e^{-w}. $$
Thus, if $K$ is a fundamental solution of $\pa_t+L_0$, then
$$(\pa_t +L) (e^{w} K) = e^{w} (\pa_t+L_0)K=e^{w} \delta_0=\delta_0 $$
so that $e^{w}K$ is a  fundamental solution of $\pa_t+L$.   You can find $K$ using Mehler's formula.<|endoftext|>
TITLE: A maximal element, where Schur gives a minimal element
QUESTION [12 upvotes]: Let me recall a result due to I. Schur, which I learnt from F. Goldberg's answer to my MO question Hadamard-like inequalites for positive definite symmetric matrices. If $H$ is a subgroup of $\frak S_n$ 
and $\chi$ is an irreducible complex character over $H$, define
$$d_\chi(S)=\frac1{\chi(e)}\sum_{g\in H}\chi(g)\prod_{i=1}^ns_{ig(i)}.$$
Then for every $S\in SPD_n$, we have
$$\det(S)\le d_\chi(S).$$
Notice that if $H=\frak S_n$ and $\chi$ is the signature, then $d_\chi$ is the determinant. Thus $\det$ is the smallest element among the $d_\chi$'s. If instead $\chi={\bf1}$, then $d_\chi$ is the permanent. If $H=(e)$, Schur's inequality is just the Hadamard inequality
$$\det S\le\prod_is_{ii}.$$
Given $n$, there are many distinct $d_\chi$'s, even though several choices of the pair $(H,\chi)$ yield the same function. For instance, there are only $11$ distinct functions if $n=3$, among $13$ pairs. 

My question is whether the permanent is the largest element among the $d_\chi$'s. In other words, is it true that for every $S\in SPD_n$, we have
  $$d_\chi(S)\le{\rm per}(S)\quad ?$$

I checked the truth of this assertion if $n=2$, $n=3$, and also in quite a complicated case of $n=4$, where $H={\frak A}_4$ and $\chi\ne{\bf1}$ is a linear character.

REPLY [17 votes]: This question is better known as the permanental dominance conjecture and is still an open problem.
According to Zhan's survey, it has been confirmed for every irreducible character  of $S_n$ for $n \le 13$. Another reference cited for this conjecture is this survey on open problems about permanents by Cheon and Wanless.

EDIT (added 12/10/2015): Incidentally, the closely related Soules's conjecture whose proof would yield the above permanental dominance (and which states that the largest eigenvalues of the Schur-product matrix of a given Hermitian semidefinite matrix $A$ equals the permanent of $A$), has been very recently shown to be false: check out this explicit counterexample!. (If the first link does not work, try this link on LAA)<|endoftext|>
TITLE: Over which fields does the Mordell-Weil theorem hold?
QUESTION [31 upvotes]: According to a well-known theorem of Mordell, the group of rational points $E(\mathbf{Q})$ of an elliptic curve $E/\mathbf{Q}$ is finitely generated. Weil generalized this theorem to abelian varieties over number fields.
Less well-known is the following generalization, due I believe to Néron : if $K$ is a field of finite type (that is, finitely generated over its prime field) and $A$ is an abelian variety over $K$, then $A(K)$ is finitely generated. There is an even more general statement, the Lang-Néron theorem, for relative field extensions which are finitely generated (see Brian Conrad's article for the precise statement and a proof of this theorem).
Q1. Are there other fields $K$ for which the group of $K$-rational points of an abelian variety over $K$ is always finitely generated?
In the other direction, there exist fields $K$ for which $A(K)$ is clearly never finitely generated whenever $\operatorname{dim}(A) \geq 1$. For example $K=\mathbf{C}$, in which case it follows from the description af abelian varieties as complex tori. If $K$ is a finite extension of $\mathbf{Q}_p$, then $A(K)$ contains a finite-index subgroup isomorphic to $\mathcal{O}_K^{\operatorname{dim} A}$, so $A(K)$ is again never finitely generated. Other examples I can think of are complete discretely valued fields and algebraically closed fields. Note that we often have the stronger result that $A(K) \otimes \mathbf{Q}$ is infinite-dimensional (except when $K=\overline{\mathbf{F}}_p$, in which case $A(K)$ is a torsion group).
Q2. Are there other fields $K$ for which the group of $K$-rational points of a non-trivial abelian variety over $K$ is never finitely generated?

REPLY [5 votes]: This is a partial answer to Q2, in particular a full answer to the question asked by François Brunault in the comments:

By the way, if K is ferile, will A(K)⊗Q be necessarily be infinite-dimensional? 

Here, A is a non-trivial abelian variety over K, and fertile = large = ample = for every geometrically integral smooth K-variety the set of K-rational points is either empty or Zariski-dense.
If K is algebraic over a finite field, the answer is NO, as already pointed out by Laurent Moret-Bailly. In all other cases, the answer is YES: 
For fields of infinite transcendence degree, this was proven in the paper On the rank of abelian varieties over ample fields with Sebastian Petersen using elementary methods.
For fields of characteristic zero, in the same paper we deduce the claim using the Mordell-Lang conjecture as proven by Faltings, Raynaud and possibly others.
For fields of positive characteristic and transcendence degree at least one, in Ranks of abelian varieties and the full Mordell-Lang conjecture in dimension one we deduce the claim from a recent result of Rössler by a similar approach.<|endoftext|>
TITLE: Inequivalent complete norms and the axiom of choice
QUESTION [10 upvotes]: Hi,
I've been wondering about the following :
Is it possible, without the axiom of choice, to have two inequivalent complete norms on a vector space?
All the examples of inequivalent complete norms I've seen rely on the existence of Hamel bases...
This is most likely well-known, but I'd be glad if someone could provide a good reference.
Thank you,
Malik

REPLY [11 votes]: No, it is not possible. There is a model due to Shelah of ZF+ dependent choice+every set of reals has the Baire property. There is a result of Garnir and Wright that implies that in such model, any two complete norms are equivalent. The Handbook of Analysis and Its Foundations by Eric Schechter has a chapter on this result and its consequences. The chapter is "The Dream Universe of Garnir and Wright". The result in that chapter is supposed to be slightly stronger than the one in the original papers, with which I'm not familiar.<|endoftext|>
TITLE: What is a good book on topological groups?
QUESTION [17 upvotes]: I am looking for a good book on Topological Groups. I have read Pontryagin myself, and I looked some other in the library but they all seem to go in length into some esoteric topics.
I would love something 250 pages or so long, with good exercises, accessible to a 1st PhD student with background in Algebra, i.e. with an introduction covering necessary background in Functional Analysis.
If possible, I would also like it covering particularly important (in my view) topics:

emphasize on locally compact groups, but both locally Euclidean and totally disconnected cases;
Pontryagin duality;
Kazhdan property T;
Tannaka reconstruction.

REPLY [3 votes]: I have just read T. Tao's 'Hilbert's Fifth Problem and Related Topics"
and I found it fantastic.<|endoftext|>
TITLE: A delicate elementary inequality
QUESTION [13 upvotes]: The following "piecewise-quadratic" inequality emerged in a joint work of Rom
Pinchasi and myself. The inequality is surprisingly delicate, and all our
attempts to simplify it made it false. By the end of the day, we were able to
prove the inequality, but the proof is unreasonably sophisticated, totalling
to about 15 pages. We would be happy to have a shorter proof.
The inequality involves the function $G$ of three real variables, defined as
follows: if $(\xi,\eta,\zeta)$ is a non-decreasing rearrangement of
$(x,y,z)$, then we let
  $$ G(x,y,z) := \begin{cases}
                   \xi\eta &\ \text{if}\ \zeta\ge \xi+\eta, \\
                   \xi\eta-\frac14\,(\xi+\eta-\zeta)^2 &\ \text{if}\ \zeta\le \xi+\eta.
                 \end{cases} $$
Thus, for instance, we have $G(9,6,7)=38$, whereas $G(7,14,6)=42$. Now
consider the function $f$ of four variables defined by
$$\begin{align*}
  f(x_0,x_1,y_0,y_1)
    &:= \min \{ 0.15s^2, x_0y_0+x_1y_1 \}  \\
    &\qquad + G(x_0,y_1,1-s) + G(x_1,y_0,1-s)  \\
    &\qquad + 0.25(1-s)^2,
\end{align*}$$
where for brevity I write $s=x_0+x_1+y_0+y_1$, and let
  $$ \Omega := \{ (x_0,x_1,y_0,y_1)\in{\mathbb R}_{\ge 0}^4\colon 1/2 \le s \le 1. \}. $$
All we want to show is that
  $$ \max_\Omega f \le 0.15. $$
(Indeed, the maximum is actually equal to $0.15$: say, we have
$f(0.5,0,0.5,0)=0.15$.)

At Pat Devlin's suggestion, here is the graph of the maximum as a function of the sum $s=x_0+x_1+y_0+y_1$. 

(source: haifa.ac.il) 
It looks nice, but does not seem to be a graph of some "simple" function; and so, there is probably no simple analytic expression for $\max f$ over all quadruples $(x_0,x_1,y_0,y_1)$ adding up to $s$.

REPLY [7 votes]: The paper where the inequality in question emerged is, finally, written (and
uploaded to the arXiv, in case anybody is interested). We were
eventually able to simplify the proof and squeeze it down to just about
six pages; indeed, the whole paper is now shorter than our original proof. I
sketch very briefly the new proof below.
We start with the identity
\begin{multline*}
  G((x+y)/2,(x+y)/2,z) = G(x,y,z) \\
     + \frac14(x-y)^2
                - \frac14 \big(\max\{|x-y|-z, 0 \} \big)^2. \tag{1}
\end{multline*}
Once stated, this is easy to verify by a careful case analysis. An immediate
consequence is that $G(x,y,z)$ can only grow if both $x$ and $y$ are replaced
with their average $(x+y)/2$.
As another preparation step, we exploit the internal symmetries of $f$ to
assume, without loss of generality, that
  $$ x_0+x_1 \ge y_0+y_1 \tag{2} $$
and also
  $$ x_0+y_0 \ge x_1+y_1. \tag{3} $$
Our big plan is to investigate how $f$ changes under the balancing operation
which includes replacing $x_0$ and $y_1$ with their average $(x_0+y_1)/2$
and, simultaneously, $x_1$ and $y_0$ with their average $(x_1+y_0)/2$. Using
the identity (1), we could show that either $f$ is non-decreasing under such
balancing, or, under the assumptions (2) and (3), we have
\begin{align*}
  x_0 &\ge y_1+(1-s), \tag{4} \newline
  y_0 &\ge x_1+(1-s), \tag{5}
\end{align*}
and
  $$ 3(x_0+y_0)+(x_1+y_1) \ge 2. \tag{6} $$
The precise meaning of being non-decreasing under balancing is that
  $$ f(x_0,x_1,y_0,y_1) \le f(z_0,z_1,z_1,z_0), $$
where $z_0=(x_0+y_1)/2$ and $z_1=(x_1+y_0)/2$. Consequently, in this case the
problem reduces to maximizing a function of just two variables, which is a
feasible task.
Now, if $f$ is decreasing under balancing, then, in view of (4) and (5) and
by the definition of the function $G$, we have
  $$ G(x_0,y_1,1-s) = y_1(1-s) \ \text{and}\ G(x_1,y_0,1-s)=x_1(1-s). $$
Hence,
\begin{multline*}
  f(x_0,x_1,y_0,y_1) = \min\{0.15s^2,x_0y_0+x_1y_1\} \newline
                                             + (x_1+y_1)(1-s) + 0.25(1-s)^2.
\end{multline*}
The expression in the right-hand side is can only increase if $x_0$ and $y_0$
are both replaced with their average, and, simultaneously, $x_1$ and $y_1$
are replaced with their average. Consequently, we can assume that $x_0=y_0$
and $x_1=y_1$. This, again, reduces the problem to maximizing a function of
two variables, which takes some two more pages to accomplish.<|endoftext|>
TITLE: How much do universes matter in topos theory?
QUESTION [17 upvotes]: Suppose we fix two Grothendieck universes $\mathcal{U} \in \mathcal{V}.$ Then one has that $\mathcal{U}$-$\mathbf{Set},$ the category of $\mathcal{U}$-small sets, is a locally $\mathcal{U}$-small, $\mathcal{V}$-small category. Grothendieck universes were used often by, well, Grothendieck, in his work with topoi. Part of the reason was to work with "large sites". One important example is that for a given topos $\mathcal{E},$ it carries its canonical topology. Grothendieck universes can be used to make sense of the statement that $\mathcal{E}$ is equivalent to sheaves over itself with respect to this topology, but lets look at another idea. If $\mathcal{E}=Sh\left(C,J\right)$ is a sheaf topos, we will instead of considering sheaves  with values in the large category of all sets, consider only sheaves with values in $\mathcal{U}$-$\mathbf{Set}.$ And this is a $\mathcal{U}$-topos. Lets take this idea and run with it:
There is a functor from $\mathcal{U}$-topoi to $\mathcal{V}$-topoi which sends a $\mathcal{U}$-topos $\mathcal{E}$ to $$Sh\left(\mathcal{E},\mathcal{V}\mbox{-}\mathbf{Set}\right),$$ the $\mathcal{V}$-topos of sheaves on $\mathcal{E}$ with its canonical topology, considered as a $\mathcal{V}$-small site.
This easily generalizes for $n$-topoi for any $n$. Are these functors full and faithful?
For $n=0$ the answer is yes, and remains yes for $n=1$ if we restrict to localic topoi, for topological reasons-  but is it true for higher $n$?
(I foresee a possible complication for $\infty$-topoi which are not equivalent to sheaves on a site.)

REPLY [24 votes]: The change-of-universe construction is faithful but not full. 
For example, let X be the topos of sets and let Y be the classifying topos for abelian groups.
The category of geometric morphisms from X to Y is equivalent to the category of abelian groups. If you pass to a larger universe, you get more abelian groups.<|endoftext|>
TITLE: Number of Normal subgroups In a p-Group
QUESTION [8 upvotes]: Dear all,
Does someone know of any paper/method that enables us counting/estimating the number of normal subgroups of some p-group of order $p ^n $ ($ n$ is some natural number ? ) .
Is there anyway we can count the maximal subgroups it has (i.e.- the groups of order $p^{n-1} $ ? ) ?
Thanks in advance

REPLY [4 votes]: The wikipedia article on p-groups reminded me that

Every normal subgroup of a finite
  p-group intersects the center
  nontrivially.

This implies immediately that minimal normal subgroups of a p-group $G$ will be central. This fact can be used to prove the statement that Wei Zhou made:

A $p$-group of maximal class and size $p^n$ has the least number of normal subgroups of all groups of order $p^n$.

(If I'm thinking straight this number is $n+1$ and the bound is also achieved by the cyclic group of order $p^n$.)
It seems to me that one might be able to prove something a little stronger using an inductive argument: counting the minimal normal subgroups in the center $Z$, and then counting the normal subgroups in $G/Z$, and then putting these two numbers together... It's that last bit that's going to be tricky though. If the center is cyclic, then everything is fine but when it's not cyclic, eek...<|endoftext|>
TITLE: Is $x \, \tan(x)$ integrable in elementary functions?
QUESTION [18 upvotes]: I'm teaching Calculus and my students asked me to calculate the integral of $x \, \tan(x)$.
I spent quite a lot of effort to do this, but I'm now even not sure if the integral could be presented in elementary functions.
Does anybody know how to calculate it, or otherwise prove it is not integrable in elementary functions?

REPLY [3 votes]: $\int x \tan x \,dx=\frac{1}{2} i \text{Li}_2\left(-e^{2 i x}\right)+\frac{i x^2}{2}-x \log\left(1+e^{2 i x}\right)+C $
Where $\text{Li}_2$ is the dilogarithm function<|endoftext|>
TITLE: Estimates for Bezout coefficients
QUESTION [10 upvotes]: The answer to my question is probably well-known, but I was unable to find a reference. 
The Bezout's identity states that for any positive non-zero integers $a_1, \ldots , a_n$ there exist integers $x_1, \ldots , x_n$ such that 
$$
a_1x_1+ \cdots + a_nx_n=gcd(a_1, \ldots, a_n) .
$$
What is the best estimate for $|x_1|+\cdots + |x_n|$ in terms of $|a_1|+\cdots +|a_n|$? 
More precisely, we define 
$$
b(a_1, \ldots , a_n) =\min\limits_{a_1x_1+ \cdots + a_nx_n=gcd(a_1, \ldots, a_n)} (|x_1|+\cdots + |x_n|) 
$$
and 
$$
f(k)= \max\limits_{|a_1|+\cdots +|a_n|\le k} b(a_1, \ldots, a_n) .
$$
What is known about the growth of $f(k)$?
Here is a very particular question. It is not hard to show that $f(k)=O(k^2)$. Is there any better estimate? Does $f(k)=O(k\log k)$ hold?

REPLY [9 votes]: We can prove $b(a_1,\ldots,a_n) \leq a_1+\cdots+a_n$ (and thus $f(k) \leq k$) by elementary means as follows.
We may assume $\operatorname{gcd}(a_1,\ldots,a_n)=1$ and $1 < a_1< \cdots <a_n$.
Start with any Bézout identity $x_1 a_1+ \cdots + x_n a_n = 1$. Using the transformations $x_n \leftarrow x_n + ka_1, x_1 \leftarrow x_1-ka_n$, we may ensure $|x_n| \leq a_1$.
Similarly, we ensure $|x_{n-1}| \leq a_1$, but we choose the sign of $x_{n-1}$ so that $x_{n-1} a_{n-1}$ and $x_n a_n$ have opposite signs. In this way $|x_{n-1}a_{n-1}+x_n a_n| \leq \max(|x_{n-1} a_{n-1}|,|x_n a_n|) \leq a_1 a_n$.
Repeating the process, for each $n-2 \geq k \geq 2$, we change $x_k$ so that $|x_k| \leq a_1$ and $x_k a_k$ and $\sum_{j=k+1}^n x_j a_j$ have opposite signs. Thus $|\sum_{j=k}^n x_j a_j| \leq \max(|x_k a_k|,|\sum_{j=k+1}^n x_j a_j|) \leq a_1 a_n$.
At the end we have $|x_1 a_1| \leq 1+ |\sum_{j=2}^n x_j a_j| \leq 1+a_1 a_n<a_1(1+a_n)$ so $|x_1| \leq a_n$ and finally $|x_1|+\cdots+|x_n| \leq a_n+(n-1)a_1 \leq a_1+\cdots+a_n$.
I suspect that there are better bounds when $n$ is big.<|endoftext|>
TITLE: $K3$ surfaces admitting finite non-symplectic group actions are projective
QUESTION [7 upvotes]: I have read somewhere that "$K3$ surfaces admitting finite non-symplectic group actions are projective". Could someone remind me of a proof?

REPLY [9 votes]: A sketch of the proof goes as follows.
Let $G$ be a finite group of non-symplectic automorphisms on a $K3$ surface $X$. Since $G$ is non-symplectic, there exists $g \in G$ such that $g \omega \neq \omega$, where $\omega$ is the holomorphic $2$-form on $X$.
Then, setting $Y:=X/G$, one has $q(Y)=p_g(Y)=0$, since $q(X)=0$ and by the previous remark the holomorphic $2$-form $\omega$ does not descend to the quotient.
From this, one proves that either $Y$ is rational (i.e, bimeromorphic to $\mathbb{P}^2$) or the minimal desingularization of $Y$ is an Enriques surface. In any case, $Y$ is a compact algebraic surface, so there exists an ample divisor $H$ on $Y$.
Finally, the quotient $\pi \colon X \longrightarrow Y=X/G$ is a finite holomorphic map since $G$ is a finite group. It follows that $\pi^*H$ is an ample divisor on $X$, hence $X$ is projective.
Remark 1. As pointed out by Rita in her comment, one must be a bit careful since $Y$ can be singular. I skip the details, which can be found in Frantzen's dissertation.
Remark 2. One also proves that $G$ is a symplectic grup of automorphism if and only if the minimal desingularization of $X/G$ is again a $K3$ surface. The reason is that in this case $g \omega=\omega$ for all $g \in G$, so $\omega$ descends to the quotient.<|endoftext|>
TITLE: question about higher geometric stacks
QUESTION [6 upvotes]: I have a naive question I am asking. Given a higher geometric stack X in the sense of Simpson, Toen etc is it true that there is an affinization Spec Gamma(O_X) such that Hom(X, Spec(A))= Hom(A,Gamma(O_X)) for every affine scheme Spec(A)? Or does this require some more hypotheses. I have very much a hard time finding this out.

REPLY [9 votes]: Yes -- affinization is defined (as you wrote) as the left adjoint to the inclusion  of affine schemes into higher stacks.
This left adjoint exists by the ($\infty$-categorical) adjoint functor theorem, since the inclusion of affines into higher stacks preserves all limits (though it certainly changes colimits). Some references for this or closely related notions: Toen's Affine Stacks (here) and Lurie's DAG VIII (available here), where the relevant notion is called "coaffine stacks". There's also a less professional and more informal discussion (in the derived context) in Section 3.2 here.<|endoftext|>
TITLE: Fake versus Exotic
QUESTION [32 upvotes]: Without recourse to the Disc Theorem (or its progeny), is it true that all known examples of exotic differentiable structures on 4-manifolds would be fake rather than exotic?
Terminology (perhaps non-standard):
(1) By "Disc Theorem" I mean the statement, contained in [1], that kinky handles are not only homotopic to standard 2-handles (as proved by Casson [2]), but homeomorphic to standard 2-handles,
(2) The "progeny" is all extensions of the Disc Theorem to non-simply connected settings (including gropes, capped gropes, etc), 
(3) A smooth 4-manifold is "exotic" if it is homeomorphically equivalent to, but smoothly inequivalent to a standard 4-manifold,
(4) A smooth manifold is "fake" if it is homotopically equivalent to, but smoothly inequivalent to a standard 4-manifold. 
The motivation for this question is the same as that of the earlier Question, namely: 

The classification [2] of topological 4-manifolds, which follows from the Disc Theorem, is now 30 years old and an easier version of the proof has not emerged. In contrast, Donaldson's invariants have been followed by more easily computed invariants. This asymmetry is an unsatisfactory state of affairs for such a far-reaching topological result, particularly as it is so regularly used in proof-by-contradiction arguments against results in smooth 4-manifold theory.
As the Bing topologists familiar with these arguments retire, the hopes of reproducing the details of the proof are fading, and with it, the insight that such a spectacular proof affords.
It may be possible to refine the proof under the assumption of more regularity to gain more control over the resulting infinite towers - and perhaps get Hoelder maps rather than homeomorphisms, for example.

References: 
[1] Freedman, M. H. (1982), "The topology of four-dimensional manifolds", Journal of Differential Geometry 17 (3), 357-453.
[2] Casson, A. J. (1986), "Three lectures on new-infinite constructions in 4-dimensional manifolds", A la recherche de la topologie perdue, Progr. Math., 62, Boston, MA: Birkhauser Boston, 201-244.

REPLY [7 votes]: The answer appears to be "Yes" - without recourse to the Disc Theorem, all known examples of exotic differentiable structures on 4-manifolds would be fake rather than exotic. 
Casson's result is that his infinite kinky handle construction produces objects that are homotopic to standard 2-handles and so, a priori, leads only to fake 4-manifolds. Since fakeness is not such an unusual phenomenon (homotopies are weak creatures), one could interpret the result as something of a failure - perhaps this explains why the result languished unpublished for almost a decade.
On the other hand, the existence of exotic 4-manifolds, with all of their strangeness, follows solely from the central claim of the disc theorem: that Casson handles are homeomorphic to standard 2-handles. This introduces extraordinary dichotomies in 4-dimensions, as it contradicts many different subsequent smooth results. For example, it leads to:

Small exotic structures: disc theorem versus smooth h-cobordism theorem 
Large exotic structures: disc theorem versus smooth connected-sum-splitting  
Topologically slice knots: disc theorem versus smooth knot theory

All of this is deeply unsettling. 
As an attempt at an alternative geometrization of these paradigms, I have just posted a preprint to the arxiv (joint work with Wilhelm klingenberg) utilizing a neutral Kaehler structure on 4-manifolds. While the main result is the extension of a differential geometry result of 1846, this is essentially a byproduct of our consideration of the disc theorem.  The topological motivation is more fully explained in the video:
A global version of a classical result of Joachimsthal and the slice problem for knots
In the near future I will post further details of an alternative construction of Casson handles with more geometric control of the boundary in order to explore these questions directly.<|endoftext|>
TITLE: Relations in a particular subgroup of the braid group.
QUESTION [6 upvotes]: I think this should be a 10 minute exercise in a decent computer algebra package - unfortunately I'm hopelessly ignorant of such things, so I'm putting it up here in the hope that someone will be kind enough to do it for me...
Here's the question: partition $n$ into two pieces, $n= p+q$, and let $S_p\times S_q \subset S_n$ be the associated Young subgroup. Now consider the braid group $B_n$. I'm interested in the subgroup of $B_n$ consisting of braids that preserve this partition of their endpoints, i.e.
$$ B_{p,q} := B_n \times_{S_n} (S_p\times S_q )$$
I can write down generators for $B_{p,q}$, namely $s_1,.., s_{p-1}, s_p^2, s_{p+1}, ..., s_{n-1}$ where the $s_i$ are the standard generators of $B_n$. My question is what are the relations? 
Obviously I need the usual braid relations on each piece of the partition, but are there any others? I'd be happy to extrapolate heuristically from low values of $p$ and $q$.

REPLY [7 votes]: You probably already noticed that, but $B_{p,q}$ is the fundamental group of
$$
X_n/(S_p \times S_q)
$$
where $X_n$ is the configuration space of $n$ points in the complex plane. Ths may help to guess some facts about these groups.
So far I know these group are usually called "mixed braid groups" in the litterature, though this name is sometimes used for the group of braids with the $p$ first strands fixed. Anyway, a presentation of it is obtained in S. Manfredini Some subgroups of Artin's braid group (available here). Results on their relation with the representation theory of $B_n$, as well as references you may find interesting, can be found in Bellingeri, Godelle, Guaschi, Exact sequences, lower central series and representations of surface braid groups (arXiv).<|endoftext|>
TITLE: De Rham representatives of the cohomology classes in $H^*(SU(3))$
QUESTION [9 upvotes]: I would like to know differential forms representing the cohomology classes of $SU(3)$. I know that there exist a unique bi-invariant form in each class, but I'm not highly motivated by simply putting $R_g^*\omega = \omega = L_g^*\omega$ and solving it by hand in coordinates. Is there a way to do it less brutally ? On a side note, what is the topology of $SU(3)$, or where can I find an answer ? I have looked in Strickland's Bestiary of Topological Objects, but didn't find the section on unitary groups very enlightening.

REPLY [11 votes]: Relying on the nice answer by Liviu Nicolaescu on the particular question, let me comment on a little more general setup. My point is that "many" things (cohomology classes, invariant forms/vector fields, invariant polynoms, etc...) can be written very explicitly and down-to-earth for classical groups. Such technique is standard in math.physics, but to the best of my knowledge is not presented systematically in basic math. textbooks on Lie groups, where it should be (imho). 
Claim "Everything" (cohomology classes, invariant forms/vector fields, invariant polynoms, etc...) for A,B,C,D groups can be written as traces, or determinants or other linear algebra expressions applied to some explicitly given matrices ("Lax matrices").
This is sometimes called "matrix" (or "Leningrad") notations which were developed by L.D.Faddev's school and led to discovery of quantum groups.
"Examples first" 
Consider the group GL(2) embed it into Mat(2) so we have natural coordinates on this linear space:  "x,y,z,u". Consider their de-Rham differential, i.e. 1-forms: "dx,dy,dz,du".
And define the following matrices:
$$
M=\begin{bmatrix}
 x &   y \\\\
 z &   u 
\end{bmatrix}, ~~~dM= 
\begin{bmatrix}
 dx &   dy \\\\
 dz &   du 
\end{bmatrix}
$$
Pay attention on some psychologically important detail: the elements of these matrices are functions  and 1-forms, but NOT complex numbers. 
Lemma 1 (Invariant 1-forms): take any 2x2 number valued matrix "K", then 
$Tr(KM^{-1}dM)$ - left invariant 1-form on the group GL(2);
$Tr(KdM M^{-1})$ - right invariant 1-form on the group GL(2).
Changing matrix "K" we obtain the whole space of left/right invariant 1-forms.
The map K->1-forms correspond to identification of dual to Lie algebra with invariant forms. 
Proof: the idea is very simple action of group is M-> GM, so when you have (GM)^{-1}dGM=,
(M)^{-1}(G)^{-1}GdM=(M)^{-1}dM.
I hope you would agree that these are very explicit coordinate-written
expressions which "everyone can understand".
Lemma 2 (Cohomology classes) The group GL(2), is contractible to U(2), not SU(2) and has basic generators of cohomology classes of degrees 1,3. 
The generators of cohomology classes can be written explicitly:
$ Tr (M^{-1} dM)^k$, k =1,3 .
One can check that these are closed  and invariant forms, which by some general reasons is enough to ensure that they are generators of the cohomology.. 
Again, I hope you would agree that these are very explicit coordinate-written
expressions which "everyone can understand".
Notation Let us give similar description for invariant vector fields and get in contact with beautiful story of Capelli identity.
Let introduce vector fields $\partial_{x},\partial_{y},  \partial_{z},\partial_{u}$
and form a matrix:
$$
\partial_M=\begin{bmatrix}
 \partial_x &   \partial_y \\\\
 \partial_z &   \partial_u 
\end{bmatrix}
$$
Lemma 3 (Invariant vector fields) take any 2x2 number valued matrix "K", then 
$Tr(KM \partial_M^t )$ - right invariant vector field on the group GL(2).
$Tr(K\partial_M^t M)$ - left invariant field on the group GL(2);
Capelli's story, Howe duality, Langlands correspondence, Talalaev's formula The 2x2  matrix $ E = M (\partial_M)^t$ is a matrix with non-commutative elements. Nevertheless for the purpose of some Lie algebra theory one wants to calculate its determinant. Actually this can be done:
Capelli identity:
$ det^{column} (E +diag(1,0) ) = det(M) det(\partial_M)  $
One of  the points of Capelli's story is that this determinant is generator of the center of U(gl(2)), another point related to Howe's duality is that we can express it  via left or right invariant vector fields and will get essentially the same result (up to details). It is related to the fact that GL(n) acting from the left on Mat_n is Howe's dual to the  GL(n) acting from the right - and hence images of the centers for the two action must coincide.
There are some modern developments around this. Let us  introduce auxiliary variable "z"
and define a new matrix: 
$$ T= (Id \partial_z -E/z)$$
the point is that  it will be Manin matrix - class of matrices which are "slighly non-commutative", but behave like commutative ones. And toy model for Talalaev's theorem states
$$det(T)$$ will generate the center of U(gl(2)). "Non-toy" is similar result for loop algebra for gl(n). Moreover the differential operator in "z" is related to the so-called GL(n)-oper which is Langlands dual to the corresponding irrep of loop algebra. 
Sorry for self-advertisement: one may look at http://arxiv.org/abs/0711.2236 "Manin matrices and Talalaev's formula" for further info.<|endoftext|>
TITLE: approximate uncertainty principle for finite abelian groups
QUESTION [8 upvotes]: Edit: we cannot find such an example. It would imply a negative solution to the KS${}_2$ conjecture which has now been proven by Marcus, Spielman, and Srivastava in this paper.
In fact, their solution implies that there always exists $S$ such that any $f$ supported on $S$ (or $S^c$) satisfies $\|\frac{1}{2}\hat{f} - \hat{f}|_T\|_2 \leq O(\sqrt{\epsilon})\cdot\|\hat{f}\|_2$.

Can we find, for every $\epsilon > 0$, an example of the following?
$\bullet$ a finite abelian group $G$
$\bullet$ a small subset $T$ of the dual group $\hat{G}$ (meaning $|T| \leq \epsilon |\hat{G}|$)
such that for any subset $S$ of $G$, there is a nonzero complex valued function $f$ supported either on $S$ or $S^c$ whose Fourier transform is approximately supported on $T$ (meaning $\|\hat{f} - \hat{f}|_T\|_2 \leq \epsilon \|\hat{f}\|_2$).
It seems unlikely, but I don't know any version of the uncertainty principle that would forbid this.

Two comments: (1) This arose out of discussions I had with Chuck Akemann about the Kadison-Singer problem. It wouldn't, as it stands, imply a negative solution to Kadison-Singer, but it would be close and (I believe) could probably be easily converted into a full negative solution. (2) Since either $S$ or $S^c$ has cardinality at least $|G|/2$, one could consider assuming $|S| \geq |G|/2$ and demanding that $f$ be supported on $S$. But I know that this makes the problem too hard. (Given $T$, I can always find an $S$ with $|S^c|/|G| = O(\sqrt{\epsilon})$ such that no nonzero function supported on $S$ has Fourier transform approximately supported on $T$.)

REPLY [7 votes]: The restricted isometry property, or RIP (formerly known as the uniform uncertainty principle, or UUP, as per your suspicion that uncertainty principles should be relevant) for random Fourier measurements prohibits $\varepsilon$ from being smaller than about $\log^{-4} |G|$; this property was first proven (for $G$ a cyclic group, and with exponent 6 instead of 4) by Candes and myself, and then (for arbitrary abelian $G$, and with exponent 4) by Rudelson and Vershynin.  If one takes $S$ to be a random subset of $G$ of density $1/2$, then for any sufficiently sparse $f$ (of sparsity less than $c |G|/\log^4 |G|$ for some small $c$), the Fourier energy of $\hat f$ will be split more or less equally between $S$ and its complement thanks to the RIP, and so the situation described in your post will not occur.
If one takes gaussian measurements instead of Fourier ones, one only needs to oversample by a constant factor (see Lemma 4.1 of the previously mentioned paper of Candes and myself and the remark at the end of the proof), so it is conceivable that one can obtain an analogous result in the Fourier setting and give a negative answer to your question for some sufficiently small $\varepsilon$ independent of the size of the group.  But unfortunately this is probably outside of reach of the technology described in the above papers.  (But there has been a number of advances in that area since then, which I have not followed as closely.  For instance, there has been a slight improvement to the Rudelson-Vershynin bound obtained recently by Cheraghchi, Guruswami, and Velingker, although for the problem at hand, it does not appear to lower the exponent $4$ further.)
ADDED LATER: Actually, on thinking about it a bit more, the RIP is probably too strong a property for this purpose (it creates a set S that becomes a counterexample for all sparse T, whereas for your problem it would suffice to find a different counterexample S for each sparse T).  The situation is then a bit closer to my original paper with Candes and Romberg which only needed an oversampling of order $\log |G|$ or so, though we didn't phrase our analysis in an easily portable form and one would have to look through the argument in detail to check if the bounds there indeed give a negative answer to your question for $\varepsilon < c/\log |G|$.  I still don't know how to get rid of the final logarithm, though.<|endoftext|>
TITLE: Open problems about CMC hypersurfaces with symmetries?
QUESTION [7 upvotes]: Recently, Andrews and Li announced a complete classification of CMC ($H=const.$) tori in $S^3$, confirming a conjecture of Pinkall and Sterling. Their main result is that any such torus is rotationally symmetric, and this reduces the problem to a previous classification due to Perdomo, also studied by Hynd, Park and McCuan. [As a side note, some of the techniques used are similar to Brendle's solution of the Lawson Conjecture, that states that the only minimal ($H=0$) torus in $S^3$, up to congruences, is the Clifford torus $S^1(\tfrac{1}{\sqrt2})\times S^1(\tfrac{1}{\sqrt2})$.] It is also known that there are CMC tori in $S^3$ that are not congruent to the family $S^1(r)\times S^1(\sqrt{1-r^2})$ of Clifford tori, so, in a certain sense, their only (continuous) symmetries are encoded in an isometric circle action, as confirmed by Andrews' and Li's result.
Given this context, my question is:

Are there  open questions, conjectures, etc., regarding existence or non-existence of CMC hypersurfaces with symmetries in highly symmetric compact manifolds, analogously to the above case of CMC tori in $S^3$? Actually, I'd be particularly interested in questions regarding non-minimal CMC embeddings ($H=const\neq 0$).

To be (slightly) more precise, I am interested in knowing if there are questions of the form "Let $(M,g)$ be a compact manifold with many symmetries (i.e., a "big" Lie group acts isometrically), and consider embeddings of some manifold $N$ into $M$. Then if $N\subset M$ is CMC, it must have certain symmetries (and perhaps cannot have other symmetries)." In other words, are there any analogues of (or questions similar to) the conjecture of Pinkall and Sterling that CMC tori in $S^3$ are rotationally symmetric, but for other (perhaps higher dimensional) compact manifolds? Perhaps some of these questions are disguised as questions about Mean Curvature Flow?
Notice also that if $N\subset M$ is an orbit of an isometric group action, then it is automatically CMC. However, there could be other CMC embeddings of $N$ into $M$, not congruent to any orbits (e.g., with less symmetries, as the tori above). Should one expect any general behavior for this other CMC embeddings? Are there any known results, questions or conjectures in this direction?

I apologize for the somewhat vague question, but none of the above references seems to risk any further conjectures or indicates natural extensions of their results. Also, a quick google search and inverse citations on MathSciNet don't give much in terms of open questions. I was wondering if there is a reason behind it... Any references to recent survey-type papers related would also be very appreciated!

REPLY [6 votes]: From my understanding there are two main directions of research related to your question.
1) There is a lot of interest in seeing how much of the classical theory of CMC and minimal surfaces that is known to hold in the three-dimensional space forms continues to hold in the other Thurston geometries (or more generally -- three dimensional metric lie groups).  W. H. Meeks and his collaborators have been pretty active recently in this area.  See for instance this survey article by Meeks and Perez -- which contains a section on open problems.  
2) The other main area -- is to try and understand the situation in geometries arising mathematical relativity.  The starting point here is an observation of Bray (I believe it is in his thesis) that the solution to the  (appropriately formulated)  isoperimetric problem in the Schwarzschild metric are rotationally invariant spheres.  Huisken has recently connected this to important questions in mathematically general relativity and so has attracted a lot of attention recently.  I refer you to this preprint of Brendle and also this preprint of Brendle and Eichmair.<|endoftext|>
TITLE: Maximum of a function of one variable
QUESTION [16 upvotes]: Let $D$ be a circular quadrilateral (that is a Jordan region whose boundary consists of 4 arcs
of circles all orthogonal to the unit circle) whose interior angles are all equal to 0, the vertices lie on the unit circle,
and $D$ is inside the unit disc.
Suppose also that $D$ is symmetric with respect to the real and imaginary axes, and has
one vertex $z_1=\exp(i\theta)$ where $\theta\in (0,\pi/2)$. 
Then this number $\theta$ determines such a $D$ completely.
Let $f$ be the inverse of the Riemann map, so that $f$ is the conformal map from the unit disc onto $D$, $f(0)=0$ and $f'(0)>0$.
Is it true that maximum $f'(0)$ is achieved when $\theta=\pi/4$ ?
There is a strong computer evidence for this, as well as the general considerations
(where else can the maximum be?).

REPLY [2 votes]: Alexandr Solynin told me that he solved this problem (even the more general one, for
hyperbolic n-gons with all zero angles) in 1993.
A. Solynin, Some extremal problems for circular polygons, (Russian)
Zapiski Nauchnyh Seminarov POMI 206, 1993, 127-136.
English translation is in Journal of Math Sci. 80 4, 1996, 1956-1961.<|endoftext|>
TITLE: Multivariate Hensel's Lemma, but with only one polynomial
QUESTION [5 upvotes]: One version of Hensel's Lemma is the following statement: 
Let $R$ be a commutative ring with a unit. Given a polynomial $Q\in R[X]$ and a root $\alpha$ of $Q$ modulo some ideal $I$ (i.e. $Q(\alpha) \in I$), assuming some non-degeneracy conditions (e.g. $Q$ is square-free), then for every $t > 1$, there exists $\beta_t \in R$ such that $\beta_t = \alpha \mod I$, and $Q(\beta_t) \in I^t$, and furthermore, $\beta_t$ is unique. 
The multidimensional generalization of Hensel's Lemma is often presented as:
Given $f_1,\ldots,f_n$ in $R[X_1,\ldots,X_n]$ and a simultaneous root $\alpha \in R^n$ modulo an ideal $I \subset R$ (i.e. $f_i(\alpha) \in I$ for all $i$), assuming some non-degeneracy conditions (e.g. $\det J(\alpha)$ is a unit), there exists $\beta_t \in R^n$ such that $\beta_{t,j} = \alpha_j \mod I$ for all $j$, and $f_i(\beta_t) \in I^t$ for all $i$.
Here, $J(\alpha)$ denotes the evaluation of the Jacobian of $f_1,\ldots,f_n$ on $\alpha$. 
My question is: is there an intermediate generalization in between the univariate case and the multivariate case above where we only consider one polynomial $Q\in R[X_1,\ldots,X_n]$, and we simply want to lift roots of $Q$ modulo an ideal $I$ to roots of $Q$ modulo $I^t$? It seems intuitively like an easier thing to do (we don't require simultaneous solutions to a system of polynomial equations). Does this intermediate generalization exist, and if so, what non-degeneracy conditions would we require?
Thank you!

REPLY [6 votes]: Yes. If the ideal generated by $(I,dQ/dX_1,..,dQ/dX_n)$ is the unit ideal at some mod-$I$ solution of $Q$, one can lift to a mod $I^t$ solution of $Q$.
Proof: It is clear that we merely need to check the induction step, that if we have a mod $I^t$ solution we can get a mod $I^{t+1}$ solution. Suppose $Q(X_1,..,X_n)$ is in $I^t$. Then, for $a_1,...,a_n \in I^t$
$Q(X_1+a_1,...,X_n+a_n)=Q(X_1,...,X_n)+a_1 \frac{dQ}{dX_1}  + .... + a_n \frac{dQ}{dX_n} + \epsilon$
where $\epsilon \in I^{t+1}$.
Choose $a_1,\dots, a_n$ such that
$a_1 \frac{dQ}{dX_1}+ \dots + a_n \frac{d Q}{d X_n}=- Q(X_1,...,X_n)$ mod $I^{t+1}$.
This is always possible since we can choose the sum to equal any element in:
$I^t \frac{dQ}{dx_1} + \dots + I^t \frac{dQ}{dx_n} + I^{t+1} = I^t\left(\frac{dQ}{dx_1} + \dots + \frac{dQ}{dx_n} + I\right)=I^t (1)=I^t$.
Then $Q(X_1+a_1,\dots, X_n+a_n) \in I^{t+1}$.<|endoftext|>
TITLE: If each strict subgroup of G is free, must G be free or cyclic of prime order ?
QUESTION [13 upvotes]: If each strict subgroup of a group G is free, must G be free or cyclic of prime order ?

REPLY [26 votes]: No. There is a variation of Tarski monster: a nonabelian group whose each proper nontrivial subgroup is infinite cyclic, see the book of Olshanskii.
Concerning Misha's comment. For any countable family of countable involution-free groups $G_1,G_2,\dots$, there is a group $H$ containing all $G_i$ as proper subgroups such that each proper subgroup of $H$ is either infinite cyclic or a conjugate of a subgroup of some $G_i$. This is Obraztsov's embedding theorem.<|endoftext|>
TITLE: on a Deformation long exact sequence of moduli space of stable maps
QUESTION [7 upvotes]: I am reading the book "mirror symmetry" by Hori,Katz,Klemm,etc. And I want to understand the following Deformation long exact sequence
\begin{align}
 0 & \to Aut(Σ, p_1, . . . , p_n, f)\to Aut(Σ, p_1, . . . , p_n) &\newline \to Def(f) &→ Def(Σ, p_1, . . . , p_n, f) → Def(Σ, p_1, . . . , p_n)  &\newline \to 
Ob(f) &\to Ob(Σ, p_1, . . . , p_n, f) \to 0
\end{align}
it connects three deformation theory:
1. deformation of stable curves
2. deformation of maps(with fixed source)
3. deformation of stable maps(with possible changing source curves)
And my understanding goes as follows:
Let $\mathscr{X}=M_{g,n}$ be the moduli stack of algebraic curves(genus $g$, n-marked point), and let $\mathscr{Y}=M_{g,n}(X,\beta)$ be the moduli stack of stable maps. Then there is a natural "forgetful" morphism:
$\pi : \mathscr{Y} \to \mathscr{X}$
by forgeting the "map".  
We have a distinguished triangle of cotangent complexes in the derived category $D^{-} (\mathscr O_{\mathcal{Y}})$: 
\begin{equation}
 \pi^* L_{\mathscr{X}}\to L_{\mathscr Y}\to  L_{\mathscr{Y}/\mathscr{X}}\to \cdot
\end{equation}
Now apply $R\mathscr{Hom}$, we have a long exact sequence:
\begin{align}
\mathscr Ext ^0 (L_{\mathscr Y/\mathscr X },\mathcal O_{\mathscr Y }) & 
\to \mathscr Ext^0 (L_{\mathscr Y}, \mathcal O_{\mathscr Y} ) \to \mathscr Ext^0 (\pi^* L_{\mathscr X},\mathcal O_{\mathscr Y} )& \newline
\to \mathscr Ext ^1 (L_{\mathscr Y/\mathscr X },\mathcal O_{\mathscr Y }) & \to \mathscr Ext^1 (L_{\mathscr Y}, \mathcal O_{\mathscr Y} ) \to \mathscr Ext^1 (\pi^* L_{\mathscr X},\mathcal O_{\mathscr Y} )& \newline
\to \mathscr Ext ^2 (L_{\mathscr Y/\mathscr X },\mathcal O_{\mathscr Y }) & \to \mathscr Ext^2 (L_{\mathscr Y}, \mathcal O_{\mathscr Y} ) \to \mathscr Ext^2 (\pi^* L_{\mathscr X},\mathcal O_{\mathscr Y} )
\end{align}
My questions:
 (1). is it an exact sequence of sheaves on $\mathscr Y$ with the first long exact sequence as its stalks?
 (2). If (1) is true, then how to see $\mathscr Ext^i (\pi^* L_{\mathscr X},\mathcal O_{\mathscr Y} )$ (i=0,1,2) corresponds to Aut,Def,Ob of curves? And why the two ends of the exact seqence vanishes?

REPLY [2 votes]: I don't believe that this is correct. The easiest way to see this is to look at your second question: The automorphisms/deformations/obstructions of a curve come from $H^i(C, T_C)$, i.e. they are the sheaves
$R^i p_*\omega_{U/\overline{\mathcal{M}_{g,n}}}^\vee$
where $p : U \to \overline{\mathcal{M}_{g,n}}$ is the universal family, and $\omega_{U/\overline{\mathcal{M}_{g,n}}}$ the relative dualizing sheaf. But these do not depend on $\overline{\mathcal{M}_{g,n}}(X, \beta)$ !
In the end, I think the issue is that you have the wrong exact sequence. What you want (to produce the relative obstruction theory) is the complex
$R^i p_*f^*T_X$
where the maps $p, f$ arise in the universal diagram 
$\overline{\mathcal{M}_{g,n}}(X, \beta) \longleftarrow_p U \longrightarrow_f X$
It is not obvious to me that your sheaves should be the same as these ones.<|endoftext|>
TITLE: A question on invariant theory of $GL_n(\mathbb{C})$.
QUESTION [12 upvotes]: Let $\rho$ denote the irreducible algebraic representation of $GL_n(\mathbb{C})$ with the highest weight $(2,2,\underset{n-2}{\underbrace{0,\dots,0}})$. 
Let $k\leq n/2$ be a non-negative integer. How to decompose into irreducible representations the representation $Sym^k(\rho)$? 
More specifically, I am interested whether $Sym^k(\rho)$ contains the representation with the highest weight $(\underset{2k}{\underbrace{2,\dots,2}},\underset{n-2k}{\underbrace{0,\dots,0}})$, and if yes, whether the mutiplicity is equal to one.
A a side remark, the representation $\rho$ has a geometric interpretation important for me: it is the space of curvature tensors, namely the curvature tensor of any Riemannian metric on $\mathbb{R}^n$ lies in $\rho$.

REPLY [15 votes]: The plethysm $\mathrm{Sym}^k \rho$ contains the irreducible representation with highest weight $(2,\ldots,2,0,\ldots,0)$ exactly once. It looks like a tricky problem to say much about its other irreducible constituents.
Let $\Delta^\lambda$ denote the Schur functor corresponding to the partition $\lambda$, and let $E$ be an $n$-dimensional complex vector space. Using symmetric polynomials (or other methods) one finds
$$\mathrm{Sym}^2 (\mathrm{Sym}^2 E) = \Delta^{(2,2)}E \oplus \mathrm{Sym}^4 E.$$ 
Therefore
$$ \mathrm{Sym}^k \mathrm{Sym}^2 \mathrm{Sym}^2 E \cong \sum_{r=0}^k \mathrm{Sym}^r (\Delta^{(2,2)}E) \otimes \mathrm{Sym}^{k-r} (\mathrm{Sym}^4 E) .$$
The irreducible representations contained in the $r$th summand are labelled by partitions with at most $2r+(k-r) = k+r$ parts. So to show that $\mathrm{Sym}^k(\Delta^{(2,2)}(E))$ contains $\Delta^{(2^{2k})}E$, it suffices to show that $\Delta^{(2^{2k})}E$ appears in
$\mathrm{Sym}^k \mathrm{Sym}^2 \mathrm{Sym}^2 E$.
Let $U = \mathrm{Sym}^2 E$. There is a canonical surjection
$$ \mathrm{Sym}^k (\mathrm{Sym}^2 U ) \rightarrow \mathrm{Sym}^{2k} U. $$
given by mapping $(u_1u_1')\ldots (u_ku_k') \in \mathrm{Sym}^k (\mathrm{Sym}^2 U )$ to $u_1u_1'\ldots u_ku_k' \in \mathrm{Sym}^{2k} U$. Therefore $\mathrm{Sym}^k (\mathrm{Sym}^2 U )$ contains
$ \mathrm{Sym}^{2k} U = \mathrm{Sym}^{2k} (\mathrm{Sym}^2 E)$. It is well known that
$$ \mathrm{Sym}^{2k} (\mathrm{Sym}^2 E) = \sum_{\lambda} \Delta^{2\lambda}(E) $$
where the sum is over all partitions $\lambda$ of $2k$ and $2(\lambda_1,\ldots,\lambda_m) = (2\lambda_1,\ldots, 2\lambda_m)$. Taking $\lambda = (1^{2k})$ we see that $\Delta^{(2^{2k})}E$ appears.
It remains to show that the multiplicity of $\Delta^{(2^{2k})}E$ in $\mathrm{Sym}^k (\Delta^{(2,2)}E)$ is $1$. We work over $\mathbb{C}$, so there is a chain of inclusions
$$ \mathrm{Sym}^k (\Delta^{(2,2)}(E)) \subseteq \mathrm{Sym}^k (\mathrm{Sym}^2 E \otimes \mathrm{Sym}^2 E) \subseteq (\mathrm{Sym}^2 E)^{\otimes 2k}.$$ 
By the Littlewood–Richardson rule (or the easier Young's rule), the multiplicity of $\Delta^{(2^k)}E$ in the right-hand side is $1$.<|endoftext|>
TITLE: Getting a bound on the coefficients of the factor polynomial
QUESTION [8 upvotes]: Suppose $f(x):=a_0+a_1x+\cdots+a_nx^n$ is a polynomial in $\mathbb{Z}[x]$ and $|a_i|\leq M$ for each $i=0,\ldots ,n.$ Now suppose $g(x)$ is a factor of $f(x)$ in $\mathbb{Z}[x]$, then is it possible to get a bound on the coefficients of $g(x)$ in terms of $M$ i.e. if $g(x)=\sum_{i=0}^mb_ix^i$ then does there exist some $M^\prime $, which depends only on $M,n$ and $m$, such that $|b_i|\leq M^\prime$ for all $i=0,\ldots ,m$ ?

REPLY [9 votes]: Gelfond's inequality is probably what you want here; see for example my book with Hindry, Diophantine Geometry: An Introduction, Proposition B.7.3. I'll state it for polynomials in $\mathbb{Z}[X_1,\ldots,X_m]$, although there's a version that's true over $\overline{\mathbb{Q}}$. The statement uses the projective height, so for a polynomial $f$ with coefficients $a_i\in\mathbb{Z}$, we let $$H(f) = \frac{\max|a_i|}{\gcd(a_i)}.$$ Then
Proposition B.7.3 (Gelfond's inequality) Let $f_1,\ldots,f_r\in \mathbb{Z}[X_1,\ldots,X_m]$, and for $1\le i\le m$, let $d_i$ denote the $X_i$ degree of $f_1f_2\cdots f_r$. Then
$$
  H(f_1)H(f_2)\cdots H(f_r) \le e^{d_1+\cdots+d_m}H(f_1f_2\cdots f_r).
$$
For the OP's question, we have $f$ is divisible by $g$, say $f=gg'$, so
$$
  H(g) \le H(g)H(g') \le e^{\deg f}H(gg') = e^{\deg f}H(f).
$$

REPLY [7 votes]: Completely explicit bounds are given by Granville (bounding the coefficients of the divisor of a given polynomial, 1990, Monat. Math).<|endoftext|>
TITLE: What is known about a 3-manifold $M$ when its fundamental group is linear?
QUESTION [6 upvotes]: Suppose we have a 3-manifold $M$ and its respective fundamental group $\pi_1(M)$. An important question about its fundamental group is to ask if it is linear, i.e. they are isomorphic to a subgroup of the Lie group $GL(n,\mathbb{C})$. What is known about the manifold in such a case, i.e. what are the implications given the earlier question is correct? I am looking for references on any property, as asking for a specific one is too much to ask for.
Your answers are greatly appreciated.

REPLY [11 votes]: If you are considering compact 3-manifolds, then it is conjectured that the fundamental groups are always linear, so there should be no restriction on the topology.
One may as well consider 3-manifolds with indecomposable fundamental group. Then the only remaining case to consider is graph manifolds with a non-trivial JSJ decomposition and which do not admit a non-positively curved Riemannian metric. See the papers of Yi Liu and Przytycki-Wise.
For noncompact 3-manifolds, the issue of linearity of the fundamental group is a wide-open problem. I don't know of any evidence against it though.<|endoftext|>
TITLE: Symmetries and faces of the associahedron
QUESTION [10 upvotes]: The dihedral group of order $2n+2$ acts on $K_n$, the ($n-2$)-dimensional associahedron.  Are there any other symmetries?  References?

Does the answer to 1 change if we restrict to just the 1-skeleton of $K_n$?  References?

It is "obvious" that any simple circuit (simple closed walk, simple closed path, whatever terminology you prefer) of length 4 or 5 is a 2-dimensional face of $K_n$.  Is this true?  Proof?  Reference?

REPLY [3 votes]: I have just seen this question, while looking for something else. The answer to 1 is indeed "no", and an explicit proof appears in Lemma 2.2 of Ceballos, Santos, and Ziegler - Many non-equivalent realizations of the associahedron (not surprisingly, it follows the same ideas as Patricia's).<|endoftext|>
TITLE: Surfaces  in a 3-manifold with the same Gaussian curvature with respect to  two ambient conformal metrics
QUESTION [9 upvotes]: Let $M$ be a 3-smooth manifold and $g_{1}$ and $g_{2}$ two conformal metrics on $M$. Consider an immersed surface S in $M$ and let $K_{1}$ and $K_{2}$ be the Gaussian curvatures of $S$ with respect to the induced ambient metrics $g_{1}$ and $g_{2}$.
Question : Has anyone already studied the problem of finding examples of surfaces S such that $K_{1}=K_{2}$?
For the particular case where $M$ is the upper half-space and $g_{1}$ and $g_{2}$ are the usual euclidean and hyperbolic metrics there is a way to find lots of non-trivial examples besides the obvious ones (such as horizontal planes or properly placed spheres). 
Can this problem be formulated in terms of some geometric flow? (in the sense that we can start with an intial immersed surface $S_{0}$ and via some evolution in time $S_{t}$ converges to an immersed surface such that $K_{1}=K_{2}$? )
thanks for any help on this!

REPLY [12 votes]: I don't know the answers to your questions, i.e., I don't know whether there has been any work already done on this problem, nor do I know whether there is any kind of 'heat flow' argument for constructing solutions.  
However, I suspect that the latter, if possible, is not going to be completely straightforward.  After all, in the example I mention in my comment above, that of the $3$-sphere with $g_1$ being the standard metric with constant sectional curvature $1$ and $g_2=\lambda g_1$ for some constant $\lambda >1$, the surfaces of interest are the flat surfaces, of which there are many in the $3$-sphere, even compact ones near the Clifford torus.  Moreover, because this is a hyperbolic problem in this case, one doesn't have good regularity for the 'stationary' solutions, so finding them by 'heat equation' methods appears to be a doubtful proposition.
In the general case (I'll assume that $M$ is oriented, for simplicity), one can understand the nature of the equations by using the structure equations on the oriented orthonormal frame bundle:  Let $\pi:F\to M$ be the orthonormal frame bundle with respect to $g_1$. One has the tautological forms $\omega_i$ such that $\pi^*g_1 = {\omega_1}^2+{\omega_2}^2+{\omega_3}^2$ and the corresponding connection forms $\omega_{ij}=-\omega_{ji}$ satisfying the first structure equations
$$
d\omega_i = -\omega_{ij}\wedge \omega_j
$$
and the second structure equations
$$
d\omega_{ij} = -\omega_{ik}\wedge \omega_{kj} + \Omega_{ij} 
=-\omega_{ik}\wedge \omega_{kj} + \tfrac12R_{ijkl}\ \omega_k\wedge\omega_l\ .
$$
Now, $F$ has a map to the unit sphere bundle $\nu:F\to\Sigma(M)$ defined by sending an orthonormal frame $(p;e_i)$ to the unit vector $(p,e_3)$.  The $1$-form $\omega_3$ is the $\nu$-pullback of a well-defined $1$-form on $\Sigma(M)$, which, by abuse of notation, I write as $\omega_3$.  Similarly, $\omega_1\wedge\omega_2$ is the $\nu$-pullback of a well-defined $2$-form on $\Sigma(M)$, which I denote by the same symbol.  Meanwhile, although $\omega_{12}$ is not the $\nu$-pullback of any $1$-form on $\Sigma(M)$, its exterior derivative $d\omega_{12}= \omega_{31}\wedge\omega_{32} + \Omega_{12}$ is the $\nu$-pullback of a well-defined $2$-form on $\Sigma(M)$.
Now, if $S\subset M$ is any oriented surface in $M^3$, it has a Gauss map $\gamma:S\to \Sigma(M)$ given by sending a point of $S$ to its oriented normal vector.  This map satisfies $\gamma^*\omega_3 = 0$, $\gamma^*(\omega_1\wedge\omega_2) = dA$, and $\gamma^*(d\omega_{12})=K\ dA$.
Now, let $g_2 = \lambda g_1$ and set $\lambda = e^{2\mu}$ for some $\mu$.  Then the orthonormal frame bundle for $g_2$ can be compared with the orthonormal frame bundle of $g_1$ in the obvious way, and, under that identification, one has
$$
\omega^\ast_i = e^\mu\ \omega_i,\qquad\text{and}\qquad
\omega^\ast_{ij} = \omega_{ij} +\mu_j\omega_i - \mu_i\omega_j\ ,
$$
where I have decorated the forms associated to $g_2$ with an asterisk.  In particular, one has, computing modulo $\omega_3$ (which is the same as computing modulo $\omega^\ast_3$),
$$
d\omega^\ast_{12} \equiv \omega_{31}\wedge\omega_{32} 
-\mu_3(\omega_{31}\wedge\omega_2 + \omega_1\wedge\omega_{32})
+ (e^{2\mu}R^\ast_{1212}+{\mu_3}^2)\ \omega_1\wedge\omega_2\ .
$$
The condition that the surface $S$ have the same Gauss curvature with respect to $g_1$ and $g_2$ is then expressed as the condition that its Gauss map $\gamma$ pull back the $2$-form $\Upsilon = d\omega^\ast_{12}-e^{2\mu}d\omega_{12}$ to vanish, since $\gamma^\ast\Upsilon=K^\ast\ dA^\ast - e^{2\mu}K\ dA = e^{2\mu}(K^\ast-K)\ dA$.  Now, one computes, using the above formulae, that
$$
\Upsilon = (1{-}e^{2\mu})\omega_{31}\wedge\omega_{32} 
-\mu_3(\omega_{31}\wedge\omega_2{+}\omega_1\wedge\omega_{32})
+ \bigl(e^{2\mu}(R^\ast_{1212}{-}R_{1212}){+}{\mu_3}^2\bigr)\omega_1\wedge\omega_2\ .
$$
The exterior differential system generated by $\omega_3$, $d\omega_3$ and $\Upsilon$ is a Monge-Ampère system on the $5$-manifold $\Sigma(M)$.  Its type depends on the sign of the quantity
$$
\Delta = (1-e^{2\mu})(R^\ast_{1212}-R_{1212}) - {\mu_3}^2.
$$
Where $\Delta>0$ it is hyperbolic, where $\Delta<0$ it is elliptic, and where $\Delta$ vanishes, it is degenerate.
Note about P. Roitman's example:  It is interesting to note that this example is not elliptic everywhere.  In fact, using the usual coordinates $x^1,x^2,x^3$ on $M=\mathbb{R}^3$ and letting $u = (u^1,u^2,u^3):\Sigma(M)\to S^2$ be the projection onto the unit $2$-sphere,
one has the formulae for the metrics he mentions in the form
$$
g_1 = (dx^1)^2+(dx^2)^2+(dx^3)^2\qquad\text{and}\qquad
g_2 = \frac{(dx^1)^2+(dx^2)^2+(dx^3)^2}{(x^3)^2}.
$$
Then the above formula for $\Delta$ works out to be
$$
\Delta = \frac{1-(x^3)^2-(u^3)^2}{(x^3)^2}.
$$
In particular, the equation is elliptic in the entire region $x^3 >1$, but in the slab $0 < x^3 < 1$, only the solution surfaces whose normals are 'close enough' to vertical are defined by an elliptic equation.  In other words, there are hyperbolic solutions in this 'boundary region'.  It might be interesting to see what these non-analytic solutions look like.
Added Comment 1:  This example has degenerate solutions (i.e., solutions whose $1$-graphs lie in the locus $\Delta=0$) depending on an arbitrary function of one variable:  In fact, one can parametrize them in the form 
$$
x^1=a(s)+\cos(s)\bigl(t{-}\tanh t \bigr), \quad
x^2=b(s)+\sin(s)\bigl(t{-}\tanh t \bigr), \quad
x^3 = \text{sech}\ t
$$ 
where $a$ and $b$ are functions of $s$ that satisfy $a'(s)\cos(s)+b'(s)\sin(s)=0$.
Choosing $a$ and $b$ to be non-analytic produces non-analytic surfaces.
Added Comment 2:  It turns out that there is a Lagrangian $\Lambda_E$ for surfaces of elliptic type (i.e., ones for which $\Delta<0$) in the upper half space such that the elliptic integral surfaces of the above system are extrema of $\Lambda_E$, and there is a Lagrangian $\Lambda_H$ for surfaces of hyperbolic type (i.e., ones for which $\Delta>0$) in the upper half space such that the hyperbolic integral surfaces of the above system are extrema of that $\Lambda_H$.  Each of these Lagrangians blows up along the locus $\Delta=0$ in $\Sigma(M)$, and it appears that the degenerate solutions listed in the Comment 1 above are not the extrema of any Lagrangian.<|endoftext|>
TITLE: Independence of Leibniz rule and locality from other properties of the derivative?
QUESTION [16 upvotes]: The following is meant to be an axiomatization of differential calculus of a single variable. To avoid complications, let's say that $f$, $g$, $f'$, and $g'$ are smooth functions from $\mathbb{R}$ to $\mathbb{R}$ ("smooth" being defined by the usual Cauchy-Weierstrass definition of the derivative, not by these axioms, i.e., I don't want to worry about nondifferentiable points right now). In all of these, assume the obvious quantifiers such as $\forall f \forall g$.
Z. $\exists f : f'\ne 0$
U. $1'=0$
A. $(f+g)'=f'+g'$
C. $(g \circ f)'=(g'\circ f)f'$
P. $(fg)'=f'g+g'f$
L. The value of $f'(x)$ is determined by knowing $f$ in any neighborhood of $x$.
Axioms A and P hold in any differential algebra. C and L mean that we're talking about something more specific than a differential algebra; they're meaningful only because we're talking about a ring of functions.
I could choose to omit U, since it can be proved from the others. I would prefer to keep U and omit P. Is P superfluous, or can anyone find a model in which P fails?
Likewise, is L independent of the others?
What seems to be tricky is to rule out models of the general flavor of $f'(x)=f^{CW}(x-17)$, where $f^{CW}$ is the usual Cauchy-Weierstrass derivative of $f$.
Are there models that are not the same as CW? Is this a nice axiomatization? Could it be improved?
[EDIT] Tom Goodwillie didn't say so explicitly, but his answer, along with one of my comments below his answer, shows that Z, A, and C suffice, so U, P, and L are not needed.
It looks like you can also take P as an axiom and recover the standard derivative, i.e., either P or C can be proved from the other: Do these properties characterize differentiation?

REPLY [15 votes]: If you grant that $c'=0$ when $c$ is a constant, you can argue as follows:
EDIT: Actually $c'=0$ follows from $1'=0$ using the chain rule, since $c=c\circ 1$.
Let $I(x)=x$. Since $I=I\circ I$, the chain rule gives $(I')^2=I'$, so $I'$ is the characteristic function of a set $A$ of real numbers. Let $T_c(x)=x+c$. Then $T_c'=I'+c'=I'$, and the chain rule applied to $T_c=I\circ T_c$ then implies $I'=(I'\circ T_c)I'$. That means that the set $A$ is either all or nothing. But you can't have $I'$ identically $0$ because that would imply $f'=(f\circ I)'=0$ for all $f$. So $I'=1$ as expected.
Now consider linear functions. If $L_m(x)=mx$, then since $L_m(x+a)-L_m(x)$ is constant, $L_m'(x+a)-L_m'(x)$ is zero. Thus $L_m'$ is a constant depending on $m$, say $h(m)$. The map $h$ is an additive homomorphism, and by the chain rule it is also multiplicative. We also know $h(1)=1$. This implies that $h(m)=m$ for all $m$. (A ring map from reals to reals preserves squares, therefore preserves ordering, therefore is continuous ...)
So $f'$ is what it should be when $f$ is polynomial of degree at most one. Now let $S(x)=x^2$. From $S(x+t)=S(x)+2tx+t^2$ we get $S'(x+t)=S'(x)+2t$, therefore $S'(x)=S'(0)+2x$. But $S'(0)=0$ using $S(-x)=S(x)$. So the derivative of squaring is what it should be.
Now the special case of Leibniz that says $(f^2)'=2ff'$ follows by the chain rule. The general case follows by expressing $fg$ in terms of $f^2$, $g^2$, and $(f+g)^2$.
EDIT: This was all about global functions. But it can be extended. Let me spell out what I hope your axioms are: $f'$ is defined when $f$ is a $C^\infty$ real function whose domain is an open set $U\subset \mathbb R$, and $f'$ is another such function with the same domain. The axioms are
(U) $1_U'=0_U$ where $1_U$ is the constant function on $U$.
(A) $(f+g)'=f'+g'$ where $f$ and $g$ (and $f+g$) have the same domain.
(C) $(f\circ g)'=(f'\circ g)g'$, when $f$ has domain $U$ and $g$ has domain $V$ and $g(V)\subset U$, so that $f\circ g$ and $f'\circ g$ have domain $V$.
(Z) For every nonempty $U$ there is some $f$ with domain $U$ such that $f'$ is not identically zero.
The arguments that I gave above can be adapted to show then that:
$c_U'=0$ for any constant function on any $U$.
$I_U'=1$ where $I_U$ with domain $U$ is defined by $I_U(x)=x$. (Here you have to mess around with compositions $I_U\circ (I_V+c_V)$.)
So in the end you get the desired localization property, too: the operator commutes with restriction from $U$ to $V\subset U$ by the chain rule, because restriction is composition with $I_V$.<|endoftext|>
TITLE: Are two elements of a group determined up to simultaneous conjugacy by the conjugacy classes of all of their products?
QUESTION [7 upvotes]: Let $G$ be a group (if it helps, assume that $G$ is a Lie group or finite). Is a pair of elements $(g, h) \in G \times G$ determined up to simultaneous conjugacy by the conjugacy class of every element $w(g, h) \in G$, where $w$ runs over all words in the free group on two generators? 
If $G$ is finite, can we bound the length of the words $w$ needed in terms of $|G|$? 
If the answer to the above question is positive, let $\pi$ be a second group (if it helps, assume that $\pi$ is finitely presented). $G$ acts on the set $\text{Hom}(\pi, G)$ by pointwise conjugation. Is an element $\phi \in \text{Hom}(\pi, G)$ determined up to conjugacy by the conjugacy class of every element $\phi(w)$ where $w \in \pi$? (The above is the special case $\pi = F_2$.)

REPLY [5 votes]: Just saw this old question thanks to the link from
the new 
MO question #212929.  Another flavor of counterexample:
let $G$ be the $ax+b$ group over some field $k$, and let
$t(b) \in G$ be the transformation $x \mapsto x + b$.
Then the $t(b)$ with $b \neq 0$ are all conjugate, but
a pair $(t(b_1),t(b_2))$ is simultaneously conjugate with
$(t(b'_1),t(b'_2))$ iff $b_1/b_2 = b'_1/b'_2$.
Suppose, then, that $b_1/b_2 \neq b'_1/b'_2$, 
and that both quotients are irrational
(i.e. if $mb_1=nb_2$ or $mb'_1=nb'_2$ for some integers $m,n$ 
then $m=n=0$ in $k$).  Then the pairs
$(g,h) := (t(b_1),t(b_2))$ and $(g',h') := (t(b'_1),t(b'_2))$
are not related by simultaneous conjugacy, but cannot be
distinguished by the conjugacy class of any $w(g,h)$.
The same is true if we take $G = {\rm SL}_2(k)$
or $G = {\rm PSL}_2(k)$ and work in its $ax+b$ subgroup
$({* \; * \atop 0 \; *})$ [with $t(b) = ({1 \; b \atop 0 \; 1})$].
Indeed that was the example that first came to mind, 
suggested by the restriction to diagonalizable elements in
MO 212929
(NB the $t(b) \in {\rm SL}_2(k)$ with $b\neq 0$ are not diagonalizable).
In the smallest finite examples of this kind, $k$ is the 4-element field, and
$b_1/b_2$, $b'_1/b'_2$ are the two elements of $k$ other than $0$ and $1$.
Then the $ax+b$ group is isomorphic with $A_4$, and 
${\rm SL}_2(k) \cong {\rm PSL}_2(k) \cong A_5$,
and in each case we can take
$g=g'=(12)(34)$, $h=(13)(24)$, and $h'=gh=(14)(23)$. 
[The pairs $(g,h)$ and $(g',h')$ become conjugate in $S_4$ and $S_5$,
which are the extensions of the $ax+b$ group and ${\rm SL}_2(k)$
by the Galois automorphism of $k$ that also switches the two
irrational elements of $k$.]<|endoftext|>
TITLE: Hochschild (co)homology and representation theory
QUESTION [7 upvotes]: Dear members of Mathoverflow,
I just discovered the notion of Hochschild (co)homology. I understand well the formalism however I am wondering about the meaning of this (co)homology for representation theory. 
I consider an algebra $\mathcal{A}$, such as $\mathcal{U}(su(n))$, a finite dimensional representation space $V$ for $\mathcal{A}$ (they are well known in the case of $\mathcal{U}(su(n))$) and the $\mathcal{A}$-bimodule $\mathcal{M}=\mathrm{End}(V)$.
In this example, what would be the interpretation of Hochschild (co)homology from the point of view of representation theory ? To what is it an obstruction ?
I can compute things in simple cases such as $\mathcal{U}(su(2))$ but I do not the global interpretation emerge in this case...
Thank you in advance,
Damien.

REPLY [2 votes]: If $A$ is a $k$-algebra and $V$, $W$ are $A$-modules, then $Hom_k(V,W)$ is an $A$-bimodule in a natural way (acting contravariantly in $V$ and covariantly in $W$).   It is well known (can be found in Cartan-Eilenberg) that $n^{th}$-Hochschild cohomology of $Hom_k(V,W)$ is $Ext_A^n(V,W)$.  Your case of $End(V)$ would then give $Ext_A(V,V)$.<|endoftext|>
TITLE: What do cluster algebras tell us about Grassmannians?
QUESTION [33 upvotes]: One of the first examples of a cluster algebra given in Fomin and Zelevinsky's original paper is the homogeneous coordinate ring $\mathbb{C}[G_{2,n}]$ of the Grassmannian of planes in $\mathbb{C}^n$. It has also been shown by Scott that $\mathbb{C}[G_{k,n}]$ carries a cluster algebra structure for any $k$.
When I'm explaining cluster algebras to somebody for the first time, I usually start with these examples, because they're the ones I spend the most time thinking about. Another reason is that the $\mathbb{C}[G_{2,n}]$ case can be nicely visualized by identifying clusters of Plücker coordinates with triangulations of the $n$-gon.
A common response to this example is "that's very pretty, but what does it tell us about $\mathbb{C}[G_{2,n}]$ as an algebra?", which is a reasonable question I don't feel I can answer well, if at all. As I'm probably going to be giving several talks about cluster algebras in the next few months, it would be nice to have a good answer to this.
Essentially all I can say so far is that the cluster monomials form a distinguished linearly independent set, and in the case of $\mathbb{C}[G_{2,n}]$ they are even a basis, but this isn't hugely satisfactory.
So, my question is:

Are there any results about
  $\mathbb{C}[G_{k,n}]$ proved using the
  cluster algebra structure that weren't
  known before this additional structure
  was discovered?

If there are better answers for different algebras then I would also be interested, but for the purposes of asking a hopefully-not-too-vague question, I'll stick to Grassmannians.

REPLY [17 votes]: One simple answer is to talk about the totally positive part of $(G_{k,n})_{> 0}$, the part of the Grassmannian where all the maximal minors (=Plücker coordinates) are real and positive.  Naively, if you want to test whether a point is in the totally positive part, you would check all the Plücker coordinates.  Since there is a cluster structure which includes all the Plücker coordinates as cluster variables, though, it suffices to check positivity with respect to any single cluster of them.
This might be a bit more geometrical than you were looking for, but I think it's a nice, and easy-to-explain, idea, which is also consistent with the roots of the development of cluster algebras.<|endoftext|>
TITLE: (The) missing Moore graph(s) - uniqueness
QUESTION [7 upvotes]: In the related literature one often sees the phrase "The missing Moore graph" which (to me) tacitly implies that the missing Moore graph (if exists) is unique.
Is there a result of this type or is or is this just a limitation of words that do not express the fact that there could be nonisomorphic graphs of diameter 2 and degree 57?

REPLY [14 votes]: The uniqueness of a Moore graph of degree 57 and diameter 2 is not known. See, for instance, this paper -
http://www.sciencedirect.com/science/article/pii/S0024379509003735
- where they refer to `the missing Moore graph(s)' to indicate this fact.
Other discussion can be found here:
http://symomega.wordpress.com/2009/09/11/i-want-more-moore-graphs/
Edit: I have browsed the paper of Macaj and Siran linked to above, the main result of which says that if a Moore(57,2)-graph exists then its automorphism group has order at most 375. Let me give a relevant quote from the paper:

On the other hand, in the study of
  possible actions of groups of order
  375 with 10 orbits we found hundreds
  of matrices satisfying conditions of
  Lemma 5 which we were not able to
  exclude by our techniques.

(The `matrices satisfying conditions of Lemma 5' are adjacency matrices for particular partitions of a putative Moore(57,2)-graph.) In other words, in theory there could be many different Moore(57,2)-graphs even if you just restrict to those with an automorphism group of order 375.<|endoftext|>
TITLE: A graph parameter possibly related to treewidth
QUESTION [10 upvotes]: (Cross-posted from the Theoretical Computer Science StackExchange site after there was no conclusive answer after a week.)
I am interested in graphs on $n$ vertices which can be produced via the following process.

Start with an arbitrary graph $G$ on $k\le n$ vertices. Label all the vertices in $G$ as unused.
Produce a new graph $G'$ by adding a new vertex $v$, which is connected to one or more  unused vertices in $G$, and is not connected to any used vertices in $G$. Label $v$ as unused.
Label one of the vertices in $G'$ to which $v$ is connected as used.
Set $G$ to $G'$ and repeat from step 2 until $G$ contains $n$ vertices.

Call such graphs "graphs of complexity $k$" (apologies for the vague terminology). For example, if $G$ is a graph of complexity 1, $G$ is a path.
I would like to know if this process has been studied before. In particular, for arbitrary $k$, is it NP-complete to determine whether a graph has complexity $k$?
This problem appears somewhat similar to the question of whether $G$ is a partial $k$-tree, i.e. has treewidth $k$. It is known that determining whether $G$ has treewidth $k$ is NP-complete. However, some graphs (stars, for example) may have much smaller treewidth than the measure of complexity discussed here.

REPLY [4 votes]: Following Felix's suggestion, and corroborating Ashley's characterization,
here are five examples of graphs grown from
$K_2$, i.e., $E=\lbrace (1,2) \rbrace$:

     


The vertices are numbered in the order in which they were added.

Here is just one more, this time grown from $K_5$, to give a gestalt view of the influence of
the start $G$:<|endoftext|>
TITLE: The significance of modularity for all Galois representations
QUESTION [9 upvotes]: On pg. 1 of the slides of a talk, Henri Darmon wrote:

Question: What is an interesting Diophantine equation? 
A “working definition”. A Diophantine equation is interesting
  if it reveals or suggests a rich 
  underlying mathematical structure.

One can adopt the perspective that the interesting elliptic curves over $\mathbb{Q}$ are those that are modular and view the fact that all elliptic curves over $\mathbb{Q}$ are modular as being of minor significance (in the sense that if some weren't, they wouldn't be so interesting). 
I realize that my raising this perspective may come across as an affront to some of the celebrated research of recent times, and would hasten to emphasize that I'm asking my questions here in good faith with a view toward learning more.

What would we lose if we decided to
  focus only on those Galois
  representations that are attached to
  automorphic forms and ignore the
  possibility that some do not?

One thing that one would lose is Wiles' proof of Fermat's last theorem. Until recently, my attitude had been that the Frey Curve construction is a curiosity and that a "morally right" proof would come from the $abc$-conjecture applied to sufficiently large exponents together with the theory of arithmetic of cyclotomic fields to rule out the possibility of nontrivial solutions to the equations with smaller exponents. However, recently I came across slides from a talk by Minhyong Kim in which Kim wrote (pg. 29):

The idea of encoding points into
  'larger' geometry is a common one in
  Diophantine geometry, as when
  solutions
$a^n + b^n = c^n$
to the Fermat equation are encoded
  into the elliptic curves
$y^2 = x(x - a^n)(x + b^n)$.
The geometry of the path torsor
  $\pi_1(X(\mathbb{C}); b, x)$ is an
  extremely canonical version of this
  idea.

I find the idea that the Frey curve construction is canonical to be fascinating! It raises the possibility that the proof of Fermat's last theorem coming from the study of Frey equations is morally right. [Edit: As KristianJS aptly points out, I misread Kim's quote. So I'd recur to my remark above about my impression on what a "morally right" proof of Fermat's Last Theorem would look like.]
Anyway, I'd be very interested in further examples concerning the significance of all L-functions attached to Galois representations (of suitable type) being modular.
[Added: If I remember correctly, In "The Map of My Life" Shimura wrote that he was more interested in the fact that suitable cusp forms correspond to elliptic curves than in the converse. This seems relevant. However, I cannot find the quotation and may be misremembering. I would welcome a reference from anybody who remembers this.]
[Added: I just asked another question that touches on material that may provide a partial answer to this question.]

REPLY [7 votes]: I want to make a couple of comments about the premise of this question. First, the OP asks what the consequences would be if we "focus only on those Galois representations that are attached to automorphic forms and ignore the possibility that some do not".  But of course it is certainly the case that there exist Galois representations that are not attached to automorphic forms (Galois representations that are ramified at infinitely many primes, for instance). 
Second, the OP writes about "Galois representations that are attached to automorphic forms" as though this phrase is well-defined.  But do you mean the automorphic forms that conjecturally believed to have Galois representations attached to them, or the automorphic forms that are currently known to have Galois representations attached to them?  The latter is a (growing but proper!) subset of the former; and note again that the former is not "all automorphic forms" (see e.g. the paper of Buzzard-Gee in the 2011 LMS Durham Symposium).
So, there's a conjectural map from (some) automorphic forms to (some) Galois representations, and this has been understood for decades to have significant arithmetic consequences for representations in the image of the map.  The most generous way I can think to interpret the question is "What would happen if we satisfied ourselves with statements of the form "If $\rho$ is in the image of this map in an instance where the map has been constructed...", without attempting to characterize the image in some easily checkable way?", and the answer is simply that we would lose most or all of the arithmetic consequences.<|endoftext|>
TITLE: Right Angled Artin Group Reference request
QUESTION [7 upvotes]: The following should be true: every normal subgroup of a non-Abelian Right Angled Artin Group should contain a free group on two generators. Is there a standard reference one can cite for this?

REPLY [13 votes]: Igor, I assume the question is about centerless RAAGs. Then it follows from the classical result of A. Baudisch (see MR0634562): every 2-generated subgroup of a RAAG is abelian or $F_2$. 
Indeed if $N$ is a non-trivial normal subgroup in a RAAG $G$, take any $x\in N$. Since $G$ is centerless  there exists $g\in G$ that does not commute with $x$. Then $\langle x,g\rangle\cong F_2$ and hence $\langle x, x^g \rangle \cong F_2$.<|endoftext|>
TITLE: graphs with independence number = Shannon capacity
QUESTION [14 upvotes]: For $G$ a graph, let $\alpha(G)$ be its independence number and $\Theta(G)=\lim_n \sqrt[n]{\alpha(G^{\boxtimes})}$ its Shannon capacity, where $\boxtimes$ denotes strong product.
Consider graphs $G$ and $H$ satisfying $\alpha(G)=\Theta(G)$ and $\alpha(H)=\Theta(H)$. For example, $G$ and $H$ could be perfect, but the more interesting situations arise when neither of them is perfect.
Question: Does this assumption imply
(1) $\alpha(G\boxtimes H) = \alpha(G)\alpha(H)$ ?
(2) $\Theta(G\boxtimes H) = \Theta(G)\Theta(H)$ ?
(3) $\Theta(G + H) = \Theta(G) + \Theta(H)$ ?
Here, $G+H$ stands for the disjoint union of $G$ and $H$.
If my reasoning is correct, then (1) and (2) are equivalent and imply (3).
As far as I can see, neither the work of Haemers nor the results of Alon have anything directly to say about these questions. But then again, I am not an expert on this, so I might have missed something obvious.
Edit (see Will Traves' answer): Actually, I am specifically interested in those $G$ and $H$ which are well-covered.
Edit: The paper is here.

REPLY [5 votes]: By now, we have been able to resolve this question, and our revised paper contains a proof showing that the answer to all three questions is negative in a very strong sense. Let me provide a brief summary here and refer to the paper for more details.
What we show is this: there exist graphs $G$ and $H$ with $\alpha(G)=\Theta(G)$ and $\alpha(H)=\vartheta(H)$ which violate all three desired inequalities. Note that the property $\alpha(H)=\vartheta(H)$ is even stronger than the required property $\alpha(H)=\Theta(H)$.
The construction crucially relies on the definitions and results on hypergraphs discussed in our paper and on results of Haemers. We have partially translated it into pure graph-theoretic terms, but it does not seem possible to do so completely.
The counterexample turns out to be the following:

$G$ is a $108$-regular graph on $220$ vertices. The vertices correspond to the $3$-element subsets of $\{1,\ldots,12\}$ and two such vertices are adjacent whenever the subsets intersect in exactly one element. This graph was considered by Haemers, who showed that $\alpha(G) = \Theta(G) < \vartheta(G)$, and this property is the most important ingredient for us.
$H$ is a graph constructed from the complement of $G$ which turns out to have $1131460$ vertices.

I suspect that a similar construction can be carried out starting with any graph $G$ having the property that $\alpha(G) = \Theta(G) < \vartheta(G)$, but we haven't checked all the details of this.<|endoftext|>
TITLE: Anabelian geometry study materials?
QUESTION [24 upvotes]: I want to study anabelian geometry, but unfortunately I'm having difficulties in finding some materials about it. If you could offer me some books/papers/articles I would be glad.

REPLY [3 votes]: Florian Pop, Lectures on Anabelian phenomena in geometry and arithmetic (pdf)
Yuri Tschinkel, Introduction to anabelian geometry, talk at Symmetries and correspondences in number theory, geometry, algebra, physics: intra-disciplinary trends, Oxford 2014 (slides pdf)<|endoftext|>
TITLE: Jacobians defined over smaller fields
QUESTION [7 upvotes]: Let $L/K$ be an extension of number fields.
Let $X$ be a curve over $L$ which can not be defined over $K$. Let $J(X)$ be the Jacobian of $X$ over $L$.
In general, the Jacobian $J(X)$ probably doesn't admit a model over $K$. But one could imagine that this does happen sometimes.
Question. Does there exist an example where $J(X)$ admits a model over $K$?
In other words, can "the" field of definition of $J(X)$ be smaller than "the" field of definition of $X$?

REPLY [3 votes]: As Piotr has suggested, Torelli essentially answers the question if the canonical principal polarization is included in the data. More accurately, this is a version of Torelli due to Serre: if $(A,\lambda)$ is a ppav of dimension at least $2$ over a field $K$ and if over some extension $L$ of $K$ there is a curve $C$ such that $(A,\lambda)_L$ is $L$-isomorphic to the Jacobian of $C$, then there is a curve $C_0$ over $K$ such that $C$ is isomorphic to $(C_0)_L$ and some quadratic twist of $(A,\lambda)$ is $K$-isomorphic to the Jacobian of $C_0$. Moreover, if $C$ is hyperelliptic then no twist is necessary.<|endoftext|>
TITLE: Lifting the Frobenius to the absolute Galois group of the $p$-adics
QUESTION [6 upvotes]: I tend to ask questions on mathstackexchange, because I don't feel adequate yet for mathoverflow. I had previously asked this question there (I have now deleted it), where it was quite popular, but it didn't seem like anybody knew the answer. I thought that perhaps this question should graduate to mathoverflow. Here it is, as it had originally appeared:
I've been playing around with the $p$-adics, and I wondered about the structure of their Galois group.
We have the short exact sequence:
$$1\rightarrow Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p^{un})\rightarrow Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)\rightarrow Gal(\mathbb{Q}_p^{un}/\mathbb{Q}_p)\rightarrow 1$$
I as wondering if this exact sequence is split. I.e., is it true that you can lift the Frobenius automorphism of $Gal(\mathbb{Q}_p^{un}/\mathbb{Q}_p)$ to $Gal(\bar{\mathbb{Q}}_p/\mathbb{Q}_p)$? How would the lifting of this automorphism act on individual elements of $\bar{\mathbb{Q}}_p$?

REPLY [4 votes]: As others have mentioned, the exact sequence is continuously split.  Furthermore, any lift of Frobenius will act as you expect on the subfield $\mathbb{Q}_p^{nr}$, i.e., the extension you get by adjoining all prime-to-$p$ roots of unity.  Indeed, you can form a $\mathbb{Q}_p$-basis of $\mathbb{Q}_p^{nr}$ using roots of unity, and then the action is determined by $p$-th powers on the basis elements.
As far as other elements are concerned, the set of lifts is a torsor under $\operatorname{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p^{nr})$, so you have a large amount of freedom.  For example, your typical $S_3$-extension whose intersection with $\mathbb{Q}_p^{nr}$ is quadratic will have three Frobenius lifts.<|endoftext|>
TITLE: Erratum for Fulton and Harris
QUESTION [14 upvotes]: I am currently using Fulton and Harris for a course on representation theory, and I have noticed that there are a few errors throughout the book. A search on google with the keywords "Errata for Fulton and Harris" doesn't come up with anything. 
I believe the book is widely used in many representation theory courses, and it would be helpful if a list of errata could be compiled. Does anyone know of any list of errata for this?
Thanks.
Errors I have found: On page 150 (in the middle) the line "...so the representation $W = \Bbb{C}\cdot x^2 \oplus \Bbb{C} \cdot xy \oplus \Bbb{C} \cdot y^2 = W_{-2} \oplus W_0 \otimes W_2$ is the..." should have a direct sum in place of tensor product in between $W_0$ and $W_2$.

REPLY [3 votes]: I may have found another error. In exercise 14.37, they ask to show that the Killing Form on $SO(n,\mathbb{R})$ is positive definite. But the formula for the Killing form on the underlying Lie algebra, which is $so(n,\mathbb{R})$ consisting of skew-symmetric matrices, is $(n-2)tr(XY)$. Firstly, for $n=2$, that isn't even definite. Even when $n>2$, it isn't positive.  For any skew-symmetric matrix, its square is negative-semidefinite, which can be verified with any simple example. Thus, $(n-2)tr(X^2)\leq 0$ for skew-symmetric matrices. Am I going crazy?<|endoftext|>
TITLE: Tensor powers of the standard representation
QUESTION [14 upvotes]: Consider $V_{(n-1, 1)}$, the $n-1$ dimensional irreducible representation of $S_n$, i.e. the "standard" or "defining" representation.  Is there a nice formula for how the $k$-th tensor power of $V_{(n-1, 1)}$ decomposes into irreps?

REPLY [6 votes]: The problem has been solved in the reference indicated in the comment
by Vasu Vineet, namely:
"Combinatorial Operators for Kronecker Powers of Representations of Sn" by
Alain Goupil and Cedric Chauve. However, one cannot say that the formulas in Propositions 1 and 2 of this paper are "nice".<|endoftext|>
TITLE: Certain compact subset of $L_1$
QUESTION [8 upvotes]: Let $(\Omega,\Sigma, \mu)$ be a probability measure and $X$ a Banach space. I am interested in subsets $F\subseteq L_\infty (\mu,X)$ that satisfy these two compactness conditions: 

$F$ is a norm-compact subset of $L_1(\mu, X)$; and
For any sequence $f_n$ in $F$ there exists a set $E\in \Sigma$, $\mu(E)=1$, such that each $L_1$-norm convergent subsequence of $f_n$ converges pointwise on $E$. 

An example of such a set is any compact subset $F$ of $L_\infty (\mu,X)$. To see this suppose that $f_n$ is a sequence in $F$. There is $E\in \Sigma$, $\mu(E)=1$, such that for any $m,n$ we have $\|f_m(\omega) - f_n(\omega)\|\leq \|f_m-f_n\|_\infty$ for all $\omega\in E$. Now if $f_m$ is an $L_1$-convergent subsequence of $f_n$, then it must be converging in $L_\infty$, and converging pointwise on $E$. 
More generally, the compactness
   condition is satisfied by 

$\star$
$F\subseteq L_\infty (\mu,X)$ and for every
$\epsilon>0$ there exists $E\in
   \Sigma$, $\mu(E)>1-\epsilon$ with
$$
\{f\chi_E: f\in F\} 
$$ 
is compact in $L_\infty (\mu,X)$. 

An example of an $L_1$-compact set that does not satisfy the compactness conditions  2 or  $\star$ is the set of monotone step functions $f\colon [0,1]\to \{0,1\}$. 
Now for my question: Does there exist a set satisfying 1 and 2 but not satisfying the $L_\infty$ compactness condition $\star$? 
I don't have an answer to this for the case $X=R$ and $\Omega=[0,1]$.

REPLY [3 votes]: The answer is that (1) and (2) implies $\star$ (this resolves a nice problem in a game theory paper I'm working on though the final unresolved problem has to do with decomposable Banach spaces, which I'll ask in a different question). 
Assume (1) and (2). Let $P=\{f_n\}$ be a sequence (actual versions of equivalence classes) in $X$ such that each $f\in X$ is the $L_1$ limit of some subsequence of $f_n$. Let $E$ be the measurable set in (2). 
By (2), $P$ has compact closure $\overline{P}^p$ in the product topology $X^{E}$. Notice that   $X= \overline{P}^{L_1}=\overline{P}^p$ also by (2) taken as equivalence classes.  Thus, we can now quickly show by means of Egorov's theorem that for every $\epsilon>0$ there is $F\in \Sigma$, $\mu(F)>1-\epsilon$ such that $\overline{P}^p$ is compact in the topology of uniform convergence (on $F$). We have the result.<|endoftext|>
TITLE: Number of elements in the set $\{1,\cdots,n\}\cdot\{1,\cdots,n\}$
QUESTION [53 upvotes]: Let $A_n=\{a\cdot b : a,b \in \mathbb{N},  a,b\leq n\}$. Are there any estimates for $|A_n|$? Will it be $o(n^2)$?

REPLY [15 votes]: Let me give here an answer with a quick argument why it is $o(n^2)$. I do not know whether it is the same as Erdős original proof. UPD: it really is, and is mentioned above in a comment by Kevin P. Costello.
Most numbers from 1 to $n$ have $\log \log n (1+o(1))$ prime divisors (counted with multiplicity) by Erdős-Kac theorem. Then most their products have $2\log \log n (1+o(1))$ prime factors, while most numbers from 1 to $n^2$ have again $\log \log n (1+o(1))$ prime factors. It proves that products of two numbers from 1 to $n$ are rare in $\{1,2,\dots,n^2\}$<|endoftext|>
TITLE: The complete list of continued fractions like the Rogers-Ramanujan?
QUESTION [14 upvotes]: I have two questions about q-continued fractions, but a little intro first. Given Ramanujan's theta function,
$$f(a,b) = \sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}$$
then the following,
$$A(q) = q^{1/8} \frac{f(-q,-q^3)}{f(-q^2,-q^2)}$$
$$B(q) = q^{1/5} \frac{f(-q,-q^4)}{f(-q^2,-q^3)}$$
$$C(q) = q^{1/3} \frac{f(-q,-q^5)}{f(-q^3,-q^3)}$$
$$D(q) = q^{1/2} \frac{f(-q,-q^7)}{f(-q^3,-q^5)}$$
$$E(q) = q^{1/1} \frac{f(-q,-q^{11})}{f(-q^5,-q^7)}$$
are q-continued fractions of degree $4,5,6,8,12$, respectively, namely,
$$A(q) = \cfrac{q^{1/8}}{1 + \cfrac{q}{1+q + \cfrac{q^2}{1+q^2 + \ddots}}},\;\;B(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \ddots}}}$$
$$C(q) = \cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \ddots}}},\;\;\;\;D(q) = \cfrac{q^{1/2}}{1 + q +\cfrac{q^2}{1+q^3 + \cfrac{q^4}{1+q^5 + \ddots}}}$$
$$E(q) = \cfrac{q(1-q)}{1-q^3 + \cfrac{q^3(1-q^2)(1-q^4)}{(1-q^3)(1+q^6)+\cfrac{q^3(1-q^8)(1-q^{10})}{(1-q^3)(1+q^{12}) + \ddots}}}$$
The first three are by Ramanujan, the fourth is the Ramanujan-Gollnitz-Gordon cfrac, while the last is by Naika, et al (using an identity by Ramanujan). Let $q = e^{2\pi i \tau}$ where $\tau = \sqrt{-n}$ and these can be simply expressed in terms of the Dedekind eta function $\eta(\tau)$ as,
$$\tfrac{1}{A^4(q)}+16A^4(q) = \left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8$$
$$\tfrac{1}{B(q)}-B(q) = \left(\tfrac{\eta(\tau/5)}{\eta(5\tau)}\right)+1$$
$$\tfrac{1}{C(q)}+4C^2(q) = \left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3$$
$$\tfrac{1}{D(q)}-D(q) = \big(\tfrac{1}{A(q^2)}\big)^2$$
$$E(q) = \;???$$
Question 1: Does anybody know how to express $E(q)$ in terms of $\eta(\tau)$? (It's SO frustrating not to complete this list. I believe there might be a simple relationship between orders 6 and 12, just like there is between 4 and 8.) This cfrac can be found in "On Continued Fraction of Order 12", but the authors do not address this point.
Question 2: Excluding these five and the Heine cfrac which gives $\eta(\tau)/\eta(2\tau)$, are there any other q-continued fractions which yield an algebraic value at imaginary arguments?

REPLY [3 votes]: Based on Elkies' answer and an email by Michael Somos, we can give an alternative expression to my Question 1. If a sum is used, instead of a difference,
$$u=\frac{1}{E(q)}+E(q)$$
then,
$$\frac{u(u-4)^3}{(u-1)^3} = \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8$$
Or more simply, we can relate it to the cubic continued fraction $C(q)$ as,
$$u=\frac{1}{E(q)}+E(q) =\frac1{C(q)\,C(q^2)}$$
Since $\displaystyle C(q)=\frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}$, this implies,
$$u= \frac{\eta(4\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(12\tau)}$$
an eta quotient also mentioned by Elkies in his answer.<|endoftext|>
TITLE: Cut-distance between two Erdos-Renyi random graphs
QUESTION [6 upvotes]: Consider two Erdos-Renyi random graphs $G_1,G_2$ on $n$ nodes, with the edges in each graph generated independently at random with probability $1/2$. My question is about the cut-distance between these two graphs. Recall that the cut-distance is the maximum over all cuts of the difference between the cut values of the two graphs. Using a standard Chernoff bound argument for each cut, followed by a union bound to maximize over all cuts, I get that the cut-distance between $G_1,G_2$ is $O(n^{3/2})$ with large probability. Is this the tightest bound one can show? I'm wondering if it is possible to get this distance down to, for example, $O(n)$?

REPLY [5 votes]: (To answer Anthony, I'm taking the two graphs to be on the same vertex set.)
This is an argument, without details, that the answer is $\Omega(n^{3/2})$.
Generate the graphs two vertices at a time. As each pair of vertices is generated and their adjacencies with the previous vertices are chosen randomly, put one vertex on each side of the cut. Choose which vertex goes on which side so as to maximize the number of edges of $G_1$ crossing the cut minus the number of edges of $G_2$ crossing the cut.
Let $X_i$ be the number of edges of $G_1$ across the cut, minus the number of edges of $G_2$ across the cut, after $i$ pairs of vertices have been generated.  The differences $X_{i+1}-X_i$ are independent and the expectation of $X_{i+1}-X_i$ is, I believe, $\Omega(i^{1/2})$ (maybe someone will prove this for us).  So with high probability, using some standard concentration inequality, $X_{n/2} = \Omega(n^{3/2})$.
Probably the constant can be improved by generating one vertex at a time and deciding which side of the cut to put it.  But then the differences are not independent so the analysis is trickier.<|endoftext|>
TITLE: Arbitrary small positive lower semi continuous functions
QUESTION [6 upvotes]: This question is a generalization of the question posed in this page to lower semi continuous functions. so let me describe the Question in the following way.

Def:  Let $(X,\tau)$ be a Tychonoff Topological space. we say that this space has an arbitrary small lower semi continuous function, if the following statement is true for it:
Statement:For each $x\in X$ consider an arbitrary positive real number $\epsilon_x>0$. then there exists  a lower semi continuous real valued function $f:X\rightarrow \mathbb{R}$   with the following property:
$$\forall x \in X $$  $$ 0< f(x) < \epsilon_x$$

Question:Can we characterize the spaces which has the above statement as it's property?

REPLY [4 votes]: Since we are dealing with upper and lower semicontinuous functions instead of continuous functions, it is fruitful to consider all topological spaces instead of just completely regular spaces. The following theorem characterizes all such spaces, and $2\rightarrow 1$ incorporates François Dorais's idea in his answer to the previous question. As before, the answer involves the first measurable cardinal, and all such $T_{1}$-spaces are discrete if there does not exist a measurable cardinal. Since a space has arbitrarily small lower semicontinuous functions if and only if it has arbitrarily large upper semicontinuous functions, it suffices to characterize the spaces with arbitrarily large upper semicontinuous functions.
Theorem: Let $X$ be a topological space. Then the following are equivalent.

For every $x\in X$, the neighborhood filter $\mathcal{N}(x)$ of $x$ is the intersection of finitely many $\sigma$-complete ultrafilters.
For every countable partition $P$ of $X$ there is an open cover $\mathcal{U}$ of $X$ such that
for each $U\in\mathcal{U}$ there are $A_{1},...,A_{n}\in P$ with $U\subseteq A_{1}\cup...\cup A_{n}$.
For every countable partition $P$ of $X$ there is a countable open cover $\mathcal{U}$ such that
for each $U\in\mathcal{U}$ there are $A_{1},...,A_{n}\in P$ with $U\subseteq A_{1}\cup...\cup A_{n}$.
If $\epsilon_{x}\in\mathbb{R}$ for all $x\in X$, then there is an upper semicontinuous function
$f:X\rightarrow\mathbb{R}$ with $f(x)\geq\epsilon_{x}$ for $x\in X$. 
If $n_{x}\in\mathbb{N}$ for all $x\in X$, then there is an upper semicontinuous function
$f:X\rightarrow\mathbb{N}$ with $f(x)\geq n_{x}$ for $x\in X$.

Proof:
$1\rightarrow 2$. Assume that every neighborhood filter $\mathcal{N}(x)$ is the intersection of finitely many $\sigma$-complete ultrafilters. Let $x\in X$ and assume that
$\mathcal{N}(x)=\mathcal{M}_{1}\cap...\cap\mathcal{M}_{n}$ where $\mathcal{M}_{1},...,\mathcal{M}_{n}$ are $\sigma$-complete ultrafilters on $X$. Let $P$ be a countable partition of $X$. Then there are $R_{1},...,R_{n}\in P$ where $R_{i}\in\mathcal{M}_{i}$ for $1\leq i\leq n$, so $R_{1}\cup...\cup R_{n}\in\mathcal{M}_{1}\cap...\cap\mathcal{M}_{n}
=\mathcal{N}(x)$, so if $U_{x}=(R_{1}\cup...\cup R_{n})^{\circ}$, then $\{U_{x}|x\in X\}$ is the required open cover of $X$.
$2\rightarrow 5$ Assume $n_{x}\in\mathbb{N}$ for $x\in X$, and let $A_{n}=\{x\in X|n_{x}=n\}$. Then $\{A_{n}|n\in\mathbb{N}\}$ is a countable partition of $X$. Define a function $f:X\rightarrow\mathbb{N}$ where if $x\in X$, then $f(x)$ is the smallest
natural number such that $x\in(A_{1}\cup...\cup A_{f(x)})^{\circ}$. Then one can see that $f(y)\leq f(x)$ whenever $y\in(A_{1}\cup...\cup A_{f(x)})^{\circ}$. Therefore the function $f$ is upper semicontinuous. Furthermore, we must have $f(x)\geq n_{x}$. Therefore, $f$ is the required function.
$5\rightarrow 4$. This is trivial.
$4\rightarrow 3$. Assume that $P=\{A_{n}|n\in\mathbb{N}\}$ is a countable partition of $X$. Let $n_{x}=n$ whenever $x\in A_{n}$. Then there is an upper semicontinuous function $f:X\rightarrow\mathbb{R}$ with $f(x)\geq n_{x}$ for all $x\in X$. Then $\{f^{-1}(-\infty,n)|n\in\mathbb{N}\}$ is a countable open cover of $X$. However, if $n\in\mathbb{N}$ and $x\in f^{-1}(-\infty,n)$, then $n_{x}\leq f(x)<n$, so $x\in A_{n_{x}}\subseteq A_{0}\cup...\cup A_{n-1}$. Therefore $f^{-1}(-\infty,n)\subseteq A_{0}\cup...\cup A_{n-1}$.
$3\rightarrow 2$. This is trivial.
$2\rightarrow 1$. Let $x\in X$. We claim that the neighborhood filter $\mathcal{N}(x)$ is $\sigma$-complete. Assume that $\{A_{n}|n\in\mathbb{N}\}$ is a descending sequence of neighborhoods of $x$. Let $P=\{A_{0},A_{0}\setminus A_{1},...,A_{n}\setminus A_{n-1},...,\bigcap_{n}A_{n}\}$. Then since $P$ is a partition of $X$, there is some $N$ where $A_{0}^{c}\cup(A_{0}\setminus A_{1})\cup...\cup(A_{N-1}\setminus A_{N})\cup
\bigcap_{n}A_{n}\in\mathcal{N}(x)$. Therefore $A_{0}^{c}\cup(A_{0}\setminus A_{1})\cup...\cup(A_{N-1}\setminus A_{N})\cup
\bigcap_{n}A_{n}=A_{N}^{c}\cup\bigcap_{n}A_{n}\in\mathcal{N}(x)$, so
$A_{N}\cap(A_{N}^{c}\cup\bigcap_{n}A_{n})=\bigcap_{n}A_{n}\in\mathcal{N}(x)$ as well. Therefore $\mathcal{N}(x)$ is $\sigma$-complete. We now claim that the Boolean algebra $P(X)/\mathcal{N}(x)$ is finite. Otherwise, there is a countable partition $\{a_{n}|n\in\mathbb{N}\}$ of $P(X)/\mathcal{N}(x)$. However, one can easily
show that this implies there is a countable partition $\{A_{n}|n\in\mathbb{N}\}$ of $P(X)$ where $a_{n}=A_{n}/\mathcal{N}(x)$ for all $n$. However, for all natural numbers $n$, we have $(A_{1}\cup...\cup A_{n})/\mathcal{N}(x)=a_{1}\vee...\vee a_{n}\neq 1$, so $A_{1}\cup...\cup A_{n}\not\in\mathcal{N}(x)$. This is a contradiction. Therefore, $P(X)/\mathcal{N}(x)$ is finite, so $\mathcal{N}(x)$ is the intersection of finitely many $\sigma$-complete ultrafilters.
QED
If there does not exist a measurable cardinal, then these spaces are fairly trivial. In fact, these spaces are essentially the pre-ordered sets $X$ where if $x\in X$, then $\{y\in X|y\leq x\}$ is finite. The $T_{1}$ such spaces are the partially ordered sets $X$ where if $x\in X$, then $\{y\in X|y\leq x\}$ is finite. If $X$ is a preordered set, then the collection of lower sets forms a topology.
On the other hand, if we assume that there is a measurable cardinal, then these spaces are much more interesting. Therefore in the remainder of this discussion, assume that there is a least measurable cardinal $\mu$. If a space has arbitrarily small lower semicontinuous functions, then the intersection of less than $\mu$ open sets is open. In particular, every regular space with arbitrarily large upper semicontinuous functions is a $P$-space. Thus, every regular space with arbitrarily large upper semicontinuous functions is zero-dimensional.
If $\mathcal{U}_{x}$ is a $\sigma$-complete ultrafilter on $X$ for each $x\in X$ and
$\mathcal{V}$ is also a $\sigma$-complete ultrafilter, then let $\sum_{x\in X}^{\mathcal{U}}\mathcal{U}_{x}$ be the subset of $P(X)$ where $R\in\sum_{x\in X}^{\mathcal{U}}\mathcal{U}_{x}$ if and only if $\{x\in X|R\in\mathcal{U}_{x}\}\in\mathcal{U}$.` Clearly $\sum_{x\in X}^{\mathcal{U}}\mathcal{U}_{x}$ is a $\sigma$-complete ultrafilter on $X$.
We can even describe spaces with arbitrarily large upper semicontinuous functions combinatorially in terms of the convergent ultrafilters as follows.
Now assume that $X$ is a set and $M_{x}$ is a collection of finitely many $\sigma$-complete ultrafilters on the set $X$ for each $x\in X$. Then call the system $(M_{x})_{x\in X}$ additive if 

each $M_{x}$ contains the principal ultrafilter $\{R\subseteq X|x\in R\}$ and
If $\mathcal{U}\in M_{x}$ and $\mathcal{U}_{y}\in M_{y}$ for $y\in Y$, then
$\sum_{y\in X}^{\mathcal{U}}\mathcal{U}_{y}\in M_{x}$ as well.

It turns out that if $X$ is a set and $M_{x}$ is a collection of finitely many $\sigma$-complete ultrafilters on $X$ for each $x\in X$, then the system $(M_{x})_{x\in X}$ is additive if and only if there is a topology on the set $X$ such that $\mathcal{N}(x)=\bigcap M_{x}$ for each $x\in X$.
We conclude that one may consider topological spaces with arbitrarily small lower semicontinuous functions as additive systems $(M_{x})_{x\in X}$ where each $M_{x}$ is a finite set of $\sigma$-complete ultrafilters on $X$. Furthermore, the Hausdorff spaces with arbitrarily small lower semicontinuous functions correspond to the additive systems $(M_{x})_{x\in X}$ where $M_{x}\cap M_{y}=\emptyset$ whenever $x,y$ are distinct points in $X$.<|endoftext|>
TITLE: Irreducible representation decomposition of tensor on manifold with metric
QUESTION [6 upvotes]: I'm aware that for some tensor product space, Schur-Weyl duality lets me decompose the space into irreducible representations by looking at irreps of the symmetric group. The simplest example is $V\otimes V\cong \mathrm{Sym}^2(V)\oplus \Lambda^2(V)$. In physicists' notation (warning, I am a physicist), we would write $T_{ab} = T_{(ab)} + T_{[ab]}$.
However, when talking about tensors on a manifold $M$ with metric $g$, we can also take traces and so further reduce symmetric products into a trace and trace-free part, i.e. $T_{ab} = \frac{1}{d}g_{ab}T + [T_{ab}]^{\tiny{STF}} + T_{[ab]}$ where $T\equiv g^{ab}T_{ab}$, $d$ is the dimension of the manifold, and $[T_{ab}]^{\tiny{STF}} = T_{(ab)}-\frac{1}{d}g_{ab}T$. This seems to be relevant because because $g$ is an invariant symbol of $SO(p,q)$ where $(p,q)$ is the signature of $g$. There will also be an alternating tensor of highest rank which will be an invariant symbol, and this could also appear in the decomposition.
Schur-Weyl doesn't seem to say anything about metrics, trace/trace-free decompositions, etc. What is the general decomposition here? Is there an algorithm to follow?

REPLY [3 votes]: There is a "Schur-Weyl theory" for representation of $O(n)$ and $Sp(n)$. The group algebra of the symmetric group is replaced by the Brauer algebra. Basically, you first decompose your tensor product with respect to "the number of traces the vectors contain" and then for each such part you can use the classical $GL$ decomposition. 
The relevant representation theory over complex numbers is treated in a book by Wallach and Goodman Symmetry, Representations, and Invariants. In particular look at the appendix F on the linked web page. Alternatively, you can look into the older version of this book which was published under the name Representations and Invariants of the Classical Groups. 
You should be careful when dealing with representation of noncompact real groups.<|endoftext|>
TITLE: Motivic generalisation of Neron-Ogg-Shaferevich criterion
QUESTION [15 upvotes]: Given a variety $X$ over $\mathbb{Q}$ with good reduction at $p$, proper smooth base change tells us that its $l$-adic cohomology groups are unramified at $p$ (and I'd guess some $p$-adic Hodge theory tells us its p-adic cohomology is crystalline).
My question is to what extent it's possible to find a converse to this statement. More precisely, I have yet to see a counterexample to the following "conjecture" (though I still suspect it's wrong).
"Conjecture": Let $K$ be a number field, $p$ and $l$ primes, and $V$ a geometric (say, coming from the variety $Y$) $l$-adic representation of $G_K$ that is unramified/crystalline at $\mathfrak{p}|p$. Then there exists a smooth proper variety $X$ such that $X$ has good reduction at $\mathfrak{p}$ and $V$ can be cut out of the cohomology of $X$.
From googling around, the things I know so far are (at least for $l \not= p$):

If $Y$ is an abelian variety, the classical Neron-Ogg-Shafarevich condition means that $Y$ itself is a witness to the conjecture.
We can take torsors for abelian varieties with no $K$-rational points, and these can have the same representations, but fail to have good reduction (in this paper http://arxiv.org/abs/math/0605326 of Dalawat).
There exist curves which have bad reduction, but whose Jacobians have good reduction.

If anyone knows any more about this story I'd be interested to hear. Ultimately I guess it would be nice to have a definition for when a motive is unramified/has good reduction, and cohomologically this surely has to mean unramified/crystalline, but it would be nice if this could always be realised "geometrically".
Thanks,
Tom.

REPLY [4 votes]: Unfortunately I don't know much about motives in genereal, but this might be relevant to your question. One result of my thesis, that I am currently writing, is to prove Neron-Ogg-Shafarevich for 1-motives. The proof is not particularly difficult and it ultimately reduces to the corresponding results for the components of the 1-motive. I will describe below what good reduction means in this particular case.
A 1-motive $M = [u\colon Y\to G]$ over a scheme $S$ consists of a group scheme $Y$, which is locally etale isomorphic to $\mathbb{Z}^r$, a group scheme $G$ which is an extension of an abelian scheme $A$ by a torus $T$ and a homomorphism $u\colon Y\to G$. If $S$ is the spectrum of a field $K$, this means that $Y$ is a free finitely-generated $\mathbb{Z}$-module with a continuous action of the absolute Galois group $\Gamma_K$ and that $u$ is a $\Gamma_K$-equivariant homomorphism $u\colon Y\to G(\bar K)$.
If $R$ is a complete discrete valuation ring with a fraction field $K$ we say that a 1-motive $M$ over $K$ has good reduction if there exists a 1-motive $\widetilde{M}$ over $R$ whose generic fiber is isomorphic to $M$. This is equivalent to the following:

$G$ has good reduction $\widetilde{G}$ over $R$, which is equivalent to saying that both $A$ and $T$ have good reduction;
The action of $\Gamma_K$ on $Y$ is unramified;
The image of $u(Y)$ is contained in the set of those points in $G(K')$ which can be reduced, where $K'/K$ is some finite field extension. Equivalently, $u(Y)$ is contained in the maximal compact subgroup of $G(K')$;

With this definition, the criterion of Neron-Ogg-Shafarevich is as follows: Let $l,p$ be primes, with $l\neq p$. A 1-motive $M/\mathbb{Q}$ has good reduction mod p if and only if the Tate module $T_l(M)$ is unramified at $p$. For general number fields replace $p$ by a prime ideal.
If you want to learn more about reduction of 1-motives you can look at M. Raynaud's paper 1-Motifs et Monodromie Géométrique.<|endoftext|>
TITLE: Unboundedness of primes in bounded arithmetic
QUESTION [14 upvotes]: Wilkie's well known question asks whether $I\Delta_{0}$ proves the unboundedness of primes. We know that by adding a sentence to $I\Delta_{0}$ which says "the exponential function is total", it is possible to prove the unboundedness of primes. This sentence is $\Pi_{2}$. Suppose $\Pi_{1}\text{-Th}(\mathbb{N})$ denotes the set of all $\Pi_{1}$ sentences that are true in $\mathbb{N}$. My question is:
Is it known that $I\Delta_{0} 
+\Pi_{1}\text{-Th}(\mathbb{N})$ proves the unboundedness of primes?

REPLY [8 votes]: Yes, for trivial reasons: $\forall x>0\,\exists y\le2x\,(y>x\land\mathrm{Prime}(y))$ is a true $\Pi^0_1$ sentence which implies the unboundedness of primes. (Of course, weaker bounds than Bertrand’s postulate would also do, such as $y\le x^2$.)<|endoftext|>
TITLE: What is knot contact homology?
QUESTION [8 upvotes]: Recently, it was conjectured by the paper of Aganagic and Vafa that the $Q$-deformed $A$-polynomials can be identified with the augmentation polynomials of the knot contact homology. The $Q$-deformed $A$-polynomial of a knot $K$ can be obtained by finding the difference equation of minimal order for the colored HOMFLY polynomials of the knot $K$. This conjecture seems to hold true for torus knots and twist knots. However, I do not understand what  the knot contact homology is. 
First of all, the knot contact homology describes knot invariants as invariants of the Legendrian submanifolds in the contact manifold. A knot is realized by an intersection of the cosphere bundle $ST^∗M$ of a 3-manifold $M$ with the unit conormal bundle $\Lambda_K$ where $ST^∗M$ admits a contact structure.

1) Is there any way to visualize an intersection of  $ST^∗M$ with  $\Lambda_K$?

The knot contact homology is constructed by the Legendrian differential graded algebra (DGA)

2) Why do you need DGA to obtain homology theory invariant under Legendrian isotopy?

From the combinatorial definition (Appendix B of the paper),  I cannot see the reason why this is isomorphic to Legendrian DGA.  Although the differentials are determined by the braiding data of a knot, it seems to me that there is no contact structure involved.

3) Could the isomorphism between the two DGA be explained in layman's terms?

I do not understand what the augmentation polynomials of the knot contact homology are. 

4) Is there any relation between augmentation polynomials and Porincare-Chekanov polynomials?

In addition, 

5) I would like to know if there is an explicit connection of knot contact homology to other knot homologies such as Khovanov-Rozansky and HOMFLY homology.

REPLY [5 votes]: Check out these lecture notes (and the references listed therein) http://www.renyi.hu/~cast2012/ng-cast.pdf
1) Visualize elements of $ST^{*}\mathbb{R}^{3}$ as unit length vectors -- not necessarily based at the origin -- in $\mathbb{R}^{3}$.  The projection $\pi$ onto $\mathbb{R}^{3}$ send each vector to its basepoint. The unit conormal bundle bundle $\Lambda_{K}$ consists of those vectors orthogonal to $K$.  If you apply the time-$\epsilon$ Reeb flow (=geodesic flow) to $\Lambda_{K}$ for $0<\epsilon$ very small and then apply $\pi$, you will get the boundary of a tubular neighborhood of $K$ in $\mathbb{R}^{3}$.
2) I take it that this question means ``Why can't you define homology groups by counting Reeb chords and holomorphic disks (which are easier to work with than DGAs)?''  You can only define homology groups by counting holomorphic strips (disks with 2 boundary punctures) if you can say that the only way that strips break (in a Gromov limit) is into a pair of strips.  This if often possible for Lagrangian intersection problems by imposing various geometric constraints which rule out undesireable degenerations of homolorphic disks.  In this Legendrian case, this does not work if you are to obtain a Legendrian isotopy class invariant.  Play around with some examples of Legendrian knots as in Etnyre-Ng-Sullivan's paper and this will become clear.
3) Not really: (Mild) You have to first identify $ST^{*} \mathbb{R}^{3}$ with the 1-jet space $ J^{1}S^{2}=\mathbb{R}\times T^{*}S^{2}$ of $S^{2}$.  This isn't too hard to intuit (the zero section of the 1-jet space is identified with $\pi^{-1}$ of the origin in $\mathbb{R}^{3}$).  (Medium) Now think about what the conormal bundle $\Lambda_{\mu}$ of the unknot $\mu$ looks like in this 1-jet space.  As an approximation, the projection of this torus onto $S^{2}$ is obtained by applying the geodesic flow to a fiber of the unit tangent bundle of $S^{2}$  for time $t\in[0,2\pi]$. (Spicy) If $K$ is braided about $\mu$, then $\Lambda_{K}$ will be braided about $\Lambda_{\mu}$ in $J^{1}S^{2}$. It's well-known that any knot is braided about the unknot.  (Habanero) Use a direct limit argument and Ekholm's Morse flow trees technique to read off the holomorphic disks from the braiding data.
4) They're both defined by looking at the representations of the DGAs.
5) Ng's papers describe how knot contact homology encodes various knot polynomials.  It is conjectured that you can use it to obtain the HOMFLY-PT polynomial.  As for knot homology theories, I'm sure that people have thoughts on this but as far as I'm aware no conjectures have been anounced.<|endoftext|>
TITLE: Seeking a seemingly missing reference of Casson
QUESTION [6 upvotes]: Floer's paper, An Instanton-Invariant for 3-Manifolds, makes reference to Casson's construction of a topological invariant for homology 3-spheres. He literally references it by placing the bibliographic-tag [C] in one of his sentences... but he doesn't actually list it in the bibliography!
I subsequently search around, and cannot find where Casson originally defines his invariant; only locate papers that talk about his invariant. Does this "paper" not exist? I would think that such an important construction has a foundational paper in existence.

REPLY [9 votes]: According to the following paper, this invariant's introduction is sourced as: A. Casson, Lecture notes, MSRI Lectures, Berkeley, 1985.
The first published discussion appears to be found in: S. Akbulut & J. McCarthy, Casson's invariant for oriented homology 3-spheres, an exposition, Math. Notes, No. 36, Princeton Univ. Press, Princeton, 1990.

"I don't know if he [Casson] has ever written his notes. I attended his lectures at MSRI then I went back to MSU and tried to fill in the details together with McCarthy, those notes are the result." S. Akbulut

"Here is a pdf of hand-written notes of T. Cochran's. Page two includes annotated notes by Boyer, and I got the notes from my friend and colleague, A. Nicas." H. Boden<|endoftext|>
TITLE: Has any attempt been made to classify finite groupoids?
QUESTION [20 upvotes]: I recently stumbled upon the Mathieu groupoid and I found them fascinating.
It appears as a subset of $S_{13}$ which is not closed under multiplication, but it turns out to be a groupoid with 13 objects. The "first" sporadic finite simple group $M_{12}$ appear as the Automorphism of a point. One can further find $M_{11}$ from this.
Now that people classified finite simple groups with decades of effort, I'm wondering if there has been any attempt to classify finite groupoids. 
(Maybe some results like this is already implied by the classification of finite simple groups, but the groupoid $M_{13}$ seems so amazing, so I got curious.)

REPLY [11 votes]: Everything that's been written so far about the classification of finite groupoids reducing to the classification of finite groups is true but, I think, misleading. In order to actually produce a list of finite groups from a finite groupoid $X$ you need to choose a basepoint in each connected component of $X$. For example, to produce $M_{12}$ from $M_{13}$ you need to choose one of $13$ points. 
There are various reasons it's undesirable to make such choices, with maybe the most practical one being that they are often unavailable once you introduce extra structure. For example, given a group $G$ you might want to study groupoids equipped with a $G$-action. These are strictly more interesting than disjoint unions of groups equipped with a $G$-action, and the reason is precisely that a $G$-groupoid need not have any $G$-invariant basepoints. 
This is in fact an issue in the present example: 

The Mathieu groupoid $M_{13}$ is equipped with an action of $\text{SL}_3(\mathbb{F}_3)$ which does not fix any points. When you pick a basepoint and get $M_{12}$ back you lose this action.

Another example of this phenomenon is the following: the configuration space $\text{Conf}_k(\mathbb{R}^2)$ of $k$ ordered points in the plane is naturally an Eilenberg-MacLane space $K(P_k, 1)$, where $P_k$ is the pure braid group. On the other hand, $\text{Conf}_k(\mathbb{R}^2)$ clearly has an action of $S_k$ on it given by permuting points. This action does not fix any basepoints, and so it does not induce an action on $P_k$. 
It does induce an action on a groupoid equivalent to $P_k$ with $k!$ points given by taking the fundamental groupoid of $\text{Conf}_k(\mathbb{R}^2)$ at a set of basepoints which are setwise fixed under the action of $S_k$. A more complicated choice of basepoints setwise fixed under the action of $S_k$ can be used to give a model of the little disks operad as an operad in groupoids. This operad cannot be simplified to an operad in groups, despite the fact that all of the groupoids involved are connected, precisely because of basepoint issues (the basepoints must be chosen not only compatibly with the $S_k$-actions but with operadic composition). 
Edit: One way to measure the interestingness of the $S_k$ action on $\text{Conf}_k(\mathbb{R}^2)$ is that it does not even fix a basepoint in the homotopy coherent sense: equivalently, there is a corresponding short exact sequence
$$1 \to P_k \to B_k \to S_k \to 1$$
that does not split. On the other hand, as pointed out in the comments, any extension of $\text{SL}_3(\mathbb{F}_3)$ by $M_{12}$ must be trivial.<|endoftext|>
TITLE: Exact DG Poisson algebra
QUESTION [13 upvotes]: A symplectic manifold gives rise to a Poisson algebra.  If the symplectic form is exact,
how is this revealed in the algebra?

REPLY [6 votes]: It seems to me that there is a (quasi-)isomorphism between the de Rham algebra and the dg algebra of polyvector fields equipped with the differential $[\pi,-]$ (where $\pi$ is the Poisson structure corresponding to the symplectic form). 
Through this isomorphism the equation $d\omega=0$ is sent to $[\pi,\pi]=0$, and the equation $\omega=d\lambda$ is sent to $\pi=[\pi,V]$, where $V$ is a vector field. 
On the level of the Poisson algebra of functions it tells you that for any two functions $f,g$, we have (up to a sign)
$$
\{f,g\}=V(\{f,g\})-\{V(f),g\}+\{f,V(g)\}
$$
Algebraically you can say that there is a derivation $V$ for the product such that the Poisson bracket is its own derived bracket w.r.t. $V$.<|endoftext|>
TITLE: Constructible sets in Hausdorff spaces
QUESTION [9 upvotes]: In an attempt to write a proof by contradiction, I end up with a space $X$ with the following properties:
(0) $X$ is nonempty,
(1) $X$ is Hausdorff,
(2) $X$ has no isolated points,
(3) every subspace of $X$ is constructible (finite union of locally closed subsets).

Is this indeed a contradiction? 

It would suffice to know that any $X$ with properties (1) and (2) has a dense subset with dense complement: such a set cannot be constructible unless $X=\emptyset$. [Edit: there are counterexamples to this, see the comment by Yves Cornulier]

REPLY [6 votes]: No, it is not a contradiction.
A space $X$ is called submaximal if every subset of $X$ is locally closed (and hence constructible). It is easy to see that spaces with only finitely many non-isolated points are submaximal. Finding a submaximal Hausdorff space with no isolated points seems harder but there are plenty of those too.
It is a nice exercise to show that any maximal space is in fact submaximal. A Hausdorff space $(X,\tau)$ is called maximal if it has no isolated points but any finer topology $\tau' \supset \tau$ has isolated points. With a standard use of Zorn´s lemma you can show that any topology without isolated points is contained in a maximal topology. This gives you a way of "constructing" a lot of submaximal Hausdorff spaces with no isolated points. Some of them are even Tychonoff spaces (see E. van Douwen, "Applications of maximal topologies", Topology Appl., 1993), although these are a lot harder to get.<|endoftext|>
TITLE: K(r)-localization and monochromatic layers in the chromatic spectral sequence
QUESTION [31 upvotes]: While preparing some lecture notes, I had a basic point of confusion come up that I haven't been able to settle.
The $BP$-Adams spectral sequence (or $p$-local Adams-Novikov spectral sequence) for the sphere begins with $E_2$-page $$E_2^{*, *} = \operatorname{Ext}^{*, *}_{BP_* BP}(BP_*, BP_*)$$ and converges to $\pi_* \mathbb{S} \otimes_{\mathbb{Z}} \mathbb{Z}_{(p)}$.
There are a variety of cool periodicities visible in this $E_2$-page, which we can organize via the following secondary spectral sequence.  There is an ascending chain of $(BP_* BP)$-invariant ideals for $BP_*$ given by $I_r = (p, v_1, \ldots, v_{r-1})$, connected to one another by the short exact sequences $$0 \to BP_* / I_r^\infty \to v_r^{-1} BP_* / I_r^\infty \to BP_* / I_{r+1}^\infty \to 0.$$  The quotient $BP_* / I_r^\infty$ is thought of as the closed substack of the moduli of formal groups detected by the ideal sheaf corresponding to $I_r$ together with its formal neighborhood inside the parent stack.  Applying $\operatorname{Ext}$ and stringing the resulting long exact sequences together, one arrives at the (trigraded) chromatic spectral sequence (CSS): $$E_1^{r, *, *} = \operatorname{Ext}^{*, *}_{BP_* BP}(BP_*, v_r^{-1} BP_* / I_r^\infty) \Rightarrow \operatorname{Ext}^{*, *}_{BP_* BP}(BP_*, BP_*).$$  Much of the fun in chromatic homotopy theory after this point comes from identifying the groups in this $E_1$-page as other sorts of things, like certain group cohomologies.
Shifting gears somewhat, Bousfield localization at the Johnson-Wilson $E(r)$-theories and the Morava $K(r)$-theories is meant to perform the same organization at the level of homotopy types.  The spectra $E(\infty)$ and $E(0)$ correspond to $BP$ and to $H\mathbb{Q}$ respectively, so the sequence of localization functors $L_{E(r)}$ are meant to interpolate between rational homotopy theory and the sort of homotopy theory visible to the $p$-local Adams-Novikov spectral sequence.
There are two ways to study these functors as $r$ increases.  First, there there is a natural map $L_{E(r)} X \to L_{E(r-1)} X$.  Its homotopy fiber detects the difference between these two spectra, denoted $M_r X$ and called the $r$th monochromatic layer of $X$.  Second, there is a pullback square, dubbed chromatic fracture:
$$\begin{array}{ccc} L_{E(r)} X & \to & L_{K(r)} X \\ \downarrow & & \downarrow \\ L_{E(r-1)} X & \to & L_{E(r-1)} L_{K(r)} X. \end{array}$$  In both of these situations, you can hope to inductively study the filtering spectra $L_{E(r)} X$ by studying the "filtration layers", which are either $M_r X$ or $L_{K(r)} X$ depending upon your approach.

My question is: How exactly do these two approaches connect to the chromatic spectral sequence?

I suspect that the CSS for $L_{E(R)} X$ looks like the CSS for $X$, after quotienting out  the information in $r$-degrees $r > R$.  I also suspect that the CSS for one of the two of $M_R X$ and $L_{K(R)} X$ looks like that for $L_{E(R)} X$, after additionally quotienting out the information in $r$-degrees $r < R$.  However, I can't seem to make the pieces line up.  For instance, Prop. 7.4 of Hopkins, Mahowald, and Sadofsky's Constructions of elements in Picard groups suggests that this description holds for $L_{K(R)} \mathbb{S}$, as that statement matches their Adams-Novikov spectral sequence converging to $\pi_* L_{K(R)} \mathbb{S}$ --- just as one would expect from a collapsing chromatic spectral sequence.  On the other hand, the bottom corner of the fracture square is of the form $L_{E(R-1)} L_{K(R)} X$, and this description seems to say that its CSS is empty, which doesn't sound right.
I'd appreciate someone setting me straight about this.  Thanks!

REPLY [18 votes]: I have at least a partial answer to my question.  It's fairly complicated, and pieces of it are written down in a variety of places, so I'm going to do what I can to be thorough.  Before we do anything involving spectral sequences at all, it will turn out to be useful to have a certain pair of families of $BP_* BP$-comodules at our disposal, defined by the formulas $$N_r^s = BP_* / \langle p, \ldots, v_{r-1}, v_r^\infty, \ldots, v_{r+s-1}^\infty \rangle,$$ $$M_r^s = v_{r+s}^{-1} BP_* / \langle p, \ldots, v_{r-1}, v_r^\infty, \ldots, v_{r+s-1}^\infty\rangle = v_{r+s}^{-1} N_r^s.$$  In fact, these formulas even make sense on the level of spectra, since $N_0^0$ can be taken to be $BP$, $M_r^s$ appears as a mapping telescope built out of $N_r^s$, and there are cofiber sequences (/ short exact sequences) $$N_r^s \xrightarrow{\cdot v_{r+s}^\infty} M_r^s \to N_r^{s+1},$$ $$N_r^r \xrightarrow{\cdot v_r} N_r^r \to N_{r+1}^{r+1}.$$  The $BP_* BP$-comodules are recovered by taking homotopy groups.
The most fundamental of all the spectral sequences in play was brought up by Drew in the comments above.  The chromatic tower is a tower of fibrations $$\cdots \to L_{E(n+1)} \mathbb{S}^0 \to L_{E(n)} \mathbb{S}^0 \to \cdots \to L_{E(0)} \mathbb{S}^0,$$ and the fibers of these maps define the monochromatic layers.  Applying $\pi_*$ to the diagram produces a spectral sequence of signature $$\pi_* M_r \mathbb{S}^0 \Rightarrow \pi_* \mathbb{S}^0_{(p)}.$$  To study this spectral sequence, there are two reasonable-sounding things to do involving the homology theory $BP_*$:

Apply $BP_*$ to the chromatic tower diagram and study the resulting spectral sequence.
Use the $BP$-Adams spectral sequence to compute $\pi_* M_n \mathbb{S}^0$ from $BP_* M_n \mathbb{S}^0$.

These both turn out to be relevant, and they both rest upon a certain input, computed by Ravenel.  Namely, he shows how to compute $BP_* L_{E(r)} \mathbb{S}^0$ and the surrounding pieces:

Theorem 6.2 (Ravenel, Localizations with respect to certain periodic cohomology theories): The going-around maps $N_0^{s+1} \to \Sigma N_0^s$ compose to give a map $\Sigma^{-s-1} N_0^{s+1} \to N_0^0 = BP$.  The cofiber of this map can be identified as $$\Sigma^{-s-1} N_0^{s+1} \to BP \to L_{E(s)} BP.$$  Moreover, the rotated triangle $BP_* \to \pi_* L_{E(s)} BP \to \pi_* \Sigma^{-s} N_0^{s+1}$ is split short exact.  (There's an exception in the bottom case, where $BP_* L_{E(0)} \mathbb{S}^0 = BP_* \otimes \mathbb{Q}$.)

Applying the octahedral axiom to the pair $\Sigma^{-s-1} N_0^{s+1} \to \Sigma^{-s} N_0^s \to BP$ and then applying $BP_*$-homology gives the calculation $$BP_* M_s \mathbb{S}^0 = \Sigma^{-s} M_0^s.$$
Now we can address 1. and 2.:

If we delete the boring $BP_*$ summands in $BP_* L_{E(n)} \mathbb{S}^0$, then the exact couple coming from applying $BP_*$-homology to the chromatic tower just falls apart into a string of short exact sequences of $BP_* BP$-comodules.  Now, we know that $H^{*, *} N_0^0$ is the input to the $BP$-Adams spectral sequence computing $\pi_* \mathbb{S}^0$, and applying $H^{*, *}$ to this diagram of short exact sequences of $BP_* BP$-modules yields a spectral sequence of signature $$E_1^{r, *, *} = H^{*, *} M_0^r \Rightarrow H^{*, *} N_0^0.$$  This is the usual chromatic spectral sequence, as stemming from algebraic considerations.
Applying the $BP$-Adams spectral sequence to compute $\pi_* M_r \mathbb{S}^0$ begins with the computation of $\operatorname{Ext}^{*, *}_{BP_* BP}(BP_*, BP_* M_r \mathbb{S}^0)$, which we now know to be isomorphic to $H^{*, *} M_0^r$.  (In fact, this spectral sequence is supposed to collapse for $p \gg n$, e.g., $p \ge 5$ for $n = 2$.)

Another part of this whole story is how the $K(r)$-local sphere plays into this picture.  Now, there is also an Adams-type spectral sequence computing $\pi_* L_{K(r)} \mathbb{S}^0$, and it has signature $$\operatorname{Ext}^{*, *}_{\Gamma}(K(n)_*, K(n)_*) \Rightarrow \pi_* L_{K(n)} \mathbb{S}^0,$$ where $\Gamma = K(n)_* \otimes_{BP_*} BP_* BP \otimes_{BP_*} K(n)_*$.  Morava's change of rings theorem states that the map $M_r^0 \to K(n)_*$ induces an isomorphism between the sheaf cohomology groups $H^{*, *} M_r^0$ and the $\operatorname{Ext}$-groups in the Adams-type spectral sequence.
The difference, then, between the monochromatic sphere and the $K(r)$-local sphere is recorded in the index swap $M_0^r$ and $M_r^0$ --- i.e., whether the generators below $v_r$ are taken to be zero or to be torsion.  The difference in these two situations is of course itself recorded as a spectral sequence: the inclusions $$v_{r+s}^{-1} BP_* / \langle p, \ldots, v_{r-1}, v_r^j, v_{r+1}^\infty, \ldots, v_{r+s-1}^\infty \rangle \xrightarrow{\cdot v_r} v_{r+s}^{-1} BP_* / \langle p, \ldots, v_{r-1}, v_r^{j+1}, v_{r+1}^\infty, \ldots, v_{r+s-1}^\infty \rangle$$ all have cofiber $M_{r+1}^{s-1}$, regardless of choice of index $j$.  Applying $H^{*, *}$ to the string of inclusions (extended to cofiber sequences) yields the $v_r$-Bockstein spectral sequence, of signature $$H^{*, *} M_{r+1}^{s-1} \otimes \mathbb{F}_p[v_r] / v_r^\infty \Rightarrow H^{*, *} M_{r}^{s}.$$  So, there is a length $r$ string of $v_*$-Bockstein spectral sequences beginning with $H^{*, *} M_r^0$ and concluding with $H^{*, *} M_0^r$.
Some things not included in this answer are: 

What happens when analyzing the chromatic tower of spaces other than the sphere spectrum?
What is the relevance of the corner space $L_{E(r-1)} L_{K(r)} \mathbb{S}^0$ in terms of the chromatic spectral sequence?
What parts of this story can be made sense of mutatis mutandis when replacing $BP$ with other spectra in the same family, like $E(r)$?  My expectation (as the comments reveal) is that there should be an analogue of the algebraic chromatic spectral sequence for $E(R)$-homology, which is the truncation of the usual one for $r \le R$.  (In October I even thought I knew how to prove this, but I've since forgotten.  This is the least interesting question of the bunch.)<|endoftext|>
TITLE: Applications of Frobenius theorem and conjecture
QUESTION [35 upvotes]: A theorem of Frobenius states that if $n$ divides the order of a finite group $G$, then the number of solutions to $x^n = 1$ in $G$ is a multiple of $n$. Frobenius conjectured that if the number of solutions is exactly $n$, then the set of solutions form a characteristic subgroup of $G$. The conjecture was eventually proved in the 90's, and the full proof uses the classification of finite simple groups.
The theorem feels a bit isolated for me.. I'm not sure how the conjecture fits into a wider context, either. What is their importance, if any? Are there any good examples of applications of the theorem or the conjecture? If I'm interested in finite groups, why should I care about the theorem or the conjecture, other than that they are kind of neat?
One example I know is that if $G$ has every Sylow subgroup cyclic, then with Frobenius theorem we can show that the Sylow subgroup corresponding to the largest prime divisor of $G$ is normal. Also (this one is too easy, but I like it) for any prime $p$, the number of elements satisfying $x^p = 1$ in the symmetric group $S_p$ is $(p-1)! + 1$, so Frobenius theorem implies $(p-1)! \equiv -1 \mod p$.

REPLY [12 votes]: An application of this theorem to $p$-adic analysis is the $p$-integrality of the coefficients of the Artin-Hasse exponential
$$
{\rm AH}_p(X) = e^{X + X^p/p + X^{p^2}/p^2 + \cdots}.
$$
That means when this series is expanded as $\sum_{k \geq 0} a_kX^k$, the coefficients $a_k$ don't have their denominator divisible by $p$. This is obvious for $k = 0$ since $a_0 = 1$.  We have $a_k = 1/k!$ for $1 \leq k \leq p-1$, so $a_1,\dots,a_{p-1}$ are all $p$-integral. And $a_p = ((p-1)!+1)/p!,$ which is $p$-integral by Wilson's theorem. To handle the general case we'll use the Frobenius theorem.
For $k \geq 1$, a non-obvious combinatorial/probabilistic formula for the coefficients is 
$$
a_k = \frac{\#\{g \in S_k : g \text{ has } p\text{-power order}\}}{k!}.
$$
Using the Frobenius theorem with $n$ being the largest power of $p$ dividing $|S_k|$, the numerator of $a_k$ is divisible by the largest power of $p$ in the denominator. Hence each $a_k$ is $p$-integral.
There are simpler proofs of the $p$-integrality of the coefficients of this series. See the Wikipedia page on the Artin-Hasse exponential.<|endoftext|>
TITLE: Reference request: parabolic PDE
QUESTION [14 upvotes]: I want to learn about parabolic PDE and it seems to me that there is no established reference as far as where one should look if one wants to learn the subject from basics.
I think I have a firm grip on elliptic PDE after going through the first part of Gilbarg and Trudinger + some Monge-Ampere stuff. But that concludes my PDE background at this moment.
Can someone provide me with a good textbook for parabolic PDE? Any chunk of information will be appreciated.
EDIT: I would like to learn about parabolic PDE arising in geometry, mostly the Kahler-Ricci flow and related questions. But since I am new to this approach perhaps a more broad introduction would be appropriate.

REPLY [19 votes]: I have to kindly dissent from Deane Yang's recommendation of the books that I coauthored. The reason being that the question by The Common Crane is about basic references for parabolic PDE and he/she is interested in Kaehler--Ricci flow, where many cases can be reduced to a single complex Monge-Ampere equation, and hence the nature of techniques is quite different than that for Riemannian Ricci flow. 
I would second user23078's recommendation of Avner Friedman's Partial Differential Equations of Parabolic Type (beautifully and carefully written), and Andras Batkai's recommendation of Nick Krylov's book Lectures on Elliptic and Parabolic Equations in Sobolev Spaces, AMS Graduate Studies in Mathematics Volume 96 [11-18-2013: this is a correction: earlier I quoted the wrong book [Hoelder spaces]] (my colleague Lei Ni will use this book next quarter for the second part of the graduate PDE course at UCSD). To the books mentioned in the above posts on elliptic and general PDE, I would add the book by Qing Han and Fanghua Lin titled Elliptic Partial Differential Equations (in my opinion, it is a really good book; I've been using it in the PDE graduate class this quarter).
For many aspects of Ricci flow, perhaps the same may be said for Kaehler--Ricci flow, one does not need to know much parabolic PDE per se (although knowing more always helps). A fair percentage of estimates in Ricci flow are what one could call Bochner formulas. In this sense, it may be helpful to see some classic examples. A few that come to mind (perhaps not representative), are:
(1) The Bochner formula for 1-forms on a closed Riemannian manifold, which shows that there are no harmonic 1-forms if $\operatorname{Ric}>0$. 
(2) The Weyl estimate used in the Weyl embedding problem; see e.g. Nirenberg's paper.
(3) Lichnerowicz formula for estimating $\lambda_1$ when $\operatorname{Ric}$ is bounded from below by a positive constant.
(4) For modern geometric analysis, the works of S.T. Yau and his coauthors on the many applications of PDE (especially the maximum principle) to problems in geometry. For parabolic equations, his paper with Peter Li on differential Harnack estimates for heat-type equations is an absolute must read. Anecdotally, I was once in Nick Krylov's house in Minnesota and I wandered into his basement and I went into his study. There, sitting prominently on the center of his table, was Li and Yau's paper. I thus learned a trade secret.
(5) Regarding Kaehler--Ricci flow, I would look into the specific techniques and calculations that workers in the field use and perform. For some of the references they quote, they may just use the techniques; for other references that they quote, they may just use the results. It may be more efficient to focus on those aspects that are most pertinent to what you are studying.
The above comments only refer to a minuscule, albeit fundamental, part of the landscape.<|endoftext|>
TITLE: Parameter space for complete intersections and their discriminant
QUESTION [11 upvotes]: Consider globally complete intersections in $\mathbb{P}^n$, of codimension $k$, of some fixed multi-degree $(d_1,\dots,d_k)$. 

Is there some nice (i.e. "explicit") parameter space for them?
(even if all the $d_i$'s are distinct one cannot take just the product of linear systems :)
Or, does the corresponding locus in the Hilbert scheme admit some nice/explicit description?
(a relevant question on this locus in Hilbert scheme)
Consider the discriminant in the parameter space of complete intersections, i.e. the locus parametrizing singular c.i.'s. I need standard results: it is an irreducible, reduced hypersurface, singular in codimension one etc. (I guess many of these properties do not depend on the particular choice of parameter space.) Where can I read about this? 
Somehow I did not find this in Gel'fand-Kapranov-Zelevinsky. And the book of Looijenga treats the local case only.

a somewhat related question

REPLY [10 votes]: The description of the Hilbert scheme of complete intersections (obtained by taking in an iterative way open subsets of grassmannian bundles, as explained in the answer above) may be found in part 2.2 of my thesis http://www.math.ens.fr/~obenoist/articles/Thesefinale.pdf. If the distinct degrees are $\delta_1<\dots<\delta_r$, at the $i$-th step you construct a grassmannian bundle corresponding to choosing the equations of degree $\delta_i$ up to multiples of the equations of lower degree, and you remove the closed locus where your equations do not define a complete intersection.
Note that this space is smooth. It is not complete unless $k=1$. However, the construction above provides an explicit compactification if $d_1\leq d_2=\dots=d_k$. This compactification is studied in part 2 of arXiv:1111.1589.
The Picard group of this Hilbert scheme is easy to describe (its rank increases by one with each grassmannian bundle and it is not changed when removing the small closed loci).The class of the discriminant in this Picard group (i.e. the degrees of the discriminant) is calculated in arXiv:1009.0704 Théorème 1.3, building on Gel'fand, Kapranov and Zelevinsky's work.<|endoftext|>
TITLE: non-artificial examples of non-smooth and non-admissible representations of GL_2
QUESTION [10 upvotes]: Let $F$ be a finite degree extension over $\mathbf{Q}_p$ and consider the locally profinite group $G:=GL_2(\mathbf{Q}_p)$. 
P1: Give an interesting example (non-artificial one, i.e., one that arises in real life for a representation theorist) of a non-smooth representation $\rho$ of $G$ on a topological $\mathbf{C}$-vector space $V$. 
So here I would like the representation $\rho:G\rightarrow Aut(V)$ to be at least continuous in the following sense: for every $v\in V$ I want the orbit map $\pi^v:G\rightarrow V$, $g\mapsto \pi(g)(v)$ to be continuous.
P2: Give an interesting example of a smooth representation of $G$ on a topological $\mathbf{C}$-vector space which is not admissible. 
added: Is it possible to construct such representations by inducing over an appropriate closed subgroup of $G$?

REPLY [6 votes]: Another example, this time of a natural representation which is smooth, but not admissible. The group $G$ acts naturally on its Bruhat-Tits tree $X$. Let $C(X)$ be the space of complex-valued functions on (the set of vertices of) $X$ with finite support. Then $G$ has a natural
representation on $C(X)$, which is smooth (the stabilizer of a point is a maximal compact subgroup is a compact maximal, the stabilizer of a function is an intersection of finitely many such maximal compact subgroup hence is open) but not admissible (for example, the space of invariants in $C(X)$ by a compact maximal is the underlying space of the unramified (or spherical) Hecke algebra of $G$, which has infinite dimension over $\mathbb C$.<|endoftext|>
TITLE: A q,t-extension of Plancherel Measure thru Yang-Mills Theory ?
QUESTION [8 upvotes]: Buried in the physics paper by Nekrasov and Okounkov, a strange identity is proven:
$$ \prod_{n > 0} (1 - q^n)^{\mu^2-1} = \sum_{\mathbf{k}} q^{|\mathbf{k}|} \prod_{\square \in k} \left( 1 - \frac{\mu^2}{h(\square)^2}\right) $$
where the left side is a q-series and the right side is the sum over all partitions.  Ihis was proven by physical considerations, evaluating the Yang-Mills partition function in 2 different ways.
The partitions could index representations of the permutation group $S_n$.    We can define measure on partitions, $\mathrm{Irr}(S_n)$ by
$$ \mathbb{P}\_{\mu, t} (\mathbf k) = \prod_{n \geq 1} (1-t^n)^{1-\mu^2} q^{|\mathbf{k}|} \prod_{\square \in k} \left( 1 - \frac{\mu^2}{h(\square)^2}\right) $$
In fact, 3 years later Alexei Borodin explains this formula interpolates between  uniform and Plancherel measures on partitions.

Can this be extended to a q,t-deformation of uniform measure on the permutation group?  Maybe through something similar to Robinson-Schensted correspondence.

REPLY [4 votes]: Recently there is An insertion algorithm associated with q-Whittaker functions 
defining a q-weighted form of the Robinson-Schensted algorithm.  
The MacDonald processes contain many examples of "integrable" combinatorial processes on Gelfand-Tsetlin patterns (see this talk).

q-Whittaker processes
Hall-Littlewood processes
Random matrices
Whittaker processes
Kingman partition structures
schur processes

Many of these feel nature from the point of view of representation theory of symmetric groups.  Macdonald functions are a q,t-deformation of this.<|endoftext|>
TITLE: A question on minimal idempotent ultrafilter on N^2
QUESTION [5 upvotes]: Is there some minimal idempotent ultrafilter  $q \in \beta( \mathbb{N}^2)$ (with respect to the law $"+"$) such that any $A \in q$ is a subset of $\mathbb{N} \times \{ 0 \} $ ? 
(See for example http://www.math.osu.edu/~bergelson.1/vbkatsiveli20march03.pdf for definitions).
Motivation : Van der Waerden theorem can be quickly deduced from a negative answer to this question.

REPLY [6 votes]: Since $(\mathbb N-\{0\})^2$ is a 2-sided ideal in $\mathbb N^2$, it follows that $\beta((\mathbb N-\{0\})^2)$ is a 2-sided ideal in $\beta(\mathbb N^2)$ and therefore contains all of the latter semigroup's minimal idempotents.<|endoftext|>
TITLE: What's the use of Malgrange preparation theorem?
QUESTION [18 upvotes]: The Malgrange preparation theorem,which is the $C^{\infty}$ version of the classical Weierstrass preparation theorem,says that if $f(t,x)$ is a $C^{\infty}$ function of $(t,x)\in\mathbb{R}^{n+1}$ near $(0,0)$ which satisfies 
$$
f=\frac{\partial{f}}{\partial{t}}=\cdots =\frac{\partial^{k-1}{f}}{\partial{t^{k-1}}}=0\quad \frac{\partial^{k}{f}}{\partial{t^{k}}}\ne 0\quad\text{at}(0,0)
$$
Then there exists a factorization
$$ 
f(t,x)=c(t,x)(t^{k}+a_{k-1}(x)t^{k-1}+\cdots +a_{0}(x))
$$
where $a_j$ and $c$ are $C^{\infty}$ functions near $0$ and $(0,0)$ respectively,$c(0,0)\ne 0$ and $a_{j}(0)=0$.As a corollary, there is a division thereom just like the Weierstrass formula.However, unlike the analytic case,this factorization is not unique.The result is said to be highly non-trival even when $k=1$,the difficulty is then the zeros may be lost,For example,$t^{2}+x$ has two real zeros when $x<0$ but none when $x>0$.The proof can be seen in Theorem 7.5.6 in Hormander's The Analysis of linear partial differential operators.
My question is What's the use of Malgrange preparation theorem in mathematics?Is this a verey useful formula in analysis ? Can anyone take some examples to apply this theorem?(In hormander's book,this is used in the method of Stationary Phase).
A quick google search shows that there is also a algebraic version which can be restated as a theorem about modules over rings of smooth, real-valued germs.

REPLY [2 votes]: The origin of the differential preparation theorem as well as its relation with distribution division problem (solved independently by Lars H\"ormander and Stanis\l aw \L ojasiewicz; a version is mentioned in Denis Serre's answer) is nicely explained by Malgrange himself in the paper MR2065138 (2006i:46039) Malgrange, Bernard(F-GREN-F) Idéaux de fonctions différentiables et division des distributions. (French) [Ideals of differentiable functions and division of distributions] With an Appendix: "Stanisław Łojasiewicz (1926–2002)''. Distributions, 1–21, Ed. Éc. Polytech., Palaiseau, 2003<|endoftext|>
TITLE: Algorithm for reducing words in a Coxeter group
QUESTION [8 upvotes]: Let $W$ be a Coxeter group with set of simple reflections $S$. Suppose that I have chosen a preferred reduced decomposition for every element of $W$. Given an arbitrary word in the alphabet $S$, is there an algorithm for reducing this word to my chosen decomposition using the Coxeter relations? That is to say, an algorithm which at each step replaces a subword of the form $s_i s_k s_i \dots$ by $s_k s_i s_k \dots$, or which removes an occurrence of $s_k^2$.
I would be happy with an answer in the following situation: $W=\Sigma_n$ is the symmetric group on $n$ letters and the preferred decomposition is given by taking the lexicographically smallest (or largest) reduced decomposition.

REPLY [11 votes]: An algorithm for reducing arbitrary words in any Coxeter group to their lexicographically least reduced representatives is described in the paper:
B. Brink and R.B. Howlett,
``A finiteness property and an automatic structure for Coxeter groups'',
Math. Ann. 296, (1993), 179-190.
It doesn't do exactly what you asked for, in that it doesn't do the reductions using only the defining Coxeter relations, but it uses longer substitutions, which can of course be derived from the defining relations, and are always length or lexicographically decreasing. It is the reduction algorithm based on the automatic structure of Coxeter groups, and runs in quadratic time in general. It has been implemented (for example) in the Computer Algebra System Magma.<|endoftext|>
TITLE: Extension of covering map
QUESTION [5 upvotes]: The question is the following: suppose I have manifolds $N$ and $M$ both with boundary, and I have a covering map $\phi$ from $\partial N$ to $\partial M.$ The question is: when is there a covering map $\tilde{\phi}: N \rightarrow M$ which restricts to $\phi?$ When $N, M$ are surfaces, the covering map of the boundaries is given by its monodromy, and there is a $\mathbb{Z}/2 \mathbb{Z}$ obstruction (the sign of the monodromy permutation should be $+1.$ (put a different way: each component of the boundary gets mapped to a collection of cycles. The covering can be extended if and only if the sum of the numbers of even cycles over all boundary components is even). This was proved by Husemoller and Gleason in Husemoller's thesis in the late fifties. I know of no general result in higher dimensions (even in dimension three), from which I conjecture that the question is known to be hard.

REPLY [6 votes]: There's an algorithm to tell if such a map exists if $N$ and $M$ are compact 3-manifolds. Given a covering map $\partial N\to \partial M$ of degree $n$, any extension to a covering $N\to M$ must be a covering of the same degree $n$. Construct all covers of $M$ of degree $n$ (this amounts to finding all reps. of $\pi_1 M$ to the symmetric group $S_n$), and determine for which of these covers there is a lift of the maps $\partial N\to \partial M$. This reduces the question to whether given a homeomorphism $\partial N \to \partial M$ extends to a homeomorphism $N\to M$ by replacing $M$ with the $n$-fold covers together with the lifts of the boundary maps (in fact this part of the argument works in any dimensions). So we will now restrict to the case that the map $\partial N\to \partial M$ is a homeomorphism.
I'll now assume that $N$ and $M$ are 3-dimensional with non-trivial boundary (otherwise you're just asking for the homeomorphism problem), and irreducible (one may reduce to the irreducible case in the usual fashion using the Kneser-Milnor decomposition). Then $N$ and $M$ are Haken.
Put a boundary pattern in $\partial N$ which has no non-trivial automorphisms, and use the homeomorphism to $\partial M$ to transfer the boundary pattern to $\partial M$.
By Theorem 6.1.6 of Matveev's book, there is an algorithm to tell if there is a homeomorphism of $N$ with $M$ which preserves the boundary pattern. This algorithm will then tell if the homeomorphism $\partial N\to \partial M$ extends to a homeomorphism $N\to M$ since the boundary pattern has no automorphisms.
I don't expect this homemorphism extension problem to be solvable in higher dimensions, unless possibly one restricts to some very special class of manifolds.<|endoftext|>
TITLE: The main idea in the proof of Artin's vanishing
QUESTION [7 upvotes]: Does anybody know an easy explanation of the proof of Artin's vanishing theorem (that the etale cohomology of an affine variety of dimension $n$ over an algebraically closed field vanishes in degrees $>n$, or of any other version of this statement)? I have found some proofs; all of them are step by step, and it is not clear to me which of these steps are the most important ones. So, what is the central idea here?

REPLY [14 votes]: I was curious myself after learning this result sometime ago from Lazarsfeld's book on positivity (he calls it the Artin-Grothendieck theorem). The corresponding statement for smooth varieties over the complex numbers and singular cohomology (theorem of Andreotti-Frankel) follows from the fact that Morse theory shows that the variety is homotopy equivalent to a CW-complex with no cells in dimensions $>n$. 
The etale cohomology counterpart works more generally for constructible sheaves. This is probably not very helpful, but here is a sketch of the argument from Lazarsfeld (he uses constructible sheaves in the complex topology, but it should adapt to etale sheaves):

Reduce to the affine space by choosing a finite map $X\to \mathbb{A}^n$ using Noether normalization (already here it is crucial to work with constructible sheaves, not constant sheaves, so we clearly gain something from generalization),
Prove the result for $\mathbb{A}^1$,
Prove the result for $\mathbb{A}^n$ by induction on $n$ using the Leray spectral sequence. Here the crucial observation is that if we choose a sufficiently generic linear projection $\mathbb{A}^n\to \mathbb{A}^{n-1}$, then the stalks of the higher direct images will compute cohomology on the fibers.

So all in all it is a typical example of devissage, which I usually to dislike but slowly learn to appreciate. I think from the outline it is clear which are the key ideas, but I would still really like to see a conceptual proof.<|endoftext|>
TITLE: Integer solutions of x^n + y^n = z^{n-1}
QUESTION [7 upvotes]: This is related to another question
I am interested in the non-trivial integer solutions of
$$ x^n + y^n = z^{n-1} $$
for $n \ge 4$. A solution is trivial if $xyz=0$ or $x = \pm y$.
There are infinitely many rational solutions to $x^n + y^n = (x+y)^{n-1}$ parametrized in the linked question.
For $n=5$ parametric solutions are $(-121 \cdot 2^{{\left(4  k + 3\right)}}, 363 \cdot 2^{{\left(4  k + 3\right)}},11^3 \cdot 2^{5k+4}) $
For $n > 5$ couldn't find any solution so far.

Q1. Are there non-trivial solutions for $n > 5$?
Q2. Are there $n$ for which non-trivial solutions don't exist?
Q3. Is it possible for some $n > 5$ to find solutions without searching? Parametrizing all solutions
(this might settle a case of Fermat-Catalan Conjecture)?

Computationally the fastest way I found so far is pari's "t=thueinit(x^5+1,1);sol=thue(t,a^4);"
though iterating over the divisors is another option.
There are congruence conditions mod $\varphi^{-1}(n)$.

REPLY [32 votes]: Take any $a,b$ and set $c=a^n+b^n$. Then the triple $(ac^{n-2},bc^{n-2},c^{n-1})$ is a solution of your equation.
Conversely, if $(x,y,z)$ is a solution and $d$ is their gcd, so $(x,y,z)=(ad,bd,cd)$, then you get $d(a^n+b^n)=c^{n-1}$. One of the solutions is presented above (with $d=c^{n-2}$). But there also exist smaller solutions --- they appear as soon as $a^n+b^n$ is not square-free.<|endoftext|>
TITLE: Cyclotomic polynomials with coefficients $0,\pm1$
QUESTION [23 upvotes]: Let me begin with what looks like a joke. According to a Bourbaki member, the following conversation occurred during a meeting dedicated to polishing the but-last version of an Algebra Bourbaki volume:
(a Bourbaki member) Why not state explicitly that the coefficients of cyclotomic polynomials are $0,\pm 1$ ?
(another member) Because it's false.
Here is what I am aware: if $n$ has at most two distinct odd prime factors, then the coefficients of  $\Phi_n(X)$ are $0,\pm1$ (Migotti, 1883). In other words, this holds true for $n=2^mp^kq^\ell$, where $p,q$ are primes. On the other hand, it is false for $n=105=3\cdot5\cdot7$, because the coefficient of $X^7$ (or of $X^{41}$ as well) is $-2$.

My question is whether there is a complete characterization of those $n$ for which the coefficients of  $\Phi_n(X)$ are $0,\pm1$ ? If not, are there other infinite lists of cyclotomic polynomials with this property?

REPLY [9 votes]: quid is correct; there are no complete characterizations of the known such numbers. In my work, now posted to the arXiv at http://arxiv.org/abs/1207.5811, I found the family he cites as well as the first fully general one for products of four primes, both listed in the abstract.<|endoftext|>
TITLE: Axiomatizing Gross-Zagier formulae
QUESTION [22 upvotes]: Let $\pi$ be a global automorphic representation of some reductive group over a number field, and let $L(\pi,s)$ denote its L-function. Assume $L(\pi,s)$ extends meromorphically to the complex plane and satisfies a functional equation of the form 
$$
L(\pi,s)= \varepsilon(\pi,s) L(\pi^\star,1-s),
$$ 
where $\pi^\star$ denotes the contragredient dual of $\pi$.
Assume $L(\pi,1/2)=0$  and $L'(\pi,1/2)\ne 0$.
Question 1:
Under what circumstances do we expect the existence of an algebraic null-homologous cycle $D$ on some variety $V$ for which its height $h(D)$ is meaningful and well-defined, and the equality (up to a non-zero, well-understood, fudge factor)
$$
L'(\pi,1/2) \overset{\cdot}{=} h(D)
$$
holds?
Question 2:   Assume the answer to the first question is expected to be "yes" for a given $\pi$, and assume we are also given a good, conjectural, candidate for $D$. Is it possible to axiomatize what one needs to show about the L-function $L(\pi,s)$, the height pairing $h$ and the cycle $D$ in order to prove the above Gross-Zagier formula?
My feeling is that the answer to Q1 should be yes at least when $\pi$ is self-dual, that is to say, $\pi \simeq \pi^\star$. But I do not know whether such a formula is to be expected also for non self-dual $\pi$.
Let me also clarify that I am not asking about the difficulties of constructing a suitable candidate for $D$. Not because it is an uninteresting question, but rather to focus the discussion. In the first question I just ask for whether there exists a cycle satisfying the Gross-Zagier formula, but I am not asking who $D$ is. In the second question I am assuming $D$ is given, and I am asking what properties it should satisfy, but I again don't care who $D$ is.
Example 1: Let me explain the basic scenario I have in mind. Let $E/\mathbb{Q}$ be an elliptic curve and $K$ an imaginary quadratic field. If the pair $(E,K)$ satisfies the Heegner hypothesis, then the order of vanishing of the (self-dual) L-function $L(E/K,s)$ at its central critical value $s=1$ is odd, and Gross-Zagier proved that there exists a certain (Heegner) point $P_K\in E(K)$ such that
$$
L'(E/K,1) \overset{\cdot}{=} h(P_K).
$$
Here $P_K$ plays the role of $D$ in the general question. And we are evaluating the L-function at $s=1$ instead of $s=1/2$ just because we re-normalized it so that the functonal equation relates the values at $s$ and $2-s$. 
And let me explain now some examples in which I do not know the answers.
Example 2: Let $E/\mathbb{Q}$ be an elliptic curve of prime conductor $p$ and $K$ a real quadratic field in which $p$ remains inert. Then the order of vanishing of the (self-dual) L-function $L(E/K,s)$ at its central critical value $s=1$ is odd. Henri Darmon has constructed a point $P_K\in E(K_p)$, rational over the completion of $K$ at $p$, which he conjectures to be actually rational over $K$. I am not asking how to prove this statement here, but rather: assume as a black box that $P_K$ indeed lies in $E(K)$. What one would need to know about $L(E/K,s)$ and $P_K$ in order to prove that $L'(E/K,1) \overset{\cdot}{=} h(P_K)$?
Example 3: Let $E/\mathbb{Q}$ be an elliptic curve. Let $\chi$ be a Dirichlet character and $\mathbb{Q}_\chi$ be the abelian extension of $\mathbb{Q}$ cut out by $\chi$. Assume $L(E,\chi,1)=0$ and $L'(E,\chi,1) \ne 0$. The conjecture of Birch and Swinneton-Dyer predicts the existence of a non-zero point $P_{\chi} \in E(\mathbb{Q}_{\chi})\otimes \mathbb{C}$ lying in the $\chi$-eigenpart of the Modell-Weil group under the Galois action. 
Do we expect the equality $L'(E,\chi,1) \overset{\cdot}{=} h(P_\chi)$ to hold up to some conceptually well-understood fudge factor? (Note that if we assume both sides to be non-zero, the formula obviously holds by setting the fudge factor to be $L'(E,\chi,1)/h(P_\chi)$, and this is not what one would call a well-understood fudge factor!) 
Example 4: Let $f\in S_2(N,\chi)$ be a (cuspidal, normalized) newform of weight $2$, level $N$ and nebentype character $\chi$. Then the field $\mathbb{Q}_f$ generated by the fourier coefficents of $f$ is a finite extension of $\mathbb{Q}$, say of degree $d$. The Eichler-Shimura construction yields an abelian variety $A_f/\mathbb{Q}$.
On the geometric side, we again have a natural construction of Heegner points: $A$ is a quotient of the jacobian $J_1(N)$ of $X_1(N)$. Given an imaginary quadratic field $K$, the theory of complex multiplication allows us to construct Heegner points $P$ on $X_1(N)$ which are rational over a suitable abelian extension $H/K$. This has been extensively studied for $X_0(N)$, in which case $H$ is a ring class field. But is also well-known for $X_1(N)$, where $H$ is no longer anticyclotomic; it contains for instance the abelian extension of $\mathbb{Q}$ cut out by $\chi$. 
In any case, one can construct a Heegner point $P_K\in A(K)$ by tracing down $P$ from $H$ to $K$. And if $\psi$ is a character of $\mathrm{Gal}(H/K)$, one can also define
$$
P_\psi = \sum_{\tau\in \mathrm{Gal}(H/K)} \psi^{-1}(\tau)P^\tau \in E(H)\otimes \mathbb{C},
$$ 
which lies in the $\chi$-eigenpart of $E(H)\otimes \mathbb{C}$.
On the L-function side, $L(A/\mathbb{Q},s)$ factors as
$$
L(A/\mathbb{Q},s) = \prod L(f^\sigma,s)
$$ 
where $\sigma$ ranges over the $d$ different embeddings of $\mathbb{Q}_g$ into $\mathbb{C}$.
While $L(A/\mathbb{Q},s)$ is self-dual, each of the individual factors $L(f^\sigma,s)$ is self-dual if and only if $\chi$ is the trivial character. If $f^\star$ denotes the modular form obtained from $f$ by complex conjugating its fourier coefficients, then the functional equation of $L(f,s)$ relates it to $L(f^*,2-s)$.
A similar discussion holds for the base change of $A$ to $H$. The L-function of $A\times H$ is self-dual, but it factors as the product of L-series of the type $L(f/K,\psi,s)$ where $\psi$ ranges over the characters of $\mathrm{Gal}(H/K)$. Each of the individual L-functions are not always self-dual (regarding $\chi$ and $\psi$ adelically over $\mathbb{Q}$ and $K$ respectively, $L(f/K,\psi,s)$ is self-dual if and only if the restriction of $\psi$ to the ideles of $\mathbb{Q}$ is the inverse of $\chi$.)
Gross-Zagier formulas are proved in the self-dual setting by Zhang and his collaborators, and also by Howard. And Olivier reminded us that such a formula is not to be expected if we insist to use the point $P_\psi$. So the question is: for arbitrary pairs $(\chi,\psi)$,  does there exist a point $P\in (E(H)\otimes \mathbb{C})^{\psi}$ for which $L'(f/K,\psi,1)\overset{\cdot}{=} h(P)$  up to a well-understood non-zero fudge factor?

REPLY [4 votes]: Here is a more detailed version of my comment above.
I will consider only the setting of your Example 3, namely $E/\mathbf{Q}$ is an elliptic curve and $\chi$ is a Dirichlet character such that $L(E \otimes \chi,1)=0$ and $L'(E \otimes \chi,1) \neq 0$.
Let $m \geq 1$ be an integer. The base change $L$-function $L(E \otimes \mathbf{Q}(\zeta_m),s)$ is the product of twisted $L$-functions $L(E \otimes \chi,s)$ where $\chi$ runs through the characters of conductor dividing $m$. The BSD conjecture for the base change $E \otimes \mathbf{Q}(\zeta_m)$ can be refined for each factor $L(E \otimes \chi,s)$. Roughly, the idea is that each arithmetical invariant appearing in the conjecture for $E \otimes \mathbf{Q}(\zeta_m)$ should factor in a way that reflects the decomposition of the motive $h^1(E \otimes \mathbf{Q}(\zeta_m)) = \bigoplus_{\chi} h^1(E \otimes \chi)$. The motive $M_\chi = h^1(E \otimes \chi)$ has dual $M_{\overline{\chi}}$, so it is self-dual only when $\chi$ is quadratic.
Assume $L(E \otimes \chi,s)$ vanishes at order one at $s=1$. The main invariant to consider is the discriminant of the Néron-Tate height pairing $\langle,\rangle$ on $E(\mathbf{Q}(\zeta_m)) \otimes \mathbf{R}$. We can extend this pairing to a positive definite hermitian form on $V=E(\mathbf{Q}(\zeta_m)) \otimes \mathbf{C}$. There is a decomposition of $V$ into isotypical components $V_\chi$ which are pairwise orthogonal with respect to the pairing. In this particular case we indeed expect $L'(E \otimes \chi,1) \sim \langle P_{\chi},P_{\chi} \rangle$ where $P_\chi$ is a generator of $V_\chi$. Usually $P_\chi$ takes the form $\sum_{\sigma} P^{\sigma} \otimes \chi(\sigma)$ for some $P \in E(\mathbf{Q}(\zeta_m))$.
In order to convince you this is reasonable, here is an example I computed using Magma. The rank $0$ elliptic curve $E=X_0(20):y^2 = x^3 + x^2 + 4x + 4$ acquires two independent points of infinite order over the cubic extension $\mathbf{Q}(a)$ with $a=2\operatorname{cos}(2\pi/9))$. Letting $P=(a+1,2a+3)$, we check numerically that $L'(E \otimes \chi,1) \sim \Omega_E \cdot \langle P_\chi,P_\chi \rangle$. The fudge factor appears to be an algebraic number of degree 6.
E:=EllipticCurve("20a1");
G:=FullDirichletGroup(9);
chi:=(G.1)^2; // Character of conductor 9 and order 6
LEchi:=TensorProduct(LSeries(E),LSeries(chi)); // L-Series L(E \otimes \chi,s)
Evaluate(LEchi,1);
//2.22329881577004394873515961159E-30
LEchi1:=Evaluate(LEchi,1:Derivative:=1);
LEchi1;
//2.78851510267155729197040153856 + 0.491690448714030907428058875920*$.1

Q<x>:=PolynomialRing(Rationals());
K<a>:=NumberField(x^3-3*x+1); // Cubic sufield of Q(zeta_9)
EK:=BaseChange(E,K);

G,_,map:=AutomorphismGroup(K); // Galois group of K
P:=[EK![(map(g))(a)+1,2*(map(g))(a)+3] : g in G]; // Set of conjugates of P=[a+1,2a+3] with a=2cos(2pi/9)
M:=HeightPairingMatrix(P);

C:=ComplexField();
PchiPchi:=C!M[1,1] + C!chi(2)*C!M[1,2] + C!chi(4)*C!M[1,3];

ratio:=LEchi1/(Periods(E)[1]*PchiPchi);
PowerRelation(ratio,6);
//27*x^6 - 9*x^3 + 1

I'm not sure about the situation for general abelian varieties $A_f/\mathbf{Q}$, but I guess some of the above might extend.
Of course, the real question seems to be whether or not these points $P_\chi$ are related in some way to Heegner points!<|endoftext|>
TITLE: References for Eilenberg-Zilber shuffle product
QUESTION [17 upvotes]: Most of the treatments I can find in the literature for the Eilenberg-Zilber shuffle product approach it from the point of view of simplicial sets (including the original Eilenberg-MacLane paper). I was wondering if anyone knows of a reference written directly in the language of singular or PL chains. By this I mean that one should show that the (p,q)-shuffles each produce a distinct embedded p+q simplex in $\Delta^p\times \Delta^q$ and that, collectively, this produces a triangulation of $\Delta^p\times \Delta^q$. Furthermore, this should be a chain map, in the sense that the boundary of the triangulation is compatible with the same triangulation process applied to the boundaries (or in other words, the map $S_p(X)\otimes S_q(X)\to S_{p+q}(X\times Y)$ obtained by applying this process is a chain map). Dold constructs this map explicitly in his book, but it's in an exercise and he doesn't provide any proofs of properties (he does helpfully indicate with an asterisk that it is a challenging exercise to do so!)
The reason I ask is that I'm working on some lecture notes (for publication) for which I need to have an explicit construction of the homology cross product at the chain level, but the intended audience is not expected to know about simplicial sets. 
If what I'm looking for does not exist, can anyone recommend the simplicial set treatment that provides the best description of the connection to the explicit geometric triangulation side of things?

REPLY [2 votes]: This is very late, I know, but I just stumbled across this question as I happen to be thinking about these things myself. There is a nice discussion of the simplicial cross product in Hatcher's "Algebraic Topology" text, on pp. 277-278. There Hatcher refers to Eilenberg and Steenrod's "Foundations of Algebraic Topology" textbook, p. 68, for a proof of precisely the fact you mention. I have not yet waded through their proof, but this would seem to be the canonical reference.
In fact they seem to be proving something more general, that the shuffles induce a natural $\Delta$-complex structure on a product of (finite) $\Delta$-complexes.<|endoftext|>
TITLE: The function $\sum_{0}^{\infty} x^n/n^n$
QUESTION [28 upvotes]: The function $F(x) = \sum_{0}^{\infty} x^n/n^n$ may be familiar to many readers as an example sometimes used when teaching tests for absolute convergence of entire functions defined by power series. I know of no name for it, nor any use for it aside from pedagogical, so this is a pure curiosity question which I hope is acceptable.
The function seems to have one real zero around $x = -1.40376$; a single extremum, a minimum around $x = -5.71837$; and then to approach the $x$-axis from below asymptotically as $x$ goes to negative infinity. Is this true?

REPLY [11 votes]: There is a paper of G. H. Hardy, where this function is studied in great detail:
G. H. Hardy, On the integral function $ \Phi_{ a,\alpha,\beta}(z)=\sum x^n/(n+a)^{\alpha n+\beta}$, Quarterly J. Math., 5 (1906) 37, 369-378. (Collected papers of G. H.Hardy, vol. IV, p. 128).<|endoftext|>
TITLE: A "geometric" insight into a proof of Krull intersection theorem?
QUESTION [7 upvotes]: I an not 100% sure that this question has an answer, but still I would like to ask it.
There is a short an simple proof of Krull intersection theorem (for example page 12 in http://www.jmilne.org/math/xnotes/CA.pdf or page 154 http://math.uga.edu/~pete/integral.pdf ).
As far as I got this short proof is very recent (at least in Pete Clark's notes there is a reference to an article of 2004). 
I would like to know if this proof can be stated in a more "geometric" language. Even though I seem to understand the proof it looks to me a bit as a trick...

REPLY [13 votes]: EDIT: This is a geometric interpretation of the statement of the Krull Intersection Theorem, and not the proof, as the OP asked for.  Here it is anyways.
Ok, so let's work in the setting of a Noetherian local ring $(R, \mathfrak{m})$.  Then the Krull intersection says that $\bigcap_{n \geq 1} \mathfrak{m}^n = (0)$.  (You can remove the local hypothesis if you assume that $R$ is a domain, and then $\mathfrak{m}$ can be any ideal).
The easiest geometric interpretation of this statement that I can think of is something like the following.  
There is no hypersurface passing through a point (or subvariety) of a variety $X$ with infinite order of vanishing through that point/subvariety.
Or, 
The only function which vanishes to arbitrarily high order at a point is the zero function.
The interpretation should be pretty easy.  Given an $f \in R$, we can measure the order of vanishing of $f$ by asking what's the biggest power $n$ of $\mathfrak{m}$ such that $f \in \mathfrak{m}^n$ but such that $f \notin \mathfrak{m}^{n+1}$.  Given now a scheme $X$, go look at a stalk of some (possibly non-closed) point.
Obviously, this result shows up a lot when studying completion, so it has geometric applications as well.<|endoftext|>
TITLE: Half integral weight Hecke operators
QUESTION [6 upvotes]: I would like to find a source giving the exact formula for the product of two Hecke operators $T_{\kappa}(n^2)$ and $T_\kappa(m^2)$ of half integral weight. That is, $\kappa \in \frac 12 \mathbb{Z} - \mathbb{Z}$. I am searching for a formula which is similar to
$$
T_k(m)T_k(n) = \sum_{d|(m,n)} d^{k-1}T_k\left(\frac{mn}{d^2}\right)
$$
in the case of full integral weight $k$. 
I suspect the answer to be exactly the same, since double coset decompositions in $\operatorname{GL}_2(\mathbb{R})^+$, and in its double cover should correspond to each other in a one-to-one fashion. The only difference being that the half integral Hecke operators are supported only on the squares.
Am I missing anything?

REPLY [4 votes]: I think you can find the answer here.<|endoftext|>
TITLE: Reference request: the "Kauffman bracket skein category"? 
QUESTION [7 upvotes]: There should be a category $3\text{CobTang}$ whose 

objects are some kind of surfaces with a finite set of marked points         
morphisms $M : S \to T$ are some kind of $3$-dimensional cobordisms $M$ between surfaces together with a choice of (framed) tangle in $M$ beginning at the marked points in $S$ and ending at the marked points in $T$. 

Composition is given by gluing. Taking free $\mathbb{Z}[q, q^{-1}]$-modules on the morphisms in this category and quotienting by the skein relations defining the Kauffman bracket should give a category (the "Kauffman skein bracket category") $K$ in which

the Temperley-Lieb category,
skein algebras of surfaces, and
skein modules of $3$-manifolds

all sit as subcategories. The first bullet is the subcategory where $S, T$ are both, say, the open disc with some marked points and $M$ is a cylinder. The second bullet is the subcategory where $S, T$ are both a fixed surface with no marked points and $M$ is a cylinder. The third bullet is the subcategory where $S, T$ have no marked points. 
There is also a forgetful functor $3\text{CobTang} \to 3\text{Cob}$ given by forgetting the marked points and the tangle, and this gives a span of categories $3\text{Cob} \leftarrow 3 \text{CobTang} \rightarrow K$. 

(Where) does this construction appear in the literature?

It seems related to a TQFT except that we only get a span and not a functor. In addition, the Temperley-Lieb category quantizes the representation theory of $\text{SU}(2)$ while skein algebras quantize character varieties. So: 

What does $K$ quantize?

REPLY [10 votes]: I'm not sure where in the literature this exact construction occurs, but it is certainly well-known.  I think some skein-module-centric papers consider non-empty boundary conditions on the skein modules (which would be equivalent to what you describe), so you might try looking for that.  You could also have a look at the (incomplete, a little out of date) "TQFTs" notes on my web page.

There is a standard way to view this as a functor $\mathrm{3Cob}\to K$ instead of a span.  Assign to a surface $Y$ the category $A(Y)$ whose objects are finite collections of framed points in $Y$ and whose morphisms are (finite linear combinations of) framed tangles in $Y\times I$ modulo Kauffman bracket relations.  Then given a 3-dimensional bordism between surfaces $Y_1$ and $Y_2$ we can construct an $A(Y_1)$-$A(Y_2)$ bimodule.  These assignments give a functor from $\mathrm{3Cob}$ to the category whose objects are linear categories and whose morphisms are bimodules.
The above construction can be extended "all the way down", assigning 2-categories to 1-manifolds and 3-categories to 0-manifolds, with gluing of 1- or 2-manifolds corresponding to the appropriate categorified tensor product over a 3- or 2-category.
To extend the construction "all the way up", assigning an intertwiner of bimodules to a 4-dimensional bordism, one needs to assume that the deformation parameter $q$ is a root of unity and take the usual quotient.  The resulting fully extended 3+1-dimensional TQFT is a disguised version of the Witten-Reshetikhin-Turaev TQFT based on the Kauffman bracket at that root of unity.

Finally, I'll make a small but important quibble with your statement that the Kauffman bracket quantizes the representation theory of $SU(2)$.  As it's name/notation suggests, $Rep(U_q(SU(2)))$ quantizes the representation theory of $SU(2)$.  The Kauffman bracket differs from $Rep(U_q(SU(2)))$ by a sprinkling of $\pm 1$'s.  By tweaking the signs like this, one gains the convenience of a trivial Frobenius-Schur indicator.  (This means that one can work with unoriented tangles.)  But one loses some positivity properties.  In the short run the Kauffman bracket is more convenient, but in the long run $Rep(U_q(SU(2)))$ is the nicer object.  For example, $Rep(U_q(SU(2)))$ can be categorified (Khovanov homology), while the Kauffman bracket can only be categorified up to ambiguous $\pm$ signs.<|endoftext|>
TITLE: Any algebraic substitute for Morse theory (and homology) in arbitrary characteristic? 
QUESTION [19 upvotes]: As far as I know, Morse theory yields much information on the topology of smooth manifolds; in particular, it can be used to prove Artin's vanishing (that the singular cohomology of smooth complex variety of dimension n vanishes in degrees >n). My question is: are there any ideas how to extend any of the consequences of Morse theory to (the study of the etale homology of) algebraic varieties (over fields of arbitrary characteristic)? In particular, what is the relation between Morse homology and Lefschetz pensils?

REPLY [20 votes]: A Morse function is  a map of a manifold to the real line locally equivalent to: 
$$f(x_1,\ldots, x_n)=-x_1^2\ldots -x_k^2+ x_{k+1}^2+\ldots+x_n^2$$
 for some $k$. In other words,
for which the singularities are as simple as possible.
While a Lefschetz pencil is a map of a smooth projective variety to the projective line local analytically given by $f=x_1^2+\ldots+x_n^2$.
So in this sense, they are very analogous. There are differences, however.  Given
a Morse function $f:X\to \mathbb{R}$, the collection of the above numbers $k$, called indices,
determine the homotopy type.[the number of cells in a complex homotopic to $X$].  For "Artin's" vanishing, it is enough to choose an $f$, where these indices are bounded by dimension. (Details can be found in the first few pages
of Milnor's Morse theory.)
I'm not aware that there is any complete substitute on the other side, given say a Lefschetz pencil $f:X\to \mathbb{P}^1$. However, the pencil does give a way to calculate the (etale) cohomology of $X$ in terms of the critical points of $f$ and the monodromy, or more formally in terms of the direct images $R^if_*\mathbb{Q}_\ell$. And this is quite powerful.

Perhaps it would be more instructive give a Morse-like pseudo-proof of Artin's theorem.
By "pseudo" I mean that there is step which I can't justify without a lot more effort than this is worth. Let $X\subset \mathbb{A}^n$ be an irreducible affine variety of dimension  $n$ over an algebraically closed field. By generic projection, we get a nonconstant morphism
$f:X\to \mathbb{A}^1$ which will play the role of our Morse function. Suppose that (*) the direct images $R^jf_*\mathbb{Q}_\ell$ were constructible and commuted with base change.
Then by induction, $R^jf_*\mathbb{Q}_\ell=0$ for $j>n-1$. From the Leray spectral
sequence, we need only prove that $H^i(\mathbb{A}^1,F)=0$ for $i>1$ and any constructible sheaf $F$,
but this is easy. Note that if $f$ were proper, (*) would be automatic. In general,
one could get around it by generic base change [SGA 41/2, p 236] and devissage.<|endoftext|>
TITLE: Uniqueness of analytic continuation in rigid analytic geometry
QUESTION [10 upvotes]: In classical complex analysis it is easy to prove that a meromorphic function has at most one analytic continuation (on an open connected subset of $\mathbb C$, say).
The problem of non-uniqueness of analytic continuation is one of the reasons why it is not possible (if one wants a good theory) to translate the complex theory to the $p$-adic case without some modifications, and so it is one of the motivation for introducing rigid analytic variety. However, I am not able to find a precise statement that explains under which hypothesis the uniqueness of analytic continuation holds for rigid analytic varieties. So the question is the following.
Let $k$ be a non-archimedean field and let $X$ be a connected rigid analytic space over $k$. Let $f \colon X \to k$ be a rigid analytic function that vanishes on $Y$, that is an admissible subdomain of $X$. It is always true that $f$ vanishes on $X$? If this is not the case, under which assumptions it is true?
Any references will be very appreciated!

REPLY [3 votes]: Let me just complement xbnv's answer with a mild generalization.  If $X$ is an irreducible rigid space (and let's suppose we're dealing with reduced spaces from the outset), then its normalization $\tilde{X}$ is connected and normal and is equipped with a finite surjective map $\tilde{X}\to X$.  Using xbnv's answer, it follows that the statement holds for $X$ as well.
In short, if you replace "connected" by "irreducible" then you're good.<|endoftext|>
TITLE: Euler characteristic and universal cover
QUESTION [26 upvotes]: Let $M$ be a compact manifold, let $\tilde{M}$ be its universal cover, and suppose that the Euler characteristic $\chi(\tilde{M})=0$.
My question is: does this imply that $\chi(M)=0$?
This is clear if $\pi_1(M)$ is finite, but I am interested in the case $|\pi_1(M)|=\infty$.
It might not feel right, but I can't think of any counterexample, either.
Thank you very much in advance!

EDIT: I was rightfully asked what I mean by Euler characteristic of the (non compact) manifold $\tilde{M}$. My answer right now is: the one you want!
What I am thinking of, is $\chi(\tilde{M})=\sum_i (-1)^i\dim H_i(\tilde{M},k)$, with $k=\mathbb{Q}$ or $\mathbb{R}$, and $H_i$ are either the usual or the compactly supported cohomology groups.
In my case, $\tilde{M}$ retracts to a compact Lie group.

REPLY [9 votes]: Here is sketch proof that $\chi(\tilde{M})=0$ implies that $\chi(M)=0$ for fundamental groups with certain finiteness properties.
The idea is to adapt the usual spectral sequence proof that the Euler characteristic is multiplicative on fibrations. Let $k$ be a field of characteristic zero, and let $\pi=\pi_1(M)$. The Cartan-Leray spectral sequence of the regular cover $\tilde M\to M$ has $E^2 = H_p(\pi;H_q(\tilde M;k))$ and converges to $H_{p+q}(M;k)$. But we can also go one page back and start at $E^1 =F_p\otimes_{k[\pi]} H_q(\tilde M;k)$, where $F_\bullet$ is a free resolution of $k$ by $k[\pi]$-modules. (Reference: K. Brown, Cohomology of groups, VII.5 and VII.7.) Now in a spectral sequence the Euler characteristic of each page is the Euler characteristic of the previous one, so it suffices to check that $\chi(E^1)=0$ when $\chi(\tilde M)=0$. 
If the trivial module $k$ admits a finite free resolution by $k[\pi]$-modules, we can choose each $F_p$ to be a finitely generated free $k[\pi]$ module, so $F_p$ is a direct sum $k[\pi]^{n(p)}$ for some $n(p)$. I think we then get $F_p\otimes_{k[\pi]} H_q(\tilde M;k)\cong k^{n(p)}\otimes_k H_q(\tilde M;k)$ as vector spaces, and the result follows in the usual way since the Euler characteristic of a tensor product of finite dimensional graded vector spaces is the product of their Euler characteristics.<|endoftext|>
TITLE: Hahn-Banach theorem with real extended valued function
QUESTION [8 upvotes]: Hello to everyone,
My problem is the following: I have this version of the Hahn-Banach theorem:
Let V be a vector space and let $p:V\rightarrow \mathbb{R}$ be any
convex function. Let $W$ be a vector subspace of $V$ and let $L:W\rightarrow \mathbb{R}$ be a linear functional dominated by $p$ on $W$. Then there is a (not generally unique) linear extension $\hat{L}$ of $L$ to $V$ that is dominated by $p$ on $V$.
Furthermore $\hat{L}_{|U}=L$.
Does the theorem still hold when $p:V\rightarrow(-\infty,+\infty]$ ? Is someone able to give me a proof or to provide a counter-example that show that the theorem does not hold? And if it does not hold, it is possible to add some conditions that make it still true?
To put it in another way, is the same true if we further relax the hypothesis on $p$ and allow it to be real extended with nontrivial domain, i.e. $\lbrace x\in V: p(x) \in \mathbb{R} \rbrace \neq \emptyset $ ?
Thanks to everyone in advance for helping me.

REPLY [11 votes]: The Hahn-Banach theorem is wrong for extended real $p$: For $V=\mathbb R^2$ let
$p$ be the Minkowski functional of $A= \mathbb R \times (0,\infty)$
(so that $p(x,y)=0$ if $y>0$ and $p(x,y)=\infty$ if $y\le 0$),
$W= \mathbb R \times \lbrace 0 \rbrace$, and $L: W\to\mathbb R$ defined by $L(x,0)=x$.
Then $L$ is $p$-dominated but there is no $p$-dominated linear extension.<|endoftext|>
TITLE: Inner Automorphisms of Matrix Algebras
QUESTION [6 upvotes]: Let $\mathbb R$ be the field of real numbers and $\mathbb C$ the field of complex numbers. 
It is  well known that that $\mathbb C$ can be embedded in $M_2(\mathbb R)$.
This embedding can be extended in the obvious way to an embedding function $\varphi : M_n(\mathbb C)\rightarrow M_{2n}(\mathbb R)$. 
My question is: consider the embedding $\varphi :M_2(\mathbb C)\rightarrow M_{4}(\mathbb R)$, it is possible to find an inner automorphism $\Psi : M_4(\mathbb R)\rightarrow M_{4}(\mathbb R)$ such that the intersection
$$\varphi (M_2(\mathbb C))\cap \Psi(\varphi (M_2(\mathbb C)))$$
is the scalar matrices in $M_4(\mathbb R)$?

REPLY [7 votes]: This is an interesting problem. (According to V.I.Arnold, Misha Gromov said once: "There is only one way to find out whether a problem is good or not --- it simply must be solved!") First, let us reformulate it.
Given a finite-dimensional $\Bbb R$-linear space $V$ (in the original problem, $V$ is $4$-dimensional), denote by $M:=\text{Lin}_\Bbb R(V,V)$ the algebra of all $\Bbb R$-linear endomorphisms of $V$ and by $G:=\text{GL}V$, the group of $\Bbb R$-linear automorphisms of $V$. An $\Bbb R$-linear map $i\in M$ is said to be a complex structure on $V$ iff $i^2=-1$. An $\Bbb R$-linear map $h\in M$ is  $\Bbb C$-linear with
respect to a complex structure $i$ on $V$ iff $h$ commutes with $i$.
Problem. Is it possible to find $2$ complex structures $i_1,i_2$ on $V$ such that any $\Bbb C$-linear map with respect to both $i_1,i_2$ is just a multiplication by a real number? (Two complex structures on $V$ are conjugated by means of an $\Bbb R$-linear isomorphism because $\Bbb C$-linear spaces of the same dimension are isomorphic. Therefore, this problem is equivalent to the original one.) In other words, we are looking for $g\in G$ such that the centralizer of $i$ and $gig^{-1}$ in $M$ is trivial.
Remark. If such a $g$ exists, a generic $g$ also works. This means that, under the assumption of existing $g$ in question, there exists also
$g\in M\setminus X$ with trivial centralizer of $i$ and $gig^{-1}$ for any proper Zariski closed $X\subset M$ (i.e., $M\ne X$ is given in $M$ by polynomial equations). Indeed, the condition on $g$ that the centralizer is nontrivial is
Zariski closed (it is equivalent to vanishing some determinants) and a linear space over an infinite fieald cannot be a union of finitely many proper Zariski closed subsets (a well-known and easy fact).
Denote by $M_0$ the centralizer of $i$ in $M$ and let $a\in G$ be induced by the complex conjugation in a decomposition $V=\Bbb C\otimes_\Bbb RV_0$ compatible with $i$, $a^2=1$. Then $M=M_0\oplus aM_0$ is a $\Bbb Z/2$-grading, where
$M_0=\{m\in M\mid imi^{-1}=m\}$ and $M_1:=aM_0=\{m\in M\mid imi^{-1}=-m\}$.
Answer. Suppose that $g=m+am_0\in G$ provides a trivial centralizer of $i$ and $gig^{-1}$ in $M$, where $m,m_0\in M_0$. By Remark, we can assume $m_0\in G$. As $m_0im_0^{-1}=i$, we can replace $g$ by $gm_0^{-1}$, i.e., we can assume $m_0=1$. By Remark, we can assume that $am+1\in G$ and that $(ma)^2$ does not belong to the centre of $M$. It follows from $a^2=1$ that $(am+1)^{-1}a(m+a)=1$. In other words, $g^{-1}=(am+1)^{-1}a$. Hence,
$$gig^{-1}=i(m-a)(am+1)^{-1}a=$$
$$=i\big((m+a)(am+1)^{-1}a-2a(am+1)^{-1}a\big)=i\big(1-2(ma+1)^{-1}\big).$$
Therefore, the centralizer of $i$ and $gig^{-1}$ coincides with that of $i$ and $ma$. The latter contains $(ma)^2\in M_0$. A contradiction.<|endoftext|>
TITLE: On simple factors of modular jacobians: endomorphism ring and simplicity of mod p reduction
QUESTION [8 upvotes]: Let $A_f$ be the abelian variety over $\mathbf{Q}$ arising as a $\mathbf{Q}$-simple factor of the Jacobian $J_0(N)$ of the modular curve associated to a normalized newform $f$ of weight $2$ on the congruence subgroup $\Gamma_0(N)$. There is a piece $\mathbf{T}_f$ of the Hecke ring acting on $A_f$. Such a ring is an order in the number field generated by the Fourier coefficients of $f$, whose degree equals the dimension of $A$. It is a result of Ribet (I think) that the natural map $\mathbf{T}_f\otimes\mathbf{Q}\rightarrow\textrm{End}(A)\otimes\mathbf{Q}$ is an isomorphism, where $\textrm{End}(A)$ is the ring of $\mathbf{Q}$-enndomorphisms of $A$ (I couldn't attach the index to End..).
I have two questions, please:
1) Can we find an abelian variety $A_f'$ over $\mathbf{Q}$, which is $\mathbf{Q}$-isogenous to $A_f$, and such that $\textrm{End}(A_f')$ is the maximal order of $\textrm{End}(A_f')\otimes\mathbf{Q}=\textrm{End}(A_f)\otimes\mathbf{Q}$? (Notice that in the one dimensional case $\textrm{End}(A_f)$ is already maximal, being it the ring of integers)
2) Does the fact that $A_f$ has such a large endomorphism ring imply that the mod $p$ reduction of $A_f$ be simple over the prime field with $p$ elements? Here $p$ is a prime of good reduction for $A_f$.
Thanks!

REPLY [5 votes]: I found a reference for the first question in the following PhD thesis :
J. Wilson, Curves of genus 2 with real multiplication by a square root of 5
The answer is yes. This is a consequence of the following general fact about abelian varieties (see Prop 2.5.4 in the thesis).
Let $A$ be an abelian variety over a field $k$. Let $R$ be an order in a number field $F$. Assume that $R$ embeds into $\operatorname{End}_k(A)$. Then there exists an abelian variety $B/k$ which is $k$-isogenous to $A$ and such that $\mathcal{O}_F$ embeds into $\operatorname{End}_k(B)$.
The idea is to take $B=A/G$ with $G=(n\mathcal{O}_F) A[n^2]$, where $n$ is the index of $R$ in $\mathcal{O}_F$.
The thesis also contains interesting examples of varieties $A_f$ with Hecke field $K_f=\mathbf{Q}(\sqrt{5})$. These are natural examples to try for Question 2 (although I have no idea how to compute the reduction of $A_f$ mod $p$).
EDIT. The answer to Question 2 is negative in general. There are newforms $f$ of weight $2$ on $\Gamma_0(N)$ such that $A_f$ splits over $\overline{\mathbf{Q}}$. This happens for example when $f$ has extra-twist. The first example appears at level $N=63$, see Table 1 p. 13 in
MR1933828 (2003i:11078)  González-Jiménez, Enrique ;  González, Josep. Modular curves of genus 2. Math. Comp.  72  (2003),  no. 241, 397--418 (electronic).
Assume $A_f \sim E_1 \times E_2$ where everything is defined over some number field $K$. If $p$ is a prime of good reduction for $A_f$ which splits totally in $K$, then $A_f$ mod $p$ is $\mathbf{F}_p$-isogenous to a product of elliptic curves over $\mathbf{F}_p$, so it is not simple.<|endoftext|>
TITLE: Good even grading and principal Levi type
QUESTION [5 upvotes]: Let $\mathfrak g$ be a simple Lie algebra over $\mathbb C$ and let $e$ be a nilpotent element in it. In the theory of finite W-algebras one often encounters the following two conditions:
1) $e$ is principal in some Levi subalgebra $\mathfrak l$ of $\mathfrak g$.
2) There exists a good even grading on $\mathfrak g$ (recall that a grading on $\mathfrak g$
is called good for e if $e \in \mathfrak g_2$ and the linear map
$ad~ e : \mathfrak g_j → \mathfrak g_{j+2}$
is injective for $j \leq −1$ and surjective for $j\geq −1$).
My question is this: is there any relation between these conditions, or are they completely independent?

REPLY [4 votes]: If $e$ is principal in a proper Levi subalgebra whose Dynkin diagram involves a component of type
$A_k$ with $k$ odd, then e is not even. This is very easy to see by writing down an explicit $sl_2$-triple containing $e$. Occasionally, one can find another good grading for such $e$ (if it is Richardson) but this is quite rare outside type $A$. Also, if $e\ne 0$ is rigid in the sense of Lusztig-Spaltenstein then $e$ is never even (nor Richardson) and for $g$ exceptional all rigid nilpotent orbits except two (I think) are principal in Levi subalgebras. From the representation-theoretic viewpoint, rigid nilpotent elements give rise to the most interesting finite $W$-algebras.<|endoftext|>
TITLE: heegard diagram 
QUESTION [7 upvotes]: It seems like there is an algorithm to find the Heegard diagram of a 3 manifold obtained by surgery on a link. Also someone told me I can find it in the Gompf and Stipciz's book. But I could not find it. Can anyone help?

REPLY [14 votes]: (1) Choose a planar presentation of your link, approximately in the plane of the blackboard.  Let $N$ be a tubular neighborhood of the link in this position.
(2) For each crossing $c$ of the presentation, add a 1-handle $T_c$ to $N$ which is perpendicular to the blackboard and connects the upper and lower parts of the crossing.  Let $H$ be the union of $N$ and all the 1-handles $T_c$.
(3) $S^3 \setminus H$ is a handlebody.  The corresponding set of Heegaard curves on $\partial H$ bijects with the complementary regions of the (flattened) planar diagram of the link.
(4) $H$ is, of course, also a handlebody.  For the corresponding Heegaard curves take the surgery curves on $N$ (this involves a choice to make them disjoint from the attaching disks of the $T_c$) union the obvious small, disk-bounding curves on the boundary of each $T_c$.  (Correction added later: Actually, if there are $k$ components of the link then one should omit $k-1$ of the $T_c$ curves.  The omitted crossings should be minimal with respect to connecting the components of the link.  Also, I'm assuming that the planar projection is a connected graph.)<|endoftext|>
TITLE: Optimum Tournament Strategy
QUESTION [7 upvotes]: Consider a symmetric N-player game in which all players partition one total unit of
energy among individual games.  The probability of winning each game is simply proportional to the spent energy (player #1 wins with probability $\frac{E_1}{E_1+E_2+...+E_N}$).
The winner is the first player to win G games.
Before each game, players know both "how much energy each person has left" and
"how many games each person has won" to choose the energy to spend in the next
game.  (I'm additionally interested in game theory if these are not both known,
but that's a bonus.)
A full game-tree solution for all cases would be nice, but maybe too much to
ask for...instead, how much energy should be spent on the first game if N=2 and
G=4 (World Series)?
[this is a tangent from Flipping coins on a budget

REPLY [4 votes]: I will address the case $N=2$.
The best strategy is the boring one of distributing your energy evenly. This can be proved inductively in the number of games left. If the players need $a$ and $b$ games to win, respectively, then we say the number of games left is $a+b-1$. 
It is trivial for $1$ game left, and a simple calculation with $2$ games left, where one side needs to win both games and the other side needs to win either. 
Suppose it is true for $g-1$ games, and let there be $g$ games left. Suppose your opponent distributes his/her energy evenly. Consider the strategy of using $x/g + \delta$ energy on the next game. By induction, regardless of the result, the correct strategy for the last $g-1$ games is to distribute your energy evenly. Choosing $x/g + \delta$ energy for the first game and $x/g - \delta/(g-1)$ equity for the remaining games is equivalent to choosing $x/g + \delta$ energy for the last game and $x/g - \delta/(g-1)$ for the first $g-1$ games. But by induction, after the first game where you use the equity $x/g - \delta/(g-1)$, your equity is less than or equal to the choice of distributing your energy evenly among the last $g-1$. Thus, for any $x/g + \delta$, the equity of that choice is less than or equal to the choice of $x/g + \delta(\frac{-1}{g-1})^n$. By the continuity of winning chances and using $|\frac{-1}{g-1}| \lt 1$, the equity using $x/g+\delta$ on the first choice is at most the equity of using $x/g$. So, by induction, it is optimal to distribute your energy equally.
In the first game of a best-of-seven series, spend $1/7$ of your energy.<|endoftext|>
TITLE: partly obscured Rubik's cube
QUESTION [8 upvotes]: I just came back from a beach which features a large Rubik's cube (2m high).  The base of the cube is not visible and the top is not coloured. The four vertical sides are each divided $3\times 3$ into coloured squares as they should be.  I was idly wondering: how can I tell if the patterns appearing on those four sides are actually possible for a real cube?  Clearly there are necessary conditions like having at most 6 different colours and at most 9 squares of each colour, but is that sufficient?  Can the legal configurations be characterized?

REPLY [10 votes]: Is this your cube? 
http://www.redbubble.com/people/yolanda/works/6358896-rubiks-cube-at-maroubra
If so, it is not solvable. The dark blue (grey?) and white centred faces are opposite each other, but there is a corner with both colours.
Edit: the linked site's interface changed slightly, so now the two images are on different pages. See here http://www.redbubble.com/people/yolanda/works/6351863-surfing-fun-at-maroubra-beach for the alternate view showing the corner.<|endoftext|>
TITLE: What's "bad" about unstable sheaves?
QUESTION [23 upvotes]: To construct a (coarse or fine) moduli space that is separated, one usually throw away some class of the object in question. For moduli of sheaves people talk about (semi-)stability. A coherent sheaf $E$ on a scheme $X$ is (semi-)stable if it is pure and for all subsheaf $F$ we have $p(F) <(\leq)\text{ } p(E)$, where $p$ is the reduced Hilbert polynomial with respect to some fixed polarization.
My question is, what's the $bad$ property of unstable sheaves so that one has to throw it away to get a $good$ moduli space? Or is this definition merely there to make GIT work?

REPLY [4 votes]: The standard explanation is that if we want to work on the level of the coarse moduli space and not on the level of the stack itself, we need to tame the automorphism groups of objects (as Peter pointed out) and have some control over the maps between objects in your moduli. Semi-stable objects, Harder-Narasimhan and Jordan-Holder filtrations are exactly the notions supplying the rigidity for the morphisms between objects.
In this way it seems that the need to impose certain stability conditions spans  beyond moduli of sheaves, e.g. (just one other case I am familiar with) to have a good moduli space one considers (semi)-stable representations of quivers.
A. Rudakov axiomatises the situation in the case of an arbitrary abelian category, see his paper "Stability for an Abelian Category".<|endoftext|>
TITLE: When should I publish my results?
QUESTION [28 upvotes]: I just finish my masters thesis, and I was told by one of the referees that I should publish some of the results in there. However I feel that I mainly repeat arguments by other people in a slightly more general form.
I do feel that I improved certain results. I also approach these results in a different way (thanks to several decades of mathematical development). However the ideas behind the proofs are not mine, and I feel that all I do is rehash the original idea.
Of course for a masters thesis this is fine, I do feel very happy with what I did. However I don't know whether or not I should write those into a paper form.
I was hoping to get a general advice from those more experienced than me. When do you publish a result? Especially as a young mathematician-to-be.

REPLY [16 votes]: I think that publishing a reappraisal and generalization of older work in modern terms can be very useful, provided you clearly label it as such in your paper. Depending on the field, there might be some journals who would gladly publish this kind of paper and others who wouldn't consider it, so you might want to consult your advisor about where to submit to.<|endoftext|>
TITLE: random walk returning probability
QUESTION [7 upvotes]: Consider a two-dimensional random walk, but this time the probabilities are not 1/4, but some values p_1, p_2, p_3, p_4 with $\sum_{i=1}^4 p_i=1$. For example, from (0,0), it goes to (1,0) with p_1, goes to (0,1) with p_2 etc.
The question is how to compute the probability x of going back to (0,0), starting from (0,0). In general, this probability is not 1.  
Thanks.

REPLY [2 votes]: A special case where we can avoid one of the two sums in James Martin's answer is when the "determinant" $p_1p_3-p_2p_4$ of the four probabilities is zero. Then we can treat the random walk as a "sum" of two independent one-dimensional random walks, one taking steps $(+1/2, +1/2)$ or $(-1/2,-1/2)$, the other taking steps $(+1/2,-1/2)$ or $(-1/2,+1/2)$. The original walk is back at the origin precisely when both these 1D-walks are.
If the transition probabilities of the two component walks are $p, (1-p)$ and $q, (1-q)$ respectively, then the expected number of visits to the origin (including the start) is $$\sum_{n=0}^\infty \binom{2n}{n}^2 p^nq^n(1-p)^n(1-q)^n.$$
In terms of the original probabilities, $$pq(1-p)(1-q) = (p_1+p_2)(p_2+p_3)(p_3+p_4)(p_4+p_1).$$
I'm no expert on this type of sum, but according to Maple, $$\sum_{n=0}^\infty \binom{2n}{n}^2 x^n = \frac2{\pi} EllipticK(4\sqrt{x}).$$
The probability of never returning is the reciprocal of this (with $x=pq(1-p)(1-q)$), in other words 
$$\frac{\pi}{2EllipticK(4\sqrt{(p_1+p_2)(p_2+p_3)(p_3+p_4)(p_4+p_1)})}.$$
An obvious question is whether this holds also when $p_1p_3 \neq p_2p_4$. Off the top of my head I don't see why it couldn't.<|endoftext|>
TITLE: Does $C'\left(\frac{5}{11}\right)$ imply exponential growth?
QUESTION [5 upvotes]: I came across this rather week small cancellation condition $C'\left(\frac{5}{11}\right)$ of a group $G$. It has been proved that $C'\left(\frac16\right)$ is enough for $G$ to contain free subgroups. I was therefore wondering if $\frac{5}{11}$ is maybe enough to still have exponential growth. 
Does anyone know of any related papers or results?

REPLY [11 votes]: Every finitely presented group has presentation satisfying $C'(1/5)$. Note that $1/5 < 5/11$. See the book by Olshanskii's book "Geometry of defining relations of groups".<|endoftext|>
TITLE: Example of a diophantine application of an open image theorem
QUESTION [6 upvotes]: I'm an applied model theorist, and open image theorems are important in the mathematical structures I study (they limit the number of types of elements being realised, and therefore keep things model theoretically nice e.g. stable). 
So I have some idea as to why these open image theorems should hold from a model theoretic viewpoint, and I know that these are regarded as important theorems, but I don't think I've ever come across a diophantine application of an open image theorem in the literature and I'd like to see one.
I'm most familiar with Serre's open image theorem for elliptic curves so an example in this context would be ideal.

REPLY [5 votes]: Here's an application to independence of Heegner points. (But if you search on MathSciNet for papers that reference Serre's two results, I expect you'll find a very large number of applications.)
Let $E/\mathbb{Q}$ be an elliptic curve with no CM, and let $\Phi:X_0(N)\to E$ be a modular parametrization. (Wiles et.al. show that $\Phi$ exists for all such $E$.) The modular curve $X_0(N)$ has special points called Heegner points associated to pairs $(C,\Gamma)$, where $C$ is a CM elliptic curve and $\Gamma\subset C$ is a cyclic subgroup of order $N$. More precisely, we can associate to each imaginary quadratic field $K$ (satisfying some conditions) a Heegner point $x_K\in X_0(\overline{\mathbb{Q}})$ associated to the maximal order in $K$.
Theorem [1] Let $K_1,\ldots,K_r$ be distinct imaginary quadratic fields such that the odd parts of their class numbers are sufficiently large. Then the points $\Phi(x_{K_1}),\ldots,\Phi(x_{K_r})$ are linearly independent in the group $E(\overline{\mathbb{Q}})$.
The proof uses Serre's image of Galois theorem in a crucial way. Not simply that the image of Galois is open in each $\hbox{Aut}(T_\ell(E))$, but also that it is surjective for almost all $\ell$.
[1] M. Rosen, JH Silverman, On the independence of Heegner points associated to
distinct quadratic imaginary fields, Journal of Number Theory
127 (2007), 10-36.<|endoftext|>
TITLE: Genus 2 curves vs Abelian surfaces
QUESTION [6 upvotes]: In the Satake compactification of abelian surfaces we have the following degeneration of a family of abelian surfaces in $\mathbf{H}_2$
$lim_{t \to \infty}\begin{pmatrix} it & b \\\ b & \tau\end{pmatrix} = \tau.$
Since we have that $M_2$ is an open of $A_2$, it is natural to look for a family of genus 2 curves depending on $t$ which gives the previous family of period matrices.
Can you describe explicitely such a family of genus 2 curves?

REPLY [3 votes]: The analytic solution is easy enough to describe: Compute the gradients of the Theta function at the six odd 2-torsion points, and projectivize these gradients. You now have six points on the projective line. This are the 6 Weierstrass points of the curve, alternatively there is the "Rosenhaon normal form", which expresses the Weierstrass points in terms of (quotients and products of) values of the theta function at even 2-torsion point.
In light of these two construction, and since theta functions involve non algebraic functions, I doubt the existence of a an algebraic expression for the family of curves you want.<|endoftext|>
TITLE: A line bundle that does not admit a G-linearisation
QUESTION [18 upvotes]: I have been thinking about quotients lately and pondered the following:
Let $G$ be a connected linear algebraic group and $X$ a $G$-variety where the action is the morphism $\sigma:G\times X\rightarrow X$. Let $p:L\rightarrow  X$ be a line bundle on $X$. 
A $G$-linearisation of $L$ is an action of $G$ on $L$ such that $p(g\cdot l)= g\cdot p(l)$, for $l\in L, g\in G$, and which restricts to a linear isomorphism $L_{x}\rightarrow L_{g\cdot x}$ on the fibres. This last condition can be expressed as saying that there is an isomorphism $L\rightarrow g^{\ast}L$, for each $g\in G$ (here $g^{\ast}L$ is the pullback bundle by the automorphism $g$ of $X$). In fact, since $G$ is connected, a $G$-linearisation of $L$ exists if and only if there is an isomorphism $p_{2}^{\ast}L\rightarrow \sigma^{\ast}L$ of bundles on $G\times X$, with $p_{2}$ the projection to $X$.
It is known (Corollary 7.2, p.109, 'Lectures on Invariant Theory' - Dolgachev) that if $X$ is normal then for any $L$ there is some power of $L$ that admits a $G$-linearisation.
Question 1: Can someone provide an example of a non-normal $G$-variety $X$ and a line bundle $L$ for which no power $L^{n}$ admits a $G$-linearisation?
Question 1': If no such example can exist can someone point me towards the literature (if any) where this question is addressed?
The existence result for normal $X$ relies on that fact that there is an exact sequence
$ 0\rightarrow K \rightarrow Pic^{G}(X)\rightarrow Pic(X) \rightarrow Pic(G)$
and that $Pic(G)$ is finite. Here $K$ is the group of rational characters of $G$ and $Pic^{G}(X)$ is the group of line bundles with a $G$-linearisation (or line $G$-bundles in Dolgachev's terminology).
Question 2: Can we extend the exact sequence
$0\rightarrow K \rightarrow Pic^{G}(X) \rightarrow Pic(X)$
to the right for arbitrary $X$ and in a 'canonical' manner? (i.e., is this exact sequence the tail of a canonical long exact sequence for any $G$-variety X?) 
Question 2': If so, what groups appear? Do they have any 'down-to-earth' interpretations? (e.g., we have $Pic(G)$ appearing for normal $X$).
Thanks in advance and apologies if this is standard material in GIT - I only have a copy of Dolgachev's notes at hand and these questions are not addressed.

REPLY [18 votes]: Let me give an example showing that the normality hypothesis is necessary.
Let $Y=\mathbb{P}^1$ with natural $G=\mathbb{G}_m$-action. Let $X$ be the $G$-variety obtained by glueing transversally the two fixed points $0$ and $\infty$. Consider the line bundle $\mathcal{O}(l)$ with $l\neq 0$ on $Y$ and glue the fibers over $0$ and $\infty$ using any linear isomorphism to obtain a line bundle $\mathcal{L}$ on $X$. Suppose that $\mathcal{L}$ has a $G$-linearisation. Pulling it back to $Y$, we obtain a $G$-linearisation of $\mathcal{O}(l)$ on $Y$ such that $G$ acts on the fibers over $0$ and $\infty$ with the same character. However, the description of the $G$-linearisations of $\mathcal{O}(l)$ when $l\neq 0$ shows that this is not possible (more precisely, for any $G$-linearisation of $\mathcal{O}(l)$, the characters through which $G$ acts on the fibers over $0$ and $\infty$ differ by the character $t\mapsto t^l$). This argument shows moreover that no multiple of $\mathcal{L}$ has a $G$-linearisation.
Note that it follows that there is no ample $G$-linearised line bundle on $X$.
As for the second question, the natural map to study is more likely to be $Pic^G(X)\to Pic(X)^G$ where $Pic(X)^G$ denotes the group of line bundles whose class in $Pic(X)$ is $G$-invariant. When $X$ is normal and proper, its Picard group is an extension of a discrete group by an abelian variety so that if $G$ is linear connected, $G$ acts necessarily trivially on $Pic(X)$ and $Pic(X)^G=Pic(X)$. However, when $X$ is not normal, this is not the case anymore (for instance in the above example). 
Moreover, the arguments in Dolgachev's notes extend to show that, if $X$ is an integral proper variety over an algebraically closed field endowed with an action of a connected linear algebraic group $G$, there is an exact sequence : $$0\to K\to Pic^G(X)\to Pic(X)^G\to Pic(G).$$
A more general statement, that does not assume $X$ proper or $G$ affine, may be found in [Brion, On linearization of line bundles, arXiv:1312.6267, Proposition 2.10].<|endoftext|>
TITLE: Stationary, ergodic measures from the structuralist point of view
QUESTION [5 upvotes]: Stationary, ergodic measures are a class of objects very familiar to probabilists. In a sense, these are the weakest generalization of the classic case of independent, identically distributed random variables. I am working on a model from mathematical physics involving such measures, and I would like to express them in the language of Dmitri Pavlov's localizable measurable spaces.
Let $Q = \mathbb Z^d$, and let $X$ be a finite set. Consider the space $S = X^Q$, equipped with the product topology. Since $X$ is finite, $X^Q$ is compact. 
Elements of $S$ are spin fields, and there is the natural isomorphism from $S$ to $S' = C(Q,X)$. There is probably a nice name for the topology on $S'$ induced by the product topology $S$, but I don't know what it is.
Question 1: The product topology on $S = X^Q$ corresponds to what topology on $S' = C(Q, X)$?
Let $G \cong \mathbb Z^d$ be the translation group of $Q$. This naturally acts on $S$. i.e., if $\tau_v \in G$, then $\tau_v(s)(q) := s(v + q).$
Let $\mathcal B(S)$ denote the Borel $\sigma$-algebra of $S$. Following the perspective of Dmitri Pavlov, I would like to turn $S$ into a localizable measurable space so that I can consider families of measures on $S$. 
To do this, I need a natural $\sigma$-ideal $\mathcal N(S)$ which is closed under the action of $G$, and for which the quotient $\mathcal B(S) / \mathcal N(S)$ is a complete Boolean lattice. 
Question 2: Does there exist a natural $\sigma$-ideal so that $(S, \mathcal B(S), \mathcal N(S))$ is a localizable measurable space?
Maybe this is too general; the classical existence of many measures on $S$ suggets that there are lots of such $\sigma$-ideals! (take any complete measure, and let $\mathcal N$ be its collection of null sets)
However, I want to build ergodicity into the definition too. We say that $\mathcal N$ is an ergodic $\sigma$-ideal when $$A \triangle \tau^{-1} A \in \mathcal N \mathrm{~for~all~} \tau \in G \mathrm{~implies~} A \in \mathcal N \mathrm{~or~} S-A \in \mathcal N.$$ That is, if a measurable set $A$ is effectively translation-invariant, then it must have either zero or full measure.
Supposing that such an ergodic $\sigma$-ideal exists, let $M(S)$ denote the space of real-valued ergodic measures on $(S, \mathcal B(S), \mathcal N(S))$. Measures push-forward, so the group $G$ naturally acts on $M(S)$. We say that a measure is stationary if it is invariant under translations. i.e., $\mu = \tau_* \mu$ for all $\tau \in G$.
Question 3: What is the structure of the space of stationary, ergodic measures?

REPLY [4 votes]: I'm not familiar enough with the notion of localisable measurable space in Pavlov's sense to say anything too authoritative, but I can make the following comments which apply to at least the case $d=1$ (as a dynamicist working mostly with group actions generated by a single continuous map $f\colon X\to X$ this is the natural setting for me):

An ergodic measure is uniquely defined by its collection of null sets, in the following sense:  if $\mu$ is an ergodic stationary measure and $\nu\ll \mu$ is also ergodic and stationary, then $\nu=\mu$.  This is just because the Radon-Nikodym derivative $d\nu/d\mu$ is an invariant function and hence a.e.-constant.
As discussed in this other question, there are many examples of dynamical systems (including in particular full shift spaces, which is the dynamicist's way of talking about the set of spin fields when $d=1$) for which the space of stationary ergodic measures is the set of extreme points of a Poulsen simplex.  In particular it is path-connected and dense in its convex closure.<|endoftext|>
TITLE: ordered exponential of unbounded operators
QUESTION [10 upvotes]: Let $H$ be a Hilbert space,
and let $A_t$ be a family of unbounded positive (self-adjoint) operators on $H$ parametrized by $\mathbb t\in R_{\ge 0}$. Consider the ordinary differential equation
$$
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\frac{d}{dt} E_t = -A_tE_t
\quad\qquad\qquad\qquad\qquad\qquad\qquad(1)
$$
that defines the ordered exponential of the family $A_t$.
If $A_t=A$ is independ of $t$, then the solution of the ODE is the usual exponential $E_t=e^{-tA}$.
Note that the above operators $E_t$ are bounded.
I suspect that, if I put appropriate hypotheses on $A_t$, (such as having a common dense domain, depending continuously on $t$, whatever that might mean, etc.) the solution of (1) will also be bounded. Intuitively, it's kind of clear:
$$
E_t = \lim_{N\to\infty} \Big(e^{-\frac t N  A_t} \cdot e^{-\frac t N A_{t(1-1 /N)}} \cdot e^{-\frac t N A_{t(1-2 /N)}}\cdots  \cdot e^{-\frac t N A_{t(3 /N)}} \cdot e^{-\frac t N A_{t(2 /N)}} \cdot e^{-\frac t N A_{t(1 /N)}}\Big)
$$
and each little exponential in the above product has norm $\le 1$.
Q:
Which properties should I impose on the family $A_t$ in order for the solution of (1) to be well defined and bounded?

REPLY [2 votes]: There are zillions of articles and books on this topic, for more and more general families of operators, where usually you can choose a specific trade-off between regularity of solutions, variability of domains of the operators, and smoothness of the dependence on time.
In the "variational" setting you consider, it is probably most efficient to introduce a weak formulation based on a family of quadratic forms. 
In this case, there is a classical theory by Lions (see e.g. Chapt. 3 of Equations Differentielles Opérationelles et Problèmes aux Limites, Springer 1961), which essentially says that if the family of forms is equi-continuous and equi-coercive, and the dependence on time is merely measurable, then you have well-posedness (in a certain weak sense) of your equation (1) (even if you add an inhomogeneous term in the equation), as well as boundedness. Also Chapt. 4 of Tanabe's Equations of Evolutions (Pitman 1979) is a good reference.<|endoftext|>
TITLE: spherical orthoscheme content above 4 dimensions 
QUESTION [5 upvotes]: I know how to compute the content of orthoschemes in 3- and 4-dimensional spherical space from dihedral angles using Schlafli series computations. Can anyone direct me to a textbook description of the general computation in 5- or higher dimensional spherical space? I understand it probably involves iterated integrals, but I would like to see a detailed example such as might be given in a textbook.
It would also be helpful to know of any off-the-shelf software that performs such computations. Also, any tabulated listing of contents of 5-D and/or higher spherical orthoschemes for various dihedral angles would be helpful.
I am a self studying enthusiast rather than a mathemetician and have taken no courses on this subject. A pointer to a comprehensive textbook would be perfect.

REPLY [3 votes]: There is no software I am aware of. Nor is there any textbook. For hyperbolic orthoschemes (a very similar subject) you should check out Ruth Kellerhals' articles (there is a very detailed one in GAFA in 1995).<|endoftext|>
TITLE: Did Gauss know Dirichlet's class number formula in 1801?
QUESTION [24 upvotes]: Let $h_d$ be the number of $SL_{2}(\mathbb{Z})$ classes of primitive binary quadratic forms of discriminant $d$. It's natural to impose the hypothesis that $d$ is not at square, as we do below. 
In Carl Ludwig Siegel's paper titled The Average Measure of Quadratic Forms With Given Discriminant and Signature Siegel cites two formulae given by Gauss in Disquisitiones Arithmeticae: 
(a) $\displaystyle\sum\limits_{d= -N }^1 h_d  \sim  \frac{\pi}{18 \zeta(3)}N^{3/2}$ 
(b) $\displaystyle\sum\limits_{d = 1}^N h_d \log{\epsilon}_d  \sim  \frac{{\pi}^2}{18 \zeta(3)}N^{3/2}$ 
Where $N > 0$ and $\epsilon_{d} = \frac{1}{2}(t + u \sqrt{d})$ where $(t,u)$ is the smallest positive solution to $t^2 - ud^2 = 4$.
(Actually, Gauss restricts to consideration to binary quadratic forms with even middle coefficient correspondingly arrives at different formulae, but they're essentially the same as those above).
Siegel gives two proofs of these formulae: one proceeding from Dirichlet's class number formula together with character sum estimates due to Polya and Landau, and one via a direct lattice point counting argument. 
In light of the facts that (i) I haven't heard anyone say that Gauss's was the one to discover the class number formula and (ii) the character sum estimates seem outside of the scope of Gauss's work, I imagine that his argument was via lattice point counting. Do we have any evidence otherwise? (I checked Gauss's book and he doesn't describe his methods there.)

REPLY [25 votes]: In 1801, Gauss certainly was aware of the general procedure to obtain the class
number formula (or asymptotic results) via counting lattice points. As a matter of fact, the approach using lattice points in general, and Gauss's circle problem in particular, can already be found in Legendre's Essai sur la Théorie des Nombres in 1798, in connection with his approach to the three-squares theorem. 
There do exist a couple of posthumous papers by Gauss on this topic, which can be found in his collected works as well as in Maser's German translation of the Disquisitiones (but not, unfortunately, in the English translation). In fact Gauss attempted twice to publish his proof of the class number formula; the first attempt begins with the sentence "33 years have passed since the principles of the wonderful connection, to which this memoir is dedicated,
was discovered, as I have remarked at the end of the Disquisitiones". Here Gauss refers to the last paragraph of the Disquisitiones, where he reports to have discovered the analytic solution to a problem stated in articles. 306 and 302. The second version of his manuscript begins with the same sentence, except that the 33 years have been replaced by 36 years.
In any case what this means is that the question in your title should be answered with a firm "yes".<|endoftext|>
TITLE: which homogeneous polynomials split into linear factors?
QUESTION [13 upvotes]: Let $R$ be the set of homogeneous polynomials of degree $n$ in $d$ variables over $\mathbb{C}$. When $n>2$, the set of elements of $R$ that split into a product of linear factors forms a proper subset $S$ of $R$. 

Is $S$ an algebraic variety, or something almost as nice?
If so, how can $S$ be described implicitly, in terms of the original coefficients, without using a factorization algorithm? In other words, is there a finite set of polynomials in the coefficients which vanish if and only if $p\in S$ (or something almost as nice)?
If so, how do I compute these polynomials for each fixed $d$ and $n$?
Are there any other shortcuts for checking whether a homogeneous polynomial splits into linear factors?

REPLY [16 votes]: 1) This is the Chow variety of degree $n$ zero cycles in $\mathbb{P}^{d-1}$.
2) Yes, this collection of polynomials can be bundled together into the Brill form or covariant.
3) Rather explicit descriptions of the Brill equations can be found in the book
by Gelfand, Kapranov and Zelevinsky on resultants. There is also a paper by Rota
and Stein. But first check out Emmanuel Briand's page and in particular the articles "Covariants decomposing on totally decomposable forms" and "Brill's equations for the subvariety of factorizable forms"
and if you read French (or German) the translation of the original article by Gordan (respectively the article itself).
As an aside, analogues of the Brill equations for the variety of forms which are powers of forms of degree dividing $n$ have been given recently in my paper with Chipalkatti
"On Hilbert covariants".

REPLY [8 votes]: There are Brill's equations. Look for them at the book by Gelfand, Kapranov, and Zelevinski. In general Brill's equations do not generate the ideal of totally decomposable polynomials, see this paper.<|endoftext|>
TITLE: Point modules of quantum projective space $\mathbb{P}^n$
QUESTION [9 upvotes]: Let $A$ be a quantum $\mathbb{P}^n$ defined by 
$$
A=\mathbb{C}\langle x_1,x_2,\dots,x_{n+1}\rangle/(x_ix_j-r_{ij}x_jx_i)_{1\le i < j\le n+1}.
$$
I would like to know the set $X$ of isomorphism classes of point modules for $A$. Here a point module is a cyclic graded right $A$-module $M$ such that each graded piece of $M$ is one-dimensional. 
The set $Y$ of isomorphism classes of of point modules for the quantum $\mathbb{P}^2$
$$
\mathbb{C}\langle x_1,x_2,x_3\rangle/(x_ix_j-r_{ij}x_jx_i)_{1\le i < j\le 3}
$$
is known to be $\mathbb{P}^2$ or a union of three lines in $\mathbb{P}^2$. 
It seems known that $X$ is projective for $n=4$, but does anyone know explicit description of $X$? I tried to compute $X$ in a similar manner as the quantum $\mathbb{P}^2$ case, but it seems quite complicated. I would also appreciate it if someone give me a good description for higher dimensional $n$ case (especially $n=3,4$). 
Thank you very much.

REPLY [8 votes]: $X$ is either isomorphic to $\mathbb{P}^n$ or is the union of some faces of the fundamental $n$-simplex (on the points $v_i = [\delta_{1i} : ... : \delta_{n+1 i}]$) containing all $\mathbb{P}^1$'s making up the $1$-faces. The generic case corresponds to the collection of all these $\mathbb{P}^1$'s.
One proves this by induction, the essential reduction being that either $X$ is contained in the collection of hypersurfaces $x_1.x_2....x_{n+1}=0$ or $A$ is the twisted homogeneous coordinate ring of (ordinary commutative) $\mathbb{P}^n$ (and hence $X \simeq \mathbb{P}^n$).
Indeed, let $x_i$ act as a non-zero divisor on the point module $P$ then $P$ corresponds to a one-dimensional representation of the degree-zero part of the graded localization of $A$ at the normalizing element $x_i$. This degree zero part is a quantum polynomial ring in $n$ variables. Either these variables all commute in which case $A$ is the claimed twisted coordinate ring or one of the new variables $y_j = x_jx_i^{-1}$ vanishes on the one-dimensional representation and hence $x_j$ vanishes on $P$.
If $X \subset \mathbb{V}(x_1.x_2... x_{n+1})$ then one gets all point-modules on which $x_i$ vanishes by looking at the point-modules of the quotient $A/(x_i)$ which is again a quantum $\mathbb{P}^{n-1}$ and one gets the claimed $X$ by induction.
The induction starts with $n=1$ in which case $X=\mathbb{P}^1$ (as $A$ is a twisted coordinate ring) and $n=2$ in which case $X$ is either $\mathbb{P}^2$ or the triangle $x_1x_2x_3=0$. So for $n=3$ one has $X$ either isomorphic to $\mathbb{P}^3$ or the union of the point-modules of the $4$ quantum $\mathbb{P}^2$'s determined by the quotients $A/(x_i)$, each of these giving either a triangle or a $\mathbb{P}^2$. Etc.<|endoftext|>
TITLE: Generic Extensions and $L(V_{\lambda+1})$
QUESTION [10 upvotes]: Suppose $\lambda$ is a strong limit cardinal of cofinality $\omega$ and for $A$ a transitive set, define $L(A)$ in the usual fashion by setting 
$$L_0(A)=A;$$
$$L_{\alpha+1}(A) = L_\alpha (A)\cup \mathcal P_{Def}(L_\alpha(A));$$
$$L(A)=\bigcup_{\alpha\in Ord} L_\alpha (A).$$
In Woodin's longer article "The Continuum Hypothesis" (in LNL 19, Logic Colloquium 2000), the following facts are stated regarding $L(V_{\lambda+1})$: 
(1) If $c$ is Cohen generic over $V$ then very likely $$(L(V_{\lambda+1}))^{V[c]}\neq L(V_{\lambda+1})[c].$$
(2) On the other hand, if $G\subset Coll(\omega_1,\mathbb{R})$ is $V$-generic then  $$(L(V_{\lambda+1}))^{V[G]}= L(V_{\lambda+1})[G].$$ 
Can anyone give a (sketch of) proof of either (1) or (2)? Are these results given only in the context of a non-trivial elementary embedding $j:L(V_{\lambda+1})\prec L(V_{\lambda+1})$ with $crit(j)<\lambda$?
More generally, for a partial order $\mathbb{P}$ and a $G\subset \mathbb{P}$ which is $V$-generic, which properties of $\mathbb{P}$ are sufficient to ensure the equality $$(L(V_{\lambda+1}))^{V[G]}= L(V_{\lambda+1})[G]$$ holds? Fails? Is this even known?

REPLY [9 votes]: They are immediate consequences of Theorem 175 of W. Hugh Woodin, "Suitable extender models II: beyond $\omega$-huge", J. Math. Log., vol. 11 (2011), no. 2, pp. 115–436, which says 

If there is a (proper) elementary embedding $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ with $\text{crit}(j)<\lambda$ and if $G\subset \mathbb{P}$ is $V$-generic for some poset $\mathbb{P}\in V_\lambda$, then $V_{\lambda+1} \in L_\lambda(V_{\lambda+1})^{V[G]}$ if and only if $(\lambda^\omega)^V = (\lambda^\omega)^{V[G]}$.

This theorem can be proved using the Large Perfect Set Theorem for subsets of $V_{\lambda+1}$ in $L_\lambda(V_{\lambda+1})$, which says assuming there is a (proper) elementary embedding $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ with $\text{crit}(j)<\lambda$, those subsets of $V_{\lambda+1}$ either has size $\leq \lambda$ or contains a large perfect subset, i.e. a homeomorphic copy of $\lambda^\omega$.
Cramer recently improves the Large Perfect Set Theorem to all subsets of $V_{\lambda+1}$ in $L(V_{\lambda+1})$, so $L_\lambda(V_{\lambda+1})$ can be replaced by $L(V_{\lambda+1})$.<|endoftext|>
TITLE: when is a locally homeo a covering map?
QUESTION [14 upvotes]: Let $X$ and $Y$ be locally comapct Hausdorff spaces, and $f:X\to Y$ be a surjective local homeomorphism.
When is $f$ a covering map?
It is well-known that when $f$ is proper, $f$ is a covering map.
However I think the properness is too strong because there are many example of non-proper covering map (e.g. the universal covering of a space whose fundamental group is not finite is not proper).
Is there any other condition?

REPLY [3 votes]: Here is a proof that I wrote for my differential geometry class, of the fact that if $f:X\to Y$ is a surjective local homeomorphism between semilocally simply connected topological spaces (e.g., manifolds), which satisfies the 'path-lifting property', then $f$ is a covering map. You can also look at do Carmo's book "Differentiable curves and surfaces", page 383.<|endoftext|>
TITLE: Interpolating a sum of binomial coefficients using a sin function
QUESTION [8 upvotes]: While studying a problem about orthogonal polynomials I encountered the following
 expressions
\begin{equation}
f(n)=\sum_{k=0}^{n}(-1)^k\binom{n+k}{2k} \frac{1}{k+1}\binom{2k}{k}
\end{equation}
and
\begin{equation}
g(n)=\sum_{k=0}^{n-1}(-1)^k\binom{n+k}{2k+1} \frac{1}{k+2}\binom{2k+2}{k+1}
\end{equation}
I can prove that $f(n)=0$ for all integers $n\geq 1$ and $g(n)=0$ for all integers $n>1$ using properties of orthogonal polynomials. However,  I would like  to find an elementary proof, which might also  be more illuminating.
While doing some experiments with calculations using Mathematica, I defined the functions $f(n)$ and $g(n)$ with the above formulas but forgot to specify that $n$ is an integer value.
When I typed $f(n)$ and $g(n)$ for a generic variable $n$, I guess that Mathematica "assumed" that the variables involved were real and provided the following simple formulas:
\begin{equation}
f(x)=\frac{\sin \pi x}{x(x+1)\pi}
\end{equation}
and
\begin{equation}
g(x)=-\frac{2\sin \pi x}{(x+1)(x-1)\pi}
\end{equation}
which makes it evident that these functions are zero for all positive (and negative) integers with the possible exceptions of $0,1,-1$.
Why are these equalities true?
I realize it might have to do, perhaps, with properties of the Euler Beta and Gamma functions, but I know too little about these functions to figure out a proof along these lines. Can anybody help? Thank you!

REPLY [12 votes]: The two identities are both special cases of Vandermonde's theorem (also called the Chu-Vandermonde theorem), which is the most well-known binomial coefficient identity after the binomial theorem.
The first sum may be written
$$f(n) = \frac 1n \sum_{k=0}^n (-1)^k\binom nk \binom {n+k}{k+1}.$$
Applying the identity  $\binom{-c}m = (-1)^{m}\binom{c+m-1}{m}$, we get
$$f(n)= -\frac 1n \sum_{k=0}^n \binom nk \binom {-n}{k+1},$$
and reversing the order of summation gives
\begin{equation*}
f(n) =-\frac 1n \sum_{k=0}^n \binom nk \binom {-n}{n-k+1}.
\end{equation*}
Vandermonde's theorem is often written
\begin{equation*}
\sum_{k=0}^m \binom ak \binom b{m-k} = \binom{a+b}m.
\end{equation*}
There are many simple proofs of Vandermonde's theorem. For example, we can prove this formula by equating coefficients of $x^m$ in $(1+x)^a (1+x)^b = (1+x)^{a+b}$. 
Setting $a=n$, $b=-n$, and $m=n+1$ in this formula (and observing that the $k=n+1$ term vanishes) and using the last formula for $f(n)$ shows that for $n>0$, 
$f(n) =- \frac 1n\binom 0{n+1}=0.$
Similarly, $g(n)$ can be evaluated by Vandermonde's theorem.
I believe that Pietro's indefinite summation approach is correct, but I don't think that this is what Mathematica is doing. There is a nonterminating generalization of Vandermonde's theorem, called Gauss's theorem (see, e.g., http://mathworld.wolfram.com/GausssHypergeometricTheorem.html) that evaluates the sum $\sum_{k=0}^\infty (-1)^k \binom ak \binom {b+k}{m+k}$ in terms of gamma functions when it converges, where $a$ and $b$ are arbitrary and $m$ is an integer. (Actually Gauss's theorem is a little more general than this.) In the particular case $a=n$, $b=n$, $m=1$, the reflection formula for the gamma function can be applied to give Stefano's formula for $f(x)$ in terms of $\sin \pi x$, and similarly for $g(x)$. Mathematica is probably applying Gauss's theorem, since it knows how to convert binomial coefficient sums to hypergeometric series and it knows how to evaluate them in cases like this one.<|endoftext|>
TITLE: Proofs for doubly ruled surfaces
QUESTION [7 upvotes]: Hello,
I am interested in proofs for why the only irreducible doubly ruled surfaces in ${\mathbb R}^3$ are the one sheeted hyperboloid and the hyperbolic paraboloid. While many books and papers state that this is "well known", I could hardly find any sources that give more details. I only found the following two:

In the book "Mathematical Omnibus: Thirty Lectures on Classic Mathematics" by Fuchs and Tabachnikov there is a proof relying on rather unusual tools. The proof heavily relies on the property that the neighborhood of every (non-singular) point behaves similarly to a plane.
Various places state that we can take three lines from one generating family, and these should intersect every line of the second family. I am not sure how to prove such a claim, and couldn't find a reference with more details (it does seem much simpler in the complex projective space, where one could rely on plucker coordinates).

Could anyone provide references to proofs of this property? Or describe a proof different from the one I mentioned in item 1?
Many thanks!
Adam

REPLY [4 votes]: Here's a sketch of an argument for approach 2, under the mild hypothesis that the lines in the family of doubly ruled lines varies continuously, which I think is intuitively clear, but requires some justification.
Take a point $p$ on a doubly ruled surface $\Sigma$, and take two lines $l_1, l_2$ going through $p$. For any nearby points $p_1  \in l_1, p_2 \in l_2$, there are lines $l_2'$ intersecting $l_1$ in $p_1$, and $l_1'$ intersecting $l_2$ in $p_2$. But since $l_1$ and $l_2$ intersect on the surface $\Sigma$, and since the family varies continuously by hypothesis, $l_1'$ and $l_2'$ must intersect when $p_1,p_2$ are close enough to $p$ by general position on $\Sigma$. Similarly, for any point $q$ near $p$, there are lines $l_1'$ and $l_2'$ going through $q$ meeting $l_2$ and $l_1$ respectively by continuity of the family and general position on $\Sigma$.
Now, take points $p_1, p_1'$ on $l_1$ near $p$, together with lines $l_2'$ and $l_2''$ meeting $l_1$ in $p_1$ and $p_1'$ respectively. Then $l_1'$ must meet both $l_2'$ and $l_2''$ for points $p_2$ near $p$ on $l_2$. Thus, one sees a neighborhood $p\in U\subset \Sigma$ such that any point in $U$ lies on a line meeting these three lines $l_2,l_2',l_2''$. 
If any pair of these lines is coplanar (i.e. intersect or are parallel), then the portion of surface $U$ near $p$ must be planar. Otherwise, one has 3 skew lines. 
Now, I claim that for 3 skew lines, there is a unique surface of lines meeting all three lines, which is either a hyperbolic paraboloid or hyperboloid. This is proved by Hilbert-Cohn Vossen, but I'll give a sketch of the proof. The uniqueness follows because for any point $p$ in $l_2$, the plane spanned by $p$ and $l_2'$ meets $l_2''$ uniquely in a point $q$, and thus every point on $l_2$ goes through a unique line $\overline{pq}$ meeting $l_2'$ and $l_2''$. 
Three skew lines in $\mathbb{R}^3$ give three projective lines in $\mathbb{RP}^3$ which do not intersect. I claim that this configuration is unique up to projective transformation. 
Three skew projective lines in $\mathbb{RP}^3$ correspond to three planes $P_1,P_2,P_3$ in $\mathbb{R}^4$ meeting pairwise only in the origin. I claim $GL_4(\mathbb{R})$ acts transitively on such configurations. Take basis vectors for $P_1, P_2$, then these form a basis for $\mathbb{R}^4$. Thus up to linear transformation, we may assume $P_1=\{(1,0,0,0),(0,1,0,0)\}, P_2=\{(0,0,1,0),(0,0,0,1)\}$. Now, the subspace $P_3$ has a basis of two vectors, such that the first two coordinates are linearly independent in $P_1$, and the second two coordinates are linearly independent in $P_2$ since we assume that the planes intersect only in the origin. We may take linear transformations in $GL_2(\mathbb{R})\times GL_2(\mathbb{R})$ stabilizing $P_1, P_2$ and sending these two vectors to $\{(1,0,1,0),(0,1,0,1)\}$, and thus we have normalized the three planes, so that the action is transitive. The corresponding action of $PGL_4(\mathbb{R})$ is thus also transitive on skew projective lines. 
Take a piece of a hyperbolic paraboloid, and take 3 skew lines lying on it. The three skew lines uniquely determine the paraboloid, since it is the surface of lines meeting the three skew lines. Then there is a projective transformation taking $l_2,l_2',l_2''$ to these three skew lines, and therefore sending a portion of $\Sigma$ near $p$ to a hyperbolic paraboloid by uniqueness. 
In fact, when we compactify a hyperbolic paraboloid or a hyperboloid in $\mathbb{RP}^3$, we get a 2-torus with two foliations by projective lines. So up to projective transformation, there is only one such surface. 
Now the surface $\Sigma$ has patches which are hyperbolic paraboloids or hyperboloids. But one can see that each such surface lies in a unique doubly ruled torus in $\mathbb{RP}^3$, so $\Sigma$ must be identically such a surface intersected with $\mathbb{R}^3$.<|endoftext|>
TITLE: Clean introduction to toric varieties for an undergraduate audience
QUESTION [6 upvotes]: I will be giving a talk to a (primarily) undergraduate audience on certain relatively concrete computations with toric varieties and their blowups. The talk is short, about 20 mins. As I result I need to introduce toric varieties (smooth, projective) and I would like to have my audience understand that the combinatorics of a fan/polytope can be used to understand the geometry of the toric variety.
When I've given similar talks to similar audiences in the past, I have been very hand-wavey just saying "there's a copy of $(\mathbb{C}^*)^n$ inside my variety and as a consequence we have this magical polytope encoding all this information." 
The Question: How would you motivate and introduce toric varieties quickly? Is there some classical hands-on example that you find particularly illustrative?

REPLY [12 votes]: David Cox has some nice expositions on toric varieties on his web page here. Cox is also one of the authors of the book "Toric Varieties", which is a very readable, yet comprehensive introduction to toric varieties. The first chapter here should provide you with enough motivation and examples for your talk. Then there is also chapter 1 in Fulton's book, which is the classic reference on the subject.
As for the motivational examples, you should look for examples that show the real power of toric varieties: That abstract algebro-geometric constructions can uaually be viewed very concretely by working with the defining combinatorial data (e.g. the fan). Some of these examples might do the trick:
1) The quadric surface $\{xy-zw=0\}$ in $\mathbb P^3$ and its affine cone in $\mathbb A^4$
2) The singular quadric $y^2=zw$ in $\mathbb A^3$. 
3) Hirzebruch surfaces 
4) Toric blow-ups and subdivisons of the fan
In the basic examples 1)-3), it is straightforward to write out the action of the torus, and see directly how monomials in the coordinate ring relates to the lattice points in the dual cones. Also, in the projective examples you can see how gluing the affine toric varieties works in terms of the fan data. 
These examples demonstrate typical features of toric varieties, for example that their ideals are generated by binomials and their Chow ring is generated by the torus invariant subvarieties. 
I like the Hirzebruch surface example because you can somehow 'see' the $\mathbb P^1$-bundle structure in the defining polytope and it is intuiticely clear that any toric surface is a blow-up of either $\mathbb P^2$ or a Hirzebruch surface. Moreover, I think it's pretty cool that you can view birational morphisms of toric varieties (e.g., resolution of singularities) as subdivisions of the defining fans. The example in this MO thread illustrates this. Another interesting example is the affine cone $Z(xy-zw=0)\subset \mathbb A^4$ which gives a nice combinatorial interpretation of the Atiyah flop.<|endoftext|>
TITLE: Jets of Equivariant Vector Bundles
QUESTION [5 upvotes]: Let $M$ be a (compact) $G$-homogeneous space with fibre group $H$, and let ${\cal E}$ be a $G$-equivariant $k$-dimensional vector bundle over $M$ with corresponding representation $\pi:H \to $R$^k$. What I would like to know is whether all the jet bundles  (see here for a definition of jet bundle) of $\cal E$ are also $G$-equivariant, and if so, can one construct their corresponding representations from $\pi$?

REPLY [2 votes]: See chapter IV and  section 32 (and others) of the book:
Ivan Kolár, Jan Slovák, Peter W. Michor: Natural operations in differential geometry. Springer-Verlag, Berlin, Heidelberg, New York, (1993)
(pdf).<|endoftext|>
TITLE: References request on the algebraic geometry of projective homogeneous spaces
QUESTION [9 upvotes]: Hi everybody.
Let $G/P$ be a complex projective homogeneous variety with $G$ a simple Lie group and $P$ a parabolic subgroup.
I believe that it is possible to 

(1) describe ${\rm Pic}(G/P)$
(2) characterize the ample line bundles and 
(3) express the canonical class of $G/P$ 

in terms of the nodes corresponding to P in the Dynkin diagram of $G$.
Are there some canonical (or a least some good) references where this is explained?
This is certainly "well-known by the experts" but I'm not one of them... 
Thanks to those who will answer.

REPLY [2 votes]: In addition to the references already mentioned, let me recommend Dennis Snow's excellent notes on homogeneous vector bundles. They cover everything you ask for (and much more):

For a description of ${\rm Pic}(G/P)$ in terms of weights, see Theorem 6.4.
Ample line bundles are characterized in Theorem 6.5.2.
The canonical class is described on page 37, right above Definition 10.2.<|endoftext|>
TITLE: Is there a "geometric" intuition underlying the notion of normal varieties?  
QUESTION [69 upvotes]: I first got concious of the notion of normal varieties around 3 years ago and despite the fact that by now I can manipulate with it a bit, this notion still puzzles me a lot.
One thing that strikes me is that the definition of normality is so entirely algebraic.
From my common sense understanding the notion of normal varieties restricts the class of spaces that we consider to more-less reasonable ones. It looks to me that this definition is analogous to the definition of pseudo-manifold. At least the obvious similarity is that in both cases the set of non-singular points is connected.
Normality pops up everywhere and its definition is very short. But it is hard for me to imagine that a differential topologist or differential geometer could come up with such a definition.
Why is the notion of normatilty is so omnipresent? 
What is "geometric" meaning of normality?  
Maybe a more concrete question would be like this. Suppose $X$ is an irreducible algebraic subvariety in $\mathbb C^n$ with singularities in co-dimension $2$. Can one somehow looking on singularities, their stratification and the way $X$ lies in $\mathbb C^n$ say if it is normal or not? 
Added. Who was the person who invented this notion? 
I would like to thank everybody for useful comments and links.

REPLY [13 votes]: An excellent non-algebraic meaning (using analysis) of normality is found in Kollar's article in the Bulletin of AMS (1987).
Restrict to irreducible varieties $X$ so we can talk of function fields.
A point $x_0\in X$ is considered normal whenever a rational function exhibits decent behaviour in a neighbourhood of $x_0$ then it finds a place in the local ring of $X$ at $x_0$.
Decent behaviour here is: If $f\in K(X)$  and if  $|f(x)|$ remains a bounded function  as $x$  approaches $x_0$ by  paths lying in $X$, then $x_0$ should be good enough to admit $f$ in its local ring. 
This survey article of Kollar is about Mori's Fields-medal winning work on 3-folds. But it starts from the scratch defining what  an algebraic variety is. It is a great  source to learn the meanings of fundamentals objects of algebraic geometry. (for example  Kollar explains why we have to deal with line bundles when we study projective varieties).<|endoftext|>
TITLE: Contractibility of the space of collars
QUESTION [7 upvotes]: I'm looking for a reference or a proof of the following statement :
Let $M$ be a compact smooth manifold with boundary.  Then the space of embeddings $\partial M\times[0,1]\to M$ inducing the identity $\partial M\times\{0\}\to \partial M$ is contractible.

REPLY [9 votes]: The theorem you're looking for is proven in Cerf's dissertation. 
J. Cerf, Topologie de certains espaces de plongements, Bull S.M.F., tome 89 (1961) 227-380. 
This is for the case you mention, when you look at the space of $C^k$-smooth embeddings with the $C^k$-Whitney/weak topology.   Cerf proves a lot of other similar foundational results about embedding spaces, tubular neighbourhoods, collars, restriction maps and so on. 
To set up the proof, just take your favourite proof that collars are unique up to isotopy.  These proofs can all be souped-up to work at the embedding space level, you just have to make smart choices for your maps.   
For example, consider the space of smooth embeddings $f : [0,1] \to [0,\infty)$ with $f(0)=0$.  Here is a homotopy:
$$F(t,x) = \frac{f((1-t)x)}{1-t} \text{ for } t < 1$$
and
$$F(1,x) = f'(0)x$$
Notice that this is an isotopy from $f$ to a linear map.  The space of linear embeddings is contractible. 
The proof for general collar neighbourhoods is not much more sophisticated than the above, but you have the complication of not having such nice coordinates in your manifold.  You can make up for that in various ways, one being embedding the manifold in Euclidean space and using tubular neighbourhoods.  Another would be to use flows.<|endoftext|>
TITLE: On the representation of a (real) square matrix as a product of two symmetric matrices
QUESTION [5 upvotes]: (For this question, all matrices are real).
According to the ancient paper "Über die Darstellbarkeit einer Matrix als Produkt yon
zwei symmetrischen Matrizen, als Produkt yon zwei
alternierenden Matrizen und als Produkt yon einer symmetrisehen and einer alternieenden Matrix"  by H. Stenzel in Göttingen (1922)
which I cannot really read fully, since it is in German, any square matrix can be written as a product of two symmetric matrices (one of which is non-singular). If we strengthen the conditions, such that one of the factors must be positive-(semi)definite, what can we say? Any way of characterizing square matrices which can be written as a product of a symmetric and a symmetric positive-semidefinit matrix?
If $A$ is symmetric positive-definite and $B$ is symmetric, then the product $AB$ is similar to a symmetric matrix, so has real eigenvalues. So if any square matrix could be written such, all square matrices would have real eigenvalues, which is absurd.  So there must be some restriction.
A version of the paper is here.
And, any more recent references for this problem?

REPLY [3 votes]: The only restriction is to be diagonalizable with real eigenvalue. For if $M=PDP^{-1}$ with $P,D$ real and $D$ diagonal, then $M=AB$ with $B=P^{-T}DP^{-1}$ symmetric and $A=PP^T$ is PSD. And conversely, such a product is similar to the symmetric matrix $A^{1/2}BA^{1/2}$, hence is diagonalizable with real eigenvalues.
About the fact that every square real matrix is the product of two Hermitian matrices (complex counterpart of what you mention), see Factorization of a real matrix into Hermitian x Hermitian. Is it stable ? .

REPLY [3 votes]: I don't have an answer, but slightly more restrictions.
If $A$ is symmetric positive semidefinite and $B$ is symmetric, $AB = A^{1/2} A^{1/2} B$ has the same characteristic polynomial as the symmetric matrix $C = A^{1/2} B A^{1/2}$, and in particular has real eigenvalues.  If $u$ is an eigenvector of $C$ for nonzero eigenvalue $\lambda$ then $A^{1/2} u \ne 0$ and $AB A^{1/2} u = \lambda A^{1/2} u$, so $A^{1/2}$ maps the eigenspaces 
of $C$ for nonzero eigenvalues to the corresponding eigenspaces of $AB$.
In particular, the geometric and algebraic multiplicities of a nonzero eigenvalue of $AB$ must be equal.  However, this may not be the case for eigenvalue $0$, e.g. 
$$ \pmatrix{0 & 1\cr 0 & 0\cr} = \pmatrix{1 & 0\cr 0 & 0\cr} \pmatrix{0 & 1\cr 1 & 0\cr}$$
EDIT:
On the other hand, $AB$ can't have a Jordan block of size greater than $2$ for eigenvalue $0$.  In fact, suppose $(AB)^3 v = 0$.
Since $\text{Ker}(A) = \text{Ker}(A^{1/2})$, this says $ A^{1/2} B A B A B v = (A^{1/2} B A^{1/2})^2 A^{1/2} B v = 0$, and since $\text{Ker}((A^{1/2} B A^{1/2})^2) = 
\text{Ker}(A^{1/2} B A^{1/2})$ we have
$(A^{1/2} B A^{1/2})A^{1/2} B v = A^{1/2} B A B v = 0$ and thus $(AB)^2 v = 0$.
Since being of the form $AB$ with $A$ and $B$ symmetric and $A$ positive definite is a similarity invariant, i.e. for any matrix $AB$ of this form and any invertible $S$,
$SABS^{-1} = (SAS^T)((S^{-1})^T B S^{-1})$, we see that a necessary and sufficient
condition is that the eigenvalues are real and the only  Jordan blocks of size greater than $1$ are
 $\pmatrix{0 & 1\cr 0 & 0\cr}$.<|endoftext|>
TITLE: Is $\mathbb{H}P^\infty_{(p)}$ an H-space?
QUESTION [26 upvotes]: Put $X=\mathbb{H}P^\infty$ (so $X$ classifies quaternionic line bundles, and $\Omega X=S^3$). There is no obvious reason for $X$ to be an H-space, because the tensor product of quaternionic vector spaces is not naturally a quaternionic vector space.  Below I will prove that there is no nonobvious H-space structure.  However, the obstruction that I use has order $12$ and so vanishes if we localise at a prime $p>3$.  My guess is that $X_{(p)}$ is not an H-space for any prime $p$; does anyone know a proof of that?
Note that $H^*(X)=\mathbb{Z}[y]$ with $|y|=4$, and this has a Hopf algebra structure given by $\psi(y)=y\otimes 1+1\otimes y$, which is compatible with all Steenrod operations.  Thus, there do not seem to be any primary obstructions.
However, if $X$ were an H-space then $S^3=\Omega X$ would have two commuting binary operations with the same identity and so (by a standard argument) they would be the same and would be commutative.  However, it is known that $S^3$ is not homotopy commutative: the commutator map $S^6=S^3\wedge S^3\to S^3$ is the standard generator $\nu'$ of $\pi_6(S^3)\simeq\mathbb{Z}/12$.

REPLY [9 votes]: Sorry for dredging up this question, but here is another argument (at least for $p$ odd, but maybe you don't need this) that came up while thinking about an unrelated problem. If $\mathbf{H}P^\infty_{(p)}$ was a H-space, then all Whitehead products must vanish, so it suffices to establish that there's a nontrivial Whitehead product. One that does not vanish is the following: take the element $\iota:\mathbf{H}P^1\hookrightarrow \mathbf{H}P^\infty$ in $\pi_4(\mathbf{H}P^\infty)$; then, the $(p+1)/2$-fold Whitehead product $[\iota, \cdots, \iota]\in \pi_{2p+1}(\mathbf{H}P^\infty)$ is nonzero. Under the isomorphism $\pi_{2p+1}(\mathbf{H}P^\infty) \cong \pi_{2p}(S^3)$, it's precisely the unstable representative for $\alpha_1$, aka the first nontrivial attaching map in $\mathbf{C}P^{p}_{(p)}$.<|endoftext|>
TITLE: A weird function related to the denominators of rational squares
QUESTION [6 upvotes]: Between any consecutive integers $a$ and $a+1$ there are infinitely many rational squares of the form $t^2 / s^2$. I have been working to understand the following question: How small can $t$ and $s$ be? That is, let $\sigma (a)$ denote the least natural number $s$ for which there exists a natural number $t$ such that $(s^2)a < t^2 < (s^2)(a + 1)$. What are the properties of the function $\sigma (a)$?
It's not hard to calculate $\sigma (a)$ numerically, and the graph of the function is weird.  Full of crazy fluctuations but bounded by a smooth curve both above and below.  I've been working on this for a while now and have found some nice partial results, including sharp formulas for the lower and upper bounds, a criterion for when the upper bound is attained, and several special cases. (I'll be happy to share those with anybody who asks.)  I am at the point where I think I need to find out if anybody else has worked on this sort of thing before. Standard websearches haven't turned anything up. Does anybody know if this question has been previously investigated?

REPLY [2 votes]: I found this on the OEIS, but it doesn't list much information, so I don't know whether it's been studied before.  One way to look at it is that you're looking for the rational number with the smallest denominator between $\sqrt{n}$ and $\sqrt{n+1}$.  There are algorithms that use continued fractions to calculate the "best" rational in any given interval; see here.  Square roots have particularly nice continued-fraction representations, so you might even be able to get some sort of formula for $\sigma$.<|endoftext|>
TITLE: Motivation behind the construction of Deligne and Lusztig
QUESTION [20 upvotes]: If $G$ is a connected reductive group over a finite field $\mathbb{F}_q$ and $T$ is a maximal torus in $G$, the famous construction of Deligne and Lusztig (Annals of Math, 1976) associates representations of $G(\mathbb{F}_q)$ to $1$-dimensional representations of $T(\mathbb{F}_q)$. These representations come from the cohomology of the Deligne-Lusztig variety associated to $G$ and $T$, which admits commuting actions of the groups $G(\mathbb{F}_q)$ and $T(\mathbb{F}_q)$. According to various remarks in the Deligne-Lusztig article, two of their sources of motivation were as follows:
1) The conjecture of Macdonald that a construction of this sort should exist.
2) The example of the Drinfeld curve: if $G=SL_2$ and $T$ is the unique (up to conjugacy) non-split maximal torus in $G$, then $T(\mathbb{F}_q)$ can be identified with the kernel of the norm homomorphism
$$\mathbb{F}_{q^2}^\times\to\mathbb{F}_q^\times$$
and $G(\mathbb{F}_q)$ and $T(\mathbb{F}_q)$ both act naturally on the curve $X$ given by the equation $x^qy-xy^q=1$ in the affine plane over $\overline{\mathbb{F}}_q$. (The group $T(\mathbb{F}_q)$ acts by dilations.) The (first) $\ell$-adic cohomology of $X$ realizes all cuspidal irreducible representations of $SL_2(\mathbb{F}_q)$.
In their 1976 paper Deligne and Lusztig give two different constructions of (what later became known as) Deligne-Lusztig varieties (and proved that they are equivalent). The Drinfeld curve turns out to be a special case. However, to me it seems like the jump from the example of the Drinfeld curve (and MacDonald's conjecture) to either of the two general constructions of Deligne-Lusztig is absolutely giant. I was wondering if someone has some additional insight into how the construction was invented.

REPLY [24 votes]: Here is a metaphor which is probably well-known. (I'm repeating what Nick is saying with mild variation.)
When one considers representations of $SL_2(\mathbb{R})$ there are two fundamental homogenous spaces: $\mathbb{P}^1(\mathbb{R})$ and the upper half-plane $\mathbb{H}$. Sections of line bundles ($L^2$ or holomorphic) on these two spaces realize all the admissible representations of $SL_2(\mathbb{R})$.
One can look at this slightly differently: one considers the fundamental homogenous space for the complexified group i.e. $\mathbb{P}^1(\mathbb{C})$ and one decomposes into its rational points $\mathbb{P}^1(\mathbb{R})$ and its complement (the upper and lower hemi-spheres). These two spaces are just (complex conjugate) incarnations of $\mathbb{H}$.
Now do the same thing for $SL_2(\mathbb{F}_q)$: one has the natural action of $SL_2(\mathbb{F_q})$ on $\mathbb{P}^1$ and one can consider the rational points $\mathbb{P}^1(\mathbb{F}_q)) = SL_2(\mathbb{F}_q)/B$ and the complement $X := \mathbb{P}^1 \setminus \{ 0, 1, \dots, p-1 \}$ (an affine algebraic curve). Pursuing the above analogy we might be tempted to call $X$ the "upper half plane" in this context.
If one is optimistic then one might hope that global sections / cohomology of local systems on $X$ realise interesting representations of $SL_2(\mathbb{F}_q)$, which is indeed the case. In this picture the Drinfeld curve emerges as a $T(\mathbb{F}_q)$ (your notation) cover of $X$ which affords a family of interesting local systems via direct image. (Indexed by the characters of $T(\mathbb{F}_q)$.)
In this point of view (local systems on $X$, rather than isotypic components in the cohomology of a cover) the Drinfeld curve loses some of its significance.
Note that there are two elements of the Weyl group: 1 and $s$. We may see $\mathbb{P}^1(\mathbb{F}_q)$ and $X$ as points in relative position $1$ (i.e. equal) and $s$ (i.e. not equal) with their image under Frobenius. Now it is perhaps clearer how to generalise.
(A side remark to be taken with a grain of salt: someone told me once that Drinfeld had the real case in mind when defining Drinfeld space, which is the analogue for p-adic groups. The finite field case was a useful testing ground.)<|endoftext|>
TITLE: Algebraic integers in skew fields
QUESTION [6 upvotes]: Hi everyone,
let $D$ be a skew field, which is finite dimensional over its center $k$. Assume that $k$ is a number field, and let $\mathcal{O}_D$ be the set of elements $z\in D$ which are roots of a monic polynomial with coefficients in $\mathcal{O}_k$ . Is $\mathcal{O}_D$ a subring of $D$ ?
Thanks!
G.

REPLY [10 votes]: No, the "integral" elements are not a subring. There are at least two ways to understand this.
In the case of $D$ being the integral Hamiltonian quaternions, or even the Hurwitz integers therein (adjoining $(1+i+j+k)/2$ to give a maximal subring), we can easily conjugate ourselves to another subring: using the model of $D$ inside two-by-two complex matrices spanned over $\mathbb R$ by $\pmatrix{1 & 0 \cr 0 & 1}$, $\pmatrix{i & 0 \cr 0 & -i}$, $\pmatrix{0 & 1 \cr -1 & 0}$, and $\pmatrix{0 & i \cr i & 0}$, conjugating by diagonal elements $\pmatrix{a+bi & 0 \cr  0 & a-bi}$ where $a\pm bi$ have some odd Gaussian prime factors not in common will move us out of the Hurwitz integers.
This is a direct manifestation of the point that subrings of $D$'s finitely-generated as $\mathbb Z$-modules are obtained by taking products of maximal compact subrings of all the $D\otimes_{\mathbb Q} \mathbb Q_p$ and intersecting with $D$. At primes $p$ where $D$ becomes the matrix algebra, there is by-far not a unique such maximal subring. Approximating this globally gives the kind of counterexample in the previous paragraph.<|endoftext|>
TITLE: State of the art for integral models of PEL type Shimura varieties with deep level structure
QUESTION [7 upvotes]: The theory of PEL type Shimura varieties is nowadays well developed, but it is not easy to be updated with the latest results. Here I am particularly interested in integrals models. Let me describe what I understand.
Let $B$ be a (semi)simple algebra over $\mathbb Q$, of finite dimension. We suppose that $B$ is endowed with a positive involution $\ast$. Let $\mathcal O_B$ be an order of $B$, preserved by $\ast$. Let $(V,\Psi)$ be a finitely generated symplectic left $(B, \ast)$-module. Let $h \colon \mathbb C \to End_{B_{\mathbb R}}(V_{\mathbb R})$ be an $\mathbb R$-algebras homomorphism that gives an Hode strucure of type $(1,0),(0,1)$ on $V_{\mathbb R}$. We can now define a Shimura datum $(G,X)$ in the usual way. We obtain in particular a family of complex varieties $S_K$ parametrized by compact open subgroup $K \subseteq G(\mathbb A_f)$. We will assume that $K$ is 'small enough', in particular these varieties are moduli spaces of abelian varieties with additional (PEL) structure. It turns out that there is a number field $E$, called the reflex field, such that $S_K$ admits a canonical model defined over $E$.
In arithmetic it is very interesting to consider integral model of $S_K$. We fix a rational prime $p$. We assume that there is a lattice $\Lambda \subseteq V_{\mathbb Q_p}$ that is self-dual for $\Psi$ and we fix a compact open, small enough, subgroup $K^p \subseteq G(\mathbb A_f^p)$. Assuming that $B$ splits over an unramified extension of $\mathbb Q_p$, we have that $G(\mathbb Q_p)$ admits an hyperspecial subgroup, that we denote $K_{0,p}$. We assume that $B$ is of type A or C (another question is what can be done in the case D). It is well known that $S_{K^pK_{0,p}}$ admits a canonical integral model over $\mathcal O_E$, that is smooth over $\mathcal O_E \otimes Z_p$ and solves a very reasonable moduli problem. This goes back to Kottwitz.
Question 1 What can be done without the unramifiedness assumption? Of course in this case we do not have an hyperspcial subroup of $G(\mathbb Q_p)$, so we need a level strucure also at $p$. Rapoport and Zink have defined some integral models that satisfy a moduli problem, but it seems that their models are not even flat over the base.
Let me go back to the unramified case. For different reasons, it is interesting to consider level structures at $p$ (for example of type $\Gamma_1(Np^m)$ or $\Gamma_1(N) \cap \Gamma_0(p^m)$ in the case of modular curves). Now there is no hope for a smooth model, but of course one wants a good integral models. I am in particular interested in Iwahoric (or, more generally, parahoric) level structure. In the Siegel case, for example, we have good integral models.
Question 2 Under which assumptions we have good (flat and with a moduli interpretation) model of Shimura varieties with Iwahoric level structure at $p$? We have the models of Rapoport and Zink, but I do not if they are flat in general. Some cases are studied by Görtz (http://arxiv.org/abs/math/9912064 and http://arxiv.org/abs/math/0011202), but it seems that the general case is open (here I am always assuming that $G$ is quasi-split).
In general, I am interested in various condition 'at $p$' one have to put in order to obtain good integral models of PEL type Shimura varieties.
Thank you!

REPLY [7 votes]: Here are some remarks:
Rapoport's Guide paper which you mention in your comment certainly is a good starting point. There is also a survey paper by Pappas, Rapoport and Smithling with many more recent results; it focusses on the "local model side", but the understanding of the local model more or less is enough to understand the Shimura variety side.
Generally speaking, when the group splits over an unramified extension of $\mathbb Q_p$, the original Rapoport-Zink model is flat (types $A$ and $C$). (Re your comment: The general case can be reduced to the case $B=F$ by a Morita equivalence argument.)
If the group does not split over an unramified extension, the original (now called: naive) local model is typically not flat. Pappas and Rapoport have studied this case in a series of papers. (Of course one can pass to the flat closure in order to obtain a flat model; then the question becomes giving a moduli description of the flat closure, and maybe describe its geometry.)
For type $D$ less is known, but see the papers of Smithling for some results.
For level structures deeper than Iwahori, much less is known in general: In special cases (such as the case considered by Harris and Taylor, or, of course, for modular curves) one can obtain a nice model with a good geometric description. Haines and Rapoport have a paper on the $\Gamma_1(p)$ case. Sometimes it is also enough to work with some "abstractly defined" models (e.g. Drinfeld level structures and flat closure as in Mantovan, On the cohomology of certain PEL type Shimura varieties, Duke Math. J. 129 (2005), no. 3, 573--610).<|endoftext|>
TITLE: kapranov's realization of $\overline{M}_{0,n}$ over other fields
QUESTION [6 upvotes]: Kapranov gave a very nice desciption, over $\mathbb{C}$ of the moduli space of stable pointed rational curves $\overline{M}_{0,n}$ as a series of blow-ups of $P^{n-3}$. Does this, or a similar result, hold over other fields? e.g. positive characteristic, non algebraically closed, etc.
ps
I am afraid one could only dream of this, over non alg closed fields...

REPLY [6 votes]: It seems to me Kapranov's methods are purely algebraic and that his description works verbatim over $\mathrm{Spec}(\mathbf Z)$.<|endoftext|>
TITLE: The highest root of an ADE quiver
QUESTION [9 upvotes]: Let $\Gamma$ be a finite subgroup of $SL_2({\mathbb C})$, and $Q$ the set of
its irreducible representations. McKay makes $Q$ into a directed graph by
having $V \to W$ if $W \leq V \otimes {\mathbb C}^2$, where the latter comes
from the natural action of $\Gamma$ on ${\mathbb C}^2$. (But since 
${\mathbb C}^2 \otimes {\mathbb C}^2$ has an $SL_2$-invariant hence $\Gamma$-invariant
vector, the directed graph is actually undirected: each edge comes with its reverse.)
In this way we get a graph with a vector space at each edge. 
McKay observes that the graphs so arising are exactly the simply-laced
affine Dynkin diagrams, with the trivial rep as the affine node.
(In particular, the extra symmetry of the affine diagram comes here
from $(\Gamma / \Gamma')^*$, which is therefore identifiable with $Z(G)$,
for $G$ the corresponding simply-connected Lie group. I wonder if there's
some larger correspondence there... but that's not my question.)
If we toss that node, and orient the edges (i.e. throw out half), we
can look at the "roots", or indecomposable representations, of the
resulting quiver.  McKay observes further (in effect) that the largest such quiver
representation has the same-dimensional vector spaces as in the first
construction. But now, since it's a quiver representation, there are
maps between the spaces. So my question:

In McKay's construction, the vertices of a Dynkin diagram are labeled 
  by nontrivial irreps $\{V\}$ of $\Gamma$. Given an orientation on the diagram
  and an edge $V \to W$, is there a natural linear map $V \to W$,
  such that the result is the largest indecomposable quiver
  representation? 

Obviously these maps aren't $\Gamma$-equivariant. The natural map is
$V \otimes {\mathbb C}^2 \to W$, so maybe these other maps correspond
to choosing a vector, or a list of vectors, in ${\mathbb C}^2$.
So a more specific version of the question:

If $\vec x$ is a generic vector in ${\mathbb C}^2$, e.g.
  with no $\Gamma$-stabilizer, do the resulting composite maps
  $V \cong V \otimes \vec x \hookrightarrow V \otimes {\mathbb C}^2 \twoheadrightarrow W$
  give the largest indecomposable? If so, what if $\vec x$ isn't generic?

Feel free to add tags; I couldn't think of anything other than rt.representation-theory.

REPLY [3 votes]: Part of what makes this question more interesting outside type $A$ is that the highest root can't be projective or injective.  A test case to consider is $D_4$, say with all three arrows pointing to the central vertex. A representation of $D_4$ with dimension vector (1,1,1,2) will be indecomposable provided you don't choose any zero maps or have any two of the maps be multiples.  
In your setting, this translates into saying that you shouldn't choose the zero vector at any of the three vertices, and they shouldn't map to the same line in the 2-dimensional vector space over the central vertex either.  It's not quite clear to me how to express this as a condition on the choice of vectors at the three vertices, but independent generic choices would work.  
It's a general fact about quiver representations that if the dimensions come from a real root, then if you choose the maps generically, you will get the indecomposable representation. So one way to view the question is whether choosing a vector $v$ for each vertex is sufficiently generic. 
For $D_n$ with $n>4$, I find it plausible that someone a bit more comfortable with representation theory of finite groups than I am, could convince themselves that this works fine as well.  The situation for $E$ seems more complicated.  

The following is not an answer to the question, but if it's unfamiliar to you, it might be of interest.  This setup is called the "algebraic McKay correspondence", and is due to Auslander.  
Let $S=\mathbb C[x,y]$.  Let $R=S^G$.  As an $R$-module, $S$ is a direct sum of the maximal Cohen-Macaulay $R$-modules.  The maximal CM $R$-modules can also be constructed from the irreps of $\Gamma$ as follows: for $V$ an irrep, take $(V\otimes_R S)^G$.  
The endomorphism ring of $S$ as an $R$-module is the preprojective algebra of the corresponding affine type. (If we throw away the node for $R$, we get the finite type.)  
From this point of view, the arrows of the McKay quiver come with maps for free.  In the $A_n$ case, they are just multiplication by $x$ and $y$, so the preprojective relation is just the fact that $xy=yx$.  
However, I don't see how to get back to finite-dimensional $\mathbb C$-vector spaces in a natural way now.<|endoftext|>
TITLE: maximal order of elements in GL(n,p)
QUESTION [21 upvotes]: I am looking for a formula for the maximal order of an element in the group $\operatorname{GL}\left(n,p\right)$, where $ p$ is prime. 
I recall seeing such a formula in a paper from the mid- or early 20th century, but could not find again this reference. I will be grateful for any hint.

REPLY [46 votes]: Well, by Cayley--Hamilton, each matrix $A\in {\rm GL}(n,p)$ generates an at most $n$-dimensional subalgebra ${\mathbb F}_p[A]\subseteq M(n,p)$ thus containing at most $p^n-1$ nonzero elements. Hence the order of $A$ cannot exceed $p^n-1$.
On the other hand, consider a degree $n$ monic polynomial $P_n$ whose root is a generator $\xi$ of ${\mathbb F}_{p^n}^*$. Then a matrix with $P_n$ as its characteristic polynomial has order at least $p^n-1$ since $\xi$ is its eigenvalue.
ADDENDUM. if you wish the order to be the power of $p$, then the answer is $d=p^{\lceil \log_p n\rceil}$. Since the order of $A$ is divisible by the multiplicative orders of its eigenvalues, all the eigenvalues should be $1$. Hence the characteristic polynomial is $(x-1)^n$, so $A^d-I=(A-I)^d=0$.
On the other hand, if $A=I+J$ is the Jordan cell of size $n$ (with eigenvalue 1), then $A^{d/p}=I^{d/p}+J^{d/p}\neq I$, but $A^d=I+J^d=I$.
NB. The subgroup of all (upper-)unitriangular matrices is a Sylow $p$-subgroup in ${\rm GL}(n,p)$. So you may concentrate on it when looking at the elements of this kind.
ADDENDUM-2 (much later). This is to answer the question in the comments about the maximal order of an element $f\in AGL(n,q)$, where $q$ is a power of $p$. Write $f(x)=Ax+b$.
If $1$ is not an eigenvalue of $A$, then $f$ has a fixed point (the equation $f(x)=x$ has a solution), so we may regard it as an element of $GL(n,q)$, and the maximal order of $f$ is again $q^n-1$.
So we are concerned with the case when the minimal polynomial $\mu(x)$ of $A$ vanishes at $1$, say $\mu(x)=(x-1)^k\nu(x)$, where $\nu(1)\neq 0$. Then $A$ is similar to a block-diagonal matrix with blocks having minimal polynomials $(x-1)^k$ and $\nu(x)$ (in view of $\mathbb F_q^n=\mathop{\mathrm {Ker}}(A-I)^k\oplus\mathop{\rm Ker}\nu(A)$). So the order $d$ of $A$ does not exceed $p^{\lceil\log_p k\rceil}(q^{n-k}-1)$ if $n<k$, and $p^{\lceil\log_p n\rceil}$ otherwise. If $n>3$ (or $n=3$ and $q>2$), one may easily see that this bound does not exceed $q^{n-1}-1$. So $f^d$ is a translation, which yields that $f^{pd}=\mathord{\rm id}$, and $pd\leq p(q^{n-1}-1)<q^n-1$. Thus the maximal order of $f$ in these cases is still $q^n-1$.
We are left with the cases $n=1$, $n=2$, or $n=3$, $p=2$. When $n=1$, the answer is obviously $\max(p,q-1)$. When $n=2$, the only case left is $d=p=q$ achieved when $A$ is similar to the Jordan cell $\begin{pmatrix}1&1\\0&1\end{pmatrix}$. In this case, $f^p(x)=x+(A^{p-1}+\dots+I)b=x$ unless $p=2$, when $f^2(x)=x+\begin{pmatrix}0&1\\0&0\end{pmatrix}b$. So the answer is still $q^2-1$ when $q>2$, and $4$ otherwise.
Finally, if $n=3$ and $q=2$, then the order of $A$ having eigenvalue $1$ exceeds 3 only if $A$ is similar to $3\times 3$ Jordan cell (then $d=4$); but in this case $f^4=\mathord{\rm id}$. So this is not an exception.
Summing up, the only cases when the order may be greater than $q^n-1$ are: (1) $n=1$, $q=p$ (the maximal order is $p$), and (2) $n=2$, $q=2$ (the maximal order is $4$).<|endoftext|>
TITLE: Manifolds with prescribed fundamental group and finitely many trivial homotopy groups
QUESTION [13 upvotes]: Fix $G$, a finitely generated presented group. 
It is known that for every $k > 3$ there is a closed $k$-manifold whose fundamental group is $G$. Similarly, there is a topological space with fundamental group $G$ and all higher homotopy groups trivial. 
However, even for simple examples such as when $G \cong \mathbf{Z}_2$, such a topological space is not a manifold. It seems like the problem with these spaces really lies in the infinite constructions process adding in cells of arbitrarily high dimension. So instead if we only require the first $n$ homotopy groups to be trivial can we still work with manifolds. That is,

Is it true that for each $n > 1$ there is a closed manifold $M$ such that $\pi_1(M) \cong G$ and for $1 < i \leq n$, $\pi_i(M)$ is trivial?

Note that if we allow $M$ to be a non-compact manifold / a manifold with boundary then the answer is yes. This follows as we can always find a finite simplicial complex $X$ whose fundamental group is $G$. By correctly adding $i$-cells (for $1 < i \leq n$) we obtain a simplical complex $X'$ with $\pi_1(X') \cong G$ and for $1 < i \leq n$, $\pi_i(M)$ trivial. By embedding $X'$ in a suitably high dimensional Euclidean space and taking an closed / open regular neighbourhood we obtain $M$, a non-compact manifold / manifold with boundary with the required properties.
Assuming that the answer to the first question is yes, can we also get manifolds of almost any dimension that we like?

REPLY [17 votes]: No, the answer is negative in general (if you require $M$ to be compact). $M$ comes with a map $M \to BG$ that is, by definition, $n+1$-connected (iso on $\pi_i$ for $i=0,...,n$, epi on $\pi_{n+1}$). You can turn it into a weak equivalence by attaching cells of dimension $\geq n+1$. From that you see, that there is a model for $BG$ having finite $n$-skeleton. This is a special property of a group that is called $F_n$ (for more information, see http://berstein.wordpress.com/2011/03/16/morse-theory-finiteness-properties-and-bieri-stallings-groups/). Finitely presented groups are $F_2$ and you find that a necessary condition on your $G$ is that it is of type $F_n$. The are concrete examples of groups that are $F_i$ but not $F_{i+1}$ for each $i$, which are discussed in same blog post (on page 423 in Hatcher's AT, you find the same examples in a slightly different context).
On the other hand, let $G$ be $F_n$ and let $K$ be the $n$-skeleton of $BG$; a finite complex. Then I claim there is a closed manifold $M$ with the desired properties. $M$ can be chosen of arbitrary dimension $d \geq 4,2n+1$ and to be stably parallelizable. Start with a sphere $S^d \to K$ and do surgery on $S^d$ to get rid of the homotopy groups in low dimensions. The precise formulation is for example Proposition 4 in Kreck's paper "Surgery and duality".
So we can say that a necessary and sufficient condition is that $G$ is of type $F_n$.
Caveat: I might have confused $n$ and $n+1$ at various places. 
If you want to have $dim M \leq 2n$, you meet a new obstruction enforced by Poincare duality and things become really difficult.<|endoftext|>
TITLE: The ABC of categories: ABstract vs Concrete
QUESTION [7 upvotes]: From Wikipedia:

A concrete category is a category that is equipped
with a faithful functor to the category of sets.
This functor makes it
possible to think of the objects of the category as sets with
additional structure, and of its morphisms as structure-preserving
functions.

This definition gives rise to the well-established dichotomy between concrete and abstract categories.
The examples of abstract  (= non-concrete) categories I've heard of come in two flavours:

categories with structured sets as objects, and some other sort of structure-preserving maps than standard homomorphisms, e.g. interpretations. (I understand that - basically - interpretations are structure-preserving maps, just involving re-definitions of individuals (as n-tupels) and relations.)

categories with structured set as objects, and equivalence classes of structure-preserving maps as morphisms, e.g. the homotopy category of topological spaces or much simpler: the category $C'$ which collapses all arrows $X \rightarrow Y$ of a (concrete) category $C$ into one (what's its name?)


[Side remark: The trivial equivalence relation on morphisms: $f \sim g :\equiv f = g$ leaves a given category $C$ unchanged.]
Given such a vast variety of possible definitions of structure-preserving maps and equivalence relations between them, I wonder why only classical homomorphisms and the trivial equivalence relation give rise to so-called concrete categories?
The other way around:

What is an example of an abstract category (in the
standard sense) with structured sets as objects, such that we cannot think of its morphisms as some equivalence class
of some sort of structure-preserving maps?

REPLY [7 votes]: This article by Ivan di Liberti and Fosco Loregian delves deeply into the relationship between concreteness and homotopy:
Homotopical algebra is not concrete. On the concreteness of certain model categories
It is introduced by a quotation from Freyd:<|endoftext|>
TITLE: About the definition of Borel and Radon measures
QUESTION [19 upvotes]: I am trying to understand the notion of Radon measure, but I am a little bit lost with the different conventions used in the litterature. 
More precisely, I have a doubt about the very definition of Borel measure. 
Suppose that $(X,\mathcal{B},\mu)$ is a measure space, where $X$ is a topological space. 
I have find two different definitions for "$\mu$ is a Borel measure":
-Def 1 : $\mu$ is a Borel measure if $\mathcal{B}$ contains the Borel $\sigma$-algebra of $X$,
-Def 2:  $\mu$ is a Borel measure if $\mathcal{B}$ is exactly the Borel $\sigma$-algebra of $X$.
The same thing happens for the notion of Radon measure, as it can be either considered as Borel measure in the sense of Def 1, or in the sense of Def 2. 
Of course, Def 1 gives a more general notion of Borel or Radon measure. For example 
the Lebesgue measure (defined on the Lebesgue $\sigma$-algebra of $\mathbb{R}^n$) is Radon in the sense of Def 1, but not in the sense of Def 2.
Are there (other) reasons as to why one may prefer Def 1 to Def 2 or vice versa ? 
Apparently, Def 2 makes it quite difficult to have a "complete Radon measure", which makes me think that it is a little bit artificial or restrictive. But maybe many results hold only for Radon measures in the sense of Def 2, without possible extension to Radon measures in the sense of Def 1 ? Or maybe there is a trivial way to transfer any result involving a Borel measure in the sense of Def 2 to a result involving a Borel measure in the sense of Def 1 ?
A related question is the following : if $\mu$ is Radon in the sense of Def 2, will its completion be Radon in the sense of Def 1 ? Same question when you replace "Radon" by "inner regular", "outer regular", and "locally finite".

REPLY [22 votes]: From a geometric measure theory perspective, it is standard to define Radon measures $\mu$ to be Borel regular measures that give finite measure to any compact set. Of course, their connection with linear functionals is very important, but in all the references I know, they start with a notion of a Radon measure and then prove representation theorems that represent linear functionals by integration against Radon measures.
Here are some examples:
$\color{blue}{I:}$ Evans and Gariepy's Measure Theory and Fine Properties of Functions states it this way:

A [outer] measure $\mu$ on $X$ is regular if for each set $A \subset X$ there exists a $\mu$-measurable set $B$ such that $A\subset B$ and $\mu(A)=\mu(B)$.
A measure $\mu$ on $\Bbb{R}^n$ is called Borel if every Borel set is $\mu$-measurable.
A measure $\mu$ on $\Bbb{R}^n$ is Borel regular if $\mu$ is Borel and for each $A\subset\Bbb{R}^n$ there exists a Borel set $B$ such that $A\subset B$ and $\mu(A) = \mu(B)$.
A measure $\mu$ on $\Bbb{R}^n$ is a Radon measure if $\mu$ is Borel regular and $\mu(K) < \infty$ for each compact set $K\subset \Bbb{R}^n$.

$\color{blue}{II:}$ In De Lellis' very nice exposition of Preiss' big paper, he doesn't even define Radon explicitly, but rather talks about Borel Regular measures that are also locally finite, by which he means $\mu(K) < \infty$ for all compact $K$. His Borel regular is a bit different in that he only considers measurable sets -- $\mu$ is Borel regular if any measurable set $A$ is contained in a Borel set $B$ such that $\mu(A) = \mu(B)$. (I am referring to Rectifiable Sets, Densities and Tangent Measures by Camillo De Lellis.)
$\color{blue}{III:}$ In Leon Simon's Lectures on Geometric Measure Theory, he defines Radon measures on locally compact and separable spaces to be those that are Borel Regular and finite on compact subests.
$\color{blue}{IV:}$ Federer 2.2.5 defines Radon Measures to be measure a $\mu$, over a locally compact Hausdorff spaces, that satisfy the following three properties:

If $K\subset X$ is compact, then $\mu(K) < \infty$.

If $V\subset X$ is open, then $V$ is $\mu$ measurable and
$\hspace{1in} \mu(V) = \sup\mu(K): K\text{ is compact, } K\subset V$

If $A\subset X$, then
$\hspace{1in} \mu(A) = \inf\mu(V): V\text{ is open, } A\subset V$


Note: it is a theorem (actually, a Corollary 1.11 in Mattila's Geometry of Sets and Measures in Euclidean Spaces) that a measure is a Radon a la Federer if and only if it is Borel Regular and locally finite. I.e {Federer Radon} $\Leftrightarrow$ {Simon or Evans and Gariepy Radon}. (I am referring of course to Herbert Federer's 1969 text Geometric Measure Theory.)
$\color{blue}{V:}$ For comparison, Folland (in his real analysis book) defines things a bit differently. For example, he defines regularity differently than the first, third and fourth texts above. In those, a measure $\mu$ is regular if for any $A\subset X$ there is a $\mu$-measurable set $B$ such that $A\subset B$ and $\mu(A) = \mu(B)$. In Folland, a Borel measure $\mu$ is regular if all Borel sets are approximated from the outside by open sets and from the inside by compact sets. I.e. if
$\hspace{1in}\mu(B) = \inf \mu(V): V\text{ is open, } B\subset V$
and
$\hspace{1in}\mu(B) = \sup \mu(K): K\text{ is compact, } K\subset B$
for all Borel $B\subset X$.
Folland's definition of Radon is very similar to Federer's but not quite the same:
A measure $\mu$ is Radon if it is a Borel measure that satisfies:

If $K\subset X$ is compact, then $\mu(K) < \infty$.

If $V\subset X$ is open, then
$\hspace{1in} \mu(V) = \sup\mu(K): K\text{ is compact, } K\subset V$

If $A\subset X$ and $A$ is Borel then
$\hspace{1in} \mu(A) = \inf\mu(V): V\text{ is open, } A\subset V$


... and by Borel measure, Folland means a measure whose measuralbe sets are exactly the Borel sets.
Discussion: Why choose one definition over another?  Partly personal preference -- I prefer the typical approach taken in geometric measure theory, starting with an outer measure and progressing to Radon measures a la Evans and Gariepy or Simon or Federer or Mattila. It seems, somehow, more natural and harmonious with the Caratheodory criterion and Caratheodory construction used to generate measures, like the Hausdorff measures.
With this approach, for example, sets with an outer measure of 0 are automatically measurable.
Another reason not to use the more restrictive definition 2 (in the question above) it makes sense to require that continuous images of Borel sets be measurable. But all we know is that continuous maps map Borel to Suslin sets. And there are Suslin sets which are not Borel! If we use the definition of Borel regular, as in I,III and IV above, then Suslin sets are measurable. There is a very nice discussion of this in section 1.7 of Krantz and Parks' Geometric Integration Theory -- see that reference for the definition of Suslin sets. (Krantz and Parks is yet another text I could have added to the above list that agrees with I, III, and IV as far as Radon, Borel regular, etc. goes.<|endoftext|>
TITLE: The Gauss circle problem on a hexagonal lattice
QUESTION [11 upvotes]: Take an infinite hexagonal lattice (or equivalently, an equilateral triangular lattice), with unit spacing between the closest lattice point pairs, and draw a disc of radius $r$ centered on a lattice point at $(0, 0)$.  Let $N(r, hex)$ denote the number of hexagonal lattice points at coordinates $(a, b)$ s.t. $(a^2 + b^2) \leq r^2$, i.e. the number of lattice points on or within the aforementioned disc of radius $r$.  
Are there any literature references for approximations to $N(r, hex)$ (I haven't been able to find any through a Google search)?  What is an exact counting solution for $N(r, hex)$?  
Using the exact counting solution for the $Z^2$ integer lattice, (http://mathworld.wolfram.com/GausssCircleProblem.html) I suppose we can guess a lowerbound for the hexagonal lattice of:
Lowerbound $N(r, hex) = 1 + Floor[\frac{r}{2}] + 4*\sum^{Floor[\frac{r}{2}]}_{i=1} Floor[((\frac{r}{2})^2-i^2)^{\frac{1}{2}}] + 2*Floor[r]$
Where we simply overlay the $Z^2$ lattice with (closest) nearest-neighbor spacing $2$ on top of an $A_2$ hexagonal lattice with (closest) nearest-neighbor spacing $1$, and add an additional $2*Floor[r]$ correctional term.
[10/13/12] The OEIS sequences are extremely helpful, but after searching the literature for awhile, I'm still having difficulty finding an exact (counting) solution for the number of lattice points within a circle of real number radius $r$.  Any references would be very much appreciated!  
[10/14/12] Still no luck finding a reference in the literature.  Surely someone has looked at this problem for, say, graphene and other molecular or atomic lattices where one would like to have a precise atom count a certain physical distance away from one atom?

[10/19/12] I managed to find the exact OEIS sequence I was looking for: http://oeis.org/A053416
However, I'd still like to find an exact counting solution, like the one presented above the $Z^2$ integer lattice.

REPLY [4 votes]: By using a transformation, it is possible to give a counting formula for the hexagonal lattice, similar to the formula on MathWolf. Here are the details.
Suppose your lattice is given by $\Lambda=T(\mathbb{Z}^2)$, where $T$ is an invertible 2x2 matrix.
Let $E$ be an ellipse, and let $N_E(R)$ denote the number of points of $\mathbb{Z}^2$ that are in the dilate $R\cdot E$. That is,
$$
N_E(R)=|R\cdot E\cap\mathbb{Z}^2|,
$$
where the vertical bars denote cardinality.
Now if we let $D$ be the unit disk, you wish to count
$$
|R\cdot D\cap \Lambda| = | R\cdot D\cap T(\mathbb{Z}^2)| = |R\cdot T^{-1}(D)\cap\mathbb{Z}^2| = N_E(R),
$$
where $E=T^{-1}D$ is an ellipse.
If
$$
T=\begin{pmatrix} a & b\\ c& d \\ \end{pmatrix}
$$
then the boundary of $R\cdot E$ is given by the equation
$$
(ax+by)^2+(cx+dy)^2=R^2.
$$
In general, the explicit expression for $y$ is fairly messy.
But for a hexagonal lattice, $a=1,b=0,c=1/2,d=\sqrt{3}/2$, so we have
$$
x^2+\frac{(x+\sqrt 3 y)^2}{4} = R^2,
$$
hence 
$$
y=\frac 1{\sqrt 3}[ -x\pm \sqrt{4R^2-x^2}].
$$
This explicit expression let's us write down a counting expression like the one on MathWorld.
Since $E$ is symmetric under $(x,y)\mapsto (-x,-y)$, it suffices to count the number of lattice points above the $x$-axis. The number of lattice points on the axes can be counted fairly easily. It is slightly tricky to count the number of lattice points within the ellipse. In the first quadrant, we may count as usual, but in the second quadrant, we must subtract lattice points that are below the ellipse:
$$
N(R)=1 + 2\lfloor \sqrt{4/5}R\rfloor + 2\lfloor \sqrt{4/3}R\rfloor + 2\sum_{i=1}^{\lfloor \sqrt{4/5}R\rfloor} \left\lfloor \frac 1{\sqrt 3}[ - i+\sqrt{4R^2-i^2}]\right\rfloor +2\sum_{i=1}^{\lfloor 2R\rfloor}\left\lfloor \frac 1{\sqrt 3}[ i+\sqrt{4R^2-i^2}]\right\rfloor -2\sum_{i=\lceil\sqrt{4/5}R\rceil}^{\lfloor 2R\rfloor}\left\lfloor \frac 1{\sqrt 3}[ i-\sqrt{4R^2-i^2}]\right\rfloor
$$<|endoftext|>
TITLE: A question about Q?
QUESTION [5 upvotes]: Let A=$\{a_n : n\in \omega \}\subset 2^{\omega\times\omega}$ be nonempty countable without isolated points (i.e. homeomorphic to $\mathbb{Q}$), and satisfy $ \forall n\in \omega \exists^\infty m|\{k:a_n(m,k)=1\}|=\omega $.
Does there exist $ a\in cl(A)\setminus A$ satisfying $\exists^\infty m|\{k:a(m,k)=1\}|=\omega $?

REPLY [3 votes]: I think the answer is no, as shown by the following set $A$ (whose elements I will describe as subsets rather than binary functions).
For any map $\sigma:\omega\to\omega+1:=\omega\cup\{\omega\}$ define its hypograph
$$\mathrm{hypo}(\sigma):=\{(m,k)\in\omega\times\omega: k < \sigma(m)\}\subset\omega\times\omega\\ .$$
An increasing map $\sigma:\omega\to\omega+1$   takes the value $\omega$ if and only if its hypograph contains a subset of the form $[n,\omega)\times \omega$, thus, if and only if it has the property (P) stated in the question (there exist infinitely many $m$ such that there exist infinitely many $k$ such that $(m,k)\in \mathrm{hypo}(\sigma)$; which is indeed verified quite in a strong way). 
Let $A$ be the set of all  hypographs of all  increasing maps that take  the value $\omega$. Clearly, the set $A$ is countable, with no isolated points; and all its elements enjoy  property (P). A point in $\mathrm{cl}(A)\setminus A$ is exactly the hypograph of an increasing $\omega$-valued map, which never satisfies  property (P).<|endoftext|>
TITLE: Generalizing Feferman - Levy
QUESTION [7 upvotes]: The Feferman - Levy model makes $\aleph_1$ singular by a cardinal collapse $\aleph_1 = \aleph_{\omega}^L$.  Unless I've got something wrong, the same thing would work to make any well-orderable cardinal $\alpha$ cofinal in its well-ordered cardinal successor.  Is that right?
The Feferman -Levy model also makes the continuum a countable union of countable sets.  Does that generalize to Beth numbers, in the sense of successive power sets starting with $\omega$?  For each finite $n$, are there models where $\beth_{n+1}$ is a union of $\beth_{n}$ many sets each smaller than or the same size as $\beth_{n}$?

REPLY [3 votes]: Posting this as an answer at Colin's request. The second paragraph of the question is addressed at this other MO question. 
The answer to the question in the first paragraph is delicate, it depends on how much of the ground model we decide to preserve: $\alpha$ and $\alpha^+$? $\rm{cof}(\alpha)$ and $\alpha^+$? Only $\alpha^+$? The issue is that a straight generalization of Feferman-Lévy must fail just based on consistency strength considerations, because Jensen's covering lemma gives us that preserving a singular and collapsing its successor requires large cardinals (even in $\mathsf{ZF}$). For the covering lemma, see:

Devlin, Keith I.; Jensen, Ronald B. Marginalia to a theorem of Silver. In $\models$ISILC Logic Conference (Proc. Internat. Summer Inst. and Logic Colloq., Kiel, 1974), G. H. Müller, A. Oberschelp, and K. Potthoff, eds., pp. 115–142. Lecture Notes in Math., Vol. 499, Springer, Berlin, 1975. MR0480036 (58 #235) 
Mitchell, William J. The covering lemma. In Handbook of set theory. Vols. 1, 2, 3, Kanamori, Foreman, eds., pp. 1497–1594, Springer, Dordrecht, 2010. MR2768697<|endoftext|>
TITLE: A sequence that tell us if a holomorphic function of several variables is identically zero
QUESTION [5 upvotes]: Is there any sequence $ \{ Z_{\nu} \}_{\nu \in \mathbb{N}}$ in $\mathbb{C}^{n}$, $Z_{\nu} \rightarrow 0$, such that any holomorphic function in $\mathbb{C}^{n}$ which vanishes in $Z_{\nu}$ for all $\nu \in \mathbb{N}$ is identically zero?
Thank you!

REPLY [4 votes]: Start from a countable dense subset $S$ of the unit ball, and take a sequence $Z_\nu$ such that for all $s\in S$, one has $\nu Z_\nu=s$ infinitely often. Then, any holomorphic function in $\mathbb{C}^n$ that vanishes along the sequence $Z_\nu$ has in particular a subsequence of zeros accumulating to the origin that belong to the complex line generated by any $s\in S$. Therefore it vanishes identically on that complex line, by the principle of isolated zeros in one variable. Since the union of these lines is dense in $\mathbb{C}^n$, the function is identically zero by continuity.<|endoftext|>
TITLE: An application of Mobius Inversion in a paper of Shintani
QUESTION [5 upvotes]: I've been reading about Shintani zeta functions and in particular with respect to finding the density of cubic discriminants as in the theorem of Davenport-Heilbronn. In Shintani's paper "On zeta-functions associated with the vector space of quadratic forms" [Tokyo Univ. J. Fac. Sci Sect. 1A Math 1975], in the proof of Theorem 4, Shintani writes

Hence, we have (by earlier results in this paper and in a previous paper):  $$\sum_{nk^4 \leq x} h_r(n) = 2^{-1}\zeta(2)\zeta(4)x + O(x^{2/3 + \epsilon}) \qquad (x \to +\infty, \forall \epsilon > 0)$$ An application of the Mobius inversion formula now yields that
  $$ \sum_{n \leq x} h_r(n) = 2^{-1}\zeta(2)x + O(x^{2/3 + \epsilon})$$

When I think of Mobius Inversion, I think of two things: $$ \begin{align*} g(n) &= \sum_{d \mid n} f(n/d) \iff f(n) = \sum_{d \mid n} \mu(n)g(n/d) \quad \text{or}\\\\ g(x) &= \sum_{n \leq x}f(n/x) \iff f(x) = \sum_{n \leq x} \mu(n)g(n/x) \end{align*}$$
But I don't see how I can use these here. Unfortunately, I also know that there are many things that might be called Mobius Inversion. This is one of those steps that taunts me. Qualitatively, we remove the fourth-power condition and end up losing a factor of $\zeta(4)$, and that feels very reasonable.
Further, in playing with it for a while, I re-stumbled upon the fact that $\displaystyle \dfrac{1}{\zeta(s)} = \sum_n \dfrac{\mu(n)}{n^s}$ (easy, but which I did not originally remember) and thus that $\displaystyle \dfrac{1}{\zeta(4)} = \sum \dfrac{\mu(n)}{n^4}$.
To make my question explicit - how can we arrive at the second equation from the first?

REPLY [13 votes]: Let $g(n)$ denote the indicator function for the fourth powers. Then your sum equals 
$$\sum_{n\leq x}\left(h_{r}*g\right)(n),$$
where $*$  denotes Dirichlet convolution. We may rewrite the given asymptotic as 
$$\sum_{k\leq x}g(k)\sum_{n\leq\frac{x}{k}}h_{r}(n)=2^{-1}\zeta(2)\zeta(4)x+O\left(x^{2/3+\epsilon}\right),$$
noticing that this is of the form
$$G(x)=\sum_{n\leq x}\alpha(n)F\left(\frac{x}{n}\right).$$
Mobius inversion tells us that 
$$F(x)=\sum_{n\leq x}\alpha^{-1}(n)G\left(\frac{x}{n}\right),$$ 
where $\alpha^{-1}$  is the multiplicative inverse of $\alpha$ with respect to Dirichlet convolution.  Applying Mobius inversion to our sum, we have that 
$$\sum_{n\leq x}h_{r}(n)=\sum_{j^{4}\leq x}\mu(j)\sum_{n\leq\frac{x}{j^{4}}}\left(h_{r}*g\right)(n)$$
which equals 
$$2^{-1}\zeta(2)\zeta(4)x\sum_{j^{4}\leq x}\frac{\mu(j)}{j^{4}}+O\left(x^{2/3+\epsilon}\left(\sum_{j^{4}\leq x}\frac{1}{j^{4}}\right)\right)$$ 
$$=2^{-1}\zeta(2)x+O\left(x^{2/3+\epsilon}\right).$$ 
Notice that the Dirichlet inverse to the function $g(n)$, the indicator function for the fourth powers, is in some sense the mobius function on fourth powers.<|endoftext|>
TITLE: A linearly orderable monoid which does not embed into a linearly orderable group
QUESTION [7 upvotes]: It is known (after an example of A.I. Mal'cev) that there exist cancellative semigroups which do not embed into a group. On the other hand, it is not difficult to see that every linearly orderable semigroup (is cancellative and torsion-free, and) embeds into a linearly orderable monoid (see here for terminology and motivations). Then, my question is:


Is it known whether or not a linearly orderable monoid embeds into a linearly orderable group?


The answer is surely yes in the commutative setting (by the construction of the Grothendieck group). But I've no clue about the non-commutative case. Thank you in advance for any hint.

REPLY [4 votes]: There is a prominent example in the non-commutative setting.
[Edit (YCor): here linearly ordered monoid is interpreted as a monoid endowed with a total ordering satisfying: $a\le b,c\le d$ imply $ac\le bd$. As mentioned in the comments, this does not imply that left/right multiplication preserve the strict ordering.]
Take some set $O$ of ordinals which is closed under addition and
contains $0$, $1$ and $\omega$, where addition is defined the way Cantor did it.
$(O,+)$ has a first-order definable order:
$x \leq y$ if and only if there is a $z$ such that $y = x+z$.
It is easy to see that $x \leq y$ implies $x+z \leq y+z$ and $z+x \leq z+y$.
However, there is no group $(G,+)$ having $(O,+)$ as a subgroup.
The main reason is that adding $\omega$ cannot be inverted:
$$0+\omega = \omega, \quad 1+\omega = \omega.$$
If $(G,+)$ is a group extending $(O,+)$, then there would be an inverse $g$ to $\omega$ in $G$ and $0 = \omega+g = (1+\omega)+g = 1+(\omega+g) = 1+0 = 1$, a contradiction.<|endoftext|>
TITLE: Eigenvalues of infinite matrices
QUESTION [8 upvotes]: I am trying to find some literature on infinite matrices because I want to know how to get the eigenvalues of infinite matrices. Seriously, it seems there are very few references available. Can someone tell me how to determine the spectrum of infinite matrices?

REPLY [16 votes]: One simple example with a special matrix, which has somehow "a continuum" as eigenvalue...
Consider some function $ f(x) = K + ax + bx^2 + cx^3 + ... $ having a nonzero radius of convergence. Then think of the infinite matrix of the form
$$ \small  \begin{bmatrix}
  K & . & . & . &  \cdots  \\\ 
  a & K & . & . &  \cdots \\\ 
  b & a & K & . & \cdots  \\\ 
  c & b & a & K & \cdots \\\ 
 \vdots & \vdots & \vdots& \vdots  & \ddots 
\end{bmatrix} $$ 
From the properties of finite matrices we would expect, that K is an eigenvalue. But consider a type of an infinite vector
$$ V(x) = [1,x,x^2,x^3,x^4,\ldots ] $$ with a scalar parameter $x$ from the range of convergence, then 
$$ V(x) \cdot F = f(x) \cdot V(x) $$
This means also: any vector $V(x)$ is an eigenvector of the matrix F and corresponds to the eigenvalue $f(x)$. If now $f(x)$ is entire, for instance the exponential function $ f(x)=\exp(x)$, then any value from the complex plane (except $0$ because $\exp(x)$ is never $0$) "is an eigenvalue" of F contradicting the "naive" extrapolation from the finite truncation of the matrix ...<|endoftext|>
TITLE: Can the level set of a critical value be a regular submanifold?
QUESTION [10 upvotes]: I know that there is a theorem that a non-empty level set of a regular value of a smooth function $f:M\rightarrow\mathbb{R}$ on a smooth manifold is a regular submanifold (or embedded submanifold) of codimension 1.
Now I wonder if there is also a condition in which the converse holds true. I mean suppose $c\in\mathbb{R}$ is a critical value of $f$ is there a condition under which you know the level set of $c$ is not a regular submanifold. 
If $g:M\rightarrow\mathbb{R}$ is defined by mapping all of $M$ to $0$ then it seems to me that $0$ is a critical value of $g$. Now $g^{-1}(0)=M$ so definitely a regular submanifold of $M$. So I know that it is at least not true without an extra condition. 
Also if the function is second degree polynomial from $\mathbb{R}$ to $\mathbb{R}$ then the level set of a critical value is just a point which would again be a regular submanifold. 
If there is no such general condition on the function $f$ or the critical value then how in general does one go about showing that a critical level set is or is not a regular submanifold?
Thanks in advance

REPLY [12 votes]: Suppose that $0$ is a regular value of $f$. Then $0$ is a critical value of $g=f^2$, yet the level set $g^{-1}(0)$ is a regular submanifold. In this case all the points on $g^{-1}(0)$ are critical points of $g$.
The general answer is difficult.     You need to assume something about $f$. A natural assumption would be that the critical points of $f$ are isolated and of finite type. Near such points one can find local coordinates  so that in these coordinate $f$  looks like a polynomial.  (This is  a generalization of the classical Morse lemma due to Tougeron.)    In such cases  you need to understand   the zero sets  of real polynomials which can be challenging. A nice place to consult for such issues is the book by Arnold, Gussein-Zade and Varchenko on   singularities of differentiable mappings.<|endoftext|>
TITLE: Number of monomials in three variables of quasihomogeneus degree d
QUESTION [5 upvotes]: I am interested in computing the number of natural solutions $(x_0,y_0,z_0)$ of the equation $w_0x+w_1y+w_2z=d$, with $w_0,w_1,w_2$ and $d$ natural numbers and $\gcd (w_i,w_j)=1$ . Or equivalently, the number of monomials in three variables of quasihomogeneus degree d and weights $(w_0,w_1,w_2)$. Does anybody now any formula in terms of $w_0,w_1,w_2$ and $d$ or any reference in the literature? 
With the previous notations I really need to compute $a_{\alpha+w_0w_1w_2}-a_{\alpha}$ being $\alpha< w_0w_1w_2$. I do not know if there are some kind of relations among the terms of this series...I want to look at the properties of these series because I need an explicit expresion in terms of the weights and $\alpha$.

REPLY [5 votes]: There is an asymptotic formula: If $w_0,...,w_r$ are nonnegative integers such that $gcd(w_0,...,w_r)=1$, then for large $n$: 

$$a_n := |\lbrace (k_0,...,k_r) \in \mathbb{N}_0^r \mid \sum_i k_iw_i = n \rbrace| = \frac{n^r}{r!\;w_0 \cdots w_r} + O(n^{r-1})$$

For, as in Francesco's answer we have 
$$\sum_{n \ge 0} a_n t^n = \frac{1}{(1-t^{w_0})\ldots (1-t^{w_r})}$$
(let me know if you need more details for this step). Now the result follows from Robert Israel's answer of the following question: 
Asymptotics for the coefficients of a rational function 

Moreover, it's known that for large $n$, $a_n$ is a "polynomial with periodic coefficients" in $n$, i.e. there are polynomials $f_i$ over $\mathbb{Q}$ of degree $r$ such that $a_{nw+i} = f_i(n)$ for $0 \le i < w$ and all large enough $n$, where $w := lcm(w_0,...,w_r)$. 
This is just Excercise 10.12 in [Eisenbud: Commutative Algebra].<|endoftext|>
TITLE: Is every ''group-completion'' map an acyclic map?
QUESTION [32 upvotes]: I start with a longer discussion which will result in a precise version of the question. A am puzzled about an issue with the 
Quillen plus construction. I have seen outstanding experts being confused about this point. 
There are the following different ways of calling a map $f:X \to Y$ a homology equivalence:

$f_*:H_*(X;\mathbb{Z}) \to H_*(Y;\mathbb{Z})$ is an isomorphism ("weak homology equivalence").
For each abelian system of local coefficients $A$ on $Y$ ($\pi_1 (Y)$ acts through an abelian group), the induced map 
$H_* (X;f^* A) \to H_* (Y;A)$ is an isomorphism ("strong homology equivalence").
For each system of local coefficients $A$ on $Y$, the induced map 
$H_* (X;f^* A) \to H_* (Y;A)$ is an isomorphism ("acyclic map").

The third condition is equivalent to each of
3'. The homotopy fibres of $f$ are acyclic.
3'''. $f$ is can be identified with the Quillen plus construction.
EDIT: Before, I included the statement ''3''. $f$ is weak homology equivalence, $\pi_1 (f)$ is epi and $ker(\pi_1 (f))$ is perfect.'' This is false (does not imply the other two conditions); in my answer to Spaces with same homotopy and homology groups that are not homotopy equivalent?
I gave an example of a weak homology equivalence that is even an isomorphism on $\pi_1$, but whose homotopy fibre is not acyclic.
END EDIT
The implications $(3)\Rightarrow (2)\Rightarrow ( 1)$ hold. If all components of $Y$ are simply connected, then all these notions coincide; if $\pi_1 (Y)$ is abelian (each component), then $(2)\Rightarrow(3)$. In that case, $\pi_1 (X)$ is quasiperfect (i.e., its commutator subgroup is perfect). If $\pi_1 (Y)$ is nonabelian, then $(2)$ does not imply $(3)$ (take the inclusion of the basepoint into a noncontractible acyclic space).
Even if $Y$ is an infinite loop space, a weak homology 
equivalence does not have to be strong: Take $X=BSL_2 (Z)$, $Y=Z/12$. The abelianization of $SL_2 (Z)$ is $Z/12$, and the map $SL_2 (Z) \to Z/12$ is a weak homology equivalence.
The kernel, however, is a free group on two generators.
Now, many cases of such maps arise in the process of ''group completion''. Here are some examples

$X=K_0 (R) \times BGL (R)$ for a ring and $Y=\Omega B (\coprod_{P} B Aut (P))$ ($P$ ranges over all finitely generated projective $R$-modules).
The commutator subgroup is perfect due to the Whitehead lemma.
$X=\mathbb{Z} \times B \Sigma_{\infty}$; $Y=QS^0$. The alternating groups are perfect.
$X=\mathbb{Z} \times B \Gamma_{\infty}$ (the stable mapping class group); $Y$ the Madsen-Weiss infinite loop space. 
Here there is no problem, $\Gamma_g$ is perfect for large $g$.
$X=\mathbb{Z} \times B Out(F_{\infty})$ (outer automorphisms of the free group), $Y=Q S^0$. Galatius proves that this is a weak 
homology equivalence and he states implicitly this map is a strong homology equivalence.

I explain why I am interested: if you only look at the homology of $X$ and $Y$, this is only an aesthetical question. I want to take 
homotopy fibres and as explained above, the distinction is essential and a mistake here can ruin any argument.
In all these cases, there is a topological monoid $M$ (for example $\coprod_{P} B Aut (P)$) and $X$ is the limit $M_{\infty}$ obtained by 
multiplying with a fixed element. There is an identification $\Omega BM$ with $Y$ that results from geometric arguments and 
does not play a role in this discussion.
There is a map $\phi:M_{\infty} \to \Omega BM$, which is the subject of the ''group-completion theorem'', see
the paper "Homology fibrations and the ''group completion'' theorem" by McDuff-Segal. The map arises from letting $M$ act on 
$M_{\infty}$ and forming the Borel construction
$EM \times_M M_{\infty} \to BM$. The point-preimage is $M_{\infty}$, the space 
$EM \times_M M_{\infty}$ is contractible and so the homotopy fibre is $\Omega BM$. $\phi$ is the ''geometric-fibre-to-homotopy-fibre'' map.
What Segal and McDuff prove is that if the action is by weak homology equivalences, then $\phi$ is a weak homology equivalence.
This is what is typically used to established the above results. To prove that 1,2,3 are strong homology equivalences, one can invoke an extra 
argument which is specific to each case.
Now, in McDuff-Segal, I find the claim (Remark 2) that 
their methods give that $M_{\infty} \to \Omega BM$ is a strong homology equivalence and I want to understand this.
I convinced myself that the whole argument goes through with strong homology equivalences (and the corresponding notion of 
"strong homology fibration"). Proposition 2 loc.cit. then has the assumption that $M$ acts on $M_{\infty}$ by strong homology equivalences
(one needs the notion of homology equivalences one wants to prove in the end - which I find plausible). 
This amounts, say in example 4, to prove that the stable stabilization map $B Out(F_{\infty}) \to B Out(F_{\infty})$ is a homology 
equivalence in the strong sense. For "weak homology equivalence", one invokes the usual homology stability theorem (Hatcher-Vogtmann-Wahl). 
But it seems that for the map being a strong homology equivalence, one needs a stronger homological stability result. I can imagine how the homological stability arguments can be modified to include abelian coefficient system, but that is not a satisfying solution.
Here are, finally, some questions: 

McDuff and Segal refer to ''argument by Wagoner'' in his paper ''Delooping classifying spaces in algebraic K-Theory''. I am unable to
find an argument in Wagoners paper that proves under general assumptions quasiperfectness. What argument do McDuff and Segal refer to?
If $M$ is a topological monoid and if $M_{\infty} \to \Omega BM$ is a weak homology equivalence, is it always a strong homology equivalence?
If not, do you know a counterexample?
If 2 is not true, is there a useful general criterion to prove that the group completion map is acyclic (besides the trivial case
$H_1 (M_{\infty})=0$ and besides proving quasiperfectness of $\pi_1 (M_{\infty})$ by hands).

A related, but not central question:

What are good counterexamples to the ''group-completion'' theorem in general that explain why the hypothesis is essential?

REPLY [17 votes]: I think I have been able to reproduce the "argument by Wagoner" (perhaps it was removed from the published version?). It certainly holds in more generality that what I have written below, using the notion of "direct sum group" in Wagoner's paper (which unfortunately seems to be a little mangled).
Let $M$ be a homotopy commutative topological monoid with $\pi_0(M)=\mathbb{N}$. Choose a point $1 \in M$ in the correct component and let $n \in M$ be the $n$-fold product of 1 with itself, and define $G_n = \pi_1(M,n)$. The monoid structure defines homomorphisms
$$\mu_{n,m} : G_n \times G_m \longrightarrow G_{n+m}$$
which satisfy the obvious associativity condition. Let $\tau : G_n \times G_m \to G_m \times G_n$ be the flip, and
$$\mu_{m,n} \circ \tau : G_n \times G_m \longrightarrow G_{n+m}$$
be the opposite multiplication. Homotopy commutativity of the monoid $M$ not not ensure that these two multiplications are equal, but it ensures that there exists an element $c_{n,m} \in G_{n+m}$ such that
$$c_{n,m}^{-1} \cdot \mu_{n,m}(-) \cdot c_{n,m} = \mu_{m,n} \circ \tau(-).$$
Let $G_\infty$ be the direct limit of the system $\cdots \to G_n \overset{\mu_{n,1}(-,e)}\to G_{n+1} \overset{\mu_{n+1,1}(-,e)}\to G_{n+2} \to \cdots$.
Theorem: the derived subgroup of $G_\infty$ is perfect.

Proof: Let $a, b \in G_n$ and consider $[a,b] \in G'_\infty$. Let me write $a \otimes b$ for $\mu_{n,m}(a, b)$ when $a \in G_n$ and $b \in G_m$, for ease of notation, and $e_n$ for the unit of $G_n$.
In the direct limit we identify $a$ with $a \otimes e_n$ and $b$ with $b \otimes e_n$, and we have
$$b \otimes e_n = c_{n,n}^{-1} (e_n \otimes b) c_{n,n}$$
so $b \otimes e_n = [c_{n,n}^{-1}, (e_n \otimes b)] (e_n \otimes b)$. Thus
$$[a \otimes e_n, b \otimes e_n] = [a \otimes e_n, [c_{n,n}^{-1}, (e_n \otimes b)] (e_n \otimes b)]$$
and because $e_n \otimes b$ commutes with $a \otimes e_n$ this simplifies to
$$[a \otimes e_n, [c_{n,n}^{-1}, (e_n \otimes b)]].$$
We now identify this with
$$[a \otimes e_{3n}, [c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}]$$
and note that $a \otimes e_{3n} = c_{2n,2n}^{-1}(e_{2n} \otimes a \otimes e_{n})c_{2n,2n} = [c_{2n,2n}^{-1}, (e_{2n} \otimes a \otimes e_{n})]\cdot (e_{2n} \otimes a \otimes e_{n})$.
Again, as $(e_{2n} \otimes a \otimes e_{n})$ commutes with $[c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}$ the whole thing becomes
$$[a,b]=[[c_{2n,2n}^{-1}, (e_{2n} \otimes a \otimes e_{n})], [c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}],$$
a commutator of commutators.<|endoftext|>
TITLE: Matrices that are > 1 in a sense
QUESTION [5 upvotes]: How can I characterize the class of square matrices such that:
$ ||MN||_F \ge ||M||_F $?
In other words, when multiplied, they always give "bigger" products.
The norm is the Frobenius norm, which is the same as the Euclidean norm of the vectorization of the matrices.
Note:
if we recast this in vectors, we are asking for a class of vectors such that:
$ || U \otimes V|| \ge ||V|| $
where $\otimes$ is the usual matrix product by turning the vectors into matrices and converting the result back to a vector.  In other words, if vec means vectorization, and mat means matrixization, so $mat(X) = vec^{-1}(X)$, then: 
$ U \otimes V = vec(mat(U) \cdot mat(V)) $
Note:
the Frobenius norm can be defined $||M|| = \sqrt{trace M^\ast M}$, so what we want is: 
$ tr (MN)^\ast (MN) \ge tr M^\ast M$ 
$ tr N^\ast M^\ast M N \ge tr M^\ast M$ 
$ tr M^\ast M N N^\ast \ge tr M^\ast M$ 
$ tr N N^\ast M^\ast M \ge tr M^\ast M$ 
using the cyclic property of trace.
Note:
the Frobenius norm is also the $\sqrt{\mbox{sum of squares of entries}}$.  So the requirement can also be spelled out as: 
$ \sqrt{ \sum_{ij} (\sum_k m_{ik} n_{kj})^2 } \ge \sqrt{ \sum_{ij} m_{ij}^2 } $ 
$ \sum_{ij} (\sum_k m_{ik} n_{kj})^2 \ge \sum_{ij} m_{ij}^2 $
But I don't know how to proceed further from this...

REPLY [5 votes]: $\def\vec#1{\mathbf{#1}}\def\tr{\mathop{\mathrm{tr}}}$ I'll develop the answer suggested in the comments for the sake of clarity. I'm assuming that you want conditions for $N$ such that $\forall M: \| MN \|_F \geqslant \| M \|_F~$(where $\|\ast\|_F$ is the Frobenius norm).
I will generally consider the squares of the Frobenius norm, as the inequality is preserved under squaring. Consider an operator $M$ with singular value decomposition $$ M = \sum_j s_j \; \vec q_j \vec r_j^\ast \;,$$
where $\vec q_j$ and $\vec r_j$ are the orthonormal sets of left- and right-singular vectors, and where the singular values are a decreasing sequence of non-negative reals, $s_1 \geqslant s_2 \geqslant \cdots \geqslant 0$.
Then the Frobenius norm of $M$ is just the Euclidean norm of the vector $\vec s$ of singular values, by
$$ \| M \|_F^2 \;=\;\tr(M M^\ast) = \tr\left( \sum_j \sum_k s_j s_k \; \vec q_j^{\phantom \ast} \vec r_j^\ast \vec r_k^{\phantom \ast} \vec q_k^\ast \right) = \;\sum_j s_j^2 \;.$$
Consider what happens when we multiply on the left by $M$: the square of the Frobenius norm is
$$\begin{align*} \| MN \|_F^2 \;&=\;\tr(MN N^\ast M^\ast) \;=\;\tr(N^\ast M^\ast MN)
\\\\&= \sum_j \sum_k s_j s_k \tr\left( N^\ast \vec r_j^{\phantom \ast} \vec q_j^\ast \vec q_k^{\phantom \ast} \vec r_k^\ast N \right)
\\\\&= \;\sum_j s_j^2 \tr\left( N^\ast \vec r_j^{\phantom \ast} \vec r_j^\ast N \right) 
\\\\&= \;\sum_j s_j^2 \tr\left( \vec r_j^\ast N N^\ast \vec r_j^{\phantom \ast} \right)
\\\\&= \;\sum_j s_j^2 \bigl\| N^\ast \vec r_j^{\phantom \ast} \bigr\|_F^2
\;,\end{align*}$$
using the cyclic property of the trace on the second and second-to-last lines, and the fact that the trace of a scalar is just the scalar itself (which happens in this case to be the inner product of a vector with itself, or the Euclidean-norm-square of that vector).
We want the value on the last line above to be larger than $\| M \|_F^2$ no matter what the right-singular vectors $\vec r_j$ happen to be, or what the singular values $s_j$ are. In particular, it must be larger even if $s_1$ is the only non-zero singular value (that is, even if $M$ is a rank one operator); so we may as well reduce to that special case — we require $\| N^\ast \vec r \|_F \geqslant 1$ for all unit vectors $\vec r$. If you consider the singular value decomposition of $N^\ast$,
$$ N^\ast = \sum_k c_k \; \vec a_k \vec b_k^\ast \;,$$
this means in particular that the smallest singular value $c_n$ must be at least $1$; otherwise, we would have $\| N^\ast \vec b_n \|_F = c_n \| \vec a_n \| < 1$.
We have almost shown what was stated in the comments. Note that we can easily obtain the singular value decomposition of $N$ from that of $N^\ast$:
$$ N = \left( \sum_k c_k\; \vec a_k \vec b_k^\ast \right)^\ast = \sum_k c_k\; \vec b_k \vec a_k^\ast \;;$$
then the singular values of $N$ must also be at least $1$. Also, because all of the singular values of $N$ are positive, it is invertible; and we can easily show
$$ N^{-1} = \sum_k c_k^{-1} \;\vec a_k \vec b_k^\ast \;.$$
Then the maximum singular value of $N^{-1}$ is at most $1$, or equivalently
$$ \Bigl\| N^{-1} \Bigr\|_\infty \leqslant\; 1\;, $$
where $\| \ast \|_\infty$ is the uniform norm on operators:
$$ \| A \|_\infty = \sup\; \Bigl\{ \| A \vec v \| \;:\; \vec v \in \mathop{\mathrm{dom}}(A) \text{ and } \|\vec v\| = 1 \Bigr\}. $$
(For operators on finite-dimensional vector spaces, the supremum can be replaced with a maximum; then the uniform norm is essentially the largest singular value by definition.) This is just another way to formulate the criterion, and (because the uniform norm is a useful operator norm in its own right) possibly the most useful way to present it succinctly. It is easy to see that $N$ being invertible and $\| N^{-1} \|_\infty \leqslant 1$ are both necessary and sufficient conditions: if $N^{-1}$ shrinks all vectors, then $N$ stretches all vectors, and in particular the right-singular vectors of any matrix $M$.<|endoftext|>
TITLE: Kodaira-Spencer map as a "differential"
QUESTION [5 upvotes]: Using the laguage of derived category, the Kodaira-Spencer map
$\kappa(x) : Ext^1_X(k(x), k(x)) \rightarrow Ext^1_X(\mathcal F_x, \mathcal F_x)$
can be described as a Fourier-Mukai transformation $\Phi_{\mathcal F}$.
$\Phi_{\mathcal F} : Hom_{D^b(X)} (k(x), k(x)[1]) \rightarrow Hom_{D^b(X)}(\mathcal F_x, \mathcal F_x[1]) $
Here, $x \in X$ is a point of a smooth projective variety over an algebraically closed field. 
$\mathcal F$ is a coherent sheaf on $X \times X$ which is flat over the first factor, and $\mathcal F_x$ means a $\mathcal {i}^*_{x\times X} \mathcal F$. 
(which is equal to $\Phi_{\mathcal F} (k(x))$)
Of course, KS map is not defined as a differential of some morphism, but please consider the following argument ;

Suppose $\mathcal F_x$  is concentrated at $x$ for all $x \in X$. It means the map $f:x \mapsto \mathcal F_x$ is injective. Hence $\kappa(x) := df(x)$ is injective at generic point. 

(see D.Huybrechts, Fourier-Mukai transform in algebraic geometry, Ch7, Prop7.1)
Here, the definition of $f$ is a nonsense. It is not even clear where the image of $f$ lives. But the argument is quite persuasive. And it is even geometrically intuitive, because it describes a KS map as a differential of some "function". 
I think this must be a shadow of rigorous mathematical contents(probably a deformation theory), but I failed to make it so. Could someone explain to me what's going on? Thanks in advance.

REPLY [4 votes]: You can view the sheaf $\mathcal{F}$ on $X \times X$ instead as a map from $X$ (the first factor) to the moduli stack $M$ of sheaves on $X$ (the second factor).  This induces a map on the tangent spaces.  The tangent space to $M$ at a sheaf $F$ is the space of first order deformations of $F$, which is $\mathrm{Ext}^1(F, F)$.
Using the exact sequence
$0 \rightarrow I \rightarrow \mathcal{O}_X \rightarrow k(x) \rightarrow 0$
you get an identification between $T_x X = \mathrm{Hom}_{\mathcal{O}_X}(I, k(x)) = \mathrm{Ext}^1(k(x),k(x))$.  Therefore the differential gives a map
$T_x X = \mathrm{Ext}^1(k(x),k(x)) \rightarrow T_{\mathcal{F}_x} M = \mathrm{Ext}^1(\mathcal{F}_x, \mathcal{F}_x)$.<|endoftext|>
TITLE: Norm concentration of trigonometric polynomials - Uncertainty principle
QUESTION [7 upvotes]: Hi all, 
I am interested in the following question (which is quite similar to one I posed a long while ago): Let $P_{N}(t)=\underset{k=-N}{\overset{N}{\sum}}c_{k}e^{ikt}$ be a unit norm trigonometric polynomial, we look at it as a function of $L^{2}\left(\mathbb{T}\right)$.
I'd like to find a direct proof to the fact that there exists $\varepsilon>0$ such that for every $N$ and every such polynomial $P_{N}$ we have $\underset{E}{\int}|{P_{N}( t)}|^{2}dt\leq1-\varepsilon$ whenever $E\subset\mathbb{T}$ of measure $|E|=\frac{c}{N}$, and   $c>0$ is some absolute constant. I would be happy with a proof only in the case $E$ is an interval, if it is any different than the general case.
To rephrase the statement; one cannot concentrate the norm of a trigonometric polynomial of degree $N$ on an interval (or any measurable set) of length (measure) of the order of magnitude $\frac{1}{N}$.
Let me comment that there is a result by Nazarov which implies this but it is way too general for my purposes.

REPLY [3 votes]: Let's talk about algebraic polynomials (just multiply by $z^N$). Let $P$ be a polynomial of degree $N$. Let $\max|P|=1$ and assume that this maximum is attained at some point $p$. Take the disk $D$ of radius $10C/N$ centered at $p$. Note that $z^{-N}P(z)$ satisfies the maximum principle in the complement of the unit disk and $P(z)$ satisfies the maximum principle in the disk itself, so $|P|<e^{10C}$ in $D$. Rescale $D$ to the unit disk $\mathbb D$ and divide by $e^C$. You'll get a bounded by $1$ function $F$ that is $e^{-10C}$ at the center. 
Now it is the usual story about subharmonicity of the logarithm of an analytic function. Let $H$ be any closed set on the circular arc passing through the origin on which $F$ is very small. Suppose that the length of $H$ is $1/2$ or more. Consider the function
$$
U(z)=\int_H\log \frac{|z-w|}{|1-\bar wz|}d\ell(w)
$$
It is easy to see that $U\ge -A$ in $\mathbb D$, harmonic in $\mathbb D\setminus H$, $U=0$ on the unit circumference, and $U(0)\le -a$ for some absolute $a,A>0$. If $|F|\le e^{-10CA/a}$ everywhere on $H$, then $\log|F|\le (10C/a)U$ on the boundary of $\mathbb D\setminus H$ and, thereby, $\log|F(0)|\le -10C$, which is not the case. Thus, the minimum of $|F|$ over every set $H$ of length $1/2$ or more is bounded from below. Coming back to the original problem, we see that $|P|$ is bounded from below by some constant depending on $C$ on at least half of the arc of length $10C/N$, so we have plenty of noticeable values outside any set of measure $C/N$.
This trick is pretty old and goes back to Bernstein. There is also another approach due to Remez (moving zeroes and looking at the level sets). You are 100% right when saying that you do not need the Turan type bounds and the related fancy techniques for this problem. :)<|endoftext|>
TITLE: Definitions of Reductive and Semisimple Groups
QUESTION [12 upvotes]: I'm a graduate student.  I've been reading Knapp's two books Representation Theory of Semisimple Groups and Lie Groups Beyond an Introduction.  He seems to give wildly different definitions for the basic terms, 'reductive' and 'semisimple', and it leaves me unclear on what to think when I read 'reductive group' or 'semisimple group' alone in the literature.
To be more specific, Knapp doesn't actually define "reductive" in Representation Theory; he defines "linear connected reductive group".  But in his other book, one of the reasons he says we need to introduce reductive groups is to deal with the disconnected groups arising from our analysis.  Their disconnectedness is "not too wild", and so much of the structure theory developed for semisimple groups goes through.  The definitions in Lie Groups make more sense to me.  
I'm similarly confused about whether semisimple groups have finite center, because Knapp seems to disagree with himself on that.  
How should I think about what these terms mean?  Is there an authoritative definition.  The definitions in Representation Theory seem most strange to me, but the resulting theory he develops is very familiar from everything else I've read.  I'm curious why Knapp chose such strange definitions for his Representation Theory in the first place.  Does the substance of his book go through if I replace his definitions with something more standard.  (I'm aware this last bit is rather vague.)

REPLY [4 votes]: In addition to other answers, especially if one is interested in (infinite-dimensional) representation theory of Lie groups, a common technical issue is that the maximal compact subgroup should meet all the connected components of the ambient groups. This is part of what often gets called "the Harish-Chandra class"' definition for semi-simple or reductive, and is (apparently!) often sufficient to cope with disconnectedness of the ambient non-compact group.
And, reiterating a point made in the other answers, one always must pay attention to context to know the precise assumptions, e.g., about connectedness and/or what compromises suffice for the mechanisms to succeed without assuming the ambient group is connected.<|endoftext|>
TITLE: Definition of continuous functions in order theory
QUESTION [5 upvotes]: If we have a complete partial order (i.e. directed complete) I find frequently the following definition of a continuous function. A function $f:A\to B$ where $A$ and $B$ are cpos is called continuous if it maps the suprema of directed subsets of $A$ (if exist) to the corresponding suprema of directed subsets of $B$.
In complete lattices I would define continuous functions as functions which preserve suprema and infima (since both exist in a complete lattice for any subset).
Since complete lattices are cpos the following question arises: Are both definitions consistent? The requirement that all suprema and infima are preserved is stronger than the requirement that only suprema of directed sets are preserved. Therefore it might be possible that both definitions are different. Or are they equivalent?

REPLY [6 votes]: To make some sense of the various possibilities for maps in order theory, it is best to look at structures in the sense of algebra, rather than properties of maps.
According to algebra, when dealing with structured sets, the corresponding notion of homomorphism should be a map which preserves structure. And then category theory teaches us that the morphisms are just as important as the object (in fact, they are more important).
For example, it may seem odd to distinguish between posets which have all suprema and posets which have all suprema and infima. After all, any poset which has all suprema has the infima too. But the difference matters when we look at the two categories:

$\mathbf{SupLat}$: the objects are posets with arbitrary suprema (which are just complete lattices), the morphisms are maps which preserve suprema.
$\mathbf{SupInfLat}$: the objects are posets with arbitrary suprema and inifma (which again are just complete lattices), the morphisms are maps which presrerve suprema and infima.

As was already pointed out in Zhen's answer, there are maps on complete lattices which preserve suprema but not infima, so the distinction is meaningful.
You ask about a clear definition of continuity in complete lattices. Continuity is about topology, so we should look at ways of topologizing complete lattices, or more generally posets, as then it will be clear what the continuous maps are. Among all the possibilities, it is probably desirable to restrict to those that allow us to recover the partial order from the topology by passing to the specialization order. The strongest topology with this property is the Alexandrov topology for which continuity coincides with monotonicity. A very reasonable choice of topology induced by a partial order might be
the Scott topology which leads to the concept of a Scott continuity: for reasonable posets a map is Scott-continuous when it preserves directed suprema. You may entertain yourself by figuring out whether there is a topology for which the continuous maps are those that preserve suprema and infima.
I cannot tell you which topology is the right one for you. That depends on what you are doing. I hope it is at least clear that the question should be framed in the context of algebra and category theory (morphisms preserve structure) and that continuity is about topology (so we should topologize partial orders).<|endoftext|>
TITLE: Representations of reductive groups over local fields through parahoric induction
QUESTION [5 upvotes]: Let me take $G$ to be a simple (connected) split reductive group over a local field $K$. One way I might go about constructing a (smooth, admissible) complex representation $\sigma$ of $G$ is as follows:

pick a parahoric subgroup $P$ of $G$, with pro-unipotent radical $U$,
form the quotient $P/U$, which is a (connected) reductive group over a finite field, 
write down a representation $\overline{\rho} \colon P/U \to \mathrm{GL}_n(\mathbb{C})$,
inflate this to a representation $\rho \colon P \to \mathrm{GL}_n(\mathbb{C})$,
induce $\rho$, giving $\sigma = \mathrm{ind}_P^G(\rho)$ (compact induction).

How does this procedure work out? Excepting that I might have to take possibly smaller $U_0 \subset U$, is this expected to produce all the (smooth admissible irreducible complex) representations of $G$? When do I get something irreducible, or a supercuspidal? What is the proper formulation when $G$ is not assumed to be split over $K$?

REPLY [5 votes]: This procedure allows to construct all "level $0$" irreducible representations of G. They appear as subquotients of your compactly induced representations.  Here "level $0$ means that the representation has a non-zero fixed vector under the pro-unipotent subgroup of some parahoric. The answers to your questions are is the following paper:
Morris, Lawrence Level zero $\bf G$-types. Compositio Math. 118 (1999), no. 2, 135–157
If you use smaller groups $U_o \subset U$, then you can indeed get any irreducible representation as subquotient of a compactly induced representation. However when the compactly induced representation is irreducible it is automatically supercuspidal (see e.g. Bushnell-Henniart for a proof of that). All explicitely known supercuspidal representations
are indeed obtained by compact induction. But it is still conjectural that they all are. 
In general a compactly induced representation from an irreducible representation of a compact open subgroup splits in two part. An admissible part which is a finite sum of supercuspidal representations and a non admissible part which contains non-supercuspidal as irreducible subquotients. 
To describe the non supercuspidal representations by compact open data, a good point of view is that  of "types". You may read Bushnell and Kutzko's papers on that subject.<|endoftext|>
TITLE: $\Lambda$-Ring Structures on $\mathbb A^2$
QUESTION [18 upvotes]: A $\Lambda$-ring structure on a torsion-free ring over $\mathbb Z$ is a commuting family of endomorphisms $\psi_p$ satisfying $\psi_p(x) \equiv x^p$ mod $p$.
One $\Lambda$-ring structure on $\mathbb Z[x]$ is defined by $\psi_p(x)=x^p$.
Another can be defined in terms of the Chebyshev polynomials of the first kind, where $\psi_p(y)=2 T_p(y/2)$. One can view this $\Lambda$-ring as the quotient of the toric $\Lambda$-ring $\mathbb Z[z,z^{-1}]$  where $\psi_p(z)=z^p$ by the $\psi$-equivariant automorphism $z\to z^{-1}$, under the identification $y=z+z^{-1}$.
A result of F.J.-B.J. Clauwens in  Commuting polynomials and $\lambda$-ring structures on $\mathbb Z[x]$ shows that these are the only two $\Lambda$-ring structures on $\mathbb Z[x]$ up to isomorphism, but according to the paper Lambda-rings and the field with one element by James Borger, it is not even known whether there are finitely many $\Lambda$-structures on $\mathbb Z[x,y]$ up to isomorphism.

How many non-isomorphic $\Lambda$-ring structures on $\mathbb Z[x,y]$ are known?

All the $\Lambda$-ring structures I know how to construct come from quotients of toric varieties by $\psi$-equivariant group actions.

Are there any $\Lambda$-ring structures on $\mathbb Z[x,y]$ known which are not quotients of toric varieties by group actions? Are there any known which behave very differently from such quotients?

REPLY [10 votes]: This answer contains a list of all the $\Lambda$-ring structures that I have found. If anyone can find any others, they should edit this list.

The two $\Lambda$-ring structures on $\mathbb A^1$ give three product $\Lambda$-ring structures on $\mathbb A^2 = \mathbb A^1 \times \mathbb A^1$.
One of these product structures, $\psi_p(x)=x^p$ and $\psi_p(y)=y^p$, has a symmetry of order two switching $x$ and $y$. The ring of invariants of this symmetry is another structure.
One can look at the ring of invariants of $\mathbb Z[x,x^{-1},y,y^{-1}]$ with $\psi_p(x)=x^p$, $\psi_p(y)=y^p$ under the action of a subgroup of $GL_2(\mathbb Z)$, where $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)$ sends $x$ to $x^ay^c$ and $y$ to $x^by^d$. If that subgroup is $D_3$, $D_4$, or $D_6$, one gets a $\Lambda$-ring structure on $\mathbb Z[x,y]$. These can also be viewed as the $K$-theory $\Lambda$-rings of the simple Lie groups $SL_3$, $SP_4$, and $G_2$, respectively.

That is a total of $7$ structures.

Here is a proof sketch that they are distinct: These can all be expressed as $\psi_p$-stable subrings of the $\Lambda$-ring  $\mathbb Z[x,x^{-1},y,y^{-1}]$ with $\psi_p(x)=x^p$, $\psi_p(y)=y^p$, with the induced $\Lambda$-ring structure. It is easy to check that they are not isomorphic as subrings. Thus it sufficies to recover the embedding into that ring. 
Extend $\psi_p$ to $\psi_n$ the obvious way: so that $\psi_n \circ \psi_m = \psi_{nm}$. Then, for any element $\alpha\in \mathbb Z[x,x^{-1},y,y^{-1}]$, the sequence $\psi_n(\alpha)$ is a sum of geometric progressions and so satisfies a finite linear recurrence relation. So if $\alpha$ is a generic element of a $\Lambda$-subring $R$ of dimension $2$, the recurrence relation is defined over $R$, and the roots generate $\mathbb Z[x,y,x^{-1},y,y^{-1}]$. By picking a generic element, we can determine, purely from the $\Lambda$-ring structure on $R$, its embedding, and thereby distinguish the different rings.

EDIT: In fact, the first two constructions suffice to produce infinitely many $\Lambda$-rings! Consider the product ring $\mathbb Z[x,y]$ where $\psi_p(x)=x^p$ and $\psi_p(y)=2T_p(y/2)$. Then the subring $\mathbb Z[x,yx^n]$ is a sub-$\Lambda$ ring for each $n\geq 0$, and none of these are isomorphic. To check that they are non-isomorphic, one can recover $x$ as the unique solution to $\psi_2(x)=x^2$, then invert $x$, then recover $y$ as the unique solution to $\psi_2(y)=y^2-2$ that generates the ring along with $x$ and $x^{-1}$. Thus the embedding of the ring into $\mathbb Z[x,x^{-1},y]$ is unique, so the embedding into $\mathbb Z[x,y]$ is unique, and because they are distinct as subrings, they are distinct.
Similarly, inside $\mathbb Z[x,y]$ where $\psi_p(x)=x^p$ and $\psi_p(y)=y^p$, there is the subring $\mathbb Z[x^{n+1}y^n+x^ny^{n+1},xy]$ which is a sub-$\Lambda$-ring and depends on $n$ for similar reasons.
However these constructions are birationally equivalent to previous defined constructions. So perhaps it is better to consider this up to birational equvialence!<|endoftext|>
TITLE: Which forcings preserve (some) determinacy?
QUESTION [13 upvotes]: The question is exactly as in the title. I'm interested in general in all questions of the form "which forcings preserve property P?" for any P, but determinacy assumptions occupy a special place in my interests. More specifically, what results are known of the form "the forcings which preserve $\Gamma$-determinacy are exactly the following: . . ." for $\Gamma$ some reasonably natural pointclass? I am, for the purposes of this question, taking my base set to be $\mathbb{N}$ (that is, all payoff sets are subsets of $\mathbb{N}^\mathbb{N}$).
What I've been able to figure out so far (which is not very much):

Continuum-closed forcings preserve all determinacy assumptions. This is just because continuum-closed forcings add no new sets of reals - hence no new payoff sets, no new points in old payoff sets, and no new strategies.
Countably closed forcings preserve PD (projective determinacy). A countably closed forcing adds no new reals, and hence preserves the truth of analytic formulas with parameters (=definitions of projective sets), and since no new reals are added, no new strategies are added either. (As Andreas points out below, "countably closed" can be replaced by "adds no new reals.")
Countable closure is not enough to guarantee preservation of AD. The usual construction of a non-determined game can be reformulated as a countably closed forcing construction over a model of ZF; and even if the ground model satisfies AD, the generic extension will have a non-determined game.

I have tried to figure out whether either of these results reverse, but I've had no success here. The way I would attempt to phrase such a reversal would be something like the following:
(*) If $\mathbb{P}$ is some poset without property $P$, then there is a transitive model of set theory $W$ containing $\mathbb{P}$ and satisfying $\Gamma$-determinacy such that forcing with $\mathbb{P}$ over $W$ does not preserve $\Gamma$-determinacy.
[EDIT: As Francois points out, this isn't a good way to phrase a reversal statement, and it's not clear what a good way would be. So as an additional question, how can this idea be phrased in a non-silly way? Or is there good reason to believe that this can't be done?]
So, in addition to the main question, I have the following subquestions:
(i) Are any results along the lines of (*) known?
(ii) What methods seem like they could be useful for proving results along the lines of (*)?
(iii) For that matter, is my reasoning in the bullet points above correct? It seems straightforward enough, but I've been very wrong about these sorts of things before.
Thanks in advance!
[EDIT: I forgot to mention this initially, but for the purposes of this question I'm assuming the consistency of arbitrary large cardinals, although I am very interested in how much large cardinal strength any answers require.]

REPLY [6 votes]: Bumping to mention/advertise a recent development:
Today, William Chan and Stephen Jackson posted this paper to the arxiv. They proved that a broad class of forcings (under reasonable hypotheses) always kill determinacy. Specifically, they showed

Theorem 3.2: ZF + AD proves that every well-orderable forcing of cardinality $<\Theta$ forces $\neg$AD

and

Theorem 5.6: ZF + AD + "$\Theta$ is regular" proves that every nontrivial forcing which is a surjective image of $\mathbb{R}$ forces $\neg$AD, as does ZF + AD$^+$ + $\neg$AD$_\mathbb{R}$ + V=L$(\mathcal{P}(\mathbb{R}))$.


Note that we generally don't have $L(\mathbb{R})^V[g]=L(\mathbb{R})^{V[g]}$ (which would of course contradict these results).

Below I'll give a tissue-thin, hopefully-not-too-inaccurate description of their paper.

The arguments in the paper generalize the existing proofs of "determinacy-destruction" for a wide class of forcings under various assumptions (specifically: Cohen forcing, $Col(\omega_1,\omega_2)$, and $Col(\omega,\omega_1)$) from prior analysis by Ikegami and Trang:

Ikegami and Trang initiated the study of the preservation of AD under forcing. They showed that many forcings, such as Cohen forcing, can never preserve AD. They also showed that if one is working with natural
  models of AD, i.e. models satisfying ZF + AD+ + V = L$(\mathcal{P}(\mathbb{R}))$, then any forcing which preserve AD must
  preserve $\Theta$, where $\Theta$ is the supremum of the ordinals which are surjective images of $\mathbb{R}$. They also showed
  that the consistency of ZF + AD+ + $\Theta>\Theta_0$ implies the consistency of ZF + AD and there is a forcing which
  preserve AD and increases $\Theta$. Thus necessarily this forcing must disturb $\mathcal{P}(\mathbb{R})$ by adding a new set of reals

(Unfortunately, it looks like this work hasn't appeared yet, and Ikegami and Trang don't appear in the bibliography. My quick impression, though, is that the Chan/Jackson paper subsumes their results.)
Chan and Jackson isolate a common combinatorial aspect of these examples:  the ground club property at $\kappa$ (Definition 2.9), that any new club in $\kappa$ contains a ground club in $\kappa$. The ground club property provides a connection between Ramsey properties and reals (slightly rephrased):

Lemma 2.10: If $V[G]$ is a generic extension via a forcing with the ground club property at $\kappa$ and $V[G]\models\kappa\rightarrow(\kappa)^\omega_2$, then $\mathbb{R}^V=\mathbb{R}^{V[G]}$.

From this they quickly deduce Theorem 3.2 above. (A different argument, joint with Goldberg, improves the theorem by replacing "cardinality $<\Theta$" with "adds a real" - Corollary 3.5.)
This lemma is also fundamental to the proof of Theorem 5.6, but the argument there is much more complicated, and relies on an analysis of the behavior of $\Theta$ after forcing, building off of a previous result by Ikegami and Trang. In particular, they show that (in ZF + AD) any nontrivial forcing which is a surjective image of $\mathbb{R}$ and preserves AD must add a real (Fact 4.4), and so must preserve $\Theta$ (Lemma 4.3 - which surprised me quite a bit).

I hope I haven't misrepresented anything here; regardless of my understanding, this is certainly a very relevant paper.<|endoftext|>
TITLE: Small primes in arithmetic sequences
QUESTION [13 upvotes]: Fix an integer $a>1$. For $n \geq 1$ an integer, let $\pi_{n,1}(an)$ the number of primes 
$p \leq an$ such that $p \equiv 1 \pmod{n}$, and $\pi(an)$ the number of all primes $p \leq an$. Let
$$Q_a(n) = \frac{\pi_{n,1}(an)}{\pi(an)} \phi(n),$$ 
where $\phi(n)$ is Euler's phi function.
If instead of fixing $a$ we fix $n$ and let $a$ goes to infinity, then by 
Dirichlet's theorem that $\lim_{a \rightarrow \infty} Q_a(n) = 1$. If we don't fix $n$ but let  it goes to infinity fast enough relatively to $a$, for example $n=a^{1+\epsilon}$ with $\epsilon>0$ under (GRH), then one can prove that the limit is still 1 by some effective version of Dirichlet. But I am interested here in the case where $a$ is fixed. In this case, it is clear that $Q_a(n)$ varies to widely to have a limit when $n \rightarrow \infty$. 
Hence let us tame $Q_a(n)$ by considering, following Cesaro,
$C_a(n) = \frac{Q_a(1)+\dots+Q_a(n)}{n}$.

Does $C_a(n)$ have a limit when $a$ is fixed and $n$ goes to infinity ? If so what is this limit ?

I have made some sage computations for different values of $a$ ($a=2$ to $10$) and $C_a(n)$ 
seems to have a tendency to grow very slowly, though it is not clear if it is toward a finite limit or $+\infty$ -- or if the whole thing is just an artefact.
My motivation is trying to understand (if only conjecturally), in the simplest case I can think of, what happens to the effective Chebotarev density theorem beyond the version I can find in the literature. I'll appreciate any answer, be it unconditional,
based on a conjecture like GRH, or even purely heuristic.
Comments (added on October 16th): I am interested in the question above for any integer $a$ but in my research a similar question arose with $a=8$. Actually I don't think the answer will really change of nature with $a$, so we can focus on the case $a=2$. In this case,
the only number congruent to $1$ modulo $n$ between $1$ and $2n$ which is susceptible to be prime is $n+1$, hence $Q_2(n)=0$ if $n+1$ is not prime, while when $n+1$ is a prime $p$,
$$Q_2(n)=\frac{\phi(p-1)}{\pi(2 (p-1))}.$$
By the way, this illustrates the fact that $Q_2(n)$ does not have a limit when $n \rightarrow \infty$: $0$ is obviously the inf.lim., but $+\infty$ is the sup.lim. : think on $n=p-1$
being for example the higher prime I'm a Sophie Germain's pair of prime, so that $\phi(n)=(p-1)/2-1$ and $Q_2(p-1) \sim \log p / 4$ which goes to inanity if one choses an infinite sequence of Germain's primes (which is widely expected to exist -- at this stage, I am perfectly happy to use any conjecture even if one can do otherwise). 
Back to the question, one has:
$$C_2(n) = \frac{1}{n} \sum_{1 < p \leq n,\  p \ \rm prime} \frac{\phi(p-1)}{\pi(2p-2)}.$$
Note that since $Q_2(n)$ is non-negative, one can replace it by an equivalent, so
$$C_2(n) \sim D_2(n) := \frac{1}{n} \sum_{1 < p \leq n,\  p \ \rm prime} \frac{\phi(p-1) \log(p)}{2p},$$
and the question begins to look like a complicated version of a question already asked several times on mathoverflow about the Cesaro average of Euler's $\phi(n)$, which behaves much more smoothly than $\phi(n)$ itself. Experimentally, here is what I get for $C_2(n)$ for $n=2^k$, $k$ running from $1$ to $23$:
0.750000000000000 
0.500000000000000 
0.300000000000000
0.254482323232323
0.204751427085986
0.182394996041895
0.174044947095252
0.177096489596196
0.177412757367371
0.175004984083009
0.175280949354989
0.176774240882088
0.177008402332853
0.178414103595542
0.178516411591865
0.179091173423042
0.179809089385918
0.180252447106263
0.180775697751659
0.181112338150868
0.181529153981739
0.181858564625316
0.182136158910456
To me it looks like $C_2(n)$ has a limit or perhaps goes to infinity a little bit slower than $\log n$, but I am really not well-trained in the difficult art of divination of limits of
sequences from their first terms... What do you think?

REPLY [12 votes]: When $a=2$, the sum you want the asymptotics of is basically
$$
\frac12 \frac{\log n}n \sum_{p\le n} \frac{\phi(p-1)}{p-1} \sim \frac12 \frac1{\pi(n)} \sum_{p\le n} \frac{\phi(p-1)}{p-1},
$$
which is half the average value of the multiplicative function $\phi(n)/n$ on shifted primes $p-1$. The heuristic for evaluating this constant is as follows: recall that
$$
\frac{\phi(p-1)}{p-1} = \prod_{q\mid(p-1)} \bigg( 1-\frac1q \bigg).
$$
For every fixed prime $q$, a proportion $1/(q-1)$ of primes $p$, namely those congruent to $1$ (mod $p$), will have $\phi(p-1)/(p-1)$ containing a factor of $1-1/q$; the others, a proportion $(q-2)/(q-1)$ of the primes, simply have the factor 1 instead. Heuristically, all these contributions are independent, and so the average value of $\phi(p-1)/(p-1)$ should be the product of the averages for each $q$, which are
$$
\frac1{q-1} \bigg( 1-\frac1q \bigg) + \frac{q-2}{q-1}1 = 1-\frac1{q(q-1)}.
$$
Therefore we predict that
$$
\frac12 \frac1{\pi(n)} \sum_{p\le n} \frac{\phi(p-1)}{p-1} \to \frac12 \prod_q \bigg( 1-\frac1{q(q-1)} \bigg) \approx 0.186978.
$$
This can be proved without much difficulty, using the method outlined in Terry's comment. EDITED TO ADD:
$$
\sum_{p\le n} \frac{\phi(p-1)}{p-1} = \sum_{p\le n} \sum_{d\mid(p-1)} \frac{\mu(d)}d = \sum_{d\le n} \frac{\mu(d)}d \sum_{p\le n, p\equiv1\pmod d}1 = \sum_{d\le n} \frac{\mu(d)}d \pi(n;d,1),
$$
where $\pi(x;q,a)$ is the number of primes $p\le x$ with $p\equiv a\pmod q$.
Now one can use the prime number theorem in arithmetic progressions to get an asymptotic formula for $\pi(n;d,1)$ when $d$ is small and the Brun-Titchmarsh theorem to get an upper bound for $\pi(n;d,1)$ when $d$ is large. The result will be the same, in the limit, as what you get if you simply plug in $\pi(x)/\phi(d)$ for $\pi(x;d,1)$:
$$
\frac12 \frac1{\pi(n)} \sum_{p\le n} \frac{\phi(p-1)}{p-1} \sim \frac12 \sum_{d\le n} \frac{\mu(d)}{d\phi(d)}
\sim \frac12 \sum_{d=1}^\infty \frac{\mu(d)}{d\phi(d)} = \frac12 \prod_p \bigg( 1 + \frac{-1}{p\phi(p)} + \frac0{p^2\phi(p^2)} + \cdots \bigg).
$$<|endoftext|>
TITLE: Maximum entropy priors in infinite dimensional spaces
QUESTION [8 upvotes]: Is there an extension of maximum entropy probability distributions for function spaces?
For $\mathbb{R}^n$ and discrete spaces, there is much literature about this problem under names such as "non-informative priors", "maximum entropy distributions", "Jeffrey's priors", and the like.
There is an extension to locally compact topological groups, where the Haar measure $U$ takes the place of the Lebesgue measure, and one looks for measures $P$ minimizing the information divergence,
$$D(P||U):= \begin{cases}
\int log \frac{dP}{dU} dP, & \text{ if } P\ll U; \\\
\infty, & \text{else.}\end{cases}$$
However, I've found little about this in the infinite dimensional setting. Can the concept of maximum entropy priors be generalized to (some class of) function spaces, or is the idea of entropy fundamentally incompatible with spaces that are not locally compact?
Notes,

This is a repost from this statistics stackexchange thread which got no answers.
There is an infinite dimensional generalization of the "min- entropy" $H_\infty$, though this is a different concept from the "minimum entropy" $\text{min } H$.

REPLY [5 votes]: There is work on infinite-dimensional exponential families of measures which might be what you are looking for.
There are these possible references:

Scricciolo (2006). Convergence rates for Bayesian density estimation of infinite-dimensional exponential families
Rivoirard and Rousseau (2012). Posterior concentration rates for infinite dimensional exponential families
Deuschel (1987). Infinite-dimensional diffusion processes as Gibbs measures on $C \left[ 0, 1 \right]^{Z^d}$

The first paper provides such distributions on Sobolev spaces. Also, the references in that paper are helpful about using infinite-dimensional entropy optimizing-based measures in order to construct priors for use in Bayesian non-parametric statistics.
Edit: I added an additional reference that is also relevant to the topic.<|endoftext|>
TITLE: Stabilizer in automorphism group of free group of a certain finite-index subgroup
QUESTION [6 upvotes]: The following question arose in my research.  I'd be interested in an answer to it, but I'd also be interested in general techniques for solving this kind of problem (or, even better, pointers to computer programs that can solve it automatically).
Consider the group $G = \text{Aut}(F_2)$.  Of course, $G$ acts on $F_2$.  Let $x$ and $y$ be the generators for $F_2$, and let $K \subset F_2$ be the normal closure of the set $\{x^4, x^2 y^{-2}, y^{-1} x y x\}$.  It is standard that $F_2/K$ is the $8$-element group of quaternions.  
Question : What are generators for the subgroup $\{\text{$g \in G$ $|$ $g(K) = K$}\}$ of $G$?

REPLY [10 votes]: Here's a short proof by hand that $K$ is characteristic in $F_2$.
The natural map $Q_8\to V=\mathbb{Z}/2\oplus\mathbb{Z}/2$ induces $\eta:F_2\to H_1(F_2,\mathbb{Z}/2)\cong V$.  The kernel $N=\ker\eta$ is therefore characteristic and contains $K$ with index two.
Suppose now that $\phi$ is an automorphism of $F_2$ that does not preserve $K$.  Then $\phi$ descends to an automorphism of $V$ which does not lift to an automorphism of $Q_8$.  But it's easy to check that every automorphism of $V$ does, in fact, lift.<|endoftext|>
TITLE: Does ZF bound countable unions of countable sets?
QUESTION [6 upvotes]: ZF proves that whenever a countable union of countable sets can be well ordered then its cardinality is at most $\aleph_1$.  But what if it cannot be well ordered?  The Feferman-Levy model shows the continuum can be a countable union of countable sets.   Is there a ZF proof that a countable union of countable sets must be smaller than or equal to that in size?

REPLY [8 votes]: In Jech The Axiom of Choice, problem 14 on chapter 5 states:

Let $M$ be a transitive model of ZFC, there exists $M\subseteq N$ with the same cardinals as $M$ [read: initial ordinals] and the following statement is true in $N$:
For every $\alpha$ there exists a set $X$ such that $X$ is a countable union of countable sets, and $\mathcal P(X)$ can be partitioned into $\aleph_\alpha$ nonempty sets.

Note that $\mathcal P(X)$ can only be mapped onto set many ordinals, so this ensures us that there is indeed a proper class of cardinalities which can be written as countable unions of countable sets.
Jech adds and points out that this $N$ is not an inner model of any transitive model of ZFC, as that would imply $2^{\aleph_0}$ can be partitioned into any number of sets. The reference given is to Morris from 1970:

D. B. Morris, A model of ZF which cannot be extended to a model of ZFC without adding ordinals, Notices Am. Math. Soc. 17, 577.


I haven't sat down to verify all the details, but it should probably work:
Using an Easton symmetry-like class forcing, we may obtain model add for every regular $\kappa$ a countable collection of countable subsets of $2^\kappa$ whose union is not countable, let us denote this union $A_\kappa$. Genericity arguments should give us that for $\kappa\neq\kappa'$, $|A_\kappa|$ is incomparable with $|A_{\kappa'}|$ - namely there are no injections between those two sets.
This shows that there is a proper class of different sets which can be represented as countable unions of countable sets. So no upper-bound is possible.<|endoftext|>
TITLE: Is it true that the orbit space of a free finite group action on a CW-complex is also a CW-complex?
QUESTION [15 upvotes]: Suppose a finite group G acts freely and continuously on an n-dimensional CW-complex X. Then can we conclude that the orbit space of this action is still an n-dimensional CW-complex? (or homotopy equivalent to an n-dimensional CW-complex?)  In particular, we do not assume G acts cellularly on X.

REPLY [10 votes]: Lemma If $X$ is a countable locally finite CW-complex and $G$ acts freely and properly discontinuously on $X$, then $X/G$ is homotopy equivalent to a CW-complex.
Proof Any metrizable ANR is homotopy equivalent to a CW-complex
(I am not sure who proved it first but see Theorem 3.6.1 here. Since $X$ is countable and locally finite, it is a metrizable separable ANR. As Misha remarks in comments averaging the metric over the group action implies that $X/G$ is metrizable. Also a countable dense subset of $X$
projects to a countable dense subset of $X/G$. Finally, if a metrizable separable space is locally ANR, it is an ANR (see Borsuk's "Theorey of Retracts", Corollary 10.4, Chapter IV). It follows that $X/G$ is a metrizable ANR as desired.
 Remark In seeing whether $X/G$ is homeomorphic to a CW-complex, even the case when $X$ is a PL manifold is unclear. The difficulty is that it seems unknown which topological manifolds are homeomorphic to CW-complexes (Kirby-Siebenmann prove this for compact manifolds of dimension $\ge 6$ (or maybe $\ge 5$?, but certainly not $4$).
So there might exist manifolds not homeomorphic to CW-complexes but whose finite covers
are PL.

REPLY [3 votes]: The 3-sphere gives an example of an action with fixed points. If one takes the solid Alexander horned sphere, then Bing proved that its double is homeomorphic to the 3-sphere. So the quotient of the involution acting on $S^3$ is the solid Alexander horned sphere. However, the solid horned sphere is not homeomorphic to a CW complex. This follows from the answer to this question on the Alexander horned sphere. If the solid Alexander horned sphere were a CW complex, then one could attach the exterior 3-ball to get a CW structure on $S^3$ with the Alexander horned sphere being the boundary of the closure of a 3-cell, which is a contradiction to the other question.<|endoftext|>
TITLE: Functional/variational derivative and the Leibniz rule
QUESTION [7 upvotes]: I am currently trying to understand the BV-formalism, which makes heavy use of the functional derivative.
Let us consider the functional derivative, as defined in for example its Wikipedia article.
Let $F$ be a functional, i.e. a map from, say, $C^\infty(\mathbb{R})$ to $\mathbb{R}$, and suppose it may be written as $F[\phi] = \int f\big(x,\phi(x),\phi'(x),\dots,\phi^{(n)}(x)\big)\\,dx$ for some function $f$ which depends on the derivatives of $\phi$ up to order $n$. Then the functional derivative of $F$ is $\displaystyle \frac{\delta F}{\delta \phi} = \sum_{i=1}^n(-1)^i\frac{d^i}{dx^i}\frac{\partial f}{\partial \phi^{(i)}}$.
Now, my background is that of differential equations and differential geometry, i.e. jet spaces and variational calculus and the like. In that area, the latter operator, $\sum_{i}(-1)^i\frac{d^i}{dx^i}\frac{\partial}{\partial \phi^{(i)}}$, is well known; it is called the variational derivative. Summarizing, then, we seem to have that the functional derivative of a functional is the variational derivative of (one of its) densities.
Since the variational derivative involves lots of derivatives, it certainly does not satisfy the Leibniz rule, i.e. it is not a derivation. In various places, however, I've come across the statement that the functional derivative does satisfy the Leibniz rule. (That already seems unexpected to me: how can an operator which is so intimately connected to a decidedly non-derivation be a derivation?) There are various ways to prove it, but I would like to understand this fact in terms of the variational derivative, if possible. So: how can the Leibniz rule of the functional derivative related to variational derivative; can the former be expressed somehow in terms of the latter?

REPLY [4 votes]: Connection of functional derivative with variational derivative: $\frac{\delta}{\delta\phi(x)} F[\phi] = \frac{\delta F[\phi]}{\delta\phi}(x)$. Note that the variational derivative carries an extra coordinate variable dependence. It helps to make it explicit when there is similar confusion.
Functional derivative Leibniz rule: $\frac{\delta}{\delta\phi(x)} F[\phi] G[\phi] = \frac{\delta F[\phi]}{\delta\phi}(x) G[\phi] + F[\phi] \frac{\delta G[\phi]}{\delta\phi}(x)$.
Special case: $F_x[\phi] = \phi(x)$, $G_{i,y}[\phi] = (\partial_i\phi)(y)$, and $$\frac{\delta}{\delta\phi(z)} F_x[\phi] G_{i,y}[\phi] = \delta(x-z) (\partial_i\phi)(y) - \phi(x) \frac{d}{dz_i}\delta(y-z)$$.
Notice the distributional coefficients in the derivatives. There is no way to get away from them if you wish to consider $\phi(x)$ and such as functionals in their own right.
If you are interested in the BV formalism in the physics formalism, where the distinction between the functional and variational derivatives is barely remarked, I recommend the reviews by Henneaux and by Gomis, París and Samuel: doi:10.1016/0920-5632(90)90647-D, doi:10.1016/0370-1573(94)00112-G. If you are interested in the BV formalism purely from the point of view of jets, without bringing functionals into the picture, other than peripherally, I recommend the early paper of McCloud and this sequence of papers by Barnich, Brandt and Henneaux: arXiv:hep-th/9307022, arXiv:hep-th/9405109, arXiv:hep-th/9405194, arXiv:hep-th/0002245. If you are more interested in the BV formalism more from the functional point of view, with the appropriate level of functional analysis included, and with jets appearing only peripherally, I recommend the papers by Fredenhagen and Rejzner, as well as Rejzner's thesis: arXiv:1101.5112, arXiv:1110.5232, arXiv:1111.5130.<|endoftext|>
TITLE: Projections onto $n$-codimensional subspaces of a Banach space: norms.
QUESTION [9 upvotes]: Hello, I'd like some help to find an answer I've been looking for since this morning. 
Let $X$ be a Banach space and let $Y$ be an $n$-codimensional subspace of $X$. Let $P$ be a projection from $X$ onto $Y$. Which is the best estimate for the norm of $P$? I found this information in an article by Bohnenblust as far as $n=1$ is concerned (that is, there always exists a projection $P$ such that $\|P\|\leq 2+\varepsilon$), but nothing satisfactory when the codimension increases. 
Thank you.

REPLY [10 votes]: In many books$^*$ you can find the result that there is a projection of norm at most $\sqrt{n}$ onto any $n$ dimensional subspace of a Banach space. For reflexive spaces, this gives immediately that every $n$ codimensional subspace is the range of a projection that has norm at most $\sqrt{n} +1$. For non reflexive spaces, by using the principle of local reflexivity (which also is in many books), you get for any $\epsilon > 0$ the estimate $\sqrt{n} +1 + \epsilon $.
$$
$$
$*$ See, for example, Albiac and Kalton, ``Topics in Banach space theory", Theorem 12.1.6.  In this book you can also find the principle of local reflexivity.<|endoftext|>
TITLE: Bounding the minimal maximum norm of a solution of a linear system.
QUESTION [5 upvotes]: I would be grateful for pointing me out a reference to some general bound on the $\ell_{\infty}$ norm of a solution of a linear system. To be specific, suppose that we have an underdetermined linear system $Ax = b$. What I want to know, is some general upper bound on the minimum $\ell_{\infty}$ norm of the solution. 
In other words: I want to know that I can find a solution of the system with a norm not greater than... 
Bound should be in terms of: the number of variables, the number of equations (we can assume that this number is much smaller if it helps) and of course of $A$ and $b$.
For example, if we have $n$ variables and one equation: $a_1x_1 + a_2x_2 + \ldots + a_nx_n = b$, then the minimal norm is given by $\frac{|b|}{|a_1|+|a_2| + \ldots + |a_n|}$. I suspect that for an arbitrary number of equations, smaller than $n$, there should exist some nice estimation of the norm. I have searched in the web for such a result on my own, but have only managed to find a papers concerning the problem of finding such solution algoritmically and no explicit bounds were mentioned.

REPLY [2 votes]: The question is easier with $L^2$ norm instead of $L^\infty$ norm -- obviously the former upper bounds the latter. Now, you are trying to minimize the norm of $x$ subject to the constraints that $x \cdot v_i = b_i.$ Using Lagrange multipliers, we see that we need $x = \sum \lambda_j v_j.$ Plugging into the original equation, we get $\mathbf{\lambda} V^t V = b.$ Since $V^t V$ is symmetric, it can be (orthogonally) diagonalized) to the matrix of the squares of singular values of $V.$ Let $\sigma_0$ be the smallest such singular value. This upper bounds all $\lambda$s by $b_{\max}/\sigma_0^2,$ and then we get an upper bound on the $L^2$ norm of $x$ in the obvious way.<|endoftext|>
TITLE: Determinacy and definable ultrafilters
QUESTION [10 upvotes]: It is a simple consequence of AD that there are no non-principal ultrafilters on $\omega$: for $U$ an ultrafilter on $\omega$, consider the game $G_U$ where players I and II play natural numbers $x_0$ < $y_0$ < $x_1$ < $y_1$ < . . .
Let $$R_I=\lbrace 0, 1, 2, . . . , x_0\rbrace\cup\lbrace y_0+1, y_0+2, . . , x_1\rbrace\cup . . .$$ and $$R_{II}=\lbrace x_0+1, x_0+2, . . . , y_0\rbrace\cup\lbrace x_1+1, x_1+2, . . , y_1\rbrace\cup . . .$$ Then either $R_I$ or $R_{II}$ is in $U$. Say that I wins if $R_I\in U$, and II wins otherwise. Then if $U$ were non-principal, neither player can have a winning strategy, by a strategy-stealing argument.
Bringing this proof down into the AC world, we have that, for instance, projective determinacy implies that no non-principal ultrafilter on $\omega$ is projective. The hypothesis here can't be removed completely: assuming $V=L$, there are projective non-principal ultrafilters.
My question is whether this reverses. Does "every projective ultrafilter is principal" imply PD over ZFC? If not, then what are the consequences of this claim - and what is the consistency strength of ZFC+"every projective ultrafilter is principal?"

REPLY [10 votes]: The answer is no. The strength of the theory "ZFC+ there is no projective nonprincipal ultrafilter" is at most that of an inaccessible cardinal, which is strictly weaker than PD. 
The reason is that if there is an inaccessible cardinal, then in the Solovay model, the $L(\mathbb{R})$ of the Levy collapse $V[G]$, every set is Lebesgue measurable. It follows that every projective set in $V[G]$ is Lebesgue measurable, and consequently, there can be no nonprincipal projective ultrafilters in $V[G]$, since the existence of a nonprincipal ultrafilter implies the existence of a non-measurable projective set, namely, the ultrafilter itself, as I explain in my answer to a remark of Connes.
Since an inaccessible cardinal has weaker consistency strength than PD, which is equiconsistent with infinitely many Woodin cardinals the assertion that every real is in an inner model with an arbitrarily large finite number of Woodin cardinals (see Andres's comment), we cannot make a model of PD this way (unless our theories are inconsistent). 
Perhaps there may be a way to get rid of the inaccessible, by using the property of Baire instead of Lebesgue measurability. [Update:] And indeed, by a result of Shelah, it is equiconsistent with ZFC that every projective set has the property of Baire, and in that model, there can be no projective nonprincipal ultrafilters, since they would need to be meager and this leads to a contradiction. (See comments by Asaf and Andreas.) So the theory "ZFC + there is no projective nonprincipal ultrafilter" is equiconsistent with ZFC.<|endoftext|>
TITLE: Trace of a functor (or dimension of a category) in extended 2d TQFTs
QUESTION [9 upvotes]: In an extended 2d TQFT $Z$, a point (with orientation + or -) is assigned a category $Z(+)$ or $Z(-)$. This category should be as close to a vector space as possible: $\mathbb{C}$-linear, monoidal, etc. $Z( + \cup +)$ should be something like $Z( + ) \otimes Z( + )$, the empty set of points should get the unit category for this tensor operation, Vect$_\mathbb{C}$, and $Z(+)$ and $Z(-)$ should be dual.
If we consider a circle as broken up into two opposite U shapes, these properties tell us that $Z(S^1)$ (a monoidal $\mathbb{C}$-linear functor $Z(empty)\rightarrow Z(empty)$, ie. $V\otimes -$ for some vector space $V$) is something like the dimension of $Z(+)$ or the trace of the identity functor.
Can we make sense of this enough to compute it for some simple categories? Eg. the category of $\mathbb{C}$-representations of a finite group?
I'm sure that this wouldn't be hard to answer if I knew more about what the tensor product should be when I write $Z(+)\otimes Z(+)$. All I know about this operation is that the unit should be the category of $\mathbb{C}$-vector spaces.
How about for higher dimensional TQFTs? Does someone know a good reference?
Thanks.

REPLY [11 votes]: I won't be able to give any references, so I hope some more experts can help me out, as there is much work on traces of functors.  In general, there are two reasonable notions of "trace" of a functor, and they can be different.  Throughout, I let $F$ denote an endofunctor of a nice-enough ($\mathbb C$-linear, etc.) category $\mathcal C$, and $\DeclareMathOperator\id{id}\id$ the identity functor.
Then one notion of "trace" is:
$$ \operatorname{trace}(F) = \hom(\id,F)$$
where the $\hom$ is taken in the (monoidal) category $\operatorname{End}(\mathcal C)$ of endofunctors of $\mathcal C$, i.e. it is the space of natural transformations.  For this definition to make sense, we need only that $\mathcal C$ is small enough for $\operatorname{End}(\mathcal C)$ to be locally small (otherwise, for generic categories, the hom spaces between functors can be proper classes), or at least for $\hom(\id,F)$ to be small.  In the $\mathbb C$-linear setting, one expects that $\hom(\id,F) \in \mathrm{Vect}$, and in fact it is a $\hom(\id,\id)$-module.  Note that $\hom(\id,\id)$ is always an algebra.  In fact, since $\operatorname{End}(\mathcal C)$ is a monoidal category, $\hom(\id,\id)$ is always a commutative algebra.
For example, when choose a $\mathbb C$-algebra $A$, and let $\mathcal C$ denote the category of left $A$-modules.  Then $\hom(\id,\id)$ is the center of $A$.
There is an important generalization: work not with categories but $(\infty,1)$-categories.  Then one can set $\mathcal C$ to be an appropriate "derived" category of chain complexes of $A$-modules, and $\hom(\id,\id)$ is then the Hochschild cochain complex of $A$.
There is another important notion of "trace", which is given by an end (or is it a coend?) of the functor $\hom(-,F-)$.  This notion is slightly closer to the idea of "adding up the diagonal entries of a matrix for $F$".  In the $A$-module case, this version gives $\operatorname{trace}(\id) = A / [A,A]$, where $[A,A]$ is the subvector space of commutators (and not an ideal or anything), so that the quotient is simply a vector space (with a distinugished element, namely the image of $1\in A$).  In the derived setting, one gets the Hochschild chains of $A$.
The two constructions must give canonically-the-same answer if $\mathcal C$ is the image of an oriented but otherwise unframed 2-TQFT.  But if you work with framed TQFTs, they can give different answers.  Recall that a 2-framing of a 1-manifold $S$ is a framing of $S\times \mathbb R$, and that a framing of an $n$-manifold is a collection of $n$ vector fields which are at every point linearly independent.
The first "trace" corresponds to the circle with "outward" framing, i.e. it has a "2-framing" inherited from embedding the circle as a simple closed curve in $\mathbb R^2$.  The second trace corresponds to the "product" framing, i.e. the framed circle where one of the two vector fields is parallel to the circle and the other is orthogonal.
When thought of in this geometric picture, the "Deligne conjecture" that Hochschild chains has a homotopy-$S^1$-action becomes natural, and Hochschild cochains have their $E_2$-algebra structure coming from embedding two disks into a larger disk.
Actually, if you have a complete framed 2-dimensional TQFT which assigns $\mathcal C$ to a point, then the two notions of trace must agree for $\id$, at least in dimension.  I mean, if you look at the torus (with its unique framing), the value of the torus must be the dimension of each $\operatorname{trace}(\id)$, by cutting the framed torus into an annulus in two different ways.  A framed 2-TQFT does not pick out a chosen isomorphism between the two different traces, and I believe that any choice of such an isomorphism is pretty much enough to extend the framed TQFT to an unframed one.  Algebras with such a choice are called "Calabi–Yau", at least by some people, because the data of such an isomorphism is roughly the same (when $A$ is commutative) as a trivialization of the canonical line of $\operatorname{Spec}(A)$.<|endoftext|>
TITLE: C*-algebras with no nontrivial endomorphisms
QUESTION [12 upvotes]: Pick a C*-algebra $A$ and call a (*-)endomorphism $\alpha:A\to A$ nontrivial if it is injective and  $\alpha(A)\neq A$. 
Question: Do there exist infinite dimensional C*-algebras with no nontrivial endomorphisms? 
I'm particularly interested in the case of simple C*-algebras, but any example would do. In the commutative case the first part of this boils down to the following.
Question: Does there exist a locally compact Hausdorff space $X$ (with infinitely many points) for which every continuous surjection $X\to X$ is automatically injective?
I'm not sure such a space can exist, but it would necessarily be quite exotic: topological manifolds will not suffice (just work locally).

REPLY [3 votes]: Trying to add information to Bill Johnson's Answer, I'd just like to say that the paper he mentioned by Cook is available for free on-line.  Hopefully, below is a working link to the Cook paper in PDF format:
matwbn.icm.edu.pl/ksiazki/fm/fm60/fm60123.pdf
David Bernier<|endoftext|>
TITLE: Reference for representations of quaternion group
QUESTION [8 upvotes]: To satisfy a referee, I need a reference for the following well-known fact (which is not hard to prove, but it seems silly to prove it in a paper).  I can't find it in either Serre or Fulton-Harris's books on representation theory.  Can anyone provide me a reference?
Let $Q$ be the $8$-element quaternion group, so we have a central extension
$$1 \longrightarrow \mathbb{Z}/2 \longrightarrow Q \longrightarrow (\mathbb{Z}/2)^2 \longrightarrow 1.$$
Then the following constitute a complete list of irreducible representations of $Q$ over $\mathbb{R}$.

One-dimensional representations that factor through $(\mathbb{Z}/2)^2$.
A four-dimensional representation $W$ obtained from the left action of $Q$ on the real quaternions (viewed as a $4$-dimensional real vector space).

I remark that the representation $W$ is irreducible but not absolutely irreducible -- when we extend the field of scalars to $\mathbb{C}$, it breaks up into two copies of the (unique) two-dimensional irreducible complex representation.

REPLY [2 votes]: Here is a stupid proof which, I think, would not elicit any further questions:
Let $\left[1\right]$, $\left[-1\right]$, $\left[i\right]$, $\left[-i\right]$, $\left[j\right]$, $\left[-j\right]$, $\left[k\right]$, $\left[-k\right]$ be the elements of $Q$ which correspond to $1$, $-1$, $i$, $-i$, $j$, $-j$, $k$, $-k$ under the canonical embedding $Q\subseteq \mathbb H^{\times}$. Then, the $\mathbb R$-linear map $\mathbb H\oplus \mathbb R\oplus \mathbb R\oplus \mathbb R \oplus \mathbb R$ which sends every $\left(a+bi+cj+dk,e,f,g,h\right)$ to
$\dfrac{\left[1\right]-\left[-1\right]}{2}\left(a+b\left[i\right]+c\left[j\right]+d\left[k\right]\right) $
$+ \dfrac{\left[1\right]+\left[-1\right]}{2}\left(\left(e+f+g+h\right)\left[1\right]+\left(e-f-g+h\right)\left[i\right]+\left(e+f-g-h\right)\left[j\right]+\left(e-f+g-h\right)\left[k\right]\right)$
(for all reals $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$) is easily seen to be an $\mathbb R$-algebra isomorphism. Now, the representation theory of a direct product of algebras is well-known, and so is the representation theory of fields.<|endoftext|>
TITLE: Is the fundamental group functor a left-adjoint?
QUESTION [33 upvotes]: Theorem 1B.9 in Hatcher's Algebraic Topology says that for a (pointed) connected CW complex $X$ and group $G$, there is a bijection $\text{Hom}(\pi_1(X), G) \cong [X,K(G,1)]$, where $\pi_1(X)$ is the first fundamental group of $X$, and $K(G,1)$ is the first Eilenberg-MacLane space of $G$. I guess he is describing an adjunction of functors here, between the category of homotopy classes of maps between connected pointed CW complexes and groups.
This surprised me. If $\pi_1$ is a left-adjoint functor, then we should conclude that it is cocontinuous, i.e. takes pushouts to pushouts. But I had understood the van Kampen theorem to say something like "$\pi_1$ takes certain pushouts in $\text{hTop}_*$ to pushouts in groups". For example, van Kampen requires the morphisms to be inclusions, among other things. Presumably then not all pushouts are preserved under $\pi_1$, for example if the maps are not inclusions.
I tried to come up with a pushout of non-injective pointed topological spaces which would give a counterexample to van Kampen, but I could not. Is there one? Can you give one?
And if there is one, why doesn't that contradict the status of $\pi_1$ as a left-adjoint? And if there isn't one, then why can't the hypotheses of the van Kampen theorem be weakened?

REPLY [13 votes]: This question is related to the paper of P. Olum, Non-abelian cohomology and van Kampen's theorem.
Ann. of Math. (2) {68} (1958) 658--668. He defines nonabelian singular cohomology $H^1(X,A;G)$ of a pair of spaces with coefficients in an in general nonabelian group, and verifies that $H^1(X,x;G) \cong Hom(\pi_1(X,x),G)$ if $X$ is pathconnected and $x \in X$. 
If $A,B$ are subspaces of $X$, then under the assumption that 
$H^1(A \cup B; G) \cong H^1(S(A) \cup S(B);G) $ he obtains a Mayer-Vietoris type sequence
$$\matrix{H^0(A\cap B,x;G)& \to & H^1(A \cup B,x;G)& \to &  H^1(A,x;G) \cr
&&\downarrow&&\downarrow\cr
&&H^1(B,x;G) & \to &H^1(A \cap B,x;G) }$$ and proves exactness conditions which imply that if $A,B,A \cap B$ are pathconnected and $x \in A \cap B$ then we obtain the usual pushout diagram of the standard Seifert-van Kampen Theorem. This result is put in the context of groupoids in R. Brown, P.R. Heath, K.H. Kamps, ``Groupoids and the
Mayer-Vietoris      sequence'', J. Pure Appl. Alg. 30 (1983)
109-129.
Later: with regard to  the question of $\pi_1$ as a left adjoint, one can say that $\pi_1$  as a functor from Simplicial Sets to Groupoids is a left adjoint to the nerve functor, and some have assumed that all van Kampen type theorems are of this kind of depth, i.e. not much. 
However more work is needed to formulate and prove the case of the fundamental groupoid  with a set of base points, which require connectivity conditions on intersections of the sets of the cover; and such simple adjointness arguments have not touched the higher homotopy Seifert-van Kampen Theorems, which require more complex connectivity assumptions, and so imply of course that they solve only some problems.<|endoftext|>
TITLE: Structure of units in a maximal order
QUESTION [12 upvotes]: Hello,
my question is simple: do we have a "Dirichlet's unit theorem" for the group of units of a maximal order of a central division algebra ?
In other words: let $k$ be a number field, let $D$ be a central division $k$-algebra (i.e. a skew field with center $k$), and let $\Lambda$ be a maximal order over $\mathcal{O}_k$.
Is $\Lambda^\times$ a finitely generated group ? what is known about its group structure ?
I browsed the web and looked at Reiner's "Maximal orders" but didn't find anything.
I'm happy to assume that $D$ satisfies Eichler's condition if necessary.
In fact, my original question is even more precise: $k/\mathbb{Q}$ is quadratic imaginary, $D$ carries a unitary involution $\tau$ (which therefore restricts to complex conjugation on $k$), and I am interested in the structure of UNITARY units in a maximal order $\Lambda$.
If anyone knows any results/references, I would be happy to know them.
Thanks in advance!
Greg

REPLY [5 votes]: Hello Greg!
I'm not a specialist on the subject (and might have misunderstood something), but I did search on this issue while ago so here are my impressions.
I think that the answer to the first general question is negative (at least in strong sense).
Dirichlet's theorem describes the unit group algebraically almost completely in terms of signature.
In most cases (and in particular in the case you are interested in) the unit group of a maximal order of 
a division algebra is a very complicated object and as far as I know there is no general theorem that gives a good 
idea of the algebraic structure of this group.
An example of  troubles one encounters in division algebras: "Presentations of the unit group of an order in a
non-split quaternion algebra" Capi Corrales,a, Eric Jespers, Guilherme Leal,
and Angel del Riod, Advances in Mathematics 186 (2004) 498–524.
Probably the best overall sources on the subject are  Ernst Kleinert's book (Units in skew fields) and a survey article(Units of classical orders,
Enseigment mathematique, 1994).  One of his central themes is consideration of  what should an analogue of Dirichlet unit theorem look like in a
division algebra.  So these references are probably quite a good answer to your general question. 
While the algebraic side of Dirichlet's theorem seem to be quite hard to generalize,
there is also the geometric side, which describes how "dense" the unit group is geometrically if we consider the ring of algebraic integers as a lattice through the usual Minkowski embedding. 
In the case you are interested the unit group has a subgroup of finite index (the norm 1 group), which is a co-compact subgroup in $SL_n(C)$. 
The "density" of this norm 1 group  is decided by algebraic invariants of the division algebra. So in this sense we can generalize the geometric 
side of Dirichlet's theorem. Here the key word is point counting in Lie groups. There is a recent book on the subject:
The Ergodic Theory of Lattice Subgroups, A. Gorodnik and  A. Nevo, Princeton University Press, 2010.
If you are interested in for example of  the number of unitary units (if I understood what you are asking it is indeed a finite number) I don't think this approach helps much.
Actually even   the original Dirichlet's theorem does not directly tell much about the roots of unity part, except that it exist and is generated by a single element.<|endoftext|>
TITLE: The Hodge  numbers of a covering
QUESTION [5 upvotes]: Let $X$ be a Kahler manifold and $Z\subset X$ 
be a smooth hypersurface. How to compute the Hodge 
diamond of the double covering $Y\to X$ ramified 
over $Z$? (And what I have to know? Would the map 
$H^*(X)\to H^*(Z)$ be enough?) 
P.S. I tried the Gysin sequence, but it looks 
like there are many loose ends.

REPLY [5 votes]: This started as a competing answer, but now it is just a computation of what Donu has stated already. It might still be useful for some.
First let's introduce some notation: $\pi:Y\to X$ is the double cover and $Z'=(\pi^*Z)_{\mathrm{red}}$ is the reduced pre-image of $Z$.
1
In the situation of the question we have that
$$
\pi_*\Omega_Y^p(\log Z')\simeq \Omega_X^p(\log Z) \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \tag{$\star$}
$$ 
where $\mathscr L$ is (as Donu already said) the anti-invariant part of the direct image of $\mathscr O_Y$ to $X$ under the
natural $\mathbb{Z}/2$ action.
Since $\pi$ is finite, all higher direct images vanish and hence we have a similar isomorphism for cohomology:
$$
H^q(Y,\Omega_Y^p(\log Z'))\simeq H^q(X, \pi_*\Omega_Y^p(\log Z'))\simeq 
H^q(X, \Omega_X^p(\log Z) )\oplus H^q(X, \Omega_X^p(\log Z) \otimes \mathscr L^{-1})
$$ 
by (3.22) of Esnault-Viehweg, Lectures on Vanishing Theorems.
2
If one is interested in Hodge numbers of the open manifolds $X\setminus Z$ and $Y\setminus Z'$, then this should be good. Otherwise we need to connect these to the non-logarithmic sheaves. For that probably the best tool is the following short exact sequence:
$$
0 \to \Omega_X^p \to \Omega_X^p(\log Z)\to \Omega_Z^{p-1} \to 0.
$$
(The existence of this short exact sequence is a simple exercise, or can be found in (2.3) of ibid.
There is of course an equivalent one on $Y$ with $Z'$:
$$
0 \to \Omega_Y^p \to \Omega_Y^p(\log Z')\to \Omega_{Z'}^{p-1} \to 0.
$$
Aha!
Until this point I thought that I was going to get a different answer than Donu and that was the main reason I even started writing, but now it seems that I might get from this what Donu stated.
The point is, $\pi$ induces an isomorphism $Z'\to Z$ and hence the right hand side of the two short exact sequences are the same. So if we add $\Omega_X^p(\log Z) \otimes \mathscr L^{-1}$ to the first short exact sequence and push-forward the second short exact sequence, then we get
$$
0 \to \Omega_X^p \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \to \Omega_X^p(\log Z)\oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big)\to \Omega_{Z}^{p-1} \to 0.
$$
and 
$$
0 \to \pi_*\Omega_Y^p \to \pi_*\Omega_Y^p(\log Z')\to \pi_*\Omega_{Z'}^{p-1} \to 0.
$$
Now, since $\pi|_{Z'}:Z'\to Z$ is an isomorphism and by $(\star)$ we get that these two short exact sequences are the same, so we have 
$$
\pi_*\Omega_Y^p\simeq \Omega_X^p \oplus \big( \Omega_X^p(\log Z) \otimes \mathscr L^{-1} \big) \tag{$\star$}
$$ 
and we get 
$$
H^q(Y,\Omega_Y^p)\simeq H^q(X, \pi_*\Omega_Y^p)\simeq 
H^q(X, \Omega_X^p )\oplus H^q(X, \Omega_X^p(\log Z) \otimes \mathscr L^{-1})
$$
as stated by Donu.<|endoftext|>
TITLE: What's the idea behind Carleman estimates?
QUESTION [18 upvotes]: A standard Carleman-type estimate is of the form
$$
\sum_{|\alpha|<m}{\tau^{2(m-|\alpha|-1)}\int{|D^{\alpha}u|^{2}e^{2\tau\phi}}dx}\leq K\int{|Pu|^{2}e^{2\tau\phi}dx},\quad u\in C_{0}^{\infty},
$$
where $\phi$ is some weight function. This formula turns to be very useful in the study of uniqueness of Cauchy problems, and many mathematicians have considered this (such as Calderon, Hormander, Kenig, Sogge, and Tataru...)
For a first look at this inequality, I'm wondering whether the weight fuction has an essential role, and furthermore, what's the original idea of it? Are there some very simple but illuminating examples that show the reasonability of the Carleman estimates?
One example  in my mind is the first order operator $P=D+ix$, where $D = \frac{1}{i} \frac{d}{dx}$.
It's easy to see that $P^*=D-ix$, and 
$$
P^*P-I=PP^*+I=-\frac{d^2}{dx^2}+x^2,
$$
which is the so-called harmonic oscillator. Here we have 
$$
2\|u\|_{L^2}\leq \|Pu\|_{L^2},\quad u\in C_{0}^{\infty}.
$$
But in this simple example, there is no need to use a weight function.
From the proof I guess the decomposition $P=\frac{P+P^*}{2}+\frac{P-P^*}{2}$
may be one of the general ideas.

REPLY [21 votes]: The weight function $\phi$ indeed plays an (I would even say the) essential role in Carleman  estimates, and constructing one such that the estimate holds is one of the major challenges in proving and applying Carleman estimates. I will give a very informal overview, and refer to the references below for the technicalities.
The original motivation for introducing Carleman estimates was to prove unique continuation theorems, which can informally be stated as follows: Given a partial differential operator $P$, an oriented hypersurface $\Sigma$, and a function $u$ satisfying $Pu=0$, then if $u=0$ (locally) on one side $\Sigma^+$ of $\Sigma$, it vanishes on the other side $\Sigma^-$ as well. This can be interpreted in the sense that the complete information about $u$ in $\Sigma^-$ can be retrieved from information in $\Sigma^+$, i.e., the complete information flows across $\Sigma$. For linear hyperbolic equations, such information flow is described by the characteristics, and the idea behind Carleman estimates is to extend this concept to general (pseudo-)differential operators via microlocal analysis. This leads to considering so-called bicharacteristic rays, which need to satisfy the geometric property that all such rays passing near $\Sigma$ in $\Sigma^+$ must cross into $\Sigma^-$. Furthermore, solutions to $Pu=0$ should not decay exponentially along such rays towards $\Sigma$ (otherwise the information gets lost before it crosses $\Sigma$). A surface satisfying these conditions is called strongly pseudo-convex (with respect to $P$). 
To make these conditions precise, one considers $\Sigma$ as the level set $\{\phi(x)=0\}$ (and $\Sigma^+$ as $\{\phi(x)>0\}$). The pseudo-convexity conditions above then become positivity conditions for the derivative of $\phi$ along the bicharacteristic flow (expressed using Poisson brackets). The connection is now that if $\Sigma=\{\phi(x)=0\}$ is a strongly pseudo-convex surface, then $e^{\lambda\phi}$ is strongly pseudo-convex function (in the sense of these positivity conditions) for sufficiently large $\lambda$. The parameter $\lambda$ is tied to the decay rate along the bicharacteristic rays, and hence the Carleman weight can be seen as accounting for this decay: The weight vanishes as the information flow across $\Sigma$ decays. The weight thus localizes the estimate to that part of $\Sigma^-$ where you still have usable information from $\Sigma^+$.
For example, for elliptic operators of second order, every smooth surface is strongly pseudo-convex (the decay condition is always satisfied); similarly, for hyperbolic operators with constant (real) coefficients, every non-characteristic surface is strongly pseudo-convex. In particular, for the wave operator $Pu=u_{{{tt}}} - c^2\Delta u$, any convex surface is strongly pseudo-convex, as well as the zero level sets for $\phi = x^2 - \beta t^2$ for $\beta < c^2$ (this bound is sharp).
[1] A good overview of unique continuation using Carleman estimates (along the lines I've given) is Daniel Tataru, Unique Continuation Problems for Partial Differential Equations, The IMA Volumes in Mathematics and its Applications Volume 137, 2004, pp 239-255. 
[2] Carleman estimates are also useful for inverse problems for partial differential equations; this side is discussed in Victor Isakov, Inverse Problems for Partial Differential Equations, 2nd ed., 2006, Springer. (The example for the wave equation is Theorem 3.4.1.)
[3] David Dos Santos Ferreira has written a habilitation thesis on Carleman estimates.<|endoftext|>
TITLE: Locally Compact Quantum Groups application
QUESTION [12 upvotes]: Recently Mathematicians in harmonic analysis become more and more interested in Locally compact quantum groups and try to transfer concepts from abstract harmonic analysis to the setting of locally compact quantum groups.
I know that the first goal of defining the category of LCQGs was extending the Pontryagin duality for locally compact abelian groups. In fact this leads me to study LCQGs and do research in this area.
But after a while I wonder whether there exists any application for locally compact quantum groups in the real world for example physics.

REPLY [6 votes]: Locally compact quantum groups can be used to construct quantum channels, see the recent preprint http://arxiv.org/abs/1210.2738 by Jason Crann and Matthias Neufang.<|endoftext|>
TITLE: Second homotopy groups of 3-complexes and Fenn's spiders.
QUESTION [5 upvotes]: Let $X$ be a finite CW complex then with one zero cell. Then (up to homotopy) the two skeleton of X is the same as a group presentation, via the Cayley complex construction. For a while I had been searching for some planar description of the second homotopy group, which would allow a concrete combinatorial description of the fundamental 2-groupoid of X (up to equivalence).
I found many discussions close to what I needed, before stumbling on the (IMHO beautiful) book "Techniques of geometric topology" by Roger Fenn. In Chapter 2 he gives a description of $\pi_2(X)$ of a 3-complex in terms of certain diagrams modulo local relations. Each relation in the 2-complex gives a "relation spider" and the second homotopy group of $X$ is the group of isotopy classes of planar diagrams generated by these spider diagrams modulo certain "universal" local relations (analogous to $gg^{-1} = 1$ in the $\pi_1$ case) and relations given by the 3-cells of $X$. (The spider diagrams are roughly dual to Van Kampen diagrams.)
My questions are:
1) are spider diagrams Fenn's invention? Perhaps this way of thinking about $\pi_2$ was folklore?
2) what are other sources describing $\pi_2$ (or even better the fundamental 2-groupoid) concretely (ideally diagrammatically) for small dimensional complexes?
I am aware that all of this can be viewed as a concrete example (for $n = 2$) of the dictionary between n-groupoids and n-types. However because of the applications I have in mind I am only looking for "concrete" sources!

REPLY [3 votes]: The web site 
Homological algebra programming  by Graham Ellis gives methods of constructing resolution of groups; the basic idea is to construct inductively a universal cover of a $K(G,1)$ together with a contracting homotopy, each inductive step gives another "home" for a contracting homotopy. This method is a higher dimensional version of constructing a tree in a Cayley graph, and is more computational than the traditional "killing kernels".<|endoftext|>
TITLE: State of the art on a question on the existence of dualizing complex
QUESTION [8 upvotes]: Let A be a noetherian ring and D(A) be the derived category of modules on A.
Recall that a dualizing complex for A is an object R in D(A)  of finite injective dimension, with cohomology of finite type and such that the natural morphism of functors
$ Id \longrightarrow R\mathcal{H}om(R\mathcal{H}om(., R), R )$
is an isomorphism of functors.
In the book "Residues and Duality" (R. Hartshorne) (V.10), it is presented as an open 
problem to know if a noetherian local domain of dimension 1 admits a dualizing complex.
What is the state of the art on this question ?

REPLY [15 votes]: Actually, more has been determined since Sharp's work in the 1980s.  It is easy to see that a homomorphic image of a Gorenstein ring of finite Krull dimension has a dualizing complex.  Sharp conjectured that this is the only way for a Noetherian ring to have a dualizing complex.  In a pair of papers (2000 and 2002) in Transactions of the AMS, Takesi Kawasaki showed that Sharp's conjecture is true.  That is, a Noetherian ring has a dualizing complex if and only if (1) it has finite Krull dimension and (2) it is the homomorphic image of a Gorenstein ring.<|endoftext|>
TITLE: Embedding of a scheme into a regular scheme
QUESTION [6 upvotes]: Hello,
Is it true that if I have a scheme $X$ which is, say, Noetherian, of finite Krull dimension, and semi-separated (intersection of two open affines is again open affine), then I can find a locally closed embedding of it into a scheme of the same type, which is, in addition, regular?
I ask just of curiosity, or minor desire not to say "quasi-projective" when I want to have enough locally free objects.
Thank you,
Sasha

REPLY [11 votes]: A noetherian regular scheme is universally catenary (Matsumura, 14.B, 16.D), so any subscheme of a regular scheme is universally catenary. 
But there are affine noetherian schemes (integral of dimension $2$) which are not universally catenary. See an example of Nagata in Matsumura, 14.E, or a slightly simpler one in EGA IV.5.6.11 (it consists in identifying two points of respective codimension 1 and 2 in an affine regular scheme of dimension 2).<|endoftext|>
TITLE: Prove that a particular polynomial sequence is a Sturm sequence
QUESTION [7 upvotes]: Let us define the following sequence of polynomials for every two non-negative integers $i,d$: $$s_i^d(w)=\sum_{j=0}^{d+1} (-1)^j {d+1\choose j} (j+1)^i w^{d+1-j}.$$
Conjecture: The sequence $\{s_d^d(w),s_d^{d-1}(w),\dots,s_d^0(w)\}$ is a Sturm sequence.
An easy corollary of this conjecture is that:
Corollary: For $i\geq d$, the roots of the polynomial $s_i^d(w)$ are all real, simple (and positive).
Any idea how to prove either the Conjecture or (directly) the Corollary? Or, in the worst case, the Corollary for the special case $i=d$?
Note that:

$s_0^d(w)=(w-1)^{d+1}$
$s_i^0(w)=w-2^i$
${d\over dw} s_i^d(w)=(d+1)s_i^{d-1}(w)$.

REPLY [7 votes]: This is just a partial answer, but anyway:
( I use $s[n,k]$ for $s^n_k$ since I just copy-pasted from Mathematica:
I do not know if this helps, or if you already know, but you have the recurrence
$$s[n- 1,k + 1] =  s'[n, k] - x \cdot s'[n - 1, k]$$
Note, by induction, we assume $s[n,k]$ and $s[n-1,k]$ to have interlacing roots.
By thm 1.47 in http://arxiv.org/abs/math/0612833 (look into this work),
we know that the derivatives have interlacing roots as well.
Now, the recursion is quite similar to many other that appear in the link above,
so it might not be too hard to prove stability.
EDIT: 
I show/sketch below that the operator $f \mapsto A f - x f'$
always produce roots interlaced (and to the right) of the roots of $f$,
provided $A>\deg f$ and that all roots of $f$ are positive. 
(We may assume leading term of $f$ has positive coefficient).
Clearly if f has a root of multiplicity m, then the result will have the same root with multiplicity m-1, so the problem is essentially to ensure that no new multiple roots may appear.
Assume now we have two consecutive roots $0\le a \le b$ of $f$ and that $f$ is positive between these.
The derivative of $f$ this first positive, and then negative in $[a,b]$, 
thus $-x f'$ is first negative and then positive.
Hence, $A f - x f'$ is negative in $a$, and positive in $b.$ 
Thus, we have a root of $A f - x f'$ in the interval $[a,b]$.
(Mutatis mutandis for the case when f is negative between $a$ and $b$).
Now, if the degree of $f$ is even, $f$ is positive to the right of its largest root.
The derivative is also positive here (since even degree poly), so $x f'$ is positive.
Hence, $A f - x f'$ is negative in this point. However, notice this is a polynomial with even degree, and with a leading coefficient! Hence, it must eventually cross the real line and grow to infinity.
(Mutatis mutandis for the case when f is odd).
This proves that $f \mapsto A f - x f'$
produces interlacing roots, and eventual multiplicities are decreased, no new multiple roots can be introduced.<|endoftext|>
TITLE: Derivatives through random variables?
QUESTION [5 upvotes]: Suppose I have some random variable X with probability distribution P(.;theta). Suppose I have a single sample x from this distribution.
Does it make any sense to ask for the derivative of x with respect to theta?
I believe it does not, but am having trouble convincing some of my colleagues.

REPLY [5 votes]: What might make sense is if $X$ is a function $g(U,\theta)$ of some underlying random variable $U$ (with distribution not depending on $\theta$) and $\theta$, where the differentiable function $g$ is chosen so that $g(U,\theta)$ has the given (continuous) distribution.  Then you could say $\dfrac{dX}{d\theta} = \dfrac{\partial g}{\partial \theta}(U,\theta)$.  Of course this depends on the "implementation", i.e. the choice of $g$ and $U$, rather than just on the distribution of $X$.<|endoftext|>
TITLE: Categorical properties of metric Boolean algebras
QUESTION [5 upvotes]: According to Kolmogorov ("Algèbres de Boole métriques complètes", VI Zjazd Matematykòw Polskich, 1948, english translation Phil. Studies, 1995, 77, 57-66), a Boolean algebra $(B, \wedge, \vee,-,1,0)$ is metric if it is given together with a map $ B\xrightarrow\mu R $ to the reals such that: 
1) $[x \wedge y=0] \Rightarrow [\mu(x \vee y)=\mu (x) + \mu (y)]$
2) $ [x \neq 0] \Rightarrow [\mu (x)>0] $.
In facts, if $B \times B \xrightarrow+ B$ denotes symmetric difference, the composite $ B \times B \xrightarrow+ B \xrightarrow\mu R$ is a metric.
Denoting: Bool the category of Boolean algebras, Met the category of metric spaces with non-expansive maps and MBool the category of metric Boolean algebras with non-expansive Boolean homomorphisms: 

what are the main differences between the categories MBool and Bool ?
the two forgetful functors MBool**$\rightarrow$ *Bool and MBool*$\rightarrow$ **Met have left adjoints ?

REPLY [2 votes]: I have a feeling that uncountable products in $MBool$ do not exist in general. One thing is for sure: if $MBool$ has uncountable products, then the forgetful functor $MBool \to Bool$ cannot preserve them (whence the forgetful functor couldn't have a left adjoint.) 
Indeed, if $2$ is the 2-element Boolean algebra with its unique measure $\mu$. If the uncountable product $(2, \mu)^{\omega_1}$ existed in $MBool$, where $\omega_1$ is the first uncountable ordinal, and if this product is calculated as it would be in $Bool$ (which is the set-theoretic infinite product), then we could construct an uncountable strictly decreasing chain of elements (uncountable tuples) 
$$(1, 1, 1, \ldots) > (0, 1, 1, \ldots) > (0, 0, 1, \ldots) > \ldots$$ 
and there would have to be a corresponding uncountable strictly decreasing chain of positive real numbers 
$$\mu(1, 1, 1, \ldots) > \mu(0, 1, 1, \ldots) > \mu(0, 0, 1, \ldots) > \ldots$$ 
by the axioms on metric Boolean algebras. This is impossible by cofinality considerations. 
Regardless of whether there are infinite products in $MBool$, there is a trivial reason why the forgetful functor $U: MBool \to Bool$ cannot have a left adjoint $F$ under the axioms given for metric Boolean algebras. Suppose WLOG that $\mu(1) = 1$ where the $1$ on the left is the top element of the putative free metric Boolean algebra $FB$. Then define a different metric $\mu'$ on $UFB$ by $\mu'(b) = r\mu(b)$ where $r > 1$, giving a different metric Boolean algebra $B'$. Then there is no nonexpansive metric Boolean algebra map $FB \to B'$ which extends the Boolean algebra embedding $i: B \to UB'$ along $i: B \to UFB$. 
One might think this trivial objection could be remedied by adding an extra axiom like $\mu(1) = 1$ (so we are working with probability measures in a sense), but it seems highly doubtful even in that case that a left adjoint out of $Bool$ would exist. As noted above, it could only exist if $MBool$ lacked many limits like uncountable products, and this would rule out a straightforward application of an adjoint functor theorem. 
I am not sure about a left adjoint out of $Met$.<|endoftext|>
TITLE: Publication and Career as a fresh Ph.D
QUESTION [23 upvotes]: This may be a little off topic.
But as new Ph.D in geometry/topology area, I have a feeling that it is relatively harder to publish a descent paper. However after seeing some peers who study PDE or Geometric analysis have a lot of paper, like 5-6 papers when graduate and after 3 years posdoc, they will had around 12-15papers, and also have  relatively higher citation on mathscinet, I am a little depressed. Even some professors in some good university (rank 50+) can have such publication when admitted as an AP in the geometry/topology area. 
So should I publish some 'small' results to catch up peers during the beginning of my career or bet on Bigger results, which may not succeed within 1 or 2 years. (sometimes I found the knowledge got from graduate school is far away from enough, or a new tool need to be studied in order to understand something new, this study could take times).

REPLY [5 votes]: The question you ask, I think is fair and good, and I wish you to find the right answer. Probably the question is in the mind of many young researchers, although the answer should be individual, I think it is relevant for MO. But actually  I am not in good position to give advises on your main question, 
so let me comment on the following " ...bet on Bigger results, which may not succeed..."
I do not think it is good to "bet on Big results", I mean you may try to invest your time to studying some big conjecture or modern popular technique, but when one is young  it is difficult (imho) to estimate properly ratio  (efforts spent/result obtained).
The way which I choose for myself (actually much later when I got PhD) is the following:
I start working on the subject when:
a)  I am 100% sure that I will get "some" (may be "bad") result, which can be obtained in reasonable amount of time - say 2-3 months  
b)  I hope for some "big result" if I would be lucky/clever/whatever. I mean I see some "big problem", such that 100% guaranteed result is some "partial case" of it.
To the best of knowledge similar way is used by many people.<|endoftext|>
TITLE: what is the probability that a scissor became the champion?
QUESTION [20 upvotes]: Here is a question from one of my students:
suppose 8 players are in an elimination match. The players are marked with marked with either R (for rock), P (for paper) or S (for scissors). If two players are marked with the same letter, then one is picked as winner of this round. An example of this sort of matches is:
First round：R-P S-S R-S P-S
Second round： P-S R-S
Third round： S-R
Champion：    R
The question is：Given r+p+s=8, for the match that there are r many R, p many P and s many S, if the table for the elimination match is assigned randomly, what is the probability that a S became the champion?
A more general question is: for a match with 2^n players, and given r+p+s=2^n, how much chance  can some S wins the champion?
My answer did not meet his satisfaction. What I know is, firstly for small n, one can list all  possibilities (there are (2^n)! many, if we make a distinction between two rocks etc.)and find the answer; secondly, one may use a computer program to solve the general question by inputting r,p and s; thirdly, one can also do some induction to find a pattern for the general question; and lastly, if the number r,p and s are assigned randomly and we let n goes to infinity, and the answer should be a third.
So can we have some smart ways to figure out these questions? Or can we reduce them to other known questions? I guess graph theory may help but my knowledge is very limited there.

REPLY [16 votes]: Here is a generating function approach. A $P$-tree $T$ of length $n$
is a complete binary tree of length $n$ with vertices labelled $p,r,s$
such that the root is $p$, the children of $p$ are either $p,p$ or
$p,r$, the children of $r$ are either $r,r$ or $r,s$, and the children
of $s$ are either $s,s$ or $s,p$. Similarly define $R$-trees and
$S$-trees. The weight $w(T)$ is the product of all leaf
vertices. Let $P(n)=\sum_T w(T)$, summed over all $P$-trees of length
$n$, and similarly define $R(n)$ and $S(n)$. Thus
  $$ P(0)=p,\ \ \ P(1)=p^2+2pr, $$
  $$ P(2)=p^4+4p^3r+6p^2r^2+4sp^2r+4pr^3+8pr^2s. $$
Clearly
   $$ P(n+1) = P(n)^2+2P(n)R(n), $$
and similarly for $R(n+1)$ and $S(n+1)$. We can eliminate two of the
functions from these three recurrences and obtain a recurrence for
each function alone. For $P(n)$ the recurrence is the following,
writing e.g. $p3$ for $P(n+3)$:
$$ 0 = 4 p2^4 p0^4+p2^2 p1^6-p2 p1^8-8 p1^5 p0^2 p3+16 p1^4 p0^4 p3 $$
$$ -8 p1^3 p0^6 p3+4 p1^2 p0^4 p2^3-16 p1 p0^6 p2^3+12 p1^7 p0^2 p2 $$
$$ -18 p1^6 p0^4 p2+12 p1^5 p0^6 p2-4 p2^2 p1^5 p0^2+12 p0^6 p2^2 p1^3$$
$$ -6 p2^2 p1^4 p0^4+9 p1^2 p0^8 p2^2-9 p2 p1^4 p0^8. $$
Is there a simpler recurrence? The highest term is $p3=P(n+3)$, so the
initial conditions are $P(0),P(1),P(2)$. The coefficient of $p3$ is
 $$ -8 p1^5 p0^2+16 p1^4 p0^4-8 p1^3 p0^6. $$
Thus there is a "Laurent phenomenon" behavior, since it is not at all a
priori clear why we get an answer that is a polynomial (with
nonnegative coefficients). For more on the Laurent phenomenon, see
e.g. http://arxiv.org/abs/math/0104241.<|endoftext|>
TITLE: Algorithm to test for discrete or quasi-Fuchsian subgroups of PSL(2,C)
QUESTION [14 upvotes]: Let $\Gamma = \pi_1(S)$ denote the fundamental group of a compact surface $S$ of genus $g>1$.
Given a representation $\rho : \Gamma \to \mathrm{PSL}(2,\mathbb{C})$, specified by matrix representatives for the images of a fixed generating set, is there an algorithm to answer either of the following questions?
1) Is the image $\rho(\Gamma)$ a quasi-Fuchsian group?
2) Is the image $\rho(\Gamma)$ discrete?
There are a number of related situations where I am aware of algorithms of this type, but   all are limited to special classes of two-generator subgroups of $\mathrm{PSL}(2,\mathbb{C})$.   For example:

For two-generator subgroups of $\mathrm{PSL}(2,\mathbb{R})$ there is the Gilman-Maskit algorithm which tests for discreteness.  There are some related sufficient conditions for discreteness in the two-generator case in $\mathrm{PSL}(2,\mathbb{C})$.
For punctured torus groups (i.e. representations of $\mathbb{F}_2$ where $abab^{-1}$ maps to a parabolic element) in $\mathrm{PSL}(2,\mathbb{C})$ there is a method of Komori, Sugawa, Wada, and Yamashita based on simultaneously testing Jorgensen's inequality (attempting to find a certificate that the group is not discrete) while also trying to find a Ford fundamental domain (of a type that would give a certificate that the group is quasi-Fuchsian).
Also in the punctured torus case, Bowditch has a conjectural characterization of quasi-Fuchsian groups in terms of a certain subset of the infinite trivalent tree of "generating triples", which is easy to test algorithmically.  As in the previous case this can be combined with Jorgensen's inequality to get a heuristic test for discreteness.

Based on these cases I would especially like to know about methods for discreteness or quasi-Fuchsian testing that apply in the compact surface case without assuming that the representation maps into $\mathrm{PSL}(2,\mathbb{R})$.

REPLY [7 votes]: Indeed, the answer to the discreteness problem turns out to depend on the computability model. In the Real-RAM or BSS (Blum-Shub-Smale) model of computations over the reals, the answer is negative: The discreteness problem (for finitely generated nonelementary subgroups of $PSL(2,C)$) is undecidable, see here, as in the case of the Mandelbrot set $M$ (due to Blum, Shub and Smale). On the other hand, in the bit-computability model, the discreteness problem for finitely generated nonelementary subgroups of $PSL(2,C)$, turns out to be decidable. Interestingly, computability of the Mandelbrot set $M$ in the bit-computability model is still an open problem. However, $M$ is bit-computable, conditional on some major conjectures in holomorphic dynamics (say, local connectivity of $M$). This conditional result is due to Peter Hertling.<|endoftext|>
TITLE: Tarski monster groups: for which primes they don't exist?
QUESTION [14 upvotes]: What can be said about the set of primes $p$ for which it is proven that an infinite group with all non-trivial proper subgroups cyclic of order $p$ doesn't exist? Specifically, what is the largest such $p$ (say $p_0$)? All I could find in the literature is $p_0\le 10^{75}$, but I admit I didn't look very far...

REPLY [8 votes]: The largest known prime for which existence of Tarski monster is not known is $997$, see Adyan, S. I.; Lysënok, I. G. Groups, all of whose proper subgroups are finite cyclic.  Izv. Akad. Nauk SSSR Ser. Mat. 55 (1991), no. 5, 933--990; translation in Math. USSR-Izv. 39 (1992), no. 2, 905–957. There are currently no methods of proving that for a given prime $p$ a Tarski monster does not exist except for showing that all finitely generated groups of exponent $p$ are finite. This gives hope that the bound $997$ can be lowered to $665$ (or below $300$ assuming the recent Adian's announcement is correct).<|endoftext|>
TITLE: Is a simply connected Ricci-flat Kaehler manifold a Calabi-Yau manifold? 
QUESTION [6 upvotes]: Hi,
I have the following question: Let $(M,\omega, J)$ be a simply connected Kaehler manifold with Ricci-flat Kaehler metric. How can one show that $M$ is a Calabi-Yau manifold. By Calabi-Yau manifold I mean that there exists a holomorphic $(n,0)-$form $\Omega$ such that the following equation is satisfied: $\frac{\omega^{n}}{n!} = (-1)^{\frac{n(n-1)}{2}}(\frac{i}{2})^{n} \Omega \wedge \bar{\Omega}$. Should one put the assumption on $M$ to be compact? But what kind of compactness? With or without boundary? Is this necessary? Does this also work without any compactness assumption? Where can I find a proof of this? Is there any reference?
Or is this question too trivial? I hope that someone has the answer and also hope for a lot of replys. Thanks in advance.
Miguel B.

REPLY [4 votes]: José is correct, with the caveat that Gunnar mentioned - you need simple-connectedness to know that reduced holonomy = holonomy. Below I expand a bit more on the details. [Thanks to Tim Perutz for catching errors in the initial version of this answer.]
Notice that the OP did not ask for $\Omega$ to be parallel or even closed. The following is true: If $(M, J, g, \omega)$ is Ricci-flat Kaehler, then the image of the first Chern class $c_1 (M)$in $H^2 (M, \mathbb R)$ vanishes, so that if $\pi_1(M) = 0$, then $H^2(M, \mathbb Z)$ has no torsion, and thus the canonical bundle $\Lambda^{n, 0} (M)$ is topologically trivial. So there exists a nowhere vanishing smooth $(n,0)$-form $\Omega$ that trivializes the canonical bundle. By consideration of type, $\Omega \wedge \overline \Omega$ is a nonvanishing $(n,n)$-form, so by rescaling $\Omega$ by a nowhere vanishing complex valued function, one gets for "free" the identity that
$$ \frac{\omega^n}{n!} = (-1)^{\frac{n(n-1)}{2}} \Omega \wedge \overline \Omega.$$
Since $\Omega$ is type $(n,0)$ and the complex structure is integrable, then $\Omega$ will be  holomorphic (and thus the canonical bundle is holomorphically trivial) if and only if it is closed. Since $M$ is Ricci-flat, the Bochner theorem tells you that an $(n,0)$ form is closed if and only if it is parallel, which would give you holonomy contained in $SU(n)$.
Compactness is needed to go the other way: Yau's theorem says that if $M$ is compact Kaehler and $c_1 (M) = 0$, then there exists a unique Ricci flat Kaehler metric in each Kaehler class. There are noncompact examples where uniqueness fails. I don't know as much as I should about the literature on existence in the noncompact case, but the papers of Tian-Yau should have the answer.
A good elementary reference is Chapter 6 of Compact Manifolds with Special Holonomy by Dominic Joyce.<|endoftext|>
TITLE: Concentration results for inner products of two independent random gaussian vectors
QUESTION [7 upvotes]: Hi,
I wanted to know if there are standard results on concentration of absolute
value of inner products of two random vectors. Thus if $X, Y \in R^m$ are two
independent random vectors with each entry distributed as $\mathcal{N}(0, 1/m)$, then
how can we bound the following probability expression: $P ( | X^T Y |  > \epsilon )$ ? 
Here, $\epsilon > 0$ is a given constant that is small.

REPLY [9 votes]: An alternative method is to exploit the rotational invariance of the Gaussian.  You can write 
$$X^T Y = |X| \left( \left(\frac{X}{|X|}\right)^T Y \right).$$
Because $Y$ is rotationally invariant, the inner product is now independent of $X$, and in fact just has distribution $N(0,1/m)$.  Now let $C>1$ be an arbitrary parameter.  We can bound the probability $X^T Y > \epsilon$ by the probability one of the following two events occur.  

$ \left(\frac{X}{|X|}\right)^T Y \geq \frac{\epsilon}{C}$.  Assuming $ \epsilon \sqrt{m}/C$ tends to infinity, this  occurs with probability $\Phi (\frac{\epsilon \sqrt{m}}{C})=(1+o(1)) \sqrt{\frac{m}{2 \pi}} \exp(-\frac{\epsilon^2 m}{C^2})$.
$|X| \geq C$.  The norm of a Gaussian vector is well studied, and it is standard (see, for example Chapter 2 of these notes, that $|X|$ is tightly concentrated around its expectation.  For example, applying Corollary 2.3 of the linked notes gives that the probability this occurs is at most $\exp(-\frac{1}{4} (1-\frac{1}{C^2})^2 m)$ 

For $\epsilon$ bounded away from $0$ you can choose $C$ to optimize the sum of the two terms getting a bound that is exponential in $m$ but with a non-optimal exponent.  If $\epsilon$ is tending to $0$ with $m$, then the first term is dominant.  That term remains small so long as $\epsilon$ is much larger than $\sqrt{\frac{\log m}{m}}$.<|endoftext|>
TITLE: Examples of 3-manifolds with RFRS fundamental group
QUESTION [10 upvotes]: I'm wondering if anyone knows how to construct hyperbolic 3-manifolds whose fundamental group is RFRS. Clearly the recent work of Agol, Wise, etc. says that such manifolds are abundant, and in particular present in every commensurability class. But how do you construct examples?
The only examples of RFRS manifolds that I'm aware of are torus knot complements (thanks to Stefan Friedl for pointing this out), but these are of course non-hyperbolic.

REPLY [11 votes]: Agol's original paper on RFRS gives a nice short proof that the fundamental group of any manifold which also happens to be a finite-index subgroup of your favourite right-angled reflection group is RFRS.  (Sketch proof: watch how loops bounce off the mirrors!)
So take your favourite right-angled reflection group $\Gamma$ in $\mathbb{H}^3$ and take the commutator subgroup $[\Gamma,\Gamma]$ (which is always torsion-free, as all the torsion injects into $H_1$).  This will give you an example.<|endoftext|>
TITLE: Level lowering for weight 1 forms
QUESTION [5 upvotes]: I'm interested in knowing to what extent is level lowering known to hold in weight 1.  Specifically, let's say I have an eigenform $f$ in $S_1(N,\chi)$ and a prime $p$ which doesn't divide $N$.  Let's assume that the mod $p$ Galois representation attached to $f$ is unramified at some prime $q$ dividing $N$.  Is there then an eigenform $g$ in $S_1(N/q,\chi)$ with the same mod $p$ Galois representation as $f$?
I think I can show this if $p>5$, but the argument relies on Artin's conjecture and feels like overkill.  Let me sketch the argument, and then ask some more specific questions at the end.
Here's the argument: if $f$ is CM, check directly that no such $q$ can exist.  So the projective image of $\overline{\rho}_f$ must be $A_4$, $S_4$, or $A_5$.  In particular, if $p>5$ then one can lift $\overline{\rho}_f$ to a $p$-adic representation with the same image.  In particular, the ramification properties are unchanged by the lift.  Take this $p$-adic representation and view it as a complex representation (i.e. an Artin representation).  By Artin's conjecture (which I guess is now a theorem in this 2-dimensional case by Serre's conjecture), there is a weight 1 eigenform $g$ giving rise to this representation.  But since the representation is unramified at $q$, we can take $g$ to have level $N/q$, and we're done.
So questions:
0) Does this argument look okay?
1) Is there a more direct argument that doesn't rely on Artin's conjecture to achieve this.  (I'm ultimately interested in the Hilbert modular case, and so I don't want to be using relying on Artin's conjecture).
2) Is there a way to handle $p=2,3,5$ even assuming Artin?  Is this even true with $p=2$??

REPLY [2 votes]: Everything you wish for is true for modular forms over $\mathbb Q$ even at $p=2$, as it follows from refined forms of Serre's conjecture; here I am assuming of course that $\bar{\rho}$ is absolutely irreducible (I think that you meant to include this explicitly in your set-up, of course otherwise level-lowering can fail). In particular your argument is correct. I don't think there is a significantly more direct general argument incorporating $p=2$ as it is my understanding that it is only with Khare-Wintenberger's proof that the last cases of "Weak Serre implies Refined Serre" were proved. 
Over totally real field, this is the whole business of Serre's weight; a topic which has seen exponential development these last 10 years. I am vey far from the most superficial understanding of the current literature but I guess Buzzard-Diamond-Jarvis is a good place to start. Other key players are Toby Gee, Matt Emerton and Florian Herzig. I don't think much is actually known about level-lowering in weight 1, though I could very well be wrong.<|endoftext|>
TITLE: Discrete Morse function from smooth one
QUESTION [13 upvotes]: Suppose I have a smooth manifold $M$ and a smooth Morse function $f$ on $M$.  Is there a standard way to replace $M$ with a cell complex and $f$ with a discrete Morse function such the resulting homology (via discrete Morse theory) is the same as that of $M$?
My question is motivated by a situation where I know the critical points of $f$, but not how they are connected by gradient flow lines.  So I am interested in a solution which doesn't depend on that information.  I imagine something like the following: Triangulate so that every critical point of $f$ lies on a simplex of the appropriate dimension, then define a discrete vector field by [some process].  This vector field has no nontrivial closed paths [for some reason].  By [some observation], the resulting homology is isomorphic to that of $M$.  
For an expert in Morse theory, does this even sound plausible?  Are there "well-known" methods or results which would fill in the gaps?  Even better, does such a result exist somewhere already?

UPDATE:  Mainly I am interested in computing the homology of $M$ without having complete information about the gradient flow of $f$.  In particular, I have a specific smooth function on $\mathbb{R}^9$ coming from some data analysis.  I would like to find the homology of the region $M = $ { $ f \le C $ }.  I have a strategy for finding the critical points of $f$, but determining how they are connected by flow lines seems problematic.  Converting to a discrete problem would, I hope, provide a way around this.
I'm also not worried about pathologies: $f$ has finitely many critical points, and $M$ is compact.

UPDATE 2:  Thanks for the comments everyone!  I will now go to think some more.

REPLY [4 votes]: This is a rapidly developing area, and there are many short-cuts if all you want to do is compute the homology of sub-level sets of $f$. To answer your main question, as Liviu has already mentioned: there is no standard way. However, you should be able to compute the homology you desire as follows.
My impression is that you have three jobs, in chronological order:

Construct a simplicial complex $X_M$ homologically faithful to $M$.
Construct a discrete Morse function $\mu:X_M \to \mathbb{R}$ which approximates $f$, and
Compute homology of everything in sight.

First, the easiest way to build a simplicial approximation if you know your $M$ is to embed it in some suitable $\mathbb{R}^n$ and sample the hell out of it. Since you are working on data analysis, this should not be too drastic a step. Given a point sample $P$ coming from a submanifold of Euclidean space, for each radius $\epsilon$ you can construct a Cech complex of radius $\epsilon$ around $P$. Precise bounds on how many points $P$ should have and how large $\epsilon$ can be in order for the Cech complex to recover the homology of $M$ with high confidence are available in the work of Niyogi, Smale and Weinberger here in the case when $P$ is uniformly sampled. These bounds are in terms of the injectivity radius of the embedding of $M$ into Euclidean space, and of course once these bounds are satisfied it doesn't hurt to add your known critical points to $P$. You have your homologically faithful Cech complex $X_M$.
Next, for 2, you can easily infer a discrete Morse function on an entire simplicial complex just from knowing its values on the vertices using the work of King, Knudson and Mramor. You may be required to perturb $f$ slightly so that its restriction to $P$ is injective, but this is easy and generically true. You have $\mu$!
And finally, I have written software to handle 3 if you already have a $\mu:X_M \to \mathbb{R}$: you can input a filtered simplicial complex and compute not just homology at each sub-level set of $\mu$ but the persistent homology across all level-sets in the case of field coefficients. Meaning, instead of just knowing the homology of the subcomplexes $X_M^c$ consisting of all simplices with $\mu$-value less than or equal to $c$, you also recover the morphism on homology groups induced by including $X_M^c$ into $X_M^d$ whenever $c \leq d$.
All the best with your computations.<|endoftext|>
TITLE: Self-Intersection of closed curves
QUESTION [9 upvotes]: Supoose I have a closed curve $\gamma$ in the plane such that for any isometry $g$ of $\mathbb{E}^2,$ such that $g(\gamma)\neq \gamma,$ $\gamma$ intersects $g(\gamma)$  in at most two points. It should be an easy corollary of the four-vertex theorem that $\gamma$ is a circle, but I am not quite seeing it.

REPLY [6 votes]: Igor Pak's argument takes care of the general convex case. So to complete the proof (without assuming any smoothness) it remains to consider the nonconvex case. To this end it is enough to note that if $\gamma$ is not convex, then it must intersect some line $L$ at least $4$ times. Let $g(\gamma)$ be the reflection of $\gamma$ with respect to $L$. Then $\gamma$ and $g(\gamma)$ intersect at least $4$ times.<|endoftext|>
TITLE: Lost soul: loneliness in pursing math.  Advice needed.
QUESTION [88 upvotes]: This is a atypical question for the forum.  I'd like to get some advice on whether I should keep pursuing Math in the traditional route, i.e. get a PhD, do research & teach, etc.
Due to financial constraints, I worked almost full time and did not really explore much during my undergrad years.  I got my BS in Finance in 2.5 years from one of the top schools in the country.  Then I worked for a couple years in the investment industry, and decided it is not what I wanted - very flat learning curve and to be a leader in the industry, it's really not how much you know, but how well you can sell.  So after I saved enough money to live frugally for a few years, I quit this industry and see if I can find another career path. 
I then went to take various classes at community college to explore, while working part-time on and off.  In about a year or so, I decided to do math.  I liked the subject very much, and seemed capable in it.  Later, I did almost two more years of Math at a state university, all the way to advanced real analysis and advanced linear algebra.  I also participated in some research with professors.  I struggled in some classes, but overall I got As and Bs.  Everyone expects me to pursue a PhD.
This is my struggle: although I like the subject and mostly enjoy the thought-process, I hate the lifestyle.  I was either studying or in the lab, because in order to do well, I have to do it all the time, including almost all weekends.   I feel very very lonely.  A few years ago, I had a very busy undergraduate schedule, but I could predict how long a homework/project would take and schedule time to make and keep friends accordingly.  Now, I don't know if I can be done with a proof in 3 hours or 3 days! I feel bad if I abandon my unfinished work and just go out.  I do try to talk to my current peers and professors, but they are mostly about the coursework or the projects we or they are working on.  I look at my professors, they are almost always there to do research, day and night, weekday and weekends, even the ones with family. The more leisurely ones are the ones who either got their professorship or published some great papers a while ago.  Otherwise, everyone else is very busy.  I start to doubt that if everyone else is okay with it, it's me who don't love Math enough to infiltrate it into most parts of my life. After all, they are satisfied with seeing their friends once in a while, have a little chit-chat here and there, but I am very unhappy.  I also realize that while I have to spend hours on a proof, some people get it very quickly. Maybe I am just incapable.  People also work on a very irregular schedule - sometimes they come in at 2am and leave at noon, sometimes vice-versa.  I don't mind when they work, but somehow the social activity, if any, is held very late at night, out of a spontaneous mode... "oh, my brain can't function, let's watch a movie/play frisbee/poker!" at midnight!  And sadly, I don't want to be out at midnight.  Again, other people seem to be fine with it as well.
Should I continue this path?  If I do it, I don't want to do it in a rush way; so it'll take me another 2 years to do a master; and another 4 years to do a PhD, plus many more years in this research environment.  (I also would like to learn how you cope with the financial challenges to support a family if you are pursuing the research route.  I am afraid that my research is not good enough to get good funding; and positions at universities are very very competitive.)
Should I somehow engage it part-time only?  I think my professors will welcome me to continue doing research for them, even just on the weekend.  However, without formerly enrolling in school, my work will be mostly unpaid since their funding comes from NASA and other institutions, and I don't have a PhD to qualify me as a visiting scholar to let me access school's supercomputers legally.  Also, I feel like I won't grow much in this part-time manner.
I had considered doing this on my own, at my own pace, but I am not at a level where I can do/learn on myself yet.  I need someone to  discuss and check if I understand something correctly.  I much prefer doing it in person than digging through online forums.
This is long, and I appreciate your time in reading and I really look forward to some guidance or experience-sharing. Thank you.
[Technical comments:] Question asked by Flora. Originally tagged mp.mathematical-physics , ca.analysis-and-odes , ra.rings-and-algebras,  to provide more focused mathematical context.   [end technical comments, by quid].

REPLY [23 votes]: Dear Flora,
Your question hits so close home. I am now an old math professor and I had to answer this question so often!
When I was young, younger students would approach me with this question. I would say "If you have to ask, don't." This answer was given at the tone of saying "If you are not cut for it, don't even think about it." The arrogance was intended. I was young.
Over the years however I observed through my students that "being cut for it" is mostly about having the motivation for it. Talent is a fertile soil which nourishes any seed you plant. Math is no exception. 
Nowadays when faced with this question, I again give the same answer but my message is now more humane. I talk about wanting to do it, will power, and passion. 
I also ask "Where do you want to see yourself in ten years from now?" Admittedly this is difficult to answer. But the actual answer to your question will probably be guided by your answer to this particular question. Finding out your life's passion is the most difficult thing in life, in my opinion. Once that is settled, the rest is easy.
So, in summary, my answer to your question is this: re-evaluate your passion and follow where it leads you.
Best of luck with your decision.
--Sinan Sertoz<|endoftext|>
TITLE: Start with a topological group, take the meet of the two uniformities, and take the topology. Is the result again a topological group? [xpost from math.SE]
QUESTION [12 upvotes]: And what else can be said, if so?
(Original math.SE post)
In more detail: Say $(G,\mathscr{T})$ is a topological group. It has a left uniformity $\mathscr{L}$ and a right uniformity $\mathscr{R}$.  (It also has a two-sided uniformity $\mathscr{U}$, which is the join of the two.)
Now, uniformities on a given set form a complete lattice, so we can also consider the meet of the two, $\mathscr{V}$.  However, the meet of two uniformities that yield the same topology does not necessarily again yield the same topology, so it's possible that $\mathscr{T}'$, the topology coming from $\mathscr{V}$, is coarser than our original topology $\mathscr{T}$.
(Obviously, this does not happen if the group is balanced, i.e. $\mathscr{L}=\mathscr{R}$; it also does not happen if $\mathscr{T}$ is locally compact, since the meet of two uniformities yielding the same locally compact topology does again yield the same topology.  Actually, I don't know an actual case where this does happen, so I guess a first question I can ask is, are there any actual examples of this?)
So my question is, is $(G,\mathscr{T}')$ again a topological group?  Obviously inversion is continuous, since $\mathscr{V}$ makes inversion uniformly continuous, but it's not clear what would happen with multiplication.
If it is a topological group, then we can ask things like, how does $\mathscr{V}$ compare to $\mathscr{L}'$, $\mathscr{R}'$, $\mathscr{U}'$, and $\mathscr{V'}$?  (Well, obviously it's coarser than the last of these.)  And considering $\mathscr{T} \mapsto \mathscr{T}'$ as an operation on group topologies on $G$, what happens when we iterate it?  When we iterate it transfinitely?

REPLY [5 votes]: The meet of the left- and right- uniformities is called the Roelcke uniformity, as Todd Eisworth mentions. The topology it generates is the original topology (the same is true for the join of the two uniformities).
One way to see it is as follows: a fundamental system of entourages for the Roelcke uniformity is given by sets of the form $\{(x,y) \colon y \in VxV \}$, for $V$ a neighborood of the neutral element. If $U$ is an open neighborhood of some $g \in G$, then by joint continuity of the group operations there exists an open $V$ containing the neutral element and such that $VgV \subseteq U$, which shows that $U$ is open for the topology induced by the Roelcke uniformity.
So the topology induced by the Roelcke uniformity is finer than the original one, and it is clearly coarser.<|endoftext|>
TITLE: Showing non-expansion for $x\rightarrow x+1, x\rightarrow 2x.$
QUESTION [8 upvotes]: Construct a graph having $V=\mathbb{Z}/p\mathbb{Z}$ as its set of vertices and $\{\{x,x+1\}: x \in V\}\cup \{\{x,2x\}: x \in V\}$ as its set of edges. This graph is not an expander - quite unsurprisingly, since it is induced by a solvable group of actions.
Question: what is the simplest way to show that this graph is not an expander?
An obvious strategy is to construct a set $A$ such that $|A \cup (A+1) \cup 2A| < (1+\epsilon) |A|$
(for $\epsilon$ arbitrary and $p$ large enough in terms of epsilon). How to construct a set A is less obvious.
Two possible constructions:
(a) If $p = 2^n+1$, or more generally $p = 2^n+O(1)$, then $A = $(reductions modulo $p$ of itnegers between $0$ and $p!$ with more $0$'s than $1$'s in their binary expansion) should work.
(b) For general p, J. Cilleruelo points out to me that the set $A$ constructed by Gonzalo Fiz in Proposition 3.2 of http://arxiv.org/abs/1203.2659 (based on a Lemma of Rokhlin´s)
should give an answer, at least if 2 is replaced by 4 (or any other constant square).
Any other proposals? I'd like something that can be shown quickly to work in a survey or in a class.

REPLY [3 votes]: Just noticed this.  The problem has been already solved in two ways,
but not the generalization suggested by H A Helfgott where the multiplier $2$
is replaced by an arbitrary constant (though the solution may be
implicit in Alain Valette's answer).  In fact, for any fixed $r$, $s$
and any $a_j$ ($1 \leq j \leq r$) and $m_k$ ($1 \leq k \leq s$)
the graph of degree $2(r+s)$ on ${\bf Z} / p {\bf Z}$ obtained by joining
every $x \bmod p$ to the $2(r+s)$ residues $x \pm a_j$ and $m_k^{\pm 1} x$
is not an expander as $p \rightarrow \infty$, no matter how the
addends $a_j$ and multiplicands $m_k$ are chosen.  This is basically
what I gave as the "Exercise" at the end of my accepted answer to
question
MO.125251 on the analogous graphs on finite fields of $2^n$ elements.
(The exercise concerned only $r=s=1$, but a comment noted that the
same argument applies in for any fixed $r,s$).
To spell it out: Let $A = \sum_j A_j$ and $M = \sum_k M_k$ be the
corresponding operators on ${\bf C}^p$, so that $A_j$ (respectively $M_k$)
sends the unit vector $e_x$ to $e_{x+a_j} + e_{x-a_j}$
(resp. $e_{m_k x} + e_{m_k^{-1} x}$
The all-$1$ vector ${\bf 1}$ has eigenvalue $2(r+s)$ for $A+M$.
We show that there is no spectral gap by finding a vector $v$
orthogonal to ${\bf 1}$ such that the Rayleigh quotient
$\langle v, (A+M) v \rangle / \langle v, v \rangle$ is $2(r+s) - o(1)$
as $p \rightarrow \infty$, uniformly over all choices of $a_j$ and $m_k$.
(The inner product is
$\langle v, w \rangle = \frac1p \sum_{x \in {\bf Z}/p{\bf Z}} v_x \overline w_x$.)
Fix $N$.  We'll construct $v$ such that
$\langle v, (A+M) v \rangle / \langle v, v \rangle > 2(r+s) - 2s/N - o(1)$.
For $u \in {\bf Z} / p{\bf Z}$ let $\chi_u$ be the character
$x \mapsto e^{2\pi i u x/p}$ on ${\bf Z}/p{\bf Z}$, so that the
$\chi_u$ form an orthonormal basis for ${\bf C}^p$.  Set
$$
v = \sum_c \chi_{m^c u_0}
$$
for some nonzero $u_0 \in {\bf Z} / p{\bf Z}$ to be chosen later;
here the sum extends over all integer vectors $c = (c_1,c_2,\ldots,c_s)$
with each $c_k \in [1,N]$, and $m^c := m_1^{c_1} m_2^{c_2} \cdots m_s^{c_s}$.
Then
$$
\langle v, M v \rangle \geq \frac{N-1}{N} 2s \langle v, v \rangle
$$
(the inequality may be strict if there are small multiplicative
relations among the $m_k$).  Now each $\chi_u$ is an eigenvector for $A$
with eigenvalue $2\sum_{j=1}^r \cos (2 a_j u/p)$.
We want to choose $u_0$ so that each of the $N^r s$ multiples
$a_j m^c u_0$ is $o(p)$, so that $A \chi_{m^c u_0}$ has eigenvalue
$2r-o(1)$ and $\langle v, A v \rangle = (2r-o(1)) \langle v, v \rangle$.
By a standard pigeonhole argument such $u_0$ exists that makes
each $a_j m^c u_0 / p$ within $O(p^{-1/N^r s})$ of an integer,
uniformly over choices of $a_j$ and $m_k$.
This gives the desired estimate as $p \rightarrow \infty$.
Since $N$ can be taken arbitrarily large, we are done.
The special case $r=s=1$, $m_1=2$ is a bit easier:
we can always take $u_0=1$ and obtain a bound $O(2^N/p)$
that decreases faster than $O(p^{-1/N})$.<|endoftext|>
TITLE: Ising model on a cycle
QUESTION [8 upvotes]: The Ising model on $\mathbb{Z} / 2d\mathbb{Z}$ gives to the configuration $x=(x_0, \ldots, x_{2d-1}) \in \{-1,+1\}^{2d}$ a probability proportional to $\exp\\big(\beta  \sum_i x_ix_{i+1} \\big)$. The Gibbs sampler with block updates is a Markov chain $X_k$ that evolves on the set of such configurations and updates the odd (resp. even) indices conditionally on the even (resp. odd) indices with probability a half.
It seems like a relatively straightforward application of the path coupling [1] approach (two configurations are neighbours if they agree on all odd or all even coordinates; distance between two neighbours is $1+H(x,y)/d$ where $H$ is the Hamming distance) shows that the mixing time of the Gibbs sampler stays bounded as the size $d$ of the system goes to infinity, which looks rather surprising. Any intuition behind that? If this is already written somewhere, any reference concerning this (or similar) result?

[1] Chapter 14 of Markov Chains and Mixing Times by D. Levin, Y. Peres and E. Wilmer

REPLY [7 votes]: This is not so surprising, and is related to the lack of phase transition in the one dimensional Ising model.
Consider first why the mixing time might be large. If $\beta$ is very high, and we start with a configuration where half the circle is + and half -, it will take a fairly long time for the chain to converge to one of the extreme states. (Roughly $d^2$, as the interface will  perform a random walk.)
However, if $\beta$ is fixed and $d$ is large, then at every step the process will create islands of the opposite sign, at distance of order $e^{4\beta}$, regardless of $d$. Notethat the stationary distribution also has a finite correlation length.
Finally, another way to see the bounded mixing time is by a coupling argument. The simplest local coupling, will create agreement with some density, and segments of agreement will grow at positive rate, so after roughly at most $e^{4\beta}$ steps any two starting configurations will couple. This bound can be improved.<|endoftext|>
TITLE: Convex PBW bases
QUESTION [5 upvotes]: Given a reduced expression for the longest word $w_0$ in the Weyl group of $\mathfrak{g}=\mathfrak{n}^+\oplus\mathfrak{h}\oplus{n}^-$, one obtains a convex ordering on the set  of positive roots, $\beta_1<\ldots<\beta_N$. Using the braid group action we get (divided powers of) root vectors $E_{\beta_i}^{(n_i)}$ in $U_q(\mathfrak{n}^+)$ and the set of elements of the form 
$$E(\underline{n})=E_{\beta_N}^{(n_N)}\cdots E_{\beta_1}^{(n_1)},\;\;\;\underline{n}=(n_N,\ldots,n_1)\in\mathbb{Z}^N_{\geq0}$$
forms a basis for $U_q(\mathfrak{n}^+)$.
I'm wondering if anyone can point me to an elementary proof of the following fact (the shorter the better):
$$\overline{E(\underline{n})}=\sum_{\underline{m}}\lambda_{\underline{n}\underline{m}}E(\underline{m})$$
where $\lambda_{\underline{n}\underline{n}}=1$ and $\lambda_{\underline{n}\underline{m}}=0$ if $\underline{m}>\underline{n}$ with respect to some total ordering. (I think the lexicographic ordering on $\mathbb{Z}_{\geq0}^N$ or maybe the opposite ordering works,  but if not I'd like to know what does).
By elementary, I mean that I want to avoid the geometry of quiver varieties. Also, I know that Leclerc has a proof of this fact in his paper on quantum shuffles, but I'm hoping for something less involved.

REPLY [3 votes]: If by Leclerc, you mean his paper "Dual canonical bases, quantum shuffles and q-characters", then his Lemma 37 is not for a general convex order, but only for those that come from Lyndon words. And if you mean a different Leclerc paper, I'd like to know which you have in mind.
I know two proofs of this fact.
One relies on Proposition 1.9 in Lusztig's "Braid group action and Canonical Bases", together with standard facts about the T_i's, where standard means that you can read about them in Lusztig's book. From here, now that Lusztig has done the hard work for us, it is not hard (but may be a pain to typeset a proof). You might want to look at Beck and Nakajima's paper "Crystal Bases and two-sided cells of Quantum Affine algebras". In particular Section 2.6 is useful since it contains a restatement of Lusztig's result with all the relevant definitions on the same page, and then later on, (Prop 3.36) they prove an analogous upper triangularity result in the affine case under consideration.
EDIT(20/12/12): Another proof of this fact is to consider the dual picture, where the uppertriangularity is equivalent to bar-invariance of the root vectors together with the Levendorskii-Soibelman formula. Originally here I claimed that
the uppertriangularity of the bar involution is a Corollary of Theorem 3.1 of my own paper "Finite Dimensional Representations of Khovanov-Lauda-Rouquier Algebras I: Finite Type." Since the proof of said theorem invokes the Levendorskii-Soibelman formula, I cannot call this an independent proof. There is a discussion of this in arXiv:1210.6900.
I would be happy to see a quiver variety proof, the only one I know of is Prop 7.9 in Lusztig's "Canonical Bases arising from Quantized Enveloping Algebras", where, as in Leclerc, not all convex orders are considered.<|endoftext|>
TITLE: Well defined Tensoring of spectral triples 
QUESTION [5 upvotes]: Hi,
I have a misunderstanding that I am hoping is really quite trivial. I will give my question directly and context below for those that need/want it.
Question: In connes standard model he takes the finite algebra input $\mathcal{A}_F = \mathbb{C}\oplus\mathbb{H}\oplus M_3(\mathbb{C})$. I am a little confused how connes takes his spectral tensor products over $\mathbb{C}$ to form an almost commutative triple. Connes represents the quaternions $\mathbb{H}$ as $2\times2$ complex matrices, but the quaternions are a real algebra. It is impossible to centralize the complex numbers as a sub algebra of the quaternions. For this reason I am wondering how his tensor product over $\mathbb{C}$ is well defined ('prior' to representation).
Context:
In Connes almost commutative models he forms his almost commutative algebra as $\mathcal{A}_{ac}= C^\infty(M)\otimes_C \mathcal{A}_F$,
and what I am going to call his almost commutative 'cyclic coboundary operator' is given as $\delta_{ac} = \delta\otimes_C\mathbb{I} + c\otimes_C \delta_F$, where c is a Hochschild cycle. These are both represented faithfully on some hilbert space $\mathcal{H}_{ac} = \mathcal{H}\otimes\mathcal{H}_F$.
The representation of the 'cyclic coboundary operator' is then given by the unfluctuated almost commutative dirac operator $D_{ac} = D\otimes_C\mathbb{I} + \gamma_5\otimes_C = i\gamma^\mu \otimes_C\partial_\mu\mathbb{I} + \gamma_5\otimes_C D_F$ (notice that here we really require the tensor product to be over C and not R).
The fluctuated dirac operator is given by $\tilde{D}_{ac} = D_{ac} + A + JAJ^*$ 
Where A is a representation of a one form given by $A = \sum_i \pi_{\alpha_i} [D_{ac},\pi_{\beta_i}] $. A general 'one form' is given by given by a sumation of elements $\alpha\delta_{ac}\beta = (a\otimes X)(\delta_{ac} = \delta\otimes_C\mathbb{I} + c\otimes_C \delta_F)(b\otimes Y)$, with $\alpha,\beta\in\mathcal{A}_{ac}$. 
It seems that forming well defined fluctuated operators in an for an almost commutative geometry requires tensor products to be over C but I am confused how this is done when a real finite algebra is chosen.

REPLY [6 votes]: There's a reasonably convenient abuse of notation in play. When dealing with almost-commutative spectral triples, $C^\infty(M)$ means $C^\infty(M,\mathbb{C})$ except when forming $C^\infty(M) \otimes A_F$ for $A_F$ real, in which case $C^\infty(M) \otimes A_F$ really means $C^\infty(M,\mathbb{R}) \otimes_{\mathbb{R}} A_F$.<|endoftext|>
TITLE: Enlarging a tetrahedron with integer edge lengths
QUESTION [6 upvotes]: Given a tetrahedron with all edges having integer length, is it always possible to increase all of the edge lengths by one?
More precisely: Let $P_1, P_2, P_3, P_4$ be four distinct non-coplanar points in $\mathbb{R}^3$, such that $d(P_i, P_j) \in \mathbb{Z}$ for all $i,j$, Must there exist non-coplanar points $Q_1, Q_2, Q_3, Q_4$ in $\mathbb{R}^3$ such that $d(Q_i, Q_j) = d(P_i, P_j) + 1$ for all $i < j$?
I don't know if the integer length condition is necessary. My intuition is that if all of the edges of a tetrahedron are increasing at the same rate, then the tetrahedron will approach a regular tetrahedron in the limit, so it should not become degenerate. But I have no proof of this.

REPLY [3 votes]: It's possible to completely characterize when a six-tuple $(a, b, c, d, e, f)$ forms the lengths of edges of a (non-degenerate) tetrahedron. Namely, by a quick google search I found for example a paper Edge lengths determining tetrahedrons by Wirth and Dreiding, Elem. Math. 64 (2009) 160 – 170.
By their Lemma 2.1, such a 6-tuple (taken as determining not only the lengths of the edges, but also the positions of the edges in the tetrahedron) actually comes from a tetrahedron if and only if each face triangle satisfies triangle inequality and there is a vertex such that the 3 angles on faces around it are all acute and satisify the triangle inequality.
Now we can try to check that if $(a, b, c, d, e, f)$ come from a tetrahedron, then so does $(a+1, b+1, c+1, d+1, e+1, f+1)$: for the faces, just note that adding 1 to the edge lengths makes the triangle more equilateral, and so the triangle inequality will again hold (if $a < b+c$, then $(a+1) < (b+1)+(c+1)$). 
EDIT: Ok, for the angles it's more tricky, as Igor Rivin says in his comment.
First, seeing that acute angles remain acute is easy, just from the fact that the angle opposite side $x$ in a triangle with sides $x, y, z$ is acute iff $x^2 < y^2+z^2$ (by the cosine rule).
As Anton Petrunin pointed out, the counterexample was nonsense. Sorry.  For the rest we'll need some lower bound on the lengths of edges, the statement in fact does not hold with short edges. 
 Here is a counterexample: Label the edge lengths so that the faces are $abc$, $dec$, $dbf$ and $aef$. We'll be looking at the vertex $V$ where $a, b, f$ meet. Choose some small $\varepsilon$ and set $a=b=\varepsilon$, $c=\varepsilon^2$, and choose $d, e, f$ very large ($> \varepsilon^{-1}$) so that the angles at $V$ in triangles $dbf$ and $aef$ are small (and the edges come from a tetrahedron). However, after I increase all edge lenths by 1, the triangles $(d+1),(b+1),(f+1)$ and $(a+1),(e+1),(f+1)$ have almost the same shape as $dbf$ and $aef$, hence their angles at $V$ have almost not changed. But the triangle $(a+1),(b+1),(c+1)$ is now almost equilateral, and so its angle at $V$ is almost $\pi/3$. Thus the triangle inequality between angles at $V$ has no chance of holding. 
It seems intuitively clear that if we assume that all the sides are $\geq C$ for a constant $C$, than the bigger $C$ is, the less will the angles change when we enlarge the sides by 1. But to make this into a proof of the statement, we'd need some quantitative estimate on how much the angles can change depending on the size of the angle (and $C$). It just seems to me like very technical and not very enjoyable calculations, so I'll be happy to leave them to someone else :) At least, then the solution of the integral problem would follow just by checking the finitely many remaining cases.<|endoftext|>
TITLE: On totally nonnegative Grassmannian
QUESTION [5 upvotes]: I was reading Postnikov's paper [TOTAL POSITIVITY, GRASSMANNiANS, AND NETWORKS][1] when I came across the definition of the totally nonnegative Grassmannian $Gr_{kn}^{tnn} \subset Gr_{kn}$ as the quotient 
$Gr_{kn}^{tnn}  = GL_k^{+}\backslash Mat^{tnn}_{kn}$,
where
$Mat^{tnn}_{kn}$ is the set of real $k\times n$-matrices  $A$
of rank $k$ with nonnegative maximal minors $\Delta_I(A)\geq 0$
and $GL_k^+$ is the group of $k\times k$-matrices with
positive determinant.
My question is, what is the meanig of the quotient here? And how do we visualize this quotient geometrically?
I understand that the grassmannian $Gr_{kn}$ is the space of k-subspaces of n-dimensional linear space $V$.

REPLY [11 votes]: There are two issues here:

You can represent a $k$-subspace of $V$ by a $k \times n$ matrix whose rows are a basis for the subspace, but this way of representing the subspace is not unique, since your subspace has many different bases.  The usual way to handle this is to consider cosets of $k\times n$ matrices under the left action of $GL_k$, since two matrices will be in the same coset iff one is obtained from the other by row operations that preserve the span of the rows.  The notation $GL_k \setminus Mat_{kn}$ (where I've left off the + from $GL_k$ and the tnn from $Mat_{kn}$) means the left cosets of the $k\times n$ matrices under the action of $GL_k$.
Postnikov is restricting to those cosets which include a $k\times n$ matrix that is "totally nonnegative", which means that all of its minors are nonnegative.  (Warning: people often study totally positive matrices where all minors must actually be positive, but Postnikov, Lusztig and others have been considering real matrices where minors can be either 0 or positive.)  The "tnn" is notation for this.  And in this case he takes $GL_k^+$ cosets instead of $GL_k$ cosets, since he only needs to identify totally nonnegative $k\times n$ matrices with each other. 

Edit: I forgot to say anything about how to visualize this space.  People believe it is a ball, but have not so far proven this (to my knowledge).  It is the image of a map from a polytope Postnikov introduced in the paper you are reading.  The restriction of this map to the interior of the polytope is a homeomorphism, but many points on the boundary of the polytope get identified with each other.  My impression from talking with people who've worked with this space is that they can visualize it to varying degrees, but mainly I think try to understand it by trying to understand this map really well.  Another paper you might find helpful is:
MR2525057 Reviewed Postnikov, Alexander; Speyer, David; Williams, Lauren Matching polytopes, toric geometry, and the totally non-negative Grassmannian. J. Algebraic Combin. 30 (2009), no. 2, 173–191. (Reviewer: T. Oda) 20G20 (05B35 13F60 14M25 52B70)
This proves it has a CW decomposition and relates this space you are considering to the totally nonnegative part of a toric variety.  Perhaps others here will have more to say about how to visualize this space.<|endoftext|>
TITLE: Location of Archimedes' grave in Syracuse (math/archaelogy trivia)
QUESTION [20 upvotes]: This is really a question for our archaelogist friends, but I could not find an "archaelogy overflow" and some mathematicians might find it interesting. 
In a few weeks I am giving a talk in which I would like to mention Archimedes' epitaph, which is reported to show a sphere and a cylinder.  (Archimedes proved that the surface area of the sphere is equal to the lateral surface area of the cylinder with the same height and radius.)  I realize that it is not possible to discover where Archimedes' epitaph was, or what exactly the representation was. (That is, whether it was inscribed, a sculpture, etc.)  However, I would like to know whether there a location of Archimedes' grave which is consistent with
the description given by the Roman politician Cicero and also the 1925 century excavation report of the Italian archaelogist Paolo Orsi.   In particular, is the description  most consistent with these descriptions,  that his tomb would have been under present location of the shopping mall "I Papiri"?  http://www.centrocommercialeipapiri.it/negozi.php
Here are the two sources:   (I realize that taking the word of a Roman politician at face value is a bit dangerous:)
Cicero describes Archimedes as being buried in a tomb to the west of the city:
"But from Dionysius’s own city of Syracuse I will summon up from the dust—where his measuring rod once traced its lines—an obscure little man who lived many years later, Archimedes. When I was questor in Sicily [in 75 BC, 137 years after the death of Archimedes] I managed to track down his grave. The Syracusians knew nothing about it, and indeed denied that any such thing existed. But there it was, completely surrounded and hidden by bushes of brambles and thorns. I remembered having heard of some simple lines of verse which had been inscribed on his tomb, referring to a sphere and cylinder modelled in stone on top of the grave. And so I took a good look round all the numerous tombs that stand beside the Agrigentine Gate. Finally I noted a little column just visible above the scrub: it was surmounted by a sphere and a cylinder. I immediately said to the Syracusans, some of whose leading citizens were with me at the time, that I believed this was the very object I had been looking for. Men were sent in with sickles to clear the site, and when a path to the monument had been opened we walked right up to it. And the verses were still visible, though approximately the second half of each line had been worn away."        
(from Cicero (106-43 BC), Tusculan Disputations, Book V, Sections 64-66, translation provided on http://www.math.nyu.edu/~crorres/Archimedes/Tomb/Cicero.html)
The Italian archaelogist Orsi describes the exacavation of a Hellenistic necropolis from the 7th-1st century BCE to the west of the city.
"La necropoli del Fusco si estende per circa 2km lungo la rotabile di Floridia,
a partire dall'ex-osteria Rejna, attraverso le proprieta Corvaia de Gargallo (trappeto di S. Nicola e Tor di Conte)  sino alla contra. Canalicchio prop. Carpinteri. All'inizio orientale di questa zona io ebbi la venture di scoprire un groppo di sepolcri, ancora in gran parte intatti, della fne del Sec. VIII e del VII, e po via via, procedendo ad occidente, quelli del VI e del V e piu la ancora gruppi del I-VI sec. av. Cr. 
Rough Google-assisted translation:  "The necropolis of Fusco stretches for about 2km along the Floridia carriageway (edit: not railway) including former tavern-Rejna through the property Corvaia de Gargallo (olive mill of St. Nicholas, and Tor di Conte) to the district Canalicchio prop. Carpinteri. At the east of this area I discovered a cluster of tombs, still largely intact, from the end of the eighth century and the seventh, and little by little, moving to the west, those of VI and V and still more the groups from I to VI century BCE."
(in NOTIZIE DEGLI SCAVI DI ANTICHITÀ - n. 4-5-6 (1925)
available at http://periodici.librari.beniculturali.it/visualizzatore.aspx?anno=1925&id_immagine=9999505&id_periodico=9817&id_testata=31
The reason that I suggest that the location most consistent with descriptions is the shopping
mall is that (1) Necropolis del Fusco is apparently the largest Hellenistic necropolis in
Syracuse, and so the most likely location of Archimedes tomb (2) it was, as in Cicero's
description, to the west of the city (3) the shopping mall I Papiri is at the end of the street "Necropolis del Fusco" in Syracuse, which runs through the area described by Orsi as the location of the Fusco necropolis, and the tombs from the later centuries are reported to be those at the western end.

REPLY [9 votes]: In 1802 Johann Gottfried Seume visited Syracuse (from Germany per pedes). In his diary he mentions the place where Cicero had found Archimedes' grave: A chapel near the Greek theatre and the water pipe.

Etwas rechts weiter hinauf hat Landolina das römische Amphitheater besser aufgeräumt und hier und da Korridore zu Tage gefördert, die jetzt zu Mauleseleien dienen. Die Römer trugen ihre blutigen Schauspiele überall hin. Die Area gibt jetzt einen schönen Garten mit der üppigsten Vegetation. Weiter rechts hinauf ist das alte große griechische Theater, fast rundherum in Felsen gehauen. Rechts, wo der natürliche Felsen nicht weit genug hinausreichte, war etwas angebaut ...  Die Wasserleitung geht nahe am Theater weg; vermutlich brachte sie ehemals auch das Wasser hinein. ... Gegenüber steht eine Kapelle an dem Orte, wo Cicero das Grab des Archimedes gefunden haben will. Wir fanden freilich nichts mehr; aber es ist doch schon ein eigenes Gefühl, daß wir es finden würden, wenn es noch da wäre, und daß vermutlich in dieser kleinen Peripherie der große Mann begraben liegt.

Johann Gottfried Seume: Spaziergang nach Syrakus im Jahre 1802 - Kapitel 35
EDIT
1) According to Seume's report this ruin is not the grave of Archimedes.
2) This painting by Benjamin West, showing Cicero and the magistrates discovering the tomb of Archimedes, is very popular but not very reliable.<|endoftext|>
TITLE: Inversion of Radon transform by incomplete data: specific case
QUESTION [7 upvotes]: Let $R[f](p,t)$ denote the Radon transform of smooth function $f(x) \colon \mathbb{R}^n \to \mathbb{R}$ with compact support in $\mathbb{R}^n_+$:
$$
   R[f](p,t) = \int\limits_{x \cdot p = t} f(x) dx.
$$
In field of mathematical economics $p$ and $t$ may have meaning of prices and then be positive. Hence there arises a problem of recovery of function $f(x)$ via it's Radon transforms $R[f](p,t)$ with $p \in \mathbb{R}^n_+$ and $t > 0$. I would like to know if there are some results on this topic.

I was asked about economical interpretation of the Radon transform. A part of economical models consists of models of production. The simplest model of production is so called Leontieff model. Say, in some sector one produce some product and use $n$ other resourses. To produce $1$ unit of our product we have to use $x_1$ units of first resourse, $x_2$ units of second e.t.c., in other words, if we have $(u_1,\ldots,u_n)$ resourses we can produce $\min \left( \frac{u_1}{x_{1}},\ldots,\frac{u_n}{x_n}\right)$ units of our production. This function is called Leontieff production function. In real models there exists an effect of substitution of production factors at the microlevel and we have to modify the Leontieff production function by the appropriate neoclassic production function. But we will still talk about Leontieff model. Let $p_0$ be a price of a unit of our production and $p=(p_1,\ldots,p_n)$ be a vector of prices of resourses, $\mu$ be a measure, that describes a distribution of available production powers over production technologies (any technology is given by a vector of resourses to produce a unit vector of our production, but we can use many technologies in production, see Hauthekker-Johaneson model) Then our profit function is a function
$$
   \Pi(p,p_0) = \int (p_0 - p \cdot x)_{+} \mu(dx)
$$
and it gives us a maximal profit possible (it is a result of maximisation over distributions $\mu$ allowed by existing production powers). It's second derivative is exactly the Radon transform of measure $\mu$.

REPLY [4 votes]: I second the recommendations in the comments which suggest looking at the tomography literature and at the characterization of the problem as an ill-posed inverse problem. I, in particular, recommend these two references.
The first one contains a detailed exposition of the recovery of $f$ in the special case $n=2$ (chapter 5): 

Kabanikhin (2011). Inverse and Ill-posed problems. Theory and Applicatons

That book is quite neat in the sense that it presents a very wide and rigorous exposition of the field of ill-posed inverse problems. Thus, it constitutes an excellent entry point to the field.
There are also a lot of papers in statistics on that particular problem. The second reference that I recommend is a paper which might be of particular interest to you (since it is an econometric paper that studies inversion of Radon transforms for the purpose of some non-parametric estimation problem):

Hoderlein, Klemelä and Mammen (2010). Analyzing The Random Coefficient Model Nonparametrically

Edit: I updated a link above for the actual published paper location of the second reference. A 2007 working paper version is available at the following link.<|endoftext|>
TITLE: Uniqueness for a non-local differential equation
QUESTION [7 upvotes]: Question:Fix $\epsilon>0$.  Consider the differential equation, defined for functions $f(t,x)\in C^\infty([0,\epsilon]\times[0,\epsilon])$ defined by
$$\frac{\partial}{\partial t} f(t,x)=\frac{f(t,x)^2-f(t,0)^2}{x}, x>0,$$
$$\frac{\partial}{\partial t} f(t,0)=2f(t,0) \frac{\partial f}{\partial x}(t,0).$$
I want to know if this differential equation has uniqueness for $f>0$.  More precisely, suppose $f$ and $g$ are two $C^\infty$ functions satisfying the above equation with $f(0,x)=g(0,x)$, $\forall x$, and $f(t,x), g(t,x)>0$, $\forall t,x$.  Must $f=g$?

Comment:  The similar linear differential equation does have uniqueness.  Indeed, consider, the equation
$$\frac{\partial}{\partial t} f(t,x)=\frac{f(t,x)-f(t,0)}{x}, x>0,$$
$$\frac{\partial}{\partial t} f(t,0)=\frac{\partial f}{\partial x}(t,0).$$
If $f$ and $g$ are two solutions to the above equation, then $f=g$.  To see this, note that since the equation is linear, we may assume $f(0,x)=0$ for all $x$, and we wish to prove that $f(t,x)=0$ for all $t$.  Define $a(t)=f(t,0)$.  Then, for $x>0$, $f$ satisfies the equation
$$\frac{\partial}{\partial t} f(t,x)=\frac{f(t,x)-a(t)}{x}.$$
We can solve this equation explicitly.
$$f(t,x)=-x^{-1}\int_0^t exp((t-s)/x) a(s) ds.$$
Since $f$ is $C^\infty$, the right hand side of the above equation is bounded in $x$.  It can be shown that if such an expression is bounded as $x$ tends to $0$, $a$ must be the zero function.  This means $a(t)=f(t,0)=0$ for all $t$.  Standard uniqueness theorems for ODEs then show $f(t,x)=0$ for all for all $t,x$, completing the proof.

Comment:  I would also be interested to know about existence, etc.

REPLY [3 votes]: The answer is yes:  f=g.  I wrote up a paper with a more general result here.  The idea is the following.  If $f$ were assumed to be of Laplace transform type
$$
f(t,x) = \frac{1}{x} \int_0^\infty e^{-w/x} A(t,w)\: dw,
$$
then, by taking the inverse Laplace transform, one finds that $A(t,w)$ satisfies a PDE which is much easier to study.
But not every solution is of Laplace transform type!  What is proved, though, is that every solution is of Laplace transform type, modulo an error which can be controlled.  In particular, the above equation does not have existence:  only initial conditions $f(0,x)$ which are of Laplace transform type modulo an appropriate error give rise to solutions.
Incidentally, there is a slight error in the above question.  Where I said "we can solve this equation explicitly," I didn't include the initial condition.  One should treat $f(\epsilon,x)$ as the initial condition, and solve backwards in time.<|endoftext|>
TITLE: Is the stable homotopy category idempotent complete?
QUESTION [5 upvotes]: Is the stable homotopy category idempotent complete? I have not been able to prove it, and the proof for abelian groups seems to strongly rely on looking at elements. 
Thanks,
Jon

REPLY [14 votes]: Yes, this is a standard fact.  Given a self-map $e\colon X\to X$, we write $e^{-1}X$ for the telescope of the sequence $X\xrightarrow{e}X\xrightarrow{e}X\xrightarrow{e}\dotsb$ (constructed as the cofibre of a suitable self-map of $\bigvee_{i=0}^\infty X$).  If $e$ is idempotent, one can check that the natural map $X\to e^{-1}X\vee (1-e)^{-1}X$ is an equivalence, and that $e$ acts as the identity on the first factor and as zero on the second; in other words, we have a splitting of $e$.<|endoftext|>
TITLE: Properly "transfinitely" Euclidean domains
QUESTION [6 upvotes]: Are there integral domains which admit ordinal-valued Euclidean functions but not $\mathbb{N}$-valued Euclidean functions?

REPLY [6 votes]: Yes, they exist. Even if the problem is left open in the the papers of T. Motzkin and P. Samuel cited in Comparing different Euclidean algorithms on a Euclidean domain 
the problem is solved in 
J. Hiblot, Des Anneaux euclidiens dont le plus petit algorithms n'est pas  valeurs finies, C. R. Acad. Sci. Paris, 281 (1975), 411-414 with correction in vol 287.
M. Nagata, On Euclid algorithm. C. P. Ramanujama tribute, pp. 175 Tata Inst. Fund. Res. Studies in Math., 8, Springer, Berlin-New York, 1978.
See also: 
Pag. 195 in P. M. Cohn, Free Ideal Rings and Localization in General Rings
Rod Downey, Euclidean Domains and Euclidean Functions, http://homepages.mcs.vuw.ac.nz/~downey/ntu_13.pdf
Pete L. Clark, A note on euclidean order types, http://arxiv.org/abs/1208.0977<|endoftext|>
TITLE: Understanding groups that are not linear
QUESTION [19 upvotes]: I have a really hard time "feeling" what it means for a group to fail to be linear. Vaguely, I'd like to know how one should think about such groups. More precisely:
What are some interesting examples of groups that aren't linear?
Are there general constructions that one can use to cook up a group or a family of groups that isn't linear?
Are there general techniques that one can use to show that a given group isn't linear?
More loosely, what is it about these groups that makes them interesting?

REPLY [4 votes]: Let me give you two examples from the theory of pro-$p$ groups:

The Nottingham group contains every finite $p$-group. Thus, it is not linear over any field.
Zubkov proved that a non-abelian free pro-$p$ groups cannot be embedded into $SL_2(R)$, where $R$ is any pro-$p$ ring. This implies that any pro-$p$ subgroup of $SL_2(R)$ satisfies a pro-$p$ identity. Lubotzky and Shalev conjectured that this is true for $SL_n(R)$ for any $n$ and any pro-$p$ ring $R$. Micheal Larsen and I proved that a non-abelian free pro-$p$ group is not linear over local fields. Our proof was a fairly easy application of Richard Pink's work which enabled us to reduce the question to showing that an open subgroup of semisimple group over local fields is not free. I think we then used some result of Lubotzky, but I am sure there are many arguments to deal with that case. Let me emphasize that our result does not imply the existence of a pro-$p$ identity (for that you need to deal with all pro-$p$ rings). However, Zelmanov claims in an unpublished work that pro-$p$ identity exists if we fix $n$ and let $p$ be big enough. Also, notice that this does not imply non-linearity in general.<|endoftext|>
TITLE: Is there a notion of a chain complex with corners?
QUESTION [16 upvotes]: Roughly speaking, algebraic topology works by reducing questions about topological objects such as manifolds and cell to questions about chain complexes. 
On the topological side, although in the PL category a manifold can have a boundary and not much more, in the smooth category there is a notion of a manifold with corners, that is that every point has a neighbourhood diffeomorphic to $\mathbb{R}_{\geq 0}^n$. To go further, there is a notion of a manifold with faces, which adds an additional piece of stratified structure whose existence guarantees that each piece of the boundary has a smooth collar in the manifold (See Appendix A of Farber's Topology of Closed One-Forms).
On the algebraic side, there is a notion of the boundary of symmetric chain complex (I think due to Ranicki), which measures the chain-level failure of Poincare-Lefshetz duality.

Question: Is there a notion of a chain complex with corners or with faces that has been studied in the literature?

It's not hard to imagine how this would work, by mapping a symmetric chain complex with boundary to a boundary of a symmetric chain complex, for example; but I'm asking whether there is any literature on such structures.

REPLY [9 votes]: The chain complex n-ads in my 1992 CUP book Algebraic L-theory and topological manifolds are chain complexes with corners. They are the chain complex analogues of Wall's n-ads (which hark back to J.H.C. Whitehead).<|endoftext|>
TITLE: Is there an algorithm for writing a number as a sum of three squares?
QUESTION [11 upvotes]: By Gauss's Theorem, every positive integer $n$ is a sum of three triangular numbers; 
these are numbers of the form $\frac{m(m+1)}2$. Clearly
$$ n = \frac{m_1^2+m_1}2 + \frac{m_2^2+m_2}2 + \frac{m_3^2+m_3}2, $$
so multiplying through by $4$ and completing the squares gives
$$ 8n+3 = (2m_1+1)^2 + (2m_2+1)^2 + (2m_3+1)^2. $$
Thus writing $n$ as a sum of three triangular numbers is equivalent to writing $8n+3$ as a sum of three (necessarily odd) squares. 
My question is;
 Is there an algorithm for writing a positive integer as a sum of three squares?

REPLY [11 votes]: This problem is discussed in my paper with Rabin, Randomized algorithms in number theory,
Commun. Pure Appl. Math. 39, 1985, S239 - S256.  We give an algorithm that, assuming a couple of reasonable conjectures, will produce a representation as a sum of three squares in random polynomial time.<|endoftext|>
TITLE: Surface Laplace-Beltrami without coordinates, exterior calculus?
QUESTION [6 upvotes]: Let $f: M \rightarrow \mathbb{R}^3$ be an immersion of a surface $M$.  For pedagogical purposes (i.e., I'm teaching a class!) I am looking for an expression for the scalar Laplace-Beltrami operator $\Delta$ applied to a real function $\phi$ on $f(M)$ that:

explicitly depends on the immersion $f$,
does not rely on local coordinates, and
does not use exterior calculus.

A standard coordinate expression is
$$\Delta \phi = \frac{1}{\sqrt{g}} \partial_i (\sqrt{|g|} g^{ij} \partial_j \phi),$$
and a standard expression using exterior calculus is
$$\Delta\phi = \star d \star d \phi.$$
However, the students do not have exposure to exterior calculus, and I am discouraging the use of coordinates whenever possible (and have so far been able to get by without them).
To give a concrete example of the "style" of expression I'm looking for, consider the normal curvature in a direction $X \in TM$, which can be expressed as
$$\kappa_n(X) = -\frac{dN(X) \cdot df(X)}{|df(X)|^2},$$
where $N: M \rightarrow S^2 \subset \mathbb{R}^3$ is the Gauss map and $\cdot$ denotes the usual Euclidean inner product.  This expression uses the differential $d$ of a function, but it does not use the exterior derivative on $k$-forms (at least, not for $k>0$), nor does it use the Hodge star, nor does it rely on a coordinate system.
In English, $\Delta$ is not hard to describe: take the sum of second derivatives along orthogonal directions in the ambient space.  But after a lot of digging, I'm surprised to find there isn't a more suggestive algebraic description.
Thanks!

REPLY [4 votes]: This is just a riff on Robert Bryant's answer but thought I would throw it out there -- its the way I think about this stuff at least....
Suppose $\mathbf{H}: M\to \mathbb{R}^3$ is the mean curvature vector (i.e. locally $\mathbf{H}=-H\mathbf{n}$ were $\mathbf{n}$ is a unit normal vector field to $f(M)$ and $H=tr A$ is the mean curvature -- this is well defined even if $M$ is unoriented).  This of course depends on the immersion.
If $\phi$ is a function on $\mathbb{R}^3$ which restricts to $f(M)$ as the given function $\phi$ then we have that
$$
\Delta_{f(M)} \phi =\Delta_{\mathbb{R}^3} \phi -\nabla^2_{\mathbb{R}^3} \phi (\mathbf{n}, \mathbf{n})+\mathbf{H}\cdot \nabla_{\mathbb{R}^3} \phi
$$
Note that $\nabla^2_{\mathbb{R}^3} \phi (\mathbf{n}, \mathbf{n})$ also does not depend on choice of $\mathbf{n}$ so this is also well defined on unoriented surfaces.<|endoftext|>
TITLE: Can one ignore primes lying over $l$ in the Fontaine-Mazur conjecture? Counterexamples?
QUESTION [9 upvotes]: The Fontaine-Mazur conjecture predicts that an $l$-adic Galois representation of a number field is 'geometric' if it is unramified outside a finite set of primes and is De Rham for primes lying over $l$. Now, what happens if one forgets about the latter restriction; are there any counterexamples, and is there any (general?) way to understand that those are not geometric without using the De Rham restriction?

REPLY [12 votes]: To complete Kevin's good answer: the number of $\ell$-adic representations (up to isomorphism) of a number field $K$ is countable, since so are varieties over a $K$. On the other hand, we know by Mazur's theory of deformations that representations of the type you consider
that is, of the Galois group of the maximal extensions of a number field $K$ unramified outside a finite set of places $S$ containing places above $\ell$ and $\infty$) form multi-dimensional $\ell$-adic family (e.g. parametrized by spaces like $\mathbb{Z}_\ell^n$ for some $n>0$, hence are uncountable. Thus not only are there counter-examples to Fontaine-Mazur's conjecture without the de Rham hypothesis, but most examples of such representations are counter-examples.
Among all representations, the de Rham (or geometric) representations are expected to be dense in certain cases (e.g. representation of dimension 2 of $\mathbb Q$) but not in general (e.g.
representations of dimension $2$ of a non-totally real number field)<|endoftext|>
TITLE: Proper class forcing vs forcing with a set of conditions bigger than one's model
QUESTION [6 upvotes]: This seems like a natural question to ask, but I've not seen it discussed in my reading around (limited to Easton's paper, the third edition of Jech's Set theory and a small handful of articles). What do you get when forcing with a proper class of conditions that you don't get when forcing with, say, a set of conditions larger than one's model of set theory? Or rather, why isn't a large set of conditions enough?
Blass makes a throwaway remark in his 1984 paper The interaction between category theory and set theory 

Although this approach [reflection principles] was first proposed in connection with the problem of foundations for category theory, it is natural to use it whenever objects seem to be too large to be coded as sets. In particular, it seems to me that it should be of some use in clarifying forcing with proper classes by making the natural (regular open) Boolean algebra available even though it is superlarge.

It seems to indicate that if we accept some sort of reflection principle, use an innaccessible cardinal $\kappa$ (or similar) - hence a Grothendieck universe - and a set of forcing conditions larger than $\kappa$, then we should arrive at our goal without using a proper class of conditions.
Alternatively, cannot one (ok, this is very naive, but this is why I'm asking) consider an inaccessible in ZFC and thus cook up a model of NBG, and then work with that a la Easton - and then at the end turn around a say 'Ahah! I was working in ZFC the whole time!'
One reason I ask is that in the paper Injectivity, projectivity, and the axiom of choice, Blass gives a symmetric model of ZFA with no nontrivial injective abelian groups using an uncountable set of atoms and a base model of ZFCA whose sets were in some sense 'small' (they arise, if I understand correctly, using the cumulative hierarchy generated from $A$ in the usual sense, but only taking countable subsets of $A$ at the first stage, rather than all of $\mathscr{P}A$). However, he gives a model of ZF with no nontrivial injective abelian groups using forcing involving a proper class of conditions. (Notice that Jech-Sochor is not useful in its usual statement because a global statement about a proper class of objects is required.) Perhaps the techniques given in Blass' Theorem 3.2 have been given a general treatment by now, I do not know.

REPLY [4 votes]: The idea behind the remark quoted in the question was that, in situations ordinarily treated with proper-class forcing (e.g., Easton's theorem), the work can be transcribed rather routinely into a Feferman-style set theory (ZFC plus a constant $\kappa$ for an ordinal and axioms saying, one formula at a time, that $V_\kappa$ is an elementary submodel of the universe $V$). Just do with $V_\kappa$ what you would otherwise have done with $V$. Where (for example) Easton got arbitrary cardinal exponentiation at all regular cardinals, you'd now get arbitrary cardinal exponentiation only at all regular cardinals below $\kappa$, but that's "morally" or "intuitively" the same (and gives the same relative consistency result) because of the elementarity of $V_\kappa$ in $V$. This "large set" approach allows you to work with the framework of Boolean-valued models rather than forcing, whereas a proper-class forcing would, in general, need super-classes (yet another level higher in the cumulative hierarchy) to do this.
In both frameworks, the real issue is not whether you work with large (i.e., $\kappa$-sized or bigger) sets or with proper classes but rather what additional conditions you impose on your forcing notions (or Boolean-valued models). As Nate pointed out, you need some conditions (in either framework) to make sure you get a model of ZFC. If you just go blindly ahead (in either forcing), you could, for example, add a proper class (respectively a $\kappa$-sized family) of Cohen reals, so that the continuum will no longer be a set in your  forcing extension (of $V$, respectively $V_\kappa$).  Or you might collapse all the cardinals (resp. all the cardinals below $\kappa$).  
Of course, some people might want to sacrifice (part of) ZFC and work with such "strange" models.  If I remember correctly, Rudy Rucker once (before he turned to science-fiction writing) proposed working in the theory obtained from ZFC by deleting the power set axiom and adding Martin's Axiom for arbitrarily large collections of dense sets (so the continuum has to be a proper class).  But here again, it seems to me that it makes little difference which framework you use.
Also, I recall that Sy Friedman did some work on super-class forcing.  I don't know any of the details, but I would expect that this too can be easily recast in terms of forcing over a Feferman-style $V_\kappa$.
Finally, let me mention that, if you force in the Feferman framework, you actually have two choices for what should be the generic extension of $V_\kappa$.  One is to take the elements of rank below $\kappa$ in the generic extension of the full universe.  The other is to take the denotations of names whose rank is below $\kappa$.  The two seem to coincide in nice cases, but I don't see any reason for them to coincide in general.  (The second is what corresponds to proper-class forcing over $V$.)<|endoftext|>
TITLE: On the natural (bigraded) homotopy groups of a simplicial object in a model category
QUESTION [7 upvotes]: $\def\mc{\mathcal} \def\sm{\wedge}$
This question stems from the Goerss-Hopkins paper Moduli Problems for Structured Ring Spectra.  Let me begin by attempting to summarize the relevant framework -- this comes from the beginning of Chapter 3, on page 92.

Let $\mc{C}$ be a model category, and let
  $\mc{P}$ be a set of projectives defining
  the $\mc{P}$-resolution model stucture
  on $s\mc{C}$.  (See Question 1 below.)
  Given $X \in s\mc{C}$ and for any
  $P\in \mc{P}$, define the $(n,P)$th
  natural homotopy group of $X$ by $\pi_{n,P}(X)=\pi_n(\mbox{map}(P,X))$ (using the derived mapping space).
  This is corepresentable in
  $\mbox{Ho}(s\mc{C})$ by $P \sm
> \Delta^n/\partial\Delta^n$, which is
  defined as the pushout of the corner
  $\emptyset \otimes \ast \leftarrow P
> \otimes \ast \rightarrow P \otimes
> \Delta^n / \partial \Delta^n$.  This smash product construction actually
  fits into an adjunction $$(-) \sm K :
> \mc{C} / \emptyset \leftrightarrows s\mc{C}
> : C_K,$$ where we define the right adjoint as follows.  First, there is an adjunction $$(-)\otimes K:\mc{C} \leftrightarrows s\mc{C}:\mbox{hom}(K,-)_0 \stackrel{def}{=}
> M_K(-)$$
  (where the right
  adjoint is the $0$th object in the
  "hom object" of $s\mc{C}$ coming its
  simplicially contensored
  structure), and then we define $C_K$ via the pullback diagram
  $$\begin{array}{ccc} C_KX &
> \rightarrow & M_KX \\ \downarrow & &
> \downarrow \\ \emptyset & \rightarrow
> & M_*X = X_0. \end{array} $$ In
  particular, write $C_nX=C_{\Delta^n /
> \Lambda^n_0}X$ and $Z_nX=C_{\Delta^n /
> \partial \Delta^n}X$.  The "inclusion
  of the boundary" $\Delta^{n}
> /\partial \Delta^{n} \rightarrow
> \Delta^{n+1}/\Lambda^{n+1}_0$ induces the
  second map in a fibration sequence
  $$Z_{n+1} X \rightarrow C_{n+1}X \rightarrow
> Z_{n}X$$ in the category
  $\mc{C}/\emptyset$ whenever $X \in s\mc{C}$ is
  Reedy fibrant.  The neat result is
  that in this case, this in fact gives a
  "presentation" of the natural homotopy
  group $\pi_{n,P}(X)$ via the exact
  sequence $$ [P,C_{n+1}X] \rightarrow
> [P,Z_nX] \rightarrow \pi_{n,P}(X)
> \rightarrow 0.$$

I have a few questions about the details and heuristics here.
Question 1: I'm confused about the extra conditions on $\mc{C}$ that might be necessary.  They claim that the objects $P\in \mc{P}$ will be h-cogroup objects; this is what in particular gives a canonical map $P \rightarrow \emptyset$, which endows $\mbox{map}(P,X)$ with a basepoint.  On the one hand, the definition of the $\mc{P}$-resolution model structure (see pages 24-26) starts out with the axiom that $\mc{P}$ be closed under suspension and desuspension, which suggests strongly that we're in a stable model category.  But then it's silly to point out that the objects of $\mc{P}$ corepresent homotopy functors valued in abelian groups, since this is true of all objects.  Even more, a stable model category is automatically pointed, which would mean that $\mc{C}/\emptyset$ is a totally vacuous thing to discuss; I'd consider this to be strong evidence that somehow $\mc{C}$ isn't meant to necessarily be stable.  Of course the final application will be for $\mc{C}$ a category of spectra so this doesn't matter, but I've been pondering this long enough that the question has become of independent interest.
Question 2: Slightly further down (3.1.3 on page 95), they claim that the case $n=0$ of the result I cited yields an isomorphism $\pi_0([P,X]) \cong \pi_{0,P}(X)$ (where $[P,X]$ is a simplicial abelian group -- "simplicial" via $[P,X]_\bullet = [P,X_\bullet]$, and "abelian group" since supposedly $P$ is an (abelian, I guess) h-cogroup).  However, unless I'm making a totally stupid mistake, I'm pretty sure that $Z_0X=C_{\Delta^0/\partial \Delta^0}X = C_*X = \emptyset$.  I don't see how to make sense of this.
Question 3 (assuming $\mc{C}$ doesn't need to be stable): How should I understand taking a pullback over the initial object?  If $\emptyset=*$, then I could consider $M_KX$ as "maps from $K$ to $X$" and $C_KX$ as "based maps from $K$ to $X$".  This is really appealing; then I could view the "presentation" as saying that $\pi_{n,P}(X)$ is just $[P,-]$ applied to "based maps $S^n \rightarrow X$ mod based maps $D^{n+1} \rightarrow X$"!   (It also makes sense that "based maps" should be adjoint to "smash product".)  But as things stand, this isn't really honest.  In fact, I don't think I've ever seen anyone take a pullback over an initial object (besides in the pointed case, of course).  I can't seem to wrap my head around what $\mbox{lim}(\emptyset \rightarrow A \leftarrow B)$ should mean, besides "the last object $C$ over $B$ such that $C \rightarrow B \rightarrow A$ is trivial" (i.e. factors through $\emptyset$, obviously).  This sounds an awful lot like the fiber of a map, but I'm not convinced that this analogy makes any sense.  I'd be grateful to hear any geometric intuition that anyone can give here.

REPLY [4 votes]: Dear Aaron,
For question 1, you're correct in that all objects are cogroup objects.  This isn't necessary for the original version from Bousfield's paper--only one of the suspensions is--and there you do need to make sure that your objects you're using to define the model structure really are cogroup objects.  (You can make use of this in the stable setting too.  Paul Goerss once pointed out to me that you can restrict your cogroup objects to high-dimensional spheres to produce Postnikov towers out of the resolution structure).
For question 2, my guess is that you're running into issues because $\Delta^0 / \partial \Delta^0$, the space obtained by taking a single point and collapsing the empty set to a single point, isn't $\ast$; by convention in homotopy theory, $X / \emptyset = X_+$.<|endoftext|>
TITLE: Judging whether a finitely presented group is a 3-manifold group?
QUESTION [9 upvotes]: Given a finitely presented group $G$, how many necessary conditions do people know for $G$ to be isomorphic to the fundamental group of some closed connected 3-manifold?  (e.g. residually finite)

REPLY [8 votes]: since Henry started the shameless self-promotion, let me also do so...
Given any group $\pi$ one can study the corresponding Alexander polynomial $\Delta_\pi$ which lies
in the group ring of $H:=H_1(\pi;\Bbb{Z})/\mbox{torsion}$. 
If $\pi$ is the fundamental group of a closed 3-manifold, then the 
Alexander polynomial $\Delta_{\pi}$ is symmetric and the one-variable specializations 
have even degree.
(see F, Kim, Kitayama: Poincaré duality and degrees of twisted Alexander polynomials)
The symmetry holds also if $\pi$ is a 3-dimensional Poincare duality group, but I am not sure whether the degree condition holds in that case.
The advantage is that this condition can be checked easily, and by checking it for finite index subgroups one gets even more necessary conditions. I would guess that in practice this is a very effective way for weeding out non 3-manifold groups.
At least it allowed me to make the right bet on Ryan's example...<|endoftext|>
TITLE: Strata of K-types appearing in irreducible representations of p-adic GL(2)
QUESTION [5 upvotes]: I'm trying to use the language of strata to organize $K$-types of irreducible smooth representations of $GL(2)$ (and then hopefully prove things). Unfortunately, I'm still new to it, so I might be making some mistakes.
My main reference is the Bushnell-Henniart book. For anyone who doesn't have access to it and wants to "play along at home", this essay contains a distillation of the important ideas.
Let $k$ be a $p$-adic field with ring of integers $\mathfrak o$, $G=GL_2(k)$, $K=GL_2(\mathfrak o)$. The principal congruence subgroup of level $\varpi^N$ (or level $N$, by abuse of notation) is $K_N=1+\varpi^N M_2(\mathfrak o)$ (and $K_0=K$). Let $\psi$ be an additive character on $k$ with conductor $\mathfrak o$ and extend it to $M_2(k)$ by composing with the trace: $\psi_M(x):=\psi({\rm tr}\ x)$. To further abuse notation, we'll suppress the subscript $M$ in $\psi_M$.
We'll say that the level of an irreducible (smooth) representation $\sigma$ of $K$ is the largest $N$ such that $\sigma$ is nontrivial on $K_N$ and trivial on $K_{N+1}$. Assume $\sigma$ has level $N\ge 1$. Since $K_N/K_{N+1}$ is a finite abelian group, $\sigma|_{K_N}$ decomposes into a direct sum of characters. Using the isomorphism $K_N/K_{N+1}\simeq M_2(\mathfrak o/\varpi)$ (given by $x\rightarrow x-1$), these characters can be written in the form $\psi_a(x):=\psi\big(a(x-1)\big)$ for some $a\in M_2(\mathfrak o/\varpi)$. For our purposes, a stratum is the level $N$ and the character $\psi_a$ on $K_N$.

What strata appear in the $K$-types of irreducible representations of $G$?

My possibly-incorrect understanding is that for a representation that can be compactly-induced from $ZK$, the fundamental stratum of the representation will appear in a $K$-type of lowest level. What happens for higher levels? And what happens for representations associated to the other chain order (than $M_2(\mathfrak o)$)?
For facts about $K$-types of irreducible representations, see Casselman's Restriction paper (this version should be more easily available), or Henniart's Appendix to this paper. For the supercuspidal case, I am aware Hansen's paper (though I don't understand it). The subtext of this paragraph is that I know (in principle) the $K$-types I'm interested in, yet I am still unable to transform this into knowledge of the corresponding strata. Whether this is due to me overlooking something simple or to a more serious issue, I do not know.

As an example, let $\chi_1$ and $\chi_2$ be characters of $k^\times$, with $\chi_1$ unramified and $\chi_2$ ramified with conductor $N_0\ge 1$ (so $\chi_2$ is nontrivial on $1+\varpi^{N_0}\mathfrak o$ but trivial on $1+\varpi^{N_0+1}\mathfrak o$). Set $\chi=\chi_1\otimes\chi_2$, and let ${\rm Ind}_P^G\chi$ be the corresponding ramified principal series. Then ${\rm Ind}_P^G\chi$ contains $K$-types $\sigma_N(\chi)$ of level $N\ge N_0$, where $\sigma_N(\chi)$ is the subrepresentation of ${\rm Ind}_{P\cap K}^K(\chi)$ of level $N$ (we aren't distinguishing between $\chi$ and its restriction to $K\cap P$).
Take $g=\bigg(\matrix{a & b\cr c & d}\bigg)\in K_N$, so that $c=u\varpi^N$, with $u\in\mathfrak o^\times$, then (for example)
$$g=\bigg(\matrix{adu^{-1}-b\varpi^N&b\cr &d}\bigg)\bigg(\matrix{1&\cr \varpi^N&1}\bigg)\bigg(\matrix{ud^{-1}&\cr &1}\bigg)$$
Thus, for $v\in \sigma_N(\chi)$ and $g=\bigg(\matrix{a & b\cr c & d}\bigg)\in K_N$
$$\sigma_{N,\chi}(g)\cdot v=\chi_2(d)\sigma_{N,\chi}\Bigg(\bigg(\matrix{1&\cr \varpi^N&1}\bigg)\bigg(\matrix{ud^{-1}&\cr &1}\bigg)\Bigg)\cdot v$$
Since $\chi_2$ has conductor $N_0$, there exists $a_2$ such that $\chi_2(d)=\psi\big(a_2(d-1)\big)$.
When $d\in 1+\varpi^N$ for $N>N_0$, the character will be trivial. I'm happy with this, but I don't understand how to get the rest of the calculation to work out, though I feel it should be a straight-forward exercise.

REPLY [3 votes]: I will try to answer the question as far as I have understood it. Please comment.
Clifford's theorem
Let $G$ be a finite group. Let $H$ be a normal subgroup of $G$. The groups $G$ resp. $G/H$ act on the irreducible representations of $H$ via conjugation on $H$.

Let $\rho$ be an irreducible representation of $G$, then the restriction ( \Res_{H} \rho ) contains precisely one $G$-orbit of irreducible representations of $H$.
Let $\psi$ be an irreducible representation of $H$, and let $G_\psi$ be its stabilizer in $G$. We have a one-to-one correspondence between irreducible representations $\rho_0$ of $G_\psi$, contained in $Ind_{H}^{G_\psi} \psi$, and irreducible representations $\rho$ of $G$, contained in $Ind_{G_\psi}^{G} \psi$. The correspondence is given by
$$\rho_0 \mapsto \rho = Ind_{G_\psi}^{G} \rho_0.$$

We want to apply Clifford theory to the irreducible representations of $GL_2(o)$ of level $n \geq 1$, we follow Stasinski. The quotient $\Gamma(p^{n}) / \Gamma(p^{n+1})$ is abelian and isomorphic to the endomorphism ring $M_2(F_q)$ of $F_q \oplus F_q$ via
$$ \iota_n \colon \Gamma(p^{n}) / \Gamma(p^{n+1}) \xrightarrow\cong M_2(F_q), \qquad x \mapsto (x-1)/\pi^{n} \bmod p.$$
According to Clifford theory, the restriction
$$ Res_{\Gamma(p^{n})} \rho $$
decomposes into a $GL_2(o)$-orbit of one-dimensional representations 
$$\psi: \Gamma(p^n) \rightarrow \mathbb{C}, \qquad \psi|_{\Gamma(p^{n+1})} = 1.$$
The Pontryagin dual $\widehat{M_2(F_q)}$ is canonically isomorphic to $M_2(F_q) = \Gamma(p^{n}) / \Gamma(p^{n+1})$ via
$$     M_2(F_q)  \xrightarrow{\cong}    \widehat{M_2(F_q)}, \qquad x \mapsto \psi_x, \qquad \psi_x(y) := \psi_{p} \circ tr \left( x \cdotp \right).$$
The orbit space of the conjugation action of $GL_2(o)$ on  $\widehat{ \Gamma(p^{n-1}) / \Gamma(p^n)} $ is isomorphic to the orbit space of the action of $GL_2(F_q)$ on the ring $M_2(F_q)$ by conjugation with inverse elements.

Via induction and restriction, Frobenius reciprocity, you can figure out: Every strata with $\mathfrak{e}=1$ gives you a $GL_2(o)$-type. Every $GL_2(o)$-type occurs in the restriction of some unitarizabile $GL_2(F)$-representation.

This is a very standard fact. Perhaps use that $GL_2(F)$ unitary representation are all admissible, and that the compact induction is unitarizabile together with Frobenius reciprocity, and general decomposition theorems for unitarizabile representations into irreducibles. 
For the normalizer of the Iwahori subgroup, the relation betwenn $K$-type and strata is somewhat more difficult, but follows the same principle (Clifford theory).
Silberger has considered the story for both in odd residue characteristic (the oddness assumption seems irrelevant). He is also a little more detailed about the $K$-types regarding the last section of your question than Casselman.
Usually everything you want to show can be done via Frobenius reciprocity, the Mackey Induction-Restriction formula, Clifford theory, and coset decomposition (Bruhat over the residue field/Iwahori decomposition), of course exploited in the right fashion.
So regarding the last section, you only want to compute $Res_{\Gamma(p^{N-1})} Ind_{\Gamma_0(p^N)}^K \chi$ via Mackey IndRes formula, and you will get the conjugacy class of strata (via Clifford theory). The necessary coset space is computed via Bruhat and Iwahori decompositions.<|endoftext|>
TITLE: Injective objects in Mor(Ab)
QUESTION [8 upvotes]: Consider the abelian (Grothendieck) category $\mathcal{C} := \mathrm{Fun}(\{0<1\},\mathrm{Ab}) = \mathrm{Mor}(\mathrm{Ab})$. Objects are morphisms $(A \to B)$ of abelian groups, morphisms are commutative diagrams. Equivalently, this is the category of abelian sheaves on the Sierpinski space.
Question. How do injective objects in $\mathcal{C}$ look like?
Since injective sheaves are stable under restriction (use extension by zero), clearly $(A \to B)$ injective implies that $A$ is injective. But is this sufficient (probably not)? When $A,B$ are injective, is the same true for $(A \to B)$?

REPLY [3 votes]: I will use notation $A_0 \to A_1$ for objects of $\mathrm{Mor}(\mathrm{Ab})$.
EDIT: previously I claimed something stronger (that I can produce lifting properties in the functor category without factorizations), but I am not so sure about it.
The following is a lot more general than necessary, but I think this added generality is also useful. Let $(\mathcal{L}, \mathcal{R})$ be a weak factorization system in a category $\mathcal{C}$ with enough colimits and limits for the following to make sense. Let $J$ be a Reedy category. Then in the functor category $\mathcal{C}^J$ the "Reedy $\mathcal{L}$-cofibrations" and "Reedy $\mathcal{R}$-fibrations" form a weak factorization system. By "Reedy $\mathcal{L}$-cofibrations" I mean morphisms of diagrams $X \to Y$ such that for every $j \in J$ the latching morphism $X_j \sqcup_{L_j X} L_j Y \to Y_j$ is in $\mathcal{L}$ and dually "Reedy $\mathcal{R}$-fibrations" are morphisms $X \to Y$ such that for every $j \in J$ the matching morphism $X_j \to M_j X \times_{M_j Y} Y_j$ is in $\mathcal{R}$. The proof is exactly as in the construction of the Reedy model structures and can be found for example in Hovey's Model Categories.
Now we take $\mathcal{C} = \mathrm{Ab}$, $\mathcal{L} = $ monomorphisms and $J = [1]$. Then $\mathcal{R}$ are split epimorphisms with injective kernel. The lifting properties are easily verified while the factorizations use the fact that there are enough injectives in $\mathrm{Ab}$. If $f : A \to B$ is a map in $\mathrm{Ab}$, pick an injective hull $i : A \to \hat A$, then $f$ factors as an injection $[i, f] : A \to \hat A \oplus B$ followed by a split surjection with injective kernel $\hat A \oplus B \to B$. We consider $J$ as a Reedy category where $0$ has degree $1$ and $1$ has degree $0$. Then "Reedy $\mathcal{L}$-cofibrations" are monomorphisms again, so an object $X$ is injective if and only if the map $X \to 0$ is a "Reedy $\mathcal{R}$-fibration" i.e. when both $X_1 \to 0$ and $X_0 \to X_1$ are split epimorphisms with injective kernel i.e. when $X_0 \to X_1$ is a split epimorphism with injective source.<|endoftext|>
TITLE: What time does it take for irrational rotations to hit an interval?
QUESTION [13 upvotes]: Hi,
Consider $\theta_n = (\theta_0 + n \theta) \mod 1$, $\theta$ being an irrational number, and $\theta_0$ an uniform random variable in $(0,1)$. Is there any estimates for the time it will take this process to hit $(0,\alpha)$ ? From the ergodic theorem I know that, if I denote $N(n)$ the number of times $\theta_n \in (0,\alpha)$, then $N(n)/n \to \alpha$. What I want to know is how much time it will take for this limit to be attained.
Another way of framing this question is : is there any "central limit theorem" (or weakening thereof ; I'm mainly interested in guaranteed bounds for $P(N\geq 1)$) for ergodic processes? From what I've read, there is no general answer to this for a generic ergodic process and function f. There are some results that depend on $f$ being smooth, which it isn't here.
The same question was asked on Quantitative versions of ergodic theorem, but I haven't found anything there that relates to my question.

REPLY [10 votes]: There is a theorem of Kesten, which roughly says, that if you take $(\theta, \theta_0)$ random, and the number of times you hit $(0, \alpha)$ in the first $N$ iterations, subtract the expected $N \times \alpha$, and normalize by $\rho \times ln(n)$, the result will converge to Cauchy distribution. This can be viewed as an analogue of CLT in this case.<|endoftext|>
TITLE: A conjecture in Number Theory
QUESTION [9 upvotes]: Hi all.
I've had this idea - a conjecture in the field of Number Theory - for a few years now.
The conjecture is rather simple, as were the logical steps that I made in order to infer it, so I would have assumed that it had already been suggested in the past. Nevertheless, I have not been able to find any piece of evidence that it had (possibly due to the difficulty of "phrasing" it into Google's search engine).
I would appreciate your opinion of the following:

Are you familiar with this theorem in any way?
Is it eligible to be stated as an open conjecture in Number Theory?

My conjecture can be stated in any of the following ways:

No       set S ⊂ { (3n+2)/(2n+1) │ n∈N } exists such that ∏Si is a power of 2
No       set S ⊂ { (3n+2)/(2n+1) │ n∈N } exists such that ∏Si is integer
No multi-set S ⊂ { (3n+2)/(2n+1) │ n∈N } exists such that ∏Si is integer

Reminder:

In a set, no element can appear more than once
In a multi-set, any element may appear more than once

In simple words:
Take any group of numbers from the series {5/3, 8/5, 11/7, 14/9, 17/11, 20/13, ...}.
Calculate the product (multiplication) of these numbers - the result will never be an integer number.
Note:
In its weakest form (#1), my conjecture is sufficient for proving that there are no cyclic sequences in the '3n+1' conjecture (the proof for that is pretty simple, but I am not including it here because it is not the main purpose of my question). I believe that my conjecture also holds in its strongest form (#3).
Thank you very much for your time.

REPLY [39 votes]: Even conjecture 1. is false: For $S=\{(3n+2)/(2n+1)\;|\;n=9, 12, 14, 27, 41\}$ your product is $8$.<|endoftext|>
TITLE: Does a power series converging everywhere on its circle of convergence define a continuous function?
QUESTION [73 upvotes]: Consider a complex power series $\sum a_n z^n \in \mathbb C[[z]]$ with radius of convergence $0\lt r\lt\infty$ and suppose that for every $w$ with $\mid w\mid =r$ the series $\sum a_n w^n $ converges .
We thus obtain a complex-valued function $f$ defined on the closed disk $\mid z\mid \leq r$ by the formula $f(z)=\sum a_n z^n$.  
My question: is $f$ continuous ?  
This is a naïve question which looks like it should be answered in any book on complex analysis.
But I checked quite a few books, among which the great treatises : Behnke-Sommer, Berenstein-Gay, Knopp, Krantz, Lang, Remmert, Rudin, Stein-Shakarchi, Titchmarsh, ... .
I couldn't find the answer, and yet  I feel confident  that it was known in the beginning of the twentieth century.  
Edit
Many thanks to Julien who has answered my question: Sierpinski proved (in 1916) that there exists such a power series  $\sum a_n z^n    $ with radius of convergence $r=1$ and associated function  $f(z)=\sum a_n z^n $  not  bounded on the closed unit disk and thus certainly not continuous. 
It is strange that not a single book on complex functions seems to ever have mentioned this example.   
On the negative side, I must confess that I don't understand Sierpinski's article at all!
  He airdrops a very complicated, weird-looking power series and proves that it has the required properties in a sequence of elementary but completely obscure computations.
I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing.

REPLY [43 votes]: This answer is in response to final sentence, "I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing". In fact, it is easy to construct power series converging on the circle of convergence, but are unbounded. For example,
$$
f(z)=\sum_{n=1}^\infty\frac1{n^5(1+in^{-3}-z)}
$$
defines a function whose power series expansion has radius of convergence 1 and converges everywhere on the unit circle, but is unbounded in a neighbourhood of 1.
A method of constructing such functions is as an infinite sum
$$
f(z)=\sum_{n=1}^\infty f_n(z).
$$
Here, $f_n(z)$ are chosen to have a power series expansion converging everywhere on the closed unit ball. Let $f^{(r)}_n(z)$ denote the sum of the first $r$ terms in the power series expansion of $f_n$. We need to arrange it so that $f^{(r)}(z)\equiv\sum_nf_n^{(r)}(z)$ converges on the closed unit ball, and that $f(z)=\lim_{r\to\infty}f^{(r)}(z)$ holds. That is, we need to be able to commute the limit $r\to\infty$ with the summation over $n$. A sufficient condition to be able to do this is that $\sum_n\sup_r\lvert f^{(r)}_n(z)\rvert < \infty$, for all $\lvert z\rvert\le1$. That this allows us to commute the summation with the limit is just a special case of dominated convergence.
Next, to ensure that $f(z)$ is unbounded on the unit ball, we want to choose $f_n$ such that there exists $q_n$ in the closed unit ball with $f_n(q_n)$ large, and such that it does not get cancelled out in the summation, so that $f(q_n)$ is large and diverges as $n\to\infty$.
For example, choose positive reals $\delta_n,\epsilon_n$ tending to zero, and setting $a_n=1+i\epsilon_n$, and
$$
f_n(z)=\frac{\delta_n}{a_n-z}=\sum_{m=0}^\infty \delta_na_n^{-m-1}z^m.
$$
These are all well-defined as power series with radius of convergence greater than 1. Furthermore, the partial sums are
$$
f^{(r)}_n(z)=\delta_n\frac{1-(z/a_n)^r}{a_n-z},
$$
which are bounded by $2\delta_n/\lvert a_n-z\rvert$. As $a_n\to1$, this is bounded by a multiple of $\delta_n$ for each fixed $z\not=1$, so the dominated convergence condition is satisfied when $\sum_n\delta_n$ is finite. On the other hand, if $z=1$, then $\lvert a_n-z\rvert=\epsilon_n$, so the dominated convergence condition is satisfied everywhere whenever $\sum_n\delta_n/\epsilon_n$ is finite.
Next, $f_n(z)$ achieves its largest value on the unit ball at $q_n=a_n/\lvert a_n\rvert$, and its real part there is given by
$$
\Re f_n(q_n)=\frac{\delta_n}{\sqrt{1+\epsilon_n^2}(\sqrt{1+\epsilon_n^2}-1)}\ge\frac{2\delta_n}{\epsilon_n^2\sqrt{1+\epsilon_n^2}}.
$$
As $f_m(z)$ has positive real part for all $m$, this bound also holds for $f(q_n)$, and we get that $f$ is unbounded whenever $\delta_n/\epsilon_n^2\to\infty$. These conditions are satisfied by taking $\epsilon_n=n^{-3}$ and $\delta_n=n^{-5}$.
Alternatively, for an example closer to Sierpinski's, consider choosing a sequence $a_n\to1$ on the unit circle and positive reals $K_n$, and set
$$
f_n(z)=K_n2^{-n}\sum_{k=0}^{2^n-1}a_n^{2^n-1-k}z^k=2^{-n}K_n\frac{a_n^{2^n}-z^{2^n}}{a_n-z}.
$$
The partial sums of the power series expansion of $f_n(z)$ are bounded by $2^{1-n}K_n/\lvert a_n-z\rvert$, so the dominated convergence condition is satisfied for $z\not=1$ so long as $\sum_n2^{1-n}K_n$ is finite. Sierpinski chooses $a_n=(n^2-1+2ni)/(n^2+1)$ so that $a_n-1$ goes to zero at rate $1/n$. The dominated convergence condition is therefore satisfied whenever $\sum_n2^{-n}K_nn$ is finite.
Now, $f_n(z)$ is maximized at $z=a_n$ where $\lvert f_n(a_n)\rvert=K_n$. So,
$$
\lvert f(a_n)\rvert\ge K_n-\sum_{m\not=n}\frac{2^{1-m}K_m}{\lvert a_m-a_n\rvert}.
$$
As $a_m-a_n$ is bounded below by a multiple of $1/m^2$, the summation on the right is bounded whenever $\sum_m2^{-m}K_mm^2$ is finite, and $f(a_n)$ is unbounded if we also take $K_n$ going to infinity. Sierpinski takes $K_n=n^2$ here. Finally, in Sierpinski's example, he multiplies $f_n$ by $z^{2^n}$. This changes nothing, except to separate out the non-zero terms of the power series of $f_n(z)$, so that the power series of $f(z)$ can be written easily term by term.<|endoftext|>
TITLE: Heegner Points and Binary Quadratic Forms
QUESTION [8 upvotes]: I've been trying to read Gross' paper on Heegner points on $X_0(N)$ and I am stuck on a few details. The definition he is working with is that a heegner points is a pair $y=(E,E')$, where $E$ and $E'$ are elliptic curves admitting an isogeny that has cyclic kernel of order $N$ and where $E$ and $E'$ both have complex multiplication by the order $\mathcal{O}$ of discriminant $D$ in a quadratic imaginary field $K$. Gross goes on to explain that we may assume the lattice for $E$ is a fractional ideal $\mathfrak{a}$ and the lattice for $E'$ is $\mathfrak{b}$ such that the ideal $\mathfrak{n}=\mathfrak{a}\mathfrak{b}^{-1}$ is proper ideal of $\mathcal{O}$ such that the quotient $\mathcal{O}/\mathfrak{n}$ is cyclic of order $N$. It is the next line that I don't understand:
"Such an ideal will exist if and only if there is a primitive binary quadratic form of discriminant $D$ which properly represents $N$...". The line goes on, but this is one of the things I'm stuck on. I've tried googling some notes/papers on binary quadratic forms, but I can't find anything that helps me understand what a binary quadratic form representing $N$ has to say about an order admitting a cyclic quotient. An explanation or a good reference would be much appreciated. 
The second and, I think, more important part of my confusion is a bit later on in the same section: Gross goes on to explain that if we have such an $\mathfrak{n}$, we can construct a heegner point as follows. Let $\mathfrak{a}$ be an invertible $\mathcal{O}$-submodule of $K$ and let $[\mathfrak{a}]$ denotes its class in $Pic(\mathcal{O})$. Let $\mathfrak{n}$ be a proper $\mathcal{O}$-ideal with cyclic quotient of order $N$, put $E=\mathbf{C}/\mathfrak{a}$, $E'=\mathbf{C}/\mathfrak{a}\mathfrak{n}^{-1}$. They are related by an obvious isogeny and thus determine a Heegner point, denoted $(\mathcal{O},\mathfrak{n},[\mathfrak{a}])$. 
Next, given $y=(\mathcal{O},\mathfrak{n},[\mathfrak{a}])$, we can find the image of it in the upper-half plane by picking an oriented basis $\langle\omega_1,\omega_2\rangle$ of $\mathfrak{a}$ such that $\mathfrak{a}\mathfrak{n}^{-1}=\langle\omega_1,\omega_2/N\rangle$. Then $y$ corresponds to the orbit of $\omega_1/\omega_2$ under $\Gamma_0(N)$. Lastly, since $\tau\in K$ it follows that it satisfies $A\tau^2+b\tau+C=0$ for some integers $A,B,C$ such that $gcd(A,B,C)=1$. 
Finally, what I don't understand is that Gross claims that $D=B^2-4AC, A=NA'$ from some $A'$ and $gcd(A',B,NC)$. I don't see what the $\tau$ we cooked up has to do with the discriminant of our order. I have read a paper that defined a Heegner point to be a quadratic imaginary point in the half-plane such that $\Delta(\tau)=\Delta(N\tau)$. I have seen how this would help with part of the claim above, but I don't see why in this situation, $\Delta(\tau)=\Delta(N\tau)$. In fact, it seems that everything I'm confused about here is the fact that it seems to be the case that $$D=\Delta(\tau)=\Delta(NT),$$ where $\Delta$ denotes discriminant.
Any insight into these two questions would very appreciated.

REPLY [6 votes]: I think the following facts, which you can find in Cox's book Primes of the Form $x^2+ny^2$, will alleviate your confusion. First off, if ${\mathfrak a}=[\alpha,\beta]$ is a proper ideal of ${\mathcal O}$ then one can show that
$$ f(x,y) := \frac{N(\alpha x-\beta y)}{N{\mathfrak a}} $$
is a primitive binary quadratic form of discriminant $D = {\rm disc}(\mathcal O)$. Moreover, the map that associates such an ${\mathfrak a}$ to such an $f(x,y)$ induces an isomorphism from the class group ${\rm Pic } ({\mathcal O})$ onto the form class group $C(D)$. The inverse of this map is given by
$$ f(x,y) := ax^2+bxy+cy^2 \mapsto [a,(-b+\sqrt{D})/2] = [a,a\tau], $$
where $\tau$ is the unique point in the upper-half plane such that $f(\tau,1)=0$. It's not hard to show that we'll have ${\mathcal O} = [1,a\tau]$ for all such $\tau$ (see Addendum below). In particular, we see that ${\mathcal O}/[a,a\tau] \cong {\mathbb Z}/a{\mathbb Z}$ is cyclic.
The last piece of the puzzle is this: a positive integer $N$ is represented by a form $f(x,y)$ in $C(D)$ if and only if $N$ is the norm of some ideal in the corresponding ideal class in ${\rm Pic}({\mathcal O})$ (loc. cit., Theorem 7.7(iii)). On the other hand, $N$ is properly represented by such an $f(x,y)$ if and only if $f(x,y)$ is properly equivalent to $Nx^2+bxy+cy^2$ for some $b,c \in {\mathbb Z}$. Now the results mentioned in the preceding paragraph will take you home.
Addendum: Given a proper ideal $\mathfrak a$ of $\mathcal O$, we can recover $\mathcal O$ as the set ${\mathfrak a}^\vee = \{x \in K \mid x\mathfrak a \subset \mathfrak a \}$. This last set is easy to compute in the following special case. Let $K=\mathbb Q(\tau)$ be quadratic and suppose that $ax^2+bx+c$ is the minimal polynomial of $\tau$, where $a,b,c$ are coprime integers. Then $[1,\tau]^\vee = [1,a\tau]$ (loc. cit., Lemma 7.5).
By applying this to Gross's $\mathfrak a = [\omega_1, \omega_2] = \omega_2 [\tau, 1]$, which is a proper ideal in some order $\mathcal O$, we find that $\mathcal O = [A\tau, 1]$. Consequently,$$D = {\rm disc}({\mathcal O}) = \det \begin{pmatrix}1 & A\tau \\ 1 & A\bar{\tau} \end{pmatrix}^2 = B^2 - 4AC.$$
The assertion about $A$ can be gotten in a similar manner.<|endoftext|>
TITLE: Do all curves have Néron models
QUESTION [31 upvotes]: Let $X$ be a smooth projective geometrically connected curve over a number field $K$. Assume that $g\geq 2$.
Does there exist a Néron model $\mathcal X$ for $X$ over $O_K$?
By a Néron model, I mean a smooth model (not necessarily proper) with the "Néron universal property": for any smooth $O_K$-scheme $\mathcal Y$,
$$\mathrm{Hom}(\mathcal Y, \mathcal X) = X(\mathcal Y_K).$$
Is the smooth locus of the minimal regular model of $X$ over $O_K$ a Néron model? Does base change help? That is, does there exist a Néron model for $X_L$ after some suitable base change $L/K$?
Note that if $\mathcal{X}$ is the smooth locus of the minimal regular model of $X$ over $O_K$, we have the "Néron" property $\mathcal{X}(O_K) = X(K)$. (In fact, the image of a section in the minimal regular model lies in the smooth locus.)
This question was asked on stackexchange four months ago:
https://math.stackexchange.com/questions/153369/do-neron-models-of-hyperbolic-curves-exist

REPLY [29 votes]: Here are some observations. I include the case $g=1$ (even if $X$ has no rational point). 
Denote by $\hat{\mathcal X}$ the (proper) minimal regular model of $X$ over the $O_K$ and let $\mathcal X$ be the smooth locus of $\hat{\mathcal X}$. 

(1) If the Néron model exists, it is equal to the smooth locus $\mathcal X$ of the minimal regular model. 
(2) If the fibers of $\mathcal X$ over $O_K$ have no rational irreducible component (e.g. if $X$ has good reduction), then $\mathcal X$ is the Néron model of $X$. 
(3) (localization) If Néron models exist over DVRs, then they exist over any Dedekind domain. 
(4) (base change) Let $R$ be a DVR. Let $R'/R$ be an extension of DVR such that an uniformizing element of $R$ is also an uniformizing element of $R'$ and such that the residue extension is separable (e.g. $R'$ can be the completion of a strict henselization of $R$). If the Néron model exists over $R'$, then the Néron model exists over $R$. 
(5) You were right to not include the case $g=0$. The projective line doesn't have Néron model. 
(6) Let $Y$ be a smooth scheme over a noetherian regular scheme $S$, let $Z$ be a regular scheme, flat and of finite type over $S$ and let $f: Y\to Z$ be a morphism. Then $f(Y)$ is contained in the smooth locus of $Z/S$. In particular, the canonical map ${\mathcal X}'(O_K)\to X(K)$ is bijective if $\mathcal X'$ is the smooth locus of (any) proper regular model of $X$. 
(7) If $g=1$, then $\mathcal X$ is the Néron model of $X$. 

Proof. Sorry I can't give all details by lack of energy and because it would be pretty unreadable in MO. 
(1) Let $\mathcal N$ be the Néron model. Embedd it in a proper flat model, solve its singularity without touching to the regular locus (which contains $\mathcal N$). Then we get a proper regular model $\hat{\mathcal N}$ containing $\mathcal N$ as an open subset. The identity on $X$ extends to morphism $\hat{\mathcal N}\to \hat{\mathcal X}$. By (6), this morphism induces a morphism $\mathcal N\to \mathcal X$. 
Then $\mathcal X$ satisfies the universal Néron mapping property. By the uniqueness of Néron model, we get $\mathcal N\simeq \mathcal X$. 
(2) Let $\mathcal Y -\to \mathcal X$ be a rational map defined over $K$ with $\mathcal Y$ smooth (regular is enough). The projection $p: \Gamma\to \mathcal Y$ is birational. Let $y\in Y$ and suppose $\Gamma_p$ is not finite. By a theorem of Abhyankar (the base scheme is excellent here, otherwise, localize and pass to the completion and use (4)), the components $E$ of $\Gamma_p$ are uniruled. But $E\to \mathcal X$ is a closed immersion, so $E$ is a rational curve in a close fiber of $\mathcal X$. Contradiction. Thus $p$ is quasi-finite biratonal and surjective. As $\mathcal Y$ is normal, $p$ is an isomorphism by Zariski's Main Theorem and the rational map we consider is actually defined everywhere. So $\mathcal X$ is the Néron model. 
(3) The curve $X$ has good reduction away from finite many places. Using (2) for good reduction places and by gluing with Néron models over bad reduction places, we get a global Néron model over $O_K$. 
(4) First the formation of the minimal regular model (and its smooth locus) is compatible with such base change. So if $\mathcal X\otimes R'$ satisfies the universal Néron mapping property over $R'$, then so does $\mathcal X$ over $R$ by faithfully flat descent for the definition domain of rational maps. 
(5) Fix a model $\mathbb P^1_{O_K}$ of $\mathbb P^1_K$. They are plenty of endomorphisms of the generic fiber which don't extend to $\mathbb P^1_{O_K}$ (e.g. $[x,y]\mapsto [x, py]$). This shows that $\mathbb P^1_K$ has no proper smooth Néron model. The general case can be proved similarly with some extra works. 
(6) Sketch: Consider $Y\times_S Z\to Y$. It is enough to show that its sections have images in the smooth locus (over $Y$), then use descent of smoothness (easy). The left hand side is regular because it is smooth over the regular scheme $Z$, and the right hand side si regular because it is smooth over the regular scheme $S$. So we can reduced to the case of flat morphism of finite type $W\to Y$ between two regular schemes. Let $y\in Y$ and let $w\in W$ be its image by a section $Y\to W$. Then 
$ O_{W,w}\to O_{Y,y}$ is a surjective map of regular local rings. Its kernel is generated by {$t_1, \dots, t_d$}, a part of a system of coordinates of $O_{W,w}$. So the maximal ideal $m_w$ of $O_{W,w}$ is generated by $t_1, \dots, t_d$ and $m_y$. Thus the maximal ideal of $O_{W_y, w}$ is generated by the images of $t_1, \dots, t_d$. The flatness of $W\to Y$ implies that $W_y$ has dimension $d$ at $w$. So $W_y$ is regular at $w$. It is in fact smooth because $w$ is a rational point of $W_y/k(y)$. 
Another proof is to use $\Omega^1_{W/Y}$ and the fact that the image of a section is locally complete intersection. 
Application: if $\hat{\mathcal X'}$ is a proper regular model over $O_K$, by the valuative criterion, $\hat{\mathcal X'}\to X(K)$ is bijective. But we just saw that the LHS is $\mathcal X'(O_K)$. 
(7) We can work over a DVR $R$. If there exists a smooth $R$-scheme $\mathcal Y$ with non-empty special fiber and a morphism $\mathcal Y_K\to X$, then $\mathcal Y_K$ has a point in an étale extension of $R$. So $X$ has 
a point in an étale extension. By (4), we can thus suppose $X(K)\ne\emptyset$. So it is an elliptic curve, and Néron showed that $\mathcal X$ is the Néron model. If such $\mathcal Y$ doesn't exists, then $\mathcal X$ trivially satisfies the Néron mapping property.<|endoftext|>
TITLE: A question about the comparability of large cardinals.
QUESTION [7 upvotes]: Are there any examples of two large cardinal axioms $AX$ and $AY$, in the language of first order $ZFC$, which satisfy the following conditions.

Each of them defines a unique cardinal number - $C(AX)$ for $AX$
and $C(AY)$ for $AY$ - not like the axiom of measurable cardinals which defines a whole collection of cardinal numbers. 
If $T$ denotes the theory $ZFC+AX+AY$, then $T$ has not yet been proved inconsistent. 
$T$ proves that each of $C(AX)$ and $C(AY)$ is larger that the smallest strongly inaccessible cardinal number. 
The sentences of $T$ stating that $C(AX) < C(AY)$ and that $C(AY) < C(AX)$ are each consistent with $T$, if $T$ is consistent.

REPLY [14 votes]: Let AX = "there exists a least strongly compact cardinal", and let AY be "there exists a least whatever cardinal" (there are many possibilities for "whatever", e.g., "inaccessible  with a measurable below it" if you want to be modest).  
C(AX) = the least strongly compact, C(AY) = the least whatever, say the least inaccessible past the least measurable.
It is well known (Magidor) that the least strongly compact can be the least supercompact or the least measurable. In the first case, C(AX) is much larger than C(AY), in the second case, slightly smaller. 
(Arthur Apter has written several papers about this "identity crisis of strongly compacts".  Many more examples in the same spirit can be obtained from his results.)<|endoftext|>
TITLE: Sylow theorems for infinite groups 
QUESTION [8 upvotes]: Are there classes of infinite groups that admit Sylow subgroups and where the Sylow theorems  are valid ? 
More precisely, I'm looking for classes of groups $\mathcal{C}$ with the following properties: 

$\mathcal{C}$ includes the finite groups 
in $\mathcal{C}$ there is a notion of Sylow subgroups that coincides with the usual one when restricted to finite groups 
Sylow's theorems (or part of them) are valid in $\mathcal{C}$ 

An example of such a class $\mathcal{C}$ is given by the class of profinite groups.

REPLY [3 votes]: The best reference for this subject is the book of martyn Dixon: Locally finite groups and Sylow theory.<|endoftext|>
TITLE: Identity involving partitions coming from representations of alternating groups
QUESTION [5 upvotes]: It is not difficult to show that the number of conjugacy classes in the alternating group $A_n$ is given by

classes in the alternating group = no. of even partitions + no. of self-transpose partitions

Note that a partition is even if it is the cycle decomposition of an even permutation.
Likewise, using some Clifford theory and the representation theory of $S_n$, one can show that the number of irreducible representations of $A_n$ is given by

irreps. of $A_n$ = $\frac 12$(no. of non-self transpose partitions) + no. of self-transpose partitions

Equating these leads to the identity:

$2\times $ no. of even partitions - no. of self-transpose partitions  = no. of partitions

In his book Representations of Finite Groups, Musili refers to this as a bizarre identity.

Question. Is there a proof of this identity which does not use the representation theory of alternating groups? Better still, is there a bijective proof?

REPLY [2 votes]: This question got answered by Gjergji Zaimi and Richard Stanley in the comments. I simply reproduce their comments here as an answer:
A very simple explanation for this identity comes from the theory of symmetric functions. The ring $\Lambda$ of symmetric functions in infinitely many variables comes with an involution $\omega$, which interchanges the complete symmetric function $h_\lambda$ with the elementary symmetric function $e_\lambda$ for each partition $\lambda$.
Comparing the answers obtained for the trace of $\omega$ on homogeneous symmetric functions of degree $n$ using Schur functions and power sum symmetric functions yields the identity in question, for $\omega(s_\lambda)=s_{\lambda'}$ (giving trace as the number of self-transpose partitions) and $\omega(p_\lambda)=\epsilon(\lambda)p_\lambda$, where $\epsilon(\lambda)$ is the sign of a permutation with cycle decomposition $\lambda$ (giving trace as number of even partitions minus number of odd partitions).
Proofs of these facts concerning symmetric functions can be found in Stanley's Enumerative Combinatorics 2 (Sections 7.7 and 7.14).
A bijective proof for this identity was given by Marc van Leeuwen to the same question on  https://math.stackexchange.com/a/102293/10126; he constructs an explicit bijection between the sets of even and odd partitions which do not have distinct odd parts. Also, there is a fairly standard bijection between partitions with distinct odd parts and self-transpose partitions.<|endoftext|>
TITLE: Characterization of cocompact group action
QUESTION [15 upvotes]: Wikipedia claims the following:

In mathematics, an action of a group G on a topological space X is cocompact if the quotient space X/G is a compact space or, equivalently, if there is a compact subset K of X such that 
  the image of K under the action of G covers X.

My question is: Isn't this wrong? It is evident that the existence of such a subset K ensures cocompactness, but I am doubting the other direction. How could one possibly choose K? Taking an arbitrary transversal (or its closure) does not work, and I do not see what else could be a candidate.
By the way: Wikipedia points to a specific page in the Handbook of Geometric Topology. This page, however, contains only the definition of a cocompact space, not the claimed equivalence.

REPLY [9 votes]: Here is a counterexample in which $X/G$ is also Hausdorff.
We give $X:=(\mathbb{N} \times \mathbb{Z}) \cup \{\infty\}$ a topology similar to the Arens-Fort topology. That is, each point $(m,n) \in \mathbb{N} \times \mathbb{Z}$ is isolated and basic neighborhoods of $\infty$ are of the form $$B_{f,k}:=\{\infty\} \cup \{(m,n) : n \geq f(m) \land m \geq k\},$$ where $f:\mathbb{N} \to \mathbb{Z}$ and $k \in \mathbb{N}$. 
We let $G:=\mathbb{Z}$ act on $X$ in the natural way: $g(m,n)=(m,n+g)$ and $g(\infty)=\infty$.
Any compact subset of $X$ is finite, so its orbit cannot cover $X$. On the other hand $X/G$ is just a convergent sequence (which is compact and Hausdorff).<|endoftext|>
TITLE: How does the lack of partitions of unity affect the structure of analytic/holomorphic manifolds?
QUESTION [9 upvotes]: The standard way to define integration on a smooth manifold is to use partitions of unity, to extend to the case where the form you're integrating isn't supported on just one coordinate patch. Of course, in the analytic/holomorphic case, we don't have partitions of unity. So how do we do integration?
Furthermore, how does this affect the space $\mathcal{T}(M)$  of analytic vector fields (analytic global sections of the tangent bundle)? The usual extension lemma for smooth sections of a vector bundle depends on partitions of unity, so there doesn't seem to be any reason you should always be able to find an nonzero analytic section. Does it ever happen that $\mathcal{T}(M) = 0$?
Also in that vein - I remember needing to use partitions of unity to prove in an exercise that the space of 1-forms $\mathcal{T}^*(M)$ is actually the dual $C^\infty(M)$ module $\, \text{Hom}(\mathcal{T}(M),C^\infty(M))$ - because given a map $\mathcal{T}(M) \to C^\infty(M)$, it seems like you need some kind of extension lemma to construct a 1-form that induces it. Does this fail in the analytic case?
I imagine the answer to the previous two questions will involve sheaves, but I don't quite know enough about sheaves to frame them in the appropriate sheaf-theoretic way. Maybe per this question, I should ask if this means the sheaves are not soft and the sheaf cohomology doesn't vanish?
And in general, any good references (or just explanations you care to give) on the major differences between the smooth and analytic cases? It seems like most differential geometry books pay almost no attention to the analytic case.

REPLY [5 votes]: (1) A holomorphic manifold is also (or "can also be viewed as") a smooth manifold, and that lets you define integration.  To put it another way, you do have partitions of unity, just not holomorphic ones.  
(2) Even before you get to tangent bundles, there are well-known cases where local things can't be patched globally.  For example, on the Riemann sphere, there are non-trivial holomorphic functions in neighborhoods of every point, but the only global holomorphic functions are the constants.  In the case of the tangent bundle, it seems to me (experts please edit if I mess this up) that there are no non-zero global tangent vector fields on Riemann surfaces of genus 2 or more, though of course there are such fields locally everywhere.  
(3) You're right that this phenomenon (and related ones) are the beginning of sheaf cohomology.  
(4) I won't try to answer your question about 1-forms being the dual of tangent vectors, since it seems to mix pointwise things (the space of 1-forms $T^*(X)$) with global things in a way that I don't understand.<|endoftext|>
TITLE: Brauer group elements of order $2$
QUESTION [6 upvotes]: Let $K$ be a field and let $Q$ be a quaternion algebra over $K$. Then it is well-known that the class $[Q]$ of $Q$ in $Br(K)$ has order $2$. One can show this by constructing an explicit isomorphism $Q \otimes_K Q \cong M_2(K)$. My question is about the converse. 

Does there exist a field $K$ and a division algebra $D$ over $K$ such that the class $[D]$ of $D$ in $Br(K)$ has order $2$, but such that $D$ is not isomorphic to a quaternion algebra over $K$?

If such a $K$ and $D$ exist, then it would also be nice to see an explicit example. 
As a non-example, I believe that it follows from local and global class field theory that if $K$ is a local or global field, then every element of order $2$ in $Br(K)$ may indeed be represented by a quaternion algebra.

REPLY [4 votes]: I think what you're after is, in modern parlance, an algebra of period 2 but index strictly greater than 2.  Googling "period-index problem" gives plenty of references and examples.
I had a look on Colliot-Thélène's web page and found the following article: "Exposant et indice d'algèbres centrales simples non ramifiées" (avec  un appendice par Ofer Gabber),  L'Enseignement Mathématique 48 (2002) 127–146.  In that article, there is a nice introduction with a big list of references.  Apparently the first example of the type you're after was given by Brauer himself:  R. Brauer, "Untersuchungen über die arithmetischen Eigenschaften von Gruppen linearer Substitutionen", Zweite Mitteilung, Math. Zeitschrift 31 (1929) 733–747.  The article by Brauer is available online, but only one page at a time from a rather clunky web site, so I haven't gone into it to track down the example.<|endoftext|>
TITLE: Determining homotopy classes [T^2, RP^2] 
QUESTION [11 upvotes]: So I've been interested in computing homotopy classes of maps $T^2=S^1\times S^1$ to $\mathbb{R}\mathbb{P}^2$. So first, we can decompose $T^2$ into a cell complex with one zero cell, $S^1\vee S^1$ and a disc $D^2$ glued down to $S^1\vee S^1$ using the commutator. Since the commutator vanishes in $\pi_1(\mathbb{R}\mathbb{P}^2)$ given any map $S^1\vee S^1\rightarrow \mathbb{R}\mathbb{P}^2$, one can extend this to a map $T^2\rightarrow \mathbb{R}\mathbb{P}^2$ by the map $D^2\rightarrow T^2$, but since the gluing map is nullhomotopic, this is the same thing as a map $S^2\rightarrow \mathbb{R}\mathbb{P}^2$ and there are $\mathbb{Z}$ homotopy classes of maps like this. Therefore we can write (as a set) $[T^2,\mathbb{R}\mathbb{P}^2] = \{a,b,c,d\}\times \mathbb{Z}$ where each of the $\{a,b,c,d\}$ is a homotopy class $S^1\vee S^1\rightarrow \mathbb{R}\mathbb{P}^2$. My question is given an arbitrary map $f:T^2\rightarrow \mathbb{R}\mathbb{P}^2$ is there some sort of computation we can do in $\mathbb{R}\mathbb{P}^2$ to determine which homotopy class $f$ belongs to or at least which of the $a,b,c,d$ it belongs to?

REPLY [12 votes]: For dimension reasons, the set of homotopy classes of maps between these spaces can be computed by using their fundamental crossed module. Despite the question has already been answered twice, let me explain this approach, just for those who may like it.
A crossed module $C_*$ is a group homomorphism $\partial\colon C_2\rightarrow C_1$ together with an action of $C_1$ on $C_2$, that we write exponentially $c_2^{c_1}$, such that 
$$\begin{array}{rl}
\partial(c_2^{c_1})&=-c_1+\partial(c_2)+c_1,\\\
c_2^{\partial(c_2')}&=-c_2'+c_2+c_2'.
\end{array}$$
Here we use additive notation despite the groups may be nonabelian.
The canonical (topological) example of a crossed module is the boundary map in the long exact sequence of a pair of spaces:
$$\partial\colon\pi_2(X,Y)\longrightarrow\pi_1(X).$$
The fundamental crossed module of CW-complex $X$ is obtained in this way for the pair formed by $X$ and its $1$-skeleton.
$$\partial\colon\pi_2(X,X^1)\longrightarrow\pi_1(X^1).$$
Notice that $\ker\partial\cong\pi_2(X)$. The image of $\partial$ is normal in any crossed module, and in this case $\operatorname{coker} \partial\cong\pi_1(X)$.
A morphism of crossed modules $f_{*}\colon C_{*}\rightarrow D_{*}$ is a commutative square
$$\begin{array}{rcccl}
&C_1&\stackrel{\partial}\longrightarrow&C_2\\\
{\scriptstyle f_2}&\downarrow&&\downarrow&{\scriptstyle f_1}\\\
&D_1&\stackrel{\partial}\longrightarrow&D_2
\end{array}$$
such that $f_2(c_2^{c_1})=f_2(c_2)^{f_1(c_1)}$. Notice that any cellular map between CW-complexes $X\rightarrow Y$ induces a morphism between their fundamental crossed modules.
Two such morphisms $f_{*},g_{*}\colon C_{*}\rightarrow D_{*}$ are homotopic if there exists a map $H\colon C_1\rightarrow D_2$ such that 
$$\begin{array}{rl}
H(c_1+c_1')&=H(c_1)^{f(c_1')}+H(c_1'),\\\
\partial H(c_1)&=-f_1(c_1)+g_1(c_1),\\\
 H\partial(c_2)&=-f_2(c_2)+g_2(c_2).
\end{array}$$
If $X$ and $Y$ are CW-complexes and $X$ is $2$-dimensional the set of homotopy classes $[X,Y]$ can be computed as the set of algebraic homotopy classes between their fundamental crossed modules.
The fundamental crossed module of $T^2$ is isomorphic to $\partial\colon G'\hookrightarrow G=\langle a,b\rangle$. Here $G$ os a free group on two generators, $G'$ is its commutator subgroup, and $\partial$ is the inclusion. The action of $G$ on $G'$ is by conjugation.
The fundamental crossed module of $\mathbb{R}P^2$ is even easier: $\partial=2\cdot\varepsilon\colon \mathbb{Z}[\mathbb{Z}/2]\rightarrow \mathbb{Z}$. Here $\varepsilon\colon \mathbb{Z}[\mathbb{Z}/2]\rightarrow \mathbb{Z}$ is the augmentation map of the group ring and $\mathbb{Z}$ acts on $\mathbb{Z}[\mathbb{Z}/2]$ via the natural projection $\mathbb{Z}\twoheadrightarrow \mathbb{Z}/2$.
It is a beautiful exercise to compute the set of homotopy classes of maps between these two crossed modules. It's easy since the second one is made of abelian groups! If you're given a map $T^2\rightarrow\mathbb{R}P^2$ and you manage to find a homotopic cellular map (e.g. using the proof of the cellular approximation theorem), then you can say what homotopy class you started with by looking at the induced morphism on the level of crossed modules.<|endoftext|>
TITLE: Is there an upper bound on the dimension for irreducible representations of a continuous trace $C^{*} $-algebra?
QUESTION [7 upvotes]: The following are questions of Don Hadwin:
If $A$ is a unital continuous trace C*-algebra, is there an upper bound on the dimension of all the irreducible representations? 
It is known that all irreducible representations are finite-dimensional?

REPLY [5 votes]: There is an upper bound. Here is why (this is a simpler way of arguing than the one I used below):
In 4.5.2 of "C*-algebras", Dixmier defines a continuous trace C*-algebra as one such that
the set \[\{ a\in A^+ \mid \pi\mapsto\mathrm{Tr}(\pi(a))\mbox{ is continuous}\}\] spans a dense two-sided ideal of $A$. Here $\pi$ ranges through the spectrum of $A$. It is remarked in 4.5.2, that for every $x$ in this dense ideal the map $\pi\mapsto \pi(x)$ is continuous. If $A$ is unital, then this dense ideal must in fact be $A$ (since by getting very close to 1 it will contain invertible elements). It follows that $\pi\mapsto \mathrm{Tr}(\pi(1))$ is continuous, and thus bounded (since the spectrum of a unital C*-algebra is compact).
But $\mathrm{Tr}(\pi(1))$ is the dimension of $\pi$.

Longer argument:
Consider first the case where $A$ has a positive element $a$ such that each irreducible representation maps $a$  to a rank 1 operator. Then $a$ must be a full element (i.e., generate $A$ as a closed two-sided ideal). Thus, there exist finitely many $x_i$ and $y_i$ such that $1=\sum_{i=1}^n x_i a y_i$. This puts the bound $n$ on the rank of $\pi(1)$ for any irreducible $\pi$.
The general case is reduced to the previous case using the Fell condition. Since the spectrum of $A$ is compact it suffices to show that the dimension of its irreducible representations is locally bounded. The Fell condition for a continuous trace C*-algebra says that for each irreducible $\pi_0$ there exists a neighborhood $U(\pi_0)$ and a positive element $a\in A$ such that $\pi(a)$ has rank 1 for every $\pi\in U(\pi_0)$. We may choose $U(\pi_0)$ a closed neighborhood, since the spectrum of $A$ is Hausdorff. This entails the existence of a closed two-sided ideal $I$ such that $U(\pi_0)$ consists of the irreducible representations that factor through $A\to A/I$. Now the special case proved earlier can be applied to $A/I$ (and the image of $a$ in $A/I$).<|endoftext|>
TITLE: Direct proof of injectivity of $L_\infty$
QUESTION [8 upvotes]: I would like to know a simple proof of isometric injectivity of $L_\infty$. The proof I've found in Topics in Banach space theory. F. Albiac, N. Kalton uses two deep result. 

$L_\infty$ as commutative unital $C^*$ algebra is isometrically isomorphic to $C(K)$ for some compact $K$.
Every $C(K)$ space which is a dual space is isometrically injective.

However the proof for $\ell_\infty$ is quite simple. Let $i:X\to Z$ be isometric embedding and $T:X\to \ell_\infty$ be a bounded operator. Let $e_n:\ell_\infty\to\mathbb{C}:x\mapsto x(n)$ be coordiante functionals, then consider bounded functionals $f_n:\mathrm{Im}(i)\to \mathbb{C}:z\mapsto e_n(T(i^{-1}(z)))$ extend them by Hahn-Banach theorem to get functionals $g_n:Z\to\mathbb{C}$. The desired operator is $
\hat{T}:Z\to\ell_\infty: z\mapsto(g_1(z), g_2(z),\ldots)$
My question:
Does there exist a direct proof that $L_\infty$ is isometrically injective, a proof similar to the arguments used for the $\ell_\infty$ space? The problem in mimicking proof for $\ell_\infty$ arose from the fact that I can't find family of functionals $(E_n:n\in\mathbb{N})\subset L_\infty^*$ similar to coordinate functionals $(e_n:n\in\mathbb{N})\subset\ell_\infty^*$. 
Thank you.

REPLY [8 votes]: I am not sure whether the following would meet your requirements but I vaguely remember
having heard it in a course many years ago and I think that it is sufficiently distinct
from the above proofs to justify a brief mention.  The crucial common property of the three spaces in question---the real line, the sequence space and the function space---is that they are 
Dedekind-complete Banach lattices for which the norm is intimately connected to the order
structure---the unit ball coincides with the interval $[-1,1]$.  This fact allows one to mimic directly the standard proof of the classical Hahn-Banach theorem in the two more advanced cases.<|endoftext|>
TITLE: Examples of separable ordinary differential equations in economics
QUESTION [6 upvotes]: I'm currently teaching an integral calculus course for business students, and we're just about to discuss differential equations. They've worked hard, and I'd like to reward them with some economic applications of ODEs, but they can only handle simple separable equations.  
I'm going to frame exponential growth in terms of economic growth(among other things,) and then I'm currently planning on looking at which demand functions have constant elasticity and looking at the logistic model of a population. I might be asking for too much, but I was wondering whether anyone could suggest a separable equation that arises from a simple model(they've all taken an introduction to economics, but no more.)

REPLY [17 votes]: Suppose you maintain a pond with fish (for profit, of course, this is economics!).
When the food is abundant and there are not many fish, the population grows at a constant rate
$k>1$ (reproduction rate minus death rate), so we have $y'=ky$. This is separable. Solve it. Give a numerical example.
Conclude from the example that our assumptions are not realistic. So what is wrong with our assumptions?
Abundant food!!!
(Of course. This is economics after all:-)
The next simple assumption is that the pond can support only some maximal population, say $A$.
Which means that when the population approaches $A$ the death rate increases (starvation), so the net
growth rate is not just $k$ but $k(1-y/A)$. When $y$ is small, (or $A$ is very large) we
have almost $y'=ky$ as before. When $y$ is close to $A$, the net rate of change approaches $0$,
as it should be.  We obtain $y'=ky(1-y/A)$, another separable equation!
But this pond  brings you no profit yet. To make a profit, you have to catch some fish, say at
a constant rate. You obtain another separable equation $y'=ky(1-y/A)-c$. Discuss what happens
for various values of parameters $k,A,c$.
And so on:-)
You can go further and further with this model when time permits. Suppose that
instead of harvesting a fixed amount $c$, you gauge the population somehow, and harvest $cy$, a fixed proportion of the population. This leads to another separable equation, as well as to a useful discussion, which strategy is better, $c$ or $cy$ in terms of long term profits and in terns the pond sustainability.
Then, if time permits, you can pass to two functions and systems of equations.
The classical example is Volterra-Lottka system, which involves a slightly more complicated
ODE, but it is also separable. And its original motivation was also economics: the influence of World War I on the population
of sardines in the Mediterranean (an important economic resource for surrounding countries).
Remark. Besides fish, there are somewhat similar models of warfare (also a kind of economics btw), search on "Lanchester laws"; they lead to simple 2x2 systems of linear differential equations, and they have been compared to what happens in real wars.<|endoftext|>
TITLE: Classification of Platonic solids
QUESTION [8 upvotes]: My question is very basic: where can I find a complete (and hopefully self-contained) proof of the classification of Platonic solids? In all the references that I found, they use Euler's formula $v-e+f=2$ to show that there are exactly five possible triples $(v,e,f)$. But of course this is not a complete proof because it does not rule out the possibility of different configurations or deformations. Has anyone ever written up a complete proof of this statement?!

REPLY [22 votes]: This is a classical question.  Here is my reading of it: Why is there a unique polytope with given combinatorics of faces, which are all regular polygons?  Of course, for simple polytopes (tetrahedron, cube, dodecahedron) this is clear, but for the octahedron and icosahedron this is less clear.
The answer lies in the Cauchy's theorem. It was Legendre, while writing his Elements of Geometry and Trigonometry, noticed that Euclid's proof is incomplete in the Elements.  Curiously, Euclid finds both radii of inscribed and circumscribed spheres (correctly) without ever explaining why they exist.  Cauchy worked out a proof  while still a student in 1813, more or less specifically for this purpose.  The proof also had a technical gap which was found and patched up by Steinitz in 1920s.  
The complete (corrected) proof can be found in the celebrated Proofs from the Book, or in Marcel Berger's Geometry.  My book gives a bit more of historical context and some soft arguments (ch. 19).  It's worth comparing this proof with (an erroneous) pre-Steinitz exposition, say in Hadamard's Leçons de Géométrie Elémentaire II, or with an early post-Steinitz correct but tedious proof given in (otherwise, excellent) Alexandrov's monograph (see also ch.26 in my book which compares all the approaches).  
P.S. Note that Coxeter in Regular Polytopes can completely avoid this issue but taking a different (modern) definition of the regular polytopes (which are symmetric under group actions).  For a modern exposition and the state of art of this approach, see McMullen and Schulte's Abstract Regular Polytopes.<|endoftext|>
TITLE: Large Intersecting Subsets of a Set
QUESTION [10 upvotes]: I am sure the answer to this question is well known, but I am not able to figure it out.
Question: Let $U$ be a finite set. Let $F=(S_1,S_2,...,S_n)$ be such that:
(1) $S_i\subset U$
(2) $|S_i|=n$  
(3) $|S_i\cap S_j|\leq n/2$
Then, what is the lowerbound on $|U$|? In other words, what is the smallest $U$ for which there exists an $F$ satisfying the above conditions.
Clearly, if $U$ has size $n^2$, it easy to construct such an $F$. You can also do this with just $n^2/2$ elements in $U$. Can you do this with just $O(n)$ elements? What about $O(n^{1+\epsilon})$ for a constant $\epsilon<1$?

REPLY [3 votes]: Just to have something for all sufficiently large n, one can also take, for any small enough $\varepsilon>0$, the set $U$ to have $(2+5\varepsilon)n$ points, and then let $S'_i$ be independent random subsets of $U$ each obtained by choosing the points independently with probability $1/2-\varepsilon$. An application of Chernoff's inequality says that the size of any given $S'_i$ is concentrated close to expectation, and in particular greater than $n$. And the expected intersection of any two sets is $(1/4-\varepsilon+\varepsilon^2)(2+5\varepsilon)n$ which is smaller than $n/2$, and also by Chernoff we have good concentration. The failure probability of each of these applications of Chernoff is something like $2^{-\varepsilon^2 n}$. In particular we can certainly take a union bound over all $n+n^2$ applications (in fact we could have exponentially many $S_i$). Alternatively we can take $\varepsilon$ to be something like $\sqrt{\log n/n}$ and get $n$ sets.
So let $S_i$ be a subset of $S'_i$ of size $n$ for each $i$ and the construction is done.<|endoftext|>
TITLE: Seven pages -- too long for a research statement for postdoc application?
QUESTION [13 upvotes]: My question is rather straight forward. I am currently applying for postdoctoral positions in (pure) mathematics. I am almost done writing my research statement and it seems to come out to about seven pages. Is that too long?
A similar question has been asked here, but it seems to deal with applications for graduate school.
In my research statement I have described my previous work (which has resulted in four published/submitted papers and two currently in progress), as well as some short term and long term projects that I plan to work on. Part of the reason that the statement is turning out to be so long, is that I work in two (a priori disjoint) fields of (pure) mathematics, often applying one to the other. So it takes about two pages to describe the motivation and the necessary background.
So, again, is seven pages too long? Do you have any specific advice about preparing a statement for research on an intersection of different fields?

REPLY [32 votes]: Unsurprisingly, your research statement addresses at least two very-different audiences, and in your specific situation, maybe three or more. The usual entirely-true cliche is that most people on the postdoc-hiring committee will not themselves read beyond the first page or so of the whole thing, so you should be sure to make your big points, of course necessarily in rough terms, just on that first page. In usual scenarios, there is also the cadre of "specialists" whose strong support is necessary, if not sufficient, to get the postdoc offer, and the rest of the thing is written for them.
For people in a "new" field, or in a novel interaction of two existing "specialties", I'd think you'd want to as-quickly-as-possible make the point that there is at least some interest for specialists in both those specialties in your application/combination of them... or you may find yourself with "no constituency at all", rather than the union of the two. 
There are two sorts of "persuasion" that successful research descriptions typically have. Often enough, proving one's adherence to some orthodox school of thought/research, in effect sanctioned by famous people at the best institutions, is a useful/necessary assertion of pedigree. At the same time, in some ways opposite to the spirit of that, is making a persuasive argument that one will do interesting things with the materials described, so that it will be stimulating (not merely "prestigious") to have you around. The conflict with certifying your orthodoxy is that progress and constructive innovation typically involve not doing things as always done before, no matter by whom.
So whatever the length of your research description is probably fine, if you just prioritize things so that the first page is an adequate representation of your ideas for non-specialists, and the rest is persuasive, rather than merely "descriptive"... the latter lending itself to boredom and not-being-read.<|endoftext|>
TITLE: A continuous notion of realizability
QUESTION [6 upvotes]: I have been interested in non-classical logics, off and on, for quite a while. This question is probably very basic, and I hope it is not too low-level for MO. My question stems from an attempt to define a notion of "continuous truth" in a topological structure.
Suppose I have a topological structure $(M, \tau)$ - that is, a first-order structure $M$ with a topology $\tau$ on the underlying set. Usually - i.e., as Abraham Robinson does in "A note on topological model theory" (Fundamenta Mathematicae, 1974) - one demands that the topology and the first-order structure be compatible in some way, and then goes on to formulate a version of classical first-order logic augmented by some topological quantifiers. I'm interested in going a slightly different direction. I do want to demand that my first-order and topological structures be compatible - specifically, I will demand that the interpretations of function symbols be continuous functions from the relevant product topology to the structure, and the interpretations of relation symbols be closed sets in the relevant product topology - but I want to wind up with a non-classical logic at the end. Specifically, motivated by Kleene's notion of realizability, I will say that a topological structure $(M, \tau)$ continuously realizes a formula $\phi$ in prenex normal form if Skolem functions for $\phi$ can be found which are continuous according to $\tau$.
For example, consider any map $m: X\rightarrow Y$ between two topological spaces which is continuous and bijective but has no continuous inverse. Consider the structure $M$ with underlying set $X\sqcup Y\sqcup\lbrace o\rbrace$, topologized naturally, with a unary relation $U$ representing $X$, a unary relation symbol $V$ representing $Y$, and a unary function symbol $f$ representing $m$ which sends all $x\in X$ to $m(x)$ and all other points to $o$. Then the statement "$\forall x\exists y(V(x)\implies f(y)=x)$" is not continuously realized, although classically it should be true. 
If I understand things correctly, this is a situation generalized by the construction of first-order logic internal to a topos; however, I know virtually nothing about topos theory. My question is the following: what am I talking about? That is, what is this notion generally called, and when did it originate? Also, am I correct in thinking that it is generalized by (the logical side of) topos theory? 
(Additionally, as a minor subquestion: is my reasoning in the example two paragraphs prior correct?)
Thank you all in advance, and I hope this question is not too elementary.

REPLY [9 votes]: To see how the realizability logic works in a topological model have a look at my PhD thesis where these things are explicitly spelled out (chapter on equilogical spaces, for example). Alternaltively you could look at my notes "Realizability as the Connection between Computable and Constructive Mathematics", which are shorter and written as a tutorial.
In short, Kleene's realizability intepretation of logic works for a wide class of "computatonal models" known as (typed) combinatory algebras. Kleene's number realizability is just one such model, but there are topological models as well, for example Scott's graph model $\mathcal{P}{\omega}$ (which is closely related to equilogical spaces). In fact, the topological spaces themselves form a (large) type combinatory algebra which allows us to write down a realizability interpretation of logic in which the realizers are elements of topological spaces.<|endoftext|>
TITLE: Is there a simple test to determine whether a polytope is integral? 
QUESTION [5 upvotes]: It is known that any rational convex polytope expressed as $\{ x\in\mathbb{R}^d : Ax \ge b \}$, where $A\in\mathbb{Z}^{k\times d}$ and $b\in\mathbb{Z}^k$, can be written as the convex hull of finitely many points. 
My question is, given the above representation in terms of hyperplanes, one can determine (easily) whether the polytope is integral ---that is, whether the polytope can be written as the convex hull of points in the integer lattice.

REPLY [7 votes]: In the paper
The complexity of recognizing linear systems with certain integrality properties, Guoli Ding, Li Feng & Wenan Zang, Mathematical Programming 114, pages 321–334 (2008)
the authors prove that testing integrality is coNP-complete.<|endoftext|>
TITLE: An etale version of the van Kampen theorem
QUESTION [33 upvotes]: Let $V$ be a smooth connected algebraic variety over an algebraically closed field $k$.  Let $W_1, W_2$ be closed subvarieties of $V$ of positive codimension whose intersection $W_1 \cap W_2$ has codimension at least 2, and let $p$ be a point in $V  \backslash (W_1 \cup W_2)$.  Then we can form the four etale fundamental groups
$$ \pi_1( V, p ), \pi_1(V \backslash W_1, p), \pi_1(V \backslash W_2, p ), \pi_1(V \backslash (W_1 \cup W_2), p ).$$
There are canonical surjective homomorphisms from $\pi_1(V \backslash W_1, p)$ and $\pi_1(V \backslash W_2, p)$ to $\pi_1(V, p)$, and from $\pi_1(V \backslash (W_1 \cup W_2),p)$ to $\pi_1(V \backslash W_1, p)$ and $\pi_1(V \backslash W_2, p)$, forming a commuting square.  (The surjectivity comes from the fact that a connected finite etale cover of a smooth variety remains connected even if one removes a positive codimension piece from the base.)  So there is a canonical homomorphism from $\pi_1(V \backslash (W_1 \cup W_2),p)$ to the fibre product $\pi_1(V \backslash W_1, p) \times_{\pi_1(V,p)} \pi_1(V \backslash W_2,p)$.
My question is: is this latter homomorphism necessarily surjective also?
In the case when $k$ has characteristic zero, I believe I can deduce this from the topological van Kampen theorem, after first using the Riemann existence theorem to describe the etale fundamental group as the profinite completion of the topological fundamental group (after passing to a complex model).  In the positive characteristic case, it seems to boil down (if I understand the etale fundamental group construction correctly) to verifying the following fact: if one has two finite etale covers of $V \backslash W_1$ and $V \backslash W_2$ respectively that become isomorphic on restriction to $V \backslash (W_1 \cup W_2)$, then they can be "glued" together to create a finite etale cover of $V$ (or of $V \backslash (W_1 \cap W_2)$).  By reasoning in analogy with the topological case, this seems very reasonable to me, but I had trouble verifying it rigorously (I could glue together the covers as a prevariety, but then I couldn't establish separability to make the cover a variety again).

REPLY [30 votes]: To simplify notation, let me write $U_i$ for $V\smallsetminus W_i$, and $U_{12}$ for $U_1\cap U_2=V\smallsetminus(W_1\cup W_2)$.  
Fact: The obvious functor
$$(\mathrm{Sch}/V)\longrightarrow (\mathrm{Sch}/U_1) \times_{(\mathrm{Sch}/U_{12})} (\mathrm{Sch}/U_2)$$
is an equivalence. In other words, a $V$-scheme $X$ is the same thing as a $U_1$-scheme $X_1$, a $U_2$-scheme $X_2$, and a $U_{12}$-isomorphism of their restrictions to $U_{12}$.
This is probably somewhere in EGA1. [EDIT: all I could find was section 2.4 of EGA1, relying on (4.1.7) of Chapter 0 (glueing of ringed spaces).]
However,  we are dealing here with finite étale schemes, which happen to be affine over the base, so this boils down to the analogous statement for categories of quasicoherent sheaves, which is essentially trivial (plus the fact that ``finite étale'' is a local condition).  
If we describe the categories of finite étale covers in terms of $\pi_1$-sets, the above equivalence says that the diagram of groups 
$$(*)\qquad\begin{array}{rcl}
\pi_1(U_{12},p)=:G_{12}& \longrightarrow &G_1:=\pi_1(U_{1},p)\cr
\downarrow && \downarrow\cr
\pi_1(U_{2},p)=:G_2& \longrightarrow &G:=\pi_1(V,p)
\end{array}$$
is cocartesian. In other words, we get the usual van Kampen statement: the natural map
$$\pi_1(U_{1},p)\ast_{\pi_1(U_{12},p)}\pi_1(U_{2},p)\longrightarrow \pi_1(V,p)$$
is an isomorphism.  [EDIT: the coproduct is in the profinite category, which perhaps makes it hard to describe in general. See Will Savin's comment.]
What we want to prove is that the  map "on the other side" 
$$G_{12}\longrightarrow G_1 \times_G G_2$$
is surjective, given that all the maps in diagram ($\ast$) are surjective.
Identifying $G_i$ ($i=1,2$) with $G_{12}/N_i$, we see from the universal property of the coproduct that $G=G_{12}/N_{1}N_{2}$. [EDIT: clearly this works also in the profinite category: since $N_1$, $N_2$ are both compact normal subgroups, so is $N_1 N_2$, hence $G_{12}/N_{1}N_{2}$ is profinite].
Take any $(x_1,x_2)\in G_1 \times_G G_2$: thus we have $x_i=g_i N_i$ for some $g_i\in G_{12}$, and the fiber product condition says that $g_1=g_2 n_2 n_1$ for some $n_i\in N_i$ (recall that $N_1 N_2=N_2 N_1$). So, $(x_1,x_2)$ is the image of $g_1 n_{1}^{-1}=g_2 n_2\in G_{12}$. QED<|endoftext|>
TITLE: Fundamental groups of compact manifolds with non-negative Ricci curvature.
QUESTION [7 upvotes]: I would like to find an appropriate reference for the following statement:
Statement. Let $M$ be a compact Riemannian manifold with non-negative Ricci curvature.
Then $\pi_1(M)$ is virtually abelian.
It seems to me that the statement should follow from the article of Cheeger and Gromoll 
"The spletting theorem for manifolds of non-negative Ricci curvature"
http://intlpress.com/JDG/archive/1972/6-1-119.pdf
but since it is not stated explicitly in the article I am not 100% sure.
So, what would be a reference?

REPLY [7 votes]: The following paper has more than you want.
Wilking, Burkhard On fundamental groups of manifolds of nonnegative curvature. Differential Geom. Appl. 13 (2000), no. 2, 129–165.<|endoftext|>
TITLE: Measuring how far from being cocompact a lattice is 
QUESTION [6 upvotes]: Let $G$ be a locally compact group and $\Gamma$ a lattice (=discrete
subgroup of $G$ such that $G/\Gamma$ carries a probability measure $\mu$
that is invariant under the action of $G$ by left-multiplication).
My vague question is: "How to measure the lack of cocompactness of
$\Gamma$"?.
Edit: My question was indeed unclear. I am not looking for a criterion that says me whether a lattice is cocompact. Given a non-cocompact lattice, I am looking for a way of quantifying "how much non-cocompact" it is. I propose below such a measurement (and ask what can be said about it), but if somebody has other propositions I will be happy.
A natural such measurement in the case when $G$ has a compact generating
set is the following. One can equip $G$ with the word length metric $d$
corresponding to some compact generating set, and consider the sequence
$n \mapsto u_n=1 - \mu (B(0,n) \Gamma)$. It is easy to see that $\Gamma$
is cocompact if and only if $u_n=0$ for all sufficiently large $n$. The
speed of convergence of $u_n$ to zero in some sense should measure how
far we are from this ideal situation. This is related to the integrability properties of $w \mapsto d(1,w)$ on a suitably chosen fundamental domain $\Omega \in G$ for the action of $\Gamma$ on $G$ by right-multiplication. 
Has this been studied? What is the typical behaviour of $u_n$ in the
case of (arithmetic) lattices in Lie or algebraic groups? In a first time I would
already be happy to have an answer for $\Gamma=SL(3,\mathbf Z)$. My
guess is that number theorists may have already studied this.

REPLY [7 votes]: It seems to be that what you are asking for is roughly the measure of a neighborhood of the "cusp" of $G/\Gamma$. 
For the case of $\Omega(n) = SL(n,\mathbb{R})/SL(n,\mathbb{Z})$ there is a classical calculation of a closely related quantity. Recall that $\Omega(n)$ is the moduli space of unimodular lattices in $\mathbb{R}^n$. 
For $\epsilon > 0$, let $\Omega_\epsilon(n) \subset \Omega(n)$ denote the subset parametrizing lattices in which the shortest vector has length at least $\epsilon$. Then the complement of $\Omega_\epsilon(n)$ is compact (This is called Mahler compactness). 
I will describe below how to estimate the measure $\mu(\Omega_\epsilon(n))$. (This is similar to the quantity you wanted to compute). 
Let $f: \mathbb{R}^n \to \mathbb{R}$ be a compactly supported function. Define the "Siegel Transform" $\hat{f}: \Omega_n \to \mathbb{R}$ by
$$\hat{f}(\Delta) = \sum_{v \in \Delta'} f(v),$$
where $\Delta'$ is the set of primitive vectors in the lattice $\Delta$. The Siegel integral formula (which you can prove by unfolding) says that
$$\int_{\Omega_n} \hat{f} \\,d\mu = \frac{1}{\zeta(n)} \int_{\mathbb{R}^n} f.$$
Now take $f$ to be the characterisric function of the ball of radius $\epsilon$ around the origin. Then the integral on the left is supported on $\Omega_\epsilon(n)$, and $\hat{f} \ge 1$ on $\Omega_\epsilon(n)$.  It turns out that to compute the leading term in the asymptotics as $\epsilon \to 0$, you might as well assume that $\hat{f} = 1$ on $\Omega_{\epsilon}(n)$. You get, that as $\epsilon \to 0$, 
$$\mu(\Omega_\epsilon(n)) \sim \frac{1}{\zeta(n)} Vol(B(0,\epsilon)).$$
Similar calculations can be done for other non-uniform lattices but things get more technical. The terms to google for are "Eisenstein Series" and "precise reduction theory".<|endoftext|>
TITLE: Harmonic function with gradient of constant norm in hyperbolic 3-space
QUESTION [7 upvotes]: I investigate certain stationary charged perfect fluid solutions to the Einstein-Maxwell equations in general relativity. The classification of these solutions has led me to the following question:
Does hyperbolic 3-space admit a non-constant harmonic function (i.e., with vanishing Laplacian) that has a gradient of constant norm?
A non-constant affine function (i.e., with vanishing Hessian) would fit this description. However, hyperbolic space does not admit such a function.
T. Sakai: "On Riemannian manifolds admitting a function whose gradient is of constant norm", Kodai Math. J. 19 (1996), 39-51.

Finally, a preprint of the paper is ready.
https://www.researchgate.net/publication/291823768_ON_SOME_STRUCTURE_RESULTS_FOR_GODEL-TYPE_SPACETIMES_Preprint

REPLY [7 votes]: The main paper you want to consult is 
É. Cartan, Familles de surfaces isoparamétriques dans les espaces à courbure constante,
Annali di Mat. 17 (1938), 177–191.
In this paper, Cartan considers the problem of studying the functions $f$ defined on an open set in a space $M$ of constant curvature that satisfy two equations of the form $|\nabla f|^2 = A(f)$ and $\Delta f = B(f)$ for some functions, $A>0$ and $B$, of one variable.  He shows that the (hypersurface) level sets of $f$ are what he called isoparametric, i.e., they have all of their principal curvatures constant.
This had been considered somewhat earlier earlier by Levi-Civita for the case $n=3$ in flat space and by Segre for all $n$ (again, in flat space).  They showed that, up to rigid motion and homothety, a connected isoparametric hypersurface must be an open subset of $S^k\times \mathbb{R}^{n-k-1}$ for some $0\le k\le n{-}1$.  
Cartan showed that, when the ambient sectional curvature is negative, i.e., in hyperbolic $n$-space, then, again, a connected isoparametric hypersurface must either be totally umbilic (i.e., an open subset of either a totally geodesic hyperplane, an equidistant hypersurface, a horosphere, or a hypersphere) or be an open subset of a hypersurface of the form $S^k\times H^{n-k-1}$, i.e., it must be a totally geodesic hypersurface or horosphere or else the set of points of constant distance from a totally geodesic submanifold of lower dimension.  Thus, if you did have a solution of $|\nabla f|^2=1$ and $\Delta f = 0$ (even a local one in a neighborhood of $p$) in hyperbolic $n$-space, then $f-f(p)$ would be the oriented distance from the level set $f=f(p)$, and it is easy to calculate that no such function is harmonic, so you have a contradiction.
By the way, there is a much greater variety of isoparametric hypersurfaces in spaces of constant positive curvature.  In fact, even though Cartan found many nontrivial examples, and there has been a lot of work since then constructing more examples and classifying them, the classification of these is still not complete.<|endoftext|>
TITLE: Today's world record on the Betti numbers of Calabi-Yau three-folds.
QUESTION [13 upvotes]: What are largest betti numbers $b_2$ and $b_3$ of three-dimensional Calabi-Yau manifolds that are discovered for today?
Is there some nice reference?

REPLY [21 votes]: Since mirror symmetry exchanges the Hodge numbers $h^{1,1} = b_2$ and $h^{2,1} = \frac{1}{2}(b_3 - 1)$, it is perhaps more natural (and of course equivalent) to discuss these.  The record-holders all come from the list of hypersurfaces in toric fourfolds, constructed by Kreuzer and Skarke.  The largest value of 
$h^{1,1}$ is $491$, and the same for $h^{2,1}$.  These two manifolds (which are mirror) also hold the record for largest $h^{1,1} + h^{2,1}$, which is 502.  There is a third manifold which shares this record; its Hodge numbers are $(251,251)$, and it is also in the Kreuzer-Skarke list.
See this recent paper and this one for nice plots and a discussion of these Calabi-Yau threefolds and others with large Hodge numbers.<|endoftext|>
TITLE: Elephant populations (and Dyck words)
QUESTION [10 upvotes]: Hello,
I'm relatively new to this forum so apologies if I have tagged my
question incorrectly.
I have been in contact with a wildlife biologist recently concerning
counting elephant populations and I wonder if people could comment on
the following approach.
de Bruijn, Knuth and Rice showed that the expected height of a general Catalan tree
is $\sqrt{\pi} \sqrt{n}$.  In terms of Dyck words this translates to
saying that the expected maximum excess of Xs over Ys in a Dyck word of
length $2n$ is $\sqrt{\pi} \sqrt{n}$.
Now if we think of X as "an elephant arrives at the watering hole" and Y
as the elephant leaving it then should not the largest number seen
correspond to the above expression?  Thus we could get an estimate of
the population.  And that is my question, how good an approximation to
the population $n$ will this be?
Note that it is not a concern that an elephant revisits the site and is
doubly counted, because this will not affect the maximum seen.  However,
one flaw in the model is that it assumes that elephants arrive
independently.  Certainly this will not be the case with young calves
but there may be cliques who are fellow travellers also.  Also, perhaps,
the model assumes that the likelihood of an elephant being present at
time $t$ is equally likely for all $t$.  This may not be realistic
either.
Apologies also to those who find all of the above just elephants :-).
Thanks,
Patrick healy

REPLY [4 votes]: Did you look at the many papers by Andy Royle (https://profile.usgs.gov/aroyle/) that adress similar problems with birds and amphibians. His N-mixture model approach would probably answer your colleague's problem in a way that is widely accepted by the wildlife ecologists' community to which I belong... (a mathematician friend of mine sent me the link if you wonder what I'm doing on this forum :-))<|endoftext|>
TITLE: fpqc sheafification and localisation
QUESTION [5 upvotes]: I am slightly confused about sheafification at the moment.
I first learned sheaves defined as a subcategory of presheaves, then I was told that sheaves are also a localisation of presheaves, then I was told this was a common feature of localisations (i.e. they are often reflective), but then I was told that not every presheaf can be fpqc-sheafified.
So, what's the deal? What is the correct notion of an fpqc sheaf (localisation vs subcategory)? Or is the problem that the localisation doesn't exist (as can sometimes happen)? But then again, can't one always put a model structure on presheaves so that the homotopy category is sheaves?

REPLY [5 votes]: An fpqc sheaf is exactly what you think it is: a functor from the opposite of the category of schemes (or relative schemes) to $Set$ (i.e. a presheaf) such that the usual glueing conditions hold for fpqc covers.
You may be thinking of the fact remarked on here: http://ncatlab.org/nlab/show/fpqc+site that the collection of fpqc covers of a scheme doesn't have a cofinal set, and so one cannot just assume that the site at hand is small. 
Even though sites that people work with can be large (say $Aff$) 'nice' Grothendieck pretopologies are given by a set of covering families for each object, or at the very least have a cofinal set of covering families (this means there is a coverage given by a set of covering families, and this is enough to define sheaves, though generally weaker than a pretopology). Without this 'local smallness' condition (called WISC), the category of sheaves may not be locally small.
There is an example of a functor on schemes which admits no fpqc sheafification.
In the case of large sites without the condition WISC, the appropriate thing to consider is small sheaves, namely sheaves that are small colimits of representable sheaves.<|endoftext|>
TITLE: Is there any theorem like implicit function theorem in $\mathbb{Q}$ ?
QUESTION [5 upvotes]: My qeustion is that,
is there any theorem like implicit function theorem in $\mathbb{Q}$ ?
More precisely, let $p(\bar{x},\bar{y})$ be in $\mathbb{Z}[\bar{x},\bar{y}]$ such that in $\mathbb{Q}$, for any $\bar{a}$, there is a solution of $p(\bar{x},\bar{a})$. Then for some polynomial(or rational polynomial) $q(\bar{y})$ with $\mathbb{Q}$ coefficients, $p(q(\bar{y}),\bar{y})=0$ holds in the rational polynomial fields over $\mathbb{Q}$.
For example, $x^2+y^2=1$ does not satisfy the condition
but for $x+y=0$ it holds.
And how about the same question in p-adic field $\mathbb{Q}_{p}$?

REPLY [5 votes]: The same question in $\mathbb Q_p$ is false. For instance, if $p \neq 2$, let $\alpha \in \mu_{p-1}$ be a primitive root of unity. Then $(x^2-y)(x^2-py)(x^2-\alpha y)(x^2-p \alpha y)$ has a solution for each $y$, but you cannot make that solution a polynomial in $y$.<|endoftext|>
TITLE: Is this a "folk theorem" about analytic functions of a complex variable?
QUESTION [5 upvotes]: In a comment on question 110345 I made a claim that might be incorrect. I claimed that if
f(z) is a non-constant analytic function defined by a power series whose circle of convergence C
has a positive radius, then f(z) cannot be bounded at all points in the interior of C. But is 
this really true?  Or am I just imagining that I learned it somewhere. I could not come up with
any simple counter-examples. It sounds like some weird generalization of Liouville's theorem.

REPLY [15 votes]: You are probably misunderstanding the following folk theorem: 
If $D $ is the convergence disc of a power series converging to $f$, then there must be some singularity of $f$ on $\partial D$. In other words, you cannot continue $f$ analytically onto a larger disc.
A counterexample that is more explicit than quids example is the power series expansion of $\sqrt{1+z}$ around $z=0$, which has convergence radius $1$. The singularity at $z=-1$ is not a pole.
A hint to the proof: if $D \subset U$ is a disc in the domain of definition of a holomorphic function $f$, then the Taylor expansion around the midpoint of $D$ converges in $D$.<|endoftext|>
TITLE: Minimal number of cells of a CW complex (up to homotopy) 
QUESTION [10 upvotes]: Given a finite connected CW complex $X$, one can ask what can said about the number of its cells. 
As an example, let's estimate the number of 1-cells: There is an epimorphism $\pi_1(X_1) \to \pi_1(X)$ from the 1-skeleton which is a connected graph and whose fundamental group is a free group on a subset of the 1-cells of $X$ [Hatcher, 1A.2]. Hence we have the lower bound
$$\text{number of 1-cells } \ge \text{ minimal number of generators of } \pi_1(X)$$ 
Conversely, given a presentation of $\pi_1(X)$ with a minimal number of generators $d$, there is a CW complex $X'$ with $d$ 1-cells and $\pi_1(X') = \pi_1(X)$ [Hatcher, 1.28].
Question 1: Can $X'$ be choosen to be (cellularly) homotopy equivalent to $X$ ?
Futhermore, by taking into account the cellular chain complex, it's not hard to see that the number of $n$-cells $(n \ge 0)$ is bounded below by the fact that we need (at least) 

one $n$-cell for each direct summand of $H_n(X)$ 
one $n$-cell for each direct summand of finite order of $H_{n-1}(X)$ 

As formula: 
$$\text{number of n-cells } \ge d(H_nX) + d(H_{n-1}(X)_{tor}) =: m_n(X)\qquad(\ast)$$
where $d(\cdot)$ denotes the minimal number of generators. 
It's known that if $X$ is simply connected, then $X$ is homotpoy equivalent to a complex $X'$ having exactly $m_n(X)$ cells in each dimension [Hatcher, 4C.1]. 
Question 2: Are there other classes of CW complexes where each $X$ is homotopy equivalent to a complex $X'$ that has $m_n(X)$ cells in each dimension ?
By the estimate above, a necessary condition for such a class is $d(\pi_1X)=d(\pi_1(X)_{ab})$, e.g. $\pi_1(X)$ solvable. 
Question 3: What's the best current bound for the minimal number of cells of (not necessarily simply-connected) finite CW complexes ?

REPLY [5 votes]: To expand on my comment, there's a very general tool to manipulate CW-complexes, due to Whitehead.  It tells you when you can in effect remove a cell from a CW-decomposition via `elementary moves', usually called Whitehead Moves.   In smooth manifold theory there are parallel constructions -- people talk about "handle slides" and "handle cancellations".   This comes up in the proof of the h and s-cobordism theorems, which are the smooth-category analogue of the Whitehead moves.   Technically these moves have to do with a slightly more refined notion of homotopy-equivalence, called simple homotopy equivalence.  Provided the fundamental group of the CW-complex is trivial, simple homotopy-equivalences are in effect the same as homotopy-equivalances, but in general they're a little more fussy. 
What are the Whitehead moves?  On the 0-skeleton, it's the move where you collapse a maximal forest in the 1-skeleton. On the 1-skeleton these are moves where you can cancel a 1-cell using a 2-cell that's incident to it only once.  This is explained in detail in Marshall Cohen's "A course in simple-homotopy theory".  GTM 10 Springer-Verlag.<|endoftext|>
TITLE: A question on Ricci curvature and Ricci form. 
QUESTION [7 upvotes]: It seems to me that there are two definition of Ricci flatness; real and complex ones. The real Ricci flatness claims vanishing of Ricci curvature. To define complex Ricci flatness, one needs to introduce so called the Ricci form for a Kahler manifold $(X,g)$ defined by 
$$
Ric(\omega)=-i\partial \overline{\partial}\log\det(g_{i,\overline{j}}),
$$
where $g=(g_{i,\overline{j}})$ is a Kahler metric. A Kahler manifold$(X,g)$ is called Ricci flat if $Ric(\omega)$ vanishes. 
I wonder if the complex one coincides with real one for a Kahler manifold $(X,g)$. How Ricci curvature and Ricci form are related?

REPLY [3 votes]: If you only have an almost complex manifold with a compatible metric i.e. an almost Hermitian manifold $(M,g,J)$ then you can define the Ricci form globally by $$\rho(X,Y):=Ric(JX,Y).$$ And vanishing of $Ric$, $\rho$ are equivalent. You don't need your manifold to be complex or Kähler.<|endoftext|>
TITLE: Degeneration of varieties to simple normal crossings
QUESTION [6 upvotes]: Let $\mathcal{X}\to\Delta$, $\Delta \subset \mathbb{C}$ is the unit disk, be a smooth family of varieties whose fibers over $t\neq 0$ are smooth and the central fiber $\mathcal{X}_0$ is a nice simple normal crossing divisor (in $\mathcal{X}$).
Let $\mathcal{X}_0=\cup X_i$, and define $X_I=\cap_{i\in I} X_i$.
This assumption imposes some relations between the $N_{X_{I}}^{X_{J}}$, $J\subset I$.
For example if $\mathcal{X}_0=X_1\cup_D X_2$ ($D=X_{12}$), then $N_D^{X_1}\otimes N_{D}^{X_2}=\mathcal{O}_D$ is trivial, and conversely if these condition holds, then there is a smooth one parameter family realizing that.
Question:
Is it known under what conditions, a simple normal crossing space can be realized as the central fiber of a smooth family? Is it known if those relations are enough for the existence of such family (similar to example above)
In general, is this probelm studied in literature or not?

REPLY [11 votes]: The more modern approach to the question adressed by Friedman is via
logarithmic geometry. Most relevant for your question is the paper of
Kawamata and Nammikawa, "Logarithmic deformations of normal crossing
varieties and smoothing of degenerate Calabi-Yau varieties," Invent. Math.,
118, (1994) 395-409. However, even that paper did not use the language
of log geometry as thoroughly as possible.
Briefly put, suppose we are given a normal crossings variety $X$. The
first step is to understand when one can put a log structure on $X$ of
the correct sort,
(I'll write the log scheme as $X^{\dagger}$) along with a log smooth
morphism $X^{\dagger}\rightarrow {\rm Spec} k^{\dagger}$, where the
latter is the "standard log point", i.e., a point with associated monoid
$k^{\times}\oplus {\bf N}$, where ${\bf N}$ denote the natural
numbers. Kawamata and Namikawa show that this can be done precisely
when Friedman's d-semistability condition holds, i.e., when the local
$T^1$ sheaf
$N_D:={\mathcal Ext}^1(\Omega^1_X,{\mathcal O}_X)$ is the structure sheaf
of the singular locus of $X$.
One then applies log deformation theory, which was sketched out by
K. Kato in his original paper on logarithmic geometry, "Logarithmic structures
of Fontaine-Illusie," and fleshed out by F. Kato in
http://arxiv.org/abs/alg-geom/9406004
Kawamata and Namikawa in fact show the log deformation theory of a
normal crossings Calabi-Yau is unobstructed, using similar techniques
for proving the Bogomolov-Tian-Todorov theorem. So for d-semistable Calabi-Yau
varieties, the statement you want is true.
The advantage of using log deformation theory is that if one uses ordinary
deformation theory as Friedman did, it is not likely that the deformation
space will be unobstructed. Typically a normal crossings variety has
many locally trivial deformations which give a large irreducible component
of the deformation space, but most of these locally trivial deformations
do not smooth because they don't carry log structures. Log deformation theory
does not see these bad deformations.<|endoftext|>
TITLE: Background reading for proving irrationality of real numbers
QUESTION [5 upvotes]: I'm almost finishing my PhD in applied mathematics, but I'm planning soon (after doing a post-doc) to start seriously doing research on problems about proving irrationality of real numbers. Whenever I have a chance I train myself reading proofs of this type and collecting articles and bibliography I find on internet. In 2010 I read a beautiful proof about $e$ being irrational, this is the link:
           http://math.stanford.edu/~rhoades/FILES/irrationalityE4d.pdf  

The beauty of the proof proposed in the above link is that is applicable to many other real numbers expressed in terms of infinite series.
I read the proof that $\pi$ is irrational in Michael Spivak's calculus book. The proof uses elementary mathematics. I found it hard to see the intuition behind it (maybe reading Lambert's proof will show the intuition), but I was able to follow all the details and understand the contradiction exposed in the book.
Surprisingly I found an article where differential equations are used to prove irrationality of certain real numbers: Irrational numbers arising from certain differential equations by M. Ram Murty and V. Kumar Murty.
I would like somebody to guide me with some literature I must follow so that I can train myself in tackling this kind of problems. What kind of books are "a must read"? Articles? Books that give historical context of problems and the ideas of the proofs, the intuition behind them, books written with the "heart" showing the beauty of this subject.
Also, what area of mathematics specializes in solving problems of this kind? Is it analytic number theory? Transcendental number theory?
There are so many attractive open problems like $\pi^e$, $\pi + e$ and Euler's gamma constant (my favorite!), where irrationality is not known. I found an article written by Jonathan Sondow titled "Criteria for irrationality of Euler's constant", but I couldn't
follow the details. That's why I need the training.
I consider myself "not too bad" in real and complex analysis. I was able to follow proofs like the prime number theorem (in Stein's book on complex analysis), which requires analytic continuation of the zeta function. I know this proof by heart. Also proofs like (I found this on "An Introduction to Number Theory" by Graham Everest; I like his way of writing):     $$\sum_{p \leq N} \frac{1}{p} \geq \log\log N - 1$$
where $p$ is a prime-valued variable. I like to read about finding closed formulas too, formulas of complicated series (zeta function evaluated at even numbers), complicated definite integrals (leading to Euler's gamma constant for example)... that kind of good stuff! (I know this is connected with the irrationality proofs, based on the first link I gave, I'm sure!).

REPLY [3 votes]: To add to quid's suggestion (look around Michel Waldschmidt's webpage), I'd say specifically that Waldschmidt's slides on Schanuel's conjecture are probably worth looking at. This may be more for one's general education about transcendental number theory than any suggestion for research, since Schanuel's conjecture is widely regarded as a Holy Grail of transcendental number theory, and a proof seems to be nowhere in sight. It is, however, quite beautiful!<|endoftext|>
TITLE: For which Calabi-Yau threefolds is SYZ conjecture known to hold?
QUESTION [8 upvotes]: I would like to know for which Calabi-Yau threefolds SYZ conjecture is known to hold. I am aware of works by Gross-Wilson (Borcea-Voisin CY3s) and Ruan (quintic CY3), but they are quite classical works today and I now wonder if there are any more examples of SYZ. Of course, answer depends on what I mean by "SYZ conjecture", so I am happy with SYZ examples at any level. I also wonder what can one conclude from Gross's many works on this subject.  
I am also aware of Auroux's (and his collaborators') work on SYZ for Fano manifolds, but I am mainly interested in CY threefold case. This comment does not exclude any information about their work that might be relevant to CY3 case (I just don't know any relation due to my ignorance). 
Thank you for your information.

REPLY [12 votes]: Let me go from the weakest to strongest sense in which the conjecture should
be true.
First, at the purely topological level, it is true for any Calabi-Yau variety
with a toric degeneration whose dual intersection complex is ``simple''.
These notions are part of my program with Bernd Siebert: see our paper
http://arxiv.org/abs/math/0309070 for the definitions of toric degeneration
and simple. In http://arxiv.org/abs/math/0406171 I proved that all
Calabi-Yau varieties arising in the Batyrev-Borisov construction as
complete intersections in toric varieties have such degenerations.
The problem is that Bernd and I have been putting off writing the paper linking
the logarithmic approach to topological fibrations for years now, largely
due to lack of motivation. So there is no reference in the literature yet
for this result. I do hope we will finally complete this paper next year.
Second, at the Lagrangian level, there are the results of W.-D. Ruan you
mentioned. In addition, Castano-Bernard and Matessi in
http://arxiv.org/abs/math/0611139 showed that given an affine three-manifold $B$
with ``simple'' singularities, one can construct a symplectic six-manifold
along with a Lagrangian fibration to $B$. So one can apply this to the case
where $B$ is the intersection complex of a polarized toric degeneration of
Calabi-Yau threefolds. One expects this six-manifold to be symplectomorphic
to a general fibre of the degeneration, but there is no proof of this at the
moment.
Finally, at the special Lagrangian level, I think it is safe to say there
are no known examples on compact non-singular Calabi-Yau threefolds with
non-degenerate metric. There are some examples for non-compact
Calabi-Yau varieties, specifically toric ones, see my paper
http://arxiv.org/abs/math/0012002<|endoftext|>
TITLE: Holonomy of a Kähler manifold
QUESTION [5 upvotes]: Hi,
I have the following question: Let $(M,J, \omega)$ be a Kähler manifold (not necessary compact). We know that the holonomy group is a subgroup of $U_{n}$. Let $\Omega$ be a constant ($\nabla \Omega = 0$) holomorphic non-vanishing (n,0)-form. Can one say that the holonomy group is now cotained in $SU_{n}$ ? Is it true ? I hope a lot of answers. Thanks in advance.
Mina

REPLY [5 votes]: The answer is yes.
The holonomy principle states that a given a riemannian manifold $(M,g)$ and a point $x\in M$, the datum of a parallel tensor field of a given type is equivalent to the datum of a tensor of the same type at the point $x$ which is invariant under the action of the holonomy group.
Now, $SU(n)$ is the subgroup of $U(n)$ of operators which preserve a (complex) non zero $n$-form in $\mathbb C^n$. Thus, if your manifold is Kähler, thanks to the holonomy principle, the holonomy group of your manifold is contained in $SU(n)$ if and only if there exists a non zero parallel $(n,0)$-form on $M$.
Such a from is closed, hence holomorphic.
In other words, the holonomy group is contained in $SU(n)$ if and only if there exists a holomorphic $n$-form on $M$, which is non zero and parallel.<|endoftext|>
TITLE: Proving that a complicated function is eventually concave
QUESTION [5 upvotes]: I have a function $f:\mathbb{R}^+ \to \mathbb{R}^+$ that I want to prove is eventually concave - i.e. that there exists $\gamma _0 > 0$ such that for every $\gamma>\gamma_0$, $f(\gamma)$ is concave. I also know that $f$ is real-analytic, increasing and upper bounded, and it is therefore enough to show that $f''$ has a finite number of zeros. 
A concrete definition of $f(\gamma)$ is as follows:
$$f(\gamma)= \frac{1}{\sqrt{2\pi}}\sum_{x\in\mathcal{X},x'\in\mathcal{X}'}p_{x,x'}\int_{-\infty}^\infty{e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}\log\left(\frac{\sum_{x\in\mathcal{X}}p_{x|x'}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}{\sum_{x\in\mathcal{X}}p_{x}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}\right)dy}$$
where $\mathcal{X},\mathcal{X}'$ are two finite sets, $p:\mathcal{X}\times\mathcal{X}'\to[0,1]$ is a probability mass function and $p_{x|x'}=p_{x,x'}/\sum_{\xi\in\mathcal{X}}p_{\xi,x'}$ , $p_x=\sum_{\xi'\in\mathcal{X}'}p_{x,\xi'}$ are the conditional and marginal distributions on $\mathcal{X}$, respectively. In Information-Theoretic terms, $f(\gamma)$ is the mutual information between $X'$ and $Y_{\gamma}=\sqrt{\gamma}X+N$, where $N \sim \mathcal{N}(0,1)$  is independent of $X'$ and $X$, which are dependent and have finite alphabets. The variable $\gamma$ is the signal-to-noise ratio.
The function is known to be concave in the special case $X=X'$ ($p_{x|x'}=\delta_{x,x'}$). When $X\neq X'$, examples can be found in which $f$ is not always concave, but becomes concave eventually as $\gamma$ grows.
Intuitively, $f''$ must have finitely many zeros (=$f$ is eventually concave), as $f$ is a finite combination of "well-behaved" functions. I would like to formalize this by showing that $f$ belongs to class of functions that have a finite number of zeros, that is closed under differentiation. This will show that $f''$ also has finitely many zeros, which is just the thing we need.
There exist several such classes of functions including Hardy's class L of "orders of infinity" and the Pfaffian closure. See also this related question. However, the mathematical machinery here is a bit complicated for me, and I could not show $f$ belongs to either class due to the integration with resepect to $y$ in its definition. Perhaps this can be shown by recasting $f$ as a solution to an ODE? Or perhaps there exists another, more suitable class of functions that I am not aware of?

There is also an expression for $f''(\gamma)$ which is quite lengthy. Here it is for completeness:
$$f''(\gamma) = -\frac{1}{2\sqrt{2\pi}}\sum_{x\in\mathcal{X},x'\in\mathcal{X}'}p_{x,x'}\int_{-\infty}^\infty{e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}[\phi_X^2(y;\gamma)-\phi_{X|X'}^2(y,x';\gamma)]dy}$$
where
$$\phi_X(y;\gamma)=\frac{\sum_{x\in\mathcal{X}}x^2p_{x}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}{\sum_{x\in\mathcal{X}}p_{x}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}-\left(\frac{\sum_{x\in\mathcal{X}}xp_{x}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}{\sum_{x\in\mathcal{X}}p_{x}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}\right)^2$$
and similarly
$$\phi_{X|X'}(y,x';\gamma)=\frac{\sum_{x\in\mathcal{X}}x^2p_{x|x'}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}{\sum_{x\in\mathcal{X}}p_{x|x'}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}-\left(\frac{\sum_{x\in\mathcal{X}}xp_{x|x'}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}{\sum_{x\in\mathcal{X}}p_{x|x'}e^{-\frac{1}{2}(y-\sqrt{\gamma}x)^2}}\right)^2$$
This has some estimation-theoretic meaning, but not anything that seems useful in establishing my desired result.

REPLY [4 votes]: Well, I also couldn't understand Thierry's argument, so I decided to use the old "pushing junk down the ladder" trick from classical asymptotic analysis. Note first of all that looking at the expression for the first derivative, which is an integral of a non-negative quantity, we can easily realize that either $f'(t)$ is identically $0$, or $f'(t)\ge ce^{-Ct}$ (I prefer to use $t$ instead of $\gamma$ for typing reasons). Canceling $e^{-{y^2}/{2}}$ whenever possible and changing the variable $y$ to $\sqrt ty$, we see that up to some irrelevant factor, $f''(t)$ is the same as
$$
F(t)=\int_{-\infty}^\infty e^{-t\frac{y^2}2}\frac{\sum_k\alpha_k e^{t U_k(y)}}{\sum_j\beta_j e^{t V_j(y)}}\,dy
$$
where $U_k,V_j$ are some fixed linear functions of $y$ and the sums are finite. Note that $I(t)\le e^{Ct}$ for some $C>0$ (in your case, it has no chance to grow at all, but let's be generous and use the particular structure of your function only when we need it). Now look at $W(y)=\max_j V_j(y)$. It is a piecewise linear function. Note that the whole real line can be split into finitely many (bounded or unbounded) intervals  $J_m$, each interval containing one point where the graph of $W$ has a corner so that on each interval we have a few linear forms $V_j$ meeting at one point while all the rest stay down from the maximum of these meeting forms by some constant. Splitting at that corner point and moving the orijin to it, we see that $F(t)$ is a sum of integrals of the form
$$
I(t)=\int_0^a e^{-t\frac{y^2}2}\frac{S(t,y)}{R(t,y)+J(t,y)}\,dy
$$
where $S(t,y)$ and $J(t,y)$ are of the same form $\sum_k\alpha_k e^{t U_k(y)}$ with arbitrary linear forms $U_k$, while $R(t,y)=1+\sum_j{\beta_je^{-\lambda_j t}}$ with some positive $\lambda_j$ and $J(t,y)\le e^{-ct}R(t,y)$ on $[0,a)$.
Now it is time to push the junk term $J(t,y)$ down the ladder by writing
$$
\frac 1{R(t,y)+J(t,y)}=\frac 1{R(t,y)}\sum_{m=0}^M (-1)^m\left[\frac{J(t,y)}{R(t,y)}\right]^m+\frac 1{R(t,y)+J(t,y)}(-1)^{M+1}\left[\frac{J(t,y)}{R(t,y)}\right]^{M+1}
$$ 
If $M$ is chosen large enough, the last term we want to drop is dominated by $e^{-Ct}$ times the original expression, so, stopping far enough, we can make the error decay exponentially with as large exponent as we wish. Thus, we can get an approximation, which is a linear combination of terms of the kind
$$
T(t)=e^{bt}\int_0^a e^{-t\frac{y^2}2}\frac{e^{\lambda ty}}{1+Q(t,y)}\,dy
$$
where $Q(t,y)=\sum_j{\beta_je^{-\lambda_j t}}$ with some positive $\lambda_j$. 
Now we have to consider 2 cases.
Case 1: There is no $Q(t,y)$ (pure exponent).
Here we can shift everything to the point of maximum of $\lambda y-\frac{y^2}2$, split if necessary, and get a linear combination of terms of the kind 
$$
e^{bt}\int_0^a e^{-t\frac{y^2}2}\,dy=ce^{bt}t^{-1/2}-ce^{b't}\int_0^{\infty}e^{-t\frac{y^2}2}e^{-aty}\,dy
$$
(write $\int_0^a=\int_{0}^\infty-\int_{a}^\infty$ and shift the starting point back to the origin in the second one).
Case 2: There is a non-trivial $Q(t,y)$ in the game. Then we want to push the junk down the ladder once again writing 
$$
\frac{1}{1+Q(t,y)}=\sum_{m=0}^M (-Q(t,y))^m+\frac{(-Q(t,y))^{M+1}}{1+Q(t,y)}
$$
The purpose of the game is to get a lot of pure exponents and an exponentially decaying function $e^{\lambda t}Q(t,y)^{M+1}$ in the numerator (note that, for all we know, $\lambda$ could be large positive in the beginning). Moreover, when integrating the last term, we can add the integral from $a$ to $\infty$ because that one can be made to decay exponentially with arbitrarily large exponent by choosing $M$ large enough.
Now, with all the junk pushed ruthlessly away, we get the sum of the terms of two kinds:
$e^{bt}t^{-1/2}$ and $e^{bt}\int_0^\infty e^{-t\frac{y^2}2}f(ty)\,dy$ where $f$ is some exponentially decaying function on $(0,\infty)$. Plus, of course, we have a very quickly decaying error. Of course, we can combine the terms of each kind with the same $b$, so I'll assume that all $b$'s are different.
In the terms of the second kind, one is tempted to change variable to $ty$ getting
$$
t^{-1}\int_0^\infty e^{-t^{-1}\frac{y^2}2}f(y)\,dy
$$
At this point, we can safely write the asymptotic expansion in the inverse powers of $t$ just expanding $e^-z$ and integrating formally. Note that it will be merely an asymptotic expansion, not a convergent series. However, it is enough for us to get the asymptotics for the whole thing unless all coefficient miraculously vanish. However, the coefficients are even moments of an exponentially decaying function. Reflecting it around the origin, we get a function with all vanishing moments and exponential decay. The Fourier transform of such function will be a real analytic function with all derivatives at the origin equal to $0$. Thus, it is identically $0$, and so is the original $f$. But then we have total collapse of all terms and only the error is present, i.e., $F$ decays faster than any exponent. It is, probably, impossible in general but in your particular case, we have a clear reason: if $f''$ decays superexponentially, so does $f'$, but we ruled that out from the beginning.
Of course, if we can fix the Pfaffian argument, this "junk" method will become totally obsolete. But until Thierry replies, one can live with it :).<|endoftext|>
TITLE: Links with same Jones polynomial
QUESTION [5 upvotes]: Is there anything known about when two links have the same Jones polynomial (beyond a calculated list of small actual examples)? The first thing I would try is to compute the (formal - you would have something*split-horizontal+something*split-vertical) Jones polynomial of (4-)tangles and look there for two tangles with same Jones polynomial. Any instance would then generate an infinite example family. The snag would be, of course, that I'd first need a tangle table - my own goes to meagre 6 crossings. Aaaand the computation can't be automated that good. Still - any takers?

REPLY [5 votes]: Liam Watson generalized a construction of Kanenobu to produce infinitely many pairs of knots with the same Jones polynomial (and Khovanov homology) but distinct HOMFLYPT polynomials, so they are not mutants.  See the references below.
MR2287438 Watson, Liam.  Any tangle extends to non-mutant knots with the same Jones polynomial. J. Knot Theory Ramifications 15 (2006), no. 9, 1153–1162. 
MR2350287 Watson, Liam.  Knots with identical Khovanov homology. Algebr. Geom. Topol. 7 (2007), 1389–1407.<|endoftext|>
TITLE: About MF Atiyah and R Bott's  1983  paper
QUESTION [11 upvotes]: I am a theoretical physics major student working on string theory. I want to understand the work of MF Atiyah and R Bott, "The Yang-Mills equations over riemann surfaces" . What kinds of mathematical background does it need? books or papers? (I only learned Nakahara's book on geometry) 
Thanks in advance.

REPLY [2 votes]: Just as a reference, there was a summer course given by Alex Waldron at USTC here on "Yang-Mills Equations over Riemann surface", and his aim is to explain the proof of Donaldson's paper from the most basic differential geometry.<|endoftext|>
TITLE: Strong convergence of projections in $B(H)$
QUESTION [6 upvotes]: (I asked this question at math stackexchange 4 months ago, but received no answers)
Let $\{e_{kj}\}$ be the canonical matrix units in $B(H)$, with $H$ separable. Define projections $q_k$ by 
$$
q_k=\sum_{n=1}^ke_{nn}.
$$
Let $\{p_1,p_2,\ldots\}\subset B(H)$ be a sequence of orthogonal projections in $B(H)$ with the property that $q_kp_hq_k=q_kp_kq_k$ whenever $h\geq k$ (i.e. the sequence "fixes" the upper left corner as the index grows). 

Question: Does the sequence $\{p_k\}$ converge strongly?

(my gut feeling is that it should, but after a while thinking about it I couldn't get neither a proof nor a counterexample; it is easy to show that the sequence converges weakly so it would be enough to prove that the limit is a projection, but I got nowhere through this route either)

REPLY [7 votes]: $Q_k$ is just the orthogonal projection to the linear span $L_k$ of $e_1,\dots,e_k$, right? 
Let $u_m$ be some sequence of unit vectors. Let $P_mx=(x,u_m)u_m$. Now, $Q_kP_mQ_k x=(Q_kx,u_m)Q_ku_m=(x,Q_ku_m)Q_ku_m$, so the condition is that $Q_ku_m$ stabilize for each $k$. Let's stabilize them to $\frac 12e_1$ (just add some tails far away to get norm $1$). Take $v=e_1$ as the test vector for strong convergence. Then $P_m v=\frac 12 u_m$. Now, the tails are wild and carry a lot of norm, so, alas, $u_m$ do not converge to anything in norm...<|endoftext|>
TITLE: Constituents of induced representation
QUESTION [9 upvotes]: Let $H < G$ be finite groups with $|G:H|=n$, and let $M$ be an irreducible $FH$-module for some field $F$. Is it always true that all irreducible constituents of the induced $FG$-module $M^G$ have dimension at least $\dim M$? If not, then is it true that the composition length of $M^G$ is at most $n$?
For complex representations, these statements follow from Frobenius Reciprocity. But do they remain true for modular representations? I am particularly interested in the case when $F$ is a finite prime field.

REPLY [2 votes]: In the case that $n$ is prime and $H$ is normal in $G$. Your second questions has an affirmative answer. The composition length is $1$, $n$, or $1+d$ where $d\mid (n-1)$, in each case at most $n$. See the main theorem from:
S.P. Glasby and L.G. Kovács,
Irreducible modules and normal subgroups of prime index.
Comm. Algebra 24 (1996), no. 4, 1529–1546.<|endoftext|>
TITLE: Believing the Conjectures
QUESTION [37 upvotes]: In Believing the axioms (I and II), Penelope Maddy proposes five "rules of thumb" that she then uses to justify large cardinal axioms in set theory. These extrinsic rules are modeled after the development of set theory and the techniques of natural science. As such, applications of these rules should be found in all branches of mathematics. The most natural context for these to manifest themselves is through conjectures that are obtained by applying one of these rules of thumb in some context. I would like to hear about such conjectures (open or closed, big or small, true or false) in your area.
Maddy's five rules of thumb are:

Maximize: This is the opposite of Occam's Razor. The idea is that the universe should be as large as possible, anything that is likely to occur should actually occur.
Inexhaustibility: This is the idea that the universe is too rich to be generated by a handful of basic building blocks: there should be transcendental objects.
Whimsical identity: An object is unlikely to be the unique object satisfying a property that does not directly pertain to the object in question.
Uniformity: The richness of the universe should not localize in one particular part, similar richness should be found in all suitably large parts.
Reflection: If there is one object with a given property then there must also be a small (or otherwise simple) object with that property.

The above brief descriptions are mine. These were formulated by Maddy in the set-theoretic context, I attempted to phrase them in a way that would make sense in a lot of other contexts. Interpret them loosely: object, property, universe can be anything you want.
Note that these rules of thumb are not always good ideas and their negatives are sometimes plausible too. Although I am mainly looking for conjectures formulated in the positive sense, I think negative conjectures are also acceptable if the main reason to disbelieve the conjecture is one of the five rules above. For example, I think the Poincaré Conjecture can be understood as a negative example of the whimsical identity rule.
Standard Big List rules apply... One example per answer please! Try to include some brief context for the benefit of people outside your area.

REPLY [8 votes]: Given that you used the word "transcendental" in describing inexhaustibility, Schanuel's conjecture seems to be an obvious instance.  In effect, Schanuel's conjecture implies that numbers such as $e+\pi$ that have "no reason" to be algebraic are indeed not algebraic.  See also the conjectures of Kontsevich and Zagier about periods, which have a similar flavor.<|endoftext|>
TITLE: Which Abelian Group sequences arise as the Homology of Embedded CW Complexes?
QUESTION [7 upvotes]: Background
Let $\mathcal{A} = \lbrace A_0, \ldots, A_M \rbrace$ be an arbitrary sequence of finitely generated Abelian groups. It is well-known that a finite CW complex $X_\mathcal{A}$ may be constructed so that its $m$-th homology group $H_m(X_\mathcal{A})$ equals $A_m$ for all $0 \leq m \leq M$. In particular, one can build wedge sums of Moore spaces.
On the other hand, any CW complex is an Euclidean neighborhood retract (see Hatcher corollary A.10) and hence embeds into some Euclidean space. By a geometric complex of dimension $n$ I mean a CW complex whose minimal Euclidean embedding dimension is $n$.
Question
What $\mathcal{A}$'s arise as homology sequences for $n$-dimensional geometric complexes as a function of $n$?
Notes and Considerations
Some obstructions to having a CW complex embed in Euclidean space are outlined in the answers to this older question and may be helpful although I have not been able to use them efficiently. What one can say immediately about the sequences $\mathcal{A}$ arising from an $n$-dimensional geometric complex is that all but the first $n-1$ groups must be trivial by dimension considerations, and that the $(n-1)$-st group must be torsion-free. Alexander duality imposes constraints as well, but it is unlikely that this list of conditions characterizes the $\mathcal{A}$'s that are possible homology groups of geometric complexes. Is there a complete answer available somewhere?

REPLY [10 votes]: This is simpler than that other question, because it's only the homology and not the homotopy type that is being prescribed.
A Moore space with $H_1$ finite cyclic can be embedded in $4$-space, so by suspending this a Moore space with $H_m$ finite cyclic can be embedded in $n$-space if $1\le m\le n-3$. And of course an $m$-sphere can be embedded as long as $0\le m\le n-1$. Wedging together such Moore spaces and spheres you can achieve any sequence of finitely generated homology groups such that 
$A_m=0$ for $m\ge n$
$A_{n-1}$ and $A_{n-2}$ are torsionfree
$A_0$ is torsionfree and nontrivial.
Of course you can also achieve $A_m=0$ for all $m$ by taking $X$ to be empty. The other trivial case not satisfying the conditions above is a point embedded in $0$-space.
Conversely, the homology of any finite complex $X$ in $n$-space (other than those two trivial cases) must satisfy the conditions above.
The fact that $A_{n-2}$ is torsionfree follows from duality. Without loss of generality $X$ is a compact $n$-manifold-with-boundary in $n$-space. The map $H_1(X)\to H_1(X,\partial X)=H^{n-1}(X)$ is zero because it factors through the map $H_1(B)\to H^{n-1}(B,\partial B)$ for a closed ball $B$ containing $X$. This makes $H_1(X,\partial X)$ inject into $H_0(X)$, which is impossible if there is torsion in $H_{n-2}(X)$ and therefore in $H^{n-1}(X)=H_1(X,\partial X)$.<|endoftext|>
TITLE: An equivalence relation on group actions
QUESTION [18 upvotes]: Suppose a group $G$ acts faithfully on a set $X$, or equivalently, $G$ is a subgroup of ${\rm Sym}(X)$. 
By functoriality, $G$ acts on $P(X), P(P(X)), P(P(P(X))),$ etc. ($P(\cdot)$ means powerset.) 
Henceforth, I'll omit parentheses.
One can recover $G$ from ${\rm Fix}_G(PPPX)$ because, for example, one can encode a well-ordering of $X$ as an element of $PPX$.
Generally, one cannot recover $G$ from ${\rm Fix}_G(PPX)$.  For example, the alternating group and symmetric groups on a finite set will give the same set of fixed elements.
Write $G\sim H$ if both groups act on $X$ and ${\rm Fix}_G(PPX) = {\rm Fix}_H(PPX)$.  Equivalently, $$\forall u,v \subset X (\exists g\in G, gu=v \leftrightarrow \exists h\in H, hu=v) \ .$$
Questions 
When does $G\sim H$ imply $G=H$?  
Are there nontrivial examples of $G\sim H$ for infinite $X$? for $X$ of any infinite cardinality?  
Is there a classification of such pairs for finite $X$?  
Does this phenomenon have a name? 
Can one always recover $G$ from ${\rm Fix}_G(PPPX)$ in ZF?

REPLY [2 votes]: The group of pwop (piecewise order-preserving) permutations of $\mathbf{N}$ is equivalent to the full symmetric group (i.e., has the same orbits on the power set $2^\mathbf{N}$. These are permutations for which there's a finite partition such that on every component, the permutation is order-preserving. It is transitive on moieties (infinite subsets with infinite complement) and obviously is also transitive on finite subsets of given cardinality. You can google "transitive on moieties" to find more.<|endoftext|>
TITLE: A riemannian manifold with finitely many closed contractible geodesics
QUESTION [12 upvotes]: By a closed geodesic, I mean a smooth periodic geodesic $\mathbb{R} \rightarrow (M,g)$. I will consider them up to geometric distinction.
This means that any two closed geodesics are equivalent if they have the same image in $(M,g)$. 
Manifolds with constant curvature $\leq 0$,
by Cartan's theorem,
cannot have any closed contractible geodesics,
and every riemannian metric on $S^2$ has infinitely many closed geodesics 
(for $n\geq 3$, the analogous theorem for $S^n$ is not known).
Moreover,
if the sequence of Betti numbers of the loops space $\Omega(M)$ is unbounded and $M$ is simply-connected,
then $(M,g)$ contains infinitely many (contractible) closed geodesics.
Are there any known examples of riemannian manifolds with finitely and positively many closed contractible geodesics (or even just closed geodesics)?
There is a theorem associated with Gromov asserting that the word problem of $\pi_1 M$ is solvable if there is a metric $g$ on $M$ with only finitely many contractible closed geodesics. I was wondering if there are any non-trivial examples for this theorem.

REPLY [7 votes]: Ellipsoids with almost but not quite equal axes have exactly three simple closed geodesics.<|endoftext|>
TITLE: radius of tubular neighborhood
QUESTION [7 upvotes]: Hi there,
Is there any result about the calculation of radius of tubular neighborhood of submanifold inside a Riemannian manifold?
For example, given a simple smooth curve on R^2, what's the radius of its tubular neighborhood? (One upper bound is given by the minimal curvature, but general it is not the radius)
Maybe that is what we can expect:
if the curvature of the curve is always decreasing, then the radius of the tubular neighborhood is given by the injective radius of (left) end point, it seems true, right? 
For example, they curve is given by: $xy=1, x \in [1,+\infty)$
Thanks!

REPLY [3 votes]: As has already been pointed out, while the tubular radius bounds the curvature globally from below, the curvature information is not enough to correctly estimate this radius. Many examples can be considered, often silly ones when you allow your sub-manifold to be disconnected: consider the case of two parallel line segments in the plane and note that the embedding radius depends on their distance.
Since it is not clear what exactly you are after (computing the tubular radius of the curve $xy = 1$ is not so hard) or what information you already have, I would like to mention that your problem is solvable by calculus alone! Whether this calculus is tractable or not obviously depends on your choice of Riemannian manifold $X$ and submanifold $M \subset X$.
Given $M \subset X$, the Medial Axis $A_X(M)$ of $M$ in $X$ is defined to be the collection of all $x \in X \setminus M$ such that there are multiple solutions to the following constrained optimization problem in $X$:
Minimize $\text{dist}_X(x,m)$ subject to $m \in M$.
Here is a rough sketch of what a medial axis looks like when $X = \mathbb{R}^2$ and $M$ is the Nicolaescu horseshoe. The axis itself is in blue, and the red lines are my amateurish attempts at showing equidistant $M$-points

Once you know this medial axis, the distance from $A_X(M)$ to $M$ is precisely your tubular radius. Figuring out this distance again reduces to calculus which may be intractably hard depending on the choice of $X$ and $M$.
Update Here is a simple pictorial counterexample to the claim that if the curvature is decreasing from "left to right" then the tubular radius is the injectivity radius of the "left" endpoint. A straight line going up a lampshade suffices. If you want the curvature to decrease strictly, you can wrap the initial segment of the curve around the lower edge of the lampshade and then straighten as you go up. The point is that the curvature of the ambient manifold also plays a part in restricting the tubular radius. It is likely that your conjecture applies in Euclidean space, although I don't have an immediate proof of this.<|endoftext|>
TITLE: On a theorem of Galois
QUESTION [19 upvotes]: I am currently teaching Galois theory and this week, I mentioned the following theorem of Galois :
Let $P(x) \in \mathbf{Q}[x]$ be an irreducible polynomial of prime degree. Then $P$ is solvable by radicals if and only if the splitting field of $P$ is generated by any two roots of $P$.
I was asked by a student whether this theorem can be generalized to polynomials whose degree is composite, maybe allowing the splitting field to be generated by more than two roots. I know that the proof of Galois's theorem relies on determining the solvable subgroups of $\mathfrak{S}_p$, but I don't know enough group theory to tell what can be proved in the case where the degree of the polynomial is composite, say $pq$ where $p$ and $q$ are (possibly equal) primes.
Does such a generalization of Galois's theorem exist? Or is there a conceptual reason why such a generalization cannot hold? In the latter case, do there already exist generalizations of Galois's theorem, possibly in different directions?

REPLY [23 votes]: The question asks about the relation of the properties 1. and 3., though possibly the intended meaning of 1. was 2.:
The splitting field of $P$ is generated by two roots of $P$.
The splitting field of $P$ is generated by any two roots of $P$.
The Galois group $G$ of $P$ is solvable.

In the prime degree case, all three properties are equivalent.
For arbitrary degrees, 1. is a weak condition and so doesn't tell much about solvability of $G$. Condition 1. is also much weaker than 2. Despite its weakness, 3. does not imply 1. In fact, for each non-prime degree $n\ge6$, there is a solvable group $G$ of degree $n$ for which 1. does not hold. Indeed, it's easy to construct explicit examples for all such degrees: Let $n=rs$ with $r\ge3$, $s\ge2$. Then for suitable rational $a,b$, the polynomial $P(X)=(X^s-b)^r-a$ doesn't fulfill 1.
So let's forget about 1. Also, this example shows that 3. is far from implying 2.
The question remains whether 2. implies 3. Indeed, it comes close:
 If $n=\deg(P)\not\equiv1\pmod{120}$, and the splitting field of $P$ is not generated by a root (see Kevin Ventullo's comment above), then 2. implies 3.
The reason for this is as follows: 2. says that $G$ is a Frobenius group. By Frobenius' Theorem, $G$ is a semidirect product of a regular normal subgroup $N$ and a point stabilizer $H$, called the Frobenius complement. By Thompson's Theorem, $N$ is nilpotent, so in particular solvable. What about $H$? By an old result of Zassenhaus, $H$ is either solvable, or its series of derived subgroups terminates in $\text{SL}(2,5)$, a group of order $120$. As $n=\lvert N\rvert$, and $H$ has regular orbits on $N\setminus\{e\}$, we get what I claimed above.
As to the excluded degrees: The smallest candidate of degree $121$ exists group theoretically: $\text{SL}(2,5)$ has a regular action on the nonzero elements of $\mathbb F_{11}^2$, yielding a Frobenius group $\mathbb F_{11}^2\rtimes\text{SL}(2,5)$. By a result of Jack Sonn (see here), every finite Frobenius group is a Galois group over the rationals. Thus there is an irreducible $P(X)$ of degree $121$ for which 2. holds, but 3. does not.
Added (Answering François' question in his comment below): This minimal number of roots which generate the splitting field is the size of a so-called minimal base of the permutation group $G$. A base of a permutation group is a subset of the points the group acts on whose elementwise stabilizer is trivial. If you take the wreath product $G=C_2\rtimes C_m$, in its natural action on $2m$ points, this number is $m$. So even if the degree is a product of two primes, the minimal base size can be arbitrarily large.
Things are better if $G$ is primitive and solvable: Then the minimal base size is at most $4$. See here for more results on minimal bases.<|endoftext|>
TITLE: How much $\beta \mathbb{N}$ is homogenous?
QUESTION [6 upvotes]: Let $p,q\in \beta \mathbb{N}\setminus \mathbb{N}$. Must always the spaces $\beta \mathbb{N}\setminus \{p\}$ and $\beta \mathbb{N}\setminus \{q\}$ be homeomorphic? If no, can we for each point $p\in \beta \mathbb{N}\setminus \mathbb{N}$ find $q\in \beta \mathbb{N}\setminus \mathbb{N}$ ($q\neq p$) such that $$\beta \mathbb{N}\setminus \{p\}\approx \beta \mathbb{N}\setminus \{q\}?$$
Or maybe there exists an uncountable family $F\subset \beta \mathbb{N}\setminus \mathbb{N}$ such that the spaces corresponding to points in $F$ are mutually non-homeomorphic?

REPLY [5 votes]: As Emil pointed out in his comment, there is a correspondence between permutations of $\mathbb{N}$ and self-homeomorphisms of $\beta\mathbb{N}$, and that pretty much answers the OP´s original questions.
If we restrict ourselves to $\mathbb{N}^\ast=\beta\mathbb{N} \setminus \mathbb{N}$ the questions get more interesting (although the answers remain the same): When you mod out $\mathbb{N}^\ast$ by its group of self-homeomorphisms you get $2^\mathfrak{c}$ many orbits. This follows from Frolik´s theory of types of ultrafilters.<|endoftext|>
TITLE: (Mixed) Tate motives
QUESTION [7 upvotes]: Hi there,
 in recent times I was reading texts about motives, and I want to ask
something about Tate motives which is not clear to me (as I came across 
different definitions in different texts).
Let $V_k$ be the category of (projective) k-varieties. I read that the polynomial ring $\mathbb{Z}[\mathbb{L}] \subset K_0(V_k)$, the latter being the Grothendieck ring of $V_k$, with $\mathbb{L}$ the Lefschetz class $[\mathbb{A}_1]$, "corresponds to" mixed Tate motives generated by the Tate objects $\mathbb{Q}(m)$ in the Grothendieck ring $K_0(M_k)$, $M_k$ being the category of pure $k$-motives. In another article one rather spoke about the subring $\mathbb{Z}[\mathbb{L},\mathbb{L}^{-1}] \subset K_0(M_k)$ ($\mathbb{L}$ now the Lefschetz motive). And in yet another paper I read that mixed Tate motives are defined differently.
My question is: is one of the two first approaches indeed the correct way to see mixed Tate motives ?  Or is this a restricted way to define them ?  
I am especially interested in the connection between mixed Tate motives and $\mathbb{Z}$-varieties which are polynomial-countable. (When assuming the Tate conjecture, these varieties would have mixed Tate motives, and conversely.)
Thanks !!!

REPLY [4 votes]: There exists an abelian category of mixed Tate motives M (defined as the heart of some t-structure on the subtriangulated category generated by the $\mathbb{Q}(m)$ in the triangulated category of mixed motives defined by Voevodsky). In this category, every object has a increasing filtration whose quotients are one $\mathbb{Q}(m)$. This implies that at the K-group level, we obtain just the ring $\mathbb{Z}[\mathbb{L}, \mathbb{L}^{-1}]$ which is your second definition. 
Your first definition is not very coherent : if one take only $\mathbb{Z}[\mathbb{L}]$ then one don't have all the $\mathbb{Q}(m)$.
A remark about the restriction of the definition : the existence of the category M (and more important the fact that we know the Ext groups in M) is a much more non-trivial and poweful thing than the description at the K-group level.<|endoftext|>
TITLE: Lotteries, Turan's problem, and minimization of risk
QUESTION [13 upvotes]: Suppose I am a high-volume broker aiming to make some money on a state lottery.  In this lottery, six balls are drawn from a population of (let's say) 50, without replacement.  A ticket is a choice of a size-6 subset of 1,2,..50.
The prize structure of this lottery is such that the jackpot alone doesn't impart much value to the ticket.  But it turns out that lesser prizes are sufficiently large relative to their probability that the ticket has a positive expected value, which is why I'm buying a lot of tickets in the first place.  For instance, I can expect to get a pretty substantial return from tickets which match 4 of the 6 numbers drawn by the lottery.
There is a substantial literature related to the "Turan problem," which asks:  what is the minimum number of tickets I need to purchase in order to guarantee that one of my tickets matches 4 of the 6 numbers in the lottery?
My question is somewhat different.  Let's say I have enough capital to buy a fixed number N of tickets, large in absolute terms but small relative to 50 choose 6.  Then my expected gain is fixed.  But of course as a wise investor I may want to minimize the variance of my winnings.
Thus my question.  
If the random variable X is the number of (4 out of 6) wins among my N tickets, how small can I make Var(X) by judicious choice of ticket purchases?
(Of course, the same question applies for (k out of 6) where k=2,3,5.)
By the way, in case the setup seems unrealistic, let me add that the reason I'm asking this is that the situation described here actually happened, and I'm trying to reverse-engineer what the broker's risk-minimization strategy must have been, and assess whether it was worth it.

REPLY [4 votes]: This is not a real "answer" but an observation. Each 6-tuple has ${6 \choose 4}$ 4-tuples, so it stands to reason that once $N$ is some smallish multiple $m$ of ${50\choose 4}/{6\choose 4}$ then Bob's your uncle, and you can come reasonably close to equidistribution. This is to be contrasted with (say) the binomial expectation. The number of fours (or more) you expect is ${6\choose 4}m$, and the reasonable size $N=138180$ gives $m=9$ and 135 fours. The binomial distribution gives a variance about of slightly under $135$. I expect that one can essentially ensure about 135, plus or minus a small amount, via some covering selection.
ADDITION: In verities, the binomial model does not well model the random choices, as they have considerably higher variance, 180-185 compared to 134.8. I do not understand the theoretical concepts in toto, but an aspect is that the coverage of fours from the random sixes is already askew.
UPDATE: OK, here's the skinny on covering. I did a rather simple process. Do the following 138180 times. Pick a 4-tuple that so far has not appeared 9 times. Append to it the 2 numbers for which the resulting six minimizes the sum of the current counts of its 15 sub-fours. Accumulate the counts of 4-tuples from this six.
Then apply a bit of post-processing if you want (throw out populous sixes). This gives a set of 138180 6-tuples in which every 4-tuple appears between 7 and 11 times (the average is 135/15 or 9, with random choices of sixes the four-counts will range from 0 to 20 or more). Then simulate the ${50\choose 6}$ lotteries. These give an expectation of 135 fours, the minimum was 120 and the maximum was 149. The variance was a mere 5.8, versus 135 (binomial) or 180 (random). The binomial distro gives less than 120 a 8.9% chance (and more than 149 a 10.7% chance). As added above, the actual random distro is even worse than binomial.
I think this shows that with a (small) bit of work, some quite good variance reduction is possible. You can try to further trim the ends if desired. In the actual example, their edge was about 20-25% when free bets were included (later comments suggest 15-20% over the history). The accounting on page 7 says "12.8%" for just the cash component, but I get 425840/400000 is 6.4%. This analysis also lacks the jackpot, which I guess is equally likely to help/hurt among the big players (it is slightly chancey that only 1 of the 45 jackpots was hit, given there are 2-4 groups each buying up to 1/30 of the pool every time).
alt text http://www.freeimagehosting.net/newuploads/3tygt.jpg<|endoftext|>
TITLE: Algebraic characterisation of directed acyclic graphs
QUESTION [7 upvotes]: Any characterization based on the adjacency matrix for directed acyclic graphs (DAG)?
An undirected graph could be simply characterized by saying that its adjacency matrix is symmetric. What about a DAG?

REPLY [9 votes]: Given a finite, directed graph, it will be a DAG if and only if you can conjugate its adjacency matrix $A$ by a permutation matrix to get an upper triangular matrix.  The idea is to index the rows and likewise the columns of $A$ by the vertices of the graph.  Conjugating by a permutation matrix amounts to simultaneously permuting rows and columns in the same way, i.e. choosing a new ordering on the vertices of the graph.  A finite directed graph is acyclic if and only if you can put a total order on its vertices such that the directed edges always go from the earlier vertex to the later vertex.  This is equivalent to the adjacency matrix with respect to this vertex ordering being upper triangular.
Alternatively, one can simply raise the adjacency matrix to powers.  Having no directed cycles is equivalent to $(A^i)^1_{j,j} = (A_{j,j})^i $ for all $i,j\le |V|$.   (Note: the redundant exponent of 1 was inserted to get latex to work.)
Edit: Mariano sums this up well in his comment, saying that $A$ is nilpotent.<|endoftext|>
TITLE: convergence action on the boundary of hyperbolic groups
QUESTION [5 upvotes]: Let G be a word-hyperbolic group acting on its boundary, which is homeomorphic to $S^n$ (n-sphere), effectively. Does this imply that G acts on the boundary as a convergence group of $S^n$? 
If this is true in general, is it easy to see, at least for n = 1 or 2?

REPLY [5 votes]: Bowditch proved much more.  Namely, if a group $\Gamma$ acts properly discontinuously on a $\delta$-hyperbolic space $X$, then $\Gamma$ acts as a convergence group on $\partial X$.  See Lemma 1.11 of his paper 
B.H. Bowditch,
Convergence groups and configuration spaces,
in ``Geometric Group Theory Down Under, proceedings of a Special Year in Geometric Group Theory, Canberra, Australia'' (ed. J.Cossey, C.F.Miller III, W.D.Neumann, M.Shapiro), de Gruyter (1999), 23-54.
which is available here.  To get the result you want, consider the action of $\Gamma$ on its Cayley graph.<|endoftext|>
TITLE: Computing the limit of a certain recursively defined sequence 
QUESTION [5 upvotes]: The following is not exactly a research question (it was originated from manufacturing of exercises for calculus), and has no other motivation than explaining a phenomenon. I apologize if it is inappropriate (and will quickly remove it). 
Consider the sequence of real numbers defined recursively as follows
$$u_0:=\lambda > -1\\ ,$$ 
$$u_{n+1}=\frac{2u_n}{1+\sqrt{1+2^{-n}u_n}}\\ .$$ 
Numerical evidence suggests that $u_n$ always converges to $\log(1+\lambda)$. I imagine there should be a simple explanation. How can one prove it?

REPLY [7 votes]: Multiplying top and bottom by the conjugate and simplifying (assuming $u_n \neq 0$) we get: $$u_{n+1}=2^{n+1}(\sqrt{1+2^{-n}u_n}-1).$$ Calling $v_n:=\frac{u_n}{2^n}+1$ we have the recursion: $v_{n+1}=\sqrt{v_n}$ and therefore $v_n=v_0^{2^{-n}}$. Going back to $u$´s we have $u_n=2^n[(1+\lambda)^{2^{-n}}-1]$. Finally: $$\lim_{n\to\infty} u_n=\lim_{h\to 0}\frac{(1+\lambda)^h-(1+\lambda)^0}{h}= \ln (1+\lambda).$$<|endoftext|>
TITLE: How does one go about finding real/complex irreducible and faithful representations of PSL(2,7)?
QUESTION [10 upvotes]: It is well known that PSL(2,7) contains 168 elements. I'm looking for a method of obtaining irreducible representations (the matrices, not just the character table) which as implied by the title are:
1) over the real or complex fields
2) faithful, i.e. isomorphic to PSL(2,7)
I'd prefer an analytic expression for such matrices if this is possible.
Can the question be answered for the more general case of PSL(2,p), p prime, or even the more ambitious PSL(n,p)?
Thanks!

REPLY [9 votes]: The irreducible character degrees for this group have degree $1,3,3,6,7,8$.

To get the degree $8$ irreducible, induce a non-trivial linear character of the Sylow $7$-normalizer (a Frobenius group of order $21$).
To get the degree $7$ irreducible, induce a non-trivial linear character of either of the maximal parabolics isomorphic to $S_4$.
To get the degree $6$ irreducible, induce the trivial character of either of the parabolics isomorphic to $S_{4},$ obtaining an orthogonal representation. Then take the orthogonal complement of the $1$-dimensional fixed-point space.
To get one of the two $3$-dimensional representations, induce a non-trivial linear character of the Sylow $7$-subgroup. The already constructed (unitary) $8$-dimensional representation shows up, as does the $6$-dimensional (unitary) repesentation and the $7$-dimensional unitary irreducible representation. Take the orthogonal complement of the sum of these. This gives a $3$-dimensional unitary representation. Take the dual of that as well, and you have all (non-trivial) irreducible representations, up to equivalence.

Later edit: Note that for the irreducible characters of degree $6,7$ and $8$ the representations above may be explicitly given as real representations. The degree $8$
representation requires a little further thought. If we induce the trivial character from a Sylow $2$-subgroup, the $8$-dimensional irreducible character occurs with multiplicity $1$, the trivial character occurs once, the $7$-dimensional irreducible character does not occur, the $6$-dimensional character occurs once and the two three dimensional character each occur once. Since the permutation representation is realized over $\mathbb{R},$ it follows that the $8$-dimensional representation may be realized over the real field.
The two $3$-dimensional representations do not have real characters. To obtain a real representation of their sum, do the last procedure instead with a real irreducible two dimensional orthogonal representation of the Sylow $7$-subgroup (which is just as  group of real rotations): two copies of each of the $6,7$ and $8$ degree real orthogonal representations show up, so take the orthogonal complement of their sum, obtaining an orthogonal representation of degree $6$ which is irreducible as a real representation, but not a complex representation.<|endoftext|>
TITLE: A sudden smiley? :-)
QUESTION [48 upvotes]: This is a vague question, and I will no doubt be (properly!) chastised for posing it.
I would like to generate a set $S$ of points in $\mathbb{R}^3$—$|S|$ finite or infinite—which
has the property that, viewing $S$ under orthogonal projection along a random direction $\vec{u}$ results
in a more-or-less generic, undistinguished cloud of points. But, there is some specific projection
direction $\vec{u^*}$, where suddenly (if one were 3D-rotating the points under mouse control) the cloud resolves itself, through unlikely point alignments, to paint a recognizable image, e.g.,

           

Is this an impossible :-) hope?
Update.
Following Michael Murray's recipe, with $10,000$ points within a cube in $\mathbb{R}^3$, three different viewpoints:

     
(Somehow my analytical smiley has a Halloween evil glint!)
PS(31 Oct 2012).  Happy Halloween!

          


Another addition (23Jun2018):

          


          

(Image from John Urschel / MoMath video.)

REPLY [16 votes]: A Painting Made From Pieces of Glass<|endoftext|>
TITLE: Does ZFC prove the universe is linearly orderable?
QUESTION [16 upvotes]: It is consistent with ZFC that the universe is well-ordered, e.g. in $V=L$ where global choice holds. I also know that it is consistent that global choice fails (although I have no immediate example from the top of my head).
However one can try and ask a slightly weaker question, much like the axiom of choice implies every set can be linearly ordered, but the reversed implication fails; is it the same with global choice? The immediate answer is yes, it is consistent that the universe can be linearly ordered but the axiom of choice fails and therefore global choice has to fail too. But just how far can this be pushed?

Can we construct a model of ZFC in which the axiom of choice holds, but there is no linear ordering of the universe? Or does ZFC prove that the universe is linearly ordered?

One would expect that it would, because if we assume AC then the sets of ordinals already decide the universe in its entire form, and sets of ordinals can always be linearly ordered by: $$A\prec B\iff\min (A\triangle B)\in A.$$ (where $\triangle$ denotes the symmetric difference.)

REPLY [8 votes]: Joel nicely answers Asaf's question.
  Here I just want to add some footnotes
  to his answer.

1. Joel' argument shows that $ZFC$ cannot even prove the definable class form of axiom of choice for pairs. Historically, this was first done by with a similar argument by Easton in his 1964-thesis (printed in the Annals of Mathematical Logic).  
2. In the pre-forcing era, Mostowski had already shown that $ZFA$ + "the universe can be linearly ordered" does not imply "every set can be well-ordered" (see p.51 of Jech's book on the Axiom of Choice). 
3. As shown in the proof of Theorem 5.21 (p.71) of Jech's text, Mostwoski's argument can be transplanted to the forcing context to show that indeed $V$ can be definably linearly ordered in Cohen's so-called basic model of the failure of the axiom of choice.<|endoftext|>
TITLE: Fundamental groups and homology groups of closed subsets of the plane
QUESTION [6 upvotes]: Let $X$ be a closed subset of $\mathbb{R}^2$.  What restrictions are there on $\pi_1(X)$ and on the homology groups of $X$ (both singular and Cech)?  This is elementary if $X$ has reasonable local properties, but the example of the Hawaiian Earring shows that things can be very complicated indeed.

REPLY [10 votes]: Fundamental Group: The fundamental group of a planar set naturally injects into the first Cech homotopy group, which is an inverse limit of free groups. In particular, the algebraic restrictions gained from this fact are: the fundamental group must be locally free, fully residually free (and thus torsion free), and residually finite. See:

Fischer, H., Zastrow, A., The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655-1676, doi:10.2140/agt.2005.5.1655, arXiv:math/0512343.

Cech Homology: It follows that the first homology pro-group is an inverse system of of finitely generated free groups. The first Cech homology group $\check{H}_1(X)$ is an inverse limit of free abelian groups.
Singular Homology: This is more complicated. Certain cases are understood.
Easy case. If $X$ is locally path-connected and semilocally simply connected, then $H_1(X)\cong \check{H}_1(X)$ is free abelian.
Harder case. Suppose your planar set $X$ is path-connected but not semilocally simply connected. To simplify things, let's at least suppose $X$ is a Peano continuum (locally path-connected compact metric space). It is a general result of Katsuya Eda that for any Peano continuum $X$ (planar or otherwise) the canonical map $\phi:H_1(X)\to \check{H}_1(X)$ is surjective. See:

Katsuya Eda, Kazuhiro Kawamura, The surjectivity of the canonical homomorphism from singular homology to Cech homology, Proc. Amer. Math. Soc. 128 No. 5 (1999) pp 1487-1495, doi:10.1090/S0002-9939-99-05670-1

This result helps because it means that if know $\phi$ splits and we can identify $\ker(\phi)$, then we can "compute" $H_1(X)$.
Now, if $X$ happens to also be one-dimensional (such as the Hawaiian earring or Sierpinski carpet) then we can do exactly that. In particular, it turns out that $\check{H}_1(X)\cong \prod_{n\in\mathbb{N}}\mathbb{Z}$ and $\ker(\phi)\cong \prod_{n\in\mathbb{N}}\mathbb{Z}\Big/ \bigoplus_{n\in\mathbb{N}}\mathbb{Z}$ where the second isomorphism is purely abstract. Moreover, since $\prod_{n\in\mathbb{N}}\mathbb{Z}\Big/ \bigoplus_{n\in\mathbb{N}}\mathbb{Z}$ is algebraically compact, the homomorphism $\phi$ splits. We can conclude that for any (planar) one-dimensional Peano continuum $X$, that
$$H_1(X)\cong \prod_{n\in\mathbb{N}}\mathbb{Z}\oplus \left(\prod_{n\in\mathbb{N}}\mathbb{Z}\Big/ \bigoplus_{n\in\mathbb{N}}\mathbb{Z}\right).$$
The fact that all of these spaces have exactly the same first singular homology group tells you that abelianizing $\pi_1$ for these spaces actually kills all of the geometry remembered by $\pi_1$.
For details, see:

Katsuya Eda, Kazuhiro Kawamura, The Singular Homology of the Hawaiian Earring, Journal of the London Mathematical Society 62 Issue 1 (2000) pp 305–310, doi:10.1112/S0024610700001071 (pdf)
Katsuya Eda, Singular homology groups of one-dimensional Peano continua, Fundamenta Mathematicae 232 Issue 2 (2016) pp 99–115, doi:10.4064/fm232-2-1 (pdf)

Possibly unknown case. As of 2019, I don't think $H_1$ is known for a general 2-dimensional planar set (or Peano continuum) because path reduction is not as straightforward as in the 1-dimensional case. However, in the end, the answer is likely to be similar to the one-dimensional case.<|endoftext|>
TITLE: How Many 4-Manifolds are Symplectic?
QUESTION [16 upvotes]: As an honest question (probably with some subjectivity), how many smooth oriented 4-manifolds are actually symplectic? Can I say half (perhaps under some mild assumptions)? I ask this question because every compact smooth oriented 4-manifold with $b^2_+\ge 1$ admits a near-symplectic form, i.e. a closed 2-form which is symplectic away from a finite set of circles.
Some results that might push the percentage one way or the other:
1) Gompf has shown that any finitely presented group can be realized as the fundamental group
of a compact symplectic 4-manifold.
2) The Seiberg-Witten invariants are nonzero for symplectic 4-manifolds, and in a sense show that they are the "irreducible" basic forms of smooth 4-manifolds.
3) Every compact symplectic 4-manifold is a branched cover of $\mathbb{C}P^2$.
The responses/comments show that we can ask this question (on when can I expect my 4-manifold to be symplectic) in many different ways, each with different expectations. So I am interested in some further thoughts on Tim's and Dmitri's questions.

REPLY [3 votes]: By Donaldson, symplectic manifolds are the same as topological Lefschetz pencils. So you may ask how many 4-manifolds are topological Lefschetz pencils.<|endoftext|>
TITLE: What kind of spectral sequences come from double complexes?
QUESTION [6 upvotes]: Given a double complex in the first quadrant, one can derive from it a (homological or cohomological) spectral sequence converging to the (co)homology of the total complex of the double complex.
My question is: When is a (homological or cohomological) spectral sequence coming from a double complex?

REPLY [15 votes]: There are two different ways to understand the question: 

If I see an abstract spectral seqeunce, is there a double complex such that its spectral sequence is isomorphic to the given spectral sequence? I do not have an answer to that question and, to be honest, do not believe it is an interesting question.
For wich set of names ''$XY$''; the $XY$-spectral sequence can be derived from a double complex?

The answer is that, as a general rule (it might have exceptions), all $XY$-spectral sequences whose $E_2$-terms and $E_{\infty}$ terms are purely homological can be derived from filtered complexes; and most of them in fact from double complexes.
Examples:

The spectral sequence of a simplicial space (Segal; ''Classfying spaces and spectral sequences'') can be reformulated using a double complex (a simplicial space $X_{\bullet}$ gives rise to a simplicial chain complex $C_{\ast} X_{\bullet}$ and thus a double complex. 
The Serre spectral sequence is a special case of the above; a direct construction using a double complex was given by A. Dress, ''Zur Spectralsequenz von Faserungen''.
Special cases of 2. include the Lyndon-Hochschild-Serre spectral sequ. for group extensions; special cases of 1. include the Bousfield-Kan spectral sequ. of a homotopy colimit and some others.
The Eilenberg-Moore spectral sequence comes from a double complex.
Purely algebraic versions: Grothendieck-spectral sequence. Probably the spectral sequence of a Lie algebra extension fits into here. The Van Est spectral sequence for Lie algebra cohomology also comes from a double complex.

The Bockstein spectral sequence is a purely homological construction, it can be derived from a filtered complex; but it does not seem to stem from a double complex.
Other counterexamples are the typical spectral sequence of stable homotopy theory (Atiyah-Hirzebruch, Adams spectral sequence): they cannot be derived from filtered complexes. In fact, if $E$ is a generalized homology theory, you cannot write $E_{\ast} (X)$ of a space in a sensible way as the homology of a chain complex functorially associated with $X$.<|endoftext|>
TITLE: Conjugating the Lyness 5-cycle into a rotation of the plane
QUESTION [6 upvotes]: The Lyness 5-cycle is the map that sends $(x,y)$ to $(y,z)$ with $z=(y+1)/x$.  Leaving aside the set on which the map is not well-defined, the map is of order 5 (hence its name).  Is there an algebraic map that conjugates the map to a rotation by 72 degrees?

REPLY [3 votes]: Yes, this map is conjugate to an automorphism of $\mathbf{P}^2$. See
A. Beauville, J. Blanc, On Cremona transformations of prime order, C.R. Acad. Sci. Paris 339 (2004), no4, 257-259.
See also T. de Fernex, On planar Cremona maps of prime order, Nagoya Math. Journal, Vol. 174 (2004), 1–28, which contains a classification of planar Cremona maps of prime order up to conjugation. According to Remark 1.3.4 therein, the fact above was already known to Iskovskikh.<|endoftext|>
TITLE: Product of conjugacy classes - is there an analog of Tanaka-Krein reconstruction ?
QUESTION [7 upvotes]: Consider a  finite group G. The product of conjugacy classes can be defined in natural way just by multiplying the representatives and counting multiplicities (see e.g. MO 62088). So we get ring with a basis and structure constants are natural numbers. Similar to what one has for product of irreps.
There are many analogies between conjugacy classes and irreps in particular see this article.
Tanaka-Krein duality states that group can be reconstructed from the  tensor category of its representations which is semisimple for finite groups, and hence carries the same information as ring + basis of irreps.
Question:  Can one reconstruct a group having  (ring + basis) made of  conjugacy classes ?
If not -  what partial information (e.g. character table) one can get ? 

Question: Is there any relation between this ring and ring of irreps of the same group ? or may be some other group ?
(Remark. For abelian group they are isomorphic.)
Question: Are there any further analogies between ring of irreps and conjugacy classes except mentioned in the paper cited above ?

REPLY [2 votes]: One nice fact, I think, is that the formula of Burnside that Geoff Robinson gives above,
$$
 N^{C_z}_{C_x C_y} = \frac{|G|}{|C_G(x)| |C_G(y)|} \frac{\chi(x) \chi(y) \chi(z^{-1})}{\chi(1)} 
$$
can be understood nicely from a geometric / topological quantum field theory perspective. I think it is precisely the "Verlinde formula" for the modular category Rep(/\G), the representation category of the Drinfeld double of C[G]. The Verlinde formula says that in a general modular category, we have
$$
 N^k_{ij} = \sum_r \frac{s_{ir} s_{jr} s_{k^* r}}{s_{0 r}}
$$
where $s_{ij}$ is the S-matrix. 
More concretely, this is to say that it has a natural interpretation in terms of G-bundles on the torus. This perspective also comes out in the appendix of Zagier to "Graphs on surfaces and their applications" by Lando and Zvonkin.<|endoftext|>
TITLE: Is there a classification of or work towards a classification of countable ordered sets?
QUESTION [6 upvotes]: Is there a classification of or work towards a classification of countable ordered sets?

REPLY [6 votes]: Yes. The reference to get started is 

MR0662564 (84m:06001) 
  Rosenstein, Joseph G.
  Linear orderings. 
  Pure and Applied Mathematics, 98. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London, 1982. 

The key idea is Hausdorff's notion of scattered order. These are the ordered sets that contain no copies of the rationals. They admit a detailed classification, essentially by an analysis reminiscent of the study of Cantor-Bendixson derivatives (more precisely, we can build up all scattered ordered sets by iterating "sums" of simple orders, the last link below gives more details). 
The key result is Laver's theorem that Fraïssé’s conjecture holds: The class of countable linear ordering can be quasi-ordered by embeddablity. With this ordering, the class contains no infinite descending chain and no infinite antichain.
(So, this is not a classification in the model theoretic sense, but it is more detailed and far-reaching than simply discussing a basis, which of course consists simply of $\omega$ and $\omega^*$.)
If you want to see how Hausdorff's analysis can be used, Laver's proof is the result to study. For an extension, see:

Continuous Fraïssé Conjecture. Arnold Beckmann, Martin Goldstern, Norbert Preining. MR2470199 (2010a:03021) Order 25 (2008), no. 4, 281–298. 

Hausdorff's analysis can be continued into a study of uncountable ordered sets, but now things get much more involved, and ZFC does not settle their basic structure theory. See this answer to a related question on the sister site for more details.<|endoftext|>
TITLE: Interpreting $f^*f_*$
QUESTION [12 upvotes]: For a morphism of schemes $f: X\rightarrow Y$, one often considers the function $f^*f_*$ on sheaves. For example, this appears in adjunction for sheaves of $\mathcal{O}_X$-modules, the projection formula, and other relatively abstract places. As a somewhat vague question, I'd like to see some non-trivial examples of how to interpret $f^*f_* \mathcal{F}$, or typical examples where this is an important construction. How about the $f^*f_*$ map on sheaves in other sites, such as the etale site?
Some context:
1.) Pullback of coherent sheaves is a fairly natural operation. I'm also perfectly happy with the pushforwards and higher pushforwards. Specifically, Mumford (in "Abelian Varieties") gives some fairly general results about how to interpret stalks of such sheaves (if $f$ is proper flat and $\mathcal{F}$ is locally free or more generally flat over $Y$) in terms of cohomology groups of $\mathcal{F}$ on fibers. More generally the Leray spectral sequence gives you a context to "interpret" higher pushforwards. If the pushforward of the structure sheaf is again the structure sheaf then Stein factorization gives us information about the fibers. More generally we have the theorem of formal functions.
2.) For finite degree $n$ maps of smooth proper curves over $\bar{k}$, the related map$f_*f^*$ can be defined at the level of the Picard group, where it is the multiplication by $n$ map.
$\textbf{Question}$ How should I think about Thm III.8.8 of Hartshorne (below)? Is this just a useful technical result (and where?), or is it more than that? Are there good examples? Affine locally one can certainly describe the natural map $f^*f_*(\mathcal{F}) \rightarrow \mathcal{F}$ explicitly. Is this how one "works" with it in practice, or are there better ways (I hope)?
$\textbf{Thm}$ III.8.8: If $f:X \rightarrow Y$ is a projective map of noetherian schemes, $\mathcal{F}$ coherent on $X$ and $\mathcal{O}_X(1)$ a very ample sheaf on $X$ over $Y$, then $\forall n>> 0$, the natural map $f^*f_*(\mathcal{F}(n)) \rightarrow \mathcal{F(n)}$ is surjective.

REPLY [5 votes]: Let both $f_*$ and $f^*$ be derived pushforard and pullback. Assume that $f_*O_X = O_Y$. Then there is a semiorthogonal decomposition of the derived category 
$$
D(X) = \langle Ker f^*, f^*(D(Y)) \rangle
$$
and $f^*f_*$ is the projection onto the second component.<|endoftext|>
TITLE: Use of Hilbert Schemes in Arithmetic?
QUESTION [9 upvotes]: I'm curious about the following: What are some arithmetic application of the Hilbert Schemes?
The application of Hilbert schemes in algebraic geometry seems to be a great success, from birational geometry to enumerative geometry. However, although the Hilbert scheme is defined over $\mathbb{Z}$, I haven't seen any use of it in the arithmetic side. If you know such an example please let me know.

REPLY [12 votes]: Quite generally, whenever you ``need a moduli space'', say, polarized deformations of varieties, or spaces of morphisms, you oftentimes construct it as follows: first, you construct some family in projective space that (over-)parametrizes your data. Then, using that you've fixed some numerical invariants, you prove that the Hilbert polynomial of this projective family is constant. Then, you use the Hilbert scheme to realize this family as a subscheme of a Hilbert scheme. This is a first approximation to your moduli problem. Usually, you've overparametrized your data and need to take some appropriate stack/GIT quotient (which is usually subtle)...
Thus, in case you're interested in arithmetic moduli, say, moduli spaces of polarized Abelian varieties over $\mathbb{Z}$, moduli spaces of curves of genus $g\geq2$ over $\mathbb{Z}$, you will need the fact that the Hilbert scheme is defined over $\mathbb{Z}$.
Let me even give an application to complex geometry: when proving the existence of rational curves on (complex!) varieties, whose $K_X$ is not nef, via "bend and break", you do the following: you reduce your variety modulo positive characteristic $p$, and construct the desired rational curves on infinitely many reductions modulo $p$ using the Frobenius morphism and characteristic-$p$-methods. Then, you bound the degree of these rational curves (w.r.t. some polarization). Now, to conclude the existence of a rational curve in characteristic zero, you use the space of morphisms from $\mathbb{P}^1$ to show lifting of these curves from characteristic $p$ to characteristic zero. Here, it is essential that this space of morphisms (whose existence relies on the Hilbert scheme) is defined over some ring of integers.<|endoftext|>
TITLE: Can measures be added by forcing?
QUESTION [17 upvotes]: The Lévy-Solovay theorem says that small forcings do not create measures.  J.D. Hamkins has generalized this to a larger class of forcings called gap forcings.  I would assume this cannot be generalized to all forcings, but I cannot think of a counterexample.
Is there a forcing notion that creates a $\kappa$-complete (or even countably complete) measure $\mu$ on some uncountable cardinal $\kappa$ such that $\mu \cap V$ is not in $V$?

REPLY [4 votes]: Just some comments to complement Joel's answer: 
That forcing can destroy and then recreate measurability is due to Kunen: 

Kenneth Kunen. Saturated ideals, The Journal of Symbolic Logic 43 (1) (1978), 65–76. MR0495118 (80a:03068) 

The same is true for real valued measurability, this is due to Gitik: 

Moti Gitik, Saharon Shelah. More on Real-valued measurable cardinals and forcing with ideals, Israel Journal of Mathematics 124 (2001), 221–242 ([GiSh 582]). MR1856516 (2002g:03110)

For a while it was open whether we can destroy and then reconstruct measurability while still preserving weak compactness. In all arguments I was aware of, it is essential that weak compactness fails in the intermediate model, as what we accomplish is a (destructible) instance of failure of stationary set reflection. Joel points out that his argument actually preserves weak compactness, and can be carried out in other settings to preserve measurability or stronger properties. 
More generally, one can ask whether measurability can "appear spontaneously" by forcing, through some other means, just as we obtain saturation on the non-stationary ideal on $\omega_1$ by forcing MM, which cannot be traced back to some measure in some inner model that the forcing is reconstructing. 
For another proof of the specific result you are asking (but one that destroys weak compactness in the intermediate extension), see section 4 of my paper on RVM cardinals:

Real-valued measurable cardinals and well-orderings of the reals. In Set theory. Centre de Recerca Matemàtica, Barcelona, 2003–2004, Joan Bagaria, and Stevo Todorcevic, eds.; Trends in Mathematics, Birkhäuser Verlag, Basel, 2006, pp 83–120, MR2267147 (2007g:03064)<|endoftext|>
TITLE: Condition Number related to Root finding problems
QUESTION [5 upvotes]: Suppose we want to find the root of the equation $f(x)=\phi(x) - d = 0$, where d is a real constant and $f$ is continuously differentiable function. 
The problem is well posed if the inverse $\phi^{-1}$ exists, since in that case $\phi^{-1} (d) = x$. 
Now most numerical analysis books I read on the topic of solving nonlinear equations (for example this grad level book) open the discussion by introducing the condition number associated with the problem, in this case it happens to be. 
$$ \kappa = \frac{1}{f^{\prime}(x)} = \frac{1}{\phi^{\prime}(x)}$$
From this equation we see that the condition number $\kappa$ is large when $\phi^{\prime}(x)$ is close to zero and we are told large $\kappa$ leads to ill conditioned problems.
But so what?? I can't think of an example for which say Newton Raphson or any other root finding shceme fails when the problem is ill conditioned.
I understand the condition number measures the sensitivity of finding roots of $f$ with respect to the input datum, $d$ in this case. But what practical implications does this have? 
I can think of plenty of severely ill conditioned problems. But Newton raphson has no problems with them (I know there are situations in which Newton Raphson fails, but these are due to problems intrinsic to the algorithm and are not dependent on the condition number). Does any one know of an example where the condition number for the above problem is large and a root finding scheme fails? If not then whats the point of examining the condition number.   
I apologize in advance if this question is not of the level of the forum. I thought since the answer cannot be found in an elementary text, here was a good place to ask.

REPLY [5 votes]: In addition to the convergence speed (and radius) mentioned in Pietro Majer's answer, there is another factor: if the problem is ill-conditioned, the solution is sensitive to perturbations. 
If you make an error of magnitude $\varepsilon$ in computing the iteration or the parameters data, then the solution is perturbed by a quantity $\kappa \varepsilon$, where $\kappa$ is the condition number. In practice, you have an error of at least one part in $10^{-16}$ in most computer implementations, and often an even larger one if your function depends on measured data or approximations. So Newton's method may converge without trouble, but the solution that you get could be total garbage.<|endoftext|>
TITLE: Can bilipschitz models of hyperbolic 3-manifolds be made effective?
QUESTION [6 upvotes]: In their proof of the Ending Lamination Conjecture, Brock, Canary, and Minsky prove existence of bilipschitz models of hyperbolic 3-manifolds (homeomorphic to a surface times $\mathbb{R}$) depending only on the topology and ending invariants of the manifold.
(See The classification of Kleinian surface groups, II: The Ending Lamination Conjecture and
The classification of Kleinian surface groups. I. Models and bounds)
Their proof is non-constructive, so the bilipschitz constants cannot be computed from their proof.
Are the bilipschitz constants close to being computable from their proof? In other words is it "easy" to see what steps in their proof are non-constructive and whether these steps can be made effective?

REPLY [7 votes]: See Bowditch: link text Systems of bands in hyperbolic 3-manifolds
with an approach to the Brock-Canary-Minsky Theorem (though not through their model manifold) that is, in principle, effective. Though I am not aware of an explicit algorithmic realization.<|endoftext|>
TITLE: Effect of large cardinals on the value of $\omega_1^L$ in $L$
QUESTION [7 upvotes]: I have three three questions, the first two of which probably have the same answer and the third of which is more vague. 
For a set $A$ let $L_\alpha(A)$ be the constructible universe up to $\alpha$, built from $A$ as a set (and not a predicate). Further let $X = (B, f)$ where $B$ is a transitive set and $f$ is a bijection from $\omega$ to $B$. 
Also assume that the background universe has whatever large cardinals you would like (or that would be helpful). In particular though there is at least one inaccessible cardinal in $L$. 
(1) Suppose $L_\alpha\models ZFC$. Is it the case that $\omega_1^L = \omega_1^{L_\alpha}$?
(2) Suppose $L_\alpha(X)\models ZFC$. Is it the case that $\omega_1^{L(X)} = \omega_1^{L_\alpha(X)}$?
(3) If the answer to (1), (2) is yes, is there any simpler way for $L_\alpha$ to know that $\omega_1^{L_\alpha} = \omega_1^L$ (other than $L_\alpha\models ZFC$)?
Finally I will just make one observation to highlight why this question isn't trivial. If you replace $ZFC$ with $KP$ then there are many countable admissible sets $L_\alpha\models KP$ with countable (in $L$) ordinals $\beta\in L_\alpha$ such that $L_\alpha \models \omega_1 = \beta$. 
Thanks

REPLY [4 votes]: Andres Caicedo has answered Questions 1 and 2 as stated, but, just in case you intended $\kappa>\omega_1^L$ in Question 1 (since that's obviously necessary for the proposed conclusion), let me add that this is also sufficient for an affirmative answer to Question 1.  The point is that, if an ordinal $\alpha$ is countable in $L$ then a bijection between $\alpha$ and $\omega$ appears in the construction of $L$ well before stage $\omega_1^L$.  This is essentially contained in Gödel's proof of GCH in $L$.  It uses much less than the full strength of ZFC in $L_\kappa$; you just need enough set theory to talk sensibly about countability of ordinals.<|endoftext|>
TITLE: The existence of an elliptic curve with a specific Galois representation induced by a character
QUESTION [6 upvotes]: In Kevin Buzzard's survey article on potential modularity Buzzard writes:

Let us say that we have an elliptic
  curve $E$ over a totally real field $F$,
  and we want to prove that $E$ is
  potentially modular (that is, that $E$
  becomes modular over a finite extension
  field $F^{′}$ of $F$, also assumed totally
  real). Here is  a strategy. Say $p$ is
  a large prime such that $E[p]$ is
  irreducible. Let us write down a
  random odd $2$-dimensional mod $ℓ$
  Galois representation $\rho_{ℓ} :
> Gal(\overline{F}/F)  → GL(2,\mathbf{F}_ℓ )$ which is
  induced from a character; because this
  representation is induced it is known
  to be modular. Now let us consider the
  moduli space parametrising elliptic
  curves $A$ equipped with 

An isomorphism $A[p] \cong E[p] $
An isomorphism $A[ℓ]\cong ρ_ℓ$

This moduli problem will be
  represented by some modular curve,
  whose connected components will be
  twists of $X(pℓ)$ and hence, if $p$
  and $ℓ$ are large, will typically have
  large genus. However, such a curve may
  well still have lots of rational
  points, as long as I am allowed to
  look for such things over an arbitrary
  finite extension $F^{′}$ of $F$ !

It's not immediately obvious to me that there's an elliptic curve $A$ over some $F^{′}$ satisfying the second condition alone (never mind satisfying both conditions simultaneously). Is there a simple explanation for why there should be such an $A$? Did Professor Buzzard mean "consider the set of A such that $A[ℓ]\cong ρ_ℓ$ for some representation induced by a character" (as opposed to a particular one)?

REPLY [8 votes]: In this context, if $\rho$ is a mod $\ell$ representation of $Gal(\overline{F} / F)$, and $A$ is an elliptic curve over an extension $F' / F$, then the statement "$A[\ell] \cong \rho$" needs a little bit of interpretation, because the two sides are representations of different things: $A[\ell]$ is a mod $\ell$ representation of the subgroup $Gal(\overline{F} / F') \subset Gal(\overline{F} / F)$. So the statement is to be read as "$A[\ell]$ is isomorphic as a $Gal(\overline{F} / F')$-representation to the restriction of $\rho$". Now, the bigger $F'$ is, the weaker this condition becomes: in particular, if we take any elliptic curve $A$ over $F$ and define $F'$ to be the extension of $F$ generated by the $\ell$-torsion points of $A$ and the splitting field of $\rho$, then the statement is automatic (both sides are the trivial representation). 
(This is kind of a stupid example, but maybe you can believe now that there exist non-stupid examples as well!)<|endoftext|>
TITLE: If a compact Kahler manifold $(M,g)$ has constant scalar curvature, is the metric $g$ real analytic?
QUESTION [8 upvotes]: Hi to all!
Perhaps it is a silly question, if so i'll delete this post.
Suppose we have a compact Kahler manifold $(M,g)$ of complex dimension $m$ with constant scalar curvature with respect to its metric $g$. My question is: does the condition of constant scalar curvature imply that the metric $g$ automatically real analytic? 
When i say that the metric is real analytic i mean that in a holomorphic coordinate chart with coordinate functions 
$$(z^1,\ldots, z^m)\quad \textrm{ with } z^j=x^j+iy^j \textrm{ for }1\leq j\leq m$$ 
the coefficients $(g_{i\bar{j}})_{1\leq i,j\leq m}$ are analytic functions w.r.t. $x^k,y^k$.
Thank you in advance!

REPLY [18 votes]: It's not a silly question, but there's a standard answer, and it's a purely local result:  If the Kähler metric is $C^2$ and has constant scalar curvature, then it is real-analytic with respect to the real-analytic structure that underlies the complex-analytic structure.  The reason is that setting the scalar curvature equal to a constant is an elliptic equation for the potential of the metric that is an analytic function of its arguments in the local real-analytic coordinates that you define, and so the elliptic regularity results of Hopf and Morrey apply.<|endoftext|>
TITLE: Bochner integral of stochastic process = path by path Lebesgue integral?
QUESTION [22 upvotes]: After some helpful comments, I realized that I had to repost this question in a more systematic way. 
On a complete probability space, let $\mathcal{H}_0$ denote the Hilbert space of square integrable random variables with zero mean. A stochastic process $X$ is called a second order process if $\mathbf{E}X(t)^2 < \infty$ and $\mathbf{E}X(t) = 0$, all $t \in [0,T]$. Such a process can be regarded as a map $[0,T] \rightarrow \mathcal{H}_0$. It is called q.m. continuous if this map is continuous, i.e. $X(s) \rightarrow X(t)$ in quadratic mean as $s \rightarrow t$. One can show that each q.m. continuous process has a measurable version.
Let $X$ be a q.m. continuous second order process. We want to compute the integral $\int_0^T X(s) \mathrm{d} s$. There are two ways.
Bochner integral. Clearly, $X$ considered as a continuous map $[0,T] \rightarrow \mathcal{H}_0$ is Bochner integrable. We denote its Bochner integral by
\begin{equation}
\text{(B-)}\int_0^T X(s) \mathrm{d}s.
\end{equation}
Lebesgue integral. We may assume that $X$ considered as a map $[0,T] \times \Omega \rightarrow \mathrm{R}$ is measurable. Thus, for fixed $\omega$, the integral $\int_0^T X(s,\omega) \mathrm{d} s$ exists as a Lebesgue integral, and we denote the random variable constructed in this way by
\begin{equation}
\text{(L-)}\int_0^T X(s) \mathrm{d}s.
\end{equation}
Question. Do we have 
\begin{equation}
\text{(B-)}\int_0^T X(s) \mathrm{d}s = \text{(L-)}\int_0^T X(s) \mathrm{d}s \quad \text{a.s.?}
\end{equation}
Ideas.
Let $\lbrace t^n = t_0^n, \ldots t_{k_n}^n \rbrace$ be a sequence of partitions of $[0,T]$ with mesh going to zero. Define the simple functions
\begin{equation}
\xi_n = X(t_0^n)1[t_0^n,t_1^n] + \sum_{i=1}^{k_n-1} X(t_i^n) 1[t_i^n, t_{i+1}^n).
\end{equation}
Then one can show that for almost every $t$, we have $\xi_n(t) \rightarrow X(t)$ in $\mathcal{H}_0$, and
\begin{equation}
\int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(B-)}\int_0^T X(s) \mathrm{d}s 
\quad \text{in $\mathcal{H}_0$}, 
\end{equation}
where the integral on the left is defined in the obvious way (we omit (B-)) (to show this, one uses the fact that the covariance function $r(s,t) = \mathbf{E}X(s)X(t)$ of a q.m. continuous process is continuous). After switching to a subsequence if necessary, we may assume that
\begin{equation}
\int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(B-)}\int_0^T X(s) \mathrm{d}s 
\quad \text{$\mathbf{P}$-a.s.}, 
\end{equation}
Now, we would like to have that also 
\begin{equation}
\int_0^T \xi_n(s) \mathrm{d}s \rightarrow \text{(L-)}\int_0^T X(s) \mathrm{d}s 
\quad \text{$\mathbf{P}$-a.s.}, 
\end{equation}
But this is tricky. The sums on the left hand side are Riemann sums, i.e. 
\begin{equation}
\int_0^T \xi_n(s) \mathrm{d}s = \sum_{i=0}^{k_n-1} X(t_i^n)(t_{i+1}^n - t_i^n  ).
\end{equation}
So if we knew that the paths of $X$ are a.s. Riemann integrable, we would be done. But this is not clear. I also tried to use some approximation arguments, but couldn't do it. It seems like one needs to deduce some kind of path regularity of $X$ from the assumption of q.m. continuity, but I don't know any results in this direction.

REPLY [17 votes]: Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question.
For a probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $X\colon[0,T]\to L^p(\mathbb{P})$ ($1\le p\le\infty$) be Bochner integrable w.r.t the Lebesgue measure on $[0,T]$, and also jointly measurable as a map $(t,\omega)\mapsto X(t)(\omega)$ from $[0,T]\times\Omega$ to $\mathbb{R}$.
Then, the Bocher integral $\int_0^T X(t)\,dt$ agrees with the pathwise Lebesgue integral $\int_0^TX(t)(\omega)\,dt$ for almost every $\omega$.
First, this statement clearly holds for simple functions, which are finite linear combinations of terms of the form $X(t)(\omega)=1_{\lbrace t\in A\rbrace}1_{\lbrace\omega\in B\rbrace}$, for $A$ a Borel subset of $[0,T]$ and $B$ in $\mathcal{F}$. Now, by definition, if $X$ is Bochner integrable then, for each $n\ge1$, there is a simple $\xi_n$ such that
$$
\int_0^T\lVert X(t)-\xi_n(t)\rVert_p\,dt\le2^{-n}.
$$
The Bochner integral is given by
$$
\int_0^T\xi_n(t)\,dt \rightarrow\text{(B-)}\int_0^T X(t)\,dt.
$$
Here the limit is taken in the $L^p$ norm and, hence, also holds for convergence in probability.
Using pathwise Lebesgue integration along the sample paths of $X$ now, we can use Fubini's theorem to commute expectation, integration and summation signs.
$$
\begin{align}
\mathbb{E}\left[\int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\,dt\right]
&=\sum_{n=1}^\infty\int_0^T\mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]\,dt\cr
&\le\sum_{n=1}^\infty\int_0^T\left\lVert X(t)-\xi_n(t)\right\rVert_p\,dt\cr
&\le\sum_{n=1}^\infty2^{-n}=1 < \infty.
\end{align}
$$
In particular,
$$
\int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\,dt < \infty
$$
with probability one. Looking at any sample path for which this is finite, $\lvert X(t)-\xi_n(t)\rvert\to0$ as $n\to\infty$ for Lebesgue almost every $t$. Also, $\lvert X(t)-\xi_n(t)\rvert$ is dominated by its sum over $n$. Therefore dominated convergence applies,
$$
\int_0^T\xi_n(t)\,dt\rightarrow\textrm{(L-)}\int_0^T X(t)\,dt.
$$
This holds for almost every sample path of $X$, so the limit holds in probability. Hence the Lebesgue integral on sample paths agrees with the Bochner integral.<|endoftext|>
TITLE: A doubt about the parts of the spectrum of tensor products
QUESTION [7 upvotes]: Let $\mathcal{H}$ be any complex Hilbert space of infinite dimensional. By an operator $T$ I mean a linear bounded transformation from $\mathcal{H}$ into $\mathcal{H}$, i.e, $T:\mathcal{H}\rightarrow\mathcal{H}$ is a linear transformation, such that $\left\|Tx\right\|\leq \beta \left\|x\right\|$, for some $\beta\geq 0 $ in $\mathbb{C}$. The sets $\mathcal{N}(T)$ and $\mathcal{R}(T)$ denote the kernel and the range of $T$. Also, $\mathcal{B}[\mathcal{H}]$ is the set the of all operators defined in $\mathcal{H}$.
The spectrum of $T$ is the set $\sigma(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)\neq 0 $ or $\mathcal{R}(\lambda I - T)\neq \mathcal{H} )$, where $I\in\mathcal{B}[\mathcal{H}]$ is the identity operator (i.e., the spectrum of $T$ is the set of all $\lambda$ such that $(\lambda I - T)$ fail to have a bounded inverse on $\mathcal{R}(\lambda I - T)=\mathcal{H}$). A classical partition of the spectrum is:
i. Point Spectrum: $\sigma_{P}(T) = (\lambda \in \mathbb{C}: \mathcal{N}(\lambda I - T) \neq 0)$.
ii. Continuous Spectrum: $\sigma_{C}(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)= 0 , \mathcal{R}(\lambda I - T)^{-}=\mathcal{H}$ and $\mathcal{R}(\lambda I - T)\neq \mathcal{H} )$, where $\mathcal{R}(.)^{-}$ denotes the closure of $\mathcal{R}(.)$.
iii. Residual Spectrum: $\sigma_{R}(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)= 0 $ and $ \mathcal{R}(\lambda I - T)^{-}=\mathcal{H})$.
Let $\otimes$ denote the tensor product. Consider $T=(A\otimes B)\in\mathcal{B}[\mathcal{H}\otimes\mathcal{H}]$, where $A\in\mathcal{B}[\mathcal{H}]$ and $B\in\mathcal{B}[\mathcal{H}]$. It is well known that $\sigma(T) = \sigma(A\otimes B) = \sigma(A)\sigma(B)$. My questions are:
a. $\sigma_{C}(T) = \sigma_{C}(A)\otimes \sigma_{C}(B)$ ?
b. $\sigma_{R}(T) = \sigma_{R}(A)\otimes \sigma_{R}(B)$ ?
Thank you very much for your attention.

REPLY [7 votes]: Let me first show that there are problems with various parts of the spectrum here, hence your concerns are well-grounded.
The first is, citing Ichinose in the Transactions from 1978 here and here, that
$$\sigma_{ess}(A\otimes B) = \sigma_{ess}(A)\sigma(B)\cup\sigma(A)\sigma_{ess}(B)$$
showing that equalities like your a. and b. may not always be true. 
Further, here is an example of Ichinose showing a funny fact: Let $H=l^2$, $Ae_n=-e_{n-1}$ be the negated left-shift $(Ae_1=0)$. Let $\tau_n\to 0$, $\max |\tau_n|<1$ and define $Be_n:=\tau_n e_{n+1}$. 
Then the point spectrum of $B$ is empty, the point spectrum of $A$ is the open circle $|\lambda|<1$. There will be however many elements in the kernel of $(A+I)\otimes(B+I)$: in fact, for every $y\in H$, 
$$\sum_{n=1}^{\infty}e_n\otimes B^{n-1}y$$
belongs to the kernel of this operator. 
Hence, the point spectrum of the tensor product cannot be recovered from the point spectra of the original operators. Since the residual spectrum is the (conjugate of) the point spectrum of the adjoint, it follows that your claim b. cannot hold, i.e., in general
$$\sigma_R(A\otimes B) \neq \sigma_R(A)\cdot\sigma_R(B),$$
or even
$$\sigma_R(A\otimes B) \neq \sigma_R(A)\cdot\sigma(B)\cup\sigma(A)\cdot\sigma_R(B).$$<|endoftext|>
TITLE: What does the numerically verified part of the Riemann Hypothesis tell about prime numbers?
QUESTION [24 upvotes]: I'm curious about the following question:
As of 2005(?) the Riemann hypothesis is verified for the first 10 trillion zeroes, they are all on the critical line. Does this verification gives us any information about prime number? 
In particular, are there any results saying if all the non-trivial zeroes whose imaginary part is < N and > 0 are on the critical line, then we understand something about prime number < M, where M is a number depend on N?

REPLY [13 votes]: If you look at the explicit formula, then you can get a bound for the error term in the PNT: If 
$$
\psi(x) = \sum_{p^k\le x}\log p,
$$ 
then formula (9) in page 109 of Davenport's book (multiplicative number theory) implies that
$$
\psi(x) = x + \sum_{-T\le \gamma\le T}\frac{x^\rho}{\rho}  + O\left(1+\frac{ x\log^2(xT) }{T}\right),
$$
for every $T\ge1$, where the implied constant is completely effective. Now, if we know that all the zeroes of $\zeta$ up to height $T$ lie on the critical line, then this automatically implies that
$$
|\psi(x) - x | \le x^{1/2}\sum_{-T\le \gamma\le T} \frac{1}{\sqrt{1/4+\gamma^2}} + O\left(1+ \frac{ x\log^2(xT)}{T} \right) \ll x^{1/2}\log^2T + \frac{ x\log^2(xT)}{T},
$$
for some effective implied constants. So, in certain ranges of $x$, depending on $T$, you can get very good bounds on the size of $\psi(x)$ and therefore on how many primes there are up to $x$.<|endoftext|>
TITLE: Is there a topograph for Pythagorean triples?
QUESTION [12 upvotes]: I have been reading Allen Hatcher's notes on quadratic forms.  Naturally, we draw a picture encoding all the values of a quadratic form in a topograph.  These are build by iterating the parallelogram identity : 
$$ 2 Q(\vec{v})+2Q(\vec{w}) = Q(\vec{v}+\vec{w}) + Q(\vec{v}-\vec{w})$$
These can be found in Ch 1 of The Sensual (Quadratic) Form by John H Conway.  
They are many interesting related to Farey fractions, circle packings, Voronoi tesselations and Kleinian groups.
      
(source)
I am interested points $x,y,z \in \mathbb{Z}^3$ in the quadratic form $Q(x,y,z) = x^2 + y^2 - z^2$ vanishes.  It's known such triples exhibit a ternary tree structure.  One can multiply vector $(x,y,z)$ by any of
$$ \left[\begin{array}{ccc}1 &-2 & 2 \\\\ 2& -1& 2\\\\ 2& -2 & 3 \end{array} \right]
\text{ or }
\left[\begin{array}{ccc}1 &2 & 2 \\\\ 2& 1& 2\\\\ 2& 2 & 3 \end{array}  \right]
\text{ or }
\left[\begin{array}{ccc}-1 &2 & 2 \\\\ -2& 1& 2\\\\ -2& 2 & 3 \end{array}  \right] $$
and get another Pythagorean triple.  The result is an $\Gamma(2)$ action on the Pythagorean triples.
      
(source)
If I had to guess, the topograph would be somehow dual to the hyperbolic tessellation associated to the congruence group.  The vertices of the "topograph" would be (similar to) the Farey fractions and the relation would involve 6 numbers instead of 4.  I wonder what it could be.
      
 (source, from Wayback Machine)
  
(but see also)
Having a topograph for solutions of quadratic forms rather than values is not without precent.  It's been done for Appolonian circle packings and Markoff triples.  The topograph itself, has extension to ground fields other than $\mathbb{Q}$.

REPLY [6 votes]: I'm not sure what "topograph" means, but if it's just a fancy word for "pictoral description of features in a region" then the answer is yes.
Send your triple $(a,b,c)$ to the element $a/c+ ib/c$ in the unit circle, and apply the Cayley transform $z \mapsto -i\frac{z-1}{z+1}$ to get elements of the rational projective line.  The basic triple $(1,0,1)$ is sent to $0$, and the standard symmetry-breaking restriction specifying that $b$ is even identifies the set of Pythagorean triples with the $\Gamma(2)$ orbit of $0$, i.e., the set of rationals with even numerator.  We will call rationals with even numerator "special cusps".
Divide the upper half plane into fundamental domains for $\Gamma(2)$ - each domain will touch a single special cusp, and three non-special cusps.  The topograph is formed by dividing the half-plane into regions, where each region is the union of  all fundamental domains touching a special cusp.  You can also use the Cayley transform to get a topograph in the disc (as in your last picture above), where the special cusps are precisely the rational points on the unit circle whose $y$-coordinate has even numerator.
Unlike the quadratic form topograph, the regions only meet at edges, because the fundamental domain of $\Gamma(2)$ has no corners - this has something to do with the lack of finite order elements.  I tried drawing the topograph in a disc, but it doesn't look so great.  The regions attached to a single Pythagorean triple are so big that everything except about three regions will be smushed into a small neighborhood of the boundary circle.

Since you added some other information, like $3 \times 3$ matrices and ternary trees, I guess you want to know how they are related to topograph process.  The correspondence between the $\Gamma(2)$ action and the $3 \times 3$ matrices is an integral version of the exceptional isomorphism between $PSL_2(\mathbb{R})$ and $SO(2,1)^+$ (but perhaps $PGO(2,1)^+$ is more canonical given the process that follows).  For example, conjugation by Cayley sends $U = \binom{12}{01}$ to the Möbius transformation $z \mapsto \frac{(1+i)z + i}{-iz+1-i}$, which takes $a+ib$ to $\frac{-a-2b-2 + (2a+b+2)i}{2a+2b+3}$.  We can now view the denominator and the real and imaginary parts of the numerator as row vectors, sending $(a,b,1)^T$ to the given combinations.  This yields your second matrix with the first row negated.  The sign change doesn't affect the validity of the triples, since it just negates $x$.  A similar process with $L = \binom{10}{21}$ yields $z \mapsto \frac{(1-i)z + i}{-iz+1+i}$, taking $a+ib$ to $\frac{-a+2b+2+(-2a+b+2)i}{-2a+2b+3}$, which produces your third matrix.  Your first matrix corresponds to $U^{-1}$, and $L^{-1}$ goes to a fourth matrix that you didn't write.
The ternary tree is a quotient of a connected subset of the Cayley graph of $\Gamma(2)/\langle - I\rangle$, which is free of rank 2.  Because the subgroup generated by $L$ is the stabilizer of the cusp at 0, the cusps are in correspondence with left cosets, i.e., the set of words in $U$ and $L$ that don't end in a nonzero power of $L$.  You can write this coset space as the subset of the Cayley graph (an infinite 4-valent tree) from which two of the four branches incident to the identity have been deleted, i.e., a pair of ternary trees sharing a common root at the identity.  The entries in the two ternary trees differ by a sign, so we typically identify entries with the same absolute value.  This is described in greater length in the article by Alperin that you linked (the main difference being that Alperin uses an action by conjugation on nilpotent $2 \times 2$ integer matrices instead of Möbius transformations).<|endoftext|>
TITLE: Interesting results for open Riemann surfaces
QUESTION [9 upvotes]: As far as I know, interesting results for open Riemann surfaces are quite rare. One of them is the theorem of Gunning and Narasimhan, which asserts that every connected open Riemann surface admits a holomorphic immersion into the complex plane. Another example is given by the theorem of Behnke and Stein, which says that every connected open Riemann surface is a Stein manifold. Is there any more interesting results for open Riemann surfaces?

REPLY [4 votes]: I hope that the following result by Bishop and myself might be "interesting": Every open Riemann surface is a cover over the sphere, branched over only three points. Equivalently, every open Riemann surface is equilaterally triangulable.
See
Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles and https://arxiv.org/abs/2103.16702 .<|endoftext|>
TITLE: Linear algebra over principal rings
QUESTION [6 upvotes]: Consider an extension $R\subseteq S$ of commutative rings, and suppose that $R$ is principal (i.e., $0$ is the only zero-divisor of $R$ and every ideal of $R$ has a generating set of cardinality $1$). By means of scalar restriction we consider $S$ as an $R$-module. Let $M$ be a sub-$R$-module of finite type of $S$ containing $R$.
In this situation, Gilmer and Heinzer claim (in Remark 2 in their article "On the complete integral closure of an integral domain") that there exists an $R$-module $N$ containing $M$ such that $R$ is a direct summand of $N$.
Their argument is just the remark that $M$ "has a linearly independent module basis containing the identity of $R$".
Unfortunately, I cannot follow this argument, nor can I prove the claim in a different way. Even worse, I meanwhile have the feeling that the claim is not true.
Does someone know either a proof of this claim or a counterexample?

REPLY [6 votes]: I think that I have a counterexample. Let $R=\mathbb Z$, $S=\mathbb Q$ and $M= 1/2 \mathbb{Z}$. The induced map $R/2 R \rightarrow M/2M$ is zero. If $R$ is a direct summand of some $N$ the $R/2R\rightarrow N/2N$ is non-zero. Hence, $R\subset M\subset N$ implies that 
$R$ cannot be a direct summand of $N$, since $R/2R \rightarrow N/2N$ factors through $R/2R \rightarrow M/2M$.<|endoftext|>
TITLE: Historical questions on the term "general abstract nonsense"
QUESTION [6 upvotes]: Saunders Mac Lane reports that the contents of his 1942 paper (joint with Samuel Eilenberg), that first introduced categories, were then referred to (in the words of prominent representatives of the mathematical community of that time?) as "general abstract nonsense''. While today the term is mostly used, especially by practitioners, as an implicit recognition of deep mathematical perspectives (rather than in a derogatory sense), is it correct that the tone was actually sarcastic, to say the least, in the early days of the subject? Or is it a falsehood? Could you point me out some references (more focused on this than Mac Lane's article from the above link) in support of one or other of the two versions? Thanks in advance for any help.
Added later. According to the bibliography included in the Wiki article linked by Robert Israel in his answer below, the term general abstract nonsense is believed to have been coined by Norman Steenrod - and surely it was not intended by him as a putdown (in spite of what happens today, in some fringes of the mathematical community). On the other hand, I'm now particularly intrigued (and, I must really confess, a little bit puzzled) by P.A. Smith's, let's say, warm comments about Eilenberg and Mac Lane's General Theory of Natural Equivalences, as they are reported by Michael Barr in an old thread from the Category Theory mailing list (dating back to May 1998). 


Does anybody know if Smith's comments are taken from a letter, review, or anything else appearing in a journal, book, etc.? If so, have they ever been "revised" by Smith?

REPLY [6 votes]: Norman Steenrod is generally credited with coining the phrase.  I'd call it "tongue-in-cheek" rather than sarcastic.  See http://en.wikipedia.org/wiki/Abstract_nonsense and references there.<|endoftext|>
TITLE: Ising model - phase transition vs rapid mixing
QUESTION [14 upvotes]: Consider a graph $G=(V,E)$ and Ising model on that graph, i.e. configuration space is $\Omega=${$-1,+1$}$^V$ and energy of a configuration $s \in \Omega$ is given by:
$H(s) = -\beta \sum_{u \sim v}s(u)s(v)$
Consider:
a) the phase transition between the ordered and disordered phase
b) transition in a Markov chain simulating the dynamics (e.g. Glauber/Metropolis dynamics), from rapid mixing ($N\log N$) to exponentially slow mixing
It is generally "known" that often phase transition as in a) is accompanied by phase transition in b) ("critical slowdown" in physics parlance). Is there any formal result capturing this knowledge (e.g. theorem of the form: physical phase transitions is equivalent to critical slowing down of the appropriate Markov chain)? Or, at least, a nonrigorous argument going beyond "this seems to hold for all systems that physicists are interested in" (proving tight results about critical points is usually difficult, so a heuristic would be OK).

REPLY [3 votes]: As only a partial answer, in the paper Bierkens, Roberts, A piecewise deterministic scaling limit of Lifted Metropolis Hastings for the Curie-Weiss model, http://arxiv.org/abs/1509.00302, we have obtained a result of this nature. I am not aware of other results in this spirit. 
We showed how in the Curie-Weiss model above critical temperature, a Metropolis chain has to make steps at rate $N$ (the number of spins) in order for the sample paths of the (suitably scaled) magnetization to converge to those of a Langevin diffusion with the correct limiting invariant distribution (a Gaussian). At the critical temperature the steps have to occur at the rate $N^{3/2}$, to obtain a Langevin diffusion for the invariant distribution proportional to $\exp(-x^4/12)$ in the proper scaling. The scaling in time is perhaps not exponential at the critical temperature because this is just about the magnetization, which does not describe the full state of the system.<|endoftext|>
TITLE: Waldhausen Additivity in a More General Context
QUESTION [7 upvotes]: The following arose when I was thinking about a talk at the Midwest Topology Seminar:
Background
I want to consider a generalization of a Waldhausen-like structure on a category $C$ with 0-object $\ast$. Namely, I will have a category $\text{co}C$ consisting of "cofibrations," but I will relax the usual axioms to be:
Cof1: The isomorphisms of $C$ are cofibrations;
Cof2: The arrow $\ast \to A$ is a cofibration for any object $A$ of $C$.
Cof3': Pushouts of cofibrations along cofibrations are always defined, and all arrows in the the pushout will be cofibrations.
(Note that Cof3' is different from Waldhausen's usual axiom. In particular, quotient objects are not necessarily defined.)
For my purposes, I am happy to work with the isomorphisms as the weak equivalences.
Using the above structure, one can apply a construction due to Thomason to form a 
simplicial category $iT.C$. An object in simplicial degree $n$ is a string of cofibrations
$$
C_\bullet \quad \equiv \quad C_0 \rightarrowtail C_1 \rightarrowtail \dots \rightarrowtail C_n
$$
A morphism is filtration preserving map $C_\bullet \to D_\bullet $, 
such that each $C_k \to D_k$ is a cofibration and   each square
$$
C_{k-1} \to C_k 
$$
$$
\downarrow \qquad \quad  \downarrow 
$$
$$
D_{k-1} \to D_k 
$$
is a pushout (Note: I do not impose the condition that the vertical maps be weak equivalences in this set up.)
The simplicial structure is defined by dropping the $k$-th term of a filtration or
inserting the identity in the appropriate place.
Waldhausen showed that Thomason's construction is homotopy equivalent to the $wS.$-construction whenever the above structure arises from a bona fide Waldhausen category structure on $C$. The idea, roughly is given by mapping the filtration above
to the filtration $C_1/C_0 \rightarrowtail \cdots \rightarrowtail C_n/C_0$, provided we incorporate quotient data into the definition. (See Waldhausen LNM1126, p.334.)
My Question
My question is basically whether a version of the additivity theorem holds in this context.
To formulate this, consider the category whose objects are pushout squares of objects of $C$:
$$
A \to B 
$$
$$
\downarrow \quad \quad  \downarrow 
$$
$$
C \to D
$$
in which all displayed maps are cofibrations.  Morphisms are defined in the obvious way (natural transformations of such diagrams)
It seems to me that this diagram category is also equipped a cofibration structure as above.
Then we may consider the functor
which sends this diagram to
$$
D \vee A
$$
as well as the functor which sends the diagram to
$$
B \vee C
$$
These are both exact functors, and one can ask:
Question:
Are these  functors homotopic after passing to  $iT_\bullet$ constructions?

REPLY [6 votes]: I tried thinking about this directly, via a direct homotopy as in the paper http://www.math.uni-bielefeld.de/documenta/vol-16/14.html, but didn't get anywhere.  Suppose we pin down the definition of iT. a bit so that in this case the vertical maps are cofibrations, too; otherwise, the use of squares that "happen" to be pushouts seems unnatural.  Now try to construct the homotopy bit by bit, starting with the 0,0-cells (vertices) of the source and progressing to the 1,0-cells and the 0,1-cells.  Nothing reasonable seemed to pop up.<|endoftext|>
TITLE: Is there a notion of a “model category which admits left Bousfield localization?”
QUESTION [10 upvotes]: At a conference not too long ago I gave a talk on (left) Bousfield localization and was asked an interesting question afterwards. The question was whether I knew any examples of model categories which were not left proper, but which did admit Bousfield localization. There are some trivial examples, but I'm interested now to hear if anyone knows any interesting or surprising examples in this vein. I do have one example (from the person who asked the question), but it's from unpublished work so I'm not going to write about it here. Since localization with respect to a set of maps is the same as localization with respect to their coproduct, I'll just focus on localization with respect to a single map $f$.

Are there interesting or surprising examples of model categories which are not left proper but which admit left Bousfield localization with respect to every map $f$? What about if we restrict to some non-trivial class of maps $f$?

Also, this question got me thinking of whether one can do Bousfield localization without assuming the model category $M$ is cellular or combinatorial. It has always seemed to me that these hypotheses are there just to make the proofs work (cellularity for Bousfield's original proof and combinatoriality to simplify things even further). Because the proof that the localized model structure exists (i.e. that it is a model category) always comes down to the Transfer Principle for transferring structures across an adjunction, so there is always some drawn out analysis of pushouts. Perhaps because of this there is no hope of removing the hypotheses, but I still want to ask to see if that's what the experts think.

Are there interesting or surprising examples of model categories which are neither cellular nor combinatorial but which admit left Bousfield localization with respect to every map $f$? Or has someone studied this question and found subclasses of maps $f$ which work?

Most importantly:

Has anyone tried to classify model categories which admit Bousfield localization? Are there any theorems of the form “if $M$ admits Bousfield localization with respect to a set of maps $S$ then some other thing holds” in the literature?

REPLY [2 votes]: Here is another family of examples of non-cofibrantly generated model categories which admit left Bousfield localization with respect to certain classes of maps. Let $C$ be a proper simplicial model category and consider the strict model structure on $Pro(C)$ defined by Isaksen in 
http://arxiv.org/abs/math/0108189. 
Let $K$ be any set of fibrant objects in $C$. A map $f : X\to Y$ in $Pro(C)$ is called a $K$-local weak equivalence if it induces a weak equivalence 
$$Map_{Pro(C)}^h(Y,A) \to Map_{Pro(C)}^h(X,A)$$
for all $A$ in $K$, where $Map_{Pro(C)}^h$ denotes the derived mapping space in the model category $Pro(C)$. 
Then it is shown in Theorem 4.4 of 
http://arxiv.org/abs/math/0403451
that the underlying (essentially levelwise) cofibrations and the $K$-local weak equivalence define a (left proper simplicial) model structure on $Pro(C)$. This is a left Bousfield localization of the strict model structure on $Pro(C)$, which is not cofibrantly generated (and also not fibrantly generated). In fact, this is the left Bousfield localization of $Pro(C)$ with respect to the class of $K$-local weak equivalences (which is a certain proper class of maps defined using the small set $K$ of objects).<|endoftext|>
TITLE: Monoidal Model Categories with Suspension Functor
QUESTION [8 upvotes]: This is basically just me trying to find out what such categories are called, and where they are written about.  If I think of some model category of spectra being a "stabilization" of some model category of spaces, i.e. obtained by inverting the suspension functor, or something along those lines, what is this process called? What is the data on a model category that I need to make this happen? Is this structure available in $(\infty,n)$-categories as well? I also think about this in terms of going from R-algebras to R-modules (perhaps in some derived sense), i.e. taking a sort of tangent space or tangent category. Is there a standard framework for such things? In the case of model categories, it seems that we have some good ones for spectra, and some good ones for spaces, but I'm not clear how to go between them. I also want to do all of this in the presence of a monoidal structure which respects everything else, maybe closed, etc. etc. 
Any references or guidelines would be dearly appreciated. 
-Jon

REPLY [3 votes]: Fernando Muro's comment is right on the money. I'm interested in this type of question, too, and have read Hovey's paper closely. I'm going to summarize the paper here. I believe Marc Hoyois is correct that you don't need much machinery to make this work in $(\infty,n)$ categories (see Dylan Wilson's comment here), so I'll focus on model categories with an eye towards answering your question about hypotheses. The process of going from $M$ to the stable model structure on $Sp^{\mathbb{N}}(M,G)$ or on $Sp^{\Sigma}(M,G)$ is called stabilization.
Going from the model category of spaces to the model category of spectra, the idea is to make the endofunctor $\Sigma$ into a Quillen equivalence. So Hovey takes a model category $M$ and an endofunctor $G$, then constructs a model category $Sp^{\mathbb{N}}(M,G)$ whose objects are sequences $(X_n)$ along with structure maps $GX_n\to X_{n+1}$. You can endow this category with the projective model structure (i.e. weak equivalences and fibrations are defined levelwise) whenever $M$ is cofibrantly generated. This is the only hypothesis you need on $M$. As for $G$, you need to know it's a left Quillen functor from $M$ to $M$. 
In order to make $G$ into a left Quillen equivalence on $Sp^{\mathbb{N}}(M,G)$, you want to do Bousfield localization. So you need this category to be left proper and cellular or left proper and combinatorial (or do you? See my most recent MO question). It turns out that if you assume these properties on $M$ then they hold on $Sp^{\mathbb{N}}(M,G)$. With this assumption, you can create a stable projective model structure on $Sp^{\mathbb{N}}(M,G)$ which makes $G$ into a left Quillen equivalence. Furthermore, if $G$ was already a left Quillen equivalence then the embedding $M\to Sp^{\mathbb{N}}(M,G)$ is a left Quillen equivalence. So $Sp^{\mathbb{N}}(M,G)$ is initial in some sense with respect to the property that $G$ becomes a left Quillen equivalence. You don't have a notion of stable homotopy groups in $Sp^{\mathbb{N}}(M,G)$, but if you want one you can read section 4 of Hovey's paper where he tries to find the correct hypotheses to make this $Sp^{\mathbb{N}}(M,G)$ because more like the usually category of (topological) spectra.

For the monoidal situation you should start with a monoidal model category $M$ with a monoidal left Quillen endofunctor $G$. More generally, let $M$ be a $D$-model category and let $G$ be a left $D$-Quillen endofunctor (you recover the monoidal case for $D=M$). Then Theorem 5.7 shows that $Sp^{\mathbb{N}}(M,G)$ is a $D$-model category and $G$ is a left Quillen equivalence on it provided we know that $G(X\otimes K) = GX\otimes K$ coherently for $X\in M$ and $K\in D$, that $M$ satisfies the properties above, and that $D$ is cofibrantly generated with domains of the generating (trivial) cofibrations being cofibrant. This hypothesis on $D$ appears in a lot of Hovey's work. One small result in my thesis obtains this hypothesis from more standard hypotheses. I can edit or comment with details on that if you're really interested.

Section 6 of Hovey's paper shows you how to do symmetric spectra $Sp^{\Sigma}(M,G)$. Now $G$ is again a left $D$-Quillen endofunctor on $M$, so it must have the form $G(X) = X \otimes G(S)$ where $S$ is the unit of $M$. Now you only need to assume $M$ and $D$ are left proper and cellular, and that $G(S)$ is cofibrant. You again get the projective model structure on $Sp^{\Sigma}(M,G)$ and you can again take Bousfield localization to get the stable version. Theorem 7.11 shows you that everything works out (i.e. $Sp^{\Sigma}(D,G)$ is a monoidal model category and $Sp^{\Sigma}(M,G)$ is a $Sp^{\Sigma}(D,G)$-model category) provided that the domains of the generating (trivial) cofibrations of $M$ and $D$ are cofibrant. The category $Sp^{\Sigma}(M,G)$ is again initial and satisfies other nice properties, as you can see in Sections 8 and 9.
Last comment: I don't know what these hypotheses reduce to in the situation of $R$-modules and $R$-algebras. I'd be interested in thinking about that, especially regarding this hypothesis that makes the domains of the generating maps be cofibrant.<|endoftext|>
TITLE: Prime Power Gaps
QUESTION [11 upvotes]: In 2000, Baker, Harman and Pintz proved that there is always a prime in
the interval $(n-n^{0.525}, n)$. There are also conditional results
implying smaller intervals. Nevertheless, I could not find any
information about prime power gaps. So, what I'm asking is:

What is the asymptotically largest function $f(n)$ s.t. there is always a prime power in the interval $(f(n), n)$?

For example, Bertrand’s postulate is almost trivial in this case, since there is always a power
of $2$ in the interval $(n, 2n]$. On the other hand, the distribution of prime powers with
exponent $e>1$ is much smaller than the distribution of primes, so adding them might not change
the answer.

REPLY [3 votes]: The paper shows that for $x \gt x_0$ there are primes in the interval $[x-x^{0.525},x]$ where the value of $x_0$ could be found effectively "with enough effort."  The result fails for $x=126$ but I can't immediately rule out that $x_0=127$ suffices. The paper seems to say that for large enough $n$ the number of primes in $[n,n+n^{0.525}]$ eventually exceeds  $\frac{9n^{0.525}}{100\log{n}}.$
Call an integer power proper if it is at least a square and respectable if it is at least a cube, To expand on Gerhard's observation: If we looked for intervals which contain either a prime or a proper integer power then we can say that there is always a square in $(n-2\sqrt{n},n)$ so this eventually improves on $x-x^{.525}.$ 
There is always a respectable power in $(n-3n^{2/3},n)$ but that is pretty near best possible, so I would guess that if we said  "a prime or a respectable power" then we would not be able to get any improvement over just "prime". Likewise, there seems no  reason to think that "prime or prime power" is essentially better than "prime or prime square" nor that "prime or prime square" is essentially better than "prime."<|endoftext|>
TITLE: Best upper bound on rate for q-ary codes
QUESTION [5 upvotes]: Among the many upper bounds for families of codes in $\mathbb F _2 ^n$, the best known bound is the one by McEliece, Rodemich, Rumsey and Welch which states that the rate $R(\delta)$ corresponding to arelative distance of $\delta$ is such that:
\begin{equation*}R(\delta) \leq H_2(\frac{1}{2}-\sqrt{\delta(1-\delta)}) \end{equation*}
where H is the binary entropy function.
(A slight improvement of the above exists in the binary case, but within the same framework)
In the case of q-ary codes, i.e. codes over $\mathbb F _q ^n$, the above bound is generalized to:
\begin{equation*}R(\delta) \leq H_q(\frac{1}{q}(q-1-(q-2)\delta-2\sqrt{(q-1)\delta(1-\delta)})) \end{equation*}
My question is as follows:
For larger alphabet size q, the above bound seems to weaken significantly. In fact, observing the growth of the above bound as $q \rightarrow \infty$, we see that:
\begin{equation*} R(\delta) \leq 1-\delta+\mathcal{O}(\frac{1}{\log{q}})  \end{equation*}
Thus, it seems to get worse than even the Singleton bound $R(\delta) \leq 1-\delta$.
So which is the best bound for large alphabet size $q$?
Also, could someone direct me to references for comparisons of different bounds for larger $q$? I am able to find reliable comparisons only for $q=2$.

REPLY [9 votes]: So, the supposedly the sharpest one among all known bounds is somehow poorer than the bound you learn in Coding Theory 101 if the alphabet size $q$ approaches infinity. I think the reason you find it funny lies in what you mean by "large $q$."
You can think this way: If the parameter $q$ is large compared to the code length $n$, the Singleton bound is an extremely sharp, invincible upper bound. In fact, it's best possible in the sense that there are infinitely many nontrivial codes attaining it. I think that's why you got caught in a seemingly contradicting situation where the "best upper bound (MRRW)" fails to beat the "poor and elementary bound (Singleton)."
Here's a bit more formal explanation. The upper bounds on the information rate $R$ you're considering are asymptotic ones described by the Hilbert entropy function
$$H_q(x) = x\log_q(q-1)-x\log_qx-(1-x)\log_q(1-x)$$
and the relative distance $\delta = \frac{d}{n}$, where $d$ is the minimum distance of a code. Bounds of this kind try to understand the behavior of the information rate when the code length $n$ tends to infinity for a fixed $\delta$.
More precisely, by writing the largest possible number of codewords for a $q$-ary code of length $n$ and relative distance $\delta$ as $A_q(n,\delta n)$, we would like to know the exact value of
$$R_q(\delta) = \lim_{n\rightarrow \infty}\sup n^{-1}\log_qA_q(n,\delta n).$$
The MRRW bound you mentioned is a pretty good upper bound on $R_q(\delta)$ obtained through linear programing. Of course, the plain, simple Singleton bound can also be viewed as an asymptotic upper bound if you simply rewrite it with $\delta$, i.e., you have $R \leq 1-\delta$.
Now, what you did to the MRRW bound seem to be simply applying
$$\lim_{q\rightarrow \infty}H_q(\delta) = \delta$$
to see what it's like when "$q$ is large." The problem is that, in order to get a meaningful upper bound better than the Singleton bound, you should consider exactly what you mean by "large $q$."
As you probably already know, there are codes that meet the Singleton bound. Indeed, it's the definition of maximum separable distance (MDS) codes. The famous Reed-Solomon codes are the canonical examples of this type of code, which are conjectured to have the "largest" $n$ relative to $q$ among all possible (linear) MDS codes. More generally, algebraic geometry codes can attain
$$R \geq 1 - \delta - \frac{1}{\sqrt{q}-1}$$
when prime power $q$ is a square. Hence, if your $q$ is in the range where MDS codes may exist, no general upper bound on $R$ can beat the Singleton bound unless it's a very contrived one. If $q$ is small enough compared to $n$ such that unbelievably strong codes like algebraic codes can be realized, you don't expect any other bound can beat it.
So, if you would like a stronger asymptotic upper bound than the Singleton bound, you have to at least rule out the existence of MDS codes by making $q$ grow slow enough. Here's one theorem I know that can be used for this purpose:

If there exists a $q$-ary MDS code of length $n$ and minimum distance $d$, then $q \geq n-d+2.$

(For the proof, see L. M. G. M. Tolhuizen, On maximum distance separable codes over alphabets of arbitrary size, Proc. IEEE Int. Symp. Inf. Theory (1994) 431.)
So, by fixing the relative distance $\delta$, the inequality can be rewritten as $q \geq n(1-\delta)+2$. By this necessary condition for the existence of an MDS code, if you only allow $q$ to be always smaller than the right-hand side when driving $n$ to infinity, your favorite general upper bound may have a chance to beat the Singleton bound.
MDS codes and other very powerful error-correcting codes are extremely important and have various interesting connections to other fields both inside and outside of mathematics. So if you ask a real coding theorist, I think they know the best theorem you can exploit to rule out the existence of MDS codes and also the best methods to get sharp upper bounds on $R$ for $q$ of your favorite size and growth rate.<|endoftext|>
TITLE: Relative De Rham cohomologies
QUESTION [33 upvotes]: as far as I know, there are two main ways to have a relative version of De Rham Cohomology for a pair (M,N), where M and N are smooth manifolds and N is a closed (as a topological subspace) submanifold of M:
1) Godbillon, Elements de topologie algébrique: $\Omega^p(M,N)$ is the space of all forms on $M$ whose restriction to $N$ is zero. This is a subalgebra of $\Omega^p(M)$ so it defines a cohomological space $H^p(M,N)$.
2) Bott-Tu, Differential forms in algebraic topology: this times, $\Omega^p(M,N)=\Omega^p(M)\oplus  \Omega^{p-1}(N)$ with differential $d(\omega,\theta)=(d\omega,i^*(\omega)-d\theta)$, where $i:N\to M$ is the inclusion.
Does these two cohomologies give the same results? Otherwise, are they related and how are they related?
Bott and Tu's paragraph on the relative De Rham cohomology is very short. Does someone know a good reference on this subject? 
Thank you in advance.

REPLY [4 votes]: AG learner's is almost the answer I wanted, but really bugged me that that $\Psi$ wasn't a chain map that was homotopic to the identity.  I think I figured out how to make that happen.  Let $T$ be an open tubular neighborhood of $N$, and $r\colon R\to N$ a smooth deformation retract.     Let $f$ be a smooth bump function whose support is in $T$ and which is 1 on $N$.  We can find the formula for $\Psi$ by considering the homotopy $h(\alpha, \beta)=(fr^*\beta,0)$.  We have that $$(dh+hd)(\alpha,\beta)=(d(fr^*\beta)+fr^*(\alpha|_N-d\beta),\beta)$$ so
$$(1-dh-hd)(\alpha,\beta)=(\alpha-fr^*(\alpha|_N)-df\wedge r^*\beta,0).$$
This coincides with AG learner's formula on closed forms, and is a chain map.<|endoftext|>
TITLE: a normal form for matrices over Z[x]/(x^2-1) ?
QUESTION [13 upvotes]: We are discussing, offline, modules over the $\mathbb{Z}$-group ring of the cyclic group of order 2, which is probably better known as the quotient ring $R=\mathbb{Z}[t]/(t^2-1)$.
Is there any way to describe matrices over it, in a way similar to Smith Normal Form (SNF), or Hermite Normal Form (HNF)? That is, for $A\in R^{n\times m}$, find $X\in GL(n,R)$ and $Y\in GL(m,R)$, such that
$XAY$ is "nice", e.g. diagonal (resp. upper-triangular), like one would get if SNF (resp. HNF) was possible for $R$.
I am aware of a similar question for $\mathbb{Z}[t]$, which looks harder. One immediate observation is that $A=B+tC$, for $B,C\in \mathbb{Z}^{n\times m}$, and so one can choose $X$, $Y$ to have integer entries, so that $XAY=B'+tC'$, where $C'$ is the SNF of $B'$.

REPLY [12 votes]: There is a general concept of Hermite Normal Form developed by Kaplansky [1] for associative rings with identity. His results were revived in [Appendix to §I.4 and Notes on Chapter I, 3] and [4]. (A quick Google search shows that other recent publications revolves around Kaplansky's definition.)  
Let us suppose rings commutative for the sake of simplicity. A commutative ring $R$ with identity is said to be an elementary divisor ring if every matrix $A$ over $R$ admits a diagonal reduction, i.e., there are invertible matrices $P$, $Q$ over $R$ and elements $d_i \in R$ such that 
$PAQ = \operatorname{diag}(d_1, \dots, d_n)$ with 
$d_1 \, \vert \cdots \vert \, d_n$. A commutative ring $R$ with identity is a Hermite ring in the sense of Kaplansky, or concisely a K-Hermite ring (this is T. Y. Lam's naming), if every $1$-by-$2$ matrix admits a diagonal reduction, i.e., if for every $(a, b) \in R^2$ we can find an invertible matrix $Q$ such that $(a, b)Q = (d, 0)$ for some $d \in R$. It should be clear that a K-Hermite ring $R$ is a Bézout ring, i.e., the finitely generated ideals of $R$ are principal.  
The ring $R = \mathbb{Z}[t]/(t^2 -1) = \mathbb{Z}[C_2]$ is not a Bézout ring since the image of the ideal $(2, t - 1)$ is not principal. Therefore we cannot expect matrices over $R$ to have a diagonal reduction in the sense of Kaplansky. However $R$ is a generalized Euclidean ring in the sense of P. M. Cohn, i.e., $SL_n(R)$ is generated by transvections for every $n$ [2]. This fact is established using one of the obvious embeddings of $R$ into $\mathbb{Z}^2$ and some strong Euclidean property of $\mathbb{Z}$,  that is $\mu(\mathbb{Z}) = \frac{1}{2}$, see [Lemma 4.1, 2]. This could be a starting point to study possible "nice" reduced forms for matrices over $R$. For instance, it is not difficult to show that the following holds: for every $(a, b) \in R^2$ there exists $E \in SL_2(R)$ such that $(a, b)E = (d, d')$ with $d d' = 0$; in addition we can take $(d, d') = (1, 0)$ if $(a, b)$ generates $R$.  
In contrast, $\mathbb{Z}^2$ is trivially an elementary divisor ring. 

[1] "Elementary divisors and modules", I. Kaplansky, 1949.
[2] "Generalized euclidean group rings", K. Dennis et al., 1984.
[3] "Serre's problem on projective modules", T. Y. Lam, 2006.
[4] "Euclidean pairs and quasi-Euclidean rings", A. Alahmadi et al., 2014.<|endoftext|>
TITLE: Are there any techniques for solving a differential equation of the form $f ' (x) = f( f( x ) )$?
QUESTION [15 upvotes]: I am trying to solve the following differential equation
$$f ' (x) = f( f( x ) ),$$
but I have no idea how. I don't think the chain rule is useful for this.
Although I don't think this differential equation is solvable, I'd like to know if there is any interesting approach to solve a differential equation of this kind, or, at least, a non-trivial solution of the equation.

REPLY [5 votes]: And regarding real solutions to the question, Alex Gavrilov is completely correct. A Taylor expansion at fixed point $p$ gives us the real solution. Existence of this solution is proven in the paper which I already referenced from my another answer.
$$f(z)=\sum_{n=0}^\infty \frac{d_n (z-p)^n}{n!}$$
where $d_n$ is defined as follows:
$$d_0=p$$
$$d_{n+1}=\sum _{k=0}^n d_k \operatorname{B}_{n,k}(d_1,...,d_{n-k+1})$$
where $B_{n,k}$ are the Bell polynomials
This gives the following starting coefficients:
$$d_1=p^2$$
$$d_2=p^3+p^4$$
$$d_3=p^4 + 4 p^5 + p^6 + p^7$$
$$d_4=p^5 + 11 p^6 + 11 p^7 + 8 p^8 + 4 p^9 + p^{10} + p^{11}$$
etc.
The fixed point $p$ here serves as a parameter, which determines the family of solutions. According the linked theorem, the expansion should converge in the neighborhood of $p$ for $0 < |p| < 1 $ or $p$ being a Siegel number.<|endoftext|>
TITLE: Why are derived categories natural places to do deformation theory? 
QUESTION [23 upvotes]: It seems to me that a lot of people do deformation theory (of schemes, sheaves, maps etc) in derived category (of an appropriate abelian category). For example, the cotangent complex of a morphism $f:X\rightarrow Y$ of schemes is defined as an object in the derived category of coherent sheaves $X$. I would like to understand why derived category is an appropriate category to do deformation theory. I would appreciate it if someone could give me a good example or motivation.

REPLY [14 votes]: Perhaps one way of reading this question is "Why is it important to think about complexes when doing deformation theory?"  One must certainly accept that deformation theory is cohomological in nature:  consider the standard Čech construction of classes in $H^1(X, T_X)$ and $H^2(X, T_X)$ corresponding respectively to deformations and obstructions to deformations of a smooth scheme $X$; or consider the deformations of a regular embedding $X \rightarrow Y$, which are parameterized by $H^0(X, N_{X/Y})$ and obstructed by $H^1(X, N_{X/Y})$.
Let me recall the arguments that yield these classes.  Locally a smooth scheme has only one deformation up to isomorphism, so global deformations arise from gluing the local ones; the gluing data come from the tangent bundle, so deformations are classified by $H^1(X, T_X)$.  Similarly, solutions to an obstructed deformation problem exist locally and are locally unique up to isomorphism, so the obstruction comes from gluing.  One writes down the gluing condition and observes that it is a Čech $2$-cocycle (or observes directly that deformations form a gerbe...).
The argument for a regular embedding is similar.  One first gives a direct construction for the identification of $H^0(X,N_{X/Y})$ and the deformations of the embedding; then one shows that there are no local obstructions, so global obstructions arise entirely from gluing, and this gives a Čech 1-cocycle with values in $N_{X/Y}$.
What about a more general deformation problem?  For example, suppose $X$ is a local complete intersection.  A great deal of the arguments above goes through without modification:  locally, deformations of a local complete intersection are unobstructed, so obstructions should come from gluing.  One should be able to produce a Čech cocycle obstructing the existence of a global deformation.  But a cocycle valued in what?
The answer is the (shifted) tangent complex (dual to the cotangent complex mentioned in another answer).  This is a complex in in degrees $[0,1]$ that coincides with $T_X$ if $X$.  The relative tangent complex of a regular embedding $X \rightarrow Y$ is $N_{X/Y}[-1]$.  Thus the tangent complex recovers the two special cases discussed above.
For a complete intersection $X$ in a smooth scheme $Y$, the tangent complex of $X$ may be constructed as $[T_Y \vert_X \rightarrow N_{X/Y}]$.  If $X$ is only a local complete intersection then this construction works locally in $X$, but it's not going to glue to something global.  However, the local construction is quasi-isomorphic to the restriction of something global.  At last, the derived category appears.
All of this is vastly generalized by the cotangent complex, which works the same way as above for arbitrary schemes, without the lci restriction.  You can read more about that in Chris Brav's answer.<|endoftext|>
TITLE: Little puzzle about intersections of Schubert cells.
QUESTION [6 upvotes]: I was reading chapter 6 in the book of Harris on Algebraic geometry and came to the following puzzle.
It seems to me that every Schubert cell in a Grassmanian is obtaining by cutting the Grassmanian by a certain plane in its Plucker embedding (hope this is correct, I got this idea from Harris' book)
Now, I was thinking always, that there is a whole theory of Littlwood Richardson coefficients
that explains the intersection theory of Grassmanian and in particular explains what are intersection numbers of various cells. 
My question is as follows. Why do we have this complicated theory of Littlwood Richardson coefficients? (is this because Schubert cells don't always have expected dimension ?) Naively, if all Schubert cells were obtained by intersecting Grassmanian transversally by some planes
the only number that would pop up would be the degree of Grassmanian in its Plucker embedding.

REPLY [10 votes]: Precisely as you say, the intersection is almost always not transverse.
For example, think about $G(2,4)$. It is the quadric $p_{12} p_{34} - p_{13} p_{24} + p_{14} p_{23} =0$ in $\mathbb{P}^{\binom{4}{2}-1}$. There are two codimension $2$ Schubert cells: One of them is given by $p_{12}=p_{13} = p_{23}=0$ and the other is given by $p_{12}=p_{13}=p_{14}=0$. In each case, we are using $3$ linear equations to cut out something which is codimension $2$. If we just interesect $G(2,4)$ transversely with a codimension $2$-plane, we get something whose cohomology class is the sum of these two classes; if we intersect $G(2,4)$ with $p_{12}=p_{13}=0$ then we literally get the union of these two cycles.<|endoftext|>
TITLE: Recovering classical Tannaka duality from Lurie's version for geometric stacks
QUESTION [10 upvotes]: In Lurie's paper Tannaka Duality for Geometric Stacks, it is essentially shown that specifying a morphism of geometric objects
$$ f \colon X \to Y$$
is equivalent to giving a corresponding pullback morphism, which is a symmetric monoidal functor
$$ f^* \colon \mathrm{QC}(Y) \to \mathrm{QC}(X),$$
where $\mathrm{QC}$ indicates the category of quasi-coherent sheaves.  
In actuality, Lurie proves the following result.
Theorem 5.11. Suppose that $(S,\mathcal{O}_S)$ is a ringed topos which is local for the étale topology, and that $X$ is a geometric stack. Then the functor
$$ T \colon f \mapsto f^* $$
induces an equivalence of categories
$$ T \colon \mathrm{Hom}( (S, \mathcal{O}_s), \mathrm{Sh}(X_\mathrm{\acute{e}t})) \to \mathrm{Hom}(\mathrm{QC}(X), \mathcal{O}_S\mathrm{-Mod}).$$
Short explanation:
Here the Hom functor on the left is taken in the (2-)category of locally ringed toposes, and on the right it corresponds to the tensor functors, which are symmetric monoidal functors which also preserve finite colimits, with an additional tameness condition.
(A stack is said to be geometric if it is quasi-compact, and the diagonal morphism is representable and affine.)    
Now I don't see how to recover from this the usual Tannakian reconstruction results, for instance that given a neutral Tannakian category I can recover an algebraic group (by looking at automorphisms of the fibre functor) whose category of representations is this category I started with.
In his helpful answer, David Ben-Zvi mentions that from Lurie's result, we can find that having a faithful fibre functor from our Tannakian category $\mathcal{C}$ to $K$-vector spaces means that $\mathcal{C}$ is the category of quasi-coherent sheaves on some classifying stack $\mathbf{B}G$ (or, more accurately, on some $G$-gerbe if we are not just working over an algebraically closed field). However, I fail to see why this is so; in particular, Lurie does not seem to have a result about essential surjectivity of the above functor $T$, and I don't see how one would deduce such a thing from the previously mentioned theorem.
Sorry if this is all too trivial, David Ben-Zvi's answer left me wanting for a fuller understanding of the picture.

REPLY [15 votes]: Tannaka duality makes two complementary statements: first, that an affine algebraic group can be recovered from its category of representations (the "reconstruction problem"), and second, that certain categories are the categories of representations of an affine algebraic group (the "recognition problem"). 
Lurie's paper generalizes the first statement, but not the second. The generalization of the second statement to Lurie's setting (in fact, to a slightly more general setting) is contained in Daniel Schäppi's paper, A characterization of categories of coherent sheaves of certain algebraic stacks. In this paper, Schäppi defines the notion of a weakly Tannakian category, and shows that weakly Tannakian categories are precisely the categories of coherent sheaves for "coherent algebraic stacks with the resolution property." This result specializes to the classical recognition result on affine algebraic groups.<|endoftext|>
TITLE: Embedding of F(4) in OSp(8|4)?
QUESTION [6 upvotes]: Is the superconformal algebra in five dimensions, $F(4)$, a subalgebra of the (maximal) six-dimensional superconformal algebra $OSp(8|4)$?

REPLY [2 votes]: Apparently the answer is "no". For more details see Appeindx C.4.1 of this paper http://arxiv.org/pdf/0810.1484.pdf<|endoftext|>
TITLE: finding highest weight of dual of a representation of a semisimple lie algebra
QUESTION [10 upvotes]: If V is an irreducible representation of a semi simple lie algebra having highest weight λ then what will be the highest weight of the corresponding irreducible representation V∗ (Dual of V)?

REPLY [13 votes]: To expand my short comment, the result itself (formulated by Sasha) has been around a long time and depends only on the definitions involved.   Textbooks dealing with the highest weight classification of finite dimensional modules over $\mathbb{C}$ (which you assume but haven't specified) always say something about dual modules of simple modules but sometimes informally in exercises and sometimes more formally.    For instance, the result appears in the book by Onishchik-Vinberg (where the tough exercise is to find it) but also much more formally in the treatise by Bourbaki: see Chapter VIII, $\S7.5$.  (This statement about highest weight of the dual simple module actually carries over easily to prime characteristic as well, though the dimensions of such modules tend to shrink.)
The statement is fairly easy to visualize in terms of weight diagrams and Weyl group symmetry, since the modules are finite dimensional.    An easy step (not requiring machinery and done quite generally in Bourbaki's Chapter I) is to show that $V^*$ is simple when $V$ is: this just requires duality and finite dimensionality along with the definition of the dual action, since submodules of one correspond to quotients of the other.
In the highest weight situation, the weight diagram of $V$ is symmetric under the Weyl group as is that of $V^*$.   The weights (with multiplicity) of the latter are the negatives of the original weights, using just the definition of the dual action, so the "lowest" weight is $-\lambda$.   To get the highest weight use $w_0$ to pass from negative to positive root action, so the answer is $-w_0 \lambda$. 
P.S. Visualization aside, the algebraic step needed is to pick out a maximal vector $f^+ \in V^*$ having the indicated weight:   if $v^+ \in V$ is a maximal vector of weight $\lambda$, just take $f^+$ to be the characteristic function of $w_0 v^+$ (taking value 1 there, 0 on other weight vectors in a basis of $V$).   Here $w_0 v^+$ is shorthand for an arbitrary nonzero vector in the 1-dimensional weight space for $w_0 \lambda$.<|endoftext|>
TITLE: Proving continuity on spaces of distributions?
QUESTION [6 upvotes]: Let $\mathcal{D}'(\Omega)$ be the space of distributions on an open set $\Omega$, and $\mathcal{E}'(\Omega)$ the compactly supported ones.
When you have a linear operator $T:\mathcal{D}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ or $T:\mathcal{E}'(\Omega)\rightarrow\mathcal{E}'(\Omega)$, I often find it easy to prove that convergent sequences get mapped to convergent sequences (in the weak topology), but proving continuity with respect to the weak topology is much harder. Since these spaces are not first-countable (I think), one must work with a local basis around 0.
For example, if you have a pseudodifferential operator $P:\mathcal{E}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ which is properly supported, then it continuously extends to a (necessarily unique) operator $P:\mathcal{D}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ and maps $\mathcal{E}'(\Omega)$ continuously into $\mathcal{E}'(\Omega)$. I was able to prove the first statement, but I was able to prove the second statement only in terms of sequential continuity. Note that the weak topology of $\mathcal{E}'(\Omega)$ is not the subspace topology that it inherits from the weak topology of $\mathcal{D}'(\Omega)$ (I think), so this is not trivial.
In many textbooks involving distributions, authors don't seem to be careful about this. They give an argument proving sequential continuity and claim that it is continuous in the more general sense. Is there a general scheme for converting sequential continuity arguments to actual continuity arguments for these spaces? Or can you point me to a detailed proof of the aforementioned fact about proper PDOs?

REPLY [4 votes]: Every linear mapping $T:\mathcal D'\to \mathcal D'$ which is $\sigma(\mathcal D',\mathcal D)$-sequentially continuous is $\sigma(\mathcal D',\mathcal D)$-continuous.
Perhaps, this is the same as Peter Michor's answer, but let me try to explain that the major point is the bornologicity of $\mathcal D'$ endowed with the strong topology
$\beta (\mathcal D',\mathcal D)$ (uniform convergence on the bounded subsets of $\mathcal D$) which follows e.g. from a beautiful theorem of Laurent Schwartz (the strong dual a every complete Schwartz space is bornological).
Moreover, since $\mathcal D$ is barrelled, the strong and the weak-$*$ dual have the same bounded sets.
If now $T: (\mathcal D',\sigma(\mathcal D',\mathcal D)) \to (\mathcal D',\sigma(\mathcal D',\mathcal D))$ is sequentially continuous then every bounded sequence of $\mathcal D'$ is mapped to a bounded sequence and hence $T$ is bounded on bounded sets
of $\mathcal D'$. As $(\mathcal D',\beta(\mathcal D',\mathcal D))$ is bornological
$T:(\mathcal D',\beta(\mathcal D',\mathcal D)) \to (\mathcal D',\beta(\mathcal D',\mathcal D))$ is continuous and therefore
$T:(\mathcal D',\sigma(\mathcal D',\mathcal D''))\to (\mathcal D',\sigma(\mathcal D',\mathcal D''))$ is continuous. By reflexivity, you get the desired weak-$*$ continuity.
In general, you are perfectly right with your skepticism against careless arguments
involving sequential continuity. For example, it may very well be that for a linear partial differential operator with constant coefficients $T=P(D): \mathcal D'(\Omega) \to \mathcal D'(\Omega)$ the inverse of $P(-D): \mathcal D(\Omega) \to \mathrm{Im} (P(-D))$ is sequentially continuous without being continuous (the former condition is equivalent to $P$-convexity, the latter to strong $P$-convexity)<|endoftext|>
TITLE: Riemann zeta function at positive integers and an Appell sequence of polynomials related to fractional calculus
QUESTION [13 upvotes]: I was exploring some raising and lowering operators related to an infinitesimal generator for fractional integro-derivatives and found an Appell sequence of polynomials, i.e., an infinite sequence of polynomials for which $\frac{d}{dx}p_n(x)=np_{n-1}(x)$, that is defined by the following recursion relation:
$p_{0}(x)=1$, $p_{1}(x)=x+\gamma$, and for $n>0$
$$p_{n+1}(x)=(x+\gamma)p_{n}(x)+\sum_{j=1}^{n}(-1)^j\binom{n}{j}j!\zeta (j+1)p_{n-j}(x)$$
where $\gamma=-\frac{\mathrm{d} }{\mathrm{d} \beta }\beta !\mid_{\beta =0 }$, the Euler-Mascheroni constant, and $\zeta(s)$ is the Riemann zeta function.
They satisfy $$p_{n}(x)=\frac{\mathrm{d^n} }{\mathrm{d} \beta^n }\frac{\exp(\beta x)}{\beta !}
\mid_{\beta =0 }.$$
Explicitly,
$$p_2(x)=(x+\gamma)^2-\zeta(2)$$
$$p_3(x)=(x+\gamma)^3-3\zeta(2)(x+\gamma)+2\zeta(3)$$
$$p_4(x)=(x+\gamma)^4-6\zeta(2)(x+\gamma)^2+8\zeta(3)(x+\gamma)+3[\zeta^2(2)-2\zeta(4)]$$
$$p_5=p_1^5-10\zeta(2)p_1^3+20\zeta(3)p_1^2+15[\zeta^2(2)-2\zeta(4)]p_1+4[-5\zeta(2)\zeta(3)+6\zeta(5)]$$
Can anyone provide a reference for these polynomials or point out an interesting combinatorial interpretation?
Background: Rich associations with fractional calculus, digamma function, ladder operators
The fractional integro-derivative can be represented as an exponentiated convolutional infinitesimal generator (cf. MSE-Q125343):
$\displaystyle\frac{d^{-\beta}}{dx^{-\beta}}\frac{x^{\alpha}}{\alpha!}= \displaystyle\frac{x^{\alpha+\beta}}{(\alpha+\beta)!} = exp(-\beta R_x) \frac{x^{\alpha}}{\alpha!}$
where
$$R_xf(x)=\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=|x|}\frac{-ln(z-x)+\lambda}{z-x}f(z)dz$$
$$=(-ln(x)+\lambda)f(x)+\displaystyle\int_{0}^{x}\frac{f\left (  x\right )-f(u)}{x-u}du.$$
with $\lambda=d\beta!/d\beta|_{\beta=0}$. (Note the integrand is related to the q (Jackson) derivative, and the Pincherle derivative / commutator is $[R_x,x]=D_x^{-1}$.)
Then
$$exp(-\beta R_x) 1 =\displaystyle\frac{x^\beta}{\beta!} = exp(-\beta\psi_{.}(x)), $$
with $(\psi_{.}(x))^n=\psi_n(x)$, which implies
$$\psi_{n}(x)=(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|_{\beta=0},$$
$$L_x\psi_{n}(x)=n\psi_{n-1}(x)=-x\displaystyle\frac{d}{dx}\psi_{n}(x),$$
$$R_x\psi_{n}(x)=\psi_{n+1}(x).$$
Let $x=e^z$ and $p_n(z)=(-1)^n \psi_{n}(e^z)$. Then
$$exp(-\beta R_z) 1 =\displaystyle\frac{exp(\beta z)}{\beta!} = exp(\beta p_{.}(z)), $$
$$L_z p_{n}(z)=n p_{n-1}(z)=\displaystyle\frac{d}{dz} p_{n}(z),$$
$$R_z p_{n}(z)= p_{n+1}(z)= (z+\gamma)p_n(z)-\displaystyle\int_{-\infty}^{z}\frac{p_n\left (  z\right )-p_n(u)}{e^z-e^u} e^u du$$
with $\gamma=-d\beta!/d\beta|_{\beta=0}$, the Euler-Mascheroni constant.
Since $p_n(z)$ is an Appell sequence and, consequently, $p_n(x+y)=(p.(x)+y)^n$, umbrally, a change of integration variables $\omega=z-u$ gives
$$R_z p_{n}(z)= p_{n+1}(z)= (z+\gamma)p_n(z)-\displaystyle\int_{0}^{\infty}[p_n(z)-(p_{.}(z)-\omega)^n] \frac{1}{e^{\omega}-1}d\omega$$
from which the recursion formula follows.
In addition, using the operator formalism for Sheffer sequences, of which the Appell is a special case,
$$R_z=z-\frac{\mathrm{d} }{\mathrm{d} \beta}ln[\beta!]\mid _{\beta=\frac{\mathrm{d} }{\mathrm{d} z}=D_z}=z-\Psi(1+D_z)$$
where $\Psi(x)$ is the digamma or Psi function.
UPDATE (Nov. 16, 2012): Just found this exact sequence in the thesis "Regularized Equivariant Euler Classes and Gamma Functions" by R. Lu with a discussion of the relationships to Chern and Pontrjagin classes.

REPLY [3 votes]: Follow-up on Rupinski's and Chapoton's observations:
To nail down the identification of the $p_n(x)$ with the cycle index polynomials for $S_n$ (or the partition polynomials of the refined Stirling numbers of the first kind A036039), look at the Taylor series rep of the digamma operator for the raising / creation operator for the $p_n(z)$ basis 
$$R_z = z-\Psi(1+D_z) = z+\gamma + \sum_{n=1}^{\infty } (-1)^n\zeta (n+1)D_z^n.$$ 
This is precisely the raising operator for the cycle index polynomials as presented on page 23 of Lagrange à la Lah Part I with $c_1=z+\gamma=p_1(x)$ and $c_{n+1}=(-1)^n\zeta(n+1)$ for $n>0$
$$D^{-1}_{c_1}= :\frac{c_{.}}{1-c_{.}D_{c_1}}: = c_1+\sum_{n=1}^{\infty } c_{n+1}D_{c_1}^n.$$
Alternatively, the Newton identities extrapolated to an entire function as an infinite order polynomial using the Weierstrass factorization maneuver can be applied to see the connections to the power and elementary symmetric polynomial formalism: 
$$\exp\left (-\beta p_{.}(z)\right )=\frac{\exp\left (-\beta z  \right )}{\left (-\beta  \right )!}=\exp\left (-\beta(z+\gamma)  \right )\prod_{k=1}^{\infty }\left ( 1-\frac{\beta}{k} \right )\exp\left (\frac{\beta}{k}  \right )$$
$$=\exp\left [-(z+\gamma)\beta -\sum_{k=2}^{\infty } \frac{\zeta (k)\beta ^k}{k}  \right ]=\exp\left [ :ln(1-a\beta ) :\right ]$$ where $a^1=a_{1}=(z+\gamma)$ and $a^k=a_k=\zeta(k)$ for $k>1$, but this is precisely the umbral form of the e.g.f. for the cycle index polynomials (mod signs).
(Also there are connections to rational zeta series.)
Update (Nov. 16, 2012): The generating series appears on pg. 58 in "Hodge theoretic aspects of mirror symmetry" by L. Katzarkov, M. Kontsevich, and T. Pantev (following Lu's references).<|endoftext|>
TITLE: What ordinals are definable relations in Peano Arithmetic?
QUESTION [9 upvotes]: I am not asking which order types PA proves are well ordered.  That would be all up to $\epsilon_0$.  Rather I mean, assuming a stronger ambient theory such as Zermelo set theory, which ordinals have the order type of some relation on $\mathbb{N}$ that is defined by a formula of PA (not requiring that PA prove the relation is a well ordering).

REPLY [9 votes]: These are the recursive ordinals. The same well-order-types can be realized by recursive relations as by hyperarithmetical relations.  PA-definable, i.e., arithmetical, falls nicely between these two.  (I think you can go considerably lower, say to PTime-computable relations, and still have the same order-types.)

REPLY [8 votes]: The answer is the ordinal $\omega_1^{ck}$, named after Church and Kleene, which is defined to be the supremum of the ordinals coded by a computable relation on $\mathbb{N}$. It happens also to be the supremum of the order types of the relations coded by any arithmetic relation, that is, by any relation definable in the language of arithmetic.

REPLY [7 votes]: The computable ordinals---that is, the ordinals below $\omega_1^{CK}$---are, by definition, represented by computable relations, all of which can be represented by formulas in PA, and indeed, even by fairly simple formulas.  As Andreas points out, allowing arithmetic formulas instead of computable ones does not change the class of ordinals.<|endoftext|>
TITLE: Direct proof of "K is projective iff C(K) has the Hahn-Banach property" ?
QUESTION [7 upvotes]: An object $X$ of a given category is called projective if for each morphism $f : X \rightarrow Z$, and each epimorphism $ g : Y \twoheadrightarrow Z$, there is a morphism $h : X \rightarrow Y$ such that $f=gh$.
An ordered vector space $X$ is said to have the Hahn-Banach extension property if for each real vector space $Y$, each subspace $Z$ of $Y$, each sublinear operator $V : Y \rightarrow X$, and each linear operator $T : Z \rightarrow X$ satisfying $T \leq V_{|Z}$ pointwise, there is a linear operator $\hat T : Y \rightarrow X$ with $\hat T_{|Z}=T$ and $\hat T \leq V$.
A theorem of Gleason asserts that a compact (Hausdorff) space $K$ is projective in the category $\mathbf{CHaus}$ of compact Hausdorff spaces iff $K$ is extremally disconnected.
By results of Goodner and Nachbin, the ordered vector space $C(K)$ has the Hahn-Banach extension property iff $K$ is extremally disconnected.
It is thus known that: $K$ is projective in  $\mathbf{CHaus}$ iff $C(K)$ has the Hahn-Banach extension property $(\star)$.
But : the Hahn-Banach extension property is a kind of projectivity with reversed arrows (reflecting the contravariance of the functor which sends maps $f : X \rightarrow Y$ to $u \in C(Y) \mapsto u \circ f \in C(X)$ ), so that it is reasonable to expect a "direct proof" of $\star$. It is relatively easy to show the direction "$\rightarrow$" in the equivalence (see below), and I would be very happy if the other direction also had a "direct proof" ... Thanks in advance.
Proof of the direction "$\rightarrow$" in $\star$ : Let $K$ be a projective object of $\mathbf{CHaus}$, and $(Y, Z, V, T)$ as in the definition of the Hahn-Banach property. Following Rainwater, let $D$ be the set $K$ endowed with the discrete topology and $r : \beta D \twoheadrightarrow K$ be the (unique) continuous extension of the canonical map $D \rightarrow K$ to the Stone-Cech compactification $\beta D$ of $D$. By projectivity, there is some $i : K \rightarrow \beta D$ with $id_{K} = r \circ i$. Without loss of generality $i$ is an inclusion map. Then $r$ is a retraction of $\beta D$ onto $K$. The maps $r$ and $i$ induce (by the functor described above) maps $\tilde r : C(K) \rightarrow C(\beta D)$ and  $\tilde i : C(\beta D) \rightarrow C(K)$. As a consequence of the usual Hahn-Banach theorem, the space $C(\beta D) \simeq \ell^\infty(D)$ has the Hahn-Banach extension property ; applying this to $(Y,Z,\tilde r \circ V, \tilde r \circ T)$ yields an operator $\hat T'$ with $\hat T'_{|Z}=\tilde r \circ T$ and $\hat T'\leq \tilde r \circ V$. The desired operator is $\hat T = \tilde i \circ \hat T'$.

REPLY [5 votes]: As indicated by Theo Buehler above, Fremlin proves what I want and much more in his book. However, the proof given in this reference can be simplified a lot in my setting, so that I can answer my own question by giving a relatively simple proof :
Basically, Fremlin's proof begins by finding a projective resolution of $K$ inside the dual ball of $C(K)$. But the existence of a projective resolution in $\mathbf{CHaus}$ has a short proof (see this article), so I use it freely below.
Proof of the direction "$\leftarrow$" in $\star$ : Let $X$ be a projective resolution of $K$ in $\mathbf{CHaus}$, and let $p : X \twoheadrightarrow K$ be the corresponding map, which satisfies $p(S) \neq K$ whenever $S$ is a proper closed subspace of $X$. Then $p$ induce an isometry $\tilde p : f \in C(K) \rightarrow f \circ p \in C(X)$. Since $\mathrm{Im} \; \tilde p$ is isometric to $C(K)$, which has the Hahn-Banach extension property, we can extend the identity on $\mathrm{Im} \; \tilde p$ to a norm-one operator $T : C(X) \rightarrow \mathrm{Im} \; \tilde p$.

Let $h$ be in the unit ball of $C(X)$, and set $S=\overline{ \lbrace h \neq 0 \rbrace } \cup \lbrace Th = 0 \rbrace$. If $S$ is a proper subspace of $X$ then some $p(x) \in K$ doesn't belong to $p(S)$. Let $f \in C(K,\[ 0,1 \])$ be such that $f \circ p(x) = 1$ and $f_{|p(S)}=0$. Then $\lVert \tilde p (f) \pm h \rVert \leq 1$, so that $\lVert \tilde p (f) \pm Th \rVert = \lVert T( \tilde p (f) \pm h )\rVert \leq 1$. But for an appropiate choice of sign, the value of $\tilde p (f) \pm Th$ at $x$ exceeds $1$, a contradiction. Thus $S=X$.

Let $x_1$ and $x_2$ be distinct points in $X$, hence separated by disjoint closed sets $F_1$ and $F_2$ in $X$. Let $h \in C(X,\[ 0,1 \])$ be such that $h_{|F_1}=0$ and $h_{|F_2}=1$. Then $\overline{ \lbrace h \neq 0 \rbrace } \subset {}^c( \mathring F_1)$, so that by the point above $x_1 \in \mathring F_1 \subset \lbrace Th = 0 \rbrace$, and $Th(x_1)=0$. Similarly, $Th(x_2)=1$. Since $Th \in \mathrm{Im} \; \tilde p$, this shows that $p(x_1) \neq p(x_2)$.


We have shown that $p$ is injective, so that $p$ is a homeomorphism $X \simeq K$. Thus $K$ is projective.
PS: Thanks for the Latex,Theo.<|endoftext|>
TITLE: A "dual" universal coefficient theorem
QUESTION [5 upvotes]: Universal coefficient theorem allows us to calculate 
$H^*(X,M)$ 
from $H_*(X,Z)$.
Do we have a "dual" universal coefficient theorem that allows us to calculate $H_*(X,M)$ from $H^*(X,Z)$? 
Here $Z$ is the set of integers.

REPLY [5 votes]: The proof of Spanier's 6.5.12 starts from a free chain complex with homology of finite type and replaces it by a quasi-isomorphic free chain complex of finite type.  Then the conclusion reduces directly to application of the usual universal coefficient theorem for computing cohomology of chain complexes from homology.  The reference to EKMM should be to p. 82, which states "A reference to Adams [1] is mandatory''; [1] is "Lectures on generalized cohomology". Springer Lecture Notes 99,
1969, pp 1--138.  As recalled on EKMM p. 82, Adams shows how to deduce further spectral sequences from
those listed on that page, by duality.  Adams' (UCT 4), page 5 op cit, gives the spectral sequence that generalizes the result cited from Spanier to generalized homology/cohomology theories.<|endoftext|>
TITLE: Triviality of Associated Bundles
QUESTION [6 upvotes]: Let $P\rightarrow  M$ be a principal (right) $G$-bundle, where $G$ is a Lie group. Given a finite-dimensional representation of $G$, $V$ say, we can define the associated bundle $P\times_{G}V\rightarrow M$. This is a vector bundle over $M$ defined as the quotient of the (free, right) action of $G$ on $P\times  V$ - $(p,v)\cdot g  =(p\cdot g, g^{-1}v)$.
Hence, for a given representation $V$ of $G$ we can associate to a principal $G$-bundle $P\rightarrow M$ a vector bundle $P\times_{G} V\rightarrow M$. Moreover, this assignment is functorial and so induces a map from isomorphism classes of principal $G$-bundles to $K_{0}(M)$, the Grothendieck group of vector bundles on $M$. Call this functor (and, by abuse of notation, the map it induces) $\theta_{V}$. Furthermore, it seems (there may be problems here?) that we obtain a functor 
$\theta: Rep_{G}\rightarrow Fun(Prin_{G}(M),Vec(M))$
where the left hand side is the category of (finite dimensional) representations of $G$ and the right hand side is the category of functors from $Prin_{G}(M)$ to $Vec(M)$, the categories of principal $G$-bundles on $M$ and vector bundles on $M$ (respectively).
Question 1: Which representations induce the trivial map on iso-classes? For example, the trivial representation $T$ will always give 
$\theta_{T}(P\rightarrow M)=M\times T$
since we can choose linearly independent generating sections of $P\times_{G} T$ using triviality of $T$. My question is, are there other representations of $G$ which afford this property?
Question 2: What am I really discussing here? Is there a name for $\theta$? Do these ideas arise in some 'deeper' (or more natural) framework?
Question 3: Is this formulation useful? Are there any interesting results related to this construction? 
I have come to these conclusions as a result of thinking about associated bundles based on knowing the basic definition only and any references/comments would be appreciated. My apologies if this is standard material to topologists, or well-known to experts  - I am neither.

REPLY [4 votes]: Theorem: ''Let $G$ be a compact, connected Lie group and $f: G \to U(n)$ a group homomorphism such that for each principal bundle $P \to M$ on a manifold, the induced vector bundle $P \times_{G,f} \mathbb{C}^n$ is a trivial vector bundle. Then $f$ is the constant homomorphism.''
Proof: ''For a given $k$, there exists a compact manifold $M$ and a map $M \to BG$ that is $k$-connected, $dim (M) \geq 2k+1$. This is manufactured using surgery below middle dimensions. Applying this to the assumption, you get that $f$ induces the trivial map on cohomology $H^{\ast}(BU(n)) \to H^{\ast}(BG)$ of any degree.
Now assume $f$ is zero on real cohomology $H^{\ast}(BU(n)) \to H^{\ast}(BG)$. By Chern-Weil theory, $H^{\ast}(BG) \cong Sym^{\ast}(\mathfrak{g})^G$, the algebra of Ad-invariant symmetric polynomials on the Lie algebra. There is a symmetric polynomial of degree $2$ on $\mathfrak{u}(n)$ that is nowhere zero: take an invariant scalar product. Therefore, the assumption implies that $f$ has to be zero on the Lie algebra level; hence $f$ is constant on the unit component of $G$.''
I think this is true for nonconnected $G$ and believe the argument is similar to the one by Chris Gerig and myself to this question:
Non-vanishing of group cohomology in sufficiently high degree
But I do not have time to think this through right now.<|endoftext|>
TITLE: Is a wedge of spheres an $E_\infty$ ring spectrum? 
QUESTION [6 upvotes]: The wedge sum $\bigvee_{k \in 2 \mathbb{Z}} S^{k}$ is an $A_\infty$-ring spectrum: the connective cover is the free $A_\infty$-ring on the sphere $S^2$, if I'm not mistaken, and then one inverts the element in $\pi_2$. It is also homotopy commutative. Can it be made into an $E_\infty$-ring spectrum? 
More generally, given an $E_\infty$-ring spectrum $R$, when can $\bigvee_{k \in 2 \mathbb{Z}} \Sigma^k E$  be made into an $E_\infty$-ring?

REPLY [14 votes]: Ah, tracked it down.  Here is an argument.  I should mention that Peter once pointed me towards an original source due to McClure, or page 238, Prop. 6.1, of SLN 1176 (the $H_\infty$ book).
Suppose you had such a ring object $R$.  We examine its mod-2 homology.  This has several features:

It is the ring $\mathbb Z/2[t^{\pm 1}]$.
It has trivial action of the Steenrod operations $P_r$ (this is dual to the cohomology action), because it's a wedge of spheres.
From the $H_\infty$ book, it has Dyer-Lashof operations $Q^s$.  These satisfy $Q^{|x|} x = x^2$, and the Nishida relations
$$
P_r Q^s = \sum \binom{s - r}{r - 2i} Q^{s-r+i}P_i.
$$

In particular, these together would say 
$$
0 = P_2 Q^4 t = \binom{2}{2} Q^{2}P_0 t + \binom{2}{0} Q^{4}P_2 t = Q^2 t = t^2.
$$
If $E$ is a commutative $MU$-algebra, then you can use the map from $MU$ to its periodic version $MUP$ to produce $R \wedge_{MU}MUP$, which is 2-periodified.  Unfortunately, we know very few genuine $MU$-algebras.  Barry Walker proved that complex K-theory is one of them, based on Matthew Ando's work on studying $H_\infty$ structures on Lubin-Tate cohomology theories.  My understanding is that the problem is still open for almost all of the Lubin-Tate cohomology theories.<|endoftext|>
TITLE: Number of linear orders
QUESTION [13 upvotes]: It is well known that for every infinite cardinal $\kappa$ the number of non-isomorphic total orders of cardinality $\kappa$ is $2^\kappa$.  Who first proved this, and in what context?   Was it proved for $\kappa=\aleph_0$ first, and then for uncountable $\kappa$, or for all $\kappa$ right away? 
(This question has been unanswered at stackexchange for a long time.)

REPLY [18 votes]: Hi Martin. 
I learned what follows in J.M. Plotkin, ed., Hausdorff on Ordered sets, AMS, History of Mathematics 25, 2005.
The result for countable ordered sets is due to Cantor. More precisely, Cantor produced continuum many countable order types: 

Assign to each $x\in 2^\omega$ the type $x_0+(\omega^*+\omega)+x_1+(\omega^*+\omega)+x_2+\dots$ 

Bernstein proved (by March, 1901 or earlier) that there can be no more than continuum many. The result (both parts) appears in his 1901 dissertation, Untersuchungen aus der Mengenlehre. That there are at most continuum many types was also found independently by Hausdorff (June 27, 1901, according to his Nachlass, during a Summer course on Set theory at the University of Leipzig); it is referred to as the "Cantor-Bernstein theorem" in Hausdorff's Über eine gewisse Art geordneter Mengen, Gesellschaft der Wissenschaften zu Leipzig, Mathematisch-Physische Classe 53 (1901), 460-475. (See pg. 460) 
It is probably worth remarking that this shows Bernstein's and Hausdorff's comfort with using choice in their arguments from the beginning, as clearly this gives us a "natural" example showing that $\omega_1\le 2^{\aleph_0}$. (On the other hand, Hausdorff remains unsure at this time on whether $\mathbb R$ can be well-ordered.)
Bernstein gives two proofs of the upper bound. The first goes by identifying an ordered set $M$ with a function $f:M^2\to\{-1,0,1\}$: 

$f(a,b)=-1$ iff $a\lt b$, $f(a,b)=0$ iff $a=b$, and $f(a,b)=1$ iff $a\gt b$. 

One then just has to count such functions for $M=\mathbb N$. 
The second proof uses Cantor's result that $\mathbb Q$ is universal for countable linear orders, so one only has to count subsets of $\mathbb Q$.

The general result is due to Hausdorff, in his 1901 paper, with the published upper bound the same argument as Bernstein's first proof. (There seem to be no records of Hausdorff's original argument, or of whether it was different.) He doesn't quite use that $M$ is well-orderable, but rather that $\mathfrak m=|M|$ satisfies $\mathfrak m^2=\mathfrak m$. 
The lower bound is argued as follows: Given an infinite linearly ordered set $M$ of size $\mathfrak m=|M|$, call it graded iff no two distinct initial segments of $M$ are order isomorphic. Assume $M$ admits a graded ordering (which is clearly the case if $\mathfrak m$ is an aleph). For $m\in M$, denote by $A_m$ the set of predecessors of $m$. 
Given a set $S\subseteq M$ of size $\mathfrak m$, assign to it the ordered sum $L_S$ of the sets $\mathbb Q+A_s$, $s\in S$, and note that if $S\ne S'$ then $L_S$ and $L_{S'}$ are not order isomorphic. The result follows if $[\mathfrak m]^{\mathfrak m}=2^{\mathfrak m}$ which, again, holds if $\mathfrak m$ is an aleph. (Plotkin remarks that Hausdorff could have elaborated a bit more on why the sets $L_S$ are not isomorphic.)<|endoftext|>
TITLE: Subtract Rectangle from Polygon
QUESTION [7 upvotes]: I'm looking for an algorithm that will subtract a rectangle from a simple, concave polygon and return a remainder of polygons. If the rectangle encloses the polygon, the remainder is null. In most cases, it looks like at least one edge will be shared between the rectangle and the polygon.
I've been digging around the internet, but I've not found a good lead.
Can someone point me in the right direction?
Edit
I did some more digging and found a few polygon clipping algorithms. However, all of the ones I found finds what is inside the clipping region. I want all of the stuff outside as separate rectangles.
If I understand correctly, the Weiler–Atherton algorithm detects all of the intersections between the clipping region and the polygon. I was thinking of using that approach. I've got some tricky cases where the edge of the clipping region and the edge of the polygon are collinear.
Do I sound like I am getting closer?

REPLY [16 votes]: I suggest you investigate the CGAL manual, Chapter 19:
"2D Regularized Boolean Set-Operations".
Here is an example of polygon difference:

            
This is well-understood & explored algorithmically, but nevertheless a very delicate computation, as the term regularized indicates.  You are subtracting a rectangle, but there is nothing special about a rectangle—"degenerate" situations may occur.
You really need to consider the general situation. So I recommend you look at general resources,
such as

"Boolean operations on general planar polygons."
  M. Riveroa, F.R. Feitob.
  Computers & Graphics.
  Volume 24, Issue 6, December 2000, Pages 881–896
  (Elsevier link)

or 

"A General Polygon Clipping Library."
  Version 2.32.
  Alan Murta.
  (link)

The upperleft corner of Murta's Fig.2 illustrates a complicated polygon difference (and the
remaining images illustrate other Boolean operations).<|endoftext|>
TITLE: Length of Hirzebruch continued fractions
QUESTION [9 upvotes]: Suppose $a,b$ are two natural numbers relatively prime to $n$ and to each other. Assume $n\geq ab+1$. Suppose further that $\frac{a}{b}\equiv k \pmod{n}$ for some $k\in \lbrace 1,2,\dots, n-1\rbrace$ and $\frac{a}{b}\equiv k'\pmod{n+ab}$ for some $k'\in \lbrace 1,2,\dots, n+ab-1\rbrace$.

Question: Is there an elementary proof that the length of the continued fraction of $\frac{n}{k}$ is equal to the length of the continued fraction of $\frac{n+ab}{k'}$?

This came out of a broader result, and for this particular case I can prove it using routine toric geometry, however I would like to know of some elementary tricks to deal with continued fractions. 

Here by continued fraction I mean the Hirzebruch continued fraction 
$$\frac{n}{k}=a_0-\frac{1}{a_1-\frac{1}{a_2-\cdots}}.$$
For example, when $a=2, b=3$ and $n=17$, we get $k=12$ and $k'=16$, so the fractions are 
$$\frac{17}{12}=2-\frac{1}{2-\frac{1}{4-\frac{1}{2}}}\qquad and \qquad\frac{23}{16}=2-\frac{1}{2-\frac{1}{5-\frac{1}{2}}}.$$

REPLY [10 votes]: Lets call expansions
$$\langle
x_1,\ldots,x_m\rangle:=\cfrac{1}{x_1-{\atop\ddots\,\displaystyle{-\cfrac{1}{x_m}}}}$$
(as in Perron's book) reduced regular continued fractions (RRCF).
Probably they are older then Hirzebruch.
We'll prove more precise statement.
Theorem. If $(n,ab)=1$ and $n>ab$ then RRCF for all numbers
$$\left\{\frac{ab^{-1}\pmod{(n+kab)}}{n+kab}\right\}\qquad(k\ge
0)$$ are almost equal: they have equal length and differ only in
one partial quotient.
Remark 1. Common factors of $a$ and $b$ can be moved into $k$.
If $d=(a,b)$, $a=da_1$, $b=db_1$, then \begin{gather*} \left\{
\frac{ab^{-1}\pmod{(n+kab)}}{n+kab}\right\} =\left\{
\frac{a_1b_1^{-1}\pmod{(n+kab)}}{n+kab}\right\} \\=\left\{
\frac{a_1b_1^{-1}\pmod{(n+(kd^2)a_1b_1)}}{n+(kd^2)a_1b_1}\right\}.
\end{gather*}
So we can assume that $(a,b)=1$.
Remark 2. The proof will be given in terms of modified
continuants $K(x_1,\ldots, x_n)$ (see ``Concrete Mathematics'' for
more explanations). These polynomials are defined by initial
conditions
$$K()=1,\quad K(x_1)=x_1$$
and the following recurrence:
$$K(x_1,\ldots,
x_n)=x_nK(x_1,\ldots, x_{n-1})-K(x_1,\ldots,
x_{n-2})\qquad(n\ge2).$$ (In the usual definition minus must be
replaced by plus.) For convenience $K_{-1}:=0$ (empty RRCF is
$0$).
In terms of continuants RRCF can be written as
$$\langle x_1,\ldots,x_n\rangle=\frac{K(x_2,\ldots,
x_n)}{K(x_1,\ldots, x_n)}.$$
Continuant's properties. All these properties can be proved by
induction (or from ``Euler’s rule'').
1$^{\circ}.$ $K(x_1,\ldots, x_n)=K(x_n,\ldots, x_{1})$.
2$^{\circ}.$ \begin{gather*} K(x_1,\ldots, x_n,x_{n+1}, \ldots,
x_{m+n})\\=K(x_1,\ldots, x_n)K(x_{n+1}, \ldots,
x_{m+n})-K(x_1,\ldots, x_{n-1})K(x_{n+2}, \ldots, x_{m+n})
\end{gather*}
3$^{\circ}.$ $\begin{vmatrix}
     K(x_2,\ldots, x_{n-1})&K(x_2,\ldots, x_n) \\
     K(x_1,\ldots, x_{n-1})&K(x_1,\ldots, x_n)
\end{vmatrix}=-1$. In particular if
$$\frac{A}{a}=\left<r_1, \ldots, r_v\right>=\frac{K(r_2, \ldots,
r_v)}{K(r_1, \ldots, r_v)}$$ then
$$K(r_1, \ldots, r_{v-1})=A^{-1}\pmod{a},\qquad K(r_2, \ldots, r_{v-1})=\frac{AA^{-1}\pmod{a}-1}{a}.$$
4$^{\circ}.$ Euler's identity (see A Short Proof of Euler's Identity
for Continuants for additional arguments). (2$^{\circ}$ and
3$^{\circ}$ are special cases of this identity) $$
K(x_1, \ldots, x_{m+n})K(x_{m+1}, \ldots, x_{m+l})-K(x_1, \ldots,
x_{m+l})K(x_{m+1}, \ldots, x_{m+n})$$ $$+K(x_1, \ldots,
x_{m-1})K(x_{m+l+2}, \ldots, x_{m+n})=0.
$$
Proof of the Theorem. For a given $n$ define
$a^{-1}:=a^{-1}\pmod{n}$, $b^{-1}:=b^{-1}\pmod{n}$, $0\le a,b\le n-1$
(inverse number is always least possible nonnegative) and $t_a$,
$t_b$ such that $aa^{-1}=1+t_an$, $bb^{-1}=1+t_bn$. Let $$
\frac{A}{a}=\left\{\frac{bt_a}{a}\right\}=\left<r_1, \ldots,
r_v\right>,\qquad A^{-1}:=A^{-1}\pmod{a};$$
$$\frac{B}{b}=\left\{\frac{at_b}{b}\right\}=\left<q_1, \ldots,
q_u\right>,\qquad B^{-1}:=B^{-1}\pmod{b};$$
$$\frac{P(x)}{Q(x)}=\left<q_1, \ldots, q_u,x,r_v, \ldots,
r_1\right>.
$$
By 2$^{\circ}$, 3$^{\circ}$ and main recurrence $$
Q(x)=K(q_1, \ldots, q_u,x,r_v, \ldots, r_1)=$$
$$xK(q_1, \ldots, q_u)K(r_v, \ldots, r_1)-K(q_1, \ldots,
q_{u-1})K(r_v, \ldots, r_1)-K(q_1, \ldots, q_u)K(r_{v-1}, \ldots,
r_1)$$ $$=xab-aB^{-1}-bA^{-1}.
$$
Hence $$ Q(x)\equiv -bA^{-1}\equiv -b((bt_a)^{-1}\pmod{a})\equiv
-t_a^{-1}\equiv n\pmod{a},$$ $$Q(x)\equiv -aB^{-1}\equiv
-a((at_b)^{-1}\pmod{b})\equiv -t_b^{-1}\equiv n\pmod{b}.$$
Therefore
$$Q(x)\equiv n\pmod{ab},$$
and for some integer $x_0$ we have $Q(x_0)=n$. We know that $n>ab$. It
means that
$$x_0ab-aB^{-1}-bA^{-1}>ab,$$
so $x_0\ge 2$ and $\left<q_1, \ldots, q_u,x_0,r_v, \ldots,
r_1\right>$ is really RRCF. Choosing arbitrary $x=x_0+k$ we'll get
progression $n+kab$ as in the statement of the theorem.
Let's check the numerator $P(x)$. Final step $P(x)\equiv
ab^{-1}\pmod{Q(x)}$ follows from identity
$$bP(x)-BQ(x)=a,$$
which is a special case of Euler's identity. Nevertheless this
identity can be verified directly with help of
1$^{\circ}$--3$^{\circ}$: $$
P(x)=K(q_2, \ldots, q_u,x,r_v, \ldots, r_1)=$$
$$xK(q_2, \ldots, q_u)K(r_v, \ldots, r_1)-K(q_2, \ldots,
q_{u-1})K(r_v, \ldots, r_1)-K(q_2, \ldots, q_u)K(r_{v-1}, \ldots,
r_1)$$ $$=xaB-a\frac{BB^{-1}-1}{b}-BA^{-1},$$ $$
bP(x)-a=B(xab-aB^{-1}-bA^{-1})=BQ(x).
$$<|endoftext|>
TITLE: Provable zero-free region for any entire function that analytically is similar to zeta(s)
QUESTION [20 upvotes]: Is there an entire function $f:\mathbb C\rightarrow\mathbb C$ such that for some $\delta>0$:

$f(z)$ is bounded when $\Re z>1+\delta$
$f(z)$ is unbounded when $\Re z=1$
$f(z)$ grows polynomially in vertical strips, ie for all $\sigma$ there is $C_\sigma$ so that $|f(\sigma+i t)|\ll|t|^{C_\sigma}$
$f(z)$ does not vanish when $\Re z>\frac12$ (provably!).

Conjecturally there is a very rich family: $L$-functions, but (4) is unproven.
If you drop (2), $1+e^{-z}$ works, or $\zeta(1/2+i t)$, or Wang zeta functions
If you drop (3), the Selberg zeta functions works, or $\exp(L(s))$
Edit: Note that $\zeta(s)$ is not entire so you can instead look at $\zeta(s)(s-1)/(s+2)$ or change the question to allow finitely many singularities, where the bounds are taken away from the singularities.

REPLY [13 votes]: OK, shameless cheating, as promised.
Part 1. Let's start with something.
We need a function bounded in $\Re z>1$ and growing not too fast on each vertical line whose zeroes are somewhere on the left. The first thing that comes to mind is $1$. No zeroes anywhere in sight, beautiful control on vertical lines. All that is lacking is the unboundedness on $\Re z=1$.
Part 2. Push it up!
We now want to add some bumps on the uneventful road $\Re z=1$. It is natural to add one bump a time. We have two options for bumping: addition and multiplication. Since we want to control the zeroes without trouble, we'll use multiplication. So, we'll be looking for an infinite product.
Part 3. A tiny little bump.
Take some entire function $g$ bounded by $1$ in the right half-plane, tending to $0$ at infinity in any right half-plane, and attaining its maximum of absolute value in $\Re z\ge 1$ at $1$, where it is real and positive. Denote $g(1)=a>0$. The exact choice doesn't matter.  I'll take $g(z)=\frac{1-e^{-z}}{z}$. Put $F(z)=1+g(z)$. Now the ride along the line $\Re z=1$ is not that smooth anymore: you have to ascend to a small hill at $1$. However, at infinity everything levels to $1$ uniformly in any right half-plane. Also, if there are any zeroes, they all have non-positive real parts. 
Part 4. Amplify the bump (being naive and fair)
Just raise $F$ to a high power $N$. You'll get as huge bump as you want. The problem is that it also becomes huge well to the right of $1$.
Part 5. Discriminate against numbers with the large real part.
Replace $g(z)$ by $g(z)e^{-n^2z}$. Of course the value at $1$ will suffer enormously, but everything with real part greater than $1$ will suffer much more (which is the whole point of any true discrimination).
Part 6: Amplify with discrimination. 
Raise $F(z)=1+g(z)e^{-nz}$ to the power $N$. We'll get $(1+ae^{-n^2})^N$ at $1$ but only at most $(1+ae^{-n^2-2n})^N$ for $\Re z>1+\frac 2{n}$. Choose $N\approx a^{-1}e^{n^2+n}$. We'll get about $e^{n}$ at $1$ and at most $1+2e^{-n}$ to the right of $1+\frac 2n$. Now it is quite a bump, and it is next to invisible just a tiny bit to the right of $1$.
Part 7: Ship it up the line to satisfy the local regulations.
Replace $F(z)^N$ with $F(z-iy_n)^N$ with large $y_n$ to satisfy the polynomial growth restriction in $\Re z>-n$: let's even make $|F(z)^N-1|<2^{-n}(1+|z|)^{2^{-n}}$ in $\Re z>-n$. Remember that though our bump function is huge, it is still bounded in any right half plane and levels to $1$ at infinity there. We also have $|F(z)|^N\le 1+2e^{-n}$ when $\Re z>1+\frac 2n$ regardless of the shipment.
Part 8. Put the production and shipment of bumps on the conveyor belt with $n=1,2,3,...$, and enjoy the product.
Of course, this is as shameless, abominable, and mostly illegal as any manufacturing under loose government regulations. Every loophole that could be exploited in the formulation of the problem has been exploited. So, please, do not accept or upvote. Instead, think of how to tighten the regulations to force someone to do honest work. :)<|endoftext|>
TITLE: Where in the literature does the anticyclotomic $\mathbf{Z}_p$-extension of an imaginary quadratic field first appear?
QUESTION [10 upvotes]: If $K$ is an imaginary quadratic field, then the $\mathbf{Z}_p$-rank of $K$ is $2$, meaning that the Galois group of the compositum of all the $\mathbf{Z}_p$-extensions of $K$ in an algebraic closure $\overline{K}$ is isomorphic to $\mathbf{Z}_p^2$. It is known that the compositum is generated by two special $\mathbf{Z}_p$-extensions: the cyclotomic and the anticyclotomic $\mathbf{Z}_p$-extensions.
Both are Galois over $\mathbf{Q}$ (I believe they are the only $\mathbf{Z}_p$-extensions of $K$ which are Galois over $\mathbf{Q}$). The cyclotomic extension is abelian over $\mathbf{Q}$ and is equal to the compositum of $K$ and the cyclotomic (the only!) $\mathbf{Z}_p$-extension of $\mathbf{Q}$. The anticyclotomic extension is "generalized dihedral" over $\mathbf{Q}$, which means that the unique non-trivial element of $\mathrm{Gal}(K/\mathbf{Q})$ acts on $\mathrm{Gal}(K_\infty^{anti}/K)$ by inversion.
I learned these facts from various sources, after being told by my advisor what the anticyclotomic $\mathbf{Z}_p$-extension was (she gave me the generalized dihedral definition). My question is: when did $K_\infty^{anti}$ first appear in the Iwasawa theory literature? Does it appear in the work of Iwasawa (I'm not all that familiar with his work)? Maybe the work of Greenberg or Washington? Was the primary motivation for studying it the connection with CM elliptic curves?
My motivation for asking this is because I'd really like to read a comprehensive exposition of its basic properties (although perhaps this doesn't exist). I've picked up bits and pieces from books and papers. For example, I know that $K_\infty^{anti}$ is the unique $\mathbf{Z}_p$-extension contained in the union of the $p$-power conductor ring class fields of $K$, and the behavior of primes in these class fields can be determined by the ideal-theoretic formulation of global class field theory, but I feel like the basics are not quite as straightforward as for cyclotomic $\mathbf{Z}_p$-extensions (which somehow seem more natural to me).

REPLY [6 votes]: First time I heard  about antycyclotomic $\Gamma$-extensions  in 1972 from  Pavel Kurchanov, in connection with his paper 
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=3020&option_lang=eng
Actually, his goal was to construct elliptic curves of infinite rank over $\Gamma$-extensions. (According to Mazur's conjecture, one cannot do it over the cyclotomic extensions.)<|endoftext|>
TITLE: Disjoint stationary sets that reflect
QUESTION [11 upvotes]: Hello, I have the following question (for definitions see at the end):
Let $\kappa$ be an uncountable regular cardinal. 
Can we prove in ZFC that there exist two disjoint stationary sets $A$, $B$ such that for every limit ordinal $\alpha<\kappa$ of uncountable cofinality, both $A$ and $B$ reflect at $\alpha$?
Definitions: (1) $A$ is a stationary set on $\kappa$, if $A\subset\kappa$ and $A$  intersects every closed and unbounded set in $\kappa$. 
(2) A set is closed if it contains its limit points.
(3) A stationary set $A\subset\kappa$ reflects at $\alpha$ if $A\cap\alpha$ is stationary on  $\alpha$.

REPLY [6 votes]: In the presence of large cardinals, one can (or rather Shelah can...) force the answer to be "NO" in a very strong sense.  The place to look is Section 7 of Chapter X of Proper and Improper Forcing.
In particular, Theorem 7.4 shows that assuming the consistency of 2 supercompact cardinals, one can force that for any regular $\kappa>\omega_1$, any stationary subset of $S^\kappa_{\aleph_0}$ contains a closed copy of $\omega_1$.
This implies the answer to your question is no by the following argument: 
Step 1:  If $\kappa>\aleph_1$ is regular and $A$ reflects at all uncountable limit ordinals below $\kappa$, then so does $A\cap S^\kappa_{\aleph_0}$ (where $S^\kappa_\tau$ is the set of ordinals less than $\kappa$ of cofinality $\tau$).
Proof: Let $A_0= A\cap S^\kappa_0$, and let $A_1= A\setminus A_0$.  $A_1$ cannot reflect at ordinals of cofinality $\omega_1$, and so it must be the case that  $A_0$ reflects at all ordinals of cofinality $\omega_1$. But then $A_0$ also reflects at any place where $S^\kappa_{\aleph_1}$ reflects as well, and so $A_0$ reflects at all ordinals of uncountable cofinality below $\kappa$.
Step 2:
Assume we are in a model like that obtained by Shelah.  If $\kappa$ is a regular cardinal greater than $\aleph_1$ and $A$ is a stationary subset of $S^\kappa_{\aleph_0}$.  We know $A$ contains a closed copy $C$ of $\omega_1$, and if we set $\delta=\sup(C)$ then $\delta$ is an ordinal of cofinality $\omega_1$ where $A$ reflects but $\kappa\setminus A$ does not.  In particular, no stationary subset disjoint to $A$ can reflect at $\delta$, hence there is no way to get your "$B"$.
Edit:
A "no" answer to your question at $\omega_2$ is equiconsistent with the existence of a Mahlo cardinal.
As Joel mentioned in (an earlier version of) his answer, one can build $A$ and $B$ in $\omega_2$ from a $\square_{\omega_1}$-sequence.  The failure of $\square_{\omega_1}$ implies that $\aleph_2$ is Mahlo in $L$  (Credited to Jensen on page 453 of Jech's "Set Theory"; I don't know a better reference.)
On the other hand, Theorem 7.1 in Chapter XI (page 576) of Proper and Improper forcing tells us that from a Mahlo cardinal, we can force ZFC+GCH + "every stationary subset of $S^{\omega_2}_{\omega}$ contains a closed copy of $\omega_1$, which we argued above gives a "No" answer.
Note that what Shelah is really showing is the consistency of the following statement:
"If $S$ is a stationary subset of $S^{\omega_2}_{\omega}$ that reflects at every member of $S^{\omega_2}_{\omega_1}$, then $S^{\omega_2}_{\omega}\setminus S$ is non-stationary,"
while the original question is equivalent to asking of $S^\kappa_\omega$ can be partitioned into two disjoint stationary sets, each of which reflects at every ordinal in $S^\kappa_{\omega_1}$.<|endoftext|>
TITLE: Varieties with infinitely many etale covers and rational points
QUESTION [9 upvotes]: Let $X$ be a (smooth projective geometrically connected) variety over a field $k$. 
Consider the set Et$(X,k)$ of finite etale covers $Y\to X$ over $k$, with $Y$ geometrically connected over $k$.
Assume Et$(X,k)$ is infinite. Consider the following question:
Does $X$ have a $k$-rational point?
The answer should be negative in general. In fact, I think one can construct a surface with infinitely many etale covers but no rational points by taking the product of two curves $C$ and $D$ over $k$, where $C$ has infinitely many etale covers and a rational point, but $D$ doesn't have any rational points. Then $C\times D$ has no rational points, but infinitely many covers.
What if $X$ is a curve? Is $X(k)$ non-empty?
Note that the converse is true if we consider curves of positive genus. That is, if $X$ is a curve of positive genus over $k$ with a $k$-rational point, then it has infinitely many etale covers. 
I'm mainly interested in the characteristic zero case, but comments on the situation in positive characteristic would also be interesting.

REPLY [7 votes]: Maybe I misunderstand something, but don't all curves have etale covers? Embed $X$ in $J^1$ (divisors of degree $1$ modulo linear equivalence). Then $J^1$ is a torsor for the Jacobian $J$ and since $J$ has etale covers, e.g. coming from multiplication by an arbitrary $n$, $J^1$ does too. Certainly, for those curves with a rational divisor of degree one, they have covers, as $J^1$ is isomorphic to $J$.
EDIT: Upon further reflection, I guess it's not true that $J^1$ always has covers, as it may not be in the divisible part of the Weil-Chatelet group of $J$. But there definitely exist curves with no points having divisors of degree one, and therefore covers of arbitrarily large degree. 
However, you question is a good one and you might be heading in the direction of Grothendieck's section conjecture: For finitely generated fields $k$, $X(k)$ is non-empty if and only if there is a section $G_k \to \pi_1(X)$ of the canonical projection $\pi_1(X) \to G_k$, where $G_k$ is the absolute Galois group of $k$ and $\pi_1$ is the etale fundamental group.<|endoftext|>
TITLE: Hypersurfaces containing no lines
QUESTION [5 upvotes]: Let $k = \bar{k}$ a fixed field. I would like to know if there exist hypersurfaces $X \subset \mathbb{A}_k^n$ that contain no lines. By line I really mean line, and not just rational curve.
I haven't put any restrictions on $X$, but it's still not clear to me that such things exist. Most likely they form a nonempty open subset of some Hilbert scheme if the degree is large enough.

REPLY [7 votes]: First, consider the projective space $P^n$ instead of affine --- if a hypersurface in affine space contains a line then its closure contains the closure of the line which is a line in the projective space. Consider the Grassmannian $G = Gr(2,n+1)$ parameterizing those lines and let $U$ be the tautological rank 2 subbundle on it. The equation of a hypersurface of degree $d$ gives a section of the vector bundle $S^dU^*$ on $G$ (and vice versa). Lines on the hypersurface are parameterized by the zero locus of the corresponding section. When 
$$
r(S^dU^*) = d + 1 > 2(n-1) = \dim G
$$
the zero locus of a general section is empty, since the vector bundle is generated by global sections.<|endoftext|>
TITLE: Reference request:  Forcing Axiom for the class of Axiom A posets
QUESTION [7 upvotes]: Does anyone know of a paper which asks, or conjectures about, the following question?
QUESTION:  Is the Forcing Axiom for Axiom A posets equiconsistent with the Proper Forcing Axiom (PFA)?
Axiom A posets are a subclass of the proper posets, defined by Baumgartner around the same time as the discovery of proper forcing by Shelah.  Many consequences of PFA (e.g. those involving finite iterations of c.c.c. and $\sigma$-closed posets) are consequences of the Forcing Axiom for Axiom A.

REPLY [2 votes]: this may help (building on earlier work by Ishiu and Weinert):
Bounded forcing axioms and Baumgartner's conjecture, by Aspero, Friedman, Mota & Sabok.

We study the spectrum of forcing
  notions between the iterations of
  σ-closed followed by ccc forcings and
  the proper forcings. This includes the
  hierarchy of α-proper forcings for
  indecomposable countable ordinals α as
  well as the Axiom A forcings. We focus
  on the bounded forcing axioms for the
  hierarchy of α-proper forcings and
  connect them to a hierarchy of weak
  club guessing principles. We show that
  they are, in a sense, dual to each
  other. In particular, these weak club
  guessing principles separate the
  bounded forcing axioms for distinct
  countable indecomposable ordinals. In
  the study of forcings completely
  embeddable into an iteration of
  σ-closed followed by ccc forcing, we
  present an equivalent characterization
  of this class in terms of
  Baumgartner’s Axiom A. This resolves a
  well-known conjecture of Baumgartner
  from the 1980’s.<|endoftext|>
TITLE: Simple Tamagawa number calculations
QUESTION [20 upvotes]: As is well known, Euler proved the Basel identity $\displaystyle\sum\limits_{i=0}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6}$. By far the most illuminating explanation of this fact that I've seen is as follows:

By another theorem of Euler we can
  rewrite the identity as $\displaystyle
 \frac{{\pi}^2}{6}\cdot \prod_{p}
 \left(1 - \frac{1}{p^2}\right) = 1$,
  where $p$ ranges over all primes. The
  first term is the normalized volume of
$\displaystyle SL(2,\mathbb{R})/SL(2,
 \mathbb{Z})$ 
and the term corresponding to $p$ in
  the product is the normalized volume
  of 
$SL(2, \mathbb{Z}_p)$. 
With these replacements, the right
  hand side can be written as the
  normalized volume of 
$\displaystyle
 SL(2,\mathbb{\mathbb{A}_{\mathbb{Q}}})/SL(2,
 \mathbb{Q})$. 
where $\mathbb{A}_{\mathbb{Q}}$
  denotes the adeles of $\mathbb{Q}$ But
  this last volume is equal to 1: this
  is a special case of the Weil
  Conjecture on Tamagawa Numbers.

I've been fascinated by this result for years, but have never been able to understand a proof of it (from the adelic perspective) even in cases as simple the one above. In this way, my question contrasts with that of Ben Weiland who asked about the theorem in more general settings. 
I tried reading André Weil "Adeles and algebraic groups" with a view toward learning a proof but found the book unintelliglbe. I gathered that the idea of the proof is to show that the nonzero volume of some object is equal to the Tamagawa number multiplied by the original volume but beyond that understood nothing. My impression is that Marie-France Vigneras' book titled Arithmetique des algebres de quaternions has this material, but I don't read French.

What are some lucid sources that you
  would recommend for learning proofs of
  the some of the first few cases
  (including the case above) of the Weil
  Tamagawa Number conjecture from an
  adelic perspective?


1 I learned this material from Yuri Manin's "Reflections on Arithmetical Physics" and Maclachlan and Reid's "The Arithmetic of Hyperbolic 3-Manifolds"

REPLY [4 votes]: I have only read Paul Garrett's paper. I find the use of Poisson summation rather mysterious, but it seems to me that it applies mutatis mutandis to the adelic case.
Very short version: the generic orbit of $SL(V)$ on $V$ is open. If we can compute volumes for $V$ and the stabilizer $N$, then we can do so for $SL(V)$.
It seems to me that most of the argument works for pretty general locally compact ring $A$ containing a lattice $B$. I see three steps:
1. We must choose Haar measures. First, scale Haar measure on $A$ so that $\mathop{vol}(A/B)=1$; this is important for Poisson summation. Choose invariant differential forms defined over $\mathbb Z$ for the three groups; these are unique up to sign. The differential forms lift Haar measure from $A$ to $G(A)$. Because the product of the differential forms on $V$ and $N$ is that of $SL(V)$, so are the Haar measures multiplicative. This gives the best normalization for understanding the large group in terms of the small groups, but if you care about other measures (eg, the measure on hyperbolic space), you may need a correction factor. For number rings, often a power of the discriminant is needed.
2. We need to understand the orbits of $SL(V)(R)$ on $V(R)$. In particular, if $R$ is a field, the action is transitive away from the origin. For any ring, $SL(V)(R)$ is transitive on the Zariski open $(V-0)(R)$. This is not true for general quotients. For example, $PGL_2$ is the quotient variety of $SL_2$, but $SL_2(\mathbb Q)$ does not surject onto $PGL_2(\mathbb Q)$. Thus for any representation of $SL_2$ that factors through $PGL_2$, such as the adjoint representation, $SL_2(\mathbb Q)$ is not transitive on the $\mathbb Q$-points of the algebraic orbit.
3. We need to understand the other orbits. They reflect the ideals of $R$. There is a tradeoff between the cases $\mathbb Z\subset\mathbb R$ and $\mathbb Q\subset\mathbb A$, in that we can't have both rings be fields. However, the complement for the locally compact ring has measure zero and is unimportant, while the counting measure on the discrete ring means that every point counts. This is an advantage of the adelic case and explains its simpler answer.
Finally, we have the argument. We take a Schwartz class function $f:V(A)\to\mathbb R$, and compute 
$$\int_{G(A)/G(B)}\sum_{x\in V(B)}f(gx)\,dg$$
One way to compute it is to break $V(B)$ into pieces according to the orbits of $SL(V)$. The zero orbit contributes $\mathop{vol}(G(A)/G(B))f(0)$. In the PID case, the other orbits are all scaled versions of the open orbit. Summing the scaling factors yields the zeta function of $B$ evaluated at the dimension of $V$. If $N$ is the stabilizer of $v$, then using point (1), the contribution of the open orbit is
$$\int_{G(A)/N(B)}f(gv)\,dg=\mathop{vol}(N(A)/N(B))\int_{(V-0)(A)}f(x)\,dx
=\mathop{vol}(N(A)/N(B))\hat f(0)$$ 
Thus the original integral is $\mathop{vol}(G(A)/G(B))f(0)+\zeta \mathop{vol}(N(A)/N(B))\hat f(0)$. Using Poisson summation for $V(B)\subset V(A)$ allows us to interchange $f$ with its Fourier transform and conclude that 
$\mathop{vol}(G(A)/G(B))=\zeta \mathop{vol}(N(A)/N(B))$. In the adelic case, the factor of $\zeta$ is always $1$ and this shows by induction that $\tau(SL(V))=1$. But the argument also explains where the factor of $\zeta(s)$ comes from in the case of $\mathbb Z\subset \mathbb R$ and extends to other number rings.<|endoftext|>
TITLE: Can there exist Chow motives/motivic cohomology for compact Kähler manifolds?
QUESTION [14 upvotes]: Can there exist a 'reasonable' extension of the (higher) Chow groups of complex smooth projective algebraic varieties to functors on the category of compact Kähler manifolds? Are there any obstructions for the existence of such a ('nice') extension? In particular, could there exist some 'Chow motives for compact Kähler manifolds'?
If one tries to mimick the usual 'algebraic' definitions, then one should define an analogue of algebraic cycles for compact Kähler manifolds. Is it reasonable to consider subsets that are images of compact Kähler manifolds with respect to birational morphisms?
Upd. Which (GAGA?) statements could help here? I would be deeply grateful for any references!
In particular, does there exist a good exposition of GAGA that includes the following statement: let X and Y be projective complex varieties and let $ϕ: X_h \to Y_h$ be a
morphism of analytic spaces, then there is a unique morphism $f : X \to Y$ such
that $f_h = ϕ$.

REPLY [12 votes]: Hi Mikhail,
I honestly don't have a good answer, but I'll share my thoughts on this and related things, since this is potentially quite interesting.

I don't see a problem in formally defining the Chow group of a compact Kähler manifold
as the group of cycles modulo rational equivalence.
However, if you want to compose correspondences, then you would need a ring structure,
and this might be difficult. Cycles on a Kähler manifold are probably more rigid than on an algebraic variety, so proving a moving lemma could be problematic.
In the algebraic case, $CH^*(X)\otimes \mathbb{Q}$ is the same as the associated graded
of the Grothendieck group of algebraic vector bundles $K^0(X)\otimes\mathbb{Q}$. I doubt
that this isomorphism would work in the analytic case because of the failure to have
global resolutions, but you could simply use $K^0(X)$ directly. This would give a ring, and therefore a category of $K^0$-correspondences...
You could also consider cycles modulo homological equivalence, and I don't see a problem in getting everything to work. So that you should be able construct a category of pure
homological motives for Kähler manifolds. However, unlike that algebraic case, I don't
think that one should expect this to be semisimple and abelian. Part of my pessimism stems from the fact the Hodge conjecture is known to fail in this setting (Voisin).

I don't really want to say much more here, but I think you have my email in case you want to discuss this further.<|endoftext|>
TITLE: Can there be a global linear ordering of the universe without a global well-ordering of the universe?
QUESTION [15 upvotes]: This question arose in the answers to Asaf Karagila's
question Does ZFC prove that the universe is linearly orderable?. The answer there was that one can have a ZFC model with no global linear ordering of the universe (and consequently also no global well-ordering). The question here is whether we can separate these two principles from each other.
Question. Is it consistent with ZFC that there is a global
linear ordering of the universe but no global well-ordering of the
universe?
More specifically, there are two forms of the question, depending
on whether one requires the classes to be definable classes as in
ZFC, or whether one allows classes in the sense of
Gödel-Bernays set theory.

If ZFC is consistent, is there a ZFC model with a definable
linear ordering of the universe, but no definable (with parameters)
well-ordering of the universe?
If ZFC is consistent, is there a model of GB+AC with a class
linear ordering of the universe, but no class well-ordering of the
universe?

The answer to the other question showed that there can be models
of ZFC having no definable linear ordering of the universe,
because one can make a class forcing extension which adds generic
sets in a homogeneous manner, which prevents any definition from
ordering them. Can we somehow modify the construction to allow a linear order, but no well-order? I suspect that we can, but I also suspect it will be easier to do this with GB classes than to make them definable.

REPLY [2 votes]: I would like to record here my observation that one natural way of
proceeding does not actually succeed.
Specifically, there is a very natural class forcing $\mathbb{P}$
to add a generic linear ordering of the universe, without adding
any sets, as follows: conditions in $\mathbb{P}$ are simply
set-sized linear orderings of any piece of the universe. We order
these conditions by $p\leq q\iff p$ includes $q$ as a suborder.
This forcing is $\kappa$-closed for any $\kappa$ and hence adds no
new sets. Meanwhile, it is dense for any set to be added to the
field of the order, and so the generic filter $G$ provides a
generic linear ordering of the universe. Basically, this forcing builds up a linear ordering of the universe by fitting each set generically into the order.
This forcing notion is exactly analogous to the forcing
$\mathbb{Q}$ of global AC, where one uses conditions that are
well-orderings of any part of the universe, ordered by
end-extension. That forcing also adds no new sets, and the generic
filter is a global well-ordering. Thus, every model of ZFC can be
extended to a model of GBC = GB + global AC without adding any
sets. This is why GBC is conservative over ZFC for first order
assertions.
Meanwhile, my observation is that adding the global linear order
via $\mathbb{P}$ will also create a global well-order. To see
this, suppose that $G\subset\mathbb{P}$ is $V$-generic. For any
condition $p$, a set-sized linear order, there is some large
$\kappa$ such that the field of $p$ does not mention any unbounded
subset of $\kappa$. Let $\lhd$ be a well-ordering of the unbounded
subsets of $\kappa$, and let $q$ extend $p$ by placing a copy of
$\lhd$ above the field of $p$. Thus, $q$ has as a suborder a
well-ordering of all the unbounded subsets of $\kappa$. So by
density, the generic linear ordering given by $G$ must also have,
for arbitrarily large $\kappa$, a suborder that is a well-ordering
of all of the unbounded subsets of $\kappa$. This is enough to
define a global well-ordering of the universe, since we can say
that $X$ preceeds $Y$ if the transitive closure of $X$ has smaller
cardinality than $Y$, or they have the same size transitive
closure, but for the smallest $\kappa$ having in $G$ a
well-ordering of the unbounded subsets of $\kappa$, such that $\kappa$ is large enough to code these sets, that $X$ is
coded by an earlier such set than $Y$ is.
So the naive attempt to add a global linear order without adding a
global well-ordering doesn't work.<|endoftext|>
TITLE: How to compute all irreducible representations of a finite group ? (how GAP is doing this?)
QUESTION [11 upvotes]: Let us "take" a finite group G. Here "take" I mean any type of group-theoretic description you prefer: e.g. as an explicit subset of GL (or other group) or Cayley table, whatever.
Question: How  should we compute representation theoretic information about the group ? 
By "representation theoretic information" I mean
a) number of irreps over complex numbers
b) their dimensions
c) their characters 
d) matrices defining the representations
Any other kind of comments on further questions like modular or rational representations is also welcome. 
What complexity should we expect from these algorithms ? And, are the current algorithms best possible or developing group theory may lead to more effective algorithms ?

Let me clarify my question by example.
For example - number of irreps = number of conjugacy classes. So if I would be asked
 question a) (i.e. compute the number of irreps) I would do the following. Just take element - compute its conjugacy class, take element, out this conjugacy class - compute its conjugacy class and so on. Finally I get the number of conjugacy classes and hence I get number of irreps.
So pay attention a) I assume that all operations like "take element", "are two elements equal", "take inverse of an element" has same complexity say "1".
b) this algorithm is based on the not so trivial fact that number of irreps = number of conjugacy classes. 
Subquestion are there less complexity algorithms ?

REPLY [16 votes]: You seem to be asking for a description of the the algorithms in computational representation theory. I think that is far too broad a question for this site and you need to be more specific. There is a recently published book on this topic, which you might like to look at:
Representations of Groups A Computational Approach
    Klaus Lux, Herbert Pahlings
    Cambridge University Press, 2012.
    Series: Cambridge Studies in Advanced Mathematics(No. 124)
As for the algorithms used by GAP, you can always look at source code! For example, for complex representations, the general approach is to compute the conjugacy classes and character table first, using the Dixon-Schneider algorithm for the latter, and a method of Dixon for getting the matrices of the representation from the character, which works if the group has a subgroup for which the restriction of the character has a linear constituent with multiplicity 1. For modular representations the Meataxe algorithm mentioned by Benjamin Steinberg is more effective.
The method you describe for finding conjugacy classes involving choosing random elements until you have covered all classes works well for groups that are not too large do not have very small conjugacy classes that are hard to find. For larger groups, there are better methods - I wrote something about this recently in
https://math.stackexchange.com/questions/198353/algorithm-to-find-conjugacy-classes-of-subgroups-elements-in-matrix-groups/198425#198425groups
BTW, I have no idea what you mean by "are there less complexity algorithms?"

REPLY [8 votes]: If the group is given by its multiplication table then there are polynomial-time algorithms for all the tasks you mentioned (over $\mathbb{C}$; these don't handle modular representations).
Babai and Ronyai. Computing irreducible representations of finite groups.  Math. Comp. 55 (1990), no. 192, 705–722.<|endoftext|>
TITLE: Where does Chu fit in with "Nice Categories of Spaces"?
QUESTION [7 upvotes]: Hi,
Here is a page about categories of spaces with "Nice" properties.  When we consider Chu spaces, I believe it was chosen to have certain properties that were "nice" at least to Pratt.  How does Chu fit in with the notion of "nice" that is outlined in the NLab site above?

REPLY [6 votes]: As one of the principal authors of the nLab page on "nice category of spaces" mentioned by the OP, let me quote: 

Within the nLab, “nice category of spaces” is a general but inexact term referring to nice or convenient properties one would like a category of spaces to have for some purpose (“space” here connoting something along topological lines)

I say this to correct the impression of Vaughan Pratt (seeing that he mentions my name in this regard, which was a surprise) that 'nice' in the nLab sense implies cartesian closure; for that, we use the more specific terminology "convenient category of topological spaces", after Ronnie Brown and Norman Steenrod who made that phrase famous. 'Nice' in the nLab sense is applied in a much more liberal and inclusive sense, as ought to be clear from the quote. Indeed, a number of examples given on the nLab page are not cartesian closed at all. (Many of the examples given however do emphasize various exactness properties.) 
Thus, there is no reason that categorical self-duality (or $\ast$-autonomous categorical structures) couldn't also be considered 'nice'. As the article says, 'nice' really only means nice for some purpose at hand. 
However, the main thrust of the nLab article is toward "spaces" in a topological sense. Chu spaces are, to my mind, "spaces" much more general than that. It's okay to call them "spaces" of course -- but consider that classical Chu spaces also include structures like $\mathbb{Z}_2$-vector spaces which would be a stretch to consider specifically topological. So Chu spaces go a bit beyond the types of structures originally envisioned for that article. Thus, the discrepancy isn't really centered on the word 'nice', but more on the word 'space'. 
If anyone wishes to discuss (or dispute!) the contents of an nLab article, the best and most proper place to do so would be on the nForum.<|endoftext|>
TITLE: Is $e^p\in\mathbb{Q}_p$ known to be transcendental?
QUESTION [10 upvotes]: $\sum\limits_{n=0}^{\infty}\dfrac{1}{n!}$ doesn't converge in $\mathbb{Q}_p$, however, $e^p:=\sum\limits_{n=0}^{\infty}\dfrac{p^n}{n!}$ does converge for $p\neq 2$. So my question is,

Are $e^p\in\mathbb{Q}_p$ for $p\neq2$ (and $e^4\in\mathbb{Q}_2$) known to be non-algebraic numbers?

REPLY [14 votes]: According to the last paragraph in Section 3 of the paper "Transcendental numbers in the p-adic domain" by William W. Adams (Amer. J. of Math., Vol. 88, 1966):
http://www.jstor.org/discover/10.2307/2373193?uid=3738736&uid=2&uid=4&sid=21101389828747
the answer is yes. More specifically, they prove that if $a \in \mathbf{Q}_p$ is a non-zero element, algebraic over $\mathbf{Q}$, such that $\left| a \right| < p^{-1/(p-1)}$, then $\exp(a) \in \mathbf{Q}_p$ is transcendental over $\mathbf{Q}$ (ibid.).<|endoftext|>
TITLE: Good Computer Package for Calculating Inverse of a Formal Power Series?
QUESTION [5 upvotes]: This might be a question people already asked or is obvious to experts, or is not appropriate for this forum, if so, I apologize. I am trying to calculate things like $z/(e^z-1)$, or find the inverse of $x=z+ 2z+5z^2+\cdots$, expand as power series. What is the best (or favorite) software package you use to do stuff like this?

REPLY [5 votes]: Okay, I've gathered the answers from the comments in alphabetical order.

flint has the fmpz_poly_revert_series command (A web search revealed a reference pdf, but no web page).  Some benchmarks seem to indicate that flint is unusually fast.
Maple has the reversion command.
Mathematica has the InverseSeries command.
Pari/GP has the serreverse function.
SAGE has the reversion command.<|endoftext|>
TITLE: Is the hyperspace of the Hilbert cube homeomorphic to the Hilbert cube
QUESTION [5 upvotes]: Question: Is the hyperspace of the Hilbert cube $H=[0,1]^\mathbb {N}$ homeomorphic to $H$?
Remarks and definitions: 
1) The Hilbert cube $H$ is a compact metric space, where the metric is given by the $\ell_2$-norm of sequences. A classical theorem on metric spaces says that every compact metric space is isometric to a closed subspace of $H$. 
2) The hyperspace of a metric space $X$ is the metric space of all non-empty compact subsets of $X$ given by the Hausdorff metric. Another classical theorem on metric spaces says that the hyperspace of a compact metric space is again a compact metric space.
Combining 1) and 2) shows that the hyperspace of the Hilbert cube is isometric to a closed subspace of the Hilbert cube. So my question asks whether we also can get a homeomorphism 
(can we even get both spaces isometric?).

REPLY [3 votes]: Rule out isometry ... In the hyperspace of $\mathbb R$, let $A=\{0,1,2\}$, $B=\{0\}$, $C=\{2\}$ and $M = \{1\}$.  Then points $A, B, C$ all have distance $2$ from each other; they are the vertices of an equilateral triangle.  But $M$ has distance $1$ from each of $A, B, C$, so $M$ is the midpoint of each of the three sides of that triangle.  Not possible in Euclidean geometry.<|endoftext|>
TITLE: Algorithm  to check is representation irreducible ? Algorithm to decompose the reducible one ?
QUESTION [14 upvotes]: Question 1 Given a representation of a finite group, what algorithm can be used to check is it irreducible or not ? 
(Main case - complex numbers, comments on other cases are also welcome. "Given" means finite set of matrices is given).
Question 2 Given a representation of a finite group, what algorithms can be used to decompose it to the direct sum of irreducibles) ? 

For the question 1 I would do the following: rep is irrep if its commutant consists of  scalar matrices. So I can try to find matrices commuting with  all elements of the group and look whether I get only scalar matrices.
Are there more effective ways to do it ?

Related question: How to compute all irreducible representations of a finite group ? (how GAP is doing this?)

REPLY [3 votes]: Here's a simple algorithm that was proposed in Dixon's 1970 paper: http://www.ams.org/journals/mcom/1970-24-111/S0025-5718-1970-0280611-6/S0025-5718-1970-0280611-6.pdf
It's split into 2 parts:
1. Construct a non-scalar commuting matrix $H$ (if possible)
Dixon presents two variants for part 1. I'll just mention the simpler one here. Let $\rho:G \to \text{GL}(n,\mathbb{C})$ be a (unitary) representation. For $r,s = 1,2,\dots,n$, define
$$
H_{rs} = \begin{cases}
E_{rr} &\text{if } r = s \\
E_{rs} + E_{sr} &\text{if } r > s \\
i(E_{rs} - E_{sr}) &\text{if } r < s,
\end{cases}
$$
where $E_{rs}$ is the $n \times n$ matrix with 1 in the $(r,s)^{th}$ entry and 0 everywhere else. Then $\{H_{rs}\}_{r,s=1}^n$ forms a Hermitian basis for the $n \times n$ matrices. 
Now for each $r,s$, compute the sum
$$
H = \frac{1}{|G|} \sum_{g \in G} \,\, \rho(g)^* \, H_{rs} \, \rho(g)
$$
Observe that $H$ commutes with all $\rho(g)$.
If $\rho$ is irreducible, then $H$ is a scalar matrix for all $r,s$. Otherwise, there will be some $r,s$ such that $H$ is non-scalar (because $\{H_{rs}\}_{r,s=1}^n$ forms a basis). In this case, proceed to part 2:
2. Use the eigenspaces of $H$ to decompose $\rho$
Let $H = UJU^*$ be the Jordan decomposition of $H$, where $U$ is unitary. Then
$$
U^* \rho(g) U
$$
will have the same block-diagonal form for all $g \in G$, yielding a decomposition of $\rho$.
Of course this assumes that the $n$ and $|G|$ are small enough that one can easily run through all the $H_{rs}$ and compute the Jordan decomposition. There are more sophisticated and efficient methods, but I like the simplicity of this one.
I've written an implementation of this algorithm in Sage.<|endoftext|>
TITLE: Smallest real closed field realizing all cuts of the rational numbers
QUESTION [5 upvotes]: Let $K$ be a real closed field of transcendence degree 1 over $\mathbb{R}$.
It is not difficult to see that $K$ has the following "minimality property": Whenever $L$ is a real closed field that realizes all cuts of the rational numbers, then
$K$ embeds into $L$.
Question: Is $K$ up to isomorphism uniquely determined by this minimality property?
Terminology:
By a cut of $\mathbb{Q}$ I mean a pair $(L,R)$ of subsets of $\mathbb{Q}$
with the property that $l < r$ for all $l\in L$, $r\in R$ and such that $R\cup L=\mathbb{Q}$
The question asks whether $K$ can be identified parallel to the way one can define $\mathbb{R}$:
The real field $\mathbb{R}$ is up to isomorphism the unique real closed field that embeds into any real closed field which realizes every non-principal cut of $\mathbb{Q}$; a cut $(L,R)$ is non-principal if $L$ does not have a supremum in $\mathbb{Q}\cup \{\pm\infty\}$

REPLY [4 votes]: Yes.  If $F$ is another such field, then $F$ would embed over ${\mathbb R}$  into $K$ so
$F$ would also be a transcendence degree 1 extension of $\mathbb R$.  We can find $s\in K$ and $t\in F$ with ${\mathbb R} < r,s$.  The ordered fields ${\mathbb R}(s)$ and ${\mathbb R} (t)$ are 
isomorphic.  Thus their real closures $K$ and $F$ are isomorphic.<|endoftext|>
TITLE: Egg-ovoid rolling down an inclined plane
QUESTION [14 upvotes]: I am seeking a mathematical analysis of an egg-ovoid rolling down an inclined plane, 
for pedagogical reasons.
It is well-known folk lore that the shape of an egg prevents it from rolling away from
the mother's nest, e.g., the article
"Why are eggs egg-shaped?" (link), or,
a bit more formally,
"The Mathematics of Egg Shape" (PDF download) by Yutaka Nishiyama, from which I
copied this figure (and the one below):
           
If anyone has seen a mathematical analysis of this phenomenon, more precise than observing
that an egg-ovoid fits inside a cone, and a cone rolls a circular arc, I would appreciate
a pointer.  Thanks!

REPLY [4 votes]: This paper https://jeb.biologists.org/content/221/19/jeb178988 contains an experimental investigation of egg rolling. Theoretically, it seems "the relationship of egg shape to egg movement (e.g. rolling) is an understudied topic".<|endoftext|>
TITLE: Functions that can be computed faster simultaneously than expected
QUESTION [6 upvotes]: The following is an elementary question about circuit complexity. It is different from the kind of thing I have seen discussed, so I would be interested in any work that has been done on this kind of question. My apologies if this is easy or well known.
By a Boolean function, I mean a function from $2^n\rightarrow2^m$ where $2$  is the set with two elements $0,1$. If $f:2^n\rightarrow2^m$ and $g:2^n\rightarrow2^k$ are Boolean functions, I denote by $\langle f,g\rangle$ the function $2^n\rightarrow2^{m+k}$ whose output is the output of $f$ followed by the output of $g$. Finally, let $|f|$ be the size (i.e. number of internal nodes) of the minimal circuit computing $f$. My question is the following: Suppose $| \langle f, \langle g,h \rangle \rangle | < |f|+|g|+|h|$ . Does it follow that $|\langle x,y \rangle | < |x|+|y|$ for some pair $x,y$ chosen from $f,g,h $ ? Naively, the idea is that we can build a circuit computing $\langle f,g \rangle$ from one computing $f$ and one computing $g$, and this has size $|f|+|g|$. But if $f$ and $g$ have something "in common", perhaps we can do better. Then the question is, suppose $f,g,h$ "have something in common". Does this mean two of these functions already "have something in common"?
I would also be interested in any variations on this question, for example for considering only formulas or bounded circuits.

REPLY [6 votes]: A specific example that has a negative answer comes from Karatsuba multiplication in which one multiplies two polynomials
$$(a_0 + a_1x)(b_0 + b_1x) =: c_0 + c_1x + c_2x^2$$
$$= (a_0b_0) + (a_0b_1 + a_1b_0)x + (a_1b_1)x^2$$
$$= (a_0b_0) + (a_0b_0 + (a_0+a_1)(b_0 + b_1) + a_1b_1)x + (a_1b_1)x^2$$
If we take $\mathbb{F}_2$-coefficients (or $2$-coefficients, in your notation), this can be viewed as three Boolean functions $2^4 \to 2^1$, or their concatenation a single function $2^4 \to 2^3$.
Karatsuba's trick allows one to turn the four multiplications and one addition into three multiplications and four additions. If you're counting all operations, and not just multiplications, it looks like you have increased from five to seven operations. But if you use this trick recursively, or take the $a_i,b_j$ to be polynomials themselves, then the reduction in multiplication wins out, and you can save overall operations.
I don't have a proof that two of the coefficients cannot be optimized without the third, but the reason I believe it is the case is because the Karatsuba trick exploits the fact that our coefficients of interest live in a rank-3 subspace of a 4-dimensional space.<|endoftext|>
TITLE: (Finite) Models of two subtheories of Peano Arithmetic
QUESTION [7 upvotes]: Consider first-order theory (with identity) of  Peano Artithmetic built in the language $\{S,+,\times,0\}$ and with the following set of axioms:
\begin{align}
\neg Sx&=0\tag{1}\\\
Sx=Sy&\rightarrow x=y\tag{2}\\\
x+0&=x\tag{3}\\\
x+S(y)&=S(x+y)\tag{4}\\\
x\times 0&=0\tag{5}\\\
x\times S(y)&=(x\times y)+x\tag{6}
\end{align}
plus the full induction schema.
Let $PA^{(-1)}$ be the subtheory of $PA$ (first-order Peano Arithmetic) which has all other axioms except for (1), similarly $PA^{(-2)}$ let be the subtheory without (2). It is rather easy, but nevertheless interesting, result that both this theories have finite models, $PA^{(-1)}$ even has the degenerate one-element model.
My question is: has any research been made towards characterization of class of models of the theories above? If yes, could please someone provide me with suitable information? I am particularly interested in finite models.
EDIT: Following J.D. Hamkins advice I explicitly stated the language and the axiomatization I am interested in.

REPLY [6 votes]: Infinite models are partly classified by two theorems of  $PA^{(-1,2)}$ the subtheory of $PA$ without axioms 1 or 2. Then I will describe the finite models completely. 
In $PA^{(-1,2)}$ if axiom 1 fails then axiom 2 holds.  For proof, express failure of axiom 1 by a constant $c$ with $S(c)=0$.  Then  $PA^{(-1,2)}$ proves $\forall x (c+S(x)=x)$, which implies axiom 2.  In this case the numbers form a group under addition.
So if axiom 2 fails then axiom 1 holds.  In that case we are in $PA^{(-2)}$ and axiom 2 fails at just one number $p$, there is an interval $[0,p]$ linearly ordered by addition, and the remaining numbers satisfy $PA^{(-1)}$ with $p$ in the role of 0.  
To prove the claims in the preceding paragraph, consider the following formula $\Phi(x)$ which is meant to say successor is one-to-one up to $x$ (but so far we have not defined an order relation):
$\forall y,z,u,v (\ (y+z=x\ \&\ S(u)=S(v)=y) \rightarrow u=v)$ 
The usual proof of $x+y=0 \rightarrow y=0$ works in $PA^{(-2)}$ and the usual additive order relation is well defined on the set defined by $\Phi$. So $\Phi(0)$ and if axiom 2 fails there is some $p$ in $\Phi$ with successor not in $\Phi$ and the set defined by $\Phi$ is linearly ordered as an interval $[0,p]$. 
Induction shows for every $x$ either $\Phi(x)$ or $\exists y (x=p+y)$  while conversely $\Phi(p+y)\rightarrow y=0$.  In particular $p$ is successor to at least one number of form $p+y$.
The numbers of form $p+y$ provide an obvious interpretation of  $PA^{(-1,2)}$ with $p$ in the role of 0, and this interpretation falsifies axiom 1.  So it satisfies axiom 2.  It interprets  $PA^{(-1)}$. 
Finite models of full induction are easy. Assuming no axioms for now, the list of iterated values $0,S(0),SS(0),\dots$ is finite in any finite model so the set of all the iterated values is definable (without parameters) by a finite disjunction $x=0\vee x=S(0)\vee x=SS(0)\vee \dots$ in that model.  Assuming induction that set is the whole model.
That means (following Blass's comment) the $S$ series is eventually cyclic, so there are $m$ and $n > m$ with $0,S(0),\dots S^{n-1}(0)$ all distinct but $S^{n}(0)=S^m(0)$.
Now assume axioms 3--6 on $+$ and $\times$.  Then all classically correct equalities between numerals (terms $S^p(0)$ for any number $p$) are provable, just as in $PA$ (but the classically false ones are not refutable).  And every element of any model is named by at least one numeral of that form.
So, for any eventually cyclic $S$ series, at most one definition of $+$ and $\times$ produces a model of axioms 3--6 plus full induction.  And one does, namely using standard arithmetic below $m$ and arithmetic mod $n-m$ above $m$.  The extreme cases are helpful in understanding: if  $m=0$ then we have arithmetic mod $n$ and a model of axioms 2--6 plus induction.  If $m=n-1$ then we have arithmetic with $n-1$ as an absorbing upper bound, a model of axioms 1 and 3--6 plus induction.
To be a bit more explicit, given an $S$ series with $n$ and $m$ as above, define an equivalence relation $=_0$ by saying: $x=_0 y$ iff either $x=y$ and both are $\leq m$, or both are $\geq m$ and have $x=y$ mod $n-m$,  You just have to show this is a congruence for $S,+,\times$ which is pretty direct since $S,+,\times$ are all strictly increasing in each argument, except multiplying by 0.<|endoftext|>
TITLE: Self-intersection and the normal bundle
QUESTION [10 upvotes]: Let $X/k$ be a surface nonsingular and proper over an algebraically closed field $k$. Let $C \subset X$ be a nonsingular curve. Then it is clear that the self-intersection $(C \cdot C)_X$ is $\textrm{deg}_C ( \mathcal{N}_{X/C} )$ , basically a matter of definition in intersection theory. More generally, if $X/k$ is a proper variety of dimension $k$, and $Y \subset X$ is a cartier divisor, the the class $[Y\cdot Y] \in A_{k-2}(Y)$ is the class of the line bundle $\mathcal{O}_X(Y) \vert_{Y} = \mathcal{N}_{Y/X}$. Both of these results are fairly easy to prove. I'm asking for something a little different:
$\textbf{Question:}$ I imagine these results are "intuitively clear" at some level to geometers. Let's stick to complex algebraic varieties. In the setting of surfaces, can one explain why the normal bundle controls the number of points that divisors linearly equivalent to $C$ meet $C$? I want to say that this "follows" because we can consider the normal bundle as a "tubular neighborhood", but I don't know how to do this precisely, or how to finish the argument. How about in the higher dimensional case?

REPLY [16 votes]: I'd like to expand a bit on the excellent comments of Charles Staats and Donu Arapura.  They both suggest understanding the self-intersection number of a curve as the number of fixed points of an infinitesimal deformation of the curve, which is manifestly the degree of the normal bundle when such a deformation exists.  Here's a slightly more pedestrian route, which I think has the benefit of being rigorous and almost as intuitive.
Suppose we have two curves in a surface:  $$\iota_C: C\hookrightarrow X, \iota_D: D\hookrightarrow X.$$  If $C\cap D$ has dimension zero, the intersection number should manifestly be $$C\cdot D:=\dim\Gamma(C\cap D, \mathcal{O}_{C\cap D})=\dim\Gamma(C, \iota_C^*\mathcal{O}_D)=\dim\Gamma(C, \iota_C^*\mathcal{O}_D(D)).$$  (The twist in the last equality does nothing by our assumption on the dimension of $C\cap D$ --- I've just inserted it to simplify things slightly later on.) We'd like to write this as an Euler characteristic, to make it constant if we vary $C$ or $D$ in a flat family.  But this is easy; since $\mathcal{O}_{C\cap D}$ has zero-dimensional support, it has no higher cohomology, so its Euler characteristic equals $C\cdot D$ as defined above.  Line bundles are nice (and more importantly, are acyclic with respect to restriction), so we use the short exact sequence $$0\to \mathcal{O}_X\to \mathcal{O}_X(D)\to \mathcal{O}_D(D)\to 0$$ to rewrite this Euler characteristic as $$\chi(\mathcal{O}_X(D)|_C)-\chi(\mathcal{O}_C)=\operatorname{deg}(\mathcal{O}_X(D)|_C).$$
I think this is a reasonably intuitive motivation for the definition of the intersection number.  So to fully answer your question, one should give an intuitive reason for why $\mathcal{O}_X(D)|_D$ is $\mathcal{N}_{D/X}.$  Of course, this is just the definition of the normal bundle, but let's motivate the definition.  First, why is the conormal bundle $I/I^2$, for $I$ the ideal sheaf of a closed subvariety $V\subset X$?  Well, an element of $I/I^2$ is precisely a function on $X$ vanishing at $V$, but ignoring higher-order terms.  A section to the normal bundle precisely takes functions $f$ defined in a neighborhood of $Y$ and differentiates them--but the partial derivative should depend only on the first-order part of $f$.  So the normal bundle should be precisely $(I/I^2)^\vee$.  This is another name for $\mathcal{O}_X(D)|_D.$
I hope that was some reasonable intuition/motivation.<|endoftext|>
TITLE: Upper bound on joint Renyi entropy
QUESTION [5 upvotes]: Renyi entropy of a random pair $(X,Y)$ with probability distribution $p_{X,Y}$ is defined by
\begin{equation}
H_\alpha(X,Y) = \frac{1}{1-\alpha}\log\sum_{x,y} p_{X,Y}(x,y)^\alpha.
\end{equation}
Assume that $X$ and $Y$ have countably infinite alphabets, and observe the case $0\leq\alpha\leq 1$. 
It is known that this functional is subadditive, i.e., that $H_\alpha(X,Y)\leq H_\alpha(X) + H_\alpha(Y)$, only for $\alpha=0$ (Hartley entropy) and $\alpha=1$ (Shannon entropy). 
My question is the following: Does there exist some upper bound on the joint entropy for $0<\alpha<1$, in terms of the marginal entropies? In fact, I don't even need an explicit upper bound, answering the following simpler question would suffice: If $H_\alpha(X)<\infty$ and $H_\alpha(Y)<\infty$, does there exist a constant $h$ such that $H_\alpha(X,Y)\leq h$?

REPLY [3 votes]: The answer is no. Renyi entropy can even be unbounded and discontinuous over the set of probability distributions with given marginals (but not always, this depends on the marginals). You can find detailed proofs in http://arxiv.org/abs/1303.3235v1.<|endoftext|>
TITLE: Dual versions of "folding" symmetric ADE Dynkin diagrams? 
QUESTION [17 upvotes]: Start with the Dynkin diagram of an irreducible root system, typically associated with a simple 
Lie algebra over $\mathbb{C}$ or a simple algebraic group.    Most of the simply-laced ADE
diagrams admit natural symmetries of order 2:  type $A_r$ when $r \geq 2$, type
$D_r$ when $r \geq 4$, and type $E_6$.   In the case of type $D_4$, there is also a
symmetry of order 3.    There is an obvious way to "fold" each diagram by identifying those vertices which are
interchanged by the symmetry.    This yields Dynkin diagrams with two root lengths.   For
example, $E_6 \leadsto F_4$.    
What I don't see is any general theory dictating in advance how to interpret this process.
In particular, does a pair of simple roots (vertices) interchanged by the symmetry in type $E_6$ 
lead to a long or a short simple root in type $F_4$?      In effect, there are (Langlands) dual
possibilities, switching long and short roots while also switching types $B_r$ and $C_r$.   It may be difficult to formulate a single theory of folding and duality which makes all instances
look natural, but it's at least reasonable to ask:

What is the origin of the notion of "folding" for Dynkin diagrams?

It's well-known that the ADE graphs are ubiquitous in various areas of mathematics; for an
older survey see the paper by Hazewinkel et al. 
here.   The usual 
convention is to regard all vertices of the graph as corresponding to long simple roots.
In their examples, the most common type of folding occurs this way:
$A_{2r-1} \leadsto C_r, \: D_{r+1} \leadsto B_r, \:  D_4 \leadsto G_2, \: E_6 \leadsto F_4$,
where multiple simple roots are always taken to short simple roots.    Examples occur in
various parts of Lie theory: classification of real forms of complex semisimple Lie algebras,
classification of $k$-forms of reductive $k$-isotropic algebraic groups (Satake, Tits), etc.
But in the study of simple singularities of types ADE (work of Grothendieck, Brieskorn, Slodowy),
adapted to other Lie types,  the natural folding procedure is dual to the above one:
$A_{2r-1} \leadsto B_r, \: D_{r+1} \leadsto C_r, \:  D_4 \leadsto G_2, \: E_6 \leadsto F_4$,
where now multiple simple roots lead to long simple roots.   
My own motivation comes from related work on representations of Lie algebras of simple algebraic
groups in prime characteristic.   Here the most elusive non-restricted representations
belong to nilpotent elements (or orbits).   The case of regular nilpotents is easy, but 
already the subregular case is too difficult for current algebraic methods to complete in
case of two root lengths.   (For large enough primes, the deep work of Bezrukavnikov, 
Mirkovic, and Rumynin has pushed farther.)
It's instructive to think of (dual) folding in connection with Dynkin curves (Springer fibers
in the subregular case) and the detailed results of
J.C. Jantzen in Subregular nilpotent representations of Lie algebras in prime characteristic,
Represent. Theory 3 (1999), 153-222, freely available 
here.     His incomplete
dimension calculations involving two root lengths would become transparent if the folding process could be rigorously justified in advance.

REPLY [5 votes]: Let me explain how both kinds of foldings of Dynkin diagrams (i.e., $A_{2n-1} \to B_n$ and $A_{2n-1} \to C_n$) arise in the context of Lie algebras and characters of their representations.
First of all, what I will call the "standard combinatorial procedure" for folding a root system $\Phi$ according to a Dynkin diagram automorphism $\sigma$, as described by Stembridge here, will produce the Type $B_n$ diagram from a Type $A_{2n-1}$ diagram (and will produce a Type $C_n$ diagram from a Type $D_{n+1}$ diagram). The standard procedure is to produce the root system whose simple roots $\beta_{I}$ correspond to orbits $I \subseteq \Delta$ of the simple roots of the original diagram under the automorphism $\sigma$: we just take the sum of the roots in each orbit $\beta_{I} := \sum_{\alpha\in I} \alpha$. Let me call this folded diagram root system $\Phi^{\sigma}$.
However, if $\mathfrak{g}$ is the Lie algebra of $\Phi$, then the automorphism $\sigma$ acts on $\mathfrak{g}$ in an obvious way, and the fixed-point Lie subalgebra $\mathfrak{g}^{\sigma}$ has as its root system the dual of $\Phi^{\sigma}$. In this way we get the inclusions $\mathfrak{sp}_{2n}\subseteq \mathfrak{sl}_{2n}$, and $\mathfrak{so}_{2n} \subseteq \mathfrak{so}_{2n+1}$ (i.e. $A_{2n-1} \to C_n$ and $D_{n+1} \to B_n$).
But there is way to make the "standard" folded root system $\Phi^{\sigma}$ appear in the context of Lie algebras as well, namely, by considering so-called "twining characters." Let me call the Lie algebra associated to $\Phi^{\sigma}$ the "orbit Lie algebra" of $(\Phi,\sigma)$.
The set-up in which the orbit Lie algebra arises is this: we can "twist" any representation $V$ of $\mathfrak{g}$ by the automorphism $\sigma$ to get a new, twisted representation $V^{\sigma}$; if $V=V^{\lambda}$ is the highest-weight representation with highest weight $\lambda$, then $V^{\sigma} = V^{\sigma(\lambda)}$, where $\sigma$ acts on the weight lattice of $\Phi$ in the obvious way. Suppose that we choose a $\sigma$-fixed weight $\lambda$. Then we can view $\sigma$ as a map $\sigma\colon V^{\lambda}\to V^{\lambda}$ (I think this is technically defined up to scalar). The twining character of $V^{\lambda}$ is defined to be $\mathrm{ch}^{\sigma}(V^{\lambda})(h) = \mathrm{tr}(\sigma\cdot e^{h})$ for $h \in \mathfrak{h}$, just like the usual character would be $\mathrm{ch}(V^{\lambda})(h) = \mathrm{tr}(e^{h})$. The twining character formula, which is originally due to Jantzen (see the discussion at the beginning of https://arxiv.org/abs/1404.4098) but has been rediscovered by many people (e.g., https://arxiv.org/abs/hep-th/9612060, https://arxiv.org/abs/q-alg/9605046), asserts that the twining character $\mathrm{ch}^{\sigma}(V^{\lambda})$ is equal to the usual character $\mathrm{ch}(U^{\lambda})$ where $U^{\lambda}$ is the highest-weight representation of the orbit Lie algebra with highest weight $\lambda$ (note that since $\lambda$ is fixed by $\sigma$, it is naturally a weight of the folded root system $\Phi^{\sigma}$). So the upshot is that a Type $A_{2n-1}$ twining character is a Type $B_{n}$ ordinary character.
The twining characters have some interesting applications to combinatorics when considering "symmetric" versions of combinatorial objects associated to Lie algebras, which is how I became aware of them. I quote from the 2nd column of the 4th page of this paper of Kuperberg (https://arxiv.org/abs/math/9411239):

As stated in the proof, $\sigma_B$ is a Dynkin diagram automorphism. The character theory of semi-direct products arising from Dynkin diagram automorphisms is described by Neil Chriss [2], who explained to the author that although this theory is known to several representation theorists, it may not have been previously published. The group $\mathbb{Z}/2 \ltimes_{\sigma_B} SL(2a)$ has two components. The character of a representation on the identity component is just the usual character of $SL(2a)$. The character on the $\sigma_B$ component, when non-zero, equals the character of an associated representation of the dual Lie group, in this case $SO(2a + 1)$, to the subgroup fixed by the outer automorphism, in this case $Sp(2a)$. The representation associated to $V_{SL(2a)}(c\lambda_a)$ is the projective representation $V_{SO(2a+1)}(c\lambda_a)$, where $\lambda_a$ is now the weight corresponding to the short root of $B_a$, the root system of $SO(2a + 1)$. In particular, the trace of $\sigma_B$ is the dimension of $V_{SO(2a+1)}(c\lambda_a)$, as given by the Weyl dimension formula, and the trace of $\sigma_BD_q$ is the q-dimension, as given by the Weyl q-dimension formula.

The disclaimer at the beginning of this quote suggests that (at least in 1994) this folding business was not well-known or written down precisely in a canonical text.<|endoftext|>
TITLE: Lambda-operations on stable homotopy groups of spheres
QUESTION [37 upvotes]: The Barratt-Quillen-Priddy theorem says in one interpretation that there is a weak equivalence of spectra $K(FinSet) \simeq \mathbb{S}^0$. In other words K-theory groups of finite sets are the stable homotopy groups of spheres $\pi_*^s$.
If $A$ is a commutative ring, $K_0(A)$ has a simple definition as the free abelian group of projective finitely generated $A$-modules modulo exact sequences. On this group we use the exterior powers $\Lambda^k$ to get so-called Lambda-operations $\lambda^k$. These have nice properties and one can use them to alternatively construct Adams operations $\Psi^i$. This construction can be extended to all $K_n(A)$, giving $K_*(A)$ the structure of a Lambda-ring. This can found in sections II.4 and IV.5 of Weibel's book.
There is a strong analogy between finite sets and vector spaces. This tells you that an analogue of the exterior power $\Lambda^k$ is given by construction that sends a finite set $X$ to its set of $k$-element subsets ${X \choose k}$. This gives the standard Lambda-ring structure on $\mathbb{Z} = \pi_0^s$, i.e. the one on Wikipedia. 
It seems that Weibel's construction of the Lambda-operations on higher K-theory groups works in this context as well. Is this correct? If so, we get $\lambda^i$ and $\Psi^i$ on the stable homotopy groups of spheres. What is known about these? Have they been used for anything?

REPLY [18 votes]: The operation which sends a finite set $S$ to its set of $k$-element subsets, $\binom{S}{k}$, gives rise to the $k$-th stable Hopf invariant. There is additional structure in this: the set $\binom{S}{k}$ has a canonical $k$-fold covering so the operation is better viewed as a map
$$
QS^0 \to Q(B\Sigma_k)_+
$$
rather than as a map 
$$
Q S^0 \to QS^0 ,
$$ 
where for a based space $X$, the space $QX$ is $\Omega^\infty\Sigma^\infty X$
is the representing space for the stable homotopy of $X$, i.e., 
$\pi_j(QX) = \pi_j^{\text{st}}(X)$. So the operation induces a homomorphism
$$
\pi_j^{\text{st}}(S^0)  \to \pi_j^{\text{st}}((B\Sigma_k)_+) .
$$
These operations satisfy certain axioms (Cartan Formula, compatibility with transfers, etc.).
A good place to read about these operations is:
Segal, Graeme: Operations in stable homotopy theory. New developments in topology (Proc. Sympos. Algebraic Topology, Oxford, 1972), pp. 105–110. London Math Soc. Lecture Note Ser., No. 11, Cambridge Univ. Press, London, 1974.<|endoftext|>
TITLE: The geometry of crinkled aluminum foil
QUESTION [14 upvotes]: I wonder if the geometry of crinkled aluminum foil has been studied?

          


The above is a photo of foil I flattened to reuse.
It might be described as a partition into nearly-uncreased polygons,
each polygon of not too many sides, and arranged in a
rather un-Voronoi like pattern.
It superficially resembles a rugged mountain terrain seen
from a great height.
I searched a bit for some mathematical analysis
of this pattern without luck.  Has anyone seen such an analysis?
There might be some interesting mathematics here...
Update.
Here is Fig.1 from the PNAS article that jc identified, "Three-dimensional structure of a sheet crumpled into a ball," by Anne Dominique Cambou and Narayanan Menon,
slices through an equatorial plane of three crumpled spheres:

REPLY [13 votes]: Crumpled structures are certainly of great interest among some soft matter physicists; you might with this review article of Tom Witten's.  He also has a nice webpage with some nice pictures and summaries of papers of his on related topics.
My understanding is that while we have some handle on the behavior of the cone-like and ridge-like singularities that are forced by the crumpling, not much is known about how they end up distributed after crumpling, though see this nice recent PNAS article from UMass on X-ray scans of crumpled metal foil balls.
I might add more later, but these references and their references, etc. should be enough to get you started.  There are indeed many beautiful problems in the area of elasticity of thin sheets.<|endoftext|>
TITLE: Do complete non-projective varieties arise "in nature"?
QUESTION [10 upvotes]: I'm aware of the existence of complete (abstract) algebraic varieties that are not projective but, probably due to my ignorance, I have the impression that they arise only as very particular examples constructed just with the purpose of finding such an example. My question (perhaps a bit vague) is:

Are there exemples in the literature in which complete non-projective varieties appear without "being expected" from the beginning or without just being the goal of the construction or proof?

REPLY [11 votes]: Assume $X$ is a projective threefold with $n$ ordinary double points. Then $X$ has $2^n$ small resolutions of singularities. Usually, almost of all of them are non-projective.<|endoftext|>
TITLE: Partitions into parts from an arithmetic progresion
QUESTION [12 upvotes]: Fix an arithmetic progression $R=(a, a+m, a+2m, \ldots)$, and assume that $gcd(a,m)=1$.  Define $q_R(n)$ as the following coefficients:
$$\prod_{i=0}^\infty (1+ t^{a+mi}) = \sum_{n=0}^\infty q_R(n) t^n
$$
In other words, $q_R(n)$ is number integer partitions of $n$ into distinct parts from $R$.  
Problem 1.  Prove that $q_R(n)$ are increasing for $\ n\ge n(a,m)$ large enough.  
I first assumed this is either standard, well known, or easily follows from the existing results.  Now I am less sure.  My literature search gives only papers like this (A. Tripathi, "Coin exchange problem for arithmetic progressions"). 
Note that for $a=m=1$, we get the usual partitions into distinct parts and the claim follows from Euler's theorem that they are equinumerous with partitions into odd parts. 
More generally, I need to prove that all finite differences are positive for large enough $n$.  Formally, define
$$(t-1)^r \prod_{i=0}^\infty (1+ t^{a+mi}) = \sum_{n=0}^\infty q_R(n,r) t^n
$$
Problem 2.  For every $r\ge 1$, prove that $q_R(n,r)>0$ for $\ n\geq n(a,m,r)$ large enough.

REPLY [4 votes]: [This is more of a comment than an answer, but I lack the reputation.] For partitions with repetitions allowed, the analogue of your Problem #2 is solved by a very general theorem of Bateman and Erdos:
http://www.renyi.hu/~p_erdos/1956-05.pdf
Let $A$ be an arbitrary set of natural numbers. For each nonnegative integer $k$, define $p_k(n)$ so that
$$ \sum_{n=0}^{\infty} p_k(n) X^n = (1-X)^k \prod_{a \in A} (1-X^a)^{-1}. $$
They show that $p_k(n)$ is positive for all sufficiently large $n$ if and only if the following holds: There are more than $k$ elements in $A$, and if we remove an arbitrary subset of $k$ elements of $A$, the remaining elements have greatest common divisor $1$.
Unfortunately they remark that the problem of partitions into distinct parts is much harder and refer to the paper of Roth and Szekeres already mentioned in Gjergji's answer.<|endoftext|>
TITLE: Ways of choosing k items out of n with exactly one symbol in common
QUESTION [5 upvotes]: This is inspired by a card-game – the game involves cards with 8 different symbols printed on each, with the property that each pair of cards has exactly 1 symbol in common. That set us thinking – given an $n$-alphabet, what's the maximum number of $k$-sets of symbols such that each pair of $k$-sets has exactly one symbol in common? Formally:
Given $k,\ n,\ A = \{a_1, …, a_n\}$, maximise $s = s(k,n)$ such that

$∃ S=\{S_1, S_2, …, S_s\}$, and
$∀ 1 ≤ t ≤ s,\ |S_t| = k ∧ S_t - A = ∅$, and
$∀ 1 ≤ u, v ≤ s, |S_u ∩ S_v| = 1 $

Or a dual problem: to generate $m$ $k$-sets with the above property between any $2$ $k$-sets, how big must the alphabet be?

REPLY [10 votes]: The kind of object you're looking at is exactly an $(r, \lambda)$-design for $\lambda = 1$ in combinatorial design theory.
An $(r, \lambda)$-design is an ordered pair $(V, \mathcal{B})$, where $\mathcal{B}$ is a collection of subsets of finite set $V$ such that every element of $V$ appears in exactly $r$ elements of $\mathcal{B}$, and every pair of distinct elements of $V$ appears in exactly $\lambda$ elements of $\mathcal{B}$.
Usually, the cardinalitys $\vert V \vert$ and $\vert \mathcal{B} \vert$ are written as $v$ and $b$ respectively. We call the elements of $V$ points and those of $\mathcal{B}$ blocks.
To see the equivalence, call each card a point, and each symbol a block. The set $A =\lbrace a_1, \dots, a_n \rbrace$ is the set $\mathcal{B}$ of blocks here, and you say a point  is contained in block $a_i$ if the corresponding card has the symbol $a_i$ on it. Then setting $r = k$  and $\lambda = 1$, the above definition defines exactly what you described in the language of cards with symbols on them; every card has $k$ symbols (= every point appears $r$ times) and every pair of cards share exactly one symbol of $A$ (= every pair of points appear exactly once in a block in $\mathcal{B}$). The question you asked can be understood as "What's the maximum number of points in a $(k,1)$-design when the number of blocks is $n$?" To answer this, the basic relation between parameters of a nontrivial $(r, \lambda)$-design is:
$v \leq r(r-1)+1$ with equality if $r-\lambda$ is the order of a finite projective plane.
The following might also be helpful if you ask the same kind of question by fixing some parameters:
$b \geq v r^2/(r+\lambda(v-1))$,
$c_i(r-\lambda+\lambda v - \lambda k_i) \leq (r-\lambda)(r-\lambda +\lambda v)$, (c_i is the size of the $i$th block or equivalently the number of cards that have the symbol $i$).
You can find more about $(r, \lambda)$-designs in chapter "(r, \lambda)-designs" by G.H.J. van Rees in the book "Handbook of Combinatorial Designs" edited by C.J. Colbourn and J. Dinitz.
Edit: The correct definition should allow the same subset of $V$ appearing more than once in $\mathcal{B}$, so $\mathcal{B}$ shouldn't be a set but a collection.<|endoftext|>
TITLE: Density of integers $n$ whose totient $\varphi(n)$ is larger than $\alpha n$
QUESTION [5 upvotes]: Fix $0 < \alpha < 1$ a real. Let $S_\alpha$ the set of integers $n \geq 1$ such that be $\phi(n)>\alpha n$. For $x>0$, let $S_\alpha(x)$ be the number of positive integers $n$ less han $x$ 
such that $n \in S_\alpha$.

Does $S_\alpha$ has a natural density (that is, does $\lim_{x \rightarrow \infty} S_\alpha(x)/x$ exist)? If so, is that density $0$, $1$ or in between? If not, same questions for   $\liminf_{x \rightarrow \infty} S_\alpha(x)/x$ and $\limsup_{x \rightarrow \infty} S_\alpha(x)/x$.

Of course, $n \in S_\alpha$ if and only if $\prod_{p \mid n} (1-1/p) > \alpha$, where the product is over prime factors of $n$. But I am not sure where to go from here.

REPLY [11 votes]: Anonymous gave a perfect answer already. All I want to add to it is that you do not need to be Erdos or even Schoenberg to figure such things out. I'll just show how to establish the existence of density for $\alpha>0$, leaving the rest to you.
The starting point is that we know that for a pair of integers not exceeding $N$, to have a common divisor is a not very probable event and to have a large common divisor is a really rare event. Indeed, the number of pairs $(m,n)\in [1,N]\times [1,N]$ such that some fixed number $d$ divides both $m,n$ is at most $N^2/d^2$, so the number of pairs having a common divisor greater than $D$ is at most $N^2\sum_{d\ge D}d^{-2}\le N^2/(D-1)$.
Now, fix some $D$ and denote by $\Phi_D(n)$ the number of numbers $m\le n$ such that both $m,n$ are divisible by some prime $p>D$. What we just proved shows that $\frac 1N\sum_{n\le N}\Phi(n)\le \frac 1{D-1}$, so for every $\varepsilon>0$ we can be sure that the upper density of $n$ for which $\Phi_D(n)>\varepsilon n$ is less than $\varepsilon$ if $D$ is chosen large enough.
Let now $\varphi_D(n)$ be the number of $m\le n$ such that $m$ and $n$ have no common prime divisor less than $D$. Note that $\varphi_D(n)-\Phi_D(n)\le\varphi(n)\le\varphi_D(n)$ and $\Phi_D(n)$ is usually less than $\varepsilon n$. On the other hand, $\varphi_D(n)/n$ is completely determined by the remainder of $n$ modulo the product of primes less than $D$, so if we replace $\varphi$ by $\varphi_D$, the existence of density problem becomes trivial. Note that we can do it outside a set of arbitrarily small upper density with arbitrarily small error. So, the only problem may arise when the small error is not tolerable, i.e., when $\alpha$ is such that for all $\varepsilon>0$ the set of numbers $n$ with $\frac{\varphi(n)}n\in(\alpha-\varepsilon,\alpha+\varepsilon)$ has upper density bounded from below by some $c>0$ independently of $\varepsilon$. 
The last step is to show that it is impossible. Let the upper density of the set $S_\varepsilon=\{n:|\frac{\varphi(n)}n-\alpha|<\varepsilon\}$ be greater than $c$ regardless of $\varepsilon$. Then the upper density of the set $S_D=\{n:|\frac{\varphi_D(n)}n-\alpha|<2\varepsilon\}$ is at least $c/2$ if $D$ is large enough. But $S_D$ has density, so, going back, we conclude that the lower density of $S_{3\varepsilon}$ is at least $c/4$. Now let $p$ be any prime greater than $8/c$. Let $S(p)=\{n\in S_{3\varepsilon}:p\not\mid n\}$. Then the lower density of $S(p)$ is at least $c/8$. Note now that $pS(p)$ has numbers with $\frac{\varphi(n)}n\approx \alpha(1-\frac 1p)$ up to $\pm 3\varepsilon$. If we take sufficiently large finite set $P$ of primes $p>8/\delta$ (the primes to choose depend on $c$ only), we'll get the disjoint sets $pS(p)$ (the disjointness will be guaranteed if $\varepsilon$ is so small that the $3\varepsilon$-intervals around the points $\alpha(1-\frac 1p)$ do not overlap) of lower density $\frac c{8p}$. However, we can choose $P$ so that $\frac c 8\sum_{p\in P}\frac 1p>1$, which is a clear contradiction.
Needless to say, the full Erdos-Wintner theorem is much deeper than this and is certainly worth learning if you like the elementary number theory.<|endoftext|>
TITLE: Why might André Weil have named Carl Ludwig Siegel the greatest mathematician of the 20th century?
QUESTION [51 upvotes]: According to Steven Krantz's Mathematical Apocrypha (pg. 186):

As was custom, Weil often attended tea
  at [Princeton] University . Graduate
  student Steven Weintrab one day went
  about the room asking various famous
  mathematicians who was the greatest
  mathematician of the twentieth
  century. When he asked Weil, the
  answer (without hesitation) was "Carl
  Ludwig Siegel (1896-1981)."

As the title of Krantz's book suggests, the anecdote may be apocryphal. However, there are other better grounded accounts of great mathematicians expressing the highest admiration for Siegel:
(A) In The Map of My Life Shimura wrote:

I always thought that few people
  really understood my work. I knew that
  Chevalley, Eichler, Siegel, and Weil
  understood my work, and that was
  enough for me [...] Of course [Siegel]
  established himself as one of the
  giants in the history of mathematics
  long ago [...] Among his
  contemporaries, [Weil] thought highly
  of Siegel [...]

(B) In an published interview (pg. 30) Selberg said

[Siegel] was in some ways, perhaps,
  the most impressive mathematician I
  have met. I would say, in a way,
  devestatingly so. The things that
  Siegel tended to do were usually
  things that seemed impossible. Also
  after they were done, they seemed
  still almost impossible.

Why might Weil, Shimura and Selberg have been so impressed by Siegel? I should emphasize that I'm not trying to precipitate a debate about the relative standing of historical mathematicians - rather - I'm hoping to learn about aspects of Siegel's work that I might otherwise overlook. I'm also not looking for, e.g. quotations from the Wikipedia article on him, but rather, less familiar material.

REPLY [9 votes]: Recently I was digging through the meeting minutes for IAS and chanced upon a report prepared for Siegel to justify hiring him. The relevant pages are 13-18. In this report, it is stated that "we decided to propose at this juncture only one name, because it eclipses all others - that of Carl L. Siegel". 
Later on, an excerpt from a letter from Courant is given which states "the only one of his generation whose strength could be compared with that of the mathematical heroes of the preceding era". More praise is lavished on by Chevalley, where he views Siegel as "on a level with a Hilbert or a Poincare" and by Hardy where he claims that nobody questioned if "he was the equal of any mathematician in his generation, and certainly I never doubted it myself". 
With this in mind, it seems that it was in fact the consensus of the mathematical community then that Siegel was the greatest of his time, not just of Weil. Details of his accomplishments are given in the report for those interested. One thing I personally find puzzling is the lack of his influence in current areas of modern mathematics, as suppose to say Weil or Kolmogorov.<|endoftext|>
TITLE: Sets with zeta functions that are not the primes
QUESTION [11 upvotes]: Does there exist a set $S \subset \mathbb N$ such that the Dirchlet density of $S$ is well-defined and positive, the Dirchlet density of $S \cap \operatorname{PRIMES}$ is well-defined and zero, and:
$ \prod_{n \in S} \frac{1}{1-n^{-s}}$ 
has a meromorphic continuation to the whole complex plane? Can we construct it?
Motivation: How special is the set of primes, among sets of natural numbers of approximately the same size, for having a meromorphic continuation? I would guess that analytic continuation should be a very rare occurrence, but I don't have an intuition for how rare, or how hard it is to find an example if one exists.
Extra credit for a functional equation.

REPLY [6 votes]: You're asking about how special the prime numbers are as a subset of the integers. One can equally well ask how special the sequence $a_k = 1$ is when viewed as the sequence of coefficients of a Dirichlet series. I don't have anything to offer on your original question, but have read a few things about the latter question that may be of some interest:

According to Section 8 of David
Farmer's article titled Basic
Analytic Number Theory, if $f(s)
    = \displaystyle \sum_{k = 1}^{\infty} \frac{a_k}{k^s}$ where
the $a_k$ are integers, then a
sufficient condition for $f(s)$ to
admit a meromorphic continuation to $\Re{(s)} = 0$ is that:
(1) $a_k$ is of subpolynomial growth
(2) $a_k$ is multiplicative
(3) If $p$ is prime then $a_{p^m}$ is
independent of $m$
and may or may not have a natural boundary there.
According to section 9.5 of Titchmarsh's The Theory of the Riemann Zeta-function, if $a_k = 0$ when $k$ is composite and $a_{k} = 1$ when $k$ is prime then $f(s)$ (provably) has a natural boundary at $\Re{(s)} = 0$ 
I've also heard of results of the type "a Dirichlet series with $a_k$ chosen at random uniformly from $[-1, 1]$ has natural boundary $\Re{(s)} = 0$ with probability $1$," but don't know  a precise statement or a reference.<|endoftext|>
TITLE: polycirculant conjecture
QUESTION [5 upvotes]: By the polycirculant conjecture, every vertex-transitive graph is a polycirculant graph (D. Marusic 1981 and D. Jordan 1988).
There are two papers that claim to prove this conjecture:
1. A. Golubchik, "On the polycirculant conjecture", available on http://arxiv.org/abs/math.GM/0204209, April 2002.
2. E. Mwambene, "A proof of the polycirculant conjecture", available
on http://arxiv.org/abs/math/0506617, Jun 2005.
But I find some papers that proved the conjecture in special cases, after 2005. For example
(a) Every vertex-transitive graph of valency four is a polycirculant (E.
Dobson et.al 2007)
(b) All vertex-transitive locally-quasiprimitive graphs have a semiregular
automorphism (M. Giudici and J. Xu 2007).
(c) Every connected distance-transitive graph admits a semiregular
automorphism (K. Kuntar and P.Sparl 2010).
So I want to know that the polycirculat conjecture is proved or not?

REPLY [13 votes]: The Conjecture is still open. 
Lemma 5 of math.GM/0204209 is false. For example, any primitive group on a prime number of points is a counterexample.
Lemma 6 of math/0506617 is also false. Any transitive permutation group without a derangement of prime order satisfies the hypotheses and does not contain a semiregular element. (Any semiregular element has a power that is stil semiregular and of prime order.) Such groups exist, such as $M_{11}$ acting on the twelve  points.<|endoftext|>
TITLE: Green functions on Riemann surfaces
QUESTION [6 upvotes]: Let $(M,g)$ be a compact Rieamnnian surface without boundary and $\Delta_g$ be the Lapalce operator. We note $\lambda_i$ and $\phi_i$ the eigenvalues and  eigenunctions of $\Delta_g$. Let also $G_g$ be the Green function of $\Delta_g$ that is to say 
$$\Delta_g G_g( .  , y) =\delta_y -\frac{1}{vol(M)}$$
Formally, we have 
$$G_g(x,y)= \sum_{i>0} \frac{\phi_i(x) \phi_i(y) }{\lambda_i}.$$ 
My questions are: How strong is the convergence on the right hand side? Do someone have good references about this subject?

REPLY [3 votes]: You may want to clarify  your question. What do you mean by how strong is the convergence?  Are you referring to some norm, and if so, what kind of norm are you interested in? The operator $G_y$ acts between various  Sobolev spaces. In fact it is a pseudo-differentuial operator of order $-2$.   
Consider the the difference 
$$ R^L:=G_y-\sum_{0<\lambda_i\leq L}\frac{1}{\lambda_i}\phi_i(x)\phi_i(y),\;\; L\to \infty ?$$
It is also a pseudodifferential operator of order $-2$ so it defines bounded operators
$$R^L: L^{s,2}(M)\to L^{s+2,2}(M),\;\;s\in\mathbb{R}, $$
where $L^{2,s}$ denotes the Sobolev space of distributions  $s$-times differentiable with derivatives in $L^2$. Denote by $\Vert-\Vert_s$ the norm on $L^{s,2}$. We can define an operator norm 
$$\Vert R^L \Vert_s = \sup_{\Vert\psi\Vert_s=1}\|R^L \psi\|_{s+2}.  $$
Are you interested in how fast $\|R^L\|_s\to 0$ for some $s$?
You can also think of $G_y$ as a distribution on $M\times M$ are you interested  in the convergence in Sobolev spaces of distributions on $M\times M$?
Two places you can look for the Green function.
M. Taylor: Pseudodifferential operators, Chap XII, or
Berligne-Getzler-Vergne: Heat Kernels and Dirac operators, Chap 3.<|endoftext|>
TITLE: Does the moduli space of genus three curves contain a complete genus two curve
QUESTION [18 upvotes]: Inspired by the question 
Does the moduli space of smooth curves of genus g contain an elliptic curve
and its amazing answers, I ask (pure out of curiosity) whether the moduli space $M_3$ of (smooth projective connected) curves of genus $3$ contains a (smooth projective connected) curve of genus $2$. 
The existence of such a genus two curve is (Edit: stronger) than the existence of a surface $S$, a genus two curve $C$ and a smooth projective non-isotrivial morphism $S\to C$ whose fibres are genus three curves. 
If the answer is positive, how explicit can our answer be made? I'm already aware of the fact that $M_g$ contains a complete curve for all $g\geq 3$. For instance, in the  paper by Chris Zaal 
http://dare.uva.nl/document/38546
many curves of some genus (I think 513) are shown to embed into $M_3$.
Of course, by Shafarevich' conjecture, if $K(C)$ denotes the function field of $C$, there are only finitely many non-isotrivial $K(C)$-isomorphism classes of genus three curves over $K(C)$ with good reduction over $C$. I'm asking whether there exists some genus two curve $C$ such that there exists a genus three curve over $K(C)$ with good reduction over $C$. 
Edit: the arithmetic analogue also has a negative answer.
The latter (weaker) phrasing of my question allows us to formulate an arithmetic analogue of the above question. (I know that I'm considering function fields over $\mathbf{C}$ and that some of you might argue function fields over $\mathbf{F}_p$ are a better analogue of number fields.) This arithmetic analogue reads as follows. There exists a number field of "genus two" such that there exist a genus three  curve over $K$ with good reduction over the ring of integers of $K$. Here a number field of "genus two"  should be a number field of absolute discriminant $e^2$. I'll take this to mean discriminant at most $8$.
Arithmetic analogue. (Abrashkin-Fontaine) There do not exist non-zero smooth abelian schemes over the ring of integers of a number field of absolute discriminant at most 8.
There are many related questions I'd also like to ask. For example, what is the minimal $g$ such that $M_g$ contains a genus two curve? Or, what is the minimal $g$ such that $M_3$ contains a genus $g$ curve? And, finally, is there an example of a complete curve in $M_g$ which is defined over $\overline{\mathbf{Q}}$? (Edit: The answer to the last question is positive. This is explained in the comments below.)

REPLY [27 votes]: There does not exist a map of a smooth complete genus 2 curve to $M_3$. 
Such a map would give rise to a surface $S$ (of general type) which violates the Bogomolov-Miyaoka-Yau inequality $c_1(S)^2 \leq 3c_2(S)$. This inequality is equivalent to $3\sigma (S) \leq e(S)$ where $\sigma$ and $e$ are the signature and topological euler characteristic of the surface. The euler characteristic of this surface is 8 (since it is multiplicative for fiber bundles) and by the index theorem, the signature is given by 4 times the integral of $\lambda_1$ over the curve in $M_3$. Since $\lambda_1$ is ample on $M_3$, $\sigma$ must be positive and divisible by 4. 
This argument is due to Dieter Kotschick in his paper "Signatures, Monopoles, and Mapping Class Groups" (MRL vol 5, 1998).<|endoftext|>
TITLE: Non emptyness of ordinary locus for PEL type Shimura varieties
QUESTION [8 upvotes]: We let $B$ be a simple algebra over $\mathbb Q$, with the usual notations for PEL type Shimura varieties. In his paper "Ordinariness in good reductions of Shimura varieties of PEL-type" (available here http://archive.numdam.org/article/ASENS_1999_4_32_5_575_0.pdf), Wedhorn proved, under the assumption that $p$ is unramified in $B$, that the ordinary locus in (the reduction) a PEL type Shimura variety is non-empty if and only if it is dense if and only if $p$ splits completely in the reflex field.
So my question is the following:
Are there similar results, even partial, without the unramifiedness assumption?

REPLY [3 votes]: In the case of Iwahori level structure, there is the result of Stamm (cited in Wedhorn's paper) who gives an example where the ordinary locaus fails to be dense (Hilbert-Blumenthal case).
Stamm's result is reproved and generalized in
Philipp Hartwig, Kottwitz-Rapoport and p-rank strata in the reduction of Shimura varieties of PEL type,
who shows that, still in the Iwahori case (loc.cit. Theorem 1.1.1):
Assume that $\mathcal B$ is the symplectic PEL datum associated with a totally real extension $F/\mathbb Q$. Assume that there is only a single prime of $\mathcal O_F$ dividing $p$. Then the ordinary locus lies dense in $\mathcal A_F$ if and only if $p$ is totally ramified in $\mathcal O_F$.
Of course, if the ordinary locaus is dense in the Iwahori case, then the same is true for all parahoric level structures.<|endoftext|>
TITLE: Projections in Banach spaces
QUESTION [10 upvotes]: Dear All, 
I am absolutely lost in the following problem:
Let $P_s, \: s \in [0,1],$ be a uniformly bounded family of projections (idempotents) in a Banach space $X$ such that $P_s P_t = P_{{\rm min}(s,t)}$. Let $Q$ be a bounded linear operator on $X$ such that $QP_s = P_sQ$ for every $s \in [0,1]$ and the function $$s \mapsto P_s Q $$ is continuous in operator norm. Does it follow that the function must be constant (i.e. $P_sQ \equiv P_0Q$)? 
For simplicity, one can also assume that the function $s \mapsto P_s$ is strongly continuous. I can give an affirmative answer only in some trivial situations (finite dimensional case, Hilbert spaces with family of orthogonal projections) but nothing more.

REPLY [8 votes]: Here a simple example :
Let X be the cartesian product of $L^{\infty}$ and $L^{1}$ on the
interval $[0,1]$, let $P_{t}$ the canonical projection on the subspace
of functions with support $[0,t]$ and choose $Q(f_{1},f_{2}) =
(0,f_{1})$ .<|endoftext|>
TITLE: What is the Plancherel Measure for $\textrm{SL}_3(\mathbb{Q}_p)$?
QUESTION [6 upvotes]: I am looking for a description of the Plancherel Measure of $\textrm{SL}_3(\mathbb{Q}_p)$. Has this been calculated yet? I've search many places for it, but I've only found results on real/complex special linear groups and on $p$-adic general linear groups. Any help would be appreciated.

REPLY [2 votes]: This is not a complete answer but only some hints that could help you.
Bushnell, Kutzko and Henniart have shown, for a general reductive group,  that the restriction of the Plancherel measure to each block of the Bernstein decomposition may be computed via isomorphisms of Hecke algebras :
Bushnell, Colin J.; Henniart, Guy; Kutzko, Philip C. Types and explicit Plancherel formulæ for reductive $p$-adic groups. On certain $L$-functions, 55–80, Clay Math. Proc., 13, Amer. Math. Soc., Providence, RI, 2011.
In the following paper, Bushnell and Kutzko show that for certain blocks of ${\rm SL}(N)$the Hecke algebra
is isomorphic to a Iwahori Hecke algebra for some ${\rm SL}(N')$ over another field. 
Bushnell, Colin J.; Kutzko, Philip C. The admissible dual of ${\rm SL}(N)$. I. Ann. Sci. École Norm. Sup. (4) 26 (1993), no. 2, 261–280.<|endoftext|>
TITLE: Quadratic Farkas' Lemma?
QUESTION [11 upvotes]: The Farkas Lemma says that if a system of linear inequalities implies
yet another linear inequality, then this last inequality can be obtained by
taking a positive linear combination of the inequalities from the system. The
precise statement is as follows:

Let $L_1,\dotsc,L_m$ and $P$ be linear polynomials in the $n$-dimensional real variable $x=(x_1,\dotsc,x_n)$, and suppose that the set of all those $x$ with $L_1(x)\ge 0,\dotsc,L_m(x)\ge 0$ is non-empty. If $P(x)\ge 0$ for each $x$ from this set, then there exist $c_1\ge 0,\dotsc,c_m\ge 0$ with $P\ge cL_1+\dotsb+cL_m$.

For $P$ quadratic this may fail: consider, for instance, $L_1(x)=x$, $L_2(x)=1-x$, and $P(x)=x(1-x)$. I wonder, however, whether the assertion stays true if we allow summands of the form $L_iL_j$:

Suppose that $L_1,\dotsc,L_m$ are linear, and $P$ a quadratic polynomial in the
  $n$-dimensional real variable $x=(x_1,\dotsc,x_n)$. Given that $P(x)\ge 0$
  whenever $L_1(x)\ge 0,\ldots,L_m(x)\ge 0$ (and the set of all such $x$ is
  non-empty), must there exist $c_i,c_{ij}\ge 0$ with $P\ge \sum c_iL_i+\sum
c_{ij} L_iL_j$?

I was able to settle some particular cases; most notably, that where $n=1$ (one variable), and also that where $m=1$ (one constraint). Perhaps, with some effort I can also resolve the case $m=n=2$ (from which the case of $m=2$ and $n$ arbitrary will follow, if I am not mistaken).
I would expect that this is either false, or should be known. Can anybody construct a counterexample or suggest a reference?

REPLY [7 votes]: Here is a counterexample: Take $n=2$ variables $X$ and $Y$. Let $L_1,\dots,L_5$ be linear  polynomials such that
$$S:=\{ (x, y) \in {\mathbb R}^2 ~|~ L_i(x,y) \ge 0\}$$ is a pentagon inscribed in the unit circle. Furthermore set $P:=1-X^2-Y^2$. Assume we could write $P$ as the sum of a globally nonnegative quadratic polynomial $Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$. Now $P$ vanishes at the vertices of the pentagon and each $L_i$ is nonnegative at these vertices. Therefore $Q$ vanishes also at the vertices. But being a nonnegative quadratic polynomial, $S$ is a sum of squares of linear polynomials which all have also to vanish at the vertices and therefore are identically zero. This shows that $S$ is the zero polynomial. Now notice that each of the $L_i$ and $L_iL_j$ is strictly positive on at least one of the vertices of the pentagon (at which $P$ vanishes, of course). Since $P$ is a nonnegative linear combination of the $L_i$ and $L_iL_j$, this shows that $P=0$.
If the set $S$ defined by the $L_i$ has non-empty interior, then the convex cone of quadratic polynomials which can be written as a globally nonnegative quadratic polynomial $Q$ and nonnegative linear combinations of the $L_i$ and $L_iL_j$ is closed. In fact, this follows from a much more general result on truncated quadratic modules, see e.g. the book of Marshall cited below (Lemma 4.1.4). This implies that, in the above counterexample, even $P+\varepsilon$ for small $\varepsilon>0$ will fail though this polynomial is strictly positive on $S$.
However, there are a lot theorems going into the direction of what you want. You might want to have a look at the following books...

Marshall: Positive polynomials and sums of squares
Prestel: Positive polynomials
Bochnak, Coste, Roy: Real algebraic geometry
Basu, Pollack, Roy: Algorithms in real algebraic geometry
Knebusch, Scheiderer: Einführung in die reelle Algebra
Andradas, Bröcker, Ruiz: Constructible sets in real geometry

...and the following articles...

http://homepages.cwi.nl/~monique/files/moment-ima-update-new.pdf
http://www.math.uni-konstanz.de/~schweigh/publications/purestates.pdf
http://www.math.uni-konstanz.de/~schweigh/publications/sosdualsdp.pdf

Also the so-called "S-procedure" could be of interest for you.<|endoftext|>
TITLE: ANOTHER Exterior differential system on $SO(3;\mathbb R) \times \mathbb R$
QUESTION [6 upvotes]: I have another exterior differential system for one forms $U^i$, where the $\theta^i$ are a cotangent basis on $SO(3)$, i.e. they satisfy $d \theta^i = \epsilon_{ijk} \theta^j \wedge \theta^k$ for the antisymmetric tensor $\epsilon$. This is related to a previous question here.
\begin{array}{c}
 \text{dU}^1+\sqrt{3} \theta^1 \wedge U^3+\sqrt{3} \theta^2\wedge U^4 =0\\\\
 \text{dU}^2+\theta^1 \wedge U^3-\theta^2 \wedge U^4+2 \cosh(\rho) \theta^3 \wedge U^5 =0\\\\
 \text{dU}^3+\sinh(\rho)\theta^2\wedge U^5+\theta^3 \wedge U^4 =0\\\\
 \text{dU}^4-\sinh(\rho)\theta^1\wedge U^5-\theta^3\wedge U^3 =0\\\\
 -\sinh(\rho) \text{dU}^5 -\cosh(\rho) \text{d$\rho $}\wedge U^5-\theta^1\wedge U^4+\theta^2\wedge U^3 =0\\\\
 \cosh(\rho) \text{dU}^5+\sinh(\rho) \text{d$\rho $}\wedge U^5+\theta^1\wedge U^4+\theta^2 \wedge U^3-2 \theta^3 \wedge U^2 =0\\\\
 -\text{dU}^4+\cosh(\rho) \theta^1\wedge U^5+\sqrt{3} \theta^2\wedge U^1-\theta^2 \wedge U^2+\theta^3\wedge U^3 =0\\\\
 -\text{dU}^3+\sqrt{3} \theta^1\wedge U^1+\theta^1\wedge U^2+\cosh(\rho) \theta^2\wedge U^5-\theta^3\wedge U^4=0
\end{array}
FYI: this problem is related to my work on asymptotic symmetries of certain noncompact homogeneous spaces, i.e. I want to find diffeomorphisms which preserve a certain metric tensor asymptotically.
It seems like it should be possible to solve systems like this (semi-) automatically with a computer algebra package. After all, the system reduces to an overdetermined system of first order partial differential equations, for which such tools already exist... but the EDS form is so much more convenient that I would hate to rewrite everything as 1st order PDEs!
EDIT: As R.B. suggested in the comments, I forgot to mention that as in the other question, there is also a coordinate $\rho$ (of course), i.e. the one forms can be expanded as $\alpha = \alpha_a \theta^a + \alpha_\rho d\rho$ etc. Also, the summation convention is not implied in the expression for $d \theta^i$.
EDIT2: Of course a good start would be to add the third and last equations to get rid of $dU^3$ and similarly for $dU^4$ and $dU^5$. The number of unknowns will then be reduced from 20 to 7, but the system still seems quite difficult to solve... I've been hacking away at it with Mathematica and some exterior differential and wedge product functions, but it's still a mess!

REPLY [16 votes]: Jeanne's calculations give the right answer, i.e., that the solutions depend on two arbitrary functions of 2 variables.  
It turns out, though, that, with the right choice of variables, one can reduce the problem to an underdetermined system of $2$ linear equations for a pair of tensors of rank $2$ on the $2$-sphere.  (This choice of variables is suggested by examining the characteristic variety and tableau of the EDS.)
First of all, if one solves for the coefficients of the $U^i$ as the OP suggests, one does get $7$ parameters.  One way to do this is as follows:
$$
\begin{align}
U^1 &= -(\sqrt{3}/3)\bigl((v_4{-}(\cosh(\rho)-\sinh(\rho))v_1)\ \theta^1
                         -(v_2{+}(\cosh(\rho)+\sinh(\rho))v_3)\ \theta^2\bigr)\\\\
U^2 &= (\cosh(\rho)-\sinh(\rho))v_1\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_3\ \theta^2\\\\
U^3 &= (\cosh(\rho)-\sinh(\rho))v_5\ \theta^1 + v_6\ \theta^2
            + 2\sinh(\rho)v_3\ \theta^3 +v_4\ d\rho\\\\
U^4 &= v_7\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_5\ \theta^2 
            + 2\sinh(\rho)v_1\ \theta^3 +v_2\ d\rho\\\\
U^5 &= (\cosh(\rho)-\sinh(\rho))v_2\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_4\ \theta^2.\\\\
\end{align}
$$
The $v_i$ are now $7$ unknowns, and substituting this into the above $2$-forms gives an involutive differential system for the $v_i$ that is generated by five $2$-forms and has characters $s_1=5$ and $s_2=2$, with $s_i=0$ for $i>2$.  
However, examining the characteristic variety and tableau of this system suggests that one should reparametrize the first four $v_i$ as
$$
\begin{align}
v_1 &= -(2\sinh(\rho){+}\cosh(\rho))\ p_2-\cosh(\rho)\ q_2\\\\
v_2 &= (2\cosh(\rho)\sinh(\rho)+2\cosh^2(\rho)-1)(p_1{+}q_1)\\\\
v_3 &= +(2\sinh(\rho)-\cosh(\rho))\ p_1 + \cosh(\rho)\ q_1\\\\
v_4 &= (2\cosh(\rho)\sinh(\rho)-2\cosh^2(\rho)+1)(p_2{-}q_2).\\\\
\end{align}
$$
for some functions $p_1,p_2,q_1,q_2$. When one does this, one finds that the differential equations imply
$$
\begin{align}
dp_1 &= (p_0{+}p_3)\ \theta^1 + (p_4{+}p_5)\ \theta^2-3p_2\ \theta^3\\\\
dp_2 &= (p_4{-}p_5)\ \theta^1 + (p_0{-}p_3)\ \theta^2+3p_1\ \theta^3\\\\
dq_1 &= (q_0{+}q_3)\ \theta^1 + (q_4{+}q_5)\ \theta^2-1q_2\ \theta^3\\\\
dq_2 &= (q_4{-}q_5)\ \theta^1 + (q_0{-}q_3)\ \theta^2+1p_1\ \theta^3\\\\
\end{align}
$$
for some functions $p_0,p_3,p_4,p_5,q_0,q_3,q_4,q_5$.  This implies that the (complex-valued) linear form $Q = (q_1{-}iq_2)(\theta^1{+}i\theta^2)$
and the (complex-valued) cubic form $P = (p_1{-}ip_2)(\theta^1{+}i\theta^2)^3$ are 
well-defined on the $2$-sphere $S^2 = \mathrm{SO}(3)/\mathrm{SO}(2)$, where the $\mathrm{SO}(2)$-subgroup of $\mathrm{SO}(3)$ is the one that is an integral of the forms $\theta^1$
and $\theta^2$.  Thus, $P$ and $Q$ are sections of natural complex line bundles over the $2$-sphere.
One then finds that the integral manifolds of the differential system must satisfy
$$
\begin{align}
v_5 &= -\sinh(\rho)\ (p_4{+}q_5)-\cosh(\rho)\ (p_5{+}q_4)\\\\
v_6 &= (\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho){-}2)\ p_0
        +(\cosh(\rho)\sinh(\rho){-}\cosh^2(\rho){+}1)\ p_3\\\\
    &\quad +(\cosh(\rho)\sinh(\rho){-}\cosh^2(\rho){+}{\tfrac{1}{3}})\ q_0 
        +(\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho))\ q_3\\\\
v_7 &= (\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho){-}2)\ p_0
        +(\cosh(\rho)\sinh(\rho){+}\cosh^2(\rho){-}1)\ p_3\\\\
    &\quad +(\cosh(\rho)\sinh(\rho){+}\cosh^2(\rho){-}{\tfrac{1}{3}})\ q_0 
        +(\cosh^2(\rho){+}\cosh(\rho)\sinh(\rho))\ q_3\ .\\\\
\end{align}
$$
Finally, once these are substituted into the $2$-forms, one finds that the differential system reduces to a pair of second order linear differential equations on $S^2$.  Using notation that can be found in a paper of mine (TAMS 290 (1985), 259–271), this second order equation for the tensors $P$ and $Q$ can be written in the form
$$
Y(Y(P)) = \tfrac12\ Q + \tfrac23\ X(Y(Q)) - \tfrac13\ X(X({\overline{Q}})),
$$
where $X$ and $Y$ are certain first-order differential operators that are invariant under
$\mathrm{SO}(3)$ and that generalize $\partial$ and $\bar\partial$ to symmetric $(1,0)$-forms of arbitrary degree.
This is an elliptic linear equation of second order that is underdetermined (it is $2$ equations for the $4$ components of $P$ and $Q$).  It is invariant under the rotations of $S^2$ and equivalent to the original (overdetermined) EDS on $\mathrm{SO}(3)\times\mathbb{R}$.
It is conceivable that this equation admits an explicit solution in terms of a potential (which would be a section of a rank $2$ vector bundle over $S^2$), but I have not tried to check whether this is true or not.
Added Comment:  I have now checked about the possibility of a potential and, miraculously, it turns out that there is a potential:  One can show that the solutions of the above equation for $P$ and $Q$ are expressible in the form
$$
\begin{align}
P &= X(X(L))\\\\
Q &= L + (XY{+}YX)L + X(X({\overline{L}})\bigr)\\\\
\end{align}
$$
where $L=(L_1{+}iL_2)(\theta^1{+}i\theta^2)$ is an arbitrary complex-valued $(1,0)$-form on $S^2$.  Thus, components of $L$ are the two arbitrary functions of $2$ variables predicted by the theory for the general solution.  
$L$ is not quite unique.  It turns out that one can also add an expression of the form $X(a_1+ib_2)$ to $L$ where $a_1$ is the restriction to the $2$-sphere of a linear function in $\mathbb{R}^3$ and $b_2$ is the restriction to the $2$-sphere of a harmonic homogeneous quadratic function in $\mathbb{R}^3$.  This describes the (local and global) ambiguity in the potential $L$ completely.
Thus, with this construction and the above formulae, we have the complete description of the (local and global) integral manifolds of the original EDS in terms of the second and third derivatives of $L$ (which is arbitrary).<|endoftext|>
TITLE: Approximating commuting matrices by commuting diagonalizable matrices
QUESTION [20 upvotes]: Suppose the matrices $A$ and $B$ commute. Do there exists sequences $A_n$ and $B_n$ of matrices such that

$A_n \rightarrow A$, $B_n \rightarrow B$. 
Each $A_n$ is diagonalizable  and the same for each $B_n$. 
For every $n$, $A_n$ commutes with $B_n$. 

Moreover, it would be nice if the following property was additionally satisfied: if $A,B$ are real, then $A_n,B_n$ can be chosen to be real as well. 
P.S. I asked this question on math.SE a couple of days ago.

REPLY [13 votes]: This result was first proved by Motzkin and Taussky (essentially in all characteristics) depending upon how you state the result:  the set of pairs of n x n matrices (A,B) where
AB = BA and A and B have distinct eigenvalues is Zariski dense in the set of commuting pairs of matrices (which is equivalent to the irreducibility of the commuting variety; both results are in Motzkin-Taussky).  The reference is:  Pairs of matrices with property L. II, Trans. AMS 80 (1955), 387-401.   A trivial consequence of this is that any pair of commuting n x n matrices generates an algebra of dimension at most n.  Gerstenhaber proved a slightly stronger result: any pair of commuting n x n matrices is contained in a commutative n-dimensional subalgebra. See Guralnick, LAMA 31 (1992), 71-75 for a slightly easier proof and also some remarks on commuting 3-tuples (this variety is reducible for n at least 28, irreducible for n at most 10).   So one cannot in general approximate 3 commuting matrices by commuting semismple matrices.   See also Guralnick-Sethuram (Commuting pairs and triples of matrices and related varieties. Linear Algebra Appl. 310 (2000),  139–148) showing that Gerstenhaber's stronger result also follows easily from Motzkin-Taussky.<|endoftext|>
TITLE: Proper maps and transversality
QUESTION [7 upvotes]: I'll begin with the question, which is intrinsically interesting:

Let M be a manifold with some submanifold Y. Suppose that $W \rightarrow M$ is a smooth, proper map. Does there exist another map $W \rightarrow M$ homotopic to the original that is ALSO proper and transverse to the submanifold Y?

Let me note that I am definitely not assuming the manifolds are compact. 
Why I care: This question came up while thinking about the geometric description of complex cobordism given by Quillen in "Elementary proofs of some results of cobordism theory using Steenrod operations." I have a geometric description of the coboundary map in the Mayer-Vietoris sequence but as of now it relies on the answer to the above question being "yes."

REPLY [2 votes]: Regarding the coboundary map in complex cobordism, I think that this was done by Dold in "Geometric Cobordism and the Fixed Point Transfer" (at least for oriented cobordism) in 2.10.<|endoftext|>
TITLE: Intersecting 4-sets
QUESTION [12 upvotes]: Is it possible to have more than $N = \binom{\lfloor n/2\rfloor}{2}$ subsets of an $n$-set, each of size 4, such that each two of them intersect in 0 or 2 elements?
To see that $N$ is achievable, choose $\lfloor n/2\rfloor$ disjoint pairs and then take each 4-set consisting of two of the pairs.  But this is not the unique way of doing it in general.
EDIT: Patricia has provided a counterexample with $n=7$, so I'll remove odd $n$ from the question.  Is there a counterexample for even $n$?

REPLY [16 votes]: The conjectured maximum of $N = \binom{\lfloor n/2\rfloor}{2}$
is correct except for $n=7$, when the maximum is $7$, and
$8 \leq n \leq 11$, when the maximum is $14$.  The maximal
configuration is unique except for $n=12$, $13$, $15$, $16$, and $17$.
Let $L$ be the subgroup of ${\bf Z}^n$ generated by $(2{\bf Z})^n$
and the characteristic functions $e_i + e_j + e_k + e_l$ of each
4-set $\lbrace i,j,k,l \rbrace$ in our family $\cal F$ of subsets of
$\lbrace 1,2,\ldots,n \rbrace$.  Give $L$ the structure of lattice
using the inner product 
 $$ \langle x, y \rangle = \frac12 \sum_{i=1}^n x_i y_i $$
(i.e. half the usual inner product).  Then $L$ is generated by
vectors $2e_i$ and $e_i + e_j + e_k + e_l$ of norm $2$, any two of which
are either orthogonal or have inner product $1$.  Hence $L$ is an
even integral lattice, with at least $2n+16|{\cal F}|$ roots
(vectors of norm 2), namely $\pm 2 e_i$ and $\pm e_i \pm e_j \pm e_k \pm e_l$
for $\lbrace i,j,k,l \rbrace \in \cal F$.   Equality holds iff $\cal F$ contains every tetrad $\lbrace i,j,k,l \rbrace$ such that $e_i + e_j + e_k + e_l \in L$.
Now we can use the theory of root systems to partition the set of
roots of $L$ into mutually orthogonal simple root systems.  Since
$L$ contains the root lattice $A_1^n = (2{\bf Z})^n$, the only possible
components of the root system of $L$ are $A_1$, $D_{2k}$ for $k \geq 2$,
and the exceptional systems $E_7$ and $E_8$.  These contribute respectively
$0$, $\binom{k}{2}$, $7$ and $14$ tetrads to $\cal F$.  Namely, each $A_1$
corresponds to a coordinate that does not appear in $\cal F$; each $D_{2k}$
corresponds to $k$ pairs of coordinates paired in each of $\binom{k}{2}$
possible ways; and $E_7$ and $E_8$ correspond to the tetrads of the
Hamming $[7,3,4]$ and extended Hamming $[8,4,4]$ codes respectively.
It is now elementary bookkeeping to obtain the maximum configuration.
$\circ$ Except for $7 \leq n \leq 11$, the maximal $|{\cal F}|$ is $\binom{k}{2}$
for $n = 2k$ or $n = 2k+1$, attained by the $D_{2k}$ configuration.
$\circ$ For $n=7$, the maximum of $7$ is attained by the $E_7$ (Hamming) configuration,
and for $8 \leq n \leq 11$, by $E_8 \oplus A_1^{n-8}$ (extended Hamming).
$\circ$ For $n=12$ ($n=13$), the maximum of $15$ is attained by both
$D_{12}$ ($D_{12} \oplus A_1$) and $E_8 \oplus D_4$ ($E_8 \oplus D_4 \oplus A_1$).
$\circ$ For $n=15$, the maximum of $21$ is attained by both
$D_{14} \oplus A_1$ and $E_8 \oplus E_7$.
$\circ$ Finally, for $n=16$ ($n=17$),
the maximum of $28$ is attained by both
$D_{16}$ ($D_{16} \oplus A_1$) and
$E_8 \oplus E_8$ ($E_8 \oplus E_8 \oplus A_1$).
[The lattice $L$ corresponds via "construction A" to a binary linear code
generated by $\cal F$, which is doubly even by hypothesis.  Koch developed
a theory of "tetrad systems" of such codes that could be used to
give a more direct but less familiar derivation of this answer.]

REPLY [5 votes]: It looks like the conjecture is very close to be right (i.e. $N$ is an upper bound) for $n\geq 12$, as the linear programming (LP) bound equals $N$ for even $n$ even, $500\geq n\geq 12$.
When $n$ is odd, there is a gap between the LP bound and $N$, but this is most likely to do with the fact that the optimal solutions are non-integer.
The corresponding LP has just 2 variables and 6 constraints, so it should be perfectly possible to derive it by hand for general $n$.
Here one can get Sage code to solve the LP in question: e.g. (it's actually an arbitrary precision solver, it only shows the LP with the floating point...)
sage: load j.sage
sage: A,p,bd=delsarte_bound_J(130,4,[1,3], return_data=True)
sage: bd
2080
sage: binomial(130/2,2)
2080
sage: p.show()
Maximization:
x_0 + x_1 + x_2 + x_3 + x_4
Constraints:
constraint_0: 1 <= x_0 <= 1
constraint_1: 0 <= x_1 <= 0
constraint_2: 0 <= x_3 <= 0
constraint_3: -10836.0 x_0 - 8041.0 x_1 - 5246.0 x_2 - 2451.0 x_3 + 344.0 x_4 <= 0
constraint_4: -52006500.0 x_0 - 25384125.0 x_1 - 7848854.0 x_2 + 599313.0 x_3 - 39624.0 x_4 <= 0
constraint_5: -682329375.0 x_0 - 162459375.0 x_1 + 5285345.0 x_2 - 192855.0 x_3 + 8385.0 x_4 <= 0
constraint_6: -1.79043228e+11 x_0 + 1420978000.0 x_1 - 22735648.0 x_2 + 550056.0 x_3 - 17888.0   x_4 <= 0
Variables:
   x_0 is a continuous variable (min=0, max=+oo)
   ...
   x_4 is a continuous variable (min=0, max=+oo)

Unfortunately, the code will only work in an experimental version of Sage; one needs version 5.4 (with is only in "release candidate" state now), and install this ticket, which is not for faint-hearted...<|endoftext|>
TITLE: Background for Hejhal's "The Selberg Trace Formula for $PSL(2, \mathbb{R})$
QUESTION [10 upvotes]: Reposted from math.stackexchange where my question received only five views and no answers...
I'm trying to learn the Selberg trace formula, but have very little background in harmonic analysis. I was referred to Dennis Hejhal's The Selberg Trace Formula for $PSL(2, \mathbb{R})$ but just got the book and was dismayed to learn that that the author assumes familiarity with Selberg's original paper (which I don't have access to - would welcome a pointer to an online copy). 
There's much in the first few pages that I don't know. For example, the author states without proof that the spectrum of the Laplacian on a compact hyperbolic surface is discrete. He gives a reference to a 1912 book by Hilbert, but aside from the fact that I don't read German, it's not clear to me that this is the best place to learn from (in light of the fact that Hejhal's book is from 1976 and many books have been written since). 
Does anyone have a suggestion as to what to read before Hejhal's book?

REPLY [5 votes]: When I tried to read Hejhal's books around 1983, I enjoyed the book $SL_2(\mathbb{R})$ (where $SL$ stands for Serge Lang) for background.<|endoftext|>
TITLE: Pesin Entropy Formula
QUESTION [10 upvotes]: In the form that I've seen it stated, the Pesin entropy formula states that if $M$ is a compact Riemannian manifold and $f$ is a $C^{1+\alpha}$ diffeomorphism of $M$ that preserves smooth invariant measure $\mu$, then
$$
h_{\mu}(f)=\int_M \Sigma(x)d\mu(x)
$$
where $\Sigma(x)$ denotes the sum of the positive lyapunov exponents of $f$ at $x$.
$Question:$ Does the above hold if $f$ is only piecewise $C^{1+\alpha}$?
In fact I'm really interested in a specific example called the random $\beta$-transformation, which is interesting in the study of Bernoulli convolutions and $\beta$-expansions. 
This can be written as a map on $[0,1]^2$ which is piecewise linear (on four pieces) but not Markov in general. It preserves a measure $\mu$ equivalent to Lebesgue measure. I'd be really grateful to hear of a reference where the Pesin entropy formula has been pushed forward to this kind of situation.

REPLY [7 votes]: Dear Tom,
I believe that the Pesin entropy formula for maps with singularities (such as piecewise smooth maps) is discussed in the book "Invariant Manifolds, Entropy and Billiards. Smooth Maps with Singularities" of A. Katok and J.-M. Strelcyn (see its parts III and IV). 
Best, 
Matheus<|endoftext|>
TITLE: Upper bound on largest eigenvalue of a real symmetric n*n matrix with all main diagonal >0, everywhere else <=0
QUESTION [7 upvotes]: Is there a good analytic upper bound on the largest eigenvalue of a real symmetric n*n matrix with all main diagonal entries strictly positive, all other entries <=0 with typically many of them exactly 0 (so sparse)? We have upper bounds on the magnitudes of all nonzero entries. We do not know whether the main diagonal terms are more or less than minus the off-diagonal nonzero terms.
Zhan's "Extremal eigenvalues of real symmetric matrices with entries in an interval" (SIAM J. Matrix Anal. Appl. Vol. 27, No. 3, pp. 851-860) provides an answer for general n*n real symmetric matrices but can we improve/simplify this given the extra conditions?

REPLY [6 votes]: Probably, the following elementary argument using sparseness will help. Assume that $|A_{i,j}|\le c$. Let $\tau\in[0,1]$ be a proportion of non-zero entries among all entries in matrix, so that there are exactly $\tau n^2$ non-zero entries. Then
$
 \max_{k}{|\lambda_k|} \le \sqrt{\mathrm{tr}(A^*A)}= \sqrt{\sum_{i,j=1}^{n}{|A_{i,j}|^2}} \le \sqrt{\tau n^2 c^2} = \sqrt{\tau}n c.
$
For $\tau$ small this beats Zhan's estimate, which is roughly $nc$ in this context.
Gershgorin theorem is good for matrices with small off-diagonal elements.<|endoftext|>
TITLE: Strongly convergent operator sequence
QUESTION [5 upvotes]: Let $T_j$ be a sequence of compact operators on a Hilbert space $H$ which converges strongly to the identity, i.e., for each $v\in H$ the sequence
$$
\parallel T_jv-v\parallel
$$
tends to zero. Is it true that there must exist an index $j$ such that the spectrum of $T_j$ contains a non-zero number?

REPLY [6 votes]: I think the answer is no: there should be a sequence of quasinilpotent compact operators on $L^2[0,1]$ which converges in SOT to the identity map. This is roughly for the same reason one can have radical Banach algebras with compact multiplication and bounded approximate identities.
The following is an outline, as I am a bit short of time and sleep right now.
Specifically, try Volterra-type operators $T_j:L^2[0,1]\to L^2[0,1]$
$$ T_j\xi(t) = \int^t_0 f_j(s)\xi(t-s)\,ds $$
where $f_j$ is something like a heat kernel or Gaussian that ``approaches the Dirac point mass at the origin''.
Certainly if $f_j$ is continuous on $[0,1]$ then $T_j$ will be quasinilpotent and compact (just approximate $f_j$ with polynomials and use known properties of the classical Volterra operator).
So I guess we just need to arrange that $\Vert T_j \xi - \xi \Vert_2 \to 0$ for any $\xi \in C[0,1]$ (then we deduce it for all $\xi \in L^2[0,1]$ by density).
But now given such a $\xi$ it is uniformly continuous, so on small intervals of width $\delta$ it can only vary by $\epsilon$, so provided $f_j$ lives mostly on the interval $[0,\delta]$ and has total mass $1$ we should have $|T_j(\xi)(t)-\xi(t) | \leq 2\epsilon$ for all $t$, which certainly implies $\Vert T_j \xi - \xi \Vert_2 \leq O(\epsilon)$.<|endoftext|>
TITLE: series expansion of the q-Pochhammer symbol
QUESTION [6 upvotes]: The following identity arose while I was working on a recent MO question:
$-\sum_{n=1}^{\infty}\frac{1}{n}\frac{(-x)^n}{1-x^n}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^n}{1-x^{2n}}.$
I have no doubt that the identity is true, but I am not able to prove it. Can anyone help?
It is easy to prove by Taylor expansion that the left-hand-side of the identity can equivalently be written as $\sum_{n=1}^{\infty}\ln(1+x^n)$, which is the logarithm of the q-Pochhammer symbol $(-x,x)_{\infty}$, so an alternative way to pose my question is to ask for a proof of the series expansion
$\ln(-x,x)_{\infty}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^{n}}{1-x^{2n}}.$

REPLY [6 votes]: I would make a mere comment since Gjergji has already answered, but I am not allowed to make comments.

... so an alternative way to pose my question is to ask for a proof of the series expansion
  $\ln(-x,x)_{\infty}=\sum_{n=1}^{\infty}\frac{1}{n}\frac{x^{n}}{1-x^{2n}}$.

This is a corollary of Euler's theorem that the number of partitions of $n$ into distinct parts is equal to the number of partitions of $n$ into odd parts.  In terms of generating functions, Euler's theorem is just
$(-x,x)_{\infty}=\frac{1}{(x,x^2)_\infty}$, which can be easily proved by replacing the
term $(1+x^i)$ in $(-x,x)_\infty$ by $\frac{1-x^{2i}}{1-x^i}$ and cancelling all the terms in
the numerator against the corresponding terms in the denominator. By Euler's theorem,
$\ln \left( (-x,x)_{\infty}\right) =\ln\left( \frac{1}{(x,x^2)_\infty}\right)
=\sum_{i=1}^\infty \ln \left( \frac{1}{1-x^{2i-1}} \right)
=\sum_{i=1}^\infty \sum_{n=1}^\infty \frac 1 n x^{n (2i-1)}$ 
$=\sum_{n=1}^\infty \sum_{i=1}^\infty \frac 1 n x^{n (2i-1)}$ 
$
=\sum_{n=1}^\infty \frac 1 n \frac{x^n}{1-x^{2n}}
$.<|endoftext|>
TITLE: Rational curved lying in the boundary of Deligne-Mumford compactification $\bar M_g$
QUESTION [7 upvotes]: Let $\bar M_g$ be the Deligne-Mumford compactifiction of the moduli space of complex genus $g$ curves $M_g$. Is this correct that through every point of the boundary $\bar M_g\setminus M_g$ passes a rational curve that lies in the boundary $\bar M_g\setminus M_g$?

REPLY [2 votes]: The irreducible components of the normalization $\mathcal{B}^{\nu}$ of the boundary $\mathcal{B}\subset\overline{\mathcal{M}}_{g,n}$ are finite images of the moduli spaces:

$\overline{\mathcal{M}}_{g_{1},S_{1}\cup\{n_{1}+1\}}\times\overline{\mathcal{M}}_{g_{2},S_{2}\cup\{n_{2}+1\}}$, where $g_{1}+g_{2} = g$ and $S_{1},S_{2}$ is a partition of $\{1,...,n\}$,
$\overline{\mathcal{M}}_{g-1,n+2}$,

Therefore any component of the boundary can be interpred as a moduli space of curves and you question can be rewritten in this way: For which $(g,n)$ is $\overline{\mathcal{M}}_{g,n}$ uniruled?
It is well known that $\overline{\mathcal{M}}_{g,n}$ is of general type for $g\geq 24, n\geq 0$. Then it is not uniruled. In lower genus, at the best of my knowledge:

$\overline{\mathcal{M}}_{0,n}$ is rational for any $n\geq 3$.
$\overline{\mathcal{M}}_{1,n}$ is rational for any $1\leq n\leq 10$, P. Belorousski, Chow rings of moduli spaces of pointed elliptic curves, Ph.D. thesis, Chicago,
1998.
$\overline{\mathcal{M}}_{g,n}$ for $g = 2$ and $1\leq n\leq 12$, $g = 3$ and $1\leq n\leq 14$, $g = 4$ and $1\leq n\leq 15$, $g = 5$ and $1\leq n\leq 12$, http://arxiv.org/abs/math/0504249.
$\overline{\mathcal{M}}_{g,n}$ is rational for $g=6$ and $1\leq n\leq 18$, and it is unirational for $g=8$ and $1\leq n\leq 11$, $g=10$ and $1\leq n\leq 3$, $g=12$ and $n=1$, http://arxiv.org/abs/math/0701475.
In Section $7$ of A. Logan, The Kodaira dimension of moduli spaces of curves with marked points, Am. J. Math. 125 (2003), 105–138, the author determines for
$g = 2, . . . , 9, 11$ an integer $f(g)$ such that $\overline{\mathcal{M}}_{g,n}$ is unirational for $n \leq f(g)$.
$\overline{\mathcal{M}}_{g}$ is unirational for $g\leq 14$, $\overline{\mathcal{M}}_{15}$ is rationally connected, $\overline{\mathcal{M}}_{16}$ is uniruled, http://arxiv.org/abs/0805.2424. 
$\overline{\mathcal{M}}_{22}$ is of general type. The Kodaira dimension of $\overline{\mathcal{M}}_{g}$ for $17\leq g\leq 21$ and $g = 23$ is not know, http://arxiv.org/abs/0805.2424.
$\overline{\mathcal{M}}_{g,n}$ is uniruled for $g = 12$ and $n \leq 5$, $g = 13$ and $n \leq 3$, $g = 15$ and $n \leq 2$, http://arxiv.org/abs/1206.1424.<|endoftext|>
TITLE: Eigenvalues of a Symmetric Positive Semi-Definite (PSD) matrix after rank one update
QUESTION [5 upvotes]: I have a Symmetric Positive Semi-Definite matrix $A$ which i know its eigenvalue and eigenvectors. let $v$ and $u$ be a random column vector. i want to know if it is possible to have eigenvalues of matrix $A+uv^T$.
I don't need its eigenvectors, but it is required to have the most precise eigenvalues. We know that $uv^T$ is also a rank one PSD matrix.
Is there a close form to this problem?
${\bf PS:}$ As you see i found the solution, but after implementation of this i can see a very small error in result, it would be appreciated if anyone know why this is happening, because we didn't used any approximation to get the result and it should be exact.

REPLY [3 votes]: Complementing Denis's answer, you can look at the paper Some Modified Matrix Eigenvalue Problems by Golub (http://www.jstor.org/stable/2028604) where this issue is treated at some length.<|endoftext|>
TITLE: Blueprint of L-functions and need for introducing them ( Hasse-Weil L-functions ) 
QUESTION [5 upvotes]: Dear All, 
This question may appear elementary to all the experts in number theory , but forgive me. I really wanted to know how did the $L$-functions came into existence, especially the Hasse-Weil L-functions . Do they have some specific meaning in their formulation or they are just framed heuristically to build some thing else , as scaffolding . 
I do know that Zeta functions and L-functions of the curve act as spies in collecting the secret information about the local part of curves and embed that information inside them, but its really a great trouble in understanding the formulation. I referred to many books and they have started saying " Let $L(s,E)$ be the ...." in an assuming manner . 
I just wanted to know , why should one consider $$\zeta_{C/\mathbb{F_q}}(u)=\exp  \bigg(\ \sum_{n=1}^{\infty}\frac{ | C(\mathbb{F_{q^n}})|}{n} u^{n} \bigg)$$ where $C$ is a projective curve with non-negative genus over finite field  $\ \mathbb{F_q}$. Here are my pointers :

I didn't understand about the reason behind introducing exponential function on the right side .
I understood that there is some measure of points taking a ratio of the cardinality ( on R.H.S ) of the solutions, but why is the ratio needed ? I got this doubt when I looked at some other heuristic consideration $\prod\frac{N_p}{p}$ ( Where $N_p$ is the cardinality of solution set at some prime $p$ ) , why is the need to take the ratio ? Isn't it not sufficient to look at just $N_p$ ? We get the cardinality directly, why should we find the ratio again by dividing it with $p$ ? 

Similarly , why is the formulation of local part of $L$-series ( Hasse Weil L-function )  appear as $L_p(T)=1-a_pT+pT^2$ when the curve has good reduction at $p$ ( here $a_p=p+1-N_p$ and has some other formulation like $L_p(T) = 1-T$ and $1+T $ when the curve has split multiplicative and non-split multiplicative reductions at $p$ respectively , and $L_p(T)=1$ when the curve has additive reduction at $p$. 
How was  the quadratic equation on R.H.S ( i.e $1-a_pT+pT^2$ ) formulated ? Was it a scaffolding  to get some heuristic output later , or it has a specific meaning derived from something, or what ? 
Same with $1-T$ and $1+T$ . 
Please do explain me , I am sorry my learned friends, if I have wasted your time, but every book I referred starts with Let, and I thought that its just a setting . If you want me to suggest some book that does the same task of explaining what I asked, you are welcome to suggest me .
Cordially, 
Shanmukha Srinivasan.

REPLY [5 votes]: One justification for this is the Euler product expression. To find the Euler product expression for the Hasse-Weil function, you have to ask yourself what the appropriate analogue of a prime is. It turns out to be a closed point on the variety. The analogue of the size of the prime is the size of the residue field of the closed point.
Without doing any calculation you can already see why the exp makes sense, because the Euler product will be a product over points, and counting points will be a sum over points, so to turn the one into the other you need to take an exponential.
To figure out why you need to divide, you need to compute more carefully.
You want to take the product of, for each closed point $x$, of residue degree $d_x$, $1/\left(1-\left(q^{d_x}\right)^{-s}\right)$. 
Setting $u=q^{-s}$, you want to take the product of $1/(1-u^{d_x})$.
Since the product is exp of the sum of the log, you want to take the exp of the sum of minus the log of $1-u^{d_x}$. If you expand the log out as a power series, you get $\sum_x \sum_k u^{kd_x}/k$. 
Each $\mathbb F_{q^n}$-point comes from a unique closed point $x$. That point must have a degree $d_x$ dividing $n$, and each closed point of degree $d$ dividing $n$ corresponds to $d$ $\mathbb F_{q^n}$-points. 
So you can write $u^{kd_x}/k$ as coming from the $d_x$ $\mathbb F_{q^{kd_x}}$-points, divided by $kd_x$.<|endoftext|>
TITLE: On some finiteness properties for schemes
QUESTION [6 upvotes]: Consider the following properties of scheme $X$:
A: $X$ is of finite type over $\mathbb{Z}$
B: $X$ is Noetherian
C: $X$ is of finite Krull dimension
What implications are there between these three? I believe that B and C are independent of each other (although I can't find a reference right now), and it follows from EGA I, 6.3.7 that A implies B. But does A imply C?
(Apologies if this question is "trivial", but I'm not an expert in algebraic geometry.)
As an aside, I would also be interested if any of these properties can be related to some notion of cohomological dimension (not sure what kind of topologies would be relevant for this).

REPLY [8 votes]: An example of a noetherian ring of infinite dimension can be found in Nagata's Local Rings, Appendix A1, Example 1.
Edit: An interesting generalisation of Nagata's construction yielding noetherian rings of infinite dimension whose maximal ideals have prescribed heights was given by Fujita in his article Infinite dimensional Noetherian Hilbert domains, Hiroshima Math. J. 5 (1975), 181–185.<|endoftext|>
TITLE: What is geometric intuition of special Lagrangian manifolds?
QUESTION [6 upvotes]: Let $M$ be (for example) a Calabi-Yau threefold with Kaehler form $\omega$ and holomorphic 3-form $\Omega$. We say that a submanifold $L$ of $M$ is a special Lagrangian submanifold if $L$ is Lagrangian with respect to symplectic form $\omega$ and also $\mathrm{Im}\Omega|_L=0$. I would like to know geometric intuition of the latter condition. I am aware that Lagrangian condition roughly corresponds to $L$ having only position coordinate, momentum coordinate, or certain mixture of them (Lagrangian formulation of classical mechanics). I wonder if there is a good explanation about the extra condition $\mathrm{Im}\Omega|_L=0$.

REPLY [7 votes]: Let $(M,g,J,\Omega)$ be a Calabi-Yau $n$-fold. Then $\textrm{Re }\Omega$ is a calibration on $(M,g)$. Let $L\subset M$ be a real submanifold with $\dim_{\mathbb{R}}L=n$. You have the following
Proposition 
$L$ is a special Lagrangian if and only if it admits an orientation making it into a calibrated (for $\textrm{Re }\Omega$) submanifold of $(M,g)$. In that case, it is volume-minimising in its homology class.
See Propositions 10.1 and 7.1 in the "Calabi-Yau manifolds..." book by Gross, Huybrechts and Joyce.
Here $\textrm{Re }\Omega$ being a calibration means that at each point $p\in M$, and for every oriented  tangent $n$-plane $V\subset T_{M,p}$, one has $\left.\textrm{Re }\Omega\right|_V\leq vol_V$, where  $vol$ is the volume form of $g$.
Then $L$ being a calibrated submanifold for $\textrm{Re }\Omega$ means that on the tangent spaces $T_{L,p}$ of $L$ the above inequality becomes equality.<|endoftext|>
TITLE: Topological equivalence of homotopic vector fields
QUESTION [9 upvotes]: Two (tangent) vector fields $X$ and $Y$ on oriented differentiable manifolds $M$ and $N$, respectively, are topologically equivalent, if there is an orientation-preserving homeomorphism $M \to N$, which sends orbits of $X$ to orbits of $Y$, preserving the direction of the orbits. The vector fields are topologically conjugate, if there is a homeomorphism $h: M \to N$, such that $\phi^X_t = h^{-1} \circ \phi^Y_t \circ h$, where $\phi^X$, $\phi^Y$ are the flows of $X$ and $Y$, respectively. A vector field is structurally stable, if small perturbations of it result in a topologically equivalent vector field.
In dynamical systems, structural stability is an important and well-studied aspect, but one is also interested in comparing the qualitative behavior of vector fields, which are not just small perturbations from each other. In order to show the equivalence or conjugacy of vector fields in concrete cases, one can sometimes construct the homeomorphism by hand. Instead of showing that two vector fields are topologically equivalent, it is often much easier to show, that they are homotopic (as sections of the tangent bundle) via vector fields which preserve a property. In concrete examples, one might like to know, what happens to the qualitative behavior of a dynamical system, if one varies parameters. Therefore it would be nice to have a few general results that under certain conditions homotopic vector fields are topologically equivalent. I am surprised that I could not find anything in the literature addressing this.
Homotopy classes of non-singular vector fields have been studied a lot in the literature, but in dynamical systems singular vector fields play an important role. Since all vector fields are homotopic, only homotopies, which preserve certain properties of the vector fields, have the chance of giving topologically equivalent vector fields. Most importantly from a dynamical point of view, the singular sets should be preserved by the homotopy up to homeomorphism.
For example, a homotopy along structurally stable vector fields gives topologically equivalent vector fields at the endpoints. However, it might be difficult to confirm structural stability for all vector fields along the homotopy. Also, there should be weaker conditions under which homotopic vector fields are topologically equivalent. There is a theorem by Shub proving topological conjugacy for homotopic expanding endomorphisms on a compact manifold, which could be relevant, but I haven't found statement/proof in the vector field setting.

Does anybody know a reference for Shub's theorem in the vector field setting? Does anybody know results about when certain homotopic vector fields are topologically equivalent? If the local behaviour of the homotopy around the singularities is known, are there methods to deduce something about the global behaviour?

REPLY [3 votes]: I don't know much about the general setting you discuss, but a reasonable notion of homotopy between singular vector fields in the local holomorphic setting has been given, and studied, by J.-F. Mattei in a French article
Modules de feuilletages holomorphes singuliers. I. Équisingularité
The idea is to require that the 1-dimensional foliation induced by the integral curves of the vector fields $X$ and $Y$ be embedded in an integrable 2-dimensional foliation (a kind of foliated cobordism) with special assumptions about the deformation of the singularities ("equisingularity"). $X$ and $Y$ will then be topologically conjugate as germs of a vector field. As for having global results, even in the complex analytic realm, I don't know of general theorems. You'll probably need to impose rather restrictive conditions on the deformation of the local type of the singularities, making it barely usable (though it is only a guess). The point is that the generic holomorphic foliation on a compact surface is rather rigid, meaning that topological conjugacy often leads to analytical conjugacy. This is particularly the case for complex foliations of the complex projective plane, as was proved by Y. Il'Yashenko
Topology of phase portraits of analytic differential equations on a complex projective plane
(I'm afraid this link only points a Russian text, yet a translation of this paper exists somewhere, e.g. on some shelf in my out-of-reach-for-now office...)<|endoftext|>
TITLE: Cohomology Ring of the Flag Manifolds, Cartan Subalgebras, and Weyl Groups
QUESTION [6 upvotes]: I've recently read the following line in an interesting paper:

It is well-known that the cohomology ring of a flag variety $G/B$ is isomorphic to the quotient ring of the ring of polynomial functions on the Cartan sub algebra $\frak{h}$ by the ideal generated by the fundamental invariants $f_1 , . . . , f_r, r = $rank$({\frak h})$, of the Weyl group W, i.e.
  $$
H^∗(G/B,{Q}) \simeq {\text Sym}_Q{\frak h}^∗/(f_1, . . . , f_r).
$$

I would like to ask:
(1) Does this result extend to other fields, i.e. the complex and real case?
(2) What is a good understandable reference for learning about this result?

REPLY [5 votes]: (1) Assuming you are referring to the coefficient field for your cohomology theory, then yes, the result immediately extends to $\mathbb{R}$ and $\mathbb{C}$ coefficients.
(2) Two sources are Fulton's Young Tableaux (Chapters 9 and 10) and Manivel's Symmetric Functions, Schubert Polynomials, and Degeneracy Loci (Chapter 3).  Both of these sources assume a basic knowledge of and familiarity with algebraic geometry.
Edit: The original source for the result is a paper of Armand Borel, Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de groupes de Lie compacts, Ann. of Math. (2) 57, (1953), 115–207.<|endoftext|>
TITLE: Who wrote up Banach's thesis?
QUESTION [50 upvotes]: Sometime ago I read somewhere (and I don't remember where it was) that Stefan Banach--a highly creative and great mathematician--did not always write down his ideas.
Allegedly, he did not write his own thesis (but of course, all the mathematics in it came from him). Is that true? And is it known who wrote it then?

REPLY [15 votes]: There is a paper on this topic in pages 1-7 of the September 2021 issue of The Mathematical Intelligencer. The authors are Danuta Ciesielska and Krzystof Ciesielski. If I understand correctly, the aim in their paper is to set the record straight regarding the (infamous) story about the way in which S. Banach obtained his Ph. D.
I am going to share with you the main paragraphs of the Ciesielska - Ciesielski paper below: both the phrases in boldface and the sics are mine.
*** THE STORY ***

"The story goes that Banach could not be bothered with writing a thesis, since he was interested in solving problems not necessarily connected to a possible doctoral dissertation. After some time, the university authorities became impatient. It is said that another university assistant (instructed by Stanisław Ruziewicz) wrote down Banach's theorems and proofs, and those notes were accepted as a superb dissertation. However, an exam was also required, and Banach was unwilling to take it. So one day, Banach was accosted in the corridor by a colleague, who asked him to join him in a meeting with some mathematicians who were visiting the university in order to clarify certain details, since Banach would certainly be able to answer their questions. Banach agreed and eagerly answered the questions, not realizing that he was being examined by a special commission that had arrived from Warsaw for just this purpose. In some sources [11, 19, 20], this event is described only as a possible version of events. Nevertheless, in several (mainly Polish-language) books, it is presented as a fact. There is even a book on the phobias and fears of great Poles that devotes a whole chapter to Banach and this story, claiming to demonstrate that Banach was unable to deal with his own psyche and phobias, although even this story presents Banach simply as someone who did not consider the PhD a very important acquisition."

*** DEBUNKING THE STORY ***

"... good stories aside, the truth about Banach's exam should be known. Nowadays, it is possible to check the facts, since many sources have become more easily available than they were some decades ago. It is enough to look carefully at some dates and university rules to see that the proposed account could not be accurate. Banach moved to Lvov in 1920 to take up his job at the Lvov Polytechnic. On June 24 of that year, he presented his doctoral dissertation to the Philosophy Faculty of Jan Kazimierz University. The time interval of just a couple of months was definitely too short for the university authorities to have become impatient, let alone for someone else to have written a thesis on the basis of Banach's overheard comments. Moreover, in 1920, Banach had already published three research papers. Why  would he be reluctant to write a doctoral dissertation, which would be a requirement for him to keep the job?
Now let's have a closer look at the exam. According to the university rules, a PhD dissertation had to be refereed and accepted, and then two exams--in the candidate's main scientific disciplines (in Banach's case they were mathematics and physics) and in pure philosophy--were to be taken by the candidate. It turns out that the records of Banach's PhD exams have survived (they are reproduced in [22] and [26]), and we may read that Banach passed his PhD examinations in mathematics and physics. The examining board consisted of four scientists: the dean of the faculty, Zygmunt Weyberg, wo was a mineralogist; two mathematicians, Eustachy Żyliński and Hugo Steinhaus; and a physicist, Stanisław Loria. None of them was from Warsaw, and Banach knew all of them.
There is another interesting story [sic] concerning Banach's doctoral dissertation. The referees were Żyliński and Steinhaus. In October 1920, Steinhaus, who was mentoring Banach, wrote to the dean to inquire about the date of Banach's doctoral exam, for it had been four months since Banach had delivered his dissertation. The dean replied that everything was ready for the exam, but they were awaiting the referee's report (one of whom was Steinhaus himself!). Indeed, when the joint report from Steinhaus and Żyliński arrived, the exam took place immediately. Banach had submitted his dissertation on June 24, the report is dated October 30, and the exam in mathematics and physics took place on November 3. Bearing in mind that in 1920, October 30 fell on a Saturday, November 3 was therefore a Wednesday, and November 1 (Monday) is a public holiday in Poland, everything must indeed have been prepared for the exam. Banach passed this exam with a unanimous grade of 'excellent' from all four examiners.
On December 11, 1920, Banach passed the exam in philosophy (the examining board consisted of the two philosophers Kazimierz Twardowski and Mścisław Wartenberg and the dean, Zygmunt Weyberg). Banach had now fulfilled all the requirements for being granted the PhD degree, and in many sources (including a CV signed by Banach; see [19]), 1920 is given as the year of Banach's doctorate. However, the precise rules for obtaining a PhD from Austro-Hungarian times had been retained by Poland after regaining its independence (see [14]). According to those rules, the candidate was allowed to call himself a 'doctor' only after the doctoral conferment ceremony, which in the case of Banach took place on January 22, 1921. The official documents state that the   academician who conferred the degree on Banach was Kazimierz Twardowski. To a mathematician, that is surprising news indeed. Why Twardowski, who was an eminent Polish philosopher? What was his connection to Banach? Could he have been his dissertation advisor? According to the rules then in force, the conferment of a new doctorate had to be celebrated by a professor from the faculty appointed by the dean, and so there is no reason to regard Twardowski as the supervisor of Banach's thesis. By analogy, one might incorrectly claim that Steinhaus's supervisor in Göttingen in 1911 was the German botanist Gustav Albert Peter, who played the same role as Twardowski in Banach's case (for details, see [9]).
It is frequently said that Banach was not a university graduate, so the fact that he obtained a position at the Polytechnic and a university doctorate was exceptional. This is also slightly misleading. According to the rules that were  then in effect in Poland [14], four years of study at the university was enough  for one to be eligible for a PhD, but even that requirement could be relaxed. The professors of a faculty could, at their discretion, allow someone with outstanding achievements to apply for a PhD. Moreover, in those years, there was no precise definition of who counted as a university graduate. Banach had studied at the Lvov Polytechnic for precisely four years, which was enough."

*** A KERNEL OF TRUTH? ***

"Let us dig further in an attempt to discover [a kernel of truth
underneath the gossip about Banach's doctorate].
This is a good place to recall the illustrious figure of Andrzej Turowicz (1904-1989), a mathematician, priest, and monk active mostly in Kraków, but who also spent some time working in Lvov... Turowicz knew many excellent stories, abounding in colorful detail, about mathematics and mathematicians of his time. It was not unusual for participants in various meetings that he attended to ask him to share some of his anecdotes. Whenever Turowicz had himself been a witness of an event, he recounted it with great accuracy, and one could be sure that things had really happened that way, but there were also stories he had heard from others.
On November 17, 1984, the Jagiellonian University Students' Mathematics Society (see [10]) invited several mathematicians to share their memories during a special meeting. Their reminiscences were taped. Turowicz was one of the guests. He contributed the anecdote about Banach's PhD exam, beginning with the words: 'This is a story I heard from Nikodym, and I am repeating it here at Nikodym's responsibility'. Turowicz recounted this event on several occasions and always credited it to Nikodym. The same attribution is also given in [20].
It was Nikodym whose conversation with Banach was accidentally overheard by Steinhaus in Kraków. Later, Nikodym became a prominent mathematician; after World War II he emigrated to the United States...
And it turns out that it was Nikodym who was reluctant to obtain a PhD. He used to ask: 'Will it make me any wiser?' In 1924, Nikodym (aged 35), still without a PhD, and his wife Stanisława (who was also a mathematician) moved from Kraków to Warsaw. Walerian Piotrowski made a very solid investigation concerning PhDs in mathematics at Warsaw University in the interwar period (see [24, 25]). According to [25], Wacław Sierpiński decided to take the matter of Nikodym's PhD exam into his own hands. He invited Nikodym to a café and began to talk with him. After a while, the dean of the department 'accidentally' appeared in the café and joined the conversation, which quickly drifted toward mathematics. More than an hour later, Sierpiński said to Nikodym: 'Congratulations. You have just passed your PhD exam.'
In our opinion, this is the source of the urban legend about Banach's doctorate. We will never know whether Nikodym gave Turowicz a twisted account of his own PhD exam, changing the main protagonist's name in the process, or whether Turowicz missed something. Our view is that the first explanation is more likely."

These are the references to which D. Ciesielska and K. Ciesielski alluded to in those paragraphs:
[9] D. Ciesielska, L. Maligranda, and J. Zwierzyńska. Doktoraty Polaków w Getyndze. Matematyka. Analecta 28:2 (2019), 73-116.
[10] K. Ciesielski. 100th anniversay of the Jagiellonian University Students' Mathematics Society. Math. Intelligencer 17:4 (1995), 42-46.
[11] K. Ciesielski. Lost legends of Lvov 2: Banach's grave. Math. Intelligencer 10:1 (1988), 50-51.
[14] T. Czeżowski (editor). Zbiór ustaw i rozporządzeń o studiach uniwersyteckich oraz innych przepisów ważnych dla studentów uniwersytetu, ze szczególnym uwzględnieniem Uniwersytetu Stefana Batorego w Wilnie. Wilno, 1926.
[19] E. Jakimowicz and A. Miranowicz (editors). Stefan Banach. Remarkable Life, Brilliant Mathematics. Gdańsk University Press, 2010.
[20] R. Kałuża. Through a Reporter's Eyes: The life of Stefan Banach. Birkhäuser, 1996.
[22] L. Maligranda. 100-lecie doctoratu Stefana Banacha. To appear in Wiad. Mat. 52 (2020).
[24] W. Piotrowski. Doktoraty z matematyki i logiki na Uniwersytecie Warszawskim w latach 1915-1939. In Dzieje Matematyki Polskiej II, edited by W. Więsław, pp. 97-131. Instytut Matematyczny Uniwersytetu Wrocławskiego, 2013.
[25] W. Piotrowski. Jeszcze w sprawie biografii Ottona i Stanisławy Nikodymów. Wiad. Mat. 50 (2014), 69-74.
[26] J. Prytuła. Doktoraty matematyki i logiki na Uniwersytecie Jana Kazimierza we Lwowie w latach 1920-1938. In Dzieje Matematyki Polskiej, edited by W. Więsław, pp. 137-161. Instytut Matematyczny Uniwersytetu Wrocławskiego, 2012.<|endoftext|>
TITLE: Any good books on numerical methods for ordinary differential equations? 
QUESTION [12 upvotes]: I need to find some masters-level exercises about numerical methods for solving ODEs. Are there any good references?

REPLY [2 votes]: A.M. Stuart and A.R. Humphries, Numerical Analysis of Dynamical Systems.
Arieh Iserles, A First Course in the Numerical Analysis of Differential Equations.
Desmond J. Higham, Numerical Methods for Ordinary Differential Equations.<|endoftext|>
TITLE: Origin of the banana graph
QUESTION [12 upvotes]: The graph with two vertices and $n > 1$ edges connecting them has been called the "banana graph" in a number of papers. For one example, see "Feynman Motives of Banana Graphs" by Aluffi and Marcoli, Comm. in Number Theory and Physics (2009) 1-57. (The short title of this paper is "Banana Motives", which I find endlessly entertaining.)
Does anyone know who coined the term "banana graph"?

REPLY [4 votes]: I just had lunch with Oliver Schnetz and our conversation broached the topic whether any of us ever coined a name which stuck. He mentioned to be the first one to attach the word "banana" to the banana graph. As a witness serves his unpublished paper Calculation of the $\phi^4$ $6$-loop non-zeta transcendental
from 1999 which deals with "$n$-banana diagrams". Later he suggested the term to Marcolli. She modified it to "banana graph" in her paper with Aluffi from 2009.
Incidentally, Oliver mentioned also that the name sunset diagram in Carlo's answer is a misnomer. It should be sunrise diagram because "the sun never sets on Quantum Field Theory" (D. Broadhurst) (-:<|endoftext|>
TITLE: A natural way of thinking of the definition of an Artin $L$-function?
QUESTION [16 upvotes]: Emil Artin knew that given a finite extension of $L/\mathbb{Q}$, the local factor of the zeta function $\zeta_{L/\mathbb{Q}}$ at the prime $p$ should be $\displaystyle\prod_{\mathfrak{p}|p}\frac{1}{1 - N(\mathfrak{p})^{-s}}$. He also knew that if $L/K$ is a class field then $\displaystyle\prod_{\mathfrak{P}|\mathfrak{p}}\frac{1}{1 - N(\mathfrak{P})^{-s}} = \displaystyle\prod_{\chi}\frac{1}{1 - \chi{(Frob{_\mathfrak{p})}}\cdot N(\mathfrak{p})^{-s}}$ where $\mathfrak{P}$ runs over all primes in $L$ lying above $\mathfrak{p}$ and $\chi$ runs over all characters of $Gal(L/K)$. 
It's natural then to 

Define $L$-series attached to characters on $Gal(L/K)$.
Recognize that the definition makes sense whether or not $L/K$ is a class field.
In light of the fact that characters are $1$-dimensional representations of $Gal(L/K)$, ask whether there's a good definition of the $L$-series attached to a higher dimensional representation of non-abelian $Gal(L/K)$. 

But having come this far, how does one then arrive at the definition of the local factor of an $L$-series attached to a representation $\rho: Gal(L/K) \to GL_{n}(\mathbb{C})$ at a prime $\mathfrak{p}$ unramified in $K$  as 
$\displaystyle \frac{1}{\det(Id - \rho(Frob_\mathfrak{p})N(\mathfrak{p})^{-s})}$
?
To be sure

It specializes to the definition of the $L$-series attached to a character on $Gal(L/K)$.
It's well-defined (independent of which member of the conjugacy class $Frob_\mathfrak{p}$ one chooses).
One has the theorem $\zeta_{L/\mathbb{Q}} = \prod_{\rho} L(\rho, s)$ where $\rho$ ranges over irreducible representations of $Gal(L/\mathbb{Q})$, generalizing the analogous fact for characters on Galois groups of class fields.

And perhaps the three properties listed above are sufficient to uniquely determine the definition. (Maybe one needs more than the above three, I would have to think about it it.) Maybe this is how Artin discovered the definition. This line of thinking is similar to Feynman's heuristic derivation of Heron's formula. But I somehow feel as though this doesn't get at the essence of things. Is there a way of thinking about the definition of an Artin L-series that gives it more of a sense of inevitability and canonicity?
[Reposted from mathstackexchange.]

REPLY [2 votes]: Since this is near the top again, I'll add what seems natural to me. Let $F/\mathbb{Q}$ be an $S_3$ extension, and let $C$ and $K$ be the cubic and quadratic subfields. Then the $\zeta$ functions of these number fields have Euler products.
Let $p \in \mathbb{Z}$ be a prime which is not ramified in $F$. Then the corresponding Euler factor in each of the fields $\mathbb{Q}$, $K$, $C$ and $F$ is dictated by whether the Frobenius element has conjugacy class $e$, $(12)$ or $(123)$. Here is the formula for the Euler factor in each case:
$$\begin{array}{|c|c|c|c|}
\hline
& e & (12) & (123) \\
\hline
\zeta_\mathbb{Q} & (1-p^{-s})^{-1} & (1-p^{-s})^{-1} & (1-p^{-s})^{-1} \\
\hline
\zeta_K & (1-p^{-s})^{-2} & (1-p^{-s})^{-1}(1+p^{-s})^{-1}&(1-p^{-s})^{-2} \\
\hline
\zeta_C & (1-p^{-s})^{-3} & (1-p^{-s})^{-2} (1+p^{-s})^{-1}&(1-p^{-s})^{-1}(1+p^{-s}+p^{-2s})^{-1} \\
\hline
\zeta_F & (1-p^{-s})^{-6} &(1-p^{-s})^{-3}(1+p^{-s})^{-3}& (1-p^{-s})^{-2}(1+p^{-s}+p^{-2s})^{-2}\\
\hline
\end{array}$$
Looking at this table makes it natural to imagine additional rows for $L$-functions $L_1$ and $L_2$ with $\zeta_K = \zeta_{\mathbb{Q}} L_1$,  $\zeta_C = \zeta_{\mathbb{Q}} L_2$ and  $\zeta_F = \zeta_{\mathbb{Q}} L_1 L_2^2$ and Euler factors
$$\begin{array}{|c|c|c|c|}
\hline
& e & (12) & (123) \\
\hline
L_1 & (1-p^{-s})^{-1} &(1+p^{-s})^{-1} &(1-p^{-s})^{-1}  \\
\hline
L_2 &(1-p^{-s})^{-2} &(1-p^{-s})^{-1} (1+p^{-s})^{-1}& (1+p^{-s}+p^{-2s})^{-1}\\
\hline
\end{array}$$
(It especially helps that Hecke and Dirichlet already studied $L_1$.)
We now have three things associated to $S_3$, namely $\zeta_{\mathbb{Q}}$, $L_1$ and $L_2$, and more complicated things are made from them. In particular, $\zeta_F$ has 1 copy of the "trivial" $\zeta_{\mathbb{Q}}$, 1 copy of $L_1$ and $2$ copies of $L_2$. That already screams representation theory to me. Then when I notice that the degree $d$ representations gives degree $d$ polynomials in $p^{-s}$ in the denominator, what could be more natural than a characteristic polynomial?<|endoftext|>
TITLE: Unprovable statements S where the only way to prove S is to assume S
QUESTION [16 upvotes]: Motivation: Incompleteness (and various independence statements) is about unprovable statements. One natural way to make an unprovable statement provable is to assume it as a new axiom. But this feels like cheating, so people often look for "natural" axioms to add that will imply their favorite unprovable statement. The question is whether there are any statements that are unprovable in a theory where essentially the only way to get a theory which proves the statement is to assume it.
Question:
Let $T$ be a theory (e.g. ZFC). Say a statement $S$ is "minimally unprovable" in/over $T$ if $S$ is not provable from $T$ (and neither is its negation), and such that if $R$ is any statement that is provable from $T \cup \{S\}$ but not provable from $T$ alone, then there is a proof in $T$ that $R$ is equivalent to $S$.


Does every (sufficiently powerful?)
  theory have minimally unprovable
  statements?
Are there examples of a(n interesting) theory $T$
  together with a minimally unprovable
  statement $S$?

REPLY [28 votes]: The following is essentially Joel's answer and also essentially the last part of Francois's answer, but its "look and feel" seems different enough to make it worth pointing out.  The main point is that, if $S$ is minimally unprovable over $T$ then $T\cup\{\neg S\}$ is consistent and complete.  Consistency is just your requirement that $S$ is not provable from $T$.  
To establish completeness, suppose $Y$ is a sentence that is not provable from $T\cup\{\neg S\}$; I'll show that $\neg Y$ must be provable from $T\cup\{\neg S\}$.  The assumption means that $(\neg S)\to Y$ isn't provable from $T$, and therefore neither is its contrapositive $(\neg Y)\to S$.  Therefore neither is the (propositionally equivalent) formula $(S\lor\neg Y)\to S$.  Now apply the main clause in the definition of minimally unprovable, with $S\lor\neg Y$ in the role of $R$.  We've just seen that the conclusion of that clause fails, so one of the hypotheses must fail.  The first hypothesis says that $R$ is provable from $T\cup\{S\}$, which is clearly true because of the disjunct $S$ in $R$.  So the second hypothesis must fail; that is, $R$ must be provable from $T$.  But then, by propositional logic, $\neg Y$ is provable from $T\cup\{\neg S\}$, as claimed.
Now if $T$ is recursively axiomatizable and contains enough arithmetic, then $T\cup\{\neg S\}$ has the same properties and, by Rosser's improvement of Goedel's second incompleteness theorem, it cannot be both consistent and complete.  Therefore, $S$ cannot be minimally unprovable over $T$.<|endoftext|>
TITLE: Cycling through the Zeta Garden: Zeta functions for graphs, cycle index polynomials, and determinants
QUESTION [11 upvotes]: Zeta functions abound in mathematics. Audrey Terras describes in Zeta Functions and Chaos three zeta functions--the zeta fct. of a projective non-singular algebraic variety; the Artin-Mazur zeta function; and a special Reulle (aka dynamical systems or Smale) zeta function, the Ihara zeta function for a graph $G$--all can be expressed in the same basic form:
$$\zeta(u)=\exp\left ( \sum_{m\geq 1} \frac{N_mu^m}{m} \right ).$$
For graph zeta functions $\zeta(u,G_n)$ typically $N_m$ is the number of closed walks of $m$ steps (with some qualifications) on the graph $G$ with $n$ vertices and is related to the trace of the power of an edge adjacency matrix. For a vertex adjacency matrix $A_n$,  also $N_m = \operatorname{tr}[A_n^m]$ (e.g., A054878 and A092297). (Edited per draks' comment.)
You can use the general heuristic $O=KPK^{-1}\Leftrightarrow P=K^{-1}OK$ to obtain 
$$\operatorname{tr}(A)=\ln[\operatorname{det}[\exp(A)]] \Leftrightarrow \operatorname{det}(A)=\exp[\operatorname{tr}[\ln(A)]]$$
and then
$$\operatorname{det}(I-uA_n)=\exp[\operatorname{tr}[\ln(I-uA_n)]]=\exp\left( -\sum_{m\geq 1} \frac{\operatorname{tr}(A_n^m)u^m}{m} \right)$$
$$=\exp\left (-\sum_{m\geq 1} \frac{N_mu^m}{m} \right ),$$
so 
$$\zeta(u;G_n)=\frac{1}{\operatorname{det}(I-uA_n)}=\exp\left(\sum_{m\geq 1} \frac{\operatorname{tr}(A_n^m)u^m}{m} \right)=\exp\left(-:\ln(1-ua): \right).$$ where $a^k=a_k=\operatorname{tr}(A_n^k)$ for $k>0$.
This last expression is the umbral form for the exponential generating function for the cycle index polynomials (OEIS-A036039) for the symmetric group (mod signs).
The Appell sequence in MO-Q111165 incorporating the Riemann zeta function reverses the last relation in some sense:
$$\exp\left (-\beta p_{.}(z)\right )=\exp\left [-(z+\gamma)\beta -\sum_{k=2}^{\infty } \frac{\zeta (k)\beta ^k}{k}  \right ]=\exp\left [ :\ln(1-b\beta ) :\right ]$$ where $b^1=b_{1}=(z+\gamma)$ and $b^k=b_k=\zeta(k)$ for $k>1$.
For easy reference:
$$p_{0}(x)=1$$
$$p_{1}(x)=x+\gamma$$
$$p_2(x)=(x+\gamma)^2-\zeta(2)$$
$$p_3(x)=(x+\gamma)^3-3\zeta(2)(x+\gamma)+2\zeta(3)$$
$$p_4(x)=(x+\gamma)^4-6\zeta(2)(x+\gamma)^2+8\zeta(3)(x+\gamma)+3[\zeta^2(2)-2\zeta(4)]$$
These polynomials are the first few cycle index polynomials for the symmetric group. I'd like to relate each $p_n(x)$ to the characteristic polynomial of a matrix with a null main diagonal.
For example, for such a 3x3 matrix the char polynomial is
$$ \sigma^3-(a_{12}a_{21}+a_{13}a_{31}+a_{23}a_{32})\sigma+(a_{12}a_{23}a_{31}+a_{13}a_{32}a_{21}).$$
Picture a triangle with the vertices ($v$) labelled 1 to 3. Make an orbit/cycle/closed loop, or path, traversing the triangle from $v_1$ through $v_2$ and $v_3$ and then to $v_1$. Denote this path of three steps and length three by $a_{12}a_{23}a_{31}$ and assign it the "moment/transition amplitude" of $\zeta(3)$. Likewise, assign the amplitude $\zeta(2)$ to paths of two steps and length one $a_{12}a_{21}$, an amplitude of $\sigma=x+\gamma$ to a self- or null-loop, and so on. This generates $p_3(x)$. 
Similarly, consider a square with labeled vertices and edges between all pairs of vertices. With cycles/orbits/closed paths of opposing circulation considered distinct cycles, the associated 4x4 determinant generates six paths each with four steps and length four, e.g., $a_{12}a_{24}a_{43}a_{31}$, that can be assigned an amplitude of $\zeta(4)$ each and three sets of two paths of two steps and length one, e.g., $a_{13}a_{31}a_{24}a_{42}$, that can be assigned an amplitude of $\zeta^{2}(2)$. The algorithm can be continued to the other terms to generate $p_4(x)$. 
How to prove that the algorithm will work for all $p_n(x)$, i.e., that each $p_n(x)$ can be generated in the above manner from an $n$ by $n$ "adjacency" matrix?
[Nov. 15, 2013 update: Replacing $p_1(x)=x+\gamma$ by $x$ and the $\zeta(n)$ by $1$ gives the characteristic polynomials (mod signs) of the adjacency matrix of the complete n-graph (see A055137).]

REPLY [2 votes]: I think the validity of the algorithm is corroborated by the relation between the trace and determinant of $m$-dimensional square matrices $A$ inherent in the Cayley-Hamilton theorem applied to the characteristic polynomial of $A$ as explained in Wikipedia.
The relation between the $\det A$ and $(\operatorname{tr} A^k)^j$ for $k,j<m+1$ is precisely that given by the cycle index partition polynomials, and the cycle mapping is clearly shown by Mark Dominus in the link in OEIS/A036039. Substitute $\zeta(k)^j$ for $(\operatorname{tr}(A^k))^j$ in the Wikipedia entry, just as above, but how to formally prove the relation between the indices mapping above and the cycle mapping still is a mystery to me.

Edit Oct. 9, 2020
(The Wikipedia article has changed quite a lot.) I've finally written up a draft compiling some old notes of mine on this and related topics and posted it on my blog as "Appells and Roses: Newton, Leibniz, Euler, Riemann and Symmetric Polynomials." In looking for further reading material, I found Qiaochu Yuan's excellent blog post "GILA VI: The cycle index polynomials of the symmetric groups," which elucidates the combinatorial interpretation of the generating function as enumeration of cycles, an instance of Polya's counting theorem.<|endoftext|>
TITLE: Is there an analytical method of solving general square root equations?
QUESTION [6 upvotes]: Equations such as $\sqrt{x+1}+\sqrt{x+2}=x+3$ are easily solvable by squaring both sides. But if we increase an extra square root, like if trying to solve $\sqrt{x+1}+\sqrt{x+2}+\sqrt{x+3}=x+4$ we encounter a problem. Squaring both sides doesn't work now, since the L.H.S. would still end up with three terms involving square roots, rather than only one as for the previous equation.
That is, unless I'm missing something. Is there a way of solving equations such as the following analytically?
$\sqrt{P_1(x)}+\sqrt{P_2(x)}+\cdots+\sqrt{P_n(x)}=Q(x)$ where $n>2$ and $Q(x), P_1(x), \ldots, P_n(x)$ are all polynomials.

REPLY [10 votes]: Let us set $z_j=\sqrt{P_j(x)}$, so that the equation becomes a system of algebraic equations:
$$z_1+\cdots+z_n=Q(x),\qquad P_j(x)=z_j^2\qquad\forall j=1,\ldots,n.$$
Using the resultant, you may eliminate the $z_j$'s. At the end, you obtain a single polynomial equation $R(x)=0$. Now, the difficulty is that $R$ has degree $\ge5$ (except in your solvable example, where it has degree $4$), and unless your data is really exceptional, this equation is not solvable, in the sense of Galois's theory. For instance, if $n=3$, the elimination of $z_3$ gives $(z_1+z_2-Q(x))^2=P_3(x)$. Then the elimination of $z_2$ yields an equation of degree $\ge4$ and the last elimination, of $z_1$ gives an $R$ of degree $\ge8$.<|endoftext|>
TITLE: 3-D continued fractions
QUESTION [5 upvotes]: Which theorems from classical theory of continued fractions have 3-(or multi-) dimesional analogs?
Of cause classical one is a periodicity of Klein polyhedra. Probably there are some more...

REPLY [2 votes]: (1) There is 3-D analog of Vahlen's theorem, see http://link.springer.com/article/10.1007%2Fs11006-006-0018-6?LI=true
(2) 3-D isolation theorems and extemal Davenport forms (see Cassels "An Introduction to the Geometry of Numbers" and Swinnerton-Dyer, "On the product of three homogeneous linear forms" Acta Arith., 1971, 18, 371-385). The key role here play numbers $2\cos\frac{2\pi}7$, $2\cos\frac{4\pi}7$, $2\cos\frac{6\pi}7$ (3-D Golden Ratios).<|endoftext|>
TITLE: Nim game for odd number of stones
QUESTION [10 upvotes]: Consider the classical Nim game with total number of stones being odd. Then the first players wins, of course, what follows from the general description of winning positions. But is there some shorter (independent of full theory) explanation of this fact, maybe with implicit strategy or whatever?

REPLY [3 votes]: This doesn't answer the OP's question, but it occurred to me (while trying to answer the question) that there's a nice strategy-stealing argument in a somewhat artificial setting:  Suppose one pile has more than half of all the stones.  Then the first player has a winning strategy.  
It's actually (I think) a little tricky to show that the usual analysis guarantees a first-player win:  You need to show that the binary number for the dominant pile has a 1 where all the other piles' numbers have a 0.  But here's the strategy-stealing argument:
Suppose there are $n+1$ piles of size $p_0,p_1,\ldots,p_n$ with $p_0 \gt P=p_1+\cdots+p_n$.  The first player's opening move will be to take some number $1\le k\le P+1$ from pile "0."  If the second player had a winning strategy, it couldn't possibly be to take additional stones from pile 0, because the first player could have stolen that move directly.  This leaves the second player with only $P$ possible responses.  A winning strategy, if there were one, would be to assign one of these responses to each of the first player's opening moves.  (There may, in principle, be many many ways of making the assignment, but we need only imagine that the second player has decided in advance, "If the first player does this, I'll do that.")  By the pigeonhole principle, there would be two numbers, $1\le i \lt j \le P$ for which the second player's move would be the same.  There's the opening for the heist:  First take $i$ stones from pile 0, then take $j-i$ on the next move.<|endoftext|>
TITLE: Remove denominators in de Rham cohomology
QUESTION [5 upvotes]: Let $\omega = \mathrm d \eta$ be an exact rational $n$-form on $\Bbb P^n$. 
It may happen that the polar locus of $\eta$ is not included in the polar locus of $\omega$. But is it true that $\omega = \mathrm d \eta_0$, where $\eta_0$ is an $(n-1)$-form which is regular where $\omega$ is?

In other words (and considering the affine case), assume that
  $$ F = \partial_1 G_1 + \dotsb + \partial_n G_n $$
  with $F$ and $G_i$'s rational functions in $x_1,\dotsc,x_n$.
  Is it possible to choose the $G_i$ such that their denominators are a power of the denominator of $F$?


I've been unable to provide a counter example to this question whereas the following formulation seems to indicate that there exists one.
Let $X$ be the open set of $\Bbb P^n$ where $\omega$ is regular, and $Z$ the polar locus, in $X$, of $\eta$, so that $Z$ is a hypersurface of $X$.
If I'm not mistaken, a positive answer to my question is equivalent to the injectivity of the restriction map in the de Rham cohomology $$H^n(X)\to H^n(X\setminus Z).$$
And following Hartshorne's On the De Rham cohomology of algebraic varieties, we have an exact sequence
$$ \dotsb \to H^{n-1}(X\setminus Z) \to H^{n-2}(Z) \to H^n(X)\to H^n(X\setminus Z), $$
so that the injectivity of the last arrow is equivalent to the nullity of the previous one, or the surjectivity of the one before, the Poincaré residue map. Is there any reason for this residue map to be surjective? What if $Z$ is smooth, or is a hyperplane?
For $n=2$ we have to check that
$$0\to H^1(X) \to H^1(X\setminus Z) \to H^0(Z) \to 0$$
is exact. Which seems to be true … although I don't know why.

Any thought will be much appreciated, including on the specific case $n=2$.

REPLY [2 votes]: Since this question is back on the front page, here is another counterexample. Again, it is algebraic but I think it could be made rational in one more variable; it has the nice feature that all the Hodge structures involved are of Tate type.
Let $f(z) = \prod_{i=1}^d (z-z_i)$ with $z_1$, $z_2$, ..., $z_d$ distinct complex numbers. Let $X$ be the affine surface
$$xy = f(z)$$
and let $Z = \{ y=0 \}$. Then $X \setminus Z$ projects isomorphically to $\{ (y,z) : y \neq 0 \}$, and we see that $H^1(X \setminus Z) \cong \mathbb{C}$ and $H^2(X \setminus Z) = 0$. However, $Z$ is the disjoint union of $d$ copies of $\mathbb{C}$. So $H^1(X \setminus Z) \to H^0(Z)$ is not surjective, and $H^2(X) \cong \mathbb{C}^{d-1}$ does not inject into $H^2(X \setminus Z)$.
Concretely, let $\gamma$ be a path from $z_i$ to $z_j$ in $\mathbb{C}$, not passing through another root of $f$. Let 
$$S = \{ (x,y,z) : |x|=|y|=\sqrt{|f(z)|},\ xy=f(z),\ z \in \gamma \}.$$
So $S$ is a $2$-sphere.
On $X$, we have $dx/x + dy/y = f'(z)/f(z) dz$ and thus $(dx \wedge dz)/x = -(dy \wedge dz)/y$, wherever these expressions are defined. Now, $(dx \wedge dz)/x$ is defined except where $x=0$ and $-(dy \wedge dz)/y$ is defined except where $y=0$, and the locus where $x=y=0$ is codimension $2$ in the smooth variety $X$, so $(dx \wedge dz)/x$ extends to a global $2$-form on $X$. For any polynomial $g(z)$, we have $\int_S g(z) (dx/x) \wedge dz = (2 \pi i) \int_{z_i}^{z_j} g(z) dz$. 
So, for most choices of $g$, we have $\int_S g(z) (dx/x) \wedge dz \neq 0$, and $g(z) (dx/x) \wedge dz$ is a closed, non-exact, $2$-form.<|endoftext|>
TITLE: What is the correct notion of Morita Equivalence between topological groupoids
QUESTION [11 upvotes]: Hello,
Morita Equivalences occur in various categories, such as rings, operator algebras, homotopical categories, groupoids, etc. I'd like to know: What is the correct and precise definition of Morita Equivalence and what is the fundamental concept behind it. Here I want to restrict to (possibly étale, i.e. r-discrete) topological groupoids. From the literature, I have collected the following concepts (roughly stated, probably incorrect):
1) Two (topological) groupoids are Morita equiv. iff the associated stacks are isomorphic.
2) Two groupoids are Morita equiv. iff their representation categories (representation of G = groupoid morphism into the frame groupoid associated to a vector bundle over the space of objects of G) are equivalent.
3) Referring to David Roberts: Internal Categories, anafunctors and localisations http://arxiv.org/abs/1101.2363. There is a notion of canonical weak equivalence in the category of groupoids internal to TOP after choosing a certain Grothendieck pretopology. The literature somehow suggests étale surjections (étale=local homeos) as singleton covers in the topology, but I don't know the conceptual reason behind this. Anyway, those weak equivalences satisfy a sort of calculus of right fractions and the localisation can be described as a category of anafunctors. Following http://ncatlab.org/nlab/show/anafunctor, anafunctor isos are spans $G\leftarrow A\rightarrow H$ where both arrows are étale surjections on objects and fully faithful in the internal sense. This gives one possible precise definition of Morita equivalence.
4) In the context of homotopical category theory, it is a span $G\leftarrow A\rightarrow H$ such that both arrows are acyclic fibrations. But what are the weak equivalences and the fibrations? As weak equivalences we can choose the same as in 3 above and as fibrations the étale surjections on objects, then the notion of Morita equiv. would be the same in both cases. The question is, what homotopical structure do those weak equivalences and fibrations satisfy? It is not a category of fibrant objects since not every top. groupoid is fibrant in this sense. Anyway if some sort of factorization lemma is true in this category (like the one in a category of fibrant objects), this would explain the following statement which I have found several times in the literature (between the lines): Every Morita equivalence (span of acyclic fibrations) is equivalent to a single weak equivalence.
So here are a bunch of questions: What is the correct unifying concept of Morita equiv.? What is the explicit definition of Morita equiv. in the case of groupoids internal to TOP? Is it the one in 3,4? Why étale surjections? Is the last statement in 4 correct? What axioms does this homotopical structure satisfy?

REPLY [2 votes]: David C answered the question pretty thoroughly, but let me add a few more details. You mention that you want to restrict to etale groupoids. This is an important distinction. Without some sort of condition on the source and target maps you do not have the connection to topological stacks. And given this condition, it is important that this is compatible with the pretopology $J$ that you are using, in the sense that given a topological groupoid $X$ of the kind you are interested in (etale, or Hurewicz, or proper, or...) and a $J$-cover $U\to X_0$, the induced groupoid with objects $U$ and arrows $U^2\times_{X_0^2} X_1$ needs to also be of the kind you are interested in. For conditions like properness (i.e. $(s,t)$ is a proper map) this is automatic. Otherwise there is a subtle interplay between the sort of maps the source and target are, and the pretopology.
In terms of 4), given a finitely complete site with a subcanonical singleton pretopology, the category of groupoids in it where the source and target maps are covers, forms a category of fibrant objects. The fibrations are the functors $f\colon E\to B$ for which the canonical map
$$
E_1 \to E_0 \times_{f,B_0,s} B_1
$$
is a cover. Then every object is fibrant, because this reduces to the defining condition on the groupoids you are considering. In the case you mention, you are dealing with $Top$ with the pretopology of etale surjections. (This is another reason to restrict to a full sub-2-category of the 2-category of all topological groupoids). Acyclic fibrations are then fully faithful functors where the object component is a cover, and this recovers your characterisation of Morita equivalences in terms of the category of fibrant objects structure.
[Aside:
This generalises a result by Everaert, Kieboom and van der Linden, where they show that given certain conditions on the ambient category, one gets a Quillen model structure on the category of internal groupoids (this extends earlier work by Joyal and Tierney, I believe). Isomorphism in the homotopy category is the same as being equivalent in the weak 2-category of anafunctors. That anafunctors work in the absence of all finite limits is one motivation for me to consider them.
]
Notice you can weaken the condition that the map $t\circ pr_1$ is an etale surjection in David C's definition and still get the same result. It can be a map from any subcanonical singleton pretopology in which open covers are cofinal. Essentially you need to have local sections of $t\circ pr_1$, and you pick the right sort of maps for the application at hand.
I should mention there is a newer version of my paper available from my nlab page, which I hope will soon be finished being rewritten and then updated on the arXiv.<|endoftext|>
TITLE: Right adjoint to pullback functor
QUESTION [8 upvotes]: Hello everyone,
Let $\mathcal{C}$ a category which has the pullbacks and $f$:$X\longrightarrow Y$ a morphism in $\mathcal{C}$ then I know that the functor pullback $f^{\star}$:$\mathcal{C}/Y\longrightarrow\mathcal{C}/X$ has a left adjoint which is just the postcomposition by $f$.
If I add that $\mathcal{C}$ is locally cartesian closed it will be possible to show that it has a right adjoint. But I don't see how this right adjoint behaves on object and morphisms in $\mathcal{C}$.
Thanks for your help.

REPLY [16 votes]: Let's focus on the case where $C = Set$, since this will give the intuition for other cases. 
An object $p: E \to X$ in the category $Set/X$ can be thought of as an $X$-indexed set, where over every $x \in X$ there is a fiber $p^{-1}(x)$. Similarly, a morphism in $Set/X$ from $p: E \to X$ to $q: F \to X$ is a global function $h: E \to F$ which takes fibers to fibers, i.e., is an $X$-indexed family of functions $h_x: p^{-1}(x) \to q^{-1}(x)$. 
Now, for simplicity, take $Y = 1$ to be a 1-element set, where $Set/1 \simeq Set$. The pullback functor $X^\ast: Set \to Set/X$ takes a set $A$ to the $X$-indexed set where $A_x = A$ for all $x$. A morphism from $X^\ast A \to (p: E \to X)$ is thus a family of functions $h_x: A \to p^{-1}(x)$. Such families are in natural bijection with functions 
$$A \to \prod_{x \in X} p^{-1}(x)$$ 
where on the right we take the product of all fibers together. That basically gives you the right adjoint, and suggests the usual notation for this functor $\prod_X$. More formally, the set $\prod_x p^{-1}(x)$ is constructed as the set of sections $s: X \to E$ of $p: E \to X$; categorically it is the equalizer of a pair of functions 
$$Sect(p) \to E^X \stackrel{\to}{\to} X^X$$ 
which you can work out yourself; basically it's the solution set to the equation $p \circ s = 1_X$. The effect on morphisms is similarly described: $\prod_X h = \prod_{x \in X} h_x$; formally, it can be constructed by taking advantage of the universal property of equalizers. 
The situation for the right adjoint to a pullback $f^\ast: Set/Y \to Set/X$ is only slightly more complicated. Intuitively, the right adjoint $\prod_f$ sends an $X$-indexed set $p: E \to X$ to a $Y$-indexed set where for each $y \in Y$, we have 
$$(\prod_f p)_y := \prod_{x \in f^{-1}(y)} p^{-1}(x)$$ 
i.e., don't take the product of all fibers $p^{-1}(x)$, but only over those where $x$ sits over $y$ via the map $f$. Again, this can be constructed more formally by considering $Y$-indexed sets of sections, where we take families of equalizers which implement section equations; here we consider a $Y$-indexed family of diagrams of the form 
$$(f \circ p)^{-1}(y)^{f^{-1}(y)} \stackrel{\to}{\to} f^{-1}(y)^{f^{-1}(y)}$$ 
More compactly, compute the object of sections of $p$ seen as a morphism from $f \circ p$ to $f$ in the category $Set/Y$. 
Once the formal categorical details of that have been squared away, it works the same way for any locally cartesian closed category.<|endoftext|>
TITLE: Is it possible to decide in polynomial time if a poset is a subposet of another which is given ?
QUESTION [6 upvotes]: I am reading some theory on partial orders and I wonder something which perhaps has a simple answer : Given two partial orders $G_1,G_2$ (by their hasse diagrams), is it possible to know in polynomial-time if it exists a injective order-preserving map from $G_1$ to $G_2$ ? (that is to say a function $f : G_1 \rightarrow G_2$ wich is injective and such that $\forall x,y, x< y  \Rightarrow f(x) < f(y)$)
We can easily solve the problem in exponential-time (and it is in NP, of course) but I don't find neither a better algorithm neither literature about this. Is this an already-know problem and do we have something about this ?
Thanks

REPLY [7 votes]: Completing Emil's observation: Take any subgraph isomorphism problem (well known to be NP-complete).  Add a new vertex in the middle of each edge and then orient the new edges outwards from the new vertex.  That is, replace each undirected edge $x-y$ by $x\leftarrow z\rightarrow y$.  I think you get two posets (with two levels) for which the subposet problem is equivalent to the original.  So it is NP-complete.
ADDED: It makes no difference if we define "subgraph" and "subposet" as containing all relations within a given set of points (the "induced subgraph" interpretation).  Just take the smaller graph to be a clique and reduce from the NP-complete CLIQUE problem using the same construction.<|endoftext|>
TITLE: Uncertainty principle (really for Mellin, but never mind that!)
QUESTION [5 upvotes]: Is there a smooth funtion $f:\mathbb{R}\to \mathbb{C}$ such that
(a) $f(x)$ decreases faster than $e^{-e^x}$ when $x\to \infty$,
(b) $\widehat{f}(t)$ decreases faster than $e^{-|t|}$ when $t\to \pm\infty$?
Note there is no restriction on $f(x)$ for $x\to -\infty$ except that it decay fast enough for the Fourier transform $\widehat{f}$ to be well defined. The function $f_\epsilon(x) = e^{\epsilon x - e^x}$, $\epsilon>0$, decays almost a fast as $e^{-e^x}$ for $x\to \infty$ and $\widehat{f_\epsilon}(t)$ decays roughly as $e^{-|t|}$ for $t\to \pm \infty$  -- so I am really asking whether one can defeat $f_\epsilon$ on both the physical and Fourier aspect.
PS. Yes, this does come from a Mellin transform, as the double exponential probably gives away.

REPLY [3 votes]: How about
$$\frac{e^{-e^x}}{1+\epsilon x^2}$$
If you compute the Fourier transform you can shift the contour to height $\pm\pi/2$ to get an $e^{-|t|}$ times something decaying to 1, by Riemann-Lebesgue lemma
Edit.
Or you can look at a shift of your original example: $f(x+\log A)$ to get something on the lines of
$$e^{-A e^x} e^{-B x^2} e^{C x}$$<|endoftext|>
TITLE: Is there any general index theorem for manifold with boundary?
QUESTION [11 upvotes]: My understanding is Atiyah-Patodi-Singer solved the index theorem for manifold with boundary only for certain types of Dirac operators, correct?
There is still no (or no hope to get) uniform theorem for the Dirac operator associated with any Dirac bundle (in the sense of Gromov-Lawson)?

REPLY [11 votes]: The APS theorem works for any  Dirac-type operator; see  e.g. the excellent monograph  by Booss-Wojchiecowski on this topic.
More than four decades  ago, Boutet de Monvel has described a general set-up for dealing  with boundary value  problems that mimicks  the K-theoretic approach to the  index theorem on closed manifolds.   For a  modern presentation  of this point of  view I recommend this  paper by  Melo-Shrohe-Schick  arXiv: 1203.5649   and the references  therein. It  involves some noncommutative  geometry because  the symbols in the Boutet-de-Monvel calculus of elliptic boundary value  problems define elements in the $K$-theory of a noncommutative    $C^*$-algebra. In the case of closed manifolds symbols  of elliptic operators lead to elements in the $K$-theory of a commutative $C^*$-algebra.

REPLY [5 votes]: A general setup for an index theorem on manifolds with boundary, has been developed by the French school in non-commutative geometry, see e.g.
http://www.math.univ-toulouse.fr/~monthube/articles/CMcrascorrection.pdf
and the references therein.<|endoftext|>
TITLE: Ultralimit versus partial limit
QUESTION [12 upvotes]: Let $\omega$ be a nonprincipal ultrafilter on $\mathbb N$.
A standard construction gives an $\omega$-limit, say $x_\omega$, for any bounded sequence $(x_n)$ of real numbers. 
Namely, there is unique real value $x_\omega$ such that
$$\{\,n\in\mathbb N\mid  |x_\omega-x_n|<\varepsilon\,\}\in \omega$$
for any $\varepsilon>0$.
Clearly $x_\omega$ is a partial limit of $x_n$ [i.e., $x_\omega$ is a limit of a subsequence of $(x_n)$].

Question. Is it always possible to choose subsequence $(x_n)$, $n\in J$
  converging to  $x_\omega$ and such that $J\in\omega$?

REPLY [18 votes]: No, only if $\omega$ is a p-point.  
If $(A_k:k\in \mathbb N)$ is a partition of the natural numbers into $\omega$-small sets such that there is no $\omega$-large set meeting each $A_k$ in a finite set, then we can choose a sequence $x_n$ by declaring $x_n:= 1/k$ whenever $n\in A_k$.   On each set $J\in \omega$ the sequence $x_n$ does not converge to 0. 
To find a non-p-point, partition $\mathbb N$ into countably many infinite sets $A_k$, let $I$ be the ideal of sets meeting each $A_k$ only finitely often (except for finitely many $k$), and let $\omega$ be dual to any maximal ideal extending $I$. 
Then each $A_k$ is in $I$, hence null modulo $\omega$.  The partition $(A_k)$ witnesses that $\omega$ is not a p-point. 
Note that ZFC does not prove that there are p-points, but existence of p-points follows from CH (or weaker assumptions).<|endoftext|>
TITLE: Explicit examples of algebraic Hecke characters with infinite image?
QUESTION [11 upvotes]: Jerry Shurman has a lovely set of notes explaining the classical definition of Hecke characters, the idelic definition of Hecke characters, their relationship, and the classification of algebraic Hecke characters for $\mathbb{Q}$ as Dirichlet characters. He also gives a single family of examples of algebraic Hecke characters with infinite order, namely $\displaystyle \chi: \mathbb{Z}[i] \to \mathbb{C}^{\times}$ given by
$\displaystyle \chi(z) = \left(\frac{z}{|z|}\right)^{4n}$ for integers $n$.
It's clear that one has essentially the same family for imaginary quadratic number fields with class number $1$. But what about imaginary quadratic fields with higher class number? I imagine that one has one family analogous to the one above for each ideal class, but I don't know what they should look like...
What do infinite image algebraic Hecke characters for real quadratic fields look like? Because the unit group is infinite, one can't kill the unit group as above, by putting a $4$ in the exponent...

REPLY [23 votes]: The "most obvious" algebraic Hecke characters of a field $K$ are the characters of the ideal class group of $K$, which have trivial infinity-type and trivial conductor. There might be no non-trivial examples (as in the case of Q). But you can get more examples by:

beefing up the conductor (which gets you Dirichlet characters, or more generally characters of ray class groups, but never anything of infinite order)
changing the infinity-type (the restriction of $\chi$ to the connected component of the identity in $(K \otimes \mathbb{R})^\times \subseteq \mathbb{A}_K^\times$).

Over $K=\mathbb{Q}$, the infinity-types are pretty restricted: the only algebraic characters of the group of positive reals are the maps $x \mapsto x^k$, so you just get powers of the "norm" character (the character of the ideles whose restriction to $\mathbb{R}_+$ is the identity, and which sends a uniformizer at a prime $p$ to $1/p$).
Over a number field the game is more subtle. Let's first suppose $K$ is totally real of degree $n$. The infinity-type of a character looks like $z \mapsto z_1^{k_1} \dots z_n^{k_n}$ for integers $k_1, ..., k_n$, where $z_1, ..., z_n$ are the embeddings into $\mathbb{R}$, but there is a constraint that the infinity-type needs to vanish on a finite-index subgroup of the global units, and this forces the vector $k_1, ..., k_n$ to be orthogonal to the lattice in $\mathbb{R}^n$ generated by the vectors
$$ \{ (\log |u_1|, ..., \log |u_n|) : u \in \mathcal{O}_K^\times\}. $$
By Dirichlet's unit theorem this lattice has rank $r_1 + r_2 - 1 = n-1$, so its orthogonal complement has rank at most 1 -- it's spanned by $(1, ..., 1)$. This tells us that every algebraic Hecke character is just a finite-order character times a power of the norm character, which is a bit boring.
For non-totally-real fields the game gets more interesting because there are not so many units. If $K$ is a CM field of even degree $2d$, then the unit group has rank $d-1$, and you can show that the weights of algebraic Hecke characters span a lattice of rank $d + 1$ (spanned by the norm character and characters of the form $x \mapsto \sigma_i(x) / \overline{\sigma_i(x)}$ for each embedding $\sigma_i: K \hookrightarrow \mathbb{C}$). 
If $K$ is not either totally real or CM, things are more interesting still: the lattice spanned by the logs of the units has rank $r_1 + r_2 - 1$, so its orthogonal complement has dimension $1 + r_2$ over $\mathbb{R}$, but you can't find enough integer vectors in the orthogonal complement. For instance, if $K = \mathbb{Q}(\sqrt[3]{2})$ then the only possibility is the norm character, again (it is a fun exercise to check this by hand in this case). This is an instance of a theorem of [edit: Emil Artin and] Andre Weil: for any number field $K$, and any algebraic Groessencharacter of $K$, the infinity-type of the character must factor through the norm map to the maximal CM subfield of $K$ [edit: or to $\mathbb{Q}$ if there is no such subfield].<|endoftext|>
TITLE: Algebraic closure as a fibrant replacement?
QUESTION [18 upvotes]: Emil Artin's construction of the algebraic closure of a field $K$ is as follows.  Let $K_{0} = K$, and inductively let $\{x_f\}$ be a set of indeterminates indexed by the irreducible $f$ in one variable over $K_i$.  Let $\mathfrak{m}_i$ be a maximal ideal of $K_{i}[x_f]$ containing the ideal generated by all $f(x_f)$ for $f$ irreducible (one can prove this ideal is proper).  Define $K_{i+1} = K_i[x_f]/\mathfrak{m}_{i}$.  Clearly, each $K_{i+1}$ is algebraic over $K_i$ (and thus over $K_0$), and every polynomial with coefficients in $K_i$ splits in $K_{i+1}$.  In particular, the union $\bar{K}$ of the $K_i$ is algebraic over $K$ and algebraically closed.
To me, a fan of homotopy theory, this seems an awful lot like a small object argument.  We can express a solution of the polynomial $f$ over the field $K$ as a solution to a lifting problem: we want the map $0:K[x] \to K$ sending $x$ to $0$ to factor through the map $f:K[x] \to K[x]$ sending $x$ to $f(x)$, so we want a lift to a square which has $K \to 0$ on the right, $f:K[x] \to K[x]$ on the left, and $0$ on the top.  (I'd draw this here, but I don't know how.)  To construct $K_1$, we are doing something like taking the pushout of the diagram 
$$\bigotimes_f K[x_f] \stackrel{\otimes f}{\leftarrow} \bigotimes_f K[x_f] \stackrel{0}{\rightarrow} K$$
whose left arrow is a coproduct of the left sides of the aforementioned lifting problems.  The algebraic closure itself is then the colimit of these pushouts, which is doing something like giving a fibrant replacement for $K$.
As you've probably noticed, this naïve outline has quite a few problems.  The pushouts constructed above are not actually the fields $K_i$, but rather become them after a quotient by a maximal ideal.  Also, the lifting problems we are solving differ between the $K_i$: the left sides of the squares that appear are of the form $f:K_i[x] \to K_i[x]$, where $f$ is an irreducible polynomial over $K_i$.  On the other hand, the small object argument is done via a single set of generating cofibrations which appear on the left sides of squares at every stage.  
I'd like to think that, after passing from the category of $K$-algebras to something presumably more complicated, one can construct a model structure in which the algebraic closure of a field appears as its fibrant replacement.  Does anyone know if this is true?

REPLY [10 votes]: There is at least one significant reason why we don't expect algebraic closure to be a fibrant replacement. Fibrant replacements are usually "unique up to contractible choice", so that there should be no indeterminacy about what the fibrant replacement is. On the order hand, an algebraic closure $K \to \bar K$ does not determine the target uniquely because there are typically nontrivial automorphisms that fix $K$.
However, this has a relatively straightforward fix: we can consider the relative case where we have a map $K \to L$ of fields, and consider the algebraic closure of $K$ inside $L$ (alias the integral closure). This is unique. Moreover, we might hope that there's a model structure somewhere so that perhaps the fibrant replacement of $K$ is boring, but the algebraic closure of $K$ inside $L$ is a factorization $K \to K' \to L$ into an acyclic cofibration followed by a fibration (so it's fibrant replacement in the "over $L$" category).

The following describes an attempt to do so. Here are some definitions.

For a ring homomorphism $R \to S$, let $R^c \subset S$ be the set of elements in $S$ which are integral over $R$: they satisfy a monic polynomial with coefficients in $R$. This is a subring of $S$ (the integral closure of $Im(R) \subset S$).
A map $R \to S$ is a weak equivalence if $R^c$ contains all the units of $S$.
A map $R \to S$ is a cofibration if $S$ is a localization of $R^c$.
A map $R \to S$ is a fibration if $R \to S$ is a monomorphism and $Im(R) = R^c$.

Hence:

A map $R \to S$ is an acyclic cofibration if and only if $S = R^c$.
A map $R \to S$ is an acyclic fibration if and only if $R \to R^c$ is an isomorphism and $R^c$ contains the units of $S$.

We'd like to use this to define a (cofibrantly generated) model structure on the category of commutative rings with unit. If it does, then $R \to R^c \hookrightarrow S$ is always a factorization into an acyclic cofibration followed by a fibration, and factorization into a cofibration followed by an acyclic fibration also adds the units of $S$ to $R^c$. In particular, for fields this recovers the integral closure as desired.
To prove that this is a model structure, I'll write down some generating classes and apply the recognition theorem.

Let $J$ be the set containing the map $\Bbb Z[x] \to \Bbb Z$ sending $x$ to zero, together with the maps $\Bbb Z[a_1,\dots,a_n] \to \Bbb Z[a_1,\dots,a_n,y]/(y^n = \sum a_i y^{n-i})$.
Let $I$ be the union of $J$ with the map $\Bbb Z[x] \to \Bbb Z[x^{\pm 1}]$.

Pushouts along the maps in $J$ are either quotients or adjoining roots of monic polynomials, and so the closure of $J$ under pushouts, retracts, and transfinite compositions is the collection of maps $R \to S$ such that $S = R^c$. This is precisely the set of acyclic cofibrations.
Pushouts along the maps in $I$ are quotients, algebraic extensions, or localizations. The closure of $I$ under pushouts, retracts, and transfinite compositions is the collection of maps $R \to S$ such that $S$ is a localization of $R^c$. This is precisely the set of cofibrations.
Maps $R \to S$ with the right lifting property with respect to $J$ have the right lifting property with respect to $\Bbb Z[x] \to \Bbb Z$ (i.e., are monomorphisms) and with respect to adjoining solutions to monic polynomials (i.e. $R \to R^c$ is surjective). These are precisely the fibrations.
Maps $R \to S$ with the right lifting property with respect to $I$ also have the right lifting property with respect to $\Bbb Z[x] \to \Bbb Z[x^{\pm 1}]$, and hence $R$ must contain the units of $S$. These are precisely the acyclic fibrations.
As a result, it almost looks like we get a model structure on the category of commutative rings with the desired properties. However, a critical part of the recognition theorem fails. Our definition of "weak equivalence" does not satisfy 2-out-of-3 for a really stupid reason: any map $R \to 0$ is a weak equivalence.
If we restrict our attention to some class of rings and monomorphisms, we fix the 2-out-of-3 axiom but lose completeness and cocompleteness. If we restrict our attention to the lattice of subrings of a fixed large ring (like a field $L$), we still inherit the lifting and factorization axioms from the category of rings, and we do get a model structure (but it's somewhat less satisfying than if it applied to the category of rings). It seems most promising to instead ask that our weak equivalences also be monomorphisms, but my time is up on this problem for now.<|endoftext|>
TITLE: Algorithm for numerically approximating the Prokhorov metric?
QUESTION [6 upvotes]: Question: What is known about algorithms for numerically computing/approximating the Prokhorov distance between two measures?
Recall that the Prokhorov distance metrizes the topology of weak(-*) convergence of measures on separable metric spaces, and is defined as follows.

Let $\mu_1$, $\mu_2$ be finite measures on a metric space $(X,d)$. The Prokhorov distance $\rho$ between them is,
  $$\rho(\mu_1,\mu_2):=\inf \left\{ \epsilon > 0 : \mu_1(A) \le \mu_2(A^\epsilon)+\epsilon~ \text{ for all } A \in \mathcal{B} \right\},$$
  where $\mathcal{B}$ is the Borel $\sigma$-algebra on $X$ and $A^\epsilon$ is the $\epsilon$-neighborhood of $A$.

Has a constructive/algorithmic approach to the Prokhorov metric been studied in any contexts? How could one go about constructing numerical algorithms to compute it?
Note: Per asked this nearly identical question at math.stackexchange, where it got no answers even after having a bounty for a week. I'm reposting here with his/her permission.

REPLY [2 votes]: Answered Per's post in the special case of distributions on $R$.<|endoftext|>
TITLE: The use of Hall algebras in physics
QUESTION [12 upvotes]: I once read a statement (not memorized precisely) that a certain physics quantity between two states of charge $d_1$ and $d_2$ respectively could be computed by running over the states of charge $d_1+d_2$ which is the extension of the original two states. Therefore we need to consider some Hall algebras on a moduli space.
I couldn't find that literature any more, so I am not sure that this statement is correct. Could anyone help me to make clear this sort of things? Thanks a lot!
My questions are:
1) What is the basic physics setting of this story?
2) Why is this "extension" important?
3) If this is not correct, what is the correct statement/why do physicists care about Hall algebras?

REPLY [4 votes]: Let me add a small comment to Jeff Harvey's nice answer. Some years after the original BPS-algebra papers of Harvey-Moore, there was a paper by Fiol-Marino, hep-th/0006189, which analyzed the algebra of BPS-states for local compactifications where the BPS-states can be described by quivers. They gave a rather explicit realization of Harvey and Moore's "correspondence conjecture", and they also noted that the algebra of BPS-states is essentially that of a Ringel-Hall algebra.<|endoftext|>
TITLE: How does a moduli interpretation give an analytic object an algebraic structure?
QUESTION [5 upvotes]: I remember hearing this in other contexts, but I encountered it again when reading Elkies' paper "Shimura Curve Computations", where on page 10, he says that:
"We now return to the Shimura curves $\mathcal{X}(1),\mathcal{X}^*(1)$ obtained from arithmetic groups $\Gamma = \Gamma(1),\Gamma^*(1)$. These curves also have a modular interpretation that gives them the structure of algebraic curves over $\mathbb{Q}$."
I've always thought that they have the structure of algebraic curves over $\mathbb{Q}$ simply because they're compact riemann surfaces. In what sense does them having a moduli interpretation give them such a structure, and is this structure different from the usual algebraic structure that we know must exist independent of whether or not there exists a moduli interpretation?
Some references would also be appreciated.
thanks,

will

REPLY [6 votes]: Even on the level of sets, the idea that any compact Riemann surface gives rise to an algebraic curve over $\mathbf{Q}$ should feel resoundingly false. There are uncountably many compact Riemann surfaces and only countably many algebraic curves over $\mathbf{Q}$. 
I think you may be confusing one or more of the following statements:
First, although a compact Riemann surface does not necessarily give rise to an algebraic curve over $\mathbf{Q}$, an algebraic curve over $\mathbf{Q}$ does give rise to an algebraic curve over $\mathbf{C}$, simply by extending the base of your curve to $\mathbf{C}$. Then we have an equivalence of categories between Riemann surfaces with analytic maps and algebraic curves over $\mathbf{C}$ with algebraic morphisms. You can prove this with the Riemann Existence theorem, but this is true in much more generality by Serre's GAGA. For the analytic theory I like Rick Miranda's book, but there are lots of potentially great references as it's an extremely classical subject.
Then, which Riemann surfaces give rise to algebraic curves over $\mathbf{Q}$? Well, that's a complicated question, but the start of the answer is Belyi's Theorem:

An algebraic curve $C$ over $\mathbf{C}$ is isomorphic to the base change of an algebraic curve over $\overline{\mathbf{Q}}$ if and only if there exists a finite map $C \to \mathbb{P}^1$ ramified only at 3 points.

You asked for references and at least with this one, Koeck's "Belyi's Theorem revisited" http://arxiv.org/abs/math/0108222 is pretty canonical.
Moving from $\overline{\mathbf{Q}}$ to $\mathbf{Q}$ is an exercise in Galois cohomology, and although you're talking about general algebraic curves, Chapter X and Appendix B of Silverman's Arithmetic of Elliptic Curves are as good as any.
If you want to bypass all of that and just be given an algebraic curve from on high, moduli spaces are great ways to do so! All you have to do is to cook up a functor taking schemes $S$ over $\mathbf{Q}$ to isomorphism classes of certain objects over $S$ and call it a "moduli problem." If the problem is rigid - there are no nonidentity automorphisms of the objects - then by certain general nonsense your moduli problem will be representable - i.e., will give rise to a scheme over $\mathbf{Q}$. If you pick the right problem, you get an algebraic curve over $\mathbf{Q}$.
Now it's worth noting that the moduli problems that Elkies references are not quite rigid. Still they are not so far from being rigid, so we can still get algebraic curves out of them. See Edidin's article for details on this process - http://arxiv.org/abs/math/9805101
Finally I'll make a note about S. Carnahan's comment: There is a little bit of a subtle issue which people are fond of "passing over in silence" - Moduli problems give algebraic curves over $\mathbf{Q}$, which give algebraic curves over $\mathbf{C}$, which give Riemann surfaces. Which Riemann surface? Well the Riemann surfaces that Elkies works with are chosen because over $\mathbf{C}$ they form certain analytic moduli spaces - so if we start off with "the analogue over $\mathbf{Q}$" we ought to get back to that very special quotient of upper half space, right? Well, we do, but there's something to prove here and that's been mentioned in comments on this site in the past - Is there a schemetical construction for modular curves over the rationals?<|endoftext|>
TITLE: (Un)Decidability of the root existence problem for functions with bounded domain
QUESTION [5 upvotes]: The problem whether a real function $f$ has a root or not is undecidable, given that $f$ is from a class of functions including polynomials and the sine function (http://dl.acm.org/citation.cfm?id=321856). Usually, undecidability is proved by using a periodic function like sin to encode integer problems. Is there anything known about undecidability of the root existence problem for some "reasonable" class of functions with bounded domains, such as from a bounded $\Omega\subset\mathbb{R}^m$ to $\mathbb{R}^n$?

REPLY [6 votes]: Suppose there is an algorithm that decides whether a function $f\colon \Omega \to \mathbb{R}$ has a root. Then one can also compute a root of $f$ if $f$ has one.
One can see this using a standard bi-partition argument:
Cover $\Omega$ with finitely many balls of radius $1$. This is possible since the closure of $\Omega$ is compact. Then using the algorithm we can find a ball that contains a root of $f$.
Then we cover this ball with balls of radius $2^{-1}$ and again find a smaller ball which contains a root...
Iterating this process yield a sequence converging to a root of $f$ with rate $2^{-n}$ or in other words a Cauchy-real representation for a root.
Now, finding a root for a function implies Brouwer's fixed point theorem.
To see that let $\Omega$ be bounded and closed and $g\colon \Omega \to \Omega$ continuous. $g$ has a fixed-point at any root of the function $f\colon \Omega \to \mathbb{R}$, $f(x):= \lvert g(x) - x\rvert$.
For Brouwers fixed point theorem it is known that there is no algorithm to find solutions, see for instance Computable counter-examples to the Brouwer fixed-point theorem, Petrus H. Potgieter.
Thus, we can conclude that there is no algorithm which decides whether a function has a root.<|endoftext|>
TITLE: Confusing Point in Proof: Semisimple Automorphism Fixes Torus
QUESTION [6 upvotes]: I am reading a proof on p.51 of Robert Steinberg in his book "Endomorphisms of Algebraic Groups" and I am having a bit of difficulty understanding one point in the proof.
The setting is as follows. Consider an algebraic group $G$ over an algebraically closed field of arbitrary characteristic. We say that an automorphism $\sigma:G\to G$ is semisimple if there exists an embedding $G\hookrightarrow G'$ such that $\sigma$ is realised by conjugation in $G'$ by a semisimple element. The theorem is:

If $G$ is solvable and $\sigma:G\to G$ is a semisimple automorphism then there exists a maximal torus $T\subseteq G$ such that $\sigma(T) = T$.

I shall paraphrase the proof on pp.51-52 of Steinberg. Assume $G$ is connected. We choose at first an arbitrary maximal torus $T\subseteq G$, and let $U$ be the normal subgroup of $G$ of all unipotents; then we have $G = T\ltimes U$. We embed $G$ in a larger group so that $\sigma:G\to G$ is realised by conjugation by a semisimple element $s$. Then $sTs^{-1}$ and $T$ are maximal tori of $G$, so we can write $T = usTs^{-1}u^{-1}$ where $u\in G$ and by our decomposition, we can assume $u\in U$.
It then suffices to show that $us$ is actually conjugate to $s$: i.e. $us = xsx^{-1}$ for some $x\in G$, for then $\sigma$ will fix $x^{-1}Tx$. 
Then Steinberg says: "and on replacing $us$ by its semisimple part we may assume it to be semisimple." After this, he proceeds to show that if $us$ is semisimple, then it indeed is conjugate to $s$, which completes the proof.

Question: why can we assume that $us$ is semisimple? This would follow if the semisimple part could be written as $u's$ where $u' \in U$, but I cannot see why this is true at the moment.

Thanks!

REPLY [9 votes]: Consider yourself fortunate to be having only "a bit of difficulty" in reading Steinberg's concise proofs.    His style typically involves numerous reductions to get to the essential point, along with sentences crammed with dense arguments.   Though he is almost always right on target, it can be frustrating to unpack such proofs (as I learned the hard way in graduate school when giving seminar talks on two of his papers from 1963 and 1965).   
Here you are looking at his 7.6 in the well-filled AMS memoir published in 1968 but mostly developed earlier.   (I'm looking at the typeset version in his collected papers, where the page numbers are different.)    It's helpful to add some notation, which he wisely avoids as far as possible.   The essential situation involves a connected solvable algebraic group $G= T \ltimes U$ with  $T$ any maximal torus and $U$ the unipotent radical.  A basic fact is that all maximal tori of $G$ are conjugate under elements of $U$ (a step to the general conjugacy theorem).   We are given some embedding of $G$ into a possibly larger algebraic group $G'$ and a "semisimple" automorphism $\sigma$ of $G$ coming from conjugation by a semisimple $s \in G'$.  The problem is to show that $\sigma$ stabilizes some maximal torus of $G$.  
Now $T$ itself might not be $\sigma$-stable, but $us T s^{-1} u^{-1} = T$ for some $u \in U$.   Thus $us \in N_{G'}(T)$, where its Jordan decomposition in $G'$ lives.  Note too that $us \in N_{G'}(G)$ since each factor does; thus the Jordan parts of $us$ stabilize $G$ and also $T$.  In particular, it is harmless now to replace $us$ by its semisimple part in $G'$ (i.e., assume it is semisimple).    This is Steinberg's reduction.
After this reduction, the remaining point is to show that $us$ is actually conjugate in $G'$ to $s$.  Moreover, the conjugating element stabilizes $G$ and takes $T$ to a new maximal torus which will be $\sigma$-stable as desired.  Here again the argument has to be reduced, now to the case when $U$ is commutative.   Etc. 
P.S. Steinberg's last steps are also compressed (using the commutativity of $U$ in a devious way), but his reference to page 4-14 in Expose 4 of the 1956-58 Chevalley seminar 
here is again elementary. 
As Steinberg points out, a different proof (which actually uses more of the Chevalley structure theory) was given by David J. Winter in Corollary 4

here.<|endoftext|>
TITLE: Curve through a point on a variety
QUESTION [8 upvotes]: Let $k$ be a not-necessarily algebraically closed field of characteristic zero. Let $X$ be a positive-dimensional projective variety over $k$. Let $x$ be a closed point on $X$. Does there exist a curve over $k$ on $X$ which contains this point?
Variety = geometrically integral quasi-projective $k$-scheme
Curve = $1$-dimensional variety.
What about the special case where $x$ is a $k$-rational point? 
I can blow-up $X$ at $x$ and take the image in $X$ of an effective ample Cartier divisor via this blow-up and reason by induction. But I'm afraid this doesn't give me a geometrically connected curve passing through $x$.

REPLY [8 votes]: Your idea is good. Let $X'\to X$ be the blowup along $x$. Then $X'$ is projective, geometrically integral and of dimension $\dim X>1$. Embed $X'$ in some $\mathbb P^n_k$. 
When $k$ is infinite, by Jouanolou, "Théorèmes de Bertini et applications" (Progress in Maths), Corollaire 6.11 (2)+(3), there exists a hyperplane $H$ such that $H\cap X'$ is geometrically integral. 
When $k$ is finite, the existence of such a hypersurface $H$ is proved in Poonen "Bertini theorem over finite fields", Ann. Math. (2000), Proposition 2.7. 
Now the image of $H\cap X'$ in $X$ is a geometrically integral closed subscheme of $X$ passing throught $x$ of dimension $<\dim X$. By induction we find a geometrically  integral curve in $X$ passing through $x$. 
Edit In fact through any closed finite subset $Z$ of $X$, it passes a geometrically integral curve in $X$. The proof is the same, we just blowup $X$ along $Z$ instead of $x$.<|endoftext|>
TITLE: Does combinatorial formula for the Pontrjagin classes exist?
QUESTION [11 upvotes]: Gelʹfand, I. M. and MacPherson, R. D. "A combinatorial formula for the Pontrjagin classes" Bull. Amer. Math. Soc. (N.S.) 26 (1992), no. 2, 304–309. 
In the above paper the authors claimed a construction of combinatorial formula for the Pontrjagin classes. According to mathscinet review: MR1129313, it seems the construction is not really local. So the existence of such formula is still open?
Also for first Pontrjagin class, do I understand correctly that "Gabrièlov, A. M.; Gelʹfand, I. M.; Losik, M. V. A local combinatorial formula for the first Pontrjagin class" is a well know result? If so is there any English paper gave detailed constructions?

REPLY [5 votes]: That the Gabrielov, Gelfand Losik paper is "well-known" you can see from Mathscinet citation
records.
This journal is translated into English.
Here are some other papers: MR1129313, MR0410758.<|endoftext|>
TITLE: Modern version of an inequality of R. M. Gabriel for contour integrals
QUESTION [10 upvotes]: I am currently reading the 1998 article Dynamics of the Binary Euclidean Algorithm:
Functional Analysis and Operators by Brigitte Vallée, which cites a 1928 article by R. M. Gabriel for the following inequality: if $f$ is a holomorphic function defined in an open ball $U$, $\Gamma$ is a circular contour contained in $U$, $\gamma$ is a convex contour enclosed by $\Gamma$, and $p>0$, then
$$\int_\gamma |f(z)|^p|dz|\leq 2\int_\Gamma |f(z)|^p|dz|.$$
The text suggests that Vallée borrowed this reference from the 1969 PhD thesis of Howard J. Schwartz - which deals with composition operators on Hardy spaces - although I haven't seen this thesis myself. I found Gabriel's original paper somewhat dated in its terminology and presentation, and I am wondering whether a more up-to-date reference exists for this result, or whether it has been subsumed into a more general result which is now relatively well-known. Is anyone able to point me in the direction of a modern version of this inequality, or a textbook which contains this inequality? In Vallée's application and in the one which I am considering, it is sufficient to consider the case in which $\gamma$ is also circular.
Since the result holds for all $p>0$, I wonder whether the key property being used is subharmonicity rather than holomorphicity. In any case, I haven't been able to find this result either in books on Hardy spaces (the context in which it is applied by Vallée, and presumably also Schwartz) or in books on convex analysis; or if I have found it, its modern form is so far removed from Gabriel's statement that I was unable to recognise it.
Anyway, I would be very grateful if someone could direct me to a more modern reference for this result.

REPLY [3 votes]: This has been called "Gabriel's problem" by Ana Granados, who has reviewed it in Michigan Math. J. 46, 461-487 (1999). I think you'll find all the background and recent developments, together with open problems, that you might want in her overview. There are several variations on this inequality, some for holomorphic and some for subharmonic functions. The inequality you give requires a holomorphic function and was proven by Gabriel in 1946.<|endoftext|>
TITLE: When does Pontryagin duality generalize?
QUESTION [11 upvotes]: Let $T$ be a locally compact abelian (LCA) group. For any other LCA group $G$, let 
$\hom(G,T)$ be the set of continuous homomorphisms $G\to T$. With the compact-open 
topology, $\hom(G,T)$ is certainly a topological group, but is not in general locally 
compact, even if $T$ is compact. In any case there is an obvious homomorphism 
$$
  \alpha_G : G \to \hom(\hom(G,T),T) 
$$
sending $x\in G$ to the functional $\chi\mapsto \chi(x)$. 
I have three questions: 

For what LCA groups $T$ is $\hom(G,T)$ locally compact for all locally compact 
      $G$?
For what groups $T$ is $\alpha_G$ always a (topological) isomorphism?
  More generally, is there any full subcategory $\mathcal L$ of the category of LCA 
      groups for which $\hom(G,H)$ is in $\mathcal L$ whenever $G$ and $H$ are in 
      $\mathcal L$?

(Edit: I'm most interested in 1 and 2. If there is some $\mathcal L$ satisfying 3 
where the objects in $\mathcal L$ can be characterized by some interesting topological 
criterion, that would be fantastic. My question is not whether there are "ad-hoc" ways of constructing $\mathcal L$.)

REPLY [12 votes]: I gave an answer to the second question already on math.stackexchange at https://math.stackexchange.com/questions/124379/why-unitary-characters-for-the-dual-group-in-pontryagin-duality-if-g-is-not-co, but I'll put it here too: if $\alpha_G$ is an isomorphism of topological groups for all locally compact abelian groups $G$ then $T$ must be the unit circle. This explains the importance of the unit circle as a target group. It is discussed in Pontryagin's book on topological groups (see Example 72) and the first volume on abstract harmonic analysis by Hewitt and Ross (p. 424).

REPLY [9 votes]: Here's an answer for 1: given a LCA group $T$, $\text{Hom}(G,T)$ is LC for every LCA $G$ iff $T$ is a compact Lie group.
Proof: Clearly this is a sufficient condition, because it holds for connected tori, and is obviously stable under taking closed subgroup.
Conversely suppose the condition is satisfied.
Let $\mathbf{Z}^{(\mathbf{N})}$ denote the discrete free abelian group of countable rank. Then $\text{Hom}(\mathbf{Z}^{(\mathbf{N})},T)=T^{\mathbf{N}}$ is locally compact iff $T$ is compact. 
So $T$ is compact. Let now $D$ be the Pontryagin dual of $T$, so that $D$ is discrete. By the condition and Pontryagin duality, $\text{Hom}(D,G)$ is LC for all LCA $G$.
Consider a discrete abelian group $G_0$ (I'll fix it later). For $F\subset D$ finite subset, define $U_F$ as the set of homomorphisms $D\to G_0$ vanishing on $F$. So the $U_F$ form a basis of clopen neighbourhoods of 0 in $\text{Hom}(D,G_0)$. So one of those, say $U_F$ is compact. If $E$ is the subgroup generated by $F$, this shows that $\text{Hom}(D/E,G_0)$ is compact. Now to specify, let us assume from the beginning we picked the group $G_0=H^{(\mathbf{N})}$, where $H=\mathbf{Q}/\mathbf{Z}$. It is a standard verification that if $D/E$ is nonzero, then $\text{Hom}(D/E,H)$ is not trivial, and an easy consequence is that $\text{Hom}(D/E,G_0)$ is not compact, a contradiction. This shows that $D/E=0$. So $D$ is finitely generated, and thus $T$ is a compact abelian Lie group, as required.
A careful look on the proof shows that otherwise, $\text{Hom}(G,T)$ is not locally compact for the specific choice of $G=\hat{\mathbf{Z}}^{\mathbf{N}}\times\mathbf{Z}^{(\mathbf{N})}$, where $\hat{\mathbf{Z}}=\prod_p\mathbf{Z}_p$ and $\mathbf{Z}_p$ are the $p$-adics.
Edit: the answer for 2. is, as mentioned in another post, that only the circle satisfies the condition. (I initially thought it was trivially true for the 2-torus (or even the trivial group!), while it's trivially false.)
Here's a proof. Taking $G=\mathbf{R}/\mathbf{Z}$, the condition implies that $\text{Hom}(\mathbf{R}/\mathbf{Z},T)$ is nonzero; pick $f$; its kernel is finite and the quotient of $\mathbf{R}/\mathbf{Z}$ by the kernel of $f$ is also isomorphic to the circle, so the image is isomorphic to the circle. Now by Pontryagin duality, whenever $\mathbf{R}/\mathbf{Z}$ stands as a closed subgroup in a LCA group, it stands as a direct factor. So write $T=H\times \mathbf{R}/\mathbf{Z}$. Then for every $G$ we get $\text{Hom}(\text{Hom}(G,T),T)=G\times \text{Hom}(\hat{G},H)\times \widehat{\text{Hom}(G,H)}\times\text{Hom}(\text{Hom}(G,H),H)$. Since by assumption the identity of $G$ into the first factor is an isomorphism for every $G$, it follows that all other 3 factors are zero. Taking $G=\mathbf{Z}$, we deduce $\widehat{\text{Hom}(\mathbf{Z},H)}=0$, so $H=\text{Hom}(\mathbf{Z},H)=0$ by Pontryagin duality, hence $T=\mathbf{R}/\mathbf{Z}$.

REPLY [5 votes]: Piggy-backing on the excellent answer by Yves, I think $\mathbb{R}/\mathbb{Z}$ is the only answer for $T$ in question 2. 
Every compact Lie group $T$ is isomorphic to one of the form $F \times (\mathbb{R}/\mathbb{Z})^n$ for a finite abelian group $F$ and finite $n \geq 0$. A necessary condition for satisfaction of 2. is that the canonical map $\mathbb{Z} \to \hom(T, T)$ is an isomorphism. 
We then have 
$$\hom(T, T) = \hom(T, F \times (\mathbb{R}/\mathbb{Z})^n) \cong \hom(T, \mathbb{R}/\mathbb{Z})^n \times \hom(T, F)$$ 
which contains $\hom((\mathbb{R}/\mathbb{Z})^n, \mathbb{R}/\mathbb{Z})^n \cong \mathbb{Z}^{n^2}$ as an infinite summand (with a complementary summand that is finite). Thus $n = 1$. Next, if $F$ is non-trivial, we have a product decomposition 
$$\hom(T, T) = \hom(\mathbb{R}/\mathbb{Z}, \mathbb{R}/\mathbb{Z}) \times \hom(\mathbb{R}/\mathbb{Z}, F) \times \hom(F, \mathbb{R}/\mathbb{Z}) \times \hom(F, F)$$ 
for which the third and fourth factors are non-trivial, and hence $\mathbb{Z}$ cannot project onto the displayed product. Thus, $F = 0$, which completes the proof. 
Edit: I might as well promote my earlier comment as an answer to question 3, as efficiently reformulated by Yves Cornulier: a suitable full subcategory of LCA-groups that is closed under homs is the category of compactly generated abelian Lie groups. (I make no claim is that this is maximal among subcategories that are hom-closed, however.) 
Edit 2: Actually, looking back on an old $n$-Category Café discussion on related matters, John Baez mentions a paper by Michael Barr which in some sense blows my answer to 3 out of the water, and which I'm sure Daniel Miller would be interested in. Here is the abstract: 

Abstract. Let $\mathcal{G}$ denote the full subcategory of topological abelian groups consisting of the groups that can be embedded algebraically and topologically into a product of locally compact abelian groups. We show that there is a full coreflective subcategory $\mathcal{S}$ of $\mathcal{G}$ that contains all locally compact groups and is $\ast$-autonomous. This means that for all $G$, $H$ in $\mathcal{S}$ there is an “internal hom” $G \multimap H$ whose underlying abelian group is $\hom(G, H)$ and that that makes $\mathcal{S}$ into a closed category with a tensor product whose underlying abelian group is a quotient of the algebraic tensor product. Moreover a perfect duality results if we let $T$ denote the circle group and define $G^\ast = G \multimap T$. This is essentially a new exposition of work originally done jointly with H. Kleisli [Theory Appl. Categories, 8, 54–62].<|endoftext|>
TITLE: The tightest prime zipper
QUESTION [6 upvotes]: Define a prime zipper as an increasing function $f(n)$ mapping $\mathbb{N}$ into $\mathbb{N}$
with the property that, for every $n \ge 1$, there is at least one prime within the
inclusive interval $[ f(n), f(n+1) ]$.
For example, let $f(n)=2^n$. 

   


This is a prime zipper, because Bertrand's Postulate says that, for every $n$, there is a prime $p$
such that $n < p < 2n$.
What is the slowest-growing known or conjectured prime zipper? Is there a polynomial prime zipper?

REPLY [11 votes]: The slowest growing zipper will depend on the size of $p_{n+1}-p_n$ where $p_n$ is the $n^{th}$ prime number.  There are many results regarding the size of the largest prime gap.
Unconditional:  The work of Baker, Harman and Pintz shows that $$p_{n+1}-p_n \ll p_n^{0.525}$$ for some computable constant.   This means that your zipper function may be taken to be $f(n)=Cn^{40/19}$ for some constant $C$.  The $\frac{40}{19}$ appears in the exponent because $\frac{40}{19}=\frac{1}{1-0.525}$.
Conditional:
If we assume the Riemann Hypothesis, then we have $$ p_{n+1}-p_n \ll \sqrt {p_n}\log p_n,$$ and we may take $f(n)=n^2 \log n$.  Assuming Cramer's conjecture, which says that $$p_{n+1}-p_n =O\left((\log p_n)^2\right),$$ would allows us to take $f(n)=Cn(\log n)^2$ for some constant $C$.
Also see this Wikipedia article on prime gaps.
Remark: Note that finding a prime zipper which grows slower than $f(n)=Cn^{40/19}$ would imply better bounds on the largest prime gap, so your question is equivalent to asking what is the largest prime gap.
** Avoid pointless functions such as $f(n)=p_n+1$.<|endoftext|>
TITLE: Equivalent to Oriented knot complement conjecture
QUESTION [6 upvotes]: I would like to see why the following two statements in Kirby's list of problem are equivalent:
Statement 1:
If $K_1$ and $K_2$ are knots in a closed oriented 3-manifold $M$ whose complements are homeomorphic via orientation-preserving homeomorphism then there exists an orientation-preserving homeomorphism of $M$ taking $K_1$ to $K_2$.
Statement 2:
Let $M$ be an oriented 3-manifold with torus boundary.
 If $M(r_1)\cong M(r_2)$ for inequivalent slopes, then the homeomorphism is orientation-reversing.
(Here $M(r_1)$, $M(r_2)$ are the result of the Dehn filling of $M$ with slope $r_1$, $r_2$.)

REPLY [4 votes]: You are right about this equivalence. It is also worth noting why the hypotheses are interesting. These question have now become the cosmetic surgery conjecture. As stated in Ni-Wu (http://arxiv.org/pdf/1009.4720.pdf), the conjecture (Conj 1.1) is as follows:
Suppose K is a knot in a closed oriented three-manifold Y such that Y − K is irreducible and not homeomorphic to the solid torus, then for all pairs of slopes $r_1$,$r_2$. If $Y(r_1)\cong Y(r_2)$. The homeomorphism is orientation reversing.
First, there are plenty of knot complements $Y\cong S^3-K$ that are amphichiral which implies $Y(r_1) \cong Y(r_2)$ for all but 2 slopes (if $\partial Y$ is framed in the standard way these slopes should be 1/0 and 0/1). The figure 8 knot complement is a good example of this.
Finally, there are plenty of knot complements in $Y\cong S^1\times D^2$ that admit two $S^1\times D^2$ fillings and one that even admits 3 fillings. These are called the Berge-Gabai knots.<|endoftext|>
TITLE: Are there any nontrivial abelian categories with only finitely many objects?
QUESTION [35 upvotes]: The title says pretty much what I want. Of course, the abelian categories should contain at least one nonzero object. 
In particular, is there an abelian category containing only one nonzero object? On the one hand, this is equivalent to construct a ring which is the endomorphism of the nonzero object. On the other hand, this is equivalent to construct a special module by Freyd–Mitchell theorem.
This seems silly for that's not what abelian category is invented for, but I really want to know the answer.

REPLY [55 votes]: Take the category of (at most) countable-dimensional vector spaces over your favourite field. Then take the quotient by the Serre subcategory of finite-dimensional vector spaces. (And take a skeletal subcategory so that it strictly has only two objects.)
Then this is an abelian category with only one non-zero object, whose endomorphism ring is the endomorphism ring of a countable-dimensional vector space, localized at the set of endomorphisms with finite-dimensional kernel and cokernel.<|endoftext|>
TITLE: flat metrics on the 2-sphere with conical singularities
QUESTION [7 upvotes]: Consider some flat metric on $S^2$ with a fixed finite number of conical singularities $p_1,\ldots,p_n$.
What is the moduli space of such metrics up to isometry? In particular what is its dimension?

REPLY [3 votes]: Another example: any meromorphic quadratic differential on the Riemann sphere (with at most simple poles) gives such a metric. There the cone angles are always positive integer multiples of $\pi$ . 
These moduli spaces have been extensively studied from many different points of view.<|endoftext|>
TITLE: Meaning of a quote of Doubilet, Rota and Stanley on harmonic analysis and combinatorics
QUESTION [10 upvotes]: The beginning of the paper On the foundations of combinatorial theory. VI. The idea of generating function (1972) says that

Since Laplace discovered the remarkable correspondence between set theoretic operations and operations on formal power series, and put it to use with great success to solve a variety of combinatorial problems, generating functions (and their continuous analogues, namely, characteristic functions) have become an essential probabilistic and combinatorial technique. A unified exposition of their theory, however, is lacking in the literature. This is not surprising, in view of the fact that all too often generating functions have been considered to be simply an application of the current methods of harmonic analysis.

Would someone explain the meaning of the paragraph in detail, please? Especially, what is

"the correspondence",

characteristic functions as continuous analogues of generating functions, and

the "current methods of harmonic analysis"?


Since I'm not familiar with harmonic analysis, I also wonder if "more current" methods in this discipline might aid solving combinatorial problems.

REPLY [3 votes]: These two papers answer your question in the introductory paragraphs:

Norbert Wiener's "The historical background of harmonic analysis"

and

G. Mackey's "Harmonic analysis as the exploitation of symmetry--a historical survey".

The characteristic-function approach still abounds in generating series related to combinatorics in the umbral calculus / Sheffer sequences / finite operator calculus of Rota et al., where one might define the umbral variables as moments of distributions, defined by characteristic functions, and, of course, in quantum field theory and statistical mechanics with their diverse partition functions and cumulant-moment expansion theorems and associated enumerative diagrammatics, incuding Feynman diagrams, (cf. OEIS A036040 and A127671). E.g., the Laplace transform version gives
\begin{gather*}
(b_\cdot)^n = b_n = (-1)^n\left.\frac{d^n}{dt^n}\langle\exp(-tx)\rangle\right|_{t=0} \\
= (-1)^n\left.\frac{d^n}{dt^n} \int_0^\infty \exp(-tx)\operatorname{pdf}(x) dx\right|_{t=0} \\
= \int_0^\infty x^n\operatorname{pdf}(x) dx = \langle x^n\rangle,
\end{gather*}
where the characteristic function for the probability distribution function $\operatorname{pdf}(x)$ is
$$\langle\exp(-tx)\rangle = \int_0^\infty \exp(-tx)\operatorname{pdf}(x) dx .$$
There is an analogous Fourier transform characteristic function
$$\langle\exp(ixt)\rangle.$$
The gaussian distribution and the central limit theorem are key historical focal points in this appoach to probability theory, which is rife with enumerative combinatorics.
More recently, free probability theory employs the Cauchy transform to define characteristic functions for the generating functions of free moments, related to noncrossing partitions, parking functions, random matrices and the Wigner semicircular distribution—the counterpart to the gaussian distribution (cf. "A Simple Introduction to Free Probability Theory and Its Application to Random Matrices" by Xiang-Gen Xia, and A134264.)


(Update 4/1/2021) A walkthrough
First, for the correspondence, see the simple examples in this excerpt ("History of Probability (Part 5) – Laplace (1749-1827)" by Bob Rosenfeld, cf. blogpost by David Giles) of Laplace's use of generating functions (originating with Euler, but a phrase Laplace coined) to encode information on combinatorics ("putting balls in urns" as Rota once put it) and associated probabilities. A more sophisticated example is given by Ulrich in Quora. These generating functions were often generated from difference (recurrence) or differential equations or a mix thereof.
Note that Euler's integral rep for the Euler beta function is in the denominator of the ratio in Problem I of the excerpt and in this case is a convolution of probabilities (some refs state Laplace did not use convolutions). Now read Wiener's survey of the history of harmonic analysis up to the end of the second paragraph of page 59, in which convolution, or Faltung, is stressed. For another perspective on Laplace and probability generating functions, read up to page 551 of Mackey's survey.
Now jump to Loeb's "The world of generating functions and umbral calculus" and read in the section summarizing DR&S:
Thus, Doubilet, Rota, and Stanley have accomplished a tour de force —presenting a unified treatment of all known types of generating functions, and justifying our expectations as to what new types are yet to be found. [Stanley 1986]
“The explanatory paradigm based on incidence algebras is this: connected with each special algebraic operator on a ‘variety’ of generating functions is a family of
partially ordered sets.. . . The fundamental operation of convolution in the incidence
algebra reflect the algebraic operator in question on generating functions. In this way, the particular algebraic operation acquires a combinatorial interpretation.”
Rota had two careers--first in probability theory and associated operators, second in the umbral / finite operator / Sheffer polynomial calculus and associated combinatorics. Di Nardo nicely summarizes the co-evolution of these two passions of Rota in "Symbolic calculus in mathematical statistics: A review."
(Joseph Kung in "Gian Carlo Rota: A biographical memoir"  gives a broader sketch of Rota's interests and contributions, dividing his work into three periods.)
Edit 10/10/21: (Start)
From "Generalized Noncrossing Partitions
and Combinatorics of Coxeter Groups" by Armstrong

It is an observation of Rota from the 1960’s that the
classical convolution of random variables is — in some sense — the same as Möbius inversion on the lattice of set partitions. Speicher proved an analogous result in a
different setting: he showed that by restricting attention to the lattice of noncrossing set partitions, one obtains Voiculescu’s free convolution of random variables.

And from Möbius Inversion for Categories
by Leinster:

In the mid-20th century, Möbius inversion (nLab) was generalized to arbitrary posets. Several people came up with this idea, but it’s usually associated with the name of Gian-Carlo Rota, who demonstrated its importance in enumerative combinatorics.

That the associations among probability calculus and enumerative combinatorics were constantly in Rota's thinking is further supported by a tangential comment in his preface to "Combinatorial Species and Tree-like Structures" by Joyal:

Species are related to generating functions in much the same way as random variables are related to probability distributions.

(End)<|endoftext|>
TITLE: Monodromy group of 1-dimensional families of hyperelliptic curves
QUESTION [5 upvotes]: If $f: \mathcal{C} \rightarrow \mathcal{P}_{2g+2}$ is a general family of hyperelliptic curves (defined over $\mathbb{C}$), we know that the algebraic connected monodromy group 
$Mon^{0}$ of this family is equal to the full symplectic group. Now if we have a 1-parameter family of hyperelliptic curves given for example by the expression 
$y^{2}= (x-h_{1}(t))....(x-h_{2g+2}(t))$ where $t$ is the parameter in $\mathbb{P}^{1}$ and the $h_{i}$ are holomorphic functions in $t$, is the group $Mon^{0}$ of this family also equal to the full symplectic group $Sp_{2g+2}$? if not, how can one compute the monodromy group of such family?

REPLY [3 votes]: When you say "algebraic monodromy group" do you mean the etale fundamental group?  In that case, you're really asking about the image of monodromy in the symplectic group mod p.  At this point there are lots of techniques for showing monodromy is big; for instance, see Chris Hall's paper  "Big symplectic or orthogonal monodromy modulo l"
http://arxiv.org/abs/math/0608718
where he proves, among other things, that a family of the form y^2 = f(x)(x-t) has big monodromy in this sense.  That paper, the paper it cites, and the papers that cite it should give you good ideas about how to approach the computation of mod p monodromy for any particular family of hyperelliptic curves you might encounter.
On the other hand, you might have in mind the topological question about the monodromy of a family of holomorphic curves in the discrete group Sp(2g,Z), which is somewhat more subtle.  It can be the case that the monodromy surjects onto Sp(2g,Z/p) for every p but has infinite index in Sp(2g,Z); this is called the thin case.  Actually, for y^2 = f(x)(x-t) one can prove that the monodromy is big even in this stronger sense; this was first proved in unpublished work of JK Yu, and I give another argument in Example 5 of my expository paper "Superstrong Approximation in Monodromy Groups":
http://arxiv.org/abs/1210.3757
which also has some general discussion of thin and non-thin monodromy groups.
I learned the argument in that paper from Ian Agol as a result of asking an MO question about it:
The image of the point-pushing group in the hyperelliptic representation of the braid group<|endoftext|>
TITLE: Reference request for the theory of heights over function fields
QUESTION [5 upvotes]: I am looking for an article or book where the theory of heights over function fields (in any characteristic) is treated. I am especially interested in Northcott-type statements. For instance, over a function field $K$ over $\bf Q$, say, a subvariety $X$ of ${\bf P}^n_{K}$, which has a dense subset of $K$-points with bounded height, should have a model over (possibly a finite extension of) $\bf Q$. Where can I find the proof of such a statement ? When $K$ is a function field over a finite field, then one can use Hilbert schemes to get Northcott-type finiteness statements but in general, it seems that one should combine the theory of Hilbert schemes with some descent arguments. I would be grateful for any suggestions.

REPLY [5 votes]: This is discussed (using somewhat older language) in Lang's Fundamentals of Diphantine Geometry. See in particular Chapter 3, Section 3, which is called "Heights in Function Fields". There is a theorem of Neron (Theorem 3.6 on page 66) which says that bounded height implies that the associated map has bounded degree. Then one should be able to use the theory Hilbert schemes (or as Lang uses in Chapter 6, the theory of Chow coordinates) to complete the argument. See also the discussion in Chapter 6, Section 5, which is entitled "Points of Bounded Height", where he uses Neron's result to analyze points of bounded height on abelian varieties; it seems that at least parts of the argument should apply generally.<|endoftext|>
TITLE: multiplicative order of 2 mod p
QUESTION [6 upvotes]: Can anyone tell me whether or not it is true that for all odd primes p the multiplicative order of 2 modulo p is strictly less than the multiplicative order of 2 modulo p^2 ? What are some good references regarding this problem ? Thank you

REPLY [10 votes]: It is well-known that there are primes $p$ such that $2^{p-1} \equiv 1$ (mod $p^{2}$), a question which arises in connection with Fermat's Last Theorem. For such a prime $p,$ let $e$ be the smallest positive integer such that $p$ divides $2^{e}-1,$ and write (as we may) $p-1 = ed$ with $d$ an integer. Then we see easily that $2^{p-1}-1 \equiv d(2^{e}-1)$ (mod $p^{2}$). Certainly $d$ is not divisible by $p,$  so we must already have $2^{e} \equiv 1$ (mod $p^{2}$). Hence for such a prime $p,$ the multiplicative order of $2$ (mod $p$) is the same as the multiplicative order of $2$ mod $p^{2}.$ 
I see in the meantime that Francois Brunault has made a comment to similar effect, and that $1093$ is the smallest such prime<|endoftext|>
TITLE: Local fractional derivative that doesn't vanish on differentiable functions
QUESTION [5 upvotes]: Riemann-Liouville fractional derivative is a nonlocal fractional derivative that doesn't vanish in general on differentiable functions. Kolwankar-Gangal fractional derivative is local but vanishes on any differentiable function. Is there some local fractional derivative that doesn't vanish on differentiable functions in general and for which
$$
   D^{\alpha} x^{n \alpha} = \frac{\Gamma(n\alpha+1)}{\Gamma((n-1)\alpha+1)} x^{(n-1)\alpha}
$$
holds for any $x > 0$?

REPLY [6 votes]: By the theorem of Peetre, a linear operator $C^\infty(\mathbb R)\to  C^\infty(\mathbb R)$ which is local (=support non-increasing) is a differential operator (so involves only integer derivatives).<|endoftext|>
TITLE: What conjectures in anabelian geometry are false?
QUESTION [14 upvotes]: Proving things suspected to be true in anabelian geometry is usually very hard. Maybe it is easier to disprove things suspected to be false?
In particular, I am interested in false generalizations of the main conjecture to higher dimensions.

Is there a pair of non-isomorphic varieties, $X$ and $Y$, which one might naively expect to satisfy an anabelian conjecture, but which in fact have the same etale fundamental group?

Obviously I am not interested in simply-connected varieties, abelian varieties, fibrations with the same fundamental group as their base, and other obvious counterexamples. Thus an example must certainly be of dimension $>1$.
Another question is:

Are there guesses about how exactly to reconstruct the geometry of a variety from its etale fundamental group, that might seem true, but are in fact false?

I am thinking about ideas like the Section Conjecture.

REPLY [10 votes]: Mochizuki proved that the anabelian conjecture for hyperbolic curves (the hom-form) only needs the maximal pro-$p$ quotient of the fundamental groups, for some prime $p$, which gives a stronger result than originally envisioned by Grothendieck.
Y. Hoshi (a student of Mochizuki) proved that an analogous generalization of the section conjecture is false:
Existence of nongeometric pro-p Galois sections of hyperbolic curves,
  Publications of the Research Institute for Mathematical Sciences (Publ. Res. Inst. Math. Sci.) 46 (2010), no. 4, 829-848.
You can get this paper from his homepage.<|endoftext|>
TITLE: Morita semi-equivalences
QUESTION [7 upvotes]: Recall that algebras (or linear 1-categories) $A$ and $B$ are Morita equivalent if there exist bimodules $_AM_B$ and $_BN_A$ and isomorphisms $u: {}_A(M \otimes_BN)_A \to {}_AA_A$ and $v: {}_B(N \otimes_AM)_B \to {}_BB_B$ such that $u$ and $v$ satisfy the zig-zag relations.
I'm interested in what can said under the weaker hypothesis that ${}_B(N \otimes_AM)_B \cong {}_BB_B$ but ${}_A(M \otimes_BN)_A$ is not necessarily isomorphic to ${}_AA_A$.
One easy consequence is that the isomorphism classes of $B$-modules inject into the isomorphism classes of $A$-modules.  Can the $A$-modules which do not come from $B$-modules be understood in terms of the bimodule ${}_A(M \otimes_BN)_A$?
In the situation I'm mainly interested in, $A$ and $B$ are *-algebras (actually *-categories) and $_AM_B$ is the same as $_BN_A$ under the usual identification of left and right modules.

REPLY [2 votes]: This was slightly too long to be a comment, but it certainly is not a complete answer.
Since module categories behave very much like abelian groups, I expect you can say a lot by thinking in analogy with the lower-categorical-level situation: you have two abelian groups $A$ and $B$, and homomorphisms $n : A \to B$ and $m: B \to A$, such that $nm = \mathrm{id}_B$. Then you certainly can say that $mn : A\to A$ is a projection onto a direct summand isomorphic to $B$, and $\mathrm{id}_A - mn$ is a projection onto the other direct summand.
Back to module categories, you may not always be able to make sense of the difference ${_A A_A} \ominus ({_A {M\otimes_B N}_A})$.  Two best cases are: perhaps you have a natural inclusion ${_A {M\otimes_B N}_A}\hookrightarrow {_A A_A}$, in which case the difference is the quotient; or perhaps you have a natural projection ${_A A_A} \twoheadrightarrow {_A {M\otimes_B N}_A}$, in which case the difference is the kernel.
More generally, you may be in the situation where the functors $M\otimes_B$ and $N\otimes_A$ are one-sided adjoints, in which case you can use whichever of the unit or counit of the adjunction is appropriate.  Recall also that the cleanest way to take differences is to work in the derived category, in which case the difference is the cone of the (co)unit.  (In the derived category, there are many available differences, one of which is the cone of the zero map, but that almost certainly will not be useful to you.)  There probably will be some conditions on the map between $A$ and $M\otimes_B N$ coming from the request that ${_A A_A} \ominus ({_A {M\otimes_B N}_A})$ is a projection.<|endoftext|>
TITLE: The history of the geometrization of closed surfaces
QUESTION [9 upvotes]: Who first recognized that the torus supports a flat structure? 
Who first characterized the moduli space of flat structures on the torus?
Who first recognized that the closed, orientable genus 2 supports a hyperbolic structure?
Who first thought of a geometrized surface in terms of the property that for any two points $A$ and $B$ there exists neighborhoods $U$ and $V$ of $A$ and $B$, respectively, and an isometry from $U$ to $V$?

[Reposted from Math Stack Exchange.]

REPLY [2 votes]: See this Wikipedia article, and the references therein. Jeremy Gray's article is particularly enlightening.<|endoftext|>
TITLE: How does one go from Chern--Weil to cohomology classes on BGL(n,C)?
QUESTION [7 upvotes]: Let's assume we start with Chern--Weil theory in the following form:

Given a manifold $M$ and a complex vector bundle $V$ over $M$, we can equip $V$ with a $\mathfrak g\mathfrak l_n(\mathbb C)$ connection and from this connection compute a closed differential $2k$-form (from the curvature of the connection) which thus determines an element $c_k(V)\in H^{2k}(M,\mathbb C)$ (by deRham theory).  This value is independent of the connection chosen.  If $f:N\to M$ is a smooth map, then $c_k(f^\ast V)=f^\ast c_k(V)$.

I've often heard that Chern--Weil theory gives cohomology classes $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$.  However, if we take Chern--Weil theory to mean the above boxed summary, then this does not seem obvious to me unless we use the fact that we can choose $B\operatorname{GL}_n(\mathbb C)$ to be a direct limit of manifolds (namely Grassmannians).
My question is whether there is more abstract way of constructing the classes $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$ from Chern--Weil theory, using only the fact that $B\operatorname{GL}_n(\mathbb C)$ is the classifying space of complex vector bundles.  I am fine with assuming that $B\operatorname{GL}_n(\mathbb C)$ is a direct limit of finite CW complexes, but of course, Chern--Weil theory does not obviously define Chern classes for vector bundles over CW complexes.

REPLY [3 votes]: I can think of several versions, besides those that have been mentioned in the earlier answers:

You can in fact construct $B GL_n (\mathbb{C})$ as a manifold, but of course an infinite-dimensional one. Start with a countably dimensional Hilbert space $H$. Look at the Stiefel manifold $V_n (H)$ of linear embeddings $\mathbb{C}^n \to H$. Being an open subset of $H^n $, it is a secound-countable Hilbert manifold. It can be proven directly that $V_n (H)$ is contractible and that the quotient $V_n (H) \to V_n (H)/CL_n (\mathbb{C})$ is a principal bundle. The proof for the second fact is more or less the same as in the finite-dimensional case, the first fact in proven in an Eilenberg-swindly way.

Now second-countable Hilbert manifolds are a particularly simple type of infinite dimensional manifolds. They have smooth partitions of unity, and as a consequence the proof of the de Rham theorem (for example the one given in Bredon' book) can be carried out without any substantial change. 
The theory of connections on principal bundle works in the same way for Hilbert manifolds as base space (if the fibre is a finite-dimensional Lie group). So you get a Chern-Weil homomorphism in the universal case.
If you replace $GL_n (\mathbb{C})$ by any closed subgroup $G$, then $V_n (H) \to V_n (H)/G$ is a Hilbert manifold model for $BG$; and the same arguments as before work.

There exist a simplicial set model for $BG$, classifying $G$-bundles with connection. The set of $p$-simplices is the set of all triples $(P,\pi,\omega)$, where $\pi:P \to \Delta^p$ is a smooth $G$ principal bundle and $\omega$ a connection $1$-form on $P$. To turn it into a set (and to make the simplicial structure precise), you take those $P$ with $P \subset \Delta^p \times \mathbb{R}^{\infty}$ (as a manifold).

By the ordinary Chern-Weil construction, you get a simplicial differential form on this simplicial set. What do I mean by this? Observe that forms on the standard simplices assemble to a simplicial d.g.a: $q \mapsto \mathcal{A}^{\ast} (\Delta^q)$. For a simplicial set $X_{\bullet}$, you look at the set of simplicial set maps $X_{\bullet} \to \mathcal{A}^q (\Delta^{\bullet})$; which is a vector space, and for varying $q$ gives a d.g.a.; which by definition is the simplicial de Rham complex.
There are two things to be proven here: that the simplicial set I described is indeed $BG$ and that the simplicial de Rham complex computes the real cohomology. The second one you find in the book ''Rational homotopy theory'' by Felix, Halperin, Thomas. For the first part, I do not have a reference; this is folklore.<|endoftext|>
TITLE: The equality problem between conjugate group elements
QUESTION [8 upvotes]: The Novikov--Boone Theorem, which is perhaps the archetypal local unsolvability result in group theory, states existence of a finitely presented group whose word problem is recursively unsolvable. Recall that the word problem asks whether there exists an algorithm to decide whether an arbitrary word in generators of the group $G$ is trivial. Closely related is the equality problem, which asks whether there exists an algorithm to distinguish between two given words in generators of the group $G$. Clearly an algorithm solving one of these problems gives rise to an algorithm solving the other, but it's not clear to me what non-existence of an algorithm implies for subsets.

Question: Given a finitely presented group $G$ with unsolvable word problem and a subset $C$ of $G$ closed under conjugation by elements in $G$ (in the case of interest $C$ is a right coset), might it happen that the equality problem is solvable in $C$? Is there a characterization of when this happens?

In a different language, $C$ is a conjugation quandle, and I'm asking whether a conjugation quandle (with its conjugation structure) might be algorithmically better behaved than the group in which it sits (with its product structure). I'm interested also in algorithmic complexity, but first in recursive solvability of the equality problem in particular. Intuitively I can't see why the quandle couldn't computationally be simpler than the group, but I don't know a good non-trivial example when this happens.

REPLY [7 votes]: If C is a right-coset Hg then the equality problem in C is equivalent to the word problem in H.  The word problem in H can be solvable even if the word problem in G is unsolvable, as long as the membership problem for H is unsolvable.
For instance, take a group G' with unsolvable word problem and set $G= G'\times H$ where $H$ is any group with solvable word problem.
By the way, in this case the condition that Hg is conjugation-closed translates into the requirement that g is central in G'.  To construct lots of interesting examples of this form you could use Ould Houcine's paper 'Embeddings in finitely presented groups which preserve the center'.<|endoftext|>
TITLE: A fibrant-objects structure on Top
QUESTION [7 upvotes]: (Sorry for the crossposting, but I'm really interested in this question).
One can define (Paragraph 1.5, page 10) a fibrant-object structure on a suitable cartesian closed category of topological spaces $\bf Top$, called the $\pi_0$-fibrant structure:

A $\pi_0$-equivalence is a map inducing a bijection at the level of $\pi_0$
A $\pi_0$-fibration is a continuous map $p\colon E\to B$ having the RLP with respect to the map $\{0\}\to [0,1]$ including the 0:
[I'm not able to reproduce the diagram, the TeX engine seems not to accept the "array" environment]

Every property defining a fibrant structure can be easily shown in the way you see.
Now I'm interested in extending this. The natural definition for a $\pi_n$-equivalence is a map $A\to B$ inducing isomorphisms $\pi_i(A)\to \pi_i(B)$ for all $0\le i\le n$.

What should a $\pi_n$-fibration be in order to define a fibrant structure $\pi_n\text{-}\bf Top$ for all $n\in\mathbb N$?
What if we "go to the limit" (and can it be done?) $\varinjlim_n \big(\pi_n\text{-}\bf Top\big)$ of these fibrant structures? Do we recover a known fibrant structure, obtained forgetting cofibrations and mutual lifting properties of a suitable model structure, on $\bf Top$?

REPLY [2 votes]: There was a mistake in an earlier version of the paper that you mention. If you define  $\pi_0$-fibrations and $\pi_0$-equivalences the way that you did, they do not give the structure of a category of fibrant objects on $\bf Top$. The reason is that then the acyclic fibrations will not be closed under pullbacks, as they should be by (FW1) loc. cit.
A counterexample is the map $R\to S^1$ of the real line to the circle that gives the universal cover ($t\to \exp(2\pi it)$). This is a $\pi_0$-acyclic fibration. Take the pullback of this map with respect to $*\to S^1$. The pullback is the fiber of the first map at the point, which is $Z$ (integers). The map from $Z$ to the point $*$ is not a $\pi_0$-equivalence.
I remarked this to Otgonbayar Uuye and he fixed this in the new version. The main conclusion he wanted to draw from it, namely, that Schochet fibrations and homotopy equivalences form a category of fibrant objects on $C^*$-algebras, still holds. The point is to use the standard (Quillen) model structure on $\bf Top$.
As for larger $n$, we can begin with the Quillen model structure on $\bf Top$, and take its left Bousfield Localization with respect to the map $S^{ n+1}\to *$. We obtain a new model structure on $\bf Top$ (so restricting to the fibrant objects we get a fibrant objects structure), with the weak equivalences being the $\pi_n$-equivalences.
Note, however, that there are less fibrations in this model structure then in the usual one. In other words, any $\pi_n$-fibration would also be a Serre fibration. For e.g., the fibrant objects in the localized model structure would be (not all spaces, but only) spaces $X$ such that $\pi_i(X)=0$ for $i> n$.<|endoftext|>
TITLE: What is Gödel's pairing function on ordinals?
QUESTION [11 upvotes]: I find many references to Gödel's pairing function on ordinals but I have not found a definition.  What is it?

REPLY [6 votes]: Asaf and Joel have answered the question. Let me add a remark that expands the fact that it helps us prove that $\kappa\times$ and $\kappa$ have the same size. All the claims here can be verified rather easily. Multiplication and exponentiation are in the ordinal sense.
It is customary to write $\Gamma(\alpha,\beta)$ for the order type of the predecessors of $(\alpha,\beta)$ under the order than Asaf denotes $\prec$. For example, $\Gamma(\omega,\omega\cdot2)=\omega^2+\omega$.
An ordinal $\alpha$ is (additively) indecomposable iff $\alpha\gt 0$ and whenever $\beta,\gamma\lt\alpha$, then $\beta+\gamma\lt \alpha$. One can easily check that the indecomposable $\alpha$ are precisely those of the form $\omega^\beta$. Say that $\alpha$ is multiplicatively indecomposable iff $\alpha>0$ and $\beta\gamma\lt \alpha$ whenever  $\beta,\gamma\lt\alpha$. Then $\alpha$ is multiplicatively indecomposable iff it is $1$ or has the form $\omega^{\omega^\beta}$.
Ok, we can now state the remark; unfortunately I would not know who to credit for this observation, I think of it as folklore: An ordinal $\alpha$ is multiplicatively indecomposable iff it is closed under Gödel pairing, that is, $\Gamma(\beta,\gamma)\lt\alpha$ whenever $\beta,\gamma\lt\alpha$. In particular, $\Gamma(\kappa,\kappa)=\kappa$ for any infinite cardinal $\kappa$, which of course implies that $\kappa\times\kappa$ and $\kappa$ have the same size. Also, if $\kappa$ is uncountable, then there are $\kappa$ ordinals $\alpha$ below $\kappa$ such that $\Gamma(\alpha,\alpha)=\alpha$. Of course, all of this works well in $\mathsf{ZF}$ and all the definitions involved are absolute.

I prefer a different approach when verifying that $\kappa\times\kappa$ and $\kappa$ have the same size, one that (again) is absolute and goes through in $\mathsf{ZF}$, but only requires the use of additively indecomposable ordinals: One first checks that there is a (recursive) bijection $h:\omega\times\omega\to\omega$ with $h(0,0)=0$. Then, given ordinals $\alpha,\beta$, use their Cantor's normal form to write them as 
$$ \alpha= \omega^{\alpha_1}n_1 + \omega^{\alpha_2}n_2 + \dots + \omega^{\alpha_k}n_k $$
and
$$ \beta= \omega^{\alpha_1}n'_1 + \omega^{\alpha_2}n'_2 + \dots + \omega^{\alpha_k}n'_k $$
where $\alpha_1 \gt \alpha_2 \gt \dots \gt \alpha_k$ are ordinals, and $n_1,\dots,n_k, n'_1,\dots,n'_k$ are natural numbers. (Note that these representations are not unique, but at least one of $n_i$ and $n_i'$ is non-zero iff $\alpha_i$ appears as an exponent in the canonical form of $\alpha$ or $\beta$).
Now set 
$$ H(\alpha,\beta)=\omega^{\alpha_1}h(n_1,n'_1)+\omega^{\alpha_2}h(n_2,n'_2)+\dots+ \omega^{\alpha_k}h(n_k,n'_k). $$ 
Then $H$ is a bijection between $\alpha\times\alpha$ and $\alpha$ whenever $\alpha$ is indecomposable. And an easy inductive argument, appealing to the explicit proof of Schröder-Bernstein, allows us to use $H$ to argue that there is, provably in $\mathsf{ZF}$, a class function that assigns to each infinite ordinal $\alpha$ a bijection between $\alpha\times\alpha$ and $\alpha$.
(Of course, the existence of this class function can also be argued from $\Gamma$, using that there are $\kappa$ ordinals $\alpha$ below $\kappa$ with $\Gamma(\alpha,\alpha)=\alpha$, but this second approach is somewhat easier.) 
I found this argument a while ago, but then saw that Levy gives essentially the same approach in his textbook on set theory. Again, I am not sure who to credit for this construction, it seems to go back to Gerhard Hessenberg's 1906 book, "Grundbegriffe der Mengenlehre".<|endoftext|>
TITLE: Finitely-affine morphisms; cohomological dimension of schemes
QUESTION [6 upvotes]: Let $f\colon X\to U$ be a morphism of Noetherian schemes such that the scheme $U$ is affine and the scheme $X$ is separated and, e.g., quasi-projective over affine.  Let $U=\bigcup_\alpha U_\alpha$ be an affine open covering of $U$.  Suppose that for each $\alpha$ the full preimage $f^{-1}(U_\alpha)$ can be covered by at most $n$ affine open subschemes.  Is there any bound on the number of affine open subschemes covering $X$?
Of course, the bound is supposed to depend only on the number $n$ and not on the number of open subschemes in the covering $U_\alpha$ (or otherwise it is not interesting).  It is well-known that if $n=1$ then the morphism $f$ is affine and the scheme $X$ is affine.  What happens for $n\ge2$?
One approach to this question would be to notice that one has $R^if_*(\mathcal F)=0$ for any quasi-coherent sheaf $\mathcal F$ on $X$ and all $i\ge n$, hence also $H^i(X,\mathcal F)=0$ for all $i\ge n$.  Is there any way to obtain a bound on the number of open affines covering $X$ from this finiteness of cohomological dimension?
The only reference I was able to find is Hartshorne "Cohomological dimension of algebraic varieties", Annals of Math. 88, 1968 (referred to from Exercise III.4.8 in Hartshorne's "Algebraic geometry"), but this doesn't seem to answer my question.

REPLY [3 votes]: "This is too long to be a comment". 
A way to construct examples $X\to U$ with a given $n$ is to take a (quasi-)projective scheme $X\to U$ whose fibers have dimension $\le n-1$. Indeed, let $s\in U$. There are $n$ hypersurfaces in $X_s$ with empty intersection, so $X_s$ is covered by the $n$ complements of hypersurfaces (more care are needed if $X_s$ is not projective, anyway, the exercise in Hartshorne you refered to is done this way). Now lift these hypersurfaces to hypersurfaces over $O_{U,s}$, then the same arguments show that $X\times_U \mathrm{Spec}(O_{U,s})$ is covered by $n$ affine open subsets. By standard arguments, this will hold above an affine open neighborhood of $s$. So $X\to U$ satisfies your hypothesis. 
Now in a very special case, $X$ is actually covered by $n$ affine open subsets. Namely, if $U$ is the spectrum of a ring of integers or if $U$ is a regular curve over a finite field, then any projective $X\to U$ with fiber dimensions $\le n-1$ admits a finite morphism $\pi: X\to \mathbb P^{n-1}_U$. This is proved independently in a preprint of Chinburg, Moret-Bailly, Pappas, Martin Taylor, and a preprint of Gabber, Lorenzini and myself. As $\mathbb P^{n-1}_U$ is covered by $n$ affine open subsets, the same is true for $X$. 
Edit 
It would be interesting to see whether any projective curve over a Dedekind domain can be covered by two (or more, but absolutely bounded number of) affine open subsets.<|endoftext|>
TITLE: Regularity of the Maxwell equations
QUESTION [14 upvotes]: As is well-known, the Maxwell equations can be phrased vectorially as,
\begin{align}
    \nabla \cdot \mathbf E &= \frac{\rho_f}{\varepsilon}, &\text{Gauss's law,}\\\
    \nabla \cdot \mathbf B &= 0, &\text{No-name law (no monopoles),}\\\
    \nabla \times \mathbf E &= - \partial_t \mathbf B, &\text{Faraday's law,}\\\  
\nabla \times \mathbf B &= \mu \varepsilon\partial_t \mathbf E + \mu \mathbf J_f, &\text{Ampere's law}.
  \end{align}
There are many equivalent formulations, for instance in terms of potentials and Gauges. My question is related to the regularity of the solution pair $(\mathbf E, \mathbf B)$. As the equations are hyperbolic and my knowledge is largely in elliptic equations (which seem to be completely different beasts to handle... I have heard: "Partial differential equations are like a zoo, even if the animals look the same you might have to treat them differently").
Regularity questions:

What are the standard references for the regularity (of the solutions) of the Maxwell equations?
If we have the equations on domains, what is the dependence of the regularity of the solutions in terms of the regularity of the boundary?
Which formulations are most convenient to prove regularity properties for hyperbolic equations? As I have said above, there exist many equivalent ones.
Is there any work done, and what work, on the regularity questions for the Maxwell equations in a functional analytic framework? Here I mean phrasing the equations as a ordinary differential equations in a Banach space, just as we would have the analysis of the heat kernel as a convolution-type operator for the heat equation. How about harmonic analysis?
Has there been any work done on the Maxwell equations in terms of gradient flows on metric spaces (as in the work of Felix Otto et al., for the Fokker-Planck equation, sorry, the Ornstein-Uhlenbeck process)?

Before the question gets closed before it is "overly broad, rhetoric or whatever", please note that my question is mainly about the regularity for Maxwell equations and if one of the other questions can get answered or get pointed to a reference in the process, that would be nice. My background in PDE is mainly from the elliptic side, I do not have much knowledge about their hyperbolic ones, other than the trivial results.

REPLY [19 votes]: There are a few, not many, books on hyperbolic equations. You might have a look to that of S. Benzoni-Gavage and myself: Multi-dimensional hyperbolic partial differential equations. First order systems and applications, Oxford Mathematical Monographs, Oxford University Press (2007).
A basic fact of hyperbolic systems of PDEs is that the Cauchy problem is well-posed in both directions of time. Therefore the regularity of the solution cannot be improved as time increase, contrary to the parabolic case. Also, this implies that such systems cannot be recast as gradient flows; instead, some of them can be reformulated as Hamiltonian system (say, if the semi-group is reversible).
That said, there exists nevertheless some regularity properties. On the one hand, the singularities are polarized. This means that the solution is smooth along non-characteristic directions, and most of (but not all) the solution is smooth even in characteristic directions. Let me take the example of the wave equation
$$\partial_t^2u=\Delta_x u$$
in ${\mathbb R}^{1+d}$. Then the wave-front set is invariant under the bi-characteristic flow 
$$\frac{dx}{dt}=p,\qquad\frac{dp}{dt}=-x.$$
A by-product (which can be proved directly by an integral formula of the solution) is that if the initial data $u(t=0,\cdot)$, $\partial_tu(t=0,\cdot)$ is smooth away from $x=0$, then the solution is smooth away from $|x|=|t|$. However the wave-frontset approach tells you much more.
On another hand, the decay of the initial data at infinity implies some space-time integrability of the solution. These properties are not directly related to hyperbolicity. They are consequences of the dispersion. In the case of the wave equation, this is the fact that the characteristic cone $|x|=|t|$ has not flat part. Such integrability statements are know as Strichartz-like inequalities.
Finally, the ODE point of view is adopted by Klainerman, Machedon, Christodoulou and others, mixed with Strichartz inequalities, to prove the well-posedness of the Cauchy problem for semi-linear hyperbolic systems, like Einstein equations of general relativity.<|endoftext|>
TITLE: Questions about local triviality
QUESTION [5 upvotes]: I have read in the paper of Meigniez "Submersions, fibrations and bundles" 
that a smooth surjective submersion $f: E \rightarrow B$ whose fibers are all diffeomorphic to $\mathbb{R}^{n}$ is locally trivial, i.e. a fiber bundle (corollay 31).
1) Is there a counterexample for $f$ not a submersion, which is not a fiber bundle even topologically?
2) Is there a smooth family of vector spaces all isomorphic to $\mathbb{R}^{n}$ which is not a vector bundle, even topologically?
3) Is there a surjective smooth map $f: E \rightarrow B$, with $E$ and $B$ compact, such that all the fibers are pairwise diffeomorphic, but which is not a fiber bundle, even topologically? (Here of course I do not require any more that the fibers are diffeomorphic to $\mathbb{R}^{n}$)

REPLY [6 votes]: Here is a counter-example to Q1. Consider a Reeb-like foliation on the annulus $E=S^1\times\mathbb R$. Namely fix a point $p\in S^1$, let $I=S^1\setminus \{p\}$ and fix a smooth function $h:I\to\mathbb R$ which tends to $+\infty$ at both ends of the interval $I$. One leaf of the foliation is $\{p\}\times\mathbb R$, and the remaining strip $I\times\mathbb R$ is foliated by graphs of the form $y=h(x)+const$. Note that all leaves are diffeomorphic to $\mathbb R$.
There is a map $f:E\to S^1$ whose fibers are leaves of this foliation. First define $g:I\times\mathbb R\to(-\pi/2,\pi/2)$ by $g(x,y)=\arctan(y-h(x))$. Extend $g$ to the entire annulus $S^1\times\mathbb R$ by setting $g(p,y)=-\pi/2$. Now we have a continuous map $g$ from the annulus onto $[-\pi/2,\pi/2)$ whose fibers are leaves of our foliation. It remains to compose it with a continuous bijection from $[-\pi/2,\pi/2)$ to $S^1$, e.g. $t\mapsto(\cos 2t,\sin 2t)$.
The resulting map $f:E \to S^1 =: B$ is not a fiber bundle as a small neighbohood of a point on $\{p\} \times \mathbb R$ cannot have a connected inersection with a nearby leaf.
Replacing $\arctan$ by a function which converges to its asymptotic values sufficiently fast, one can make the map $f$ smooth (with zero derivatives at $\{p\}\times\mathbb R$).<|endoftext|>
TITLE: Cardinality of $\omega\uparrow^\omega\omega$
QUESTION [7 upvotes]: I was wondering what the cardinality of $\omega\uparrow^\omega\omega$ is, with $\uparrow$ being Knuth's up-arrow notation. I ask this purely out of curiosity; after finding out about set theory I feel like a child with a new toy. I'm not on the same level as everyone else on this site, and everything I've learned about set theory I've learned on Wikipedia. The answer to my question might very well be there, but after much searching and attempts to understand the language there, I'm still not sure. Is this an appropriate question for this site? Thank you.

$|\omega\uparrow^\omega\omega|=?$

REPLY [14 votes]: The Knuth arrow notation is most often defined only on the natural numbers, but the central idea of it
can be easily extended to the ordinals, for example as follows:
$$\alpha\uparrow^0\beta=\alpha\beta$$
 $$\alpha\uparrow^\eta 0=1\qquad\text{for }\eta\geq 1$$
 $$\alpha\uparrow^\eta\beta=\sup_{\eta'\lt\eta}[\alpha\uparrow^{\eta'}\sup_{\beta'\lt\beta}(\alpha\uparrow^\eta\beta')]\qquad\text{otherwise}.$$
This definition simply replaces the use of $n-1$ and $b-1$ in the usual natural number definition of the Knuth arrow with a supremum over all smaller values, which allows the definition to work with limit ordinals, which have no immediate predecessor. On finite and successor values, this definition agrees with the standard formula. 
Please note that there are other natural definitions, depending on how one treats the limit stage. which will not be the same as this one. For example, in the definition above, by $\alpha\beta$ I had intended that one should use the natural product, rather than the common product, because this achieves some nicer properties. But others may want to do things differently, and the resulting functions will differ. Nevertheless, what I say below will apply to all the natural formulations of the arrow. 
Using this definition, one can show by transfinite induction
that if $\alpha, \beta$ and $\eta$ are countable ordinals, then
$\alpha\uparrow^\eta\beta$ is also countable, because by the
induction hypothesis, this will be a countable supremum of
countable ordinals.
In particular, $\omega\uparrow^\omega\omega$ is a very large
countable ordinal, and the anwer to your question is that it has
cardinality $\aleph_0$.<|endoftext|>
TITLE: A weaker concept of graph homomorphism
QUESTION [7 upvotes]: In the category $\mathsf{Graph}$ of simple graphs with graph homomorphisms we'll find the following situation (the big circles indicating objects, labelled by the graphs they enclose, arrows indicating the existence of a homomorphism):

Speaking informally, the "obvious" structural relatedness between the two circle graphs $C_3$ and $C_4$ reduces to its two common subgraphs $P_3$ and $P_4$, with $P_3$ being a subgraph of $P_4$.
But even though there is an "undeniable" structural relatedness between $C_3$ and $C_4$, there is no single graph homomorphism between the two. This in complete contrast to the category $\mathsf{Top}$, where they are even isomorphic. (Instead of this: no arrows between $P_i$ and $C_j$!)
But why is there no graph homomorphism between $C_3$ and $C_4$? There are two interrelated reasons:

If all vertices of a graph were forced to have a self-loop, there would be a homomorphism from $C_4$ to $C_3$, since two adjacent vertices $x,y$ were allowed to be mapped onto the same vertex $f(x)=f(y)$. Furthermore, there would also be a homomorphism from $P_4$ to $P_3$.
If one insists on graphs to be loop-less (as in topological graph theory?), one might instead weaken the definition of a graph homomorphism. Instead of defining $f:G\rightarrow G'$ to be a homomorphism when $(x,y) \in E(G)$ implies $(f(x),f(y)) \in E(G')$, one might define it like this:


$f:G\rightarrow G'$ is a (weak) homomorphism when $(x,y) \in E(G)$ implies  $f(x) = f(y) \vee (f(x),f(y)) \in E(G')$.



My questions are:

In which contexts does this definition of a (weak) homomorphism between simple graphs have drawbacks other than not being standard, e.g. technical ones?
Is there a known or imaginable "problem" that might be easier to handle with weak homomorphisms (other than the missing homomorphism between $C_4$ and $C_3$ which isn't really a problem)?
Might weak homomorphisms be conceptually more appropriate, i.e. catch the "meaning" of structure preserving better?
Where can I find weak homomorphisms in the literature, maybe under another name?

REPLY [6 votes]: What you are describing is sometimes goes under the phrase reflexive vs. irreflexive graphs. What you seem to be describing are two categories:

Irreflexive graphs: the objects are sets with irreflexive symmetric relations, the morphisms are relation-preserving functions on the underlying sets.
Reflexive graphs: the objects are sets with reflexive symmetric relations, the morphisms are relation-preserving functions on the underlying sets.

It is fun to compute cartesian products in both of these and to discover the two well-known kinds of graph products. Graph theorists could benefit from a bit of category theory.
There are actually many variations on categories of graphs. Here are some others.
Graphs as relations
The categories of relations and relation-preserving functions are categories of simple graphs. We can require various additional properties. Symmetric relations give us symmetric graphs, reflexive relations give us graphs in which morphisms are allowed to squish edges, etc.
Directed graphs
This is perhaps the most straightforward example. Let $\mathcal{C}$ be the category with two objects, called $\mathsf{E}$ and $\mathsf{V}$ and two arrows $\mathsf{s}, \mathsf{t} : \mathsf{E} \to \mathsf{V}$. Then the (covariant) presheaf category $\mathbf{Set}^\mathcal{C}$ is just the category of directed graphs. An object of the category can be viewed as a pair of sets $(E, V)$ together with two maps $s, t : E \to V$.
Reflexive directed graphs
We can play with directed graphs. For example, if we add an arrow $\mathsf{r} : \mathsf{V} \to \mathsf{E}$ which is a common right inverse of $\mathsf{s}$ and $\mathsf{t}$, we obtain directed graphs with chosen loops.
Symmetric graphs
Another option is to add an arrow $\mathsf{o} : \mathsf{E} \to \mathsf{E}$ satisfying $\mathsf{t} \circ \mathsf{o} = \mathsf{s}$ and $\mathsf{o} \circ \mathsf{o} = \mathsf{o}$ (yes, I am using different kinds of circles on purposes to confuse you). The presheaf category then corresponds to directed graphs in which each edge $e : a \to b$ has an opposite $o(e) : b \to a$, and "opposite" is an involution. It may happen that a loop is its own opposite.
Graphs as monoid actions
To get more fun out of categories of graphs we can do the following. Let $M$ be the monoid of functions $\lbrace 0, 1 \rbrace \to \lbrace 0, 1 \rbrace$. There are four elements, the identity $1$, the map $t$ which swaps $0$ and $1$, and the two constant maps $0$ and $1$. The category of right $M$-actions can be seen as a category of graphs.
Suppose $(X, m : X \times M \to X)$ is a right action on $X$. This is a graph in the following sense. The vertices are those elements of $X$ that are fixed by the action of $0$, or equivalently by the action of $1$. We do not have edges, but rather "half-edges". A half-edge is an element of $X$ which is not a vertex. Each half edge $e$ has its mate $e \cdot t$, and together you can think of them as a wholesome edge (unordered). A half-edge $e$ is attached to the vertex $e \cdot 0$, while its mate is attached to $e \cdot 1$.
We almost get the usual symmetric (non-simple) graphs, except that there can be degenerate half-edges that are their own mates (and therefore necessarily loop-like).
The relevant reference for this answer is Bill Lawvere's Categories of spaces may not be generalized spaces as exemplified by directed graphs.<|endoftext|>
TITLE: Weakest assumption for pointwise convergence of Fourier series
QUESTION [5 upvotes]: This should be a quick one, but so far books, my brain, and the internet have not produced a clear answer.  Or maybe it's subtle and exposes a weakness in my understanding of FS!
Suppose $f(x)=\sum_{k\in\mathbb{Z}}c_ke^{ikx}$, whereby we mean pointwise convergence.  What properties must $f(x)$ then satisfy?  Clearly continuity is too strong (take for example an appropriately defined square wave).  $L^1[-\pi,\pi]$ seems troublesome as well, since term-by-term integration is not necessarily valid with only pointwise convergence.  
Thanks ahead for any tips!

REPLY [7 votes]: The function must be integrable in a certain sense defined by Denjoy and others. Here is an interesting survey paper on the subject:

One of the problems in the theory of trigonometric series
  $$\frac12a_0+\sum_{n=1}^\infty(a_n\cos nx+b_n\sin nx)\tag{1.1}$$ is that of suitably defining a trigonometric integral with the property that, if the series (1.1) converges everywhere to a function $f(x)$, then $f(x)$ is necessarily integrable and the coefficients, $a_n$ and $b_n$, given in the usual Fourier form. It is well known that a series may converge everywhere to a function which is not Lebesgue summable nor even Denjoy integrable (...) The problem has been solved by Denjoy [4; 5], Verblunsky [14],
  Marcinkiewicz and Zygmund [10], Burkill [1; 2], and James [8]. (...) The solutions are described, mainly in the order in which they were published, in §§2-7 below.<|endoftext|>
TITLE: A weaker version of the ABC conjecture
QUESTION [36 upvotes]: I posted this question over at Stackexchange, where a user informed me that it was probably more appropriate for Mathoverflow.  Here's to hoping that the answer is out there:
The ABC conjecture states that there are a finite number of integer triples (a,b,c) such that $\frac {\log \left( c \right)}{\log \left( \text{rad} \left( abc \right) \right)}>1+\varepsilon $, where $a+b=c$ and $\varepsilon > 0$.
I am however more interested in a weaker version of the ABC conjecture where the following inequality holds true: $\frac {\log \left( c \right)}{\log \left( a \: \text{rad} \left( bc \right) \right)}>1+\varepsilon $.  This weaker conjecture has a number of applications in music theory — specifically concerning temperament theory.  For instance, it establishes a type of intuitive complexity metric on various temperaments, and then lets us bound a finite number of these temperaments underneath a given complexity. (if you are not familiar with temperament theory, you can think of these "temperaments" as z-module homomorphisms from one free abelian group to another of lower rank)
It is easy to see that this conjecture is implied by the ABC conjecture.  However, I was wondering if this weaker version is already proven?  And if not, what is the best approach to a proof that does not rely on ABC?  I'm not very familiar with number theory so I don't know where to start.

REPLY [10 votes]: Just to point out there are infinitely many coprime solutions to
$\frac {\log \left( c \right)}{\log \left( a \: \text{rad} \left( bc \right) \right)} > 1$
Take $a=1$ and $b,c$ consecutive powerful numbers.
If $n,n+1$ are consecutive powerful numbers so are $4n(n+1),4n(n+1)+1$ so the solutions without epsilon are infinite.<|endoftext|>
TITLE: Games that never begin
QUESTION [18 upvotes]: Games that never end play a major role in descriptive set theory.  See for example Kechris' GTM.  
Question: Does there exist a literature concerning games that never begin?
I have in mind two players, Alice and Bob, making alternate selections from ${\Bbb N}$,  their moves indexed by increasing non-positive integers, the game terminating when Bob plays his move 0.
As for payoff sets and strategies, define these as for games that never end, mutatis mutandis.  
One major difference: a pair of strategies, one for Alice, one for Bob no longer determines a unique run of the game, but rather now a set of runs, possibly empty.  Even so one may still say that Alice's strategy beats Bob's if every compatible run of one strategy against the other belongs to the payoff set.
Another major difference involves the set-theoretic size of strategies.  Now Alice and Bob play every move in the light of infinite history.  So size considerations mean that certain familiar arguments, for example non-determined games from the axiom of choice, don't work in any obvious way?
Question: What payoff sets give determined games that never begin?
Edits:  By the cold light of morning, I see that I abused the words "mutatis mutandis."  A la Kechris, Alice's strategy beats Bob's if the the unique run of the game falls in the payoff set.  What I had in mind here was that Alice's strategy beats Bob's
if the run set (consistent with both strategies) is a subset of the payoff set.  Joel David Hamkins' clever answer remains trenchant, only now with the import that Alice always wins by playing a strategy with empty run sets regardless of Bob's strategy.
Joel's Alice needs an infinite memory, but if her strategy at each move consists of increasing her previous move by 1, that also necessarily produces an empty run set regardless of Bob's choice.
Possible fix 1  Alice must play an engaged strategy, a strategy that produces a nonempty run set against at least one strategy of Bob's.
Possible fix 2  The residual game at any turn only has a finite future, so one player has a winning strategy from that point on.  Call a strategy rational if it requires the player to play a winning move in the residual game whenever one exists.  Call a strategy strongly rational if it requires the player to play the least possible winning move whenever one exists.  To avoid easy disengaged strategies that don't even reference the payoff set, insist on rational, or even strongly rational strategies.
Possible fix 3  Combine the previous fixes.

REPLY [4 votes]: A slightly more tangential answer, but one which I hope is still useful: there is a well-known connection between infinite games and infintary logic. In the usual context of games with no ending, determinacy principles can be viewed as versions of De Morgan's Law for certain infinitary sentences: for $\Gamma$ a pointclass, $\Gamma$-Det is the statement that each of the disjunctions $$ \forall x_0\exists x_1\forall x_2\exists x_3 . . . . ((x_0, x_1, x_2, x_3, . . . )\not\in X) \vee \exists x_0\forall x_1\exists x_2\forall x_3 . . . ((x_0, x_1, x_2, x_3, . . . ) \in X)$$ for $X\in\Gamma$ is true. Similarly, games with no beginnings should be connected to the semantics of infinitary sentences with ill-founded strings of quantifiers. In the paper "On languages with non-homogeneous strings of quantifiers" (https://doi.org/10.1007/BF02771553), Saharon Shelah did some work on the behavior of such sentences (his semantics for these sentences is in terms of Skolem functions; it appears to avoid Joel's observation by requiring that a strategy for one player look only at moves made by the other player, but I'm not certain of this - please correct me if I'm wrong!). The main result is that "every linear string of quantifiers can be replaced by a well-ordered sequence of quantifiers," which goes some way towards reducing the study of beginningless games to the study of endless games.
However, it should be noted that non-linear "strings" (posets?) of quantifiers have also been studied (cf. "Dependence Logic"), and I have no idea what happens if we look at branching, ill-founded collections of quantifiers, or if this has been looked at in the past (although I vaguely recall a paper by either Hintikka or Vaananen on the subject, but I can't find it, so maybe it doesn't exist). I also don't know a good game-theoretic interpretation of such collections of quantifiers, but I imagine one would not be too hard to come by.<|endoftext|>
TITLE: What CASes say about the analytic rank of rank 8 elliptic curve '457532830151317a1'
QUESTION [11 upvotes]: For the rank $8$ elliptic curve with a-invariants $(0, 0, 1, -23737, 960366)$
sage 5.3 reports analytic rank $4$ in about 2.4 hours.
Almost sure this a bug, so I am interested what other CAS say on the matter of
the analytic rank. Currently testing pari's ellanalyticrank() but I have the impression
sage is several times faster than pari on this problem.

What CASes say about the analytic rank of '457532830151317a1'?

(It might be a good idea to verify my results, I ran them twice).
Session:
sage: e= elliptic_curves.rank(8)[0]
sage: e.ainvs()
(0, 0, 1, -23737, 960366)
sage: time e.analytic_rank()
4
Time: CPU 8556.68 s, Wall: 8607.92 s
sage: e
Elliptic Curve defined by y^2 + y = x^3 - 23737*x + 960366 over Rational Field
sage: e.gens()
[(-171 : 138 : 1), (-647/4 : 6025/8 : 1), (-159 : 845 : 1), (-158 : 875 : 1), (-142 : 1211 : 1), (-136 : 1293 : 1), (-120 : 1442 : 1), (166/9 : -19648/27 : 1)]
sage: e.cremona_label()
'457532830151317a1'

REPLY [14 votes]: Did you read the Sage's documentation on analytic_rank()?
   Return an integer that is *probably* the analytic rank of this
   elliptic curve.  If leading_coefficient is "True" (only implemented
   for PARI), return a tuple (rank, lead) where lead is the value of
   the first non-zero derivative of the L-function of the elliptic
   curve.

So Sage says that it is probably 4. This does not qualify as a bug. Further on, the documentation says:
   Note: It is an open problem to *prove* that *any* particular elliptic
   curve has analytic rank >= 4.

So either the latter is nonsense (well, I am not a number theorist, I don't know), or Magma just solved an open problem for you... Or maybe it didn't, and it fact no CAS can actually compute it.
EDIT: please see this discussion. Also, slide 22 of the talk by J.Cremona explains that there is currently no way to check for sure that the analytic rank $\geq 4$.<|endoftext|>
TITLE: stackification commutes with finite limits?
QUESTION [9 upvotes]: Suppose we work on the Grothendieck site $\mathcal{C}$ of all schemes in the fpqc topology. If it helps it is also fine with me to work only over affine schemes.
Let us denote the category of stacks over $\mathcal{C}$ by $Stacks(\mathcal{C})$. This is naturally a full subcategory of the category $Pre_{grpds}(\mathcal{C})$ of presheaves in groupoids over $\mathcal{C}$. This is equivalent to formulating stacks as being categories fibered in groupoids, fulfilling effective descent.
Now the inclusion $Stacks(\mathcal{C}) \to Pre_{grpds}(\mathcal{C})$ has a left adjoint called the stackification functor.
It is a classical fact that the sheafification functor on presheaves of sets commutes with finite limits. Is this also true for the mentioned stackification functor?
Precisely, I want to know if and if so, why, the stackification functor commutes with 2-categorical pullbacks, here is the precise setting.
Suppose we are given Hopf-algebroids $P_1,P_2,P$ and view them as presheaves in groupoids.
Suppose we are given two morphisms $P \to P_i$ and we construct the 2-pullback of functors $$Q = pullback(P_1 \to P \leftarrow P_2).$$
In my situation I can show that Q is equivalent to an affine scheme, and I want to conclude that the pullback of the stackified diagram is also equivalent to an affine scheme.
If I knew that stackification commuted with finite limits, this would be ok, since the (co)unit (I always mix them up) of the adjunction is an equivalence, i.e., if $X$ is a stack, and $st(X)$ is its stackification, then the natural map $X\to st(X)$ adjoint to the identity of $st(X)$ is an equivalence of stacks.
Any help and comments are appreciated, also if you could give references concerning these questions, that would be great.

REPLY [9 votes]: The answer is yes, at least for $2$-fiber products. And fortunately there is an excellent reference online: Tag04Y1 in the Stacks Project. I quote:

Lemma 8.4. Let $C$ be a site. Let $f : X \to Y$ and $g : Z \to Y$ be morphisms of fibred categories over $C$. In this case the stackification of the 2-fibre product is the 2-fibre product of the stackifications.<|endoftext|>
TITLE: Compactification of topological spaces
QUESTION [5 upvotes]: Hello,
If we take a localy compact space $X$ and we put $A=C_{b}(X)$ the $\mathbb{R}$-algebra of bounded continous functions on $X$, we have an embeding of topological space 
$$\psi:X\longrightarrow \text{Spec}_{\max}(A) $$
defined by $\psi(x)=I_{x}$, where $I_{x}=\{ f\in A / ~ f(x)=0 \}$, and where $\text{Spec}_{\max}(A)$ is the set of maximal ideal of the ring $A$ with the Zariski topology. Then we define the Stone-Čech compactification of $X$ by $\displaystyle \overline{X}^{sc}=\overline{\psi(X)}$.
We can also define the Alexandroff compactification of $X$ with the same method, we just must to take $A=C_{0}(X)$ the algebra of continous functions with zero limit at infinity.
My question is: Can we define all compactification of $X$ with the same method, in other hand if $\widetilde{X}$ is a compactification of $X$ ($X$ locally compact), then there is a sub-algebra $A$ of the algebra $C_{b}(X)$ such that $\widetilde{X}$ is homeomorphic with $\psi(X)$, where 
$$\psi:X\longrightarrow \text{Spec}_{\max}(A) $$
is defined by $\psi(x)=I_{x}$ the maximal ideal of elements $f\in A$ such that $f(x)=0$.

REPLY [6 votes]: Yes. This is called Constantinescu-Cornea theorem,
Ideale Ränder Riemannscher Flächen, Springer 1963. They deal with 2-dimensional
manifolds only, but the result is true for any locally compact space.
It is stated and proved in this generality in the book
of Marcel Brelot, On topologies and boundaries in potential theory, Springer 1971.<|endoftext|>
TITLE: Differential equations and axiom of choice
QUESTION [15 upvotes]: In the most general context, the Picard-Lindelöf theorem (aka Cauchy-Lipschitz in French) asserts the existence of a maximal solution for $\dot{x}(t) = f(t,x(t))$, i.e. of a solution $x(t)$ defined on a interval $I$ such that there exist no other solution whose restriction  to $I$ coincide with $x$. The usual proofs of this (when $f$ is such that there is no local unicity) use Zorn's lemma, or some other weaker form of choice. But is this result actually not provable in ZF?

REPLY [11 votes]: At least for scalar equations $\dot x(t)=f(t,x(t))$, that is with a nonlinearity $f\in C^0(\Omega,\mathbb{R})$, defined on an open set  $\Omega\subset \mathbb{R}^2$, the Zorn's lemma is not necessary: the order structure of $\mathbb{R}$ allows to select a preferred solution (actually, two)
Any IVP  admits  an upper and a lower solution, whose domains are maximal intervals. For  $(t_0,x_0)\in\Omega$, define, for $t\in\mathbb{R}$
$$x ^ * (t):=\sup\big\{x(t)\, : x\in C^1(\mathrm{co}(t_0,t),\, \mathbb{R}),\, \mathrm{graph}(x)\subset\Omega, x(t_0)=x_0, \dot x(s)=f(s,x(s))  \big\}\, ,$$
(where of course $\sup\emptyset=-\infty$). Define
$\mathrm{dom}(x ^  *)$ to be the connected component of $t_0$ in the set $\{ t: x ^ *(t) \in\mathbb{R} \}$. Then, $x ^ *$ is a solution of the ODE with IVC $x(t_0)=x_0$, maximally defined on the interval $\mathrm{dom}(x ^  *)$.<|endoftext|>
TITLE: Eigenvalues of nonnegative integer matrices
QUESTION [11 upvotes]: Edit
I realized that the key piece of information that I need is question 1, and so I'd like to rephrase this post:

What are the possible eigenvalues of nonnegative integer matrices?

Any answer to this question would be appreciated and checkmarked.

Original Question:
My question is related to this question  about integer nonnegative matrices but goes in a slightly different direction. Like the previous poster, my question comes from solving linear recursions (specifically, computing the discrete modulus of a product finite subdivision rule acting on a grid).
Given $A$ a square matrix with nonnegative integer entries and $v$ a column vector of the same dimension, we can use the Jordan canonical form to get a closed expression for $A^n(v)$. If we sum the entries of $A^n(v)$, we get a function of the form $f(n)=a_1 P_1(n)\lambda_1^n+...+a_kP_k\lambda_k$, where the $\lambda$'s are the distinct eigenvalues and each $P_i$ is a monic polynomial in $n$. 
My question is, what are the possible values for the $a_i$ and the $\lambda_i$? In particular:

Can the $\lambda_i$ be any algebraic integer?  
Can the $a_i$ be non-integers?
(My real question): Given an algebraic integer $\lambda$, can we construct two matrices $A_1$ and $A_2$ such that the ratio of their associated polynomials  $\frac{f_1(n)}{f_2(n)}$ (where, as above, $f_i(n)$ is the sum of the entries of $A_1^n v$) has limit $\lambda$? The limit of such a fraction will be 0 unless the 'largest terms' of each polynomial have the same magnitude; my worry is that it would be impossible to get some algebraic integers in this way because they have Galois conjugates of equal or larger size. But it seems that the column vector $v$ might allow one to 'cancel out' unwanted eigenvalues. Is this possible? Is it even possible to get $\sqrt{3}$ in this way?

Thank you for your help! The first two questions seem like they would be easily answerable by experts in matrix theory, but google searches have led to nothing. I appreciate in advance your help.
An example of $f_1(n)$
Let $A=\left[ \begin{array}{cc} 1 & 2 \\ 0 & 2
\end{array} \right]$. It can be diagonalized as $A=\left[ \begin{array}{cc}
1 & 2
\\ 0 & 1 \end{array} \right]
\left[ \begin{array}{cc} 1 & 0
\\ 0 & 2 \end{array} \right]
\left[ \begin{array}{cc} 1 & -2
\\ 0 & 1 \end{array} \right]$.  Now, let $v = \left[ \begin{array}{c} 2
\\ 3 \end{array} \right] $.
Then $A^{n} v=
\left[
\begin{array}{cc} 1 & 2
\\ 0 & 1 \end{array} \right]
\left[ \begin{array}{cc} 1 & 0
\\ 0 & 2^{n} \end{array} \right]
\left[ \begin{array}{cc} 1 & -2
\\ 0 & 1 \end{array} \right]
\left[ \begin{array}{c} 2
\\ 3 \end{array} \right] =
\left[
\begin{array}{cc} 1 & 2^{n+1} -2
\\ 0 & 2^{n} \end{array} \right]
\left[ \begin{array}{c} 2
\\ 3 \end{array} \right] =
\left[ \begin{array}{c} 3(2^{n+1})-4
\\ 3(2^{n}) \end{array} \right]$.  Adding all the entries of this matrix together, we see that
the growth function $f(n)$ is $3(2^{n+1} +
2^{n})-4=9(2^{n})-4$.

REPLY [26 votes]: I think there is a simpler approach. Let $\alpha$ be any algebraic integer, and let $A$ be an $n \times n$ integer matrix with $\alpha$ as an eigenvalue. Take $B$ to the the direct sum of 2 copies of $A$, ie a $2n \times 2n$ matrix, and let $J$ be the $2n \times 2n$ matrix with all entries $1.$ For large enough positive integers $m,$ $B +mJ$ is a matrix with non-negative integer entries, and still has $\alpha$ as an eigenvalue.
  This is because the $\alpha$ eigenspace of $B$ is (at least) $2$-dimensional, while the $0$-eigenspace of $J$ has codimension $1,$ so there is a nonzero vector $u$ with $Bu = \alpha u$ and $Ju = 0.$ Hence $(B+mJ)u = \alpha u$ for all positive integers $m.$<|endoftext|>
TITLE: An S-lemma for polynomials of degree 4 in three variables
QUESTION [6 upvotes]: Might the following be true:

Let $p,q\in\mathbb{R}[x,y,z]$ be homogeneous polynomials, with $\deg(p)\leq 4$ and $\deg(q)= 2$. Suppose $q(x,y,z)>0$ for some $x,y,z\in\mathbb{R}$. Then the following are equivalent:
1) $q(x,y,z)\geq 0\Rightarrow p(x,y,z)\geq 0$ for all $x,y,z\in \mathbb{R}$ 
2) there are nonnegative homogeneous polynomials $s,t$ with $\deg(s)\leq 4$ and $\deg(t)\leq 2$ such that $p=s+qt$.  

Background:
A) The case of the above conjecture that $\deg(p)=2$ is exactly the S-lemma for polynomials $p,q$ in three variables. 
B) The analogous statement for homogeneous polynomials $p,q$ in two variables is also known to be true.
C) By a theorem of Hilbert, a homogeneous polynomial $p\in\mathbb{R}[x,y,z]$ of degree 4 is nonnegative if and only if it is a sum of three squares of quadratic polynomials.
If the above conjecture holds up then $$\min\{p(x,y)\mid q(x,y)\geq 0\}$$ for inhomogeneous polynomials $p,q$ with $\deg(p)=4$ and $\deg(q)=2$ can be cast as a semidefinite optimization problem.

REPLY [5 votes]: There is the following counterexample:
Set $f=\mathrm{x}_1^3\mathrm{x}_3+\mathrm{x}_1^3\mathrm{x}_2+\mathrm{x}_2^2\mathrm{x}_3^2$ and $g=\mathrm{x}_1\mathrm{x}_3+\mathrm{x}_2\mathrm{x}_3+\mathrm{x}_1\mathrm{x}_2$. 
The two homogeneous polynomials satisfy condition (1): The easiest way to verify this, is to type in the following command 
Reduce[ForAll[{x1,x2,x3},Implies[x1*x3+x2*x3+x1*x2>=0,x1^3*x3+x1^3*x2+x2^2*x3^2>=0]]]
in Mathematica.
But condition (2) is violated:
Suppose we could find a polynomial $t=a_1\mathrm{x}_1^2+a_2\mathrm{x}_2^2+a_3\mathrm{x}_1\mathrm{x}_2+a_4\mathrm{x}_1\mathrm{x}_3+a_5\mathrm{x}_2\mathrm{x}_3+a_6\mathrm{x}_3^2$ ($a_1,\ldots,a_6\in \mathbb{R}$) such that $f(y)-t(y)g(y)\geq 0$ for all $y\in \mathbb{R}^3$. 
A simple calculation shows that $f-tg=\mathrm{x}_1^3\mathrm{x}_2-a_1\mathrm{x}_1^3\mathrm{x}_2-a_3\mathrm{x}_1^2\mathrm{x}_2^2-a_2\mathrm{x}_1\mathrm{x}_2^3+\mathrm{x}_1^3\mathrm{x}_3-a_1\mathrm{x}_1^3\mathrm{x}_3-a_1\mathrm{x}_1^2\mathrm{x}_2\mathrm{x}_3-a_3\mathrm{x}_1^2\mathrm{x}_2\mathrm{x}_3-a_4\mathrm{x}_1^2\mathrm{x}_2\mathrm{x}_3-a_2\mathrm{x}_1\mathrm{x}_2^2\mathrm{x}_3-a_3\mathrm{x}_1\mathrm{x}_2^2\mathrm{x}_3-a_5\mathrm{x}_1\mathrm{x}_2^2\mathrm{x}_3-a_2\mathrm{x}_2^3\mathrm{x}_3-a_4\mathrm{x}_1^2\mathrm{x}_3^2-a_4\mathrm{x}_1\mathrm{x}_2\mathrm{x}_3^2-a_5\mathrm{x}_1\mathrm{x}_2\mathrm{x}_3^2-a_6\mathrm{x}_1\mathrm{x}_2\mathrm{x}_3^2+\mathrm{x}_2^2\mathrm{x}_3^2-a_5\mathrm{x}_2^2\mathrm{x}_3^2-a_6\mathrm{x}_1\mathrm{x}_3^3-a_6\mathrm{x}_2\mathrm{x}_3^3.$
Thus $f(\mathrm{x}_1,0,1)-t(\mathrm{x}_1,0,1)g(\mathrm{x}_1,0,1)=\mathrm{x}_1^3-a_1\mathrm{x}_1^3-a_4\mathrm{x}_1^2-a_6\mathrm{x}_1$. We know that $f-tg$ is a non-negative polynomial. This is only possible if $a_1=1$, $a_6=0$, and $a_4\leq 0$. Since $t$ is non-negative and $a_6=0$, the two coefficients $a_4$, $a_5$ must also vanish. The leading term of $f(0,\mathrm{x}_2,1)-t(0,\mathrm{x}_2,1)g(0,\mathrm{x}_2,1)$ is $-a_2\mathrm{x}_2^3$, and therefore $a_2=0$. Now only $a_3$ is not determined. But it is easy to see that $a_3$ must also vanish. Thus $f-tg$ reduces to $f-tg=-\mathrm{x}_1^2\mathrm{x}_2\mathrm{x}_3+\mathrm{x}_2^2\mathrm{x}_3^2$, which is obviously not non-negative.<|endoftext|>
TITLE: Is "Notices of the American Mathematical Society" available before 1995?
QUESTION [17 upvotes]: I tried to find some paper published in Notices of the American Mathematical Society 1975

By E. Calabi, On manifolds with non-negative Ricci curvature II 

Notices of the American Mathematical Society 22 1975 A205
But when I searched at mathscinet and click Calabi, there is no such a paper listed under his name.

Also AMS's website does not have any issues from before 1995. So is it a book or something?

REPLY [10 votes]: To update the answer in the title question: Full issues of the Notices of the AMS from 1954 onwards are now available online. Just follow this link and click on "Full Issues Browsing Options and Other Notices Collections."<|endoftext|>
TITLE: Relation of SW and Donaldson Invariant
QUESTION [7 upvotes]: My question is:
I am request for the reference that Is there any relationship between the Seiberg-Witten Invariant and Donaldson's Invariant? Or the relationship between Seiberg-Witten Moduli Space and Yang-Mills Moduli space?
My question is a reference request.
My question is based on the following observation: For given compact smooth manifold M,
Consider the Seiberg-Witten Equation without perturbation, $F^+_A+(\phi \phi^*)_0$=0 $D_A^+ \phi =0$. Then every solution of the Seiberg-Witten Equation is a pair $(A, \phi)$, A is connection and $\phi$ is a spinor. However, for the reducible solution of Seiberg-Witten equation(the solution that $\phi=0$), we get the equation that $F_A^+$=0
An anti-selfdual equation of the curvature. When we change the orientation of the manifold M, all the Yang-Mills connection can be the antiself-dual. So I think in some way the Yang-Mills connection is the reducible solution of the Seiberg-Witten equation. Therefore, I believe there must be some strictly relationship between two manifold.
However, I ignore the difference of the $Spin^c$ structure and $SU(2)$, but I think with a proper choose of a Lie group homomorphism $f:SU(2)\rightarrow Spin^c$, we can map the Yang-Mills Moduli space into Seiberg-Witten Moduli space. 
Maybe there exist some related questions in MO, but not what I want. 
Evans has asked a similar question:Is there a Seiberg-Witten version of Donaldson-Thomas theory?

REPLY [13 votes]: This was conjectured by Witten in his paper in his paper Monopoles and four-manifolds.  The conjecture says that if $X$ has Seiberg-Witten simple type (meaning that $SW_X(\mathfrak{s})$ is nonzero only when the moduli space associated to $X$ is 0-dimensional) and satisfies some mild homological conditions (including $b_1(X)=0$ and $b^+(X)>1$ odd), then $X$ has Donaldson simple type, its Donaldson and Seiberg-Witten basic classes coincide, and its Donaldson series has the form
${\mathcal D}^w_X(h) = 2^{2+(7\chi(X)+11\sigma(X))/4}e^{Q(h)/2}\sum_{\mathfrak{s}} (-1)^{w^2+c_1(\mathfrak{s})\cdot w} SW_X(\mathfrak{s})e^{c_1(\mathfrak{s})\cdot h}$
where $\mathfrak{s}$ ranges over Spin^c structures on $X$ and $Q(h)$ is the intersection form on $X$.  Recall that the Donaldson series is defined as a formal power series by the sum $\sum_i D^w_X((1+\frac{x}{2})\frac{h^i}{i!})$, where $x$ is a point of $X$ and $h$ is an element of $H_2(X)$; this definition is originally due to Kronheimer and Mrowka, in Embedded surfaces and the structure of Donaldson's polynomial invariants.
There is a series of papers by Feehan and Leness proving many cases of the conjecture, although it has not been established in full.  For an overview of this program, originally proposed by Pidstrigach and Tyurin, you might try their survey article PU(2) monopoles and relations between four-manifold invariants.  Notably, their most recent paper, Witten's conjecture for many four-manifolds of simple type, proves it assuming that $c_1^2(X) \geq \chi_h(X)-3$, and before that, A general SO(3)-monopole cobordism formula relating Donaldson and Seiberg-Witten invariants was used by Kronheimer-Mrowka to prove enough cases of the conjecture to establish the Property P conjecture, in Witten's conjecture and Property P.<|endoftext|>
TITLE: Orders of Number Fields
QUESTION [5 upvotes]: Let $K$ be a number field over $\mathbb{Q}$ of degree $n$, and $\mathcal{O} \subset \mathcal{O}_K$ an order.
$\textbf{Questions:}$
$\newcommand{\Spec}{\textrm{Spec }}$
$\newcommand{\cO}{\mathcal{O}}$
1.) Is the natural map $\phi: \Spec \mathcal{O}_K \rightarrow \Spec \mathcal{O}$ flat? 
2.) How many distinct primes (can) lie under a given prime? All but finitely many local rings of $\cO$ are canonically identified with local rings of $\cO_K$. Since it is not obvious to me that $\phi$ is surjective, how many more primes are in $\cO$ than in $\cO_K$ (w.r.t. the canonical identification above).
3.) In $\mathcal{O}_K$, and dedekind domains, prime ideals can be generated by two elements. How many elements are required to generate prime ideals of $\mathcal{O}$ ? Is it possible to give an answer depending on the degree $n = [K:\mathbb{Q}]$ and the index $[\mathcal{O}_K: \mathcal{O}]$?
4.) Is every ideal $I$ of $\cO$ also a proper $\cO$-ideal as is the case for the maximal order? That is, has ring of multipliers $R$ exactly $\cO$. (The rings of multipliers $R\subset K$ is the subring of elements $\alpha$ so that $\alpha \cdot I \subset I$). Certainly $\cO \subset R$.
$\textbf{Note:}$ If need be, feel free to assume that $K$ is quadratic imaginary. I'm primarily interested in this case, but I would like to have a clearer picture of the general situation.

REPLY [6 votes]: 2) I'm not sure what you mean by "lie under". Any number of equal-characteristic primes of $\mathcal O_K$ can map to a single prime of $\mathcal O$, but that's lying over, not under. The map on primes is surjective - for any local ring of $\mathcal O$, its integral closure is a local ring of $\mathcal O_K$.
3) They can be at least $n-1$. Take $p$ a totally split prime, and consider the subring of $\mathcal O_K$ of elements that are in $\mathbb Z$, modulo $p$.  Then the primes lying over $p$ glue together into a single prime ideal, whose local ring is the inverse image of the diagonal $\mathbb F_p$ in the natural map $\mathbb Z_p^n \to \mathbb F_p^n$. If $m$ is the maximal ideal of this local ring, then $m/m^2 = \mathbb F_p^n = (R/m)^n$, so the ideal requires at least $n$ generators.
4) No. Certainly some ideals have ring of multipliers $\mathcal O_K$. in $\mathbb Z[\sqrt{-3}]$, say, the ideal $(2)$ has this property.

REPLY [4 votes]: I can't figure out how to just comment, perhaps because I am a new user.
It seems to me like the example cited in Flatness of normalization should apply to 1).

REPLY [4 votes]: 1) No. The normalization of a ring $R$ is never flat over $R$, unless $R$ was normal in the first place.<|endoftext|>
TITLE: Cogroups in the category of groups are free
QUESTION [10 upvotes]: The free group $F(S)$ on a set $S$ is a cogroup in the category of groups since $\hom(F(S),G) \cong G^S$ carries a natural group structure for every group $G$. I have read that these are the only cogroups in the category of groups. This result is attributed to Kan:

Daniel M. Kan, On monoids and their dual, Bol. Soc. Mat. Mexicana (2) 3 (1958), 52–61. MR 0111035 (22 #1900)

However I have no access to this paper, and could not find it online either. Perhaps someone knows the paper and can give me a hint how to prove the result? Thanks a lot.
Edit: Tyler's answer explains why the underlying group of any cogroup is free. I accept it because meanwhile I've found Kan's paper.

REPLY [3 votes]: This is an extended comment to Tyler's answer. The Kurosh theorem implies the equalizer is a free product of a free group with conjugates of subgroups of the free factors. But each non-trivial subgroup of a free factor is killed by one projection and not the other. Thus the equalizer is free.<|endoftext|>
TITLE: Are the only local minima of $\angle(v, Av)$ the eigenvectors?
QUESTION [19 upvotes]: Let $A$ be an invertible $n \times n$ complex matrix. For $v \in \mathbb{CP}^{n-1}$, define 
$$d(v) =  \frac{|\langle A \tilde{v}, \tilde{v} \rangle |^2}{ \langle A \tilde{v}, A \tilde{v} \rangle \langle \tilde{v}, \tilde{v} \rangle}$$
where $\langle \ , \ \rangle$ is the standard  Hermitian inner product and $\tilde{v}$ is any lift of $v$ to $\mathbb{C}^n \setminus \{ 0 \}$
So $d(v) \leq 1$, with equality precisely if $v$ is an eigenvector (by Cauchy-Schwarz).

Are the eigenvectors the only local maxima of $d$?

Motivation: If this is true, than we can prove that complex matrices have complex eigenvectors by a proof analogous to the standard proof that real symmetric matrices have real eigenvectors.

REPLY [7 votes]: Thank you for this interesting question! (Long time I was expecting the opposite answer.)
True motivation. Let $V$ be a finite-dimensional $\mathbb C$-linear space equipped with a positive-definite hermitian form $\langle-,-\rangle$. Pick a $1$-dimensional $\mathbb C$-linear subspace $p\subset V$. The orthogonal decomposition $V=p\oplus p^\perp$ provides the natural identification and inclusion in
$$\text{T}_p{\mathbb P}_{\mathbb C}V=\text{Lin}_{\mathbb C}(p,V/p)=\text{Lin}_{\mathbb C}(p,p^\perp)\subset\text{Lin}_{\mathbb C}(L,L).$$ The rule $\langle t_1,t_2\rangle:=\text{tr}(t_1\circ t_2^*)$ defines a positive-definite hermitian form on $\text{Lin}_{\mathbb C}(L,L)$ and thus induces a hermitian structure on ${\mathbb P}_{\mathbb C}V$ known as Fubini-Study. The Fubini-Study distance is well known to satisfy the inequalities $0\le\text{dist}(p_1,p_2)\le\frac\pi2$ and the identity
$$\cos\text{dist}(p_1,p_2)=\frac{\langle p_1,p_2\rangle\langle p_2,p_1\rangle}{\langle p_1,p_1\rangle\langle p_2,p_2\rangle}=:\text{ta}(p_1,p_2)$$
(see, for instance, arXiv:0702714).
Question. Are the fixed points of $A$ the only local minima of $\text{dist}(Ap,p)$, where $A$ is an arbitrary holomorphic automorphism of ${\mathbb P}_{\mathbb C}V$ ?
Answer. Suppose that a local maximum $x=p$ of $\text{ta}(Ax,x)$ is not fixed by $A$. Taking suitable representatives $p\in V$ and $A\in\text{GL}_{\mathbb C}V$, we assume
$\langle p,p\rangle=\langle Ap,Ap\rangle=1$ and $g:=\langle Ap,p\rangle\ge0$. If $g=0$, then $\langle Ax,x\rangle=0$ for all $x\in V$ sufficiently close to $p$, hence, for all $x\in V$. Taking an eigenvector $x$ of $A$, we arrive at a contradiction. So, $0<g<1$ (because the hermitian form on $V$ is positive-definite).
Since
$\dim_{\mathbb C}p^\perp+\dim_{\mathbb C}({\mathbb C}p+{\mathbb C}Ap)>\dim_{\mathbb C}V$,
there exists $0\ne w\in p^\perp$ such that
$(1-gA)w\in{\mathbb C}p+{\mathbb C}Ap$. We will show that, for some
$0\ne c\in{\mathbb C}$, an arbitrarily small deformation $p'=p+tcw$ of $p$, $t\in{\mathbb R}$, provides $\text{dist}(Ap',p')<\text{dist}(Ap,p)$. We assume $\langle w,w\rangle=1$ and write $w-gAw=ap+bAp$ for some $a,b\in{\mathbb C}$. It follows that $g\langle Aw,p\rangle+a+bg=0$.
For any $v\in p^\perp$ such that $\langle v,v\rangle=1$, we define
$$w_1(t):=\big\langle A(p+tv),p+tv\big\rangle\big\langle p+tv,A(p+tv)\big\rangle,$$
$$w_2(t):=\big\langle A(p+tv),A(p+tv)\big\rangle(1+t^2)$$
so that
$$\text{ta}\big(A(p+tv),p+tv\big)=\frac{w_1(t)}{w_2(t)}=:\varphi(t).$$
The fact that $x=p$ is a local maximum of $\text{ta}(Ax,x)$ implies $\varphi'(0)=0$ and $\varphi''(0)\le0$.
Taking derivatives with respect to $t$, we obtain
$$w'_1(t)=2\text{Re}\Big(\big(\langle Av,p+tv\rangle+\langle Ap+tAv,v\rangle\big)\langle p+tv,Ap+tAv\rangle\Big),$$
$$w'_2(t)=2\text{Re}\langle Av,Ap+tAv\rangle(1+t^2)+2\langle Ap+tAv,Ap+tAv\rangle t.$$
Therefore,
$$w_1(0)=g^2,\qquad w'_1(0)=2g\text{Re}\big(\langle Av,p\rangle+\langle Ap,v\rangle\big),$$
$$w''_1(0)=4g\text{Re}\langle Av,v\rangle+2\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2,$$
$$w_2(0)=1,\qquad w'_2(0)=2\text{Re}\langle Av,Ap\rangle,\qquad w''_2(0)=2\langle Av,Av\rangle+2.$$
The condition $\varphi'(0)=0$ is equivalent to $w'_1(0)w_2(0)-w_1(0)w'_2(0)=0$, i.e., to
$2g\text{Re}\big(\langle Av,p\rangle+\langle v,Ap\rangle-g\langle Av,Ap\rangle\big)=0$.
Since it holds for any $v\in p^\perp$, we obtain
$\langle Av,p\rangle+\langle v,Ap\rangle-g\langle Av,Ap\rangle=0$. In particular, $\langle Aw,p\rangle+ag+b=0$, hence,
$g\langle Aw,p\rangle+ag^2+bg=0$. It follows from $g\langle Aw,p\rangle+a+bg=0$ that $a=0$. So, $w-gAw=bAp$ and $\langle Aw,p\rangle=-b$.
The equality $w'_1(0)w_2(0)-w_1(0)w'_2(0)=0$ implies $\varphi''(0)=\frac{w''_1(0)w_2(0)-w_1(0)w''_2(0)}{w_2^2(0)}$. As
$$w''_1(0)w_2(0)-w_1(0)w''_2(0)=$$
$$=4g\text{Re}\langle Av,v\rangle+2\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2-2g^2\langle Av,Av\rangle-2g^2,$$
we obtain
$2g\text{Re}\langle Av,v\rangle+\big|\langle Av,p\rangle+\langle Ap,v\rangle\big|^2\le g^2\langle Av,Av\rangle+g^2$,
i.e.,
$$2\text{Re}\big(\langle Av,p\rangle\langle v,Ap\rangle\big)+\big|\langle Av,p\rangle\big|^2+\big|\langle Ap,v\rangle\big|^2+1-g^2\le\langle v-gAv,v-gAv\rangle.$$
Replacing $v$ by $uv$ with a suitable unitary $u\in{\mathbb C}$, $|u|=1$, we obtain $\text{Re}\big(\langle Av,p\rangle\langle v,Ap\rangle\big)\ge0$ and conclude that
$$\big|\langle Av,p\rangle\big|^2+1-g^2\le\langle v-gAv,v-gAv\rangle.$$
In particular, for $v=w$, we obtain $|b|^2+1-g^2\le|b|^2$. A contradiction.
So, the answer is yes.<|endoftext|>
TITLE: Beautiful examples of arc-like continua
QUESTION [6 upvotes]: A continuum is a nonempty compact, connected metric space. 
A continuum $X$ is called arc-like if, for every $\varepsilon>0$, there is a continuous and surjective function $f:X\to [0,1]$ such that $f^{-1}(t)$ has diameter less than $\varepsilon$ for every $t\in[0,1]$. 
(Equivalently, $X$ is homeomorphic to an inverse limit of arcs with surjective bonding maps.)
Arc-like continua are also called "snake-like" or "chainable" continua. For more background, see Nadler's excellent textbook 'Continuum Theory: An Introduction'.

Examples include the arc, the $\sin(1/x)$-continuum, the Knaster bucket-handle and, perhaps most famously of all, the pseudo-arc (which is the unique hereditarily indecomposable arc-like continuum).
It is easy to make a nice picture of the bucket-handle (and of the $\sin(1/x)$-continuum). As far as I know, there isn't really any good way to make a sensible picture of the pseudo-arc.
I am writing a paper that involves arc-like continua, and I would be interested to know:
Are there other interesting examples of arc-like continua that lend themselves to making nice and illuminating computer pictures?
(Of course we could combine the above examples to create new arc-like continua, but I wouldn't class this as being 'interesting'. Nadler's book has an example of a hereditarily decomposable arc-like continuum that contains no arc, but it would seem difficult to turn this into a sensible picture.)
Any pointers (or, even better, pictures!) would be appreciated. As the question is open-ended, I'm making it Community Wiki.

(In case you are interested, the main result of my paper states that there is a transcendental entire function $f:\mathbb{C}\to\mathbb{C}$ with the following property. If $X$ is an arc-like continuum with a terminal point $x_0\in X$, then there is a component $C$ of the Julia set $J(f)$ such that $C\cup\{\infty\}$ is homeomorphic to $X$. In particular, the pseudo-arc can appear as (the one-point compactification of) a component of the Julia set of a transcendental entire function.)
EDIT. The paper, Arc-like continua, Julia sets of entire functions, and Eremenko's Conjecture, is now available as preprint arXiv:1610.06278.

REPLY [2 votes]: Have you tried solenoids? Solenoids certainly seem like they would satisfy your definition. Also, Antoine's necklace is another likely candidate.
Edit: Sorry, I didn't realize you required it to be connected. Solenoids still work, but Antoine's necklace is out.<|endoftext|>
TITLE: Kernel of linear representation of Baumslag-Solitar group
QUESTION [13 upvotes]: Let $BS(m,n)$ be the Baumslag-Solitar group defined by $B(m,n) =  < a,b ~|~ b a^m b^{-1} = a^n > $, $mn \neq 0$. There is a linear representation of $BS(m,n)$ by mapping $a$ to the matrix $\left(\begin{matrix} 1&1 \cr 0&1\end{matrix}\right)$ and $b$ to the matrix $\left(\begin{matrix} \frac{n}{m}&0 \cr 0&1\end{matrix}\right)$. Denote this representation homomorphism as $f$, assume $|m| \neq |n|$, my main question is:

What is the kernel of $f$ ?

Some observations
(1) Commutator of the form $ [a, a^b], [a,a^{b^2} ], [a,a^{b^3}] \ldots$, are in the kernel. Do these elements generate the Kernel of $f$? Do they form an infinite generated  free group? .
(2) If $|m| \ne |n|$ and either $|m| = 1$ or $|n| = 1$ then $f$ is known to be injective.

REPLY [12 votes]: The kernel $K$ is a free group of infinite rank. 
To see that it is a free group, take the action of $BS(m,n)$ on the Bass-Serre tree $T$ of the usual graph of groups presentation for the usual HNN decomposition $Z*_Z$, where one of the $Z \mapsto Z$ edge-to-vertex homomorphisms is multiplication by $m$ and the other is multiplication by $n$. Each vertex stabilizer of $BS(m,n)$ is conjugate to the $\langle a \rangle$ subgroup, whose intersection with $K$ is trivial. So, all vertex stabilizers of the action of $K$ on $T$ are trivial, implying that all edge stabilizers are trivial. $K$ therefore acts properly discontinuously on the tree $T$. 
One can show that the rank is infinite by exhibiting arbitrarily long simple closed edge paths in the quotient graph $T/K$, although maybe there is a slicker way.<|endoftext|>
TITLE: Vector Bundles on normal surfaces 
QUESTION [6 upvotes]: Let $X$ be a projective normal surface over $\mathbb{C}$.  In this related question it is stated as soon as $X$ is smooth any vector bundle defined on the compliment of a codimension 2 subset extends to all of $X$.
Does this fail when $X$ has codimension 2 singularities?  If vector bundles do not always extend, is there a nice example of a surface $X$ and a vector bundles $E$ on $X - \{p_1, \dotsc, p_n\}$ that does not extend to a vector bundles on $X$?
Does anyone know of references that discuss the classification of vector bundles on non smooth but normal surfaces?  
Variant: What about when you replaces vector bundle with principal $G$-bundle for a reductive group $G$?

REPLY [21 votes]: As Piotr remarks, these kind of questions quickly lead to studying reflexive sheaves. I would add that one also better get acquainted with Serre's condition $S_2$. For more on this see
this and this MO answers.
Perhaps the first remark is that besides the singularities of the surface $X$ you also have to take into account the singularities of the sheaf you are considering. 
You are asking about normal surfaces. Normal implies $S_2$ and pretty much everything that follows is OK for $S_2$ in arbitrary dimensions.
A coherent sheaf on an open set can always be extended as a coherent sheaf on the ambient space. Furthermore if $X$ is $S_2$ and $j:U\hookrightarrow X$ is an open set such that $\mathrm{codim}_X(X\setminus U)\geq 2$, then a coherent sheaf $\mathscr F$ on $X$ such that $\mathscr F|_U$ is locally free  is reflexive if and only if $$\mathscr F \simeq j_*(\mathscr F|_U).$$
Notice that this means that if $X$ is $S_2$, then a locally free sheaf $\mathscr E$ on $U$ can be extended as a locally free sheaf to $X$ if and only if $j_*\mathscr E$ is locally free. (The point is, that since $X$ is $S_2$, $j_* \mathscr E$ is reflexive and if there is a locally free sheaf extending $\mathscr E$, then that is also a reflexive sheaf which agrees with $j_*\mathscr E$ on an open set with at least codimension $2$ complement, so they have to agree). 
This also tells you how to produce locally free sheaves that cannot be extended as a locally free sheaf: Take any reflexive sheaf that is not locally free, then take the open set where it is locally free. By the above, this sheaf on that open set is a locally free sheaf that does not have an extension as a locally free sheaf on the ambient space. Piotr's example is probably the simplest such sheaf. 
So now the question is: When is a reflexive sheaf locally free? 
In some contexts one defines the singularity set of a coherent sheaf $\mathscr F$ as the locus where it is not locally free. Then there are various results that say that the singularity set of certain sheaves have to be at least such and such codimension. Here is a short list of those:

Sample statement: If $\mathscr F$ is bluh, then the singularity set of $\mathscr F$ has codimension at least boo.

Actual statement If $X$ is smooth, we can substitute the following into the above sentence:

If bluh = coherent, then boo = $1$.
If bluh = torsion-free, then boo = $2$.
If bluh = reflexive, then boo = $3$.


You may recognize theorems that you know as simple consequences of these:

Thm 1 For any coherent sheaf there exists a non-empty open set on which it is (locally) free.

Remark You may also note that this does not require $X$ smooth, but that's essentially because the conclusion is invariant under first restricting to the open set where $X$ is smooth. 

Thm 2 A torsion-free sheaf on a smooth curve is locally free.

Remark We know that this fails if the  variety is either not smooth or has dimension at least $2$.

Thm 3 A reflexive sheaf on a smooth surface is locally free.

Remark This is the reason why the statement you mention at the start is true. If $X$ is a smooth surface, then a sheaf that's locally free on the complement of finitely many points pushes forward to a reflexive sheaf which by this theorem has to be locally free.
Piotr's example or any Weil divisor that's not Cartier shows that this fails if the surface is not factorial. (Meaning that every local ring is a UFD. This is a weaker condition than being smooth.) It is also true that a reflexive sheaf of rank $1$ on a smooth variety is always locally free, so you need to go to higher ranks to get a counter-example.
To complete the picture here is an example of a reflexive sheaf on a smooth $3$-fold that is not locally free. 
Example Consider the Euler sequence of $\mathbb P^3$:
$$
0\to \mathscr O_{\mathbb P^3}\to \bigoplus_{i=0}^3 \mathscr O_{\mathbb P^3} (1) \to 
T_{\mathbb P^3} \to 0
$$
and consider (one of) the induced maps
$$
\alpha: \mathscr O_{\mathbb P^3} (1) \to T_{\mathbb P^3}.
$$
This is clearly injective and let the cokernel of $\alpha$ be $\mathscr F$.
The original (Euler) sequence shows that $\alpha$ cannot be a vector bundle embedding and hence $\mathscr F$ cannot be locally free. I leave it for you to prove that it is reflexive. (This is not absolutely trivial, but it is a good exercise).
So, to answer your question, the statement you cite is not true for singular varieties, at least not in the way it is stated. You can cook up some versions that work by adding additional assumptions. In any case, these are likely the notions that will help you do that.
Finally, here are some references:

On reflexive sheaves I would recommend Hartshorne's series of papers Stable reflexive sheaves I-II-III.
You can read about the singularity set and other interesting stuff in Chapter 2 of Okonek-Schneider-Spindler's Vector bundles on complex projective spaces<|endoftext|>
TITLE: Representation ring of SU(n)?
QUESTION [7 upvotes]: What's the structure of representation ring of SU(n)?
What are the representations of generators?

REPLY [16 votes]: Will has summarized concisely the classical structure theory for the representation ring of $SU(n)$, but it's worth emphasizing that all of this is found in textbooks on compact Lie groups.   Doing any kind of research involving such older parts of representation theory requires some acquaintance with this kind of literature, to avoid re-inventing the wheel.   It's also a good idea to place the special example in the context of simply connected semisimple compact Lie groups.
A typical source is the Springer GTM 98 Representations of Compact Lie Groups (1985) by Brocker and tom Dieck.   They have a clear discussion of representation rings in section II.7, along with the general version of Will's answer in the setting of highest weights and fundamental representations in section VII.2.   They also provide a lot of concrete details about classical groups, etc.   (All of this material goes back to much older work of Weyl and others, and is treated in multiple sources.)<|endoftext|>
TITLE: Recursively enumerable sets as range sets of functions in Grzegorczyk-hierarchy
QUESTION [7 upvotes]: It is well known that recursively enumerable sets can be defined (among many other equivalent alternatives) as the range sets of primitive recusive functions (except for the trivial case of the empty set).
On the other hand, primitive recursive functions can be presented in an stratified way using Grzegorczyk-hierarchy.
I am wondering what it is known about whether recursively enumerable sets can be presented as the range sets of functions in some particular level of Grzegorczyk-hierarchy.

REPLY [7 votes]: Gregorczyk hierarchy is not suitable to classify functions of very low complexity. The answer here is that for any r.e. set, you can take the enumerating function from pretty much as low a complexity class as you can imagine, as long as the class can compute anything nontrivial.
For example, the function can be made uniform $\mathrm{AC}^0$, in other words, computable by a logarithmic-time alternating Turing machine with constantly many alternations (actually, 1 or 2 alternations should be enough, but I would have to check).
For another notion of simplicity, if $A$ is an r.e. set and $a_0$ is its least element, then $A$ is the range of a function $f\colon\mathbb N^k\to\mathbb N$ of the form
$$f(\vec x)=\max\{p(\vec x),a_0\}$$
where $p$ is a polynomial with integer coefficients. (This follows from the MRDP theorem.)<|endoftext|>
TITLE: Resolutions of unbounded complexes and homotopy (co)limits.
QUESTION [6 upvotes]: I want to understand once and for all what the resolution of an unbounded complex is. I've been trying to read 'Homotopy limits in triangulated categories' by Marcel Bokstedt and Amnon Neeman and can't understand at all even some of the notation. From what I've gathered, the resolution of an unbounded complex $X$ is $M$ in:
$X =$ $\underleftarrow{lim_{n}} (X_{-n}) \xrightarrow{\alpha} \underleftarrow{holim_{n}} (X_{-n}) \xrightarrow{\beta} M = \underleftarrow{holim_{n}} (I_{-n})$,
where $\alpha$ and $\beta$ are quasi-isomorphisms. 
But what's the homotopy limit in this context? The thing I can relate it to is this: 
A homotopy limit of a sequence of maps
$X_{0} \xrightarrow{f_0} X_{1} \xrightarrow{f_1} X_{2} \rightarrow \cdots$
is defined as the subspace of the product of $X_0$ with $X_{i}\times\hom(\Delta[1],X_{i})$, $i \geq 1$, of elements $(x_0,${$(x_i,\gamma_i)$}$_{0 \leq i \leq n}$) such that
$\gamma_i(0) = f(x_{i-1})$,
$\gamma_i(1) = x_i$.
This is completely over my head, as someone coming from Cartan-Eilenberg resolutions being the resolutions of bounded complexes I was expecting to see a bicomplex, I have no idea how to interpret that, does anyone know? Be nice please, I'm still learning.

REPLY [3 votes]: Let $X$ be a complex (of $R$-modules, say). A resolution (on the right, to fix ideas) is just a quasi-isomorphism $X \to P$. If we want to derive a functor $F$ on the left, we would choose for $P$ an acyclic resolution and we have that $\mathbf{L}F X = F P$. Of course, if $P$ is homologically projective (or K-projective) then it is acyclic for any $F$. The dual thing with resolutions on the left and right derived functors. The classical construction of the unbounded resolution is a tedious step by step construction. (A totalization of a Cartan-Eilenberg bicomplex will not work except under some exactness situations). So what Bökstedt and Neeman prove is that with a simple countable homotopy colimit (a cone) you get the desired resolution for modules.
For sheaves things are more delicate. If you dualize the argument in Bökstedt and Neeman, you get K-injective resolutions for modules, but the procedure does not work for sheaves because countable products are not exact over sheaves. That's where Spaltenstein work comes into the picture.
A good discussion of these kind of questions without full proofs, but pointers to the relevant literature, are the first two chapters of Lipman's "Notes on Derived Functors and Grothendieck Duality", (in Foundations of Grothendieck Duality for Diagrams of Schemes, Lecture Notes in Mathematics, no. 1960, Springer, 2009).
Actually, what these works do is to reduce the problem of deriving additive functors to a question in homological algebra, without making recourse to model categories. The model structure for the unbounded category of complexes is a delicate thing (see the papers on this topic by Hovey and collaborators). To establish a certain model structure on the derived category of a Grothendieck Abelian category is as difficult as proving that there are K-injective resolutions by homological methods.<|endoftext|>
TITLE: Groups of exponent 4
QUESTION [7 upvotes]: Is there a classification of finite nonabelian 2-groups of exponent 4?
What about, finite nonabelian 3-groups of exponent 3?

REPLY [18 votes]: There is no classification of finite groups of exponent 4. You might find this paper interesting - it contains lots of information about how the group Burnside group $B(m,4)$ grows (all $m$-generator exponent-4 groups are quotients of this group).
There is also no classification of finite groups of exponent 3. However it is known that these groups must be $2$-Engel and class three. Furthermore in this case the precise size of the corresponding Burnside group is known: $B(m,3)$ is a finite group of size $3^{m + \binom{m}{2} + \binom{m}{3}}$.

REPLY [9 votes]: What about, finite nonabelian 3-groups
  of exponent 3?

Those are all quotients of the Burnside group $B(m,3)$ for some value for $m$.<|endoftext|>
TITLE: Smallest value of largest angle in finite planar configurations
QUESTION [5 upvotes]: Does every set of $n$ points in the Euclidean plane contain three points $A,B,C$ such
that the two segments obtained by joining $A,B$, respectively $A,C$ form an angle
at least equal to $(1-2/n)\pi$ at the point $A$? (Equality is of course achieved by the
vertex set of a regular $n-$gone.)
Pietro Majer's example below can be generalized and shows that $(1-2/n)\pi$ has to be replaced by a somewhat smaller constant (at least for values of $n$ which are large enough). For his example
we have to take $(1-2/6)\pi$ instead of $(1-2/7)\pi$. Is the best possible constant
$(1-a(n))\pi$ asymptotically substantially better, ie. can $na(n)$ become for example arbitrarily large for $n$ large enough? (It is of course obvious that $a(n)$ is decreasing but how fast?)
Update: For $n=5$, one can get arbitrarily close to $(1-1/4)\pi$:
Take a right-angled isocele triangle. Split the right-angled-vertex infinitesimally
along a line parallel to the longest side of the initial triangle and add
an additional point on the symmetry axis very high above the two infinitesimal points.

REPLY [6 votes]: The key bound is $(1 - 1/n) \pi$, due to Erdős and Szekeres:

          


The above is an excerpt from this paper:

          


The Erdős-Szekeres result is in their 1961 paper, "On some extremum problems in elementary geometry.", Ann. Univ. Sci. Budapest. Rolando Eötvös, Sect. Math. 3-4, 53-62 (1961)
(PDF download link).<|endoftext|>
TITLE: volume of exceptional group orbits
QUESTION [6 upvotes]: Assume that $G$ is a compact group acting by isometries on a (compact) Riemannian manifold (M,g), with principal orbits of dimension $d>0$. For $x\in M$, let $G(x)$ denote the $G$-orbit of $x$, by $G_x$ the stabilizer of $x$ and by $G_x^0$ the identity connected component of $G_x$. Does anyone know a reference where it it proved the the function $f(x)=\vert G_x/G_x^0\vert\cdot vol_d(G(x))$ is continuous
on $M$? Here, $\vert .\vert$ denotes the cardinality, and $vol_d$ the $d$-dimensional volume (induced by the restriction of the metric $g$ on the orbit).
How about a pseudo-Riemannian extension of the above? If $(M,g)$ is pseudo-Riemannian,
then one defines $f(x)$ as above when $G(x)$ is a nondegenerate submanifold, and $f(x)=0$ otherwise. Is such $f$ continuous? 
Note that $f(x)\ne 0$ only if $G(x)$ has dimension $d$, i.e., if $G(x)$ is either a principal or an exceptional orbit. The function $v(x)=vol_d(x)$ is only continuous at points
$x$ whose orbit is principal, but it fails to be continuous at points with exceptional orbit.

REPLY [4 votes]: To my understanding, Proposition 1 in this paper of Pacini, TAMS 2003 gives exactly the proof that you ask for in the Riemannian case; namely, that the volume of orbits: $$vol\colon M\to \mathbb R, \quad vol(x)=\int_{G(x)} i^*(vol_M),$$ where $i\colon G(x)\hookrightarrow M$ is the immersion of the $G$-orbit through $x$ and $vol_M$ is the volume form of $M$, is a continuous function on $M$, vanishing exactly at singular orbits.

More precisely, he proves that:

the volume function on regular (i.e., principal or exceptional) orbits $vol\colon M^{reg}\to\mathbb R$ is a smooth function;
it has a continuous extension $vol\colon M\to\mathbb R$ that is zero on the singular points $M^{sing}=M\setminus M^{reg}$;
$vol^2\colon M\to\mathbb R$ is smooth.

Note that Pacini defines the volume of an orbit not by using the volume of the image, but rather by integrating the pull-back of the volume form. These are the same thing only if the immersion is $1$-to-$1$ (e.g., for principal orbits $G/P$). For an exceptional orbit $G/K$, the immersion is $k$-to-$1$, where $k$ is the number of sheets on the covering by a principal orbit $G/P\to G/K$, so the volume of the image is multiplied by $k$. This correction factor is precisely the cardinality of the fiber, $k=|P/K|$, as pointed out by the OP.
Regarding the second question, adapting the proof to the nondegenerate semi-Riemannian case should be straightforward.

Edit: I recently realized that also the classic paper by Hsiang-Lawson, JDG 1971 (see first few lines of page 7) cites the continuity of the volume function above in $M$ (being zero on singular points) and smoothness in the set of regular points. Although they do not provide an explicit proof, they say it is straightforward from the Slice Theorem. There are also many nice examples following that.<|endoftext|>
TITLE: Intuition for Levi-Civita connection?
QUESTION [13 upvotes]: Levi-Civita connection is usually defined as the unique connection which is torsion free and preserves metric.
Question Is there some intuitively transparent  constructive way to define   it  (or corresponding parallel transport) ?  
"intuitively transparent" is up to "good will" of the ones answering. 

PS
I remember the following construction but it is not intrinsic and it is not clear for me how to derive the formula for Christoffel symbols from it in transparent way. Nevertheless let me mention it.
Assume we have a submanifold in some Riemann manifold. To define the transport along the curve on a submanifold we can do infinitesemal translation in bigger manifold - the resulting vector may not be tangent to submanifold - so we will make a projection on the tangent space of the manifold. 
In this way starting from standard metric on R^n we can derive parallel transport and hence Levi-Civita connection on a submanifold. Expressing result in terms of submanifold's metric
may be considered as way to answer the question - but it seems to be very indirect.

REPLY [6 votes]: Here is a very visual way to think of what parallel transport is.  It is Levi-Civita's original description for surfaces.  
Take a surface embedded in Euclidean 3 space.  draw a curve on it between 2 points.  take the envelope of the tangent planes to the surface that are tangent planes to the curve between the two points.  Note that that this envelope is a developable surface (of course, not the same surface), developable surfaces  are isometric to a region in the plane, so that provides a unique way along the curve to move the tangent plane and so to say whether vectors are "parallel"<|endoftext|>
TITLE: Is there a local-global principle for integral Laurent series ? 
QUESTION [5 upvotes]: Motivation: A real number is rational iff its decimal expansion is periodic (by "periodic" I mean periodic after some steps). Similar, a p-adic number is rational iff its p-adic expansion is periodic. However, this isn't true in general for rational functions as the example 
$$\frac{1}{(1-X)^2}= \sum_{n=0}^\infty (n+1)X^n \in \mathbb{Q}(X)\subseteq \mathbb{Q}((X))$$
shows. But an inspection of the proof in the p-adic case (cf. Hasse: Number Theory, Chap. 9) shows: 

If $F$ is a finite field, then $f \in F((X))$ is rational, i.e. $f \in F(X)$, iff its Laurent series 
  is periodic.

Now let $f=g/h$ with polynomials $g,h \in \mathbb{Z}[X]$ such that the Laurent series of $f$ has  integer coefficients. Reduction modulo p yields $\bar{f}=\bar{g}/\bar{h} \in\mathbb{F}_p(X)$. Hence the coefficients of $f$ are periodic modulo p. I wonder  whether the converse is true:

Question 1: Let $f$ be a Laurent series with integer coefficients. Is $f \in Quot(\mathbb{Z}[X])$ iff the coefficients of $f$ are periodic for all primes (with period depending on the prime) ?

Suppose that the coefficients of $f$ are periodic modulo p. Hence the reduction modulo p 
is rational, i.e. $\bar{f} \in \mathbb{F}_p(X)$.  Therefore an equivalent formulation of the question is: 

Question 2: Let $f$ be a Laurent series with integer coefficients. 
  Is $f$ rational, i.e. $f \in Quot(\mathbb{Z}[X])$ iff the reduction modulo p is rational for all primes ? 


Edit: As shown by Felipe, the answer is in general negative. Since I'm mostly interested in convergent power series (with integer coefficients), I would like to ask in addition, if there are also counterexamples in this case ?

REPLY [9 votes]: No $\sum n! x^n$ is a counterexample. I think it's true for algebraic functions, though.<|endoftext|>
TITLE: The derived category of the heart of a t-structure
QUESTION [17 upvotes]: Suppose $\mathcal{D}$ is a triangulated category and that we are given a $t$-structure $(\mathcal{D}^{\leq 0},\mathcal{D}^{\geq 0})$ on $\mathcal{D}$.  The heart of the $t$-structure, $\mathcal{A}=\mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}$, is an abelian category.  Is it true in general that $\mathcal{D}=D(\mathcal{A})$ is the derived category of the heart of the given $t$-structure on $\mathcal{D}$?  If not, is there an easy example that shows why not?

REPLY [15 votes]: The most serious problem is that in general there is no natural functor from $D(A)$ to $D$. To construct one you need an additional structure on $D$. There are several approaches here. One was suggested by Beilinson and gives a notion of a filtered triangulated category. Another important approach uses derivators.  
On the other hand, if you have enough structure to construct a functor, then the criterion for it to be an equivalence is rather simple. If I remember right, the necessary and sufficient condition is that each morphism $A \to A'[n]$ in $D$ (with both $A$ and $A'$ in the heart) should be decomposable into a sequence $A \to A_1[1] \to A_2[2] \to \dots \to A_{n-1}[n-1] \to A'[n]$
with all $A_i$ being objects in the heart (in other words, the graded algebra of $Ext$'s should be 1-generated).<|endoftext|>
TITLE: Why Donaldson's Four-Six Conjecture?
QUESTION [16 upvotes]: Simon Donaldson apparently made the following conjecture: Two closed symplectic 4-manifolds $(X_1,\omega_1)$ and $(X_2,\omega_2)$ are diffeomorphic if and only if $(X_1\times S^2,\omega_1\oplus\omega)$ is deformation-equivalent to $(X_2\times S^2,\omega_2\oplus\omega)$. Here $\omega$ is a symplectic structure on $S^2$, and a deformation-equivalence is a diffeomorphism $\phi:X_1\times S^2\to X_2\times S^2$ such that $\omega_1\oplus\omega$ and $\phi^*(\omega_2\oplus\omega)$ can be joined by a path of symplectic forms.
However, where I read this did not contain any background or the original source. Where did Donaldson make this claim? And why did he make this claim? What is the motivation / are there good examples where this holds? Ivan Smith showed (through examples) that this conjecture fails when we replace $S^2$ by $\mathbb{T}^2$, so the statement itself seems pretty rigid.
[Edit] Motivation and examples come from the 1994 paper "Symplectic Topology on Algebraic 3-Folds" of Ruan and the 1997 followup "Higher Genus Symplectic Invariants..." of Ruan-Tian, which states and proves the conjecture for simply-connected elliptic surfaces!

REPLY [12 votes]: I think that YangMills is probably right that Donaldson never wrote the conjecture down. But there are some interesting circles of ideas surrounding the conjecture which deserve mention which, again he probably never wrote down, but I think motivated some of his work on symplectic manifolds: namely, the idea that one could define invariants of symplectic manifolds inductively by dimension. For instance, take a 4-dimensional symplectic (Donaldson) hypersurface in a symplectic 6-manifold. There is a sense (only an asymptotic sense) in which you can do this uniquely. Is that enough to use smooth invariants of the 4-manifold to define symplectic invariants of the six-manifold? No-one has ever succeeded, due to the complicated nature of the asymptotic uniqueness.
The question about 4/6-manifolds which Chris Gerig is asking about is probably motivated by a more concrete phenomenon: smooth (i.e. Seiberg-Witten) invariants of symplectic 4-manifolds see the same information as symplectic (i.e. Gromov-Witten) invariants; after crossing with a sphere, homeomorphic but non-diffeomorphic symplectic 4-manifolds become diffeomorphic 6-manifolds, however symplectically you can still detect their Gromov-Witten invariants by counting curves in the 6-manifold (see the early papers of Ruan). The classic example is to compare the Barlow surface (a surface of general type) and a (homeomorphic) blow-up of the projective plane. One is minimal, the other has many -1-curves and you can still see these after crossing with a sphere.
This also explains why 4 and 6 are the relevant dimensions: smooth geometry in dimension 4 and symplectic geometry in dimension 6 are both "hard" in the Gromov sense. There are elliptic PDEs whose moduli spaces can be used to distinguish exotic pairs. By contrast there's no hard smooth invariants for 6-manifolds, so the question doesn't generalise.
I guess the conjecture Chris mentions is the most optimistic extrapolation of this observation, designed to encourage people to think about the circle of ideas.<|endoftext|>
TITLE: Given the vertices of a convex polytope, how can we construct its half-space representation?
QUESTION [12 upvotes]: Let us say I have the vertices of a polytope $V = \{v_1,\dots,v_k\} \subset \mathbb R^n$. Is it possible to write $V$ as intersection of half-spaces using the information from the vertices, i.e., can I write the polytope in the form $Ax \leq b$ where $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^m$? 
The columns of $A$ are not necessarily the vertices of the given polytope. An example, consider a polytope in $\mathbb R^2_+$ with vertices $\{(0,1),(1,1),(2,0),(0,0)\}$. It can be observed that the corresponding half space representation is $Ax\le b$, where 
$$A=\begin{pmatrix} 0 & 1 \\\ 1 & 1 \\\ -1 & 0\\\ 0 & -1\end{pmatrix}$$ 
and $b = (1,2,0,0 )^T$. Thank you.

REPLY [14 votes]: The problem you identify is called the facet enumeration problem in the literature: Given the vertices, find a description of the facets.
There has been quite a bit of work on this.  For $n$ points in $d$ dimensions,
$O(n^{\lfloor d/2 \rfloor})$ is achievable, and aymptotically worstcase optimal.
But this is a theoretical result.  The work of Avis & Fukuda, to which Igor refers,
is quite practical, achieving a complexity of $O(d^{O(1)} n M)$ where $M$ is the size
of the output description.  Here is one reference:

D. Bremner, K. Fukuda, and A. Marzetta. 
  "Primal-dual
  methods for vertex and facet enumeration." 
  Discrete
  Comput. Geom., 20(3):333 – 357, 1998.
  (Citeseer link, which includes a PDF download link.)

If you are interested in software, permit me to also suggest polymake:<|endoftext|>
TITLE: Schwarz type inequality
QUESTION [7 upvotes]: a) Is true the following statement. Let $h$ be analytic in the unit disk such that $$|h(z)|\le \frac{|z|^2}{1-|z|^2},$$ then $$|h'(z)|\le \frac{2}{(1-|z|^2)^2}.$$ 
a') Is true the following statement. Let $h$ be analytic in the unit disk such that $$|h(z)|\le \frac{|z|^2}{1-|z|^2},$$ then the inequality $$|h'(z)|\le \frac{8}{\pi(1-|z|^2)^2}$$ is sharp. The inequality can be proved by using Schur test, and Riesz-Thorin convexity type theorem (Dunford & Schwartz 1958, §VI.10.11).
b) If $$|h(z)|\le \frac{|z|^2}{|1-z^2|}$$ then we have better conclusion $$|h'|\le \frac{2|z|}{(1-|z|^2)|1-z^2|}$$ and this follows by using Schwarz lemma. Namely in this case $$|H(z)|=|(1-z^2) h(z)/z^2|\le 1.$$ Then $$|H'(z)|\le \frac{1-|H(z)|^2}{1-|z|^2}.$$
As $$H'(z)=(1-z^2) h'(z)/z^2-2/z^3 h(z),$$ it follows that $$|(1-z^2) h'(z)/z^2|\le \frac{2(1-|z|^2)/|z|^3 h(z)+1-|H(z)|^2}{1-|z|^2}$$ $$\le \frac{2|H(z)|/|z| +1-|H(z)|^2}{1-|z|^2}\le \frac{2|z|^{-1}}{1-|z|^2}.$$
The question a) is related to precise estimation of norm of a Bergman projection into Bloch space and is far for being a homework.

REPLY [4 votes]: Looks like we are closing the question anyway, so I'll just provide a counterexample quickly before the final vote is cast. 
If you think a bit of what is asked and what the natural freedoms and scalings are present here, you'll see that it is enough to get an analytic $f$ in the right half-plane $x>0$ ($z=x+iy$ as usual) such that $|f|<1/x$ and $|f'(1)|>1$. Now just take something like $f(z)=\frac 1z-aze^{-\sqrt{z}}$ with sufficiently small positive $a$. I leave it to somebody else to beat $4$ in the upper bound. 
As to "motivation" in general, look up in the evening. You'll see the stars in the sky. What other motivation do you need?<|endoftext|>
TITLE: Some mid-sized ¿hyperbolic? manifolds and SnapPea
QUESTION [11 upvotes]: I've recently come across some mid-sized 3-manifolds that I think are likely hyperbolic, but SnapPea has some trouble with them.  This is related to my previous question
can you fool SnapPea?
but in this case I'm dealing with closed, orientable 3-manifolds instead of knot and link complements. 
The 3-manifolds I've come across have 11, 12 and 13 tetrahedra in their triangulations.  One of them SnapPea finds a solution to the gluing equations but it has "negatively oriented tetrahedra".  Does this mean what I think it means -- that once you've put the geometric structure on the tetrahedra, you have a tetrahedron folded-over?  If you view the gluing equations from the upper half-space model, they say that the sum of a bunch of angles should be $2\pi$.  Is this the case where one of those angles is negative? 
The other two triangulations SnapPea finds geometric structures with degenerate tetrahedra.  In the upper half-space model this is where one of the angles is zero, I believe. Is there a way to fix this, so that I could get a Dirichlet domain, drill and fill, etc?   This is my primary question. 
Here are the triangulations, both in SnapPea format and Regina format. 
Regina file, all triangulations
11-tet, SnapPea
12-tet, SnapPea
13-tet, SnapPea
With the 12-tetrahedron example, SnapPea can generate a Dirichlet Domain.  This one has what looks almost like bevelled-edges.  Is there a way to find a better basepoint to grow the Dirichlet domain from? 
12-tet Dirichlet domain http://dl.dropbox.com/u/46424505/triangulations/pic.png
edit: After working a bit with Nathan's answer, I've identified these three manifolds and got a little closer to understanding how to work SnapPea to maximal advantage. 
tri11, as Nathan mentioned, is hyperbolic. It has a fairly pretty Dirichlet domain.
Tri11 Dirichlet Domain http://dl.dropbox.com/u/46424505/triangulations/tri11.jpg
Another common name for this manifold would be the 0-surgery on the 2-component link $7a_6$ (in the Thistlethwaite table).  Similarly tri12 can be identified as Nathan says. 
After playing around with Regina a bit I found an incompressible torus in tri13 that splits tri13 into the union of an orientable $I$-bundle over the Klein bottle and a figure-8 complement.  So this answers the core of my question.

REPLY [5 votes]: That's just a long comment about tri13. I don't have Regina now, but you can get some informations by looking only at the 1-skeleton $G$ of the closed triangulation for tri13. Among the (experimentally) minimal triangulations for tri13, i.e. the ones that you find by trying to simplify it as much as you (or Regina) can, you probably find one where the 1-skeleton $G$ contains a pair of disconnecting edges, which split the $G$ into two parts $G_1$ and $G_2$. In any experimentally minimal triangulation such a pair of edges is always transverse to a torus $T$, which cut the manifold $M$ into two parts $M_1$ and $M_2$, containing $G_1$ and $G_2$ correspondingly.
According to Nathan's suggestion, the torus $T$ should separate a figure-8 knot complement (or its sibling) $M_1$ and a small Seifert space $M_2$. You can see this from the 1-skeleton $G$: the portion $G_1$ must have at least 8 vertices to be hyperbolic, and it has probably 8 or 9 (corresponding to the figure-8 knot sibling and the figure-8 knot complement), and the other portion $G_2$ has of course 5 or 4 vertices. If $G_2$ is layered, then the torus is compressible and you have a Dehn fillig of $M_1$. If (as we expect) $G_2$ is not layered, then it is certainly not a solid torus (minimal triangulations of solid tori with small number of tetrahedra must be layered, as far as I remember).
Summing up, I think that:

if a triangulation for $M$ is experimentally minimal with a reasonable number of tetrahedra and its 1-skeleton $G$ decomposes into two pieces $G_1$ and $G_2$, noone of which is layered, then you can conclude theoretically that it is not hyperbolic.

The numer of tetrahedra must be reasonable, because minimal triangulations of solid tori are classified only for small number of tetrahedra (I don't know how many...)
A fool-proof alternative is to use Matveev's Recognizer, which tells you exactly the JSJ decomposition of the manifold.<|endoftext|>
TITLE: Constructing a field from a spherical building
QUESTION [6 upvotes]: Tits proved that (sufficiently high rank) spherical buildings arise from an algebraic group and a field, so any building is some $\Delta(G, F)$.  He also showed that a building isomorphism $\Delta(G,F)\simeq\Delta(G',F')$ induces a field isomorphism $F\to F'$.
This shows that the field is somehow coded up in the isomorphism type of the building.  I'm wondering whether a construction is floating around anywhere that shows how to construct the field from the combinatorics of the building.
To help clarify what I'm after: I've seen a construction of the real field from the incidence structure or geometry of the real projective plane (pick a pair of lines, prove they're bijective, define addition via some more lines and unique intersection points which must exist, etc.).  In the end you have parametrized a projective line by matching up points on it with the underlying field.  The construction requires a choice of a few points in general position and so is given sort of "up to collineation".  I'm under the impression that I should view those results on spherical buildings as generalizations of results in projective geometry, and I'm looking for an analogous construction.  That is, a construction of a field in terms of the apartments, relations between them, etc.  In the end presumably some collection of objects is parametrized by the underlying field.  Or is this not the case?

REPLY [4 votes]: There is no uniform definition for the field of definition. In particular this means that recovering $F$ (somewhat) depends on the classification of spherical buildings of rank at least three.
The basic method to recover a field is to find a projective line embedded in a rank one residue of the building. Such a projective line then has a field associated to it, which you then call the field of definition.
If you want to recover the field for a spherical building of rank at least three combinatorially, one can do the following. Every such building contains a rank two residue isomorphic to a Moufang projective plane (the projective plane is possible because of the classification of spherical coxeter diagrams, its residue is Moufang because of results of Tits), from which you can extract a field, skew field or octonion algebra somewhat combinatorially. 
In general it is possible to extract a (skew) field from a rank two Moufang spherical building (also known as a Moufang generalized polygon) in a combinatorial way, but the algorithm is different for each class coming from the classification of Moufang polygons by J. Tits and R. Weiss.
I hope this helps.

REPLY [2 votes]: If you want to see explicitly how a field (or division ring) arises from combinatorics of the building (say, in the $A_2$ case, provided that enough extra axioms are satisfied), you can use Von Staudt's construction: He encodes algebraic operations into certain arrangements of points and lines in the abstract projective plane. You can find this construction, say, in Hartshorne's "Foundations of Projective Geometry". Von Staudt used this construction to prove his "fundamental theorem of projective geometry". Tits' theorem is a far-reaching generalization of this result, but the idea is clearest in the $A_2$ case. 
Incidentally, this encoding procedure has lead to other interesting mathematical constructions, like Mnev's Universality theorem and its generalizations.<|endoftext|>
TITLE: spectral sequence for cobordism without leaving smooth category
QUESTION [5 upvotes]: In Bott & Tu's marvelous book there is a derivation of the spectral sequence for a (smooth) fiber bundle for deRham cohomology done entirely in the realm of the smooth category. Unfortunately, as it's written there it is not clear how to generalize to other cohomology theories (they write down a double complex, and they use very much the fact that cohomology is given as the cohomology of this chain complex.)
Now, in Quillen's paper computing the complex cobordism ring, he introduces a geometric description of complex cobordism as a cohomology theory on smooth manifolds. I've been attempting, to no avail, to come up with a derivation of the Atiyah-Hirzebruch spectral sequence for a smooth fibration without using CW complex-esque techniques. My question is:

Does anyone know of a derivation of the Atiyah-Hirzebruch spectral sequence of a smooth fiber bundle for a generalized cohomology theory that does not leave the realm of manifolds?

So far, the most promising bet I have found is Segal's approach to this spectral sequence in the paper "Classifying spaces and spectral sequences." The trouble appears, however, in his use of a complex $BX_U$ for a covering $U$ of $X$. This is most definitely not a manifold, though it is homotopy equivalent to one for numerable covers. The question in this context, however, reduces to:

Does the natural filtration on $BX_U$ induce a filtration on $X$ that gives rise to the spectral sequence for a covering? If so, is there a nice description of this filtration using, say, just the data of $X$ and the numerable cover?

For this question I should be more specific about $BX_U$. This is defined as the geometric realization of the nerve of the topological category $X_U$ whose points are pairs $(x, U_{\sigma})$ where $U_\sigma$ is a finite intersection of elements of $U$ and $x \in U_\sigma$. Morphisms are inclusions $U_\sigma \subset U_\tau$. The filtration on $BX_U$ is given by looking at the images of $(NX_U)_n \times \Delta^n$ in $BX_U$, where $NX_U$ is the nerve.

REPLY [6 votes]: You can use Morse theory. But I do not know if you will be very happy with that, because in some sense you recover a CW-decomposition of your manifold from this data.  
Anyway, if you use a Morse-Smale function on the basis of your bundle and if you suppose that it is self indexing and Morse-Smale you will get a nice filtration of your basis. Then to get the Atiyah-Hirzebruch spectral sequence you proceed by pulling-back this filtration from the basis to the total space of your fiber bundle.<|endoftext|>
TITLE: Can Assumptions about forcing produce Mice?
QUESTION [5 upvotes]: This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident: 
For every partial order $\mathbb{P}$ and regular cardinal $\lambda > \omega$ we can define the following two statements 
$$ \mathcal{C}(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f: \alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$
(this is the formalized version of the statement "$\mathbb{P}$ preserves $\lambda$ is a cardinal" in the forcing language, this statement is normally certified by reasoning which does not involve the forcing relation and depends on the structure of $\mathbb{P}$-names) 
and
$$ Cof(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f:\alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda} \ (\sup(ran(f)) \le \gamma) $$
(Again a forcing language version of the statement $\mathbb{P}$ preserves $\forall \alpha \in \lambda \ (cf(\alpha) < cf(\lambda))$: we had to be careful here because we need to be able to distinguish between the two (If this is not the correct way to formalize this please let me know.))
Now, here comes the question: Does the following conjunction:
$\exists \lambda > \omega\ \exists\ \mathbb{P}$ such that

$\lambda$ is a Regular cardinal.
$\vert \mathbb{P} \vert = \lambda^{+}$
$\forall \mu \  (\mu$ is a cardinal $\implies \mathcal{C}(\mu,\mathbb{P}))$ 
$\neg Cof(\lambda, \mathbb{P})$

Imply there is an inner model with a measurable cardinal? (changed based on the answers.)
(Namba for $\omega_2$ and threading a generic square collapse cardinals; moreover if $ 0^\sharp $ exists then $\aleph_\omega^{V}$ is regular in $L$ producing a model in some sense)
Edit:
(It was not my intention to scare a lot of nice mice)
(also, mice need to be more damn direct and stop subtly hinting things.... didn't realize what was going on until just now....)

REPLY [10 votes]: Here's an argument for an affirmative answer to Joel's modified version of the question. Suppose we have a forcing that preserves cardinals but singularizes some cardinal $\lambda$ that was regular in the ground model.  Note that $\lambda$ had to be a limit cardinal, since otherwise singularizing it would collapse it down to its immediate predecessor cardinal (if not even lower).  Now let $C$ in the forcing extension be a cofinal subset of $\lambda$ of smaller cardinality $\kappa$.  I claim that C is not included in any set $D$ in the ground model of cardinality $\leq\max\{\kappa,\aleph_1\}$; in other words, I claim that $C$ is a counterexample to the assertion that the forcing extension satisfies the covering lemma over the ground model.  Indeed, suppose we had such a $D$.  Intersecting it with $\lambda$, we'd have a cofinal subset of $\lambda$ strictly smaller than $\lambda$ in the ground model, contrary to the assumption that $\lambda$ is regular in the ground model.  ("Strictly smaller" in the preceding sentence uses that $\lambda>\aleph_1$, which is why I pointed out earlier that $\lambda$ has to be a limit cardinal.)  So the forcing extension doesn't satisfy the covering lemma over the ground model.  That implies the existence of an inner model with a measurable cardinal, by an ancient result of mine --- "Small extensions of models of set theory" in "Axiomatic Set Theory" (Proc. of 1983 Boulder Conference, edited by Baumgartner, Martin, and Shelah) Contemporary Math. 31 (1984) pp. 35-39.<|endoftext|>
TITLE: Ultrapowers of operators
QUESTION [6 upvotes]: Can we prove that for each infinite dimensional Banach space $X$ and any free ultrafilter (possibly over uncountable set of indices) $\mathcal{U}$ the obvious embedding 
$$({\mathcal{L}(X)})_{\mathcal{U}}\to \mathcal{L} (X_U ) $$
is not surjective? Even when $X$ is superreflexive? $X_{\mathcal{U}}$ stands for the Banach space ultrapower of $X$ along $\mathcal{U}$.

REPLY [4 votes]: As Andreas suggests, I shall fix $\mathcal U$ to be countably-incomplete.  In fact, wlog, $\mathcal U$ will be over $\mathbb N$.  If $X$ is not super-reflexive, then you don't even get all the rank-one operators.  We know that $(X)_{\mathcal U}^* = (X^*)_{\mathcal U}$ if and only if $X$ is super-reflexive, so there is $\lambda \in (X)_{\mathcal U}^* \setminus (X^*)_{\mathcal U}$.  Choose $y=(y_n)\in (X)_{\mathcal U}$.  Let $T(x) = \lambda(x)y$ so $T$ is a rank-one map on $(X)_{\mathcal U}$.  Suppose $T=(T_n)$.  For each $n$ pick $\mu_n\in X^*$ with $\|\mu_n\|\leq 1$ and with $\lim_n \mu_n(y_n)=\lim_n \|y_n\|$ (limits over $\mathcal U$ of course).  Set $\mu=(\mu_n)$.  Then
$$ \mu(T(x)) = \lambda(x) \mu(y) = \lambda(x) = \mu((T_n)(x)) = \lim_n \mu_n(T_n(x_n)), $$
which holds for all $x$, so $\lambda = (\mu_n\circ T_n)\in (X^*)_{\mathcal U}$, contradiction.
If $X$ is super-reflexive, then I want to use some "co-ordinate" structure, so I need to think some more...<|endoftext|>
TITLE: Expected Hitting Time for Simple Random Walk from origin to point (x,y) in 2D-Integer-Grid
QUESTION [5 upvotes]: Consider a simple random walk on the lattice $\mathbb Z^2$ starting at the origin $(0,0)$ where in each step, one of the four adjacent vertices in chosen uniformly at random, i.e. with probability $1/4$. I want to know the distribution of the time that it takes to hit a certain vertex $(x,y)$. I have already done quite some research on related work but so far I was not able to come up with a paper that gives an answer to my question. I would appreciate any hint towards a paper that covers this problem or a hint on how I can calculate it myself.

REPLY [8 votes]: Oops, too long for a comment: 
The expectation of the time until you reach a given point other than the origin is infinite already in one dimension. To see this, let $T$ be the expected time until you get from 0 to 1 (in the obvious 1-dimensional setting). The first step is either to the left or to the right, and if you go to $-1$, the expected remaining time is going to be $2T$ since you have to get back to the origin and then to 1. Consequently $T = 1 + 1/2\cdot (2T)$, from which we see that finite $T$ leads to contradiction. In higher dimensions it becomes even worse. Presumably you want to modify your question and ask about something like the distribution of the time (in dimensions 1 and 2 you actually get there!) or about the expected time to get from one point to another in a finite box.<|endoftext|>
TITLE: Do sufficiently regular distances on manifolds come from riemannian metrics?
QUESTION [7 upvotes]: Hi to all!
Let $M$ be a compact smooth manifold without boundary and let $$d:M\times M\rightarrow [0,+\infty)$$ a distance on $M$ compatible with its topology. Suppose there exist $\varepsilon\in (0,+\infty)$ s.t. on the open set $\Delta_{\varepsilon}\subset M\times M$ 
$$\Delta_{\varepsilon}:=d^{-1}\left([0,\varepsilon)\right)$$
the function 
$$d^{2}:\Delta_{\varepsilon}\rightarrow [0,+\infty)$$ 
is smooth. Does a distance like this one come from a riemannian metric on $M$?
Thank you in advance.

REPLY [6 votes]: As pointed out in other answers, only length metrics can be Riemannian and there are plenty of non-length metrics satisfying your condition.
If your metric is a length metric, then yes, it is Riemannian. Indeed, for every $x\in M$ the function $f_x=d(x,\cdot)^2$ is smooth near $x$. Since it attains its minimum at $x$, its derivative at $x$ is zero. Therefore its second derivative at $x$ is well-defined as a quadratic form on $T_xM$. Define the metric tensor at $x$ as one half of this second derivative. Now from the Taylor expansion it is easy to see that for every $\varepsilon>0$ there is a neighborhood of $x$ where the original distance and the new Riemannian distance are Lipschitz equivalent with Lipschitz constant $1+\varepsilon$. Since both metrics are length metrics, this property implies that they are $(1+\varepsilon)$-Lipschitz equivalent globally. Since $\varepsilon$ is arbitrary, this means that they are equal.<|endoftext|>
TITLE: Does every orientable surface embed in $\mathbb{R}^{3}$
QUESTION [12 upvotes]: A topological surface can be pretty strange (consider, for instance, covers of $S^{2}-K$, where $K$ is a Cantor set.)  Can every orientable topological surface be topologically embedded in $\mathbb{R}^{3}$?

REPLY [15 votes]: Expanding slightly on my comment above: here is how one can get the embedding theorem from the classification theorem.
The classification theorem for non-compact surfaces (theorem 3 in http://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf; by the way, the comment links to a version that tries to charge one $30 unless one's reading this from a university account; sorry about that; this has been fixed now) states that
Every surface is homeomorphic to a surface formed from a sphere $\Sigma$ by first removing a closed totally disconnected set $X$ from $\Sigma$, then removing the interiors of a finite or infinite sequence $D_1,D_2,\ldots$ of nonoverlapping closed discs in $\Sigma - X$, and finally suitably identifying the boundaries of these discs in pairs. (It may be necessary to identify the boundary of one disc with itself to produce an odd "cross  cap.'')[...]
To get a proper embedding in $\mathbb{R}^3$ first note that one may assume that $X$ is non-empty; otherwise the surface will be compact (since we are throwing away something open and then identifying something), in which case everything is clear. Let $f:\Sigma\to [0,\infty)$ be a smooth function such that $f^{-1}(0)=X$.
Take an $x_0\in X$ and identify $\Sigma\setminus \{x_0\}=S^2\setminus \{x_0\}$ with $\mathbb{R}^2\subset\mathbb{R}^3$. First we properly embed $\Sigma-X=\mathbb{R}^2\setminus X$ in $\mathbb{R}^3$ as the graph of the function $\frac{1}{f}$.
Let $D_i,D_j$ be two disks whose boundaries are to be identified. There is an $\varepsilon(D_i,D_j)>0$ such that one can join $D_i$ and $D_j$ with a curve $\gamma(D_i,D_j)\subset \Sigma$ that misses $f^{-1}([0,\varepsilon(D_i,D_j)))$. Now delete the portion of the graph over the interiors of $D_i$ and $D_j$, attach vertical tubes ("chimneys") that reach at least as high as $1/\varepsilon(D_i,D_j)$ to the resulting boundaries, and then connect the tops of the chimneys with a horizontal tube along $\gamma(D_i,D_j)$.
If $D_k,D_l$ are two other disks that we must throw away and then glue the boundaries we proceed in a similar way but this time we may have to make the chimneys higher so that they miss the first tube. And so on. It is not hard to see that each ball centered at the origin intersects only finitely many tubes.<|endoftext|>
TITLE: Absolute Galois group of the field of Puiseux series over $\overline{\mathbb{F}}_p$
QUESTION [8 upvotes]: Let $K$ be the field of Puiseux series with coefficients in $\overline{\mathbb{F}}_p$ (the algebraic closure of the field with $p$ elements).
What is the absolute Galois group of $K$?
Thank you to anyone who could help!

REPLY [17 votes]: Let $E$ be the field $\overline{\mathbb{F}}_p((X))$. The field of Puiseux series whose exponents have denominators prime to $p$ is a subfield of $E^{sep}$, so the group you're asking about would then be the wild inertia subgroup of $Gal(E^{sep}/E)$. The group $Gal(E^{sep}/E)$ is quite complicated, and it comes up in arithmetic geometry, for example when studying the $\pi_1$ of curves. It also occurs as a closed subgroup of $Gal(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$ by the theory of the field of norms of Fontaine and Wintenberger. Its representations on $\mathbb{Z}_p$-modules are described by $\varphi$-modules'' (like $(\varphi,\Gamma)$-modules without the $\Gamma$). If you want to include Puiseux series whose exponents have denominators divisible by $p$, then you're looking at the perfection of $E$. The group does not change, as $E^{sep}$ is dense in $E^{alg}$ by a theorem of Ax.<|endoftext|>
TITLE: Index of a differential operator between trivial bundles.
QUESTION [6 upvotes]: Let $M$ be a closed parallelizable manifold and $D: \Gamma(E) \to \Gamma(F)$ an elliptic differential operator between trivial vector bundles $E,F \to M$.  The Atiyah Singer index theorem implies that the index of $D$ is zero.  Is there a way to prove this with less machinery?
By the way, this question is a cross-post from math.SE.
EDIT:
Actually I think I made a mistake in my reasoning, which was that the symbol class $[\sigma_D] \in K(TM)$ is zero (I actually didn't need triviality of $TM$).  Viewing $K$-theory as sequences of bundles the symbol class is
$$
0 \to \pi^* E \stackrel{\sigma_D}{\to} \pi^* F\to 0
$$
where $\pi: TM \to M$.  Now if $TM^+$ is the one-point compactification of $TM$ then the isomorphism $K(TM) \to \tilde K(TM^+)$ is given by extending the sequence to $TM^+$.  I thought that the extension would have to involve trivial bundles as well, from which it will follow that $\sigma_D = 0$  since for a compact space any sequences involving trivial bundles is zero in $\tilde K$.  But now I think this extension need not involve trivial bundles: $K(\mathbb R^2) \simeq \tilde K(S^2) = \mathbb Z$.  But every bundle over $\mathbb R^2$ is trivial so my argument would give $K(\mathbb R^2) = 0$.

REPLY [4 votes]: The result is wrong; the case of a point as base manifold creates counterexamples. Here is a less trivial construction in dimension $2$:
Let $M$ be a manifold and $V \to M$ be any vector bundle. There is an elliptic differential operator $D$ of order $2$ on $V$, which is self-adjoint and has thus index $0$: take a connection $\nabla$ on $V$ and put $D=\nabla^{\ast} \nabla$ (this is a Laplace type operator).
Now let $M= T^2$ and let $W \to T^2 $ be a holomorphic line bundle of degree $1$. By Riemann-Roch, the operator $\bar{\partial}_W$ has index $1$; and it goes from sections of $W$ to sections of $W$, since the canonical line bundle of a torus is trivial. Therefore one can form the composite $P:=(\bar{\partial}_W)^2$, and $P$ has index $2$.
Now let $V$ be a complex vector bundle such that $V \oplus W$ is trivial; with the operator $D$ constructed above. Consider the operator $D \oplus P$; this is an order $2$ elliptic operator on the trivial vector bundle over a parallelizable manifold and has index $2$.
I do not see how to produce an order $1$ operator of index $1$, though.
The vanishing theorems in Aityah-Singer, IoEO III, are quite optimal. My construction does not work in odd dimensions; and it is clear that the resulting trivial vector bundle has dimension at least $2$. If the dimension of the trivial vector bundle is too small, each (pseudo)differerential operator will have index $0$, as proven by Atiyah-Singer.<|endoftext|>
TITLE: Do people still use Massey Products for computations in the Adams Spectral Sequence
QUESTION [7 upvotes]: Hey everyone,
It seems to me like in the literature of the Adams Spectral Sequence, older publications (Toda, May, Tengora+Mahowald) make heavy and explicit use of Massey Products for computations.
More "recent" sources (Kahn, Milgram, Ravenel, May (again), Bruner) seem to like to make references to the Massey products, even just for the sake of naming, but use Steenrod Squares for actual computations.  The few computations I know of using Massey products can be more easily done with Steenrod Squares, but this may just be my ignorance about Massey products.  My question then is do these things still play an important computational role?
Thanks

REPLY [17 votes]: Why would anything computationally useful be obsolete?  Massey products and Toda brackets are intrinsic to stable homotopy theory.  It is guaranteed in advance that every element of $E_2$ 
of the classical Adams spectral sequence (for the homotopy groups of spheres, a similar statement holds more generally for the ASS computing maps between spectra)  above the $s=1$ line is decomposable in terms of matric Massey products.
Similarly, every element of the stable homotopy groups of spheres is decomposable in terms of matric Toda brackets built from the Hopf invariant one elements.  Drew's comments are on the mark: few people nowadays do these kinds of calculations, or know how to do them, which is a pity.  We still know relatively little about concrete calculations of stable homotopy groups.  These operations are complementary to, not in competition with, Steenrod operations and their related homotopy operations (see e.g. Bruner's contributions to $H_{\infty}$ ring spectra and their applications'').<|endoftext|>
TITLE: Projective bundle given by vanishing of a section
QUESTION [5 upvotes]: This question might be tautological. It comes from a statement in the proof of the non-emptiness of the degeneracy loci of a vector bundle homomorphism that Prof. Lazarsfeld gives in his book "Positivity in AG II" (Theorem 7.2.1)
Take a homomorphism of vector bundles $v:E\rightarrow F$ with kernel $F=\ker v$ and image  $K=Im v$ and consider the short exact sequence $$0\rightarrow N \rightarrow E \rightarrow K \rightarrow 0$$
The surjection $E^{\ast}\to N^{\ast}$ gives an embedding $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$ and we seek to realize $\mathbb{P}(N^{\ast})$ as the zero locus of the section of some vector bundle.
The projectibve bundle $\pi:\mathbb{P}(E^{\ast})\rightarrow Y$ comes with a tautological surjection $$\pi^{\ast}E^{\ast}\rightarrow \mathcal{O}_{\mathbb{P}(E^{\ast})}(1) \rightarrow 0$$
(given by the identity $E^{\ast}\rightarrow E^{\ast}$) and the composition of its dual $\mathcal{O}_{\mathbb{P}(E^{\ast})}(-1) \rightarrow \pi^{\ast}E$ with the pullback homomorphism $\pi^{\ast}v:\pi^{\ast}E \rightarrow \pi^{\ast}K$ gives a section 
$$s\in \Gamma(\mathbb{P}(E),\pi^{\ast}K \otimes \mathcal{O}_{\mathbb{P}(E^{\ast})}(1))$$
Then it is claimed that the zero-locus of this section gives precisely the subvariety $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$.
I am wondering whether this is obvious or not, but this correspondence is not apparent to me.
Thanks in advance for any insight.

REPLY [2 votes]: It's tautological. Remember that $\mathcal O_{\mathbb P(E^*)}(-1)$, viewed as an actual bundle of lines, is the tautological bundle: It is $E$ with the origin blown up. The map to $\pi^* E$ is literally the map from $E$ with the origin blown up to $E$. So the map to $\pi^* K$ is that, composed with the projection down to $K$. The section is trivial at a point if the map is trivial in the corresponding line - that is, if the point of $\mathbb P(E^*)$ corresponds to a line lying in the part of $E$ that maps to $0$ in $K$ - that is, if the line lies in $N \subset E$.
But the points with lines lying in $N \subset E$ are of course just $\mathbb P(N^*) \subset \mathbb P(E^*)$.<|endoftext|>
TITLE: The sparsest planar net that captures every unit segment
QUESTION [17 upvotes]: Let $\cal C = \lbrace C_i \rbrace$ be a collection
of rectifiable curves in the plane with the property that
every unit-length segment meets at least one curve
in at least one point.
Call such a collection $\cal C$
a needle net: any unit-length "needle" is captured by the net.
I would like to find the sparsest needle net, sparse in the
sense that the curves have minimum length per unit area.
That is, the limit of $L/A$ of the ratio of the length $L$ of the curves within a region
to that region's area $A$, as the region grows large, is as small as
possible.
For example, a regular grid of orthogonal parallel lines
separated by $\sqrt{2}/2$ is a needle net: the diagonal of
each square cell of the grid has length $1$.
If I've calculated correctly, the length of its
curves (lines) within each unit area region $L/A$ is $2 \sqrt{2}$.
See left below, where a unit-length
diagonal is highlighted in red, and the region of the plane
I used to compute $L/A$ is marked.

   

Again if I've calculated correctly, the equilateral-triangle tiling of
the plane obtained from three sets of parallel lines is less efficient,
and the packing arrangement of unit-diameter circles shown right above
is less efficient still.
Is the square-grid the sparsest needle net?
This feels like a question that has been addressed before, perhaps
in another guise.  If so, a pointer would be welcomed.
Thanks!
Update. Roland Bacher's more efficient needle net:

            

Is this the optimal net?

REPLY [11 votes]: Without error of my part, a paving with regular hexagons with sides of length $1/2$ gives 
$L/A=4\sqrt{3}/3\sim 2.3094$. This could very well be the optimal candidate.<|endoftext|>
TITLE: Cocomplete but not complete abelian category
QUESTION [40 upvotes]: This is a duplicate of the following question to which I did not receive any answer: https://math.stackexchange.com/questions/238247/complete-but-not-cocomplete-category
Let $\mathfrak C$ be an abelian, cocomplete category. If $\mathfrak C$ has a generator and colimits are exact (i.e., $\mathfrak C$ is Grothendieck) then $\mathfrak C$ is the torsion-theoretic localization of a full category of modules (by the Gabriel-Popescu Theorem) and so it is also complete. Anyway I'm not aware of any counter-example showing that a cocomplete abelian category may not be complete. So my question is: could you provide such example or a reference to a proof of the bicompleteness of cocomplete abelian categories?
My first idea was to look for counterexamples in non-Grothendieck subcategories of a Grothendieck category. After some attempt I realized the following 
Lemma. Let $\mathfrak C$ be a Grothendieck category and $\mathcal T$ a full hereditary torsion subcategory (i.e. $\mathcal T$ is closed under taking sub-objects, quotient objects, extensions and coproducts). Then $\mathcal T$ is bicomplete.
Proof. Let $T:\mathfrak C\to \mathcal T$ be the hereditary torsion functor associated to $\mathcal T$. Now, given a family {$C_i:i\in I$} of objects in $\mathcal T$ we can take the product $(P,\pi_i:P\to C_i)$ of this family in $\mathfrak C$. We claim that $(T(P), T(\pi_i))$ is a product in $\mathcal T$. Indeed, let $X\in \mathcal T$ and choose maps $\phi_i:X\to C_i$. By the universal property of products in $\mathfrak C$, there exists a unique morphism $\phi:X\to P$ such that $\pi_i\phi=\phi_i$ for all $i\in I$. Now, since $X\in\mathcal T$, there is an induced map $T(\phi):X\to T(P)$ which is clearly the unique possible map satisfying $T(\pi_i)T(\phi)=T(\phi_i)=\phi_i$. \\\
Thus there are lots of non-Grothendieck bicomplete abelian categories.
EDIT: notice that in the lemma we never use the hypothesis that the subcategory $\mathcal T$ is closed under taking extensions or subobjects. In fact, if $\mathcal T$ is just closed under taking coproducts and quotients, one defines the functor $T:\mathfrak C\to \mathcal T$  such that, for all object $X\in\mathfrak C$, $T(X)\in \mathcal T$ is the direct union of all the subobjects belonging to $\mathcal T$ (image (which is a quotient) of the coproduct of all the subobject of $X$ belonging to $\mathcal T$ under the universal map induced by the inclusions of the subobjects in $X$). Clearly $T(X)$ is fully invariant as a subobject of $X$ (by the closure of $\mathcal T$ under taking quotients and the construction of $T$) and so $T$ can be defined on morphisms by restriction. It is also clear that $T(X)=X$ if $X\in\mathcal T$ so the proof of the lemma can be easily adapted to this case.
REMARK: the new relaxed hypotheses of the lemma allow us to exclude other "exotic" examples... in particular, if you want to take the abelian subcategory of all the semisimple objects in a given Grothendieck category, this is closed under coproducts and quotients.

REPLY [27 votes]: I think I have an example.
Fix a chain of fields $k_\alpha$ indexed by ordinals $\alpha$, where $k_\alpha\subset k_\beta$ is an infinite field extension for all pairs $\alpha<\beta$ of ordinals.
First I'll define an "abelian category" which has large Hom-sets.
An object $V$ will consist of a $k_\alpha$-vector space $V(\alpha)$ for each ordinal $\alpha$, together with a $k_\alpha$-linear map $v_{\alpha,\beta}:V(\alpha)\to V(\beta)$ for each pair $\alpha<\beta$ of ordinals, such that $v_{\beta,\gamma}\circ v_{\alpha,\beta}=v_{\alpha,\gamma}$ whenever $\alpha<\beta<\gamma$.
A morphism $\theta:V\to W$ will consist of a $k_\alpha$-linear map $\theta_\alpha:V(\alpha)\to W(\alpha)$ for each $\alpha$, such that $w_{\alpha,\beta}\circ\theta_\alpha=\theta_\beta\circ v_{\alpha,\beta}$ for all $\alpha<\beta$.
Now let's say that an object $V$ is "$\alpha$-good" if, for every $\beta>\alpha$, $V(\beta)$ is generated as a $k_\beta$-vector space by the image of $v_{\alpha,\beta}$, and that $V$ is "good" if it is $\alpha$-good for some $\alpha$. If $V$ is $\alpha$-good, then any morphism $\theta:V\to W$ is determined by $\theta_\gamma$ for $\gamma\leq\alpha$, so the full subcategory $\mathfrak{C}$ of good objects has small Hom-sets.
It's straightforward to check that $\mathfrak{C}$ is an abelian category, and it has small coproducts in the obvious way, where $\left(\coprod_{i\in I} V_i\right)(\alpha)=\coprod_{i\in I} V_i(\alpha)$.
I claim that $\mathfrak{C}$ does not have all small products.
For any $\alpha$, let $P_{\alpha}$ be the ($\alpha$-good) object with
$$P_\alpha(\beta)=\begin{cases}0&\mbox{if }\beta<\alpha\\k_\beta&\mbox{if }\alpha\leq\beta\end{cases}$$
and obvious inclusion maps. Then for any object $W$, $\operatorname{Hom}_{\mathfrak{C}}(P_\alpha,W)$ is naturally isomorphic to $W(\alpha)$ (i.e., $P_\alpha$ represents the functor $W\mapsto W(\alpha)$ from $\mathfrak{C}$ to $k_\alpha$-vector spaces), and if $\alpha<\beta$ then the map
$$W(\alpha)=\operatorname{Hom}_{\mathfrak{C}}(P_\alpha,W)\to\operatorname{Hom}_{\mathfrak{C}}(P_\beta,W)=W(\beta),$$
induced by the obvious inclusion $P_\beta\to P_\alpha$, is just $w_{\alpha,\beta}$.
Suppose $W$ were the product in $\mathfrak{C}$ of of a countable number of copies of $P_0$. Since it's an object of $\mathfrak{C}$, $W$ must be $\alpha$-good for some $\alpha$.
Then 
$$W(\alpha)=\operatorname{Hom}_{\mathfrak{C}}(P_\alpha,W)=\prod_{i\in\mathbb{N}}k_\alpha$$
and for $\beta>\alpha$
$$W(\beta)=\operatorname{Hom}_{\mathfrak{C}}(P_\beta,W)=\prod_{i\in\mathbb{N}}k_\beta.$$
But then $W(\beta)$ is not generated as a $k_\beta$-vector space by the image of the natural map $w_{\alpha,\beta}:W(\alpha)\to W(\beta)$, since $k_\beta$ is an infinite extension of $k_\alpha$, contradicting the $\alpha$-goodness of $W$.<|endoftext|>
TITLE: Combinatorial meaning of the functional equation for logarithm
QUESTION [24 upvotes]: If we set $\exp(x)=\sum x^k/k!$, then $\exp(x+y)=\exp(x)\cdot \exp(y)$. In terms of coefficients it means that $(x+y)^n=\sum \frac{n!}{k!(n-k)!} x^ky^{n-k}$, i.e. just binomial expansion.
Now consider logarithm. Set $L(x):=\sum_{k>0} x^k/k$, then $L(x)=-\log(1-x)$ in a sense, and hence $$L(u+v-uv)=L(u)+L(v),$$ i.e. $\sum (u+v-uv)^n/n=\sum (u^n+v^n)/n$, or, if we pass to coefficients of $u^av^b$ ($a,b\geq 1$), we get
$$
\sum_k (-1)^k\frac{(a+b-k-1)!}{(a-k)!(b-k)!k!}=0
$$
The question is what is combinatorial meaning of this identity. Maybe, it is some exclusion-inclusion formula, as it is usual for alternating sums?

REPLY [24 votes]: As David noted, since the summands aren't in general integers, it's difficult to give a combinatorial interpretation to the formula. However, if we multiply by $a$ or $b$ we get integers and we can give a combinatorial interpretation to the identity that we obtain (though doing this destroys the symmetry between $a$ and $b$).
If we multiply by $b$, the sum may be written
$$\sum_{k=0}^a (-1)^k \binom{a+b-k-1}{a-k}\binom{b}{k}.$$
This is a special case ($m=a+b-1$) of the more general identity
$$\sum_{k=0}^a (-1)^k\binom{m-k}{a-k}\binom{b}{k} = \binom{m-b}{a},$$
which we can prove combinatorially. (Incidentally, this identity is a form of Vandermonde's theorem.)
To prove this formula, we start with a set $M$ of size $m$, with a subset $B$ of size $b$. To interpret $\binom{m-k}{a-k}\binom{b}{k}$, we choose a $k$-subset $K$ of $B$ and then choose an $(a-k)$-subset $C$ of $M-K$. The right side $\binom{m-b}{a}$ counts pairs $(K,C)$ in which $K$ is empty and $C$ is an $a$-subset of $M-B$. To prove the identity, we find an involution  on the set of pairs $(K,C)$ not of this form that changes the parity of $|K|$. The pairs $(K,C)$ to be canceled are those in which  $K\cup C$ contains at least one element of $B$. Then the involution moves the smallest element of $(K\cup C) \cap B$ from $K$ to $C$ or from $C$ to $K$.<|endoftext|>
TITLE: Adelic description of moduli of $G$-bundles on a curve
QUESTION [12 upvotes]: Let $X$ be a smooth, projective, geometrically connected curve over a field $k$ and $G$ an an affine algebraic group group over $k$ (we can put more hypotheses on $G$ if necessary). If $K$ denotes the function field of $X$ and $\mathbb{A}$ the corresponding ring of adeles with integral adeles $\mathcal{O}$, I expect that there is a bijection from $G(\mathcal{O}) \backslash G(\mathbb{A}) / G(K)$ to the set of isomorphism classes of $G$-bundles on $X$ (although I have only seen this fact stated when $G = \text{GL}_n$). Also, I think there should be an interpretation of the "class group" $G(\mathbb{A}) / G(K)$ as $G$-bundles with level structure, but I don't even know how to make this into a precise statement.
Can someone explain these things to me? I would also be happy with a reference.

REPLY [14 votes]: The one-sentence answer to this question is: use fpqc descent theory (and an "answer" which doesn't address the role of fpqc descent -- sometimes presented in the form of a reference to a paper of Beauville and Laszlo -- is missing the key technical issue in the rigorous proof when working with general $G$, as far as I know).
We make two "necessary" hypotheses (as noted in the comments to Sawin's answer) for our smooth connected affine $k$-group $G$:  ${\rm{H}}^1(K,G) = 1$ for $K = k(X)$ and ${\rm{H}}^1(k',G) = 1$ for all finite extensions $k'/k$.  For example, if $k$ is finite then this holds for any simply connected semisimple (connected) $G$ by the theorems of Harder and Lang respectively.  If instead $k$ is algebraically closed of char. 0 then it holds for any $G$ by theorems of Tsen and Springer.  If $k$ is algebraically closed of positive characteristic then it holds for any connected reductive $k$-group $G$, but Springer's theorem doesn't literally apply; see Remark 2(b) of the Drinfeld-Simpson paper mentioned in the comments to Sawin's answer.
We shall allow $X$ to be any 1-dimensional reduced and irreducible $k$-scheme of finite type, not assumed to be proper or even normal. This way we incorporate Chervov's observations about using singular curves to build in more level structure. 
We construct the desired bijection as follows.  Consider a left $G$-torsor $E \rightarrow X$ (local triviality equivalent for the fpqc and etale topologies due to the smoothness of $G$; the equivalence will be crucial later on and is false in general if we try using the Zariski topology, and officially we work with the etale topology in the definition as is traditionally done).  Since ${\rm{H}}^1(K,G) = 1$, the generic fiber $E_{\eta}$ has a $K$-point and this spreads out over a dense open $U$ in $X$.  Fix such a $U$ and trivialization $\xi \in E(U)$. 
Let $X^0$ be the set of closed points of $X$. Consider the pullback of $E$ over the completion $O^{\wedge}_x$ at $x \in X^0$.  This pullback is a smooth $O^{\wedge}_x$-scheme whose special fiber is a $G$-torsor over the finite extension $k(x)$ of $k$ and so has a $k(x)$-point due to the other vanishing hypothesis.  By smoothness (!) of $G$ (and the henselian property of $O^{\wedge}_x$) this lifts to a point $\xi_x \in E(O^{\wedge}_x)$.
For each $x \in X^0$ consider the pullbacks of $\xi$ and $\xi_x$ over $U \times_X {\rm{Spec}}(O^{\wedge}_x) = {\rm{Spec}}(K_x)$ where $K_x$ is the total ring of fractions (product of finitely many fields) of the 1-dimensional reduced complete local ring  $O^{\wedge}_x$.  These two $K_x$-points of a common $G$-torsor are related through the action of a unique $g_x \in G(K_x)$; to be precise, $\xi = g_x \xi_x$ in $E(K_x)$.  If $x \in U^0$ then clearly $g_x \in G(O^{\wedge}_x)$, so $(g_x) \in G(\mathbf{A}_X)$.
If we
change $\xi$ then we multiply every $g_x$ on the left by some common $g \in G(U)$, and if we change the various $\xi_x$'s then we multiply each $g_x$ on the right by an element of
$G(O^{\wedge}_x)$.  Finally, taking into account that we may shrink $U$ (and thereby enlarge $X - U$), we obtain an element
$$(g_x) \in G(K)\backslash G(\mathbf{A}_X)/G(O^{\wedge})$$
(where $O^{\wedge} = \prod_{x \in X^0} O^{\wedge}_x$) that depends only on the isomorphism class of $E$ over $X$.
Our problem is to show that (i) this adelic double coset determines the isomorphism class of $E$ and (ii) all double cosets arise in this way. 
The assertion (i) is proved as follows. Assume $E$ and $E'$ give rise to the same double coset, so for a Zariski-dense open $U$ in $X$ trivializing $E$ and $E'$ we have $(g'_x) = \gamma (g_x) h$ for some $\gamma \in G(K)$ and $h \in G(O^{\wedge})$ with $g'_x, g_x \in G(O^{\wedge}_x)$ for all closed points $x$ of $U$. By shrinking $U$ we may assume $\gamma \in G(U)$. We may replace $\xi_x$ with $h_x\xi_x$ for all $x \in X^0$ and replace $\xi$ with $\gamma \xi$ so that $g'_x = g_x$ for all $x$.
In other words, $E_U$ and $E'_U$ are each identified with the trivial $G_U$-torsor and has corresponding trivial $K_x$-fiber identified with the generic fiber of $G_{O^{\wedge}_x}$ via $g_x$-translation. Working one point of $X - U$ at a time, we just have to check:
${\mathbf{Claim}}$: The category of $G$-torsors over $O_x$ is equivalent to the category of $G$-torsors over $O^{\wedge}_x$ equipped with a $K$-descent on its generic fiber over $K_x$. 
The categorical aspect of this Claim is essential (i.e., we do not just consider sets of isomorphism classes).
Proof: By fpqc descent theory (the "Beauville-Laszlo step", though for us all we need was provided by Grothendieck) applied to the fpqc cover
$${\rm{Spec}}(K) \coprod {\rm{Spec}}(O^{\wedge}_x) \rightarrow {\rm{Spec}}(O_x)$$
whose fiber square is ${\rm{Spec}}(K_x)$, 
the category of affine $O_x$-schemes is equivalent to the category of affine
$O_x^{\wedge}$-schemes equipped with a $K$-structure on the generic fiber over $K_x$. Since the notion of $G$-torsor is well-behaved for the fpqc topology (and recall that fpqc $G$-torsors are automatically etale-topology torsors, due to the smoothness of $G$!!!), this equivalence specializes to the case of $G$-torsors.  QED
Now we run the game in reverse.  Pick a class in $G(K)\backslash G(\mathbf{A}_X)/G(O^{\wedge})$ represented by some $(g_x) \in G(\mathbf{A}_X)$.  Since
$g_x \in G(O^{\wedge}_x)$ for all but finitely many $x \in X^0$, we can pick a Zariski-dense open $U$ in $X$ such that $g_x \in G(O^{\wedge}_x)$ for all $x \in U^0$, so 
we may change our representative to satisfy $g_x = 1$ for all $x \in U^0$.  Applying the above Claim then enables us to extend the trivial $G_U$-torsor to a $G$-torsor $E$ over $X$ by fpqc-gluing using the elements $g_x$ for each $x \in X - U$ one at a time, and by design this $E$  gives rise to the chosen adelic double coset (if we are careful not to mix up $g_x$ and $g_x^{-1}$ for $x \in X - U$).<|endoftext|>
TITLE: Least prime primitive root
QUESTION [9 upvotes]: For $p$ a prime number, let $G(p)$ be the least prime $q$ such that $q$ is a primitive root mod $p$, that is $q$ generates the multiplicative group $(\mathbb Z/p\mathbb Z$)* . 

Is it known that $G(p)=O(p)$ ? I don't mind if the answer assumes GRH or any other standard conjecture. 

I am interested in results true for all $p$, much less (though a little bit) on results which exclude a density $0$ or other smallish set of $p$. I note that it is easier to find bounds in the literature for $g(p)$, the least integer $n$ such that $n$ is a primitive root mod $p$.
For example $g(p)=O(p^{1/2+\epsilon})$ was known unconditionally to Vonogradov in the 1930's
(we have better unconditional results since), and with GRH we have result of type $g(p)=O(log^A p)$ with $A$ is some small constant. But what are the best results we have for $G(p)$? What are the best expected results ?
I am interested by $G(p)$ and not $g(p)$ because I use this problem as a testing ground of various effective forms of Chebotarev's there, and Chebotarev provides prime numbers. The best result I can prove this way is, under GRH, is $G(p)=O(p \log^{6+\epsilon} p)$ (edited: I made a mistake on the exponent of the $\log$), using Proposition 8.3 of the book of Ram Murty and Kumar Murty "Non-vanishing of $L$-functions and applications". With the GRH version of Lagarias-Odlyzko I get only $O(p^2 \log^2 p)$.
EDIT: Here is the proof of the estimate using Murty and Murty, as GH asked.
Proposition 8.3 of Murty and Murty states that if $G$ is the Galois group of an extension $L$ of $\mathbb Q$, $D$ a union of conjugacy classes in $G$, and $M=\sum \log p$, the sum being on the primes ramified in $L$, then 
$$| \pi_D(x) - \frac{|D|}{|G|} Li\, x | < C |D|^{1/2} x^{1/2} \log(Mx),$$
where $C$ is an absolute constant, $\pi_D(x)$ the numbers of primes $p \leq x$ such that  $Frob_p \in D$.
Let us apply this to $L=\mathbb Q(\mu_p)$, $D=$ set of primitive roots in $G=(\mathbb Z/p\mathbb Z)^\ast$.
If for some real $x$, the principal term $\frac{|D|}{|G|} Li x = Li(x)/2$ is bigger than the error term $C |D|^{1/2} x^{1/2} \log(p x)$, then $\pi_D(x) > 0$ which means that $G(p)< x$.
So we write that inequality, and solve it for $x$, using $|D|=\phi(p-1)$, and replacing 
$Li(x)$ by $x/\log x$ which just changes the constant $C$. So we want:
$$ x/(\log(x) x^{1/2}) > C \phi(p-1)^{1/2} (\log p + \log x).$$
Since $\log p \log x > \log p + \log x$ except for $x$ ridiculously small,
it is enough to have
$$ x/(\log(x) x^{1/2}) >  C \phi(p-1)^{1/2} \log p \log x,$$
or, taking the square,
$$x / \log^4(x) > C^2 \phi(p-1) \log^2 p$$
which is implied by 
$$x > C' \phi(p-1) \log^2 p \log^4(\phi(p-1) \log^2 p),$$
Hence $G(p)=O(\phi(p-1) \log^2 p \log^4(\phi(p-1) \log^2 p)) = O(p \log^{6+\epsilon} p)$.

REPLY [13 votes]: For the expected behavior, see Paszkiewicz and Schinzel's paper "On the least prime primitive root modulo a prime" in Math. Comp. 71 (2002), no. 239, 1307–1321. There they examine a conjecture of Bach that
$$\limsup \frac{G(p)}{(\log p)(\log\log p)^2}=e^{\gamma}.$$
It is known that almost always $G(p)$ is bounded by a fixed power of $\log{p}$, and the word "almost" can be removed if we assume GRH. (Under GRH, we in fact have $G(p) \ll (\log{p})^6$, and one can do better as long as $p-1$ doesn't have atypically many prime factors.) The best results I know in this direction are due to Greg Martin; see 
 "The Least Prime Primitive Root and the Shifted Sieve" in Acta Arith. 80 (1997), no. 3, 277–288;
also on arxiv.
Unconditionally, I believe it's not even known that $G(p)$ is less than $p$ for all large $p$.<|endoftext|>
TITLE: Does every finite free R-algebra have a basis starting with 1?
QUESTION [6 upvotes]: Suppose $S$ is an $R$-algebra (associative, commutative, with unit...) such that $S$ is free of finite rank $n$ over $R$. Is it necessarily the case that we can find an $R$-basis $y_1, ..., y_n$ for $S$ with $y_1 = 1$?
I can prove this when $n = 2$:
Suppose $x_1, ..., x_n$ is an $R$-basis for $S$, and write $1 = \sum_{i=1}^n a_ix_i$, with $a_i \in R$. We have $S$ faithfully flat over $R$, so we have $(a_1, ..., a_n) = (a_1, ..., a_n)S \cap R = (1)$, so we can find $b_1, ..., b_n$ in $R$ such that $1 = \sum_{i=1}^n a_ib_i$. Now I use the assumption $n = 2$: the matrix $\left( \begin{array}{cc} a_1 & a_2 \\\ -b_2 & b_1 \end{array} \right)$ is unimodular, so $y_1 = a_1x_1+a_2x_2,$ $y_2 = -b_2x_1+b_1x_2$ is an $R$-basis of $S$ with $y_1 = 1.$
More generally, as long as the unimodular row $(a_1, ..., a_n)$ can be completed to a unimodular matrix over $R$, we can use the unimodular matrix to produce an $R$-basis of $S$ with $y_1 = 1$.
Since not every unimodular row can be completed to a unimodular matrix when $n \ge 3$ and $R$ is arbitrary, it seems like it may be possible to construct a counterexample, but I haven't been able to do it (specifying an algebra structure on a rank $3$ $R$-module is surprisingly tricky).

REPLY [7 votes]: As Will Sawin says in his answer, if $R$ has a module $M$ such that the module $R\oplus M$ is free of rank 3 but $M$ is not free then there is a counterexample: $R\oplus M$ with zero multiplication in $M$.
For an example, let $R$ be the continuous (or just polynomial) functions on the $2$-sphere and let $M$ be the continuous (or polynomial) tangent vector fields.
Conversely, suppose $R$ has no such module. Whenever $S$ is an $R$-algebra such that $S$ is free of rank 3, then the ring map $R\to S$, considered as an $R$-module map, has a left inverse because the composed map $R\to S\to End_R(S)$ has a left inverse. (There is an $R$-basis for the $3\times 3$ matrices over $R$ starting with the identity matrix.) Therefore as an $R$-module $S$ is the direct sum of $R$ and a free module; $S$ has a basis starting with $1$.
Note: We didn't need commutativity in $S$. We didn't need commutativity of $R$, either; we could just assume $R\to S$ is a homomorphism of associative unital rings for the last argument. But you did need commutativity of $R$ to say that if the module $M$ is such that $R\oplus M$ is free of rank $2$ then $M$ is free of rank one.<|endoftext|>
TITLE: Why is Gauss credited with this connection?
QUESTION [11 upvotes]: Let $\pi : X \to B$ be a family of compact Kähler manifolds over a smooth base $B$. We then have a local system $\mathcal R^k \pi_* \mathbb Z$ (for your favorite $k$) of abelian groups over $B$, whose fiber over a point $b$ is the cohomology group $H^k(X_b, \mathbb Z)$.
We can tensor this system by $\mathcal O_B$ and obtain a holomorphic vector bundle $E^k \to B$. This bundle is equipped with a flat connection $\nabla$, that is induced by the exterior derivative $d$ in local coordinates. This connection is called the Gauss-Manin connection of the bundle $E^k$.
Now: Why do we call this the Gauss-Manin connection, when it seems that nothing that Gauss could have worked on relates to it?

REPLY [18 votes]: I've heard from a reliable source that Grothendieck decided on this name because he wanted to include the
first person who discovered the thing and the last person who made an important contribution:-)
(Of course, Grothendieck knew that there were many in between).
However, as it frequently happens, this attribution of the first discovery is not correct.
The fact that periods of an elliptic integral satisfy a hypergeometric equation was discovered
by Legendre. Actually several important discoveries of Legendre are attributed to Gauss,
(for example, quadratic reciprocity law, and the method of least squares)
and Legendre bitterly complained about this himself. 
Also the discovery of the arithmetic geometric mean is universally credited to Gauss, while Lagrange discovered it and published 6 years earlier, and Legendre used it to compute elliptic integrals.<|endoftext|>
TITLE: How many well orderings of $\aleph_0$ are there?
QUESTION [8 upvotes]: What is known about the set of well orderings of $\aleph_0$ in set theory without choice?  I do not mean the set of countable well-order types, but the set of all subsets of $\aleph_0$ which (relative to a pairing function) code well orderings.  And I would be interested in an answer in, say, ZF without choice.  My actual concern is higher order arithmetic.
I would not be surprised if ZF proves there are continuum many.  But I don't know.
At the opposite extreme, is it provable in ZF that there are not more well orderings of $\aleph_0$ than there are countable well-order types?

REPLY [7 votes]: This is an aside that I mentioned elsewhere long ago but deserves mention here since it homes in on the counterintuition that probably led Colin to doubt the answer.
As Colin pointed out, every $R \subset \omega$ can be interpreted as a binary relation on $\omega$ through a pairing function. This leads to a partition $\mathcal{B}$ of $\mathcal{P}(\omega)$ into isomorphism classes of binary relational structures $(\omega,R)$. Every countable infinite ordinal $\alpha$ has its own isomorphism class $B_\alpha \in \mathcal{B}$ and therefore $\aleph_1 \preceq \mathcal{B}$. We can also see that $2^{\aleph_0} \preceq \mathcal{B}$ in a multitude of ways. For example, we can map each $X \subseteq \omega$ to the isomorphism class of the directed graph consisting of one directed cycle of length $n+1$ for each $n \in X$ and infinitely many isolated points to fill space. In fact, we see that $\aleph_1 + 2^{\aleph_0} \preceq \mathcal{B}$ since the ranges of these two maps are disjoint. This is all provable without the axiom of choice.
There are models of ZF in which $2^{\aleph_0}$ and $\aleph_1$ are incomparable cardinals. Solovay's model where all sets of reals are Lebesgue measurable is such an example. In such models, $\mathcal{B}$ must have cardinality strictly greater than $2^{\aleph_0}$... Yes, that's right: $\mathcal{B}$ is a partition of $\mathcal{P}(\omega)$ that has more pieces than there are elements in $\mathcal{P}(\omega)$!<|endoftext|>
TITLE: Spanning trees of plane graphs containing an edge of every face
QUESTION [8 upvotes]: I feel sure this must be known, but can I find it??
Which connected plane graphs (graphs drawn in the plane without crossings) have a spanning tree such that at least one edge of each face is in the tree?
If multiple edges are allowed, there might be simply too many faces, and other obstructions are easy to find.  But I don't know about simple graphs.

REPLY [10 votes]: A triangulation has a spanning tree with the required property if and only if its dual graph has a hamiltonian path (is traceable).
Zamfirescu constructed a 3-regular 3-connected planar non-traceable graph on 88 vertices. 
The dual of this graph is a triangulation with no spanning tree with required properties.
a reference:
Tudor Zamfirescu, Three small cubic graphs with interesting Hamiltonian properties,
Journal of Graph Theory, Vol. 4 (1980), 287-292.<|endoftext|>
TITLE: Free ultrafilters on groups and irregularity
QUESTION [5 upvotes]: Hello,
Let $G$ be an infinite finitely generated discrete group. I call an infinite set $S$ irregular iff for every $g\in G$, $g\neq 1$, we have that $S\cap gS$ is finite. For example $\{z^3|z\in\mathbb{Z}\}$ is irregular in $\mathbb{Z}$. Now my easy to state question: Does every free ultrafilter on $G$ contain at least one irregular subset?
The following is true and might probably be useful: $G$ acts freely on the space of all ultrafilters (the Stone-Cech compactification of $G$ as a discrete space).
free ultrafilters: http://en.wikipedia.org/wiki/Ultrafilter

REPLY [4 votes]: No, a free ultrafilter on the additive group of integers need not contain an irregular set.  The Galvin-Glazer proof of Hindman's theorem (which is nowadays the standard proof of that theorem) begins by showing the existence of idempotent ultrafilters $U$ on $\mathbb N$.  I won't bother to define "idempotent" here, since what I need is not the definition but the following consequence of it.  If $X$ is any set in $U$, then there is an infinite set $\{x_0<x_1<\dots\}$ of positive integers such that any finite sum of distinct $x_i$'s is in $X$.  In particular, the intersection of $X$ and $x_0+X$ contains $x_0+x_i$ for all $i>0$, and therefore $X$ is not irregular.<|endoftext|>
TITLE: Does the metric space of compact metric spaces satisfy the binary intersection property?
QUESTION [8 upvotes]: A metric space $Y$ has the binary intersection property provided that whenever a collection of closed balls in $Y$ intersects pairwise, then there is a common intersection point.

Does the metric space $M$ of compact metric spaces under the Gromov-Hausdorff distance satisfy the binary intersection property?

The motivation is simple: I have a metric space $X$ with subspace $A$ and a Lipschitz map $f:A \to M$. I'd like to know if I can extend $f$ to all of $X$ without increasing the Lipschitz constant. It turns out (see Prop 1.4 here) that this binary intersection property is one of two hypotheses that must be satisfied by $M$ if it is to admit Lipschitz extensions for arbitrary metric space pairs $(X,A)$.

REPLY [5 votes]: Sergei Ivanov showed that infinite number of balls can have problems with compactness, but there are more bad news. I can take 3 balls which intersects pairwise, but don't have any common intersection point, or non compact "intersection point".
Indeed, Let $m^1,m^2,m^3$ be subsets of $\mathbb{R}$, consisting of three points each. And distances between them are $12,7,5$ for $m^1$, $10,5,5$ for $m^2$, and $10,7,3$ for $m^3$. Then GH-distanses between $m^1, m^2, m^3$ are less or equal then 1. If intersection of balls with centers in $m^1, m^2, m^3$ and radiuses $\frac{1}{2}$ is not empty, then I can take one element $m^{o}$ from it. 
There is metric space $m^{1} \cup m^{o}$ such that Hausdorff distances between $m^{1}$ and $m^{o}$ is less then $\frac{1}{2} + ε$. For each point $A_{i}$ from $m^{1}$ I take a point $B_{i}$ from $m^{o}$ such that $|A_i B_i|$ is less then $\frac{1}{2} + 2ε$. I will call $m^{oo}$ the metric space consisting of $B_1$, $B_2$, $B_3$. 
GH-distanses between $m^{i}$ and $m^{oo}$ are less or equal then $\frac{1}{2} + 2ε$,then distances between points of $m^{oo}$ are $11, 6, 4$ plus some epsilons, but that contradicts triangle inequality.<|endoftext|>
TITLE: Does there exist an isometry between $L^p$ and $l^p$?
QUESTION [13 upvotes]: The motivation is simple, as it is trivially right when $p=2$. When considering the duality between $L^p$ ($l^p$) and $L^q$ ($l^q$) when $p$ and $q$ are conjugate in the sense that $1/p+1/q=1$, I wonder if $L^p$ and $l^p$ are the same in the sense of isometry.
I tried to use the situation when $p=2$, however I find it difficult to give a linear homeomorphism.

REPLY [32 votes]: Variants of this question show up often enough here on MO and over at math.SE that it seems worthwhile to collect some facts and links. I say isomorphic for linearly homeomorphic and isometric for isometrically isomorphic.
One main upshot is:

The family of Banach spaces $L^p, \ell^q$ for $1 \leq p, q \leq \infty$ consists of pairwise non-isomorphic spaces, except for two cases:

There is the obvious isometry between $L^2$ and $\ell^2$ and
There is a non-obvious isomorphism between $L^\infty$ and $\ell^\infty$, due to Pełczyński. However, $L^\infty$ and $\ell^\infty$ are not isometric.


Plenty of references can be found in the following threads on MO and math.SE:

Bill Johnson explains in $L^p$ vs. $L^q$ how one can tell these spaces apart for $1 \lt p \neq q \lt \infty$ using type and cotype considerations, and in the comments there is also some discussion of existence of embeddings.
A detailed explanation of the non-existence of an isomorphism between the spaces $L^p$ and $\ell^q$ for $1 \leq p,q \lt \infty$ (modulo Bill Johnson's answer in 1.) is in the answer to If $1\leq p < \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic. The main ingredient in that answer is Pitt's theorem stating that every operator $\ell^q \to \ell^p$ is compact for $1 \leq p \lt q \lt \infty$, the existence of an embedding $L^2 \to L^p$ plus a little bit of duality theory. See also

How do you prove that $\ell^p$ is not isomorphic to $\ell^q$? for a discussion how Pitt's theorem implies non-isomorphism of $\ell^p$ and $\ell^q$ and 
Inclusion of $L^p$-spaces, reloaded for a discussion of embeddings of $L^2$ into $L^p$ via Rademacher functions or Gaussians.

Pełczyński's isomorphism between $L^\infty$ and $\ell^\infty$ is discussed in Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$? and non-existence of an isometry is discussed in Isometry between $L^\infty$ and $\ell^\infty$. 

Let me finish by recommending the very nice book by Albiac and Kalton, Topics in Banach Space Theory, as an alternative to Lindenstrauss-Tzafriri. It contains a gentle introduction to the above ideas and much more.

Edit: Further links to related topics:

One crucial point used in establishing Pełczyński's isomorphism is the injectivity of $\ell^\infty$ and $L^\infty$. For $\ell^\infty$ this is a standard exercise in applying Hahn-Banach (coordinatewise) and for $L^\infty$ Bill Johnson gives a quick proof in Direct proof of injectivity of $L^\infty$. See also Direct proof of "K is projective iff C(K) has the Hahn-Banach property"? for a generalization and related results.
Complemented subspaces of $\ell_p(I)$ for uncountable $I$
Isometric embeddings of $\ell_q^m$ into $\ell_p$ and $L_p$ for $p,q\in[1,+\infty]$<|endoftext|>
TITLE: Groups becoming algebraic groups
QUESTION [6 upvotes]: Let $G$ be an algebraic variety over an algebraically closed field $k$ (any characteristic). Suppose that:
(1) the set of $k$-points has the structure of a group. 
(2) for any $g\in G$ the right-multiplication by $g$ is a morphism of algebraic varieties $G\to G$.
(3) the inverse map is a morphism $G\to G$.
Does it imply that $G$ is an algebraic group? (i.e. is the multiplication $G\times G\to G$ a morphism?)

REPLY [12 votes]: I predict that in whatever is the situation of motivating interest, you know more: for any algebraically closed field $K/k$ you likewise have a group structure on $G(K)$ functorially in $K/k$ making the translations and inversions by $G(K)$ also be morphisms on $G_K$. Under this additional condition the answer is always affirmative, by using a trick that I believe goes back to Hasse, namely base change to the function field of $G$ (or rather, in our situation, an algebraic closure thereof).  This is a Weil-style way of getting at Yoneda's Lemma.
The inversion hypothesis implies that left-translations are also morphisms, and we will use this condition instead of the inversion hypothesis (and deduce at the end that inversion is a morphism too).  In fact, in char. 0 we won't even need this condition with the left-translations (or inversion) at all, but we are using a stronger assumption on the existence of group laws on $K$-points for lots of $K/k$. This stronger initial assumption across many $K/k$ bypasses Bondarko's examples in char. 0. The near-miss examples avoiding the inversion condition in char. $p$ in the comments are not ruled out by our stronger hypothesis across all $K/k$, but they are ruled out by the left-translations being morphisms (!), so matters will be more delicate in positive characteristic. We need the left-translation condition in positive characteristic to circumvent some inseparability issues with function fields of $k$-varieties.  
Before we begin the actual arguments, note that since $k$ is algebraically closed, so $G(k)$ is Zariski-dense in $G$, for any field extension $F/k$ the subset $G(k)$ inside $G(F) = G_F(F)$ is Zariski-dense $G_F$. Hence, $F$-morphisms among $G_F$, $G_F \times G_F$, etc. are determined by their effect on $k$-points promoted to $F$-points. I mention this at the outset for peace of mind in some later discussions.
The "translation is a morphism" hypothesis ensures that $G$ is smooth, so its connected components are irreducible.  I will assume that the variety $G$ is connected, a harmless assumption since (i) for $g \in G^0(k)$ the right-translation morphism by $g$ preserves $G^0$ because it carries $e$ to $g$, (ii) inversion preserves $G^0$ since it fixes $e$, (iii) the morphism property on $G^0$ clearly implies the general case by using the "morphism" property for a few translations by points in the various connected components of $G$.  Thus, now $G$ is irreducible and so has a "function field" $k(G)$. 
By applying the functoriality in $K/k$ to $k$-automorphisms of $K$, we see via Galois descent that the initial hypothesis for algebraically closed extensions is actually valid for all perfect extensions of $k$ as well.  Let $K$ be the perfect closure of $k(G)$, and let $\eta \in G(k(G)) \subset G(K)$ correspond to the generic point. This induces a right-translation morphism $\rho_K:G_K \simeq G_K$ (computing right-translation by $\eta_K$). Since $G$ is finite type over $k$, by expressing $K$ as a directed union of finite purely inseparable extensions of $k(G)$ we obtain a finite purely inseparable extension $F$ of $k(G)$ and an $F$-morphism $\rho_F:G_F \simeq G_F$ that descends $\rho_K$ and clearly computes right-translation by $\eta_F$. The normalization $X$ of $G$ in the finite extension $F/k(G)$ is a $k$-variety finite radiciel over $G$, and $X_{\eta} = {\rm{Spec}}(F)$ over $k(G)$.  Thus, by expressing $k(G)$ as the directed union of the coordinate rings of affine opens around $\eta$ in $G$, we find such an open $U$ so that over its preimage $U'$ in $X$ (which is finite radiciel over $U$) there is a $U'$-morphism $\rho_{U'}:G_{U'} \simeq G_{U'}$ extending $\rho_F$.
The map $U'(k) \rightarrow U(k)$ is bijective, so for each $g \in U(k)$ let $g' \in U'(k)$ be the unique point over $g$.  Consider the specialization $\rho_{g'}:G \simeq G$ of $\rho_{U'}$ over $g'$.  I claim that this is the right-translation by $g$. To see this, pick $h \in G(k)$ and consider the morphism of left-translation $\ell_h:G \simeq G$ by $h$.  This carries $\eta$ to another $k(G)$-point of $G$ (over $k$) that "spreads out" to a $U$-point of $G$ whose specialization at $g \in U(k)$ is $\ell_h(g) = hg$ (so if we work instead with the $U'$-point over this via the canonical $U' \rightarrow U$ then its specialization at $g' \in U'(k)$ is also $\ell_h(g) = hg$). Now inside the group $G(K)$ we have  $\ell_h(\eta_K) = \rho_K(h_K)$, so $\rho_{U'}(h_{U'})$ is the $U'$-point obtained from applying $\ell_h$ to the canonical $U'$-point of $G$ (spreading out $\eta_F$) because this comparison of $U'$-points may be checked by working at the generic point of $U'$ (and this generic point is dominated by the canonical $K$-point, over which we have the equality $\ell_h(\eta_K) = \rho_K(h_K)$). Specializing this equality of $U'$-points at the point $g' \in U'(k)$ over $g \in U(k) \subset G(k)$ gives the equality of $k$-points $\rho_{g'}(h) = \ell_h(g) = hg$, as desired.
To summarize, we have constructed a $U'$-morphism $\rho:G_{U'} \simeq G_{U'}$ whose specialization at any $g' \in U'(k)$ is right-translation on $G$ by the corresponding 
point $g \in U(k)$. Suppose ${\rm{char}}(k) = 0$, so $U' = U$ and the composite morphism
$${\rm{pr}}_1 \circ \rho:G \times U \rightarrow G$$
on $k$-points is the group law restricted to $U(k)$ in the 2nd variable. For any $g \in G(k)$ the right translation morphism $\rho_g$ carries $U$ to another open subscheme $\rho_g(U)$, and I claim that these opens cover $G$. It suffices to check the covering property on $k$-points, so for $h \in G(k)$ we seek $g \in G(k)$ such that $h \in \rho_g(U)$, or equivalently $hg^{-1} \in U(k)$.  Pick any $u \in U(k)$ and let $g = u^{-1}h$. This proves the covering property, and the multiplication map $G(k) \times U(k)g \rightarrow G(k)$ arises from a morphism, namely the composition of ${\rm{pr}}_1 \circ \rho$ and the morphism $\rho_g$.  We conclude (in char. 0) that the group law on $G(k)$ is induced by a morphism $G \times G \rightarrow G$.  
Now we can deduce in char. 0 that inversion is a morphism (so we have a group variety), even though we never used the left-translation or inversion conditions. Indeed, the "universal right-translation" $G \times G \rightarrow G \times G$ defined by $(x,y) \mapsto (xy,y)$ between fppf $G$-schemes (via ${\rm{pr}}_2$) is a scheme isomorphism between fibers over all points in $G(k)$ and therefore is a scheme isomorphism (by fibral isomorphism criteria, adapted to the peculiarities of $k$-points when $k$ is alg. closed). This yields the morphism property for inversion for free! This was a char-free argument, but it required the composition law to be a morphism.  Anyway, char. 0 is now settled.
Assume ${\rm{char}}(k) = p > 0$, so for sufficiently large $n \ge 0$ the finite flat $n$-fold relative Frobenius morphism $G^{(1/p^n)} \rightarrow G$ of $G^{(1/p^n)}$ dominates $U'$ over $U$ (namely, pick $n \ge 0$ so that the finite purely inseparable extension $F/k(G)$ is contained inside $k(G)^{1/p^n}$). Since the initial choice of $F$ could be replaced with a finite purely inseparable extension at the outset if we wish, we may therefore assume that $U'$ is open inside $G^{(1/p^n)}$.  Using a covering and translation argument similar to characteristic 0, we arrive at a morphism
$$m_r:G \times G^{(1/p^n)} \rightarrow G$$
that recovers the given group law on $k$-points via the natural identification of $G^{(1/p^n)}(k)$ with $G(k)$ (for some $n \ge 0$). Our problem is precisely to show that $m_r$ factors (in the sense of morphisms of varieties) through the $n$-fold relative Frobenius morphism in the 2nd variable. 
Now we shall use that left-translations are morphisms too. At the cost of increasing our $n$ if necessary, we can run through the same arguments with "left" instead of "right" to arrive at another morphism
$$m_{\ell}:G^{(1/p^n)} \times G \rightarrow G$$
which recovers the given group law on $k$-points (with the same $n$).
Returning to our task of checking that $m_r$ factors through the appropriate iterated Frobenius in the 2nd variable, since that iterated Frobenius is fppf (as $G$ is smooth!) we conclude via fppf descent that it is harmless to check the existence of such a factorization after precomposing $m_r$ with an fppf morphism in the first variable.  So let's compose with the same iterated Frobenius in the first variable, arriving at a morphism
$$G^{(1/p^n)} \times G^{(1/p^n)} \rightarrow G$$
that recovers the composition law of $G(k)$ on $k$-points.  It suffices to show that this latter map factors through the $n$-fold Frobenius in its 2nd variable, but this factorization is clear: it is the composition of $m_{\ell}$ with that iterated Frobenius (as we may check by computing on $k$-points, since we're working with reduced $k$-schemes of finite type). QED<|endoftext|>
TITLE: Seeing the vertices of a polygon with rational angles
QUESTION [7 upvotes]: Given any convex polygon in the plane, is it always possible to find a point $p$ in its interior such that when we draw the line segments from $p$ to each of its vertices, the angles formed at $p$ are all (not necessarily equal) rational multiples of $\pi$?  
For a triangle $T$, it's easy to construct such a point, namely the Steiner point $p$ will do, enjoying three angles of measure $2\pi/3$ each, between $p$ and any two adjacent vertices of $T$.  But is this known in general?

REPLY [6 votes]: Consider the manifold of all $k$-sided polygons. This is $2k-3$ dimensional. Given a set of angles $\theta_1,...,\theta_k$, the manifold of polygons with those angles from a point is $k$-dimensional, since it's determined by the distances of the vertices from the point. 
You're not going to cover a $2k-3$ - dimensional manifold with countably many $k$-dimensional manifolds if $k\geq 4$. In particular, these maps are real algebraic maps, and thus aren't anything like space-filling curves.<|endoftext|>
TITLE: Bisectors in symmetric spaces
QUESTION [5 upvotes]: In William Goldman's book Complex Hyperbolic Geometry, bisector hypersurfaces play an important role. Given two points $x,y$, the bisector is the set of points equidistant from $x$ and from $y$. Do they also play an important role in the study of higher rank symmetric spaces, and do they admit concrete descriptions, say in the Siegel upper half-plane? References are especially welcome.

REPLY [3 votes]: Although I cannot address your specific question, of whether bisectors "play an important role in the study of higher rank symmetric spaces," 
I can say that
bisectors are the essence of Voronoi diagrams.
So the 
2009 paper by
Frank Nielsen and Richard Nock, entitled
"Hyperbolic Voronoi diagrams made easy"
(arXiv link), could well be relevant:

We present a simple framework to compute hyperbolic Voronoi diagrams of finite point sets as affine diagrams. We prove that bisectors in Klein's non-conformal disk model are hyperplanes that can be interpreted as power bisectors of Euclidean balls. [...]<|endoftext|>
TITLE: Relation between $\neg \square(\kappa)$ and the tree property at $\kappa$.
QUESTION [9 upvotes]: If $\kappa$ is an inaccessible cardinal then the tree property at $\kappa$ is equivalent to weak compactness of $\kappa$, which implies that $\square(\kappa)$ fails---that is, that every coherent sequence of clubs of length $\kappa$ can be threaded.
I am wondering about other implications involving square and the tree property, namely:

If $\kappa$ is an inaccessible cardinal and $\square(\kappa)$ fails, must $\kappa$ have the tree property (and therefore be weakly compact?)
If $\kappa$ is a regular cardinal, does $\neg \square(\kappa)$ imply that $\kappa$ has the tree property?
If $\kappa$ is a regular cardinal with the tree property, must $\square(\kappa)$ fail?

(By the way, I am aware of the relative consistency result that if $\square(\kappa)$ fails for some regular cardinal $\kappa$, then $\kappa$ is weakly compact in $L$ and so in particular it has the tree property in $L$.)

REPLY [5 votes]: To complete the accepted answer to my question (which addresses parts 2 and 3) perhaps I should mention here that the answer to part 1 is no: If  $\delta$ is a supercompact cardinal and $\kappa$ is the least inaccessible cardinal above $\delta$ then $\square(\kappa)$ fails but $\kappa$ is not weakly compact.<|endoftext|>
TITLE: Implications of non-negativity of coefficients of arbitrary Kazhdan-Lusztig polynomials?
QUESTION [39 upvotes]: In their seminal 1979 paper Representations of Coxeter groups and Hecke algebras (Invent. Math. 53, doi:10.1007/BF01390031),
Kazhdan and Lusztig studied an arbitrary Coxeter group $(W,S)$ and the corresponding Iwahori-Hecke algebra.   In particular they showed how to pass from a standard basis of this algebra to a more canonical basis, with the change of basis coefficients involving polynomials indexed by pairs of elements of $W$ (in the Bruhat ordering) over $\mathbb{Z}$.   Even though the evidence at the time was quite limited, they conjectured following the statement of their Theorem 1.1 that the coefficients of these polynomials should always be non-negative.  (In very special cases this is true because the coefficients give dimensions of certain cohomology groups.)
Several decades later, Wolfgang Soergel worked out a coherent strategy for proving the non-negativity conjecture, in his paper

Kazhdan–Lusztig-Polynome und unzerlegbare Bimoduln über Polynomringen. J. Inst. Math. Jussieu 6 (2007), no. 3, 501–525, doi:10.1017/S1474748007000023, arXiv:math/0403496

Now that his program seems to have been completed, it is natural to renew the question in the header:

What if any implications would the non-negativity of coefficients of arbitrary Kazhdan-Lusztig polynomials have?

It has to be emphasized that in Soergel's formulation and the following work, the non-negativity is not itself the main objective.   Instead the combinatorial framework proposed was meant to provide a more self-contained conceptual setting for proof of the original Kazhdan-Lusztig conjecture on Verma module multipliities for a semisimple Lie algebra (soon a theorem) and further theorems in representation theory of a similar flavor.   But Coxeter groups form a vast general class of groups given by generators and relations, so it is surprising to encounter such strong constraints on the polynomials occurring in this generality.
ADDED: There is some overlap with older questions related to Soergel's approach, posted here and
here.
UPDATE: It's been pointed out to me that older work by Jim Carrell and Dale Peterson involves the non-negativity condition, though their main goal is the study of singularities of Schubert varieties in classical cases.   See the short account (with a long title)

J.B. Carrell, The Bruhat graph of a Coxeter group, a conjecture of Deodhar, and rational smoothness of Schubert varieties.
Algebraic groups and their generalizations: classical methods (University Park, PA, 1991), 53–61, Proc. Sympos. Pure Math., 56, Part 1, Amer. Math. Soc., Providence, RI, 1994. https://doi.org/10.1090/pspum/056.1

The first section develops for an arbitrary Coxeter group some consequences of non-negativity of Kazhdan-Lusztig coefficients for the combinatorial study of Bruhat intervals.   For further details about the geometry, see

Carrell, J., Kuttler, J. Smooth points of T-stable varieties in G/B and the Peterson map. Invent. math. 151, 353–379 (2003). https://doi.org/10.1007/s00222-002-0256-5, arXiv:math/0005025

I'm still not sure whether such consequences of the 1979 K-L conjecture are enough to make the conjecture in itself "important".    But it's definitely been challenging to approach.

REPLY [3 votes]: Maybe I can provide a belated kind of answer to my own question, which I came across when looking for something else in the older literature.  Vinay Deodhar published a paper in 1990 here (just before family medical problems and then his own health prevented him from continuing his research).  This might have further combinatorial interest relative to Kazhdan-Lusztig polynomials for Coxeter groups outside the traditional framework of Lie theory.  But his approach depends crucially on the assumption that a given element of the Coxeter group is "good", which is implied by non-negativity of all coefficients in certain of the KL polynomials.    Deodhar seems to have expected, in line with the conjecture of Kazhdan and Lusztig, that this would always be satisfied.    
The impressive work of Elias and Williamson published in 2014 here extends earlier work of Soergel on his bimodules and thereby proves the non-negativity conjecture in general.   So Deodhar's algorithmic procedure might be worthwhile to revisit.   In any case, those who have access to Math Reviews should find it useful to track the reviews and later citations of both of these papers.<|endoftext|>
TITLE: There are two points on the Earth's surface that ... ?
QUESTION [15 upvotes]: At every moment in time, there are two points on the Earth's surface that have the same $\lbrace x, y, z, ... \rbrace$...?  
What is the strongest, most impressive statement one can make here?  The Borsuk-Ulam Theorem applies, but I am uncertain of its full implications.  Could one say that the two points are
(1) separated by a specific geodesic distance, (2) have the same temperature, and (3) have the same barometric pressure?  For example...?  I pose this question for its pedagocial import, but it clearly follows from known theorems.
To what extent do these results extend to $\mathbb{R}^d$ for $d>3$?
Thank you for your help!
           
(Wikipedia image)

REPLY [14 votes]: One of the standard generalizations is Knaster's conjecture: for every function $f: \mathbb{S}^{n-1}\rightarrow \mathbb{R}^m, m\lt n,$ and $k=n-m+1$ points $p_1, \dots, p_k \in \mathbb{S}^{n-1}$ does there always exista rotation $\rho \in SO(n),$ such that $f(\rho(p_1) = \dots = f(\rho(p_k)).$ That this is true for $k=2$ is a theorem of H. Hopf (which generalizes Borsuk-Ulam). It turns out that Knaster's conjecture is true for some $m, n$ and false for others. See this nice paper by Hinrich and Richter for more results and references.<|endoftext|>
TITLE: What goes wrong in Easton forcing if we don't just use regular cardinals?
QUESTION [5 upvotes]: Recall that Easton forcing was introduced to show that the continuum function at regular cardinals could be anything subject to 'the obvious constraints' (monotonicity etc). However, it is a handy method if one wants to add a proper class of sets. My question is why do would we now restrict to using forcing conditions only using regular cardinals (edit: if we wanted only to add class-many sets)? I've had a read through Friedman's Class forcing, and all the (nontrivial) examples given there are variants on Easton forcing, only playing with things like supports and stationarity.
I'm not interested in preserving AC, though I suspect that we lose tameness at some point, and hence axioms like powerset may fail to hold.

Hmm, let me state my actual question, which was in fact rather implicit (and everyone's comments/answers have helped me figure out how to phrase it, so thank you all).

If I try to add $F(\kappa)$ generic subsets to each cardinal $\kappa$ (by some simple class function $\kappa \mapsto F(\kappa)$, such as the identity, or constant at some given infinite cardinal), will I get a model of ZF(C)? Or is the restriction to adding subsets to only regular cardinals, as in Easton forcing, necessary?

REPLY [8 votes]: To increase the power set of a regular cardinal $\kappa$, Easton used forcing conditions that are partial functions of size $<\kappa$.  So the forcing is $\kappa$-closed and therefore adds no new subsets of any cardinals below $\kappa$.  It therefore doesn't interfere with whatever he was trying to do with the power sets of those smaller cardinals.  If he did the same thing with a singular $\kappa$, the forcing would be only cf$(\kappa)$-closed, not $\kappa$-closed.  For example, if $\kappa=\aleph_\omega$, then the union of a countable chain of conditions (each of size $<\aleph_\omega$) could have size $\aleph_\omega$ and thus fail to be a condition.  As a result, new subsets would be added at cardinals below $\kappa$ (but $\geq$ cf$(\kappa)$), thereby messing up whatever was supposed to happen with the power sets of those cardinals.
A decade later, Silver discovered that not only does Easton's method not work for singular cardinals (which Easton already knew), but there are non-trivial constraints on $2^\kappa$ for singular $\kappa$.  In particular, a singular cardinal of uncountable cofinality cannot be the first place where GCH fails.  Later, it was shown (I believe first by Magidor) that a singular cardinal of countable cofinality can be the first place where GCH fails, but a large cardinal was needed for the proof and, by a result of Jensen, large cardinals are unavoidable here.  Work of Gitik has pinned down the exact large-cardinal strength of the negation of the singular cardinal hypothesis.  
The bottom line here is that, in order to get anything like Easton's results for singular cardinals, one must use large cardinals, one must use considerably fancier forcing notions than Easton used, and even then, some manipulations of power sets of singular cardinals are outright impossible.<|endoftext|>
TITLE: What manifolds can have a (non-piecewise) linear structure?
QUESTION [7 upvotes]: By the definition I'm using, all manifolds are Hausdorff and second countable. For all non-negative integers $n$, I define $B_n$ to be $\bigl\{ \mathbf{v} \in \mathbf{R}^n : \lVert\mathbf{v}\rVert < 1 \bigr\}$.

For what manifolds does there exist an atlas of charts $c : U \to B_n$ such that the transition maps are all locally affine?

(Replacing "locally affine" with "piecewise affine" would make the answer, by definition, those manifolds for which there exists a piecewise linear structure.)

REPLY [11 votes]: If I am right, such manifolds are called affine manifolds. They are smooth manifolds together with a flat, torsion free connection.
Maybe it is worth recalling Chern's conjecture that the Euler characteristic of an affine manifold should vanish.
Kostant B. and Sullivan D. in: 
"The Euler characteristic of an affine space form is zero, Bull. Amer. Math. Soc. 81 (1975)", no. 5, 937-938
proved this conjecture in the case of the quotient of the ordinary space ${\mathbb R}^n$ by a discrete group of affine transformations.<|endoftext|>
TITLE: Subscript 0 in Reverse Mathematics
QUESTION [6 upvotes]: What does the subscript 0 mean on terms like  $\mathsf{ATR}_0$?  Does it mean the same thing in $\Pi^1_k\text{-}\mathsf{CA}_0$?
If I frame higher order analogues of these, should I change that subscript?

REPLY [5 votes]: As the other answer points out, the subscript 0 means restricted induction. However, without the subscript 0, there are two conventions:

The older convention was that the systems without the subscript 0 have the full second-order induction scheme. Thus $\mathsf{ACA}$ is the system consisting of $\mathsf{ACA}_0$ plus the full induction scheme, and the same for e.g. $\mathsf{WKL}$ vs. $\mathsf{WKL}_0$. Historically, the systems with full induction were studied first (as in Feferman's article in the Handbook of Mathematical Logic), and the systems with restricted induction were of secondary interest. The restricted systems drew more interest once it was apparent how many mathematical results they can prove, so that in the context of reverse mathematics the systems with full induction seem less natural. 
Some authors, however, use ACA to mean "Arithmetical Comprehension Axiom" and WKL to mean "Weak König's Lemma". For these authors, $\mathsf{ACA}_0$ is $\mathsf{RCA}_0$ plus "ACA". Similarly $\mathsf{WKL}_0$ means $\mathsf{RCA}_0$ plus "WKL". This terminology appears in various papers, even some published by respected proof theorists, and so you have to watch for it. Note that ACA here can be taken to be a single sentence "for all $X$ the Turing jump of $X$ exists" and similarly WKL is a single sentence. 

For higher-order analogues, it is still a question whether restricted induction or full induction is included. Kohlenbach [1] has used notation such as $\mathsf{ACA}_0^\omega$ to refer to the analogue of $\mathsf{ACA}_0 $
 formalized in arithmetic in all finite types. In this context, though, there are many different ways in which induction can be restricted. So notation like $\widehat{\mathsf{E\text{-}HA}}^\omega_\upharpoonright $ is used in the literature, where the hat and the harpoon refer to different sorts of restrictions. These notations are explained in Kohlenbach's Applied Proof Theory or in Troelstra's Metamathematical Investigations. 
1: Ulrich Kohlenbach, "Higher Order Reverse Mathematics", Reverse Mathematics 2001, Lecture Notes in Logic, 2005, ftp://ftp.daimi.au.dk/BRICS/RS/00/49/BRICS-RS-00-49.pdf<|endoftext|>
TITLE: Can pure mathematics harness citizen science?
QUESTION [42 upvotes]: Having just finished Michael Nielsen's book "Reinventing Discovery", I find myself wondering if there are ways that pure mathematics research can engage the public in the way that GalaxyZoo or Foldit have in astronomy and protein science respectively. In other words, to break up large-scale research problems into manageable chunks which can be tackled by massed ranks of amateur enuthusiasts (citizen scientists).
Let me give you three examples of what I have in mind (which each have their own drawbacks)

One obvious historically relevant possibility which comes to mind would have been getting people to check cases of the four-colour theorem. This could obviously be done by computer, but perhaps if it were done by hand(s) the proof would be less controversial. The main drawback is that the proof has already been done (by computer) and people would maybe feel like they were wasting their time because they could be replaced by a computer. A good citizen science project would make crucial use of the fact that its citizens were human.
Perhaps, given a suitably programmed piece of software for manipulating Kirby diagrams, amateurs could be let loose on checking new potential counterexamples to the smooth Poincare conjecture in dimension 4 (like the recent Nash ones which were found to not be counterexamples by Akbulut). The disadvantage would be that such software would be pretty hard to write (I imagine).
Donaldson's Lefschetz fibration theorem effectively reduces symplectic geometry in 4-dimensions to study of the mapping class groups of surfaces. In particular it would be useful to find new factorisations of the identity in mapping class groups into words of right-handed Dehn twists (these correspond to Lefschetz fibrations on closed symplectic manifolds). A piece of software allowing people to play with mapping class groups might help find such interesting factorisations. The drawback would be that this definitely feels like searching for a needle in a haystack. The beauty of GalaxyZoo is that each click you make is contributing positively to the project, even if you're not finding a new type of galaxy. With a needle-in-a-haystack problem, users would get frustrated very quickly. It's also possible that a computer would perform better.

So my question is:
Can we collectively come up with some reasonable (proto)-propositions for research projects in pure mathematics which would be amenable to citizen science? Ideally these would be both useful for pure mathematics, and intellectually engaging and rewarding for the citizens (preferably not searching for a needle in a haystack).
Edit: Just to be clear, I am interested in specific suggestions for mathematical problems which could be amenable to solution by 'citizen science' and I don't want to open up a discussion about whether this is desirable or sensible: that's not part of MathOverflow's remit. A priori, it's not even clear to me that such mathematical problems exist and I would be interested to hear if more imaginative people than me can come up with suggestions.

REPLY [4 votes]: 1.Harnessing the existing public interest on major math problems such as Turbulence and Riemann hypothesis. Provide people with software to study toy models from these problems and search for interesting structures. For example, in turbulence we have the rise of persistent structures (called coherent structures) such as the hairpin.

(Attracting (red) and repelling (blue) Lagrangian CSs extracted from a two-dimensional turbulence experiment)

Analytic Number theory is amenable to public science like having people search for particular kinds of primes by providing them with some software.

REPLY [2 votes]: There are many open bijective problems in algebraic combinatorics,
that could be attacked by amateurs.
It is basically about finding a rule to explain expansions. The data could be generated easily, or posted by a mathematician, and the public can try to explain it.
Examples:

The $e$-expansion of certain chromatic symmetric functions. This is about assigning a partition to acyclic orientations of certain graphs, to match some data.
The qt-Catalan symmetry problem. Basically finding an involution on Dyck paths interchanging two quite elementary statistics.
Inventing statistics on semi-standard tableaux explaining the qt-Kostka numbers related to modified Macdonald polynomials.<|endoftext|>
TITLE: Non-stably trivial bundle with trivial characteristic classes
QUESTION [19 upvotes]: Though it's relatively clear that the characteristic classes do not characterise a vector bundle (and after looking through some books) I could not find an example of a vector bundle which is not stably trivial but whose characteristic classes (those which may be defined*)   are all trivial. Could someone be so kind as to point out a reference for this?
*an example with trivial Stiefel-Whitney class would already be nice.

REPLY [10 votes]: There is an old reference which gives low-dimensional examples. The paper 
A. Dold and H. Whitney. Classification of oriented sphere bundles over a $4$-complex. Ann. Math.(2) 69 (1959), pp. 667--677
contains a classification of oriented vector bundles over CW-complexes of dimension $4$. Take a rank $n$ real vector bundle $\mathcal{E}$, $n\geq 5$, with trivial Stiefel-Whitney classes and trivial Pontryagin class over a $4$-dimensional CW-complex $X$. Then there is a cohomology class in $H^4(X,\mathbb{Z})$ which measures the difference between $\mathcal{E}$ and the trivial bundle, the class is $0\mod 2$ and annihilated by multiplication by $2$. Conversely, any such class can be realized by a bundle. 
Therefore, a non-trivial bundle exists on a $4$-complex if there are classes which are $0\mod 2$ and annihilated by $2$. The obvious example of such a complex is the Moore space $M(\mathbb{Z}/4,4)$ obtained by glueing a $4$-cell to $S^3$ with a degree $4$ map. Note also that rank $n\geq 5$ on a $4$-complex is in the stable range, so the bundle obtained from the Dold-Whitney paper is stably non-trivial. 

Edit: Since the question was tagged differential-geometry, I'll also give a manifold example. Take a free action of $\mathbb{Z}/4\mathbb{Z}$ on $S^5$ (which can be obtained from the embedding $S^5\subset \mathbb{C}^3$ and coordinatewise multiplication with suitable roots of unity). The quotient $X=S^5/(\mathbb{Z}/4\mathbb{Z})$, a lens space, is a $5$-dimensional smooth manifold with $H^4(X,\mathbb{Z})\cong\mathbb{Z}/4\mathbb{Z}$. This means that there is a non-trivial $SO(6)$-bundle with trivial characteristic classes on the $4$-skeleton, as follows from the Dold-Whitney paper. Since $\pi_5(BSO(6))=0$, obstruction theory implies that any $SO(6)$-bundle on the $4$-skeleton extends uniquely to a vector bundle on $X$.  Again, rank 6 on dimension 5 is stable range, so the bundle constructed this way is stably non-trivial.<|endoftext|>
TITLE: Proof without words for surface area of a sphere
QUESTION [6 upvotes]: I love the book Proofs Without Words by Roger B. Nelsen. One of the proofs I liked the most was this: Area under one arch of a cycloid is 3 times the area of the wheel that traces it. You break the cycloid in to three parts and show that each part has an area equal to that of the wheel.
I have always wondered if there is a similar proof for why the surface area of a sphere is equal to the area of a circle with radius twice that of the sphere. Is there a nice way to see this?
I would also like to know if one can show without doing any calculus that the length of the arch of the cycloid is 8 times the radius of the wheel. (Note: $\pi$ does not show up here, so this sounds tricky!)

REPLY [14 votes]: $$\frac{dy}{ds} = \frac{r}{R} \ \implies \ 2 \pi r \cdot ds =2 \pi R \cdot dy$$
I was recently watching through 3 Blue 1 Brown videos, and learned that he did this argument, and also a second proof.<|endoftext|>
TITLE: Characterization of bounded geometry - Reference-request
QUESTION [6 upvotes]: I already asked this question at stackexchange three days ago. Since I got no answer, I want to try mathoverflow now. I hope that you can help.
I'm looking for a proof of an equivalence that can e.g. be found in a paper by Shubin 'Spectral theory of elliptic operators on non-compact manifolds' (Appendix A.1.1 below Def. 1.1).
It's about manifolds of bounded geometry, where bounded geometry means here: positive injectivity radius and the curvature tensor and all its covariant derivatives are bounded.
In the paper of Shubin it is written that there are the following equivalent characterizations: (Let $(M,g)$ always have positive injectivity radius.)
(i) $(M,g)$ is of bounded geometry.
(ii) Let $r>0$ be smaller than the injectivity radius. For any $x,x′\in M$ let $y$ (resp. $y′$) be geodesic normal coordinates around  $x$ (resp. $x′$) with an r-ball as domain. If the balls of radius $r$ around $x$ and $x′$ have a nonempty intersection, then the transition function $y^{−1}\circ y$ and all its derivatives are uniformly bounded (where uniformly means here independent of $x$ and $x′$).
I know that (i) is equivalent to the following: (iii) Let $g^\alpha_{ij}$ be the representation of the metric coefficienst on a ball of radius $r$ around $x_\alpha$ with respect to geodesic normal coordinates. Then $g^\alpha_{ij}$ and all its derivatives are uniformly bounded (where uniformly means independent on $\alpha$).
I can see that (iii) implies (ii) (by looking at the geodesic flow and using Gronwall's inequality). But I have no idea how to get the converse. Is there any reference for the proof? I'm grateful for any idea for the proof as well.
Thank you in advance.

REPLY [8 votes]: I assume that by "all derivatives" you mean derivatives of every order.
Suppose that all transitions between normal coordinates have uniformly bounded derivatives within some radius $r$. For every point $p$ we have a unit radial vector field $V=V(p)$ whose derivatives are uniformly bounded at distances between, say $r/10$ and $r$ from $p$. (This holds in normal coordinates centered at $p$ and hence in normal coordinates centered at any nearby point.)
Now fix a point $q\in M$ and arrange points $p_k$, $k=1,\dots,\frac{n(n+1)}2$, where $n=\dim M$, so that the distances from $q$ to $p_k$ equal $r/2$ and the corresponding vector fields $V_k=V(p_k)$ at $q$ are generic in the sense that the symmetric tensors $V_k\otimes V_k$ are linearly independent. Then the same holds at every point in a neighborhood of $q$. Actually the points $p_k$ should have some prescribed coordinates in normal coordinate system centered at $q$, then all subsequent estimates are uniform.
We have a system of equations $g(V_k,V_k)=1$, $k=1,\dots,\frac{n(n+1)}2$. At every point near $q$ this is a nondegenerate linear system on the components of the metric tensor $g$. Hence it uniquely determines $g$ via some explicit formulae in terms of $V_k$. Since the derivatives of $V_k$ are uniformly bounded, so are the derivatives of the metric tensor.<|endoftext|>
TITLE: Fredholm theory on Fr\'echet spaces
QUESTION [8 upvotes]: Dear everybody,
In my study of the classial Fredholm theory on Banach spaces, I am interested in the corresponding Fredholm theory on Fr\'echet spaces. But it seems to me that there is
little research results in this aspect. 
Let $X$, $Y$ and $Z$ be Fr\'echet spaces. An operator $T \in L(X,Y)$
is said to be upper (resp. lower) semi-Fredholm, if its kernel space
$\mathrm{ker}(T)$ is finite dimensional and its range space $R(T)$ is closed
in $Y$. In this case, the index of $T$ is defined as
$\mathrm{ind}(T)=\dim \mathrm{ker}(T) - \dim Y/R(T)$.
Now, one of my question is: 
Suppose that $T \in L(X,Y)$ and $S \in L(Y,Z)$ are upper
semi-Fredholm, $ST$ is upper semi-Fredholm and $\mathrm{ind}(ST)=
\mathrm{ind}(S)+\mathrm{ind}(T)$? (It holds in Banach spaces case) 
If you have any opinions to this question or this topic, please communicate with me. 
Thanks!

REPLY [2 votes]: This book: Spectral Theory of Linear Differential Operators and Comparison Algebras
 By Heinz Otto Cordes (mostly available on google books)
Discusses this, as does the earlier:
Einige Klassen singularen Gleichungen, by S. Prössdorf, Akademie-Verlag,
Berlin; Mathematische Reihe, Band 46, Birkhâuser, Verlag, Basel and
Stuttgart, 1974, 353 pp.<|endoftext|>
TITLE: 2 generated arithmetic groups
QUESTION [6 upvotes]: Suppose $G({\mathbb Z})$ is a higher rank non-cocompact arithmetic group (e.g. $SL_n({\mathbb Z})$ with $n\geq 3$, or $Sp_{2g}({\mathbb Z})$ with $g\geq 2$). I have seen a result (http://arxiv.org/abs/math/0409345) which says that every finite index subgroup $\Gamma $ of $G({\mathbb Z})$ contains a smaller finite index subgroup generated by three elements. 
Does anyone know ANY example of $G({\mathbb Z})$, where three can be replaced by two? I believe Alan Reid has some result in this direction.  
[Edit] That 2 should suffice is a conjecture, attributed to Alex Lubotzky. That $3$ DO suffice for non-uniform higher rank lattices in the result mentioned in the link. What I am asking is just ONE example where 2 generators suffice.

REPLY [4 votes]: The result to which you refer is not a result but a conjecture of A. Lubotzky. Long and Reid have constructed some examples. -- the relevant preprints can be found on Alan Reid's web page. I assume that Lubotzky's conjecture is about three and not two generators because he did not want to be too ambitious -- nobody knows anything concrete, to the best of my knowledge.<|endoftext|>
TITLE: Computation of homotopy groups of spheres via Pontryagin-Thom
QUESTION [18 upvotes]: The Pontryagin-Thom construction identifies $\pi_{n+k}(S^n)$ with the group of bordism classes of framed $k$-dimensional submanifolds of $S^n$.  Before Serre's work introduced algebraic tools into the subject, this was used to calculate $\pi_{n+k}(S^n)$ for $0 \leq k \leq 2$ by Pontryagin and for $k=3$ by Rokhlin.  
Does there exist a modern exposition of these proofs anywhere?  The case $k=0$ is trivial, but as far as I can tell, the only sources for $k \geq 1$ are Pontryagin's book "Smooth manifolds and their application to homotopy theory" and Rochlin's original paper.  The book is very old-fashioned and spend way too much time developing the foundations of smooth manifold theory (I guess there was no nice source in the early 1950's), and Rokhlin's paper is unreadable (to me).

REPLY [8 votes]: I agree with you that Pontryagin's book is very hard to read!  And while there are many places that have brief sketches of what is going on (some given in the other answers), I am not aware of any other serious exposition of its contents.  To correct this, I have written a detailed modern account of Pontryagin's approach to calculating $\pi_{n+1}(S^n)$ and $\pi_{n+2}(S^n)$  in my notes "Homotopy groups of spheres and low-dimensional topology", which are available on my page of notes here.  I did not include a discussion of Rochlin's theorem, which is quite a bit harder.  The book of "A la recherche de la topologie Perdue" by Guillou and Marin (mentioned in Mike-Doherty's answer) is a good place to start for that.<|endoftext|>
TITLE: Do L-functions exist for Half-integral weight modular forms?
QUESTION [13 upvotes]: Classically, we can attach $L$-functions (with properties like, analytic continuation, functional equation) to Dirichlet characters, Hecke eigenforms, etc... 
My question is: can one attach $L$-functions (properties similar to the above) to Half-Integral weight modular eigenforms as well?  If they do exist, can someone
provide a reference for this?

REPLY [12 votes]: Since the original question asked only for analytic continuation, functional equation, I'd like to add that the Mellin transform $\sum a_n n^{-s}$ of the half integral weight modular form $\sum a_n \exp(2 \pi i n z)$ has
these two properties, but it lacks an Euler product decomposition even if the modular form is a Hecke eigenform, since the Fourier coefficients at square free indices are not multiplicatively related.
A similar phenomenon occurs for Siegel modular forms with the so called Koecher-Maaß series; here the Fourier coefficients at matrices representing maximal lattices (Kern- resp. Stammformen in Brandt's terminology) have no multiplicative relation.<|endoftext|>
TITLE: g.c.d. and Euler's totient function
QUESTION [31 upvotes]: There is this really nice paper by J.P.Serre on the congruence subgroup property for $SL_2$ for $S$-arithmetic groups (https://www.jstor.org/stable/1970630). If one looks at the proof of Proposition 3 there, Serre in fact proves the following result. 
Let $a,b \in {\mathbb N}$ be two co-prime integers, and $\phi$ be Euler's totient function. For each $x\in {\mathbb N}$ we may consider $\phi (ax+b)$. Now consider the g.c.d. of the infinite set of numbers
$$N(a,b)= g.c.d. \{ \phi (ax+b): x=1,2,3,\cdots \}.$$  Now $N(a,b)$ seemingly depends on $a,b$ but it does not much: $N(a,b)$ divides $8$. 
The proof of this uses Dirichlet's theorem on infinitude of primes. 
If ${\mathbb Q}$ is replaced by a number field $K$, and $a,b$ are co-prime integers, define $\phi (ax+b)$ to be the number of units in the quotient ring $O_K/(ax+b)$, then the analogous g.c.d. divides $2\mu _K^2$ where $\mu _K$ is the number of roots of unity in $K$. 

My question is : if I replace the linear polynomial $ax+b$  by any polynomial $P(x)=a_0+ a_1x+\cdots+ a_nx^n$, with the numbers $a_0,a_1, \cdots, a_n$ co-prime and $a_n\neq 0$, then does the corresponding g.c.d. 
  $$g.c.d \{\phi (P(x)):x=0,1,2,..\}$$ depend (i.e. is bounded by a constant dependent) only on the degree $n$ and not on the polynomial? 

The question came up in a question on discrete groups, which could be resolved, but THIS question remained. I do not have any applications for this, but I thought it was interesting on its own.  
[Edit] I should have added the link https://arxiv.org/abs/math/0409377. 
[Edit] The following paper https://arxiv.org/abs/1909.10808 answers this affirmatively (unconditionally for $n=2$ and modulo a well known conjecture in the general case). So the answer is Yes.

REPLY [7 votes]: I have made some computations which seem to corroborate the OP's conjecture, namely that for any $n$ there exists a $N$, such that for every polynomial $P$ of degree $n$, with positive integral coefficients and content 1, the quantity $$g(P):= g.c.d(\phi(P(x)),x \geq 1)$$
divides $N$.
For $n=1$, as the OP says, one can take $N=8$ as proved by Serre.
For $n=2$, it seems that one can take $N=2^4 3^2 = 144$. It seems even more
that one cannot do better, because for $P(x)=16x^2+32x+17$, I get experimentally $g(P)=16$
(this must not be hard to prove but I haven't tried), and for $P(x)=27 x^2 + 9x+1$, I get $g(P)=18$. So $144 | N$. On the other hand I have need been able to find any $P$ such that $g(P)$ was not a divisor of $144$.
For $n=3$ or $n=4$, I have failed to find any $P$ with $g(P)\geq 2$. This suggests $N=2$ in these cases.<|endoftext|>
TITLE: On intermediate transitive models for ZFC between M an M[G]
QUESTION [6 upvotes]: Let $P$ be a forcing notion. Let $B(P)$ be the boolean completion of $P$ and $i : P \rightarrow B(P)$ be the corresponding dense embedding (in $B(P)^{+}$). Let $G$ be $B(P)$-generic over $M$, the transitive ground model satisfying ZFC. 
I know that if $N$ is a transitive model of ZFC such that $M \subset N \subset M[G]$, then $N = M[D \cap G]$ for some complete subalgebra $D$ of $B(P)$. But can we say anything about $(X :=) ran(i) \cap D$ and $(Y :=) i^{-1}[D]$? Is $X$ dense in $D^{+}$? Is $Y$ the range of some complete embedding into $P$? Are there any other interesting properties about them?
Thanks in advance.

REPLY [7 votes]: It depends on the particular forcing, and in general, things may
not work out so nicely.
On the one hand, it could be that $P=\mathbb{B}^+$, in which case
for any intermediate model $N$ we have $X=Y=D^+$ and so $X$ is
dense in $D^+$ and everything you want is true.
On the other hand, consider the case where $P$ is the forcing
consisting of conditions $(s,t)$, where $s,t\in 2^{{\lt}\omega}$
are finite binary sequences of the same length $|s|=|t|$. The
Boolean completion is $\mathbb{B}=\text{Add}(\omega,2)$ the
forcing to add two Cohen reals $M[c,d]$. Consider the intermediate
extension $M[c]$ to add just the first one. In this case, $D$ is
the subalgebra consisting of conditions in $\mathbb{B}$ that do
not determine any information about the second real $d$, although
they may decide information about the first real $c$. This does
not interact well with the image of $P$ inside $\mathbb{B}$,
because conditions in the range of $P$ decide an equal number of bits for both $c$ and $d$. In
particular, $D$ contains no members of $\text{ran}(i)$ except for the
trivial condition $1$. So in this case, $X$ is not dense in $D^+$,
and $Y$ has only the trivial condition.<|endoftext|>
TITLE: For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$.
QUESTION [22 upvotes]: For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$.
I cannot find a counter-example to this. Do we know if this inequality is true? Alternatively, is this some documented problem (solved or unsolved)?

REPLY [16 votes]: Ramanujan (1919), see Eq. (18): 

$$\pi(x) - \pi(x/2) \ge 2 \quad \text{ for } x\ge 11 $$

Whence, with $x= 2p_k$ for $p_k \ge 7$, $$p_{k+2} \le 2 p_k \lt p_k+p_{k+1}, $$  and $5\le 2+3$, $7\le 3+ 5$, $11 \le 5+7$.

REPLY [11 votes]: As a matter of fact, P. L. Chebyshev knew already that for any $\epsilon > \frac{1}{5}$, there exists an $n(\epsilon) \in \mathbb{N}$ such that for all $n\geq n(\epsilon),$ 
$\pi((1+\epsilon)n)-\pi(n)>0.$
In [2], one can find a short report on the problem of determining the smallest $n(\epsilon)$ explicitly once that $\epsilon$ has been fixed.
References
[1] P. L. Chebyshev. Mémoire sur les nombres premiers. Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.
[2] H. Harborth & A. Kemnitz. Calculations for Bertrand's Postulate. Mathematics Magazine, 54 (1), pp. 33-34.<|endoftext|>
TITLE: Cohomologically trivial stacks
QUESTION [19 upvotes]: The following theorem of Serre is well-known:

A noetherian scheme $X$ is affine if and only if $H^i(X; \mathcal{F}) = 0$ for all quasi-coherent sheaves $\mathcal{F}$ on $X$ and all $i>0$. (Actually it is enough to have this for $i=1$ and all coherent ideal sheaves.)

I asked myself whether there is an extension of this theorem to (Artin/Deligne-Mumford) stacks. More precisely:

Question: Can one characterize the class of (Artin/Deligne-Mumford) stacks $X$ such that $H^i(X; \mathcal{F}) = 0$ for all quasi-coherent sheaves $\mathcal{F}$ on $X$ and all $i>0$?

It is certainly not true that affine schemes are here the only examples. For example, take a graded ring $A$ and consider $X = Spec A // \mathbb{G}_m$ (where the $\mathbb{G}_m$-action is induced by the grading). The category of quasi-coherent sheaves on $X$ is (by fpqc-descent) equivalent to that of graded $A$-modules and the global sections functor corresponds to taking the zeroth degree of such a graded module. This is clearly exact and thus all higher cohomology groups of all quasi-coherent sheaves vanish.

REPLY [12 votes]: Suppose that $\mathcal X$ is an algebraic stack with finite inertia (for example, a separated Deligne-Mumford stack); then, by a well-known result of Keel and Mori, there exist a moduli space $\pi \colon \mathcal X \to M$. The stack $\mathcal X$ is called tame when $\mathrm R^i\pi_* F = 0$ for every quasi-coherent sheaf $F$ on $\mathcal X$ and every $i > 0$. From the definition it follows easily that tame stacks with affine moduli spaces have the property you require. In characteristic 0, an algebraic stack with finite diagonal is tame if and only if it is Deligne-Mumford.
There are several different characterizations of tame stacks; see the paper "Tame stacks in positive characteristic" by Dan Abramovich, Martin Olsson and myself. Using the results in that paper, it is not hard to show that a noetherian algebraic stack with finite inertia has the property you want if and only if it is tame with affine moduli space.
[Edit:] here is a proof that if a noetherian algebraic stack $\mathcal X$ with finite inertia has the property you want it is tame with affine moduli space. Let $\mathcal X \to M$ be the moduli space. Let $\mathcal G$ be the residual gerbe over a closed point of $M$; then $\mathcal G$ is closed in $\mathcal X$, so the cohomology of each quasi-coherent sheaf on $\mathcal G$ is trivial. The moduli space of $\mathcal G$ is the spectrum of a field, so $\mathcal G$ is tame. This implies that the automorphism group of an object of $\mathcal G$ is linearly reductive. One of the results in the paper implies that an open neighborhood of $\mathcal G$ in $\mathcal X$ is tame. Since every non-empty closed subset of $M$ contains a closed point of $M$, this implies that these open neighborhoods cover $\mathcal X$, so $\mathcal X$ is tame.<|endoftext|>
TITLE: Complete anti-chain lattices and the axiom of choice
QUESTION [5 upvotes]: Hello, everyone. I'm trying to find out about lattices of anti-chains, and was wondering whether you could help me with getting to grips with a Comp. Sci. paper I'm struggling with.
I've been reading two related papers on algorithms that operate over a lattice of anti-chains. The authors give a construction for deriving the lattice of anti-chains from an underlying powerset algebra and make claims that the resulting structures are complete. I have a feeling that the authors make the hidden assumption that the powerset algebra is finite for their completeness argument (which makes sense, since the authors are ultimately concerned with finite automata), and was wondering whether someone could help me figure out whether a partial fix requires the axiom of choice. 
My interest in this is with respect to finding completeness conditions for lattices of anti-chains.
I will try to summarise the general construction given in the paper. I hope I'm not misrepresenting.

Let $(\wp(S),\subseteq, \cap,\cup)$ be a powerset algebra with glb (meet) $\cap$ and lub (join) $\cup$. An anti-chain is a subset $C$ of $S$ such that all pairs of distinct elements $c,c' \in C$ are mutually incomparable, i.e., $c \not \subseteq c'$ and $c \not \supseteq c'$. We denote the set of all anti-chains of $\wp(S)$ as ${\mathbb{C}}$.
The set of anti-chains $\mathbb{C}$ is partially ordered by the binary relation $\sqsubseteq \in \wp(\mathbb{C}\times\mathbb{C})$ defined as follows:
  $$C \sqsubseteq C' \text{ exactly if for all }c \in C\text{ there exists a }c' \in C'\text{ such that }c \sqsubseteq c'$$

So far so good. The authors go on to define binary meets and joins, here's my first suspicion that something is missing:

We can define binary meets and joins over the poset $(\mathbb{C},\sqsubseteq)$ as follows, where $\max S$ and $\min S$ denote, respectively, the maximal and minimal elements of a set $S$.
  \begin{align*} 
C \sqcap C' &= \max \{ c \cap c' ~|~ c \in C \land c' \in C'\} & 
C \sqcup C' &= \max \{ c ~|~ c \in C \lor c \in C'\} 
\end{align*}

Hold on there. This obviously works for finite lattices, but I'm not so sure about the meet in the infinite case. What if $C$ and $C'$ are infinitely big antichains, couldn't the set $\{ c \cap c' | c \in C \land c' \in C' \}$ simply lack maximal elements?
Then the authors go on to give operators for the non-binary joins and meets. In this paper one can reasonably assume that the authors simply forgot to mention that what they say only holds in finite lattices, but the other (section 2) explicitly defines meets and joins for arbitrary sets, and I don't see how they work. Here's what they say:

For a set of antichains $Q \subseteq \mathbb{C}$, we can define 
  \begin{align*}
\sqcap Q &= \max \{ \bigcap_{q \in Q} s_q ~|~ s_q \in q \} &
\sqcup Q &= \max \{ s_q ~|~ \exists q \in Q.~s_q \in q \} 
\end{align*}

So again, I think there's the problem that these maximal elements may not exist. Consider the case where the underlying powerset algebra has infinite ascending chains, e.g., consider $\wp(\mathbb{N})$ and the chain:
$$ \{1\} \subseteq \{1,2\} \subseteq \{1,2,3\} \ldots $$
Now we can derive an ascending chain of singleton anti-chains:
$$ \{\{1\}\} \sqsubseteq \{\{1,2\}\} \sqsubseteq \{\{1,2,3\}\} \sqsubseteq \ldots$$ 
But following the join operator for sets given above we get $\emptyset$ as the join of this chain, since there is no maximal element. Am I missing something?
Also, one could attempt to fix this, by requiring that the underlying lattice satifies the ascending chain condition (i.e., that it does not have ascending chains). But then I still find the definition of the meet troubling. I have a bit of trouble parsing what it means, and I can't clearly explain why. I think it is equivalent to the definition below:
$$\sqcap Q = \max \{ \bigcap_{q \in Q} f(q) ~|~ \text{ for choice function $f: Q \to \bigcup_{q \in Q} q$ s.t. $\forall q\in Q. f(q) \in q$}\}$$ 
This does require the axiom of choice doesn't it? Or does it somehow get around that by not actually being interested in the choice function itself, but just in its codomain? 
I'm happy for any suggestions / ideas / insight. Are my issues making sense or am I missing something?

REPLY [2 votes]: I claim that the lattice described above is a complete lattice if and only if $X$ is finite.
Let $X$ be a countably infinite set. Assume that the set $\{\{R\}|R\subseteq X,|R|<\infty\}$ has a least upper bound $\mathcal{A}$. Put a relation $\simeq$ on $P(X)$ where $R\simeq S$ iff there is a bijection $f:X\rightarrow X$ with $f[R]=S$. Then if $R,S\subseteq X,R\simeq S$, then $R\in\mathcal{A}$ iff $S\in\mathcal{A}$. Therefore $\mathcal{A}$ is the union of a collection of equivalence classes. However, we take note that the equivalence classes are the sets of the forms $\{R\subseteq X:|R|=n\},\{R\subseteq X:|X\setminus R|=n\},\{R\subseteq X:|R|=|X\setminus R|=\infty\}$. However, the set $\mathcal{A}$ cannot contain more than one equivalence class since $\mathcal{A}$ must be an antichain. Therefore $\mathcal{A}$ is precisely one equivalence class. The set $\mathcal{A}$ cannot be any of the equivalence classes of the form $\{R\subseteq X:|R|=n\}$ since the sets $\{R\subseteq X:|R|=n\}$ are not upper bounds of $\{\{R\}|R\subseteq X,|R|<\infty\}$.  Furthermore, $\mathcal{A}\neq\{R\subseteq X:|R|=|X\setminus R|=\infty\}$ since
$\{R\subseteq X:|R|=|X\setminus R|=\infty\}$ is not an antichain. Therefore, $\mathcal{A}=\{R\subseteq X:|X\setminus R|=n\}$ for some $n$. On the other hand, $\{R\subseteq X:|X\setminus R|=n+1\}$ is another antichain and
$\{R\subseteq X:|R|=n\}\leq\{R\subseteq X:|X\setminus R|=n+1\}<\{R\subseteq X:|X\setminus R|=n\}$, This contradicts the fact that $\mathcal{A}=\{R\subseteq X:|X\setminus R|=n\}$ is the least upper bound of $\{\{R\}|R\subseteq X,|R|<\infty\}$. Therefore the collection of antichains in $P(X)$ is not an complete lattice.
What goes wrong? If $X$ is a set, then we say $L\subseteq P(X)$ is a lower set if whenever $R\in L$ and $S\subseteq R$, then $S\in L$ as well. Let $\mathcal{L}(X)$ denote the collection of lower sets in $X$. Then $\mathcal{L}(X)$ is clearly a complete lattice.
For finite sets, the antichains are in a one-to-one correspondence with the lower sets by the following correspondences. If $A$ is an anti-chain, then one simply takes the lower set generated by $A$. If $L$ is a lower set, then the collection of all maximal elements in $L$ is an anti-chain. Therefore since $\mathcal{L}(X)$ is a complete lattice, the collection of all anti-chains also forms a complete lattice.
On the other hand, for infinite sets, there are lower sets with no maximal elements such as the collection of all finite subsets of an infinite set $X$. Therefore this one-to-one correspondence breaks down for infinite sets.<|endoftext|>
TITLE: String topology for a Lie group  
QUESTION [6 upvotes]: My question is very naive maybe, I don't have a deep knowledge about string topology. I wanted to ask (explanation or a reference) for the geometric interpretation 
of free loop space (continues maps) $\mathrm{Map}(S^{1}, G)$, when G is a topological group (Lie group). Does it classify something geometric? 
Thank you.

REPLY [4 votes]: The free loop space sits in a fibration
$$
\Omega M \to M^{S^1} \to M
$$
and in the case where $M = G$ is a Lie-group, I understand the main point of Richard Hepworths paper http://arxiv.org/abs/0905.1199 as saying that the fibration trivializes to a product $G^{S^1} \cong G \times \Omega G$. This triviality then makes it more evident how the string topology operations is a mixture of the two geometric entities: the intersection product of $\mathbb{H}_*(G)$ and the $E_1$-structure on the based loops $\Omega G$.<|endoftext|>
TITLE: What are some consequences of the Mumford-Tate conjecture?
QUESTION [10 upvotes]: Let $A/K$ be an abelian variety over a number field. On the one hand we have the singular cohomology group
$$V := H^1(A(\mathbb{C}),\mathbb{Q})$$
with respect to some fixed embedding $K \subset \mathbb{C}$, and from this we can construct the Mumford-Tate group $G_A \subset Aut_\mathbb{Q}(V)$, a reductive algebraic group over $\mathbb{Q}$, associated to the natural Hodge structure of $V$. 
On the other hand, for each prime $l$, we have the $l$-adic étale cohomology group
$$V_l := H^1(A(\overline{K}),\mathbb{Q}_l)$$ 
with respect to some fixed algebraic closure $\overline{K}$ of $K$, and from this we can construct the $l$-adic algebraic monodromy group $G_{A,l} \subset Aut_{\mathbb{Q}_l}(V_l)$, a reductive algebraic group over $\mathbb{Q}_l$, defined as the Zariski closure of the image of the $G_K$-representation acting continuously on $V_l$ (this latter representation being dual to that on the Tate module). The comparison isomorphism $V_l \cong V \otimes_\mathbb{Q} \mathbb{Q}_l$ allows us to compare the identity component $G^0_{A,l}$ of $G_{A,l}$ with $G_{A,\mathbb{Q}_l}$ (that is, $G_A$, but viewed as an algebraic group over $\mathbb{Q}_l$), and the Mumford-Tate conjecture predicts that these two groups are the same. 
Here is my question:

What are some consequences of the Mumford-Tate conjecture for the arithmetic of abelian varieties?

An example of the sort of thing I have in mind is the following recent result of David Zywina: if $A$ as above is absolutely simple, $K$ is large enough, and the Mumford-Tate conjecture for $A$ holds, then the density of good primes $v$ of $K$ where the reduction $A_v/\mathbb{F}_v$ is also absolutely simple is 1 if and only if the endomorphism ring of $A$ is commutative. (Zywina's result is actually finer than this, see the link.)
I guess I'm asking for other places in the literature where results of this flavour are proven to follow from MTC. 
(A word on my motivation: I have a problem regarding abelian varieties which I think may follow from MTC, but I'm rather stuck proving this. I think my cause would be helped if I had more tools available.)

REPLY [4 votes]: Here is a result in a somewhat different direction. Assuming MTC, Hindry and Ratazzi recently proved a bound for the torsion subgroup of abelian varieties of ${\rm GSp}$-type over arbitrary finite extensions.<|endoftext|>
TITLE: Is there an algebraic geometry analogue of the closed graph theorem?
QUESTION [29 upvotes]: In functional analysis, the closed graph theorem asserts that if a linear map $T: X \to Y$ between two Banach spaces $X, Y$ has a closed graph $S := \{ (x,Tx): x \in X \}$, then the map is continuous.  Thus, it gives a criterion for regularity of a map in terms of regularity of the graph of that map.
I am curious as to whether there is any analogous statement in algebraic geometry.  The naive formulation would be: if $T: X \to Y$ was a function (in the set-theoretic sense) between algebraic varieties $X, Y$ over an algebraically closed field $k$ whose graph $S := \{ (x,Tx): x \in X \}$ was also an algebraic variety, then $T$ would be a regular map.  (Here I will be vague as to whether I want varieties to be affine, projective, quasiprojective, or abstract.)  But this is false,  even in characteristic zero: for instance, the coordinate function $(t^2,t^3) \mapsto t$ from the cuspidal curve $\{ (t^2,t^3): t \in k \}$ to $k$ has a graph which is an algebraic variety, but is not a regular map (it is not given by a rational function in a neighbourhood of the origin).  In characteristic $p$, the inverse of the Frobenius map $x \mapsto x^p$ provides another counterexample.  Somehow the difficulty is that regular functions in $S$ need not come from pullback from regular functions in $X$, even though the vertical line test suggests that such maps should be "degree 1" in some sense.
Still, I feel like there should be some positive statement to be made here, though I was not able to find one after searching through a few algebraic geometry texts.  For instance, if one demands that $X, Y, S$ are all smooth and that the field has characteristic zero, does the claim now hold?  Ideally, I would like to only have conditions on the varieties $X,Y,S$ and not on the various maps between these varieties; for instance, I would prefer not to have to assume that the projection map from $S$ to $X$ is finite (though perhaps this is automatic?).

REPLY [26 votes]: You might be rediscovering Zariski's Main Theorem, which implies your statement in case $X$ is normal (or just weakly normal) and the projection from the graph $\Gamma$ to $X$ is proper and separable. What you really need is the map $\Gamma\to X$ to be an isomorphism, so the question is equivalent to "when is a bijective map an isomorphism"?
You already explained why we need (weak) normality (example with the cuspidal curve) and separability (the Frobenius map). To see why properness is also necessary, look at the map $\mathbb{A}^1\to\mathbb{A}^1$ sending $x$ to $1/x$ for $x\neq 0$ and sending $0\to 0$, whose graph is a union of $(0,0)$ and a hyperbola $xy=1$.<|endoftext|>
TITLE: Categorifications of Zorn's lemma
QUESTION [17 upvotes]: I'm wondering about categorifications of Zorn's lemma along the following lines.

Lemma: if $\mathbf{C}$ is a small category in which every directed diagram of monomorphisms has a cocone of monomorphisms, then there is an object $A$ such that any monomorphism $A \rightarrowtail B$ splits.

Proofsketch: If there were no such object, we could use the axiom of choice to define a function $f$ that assigns to each directed diagram $D$ of monomorphisms, a monomorphism with domain $\mathop{cocone}(D)$ that does not split. Define an ordinal-indexed diagram $D$ by setting $D(\alpha) = f\big( D(\beta) \mid \beta<\alpha \big)$. Because $D$ consists of monomorphisms that don't split, it cannot contain any cycles. But this contradicts smallness of $\mathbf{C}$.
Can this be generalised? (E.g. do we need smallness and monics?) Are other versions known? Are there similar categorical existence statements that are provable without the axiom of choice?

REPLY [7 votes]: In their book Categories and Sheaves, M. Kashiwara and P. Schapira prove the following theorem (Theorem 9.4.2) :

Let $C$ be an essentially small non empty category which admits small filtrant inductive limits. Then $C$ has a quasi-terminal object, that is, an object $X$ such that every morphism with source $X$ has a left inverse.

It seems to me that the hypothesis is stronger than yours, but the conclusion is also stronger. (What about considering the subcategory $C'$ of $C$ whose arrows are monomorphisms of $C$?)<|endoftext|>
TITLE:  A definition of non-commutative metrisable space
QUESTION [8 upvotes]: If $X$ is a compact metrisable space, a metric $d$ on $X$ can be take as an element of  $C(X\times X)$ such that
(1)  $ev_x\otimes ev_y (d)=d(x,y)\geq 0$ for all $x,y\in X$ (Non-negativity).
(2) $ev_x\otimes ev_y (d)=0$ iff $x=y$ (Identity of indiscernibles).
(3) $ev_x\otimes ev_y (d)=ev_y\otimes ev_x (d)$ for $x,y\in X$ (Symmetry).
(4) For all $x,y,z\in X$, $ev_x\otimes ev_y (d)\leq ev_x\otimes ev_z (d)+ev_z\otimes ev_y (d)$ (Triangle inequality).
Motivated by this, we can give the following definition of quantum compact metrisable space.
A unital nuclear $C^*$-algebra $A$ is called a quantum compact metrisable space if there exists an element $d\in A\otimes A$ such that
(1) $d\geq 0$(non-negativity).
(2) $\psi\otimes \phi(d)=0$ iff $\psi=\phi$ for $\psi,\phi\in P(A)$ where $P(A)$ is the pure state space of $A$(Identity of indiscernibles).
(3) $\psi\otimes \phi(d)=\phi\otimes \psi(d)$ for $\psi,\phi\in P(A)$(Symmetry).
(4) $\psi\otimes \phi(d)\leq \psi\otimes \varphi(d)+\varphi\otimes \phi(d)$ for all $\psi,\phi$ and $\varphi\in P(A)$(Triangle inequality).
I check that for $M_n(C)$, the $C^*$-algebra of $n$ by $n$ complex matrices and found that if a $d$ satisfies non-negativity, identity of indiscernibles and symmetry, then $d$ does not satisfy triangle inequality. So this means that $M_n(C)$ can only admit a semi-metric in this sense. 
My question: is there any genuine quantum compact metrisable space(i.e. noncommutative $C^*$-algebra $A$ with such a $d\in A\otimes A$ satisfying the above properties)?

REPLY [10 votes]: I would suggest that you look at the work of Rieffel on compact quantum metric spaces.  His point of view is not to directly generalize the metric by understanding it as an element of the tensor square $A \otimes A$ (NB: you have not specified which tensor product you use here), but rather he generalizes the Lipschitz seminorm associated to the metric.
As I understand it from Rieffel, it was known already to Kantorovich that the metric on a compact space $X$ is determined by the Lipschitz seminorm on $C(X)$:
$$
L(f) = \sup_{x \neq y} \{\frac{|f(x)-f(y)|}{d(x,y)}\}
$$
via the identity
$$
d(x,y) = \sup\{ |f(x) - f(y)| : L(f) \le 1\}.
$$
Anyway, this doesn't directly answer your question, but this has been a fruitful line of inquiry, and I suggest you look at Rieffel's papers to see if they have anything useful for you.  I think "Metrics on State Spaces" is a good one to start with.<|endoftext|>
TITLE: Are ranks of Jacobians over number fields unbounded?
QUESTION [7 upvotes]: Fix a number field $K$.

Is the rank of $J(K)$ unbounded, where $J$ ranges over the Jacobians of all smooth, projective, geometrically connected curves over $K$?
Does there exist an integer $g$ such that the rank of $J(K)$ is unbounded, where $J$ now ranges over the Jacobians of all smooth, projective, geometrically connected curves of genus $g$ over $K$?

I expect (perhaps naively) that the answer to 1. is "yes". Maybe one can write down an explicit family of superelliptic curves and use descent to show that their ranks are not bounded. Or else there may be a construction that, given an elliptic curve $E$ and integer $r$, produces a curve $C$ such that $\mathrm{Jac}(C)$ contains a factor $E^r$ up to isogeny. But I can't make either approach work.
On the other hand, I would be surprised if 2. were known, since it is so famously open in the case $g=1$.

REPLY [3 votes]: The answer to Question 1 is already contained in the following paper:
ANDRÉ NÉRON,
Problèmes arithmétique et géométriques rattachés à la notion de rang d’une courbe algébrique dans un corps, Bulletin de la S. M. F., tome 80 (1952), p. 101-166.
http://archive.numdam.org/ARCHIVE/BSMF/BSMF_1952__80_/BSMF_1952__80__101_0/BSMF_1952__80__101_0.pdf
The corollary to Theorem 7 (see page 155 of the article) says that for any given number field $K$, there are infinitely many curves $C$ of genus $g$ such that $\text{Jac}(C)(K)$ has rank at least $3g+5$. (With a bit more work, one can get $3g+6$. And with even more work, for $g=1$, the author gets up to rank at least $11$.)<|endoftext|>
TITLE: Does strict order-preservation of powerset curtail the candidates for violation of CH?
QUESTION [5 upvotes]: Thus, let $\mathrm{OPP}$ be the axiom that $|A|\lt|B| \Rightarrow |2^A|\lt|2^B|$ for any sets $A$ and $B$; and, for any ordinal $\alpha$, let $\mathrm{CH}_\alpha$ be the hypothesis that $\aleph_\alpha=\frak c$  (so that $\mathrm{CH}_1=\mathrm{CH}$) . Define $S$ to be the set of those ordinals $\alpha\in\frak c$ such that $\mathrm{CH}_\alpha$ does not provably (within $\mathrm{ZFC}$) violate $\mathrm{ZFC}$ (for example, it is known that $\omega\backslash${$0$}$\subseteq S$ and $\omega \notin S$); and let $S'$ be the set of those $\alpha\in\frak c$ such that $\mathrm{CH}_\alpha$ does not provably (within $\mathrm{ZFC}$) violate $\mathrm{ZFC}$ $\&$ $\mathrm{OPP}$. Clearly $S'\subseteq S$. But is $S'= S$ ? Or are any elements of $S$ known to be not in $S'$ ?
My guess is that $\mathrm{OPP}$ can't restrict the possibilities for violations of $\mathrm{CH}$ because the sets it talks about in the consequent---especially $2^B$--- are too big to be relevant; but I'm not sure of my footing here.

REPLY [5 votes]: First, let me remark that the particular way that you've posed the question has several problematic issues of formalization. One issue, noted by François, Andres and Andreas, is that it doesn't make sense to speak about proving an assertion with an ordinal parameter (one would instead want to speak of definitions of particular ordinals). Another issue is that for all we know, we may be living in a universe with ZFC + $\neg\text{Con}(\text{ZFC})$, and in this universe everything is provable, so even if we are able to resolve the first issue nevertheless the sets $S$ and $S'$ will be empty, since everything is provable. 
So let me propose a more semantic, alternative version of the question, which to my of thinking gets at the issue in which I believe you are interested. 
Question. If $\alpha$ is an ordinal and the continuum $2^{\aleph_0}$ can be $\aleph_\alpha$ in a forcing extension of the universe, then can the continuum be $\aleph_\alpha$ in a forcing extension of the universe in which also the OPP holds? 
The answer is yes, and so in this sense the OPP imposes no additional constraints on the value of the continuum. In this question and in the theorem below, I am speaking about possibly proper class forcing, and this is required, since if the OPP fails unboundedly often, it will require proper class forcing to force OPP again. 
Theorem. If the universe $V$ satisifes ZFC, then for any ordinal $\alpha$, the following are equivalent:

There is a forcing extension in which $2^\omega=\aleph_\alpha$.
There is a forcing extension in which $2^\omega=\aleph_\alpha$ and the OPP holds. 
Either $\alpha$ is a successor ordinal or $\alpha$ has uncountable cofinality.

Proof. Clearly 2 implies 1, and 1 implies 3. Suppose 3 holds, and I argue for 2. Fix any ordinal $\alpha$ as in $3$. First, we may simply force the GCH by the canonical forcing of the GCH. This forcing (which may be a proper class), is countably closed and hence preserves the property of having uncountable cofinality. So $3$ still holds about $\alpha$ in the extension with GCH. We may now simply apply Easton's theorem, using an Easton function $E$ that takes $\aleph_0$ to the current $\aleph_\alpha$, and more generally which takes $\aleph_\beta$ to $\aleph_{\alpha+\beta+1}$. (But any strictly increasing Easton function will do, and there are many variations.) Note that $\alpha+\beta=\beta$ once $\alpha\cdot\omega\leq\beta$, and so this pattern is eventually the GCH pattern.  By Easton's theorem, there is a further forcing extension in which $2^{\aleph_0}=\aleph_\alpha$ and the continuum function is given by $E$, which is strictly increasing, so the OPP holds. QED
In particular, for any ordinal $\alpha$ that you care to define, then you can provably force the continuum to become $\aleph_\alpha$ if and only if you can do so while also ensuring the OPP. 
Notice that in the proof of the theorem, the value of $\aleph_\alpha$ may have changed, during the forcing of the GCH, since this will collapse cardinals if the GCH did not already hold. So there is another version of the question, which is about cardinals, rather than about ordinals. If we start with the GCH, then a similar conclusion can be made.
Theorem. If $V$ is a model of ZFC+GCH, then for any cardinal $\delta$ the following are equivalent:

There is a forcing extension $V[G]$ in which the continuum is $\delta$. 
There is a forcing extension $V[G]$ in which the continuum is $\delta$ and the OPP holds. 
The cardinal $\delta$ has uncountable cofinality. 

The proof is essentially the same as above. The nontrivial part is 3 implies 2, which can be achieved via Easton's theorem by using a strictly increasing Easton function $E$ with the property that $E(\aleph_0)=\aleph_\alpha$. There are many choices of such $E$, such as $E(\aleph_\beta)=\aleph_{\alpha+\beta+1}$, as above. Any such $E$ will ensure the right value for $2^{\aleph_0}$ and, because it is strictly increasing, will also achieve the OPP. In this case, since we started with the GCH, one requires only set-sized forcing. 
By the way, there seems to be alternative terminology to refer to what you call the OPP. For example, in my paper, "Is the dream solution of the continuum hypothesis attainable?", I refer to the power set size axiom, denoted, PSA, and this is the same as what you call OPP. This axiom also appears in the MO question on reasonable-sounding statements that are independent of ZFC.<|endoftext|>
TITLE: Two questions about commutative theories
QUESTION [7 upvotes]: Let $\mathcal{T}$ be a commutative algebraic theory (for example sets, abelian groups, commutative monoids, but not groups etc.). References include the nlab and Borceux' Handbook of Categorical Algebra 2, section 3.10. Then $\mathsf{Mod}(\mathcal{T})$ is a monoidal category with internal homs.
Question 1. (Answered: Yes) Can we find a property of concrete categories which holds for $\mathsf{Mod}(\mathcal{T})$ if and only if $\mathcal{T}$ is commutative? In other words, does commutativity of an algebraic category not depend on the presentation? 
Question 2. (Answered: No) Let $\mathcal{T}$ be a commutative algebraic theory and $C=\mathsf{Mod}(\mathcal{T})$. Assume that $X \in C$ is a Co-$C$-algebra, i.e. we have a factorization of $\hom(X,-) : C \to \mathsf{Set}$ over $C$. Does this have to coincide with the usual factorization? This is well-known to be true in the examples I have mentioned above, for example for every abelian group $A$ there is only one natural abelian group structure on the hom-sets $\mathrm{hom}(A,B)$. 
This should be all well-known, but I don't know a reference.

REPLY [6 votes]: As for question 1: commutativity doesn't depend on the presentation of $T$. If $M = Mod(T)$ and $U: M \to Set$ is the forgetful functor, then commutativity can be formulated as saying that the monad $Ran_U U = U \circ Ran_U 1_M$ is commutative (or monoidal) in the sense of the nLab article here. Perhaps the most interesting aspect of this is that commutativity is a property, not an extra structure on a monad (where the structure of a strength constraint on an endofunctor on $Set$ is canonically given because every such endofunctor is canonically $Set$-enriched). These observations also lift to the enriched setting, provided of course that the functors involved are given as enriched functors (with respect to a base of enrichment $V$). 
(Note: $Ran_U 1_M$, which invariably exists, is just the left adjoint $F$ of $U$ if $U$ has a left adjoint. Some related discussion on the codensity monad of a general functor $U: M \to Set$ can be found in this post by Tom Leinster.) 
Edit: I had responded to question 2 earlier, but I am now editing that response out as it is superseded by Martin's second comment below, which makes the situation quite clear. Apologies for the noise.<|endoftext|>
TITLE: Block design question
QUESTION [5 upvotes]: Given fixed values for $d \leq k \leq v$.  I would like to find a set $B$ of $d$-sets of $[v]$ with the following properties:

Every $k$-set of $[v]$ contains at least one element of $B$
Every element of $[v]$ is contained in at most $r$ elements of $B$.

I would like to keep $r$ as small as possible.
This is very similar to the conditions for a balanced incomplete block design, but I only need an upper bound in condition 2. Also, I do not care about the coverage of pairs of elements of $V$.
Does this type of construction have a name? Are there any bounds on $r$ known (constructive or non-constructive)?

REPLY [4 votes]: Sounds like similar to a forbidden configuration problem in extremal set theory, hypergraph theory, and design theory. I don't know if exactly the same problem has been studied in one of those fields (or somewhere outside of mathematics as applications of this problem), but probably the following similar one has been a subject of serious reseach:
What's the maximum number of edges on a $d$-uniform (hyper)graph avoiding complete ($d$-uniform) sub-(hyper)graphs on $k$ vertices?
If $d=2$, it's just a graph avoiding $K_k$ with as many edges as possible.
You can put it in the language of sets or blocks by regarding edges as $d$-sets or blocks of size $d$. For instance, a $(v, b)$-configuration in combinatorial design theory is a set (or a list) of $b$ blocks whose union is of cardinality $v$. So, the above question is the same as asking the maximum number of blocks avoiding the $\left(k, {{k}\choose{d}}\right)$-configuration.
To relate this to your problem, consider the missing $d$-sets. Since you chose combinatorial design theory, in the remainder we go with the standard notation in the field.
Let $V$ be a finite set of cardinality $v$ and $\mathcal{B}$ a set of $d$-subsets of $V$. In other words, $V$ is the point set and $\mathcal{B}$ is the block set.
You want any $k$-subset of $V$ to contain at least one block $B \in \mathcal{B}$ as its subset(s). If $k=d$, the problem is trivial because you have to take all the ${{v}\choose{d}}={{v}\choose{k}}$ sets as blocks. The case when $d = 1$ is also trivial; you take distinct $v-k+1$ singletons for $\mathcal{B}$, and that's the optimal one. So we assume that $k > d \geq 2$.
Define $\bar{\mathcal{B}}$ as the ${{v}\choose{d}} - \vert \mathcal{B}\vert$ $d$-sets that are not in $\mathcal{B}$. Take distinct ${{k}\choose{d}}$ blocks $\bar{B}_i$ in $\bar{\mathcal{B}}$. If $\vert \bigcup \bar{B}_i \vert = k$, then all the $d$-subsets of the $k$-set $\bigcup \bar{B}_i$ are in $\bar{\mathcal{B}}$, which means that no block in $\mathcal{B}$ appears in $\bigcup \bar{B}_i$. Conversely, if there is a $k$-subset of $V$ no block of $\mathcal{B}$ appears in, then you end up with distinct ${{k}\choose{d}}$ blocks $\bar{B}_i$ in $\bar{\mathcal{B}}$ with $\vert \bigcup \bar{B}_i \vert = k$.
So, the first requirement in your problem is equivalent to avoiding the distinct ${{k}\choose{d}}$ blocks $\bar{B}_i$ in $\bar{\mathcal{B}}$ with $\vert \bigcup \bar{B}_i \vert = k$ (or equivalently avoiding complete $d$-uniform hypergraph on $k$ vertices in a $d$-uniform hypergraph).
If the size of $\bar{\mathcal{B}}$ is small, it's easy to avoid the forbidden configuration. The more blocks, the harder it becomes to not have one. Now you want to make any point in $V$ appear as few times as possible (or make $r$ as small as possible in your original phrasing). If a set of blocks avoiding the forbidden configuration has theoretically as many blocks as possible, the corresponding $\mathcal{B}$ (which you had in mind) has the smallest possible number of blocks. If every point appears as often in $\mathcal{B}$, this achieves the smallest $r$ possible. So, a lower bound on the maximum $\vert \bar{\mathcal{B}} \vert$ leads to a better-than-wild-guess kind of educated guesstimate on the best possible $r$ by assuming each point appears averagely in "optimal" $\mathcal{B}$ and dividing the number of blocks by an appropriate number. An upper bound on the maximum $\vert \bar{\mathcal{B}} \vert$ leads to a lower bound on the minimum size of $\mathcal{B}$ satisfying the first condition in your problem (and the lower bound on $r$ by assuming every point appears equally frequently in an optimal $\mathcal{B}$).
So, the problem of finding a bound on $r$ is now related to that of finding bounds on the maximum number of blocks avoiding complete (hyper)graphs.
If $d = 2$, it's a problem in graph theory. I don't know much about graph theory, but there must be nice results on this kind of forbidden subgraph, I think. So you should check the graph theory literature.
If $d \geq 3$, it's a problem in hypergraph theory (or design theory or extremal set theory). There has to be results on this somewhere, and I'm pretty sure some folks in hypergraph theory and extremal set theory can give you a very good answer. Anyway, as a design theorist, what I'd do first for the lower bound on $\vert \mathcal{B} \vert$ (while keeping in mind the requirement of each point appearing uniformly) is something like this:
Take uniformly at random $d$-subsets of $V$ with probability $p = c\cdot v^x$, where $c$ and $x$ are constants (of which nice exact values will be chosen later) and $\vert V \vert = v$.
The expected value of the number $f_k$ of forbidden configurations you get is upper bounded by ${{v}\choose{k}}\cdot p^{{k}\choose{d}}$. By Markov's Inequality, the probability that you wind up with twice as many as or more than the average is smaller than or equal to $\frac{1}{2}$. Hence, you have the probably $P(f_k \leq 2{{v}\choose{k}}\cdot p^{{k}\choose{d}}) \geq \frac{1}{2}$.
Let $t$ be the random variable counting the number of blocks and $E(t)$ its expected value. Then $E(t) = {{v}\choose{d}}p$. Because $t$ is a binomial random variable, by Chernoff's inequality, for sufficiently large $v$ we have
$P\left(t < \frac{E(t)}{2}\right) < e^{-\frac{E(t)}{8}} < \frac{1}{2}$ (if we choose $c$ and $x$ wisely). Hence, if $v$ is sufficiently large, with positive probability we end up with block set $\mathcal{B}$, where $\vert\mathcal{B}\vert > {{v}\choose{d}}p$ and there are at most $2{{v}\choose{k}}\cdot p^{{k}\choose{d}}$ forbidden configurations. Deleting one block from each forbidden configuration will give you the desired complete-graph-free block set. Hence, with positive probability you can have at least ${{v}\choose{d}}p - 2{{v}\choose{k}}\cdot p^{{k}\choose{d}}$ blocks while avoiding unwanted guys.
So, our task is now to choose $c$ and $x$ for $p=c\cdot v^x$ so that the above number is the largest. The best you can do is, I think, pick $x$ so that the two terms ${{v}\choose{d}}p$ and $2{{v}\choose{k}}\cdot p^{{k}\choose{d}}$ are of the same order, and then set $c$ so that the former becomes strictly larger; this way you get the best order in terms of $v$'s function. If it doesn't work (e.g., $c$ must be egregiously tiny to do this but the background of your problem doesn't want $v$ really large), you lower $x$ a tiny bit, and the number of blocks becomes big and positive for sufficiently (but not unrealistically) large $v$. Either way, since we picked blocks uniformly at randome, the resulting blocks are expected to have every point roughly evenly.
This is just an off-the-top-of-my-head approach, so there has to be a much better way to attack this. I didn't think about the upper bound at all, so the problem may have already be solved somewhere. I might have even botched mathematics in this post somewhere... But I hope at least this way of viewing your problem helps.
Edit: fixed typos and some inaccurate statements about bounds.
More edit: added a minor note that each point likely appears uniformly in the resulting set of blocks.
...and fixed grammar.<|endoftext|>
TITLE: What classification theorems have been improved by re-categorizing?
QUESTION [22 upvotes]: Many classification theorems (e.g. of the finite subgroups of $SO(3)$, or the finite-dimensional complex simple Lie algebras, or the finite simple groups) have some infinite lists, plus some "sporadic" or "exceptional" examples.
It was opined in Why/when classification of simple objects is "simple" ? E.g. (unknown) classification of simple Lie algebras in char =2,3...   that any such theorem only reflects our current knowledge of the subject. One might hope that better understanding would lead to an alternate way of slicing up the set of examples, with fewer sporadic cases.
Are there actually examples of this happening? I want a case where the initial classification is actually complete and correct; it's only our human description that has improved.
Two non-examples: (1) As I understand it, Suzuki found an infinite family of finite simple groups, that we now regard as twisted Chevalley groups for $G_2$ and its outer automorphism in characteristic 3. But that was before the classification was complete. (2) Killing had two root systems that turn out to both be $F_4$. So he wasn't quite correct.

REPLY [8 votes]: "Classification" might be a very strong word for this example, but I think real quadratic polynomials underwent something like the development you describe. They are classified by their discriminant: positive for two distinct roots, zero for a double root and negative for no roots ("three families"). The generalization of complex polynomials and roots simplifies this: discriminant zero for a double root and non-zero for two distinct roots ("two families"). When you specialize this to real polynomials you end up with what you started with, with the added bonus of two complex conjugate roots instead of none in the case of a negative discriminant.
Surely this is only a toy example, but it does show that broadening the viewpoint can reduce the number of cases to consider.<|endoftext|>
TITLE: Why only half-integral weight automorphic forms?
QUESTION [18 upvotes]: Why is that the theory of automorphic forms concentrates on the case of half-integral weight? I read in Borel's book "Automorphic forms on $SL_2$" (Section 18.5) that by considering the finite or universal coverings of $SL_2$, one can define modular forms with rational or even real weight. Borel mentions in passing that the case of half-integral weight is particularly important. Why?!

REPLY [2 votes]: You may want to look at the following article:
MR1809555 (2002b:11056) 
Ibukiyama, T.
Modular forms of rational weights and modular varieties.
Abh. Math. Sem. Univ. Hamburg 70 (2000), 315–339.<|endoftext|>
TITLE: A function whose fixed points are the primes
QUESTION [47 upvotes]: If $a(n) = (\text{largest proper divisor of } n)$, let $f:\mathbb{N} \setminus \{ 0,1\} \to \mathbb{N}$ be defined by $f(n) = n+a(n)-1$. For instance, $f(100)=100+50-1=149$. Clearly the fixed points of $f$ are the primes.

Is every number preperiodic? In other words, is $f(f(\ldots(f(n)\ldots))$ eventually prime?

REPLY [4 votes]: The problem may have connections with 3x+1 problem. To an integer, I can associate $(p_0,p_1,\cdots)$ the sequence of the least prime divisor of the term  of the sequence $u_{n+1}=f(u_n)$ which may have same properties as the parity function in the 3x+1 problem.
For instance : the asymptotic periodicity of $(p_n)$ is equivalent to periodicity of $u_n$ (thus convergence).
I would like to prove that for all composite integer,there exists a prime that appear infinitly many times in the sequence $(p_n)$.  Using some asymptotic estimations, it's not difficult to prove: 
$\forall N>0,\exists p | card \lbrace n \in \mathbb N | p_n=p\geq N\rbrace $.
It makes no use of the fact that $p_n$ is the least divisor of $u_n$.
It would be interesting to have two other theorems "3x+1"-like : 
The sequence $(p_n)$ determines $u_0$ and $\sum_{n=0}^{\infty}\prod_{k\leq n} \frac{p_k}{p_k +1}=u_0$ 
The sum converges since we have the inequality $\sum_{n=0}^{\infty} \prod_{k\leq n} \frac{p_k}{p_k +1}\leq u_0$.
I would be very interested by the links between the choice of a sequence $(p_n)$ and the convergence in an appropriate space of the prime sum. In the "3x+1"-equivalent would be 
$\sum \frac{2^{a_k}}{3^k}$, where $a_k=$ number of even terms before the $k^{th}$ odd term, which converges in $\mathbb{Z}_2$. In the 3x+1 problem generalized to $\mathbb{Z}_2$, the sum establish a bijection between parity functions and initial conditions in $\mathbb{Z}_2$.  
Furthermore is it true that for any k-uplet of prime number $(l_0,\cdots,l_N)$ there exists an initial condition such that $p_i=l_i, i\leq N $.<|endoftext|>
TITLE: A Question on Exterior Forms
QUESTION [6 upvotes]: For the last few days, I have been trying to answer the following algebraic question in exterior algebra. The following question appears as an algebraic step in the context of existence of solutions of a certain system of PDE. I have asked a special case of the problem in 
Link: Inequalities Involving Wedge Product (Reference Request) 
with the hope that this particular case will do what I have in mind, but this did not turn out to be the case. Any help in this direction is welcome.
QUESTION: 
Let $N$ be a linear subspace of $\Lambda^2(\mathbb{R}^n)$ satisfying
$$
\omega\wedge\omega\neq 0,\text{ for all }\omega\in N,\omega\neq 0.
$$
Is it true that there exists a $a\in\Lambda^4(\mathbb{R}^n)$, $a\neq 0$ such that
$$
\langle a;\omega\wedge\omega\rangle>0,\text{ for all }\omega\in N,\omega\neq 0?
$$
Comment:
My guess is that there will be such an $a$. But I could not prove it.

REPLY [6 votes]: For what is worth, here is a simple coordinate example in dimension 6. Take
$$
 \omega_0 = dx_1\wedge dx_2 + dx_3\wedge dx_4 + dx_5\wedge dx_6 ,
$$
$$
 \omega_1 = dx_1\wedge dx_3 + dx_2\wedge dx_4 ,
$$ 
$$
\omega_2 = dx_3\wedge dx_5 + dx_4\wedge dx_6 ,
$$
$$
\omega_3 = dx_5\wedge dx_1 + dx_6\wedge dx_2 .
$$
The sum of squares of these forms is zero, but the square of any nontrivial linear combination is not.<|endoftext|>
TITLE: Differential/difference algebraic groups as "group schemes"
QUESTION [11 upvotes]: While the common approach to algebraic groups is via representable functors, it seems that there is no such for differential algebraic groups (defined by differential polynomials). Neither the book by E. Kolchin, nor the texts by Ph. J. Cassidy contain anything like this — they work only with the groups of points over differential fields (and, naturally, don't say the words "group of points").
Concerning difference algebraic groups, i.e. defined by polynomials with some fixed endomorphism (also, I don't like the ambiguity with the notion of "difference algebraic equation"), there is no systematic treatment at all, although some of these groups are intensively studied (twisted groups of Lie type as an example).
So the question is: is there really no modern (scheme-like) exposition of the subject? If so, why?

REPLY [2 votes]: For difference algebraic groups, I think the paper
Michael Wibmer: Affine difference algebraic groups
is what you asked for.<|endoftext|>
TITLE: Solving equations in a subset of rational numbers
QUESTION [6 upvotes]: Let $S$ be a set of all positive rational numbers $x$ such that $2x^2 - 1$ is a square, excluding $x=1$.
I am interested in computing as many as possible solutions in $S$ to either the following equations:
(1) $\qquad p^2 - 1 = (q^2 - 1)\cdot r^2$
(2) ...removed...
(3) $\qquad p^2 + q^2 = 1 + r^2$
What would a reasonable computational approach for finding solutions?
For the equation (1), I know one solution: $(p,q,r)=(\tfrac{373}{23}, \tfrac{85}{41}, \tfrac{205}{23})$ -- would it help to find more solutions? 
EDIT: Equation (2) was not exactly the one I'm interested in. So I removed it.

REPLY [5 votes]: Your question is equivalent to looking for rational points on certain hypersurfaces in affine 3-space. 
The equation $2x^2 - 1 = y^2$ describes a plane conic with a rational point $(x,y) = (1,1)$. This means you can parameterize its solutions by passing a line through $(1,1)$ and finding the other point of intersection:
$$
 x = \frac{t^2 - 2t + 2}{t^2 - 2} \qquad y = \frac{-t^2 + 4t - 2}{t^2 - 2}.
$$
Now you can substitute this expression for $x$ in for $p$, $q$, and $r$ (using a different variable for $t$ each time) to arrive at an equation for a hypersurface. For example, using the variables $a,b,c$, your equation (2) becomes
$$
 (a^2 - 2a + 2)(b^2 - 2b + 2)(c^2 - 2) = (c^2 - 2c + 2)(a^2 - 2)(b^2 - 2).
$$
At this point, I think I've transformed your question, but I don't immediately see how to produce rational points on these hypersurfaces.<|endoftext|>
TITLE: Surjectivity of reduction maps of elliptic curves over Q
QUESTION [14 upvotes]: Let $E/\mathbf{Q}$ be an elliptic curve of rank $>0$. It is easy to see that there is a positive-density set of primes $p$ such that the reduction map $\mathrm{red}_p : E(\mathbf{Q}) \rightarrow \widetilde{E}(\mathbf{F}_p)$ is not surjective. Namely, for an integer $n>1$, take any rational prime $p$ that splits completely in the field $K_n = \mathbf{Q}([n]^{-1}(E(\mathbf{Q})))$, i.e. the field that results from adjoining to $\mathbf{Q}$ the coordinates of all preimages of points in $E(\mathbf{Q})$ under multiplication by $n$. Note that the $K_n$ are finite field extensions of $\mathbf{Q}$ (this is equivalent to the weak Mordell-Weil theorem). To show that these $p$ work: take $P \in E(\mathbf{Q})$ and $Q \in E(K_n)$ with $nQ=P$, then $\mathrm{red}_p(P) = n ( \mathrm{red}_p(Q))$, with $\mathrm{red}_p(Q)$ lying in $\widetilde{E}(\mathbf{F}_p)$ by the assumption on $p$, so $\mathrm{red}_p(E(\mathbf{Q}))$ lies in $n \widetilde{E}(\mathbf{F}_p)$, which is an index-$n^2$ subgroup of $\widetilde{E}(\mathbf{F}_p)$.
More generally, for any isogeny $\phi : E' \rightarrow E$ of elliptic curves over $\mathbf{Q}$ with non-trivial kernel, take any prime $p$ that splits completely in the finite field extension $\mathbf{Q}(\phi^{-1}(E(\mathbf{Q})))$ of $\mathbf{Q}$. 
My questions are in the opposite direction:

Do there exist infinitely many $p$ such that $\mathrm{red}_p$ is surjective?
Do there exist arbitrarily large sets of primes $\{ p_1, p_2, \ldots, p_m \}$ such that the combined reduction map $E(\mathbf{Q}) \rightarrow \prod_{i=1}^m \widetilde{E}(\mathbf{F}_{p_i})$ is surjective?
For a prime $p$ such that $\mathrm{red}_p$ is not surjective, is the failure of surjectivity explained by some isogeny $\phi$, by the argument sketched in the first paragraph?

Edit. The answer to 1. and 2. is obviously "no" in general by Maarten's answer. By the Gupta-Murty paper mentioned by Felipe, the answer to 1. becomes "yes" once the rank of $E$ is sufficiently large. As for 2., I would like to ask: is there even a single elliptic curve $E$ over $\mathbf{Q}$ for which question 2. has a positive answer?

REPLY [11 votes]: Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.
Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. 
Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.
A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9
Update on part 1:
I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'.
I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.
ps. Note that my counter examples to part one are not counter examples to Lang-Trotter. The reason is that the Lang-Trotter conjecture is a conjecture about the density of the of the primes such that the reduction is surjective. In my examples both the conjectured density and the actual density are both 0.
Part 3:
Let $E$ be a non CM rank 1 elliptic curve that is the only one in it's isogeny class. Then the only isogenies to $E$ are the multiplication by $n$ maps. For concreteness I let $E$ be the curve among all curves with these properties of smallest conductor. This curve $E$ is given by $y^2 + y = x^3 - x$ and $E(\mathbb Q)= \mathbb Z$ is generated by $P=(0 : -1 : 1)$. Now let $p=23$ then $\\#E(\mathbb F_p)=22$ but the order of $P$ after reduction is $11$ so that the index is $2$. This means that the obstruction cannot come from an isogeny because this would mean that it comes from some multiplication by $n$ map and hence that the index should not be squarefree.
In the article of Lang and Trotter where they state their conjecture they give a criterion in terms of $\mathbb Q(l^{-1}E(\mathbb Q))$ that is equivalent to $l$ being a divisor of the index. If you read that obstruction carefully you will realize that its really easy to cook up counter examples to part $3$, in particular using their criterion one can show that the set of $p$ such that reduction mod $p$ is a counter example has positive density for the above $E$.<|endoftext|>
TITLE: Computation of stable homotopy groups of $RP^2$
QUESTION [6 upvotes]: I would like to compute the first few stable homotopy groups of $RP^2$. 
I first thought to use the Atiyah-Hirzebruch Spectral Sequence, (see Davis & Kirk, pg. 242). Here is what I computed for the $E^2$ term of the spectral sequence:
$$E^2_{p,q}=\begin{array}{|ccc}
\mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_2 \\ 
\mathbb{Z}_2 & \mathbb{Z}_2 & \mathbb{Z}_2 \\
\mathbb{Z} & \mathbb{Z}_2 & 0 \\\hline
\end{array}$$
From this, I compute that the associated graded complex to $\pi_1^s(RP^2)$ is $\mathbb{Z}_2\oplus\mathbb{Z}_2$. (I think I made a mistake here with the local coefficients. I believe I showed the local coefficients act trivially, so it should just reduce to ordinary homology with coefficients in $\pi_q^s(S^0)$.) So either $\pi_1^s(RP^2)$ is $\mathbb{Z}_4$ or $\mathbb{Z}_2\oplus\mathbb{Z}_2$. 
On the other hand, we know that $\pi_1^s(RP^2)=\pi_2(\Sigma RP^2)$ by the Freudenthal suspension theorem. Using the evident cell structure on $\Sigma RP^2$ consisting of a single 0-cell, a single 2-cell, and a single 3-cell, we see that $\pi_1(\Sigma RP^2)=0$ by cellular approximation. So by the Hurewicz theorem $\pi_2(\Sigma RP^2)\cong H_2(\Sigma RP^2) \cong H_1(RP^2) \cong \mathbb{Z_2}$. 
Where am I going wrong using the AHSS? How does one compute $\pi_2^s(RP^2)$?

REPLY [4 votes]: Another approach will be to use Adams Spectral Sequence. This is a most suited to compute stable homotopy groups. An outline will be as follows. 
As a module over steenrod algebra $\mathbb{RP}^{2}$ looks like two cells connected with $Sq^{1}$. 
One can either compute the minimal resolution or do a May spectral sequence to compute the $E_{2}$ page of Adams Spectral Sequence. In fact, May Spectral Sequence is tailor made for such modules. All one has to do is run May SS after throwing away $h_{0}$ But after first few terms you want to probably want to see if you have any differentials in the Adams SS.
If you are going high enough to worry about Adams SS differentials, one can import the computations done for Sphere spectrum. Since $\mathbb{RP}^{2}$ is cofiber of $S \xrightarrow{2} S$ and multiplication by $2$ induces multiplication by $h_{0}$ in Adams SS. Also one has a long exact sequence 
 $Ext_{A}^{s,t}(S) \xrightarrow{h_{0}} Ext_{A}^{s+1,t+1}(S) \to Ext_{A}^{s+1, t+1}(\mathbb{RP}^{2}) \to Ext_{A}^{s+1, t}(S)$ 
where $A$ is the mod 2 Steenrod Algebra. Using this method one can import the differentials Adams SS differentials for $S$, the sphere spectrum. One can easily recover homotopy groups $\mathbb{RP^{2}}$ from the knowledge of sphere. This obviously gives the answer easily up to $50^{th}$ stem, may be more, but I did not go beyond that. `
This approach will give all the extensions as well. Also for $\mathbb{RP}^{2}$ one has to worry about what happens at prime $2$ only.  `<|endoftext|>
TITLE: Poincaré lemma in infinite dimensions
QUESTION [6 upvotes]: Hi everyone,
Is the Poincaré lemma true in infinite dimensions?
Here's a precise statement:
Let $X$ be a Banach (or maybe Hilbert) vector space, $U$ a simply connected open set in $X$. Is it true that every closed (smooth) $1$-form on $U$ is exact?
Thanks!

REPLY [10 votes]: Yes, it is, on convenient locally convex vector spaces. 
Convenient is a very weak completeness condition.
See 33.20 in: 
Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997.(pdf)<|endoftext|>
TITLE: Proper morphisms of C*-algebras / Nondegenerate representations
QUESTION [7 upvotes]: Let $A \to B$ be a proper morphism of $C^*$-algebras. A nondegenerate representation of $B$ induces a nondegenerate representation of $A$. Does the converse hold?
I.e.: let $A \to B$ be a morphism of $C^*$-algebras such that every nondegenerate representation of $B$ induces a nondegenerate representation of $A$. Does the morphism result proper? I guess not.

REPLY [6 votes]: This is true. Factoring by the kernel of the homomorphism, we may assume that $A$ is a C*-sub-algebra of $B$ and the homomorphism is just the inclusion.
So assume that $A\subseteq B$ and 
(a) Every non-degenerate representation of $B$ restricted to $A$ is non-degenerate.
Then $A$ cannot be contained in the kernel of a state $\lambda$ of $B$. Otherwise, the GNS construction
would give us a representation $\pi_\lambda\colon B\to B(H_\lambda)$  that is non-degenerate but restricted
to $A$ is degenerate (since there exists $h\in H$ such that $\lambda(a)=\langle\pi_\lambda(a)h,h\rangle$ for all $a\in B$).
So, (a) implies
(b) $A$ is not contained in the kernel of a state of $B$.
This implies that $A$ must generate $B$ as a closed left ideal, by Theorem 3.10.7 of Pedersen's "C*-algebras and their automorphims".
Now let $(a_i)$ be an approximate unit for $A$. Then for every element of the form $ab$, with $a\in A$ and $b\in B$,
we have $a_iab\to ab$. But the linear span of these elements elements  is dense in $B$. So, (b) implies
(c) Any approximate unit of $A$ is also an approximate unit of $B$.<|endoftext|>
TITLE: Salié permutations and fair permutations
QUESTION [18 upvotes]: In October 2010, I published a Monthly problem that introduced the concept of a fair permutation, which is a permutation $\pi$ such that for every $i$, either $\pi(i) > i$ and $\pi^{-1}(i) > i$, or $\pi(i) \le i$ and $\pi^{-1}(i) \le i$.  Equivalently, every cycle of $\pi$ is either a fixed point or an alternating cycle of even length, meaning that if $z$ is the largest element of the cycle, then
$$z > \pi(z) < \pi(\pi(z)) > \pi(\pi(\pi(z))) < \cdots < z.$$
I don't think anyone has studied fair permutations explicitly before.
However, 
a permutation $\sigma$ on $\lbrace1, 2, \ldots, 2m\rbrace$ is said to be a Salié permutation if for some $r\le m$, $$\sigma(1) < \sigma(2) > \sigma(3) < \cdots < \sigma(2r)$$ and
$$\sigma(2r) < \sigma(2r+1) < \sigma(2r+2) < \cdots < \sigma(2m).$$
These have been studied before, and one can show by generatingfunctionology that the number of fair permutations is twice the number of Salié permutations.  This is very suggestive and hints at a close connection.

Question: Can one construct an explicit 2-to-1 map from fair permutations to Salié permutations?

REPLY [19 votes]: Write a fair permutation in cycle form, as follows. First write all
cycles of length $>1$ in decreasing order of their smallest element,
with the smallest element of each cycle written as the leftmost
element of the cycle. Then append all the fixed points in increasing
order. An example is
  $$ (3,9,4,6)(1,7,2,10)(5)(8). $$
Now erase the parentheses. We obtain a Salié permutation. Each Salié
permutation arises in exactly two ways. If the first fixed point is
less than the element preceding it in the representation just
described (as is the case for the example above), then we can absorb
the first two fixed points into the cycle preceding them (that is, the
rightmost cycle). Otherwise, we can turn the last two elements of the
rightmost cycle into fixed points. For the example above we get the
additional fair permutation
  $$ (3,9,4,6)(1,7,2,10,5,8) $$
yielding the same Sailé permutation.
Addendum. There is a small inaccuracy above. If 1 is a fixed point, 
then rather than absorbing 1 and the fixed point $j$ following it into 
the previous cycle, we should create a new cycle $(1,j)$.<|endoftext|>
TITLE: Counterexample to Sard's theorem for a non-C1 map
QUESTION [19 upvotes]: Is there a function $f: \mathbb{R} \rightarrow \mathbb{R}$ which is differentiable but not $C^{1}$, such that the image of the points where $f'(x) = 0$ has measure bigger than 0?
If the answer is no, is it possible to find such an $f$ without requiring that it is everywhere differentiable?

REPLY [26 votes]: No, such functions do not exist. More precisely, let $f:\mathbb R\to\mathbb R$ be an arbitrary function, $\Sigma$ is the set of $x\in\mathbb R$ such that $f'(x)$ exists and equals 0. Then $f(\Sigma)$ has measure 0.
By countable subadditivity of measure, we may assume that the domain of $f$ is $[0,1]$ rather than $\mathbb R$. Fix an $\varepsilon>0$. For every $x\in\Sigma$ there exists a subinterval $I_x\ni x$ of $[0,1]$ such that $f(5I_x)$ is contained in an interval $J_x$ with $m(J_x)<\varepsilon m(I_x)$. Here $m$ denotes the Lebesgue measure and $5I_x$ the interval 5 times longer than $I_x$ with the same midpoint. Now by Vitali's Covering Lemma there exists a countable collection $\{x_i\}$ such that the intervals $I_{x_i}$ are disjoint and the intervals $5I_{x_i}$ cover $\Sigma$. Since $I_{x_i}$ are disjoint, we have $\sum m(I_{x_i})\le 1$. Therefore $f(\Sigma)$ is covered by intervals $J_{x_i}$ whose total measure is no greater than $\varepsilon$. Since $\varepsilon$ is arbitrary, it follows that $f(\Sigma)$ has measure $0$.<|endoftext|>
TITLE: How does descent theory imply a sheaf is a scheme?
QUESTION [10 upvotes]: I've noticed that often authors will comment that "descent theory" shows that some sheaf in the étale topology is actually a scheme.  I was wondering what result in descent theory actually implies this(a statement and reference for the result is what is requested in case an actual explanation would take too long).  I understand how descent theory is what allows one to define and construct stacks, but I've never seen a result which says something to this effect (for example, I haven't seen it in Vistoli's stack notes in FGA explained) and it seems to be used often in verifying that something is a DM stack.
For example, Edidin states in the proof of Prop. 2.2 of his notes on $M_g$ that "$\underline{\mathrm{Iso}}_B(e,e')$ is the étale sheaf which is the quotient of $(X\times_{X\times X} E\times_B E')$ by the free group action of $G$.  Moreover, descent theory shows that this sheaf is in fact a scheme."
In the proof of Theorem 3.2 there, he similarly writes, "Descent theory says that in this case a quotient $C=C_E/PGL(N+1)$ also exists as a scheme."  
One finds similar statements in the other standard papers about DM stacks.  I was wondering what results in "descent theory" they are referring to.

REPLY [11 votes]: When you post a question, it would be good if you include enough explanations not to force the interested reader to go search for a paper online.
Anyway, the general question is: suppose that we have an fpqc covering of schemes $Y'\to Y$ and a scheme $X' \to Y'$ with descent data. When can I conclude that $X'$ descends to a scheme over $Y$? I know of two general results in this direction: this works when $X'$ is affine over $Y'$, of when there is a relative ample line bundle $L'$ on $X'$, and the descent data can be extended to $L'$. These are covered in my notes on descent theory: the first is in 4.3.1, the second in 4.3.3. Affine descent applies to the first of the cases you mention, descent via ample line bundles to the second.<|endoftext|>
TITLE: Trasportation metric (AKA Earth-Mover's, Wasserstein, etc.) as "natural" / "induced"?
QUESTION [6 upvotes]: Context: Given a discrete finite metric space $X$ (in my case X={0,1}$^n$ with the Hamming/L$_1$ distance), I need to define the natural or canonical metric on the set of all probability distributions over $X$ (call it $D(X)$) which extends the given metric.
I have read in several places [*] that the Earth-Mover's is the natural, or induced, or most canonical extension.
I would like to know whether it is proved (or can be easily proven) that it is the only, or indeed the canonical metric that can be given to the set of distributions over $X$ so that when restricted to 
D1(X) = {d in D(X)  |  d(x)=0   for all x in X, except one} 
(the distributions whose support is just one element),
you obtain the metric of $X$.
[*]
mathoverflow: Distance metric between two sample distributions [...],
or google "EMD naturally extends the notion of distance" and you'll find several papers saying a similar thing, as a "fact", without proof or reference.

REPLY [11 votes]: Okay, so $X$ is a finite metric space and $D(X)$ is the positive part of the unit sphere of $l^1(X)$. We can consider $X$ as sitting inside $D(X)$ by identifying a point $x \in X$ with the function that is $1$ at $x$ and $0$ elsewhere.
The literal question you have asked is whether the mass transport metric on $D(X)$ is the only metric on $D(X)$ whose restriction to $X$ recovers the original metric on $X$. The answer to this question is clearly no; if you want to add a point to a metric space you generally have a lot of freedom to assign distances from it to the other points. All the more so if you are adding many new points.
But you probably meant to take the affine structure of $D(X)$ into account. $D(X)$ is the convex hull of $X$, so we can ask: if $X$ is isometrically embedded in a Banach space $E$, is the norm on its convex hull in $E$ uniquely determined? The answer is still no. For example, let $X$ have three elements, such that the distance between any two of them is $1$. We can embed $X$ as the vertices of an equilateral triangle in the euclidean plane, or we can embed it as the points $(0,1)$, $(1,1)$, and $(1,0)$ in $l^\infty_2$. The two metrics on the convex hull of $X$ aren't the same. (Look at the distance from the average of two of the points to the third.)
However, you also asked whether the mass transport metric is "canonical".  Yes, it is. It is universal in the following sense:
Theorem. Let $X$ be a metric space and let $e \in X$. Then there is a Banach space $AE(X)$ together with an isometric embedding $\iota: X \to AE(X)$ such that $\iota(e) = 0$, and such that if $f: X \to E$ is any nonexpansive map from $X$ into any Banach space $E$ with $f(e) = 0$, then there is a unique nonexpansive linear map $T: AE(X) \to E$ such that $T \circ \iota = f$.
The mass transport metric is the restriction of the metric on $AE(X)$ to the convex hull of $X$. So this theorem could be rephrased in terms of the mass transport metric being the universal metric on $D(X)$ relative to nonexpansive affine maps. In other words, it is the metric for which distances are as large as possible, given the original metric on $X$. There's a nice little book on Lipschitz algebras that covers this material, but I forget the author.<|endoftext|>
TITLE: Exceptional collections of objects in topological triangulated categories?
QUESTION [6 upvotes]: People often consider exceptional sets of objects (i.e. collections of objects satisfying certain strong orthogonality conditions: $Ext^{l}(P_i,P_j)$ should be zero for $l\neq 0$ + something else) in derived categories of coherent sheaves (over algebraic varieties;  possibly the first example corresponds to the Beilnson's description of the derived category of coherent sheaves on the projective space of dimension $n$). Are there any examples of this notion in some stable homotopy categories (in the sense of abstract model categories; one can consider the category of modules over a ring spectrum here)?

REPLY [4 votes]: A natural topological example of a [higher] category with an exceptional collection is constructible complexes on a stratified space, locally constant along contractible strata. This holds with any coefficients - e.g. you can look at sheaves of S-modules or E-modules for an $E_\infty$-ring spectrum if you prefer.
The exceptionality encodes a. the contractibility of strata, b. the absence of Exts "in the wrong direction" for extensions of constant sheaves off the strata.
I would presume you could also find examples of "derived Fano schemes" with exceptional collections - i.e. nontrivial analogs of the many (toric Fano e.g.) schemes admitting exceptional collections, but I don't know of such.<|endoftext|>
TITLE: Fastest way to factor integers < 2^60
QUESTION [19 upvotes]: I've been running a search for Mordell curves of rank >=8 for about 12 months and have identified approximately 280,000 curves in our archivable range, amongst many millions that aren't. 
For this search I have been utilizing up to 60 3Ghz+ cpus at any one time.  Right now I'm reaching the point of decreasing return and greatly increased memory requirements.  As such, I need to be a little smarter in some of the maths.
My question is, then, what is believed (or known) to be the fastest factoring algorithm for positive integers < 2^60.  
I'm not overly keen to consume further multiple GHz decades of processing power unless I can get a reasonable return on investment, for which a fast factoring algorithm would certainly help.
Any ideas are more than welcome.
EDIT:
Reading through the responses makes me realise that I should probably add that I want to keep the factoring within 64 bit arithmetic.  The Pollard Rho algorithm was interesting, but probably would exceed the 64 bit limit during its execution.
Another part of the puzzle it that, from the factorisations, I'm storing the differences of divisor pairs for each number factored.  This may, potentially, leave me with an array of about 50,000,000 values which then subsequently needs to be sorted.
I have been using Pari/GP for this project and, up till now, it's been great. My current problem is mainly with memory usage, with each Pari/GP task taking over 8GByte of memory. This is preventing my 32 core machine from being as efficient as it may otherwise have been.  Hence my intent to move to 64 bit arithmetic and 'C' code, to hopefully gain efficiencies in both time and space, thus breathing new life into an otherwise stalling project.
Update:
Thank you to all those that responded with so many good suggestions on how to proceed.
Eventually I've decided to use the flint library, as suggested by William Hart, rather than try to re-invent the wheel.  The ability of flint to work directly with 64 bit integers gives me a great advantage as regards memory usage and speed when compared to my current setup.  In particular I can now run all 32 cores on my main machine and still have memory left over, potentially giving an 8 fold improvement on processing throughput.
Kevin.

REPLY [2 votes]: A good source for highly efficient algorithms and implementations for this kind of problems is Dan Bernstein's homepage. There I found an algorithms that might be useful for weeding out all the small prime factors: 
If you have $y/(\log y)^{O(1)}\;$ integers, each with at most $(\log y)^{O(1)}\;$ bits, then you can find all the small prime factors of each integer in time $(\log y)^{O(1)}\;$ per integer.
[I didn't look at the details, so you will have to give it a try to see if 60-bit numbers are already big enough.]<|endoftext|>
TITLE: How to understand the Harish-Chandra isomorphism?
QUESTION [22 upvotes]: The Harish-Chandra isomorphism describe the center $Z(\mathfrak{g})$ of $U(\mathfrak{g})$ as invariants of $\text{Sym}^*\mathfrak{h}$ under the action of the Weyl group. (One need to twist the action or make a change of coordinate on the affine space $\mathfrak{h}^*$ for this isomorphism work.)
My question is, how does one understand this isomorphism, and what's the geometric context of it? Why should one expect something like this might be true?

REPLY [5 votes]: Depending what one means by "intuition", the computation of the behavior of the center $\mathfrak z$ of $U\mathfrak g$ on unramified principal series repns not only suggests the form of the Harish-Chandra isomorphism, but also suggests the $W$-invariance, since the intertwinings (generically isomorphisms) among principal series are exactly given by Weyl group elements (with the shift by the half-sum $\rho$ of positive roots as a result of a change-of-measure at the group level).<|endoftext|>
TITLE: Learning path for the proof of the Weil Conjectures
QUESTION [24 upvotes]: Assume you are an algebraic geometry advanced student who has mastered Hartshorne's book supplemented on the arithmetic side by the introduction of Lorenzini - "An Invitation to Arithmetic Geometry" and by Liu - "Algebraic Geometry and Arithmetic Curves".

What would be a good learning path towards the proof of the Weil Conjectures for algebraic varieties (not just curves)?
What modern references are available and in which order should be studied?

Besides the original article I and article II by Deligne and the results on rationality by Dwork, there is the book Freitag/Kiehl - "Étale Cohomology and the Weil Conjecture" and the online pdf by Milne - "Lectures on Étale Cohomology". The first title is out of stock and hard to get and the second seems to me too brief and succinct.

Is it better to master étale cohomology by itself elsewhere and then refer to the original articles? Is any further algebraic/arithmetic background necessary?

Thank you in advance for any hints on how to approach such a study program, and for any related advice towards a self-learning path in arithmetic geometry. (This question is cross posted to math.stackexchange so all kind of students and professionals can provide their advice regardless of their membership to these forums.)

REPLY [4 votes]: Here are Uwe Jannsen's lecture notes on étale cohomology http://www.mathematik.uni-regensburg.de/Jannsen/Etale-gesamt-eng.pdf and Weil I http://www.mathematik.uni-regensburg.de/Jannsen/home/Weil-gesamt-eng.pdf and Laumon's Fourier transform http://www.mathematik.uni-regensburg.de/Jannsen/home/Garben-gesamt-engproof.pdf.
For Weil II, there is also Kiehl-Weissauer.<|endoftext|>
TITLE: Free Objects in Functor Categories
QUESTION [6 upvotes]: As a bit of background, consider the category of all covariant, additive functors from a small Abelian category $C$ to Abelian groups, which I'll denote $[C,Ab]$.  First, $[C,Ab]$ is an Abelian category, and an easy argument shows that representable functors are projective objects in this category.  Indeed, if $T = \text{Hom}_C(X,-)$ is a representable functor and
\begin{equation*}
0 \to F_1 \to F_2 \to F_3 \to 0
\end{equation*}
is a short exact sequence in $[C,Ab]$ (where exactness is checked on objects of $C$), then Yoneda's lemma gives $\text{Nat}(T,F_i) \cong F_i(X)$, and the result is immediate.  I think that the reverse implication also holds (projective implies representable), but I don't remember the proof being as apparent.   We even have the Eilenberg-Watts theorems that give criteria for when additive functors (from $R$-Mod to $Ab$) are representable.  
Anyway, this is nice, but I find it a bit lacking compared to the tools we have in, say $R$-Mod for some ring $R$.  In that setting, we have such results as "A module $P$ is projective iff it is a direct summand of a free module, etc."  Or in $Grp$, we have $G$ is projective iff it is free.  My point is that we have a notion of "free object" since these are all nice concrete categories, and such a notion seems to have no nice analogue in $[C,Ab]$.  
Of course, we have results like $[C,Ab]$ being concrete over $[C,Set]$ which gives a "locally free Abelian group" type example (where $C^{op}$ here would be the category of open sets of a topological space, and $[C,Ab]$ would be presheaves of Abelian groups).  However, these are not free objects in $[C,Ab]$.  

My question is then "Can $[C,Ab]$ be reasonably thought of as concrete over another category so that we can construct free object?"  As an example, is $[R-Mod,Ab]$ concrete over the functor category $[Set,Set]$?  (of course, we'd have to juggle Grothendieck universes for this to make any sense;  The standard way being to fix some universe $\mathfrak{U}$, and say $Set$ is the category of $\mathfrak{U}$-sets, and let $\mathfrak{U}'$ be the smallest Grothendieck universe containing $\mathfrak{U}$ as an element, so that $Set$ is now $\mathfrak{U}'$-small (see Schuberts "Categories"))

REPLY [7 votes]: To amplify Qiaochu's answer (and answer S. Carnahan's question), I'd like to add that the forgetful functor $[\mathcal{C}, \textbf{Ab}]_{\textbf{Ab}} \to [\operatorname{ob} \mathcal{C}, \textbf{Set}]$ is (finitary) monadic. The easiest way to see this is to treat functors $\mathcal{C} \to \textbf{Ab}$ as a many-sorted finitary algebraic theory, with one copy of the group operation symbols for each object of $\mathcal{C}$ and one function symbol for each morphism of $\mathcal{C}$. Of course, the axioms of this algebraic theory simply express the following:

Each sort is an abelian group under the respective group operations.
The functions between the sorts are group homomorphisms.
The functions between the sorts compose the way we expect them to, so that we get a functor.
The functions between the sorts can be added pointwise the way we expect them to, so that we get an additive functor.

This presentation essentially corresponds to factorising $[\mathcal{C}, \textbf{Ab}] \to [\operatorname{ob} \mathcal{C}, \textbf{Set}]$ as the composite $[\mathcal{C}, \textbf{Ab}]_{\textbf{Ab}} \to [\mathcal{C}, \textbf{Ab}] \to [\operatorname{ob} \mathcal{C}, \textbf{Ab}] \to [\operatorname{ob} \mathcal{C}, \textbf{Set}]$. The free models of this theory can then be constructed by applying the left adjoints of each of these functors in turn, because left adjoints are unique up to unique isomorphism. (Note: the left adjoint of $[\mathcal{C}, \textbf{Ab}] \to [\operatorname{ob} \mathcal{C}, \textbf{Ab}]$ is left Kan extension along the inclusion $\operatorname{ob} \mathcal{C} \hookrightarrow \mathcal{C}$.)<|endoftext|>
TITLE: Does there exist a categorical treatment of root data(systems)?
QUESTION [5 upvotes]: What I am looking for is an abstract description of root data with their morphisms(!) plus a comparison with the categories of reductive groups over some field, Dynkin diagrams, Lie algebras, singularities...?

REPLY [7 votes]: The Isogeny Theorem does this for all isogenies between split reductive groups over any field (including non-central isogenies and exceptional isogenies in small characteristics) via the notion of "$p$-morphism" between root data ($p \ge 1$). 
Is there a motivating situation to say more? For non-split groups the "anisotropic kernel" is invisible to root systems. For split semisimple groups in positive characteristic the homomorphisms into ${\rm{GL}}_n$ seem too rich to be encoded in terms of combinatorial information in a functorial way (e.g., there is a "highest weight" theory, but dimensions of irreducibles are mysterious, and semisimplicity fails badly). Complexities due to working over a ground field that isn't algebraically closed (even char. 0) seem huge, such as for field of definition of representations, much as we know very little about ${\rm{H}}^1(k,G)$ for general $k$ and many split $G$.<|endoftext|>
TITLE: Training towards research on birational geometry/minimal model program
QUESTION [15 upvotes]: Being a not yet enrolled independently supervised graduate student in mathematics, with prospects of applying to American graduate schools hopefully in a 1-2 years' time, I have a background of having mastered Hartshorne - "Algebraic Geometry" with partial complements like Beauville - "Complex Algebraic Surfaces" and Reid - "Chapters on Algebraic Surfaces" to learn the Kodaira-Enriques classification, Kollár - "Lectures on Resolution of Singularities" to understand Hironaka's theorem, and bits of Mukai - "An Introduction to Invariants and Moduli" as introduction to GIT and moduli. Now, after studying minimal Cremona models for birational linear systems on rational surfaces from the article by Calabri and Ciliberto, I have become very interested in the general birational geometry of higher-dimensional algebraic varieties and their minimal model program. As this line of research has been very active and with important breakthroughs in recent years, I have tried to compile a large amount of fundamental articles, lecture notes, reviews and books on the subject, but I do not know which are better, which are outdated and in which order they should be studied. This is a selection of the material I think most important:
Books (added recommendations from the answers below):

Lazarsfeld - Positivity in Algebraic Geometry vol I & II, Springer 2004.
Debarre - Higher-Dimensional Algebraic Geometry, Springer 2001.
Miyaoka, Peternell - Geometry of Higher Dimensional Algebraic Varieties, Birkhauser 2004
Matsuki - Introduction to the Mori Program, Springer 2002.
Kollár; Mori - Birational Geometry of Algebraic Varieties, Springer 1998
Kollár - Rational Curves on Algebraic Varieties, Springer 2001.
Hacon; Kovács - Classification of Higher Dimensional Algebraic Varieties, Birkhauser 2010.
Kollár; Kovács - Singularities of the Minimal Model Program, CUP 2013 (to appear).
Corti (ed.) - Flips for 3-folds and 4-folds, OUP 2007.

Articles & Pre-Prints (added recommendations from the answers below):

Kollár - Exercises in the Birational Geometry of Algebraic Varieties
Andreatta - An Introduction to Mori Theory: The Case o Surfaces.
Kollár; Kovács - Birational Geometry of Log Surfaces.
Fujino - On Log Surfaces + Minimal Model Theory for Log Surfaces.
Tanaka - Minimal Models and Abundance for Positive Characteristic Log Surfaces.
Reid - Twenty Five Years of 3-folds, an Old Person's View.
Kollár - The Structure of Algebraic Threefolds: An Introduction to Mori's Program.
Kollár - Minimal Models of Algebraic Threefolds: Mori's Program.
Birkar - Lectures on Birational Geometry.
Corti; Kaloghiros; Lazić - Introduction to the Minimal Model Program and the Existence of Flips.
Corti; Hacking; Kollár; Lazarsfeld; Mustaţă - Lectures on Flips and Minimal Models.
Birkar; Cascini; Hacon; McKernan - Existence of Minimal Models for Log General Type Varieties.
Hacon; McKernan - Existence of Minimal Models for Varieties of Log General Type II.
Druel - Existence de Modèles Minimaux pour les Variétés de Type Général.
Corti - Finite Generation of Adjoint Rings after Lazić: An Introduction.
Cascini; Lazić - The Minimal Model Program Revisited.
Cascini; Lazić - New Outlook of the Minimal Model Program I.
Corti; Lazić - New Outlook of the Minimal Model Program II.
Lazić - Around and Beyond the Canonical Class.
Fujino; Mori - A Canonical Bundle Formula.
Fujino - Fundamental Theorems for the Log Minimal Model Program.
Fujino - Introduction to the Log Minimal Model Program for Log Canonical Pairs.
Birkar - On Existence of Log Minimal Models.
Birkar - Existence of Log Canonical Flips and a Special LMMP.


I would like to get any advice on how to organize a deep study course on these topics for a 1-2 years period, as it were for an independent study program preparation for a future Ph.D. thesis on birational geometry. I would really appreciate if any of the professionals or advanced students in the field could provide hints for organizing such a guide, to make good use of the time I have till then, and get already a good general background preparation on these advanced topics. 

For example, what would be a good approach to acquire the required background to work through the new book by Hacon and Kovács?

Thank you very much in advance for any hints on how to proceed.

REPLY [7 votes]: The standard answer to these type of questions on MO is "ask your (potential) PhD advisors", and I think here it applies even more so than usually. The most productive choice of topics to learn in detail depends on your choice of future research topics, and unless you want to do a PhD on your own, this will depend heavily on your advisor's skill set and interests. No PhD advisor would be equally qualified to supervise any topic in birational geometry.
Having said that, I think Lazarsfeld's books are a huge gap in your list.

REPLY [4 votes]: I would strongly agree with Arend that you should include Positivity in Algebraic Geometry I & II, by Lazarsfeld, (and that you should ask your potential advisor).
However, while you point out that some books have become outdated, it doesn't mean that they aren't worth going through. The book by Kollár and Mori I think is still the standard source for the foundations of the MMP (especially the earlier chapters).  You could also see some exercises to supplement that book by Kollár.
In particular, I would suggest that that is still the right place to start (possibly combined with Lazarsfeld's books).  Debarre's book is also a good starting text.<|endoftext|>
TITLE: Primitive subwords in a free group of rank 2
QUESTION [5 upvotes]: I am not sure yet about what I exactly need to prove, but I guess I can formulate a rough statement similar to the following:
Suppose $w\in F_2$ is a primitive word whose length is big enough. Then for every chunk of length greater than some big constant, there exists a "long" subword, contained in that chunk, which is primitive.
I suspect that the statement should be true, maybe in some similar form, and I thought about proving it by using some canonical form for primitive elements. Since the rank of the free group is just 2, I guess there should be an easy way for finding some good pattern in the primitive words, or something like that. 
Do you know any nice way of writing down primitive elements in $F_2$ that might be good in this context? 
Thank you very much.

REPLY [3 votes]: To answer your last question, check out Cohen-Metzler-Zimmerman (1981) and references there in.<|endoftext|>
TITLE: Stable equivalence and triangulated equivalence of self-injective algebras
QUESTION [5 upvotes]: Two (finite-dimensional) $k$-algebras $A$ and $B$ are said to be stable equivalent if their stable module categories $\underline{\rm mod}(A)$ and $\underline{\rm mod}(B)$ are equivalent as $k$-linear categories. If these algebras are self-injective, the stable module categories are triangulated categories. 
So the question is: if $A$ and $B$ are stable equivalent, are the categories $\underline{\rm mod}(A)$ and $\underline{\rm mod}(B)$ equivalent as triangulated categories?

REPLY [6 votes]: There are some dull counterexamples.
If $A$ is a self-injective algebra such that the square of the radical is zero, then the stable module category is a semi-simple $k$-linear category with one simple object for each non-projective simple $A$-module.
So, for example, if $A = k[x]/(x^2) \times k[y]/(y^2)$ and $B$ is the path algebra of a quiver with two vertices and two arrows $a$ and $b$ between the vertices in opposite directions, modulo the relations $ab=0=ba$, then the stable module categories of $A$ and $B$ are equivalent as $k$-linear categories (both being semisimple categories with two simple objects). However, they are not equivalent as triangulated categories, as $\Omega$ acts differently on the simple objects.
I don't think I know of any more interesting kinds of example.<|endoftext|>
TITLE: Sato-Tate conjecture for CM modular forms
QUESTION [8 upvotes]: For a non-CM holomorphic modular forms of weight $k \geq 2$, the Sato–Tate conjecture
is known to be true. Thanks to the work of Barnet-Lamb, Geraghty, Harris, and Taylor.
Do we have an analogous statement for CM modular forms as well? I mean, Is there
a precise formulation (or a proof) of the Sato-Tate conjecture for CM modular forms
of weight $k \geq 2$?

REPLY [5 votes]: Since the $L$-function of a CM modular form is just that of a Hecke character, the analogue of the Sato–Tate conjecture is much simpler to prove and follows from work of Deuring, I believe. If I remember correctly the measure one uses is (proportional to) $(1-z^2)^{-1/2}$ (and, as David points out, you only consider the primes that split in the associated imaginary quadratic field since $a_p=0$ at all inert primes).<|endoftext|>
TITLE: Explicit computations of the étale homotopy type?
QUESTION [23 upvotes]: Hi,
I'm currently trying to learn about etale homotopy for schemes as introduced by Artin-Mazur. I know that by the Artin-Mazur comparision theorem, it is possible to compute the etale homotopy type of certain class of varieties as the profinite completion of the complex points. However,  in most other cases for schemes, it seems quite cumbersome to calculate the étale homotopy type of a locally noetherian scheme say. Are there any explicit computations of the étale homotopy type that are particularly helpful for understanding the general theory? Or am I missing something here?
Sorry if my question is a bit vague.

REPLY [10 votes]: This is kind late, but here are some examples to add on which might demonstrate some of the elementary tools that one has in this subject:

As Daniel said, if $G = Gal(k_{sep}/k)$ then $BG \simeq Spec\,k_{et}$. Perhaps the easiest way to see this is the observation that if $L/k$ is a Galois extension then $L \otimes_k L \simeq \prod_{\sigma \in Gal(L/k)} Spec\,L$ so that, upon applying $\pi_0$ we get the simplicial set $BGal(L/k)$.
We can do $\mathbb{G}_m$ for $k$ being separably closed explicitly (without comparison theorems). With any scheme in which cohomology may be computed using Cech theory, the etale homotopy type may be computed using 0-coskeletal hypercovers, i.e. Cech covers in the homotopy category of hypercovers. We have a left final subcategory of Cech covers of the form $cosk_0(G_m) \rightarrow G_m$ induced by $G_m \rightarrow G_m, t \mapsto t^n$. This can be proved using, say, Riemann Hurwitz arguments (if you want to avoid comparison theorems completely). Observing that $cosk_0(G_m)_n \simeq G_m^{n+1} \simeq G_m \times \mu_n^n$, $\pi_0$ again gets us $B\mu_n$ and the etale homotopy type is the pro-space $\{B\mu_n\}$ indexed by divisibility.
Of course there are comparison theorems which might be "cheating" but here's one that is interesting. If you're a projective variety which has a model over $W(\bar{F}_p)$, then the etale homotopy type over $\mathbb{C}$ and over $\bar{F}_p$ is equivalent - this is a direct consequence of proper and smooth base change so that might tell you that the theory encodes difficult theorems in algebraic geometry.
In Friedlander's thesis (his completion of Quillens sketch of the complex Adams conjecture using etale homotopy theory), he computed the homotopy fiber of $\mathcal{E} - 0(X) \rightarrow X$ where $\mathcal{E}$ is a vector bundle of rank $r \geq 2$ and $0(X)$ is the zero section. He proved that this is (upon completion away from the residue characteristics), the $2r-1$ sphere. Roughly speaking, Friedlander does two things - he showed that for $x \rightarrow X$ a geometric point, $\mathcal{E}- 0(X)_x$ is indeed equivalent to the appropriately completed sphere. This step uses the general "six functors" type calculations for smooth pairs which can be found in Milne VI.5, say. The other step compares the fiber versus the homotopy fiber upon take etale homotopy types. This uses a comparison between the Serre spectral sequence (of fibrations between (pro)-spaces) and the Leray spectral sequence (as composition of functors).<|endoftext|>
TITLE: why Skolemization?
QUESTION [10 upvotes]: Can someone explain what is the reason for using Skolemization, I clearly understand that its the removal of existencial quantifiers, but whats the use? why is keeping existencial quantifiers not good??

REPLY [3 votes]: Somewhat related to Jason Rute's answer: suppose you give me a formula, say $\varphi\equiv\forall x\exists y\theta(x, y)$ where $\theta$ is quantifier free. Skolemization gives me a way to think about the complexity of $\varphi$ being true: how complicated must a Skolem function for $\varphi$ be?
For example, consider the formula "For every $x$ there is a $y$ such that EITHER $\Phi_x(x)$ halts in exactly $y$ many stages, OR $y=0$ and $\Phi_x(x)$ never halts." This is clearly a true sentence; however, any Skolem function for it must be equivalent to the Halting problem. So the formula is not "recursively true." 
(Note that in the above example, our underlying structure is the natural numbers with $+$ and $\times$. In general, we either need to look at copies of a given structure with domain $\omega$, or adopt some other notion of effectiveness which treats a broader class of structures.)
There are now a bunch of interesting directions one can go with this! We could try to formalize the notion of "recursively true"; depending what aesthetic we bring to the table, this could lead to an intuitionistic logic, or to something like Reverse Mathematics. We could look at set of true $\Pi^0_2$ (or beyond) sentences of arithmetic, partially preordered by "every Skolem function for $\varphi_1$ computes a Skolem function for $\varphi_2$" (or other partial preorders with the sameish motivation, for example the 'degrees of provability' http://projecteuclid.org/download/pdfview_1/euclid.ndjfl/1352383227). We could focus on some particular sentence of interest, and try to understand the family of Skolem functions associated to it. We could move one type higher - to second-order statements like "For every set $X\subseteq\mathbb{N}$, there is a set $Y\subseteq\mathbb{N}$ such that if $X$ codes an infinite linear order then $Y$ codes an infinite ascending sequence or an infinite descending sequence in $X$" and look at the descriptive set-theoretic complexity of the associated higher-order Skolem functions. Or any number of other approaches I haven't thought of!

The bottom line for me: Skolemization lets us turn a statement which may be false or true into a statement which may be false or may be true with a certain complexity; and this makes things infinitely more fascinating, at least for me!<|endoftext|>
TITLE: What is the DGLA controlling the deformation theory of a complex submanifold?
QUESTION [18 upvotes]: Let $X$ be a complex manifold, $Y\hookrightarrow X$ a complex compact submanifold.  Let $T_{X/Y}$ denote the normal bundle of $Y$ in $X$, and $\mathcal{O}(T_{X/Y})$ its sheaf of holomorphic sections. A classical result, proven by Kodaira, is that if the cohomology group $H^1(Y,\mathcal{O}(T_{X/Y}))$ vanishes, then the deformation theory of $Y$ as a complex submanifold of $X$ is unobstructed. More precisely, there exists a "maximal family" of deformations of $Y$ in $X$: this consists of a complex manifold $W$ (say with a marked point $w_0\in W$), a complex submanifold $V$ of $W\times X$, such that for each point $w\in W$ the intersection of $V$ with $w\times X$ is a complex compact submanifold of $X$, which in the case $w=w_0$ is equal to $Y$.  Furthermore, this family of submanifolds is "maximal" (or "universal") in an appropriate sense.  By saying the deformation problem is unobstructed, I mean the following: for any family as described above (maximal or not) there is a canonical injective complex linear map from the tangent space of $W$ at $w_0$ to the space of holomorphic sections of $T_{X/Y}$, i.e. the cohomology group $H^0(Y,\mathcal{O}(T_{X/Y}))$.  Given the above assumption that $H^1(Y,\mathcal{O}(T_{X/Y}))=0$, this map is an isomorphism.  Roughly speaking, one can view a holomorphic section of the normal bundle of $Y$ as a first order deformation of $Y$ in $X$, and unobstructedness means that each such first order deformation can be extended to an "honest" family of deformations.  
On the other hand, Kodaira does not explore what happens if one weakens the condition $H^1(Y,\mathcal{O}(T_{X/Y}))=0$.  For a given first order deformation $v\in H^0(Y,\mathcal{O}(T_{X/Y})$, it seems plausible that it isn't necessary for the entire group $H^1(Y,\mathcal{O}(T_{X/Y}))=0$ to vanish in order for $v$ to extend to an honest family of deformations.
$\textbf{Question 1}$: For a given first order deformation $v\in H^0(Y,\mathcal{O}(T_{X/Y}))$, is there a way to determine when $v$ extends to an honest family of deformations?  In particular, is there a map $T:H^0(Y,\mathcal{O}(T_{X/Y}))\to H^1(Y,\mathcal{O}(T_{X/Y}))$ which can be described in a reasonably explicit way such that $v$ extends if and only if $T(v)=0$?
One possible solution to this problem would be to derive a Maurer-Cartan type of equation. It's my understanding that, on general grounds, one expects the deformation theory of almost any structure to be controlled by a differential graded Lie algebra (DGLA), with (perhaps formal) deformations corresponding to solutions of the Maurer-Cartan equation. One candidate for the DGA in the above example would be the space of sections of the Dolbeault complex $\Omega^{\bullet}(Y;\Lambda^{\bullet}T_{X/Y})$ associated to the holomorphic vector bundle $\Lambda^{\bullet} T_{X/Y}$.  This comes equipped with the Dolbeault differential, but no obvious Lie bracket (at least not obvious to me).
$\textbf{Question 2}:$  Can $\Omega^{\bullet}(Y;\Lambda^{\bullet}T_{X/Y})$ be given the structure of a DGLA (i.e. equipped with a bracket compatible with the Dolbeault differential), such that formal deformations of $Y$ as a complex submanifold of $X$ are given by solutions to the Maurer-Cartan equation?  If not, is there some other DGLA controlling the deformation theory of a complex submanifold which has a "geometric" description (i.e. as the space of sections of some vector bundle)?

REPLY [7 votes]: I'd too suggest the paper by Donatella Iacono as a basic reference. I'll just add
here a few lines to complement Urs Schreiber's answer by explaining in which
sense the $\infty$-groupoids point of view clarifies what happens here (if my
memory is not playing a bad trick on my, in the paper with Elena Martinengo we do
not address the problem of deformations of submanifalds, although the needed
technology is there). So these few lines are to be read as a "How to read
Donatella Iacono's results by an $\infty$-groupoids point of view".
What one does is moving from Set-valued deformation functors to
$\infty$-groupoids valued deformation functors (these are called "formal moduli
problems" in Lurie's DAG X). One recovers the classical deformation functor from
the $\infty$-groupoid valued one simply by taking $\pi_0$. However (and this is
the reason to prefer the $\infty$-groupoid valued version), things are behaved
much more naturally in the $\infty$-groupoids setting since here one can make homotopy invariant constructions. 
More precisely, if we denote by $Def_{\mathfrak{g}}$ the $\infty$-groupoids-valued deformation functor associated with a differential graded Lie algebra $\mathfrak{g}$, then the association $\mathfrak{g}\mapsto Def_{\mathfrak{g}}$ establishes an equivalence of $(\infty,1)$ categories between differential graded Lie algebras and formal moduli problems (see Lurie's DAG X or Pridham's "Unifying derived deformation theories"). This means in particular that if the deformation problem we are interested in arises as a (homotopy) limit of simpler deformation problems for which we know differential graded Lie algebras $\mathfrak{g}_i$ governing them, then the problem we are interested in will be governed by the (homotopy) limit of the $\mathfrak{g}_i$'s.
For instance, consider the problem of $Def_{Z\hookrightarrow X}$ of infinitesimal deformations of a complex submanifold $Z$ inside a complex manifold $X$. Such a deformation is equivalently the datum of a deformation of the pair $(Z,X)$ together with the datum of a trivialization of the deformation of $X$. So if we denote by $Def_{(Z,X)}$ and by $Def_X$ the deformation functors describing infinitesimal deformations of the pair and of $X$ respectively, we see that the problem $Def_{Z\hookrightarrow X}$ we are interested in is the (homotopy) fiber of the forgetful morphism  $Def_{(Z,X)}\to Def_X$. For both $Def_X$ and $Def_{(Z,X)}$ it is simple to describe dglas governing them. They are the Dolbeault complex on $X$ with coefficients in the tangent sheaf of $X$ and the sub-dgla of this given by the kernel of the natural morphism to the Dolbeault complex on $Z$ with coefficients in the normal sheaf of $Z$ (i.e., by differential forms on $X$ with coefficients in the tangent sheaf of $X$ which, restricted to $Z$ are tangent to $Z$, thus inducing a deformation of $Z$). The dgla governing $Def_{Z\hookrightarrow X}$ will therefore be the homotopy fiber of this inclusion.
Since the model category structure on differential graded Lie algebras is induced by that on chain complexes, as a chain complex the homotopy fiber of the inclusion of the kernel above is quasi-isomorphic to the Dolbeault complex on $Z$ with coefficients in the normal sheaf of $Z$ shifted by one degree. And by the homotopical transfer of $L_\infty$-structures this means that there is a natural $L_\infty$-algebra structure on the shifted Dolbeault complex of $Z$ with values in $N_{X/Z}$, extending the chain complex structure, such that the associated deformation functor is $Def_{Z\hookrightarrow X}$. This answers Question 2 above.
In particular one recovers the well known fact that the tangent space to $Def_{Z\hookrightarrow X}$ is $H^0(Z, N_{X/Z})$: this is a degree 1 cohomology group for the shifted Dolbeault complex. Similarly one sees that $H^1(Z, N_{X/Z})$ is an obstruction space for $Def_{Z\hookrightarrow X}$.<|endoftext|>
TITLE: Automatic factors of braid groups
QUESTION [6 upvotes]: Consider the braid group $B_n$ presented in terms of the usual Artin generators $\sigma_1,\ldots,\sigma_{n-1}$. Now add the additional relations $\sigma_i^k = 1$ for $i=1,\ldots,n-1$. For lack of a better name, I'll call the resulting group $B_n^{(k)}$. $B_n^{(2)}$ is just the symmetric group $S_n$. As discussed in this mathoverflow thread, $B_n^{(k)}$ is infinite unless $\frac{1}{n}+\frac{1}{k} > \frac{1}{2}$. $B_n$ is an automatic group and thus any word $w$ in the Artin generators can be reduced to a normal form that depends only on the group element in $O(|w|^2)$ time. My question is, for fixed $k > 2$, can one reduce a word $w$ in the generators of $B_n^{(k)}$ to a normal form in $\mathrm{poly}(n,|w|)$ time, too? (It appears to me that direct simpleminded adaptation of the algorithm for braid group normal form fails.)

REPLY [7 votes]: The groups you are considering are a special class of Shephard groups, which are obtained from Artin groups by adding relators $\sigma_i^{k_i}=1$, $0\le k_i<\infty$ for every Artin generator. 
Biautomaticity of some of all these groups is conjectured in http://arxiv.org/abs/0901.0094 
and this conjecture is verified in some cases. Unfortunately for you, the results in this paper require that the Artin graph contains no triangles with label 2, which will exclude quotients of classical braid groups. Nevertheless, you may want to read the paper to see if the methods could be useful in the context of your question.<|endoftext|>
TITLE: What are first eigenfunctions of Laplacian for $CP^n$ with Fubini-Study metric?
QUESTION [10 upvotes]: I know the round $n$-sphere has $f_i=\cos(dist(e_i, x))$ as the set of first eigenfunctions for $e_i=(0, \cdots, 1, \cdots, 0)\in \mathbb R^{n+1}$. i.e. $\Delta f_i=\lambda_1 f$, where $\lambda_1$ is the first eigenvalue of $\Delta$.
So I am wondering whether there is a similar description for the 'round' complex projective spaces?

REPLY [17 votes]: Of course, Igor's answer points the way to working out the answer the OP wanted, but it may not be clear, even after you have got the eigenvalues, what the corresponding eigenfunctions are, or that they have a simple geometric interpretation analogous to the one for the sphere, as the OP asks.
The nice way to describe the first eigenfunctions on the $n$-sphere with its standard metric is to start with the standard isometric embedding $X:S^n\to\mathbb{R}^{n+1}$ and then, for each vector $v\in\mathbb{R}^{n+1}$, one defines the function $f_v:S^n\to\mathbb{R}$ by
$$
f_v(p) = v\cdot X(p).
$$
Every eigenfunction for the first nonzero eigenvalue of the Laplacian is of this form for a unique $v$.
Similarly, for $\mathbb{CP}^n$, there is a natural embedding $X:\mathbb{CP}^n\to\mathsf{H}_{n+1}$, where $\mathsf{H}_{n+1}$ is the (real) vector space of $(n{+}1)$-by-$(n{+}1)$ Hermitian symmetric matrices, given by
$$
X\bigl([v]\bigr) = \frac{v\ v^\ast}{|v|^2} = \frac{v\ v^\ast}{(v^\ast v)},
$$
where $v$ is any nonzero vector in $\mathbb{C}^{n+1}$ (thought of as column vectors of height $(n{+}1)$ and $v^\ast$ is its conjugate transpose, and $[v]\in\mathbb{CP}^n$ is the line in $\mathbb{C}^{n+1}$ that is spanned by $v$.  (With the right scaling and inner product on $\mathsf{H}_{n+1}$, this $X$ is an isometric embedding of $\mathbb{CP}^n$ endowed with the Fubini-Study metric.  In some sense, it is the simplest such.  Note that it is equivariant with respect to the natural actions of $\mathrm{SU}(n{+}1)$ on the domain and range.)
Now let $w\in\mathsf{H}_{n+1}$ be traceless and define
$$
f_w\bigl([v])\bigr) = \mathrm{tr}\bigl(\ w X\bigl([v]\bigr)\ \bigr).
$$
Then every eigenfunction of lowest nontrivial eigenvalue of the Laplacian is of the form $f_w$ for some traceless $w\in\mathsf{H}_{n+1}$.  Thus, as in the case of the $n$-sphere, the lowest eigenfunctions are essentially the components of the 'natural' isometric embedding of $\mathbb{CP}^n$.  (The reason for the 'essentially' qualifier is that, as defined, $X$ actually maps $\mathbb{CP}^n$ into the hyperplane in $\mathsf{H}_{n+1}$ that consists of the matrices of trace $1$, and one really ought to subtract the term $\tfrac1{n+1}\ I_{n+1}$ from $X$ so that it maps into the hyperplane of traceless matrices.  Then the statement is literally true.  Those who are familiar with symplectic geometry will recognize that if one then multiplies by $i$, so that the map goes into the traceless skew-Hermitian matrices, then the result is the canonical embedding of $\mathbb{CP}^n$ as a (co)-adjoint orbit of $\mathrm{SU}(n{+}1)$.)
To get something like the formula in terms of distance that held for the $n$-sphere, you just need to observe that as $[v]$ varies, the traceless elements
$$
W\bigl([v]\bigr) = X\bigl([v]\bigr) - \tfrac{1}{n+1} I_{n+1}
$$
span the space of traceless Hermitian $(n{+}1)$-by-$(n{+}1)$ matrices, so this means that this eigenspace is spanned by the functions of the form
$$
h_{[w]}\bigl([v]\bigr) = \frac{\bigl|\langle w,v\rangle\bigr|^2}{|w|^2|v|^2} - \frac{1}{n+1}
$$
as $[w]$ varies over $\mathbb{CP}^n$.  Obviously, the function $h_{[w]}$ can be expressed as a function of the distance from $[w]$ in the Fubini-Study metric, even though the generic linear combinations of such eigenfunctions aren't usually expressible in terms of distance from a point.<|endoftext|>
TITLE: Can we ascertain that there exist an epimorphism $G\rightarrow H?$
QUESTION [46 upvotes]: Let $G,H$ be finite groups. Suppose we have a epimorphism $G\times G\rightarrow H\times H$. Can we find an epimorphism $G\rightarrow H$?
A fellow graduate student asked me this question during TA sessions. Baffled, I asked this question on mathstackexchange [site][1], received some positive votes but no answer. According to him he has been running a software check on small order groups for days, and still have not find any counter example. So I venture to ask in here. It 'feels' unlikely to be true, yet we cannot find a proof or a counter example. 
This is a repost of 
https://mathoverflow.net/questions/110857/can-we-ascertain-that-there-exist-an-epimorphism-g-rightarrow-h
by the request of the moderator from meta.mathoverflow. The original post will be merged with this post.

REPLY [35 votes]: (Crossposting my counterexample from MSE here)
Let $G=Q_8\times D_8$, where $Q_8$ is the quaternion group and $D_8$ is the dihedral group of order $8$.
Let $f$ be an isomorphism $$f:G\times G =\left(Q_8\times D_8\right)\times \left(Q_8\times D_8\right)\longrightarrow \left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right).$$
Now, let $\mu$ and $\lambda$ be epimorphisms $$\begin{eqnarray*}\mu:Q_8\times Q_8&\longrightarrow&Q_8 {\small \text{ Y }} Q_8\\ \lambda:D_8 \times D_8&\longrightarrow&D_8 {\small \text{ Y }}D_8\end{eqnarray*}$$
where $A {\small \text{ Y }} B$ denotes the central product of $A$ and $B$.  Then $$\mu\times \lambda:\left(Q_8\times Q_8\right)\times \left(D_8\times D_8\right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)$$
is an epimorphism.  The key is that $D_8{\small \text{ Y }} D_8\cong Q_8{\small \text{ Y }} Q_8$, so if we take an isomorphism $$\phi:D_8{\small \text{ Y }} D_8\longrightarrow Q_8{\small \text{ Y }} Q_8,$$ we can take $H=Q_8{\small \text{ Y }} Q_8$ and form an isomorphism
$$1_H\times \phi:\left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8 \right)\longrightarrow \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8 \right)=H\times H.$$
So, all in all, we have 
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
\left(Q_8\times D_8\right) \times \left( Q_8 \times D_8 \right)& \ra{f} &\left(Q_8\times Q_8\right) \times \left( D_8 \times D_8 \right)&\\
& & \da{\mu\times \lambda} & &  & & \\
& & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(D_8 {\small \text{ Y }}D_8\right) & \ras{1_H\times \phi} & \left(Q_8 {\small \text{ Y }}Q_8\right)\times \left(Q_8 {\small \text{ Y }}Q_8\right)
\end{array}
$$
and thus an epimorphism $$f(\mu\times\lambda)(1_H\times \phi):G\times G\longrightarrow H\times H.$$
However, $Q_8{\small\text{ Y }}Q_8$ is not a homomorphic image of $Q_8\times D_8$. So this is a counterexample.
(Credit and thanks to Peter Sin for the crucial step in this answer.)<|endoftext|>
TITLE: A sentence in Shimura's "On The Periods of Modular Forms"
QUESTION [11 upvotes]: Let $K_f$ denote the number field generated by the Fourier coefficients $a_n$ of a normalized primitive holomorphic cusp form $f$.
On page 2, line 6 of the paper mentioned in the title, Shimura writes that $K_f$ is generated by $a_p$ for almost all primes $p$. In the next sentence, he says that it follows trivially from the Multiplicity one theorem. 
I don't see how it follows. I shall appreciate any comment.

REPLY [21 votes]: Suppose that a co-finite subset of the $a_p$'s generate a field $L$. Applying any element $\sigma$ of the absolute Galois group of $L$ to $f$ yields another form $g$ whose Fourier coefficients $b_p$ are equal to $a_p$ for all but finitely many $p$.  By strong multiplicity one, $g=f$ and $b_p = a_p$ for all $p$.  Therefore $\sigma$ fixes $K_f$ as well, and since $L \subset K_f$ it follows that $L=K_f$.<|endoftext|>
TITLE: Are irreducible components of a flat family flat?
QUESTION [6 upvotes]: Let $f:X\rightarrow Y$ be a flat morphism of schemes of finite type over a field $k$, and assume $Y$ is irreducible. Let $X_1, \dots, X_n$ be the scheme-theoretic irreducible components of $X$ (i.e., including embedded components). 

Is it true that each $X_i$ is flat over $Y$?
If there are counterexamples to flatness of the $X_i$, is it true at least that each of them has equidimensional fibers?

REPLY [10 votes]: No to the first question. Let $Y$ be a nodal cubic curve and let $X$ be its connected two-sheeted covering space. Each of the two components of $X$ is the normalization of $Y$.<|endoftext|>
TITLE: Smashing localizations in the category of spectra
QUESTION [10 upvotes]: Let $E$ be a spectrum. Then $E$ determines an idempotent localization functor $L_E: \mathrm{Sp} \to \mathrm{Sp}$ sending each spectrum to its $E$-localization. 
The functor $L_E$ generally does not commute with homotopy colimits. 
(It does send homotopy colimits in spectra to homotopy colimits in $E$-local spectra, though.) When $L_E$ commutes with homotopy colimits, then it is (by a formal argument) equivalent to smashing with the $E$-local sphere $L_E S$ and the localization functor is said to be smashing. 
When $E$ is a Moore space $SG$ for $G$ torsion-free, $L_E$ can be described via "arithmetic" localization: the homotopy groups get localized in the sense of commutative algebra. These are smashing localizations. But there are other examples. A (deep) theorem  of Hopkins and Ravenel  states that localization with respect to Morava $E$-theory $E_n$ is a smashing localization as well. 
(By contrast, localization with respect to torsion things tends to involve arithmetic completion, which is not smashing.)
Is it known whether there are additional examples of smashing localizations (in the $p$-local category)?

REPLY [12 votes]: Finite localizations, as defined by Miller ("Finite localizations", Boletin de la Sociedad Matematica Mexicana 37 (1992), 383–390; preprint here) are also smashing localizations. The finite localization away from the thick subcategory of objects of type $n+1$ is sometimes written as $L_n^f$ or $L\prime_n$ (unless I've messed up the indexing). One formulation of the telescope conjecture is that every smashing localization is a finite localization; in particular, if you write $L_n$ for localization with respect to $E_n$, then a consequence would be that $L_n = L_n^f$.<|endoftext|>
TITLE: Is the tangent cone of a totally convex subset again totally convex?
QUESTION [9 upvotes]: $X$ be an Alexandrov's space with lower curvature bound and $C$ be a totally convex subset, i.e. for any $x,y \in C$ and any geodesic $\gamma$ (that is a locally shortest path) connecting $x$ and $y$ we have $\gamma \subseteq C$. For $p \in C$ the tangent cone $K_pC \subset K_p X$ is thus well defined. My question is:
Is $K_pC$ totally convex as well?
It is not hard to see that $K_pC$ is convex in the sense that any unique shortest connection between points in $K_pC$ also lies within $K_pC$, solving this problem for example in the riemannian case. (In fact let $v,w \in K_pC$ together with a unique shortest geodesic $\gamma$ connecting the two points. Using the scaling invariance of the problem together with $(K_pC,0) = \lim_{\lambda \to \infty} (\lambda C,p)$ one may approximate $\gamma$ by geodesics contained in $C$.
But i think in general it might not be possible to approximate arbitrary geodesics like this).

REPLY [5 votes]: (Too long for a comment)
The question is interesting and it might be hard.
From the comments: The totally convex subset $C$ usually appears as a sublevel set of a locally Lipschitz convex function (I do not know other sources of totally convex subsets).
If $C$ is a sublevel set of a convex function for a not mimimal value $a$
then so is $K_pC$, in particular $K_pC$ is totally convex.
Related stuff. Instead of tangent cone you might consider the same question for a (noncollapsing) Gromov--Hausdorff convergence $A_n\to A_\infty$. (In particular you may think that $A=A_n=A_\infty$ for all $n$ and $C_n$ is a sequence of totally convex sets.) Here some relevant statements which might be useful. 

Any minimizing geodesic in $A_\infty$ can be approximated by minimizing geodesics in $A_n$. (Any minimizing geodesic can be approximated by unique minimizing geodesic, which is approximated by minimizing geodesic in $A_n$.)
If $A_n$ are Riemannian then any geodesic in $A_\infty$ can be approximated by geodesic in $A_n$. (You approximate a minimizing piece and then extend the approximation.) The general case would follow if the geodesic in Alexandrov space without boundary have infinite extension with probability 1 (this is not known now). 
You might consider version of definition of totally convex set with quasigeodesics instead of geodesics. In this case the answer is NO; take $A_n=A_\infty$ to be a 2-dimensional cone and the sets $C_n$ which which lie on distance $\ge 1$ from the tip, but for its limit $C_\infty$ there is a quasigeodesic which pass through the tip.<|endoftext|>
TITLE: Integral kernel for the resolvent of the laplace operator
QUESTION [7 upvotes]: Consider the Laplace operator defined in the biggest possible subset of $L^2(\mathbb{R}^2)$ and let $z \in \mathbb{C}\backslash\mathbb{R}$. Therefore $z \notin \sigma (\Delta)$ the spectrum of $\Delta$, and the resolvent $R=(-\Delta - zI)^{-1}$ is well defined and bounded in all of $L^2(\mathbb{R}^2)$.
I'm trying to find the integral kernel for this operator, that is a function $K(x,y)$ such that almost everywhere in $\mathbb{R}^2$:
$$(Ru)(x)=\int_{\mathbb{R}^2} u(y)K(x,y) dy$$
Let $f := ( - \Delta - z I)^{- 1} u$ and let $\mathcal{F}$ be the Fourier
transform. Now
$$ ( - \Delta - z I) f = u \Rightarrow \mathcal{F} ( ( - \Delta - z I) f)
   =\mathcal{F} ( u) \Rightarrow ( \xi^2 - z) \hat{f} ( \xi) = \hat{u} ( \xi)
   . $$
Solving for $\hat{f}$ and applying $\mathcal{F}^{- 1}$ we arrive at (modulo
some constant depending on your favourite definition of $\mathcal{F}$)
$$ f ( x) =\mathcal{F}^{- 1} \left( \frac{1}{\xi^2 - z}  \hat{u} ( \xi)
   \right) =\mathcal{F}^{- 1} \left( \frac{1}{\xi^2 - z}  \right) \ast u ( x)
$$
So I need to calculate
$$ \int_{\mathbb{R}^2} \frac{e^{ix \cdot \xi}}{\xi^2 - z} d \xi $$
and I'm completely stuck. Does anybody have any ideas or references? Thanks.

REPLY [2 votes]: I finally found the solution in Evans' book on Partial Differential equations,
Chapt. 4.3. Using that
$$ \frac{1}{\xi^2 - z} = \int_0^{\infty} e^{- t ( \xi^2 - z)} d t, $$
and substituting into
$$ \int_{\mathbb{R}^2}  \frac{e^{ix \cdot \xi}}{\xi^2 - z} d \xi $$
then using Fubini, completing the square in the exponential function and
evaluating a complex integral along a line $ \{ Im = const. \} $, one arrives at the solution.
Thanks everybody for the fine answers.<|endoftext|>
TITLE: Random walks on Coxeter groups
QUESTION [14 upvotes]: Let $G_N$ be the group generated by elements $a_1,\ldots,a_N$ subject to the relations $a_i^2=1$ and $(a_ia_j)^3=1$. The growth function of $G_N$ is then
$$f_N(t)=\frac{1+2t+2t^2+t^3}{1-Mt-Mt^2+\frac{M(M+1)}{2}t^3}$$
where $M=N-2$ (or at least, this seems to be the correct generalization of the $N=4,5$ known formulae, A154638 and A162740 @ OEIS). Question now: is the random walk function
$$g_N(t)=\sum_{k=0}^\infty Card(i_1,\ldots,i_k|a_{i_1}\ldots a_{i_k}=1)\cdot t^k$$ 
known? Maybe related to $f_N$? I'd be actually interested in the $N\to\infty$ behavior of $g_N$.

REPLY [9 votes]: For the case $N=3$, this is the random walk on the honeycomb lattice:
        
  (source)
Lemma 2.1 in this paper computes the number of walks of length $2n$ on
the honeycomb lattice
which return to the origin,
so $$g_3(t)=\sum_{n=0}^{\infty}\sum_{k=0}^n \binom{2k}{k}\binom{n}{k}^2 t^{2n}.$$<|endoftext|>
TITLE: Where did Nachman Aronszajn prove the existence of Aronszajn trees?
QUESTION [8 upvotes]: I understand that Nachman Aronszajn is credited for proving the existence of Aronszajn trees.  But I've not had any success finding where this was published.  Does anyone know of a specific paper or book where Aronszajn published his proof?

REPLY [9 votes]: I believe that these first appeared in the paper
Georges Kurepa, Ensembles linéaires et une classe de tableaux ramifiés (tableaux ramifiés de M.Aronszajn), Publ. Math. Univ. Belgrade, 6 (1936) 129-160.
The author attributes them to a letter that Aronszajn wrote to him (see page 132).  This paper is available here.

Edit by A. Caicedo: The following is from Stevo Todorcevic's essay introducing the papers on "Theory of partially ordered sets", part A of Selected papers of Ðuro Kurepa, Edited and with commentaries by Aleksandar Ivić, Zlatko Mamuzić, Žarko Mijajlović and Stevo Todorčević, Matematički institut SANU, Belgrade, 1996. MR1429393 (97m:01106).
The papers it refers to are 

A[35] Ensembles ordonnés et ramifiés, Publ. Math. Univ. Belgrade 4 (1935), 1-138.
A[37] Ensembles lineaires et une classe de tableaux ramifies (Tableaux ramifies de
M. Aronszajn), Publ. Math. Univ. Belgrade 6/7 (1937/38), 129-160.
[1] J. E. Baumgartner, Decomposition and embedding of trees, Notices Amer. Math. Soc. 17 (1970), 967.
[6] K. J. Devlin, Note on a theorem of J. Baumgartner, Fund. Math. 76 (1972), 255-260. MR0540759 (58 #27476).
[12] W. P. Hanf, Incompactness in languages with infinitely long expressions, Fund. Math. 53 (1964), 309-324. MR0160732 (28 #3943).


The most important publication in this group is Kurepa's thesis A[35] written under the direction of M. Fréchet. It was the first systematic study of trees and ramified partially ordered sets and of their close relationship to linear orderings. It was the source of many crucial notions and problems in this area such as, for example, the notions of Aronszajn and Souslin tree. It is the source of the problem whether inaccessible cardinals have the tree property i.e., whether they satisfy the analogue of König's infinity lemma, which was later proved by Hanf, Tarski and others ([12]) to be equivalent to the large cardinal property of weak compactness. Trees are classified in §8 A11 as "large", "étroit" and "ambigu" according to their heights and widths. The very thin and tall trees ("étroit") always have cofinal branches i.e., chains intersecting every level (Theorem $5^{bis}$). This is a fine result representing a recurring theme in applications of trees, especially in the partition calculus. This result was also a source of the problem whether the same fact is true about the class of slightly wider trees ("ambigu") i.e., the trees of height equal to
  some cardinal $\theta$ and whose levels are now only assumed to be of size $\lt \theta$ (rather than $\lt\lambda$ for some cardinal $\lambda<\theta$ as it was the case with the trees in Theorem $5^{bis}$). This is the problem known today as the problem whether $\theta$ has the tree property. For $\theta= \omega_1$ the problem was solved in June 1934 by N. Aronszajn and appeared (with an acknowledgement) as Theorem 6 of A[35]. According to the footnote on the same page, Aronszajn constructed his tree as a subtree of the tree of all $1-1$ sequences from $\mathbb Q^{\lt\omega_1}$, while Kurepa's version of the proof presented in A[35] (and also in A[37]) was to build such a tree inside the tree $\sigma\mathbb Q$ (denoted by $\sigma_0$) of all (nonempty bounded) well-ordered subsets of rationals. This is the construction most frequently used in subsequent expositions of this result. In the sequel A[37;§27] he modified his construction in order to produce an Aronszajn tree with the surprising property that it is a union of countably many antichains, thus introducing yet another remarkable notion, the notion of a special Aronszajn tree. (It is known today that some care is needed to get such a tree, as it is possible to have nonspecial Aronszajn subtrees of both $\sigma\mathbb Q$ and the set of all $1-1$ sequences from $\mathbb Q^{\lt\omega_1}$; see [1] and [6].) 

The essay goes on to describe additional work (by Kurepa and others) on the tree property.<|endoftext|>
TITLE: Who was Hermann Künneth?
QUESTION [23 upvotes]: Question as in the title:
Who was Hermann Künneth? Where can I find some biographical information beyond what is available on Wikipedia?
The well-known Künneth formula, for example in the form of exactness of the sequence 
$$
0 \to \bigoplus_{p+q = n} H_p(C) \otimes H_q(D) \to H_n(C \otimes D) \to \bigoplus_{p+q=n-1}\operatorname{Tor}_1(H_p(C),H_q(D)) \to 0,
$$
for complexes $C$ and $D$ of flat modules over a PID appears prominently in essentially every book on homological algebra and algebraic topology. Of course, Künneth formulated his insight in terms of Betti numbers, not in terms of homology groups.
Nevertheless, biographical information on its originator seems quite hard to find.
Wikipedia links to Haupt's 5 page obituary in German which mainly focuses on Künneth's mathematics with only a few lines dedicated to his life.
Any further pointers would be appreciated.

REPLY [19 votes]: For the sake of the readers who are not fluent in German, I provide a translation of the German Wikipedia page (link to the revision at the time of posting this answer):
Hermann Künneth (1892-1975) was the son of the high school ("Gymnasium", the highest form of high school) teacher Christian Künneth. Beginning with 1910, he studied mathematics at the Universität Erlangen and the Ludwig-Maximilians-Universität München with a "break" from 1914-1919 where he served in the German army; he was injured twice and was prisoner of war with the British. Künneth was member of the AMV Fridericiana Erlangen, a musically oriented fraternity. His professors in Erlangen were Ernst Sigismund Fischer, Paul Gordan, Max Noether, Richard Baldus and Erhard Schmidt.
1912 he took his first Staatsexamen to become a teacher and 1920 he took his second. In 1920 he became teacher in Bavaria, in particular at high schools ("Gymnasien") in Kronach and Erlangen. He remained in contact with the University in Erlangen, where he got in PhD under the direction of Tietze in 1922 (and was assistant (professor) beginning with 1921). The title of his thesis was "Über die Bettischen Zahlen einer Produktmannigfaltigkeit" - "About the Betti numbers of a product manifold" (where he proved the Künneth formula).
1923 he became assistant (professor) in Berlin; interestingly, he became 1923 also teacher ("Studienrat") in Kronach. As already indicated, the switched to the high school  Fridericianum in Erlangen in 1925, where he became Oberstudienrat in 1950 [this would not be a very high position at a high school these days, but I am not sure how it was then].
In 1942 he habilitated in Erlangen and was Privatdozent (a kind of freelancing professor) after that. After he retired in 1957 from his teaching job, he became associate professor in Erlangen. Otto Haupt said about this: "[he] developed an amazing and surprising scientific activity." (at the age of 65)
1964 he got the Bundesverdienstkreuz am Band (the second lowest order of the "Order of Merit of the Federal Republic of Germany"). (See this newspaper article - it says: "His chivalric personality, of clear judgement, emanates human kindness and witty humour.")<|endoftext|>
TITLE: Relating a Polynomial equation to the characteristic equation of a Hermitian matrix
QUESTION [8 upvotes]: This question arose out of mere curiosity. Given a polynomial equation and I happen to know that its roots are real (but not the roots itself). Does it mean it is the characteristic equation of a Hermitian matrix?      And if that is the case, could I just find the hermitian matrix and solve for its eigenvalues?. When I thought about it, looks like determining the hermitian matrix from the polynomial equation looks like a daunting task. Any thoughts on this?

REPLY [10 votes]: Let $K$ be a subfield of $\mathbb R$ (e.g., $K=\mathbb Q$) and $f\in K[x]$ be a monic polynomial of degree $n$ all of whose roots are real. We want to compute a symmetric matrix $A\in K^{n\times n}$ such that $f=\det(xI_n-A)$. By squarefree decomposition in $K[x]$ we can assume that $f$ has only simple roots (otherwise perform euclidean divisions to compute the squarefree decomposition, find a determinantal representation for each factor in this decomposition and use diagonal blocks).
Of course, the idea is to start off with the companion matrix $C\in K^{n\times n}$ of $f$ which represents the vector space endomorphism $$\varphi\colon K[x]/(f)\to K[x]/(f),\ \overline p\mapsto\overline{xp}$$
with respect to the canonical monomial basis. We know that $f=\det(xI_n-C)$. In particular, $C$ is similar to a triangular matrix and therefore $\text{tr}(C^k)$ is the sum of the $k$-th powers of the roots of $f$ for each $k\in\mathbb N$.
Now $C$ itself is almost never symmetric but $\varphi$ is self-adjoint with respect to the $L_2$ "scalar product" given by a measure whose mass is uniformly distributed on the roots of $f$ (we use here that the roots are simple so that this really gives a non-degenerate bilinear form). Denoting the usual scalar product on $\mathbb R^n$ by $\langle.,.\rangle$ and by $V\in \mathbb R^{n\times n}$ the Vandermonde matrix corresponding to the roots of $f$, this means that $\langle VCx,Vy\rangle=\langle Vx,VCy\rangle$ for all $x,y\in\mathbb R^n$. In other words, we have $C^TV^TV=V^TVC$.
Now the Hermite matrix $H:=V^TV$ of $f$ comes naturally into play. It is a Hankel matrix whose entries are power sums of roots of $f$. But these power sums are traces of powers of $C$ and therefore lie in $K$ and can easily be computed. The idea is to replace $V$ by another matrix $W$ with $H=W^TW$. In contrast to $V$, the matrix $W$ should have entries in $K$ and should be easily computed. But this is possible: Just take any $W\in K^{n\times n}$ with $H=W^TW$. You get such a $W$ even in a triangular form if you use the Cholesky decomposition of the positive definite matrix $H$.
Now $C^TW^TW=C^TH=C^TV^TV=V^TVC=HC=W^TWC$ and since $W$ is invertible (note that $V$ and $H$ are invertible since all roots of $f$ are simple) this shows that $WCW^{-1}$ is symmetric. So you can set $A:=WCW^{-1}$.
All this is folklore.
If you want a tridiagonal $A$ then you would have to perform Sturm's algorithm as indicated above by Denis. In fact, finding a tridiagonal $A$ is essentially equivalent to Sturm's algorithm, see the recent interesting work of Ronan Quarez:
http://arxiv.org/pdf/0811.2365v1.pdf
If the coefficients of the monic polynomial are itself polynomials in variables $y_i$ such that for each fixed real value of the $y_i$ the polynomial has again only real roots (cf. Garding's notion of hyperbolic polynomials), then you can still use the Hermite matrix to do something, see the recent article of Netzer, Plaumann and Thom:
http://arxiv.org/abs/1108.4380<|endoftext|>
TITLE: On condition when the push-forward of coherent sheaf is locally free
QUESTION [9 upvotes]: This is a result being widely used in the literature:
$f:X\rightarrow Y$ proper morphism between Noetherian schemes. $F\in Coh(X)$ flat over $Y$, if $H^i(X_y,F_y)=const$, $y\in Y$, then $R^if_\ast F$ is locally free.
The problem is that I can only find references (EGA or GTM 52, etc) of this result with a condition 'Y is reduced', which seems to be unavoidable if one try to prove it by using the canonical technique of 'Grothendieck complex'.
However, the general case is crucial in many arguments.
So my question is that: Is the general case true? When can I find the proof of the general case?
Thanks!

REPLY [15 votes]: I believe that this statement is not true. Take $Y=Spec(k[t]/t^2)$ and $X=\mathbb{P}^1_Y$. On $X$, extensions of $\mathcal{O}$ by $\mathcal{O}(-2)$ are parametrized
by : $$Ext^1_X(\mathcal{O},\mathcal{O}(-2))=H^1(X,\mathcal{O}(-2))=H^0(X,\mathcal{O})^{\vee}\simeq k[t]/t^2,$$
as a $k[t]/t^2$-module.
Let $0\to\mathcal{O}(-2)\to E\to\mathcal{O}\to 0$ be the extension corresponding to $t\in k[t]/t^2$. By construction, the map $H^0(X,\mathcal{O})\to H^1(X,\mathcal{O}(-2))$ in the long exact sequence associated to the above short exact sequence is multiplication by $t$. It follows that $H^0(X,E)$, that is the kernel of this map, is necessarily $(t)\subset k[t]/t^2$.
Hence it is isomorphic to $k[t]/t$ as a $k[t]/t^2$-module and it is not free.
As $f_*E=H^0(X,E)$ because $Y$ is affine, and $E$ is flat over $Y$ because it is a vector bundle on $X$ that is flat over $Y$, this is a counter-example to the statement.<|endoftext|>
TITLE: Projective arrows
QUESTION [7 upvotes]: We all know that a projective object in a category $\mathcal{C}$ is an object $P$ in $\mathcal{C}$ such that for every epimorphism $f: X\to Y$ in $\mathcal{C}$ and arrow $g\colon P\to Y$ there is a lift $g'\colon P\to X$ of $f$.
Let us define a projective arrow in $\mathcal{C}$ to be an arrow $g: P\to Y$ such that for every epimorphism $f: X\to Y$ there is a lift $g': P\to X$ of $f$.


What is known about such projective arrows?
What are the projective arrows in the category of groups?
Is this just some kind of limit in disguise?


(I'm using "arrow" instead of "morphism" to avoid confilct with the accepted usage of "projective morphism" in Algebraic Geometry).

REPLY [5 votes]: In addition to Fernando's answer, note that projective arrows were introduced in 1966 by A. V. Roiter (under the name of projective morphisms)
in the paper On integral representations belonging to one genus, Izv. Akad. Nauk SSSR Ser. Mat. 30 (1966), 1315-1324,
see also the English translation: Amer. Math. Soc. Transl. (2) 71 (1968), 49-59.
Roiter defines projective arrows in an abelian category, but actually works with projective arrows in the category of modules over a ring $\Lambda$.
He notices that a morphism of $\Lambda$-modules $p\colon A\to B$ is a projective arrow if and only if $p$ factors via a projective arrow that is an epimorphism.
He uses the notion of a projective arrow in his version of Schanuel's lemma:

Roiter's lemma. 
  Let 
  $$ 0\to X \to A \to U\to 0,\qquad  0\to Y \to B \to U\to 0 $$
  be two short exact sequences, where the morphisms $A\to U$ and $B\to U$ are projective arrows.
  Then $B\oplus X \simeq A\oplus Y$.

Proof: Let $W$ denote the fibered product of $A$ and $B$ over $U$.
Then $W$ is an extension
$$ 0\to Y\to W\to A\to 0.$$
Since $\varphi\colon A\to U$ is a projective arrow and $\psi\colon B\to U$ is surjective,
$\varphi$ factors as $\psi\circ s$ for some morphism $s\colon A\to B$.
We obtain a morphism $({\rm id}_A,s)\colon A\to W$, which splits the extension  $W\to A$.
Thus $W\simeq A\oplus Y$.
Similarly $W\simeq B\oplus X$, hence $B\oplus X \simeq A\oplus Y$, as required.

Proposition (Roiter).
  Assume that $\Lambda=\mathbb{Z}[\Gamma]$, where $\Gamma$ is a finite group of order $n$.
  Let $A$ be a finitely generated free abelian group on which $\Gamma$ acts.
  Let $B$ be a finitely generated $\Lambda$-module. 
  Then  for any $\Lambda$-morphism $\varphi\colon A\to B$, the morphism $n\varphi$ is a projective arrow.

Proof: Choose an epimorphism $\psi\colon S\twoheadrightarrow B$, where $S$ is a finitely generated free $\Lambda$-module,
then we have an exact sequence
$$ 0\to C\to S\to B\to 0, $$
where $C={\rm ker\,} \psi$, and we obtain the induced exact sequence
$$ 0\to {\rm Hom}(A,C)\to {\rm Hom}(A,S)\to {\rm Hom}(A,B)\overset{\delta}{\longrightarrow}{\rm Ext}_\Lambda^1(A,C).$$
Since $A$ is $\mathbb{Z}$-free, 
we have ${\rm Ext}_\Lambda^1(A,C)=H^1(\Gamma, {\rm Hom}_{\mathbb{Z}}(A,C))$.
Since $\#\Gamma=n$, we have $n{\rm Ext}_\Lambda^1(A,C)=nH^1(\Gamma, {\rm Hom}_{\mathbb{Z}}(A,C))=0$, hence $n\delta=0$,
hence $\delta\circ(n\varphi)=(n\delta)\circ\varphi=0$, 
and therefore, $n\varphi\in{\rm ker\,}\delta$.
From the exact sequence we see that $n\varphi=\psi\circ x$ for some $x\in {\rm Hom}(A,S)$.
Since $S$ is $\Lambda$-free, the morphism $\psi$ is a projective arrow, hence $n\varphi=\psi\circ x$ is a projective arrow, as required.

Corollary.
  If $\Lambda$, $\Gamma$, $n$ and $A$ are as above and $U$ is a finite $\Lambda$-module such that $nU=U$,
  then any $\Lambda$-morphism $A\to U$ is a projective arrow.<|endoftext|>
TITLE: Is the normalizer of a reductive subgroup reductive?
QUESTION [14 upvotes]: Let $G$ be a reductive algebraic group over an algebraically closed field (of characteristic zero if it matters) and $H \subset G$ a subgroup, also reductive.  Is the identity component of the normalizer of $H$ in $G$ always reductive?

REPLY [14 votes]: In positive characteristic, the normalizer of a (connected) reductive
subgroup of a (connected) reductive group is not in general reductive.
I communicated the example below in an emailed answer to a query in
April 2002 (it took me some searching in old emails to find it!)  At
the time I wrote something at the end like "I'm not aware of a good
reference" -- this remains true.
Let $k$ be alg. closed of positive characteristic. I will exhibit $H
\subset G$ reductive groups such that $C_G(H)$ is not reductive;
since $C_G(H)$ is normal in $N_G(H)$, also $N_G(H)$ is not reductive.
Here is the example. Let $n \ge 2$, and consider the adjoint
representation $$\operatorname{Ad}: \operatorname{GL}_n \to
\operatorname{GL}(\mathfrak{gl}_n)$$ where $\mathfrak{gl}_n$ is the Lie algebra of
$\operatorname{GL}_n$.
The image $H \simeq \operatorname{PGL}_n$ of $\operatorname{Ad}$ is a
reductive subgroup of $G=\operatorname{GL}(\mathfrak{gl}_n)$, and the
centralizer $C_G(H)$ identifies with the group of units of the
endomorphism ring $\operatorname{End}_H(\mathfrak{gl}_n)$.
If $(n,p) = 1$, $\mathfrak{gl}_n$ is a semisimple representation of
$\operatorname{GL}_n$ with two distinct irreducible factors, so that
$\operatorname{End}_H(L) = k \times k$. In this case $C_G(H)$ is
a 2 dimensional torus and hence is reductive.
If $(n,p) = p$, $\mathfrak{gl}_n$ is an indecomposable representation
with 3 composition factors. Thus the endomorphism ring of
$\mathfrak{gl}_n$ is a local ring. It turns out that this endomorphism
ring $\operatorname{End}_H(\mathfrak{gl}_n)$ still has dimension 2,
but it is now isomorphic to the ring $k[t]/\langle t^2 \rangle$. To exhibit a non-0
nilpotent $H$-endomorphism $t$ of $\mathfrak{gl}_n$, take the
following map: to a matrix $X$ in $\mathfrak{gl}_n$, assign the matrix
$t(X) = \operatorname{tr}(X).1$ where 1 denotes the $n\times n$
identity matrix, and $\operatorname{tr}()$ denotes the trace.  Since
$n$ is divisible by $p$, applying $t$ twice gives 0.
The unit group of this local ring is isomorphic to the product of a
$1$ dimensional torus and the additive group of the field; such a
group is not reductive. (Explicitly: the additive subgroup is precisely the set of
all automorphisms $1 + at$ of $\mathfrak{gl}_n$ with $a \in k$.)
More generally, let $H$ be any reductive group, let $V$ be a finite
dimensional $H$-module, and write $G = \operatorname{GL}(V)$. 
Suppose that $V$ is indecomposable, has
composition length 3, and that $\operatorname{soc}(V) \simeq V
/\operatorname{rad}(V)$. Then $\operatorname{End}_H(V) \simeq
k[t] / \langle t^2 \rangle$ 
and $C_G(H) =
\operatorname{End}_H(V)^\times$ is not reductive.
Here are a few more examples of such $H$ and $V$ (I'll write
$T(\mu)$ for the indecomposable tilting module with highest weight
$\mu$).

$H = \operatorname{Sp}(W)$, $V = \bigwedge^2 W$ ("exterior square"), when $p>2$ and
$\dim W \equiv 0 \pmod p$.
$H = \operatorname{SO}(W)$, $V = \operatorname{Sym}^2 W$ ("symmetric square"), when
$p>2$ and $\dim W \equiv 0 \pmod p$.
$H = \operatorname{SL}_2$, $V = T(n)$ for $p \le n \le 2p-2$.
$H = \operatorname{SL}_{\ell + 1}$, $V = T(\varpi_i +
  \varpi_\ell)$ for $1 \le i < \ell$, when $\ell + 2 - i \equiv 0 \pmod p$,
[This more-or-less interpolates the original example since when $i=1$ and $\ell + 1 \equiv 0 \pmod p$, $T(\varpi_1 +
  \varpi_\ell) \simeq \mathfrak{gl}_{\ell+1}$.]
$H = \operatorname{SO}_{2\ell}$, $V = T(\varpi_1 +
  \varpi_\ell)$ when  $\ell \equiv 0 \pmod{p}$.<|endoftext|>
TITLE:  I know that you know... 
QUESTION [28 upvotes]: A bit unsure if the following vague question has enough mathematical content to be suitable upon here. In the case, please feel free to close it.
In several circumstances of competition, a particular situation of partial information occurs, usually described as "I know that you know that I know... something". We may distinguish a whole hierarchy of more and more complicate situations closer and closer to a complete information. E.g. :

$I_0$: I know $X$, but you don't know that I know.
$I_1$: I know $X$, you know that I know, but I don't know that you know that I know.
$I_1$: I know $X$, you know that I know, I know that you know that I know, but you don't know that I know that you know that I know.
.... &c.   

For small values of $k$, I can imagine simple situations where passing from $I_k$ to $I_ {k+1}$ really makes a difference (for instance: you are Grandma Duck, and $X$ is : "you left a cherry pie to cool on the window ledge". Clearly, $I_0$ is quite agreeable position; $I_1$ may lead to an unpleasant end (for me); $I_2$ leaves me some hope, if I behave well, and so on). But, I can't imagine how passing from $I_6$ to $I_7$ may affect my strategy, or Grandma's. 

Are there situations, real or 
  factitious, concrete or abstract,  where $I_k$
  implies a different strategy than
  $I_{k+1}$ for the competitors? What
  about $I_{\omega}$ and, more generally,
  $I_\alpha$ for an  ordinal $\alpha$
  (suitably defined by induction)? How these situations are modeled mathematically?

REPLY [2 votes]: A good source for this sort of reasoning is:
J.-J. Meyer and W. van der Hoek, Epistemic Logic for AI and Computer Science, 
Number 41 in Cambridge Tracts in Theoretical Computer Science, Cambridge 
University Press, 1995.
The question of common knowledge, and of the algorithmics of epistemic modal logics, provides a lot of good material on this sort of problem.  (It is good fun to try the mathematical side of this in Popularisation activities.  I used Muddy Children with various groups with a lot of laughter and appreciation of the mathematical models.)<|endoftext|>
TITLE: Stable infinity categories vs dg-categories
QUESTION [20 upvotes]: What is the relation between dg-categories and stable $\infty$-categories?
Given a dg-category one can form its dg-nerve and get a $\infty$-category
(which will be stable if the dg-category is?). 
Can one turn a stable $\infty$-category into a dg-category or $A_\infty$-category somehow?
I have heard the statement that at least over a field of characteristic zero the theories
of stable $\infty$-categories and dg-categories are "equivalent".
What would be a precise formulation of this statement and what would be a reference?

REPLY [19 votes]: See the recent paper

Lee Cohn, Differential graded categories are k-linear stable infinity categories, arXiv:1308.2587

where a proof has been written down.  The precise statement is that the underlying $(\infty,1)$-category associated to the Morita model structure on dg-categories over $k$ (where fibrant objects are karoubian pretriangulated dg-categories) is equivalent to the $(\infty,1)$-category of karoubian stable $k$-linear $(\infty,1)$-categories.
Update: Though Cohn works over characteristic zero, Bertrand Toen told me his arguments in fact hold in arbitrary characteristic.
Also, see the new paper 

Giovanni Faonte, Simplicial nerve of an A-infinity category, arXiv:1312.2127

where it is proved that the dg-nerve functor takes pretriangulated dg-categories to stable $(\infty,1)$-categories.<|endoftext|>
TITLE: Explicit equation of Dickson invariant / quasideterminant / special orthogonal group over the integers
QUESTION [5 upvotes]: Consider $2n$ coordinates $x_1,\ldots,x_n,y_1,\ldots,y_n$ and the quadratic form $q = \sum_{i=1}^n x_i y_i$.  Now call $O(q,A)$ (orthogonal group of $q$) the group of $(2n)\times(2n)$ matrices, with coefficients in a commutative ring $A$, which preserve $q$.  (This is an algebraic group over $\mathop{\mathrm{Spec}}\mathbb{Z}$.)  When $A$ is a field of characteristic $\neq 2$, the determinant restricted to $O(q,A)$ takes its values in $\{\pm 1\}$, its kernel $O(q,A) \cap SL(2n,A)$ defines a subgroup $SO(q,A)$ of index $2$.  When $A$ is a field of characteristic $2$, the determinant is identically $1$ on $O(q,A)$ but there is still a subgroup of index $2$, which one might still denote $SO(q,A)$, defined by the so-called "Dickson invariant", also known as quasideterminant or pseudodeterminant, and it is relatively straightforward to give an explicit polynomial in the coefficients of $A$ which defines an equation of $SO(q,A)$ inside $O(q,A)$ (see, Dickson's book, Linear Groups, theorem 205 on page 206).
Now for an arbitrary ring $A$, there is still a subgroup $SO(q,A)$ of $O(q,A)$, which is the kernel of a morphism $\deg$ of $O(q,A)$ to the group $(\mathbb{Z}/2\mathbb{Z})(A)$ of idempotents of $A$ (so, when $A$ is connected, $SO(q,A)$ is a subgroup of order $2$) natural in $A$ and which coincides with $\frac{1}{2}(1-\det)$ when $2$ is invertible in $A$ and with Dickson's invariant when $2$ is zero in $A$.  This is due to H. Bass ("Commutative Algebras and Spinor Norms over a Commutative Rings", Amer. J. Math. 96 (1974), 156–206).
Concretely, this means that there exists a polynomial $\deg$ in $4n^2$ variables over $\mathbb{Z}$ such that, modulo the ideal $I$ of relations defining $O(q,A)$, we have $\deg^2 = \deg$ and $\det = 1-2\deg$.  And $I_0 := I+(\deg)$ is the ideal defining $SO(q,A)$.
My question is: can one give an explicit expression of $\deg$ (as a polynomial in $4n^2$ variables), or perhaps an explicit set of equations of $SO(q,A)$ (e.g., Gröbner basis of $I_0$ for some term order)?  (At least for the particular quadratic form $q = \sum_{i=1}^n x_i y_i$ if not in general.)

REPLY [3 votes]: Hi Gro-Tsen ! The following is the outcome of a discussion with Olivier Taïbi.
Let V be the free module of rank $2n$ over $\mathbb{Z}$ with basis $f_1,\dots,f_n,g_1,\dots,g_n$ and  equipped with the split quadratic form $q=\sum_{i=1}^nx_iy_i$. Let $C(V,q)$ be the Clifford algebra of $(V,q)$ and $C^+(V,q)$ be its even part. The orthogonal group $O(V,q)$ acts on both. Moreover, $C^+(V,q)$ is the product of two matrix algebra, hence an element of the orthogonal group either preserves these factors or exchanges them. This gives a morphism $\deg:O(V,q)\to \mathbb{Z}/2\mathbb{Z}$ that is precisely the Dickson invariant we want to compute. [For all these facts, see for instance Conrad's notes http://math.stanford.edu/~conrad/252Page/handouts/O(q).pdf, Lemma 1.4 and around.]
Now, $C^+(V,q)$ has two non-trivial projectors (one on each factor), so that an element of the orthogonal group will have Dickson invariant $0$ (resp. $1$) if and only if it preserves these projectors (resp. it exchanges them). When $n=1$, it is not difficult to compute these projectors: one is given by $e=f_1g_1\in C^+(V,q)$ (and the other is $1-e$). It is then possible to compute inductively these projectors, using for instance the formula [SGA7 II Exp. XII (1.10.1)]. When $n=2$, one of the projectors is given by $e=f_1g_1+f_2g_2+2f_1f_2g_1g_2$ (and the other is $1-e$). By induction, it is possible to give a closed formula as follows. If $I=\{i_1<\dots <i_k\}$, let $f_I:=f_{i_1}\dots f_{i_k}$ and $g_I:=g_{i_1}\dots g_{i_k}$. Then one of the projectors is given by: $$e=-\sum_{k\geq 1}(-1)^{\frac{k(k+1)}{2}}2^{k-1}\sum_{I\subset\{1,\dots,n\},|I|=k}f_Ig_I.$$
To obtain a formula for the Dickson invariant of $u\in O(V,q)$, compute $u(e)$ in $C(V,q)$. By the above, it is equal to $e$ if $\deg(q)=0$ and $1-e$ if $\deg(q)=1$. Hence, it suffices to write down $u(e)$ in the basis $(f_Ig_J)$ of $C(V,q)$: the coefficient of $Id=f_{\varnothing}g_{\varnothing}$ will be $\deg(u)$. This gives an explicit polynomial expression for $\deg(u)$ in the matrix coefficients of $u$. 
In characteristic $2$, the fact that $e=\sum_{i=1}^nf_ig_i$ allows to recover the very simple expression of the Dickson invariant in characteristic $2$. When $n=2$, it is not difficult to use this to write down by hand an expression for the Dickson invariant over $\mathbb{Z}$. If $a,b,c,\dots, n,o,p$ are the matrix coefficients of $u$ from left to right then up to bottom (sorry for the poor notations), the Dickson invariant $\deg(u)$ is given by: $$ic+mg+jd+nh+2[(ib+mf)(kd+oh)+(id+mh)(jc+ng)-(jd+nh)(ic+mg)].$$
It is also possible to obtain a (probably not very useful) closed formula from the procedure above by understanding how the $f_Ig_I$ term in the formula above contributes to the coefficient of $Id$ in $u(e)$. To write it down, let me introduce the following notation. If $I\subset\{1,\dots, n\}$ has $k$ elements, $\tilde{I}=I\cup(n+I)$ and $P_I$ is the set of partitions of $\tilde{I}$ into $k$ pairs. To every such partition $P\in P_I$, it is possible to associate a natural sign $\varepsilon(P)$. Then:
$$\deg(u)=-\sum_{k\geq 1}(-1)^{\frac{k(k+1)}{2}}2^{k-1}\sum_{I\subset\{1,\dots,n\},|I|=k}\bigg(\sum_{P\in P_I}\varepsilon(P)\prod_{\{i<j\}\in P}\sum_{k=1}^n u_{n+k,i}u_{k,j}\bigg).$$<|endoftext|>
TITLE: Properness of quotient forcing
QUESTION [8 upvotes]: It is well known that if $P$ is a proper notion of forcing and $\Vdash_P \dot{Q} \text{ is proper}$ then the iteration $P \ast \dot{Q}$ is proper. Is the converse also true, i.e
Suppose that we have a two step iteration, such that $P$ and $P \ast \dot{Q}$ are proper forcings. Can we conclude that $\Vdash_P \dot{Q} \text{ is proper}$? 
My intuition would tell me no, we could probably add in the first step a new stationary set, which gets killed after the second forcing. We have to ensure however that while doing this the iterated forcing remains proper, i.e. preserves stationary subsets of $[\lambda]^\omega$ of $V$ (for $\lambda$ an arbitrary uncountable cardinal).
If the answer to my question is negative, could at least the following be true: Suppose that $P$ and $P\times Q$ are proper. Does it follow that $\Vdash_P Q \text{ is proper}$?

REPLY [7 votes]: As you expect, the answer to the main question is negative. 
Indeed, your remarks already very nearly provide a counterexample. Namely, consider the forcing $\mathbb{P}$ to add a Cohen subset of $\omega_1$ by initial segment. This forcing is countably closed and hence proper. Let $S$ be the generic set added by $\mathbb{P}$. It is easy to see that $S$ is both stationary and co-stationary. Let $\dot{\mathbb{Q}}$ be the forcing to shoot a club $C\subset S$. That is, conditions are closed bounded subsets of $S$, ordered by end-extension. This forcing is not proper in $V[S]$, because it destroys the stationarity of the complement of $S$.
Meanwhile, consider the combined forcing $\mathbb{P}\ast\dot{\mathbb{Q}}$. Every condition in this iteration has the form $(s,\tau)$, where $\tau$ is a $\mathbb{P}$-name for a closed bounded subset of $\dot S$. By extending this condition, we can decide a little more of $\tau$, and by iterating this $\omega$ many times, we find ourselves at a condition of the form $(s,c)$ where $s$ is a bounded subset of $\omega_1$, and $c$ is a closed subset of $s\cup\{\sup(s)\}$, containing this supremum. Thus, the collection of such conditions is dense in $\mathbb{P}\ast\dot{\mathbb{Q}}$. But that dense set of conditions is countably closed, and so $\mathbb{P}\ast\dot{\mathbb{Q}}$ is proper, as desired. In fact, it is forcing equivalent to $\text{Add}(\omega_1,1)$.
A similar argument arises in an answer to Justin Palumbo's question on Cantor-Bernstein for notions of forcing, as well as in the answer to Matteo Viale's question on complete embeddings of Boolean alebras and preservation of stationarity.<|endoftext|>
TITLE: basic questions on quantum integrable systems
QUESTION [9 upvotes]: I have been learning about (classical) integrable systems lately, e.g. in the examples of a Lax pair etc. I frequently run into the term 'quantum integrable system'. May I ask a few questions:

What are quantum integrable systems? Are there examples that are not too complicated?
Why are mathematicians interested in quantum integrable systems?

I would appreciate any suitable references or papers. Thanks!

REPLY [8 votes]: To answer the question What are quantum integrable systems ? let us find common understanding what is quantization. Roughly speaking it is the following - you have some classical phase space e.g. $\mathbb{R}^{2n}$ with coordinates $q_i$ and $p_i$. Now corresponding quantum algebra of observables is an algebra with generators $\hat p_i, \hat q_i$ such that commutators $[\hat p_i , \hat q_j]=\delta_{ij}$.
Now assume you have classical hamiltonian integrable system, so you have functions $H_i(p,q)$ such that they Poisson commute.
The first aim of quantization of  integrable system is to find elements $\hat H_i( \hat p , \hat q)$,
such that their commutator will be equal to zero and their principal symbols equals to classical $H_i$.
Answer 1a: 
So a quantum integrable system corresponding to classical hamiltonian system defined by $H_i$ is
given by $\hat H_i$, with properties above.   
In general you may not consider $\mathbb{R}^{2n}$, but some symplectic manifold, and deformation quantization of algebra of functions on it. Quantization of integrable system - is again looking for $\hat H_i$ in quantum algebra corresponding to $H_i$ in algebra of functions on symplectic manifold.
Remark: Well, may you consider Poisson manifolds as well, and classical integrable systems are maximal Poisson commutative subalgebras and we may look for they lift to quantum algebras. Actually many practical examples work with Poisson manifolds g^*, for Lie algebras $\frak g$.
To the best of my knowledge the question: `Is it always possible to do so or not?', is open in general. 
And in many case it is an art to find corresponding $H_i$. Some recent research on general question can be found here. Analysis of many concrete examples is related to representation theory, quantum groups etc...
Answer 1b: Are there examples that are not too complicated?
Trivial example - take $\mathbb R^{2n}$, take $H_i = p_i$, (free motion) corresponding quantum system $\hat H_i = \hat p_i$.
One more trivial example $H_l=p_l^2 + q_l^2$  (independent Harmonic oscillators), corresponding $\hat H_i = \hat p_i ^2 + \hat q_i^2$.
Let me give non-trivial examples later - you can search for Calogero system, Toda, Gaudin model.
Answer 2:
Why are mathematicians interested in quantum integrable systems? 
1) In many cases study of quantum integrable systems is the same as certain representation theory questions - e.g. corresponding $\hat H_i$ often have a sense of centres of universal enveloping or their quantum deformations MO question about the Yangian  
2) It is related to certain "hot topics" as Langlands correspondence (e.g. http://arxiv.org/abs/hep-th/0604128 ), quantum cohomology ( e.g. http://arxiv.org/abs/1211.1287 )
3) We might want to apply math. to something "physically" sounding :)<|endoftext|>
TITLE: Integration over the orthogonal group
QUESTION [11 upvotes]: Let $O(N)$ be the orthogonal group, and $a,b,c\in\mathbb N$. The question is:
$$\int_{O(N)}U_{11}^aU_{22}^bU_{33}^cdU=?$$
This is quite a tricky question:
(1) The first thought would go to probability, because with $N\to\infty$ the variables $U_{11},U_{22},U_{33}$ become Gaussian and independent; however, there doesn't seem to be any good analytic method for computing the correlations at $N$ fixed.
(2) The second thought would go to combinatorics, and to the Weingarten function. But that doesn't work either: we spent some time with Collins and Schlenker on this question, and just got a kind of very long (and especially unusable!) formula.
So, a new point of view on all this would be probably needed.

REPLY [5 votes]: This is a response to your question (1), for an analytic method to compute the integral over $O(N)$ as a power series in $1/N$. This method was developed by Prosen, Seligman and Weidenmüller in J. Math. Phys. 43, 5135-5144 (2002) [arXiv:math-ph/0203042]. Their key result can be written, in the context of your integral, as
$$\int_{O(N)}dU\;U_{11}^{a}U_{22}^{b}U_{33}^{c}=\int d\mu\; w_{\kappa}(M)M_{11}^{a}M_{22}^{b}M_{33}^{c}+{\cal O}(1/N^{z+1})$$
The order of the approximation is $z=\mbox{Int}[(\kappa+a+b+c)/2]$. The $N\times N$ real matrix $M$ has Gaussian measure $d\mu\propto\exp(-N\;\mbox{Tr}\;MM^{T})\prod_{ij}dM_{ij}$, so the integral on the right-hand-side is simply a Gaussian integral over the matrix elements $M_{ij}$ of $M$. The orthonormality constraints of the integral on the left-hand-side are accounted for by a weight function $w_{\kappa}$ of linearly independent invariants of $M$ up to order $2\kappa$. Explicit expressions for $w_{\kappa}$ for $\kappa=1,2,3,4$ are given in the cited reference.
To lowest order, one has $\kappa=1$, when $w_{1}\equiv 1$. This amounts to treating $U_{11}$, $U_{22}$, and $U_{33}$ as independent Gaussians, which is the lowest-order approximation you mentioned in your question. Increasing $\kappa$ gives you the higher order corrections in a systematic way.<|endoftext|>
TITLE: Visualising locally flat embeddings of surfaces in R^4
QUESTION [10 upvotes]: As far as I understand it follows from the work of M. Freedman that there exist  locally flat embeddings of two dimensional surfaces in $\mathbb R^4$ that can not be smoothed in the class of locally flat embedding. (For definition of locally flat see http://en.wikipedia.org/wiki/Local_flatness)
I am curious if one can draw a realistic picture of such a surface. At least is it possible to draw an intersection of such a surface with a (linear) hyperplane in $\mathbb R^4$?
Is the Hausdorff dimension of such an intersection equals $1$?

REPLY [3 votes]: I have been thinking about this as well. My approach would be to construct a broken surface diagram or chart of an immersed disk that a classical knot bounds. For simplicity take the untwisted double of the Figure-8 knot ($4_1$) or the Conway knot (11 crossings in the table). Take any immersed disk bounded by the knot with transverse double points. Now find a pair of arcs that the double points bound and look for an immersed disk bounded by these. In my opinion, you should be able to get one or two levels down before the figure becomes impossible to read. The tricks, though, are to use a Hoffman coding for the depths of the folds and double points and draw the lines with differing colors and line thicknesses. 
This project is on my (rather long) to-do list. If I make progress, I'll let you know.<|endoftext|>
TITLE: shallow question: Why a 300 digit number is associated with "any NP-hard problem"?
QUESTION [8 upvotes]: I was reading this article (http://www.ams.org/notices/200203/fea-knuth.pdf) the other day and noticed Donald Knuth said something nontrivial: Theoretically we can compute a very large number of magnititude $10^{300}$ (or possibly even larger) and use it via taking greatest common divisor to solve any NP hard problem. He used this example to suggest the patent system is essentially absurd. 
Here is the copy of the original paragraph:
"There’s an interesting issue, though. Could you
possibly have a patent on a positive integer? It is
not inconceivable that if we took a million of the
greatest supercomputers today and set them going,
they could compute a certain 300-digit constant
that would solve any NP-hard problem by taking
the GCD of this constant with an input number, or
by some other funny combination. This integer
would require massive amounts of computation
time to find, and if you knew that integer, then you
could do all kinds of useful things. Now, is that
integer really discovered by man? Or is it something
that is God given? When we start thinking of complexity issues, we have to change our viewpoint as
to what is in nature and what is invented."
My shallow question(must have already been asked by someone) is: Why finding such a large integer would suffice to solve any $NP$-hard problem? I asked a friend majoring in algorithm design in computer science, and he could not tell me the answer as well. While he explained to me that a $NP$ hard problem can be reduced to finding the GCD, neither he nor I can understand why finding one number is suffice for all applications. I am not sure if this is more appropriate for mathoverflow or stackoverflow.

REPLY [2 votes]: No matter how you formalize the question, if we believe that this world is not a computer simulation with a program of decent length with all randomness introduced by some simple standard pseudorandom generator but a truly random chain of events, then it looks like a rare moment of true unforgivable ignorance of Donald Knuth. He had no excuse for not knowing http://en.wikipedia.org/wiki/Kolmogorov_complexity. It is actually much easier for me to believe that gods had a direct communication with him and showed him a part of the program somehow than to believe that he didn't know what he was talking about to the extent described. Well, perhaps this was the case and we are just funny configurations in some game with simple rules (it was Knuth who designed "life", wasn't it?) but I prefer to act under the alternative assumptions, at least for the next few years :).<|endoftext|>
TITLE: Is Wikipedia correct about desarguesian projective planes being self-dual?
QUESTION [12 upvotes]: I stumbled over a statement on Wikipedia http://en.wikipedia.org/wiki/Duality_%28projective_geometry%29 and would like to ask how this could possibly be true. It states the following

The projective planes $PG(2,K)$ for any division ring $K$ are self-dual.

Is this really true? Or rather: What is wrong with the following argument that (if correct) would contradict the statement:
For a finite dimensional left vector space $V$ over a division ring $K$, we have the dual left vector space $V^{\star}$ over $K^{op}$ (the opposite division ring of $K$). Now, it is certainly true that $P(V^{\star})$ is isomorphic to the dual of $P(V)$. Thus, $P(V)$ is isomorphic to its dual if and only if $P(V) \cong P(V^{\star})$. Since $V$ and $V^{\star}$ have the same dimension, does that not necessarily imply that $K$ and $K^{op}$ are isomorphic by the Second Fundamental Theorem of projective Geometry? 
In fact, isn't $PG(2,K)$ (or $PG(n,K)$ for any $n \geq 2$) being self-dual simply eqvuialent to $K \cong K^{op}$ (which is of course is not always the case). I also recall that I have read this somewhere.
So, to put it in one line: What is wrong? The statement on Wikipedia or my reasoning?

EDIT: I found some course notes in which it is also claimed that $PG(n,K)$ is self-dual iff $K$ is self-dual (see 6.1 in http://www.maths.qmul.ac.uk/~pjc/pps/pps6.pdf). This does contradict what Wikipedia says, doesn't it?

REPLY [6 votes]: I think you are right. There is a theorem that asserts that if two projective spaces $P(V,F)$ and $P(V',F')$ are isomorphic then the skew fields $F$ and $F'$ are isomorphic. Thus if $P(V,F)$ and $P(V^*,F^{op})$ are isomorphic then $F\cong F^{op}$ as skew fields. 
I recommand these course notes "Essential Concepts of Projective Geometry" available here (the author gives a lot of references):
http://math.ucr.edu/~res/progeom/
in particular look at appendix C.<|endoftext|>
TITLE: fixed point arguments in PDE
QUESTION [6 upvotes]: I was curious whether anyone knows of some examples in PDE where 
a standard fixed point argument fails to show the existence of a solution but one can apply 
one of the more advanced fixed point arguments to obtain a solution.  
By standard fixed point i mean any of the fixed point theorems in a beginning PDE/functional analysis book  and by the more advanced I mean some of these pseudo contractions versions.  
thanks

REPLY [5 votes]: The graphical minimal surface equation is a great example of a PDE where Leray-Schauder fixed point theory is applied:
$$
\left(\delta_{ij} - \frac{D_i u D_j u}{1+|Du|^2}\right)D_{ij}u = 0
$$
This represents the condition that $graph(u)$ is a minimal surface, or in other words is a critical point for the area functional.
For surfaces in $\mathbb{R}^3$, existence for the a slightly more general problem (Plateau's problem) was established by Douglass-Rado in the 1930's, using beautiful conformal methods. However, the higher dimensional problem required the introduction of the Leray-Schauder fixed point theorem. One version of this says

For $\mathscr{B}$ a Banach space, and $T : \mathscr{B} \times [0,1] \to \mathscr{B}$ continuous, compact and so that $T(x,0) = 0$ for all $x \in \mathscr{B}$. Suppose also that there is some $M > 0$ so that if $x = T(x,\sigma)$ for $(x,\sigma) \in \mathscr{B} \times[0,1]$ then $$ \Vert x \Vert_\mathscr{B} \leq M.$$
  Then, there is a fixed point for $T(\cdot, 1)$. 

One of the reasons that I find this theorem very neat is because in some sense it treats a priori estimates as just that! In other words, you never need to show a solution to some relaxed equation holds or something like that, you just have to show that if there is a solution, then you have some estimates (and of course compactness). 
To use this, we define the operator $\hat T : C^{1,\beta}(\overline \Omega) \times [0,1] \to C^{2,\beta'}(\overline\Omega) $ as the solution operator to the linear PDE, solving for some $v$ satisfying
$$
\left(\delta_{ij} - \frac{D_i u D_j u}{1+|Du|^2}\right)D_{ij}v = 0 \text{ in $\Omega$}
$$
$$
v = \sigma \varphi \text{ on $\partial\Omega$}
$$
Linear existence theory shows that this map is well defined and if we then compose it with the map $C^{2,\beta'} \hookrightarrow C^{1,\beta}$, we have a map $T: C^{1,\beta} (\overline\Omega) \times [0,1] \to C^{1,\beta} (\overline\Omega)$, which is compact because the inclusion $C^2\to C^{1,\beta}$ is.
Furthermore, if $\sigma =0$, the $0$ solution clearly works.
Thus, to prove that Leray-Schauder applies, one must show that the a priori estimate holds. This is a bit delicate, so I won't go into the details, but only remark that one needs to assume some geometric conditions on the boundary (mean convexity). If you're interested, you can find the details in Gilbarg-Trudinger, starting with 11.3 but sort of jumping around for the various bounds.<|endoftext|>
TITLE: Connection between properties of dynamical and ergodic systems
QUESTION [17 upvotes]: While studying topological and ergodic dynamics, I've got quite perplexed by the different properties a system might have (minimality, regionally recurring, transitivity, mixing, ergodic, uniquely ergodic etc.). All of which basically has the sense instability but not always equivalent.
Do any of you has a graphic summary of those properties, and in what cases do they imply each other?

REPLY [34 votes]: Edit: I've updated this answer to reflect the helpful comments made by Andres Koropecki and Ian Morris.

As the other answers mentioned, the first crucial distinction you must make is that some properties refer to a topological dynamical system $(X,T)$, while others refer to a measure-preserving dynamical system $(X,T,\mu)$.  Thus there are two different sets of definitions.  Let me attempt a sketch at some of the relationships within each set.
First suppose you have a topological dynamical system $(X,T)$.  Then four of the key properties are topological transitivity, topological mixing, minimality, and unique ergodicity.  The first three are related by

topologically mixing $\Rightarrow$ topologically transitive;
minimal $\Rightarrow$ topologically transitive.

Unique ergodicity is independent of those three properties.  The picture is the following.

Counterexamples 1-9 illustrating the strict containments are as follows.  (These may not be the simplest or the earliest counterexamples in each case, and I welcome corrections or improvements.  This is based on some quick googling for things not already in my memory, plus the helpful additions offered by commenters.)
1. $X = \Sigma_2 \times \{a,b\}$, the direct product of a full two-shift with a period-two orbit, where the dynamics is $\sigma\times S$, with $\sigma$ the shift map and $S$ the map interchanging $a$ and $b$.
2. $X=\Sigma_2$.
3. Constructed by Bassam Fayad, Topologically mixing and minimal but not ergodic, analytic transformation on $\mathbb{T}^5$, 2000.
4. Constructed by Furstenberg, Strict ergodicity and transformation of the torus, 1961.
5. An irrational flow on the torus, slowed down near a single point: see the comment below by Andres Koropecki.
6. As Ian Morris points out in the comments, the identity map on a singleton set works here.  A less trivial example was given by Karl Petersen, A topologically strongly mixing symbolic minimal set, 1970.
7. Rotation of the circle by an irrational angle.
8. Direct product of the example from 5 with a periodic orbit.  (Again as suggested by Andres in the comments.)
9. North-south map:  a map $T\colon [0,1]\to [0,1]$ with fixed points at $0,1$ and such that $T(x) < x$ for all $x\in (0,1)$. Identify the endpoints $0$ and $1$ so that this is a uniquely ergodic circle map.
A couple things are probably worth pointing out.

Terminology is not always uniform.  For example one of the papers I referenced (I think Petersen's) uses "ergodic" in place of "topologically transitive", to highlight the analogy with the measure-preserving case.  So people may sometimes use different words for the same thing.
Conversely the same word may mean different things.  There are two definitions of topological transitivity, one involving open sets ($f^n(U) \cap V \neq \emptyset$ for some large $n$) and the other involving existence of a dense orbit.  The definition involving open sets more closely mirrors the definition of topological mixing (non-empty intersection for every large $n$), while the definition with a dense orbit more closely mirrors minimality (denseness of every orbit).  The definitions are equivalent if $X$ is separable, second category, and has no isolated points.

All of the above is for topological dynamical systems, where no invariant measure is specified.  Then there are the ergodic properties: those that depend on a system preserving an measure $\mu$.  For these one has the ergodic hierarchy.

It is very often the case that one wishes to study a topological dynamical system as a measure-preserving system by equipping it with an invariant measure, and in this case it is quite reasonable to ask about the relationships between the two different classes of properties.  But this depends on which invariant measure you choose, because in general there may be very many of them.  One may ask what properties of $(X,T)$ let you pick invariant measures $\mu$ with certain nice properties, and this is a whole different story which would expand this answer far beyond the bounds of propriety.<|endoftext|>
TITLE: Langlands paper on representations of abelian algebraic groups
QUESTION [5 upvotes]: I have been working through Langlands paper Representations of Abelian Algebraic Groups, and I can't understand why one of his maps is obvious and how it helps. First I'll give the notation
Take a algebraic torus over $F$ that splits over a Galois extension $K$ of $F$, then the torus corresponds to a lattice $L$ on which $G=Gal(K/F)$ acts. Then let $\widehat{L}=Hom(L,\mathbb{Z})$. Then we define $\widehat{T}=Hom(\widehat{L},\mathbb{C}^*)$, and we define the Weil group as the extension $$0 \longrightarrow C_K \longrightarrow W_{K/F} \longrightarrow Gal(K/F) \longrightarrow 0.$$ ($C_{K}$ is the idele class group) and finally let $N = \sum_{g \in G} g$
On page 13 of his paper hes shown that there is an isomorphism $$H^{1}(W_{K/F},\widehat{T}) \longrightarrow Hom(H_{1}(W_{K/F},\widehat{L}),\mathbb{C}^{*}).$$ And  now he wants to show that that image of a continuous cocycle will be continuous. To do this he proceeds as follows. He defines $U_{K}$ as the elements on norm 1 in $C_K$ and constructs the exact sequence $$1 \longrightarrow U_K \longrightarrow C_K \longrightarrow M_K \longrightarrow 1,$$ $M_K$ being $\mathbb{Z}$ or $\mathbb{R}$. He then uses $L$ to make the sequence $$0 \longrightarrow Hom(L,U_K) \overset{\lambda}\longrightarrow Hom(L,C_K) \overset{\mu}\longrightarrow Hom(L,M_K) \longrightarrow 0$$. And now this is where I get lost, he claims there is an obvious map from $$N(Hom(L,C_K)) \cap Hom(L,U_K)/N(Hom(L,U_K))$$ to $$\hat{H}^{-1}(G,Hom(L,M_K))/ \mu \hat{H}^{-1}(G,Hom(L,C_K)).$$ Here the $\hat{H}$ means Tate groups.
Is this map obvious? and why is the second group finite? which he claims. 
Thank you

REPLY [3 votes]: The map is induced by a map $N(Hom(L, C_{K})) \cap Hom(L, U_{K})$ to $\hat{H}^{-1}(G, Hom(L, M_{K}))$, which is obtained as follows. If $z$ is in $Hom(L, U_{K})$ and $z = Nx$, where $x \in Hom(L, C_{K})$, where should we send $z$?  Well, $N(\mu(x)) = 0$.  Therefore $\mu(x)$ maps to something in $\hat{H}^{-1}(G, Hom(L, M_{K}))$ by definition of the latter group.  And that element is where we want to send $z$. 
For the finiteness question, try writing down the long exact sequence in Tate cohomology for the short exact sequence $0 \rightarrow U_{K} \rightarrow C_{K} \rightarrow M_{K} \rightarrow 0$.<|endoftext|>
TITLE: Representability of sheaf of Ext^1 of a Néron model by $\mathbb{G}_m$
QUESTION [6 upvotes]: Let's work over a trait $S=\mathrm{Spec}R$, where $R$ is a dvr with fraction field $K$, residue field $k$. Given an abelian variety $A_K$ with semi-stable reduction, let $A$ over $S$ be its Néron model and $A^{\circ}$ the neutral component. We know the sheaf $\mathscr{E}xt^1(A^{\circ},\mathbb{G}_m)$ is represented by the Néron model of the dual of $A_K$ over the category of smooth scheme over $S$, see (Mazur and Messing's LNM Universal extensions and one dimensional crystalline cohomology, chapter I section 5)
My question is: Is the sheaf $\mathscr{E}xt^1(A^{\circ},\mathbb{G}_m)$ also representable over the category of schemes over $S$?
Also we know the Poincaré biextension $W_K$ of $A_K$ and $A_K^'$ by $\mathbb{G}_m$ extends to a biextension $W$ of certain open subgroups of $A$ and $A'$ (i.e. the subgroups making the component pairing vanish),
my second question is if $W$ is represented by a scheme over $S$?

REPLY [7 votes]: First question: no. Assume, to fix ideas, that $R$ is complete with uniformizer $\pi$, $k$ is algebraically closed, and $A$ is an elliptic curve with multiplicative reduction. Denote by $\mathscr{E}$ the Ext sheaf in question. Then the restriction of $A^\circ$ to $S_n:=\mathrm{Spec\,}(R/(\pi^{n+1}))$ is isomorphic to $\mathbb{G}_{m}$, so $\mathscr{E}(S_n)$ is zero for all $n$. If $\mathscr{E}$ were a scheme, this would imply $\mathscr{E}(R)=0$ (the functor of points of a scheme "commutes with completion" for local rings), a contradiction.
Second question: yes, because $W$ is a $\mathbb{G}_m$-torsor over the product, and torsors under affine group schemes are schemes.<|endoftext|>
TITLE: 13 months and not even one report. what would you do?
QUESTION [21 upvotes]: I submitted a 24 pages paper to a good journal - say usually in the top 10-20 - of pure maths, and after 14 months from the submission I haven't received any report. The last news I had from the editor date last may. Then I tried to contact him in september but no answer. What would you do? Wait? Write again? Withdraw the paper?
It is not a VERY TOP journal (top 5 for instance), hence I think a rejection would be pretty difficult to accept after such a long time.

REPLY [12 votes]: Hey one month ago I finally received a pretty positive report, asking for some small changes!
Persistence is the key ! :)<|endoftext|>
TITLE: Infinite products of forcings
QUESTION [5 upvotes]: Suppose $M \subseteq N$ are models of ZFC such that $(ORD^\omega)^M = (ORD^\omega)^N$. Let $\langle P_n : n \in \omega \rangle$ be a sequence of countably closed partial orders in $M$, and let $\langle G_n : n \in \omega \rangle$ be a sequence of filters in $N$ such that for each $n$, $G_n$ is $P_n$-generic over $M[G_0,...,G_{n-1}]$.  Is $\Pi_{n \in \omega} G_n$ generic for $\Pi_{n \in \omega} P_n$ over $M$?

REPLY [5 votes]: This is a very nice problem, but unfortunately the answer can be negative. 
Let me describe a counterexample. Consider the forcing to add $\omega$ many Cohen subsets of
$\omega_1$. So $P_n=\text{Add}(\omega_1,1)$ adds one Cohen subset
to $\omega_1$ and the (full support) product $\Pi_n P_n$ is
$\text{Add}(\omega_1,\omega)$. Suppose that $G\subset\Pi_n P_n$ is
$M$-generic for the product forcing, and consider the two models
$M\subset N=M[G]$. Since the forcing is countably closed, it adds
no new $\omega$-sequences over $M$.
We may think of $G$ as filling in a $\omega\times\omega_1$ matrix
with $0$s and $1$s. Generically, there will be many all-zero rows, that is, rows having
zeros all the way across, so that $G(n,\alpha)=0$ for all $n$, where this is the $\alpha^{th}$ row.
Let us define $G^\ast$ to be just like $G$ in every column, except
that in any such all-zero row in $G$, we change the first bit
to a $1$ in the first column in $G^\ast$, leaving the rest of the
row all $0$s. This operation ensures that $G^\ast$ has no all-zero
rows, and thus ensures that $G^\ast$ is definitely not $M$-generic
for the product forcing. But meanwhile, I claim that this
operation does not affect the $M$-genericity of any finite number
of the columns of $G^\ast$. For this, it is an elementary exercise to see that for any dense set $D$ for the forcing in the first $n$
columns, there is a dense set $E$ in the full product, such that
the operation applied to conditions in $E$ gives a condition in
$D$. So generically, $G$ is such that $G^\ast$ will have its first $n$ factors in $D$.
Thus, every finitely many factors of $G^\ast$ are $M$-generic, but
the whole product $G^\ast$ is not $M$-generic; so it is a
counterexample.<|endoftext|>
TITLE: Examples where adding complexity made a problem simpler
QUESTION [9 upvotes]: I can think of a few situations in math where a problem becomes easier or an object becomes simpler when some complexity is added. Examples:

$S^n$ is never contractible, but $S^{\infty}$ is.
The vanishing viscosity method of PDE's.
Higher-dimensional topology as opposed to low-dimensional topology (in some specific cases)
Singular homology as opposed to simplicial homology
Cube complexes as opposed to 3-manifolds

etc.
What other examples are there where a more complex object is simpler to analyze than a 'simpler' object? I realize that you could say that if it is easier to analyze, then it is less complex, so let me restate it this way:

What examples are there where one object seems much more complicated than another, but in fact has a simpler structure?

I've been thinking about things like the Ising model for magnetic phase changes and also about Navier-Stokes; perhaps the simplifications used to derive them make them harder to analyze in the end.

REPLY [3 votes]: "The shortest path between two truths on the real line passes through the complex plane." – Jacques Hadamard.<|endoftext|>
TITLE: Adjoint Functors as Initial Objects of Some Category
QUESTION [11 upvotes]: Just as universal arrows can be characterized as initial objects of some appropriate comma category, and (co)limits can be characterized as (initial) terminal objects of the appropriate (co)cone category, is there some (and if so, what is the) appropriate category for which, given a functor with a left/right adjoint, we can characterize said adjoint as its initial/terminal object? My goal in this is to characterize [initial objects, universals, limits, adjunctions] all as examples of each other.

REPLY [15 votes]: Well, yes: the left adjoint of a functor $G: C \to D$ is the initial object in the category whose objects are pairs $(H: D \to C, \eta: 1_D \to G H)$ where $\eta$ is a natural transformation, and whose morphisms $(H, \eta) \to (H', \eta')$ are natural transformations $\theta: H \to H'$ such that 
$$\begin{array}{ccc}  & 1_D &  \\\\ {}^{ \eta} \swarrow & & \searrow {}^{\eta'} \\\\  GH & \xrightarrow{G\theta} & GH'\end{array}$$
commutes. Similarly, a right adjoint can be expressed as a terminal object in a suitable category (exercise in applying the concept of duality). 
See the discussion on comma categories in Categories for the Working Mathematician,<|endoftext|>
TITLE: Result that follows from ZFC and not ZF but are strictly weaker than choice
QUESTION [7 upvotes]: A number of results that people use that require the axiom of choice (i.e. do not follow from ZF alone) are known to actually imply the axiom of choice. Therefore, one might naturally wonder whether there are results that require choice to prove yet which, on the contrary, do not imply choice.
The most natural answer to this question is countable choice, or $\kappa$-choice for cardinals $\kappa$.
Therefore, I wonder, are there other (i.e. not equivalent to any of the above) examples of results with this property? What about which are strictly stronger or weaker than all the cardinal-choice axioms (but not equivalent to ZF or ZFC)? In particular, I am interested in results occurring naturally in algebra, geometry, etc. And whether or not there are, is there a deeper reason why results that use choice tend to imply it?

REPLY [3 votes]: Here's one that I think a lot of people don't realize: the existence of free algebras for all (possibly infinitary) algebraic theories.  In his paper "Words, free algebras, and coequalizers" Andreas Blass proved that (assuming the consistency of a proper class of compact cardinals) it is impossible to prove this in ZF.  (I don't know for sure that it is strictly weaker than AC, but it seems to me extremely unlikely to be equivalent to AC.)  I think this is especially interesting because it doesn't look like any sort of choice principle itself.<|endoftext|>
TITLE: Consequences of Legendre's conjecture
QUESTION [10 upvotes]: I am looking for a list/reference which explores the consequences of Legendre's conjecture, which states that one can always find a prime number between $n^2$ and $(n+1)^2$.

REPLY [20 votes]: I am given credit here for a conjecture that I did not make (on the maximal gaps between consecutive primes).  This is also wrong in Wikipedia (can someone please correct that?). 
What I noted, on page 24 of my paper, "Harold Cramér and the distribution of prime numbers, Scandanavian Actuarial J., 1 (1995) 12- 28" is that if one includes in Cramér's model the fact that every pth integer is divisible by $p$, for small primes $p$, then
$\limsup_{n\to \infty}  (p_{n+1}-p_n)/ (\log p_n)^2  \geq  2e^{-\gamma} .$
Cramér's conjecture is that the limsup equals 1 (which is smaller than $2e^{-\gamma}$) and so is likely to be false (even if there is not enough computational evidence yet to say that). I have not made a conjecture as to the correct value of the limsup.<|endoftext|>
TITLE: Examples of injective morphisms which are not universally injective
QUESTION [5 upvotes]: I am interested in morphisms of algebraic varieties $X\to Y$ over a field $k$, and want to have examples of injective morphisms which are not universally injective. If $k$ is not algebraically closed, it is easy to construct plenty of examples, so I am only interested in the case where $k$ is algebraically closed.
The motivation of the question is that open immersion = étale + universally injective. This latter condition is equivalent to radicial, or to the fact that it is injective under a field extension $K/k$.
I would thus be even more interested in examples with étale morphisms.

REPLY [12 votes]: Every morphism of schemes locally of finite type over an algebraically closed field that is injective on $k$-points is universally injective.
Let $f: X \to Y$ be finite, étale and injective on $k$-points, where $Y$ is integral. For any $y \in Y$, the degree of $f^{-1}(y)$ over $k(y)$ is constant, and equals the degree of $f$. If $y$ is a closed point, then $k(y) = k$, and since $k$ is algebraically closed the number of points of $f^{-1}(y)$ equals $d$. This means that $d=1$, and $f$ is an isomorphism.
The general case reduces to this. Suppose that $p \in X$; we need to show that $k(p)$ is a purely inseparable extension of $k(f(p))$ By restriction to the closures of $p$ and $f(p)$, we may assume that $X$ and $Y$ are integral, $f$ is dominant, and $k(X)$ is purely inseparable on $k(Y)$. If the dimension of $X$ is larger then the dimension of $Y$, the fibers are positive-dimensional, hence have infinitely many $k$-points. So $\dim X = \dim Y$, and $k(X)$ is finite over $k(Y)$. By restriction $Y$ we may assume that $X$ and $Y$ are normal, and $f$ is finite.
Let $Z$ be the normalization of $Y$ in the separable closure of $K(Y)$; we can factor $f$ as $X \to Z \to Y$. We have that $X\to Y$ is finite and dominant, hence surjective, so $Z \to Y$ is injective in $k$-points. Since $Z$ is generically étale over $Y$, by restricting $Y$ we may assume that $Z$ is étale over $Y$. Hence it has degree 1, which shows that $k(Z) = k(Y)$, which is what we want.<|endoftext|>
TITLE: Is there any holomorphic version of the tubular neighborhood theorem?
QUESTION [30 upvotes]: This question arised when I was studying Beauville's book 'Complex Algebraic Surfaces'. 
Castelnuovo's theorem says that a smooth rational curve $E$ on an algebraic surface $S$ is an exceptional curve iff $E^2=-1$. The proof in Beauville's book is to find a very ample divisor $H$ satisfying $H^1(S,\mathcal{O}_S(H))=0$ first, and then set $H'=H+kE$ where $k=H\cdot E$. The linear system of $H'$ gives a projective morphism from $S$ to $\mathbb{P}^n$ which contracts $E$, and then some topological arguments implies that the image of $S$ is actually smooth.
Although this proof is not difficult to understand, I still want a proof based on complex manifolds but not algebraic geometry.
Question: Is there any holomorphic version of the tubular neighborhood theorem?
I have several reasons to raise this question:

If we have some holomorphic tubular neighborhood theorem, we can identify some neighborhood $U$ of $E$ in $S$ with neighborhood $V$ of the zero section in $N_E$. Here $N_E$ is the holomorphic normal bundle of $E$. Then $E^2=-1$ easily implies $N_E\cong\mathcal{O}_{E}(-1)$, so $E$ can be contracted in $U$ directly. Thus we not only prove Castelnuovo's theorem but also generalize it to non-algebraic surfaces.
There exists a symplectic version of the tubular neighborhood theorem, so I guess the holomorphic case is also true.

Any answers or comments are welcome. I'll really appreciate your help.

REPLY [6 votes]: Perhaps it is worth saying that while in general a complex submanifold of a complex manifold does not admit a holomorphic tubular neighborhood, it does admit a special smooth tubular neighborhood; the leaves that get contracted are complex manifolds, but they do not vary holomorphically.<|endoftext|>
TITLE: Which of these relations on partial orders allows us to identify forcing equivalence? 
QUESTION [7 upvotes]: Background
This question was inspired by Justin Palumbo's excellent question Cantor Bernstein for notions of forcing.
In his question, Justin considers a relation $\lhd$ on partial orders (defined below as $\lhd_1$) which begins to capture the notion of one partial order being "stronger" than another.  He asks a natural question, namely whether this relation satisfies antisymmetry -- that is, if $\mathbb{Q}\lhd\mathbb{P}$ and $\mathbb{P}\lhd\mathbb{Q}$, does it follow that $\mathbb{P} \sim \mathbb{Q}$?  Here, $\sim$ means forcing equivalent: two partial orders $\mathbb{P}$ and $\mathbb{Q}$ are forcing equivalent if they give rise to the same generic extensions, that is, every extension $V[G]$ by $\mathbb{P}$ can be realized as an extension $V[H]$ by $\mathbb{Q}$, and vice versa (in Joel Hamkins' terminology, we could say they are forcing equivalent if they give us access to the same neighborhood of the multiverse).  
Unfortunately the given relation does not satisfy antisymmetry.  In fact, the counterexamples given in the responses to the question demonstrate that $\lhd$ falls short of capturing the idea of one partial order being stronger than another.  I'm trying to find the "right version" of $\lhd$ -- a relation on partial orders $\mathbb{Q} \lhd \mathbb{P}$ that captures our intuitive notion of '$\mathbb{P}$ is stronger than $\mathbb{Q}$'. A good test criteria is whether the relation satisfies antisymmetry. I will consider a number of (I hope) natural strengthenings  of the $\lhd$ relation (or variations -- it's not clear that all of them are strengthenings).

Consider the following relations on partial orders:

$\mathbb{Q} \lhd_1 \mathbb{P}$: Every $\mathbb{P}$ extension contains a $\mathbb{Q}$ extension. $\mathbb{Q} \lhd_1 \mathbb{P}$ if and only if whenever $V[G]$ is a forcing extension by $\mathbb{P}$, there is an $H\in V[G]$ generic for $\mathbb{Q}$ with $V[H] \subseteq V[G]$.  (this is the original definition from Justin's post).
$\mathbb{Q} \lhd_2 \mathbb{P}$: Every $\mathbb{P}$ extension contains a $\mathbb{Q}$ extension meeting an arbitrary condition. $\mathbb{Q} \lhd_2 \mathbb{P}$ if and only if whenever $V[G]$ is an extension by $\mathbb{P}$, and $q \in \mathbb{Q}$,  there is $H$ in $V[G]$ generic for $\mathbb{Q}$ with $q \in H$.  
$\mathbb{Q} \lhd_3 \mathbb{P}$: Every $\mathbb{Q}$ extension is contained in a $\mathbb{P}$ extension. $\mathbb{Q} \lhd_3 \mathbb{P}$ if and only if whenever $V[H]$ is an extension by $\mathbb{Q}$, there is an extension $V[G]$ by $\mathbb{P}$ such that $V[H] \subseteq V[G]$.  
$\mathbb{Q} \lhd_4 \mathbb{P}$:  $\mathbb{Q}$ embeds into $\mathbb{P}$.  $\mathbb{Q} \lhd_4 \mathbb{P}$ if and only if there is a complete embedding of (the Boolean algebra of) $\mathbb{Q}$ into (the Boolean algebra of) $\mathbb{P}$.
$\mathbb{Q} \lhd_5 \mathbb{P}$:  Every $\mathbb{Q}$ extension is equal to a $\mathbb{P}$ extension.  $\mathbb{Q} \lhd_5 \mathbb{P}$ if and only if whenever $V[H]$ is an extension by  $\mathbb{Q}$, there is an extension $V[G]$ by $\mathbb{P}$ such that $V[H] = V[G]$.  

Please note that I have ignored serious metamathematical considerations in giving these definitions,  but I believe each of them has a first-order definition.

Question. Which of the relations above satisfy antisymmetry?  That is, for which of the relations above do we have $\mathbb{Q}\lhd\mathbb{P}$ and $\mathbb{P}\lhd\mathbb{Q}$ implies $\mathbb{P}$ is forcing equivalent to $\mathbb{Q}$?

From the responses to Justin's question it is clear that $\lhd_1$ does not satisfy antisymmetry.  On the other hand, $\lhd_5$ does satisfy antisymmetry, but is extremely restrictive in terms of the partial orders that it can compare (for example, $Add(\omega,1)$ and $Add(\omega,1)\times Add(\omega_1,1)$ are incomparable by $\lhd_5$, although the latter is apparently more powerful than the former).
At the risk of violating the suggested guidelines for keeping questions focussed and specific, I'll include two followup questions -- feel free to ignore.

Question 2. What are the relative strengths of these relations (which ones imply which others)?

For example, we (trivially) have $\mathbb{Q} \lhd_5 \mathbb{P}$ implies $\mathbb{Q} \lhd_3 \mathbb{P}$, and $\mathbb{Q} \lhd_2 \mathbb{P}$ implies $\mathbb{Q} \lhd_1 \mathbb{P}$. 

Question 3. What is the "right" notion of $\lhd$ for partial orders, that captures our intuitive idea of one partial order being "stronger" or "more effective" than another?  

This is "too vague", but it's the question that motivated those above.

REPLY [2 votes]: I believe I have an answer to my first question.
To recap, $\lhd_1$ is not antisymmetric (as demonstrated in the answers to Justin Palumbo's question).
As Francois G. Dorais points out in a comment, $\lhd_2$ is not reflexive (and therefore not antisymmetric).
It follows directly from the definition that $\lhd_5$ is antisymmetric (it was defined to accomplish exactly this).
This leaves $\lhd_3$ and $\lhd_4$.  I will show that neither of these is antisymmetric -- the same counterexample works for both.
Let $\mathbb{R}_n$ be the product that adds a Cohen generic to each of the first $n$ cardinals.  Now let $\mathbb{P}$ be the lottery sum of the posets $\{ \mathbb{R}_{2n} \mid n \in \omega \}$.  This partial order adds Cohen generics to the first $2n$ cardinals, where $n$ is chosen generically.  Similarly, let $\mathbb{Q}$ be the lottery sum of $\{ \mathbb{R}_{2n+1} \mid n \in \omega \}$, which adds Cohen generics to the first $2n+1$ cardinals.  Observe that $\mathbb{P}$ and $\mathbb{Q}$ are not forcing equivalent, since they do not share any of the same generic extensions (extensions by $\mathbb{P}$ and $\mathbb{Q}$ always differ by a Cohen generic on at least one cardinal).  
However, any generic extension $V[G]$ by $\mathbb{P}$ can be extended to a generic extension $V[H]$ by $\mathbb{Q}$ by simply adding a Cohen generic to the 'next' ($2n+1^{th}$) cardinal.  This shows that $\mathbb{P} \lhd_3 \mathbb{Q}$.  The same argument shows that $\mathbb{Q} \lhd_3 \mathbb{P}$. Thus $\lhd_3$ is not antisymmetric.
For $\lhd_4$, observe that each $\mathbb{R}_n$ embeds completely into $\mathbb{R}_{n+1}$, and we can thus construct a complete embedding of $\mathbb{P}$ into $\mathbb{Q}$ by working term-by-term with each member of the lottery sum.  The same argument allows us to construct a complete embedding of $\mathbb{Q}$ into $\mathbb{P}$.  Thus $\mathbb{P} \lhd_4 \mathbb{Q}$ and $\mathbb{Q} \lhd_4 \mathbb{P}$, and so $\lhd_4$ is not antisymmetric either.<|endoftext|>
TITLE: How does one make sense of the $\mathbf{C}_p$-points of a rigid analytic space over $\mathbf{Q}_p$?
QUESTION [14 upvotes]: I apologize in advance if this question is terribly naive. I've just recently learned a bit of rigid analytic geometry with the hopes of understanding some basic facts about eigenvarieties. 
In the literature (e.g. in Coleman-Mazur) I've seen people talk about the $\mathbf{C}_p$-points of a rigid analytic space over $\mathbf{Q}_p$ (here $\mathbf{C}_p$ is the completion of an algebraic closure of $\mathbf{Q}_p$). Based on the case of schemes, if $X$ and $T$ are rigid spaces over $\mathbf{Q}_p$, I assume the set of $T$-valued points of $X$ is the set $\mathrm{Hom}_{\mathbf{Q}_p}(T,X)$, where the morphisms are in the category of $G$-topologized spaces over $\mathrm{Sp}(\mathbf{Q}_p)$. But $\mathbf{C}_p$ is not a $\mathbf{Q}_p$-affinoid algebra, so, strictly speaking, it's associated $G$-topologized space $\mathrm{Sp}(\mathbf{C}_p)$ is not a rigid space over $\mathbf{Q}_p$, right? I guess there is a morphism of $G$-topologized spaces $\mathrm{Sp}(\mathbf{C}_p)\rightarrow\mathrm{Sp}(\mathbf{Q}_p)$, and so one could define the $\mathbf{C}_p$-points of $X$ in the same way as I did above, just morphisms of $G$-topologized spaces $\mathrm{Sp}(\mathbf{C}_p)\rightarrow X$ compatible with the morphisms to $\mathrm{Sp}(\mathbf{Q}_p)$. 
Question: Is this what is meant by the $\mathbf{C}_p$-points of a rigid analytic space over $\mathbf{Q}_p$?
The only alternative I can think of is to define the $\mathbf{C}_p$-points of a rigid space $X$ over $\mathbf{Q}_p$ as the $\mathbf{C}_p$-valued points of the base change $X_{\mathbf{C}_p}$. Although, if my intuition drawn from schemes is to be trusted, this is probably equivalent to the definition I suggested above.

REPLY [4 votes]: I think in Coleman-Mazur, it is just to be taken in the sense of $\mathbb C_p$-points of $X_{\mathbb C_p}$. This is compatible with the more abstract definition you propose. Indeed, as $\mathrm{Sp}(\mathbb C_p)$ is just one point, it is enough to work with an affinoide space $X$ associated to an affinoide algebra $A$ over $\mathbb Q_p$. Then it is clear that the continuous homomorphisms from $A$ to $\mathbb C_p$ coincide with continuous homomorphisms from $A\widehat{\otimes}_{\mathbb Q_p} \mathbb C_p$ to $\mathbb C_p$.<|endoftext|>
TITLE: Topology of the Universal Spinor Field Bundle
QUESTION [9 upvotes]: While reading article [1] below I came across the notion of a universal spinor bundle. This is defined at the beginning of section 6 (p.14) in [1] as follows: Let $M$ be a spin manifold and $\mathcal{G}$ be a set of smooth-semi Riemannian metrics that is open in the $\mathcal{C}^\infty$-topology (for simplicity let $\mathcal{G} = \mathcal{R}(M)$, the set of all Riemannian metrics). Define $E_{(g,x)} :=\Sigma^g_x M$ to be the spinor space at $x \in M$ with respect to $g \in \mathcal{R}(M)$. The author claims that
i) The map $\pi:E \to \mathcal{R}(M) \times M$, $\psi \in \Sigma^g_xM \mapsto (g,x)$ is a fibre bundle.
ii) The space $\mathcal{S}_g$ of sections (probably smooth sections?) of $\pi^{-1}(\{g\} \times M)$ is a Frechet manifold.
iii) These spaces assemble to a Frechet fibre bundle $\mathcal{S}:=\bigcup_{g \in \mathcal{R}(M)} \to \mathcal{R}(M)$.
Prior to these claims the author refers the reader to [2,p.153ff] for more details. But unfortunately, I can't find a proof of these claims in there. Probably [2] is supposed to be a general introduction to Lagrangian field theory, which is an important subject in the rest of the section. I am however interested in the Universal Spinor Bundle in its own right. Therefore this raises the following 
Question: How are (i)-(iii) proven? More explicitely I am asking 
1) How exactly are the spaces $E$, $\mathcal{S}_g$ and $\mathcal{S}$ topologized? Equivalently, how do local trivializations look like and why are their transition functions continuous resp. smooth?
2) Are $\mathcal{S}_g$ really the smooth sections and can this construction be generalized to $L^2$-sections?
3) Can someone give a reference for more details on the Universal Spinor bundle?
4) Why does one not consider $E$ as a bundle over $\mathcal{R}(M)$? Here the later I would give the $\mathcal{C}^1$-topology (this is commmon in spin geometry).
Possible Solution: I thought about it for a while and came across the idea that the identification of the spinor bundles with different metrics as discussed in [1, Section 5] or [2, Section 2] could be helpful to construct local trivializations. But I am not sure what formal argument to use in order to show that they depend continuously on the metrics or in what sense one should define continuity here. I am also unsure, if this way of thinking is not way too complicated.

[1] Bär, Gauduchon, Moroianu - Generlized Cylinders in Semi-Riemannian and Spin Geometry, http://arxiv.org/abs/math/0303095
[2] Deligne. Quantum fields and strings
[3] Maier: Generic Metrics and Connection on Spin- and Spin-c-manifolds

REPLY [2 votes]: If you are happy with Frechet bundles here is an alternative approach.  Let $\pi \colon {\mathcal G} \times M  \to M$ be the projection and consider $\pi^{-1}(TM) \to {\mathcal G} \times M$ a  real vector bundle of rank $n$. Following the notation in the paper let $P_{GL^+} \to M$ be the 
$GL^+(n, {\mathbb R})$ bundle of oriented frames of $TM$.  The bundle of oriented frames of $\pi^{-1}(TM)$
is $\pi^{-1}(P_{GL^+})$.  As in the paper pick a lift $P_{\widetilde{GL}^+} \to M$ of $P_{GL^+}$ to $\widetilde{GL}^+(n, {\mathbb R}) \to M$ of bundles over $M$. This exists because we assume $M$ is spin. Then  $\pi^{-1}(P_{\widetilde{GL}^+})$ is a lift of $\pi^{-1}(P_{GL^+})$ to $\widetilde{GL}^+(n, {\mathbb R})$.  
If $g \in {\mathcal G}$ and $m \in M$ then $\pi^{-1}(TM)_{(g, m)} = T_m M$ so has on it an inner product defined by $g(m)$. Denote this "universal" inner product on $\pi^{-1}(TM)$ by $g$.  It will be smooth for the usual reason with Frechet manifolds which is  because if $M$ and $N$ are finite-dimensional bundles then the evaluation map
$$
M \times C^\infty(M, N) \to N
$$
is a smooth map of Frechet manifolds [1]. Again following the approach in the paper we let 
$P_{SO} \subset \pi^{-1}(P_{GL^+}) $ be the subbundle of oriented orthonormal frames for the metric 
 $g$.  Taking the pre-image of this in $\pi^{-1}(P_{\widetilde{GL}^+})$ gives us a $Spin(r, s)$
bundle over $\mathcal{G} \times M$. The associated  vector bundle to this using the spin representation  gives us  $E$ as a smooth, finite rank, Frechet vector bundle. 
Finally you want a theorem that says that when you "push-down" $E$ with $\pi$ the result is a smooth Frechet vector bundle on ${\mathcal G}$.  This seems reasonably but I'm not sure where to find it.  I can't see it in [1].
Sorry this is a bit sketchy but that reflects the sketchiness of my knowledge of Frechet manifolds.  
[1] Richard Hamilton -- The Inverse Function Theorem of Nash Moser. http://dx.doi.org/10.1090%2FS0273-0979-1982-15004-2<|endoftext|>
TITLE: Idempotent elements in matrix ring
QUESTION [5 upvotes]: Is there a relation between the idempotent elements of a ring $R$ and those of $M_{n}(R)$ - the ring of $n \times n$ matrices over $R$?

REPLY [7 votes]: Any idempotent $e$ of $R$ induces an idempotent $E=\mathop{diag}(e,\ldots,e)$ of $M_n(R)$. In fact, if $e_i$ are idempotents in $R$, then $E=\mathop{diag}(e_1,\ldots,e_n)$ is an idempotent of $M_n(R)$.
Conversely, if $R$ is nice enough, an idempotent $E$ in $M_n(R)$ can be diagonalized to $E=U^{-1} \cdot \mathop{diag}(e_1,\ldots,e_n) \cdot U$ for some $U \in M_n(R)$ and idempotents $e_i$ in $R$.
Of course, this relies crucially on $R$ being nice enough. One sufficient "nicety" condition is that $R$ is an AW*-algebra; see for example this paper.<|endoftext|>
TITLE: Name for this generalized pigeonhole principle?
QUESTION [6 upvotes]: For a set $X$, let $|X|$ denote its cardinality.
A block of a partition is a non-empty element of the partition.

Let $P$ and $Q$ be two partitions of a set $X$.
  If $|P| < |Q|$ then $P$ contains a block $B$ which intersects two distinct blocks of $Q$.

What is this principle called?
The proof is immediate for finite $X$, and is an easy consequence of the Axiom of Choice for infinite $X$.  I am interested in using this result, and would like to refer to its proper name and to cite it correctly.  Jech's Set Theory and several online sources don't seem to mention this result.  In particular the Consequences of the Axiom of Choice project doesn't seem to list it (at least not in this specific form).
I would also be interested to know whether this principle is equivalent to some known consequence of the Axiom of Choice.

REPLY [8 votes]: This is equivalent to the Weak Partition Principle (a close relative of the Partition Principle mentioned by Goldstern). The Weak Partition Principle is form 100 in Consequences of the Axiom of Choice by Howard and Rubin (the Partition Principle is form 101).
The Weak Partition Principle asserts that if there is a surjection from $X$ onto $Y$ then it is not the case that $X$ has strictly smaller cardinality than $Y$ (i.e. $X \prec Y$). In contrapositive form, if $X \prec Y$ then there is no surjection from $X$ onto $Y$.
To see that the Weak Partition Principle implies your statement, note that if $P$ contains no  block which intersects two distinct blocks of $Q$ then each block of $P$ is contained in a unique block of $Q$ and the map $P \to Q$ thus defined is necessarily a surjection. By the Weak Partition Principle, this cannot hold at the same time as the hypothesis $P \prec Q$.
For the converse, suppose that the Weak Partition Principle fails as witnessed by sets $X \prec Y$ and a surjection $p:X \to Y$. Let $Q = \lbrace p^{-1}(y) : y \in Y \rbrace$ and $P = \lbrace \lbrace x \rbrace : x \in X\rbrace$. These are two partitions of $X$ with $P \prec Q$ and every block of $P$ is clearly contained in a unique block of $Q$.<|endoftext|>
TITLE: Derived category of varieties and derived category of quiver algebras
QUESTION [9 upvotes]: I have heard that derived category of coherent sheaves $\mathrm{Coh}(X)$ on any Fano varieties $X$ may be realized as derived category $\mathrm{Coh}(\mathrm{Rep}(Q,W))$ of representation of quiver $Q$ with relations $W$. I wonder for what kind of varieties their derived category of sheaves are realized by quiver. For example, is it possible for a quintic threefold $X$? 
Thank you very much.

REPLY [11 votes]: The question is the mother of all confusions. For the starters, a quintic threefold is not Fano but rather Calabi-Yau, and there you are doomed as Aaron has explained.
In general, a triangulated category may have three types of nice gadgets:

strong exceptional collection,
exceptional collection,
tilting object.

Number 1 gives both 2 and 3. 2 or 3 do not give each other or 1. Each of these gadgets gives you a realization of your category as

derived representations of a directed quiver (with relations),
derived representations of a directed DG-quiver (with relations),
derived representations of a quiver (with relations)

correspondingly. None of these gadgets exist on Calabi-Yau-s as explained by Aaron. All of these gadgets are expected to exist on Fano-s but the reality is much tougher. While we know these gadgets on Fano-s in many partial cases, we do not know them, in general. IMHO, it is a matter of time when a smooth compact Fano without none of them is discovered (or maybe even already known!).
Having said all this, you can reasonably expect that any smooth projective variety would have its bounded derived category of coherent sheaves realized as derived representations of a DG-quiver, but this gets technical and you should consult a real expert, not just a wabbitt:-))<|endoftext|>
TITLE: When does a homeomorphism split into essentially minimal homeomorphisms?
QUESTION [6 upvotes]: Background
Suppose $X$ is a compact metric space, and that $\varphi: X\to X$ is a homeomorphism of $X$.
We say a subset $A$ of $X$ is $\varphi$-invariant if $\varphi(A) = A$. A $\varphi$-invariant set is minimal if it is closed, $\varphi$-invariant, nonempty and the smallest of all such sets. We say $(X,\varphi)$ is essentially minimal if $X$ contains a unique minimal $\varphi$-invariant set.
An orbit of $x \in X$ is the set $O_\varphi(x)= \{ \varphi^n(x) \;|\; n\in\mathbb{Z} \}$.

My Question
Now suppose that $\varphi: X\to X$ is a homeomorphism such that $X$ contains exactly two minimal $\varphi$-invariant subsets $M_1,M_2\subset X$. Moreover, $\varphi$ has the property that for all $x\in X$ either 
$M_1\subset\overline{O_\varphi(x)}$ or $M_2\subset\overline{O_\varphi(x)}$ but not both.

Does it follow that $X=X_1\dot{\cup} X_2$ splits into two clopen $\varphi$-invariant subsets?

If not, I would be thankful for a counter example.

More generally, I would like to answer the following more general question. Suppose $\varphi: X\to X$ is a homeomorphism and let $\mathcal{M}$ be the set of all minimal $\varphi$-invariant sets in $X$. Suppose also that $\varphi$ satisfies the following property:
$$\forall\; x\in X: \exists !\; M\in\mathcal{M}: M\subset\overline{O_\varphi(x)}. $$

For $M_1,M_2\in\mathcal{M}$, can one find $\varphi$-invariant open subsets $U_1,U_2\subset X$ such that $M_i\subset U_i$ for $i=1,2$ and $U_1\cap U_2=\emptyset$ ?


My motivation is, among other things, to show that certain homeomorphisms with this property can be decomposed into essentially minimal systems, i.e. $X=\bigcup E_\alpha$ and the $E_\alpha$ are closed, $\varphi$-invariant, essentially minimal and pairwise disjoint. For the case I am interested in, a positive answer to the above (more general) question would be sufficient. But since I cannot even answer the more specific question, an answer to that would already be very helpful.

REPLY [3 votes]: The answer is no.
For each pair $(n,k) \in \mathbb{Z}^+ \times \mathbb{Z}$ we define a point $p(n,k) \in \mathbb{R^2}$ as: $$p(n,k)=\left(1+\frac{1}{n},\frac{k}{n^2} \right)$$ if $|k| \leq n$, and $$p(n,k)=\left( \frac{1}{k}\cos(1/n), \frac{1}{k} \sin(1/n) \right)$$ otherwise.
We let $$X= \left\{ p(n,k): (n,k) \in \mathbb{Z}^+ \times \mathbb{Z} \right\} \cup \left\{(0,0), (1,0)  \right\}$$ viewed as a (compact) subspace of $\mathbb{R}^2$, and define $\varphi:X \to X$ by $\varphi(0,0)=(0,0)$, $\varphi(1,0)=(1,0)$ and $\varphi(p(n,k))=p(n,k+1)$.
It is easy to check that all the requirements are met (note that the two minimal $\varphi$-invariant sets are $M_1=\{(0,0)\} $ and $M_2=\{(1,0)\}$). However, if $U$ is an open subset of $X$ containing $(0,0)$ then for each $n$ there is a $k$ such that $p(n,k) \in U$; so if $U$ is also $\varphi$-invariant then it must contain every $p(n,k)$; therefore, if $U$ is also closed, it must contain $(1,0)$. In other words, the only $\varphi$-invariant clopen is $X$.<|endoftext|>
TITLE: propagation of singularities & the Schrodinger equation
QUESTION [6 upvotes]: I've been thinking about the following propagation of singularities result:
Let $X$ be a compact manifold, and let $P$ be a differential operator (of, say, order $m$) on $X$ whose principal symbol $\sigma_m(P)$ is real-valued.  Suppose that $Pu=0$.  Then the wavefront set of the solution $u$ is a union of maximally extended (null) bicharacteristics of $\sigma_m(P)$ in the co-sphere bundle $S^*X$.
Let's consider the Schrodinger operator on $X\times\mathbb{R}$:
$P=-i\partial_t+\Delta_x$.
My question is what, if anything, does the above propagation theorem tell us about solutions $u$ to the homogeneous Schrodinger equation?

REPLY [7 votes]: There are propagation of singularities results for the Schr\"odinger operator, but they are usually stated on asymptotically Euclidean manifolds. Early results appear in a paper in CMP by Zelditch in around 82 or 83 exhibit the typical weird behaviour where singularities disappear then reappear at later discrete points of time. A more general `true' propagation theorem was proved by Craig, Kappeler and Strauss, but a much sharper set of results was eventually proved by Jared Wunsch, see his papers from the late '90's.  I am not sure what the best results are on compact manifolds at this point, though there are some. For the sphere one can see the same phenomenon of solutions being singular at time 0 and then becoming smooth then instantaneously singular again at time $2\pi$, $4\pi$, and so on.<|endoftext|>
TITLE: The boundedness of the rank of twists of a fixed curve
QUESTION [14 upvotes]: It is conjectured that there are do not exist elliptic curves over $\mathbb Q$ of arbitrarily high rank. I was wondering wether someone made a similar conjecture if one restricts to a fixed $j$-invariant. If there are specific values of $j$ known such that the following question has a negative answer would also be nice to know.
Let $j \in \mathbb Q$ be fixed. Do there exist elliptic curves $E/\mathbb Q$ with $j(E)=j$ of arbitrary high rank over $\mathbb Q$?
I'm specifically interested in the value $j=0$.
Edit
I recently found an article by Bjorn Poonen called "Heuristics for the arithmetic of elliptic curves". And in that article he proves (Thm 3.7) that if elliptic curves over $\mathbb Q$ follow his heuristics then 21 is the switching point between finitely and infinitely many elliptic curves of that rank and in particular there will be only finitely many elliptic curves of rank > 21. This of course implies the boundedness of the rank of all elliptic curves over $\mathbb Q$, so the same should hold if one restricts to just one $j$-invariant.

REPLY [6 votes]: This doesn't answer your question, but I'll mention it anyway.  My thesis research shows that the average rank of $j = 0$ elliptic curves, when ordered by discriminant, is bounded by 1.5.  This follows from computing the average size of the 2-Selmer group, which is 3.  
I'll also mention that your question is much easier to answer if you replace Mordell-Weil rank with 2-Selmer rank.  In that case, Heath-Brown has two papers (The Size of the Selmer Group for the Congruent Number problem I, II) showing that for any $r$, a positive density of the curves $y^2 = x^3-D^2x$ have 2-Selmer rank $r$.  Mazur and Rubin have a paper (Ranks of Twists of Elliptic Curves and Hilbert's Tenth Problem) where they consider quadratic twists of curves with no rational 2-torsion and give some conditions for the family of twists to have arbitrary 2-Selmer rank.<|endoftext|>
TITLE: Fibrations of $SU(4)$
QUESTION [10 upvotes]: I would like to apologize in advance if my question is too simple for mathematical community here: I am physicist by education.
It is well known that for a topological group $G$ acting transitively on a space $X$ and its subgroup $H \subset G$ one can construct a principal bundle whose fibers are homeomorphic to the coset space $G/H$. A typical example would be $SO(n-1)\rightarrow SO(n)\rightarrow S^{n-1}$.
$SU(2)\times SU(2)$ is a subgroup of $SU(4)$. Would it be possible to construct a corresponding fibration and what would be the base space? If the answer is 'yes' can we say something about the Betti numbers of the base based on this fact?
Thank you very much in advance...

REPLY [8 votes]: I'm way late to the party, but with a bit more work, one can understand $SU(4)/SU(2)\times SU(2)$ even more precisely.  Namely

$SU(4)/SU(2)\times SU(2)$ is diffeomorphic to $T^1 S^5$, the unit tangent bundle of $S^5$.

The topology of the unit tangent bundles of spheres is well known.  In particular, using the Gysin Sequence, one easily sees that $H^\ast(T^1 S^5) \cong H^\ast(S^4\times S^5)$ (with integral coefficients), which recovers Robert's answer.
The "more work" is just a bit of representation theory.  There is a double covering map $$f:SU(4)\rightarrow SU(4)/\pm I \cong SO(6)$$ which is found as a real subrepresentation of $\Lambda^2 \mathbb{C}^4$, (where $\mathbb{C}^4$ denotes the standard rep of $SU(4)$).  Since $-I\in SU(4)$ lies in the subgroup $SU(2)\times SU(2)$ and $SU(2)\times SU(2)/-I = SO(4)$, $f$ induces a diffeomorphism between $SU(4)/SU(2)\times SU(2)$ and $SO(6)/SO(4)$.  For the block embedding $SO(4)\subseteq SO(6)$, the homogeneous space $SO(6)/SO(4)$ is diffeomorphic to T^1 S^5$.
Note that one gets lucky here:  there is a unique (up to conjugacy) subgroup of $SO(6)$ isomorphic to $SO(4)$.
(While we're at it, a similar argument shows $SU(4)/S(U(2)\times U(2))$ is diffeomorphic to $SO(6)/SO(4)\times SO(2)$, the Grassmanian of oriented $2$ planes in $\mathbb{R}^6$.)<|endoftext|>
TITLE: Are homeomorphic open subsets of $\mathbb{R}^n$ also diffeomorphic?
QUESTION [23 upvotes]: Let $U_1, U_2$ be open subsets of $\mathbb{R}^n$.  Both are naturally differentiable submanifold, getting the differentiable structure from $\mathbb{R}^n$.  Further, both are natural topological manifolds, as submanifolds of $\mathbb{R}^n$.
Question:
If $U_1$ and $U_2$ are homeomorphic, are they also diffeomorphic?
Of course two general topological manifolds which are homoemorphic do not need to be diffeomorphic.  But here the differentiable structure is a very special one.
The answer might depend on the dimension $n$.  For $n=1,2,3$ it is yes, as there each topological manifold has a unique differentiable structure.  For $n \geq 5$ and $U_1$ an open ball the answer is yes by the uniqueness of differentiable structures on $\mathbb{R}^n$ for $n \geq 5$.
Some special cases are:

What happens if $U_1$ (and hence $U_2$) is contractible?
What happens if $U_1$ is a ball and $n=4$?  Is there an exotic $\mathbb{R}^4$ which can be realized as an open subset of the standard $\mathbb{R}^4$?

(The question came up because I encountered a sloppy definition of a manifold.  One can view the above manifolds as being defined by only one chart.  [That of course depends on your definition of chart, if you require it to start from a ball or not.]  So the questions basically asks: How do manifolds with only one chart look like?)

REPLY [8 votes]: Adding to Andy's answer: there are lots of contractible open subset of $\mathbb R^n$ that
are not homeomorphic to $\mathbb R^n$. For example, any compact contractible manifold of dimension $n>4$ embeds into $\mathbb R^n$: the double of any compact contractible manifold is simply-connected and hence a homotopy sphere, which after removing a point becomes $\mathbb R^n$. For constructions of compact contractible manifolds, see Kervaire's paper 
Smooth homology spheres and their fundamental groups.
You mention a confusion about a "sloppy definition of a manifold" and ask which manifolds
have one chart. By a chart you seem to mean any open subset of a manifold together with a homeomorphism onto an open subset of $\mathbb R^n$, which I think is a valid definition.
Any atlas of charts whose transition functions are smooth defines a smooth structure on
your manifold, which by definition is the set of all atlases compatible the given one. 
If there is only one chart, then the (only) transition function is the identity,
which is smooth. However, this merely implies that your manifold with one chart it diffeomorphic to an open subset of $\mathbb R^n$.<|endoftext|>
TITLE: Invariant free factor of a free group
QUESTION [10 upvotes]: Let $F_n=F\ast F'$ be a free splitting of the free group $F_n$ and $\phi\in Aut(F_n)$. The free factor $F$ is said to be invariant under $\phi$ if $\phi(F)\subseteq F$.
I recently wondered if this already implies that $\phi(F)=F$, and I found a positive answer:
An exercise in Magnus-Karrass-Solitar's book on combinatorial group theory shows that if $\phi(F)\subseteq F$ then $\phi(F)$ is a free factor of $F$, and since $\phi(F)$ and $F$ both have the same rank they have to be equal. The exercise in Magnus-Karrass-Solitar's book is proved using the existence of Schreier systems.
Can anyone think of an even easier argument why $$\phi(F)\subseteq F\quad\Rightarrow\quad\phi(F)=F$$ that does not involve Schreier systems?

REPLY [2 votes]: The case of finitely generated free groups
Let me reproduce the proof that Lee Mosher pointed out.
Claim:
Let $F=A\ast B$ be a free splitting of a finitely generated free group and $\phi\in Aut(F)$. If $\phi(A)\subseteq A$ then $\phi(A)=A$.
Proof:
If $A\ast B$ is a free splitting of $F$ then so is $\phi(A)\ast \phi(B)$. Therefore, $F=\phi(A)\ast \phi(B)$ is the fundamental group of a graph of groups $X$ that consists of a single edge with trivial edge group and vertex groups $\phi(A)$ and $\phi(B)$. Let $T$ be the Bass-Serre covering tree of $X$. By construction, the group $F$ acts on $T$ with trivial edge stabilizers and with vertex stabilizers conjugate to $\phi(A)$ and $\phi(B)$ respectively. In particular, there exists a vertex $v\in V(T)$ such that $F_v=\phi(A)$, where $F_v$ denotes the point stabilizer of $v$ under the action of $F$.
The subgroup $A$ of $F$ also acts on the tree $T$ and the edge stabilizers under this action are also trivial. Our assumption $\phi(A)\subseteq A$ implies that the vertex stabilizer $A_v=F_v\cap A=\phi(A)\cap A$ is in fact given by $\phi(A)$. Now choose a fundamental domain for the action of $A$ on $T$ accordingly such that the quotient graph of groups $Y$ with underlying topological graph $A\backslash T$ has $\phi(A)$ as one of its vertex groups. Bass-Serre theory tells us that $A$ is the fundamental group of $Y$, and since $Y$ has trivial edge groups it follows that $A$ contains $\phi(A)$ as a free factor.
We haven't yet made use of the fact that we are dealing with free groups ($A$ and $B$ are free, as all subgroups of free groups are free). Since $\phi$ restricted to $A$ is an isomorphism onto its image $\phi(A)$, the two free groups $A$ and $\phi(A)$ have the same finite rank. Therefore, $A$ cannot contain $\phi(A)$ as a proper free factor and hence $\phi(A)=A$.
Does this statement also hold for arbitrary finitely generated groups?
No, not in general.  
In 'A Finitely Related Group with An Isomorphic Proper Factor Group', J. London Math. Soc. (1951) s1-26(1): 59-61, Graham Higman gives an example of a finitely presented group $G$ that is isomorphic to a proper free factor of itself. In other words, there exists a free splitting $G=G_1\ast G_2$ and an isomorphism $$\phi\colon G_1\ast G_2\stackrel{\cong}{\longrightarrow} G_1.$$ The free product of $\phi$ with its inverse $\phi^{-1}$ is then an automorphism $$\Phi=\phi\ast\phi^{-1}\colon (G_1\ast G_2)\ast {G_1}'\stackrel{\cong}{\longrightarrow} G_1\ast ({G_1}'\ast G_2)$$ which has the property that $\Phi(G_1\ast G_2)=G_1\subseteq G_1\ast G_2$, but equality does not hold.<|endoftext|>
TITLE: Can you see smoothness of the boundary of a convex body from its shadow?
QUESTION [13 upvotes]: Let $d \geq 3$ and suppose that $K \subset \mathbb{R}^{d}$ is a convex body (compact, convex, non-empty interior). Is the following true?

The boundary $\partial K$ is a $C^1$-manifold if and only if for each projection $\pi:K\rightarrow H$ to a hyperplane $H$ has the property that $\partial \pi(K)$ is a $C^1$-manifold.

I suspect the answer is true. I am most interested in the case $d=3$.

If it is true, does it hold true for $C^1$ replaced with $C^k$, $k \in \mathbb{N} \cup \{\infty\}$?

I have no familiarity with this field so I have no idea about the difficulty of the question. I would appreciate any references.

REPLY [14 votes]: The statement for $C^1$ regularity is true, but with "dimension-2  projections" instead of "codimension-1 projections''. This is even stronger, if $d\ge 3$. On the other hand, for $d=2$ the statement with "hyperplane projections" fails, since $1$ dimensional projections are just closed intervals, whose boundary is certainly smooth. Also, an analogous statement holds with "plane sections" instead of "plane projections", for analogous reasons. Here's a sketch of the proof. 
A convex function is differentiable if and only if its subdifferential is a single point. By a compactness argument, this implies that a convex function everywhere differentiable on a convex domain of $\mathbb{R}^n$  is automatically $C^1$.
The boundary of a convex body is locally the graph of a convex function. The above property then implies: if the boundary of $K$ is not $C^1$, on some point $x$ of $\partial K$ it admits two supporting hyperplanes. If we project on the plane section orthogonal to the intersection of these two hyperplanes, we find a 2-dimensional projection of $K$ with  boundary that is not smooth at $\pi x$ (because it has two supporting lines at that point). And conversely.<|endoftext|>
TITLE: Additive energy of random sets
QUESTION [5 upvotes]: Given two random sets $A,B$ in a finite field (say $x\in A$ independently and with probability $1/2$), what is known about the additive energy $E(A,B)=|\{(a,a',b,b')\in A\times A\times B\times B: a+b=a'+b'\}|$?
Equivalently, what is the distribution of the random variable 
$\|1_A*1_B\|_2$? I'm mostly interested in asap (as sharp as possible) upper bounds for the probability of it being large.

REPLY [5 votes]: Expanding on my earlier comment, the concentration has to be really quite good: at least exponential in $N$, the order of the additive group.  (Edit: and you can't get better concentration because the state space is only exponentially large.)
Recall the following variant of the Hoeffding-Azuma martingale concentration inequality.  Let $\Omega = \Omega_1 \times \cdots \times \Omega_n$ be a product space, and let $X$ be a random variable on $\Omega$ such that $X(\omega)$ changes by at most $k$ if we change a single coordinate of $\omega$.  Then $$\mathbb{P}(X \geq (1+t)\mu) \leq \exp (-2t^2\mu^2/nk^2),$$ where $\mu=\mathbb{E}(X)$.  
Apply this with $n=2N$, $\Omega$ the space of coin flips determining membership of each element of the group in $A$ and $B$, and $X=E(A,B)$.  We have $\mu = N^3/16$ and $k = O(N^2)$ (as each element of the group is in $O(N^2)$ additive quadruples), which gives $$\mathbb{P}(X \geq (1+t)\mu) \leq \exp(-ct^2N),$$ for some absolute constant $c$.<|endoftext|>
TITLE: Loop space of a category
QUESTION [16 upvotes]: This seems like it should be a "standard" thing, and I think I remember even seeing it somewhere, but I can't remember where.
Let $C$ be a small category.  Is there a category $\Lambda C$ whose nerve (or classifying space) is a model for the free loop space of the nerve (or classifying space) of $C$?  Probably the objects of $\Lambda C$ should be something like zigzags of morphisms in $C$.  A reference would be the best thing to hear.

REPLY [3 votes]: I will try to answer the question. As I said in a comment, the Thomason model structure on $Cat$ is not simplicial model structure. Let $C$ be a small category, we will view it as a topological category. Denote by $C[C^{-1}]$ the topological category where we invert all maps of $C$ such that $C\rightarrow C[C^{-1}]$ is a cofibration of topological categories, then the coherent nerve $N_{\bullet}C\rightarrow N_{\bullet}C[C^{-1}]$ induces a weak equivalence of simplicial sets. Notice that $C[C^{-1}]$ is an infinity groupoid. 
Let $C$ be a cofibrant topological category. The mapping space $map(C,D)$ in the model category of topological categories   is given by the (standard) nerve of the following $HOM(C,D)$ category :
$\underline{Objects}$ are topological functor $F:C^{op}\times D:\rightarrow Top$ such that for any $c\in C,$  $F(c)$ is equivalent to a representable functor $D(d,-)$ for some $d\in D$. 
$\underline{Morphisms}$ in this category are natural transformation $H:F\rightarrow G$ such that $F(c,d)\rightarrow G(c,d)$ is a weak equivalence for all $c\in C$ and $d\in D$.
Let $S^{1}$ a simplicial model for a circle. Let $k: sSet\rightarrow sSet$ the cocontinues Joyal functor which take $\Delta^{n}$ to the nerve of the groupoid with $n+1$ objects and only one isomorphism between any two objects.
Recall that $\mathfrak{C}: sSet\rightarrow Cat_{\Delta}$ is the left quillen adjoint to the coherent $N_{\bullet}$ betwen the joyal model structre on $sSet$ and the Bergner model structure on $Cat_{\Delta}$ 
Now $k(S^{1})$ is a simplicial set, and the cofibrant topological category $|\mathfrak{C}[k(S^{1})]|$ is an infinity groupoid and its cohenrent nerve is equivalent to $S^{1}$.
The finial result is that $HOM(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])$ is a model for $\Lambda C$, since the nerve of $HOM(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])$ is equivalent to $ map(|\mathfrak{C}[k(S^{1})]|, C[C^{-1}])\sim Map(S^{1},N_{\bullet}C[C^{-1}])\sim Map(S^{1}, N_{\bullet} C)=\Lambda N_{\bullet}C$ 
N.B. The only point that I did not explained is the construction of $C[C^{-1}]$. 
Let 1 be the category with two objets a and b and a unique morphisms from $: f:a\rightarrow b$. Let $\widehat{1}=|\mathfrak{C}[k(\Delta^{1})]|$, then $ C[C^{-1}]$ is the pushout 
$colim (\sqcup_{mor C} \widehat{1}\leftarrow \sqcup_{mor C} 1\rightarrow C  )$ i.e., for each porphism of $C$ there is a map $1\rightarrow C$.<|endoftext|>
TITLE: Does Physics need non-analytic smooth functions?
QUESTION [119 upvotes]: Observing the behaviour of a few physicists "in nature", I had the impression that among the mathematical tools they use a lot (along with possibly much more sofisticated maths, of course), there is certainly Taylor expansion. They have a quantity (function) that they need to approximate: they expand it in Taylor series, keep the order of approximation that is useful for their purposes, and discard the irrelevant terms.
Appearently, there is little preoccupation for mathematically justifying this procedure, even if the to-be-approximated quantity is not given by an explicit form which is clearly known to be analytic. As Physics clearly gets no problems from the above mathematical subtleties, this may just mean that the distinction between analytic and smooth functions is somehow irrelevant to the basic equations of physics, or rather to the approximations of their solutions that are empirically testable.

If non-analytic smooth functions are irrelevant to Physics, why is it so?
Are there equations of physical importance in which non-analytic smooth solutions actually are important and cannot be safely considered "as if they were analytic" for the approximation purposes?

Remark: analogous questions may arise about Fourier series expansions.
One possible way the practice goes might be:

Consider a (differential or otherwise) equation $P(f)=0$ usually with analytic coefficients.
Expand the coefficients in Taylor series around a point in the scale of physical interest.
Discard higher order terms obtaining an approximated equation with polynomial coefficients $\tilde{P}(f)=0$.
Make the ansatz that the solutions $f$ of interest must be analytic.
Find the coefficients of $f$ by hand or by other means.

This leaves open the question why the ansatz is mathematically justified, if the equation of interest was $P$ not $\tilde{P}$. Do analytic solutions of $\tilde{P}$ aptly approximate solutions of $P$? Edit: I understand now that these last two lines are not very well formulated. Perhaps, ignoring the $\tilde{P}$ thing, I should have just asked something like:

Given any $\epsilon>0$, does knowing the analytic solutions (i.e. knowing their coefficients, possibly up to an arbitrarily large but finite number of digits) of $P$ give all the information about all solutions of $P$ up to $\epsilon$-approximation? Are there physically well known classes of equations $P$ in which this may not happen (perhaps even up to taking very regular approximations of the coefficients/parameters of $P$ itself)?

REPLY [5 votes]: A beautiful 19th century theorem of mechanics due to Joseph Bertrand says the following. Consider the motion of a particle which is driven by a central force potential
$V$, that is, the potential $V$ is a function from the distance of the particle
to some fixed origin and the force exerted on the particle is given by $\vec F=-\mathop{\rm grad}(V)$. If (almost) all trajectories are periodic, then either $V(r)\propto 1/r$ (celestial mechanics),
or $V(r)\propto r^2$ (harmonic oscillator).
At some point of the proof one knows that all periods of all trajectories are rational multiples of a common period, and one needs to conclude that there is a common period. That part of the proof is always incorrect in the litterature I know. (In particular, Wikipedia's argument is not complete.)
It is in fact easy
to construct smooth real families of periodic functions with non-constant rational period.
The only way I can correct the proof assumes that the potential is real analytic.<|endoftext|>
TITLE: cotangent to flags as a quiver variety
QUESTION [6 upvotes]: It is easy to realize cotangent space to the flag variety $Fl=SL_n/B$ as a Nakajima quiver variety: 
consider the finite quiver of type A, the dimension vectors v=(1,2,...,n-1), w=(0,...,0,n); 
an appropriate stability condition (polarization) amounts to the condition that the arrow
from the i-dimensional space to the (i+1)-dimensional one is injective, and we end up with a complete
flag in the n-space, the arrows in the opposite direction giving a cotangent vector.
Now, if I understand correctly, the other stability conditions (of which there is n!) should produce
quiver varieties which are also isomorphic to $T^*(Fl)$. How to see this, preferably using equally explicit linear algebra? Is it explained in the literature?

REPLY [8 votes]: Consider the stability parameter lives in the Cartan subalgebra. The corresponding variety is smooth if it lies in a chamber, i.e., outside of any root hyperplanes. It is clear that the variety remains the same if the parameter stay in the same chamber. If one cross the wall, the variety is changed.
On the other hand, one can change the stability condition by what I call reflection functors. They are quiver varieties analog of reflection functors for quiver reprsentations, but behave much nicer than the original ones.
They are compatible with the Weyl group action on the space of stability parameters, i.e., the Cartan subalgebra. The dimension vector $v$ is also changed compatible with the Weyl group action, if we think $w - Cv$ as a weight.
The reflection functors give an isomorphism between quiver varieties. Therefore, for a finite type quiver varieties, as in this question. One can take any stability parameter. It can be moved to the standard one by successive applications of reflection functors.<|endoftext|>
TITLE: restriction of the cotangent bundle of an elliptic surface
QUESTION [6 upvotes]: Given an nonisotrivial elliptic fibration $f:X\rightarrow P^1$, where $X$ is smooth and $P^1$ is a projective line. Could anybody provide some information on the restriction of the cotangent bundle $\Omega^1_{X}$ to the smooth fibers of $f$, e.g. does it split to line bundles, (semi)stable...?

REPLY [9 votes]: It's semi-stable, but not stable on every fiber. 
Assuming that the characteristic is zero this sheaf does not split. 
This may be true in positive characteristic, but as Damian Rössler points out the proof below requires characteristic zero.

Claim 1. 
  Let $V\subseteq \mathbb P^1$ be an arbitrary non-empty open subset and $U=f^{-1}V$. Then 
  $$0 \to f^*\Omega^1_{\mathbb P^1}|_U \to \Omega^1_X|_U \to \Omega^1_{X/\mathbb P^1}|_U \to 0\tag{$\star$}$$
  is not a split sequence.

Proof. We may obviously assume that $f$ is smooth over $U$. 
Since $f$ is a non-isotrivial fibration, the associated Kodaira-Spencer morphism of sheaves is non-zero, that is, pushing forward the dual of the original sequence,
$$0 \to T_{X/\mathbb P^1} \to T_X \to f^*T_{\mathbb P^1} \to 0,\tag{$\star^\vee$}$$ gives an injective map
$$
\kappa: T_{\mathbb P^1}\to R^1f_* T_X.
$$
Clearly, this is injective on any non-empty open set $U$ which implies that $(\star^\vee)|_U$ cannot be split and hence the Claim is proven. $\square$
Remark 1 The map $\kappa$ is the sheaf version of the classical Kodaira-Spencer map. That is obtained by tensoring with the residue field of a point on the target to get $T_{\mathbb P^1, t}\to H^1(X_t, T_{X_t})$ for $t\in \mathbb P^1$. That may be zero even for non-isotrivial families
Remark 2 The fact that this sequence does not split does not mean that the sheaf in the middle cannot be the direct sum of two line bundles. Contemplate the Euler sequence of $\mathbb P^1$:
$$
0 \to \omega_{\mathbb P^1} \to \mathscr O_{\mathbb P^1}(-1) \oplus \mathscr O_{\mathbb P^1}(-1)\to \mathscr O_{\mathbb P^1} \to 0.
$$
On the other hand, this proof gives a little more:

Corollary (of the proof).
  $\Omega_X^1|_U\not\simeq \mathscr O_U\oplus \mathscr O_U$.

Proof. Considering the long exact sequence that $\kappa$ is sitting in we get
$$
0\to f_*T_{X/\mathbb P^1}\to f_*T_X \to T_{\mathbb P^1} \to 
R^1f_*T_{X/\mathbb P^1} \to \dots 
$$
Using that the fibers are genus $1$ curves we see that $f_*T_{X/\mathbb P^1}$ is a line bundle. Since $\kappa$ is non-trivial, it has to be injective and hence $f_*T_{X/\mathbb P^1}\to f_*T_X$ is surjective. On the other hand, if $T_X|_U$ was trivial, then $f_*T_X$ would have rank $2$ which would lead to contradiction.
$\square$
So we need a little more:

Claim 2 
  Let $V\subseteq \mathbb P^1$ be an arbitrary non-empty open subset  and $U=f^{-1}V$. Then the sheaf 
  $\Omega^1_X|_U$ does not split as a non-trivial direct sum.

Remark As I said, the fact that $(\star)|_U$ is not split does not mean that $\Omega_X^1|_U$ is not either, but more particulars about that sequence actually imply that...
Proof. Suppose $\Omega_X^1|_U=\mathscr L\oplus \mathscr M$. If this is a non-trivial decomposition, then both $\mathscr L$ and $\mathscr M$ are line bundles. We may assume that $V\neq \mathbb P^1$ and hence $f^*\Omega^1_{\mathbb P^1}|_U\simeq \mathscr O_U$ and $\Omega^1_{X/\mathbb P^1}|_U\simeq \mathscr O_U$, so we have a short exact sequence
$$
0\to  \mathscr O_U \to \mathscr L\oplus \mathscr M\to \mathscr O_U \to 0.
$$
If either of the induced maps, say $\mathscr M\to \mathscr O_U$, is trivial, then the other one, $\mathscr L\to \mathscr O_U$, is surjective and hence an isomorphism and hence $\mathscr L\simeq\mathscr M\simeq  \mathscr O_U$. However, this is impossible by the Corollary above.
So we may assume that the induced maps $\mathscr L\to \mathscr O_U$ and $\mathscr M\to \mathscr O_U$ are non-trivial. Similarly the induced maps $\mathscr O_U\to \mathscr L$ and $\mathscr O_U\to \mathscr M$ are  also non-trivial. These together imply that both $\mathscr L$ and $\mathscr L^{-1}$ as well as both $\mathscr M$ and $\mathscr M^{-1}$ have non-zero global sections and therefore, again, $\mathscr L\simeq\mathscr M\simeq  \mathscr O_U$, which is again impossible by the Corollary above.
$\square$

[This part does not require a restriction on the characteristic.]
As far as (semi-)stability goes, since $U$ is not projective, it doesn't make much sense to ask on $U$. You could ask if it is relatively semi-stable, or semi-stable on the fibers. From the short exact sequence $(\star)$ you can see that its degree on any fiber is $0$ and an argument similar to the proof of Claim 2 says that any sub-line bundle would have to have degree at most $0$. (For any sub line bundle, if the induced map to $\mathscr O$ is non-trivial, then this is clear, if it is trivial, then it has to be contained in the kernel, which is also $\mathscr O$, so again clear.) So, this is actually semi-stable on every fiber. Since it contains a copy of $\mathscr O$, it is not stable. 
For other properties, you can probably use the above short exact sequences and arguments.<|endoftext|>
TITLE: $n$-path-connected components of a variety
QUESTION [6 upvotes]: This question is motivated by the question Path Connectedness of Varieties and some funny little theorem I was trying to prove. Let $X$ be a (quasiprojective smooth connected) algebraic variety over an algebraically closed field of an arbitrary characteristic. We know from the answer above that any two points of $X$ can be connected by a curve. Can we control its genus?
More precisely, we say that the two points $x$ and $y$ are $n$-path-connected, if there is a sequence of smooth curves from $x$ to $y$ such that every curve has genus less or equal to $n$. It is an equivalence relations and I am interested in its equivalence classes that are reasonably to be called $n$-path-connected components. Let me ask three precise questions.

Is it true that there exists $n$ such that $X$ is $n$-path-connected?
How do you find such minimal $n$?
Is there an algorithm/method for computing/describing $n$-path-connected components of $X$?

PS A curve of genus $g$ is an instructive example. It is $g$-path-connected but its $(g-1)$-path-connected components are points.

REPLY [2 votes]: Here is a variation on Tom's argument (for the 1st question) that works: I will also work over ${\mathbb C}$, although, I do not think it is important. I will be proving that $n$ can be bounded depending only on dimension of ambient space and degree $d$ of the subvariety $V\subset P^N$. The proof is by induction on dimension $N$ of the ambient projective space. 
Everything is clear, if $N=1$ or if $dim(V)=1$. Suppose that we have a bound $n(d,N-1)$;  consider a smooth subvariety $V\subset P^N$. Now, given any pair of points $p, q\in V$, I can find a projective hyperplane $H\subset P^N$ intersecting $V$ transversally (algebraic geometers would call it Bertini's transversality, topologists would call it Sard's theorem). By Lefschetz, if $dim(V)\ge 2$, then $W=V\cap H$ is connected. For two generic projective hyperplanes $H_p, H_q$ through $p, q$, intersections $W_p=H_p\cap W, W_q=H_q\cap W$ are nonempty, smooth and connected. Now, we are done by induction applied to $W, W_p, W_q$ (connect $p$ to $x\in W\cap W_p$, then connect $x$ to $y\in W\cap W_q$, etc.).<|endoftext|>
TITLE: Forcing with product vs. box product
QUESTION [6 upvotes]: If we want to add one real number the simplest way to do it is to use Cohen forcing. The poset is $\lbrace p\colon n\to 2\mid n\in\omega\rbrace$ which is a countable set. We can think of this as approximating a new set by finite sets.
If we want to add $\omega$-many pairwise generic real numbers we can do it by taking the poset $\lbrace p\colon n\times m\to 2\mid n,m\in\omega\rbrace$. This forcing approximates infinitely many new sets by finite parts of finitely many of the new sets. Interestingly enough both these sets are countable and therefore forcing with one is the same as forcing with the other. 
But we can also take $\lbrace p\colon\omega\times n\to 2\mid n\in\omega\rbrace$ and approximate all new sets at each stage. This poset is not countable anymore.
If we look at these things as topological spaces then the original Cohen forcing is really just a countable dense subset of the Cantor space, and we are adding a generic point to the Cantor set. The second forcing is the product of countably many Cantor sets, it is homeomorphic to the original Cantor set. 
The third notion of forcing, though, is the box product of countably many Cantor sets. So at least topologically it is different. I also have to admit that I never saw anyone using this approach.
Is there anything wrong it? Is it at all different from the first/second approach? What do we gain/lose when we use the box-product over the usual product?

REPLY [9 votes]: Your third notion of forcing, with conditions $p:\omega\times n\to 2$ ordered by extension as $n$ increases, is the same as the forcing to add a function $\omega$ to the reals of the ground model. This forcing is equivalent to the forcing $\text{Coll}(\omega,\mathbb{R})$, to collapse the ground model continuum to be countable. 
To see this, observe that each condition $p$ is essentially the same as a finite list of reals, corresponding to the slices of $p$. Since we may extend any such condition so as to mention any given ground model real on a slice, the generic function will be a surjection from $\omega$ to the ground model reals $\mathbb{R}$. So this forcing collapses $2^\omega$ to $\omega$, and since it has size continuum, it must be forcing equivalent to $\text{Coll}(\omega,\mathbb{R})$. 
In the spirit of your question, let me mention a fourth way of adding $\omega$ many Cohen reals, namely, the full support product $\Pi_n \text{Add}(\omega,1)$. Thus, conditions are simply countable sequences of finite binary sequences, ordered by extension in the natural way. This forcing is something like your third way, except that we do not insist on having uniformly constant $n$. That is, the conditions do not need to have the same size on each coordinate. Thus, we may think of a condition $p$ as a function from the region in the plane $\mathbb{N}\times\mathbb{N}$ to $2$, where the domain consists of pairs $(n,m)$ with $m\leq f(n)$ for some function $f:\mathbb{N}\to\mathbb{N}$. (Your third poset has only constant functions $f$.) Meanwhile, this version of the poset also collapses the ground model continuum to $\omega$. To see this, consider the following operation: consider the generic function, which fills out the entire matrix with $0$s and $1$s. Start at any height $k$ in the first column, and look at the number $n_0$ of $0$s directly above it (which is finite by genericity); then go to the $n_0$th digit of the second column, get that bit $a_0$, and let $n_1$ be the number of zeros following it; and so on. This defines a certain binary sequence $a_n$, which is determined by $k$. The point now is that any given binary sequence will arise as such a sequence for some $k$ via the generic filter, since for any condition $p$, I can extend it so as to ensure that my given binary sequence is coded. Thus, the forcing adds a surjection from $\omega$ to the ground model reals, and so the forcing is equivalent to $\text{Coll}(\omega,\mathbb{R})$.

REPLY [6 votes]: Somewhat surprisingly, your third forcing collapses the cardinal of the continuum to $\aleph_0$.  That is, it adjoins a surjection from $\omega$ onto the ground model's set of reals.<|endoftext|>
TITLE: Can we promote to a Lie Group Isomorphism?
QUESTION [10 upvotes]: We regard an isomorphism of Lie groups to mean a group isomorphism which is simultaneously a diffeomorphism of the underlying smooth manifold. I'm wondering about how much rigidity is imposed by this definition.
Question: If we have maps $f, g: G\rightarrow H$, where $G, H$ are Lie groups, $f$ is an abstract group isomorphism, and $g$ is a diffeomorphism, must $G$ and $H$ be isomorphic as Lie groups? 
I think there should be a (possibly easy) counterexample, but neither I nor the professors I've asked could immediately find one. 
EDIT: assume $G$ and $H$ are connected.

REPLY [3 votes]: All you need is love:-)) Not really, just take a family of complex nilpotent Lie algebras $L_a$ depending on a complex parameter $a$. Such a family exists in dimension 7, or maybe even 6 (I do not remember the nilpotent classification).
Now consider a wild automorphism of complex numbers $\phi$. Then $L_a$ and $L_{\phi (a)}$ for a generic $a$ are isomorphic as Lie algebras over ${\mathbb Q}$ but not over ${\mathbb R}$ or ${\mathbb C}$. Love is all you need((-: the corresponding simply connected Lie groups are isomorphic as abstract groups but not as Lie groups.<|endoftext|>
TITLE: Geometric intuition behind perverse coherent sheaves? 
QUESTION [6 upvotes]: I would like to know an intuition behind perverse coherent sheaves. I am aware that it is induced by a heart of another t-structure on the derived category. Are there any better, probably more geometric, way to understand perverse coherent sheaves? 
Just in case, let us recall the definition of perverse coherent sheaves. Let $X$ be a projective threefold with at worst Gorenstein terminal singularities and $f:Y\rightarrow X$ be a crepant resolution. Define a full subcategory $\mathrm{Per}(Y/X) \subset \mathrm{D}(Y)$ consisting of objects $E \in \mathrm{D}(Y)$ satisfying the following three conditions;

$H^i(E)=0$ unless $i=0,-1$,
$R^1f_*H^0(E)=0$ and $R^0f_∗H^{−1}(E)=0$,
$Hom_Y(H^0(E),C)=0$ for any sheaf $C$ on $Y$ satisfying $Rf_∗(C)=0$.

We call the objects of $\mathrm{Per}(Y/X)$ perverse coherent sheaves.

REPLY [2 votes]: This is the definition appear in Bridgeland's paper which shows that flops of smooth 3-folds induces equivalence of derived category of coherent sheaves. From your question I think you know the word "perverse" is kind of related to t-structures. 
The main theorem of that paper, indicates that for a flop $Y\to X\leftarrow W$, $Per(Y/X)$ will be send to $Coh(W)$ under that isomorphism. In other words, these objects are sheaves on another scheme which you can construct from the data $Y\to X$! In my opinion that's pretty cool, not "perverse" at all. But as for the name, so be it.
BTW, you don't need $X$ and $Y$ to be three fold in the definition. If you check Bridgeland's paper, most of the time he work with birational morphism such that $Rf_*\mathcal{O}_Y=\mathcal{O}_X$ and fibers have dimension at most 1. (For 3-folds that's just a small resolution.)<|endoftext|>
TITLE: Does every group embed into a co-hopfian group?
QUESTION [10 upvotes]: A group $G$ is co-hopfian if every injection $f\colon G \rightarrow G$ is an automorphism, or equivalently if $G$ is not isomorphic to any of its proper subgroups.  Miller and Schupp, using small cancellation theory, showed that every countable group that does not contain elements of every finite order can be embedded in a 2-generated co-hopfian group.  I can't find a more general result.
Is it known if every countable group embeds into a finitely generated co-hopfian group?  At least, does every finitely generated group embed into a finitely generated co-hopfian group?  My instinct is to use small cancellation theory, but if a group contains elements of every finite order it doesn't seem to give enough control over the embeddings $f\colon G\rightarrow G$, as this is usually accomplished using torsion elements.

REPLY [8 votes]: Like with many problems solvable by small cancellation methods, the answer can be found by looking at A.Yu. Ol'shanskii's papers. 
Indeed, take any countable group $H$. Without loss of generality we can assume that $H$ is not virtually cyclic and has some non-trivial element of finite order. Now, let $K$ be a finitely generated torsion-free group that is not embeddable in $H$ (such $K$ exists as $H$ has only countably many f.g. subgroups, but there are uncountably many f.g. pairwise non-isomorphic torsion-free groups). 
In Theorem 2 of the paper [``Efficient embeddings of countable groups.'', Vestnik
Mosk. Univ., Ser. Matem. (1989), N 2, 28-34 (in Russian); English translation in Moscow Univ. Math. Bull. 44 (1989), no. 2, 39-49] Ol'shanskii proves a powerful embedding theorem which implies that there is a $2$-generated simple group $G$, containing both $H$ and $K$, in which every proper subgroup is either infinite cyclic or infinite dihedral or is conjugate inside $H$ or $K$ in $G$. 
Thus $H$ embeds in $G$, and it's not hard to see that $G$ is co-hopfian. Indeed, let $A < G$ be a proper subgroup such that $G \cong A$. Since $G$ is not virtually cyclic, $A$ must be contained in a conjugate of $H$ or $K$ in $G$. Since $K$ does not embed in $H$, $A$ must be conjugate inside of $K$, but $K$ is torsion-free and $H$ has torsion, giving a contradiction.
Of course this construction is flexible. For example, if $H$ is torsion-free one can take $G$ to be torsion-free, by using some other invariants instead of torsion (to ensure that $H$ is not embeddable in $K$ and $K$ is not embeddable in $H$).<|endoftext|>
TITLE: Solve equation with matrix variable
QUESTION [5 upvotes]: I want to solve a matrix $\Omega$ from a equation $\sum_k (\Omega + \Theta_k)^{-1} = Q$. The $Q$ and $\Theta, \forall k=1...K$ are known, and are positive definite matrices. $\Omega$ also has to be positive definite. all matrices are large (a few thousands of columns and rows). My questions are:
(1) Is there a closed-form solution? How do I simplify the sum of the inverse of two matrix sum?
(2) I'm OK to go for a numerical solution. But how do I define this problem? An optimization problem to minimize something like $f(\Omega) = ||\sum_k (\Omega + \Theta_k)^{-1} - Q||$? Do I need to minimize the frobenius norm, (just like minimizing the L-2 norm in a least square problem)? Considering the constraint that $\Omega$ is positive definite, can I solve it by semi-definite programming? How do I redefine the problem in a linear/semi-definite programming? I don't have much knowledge of linear programming. I would prefer a general gradient descent rather than LP. But I'm OK to use LP if I know how to do.
This problem comes from the estimation of inverse covariance matrix of multi-variate Gaussian distribution. 
EDIT: Both $\Theta_k$ and $\Omega$ are sparse, if that helps.

REPLY [5 votes]: Here is a partial solution to the first question in the original post. Let's look at the equation
\begin{equation}\tag{1}
  \sum\nolimits_{i=1}^m (X+ \Theta_i)^{-1} = Q.
\end{equation}
Lemma (Existence).    If all $\Theta_i$ are (strictly) positive definite, then (1) has a positive semidefinite solution only if $Q \preceq \sum_i \Theta_i^{-1}$.
Proof. Suppose $Q=\sum_i \Theta_i^{-1}$, then clearly $X=0$ is the solution. Since, $(X+\Theta_i)^{-1} \preceq \Theta_i^{-1}$ for any $X \succeq 0$, on summing up we see that $Q \preceq  \sum_i \Theta_i^{-1}$ must hold. Moreover, in this case if there is a solution, then it must be strictly positive definite. A little extra argument shows that in this case, there must exist a unique positive definite solution.
This lemma shows that in case $Q$ does not satisfy the requirement, the original equation has no solution, and it might be preferable to minimize $\|\sum_i (X+\Theta_i)^{-1}-Q\|_F^2$ instead.
Lemma (Bounds). Any feasible solution to (1) must lie in the set $\Omega := [0, mQ^{-1}]$.
Proof. The lower bound $X \succeq 0$ is obvious. Following an argument similar to the previous lemma, we see that $Q=\sum_i (X+\Theta_i)^{-1} \preceq \sum_i X^{-1}$, which implies that $m X^{-1} \succeq Q$, or equivalently, $X \preceq m Q^{-1}$.
Idea Now that we have a compact set $\Omega$, we just need to setup a strictly contractive nonlinear map $G : \Omega \to \Omega$. I have not proved strict contraction of the map below, but numerically it seems to work. As one might suspect from the above lemmas, the rate of convergence depends on $\|Q-\sum_i \Theta_i^{-1}\|$, so that for small values of this quantity, the iteration converges more slowly.
Suppose, that $X \succ 0$. Denote by $S^{++}$ the set of $n\times n$ strictly positive definite matrices. Then, define the nonlinear map $\mathcal{G} : S^{++} \to S^{++}$  as
\begin{equation*}
 \mathcal{G} = X \mapsto X^{1/2}\left(\sum\nolimits_{i=1}^m Q^{-1/2}(X+\Theta_i)^{-1}Q^{-1/2}\right)X^{1/2}.
\end{equation*}
TODO If I get time, I might think about proving that the above map generates convergent solutions. Or one can come up with some other fixed point iteration.<|endoftext|>
TITLE: Well-ordering with a topological property
QUESTION [5 upvotes]: Assuming the axiom of choice, is there a well-ordering of the reals such that every initial segment is closed for the usual topology? If the continuum hypothesis helps, we can also assume it.
An initial segment is a set of the form $\{x : x<y\}$ for some $y$, according to the well-ordering $<$.

REPLY [11 votes]: No.  There cannot be a strictly increasing $\omega_1$-sequence of closed sets in a topological space with a countable base.  Their complements would be unions of open sets from the basis, and it is impossible to drop elements of a countable set uncountably many times.

REPLY [9 votes]: The answer is no. Assume there is such a well-order; then some downward-closed subset $D$ (which might be an initial segment, or might be all of $\mathbb{R}$) will be of order type $2^{\aleph_0}$, i.e., the least ordinal of continuum size. Being a subset of $\mathbb{R}$, $D$ contains a countable subset that is dense in $D$ (see for example here if you are in need of proof), say $x_1 \lt x_2 \lt \ldots$ according to how the elements appear in the well-order. Since there is no countable cofinal sequence in the order type $2^{\aleph_0}$, it must be that some proper initial segment contains all the $x_i$. This initial segment is both closed and dense, hence is all of $D$. Contradiction. 
(Edited after Ramiro pointed out a glitch. Hopefully okay now.)

REPLY [6 votes]: No.
If there was such a well order, let $r$ be the first real for which the initial segment $I$ bounded by $r$ is uncountable. Since $I$ is separable (being a subspace of the reals) we can fix a countable $D \subseteq I$ dense in $I$. But $I$ has order type $\omega_1$ which is a regular cardinal (usinge some choice), so there would be an $s \in I$ that bounds $D$. This is already a contradiction because then the clousure of $D$ would be contained in the initial segment determined by $s$, which is a proper subset of $I$.<|endoftext|>
TITLE: Existence of an $R$-basis with at least one unit in it?
QUESTION [6 upvotes]: Let $F$ be a domain and let $R\le F$ be a subring such that $F$ is a free $R$-module of finite rank $n$.  
Question: Is there an $R$-basis $\lbrace e_1,...,e_n\rbrace$ of $F$ such that at least one of the basis elements is a unit in $F$ ? 
As an example consider $R = \mathbb{Z}$ and $F=\mathbb{Z}[\text{i}]$ where we can take the units $1,\text{i}$ as basis.

REPLY [2 votes]: First some general remarks. You're asking whether $S/R$ is free as a $R$-module. If $S$ is a $R$-algebra which is free of finite rank, then the map $R \to S$ splits as a map of $R$-modules. This fact was already noticed by Florian Eisele in his answer to the other MO question.
Now for an explicit counter-example to your question. Consider the ring $R={\bf Z}[x,y,z]/(x^2+y^2+z^2-1)$. It is an integral domain. It is known that there exists a $R$-module $M$ which is not free such that $R \oplus M \cong R^3$. For a nice construction, see e.g. Keith Conrad's notes. Explicitly we can take $M=\{(f,g,h) \in R^3 : xf+yg+zh=0\}$. Note that we can embed $M$ in $R^2$ by $(f,g,h) \mapsto (f,g)$, and the cokernel $R^2/M$ is a torsion module, so there exists $F \in R \backslash \{0\}$ such that $F \cdot R^2 \subset M$.
Now, we would like to construct a $R$-algebra structure on $R \oplus M$. We can do this by considering the $R$-algbera $S_0 = R \otimes_{\mathbf{Z}} \mathcal{O}$ where $\mathcal{O}$ is an order of a cubic field $K$. It is an integral domain, since the polynomial $x^2+y^2+z^2-1$ is irreducible over any field of characteristic not $2$. Let $(1,\alpha,\beta)$ be a $\mathbf{Z}$-basis of $\mathcal{O}$. Embed $R \oplus M$ in $S_0$ by $(f,(g,h)) \mapsto f+g\alpha+h\beta$. This won't be a subring of $S_0$ in general, but $S=R \oplus FM$ is a subring of $S_0$ since $(FM) \cdot (FM) \subset F^2 S_0 \subset R \oplus FM$. So we have constructed an integral domain $S$ over $R$ such that $S/R \cong M$ is not free over $R$.
I don't know whether it's possible to find a counterexample where $R \to S$ splits as a map of rings, in other words where $S=R \oplus I$ where $I$ is an ideal of $S$.<|endoftext|>
TITLE: Is it possible to approach higher categories from the point of view of the arrow functor?
QUESTION [7 upvotes]: I have not read anything thoroughly about higher categories. I am only aware that in higher categories, we have higher dimensional cells, after adjusting for the intuition that 0-, 1-, 2-dimensional cells are respectively, objects, morphisms, and commuting squares. This terminology is similar to what we from arrow construction, as it follows.

Given a category $\mathcal{C}$, the objects of an arrow category $\mathcal{C^{\rightarrow}}$ are morphisms of $\mathcal{C}$ and morphisms of $\mathcal{C}$ are commuting squares. Another way to think of an arrow category is by considering  $\mathcal{C^{\rightarrow}}$ as  $\mathcal{C}^{2}$, the category of functors from $2$ to $\mathcal{C}$, where $2$ is the category $0 \longrightarrow1$ and a morphisms of $F: 2 \longrightarrow \mathcal{C}$ and $G: 2 \longrightarrow \mathcal{C}$ is a natural transformations of $F$ and $G$; this is actually a functor $\phi:2 \times 2 \longrightarrow \mathcal{C}$. 
  Moreover, this construction is functorial, i.e., it yields an endofunctor $A$r: Cat $\rightarrow$  Cat, which I call the arrow functor.

Now, for a given category $\mathcal{C}$ (just for the purpose of this question) let $\mathcal{C} ^ {\stackrel{\rightarrow} {n}}$ denote the category which is obtained by an $n$-times iteraion of Ar, with the convention that $\mathcal{C}^{\stackrel{\rightarrow} {1}}$ is the usual arrow category and $\mathcal{C}^{\stackrel{\rightarrow} {0}}$ is $\mathcal{C}$ itself. So, for example the objects and morphisms of $\mathcal{C}^{\stackrel{\rightarrow} {2}}$ are respectively pairs of morphisms and commuting squares in $\mathcal{C}^{\stackrel{\rightarrow} {1}}$. But, as morphisms of $\mathcal{C}^{\stackrel{\rightarrow} {1}}$ are again commuting squares in $\mathcal{C}^{\stackrel{\rightarrow} {0}}$, we can think of objects and morphisms of $\mathcal{C}^{\stackrel{\rightarrow} {2}}$ respectively as commuting squares and commuting cubes in $\mathcal{C}$. Generally, objects and morphisms of $\mathcal{C}^{\stackrel{\rightarrow} {n}}$ are respectively n-dimensional and n+1-dimensional cubes in $\mathcal{C}$, where an $i$-dimensional cube is a functor $$F:2 \times 2 \times ...\times 2 \longrightarrow \mathcal{C}$$ 

Question: Can $\mathcal{C}^{\stackrel{\rightarrow}{n}}$ model higher categories? What about the category $\mathcal{C}^{\stackrel{\rightarrow}{\infty}}$ which has all $\mathcal{C}^{\stackrel{\rightarrow}{i}}$ (for $i \in \mathbb{N}$), while  as objects and functors between them as morphisms. (Not too sure about this. What would be an appropriate class of morphisms between $\mathcal{C}^{\stackrel{\rightarrow}{n}}$ and $\mathcal{C}^{\stackrel{\rightarrow}{m}}$ anyway?) Please, let me know if this construction is of any relevance to higher category theory. This might help me learning some higher category theory.

REPLY [3 votes]: (source)
@David: An advantage of strict $n$-fold categories is that they can express in a simple way the above diagram in the form of "the big square is the composition of the little squares"; namely one defines a composable array $(a_{ij})$ to be an array of elements such that each is composable with its immediate neighbours. Then by associativity and the interchange law, the composite $[a_{ij}]$ is well defined. This "algebraic inverse to subdivision", which is easily extended to higher dimensions,  is used extensively by Higgins and me in proving a Higher Homotopy Seifert-van Kampen theorem, in the paper available here. I do not know how to develop analogous methods in the globular or simplicial contexts; simplicial methods, but not of the higher categorical type,  are used by Brown and Loday to prove their version of a HHSvKT.  All these bring nonabelian colimit methods into homotopy theory. 
I'll also mention that Brown and Higgins proved a groupoid result on the equivalence between strict cubical and globular $\omega$-groupoids, and the analogous but more difficult category result was proved by Al-Agl, Brown, and Steiner. The references are on the (updated)  ncatlab page given by David. 
Later: for complete clarity, though I expect it will be clear, I just add that if $a_{ij}$ is not at the edge of the square, then its immediate neighbours are: $a_{i-1,j}, a_{i+1,j}, a_{i,j-1}, a_{i,j+1}$, and the rule for composability of two squares is that their appropriate edges agree. My papers with Higgins contain lots of 2-dimensional rewriting, using the interchange law; and the paper with Al-Agl and Steiner contains a 3-dimensional rewriting argument to obtain a braid relation.<|endoftext|>
TITLE: Fundamental motivation for several complex variables
QUESTION [18 upvotes]: I have 3 general abstract reasons to care about complex analysis in a single variable:

The laplacian is, up to a constant multiple, the only isometry invariant PDO in the plane, and so it is abstractly very important.  Holomorphic functions are intimately related to harmonic functions, so holomorphic functions are important.
Holomorphic mappings are conformal if they have nonvanishing derivatives
Many results which are real variable in nature are most easily understood in the light of complex analysis (factorization of real polynomials, radius of convergence of real power series, etc)

I currently have no such justification for several complex variables.  The connection to harmonic functions mostly breaks down.  Conformality breaks down.  I have not seen applications to real variable phenomena.  
Can anyone give me some insight into the big picture here?  Where do the phenomena studied in several complex variables (domains of holomorphy and their ilk) come up in the rest of mathematics?

REPLY [7 votes]: If those are your reasons for caring about complex analysis, then I'm not sure to what extent I can convince you to care about SCV.
But here is why I like SCV: I like multivariable calculus, and I like complex numbers.  I've always wondered what would happen if you put the two together.  And apparently pretty wild things happen (compared with either the single-variable or real-variable case).
I'm also really interested in complex geometry.  The local theory of complex manifolds involves SCV.
One final comment: Your reason (1) for caring about holomorphic functions is in fact my reason for caring about harmonic functions (for which I would otherwise desire motivation).<|endoftext|>
TITLE: Are annihilation modules in the quantum torus necessarily principal?
QUESTION [8 upvotes]: I hope that my question yields some standard fact from (noncommutative) ring theory.  In discussions with other graduate students, we have outlined some approaches to tackling the question, but haven't come up with an answer.
Notational caveats: I will work over some commutative ring that I will call $\mathbb Z$, but if you like, you may let $\mathbb Z$ denote your favorite algebraically closed field, or whatever — I am agnostic about that type of thing.  I will denote by $\mathbb Z_q$ the Laurent polynomial ring $\mathbb Z_q = \mathbb Z[q^{\pm 1}]$ in a commuting parameter $q$, but again I am reasonably agnostic: if you need to use the field of rational functions, say, then so be it.
The quantum torus: I am interested in the noncommutative (but "$q$-mutative") Laurent polynomial ring of "functions on the quantum torus", defined by:
$$ R = \mathbb Z_q\langle x^{\pm 1},y^{\pm 1}\rangle / (yx = qxy) $$
It is an "exponential" version of the Heisenberg ring $\mathbb Z[\hbar]\langle \xi,\upsilon\rangle / ([\upsilon,\xi] = \hbar)$, with $q = e^\hbar$, $x = e^\xi$, and $y = e^\upsilon$.
I am also interested in a distinguished left $R$-module:
$$ M = \mathbb Z_q\langle x^{\pm 1}\rangle $$
$$ x \triangleright f(x) = x\ f(x), \quad y \triangleright f(x) = f(qx) $$
Again, this is an exponential version of the action of differential operators on polynomials, in which $\upsilon$ acts by $\frac{\partial}{\partial \xi}$.
Fix $f\in M$.  It defines a left ideal $\operatorname{Ann}(f) \subseteq R$ consisting of those $r\in R$ such that $r\triangleright f = 0$.
My question:  (When) is $\operatorname{Ann}(f)$ principal?  I.e. (for which $f$) does there exist an element $a_f \in R$ such that $\operatorname{Ann}(f) = Ra_f$?
If it exists, then $a_f$ is determined up to units in $R$, which are precisely the monomials (unless I am mistaken).  Is there an algorithm that computes $a_f$ from $f$?
For example, I believe that $\operatorname{Ann}(x^n)$ is generated by $y - q^n$.  On the other hand, included within $\operatorname{Ann}(x-1)$ are the elements $(y-1)(y-q)$ and $x(y-q) - (y-1)$, and I don't see immediately a principal generator.
Not the main question, but an interesting topic for discussion: The classical limit $q \to 1$ gives the classical torus $R|_{q=1} = \mathbb Z[x^{\pm 1},y^{\pm 1}]$, with the symplectic structure $\omega = \frac{\mathrm dy}{y}\wedge \frac{\mathrm dx}{x}$ as the 1-jet of the noncommutativity.  If I coarsely take the classical limit before computing the annihilation ideal, then for all non-zero $f$ I get the ideal generated by $(y-1)$, because $M$ is a domain.  The more interesting thing to do is to compute $\operatorname{Ann}(f)$ in the quantum ring $R$, and then specialize $q \to 1$.  This gives the ideal generated by $a_f(q=1)$, provided $\operatorname{Ann}(f)$ is principal in $R$.  Can the vanishing locus of $\operatorname{Ann}(f)|_{q=1}$ be described in terms of the (symplectic) geometry of the function $f$?

REPLY [6 votes]: A longuish comment. 
I'll work in $R=k\langle x^{\pm1},y^{\pm1}\rangle/(yx-qxy)$ with $q\in k\setminus0$ not a root of unity.
Notice that $M\cong R/R(y-1)$. If $f\in M$ is not zero, then $M/Rf$ is finite dimensional. As $R$ is infinite dimensional and simple (this is why I picked $q$ of infinite order), $M/Rf$ must be zero, that is, $Rf=M$. 
Now we have short exact sequences $$0\to R(y-1)\to R\to M\to 0$$ with the last map being $r\mapsto r\triangleright 1_M$, and $$0\to\operatorname{ann}(f)\to R\to Rf\to 0$$ with the last map $r\mapsto r\triangleright f$, of course. The first one is a projective resolution of $M$, so $\operatorname{ann}(f)$ is projective, and Schanuel's lemma implies that there is an isomorphism of left $R$-modules $$R(y-1)\oplus R\cong \operatorname{ann}(f)\oplus R$$ We thus see that $\operatorname{ann}(f)$ is stably principal :-) (and that it can be generated by two elements)<|endoftext|>
TITLE: Ideals generated by idempotent elements
QUESTION [5 upvotes]: Let $R$ be a unitary associative ring and $J$ be an ideal of $M_{n}(R)$. We know that there is an ideal $I$ of $R$ such that $J=M_{n}(I)$.
Now there is a question.
Question: If $J$ is generated by a subset of idempotent elements of $M_{n}(R)$ say $S$,  is $I$ generated by a subset of idempotent elements of $R$, which related to $S$?

REPLY [2 votes]: I don't know if this helps. 
We readily obtain the following assertions:
(1) If the ideal $J=M_n(I)$ of $M_n(R)$ is generated by some idempotents,
then $I$ is an idempotent ideal, i.e., $I^2=I$.
(2) If $J=M_n(I)$ is generated by finitely many idempotents, then $I$
is a finitely generated ideal.
Thus in the case where $R$ is commutative and $J$ is generated by
finitely many idempotents, there exists an idempotent $e\in R$ such
that $I=Re$.<|endoftext|>
TITLE: Intersecting Family of Triangulations
QUESTION [54 upvotes]: Let $\cal T_n$ be the family of all triangulations on an $n$-gon using $(n-3)$ non-intersecting diagonals. The number of triangulations in $\cal T_n$ is $C_{n-2}$ the $(n-2)$th Catalan number. Let $\cal S \subset \cal T_n$ be a subfamily of triangulations with the property that every two triangulations of $\cal S$ have a common diagonal. 

Problem: Show that $|\cal S| \le |\cal T_{n-1}|$.

Remark: This problem was raised independently around the same time by Karen Meagher and by me.

Update (and counter-update)
A few weeks after this problem was posted Gjergji Zaimi (private communication) proposed a more general conjecture:
Conjecture:
Let $P$ be a polytope with no triangular face. Then the maximum number of vertices such
that every two vertices belongs to a common facet is attained by all vertices of
 a single facet.
The original question is the case of the associahedron. The case of the permutahedron is known- it is a result by Frankl and Deza-  and it is related to extremal combinatorics on permutations. For the cube the result is immediate but can serve as a good starting point for extremal combinatorics (Problem 1 here).
Update: Bruno le Floch showed that the more general conjecture is false: He described a quadrangulation of S^2 with 15 vertices and 13 quadrangles having 5 vertices each two on a face.   
Update Paco Santos proposed the following intermediete conjecture: 
Conjecture: For every flag simplicial polytope, the maximum size of a set of pairwise intersecting facets is achieved by the facets containing some common vertex.
A flag simplicial polytope is a simplicial polytops so that every set of vertices
that any two form an edge is the set of vertices of a face. Santos's conjecture thus asserts that the conjecture about polytopes with no triangular faces still applies for dual-to-flag polytopes.

REPLY [2 votes]: This falls slightly short of a proof but is too long for a comment.
The question inquires into the largest family of triangulations in which every pair of members shares some diagonal.
If we instead fix any single diagonal in common to every member of a family of triangulation of an n-gon, it is self-evident (from the noncrossing property of triangulations) that the number of triangulations sharing it is given by the product of the enumerations of the two families of triangulations of the ploygons so formed either side of it.
For any triangulatable n-gon, any given diagonal is a member of up to two fan triangulations, in which it can be assigned an integer the $d^{th}$ diagonal from the perimeter satisfying $1\leq d\leq \frac{n-2}{2}$. By symmetry we may choose either of the two fans with no effect upon the distance of the diagonal from the nearest edge.
$d=\frac{n-2}{2}$ shall be the distance of the central triangulation if $n$ is even and $d=\frac{n-3}{2}$ shall be the distance of the two "most central" diagonalisations if $n$ is odd.
The number of triangulations of any $n$-gon sharing the $d^{th}$ diagonal is the product of the number of triangulations $\cal T(d,n)$ either side of it:
$$\cal T(d,n)=C_d\times C_{n-d}$$
$$\cal T(d,n)=(d+1)^{-1}\binom{2d}{d}(n-d+1)^{-1}\binom{2n-2d}{n-d}$$
The ratio of one Catalan number to the next is greater than the ratio of the previous two Catalan numbers, and therefore in the choice of which diagonal to share, in order to maximise the size of the family, for any given choice $d$ of diagonal which is not at the edge, we may always identify a larger family by choosing the diagonal $d-1$ which is one position closer to the perimeter and by induction the outermost diagonal $d=1$ yields the largest family.
The size of the largest family of triangulations sharing the same diagonal is therefore given by:
$|\cal S| = C_{n-2-1}\times C_1=|\cal T_{n-1}|$
My claim which would complete the proof (and which remains to be proven) is that there is no family of triangulations in which every pair shares some diagonal, which is larger than the largest family of triangulations in which every triangulation in the family shares the same diagonal.
I suspect this is proven by the fact that the largest face on any associahedron has the same number of edges as the number of vertices of the associahedron one order smaller.  In particular I believe that each face of the associahedron represents a family of triangulations which all share the same edge.  These two statements alone would in fact be sufficient to answer the entire question in the affirmative.<|endoftext|>
TITLE: What is the universal deformation of the formal additive group $\widehat{\mathbb{G}}_a$ over $\mathbb{F}_p$?
QUESTION [12 upvotes]: Lubin and Tate show in their paper Formal moduli for one-parameter formal Lie groups that for any formal group over a field $k$ of characteristic $p>0$ with height $h<\infty$, the functor of deformations is represented by a formal scheme isomorphic to $\mbox{Spf } \mathbb{W}(k)[[u_1,\ldots,u_{h-1}]]$.  Modulo lower terms, $u_i$ is the coefficient of $x^{p^i}$ in the $p$-series of the universal deformation.  (We take $p=u_0$.)
Does this carry over for the additive group?  Certainly there is an evident deformation to $\mbox{Spf } \mathbb{W}(k)[[u_1,u_2,\ldots]]$.  In the paper, the finite height assumption is present in various results that they cite from elsewhere, so without being intimately familiar with the whole theory it's kind of hard to tell if this assumption is essential.

REPLY [18 votes]: $\DeclareMathOperator{\Ext}{Ext} \newcommand{\G}{\hat{\mathbb{G}}} \DeclareMathOperator{\Maps}{Maps} \renewcommand{\phi}{\varphi}$  The analysis of the infinitesimal deformation space of the Honda formal groups $H_n$ uses three calculations which govern the existence of square-zero deformations.  These can be phrased in terms of certain $\Ext$ groups, and here's how I think of them: we want to study the deformation of a formal group $G_0$ over a ground ring $R_0$ along a square-zero infinitesimal deformation $I \to R \to R/I = R_0$ to a formal group $G$ over $R$.  Because $I$ is square-zero and the only formal group law truncated at degree $2$ is the additive formal group, we think of $G$ as sitting in an extension $I \otimes \G_a \to G \to G_0$ sort of "over" the original extension of $R$-modules.  So, to study these extensions (their existence, their uniqueness, so on), we want to calculate some kind of $\Ext$ groups of the form $\Ext^*(G_0; \G_a)$.
This is exactly what Lubin and Tate do in their paper.  Fix a pair of formal groups $F$ and $G$; then there's a simplicial object $BF = B_*(*, F, *)$ associated to $F$ via its group structure, and the cosimplicial object $\operatorname{Maps}(BF, G)$ should be a thing whose cohomology $H^*_{LT}$ computes $\Ext^{*-1}(F; G)$.  Trying to work out exactly what this means, you'll find the following descriptions of the first few groups (written in coordinates, but this is inessential):
$$H^1_{LT}(F; G) = \{\phi: F \to G \mid \phi(x) - \phi(x + y) + \phi(y) = 0\} = \operatorname{Hom}(F, G),$$
$$H^2_{LT}(F; G) = \frac{\{\phi: F^2 \to G \mid \phi(x, y) - \phi(x, y + z) + \phi(x + y, z) - \phi(y, z) = 0\}}{\{\delta^1 \phi \mid \phi : F \to G, \delta^1 \phi(x, y) = \phi(x) - \phi(x + y) + \phi(x)\}},$$
$$H^3_{LT}(F; G) = \frac{\left\{\phi: F^3 \to G \middle| \begin{array}{c}\phi(x, y, z) - \phi(x, y, z + w) + \phi(x, y + z, w) - \\ \phi(x + y, z, w) + \phi(y, z, w) = 0\end{array}\right\}}{\{\delta^2 \phi \mid \phi : F^2 \to G, \delta^1 \phi(x, y) = \phi(x, y) - \phi(x, y + z) + \phi(x + y, z) - \phi(y, z)\}}.$$
General facts about infinitesimal deformation theory then tell you the utility of these groups: $H^1_{LT}$ tracks automorphisms of the square-zero deformation space, so it tells you whether you should expect the deformation space to be a scheme or some sort of stack; $H^3_{LT}$ tracks the obstruction to the existence of square-zero deformations; and $H^2_{LT}$ tracks the available square-zero extensions, in the sense that when $H^1_{LT} = 0$, its dimension will tell you the dimension of the tangent space of the deformation space.
So, Lubin and Tate go about computing these three things in the case that $G_0 = H_n$ is the height $n$ Honda formal group.  They know that $\operatorname{Hom}(H_n, \G_a) = 0$, and so $H^1_{LT} = 0$.  They also compute that $H^3_{LT} = 0$, so they know that the deformation space they're computing is actually a formal variety --- there are never any kind of conditions imposed on their generators $u_i$ to ensure the existence of a deformation.  The only thing left is to determine the dimension of the formal variety, and they compute $\dim \Ext^1(H_n, \G_a) = n-1$.
You ask what happens when we replace $H_n$ with $\G_a$ and instead study $H^*_{LT}(\G_a; \G_a)$.  The computation of these cohomology groups is worked out in the COCTALOS notes, and also in Theorem 4.3 and Section 8.4 of a project I worked on: [link].  Grinding through the homological algebra, you'll find that the group $H^n_{LT}$ consists of the polynomials of homogeneous degree $n$ in the dual Steenrod algebra, after assigning the degrees $|\xi_*| = 1$, $|\tau_*| = 1$, and $|P_*| = 2$.
All of these things are woefully $\infty$-dimensional.  I don't actually know what that means for the existence of your object --- and so this isn't a proper answer --- but I do know that it isn't anywhere as simple as the case Lubin and Tate analyze.
P.S.: Since I'm updating this answer anyway, Section 2.2 of the Hopkins-Lurie Ambidexterity manuscript describes computing $H^*_{LT}(\Gamma; \G_a)$ for any connected $p$–divisible group $\Gamma$ and any value of $*$.  This generalizes all but the last paragraph of this answer, since $\G_a$ is not $p$–divisible.<|endoftext|>
TITLE: Descending chain condition in noncommutative rings
QUESTION [8 upvotes]: By Hopkins Theorem it is well-known that every right (resp. left) artinian unitary ring is right (left) noetherian. Suppose that a noncommutative unitary ring R satisfies the descending chain condition on its two-sided ideals. Does R satisfy the ascending chain condition on two-sided ideals?

REPLY [5 votes]: I'm somewhat surprised this question hasn't been answered previously.  It turns out DCC on two-sided ideals does not imply ACC on two-sided ideals.
Let $V$ be a $k$-vector space with a basis of size $\aleph_{\omega}$, and put $R={\rm End}_k(V)$.  Then, by Exercise 3.16 in Lam's "A First Course in Noncommutative Rings" the two-sided ideals are of the form $0$, $R$, and $I_{\aleph}=\{x\in R\ :\ \text{the rank of $x$ is }<\aleph\}$ for each infinite cardinal $\aleph\leq \aleph_{\omega}$.  These ideals are linearly ordered and do not satisfy ACC since we have an increasing chain $\aleph_0<\aleph_1<\cdots< \aleph_{\omega}$, but these ideals do have DCC since the cardinals do.<|endoftext|>
TITLE: Spin structures on $S^1$ and Spin cobordism
QUESTION [11 upvotes]: I'm trying to understand the 2 spin structures on the circle. Since the frame bundle for the circle is just the circle itself, Spin structures on $S^1$ correspond to double covers of $S^1$. There are two choices: the connected double cover and the disconnected double cover.
From the point of view of Spin cobordism, we can view the circle as the boundary of the disk in the plane. The disk has a unique spin structure, and we can ask which spin structure this induces on the boundary.
Lawson/Michelson's "Spin Geometry" claims that this induces the spin structure coming from the double cover, but I'm having trouble seeing that. The frame bundle for the disk $D^2$ must be trivial, and thus isomorphic to $D^2\times SO(2) = D^2 \times S^1.$ There is a natural double cover given again by $D^2 \times S^1,$ and the map is just the identity on $D^2$ and $z \rightarrow z^2$ on the $S^1$ factor. 
To see what the induced spin structure on the boundary is, we must view the frame bundle of the boundary as sitting inside the frame bundle of $D^2\times S^1$ by fixing an outward normal vector field and then using it to complete any frame on $S^1$ to a frame on $D^2.$ To me, this seems to say that we view the frame bundle of $S^1$ (which is itself $S^1)$ as $S^1\times \{1\} \subset D^2 \times S^1,$ since once we fix one vector of a frame (in this case given by the normal) the other is entirely determined since we are in 2 dimensions. 
But now if we look at the inverse image of that in the double cover, we appear to get two disjoint copies of $S^1,$ i.e. the disconnected double cover. What am I doing wrong?
(This is crossposted on stack exchange as https://math.stackexchange.com/questions/245480/spin-structures-on-s1-and-spin-cobordism).

REPLY [12 votes]: As Fabian pointed out in the comments, you have to be more careful about how you trivialize $SO(D^2)$.  I'm going to use the standard coordinates $(x,y)$ on $\mathbb{R}^2$ (note that these are not global coordinates on $D^2$, but they still trivialize the frame bundle).  Thus we have a global section of $SO(D^2)$ which assigns to each point in $D^2$ the standard orthonormal basis $(e_x,e_y)$, and the action of $S^1 = SO(2)$ on $SO(D^2) \cong D^2 \times S^1$ is by counterclockwise rotation of this basis.  It is fairly clear from this picture that the principal spin bundle is $Spin(D^2) \cong D^2 \times S^1$, and the double cover $Spin(D^2) \to SO(D^2)$ is the identity on $D^2$ and the doubling map on $S^1$.
Now let's figure out how $SO(S^1) \cong S^1$ sits inside $SO(D^2)$ using this trivialization.  The boundary circle is the set of points $(\cos \theta, \sin \theta)$ in $D^2$.  The (oriented) unit tangent vector to $S^1$ at the point $(\cos(0),\sin(0)) = (1,0)$ is just $e_y$ in the frame used to trivialize $SO(D^2)$ above, and at any other point $(\cos \theta, \sin \theta)$ it is just $R_\theta e_y$ where $R_\theta$ denotes counterclockwise rotation by the angle $\theta$.  So if we denote the oriented unit vector tangent to $\theta \in S^1$ by $e_\theta$ then the embedding $SO(S^1) \to SO(D^2) \cong D^2 \times S^1$ is given by $e_\theta \mapsto ((\cos \theta, \sin \theta), \theta)$.  Finally, in this picture it is clear that the inverse image of the set $\lbrace ((\cos \theta, \sin \theta), \theta) \rbrace$ under the map $Spin(D^2) \to SO(D^2)$ is the connected double cover of $S^1$.<|endoftext|>
TITLE: What are some Applications of Teichmüller Theory?
QUESTION [19 upvotes]: I'm trying to collect some specific examples of applications of Teichmüller Theory. Here are some things I have collected thus far:

No-wandering-domain Theorem (Sullivan)
Theorems of Thurston (Classification of homeomorphism of surfaces, topological characterization of rational maps, hyperbolization theorems for special 3 manifolds)
Computer graphics. (Using the various metrics on the Teichmüller spaces as a substitute for Gromov-Hausdorff metric.)
String Theory (as elementary particles are modeled by loops, they generate a Riemann surface as they move through time.) 
Some applications to biology.(Brain morphometry)

Note: I am very sure that this is only a small fraction of what is out there, and I plan to continue to update this list.

REPLY [2 votes]: http://www.cacs.louisiana.edu/~mjin/ I saw a talk by Miao Jin where she used Teichmüller theory and Ricci flows to better store and analyse colonoscopy data. I believe her advisor at Stony Brook started a company which does that.<|endoftext|>
TITLE: topological monoid from symmetric monoidal category
QUESTION [11 upvotes]: What is the standard reference for the fact that the classifying space of a strict monoidal category is a topological monoid with respect to the operation induced by the tensor product?
EDIT: The first version of the question was stated for strict symmetric monoidal categories, but as was pointed out in the comments, a symmetry is of course not necessary to just get a monoid structure on the classifying space.

REPLY [5 votes]: Well, if you are going to reference me somewhere, I can give you something more explicit.  The cited Corollary 11.7 is only about topological monoids.  However  Theorem 4.10 of "$E_{\infty}$ spaces, group completions, and permutative categories" ( http://www.math.uchicago.edu/~may/PAPERS/13.pdf ) has the precise statement requested: "If $(\mathcal{A},\Box,\ast)$ is a strict monoidal category, then $B\mathcal{A}$ is a topological monoid with product $B\Box$."  The result goes on to say precisely what holds with respect to commutativity when $\mathcal{A}$ is permutative (= strict symmetric monoidal).<|endoftext|>
TITLE: An invariant method of stationary phase
QUESTION [10 upvotes]: The method of stationary phase is very well-known and employed in many areas of physics and mathematics, and, of course, included in various versions as theorem in textbooks, especially on pseudors and microlocal analysis.
However, it always is somewhat dependent on local coordinates and the Fourier transform, despite being a quite invariant problem. To be precise, the question would be the following.
Let $M$ be a manifold and $\phi: M \longrightarrow \mathbb{C}$ be a smooth function with values in the closed right half plane. Let $u$ be an volume density on $M$ with compact support in $M$. Determine an asymptotic expansion as $t \rightarrow \infty$ of the integral
$$ I(\phi, u, t) = \int_M e^{-\phi t} u$$
under some nondegeneracy conditions on $\phi$. 
(For example, one could require $\phi$ to be Morse or, more general, require that the set where it vanishes is a submanifold $C$ of $M$ and that at a point $p \in C$, the Hessian of $\phi$ is non-degenerate on the space $T_pM/T_pC$.)
It is well-known that in these cases $I(\phi, u, t)$ has an asymptotic expansion of the form
$$ I(\phi, u, t) = (t/\pi)^{-(n-k)/2}\sum_{j=0}^\infty t^{-j} \int_C s_j,$$
where $k$ is the dimension of $C$ and the $s_j$ are certain volume densities on $C$. In fact, they have to be certain universal terms, depending only on the $2j$-th jets of $\phi$ and $u$ at $C$. This is not stated in most textbooks. 
I wonder if it is possible to find these terms $s_j$ using Invariance theory alone. I would like if someone ever thought about this and knows a reference to this more invariant, geometric approach.

/Edit: To clarify my question: I was wondering if it is possible to determine the constants by invariance theory, i.e. some argument like "there is only one polynomial on the $2j$-jets of $u$ and $\phi$ that is invariant under coordinate transformation" or so. For the first term, this goes like this, supposed that $\phi$ is purely real:
Define the $n-k$-density $\mathrm{H}\phi$ on $C$ by setting
$$\mathrm{H}\phi[X_1, \dots, X_{n-k}] := \sqrt{\left|\det \bigl( D^2\phi[X_i, X_j] \bigr)_{ij}\right|},$$
where $D^2\phi$ is the (on $C$ well-defined) Hessian of $\phi$. Now $u/\mathrm{H}\phi$ is a $k$-density on $C$ -- this is $s_0$.
Now there should be similar characterizations of the higher $s_j$ (which obviously can get arbitrarily complicated).

REPLY [2 votes]: Check Proposition 1.2.4 from the book Fourier Integral Operators by the late great  Duistermaat. This result applies  in the case when the phase $\phi$ is Morse.  If the phase is not Morse, but the critical points    are finitely determined (finite Milnor number) then things are a bit more complicated. The vol 1 book  of Arnold-Gusein-Zade Varchenko Singularities of differentiable maps is a good source.      You can also  have a look  at the senior thesis of a Zach Lamberty, a  former student of mine.  There he deals with the $2$-dimensional case ($\dim M =2$) and he essentially works out the toric resolution trick of Arnold and comp.  for a special and quite degenerate two variable phase.<|endoftext|>
TITLE: One more question about PBW
QUESTION [8 upvotes]: Let $k$ be a commutative ring with unit and $L$ be a Lie $k$-algebra. 
Let $U(L)$ be the universal enveloping $k$-algebra of $L$ (one can define it as a quotient of the tensor algebra, as it is explained in this MO question, or one can say that $U(-)$ is left adjoint to the forgetful functor sending an associative $k$-algebra to the Lie $k$-algebra obtained by taking the same underlying $k$-module and with Lie bracket being the commutator). 
The associative $k$-algebra $U(L)$ is filtered as a $k$-algebra, and there is a canonical epimorphism $S(L)\to gr\big(U(L)\big)$. 
If this epimorphism is an isomorphism, then we say that $L$ has the PBW property. 
All the examples of Lie $k$-algebras not satisfying the PBW property I am aware of are constructed in the following way: one first finds an example of a Lie algebra for which the map $L\to U(L)$ (the unit of the adjunction) is not injective, and then it is quite clear that the PBW property can't hold. 
My question is then: 

Is there any example of a Lie
  $k$-algebra $L$ such that the map
  $L\to U(L)$ is injective which does
  not satisfy the PBW property ?

Or is it that the PBW property is just equivalent to $L\to U(L)$ being injective (it would be great, but I have no idea why this would be true - EDIT: one might want to use that $L\to U(L)$ is injective to reduce to the cas when $k\supset\mathbb{Q}$)?

REPLY [5 votes]: While I can't immediately give a precise example, it may help to focus the question a little more in order to emphasize what is actually involved.    
1) In most literature on Lie algebras (related to Lie groups or algebraic groups), the emphasis is on Lie algebras over fields or perhaps over integral domains and their fraction fields.  But there is another direction of study involving Lie rings (and often associated abstract groups).  It's possible more generally to study Lie algebras over an arbitrary commutative ring $K$ (with 1), which is done for instance in Chapter I of the Bourbaki treatise Groupes et algebres de Lie.    There the PBW theorem is developed in Section 2.7, following generalities on the universal enveloping algebra $U(L)$.  But the theorem is stated and proved under the restrictive assumption that the Lie algebra is a free module over $K$.  This leads directly to the embedding of $L$ in $U(L)$ and the existence of the usual PBW bases for the latter algebra.  Once you assume that $L$ has a basis, what you call the PBW property is then equivalent to the embedding property.      
2) Bourbaki's Exercise 9 for Section 2 provides a rather complicated construction of a Lie algebra $L$ over a commutative ring which doesn't embed in its universal enveloping algebra.   But $L$ does embed in its symmetric algebra, by definition.  This is certainly enough to defeat the PBW property, as you observe.   [But your wording "All the examples ..." is confusing and really means "All the known examples .."]    You are asking for an example of a Lie algebra which fails to be a free $K$-module and fails to have the PBW property but does embed in $U(L)$.   Given the complexity of the Bourbaki exercise mentioned above, such an example (if it exists!) is probably even more complicated.<|endoftext|>
TITLE: The diophantine equation X^2 - Y^2 - Z^2 = +- 1
QUESTION [9 upvotes]: Hi everybody. I'd like to know if the diophantine equation
(1) $$X^2 - Y^2 - Z^2 = \pm 1$$
has been studied and if the set of its solutions $(X,Y,Z)$ is known. I appreciate any reference. Thank you very much.
P.S. If instead we look at the diophantine equation
$$X^2 - Y^2 - Z^4 = \pm 1$$
surely we can solve it imposing conditions on the solution of (1) so that Z be a square. However, is there a quicker method?

REPLY [12 votes]: In general, if $Q(x,y,z)$ is a nonsingular ternary quadratic form with integral coefficients, then the integral solutions of $Q(x,y,z)=n$ fall into finitely many orbits of the integral automorphism group of $Q$, and these orbits can be effectively determined. For your case there is a shortcut as follows.
Write $a=x+y$, $c=x-y$, $b=z$, then your equation becomes $b^2-ac=\mp 1$. In the language of binary quadratic forms, this means that the form $au^2+2buv+cv^2$ has discriminant $\mp 4$. From classical theory it follows that after an invertible linear change of variables $u'=pu+qv$, $v'=ru+sv$ the form becomes $u'^2\pm v'^2$, i.e. there are $p,q,r,s\in\mathbb{Z}$ such that $ps-qr=1$ and
$$ au^2+2buv+cv^2=(pu+qv)^2\pm(ru+sv)^2. $$
Equating coefficients,
$$ a=p^2\pm r^2,\quad c=q^2\pm s^2,\quad b=pq\pm rs, $$
i.e. 
$$ x=(p^2\pm r^2+q^2\pm s^2)/2,\quad y=(p^2\pm r^2-q^2\mp s^2)/2,\quad z=pq\pm rs. $$
To summarize, all integer solutions of $x^2-y^2-z^2=\pm 1$ are of this form, with integers $p,q,r,s$ satisfying $ps-qr=1$. Using congruences one can easily distinguish integer solutions from half-integer ones.
As references, I suggest Rose: A course in number theory, and Cassels: Rational quadratic forms.<|endoftext|>
TITLE: Baum-Connes-like "conjecture" for $l^p$-spaces
QUESTION [8 upvotes]: Let $G$ be a (discrete) group. For the Baum-Connes conjecture, one looks at the reduced group $C^{\star}$-algebra: Look at the Hilbert space $l^2(G)$ and the representation of $G$ on this Hilbert space given by left multiplication. The norm-closure of the resulting $\mathbb{C}G$-representation in $B(l^2(G))$ is the reduced group $C^*$-algebra.
For any $p \geq 1$, we can pretty much do the same: Look at $l^p(G)$. We still have a representation of $G$ on $B(l^p(G))$ by left multiplication and hence obtain a kind of reduced Banach group algebra for $l^p$; lets call it $B^p(G)$.
There also should be an assembly map
$K_*(E_{Fin}G) \rightarrow K_*(B^p(G))$
as in the Baum-Connes conjecture. For $p = 1$, we have $B^1(G) = l^1(G)$ and we obtain the Bost assembly map. I have no reason to believe that for arbitrary p such an assembly map might be an isomorphism, but was wondering whether such group Banach algebras, and maybe even the assembly maps, have been considered anywhere in the literature.

REPLY [10 votes]: It is likely that in cases where I proved Baum-Connes without coefficients (i.e. reductive groups over local fields and some discrete groups with RD), some variant of the Schwartz or Jolissaint algebra will be dense and stable under functional calculus in the algebra you call $B^p(G)$. This would imply the BC conjecture for it. 
In the case of Schwartz algebras, some arguments like this (with almost the right L^p estimates)
are given in the last section of my paper in Inventiones. 
The big limitation of this $L^p$ variant of the Baum-Connes conjecture is that when you consider coefficients in a $G$-$C^*$-algebra A, $L^p(G,A)$ 
can be defined only in a naive way and cannot be a  $A$-Hilbert module as in the case where p=2.<|endoftext|>
TITLE: Is a domain all of whose localizations are noetherian itself noetherian ? 
QUESTION [25 upvotes]: Is a domain $D$, all of whose localizations $D_P$ for $P \in Spec(D)$ are noetherian, itself noetherian ? 
The question is motivated by proposition 11.5 of Neukirch's Algebraic Number Theory: 

Let $\mathfrak{o}$ be a noetherian integral domain. $\mathfrak{o}$ is a Dedekind domain if and only if, for all prime ideals $\mathfrak{p}\neq 0$, the localizations $\mathfrak{o}_\mathfrak{p}$ are discrete valuation rings. 

If the question above has a positive answer, this proposition would give an unconditioned  (i.e. without precondition "noetherian") characterization of Dedekind domains by a local property. 
By googling I found a counterexample for a ring with zero-divisors: 
https://math.stackexchange.com/questions/73421/a-non-noetherian-ring-with-all-localizations-noetherian
But I couldn't find a counterexample for a domain.

REPLY [13 votes]: All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for 

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian. 

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=\mathbb{Z} \setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.  
Remark. The foregoing shows that $\dim D=2$.<|endoftext|>
TITLE: Integrating powers without much calculus
QUESTION [15 upvotes]: I'll jump into the question and then back off into qualifications and context

Using the definition of a definite integral as the limit of Riemann sums, what is the best way (or the very good ways) to establish the results $\int_a^bx^pdx=\frac{b^{p+1}-a^{p+1}}{p+1}$ without building a general theory of integrals?

Context: For better or worse, a common sequence in teaching integral calculus for the first or second time is to define the definite integral as a limit of Riemann sums. One then notes that this (like the limit definition of derivatives) is effective for proving theorems but not very practical for specific calculations. So soon one demonstrates or implies that the definition is valid and then gets to the Fundamental Theorem of Calculus. However it is traditional to use the definition to evaluate $\int_a^bx^pdx$ for $p=0,1,2$ and perhaps $p=3$ using the lovely sum of cubes formula.
It might be tempting to prove the formula in greater generality without using the fundamental theorem (most students are less excited at this prospect than one might expect!). In my early days as a TA I came up with an approach which I thought was great. The students were not impressed and I have not used it since. Anyway, I have not seen it elsewhere although I am confident it is nothing novel.  I am not going to reveal it right away just to see if it shows up. I realize that is questionable manners here on MO but I will put it up in a day or two, I just want to see what shows up first.
Here is a very brief sketch of two approaches I have seen:
‍1) Let $S_p(n)=\sum_{k=0}^nn^p$. The explicit formulas for $p=0,1,2$ and also $p=3$ are attractive and not bad to prove by induction. Given an explicit formula for $S_p$ one easily evaluates the usual equal subinterval sums for $\int_0^1x^pdx$ and then extends to $\int_a^b$. But $S_p$ gets more tedious for larger $p$. A clever method of Pascal allows one to use strong induction, the binomial theorem and telescoping sums to derive an explicit formulas for $S_{p}$ for larger values of $p$, limited only by one's patience and stamina:
Take $(k+1)^{p+1}-k^{p+1}=\sum_1^{p+1}\binom{p+1}{j}k^{p+1-j}$ and sum for $k$ from $0$ to $n$ to get $(n+1)^{p+1}-0^{p+1}=\sum_1^{p+1}\binom{p+1}{j}S_{p+1-j}(n)$. Since we know everything except $S_p$, the rest is algebra! This is quickly unpleasant and the final results are not as aesthetic as the first cases. HOWEVER, for the desired application we only need to establish that $S_p(n)=\frac{n^{p+1}}{p+1}+\frac{n^p}{2}+O(n^{p-1})$. That is not hard and shows that $n$ subdivisions yield  $\frac{1}{p+1}-\frac{1}{2n} \lt \int_0^1x^pdx \lt \frac{1}{p+1}+\frac{1}{2n}$.
Notes: This is valid for $p$ a non-negative integer. Knowing enough about Bernouli numbers allows explicit formulas but I am interested in fairly elementary methods. I think that one involves the series expansion for $e^x$.
‍2) Due to Fermat: Partition into subintervals using points forming a geometric rather than arithmetic progression. I've seen this in two forms:
2.1) Choose $0 \lt a \lt b$ and divide using $a \lt ar \lt ar^2 \lt \cdots\lt ar^N=b$ so $r=\sqrt[N]{b/a}$. The widths of
the intervals form a geometric progression of common ratio $r$. The values of $x^p$ at the points of division form a geometric progression of common ratio $r^p$. Thus the sum of rectangle areas using left endpoints gives as a lower bound for $\int_a^bx^p$ the geometric series with $N$ terms,  first term $a^{p+1}(r-1)$ and ratio $r^{p+1}$.  Using righthand enpoints gives a similar upper bound with first term $a^{p+1}(r-1)r^p$. With very little effort one arrives at
$$ \frac{(b^{p+1}-a^{p+1})(r-1)}{r^{p+1}-1}  \lt \int_a^bx^pdx \lt  \frac{(b^{p+1}-a^{p+1})(r-1)r^p}{r^{p+1}-1}.$$
If $p+1$ is a positive integer we have
$$ \frac{b^{p+1}-a^{p+1}}{1+r+r^2+\cdots+r^p} \lt \int_a^bx^pdx \lt \frac{(b^{p+1}-a^{p+1})r^p}{1+r+r^2+\cdots+r^p}.$$
Now let $N$ go to infinity sending $r$ to $1$ and squeezing to $\int_a^bx^pdx=\frac{b^{p+1}-a^{p+1}}{p+1}$.
This particular approach requires $0 \lt a$. It is an easy extra step to extend the result to  rational values of $p$ (except for the challenging $p=-1$) using $\frac{r^{p/q}-1}{r-1}=\frac{u^p-1}{u-1}/\frac{u^q-1}{u-1}$ for $u=r^{1/q}$.
2.2) Similar except now divide the interval $[0,b]$ using a value $0 \lt r \lt 1$ and infinitely many points $ \dotsb br^3 \lt br^2 \lt br \lt b$. Now one has infinite geometric series and the rest proceeds similarly to before letting $r$ increase to $1$.
So that is the flavor of what I am asking about. I do not think this is a big list question unless there are a large number of approaches I have not seen.
CONTINUED To recap, we already know the answer, $\frac{b^{p+1}-a^{p+1}}{b-a}$, which we want for the area $A$ of the region under $x^p$ for $a \le x \le b$, we just want to prove it.  (Assume for ease that $0 \lt a$.) A partition $P$ of $[a,b]$  is a sequence $a=x_0 \lt x_1 \lt \cdots \lt x_n=b$. The mesh $m(P)$ of $P$ is $\max(x_{i}-x_{i-1}).$ (There is rarely a reason to have unequal intervals, but Fermat gave one.) We use the sub-intervals, in two ways,  as the bases of an assemblage of rectangles with heights determined by the endpoints.  Since $x^p$ is monotonic, one is covered by the region and the other covers it. So the two areas provide a lower and an upper bound.

$$ \sum_1^nx_{i-1}^p(x_i-x_{i-1}) \lt A \lt \sum_1^nx_{i}^p(x_i-x_{i-1}).$$ If we manage to compute or bound these bounds and show that, when the mesh goes to zero, they have a common limit (the one we expect), we are done. The actual bounds we compute are of value only for the interesting, but secondary, topic of speed of convergence. And anyway, if $m(P) \lt \epsilon$ then the difference between the two bounds is less than $(b-a)(b^p-(b-\epsilon)^p),$ which converges to zero. (For $p \lt 0$ use $a-(a+\epsilon)^p$.)
So I propose to instead assign to each sub-interval $[u,v]$ the height $h(u,v)=\frac{v^{p+1}-u^{p+1}}{(p+1)(v-u)}$ and "compute" $\sum_1^nh(x_{i-1},x_i)(x_i-x_{i-1})$ which immediately collapses to, of course, $\frac{b^{p+1}-a^{p+1}}{p+1}$.

Establishing that this has any relevance  requires showing that the height $h(u,v)$ is between $u^p$ and $v^p$. This is easy in practice if one simplifies. If you simplify first, then the whole thing looks like magic until you see what was done.
So for $p=5$, obviously $$u^5 \lt \frac{v^5+v^4u+v^3u^2+v^2u^3+vu^4+u^5}{6} \lt v^5.$$ OK, so what? Why not use the average, the geometric mean or $\left(\frac{u+v}{2}\right)^5$? Well,  $(v-u)h(u,v)=\frac{v^6-u^6}{6}$ so $\sum_1^nh(x_{i-1},x_i)(x_i-x_{i-1})$ collapses to $\frac{b^6-a^6}{b-a}$.
About as easily
\begin{gather*}
\frac{1}{\sqrt{v}} \lt \frac{2}{\sqrt{u}+\sqrt{v}} \lt \frac{1}{\sqrt{u}} \\
\frac{1}{v^2} \lt \frac{1}{uv} \lt \frac{1}{u^2} \\
\frac{1}{v^4} \lt \left( \frac{1}{v^3u}+\frac{1}{v^2u^2}+\frac{1}{vu^3}\right)/3 \lt \frac{1}{u^4}.
\end{gather*}
It is slightly more fun to show that
$$\sqrt{u} \lt \frac{2(v+\sqrt{vu}+u)}{3(\sqrt{v}+\sqrt{u})} \lt \sqrt{v}.$$
SO:

Is this line of argument valid? Is it interesting? Have you seen it before?

To its credit I'll say that it does not show preference to any particular partition and uses nothing more complex than the two historic treatments above (although maybe it benefits from a modern frame of reference). Also, rather than carefully converging to the correct answer as the partition evolves, it just starts there and stays unaffected. I don't immediately see that it can be applied to any other definite integrals. But the case of $x^p$  has a certain primary importance.

REPLY [4 votes]: I think it is also worth looking at the case $p=-1$, in the spirit of your context. It also provides a nice way to introduce the logarithm and the exponential function (quite closely  to the historical development). 
Premise: Suppose we have just proved the first elementary facts about the Riemann integral, including integrability of monotonic functions and the bound $\int_a^bf(t)dt\le(b-a)\sup_{a\le t\le b} f(t)$. Comparing Riemann sums, it also follows easily the formula of linear change of variable:  $\int_{\lambda x}^{\lambda y}f(t)dt=\lambda\int_x^yf(\lambda t)dt$, also valid for negative $\lambda$ due to the convention $\int_b^a=-\int_a^b$. 
Then we may look at the function $L(x):=\int_1^x{dt\over t}$, for $x>0$. By a linear change of variable we find a logarithmic law:
$$L(xy)=\int_1^x{dt\over t}+\int_x^{xy}{dt\over t}=L(x)+L(y).$$
It is clear that $L$ is strictly increasing, in fact unbounded, and it is continuous (bounding the integral), therefore a homeomorphism $\mathbb{R}_+\to\mathbb{R}$.
Also
$$L\big(1+{x }\big)=\int_1^{1+x }{dt\over t}={x }(1+o(1)),\quad (x\to0).$$
Hence for any $x\in\mathbb{R}$
$$ L\Big(\big(1+{x\over n }\big)^n\Big)=nL\big(1+{x\over n }\big)=x+o(1),\quad (n\to\infty).$$
One concludes that $\big(1+{x\over n }\big)^n$ converges to the inverse function of $L$, for any $x\in\mathbb{R}$ (thus a strictly increasing homeo $E:\mathbb{R}\to\mathbb{R}_+$ with the exponential law $E(x+y)=E(x)E(y)$).<|endoftext|>
TITLE: Monic polynomial with integer coefficients with roots on unit circle, not roots of unity?
QUESTION [24 upvotes]: There are certainly non-monic polynomials of degree 4 with all roots on the unit circle, but no roots are roots of unity; $5 - 6 x^2 + 5 x^4$ for example.
Now, for a monic polynomial of degree $n$, this is impossible (I think).
So, my question is, given a monic polynomial with integer coefficients of degree $n$,
what is the maximal number of roots that can lie on the unit circle, and not be roots of unity?
For example, $1 + 3 x + 3 x^2 + 3 x^3 + x^4$ has two roots on the unit circle, and two real roots.

REPLY [4 votes]: This is an adelic comment based on algebraic dynamics. Each such polynomial $f(x)\in\mathbb{Z}[x]$ induces an algebraic dynamical system (i.e. an automorphism of a compact abelian group). If $f(x)$ has no roots that are roots of unity, then the system is known to have strictly positive entropy. The entropy is calculated by adding up the logs of the absolute values of the eigenvalues. If they are all on the unit circle then you'd get zero, which is wrong. So where is positive entropy coming from? It's from the $p$-adic eigenvalues of $f(x)$ (see Automorphisms of solenoids and p-adic entropy, Lind and Ward, Ergodic Th. & Dynam. Syst. 8 (1988), 411-419).
In other words, if $f(x)$ has all of its (complex) roots on the unit circle (and they are not roots of unity), then it must have $p$-adic roots having $p$-adic norm strictly greater than one. For the polynomial mentioned in the first sentence, there is a 5-adic root with norm > 1. In some sense, this is the correct (adelic) formulation of Kronecker's theorem mentioned above.<|endoftext|>
TITLE: Dependence of the blow-up time of existence of an ODE with respect to initial condition.
QUESTION [7 upvotes]: Let $V$ be a smooth vector field on $\mathbb{R}^n$. Assume that the maximal solution to the Cauchy problem $x'=V(x), x(0)=x_0$ exist only for $t\in [0,T)$, where $T$ is finite, denote this time by $T(x_0)$. Is $T$ continuous with respect to $x_0$ ? Is it $C^1$ ?

REPLY [6 votes]: No, but you can say it is lower semicontiuous, even wrto the initial time (that is, the optimistic situation: perturbing a little the initial data the existence is ensured almost up to $T$ , and could even be much greater) .
Precisely, given a Banach space $E$, an open set $\Omega\subset \mathbb{R}\times E$ and  $f:\Omega\subset E\times \mathbb{R}\rightarrow E$ in the Cauchy-Lipshitz-Picard hypoteses, for any $(t_0,x_0)\in \Omega$
the Cauchy problem $$u(t_0)=x_0$$ for the ODE  $$\dot u =f( t, u(t))$$ 
admits a maximal solution defined in an interval $\big(\tau_*(t_0,x_0), \tau^*(t_0,x_0)\big)\subset\mathbb{R}$, where 
$$\tau _ *:\Omega\to [-\infty,0 ) $$ 
is upper semicontinuous and 
$$\tau ^ *:\Omega\to (0,+\infty]$$ 
is lower semicontinuous. This amount to saying that: the domain of the "general solution" $\xi:\Xi\subset \Omega\times\mathbb{R}\rightarrow E$ defined as $\xi(t_0,x_0,t):=u(t)$ with the solution $u(t)$ of the above Cauchy problem, that is the set
$$\Xi:=\{ (s,x,t)\in \Omega\times\mathbb{R} \, :\, \tau _ * (s,x) < t < \tau ^ *(s,x) \}$$
(that is the zone between the graph of $\tau _ * $ and the graph of $\tau ^ *$), is an open set.<|endoftext|>
TITLE: Morphisms between formal dg-algebras
QUESTION [7 upvotes]: Suppose we are given a map $f:A \rightarrow B$ between two dg-algebras which are formal.
Is the map $f$ also "formal" in some sense?
More precisely can we find isomorphisms $\phi_A:A\rightarrow H^\bullet(A)$ and $\phi_B:B\rightarrow H^\bullet(B)$ in the derived category of dg-algebras such that 
$$H^\bullet(f) \circ \phi_A= \phi_B \circ f $$
holds in the derived category of dg-algebras?
I fear that this may be wrong in general. What would be a counter example? Are there criteria which ensure that $f$ is "formal" in this sense?

REPLY [7 votes]: Let me give you an example which has a topological flavour. Let us consider, the De-Rham complex of differential forms on a sphere $S^n$, we denote it $A^*(S^n)$, it is a formal commutative differential graded algebra. Let us look at the morphisms of commutative differential graded algebras between $A(S^2)$ and $A(S^3)$ into the derived category.
A good way to compute them is to take a cofibrant replacement of $A(S^2)$, such a cofibrant resolution is given by the algebra $(\mathbb{R}[x_2]\otimes\Lambda(x_3),D)$ where the differential satisfies $D(x_2)=0$ and $D(x_3)=x_2^2$.
And as $A(S^3)$ is formal you can replace it by its cohomology algebra which is the exterior algebra $\Lambda(y_3)$. It is easy to see that a morphism of algebra $\phi$ between these two algebras is completely determined by the image of $x_3$ thus by a real number $\lambda$ such that $\phi(x_3)=\lambda.x_3$. We have proved:
$$[A(S^2),A(S^3)]_{CDGA}\cong\mathbb{R},$$
while $H(\phi)=0$.
This example has a topological flavour because it is a way (not the best of course) to show that $\pi_3(S^2)\otimes \mathbb{R}\cong \mathbb{R}$. 
People in rational homotopy theory have studied the concept of formality for morphisms:

Vigué-Poirrier, Micheline
Formalité d'une application continue. 
C. R. Acad. Sci. Paris Sér. A-B 289 (1979), no. 16
Félix, Yves; Tanré, Daniel
Sur la formalité des applications.
Publ. U.E.R. Math. Pures Appl. IRMA 3 (1981), no. 2, exp. no. 1, 45 pp. 
Oprea, John F.
DGA homology decompositions and a condition for formality.
Illinois J. Math. 30 (1986), no. 1, 122–137. 

Edit: if you want to play with dgas, then the only thing that you have to change in the example above is the cofibration resolution of the algebra $A(S^2)$. You start with $(T(x_2,x_3),D)$ the tensor algeba on $2$ generators with differential $D(x_2)=0$ and $D(x_3)=x_2\otimes x_2$ it has the same cohomology as $A(S^2)$ until the differential degree $5$ where you have created a new cycle :$[x_3,x_2]=x_3\otimes x_2-x_2\otimes x_3$ which is not a boundary. Then you have to add a generator $x_4$ to kill this cycle i.e. you put $D(x_4)=[x_3,x_2]$. And so on and so forth, it is easy to check that all the generators you add to $(T(x_2,x_3),D)$ in order to build a cofibrant resolution of $A(S^2)$ will have degree at least $4$.
With this resolution in hand you check that a homotopy class of morphism of dgas is also completely determined by the image of $x_3$.<|endoftext|>
TITLE: Transcendentality of all irrationals in the Cantor set
QUESTION [11 upvotes]: Hi, I am a student researcher trying to prove that all irrationals within the Cantor set are transcendental. This is grounded, intuitively, in Cantor set members' being non-normal; since algebraic numbers are widely believed to be normal, this implies the transcendentality of the irrationals in the Cantor set. Now, I am at a dead end in trying to prove this, and I would appreciate any pointers (I will give you full credit in whatever end product comes out of this).
I tried recasting the problem in the following way: if I can prove that any irrational algebraic number must have a $1$ in its ternary expansion, then the result I'm after follows. With this, I realized that if an irrational algebraic number has at least one $1$ in its ternary expansion, then it must have infinitely many of them (because irrational algebraics are a field, and otherwise I could add an appropriate irrational algebraic number and get another irrational algebraic without any $1$'s in their ternary expansions -- a contradiction). But I can't really get anywhere past this (tried playing with the field properties, too). Mostly, I just lack/do not know of the tools used in this kind of problem.
Any help very much appreciated!

REPLY [14 votes]: This question was asked by Mahler ("Some suggestions for further research", Bull. Austral. Math. Soc. 29 (1984), no. 1, 101–108). 
See Adamczewski, Bugeaud, "On the decimal expansion of algebraic numbers" (2005) for some things that are known. I have confirmed with a colleague that this is still very much a (widely) open problem. It does not appear that this problem would be appropriate for undergraduate research. 
Of course this is not to say that learning about it would not be beneficial, or that there could not be related research problems that are more tractable. I suggest that you should seek the guidance of an experienced researcher in the field.<|endoftext|>
TITLE: Similarity between Cauchy-Riemann eqs and Hamilton equations.
QUESTION [5 upvotes]: I would like to see if this idea has any applications:
So CR equations are given by:
$$ \frac{\partial u}{\partial x} =\frac{\partial v}{\partial y} ; \ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$
And Hamilton equations are given by:
$$ \dot{p}=-\frac{\partial H}{\partial q} ; \ \dot{q}= \frac{\partial H}{\partial p}$$
Now these conjugate equations ask for an analogy between them, so I tried the next thing:
I am trying to find $u(p,q),v(p,q)$ s.t:
$$\dot{p}=-\frac{\partial H}{\partial q}=\frac{\partial u}{\partial p}=-\frac{\partial v}{\partial q}$$
$$\dot{q}= \frac{\partial H}{\partial p} = \frac{\partial u}{\partial q} =\frac{\partial v}{\partial p}$$
So we can see that we can find a complex-analytic representation of the hamiltonian, $H$:
$$f(p,q)=u(p,q) + \imath (H(p,q)+c)$$
The only restriction is for $u(p,q)$ which by integration we can find it.
Does this represnetation has any applications in mathematical physics or complex analysis? I guess it's already known.
Thanks in advance, Alan.

REPLY [2 votes]: Doesn't this only work if $H(p,q)$ is a harmonic function in the plane?
If that's the case, we have 
$\{u,H\}=\frac{\partial H}{\partial p}\frac{\partial u}{\partial q}-\frac{\partial H}{\partial q}\frac{\partial u}{\partial p}=|\nabla H|^2=-|\nabla u|^2$
where $\{\cdot,\cdot\}$ is the Poisson bracket. Since $\frac{du}{dt}=\{u,H\}$ this tells us how fast $u$ is changing.<|endoftext|>
TITLE: Local finality condition (for re-indexing parameterized colimits)
QUESTION [6 upvotes]: I'm in need of a condition that is analogous to the "finality" condition in the following lemma:
Lemma: A functor $F\colon A\to B$ is final if and only if for any functor $x\colon B\to Set$, the natural map $colim (xF)\to colim(x)$ is an isomorphism.
This lemma could be taken instead as a definition of final functor, but finality is more easily recognized by whether all slice categories of a certain kind are non-empty and connected. I want a recognition principle for a more general kind of finality, which I'm calling local finality.
The more general context requires a bit of notation. If $A$ is a category, write $A-Set$ for the category of functors $A\to Set$. If $F\colon A\to B$ is a functor, write $\Delta_F\colon B-Set\to A-Set$ for the ``composition with $F$" functor, and write $\Sigma_F$ for its left adjoint and $\Pi_F$ for its right adjoint (these three are also sometimes denoted by $F^*, F_!$, and $F_*$ respectively).
The following lemma (obviously) holds for some appropriate definition of locally final. 
Lemma: Suppose that we have a commutative diagram $A\xrightarrow{F}B\xrightarrow{x}C$ and let $G:=xF$. Then $F$ is locally final if and only if the natural map $\Sigma_G\Delta_F\to\Sigma_x$ is an isomorphism.
Is there a nice recognition principle for this kind of ``local finality"? I have a big messy condition obtained by following my nose, but it's of no use. Any help would be greatly appreciated.
Thanks!

REPLY [7 votes]: Since nobody has said so, I will mention that the notion you describe is a particular case of the known --- but perhaps obscure --- concept of Guitart exact square. One can read about it in the nlab page and in an article by Maltsiniotis. Even though the latter article aims to generalize exact squares to a homotopical context, it still gives a good, if somewhat skewed, overview of the concept.
To justify the relative usefulness of exact squares, let me state that instances of that notion characterize: fully faithful functors, (co)final functors, initial functors, absolute Kan extensions, among other concepts (including my personal favourite, absolutely dense functors).
For completeness, I will summarize some characterizations of the notion of exact square. A square of small categories, functors, and natural transformations:
$$ \begin{matrix} 
A & \overset{U}{\longrightarrow} & B \\
\llap{\scriptstyle L}\Big\downarrow  & \big\Downarrow\rlap{\scriptstyle\alpha} & \Big\downarrow\rlap{\scriptstyle R} \\
A' & \underset{D}{\longrightarrow} & B'
\end{matrix} $$
(i.e. a natural transformation $\alpha:R\circ U\to D\circ L$) is called exact if any of the following equivalent conditions hold:

The natural 2-cell induced by $\alpha$ in
$$ \begin{matrix} 
\textrm{Set}^A & \overset{\Delta_U}{\longleftarrow} & \textrm{Set}^B \\
\llap{\scriptstyle \Sigma_L}\Big\downarrow  & \big\Downarrow & \Big\downarrow\rlap{\scriptstyle \Sigma_R} \\
\textrm{Set}^{\smash{A'}} & \underset{\Delta_D}{\longleftarrow} & \textrm{Set}^{\smash{B'}}
\end{matrix} $$
is an isomorphism. That is, the induced natural transformation $\Sigma_L\circ\Delta_U \to \Delta_D\circ\Sigma_R$ is an isomorphism.
For every $x\in A'$, the naturally induced functor on over-categories
$$ A/x=L/x \longrightarrow R/(D(x))=B/(D(x)) $$
is (co)final.
For every $x\in B$, the naturally induced functor on under-categories
$$ x/A=x/U \longrightarrow (R(x))/D=(R(x))/A' $$
is initial.
For each object $x\in B$ and $y\in A'$, and each arrow $f:R(x) \to D(y)$ in $B'$, the category of factorizations $C_{x,y,f}$ is connected (which I, Karol, and a surprisingly large/vocal set of mathematicians take to mean non-empty). Here, the category $C_{x,y,f}$ is defined by:

the objects of $C_{x,y,f}$ are triples $(z,g,h)$ where $z$ is an object of $A$, $g:x\to U(z)$ is a morphism in $B$, and $h:L(z)\to y$ is a morphism of $A'$, such that $D(h)\circ\alpha_z \circ R(g)=f$;
a morphism $(z,g,h)\to (z',g',h')$ in $C_{x,y,f}$ is an arrow $k:z\to z'$ such that $g'= U(k)\circ g$ and $h=h'\circ L(k)$.


For all objects $x\in B$ and $y\in A'$, the natural map from the coend
$$ \int^{a\in A} B(x,U(a))\times A'(L(a),y) \longrightarrow B'(R(x),D(y)) $$
is an isomorphism of sets.


Before proceeding, observe that condition 4 is simply a restatement of conditions 2 and 3. In fact, the categories of factorizations $C_{x,y,f}$ defined in 4 are exactly the categories whose connectedness must be checked to ensure that the functor in condition 2 is cofinal (respectively, that the functor in condition 3 is initial). More precisely, the categories $C_{x,y,f}$ are the under-categories $a/F$ of the functor $F$ in condition 2 (respectively, the over-categories of the functor in condition 3) for objects $a$ in the codomain of $F$. I feel this both motivates and gives a nice way to remember the definition of $C_{x,y,f}$.
Then $A\overset{F}{\rightarrow}B\overset{x}{\rightarrow}C$ is locally final (in the sense David Spivak states) if and only if the the square
$$ \begin{matrix} 
A & \overset{F}{\longrightarrow} & B \\
\llap{\scriptstyle G}\Big\downarrow  & \big\Downarrow\rlap{\scriptstyle\textrm{id}_G}\ & \Big\downarrow\rlap{\scriptstyle x} \\
C & \underset{\textrm{id}_C}{\longrightarrow} & C
\end{matrix} $$
(filled by the the identity 2-cell on $G=x\circ F$) is exact. Under this interpretation, condition 4 above is exactly the condition given in Karol Szumiło's answer.
Addendum: To finish off, here are a few further equivalent characterizations of the exactness of the original square ($\alpha:R\circ U\to D\circ L$) drawn at the top of this answer:

For any cocomplete category X, the natural 2-cell induced by $\alpha$ in
$$ \begin{matrix} 
X^A & \overset{\Delta_U}{\longleftarrow} & X^B \\
\llap{\scriptstyle \Sigma_L}\Big\downarrow  & \big\Downarrow & \Big\downarrow\rlap{\scriptstyle \Sigma_R} \\
X^{\smash{A'}} & \underset{\Delta_D}{\longleftarrow} & X^{\smash{B'}}
\end{matrix} $$
is an isomorphism. Note that for $X=\textrm{Set}$, we recover condition 1 above.
The natural 2-cell induced by $\alpha$ in
$$ \begin{matrix} 
\textrm{Set}^A & \overset{\Pi_U}{\longrightarrow} & \textrm{Set}^B \\
\llap{\scriptstyle \Delta_L}\Big\uparrow  & \big\Uparrow & \Big\uparrow\rlap{\scriptstyle \Delta_R} \\
\textrm{Set}^{\smash{A'}} & \underset{\Pi_D}{\longrightarrow} & \textrm{Set}^{\smash{B'}}
\end{matrix} $$
is an isomorphism.
For any complete category X, the natural 2-cell induced by $\alpha$ in
$$ \begin{matrix} 
X^A & \overset{\Pi_U}{\longrightarrow} & X^B \\
\llap{\scriptstyle \Delta_L}\Big\uparrow  & \big\Uparrow & \Big\uparrow\rlap{\scriptstyle \Delta_R} \\
X^{\smash{A'}} & \underset{\Pi_D}{\longrightarrow} & X^{\smash{B'}}
\end{matrix} $$
is an isomorphism. Note that for $X=\textrm{Set}$, we recover the preceding condition.
The opposite square
$$ \begin{matrix} 
A^{\textrm{op}} & \overset{L^{\textrm{op}}}{\longrightarrow} & {A'}^{\textrm{op}} \\
\llap{\scriptstyle U^{\textrm{op}}}\Big\downarrow  & \big\Downarrow\rlap{\scriptstyle\alpha^{\textrm{op}}} & \Big\downarrow\rlap{\scriptstyle D^{\textrm{op}}} \\
B^{\smash{\textrm{op}}} & \underset{R^{\textrm{op}}}{\longrightarrow} & {B'}^{\smash{\textrm{op}}}
\end{matrix} $$
is exact. Note that the preceding 3 conditions and condition 1 applied to this opposite square give diagrams with categories of presheaves (contravariant functors) on $A$, $B$, $A'$, and $B'$, instead of categories of covariant functors on those categories. In fact, a common (equivalent) definition of exactness is the analog of condition 1 for presheaves: the 2-cell in the square
$$ \begin{matrix} 
\widehat{A} & \overset{\Sigma_{U^{\textrm{op}}}}{\longrightarrow} & \widehat{B} \\
\llap{\scriptstyle \Delta_{L^{\textrm{op}}}}\Big\uparrow  & \big\Downarrow\rlap{\scriptstyle} & \Big\uparrow\rlap{\scriptstyle \Delta_{R^{\textrm{op}}}} \\
\widehat{A'} & \underset{\Sigma_{D^{\textrm{op}}}}{\longrightarrow} & \widehat{B'}
\end{matrix} $$
is an isomorphism.<|endoftext|>
TITLE: What is "Teichmüller Theory" and its history?
QUESTION [36 upvotes]: What is "Teichmüller Theory"? What part has been worked out / foreseen  by O. Teichmüller himself and what is further development? Is there some current work which might be considered as continuation/completion of this theory? 
Background The question might be seen as too naive and can be answered by google  or Wikipedia, but I have it in mind for a long time and do not think  that it is that much simple. Let me explain what is puzzling me: Teichmüller space is very close to moduli space of Riemann surfaces ("The Teichmüller space is the universal covering orbifold of the (Riemann) moduli space.") and reading some sources make me expression that  "Teichmüller Theory" is  everything which is related to the moduli space of Riemann surfaces. Is it really like this ? If it is true it does not seem to me good name since "theory" should be something not so diverse as current research on moduli spaces of Riemann surfaces.
On the other hand what I heard about the contribution of Teichmüller himself - it is introduction of the Teichmüller metric by means of quasiconformal maps. (See Wikipedia).
It is beautiful result, but it is kind of "theorem", not "theory", so probably there is something more? which I am missing?
The question might be considered as background to  What are some Applications of Teichmüller Theory?

REPLY [46 votes]: First of all, let me recommend a book: J. Hubbard, Teichmüller theory, vol. 1.
Let me try to list briefly Teichmüller's own contribution to Teichmüller theory.
Bers's papers of 1960-s are good primary sources. The few papers of Teichmüller himself
that I read are also exciting, but my poor knowledge of German does not allow me to
read all of them. 
Perhaps the main contribution is the introduction of Teichmüller's space
(instead of the much more complicated moduli space). It is simply connected!
The second main contribution is the definition of the Teichmüller metric on this space.
The metric is defined using
a solution of an extremal problem: finding a quasiconformal homeomorphism in a
homotopy class with smallest dilatation. Such problems in plane domains were first
considered by Grötzsch. Teichmüller's contribution was
a) considering them on compact Riemann surfaces, and
b) describing the extremal map in terms of a certain quadratic differential.
He also established existence and uniqueness of the extremal mapping with a very
original argument.
Teichmüller distance is defined as $(1/2)\log K$, where $K$ is the extremal
dilatation. 
Teichmüller died young (he was killed or MIA in the Eastern front, somewhere near Kiev
in 1943), and many of his principal papers contain a lot of heuristic arguments.
The subject was developed by Ahlfors and Bers in 1950-s. They rigorously introduced
the analytic structure on Teichmüller spaces, and proved in particular that the Teichmuller
space of surfaces of genus $g>1$ is isomorphic to a domain in $C^{3g-3}$ which is
homeomorphic to $R^{6g-6}$.
They identified the cotangent space as a space of quadratic differentials. Later Royden proved
that the Teichmüller distance coincides with the Kobayashi distance.
The crucial technical tool, existence and analytic dependence on parameters of 
the homeomorphic solution of the Beltrami equation with $L^\infty$ norm of the coefficient less
than 1,
which people call sometimes the "Measurable Riemann theorem", was not available in
Teichmüller's time. It was published for the first time by Boyarski in 1955.
This more or less constitutes the original "Teichmüller theory".
Later the meaning of the term substantially expanded, to include almost everything
about the moduli spaces. 
EDIT. The good news is that all principal papers of Teichmüller are now available in English:
MR3560242
Handbook of Teichmüller theory. Vols. IV,V,VI. 
European Mathematical Society (EMS), Zürich, 2016. (Each volume contains translations of several papers of Teichmüller)
And with comments.<|endoftext|>
TITLE: Is it known if the absolute Galois group is "divisible"?
QUESTION [9 upvotes]: The definitions of a divisible group that I have seen all seem to assume abelian is an a priori property of the group. My question is as to whether or not it is known that--given a non-torsion element $\tau\in\mathbf{G}$* and $n\in\mathbb{N}$, do we know if $\exists \tau'\in\mathbf{G}_\mathbb{Q}$ such that $\tau=\tau'^n$?
The motivation is that--if this never happens--then the fields $\overline{\mathbb{Q}}^\tau$ should be sort of "maximal" subfields of $\overline{\mathbb{Q}}$ since fixing by the generator would be equivalent to fixed by any power of it.
Edit: anon has noted that quotients of divisible groups are divisible, so we can lay to rest that $\mathbf{G}_\mathbb{Q}$ is not divisible. Can we tell if the opposite is true? I.e. the answer to "can I find such a $\tau '$ given $n$" is "no", but how about the somewhat interesting and related question:  If I have a cyclic subgroup of the absolute Galois group, $C=\langle\tau\rangle$ can we find a maximal (with respect to inclusion), cyclic group containing $C$? Of course this is equivalent to a minimal field in a chain, so it seems like if that is so something interesting must be going on.
The other related question which deals with the original spirit of the problem is:  are there $\tau$ such that $\langle \tau^n\rangle$ fixes $k\subseteq\overline{\mathbb{Q}}$ then $\langle\tau\rangle$ fixes $k$? i.e. is $k=\overline{\mathbb{Q}}^\tau$ equivalent to $k=\overline{\mathbb{Q}}^{\tau^n}$ possible for some $\tau$ non-torsion? If so can such elements be characterized?
The immediate observation is that, by the FTGT, if we identify $\langle\tau\rangle\cong\mathbb{Z}$ and write $[n]$ for the index $n$ subgroup, that--in the topology of $\mathbf{G}_\mathbb{Q}$--necessarily $\overline{[n]}=\overline{[1]}$. I.e. the monothetic group $\overline{[1]}$ has every element is a generator. If $\overline{[1]}$ were connected, of course this would be trivial to check since the set of generators has Haar measure 1 in this case, and a finite index subgroup has positive Haar measure, and so contains a generator, but the topology here is totally disconnected so it is not easy to see one way or the other.
*My TeX was not rendering properly when I wrote $\mathbf{G}_\mathbb{Q}$ here, so I just included it as a sidenote rather than have it look like a mess.

REPLY [3 votes]: Here is a result of Haran that might be relevant :
For almost all $\sigma$ in the absolute Galois group of $\mathbb{Q}$ the fixed field $\bar{\mathbb{Q}}^\sigma$ of $\sigma$ in $\bar{\mathbb{Q}}$ has no proper cofinite subextensions. (I.e. $L\subsetneq \bar{\mathbb{Q}}^\sigma$ implies  $[\bar{\mathbb{Q}}^\sigma:L]=\infty$.)
Here almost all is in the sense of the Haar measure on the absolute Galois group.<|endoftext|>
TITLE: Are there countably many diffeomorphism classes of finite radius balls of complete Riemannian manifolds?
QUESTION [5 upvotes]: Suppose $M$ is a smooth complete Riemannian manifold and $x$ is a point in $M$.  For any positive radius $r$ we consider the open ball $B(x,r)$ centered at $x$ with radius $r$.
If we ignore the Riemannian structure on $B(x,r)$ and consider it only as a smooth manifold.  Are there only a countable number of diffeomorphism classes it can belong to?
An afirmative answer includes as a particular case the fact that there are only countably many compact manifolds up to diffeomorphism.  This follows from Cheeger's finiteness theorem because on any such manifold has bounded curvature, volume, and injectivity radius.
It is well known that there are uncountably many diffeomorphism classes of smooth manifolds (even homeomorphic to $\mathbb{R}^4$; Are there uncountably many surfaces?).
My motivation for this is that I'm trying to study a particular concept of a random complete  Riemannian manifold with a distinguished point.  The starting point is to topologize the space of such manifolds (and I need something stronger then Lipschitz) and things would be simpler if the answer to this question was yes (and I had a short proof or reference :) ).   Also, the question seems interesting in its own right.

REPLY [4 votes]: The answer to this question should be negative for $M=S^2$. I will sketch a tentative proof of this claim that would use a couple of simple lemmas.
Lemma 1. The set of closed totally disconnected subsets $X_{dis}$ in the interval $[0,1]$ modulo diffeos of $[0,1]$ is uncountable.
Lemma 2. The set of surfaces $S^2\setminus X_{dis}$ where $X_{dis}$ as above is uncountable (we consider here $X_{dis}$ lying in a segment $I\cong [0,1]$ in $S^2$, $I\subset S^2$.)
Lemma 3. For any $X_{dis}\subset [0,1]$ there is a smooth non-negative function $f$ on $[0,1]$ whose set of zeros is $X$.
All these lemmas are very straightforward (Lemma 2 requires a bit of thinking but it should be correct). I want to use them to construct for any $X_{dis}\subset S^2$ a metric on $S^2$ such that for a certain point $O\in S^2$ the ball $B(O,1)$ is $S^2\setminus X_{dis}$. In order to do this start with a metric on $S^2$ such that $B(O,1)$ is $S^2\setminus [0,1]$, where $[0,1]$ is an interval smoothly embedded in $S^2$.  Note that the equidistants emanating from $O$ come to $[0,1]$ from two sides. So just "speed up" them from one side so that they hit all the points from $[0,1]\setminus X_{dis}$ before the "other side" of equidistant hits them; at the same time do so that all points of $X_{dis}$ are hit at the same time from both sides. (to do this explicitly I would use something like Lemma 3).
I am afraid that this sketch is not so well phrased, but I hope that the idea is more-less clear.<|endoftext|>
TITLE: Hausdorff group topologies on finitely generated groups
QUESTION [6 upvotes]: Suppose $G$ is a finitely generated Hausdorff topological group. Must $G$ be first countable (or perhaps a sequential space)? What if we restrict to the abelian case?
I wonder if this is even true for the additive group of integers $\mathbb{Z}$. There certainly are non-discrete, Hausdorff group topologies on $\mathbb{Z}$ where a basis at $0$ consists of subgroups (such as that used in Furstenberg's proof of the infinitude of primes). On the other hand, determining if there is a Hausdorff group topology that makes a given sequence converge to $0$ is non-trivial. For instance, it is known that the sequence of squares $n^2$ can't converge to $0$ in any Hausdorff group topology and that if there is a Hausdorff group topology on $\mathbb{Z}$ such that the sequence of primes $2,3,5,...,p,..$ converges to $0$, then the twin prime conjecture is false.

REPLY [3 votes]: It is not the answer but an important point, too long for a comment. There is no Hausdorff group topology on Z so that the sequence of prime numbers $2, 3, 5 ...$ converges to zero. If there were such topology, there would be finitely many pairs of primes of each given gap $k$. But that contradicts Zhang's Gap Theorem<|endoftext|>
TITLE: Is this height-transcendence-degree inequality true without AC ?
QUESTION [5 upvotes]: Let $R$ be a $k$-algebra ($k$ a field) and a domain of finite Krull dimension. In 
$\quad$ Krull dimension less or equal than transcendence degree? 
it is shown that 
$$\text{Krull-dim}(R) \le \text{trans.deg}_k Quot(R).\tag{*}$$
From the paper [1]  I learned the inequality 
$$\text{height}(P) + \text{trans.deg}_k Quot(R/P)  \le \text{trans.deg}_k Quot(R)\tag{**}$$
for all prime ideals $P \subseteq R$. Obviously $(\ast\ast)$ strengthens $(\ast)$. 
The paper gives as reference for $(\ast\ast)$ a combination of two results: First, in [2, Chap. IV, §3, Cor. 1] the inequality is proved for valuation rings. The general case then follows by an embedding theorem for domains into valutation rings [3, Chap. I, (11.9)]. 
However, the proof of the embedding theorem uses Zorn's Lemma. Since $(\ast)$ can be proved without the Axiom of Choice, I wonder: 
Question 1: Is $(\ast\ast)$ also true without assuming the Axiom of Choice ? 
Question 2: Are there alternative references for $(\ast\ast)$ than whose given by Wadsworth ? 
References: 

Wadsworth: The Krull dimensions of tensor products of commutative algebras over a field. J. London Math. Soc. (2), 19(1979), 391-401 
Zariski, Samuel: Commutative Algebra II 
Nagata: Local Rings

REPLY [3 votes]: Here is a proof of (**) by induction on the height of $P$.
If $P=0$, the inequality (**) is obvious. Let $P$ be a prime ideal of $R$ of height $d \geq 1$, and consider a chain of prime ideals $0=P_0 \subset P_1 \subset \ldots \subset P_d = P$ of length $d$ in $R$. The domain $R/P_1$ has finite Krull dimension, and the prime ideal $P/P_1$ has height $d-1$ so by the induction hypothesis
\begin{equation*}
{\rm height}(P/P_1) + {\rm trdeg}_k (R/P) \leq {\rm trdeg}_k(R/P_1).
\end{equation*}
It remains to prove that ${\rm trdeg}_k(R/P_1) \leq{\rm trdeg}_k(R)-1$. Let $x_1,\ldots,x_e$ be elements of $R$ whose images in $R/P_1$ are algebraically independent over $k$. Let $y$ be any element of $P_1 \backslash \{0\}$. Consider the map $\phi : k[X_1,\ldots,X_e,Y] \to R$ sending $X_i$ to $x_i$ and $Y$ to $y$. If $Q = \sum_j Q_j(X_1,\ldots,X_e) Y^j$ lies in the kernel of $\phi$, then reducing modulo $P_1$ gives $Q_0(x_1,\ldots,x_e)=0$, so that $Q_0=0$. Since we are working in a domain and $y \neq 0$, an easy induction gives $Q=0$. Thus $x_1,\ldots,x_e,y$ are algebraically independent over $k$, which concludes the proof.<|endoftext|>
TITLE: Scheme defined over $\mathbb{Z}$
QUESTION [6 upvotes]: I'd like to check a definition:
If $X$ is a scheme, what does it mean to say that $X$ is "defined over $\textrm{Spec }\mathbb{Z}$"? Is this a precise statement? Certainly this statement requires that $X$ is finite type over $\textrm{Spec }\mathbb{Z}$.
If $X$ is a projective or affine variety over $\textrm{Spec }\mathbb{C}$ (with choice of embedding) we can ask if the coefficients of the equations defining it are integers, and maybe call such a scheme to defined over $\textrm{Spec }\mathbb{Z}$. Other than not being "coordinate free", it is also easy to get schemes that are better said to be "defined over $\textrm{Spec }\mathbb{Z}[1/N]$" (in particular, there are lots of examples when a construction is defined over the latter scheme, and a goal is to make it defined over $\textrm{Spec }\mathbb{Z}$).
$\textbf{Question:}$ If $X$ is a scheme, does the phrase "defined over $\textrm{Spec }\mathbb{Z}$" mean that the structure map to $\textrm{Spec }\mathbb{Z}$ is fppf? (or is the base change to say $\textrm{Spec }\mathbb{Q}$ of a scheme with such a structure map).

REPLY [3 votes]: A scheme $X$ is defined over $\mathbb Z$ if there is a scheme $X_0$ over $\mathbb Z$ (usually of finite type) such that $X$ is obtained from $X_0$ by some base change $S \to Spec(\mathbb Z)$.  For example, if $X$ is a scheme over a field $K$ then $X$ should be isomorphic to $X_0\times_{\mathbb Z}K$.<|endoftext|>
TITLE: Is every Fibonacci number "Fibonacci-prime"?
QUESTION [6 upvotes]: Is it true that if a Fibonacci number $F_{n}$ divides the product of two Fibonacci numbers, then it must divide at least one of them? 
Is it true that for all $n \ne 1,2,6,12$, there exists a prime divisor $p$ of $F_{n}$ such that the entry point (first appearance as a divisor in the Fibonacci sequence) of $p$ is at position $n$?

I can see that the second statement implies the first (at least for $n \ne 1,2,6,12$).
Also, Carmichael's theorem says that for all $n \ne 1,2,6,12$, there exists a prime divisor $p$ of $F_{n}$ such that $p \nmid F_k$ for (positive integer) $k \lt n$, and the second statement is equivalent to a sort of counterpart going in the other direction: 
for all $n \ne 1,2,6,12$, there exists a prime divisor $p$ of $F_{n}$ such that $p \nmid F_k$ for $k \gt n$ save only for the obvious exceptions where $k$ is a multiple of $n$ (since $F_{n}\mid F_{k}$ when $n \mid k$).
These seem like natural questions to ask but I haven't seen them addressed.

REPLY [14 votes]: As François Brunault said in his comment, your #2 is true: Carmichael's theorem, which establishes the existence of a primitive prime divisor of $F_n$ for every $n\ne1,2,6,12$, together with $\gcd(F_m,F_n) = F_{\gcd(m,n)}$, shows that no Fibonacci numbers are divisible by a primitive prime divisor of $F_n$ except $F_{kn}$.
This gives an answer to #1 as well: suppose that $F_n \mid F_a F_b$, and let $p$ be a primitive prime divisor of $F_n$. Then $p \mid F_a$ or $p\mid F_b$, and so one of $a$ or $b$ must be a multiple of $n$, so that $F_n\mid F_a$ or $F_n\mid F_b$. This proof doesn't work for $n=1,2,6,12$, but $n=1,2$ are trivial, and the cases $n=6,12$ follow by considering the (periodic) Fibonacci sequence modulo $8$ and $144$, respectively. (In the latter case it may be easier to examine the periodic sequence $\gcd(F_n,144) = F_{\gcd(n,12)}$.)
Note that this proof gives the generalization that if $F_n$ divides the product of any finite number of Fibonacci numbers, then it must divide one of them - except for $n=6$ and $n=12$ for which this is false!<|endoftext|>
TITLE: Equations of the secant variety
QUESTION [8 upvotes]: Let $X\subset\mathbb{P}^N$ be an irreducible nondegenerate (i.e. not contained in a hyperplane) projective complex algebraic variety, and let $\mathrm{Sec}(X)$ be the secant variety of $X$ (i.e. the union of all secant lines of $X$).
Further suppose that the homogeneous ideal of $X$ is generated by forms $F_1,\ldots,F_m$ of degree two.
How do I write the equations of $\mathrm{Sec}(X)$ as a function of $F_0,\ldots,F_m$ ?
Let us focus attention on the simplest case, i.e. when $\mathrm{Sec}(X)$ is a cubic hypersurface. Then we have 
$\mathrm{Sec}(X)=V(G(x_0,\ldots,x_N))$, where $G(x_0,\ldots,x_N)=\sum_{i=1}^m L_i(x_0,\ldots,x_N) F_i(x_0,\ldots,x_N)$. 
The linear forms $L_0,\ldots,L_N$ depend only by $F_0,\ldots,F_N$, but how?
Thanks.

REPLY [8 votes]: If  $I(X)$, the ideal of $X$ is empty  in degrees less than $d$, then there can be no equations
of the secant variety until degree $d+1$, and the ideal in degree $d+1$ consists
of all polynomials $P$ such that all partials of $P$ are in $I_d(X)$. There is a similar
description for the ideal of the secant variety in higher degrees which I call "multi-prolongation",
but one does not know when one has generators for the ideal by this method, and it
becomes very difficult to compute. In your case,
since you have a hypersurface the termination problem does not arise - your cubic is the unique cubic all of whose partial derivatives are in the ideal of $X$.<|endoftext|>
TITLE: On equation f(z+1)-f(z)=f'(z)
QUESTION [20 upvotes]: Original Problem
If $f$ is an entire function such that
$$ f(z+1)-f(z)=f'(z) $$
for all $z$.
Is there a non-trivial solution? ($f(z)=az+b$ is trivial)
And here is something uncertainty
If we use Fourier transform, how to define it to ensure any entire function has a FT? 
Classical FT is defined by
$$ \mathcal{F}[f] = F(\xi) = \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{+\infty}f(z)\mathrm{e}^{-\mathrm{i} \xi z} \mathrm{d} z. $$
This only work for $f \in L^1(\mathbb{R})$. (If improved, it can work for $f \in L^2(\mathbb{R})$.)
I know $\mathcal{F}[\mathrm{e}^{sz}] = \sqrt{2 \pi} \delta(\xi - \mathrm{i}s)$, but I'm not sure about a general definition.

REPLY [51 votes]: Linear functional equations can be solved with Fourier transform.
Let $\lambda_k$ be the roots of the equation $e^\lambda-1=\lambda$. There are infinitely many
such roots. Then 
$$f(z)=\sum_k a_ke^{\lambda_k z}$$
is a solution.
Here the sum can be finite, and $a_k$ arbitrary, or the sum can be infinite, and $a_k$ tend to
zero with such speed that the series converges.
The trivial solution is covered, if you interpret the answer correctly.
The equation $e^\lambda-1=\lambda$ has a DOUBLE root at $0$. For the case of a double root
one includes not only the term $e^{\lambda z}$ but also the term $ze^{\lambda z}$.
Similarly for multiple roots $z^ke^{\lambda z}$. So the root $\lambda=0$ exactly covers
the solution $az+b$.
EDIT: Actually all entire solutions can be represented in this way, for the proof I refer to
Gelfond, Calculus of finite differences, MR0342890, Chap. 5 Sect 7, Thm II and Corollaries.
He gives the equation $f'(z)=f(z-1)$ as an example, but your equation is treated similarly.
EDIT2: The problem can be generalized as follows: Let $\omega$ be a distribution with bounded
support. Equation $f\star w=0$, where $\star$ is the convolution, is called a  convolution equation, and its solutions are called mean-periodic functions (fonctions moyenne-periodiques). The case we are discussing is $\omega(x)=\delta(x+1)-\delta(x)-\delta'(x)$
where $\delta$ is the delta function. The theory of mean periodic functions was created by
Delsarte and Schwartz in 1940-s. Schwartz has a general result that all mean-periodic functions
can be obtained as limit of exponential sums, like in this problem.<|endoftext|>
TITLE: Sufficient Condition for Exponential Decay in Chernoff Bound (Large Deviations)
QUESTION [6 upvotes]: Let $X_i$ ($i=1,...,n$) be a sequence of independent and identically distributed random variables.  Denote $\mu=\mathbb{E}[X_i]$ and  $S_n=\frac{1}{n}\sum_{i=1}^nX_i$.  This question concerns the tail probability $\Pr[S_n \ge \mu+\epsilon]$, where $\epsilon>0$ is a constant.
In the case that $|X_i|$ is bounded with probability one, Hoeffding's Inequality states that the tail probability decays to exponentially fast in $n$ for any $\epsilon > 0$.  [EDIT: The paper http://arxiv.org/pdf/1209.6396v1.pdf (Theorem 1.1) used to state a theorem which made it seem that finite variance was sufficient, but this is not true, and the paper has now been edited to state the sufficient condition more precisely].
A more general sufficient condition is that each variable is sub-Gaussian - see Exercise 6 of http://terrytao.wordpress.com/2010/01/03/254a-notes-1-concentration-of-measure/
With no assumptions at all on $X_i$, the probability may fail to decay exponentially.  For example, under the Cauchy distribution (which has no first or second moment), the probability is bounded away from zero.  However, is there a sufficient condition which is less restrictive than that of boundedness?
This question is related to Large Deviations Theory and the Chernoff bound.  The following upper bound is straightforward:
$$\Pr[S_{n}\ge\mu+\epsilon]=\Pr[e^{\tau\sum_{i=1}^{n}X_{i}}\ge e^{\tau n(\mu+\epsilon)}]\le\frac{\mathbb{E}[e^{\tau\sum_{i=1}^{n}X_{i}}]}{e^{\tau n(\mu+\epsilon)}}=\frac{\mathbb{E}[e^{\tau X_{1}}]^{n}}{e^{\tau n(\mu+\epsilon)}}$$
where $\tau$ is an arbitrary positive constant, and the inequality is Markov's inequality.  For the optimal $\tau$, and under some technical assumptions, one can obtain a lower bound with the same exponential behavior (see "Large Deviations Techniques and Applications" by Dembo/Zeitouni).
Is there a simple sufficient condition for the above Chernoff bound to decay exponentially, which is stronger than boundedness or sub-Gaussian?
[EDITED SINCE ORIGINAL POST - The original answer of "no" was in response to the questions "Does a bounded absolute first moment suffice?".  I certainly agree with this answer of no, but I am still interested in finding a condition as general possible]

REPLY [6 votes]: Note that Cramer's theorem applies in dimension 1 w/out any moment assumptions, but of course the rate function may vanish (for this version see theorem 2.2.3 in Dembo-Zeitouni). So the question is equivalent to the question ``when does the rate function vanish'' (the rate function is $\sup_{\lambda\in R}
(\lambda x-\Lambda(\lambda))$, where $\Lambda(\lambda)= \log E(e^{\lambda X_1}))$. So if 
$\Lambda(\lambda)=\infty$ for all $\lambda\neq 0$, clearly there is no exponential convergence, not just a failure of the upper bound.<|endoftext|>
TITLE: A "holomorphic" Peano curve?
QUESTION [14 upvotes]: A Peano curve is a continuous map $[0,1]\to [0,1]^2$ whose image is the whole square. 
I would like to know if on can obtain "holomorphic" Peano curves. Namely, is it possible to find a continuous map $\phi$ from the unit disk $|z|\le 1$ to $\mathbb C^1$  such that 
$\phi$ is holomorphic for $|z|<1$ and the image of the boundary $|z|=1$ has non-empty interior in $\mathbb C^1$ under the map $\phi$.

REPLY [17 votes]: Here it is:
MR0015154
Salem, R.; Zygmund, A.
Lacunary power series and Peano curves. 
Duke Math. J. 12, (1945). 569–578.<|endoftext|>
TITLE: Number of irreducible and connected components constant in flat families
QUESTION [6 upvotes]: A) Let $f:F\rightarrow S$ be a flat proper morphism of schemes with geometrically normal fibers.  Then supposedly the number of $\textbf{connected}$ components of the geometric fibers is constant.  Why is this?  Without some kind of vanishing of cohomology or information on the base, I don't see why this is true.  
B) Furthermore, supposedly if $F$ is now a flat proper morphism with reduced, connected, nodal curves as geometric fibers, then there is a Zariski open subset of $S$ on which the fibers all have the same number of $\textbf{irreducible}$ components.  Why is this?
Finally, how far can these results be generalized?  For example, is B) true for any flat proper morphism?

REPLY [3 votes]: Part (A) is precisely the content of Theorem 4.7, (iii) in
Deligne, Pierre; Mumford, D., The irreducibility of the space of curves of a given genus, Publ. Math., Inst. Hautes Étud. Sci. 36, 75-109 (1969). ZBL0181.48803.<|endoftext|>
TITLE: Are these numbers irrational and/or transcendental?
QUESTION [6 upvotes]: I am curious to know if the following number is irrational or transcendental:
$$\displaystyle A = \sum_p 2^{-p},$$
where the sum is over all positive primes. A similar question can be asked for any number $k$ other than 2.
Edit: In retrospect and as pointed out by some comments below that it is trivial that $A$ is irrational. The transcendence question seems still relevant.
If we define a random number 
$$\displaystyle b = \sum_n 2^{\rho(n)},$$
where $\rho(n)$ is a Bernoulli random variable defined to be $\rho(n) = -n$ with probability $1/\log n$ and $-\infty$ with probability $1 - 1/\log n$ (so that $2^{\rho(n)} = 0$), for say $n \geq 2$ and define $\rho(1) = -1$. 
Can it be proved that say $n$ is transcendental with probability 1?

REPLY [11 votes]: The number $A$ is irrational, since the characteristic function of the set of prime numbers is not eventually periodic. 
It definitely should be transcendental as it is not a base $2$ normal number, ie the asymptotic frequency of digits is not the same (and algebraic numbers are conjectured to be normal). But this should be open (as also in the link mentioned by Theo Buhler, while I was composing this).
This conjecture would also answer the question on $b$, but Pietro Majer gave a direct argument already; it might however be good to keep this in mind for variations of the question. 
Now, since the above is a bit terse some information around this. There is an interesting rather recent paper on more or less precisely this type of problem by Bailey, Borwein, Crandall, Pomerance 
"On the binary expansions of algebraic numbers" 
Among others they show that the binary distribution of a real algebraic number cannot be too extremely biased more precisely both digit have to occur (up to $N$) at least order of $N^{1/d}$ times where $d$ is the degree of the number. They also mention that thus the degree of 
$$\sum_p \frac{1}{2^{p^k}}$$ is at least $k+1$ (with the convention that the degree of a transcendental is infinite). 
There is more recent work on this by Adamczewski and Faverjon on this, too. 
Finally, a good overview on this type of problem is given in this paper of Adamczewski and Bugeaud (see in particular Chapter 6 on lacunary numbers, that is precisley those of the form considered here, ie the digits are the characteristic function of some set).<|endoftext|>
TITLE: Normal subgroups of $SL_2$ of a polynomial ring
QUESTION [6 upvotes]: What is known about normal subgroups of $SL_2(\mathbb{C}[X])$? Can one hope for a congruence subgroup property, i.e. that every (non-central) normal subgroup contains the kernel of the reduction modulo some ideal of $\mathbb{C}[X]$?

REPLY [9 votes]: [EDIT] These groups have been studied for a long time from various viewpoints, so there is a long paper-trail.  I'd emphasize however that working over the complex numbers is usually similar to working over an arbitrary infinite field.
Finite fields on the other hand occur more often in arithmetic contexts. 
Concerning normal subgroups, the problem looks essentially hopeless, just as in the closely parallel situation of $\mathrm{SL}_2(\mathbb{Z})$ (or its quotient the modular group) first studied over a century ago.    There the congruence subgroup problem has a strongly negative answer, in a sense eventually made precise by Serre.   In the modern study of the congruence subgroup problem (Bass-Milnor-Serre and beyond), the most manageable situation involves higher rank algebraic groups.   But Serre did obtain almost definitive results for rings of integers (and closely related rings of arithmetic interest).    His paper is now available online through JSTOR:  Le probleme des groupes de congruence pour SL2.   Ann. of Math. (2) 92 1970 489–527.
While the congruence subgroup problem is studied mainly in the arithmetic setting, the normal subgroup problem is complicated in a similar way because the matrix group (in your case over a polynomial ring in one variable) has a huge number of such subgroups.  In the case of the modular group, these arise from the group-theoretic structure as an amalgamated free product of two small cyclic groups.  The analogue in your question is worked out in the paper by Hirosi Nagao,  *On $\mathrm{GL}(2, K[x])$,  J. Inst. Polytech. Osaka City Univ. Ser. A 10 (1959) 117–121.   In most of the literature, attention is focused instead on the somewhat better behaved classical (or Chevalley) groups of higher Lie rank than 1.   
While the cumulative literature on matrix groups over rings of various types is enormous and spreads out into algebraic $K$-theory, I'll mention a few of the many people involved over the years:   E. Abe, W. Klingenberg, A.W. Mason, A.A. Suslin, L.N. Vaserstein, N.A. Vavilov.   Here is a randomly chosen paper:  A.W. Mason, Anomalous normal subgroups of $\mathrm{SL}_2(K[x])$.  Quart. J. Math. Oxford Ser. (2) 36 (1985), no. 143, 345–358.  Vavilov has written many papers on general structure theory, in both Russian and English, including some long surveys.   
P.S. Though I'm not a specialist on structure of linear groups over rings, all  evidence I've seen confirms that determining the normal subgroups of groups $\mathrm{SL}_2(K[x])$ (or $\mathrm{GL}_2(K[x])$) when $K$ is a field is basically hopeless.   Even though the ideal structure of the ring itself is reasonable (as for $\mathbb{Z}$), Nagao's old work already shows how far that is from locating all normal subgroups.   Much of the literature is covered in the 1989 Springer book by Hahn-O'Meara The Classical Groups and $K$-Theory (Chapter 4).  The papers by Costa-Keller get some mileage in special cases from the refined notion of "radix", but for the rings here every radix is actually an ideal unless $K$ has characteristic 2.<|endoftext|>
TITLE: Abstract characterization of polygonizations
QUESTION [5 upvotes]: Consider a polygonization of the plane by convex polygons of a given minimal size that meet edge-to-edge and vertex-to-vertex.

What's the “official” name of such a polygonization?
Such polygonizations of the plane induce infinite graphs.

How can such abstract graphs be characterized?

Somehow like this: “A graph is induced by a polygonization of the plane iff it is infinite, planar, 3-vertex-connected, and P.” (The question asks for property P, since infinite, planar and 3-vertex-connected those graphs obviously are.)

Is it true, that the graphs that are
  induced by a polygonization of the
  sphere are exactly the polyhedral
  graphs which in turn are exactly the finite planar 3-vertex-connected graphs?

Finally I want to know:

Can the graphs be characterized that are induced by a polygonization of any surface?

For the record: I asked this question at MSE before but it didn't earn a lot of interest.

REPLY [2 votes]: For the OP's claim re *infinite*polyhedral graphs, the answer is yes, this is true, and a proof is in my paper:
Rivin, Igor. "Combinatorial optimization in geometry." Advances in Applied Mathematics 31.1 (2003): 242-271.
Basically, you can construct a circle packing with any prescribed (three-connected) combinatorics. What you lose when you go from finite to infinite is uniqueness, in a spectacular way: it should be true that one can get the carrier of the packing to be any Jordan domain.<|endoftext|>
TITLE: Number of II${}_1$ factors
QUESTION [5 upvotes]: McDuff proved that there exist continuum many non-isomorphic (separable) II${}_1$ factors. I would like to politely ask whether it is known/open if one can find $2^{\mathfrak{c}}$ (or at least $\mathfrak{c}^+$) many such factors.
My feeling is that this is not possible to construct more than $\mathfrak{c}$ separable von Neumann algebras by a simple cardinality argument. The ball of $B(H)$ for $H$ separable, is metrisable under the ultraweak topology, so it has at most $\mathfrak{c}$ ultraweakly closed subsets. So we cannot have more than $\mathfrak{c}$ different balls, and consequently, have more than $\mathfrak{c}$ non-isomorphic algebras. Is this correct?

REPLY [6 votes]: Your argument is correct.  An alternate and more "intrinsic" argument is to look at the predual, which is a separable Banach space.  There are only continuum many separable Banach spaces, since they are determined by the metric on a countable dense subset that is a $\mathbb Q[i]$-vector space.  Thus there are only continuum many separable von Neumann algebras, when considered just with their structure of dual Banach spaces.  As Nik points out in the comments, the dual Banach space structure may not determine the algebra structure.  However, given any such dual Banach space, you can fix a countable weak*-dense subset, and then any algebra structure will be determined by what it does on that countable set, so you again have only continuum many possibilities.<|endoftext|>
TITLE: Where are the second- (and third-)generation proofs of the classification of finite simple groups up to?
QUESTION [53 upvotes]: According the the Wikipedia page, the second generation proof is up to at least nine volumes: six by Gorenstein, Lyons and Solomon dated 1994-2005, two covering the quasithin business by Aschbacher and Smith in 2004, and one by Aschbacher, Lyons, Smith and Solomon in 2011. However, this latter book is really just the second part of an outline of the proof, the first part of which was written by Gorenstein in the 80s (the reason for the delay is, of course, that the quasithin case hadn't actually been settled at the time of the announcement of completion). Hence the last update on the second-generation proof is really 2005.
With the recent formal proof in Coq of the Odd-order Theorem, it would be good to know where the traditional proof is up to.
EDIT 6 August 2013: Any news as to the completion of that seventh volume as mentioned in the comments?
EDIT 29 September 2016 Just a bump to this question in case people know more about where the progress is at. Books 7 and 8 should probably have made some progress since I asked this originally.

REPLY [30 votes]: There is an interesting review by Ron Solomon of a paper in this area, which has been featured on the Beyond Reviews blog. In particular, he outlines the broad tactics that people are using in CFSG II, and some of the content that will be going into volume 7.
Also, Inna Capdeboscq apparently gave an outline of volume 8, or at least a chunk of it, at the Asymptotic Group Theory conference in Budapest. This was mentioned by Peter Cameron on his blog, sadly with no detail! If anyone can get a whiff of what she said, I would be grateful.
EDIT 15 October 2016 I emailed the group-pub mailing list and was told second-hand that Ron Solomon 'has hopes' volume 7 will be submitted next year.
EDIT 27 March 2018 Thanks to Timothy Chow in a comment on another answer, here is the link to the published version of Volume 7. So now the countdown to Volume 8 starts...
EDIT 22 June 2018 Even better news: Volume 8

...is near completion and promised to the AMS by August 2018. The completion of Volume 8 will be a significant mathematical milestone in our work. (source)

Also (from the same article):

We anticipate that there will be twelve volumes in the complete series [GLS], which we hope to complete by 2023.


Considerable work has been done on this problem [the bicharacteristic case], originally by Gorenstein and Lyons, and more recently by Inna Capdeboscq, Lyons, and me. We anticipate that this will be the principal content of Volume 9 [GLS], coauthored with Capdeboscq.


When p is odd, there is a major 600-page manuscript by Gernot Stroth treating groups with a strongly p-embedded subgroup, which will appear in the [GLS] series, probably in Volume 11. There are also substantial drafts by Richard Foote, Gorenstein, and Lyons for a companion volume (Volume 10?), which together with Stroth’s volume will complete the p-Uniqueness Case.


It would be wonderful to complete our series by 2023, the sixtieth anniversary of the publication of the Odd Order Theorem. Given the state of Volumes 8, 9, 10, and 11, the achievement of this goal depends most heavily on the completion of the e(G) = 3 problem. It is a worthy goal.

EDIT Mar 2019 Volume 8 has been published. The page listing the available volumes, along with links to more details is here.
The summary of this volume is as follows:

This book completes a trilogy (Numbers 5, 7, and 8) of the series The Classification of the Finite Simple Groups treating the generic case of the classification of the finite simple groups. In conjunction with Numbers 4 and 6, it allows us to reach a major milestone in our series—the completion of the proof of the following theorem:


Theorem O: Let G be a finite simple group of odd type, all of whose proper simple sections are known simple groups. Then either G is an alternating group or G is a finite group of Lie type defined over a field of odd order or G is one of six sporadic simple groups.


Put another way, Theorem O asserts that any minimal counterexample to the classification of the finite simple groups must be of even type. The work of Aschbacher and Smith shows that a minimal counterexample is not of quasithin even type, while this volume shows that a minimal counterexample cannot be of generic even type, modulo the treatment of certain intermediate configurations of even type which will be ruled out in the next volume of our series.

EDIT February 2021 Volume 9 has now been published.  From the preface:

This book contains a complete proof of Theorem $\mathcal{C}_5$, which covers the “bicharacteristic” subcase of the $e(G) \ge 4$ problem. The outcome of this theorem is that $G$ is isomorphic to one of the six sporadic groups for which $e(G)\ge 4$, or one of six groups of Lie type which exhibit both characteristic 2-like and characteristic 3-like properties. Finally, in Chapter 7, we begin the proof of Theorem $\mathcal{C}_6$ and its generalization Theorem $\mathcal{C}^∗_6$, which cover the “$p$-intermediate” case. $\ldots$ In the preceding book in this series, we had promised complete proofs of Theorems $\mathcal{C}_6$ and $\mathcal{C}^∗_6$ in this book, but because of space considerations, we postpone the completion of those theorems to the next volume.

EDIT September 2021
In response to a question from Hugo de Garis, Ron Solomon sent the following email in January 2021:

Vol. 9 is already submitted, accepted and scheduled for publication. It should be published early this year.
As for the rest, my best guess now is that there will in fact be 4 further volumes, not 3.  A roughly 800 pages manuscript on the Uniqueness Theorem has been completed by Gernot Stroth.  With some additional material, it will fill 2 further volumes. This could probably be readied for publication by a year from now.  However, our team (Inna Capdeboscq, Richard Lyons, Chris Parker and myself) are currently focussing on the remaining work to be done for the other two volumes.  It is difficult to estimate how long this will take. With luck we might have a first draft completed this calendar year, but it might take longer.
It is safe to say that the remaining volumes will not all be published before 2023.  I hope it is also safe to say that they will all be published no later than 2025.

(Emphasis added)
EDIT 29 Dec 2021
Richard Lyons maintains an erratum for the whole published second generation CFSG on this page: https://sites.math.rutgers.edu/~lyons/cfsg/
EDIT 05 Apr 2022
In response to a further question from de Garis (see the page linked above), Solomon wrote (in March 2022):

We have been working on the theorems for both Volumes 10 and 11.  Just in the past few weeks, we have decided to concentrate on the completion of Volume 10.  This is proceeding very well and we should be able to submit Volume 10 for publication this year, I believe.  I fear that I may have been a bit overoptimistic in predicting the completion of all the volumes by the end of 2024.<|endoftext|>
TITLE: Mutually tangent ellipsoids in 3 space
QUESTION [9 upvotes]: I recently heard a claim that for any n, it is possible to arrange n ellipsoids in 3 space such that each pair of ellipsoids is kissing.  Is this true, and if so, how?
Edit: By kissing, I mean that I would like the interiors of the ellipsoids to be disjoint, but each pair of ellipsoids should intersect at a point.

REPLY [5 votes]: I've never heard of this before, and it sounds quite counterintuitive. If one is granted that this is true, then one can try to work backwards and apply a bit of dimensional analysis to understand how this can be. 
First, note that ellipsoids are quadrics, and there are a $9$ dimensional space of quadrics. Also, note that ellipsoids are invariant under affine (or more generally projective) transformations, so if we have one such configuration, we will expect to get at least a $15$-dimensional family of kissing ellipsoids. 
On the other hand, there are degenerate quadrics for which this is true. Consider $n$ mutually non-parallel lines in the plane. We can consider these as degenerate ellipsoids, with two axes $0$ radius and one axis $\infty$ radius. They the are each mutually tangent, since they each meet in a point. There is an $2n+3$-parameter family of these. One could hope to "regenerate" families of $n$ tangent ellipsoids from these, but there might be restrictions on when this is possible. 
For $n=1$, there is a $9$ dimensional family.
For $n=2$, there is a $17=9+8$ dimensional family. We have $9$ dimensions for the first ellipsoid. Choosing any other ellipsoid with center disjoint from the first, we may rescale it uniquely to be tangent to the first ellipsoid. So this gives us $8$ dimensions for the second ellipsoid. 
For $n=3$, let's assume that one of the ellipsoids is a sphere. Then its center is equidistant from the other two ellipsoids. The space of points equidistant from two kissing ellipsoids is $2$-dimensional. So we get a $17+2=19$-dimensional space of 3 tangent ellipsoids, with one round. We may change the round sphere by an affine transformation based at its center to get any ellipsoid with the same center, and we have a $6$-parameter family of such affine maps (we have $3$ dimensions for the major axis, $2$ for the minor axis, and $1$ for the third axis). However, this over-counts, since a similarity (all $3$ axes equal) will take the sphere to a sphere, so this gives us $19+6-1=24$ dimensions (or we may take a $5$-parameter family of volume-preserving affine transformations to eliminate repetitions). 
For $n=4$, let's again assume that one of the ellipsoids is a sphere. The space of points equidistant from $3$ ellipsoids is $1$-dimensional. As above, we may modify the sphere by a $5$-parameter family of volume-preserving affine transformations to get $24+1+5=30$ dimensions of 4 mutually tangent ellipsoids. 
For $n=5$, we expect finitely many points which are equidistant from $4$ mutually tangent ellipsoids, so the computation gives $35$ dimensions. 
At this point, one expects adding the next sphere will cut down on the dimension of the space of 5 tangent ellipsoids which have an equidistant point. If we continue the sequence $9,17,24,30,35$, then the next term ought to be $39$ dimensions ($42, 44, 45, ?$). Of course, this trend couldn't possibly continue if one expects to have $n$ tangent ellipsoids for all $n$, so there must be some non-generic phenomenon creating the incidence.<|endoftext|>
TITLE: Homology of classifying space of spin group BSpin(n)
QUESTION [8 upvotes]: While dealing with $BO(n)$, $BSO(n)$ and $BSpin(n)$ with the universal coefficient theorem and Künneth formula, I came to have the following question:
The universal coefficient says $H^n(X;M)\cong \hom(H_{n}(X;\mathbb{Z}),M)\oplus  {\rm Ext}^{1} (H_{n-1}(X;\mathbb{Z},M))$ for a $\mathbb{Z}$-module $M$.
When $X=BSpin(n)$, we know that $H^4(BSpin(n);\mathbb{Z})\cong \mathbb{Z}$ and it seems likely that once we know what $H_p(BSpin(n);\mathbb{Z})$ would be for $p=3,4$ we might be able to retrieve this isomorphism with the aid of universal coefficient theorem.
So what would be $H_p(BSpin(n);\mathbb{Z})$ at least for $p=0,1,2,3,4$? 
(I have to say that the question is not about how to prove the isomorphism $H^4(BSpin(4);\mathbb{Z})\cong \mathbb{Z}$.)
Thanks!

REPLY [8 votes]: Consider the classifying space $EG$ of a given group $G$. In your case, $G={Spin}(n)$. We have a fibration $G\to EG\to BG$ where $EG$ is contractible and $G$ acts freely on it. Therefore, the long exact sequence in homotopy groups tells you that $\pi_j(BG)\cong \pi_{j-1}(G)$. But $G={Spin}(n)$ is a Lie group which is simply connected. It is also classically known that $\pi_2(G)=0$ for finite dimensional Lie groups $G$. And, we also know that $\pi_3({Spin}(n))=\mathbb{Z}$. This implies that $\pi_j(B {Spin}(n))=0$ for $j=0,1,2,3$. Moreover, $\pi_4(B {Spin}(n))\cong H_4(B{Spin}(n);\mathbb{Z})$ by Hurewicz theorem. This is also isomorphic to $\pi_3({Spin}(n))$. The universal coefficient theorem now tells you that $H^4(B{Spin}(n);\mathbb{Z})=\mathbb{Z}$.<|endoftext|>
TITLE: C^\infty versus semiclassical wavefront sets
QUESTION [5 upvotes]: Zworski states that if $u$ is a compactly supported distribution, independent of the semiclassical parameter $h$, then the relationship between the $C^\infty$ and semiclassical wavefront sets of $u$ is given by:
$WF_h(u)=(\mathrm{supp}(u)\times\{0\})\cup WF(u)$.
Would someone please explain this relationship?  Thanks!

REPLY [2 votes]: Heuristically, this result follows because the semiclassical wavefront set measures oscillations at frequency $\frac{\xi}{h}$. To understand this, observe that since $u$ does not depend on $h$, if $u$ is smooth, it has small high frequency oscillations, and hence, for $\xi\neq 0$, and hence points where $u$ is smooth and $\xi\neq 0$ do not appear in $WF_h(u)$.  However, where $u$ is not smooth, $u$ has high frequency oscillations and so these points do appear.
Below you will find a proof of the statement.
To see that $WF_h(u)\cap \mathbb{R}^d\times \{0\}=$ supp $(u)\times\{0\}$ observe that 
$$\mathcal{F}_h(\chi u)(0)=\langle u, \chi\rangle \neq 0$$
for $\chi$ supported near $x_0\in $ supp $(u)$ and
$$\mathcal{F}_h(\chi u)(0)=0 $$
for $\chi$ supported near $x_0\notin $ supp $(u)$. 
To see the second part of $WF_h(u)$, simply use the characterization of $WF(u)$ that for $\xi\neq 0$, $(x_0,\xi_0)\notin WF(u)$ if and only if for $\chi$ supported sufficiently close to $x_0$, all $N>0$ and $\xi$ in a conic neighborhood of $\xi_0$, 
$$|\mathcal{F}(\chi u)(\xi)|\leq C_N\langle \xi \rangle ^{-N}.$$
Thus, $(x_0,\xi_0)\notin WF(u)$ if and only if for all $N>0$ and all $\xi$ in a neighborhood of $\xi_0$,
$$|\mathcal{F}_h(\chi u)(\xi)|\leq C_N\langle \xi/h\rangle^{-N}.$$
But, since $\xi_0\neq 0$, this gives $(x_0,\xi_0)\notin WF(u)$ if and only if for all $N>0$ and $\xi$ in a neighborhood of $\xi_0$,
$$|\mathcal{F}_h(\chi u)(\xi)|\leq C_Nh^N$$
and hence if and only if $(x_0,\xi_0)\notin WF_h(u)$.<|endoftext|>
TITLE: Fuglede-Kadison determinants in $L(\mathbb{F}_2)$
QUESTION [5 upvotes]: Given $M$ a finite von Neumann algebra with trace $\tau$, $T\in M$ invertible.
The Fuglede-Kadison determinant is defined as 
$\Delta(T)=e^{\tau(log|T|)}$, 
where $|T|=(T^*T)^{\frac{1}{2}}$, and $\tau(log|T|)=\int_{0}^{||T||}log(t)d\mu_{|T|}(t)$, and the probability measure $\mu_{|T|}$ is defined on spectrum(|T|) by requiring $\int_{spec(|T|)}fd\mu_{|T|}=\tau(f(|T|))$ for all $f\in C(spec(|T|))$. 
See the reference Determinant theory in finite factors
Especially, let $M=L(\mathbb{F}_2)$ be the free group factor associated to the free group $\mathbb{F}_2$ on two generators $a, b$, and with the canonical trace $\tau$, my question is :
1, Are there any references for the study of the determinant in the case $M=L(\mathbb{F}_2)$ ?
Especially, I also want to know 
2, Are there any nontrivial computable examples in this case, i.e., what does $\Delta(T)$ looks like for $T\in \mathbb{C}\Gamma$, invertiable ?
Note: This question is motivated by the paper Li, 2012.

REPLY [7 votes]: The spectral measures for self-adjoint elements in $\mathbb C F_2$ are very special. In particular, it is known that non of the elements in $\mathbb C F_2$ has a kernel when acting via the left-regular representation on $\ell^2 F_2$. This was shown by Peter Linnell using index-theoretic methods in
Linnell, Peter,
Division rings and group von Neumann algebras.
Forum Math. 5 (1993), no. 6, 561–576. 
(It also follows from another of Linnell's papers and the fact that $F_2$ is left-orderable.)
It is also known that Novikov-Shubin invariants are always positive. This together implies that $\Delta(T) \neq 0$ if $T\neq 0$. 
The main advantage is that $\sum_{n} \tau(T^n)z^n$ is an algebraic power series for any $T \in \mathbb C F_2$. This implies the result about Novikov-Shubin invariants (and much more). It was proved in
Sauer, Roman, 
Power series over the group ring of a free group and applications to Novikov-Shubin invariants. High-dimensional manifold topology, 449–468, World Sci. Publ., River Edge, NJ, 2003
Starting with the computations in this paper (and using the Stieltjes-Inversion formula), you can also make concrete computations for the determinant of specific elements.<|endoftext|>
TITLE: Tangent space of the stack $\overline{\mathcal{M}}_{g,n}(X,\beta)$. 
QUESTION [9 upvotes]: Let $\overline{\mathcal{M}}_{g,n}(X,\beta)$ be the moduli stack of stable maps $f$ from genus $g$, $n$-marked curve $C$ to a variety $X$ i nth curve class $\beta \in H^2(X,\mathbb{Z})$. I would like to know why the hyperext group
$$
Ext^1_C(f^*\Omega_X\rightarrow \Omega_C(\sum_{i=1}^n),\mathcal{O}_C)
$$
represents tangent space of the moduli and the obstruction lies in
$$
Ext^2_C(f^*\Omega_X\rightarrow \Omega_C(\sum_{i=1}^n),\mathcal{O}_C)?
$$
I am aware that the deformation-obstrcution of a map $f:C\rightarrow X$ with a fixed curve $C$ is governed by $H^i(C,f^*T_X)$ for $i=1,2$ and the automorphism-deformation of a $n$-marked points $(C;p_1,\dots,p_n)$ is governed by $Ext^i_C(\Omega_C(\sum_{i=1}^n),\mathcal{O}_C)$ for $i=0,1$. However, I don't know how to combine them into one package in the hyperext groups above. 
I would appreciate it if someone could kindly explain how to obtain and understand the hyperext groups above.

REPLY [5 votes]: Let $L$ be the complex $[ f^\ast \Omega_X \rightarrow \Omega_C(D) ]$, concentrated in degrees $[-1,0]$ on $C$.
One way to understand the obstructions is to understand the deformations for affine curves first.  Observe first that there is a canonical equivalence of categories between the category of deformations of a pointed map $f : C \rightarrow X$, with $C$ not-necessarily proper, and the category of extensions
$G(C) = \mathrm{Ext}(L, \mathcal{O}_C)$.
The meaning of the category on the right is the category of extensions $E$ of $\Omega_C(D)$ by $\mathcal{O}_C$ together with a trivialization of the induced extension of $f^\ast \Omega_X$ by $\mathcal{O}_C$.  One checks explicitly that the category of extensions of $\Omega_C(D)$ by $\mathcal{O}_C$ is canonically equivalent to the category of deformations of the pointed curve $C$ and that splitting the induced extension of $f^\ast \Omega_X$ is the same as extending the map $f : C \rightarrow X$ to a map from the deformed curve to $X$.
Note that $G$ is a commutative group stack (what is sometimes called a "Picard stack").  It is easy to check that $G$ is non-empty when $C$ is an affine curve (use the fact that deformations of nodal curves are unobstructed and the formal criterion of smoothness for $X$), which means that any obstructions to deforming a proper curve come from the failure of local deformations to glue.  There is a standard way to compute these local obstructions---e.g., by Cech cohomology, noting that mutatis mutandis, Cech cohomology still works for group stacks---and this yield a class in $H^1(C, G)$.  Cohomology of a commutative group stack is the same as cohomology of the associated complex, so we get an obstruction in
$H^1(C, L^\vee[1]) = H^2(C, L^\vee) = \mathrm{Ext}^2(L, \mathcal{O}_C)$.<|endoftext|>
TITLE: Norm of inverse confluent Vandermonde matrix
QUESTION [6 upvotes]: Let $\{x_1,\dots,x_n\}$ be pairwise distinct complex numbers and $l_1+l_2+\dots+l_n=N$. The $N\times N$ confluent Vandermonde matrix is defined as
$$V=
\begin{bmatrix}
v_{1,0}&v_{2,0}&\dots&v_{n,0}\\\\
v_{1,1}&v_{2,1}&\dots&v_{n,1}\\\\
\vdots\\\\
v_{1,N-1}&v_{2,N-1}&\dots&v_{n,N-1}
\end{bmatrix}$$ where $v_{j,k}=\begin{bmatrix}x_j^k,&kx_j^{k-1},&\dots&k(k-1)\times\dots\times (k-l_j+1) x_j^{k-l_j+1}\end{bmatrix}$. Let $\|\cdot\|$ denote the row sum matrix norm. In some applications (e.g. interpolation, signal processing) one would like to estimate the quantity $\|V^{-1}\|$.
Gautschi [1] has shown that for $l_1=\dots=l_n=2$ one has
$$ \|V^{-1}\| \leq \max_{1\leq \lambda\leq n} \beta_{\lambda} \prod_{\nu=1,\nu\neq\lambda}^n \biggl(\frac{1+|x_{\lambda}|}{|x_{\nu}-x_{\lambda}|}\biggr)^2$$ where $\beta_{\lambda}=\max\biggl(1+|x_{\lambda}|,1+2(1+|x_{\lambda}|)\sum_{\nu\neq\lambda}{1\over |x_\nu-x_\lambda|}\biggr)$.
I am interested in a somewhat cruder estimates, as follows: if $|x_j|\leq 1$ and $|x_i-x_j|\geq \delta$, then for the above case we have $$\|V^{-1}\| \leq C n 2^N\delta^{-N+1}\qquad (*)$$ for some absolute constant $C$.
Is it true that something like $(*)$  holds for the general configuration
$\{l_1,\dots,l_n\}$?
EDIT: using [2], this seems to boil down to the following. Let
$$ h_j(x)=\prod_{i \neq j}(x-x_i)^{-l_i}.$$ For $t=0,1,\dots,l_j$ evaluate $h_j^{(t)}(x_j).$
[1] W.Gautschi, "On Inverses of Vandermonde and Confluent Vandermonde matrices II", Numerische Mathematik 5, 425-430, 1963.
[2] R.Schapelle, "The Inverse of the Confluent Vandermonde Matrix", IEEE Trans. on Automatic Control, October 1972, pp.724-725.

REPLY [2 votes]: For $j=1,\dots,n$ and $k=0,1,\dots,l_j-1$ denote by $u_{j,k}$ the row with index $l_1+\dots +l_{j-1}+k$ of the matrix $V^{-1}$. By using a generalization of the Hermite interpolation formula (see [3]), in [2] it is shown that the elements of $u_{j,k}$ are the coefficients of the polynomial
$$ {1\over k!} \sum_{t=0}^{l_j-1-k} {1\over t!} h_j^{(t)}(x_j) (x-x_j)^{k+t} \prod_{i\neq j} (x-x_i)^{l_i}  $$
Now thanks to the answer to this MSE question, one has
$$
|h_j^{(t)}(x_j)|\leq N(N+1)\cdots (N+t-1)\delta^{-N-t}.
$$
The sum of absolute values of the coefficients of the polynomials $(x-x_j)^{k+t} \prod_{i\neq j} (x-x_i)^{l_i}$  is at most (see [4, Lemma])
$$
(1+|x_j|)^{k+t} \prod_{i\neq j}(1+|x_i|)^{l_i} \leq 2^{N-(l_j-k-t)}.
$$
So now
$$
\|u_{j,k}\| \leq {1\over k!}\sum_{t=0}^{l_j-1-k} {1\over t!} {N(N+1)\cdots (N+t-1) \over {\delta^{N+t}}}2^{N-l_j+k+t}\\\\
\leq \biggl({2\over \delta}\biggr)^N {1\over {2^{l_j-k}k!}}\sum_{t=0}^{l_j-1-k} {l_j-1-k \choose t} {N(N+1)\cdots(N+t-1)\over (l_j-k-t)\cdots(l_j-k-2)(l_j-k-1)} \biggl({2\over \delta}\biggr)^t\\\\
\leq \biggl({2\over \delta}\biggr)^N {1\over {2^{l_j-k}k!}} \biggl(1+{2N\over \delta}\biggr)^{l_j-1-k}\\\\
=\biggl({2\over \delta}\biggr)^N {2\over k!} \biggl({1\over 2}+{N\over\delta}\biggr)^{l_j-1-k}.
$$
[3] A.Spitzbart, "A Generalization of Hermite's Interpolation Formula", The American Mathematical Monthly, Vol.67 No.1, p.42-46, 1960.
[4] W.Gautschi, "On Inverses of Vandermonde and Confluent Vandermonde matrices", Numerische Mathematik 4, p.117-123, 1962.<|endoftext|>
TITLE: algebraic groups and their Lie algebras
QUESTION [5 upvotes]: I have probably a stupid question about representations of algebraic groups:
Let $G$ be an algebraic group and $L$ be a Lie algebra of $G$. What is the connection between
categories of representations of $G$ and $L$ (are they equivalent)?
Now, let $V$ be an irreducible representation of $G$, how to prove that $V$ is also 
an irreducible representation of $L$ (in case it is true)?   It should be true and I believe,
proof is easy, but still, I do not see that and I am looking for a nice proof of that. 
And vice versa, let $V$ be an irreducible representation of $L$, is it also irreducible representation of $G$ ?

REPLY [6 votes]: The question is out of focus, I think.  If it's really about algebraic groups rather than Lie groups, that should be made clear.   (And a tag should be added either way.)    As Marc Palm indicates, there are well understood analytic pathways connecting representations of Lie groups and representations of their Lie algebras, though of course the details are quite nontrivial and restrictions are needed.
For linear algebraic groups in characteristic 0, the connections are much more problematical.   Chevalley realized this in his early book (1951), where he took some steps using a sort of formal exponentiation process to get back from the Lie algebra to the algebraic group.   In modern form some of this is written down in later textbooks (Borel for instance).  Probably the most comprehensive source is Section 6 of Chapter II in Demazure-Gabriel Groupes algebriques (optimistically labelled 
"tome I").  Even here one sees that not so much can be said outside the framework of semisimple groups and their Lie algebras.   Note that the concept of "simply connected" isn't even meaningful for most algebraic groups.
In prime characteristic, almost nothing works along the lines of your question, even if you limit to simply connected semisimple groups.    Here there is lots of literature, but for example it's long been known that the irreducible representations of the group and its Lie algebra diverge drastically.

REPLY [5 votes]: A graph argument settles this issue very nicely, as follows.
Consider a linear algebraic group $G$ over a field $k$ of characteristic 0, and let $\mathfrak{g}$ be its Lie algebra.  For a finite-dimensional $k$-vector space $V$, let $f:\mathfrak{g} \rightarrow \mathfrak{gl}(V)$ be a representation.  Provided that $G$ is connected, since ${\rm{char}}(k) = 0$ clearly there is at most one $k$-homomorphism $\rho:G \rightarrow {\rm{GL}}(V)$ such that ${\rm{Lie}}(\rho) = f$, so assuming $G$ is connected we seek conditions under which such a $\rho$ always exists.
Let $\mathfrak{h} \subset \mathfrak{g} \times \mathfrak{gl}(V)$ be the graph of $f$.  Clearly if $\rho$ is to exist and  the $k$-subgroup $H \subset G \times {\rm{GL}}(V)$ is its graph then $H$ is connected and $\mathfrak{h} = {\rm{Lie}}(H)$, so $H$ is uniquely determined (as a $k$-subgroup of $G \times {\rm{GL}}(V)$) since ${\rm{char}}(k) = 0$.  So we seek conditions under which the Lie subalgebra $\mathfrak{h}$ of $\mathfrak{g} \times \mathfrak{gl}(V)$ "exponentiates" to a connected closed $k$-subgroup $H$ (and then we need conditions to ensure ${\rm{pr}}_1:H \rightarrow G$ is an isomorphism).
Assume $G$ is its own derived group, so $\mathfrak{g}$ is its own derived subalgebra (as ${\rm{char}}(k)=0$) and hence the same for $\mathfrak{h}$. Now comes the crucial point: it is a general fact over fields $k$ of char. 0 (see Cor. 7.9 in Ch. II of Borel's textbook on algebraic groups) that the derived subalgebra of any Lie subalgebra of a linear algebraic group over $k$ "exponentiates" to a connected closed $k$-subgroup.  So $\mathfrak{h} = {\rm{Lie}}(H)$ for a unique connected closed $k$-subgroup $H \subset G \times {\rm{GL}}(V)$.  The necessary and sufficient condition for $f$ to arise from some $\rho$ is that ${\rm{pr}}_1:H \rightarrow G$ is an isomorphism.  
This projection has Lie algebra map $\mathfrak{h} \rightarrow \mathfrak{g}$ that is the analogous projection which is visibly an isomorphism (due to the definition of $\mathfrak{h}$ as the graph of $f$), so $H \rightarrow G$ is an isogeny.  As such, its kernel is etale (since ${\rm{char}}(k) = 0$) and hence central (since $H$ is connected), so it is a finite central $k$-subgroup of $H$. Thus, we just need that $G$ admits no nontrivial isogenous (smooth) connected central extension.  This is automatic when $G$ is assumed to be semisimple and simply connected (in the sense of algebraic groups).
So we win whenever $G$ is a connected semisimple $k$-group that is simply connected. We also win whenever $G$ is a unipotent $k$-group (by entirely different arguments), but presumably you're not interested in that case.<|endoftext|>
TITLE: Tensor product of $\mathcal{D}$-modules and constructible sheaves
QUESTION [8 upvotes]: The Riemann-Hilbert correspondence, as proved by Kashiwara and Mebkhout, says that for X a smooth algebraic variety over $\mathbb{C}$ there is an equivalence of triangulated categories
$D^b_c(X,\mathbb{C})\cong D^b_\mathrm{rh}(\mathcal{D}_X)$
between the bounded derived category of complexes of $\mathbb{C}$-modules on $X$ with constructible cohomology sheaves, and the bounded derived category of complexes of coherent $\mathcal{D}_X$-modules with regular holonomic cohomology sheaves.
Moreover, this equivalence respects the 6 operations $f^* , \mathbf{R}f_* , f^!, \mathbf{R}f_!, \boxtimes, \mathbb{D} $ of usual and extraordinary direct and inverse images, exterior tensor product and duality.
$\mathbf{Question:}$ Does the Riemann-Hilbert correspondence also preserve the interior tensor product? 
On the constructible side, the interior tensor product is $\Delta_X^*(-\boxtimes-) $ where $\Delta_X$ is the diagonal immersion, but on the holonomic side the interior tensor product is $\Delta^!_X(-\boxtimes-)[d_X]$. So its seems to a novice like me that we're getting different operations. Or is there some comparison between $f^!$ and $f^*$ for a closed immersion $f$ that I've missed?

REPLY [10 votes]: This is correct. Verdier duality does not preserve tensor products in general. Another point of view is that each of these categories has two versions of tensor product, $\otimes ^\ast$ and $\otimes ^!$, which are interchanged by Verdier duality. It just happens that for $D$-modules the shriek version (or a shift of it) is most natural to define, whereas for constructible sheaves the $\ast$-version is more natural.
In Bernstein's notes on $D$-modules on page 28, he talks about the ``!-tensor product'', and remarks that it is just a shift of the naive tensor product of $D$-modules (over $\mathcal O$).<|endoftext|>
TITLE: The mean curvature of a hypersurface
QUESTION [5 upvotes]: It is well known to geometric analyst that the scalar curvature of a Riemannian manifold can be decomposed to two parts: one part has a divergence structure and the other part consists of lower order terms. My question is for the mean curvature of a hypersurface in a Euclidean space, dose it have a similar structure? Has any author considered a decomposition of this kind? Any reference will be useful to me. Thanks in advance.

REPLY [4 votes]: I was unable to post an image without a big song and dance. If you go to MY STUFF and open the Michigan Math J. 1991, page 256 has the formula when the hypersurface is given as the level set of a smooth function. It is just a rotated version of the graph formula, as it must be. Meanwhile, the mean curvature is just a dimension constant times the divergence of the (oriented) unit normal, where it does not matter whther you take the divergence on the manifold or extend the unit normal field off the manifold and use the ambient divergence. All the same. I do not see how   you are going to separate out first and second order derivatives. 
Wait, I can try to typeset:
$$   n H = \frac{1}{|\nabla F|} \; \sum_{i=1}^{n+1}   \sum_{j=1}^{n+1} \left( \delta_{ij} - \frac{F_i F_j}{|\nabla F|^2} \right) F_{ij}   $$
That actually looks correct. Good for me. Level set of the function $F$ and $H$ is the mean curvature with one of the choices of unit normal field.<|endoftext|>
TITLE: Do distinct idempotent measures on finite binary systems have distinct supports?
QUESTION [5 upvotes]: Suppose that $(S,*)$ is a finite set equipped with a binary operation.  Extend the binary operation to the vector space $V$ with basis $S$.
The set of probability measures on $S$, viewed as a compact convex subset of $V$ is closed under $*$
and, since $*$ is continuous, there are idempotent measures on $S$.
Must two idempotent measures on $S$ have distinct supports?
I am also interested in the more general question where the assumption of finiteness is dropped and one considers the extension (by convolution) of $*$ to the family of all finitely additive measures on $S$ (in that context, define the support of a measure to be all subsets of $S$ with positive measure).

REPLY [4 votes]: No if I understood. Take the two element left zero semigroup. All measures are idempotent. 
Added a left zero semigroup is one satisfying the identity xy=x 
Added A finite semigroup $S$ satisfies that distinct idempotent measures have distinct support iff for all idempotents $e,f\in S$ one has $SeS=SfS$ implies $e=f$.
A finite semigroup has a unique idempotent measure with full support iff it is a finite group.
Here is how the proof goes. Suppose $P$ is an idempotent measure on $S$ and assume the support of $P$ is $S$.
Claim 1: $S$ contains no ideal.
Proof.  Obviously every state is recurrent for both the right random walk an left random walk on $S$ driven by $P$ because $P$ is idempotent and the support is $S$ (and hence $P$ is stationary for these walks).  If $I$ was a proper ideal, then the states in $S\setminus I$ would have to be transient because they fall into $I$ with positive probability.  Thus $S$ has no proper ideals.
By Rees's theorem, if $e$ is an idempotent of $S$ then $eSe$ is a group $G$, there are sets $A$ and $B$, and a mapping $P\colon B\times A\to G$ such that $S\cong A\times G\times B$ with multiplication $$(a,g,b)(a',b',b') = (a,gP(b,a')g',b).$$
It is trivial to check that if $Q$ is a measure on $A$, $U$ is uniform measure on $G$ and $R$ is a measure on $B$, then $Q\times U\times R$ is an idempotent measure on $S$.  The fact that all idempotent measures $A\times G\times B$ are of this form is an easy calculation using  that the only idempotent measure on a finite group with global support is the uniform measure.    The key point is to first show that the measure is uniform on subsets of the form $\{a\}\times G\times \{b\}$ using the group result.<|endoftext|>
TITLE: There are $n$ horses. At a time only $k$ horses can run in a single race. What is the minimum number of races required to find the $m$ fastest horses?
QUESTION [28 upvotes]: The following question was asked and not (yet) answered at Math Stack Exchange.

There are $n$ horses. At a time only $k$ horses can run in the single race. What is the minimum number of races required to find the top $m$ fastest horses? Please explain your answer.

The $n = 25, k = m = 5$ case was a Google interview question and there are various answers on the web. But I am not sure what the right answer should be for this. Any ideas?

REPLY [14 votes]: A trivial lower bound is $n/k$, since clearly every horse must race to obtain the answer.
I think we can get $O(n/k)$ upper bound, by adapting the median-of-medians selection algorithm.
First, note that, up to a constant factor, this problem is equivalent to finding the $m$th best horse. For the reduction one way, simply find the $m$ best horses and the $m-1$ best horses and see which horse mysteriously disappeared. For the reduction the other way, find the $m$th best horse, then race every other against it to find which are better and which are worse. (In fact, you don't need to do this - it's not hard to check that if you've found the $m$th best horse through repeated racing, you already know which are better and which are worse.)
For $k=2$, the median-of-medians algorithm is known to give an $O(n)$ time bound, as was pointed out by Ralph Furmaniak. We will also use this to handle $k\leq 4$.
So assume $k\geq 5$ and is odd. Let $T(n)$ be the time it takes to find the $m$th horse among $n$ horses. Then divide the horses into groups of $k$, race them in time $n/k$, and take the median of each. Then select the median of those medians in time $T(n/k)$, and pivot on it in time $n/(k-1)$. (to pivot, the pivot horse must race each other horse only once. This allows us to remove at least $n(k+1)/4k$ of the horses, so we can find the $m$th horse in time $T(n (3k-1)/4k)$. So we get:
$$T(n) \leq T\left(\frac{n}{k}\right) + T\left(  \frac{n(3k-1)}{4k}\right) + \frac{n}{k}+\frac{n}{k-1} $$
By induction,
$$T(n) \leq \frac{4k}{k-3} \cdot \left( \frac{n}{k}+\frac{n}{k-1} \right) = O \left(\frac{n}{k}\right)$$
In fact we obtain an explicit constant of $8+o(1)$.
This ignores non-unique divisibility, which should only lead to a small error term unless $k$ is very large as a fraction of $n$.<|endoftext|>
TITLE: Is the von Neumann algebra associated to a unitary representation of an amenable group always injective?
QUESTION [5 upvotes]: I should be tarred and feathered for not knowing at least the status of the following question. 

Question: Let $\Gamma$ be a discrete amenable group. If $\pi:\Gamma \rightarrow B(\mathcal{H})$ is a unitary representation of $\Gamma$ on a separable Hilbert space $\mathcal{H}$, is the von Neumann algebra $\pi(\Gamma)''$ necessarily injective?

Flippantly one imagines that the answer to this question is yes, by Theorem 2.2 of Bekka's paper on amenable representations. But this result only says that the images of group elements are in the centralizer of a non-normal state...it isn't immediately clear why the entire von Neumann algebra should lie in the centralizer of such a state. If one tries to sidestep this by looking at a proof using almost invariant vectors, one is busted by the fact that a representation that is "$H$-amenable" isn't necessarily amenable in Bekka's sense. 
EDIT: Makoto's nice answer below provided me with some closure. I'm still worried that I can't see a more or less direct way to this result from Connes's '76 paper on the classification of injective factors. If this paper can, in a more or less direct and self-contained way, be used to resolve the question, please feel free to include another answer.

REPLY [8 votes]: This is because you get a representation of the maximal(=reduced by amenability) C*-algebra $C^\ast(\Gamma)$ on $H$, and the image has to be a nuclear algebra because it's a quotient of nuclear one.  Then, $\pi(\Gamma)''$, being a weak*-closure of a nuclear C*-algebra, has to be injective.<|endoftext|>
TITLE: Open problems in Birational Geometry, after BCHM
QUESTION [30 upvotes]: Rencently a breakthrough was made in the context of the Minimal Model Program by the work of Birkar-Cascini-Hacon-McKernan. They proved that the canonical ring of a smooth or mildly singular projective algebraic variety is finitely generated.
Since I'm a master student and so I have no a wide view of the subject (I'm not an expert), I would like to know what are the main open problems in this direction (I mean, in the framework of the Mori Program). More generally, right now what are the driving forces, the big open questions in birational geometry?
Feel free to close this question, if too generic for the purposes of the site.
Thanks in advance.

REPLY [19 votes]: Let me add my answer.
In characteristic 0.

Abundance conjecture.
This is probably universally accepted as the most important question for the current minimal model program after BHCM's proof of finite generation. I want to remark that n-dimensional log canonical MMP follows from n-dimensional klt MMP and n-1 dimensional log canonical MMP. So there is no new difficulty for log canonical pairs at least for MMP other than abundance for klt.

General type.
In fact, after BCHM, Koll'ar and many other people's work, I think we have a pretty good understanding of the rough classification of general types, namely, we know they form a moduli space which has a pretty reasonable compactification.

Fano and singularities.
The boundedness of Fano, namely the BAB-conjecture is one of the most fundamental question about singular Fano. We only know special cases. In philosophy, there is a local-to-global principle, and correspondingly, we should consider certain boundedness of singularities. The famous challenge there is Shokurov's ACC of mld conjecture. Note this should be substantially harder than ACC of log canonical thresholds which now is a theorem. Another one is the semi-continuity of mld conjecture. And Shokurov proved these two conjectures together imply termination of flips.

Calabi-Yau.
There are two fundamental problems there to me. One is the finiteness of the topological type,
Another one is Kawamata-Morrison Cone conjecture. Each of them is still out of reach even in dimension 3. This is a not a big surprise, since we are still quite lack of understanding of CY in dimension 3. On the other hand, if you apply local-to-global principle here, then we should consider semi-log-canonical CY. And Koll'ar's recent examples point out that the slc picture is even more complicated.


In characteristic p.
In characteristic p, the most important question to me is the resolution of singularity. The next thing is how many results in characteristic 0 MMP still hold in characteristic p.
Without vanishing type theorem, we can still formulate most of those results, but then it really puts a question mark on whether we should still believe them.
My feeling is although there are many substantial work there recently, it's still not clear what the picture should be.<|endoftext|>
TITLE: The integers as a sequential but non-first countable topological group
QUESTION [7 upvotes]: Completely unaware of the Bohr topology, I recently asked whether or not there was a Hausdorff group topology on the integers $\mathbb{Z}$ which made the group fail to be first countable. For me, this topological group is a bit extreme since there are no non-trivial convergent sequences. I'm very interested to know if there is a sequential example.
If $\mathbb{Z}$ is given a Hausdorff group topology which makes it a sequential space, must it be first countable?

REPLY [2 votes]: Consistently, you can get even more, as noted by Hrusak and Ramos-Garcia in this paper: http://www.matmor.unam.mx/~michael/reprints_files/precompact-groups.pdf

There are consistent examples of Fréchet-Urysohn non-first-countable Hausdorff group topologies on $\mathbb{Z}$.

Fréchet-Urysohn means that for every non-closed set $A$ and point $x \in \overline{A} \setminus A$ there is a countable sequence inside $A$ converging to $x$. Fréchet-Urysohn is the same as every subspace is sequential. Malykhin's problem asks for an example of a countable non-metrizable Fréchet-Urysohn topological group. The consistency of a positive answer to it has been known for some time, and recently Hrusak and Ramos-Garcia came up with a proof of the consistency of a negative answer, thus establishing its independence from ZFC.

Take a family of $\omega_1$ many distinct characters on $\mathbb{Z}$ separating the points of $\mathbb{Z}$ and consider the coarsest topology making each of those characters continuous. This is a Hausdorff group topology on $\mathbb{Z}$ with no countable local base and a base of cardinality $\omega_1$. The latter implies that it is Fréchet-Urysohn in any model of ZFC+$\mathfrak{p}>\omega_1$  (see below).

Call a family $\mathcal{F}$ of infinite subsets of a countable set strongly centered if every finite subfamily of $\mathcal{F}$ has infinite intersection. We say that a set $S$ is a pseudointersection of the family $\mathcal{F}$ if $S \setminus F$ is finite for every $F \in \mathcal{F}$. Now $$\mathfrak{p}:=\min \{|\mathcal{F}|: \mathcal{F} \mbox{ is a strongly centered family without an infinite pseudointersection} \}$$
Since $\mathcal{F}$ is a family of subsets of a countable set we have $\mathfrak{p} \leq \mathfrak{c}$. Moreover, you can cook up an infinite pseudointersection of a given countable family of infinite subsets of a countable set by an easy diagonalization, so $\mathfrak{p} \geq \omega_1$. It is known that $\mathfrak{p}>\omega_1$ is consistent. For example, under Martin's Axiom we have $\mathfrak{p}=\mathfrak{c}$, and hence it suffices to take a model of Martin's Axiom plus the negation of the Continuum Hypothesis.

Every countable topological space with a local base of cardinality $<\mathfrak{p}$ at every point is Fréchet-Urysohn.
Proof: Let $A \subset X$ be a non-closed set and $x \in \overline{A} \setminus A$. Let $\{U_\alpha: \alpha < \kappa \}$ enumerate a local base at $x$, where $\kappa < \mathfrak{p}$. Then $\mathcal{F}=\{U_\alpha \cap A: \alpha < \kappa \}$ is a strongly centered family of subsets of the countable set $A$. Since $\mathcal{F}$ has cardinality smaller than $\mathfrak{p}$ we can fix an infinite pseudointersection $S \subset A$ of $\mathcal{F}$. Now $S$ is a sequence inside $A$ which converges to $x$.<|endoftext|>
TITLE: Non-rigorous reasoning in rigorous mathematics
QUESTION [34 upvotes]: I was wondering what role non-rigorous, heuristic type arguments play in rigorous math. Are there examples of rigorous, formal proofs in which a non-rigorous reasoning still plays a central part?
Here is an example of what I am thinking of. You want to prove that some formula $f(n)$ holds, and you want to prove this by induction. Based on heuristic arguments, you conjecture what the correct formula is. Then you prove it by induction. But, if you had just given the induction proof on its own, then you would have to pluck this mysterious formula out of thin air.
I am interested in situations in which there is a heuristic argument which is valid and can be formalized. I am more interested in cases in which there is a heuristic argument and a separate (or complementary) rigorous argument, but the heuristic argument is more enlightening and more explanatory.

REPLY [7 votes]: To me, the quintessential example of the employment of non-rigorous reasoning to arrive at rigorous proofs is Archimedes' On method, where he employs a "mechanical" method, usually called the method of indivisibles, as a heuristic, to arrive at results which he then proves with the method of exhaustion. Even better, Archimedes acknowledges and explains this way of proceeding in the work itself, where he draws the distinction between discovering new results on the one hand (θεωρεῖν), and the subsequent rigorous proving of the results (ἀποδεικνύναι). 
Somehow, this feels remarkably similar to Grothendieck's description of his theory of motives as an instrument of discovery.<|endoftext|>
TITLE: Determinantal formula for the nullspace of a singular matrix
QUESTION [9 upvotes]: In June 2012, Bill Press and Freeman Dyson published a remarkable paper on the iterated prisoner's dilemma. A key step in their derivation is a simple fact from linear algebra that I feel I should have been explicitly aware of all my life but wasn't: If $M$ is a singular matrix, then we can write down explicit determinantal formulas for some of the vectors in its nullspace.  Simply observe that $$M\cdot Adj(M) = (\det M) I = 0$$ where $Adj(M)$ is the adjugate matrix, and observe that the columns of $Adj(M)$ are in the nullspace of $M$.
Now the columns of $Adj(M)$ don't necessarily span the nullspace of $M$ (for a trivial example, take $M=0$).  My question is:

Can one write down determinantal formulas for a spanning set of the nullspace of an arbitrary singular matrix $M$?  (Perhaps one needs to split into cases and give different formulas in different cases; that's O.K.)

I think that part of the reason I don't already know the answer to this question is that in today's computer age we are obsessed with efficient algorithms, and so we have learned to despise Cramer's rule.  However, Press and Dyson's paper illustrates that sometimes this kind of formula can yield important insights.
I should mention that part of the motivation for my question is that Press and Dyson tacitly assume in their paper that for the matrix $M$ they are interested in, the nullspace has dimension one, which strictly speaking is not always true.  I think that if the answer to my question is yes, then it should be easy to patch up this gap in their analysis.

REPLY [11 votes]: If the rank of $M$ is $k$ then $\Lambda^kM$ is a ${n\choose k}\times{n\choose k}$ matrix (consisting of al $k\times k$ minors of $M$) of rank $1$. Any of its nonzero rows is a point on $Gr(k,n)$ in its Plucker embedding. This point gives equations of the nullspace.
For example, if $k=1$ then any nonzero row of $M$ gives the equation of the nuspace.
In another example, $k=n-1$, you take $\Lambda^{n-1}M = Adj(M)^T$ and its nonzero rows give you a point of $Gr(n-1,n)$ which is isomorphic to $P^{n-1}$ and under this isomorphism the point gives you the nullspace.<|endoftext|>
TITLE: Profinite groups as étale fundamental groups
QUESTION [25 upvotes]: Does every profinite group arise as the étale fundamental group of a connected scheme?
Equivalently, does every Galois category arise as the category of finite étale covers of a connected scheme?
Not every profinite group is an absolute galois group of a field (the only finite ones have order $1$ or $2$ by Artin-Schreier). Therefore we cannot restrict to spectra of fields. Perhaps one first has to check if every finite group arises as a fundamental group of a scheme. I don't even know enough examples to answer this question for cyclic groups. At least order $3$ is possible (see here,  Remark 2).
If the answer turns out to be no, then I would like to know which profinite groups arise as fundamental groups.

REPLY [6 votes]: Every finitely presented group occurs as the fundamental group of an irreducible complex variety (see below), so (at least) you get the completions of all those.
Simpson, Carlos. Local systems on proper algebraic $V$-manifolds. Pure Appl. Math. Q. 7 (2011), no. 4, Special Issue: In memory of Eckart Viehweg, 1675--1759. MR2918179<|endoftext|>
TITLE: Necessary and sufficient conditions for a sum of idempotents to be idempotent
QUESTION [13 upvotes]: Given: a finite list of $n$-by-$n$ idempotent complex matrices $E_1, E_2, \ldots, E_k$.
If all pairwise products $E_i E_j$ (with $i \neq j$) are zero, it is trivial to show the sum $E_1 + E_2 + \cdots + E_k$ is idempotent.
The converse, while true, is not so trivial: If the sum is idempotent, all pairwise products are zero.
I have been told that the converse is well known among
statisticians but I cannot find a reference. Do you know one?

REPLY [12 votes]: The key point is that the image of the sum of the idempotents is necessarily the direct sum of the images of the individual idempotents.
Suppose that $E_1,\ldots,E_k$ are idempotent and $F = \sum E_i$ is idempotent.  Let $R_i$ and $K_i$ be the image and kernel of $E_i$, respectively, and let $R$ be the image of $F$.
The trace of an idempotent equals its rank, so $$\dim R = \mathrm{tr}(F) = \sum \mathrm{tr}(E_i) = \sum \dim R_i.$$  Furthermore $R$ is a subspace of $\sum R_i$ and $\sum R_i$ has dimension at most $\sum \dim R_i$, with equality iff this sum is direct, so we actually have $$R = R_1\oplus \cdots \oplus R_k.$$ Consider a vector $v_1\in R_1$.  By definition $$(v_1,0,\ldots,0) = Fv_1 = (E_1v_1,E_2v_1,\ldots,E_k v_1)$$ with respect to this decomposition.  In other words, $R_1\subset K_i$ for all $i\neq 1$, or $E_i E_1 = 0$ for all $i\neq 1$.  It follows similarly that $E_i E_j = 0$ for all $i\neq j$.<|endoftext|>
TITLE: Research in topology for a master student
QUESTION [8 upvotes]: I hope here is the best place to ask this, I will begin my master degree very soon, I've already attended the regular undergraduate courses included Real Analysis, Analysis on manifolds, Abstract Algebra, Field Theory, point-set topology, Algebraic Topology, etc...
I like very much algebraic topology and I found it really beautiful, I would like to know which areas of algebraic topology are the most interesting to begin to work with and which books I can study with my background in order to get the prerequisites to begin to study this subject. 
I want as soon as possible has a "taste" of a current research field in algebraic topology, and I know that an algebraic topologist can give me a "shortest way" while I attend the regular courses of my master degree.
Thank you

REPLY [3 votes]: For your first research problem, I recommend that you find an adviser in your department.
If there is no algebraic topologist in your department, find some other adviser, and
ask to suggest an interesting problem.
It is very unlikely that, as a master student, you will be able to find and solve
a reasonable research
problem yourself, without a help from an experienced adviser.
In this site, people can give you only reading recommendations, and this is probably not enough
to begin your own research.
But of course, there were rare exceptions in history when self-taught mathematicians
did good research.
Here is an outstanding problem in algebraic topology on this site:
fedja (mathoverflow.net/users/1131), Two commuting mappings in the disk, Two commuting mappings in the disk (version: 2009-11-25)
To understand the statement of the problem, little knowledge is required.
What does one need to learn to solve this problem, nobody knows:-)
I wish you luck.<|endoftext|>
TITLE: What are the Dirac operators on $S^1$?
QUESTION [8 upvotes]: This is crossposted at stack exchange as https://math.stackexchange.com/questions/248391/dirac-operators-on-s1.
I am trying to understand the Dirac operators associated to the 2 spinor bundles on $S^1.$ I have been getting very confused about why one bundle has nontrivial harmonic spinors and the other doesn't.(Harmonic spinors are solutions $s$ to the equation $Ds = 0$ where $D$ is the Dirac operator and $s$ is a section.)
Here is my argument (which must be wrong somewhere). We have 2 spin structures given by the connected 2-fold covering and the disconnected 2-fold covering. Since the tangent bundle $TS^1$ is trivial, we can choose the trivial connection on it given by $f \rightarrow df.$ When considered as a connection on the principal bundle of frames (also isomorphic to $S^1$), i.e. as a Lie algebra valued one form on $S^1,$ it must be the zero form.
Ok, so now given either spin structure, the connection must lift to the $0$ connection. Furthermore, any complex line bundle over the circle is trivial, so both cases look exactly the same, and the Dirac operator appears to be $f \rightarrow i\frac{df}{dx}.$
However, I am told that in the case of the connected double cover we should have an additional condition on our $f,$ namely that it should satisfy $f(-x) = -f(x).$ With this extra condition, there cannot be harmonic spinors on the spinor bundle associated to the connected spin structure. Where have I gone wrong?

REPLY [9 votes]: This one is tricky and extremely confusing. The bundle of spinors  is a complex line bundle so it is trivializable.   You detect the   spin structure only if you look at  the Dirac operator.     For one  spin structure the Dirac operator has  a kernel, for the other, it does not. 
For a more detailed  discussion on the pathological case of spin structures on $S^1$ I  recommend you to have a look at the discussion on spin structures on page 150-151 of these notes. There I discuss Milnor's point of view on  this subject, but even Milnor in  his nice little paper  on this subject (see the precise  reference in the above notes) is not very careful about this issue. 
Addendum.    Over   $S^1$ we have two real line bundles, $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ the trivial line bundle $\underline{\bR}$ and the nontrivial line bundle $\tilde{\underline{\bR}}$ with  nontrivial first Stieffel-Whitney class $w_1$. Note that $\tilde{\underline{\bR}}$ can be identified with the tautological (real) line bundle over the projective line $\bR\mathbb{P}^1$.  These line bundles are equipped with natural metrics and compatible connections $\nabla$ and respectively $\tilde{\nabla}$. Something miraculous happens. Although these two real lines bundles are not isomorphic, the  complex line bundles $\underline{\bR}\otimes\bC$ and $\tilde{\underline{\bR}}\otimes\bC$ are isomorphic  as  complex line bundles over $S^1$ (duh!).   The connections $\nabla$ and $\tilde{\nabla}$ induce connections on the complexifications  $\underline{\bR}\otimes\bC$ and $\tilde{\underline{\bR}}\otimes\bC$ and we obtain two Dirac operators $\newcommand{\ii}{\boldsymbol{i}}$  $D=-\ii\nabla_\theta$ and $-\ii\tilde{\nabla}_\theta$ on the same  line bundle.   However, the isomorphism $\Phi$ that maps $\underline{\bR}\otimes\bC$ to  $\tilde{\underline{\bR}}\otimes\bC$ does not map the complexification of $\nabla$ to the complexification of $\tilde{\nabla}$ so that the operators $D$  and $\tilde{D}$ are not conjugate to each  other.   The operator $D$ is $-\ii\frac{d}{d\theta}$,  and
$$ \Phi^{-1}\tilde{D}\Phi =-\ii\frac{d}{d\theta}+\frac{1}{2}= D+\frac{1}{2}. $$<|endoftext|>
TITLE: total variation distance between two solutions of SDE
QUESTION [7 upvotes]: Suppose we have two stochastic differential equations with the same initial conditions:
$$d X_t^1= b_1(t,X_t^1)dt + dW_t$$
$$d X_t^2= b_2(t,X_t^2)dt + dW_t,$$
$X_0^1=X_0^2=x_0$; $W_\cdot$ is a standard one-dimensional Brownian motion, the functions $b_1$ and $b_2$ are uniformly bounded and Lipschitz.
Using Gronwall's inequality one can easily derive bounds on $|X_t^1-X_t^2|$ and hence it is possible to estimate the Wasserstein distance between $X_t^1$ and $X_t^2$. 
My question is, how one can estimate the total variation distance between $X_t^1$ and $X_t^2$? Intuitively, because the processess $X_t^1$ and $X_t^2$ start from the same point, the total variation distance between them should be small (at least for small $t$). So, is it possible to obtain the bounds of the form
$$d_{TV} (X_t^1,X_t^2)\le \phi(t) $$
for some function $\phi$? (here $d_{TV}$ means the total variation distance between two measures)
Thanks!

REPLY [6 votes]: Such a bound can be derived with Girsanov's theorem and Pinsker's inequality. Let $X_t = x_0 + W_t$. Supposing $b$ has linear growth in $x$, we may define measures $P_i$, $i=1,2$ by
$\frac{dP_i}{dP} = \exp\left(\int_0^Tb_i(t,X_t)dW_t - \frac{1}{2}\int_0^T|b_i(t,X_t)|^2dt\right)$.
Then, under $P_i$, $W^i_t := W_t - \int_0^tb_i(s,X_s)ds$ is a Brownian motion, and so $X$ is a weak solution of SDE (i). Let $P^i \circ X^{-1}$ denote the $P^i$-law of the entire process, a measure on the space of continuous functions. Then
$\frac{dP_2}{dP_1} = \exp\left(\int_0^T(b_2(t,X_t) - b_1(t,X_t))dW^1_t - \frac{1}{2}\int_0^T|b_2(t,X_t) - b_1(t,X_t)|^2dt\right)$.
Let $\mathcal{H}(\cdot | \cdot)$ denote the relative entropy. By Pinsker's inequality, 
$
\begin{align}
d_{TV}^2(P_1 \circ X^{-1}, P_2 \circ X^{-1}) &= d_{TV}^2(P_1, P_2) \le 2\mathcal{H}(P_1 | P_2) \newline
 &= -2\mathbb{E}^{P_1}\left[\log \frac{dP^2}{dP^1}\right] \newline
 &= \mathbb{E}^{P_1}\left[\int_0^T|b_1(s,X_s) - b_2(s,X_s)|^2ds\right] \newline
 &= \mathbb{E}^{P}\left[\frac{dP_1}{dP}\int_0^T|b_1(s,X_s) - b_2(s,X_s)|^2ds\right]
\end{align}
$
since the stochastic integral term is a true martingale. This is actually a much stronger control than you requested, since it is easy to see that $d_{TV}(P_1 \circ X_t^{-1}, P_2 \circ X_t^{-1}) \le d_{TV}(P_1 \circ X^{-1}, P_2 \circ X^{-1})$ for any $t$. If you have a uniform bound on $|b_1 - b_2|$, you have a good bound on the $TV$ distance between your processes. You can probably get away with an $L^2$ bound, but you'll then need to fuss with the $dP_1/dP$ term a bit. Hope this helps!
EDIT 1: It's interesting to note that this approach breaks down if you have two different volatility coefficients in your SDE, because the laws of the processes are then singular. But you could still bound the TV distance between the time-$t$ laws, as requested, probably via Malliavin calculus.
EDIT 2: I should elaborate on the first of the string of equalities above. Since $X$ and $W$ generate the same $\sigma$-fields, $dP^i/dP$ is $X$-measurable, and so
$\frac{dP^i \circ X^{-1}}{dP \circ X^{-1}} (X) = \mathbb{E}\left[\frac{dP^i}{dP}|X\right] = \frac{dP^i}{dP}$.
From this it is clear that the TV-distances above are the same, from the formula $d_{TV}(\mu,\nu) = \int d\lambda|d\mu/d\lambda - d\nu/d\lambda|$ for $\mu,\nu \ll \lambda$.<|endoftext|>
TITLE: What notion captures the 'class' of all classes?
QUESTION [14 upvotes]: In ZFC there is no set that is the set of all sets, for this we introduce the notion of class. But then what is the 'class' of all classes, is it actually a class? Do we apply the same idea again? But then at what stage do we stop? Does this show that classes are not the right notion to go beyond sets, but more of an ad-hoc solution?
Further, within foundational category theory, we have the notion of grothendieck universes, if i recall rightly, this is equivalent to introducing an axiom that an inaccessible cardinal exists. Does this subsume, or is equivalent to the notion of classes?
Finally, what is the formalism that uses classes to extend ZFC, is the NBG?

REPLY [9 votes]: To give a very specific  (quite formalistic, and possibly very wrong - depending on your or even my beliefs) answer to some of your questions: 

In ZFC there is no set that is the set of all sets, for this we introduce the notion of class.  

I don't, because I use ZFC.  Whenever I say "class", I mean "formula".  (Today.  I may change my mind tomorrow.)  You may use NBG, of course. 

But then what is the 'class' of all classes.

No such thing, in ZFC.  (Well, there is the set of all formulas.  But that is not what you mean.)  No such thing in NBG either.   Try KM. 

Do we apply the same idea again? But then at what stage do we stop? 

It depends on what you mean by "we".   I stopped at ZFC.  You may go as far as you want. 
You may even use a type-theoretic approach, in which there are infinitely many levels of this
hierarchy. However, once you have countably many levels, you may ask how many levels there are.  But now the pictures looks somewhat similar to a universe $V_{\delta+\omega}$, which ZFC handles very well. 
I seem to recall that Fraenkel-Bar Hillel-Levy, "Foundations of Set Theory", has an enlightening and more detailed discussion of this topic.<|endoftext|>
TITLE: Kähler form on complex Lie group
QUESTION [6 upvotes]: Hallo,
Let $G$ be a semi-simple, compact Lie Group. Consider its complexification $G_{\mathbb{C}}$. Does there exist a Kähler structure on $G_{\mathbb{C}}$ which is $G$-invariant (maybe in a neighbourhood of $G$ in $G_{\mathbb{C}}$)?
hapchiu

REPLY [13 votes]: Yes, such a Kähler form always exists:  Embed $G$ as a matrix group in $\mathrm{SU}(n)$ for some $n$ and then let $G_\mathbb{C}\subset \mathrm{SL}(n,\mathbb{C})\subset M_n(\mathbb{C})$ be the complexification.  Choose a Kähler form on this latter vector space, pull it back to $G_\mathbb{C}$ and then, using the compactness of $G$, average its pullbacks under left and right multiplications.  This will yield a Kähler form on $G_\mathbb{C}$ that is $G$-invariant.<|endoftext|>
TITLE: Are grothendieck universes enough for the foundations of category theory?
QUESTION [11 upvotes]: Grothendieck universes are equivalent to ZFC+a strongly inaccessible cardinal. This is low on the large cardinal axiom list. Is it enough to place category theory on a firm foundational basis, and how about higher category theory, does it remain enough?

REPLY [14 votes]: Mike Shulman wrote a nice expository paper on set theoretical foundations for category theory
http://arxiv.org/abs/0810.1279
In Section 6 he explains the difficulties of working with large categories using just ZFC, and he discusses various ways to deal with these size issues. Some of these do not assume the existence of an inaccessible cardinal.<|endoftext|>
TITLE: Numerical evaluation of the Petersson product of elliptic modular forms
QUESTION [8 upvotes]: It is known how to compute the Fourier expansion of elliptic modular forms using modular symbols, and it is known how to get numerical evaluations of $L$-functions of various type ; it's possible to get explicit values on those matters with sage already.
It is also known that when an Eisenstein series is involved, it's possible to relate the Petersson scalar product to $L$-functions, and hence to evaluate them.
I have seen that sage bug about various pairings for modular forms, but it looks more like it's about the pairing between modular forms and modular symbols than the Petersson scalar product.
My question is: does there exist general formulas to compute the Petersson scalar product of two elliptic modular forms numerically?
EDIT(2012-12-23): I insist on the numerically: having an expansion with estimates on the order of the error with constants which depends on this or that (I'm thinking about those which can be found in chapter 5 of Iwaniek's "Topics in classical automorphic forms" for example) is very nice from a theoretical point of view, but doesn't help when one wants to actually compute with specific forms and to a given precision. In fact, I want to compute various things with the Petersson scalar product, so this question is to check whether I can directly work on them or if I should write something about the matter before.

REPLY [8 votes]: Let
$f(z) = \sum a(n) e(n z)$
and
$g(z) = \sum b(n) e(n z)$
be holomorphic modular forms
of weight $k \in 2 \mathbb{N}$ on $\Gamma =
\operatorname{SL}_2(\mathbb{Z})$
whose product decays rapidly.
Then
\begin{equation}
\int_{\Gamma \backslash \mathbb{H}} y^k \overline{f(z)}g(z) ~ \frac{dx ~ d y}{y^2} =
2
\sum_{n \in \mathbb{N} } \frac{  \overline{a(n)} b(n) }{  n^{k-1} } \sum_{d \in \mathbb{N}} \Phi(4 \pi d \sqrt{n}),
\end{equation}
where
\begin{equation}
\Phi(y)
= 2( \frac{y}{8 \pi})^{k-1}
(y K_{k-2}(y) - K_{k-1}(y)).
\end{equation}
Note that $\Phi(y)
\asymp_k
y^{k-1/2}
e^{-y}$ for $y \gg 1$.
For a general finite index subgroup
 $\Gamma$
 of $\operatorname{SL}_2(\mathbb{Z})$,
a correct formula
may be obtained
by
summing the RHS over the cusps
$\mathfrak{a}$ of $\Gamma$,
weighted by the width $w$ of $\mathfrak{a}$,
and taking for $a(n), b(n)$ ($n \in w^{-1} \mathbb{N}$) the Fourier coefficients at
$\mathfrak{a}$.
A reference
for
a general form of such identities
is Theorem 5.6 (p.24)
in my paper Evaluating modular forms on Shimura curves;
see also Example 5.7,
Remark 3.5,
and the discussion of Sections 5.3--5.6,
which includes a detailed comparison with the other approaches
mentioned in this thread that I will summarize briefly here.
Pros: no need to Hecke-decompose, unlike the "symmetric square"
approach; converges rapidly to the correct value,
unlike the vanilla "equidistribution of horocycles" approach;
generalizes to non-holomorphic forms
lacking a straightforward Hecke decomposition (e.g.,
certain theta series),
although perhaps this feature
is not important for your purposes.
Cons: requires the Fourier expansion at every cusp,
unlike either approach just mentioned
(although the "symmetric square" approach
is not devoid of such subtlety,
since it requires one to compute
the conductor and bad Euler factors
of the symmetric square of a newform).
Another approach (specific to the holomorphic case)
would be to exploit the connection with period polynomials,
for which
a search just now turned up
this article.
One variant of that method also requires knowing Fourier expansions at every cusp, and reduces the problem to evaluating a class of incomplete gamma functions some of which reduce to K-Bessel functions as above; another requires only that one be able to compute periods of a cusp form $f$ over split geodesics in $\Gamma \backslash \mathbb{H}$, which can apparently be done using the Fourier expansion at only one cusp.  Moreover, one can speed up the computation when $f = g$ is an eigenform by exploiting certain rationality results.<|endoftext|>
TITLE: Crossed product of a C*-algebra by a subgroup
QUESTION [6 upvotes]: Let $A$ be a unital $C^*$-algebra, let $G$ be a compact group, let $\alpha:G\to\mbox{Aut}(A)$ be a continuous action, and let $H$ be a closed subgroup of $G$. Is there any relationship between the crossed products $A\rtimes_\alpha G$ and $A\rtimes_{\alpha|_H}H$?
I really only need this for $G=\mathbb{T}$ the unit circle, and $H=\mathbb{Z}_n$ (identified with the $n$-th roots of unity in $\mathbb{T}$). For these groups, and in the case of the trivial action of the circle on $A$, $A\rtimes \mathbb{Z}_n \cong A\otimes \mathbb{C}^n$ is a corner of $A\rtimes \mathbb{T} \cong A\otimes c_0(\mathbb{Z})$, but I don't know if this is true in general.

REPLY [5 votes]: You always have an injective $*$-homomorphism from $A\rtimes H$ into the multiplier algebra of $A\rtimes G$ (the reason is that you can view functions on $H$ as measures on $G$ which are supported on $H$). If $H$ is open in $G$ (a rather unfrequent situation, as you know), then $A\rtimes H$ sits as a $C^*$-subalgebra in $A\rtimes G$.<|endoftext|>
TITLE: Eversion of the 6-sphere in 7-space
QUESTION [21 upvotes]: Say that $S^n$ "admits eversion" if the inclusion $S^n \rightarrow \mathbb{R}^{n+1}$ is regularly homotopic to the antipodal map (where a "regular" homotopy is a continuous path through immersions).
Smale proved that $S^2$ admits eversion by defining an appropriate algebraic invariant corresponding uniquely to regular homotopy classes, and noted that the group this invariant lives in is trivial.  Many people didn't believe it until someone made a movie illustrating an explicit eversion. 
It can be shown that $S^n$ admits eversion if and only if the tangent bundle of $S^{n+1}$ is trivial.  That is, the only spheres which admit eversion are $S^0$, $S^2$, and $S^6$.  
My question is: does anyone know of an explicit eversion of $S^6$ in $\mathbb{R}^7$?

REPLY [11 votes]: (That's my first post on mathoverflow. Henceforth and unfortunately I am not allowed to post comments (this needs reputation 50), so part of the present post in the answer box would better fit in the comments, sorry for this.)
Citing Sullivan's article "The Optiverse" and Other Sphere Eversions, available on the web as of today:

Models of [the Morin-Apéry] eversion were made by Charles Pugh, and Nelson Max digitized these models and interpolated between them for his famous 1977 computer graphics movie "Turning a Sphere Inside Out".

This might be the first movie. You can find it on Youtube today. The Shappiro eversion was as far as I know only published in still pictures (in Scientific American) when the movie above was realized.
The Morin-Apéry eversion was proposed after Shappiro's, which is the first realization (still according to Sullivan's article).
There is also the Geometry Center's movie outside-in, with a decisive contribution by Thurston. As already noted by Igor Rivin, this uses corrugations, in the lines of ideas of what is called the h-principle since Gromov. Note that I am not sure how deep goes the analogy between $C^1$ isometric embeddings (that cannot be taken $C^2$) and those eversions that can be chosen to be $C^k$.
John Sullivan's article presents a movie that he realized in collaboration with Rob Kusner, Ken Brakke, George Francis, and Stuart Levy. It is called the optiverse. It is done by taking optimal path with respect to the Willmore energy (intergal of the square of the mean curvature). You could probably use this idea again (with considerably more computation power needed). But it is not clear how to exploit the output (a moving 6D mesh in $\mathbb{R}^7$).
The last movie I know of is called the Holiverse (see arXiv and/or Youtube).
There is of course no hope for a 7D movie, but it would still be interesting to have more information on the $S^6$ eversion. Let me introduce a question closely related to yours: Morin proved that the minimal number of generic topological transitions in an $S^2$ eversion is 14. What would the minimal number be for $S^6$?<|endoftext|>
TITLE: How does "modern" number theory contribute to further understanding of $\mathbb{N}$?
QUESTION [73 upvotes]: I hope this question is appropriate for MO. It comes from a genuine desire to understand the big picture and ground my own studies "morally".
I'm a graduate student with interest in number theory. I feel like I'm in danger of losing the big picture as I venture a bit deeper and reflect on where I am at. My fundamental is this: I care about the natural numbers - and thus naturally care about the Riemann Zeta function. Number theorists have embarked on various adventures in studying generalized integers (rings of integers of Q-extensions), and their associated zeta functions, and beyond (e.g. Langlands program). Some mathematicians seem to be interested in these generalized integers and zeta functions for their own sake. I am not.
Given my passion for $\mathbb{N}$ and zeta, why should I study these other objects? I understand that philosophically to understand an object it's good to understand its context, and its similarities and differences to its brothers and cousins. This principle makes a lot of sense.
But specifically, what new understandings of $\mathbb{N}$ and zeta have we gained thus far by studying these more general systems? Are there clearly articulated reasons why we can hope to bring back more "treasure" from these more general searches that may shed light on $\mathbb{N}$ in particular? I worry sometimes that number theory is becoming divorced from its original "ground", though I believe (and hope) this feeling derives mainly from ignorance.
EDIT: My question was probably not written very well. I am aware of some of the benefits of studying solutions of polynomials in ring extensions (e.g. solving cases of FLT). My concern is with the broad scope of number theory research today, particularly in the land of generalized zeta-functions and Langlands program. I am uncomfortable (in my ignorance, I admit!) with the apparent lack of a clear connection to the "natural" concerns of number theorists prior to the mid 20th century.
I hope that my question is taken in the spirit of a naive apprentice asking masters for motivation, and a layout of the land of modern research as it connects to concerns that used to be universal.

REPLY [8 votes]: As yet-another comment/question, in addition to the several excellent responses: the question in itself is entirely reasonable, indeed, and is not obviously answered by contemporary professional literature, nor by our current "standard texts". One should rightly suspect/worry that there's been some sort of swindle perpetrated...
And, indeed, the monuments and icons of past times do seem to assume a life and mythology of their own, which all too often has dropped explicit or pointed references to previous history.
So, indeed, one should ask oneself these questions...
And, simultaneously, it would seem fair to ask what all this chatter is about while we've not only not proven RH, but not even proven a zero-free strip!?! Cart before the horse? :)
Probably the genuine explanation in human terms is that the immediate description of things misled us as to what was a cart, and what was a horse, and... maybe, what was "before".
So, yes, while we seem to have made frustratingly little progress on "old issues", and we make only tenuous analogies in further generalizations... what is the alternative?
That is, we do what we can. And we do have some pressure to "hype" it to funding agencies, to Deans, and to each other. Yes, this seems to claim remarkable progress... while we've not really made so much, by more sober standards. :)  But it is not easy to construct a better model of reporting progress.
At some point, then, one observes that it'd be extremely unwise to decide to prove RH for a PhD thesis, or postdoc project, ... or maybe anything else... despite the extreme desirability of progress on exactly such a thing.
One true operational point is that people have to "make a living", which currently requires reporting incredible progress to department heads and deans. Thus, ... noise.
The notion of "deconstruction" of any human activity (that is, acknowledging that it is a human activity, rather than the human-independent, context-independent "absolute" that has a sort of pernicious charm ... for humans) greatly clarifies the seeming paradoxes of human activities, in many cases. :)<|endoftext|>
TITLE: Iterating Random Matrix Operations
QUESTION [9 upvotes]: Consider the following probability measure on the integers concentrated around $0$: the probability of drawing $0$ is $\frac{1}{2}$, of drawing ($1$ or $-1$) is $\frac{1}{4}$ split evenly among the two choices, of drawing ($2$ or $-2$) is $\frac{1}{8}$ and so forth. Call this measure $\mu$.
Now consider an $n \times n$ matrix $M$ whose entries have been drawn in a $\mu$-iid fashion. Define a $\mu$-random row operation on $M$ as follows: pick two numbers $p,q$ randomly according to the uniform distribution on $\lbrace 1,2,\ldots,n \rbrace$ and also pick an integer $m$ according to $\mu$. Then, we perform the row operation that replaces the $p$-th row of $M$ by itself plus $m$ times the $q$-th row. Similarly define a $\mu$-random column operation.

What does the distribution of entries in $M$ look like as the number of $\mu$-random row and column operations goes to $\infty$?

Background
I should confess that this question does not arise from my own research, but rather from watching the progress of some undergraduates taking linear algebra who have not quite mastered the Gaussian elimination algorithm yet. Their strategy for row reduction seemed to involve blindly performing row operations without caring for pivots or GCD's or anything of the sort. In order to incur a minimal computational burden, they seemed to prefer small scalar multiples of rows, so I chose $\mu$ to be concentrated on the small integers. What are the odds that they will get lucky with this blind strategy and end up with a matrix largely populated with zeros?

REPLY [4 votes]: I would also mention a paper by H. Furstenberg and H. Kesten, "Products of random matrices", 1960.
In fact, a general idea is that when you are making a product of random matrices, unless there is a degeneracy (an invariant finite union of planes), almost surely you will see the rows to become more and more aligned (that is, the projectivization of your dynamics collapses almost all the projective space, including the points corresponding to the base vectors $e_i$, to one "wandering" point).
And quite naturally it means, that if the matrices are in the $SL(n,\mathbb{R})$ group, the contraction in all the directions but one means expansion in the last one: the elements of the product grow exponentially.
Now you multiply your matrix M on both sides by such matrices. It seems rather clear (though, perhaps, you need some cleanup for it to become a really formal proof) that what you will have is still a matrix with exponentially big entries, almost-aligned (angle close to zero) rows and almost-aligned columns.
Poor students! If they don't hit the answer quickly, they wander off to infinity in  $SL(n,\mathbb{Z})$...<|endoftext|>
TITLE: Particles chasing one another around a circle
QUESTION [18 upvotes]: Two particles start out at random positions on a unit-circumference circle.
Each has a random speed (distance per unit time) moving counterclockwise uniformly distributed
within $[0,1]$.  How long until they occupy the same position?  In the example below,
the red particle catches the green particle at $t=5.9$, i.e., nearly six times around the circle:

          


The distribution of overtake-times is quite skewed, indicating perhaps
the mean could be $\infty$.  For example, in one simulation run, it
took more than $3$ million times around the circle before one particle finally caught
the other. So I don't trust the means I am seeing (about $25$).

What is the distribution of overtake-times?

I was initially studying $n$ particles on a circle, but $n=2$ seems already somewhat interesting...
Update (2Dec12).
Alexandre Eremenko concisely established that the expected overtake-time (the mean)
is indeed $\infty$.
But I wonder what is the median, or the mode?  Simulations suggest the median is about $1.58$ and
the mode of rounded overtake-times is $1$, reflecting a distribution highly skewed toward rapid overtake. (The median is suspiciously close to $\pi/2$ ...)
Update (3Dec12). Fully answered now with Vaughn Climenhaga's derivation of the
distribution, which shows that the median is
$1 + \frac 1{\sqrt{3}} \approx 1.577$.

REPLY [17 votes]: To answer your questions about median and mode, one can take Alexandre's answer a little further and compute the exact distribution function for the overtake-times.
Note that the overtake-time doesn't depend on $v_1,v_2$ directly, but only on their difference.  Call the difference $v$.  Now $v$ is the difference of two uniformly distributed random variables on $[0,1]$, so it is supported on $[-1,1]$ with probability density function $1-|v|$.  Moreover, since $\theta$ is uniformly distributed we can without loss of generality identify the cases $(v,\theta)$ and $(-v,1-\theta)$ and reduce everything to the following set-up:

$v$ is distributed on $[0,1]$ with density function $2(1-v)$.
$\theta$ is uniformly distributed on $[0,1]$.
The overtake-time is $t=\theta/v$.

Now we can compute the cumulative density function for the overtake-time.  Indeed, we have $P(t<T) = P(\theta/v<T) = P(\theta < Tv)$, which we can get by the following integral:
$$
P(t<T) = \int_0^1 2(1-v) P(\theta < Tv | v) \,dv.
$$
The probability $P(\theta < Tv | v)$ is given by the function $f(\theta,v) = \max(Tv,1)$.  Thus for $T\leq 1$, we have $f(\theta,v)=Tv$ for all $v\in[0,1]$, so integrating gives $P(t<T) = T/3$, while for $T\geq 1$, we integrate and find
$$
P(t<T) = \int_0^{1/T} 2(1-v)Tv\,dv + \int_{1/T}^1 2(1-v)\,dv = 1-\frac 1T + \frac 1{3T^2}.
$$
So in the end the cumulative density function for the overtake-time is
$$
P(t<T) = \begin{cases} 
\frac T3 & T\leq 1, \\
1 - \frac 1T + \frac 1{3T^2} & T \geq 1.
\end{cases}
$$
The term $1/T$ in the last expression will give you the infinite mean, since upon differentiating the CDF you'll get a term $1/T^2$, which upon multiplying by $T$ and integrating to get the mean you end up integrating $1/T$ from $1$ to $\infty$.
As for the median, it looks as though any proximity to $\pi/2$ is just a red herring, because solving for $P(t<T) = 1/2$ yields $T=1 + \frac 1{\sqrt{3}} \approx 1.57735\dots$.<|endoftext|>
TITLE: Is there a composite number that satisfies these conditions?
QUESTION [13 upvotes]: We know that if $q=4k+3$ ($q$ is a prime), then $(a+bi)^q=a-bi \pmod q$ for every Gaussian integer $a+bi$. Now consider a composite number $N=4k+3$ that satisfies this condition for the case $a+bi=3+2i$. I use Mathematica 8 and find no solution less than $5\cdot 10^7$. Can someone find a larger number for the condition, and can this be used for a primality test?

REPLY [5 votes]: As Mr. R. Gerbicz pointed out in the mersenne forum an eventual counterexample for the base 3+2i must be 13-PRP (just multiply the equation with its conjugate). The first point to check is to make a list of pseudoprimes base 13 which are 3 (mod 4). I checked them to 10^10 and there is no counterexample which pass this test (a couple of them which are 1 (mod 4) passes the complex base test, but none of the 3 (mod 4)). However, the general opinion is that this test is a "hidden" multi-base PRP test, or a (n-1)(n+1) combined test, and as Mr. Tom Womack pointed out in that thread, if a couterexample exists, it must be HIGH (somewhere in 10^30 or so).<|endoftext|>
TITLE: Groups with an automorphism of order two fixing only two elements
QUESTION [8 upvotes]: It is well known that a finite group admitting an automorphism of order 2 that fixes only the identity is abelian and has odd order. Moreover, the automorphism is inversion.
Is anything known about finite groups admitting an automorphism of order 2 that fixes only the identity and one other element?

REPLY [6 votes]: MacKay [On the structure of a special class of $p$-groups, Quart. J. Math. Oxford Ser (2) 38, 489-502] and, indipendently, Kiming [Structure and derived length of finite $p$-groups possessing an automorphism of $p$-power order having exactly $p$ fixed points, Math. Scand. 62, 153-172] showed that if a finite $p$-group $G$ admits an automorphism of order $p^n$ with exactly $p$ fixed points, then $G$ contains a subgroup $H$ of index bounded by a function of $p$ and $n$ which is nilpotent of class at most 2 (and $H$ is abelian if $p=2$).<|endoftext|>
TITLE: Non-definable elements vs indiscernible elements
QUESTION [7 upvotes]: Let $\Sigma$ be a one-sorted first-order signature, let $A$ be a $\Sigma$-structure, and let $B \subseteq A$ be a $\Sigma$-substructure. Fix a class $\mathcal{L}$ of formulae over $\Sigma$. We say an element $a$ in $A$ is $\mathcal{L}$-definable over $B$ just if there is some formula $\phi (x, \vec{y})$ in $\mathcal{L}$ such that $A \vDash \phi [a, \vec{b}] \land (\phi [x, \vec{b}] \to (x = a))$ for some finite sequence $\vec{b}$ of elements of $B$; and we say two elements $a, a'$ in $A$ are $\mathcal{L}$-indiscernible over $B$ just if $A \vDash \phi [a, \vec{b}] \leftrightarrow \phi [a', \vec{b}]$ for all formulae $\phi$ in $\mathcal{L}$ and all finite sequences $\vec{b}$ of elements in $B$.
Question. What conditions can we put on $A$, $B$, $\mathcal{L}$, and/or $\Sigma$ so that that an element $a$ is not $\mathcal{L}$-definable over $B$ if and only if there exists $a'$ such that $a$ and $a'$ are $\mathcal{L}$-indiscernible over $B$?
According to Hodges [Model theory, Lem. 4.1.3], if $\mathcal{L}$ is the class of $L_{\omega_1, \omega}$ formulae over $\Sigma$ and $A$ is countable, then non-definability and indiscernibility coincide in the sense described above. However, the proof of the lemma relies entirely on Scott's theorem [Cor. 3.5.4], and it is not clear to me whether this can be generalised. (Hodges [Shorter model theory, just before Thm 3.2.5] says that there is no known satisfactory analogue for uncountable cardinalities.)
I am primarily interested in the case where $\mathcal{L}$ is the class of regular formulae over $\Sigma$, i.e. the smallest class containing the atomic formulae (not including $\bot$) and closed under finite conjunctions (including $\top$) and existential quantification. I do know that there are non-trivial examples when $\mathcal{L}$ is the class of equations: the fundamental theorem of Galois theory implies that non-definability and indiscernibility coincide when $A$ is a Galois extension of $B$.

REPLY [5 votes]: Here is one fairly general thing to say:

If $A$ is $\kappa$-saturated and $B$ has size less than $\kappa$, then every element $a\in A$ that is not first-order definable in $A$ using parameters in $B$ will be part of an indiscernible pair, and conversely.

The reason is that if $a$ is not definable in $A$ over $B$, then every finite collection of assertions $\varphi(a,\vec b)$ true in $A$ are also true of some $x\neq a$. So by saturation, that type will be realized by some $a'$, which will be indiscernible from $a$ in $A$ over $B$. Conversely, an element $a$ that is part of an indiscernible pair can never be definable. 
There are many weakenings and refinements of the saturation property, and I would expect that the model theorists will be able to say much more.<|endoftext|>
TITLE: Integers represented by the polynomial $a^2+b^3+c^6$
QUESTION [10 upvotes]: Can every sufficiently large integer be written in the form $a^2 + b^3 + c^6 $ for some non-negative integers $a,b$ and $c$?

REPLY [18 votes]: This is question 3 on page 146 of the second edition of The Hardy-Littlewood Method by R. C. Vaughan, with $b \geq 0.$ 

          


                (Image added by J.O'Rourke)
The answer is No, by a simple volume argument. It is not even necessary to know the exact constant, just that the number of lattice points with $x,y,z \geq 0  $ and $$  x^2 + y^3 + z^6 \leq N$$ is less than $CN,$ with a constant $0 < C < 1.$
I'm in question 5 on the same page. 
Image hosting did not work. Please see: VAUGHAN PDF 
Oh, well. The relevant calculation is the sum of the reciprocals of the exponents, in case the polynomial is the sum of distinct monomials in the different variables. You might think that $$  x^2 + y^2 + z^9$$ ought to represent all large numbers with $z \geq 0.$ There are no local obstructions. But this is not the case. We can think of the exponent in this volume calcultaion as $10/9.$
It is reasonable to ask, how large an exponent reciprocal sum can we get and still fail to represent large integers? The best I have is $4/3,$ and in the simplest form we require coefficients, as in
$$  x^2 + 27 y^2 + 7 z^3.  $$ This is a version of the $ 4 x^2 + 2 x y + 7 y^2 + z^3,  $ which is more natural but looks less diagonal.<|endoftext|>
TITLE: Finite order arithmetic and ETCS
QUESTION [9 upvotes]: I'm looking for a reference to the statement that Lawvere's Elementary Theory of the Category of Sets (ETCS) is equal in proof-theoretic strength to finite order arithmetic.  The person who informed me of this said it was well-known in certain circles, but he couldn't think of a reference.
Actually, all I need is a reference to one half of the equivalence: that anything provable in finite order arithmetic is provable in ETCS.  The story: I've been looking at Colin McLarty's paper A finite order arithemetic foundation for cohomology, which shows that nothing stronger than finite order arithmetic is needed anywhere in EGA or SGA.  I want to state that nothing stronger than ETCS is needed anywhere in EGA or SGA.  To back that up with references, I therefore need something that relates ETCS to finite order arithmetic.

Edit This question has generated lots of discussion about McLarty's paper.  I'm genuinely interested in that discussion, but I'd also like to emphasize that it's peripheral to my question, which is simply a reference request: where can I find it stated/proved that ETCS is equal in strength to finite order arithmetic?

Further edit Maybe I can make this question more transparent to experts in non-categorical set theory.  ETCS is well-known to have the same strength as the membership-based theory known as "bounded Zermelo with choice" or "restricted Zermelo with choice".  (One reference: Mac Lane and Moerdijk, Sheaves in Geometry and Logic, Section VI.10.)  The axioms are extensionality, empty set, pairing, union, power set, foundation, restricted comprehension, infinity, and choice.  Here "restricted comprehension" means that we only consider formulas that are restricted in the sense that all quantifiers are of the form "$\forall x \in y$" or "$\exists x \in y$".

REPLY [3 votes]: Ah, Thomas Forster's 1998 paper

Forster T. (1994) Weak systems of set theory related to HOL. In: Melham T.F., Camilleri J. (eds) Higher Order Logic Theorem Proving and Its Applications. HUG 1994. Lecture Notes in Computer Science, vol 859. Springer, Berlin, Heidelberg. doi:10.1007/3-540-58450-1_43

is available on-line at various places including here.
He says it is proved in

Jensen RB, On the consistency of a slight (?) modification of Quine's NF, Synthese 19 1969 pp 25--63, doi:10.1007/978-94-010-1709-1_16

Lake J, Comparing Type theory and Set theory, Zeitschrift fur Matematische Logik 21 1975 pp  355-56. doi:10.1002/malq.19750210144


For a fanatically detailed proof and discussion see

Mathias, A. R. D. The strength of Mac Lane set theory. Ann. Pure Appl. Logic 110 (2001), no. 1-3, 107–234,
doi:10.1016/S0168-0072(00)00031-2, author pdf<|endoftext|>
TITLE: Why does the Solovay-Tennenbaum theorem work?
QUESTION [6 upvotes]: I have decided to learn iterations at long last, and I am reading through Jech's Set Theory for now. The standard example after explaining what is an iteration with finite support is the following theorem (quoted from Jech, Theorem 16.13):

Theorem (Solovay-Tennenbaum). Assume GCH and let $\kappa$ be a regular cardinal greater than $\aleph_1$. There is a c.c.c. notion of forcing $P$ such that the generic extension $V[G]$ by $P$ satisfies Martin's Axiom and $2^{\aleph_0}=\kappa$.

I believe that I understood the proof mechanically, but even though I feel that I understand it in a more profound way - I have this feeling that I really didn't get the point. It feels like magic. 
What I managed to understand is that we have enumerations of all isomorphism classes of small posets in each stage, and generically we reach them all. Alas, like any new forcing argument, this too feels like some sort of witchcraft.
I also understand completely why $2^{\aleph_0}=\kappa$, and I have to admit that this is one clever argument, but I can't really see the full intuition behind it. I just see that it works, and I can reason why it works, by MA there is a generic for any $<\kappa$ dense sets, but this means that we have to have at least $\kappa$ dense subsets in the universe so $2^{\aleph_0}\geq\kappa$, where the other side of the inequality follows from the fact that $|P|\leq\kappa$ and $P$ is c.c.c.

Can anyone shed some intuition behind the proof? Some insights about why it works?

REPLY [9 votes]: Let me add a couple of comments to Joel's and Martin's answers.  They both pointed out that you get $\kappa$ reals for a much simpler reason than having forced MA, namely that at $\kappa$ stages you chose to add a Cohen real. You might ask what happens if you never choose to add a real.  Indeed, that's the original Solovay-Tennenbaum situation, where each step was forcing with a Suslin tree, which doesn't add reals.  Nevertheless, you get new reals anyway.  Cohen reals appear at the limit stages of cofinality $\omega$ in any finite-support iteration.  
I think of MA as a compatibility result.  There are lots of things that would naturally be done by ccc forcing --- killing a Suslin tree, adding a single measure-zero set that covers all the ground model's (coded) measure-zero sets (amoeba forcing), adding Cohen reals, etc.  The main point of MA is that you can do all these things together, and mixing them together won't produce disasters.  Notice that, in other circumstances, mixing innocuous-looking forcings can produce disasters, expecially collapsing cardinals. For example, if you specialize an Aronszajn tree and I add a path through it, then between us we've collapsed $\aleph_1$.  Similarly if you shoot a club through a stationary co-stationary set and I shoot a club through its complement.  The key technical fact underlying MA is that a finite-support iteration of arbitrary ccc forcings is still ccc.  So you can mix these forcings at will, without any risk of collapsing cardinals.<|endoftext|>
TITLE: The Dedekind eta function in physics
QUESTION [22 upvotes]: This interesting little fellow (a nice introduction is the video "Mock Modular Forms are Everywhere" by Cheng and Felder) popped up in some operator algebra (Witt / Virasoro Lie algebra) I was exploring, so I've been curious about where else the Dedekind $\eta$-function makes a cameo appearance and found that in physics it's related to the

statistical  parameters of solvable Ising models

(See "The Reasonable and Unreasonable Effectiveness of Number Theory in Statistical Mechanics" by G. Andrews and "Introduction to Exactly Solvable models in Statistical Mechanics" by C. Tracy.)
The difference between the average local occupation densities of two sub-lattices of a hard hexagon model of a lattice gas given on pp. 368-371 of Tracy is $R(\tau)=\frac{n(\tau)\eta(5\tau)}{\eta^2(3\tau)}$.

partition functions (statistical mechanics variety) for colored bosons moving on a
line  (1/24 is the associated Casimir energy) and one-color fermions

operator traces (characters) for the infinite dimensional Lie algebras $\widehat{su}_n$,
equivalent to 2-Dim current algebras

partition function of a microscopic black-hole in a 5-Dim D-brane

string theory guage corrections


(For 2-5, see "Nucleon Structure, Duality and Elliptic Theta Functions" by W. Scott. For item 2, see also "Vertex Operators and Modular Forms" by G. Mason and M. Tuite.)
From pg. 39 of "Fivebrane instantons ..." and on pg. 11 of "D3 instantons ...,"
a correction to the field basis (of the RR axion dual to D3-branes) in type IIB string perturbation theory related to the action of S-duality in ten dimensions:
$\tilde{c_a} \mapsto \tilde{c_a}-\tilde{c}_{2,a}\:\epsilon(g)$ where, with $g=\binom{a\:\:b}{c\:\:d}$,
$$\exp(2\pi i \epsilon(g))=\frac{\eta\left [ \frac{a\tau+b}{c\tau+d} \right ]}{\left ( c\tau+d \right )^{\frac{1}{2}}\eta(\tau)}.$$

partition function in 2+1 dimensions and vanishing chemical potential of
non-relativistic fermions in a constant magnetic field

("Nonrelativistic Fermions in Magnetic Fields: a Quantum Field Theory Approach" by O. Espinosa, J. Gamboa, S. Lepe, and F. Mendez)

physics of gauge theories and the Dirac operator

(See "The Logarithm of the Dedekind $\eta$ Function" by M. Atiyah.)
Michael Atiyah even goes so far as to say, "It seems therefore timely to attempt to survey the whole development of the theory of $\log(\eta)$, putting results in their natural order and in the appropriate general context. This is the aim of the present paper, in which the emphasis will be strongly geometrical. In a sense we shall show that the latest ideas from physics [circa 1987] provide the key to a proper understanding of Dedekind's original results."
[ 7.5 Edit 1/18/21: The lecture "A geometric approach to the modular flow on the space of lattices" by Bruce Bartlett interleaves topics presented by Atiyah above and Ghys below.]

knots and dynamics

(See "Knots and Dynamics" by E. Ghys, and Chapter 2 A New Twist in Knot Theory in Dana MacKenzie's book What's Happening in the Mathematical Sciences Vol. 7.)
Ghys presents the equation  $$24\log\eta\left(\frac{a\tau+b}{c\tau+d}\right)=24\: \log(\eta(\tau))+6\: \log(-(c\tau+d)^{2})+2\pi i\:\mathfrak{R}\left(\binom{a\: b}{c\: d}\right)
 $$
where $\mathfrak{R}$ is the Rademacher function, which he relates to the linking number
between two knots related to modular/Lorenz flow: “For every hyperbolic element $A=\binom{a\: b}{c\: d}$ in $PSL(2,Z)$, the linking number between the [modular/Lorenz] knot $k_A$ and the trefoil knot $l$ is equal to $\mathfrak{R}(A)$ ....”

string/brane partition functions, propagators, and metrics

In "String Theory" by S. Nibbelink, $\eta$ occurs in the denominator of string partition functions for fermionic and bosonic zero modes (pp. 163-7).
A coefficient in the 10-dim metric for a 7-brane is given as $e^{\phi}=\tau_2 \eta^2\bar{\eta}^2|\prod_{i=1}^{k}(z-z_i)^{-\frac{1}{12}}|^2$ on pg. 493 of "Supergravity vacua and solitons" by G. Gibbons.
In what other contexts in physics does the Dedekind $\eta$ function take a bow?
(Edit) Moreover, since this is a community wiki and not a test question with one best answer but an attempt to come to a better understanding of the $\eta$-function and associated math and physics, I invite people to expand on any of the items with specifics (e.g., exact formulas), more references, and/or insightful commentaries (e.g., what you believe are important aspects of the references).
Other appearances: In Gliozzi's "The Infrared Limit of QCD Effective String" on pg. 14; Panero's "A numerical study of confinement in compact QED" on pg. 4; Zahed's "Holographic Pomeron and Primordial Viscosity" on pg. 1; Caselle and Pinn's "On the Universality of Certain Non-Renormalizable Contributions in Two-Dimensional Quantum Field Theory" on pg. 3; Billo, Casselle, and Pellegrini's "New numerical results and novel effective string predictions for Wilson loops" on pg. 6 and 15; and Basar, Kharzeev, Yee, and Zahed's "Holographic Pomeron and the Schwinger Mechanism" on pg. 7.

REPLY [7 votes]: The Dedekind eta function shows up in three-dimensional quantum gravity: http://arxiv.org/abs/0712.0155 (Alexander Maloney, Edward Witten, Quantum Gravity Partition Functions in Three Dimensions). On page 17 a basic partition function $Z_{0,1}$ of the theory is calculated as $$Z_{0,1}(\tau)=\frac{1}{|\eta(\tau)|^2}|\bar q q|^{-(k-1/24)}|1-q|^2.$$ It also appears in the calculation of supergravity partition functions in sec.7.
The Dedekind eta function also enters in (supersymmetric) physics through mock modular forms: http://arxiv.org/abs/1208.4074 (Atish Dabholkar, Sameer Murthy, Don Zagier, Quantum Black Holes, Wall Crossing, and Mock Modular Forms).
A good review of mock modular forms is http://mathcs.emory.edu/~ono/publications-cv/pdfs/114.pdf (Ken Ono, Unearthing the visions of a master: harmonic Maass forms and number theory).
I first heared about Dedekind Eta Function in the physics context via Freeman Dyson's lovely essay "Missed opportunities": http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183533964<|endoftext|>
TITLE: Is this a Julia set (and if so, for which function family is it the Julia set)?
QUESTION [9 upvotes]: Consider the function family given by $f_\lambda(z) = z - p_\lambda(z)/p_\lambda'(z)$ where $p_\lambda(z) = (z^2 - 1)(z - \lambda)$.  Every attracting cycle and every rational neutral cycle of $f_\lambda$ attracts the one critical point of $f_\lambda$, which is $\lambda/3$ (see this other MathOverflow question and Alexandre Eremenko's answer for context).
If the sequence of iterates, $f^n_\lambda(\lambda/3)$, converges to a fixed point of $f_\lambda$ then the fixed point is one of the roots of $p_\lambda$, that is $1$, $-1$, or $\lambda$.  If we mark the points $\lambda$ in $\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to each of these fixed points then the result is an image that is reminiscent of a Newton basin fractal.  Let $K$ be the set of values $\lambda\in\mathbb{C}$ such that $f^n_\lambda(\lambda/3)$ converges to a fixed point of $f_\lambda$.  My question is:

Is the border, $\partial K$, a Julia set, and if so then for what function family $g$ does $J(g) = \partial K$?

I suspect that the answer is "yes" and that $g$ is a family of iterates of a rational function.  In the attached images, the very bright green points are the ones nearest to the border $\partial K$.
A note about the little Mandelbrot sets visible in two of the attached images:  In these images the white points indicate parameters $\lambda$ such that $f^n_\lambda(\lambda/3)$ converges to a rational neutral cycle of period greater than $1$.  The white points indicate the bifurcation locus of $f_\lambda$ ($J(f_\lambda)$ is not continuously determined, in the sense of the Hausdorff metric, by the parameter at these points).  The white points appear to be composed of many scaled and rotated copies of the boundary of the Mandelbrot set and I believe they are contained by $\partial K$.  Perhaps that is a useful thing to know when searching for $g$.

REPLY [5 votes]: The set in question is the bifurcation locus of the family $f_{\lambda}$. It is hence the set of non-normality of the family
$$\bigl(\lambda\mapsto f_{\lambda}^n(\lambda/3)\bigr)_{n\in\mathbb{N}};$$
see Theorem 4.2 in McMullen's book "Complex Dynamics and Renormalization".
As has been pointed out, you cannot expect the set to exactly coincide with the Julia set of a rational function. It should be possible to prove this formally. Indeed, your set has complementary components bounded by analytic curves, or regions bounded by curves analytic except at a single cusp. Such curves cannot bound Fatou components of a rational map, unless the map is a Blaschke product and the Julia set is itself a circle.
(I cannot seem to find a reference for this fact right now. However, the boundary of a Siegel disk may be smooth, but cannot be analytic anywhere; otherwise, the conjugacy to a rotation would extend beyond the boundary. On the other hand, boundaries of attracting basins are even known to have Hausdorff dimension strictly greater than one; see Przytycki, "On hyperbolic Hausdorff dimension of the boundary of a basin of attraction for a holomorphic map and of quasirepellers". Showing that the boundary cannot be analytic is much easier, both for attracting and parabolic basins.)<|endoftext|>
TITLE: expository papers related to quantum groups 
QUESTION [23 upvotes]: Hello all, 
I know basic representation theory(finite groups, lie groups&lie algebras) and I want to get a flavor of quantum groups (why they are useful, important results etc) and other related things like the Yang-Baxter equation. Can someone suggest me some good expository articles? Thank you.

REPLY [5 votes]: This is mostly meant as a reminder that the original question asks about articles and not books or research papers.    Of the latter there are a huge number by now.  The notion of "quantum group" has multiple aspects, not easily covered by a single exposition, but it's reasonable to look for a fairly brief guide to what is going on in the subject (and why).    Some of the lecture notes and other expositions mentioned here should be useful, at least within their own defined limits, but the large books such as Chari-Pressley and the research monographs such as Lusztig's book go far beyond the scope of "article".     (Jantzen's more introductory textbook on the other hand is a fairly elementary introduction to one important line of work, though not to all possible ones.)
There isn't actually a precise mathematical definition of "quantum group", which is definitely a problem with the kind of free-flowing discussion in the answers here.   Hopf algebra theory is more narrowly defined, but even here there are too many directions to encompass in a survey article or set of lecture notes.<|endoftext|>
TITLE: Suslin algebras
QUESTION [6 upvotes]: A Suslin algebra is a generalization of a Suslin tree: it is a ccc, $(\omega,\infty)$-distributive boolean algebra.  Jech proved that every Suslin algebra has size at most $2^{\omega_1}$.  A proof is given in the current edition of his book "Set Theory."  He also mentions that it is consistent to have Suslin algebras of size $2^{\omega_1}$ but does not sketch an argument.  I have been having trouble finding a proof of this fact.  I looked in the older edition of Jech's book, and he says a little more about it there, but still no proof.  I would be very grateful if someone could point me to a paper that has a construction of a large Suslin algebra, or sketch a proof here in an answer.  I am also curious about the following:
1) Is it consistent to have Suslin trees but no Suslin algebras of density larger than $\omega_1$?
2) Can the existence of large Suslin algebras be proved from several diamonds, say $\Diamond_{\omega_1}$ + $\Diamond_{\omega_2}(cof(\omega_1))$?
Thanks!

REPLY [2 votes]: Jech, Thomas J.
Some combinatorial problems concerning uncountable cardinals. 
Ann. Math. Logic 5 (1972/73), 165–198. 
Section 5 contains the forcing for arbitrarily big Suslin algebras.  See also:
Scharfenberger-Fabian, Gido
Chain homogeneous Souslin algebras. (English summary) 
MLQ Math. Log. Q. 57 (2011), no. 6, 591–610.
According to Gido, his paper, Jech's, and Jensen's result mentioned by Dorais are the only known contsructions of "big" Suslin algebras.  Jech's forcing is thus the only thing that makes one bigger than $\aleph_2$.
Note that Jech does not explicitly prove distributivity in his paper, but it is not hard to show given what is there.<|endoftext|>
TITLE: What is the importance of defining the classes $C^k$ of functions ?
QUESTION [7 upvotes]: What is the importance of defining the classes $C^k$ of functions ? Wouldn't be more natural defining the classes of functions differentiable k-times, with k=0,1,2,...
Why the continuity of the last derivative is so important to justify a specific definition ?

REPLY [7 votes]: Among the properties that make $C^k$ a nice category, and that have no analog for just differentiable maps, I'd list these (some of whom have already mentioned in the comments):
1. The total differential theorem. To check that a function of several variables is $C^k$ becomes a matter of partial differentiability in each single variable (and continuity of the partial differentials in all of them).
2. The local inverse theorem. The core of local differential calculus: checking that a local property holds locally at a point just requires verifying that the analogous property holds for the differentials at that point in the corresponding linear category. This holds for being a local diffeo, a local immersion  or submersion; for transversality of mappings, &c. 
3. Characterization. (Converse of Taylor formula). A map is $C^k$ if and only if it has a polynomial expansion of order $k$ "locally uniformly", with continuous coefficients.
4. Whitney extension theorem. With a natural notion of $C^k$ smoothness on closed sets, every smooth map on a closed set extends on the whole space.
5. Ordinary Differential Equations.   $C^k(I)$ is a convenient space for well-posed ODE, starting with the problem of finding an antiderivative, given by the Fundamental Theorem of Calculus. Compare with the difficulty of characterising derivatives of everywhere differentaible functions (see this MO question). Note however that $C^k$ is not so a convenient setting for PDE's, e.g. in potential theory.
6. Functional Analysis. For any open set $\Omega\subset\mathbb{R}^n$ the class $C^k(\Omega)$ is a Fréchet space, embedded as a closed subspace into  $C^0(\Omega,\mathbb{R}^N)$ taking a function into its $k$-jet. Something similar also works for the class of $C^k$ maps between $C^k$ manifolds, that can be embedded in the space of    $C^0$ maps in the $k$-jets.<|endoftext|>
TITLE: Multiplication of Cauchy and Dedekind real numbers
QUESTION [8 upvotes]: In Michael Dummett's book "Elements of Intuitionism", the product of real numbers is defined as follow:
$x\cdot y= \{ \langle r_n\rangle \cdot \langle s_n\rangle$ | $\langle r_n\rangle\in x , \langle s_n\rangle\in y \}$, where  $\langle r_n\rangle ,\langle s_n\rangle$ are Cauchy sequences of rational numbers, and $\langle r_n\rangle \cdot \langle s_n\rangle=\langle r_n\cdot  s_n\rangle$.
This definition is valid iff we can prove intuitionistically $\{ \langle r_n\rangle \cdot \langle s_n\rangle$ | $\langle r_n\rangle\in x , \langle s_n\rangle\in y \}$ is indeed a real number, i.e. $\{ \langle r_n\rangle \cdot \langle s_n\rangle$ | $\langle r_n\rangle\in x , \langle s_n\rangle\in y \}$ is closed under "equivalent" (we say that $\langle r_n\rangle$ is equivalent to $\langle s_n\rangle$ if for any natural number $k$, we can find a natural number $n$ such that $|r_m-s_m|<2^{-k} $ for all $m>n$ ). This is to say, we must prove the following proposition intuitionistically:
Let $\langle r_n\rangle , \langle s_n\rangle , \langle t_n\rangle $ be Cauchy sequences of rational numbers, and $\langle t_n\rangle $ is equivalent to $\langle r_n\rangle \cdot \langle s_n\rangle$. Then we can construct two Cauchy sequences $\langle r_n'\rangle , \langle s_n'\rangle $ of rational numbers such that
 (1)  $\langle r_n'\rangle$ is equivalent to $\langle r_n\rangle$;
 (2)  $\langle s_n'\rangle$ is equivalent to $\langle s_n\rangle$;
 (3)  $\langle t_n\rangle =\langle r_n'\rangle \cdot \langle s_n'\rangle$.
But  Michael Dummett doesn't justify his definition, and I find it's very difficult to prove the above proposition intuitionistically. Could you help me?

REPLY [8 votes]: Whilst the definition of addition of Cauchy or Dedekind real numbers is "obvious", multiplication is rather more tricky.  Unfortunately, most accounts, including [RD], leave it as an "exercise for the reader", without even giving a hint about what the problem is, so the questioner is right to ask about this.  The difficulty is intrinsic to multiplication: the only difference in the intuitionistic setting is that we must do the job properly, instead of bodging it by treating positive, zero and negative numbers separately.
The point is that, if you want to achieve precision $\epsilon$ in the product of two numbers, one of which is bounded by $B$, then the other must be given within $\epsilon/B$.
[MD] is not quoted verbatim in the Question, but it is close enough, whilst the accounts in [AH] and [AT] are essentially the same.
[TD] gives a more general account of uniform continuity, taking explicit account of the modulus of convergence of Cauchy sequences (the function that says how far down the sequence you have to go to get a desired accuracy). This is needed elsewhere in constructive analysis.
[BB] has by far the clearest treatment that I have seen of the arithmetic of Cauchy reals, building the modulus into the definition.  (I admire this book for its "can do" attitude, not dwelling on the counterexamples.)  It gives the explicit (but snappy) proof of correctness for multiplication.
[BT] defines multiplication for Dedekind reals and proves correctness.  It shows how Dedekind reals are the limiting case of intervals and also considers "back-to-front" (Kaucher) intervals, which are related to existential quantification just as ordinary intervals are related to universal quantification.
[JC] defines multiplication in a completely novel fashion for Conway (surreal) numbers. This is adapted to multiplication of real numbers in a topos in [PJ].
Since [MD,AH,AT] do not give the explicit answer to the Question, here it is. 
As above, $\langle r_n\rangle$ is a Cauchy sequence if  $\forall k.\exists n.\forall m.|r_{n+m}-r_n|\lt 2^{-k}$.
We write $\alpha(k)$ for such an $n$ for each given $k$; this is the modulus of convergence.
In particular, with $k=0$, for any Cauchy sequence $\langle r_n\rangle$ there are integers $N=\alpha(0)$ and $K=\log_2(r_N)$ such that $\forall m.-2^{K}\lt r_N-1\lt r_{N+m}\lt r_N+1\lt 2^{K}$.
Let $M$, $L$ and $\beta$ be the corresponding integers and modulus for the Cauchy sequence $\langle s_n\rangle$.
Now, given $h$, let $k\geq h+L+1$, $l\geq h+K+1$, $n\geq\alpha(k)$ and $n\geq\beta(l)$.  Then, for all $m$,
$$\begin{eqnarray}
  |r_{n+m} s_{n+m} - r_n s_n|
   &\leq&  |r_{n+m}| |s_{n+m} - s_n| + |r_{n+m} - r_n| |s_n| \\
   &\lt&   2^K 2^{-l} +  2^{-k} 2^L \leq 2^{-h}.
\end{eqnarray}$$
Hence $\langle r_n s_n\rangle$ is a Cauchy sequence with modulus
$\gamma(h)=\max(\alpha(h+L+1),\beta(h+K+1))$.
We can avoid considering equivalence of sequences explicitly, by observing that two Cauchy sequences are equivalent iff they are both subsequences of the same Cauchy sequence. Rationals are represented by constant Cauchy sequences and the new operation for them agrees with multiplication of rationals.
This argument amounts to saying that the new operation is continuous with respect to the Euclidean topology. Also, the rationals are dense amongst Cauchy reals. Hence the new operation is the unique continuous extension and it follows that it obeys the usual algebraic laws for multiplication.
[BT] Andrej Bauer and Paul Taylor,
  The Dedekind Reals in Abstract Stone Duality,
  in Mathematical Structures in Computer Science, 19 (2009) 757-838.
[BB] Errett Bishop and Douglas Bridges, Foundations of Constructive Analysis,
  Grundlehren der mathematischen Wissenschaften, Springer-Verlag, 1985.
[JC] John Horton Conway, On Numbers and Games,
  Number 6 in London Mathematical Society Monographs. Academic Press, 1976.
  Revised edition, 2001, published by A K Peters, Ltd.
[RD] Richard Dedekind, Stetigkeit und irrationale Zahlen, Braunschweig, 1872.
  Reprinted in [DW], pages 315–334; English translation, Continuity and Irrational Numbers, in [DE].
[DE] Richard Dedekind, Essays on the theory of numbers, Open Court, 1901;
   English translations by Wooster Woodruff Beman; republished by Dover, 1963.
[DW] Richard Dedekind. Gesammelte mathematische Werke, volume 3. Vieweg, Braunschweig, 1932; edited by Robert Fricke, Emmy Noether and Oystein Ore; republished by Chelsea, New York, 1969.
[MD] Michael Dummett, Elements of Intuitionism,  Oxford University Press, 2000.
[AH] Arend Heyting, Intuitionism, an Introduction, Studies in Logic and the Foundations of Mathematics, North-Holland, 1956. Third edition, 1971.
[PJ] Peter Johnstone, Topos Theory, London Mathematical Society Monographs 10, Academic Press, 1977.
[AT] Anne Troelstra, Principles of Intuitionism, Lectures presented
at the Summer Conference on Intuitionism and Proof Theory (1968) at
SUNY at Buffalo, NY, Lecture Notes in Mathematics 95, Springer-Verlag, 1969.
[TD] Anne Sjerp Troelstra and Dirk van Dalen, Constructivism in Mathematics, an Introduction,
   Number 121 and 123 in Studies in Logic and the Foundations of Mathematics, North-Holland, 1988.
If you know of other explicit accounts of multiplication for Cauchy or Dedekind reals then please give the references in comments below.<|endoftext|>
TITLE: Approximation power of wavelets
QUESTION [5 upvotes]: The Wikipedia article on Wavelet Transform states that:

Wavelet compression is not good for all kinds of data: transient signal characteristics mean good wavelet compression, while smooth, periodic signals are better compressed by other methods, particularly traditional harmonic compression (frequency domain, as by Fourier transforms and related).

What is the precise quantification of this statement? For the smooth periodic signals we have for example: if the signal is of class $C^d$ of $d$-times continuously differentiable periodic functions, then the partial Fourier sum of order $N$ gives uniform approximation error at most $N^{-d-1}$. What would be the corresponding characterization for wavelet series?
To make the question more concrete: if the signal is piecewise $C^d$-periodic, with, say, one discontinuity, what approximation error do I get with $N$ wavelet coefficients (the Fourier approximation gives only $N^{-1}$ away from the discontinuity)?  Conversely, how many coefficients do I need to get an accuracy $\epsilon$?

REPLY [5 votes]: The relationship between wavelets and smoothness spaces is characterized by "Jackson" and "Bernstein" inequalities. Jackson inequalities look something like 
$$||f-\pi_k f||_{L_p} \le C_1(k,p) |f|_{\text{smoothness seminorm}},$$
where $\pi_k$ is a projector onto the $k$'th level approximation space (space spanned by all wavelets up to resolution level $k$) and $1 \le p \le \infty$. 
Bernstein inequalities look something like
$$|u|_{\text{smoothness seminorm}} \le C_2(k,p)||u|| _{L_p},$$
where $u$ is any function in the $k$'th level approximation space.
The exact results vary based on the domain, choice of wavelets, and smoothness space of interest. Usually $C_1 \approx 2^{-k d}$, where $d$ is the level of smoothness in the seminorm, and $C_2 \approx 2^{k/p}$, but you'll have to look to the literature for your situation. These sort of results are often proved by breaking space up into a bunch of dyadic cubes and then applying polynomial approximation theory on them.
For a well written introduction, see DeVore's explanatory paper.<|endoftext|>
TITLE: Sum of commuting semisimple operators
QUESTION [11 upvotes]: Let $V$ be a finite dimensional vector space over a field $K$. An operator $T:V\to V$ is called semi-simple if every $T$-invariant subspace of $V$ has a $T$-invariant complement(for algebraically closed fields these are exactly  diagonalizable operators). Is it true that the sum (or the product) of two commuting semi-simple operators is semi-simple?

REPLY [10 votes]: The answer is still No, but for not that obvious reason. To show that, let us start with a positive claim.
1. FIrst of all, an operator $T$ is semisimple iff all the factors in the prime expansion of its minimal annihilating polynomial $\mu$ are distinct. Actually, the algebra $K[T]$ is isomorphic to $K[X]/(\mu)$; so, if there are no multiple factors, then this algebra is a direct sum of fields, and each its finitely generated module is semisimple. Otherwise, if $\mu=p^2q$ with $p$ nonconstant, then the annihilator space of $p(T)$ has no complement.
2. Now assume that the extension of $K$ generated by all the roots of characteristic polynomials of $T_1$ and $T_2$ is separable. Then they should be diagonalizable over this extension by the reasons of the minimal polynomial. Since they commute, they are simultaneously diagonalizable. Hence their sum and product are also diagonalizable, and their minimal polynomials have no multiple roots (from separability!) and hence no multiple factors. So $T_1+T_2$ and $T_1T_2$ are both semisimple.
3. And here is a counterexample for non-separable case. Let $K=F_2(t)$ be the field of rational fractions over $F_2$. Set $T_1=\begin{pmatrix}0&0&1&0\cr 0&0&0&1\cr t&0&0&0\cr 0&t&0&0\end{pmatrix}$ and $T_2=\begin{pmatrix}0&1&0&0\cr t&0&0&0\cr 0&0&0&1\cr 0&0&t&0\end{pmatrix}$; their common  minimal polynomial is $X^2-t$, so they are semisimple (this polynomial is irreducible although not separable). But their sum is $S=T_1+T_2=\begin{pmatrix}0&1&1&0\cr t&0&0&1\cr t&0&0&1\cr 0&t&t&0\end{pmatrix}$ with $S^2=0$, and their product is $P=T_1T_2=T_2T_1=\begin{pmatrix}0&0&0&1\cr 0&0&t&0\cr 0&t&0&0\cr t^2&0&0&0\end{pmatrix}$ with the minimal polynomial $(X+t)^2$. Hence both are not semisimple. One may present a direct example of a spaces that cannot be complemented as $\langle e_2+e_3,e_1+te_4\rangle$ in both cases.
In fact, this example was obtained from the action of algebra $K[X,Y]/(X^2-t,Y^2-t)$ on its regular module; $T_1$ and $T_2$ correspond to $X$ and $Y$, respectively.<|endoftext|>
TITLE: Non-uniform complexity of the halting problem
QUESTION [5 upvotes]: This question is approximately cross-posted from Theoretical Computer Science Stack Exchange: https://cstheory.stackexchange.com/questions/14445/complexity-of-the-halting-problem
What can be said about the non-uniform circuit complexity $C(n)$ of the halting problem? Obviously it is $O(2^n)$ as any other decision problem. However I don't know any non-trivial bound.
A related measure of complexity is the time-complexity $T(n)$ with infinite advice (i.e. we allow an extra-tape in our Turing machine which carries an infinite-amount of information in the initial state). This also gives the obvious upper bound $O(2^n)$ by storing an infinite look-up table on this tape.
The two measures of complexity are related as follows. Consider $R_n$ a family of circuits solving a decision problem $S$. Then we can construct an infinite-advice program $H$ for solving $S$ by encoding $R_n$ as advice. This yields
$$T(n) = O(n C(n) \ln C(n))$$
On the other hand if we have $H$ an infinite-advice program solving $S$, we can construct the circuit $R_n$ representing the computation process of $H$ on an input of size $n$. The size of this circuit is the product of the spatial complexity by the temporal complexity so
$$C(n) = O(T(n)^2)$$
Note that if the halting problem is in $P/poly$ i.e. $C$ is polynomial, then $NP \subset P/poly$ (which implies $PH = \Sigma_2$). To see this consider $S \subset \lbrace 0,1 \rbrace^*$ a decision problem in $NP$ and $V$ a verifier program for $S$. Deciding whether $x \in S$ is equivalent to solving the halting problem for the following program $Q_x$: "Loop over all $p \in \lbrace 0,1 \rbrace^*$, halt if $V(x,p) = 1$". The size of $Q_x$ is the same as the size of $x$, up to a constant. Therefore if we can solve the halting problem for $Q_x$ in polynomial time with polynomial advice, we can decide $x \in S$ in polynomial time with polynomial advice
If the halting problem is in $coNP/poly$ then $NP \subset coNP/poly$. This is due to reasoning similar to above i.e. an existential quantifier can be replaced by a universal quantifier at the cost of requiring polynomial advice. I think this also implies some kind of collapse of the polynomial hierarchy
It is possible to construct a specific infinite-advice algorithm of optimal complexity, analogous to Levin search for $NP$ problems. As opposed to the case of $NP$, there is no way to verify correctness of solutions, on the other hand it is possible to restrict the dovetailing only to valid programs. This is done by encoding all programs which solve the halting problem together with their respective infinite advice sequences in the infinite advice of our algorithm. The penalty incurred by using this encoding is at most polynomial, hence the resulting algorithm has complexity which is optimal up to a polynomial

REPLY [11 votes]: Every r.e. language is polynomial-time reducible to the halting problem. Since there are computable languages (indeed, in $\Delta^E_3$) having the maximum possible circuit complexity for every length $n$ (which is asymptotically $2^n/n$), the halting problem also has exponential circuit complexity. The exact complexity will depend on the particular representation of algorithms in the definition of the halting problem, and specifically, on the complexity of the reduction function which hardwires an input string into a fixed algorithm. In the most obvious representations, this function blows up the input length only linearly and can be made computable by linear-size circuits, hence we get $2^{\Theta(n)}$ as the circuit complexity of the halting problem. If the halting problem is formulated directly for algorithms accepting an input, the reduction function increases the input length by an additive constant and has essentially constant complexity, so the circuit complexity of the halting problem is $\Theta(2^n/n)$ in such a formulation.<|endoftext|>
TITLE: Examples of Clifford modules
QUESTION [17 upvotes]: For a Riemannian manifold $M$, a Clifford module is a bundle over $M$ that is, fiberwise, a representation of the Clifford algebra bundle $\mathrm{Cl}(TM)$ and has a connection compatible with this action.  The canonical examples of Clifford modules are

$\Lambda^* M$ (in two ways, with the grading formed by the degree of differential forms or with the hodge star).
The spinor bundle associated to a spin or spin$^c$ structure.
The Dolbeault complex $\Lambda^{0,*} TM$ for a complex manifold (which is really a special case of a spin$^c$ structure).

Are there other examples of Clifford modules that come up naturally?

REPLY [4 votes]: The following example is very specific (but generalizes to hypersurfaces in spin-manifolds). Consider an immersion $f\colon \Sigma\to\mathbb R^3,$ and the Riemannian metric induced by the immersion. Identify $\mathfrak{su}(2)=\mathbb R^3$. Thus, vectors in $\mathbb R^3$ act on $\mathbb C^2,$ which then is the spinor bundle for the euclidean 3-space. In this regard, tangent vectors to $\Sigma$ act on the trivial $\mathbb C^2$ bundle over $\Sigma.$ This bundle is equipped with the canonical trivial connection $d$ which is compatible with the Clifford action.
As a bundle with Clifford action, it can be identified with the spinor module,
but the connection $d$ is in general not the spinor connection of the induced Riemannian metric. In fact, the difference between these two connections is determined (and determines) the second fundamental form of the surface. A classical consequence of the above is the Weierstrass representation for minimal surfaces in euclidean 3-space.<|endoftext|>
TITLE: Matrices over Finite Prime Fields
QUESTION [5 upvotes]: Let $p$ be an odd prime and $\mathbb Z_p$ be the prime field of order $p$. Consider the matrix ring $R=M_n(\mathbb Z_p)$. Is there any method to count the solutions of the equation (in the ring $R$)
$$X^2=I.$$
Where $I$ is the identity matrix?

REPLY [2 votes]: Because $p$ is different from $2$, the solutions are the same as direct sum decompositions $\mathbb{Z}_p^{\oplus n} = E_{+1} \oplus E_{-1}$ as $\mathbb{Z}_p$-vector spaces.  You can index these by the dimensions, say $r$ and $n-r$.  For each, the number of solutions is
$$
\frac{(p^n-1)(p^n-p) \cdot \dots \cdot (p^n-p^{r-1})}{(p^r-1)(p^r-p) \cdot \dots \cdot (p^r-p^{r-1})} \cdot p^{r(n-r)}
$$
So the final answer is the sum over $r$ from $0$ to $n$ of this factor.

REPLY [2 votes]: There are $\displaystyle \sum_{i=0}^n \frac{N(n)}{N(i)N(n-i)}$ solutions to your problem, where 
$N(r)$ is the number of elements in $GL_r(\mathbb{F}_p)$.
Any such $X$ is invertible, and $X\in GL_n(\mathbb{F}_p)$ is a solution if and only if it is conjugate in $GL_n(\mathbb{F}_p)$ to a diagonal matrix with the first $i$ entries equal to $1$ 
and last $n-i$ entries equal to $-1$. This explains the number of summands. Now $g\in GL_n(\mathbb{F}_p)$  commutes with such diagonal matrix if and only if it preserves both eigenspaces. Hence, there are $\frac{N(n)}{N(i)N(n-i)}$ elements in the conjugacy class. 
The formula can be made even more explicit by substituting a formula for $N(r)$.<|endoftext|>
TITLE: Relationship between eigenvalues of $A-B$ and eigenvalues of $A^2-B^2$
QUESTION [10 upvotes]: Let us suppose that $A_{n}$ and $B_n$ are sequences of positive definite matrices satisfying
$$c \leq \lambda_{\min}(A_n)\leq \lambda_{\max}(A_n)\leq C$$
and
$$c \leq \lambda_{\min}(B_n)\leq \lambda_{\max}(B_n)\leq C$$
where $\lambda_{\min}$ and $\lambda_{\max}$ are minimal and maximal eigenvalues. Then, what's the relationship between $\|A_n-B_n\|_2$ and $\|A_n^2-B_n^2\|_2$? Is it true that 
$$\|A_n-B_n\|_2 \leq \text{constant} \cdot \|A_n^2-B_n^2\|_2$$
where the $2$-norm of a matrix $M$ is the maximum of the absolute value of its minimal eigenvalue and the absolute value of its maximal eigenvalue.

REPLY [5 votes]: The norm you consider is usually called operator norm, or the norm subordinated to the $\ell^2$ norm over ${\mathbb C}^n$ (or ${\mathbb R}_n$). The correct inequality for positive Hermitian matrices $A$ an $B$ is
$$\|A-B\|_2\le\sqrt{\|A^2-B^2\|_2}.$$
See Exercise 110 of my additional list accompagnying my book Matrices, GTM 216 (Springer-Verlag).<|endoftext|>
TITLE: Partitions comprised only of divisors
QUESTION [6 upvotes]: How many of the partitions of a natural number $n$ are comprised only of its divisors? That is, if $$p(n)=\sum_{\sum_{1}^n kj_k=n:j_k\geq 0} 1_{\[j_1,j_2,...\]},$$
is the ordinary partition function (i.e. the total number of partitions of $n$), then I want to know something about the counting function $$s(n)=\sum_{\sum_{d|n}dj_d=n:j_d\geq 0}1_{[j_1,...,j_d,...]}.$$
I would be happy to hear of anything that is known about this function, but I am particularly interested in (a) its generating functions, and (b) a bijection between this restricted partition and another (hopefully more intuitive to count) restricted partition. Any insights would be welcome. 
I should add that google finds a number of papers that study "partitions of $n$ into divisors of $m$", e.g. Gupta, 1970s, but those methods reduce to rather vacuous statements when evaluated at $m=n$.
Thanks!

REPLY [8 votes]: Bounds for this partition function were given by the editors of The American Mathematical Monthly, Paul Erdos, and Andrew Odlyzko in the March 1992 issue, p. 277, as a solution to Advance Problem number 6640. The bounds they prove are:
$$
({\tau(n)}/{2}-1)(\log n + O({\log n}/{\log \log n}))
\leq 
\log s(n)
\leq
({\tau (n)}/{2})\log n + O(\log \log n),
$$
where $\tau(n)$ is the number of divisors of $n$.<|endoftext|>
TITLE: Experimental mathematics: how are floating point equations discovered/converted to exact equations?
QUESTION [7 upvotes]: the 2005 AMS article/survey on experimental mathematics[1] by Bailey/Borwein mentions many remarkable successes in the field including new formulas for $\pi$ that were discovered via the PSLQ algorithm as well as many other examples. however, it appears to glaringly leave out any description of the crucial step. it goes from discussing large accuracy floating point operations/formulas to stating the exact theoretical formulas with no discussion of the intermediate step(s).

suppose a floating point formula is found experimentally that holds for many finite points of the formula to a high degree of precision. how is it proven that the abstract algebraic formula which does not use floating point arithmetic with finite accuracy is correct?

this seems to relate to induction. of course a formula can hold for a finite number of points or finite precision but then fail for more points or "infinite" precision. is there any discussion that focuses on this step/conversion/aspect? of course a virtually identical/analogous issue arises in statistics with curve fitting and the danger of "overfitting".
[1] Experimental Mathematics: Examples, Methods and
Implications by Bailey/Borwein, notes of the AMS, 2005

REPLY [13 votes]: In most situations, floating point approximations can not prove a closed-form formula.  An exception: if the value of an expression is known to be a member of a given discrete set, an accurate enough approximation can discriminate between members of that set.  But once discovered by numerical methods, the BBP formula and similar ones are proven by standard techniques.  In the BBP case, note that
$$ \sum_{k=0}^\infty \frac{z^{8k+j}}{8k+j} = \int_0^z \sum_{k=0}^\infty t^{8k+j-1}\ dt
= \int_0^z \frac{t^{j-1}}{1-t^8}\ dt$$
which can be evaluated using partial fractions, and the results combined and simplified to obtain $\pi$.<|endoftext|>
TITLE: When is the kernel of the etale fundamental group in a fibration abelian?
QUESTION [8 upvotes]: Let $X \to Y$ be a smooth proper morphism. Let $y$ be a geometric point of $Y$. Is the kernel of the natural map of etale fundamental groups $\pi_1^{et}(X_y) \to \pi_1^{et} (X)$ abelian?
This is true for analytic fundamental groups, by the long exact sequence of homotopy groups. 
It is not obviously true for etale fundamental groups of schemes over $\mathbb C$, since the profinite completion functor is not exact.
This article provides a version of the long exact sequence of etale homotopy, that specifically avoids saying this about $\pi_1$. Thus I suspect there is some kind of obvious counterexample to at least the most bold possible version of this question. If so, what is that counterexample? Is any special case of this statement true?

REPLY [4 votes]: Regarding the suggestion in my comment, there is a subtle point regarding my proposed pencil of curves; the Euler identity fails, so one has to check by hand that the singular members occur only for $t=0$ and $t=\infty$.  In fact, this is not always the case.  However, I checked this in one case.
Let $k$ be an algebraically closed field of characteristic $2$.  Let $a,b\in k$ be elements such that none of $a,b,a+b$ equals $0$.  Denote $\mathbb{P}^1_k$ by $\overline{Y}$, and let $[s,t]$ be homogeneous coordinates on $\mathbb{P}^1_k$.  Let $[u,v,w]$ be homogeneous coordinates on $\mathbb{P}^2_k$.  Let $\overline{X}$ be the Cartier divisor in $\mathbb{P}^1_k \times_k \mathbb{P}^2_k$ with defining equation $f(s,t;u,v,w) = suvw(u+v+w) - t(au+bv-(a+b)w)^4$.  Consider the projection morphism $\overline{\pi}:\overline{X}\to \overline{Y}$.  By direct computation, the only singular points of the morphism occur where $([s,t],[u,v,w])$ equals one of 
$$
([1,0],[1,0,0]), ([1,0],[0,1,0]), ([1,0],[0,0,1]), 
$$
$$
([1,0],[1,1,0]), ([1,0],[1,0,1]), ([1,0],[0,1,1]), ([0,1],[u,v,w]).  
$$
The tricky coordinates, of course, are $[u,v,w]=[1,1,1]$.  However, the conditions on $a$ and $b$ guarantee this is contained in the member $[s,t]=[0,1]$ of the pencil.  So this is a pencil to which my comments above apply.  Define $Y\subset \overline{Y}$ to be the open complement of $\{[1,0],[0,1]\}$.  Define $X\subset \overline{X}$ to be the inverse image of $Y$ under $\overline{\pi}$.  Finally, define $\pi:X\to Y$ to be the restriction of $\overline{\pi}$.  Then $\pi$ is a proper, smooth morphism.
To summarize the comments, by Deligne-Fulton the tame fundamental group of the open complement $U:= \mathbb{P}^2 \setminus Z(uvw(u+v+w)(au+bv-(a+b)w))$ is Abelian.  This is also a dense open subset of $X$.  Thus the induced map from the étale fundamental group of $U$ to the étale fundamental group of $X$ is surjective.  In particular, the tame fundamental group of $X$ is also Abelian.  The geometric generic fiber $X_{\overline{\eta}}$ of $\pi$ is a smooth, genus $3$ curve.  Therefore its tame fundamental group is non-Abelian.  So the kernel of the homomorphism of étale tame fundamental groups $\pi_1^t(X_{\overline{\eta}})\to \pi_1^t(X)$ is non-Abelian.  
$\textbf{Edit}.$  I am a little worried now about the distinction between the tame fundamental group and the full étale fundamental group in this argument.  Take any finite group $G$ of order prime to $p$.  There is an injective homomorphism into a finite symmetric group $S_{p^r}$.  Yet the maximal prime-to-$p$ quotient of $S_{p^r}$ is just $S_2$ (or trivial if $p$ equals $2$).  So perhaps the kernel of the map of tame fundamental groups is much larger than the kernel of the map of full étale fundamental groups.
$\textbf{Second Edit}.$  I looked in the article of Friedlander that Will Sawin linked.  Friedlander also needs to work with completions of the higher étale homotopy groups at primes different from the characteristic; in particular, this factors through the maximal prime-to-$p$ quotient.  My example above also works if we look at the maximal prime-to-$p$ quotients of the étale fundamental group, or if we look at an $\ell$-completion for $\ell$ a prime different from $p$, since both of these factor through the tame fundamental group.  So I feel that this example does explain why Friedlander did not state his theorem for $\pi_1$ as well as for $\pi_n, n\geq 2$.<|endoftext|>
TITLE: What are the relations between conjugates and commutators?
QUESTION [17 upvotes]: The following algebraic structure came up when I was thinking about invariants of coloured knots. The elements are all elements of a noncommutative free group $F$, and the operations are: 

$a^b= b^{-1}ab$, taking the conjugate in $F$.
$[a,b]= aba^{-1}b^{-1}$, taking the commutator of two elements in $F$.

And that's all. (If I were allowing only conjugation then this structure would be a conjugation quandle, but I'm also allowing to take commutators, but not to take products- the group product is not part of the structure). Are such structures at all studied or known?

Question: Is the full set of relations in this structure known (in the sense of universal algebra)? Is there a proof in the literature?

REPLY [10 votes]: There is a notion of multiplicative Lie algebras introduced here: Ellis, Graham J.
On five well-known commutator identities. J. Austral. Math. Soc. Ser. A 54 (1993), no. 1, 1–19. The signature there does include multiplication, though. The problem of finding axioms was solved there (I think somebody  finally proved that the five standard commutator identities suffice). If the product operation is removed from the signature, the correct first question would be if the class is first order axiomatizable. It is not clear. One can write a bunch of axioms which certainly hold, but this list is not complete:

$[x,x]=[y,y]$ (call this element 1)
$[x,1]=[1,x]=1$
$[[x,y], [y,x]]=1$
$[x,y]=1, [x,z]=1 \to [x,[y,z]]=1$
$[x,y]^z=[x^z,y^z]$
$x^z=x \leftrightarrow [x,z]=1$
$\exists x,y, [x,y]\ne 1 \to \exists x \exists a, [x,a]=1, a\ne 1, a\ne x$

(The last axiom follows from the fact that if all elements of a group are of order 2, then the group is Abelian.) 
I think it is clear that the class cannot be axiomatized only by universal formulas (because it is not closed under taking subalgebras). It is closed under taking ultraproducts, which is a good news.<|endoftext|>
TITLE: Quotients of Cantor cubes onto spaces 
QUESTION [5 upvotes]: Let $\lambda$ be an infinite cardinal. Consider the Cantor cube $\Delta_\lambda = \{0,1\}^\lambda$. It is a standard fact in topology that the topological weight (= minimal cardinality for a basis) of $\Delta_\lambda$ is $\lambda$. Let $S$ be a zero-dimensional compact space of weight $\lambda$ and suppose $s\colon \Delta_\lambda\to S$ is a continuous suriection. Does there exists a closed subspace $D$ of $\Delta_\lambda$, which is homeomorphic to $S$ such that $p|_D$ is a homeomorphism?

REPLY [5 votes]: The following construction is due to Pashenkov (see "Extensions of compact spaces", Soviet Math. Dokl., 1974):
Let $X=2^\omega$ and $Z=X \times 2^X$ with product topologies everywhere. Define an equivalence relation on $Z$ by $$(x_1,y_1) \sim (x_2,y_2) \Longleftrightarrow x_1=x_2 \land y_1(x)=y_2(x) \mbox{ for all } x \in X \setminus \{x_1\}.$$
Let $\hat{Z}$ be the quotient space $Z / \sim$ and let $s: Z \to \hat{Z}$ denote the quotient map. Among other things, Pashenkov shows that $\hat{Z}$ is zero-dimensional of weight $2^{\aleph_0}$ and that $s$ is an irreducible map (i.e. there is no proper closed subset of $Z$ on which $s$ is still onto $\hat{Z}$).
Thus we get a counterexample to your question by taking $\lambda=2^{\aleph_0}$, $S=\hat{Z}$ and noting that $Z$ is homeomorphic to $2^\lambda$.<|endoftext|>
TITLE: If d/dx is an operator, on what does it operate?
QUESTION [21 upvotes]: If $\frac{d}{dx}$ is a differential operator, what are its inputs?  If the answer is "(differentiable) functions" (i.e., variable-agnostic sets of ordered pairs), we have difficulty distinguishing between $\frac{d}{dx}$ and $\frac{d}{dt}$, which in practice have different meanings.  If the answer is "(differentiable) functions of $x$", what does that mean?  It sounds like a peculiar hybrid of mathematical object (function) with mathematical notation (variable $x$). 
Does $\frac{d}{dx}$ have an interpretation as an operator, distinct from $\frac{d}{dt}$, and consistent with its use in first-year Calculus?

REPLY [20 votes]: Not sure why this question is back on the front page, but I just wanted to add that the situation seems to be clarified by temporarily generalising to higher dimensions and to curved spaces, i.e., by taking a differential geometry perspective.
Firstly, a quick reminder of the concept of a dual basis in linear algebra: if one has an $n$-dimensional vector space $V$ (let's say over the reals ${\bf R}$ for sake of discussion), and one has a basis $e^1,\dots,e^n$ of it, then there is a unique dual basis $e_1,\dots,e_n$ of the dual space $V^* = \mathrm{Hom}(V,{\bf R})$, such that $e_i(e^j) = \delta_i^j$ for all $i,j=1,\dots,n$ ($\delta_i^j$ being the Kronecker delta, and where I am trying to choose subscripts and superscripts in accordance with Einstein notation).  It is worth pointing out that while each dual basis element $e_i$ is "dual" to its counterpart $e^i$ in the sense that $e_i(e^i) = 1$, $e_i$ is not determined purely by $e^i$ (except in the one-dimensional case $n=1$); one must also know all the other vectors in the basis besides $e^i$ in order to calculate $e_i$.
In a similar spirit, whenever one has an $n$-dimensional smooth manifold $M$, and (locally) one has $n$ smooth coordinate functions $x^1,\dots,x^n: M \to {\bf R}$ on this manifold, whose differentials $dx^1,\dots,dx^n$ form a basis of the cotangent space at every point $p$ of the manifold $M$, then (locally at least) there is a unique "dual basis" of derivations $\partial_1,\dots,\partial_n$ on $C^\infty(M)$ with the property $\partial_i x^j = \delta_i^j$ for $i,j=1,\dots,n$.  (By the way, proving this claim is an excellent exercise for someone who really wants to understand the modern foundations of differential geometry.)
Now, traditionally, the derivation $\partial_i$ is instead denoted $\frac{\partial}{\partial x^i}$.  But the notation is a bit misleading as it suggests that $\frac{\partial}{\partial x^i}$ only depends on the $i^{th}$ coordinate function $x^i$, when in fact it depends on the entire basis $x^1,\dots,x^n$ of coordinate functions.  One can fix this by using more complicated notation, e.g., $\frac{\partial}{\partial x^i}|_{x^1,\dots,x^{i-1},x^{i+1},\dots,x^n}$, which informally means "differentiate with respect to $x^i$ while holding the other coordinates $x^1,\dots,x^{i-1},\dots,x^{i+1},\dots,x^n$ fixed".  One sees this sort of notation for instance in thermodynamics.  Of course, things are much simpler in the one-dimensional setting $n=1$; here, any coordinate function $x$ (with differential $dx$ nowhere vanishing) gives rise to a unique derivation $\frac{d}{dx}$ such that $\frac{d}{dx} x = 1$.
With this perspective, we can finally answer the original question.  The symbol $x$ refers to a coordinate function $x: M \to {\bf R}$ on the one-dimensional domain $M$ that one is working on.  Usually, one "simplifies" things by identifying $M$ with ${\bf R}$ (or maybe a subset thereof, such as an interval $[a,b]$) and setting $x$ to be the identity function $x(p) = p$, but here we will adopt instead a more differential geometric perspective and refuse to make this identification.  The inputs to $\frac{d}{dx}$ are smooth (or at least differentiable) functions $f$ on the one-dimensional domain $M$.  Again, one usually "simplifies" things by thinking of $f$ as functions of the coordinate function $x$, but really they are functions of the position variable $p$; this distinction between $x$ and $p$ is usually obscured due to the above-mentioned "simplification" $x(p)=p$, which is convenient for calculation but causes conceptual confusion by conflating the map with the territory.
Thus, for instance, the identity
$$ \frac{d}{dx} x^2 = 2x$$
should actually be interpreted as
$$ \frac{d}{dx} (p \mapsto x(p)^2) = (p \mapsto 2x(p)),$$
where $p \mapsto x(p)^2$ denotes the function that takes the position variable $p$ to the quantity $x(p)^2$, and similarly for $p \mapsto 2x(p)$.
If one also had another coordinate $t: M \to {\bf R}$ on the same domain $M$, then one would have another differential $\frac{d}{dt}$ on $M$, which is related to the original differential $\frac{d}{dx}$ by the usual chain rule
$$ \frac{d}{dt} f = \left(\frac{d}{dt} x\right) \left(\frac{d}{dx} f\right).$$
Again, for conceptual clarity, $t, x, f: M \to {\bf R}$ should all be viewed here as functions of a position variable $p \in M$, rather than being viewed as functions of each other.<|endoftext|>
TITLE: Hormander's bracket condition for the adjoint of an operator
QUESTION [5 upvotes]: Let $X_0, X_1, \dots, X_k$ be smooth vector fields over ${\mathbb R}^n$, and let us consider the operator
$$
L = \sum_{i=1}^k X_i^2 + X_0~.
$$
Here, I assume that Hörmander's bracket condition is satisfied, that is to say that the Lie algebra generated by $X_0, X_1, \dots, X_k$ has full rank at every point in ${\mathbb R}^n$. This implies that $L$ is hypoelliptic. Now the question is: what are the conditions (if any) for the formal adjoint $L^\dagger$ of $L$ to be also hypoelliptic?
Judging by the answer to this question it appears that this is trivially true since Hörmander's condition "does not change by taking adjoints". What this means is unclear to me. We have indeed that $$
L^\dagger = \sum_{i=1}^k X_i^2 - X_0 + f~,
$$
where $f$ is a smooth scalar function, right? If there were no $f$, then this would indeed be trivial, since we find again the same Lie algebra. Since all the treatments of Hörmander's theorem that I have seen discuss operators without such an $f$ (except Hormander's paper of 1967), I guess there is an easy way to get rid of this $f$. Do you know how? Or do you have another argument to show that $L^\dagger$ is also hypoelliptic?
Thanks a lot.

REPLY [6 votes]: The hypoellipticity result is more precise:
you have
$$
Lu \in H^s_{loc}\Longrightarrow u\in H^{s+2-\delta}_{loc}\quad\text{ for some $\delta\in [0,2)$,}
$$
and that $\delta$ is linked to the number of brackets of the $X_j$ needed to generate the full tangent space. For instance, in the elliptic case, where $X_0=0$ and the $(X_j)_{1\le j\le n}$ span the tangent space, we have $\delta=0$. Now if  $f\in C^\infty$,
$$
u\in H^s_{loc},\quad Lu+fu \in H^s_{loc}\Longrightarrow Lu \in H^s_{loc}\Longrightarrow u\in H^{\epsilon+s}_{loc}
$$
with $\epsilon=2-\delta>0$.
We assume now $Lu+fu\in C^\infty$ and we want to prove $u\in C^\infty$:
since it is a local result, we may assume for some $s$ that
$
u\in H^s_{loc}
$.
Since $Lu+fu\in H^s_{loc}$, we get from above 
$u\in H^{s+\epsilon}_{loc}$. 
Since $u, Lu+fu\in H^{s+\epsilon}_{loc}$ we get 
$u\in H^{s+2\epsilon}_{loc}$ 
and so on: $u$ is smooth.
P.S. The weird typesetting is due to MathJax.<|endoftext|>
TITLE: What is the geometry of an undecidable diophantine equation?
QUESTION [54 upvotes]: As an arithmetic algebraic geometer of the highest moral fiber, I am trained to look at Diophantine equations in terms of the geometry of the corresponding scheme. For instance, if the Diophantine equation comes from a curve, I know that I should compute the genus, and do various things depending on if the genus is $0$, $1$, or $\geq 2$.
But I know very little about what the limits of this geometric approach are. I only know that there exist undecidable Diophantine equations, or families of Diophantine equations. I do not know what their geometry is like!
Do undecidable Diophantine equations, or families of equations, have interesting geometric properties? Can we compute basic geometric invariants like the Hodge diamond, Kodaira dimension, etc.? Are they pathological in every way, or do some of them have properties that might give a naive geometer hope about finding solutions? What would Noam Elkies try to do if he were asked to solve them and did not know they were undecidable, and why would he be stymied?

REPLY [16 votes]: This is just a long comment, rather than an answer to Will's question. There are "standard" algorithmically unsolvable problems in combinatorial group theory: Word problem, conjugacy problem, triviality problem for the group, isomorphism problem for a pair of groups, etc. The most fruitful line of arguments relating logic to geometry in group theory is to find geometric conditions on finitely-presented group(s) which are sufficient for solvability of these problems. It turned out that, appropriately defined hyperbolicity is a geometric condition implying algorithmic solvability of all the "standard" problems. Furthermore, it turned out that hyperbolicity is, probabilistically speaking, "generic". Moreover, one of the oldest algorithms for solving word problem (Dehn's algorithm/linear isoperimetric inequality) is one of many equivalent definitions of hyperbolicity. In view of this, I would be looking for algebro-geometric features which are sufficient for algorithmic solvability of Diophantine equations, rather than the other way around. However, since I am neither a number-theorist nor a logician, I do not know what they could be.<|endoftext|>
TITLE: Is there a Montgomery's conjecture for Dirichlet characters and Artin representations ?
QUESTION [15 upvotes]: Edit: as GH noticed, the way I tried to state Montgomery's conjecture is wrong. There were some mistakes in the references I used, which compounded with some mistakes of mine, gave a very poor post. Let me try again, hoping that the questions make more sense now.
Second Edit: I have added a bounty to the question. I would be very happy to give it for an answer to just the first greyed question. I am not comfortable with explicit formulas, but I wonder if the conjecture stated here might be translated into conjecture about the distribution of $0$'s on the critical line (assuming GRH) for Dirichlet L-function, uniformly in the character. This in turn should be known (conjecturally I mean) to specialists...

The Montgomery's conjecture I am talking about is the following: 
Let $q \geq 1$ an integer, $a \geq 1$ an integer such that $(a,q)=1$, and
let $$\psi(q,a,x) = \sum_{p^\alpha \leq x, p^\alpha \equiv a[q]} \log p.$$ 
Then for all $\epsilon>0$, one has
$$(1) \ \ \ \ \ \ \ \ \ \ \ \ \psi(q,a,x)=\frac{x}{\phi(q)} + O(x^{1/2+\epsilon}q^{-1/2})$$
uniformly for $x>q$ (that is, the implied constant depends only of $\epsilon$).
Montgomery himself states a slightly stronger conjecture in [M], with the error term in 
$O((x/q)^{1/2+\epsilon})$ and forgets to assume $x>q$. Without the latter condition, the conjecture is
trivially false, as noted by GH (see the comments below). Even with the restriction $x>q$, Montgomery's initial conjecture is false, as 
proven in [FG], who proposes the slightly weaker statement above as a replacement: see [FG, conjecture 1(b)].
The conjecture is also given in {IK,17.5] but without the restriction $x>q$.
Since Dirichlet, results on primes in arithmetic progressions are proved using Dirichlet characters.
Hence let $\chi$ be a non-principal Dirichlet character of conductor $q$. Define (as usual)
$$\psi(\chi,x) = \sum_{p^\alpha < x} \chi(p^\alpha) \log p.$$
One has $\psi (\chi,x)=\sum_{a \pmod{q}} \chi(a) \psi(q,a,x)$ and $\psi(q,a,x) = \frac{1}{\phi(q)} \sum_\chi \chi(a)^{-1} \psi(\chi,x)$.


Is it reasonable to conjecture the following estimate, uniformly in the non-principal character $\chi$ and the integer $q$, for $x>q$, 
    $$(2)\ \ \ \ \ \ \ \ \ \ \ \ \ \psi(\chi,x) = O(x^{1/2+\epsilon} q^{-1/2})$$? Has this been conjectured somewhere?


Note that the Montgomery's estimate (1) for $\psi(q,a,x)$ follows immediately from (2)
by summing over $a$ invertible mod $q$ and dividing by $\phi(q)$. 
However, it is not clear that (1) implies (2)
as the error term seems to get multiplied by $\phi(q)$. Yet, I have made some numerical computation, and the estimate for Dirichlet character seems to hold.
Now assuming that the first question has a positive answer, I want to generalize it boldly to Artin's representations. Let $\rho : Gal(K/\mathbb Q) \rightarrow Gl_d(\mathbb C)$ be an irreducible, non-trivial representation of dimension $d$, Artin's conductor $q$ and let $\chi$ be its character. In this context, I define (also as usual):
$$\psi(\chi,x)=\sum_{p^\alpha < x} \chi(Frob_p^\alpha) \log p.$$

What would be a reasonable conjectural estimate for $\psi(\chi,x)$, in terms of $q$, $d$, and $x$, giving back the above conjecture when $d=1$? Has anyone formulated such a conjecture?

Perhaps a starting point is the estimate one gets by usual methods assuming both the GRH for the Artin $L$-function $L(s,\rho)=L(s,\chi)$ and the holomorphy of its $L$-function everywhere, that is, Artin's conjecture for $\rho$ (cf. [IK, Theorem (5.15)]):
$$\psi(\chi,x)=O(x^{1/2} \log(x) (d \log x + \log q)).$$
Can we replace the estimate by $O(x^{1/2+\epsilon} d / q^{1/2})$ for example ? I am at a loss even to guess a reasonable formula.
My motivation is trying to understand the best error terms in effective Chebotarev theorem.
Any conjecture as asked will lead to a new estimate for Chebotarev, and I have a
nice collection of examples to test those versions of Chebotarev against.
References: 
[FG] Friedlander, Granville, Limitations of the equi-distribution of primes I, Annals of Maths vol. 129, no2, 1989
[IK] Iwaniec, Kowalski, Analytic Number Theory  
[M] Montgomery, Problems about Prime Numbers, in PSPM XXVIII, AMS.

REPLY [5 votes]: The basic heuristic behind the scenes in all of these conjectures is that there is optimal "square-root" cancellation in the various error terms.  The estimate (2) is surely not true; the explicit formula links this to a sum over zeros of the related Dirichlet L-function, and it has zeros on the half-line. Instead, one may hope that $$\psi(\chi, x) = \sum_{n < x} \chi(n) \Lambda(n) = O(x^{1/2 + \varepsilon}).$$  This is square-root in the number of terms, with an extra $x^{\varepsilon}$ thrown in for safety.  It's much more subtle to figure out powers of logs.
To get to Montgomery's conjecture, you use the formula $$\psi(q,a,x) = \frac{1}{\phi(q)} \sum_{\chi} \overline{\chi}(a) \psi(\chi, x),$$
and imagine that there is square-root cancellation in the sum over $\chi$.  You have roughly $q$ terms, are dividing by roughly $q$, and so in all this amounts to having an error of size $x^{1/2+\varepsilon} q^{-1/2 + \varepsilon}$.<|endoftext|>
TITLE: Persistent homology of Gaussian fields in Euclidean space
QUESTION [23 upvotes]: If you generate points in $\mathbb R^n$ via a process that respects a Gaussian normal distribution, then compute the persistent homology / barcodes, to my eye something fairly regular seems to be happening, with the barcodes tending towards something like a "wing" shape, fat in lower dimensions, thinning out towards dimension $n$. 
Has anyone proven any theorems that describe the asymptotic "shape" of the barcodes? 
Ideally I'd like a test so that I can look at some barcodes and say "that's typical of a Gaussian normal distribution".  
The closest thing I've been able to find is experiments and results on the expected Euler characteristic of the persistent homology, in the following two references (arXiv links): Persistent homology for random fields and complexes, Euler integration of Gaussian random fields and persistent homology.
Edit: 
I did a very rough computation to try and get some kind of guess as to what the distribution of barcodes should look like.  So I made a very coarse estimate based on a distribution of points that is roughly `locally cubical' and approximately respecting a normal distribution.
The density is given by:
$$\mu = N e^{-r^2}$$
where $r$ is the distance from the origin.  Then if $\epsilon$ is the parameter for persistent homology, it appears that $H_0$ is rank approximately
$$N \int_{\sqrt{\ln(N\epsilon^{1/n})}}^\infty r^{n-1}e^{-r^2} dr$$
and $H_k$ for $k \in \{1,2,\cdots,n-1\}$ has rank approximately
$$ {n \choose k+1}\frac{(\sqrt{\ln(N\epsilon^{1/n}/\sqrt{k}))}^{n-2}}{4\sqrt{k}\epsilon^{1/n}} $$
These are fairly coarse estimates, and in no way rigorous.  But if something like this is actually true it seems to be saying that for $N$ large and $n \geq 3$, the $H_0$ betti number tends to some asymptote (dependent on $\epsilon$), and $H_1, \cdots, H_{n-1}$ are non-trivial but small.  So most of the points in the distribution are in a giant homology `black hole' at the centre and persitent homology sees the thin crust around the outside. 
I'd be curious if people have done other similar guestimates (or better) and if they had similar-looking results.

REPLY [9 votes]: Adler, Bobrowski and Weinberber's "Crackle: The Persistent Homology of Noise" is an answer to my question.  I have not read it closely yet but it appears to confirm the guess in the question, and provide answers for other distributions as well. 
Although this paper does not target my question directly it gives a more quantitative answer to a nearby question, that of the length of the largest barcode for certain types of random point clouds. Maximally persistent cycles in random geometric complexes.<|endoftext|>
TITLE: Subspaces of End(V) that can fix any vector
QUESTION [8 upvotes]: Suppose V is a finite-dimensional vector space and I have a linear subspace of its endomorphisms
$$W \subseteq \mbox{End}(V).$$
How can I easily check if every vector of $V$ is fixed by some element of $W$?  I would also be interested in any nice conditions on $W$ that imply 
$$\forall v \in V: \;\;\; v \in Wv,$$
even if they aren't biconditional.

If the identity matrix happens to be in $W$, then the condition is satisfied trivially.  However, it is easy to find examples where W omits the identity but can still fix any vector.  For example, the two-by-two traceless matrices have this property.
Unfortunately, $W$ can be large without satisfying the property.  For example, take all matrices which have a zero in the upper left.  These can never fix the first basis vector.


Side note:
In my particular situation, $W$ happens to be closed under multiplication.  If this turns out to be a helpful condition, please don't hesitate to use it.

REPLY [5 votes]: For subspaces that are stable under multiplication, there's an even more profound
result that holds for all fields : 

Given a finite-dimensional vector space $V$ and a linear subspace $W$ of $End(V)$
  that is stable under multiplication but does not contain the identity, 
  there are linear subspaces $V_1$ and $V_2$ of $V$ such that $V_1$ is a proper subspace of
  $V_2$ and every operator in $W$ maps $V_2$ into $V_1$. 
  In particular, every operator in the reflexive closure of $W$ must map $V_2$ into $V_1$, 
  whence the reflexive closure of $W$ consists only of singular operators!

The proof is essentially similar to the one I have outlined in my first answer. 
The key is a theorem of Wedderburn which states that a subalgebra of $End(V)$
is either reducible or simple. If we have a linear subspace $W$ which is stable under multiplication, then we note that $W +Span(id)$ is a subalgebra of $End(V)$ and $W$
is a two-sided ideal of it: thus, either $W$ is irreducible, or $W$ is a simple algebra,
or $W$ is limited to the zero operator and $V$ has dimension $1$. 
Using this repeatedly, one finds a basis in which all the operators in $W$ are represented by block-upper-triangular matrices, and, on the diagonal, each block space is an algebra 
or consists of the $1 \times 1$ zero matrix. 
If there is at least one block of the latter time, then we have found the subspaces $V_1$ an $V_2$ we were looking for. 
If we now assume that all the diagonal block spaces
are algebras, then we prove that $W$ contains a non-singular matrix, which yields, 
as explained in some answers, that $W$ contains the identity. 
To do this, it suffices to prove the following lemma :

Let $A_1,\dots,A_n$ be algebras, and $H$ be a linear subspace of 
  $A_1\times \cdots \times A_n$ that is stable under multiplication and projects surjectively
  onto each $A_i$. Then, $H$ contains the unit element of $A_1\times \cdots \times A_n$. 

This is proved by induction over n. The case $n=1$ is trivial. Assume that this holds for
$n-1$, with $n \geq 2$ fixed. Then, we write 
$A_1\times \cdots \times A_{n-1}=B$; by induction, $H$ contains $(1_B,x)$ for some
$x \in A_n$. Then, as some $b \in B$ satisfies $(b,1_{A_n}) \in B \times A$, we see that
$(1_B,1_{A_n})=(1_B,x)+(b,1_{A_n})-(1_B,x)*(b,1_{A_n}) \in H$.<|endoftext|>
TITLE: The Riemann zeros and the heat equation
QUESTION [30 upvotes]: The Riemann xi function $\Xi(x)$ is defined, with $s=1/2+ix$, as
$$
\Xi(x)=\frac12 s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)=2\int_0^\infty \Phi(u)\cos(ux) \, du,
$$
where $\Phi(u)$ is defined as 
$$
2\sum_{n=1}^\infty\left(2\pi^2n^4\exp(9u/2)-3\pi n^2\exp(5u/2)\right)\exp(-n^2\pi\exp(2u)).
$$
This arises from integration by parts after writing $\Xi$ as the Mellin transform of the theta function, and then a change of variables from multiplicative to additive notation.  In 1950 de Bruijn (building on work of Polya)  introduced a deformation parameter $t$:
$$
\Xi_t(x)=\int_0^\infty \exp(t u^2)\Phi(u)\cos(ux)\, du,
$$
so that for $t=0$, $\Xi_0(x)$ is just $\Xi(x)/2$.  
de Bruijn proved the following theorem about the zeros in $x$:  
(i)  For $t\ge 1/2$, $\Xi_t(x)$ has only real zeros.
(ii) If for some real $t$, $\Xi_t(x)$ has only real zeros, then $\Xi_{t^\prime}(x)$ also has only real zeros for any $t^\prime>t.$  
In 1976 Newman showed that there exists a real constant $\Lambda$, $-\infty<\Lambda\le 1/2$, such that
(i) $\Xi_t(x)$ has only real zeros if and only if $t\ge\Lambda$.
(ii) $\Xi_t(x)$ has some complex zeros if $t<\Lambda$.
The constant $\Lambda$ is known as the de Bruijn-Newman constant.  The Riemann hypothesis is the conjecture that $\Lambda\le 0$.  Newman made the complementary conjecture that $\Lambda\ge 0$, with the often quoted remark

"This new conjecture is a quantitative
  version of the dictum that the Riemann
  hypothesis, if true, is only barely
  so."

Given the significance of the de Bruijn-Newman constant $\Lambda$, much work has gone into estimating lower bounds, and the current record (Saouter et. al.) is
$
-1.14\times 10^{-11}<\Lambda.
$
A breakthrough occurred in the work of Csordas, Smith and Varga, "Lehmer pairs of zeros, the de Bruijn-Newman constant, and the Riemann Hypothesis", Constructive Approximation, 10 (1994), pp. 107-129.
They realized that unusually close pairs of zeros of the Riemann zeta function, the so-called Lehmer pairs, could be used to give lower bounds on $\Lambda$.  The idea of the proof is that the function $\Xi_t(x)$ satisfies the backward heat equation
$$
\frac{\partial \Xi}{\partial t}+\frac{\partial^2 \Xi}{\partial x^2}=0,
$$
from which they are able to draw conclusions about the differential equation satisfied by the $k$-th gap between the zeros as the deformation parameter $t$ varies.  
They mention this PDE in a rather offhand way, as a remark on an alternate proof to one of the lemmas.  In fact, it does not seem to be well known that the de Bruijn-Newman constant can be interpreted as a time variable in a heat flow.  Is this well known?  Or put more concretely, does anyone have a citation prior to 1994 which mentions this fact?

Update:  Tao and Rodgers have a proof of the Newman conjecture on the arXiv.

REPLY [7 votes]: Q. Does anyone have a citation prior to 1994 which mentions this fact?

1988: Numer. Math. 52, 483-497 (the differential equation is given in a slightly different form on page 493).<|endoftext|>
TITLE: Projective objects in the category of chain complexes 
QUESTION [16 upvotes]: Excercise 2.2.1 in Weibel ("An Introduction to Homological Algebra") states that an object $P$ in the category of chain complexes over an abelian category is projective if and only it is a 
split exact complex of projectives. 
I was able to solve the only-if-part but I have touble with the if-part and would be glad if 
someone can give me some help. This is no homework! 
What have I a tried so far ? Given an epimorphism $\pi: X \to Y$ and a morphism $f: P \to Y$, it has to be shown that there is a morphism $g: P \to X$ s.t. $\pi \circ g=f$. 
Weibel hints to consider the special case  $0 \to P_1 \cong P_0 \to 0$. It's easy to construct $g$ in this case: $\pi$ epi means that each $\pi_i:X_i \to Y_i$ is epi. By projectivity of $P_1$ there is a hom. $g_1: P_1 \to X_1$ s.t. $\pi_1 \circ g_1 = f_1$. If $d^P$ resp. $d^X$ denotes the differential in $P$ resp. $X$, set $$g_0 := d^X_1 \circ g_1 \circ (d^P_1)^{-1}: P_0 \to X_0,\qquad g_i = 0: P_i \to X_i\; (i\neq 0,1)$$
Then $g=(g_i): P \to X$ is a morphisms s.t. $\pi \circ g=f$.
But I have no idea how to generalize this procedure to the general case where $d^P$ can not be expected to be an isomorphism.

REPLY [6 votes]: The question as asked has been answered, but to understand where bounded below enters the picture, it is 
helpful to think model categorically (as in Dwyer and Spalinsky
or, more recently, chapter 18 of More Concise Algebraic Topology,
by Kate Ponto and myself).  With the usual model structure (there
are others in the latter reference), a chain complex is acyclic
and cofibrant if and only if it is a projective object.   If it
is cofibrant (not necessarily acyclic) then it is degreewise
projective.  If it is degreewise projective and bounded below,
then it is cofibrant.  However, it can be acyclic and degreewise projective
and yet not cofibrant if it is not bounded below. There is a nice
example in the paper [K] that TJ refers to:  work over the ring
$Z/4$ and take all $P_n$ to be free on one generator, with all 
differentials given by multiplication by $2$:  $P$  is acyclic and
degreewise free, but it is not cofibrant and not a projective object.
Split exactness rules out such examples and is automatic when $P$ 
is exact, degreewise projective, and bounded below.
Incidentally, the role of $R$-split exactness becomes really interesting
model theoretically when $R$ is commutative and not a field and one considers 
model structures on DG modules over a DG $R$-algebra.  There are (at least) six 
different interesting projective type model structures, and the usual one is arguably not the most useful one (this is a shameless
advertisement for a paper in the writing stage by Tobi Barthel, Emily Riehl, 
and myself).<|endoftext|>
TITLE: Average orders of multiplicative functions
QUESTION [12 upvotes]: For a multiplicative function $f$ and $x>0$ let $$S_f(x)= \sum_{n \leq x} f(n).$$
Studying sums of this type is a favourite pastime of analytic number theorists. I'm trying to understand what kind of behaviour can occur for such sums. In particular, my question is the following.


Does there exist a multiplicative function $f$ and a constant $c_f>0$ such that $$S_f(x) \sim c_f\frac{x}{\log x},$$ as $x \to \infty$?


Here is some motivation for how I came across this problem. Analytic number theorists often study sums of the above type by studying the analytic properties of associated Diriclet series
$$L(f,s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}.$$
Here, if $L(f,s)$ has a pole of order $r>0$ at $s=1$ and is well-behaved for $\text{Re}( s) >1$,
then one can often show (using e.g. a Tauberian theorem such as Perron's formula) that we have an asymptotic formula
$$S_f(x) \sim c_f x (\log x)^{r-1}.$$
More generally there is the Selberg-Delange method, here one works with complex powers $\zeta^z(s)$ of the Riemann zeta function. This method, when it works, will give an asymptotic formula of the shape $$S_f(x) \sim c_f x (\log x)^{z-1}.$$ In particular, it does not seem that one can obtain an asymptotic formula like the one I am seeking using this approach. 
Note that one cannot use something like the prime number theorem to construct an example of the shape I am looking for, since $\pi(n)$ is not a multiplicative function!

REPLY [8 votes]: I think an explicit multiplicative function that should do the job for you is this one:
$f(2^n)=2^n/\big((n+1)\sqrt{\log(n+e)}\big)$, $f(3^n)=3^n/\big((n+1)\sqrt{\log(n+e)}\big)$, $f(p^n)=0$ for $n\ge 1$ and primes $p\ge 5$. The $+1$'s and $+e$'s are just there to make the formula make sense for $n=0$ also.
Here's an outline of why:
It suffices to show that for all $\epsilon>0$, for all sufficiently large $N$ one has
$$
\sum_{N < 2^k3^l\le (1+\epsilon) N} \frac{1}{(k+1)\sqrt{\log(k+e)}}\cdot \frac{1}{(l+1)\sqrt{\log(l+e)}}\sim c\epsilon /\log N
$$
Taking logs, the summation range is essentially $\log N\le k\log 2+l\log 3\le \log N+\epsilon$.
For large $N$, there are approximately $\epsilon\log N/(\log 2\log 3)$ pairs $(k,l)$ in the range. These are reasonably uniformly distributed in the band of $\mathbb R^2$, $x\log 2+y\log 3\in [\log N,\log N+\epsilon]$. Hence (with a bit of work making sure the ends of the integral don't dominate), the sum is close to 
$$
\frac{\epsilon}{\log 2}\int_{0}^{\log N/\log 3}
\frac{1}{(y+1)\sqrt{\log (y+e)}}\frac{1}{(x+1)\sqrt{\log(x+e)}}\ dx,
$$
where $y=(\log N-x\log 3)/\log 2$.
In the first half of the range, the integral is something like
$$
\frac{\epsilon}{\log N\sqrt{\log\log N}}
\int_0^{\log N/(2\log 3)} \frac{1}{(x+1)\sqrt{\log(x+e)}}\ dx
$$
which is $\sim c\epsilon/\log N$. Similarly for the second half of the range.
If you prefer your multiplicative functions to be integer-valued, of course, you can just take the floor of everything.<|endoftext|>
TITLE: The optimal constant in Vitali covering lemma
QUESTION [17 upvotes]: Let me restate Vitali covering lemma.
Let $\{B_i\}_{i\in F}$ be a finite collection of balls in the $\mathbb{R}^n$. Then there is $S\subset F$ such that the balls $\{B_i\}_{i\in S}$ are disjoint and
$$\mathop{\rm vol}\left(\bigcup_{i\in S}B_i\right)\ge \tfrac1{3^n} \mathop{\rm vol} \left(\bigcup_{i\in F}B_i\right),$$

Question. Can one make the constant better? In particular, is it OK to change $\tfrac1{3^n}$ to $\tfrac1{2^n}$ in the formulation?

It is likely that the answer is well known, but googling did not help.
Clearly the greedy algorithm which is used in the standard proof of Vitali covering lemma cannot give anything better than $\tfrac1{3^n}$.
One cannot make it better than $\tfrac{1}{2^n}$. (Consider a large collection of balls of the same radius such that each contains a fixed point.)
I see that for $n=1$ the optimal constant is $\tfrac1{2}$, 
but for higher dimensions I do not see a proof.

REPLY [6 votes]: It is not hard to see that we can improve the constant a bit. Denote the ball with the biggest radius by B and its radius by r.
If there are two disjoint balls intersecting B with radius at least 0.99r, than put them in S, this will be "locally better" than $\frac 1{3^n}$ as the blow-up of the balls intersect.
If there are no two such balls, then put B into S, this is again "locally better" than $\frac 1{3^n}$. Optimizing the constants will probably give something reasonable, but not 2, which I find to be a very nice conjecture.<|endoftext|>
TITLE: Lattice in a certain Lie group
QUESTION [9 upvotes]: Let $G_n$ be the Lie group consisting of $n \times n$ upper triangular matrices of determinant $1$ with real entries.  In other words,
$$G_n = \{\text{$\left(\begin{matrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n}  \\ 
0 & a_{22} & a_{23} & \cdots & a_{2n} \\
0 & 0 & a_{33} & \cdots & a_{3n} \\
\vdots & \vdots & \vdots & & \vdots \\
0 & 0 & 0 & \cdots & a_{nn} \end{matrix}\right)$ $|$ $a_{ij} \in \mathbb{R}$ and $a_{11} \cdots a_{nn} = 1$}\}.$$
Does $G_n$ contain a lattice (i.e. a discrete subgroup of finite covolume)?  The obvious thing to try is to take the subgroup consisting of matrices with integer entries, but that does not work (though it would work if we were working with strictly upper triangular matrices).

REPLY [14 votes]: @Edward, here is a short proof which I wrote awhile ago for my notes on group theory. 
Lemma. If a 2nd countable locally compact group $G$ contains a lattice $\Gamma$ then $G$ is unimodular. 
Proof. For arbitrary $g \in G$ consider the push-forward $\nu=R_g(\mu)$ of the (left) Haar measure $\mu$ on $G$; here $R_g$ is the right multiplication by $g$:
$$
\nu(E)= \mu(Eg).
$$
Then $\nu$ is also a left Haar measure on $G$. By the uniqueness of Haar measure, $\nu= c\mu$ for some constant $c > 0$. The lattice $\Gamma \le G$ has a fundamental domain $D\subset G$, i.e., a measurable subset of $G$ such that
$$
\bigcup_{\gamma\in \Gamma} \gamma D= G, \quad \mu(\gamma D\cap D)=0, \quad \forall \gamma\in \Gamma \setminus 1.
$$
(Proof of existence of such domain uses 2nd countable assumption.) In particular, $0<\mu(D)<\infty$, since $\Gamma$ is a lattice. Then $Dg$ is again a fundamental domain for $\Gamma$ and, thus, $\mu(D)=\mu(Dg)$. Hence, $\mu(D) = \mu(Dg) = c\nu(D)$. It follows that $c=1$. Thus, $\mu$ is also a right Haar measure. qed
Another proof could be found in Chapter I of Raghunathan's book "Discrete subgroups of Lie groups".<|endoftext|>
TITLE: On well separated point sets in the plane
QUESTION [9 upvotes]: Let us say that a finite set $A$ in the plane is $1$-separated if:
1) it has an even number of points;
2) no open ball of diameter $1$ contains more than $|A|/2$ points.
For a $1$-separated set $A$ define $G(A)$ to be a graph where two points $x,y$ in $A$ are joined by an edge iff the distance between them is at least $1$.

Question: can one find a finite set of graphs $G _ 1,\dots,G _ n$ such
  that any $1$-separated set $A $ can be
  partitioned into non-empty
  $1$-separated sets $A _ 1,\dots,A _ k$
  such that $G(A _ i)$ is isomorphic to
  one of the $G _ j$'s?

Comment: The definition makes sense on the real line (the ball of diameter $1$ is replaced by an interval of length $1$). In that case we can take $n=1$ and $G_1$ to be a graph on two vertices joined by an edge (that is, $G(A)$ contains a matching).

REPLY [3 votes]: I think Domotorp is correct. Take a regular $(2n-1)$-gon such that its longest diagonal is 1, along with its center. Then $A$ cannot be partitioned, and $G(A)=C_{2n-1}$.<|endoftext|>
TITLE: $A_\infty$-categories and their equivalent dg-categories: the case of $\mathcal{RH}om(\mathcal A,\mathcal B)$
QUESTION [6 upvotes]: Hello everyone,
I'm currently working on dg-categories, in particular I'm looking for some convenient characterization of the dg-category $\mathcal{RH}om(\mathcal A,\mathcal B)$, for two given dg-categories $\mathcal A$ and $\mathcal B$, at least in cases when $\mathcal A$ is "easy". For example, if $\mathcal A$ is the category $\Delta^1$ with two objects $0$ and $1$ and freely generated over $k$ (a fixed ground commutative ring) by one nontrivial morphism $0 \to 1$, then $\mathcal{RH}om(\mathcal A,\mathcal B) = \mathcal{M}or(\mathcal B)$ (equality up to quasi-equivalence, I suppose), where $\mathcal{M}or(\mathcal B)$ is the dg-category of morphisms of $\mathcal B$ as defined in Drinfeld's article "dg quotients of dg categories" (Drinfeld assumes $\mathcal B$ pretriangulated, but it is not really necessary).
In Keller's survey "on differential graded categories" it is stated that $\mathcal{RH}om(\mathcal A,\mathcal B)$ is (quasi equivalent to) the category of strictly unital $A_\infty$ functors from $\mathcal A$ to $\mathcal B$, both viewed as $A_\infty$-categories.
Moreover, it seems a well-known fact that a given $A_\infty$-category is equivalent - in the $A_\infty$ sense - to some dg-category. Drinfeld himself (first paragraph of Appendix IV of "Dg quotients of dg categories") sketches a procedure to associate a dg-functor to a $A_\infty$-functor between dg-categories, upon changing the source dg-category, in a simple case: as far as I can understand, the idea is to formally add morphisms and coboundary relations to the source category, in a smart way.
My question is the following: how do you explain the above procedure, in general? Namely, what morphisms and what differentials I really have to add, keeping Drinfeld's example in mind? I am looking for something as elementary as possible, and living solely in the "world of dg categories". Even an explanation in some simple situations would be of great help. A caveat: I know nearly nothing about $A_\infty$-categories!
Thanks in advance; I hope everything above is clear enough.

REPLY [2 votes]: I don't quite know what kind of properties do you want the DG-category $RHom(A,B)$ to have, but there is a natural nice DG-category which may work. Namely, the category of right quasi-representable h-projective $(A^{op}\otimes B)$-DG-modules.<|endoftext|>
TITLE: A conjecture on solvablity of finite groups
QUESTION [12 upvotes]: Suppose $G$ is a finite group and $A$ an abelian subgroup. Suppose for some natural number $n\geq 2$, elements of $\gamma_n(G)$ have the form $[a, x]$ where $a\in A$ and $x\in G$. Then $G$ is solvable.

REPLY [14 votes]: I claim that the conjecture is false, and there is a counterexample with $G=S_5$, $n=2$, and $A$ cyclic of order 6. In this case $\gamma_2(G)=[G,G]=G'=A_5$ has order $5!/2=60$. Take $A$ to be the abelian group $A=\langle(1,2,3),(4,5)\rangle$, but any conjugate of $A$ will also work. A simple computation shows that
the set $C:=\{[a,x]\mid a\in A,\ x \in G\}$ also has 60 elements. Since $C\subseteq A_5$, this proves that $C=A_5$. However, $G$ is clearly not solvable.
What a pity, the motivating idea for the question was attractive.<|endoftext|>
TITLE: What is the free monoidal category generated by a monoid?
QUESTION [8 upvotes]: In several places in a segment on  cohomology (for example, here (PDF)) in John Baez's online lecture notes for a course in 2007 on quantum gravity, much is made of the fact that the simplex category $\Delta$ (category of finite ordinals) is the "free monoidal category on a monoid". Supposedly this is the reason that this category is the initial object in the category of monoidal categories with monoid object.
But what does this mean? If it's defined in those lecture notes, I can't find it. Under some mild assumptions on the tensor product in a monoidal category $C$, we have a free monoid functor $F\colon C\to Mon(C)$, the adjoint of the forgetful functor, which takes objects $X\mapsto \coprod^\infty_{n=0} X^{\otimes n}.$ One says that $F(X)$ is the free monoid object generated by $X$. Can this be what Baez means? Not really, this is generated by an object, not a monoid. What if we take $X$ to be the underlying object of a monoid object? Still not right, the fact that the object generating the free monoidal object had its own monoidal structure seems to have no effect on the resulting free monoid object. And anyway we're looking for a monoidal category, but we only got a monoid object.
What if we take $C=\mathbf{Cat}$, the category of categories? Then we have $Mon(\mathbf{Cat})=\mathbf{MonCat}$ is the category of monoidal categories, and the free functor $F\colon \mathbf{Cat}\to \mathbf{MonCat}$ does give us a monoidal category. Can this be what Baez means? If $C$ is a category, then $F(C)$ is a category whose objects (resp. morphisms) are tuples of objects (resp. morphisms) in $C$. Now it makes better sense why we can take the free monoidal category generated by a monoid: in this context, a monoid $M$ is just a category with one object, a monoid object in $\mathbf{Set}$, and an object of $\mathbf{Cat}.$ So a tuple of a single object is just a natural number. This is starting to look like $\Delta$, at least the objects are right.
But what are the morphisms? They are tuples of morphisms in the original category, which was a category with one object, but apparently no restrictions on the morphisms. If $M$ was a nontrivial monoid, say $x,y\in M$ with $x\neq y$, then the object 1 in the free monoidal category $F(M)$ will have two distinct endomorphisms. However in $\Delta$, there is only one morphism $1\to 1.$ So $\Delta$ is not the free monoidal category generated by a monoid, unless that monoid is the trivial monoid. Is that what Baez means here? $\Delta$ is the free monoidal category generated by the trivial monoid viewed as a category with one object (and only the trivial arrow)?
He develops the idea also in his This Week's Finds column in 2001. There he says "free monoidal category on a monoid object", and the lack of specificity makes me think that not only does it not matter what monoid you start with, it doesn't even matter what monoidal category you take your monoid object from: you'll still get $\Delta$. But that's just not what I'm seeing. Can someone set me straight?

REPLY [8 votes]: I think you are interpreting it as "(free monoidal category) on a monoid", while it's really "free (monoidal category on a monoid)", or rather "free (monoidal category with a monoid)".
Now to construct it you have have to find which morphisms should exists as a consequences of the axioms, i.e. which morphisms you are sure to see whenever you have a monoid in a monoidal category. Clearly, if $X$ is an object in a monoidal category $C$, the only morphisms between $X^{\otimes n}$ and $X^{\otimes m}$ that exists whatever $(C,X)$ are, are the identity when $m=n$. So the free monoidal category is just $\mathbb{N}$ whith $\hom(m,n)=${id} if $m=n$, and is empty otherwise.
But if $X$ is a monoid, you get by definition a morphism $X \otimes X \rightarrow X$ and a morphism $I \rightarrow X$ where $I$ is the identity object. These morphisms clearly extends to morphisms $\delta_i:X^{\otimes n+1}\rightarrow X^{\otimes n}$ and $\sigma_i: X^{\otimes n}\rightarrow X^{\otimes n+1}$ just by tensoring them with identity morphisms, and then you can compose them to get maps $X^{\otimes m}\rightarrow X^{\otimes n}$. Finally, it's easy to see that the commutation relation between these morphisms imposed by the axioms of a monoid are precisely those satisfied by the simplicial maps in $\Delta$.<|endoftext|>
TITLE: ODE's without a Lipschitz condition
QUESTION [17 upvotes]: When teaching ODE's earlier this semester, one of my students asked the following question for which I didn't know the answer (and none of the textbooks I consulted seem to discuss it).  It is standard that if $f(x,y)$ is Lipschitz, then the ODE $y'=f(x,y)$ can be solved uniquely for any initial condition (at least locally).  The student asked for an example of a (non-Lipcshitz) $f(x,y)$ such that there is no solution.  Of course, one can give stupid examples (eg functions $f(x,y)$ that are so non-continuous that they cannot be the derivatives of anything).  However, I was unable to find a counterexample with $f(x,y)$ continuous.  Can someone provide one?

REPLY [5 votes]: I think it is worth mentioning that although it is true that Lipschitz  continuity guarantees unique solutions, there is a weaker regularity condition for $f$ which also gives uniqueness. The buzzword is "Osgood condition" and the theorem is given in Andrey Rekalo's answer to this MO question on "Existence/Uniqueness of solutions to quasi-Lipschitz ODEs".<|endoftext|>
TITLE: Cobordism of orbifolds?
QUESTION [27 upvotes]: Is it possible to setup classical cobordism theory in the context of orbifolds?  For example, let's consider the free abelian group generated by oriented smooth orbifolds and quotient by those which are null cobordant (that is, they are the boundary of a smooth oriented orbifold with boundary).  Is the resulting ring expressible as the homotopy groups of some Thom spectrum?  Can we prove this just by smoothly embedding an orbifold in $\mathbb R^n/S_n$ and following the classical proof?
My motivation is that in Gromov--Witten theory, the moduli space of stable maps $\bar M_{g,n}(X,A)$ is a smooth oriented orbifold (assuming it is cut out transversally, and assuming we have smooth charts for gluings) and it is defined up to cobordism.  Thus instead of taking its fundamental class and pushing forward to $H_\ast(\bar M_{g,n}\times X^n)$ to get Gromov--Witten invariants, we could consider the class it represents in the generalized cohomology theory which we might call "oriented orbifold cobordism" of $\bar M_{g,n}\times X^n$, and get a slightly more refined invariant.

REPLY [6 votes]: Here are a couple papers on oriented orbifold cobordism. The first gives rational invariants and generators and also shows that rationally odd dimensional orbifolds bound.
 The second paper develops machinery for handling the torsion (all dimensions) and applies that to show that every oriented three-orbifold bounds.
K.S. Druschel.  Oriented Orbifold Cobordism, Pacific J. Math.,  164(2) (1994), 299-319.
K.S. Druschel. The Cobordism of Oriented Three Dimensional Orbifolds, Pacific J. Math., bf 193(1) (2000), 45-55.<|endoftext|>
TITLE: Conjectures in Grothendieck's "Pursuing stacks"
QUESTION [30 upvotes]: I read on the nLab that in "Pursuing stacks" Grothendieck made several interesting conjectures, some of which have been proved since then.  For example, as David Roberts wrote in answer to this question,

Grothendieck conjectured, and Cisinski proved, that the class of weak equivalences in the Thomason model structure was the smallest basic localizer.

I am interesting in knowing what other conjectures made in PS have turned out to be true, or  other "ideas" that have been successfully realized/formalized.  Ideally it would be nice to include references to the relevant papers.

REPLY [18 votes]: I agree with Tim that calling Pursuing Stacks  a "letter to Quillen" is erroneous, especially as Quillen never replied.
Grothendieck also wrote: "This is written in English in response to a correspondence in English." At one stage he planned more volumes in French but it seems got diverted from this. I hope the following will be of help,  in addition to David Robert's answer, to put the situation with regard to models for homotopy theory in context. 
In a letter dated 02/05/1983 Alexander Grothendieck
wrote to me: "Don't be surprised by my supposed efficiency in
digging out the right kind of notions--I have just been following,
rather let myself be pulled ahead, by that very strong thread
(roughly: understand non commutative cohomology of topoi!) which I
kept trying to sell for about ten or twenty years now, without
anyone ready to ``buy'' it, namely to do the work. So finally I got
mad and decided to work out at least an outline by myself."  
But this question is about the homotopy theory of categories and the related question of "Why simplicial sets?", for which see also the good answers in March 2011 to Is there a high-concept explanation for why "simplicial" leads to "homotopy-theoretic"?.
Dan Kan's first contribution to combinatorial homotopy was in terms of cubical sets. When he went to Princeton the disadvantages of cubical sets were found: cubical groups did not satisfy the extension property, and the geometric realisation of the cartesian product of cubical sets had the wrong homotopy type, while as shown by Moore and Milnor respectively, the situation was fine for simplicial sets. So they did not attempt to refine the cubical theory. In my doctoral studies at Oxford, 1956-59, the exposition was all simplicial, especially when Michael Barratt came back from Princeton in 1957.
However the 1960 book by Hilton and Wylie on algebraic topology was cubical, as were some 1962 notes of Federer from Brown University, a later book by Massey, and cubical sets continued to be found useful in various places. Our 2011 book on "Nonabelian algebraic topology" is almost entirely cubical, because of its emphasis on the use of higher homotopy Seifert-van Kampen theorems.  

My 1965 intuition for using cubical sets was based on generalising the van Kampen theorem to higher dimensions. It seemed entirely reasonable that the above diagram could be expressed as: the big square is the composition of the little squares. C. Ehresmann's 1965 book on "Categories structuree"  gave a definition of double categories which expressed this nicely. Indeed, I have answered this on mathoverflow as using matrix notation where $(a_{ij})$ denotes a composable array and $[a_{ij}]$ denotes the composite. So one has an easy definition of the $n$-fold cubes of the nerve of an $n$-fold category, except that there seems currently no name for the cubical type geometry underlying an $n$-fold category.   Note that composable sequences of morphisms in a category are used in describing the nerve of a category, but it seems more difficult, at least for me,  to define multiple compositions in simplicial or globular terms, although the simplicial nerve of an $n$-fold category is easily defined as an $n$-fold simplicial set.
By contrast, the singular cubical complex of a space, or filtered space, is ideally suited for the description of multiple compositions, using an array notation. I have already explained this in answer to this mathoverflow question. 
So in considering what category in which to work the question of "what should be adequacy and convenience?" is crucial. It is as reasonable to ask this for combinatorial models of homotopy theory as it was in 1963 to ask it for categories for topology in  my paper Ten topologies.
A property that  was also required for the conjectured proof of a putative higher van Kampen theorem using homotopy classes of maps  was the notion of "commuting cube", and that "any composition of commuting cubes is commutative". Chris Spencer and I found that the notion of "connection" for a double groupoid was good for this, and that it allowed  an equivalence between crossed modules and single pointed edge symmetric  double groupoids with connections. Then Philip Higgins and I found in 1974 the construction of the homotopy double groupoid of a pair $(X,A,x)$ of pointed spaces using homotopy classes of maps $I^2 \to X$ which take the edges to $A$ and the vertices to $x$. This gave the first homotopy fundamental double groupoid, which enabled the proof of a 2-dimensional van Kampen theorem, including the usual theorem for the fundamental group as a special case, not just an implication. 
There are grounds for suggesting that simplicial sets are convenient,  but are not entirely adequate, since they cannot  easily express multiple compositions. On the other hand, cubical sets with connections are adequate for this test, but not entirely convenient! Andy Tonks proved that cubical groups with connection are Kan complexes. One reason for inconvenience    is that although they have been shown to form a strict test category in the sense of Grothendieck, in the paper given here,  the geometric realisation of the categorical  product is only of the homotopy type of the product of the realisations, not   actually homeomorphic to the product as in the case of simplicial sets. The latter homeomorphism property implies that in the right convenient category, the geometric realisation of a simplicial group is a topological group. 
The cubical setup is also not sufficient to describe the geometry underlying $n$-fold categories, and indeed there seems currently no name for such a structure in which cubes have different types of faces in different directions. Yet Grothendieck remarked to me on my affirming Loday's theorem, that (strict) $n$-fold groupoids model weak homotopy $n$-types: "That is absolutely beautiful!" 
So it as well not to assume we have the final story, and to investigate options! 
January, 2015: This answer is related to my answer to 
https://math.stackexchange.com/questions/1112107/why-does-seifert-van-kampen-not-hold-with-n-th-homotopy-groups/
November, 2016 There is more discussion in this preprint Modelling and Computing Homotopy Types: I.<|endoftext|>
TITLE: Index theorems and  orientability
QUESTION [6 upvotes]: Given a Dirac operator $D$ acting on some Clifford bundle $\mathcal{E}$ over a compact, even-dimensional, oriented manifold $M$, the Atiyah-Singer index theorem states that its index is given by pairing some characteristic class (which are elements of $H^n(M, \mathbb{R})$) with the fundamental class of the manifold, i.e.
$$ \mathrm{ind}(D) = \langle \hat{A}(M) \wedge \mathrm{ch}_{\mathcal{E}/S}(\mathcal{E}), [M] \rangle $$
in the case of a Dirac operator. In this case, we could use Chern-Weil theory, such that the characteristic classes are (equivalence classes) of differential forms, which can be integrated over the manifold, as it is oriented.
However, the theorem is true in the non-oriented case as well. In fact, the characteristic class above can be interpreted as a volume density, which can be integrated over $M$. 
Of course, I am aware that one can just go to the oriented double cover and use that fact that the index is multiplicative with respect to Riemannian coverings, so the "non-oriented Atiyah-Singer" follows easily from the regular one. However, I find this somewhat unsatisfactory.
Isn't there some (co-)homology theory in which the terms on the right side of the equation above can be expressed? I know that one often uses $\mathbb{Z}_2$-valued (co-)homology to deal with non-oriented manifolds; however, the index can be any integer, not just zero or one, so that this does not seem useful here.

REPLY [7 votes]: The trick, rather, is to write $\text{Ind}(D)$ in terms of the (co-)homology of (the total space of) $T^\ast M$, which is always orientable, viz,
$$
\text{Ind}(D) = \int_{T^\ast M} \text{ch}[\sigma_m(D)] \smile \text{Td}(T^\ast M \otimes \mathbb{C})
$$
where $\text{ch}[\sigma_m(D)] \in H^{\text{even}}(T^\ast M)$ is the Chern character of the symbol class $\sigma_M(D) \in K(T^\ast M)$ of $D$ and $\text{Td}(T^\ast M \otimes \mathbb{C})$ is the Todd genus of $T^\ast M \otimes \mathbb{C}$. For an explanation, see 
Intuitive explanation for the Atiyah-Singer index theorem 
and especially Paul Siegel's excellent answer. In particular, written this way, the Atiyah--Singer index theorem can be viewed as the translation into (co-)homological terms of a purely $K$-theoretic statement.<|endoftext|>
TITLE: $\int^{\infty}_{0}x^{r +s- 1}(1 + x)^{-s}(1 + x^2)^{-\frac{rm}{2}}dx$
QUESTION [11 upvotes]: I'm trying to solve the integral 
$\int^{\infty}_{0}x^{r +s- 1}(1 + x)^{-s}(1 + x^2)^{-\frac{rm}{2}}dx$, 
where $s$, $r$ and $m$>1 are positive integers.
My question is whether a closed form solution for this integral exists. By closed form here I mean an expression in terms of a finite number of special functions. There is some hope for a closed form solution I think given the output below from symbolic software, but I have been unable to find a formula yet.
To give a bit of context the integral is the $s$-th moment of the real random variable $x/(1+x)$, $x>0$, when the probability density function of $x$ is, up to a constant that I have omitted for simplicity, is $x^{r - 1}(1 + x^2)^{-\frac{rm}{2}}$. All these moments exist. I am trying to understand whether they are expressible in terms of a finite number of special functions.
I was not able to get help from Mathematica or Maple. Both Mathematica and Maple give me a solution containing a $\Gamma$ function evaluated at negative integers. More precisely, if I input in Mathematica:
Integrate[
 x^(r + s - 1) (1 + x)^-s (1 + x^2)^(-r m/2), {x, 0, \[Infinity]}, 
 Assumptions -> {r \[Element] Integers, s \[Element] Integers, 
   m \[Element] Integers, r > 0, s > 0, m > 1}]
I obtain:
(1/Gamma[s])
 Gamma[(-1 + m) r] Gamma[
   r - m r + s] HypergeometricPFQ[{(m r)/2, -(r/2) + (m r)/2, 
    1/2 - r/2 + (m r)/2}, {1/2 - r/2 + (m r)/2 - s/2, 
    1 - r/2 + (m r)/2 - s/2}, -1] + (1/(
 2 Gamma[(m r)/
   2]))(-s Gamma[1/2 (-1 + (-1 + m) r - s)] Gamma[
     1/2 (1 + r + s)] HypergeometricPFQ[{1/2 + s/2, 1 + s/2, 
      1/2 + r/2 + s/2}, {3/2, 3/2 + r/2 - (m r)/2 + s/2}, -1] + 
   Gamma[1/2 ((-1 + m) r - s)] Gamma[(r + s)/
     2] HypergeometricPFQ[{1/2 + s/2, r/2 + s/2, s/2}, {1/2, 
      1 + r/2 - (m r)/2 + s/2}, -1])
which is:
$\frac{\Gamma(r(m-1))\Gamma(s-r(m-1))}{\Gamma(s)}{}_3 F_2\left(  \frac
{mr}{2},\frac{r(m-1)}{2},\frac{1+r(m-1)}{2};\frac{1-s+r(m-1)}{2},1+\frac
{s-r(m+1)}{2};-1\right)$
$ -\frac{s\Gamma\left(\frac{r(m-1)-1-s}{2}\right)\Gamma\left(  \frac{1+r+s}{2}\right)  }
{2\Gamma(\frac{mr}{2})}{}_3 F_2\left(  \frac{1+s}{2},1+\frac{s}{2}
,\frac{1+r+s}{2};\frac{3}{2},\frac{3+s-r(m-1)}{2};-1\right)$
$   +\Gamma\left(  \frac{r(m-1)-s}{2}\right)  \Gamma\left(  \frac{r+s}
{2}\right){}_3 F_2\left(  \frac{1+s}{2},\frac{r+s}{2},\frac{s}{2}
;\frac{1}{2},1+\frac{s-r(m-1)}{2};-1\right)$
(same solution for Maple) As you can see, there is a $\Gamma$ function evaluated at $s-r(m-1)$ which can be a negative integer, so it seems to me that Mathematica and Maple are giving a solution that holds for most real values of $(r,m,s)$ but not necessarily for the values I'm interested in ($r,m,s$ are positive integers in my problem)
I'm not too sure my question is suitable here, but I did not have any luck at math.stackexchange (https://math.stackexchange.com/questions/234989/int-infty-0xr-s-11-x-s1-x2-fracrm2dx) so I thought I'd try and see if I can get some help here.

REPLY [3 votes]: Let $I(a,b,m) = \displaystyle \int_0^{\infty} \dfrac{x^{a+b-1} dx}{(1+x)^{b}(1+x^2)^{am/2}}$.
Setting $x = \tan(\theta)$. We then get that
$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \left(1+\tan^2(\theta) \right)^{am/2}}$$
$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \sec^2(\theta) d \theta}{(1+\tan(\theta))^b \sec^{am}(\theta)}$$
$$I(a,b,m) = \int_0^{\pi/2} \dfrac{\tan^{a+b-1}(\theta) \cos^{am-2}(\theta) d \theta}{(1+\tan(\theta))^b}$$
Let us use the following short hand notation. $c = \cos(\theta)$ and $s = \sin(\theta)$. We then get that
$$I(a,b,m) = \int_0^{\pi/2} \dfrac{s^{a+b-1} c^{am-a-1} d \theta}{(s+c)^b}$$
Hence, we are interested in evaluating integral of the form
$$J(p,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r}$$
First some observations:
$$J(p,q,r) = J(q,p,r) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$
$$J(p,q,0) = \dfrac{\beta((p+1)/2,(q+1)/2)}2$$
$$J(p,q,-1) = \int_0^{\pi/2} s^p c^q (s+c) d \theta = J(p+1,q,0) + J(p,q+1,0)$$
Further
$$J(p,q,r+2) = \int_0^{\pi/2} \dfrac{s^p c^q d \theta}{(s+c)^r (1+2sc)}$$
$$J(p,q,r+2) = \sum_{k=0}^{\infty} \int_0^{\pi/2} \dfrac{s^p c^q (-2sc)^k d \theta}{(s+c)^r}$$
$$J(p,q,r+2) = \sum_{k=0}^{\infty}(-2)^k J(p+k,q+k,r)$$
Hence, you can use this to compute $J(p,q,r)$ for all $r$ since these can be obtained as we know the values for $J(p,q,0)$ and $J(p,q,-1)$. (For even $r$, $J(p,q,r)$ will eventually depend on $J(p,q,0)$ and for odd $r$, $J(p,q,r)$ will eventually depend on $J(p,q,-1)$)

Few other relations, which might be of use.
We have $$\lim_{p \to \infty} J(p,q,r) = \lim_{q \to \infty} J(p,q,r) = \lim_{r \to \infty} J(p,q,r) = 0$$
Note that
$$J(p+2,q,r) = \int_0^{\pi/2} \dfrac{s^p c^q (1-c^2) d \theta}{(s+c)^r}$$
Hence, we get that
$$J(p+2,q,r) + J(p,q+2,r) = J(p,q,r)$$
Setting $p=q$, and making use of $(\star)$, we get that
$$J(p+2,p,r) = \dfrac{J(p,p,r)}2$$
Note that $$J(0,0,r) = \int_0^{\pi/2} \dfrac{d \theta}{(s+c)^r} = \int_0^{\pi/2} \dfrac1{2^{r/2}} \dfrac{d \theta}{\sin^r(\theta + \pi/4)}$$<|endoftext|>
TITLE: SAT and Arithmetic Geometry
QUESTION [16 upvotes]: This is an agglomeration of several questions, linked by a single observation: SAT is equivalent to determining the existence of roots for a system of polynomial equations over $\mathbb{F}_2$ (note though that the system is represented in non-trivial manner). The reason it is OK to consider more than one equation is because the conjunction of the conditions $f_i(x_1 ... x_n) = 0$ is equivalent to the single condition $\prod_i (f_i(x_1 ... x_n) + 1) + 1 = 0$.

This reminds of the solution of Hilbert's 10th problem, namely that it is undecidable whether a system of polynomial equations over $\mathbb{Z}$ has roots. Is there a formal relation? Can we use the undecidability over $\mathbb{Z}$ to provide clues why the problem is hard over $\mathbb{F}_2$ (that is, $P \ne NP$)? What is known about decidability and complexity for other rings? In particular, what is known about complexity over $\mathbb{F}_p$ for p prime > 2?
The system of polynomial equations defines an algebraic scheme. Is it possible to find algebro-geometric conditions on this scheme, s.t. something can be told about the complexity of SAT restricted to such schemes?
The solutions of our system of polynomial equations are the fixed points of the Frobenius endomorphism on the corresponding variety over $\bar{\mathbb{F}}_2$. There is a variant of Lefschetz's fixed-point theorem which relates the existence of such points to $l$-adic cohomology. Can this be used to provide some insight on P vs. NP?

REPLY [6 votes]: I'm going to address just one aspect of your question, which has been restated in some of the answers and comments, about whether there is any formal connection between undecidability and computational complexity.
This is certainly a very tempting idea, and you could argue that one of the main reasons to conjecture that the polynomial hierarchy does not collapse (this is a standard generalization of the conjecture that P≠NP) is that the infinitary analogue of the polynomial hierarchy, namely the arithmetical hierarchy, is known not to collapse.  Of course we know, for example from the Baker–Gill–Solovay theorem, that we can't naively carry over all intuitions from computability theory over to complexity theory, but the analogy remains in the back of the minds of many people working in this subject.
Perhaps closer to your question are two papers by Michael Freedman, Limit, logic, and computation and K-SAT on groups and undecidability.  The latter paper in particular notes that 2-SAT is polytime solvable and 3-SAT is NP-complete, and constructs infinitary analogues of these two problems that are respectively decidable and undecidable.  Your observation about the analogy between the MRDP theorem and the hardness of solving systems of equations over $\mathbb{F}_2$ could be thought of as being in the same vein.  Unfortunately, while intriguing, so far these analogies don't seem to have yielded any significant insights into the P≠NP problem.<|endoftext|>
TITLE: Notes for Bott's 1963 lectures on Morse theory
QUESTION [8 upvotes]: Would anybody happen to know where I could obtain a scanned version of
Lectures on Morse theory - [revised and expanded version of notes of lectures delivered at Professor R. Bott's topology seminar at Harvard in February and March of 1963], taken by Richard S Palais?
As far as I am aware, these notes were never published.

REPLY [2 votes]: There actually is a Morse theory book by Bott, but it's not the one you cite.  Rather, it's the 1960 volume Morse theory and its application to homotopy theory with the attribution "Lectures by R. Bott / Notes by A. van de Ven".  It's a little red book, but my copy (given to me by Bott himself!) is in a box in the rafters of my parents' garage at the moment, so I can't say anything else about it.  Content-wise, it's similar to Milnor's Morse Theory, but a bit more terse.<|endoftext|>
TITLE: Homotopy equivalent Morse functions
QUESTION [5 upvotes]: my question is the following: given a smooth manifold $M$, take a homotopy of maps $f_{t}:M \rightarrow \mathbb{R}, \quad t \in [0,1]$ such that every $f_{t}$ is a Morse function. Do $f_{0}$ and $f_{1}$ have the same critical points' structure?
In that case, is it possible to generalize the result to Morse-Bott functions?

REPLY [7 votes]: To see why the answer is yes for the first question use the implicit function theorem.   For any $t_0$  you can   find $\newcommand{\ve}{\varepsilon}$ $\ve >0$ and  smooth maps
$$\gamma_1,\dotsc, \gamma_N:(t_0-\ve,t_0+\ve)\to M$$
 such that  for any $t\in (t_0-\ve,t_0+\ve)$ the points $\gamma_1(t),\dotsc, \gamma_n(t)\in M$ are pairwise distinct and the  critical set of $f_t$ consists precisely of these points. Clearly the index of $\gamma_i(t)$ is independent of $t$, for any $i$.
If we denote  consider the set
$$ C:=\bigl\lbrace (t,x)\in [0,1]\times M;\;\;df_t(x)=0\bigr\rbrace, $$
then the above   fact shows that the natural projection
$$ C\ni (t,x)\mapsto t\in [0,1] $$
is a covering map. 
As for the   second question, I will give  an example. Denote by $\DeclareMathOperator{\Gr}{Gr}$ by ${\Gr}_k(V)$  the Grassmannian of $k$-dimensional subspaces of the complex Hermitian  space $V$. For any selfadjoint operator  $A: V\to V$ we have a smooth function
$${\Gr}_k(V)\ni L \mapsto f_A(L)= {\rm Re}\; {\rm tr}(AP_L)\in \mathbb{R}, $$
where $P_L$ denotes the orthogonal projection  onto $L$.  This function is Morse-Bott for any $A$, but it is  Morse if and only if the  eigenvalues of $A$ are  simple, i.e.,  pairwise distinct.     You can now easily construct a smooth family  of selfadjoit operators $A_t$, $t\in [0,1]$ such that $A_0=\boldsymbol{1}_V$ but $A_t$ has simple eigenvalues for any $t>0$.<|endoftext|>
TITLE: rationalized K-Theory of the group ring of finite cyclic groups
QUESTION [8 upvotes]: I am interested in calculating the rationalized algebraic K-Theory groups of the group ring of $\mathbb Z/n$, that is $K_i(\mathbb Z[\mathbb Z/n])\otimes \mathbb{Q}$ for any natural number $n\geq 2$. What is known about the rationalized K-Theory of $\mathbb Z[\mathbb Z/n]$ for an arbitrary $n$?

REPLY [10 votes]: The rationalized K-theory of finite groups is known: Let $G$ be a finite group and $n \ge 2$. 
$$K_n(\mathbb Z G)\otimes \mathbb Q =
\begin{cases}
\mathbb{Q}^r & n\equiv 1(4) \newline 
\mathbb{Q}^c & n\equiv 3(4) \newline 
0 & n \text{ even}
\end{cases}$$
where $r$ is the number of irreducible real representations of $G$ and $c$ of them are of complex type (see Theorem 2.2 here). 
Now let $G=\mathbb{Z}/n$. If $n$ is odd then $r=(n+1)/2$ and $c=(n-1)/2$. If $n$ is even, then $r=(n+2)/2$ and $c=(n-2)/2$.<|endoftext|>
TITLE: Free group actions on varieties and algebras of coinvariants
QUESTION [8 upvotes]: Suppose $k$ is an algebraically closed field of characteristic zero and $A$ is a finitely generated commutative associative reduced $k$-algebra.
Suppose the group $\mathbb{Z}_2$ acts on $A$ in such a way that the induced action on maxSpec $A$ is free.  (Side note: This is equivalent to the induced grading $A = A_0 \oplus A_1$ being a strong grading, i.e., $A_1^2 = A_0$.)
Let $S^2 A = (A \otimes A)^{S_2}$, where $S_2$ acts by swapping the factors in the tensor product.  The action of $\mathbb{Z}_2$ on $A$ induces a diagonal action of $\mathbb{Z}_2$ on $A \otimes A$.  Since this action commutes with the $S_2$ action, we have an induced action of $\mathbb{Z}_2$ on $S^2 A$.
Now let $(S^2 A)_{\mathbb{Z}_2}$ be the corresponding algebra of coinvariants. By definition, this is the quotient of $S^2 A$ by the ideal generated by elements of the form $(b - g \cdot b)$ for $g \in \mathbb{Z}_2$ and $b \in S^2A$.
My question is: Is $(S^2 A)_{\mathbb{Z}_2}$ isomorphic to $A^{\mathbb{Z}_2}$ (as an algebra)?
My reason for thinking this is true is the following intuition:

$S^2 A$ is the coordinate ring of the set of unordered pairs of points of $A$.
Thus $(S^2 A)_{\mathbb{Z}_2}$ is the coordinate ring of set of unordered pairs of points of $A$ that are invariant under the $\mathbb{Z}_2$-action.
Since the action of $\mathbb{Z}_2$ is free, unordered pairs of points of $A$
that are invariant under the $\mathbb{Z}_2$ action are just $\mathbb{Z}_2$-orbits.
Thus $(S^2 A)_{\mathbb{Z}_2}$ should be isomorphic to $A^{\mathbb{Z}_2}$, which is precisely the coordinate ring of the variety of $\mathbb{Z}_2$-orbits.

I think the above reasoning is rigorous on the level of maximal ideals.  But I'd like to know that the rings are actually isomorphic.

REPLY [3 votes]: You are right, they are isomorphic.
Denote by $g$ the generator of $\Bbb Z_2$. The map given by $x\otimes y\mapsto x^gy$ is obviously a homomorphism
$h:A\otimes A\to A$ of algebras. Since $S^2A=S^2A_0+S^2A_1+K$, where $K$ is spanned by $a_0\otimes a_1+a_1\otimes a_0$,
$a_i\in A_i$, and $A_1^2=A_0$, we conclude that $h(S^2A)=A_0$ (even $h(S^2A_1)=A_0$) and that $K$ generates in $S^2A$ the ideal $I$
such that $S^2A/I$ is the algebra of coinvariants.
In order to show that the kernel of $h$ on $S^2A$ equals $I$, we observe that, providing
$\sum_ia_{0i}a_{1i}b_{1i}=\sum_jc_{1j}d_{1j}$, we have
$$\sum_i(a_{0i}\otimes a_{1i}b_{1i}+a_{1i}b_{1i}\otimes a_{0i})+\sum_j(c_{1j}\otimes d_{1j}+d_{1j}\otimes c_{1j})=$$
$$=\sum_i(a_{0i}\otimes a_{1i}+a_{1i}\otimes a_{0i})(1\otimes b_{1i}+b_{1i}\otimes1)-$$
$$-\sum_i(1\otimes a_{0i}b_{1i}+a_{0i}b_{1i}\otimes1)(1\otimes a_{1i}+a_{1i}\otimes1)+$$
$$+\sum_j(1\otimes c_{1j}+c_{1j}\otimes1)(1\otimes d_{1j}+d_{1j}\otimes1)\in I,$$
where $a_{0i}\in A_0$ and $a_{1i},b_{1i},c_{1j},d_{1j}\in A_1$.
Let $s_i\in S^2A_i$ be such that $h(s_0+s_1)=0$. We can write $s_0=\sum_i(a_{0i}\otimes a_{1i}b_{1i}+a_{1i}b_{1i}\otimes a_{0i})$
and $s_1=\sum_j(c_{1j}\otimes d_{1j}+d_{1j}\otimes c_{1j})$ for suitable $a_{0i}\in A_0$ and $a_{1i},b_{1i},c_{1j},d_{1j}\in A_1$
because $A_0=A_1^2$. From $h(s_0+s_1)=0$, we obtain $\sum_ia_{0i}a_{1i}b_{1i}=\sum_jc_{1j}d_{1j}$ (the characteristic of $k$ is
different from $2$). By the above idenity, $s_0+s_1\in I$.<|endoftext|>
TITLE: Is a smooth cubic threefold diffeomorphic to a rational threefold?
QUESTION [7 upvotes]: A theorem of Clemmens and Griffiths states that a smooth hypesurface in $\mathbb CP^4$ of degree three is not rational. I would like to know if nevertheless it is diffeomorphic (as a smooth real $6$-dimensional manifold) to a rational complex three-dimensional variety?

REPLY [7 votes]: Let me work out the idea of Ulrich, and deduce that the answer to the question is negative, namely a cubic three-fold is not diffeomorphic to any rational variety.
The idea of Urlich is that if a rational threefold is diffeomorpic to a cubic then it is a Fano with $Pic=\mathbb Z$. So we just have to check that cubic is not diffeomorphic to any 
other type of three dimensional Fanos with $Pic=\mathbb Z$.
There exist $17$ families of Fano threefolds with $Pic=\mathbb Z$ 
($\mathbb CP^3$, quadric, five Fanos of index two including cubics, and $10$ Fanos of index $1$). The description can be found here one page two:
http://www.math.u-psud.fr/~amerik/articles/obzor-fv.pdf
Let $H$ be the hyperplane section of the cubic, then $H^3$ equals 
$3$, i.e., the degree of the cubic. So the cubic has the following property:
if we take all classes in integral second cohomology of the cubic then 
their cubes are of the form $3\cdot n^3$ where $n\in \mathbb Z$.
Checking the list of $17$ Fanos we see that the cubic is the only one 
with this property. So it is not diffeomorphic to any other Fano three fold.<|endoftext|>
TITLE: What is the algebraic geometry version of the spheres?
QUESTION [29 upvotes]: In topology the spheres $S^n$ are the "simplest" closed manifolds, and they are like "Dirac's delta at $n$" for (reduced) cohomology groups. Furthermore they are boundaries of the simplest compact manifolds-with-boundary, i.e. the disks $D^{n+1}$, which are contractible.  And $S^{n}$ is obtained by glueing two copies of $D^{n}$ along their boundary $S^{n-1}$.
My question is:

Are there some objects of algebraic geometric nature that somehow reproduce the same pattern, or that are considerable as the equivalent of spheres from topology?
More generally, are there "homology spheres" for some homology theory like -say- Chow groups? What about an "algebraic Poincaré conjecture"?

If they do exist, I don't expect them to be standard varieties or schemes, otherwise they probably would have made their appearence "classically".

REPLY [11 votes]: To expand on Tom Goodwillie's answer:
a precise definition of "motivic sphere" would be
$$S^{p,q} = \big( \Delta^{p-q} / \partial \Delta^{p-q} \big) \wedge \big( \bigwedge^q \mathbb{G}_m\big)$$
which you can interpret as the $q$-fold smash product of the multiplicative group $\mathbb{G}_m$ smashed with a $(p-q)$-dimensional simplicial sphere. This smash product makes sense in a category of simplicial presheaves on smooth schemes (which is what you work with in A¹-homotopy theory).
The indices can be explained from looking at the motive $M(S^{p,q}) = \mathbb{Z}(q)[p]$ or at the realizations, where the complex realization gives $S^p$ and the real realization gives $S^{p-q}$. More about this can be read in the Morel-Voevodsky paper.
Now these are not algebraic varieties, by definition, so a good question is

For integers $p$ and $q$, does an algebraic variety $X$ exist which is A¹-weakly equivalent to $S^{p,q}$?

We'll also say "$X$ is a $S^{p,q}$".
From looking at realizations you can already exclude $p,q$ negative.
One positive example is (by definition) $\mathbb{G}_m$, which is a $S^{1,1}$.
As Tom Goodwillie explained, $\mathbb{A}^n / (\mathbb{A}^n \setminus 0)$ is a $S^{2n,n}$ (which some people shorten to "motivic $2n$-sphere") and $\mathbb{A}^n \setminus 0$ is a $S^{2n-1,n}$.
There is not much known beyond that, I suppose.

There are even more varieties that could count as algebraic versions of spheres. An interesting feature of algebraic geometry is that you can look at many fields of definition. Projective quadrics over the complex numbers all look the same (in each dimension), as they are isomorphic (given the same dimension) to the split quadrics $Q_{2n}^{split} = \{\sum_{i=0}^n x_iy_i = 0\}$ or $Q_{2n+1}^{split} = \{\sum_{i=0}^n x_iy_i = z^2\}$ as well as to the anisotropic quadrics $Q_{m} = \{\sum_{i=0}^m z_i^2\}$ (by change of basis). Over smaller, non-quadratically closed fields, these quadrics are no longer isomorphic.
Now look at the affine quadrics inside the projective quadrics (by removing a suitable hyperplane section). Conjecturally, the split ones are motivic spheres (at least they have the right motive), while the anisotropic ones aren't. You can consider these forms of motivic spheres.<|endoftext|>
TITLE: Is the tensor product of polyhedra a polyhedron?
QUESTION [18 upvotes]: Conventions: A polytope in a finite-dimensional $\mathbb R$-vector space $V$ is defined to be a convex hull of finitely many points in $V$. A polyhedron in a finite-dimensional $\mathbb R$-vector space $V$ is defined to be an intersection of finitely many closed halfspaces in $V$ (that is, the set of solutions of a system of finitely many non-strict linear inequalities on a vector in $V$). It is known that the polytopes are exactly the bounded polyhedra.
Note that, for me, $\mathbb R$ really can mean any ordered field, like $\mathbb Q$ or $\mathbb Q\left[\sqrt{2}\right]$ or many others. Every claim stated below holds when $\mathbb R$ is replaced by any ordered field, and an answer that makes use of special properties of $\mathbb R$ is welcomed but won't be considered final.
Background: The decomposition theorem for polyhedra yields the following facts as easy consequences:
1. If $f:V\to W$ is an $\mathbb R$-linear map between finite-dimensional $\mathbb R$-vector spaces, and $P$ is a polyhedron in $V$, then $f\left(P\right)$ is a polyhedron. (The same statement holds with "polyhedron" replaced by "polytope", but that is a triviality.)
2. If $f:V\to W$ is an $\mathbb R$-linear map between finite-dimensional $\mathbb R$-vector spaces, and $P$ is a polyhedron in $W$, then $f^{-1}\left(P\right)$ is a polyhedron. (This one is obvious, but just mentioned here for the sake of "symmetry".)
3. If $P$ is a polyhedron in a finite-dimensional $\mathbb R$-vector space $V$, and $Q$ is a polyhedron in a finite-dimensional $\mathbb R$-vector space $W$, then $P\times Q$ is a polyhedron in $V\times W$. (The same holds for polytopes.)
4. If $P$ and $Q$ are two polyhedra in one and the same finite-dimensional $\mathbb R$-vector space, then the Minkowski sum $P+Q$ and the intersection $P\cap Q$ are polyhedra as well. (Again, the same holds for polytopes.)
5. If $P$ is a polyhedron in a finite-dimensional $\mathbb R$-vector space $V$, and $Q$ is a polyhedron in a finite-dimensional $\mathbb R$-vector space $W$, then $\left\lbrace f\in\mathrm{Hom}_{\mathbb R}\left(V,W\right) \mid f\left(P\right) \subseteq Q\right\rbrace$ is a polyhedron in $\mathrm{Hom}_{\mathbb R}\left(V,W\right)$. (This is inspired by Definition 9.16 in Günter M. Ziegler, Lectures on Polytopes, 1995.)
Question:
6. If $P$ is a polyhedron in a finite-dimensional $\mathbb R$-vector space $V$, and $Q$ is a polyhedron in a finite-dimensional $\mathbb R$-vector space $W$, then is it true that the convex hull of the set $\left\lbrace p \otimes q \mid p\in P,\ q\in Q \right\rbrace $ is a polyhedron in $V\otimes W$ ?
This holds for polytopes, and follows in that case from §2.5 of Lawrence Valby, A Category of Polytopes (caveat: my convex hull is not his $P\otimes Q$, but rather the image of his $P\otimes Q$ under a surjection which keeps the $p_iq_j$ coordinates and forgets the $p_i$, $q_j$ and $1$ coordinates); but the argument there does not generalize to polyhedra. On the other hand, I am at a loss when trying to find a counterexample. Any ideas?

REPLY [11 votes]: Not always! However, the closure is a polyhedron.

Not always: Take $P = \{a | 0 \leq a \leq 1\}$. Take $Q= \{ b,c | b\geq 0, c=1\}$.  Then under the map $x=ab$, $y=ac$. Since $P$ is the convex hull of $(0)$ and $(1)$, and $Q$ is the ray starting at $(0,1)$ and going in direction $(1,0)$, $P \otimes Q$ is the convex hull of $(0,0)$ and the ray starting at $(0,1)$ and going in direction $(1,0)$, which is: 
$P \otimes Q = \{x,y | 0\leq x, 0 \leq y \leq 1, (y>0 \vee x=0) \}$
This not a polyhedron because it is not closed.

The closure is: Define $B$ to be the following convex body. First we show that $B$ is a polyhedron. Next we will show that $cl(P \otimes Q)=B$.
$B= \operatorname{conv}(P_p \otimes Q_p + ((P_p+P_c) \otimes Q_c) + (P_c \otimes (Q_p+Q_c)) + (P \otimes Q_l) + (P_l \otimes Q)) $
Since $\operatorname{conv}(A + B) = \operatorname{conv}(A) + \operatorname{conv}(B)$, 
$B= \operatorname{conv}(P_p \otimes Q_p) + \operatorname{conv}(((P_p+P_c) \otimes Q_c) + (P_c \otimes (Q_p+Q_c)) )+ \operatorname{conv}((P \otimes Q_l) + (P_l \otimes Q))) $
The first part is clearly a polytope. The last part is clearly a linear subspace. The cone is the tricky but one can view a cone as the convex hull of finitely many rays. A ray tensor a polytope is the convex hull of finitely many rays, thus a cone. A ray tensor a cone is the convex hull of finitely many rays, thus a cone again. Taking the convex hull of different cones could produce more linear subspaces but will not take you out of the world of polyhedra. (Dima might be able to produce a better argument than this?)
Next we show that $cl(P \otimes Q) \subseteq B$. As a polytope, it is convex and closed, so it is enough to show that any element in $P$ tensor an element in $Q$ is in $B$. But this is obvious - just split that element into a sum.
Finally we show that $B \subseteq cl(P \otimes Q)$. Since $P \otimes Q$ is convex, $cl(P\otimes Q)$ is convex, so need to show that a sum of an element in $P_p \otimes Q_p$, an element in $(P_p+P_c)\otimes Q_c$, etc. is in $B$. Assume for simplicity we merely need to add $a \otimes b$ in $P_p \times Q_p$ to $c \otimes d$ in $(P_p+P_c) \otimes Q_c$. (To get the general caes, you just repeat the argument). Let $e$ be any element of $Q_p$. Then we notice that  
$\lim _ {\lambda \to 0} \left((1-\lambda) \left[a \otimes b\right] +  \lambda \left[c \otimes \left(\frac{d}{\lambda}+ e \right)\right]\right)= a\otimes b+ c\otimes d $
$a \otimes b$ and $c \otimes \frac{d}{\lambda}+ e$ are in $P \otimes Q$ as long as $\lambda>0$, so a convex combination of them is as well, so the limit as $\lambda \to 0$ is in $cl(P \otimes Q)$. The key fact is that $Q_c$ is closed under multiplication by positive real numbers. Since $P_c$, $Q_l$, and $P_l$ are as well, we can apply this trick again to get an arbitrary sum.<|endoftext|>
TITLE: How to solve a system of linear equations without storing the matrix?
QUESTION [7 upvotes]: I have a procedurally defined Hermitian matrix $M$, i.e. I can get any matrix element by calling a black box function (e.g. a library function), and a vector $Y$. And I have to solve a system of linear equations:
$M\cdot X=Y$.
But $Y$ is such that having $n$ elements, it takes about half of available RAM, another half would be for $X$, so if I try to store $M$, it'll take $n^2$ space, i.e. even if I double the RAM space, this will still place $n-2$ matrix columns/rows into swap.
At the same time, all the algorithms which I've found need saving large amounts ($\geq n^2$) of data while solving the system.
So, the question: are there any algorithms to solve systems of linear equations which don't require me to store the matrix, and still are fast enough (maybe not $O(n^3)$, but at least not much slower)?

REPLY [6 votes]: This looks like a situation where the Kaczmarz method could work.
What you do to maintain an approximate solution and then project cyclically onto the hyperplanes which are given by the $k$-th equation. More precisely: If you have the $m$-the iterate $X^m$ and use the $k$-th equation, then you have the next iterate
$$X^{m+1} = X^M + \frac{Y_k - a_k^T\cdot X^m}{\|a_k\|^2}a_i$$
where $a_k$ is the $k$-th row of $A$ and $Y_k$ is the $k$-the entry of $Y$.
Hence, you only need one row of $A$, one entry of $Y$ and the current iterate $X^m$ to perform one iteration, i.e. $2n+1$ space. Also the iteration complexity is very low (it's $\mathcal{O}(n)$) but you usually need a lot of iterations.
This methods is widely used in discrete tomography and is also an instance of the "Projection onto convex sets" method.
Recently it has been shown by Strohmer and Vershynin that a randomized version of this method has favorable convergence properties (when you pick each column with a probability proportional to its norm). Also "block iterative" versions work, i.e. you take a hyperplane of higher codimension to project on. So, if you have some memory left, you could also take some more rows of $A$ at once... See, e.g. here.<|endoftext|>
TITLE: Rational smooth complex projectives three fold with non-rational deformation
QUESTION [8 upvotes]: This question is prompted by a great talk of Beauville:
http://www.mathnet.ru/php/presentation.phtml?presentid=5821&option_lang=rus
The talk is called "Luroth problem". In this talk Beauville considers in particular Fano three-folds and says how one can prove that some of them are not rational. 
Still I was not able to figure out the following:  is there any example of a rational (smooth of course) complex projective three fold that admits a deformation that is not rational? If yes what is the simplest example?

REPLY [4 votes]: A conjecture of Iskovskikh says that this never happens. To be more precise, it says that if there is a family of smooth projective threefolds with general threefold nonrational then all these threefolds are nonrational. 
The conjecture is not proved. On one hand it is not clear how this can be proved, on the other hand no counterexample known.<|endoftext|>
TITLE: Groups that do not exist
QUESTION [65 upvotes]: In the long process that resulted in the classification of finite simple groups, some of the exceptional groups were only shown to exist after people had computed (most of) their character tables and other such precise information which usually can only be attached to things that exist. Maybe someone familiar with the details can tell me/us:

Was there at some point a finite simple group conjectured to exist that turned out not to exist in the end?

If so, this part of the story is much less told than the successful part! It would be interesting to know if for such a non-existent group, say, the character table was computed, and so on... 
I am asking because I just read this question on Math.SE and it reminded me I have always wanted to know this.

REPLY [15 votes]: I believe that at some point there was a conjecture (by whom, I don't recall) that Janko's smallest group, of order $175,560=11(11^2-1)(11^3-1)/(11-1)$, should be the first of an infinite sequence of finite simple groups with a Lie-type group-order formula. Such groups turned out not to exist.<|endoftext|>
TITLE: An algebra of "integrals"
QUESTION [40 upvotes]: When discussing divergent integrals with people, I got curious about the following:
Is there an $\mathbb{R}$-algebra $A$ together with a map (could be defined on just a subspace)
$$\int_0^{\infty}: C^{\infty}(\mathbb{R})--\to A$$
which behaves like integration, and is defined even for some function whose integration is divergent in the usual sense? Or,

Can we find some universal target of integration which is like the module of Kahler differentials as the universal target of derivation?

In other words, is there an $\mathbb{R}$-algebra $A$ together with a map $T: C^{\infty}(\mathbb{R})\to A$ such that (here is a list of plausible properties)
$$T(f(x))=\int_0^{\infty}f(x)dx\in\mathbb{R},\text{ if the RHS converge}.$$
$$T(f(x))-T(f(x+a))=\int_0^af(x)dx \text{ for any }f(x).$$
$$T(f(x))=aT(f(ax)) \text{ for } a>0.$$
(Edit: I removed the 4th, which is included in the first.)
Or for some reasons such an $\mathbb{R}$-algebra could only be $\mathbb{R}$? I tried to construct a ``free'' algebra of some kind but it is not clear to me what I got. (from the conditions above there are too many generators and relations, and there are even things from "integration by parts" given the last rule, I'm not sure what the quotient of the generators by the relations gives.)
(Edit: people asked why $A$ need to be an algebra, I don't have a good reason for that. I just want to see if one can extend the definition of integration so that they land in some vector space with a certain structure. The most naive thing I can think about is that the result of an integral is a certain kind of "number", and we add, subtract, multiply numbers.)

REPLY [7 votes]: I am currently working on a similar system. But your properties (2) and (3) would not work and need change.
Instead, the following properties would work much better:
$$\int_a^c f(x) dx=\int_a^b f(x)dx+\int_b^c f(x)dx\tag{1}$$
$$\int_a^b (f(x)+g(x)) dx=\int_a^b f(x)dx+\int_a^b g(x)dx\tag{2}$$
$$\int_a^b c f(x) dx =c \int_a^b f(x) dx\tag{3}$$
$$\int_{-\infty}^0 f(x) dx=\int_0^\infty f(-x) dx\tag{4}$$
where $a,b,c,f(x)$ and $g(x)$ take values from ℝ ∪ {−∞, +∞}.
Thus, any integral
$$T=\int_a^b f(x) dx$$
represents an "extended" number.
The integrals which are regularizable by a stable method like Cesaro or Abel (as opposed to Ramanujan or Dirichlet) are taken to be equal to their regularized sum:
$$\int_0^\infty f(x)\,dx=\lim_{\epsilon\to 0}\int_0^\infty e^{-\epsilon x}f(x) \, dx\tag{6}$$
Two integrals $\int_0^\infty f(x) dx$ and $\int_0^\infty g(x) dx$ are thus equal if
$$\lim_{\epsilon\to 0}\int_0^\infty e^{-\epsilon x}(f(x)-g(x)) \, dx=0$$
We also can equate some divergent series to the integrals:
$$\sum_{k=0}^\infty f(k)=\int_{-1/2}^\infty\sum_{k=0}^\infty\operatorname{rect}(x+k)f(k)dx$$
In our notation we will consider by definition $$\sum_{k=n}^\infty f(k)=\sum_{k=0}^\infty f(k)-\sum_{k=0}^{n-1}f(k)$$
Now, we postulate that the regularized value or the integral or corresponding series represents the regular part of an extended number, while the rest is the irregular part. Among the suitable regularization methods are Cesaro, Abel, Ramanujan, Borel, Dirichlet regulaization and some others (they agree with each other when applicable). We will denote the regularized value of an extended number $w$ as $\operatorname{reg} w$
Particularly, very useful would be the Faulhaber's formula for Ramanujan's summation of analytic functions:
$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n \tag{7}$$
We will use the following symbols for the three most key integrals and series:
$$\omega_+=\sum_{k=0}^\infty 1$$
$$\omega_-=\sum_{k=1}^\infty 1=\omega_+-1$$
$$\tau=\int_0^\infty dx=\omega_+-1/2=\omega_-+1/2$$
(this can also be formally interpreted as $\tau=\pi\delta(0)$ due to Fourier transform).
By interpreting formula (7) as a Taylor series, we come to a formula which allows to generalize the analytic functions to extended numbers (at least in the sense of determining the regular part of the result):
$$\operatorname{reg} f'(\omega_-+z)= \Delta f(z)\tag{8}$$ and in particular, to the generalizations of powers of our key series:
$$\operatorname{reg}\omega_-^n=B_n\tag{9}$$
$$\operatorname{reg}\omega_+^n=B^*_n\tag{9a}$$
where $B^*$ are the second Bernoulli numbers (those which have $B^*_1=1/2$ ).
A more general formula reveals the role of the Hurwitz zeta function:
$$\operatorname{reg}(\omega_-+x)^a= B_a(x)=-a\zeta(1-a,x)$$
Based on formula (7) we even can derive an expression for a derivative of an analytic function which does not use limits:
$$f'(x)=\operatorname{reg}(f(\omega_++x)-f(\omega_-+x))=\operatorname{reg} \Delta f(\omega_-+x)$$
which works for any regular $x$.
Also, since many series expansions of trigonometric functions use Bernoulli numbers, we can interpret them as regular parts of similar series but involving extended numbers. This way, and using formula (8) we can obtain the following relations:
$$\operatorname{reg}\sin (a\omega_-+x) =  \frac{a}{2} \cot \left(\frac{a}{2}\right) \sin x -\frac{a}{2} \cos x$$
$$\operatorname{reg}\cos (a\omega_-+x) = \frac{a}{2} \csc \left(\frac{a}{2}\right) \cos \left(\frac{a}{2}- x \right)$$
$$\operatorname{reg}\ln (\omega_-+z)=\psi(z)$$
$$\operatorname{reg} e^{z\omega_-}=\frac{z}{e^{z}-1}$$
Particularly, $$\operatorname{reg}\sin \omega_-=-1/2;$$ $$\operatorname{reg}\sin \omega_+=1/2;$$ $$\operatorname{reg}\ln \omega_+=-\gamma;$$ $$\operatorname{reg} e^{\omega_-}=\frac{1}{e-1};$$ $$\operatorname{reg} e^{\omega_+}=\frac{e}{e-1}.$$
Another notable thing is the possibility to express trigonometric functions via inverse trigonometric or logarithms:
$$\cot x=\operatorname{reg}\frac1{\pi }\ln \left(\frac{\omega _+-\frac{x}{\pi }}{\omega _-+\frac{x}{\pi }}\right) = \operatorname{reg}\frac2z \cos (2x\omega_\pm)$$
$$\tan x=\operatorname{reg} \frac1\pi\ln \left(\frac{\tau +\frac{x}{\pi }}{\tau -\frac{x}{\pi }}\right)$$
$$\coth x=\operatorname{reg}\frac{1}{\pi} \operatorname{arccoth}\left(\frac{\pi  \omega _+}{x}\right)+\frac1x=\operatorname{reg}\frac1x \cosh (2 x\omega_\pm)$$
Following from Faulhaber's formula (for $ n\ge0 $ ),
$$ \int_0^\infty x^n dx=\frac{\left(\tau +\frac{1}{2}\right)^{n+2}-\left(\tau -\frac{1}{2}\right)^{n+2}}{(n+1)(n+2)}=\frac{\omega _+^{n+2}-\omega _-^{n+2}}{(n+1)(n+2)} $$
Interpreting Fourier transform formally, for even $ n $ we also have
$$ \int_0^\infty x^n dx=i^n\pi\delta^{(n)}(0) $$
For $ n>1 $
$$ \int_0^\infty \frac1{x^n} dx=\frac1{(n-1)!}\int_0^\infty x^{n-2} dx=\frac{\omega _+^{n}-\omega _-^{n}}{(n-1)n!} $$
Particularly,
$$\int_0^\infty 1 dx =\tau$$
$$\int_0^\infty x dx=\frac{\tau^2}2+\frac1{24}$$
$$\int_0^\infty x^2 dx=\frac{\tau^3}3+\frac{\tau}{12}$$
We introduce generalized limits in the following way:
$$
\operatorname{gen}\lim_{x\to u^+}f(x)=f(a)-\int_u^a f'(x)dx
$$
where $a>u$ and
$$\operatorname{gen}\lim_{x\to u^-}f(x)=f(a)+\int_a^u f'(x)dx$$
where $a<u$
This value can serve as a measure of the growth rate of a function.
Particularly, for $ n\ge0 $
$$ \operatorname{gen}\lim_{x\to\infty}x^n=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{n+1} $$
For odd $ n $ ,
$$ \operatorname{gen}\lim_{x\to\infty}x^n=i^{n-1}\pi n\delta^{(n-1)}(0) $$
And
$$ \operatorname{gen}\lim_{x\to 0^+} \frac1{x^n}=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{(n+1)!} $$
Following Urs Graf, p.36, for odd $ n $ ,
$$ \operatorname{gen}\lim_{x\to 0^+} \frac1{x^n}=\frac{i^{n-1}\pi\delta^{(n-1)}(0)}{(n-1)!} $$
For instance,
$$\int_{0^+}^\infty \frac1{x^2}=\tau$$
This, combined with the delta function form of $\tau=\pi\delta(0)$ can explain why logarithm has its imaginary part represented as a step function (which is integral of delta function).
The Cauchy principal value of an analytic function at a pole corresponds to the regular part of its generalized limits at the pole, while the order of the pole corresponds to the order of polynomial of $\tau$ of the generalized limit. Thus,
$$\operatorname{gen}\lim_{x\to0^\pm}\Gamma(0)=-\gamma\pm\tau$$
$$\operatorname{gen}\lim_{x\to{-1}^\pm}\Gamma(x)=\gamma-1\mp\tau$$
$$\operatorname{gen}\lim_{x\to{-2}^\pm}\Gamma(x)=\frac{3}{4}-\frac{\gamma }{2}\pm\frac\tau 2$$
$$\operatorname{gen}\lim_{x\to{-3}^\pm}\Gamma(x)=\frac{\gamma }{6}-\frac{11}{36}\mp\frac\tau 6$$
$$\operatorname{gen}\lim_{x\to1^\pm}\zeta(x)=\gamma\pm\tau$$
For Bernoulli polynomials we have these noticeable formulas:
$$\omega_-^n=\operatorname{gen}\lim_{x\to\infty}B_n(x)$$
$$\omega_+^n=\operatorname{gen}\lim_{x\to\infty}B_n(x+1)$$
And more generally,
$$(\omega_-+a)^n=B_n(a)+n\int_0^\infty B_{n-1}(x+a)dx$$
Moreover, generalizing the above formulas we can write down the general formula for conversion of divergent integral representation to omega-tau representation:
$$\int_0^{\infty } f(x) \, dx=\operatorname{reg}\int_0^{\infty } f(x) \, dx+\int _{\omega_-}^{\omega_+} \int _0^x f(t) dtdx$$
The integral from $\omega_-$ to $\omega_+$ should be understood as the difference of the antiderivative on those values.
Here is a table of some extended numbers and their representations in different forms:
$$
\begin{array}{cccccc}
 \text{Delta form} & \text{In terms of } \tau, \omega_+,\omega_- & \text{Regular part} & \text{Integral or series form} & \text{Generalized limit form} \\
 \pi  \delta (0) & \tau  & 0 & \int_0^{\infty } \, dx;\int_0^{\infty } \frac{1}{x^2} \, dx & \operatorname{gen}\lim_{x\to\infty}x;\operatorname{gen}\lim_{x\to0^+}\frac1x \\
 \pi  \delta (0)-\frac{1}{2} & \omega _-;\tau-\frac{1}{2} & -\frac{1}{2} & \sum _{k=1}^{\infty } 1 & \operatorname{gen}\lim_{x\to\infty} (x-1/2) \\
 \pi  \delta (0)+\frac{1}{2} & \omega _+;\tau+\frac{1}{2} & \frac{1}{2} & \sum _{k=0}^{\infty } 1 & \operatorname{gen}\lim_{x\to\infty} (x+1/2) \\
 2 \pi  \delta (i) & e^{\omega_+}-e^{\omega_-}-1 & 0 & \int_{-\infty }^{\infty } e^x \, dx & \operatorname{gen}\lim_{x\to\infty} e^x \\
& \frac{\tau ^2}{2}+\frac{1}{24};\frac{\omega_+^3-\omega_-^3}6 & 0 & \int_0^{\infty} x \, dx;\int_0^\infty \frac2{x^3}dx & \operatorname{gen}\lim_{x\to\infty}\frac{x^2}2;\operatorname{gen}\lim_{x\to{0^+}} \frac1{x^2}\\
 & \frac{\tau ^2}{2}-\frac{1}{24} & -\frac1{12} & \sum _{k=0}^{\infty } k & \operatorname{gen}\lim_{x\to\infty} \left(\frac{x^2}2-\frac1{12}\right) \\
 -\pi \delta''(0) &\frac {\tau^3}3 +\frac\tau{12};\frac{\omega_+^4-\omega_-^4}{12}& 0 & \int_0^\infty x^2dx;\int_0^\infty\frac6{x^4}dx&\operatorname{gen}\lim_{x\to\infty}\frac{x^3}3;\operatorname{gen}\lim_{x\to{0^+}} \frac2{x^3}\\
\pi^2\delta(0)^2-\pi\delta(0)+1/4&\omega_-^2&\frac16&2 \int_0^{\infty } \left(x-\frac{1}{2}\right) \, dx+\frac{1}{6}&\operatorname{gen}\lim_{x\to\infty}B_2(x)\\
\pi^2\delta(0)^2+\pi\delta(0)+1/4&\omega_+^2&\frac16&2 \int_0^{\infty } \left(x+\frac{1}{2}\right) \, dx+\frac{1}{6}&\operatorname{gen}\lim_{x\to\infty}B_2(x+1)\\
\pi^2\delta(0)^2&\tau^2&-\frac1{12}&\int_{-\infty}^{\infty } |x| \, dx-\frac{1}{12}&\operatorname{gen}\lim_{x\to\infty}B_2(x+1/2)\\
&\ln \omega_++\gamma&0&\int_1^\infty \frac{dx}x;\sum_{k=1}^\infty \frac1x -\gamma&\operatorname{gen}\lim_{x\to\infty}\ln x\\
 -\pi\delta''(0)-\frac14 \pi\delta(0);\pi^3\delta(0)^3&\tau^3&0&\int_0^\infty \left(3x^2-\frac1{4}\right)dx&\operatorname{gen}\lim_{x\to\infty}B_3(x+1/2)\\
\frac{2\pi\delta(i)+1}{e-1}&e^{\omega_-}&\frac1{e-1}&\frac1{e-1}+\frac1{e-1}\int_{-\infty}^\infty e^x dx&\operatorname{gen}\lim_{x\to\infty} \frac{e^x+1}{e-1}\\
\frac{2\pi\delta(i)+1}{1-e^{-1}}&e^{\omega_+}&\frac1{1-e^{-1}}&\frac1{1-e^{-1}}+\frac1{1-e^{-1}}\int_{-\infty}^\infty e^x dx&\operatorname{gen}\lim_{x\to\infty} \frac{e^x+1}{1-e^{-1}}\\
\end{array}
$$

The system forms an integral domain which can be extended to form a field.
Still, there are many white areas, particularly, there is no straightforward way to transform a divergent integral into an expression in terms of $\omega_\pm$ and vice versa. There is no formula to construct an integral representing a product of two integrals, etc.<|endoftext|>
TITLE: The Surreal numbers satisfy all the field axioms except that its elements constitute a proper class. Is it safe to call it a field?
QUESTION [15 upvotes]: Do all the field theorems apply to surreal numbers? If fields were redefined so that their elements were allowed to come from an arbitrary class, would the theory look different to an algebraist?

REPLY [15 votes]: First, let me say that the set/class issue is not a problem to deal with properly, and so one shouldn't be very worried about it. It is true as you say that the surreal numbers No are a
proper class, and they do not form a set. So in a purely technical sense, they are not a field. But nevertheless, they do satisfy all the field axioms and have all the usual kinds of structure that one would want in a field, and so one can correctly describe them as a proper class field, or Field as Conway writes it, in much the same way that the class of all ordinals is regarded as a proper class well-order. Although this set/class issue may seem mysterious or irritating, in practice it is a routine matter
to handle correctly for those familiar with the set/class
distinction.
But since you seem particularly interested in what might go wrong, let me suggest on the negative side, one issue that could make a difference is that when dealing with the surreal field No, one will want to strengthen the background set theory from ZFC to GBC, which includes the global axiom of choice, the assertion that there is a proper class well-ordering of the universe. The reason is that it is consistent with ZFC that the surreal numbers do not admit any proper class well-ordering, and actually, the assertion that they have a definable such well-ordering is equivalent to the set-theoretic axiom known as V=HOD, as I proved in my answer to David Feldman's question on a Definable map from all the ordinals to the surreal numbers. So if one wants to undertake algebraic constructions requiring one to have a well-ordering of the field itself, such as finding a proper class maximal ideal inside a particular subring of No, then there could be difficulties undertaking such a construction in ZFC as opposed to GBC. But nevertheless, the theory GBC is conservative over ZFC and one may thereby freely assume the global axiom of choice. (This is used in the various arguments showing that No is universal for class-sized objects, such as the assertion that every class order embeds to a suborder of No.) In particular, in GBC one has a well-ordering of the entire universe, including the surreals, and this situation would address such issues. Much of the theory undertaken by Ehrlich on the surreals, for example, works in GBC as the background theory. Beyond this issue, even in GBC one does not have any sense of a well-ordering of the (meta-class) collection of all class-sized subrings of No, if this were desired for any algebraic construction, and this is the kind of issue that would arise with the set/class issue. 
But meanwhile, there is also a positive answer. The situation is that if one wants set versions of the surreal numbers, they are abundantly available in increasingly powerful and accurate approximations, which are well-understood and studied. Specifically, we have numerous set-sized
approximations to the surreal numbers, simply by considering the
set of surreal numbers born before a given ordinal birthday. For
any ordinal $\lambda$, let $\text{No}(\lambda)$ be the set of surreal numbers
born before $\lambda$. One should regard $\text{No}(\lambda)$ as
the version of the surreal numbers as constructed inside the
set-theoretic universe $V_\lambda$, which can satisfy increasing
fragments of our set theory, as $\lambda$ is chosen to exhibit
increasingly strong closure properties.
Philip Ehrlich mentioned in his recent talk at the CUNY Logic Workshop that 
he and Lou van den Dries prove in their
article Fields of surreal numbers with exponentiation the following facts:

$\text{No}(\lambda)$ is an additive subgroup of No if and only
if $\lambda=\omega^\alpha$ for some ordinal $\alpha$; that is, if
and only if $\lambda$ is additively indecomposable.
$\text{No}(\lambda)$ is an additive subring of No if and only
if $\lambda=\omega^{\omega^\alpha}$ for some ordinal $\alpha$; that is, if and only if $\lambda$ is multiplicatively indecomposable.
$\text{No}(\lambda)$ is a subfield of No if and only
if $\lambda$ is an $\epsilon$-number, that is, if and only if
$\lambda=\epsilon_\alpha$ for some $\alpha$.

These facts are proved by giving a careful analysis of exactly how long it takes to add the inverse of a given surreal number, based on its birthday, and so when the ordinal $\lambda$ is closed under those waiting times, then the resulting $\text{No}(\lambda)$ contains the requisite inverses. 
Ultimately, we obtain set-sized approximations $\text{No}(\lambda)$ to the surreals by
truncating at sufficiently powerful ordinals. Indeed, for any particular natural number $n$, there will be a closed unbounded proper class of ordinals $\lambda$ such that $\text{No}(\lambda)$ has all the same $\Sigma_n$-expressible properties as the full class of surreal numbers No. This can be proved as an immediate consequence of the reflection theorem. So in fact, No is the union of a proper class chain of increasingly elementary subfields $\text{No}(\lambda)$. 
One can think of the situation as corresponding to the small/large distinction that one finds in category theory with the use of Grothendieck universes, as in Daniel's answer. But in fact one doesn't need a whole Grothendieck universe just to have a subfield, since a mere epsilon number suffices in comparison with an inaccessible cardinal (which are all epsilon numbers).<|endoftext|>
TITLE: Pullback map in homology
QUESTION [8 upvotes]: I'm interested in a concrete description of the "wrong way maps" in homology/cohomology.
$\textbf{Question 1:}$ Let $X, Y$ be compact smooth manifolds of dimensions $n, m$ respectively, and $\phi: X \rightarrow Y$ be a surjective $C^\infty$ map. Using the isomorphism with de Rham cohomology we can define a pullback map on homology classes. If $[C] \in H_i^{cellular}(Y, \mathbb{R})$ is a homology class in $Y$, then is the pullback $\phi^*([C]) \in H^{cellular}_{m-n+i}(X, \mathbb{R})$ given by $[\phi^{-1}(C)]$ if $C$ does not contain any of the critical values of $\phi$?
(In the case I'm most interested $X,Y$ are complex algebraic varieties and $\phi$ is a smooth map.)
$\textbf{Question 2:}$ Is the same thing true with $\mathbb{Z}$ coefficents, since you can't use comparison with de Rham cohomology. Is question 1 vastly more general?
(Sorry if this question is too easy. This isn't really my area - I am an arithmetic algebraic geometer.) References will be greatly appreciated.

REPLY [8 votes]: to Leonard-
An important example of homology pull-back is given by the intersection product. Let $M$ be a closed oriented manifold then we define this product as:
$$H_i(M)\otimes H_j(M)\stackrel{\times}{\rightarrow} H_{i+j}(M\times M)\stackrel{\Delta^{*}}{\rightarrow} H_{i+j-d}(M)$$
where $\Delta^*$ is the homology pull-back of the diagonal $\Delta:M\rightarrow M\times M$. 
The intersection product was first introduced by means of geometric transverse intersection of cycles. Bredon "Geometry and Topology" explains it and makes the relationship with the cup product which is its Poincaré dual. He does not explain how to do when the cycle is not representable by a manifold but this is a good starter.
Fore general cycles, as I explained in my other answer either you play with simplexes viewed as manifolds with corners or you play with manifolds with singularities. M. Kreck in his book "Differential algebraic topology" did it for a particular type of manifolds with singularities and he proves all the tranversality results you want in order to define homology pull-backs. M. Goresky in his PhD thesis studied it in the case of stratified pseudomanifolds and also build a geometric homology and a geometric cohomology thanks to what he called Whitney chains and cochains. Here you need transversality for stratified pseudomanifolds (it works inductively strata by strata). This is in the smooth world, if you are more a PL-guy then you should have a look at: 
S. Buoncristiano, C. P. Rourke and B. J. Sanderson "A geometric approach to homology theory", Cambridge Univ. Press, Cambridge, Mass., 1976    
Another good reference is: 
Jakob, Martin
"Bivariant theories for smooth manifolds."
Papers in honour of the seventieth birthday of Professor Heinrich Kleisli (Fribourg, 2000).
Appl. Categ. Structures 10 (2002), no. 3, 279–290. 
It is another geometric approach to homology theories where homology pull-backs are geometric pull-backs. In fact this all goes back to 
Quillen, D.
"Elementary proofs of some results of cobordism theory using Steenrod operations."
Advances in Math. 7 1971 29–56 (1971). 
I like and recommand Jakob's paper because it is related to the framework of 
Fulton, W.; MacPherson, R. "Categorical framework for the study of singular spaces." Mem. Amer. Math. Soc. 31 (1981), no. 243
which explains how to mix homology covariant and contravariant morphisms in order to have a good framework to get Riemann-Roch types theorem.
To conclude this geometric interpretation of homology pull-backs certainly goes back to the first days of homology in algebraic topology before cohomology was invented.<|endoftext|>
TITLE: Endomorphism ring of Drinfeld module of rank 1
QUESTION [7 upvotes]: That the endomorphism ring of a Drinfeld $A$-module of rank 1 is isomorphic to $A$ appears as a corollary to the more general result that a Drinfeld module of rank $r$ has endomorphism ring of rank at most $r^2$ (a result obtained by analyzing the map 
$\text{Hom}(\phi,\psi)\otimes_A A_{\mathfrak{p}}\to\text{Hom}(T_{\mathfrak{p}}(\phi),T_{\mathfrak{p}}(\psi))$).  Without using this more general result or the uniformization theorem for Drinfeld modules, is there a way to get a more direct handle on $\text{End}(\phi)$ when $\phi$ is of rank 1?

REPLY [3 votes]: We can construct a new object from $O:=\text{End}(\phi)$ by looking at $\psi:O\to \mathscr{F} \{ \tau \}$ defined by $\psi_f = f$.  This thing is not automatically a Drinfeld module, but it is close.
We can either relax our definition of Drinfeld modules or use an isogeny to assume that $\psi$ is a Drinfeld module (See the proof of Proposition 4.7.17 in Basic Structures of Function Field Arithmetic by Goss and Explicit Class Field Theory of Global Function Fields by Hayes).
Now, by looking at the torsion $\psi[a]$ for some $a\in A$ (we need to be a little careful here if $\psi$ is not really a Drinfeld module), we can see that the rank of $\psi$ is less than that of $\phi$ and also that $O$ is equal to $A$.
EDIT: I forgot to say that if $\text{End}(\phi)$ is not commutative you may have to look at a commutative subring, just like in Proposition 4.7.17 of Goss' book.<|endoftext|>
TITLE: Measurability issues in the proof of Fujisaki, Kallianpur and Kunita for stochastic filtering
QUESTION [5 upvotes]: I'm currently looking over the proof(s) of the theorem of Fujisaki, Kallianpur and Kunita regarding the MRT-like characterisation of square integrable random variables measurable with respect to the augmented observation filtration. Unfortunately I'm seeing problems with respect to measurability in the two proofs I'm aware of:
The proof in Fundamentals of Stochastic Filtering (Crisan and Bain) which most of my intuition about the ideas of the proof revolves around: http://books.google.co.jp/books?id=hE3KF5Wf6ecC&pg=PA18&lpg=PA18&dq=path+regularity+for+optional+projection&source=bl&ots=XoOtAHQ6YG&sig=z1b6Zp_P3c7ExNl2DqpNHGHB54M&hl=en&sa=X&ei=asjCUIjYEO2tiQfj-YCwBA&ved=0CCoQ6AEwAA#v=onepage&q=path%20regularity%20for%20optional%20projection&f=false
The original proof: http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.ojm/1200693535
For the first proof, my understanding breaks down around the bottom of page 36 and the start of page 37, especially where they state that $\int_0^t \|\bar{\pi}_s(h)\| ds$ is continuous.  However, as I understand this process is only equal to a continuous process with respect to indistinguishability, which is not sufficient for the $T^n$ defined later to be $\mathcal{Y}_t^o$ stopping times.
For the original proof (which I've admittely skimmed), the authors seem to suggest defining $\hat{h}$ to be the optional projection of $h$ with respect to the raw (unaugmented) filtration of the observation process $z$ (which I believe can be done by a theorem of Dellacherie and Meyer though I have not read the proof). Later on we would like to define stopping times $T_n$ such that $T_n$ is the hitting time of $|\int_0^t \hat{h}_s \ ds |$ of the level n. For these to be stopping times with respect to the raw observation filtration, we'd need $\int_0^t \hat{h}_s \ ds $ to be continuous, but this can only be done if we have some path regularity of $\hat{h}_s$. In the case of taking the optional projection with respect to an augmented, right continuous filtration, given certain integrability conditions on $h$ we can choose a cadlag version of $\hat{h}$, but in the unaugmented filtration case, as far as I know, this is not possible.
As it stands, I only feel I can prove the theorem in the case $h_t = h(X_t)$ is surely bounded by a deterministic function $K_t$, in which case we don't need to apply any probabilistic stopping arguments.
As this is quite a fundamental result in the field of stochastic filtering, these issues are quite worrying to me. I'm hoping that someone more experienced in this area could help clear things up. My impression so far is that the literature seems rather fraught with problems of this nature (the proof of Lemma A.24 in the book linked, required for the proof of the same theorem, being another example: https://math.stackexchange.com/questions/168566/stopped-filtration-filtration-generated-by-stopped-process).

REPLY [2 votes]: Having looked at the proof in Diffusions, Markov Processes and Martingales Vol. 2 (Rogers and Williams) I believe I have resolved the issue. We define $\mathcal{Y}_t$-stopping times $T_n$ using the cadlag version of $\pi_t(h)$. By continuity of $\int_0^t \pi_s(h) ds$ these are $\mathcal{Y}_{t-}$ stopping times. We then can find $\mathcal{Y}_t^o$-stopping times $S_n$ equal to $T_n$ almost surely. This enough for the proof in the book linked.<|endoftext|>
TITLE: Walls of CAT(0) cube complex sufficiently far apart implies intersection of stabilizers finite
QUESTION [12 upvotes]: I was reading through Agol's paper on the Virtual Haken Conjecture and I came across a claim whose proof I am after. It seems to boil down to the following claim about the hyperplanes and their stabilizers under the action of a hyperbolic group on a CAT(0) cube complex:
Claim: Suppose $G$ is a (word-)hyperbolic group acting properly discontinously cocompactly and faithfully on a $CAT(0)$ cube complex $X$. Then there exists an $R>0$ so that if two walls(hyperplanes) $W,W' \subset X$ are such that $d(W,W')>R$ then $G_W \cap G_{W'}$ is finite ($G_W \subset G$ denotes the stabilizer of $W$ i.e the elements that send $W$ to itself)
Note: Certain easy examples show that the condition that $G$ be hyperbolic is necessary, e.g $\mathbb{Z}^2$ acts on the cube complex $\mathbb{R}^2$ and the stabilizer of any two horizontal hyperplanes is $\mathbb{Z}$ which is infinite. A similair example shows that cocompactness is also a necessary assumption.

REPLY [13 votes]: Here's a proof.
Lemma: Suppose $G$ is a (word-)hyperbolic group acting properly discontinously, cocompactly and faithfully on a C⁢A⁢T⁢(0) space $X$.  Then there is a uniform bound $R_0$ on the width $R$ of isometrically embedded flat strips $\mathbb{R}\times [0,R]$ in $X$.
Proof: If not then, by cocompactness, there exist nested flat discs of diameter tending to infinity.  Their union is an embedded copy of $\mathbb{R}^2$, which contradicts hyperbolicity. QED 
Now suppose that walls $W,W'$ have stabilizers with infinite intersection.  Then that intersection is an infinite word-hyperbolic group (since it is quasiconvex in $G$) and so contains an element $\gamma$ of infinite order.  Because $W,W'$ are convex and so themselves CAT(0), each contains an axis $l_W,l_{W'}$ for $\gamma$.  By standard facts about CAT(0) spaces (see Bridson--Haefliger), any two axes bound a flat strip.  Therefore, by the lemma, $l_W$ and $l_{W'}$ are at distance at most $R_0$, as claimed.<|endoftext|>
TITLE: Arithmetic strength of Peano + the Howard ordinal
QUESTION [8 upvotes]: Consider the theory $\mathrm{PA}+\mathrm{BHO}$ consisting of first-order Peano arithmetic ($\mathrm{PA}$) enriched by an axiom scheme which allows well-founded induction up to any ordinal less than [a standard recursive presentation of] the Bachmann-Howard ordinal.  More precisely, let $\prec$ be the explicit well-ordering on $\mathbb{N}$ which results from, say, this description and which has order type of the Bachmann-Howard ordinal, and add to Peano's axioms an axiom scheme ($\mathrm{BHO}$) which for every formula $\phi$ and every $n$ asserts that
$(\forall m\prec n((\forall p\prec m(\phi(p)))\Rightarrow\phi(m)))\Rightarrow\forall m\prec n(\phi(m))$
These are supposed to be theorems of Kripke-Platek set theory ($\mathrm{KP}$), whose proof ordinal is the B-H ordinal, so—unless I badly messed things up—all theorems of $\mathrm{PA}+\mathrm{BHO}$ are arithmetical consequences of $\mathrm{KP}$.  My question is whether the converse holds: can every arithmetical theorem of $\mathrm{KP}$ be proved in $\mathrm{PA}+\mathrm{BHO}$?  If not, what would a counterexample be?
The naïve idea I have is to somehow define the constructible hierarchy (up to and excluding the Bachmann-Howard ordinal), or α-recursion, in $\mathrm{PA}+\mathrm{BHO}$ but I can't imagine how it would work without some kind of second-order language.  I'm not asking for details, just a general idea of how things might work (if they do).
More generally, a broader question should be something like: in what respect can "artificially" increasing the proof ordinal of some theory give it the same arithmetical strength as some stronger theory with that proof ordinal.  So if there's some way to modify my question to give this, I'd like to know.

REPLY [9 votes]: The answer is yes, using the ordinal analysis of KP. See Pohlers' A Short Course in Ordinal Analysis for why the usual ordinal analyses are profound, that is, they imply conservativity of the analyzed theory over PA + TI($\prec_i$) for arithmetical sentences. Here TI($\prec_i$) is the scheme of transfinite induction along all proper initial segments of the primitive recursive well-ordering $\prec$ that measures the proof-theoretic ordinal of the theory.<|endoftext|>
TITLE: Conditions on a unit vector field to be the Gauss map of some surface immersed in R^3?
QUESTION [6 upvotes]: Let $U$ be a bounded domain in $R^2$ and let $n : U \to S^2$. Which (necessary/sufficient) conditions must $n$ satisfy in order that there exist an immersion $f : U \to R^3$ such that $n(x)$ is the normal to $f$ at the point $f(x)$, for all $x\in U$ ? The ideal answer would be of the form: "such $f$ exists if and only if $n$ satisfies the following PDE: ..."

REPLY [5 votes]: There is no PDE. Let $n: U\to S^2$ be a given smooth map, and suppose that the Jacobian is not $0$
at some point. Let $f$ be the composition of $n$ with the standard embedding of $S^2\to R^3$ as the
unit sphere. Then the Gauss map of this $f$ is exactly $n$. So there cannot be any local condition,
that is no PDE.

REPLY [5 votes]: As Alexandre Eremenko points out, there's no PDE that $n$ would have to satisfy, at least for local solvability, which is believable when you think of it in heuristic terms:  There are $3$ unknowns involved in specifying an immersion $f:U\to\mathbb{R}^3$ and only $2$ arbitrary functions needed to specify $n:U\to S^2$, so it's an underdetermined problems and there 'ought' to be many local solutions, in particular, there shouldn't be any equations that $n$ would have to satisfy in order to be the Gauss mapping of some immersion.
More explicitly, you can see how to write down all the possible $f$s for a given $n$ in the 'generic' case in which $n:U\to S^2$ is itself an immersion. [As Eremenko points out, in this case, a particular solution is just to take $f = n$.]  
Here's how to get the 'general' solution:  Let $x$ and $y$ be the standard coordinates on $\mathbb{R}^2$ and note that the assumption that $n$ be an immersion is equivalent to the condition that $n_x$ and $n_y$ be linearly independent on $U$.  Now, consider a possible solution $f$ and set $c = f\cdot n$, so that $f = c\ n + g$, where $g:U\to\mathbb{R}^3$ satisfies $n\cdot g = 0$.  Then one has
$$
0 = n\cdot df = n\cdot (c\ dn + n\ dc + dg) = dc + n\cdot dg = dc - g\cdot dn + d(g\cdot n)
= dc - g\cdot dn
$$
(note that I have used that $n\cdot dn = \tfrac12d(n\cdot n) = 0$).  This latter equation uncouples as the pair of equations 
$$
g\cdot n_x = c_x\qquad\text{and}\qquad  g\cdot n_y = c_y\ ,\tag{1}
$$
which, together with $g\cdot n = 0$, determines $g$ completely in terms of $c$.  
Conversely, if we start with any function $c:U\to \mathbb{R}$ and define $g:U\to\mathbb{R}^3$ to be the unique vector field that satisfies $g\cdot n = 0$ and the equations (1), then setting $f = c\ n + g$ will yield a mapping $f:U\to\mathbb{R}^3$ with the property that $n\cdot df = 0$.  Thus, wherever $f$ is an immersion, its Gauss map will be $n$.  It is easy to see that for any $p\in U$, the generic choice of $c$ on a neighborhood of $p$ will give an $f$ that is an immersion, so there are many local solutions to this problem, and they essentially depend on one arbitrary function (i.e., $c$) of $2$ variables.  
How to determine the conditions that $c$ defined on $U$ should satisfy such that the resulting $f$ will be an immersion everywhere on $U$ is a global question that doesn't have anything to do with solving PDE.
In the case that $n$ is not an immersion, there are singularity issues where $dn$ changes rank, but the cases when the rank is constant are easy to deal with.<|endoftext|>
TITLE: abstract evolution equations
QUESTION [8 upvotes]: Hi
Whenever I read a book on evolution equations, they set up, say the parabolic PDE
$$\dot{y} = Ay + f$$
in abstract function spaces (eg. $L^2(0,T;V)$ and $L^2(0,T;V^*)$). In examples, they always pick $V = H^1$. Can anyone give me an example where $V$ is chosen to be something other than $H^1$. Does it always need to be something like $H^k$? What's the point of calling this "abstract" if we only look at Lebesgue/Sobolev spaces (and/or $C^k$ spaces)? What's an example of a truly abstract equation?
Thanks

REPLY [4 votes]: In my opinion, the adjective "abstract" refers to the fact that one define mild solution by applying a density argument, and these mild solution are not classical solutions. Let us go back to the context of Hille-Yosida theorem, explained in Liviu's answer. If $u_0\in D(A)$, there is nothing abstract, the solution is continuous differentiable with respect to time, with values in the pivotal Hilbert space $H$ (often $H=L^2$). But if the operator is maximal monotone, the fact that $S_t:u_0\mapsto u(t)$ is linear and contracts the $H$-norm allows us to extend $S_t$ to data $u_0\in H$; in this case, the solution is not time-differentiable and teh solution is not more than continuous as an $H$-valued function. One speaks of a continuous semi-group. An even more important fact is that the non-homogeneous equation
$$\frac{du}{dt}=Au+f$$
can be solved, again in an abstract sense, by applying the Duhamel principle
$$u(t)=S_tu_0+\int_0^tS_{t-s}f(s)ds.$$
Of course, this linear theory has a non-linear counterpart. But then something new happens. While functional analysis often yields simultaneously existence and uniqueness in the linear case (well-posedness, including the continuous dependence upon data), the mild solutions of non-linear evolution equations are not always unique. A striking example is that of the Navier-Stokes equation governing the motion of an incompressible fluid in the three-dimensional domain. The gap between existence and uniqueness is worth a Million Dollars.<|endoftext|>
TITLE: Sieve of Erathostenes: removing consecutive items
QUESTION [5 upvotes]: This question comes from https://stackoverflow.com/questions/13747873/why-does-this-prime-function-work, where somebody wrote a standard Sieve of Erathostenes algorithm --- with a bug.  However, the bug does not have any bad effect up to 1'000'000.  At this point, it would be interesting to know if we can find either a proof or a counter-example.
The question is: let L(n) denote the smallest prime factor of n.  Is there a prime number p and two integers b > a >= 2, with:

L(ap) = L(bp) = p
for any m such that ap < m < bp, L(m) < p.

The question and comments in question in prime numbers let us think that it might be false in general.  However I don't really see how to draw that conclusion so far.
EDIT: reformulated the problem (now close to Gerhard Paseman's formulation).

REPLY [6 votes]: There are more efficient ways to sieve, however the question as asked is interesting. I think that there will be a failure for $p=73$ at about $3.08 \times 10^{27}$. First a reformulation which I think is equivalent:

Choose an odd prime $p$ and color the integers (starting at $2$) according to least prime divisor: Color $n$ white, green or red according as the least prime divisor of $n$ is $\lt, =,$ or $ \gt p.$
Q: Is it the case that , for every $p$, there is always at least one red integer between each green integer and the next?

The colors relate to a simple idealized  implementation of the Sieve of Erasthostenes on all integers greater than one. The white integers are already struck out by the time we get to the prime $p$. The green will be newly struck out using the prime $p$ and the red survive this step still unstruck.
The first green integer is $p$ itself and the next is $p^2.$ In between are a healthy number $m_p$ of red integers, all of which are prime (one would expect $m_p \approx \frac{p^2-2p}{2\ln{p}} .$) $p^3$ is also green. The green integers from $p^2$ up to $p^3$ are the $m_p$ numbers $pq$ where $q$ is one of the primes just mentioned. The red numbers from $p^2$ up to $p^3$ are the many primes in this range along with the $m_p(m_p+1)/2$ products $qq'$ where $q \le q'.$  All this does not rule out there being two green integers much further out with no red number between them. The entire pattern of white, green and red has period $p$#, the product of the primes up to $p$ although that is quickly very huge. The next green number after $n$ is $n+2pj$ for some $j \ge 1.$ So the first question is "can there be a run of at least $2p+1$ consecutive integers, all white and green?" Here the OEIS items Largest number of consecutive integers such that each is divisible by a prime <= the n-th prime and Smallest number which begins the maximal number of consecutive integers divisible by one of the first n prime numbers along with their linked data tables become handy. The first prime for which this can happen is $p=67$ where it happens for the $153$ integers starting at $s=7714600835154917969172.$ However we need a run which actually contains two green numbers and no red ones so the length alone is not enough. As it happens $s+61$ is a green so the next possible green number is $t=s+61+2*67$ which already past the red number $s+154$ (and it turns out that $t$ is a multiple of $9\cdot17$). 
But on further examination it seems that for $p=73$ we have two green numbers
$s=3084626641924131277081102913 =73 \cdot 483733 \cdot 169372703 \cdot 515739756619 $
$s+2\cdot 73=73 \cdot 1033 \cdot 40905285071067528770851$
and everything between them white because of the following minimal prime factors.
$73, 2, 5, 2, 3, 2, 19, 2, 41, 2, 3, 2, 5, 2, 17, 2, 3, 2, 53, $
$2, 29, 2, 3, 2, 31, 2, 7, 2, 3, 2, 13, 2, 5, 2, 3, 2, 23, 2, $
$11, 2, 3, 2, 5, 2, 19, 2, 3, 2, 17, 2, 47, 2, 3, 2, 7, 2, $
$13, 2, 3, 2, 11, 2, 5, 2, 3, 2, 37, 2, 7, 2, 3, 2, 5, 2,$
$ 43, 2, 3, 2, 29, 2, 67, 2, 3, 2, 71, 2, 31, 2, 3, 2, 41, $
$2, 5, 2, 3, 2, 7, 2, 61, 2, 3, 2, 5, 2, 11, 2, 3, 2, 13, 2,$
$ 7, 2, 3, 2, 59, 2, 17, 2, 3, 2, 19, 2, 5, 2, 3, 2, 11, 2, $
$23, 2, 3, 2, 5, 2, 13, 2, 3, 2, 7, 2, 37, 2, 3, 2, 47, 2, 73$
updates As @Gerhard points out, I was too hasty. I gap of $2p+1$ can not happen for $p=53,59,61$ nor for $p \le 41$ but does happen for $p=43,47.$  I leave the examination of those possible cases to someone else. The case I discuss for $p=67$ is the first attaining the maximal length for that $p$ of $153.$ But for all I know there is an earlier  gap of length $135 \le \ell \le 152$  which has two green numbers with no red between them (or a later gap of some length $135 \le \ell \le 153$). Similarly for $p=73.$  As pointed out, The example given is likely not a counterexample for the original SE question. I almost feel as if a counterexample to that would require the larger number to be a power of $p$ or perhaps $p^aq^b$ with $ab$ large. I was only able to give the answer I did because the particular data was there and turned out to work.The universe is infinite and our horizon is small.<|endoftext|>
TITLE: What is the structure of the space of solutions of a non linear ODE?
QUESTION [11 upvotes]: As is well known, the space of solutions of a linear ODE with, say, $\mathcal{C}^\infty$ coefficients on $\mathbb{R}$ is a finite dimensional affine space (a vector space, in the homogeneus case).

What is the structure of the space of solutions of a non linear ODE? In which cases does it have a "natural" structure of finite dimensional smooth manifold? 

For the question to make sense, some conditions on the form of the equation must be put. Let's assume the ODE is of the form
$$F(t,u(t),u'(t),u''(t),\dots,u^{(n)}(t))=0$$
where $F:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}$ is a smooth function.

REPLY [8 votes]: There is yet another direction in which one could take the question, one that makes more contact with the linear and affine structures that one observes in the standard treatment of linear equations (both homogeneous and inhomogeneous).  For example, consider the Riccati equation
$$
u'(t) = a_0(t) + a_1(t)u(t) + a_2(t)u(t)^2,\tag{1}
$$
which is nonlinear (when $a_2$ is not zero).  A standard fact that is proved in ODE courses is that there is a 'trick' for 'linearizing' this equation, yielding the result that the general solution can be written in the form
$$
u(t) = \frac{c_1b_1(t)+c_2b_2(t)}{c_1b_3(t)+c_2b_4(t)}
$$
for some functions $b_i(t)$ (defined on the entire interval on which the $a_i$ are defined) and constants $c_1$ and $c_2$ (not both zero).  Obviously, only the ratio of $c_1$ to $c_2$ matters for specifying $u$, so the ($1$-dimensional) space of solutions to $(1)$ is 'naturally' diffeomorphic to the real projective line $\mathbb{RP}^1$.  Note that this representation works even when the solution $u$ has a 'pole'.  (For example, consider the Riccati equation $u' = 1 + u^2$, which has $u(t) = \tan t$ as a solution.)  
Sophus Lie developed a theory that associated a type of ODE to each homogeneous space of what we now call Lie groups, one that generalized the theory of linear ODE (both homogeneous and inhomogeneous) and Riccati equations in their various forms (which correspond to vector spaces, affine spaces, and projective spaces, thought of as homogeneous spaces of appropriate Lie groups).  
It's possible that the OP is interested in learning more about the theory of ODE (and PDE) of Lie type, which is what this theory developed into.  For this theory, putting a (possibly non-Hausdorff) manifold structure on the space of solutions is a necessary first step, but it is only the beginning.<|endoftext|>
TITLE: What are the simple Lie superalgebras of type E?
QUESTION [12 upvotes]: Background
Simple finite dimensional Lie superalgebras over $\Bbb C$ have been classified. There are the Cartan type superalgebras (algebras of purely odd vector fields), two strange families P(n) and Q(n), and the basic superalgebras:

the A series $sl(m,n)$ ($psl(m,m)$ for $m=n$)
the B, C, D series $osp(m,n)$
a family D(2,1;$\alpha$) of deformations of D(2,1)
the exceptional Lie algebras F(4) and G(3)

Basic Lie superalgebras have roots, Cartan matrices and Dynkin diagrams, and their Dynkin diagrams look quite similar to usual Dynkin diagrams of simple Lie algebras. A noticeable difference is that the Dynkin diagram is not unique anymore (so that diagrams of type C and D for example describe the same Lie superalgebras).
Question
Why is the E series missing? More precisely, are there known Lie superalgebras that are defined similarly to the Lie algebras $E_6$, $E_7$, $E_8$ (e.g. via octonions as in the classical case) and why are they not simple or finite dimensional?
Remarks
A natural place to look for Lie superalgebras of type E is the class of Kac-Moody superalgebras (KMSA). The simplest infinite dimensional Kac-Moody superalgebras are the affine ones. Their construction is similar to the classical one, and there are no examples that "look like" algebras of type E. Expanding the search to KMSA of finite growth only gives a few exceptional families, that again are not similar to algebras of type E (see arXiv:0810.2637).
The next simplest class of infinite dimensional KMSA is that of hyperbolic KMSA, i.e. those such that removing a root from (any of) its Dynkin diagram(s) yields a Dynkin diagram of a finite dimensional or affine KMSA. There have been at least three attempts at classification of hyperbolic KMSA: 1, 2, 3.
Some diagrams of type E indeed appear, however they are just listed, and the resulting KMSA are only described by their Cartan matrix. Have these algebras been described in more detail somewhere?

REPLY [2 votes]: I think Kac - Lie superalgebras is a useful reference.
Exceptional Lie superalgebras G(3) and F(4) can indeed be expressed in terms of octonions:
Sudbery - Octonionic description of exceptional Lie superalgebras. See also Elduque - Quaternions, octonions and the forms of the exceptional simple classical Lie superalgebras.
The infinite-dimensional case is considered in Kac - Classification of infinite-dimensional simple groups of supersymmetries and quantum field theory.<|endoftext|>
TITLE: Homotopy $\pi_4(SU(2))=Z_2$
QUESTION [11 upvotes]: I am a physics student, recently I read a paper using Homotopy $\pi_4(SU(2))=Z_2$, I guess mathematicians have some visualization or explanation of this result. So I come here ask for help. 
CROSS-POST from https://physics.stackexchange.com/questions/46284/homotopy-pi-4su2-z-2

REPLY [12 votes]: This is really just a handwavy but perhaps more "visual" description of Pontryagin's result as cited by solbap in the comments above.  Though I've written a huge block of text, there are some reasonably concrete three-dimensional pictures that you can build up in your head in this case, but it does take quite a bit of practice.
First, I assume that you are familiar with Pontryagin's construction relating the homotopy classes of maps to the k-sphere with framed (co-)bordism classes of codimension k submanifolds.
Check out Milnor's book Topology from the Differentiable Viewpoint if you're not familiar with this.  Because your user profile says that you are interested in condensed matter physics, I'll add that this idea is used in the case of $k=2$ to draw some nice pictures of "homotopies around defects" in this paper of Teo and Kane.
Warmup, $\pi_3(S^2)$
As a warmup, let's try to visualize homotopy classes of maps from $S^3$ to $S^2$, i.e. the situation of the Hopf fibration.  Pontryagin's construction says that we should be looking at bordism classes of framed codimension 2 submanifolds in $S^3$.  3-2=1, so we should be looking at 1-dimensional submanifolds, i.e. links in $S^3$.  Here we have framed links in $S^3$ which can be visualized by drawing each component of the link with another parallel copy that winds around it, much like a ribbon.
You should convince yourself that all components in these framed links can be merged together into a single unknot with some integer framing.  Thus what matters ultimately is the classification of possible framings.  Imagine taking a 2D slice of $S^3$ transverse to a point $p$ of the framed link and placing the point $p$ at the origin of that plane.  Then the framing at that point is just a choice of the $x$- and $y$- axes (i.e. a 2-dimensional frame).  As we carry this plane along the original unknot, this choice of axes can rotate in that plane and so the classification of framings is naturally an integer. You may check that the inverse image of the North pole of the Hopf fibration is an unknot, and the inverse image of any other point on the sphere is an unknot which is linked once with it.  Finally, you should see how you can build up any other homotopy class from "adding" Hopf fibrations together by putting multiple copies of this framed unknot together (possibly with opposite orientations), which gives a visualization of the group structure on the set of homotopy classes.
In this way you get a visualization of $\pi_3(S^2)$ by means of some pictures of framed circles.  I can't resist here adding a link to this paper of DeTurck et al which gives some beautiful illustrations and description of the homotopy classes of maps from $T^3$ to $S^2$ with this tool.
$\pi_4(S^3)$
Now, you are interested in the case of homotopy classes of maps from $S^4$ to $S^3$. In this case you are now looking at framed links in $S^4$.  You can still arrange for the link to become a single framed unknot by a sequence of bordisms.  However, the framing can no longer be drawn with simply just a single parallel knot.  Consider taking a 3-dimensional slice transverse to a point $p$ on the link in $S^4$ and let us place $p$ at the origin of our $R^3$ that we sliced with.  In $S^4$, the framing of the link yields a choice of a 3-dimensional frame in this $R^3$ slice.  And just as the relevant topological invariant of the framing in $S^3$ was how this frame rotates as we travel along the $S^1$ corresponding to our link component, leading to an element of $\pi_1(S^1)$ (the winding number), in $S^4$, we must now track how this 3-d frame rotates as we follow the $S^1$ of the link component.  But now we are considering a continuous loop of choices of 3-dimensional orientations, i.e. an element of $\pi_1(SO(3))$, which is well known to be $\mathbb{Z}/2\mathbb{Z}$.
With this key ingredient of the 3-dimensional framing, hopefully you can see that $\pi_4(S^3)=\pi_1(SO(3))=\mathbb{Z}/2\mathbb{Z}$.<|endoftext|>
TITLE: Polyhedra that combinatorially shadow a sequence
QUESTION [8 upvotes]: Let $P$ be a polyhedron in $\mathbb{R}^3$.
Say that $P$ combinatorially shadows a sequence of natural numbers $S$ if
there is a continuous rotation of $P$ such that its orthogonal-projection
shadows are polygons whose number of sides coincide with the elements of $S$
in order.  For example, $S$ might be the odd primes: $S=(3,5,7,11,13,\ldots)$,
and we want the shadows to be a triangle, then a pentagon, then a septagon,
then a hendecagon, etc.
Q0. Let $S$ be an increasing sequence of natural numbers,
whose first element is $\ge 3$.
Does there exist a polyhedron $P$ that combinatorially shadows $S$?
I think the answer here is Yes, illustrated just for $S=(3,5)$ with this
example:

      


The generalization is that the needed increase above the previous element
in the sequence is achieved by bumping out near the centroid of an
appropriate face (the centroid $c$ 
of face $(1,2,3)$ in the
above example is bumped out to $\lbrace a, b \rbrace$),
shallow enough to be hidden for the previous element (middle image), but sufficent so
that a rotation will simultaneously expose the additional vertices (right image).
So if I am correct here, there is a prime polyhedron that realizes the odd primes—either
up to any given prime, or all odd primes if an infinite number of faces are countenanced.
Correction (9Dec12): I now think the above sketch fails to allow many vertices to appear in the shadow
simultaneously.  Better is to split existing vertices into two ...
[remaining bad idea deleted]. 23Dec12: Now I believe the construction posted in a separate
answer settles Q0 (positively).
My question concerns arbitrary—not necessarily increasing—sequences:
Q1. Let $S$ be an arbitrary sequence of natural numbers,
each $\ge 3$.
Does there exist a polyhedron $P$ that combinatorially shadows $S$?
Ideas, even half-baked, or pointers to relevant literature welcomed!  Thanks!

REPLY [4 votes]: This answers my Q0.
The polyhedron is the convex hull of these faces:

          


When rotated, the shadow has $3, 5, 7, 11, 13, 17$ vertices:

          


Clearly the same
design (not dissimilar from Gerhard's suggestion)
suffices to capture any sequence with $s_{i+1}-s_i \ge 2$.
This suffices for a prime polyhedron, my original (recreational) goal.

          


It is not difficult to capture sequences that also sometimes increase by just $1$,
by arranging for rotation to expose a (thin) triangle face.
Thus the answer to Q0 is Yes.<|endoftext|>
TITLE: Which von Neumann algebras have inner permutation of tensor factors?
QUESTION [13 upvotes]: My favorite model of quantum probability is by von Neumann algebras, i.e., a quantum measurable space is a von Neumann algebra and a quantum distribution is a normal state.  Then, one important new phenomenon in quantum probability is the existence of bosons and fermions.  Another important new phenomenon, of which bosons and fermions are a special case, is that every automorphism of the von Neumann algebra $L(H)$, the bounded operators on a Hilbert space $H$, is inner.  This is a quantum Noether's theorem that says that if the algebra of observables is $L(H)$, then every symmetry can also be used as an observable.  The connection between these two phenomena is that if you have two physical systems with the same algebra $L(H)$, then the symmetry $x \otimes y \mapsto y \otimes x$ of $L(H) \otimes L(H)$ is inner if you use the von-Neumann-completed tensor product.  (Clearly, since it is directly given by an operator on $H \otimes H$.)  So, you can measure whether the joint system is in a bosonic or fermionic state and (in this abstracted, simplified model) it has to be one of the two.
So my question is, which von Neumann algebras $M$ have the property that $x \otimes y \mapsto y \otimes x$ is an inner automorphism of $M \otimes M$?  I thought that I had learned that $L(H)$ is the only von Neumann algebra for which every automorphism is inner, but from the comments it seems that that is not true.  What about this automorphism in particular?

REPLY [18 votes]: Here's an argument showing that in the ${\rm II}_1$ case the flip automorphism is never inner. 
Let $M$ be a type ${\rm II}_1$ factor and $\tau$ its trace, so that $M \subset L^2(M, \tau)$, and $M$ acts standardly on $L^2(M, \tau)$. Suppose that the flip automorphism is implemented by a unitary $U \in \mathcal U(M \overline \otimes M)$. I'll reach a contradiction by showing that $U$ is orthogonal to every vector in the dense subspace of $L^2(M \overline \otimes M, \tau \otimes \tau)$ spanned by vectors of the form $v \otimes w$ where $v, w \in \mathcal U(M)$.
Fix $v, w \in \mathcal U(M)$, and $\varepsilon > 0$. Take $n \in \mathbb N$ such that $2^{-n} < \varepsilon$, and take a partition of unity $\{ p_k \}_{k = 1}^{2^n} \subset M$ such that each $p_k$ is a projection of trace $2^{-n}$. Then 
$$
| \langle U, v \otimes w \rangle | = \left| \sum_{k = 1}^{2^n}  (\tau \otimes \tau) ((p_k \otimes 1)(v^* \otimes w^*) U (p_k \otimes 1) ) \right|
$$
$$
\leq \sum_{k = 1}^{2^n} | (\tau \otimes \tau) ((v^* \otimes w^*) U (p_k \otimes vp_kv^*) ) |
$$
$$
\leq \sum_{k = 1}^{2^n} (\tau \otimes \tau)(p_k \otimes vp_kv^*)
$$
$$
= \sum_{k = 1}^{2^n} \tau(p_k)^2 = 2^{-n} < \varepsilon.
$$
Update: I've recently come across the 1975 paper of Sakai "Automorphisms and tensor products of operator algebras" where he proves that the flip automorphism for a von Neumann algebra $M$ is inner if and only if $M$ is a type ${\rm I}$ factor. His proof is roughly as follows:
For the type ${\rm II}_1$ case he proceeds as I did above by showing that the unitary $U$ would have to be orthogonal to every vector of the form $v \otimes w$. His argument for this is not as direct as the one above, but the argument I gave above is based on techniques of Popa which came later.
For the type ${\rm II}_\infty$ case he writes $M$ as $N \overline \otimes \mathcal B(\mathcal H)$ where $N$ is a type ${\rm II}_1$ factor and then shows with a simple argument that if the flip automorphism is inner on $M$ then it must also be inner on $N$ which it cannot be by the arguments above.
For the type ${\rm III}$ case he first writes $M$ as $N \overline \otimes \mathcal B(\mathcal H)$ where $N$ is type ${\rm III}$ and countably decomposable. Next he shows that if the flip is inner on $N$ then $N$ has trivial outer automorphism group. Indeed, if $\sigma$ denotes the flip and $\rho \in {\rm Aut}(N)$ then since $\sigma$ is inner, and ${\rm Inn}(M)$ is a normal subgroup, we must have $\tilde \rho = \sigma (\rho^{-1} \otimes {\rm id}) \sigma (\rho \otimes {\rm id})$ is also inner. Restricting $\tilde \rho$ to $N \otimes \mathbb C$ we then see then that there is a unitary $V \in N \overline \otimes N$ such that $(a \otimes 1)V = V(\rho(a) \otimes 1)$ for all $a \in N$. If we then consider the normal conditional expectation $E$ from $N \overline \otimes N$ to $N \otimes \mathbb C$, then there exists some operator $x \in \mathbb C \otimes N$ such that $E(Vx) \not= 0$, and we then have $a E(Vx) = E(Vx) \rho(a)$ for all $a \in N$. By conjugating this formula it then follows easily that $E(Vx)E(Vx)^* \in \mathcal Z(N) = \mathbb C$ and also $E(Vx)^*E(Vx) \in \mathbb C$, hence $E(Vx)$ is a non-zero scalar multiple of a unitary showing that $\rho$ is inner. Tomita-Takesaki theory though gives continuum many outer automorphism of $N$, a contradiction.<|endoftext|>
TITLE: How should a number theorist learn a modest amount of algebraic geometry?
QUESTION [32 upvotes]: A little bit vague, but I hope useful for the entire community.
I am, by training, an analytic number theorist. I have managed to learn some algebraic geometry, by reading parts of Silverman's Arithmetic of Elliptic Curves, the beginning of Ravi Vakil's notes, and a very little bit of Hartshorne (and doing a least a few of the exercises).
It has become apparent to me that it would be very useful to learn more -- the subject often comes up in reading, seminars, and discussions with colleagues, and besides I find what little I know to be fascinating.
Unfortunately I don't have the time to, say, read Hartshorne (or EGA) and do all the exercises. That said, how would MO readers recommend that I go from knowing a little bit of algebraic geometry to a modestly bigger bit of algebraic geometry? 
My goals, roughly speaking, are:

To have a broader base in the subject, and understand the basic definitions and examples better. (Naturally I am especially interested in the side of the subject that relates to arithmetic, and less so in the complex geometric side.)
To be better prepared to read books related to some of my research interests and those of my collaborators (Borel's Linear Algebraic Groups, Mumford, Fogarty, and Kirwan's Geometric Invariant Theory, etc.)
To better know where to look if I run across an algebro-geometric argument in the literature.
To enjoy myself, especially since I hope to persuade colleagues and grad students to join me.

There seem to be various books which are well-suited to this, e.g., Mumford's Red Book, Eisenbud and Harris's Geometry of Schemes, Harris's Algebraic Geometry: A First Course, and many others. Vakil's notes are also excellent, but perhaps for someone with ambitions of learning the subject more thoroughly. For the most part, I'm not very familiar with the virtues and drawbacks of these books. 
Can MO readers recommend a roadmap for learning at least a little bit more about this subject, even as I am obliged to keep most of my attention elsewhere?
Thank you very much!

REPLY [14 votes]: I'm not even a number theorist, but when I was in grad school I took a one-semester course from Hartshorne, using his book.  I felt that it carried a powerful message, more abstract and more general than what I needed for any purpose, but still really interesting.  We only made it to the beginning of the third chapter, but that was enough.  In fact, without entirely realizing it, I was taking a lot of hard commutative algebra results on faith in the homework, but somehow it was okay.
So I would ask whether Hartshorne is exactly what you want; it exactly explains what you say is missing from your experience.  You don't have to and shouldn't read it page by page, confirming every assertion.  You can learn from this book using the spiral method.  I do not know "Geometry of Schemes", although a book with that title and those authors sounds good too.  What I can say is that Hartshorne is especially enthusiastic about Grothendieck's perspective on algebraic geometry.  Again, even though I had no career reason to care, I ended up wanting to prove things Grothendieck style: using schemes, with the same argument in characteristic 0 and positive characteristic, with proper schemes as a replacement for projective varieties, etc.<|endoftext|>
TITLE: Does a weaker condition than vanishing derivative imply a function being constant?    
QUESTION [8 upvotes]: I learned this question from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does 
$$g(x):=\lim_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$
exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous?
Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?
Any help is appreciated. 
Edit: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.

REPLY [10 votes]: The ordinary proof that 0 derivative implies constant rests on the mean value theorem.  Your function $g(x)$ is the "symmetric derivative" of $f$, so we should look for a "quasi-mean value theorem" for symmetric derivatives.  A quick google search came up with the paper 
http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-27/No-3/27%283%298-5%28279-301%29.pdf
Here it is shown (Theorem 2) that if $f$ is continuous on $[a,b]$ and symmetric differentiable on $(a,b)$ with symmetric derivative $f^s$, then there are points $\xi,\eta\in (a,b)$ with 
$$f^s(\eta) \leq \frac{f(b)-f(a)}{b-a} \leq f^s(\xi).$$
Clearly then if $f^s = 0$ we must have that $f$ is constant.<|endoftext|>
TITLE: rational homotopy of a manifold
QUESTION [10 upvotes]: Given a finite dim rational homotopy type satisfying Poincaré duality,
what is the best reference to when it is the rational homotopy type of a fin dim manifold?

REPLY [20 votes]: It sounds like you are asking about the Sullivan-Barge Theorem. The original references are:
J. Barge, Structures différentiables sur les types d'homotopie rationnelle simplement connexes, Ann. Sci. École Norm. Sup. (4) 9 (1976), no.4, 469–501.
D. Sullivan, Infinitesimal computations in topology, Inst. Hautes Études Sci. Publ. Math. (1977), no.47, 269–331 (1978).
You'll find a clean statement in Chapter 3 of this book, but to paraphrase: Suppose you have a simply-connected Sullivan algebra whose cohomology $H^\ast$ is a Poincaré duality algebra of formal dimension $n$. Then it can be realised by a closed simply-connected manifold if, and only if, one of the following holds:

$n$ is not of the form $4k$;
$n$ is of the form $4k$, the signature is zero and the quadratic form on $H^{2k}$ is equivalent over $\mathbb{Q}$ to one of the form $\sum \pm x_i^2$;
$n$ is of the form $4k$, the signature is nonzero, the quadratic form on $H^{2k}$ is equivalent over $\mathbb{Q}$ to one of the form $\sum \pm x_i^2$, and one can find a sequence of classes $p_i\in H^{4i}$ (the Pontrjagin classes) such that the corresponding Pontrjagin numbers satisfy certain necessary congruences.

In other words, the conditions which are necessary for realization by a smooth manifold are also sufficient.<|endoftext|>
TITLE: Why is the transfer map Tate-dual to restriction ? 
QUESTION [5 upvotes]: In  one of their papers (before Theorem 7.2), Benson and Carlson state that the transfer map is Tate-dual to the restriction homomorphisms (also see Remark 1.3 of this recent paper).  
More precisely: If $H \le G$ are finite groups and $k$ a field those characteristic divides the order of $H$, then the there should be a commutative diagramm
$$\begin{array}{ccc}
\hat{H}^{-s-1}(G,k) & \cong & \text{Hom}_k(\hat{H}^sS(G,k),k) \newline
res^G_H \downarrow &  & \downarrow (tr^G_H)^\ast \newline
\hat{H}^{-s-1}(H,k) & \cong & \text{Hom}_k(\hat{H}^sS(H,k),k)
\end{array}$$
where the horizontal isomorphisms are Tate duality.
Does anyone know a reference with a proof or can provide a proof of this statement ? Thanks in advance.

REPLY [4 votes]: I don't know of a reference, but the duality in question can be proved by results from Brown's book on group cohomology. I'll show the case $s\ge 0$. First note that for each integer $j$ there is an isomorphism 
$$\psi: \hat{H}^j(G,k) \xrightarrow{\sim} \text{Hom}_k(\hat{H}_j(G,k),k)$$
(Brown, VI.7.2) and for $s\ge 0$ there is an isomorphism $$\varphi: \hat{H}_{-s-1}(G,k) \xrightarrow{\sim} H^s(G,k)$$
(Brown, VI.4). Denote the $k$-dual of a vector space or of a homomorphism by $(-)^\ast$. Tate duality is then the composition 
$$t = (\varphi^{-1})^\ast\circ \psi: \hat{H}^{-s-1}(G,k) \xrightarrow{\sim} \hat{H}_{-s-1}(G,k)^\ast  \xrightarrow{\sim} H^s(G,k)^\ast.$$
Hence we have to show the commutativity of the diagramm
$$\begin{array}{ccccc}
\hat{H}^{-s-1}(G,k) & \xrightarrow{\psi} & \hat{H}_{-s-1}(G,k)^\ast  & \xleftarrow{\varphi^\ast} & H^s(G,k)^\ast \newline
{\scriptstyle \widehat{res}} \downarrow & & \downarrow \scriptstyle res^\ast & & \downarrow \scriptstyle tr^\ast\newline  
   \hat{H}^{-s-1}(H,k) & \xrightarrow{\psi} & \hat{H}_t(H,k)^\ast & \xleftarrow{\varphi^\ast} & H^s(H,k)^\ast
\end{array}$$
($t$ stands for $-s-1$ which the editor doesn't accept!?) The commutativity of the left hand square follows right from the definition of the maps and the right hand square commutes if we can show the commutativity of the following square: 
$$\begin{array}{ccc}
\hat{H}_{-s-1}(G,k) & \xrightarrow{\varphi} &  H^s(G,k)\newline 
{\scriptstyle \widehat{res}} \uparrow  & & \uparrow \scriptstyle tr^G_H \newline 
\hat{H}_{-s-1}(H,k)  & \xrightarrow{\varphi} & H^s(H,k) 
\end{array}\tag{1}$$
In order to describe $\varphi$ on chain level, let $P \to k$ be a projective resolution over $kG$ and let $F$ be a complete resolution such that $F_i=P_i$ and $F_{-i-1} = P_i^\ast$ for $i \ge 0$. Then $\varphi$ is induced by the composition 
$$\varphi: F_{-i-1}\otimes_{kG}k=P_i^\ast \otimes_{kG}k \xrightarrow{\alpha\otimes id} \text{Hom}_{kG}(P_i,kG) \otimes_{kG} k\xrightarrow{\beta}\text{Hom}_{kG}(P_i,k)$$
where $\alpha(f)(x)=\sum_{g \in G}f(g^{-1}x)g$ (Brown, VI.3.4) and $\beta(f \otimes a)(x)=f(x)a$ (Brown, I.8.3). Hence 
$$\varphi(f \otimes a)(x)=\sum_{g \in G}f(g^{-1}x)(ga)=\sum_{g \in G}f(g^{-1}x)a=tr^G_E(f)(x)a\tag{2}$$
where $f \in P_i^\ast, a \in k, x \in P_i$ and $E=\{1\}$. 
On chain level $(1)$ is given by the diagramm 
$$\begin{array}{ccc}
P_i^\ast \otimes_{kG} k & \xrightarrow{\varphi_G} & \text{Hom}_{kG}(P_i,k) \newline 
{\scriptstyle \kappa} \uparrow  & & \uparrow \scriptstyle tr^G_H \newline 
P_i^\ast \otimes_{kH} k & \xrightarrow[\varphi_H]{} & \text{Hom}_{kH}(P_i,k) \newline 
\end{array}\tag{3}$$
where $\kappa(f \otimes_H a)=f \otimes_G a$. With $f,a,x$ as above, we obtain 
$$(tr^G_H \circ \varphi_H)(f \otimes_H a)(x)=\sum_{g \in G/H}\varphi_H(f\otimes_H a)(g^{-1}x)
\overset{(2)}{=}\sum_{g \in G/H}tr^H_E(f)(g^{-1}x)a$$
$$\qquad=tr^G_H(tr^H_E(f))(x)a=tr^G_E(f)(x)a$$
$$\qquad\qquad=\varphi_G(f \otimes_G a)(x)=(\varphi_G \circ res)(f \otimes_H a)(x)$$
Thus the commutativity of $(3)$ is shown. QED<|endoftext|>
TITLE: What is the support of the Whittaker function of a new vector on GL(2)?
QUESTION [7 upvotes]: Let $W$ be the normalized Whittaker function associated to a new vector in an irreducible generic representation $\pi$ of $G=GL_2(k)$, where $k$ is a $p$-adic field. Let $c$ be the conductor of $\pi$, meaning that $c$ is the smallest integer with $W(gk)=W(g)$ for all $k\in K_1(c)$ where $K_1(c)$ is the subgroup of $K=GL_2({\mathfrak o})$ with bottom row congruent to $(0,1)$ modulo $\mathfrak p^c$ (with the convention that $K_1(0)=K$).


What is the support of $W$ restricted to $K$?


Clearly, the support contains $K_0(c)$, the subgroup of upper triangular matrices modulo $\mathfrak p^c$, since $W$ won't vanish on the center of $G$. I imagine it will depend on further information about $\pi$ beyond the conductor, but I'm having trouble finding (or proving) much of anything definitive.

REPLY [3 votes]: You should find all ingredients needed for your calculation in the following very useful notes by Ralf Schmidt:
Some remarks on local newforms for GL(2).
J. Ramanujan Math. Soc. 17 (2002), 115-147 
It is available on Ralf Schmidt's webpage.<|endoftext|>
TITLE: Non-generating sets in a free group.
QUESTION [10 upvotes]: Let $F_n$ be the free group generated by $x_i$, for $1\leq i\leq n$. Let $a_i$ be some elements of $F_n$, also for $1\leq i\leq n$. Is there a nice way to tell when the list $\{a_i^{-1}x_ia_i\}$ does not generate $F_n$?
For insufficient reasons partially related to a talk I gave once in Hamburg (video and handout there), and to a paper I'm not done writing (PDF there), I expect that there might be a way to construct out of the $a_i$'s a conjugacy class in $F_n$ (or perhaps in some completion of $F_n$), whose non-triviality implies that $\{a_i^{-1}x_ia_i\}$ do not generate $F_n$. Does this to anyone make sense?

REPLY [6 votes]: Stallings' folding algorithm (described by Agol) is probably the best way of doing this, but I thought I'd mention an older algorithm which is also useful, due to Whitehead.
Let $\{w_k\}$ be a collection of words in $F_n$.  The Whitehead graph is defined as follows.  There are $2n$ vertices, labelled $x_i^{\pm}$.  Whenever $x_ix_j$ appears as subword of some $w_k$, an edge is added from $x_i^+$ to $x_j^-$.  Similarly, when $x_i^{-1}x_j$ appears as a subword, an edge is added from $x_i^-$ to $x_j^-$, and so on.
A better description of the Whitehead graph for a topologist is as follows.  Realize $F_n$ as the fundamental group of a handlebody $U_n$. Realize the generators $x_i$ by finitely many properly embedded discs $D_i$, which cut the interior of $U_n$ into a ball.  Realize the words $w_k$ by an embedded 1-dimensional submanifold of $U_n$; after a homotopy, we may assume that the $w_k$ are `pulled tight' with respect to the discs $D_i$.
As mentioned already, cutting along the discs $D_i$ gives a 3-ball $B$, and each disc $D_i$ lifts to a pair of discs $D_i^{\pm}$ in the boundary of $B$.  The submanifold $w_k$ becomes a set of intervals joining these discs.  This is precisely the Whitehead graph, with the disc $D_i^{\pm}$ corresponding to the vertex $x_i^{\pm}$.
Here's a picture from one of my papers.
The Whitehead graph of b^{-1}aba^{-2}. http://www.freeimagehosting.net/6wb7s
Whitehead used this construction to produce an algorithm that computes shortest representatives for sets of words $\{w_k\}$ under the action of the automorphism group of $F_n$: if the set of words is not of shortest length then one can perform a `Whitehead move', which replaces one cutting disc with a better one, and reduce the length.  In particular, his algorithm can be used to recognise generating sets.  In fact, he proved the following very useful lemma. 
Whitehead's Lemma: If  $F_n$ admits a free splitting $A*B$ in which every $w_k$ is conjugate into either $A$ or $B$ then the Whitehead graph is either disconnected or has a cut vertex.
In particular, if the Whitehead graph is connected with no cut vertices then the $w_k$ do not generate.  If the Whitehead graph is disconnected or does have a cut vertex then you can perform Whitehead moves until you find the answer.
For further details, see §I.4 of Lyndon and Schupp for the combinatorial approach and a paper of Stallings (click here for a dvi) for the topological approach.<|endoftext|>
TITLE: At what point does number theory stop playing with finite rings?
QUESTION [12 upvotes]: Basic results in number theory, like the Chinese remainder theorem, the Euclidean algorithm and Euler's theorem, are really about finite structures, namely the rings $\mathbb{Z}/n\mathbb{Z}$ for suitable $n$, or perhaps other finite rings (such as quotients of extensions of $\mathbb{Z}$). Such results can be treated without a worry by a finitist. It is ingredients like this that are sufficient for, say, implementation of the RSA encryption scheme. However, at some point number theoretic results became statements (much more powerful and interesting statements!) about things like rings of adeles.
My question could be taken two ways: 

What are the most powerful results in number theory which one can state and prove using the machinery of finite rings?

or

At what point in time did number theory move from considering finite rings to more analytic objects, and by what time was this move 'complete'?

For a reference post, one can point to the Lasker–Noether theorem (1921), which can be taken to be a statement about finitistic objects (and ignoring the possibility that infinite objects exist), namely finite modules $M$ for a ring $R$ (possibly infinite, in which case one could think of it as a finite ring by quotienting by the kernel of $R\to End(M)$).
My motivation for asking about this question is not to be controversial or obtuse, but to get an idea about how far a finitist might get in proving theorems in number theory. Much as reverse mathematics finds the precise strength of a subsystem of second-order arithmetic that is necessary to prove an analytic result (for example, the intermediate value theorem, the Heine–Borel theorem or the Bolzano–Weierstrass theorem), one could try to see how strong theorems in classical number theory are. This of course is something far outside the scope of a single MO question.

REPLY [10 votes]: Dirichlet's Theorem on Primes in Arithmetic Progressions, proved in 1837,
 needing real-analytic methods could possibly be the first major candidate for a
 number-theoretic result departing from finite methods. (This was proved 50 years earlier than Prime Number Theorem).<|endoftext|>
TITLE: What happened to Emmy Noether's *Zukunftsphantasie* ?
QUESTION [27 upvotes]: Recenly I came across Peter Roquette's article On the history of Artin's $L$-functions and conductors (23 July 2003) in which he talks about some letters from Emil Artin and Emmy Noether to Helmut Hasse in the early 1930s.
Artin is trying to give the definitive form to the definition of his $L$-functions (to include ramified and archimedean places), and has proved what Hasse calls the Führerdiskriminantenproduktformel : for a finite galoisian extension $L|K$ of number fields with group $G=\mathrm{Gal}(L|K)$, the discriminant $\mathfrak{d}$ of $L|K$ can be decomposed as the product
$$
\prod_{\chi}\mathfrak{f}(\chi,L|K)^{\chi(1)}
$$
extending over all characters $\chi$ of $G$, where $\mathfrak{f}(\chi,L|K)$ denotes the conductor of $\chi$ (as defined by Artin).
Emmy Noether writes to Hasse that she is looking for a decomposition formula for
the different $\mathfrak{D}$ of $L|K$ which would yield Artin’s product formula for the 
discriminant $\mathfrak{d}$ after applying the norm map $N_{L|K}$. Perhaps this is what she calls her Zukunftsphantasie (a fantasy for the future).
Question.  Is there such a decomposition of the different $\mathfrak{D}$ ?

REPLY [11 votes]: I am not sure if I am interpreting the question correctly: is it whether it is possible to assign, naturally, an ideal $I_\chi\subset O_L$ to each $\chi$ in such a way that $N_{L|K} I_\chi=f(\chi,L|K)^{\chi(1)}$? Then the answer is No, even without the word "naturally", even for cyclotomic fields and even locally:
If $K={\mathbb Q}$ and $L={\mathbb Q}(\zeta_{12})={\mathbb Q}(i,\sqrt{3})$, then ${\rm Gal}(L/K)=C_2\times C_2$ has four 1-dimensional characters, of conductor $1$, $3$, $4$ and $12$. However, $3$ and $12$ are not norms of ideals from $L/K$, because there is a unique prime ${\mathfrak p}|3$ of $O_L$ with $e=f=2$, so $N_{L|K}{\mathfrak p}=3^2$, and the norm of any ideal has even valuation at $3$. 
Another way of saying this is that the different here is ${\mathfrak D}={\mathfrak q}^2{\mathfrak p}$ (with ${\mathfrak q}|2, {\mathfrak p}|3$) and the discriminant is ${\mathfrak d}=2^43^2$, and while it is possible to split the 3-part of the discriminant into two parts for the two ramified characters, it is not possible for the different. So perhaps Noether's question means something else?
(The observation is joint with my brother.)<|endoftext|>
TITLE: Convex cones and self-duality
QUESTION [14 upvotes]: Consider the Euclidian space $E_n={\mathbb R}^n$, with standard scalar product
$$x\cdot y=x_1y_1+\cdots+x_ny_n.$$
A closed convex cone $\Gamma\subset E_n$ defines an order by $y\ge x$ iff $y-x\in\Gamma$. An order is compatible with the Euclidian structure if 

$x,y\in\Gamma$ implies $x\cdot y\ge0$,
conversely, if $x\in\Gamma$ implies $x\cdot y\ge0$, then $y\in\Gamma$.

Cones satisfying these properties are usually called self-dual. Examples of self cones are $({\mathbb R}^+)^n$, a circular cone with an appropriate aperture angle (which depends on $n$), and the cone of semi-positive definite symmetric $d\times d$ matrices if $n=\frac{d(d+1)}2$. Self-dual cones are also present in the theory of Jordan algebras.
I have two questions.

If $n=2$, the angle of a cone and of its dual are related by the formula $\alpha+\beta=\pi$. In particular, a self-dual cone has angle $\frac\pi2$. In dimension $n=3$, there is no such formula. If the cone is circular, its solid angle $\Omega$ and $\Omega'$, that of the dual cone are related by
  $$\left(1-\frac{\Omega}{2\pi}\right)^2+\left(1-\frac{\Omega'}{2\pi}\right)^2=1$$
  But for the positive orthant, the left-hand side above equals $\frac98$. Is it true that for every convex cone, the solid angles of the cone and of its dual are constrained by
  $$\left(1-\frac{\Omega}{2\pi}\right)^2+\left(1-\frac{\Omega'}{2\pi}\right)^2\ge1?$$
  In particular, what are the possible values for the solid angle of a self-dual convex cone ? Is there a similar inequality (with equality for circular cones) in higher dimension ?

The side question is whether the set of self-dual convex cones form a compact metric space, where we may take the Hausdorff metric on the intersections with the unit sphere. I should bet so.

REPLY [10 votes]: For convex figure $\Sigma$ in $\mathbb S^2$,
the isoperimetrical inequality should look like
$$\left(\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}\right)^2+\left(1-\frac{\mathop{\rm area}\Sigma}{2\cdot\pi}\right)^2\ge 1.$$
If $\Sigma$ and $\Sigma'$ are the intersections of $\mathbb S^2$ with your cones
then by Crofton formula we get
$$\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}+\frac{\mathop{\rm area}\Sigma'}{2\cdot\pi}=1$$
Hence te result.
P.S. The extreme values should be for round cone and positive octant, but I do not see a proof in higher dimensions.<|endoftext|>
TITLE: Branch locus of a 6:1 cover of the grassmannian G(1,3)
QUESTION [5 upvotes]: Given a general quartic surface $S$ in $\mathbf{P}^3$, there is a natural 6:1 surjective map
$\phi: Hilb^2(S) \to G(1,3)$ sending $\{P,Q\}$ to the line through them in $\mathbf{P}^3$.
Can you describe the branch locus of $\phi$ in terms of Schubert classes?

REPLY [12 votes]: Since you are interested in a divisor, you only need to know its degree, that is its intersection with a line. A generic line on $Gr(1,3)$ is given by the set of all lines contained in a plane $P$ and passing through a point $Q$. So, you want to know how many tangents to $S$ pass through $Q$ and lie in $P$. 
Consider the intersection $S_P = S \cap P$.
Since $P$ is generic $S_P$ is a quartic curve. The number of tangents passing through generic point is nothing but the degree of the projectively dual curve which is known to be $d(d-1) = 4\cdot 3 = 12$. 
So, the answer is that the branch locus is given by $12\sigma_1$ (honestly, I don't remember whether the standard notation for the Schubert class of codimension 1 is $\sigma_1$ or not).<|endoftext|>
TITLE: Does this group act geometrically on a Median space?
QUESTION [12 upvotes]: Let $G$ be the semidirect product of $\mathbb{Z}^2$ with $\mathbb{Z}/6$ where $\mathbb{Z}/6$ acts by the order 6 element of $SL_2(\mathbb{Z})$. We can think of this group as the group of order preserving isometries of the tesselation of $\mathbb{R^2}$ with regular triangles.
Does this group acts properly, isometrically and cocompactly on a median space??
Let for two points in a metric space $[x,y]=\{z|d(x,z)+d(z,y)=d(x,y)\}$. If $X$ is a geodesic metric space than this is just the set of all points lying on some geodesic from $x$ to $y$. $X$ is called a median space if for every triple of points $x,y,z$ we have that $[x,y]\cap[x,z]\cap[y,z]$ consists of exactly one point - the median of $x,y,z$. Examples for median spaces are trees and $\mathbb{R}^n$ with the $l^1$- metric. 
The motivation is that the one skeleton of a CAT(0) cube complex is a median graph. If a group acts geometrically on this CAT(0)-cube complex it also acts that way on that graph.
For example this group acts properly and isometrically on $\mathbb{R}^3$. This gives a proper and isometric action on a median space, but this action is not cocompact.
So I was wondering whether there is a better action. The problem seems to be that the automorphism of $\mathbb{Z}^2$ does not extend to a cube-complex automorphism of $\mathbb{R}^2$, but I could  not make this precise.

REPLY [5 votes]: It is known that the group you mention does not act geometrically on a CAT(0) cube complex. See for instance this answer for a possible argument, based on Lemma 16.12 in Wise's monograph The structure of groups with a quasiconvex hierarchy. I think you can adapt the proof of this lemma in order to prove the following statement:
Proposition: Let $G$ be a group acting properly and cocompactly on a median metric space $(M,d)$ of finite rank. Assume that $G$ contains a finite-index subgroup $H \simeq \mathbb{Z}^n$, $n\geq 2$. Then $G$ acts properly and cocompactly on $(\mathbb{R}^n, \ell^1)$.
Sketch of proof. As shown by Bowditch in Some properties of median metric spaces, there exists a CAT(0) metric $\sigma$ on $M$, and, if I understand the construction correctly, this metric satisfies the following properties: (1) any isometry of $(M,d)$ induces an isometry of $(M,\sigma)$; (2) the metrics $d$ and $\sigma$ are biLipschitz equivalent; (3) halfspaces of $M$ are $\sigma$-convex.
As a consequence, the flat torus theorem can be applied, and we find a $G$-invariant and $\sigma$-convex subspace $\Sigma \subset M$ which is $\sigma$-isometric to $\mathbb{R}^n$. 
Now, we consider the structure of measured wallspace of $\Sigma$ induced by the walls of $M$. By $\sigma$-convexity of $\Sigma$ and of the walls, we must have $m$ families of parallel hyperplanes $\mathbb{R}^{n-1}$ in $\Sigma$. (Here, we ignore the collections of hyperplanes which lie in the neighborhood of a single hyperplane. As a consequence, $m \leq n$ since otherwise it would be possible to embed coarsely $\mathbb{R}^{n+1}$ into $\Sigma \simeq \mathbb{R}^n$.) 
Let $F$ denote the median space associated to the previous wallspace. Then $F$ decomposes as the $\ell^1$-product of $m$ (discrete or continuous) unbounded lines. Up to replacing discrete lines with continuous lines, we may suppose that $F$ is $(\mathbb{R}^m,\ell^1)$.
Notice that, because $G$ acts properly on $\Sigma$, necessarily it also acts properly on $F$. So we must have $m \geq n$. But we already know that $m \leq n$, so $m=n$.
So far, we have proved that $G$ acts properly on $(\mathbb{R}^n,\ell^1)$. As $G$ is virtually $\mathbb{Z}^n$, we conclude that $G$ acts geometrically on $(\mathbb{R}^n,\ell^1)$.
$\square$
In your specific example, the question is now: does $\mathbb{Z}^2 \rtimes \mathbb{Z}_6$ act geometrically on $(\mathbb{R}^2,\ell^1)$? The same argument as the one followed here works. A presentation of the group is
$$T=\langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ac)^3=1 \rangle.$$
Notice that $\mathrm{Isom}(\mathbb{R}^2, \ell^1)= (\mathbb{R}\rtimes \mathbb{Z}_2)^2 \rtimes \mathbb{Z}_2$ does not contain elements of order three, so, for any action $T \curvearrowright (\mathbb{R}^2, \ell^1)$ by isometries, the elements $ab$, $bc$ and $ac$ must be trivial. Consequently, such an action must factorise through the quotient $T \twoheadrightarrow \mathbb{Z}_2$ sending all the generators to $1$. A fortiori, the action cannot be geometric (and even proper).
Remark: The argument above does not completely answer the question as the median metric space is supposed to have finite rank. I do not know what happens if the rank is infinite, but it seems reasonable to think that the same conclusion holds.<|endoftext|>
TITLE: jacobians with non-abelian complex multiplication
QUESTION [9 upvotes]: Hi friends,
I am looking for examples of curves over a number field such that their jacobians are CM abelian varieties by a field whose Galois group is non-abelian. Does anybody know how to produce such examples out of a curve with the action of a finite non-abelian group G?
Thanks a lot!

REPLY [19 votes]: I think a large group of automorphisms of a curve $C$ only forces 
presence of roots of unity in the endomorphism algebra 
$\text{End}^0(\text{Jac}\ C)$, so this way is likely to produce 
products of abelian varieties with CM by abelian fields. There is a lot
of literature on such examples, e.g. on Fermat curves $x^n+y^n=z^n$, but
they are probably not what you want.
Non-abelian CM seems quite hard to construct explicitly. 
It is easiest to work computationally with curves over ${\mathbb Q}$, 
and usually in genus 1 or 2.
For $g=1$ all CM fields are imaginary quadratic, and for
$g=2$ they are either abelian quartic or have dihedral Galois group $D_8$.
However, there is a theorem of Shimura that says that $D_8$ examples 
do not exist over ${\mathbb Q}$!
To construct such an example over a number field, one can use an approach
of van Wamelen in

`Examples of genus 2 CM curves defined over the rationals'.
This is a very nice paper, where he first shows how to construct such Jacobians 
over ${\mathbb C}$ as lattices (sections 2-3). Then he turns to constructing them as genus 2 hyperelliptic curves over ${\mathbb Q}$, which restricts his examples to abelian Galois groups.
(There is a reference to the aforementioned theorem of Shimura on the last page 
of the paper.) 
One of van Wamelen's examples is carefully worked through in Magma, 
in the chapter on 

hyperelliptic curves, towards the end in the `From Period Matrix to Curve' 
section. A direct link to it is currently
here
but these tend to change with new releases. 
It can be adapted to construct a $D_8$ example as well, as follows.
Change the field to ${\mathbb Q}(\sqrt{\sqrt{2}-2})$ in that example 
(Galois group $C_4$) by ${\mathbb Q}(\sqrt{2\sqrt{2}-5})$ (Galois group $D_8$).
Then going through that code (plus a little work to get the coefficients as algebraic numbers) gives a hyperelliptic curve with Igusa-Clebsch invariants 
$$
  [36,45(\sqrt{17} + 1)/2,(729\sqrt{17} - 783)/2,-4\sqrt{17} + 36].
$$
The function HyperellipticCurveFromIgusaClebsch then constructs the curve
in the usual form $y^2=$hexic over ${\mathbb Q}(\sqrt{17})$, 
but it has awful coefficients. The example done
in Magma is simplified further using ReducedModel(C: Al:="Wamelen"), but this
function is only implemented over ${\mathbb Q}$. 
The field ${\mathbb Q}(\sqrt{17})$ has class number $1$, and the same 
ReducedModel algorithm would work over this field, I suppose, 
but this is quite a bit of work to implement it properly. 
And, if I am not mistaken, in genus 2 this is possibly one of the simplest
examples.
In any case, if you are happy to have examples as lattices in ${\mathbb C}^g$
rather than equations,
then van Wamelen's paper is a good place to look, I think.<|endoftext|>
TITLE: A matrix inequality involving the Hilbert-Schmidt norm
QUESTION [6 upvotes]: This question comes from a problem in PDEs on which I'm currently working. Let $a$ be a $3\times 3$ matrix, real symmetric and positive definite. Denote with $\|a\|^2 _ 2=\sum a_{ij}^2$ the square of the Hilbert-Schmidt norm and consider the quantity
$$ Q(v)=  2\|a\|^2 _2 + [trace(a)-3(av,v)]^2
   -6[2 |av|^2 -(av,v)^2] $$
where $v$ is an arbitrary unit vector. If $a=I$ is the identity, the quantity $Q$ is identically zero. 
QUESTION: are there other matrices $a$ such that $Q(v)\ge0$ for all unit vectors $v$, or is this condition equivalent to $a=I$?
At least, it would be helpful if the matrix wizards around here could suggest ways to handle the HS norm, which is unfamiliar to me, and estimates relating trace, HS norm, operator norm and the numerical range of a matrix, which could possibly be of use here.

REPLY [8 votes]: Suppose $Q$ is such a form. Write that the mean value of $Q$ over the unit sphere is non-negative. You obtain
$$-\frac12\sum_ia_{ii}^2+\frac12\sum_{i < j}a_{ii}a_{jj}-9\sum_{i < j}a_{ij}^2\ge0.$$
This implies that $a=\lambda I_3$.<|endoftext|>
TITLE: Examples of sphere bundles
QUESTION [10 upvotes]: For certain values of $k$ it is known that $\mbox{Diff}(S^k)$ is not homotopy equivalent to $O(k+1)$. So there are sphere bundles that do not arise from vector bundles.
Since I've never (knowingly) come across such a sphere bundle I'm interested in seeing some enlightening examples of sphere bundles which do not come from vector bundles.
Thank you for any contribution.

REPLY [5 votes]: As far as I know the only explicitly-described such bundles are in Hatcher's paper:

Hatcher.  Concordance spaces, higher simple-homotopy theory, and applications.  Algebraic and geometric topology (Proc. Sympos. Pure Math., Stanford Univ, Stanford Calif 1976), Part 1, pp. 321. 

I think in Igusa's Higher Franz Reidemeister Torsion book there might also be these examples, although I don't have the book at home with me so I can't check.  But that seems likely as Igusa has also developed these examples.<|endoftext|>
TITLE: Deligne's letter to Looijenga from 1974
QUESTION [7 upvotes]: Hello,
I wonder if anyone has a copy of Deligne's letter to Looijenga from 1974 mentioned as reference [26] in Bessis' paper Finite complex reflection arrangements are $K(\pi,1)$ from 2006, see http://arxiv.org/abs/math/0610777, and is willing to share it / make it publicly available.
We were recently proving some results generalizing particular aspects of things that are supposed to be proven in there, and we were asked about the historical background. Unfortunately, we cannot really provide an answer since we don't have access to the letter...

REPLY [13 votes]: Eduard Looijenga provided a scanned version of the letter, which now can be found at
http://homepage.univie.ac.at/christian.stump/Deligne_Looijenga_Letter_09-03-1974.pdf (outdated)
http://homepage.rub.de/christian.stump/Deligne_Looijenga_Letter_09-03-1974.pdf (2020-01-24)
Many thanks for making it publicly available!
I also added a section on Deligne's hand-written letters to his wikipedia page, though I am not quite sure this is the right place to collect his publicly available letters.<|endoftext|>
TITLE: Which local ringed spaces are schemes?
QUESTION [13 upvotes]: (This was originally asked on math.stackexchange, but didn't get any responses. I figured it might be worthwhile to move it here and try again.)
This paper gives a proof that the underlying topological spaces of affine schemes are precisely the spectral spaces (compact, T0, sober, and the compact open subsets are closed under finite intersection and generate the topology; equivalently, they are the inverse limits of families of finite T0 spaces).
Instead of starting with a bare topological space, suppose we have a locally ringed space. If it is an affine scheme, then the underlying space is spectral, the supports of sections generate the topology, every restriction map of the structure sheaf to a distinguished open set is a localization of rings, etc. 
Does anyone know whether we can impose restrictions (such as the ones I have just listed, and probably together with others) to guarantee that a given locally ringed space is actually an affine scheme?
This poster asked something similar - whether locally ringed spaces had been classified in some way. This is not exactly what I'm looking for, but it would be interesting to know in any case.

REPLY [16 votes]: I think the easiest condition is the fact that the natural morphism
$$
(X, \mathcal O_X) \to \operatorname{Spec}(\mathcal O_X(X))
$$
is an isomorphism.<|endoftext|>
TITLE: Localizations of non-nilpotent spaces
QUESTION [8 upvotes]: For simplicity let's talk about $p$-localizations of spaces for a fixed prime $p$. Every space $X$ has a well-defined $p$-localization which can be constructed by the small object argument and which becomes a fibrant replacement in the $p$-local model structure on the category of spaces. It is well-known that nilpotent spaces have nice enough Postnikov towers and we can localize such spaces by taking the Postnikov tower, localizing step by step and putting it back together by taking the limit of the resulting tower of fibrations. My question is:

Is there an example of a non-nilpotent space $X$ whose $p$-localization we can explicitly describe?

I leave the meaning of "explicitly" ambiguous. I would be interested in any construction not using the small object argument.
Here's my stab at a possible example. For a group $G$ we define its lower central series by setting $G_0 = G$ and $G_{n + 1} = [G_n, G]$ and we can also continue transfinitely by setting $G_\beta = \bigcap_{\alpha < \beta} G_\alpha$ for limit ordinals $\beta$. The group $G$ is nilpotent if this construction terminates at the trivial subgroup at a finite stage. It is called hypocentral if it terminates at the trivial subgroup at some not necessarily finite stage. According to Wikipedia it is a result of Malcev that there are hypocentral groups with arbitrarily long lower central series.

If we start with a hypocentral group $G$ and convert its lower central series into a (transfinite) tower of fibrations (whose limit is a $K(G, 1)$), $p$-localize it step by step and take the limit of the resulting tower of fibrations, do we obtain the $p$-localization of $K(G, 1)$?

REPLY [5 votes]: As you know extending $P$-localization functors to non-nilpotent spaces is a very delicate matter. You can do it by homotopical localization techniques. if you do it in this way you get functorial localization which work for non-nilpotent spaces "but" it is the aim of your question it is not "very explicit". 
Now let me be very naive and let us go back to D. Sullivan's MIT notes "Geometric topology, localization, periodicity and Galois symmetry"  chapter 2 (these notes are really beautiful to read).
D. Sullivan first explains how to do $P$-localization when we have a CW-complex, of course he also explains that this CW-procedure gives a "nice localization" when the CW-complex is 1-connected (in that case it is our modern $P$-localization). He proceeds on the skeleton of the CW-complex, localizing cell by cell. 
Thus you can apply Sullivan's telescope construction to a circle and then to a wedge of circles and any CW-complex you want. The starting point being the localization of spheres through a telescope of maps $S^k\stackrel{\times p}{\rightarrow }S^k$ (you invert degree $p$-maps).
The point is that you can argue that this naive construction on non-nilpotent CW-complexes is not what you expected as a $P$-localization. D. Sullivan also explains how to do it when we have a Postnikov tower.   
Let me add that there is another interesting way to $P$-localize which is due to C. Casacuberta and G. Peschke:
"Localizing with respect to self-maps of the circle" Trans AMS 339 (1993) 117 – 140.  
This $P$-localization is as "explicit" as a homological Bousfield's localization but the homotopy related to it is very interesting. In that setting, a space is $P$-local if and only if the $p$-power map on the loop space    $\Omega X$ is a self homotopy equivalence if $p\notin P$.<|endoftext|>
TITLE: Exact arithmetic for real algebraic numbers
QUESTION [6 upvotes]: There was a reply to a question (that I can't find) which mentioned SARAG (Some Algorithms
in Real Algebraic Geometry) see http://perso.univ-rennes1.fr/marie-francoise.roy/bpr-ed2-posted2.html. This is a package for Maxima.
I looked into this and came across the Thom encoding of real algebraic numbers. If I have understood correctly, you are given a square-free polynomial with, say integer coefficients, and a real root. Then evaluate the derivatives of the polynomial at the root to get a sequence of signs. The polynomial together with the sign sequence is an encoding of the root.
My question is whether you can implement the operations of an ordered field on these encodings? The operation I not clear on is the ordering. A specific question is: given an encoding as above and a polynomial; evaluate the polynomial at the real number. Is there an algorithm for deciding if this is positive, negative, (or zero)?
I assume it is possible to evaluate this numerically using interval arithmetic and that if you did this with sufficient accuracy you would eventually arrive at a solution. However this seems clumsy and I am asking for something more effective.
As an illustration of why this could be useful. There was a question on the word problem in a Coxeter group (which again I could not find). One solution to the word problem is to look at the image of a fixed vector in the geometric representation. Then you can determine the descent set of a word by looking at which entries are negative. This is fine theoretically but not if you work with fixed precision real numbers as the entries will rapidly become very small in absolute value. A positive solution to my question would make this an effective algorithm.

REPLY [6 votes]: In the reference you posted there is a link to an online version of the monograph by S.Basu, M.-F.Roy, and R.Pollack, where algorithms like this are described (cf. Sect. 10.4, Algorithm 10.15). 
The technique there is very general, and applies to non-Archimedean real closed fields, e.g. to fields of Puseaux series w.r.t. to an infinitesimal. There you don't have anything like "numeric evaluation".
Having said this, I wonder whether Thom encoding is really better for the range of problems like "straight" arbitrary precision computations with algebraic reals. For the latter, isolating roots by rational numbers works as well, and is faster, in theory, according to the reference. 
Thom encodings really shine when one has a parametric (e.g. multivariate),  or/and a non-Archimedean, setting. 
Computation with isolating intervals is implemented e.g. 
in Sage.
sage: s=sqrt(AA(2))
sage: s.numerical_approx(prec=1000)
1.4142135623730950488016887242096980785696718753769480731766797379907324784...
sage: s>AA(0)
True
sage: t=s-sqrt(AA(3))
sage: t.minpoly()
x^4 - 10*x^2 + 1
sage: t.sign()
-1<|endoftext|>
TITLE: Discretization of a complete manifold
QUESTION [9 upvotes]: Suppose $M$ is a complete Riemannian manifold with very large injectivity radius (say larger than $100$) and $\left\lbrace x_i: i \in I\right\rbrace$ is a maximal $1$-separated subset of $M$.
Is diffeomorphism class of $M$ determined by the (possibly infinite) distance matrix $(d(x_i,x_j))_{i,j \in I}$?
Suppose now that we are only given the information of which points $x_i,x_j$ are at distance less than $2$.  Is this enough to determine $M$ up to diffeomorphism?

REPLY [12 votes]: My knowledge of this subject is obsolete, but anyway here are some partial answers.
If you have a Ricci curvature bound in addition to the injectivity radius, then  you can recover the diffeomorphism type, e.g. with harmonic coordinates, see  M. Anderson, Convergence and rigidity of manifolds under Ricci curvature bounds, Invent. Math., 102 (1990), no. 2, 429-445. But 100 should be replaced by some radius depending on the dimension and the Ricci curvature bound.
Correction. It is the constant 1 that would depend on the Ricci curvature bound. (I did the rescaling argument wrong.)
Without curvature bounds things get more complicated. It is very easy to recover homotopy type, In fact, you don't need injectivity radius, only a contractibility function: every ball of radius $r\le r_0$ is contractible within a ball of radius $f(r)$. Given this, you can construct homotopy equivalence from an almost isometry defined on a net: extend it step by step to skeletons of a cell decomposition, just make sure that $n$ iterations of $f$ (or maybe $2f$) do not grow bigger than your constant 100. 
If the dimension is not 3 (and maybe for 3 as well, since the Poincare conjecture is now solved), you can recover homeomorphism type under a pre-compactness assumption (in your set-up, this pre-compactness boils down to something like that every ball of radius 2 is covered by a bounded number of balls of radius 1, and then your radius 100 depends on this number). This follows from the arguments in Grove-Petersen-Wu, Geometric finiteness theorems via controlled topology. Invent. Math. 99 (1990), no. 1, 205-213.
Correction. Again, the pre-compactness would translate into something more complicated after rescaling the injectivity radius to 1.
They also use only local contractibility function, and in this more general context diffeomorphism stability fails in dimension 4. Of course this does not answer your question since the injectivity radius assumption is so much stronger.  I don't know how essential the pre-compactness is. They certainly need it for their result (which is finiteness of topology types) but perhaps it may be relaxed if you only need stability.<|endoftext|>
TITLE: Why pullback only defined up-to-isomorphism but nevertheless presented as functor?
QUESTION [19 upvotes]: I am reading "Category Theory" (2nd ed.) of Awodey, and I'm stuck at page 96 (proposition 5.12) when pullbacks are presented as functors:
The pullback under question corresponds to this square:
$$\begin{matrix}
C' \times_C A & \xrightarrow{h'} & A \\[1ex] 
\downarrow \rlap{\scriptstyle{\alpha'}} & &  \downarrow\rlap{\scriptstyle{\alpha}} \\[1ex] 
C' & \xrightarrow{h} &  C
\end{matrix}$$
Here is the statement of Awodey's book that I do not understand:

Pullback is a functor. That is, for fixed $C' \rightarrow_h C$ in a category $\mathbf{C}$ with pullbacks, there is a functor
$h^* : \mathbf{C}/C \rightarrow \mathbf{C}/C'$
defined by
$(A\rightarrow_\alpha C) \mapsto (C'\times_C A \rightarrow_{\alpha'} C')$
where $\alpha'$ is the pullback of $\alpha$ along h

The problem that I see is that, given initially:
$$\begin{matrix}
& & A \\[1ex]
& & \downarrow \rlap{\scriptstyle{\alpha}} \\[1ex]
C' & \xrightarrow{h} & C
\end{matrix}$$
there can be several pullbacks on it, for example, in addition to $(\alpha',h')$, there could be $(\alpha_2',h_2')$, and the unique condition is that there exists an isomorphism $i$ such that $\alpha_2' = \alpha\circ i$ and $h_2' = h'\circ i$.
Worse, given a pullback $(\alpha',h')$, one can build as many as pullbacks as there exist isomorphisms, as given any isomorphism $j$ (with domain $C' \times_C A \rightarrow_{h'}$) the two arrows $(\alpha'\circ j,h' \circ j)$ form a new pullback.
So, how could we build a functor if the image arrow is only defined up to an arbitrary isomorphism?

REPLY [13 votes]: Why has no one mentioned anafunctors? They were invented by Makkai for the express purpose of expressing universal constructions as functor-like things (I was going to say 'objects') without having to make choices. 
The definition is as follows: Let $C$ and $D$ be categories. An anafunctor from $C$ to $D$ is a span $C\leftarrow U \to D$ where $U \to C$ is fully faithful and surjective on objects.
That's it.
Of course, it is a bit more tricky to show that there are the arrows of a (weak) 2-category of categories, and that category theory works perfectly well using this notion of categories; for the present purposes, the references at the above link should satisfy the most curious.
In the case of the pullback 'functor' $h^*$ we have an anafunctor
$\mathbf{C}/C \leftarrow P \to \mathbf{C}/C'$ where $P$ is the category with objects pullback squares with bottom arrow $h$ and morphisms the canonical thing which makes $P \to \mathbf{C}/C$ fully faithful: commutative triangles involving the other leg of the cospan involving $h$ - in other words, arrows in $\mathbf{C}/C$. This induces a canonical arrow in $\mathbf{C}/C'$ by the universal property of pullbacks, and then the functor $P\to \mathbf{C}/C'$ just forgets the original pullback square, and keeps the arrow with codomain $C'$. This is a functor by using the universal property of pullbacks a couple of times.
If one has a way of choosing a particular pullback square for each object of $\mathbf{C}/C$, say by some sort of canonical construction as in the category of (ZF-)sets (Kuratowski pairs and subsets), or by judicious amounts of the axiom of choice, then one can find a section on objects of $P \to \mathbf{C}/C$, and this gives you a canonical section of the functor $P \to \mathbf{C}/C$, which then gives a pullback functor $h^*\colon \mathbf{C}/C \to \mathbf{C}/C'$.
A similar sort of anafunctor exists for any universal construction, which is defined up to isomorphism by some universal property.<|endoftext|>
TITLE: A more general form of Grauert's Theorem on Higher Direct Image Sheaves?
QUESTION [6 upvotes]: When Grauert's Theorem is presented in Hartshorne, the statement goes as follows:
Let $f:X\rightarrow Y$ be a projective morphism of noetherian schemes, and $\mathcal F$ a coherent sheaf on $X$, flat over $Y$.  Assume that $Y$ is integral and for some $i$, the function $h^i(y,\mathcal F)=\dim_{k(y)} H^i(X_y,\mathcal F_y)$ is constant on $Y$.  Then $R^if_*(\mathcal F)$ is locally free on $Y$ and for every $y$ the natural map 
$$R^if_*(\mathcal F)\otimes k(y)\rightarrow H^i(X_y,\mathcal F_y).$$
I was wondering if the assumption that $Y$ be integral can be removed.  Certainly the proof in Hartshorne uses the integrality condition.
I know that the projective requirement can also be made more general in allowing proper morphisms.  I have seen a statement to this effect (in allowing properness and any Noetherian scheme $Y$ as a base) in Nitsure's notes "Construction of Hilbert and Quot schemes" (Part 2 of "FGA:Explained"), and indeed part (3) of Theorem 5.10 there says precisely the statement above with properness replacing projective and without the requirement that the base be integral, but I was wondering if that was accurate.  The reference given for the proof is the above result and proof in Hartshorne, which doesn't cover the general case.
In short, I would like to know whether this generalization (1) is indeed true, (2) if Hartshorne's proof can be easily modified, and (3) if not, is there a good reference (EGA is acceptable, but not preferred).

REPLY [2 votes]: The standard proof (with only reduced hypotheses) is Theorem 28.1.5 in the June 11, 2013 version of the notes at http://math216.wordpress.com/<|endoftext|>
TITLE: "Softness" vs "rigidity" in Geometry 
QUESTION [35 upvotes]: According to common wisdom, there are structures in Geometry that have a more "topological" flavor, others that are more "geometrical", and others that are halfway between. Usually, geometries${}^*$ that are closer to the topological end of the (possibly non totally ordered) spectrum are referred to as being "soft", whereas geometries closer to the geometric end are considered "rigid". For example, topological, differentiable and symplectic manifolds are considered soft, while Riemannian and complex analytic manifolds are considered rigid. And algebraic varieties are even more rigid geometric objects than analytic manifolds.
I've noted there's also a tendency in deformation theory to use the term "rigid" in the opposite way: rigid is something you cannot deform ${}^{**}$, so e.g. compact differentiable manifolds are rigid in this sense (thanks to a well known theorem of Ehresmann), whereas usually algebraic varieties admit nontrivial deformations so they are usually not "rigid".
Also, near the ends of the spectrum (homotopy theory, geometric topology; finite geometries, arithmetic geometry) a more combinatorial/discrete/algebraic approach seems to prevail, while in the middle of the spectrum (metric spaces, Riemannian geometry, analytic and complex algebraic geometry) there seems to be a more "continuous" character.

How to make the above distinctions topology vs geometry and/or softness vs rigidity formally more rigorous?
What are other features of the softness vs rigidity phenomenon?

Let's start with some loose ideas about what I have the impression to be features typical of topology vs geometry.

Local invariants. Is the structure at a point distinguishable from the other points or from points of other spaces, or do they all locally look the same? In the case of topological, differentiable and symplectic manifolds there are no local invariants. This also happens for complex analytic manifolds, that we most often regard as instances of a rigid geometry, so it is certainly not a sufficient characterization. For Riemannian manifolds the curvature is a nontrivial local invariant (even a punctual one, in the sense that, even if its definition involves a neighbourhood, differences can be checked pointwise). In algebraic geometry we must be more precise as for the meaning of "locally": locally (w.r.t. some Grothendieck topology) or infinitesimally locally or formally locally? All smooth varieties over a field formally locally look like affine space, but look different locally in the Zariski topology. A condition often put on principal bundles is local isotriviality (i.e. local triviality in the étale topology); this subtlety doesn't appear for vector bundles (aka locally free sheaves).

Robustness vs deformability If you perturb the structure in some way, the resulting structure stays isomorphic. The perturbation can be a deformation in the sense of deformation theory, or just picking a close enough datum (e.g. a Riemannian metric close to the original one in the $\mathcal{C}^k$ topology). By Ehresmann, compact differentiable manifolds are invariant under deformations; but noncompact differentiable manifolds are not, indeed starting from dimension $4$ there may be fibrations with homeomorphic non diffeomorphic fibers, so in some sense it's a less purely topological feature. In analytic and algebraic geometry, projective spaces are invariant under deformation; and I have the impression that it also happens for combinatorially/algebraically defined varieties (e.g. toric).

Discreteness of moduli. This is somehow the global version of the previous point. A "deformation invariant" structure need not have trivial moduli, but while topological structures tend to have discrete moduli spaces (in whatever sense we intend "moduli space"), when moduli are nondiscrete it means there is something geometric going on. For example, if I remember correctly, line bundles on toric varieties have discrete moduli (the Jacobian is trivial), maybe because they depend only on the combinatorics of the orbits, which is -with some stretch of the meaning- a topological thing.

Homogeneity. In many topological categories the "generic" object tend to have a transitive group of isomorphisms (even $n$-transitive sometimes). This happens for topological, differentiable, and (differentiable or real analytic) symplectic manifolds. Of course transitivity of automorphisms also happens for homogeneous spaces in categories of rather "rigid" objects (Riemannian, analytic, algebraic homogeneous spaces), but they are very special objects, not a "random" representative of their category. Anyway they share with more topological categories the possibility of being described combinatorially/algebraically (Schubert cells of the Grassmannian, Lie algebras).

Obstructions. In some soft categories we have partitions of unity, which often allow us to patch local data together to obtain a globally defined thing from locally defined things; in more rigid categories this doesn't hold. Also, in rigid categories there are extension problems.


I would say, when two categories of geometric objects are to be compared in softness/rigidity, we are in the following situation. We have categories $\mathcal{C}$, $\mathcal{C}'$ concrete over some base category $\mathcal{S}$ (the latter can be sets, or topological spaces, or any fancy thing like presheaves of simplicial sets over a site). There is a "forgetful functor" $\mathbb{U} :\mathcal{C}\to\mathcal{C}'$ which commutes with the concretizations. We can say (mind that I'm only giving a vague suggestion and I have no idea of a more precise answer) that $\mathcal{C}'$ is obtained by "putting some geometric structure" on objects of $\mathcal{C}$ if one or more of the following holds:

$\mathbb{U}$ can make local invariants disappear (e.g. forgetting a Riemannian metric on a manifold).
$\mathbb{U}$ can turn a (nontrivially) deformable object into a deformation-rigid one.
If there is some notion of moduli spaces $\mathcal{M},\mathcal{M}'$ for objects of $\mathcal{C},\mathcal{C}'$, then the map $\mathcal{M}\to\mathcal{M}'$ induced by $\mathbb{U}$ has nondiscrete fibers.
$\mathbb{U}(\mathrm{Aut}(X))\subset\mathrm{Aut}(\mathbb{U}X)$.

${}^*$ (Unfortunately, the term "geometry" is, ambiguously, used both in the general sense of "pertaining to spaces of any kind" and in the more localized sense of "pertaining to structures that are metric or anyway more rigid than topological ones". E.g. the term "differential geometry" is sometimes used as a general header including differential topology, sometimes as a synonym of Riemannian geometry. I hope in the question the ambiguities will be cleared by the context)
${}^{**}$ (Rather than "rigidity", I think a more appropriate term would be "elasticity" or "flexibility", since after all one can fit a rigid variety as special fiber of a family, it's just that the deformed neighbours -i.e. the other fibers- turn out to be isomorphic to the variety you started with)

REPLY [10 votes]: Dick Lashof once told me that "differential topology is about first derivatives; differential geometry is about second derivatives".

REPLY [9 votes]: Here is very rigorous way to think about it.
Doing geometry is like a walking on a tightrope --- you can make few steps and then you fall into algebra (right side?) or into topology (left side?); in any case, you have to start all over again.<|endoftext|>
TITLE: how to find/define eigenvectors as a continuous function of matrix?
QUESTION [26 upvotes]: I asked this (with background) here
https://stats.stackexchange.com/questions/38494/principal-component-analysis-bootstrap-and-probability-of-eigenvalue-collision
but did not really get any answers. See that post for the background.
Let $D$ be some open set in the plane, say.  Not really important where the set $D$ sits, but it shoud not be only a line/curve. Suppose we have defined a continuous function on $D$
$$
f \colon   D \mapsto \text{Sym}^n
$$
where $\text{sym}^n$ is the set of (real) symmetric $n \times n$  matrices. How can I define the eigenvectors of $f(x), x \in D$ as a continuous function on $D$? How can I calculate this?
And how can I deal with eigenvalue collisions? A simple example clarifying this point (and defined on a curve):  Let
$$
f(t) =\left( \begin{matrix} 1+t & 0
\cr
   0 &   1-t \end{matrix}\right)
$$
Then the largest eigenvalue is
$$ 
\lambda_1(t) = 1+ |t|
$$
but the eigenvector corresponding to the largest eigenvalue cannot be defined as a continuous function:
$$
   v_1(t) = \begin{cases} e_2 & t\le 0        \cr
                           e_1 & t > 0 \end{cases}
$$
So what I want is to look at the two eigenvalue functions $1+t, 1-t$ and follow the eigenvectors corresponding to each one, which obviously can be done in a continuos (constant!) manner.
ADDED after the answer by Anthony Quas:
Is it possible to give some further conditions, under which a solution is possible?
Differentiability? Or, if the matrices are realizations of some random field of matrices, can something be said about the probability some continuous selection is possible?

REPLY [2 votes]: I think this might salvage the situation: Assume that you do have a path parameterized by $t$ and that it is stationary for a while any time that the multiplicity of an eigenvalue increases. In the case $n=2$ the eigenvectors are (usually) perpendicular so we could represent them as four points on the unit circle separated by $\frac{\pi}{2}$ radians . Imagine time as an axis so the eigenvectors form  four black paths traveling up a cylinder. Any time the matrix becomes a scalar multiple of the identity matrix you suddenly have the solid unit circle. As long as this is a band of some width you can arrange to leave the band in the appropriate configuration.
With larger $n$ and even more general matrices I think it would be about the same. So one point is that it can not be a function of merely where you are, but also where you were and where you will be next.
A related problem is constructive versions of the Fundamental Theorem of Algebra ( cribbed from a paper by Fred Richman which I recommend.) Let $\mathbb{A} \subset \mathbb{C}$ be the field of algebraic numbers (roots of polynomials with rational coefficients) Consider degree $n+1$ monic polynomials  $x^{n+1}+\sum_0^na_iz^i$ They can be parameterized by their coefficient vectors $\mathbf{a}=(a_0,a_1,\cdots,a_n) \in \mathbb{A}^n$ and by their "list" of roots $\boldsymbol\alpha=(\alpha_0,\cdots,\alpha_n) \in \mathbb{A}^n$ ordered somehow. There is an obvious continuous map (uniformly bicontinuous on bounded sets) in one direction $\boldsymbol\alpha \to \prod(z-\alpha_i)$ i.e. extract the coefficients using elementary symmetric functions. Is there a continuous mapping in the other? Read the paper (which gets into Dedikind cuts, extension to all of $\mathbb{C}$ and other matters.) As I recall, the correct target in the space of roots should instead be multisets of algebraic numbers with an appropriate metric. A motivating example is $z^2-b$ with $b$ real. For $b$ close to $0$ we have a sudden shift from the two roots spanning a horizontal line to a vertical one.<|endoftext|>
TITLE: Explicit Computations of Examples in Spin Geometry
QUESTION [10 upvotes]: I have been trying to learn about spin geometry, Dirac operators, and index theory by reading Lawson/Michelson's "Spin Geometry" and Friedrich's "Dirac Operators in Riemannian Geometry." Both are abstract, and basically no explicit examples are worked all the way through.
For example, I have been trying to find the spinor bundles, Dirac operators, and various indices for relatively simple manifolds: spheres and tori. However often these computations are detailed and even when I get to the end, it's not clear that I've done it correctly. 
Is there another book, or perhaps online notes, which have a bunch of examples worked through in detail so that I can make sure what I'm doing is correct and also have a bank of examples to look at as I progress?

REPLY [6 votes]: Appendix A to Chapter 9 of the book Elements of Noncommutative Geometry by Gracia-Bondia, Varilly, and Figueroa is titled "Spin geometry of the Riemann sphere".  It is 15 pages long and goes into quite some detail.  (Some might call that level of detail excruciating, but YMMV.)
As Paul Siegel notes, computations on homogeneous spaces can be done quite effectively using representation theory.  Some years ago, in the course of learning about that approach, I wrote up an account of the construction of the spinor bundle, Dirac operator, etc on $S^2$, viewed as the homogeneous space $SU(2)/U(1)$.  If you're interested, email me (you can find my email address at my website, linked in my profile) and I can send it to you.<|endoftext|>
TITLE: The BCH series in terms of Lyndon words
QUESTION [24 upvotes]: Recently I did some explicit computations that involved the BCH series, $\log(e^x e^y)$. Here $x$ and $y$ are non-commuting variables, and the BCH series lives in the graded completion $FL(x,y)$ of the free Lie algebra generated by $x$ and $y$.
Mostly by chance I found that when BCH is written in the Lyndon basis of $FL(x,y)$, the number of Lyndon words that occur in its degree $n$ piece is {2, 1, 2, 1, 6, 5, 18, 17, 55, 55, 186, 185, 630, 629, 2181, 2181, 7710, 7709, 27594, 27593, 99857, 99857}, for $n$ running from 1 to 22.
There is an obvious pattern in this sequence - it seems that the odd-numbered terms are almost equal to the even-numbered terms that follow them, with a decline of one in 2/3 of the times, and with precise equality in the remaining 1/3 of the times. I have no idea why this is so. Perhaps you do?
Why care? The truth is that I'm curious but I don't care much; I just stumbled upon this by chance. Yet Lyndon words are a very effective tool for computations in free Lie algebras, and the BCH formula appears in many of these computations. The fact that there is some unexpected symmetry in the Lyndon word description of BCH suggests that BCH contains less information than one might think, possibly leading to some computational advantage. Though in (my) reality, the computational bottlenecks are anyway elsewhere.
Some further details and observations are at http://drorbn.net/AcademicPensieve/2012-12/nb/BCH-Lyndon_Question.pdf.

REPLY [5 votes]: This pattern of vanishing coefficients in the expansion of the Baker-Campbell-Hausdorff formula in the basis of Lyndon words is explained in section IV.C of the article An efficient algorithm for computing the Baker–Campbell–Hausdorff series
and some of its applications of Fernando Casas and Ander Murua.<|endoftext|>
TITLE: Exponent of Sylow $p$-subgroup of classical groups over a field of characteristic $p$ 
QUESTION [7 upvotes]: Let $G$ be a classical group of dimension $n$ over $GF(q)$ where $q=p^f$ is a prime power, and $P$ be a Sylow $p$-subgroup of $G$. What is the maximal order of elements, i.e. the exponent, of $P$? 
For $G=GL(n,q)$, it can be easily seen that the exponent of $P$ is the least power of $p$ greater than or equal to $n$. This gives a upper bound for exponents of classical groups of dimension $n$ over $GF(q)$.

REPLY [3 votes]: For some classical groups (arising from simple algebraic groups), the answer is given in Proposition 0.5 in a paper by Donna Testerman: $A_1$-type overgroups of elements of order $p$ in semisimple algebraic groups and the associated finite groups, J. Algebra, 177 (1995), no. 1, 34--76. I quote this proposition: 
Let $G$ be a simple algebraic group defined over an algebraic closed field of arbitrary characteristic $p>0$ and $\sigma$ be a surjective endomorphism of $G$ such that $G_\sigma$ is finite. Then the exponent of a Sylow $p$-subgroup of $G_\sigma$ is the least power of $p$ greater than the height of the highest root in the root system of $G$. 
The height of the highest root is equal to $h-1$, where $h$ is the mentioned Coxeter number, and so is $n$, $2n-1$, $2n-1$ and $2n-3$ in $A_n$, $B_n$, $C_n$ and $D_n$ respectively (the Coxeter number in type $A_n$ is $n+1$ not $n$).<|endoftext|>
TITLE: Recovering Sidon sets from difference sets, part 2.
QUESTION [6 upvotes]: This is inspired by a recent question. A set $A \subset \mathbb{Z}/n\mathbb{Z}$  with $|A|=m$ is a Sidon set if all the pairwise sums of distinct elements are unequal:  $A+A=\{a+a' \mid a,a' \in A, a \ne a'\}$ has $\binom{m}2$ elements. Since $a+b=c+d$ implies $a-c=b-d$ it is equivalent to require that $\{a-a'\mid a,a' \in A\}$ has $m(m-1)+1$ elements (including 0.)
It is clear that when $\lambda$ is coprime to $n$, $B=\lambda A$ is also a Sidon set and $B-B=\lambda(A-A).$ I am going to restrict to the case that $n$ is prime (although more general rings could be of interest.) My question is if the converse is true (I'll explain why I call this the converse):

Suppose that $n$ is prime, $A,B \subset  \mathbb{Z}/n\mathbb{Z}$ are Sidon sets and $B-B=A-A.$ Must it be the case that either

There are  $\lambda=\pm 1,k$ with $B=\lambda A+k$? OR

$A-A=B-B=\mathbb{Z}/n\mathbb{Z}$ and there are  $\lambda,k$ with $B=\lambda A+k$?



I can report that this is true in all the cases I checked which were most of the possibilities with $n \le 29$ and $m \le 5$ as well as $n=31$ and $m=6.$ However this may simply be the Strong Law of Small Numbers.
In that last case (as well as $n=13$ and $m=4$) we have $A-A= \mathbb{Z}/n\mathbb{Z}$ so it is not a surprise that there are choices of $\lambda \ne \pm 1$ with $\lambda(A-A)=A-A.$  In my limited explorations there were no other examples of $\lambda(A-A)=A-A. $  I think that this is really the only way this could happen (if not, change case 2 to simply have $\lambda(A-A)=A-A$.)
Let $H$ be the multiplicative subgroup generated by $\lambda.$ Then $\lambda(A-A)=A-A$ would require that $A-A$ is a union of translates of $H$  (along with $0$). This might be the start of a proof.
I called the question a converse because we could ask if $B-B=\alpha(A-A)$ requires that $B$ is a translate of a dilation of $A$, but (since I restricted to $n$ prime) there is an inverse with $\alpha \beta=1$ and then $C-C=A-A$ for $C=\beta B.$

REPLY [2 votes]: N0. It is NOT the case that it is true in general. It is true for all cases with both  $m \le 5$ and prime $n \lt 100$. It is also true for $m=6$ and $n = 31$. But actually it is false  ( for $m=6$) over $\mathbb{Z}$ and also over $\mathbb{Z}/{n}\mathbb{Z}$ for any $n \ge 36$, prime or composite.
The sets $$A=\{0,  1, 4, 10, 12, 17  \} \text{ and } B=\{  0, 1, 8, 11, 13, 17\}$$  are Sidon sets with $A-A=B-B$ in $\mathbb{Z}$. So the only restriction on $n$ is that they are Sidon sets. Since the elements of $A-A=B-B$ range from $-17$ to $17$, they are Sidon sets with $A-A=B-B$ in $\mathbb{Z}/{n}\mathbb{Z}$ for any modulus, prime or not, starting with $n=35.$ It does not work for $n=32,33,34$. It does also work for $n=31.$ However for $n=31$ we have $B=11A.$ Also, for $n=35$ we have $B=17A-11.$ These are the only cases where an affine map takes $A$ to $B$.
I found this example looking in $\mathbb{Z}/{n}\mathbb{Z}$ but it turns out to be known.
I think it could be decided if there are any exceptions for $m=4,5$ (other then the special case $m=4$ and $n=13$) however the method seems tedious. Here is a sketch: We may assume that $A=\{0,1,a,b\}.$ If $B-B=A-A$ then $s-r=1$ for some $r,s\in B.$ We can translate to have $r=0,s=1$ so that $B=\{0,1,w,x\}.$ Further more, $w,x$ are elements of $A-A.$ Three of the 13 elements ($-1,0,1$) are ruled out, leaving $\binom{10}{2}=55$ possible cases. Many of these (like $w=-x$, $w=1-x$ and $w=x-1$) are immediately seen to be impossible. The rest come in pairs $\{0,1,w,x\}$ and $\{0,1,1-w,1-x\}$ which are reflections of each other.
If we try $B=\{0,1,1-a,a-b\}$ then $A-A$ and $B-B$ overlap in $9$ elements leaving $\{b,-b,1-b,b-1\}$ unmatched in $A-A$ and $\{1-2a+b,-1+2a-b,a-b-1,1-a+b\}$ in $B-B$ If $b=1-2a+b$ then $n=2a-1$ and $B$ is seen to be a translated reflection of A. We can't have  $b=1-a+b$ because $a  \ne 1.$ The other two possibilities for $b$ also fail.
But if we try $B=\{0,1,-a,b-1\}$ there are $6$ unmatched expressions in each of $A-A$ and $B-B$ and running through the possible matchups of pairs one case is $b=-a-1,a-1=1-b-a$ which comes out to $a=3,b=-4$ which leaves the remaining elements as $\pm2,\pm 4,\pm 7$ for $A-A$ and $\pm2, \pm4, \pm6$ for $B-B$. So this sub-sub-case is possible for, but only for, $n=13$ where, it can be checked, $B$ is a translated diolation of $A$.<|endoftext|>
TITLE: Topological Conformal Field Theories.
QUESTION [14 upvotes]: Hi there!
My question is simple, and I hope you don't misunderstand it. Why should one care about Topological Conformal Field Theories? What are the motivations to study it? I mean, for example: Topological Quantum Field Theories are important because they yield invariants of manifolds and non-trivial representations of the Mapping Class Group of a surface.
Thank you!

REPLY [16 votes]: The basic problem here is the naming rather than the object itself. The name TCFT as far as I understand indicates that it's a topological field theory, whose origin is in conformal field theory. Namely there is a construction (a "topological twist") starting from an $N=2$ supersymmetric conformal field theory in two dimensions, that produces a topological field theory (in fact two, the A- and B-twists). This is a special case of a general theory of topological twists of SUSY quantum field theories, which is by far the predominant source of topological quantum field theories as far as I know, including many of the most interesting ones coming from SUSY gauge theories in four dimensions (these are sometimes called "TFTs of Witten type" as opposed to the very rare Schwarz type, like Chern-Simons theory, which come with a manifestly topological formulation).
Now when we say "TFT" here it is at a more refined chain level than the classical Atiyah-Segal axiomatic definition --- a synonym for TCFT in the sense of say Costello's beautiful paper on the subject is differential graded TFT. This means roughly that the theory is topological on a derived level -- its outputs are topologically invariant up to coherent homotopies. (This is the kind of refined topological invariance one always gets out of twisting SUSY field theory.) 
What makes this very confusing initially is it seems conformal structures on a Riemann surface are playing an essential role: TCFT is defined in terms of chains on moduli spaces of complex structures.
However this is a red herring (unless you are interested in the CFT origin of the construction) -- the moduli of complex structures is just playing the role of a nice model for the classifying space $BDiff(\Sigma)$ of topological surfaces, and everything can be said purely topologically (as it is in Hopkins-Lurie's work on the Cobordism Hypothesis). So really we are just defining a TFT on the chain level, in families (ie universally over moduli of topological surfaces).
(This is a perspective I learned from Segal and Teleman and Freed and Costello, see Lurie's manuscript on TFTs for the contemporary perspective.)<|endoftext|>
TITLE: orderings of the field R((x, y))
QUESTION [6 upvotes]: I don't know much about the theory of ordered fields.  But I know that, for the real fields 
$\mathbb{R}(y)$, $\mathbb{R}((x))(y)$, and $\mathbb{R}((x))((y))$,
we can explicitly determine all the orderings of the field.
My question: Can we determine all the orderingds of the field $\mathbb{R}((x, y))$?
Can anyone give some brief explanations or a reference?

REPLY [2 votes]: A full description of the orderings of R((x,y)) is given in the paper
Alonso, M. E.(E-MADC); Gamboa, J. M.(E-MADC); Ruiz, J. M.(E-MADC)
On orderings in real surfaces.
J. Pure Appl. Algebra 36 (1985), no. 1, 1–14
In fact this paper describes all orderings of R[[x,y]] in terms of analytic half branches at the origin and the non-algebraic ones correspond to the orderings of R((x,y))<|endoftext|>
TITLE: A problem on convex geometry
QUESTION [13 upvotes]: Consider a convex body $K \subset \mathbb{R}^n$ containing the origin in its interior. Although the body is not necessarily symmetric, let us say that two points in its boundary $\partial K$ are antipodal if the origin lies in the segment that joins them.
Question 1. Assuming that $n$ is odd, does there always exist a pair of parallel affine hyperplanes that support $K$ at a pair of antipodal points?
If the boundary of $K$ is $C^1$ this is true and here is the simple proof: For every point $x \in \partial K$ consider the tangent hyperplane at $x$ and the tangent hyperplane at the point $\bar{x}$ antipodal to it. Translate this second hyperplane so that it passes through $x$ and consider its intersection with $T_x \partial K$. If the hyperplanes are not parallel, we obtain an $(n-2)$-plane tangent to $\partial K$ at $x$. If the tangent hyperplanes at antipodal points are never parallel, we get a continuous field of tangent hyperplanes in the tangent bundle of $\partial K$. Since $\partial K$ is an even-dimensional sphere, this is impossible. Q.E.D.
What happens if the boundary is not smooth and we allow some points to be points of support for multiple support hyperplanes?
Question 2. Assume $n$ is even. Does there exist some convex body $K$ (with a possibly smooth boundary) for which the support planes at antipodal points are never parallel?
These questions came up in a conversation with Constantin Vernicos who convinced me that we didn't really need the answers for what we were doing. Still, I remain curious about the subject.

REPLY [9 votes]: Since Petya answered question 2, let me answer question 1. Set $S=\partial K$ and assume at first that $S$ is smooth. Let $f,g:S^{n-1}\to S^{n-1}$, where $S^{n-1}$ is the unit sphere, be the maps that take an element $x$ to the outward unit normal to $S$ at the points of intersection of $S$ with the half-lines starting at the origin and going in the direction of $-x$ and $x$ respectively. These maps are homeomorphisms and they are homotopic to minus identity and identity respectively. So the map $x\mapsto g^{-1}(-f(x))$ has degree 1, and so it has a fixed point by the Lefschetz theorem (here we use the fact that $n$ is odd, so $n-1$ is even). Any such fixed point gives you a pair of opposite points on $S$ with parallel tangent planes.
upd: the above argument was essentially given in the posting. Let me explain how one can deduce the general case from it. One can approximate (e.g., with respect to the Hausdorff metric) an arbitrary compact convex $K$ with a sequence $(K_i)$ of smooth convex bodies that all have the origin in their interiors. For each $i$ let $a_i,b_i\in\partial K_i$ be opposite points with parallel tangent planes. By choosing a subsequence we may assume that there are $a=\lim a_i,b=\lim b_i$, which will again be opposite points in $\partial K$. By choosing a subsequence again we may assume that $\lim T_{a_i}\partial K_i$ and $\lim T_{b_i}\partial K_i$ also exist. These will be the required support planes.<|endoftext|>
TITLE: Does Bourbaki's (and Grothendieck's) approach to mathematics survive today?
QUESTION [5 upvotes]: I am curious if the "Bourbaki's approach" to mathematics is still a viable point of view in modern mathematics, despite the fact that Bourbaki is vilified by many.
Even more specifically, does anyone actively approach mathematics from the more "yielding" point of view famously practiced by Grothendieck? Which, or what type of, research areas are welcoming to (or practicing) Grothendieck's approach to mathematics?
Motivation:
To me, there is a deep question regarding motivation of mathematicians over time which is addressed by this viewpoint. An emphasis on resolving hard technical problems is quite depressing, generally, whereas the idea of finding a general framework which presents a natural and explanatory solution through the development of a vast theory seems very motivating. In such a view, the open problem only serves to motivate a better development of the general theory surrounding the core difficulty, bringing into focus a clearer picture of the essential issue at hand.
It seems to me that carefully developing a general (sometimes axiomatic) theory is analogous to performing scientific experiment. One is not looking to be clever, but instead is filling in data which may, when examined later, reveal clear and natural answers to mathematical questions. Obviously such an approach can be exhausting, in that one must spend much more time to fill in an entire picture than to, at some point, jump to a resolution of a particular question. On the other hand, It may be possible to persevere longer at such a task, as one is not so sensitive to one's loss of quickness or cleverness and can simply engage the task at hand.
Is this viewpoint valid?

[Edited (Dec. 17, 2012) by A. Caicedo, following suggestions here. Question originally asked by user
curious1.]

REPLY [13 votes]: The following appears in "Reminiscences of Grothendieck and his school", published in Notices of the AMS:

Bloch: I wonder whether today such a style of mathematics could exist.
Illusie: Voevodsky’s work is fairly general. Several people tried to imitate Grothendieck, but I’m afraid that what they did never reached that “oily” character dear to Grothendieck.

I am not completely sure what Illusie meant by "oily", but this seems to be a hint:

Illusie: To him no statement was ever the best one. He could always find something better, more general or more flexible. Working on a problem, he said he had to sleep with it for some time. He liked mechanisms that had oil in them. For this you had to do scales, exercises (like a pianist), consider special cases, functoriality. At the end you obtained a formalism amenable to dévissage.<|endoftext|>
TITLE: How does one prove that the complete  intersection of a quadric and a cubic of $\mathbb P^5$ is unirational?
QUESTION [15 upvotes]: The question is stated in the title, but I would like to add some motivation.
I've been teaching a course on complex tori and abelian varieties this semester and I would like to end it by showing some significant application of abelian varieties in algebraic geometry. I've come across a very beautiful  recent proof by Beauville that a certain specific sextic threefold as in the title is not rational and I have decided give an outline of it in my last lecture. Beauville refers to a paper of Enriques of 1912 for the proof of unirationality. I've got Enriques paper but I cannot make sense of it, so I'm looking either for another reference or for a sketch of proof.
Just in case it helps, here's  what I've been able to understand from Enriques' proof. Let $V_6=Q_2\cap C_3$ be the threefold, where $Q_2$ is a smooth quadric and $C_3$ a smooth cubic.  Let $P\in V_6=Q_2\cap C_3$ be a point. Then  $Q_2$ contains two families of planes through $P$, each parametrized by a $\mathbb P^1$. If $H$ is such a plane $H\cap C_3$ is a plane cubic $C_H$ and we can associate with $P$ the residual intersection $Q_H$ with $C_H$ of the tangent line to $C_H$ at $P$. As $H$ varies in one of the families of planes through $P$, $Q_H$ describes a rational curve $K$ in $V_6$. Of course, as $P$ varies, the curves $K$ fill up $V_6$. At this point, Enriques just claims that the curves $K$ thus defined ``correspond to the lines through a point in $\mathbb P^3$''.

REPLY [9 votes]: You can look at the short paper by Conte, Marchisio end Murre On the k-unirationality of the cubic complex (2007). 
It contains a proof of the unirationality of $V_6$ over a field $k$ of any characteristic $\neq 2,3$, under the assumption that $V_6$ has a $k$-rational point $p$ and that one of the two planes through $p$ on the quadric is also rational over $k$.
In the introduction, the authors write "we follow closely Enriques construction, our only contribution being to fully explain and justify his statements".<|endoftext|>
TITLE: Balls and bins variation
QUESTION [10 upvotes]: How many balls have to be thrown uniformly at random into $m$ bins, such that with high probability $n_1, n_2, \dots, n_m$ are distinct numbers, where $n_i$ is the number of balls in bin $i$ ?
Is there anything known about this problem? A trivial lower bound is $m(m-1)/2$, as we need $m$ distinct values.

REPLY [4 votes]: I think you have a good answer but I wanted to add a few computational observations. For $m=2$ it is obviously best to have an odd number of balls! But when $n=2k$ the chance of an even split is $$\frac{\binom{2k}{k}}{2^{2k}}\approx\frac{1}{\sqrt{\pi k}}$$ So figure out how high you want the probability of distinct bin numbers to be.
For $m=3$ it seems to actually be an advantage to have the number of balls be a multiple of three. There might be an easy explanation why,  but I don't see it. (But see the comments) Here are the chances of distinct bin numbers for $m=3$ and various $n$
$[20, .67531], [21, .75929], [22, .69015], [23, .69685], [24, .77016], [25, .70904], $$[121, .86695], [122, .86750], [123, .88141], [124, .86857], [125, .86909], [126, .88267], $$[1021, .95414], [1022, .95416], [1023, .95580], [1024, .95420], [1025, .95422], [1026, .95586]$$[3000, .97379], [3001, .97324], [3002, .97325]$
The last figures suggest to me that, with $m=3$ bins, IF you plan to throw in around $m_0=3000$ balls and then add $100$ more, one at a time THEN the chance of having a tie right after ball number $m_0$ is about $\frac{1}{40}$ but we would also expect (before throwing in the first ball) to have about 2 ties over the next $100$ balls. That reasoning may be too sloppy. There will be ties all along the way, they become less frequent but when they do occur it is probably a clump of several.
When the number of bins gets bigger the chances of a tie increase.<|endoftext|>
TITLE: Strange pattern in rounding errors?
QUESTION [6 upvotes]: This will look at first like a posting about trigonometry, then maybe about statistics, then finally about peculiarities of either

a certain random process; or
the pseudorandom number generator that I'm using; or
other (specify!).

My question is: Which is it?  And what's actually going on?  I suspect it's the second alternative, but I'm not at all confident about that.
In the course of doing a bit of amateur cartography, I derived this little trigonometric relation:
If $(\cos\alpha,\sin\alpha)$ is in the right half of the unit circle (in other words, $\cos\alpha>0$), and $$\tan\gamma=\dfrac{\sin\alpha\sin\beta}{\cos\alpha+\cos\beta},$$ and $\cos\gamma$ is also positive, then
$$\tan\dfrac\gamma2=\tan\dfrac\alpha2\cdot\tan\dfrac\beta2.$$
Numerical evidence bore out what I had derived, so now I should live happily ever after.
(And I was moderately intrigued by the resemblance to the simpler and more familiar tangent half-angle formula $\dfrac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}=\tan\dfrac{\alpha+\beta}{2}$.)
But then I asked what happens in the left half of the circle, where the cosine is negative.  The answer turns out to be
$$
-\cot\frac\gamma2 = \tan\frac\alpha2\cdot\tan\frac\beta2.
$$
[But see the "later note" below.]
But instead of deriving this by massaging trigonometric identities I got lazy and did some "experimental mathematics".  Using R, I entered these commands:
a <- pi/180*(runif(1000)*(177 - 93) +93)
b <- pi/180*(runif(1000)*(177 - 93) +93) 
c <- atan( sin(a)*sin(b)/(cos(a)+cos(b) ))
u <- -1/(tan(a/2)*tan(b/2))
coefficients(lm(tan(c/2) ~ u))

(Intercept)           u  
         0           1  
anova(lm(tan(c/2) ~ u))

Analysis of Variance Table 
Response: tan(c/2) 
           Df Sum Sq Mean Sq    F value    Pr(>F) 
u           1 29.222  29.222 2.1747e+34 < 2.2e-16 * 
Residuals 998  0.000   0.000 
--- 
Signif. codes:  0 ‘*’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1  
Warning message: 
In anova.lm(lm(tan(c/2) ~ u)) : 
ANOVA F-tests on an essentially perfect fit are unreliable 
plot(u,residuals(lm(tan(c/2) ~ u)))

So this puts $a$ and $b$ between $\pi/2$ and $\pi$, or more precisely, between $93\cdot\frac{\pi}{180}$ and $177\cdot\frac{\pi}{180}$, and chooses 1000 such pairs $(a,b)$ independently, and they're uniformly distributed in that region.  Then it sets $c=\arctan(\sin a\sin b/(\cos a+\cos b))$, and $u=-1/(\tan a\tan b)$.  Then we look at coefficients from a simple linear regression of $\tan(c/2)$ on $u$, and the software reports $0$ for the intercept and $1$ for the slope.  An analysis of variance gives $0$ as the sum of squares of residuals, so it seems we have a perfect fit.
Finally, I plotted $u$ on the horizontal axis and the residuals on the vertical axis, and I got the following!  If real numbers rather than approximations could be used, they would of course all be $0$, so this is about rounding errors, but still I wouldn't expect to see a pattern like this.  I tried it a dozen or so times with pretty much the same result, and I tried it with the angles in the first quadrant and the identity that holds in that quadrant, with the same result again.

LATER NOTE: Well, haste makes waste, I guess.  I should have let $\gamma$ be the "other value of" the arctangent once I moved into the second quadrant, i.e. $\gamma=\arctan(\cdots\cdots\cdots)+\pi$ as soon as the argument to the arctangent function was more than $\pi/2$.  That way we still have the identity $\tan\frac\gamma2=\tan\frac\alpha2\cdot\tan\frac\beta2$.  However, this doesn't upset the main point of this question.  As I said, this already works in the first quadrant; I simply hadn't yet noticed it because at that point I was still doing things intelligently rather than numerically.

REPLY [3 votes]: Apparently it's something to do with the "residuals" function in R.
If you do this 
h <- fitted.values(lm(tan(c/2)~u))
plot(u,h-tan(c/2),ylim=c(-2e-16,2e-16),cex=0.1)

instead (with the previous code being the same) you get the following:
 (source)<|endoftext|>
TITLE: Is there an algebraic "derived mapping space" construction that encompasses both Hochschild homology and loop spaces of non-simply-connected spaces?
QUESTION [15 upvotes]: I'm looking for directions to the literature that might contain fairly explicit constructions that might be called (the algebra of functions on) the "derived mapping space" from a simplicial set to a simplicial (affine) scheme.  To make the question reasonably self-contained and to give a sense of my background and current understanding, I will begin with some general abstract nonsense, and then point to a construction that I have found in the literature and that does not work to my satisfaction.
A little abstract nonsense
Let $C$ and $D$ be categories.  In a bit, I will give them specific values, but for now I will ask only that $D$ be small, and that $C$ have any necessary limits and colimits.  A (generalized) $D$-object in $C$ is a presheaf on $D$ valued in $C$, i.e. a functor $X : D^{\mathrm{op}}\to C$.  Each $d\in D$ determines (and is determined by) a $D$-object in $\mathrm{SET}$, by the usual Yoneda embedding $d \mapsto \operatorname{hom}_D(-,d)$.  It will be convenient for me to denote the presheaf $\operatorname{hom}_D(-,d)$ by $[d]$, and given $k\in D$ and $X : D^{\mathrm{op}}\to C$, I will write $X_k$ for $X(k)$.
For $x\in C$ and $s\in \mathrm{SET}$, there is an object $\operatorname{maps}(s,x) = x^s = \prod_s x \in C$, which is the $s$-fold cartesian product of $x$ with itself.  Now, let $X : D^{\mathrm{op}} \to C$ be a $D$-object in $C$, and $S: D^{\mathrm{op}} \to \mathrm{SET}$ a $D$-set.  Then there is an object $\operatorname{hom}_D(S,X) \in C$, which is built as a certain limit ranging over the objects $\operatorname{maps}(S_k,X_k)$ for $k\in D$.  Even better, the categories of $D$-sets and $D$-objects in $C$ have products — the ("categorical") cartesian product of functors is constructed by taking the product for each — and so we can define an enriched hom by:
$$ \underline{\operatorname{hom}}_D(S,X) : D^{\mathrm{op}} \to C, \quad d \mapsto \operatorname{hom}_D(S \times [d],X). $$
Finally, there is one more, much more naive "mapping space" between $D$-objects, which I will denote by $\operatorname{maps}(S,X) : D \times D^{\mathrm{op}} \to C$, sending $(d,k) \mapsto \operatorname{maps}(S_d,X_k)$.
A little concrete nonsense
I will be interested in the situation where $D = \Delta$ is the category of finite nonempty totally-ordered sets (and monotonic maps).  It has a skeletalization with objects indexed by the natural numbers, given by $[n] = \lbrace 0 < \dots < n \rbrace$.  Note that the $\Delta$-set $[0]$ is terminal, so $\underline{\operatorname{hom}}_D(S,X)_0 = \operatorname{hom}_D(S,X)$.  An object $X : D^{\mathrm{op}} \to C$ determines, among other data, two maps $X_1 \rightrightarrows X_0$, corresponding to the two inclusions $[0] \rightrightarrows [1]$.  By definition, $\pi_0(X) \in C$ is the coequalizer of the two arrows $X_1 \rightrightarrows X_0$.
Fix a commutative ring $\mathbb K$.  I will not be upset if you would like to make further assumptions on $\mathbb K$, e.g. that $\mathbb K \supseteq \mathbb Q$, or that $\mathbb K$ is an algebraically closed field.  I believe that I am primarily interested in the following two values for $C$, but I am open to being convinced otherwise:

$C = \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$ is the category of affine schemes over $\mathbb K$.
$C = \mathrm{Mod}_{\mathbb K}$ is the category of $\mathbb K$-modules.

There is a well-known contravariant forgetful functor $\mathcal O$ from $\mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$ to $\mathrm{Mod}_{\mathbb K}$.
I will also take inspiration from the case $C = \mathrm{Top}$ of nice enough topological spaces.
There is a further functor $\operatorname{ch}: \Delta\mathrm{Mod}_{\mathbb K} \to \mathrm{DGMod}_{\mathbb K}$ (the category of homologically-graded chain complexes of $\mathbb K$-modules) which sets $\operatorname{ch}(X)_k = X_k$ with differential a certain well-known alternating sum.  There is a standard symmetric monoidal structure on $\mathrm{DGMod}_{\mathbb K}$ which sums the homological degrees, and for this structure $\operatorname{ch}$ is not strongly monoidal, but there is a canonical Eilenberg–Zilber map $\operatorname{ch}(X) \otimes \operatorname{ch}(Y) \to \operatorname{ch}(X \otimes Y)$, which sums over all $(k+\ell)$-simplices in a product of a $k$-simplex with an $\ell$-simplex (closely related is the fact that for simplicial sets, the geometric realization of a product is homeomorphic to the product (in the category of compactly-generated spaces) of geometric realizations), making $\operatorname{ch}$ into a "lax symmetric monoidal functor".  The Eilenberg–Zilber map is a quasi-isomorphism, and one choice of quasi-inverse is the (non-symmetric) Alexander–Whitney map; if $\mathbb K \supseteq \mathbb Q$, there are other more symmetrical choices.  In any case, the Eilenberg–Zilber map means that any simplicial commutative algebra determines canonically a dg commutative algebra.  Of course, when $C = \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$, a simplicial affine scheme $X : \Delta^{\mathrm{op}} \to \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$ determines a cosimplicial commutative algebra $\mathcal{O}(X)$, and so $\operatorname{ch}(\mathcal{O}(X))$ is not quite a dgca (the Alexander–Whitney map makes it into a dga).  Anyway, this all won't matter much for me.

What I wanted to mention about all this is that if $X : \Delta^{\mathrm{op}} \to \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$ is a simplicial affine scheme, then $\operatorname{H}_\bullet(\operatorname{ch}(\mathcal{O}(X)))$ is canonically a graded commutative algebra (supported in nonpositive homological degrees; and there is more algebraic data in the form of Massey products) and
$$ \operatorname{H}_0(\operatorname{ch}(\mathcal{O}(X))) = \mathcal{O}(\pi_0(X)). $$

Examples
Let $A$ be a commutative $\mathbb K$-algebra, with corresponding affine scheme $X = \operatorname{spec}(A) \in \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$.  If you want, you can extend $X$ to a constant functor $X : \Delta^{\mathrm{op}} \to \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$.  Let $S^1$ denote the simplicial set generated by one nondegenerate $0$-simplex and one nondegenerate $1$-simplex.  Then $\operatorname{maps}(S^1, X)$ is a cosimplicial affine scheme (or simplicial cosimplicial, but constant in the simplicial direction), and so $\mathcal{O}(\operatorname{maps}(S^1, X))$ is a simplicial commutative algebra.  By definition,
$$ \operatorname{HH}_\bullet(A) = \operatorname{H}_\bullet(\operatorname{ch}(\mathcal{O}(\operatorname{maps}(S^1, X)))) $$
is the Hochschild homology of $A$.  The complex $\operatorname{ch}(\mathcal{O}(\operatorname{maps}(S^1, X)))$ can be alternately defined by making a certain choice of resolution of $A$ as an $(A\otimes A)$-module, and using this resolution to construct the derived tensor product $A \otimes_{A\otimes A} A$.

Let $G$ be an affine algebraic group over $\mathbb K$ (e.g. a finite group).  There is a well-known simplicial affine scheme $X = \mathrm{B}G$ whose space of $k$-simplices is $G^k$, with boundary maps that encode the multiplication.  Let $M$ be a simplicial set, and I am primarily interested in the case that $M$ is a simplicial finite set describing the homotopy type of a finite-dimensional compact manifold.  The simplicial affine scheme $\underline{\operatorname{hom}}_\Delta(M,\mathrm{B}G)$ is the space if $G$-local systems on $M$.  In particular, $\pi_0(\underline{\operatorname{hom}}_\Delta(M,\mathrm{B}G))$ is the character variety of $M$.

My question
I am looking for a general construction, of the flavor above, that incorporates both examples.  More specifically, the construction should:

input a simplicial (finite) set $M$ and a a simplicial affine scheme $X$ over $\mathbb K$
output a chain complex $V(M,X)$ over $\mathbb K$, supported in both directions, that deserves to be thought of as a "derived space of global functions on the space of maps from $M$ to $X$"
have good functoriality and monoidality properties in both variables (implying for instance that $V(M,X)$ has a strongly-homotopy commutative dg algebra structure, coming from various diagonal and Eilenberg–Zilber-like maps)
if $X = \operatorname{spec}(A)$ is a constant simplicial scheme, then $V(M,\operatorname{spec}(A))$ is the generalized Hochschild homology of $A$ determined by $M$
$\operatorname{H}_0(V(M,X)) = \mathcal{O}(\pi_0(\underline{\operatorname{hom}}_\Delta(M,X)))$

Some near misses
The problem seems to be when $X$ is not "simply connected".  In particular, I have not come across a construction that works even when $X = \mathrm{B}G$ for $G$ a finite simple group.
Greg Ginot and collaborators (see e.g. Higher order Hochschild cohomology, Derived Higher Hochschild Homology, Topological Chiral Homology and Factorization algebras, and A Chen model for mapping spaces and the surface product) have extended work by Pirashvili defining the generalized Hochschild homology.  Let $A$ be a cdga over $\mathbb K \supseteq \mathbb Q$ and let $M$ be a simplicial set.  Then there is a simplicial cdga $\int_M A = \mathcal{O}(\operatorname{maps}(M,\operatorname{spec}(A)))$ with good functoriality and monoidality properties, which agrees up to quasi-isomorphism with Lurie's "topological chiral homology."  
By definition, a quasi-isomorphism of cdgas is a morphism that induces isomorphisms on homology.  One of the things that Ginot et al prove is that a quasi-isomorphism $A \to B$ induces a quasi-isomorphism $\int_M A \to \int_M B$.  Thus in particular when $A = \mathcal{O}(\mathrm{B}G) = \operatorname{Ext}_G(\mathbb K,\mathbb K)$, for any meaning of this, and $G$ is a finite simple group, then the canonical map $\mathbb K \to A$ is a quasi-isomorphism, and so the chain complex $\int_M A$ will never contain data.  So this construction fails my last condition, e.g.: $\pi_0(\underline{\operatorname{hom}}_\Delta(S^1,\mathrm{B}G)) = G/G^{\mathrm{conj}}$ and  $\mathcal{O}(\pi_0(\underline{\operatorname{hom}}_\Delta(S^1,\mathrm{B}G))) = \mathcal{O}(G)^G$ is the algebra of class functions on $G$, whereas $\operatorname{H}_0(\int_{S^1}\mathcal{O}(\mathrm{B}G)) = \mathbb K$.
Ben-Zvi and Nadler have discussed loop spaces and connections their relationships to Hochschild homology and representations.  They run into what I believe are related issues, but work primarily with not the space of loops in a derived scheme, but rather the infinitesimal neighborhood of the constant loops within that space.  I should also mention that for my particular application, I really am looking for an explicit one-categorical construction (akin to the Pirashvili-style work), rather than quickly moving to model or $\infty$ categories.
Finally, perhaps the result I should have started with is one I learned from a review by Loday (original references are included there).  Suppose that $M$ is a simplicial approximation of an $n$-dimensional manifold, and that $X$ is a simplicial set which is $n$-connected, in $\pi_{\leq n}(X)$ is trivial.  (So $1$-connected means connected simply-connected.)  I can build a cosimplicial simplicial set $\operatorname{maps}(M,X)$, and a simplicial set $\underline{\operatorname{hom}}_\Delta(M,X)$, as discussed above.  There is a _free $\mathbb K$-module_ functor $\mathbb K : \mathrm{SET} \to \mathrm{Mod}_{\mathbb K}$, and with is I get a cosimplicial simplicial $\mathbb K$-module $\mathbb K\operatorname{maps}(M,X)$ and a simplicial $\mathbb K$-module $\mathbb K\underline{\operatorname{hom}}_\Delta(M,X)$.  Of course, given a cosimplicial simplicial $\mathbb K$-module, I can apply the "alternating sum of boundaries" functor $\operatorname{ch}$ to get a bicomplex, which I can then totalize.  Unless I have made a mistake, I believe the statement is that under the conditions on $M$ and $X$, the canonical map of chain complexes between $\operatorname{ch}(\mathbb K\operatorname{maps}(M,X))$ and $\operatorname{ch}(\mathbb K\underline{\operatorname{hom}}_\Delta(M,X))$ is a quasi-isomorphism.  This example specifically does not include classifying spaces of finite groups.
 Final comments and examples
Truth be told, I am most interested in the case $X = \mathrm B G$ for $G = \mathrm{SL}(2)$ and $M$ a simplicial approximation of a three-manifold.  Note that the topological space $\mathrm{B}(\mathrm{SL}(2,\mathbb C))$ is $3$-connected (it is homotopy equivalent to $\mathrm{B}(\mathrm{Spin}(3,\mathbb R))$), but I don't have a good sense about notions like "3-connected" for algebraic stacks.  And, besides, I would like a robust construction.
Let me end with an example that does work.  A much easier category than $\mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$ is the category $\mathrm{CCog}_{\mathbb K}$ of cocommutative coalgebras (or "cogebres" in French, hence the name).  A group object in $\mathrm{CCog}_{\mathbb K}$ is a cocommutative Hopf algebra, and a good example is the universal enveloping algebra $U\mathfrak g$ of a Lie algebra $\mathfrak g$.  Then $\mathrm B U\mathfrak g = \mathrm B \mathfrak g$ is a simplicial cocommutative coalgebra, which is quasi-isomorphic to the dg cocomutative coalgebra $\mathrm{CE}(\mathfrak g)$ of Chevalley–Eilenberg cochains with trivial coefficients.  I believe that it _is_ true that Hochschild homology of $\mathrm{CE}(\mathfrak g)$ is the Chevalley–Eilenberg cochain complex with coefficients in $U\mathfrak g$, as it should be if you think about loop spaces.

REPLY [12 votes]: There is a standard answer to your question, unless I'm missing something, that I learned from the great survey by Toen and satisfies everything you want. However to satisfy your requirements - ie to get the correct Hochschild complex - you must (explicitly or implicitly) embed simplicial schemes into a larger world of derived algebraic geometry, ie change from rings to simplicial rings (cosimplicial affine schemes) or some variant (homological cdgas in characteristic zero for example).
The construction is "the obvious one" - given any finite simplicial set think of it as a constant functor on your category of affines and construct the mapping space into your given target $X$.
Then pass to functions on the mapping space $X^M$, ie (derived) global sections of the structure sheaf.
When $M=S^1$, then say $BG^{S^1}=G/G$ the adjoint quotient stack. For any (derived) ring $R$ we have 
that functions on $Spec(R)^{S^1}$ is Hochschild chains of $R$ (very close to the definition).
Much stronger things are true about these loop spaces, eg for any perfect stack (a very large class of well behaved stacks with affine diagonal map, eg any quasicompact separated scheme) and any finite simplicial set we have $QC(X^M)\simeq QC(X)\otimes M$ ($QC$=quasicoherent sheaves).
$BSL_2$ is such a perfect stack (in characteristic not 2 I think) and $BSL_2^M$ is a great object to consider, with $QC(BSL_2^M)$ being functions on the derived stack of $SL_2$ local systems on your space $M$. Not sure what else you want to hold but this is a very robust nice functorial monoidal construction which has already seen quite a lot of applications.<|endoftext|>
TITLE: Rational orthogonal matrices
QUESTION [11 upvotes]: ``everybody knows'' that an integral orthogonal matrix is a signed permutation matrix, so there are exactly $2^n n!$ such matrices in $O(n).$ Now, what if we ask for the enumeration of elements of $O(n$ with rational elements, where all the denominators are equal to $q$ (where $q$ is an arbitrary integer, though if the question is much easier for $q$ prime, that's fine too...)? I would suspect this has been studied... (one can ask the same question for arbitrary algebraic groups/number fields, of course).

REPLY [3 votes]: Boris Venkov asked this question, once, in the coffea room of the Maths Department of the university of Geneva. 
Here are some things that relate to the case $q=2^m$.
Let $I_n$ be the 'euclidean' $\mathbf Z$-module of rank $n$, and $A\mapsto G_n(A)$ the group it defines.
Let us write $a(n,m)$ for the number of matrices in $G_n(\mathbf Z[\frac{1}{2}])$ all of whose coordinates have $2$-valuation at least $-m$. 
First note that $G_n(\mathbf Z[\frac{1}{2}])$ is finite for $n<5$.
In fact, one even has $G_n(\mathbf Z[\frac{1}{2}])=G_n(\mathbf Z)\simeq\mathbf Z/2^n\rtimes S_n$ for $n<4$ and $G_4(\mathbf Z[\frac{1}{2}])=Aut(D_4)=G_4(\mathbf Z)\rtimes \mathbf Z/3$, where $D_4$ denotes the sublattice of elements of even length in $I_4$. The $\mathbf Z/3$ is generated by 
$$ \begin{bmatrix}
-1/2 & 1/2 & 1/2 & 1/2\\\
-1/2 & 1/2 & -1/2 & -1/2\\\
-1/2 & -1/2 & 1/2 & -1/2\\\
-1/2 & -1/2 & -1/2 & 1/2
\end{bmatrix}$$
Thus in this case, you have $a(4,0)=384=2^7*3$ and $a(4,1)=1152=2^7*3^2$.
Next, note that $G_5(\mathbf Z[\frac{1}{2}])$ is an amalgamated sum : $A\star_C B$ with $A=G_5(\mathbf Z)$, $B=Aut(I_1)\times Aut(D_4)$ and $C=A\cap B\simeq \mathbf Z/2\times G_4(\mathbf Z)$. 
This decomposition comes from the action of $G_5(\mathbf Z[\frac{1}{2}])$ on the Bruhat-Tits Building of $G_5(\mathbf Q_2)$ : a tree whose vertices are 5-valent and 3-valent.
The number $\frac{a(5,m)}{\vert G_5(\mathbf Z)\vert}$ counts the $5$-valent vertices that are at (combinatorial) distance less than $2m$ from a fixed one. Thus 
$\frac{a(5,m)}{\vert G_5(\mathbf Z)\vert}=10*\frac{8^m-1}{7}+1$.
Things are becoming more complicated in higher dimensions since the Bruhat-Tits building is not a tree anymore ($n\geq 6$) and the action of $G_n(\mathbf Z[\frac{1}{2}])$ on the vertices of a same type (say corresponding to unimodular lattices) is not transitive anymore ($n\geq 9$).
You may adapt the same method to obtain similar results for small values of $n$ and $q=p^m$ : the Bruhat-Tits building is a tree for $n=3$, the action of $G_3(\mathbf Z[\frac{1}{p}])$ is transitive on one type of vertices (but rarely on maximal simplices) ...<|endoftext|>
TITLE: What graph parameters are determined by parameters for strongly regular graph
QUESTION [7 upvotes]: Say two graphs are not isomorphic but are both strongly regular with the same set of parameters.  Are there any parameters (other than the usual such as order, degrees, eigenvalues and multiplicities, etc.) that are determined, e.g., independence number, chromatic number, etc.?
Thanks for any help

REPLY [3 votes]: The number of cycles of length 3,4,5 are determined.  If the girth is 4, the number of 6-cycles is determined too.<|endoftext|>
TITLE: Is a reductive adelic group a Type I group?
QUESTION [19 upvotes]: I foresee that to experts of automorphic forms this question will sound unimportant or useless or even not worthy of an answer; but none of these are going to stop me from asking it! 
The question is simple: let $G$ be a reductive (or even semisimple) algebraic group over $\mathbb Q$. Is it true that the adelic group $G(\mathbb A)$ is of Type I (i.e., direct integral decomposition of its unitary representations is unique)? And if the answer is negative, then how do people actually get around it in the study of multiplicity of automorphic forms in $L^2(G(\mathbb Q)\backslash G(\mathbb A))$?

REPLY [3 votes]: Freitag and van Dijk proved that the adelic points of a reductive group over a global field is trace class (Theorem 2.3), while every trace class group is of type I (Theorem 1.7). So the answer is yes (again).<|endoftext|>
TITLE: Mathematician trying to learn string theory
QUESTION [56 upvotes]: I'm a mathematician. I want to be able to read recent ArXiv postings on high energy physics theory (String theory) (and perhaps be able to do research). I want to understand compactifications, Dualities, D-branes, M-branes etc. What's the easiest way to do so provided I have the necessary knowledge in algebraic geometry, algebraic topology, analysis and differential geometry?

REPLY [16 votes]: I want to understand compactifications, Dualities, D-branes, M-branes
etc.

What makes all this hard to learn in a systematic way is that the theory itself is still incomplete and proceeds in parts via educated guesswork.
In principle perturbative string theory is well defined: This says to pick a 2d super-conformal field theory of central charge 15, collect its n-point functions into one formal power series, interpret this as a loop-wise finite, hence normalized, scattering matrix known from QFT, then study this.
In principle "compactifications, Dualities, D-branes" (though not M-branes) are all well defined parts of this scheme. For instance D-branes are the algebraic data that defines the 2d SCFT at the boundary of the Riemann surfaces, Dualities are correspondences between different 2d SCFTs that miraculously yield isomorphic scattering matrices this way, and compactifications are decompositions of a 2d SCFT as a tensor product with some "finite" factor for instance a rational CFT. The target spacetime geometry that is being compactified thereby is to be read off from the CFT in a way generalizing how spectral Connes-style NCG reads off target geometry from algebraic data. (See the exposition at Spectral Standard Model and String Compactifications)
The problem at this point is that, while well defined, it is mathematically so hard. The only 2d CFTs that have mathematically been constructed as full CFTs are the "rational" ones, which only describe certain compact factors. For the others there is Segal's axiomatics, but essentially no examples. What one has is the local description in terms of vertex operator algebras. For these many examples are available, but not enough for the purposes of the physicists.
This is the first point where physics parts with a systematic mathematical development. Namely while it is hard to really construct full 2d SCFTs, the folk lore of the path integral allows to think of solutions to supergravity equations of motion as inducing 2d SCFTs via quantization of the "nonlinear sigma-model" with these spacetimes as their targets. So now instead of deducing effective target space geometry from the SCFT algebra, one prescribes classical target space geometry to which one believes SCFTs may be associated. It is from here on that much of string theory is now phrased in terms of classical geometry with some extra stringy effects sprinkled in (modular invariance, anomaly cancellation, brane instanton effects).
The uncertainty as to how well this sigma-model construction is under control is the cause of the discrepancy in the perception of how many string vacua are known: A perturbative string vacuum is equivalently that 2d SCFT which gives the scattering matrix, so in principle the "landscape of perturbative string vacua" is the moduli space of 2d SCFTs. But since this is not understood, what people instead scan is the space of target space geometries that are thought to induces 2d SCFTs. This is a subtle business, where one imagines one may incrementally approximate that 2d SCFT by adding alpha-prime corrections, cancelling anomalies, etc. Therefore one finds authors who worry that too many string vacua are known, and other authors who worry that too few string vacua are known.
If this were all there is to it, the solution would in principle be straightforward: mathematicians/mathematical physicists should simply sit down and find means to rigorously construct 2d SCFTs and to understand the moduli space that they form. That would be the mathematics of perturbative string theory, and all the answers as to "compactifications, Dualities, D-branes" would be encoded in there, under some well defined dictionary.
But now there is also the "M-branes", and that's indicative of the real problem: Since the string perturbation series is just a non-converging formal power series, as for a normalized perturbative QFT, it ought to be but the Taylor expansion of some non-perturbative theory about some points of its cofiguration space.
There are many compelling hints for this "theory formerly known as String" but so far they are just hints.

We still have no fundamental formulation of “M-theory” - the
hypothetical theory of which 11-dimensional supergravity and the five
string theories are all special limiting cases. Work on formulating
the fundamental principles underlying M-theory has noticeably waned.
[...]. If history is a good guide, then we should expect that anything
as profound and far-reaching as a fully satisfactory formulation of
M-theory is surely going to lead to new and novel mathematics.
Regrettably, it is a problem the community seems to have put aside -
temporarily.
But, ultimately, Physical Mathematics must return to this grand issue.
(G. Moore, "Physical Mathematics and the Future", talk at Strings 2014)

The way this is presently being studied is the same mix of classical target spacetime geometry with conjectured "extra effects" sprinkled in. For instance a "black M-brane" is to first approximation a solution to the equations of motion of 11d supergravity which preserves half of the global supersymmetry. By analogy with the D-branes and some other arguments, coincident such M-branes should "carry" a non-abelian gauge SCFT on their worldvolume. There is presently no way to derive this from anything, but superconformal invariance imposes enough constraints that a classical action functional could finally be guessed (the BLG/ABJM model). Now the search is on for the analogue on the M5-brane. Nothing definite is known, there are hints, and whatever one finds will justify itself by plausibility arguments. Because none of this can be derived from first principles
There are presently no first principles for full string theory, aka M-theory. There is no mathematics of M-theory to be learned. Instead, the mathematics of M-theory is waiting to be found.
This may not be as bad as it may sound. Maybe "M-theory" is easier to deduce following mathematical principles, than the historical route of the perturbtive string. I just wrote an exposition of such a point of view over at PF-Insights:
Super p-Brane Theory emerging from Super Homotopy Theory
So it seems not out of the question that, conversely, it will in the end be the physicists who will learn M-theory from the mathematicians.<|endoftext|>
TITLE: On multi-dimensional real trees
QUESTION [5 upvotes]: A real tree is a metric space $(M,d)$ satisfying the following two conditions: 
(1) for every $x,y\in M$, there is an unique isometry $\phi$ from the closed interval $[0,d(x,y)]$ onto $M$ such that $\phi(0)=x$ and $\phi(d(x,y))=y$; and
(2) any one-to-one continuous mapping $f:[0,1]\rightarrow M$ has the same range as the isometry $\phi$ associated to the points $x=\phi(0)$ and $y=\phi(1)$.
Metric spaces that can be isometrically embedded into a real tree are exactly the ones who satisfy the four point condition, that is, metric spaces $(N,\rho)$ such that 
$$\rho (a,b)+\rho (c,d)\leq max (\rho (a,c)+\rho (b,d),\rho (a,d)+\rho (b,c)) ,a,b,c,d\in N.$$ 
My question is: what is the most natural/useful way of defining an n-dimensional version of a real tree? Is there any hope to obtain a nice characterization of subspaces, in the spirit of the four point condition?
I would be please to see an answer to this question even for two dimensions.

REPLY [4 votes]: See the paper "Rigidity of quasi-isometries for symmetric spaces and Euclidean buildings" MR1608566 by Kleiner and Leeb for a theory of $\mathbb{R}$-buildings that generalizes the theory of $\mathbb{R}$-trees. They use this theory for classifying, up to quasi-isometry, all symmetric spaces of noncompact type whose deRham factors all have rank $\ge 2$. I do not believe they address any characterization such as the 4 point condition.<|endoftext|>
TITLE: How do you compute the primes of bad reduction?
QUESTION [7 upvotes]: Suppose that I am given a subscheme $Y$ of $\mathbf{P}^n_{\mathbf{Z}}$, flat over $\operatorname{Spec}\mathbf{Z}$ and with smooth generic fiber $Y_{\mathbf{Q}}$, defined by the vanishing of some homogeneous polynomials
$$
F_1, \ldots, F_k \in \mathbf{Z}[X_0,\ldots,X_n].
$$
How does one determine the set $S$ consisting of primes $p$ such that the fiber $Y_{\mathbf{F}_p}$ is non-smooth? 
The way I'm going about the problem seems a bit crude to me, and has some problems. In particular, I can't prove that my algorithm solves the problem.
For simplicity suppose that $k=1$, and write $F=F_1$. In a nutshell, I compute the derivatives $f_i := \frac{\partial F}{\partial X_i}$, and determine an integer $N$ such that $N$ is contained in the ideal $I = (F, f_1, \ldots, f_n) \subset \mathbf{Z}[X_0,\ldots,X_n]$  by repeatedly taking resultants of polynomials in $I$. Then at least for all $p$ not dividing $N$, the fiber $Y_{\mathbf{F}_p}$ is smooth, since for such $p$ the ideal $I_p \subset \mathbf{F}_p[X_0,\ldots,X_n]$ generated by the reduced polynomials $\widetilde{F}$, $\widetilde{f}_1$,$\ldots$,$\widetilde{f}_n$ contains $1$.
One problem is that it's quite expensive computationally. I'm using resultants to eliminate the variables $X_i$ one by one, say in the order $i=0,1,\ldots,n$. Then every time I've eliminated one of the $X_i$s, I get a lot of polynomials that don't contain $X_0$ up to $X_i$, and I take them all into the next elimination round. (You have to be careful when the resultant comes out $0$, and perhaps there are some other subtleties that I'm forgetting, but this is basically it.) 
Another problem with this is that the final outcome seems to depend on the order in which I take the $i$s. (This may have something to do with singularities being located at infinity with respect to one of the $X_i$s [edit: on second thought, this doesn't make any sense; see comment by François], but I really can't see the geometry of what I'm doing clearly enough to be confident about this.) And I don't see why my method necessarily gives me the whole $S$ - instead of some upper bound on it - even when all orderings of the $X_i$ are taken into account. 
My questions: 

Does my method give the whole set $S$? 
If yes, why? If no, what method does work?
Regardless of the answers to 1. and 2., is there a faster or more natural way of finding the primes of bad reduction?

REPLY [4 votes]: Assume for simplicity that $Y$ has pure relative dimension $d$. By considering the standard affine cover of $\mathbf{P}^n_{\mathbf{Z}}$, one easily reduces to the case where $Y=V(F_1,\ldots,F_k)$ is a closed subscheme of $\mathbf{A}^n_{\mathbf{Z}}$. Then the special fiber $Y_p = V(F_{1,p},\ldots,F_{k,p}) \subset \mathbf{A}^n_{\mathbf{F}_p}$ is smooth if and only if for every $x \in Y_p(\overline{\mathbf{F}}_p)$, the Jacobian matrix $(\frac{\partial F_{i,p}}{\partial X_j}(x))$ has rank $n-d$. Note that $Y_p$ is $d$-dimensional by the flatness assumption, so all geometric tangent spaces have dimension $\geq  d$, which implies that the rank of the Jacobian matrix is everywhere $\leq n-d$. Thus $Y_p$ is smooth if and only if the ideal generated by $F_{1,p},\ldots,F_{k,p}$ together with all $(n-d)$-minors of the Jacobian matrix is equal to $\mathbf{F}_p[X_1,\ldots,X_n]$.
Now consider the ideal $I$ of $R=\mathbf{Z}[X_1,\ldots,X_n]$ generated by $F_1,\ldots,F_k$ together with all $(n-d)$-minors of the Jacobian matrix. Since the generic fiber of $Y$ is smooth, we have $I \cap \mathbf{Z}=N\mathbf{Z}$ for some integer $N \geq 1$, and the prime factors of $N$ are precisely the primes of bad reduction of $Y$.
Proof. If $p$ doesn't divide $N$ then $(I,p)$ contains $N$ and $p$, so $(I,p)=R$ and $Y_p$ is smooth. If $p$ divides $N$, write $N=p^k M$ with $p$ not dividing $M$. If we had $(I,p)=R$, then we would also have $(I,p^k)=R$. Then $M \in (MI,p^k M)=(MI,N) \subset I$, a contradiction. Thus $Y_p$ is not smooth.
You can compute the integer $N$ using Gröbner bases for polynomials over $\mathbf{Z}$, see e.g. http://magma.maths.usyd.edu.au/magma/handbook/text/1112#12189 for the Magma implementation. I'm not expert in Gröbner bases, so I would appreciate if someone could confirm whether it is always possible to find $I \cap \mathbf{Z}$ when a set of generators of $I$ is given, and whether the Magma implementation works in all cases.<|endoftext|>
TITLE: Simple lower bounds for Bell numbers (number of set partitions)?
QUESTION [5 upvotes]: The $n$-th Bell number $B_n$ represents the number of distinct partitions of a set with $n$ distinguished elements.
It can be expressed as the infinite sum $B_n = (1/e)\sum_{k=1}^{\infty} (k^n/k!)$, which is also the $n$-th moment of a Poisson distribution with mean $1$.
The first few values are known precisely; the Bell numbers form OEIS sequence A000110.  There are also several asymptotic expressions, but for an application I need lower bounds.
Write $\log x$ for $\log_2 x$.

For $n \ge 5$, is it true that $B_n \ge (n/\log n)^n$?

Denote the set with elements $1,2,\dots,n$ by $[n]$.
Since a partition of $[n]$ has at most $n$ blocks (equivalence classes), each partition of $[n]$ can be obtained via some function $f$ mapping $[n]$ to $[n]$, by regarding $f(i)$ as a name for the partition containing the number $i$.  Many different names are possible for the same partition, so $B_n < n^n$, as an easy but crude upper bound.
It is possible to show tighter bounds.  For convenience, when $n \ge 2$ we can express $B_n$ in terms of another sequence $c_n$ as $B_n = \left(\frac{c_n n}{\log n}\right)^n$.
It seems that $\log c_n \ge -1.5$ for all integers $n \ge 2$, again by just counting functions (although a slightly more involved argument is required than for the trivial upper bound).  Moreover, for any $\epsilon > 0$, this argument then also shows that $\log c_n \ge -(1+\epsilon)$ for all large enough $n$ (where the threshold for $n$ grows as $\epsilon$ becomes smaller).
By a result of Berend and Tassa, it already follows that $\log c_n < 0.1924$ for all positive $n$, and they state that from an asymptotic argument of de Bruijn it follows that $\log c_n > -0.914$ for all large enough $n$.
My question is then whether the stronger bound $\log c_n \ge 0$ holds for $n \ge 5$.  Note that the desired inequality fails for $n \le 4$, but can be verified numerically for small values $5\le n \le 26$ via the table of Bell numbers, and for slightly larger values (up to about 100) via computer algebra by computing finite partial sums.
Consider a function $f \colon [n] \to [n]$.  This induces a partition via the equivalence relation $\equiv_f$ defined as $i \equiv_f j$ iff $f(i) = f(j)$.  As above, the function does not uniquely determine the induced partition.  Another way to think about the question is then: can every function $[n] \to [\lceil n/\log n \rceil - 1]$ be mapped to a unique partition, for $n \ge 5$?  (But note that this is a slightly different requirement, due to rounding.)

Daniel Berend and Tamir Tassa, Improved Bounds on Bell Numbers and on Moments of Sums of Random Variables, Probability and Mathematical Statistics 30(Fasc. 2), 185–205, 2010.
Donald Knuth, The Art of Computer Programming, Vol. 4A, Section 7.2.1.5.  Addison-Wesley, 2011. (ISBN 0-201-03804-8; this section also appears in Fascicle 3, 2005, ISBN 0-201-85394-9)

REPLY [2 votes]: On further reflection, it seems the answer is no.
By considering the most significant terms in the asymptotic analysis of de Bruijn, and arguing that they dominate the other terms for large enough $n$, it seems possible to show that for every $\epsilon > 0$, there is some threshold $n_0 = n_0(\epsilon)$ such that
$$-0.9139\dots < \log c_n < -0.9139\dots + \epsilon$$
for all $n \ge n_0$.  Hence my proposed inequality cannot be true.
(Here the mysterious constant $-0.9139\dots$ is just $\log_2\;\log_2 e - \log_2 e = -1/\ln 2 - \ln\;\ln 2/\ln 2$.)
It would be interesting to establish precisely for which range of $n$ the simple expression $\log c_n \ge 0$ does hold.<|endoftext|>
TITLE: From Shortest Paths to Manifold Structure
QUESTION [8 upvotes]: I'm relatively green in the differential geometry area, so my apologies if what I'm asking is ill-posed and/or not research-level.
I have a situation where I know the shortest path between any two points in the plane. Is there a way to reconstruct a corresponding 2D-manifold such that the shortest path between points on the manifold with respect to the Euclidean metric projects to the given shortest path on the plane? I'm only interested (for the moment) in the nicest cases (i.e. dense, locally compact, analytic, etc.).
I can only think of the answer in the trivial example: if the shortest path is always a straight line, then the corresponding manifold is just the plane. But what is the calculation that shows this must be true?
And what if the shortest path between two points is given by the unique exponential curve $y=C_1+C_2e^x$ through those points? What if it's the unique monic parabola through the points?

REPLY [18 votes]: NB: I'm assuming from the way you worded the question that your '2D manifold' is supposed to be a surface in $xyz$-space and that the 'projection' is the projection to the $xy$-plane.  You may have had a more general situation in mind, such as a surface in $\mathbb{R}^n$ for $n>2$ or a more general projection than the obvious one.  In that case, the answer would change, but there is still a way of finding the answer.  Let me know if you want to know about these more general situations.
You are asking a special case of an inverse problem in the calculus of variations, and there is a standard method for solving this problem.  Note that, it is not true that every 2-parameter family $\mathscr{F}$ of curves in the plane (or an open domain in the plane) that has the property that there is a unique curve in the family passing through any given point in any given direction is the set of geodesics of some Riemannian metric, so it could easily be that most of the curve families $\mathscr{F}$ you might consider aren't the projections of geodesics of a metric of the form that you describe, i.e., a metric of the form
$$
g = \sqrt{dx^2+dy^2 + df^2}
$$ 
where $f = f(x,y)$ is a function on the given domain in the plane.
The reconstruction method involves computing the Euler-Lagrange equation for the above $g$ as a Lagrangian for curves in the plane and writing it in the form 
$$
y'' = F(x,y,y')
$$
for a certain function $F$.  Then you set this $F$ equal to the equation that defines your curve family.  For example, for the family $y = c_1 + c_2 e^x$, you'd have $y'' = y'= F(x,y,y')$.  This will give you an overdetermined equation for $f$ and then you apply the usual theory of overdetermined systems to determine whether or not there is a solution.  
Here, more explicitly, is what this amounts to:  If your curve family is described by a second order equation of the form $y'' = G(x,y,y')$, then any function $f(x,y)$ on a domain such that the above metric has the curve family as geodesics must satisfy the nonlinear equation
$$
\begin{align}
(1{+}f_x(x,y)^2{+}f_y(x,y)^2)G(x,y,y') 
&= - f_y(x,y)f_{xx}(x,y) \\
&\ \ \ \ - \bigl(2f_y(x,y)f_{xy}(x,y){-}f_x(x,y)f_{xx}(x,y)\bigr)y'\\
&\  \ \ \ +\bigl(2f_x(x,y)f_{xy}(x,y){-}f_y(x,y)f_{yy}(x,y)\bigr)(y')^2\\
&\ \ \ \ +f_x(x,y)f_{yy}(x,y)(y')^3.
\end{align}
$$
Note that, one thing you'd know right off the bat is that if your defining equation for the curve family isn't of the form $y'' = G(x,y,y')$ where $G$ is at most cubic in $y'$, then there will be no solution.  This is only a first necessary condition, though, it is not sufficient.  To see more conditions, suppose that the curve family is the set of solutions of
$$
y''=G(x,y,y')=a_0(x,y)+a_1(x,y)\ y'+a_2(x,y)\ (y')^2+a_3(x,y)\ (y')^3.\tag{1}
$$
Substituting this into the above equation and equating like powers of $y'$ will yield $4$ second order equations for $f$ in terms of the functions $a_i$, namely
$$
\begin{align}
- f_y\ f_{xx} &= a_0\ (1{+}{f_x}^2{+}{f_y}^2)\\
-2f_y\ f_{xy}{+}f_x\ f_{xx} &= a_1\ (1{+}{f_x}^2{+}{f_y}^2)\\
 2f_x\ f_{xy}{-}f_y\ f_{yy} &= a_2\ (1{+}{f_x}^2{+}{f_y}^2)\\
f_x\ f_{yy} &= a_3\ (1{+}{f_x}^2{+}{f_y}^2).
\end{align}\tag{2}
$$
For most choices of $a_i$ these equations are incompatible.  
Thus, for example, if $G\equiv0$, $f$ must satisfy the four second order equations
$$
f_y\ f_{xx} = f_x\ f_{xx}-2f_y\ f_{xy} = f_y\ f_{yy}-2f_x\ f_{xy} = f_x\ f_{yy} = 0.
$$
Of course, this implies that, at any point where $(f_x,f_y)$ is nonzero, one must have $f_{xx}=f_{xy}=f_{yy}=0$, so the only solutions to these equations are to have $f$ be a linear function of $x$ and $y$ (and all linear functions are solutions). In the case in which $G\equiv 2$ (the monic quadratic polynomials), the equations become
$$
f_y\ f_{xx} + 2(1{+}{f_x}^2{+}{f_y}^2) = f_x\ f_{xx}-2f_y\ f_{xy} 
= f_y\ f_{yy}-2f_x\ f_{xy} = f_x\ f_{yy} = 0,
$$
and it is not difficult to see that there are no solutions to this overdetermined system.
In the case in which $G\equiv y'$ (your other case), the equations become
$$
f_y\ f_{xx}  = f_x\ f_{xx}-2f_y\ f_{xy} + 2(1{+}{f_x}^2{+}{f_y}^2) 
= f_y\ f_{yy}-2f_x\ f_{xy} = f_x\ f_{yy} = 0,
$$
and it is easy to see that there are no solutions to this system either.
One useful observation for solving system (2) is that it implies the first order equation
$$
a_0\ {f_x}^3 + a_1\ {f_x}^2f_y + a_2\ f_x{f_y}^2 + a_3\ {f_y}^3=0,\tag{3}
$$
which shows that, when the functions $a_i$ are not all zero, the gradient of $f$ must point in one of three possible directions.  Differentiating (3) and plugging this back into the system (2) and doing a bit more work, one can derive the necessary and sufficient conditions on the $a_i$ that there exist a solution $f$.<|endoftext|>
TITLE: Hamiltonian polar action with Lagrangian section
QUESTION [6 upvotes]: I am looking for examples of Hamiltonian polar isometric actions of a compact Lie group on a Kahler-Einstein (or perhaps just Kahler) manifold, that admits a Lagrangian section.

Recall that an isometric action on $M$ is polar if there exists a submanifold $\Sigma\subset M$, called section, that meets all orbits orthogonally. Any section is automatically a totally geodesic submanifold, and there may be many different sections on $M$. The examples I am looking for should have a section $\Sigma\subset M$ that is also a Lagrangian submanifold, i.e., $\omega|_\Sigma=0$ and $\dim M=2\dim \Sigma$.

So far, the only example I can see is the standard torus action of $T^n$ on $\mathbb C P^n$, with the Fubini metric.

This action is Hamiltonian and isometric. It is also polar (in fact, by a Theorem of Podesta-Thorbergsson, if a torus $T^n$ acts on a compact Kahler manifold $M^{2n}$ of complex dimension $n$ and positive Euler characteristic, then this action is polar). Finally, the usual totally real embedding of $\mathbb R P^n$ into $\mathbb C P^n$ gives a Lagrangian section for the action.

With such strong hypothesis I would imagine that such actions are perhaps even classified, but I could not find anything in the literature about Lagrangian sections. Most classification results have to do with coisotropic actions, where it is required that the orbits satisfy $\omega|_{G(x)}=0$; while I am interested in an "orthogonal" version of that, i.e., I want the section to be coisotropic (even more, Lagrangian).

REPLY [4 votes]: I don't know what the global classification might be, but examining the structure equations for such a structure shows that the 'local' classification is reasonable.
Let $(M,g,\omega)$ be a Kähler manifold (of real dimension $2n$) with a symplectic and isometric polar action $G\times M\to M$ and let $\Sigma^n\subset M$ be a Lagrangian section.  (Since the consideration will be local, this can be just a local section for now.)  Because the section has dimension $n$, the generic orbits (which meet $\Sigma$ orthogonally) must be of (real) codimension $n$ and must, themselves be Lagrangian. Let us restrict our attention to the open subset of $M$ consisting of $n$-dimensional $G$-orbits.  
As the OP observes, $\Sigma$ must be totally geodesic.  If $m\in\Sigma$ is fixed and $G_m\subset G$ is the stabilizer of $m$, then $G_m$, which preserves the $n$-dimensional orbit $G\cdot m$ and hence its tangent space, must preserve the orthogonal to this tangent space, i.e., $T_m\Sigma$.  Hence, $G_m$ must preserve $\Sigma$ since the image of $\Sigma$ under an element of $G_m$ must be a totally geodesic Lagrangian submanifold that is tangent to $\Sigma$ at $m$.  Thus the images of $\Sigma$ under $G$ define a Lagrangian foliation of $M$ that is transverse to the Lagrangian foliation defined by the orbits of $G$.
Following along this kind of argument, one sees that, locally, one can choose local holomorphic coordinates $z = (z^i) = x + i\ y$ so that a Kähler potential $\phi$ for $\omega$ can be found that is a function of $x$ alone and the action of $G$ is just translation in the $y$-directions.  In particular, $G$ is abelian, and, in order to make the group compact, one has to take $y\in \mathbb{R}^n$ to be defined modulo a lattice $\Lambda\subset\mathbb{R}^n$.  Locally, the section $\Sigma$ can be taken to be defined by $y=0$.
In order for the potential $\phi = \phi(x)$ to define a Kähler metric, $\phi$ has to be convex relative to the affine structure on $\mathbb{R}^n$.  Then, in order for this to define a Kähler-Einstein metric, $\phi$ has to satisfy a Monge-Ampère equation.  
I believe that there is a section in a Chapter of Besse's Einstein manifolds that describes these metrics (which, I think, were first seriously investigated by Calabi).  However, I don't have a copy of the book here at home with me this weekend.
Note that this is just a local classification on a dense open set.  Of course, to deal with the case of $M$ compact, one is dealing with the case of toric Kähler manifolds, but I'm not sure which toric Kähler manifolds admit a polar section.  However, I would imagine that this must be well-studied; one should look for more information in the toric literature. In particular, I recommend two papers (out of many possibilities):
V. Guillemin, Kähler metrics on toric varieties, J. Diff. Geom. 40 (1994), 285–309. MR
95h:32029
and 
M. Abreu, Kähler geometry of toric varieties and extremal metrics, Internat. J. Math. 9 (1998), 641–651. MR 99j:58047<|endoftext|>
TITLE: If $X$ is a simplicial complex, is there a characterization of the links of its vertices that is equivalent to the statement "$|X|$ is a manifold"?
QUESTION [19 upvotes]: We have a characterization when we want $|X|$ to be a PL-manifold, in particular that the links of all the vertices are themselves  (PL) spheres. If we are in the category of PL- spaces then this is a necessary and sufficient condition. If however, we leave the PL category, then we get simplicial complexes that are topologically spheres, but not PL spheres. For instance if we take the double suspension of a homology sphere, then we get a  sphere. If our homology sphere has a triangulation (and is not itself a sphere), then the double suspension also has a triangulation. The links of the suspension points are not spheres, but these links do have the homotopy type of spheres. Furthermore, the links of these suspension points are also pure simplicial complexes (pure meaning that that all top dimensional simplicies have the same dimension).
I have drawn the following conclusion:  If $X$ is a simplicial complex, |X| is a topological manifold of dimension $N$, then the links of all of the vertices are pure simplicial complexes of dimension $N-1$ and have the homotopy type of the $N-1$ sphere.

What is the converse to this conclusion? Namely what extra conditions must be put on the links of the vertices other than purity and being homotopy sphere that will be equivalent to $|X|$ is a manifold? 

note: This question is similar to When are (finite) simplicial complexes (smooth) manifolds? . However, the I did not find the answer there. This is also a slightly different question.

REPLY [11 votes]: In dimensions $\geq 5$,  Theorem 1.5 of Galewski and Stern implies that if the links of the vertices of a homology manifold of dimension $\geq 5$ are simply-connected, then it is a manifold. 
I'm not sure if this is equivalent to your condition. Certainly your condition implies that the links of vertices are simply-connected and are homology spheres. But to be a homology $n$-manifold, the links of the barycentric subdivision must be $n-1$-homology spheres as well, and I'm not sure whether that is implied by your conditions. 
In $1, 2, 3$ and (I believe) $4$ dimensions, I think it's necessary and sufficient to have links be $n-1$-spheres.<|endoftext|>
TITLE: The non-traveling mathematician problem
QUESTION [17 upvotes]: This is a career question. I have just begun a research postdoc position in Southern California. It has been hard, but I've enjoyed teaching my first graduate courses and working on research and publishing.
From talking to other mathematicians, I've realized more and more that traveling to conferences a lot is an important part of being a research mathematician. But I don't want a job where I have to be gone from my wife and children on a regular basis. So research positions seem out of the question. 
So what are my options? I learned the hard way applying for jobs last year that companies aren't looking for bright young mathematicians who could learn stats, programming, etc. quickly when there are people with bachelors and masters in these areas who already have real-life experience. I know how to program in a few languages.
Thus, the clearest option seems to be a university position with a focus on teaching over research. But, it may be that such schools also would require regular travel. That's one reason I'm asking this question.
So,
1. What are the best options for a job not requiring travel given that I am a mathematician (in a non-applied area) with a great teaching record, an okay publishing record, and some computer science background, and
2. What can I do now so that I can get such a job when I apply in a year or two? No matter what, I'll work on publishing and teaching now because that's why I was hired; but what else can I work on?
Thank you your help!

REPLY [29 votes]: Teaching/research job in any University (in a research oriented department or teaching-oriented
department) DOES NOT require a lot of travel. Invitations to conferences or seminar talks come
indeed but this does not mean you have to go, if you don't want to.
Traveling to two conferences per year (usually one week or less) and/or 1-2 seminar talks per year
(for one day) will be sufficient. Even if you do not travel at all, this is not going to hurt
seriously your research career. People travel a lot because they like to travel:-)

REPLY [7 votes]: Have you ever considered the NSA or other government contractors? The standard way to get a job as a mathematician at the NSA is to apply for one of their development programs, which lets you tour around through several different groups within the NSA (but all would be in Maryland, probably) for 3 years and then settle down into one you like. They love math phds with some computer science background, and they pay more for those with a phd than for those coming out of bachelor's (so you wouldn't have "wasted your time" getting the degree; it would show up in your salary and promotion prospects).
Note that in order for this to work for you, you have to be a US citizen and you have to be able to pass a fairly invasive background check to get a top secret clearance. From when you apply to when you start working there might be 9 or 10 months. The development programs are called things like "The Applied Math Program" but I don't believe the math is any more applied than graph theory (though I can't know for sure). I know several pure mathematicians who are perfectly happy there. They don't get to research whatever they like, but they do get to choose problems from a list of problems and they seem to be pretty interested in what they do.
The other benefit of the government route is that there are contractors all over the place you could also go work for. There's even one in San Diego. It's called IDA (Institute for Defense Analyses). You can apply for those at the same time as you're applying for the NSA, or you could apply after the development program was done. Tread carefully here, I imagine the NSA doesn't like to lose people they've trained to other agencies, any more than any job would like losing people they've trained to a higher paying job. I only mention this in case living in CA is important to you.
Good luck!<|endoftext|>
TITLE: 4-dimensional h-cobordisms
QUESTION [17 upvotes]: I would like to know the state of the art concerning the following two questions.
1) Does there exist a smooth 4-dimensional h-cobordism (so between closed 3-manifolds) with non-vanishing Whitehead torsion ?
2) Does there exist a smooth 4-dimensional s-cobordism (that is, with vanishing Whitehead torsion) which is not diffeomorphic to a product cobordism ?
Thank you !

REPLY [7 votes]: The answer to question 1 is no in the orientable case .Every topological 4-dim.h-cobordism is an s-cobordism.
I think that the argument in the paper S.Kwasik,R.Schultz:Vanishing of Whitehead torsion in dimension 4,Topology,vol.31(1992),pp.735-756 should also give the non-orientable case.
The orientable case is the main theorem on p.736.It was proved for geometric 3-manifolds,but now we know that all3-manifods are geometric.<|endoftext|>
TITLE: Homotopy Transfer Theorem for Differential Graded Associative Algebras
QUESTION [7 upvotes]: As in Algebra+Homotopy=Operad by Bruno Vallette, let $A$ with multiplication $\nu$ be a differential graded associative algebra equipped with degree +1 map $h$ and let $H$ be a chain complex such that there exist chain maps $i$ and $p$ such that

and I work in characteristic 2 to make everything easier. Define
$$\mu_2=p\circ\nu\circ(i\otimes i):H\otimes H\to H,$$
and in general,

is a degree $+(n-2)$ map, where $PBT_n$ means binary trees with $n$ nodes and Vallette's summand on the right is an example of a summand for $n=5.$ For $f\in\hom(H^{\otimes n},H),$ define $\partial f=d\circ f + f\circ d_{H^{\otimes n}}$; remember that we are in characteristic 2.
One way to visualize $\partial\mu_n$ is that the one term decorates the leaves with $d$'s, as $d$ is a derivation for the raw tensor product, and the other puts a $d$ at the root, which then propagates upwards, as $d$ is a derivation for $\nu.$
The Homotopy Transfer Theorem for Differential Graded Associative Algebras is that $H$ equipped with the $\mu_n$ is an $A_\infty$ algebra, which means precisely that
.
All images have been directly screencapped from Vallette's paper. He writes that it should be an "easy and pedagogical" exercise to prove this theorem, but I'm getting caught in the thicket even in this characteristic 2 case where there are far fewer $\pm$'s to keep track of. I was wondering if anyone could provide me with any insights as to how to proceed without trees popping up all over the place occluding the forest.

REPLY [12 votes]: There is a systematic graphical notation that allows the tracking of signs, which I will mention at the end of this answer.  But before doing so, let me outline the situation when $2 = 0$.
Note first that since $(A,\nu)$ is strictly associative, $\mu_0 = 0$ and $\mu_1 = d_H = d : H \to H$.  By convention, if I have multilinear maps $f: H^{\otimes k}\to H$ and $g: H^{\otimes l}\to H$, then I will write $f\circ g : H^{\otimes(k+l -1)} \to H$ for:
$$ (f\circ g) (x_1\otimes \cdots \otimes x_{k+l-1}) = \sum_{i=1}^k f\bigl(x_1\otimes \dots \otimes x_{i-1} \otimes g(x_i\otimes \dots \otimes x_{i+l-1}) \otimes x_{i+l} \otimes \cdots \otimes x_{k+l-1}\bigr) $$
This has a useful graphical notation, wherein the composition is the sum over all rooted planar trees with an $f$ at the bottom node and precisely one $g$ at one of the upper nodes.

Then axiom to be an $A_\infty$-algebra in characteristic $2$ is
$ 0 = \sum_{j=0}^{n+1} \mu_j\circ \mu_{n+1-j} $, or, since $\mu_0 = 0$ and $\mu_1 = d$:
$$ [d,\mu_n] = \sum_{j=2}^{n-1} \mu_j \circ \mu_{n+1-j} $$
The right-hand side is a sum over all rooted planar trees with $n$ leaves and precisely two nodes, each of which has at least two branches from it.

To check this, the first thing to convince yourself is that the operator $[d,-] : f \mapsto d\circ f + f\circ d$ is a derivation of composition and tensor, so that to apply $[d,-]$ to some large diagram, you sum all diagrams you get by replacing one component of your original diagram by $[d,-]$ of it.  Note also that $[d,-]$ comutes with (i.e. annihilates) $i$, $p$, and $\nu$.  So when you work out $[d,\mu_n]$, you get a sum over diagrams that look like $\mu_n$ (i.e. planar rooted trees with $n$ leaves, each node has two branches, and interior edges labeled by $h$), except one of the interior edges has been replaced by $[d,h] = \mathrm{id}_A + ip$.
Now, it should be completely clear that the diagrams where the $h$ is replaced by an $ip$ are precisely the diagrams appearing in $\sum_{j=2}^{n-1} \mu_j \circ \mu_{n+1-j}$.  (If this is not clear, let me know, and I will try to make it clearer.)
Finally, we must dispense with the diagrams in which an $h$ is replaced by an $\mathrm{id}$.  For any such diagram, consider contracting it along the offending $\mathrm{id}$ vertex, to produce a node with three branches.  Except the resulting diagram with the trivalent vertex can be produced in two ways, corresponding to the two planar ways of blowing up a rooted node with three branches into two two-branch nodes.  So, after sorting all of your offending diagrams into such pairs, you get a sum of diagrams that looks like a $\mu_n$-type sum, except one vertex is has three branches.  What is this vertex labeled by?  Why, $\nu \circ \nu$, of course, which is a sum of two terms.  On the other hand, $\nu \circ \nu = 0$ by the associativity law for $(A,\nu)$.

In characteristic not equal to $2$, the exact same argument works, but you must find a good convention / notation for signs.  The best notation that I know is as follows.  It should, of course, already by understood that the solid "$H$" or "$A$" edges extend to "infinity" at the top and bottom of the page.  You should additionally draw diagrams with some other color of edge (I usually used "dashed") that records the degrees of operators — so this "dashed" edge should carry an arrow denoting its direction.  A vertex that raises homological degree by $n$ is required to receive $n$ dashed edges, and a vertex that lowers homological degree by $n$ is required to emit $n$ dashed edges.  Free dashed edges are sent off to "infinity" at (say) the left-hand side of the page, and the order from top to bottom that the free dashed edges arrive is important.  Just as you cannot add diagrams whose numbers of input and output "$H$" strands mismatch, you similarly must have the same sequence of dashed edges.  (In categorical language, what I'm saying is that you only work with "global" elements of endomorphism spaces, which is to say actual morphisms in the category of homologically-graded abelian groups, but that you give yourself access to the objects of this category which are lines in degree $\pm 1$.)
Now whenever two edges cross, something happens with signs.  The notation basically takes care of this, but if you ever insist on working with "homogeneous elements" (which is a bad habit — it's better to work more categorically) the convention is that as an element runs down the "wire" of a solid edge, when it passes through a dashed edge it remains unchanged if it is of even degree and changes sign if it is of odd degree.  The notations that do matter are: 

A closed dashed circle can be removed for a factor of $-1$.
A dashed crossing can be resolved for a factor of $-1$.  (Since dashed edges are directed, any dashed crossing has a unique resolution.)

Resolving a crossing for a sign http://math.berkeley.edu/~theojf/conventions1.gif
Loops, etc., cost signs http://math.berkeley.edu/~theojf/conventions2.gif
For example, the operator $d$ emits a dashed edge, and the operator $h$ receives a dashed edge.  Thus an equation like "$\mathrm{id} - ip = dh + hd$" is nonsense: the left-hand side has no dashed edges running to infinity, whereas on the right-hand side the first summand emits an edge and then receives one, and the second summand does those in the opposite order.  To sum the two terms on the right-hand side you have to at least get them into their edges-at-infinity into the same order, which you can do by adding a crossing (but remember that resolving that crossing changes a sign).  To make the two sides agree, you get connect up the two dashed edges, and again you should think a moment about signs.  The correct right-hand side to "$\mathrm{id} - ip = dh + hd$" is: 
the commutator http://math.berkeley.edu/~theojf/conventions3.gif
Yes, the sign is correct.  (Incidentally, I'm lifting these images from my thesis, which works out a slightly different question, and so the colors and numbers are not for this post.)
Anyway, I'll leave it as an exercise to write out diagrams for $A_\infty$ algebras in this notation, and to get all the signs right.  (Hint: there should be no "weird" signs.)  Part of the reason that I'll leave it as an exercise is that there's not really a unique correct answer: you make a sign convention, and work with it.<|endoftext|>
TITLE: Elementary proof of Mordell's theorem
QUESTION [10 upvotes]: My question concerns a comment in Silverman-Tate's book "Rational Points on Elliptic Curves" (second printing). On page 76 they begin their proof of 'lemma 4' to the proof of Mordell's theorem, which is to say that the subgroup $2C(\mathbb{Q})$ in $C(\mathbb{Q})$, the group of rational points for some elliptic curve $E/\mathbb{Q}$, has finite index. They make the comment
"Unfortunately, we do not know how to prove Lemma 4 for all cubic curves without using some algebraic number theory, and we want to stick to the rational numbers. So we are going to make the additional assumption that the polynomial $f(x)$ has at least one rational root, which amounts to saying that the curve has at least one rational point of order two. The same method of proof works in general if you take a root of the equation $f(x) = 0$ and work in the field generated by that root over the rationals. But ultimately we would need to know some basic facts about the unit group and the ideal class group of the field, topics which we prefer to avoid."
My question is has this situation been improved? Can Mordell's theorem be proved in generality without resorting to algebraic number theory?

REPLY [10 votes]: Not sure if this answers your question, but here's a thought. 
The known proofs that $E(K)/mE(K)$ is finite are non-effective because they embed $E(K)/mE(K)$ into a larger group that is shown to be finite. Let's call that larger group $S^{(m)}$. The proof that $S^{(m)}$ is finite is effective, and indeed it yields a very nice upper bound for the order of $S^{(m)}$. However, as far as I know, this bound always involves (at least) the $m$-part of the class number of $K(P)$, where $P$ is an $m$-torsion point of $E$. So the proof requires algebraic number theory in the sense that one needs to know that the class group (or at least the $m$ part) of some extension field of $K$ is finite, unless there is a rational $m$ torsion point. So proofs along these lines would seem to require algebraic number theory and ideal class groups, either explicitly or disguised in some way.<|endoftext|>
TITLE: Density of the "multiplicative odd numbers"
QUESTION [10 upvotes]: I am interested in the set $A$ of all positive integer numbers such that when factored into primes, the sum of the exponents is odd (I think of $A$ as the multiplicative odd numbers).
I want to know if it has positive upper density, more precisely
$$\bar d(A):=\limsup_{n\to\infty}\frac{|A\cap[1,n]|}n$$
I think I read somewhere that it has density $1/2$ (and the $\lim$ exist, not just the $\limsup$), but I would be happy with a proof that $\bar d(A)>0$.

REPLY [12 votes]: Your sequence is http://oeis.org/A026424.
Define the zeta density of the set of integers $A$ as 
$$ d(A) = \lim_{x \to 1+} \frac1{\zeta(x)}  \sum_{k \in A} k^{-x}. $$
Then from results given at the OEIS, this is
$$ d(A) = \lim_{x \to 1+} {1 \over \zeta(x)} \left[ {\zeta(x)^2 - \zeta(2x) \over 2\zeta(x)} \right]. $$
After some simplifcation this is
$$ \lim_{x \to 1} \left( {1 \over 2} - {\zeta(2x) \over 2 \zeta(x)^2} \right) $$
and recalling that $\lim_{x \to 1} \zeta(x)$ is infinity while $\zeta(2) = \pi^2/6$, this is $1/2$.   
Now, if a set has a natural density, then it has a zeta density, and the two densities are equal; see for example Chapter 2 of Diaconis' PhD dissertation.  So we can conclude that if your set has a natural density, then that natural density is $1/2$.<|endoftext|>
TITLE: Software Reference Request:  Three Dimensional Staircase Visualizers for Monomial Ideals
QUESTION [7 upvotes]: Is anyone familiar with software that already exists to visualize monomial ideals in three variables as their staircase diagrams, in the sense of Combinatorial Commutative Algebra?  It looks like their graphics were done using TikZ, but after a reasonable search, I wasn't able to find any pre-written packages or software that would automatically produce a picture of the staircase diagram of a monomial ideal.
It was easy enough to write a few lines of code in Macaulay2 to produce the TikZ code for the two-dimensional case, but I wasn't looking forward to writing code to do the same thing in the three variable case if this already existed somewhere.  If there was code to draw these things using ideals encoded in Macaulay2, Singular, or Sage format, I would love that -- but if there's any pre-done visualizer for three variable monomial ideals, I would happily compromise on format.
Alternately, if someone has software (or knows a good reference to software) which produces visualizations of the Buchberger graph for one of these monomial ideals, I'd be willing to settle for that too.

REPLY [2 votes]: Jang Soo Kim's Plane Partition code produces staircase diagrams in tikz. Similar diagrams can be produced in Macaulay2 using the Visualize package.<|endoftext|>
TITLE: Combinatorial interpretation of ${i\choose n}$, where $i^2=-1$
QUESTION [45 upvotes]: At MIT all departments have numbers, and math is 18. Last year MIT
math majors produced a tee shirt that said ${i\choose 18}$ ("I choose
18") on the front, and on the back
   $$ \frac{34376687+1499084559i}{14485008384}. $$
With the more natural denominator $18!$ this is
   $$ \frac{15194495654000+662595375078000i}{18!}. $$
This suggests the question: for any $n\geq 1$ find a "nice"
combinatorial interpretation of the real and imaginary parts of
$i(i-1)(i-2)\cdots (i-n+1)=f_n+ig_n$. It is easy to express $f_n$ and
$g_n$ as certain alternating sums of Stirling numbers of the first
kind, but I don't consider this "nice." The $g_n$'s seem to alternate
in sign beginning with $n=5$. The $f_n$'s alternate in sign up to
$n=17$ and then seem to alternate in sign beginning with $n=18$. It is
curious that $i(i-1)(i-2)(i-3)=-10$, a real number. One could ask the
same question with $i$ replaced by any Gaussian integer $a+bi$. One
can also ask about the asymptotic rate of growth of $f_n$ and
$g_n$. Clearly $f_n^2+g_n^2\sim C\cdot (n-1)!^2$, so one would expect
$f_n$ and $g_n$ to be roughly of the size of $(n-1)!$.

REPLY [24 votes]: There is no need to reinvent the wheel by estimating
$\prod_{k<n}(1+\frac1{k^2})$.
The asymptotic formula for $f_n + i g_n$  follows readily from
Stirling's approximation (as I already noted in my comment to
the original question), and indeed the same is true for
the asymptotics as $n \rightarrow \infty$ of $w \choose n$
for any $w \in {\bf C}$; the answer is simply
$$
\phantom{*0000000000000000000}
{w \choose n} =
 \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) \frac{(-1)^n}{\Gamma(-w)} n^{-w-1}
\phantom{0000000000000000000}(*)
$$
(and the $O(1/n)$ can be refined to an asymptotic series in powers of $1/n$).
Note that this gives zero precisely for the values $w=0,1,2,3,\ldots$
for which $-w$ is a pole of $\Gamma$, which are also the $w$ for which
${w \choose n} = 0$ for sufficiently large $n$.
For $w=i$, we recover the observed behavior:
$\Gamma(i)$ is a complex number of absolute value
$(\pi / \sinh \pi)^{1/2}$ [in general
$$|\Gamma(it)| = (\Gamma(it)\Gamma(-it))^{-1/2} = 
\left(\frac \pi {t \phantom. \sinh \pi t} \right)^{1/2}
$$
for real $t \neq 0$],
and $n^{-w-1}$ is a complex number of absolute value $1/n$ that goes
once around the origin when $n$ increases by a factor $e^{2\pi}$.
Thus each of $\lbrace f_n \rbrace$ and $\lbrace g_n \rbrace$
alternates in sign outside an infinite sequence of exceptions
that's asymptotically a geometric sequence with common ratio $e^\pi$.
To prove $(*)$, write
$$
{w \choose n} = \frac{(-1)^n}{n!} \prod_{k=0}^{n-1} (k-w)
= \frac{(-1)^n}{n!} \frac{\Gamma(n-w)}{\Gamma(-w)}
= \frac{(-1)^n}{n\Gamma(-w)} \frac{\Gamma(n-w)}{\Gamma(n)}.
$$
Now we understand $(-1)^n/n$,
and the factor $1 / \Gamma(-w)$ is constant,
so we're left with $\Gamma(n-w) / \Gamma(n)$.
We apply the following form of Stirling's formula:
there exists a constant $\varpi>0$ (known to equal $2\pi$,
but we shall not need this) such that
$$
\Gamma(z) = \bigl(1 + O(|z|^{-1}\bigr) z^z e^{-z} \sqrt{\varpi/z}
$$
holds as $|z| \rightarrow \infty$ in the right half-plane,
where $z^z = \exp (z \log z)$ and $\sqrt{\varpi/z}$
are defined using the principal branches of $\log z$ and $\sqrt z$.
This gives
$$
\frac{\Gamma(n-w)}{\Gamma(n)} = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr)
  \frac{(n-w)^{n-w} e^{-(n-w)} (\varpi/(n-w))^{1/2}}
    {n^n e^{-n} (\varpi/n)^{1/2}}
$$ $$
 = \Bigl(1+O\bigl(\frac1n\bigr)\Bigr)
\frac{(n-w)^{n-w}}{n^n} e^w \left(1-\frac{w}{n}\right)^{-1/2}.
$$
Now the last factor is $1 + O(1/n)$; the factor $e^w$ is constant;
and
$$
(n-w)^{n-w} = (n-w)^{-w} (n-w)^n
= \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{-w} (n-w)^n.
$$
So we're left with
$$
\frac{\Gamma(n-w)}{\Gamma(n)} =
\Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{-w} e^{-w} \left(1 - \frac{w}{n}\right)^n
= \Bigl(1+O\bigl(\frac1n\bigr)\Bigr) n^{-w}.
$$
This completes the proof of $(*)$ (and the cancellation in the last step
leads me to suspect that even this use of Stirling is more complicated
than necessary).<|endoftext|>
TITLE: Morphisms between Verma modules
QUESTION [14 upvotes]: Let $\mathcal{O}_0$ be the principal block of the BGG category $\mathcal{O}$ for a finite dimensional simple Lie algebra over $\mathbb{C}$. For an element $w$ in the Weyl group $W$, let $\Delta_w$ denote the Verma module with highest weight $w_0w^{-1}\cdot 0$, where $w_0\in W$ is the longest element, and $\cdot$' denotes the dot-action.
It is well known that
$\mathrm{dim}(\mathrm{Hom}(\Delta_v, \Delta_w)) \leq 1$
This fact is pretty straightforward to prove algebraically. However, I do not know how to see this topologically. Namely, I do not know how to prove this via the interpretation of Verma modules as perverse sheaves on the flag variety.
I would be grateful if someone could explain how to see this fact topologically.
Added later: In response to Jim Humphreys comment let me add some motivation:

In this regard I think of category $\mathcal{O}$ as a "toy example". I would like to know what sort of generality this fact holds for. For instance, is the corresponding statement true for perverse sheaves smooth along a stratification given by affine spaces? The latter is certainly a highest weight category, computations in it can be undertaken topologically, etc. So as a starting point I would like to understand the topological reason for its truth for the "toy example".
In the same vein as 1) I would like to know whether this truly is a "geometric" fact, i.e., does it hold if I consider my sheaves with coefficients in a commutative ring say?
Computing extension groups of Verma modules is an old problem. If there is any hope for doing this topologically, I would think a reasonable place to start would be to compute $\mathrm{Ext}^0$ topologically!
In the same vein as 3). One can see that the extensions of Verma modules is given by (compactly supported) cohomology (appropriately shifted) of intersections of Schubert cells with opposite Schubert cells. This is related to my earlier questions:

Intersection of plus/minus cells in Bialynicki-Birula decomposition
A cohomology computation request.

The above fact about homomorphisms between Vermas translates to the lowest non-vanishing cohomology (compactly supported) being one dimensional. These are smooth affine varieties, but (at least in low ranks) their Betti numbers satisfy a curious "Poincare duality"/palindromic type phenomenon. This phenomenon is even more starkly visible if one further looks at the Hodge numbers. Amusingly, since these varieties are smooth and irreducible, one immediately gets that the highest non-vanishing extension group (when it is possible to have morphisms between the Vermas) is one dimensional and concentrated in the "right" degree. This latter fact can also be shown algebraically, but requires a careful argument using translation functors (which can also be done geometrically without ever knowing anything about the intersections, but now I am digressing). Anyway, a topological reason as in my question may hopefully give some insight as to whether this palindromic phenomenon is a low rank coincidence or has any hope for holding in general.

Apologies if any of the reasons above are too vague/ranting, I didn't want to throw in all of that in my original question in case the answer was something blatantly obvious that I had been overlooking.

REPLY [4 votes]: It may be helpful to look at the 1987 paper by Kari Vilonen and his student Ren Mirollo here.  They study BGG reciprocity on perverse sheaves in some generality, working over a field.   Here the intermediate (or standard) objects corresponding to Verma modules in the classical BGG category are straightforward to define, using an assumed Whitney stratification like that of the flag variety.   The functors $\mathrm{Ext}^i$ then come into play.
Taken in isolation, the original Verma modules are easy enough to construct: they played a technical role for Chevalley and Harish-Chandra in getting uniform existence proofs for finite dimensional simple modules in the Lie algebra framework.   But arbitrary simple quotients of Verma modules are quite elusive, requiring deeper ideas.   This shows up again in the setting of perverse sheaves.   
As in your question, Mirollo-Vilonen were proceeding in a somewhat speculative way.   To work over suitable rings, you'd probably have to go further into the work of Mirkovic-Vilonen on geometric Langlands.  
ADDED: Concerning Geordie's comment on the case of parabolic Verma modules (relative to "partial flag varieties"), it's clear from the algebraic study already done that Hom spaces get much more complicated than for ordinary Verma modules.    (See Section 9.10 of my 2008 AMS book on the BGG category for some details and references.)   As far as I know, none of this has been effectively translated into the language of perverse sheaves along the lines of Mirollo-Vilonen.<|endoftext|>
TITLE: Polynomial maps between noncommutative groups
QUESTION [12 upvotes]: Below I will give some definitions.  My question is: do these appear in the literature, and if so, under what name?
Let $G$ and $H$ be groups that may not be commutative.  For $y\in G$, define $R_y:G\to G$ by $R_y(x)=xy$.  Let $f$ be a function from $G$ to $H$.  Define $\delta_n(f):G^n\to H$ by \begin{align*}
 \delta_0(f) &= 1 \\\\
 \delta_1(f)(x) &= f(1) f(x)^{-1} \\\\
 \delta_2(f)(x,y) &= f(1) f(x)^{-1} f(xy) f(y)^{-1} \\\\
 \delta_3(f)(x,y,z) &= f(1) f(x)^{-1} f(xy) f(y)^{-1} f(yz) f(xyz)^{-1} f(xz) f(z)^{-1} 
\end{align*}
and in general
$$ \delta_{n+1}(f)(x_1,\dotsc,x_{n+1}) = 
    \delta_n(f)(x_1,\dotsc,x_n)
    \delta_n(f\circ R_{x_{n+1}})(x_1,\dotsc,x_n)^{-1}.
$$
This has one term for each subset $J\subseteq\{1,\dotsc,n\}$, with exponent $(-1)^{|J|}$.  The order of the terms corresponds to the Binary Reflected Gray Code (see Wikipedia, for example).  One could imagine using other orders such as lexicographic, but the BRGC order seems to do the right thing for the examples that I am considering.
I'll say that $f$ is polynomial of degree at most $n$ if $\delta_{n+1}(f)$ is the constant function with value $1$.  Clearly $f$ is polynomial of degree at most $0$ iff it is constant, and it is polynomial of degree at most $1$ iff it is a constant times a homomorphism.  The commutative case is fairly well-known, and is consistent with the usual meaning of 'polynomial' for maps $\mathbb{Z}^p\to\mathbb{Z}^q$.   I know of a 1971 paper by Andreas Dress, but it would not surprise me if there were earlier references.  However, I have never seen the noncommutative case.  
Even in the commutative case, it takes some work to prove that any composite of polynomial maps is polynomial.  I do not know whether that holds in the noncommutative case.

REPLY [5 votes]: Polynomial mappings of groups were investigated by Leibman in connection with density Ramsey theory.
Definition. Let $G$ be a group and $f:\mathbb{Z}\to G$ any map. The discrete derivative of $f$ is the map $D_a f(b) = f(b+a) f(b)^{-1}$. The map is called polynomial of degree $d$ if all its $d+1$-th discrete derivatives vanish identically (i.e. equal $1_G$).
One could replace $\mathbb{Z}$ by a more general group here; I stick to the integers for simplicity.
Theorem Assume that $G$ is nilpotent. Then the polynomial mappings form a group under pointwise operations.
Different proofs can be found in Leibman's article, an appendix in an article by Green, Tao, and Ziegler and in an article that I wrote.
The appendix contains probably the most systematic approach based on Host-Kra cube groups, in particular it is shown that the composition of polynomial maps between nilpotent groups is also polynomial.
Note that the theorem does not say that the polynomial mappings of a given degree form a group, to recover a result of this kind one has to refine the notion of degree, see the references above.
The theorem fails already for the dihedral group $D_3$ that is the smallest non-nilpotent group. Indeed, let δ be a rotation and σ a reflection in $D_3$ so that δ³ = σ² = (σδ)² = 1. The sequences (…, σ, 1, σ, 1,…) and (…, σδ, 1, σδ, 1,…) are polynomial (of degree 1) but their pointwise product (…, δ, 1, δ, 1,…) is not polynomial (of any degree).
I am not sure whether this presents an obstruction to polynomiality of composition of polynomial maps between arbitrary groups.<|endoftext|>
TITLE: Terminology question for real K-theory
QUESTION [5 upvotes]: This is a terminology question. Answers will help me satisfy a referee but I'm also genuinely interested. Consider the following two things that you could define for a topological space X:
(1) The Grothendieck ring of the category of real vector bundles over X;
(2) the set $[X, \mathbb{Z} \times BO]$  of homotopy classes of maps from $X$ to $\mathbb{Z} \times BO$ where $BO$ is the classifying space for the infinite orthogonal group.
It is well known that (1) and (2) are in bijection when $X$ is, say, a finite CW-complex. This can fail when $X$ is not compact though, and I think it also fails when $X$ is a nested sequence of circles in the plane with the subspace topology. Anyway, the two are different in general.
Now consider the following names and notation :
(A) the real K-theory of $X$;
(B) the real topological K-theory of $X$;
(C) the K-theory of real vector bundles over $X$;
(D) $K(X)$;
(E) $KO(X)$;
(F) $K_\mathbb{R}(X)$.

Which would you pair to which? and what is your favorite for (1) and for (2)?

For example I don't expect anyone to answer that C2 is reasonable. I'm curious whether E1 or E2 will appear. Some other notation may also be prefered, I think I saw $\mathbf{KO}(X)$ once, meaning (1), while $KO(X)$ was for (2). Or perhaps the other way around. There are also alternative names like "the Atiyah real K-theory", not sure meaning what.
Thanks!
There may not be any universal convention, but at least we can have a vote of sorts.

REPLY [8 votes]: One standard notational convention is that 
(2) = (E) for unbased spaces $X$, where the 
brackets mean unbased homotopy classes of maps.
This is consistent with the classification 
theorem that equivalence classes of $n$-plane 
bundles over $X$ are classified as $[X,BO(n)]$.
The classification works for general CW complexes
$X$, not just finite ones.  The restriction to 
finite CW complexes to prove that (1) = (2) comes 
in showing that the Grothendieck group, restricted
to elements of virtual dimension zero, is isomorphic
to $[X,BO]$.  
However (E) is perhaps more usually used for the entire 
generalized cohomology theory whose zeroth term is (2).  Then, in modern algebraic topology, it has become standard to 
focus on reduced cohomology theories defined on based spaces $X$,
and the bracket $[X,Y]$ is then understood to mean homotopy classes of based
maps.  In early literature, $[X,Y]_*$ meant based homotopy classes,
but the modern literature prefers confusion, leaving to context
which is meant.  Thus for unbased spaces,
$$KO(X) = KO^0(X) = [X,BO\times \mathbf{Z}] = [X_+, BO\times \mathbf{Z}]_*.$$
The complex analogue is 
$$K(X) = K^0(X) = [X, BU\times \mathbf{Z}] = [X_+, BU\times \mathbf{Z}]_*,$$
so you absolutely must not use (D). Here $X_+$ is not the one-point 
compactification of Paul's answer in general, but rather the union of
$X$ and a disjoint basepoint.  This is the general way to describe 
unreduced cohomology in terms of reduced cohomology.  It has gradually become 
more standard in algebraic topology to write $KO(X)$ rather than the historical
$\widetilde{KO}(X)$ for $[X,BO\times \mathbf{Z}]_*$
when $X$ is based.  One point crucial to modern algebraic topology is that
generalized cohomology theories are represented by $\Omega$-spectra $E$, which 
are sequences of based spaces $E_n$ and based (weak) homotopy equivalences
$E_n\longrightarrow \Omega E_{n+1}$; of course loop spaces only make sense in 
the based context.   
Adams used $K_{\mathbf R}(X)$ and $K_{\mathbf C}(X)$
as synonyms for $KO(X)$ and $K(X) = KU(X)$.  That is the
context of your (F), but it never really caught on. One
can use (B) instead of (A) for emphasis when necessary,
to distinguish from algebraic $K$-theory.  By abuse,
(C) is also sometimes used as synonymous with (A), (B),
(E), and (F), although it ought more logically be paired
with (1).  People sometimes use 'orthogonal' rather
than `real' to avoid confusion with Atiyah's Real $K$-theory
$KR$ as in Mark's answer.  Atiyah's real vector bundles are 
often called Real vector bundles to avoid confusion, 
which works Really badly in talks.  
In answer to Tom and Pierre, the nice paper in mind is 
``Vector bundles over classifying spaces of compact Lie groups''
by Stefan Jackowski and Bob Oliver. 
It exploits the Sullivan conjecture to study (1) when $X=BG$
for a compact Lie group $G$.<|endoftext|>
TITLE: Numerical calculation of Arnold tongue
QUESTION [5 upvotes]: Hello.
I am working on investigation of family of dynamical systems on the torus
$$\dot{x}=\cos(x)+b\cos(t)+a$$
$$\dot{t}=1$$
and it's Poincare map $$P:(x,0) \rightarrow (P(x),2\pi=0)$$
I need to find Arnold tongues of map $P$. I tried simple calculation of solution using  Runge-Kutta formulas, then iterating and checking rotation number, but it's not working effectively. Arnold tongues was first calculated in 1970s so maybe there is effective algorithm of doing it?

REPLY [8 votes]: There is a good way to compute rotation number of a circle homeomorphism (this was the way Poincaré thinked of it): you calculate the rotation number buy its continued fraction in a direct way.
You start from a point $x$ and $f(x)$: this gives you a decomposition of the circle into points that are on the right side of $x$ (in $]x,f(x)[$)  and points which are on its left side (in $]f(x),x[$). You look at $f^2(x)$ and you write $R$ if it is on the right side of $x$, $L$ otherwise. Iteranting $f$ you find a sequence of $R$'s and $L$'s. If you get $LLLLR$, for example, you record 4 (this is the number of $L$'s) and you approximate the rotation number of $f$ by $1/4$.
Renormalizing $f$, you iterate this process finding $\rho=[0,a_1,a_2,\ldots,a_k]$.
I won't be more precise here.
Every detail is very well explained in de Melo & van Strien's One-Dimensional Dynamics, section I.1.
You can find a paper by Bruin (Numerical determination of the continued fraction expansion of the rotation number) in which he compares different methods on Arnold tongues.
EDIT[update]:
Recently, I wrote for myself some sage lines implementing the algorithm I described you. 
This is my second version, now working for rational numbers too : I was originally interested only in irrational rotation numbers (comments are welcome to improve it!).
L=8 #length for cf-expansion, depending on your computer, 8 or 9 suggested for a try run
A=100000 #maximum size of single element of the sequence

def partfrac(x):
    return x-floor(x)

##### computing rational approximations given continued fraction expansion
# input b=a continued fraction expansion
# input l=L length of computed expansion
def rational_approximation(b,l):
    p=[0,1]
    q=[1,b[1]]
    for i in range(1,l+1):
        p.append(b[i+1]*p[i]+p[i-1])
        q.append(b[i+1]*q[i]+q[i-1])
    return simplify(p[l+1]/q[l+1])


#computing rotation number of a given circle map f
def rotation(f):
    a=[0]
    orbit=[]
    orbit.append(partfrac(f(0)))
    if orbit[0]==0 :
        print 'map with a fixed point'
        return 0

    def shift(x):  #set f(0) as the origin + 1
        if partfrac(x)>orbit[0]:
            return partfrac(x)-1
        return partfrac(x)

    def first_return(p,pre_p,y):
        x=shift(f(y))
        while x<pre_p or x>p:
            x=shift(f(x))
        return x

    a.append(1)
    x=orbit[0]

    if shift(f(orbit[0]))==0:
        print 'map with periodic point of order 2'
        return 1/2

    if shift(f(orbit[0]))<0:
        while shift(f(x))<0:
            a[1]=a[1]+1
            x = shift(f(x))
            if a[1]>A:
                print 'approximatively 0'
                return 0
            if shift(f(x))==0:
                print 'periodic point'
                a[1]=a[1]+1
                return 1/a[1]
        orbit.append(shift(x))
        z = shift(f(x))
        a.append(0)
        while z>0:
            y = z
            z = first_return(shift(orbit[0]),shift(orbit[1]),z)
            a[2]=a[2]+1
            if a[2]>A:
                print 'approximatively rational'
                return 1/a[1]
            if z==0:
                print 'periodic point'
                a[2]=a[2]+1
                return rational_approximation(a,1)
        orbit.append(y)

    if shift(f(orbit[0]))>0:
        def shift(y):  #set f(0) as the origin
            if partfrac(y)>=orbit[0]:
                return partfrac(y)-1
                return partfrac(y)
        orbit.append(orbit[0]-1)
        a.append(0)
        while shift(f(x))>0:
            a[2] = a[2] + 1
            x = shift(f(x))
            if a[2]>A:
                print 'approximatively rational'
                return 1/a[1]
            if shift(f(x))==0:
                print 'periodic point'
                a[2]=a[2]+1
                return rational_approximation(a,1)
        orbit.append(shift(x))    
        z = shift(f(x))

    for i in range(1,L):
        a.append(0)
        if shift(orbit[i+1])<shift(orbit[i]):
            while z>0:
                y = z
                z = first_return(shift(orbit[i]),shift(orbit[i+1]),z)
                a[i+2]=a[i+2]+1
                if a[i+2]>A:
                    print 'approximatively rational'
                    return rational_approximation(a,i)
                if z==0:
                    print 'periodic point'
                    a[i+2]=a[i+2]+1
                    return rational_approximation(a,i+1)
        if shift(orbit[i+1])>shift(orbit[i]):
             while z<0:
                 y = z
                 z = first_return(shift(orbit[i+1]),shift(orbit[i]),z)
                 a[i+2]=a[i+2]+1
                 if a[i+2]>A:
                     print 'approximatively rational'
                     return rational_approximation(a,i)
                if z==0:
                     print 'periodic point'
                     a[i+2]=a[i+2]+1
                     return rational_approximation(a,i+1)
         orbit.append(y)

    print a
    return rational_approximation(a,L)<|endoftext|>
TITLE: tiling a rectangle with the smallest number of squares
QUESTION [33 upvotes]: This is based on another thread. For $m,n\in \mathbb N$, let $f(m,n)$ be the minimum number of squares with integer sides needed to tile a $m\times n$ rectangle. Recently, a table of values for $n\le m\le 85$, obtained by what seems to be a brute force search, has been put online here.
The table looks quite fuzzy, but if we restrict it to values of coprime $m,n$ such that $2n\ge m\ge n$, there is surprisingly little fluctuation. For convenience, the following table gives $f(m,n)$ in reverse order, showing in row $n$ the values for $m=n-1,n-2,...,\lbrace n/2\rbrace$ but putting "o" wherever $(m,n)>1$.
(The number following $m$ is $ g(m):=\frac{\log(m\sqrt{5})}{\log(\phi)}$ where $\phi=\frac{\sqrt{5}+1}2$, see below.) 
 3 :  3.955  [ 3]
 4 :  4.553  [ 4]
 5 :  5.016  [ 5, 4]
 6 :  5.395  [ 5, o]
 7 :  5.716  [ 5, 5, 5]
 8 :  5.993  [ 7, o, 5]
 9 :  6.238  [ 7, 6, o, 6]
10 :  6.457  [ 6, o, 6, o]
11 :  6.655  [ 6, 7, 6, 6, 6]
12 :  6.836  [ 7, o, o, o, 6]
13 :  7.002  [ 7, 6, 7, 7, 6, 6]
14 :  7.156  [ 7, o, 7, o, 7, o]
15 :  7.299  [ 7, 8, o, 7, o, o, 8]
16 :  7.433  [ 7, o, 8, o, 7, o, 7]
17 :  7.559  [ 8, 8, 7, 8, 7, 7, 7, 8]
18 :  7.678  [ 8, o, o, o, 7, o, 7, o]
19 :  7.791  [ 7, 9, 7, 7, 7, 7, 7, 7, 7]
20 :  7.897  [ 9, o, 7, o, o, o, 7, o, 8]
21 :  7.999  [ 8, 7, o, 9, 8, o, o, 7, o, 7]
22 :  8.095  [ 8, o, 8, o, 8, o, 8, o, 8, o]
23 :  8.188  [ 8, 8, 8, 9, 8, 8, 8, 8, 8, 8, 8]
24 :  8.276  [ 8, o, o, o, 9, o, 8, o, o, o, 7]
25 :  8.361  [ 8, 8, 8, 8, o, 8, 8, 9, 8, o, 8, 8]
26 :  8.442  [ 8, o, 8, o, 8, o, 8, o, 9, o, 8, o]
27 :  8.521  [ 8,10, o, 8, 8, o, 8, 8, o, 8, 8, o, 8]
28 :  8.596  [ 8, o,10, o, 9, o, o, o, 8, o, 8, o, 8]
29 :  8.669  [ 9, 8, 8,10,10, 9, 9, 8, 9, 8, 8, 8, 9, 8]
30 :  8.740  [ 9, o, o, o, o, o, 9, o, o, o, 8, o, 9, o]
31 :  8.808  [ 8, 8, 8,10, 8, 8, 8, 8, 8, 8, 9, 8, 8, 8, 8]
32 :  8.874  [ 9, o, 8, o, 8, o, 9, o, 9, o, 8, o, 8, o, 9]
33 :  8.938  [ 9, 9, o, 9, 8, o, 8,10, o, 9, o, o, 8, 8, o, 9]
34 :  9.000  [ 9, o, 9, o, 9, o, 9, o, 8, o, 9, o, 8, o, 8, o]
35 :  9.060  [ 8, 9,10, 8, o, 9, o, 9, 8, o, 8, 9, 9, o, o, 8, 9]
36 :  9.119  [ 9, o, o, o,10, o,10, o, o, o, 9, o, 9, o, o, o,10]
37 :  9.176  [ 9, 9, 9, 9, 8,10, 9, 8, 9, 9, 9, 9, 8, 9, 8, 9, 8, 8]
38 :  9.231  [ 9, o, 9, o, 9, o, 9, o,10, o, 9, o, 9, o, 9, o,10, o]
39 :  9.285  [ 9, 9, o, 9, 9, o,10, 9, o, 9, 9, o, o, 9, o, 9, 9, o, 9]
40 :  9.338  [ 9, o, 9, o, o, o, 9, o, 9, o, 8, o, 9, o, o, o, 9, o, 8]
41 :  9.389  [ 9, 9, 9, 9,11, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9]
42 :  9.439  [ 9, o, o, o, 9, o, o, o, o, o,10, o,10, o, o, o,10, o,10, o]
43 :  9.488  [ 9, 9, 9, 9, 9, 9,10,10, 9, 9, 9, 9, 9, 9, 9, 9,10, 9, 9, 9, 9]
44 :  9.536  [ 9, o, 9, o, 9, o, 9, o, 9, o, o, o, 9, o, 9, o, 9, o, 9, o, 9]
45 :  9.582  [10, 9, o,10, o, o, 9, 9, o, o, 9, o, 9, 9, o,10, 9, o, 9, o, o, 9]
46 :  9.628  [ 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o, 9, o]
47 :  9.673  [ 9, 9, 9,11, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9,10, 9, 9, 9, 9, 9, 9, 9, 9]
48 :  9.716  [10, o, o, o, 9, o, 9, o, o, o,10, o,10, o, o, o, 9, o, 9, o, o, o, 9]
49 :  9.759  [ 9, 9, 9, 9,10,10, o, 9, 9, 9, 9, 9, 9, o, 9,10,10, 9, 9,10, o, 9, 9, 9]
50 :  9.801  [10, o, 9, o, o, o, 9, o, 9, o, 9, o, 9, o, o, o, 9, o, 9, o, 9, o, 9, o]
51 :  9.842  [ 9,10, o, 9,10, o,10, 9, o,10,10, o, 9,10, o, 9, o, o, 9,10, o,10,10, o, 9]
52 :  9.883  [10, o,11, o, 9, o,10, o,10, o, 9, o, o, o, 9, o,10, o, 9, o, 9, o,10, o,11]
53 :  9.922  [10, 9, 9,11, 9,10,10, 9,10,11,10, 9,10, 9,10,10,11,10, 9, 9, 9, 9, 9,11,11, 9]
54 :  9.961  [10, o, o, o, 9, o,10, o, o, o, 9, o, 9, o, o, o, 9, o, 9, o, o, o, 9, o,10, o]
55 :  9.999  [10,10,10,10, o, 9, 9, 9, 9, o, o, 9,10, 9, o, 9, 9, 9,10, o, 9, o,10, 9, o, 9, 9]
56 :  10.03  [ 9, o, 9, o,10, o, o, o, 9, o,10, o,11, o,10, o,10, o, 9, o, o, o,10, o, 9, o, 9]
57 :  10.07  [10,10, o,10, 9, o,10,10, o,10, 9, o, 9,10, o,10,10, o, o, 9, o,10, 9, o,10, 9, o,10]
58 :  10.11  [10, o,10, o,10, o,10, o,10, o,11, o,10, o,10, o,10, o,11, o,10, o,10, o,10, o,10, o]
59 :  10.14  [10,10, 9, 9,10,11, 9, 9, 9,10,10, 9, 9,10, 9, 9, 9,10, 9, 9,10, 9,10, 9, 9, 9, 9, 9,10]
60 :  10.18  [10, o, o, o, o, o,11, o, o, o,10, o,10, o, o, o,11, o, 9, o, o, o,10, o, o, o, o, o, 9]
61 :  10.21  [10, 9,10,10,10,10, 9, 9,10,10, 9,11,10,10,10,10, 9, 9,10,10, 9,10, 9, 9,10, 9,10, 9, 9, 9]
62 :  10.24  [10, o,10, o,10, o,10, o,10, o,10, o,11, o,11, o,10, o,11, o,10, o,10, o,10, o,10, o,10, o]
63 :  10.28  [10,10, o,10,10, o, o,10, o,10,10, o,10, o, o,10,10, o,10,10, o,10,10, o,10,10, o, o,10, o,10]
64 :  10.31  [10, o,10, o,10, o,10, o,11, o,11, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10]
65 :  10.34  [10,10,10,10, o,10,10,10,10, o,10,10, o,10, o,11,10,10,10, o,10,10,10,10, o, o,10,10,11, o,10,10]
66 :  10.37  [10, o, o, o,10, o,10, o, o, o, o, o,10, o, o, o, 9, o,10, o, o, o,10, o,10, o, o, o, 9, o, 9, o]
67 :  10.40  [10,10,10,10,10,10,11,10,10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,10,10,10,10]
68 :  10.44  [10, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o, o, o,10, o,10, o,10, o,10, o,10, o,10, o,11, o,10]
69 :  10.47  [10,10, o,10,11, o,10, 9, o,10,10, o,10,11, o,11,11, o,10,11, o,10, o, o,10,10, o, 9, 9, o,10, 9, o, 9]
70 :  10.50  [11, o,10, o, o, o, o, o,10, o,10, o,10, o, o, o,11, o,10, o, o, o,10, o, o, o,10, o,10, o,10, o,10, o]
71 :  10.53  [10,10,10,12,10,10,10,10,10,11,10,12,10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,10,10,10]
72 :  10.55  [10, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,10]
73 :  10.58  [10,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10]
74 :  10.61  [10, o,10, o,10, o,10, o,10, o,10, o,11, o,11, o,10, o,11, o,10, o,10, o,10, o,10, o,11, o,10, o,10, o,10, o]
75 :  10.64  [10,11, o,10, o, o,10,10, o, o,10, o,10,10, o,10,10, o,10, o, o,10,11, o, o,10, o,10,10, o,10,10, o,10, o, o,10]
76 :  10.67  [10, o,10, o,10, o,11, o,10, o,10, o,10, o,10, o,10, o, o, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10, o,10]
77 :  10.69  [10,10,10,10,10,11, o,10,10,12, o,10,11, o,10,11,12,10,10,10, o, o,10,12,12,10,10, o,10,10,10,10, o,10, o,12,10,10]
78 :  10.72  [11, o, o, o,10, o,10, o, o, o,10, o, o, o, o, o,11, o,10, o, o, o,11, o,11, o, o, o,11, o,10, o, o, o,11, o,10, o]
79 :  10.75  [11,10,10,11,11,10,10,10,10,11,11,10,10,10,10,10,10,11,10,10,10,10,10,10,10,10,11,11,10,10,10,11,10,10,10,11,10,10,10]
80 :  10.77  [10, o,10, o, o, o,10, o,10, o,12, o,10, o, o, o,10, o,11, o,11, o,10, o, o, o,10, o,11, o,10, o,10, o, o, o,10, o,10]
81 :  10.80  [10,10, o,12,11, o,10,10, o,10,10, o,10,12, o,10,10, o,10,10, o,10,10, o,10,10, o,11,11, o,10,11, o,10,10, o,10,10, o,10]
82 :  10.82  [10, o,11, o,11, o,11, o,10, o,11, o,11, o,10, o,10, o,11, o,10, o,10, o,10, o,10, o,11, o,10, o,11, o,10, o,10, o,10, o]
83 :  10.85  [10,10,10,10,10,11,10,10,10,10,10,10,10,10,11,10,10,11,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,10,10,10,10,10,10]
84 :  10.87  [10, o, o, o,10, o, o, o, o, o,10, o,10, o, o, o,10, o,10, o, o, o,11, o,10, o, o, o,10, o,10, o, o, o, o, o,10, o, o, o,10]
85 :  10.90  [10,10,11,10, o,12,11,10,10, o,10,12,10,10, o,11, o,10,10, o,10,11,11,10, o,10,11,11,10, o,10,10,10, o, o,10,11,10,10, o,10,10]


For a Fibonacci rectangle, we obviously have $f(F_{k+1},F_k)\le k$, and it seems straightforward to show that this bound is sharp. But is this really trivial?

It looks like under the above restrictions on $m$ and $n$, the values of $f(m,n)$ are very close to $g(m)$, more precisely $$\boxed{\lfloor g(m)\rfloor -1\le f(m,n)\le \lceil g(m)\rceil +1}.$$ Is it possible that in the minimal tilings, patterns like in a 'Fibonacci rectangle tiling' occur frequently?
Note that it is already known or at least plausible from the article quoted in the first thread that $f(m,n)\sim g(m)$.
What about $f(m,n)$ if the rectangle sides are not coprime? Obviously $f(km,kn)\le f(m,n)$ for $k\in \mathbb N$. In the range of the table, there is equality everywhere.

Is anything known concerning the conjecture $f(km,kn)= f(m,n)$?
Moreover, does there even exist a minimal tiling of a $km\times kn$ rectangle such that not all square sides are multiples of $k$?

For $m> n$, let’s call a $m\times n$ rectangle reducible if $f(m,n)=f(n,m-n)+1$. This means there is a minimal tiling such that the biggest square has side $n$. (There may exist other minimal tilings also, though I'd rather doubt it).
Using the same order as above, I have kept in row $m$ only those $n$’s for which the $m\times n$ rectangle is reducible and put "o" where it is not:
 3 : [ 2]
 4 : [ 3]
 5 : [ 4, 3]
 6 : [ o, 4]
 7 : [ o, 5, 4]
 8 : [ o, 6, 5]
 9 : [ o, 7, 6, 5]
10 : [ o, 8, 7, 6]
11 : [ o, 9, 8, 7, 6]
12 : [ o, o, 9, 8, 7]
13 : [ o, o,10, 9, 8, 7]
14 : [ o, o,11,10, 9, 8]
15 : [ o, o,12,11,10, 9, 8]
16 : [ o, o,13,12,11,10, 9]
17 : [ o, o, o,13,12,11,10, 9]
18 : [ o, o, o,14,13,12,11,10]
19 : [ o, o, o, o, o,13,12,11,10]
20 : [ o, o, o,16,15,14,13,12,11]
21 : [ o, o, o,17,16,15,14,13,12,11]
22 : [ o, o, o,18,17,16, o,14,13,12]
23 : [ o, o, o,19,18,17,16, o,14,13,12]
24 : [ o, o, o, o,19,18,17,16,15,14,13]
25 : [ o, o, o, o,20,19,18,17,16,15,14,13]
26 : [ o, o, o, o, o,20,19,18,17,16,15,14]
27 : [ o, o, o, o, o,21,20,19,18,17,16,15,14]
28 : [ o, o, o, o,23,22,21,20,19,18,17,16, o]
29 : [ o, o, o, o,24,23, o,21,20,19,18,17,16,15]
30 : [ o, o, o, o, o,24,23,22,21,20,19,18,17,16]
31 : [ o, o, o, o, o, o, o, o, o,21,20,19,18,17,16]
32 : [ o, o, o, o, o,26,25,24,23,22,21,20,19,18,17]
33 : [ o, o, o, o, o,27, o,25,24,23,22,21,20,19,18,17]
34 : [ o, o, o, o, o, o,27,26, o,24,23,22,21,20,19,18]
35 : [ o, o, o, o, o, o,28,27, o,25,24,23,22,21,20,19,18]
36 : [ o, o, o, o, o, o, o,28,27,26,25,24,23,22,21,20,19]
37 : [ o, o, o, o, o,31, o, o,28,27,26,25,24,23, o,21,20,19]
38 : [ o, o, o, o, o, o, o, o,29, o,27,26,25,24,23,22,21,20]
39 : [ o, o, o, o, o, o,32, o,30,29,28,27,26,25,24,23,22,21, o]
40 : [ o, o, o, o, o, o, o,32, o,30, o,28,27,26,25,24,23,22,21]
41 : [ o, o, o, o, o, o, o, o, o,31,30,29, o,27,26,25,24,23,22,21]
42 : [ o, o, o, o, o, o, o,34,33,32,31,30,29,28,27,26,25,24,23,22]
43 : [ o, o, o, o, o, o, o,35, o, o,32,31, o,29,28,27,26,25, o,23,22]
44 : [ o, o, o, o, o, o, o,36, o,34,33,32,31, o,29,28,27,26,25,24,23]
45 : [ o, o, o, o, o, o, o, o,36,35, o,33,32,31,30,29,28,27,26,25,24,23]
46 : [ o, o, o, o, o, o, o,38, o,36,35,34,33,32,31, o,29,28,27,26,25,24]
47 : [ o, o, o, o, o, o, o, o, o, o, o, o,34,33,32,31, o,29,28,27,26,25,24]
48 : [ o, o, o, o, o, o, o, o,39,38,37,36,35,34,33,32,31,30,29,28,27,26,25]
49 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,35,34,33,32,31,30,29,28,27,26,25]
50 : [ o, o, o, o, o, o, o, o, o,40, o,38,37,36,35,34, o,32,31,30,29,28,27,26]
51 : [ o, o, o, o, o, o, o, o, o,41,40,39, o,37,36,35,34,33,32,31,30,29,28,27,26]
52 : [ o, o, o, o, o, o, o, o, o, o, o,40,39,38, o,36,35,34,33,32,31,30,29,28,27]
53 : [ o, o, o, o, o, o, o, o, o,43, o, o,40, o,38,37,36,35,34,33,32,31, o,29,28,27]
54 : [ o, o, o, o, o, o, o, o, o, o, o,42, o,40,39,38,37,36,35,34,33,32,31,30, o,28]
55 : [ o, o, o, o, o, o, o, o, o,45,44, o, o, o,40, o, o,37, o,35,34,33,32,31,30,29,28]
56 : [ o, o, o, o, o, o, o, o, o,46, o,44,43,42,41,40,39,38,37,36,35,34,33,32,31, o,29]
57 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,43, o,41,40,39,38,37,36,35, o,33,32,31,30,29]
58 : [ o, o, o, o, o, o, o, o, o,48,47,46,45, o,43,42,41,40,39,38,37,36,35,34, o,32, o,30]
59 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,45, o, o, o,41,40, o, o,37,36,35,34,33,32,31,30]
60 : [ o, o, o, o, o, o, o, o, o, o, o,48,47,46,45,44,43,42, o,40,39,38,37,36,35,34,33,32,31]
61 : [ o, o, o, o, o, o, o, o, o, o, o,49, o,47,46, o, o, o, o,41,40,39, o,37,36,35,34,33,32,31]
62 : [ o, o, o, o, o, o, o, o, o, o, o, o,49, o,47, o,45, o,43,42,41,40,39,38,37,36,35,34,33,32]
63 : [ o, o, o, o, o, o, o, o, o, o, o,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37,36,35,34,33,32]
64 : [ o, o, o, o, o, o, o, o, o, o, o,52, o,50,49,48, o,46,45,44,43,42,41,40,39,38,37,36,35,34,33]
65 : [ o, o, o, o, o, o, o, o, o, o, o, o,52, o,50,49,48,47,46,45,44,43, o,41,40,39,38,37,36,35,34,33]
66 : [ o, o, o, o, o, o, o, o, o, o, o,54, o, o,51,50, o,48,47,46, o,44,43,42,41,40,39,38,37,36,35,34]
67 : [ o, o, o, o, o, o, o, o, o, o, o,55, o, o, o,51, o,49,48,47,46,45,44,43, o,41,40,39,38,37, o, o,34]
68 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,54, o,52,51, o,49,48,47,46,45,44,43,42,41,40,39,38,37,36,35]
69 : [ o, o, o, o, o, o, o, o, o, o, o,57, o,55,54,53,52,51,50,49,48,47,46, o,44, o,42, o,40,39,38,37,36,35]
70 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,56,55,54, o, o,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37,36]
71 : [ o, o, o, o, o, o, o, o, o, o, o,59, o, o, o,55,54, o,52, o,50,49,48,47,46,45,44,43, o,41,40,39,38,37,36]
72 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o,57,56, o,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38,37]
73 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,55,54,53,52, o,50,49,48,47,46,45,44,43, o,41,40,39,38,37]
74 : [ o, o, o, o, o, o, o, o, o, o, o,62, o, o,59, o, o,56, o,54,53,52, o,50,49,48,47,46,45, o,43,42,41,40,39,38]
75 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o,60, o, o,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39,38]
76 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,58,57, o,55,54, o,52,51,50,49,48,47,46,45,44,43,42,41,40,39]
77 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,63, o, o,60, o, o,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41,40,39]
78 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,64, o, o,61,60,59,58,57,56,55,54, o,52,51,50,49,48,47,46,45,44,43,42,41, o]
79 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,61, o, o, o, o, o,55, o,53, o,51,50,49,48,47,46,45,44,43,42,41,40]
80 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o,65,64, o, o, o,60, o, o, o,56,55,54,53,52,51,50,49,48,47,46,45,44,43,42,41]
81 : [ o, o, o, o, o, o, o, o, o, o, o, o, o,67, o, o, o,63, o, o,60,59, o,57,56,55,54, o,52,51,50,49,48,47,46,45,44,43,42,41]
82 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,63,62,61,60, o,58, o, o,55,54, o,52, o,50,49,48,47,46,45,44,43,42]
83 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,59, o,57,56,55, o,53,52,51,50,49,48,47,46,45,44,43,42]
84 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o,69,68, o,66, o,64,63,62,61,60,59,58,57,56,55,54,53,52,51,50,49,48,47,46, o,44,43]
85 : [ o, o, o, o, o, o, o, o, o, o, o, o, o, o, o, o,68, o, o,65, o,63,62,61,60,59, o,57,56,55,54,53,52,51,50,49,48,47,46,45,44,43]

The general tendency is clear, but the overall situation looks rather irregular. For some values of $m$, there are much more 'holes' than for others. Any ideas why?

REPLY [10 votes]: For rectangles with maximum size 760 or less, there is one remaining possible counterexample. The 17-square tiling of the 697x611 is not proven minimal.

16 1394 1222 723 671 120 551 499 155 69 1 32 87 39 31 8 55 47 344
17 697 611 51 51 119 119 119 119 119 34 68 34 41 82 82 492 41 205 205   
I've put 4944 more at Possible Counterexamples to the Minimal Squaring Conjecture.
My original example shows two known ways to divide a 2(7125×7081) rectangle into 20 squares.  
 
The smallest known dissection of a (7125×7081) rectangle needs 21 squares.<|endoftext|>
TITLE: This is not a category. What is it?
QUESTION [10 upvotes]: EDIT The question was based on an error, as it turns out. In fact my example is a category (and therefore a groupoid), by Eric Wofsey's argument. I can't remember why I thought it wasn't, and I feel a little silly, but I am very glad that it is. 
In the nerve of a small category an $n$-simplex is determined by $n+1$ objects and $n$ arrows: $$x_0\to x_1\to x_2\to\dots \to x_n
$$
What do you call a simplicial set in which an $n$-simplex is determined by a different part its $1$-skeleton, as follows:
$$
(x_0\to x_1\ ,\ x_0\to x_2\ ,\ \dots\ ,\ x_0\to x_n)\ ?$$
You could say that instead of a composition law for arrows you have a certain kind of decomposition law.
Have people run into this before? What is it called? Do you know any good ways of thinking about it?
I have run into something formally analogous to this (with cosimplicial commutative rings rather than simplicial sets), and I'm trying to get to know it better.
EDIT  I see that I did not say all of what I meant. I should have said that given 
$$
(x_0\to x_1\ ,\ x_0\to x_2\ ,\ \dots\ ,\ x_0\to x_n)\ $$
(i.e. a map into $X$ from the $1$-dimensional object indicated by the above) there is a unique map $\Delta^n\to X$ extending it. By the way, the dual thing is also true in my examples: given 
$$
(x_0\to x_n\ ,\ x_1\to x_2\ ,\ \dots\ ,\ x_{n-1}\to x_n)\ $$
(i.e. a map into $X$ from the $1$-dimensional object indicated by the above) there is a unique map $\Delta^n\to X$ extending it. (In fact, there is also an arrow-reversing involution $X_n\to X_n$ for all $n$, if you know what I mean.)

REPLY [2 votes]: One of the intuitions of (simplicial) T-complexes is that you can make the simplicial singular complex $SX$ of a space into a natural Kan complex, by making choices of the appropriate retractions in the  models. But of course these choices are not unique, and what is unclear is what are the relations among these natural fillers! The idea has something to do with "composition" since in a Kan complex the filler of a horn makes the remaining face in some sense the composition of the other faces; the filler is a "program" for determining this composition. 
Keith Dakin (1977) then axiomatised the idea that these natural fillers are in some sense "thin". So the axioms are: 

Degenerate elements are thin. 
Every horn has a unique thin filler. 
If every face but one of a thin element is thin, then so also is the remaining face. 

A T-complex is of rank $\leq n$ if every element of dimension $>n$ is thin. 
T-complexes of rank $1$ are equivalent to groupoids, and (Dakin) those of rank $2$ are equivalent to crossed modules over groupoids. Nick Ashley (1978) completed the job by showing, among other things, that  $T$-complexes are equivalent to crossed complexes, and giving the topological example, the fundamental $T$-complex of a filtered space. 
These theses are available from here.<|endoftext|>
TITLE: Classification of quasi-split unitary groups
QUESTION [7 upvotes]: Let $U$ be a unitary group defined with respect to an extension $E/F$ of non-archimedean local fields, and assume it is realised with respect to a pair $(V,q)$, where $V$ is an $n$-dimensional vector space over $E$ and $q$ is a hermitian form on $V$. By a decomposition theorem, $V$ decomposes as a sum of hyperbolic planes and another (possibly trivial) hermitian space of dimension at most 2. My question is as follows:


If $n$ is odd, this other hermitian space is in fact a line, and in that case $U$ is quasi-split. If $n$ is even, then $U$ is quasi-split if and only if this other space is trivial (that is, $V$ is really a sum of hyperbolic planes). Where could I find a reference for this characterisation of quasi-splitness?


This is often use in many papers, but I have never seen a reference for it. Any help would be greatly appreciated.

REPLY [8 votes]: There is a text by Scharlau about "Hermitian...". Also the older book by O'Meara.
The point is that, first, over non-archimedean local fields a quadratic form in five or more variables has an isotropic vector. In case the residue characteristic is not two, this has a reasonably elementary direct proof. Then note that a hermitian form is a (special type of) quadratic form in twice as many variables. Thus, there is no anisotropic hermitian form in more than two variables.
Edit: in response to questioner's comment, "quasi-split" means (reductive and) a "Borel subgroup" defined over the field. Then "Borel subgroup" means parabolic subgroup that remains minimal under extending scalars. If the whole space were decomposable as hyperbolic planes and an anisotropic two-dimensional space, any quadratic extension would produce a "smaller" parabolic (the Borel, here, because the minimal parabolic is still next-to-Borel).
About algebraic groups, J. Tits' article in Corvallis is good, also the book Platonov and Rapinchuk, "Algebraic groups and number theory". Both these pay attention to such rationality properties, while many classics (such as Borel's "Linear algebra groups") emphasize the algebraically closed groundfield case.

REPLY [4 votes]: As pointed out in comments, there is a detailed survey by Tits from the algebraic group viewpoint in his AMS proceedings paper, along with quite a few references to earlier literature.    From the viewpoint of groups over local fields, it's worth looking closely at his table of "indices" at the end of the paper.   There he shows concisely which labelled Dyhkin diagrams can occur over various kinds of fields.   In particular, you are interested in the twisted type $^2 \! A_n$ with $n$ even.    He indicates that in the local field case this diagram can occur only relative to a quadratic extension and corresponds then to a quasi-split special unitary group.  In his set-up, "quasi-split" corresponds to the case where $n$ is twice the relative rank $r$ and the Dynkin diagram is folded accordingly.  (In general, various special unitary groups over division algebras are possible.) 
While Tits does not spell out all the details of how such groups are constructed, he does provide a nice overview of the basic algebraic group theory leading to such a classification list.    (Some improvements are contained in his 1971 paper Représentations linéaires irréductibles d'un groupe réductif sur un corps quelconque (MR).)  Also, student of his at Bonn named Martin Selbach elaborated further on the theory:
Klassifikationstheorie halbeinfacher algebraischer Gruppen.
Diplomarbeit, Univ. Bonn, Bonn, 1973.
Bonner Mathematische Schriften, Nr. 83.
Mathematisches Institut der Universitat Bonn, Bonn, 1976. v+140 pp. 
P.S. For the characteristic 0 theory, presented in a different style, you might try the lecture notes by I. Satake (which I haven't looked at in a long time): Classification theory of semi-simple algebraic groups. With an appendix by M. Sugiura. Notes prepared by Doris Schattschneider. Lecture Notes in Pure and Applied Mathematics, 3. Marcel Dekker, Inc., New York, 1971. viii+149.   Satake also used labelled Dynkin diagrams, but with different conventions than Tits ("Satake-Tits diagrams").<|endoftext|>
TITLE: Schemes with no nonconstant maps to lower dimensional schemes 
QUESTION [12 upvotes]: Fix an algebraically closed field $k$ (arbitrary characteristic), all schemes will be of finite type over $k$.
(Property *): I'm interested in (classes of) examples of schemes $X$ (irreducible, of dimension $n$) so that any morphism of schemes $\phi: X \rightarrow Y$ with $\dim Y < n$ is constant.
There are two examples I know of. Projective spaces $\mathbb{P}^n_k$ have this property and simple abelian varieties do too. (One may also put arbitrary non-reduced structures on these, see below).
$\textbf{Claim}$: More generally, if $X$ is a proper irreducible scheme so that every effective divisor is ample (so proper=projective), then $X$ has property (*). (Projective spaces have this property by definition, and simple abelian varieties do too by a general result from Mumford's book.)
Proof: (Eisenbud-Harris give a similar argument for the case of projective space). Let $X$ be as in the theorem and $\phi: X \rightarrow Y$ be a morphism with $Y$ of smaller dimension than $n = \dim X$. Without loss of generality, we may assume $\phi$ is surjective (hence we can pullback cartier divisors). Choose an effective Cartier divisor $D$ and a point $p \not \in |D|$ the support of $D$, but in the image of $\phi$ (since $\phi$ surjective). The pullback of an effective Cartier divisor is also effective, Cartier - hence ample. The pullback of the point will contain a complete curve (hence these two subschemes of $X$ meet by the Nakai-Moishezon criteria - contradicting that $p \not \in |D|$. $\square$
 Easy observations 
(1) Having property $*$ is not stable under blowing-up (Blow up of $\mathbb{P}^2$ is $\mathbb{P}^1 \times \mathbb{P}^1$.)
(2) If a scheme $X$ satisfies the claim, then by definition, so does $X_{red}$. Further, any thickening of $X$ has property $*$.
$\textbf{Proof: }$ To check that an $X$ satisfying the claim satisfies $*$, we did calculations in the intersection ring. This is invariant under changing the non-reduced structure. $\square$
 Questions 
Main question: Are there other (families of?) examples of schemes satisfying (*)?
(1) Does every scheme (no finiteness conditions!) have a dense affine open subset? This came up when I was thinking about this, and I realized I can't prove it offhand. Certainly it is true for irreducible schemes, and suffices to show it for connected schemes.
(2) Do you suspect that the only examples also satisfy the claim above? That is, have every effective divisor ample?
(3) Certainly all examples of schemes satisfying $*$ must be connected. Are there connected, but not irreducible examples?
I thought this was a little interesting (and admittidenly, I have no applications in mind.)

REPLY [8 votes]: The point is not that every effective divisor should be ample, but that every semi-ample (=some multiple is basepoint-free) should have maximal Kodaira-Iitaka dimension. Actually this is an equivalent condition, but it is really just a reformulation of the condition you are looking for. Semi-ample divisors that are not ample lie on the boundary of the ample cone, so a sufficient condition would be something along the lines that the boundary of the ample cone does not contain any (integral) non-zero divisors. 
One probably has to make some assumptions so this would make sense. Also, this is sufficient, but not necessary. That is, if this condition on the cone holds, then your desired condition holds, but your desired condition may hold even without this. In particular, Mumford (first and then others) gave examples of nef but not semi-ample divisors. Those divisors would lie on the boundary as well. Anyway, I don't have time to add more details right now, but I will try to do it later. Alternatively, you can consider this a hint and work it out yourself. (Mumford's example is a ruled surface, so that does not work, but I would expect that one should be able to construct an example that satisfies your condition and still has a nef but not semi-ample divisor).
Sasha's example of a K3 surface falls into this category. If there are no elliptic pencils, which is equivalent to there being no smooth elliptic curve on the surface, then the only non-trivial morphisms of the surface are the contraction of some $(-2)$-curves, but those morphisms are birational, so the target has the same dimension.
For a reducible example take two varieties of the same dimension, both of which has Picard number one and who intersect each other. For instance two intersecting planes. If the morphism maps to something lower dimensional, it has to be constant on both planes but they have to map to the same point.<|endoftext|>
TITLE: Equivalences of Internal Categories
QUESTION [8 upvotes]: I am trying to understand double categories and their relatives a little better.  I do not understand much, so I apologize in advance if my question is too naive for this website.
A functor $F:X \rightarrow Y$ of ordinary $2$-categories is an equivalence if and only if:
(1) It is essentially surjective.
(2) It is fully faithful, in the sense that it induces equivalences of Hom categories.
Are there analogues of conditions (1) and (2) that detect equivalences of double categories?
To ask my question in a different way, section A.3.2 of Lurie's Higher Topos Theory constructs a model structure on $S$-enriched categories where $S$ is a monoidal model category satisfying some nice properties (an alternative set of properties were recently discovered by Berger and Moerdijk).  In Lurie's model structure, an $S$-enriched functor $F:X \rightarrow Y$ is an equivalence if and only if
(1) $Map_X(x,x') \rightarrow Map_Y(F(x),F(x'))$ is always a weak equivalence in $S$
(2) $F$ induces an essentially surjective map of homotopy categories $Ho(X) \rightarrow Ho(Y)$.  Here, arrows from $x$ to $x'$ in $Ho(X)$ are homotopy classes of morphisms from the unit of $S$ to $Map_X(x,x')$.
Suppose I have a nice model category $S$, for some definition of nice I am willing to determine.  Can I expect there to be a model structure on category objects in $S$, analogous to the above model structure on $S$-enriched categories?  What would be the analogues of conditions $(1)$ and $(2)$?

REPLY [4 votes]: As David says, the right notion of equivalence of double categories depends on what sort of double categories you're looking at, and what you want to use them for.  You might also be interested in Theorem 7.7 of this paper.<|endoftext|>
TITLE: orbit space of a topological manifold
QUESTION [11 upvotes]: Given a compact Lie group G acting freely on a topological manifold M, is it true that the orbit space M/G is also a topological manifold? If so, why?

REPLY [15 votes]: The answer is no. Bing constructed a space $X$ (called the dogbone space) so that $X$ is not a manifold, but $X\times R$ is homeomorphic to $R^4$. In particular, $M^4=X\times S^1$ is a $4$-manifold (since its universal cover is $R^4$) and $X$ is the quotient of $M^4$ by free $S^1$-action.<|endoftext|>
TITLE: larger coherent family of functions
QUESTION [7 upvotes]: One way to construct an Aronszajn tree is to build a sequence of functions $\langle e_\alpha : \alpha < \omega_1 \rangle$ such that $e_\alpha$ is an injection from $\alpha$ to $\omega$, and for any $(\alpha, \beta)$, $e_\alpha$ agrees with $e_\beta$ except on a finite set.
Is the following generalization possible?  There is a family of functions {$e_x : x \in [\omega_2]^\omega$} such that each $e_x$ is an injection from $x$ to $\omega$, and for any $x,y \in [\omega_2]^\omega$, {$\alpha : e_x(\alpha) \not= e_y(\alpha)$} is finite.

REPLY [6 votes]: To add to Joel's answer; you can construct such a family, so long as you're willing to relax the injectivity requirement to just "finite-to-one".  See P. Koszmider, "On Coherent Families of Finite-to-One Functions", JSL 58 (1993) no. 1, 128-138.  It's in JSTOR.
Koszmider proves that you can construct such a family (with functions that are finite-to-one) on any $[\omega_n]^\omega$ in ZFC, and any $[\kappa]^\omega$ ($\kappa$ an infinite cardinal) assuming $V = L$.<|endoftext|>
TITLE: Counting spanning trees when blowing up vertices
QUESTION [9 upvotes]: I have a cubic graph $G$ with $\tau(G)$ spanning trees. Now I replace each vertex in $G$ by a triangle giving me a new graph $G'$ - this operation is sometimes referred to as blowing up the vertex to a triangle or truncating the vertex. I want to find a formula for the number of spanning trees in $G'$.
For example, the complete graph on 4 vertices $K_4$ has 16 spanning trees, and blowing up each vertex in $K_4$ gives a graph on 12 vertices with 6000 spanning trees. If it is not possible to give a formula for blowing up the vertices in a general cubic graph, a formula for the specific case where the starting graph is $K_4$ is much appreciated.
As an example of how the operation works, I have provided a drawing of the graph obtained from performing the operation once on $K_4$.
(source)

REPLY [11 votes]: The resulting graph is the linegraph of the subdivision graph of $G$.  This survey paper of Bojan Mohar tells how to obtain the Laplacian spectrum of the linegraph of a semiregular graph and the subdivision graph of a regular graph.
Let's generalize.  Let $G$ be a regular graph of $n$ vertices, degree $d$, and therefore $m=nd/2$ edges.  Let $\mu(G,x)$ denote the characteristic polynomial
of the Laplacian matrix, and let $\kappa(G)$ be the number of spanning trees.
The blowup $B(G)$ of $G$, formed by replacing each vertex by a $d$-clique, is
the linegraph of the subdivision graph of $G$.  Using Theorems 3.8 and 3.9 in the
survey paper of Mohar, we find
$$ \mu(B(G),x) = (-1)^n (x-d)^{m-n} (x-d-2)^{m-n} \mu(G,x(d+2-x)). $$
We know that $\kappa(G) = n^{-1} (-1)^{n-1} \mu'(G,0)$.
Differentiating and using $\mu(G,0)=0$, we find 
$$ \kappa(B(G)) = d^{m-n-1} (d+2)^{m-n+1} \kappa(G). $$
For the $k$-fold blowup, we have
$$ \kappa(B^k(G)) = d^{d_k(m-n)-k} (d+2)^{d_k(m-n)+k} \kappa(G), $$
where $d_k=1+d+\cdots+d^{k-1}$.
For $d=3$, I believe this gives
$$ \kappa(B^k(G)) = (5/3)^k 15^{(3^k-1)n/4} \kappa(G). $$<|endoftext|>
TITLE: Positive results coming from paradoxes
QUESTION [7 upvotes]: Many examples comes to mind, the most famous being the Gödel's theorems viewed as formalisations of the Liar's paradox. I just realised that the proof of non-calculability of Kolmogorov complexity is a positive rewriting of Berry's paradox. My question (perhaps to be made into collective mode) is a) what are the best examples you know ? b) (more important) is there some explication of this productivity of paradoxes (or, conversely, do you know of paradoxes with no interesting follow-up) ?

REPLY [5 votes]: Zeno's paradox on Achilles and the tortoise is related to the formula for infinite geometric sums and more generally to the idea that infinite sums can lead to finite outcomes. 
It is quite remarkable how relevant 17th century calculus is to Zeno's three paradoxes. In fact, it looks that in a different universe these paradoxes could have started calculus. Terry Tao remarked on some post I made about it: "Zeno's arrow paradox can be reinterpreted in the light of the theory of differential equations that the equations of motion must be second-order in time rather than first-order, since one has to specify initial velocity in addition to initial position in order to have a well-posed system. So the arrow paradox may well be the earliest precursor of Newton's famous equation F=ma..."<|endoftext|>
TITLE: $\Delta^1_2$-well ordering vs $\Delta^1_3$
QUESTION [8 upvotes]: It is a classical result that if $0^{\sharp}$ exists, then there is a model of $ZFC$ in which there is a $\Delta^1_3$ well ordering of reals but no $\Delta^1_2$-well ordering.
My question is: Is $0^{\sharp}$ necessary to prove this? Or does the statement $\mathbf{CON}(ZFC+\mbox{there is a }\Delta^1_3\mbox{ well ordering of reals but no }\Delta^1_2 )$ implies the consistency of the existence of $0^{\sharp}$?

REPLY [7 votes]: Hi Yu,
No, your statement is equiconsistent with $\mathsf{ZFC}$. In 

Leo Harrington. Long projective wellorderings, Annals of Mathematical Logic 12 (1977) 1-21, MR0465866 (57 #5752).

it is shown that it is equiconsistent with $\mathsf{ZFC}$ to have a boldface $\Delta^1_3$ well-ordering and Martin's axiom. But the existence of boldface $\Delta^1_2$ well-orderings implies that the reals are the reals of $L[r]$ for some real $r$ (this is a result of Mansfield), and this fails under Martin's axiom. 
(The boldface in Leo's result can be made lightface at the cost of more complicated coding techniques. This was shown by Sy Friedman, see his book on "Class forcing". If we do not care about also having $\mathsf{MA}$, Leo's paper also shows how to obtain from $\mathsf{ZFC}$ models with lightface $\Delta^1_3$ well-orderings.)
Going beyond $\mathsf{MA}$, recently, Sy and I proved that if $\mathsf{BPFA}$ holds and $\omega_1=\omega_1^L$, then there is a lightface $\Delta^1_3$ well-ordering of $\mathbb R$. Curiously, it is still open whether starting with $L$ and using the standard forcing for $\mathsf{MA}$ ($+2^\omega=\omega_2$) results in a model with a definable well-ordering of $\mathbb R$.  
On the other hand, there is a connection with sharps: Assume that there are no inner models with a strong cardinal (or there is a measurable, but there are no inner models with a Woodin cardinal). If all reals have sharps, and there is a $\Delta^1_3(r)$ well-ordering (for some real $r$), then the reals are the reals of the core model  $K_r$. This was first noted by Welch. 

Some references:

Mansfield result was nicely reproved by Alekos: Alexander Kechris. The perfect set theorem and definable wellorderings of the continuum, J. Symbolic Logic 43 (1978), no. 4, 630–634, MR0518668 (80b:03067).
Sy D. Friedman. Fine structure and class forcing, de Gruyter Series in Logic and its Applications, 3. Walter de Gruyter & Co., Berlin, 2000, MR1780138 (2001g:03001).
Andrés E. Caicedo, Sy D. Friedman. $\mathsf{BPFA}$ and projective well-orderings of the reals, J. Symbolic Logic 76 (2011), no. 4, 1126–1136, MR2895389 (2012m:03123).
For Welch's result, see: Ralf Schindler. Coding into $K$ by reasonable forcing, Trans. Amer. Math. Soc. 353 (2001), no. 2, 479–489, MR1804506 (2002c:03083).<|endoftext|>
TITLE: Where does the notion of pseudoholomorphic curve come from?
QUESTION [8 upvotes]: I wonder, why we consider the notion of pseudoholomorphic curve: By definition a pseudoholomorphic curve in an almost complex manifold $X$ is a smooth map $f: C \rightarrow X$
from a Riemann surface $C$ into $X$ such that $df \circ J=J \circ df$ for the respective almost complex structure $J$. Why does it make sense to look at this definition? Is it just because the Cauchy-Riemann differential equations $\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$, $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$
are invariant under the symmetry $x \rightarrow y$, $y \rightarrow -x$ and $u \rightarrow v$, $v \rightarrow -u$ and therefore we have that this relation holds for a holomorphic map $f$ or is there any other reason?

REPLY [2 votes]: The work of Floer in proving the Arnold conjecture (at first in the monotone setting) interpreted pseudoholomorphic curves as a generalization of the flowlines in Morse theory.  In analogy with Morse homology, a count of pseudoholomorphic curves provides the differential in what has become known as a Floer homology.
This work predated Gromov's paper and has been seminal in its own way, spawning a variety of Floer homologies (such as the popular Heegaard-Floer homology) which provide strong invariants in low-dimensional topology.
If you're willing to accept that symplectic manifolds are reasonable things to study, then Floer's approach to the Arnold conjecture shows pseudoholomorphic curves arising very naturally.  Salamon has a nice paper explaining this carefully http://www.math.ethz.ch/~salamon/PREPRINTS/floer.pdf .<|endoftext|>
TITLE: Symmetric products of smooth non-proper curves over generalized Jacobians
QUESTION [8 upvotes]: Does anyone know a written reference for the following fact?
For large n, $\operatorname{Sym}^n X \to \operatorname{Jac}^nX$ is a vector bundle, where $X$ is a smooth, non-proper curve, and $\operatorname{Jac}X$ is its generalized Jacobian, so $\operatorname{Jac}^nX = \operatorname{Pic}^n X^+$ where $X^+$ is the one-point compactification given by the quotient of the smooth compactification $Xc$ by $Xc -X$.
(I know how to prove it-- I would like to be able to cite a reference for it.)

REPLY [5 votes]: I think I should have said "affine bundle" instead of "vector bundle." I still haven't found a reference, but I wrote a proof in the appendix of: 
http://www.math.harvard.edu/~kwickelg/papers/delta2real.pdf<|endoftext|>
TITLE: Integral cohomology operations related to Landweber-Novikov
QUESTION [7 upvotes]: Let $U^* \rightarrow H^* $ be the homomorphism describing the complex orientation of $H^* $ from complex cobordism. Let $t_1, t_2, ...$ be indeterminates. 
My question is: Does there exist an integral cohomology operation $H^* (X) \rightarrow H^*(X)[\mathbf{t}]$ that makes the following diagram commute for any $X$? If so can we describe it in a way that doesn't involve complex cobordism?

Here $s_{\mathbf{t}}$ is the total Landweber-Novikov operation.
In particular, since the Landweber-Novikov operation satisfies the Riemann-Roch type formula for proper, complex-oriented maps $$ s_{\mathbf{t}} f_{*}x = f_{*}(c_{\mathbf{t}}(\nu_{f}) \cdot s_{\mathbf{t}}x)$$
(where $\nu_{f}$ is the virtual class $1-\nu_i$ and $\nu_i$ is the stable normal bundle of the proper, complex-oriented map $f: Z \rightarrow X$) we would expect ths operations to satisfy something similar. 
EDIT: Whoops, original title didn't have to do with the question :) But it did have to do with the motivation behind asking it! Maybe another time...

REPLY [5 votes]: This is more of a comment than an answer. I just wanted to point out that such an operation, if it were to exist, cannot be stable.
This follows from the answer to Integral cohomology (stable) operations, which asserts that the set of stable cohomology operations in integral cohomology of a given degree $2k>0$ is finite, and the observation that $s_{(k)}(cf_1) = cf_1^{k+1}$, where $cf_1\in U^2(BU)$ is the first universal Conner-Floyd-Chern class.
To see this, suppose that there is a stable operation $\sigma_{(k)}:H^\ast(-)\to H^{\ast+2k}(-)$ which extends the stable operation $s_{(k)}:U^\ast(-)\to U^{\ast+2k}(-)$. Then we would have
$$
\sigma_{(k)}(c_1)= c_1^{k+1},
$$ where $c_1\in H^2(BU)$ is the first universal Chern class. But since the sum of stable operations is stable, we would have
$$
n\sigma_{(k)}(c_1)=0
$$
for $n$ sufficiently large, contradicting the fact that $nc_1^{k+1}\neq 0$.<|endoftext|>
TITLE: Is an eigenvector of a Hecke operator automatically an eigenform?
QUESTION [8 upvotes]: By definition, an eigenform is a simultaneous eigenvector for all the Hecke operators $T_n$. Suppose I know that $f$ is an eigenvector of a particular Hecke operator, say $T_N$, does it follow that $f$ is also an eigenvector for all the other Hecke operators? Or are there counter-examples to this?

REPLY [9 votes]: Following the comments, here is perhaps the simplest counterexample (once you know the Breuil-Conrad-Diamond-Taylor modularity theorem).  The curves $y^2 + y = x^3$ and $y^2 + y = x^3 + 2x$ both reduce mod 2 to the same smooth curve, so their Hecke eigenforms have equal $T_2$ eigenvalues.
However, the curves over $\mathbb{Q}$ are non-isomorphic, since only the first curve has vanishing $j$-invariant.  This means the Hecke eigenforms have different coefficients for some prime $p$, and hence different eigenvalues for $T_p$ (in fact, by counting the mod 3 solutions, you can see that $p=3$ works).
If you add the eigenforms, you get a (non-normalized) cusp form that is an eigenfunction for $T_2$, but not for $T_p$.
Edit I might as well say what I know about the level 1 case.  It is "classically" known that Hecke operators are self-adjoint with respect to the Petersson inner product on cusp forms, and they commute with each other.  By the spectral theorem, the Hecke operators are therefore simultaneously diagonalizable, i.e., there is a basis of the space of cusp forms made out of eigenforms.
Now, let us assume all of the eigenvalues of all of the Hecke operators are distinct (in contrast to the counterexample above).  If you have a form $f$ for which some $T_n$ acts by a scalar, then $f$ cannot be a non-trivial linear combination of some basis of eigenforms, i.e., it is necessarily an eigenform.  This multiplicity one property is not known in level one in general, but it is known to be true for $T_2$, as I learned in Cardinal Wolsey's brilliant answer to your previous question (the answer was deleted, it seems, because you didn't accept it within 4 days).
More generally, the multiplicity one property is implied by a conjecture of Maeda, which asserts that Hecke operators act irreducibly on $S_k(SL_2(\mathbb{Z}))$ for each $k$.  In fact, David Loeffler mentioned, in a comment on the previous question, that there is a lot of computational evidence behind an even stronger assertion, namely that the eigenvalues generate as-big-as-possible Galois extensions over $\mathbb{Q}$.<|endoftext|>
TITLE: The Unreasonable Effectiveness of Physics in Mathematics. Why ? What/how to catch?
QUESTION [51 upvotes]: Starting from 80-ies the ideas either coming from physics, or by physicists themselves (e.g. Witten) are shaping many directions in mathematics. It is tempting to paraphrase E. Wigner, saying about  "The Unreasonable Effectiveness of Physics in Mathematics".   
What can be the reasons for it ? 
Do physicists have some tools/ideas/techniques which allow them to make insights, which
are not seen for mathematicians? Or it is just because Witten&K are very ... very smart ?
If, yes, what are these tools/ideas ? How to learn/absorb/(put into math. framework) them? 
What can be the further applications of these ideas ? 

Being a mathematician, but working in physicists surrounding for many years,
I have thought on this questions for quite a while. The recent MO question 
Mathematician trying to learn string theory prompts me to ask it here.
I would think, that yes, there are such "ideas". But  from some outstanding mathematicians I've heard an opposite opinion.
My vague feeling is that quantum field theory and string theory it is something like
an analysis/differential geometry on infinite-dimensional manifolds. But these manifolds are not abstract, say Banach modeled manifolds, which theory is not so rich, but kind of maps from one finite-dim. manifold to another, which has certain specific structures which are not fully revealed by mathematicians. E.g. vertex operator algebras, arise from maps of circle to manifold, if we map not circle but something higher dimensional there should be something more complicated. Another issue is about Feynman integral, which
 allows physicist to use integration techniques in geometric problems, it is not well-defined mathematically, but it might be it cannot be defined in very general form of infinite-dimensional integrals, but again physicist have an intuition where it can be defined, where cannot, and proper mathematical theory should first clarify the setup
where it exists, rather than trying to build general theory which might not exist.
These words are probably very vague, so might be answers help to me clarify.

I think everybody knows the influence of physics happened from 80-ies, but
for completeness  let me mention just a few.
Donaldson used instanton moduli spaces in his study of 4-folds.
Faddeev, Drinfeld et. al. created quantum groups
Representation theory of infinite-dimensional algebras have been large influenced by conformal field theory developments.
Witten's contributions are numerous his Fields Medal says more than I can say.
Mirror Symmetry, quantum cohomology etc...
The works of Fields Medalist Kontsevich and Okounkov are largely influenced by physics.
So on an so forth...

REPLY [4 votes]: If you accept Max Tegmark's view that "our physical world is an abstract mathematical structure" (per https://arxiv.org/abs/0704.0646), then it would follow that the "effectiveness" of physics in mathematics is not unreasonable at all. It's exactly what you would expect.

REPLY [2 votes]: It is my experience that physics motivates and allows the consideration of objects, with some degree of rigor, whose nature would be very difficult to understand mathematically with no motivation and prior training.  
Things like manifolds and line integrals are early examples of objects whose involvement in physical theories allowed me to understand their workings through context long before I had developed sufficient mathematical language and machinery to really articulate clearly what it was I understood. 
I spent the last semester and the beginning of this summer constructing $\mathbb{R}$ (and larger real-closed fields up to the Surreals) directly out of cuts in a subclass of $O_n\times O_n\times O_n\times O_n~\;$  using set theoretical axioms (https://arxiv.org/abs/1706.08908*). Two of my math professors were reviewing the paper with me once a week, an analyst and an algebraist by training, and both of them repeatedly wondered what made me think of trying something like this.  I was motivated by the fact that infinities seem to be inextricably working their way into the leading edge of problems in physics -- for example, it is my understanding that one of the disagreements between GR and QM is the existence some nasty divergent series describing a particles position and momentum states in a quantized setting, something that is typically handled through renormalization but which can't be here.
I figured that maybe these series would converge to some value in a larger, denser real-closed field, and I wanted to build those fields in the same way that $\mathbb{R}$ can be built out of $\mathbb{N}$ since so much of quantum mechanics relies on analysis together with indexing over some well defined discrete ordered subset. To do so I constructed larger discrete ordered rings that contained the integers out of $\delta$-numbers greater than $\omega$, then created fields of fractions for them and 'cut up' the fields to real-close them. I also algebraically closed all of these fields so we have an analogous 'non-Archimedean' version of $\mathbb{C}$ to work with. All of this was motivated by happenings in physics, or as another answer elegantly put it 'considerations of Nature'.
*currently under review for publication<|endoftext|>
TITLE: Easy functions?
QUESTION [21 upvotes]: Let $f$ be an analytic function, and suppose that we want to compute
$f(x)$. The input consists of the digits of $x$ and the output of
a rational number approximating $f(x)$. A function $f$ is called easy if
there is
an algorithm which computes $f(x)$ with accuracy $2^{-n}$
using $n^{1+o(1)}$
arithmetic operations.
It is known that elementary functions like $e^x,\log x$ are easy.

Is it known (proven) about any reasonable function
  that it is hard (not easy)?

For an algorithm, using the AGM, showing that $e^x$ is easy, a reference is
D. Newman, Rational approximation versus fast computer methods,
Lectures on approximation and value distribution, pp. 149.174,
Sém. Math. Sup., 79, Presses Univ. Montréal, Montreal, Que., 1982.
EDIT1: The same paper contains a proof that multiplication is easy (fast multiplication), 
and if $f$ is easy then
the inverse function is easy (Newton's method).
EDIT2: I understand that with our present knowledge we cannot compute Euler's constant
efficiently. But I don't know a proof that this is impossible.
Remark. I am mostly interested in analytic functions, even "special functions". Are they all easy?

REPLY [3 votes]: If we consider constant functions (trivially analytic...), then we could change the original question to: Are there "reasonable" real numbers that are not computable in quadratic time?
As there is a time hierachy theorem on the real numbers (Norbert Th. Müller: Subpolynomial Complexity Classes of Real Functions and Real Numbers. ICALP 1986: 284-293), there exist numbers in qubic time that are not in quadratic time. Whether there numbers are as "reasonable" as $\pi$ or $e$, however, is another question...<|endoftext|>
TITLE: Energy quantization for $J$-holomorphic spheres
QUESTION [16 upvotes]: Let $(\mathbb{CP}^1, j, g_{\text{FS}})$ be the complex projective line with the standard complex structure and the Fubini-Study metric and let $(M,J,\omega,g)$ be an almost Kähler compact manifold ($\omega$ is closed, $J$ is not neccesarily integrable). Using energy estimates, it can be shown that there is a constant $\hbar > 0$ that depends on $J$ and $g$ (and $\omega$) such that any non-constant $J$-holomorphic sphere $u : \mathbb{CP}^1 \rightarrow M$ satisfies
$$ E(u) = \int_{\mathbb{CP}^1} |du|^2 \mathrm{dvol}_{\text{FS}} \geq \hbar. $$
That is, a non-constant $J$-holomorphic sphere must have an energy of at least $\hbar$. A more elaborate analysis can show that the set of possible "energy levels"
$$ \{ E(u) \}_{u \; \text{is } J-\text{holomorphic}} $$
is a discrete set. This phenomenon also occurs for two-dimensional harmonic maps and presumably in other settings which I'm less familiar with.

Now, this smells in some sense like a "quantization". Smooth spheres $u: \mathbb{CP}^1 \rightarrow M$ are much less rigid with respect to energy - we can pertube them a little, changing the energy a little. However, once we impose some conditions on the maps (being $J$-holomorphic, harmonic), we get a discrete "spectrum", finite dimensional moduli spaces, etc.
Is there any physical interpretation behind this phenomenon? Any hueristic that explains intuitively why this should happen (maybe as some sort of a "quantization")? 
Is there a theory in which the set of possible energy levels have some physical meaning, analogous to the a spectrum of some self-adjoint operator in quantum mechanics? Can one think of the set of possible energy levels as the "spectrum" of the non-linear $\bar{\partial}_J$ and learn interesting things on the (almost complex) geometry of $M$ from it?

REPLY [10 votes]: There is indeed something that links J-holomorphic curves, harmonic forms and energy eigenstates in quantum mechanics: a variational principle. That is to say, they can all be characterized by the extremization of a certain functional. This offers an intuitive understanding of their quantized behaviour: if you draw a random curve on a sheet of paper, the local minima will typically be a discrete set. In fact I would say the variational principle is a deeper link than the quantization, but let me first get into more detail:

A J-holomorphic curve is usually of the form $u: (\Sigma_g,j) \to (M,J,\omega)$ with the domain a Riemann surface, and the codomain a symplectic manifold with an $\omega$-tame $J$, such that $u$ is J-holomorphic (i.e. $\bar \partial_J (u) := \mathrm d u + J \circ \mathrm d u \circ j = 0$). Let us for the moment allow general smooth maps $u$, then it's straight-forward to show that if $J$ is in fact $\omega$-compatible:$\qquad \qquad \qquad \boxed{E(u)} := \frac{1}{2} \int |\mathrm d u|^2 \; \mathrm d \textrm{vol} = \boxed{\int |\bar \partial_J u|^2 \; \mathrm d \textrm{vol} + \int u^* \omega}$
Since J-holomorphic curves are exactly those with $\bar \partial_J u = 0$, we see that they extremize this energy functional, leaving the usual topological term $\int u^* \omega$.
Another analytic object that gives rise to quantization are harmonic forms: for example Hodge theory tells us (when) they represent the cohomology of our manifold (or relatedly there's the quantization of harmonic maps, as you mention). To understand them from a variational perspective: suppose we have a de Rham cohomology class $[\alpha] \in H^k(M)$ and let's look for the representative that minimizes the $\int g(\cdot, \cdot)$-norm. It should be such that for all $\beta \in \Omega^{k-1}(M)$ we have

$\qquad \qquad \qquad \frac{\mathrm d}{\mathrm dt} \langle \alpha + t\; \mathrm d \beta, \alpha + t \;\mathrm d \beta \rangle \big|_{t=0} = 0$

I.e. we want $\langle \alpha , \mathrm d \beta \rangle = \langle \delta \alpha, \beta \rangle = 0$ for all $\beta$, which means $\boxed{\delta \alpha = 0}$. Since $\alpha$ is a cohomology representative, it's closed $\boxed{\mathrm d \alpha = 0}$. These conditions are equivalent to being harmonic: $\boxed{\Delta \alpha = 0}$. The harmonic representative of a cohomology class is the one that extremizes the norm!
Finally, one formulation of quantum mechanics is in terms of path integrals. Classical point particles correspond to extrema of the classical action, but in fact there's a variational principle for the quantum wave functions such that the energy wavefunctions come out. I might have to come back at a later time to flesh this out.

So a variational principle is present. The reason I would give this more importance than the quantization, is because the latter depends on our boundary conditions. For example, if our manifold is not compact, then there is no quantization for $E(u)$, or no Hodge theory. Or similarly if instead of a quantum wave function in a potential well we have one in free space, energy won't be quantized. Yet even in these cases, the variational principle can still make sense.
But is this an answer? Or are we left with the question:

Why should the J-holomorphic curve (et cetera) be characterized by a
  variational principle?

To me this seems like a very interesting question to which I sadly have no answer!<|endoftext|>
TITLE: Status of the Isomorphism problem for automatic groups?
QUESTION [7 upvotes]: I only ask because I don't know how to look for the answer.

REPLY [7 votes]: It's still an open problem.  The isomorphism problem for hyperbolic groups (a much smaller class) was only solved recently by Dahmani and Guirardel (see here), following work of Sela.
In the same vein, the conjugacy problem is also still open for automatic groups.  It has been solved for biautomatic groups, but it is also still open whether all automatic groups are biautomatic.<|endoftext|>
TITLE: Where do the Kähler Identities first appear?
QUESTION [7 upvotes]: The Kähler identities (sometimes known as the Hodge identities) are an important collection of relationships between operators on the exterior algebra of a Kähler manifold. These relationships generalise to hermitian manifolds, sections of hermitian holomorphic vector bundles, and many other situations. 
I know that the notion of a Kähler metric was introduced by Kähler himself in 1933 and that the Kähler identities were first generalised to hermitian manifolds by Demailly in 1985 (although he mentions that the ideas were present in a paper by Griffiths in 1966). 

Can anyone fill in the historical gap and tell me where or when the Kähler identities first appeared?

REPLY [7 votes]: As you hinted yourself, they were indeed discovered by W.V.D. Hodge. They appear explicitly in his 1941 book "The Theory and Applications of Harmonic Integrals", which you can find here, see e.g. the lemmas on pages 172 and following. They were discovered by him some times in the 30s, see e.g. this 1935 paper of his.
Hodge's notation is a bit different from the modern one. The first place that I know where they appear in the modern form is A.Weil's book "Introduction a l'étude des variétés kählériennes" (1957), available here, Theoreme 1, p.42. The modern notation was probably introduced by Weil, since Kodaira in 1952 called the operator $\Lambda$ the "Hodge-Weil operator".<|endoftext|>
TITLE: I'll admit it: I don't understand the definition of the Easton product.
QUESTION [14 upvotes]: I'm teaching myself bits and pieces of forcing at the moment, for the purposes of translating them into sheaf-theoretic versions. I'm trying to write down what I feel is a cleaner description of the Easton product of forcing posets, by which I mean a global description rather than one in terms of elements like
$$
\left|{(\kappa,α,β) ∈ dom(p) : \kappa\leq \lambda}\right| \lt \lambda \qquad              (1)
$$
for $p$ in the Easton product $\prod^E P(\kappa)$, which is not very useful at the level of the category of sets if it is not the category of ZF(C)-sets.
I'm confident I understand what this condition does when you do the forcing, namely I think it puts a bound on how many sets are added to any given $\lambda$, else you might get silly things like a proper class of sets added to some 
set.
However, at the risk of embarrassing myself, and in the interest of educating others, here is my best guess for what this condition translates to. Consider, as Jech does (Set theory, 3rd edition), first the Easton product over some set $A$ of regular cardinals. For $\kappa\in A$, let $P(\kappa)$ be the set of functions $p:D(p)\to 2$ where $D(p)\subset \kappa\times B(\kappa)$ has cardinality less than $\kappa$. Here $B(\kappa)$ is some cardinal, not necessarily
given by an Easton function on regular cardinals, and forcing using the 'usual order'
on this poset will add $B(\kappa)$ subsets to each $\kappa\in A$. There is a distinguished
element $\top$ of each $P(\kappa)$, namely the unique function $\emptyset \to 2$.
Let $P=\prod P(\kappa)$ be the product over $\kappa\in A$. The Easton product  is the subset 
consisting of those $p\in P$ such that the support condition (1) holds. So what
does this mean? An element $p\in P$ is a collection of functions $p_\kappa$, one for each $\kappa\in A$. Then let $supp(p) \subset A$ be the set of those $\kappa\in A$ such that $p_\kappa\not=\top$.
Analogously, define $supp_\lambda(p) \subset A$ to be the set of those $\kappa\leq \lambda$ such that $p_\kappa\not=\top$. This last definition is where I am the most unsure, as the definition in Jech really involves $supp(p)\cap \lambda$, which doesn't make sense from a
structural set theory point of view unless $\lambda$ is viewed as a subset of $A$ (even though it makes perfect sense in a material set theory such as ZFC).
Now assuming the definition of $supp_\lambda(p)$ is correct, the support condition on $p$ as first stated in Jech is that
$$
\forall \text{ regular } \lambda, \left|supp_\lambda(p)\right| \lt \lambda.\qquad               (2)
$$
Apparently it's enough to enforce (2) whenever $\lambda$ is weakly inaccessible, (though Friedman just says 'inaccessible'), and that's fine by me. However, this doesn't look the same as the condition (1), which has me slightly worried. The condition (1) imposes a condition on the size of the domain of $p$ as well as (2), and in fact implies that the domain of $p_\kappa$ is smaller than $\kappa$ for all $\kappa$.
So where has my reasoning gone wrong? At this point I really just want to understand the set underlying the Easton product, rather than any intricacies of class forcing.

REPLY [17 votes]: Here is a general way to think about product forcing, which may be
helpful. One has forcing notions $\mathbb{Q}_\gamma$ for each
$\gamma$ in a class $D$ of ordinals. (In your example, $D$ is the
class of regular cardinals, and
$\mathbb{Q}_\gamma=\text{Add}(\gamma,E(\gamma))$ for the Easton
function $E$.) One wants to define the notion of product forcing
$\Pi_\gamma\mathbb{Q}_\gamma$.
The idea is that fundamentally different product forcing notions
are obtained by allowing different kinds of support in the
product. This is exactly the same issue as the difference between
direct sums and direct products in algebra. For each ideal $I$ on
$D$, one may use that ideal as collection of allowed supports, so that the product $\Pi_\gamma\mathbb{Q}_\gamma$ with supports
in $I$ consists of the functions $p:\gamma\mapsto
p(\gamma)\in\mathbb{Q}_\gamma$, such that $\{\gamma\mid
p(\gamma)\neq\top\}\in I$. That is, we insist that the support of
a condition $p$ is in the ideal.  It is very convenient for the collection of allowed supports to be an ideal, since we often want to manipulate conditions, either by weakening them on each coordinate or by combining two conditions that are compatible on each coordinate, and this leads us to want an ideal, which is closed under subsets and finite unions. Several natural cases arise:

If we use the ideal of all finite sets, then this product is
called the finite-support product, and this corresponds to taking
what in set theory is called the direct limit at each limit stage,
and this can be characterized in terms of a nice universal
property of how the product relates to the factors
$\mathbb{Q}_\gamma$. This is like the direct sum.
If we use the ideal of all countable sets, the product is called
a countable-support product.
If we use the (improper) ideal of all subsets of $D$, then the
product is called the full-support product, and this corresponds
to using inverse limits at every stage. This also has a nice
characterization by a universal property, like the direct product.
The Easton support ideal consists of sets $A$, which are bounded
below every inaccessible cardinal. This ideal corresponds to
taking direct limits at every inaccessible cardinal stage, and
inverse limits at all other limit stages.

This ideal perspective makes sense whether one speaks of product
forcing or iterated forcing. Although Easton used his support with
product forcing, there are now numerous applications of Easton
forcing with iterated forcing.
Each of these ideals and others finds numerous applications in
forcing. For example, the finite support iteration of ccc forcing
is necessarily ccc. The countable support iteration of proper
forcing remains proper.
Note that the Easton ideal is exactly the same as the improper
ideal of all sets, if there should happen to be no inaccessible
cardinals. Easton realized, however, that he could prove his
theorem on the continuum function, without splitting into cases as
to the existence of inaccessible cardinals, by using this hybrid
support. Easton support is important because with it, one is often
able to show that cardinals are preserved, when they would not be
preserved with a full-support iteration. The insertion of direct
limits at inaccessible stages enables one to prove better chain
conditions, which are the means by which one shows that cardinals
are not inadvertently collapsed by the forcing.
Lastly, let me say that the particular confusion about
inaccessible versus regular cardinals in the Jech presentation is
resolved when one realizes that he is forcing only at cardinal
stages. And so if $\lambda$ is a successor cardinal
$\lambda=\delta^+$, then there at most $\delta$ many cardinals
below $\lambda$, and so the support will have size less than
$\lambda$, for free. So the only operative restriction there
occurs at inaccessible cardinals. (My belief is that this
presentation in the book could be improved to explain this better,
since numerous students have had exactly this confusion.)<|endoftext|>
TITLE: What is $\omega_1^{CK}(\mathsf{Ord})$?
QUESTION [17 upvotes]: We know that if $\alpha<\omega_1^{CK}$ then there is some recursive $R$ such that $(\omega,R)$ has order type $\alpha$.
Let's consider now the ordinals, $\mathsf{Ord}$ with their natural order. This is a well-ordered class, and it behaves very much like $\omega$ (in fact $\mathsf{Ord}^{V_\omega}=\omega$). We can define new order types on $\mathsf{Ord}$ whose order type is strictly larger.
For example $\leq_0=\lbrace(\alpha,\beta)\mid 0<\alpha\land(\alpha<\beta\lor\beta=0)\rbrace$ would be a well-ordering of $\mathsf{Ord}$ which is isomorphic to $\mathsf{Ord}+1$.
We can do this with any set of ordinals, or even a class of ordinals: put all the limit ordinals strictly above all the successor ordinals (and zero), and in each part order by the natural order. We can even go further and define the following ordering:

Zero and successor ordinals first, order by the natural order.
Any limit ordinal is greater than zero and all the successors.
Given two limit ordinals with different cofinality then the order then by their cofinality.
Given two limit ordinals with the same cofinality then order them as usually.

And of course we can proceed more and more. And define even greater and more complicated order types (using parameters, of course).
My question is what sort of supremum these class-orders have? But of course this is too broad. Clearly this is going to depend on the universe itself, or rather the model. So let me put some constraints on this question.
Let $(M,\in)$ be a countable transitive model of ZFC, and suppose that $\mathsf{Ord}^M=\alpha$. Can we compute this $\omega_1^{CK}(\mathsf{Ord}^M)$ from $\alpha$ and $M$? Can we at least bound it from below and above (of course $\omega_1$ is an upper bound, but a more reasonable bound that is)?
I feel that my question is still quite broad, but is there anything intelligible we can say on those ordinals? Does the fact that $M$ is countable make any difference in the computation?
Two more points which crossed my mind as relevant, and perhaps helpful, to the above question. 

What happens if we add large cardinals assumptions into $M$. If for example there is a measurable in $M$, $\omega_1^{CK}(\mathsf{Ord}^M)>\omega_1^{CK}(\mathsf{Ord}^{L^M})$ (to make things interesting $M=L[U]$ maybe)?
What happens between extensions which don't add ordinals? In particular what about forcing extensions? Can we change this ordinal by forcing? Or maybe give some intelligible conjecture as to when an inner model of $M$ has the same $\omega_1^{CK}(\mathsf{Ord})$?

REPLY [8 votes]: Bumping an old question, I believe that Andreas' bound of the next admissible ordinal is in general (and I suspect always) not sharp. The crucial point is not the theory of the model in question, but rather its closure properties - in particular, the bound of the next admissible fails to be sharp whenever $M^\omega\subseteq M$. (It's possible of course that large cardinals will be relevant in case $M^\omega\subseteq M$, but I actually think the failure of sharpness will hold unconditionally.)
Part of the point of this answer is that the notation in the OP winds up being highly misleading. So let me notationally start from scratch:

I'll write "$\Delta(M)$" for the supremum of the definable-over-$M$ well-orderings of subsets of $M$. 
For $M$ a transitive model of KP + Infinity (= KP$\omega$), I'll write "$Ad(M)$" for the smallest transitive model of KP$\omega$ with $M$ as an element.
I'll write "$\omega_1^{CK}(M)$" for $Ad(M)\cap Ord$.


(This MSE answer of mine contains the argument below in more detail, and this old MO answer of mine is also relevant:)
The key observation is that if $M^\omega\subseteq M$ then any $M$-definable ordering of a subset of $M$ which is ill-founded has a descending sequence in $M$. This means that the set $\mathcal{W}_M$ of formulas-with-parameters-in-$M$ which define in $M$ well-orderings of subsets of $M$ is computable in $Ad(M)$: this is because, given a candidate $\varphi$, we can just search in $Ad(M)$ for either an isomorphism between $\varphi^M$ and some ordinal or a descending sequence through $\varphi^M$ (in fact, all we're really using here is $M^\omega\subseteq Ad(M)$). This means that $$\Delta(M)<\omega_1^{CK}(M)$$ whenever $M$ is sufficiently closed.
As I said above, I strongly suspect that this upper bound is never sharp, but I don't see an argument for that at present. Annoyingly, there seems to be a paucity of candidates for better bounds. The height $\eta_M$ (my notation) of the smallest $E$-closed set containing $M$ as an element is a possible candidate, being in general (even always, I think) smaller than $Ad(M)$ when $M$ is admissible, but the plausibility of $\eta_M=\Delta(M)$ is due entirely to my ignorance of $E$-recursion.<|endoftext|>
TITLE: Complexity of a matching problem on the grid $\mathbb Z^2$
QUESTION [5 upvotes]: Given $2n$ integral points of $\mathbb Z^2$, is there a polynomial algorithm which gives 
a matching consisting of $n$ non-intersecting straight vertical or horizontal segments between pairs of points if such a matching exists? (Not all segments have to be vertical
or horizontal, there can be $a$ vertical and $n-a$ horizontal segments, but two
distinct segments do never intersect.)
Examples: (1) If the number of points is even in every row (or column) one can simply 
pair points row by row.
(2) No such matching exists for the four points $\pm (1,0),\pm (0,1)$.
There is also an obvious generalization to $\mathbb Z^d$ for $d\geq 3$.
PS: There is always a solution using piecewise linear paths with vertical or horizontal
steps making at most one quarter of a turn (ie. either straight horizontal or vertical 
segments or L-shaped (and its rotations) paths).

REPLY [6 votes]: I believe the full details, along the lines of what domotorp posted, are provided in "Reconstructing sets of orthogonal line segments in the plane" by Rendl and Woeginger.  From the abstract,
"We show that reconstructing a set of $n$ orthogonal line segments in the plane from the set of their vertices can be done in $O(n log n)$ time, if the segments are allowed to cross. If the segments are not allowed to cross, the problem becomes NP-complete."
The authors give a reduction to planar 3-SAT that uses points on an integer lattice, which I believe is the setting you're interested in.<|endoftext|>
TITLE: Sane bound on number of moves for Maker-Breaker game on $\mathbb R^2$ for $\{0,1,2,3,4\}$
QUESTION [24 upvotes]: The description below comes from 

József Beck. Combinatorial games. Tic-tac-toe theory, Encyclopedia of Mathematics and its Applications, 114. Cambridge University Press, Cambridge, 2008, MR2402857 (2009g:91038).

Given a finite set $S$ of points in the plane $\mathbb R^2$, consider the following game between two players Maker and Breaker. The players alternate, each time picking one (previously unselected) point in $\mathbb R^2$, with Maker moving first. Maker's goal is to build a congruent copy of $S$, while Breaker's goal is to prevent this from happening. If at any finite stage Maker's goal is achieved, the game ends, and Maker wins. Otherwise, Breaker wins.
For example, denote by $A(n)$ the set consisting of $n$ points in a row in arithmetic progression, with common difference one. 

Maker has a winning strategy, in two moves, if $S=A(2)$.
Maker has a winning strategy, in three moves, if $S=A(3)$. 
Maker has a winning strategy, in at most five moves, if $S=A(4)$ (begin by playing the vertices of an equilateral triangle $ABC$ with side length $1$, such that at least two of the lines it determines, say $AB$ and $AC$, have no points played so far by Breaker).

Beck proves a remarkable theorem in the book (Theorem 1.1): For any finite $S$, Maker has a winning strategy. The proof is an elegant generalization of a theorem of Erdős and Selfridge:

First, one shows that (for any $n$) if $(V,\mathcal F)$ is an $n$-uniform hypergraph with 
$$ \frac{|\mathcal F|}{|V|}>2^{n-3}\Delta_2(\mathcal F), $$ 
where 
$$ \Delta_2(\mathcal F)=\max_{x\ne y\in V}|\{A\in\mathcal F\mid \{x,y\}\subseteq A\}| $$
then, in the game where Maker and Breaker alternate picking distinct elements of $V$, Maker can ensure to pick all the elements in some $A\in\mathcal F$.
Second, one shows that for any $S$, there are finite sets $X$ in the plane that contain "many" congruent copies of $S$. "Many" is formalized so that the inequality above holds, where $V=X$ and $\mathcal F$ is the collection of congruent copies of $S$ among the points in $X$. The sets $X$ obtained this way tend to be very large. 

The proof of the "Erdős-Selfridge result" goes by considering a "weighed" characteristic function that counts at each stage of the game the number of sets $A\in\mathcal F$ that have not been eliminated yet by the moves of Breaker, and having Maker play so that the value of this function is maximized at each stage. This ensures that, once all points of $X$ have been played, the function is still positive.
This elegant argument unfortunately produces ridiculously large bounds, due to its great generality. If $S=A(5)$, the number of moves needed to ensure Maker's victory following this approach is estimated to be about $309^{44}\approx 3.6\times 10^{109}$. For $|S|\ge10$, Beck tightens the argument somewhat, to show that $2^{2^{|S|^2}}$ moves suffice.
My question:

For $S=A(5)$, can we find a more decent bound on the number of moves? 

My requirement on what counts as "decent" is very loose. I expect the bound above is much larger than needed. I would be happy to be proved wrong, of course, by obtaining large lower bounds. (Additional) references in the literature are also welcome. The following is from pg. 34 of Beck's book:

The wonderful thing about Theorem 1.1 is that it is strikingly general. Yet there 
  is an obvious handicap: these upper bounds to the Move Number are all ridiculously 
  large. We are convinced that Maker can build [the set $S=A(5)$] in (say) less than 1000 moves, but do not have the slightest idea how to prove it. The problem is that any kind of brute force case study becomes hopelessly complicated.

REPLY [19 votes]: Eleven moves suffice.
François Brunault commented that the maker can get two moves to start on some hexagonal lattice (a lattice generated by unit vectors with an angle of $60$ degrees). In fact, by the fourth move, the maker can get three moves to start in a hexagonal lattice, and can choose these to be the vertices of an equilateral triangle of side $1$ so that the breaker has not played in this lattice yet. 
Proof: Let the maker's first play be the origin $A$. Rotate the coordinates after the breaker's first move so that this play be ignored. Let the maker's second move be $B = (2x,0)$ where $x$ is a transcendental smaller than $1$. There are $4$ hexagonal lattices $H_{A,C}, H_{A,D}, H_{B,C}, H_{B,D}$ containing one element of $\lbrace (0,0), (2x,0)\rbrace$ and one element of $\lbrace C=(x,\sqrt{1-x^2}),D=(x,-\sqrt{1-x^2}) \rbrace$, the points of distance $1$ from the maker's first two moves. These hexagonal lattices have the property that the breaker hasn't played in any of them yet, and their pairwise intersections are empty or a point in $\lbrace A,B,C,D\rbrace.$ So, either the second play of the breaker misses both $H_{A,C}$ and $H_{B,C}$ or both $H_{A,D}$ and $H_{B,D}$. By symmetry, we can assume the second play of the breaker misses $H_{A,C}$ and $H_{B,C}$. Let the third play of the maker be $C$. This gives the maker an unopposed pair of points in both $H_{A,C}$ and $H_{B,C}$. The third play of the breaker can only be in one of these lattices. On the fourth move of the maker, the maker can play in the other to make an unopposed equilateral triangle of side length $1$. $\blacksquare$
Next, even if the maker is constrained to play in this lattice, the maker can force $5$ in a row by move $11$. Note that if the maker has an "open $3$" of $3$ points in row with two open spaces on either side, $- - \circ \circ \circ - -$, then the breaker has to respond immediately either just to one side or the other, or else the maker can make $5$ in a row in $2$ more moves. To avoid an explosion of cases, we'll let the breaker play both sides of an open $3$. Perhaps without this, the maker could force $5$ in a row in fewer moves. 
We'll show that whatever the breaker's fourth move is, the maker can still force $5$ in a  row by the eleventh move. By symmetry, we can assume the breaker's fourth move is in a sixth of the lattice between two of the perpendicular bisectors of the triangle's sides. We'll consider two possible moves within this sixth individually, and then all others.
o o x
 o


   5
o o x
 o

    x
   5
o o x
 o 
x

     x
6   5   
 o o x
  o
 x

x     x
 6   5
  o o x
   o
  x x

x     x
 6 7 5
  o o x
   o
  x x    

 x     x
x 6 7 5 x
   o o x
    o
   x x 

 x     x
x 6 7 5 x
   o o x
  8 o
   x x 

 x   x x
x 6 7 5 x
   o o x
  8 o
 x x x 

 x   x x
x 6 7 5 x
   o o x
  8 o 9
 x x x 

This last play makes two open $3$s, with $7$ and with $8$, so with this $4$th move by the breaker, the maker can construct $5$ in a row by move $11$. 
In the next sequence, I'll show the breaker's response immediately, again letting the breaker block both sides of open $3$s.
o o
 o x

    x
 o o
  o x
 5
x

       x
x 6 o o x
     o x
    5
   x

 x     x
x 6 o o x
   7 o x
    5
   x x

 x x   x
x 6 o o x
   7 o x
    5 8
  x x x

 x x   x
x 6 o o x
   7 o x
  9 5 8
   x x x

Again, the maker's 9th move creates two open $3$s, through the $7$ and through the $5$ and $8$, so with that choice of $4$th move by the breaker, the maker can construct $5$ in a row by move $11$.
Next we let the breaker's $4$th move block every other possibility in that sixth of the lattice, which will cover all of those possibilities simultaneously.
     x x x x x
o o . x x x x x
 o . x x x x x
. . . x x x x x
 . . . . x x x

       x x x x x         
5 o o . x x x x x
   o . x x x x x
  . . . x x x x x
   . . . . x x x

This play technically does not make an open $3$ since only one space to the right is open. So, the breaker does not have to respond to either side immediately. The other possibility is $2$ to the left. We will let the breaker play in all $3$ positions simultaneously.
           x x x x x
x x 5 o o x x x x x x
       o . x x x x x
      . . . x x x x x
       . . . . x x x

         x x x x x x
x x 5 o o x x x x x x
       o . x x x x x
      6 . . x x x x x
     x . . . . x x x

   x     x x x x x x
x x 5 o o x x x x x x
     7 o . x x x x x
      6 . . x x x x x
     x x . . . x x x

   x   x x x x x x x
x x 5 o o x x x x x x
     7 o . x x x x x
    8 6 . . x x x x x
   x x x . . . x x x

   x   x x x x x x x
x x 5 o o x x x x x x
     7 o . x x x x x
    8 6 9 . x x x x x
   x x x . . . x x x

Again, the $9$th move creates two open $3$s, so the maker can construct $5$ in a row by move $11$.<|endoftext|>
TITLE: Dense subgroups of Lie Groups
QUESTION [6 upvotes]: SETUP: Let $G$ be a connected Lie group, and $H\subset G$ be a FINITELY GENERATED dense subgroup.
I am interested in knowing what kind of information one can infer on the complexity of $H$.
I am especially interested in the case in which $G$ is simply connected, non compact, and non diffeomorphic to $\mathbb{R}^n$. After some research online, the only result I found in this direction is in "On dense free subgroups of Lie groups", by Breuillard, E. and Gelander, T.. Here the authors prove that if $G$ is not solvable, and $H\subset G$ is finitely generated and dense, then it contains a free group of rank $r=2\dim G$.
Does anyone have other references of result in this direction? I hope to find results of the type "such a group $H$ needs to be at least this complicated".
In the case I am interested in, $H$ is the fundamental group of a compact manifold, so I have an "upper bound" on the complexity of "H". Now I want a "lower bound", if this makes any sense.
Thank you in advance!

REPLY [3 votes]: I recently completed a preprint with Michael Larsen in this direction. Here is a link to the paper:
https://arxiv.org/abs/1312.7294
Here is the abstract:
When does Borel's theorem on free subgroups of semisimple groups generalize to other groups? We initiate a systematic study of this question and find positive and negative answers for it. In particular, we fully classify fundamental groups of surfaces and von Dyck groups that satisfy Borel's theorem. Further, as a byproduct of this theory, we make headway on a question of Breuillard, Green, Guralnick, and Tao concerning double word maps.<|endoftext|>
TITLE: Hodge numbers of reduction mod $p$
QUESTION [6 upvotes]: Let $X$ be a projective variety defined over a number field $K$, and $p \in \textrm{Spec }\mathcal{O}_K$ a maximal ideal, so that reduction mod $p$ makes sense, and the resulting scheme (mod $p$) $\bar{X}$ is smooth over the relevant finite field. Assume that $X$ smooth over $K$.
1.) In the case that $X$ is a curve, is there a short argument to show that the geometric genus of $X$ and of $\bar{X}$ are the same? Certainly if $X$ is a plane curve this is clear.
2.) The hodge numbers $h^{p,q}_X = \dim H^p(X, \Omega^q)$ make sense in all characteristics. Are the hodge numbers preserved under reduction mod $p$, that is, $h^{p,q}_X = h^{p,q}_\bar{X}$?
3.) The Weil conjectures tell us that we can recover the Betti numbers of $X$ (considered as a complex manifold) from the zeta function of $\bar{X}$. There are many smooth projective varieties that have reduction $\bar{X}$ mod $p$ and the Weil conjectures tell us that all of them have the same Betti numbers. Can one prove this without using the Weil conjectures, perhaps with Etale cohomology?
4.) More generally, if $\mathcal{L}$ is a locally free sheaf on $X$, and $\bar{\mathcal{L}}$ denotes the reduction mod $p$, I would guess that the numbers $\dim H^p(X, L)$ and $\dim H^p(\bar{X}, \bar{L})$ don't match up - but I don't have a good example.
I am interested in proofs (not using the Weil conjectures if possible).

REPLY [13 votes]: As already pointed out, the Hodge numbers may go up under reduction modulo $p$. On the other hand, let me also point out that the situation can be controlled: 
1.) For all $p$, where $\overline{X}_p$ is smooth, the $\ell$-adic Betti numbers of $X$ and $\overline{X}_p$ are the same.
2.) Now, by the universal coefficient formula relating crystalline and deRham cohomology, we have for all $i$ short exact sequences
$$
 0 \to H^i_{cris}(\overline{X}_p/W)/p\to 
$$
$$
H^i_{dR}(\overline{X}_p/k_p)\to {\rm Tor}_1^{W(k_p)}(H_{cris}^{i+1}(\overline{X}_p/W),k_p)\to 0
$$
where $k_p={\cal O}_K/p$. Now, if $\overline{X}_p$ has torsion-free crystalline cohomology, then the term on the right is zero, and the term on the left is a $k_p$-vector space of dimension equal to the $i$.th $\ell$-adic Betti number.
Then, the Fr\"olicher spectral sequence relating Hodge- and deRham-cohomology degenerates at $E_2$, we have that $\sum_{p+q=i}h^{p,q}$ is equal to the $i$.th $\ell$-adic Betti number. Thus, simply for dimension reasons, the Hodge numbers of $X$ and $\overline{X}_p$.
The upshot is that torsion in crystalline cohomology of $\overline{X}_p$ detects and controls the differences in Hodge numbers of $X$ and $\overline{X}_p$. For almost all $p\in {\rm Spec} {\cal O}_K$, the reduction $\overline{X}_p$ will be smooth and will have torsion-free crystalline cohomology.<|endoftext|>
TITLE: On similar concepts in mathematics whose similarity is a non-trivial fact.
QUESTION [7 upvotes]: Recently, while undertaking a study of commutative algebra, I learned three concepts: (i) a local ring, (ii) a regular local ring and (iii) a regular ring.
At the end, I found myself asking this seemingly naïve question: Are regular local rings the same objects as local rings that are regular? At first, I thought, "My mind must be acting stupid again." However, upon further analysis, it turned out that the answer to my question was non-trivial after all.
One direction, namely proving that a local ring that is regular is actually a regular local ring, is not very hard to establish. Indeed, it can be assigned as a homework problem in an undergraduate abstract algebra course. The key observation is that for a local ring $ (R,{\frak{m}}) $, upon localization at $ {\frak{m}} $, we obtain $ R_{\frak{m}} = R $. This is because $ R \setminus {\frak{m}} $ is precisely the set of units of $ R $. Hence, by the definition of regular ring, we see that $ (R,{\frak{m}}) $ is a regular local ring.
The other direction is a well-known (in my opinion, highly) non-trivial result in homological algebra, which states that the localization of a regular local ring at any prime ideal is still a regular local ring. By the definition of regular ring once again, regular local rings are therefore local rings that are regular.
I am wondering, are there any pairs of concepts in other areas of mathematics that look so similar that their similarity may be mistaken for tautology but, in reality, can only be established by a hard proof?

REPLY [6 votes]: Here is a possible example that I have been speculating about, on and off, since around 2006: it might not fit the OP's question, and I may be talking rubbish, so I'd welcome corrections from the true connoisseurs (as opposed to this dilletant).
Call a short exact sequence of Banach spaces and bounded linear maps naively exact if it is exact as a s.e.s. of vector spaces (this is just to get round the fact that category-theoretic cokernels and epimorphisms in Ban don't behave as they do in Vect). Call a Banach space $E$ homologically flat if $E\widehat{\otimes}\underline{\quad}$ preserves the short naively exact sequences of Banach spaces, where the tensor is the projective tensor product a.k.a. the "right one from the POV of closed monoidal categories".
Now if I understand the literature correctly, the homologically flat Banach spaces are known: they are precisely the ${\mathcal L}^1$-spaces of Lindenstrauss and Pelczynski, which are roughly speaking those where each fin-dim subspace $E$ is contained in a fin-dim subspace $F$ which is uniformly isomorphic to $\ell_1^{|F|}$.
As $\ell_1^n$ is the "free Banach space" on an $n$-element set, this result could be airily if imprecisely captured by saying "the flat Banach spaces are those built locally from finitely generated free Banach spaces". But now this sounds awfully like the Govorov-Lazard theorem from ring theory: a module over a fixed commutative ring $R$ is flat iff it is a colimit of finitely generated free $R$-modules.
As far as I know the Banach-space result is proved using duality (duals of flat spaces are injective, and the injective Banach spaces admit a description as ${\mathcal L}^\infty$-spaces) and doesn't seem to have a proof along the lines of the GL-theorem. But one might hope to find a proof that made use of a meaningful analogy between the category of Banach spaces and bounded linear maps, and categories of modules...
(Of course the module-categories are abelian, while Ban is not, but there may be some way to exploit the "embedding" of Ban into an abelian envelope, cf. work of Noël, Waelbroeck and in greater generality Schneiders. Calling Theo Bühler... calling Theo Bühler...)<|endoftext|>
TITLE: Connection of X(n) spectra to formal group laws
QUESTION [13 upvotes]: In the proof of the Nilpotence Theorem, or at least in Ravenel's account of it in his Orange Book, a sequence of spectra are used, denoted $X(n)$ with $X(0)=\mathbb{S}$ and and $X(\infty)=MU$ such that $\langle X(n)\rangle\geq\langle X(n+1)\rangle$.  These are the Thom spectra associated to the map $\Omega SU(n)\to BU$.  They are homotopy equivalent to $MU$ up through degree 2n-1.  They all have associated Hurewicz maps $h(n):\pi_\ast(R)\to X(n)_\ast(R)$ for $R$ a finite ring spectra.  We are interested, for the Nilpotence Theorem, in determining when $h(n)(\alpha)=0$, ultimately for $n=0$. To this end, Ravenel proves that if $h(n+1)(\alpha)=0$ then $h(n)(\alpha)$ is nilpotent. 
So, from this theorem we know then that any of the $X(n)$ spectra detect nilpotence just as well as $MU$.  The nice thing of course about $MU$ (or one of the many nice things) is that it has at least one other interpretation (i.e. aside from its geometric interpretation as a Thom spectrum).  This is of course that $MU_\ast$ determines formal group laws over rings, and you have all of this amazing stuff happen. Do you have any similar such interpretations for these $X(n)$ spectra, or are they ONLY geometric in nature? Or I guess, to rephrase, does anyone KNOW of any other ways of thinking about these things?
Thanks!

REPLY [19 votes]: Although I'm not aware an interpretation of $X(n)$ in terms of formal group laws (aside from 
the ones in Eric's and Dylan's comments), I would like to point out that the nilpotence theorem is fundamentally a geometric fact and is proved that way. The nilpotence theorem has a number of corollaries to the effect that the formal group perspective gives a very good description of the global structure of the stable homotopy category, but its proof requires explicit, computational facts about very specific spectra. 
For example, one way to phrase the nilpotence theorem is that, for any connective  spectrum $X$, the $E_\infty$-page of the Adams-Novikov spectral sequence (drawn with $t-s$ horizontally and $s$ vertically) has a "vanishing curve" which is asymptotically flat: that is, has slope tending to zero as $t-s \to \infty$. That is, the maximum possible Adams-Novikov filtration of an element in $\pi_k X$ grows in $k$ by some function which is $o(k)$. (Actually determining what this function is seems to be an open problem; see Hopkins's enjoyable talk about Ravenel's work for some discussion.) If you know that there is such a vanishing curve, you can get the nilpotence theorem in its first form (about ring spectra) at once by noting that an element in $\pi_* R$ for any ring spectrum which is killed by the $MU$-Hurewicz is detected in the ANSS in positive filtration and its powers live on  a line of positive slope, which has to overtake the asymptotically flat vanishing curve. This is something that's very much specific for $MU$. With mod $2$ (or even integral) homology and with the sphere spectrum, the image of $J$ elements already rule out such a vanishing line in the classical ASS. 
It's important, however, that you don't get the vanishing curve at $E_2$, which is the part that comes from algebra. The $E_2$-term (which is the cohomology of $M_{FG}$) has lots of non-nilpotent elements, for instance the element corresponding to $\eta$. A quick way to see this is to consider the map 
$$B \mathbb{Z}/2 \to M_{FG}$$
(which corresponds geometrically to the map 
$S^0 \to KO$). The element $\eta$ is not nilpotent for $B \mathbb{Z}/2$, and therefore it can't be nilpotent in $H^*(M_{FG})$,  although $\eta^3$ is killed by a $d_3$ differential in the spectral sequence for $KO$. The nilpotence theorem is somehow saying something global about the structure of such spectral sequences, that there have to be a lot of differentials, which create this vanishing line. So at some level, it's saying what appears to be the opposite:  that homotopy can't look too much like algebra. 
(I wrote some notes on the spectral sequence for $KO$, and ultimately for $TMF$ at the prime $3$, which I mention because working through these spectral sequences helped me appreciate some of this technology. In fact, it's even better: you get flat vanishing lines at finite stages; this is related to the Hopkins-Ravenel smashing theorem which states that this happens for any $E(n)$-local spectrum.) So there's no nilpotence theorem for $M_{FG}$. (As a related note: there's no thick subcategory theorem for the derived category of perfect modules on $M_{FG}$.)
Let me try to summarize the key inductive argument in Devinatz-Hopkins-Smith's paper, which is to prove: 
Theorem: If $R$ is a connective, associative ring spectrum and $\alpha \in \pi_* R$ is such that $X(n+1)_* \alpha =0$, then $X(n)_* \alpha $ is nilpotent. 
More generally, you can ask when something like this is true: 
Question: If $R \to R'$ is a morphism of ring spectra and $R$ "detects nilpotence," then when does $R'$? 
For example, if $R'$ and $R$ are Bousfield equivalent, then it's easy to get the result, but the $X(n)$ are not Bousfield equivalent. However, they do turn out to Bousfield equivalent on "telescopes of connective spectra," which turns out to be all you need. 
In particular, if $R'$ annihilates a localization $\alpha^{-1} T$ for $T$ a connective ring spectrum (which is to say that $\alpha$ is nilpotent in $R'$-homology), then so does $R$. 
That's what D-H-S show, and their argument is summarized in:
Axiomatic nilpotence theorem:
Suppose $R'$ is obtained from a filtered colimit of spectra $G_k$ such that: 

The $G_k$ have good $R'$-based Adams spectral sequences: that is, in the  $E_\infty$-page of the $R'$-based ASS for $G_k \wedge X$ (for any connective $X$), there is a vanishing line of slope $\epsilon_k$ which tends to zero as $k \to \infty$. 
Each $G_k$ is Bousfield equivalent to $R$. 

D-H-S check these conditions for $X(n) \to X(n+1)$. The first step is something that they do purely algebraically (at $E_2$) using a series of May-type spectral sequences, and it's the piece of the argument which you might be able to get using facts about formal groups. But the second part -- which is way, way harder -- seems to require some geometry and facts about concrete things like $E_2$-algebras and Thom spectra and partial James constructions. I suspect that, at some level, the use of such geometry is an inescapable feature of the whole business.<|endoftext|>
TITLE: Triality of Spin(8)
QUESTION [26 upvotes]: Among simple Lie groups, $Spin(8)$ is the most symmetrical one in the sense that $Out(Spin(8))$ is the largest possible group. A description of this outer automorphism groups is as follows. $Spin(8)$ has 2 half-spinor representations and one vector representation (coming from standard representation of SO(8)) all of them of dimension 8. In general one can pre-compose a representation of a Lie group with an automorphism of the group to get another representation. In this special case $Out(Spin(8))$ permutes above three representations and this we get an isomorphism of $Out(Spin(8))$ and $S_3$.
My questions:
1- Is there any way to see an automorphism explicitly which interchanges the vector representation and a half-spinor representation? (It is easy to see there exists such an automorphism. But I'd like to have an explicit construction of such an automorphism.)
2- Apparently there is a 27-dimensionla Jordan algebra which has $Aut(Spin(8))$ as a subgroup of its automorphism group. Can anyone explain what is this Jordan algebra and how should I think about $Aut(Spin(8))$ acting on it?

REPLY [14 votes]: In addition to the above answers involving spinors and/or octonions, you might be interested in Cartan's original construction of the triality automorphisms, which is very explicit and takes just a couple of pages in his beautiful little paper Le principe de dualité et la théorie des groups simples et semi-simples (Bull. Sc. Math 49 (1925), 361–374).  The description is in Section 5 of that article, mainly on pages 368 and 369, though you'll probably be interested in the very geometric construction that he makes to interpret the outer automorphisms in the concluding sections 6 and 7 of that paper.
The main difference from what has been said in answer so far is that, instead of constructing spinors (which, of course, he already knew how to do at that point), Cartan works on the centerless simple group $G = \mathrm{SO}(8)/\lbrace\pm\mathrm{I}\rbrace$  (which he calls the 'adjoint group' of $\mathrm{SO}(8)$), since the triality automorphisms actually do act on $G$ and not just on its Lie algebra.  He then uses a clever choice of notation to write down an explicit action of $S_3$, the symmetric group on $3$ letters, on the Lie algebra of $G$ in such a way that these automorphisms are Lie algebra automorphisms that preserve the obvious maximal torus and yet are outer because they don't come from the Weyl group.
Another great source, of course, is Chevalley's beautiful little book The algebraic theory of spinors (1954), the last chapter of which is all about triality.  He takes pains to do everything over general fields as well, so it's quite a useful treatment.
Finally, Cartan also wrote about the tie of triality with $\mathrm{F}_4$ acting irreducibly on the $26$-dimensional space of traceless $3$-by-$3$ Hermitian octonian matrices in Section V of his paper Sur des familles remarquables d'hypersurfaces isoparamétriques dans les espaces sphériques (Math. Zeitschrift 45 (1939), 335–367).  The relevant passage is on pages 354 and 355, where he explains the construction.<|endoftext|>
TITLE: What's the maximum entropy probability distribution given bounds [a,b] and mean?
QUESTION [13 upvotes]: What is the continuous probability distribution that maximizes entropy, given only the bounds of the random variable [a,b] and the mean mu of the probability distribution?
For example:

if a=0, b=1, and mu=0.5, it should return a U[0,1].
if a=10, b=20, and mu=20, it should return Dirac delta at x=20.
if a=0, b=1, and mu=0.8 it should return ... ?

I imagine the general solution will be based on the Beta distribution with some alpha and beta parameters expressed in terms of a, b, and mu, but I don't know.
Many thanks!

REPLY [12 votes]: You can maximize the entropy using standard calculus of variations; you need to take into account the constraint that the probability distribution is properly normalized and that the mean is known, using Lagrange multipliers. You then find that the probability distribution is of the form:
$$p(x) = \frac{\alpha e^{\alpha x}} { (e^{\alpha b}-e^{\alpha a})},  x\in [a,b]$$
for $\alpha$ the unique solution to
$$\mu = \frac{ \int_a^b  \alpha  x e^{\alpha x} dx}{(e^{\alpha b} - e^{\alpha a})} = \frac{b e^{\alpha b} - a e^{\alpha a}}{(e^{\alpha b}-e^{\alpha a})} -\frac{1}{\alpha}$$<|endoftext|>
TITLE: If $M_n(R)$ and $M_m(R)$ satisfy the same polynomial identities is it true that $m=n$?
QUESTION [7 upvotes]: Let $K$ be an inifinite field of characteristic different from 2.
The well-known Amitsur-Levitzki theorem states that the algebra $M_n(K)$ satisfy the standard polynomial identity of degree $2n$, 
$$s_{2n}(x_1,\dots,x_{2n})=\sum_{\sigma\in S_{2n}}(-1)^{\sigma}x_{\sigma(1)}\cdots x_{\sigma(2n)}$$
Moreover, it does not satisfy any other identity of degree less than $2n$.
In particular, if $m < n$, $s_{2m}$ is an identity for $M_m(K)$ and is not an identity for $M_n(K)$.
My question is the following: 

If $R$ is a unitary associative noncommutative $K$-algebra that satisfy a polynomial identity, is it true that if $m < n$ then there is an identity of $M_m(R)$ which is not an identity for $M_n(R)$?

In the language of T-ideals, is the inclusion $T(M_n(R))\subset T(M_m(R))$ a proper one?
Equivalently, the PI-equivalence of $M_n(R)$ and $M_m(R)$ imply that $m = n$?
Of course, if the condition that $R$ is a unitary algebra is removed, nilpotent algebras can give counter-examples.

REPLY [3 votes]: This is not true in general. For example, if $R$ is a free associative algebra of rank $>1$, then $M_n(R)$ does not have nontrivial identities for any $n$. 
Moreover, I believe there are examples of algebras $R$ such that much stronger condition holds: $M_n(R)$ is isomorphic to $M_m(R)$ for some $m \ne n$. This reminds cancellation problems in the commutative setting, though I cannot provide examples of such algebras.
This is true if, for example, if $R$ is finite-dimensional and prime (even not necessary associative). Then we can note that $M_n(R)$ is prime too, pass to the algebraic closure of the base field, and invoke theorem of Razmyslov that finite-dimensional prime algebras over algebraically closed fields are determined by their identities (Yu.P. Razmyslov, Identities of Algebras and Their Representations, AMS, 1994 (translation from Russian), around p. 30).
EDIT:
I was not careful enough when reading the question, sorry. The question explicitly asks for situation when $R$ is a PI algebra, so the example with a free algebra obviously does not qualify. 
I still think this is not true in the whole generality: probably just the condition of being unitary is to weak, one should demand something like (semi)primeness. 
I can think of two approaches. First, there is a lot of works about identities of tensor product of algebras, and of $M_n(R)$ in particular (typical results: if $R$ satisfies the standard identity of degree $k$, then $M_n(R)$ satisfies the standard identity of some given degree in terms of $n$ and $k$, see e.g. M. Domokos, Eulerian polynomial identities and algebras satisfying a standard identity, J. Algebra 169 (1994), N3, 913-928 DOI:   10.1006/jabr.1994.1317). Second, perhaps one can do something along the lines of Sections 4 and 5 of arXiv:0911.5414.<|endoftext|>
TITLE: How would set theory research be affected by using ETCS instead of ZFC?
QUESTION [33 upvotes]: In "Rethinking Set Theory", Tom Leinster argues in favor of teaching axiomatic set theory via Lawvere's Elementary Theory of the Category of Sets with 10 axioms (but phrased in a way that requires no knowledge of category theory) which uses sets and functions as primitive elements compared to ZFC which uses sets and elements as primitives. (although if you really insist you can reintroduce elements as primitives at the expense of more clauses). This is weaker than ZFC but can be made equivalent to ZFC (or equiconsistent according to François G. Dorais) with the addition of a hardly ever needed 11th axiom.
http://golem.ph.utexas.edu/category/2012/12/rethinking_set_theory.html
How would set theory research be affected by using ETCS instead of ZFC?
Is ETCS less cumbersome than ZFC or more, or does it not matter? Does it make proofs longer/shorter or easier/harder?
Would automated theorem provers work better with ETCS?
To clarify, the question is not about the strength of ETCS: assume whatever is needed to bring ETCS up to strength with ZFC. The question is about practicalities of working with the ETCS axioms. I guess most mathematicians don't directly work with axioms of set-theory at all. So the question is: if your work does involve working directly with such axioms then what difference (if any) does it practically make to you as a set-theory researcher to use a different set of axioms that have equal strength. If I understand correctly the goal of Tom Leinster's paper was to reflect actual (non-set-theoretical)-mathematical practice but my question is about what set-theorists would do if they used these axioms. Assume also that for the purposes of this question ETCS refers specifically to the rephrasing used in this paper that is free of category and topos terminology. Perhaps this rephrasing should be renamed ETS. Adding replacement would give the name ETSR.

REPLY [8 votes]: No one has really addressed the question about automated theorem provers.  I'm going to take this to refer more generally to computer-assisted mathematics, since fully automated theorem provers are still of very limited usefulness to a mathematician.  There are a lot of good arguments that the best foundational system for these is neither ZFC (or its relatives) nor ETCS (or its relatives), but some variety of type theory.
This is mainly because type theory is so closely connected with programming and has good computation behavior, thus meshing very well with computers.  Since the question is not about type theory, I won't say much about it; I just want to point out that ETCS is more like type theory than ZFC is.  So although I kind of doubt that actually using ETCS in a computer proof assistant would be much better than using ZFC, learning to think in ETCS and do mathematics therein will probably stand you in better stead when you try to use a proof assistant based on type theory.<|endoftext|>
TITLE: The Quillen model structure on simplicial sets as a Bousfield localization
QUESTION [7 upvotes]: Starting with the trivial model structure on the category of simplicial sets (that is the weak equivalences are exactly the isomorphisms and the cofibrations and fibrations are arbitrary maps), is it possible to get the usual Quillen model structure on simplicial sets by performing a number of explicit left and right Bousfield localizations (e.g. by localizing along the inclusions of horns into simplices)?

REPLY [8 votes]: Such sequence of localizations/colocalizations does not exist, since homotopy is not concrete. 
Suppose, for contradiction, that such sequence is constructed, then we obtain a corresponding sequence of reflections/coreflections on the level of homotopy categories, producing a fully faithful embedding of $\mathrm{Ho}(sSet_{\mathrm{standard}})\rightarrow \mathrm{Ho}(sSet_{\mathrm{trivial}})\cong sSet$. Next, it is possible to construct a faithful functor $sSet\rightarrow Set$, say, by sending every simplicial set into the product of the sets of its simplices, obtaining a contradiction with Freyd's theorem.<|endoftext|>
TITLE: Integral transform and $\frac{1}{n!}$.
QUESTION [6 upvotes]: Probably this is a trivial question, but I am unable to find an answer: is there a function $v(x)$ such that 
$$
\int_{0}^\infty x^n e^{v(x)} dx =\frac{1}{n!}
$$
for all positiv integer n?

REPLY [4 votes]: Suppose you are not wedded to the interval $(0,\+\infty)$.  
If $C$ is the unit circle in the complex plane, oriented in the counter-clockwise direction as usual, then
$$
\frac{1}{2\pi i}\oint_C z^n \frac{e^{1/z}}{z}\;dz = \frac{1}{n!}
$$
for $n=0,1,2,\dots$<|endoftext|>
TITLE: About a possible generalization of Green-Tao's theorem
QUESTION [5 upvotes]: Hello,
Let's say that an integer $n$ is $k$-primal if $k$ is its smallest primality radius (i.e non negative integer $r$ such that both $n-r$ and $n+r$ are primes).
I think that for every positive integer $m$ and every non negative integer $k$, there exists an arithmetic progression made of $m$ $k$-primal integers.
Has such a generalization of Green-Tao's theorem been considered so far?
If not, is there a heuristics that would make it quite likely?
Thanks in advance.

REPLY [5 votes]: I am sure many people already know this but just I wanted to mention that the Green-Tao theorem is special case of a more general conjecture of Erdos who asserted that if $(a_n)_{n=1}^\infty \subset \mathbb{N}$ such that $\sum_n \frac {1}{a_n} = \infty$, then the sequence $(a_n)$ contains arbitrarily long arithmetic progressions.<|endoftext|>
TITLE: Publishing a bad paper?
QUESTION [52 upvotes]: First, I apologize if mathoverflow is a bad fit for this question, but it is the only place where I can think to get advice from professionals given my circumstance. I'm also sorry about any vagueness in my post since I need to make sure I maintain anonymity - anonymity is also why I can't get advice from folks in my department.
The short story is that I am a graduate student in a math related field and have been reached out to by a faculty member for writing up a paper about something we discussed to be submitted to a low ranking Mathematics journal. The material is very basic, coming very close to being trivial observations, and happens to be of, I suspect, no interest to anyone anywhere. Anyone who cared to prove what we proved would probably be able to do so within a day at most, if the results aren't already well known. At best, I think the results make good homework problems. 
Despite this, the faculty member seems excited about it. He works in a field that another faculty member has described as "toxic" to getting a job in academics, which is my ultimate goal, and advised me that I would probably be smart to leave papers in this research topic off my C.V. unless it is accepted to a top-tier journal. 
I'm in the strange position of working on this paper because I don't want to alienate him or offend him, but I'm hoping that the paper is rejected just so that my name isn't attached to the paper. This will be my first article submitted for publication, and I'm a little uneasy about even having random editors - who I conceivably could run into in the future - viewing the work.
So, my question to you: is this sort of thing worth getting worked up about? Should I just go along with it, figuring that it is highly unlikely that it will negatively impact my career, or is there some legitimate concern? Is there a chance that publishing trash might help me just because it increases my publication rate?

REPLY [12 votes]: I am tempted to attack some bits of the responses you have gotten so far! Instead I will tell you that I believe the answer is very simple. 


Put yourself in the professors shoes. (I have to assume you have the ability to do that, even if it requires great effort.)
Now, do what you would want a student to do if the tables were turned.


As simple (and as old hat) as this sounds, it is a cure for many ills that is far too rarely practiced.
(Another relevant fact: scientists - mathematicians included - almost always take themselves and what they do far too seriously!)<|endoftext|>
TITLE: Statements which were given as axioms, which later turned out to be false.
QUESTION [16 upvotes]: I know that early axiomatizations of real arithmetic (in the first half of the nineteenth century) were often inadequate. For example, the earliest axiomatizations did not include a completeness axiom. (For example, there is no completeness axiom in Cauchy's Cours d'Analyse).
I also know that mathematicians in the early nineteenth century had some false beliefs about real analysis which were later uprooted as the process of rigorization continued. For example, it was widely believed that a continuous real function must be differentiable except at isolated points - a claim which Weierstrass refuted in the 1870s by defining a function that is continuous everywhere but differentiable nowhere. (http://en.wikipedia.org/wiki/Weierstrass_function).
What I'd like to know is whether at any point an "axiom" was proposed for real arithmetic, which subsequently turned out to be false.
I'd also like to hear about such cases from other branches of mathematics.
However, I'm not so interested in cases of inconsistent sets of axioms (e.g. Gottlob Frege's Grundgesetze, or Church's first formulation of the lambda calculus).
The best example I've found so far is Leibniz's "principle of continuity", according to which "what is true up to the limit is true at the limit". Apparently this was sometimes called an "axiom" although it is obviously not true in general. (I'm getting this from Lakatos's Proofs and Refutations, pg. 128). I'm not entirely happy with this example, because it's so obviously false that I can't believe that it was accepted, except as a heuristic.
Thanks in advance!

REPLY [5 votes]: Perhaps one of the earliest examples would be with the Pythagoreans, who held that any two magnitudes were commensurable, measured as integer multiples of a smaller common unit, a belief that was connected with their mystical religious views and also with their mathematical theory of musical harmony. The Pythagoreans were shocked by the discovery of incommensurable numbers, such as $\sqrt{2}$.
But it may be anachronistic to refer to the fundamental Pythagorean beliefs or principles as "axioms".<|endoftext|>
TITLE: Spectra of elements of a Banach algebra and the role played by the Hahn-Banach Theorem.
QUESTION [16 upvotes]: This problem was posed on Math StackExchange some time ago, but it did not garner any solutions there. I think that it is interesting enough to be posed here on Math Overflow, so here it goes.
Let $ \mathcal{A} $ be a unital Banach algebra over $ \mathbb{C} $, with $ \mathbf{1}_{\mathcal{A}} $ denoting the identity of $ \mathcal{A} $. For each $ a \in \mathcal{A} $, define the spectrum of $ a $ to be the following subset of $ \mathbb{C} $:
$$
{\sigma_{\mathcal{A}}}(a) \stackrel{\text{def}}{=} \lbrace \lambda \in \mathbb{C} ~|~ \text{$ a - \lambda \cdot \mathbf{1}_{\mathcal{A}} $ is not invertible} \rbrace.
$$
With the aid of the Hahn-Banach Theorem and Liouville's Theorem from complex analysis, one can prove the well-known result that $ {\sigma_{\mathcal{A}}}(a) \neq \varnothing $ for every $ a \in \mathcal{A} $. All proofs that I have seen of this result use the Hahn-Banach Theorem in one way or another (a typical proof may be found in Walter Rudin's Real and Complex Analysis). Hence, a natural question to ask would be: Can we remove the dependence of this result on the Hahn-Banach Theorem? Is it a consequence of ZF only? Otherwise, if it is equivalent to some weak variant of the Axiom of Choice (possibly weaker than the Hahn-Banach Theorem itself), has anyone managed to construct a model of ZF containing a Banach algebra that has an element with empty spectrum?

REPLY [4 votes]: There is a proof without using even complex analysis, and at first glance without using AC. We replace Cauchy integral formula by clever finite sums. Namely, we define the spectral radius as $r(a):=\lim \|a^n\|^{1/n}$ (the limit exists by Fekete lemma), by scaling it suffices to consider two cases: $r(a)=0$ and $r(a)=1$. In the first case $a$ is not invertible, since if $ab=1$, then $a^nb^n=1$ and $\|a^n\|\cdot \|b\|^n\geqslant 1$, thus $r(a)\geqslant 1/\|b\|$. If $r=1$, we prove that there exist $\lambda$ on the unit circle such that $a-\lambda$ is not invertible. Assume the contrary, then by continuity of the inverse and compactness of the circle we get $\|(a-\lambda)^{-1}\|\leqslant M$ for all $\lambda$ on the unit circle and certain $M>0$. Now we use the rational functions identity $$n^2 a^{n-1}=(1-a^n)^2\sum_{w:w^n=1}w(a-w)^{-2}$$
and take the spectral radius of both right and left hand sides. By using the easy equalities and inequalities $r(u^n)=(r(u))^n$, $r(uv)\leqslant r(u)\cdot \|v\|$, $r(1+u)\leqslant 1+r(u)$ for commuting $u,v$ we see that $r(n^2a^{n-1})=n^2$ while $r(RHS)=O(n)$. A contradiction.
I read this proof in the paper of E. Gorin, who says that it is a simplification of the proof in [Rickart C. E. General Theory of Banach Algebras. Princeton NJ, D. van Nostrand (1960).]<|endoftext|>
TITLE: Can Liouville's number be expressed as a physical ratio in the sense that $\pi$ is?
QUESTION [5 upvotes]: Quadratic irrational numbers are perhaps the most basic examples of irrational numbers that arise as basic physical ratios: think of $\sqrt{2}$ as the distance between the corners of a square to the length of a side for example.
Similarly, the number $\pi$ is the ratio of the circumference of a circle to its diameter. Even $e$ can be realized in this sense as the distance $a$ such that $\int_{1}^{a}\frac{1}{x}dx=1$. Are there similar understandings for other irrationals?

Can Liouville's number be expressed as the ratio of two measurements of a geometrical object?

I will add my inspiration for asking this question in the hopes of encouraging more discussion that can help me focus these ideas. While pondering the argument for the existence of two irrational numbers $a$ and $b$ such that $a^b$ is rational using $\sqrt{2}^\sqrt{2}$ and the "law of the excluded middle", I wondered at the linked wikipedia article's note that an "intuitionist" would not accept this argument as proof. I feel like the only way to reject this argument as proof would be to question the assumption that $\sqrt{2}^\sqrt{2}$ exists as a real number, and in fact must be either rational,  or irrational. (Any intuitionists want to way in?)
This led me to ask myself, how do I know that $\sqrt{2}$ exists? I of course immediately answered that it was the ratio of the diagonal of a square to the length of a side! This led me to consider the "existence" of other irrational numbers as "some sort" of "physical ratio."
In considering this I would (for now) accept the description of $e$ above as being a "physical ratio."
Can anyone help me define what I mean by these "physical ratios?" I suppose I am looking for some argument for the "existence" of otherwise incomprehensible numbers, and whether or not every irrational could be explained in such a way.
I of course must note that this is incredibly vague and that the only one who can truly decide what I mean is myself, but any advice would be great!

REPLY [11 votes]: The answer is probably no.
In his paper Transcendence of Periods: the State of the Art (Pure and Applied Mathematics Quarterly, Volume 2, Number 2, p. 435-463, 2006), Michel Waldschmidt conjectures that no period is a Liouville number (see questions 2 & 3 in the introduction). This expectation is supported by the main advances of transcendental number theory in 20th century, notably Baker's theory of linear forms in logarithms and its extensions to commutative algebraic groups.
Here, the word period has to be understood in the sense of Kontsevich and Zagier: 
a complex number whose real and imaginary parts
are values of absolutely convergent integral of rational functions with rational
coefficients, over domains in $\mathbf R$
given by polynomial inequalities with rational coefficients. (The preceding definition is quoted from the paper Periods and elementary real numbers by Masahiko Yoshinaga, 
who was apparently the first to prove that periods belong to the field of elementary complex numbers, those whose real and imaginary parts can be effectively approximated by Cauchy sequences of rationals.)<|endoftext|>
TITLE: Reference: Finite $p$-Groups
QUESTION [7 upvotes]: Hall and Blackburn made important contributions in the study of regular $p$-groups and $p$-groups of maximal class. From their work, one can understand that in the classification of groups of order $p^n$, we must have to make two main cases: $p\leq n$, and $p>n$. 
With this interest, I am searching more and more material to study small $p$-groups, and their classification. The books I referred are that of Berkovich (Groups of prime power order) and of Leedham-Green, McKay (Structure of groups of prime power order). 
Beside these two main references, can one suggest other books/notes which contains study of $p$-groups of maximal class and regular $p$-groups?
(The book of Berkovich mentions one book in bibliography, that of A. Mann-Finite $p$-groups; but I couldn't find this book. Is this book or notes published?)

REPLY [13 votes]: It's true that at one time I thought of writing such a book, and even wrote a few chapters, but at the moment I'm not sure if I'll ever finish it, so I'm telling people asking about it not to hold their breath.
It should be noted that the book Structure of Groups... mentioned in the question has two authors: Charles R. Leedham-Green and Sue McKay.
Avinoam Mann<|endoftext|>
TITLE: Geodesic cuffs of pairs of pants in a hyperbolic manifold- why are they disjoint?
QUESTION [9 upvotes]: I'm trying to understand Kahn-Markovic's celebrated Immersing almost geodesic surfaces in a closed hyperbolic three manifold. There is something probably quite basic which I can't figure out.
We have $M^3= \mathbb{H}^3/\mathcal{G}$ a closed hyperbolic $3$-manifold, where $\mathcal{G}$ is a Kleinian group. Let $\Pi^0$ be a topological pair of pants with cuffs $C_0$, $C_1$, and $C_2$, and let $\rho\colon\thinspace \pi_1(\Pi^0)\to \mathcal{G}\subset PSL(2,\mathbb{C})$ be a faithful representation. Let $\gamma_i$ denote the geodesic in $M$ that represents the conjugacy class of $\rho(C_i)$ in $\mathcal{G}$ for $i=0,1,2$ (these become cuffs for pairs of pants which Kahn-Markovic construct inside $M$ and glue together to form an immersed almost geodesic surface). 

Question: Why are $\gamma_0$, $\gamma_1$, and $\gamma_2$ disjoint?
It is essential to the construction that the cuffs indeed be disjoint, because if they intersect, reduced complex Fenchel-Nielsen coordinates don't exist (there is no "foot"). In fact, Kahn and Markovic need many pairs of pants inside $M$ to coexist, so not understanding why the cuffs are non-intersecting even for a single pair of pants is particularly frustrating.

REPLY [6 votes]: This is an extended comment:
As Misha says, the geodesics might be immersed in general, and in fact the pants and the surfaces they construct will in general be highly immersed (lots of self-intersections). 
If you'd like to visualize embedded geodesics, you can imagine the lifts to the unit tangent bundle, or for a given pair of pants, it will lift to an embedded pants in some covering space corresponding to the image of the fundamental group $\rho(\pi_1(\Pi^0))$. To understand the feet of $\gamma_i$, then you can work in this covering space. The feet will be at the points of the shortest geodesics connecting the three boundary components in pairs (seams). There's a canonical involution sending $\gamma_i$ to its inverse, fixing the seams, and thus the feet lie equally spaced about each geodesic. Then project the whole picture back down into $M$ to get the feet.<|endoftext|>
TITLE: Constructive proof of "Projective implies proper"
QUESTION [12 upvotes]: For every ring $A$, the structural morphism of schemes $\pi_A : {\bf P}^n_{A} \to {\rm Spec}{A}$ is a closed map. The usual proof of this fact is not constructive : given equations of a closed subset $Z$ of ${\bf P}^n_{A}$, it doesn't produce equations for $\pi_A(Z)$.
In the case $A$ is a polynomial ring over an algebraically closed field $k$, this result is none other than the fundamental theorem of elimination theory : the image of a Zariski-closed subset of ${\bf P}^n(k) \times k^m$ under the second projection is a Zariski-closed subset of $k^m$. The first proofs of this theorem (Cayley, Kronecker, Sylvester) used resultants and thus were constructive.
In fact, the proof using elimination theory is universal in the following sense. Given integers $n,r \geq 1$, $d_1,\ldots,d_r \geq 1$, consider the universal homogenous polynomials $P_1,\ldots,P_r$ of degree $d_1,\ldots,d_r$ in the indeterminates $T_0,\ldots,T_n$, having coefficients in the polynomial ring $\widetilde{A} = \mathbf{Z}[Y_{i,\alpha} : 1 \leq i \leq r]$, where the indeterminates $Y_{i,\alpha}$ are the coefficients of $P_i$. Then there exists an explicit "resultant system" $R_1,\ldots,R_s \in \widetilde{A}$ such that $\pi_{\widetilde{A}}(V_+(P_1,\ldots,P_r))=V(R_1,\ldots,R_s)$. This means that specializations of $P_1,\ldots,P_r$ in some algebraically closed field $k$ have a common root in ${\bf P}^n(k)$ if and only if the corresponding specializations of $R_1,\ldots,R_s$ all vanish. Of course $s$ has to depend on $n,r,d_i$, but everything is explicit (at least from a theoretical point of view).
Now let $A$ be any ring and let $I=(f_1,\ldots,f_r)$ be an homogenous ideal of finite type of $A[T_0,\ldots,T_n]$. Then the resultant system above specialized at $f_1,\ldots,f_r$ provides explicit equations for $\pi_A(V_+(I))$ (this can be seen by studying the geometric fibers of $\pi_A$). In particular if $A$ is noetherian, then the map $\pi_A$ is closed, and we have a constructive proof for that. But in general, a closed subset of ${\bf P}^n_A$ need not be defined by finitely many equations. This raises the following questions :

Is there a way to prove that the map $\pi_A$ is closed for every ring $A$, by some clever reduction to the noetherian case?
If $Z$ is a closed subset of ${\bf P}^n_A$, given to us by infinitely many explicit equations $(f_i)_{i \in I}$, is there a way to produce explicit equations for $\pi_A(Z)$? In other words, is there a constructive proof of the fact that $\pi_A$ is closed?
Regarding question 2, an obvious thing to do is to look at all finite subfamilies $(f_i)_{i \in J}$, where $J$ is a finite subset of $I$, and to consider the associated resultant systems. Are all these equations sufficient to define $\pi_A(Z)$?

EDIT. Will Sawin has proved that the answers to all these questions is yes. Following Daniel Litt's comment, we can also consider $\pi_A(Z)$ as a closed subscheme of $\operatorname{Spec} A$, namely the closed subscheme defined by the kernel of the morphism $A \to \mathcal{O}_Z(Z)$. Do the resultant systems generate this ideal of $A$?

REPLY [7 votes]: 3, and thus 2 and 1: yes. By checking equality between the two sets at each point, we reduce to the case where the base is a point. But points are always Noetherian schemes, and the statement is obviously true for Noetherian schemes.
Edit: I was just reminded of this question and I realize that I now know the answer. If an element is in the kernel of the morphism $A \to \mathcal O_Z(Z)$, then it is in the kernel of the map to $\mathcal O_Z$ restricted to each of the $n+1$ standard affine open sets in $\mathbb P^n_A$. In each of those sets, it is in the ideal generated by the $f_i$. Of course if you are in the ideal generated by the $f_i$ then you are in the ideal generated by finitely many of them. Taking a finite union over the $n+1$ different affine opens, we get a finite set of relations that proves this element is in the kernel. So finite elimination theory generates infinite elimination theory.<|endoftext|>
TITLE: Torsion in cohomology of smooth manifolds
QUESTION [10 upvotes]: I've been interested in the possible (singular) cohomology groups of complex projective algebraic varieties, and there are lots of theorems that give various restrictions on these (Hodge decomposition, Lefshetz theorem, ... ). I realized that I would like to know more about the what is true for smooth manifolds hence my questions:
1.) One can construct CW complexes that have prescribed (reduced) homology groups (coeffs in $\mathbb{Z}$), these are the Moore spaces. However, they aren't even topological manifolds in general. Can one construct compact oriented smooth manifolds that have prescribed singular cohomology groups $H^i(X, \mathbb{Z})$, provided that after we remove torsion our sequence of groups satisfy Poincare duality? Should one expect that this is "generally possible" but it may be hard to actually construct examples? 
2.) If $X$ is a compact oriented smooth manifold, is there any regularity in the torsion subgroups of it's cohomology: $H^i_{sing} (X, \mathbb{Z})$? (Eg, poincare duality gives regularity between the various torsion free parts.) How about if $X$ is a nonsingular complex projective variety?
3.) For $X$ a smooth oriented manifold, it seems like compactly supported cohomology contains more information than ordinary cohomology. Can one recover ordinary cohomology $H^i_{sing}(X, \mathbb{Z})$ from compactly supported cohomology $H^i_c(X, \mathbb{Z})$? How about if we take coefficents in $\mathbb{Q}$?
I'd love to see "typical", or common examples where various phenomena appears.

REPLY [10 votes]: Here are some facts.
If  
$$ P(t) = a_0+a_1t+\cdots + a_{2k} t^{2k} $$
is a polynomial  with  nonnegative  integral coefficients such that
$$ a_0=a_{2k}=1,\;\;a_j=a_{2k-j},\;\;\forall j $$
and $\newcommand{\bZ}{\mathbb{Z}}$
$$ a_k\in 2\bZ, $$
then there exists a smooth, compact, connected, oriented  manifold $M$ of dimension $2k$  whose Poincare polynomial $P_M$ is the above polynomial $P$, i.e., 
$$b_j(M)=a_j,\;\;\forall j. $$
The manifold $M$ can be found  by taking connected sums of  products of spheres $S^{k_1}\times \cdots \times S^{k_m}$. The result is  sharp in the following sense.   There do not exist  oriented smooth manifolds whose Poincare polynomials  are
$$1 +t^6 +t^{12},\;\; 1+ t^{10}+ t^{20}. $$
This last  fact  was observed by Serre and follows from Hirzebruch's signature theorem.

REPLY [5 votes]: 1) You can pick the homology below the middle dimension quite arbitrarily. More precisely, given a finite complex $K$ and a number $n$, there exists a closed, parallelizable $2n$-dimensional manifold $M$ and an $n$-connected map $f:M \to K$. You begin with a constant map $S^{2n} \to K$ and make it more and more connected by surgeries.
2) as you said, a necessary condition for the homology of a manifold is Poincare duality. If you have a finite complex $X$ that satisfies Poincare duality, the question of whether there is a smooth manifold homotopy equivalent to $X$ is a basic problem in surgery theory. If $X$ is simply connected, this has largely been solved by Browder. The answer is that if $X$ is odd-dimensional, there is such a manifold; and if the dimension is divisible by $4$, there is a manifold precisely if there is a stable vector bundle on $X$ such that the Hirzebruch signature formula holds with this bundle. In dimensions $2,6,10,\ldots$, there is a subtle problem with the "Kervaire invariant". And: I forgot to say that the dimension has to be at least $5$. For nonsimplyconnected complexes, Wall gave at least a theoretical answer.
3) Poincare duality for integral coefficients (and closed oriented $M$) says that $H_i (M) \cong H^{n-i}(M)$. The universal coefficient theorem implies that the torsion subgroups (for each space with finitely generated homology) are $T H^{i+1} = T H_i$ (abstract isomorphism). Combined, these two results tie the torsion subgroups of cohomology together.
4) I would not say that compactly supported cohomology contains more information than ordinary cohomology - they contain different information. With rational coefficients, you have an isomorphism $H^i(M) \cong (H^{n-i}\_{c}(M))^{\ast}$; the other isomorphism $H^{i}_{c}(M) \cong (H^{n-i}(M))^{\ast}$ holds iff the cohomology vector space are finitely generated.<|endoftext|>
TITLE: Tensor rank of anti-symmetric tensor
QUESTION [6 upvotes]: Let $V$ be a vector space of dimension $n$. Let us consider $V^{\otimes n}=V\otimes V \ldots \otimes V$. This vector space contains one dimentional vector space $\wedge^n V$. My question is does it something is known about the tensor rank of the vector   $\wedge^n V$?
More formally let $e_1, e_2,\ldots e_n$ be a basis for $V$ than the question is what does it known about the tensor rank of:$$ T=\sum_{\sigma \in S_n}(-1)^{sign(\sigma)} e_{\sigma(1)}\otimes e_{\sigma(2)} \otimes \ldots \otimes e_{\sigma(n)}.$$
The trivial upper bound on the tensor rank of this form is $n!$. Does it know any better uper bound? 
As far as I know without $(-1)^{sign(\sigma)}$(i.e. for a symmetric form) it know upper bound of $2^n$.

REPLY [3 votes]: The tensor ranks of determinants and permanents are currently not known. In the $3 \times 3$ case it is known: the $3 \times 3$ determinant has tensor rank $5$ and the $3 \times 3$ permanent has tensor rank $4$. Here, $5$ is better than the naive $n!=3!=6$. Upper bounds are known. An identity due to Derksen shows that the $n \times n$ determinant has tensor rank at most $\left(\dfrac{5}{6}\right)^{\lfloor n/3 \rfloor} n!$, instead of $n!$. An identity due to Glynn shows that the $n \times n$ permanent has tensor rank at most $2^{n-1}$.
To be clear, perhaps I should have said first that the tensor $T$ described in the question can be identified with the determinant of a generic $n \times n$ matrix (each $e_i$ stands for the $i$th column [or $i$th row]). And without the signs, the resulting tensor is known as the "permanent". The original question as I understand it is asking about an upper bound for the tensor rank of the determinant and permanent.
For some context about what is the meaning of tensor ranks of determinant and permanent, and why comparisons between those particular things might be of interest, see for example Landsberg, Geometric complexity theory: an introduction for geometers, 2015 (MR3343444).
Derksen's identity for the $3 \times 3$ determinant is in his article Derksen, On the nuclear norm and the singular value decomposition of tensors, 2016 (MR3494510). It may be interesting to see the identity explicitly here. First of all the $3 \times 3$ determinant, as noted in the original question, is the following tensor in $(\Bbbk^3)^{\otimes 3}$:
$$
\begin{multline}
  e_1 \otimes e_2 \otimes e_3 + e_2 \otimes e_3 \otimes e_1 + e_3 \otimes e_1 \otimes e_2 \\
  - e_1 \otimes e_3 \otimes e_2 - e_2 \otimes e_1 \otimes e_3 - e_3 \otimes e_2 \otimes e_1 .
\end{multline}
$$
(This corresponds to the familiar-to-undergraduates method of computing $3 \times 3$ determinants by adding the "downward" diagonals and subtracting the "upward" diagonals.) Derksen observed that this same tensor is equal to:
$$
\begin{multline}
  \frac{1}{2} \Big( (e_3+e_2) \otimes (e_1-e_2) \otimes (e_1+e_2) + (e_1+e_2) \otimes (e_2-e_3) \otimes (e_2+e_3) \\
  + 2 e_2 \otimes (e_3 - e_1) \otimes (e_3 + e_1) + (e_3-e_2) \otimes (e_2+e_1) \otimes (e_2-e_1) \\
  + (e_1-e_2) \otimes (e_3+e_2) \otimes (e_3-e_2) \Big).
\end{multline}
$$
This uses $5$ terms. (I suppose we had better require $2^{-1} \in \Bbbk$.) By Laplace expansion, all $n \times n$ determinants can be expanded using "blocks" of $3 \times 3$ minors, and each "block" has this gain of rank $5$ instead of $6$, resulting in the above mentioned upper bound for tensor rank of determinant.
As of October 2017 this is, to my knowledge, the best known upper bound for tensor rank of the determinant.
It turns out that it has been known for quite some time that the maximum rank of any tensor in $(\Bbbk^3)^{\otimes 3}$ is $5$, which forces the existence of some rank $5$ expression for determinant. So an identity like Derksen's could have been observed a long time ago (decades ago). In stating this I have no intention whatsoever of taking anything away from Derksen, who was the one who finally considered this and found the above explicit expression. (But when I finally understood what Derksen had managed to do, I did spend a frantic week looking at every upper bound for rank I could find, and sadly for me, the upper bound of $5$ in $(\Bbbk^3)^{\otimes 3}$ turned out to be the only meaningful one for determinants or permanents. Oh well.)
The upper bound of $2^{n-1}$ for the tensor rank of permanent is by an identity due to Glynn, in Glynn, The permanent of a square matrix, 2010 (MR2673027). As of October 2017 it is, to my knowledge, the best known upper bound for the tensor rank of the permanent.
The lower bounds for the tensor ranks of the $3 \times 3$ determinant and permanent is in Ilten, Teitler, Product ranks of the $3 \times 3$ determinant and permanent, 2016 (MR3492642). That article presents the results in terms of "product rank" rather than tensor rank, but the article includes an explanation of the relationship between these notions of rank, including showing how to get the results about tensor ranks of determinants and permanents.
So we know the tensor ranks of $3 \times 3$ determinants and permanents (the tensor ranks are $5$ and $4$, respectively). But that's it; for $4 \times 4$ it is open. Glynn's identity shows that the $4 \times 4$ permanent has tensor rank at most $8$, and Derksen's identity shows that the $4 \times 4$ determinant has tensor rank at most $20$ (via Laplace expansion across the top row, into a sum over $4$ determinants of $3 \times 3$ minors, each with rank $5$). I'm not sure what exactly are the current best lower bounds, but I believe they are something like this. A result of Shafiei shows that the $4 \times 4$ permanent has Waring rank at least $\frac{1}{2}\binom{8}{4} = 35$; this bound for Waring rank implies the tensor rank is at least $\lceil \frac{35}{8} \rceil = 5$. A result of Derksen and myself shows that the $4 \times 4$ determinant has Waring rank at least $\binom{8}{4}-\binom{6}{3}=50$, hence tensor rank at least $\lceil \frac{50}{8}\rceil = 7$. (This business with Waring ranks and tensor ranks is explained in my paper with Harm Derksen, and also in my paper with Nathan Ilten, mentioned above.) There do exist other lower bound results, I'm just not sure right now if any of them give stronger bounds in these examples.
In summary, the $4 \times 4$ permanent has tensor rank between $5$ and $8$ (inclusive) and the $4 \times 4$ determinant has tensor rank between $7$ and $20$ (inclusive).
Regarding @aginensky's comment, it's not only the tensor ranks (and product ranks) which are completely up in the air; the Waring ranks, equivalently symmetric ranks when considered as symmetric tensors, are also completely unknown, even for the $3 \times 3$ case!<|endoftext|>
TITLE: Collection from Replacement in ZFC-extensionality
QUESTION [5 upvotes]: My question is whether the axiom of extensionality is required to show that the schema of collection follows from the schema of replacement in the usual Zermelo-Fraenkel environment with choice. In other words: Is the schema of collection a theorem schema in Zermelo-Fraenkel set theory with choice minus the axiom schema of extensionality?

REPLY [4 votes]: Collection is not provable in ZFC minus extensionality, a simple countermodel is described in https://mathoverflow.net/questions/54328 . (That the model cannot provably satisfy collection follows from Gödel’s theorem. For a specific instance of collection which fails, let $\bar\omega$ denote one of the many representations of $\omega$ in the model, and $\bar0\in\bar\omega$ the corresponding empty set: then the model satisfies “for every $n\in\bar\omega\smallsetminus\{\bar0\}$, there exists a function $f$ with domain $n$ such that $f(\bar0)=\bar\omega$, and $f(x)\in f(y)$ whenever $x\in y\in n$”, but there is no set collecting such functions for every $n\in\bar\omega\smallsetminus\{\bar0\}$.)<|endoftext|>
TITLE: distance between powers of 2 and powers of 3
QUESTION [27 upvotes]: I expect this is a classical question, so feel free to point me to classical answers: what is the fastest-growing function $f(t)$ for which we know that
$$
|2^t - 3^{t'}| \ge f(\min(t,t')) \;?
$$
In particular, do we know that the gaps between powers of 2 and powers of 3 get exponentially large as $t,t'$ increase?  Do we know anything like this for any other pair of integers besides 2 and 3?

REPLY [4 votes]: Just to satisfy the curiosity of @FelixGoldberg and other cursory readers. Here is a heuristic which pointed me to try to use $C=1.06$ for an example.     
We look at the distances     
$$\left|1-{3^b\over2^a}\right| \overset{???}\ge { 1\over (e \cdot a)^C} $$ 
with $a \gt b$ and $2^a \gt 3^b$ (fixing the $\max()$-terms.        
In the table $w=\log_2\left|1-{3^b\over2^a}\right| $ and $u=-\log_2 (e a)$.    The quotient $w/u$ should give an impression of the missing factor $C$, and in this table for all except $2$ cases ( idx=15,idx=21 ) a value of $C=1.06$ suffices to make the inequality true.        
The table reports the cases according to the continued fraction of $ß=\log_23)$ so only the best possible approximants (with $2^a \gt 3^b$) are displayed (the convergents, each second of them)
  idx   b     a     log2(b)  log2(a)      w       u         w/u      1.06*u
----------------------------------------------------------------------------
   3      1      2  0.E-201  1.00000  -2.00000  -2.44270  0.818768  -2.58926
   5      5      8  2.32193  3.00000  -4.29956  -4.44270  0.967782  -4.70926
   7     41     65  5.35755  6.02237  -6.45514  -7.46506  0.864714  -7.91297
   9    306    485  8.25739  8.92184  -9.93479  -10.3645  0.958537  -10.9864
  11  15601  24727  13.9294  14.5938  -15.7461  -16.0365  0.981894  -16.9987
  13  79335  SSSSS  16.2757  16.9401  -18.0579  -18.3828  0.982323  -19.4858
  15  NNNNN  SSSSS  17.5397  18.2042  -23.8860  -19.6469   1.21576  -20.8257
  17  NNNNN  SSSSS  23.3620  24.0265  -26.2877  -25.4692   1.03214  -26.9973
  19  NNNNN  SSSSS  27.3572  28.0217  -29.0580  -29.4644  0.986209  -31.2322
  21  NNNNN  SSSSS  28.5666  29.2311  -33.1373  -30.6738   1.08031  -32.5142
  23  NNNNN  SSSSS  32.6169  33.2814  -36.5236  -34.7241   1.05182  -36.8075
  25  NNNNN  SSSSS  37.0009  37.6654  -40.0173  -39.1081   1.02325  -41.4546
  27  NNNNN  SSSSS  42.2986  42.9630  -43.7861  -44.4057  0.986046  -47.0701
  29  NNNNN  SSSSS  43.3957  44.0601  -46.1400  -45.5028   1.01400  -48.2330
  31  NNNNN  SSSSS  48.6152  49.2797  -49.4134  -50.7224  0.974193  -53.7657
  33  NNNNN  SSSSS  52.3527  53.0172  -53.0620  -54.4599  0.974331  -57.7275
  35  NNNNN  SSSSS  56.8562  57.5206  -58.9521  -58.9633  0.999810  -62.5011
  37  NNNNN  SSSSS  58.4640  59.1284  -62.5155  -60.5711   1.03210  -64.2054
  39  NNNNN  SSSSS  62.0089  62.6734  -65.0073  -64.1161   1.01390  -67.9630
  41  NNNNN  SSSSS  64.5731  65.2376  -66.4207  -66.6803  0.996108  -70.6811
  43  NNNNN  SSSSS  66.0744  66.7389  -67.9931  -68.1815  0.997236  -72.2724
  45  NNNNN  SSSSS  67.4786  68.1431  -73.2504  -69.5858   1.05266  -73.7609
  47  NNNNN  SSSSS  74.7217  75.3861  -81.0514  -76.8288   1.05496  -81.4386
  49  NNNNN  SSSSS  80.5354  81.1999  -82.0659  -82.6426  0.993022  -87.6011


The Pari/GP script is
fmt(200,8) \\ internal precision 200 dec digits, user-procedure
{e=exp(1);l3=log(3);l2=log(2);ld3 = l3/l2;
cf = contfrac(ld3);
cvgts= mkContFracConvergents(cf,50) ; \\ user-procedure
listlogs=vectorv(50);ix=0;
forstep(i=3,50,2,
          a=cvgts[1,i];      \\ ===> a > b  and also 2^a > 3^b
          b=cvgts[2,i];   
          ix++; listlogs[ix]=[i,
                 if(b<100 000,b,'NNNNN),      if(a<100 000,a,'SSSSS),
                 log(b)/l2,                   log(a)/l2,
                 w=log((1.0-2^(ld3*b-a)))/l2, u=-log(e*a)/l2,
                 w/u      ,                   1.06*u];
        );
 listlogs=Mat(VE(listlogs,ix))}

Remark: a bit more introduction and tables and graphs for $b \to 10^{10800} \approx 2^{36000} $ can be found at my pages . Note, that I use $N$ for what we use $b$ here, and $S$ for what we use $a$ here, thus discussing $2^S-3^N$.

update A better visualization of the properties of selecting some constant $C=1+\epsilon$ using up to $b =10^{1000}$ taken from the convergents of the continued fraction of $\log(3)/\log(2)$
I show how empirically the values of $C(b)$ were when $a,b$ are inserted in the formula and $C(b)$ is computed. The image shows, that the empirical $C(b)$ are except in two cases smaller than $C=1.06$ and moreover, that possibly we can choose any $C=1+\epsilon$ and getting only finitely many cases where not $C(b) \le C$
Legend: In the picture I used my standard-notation  $N$ for $b$ here and $S$ for $a$ here.<|endoftext|>
TITLE: Intuitive pictures in characteristic p
QUESTION [27 upvotes]: This is a tough one, but does anyone know of any images that recall characteristic p geometry (over algebraically closed fields) in some sense? It is not enough if it is some picture that can be also understood solely in characteristic 0.
A quick search through the literature has proved fruitless.
I have been thinking for a while of asking this question, but I never had a pressing need rather than my own curiosity. However, now I am trying to improve a poster by including pictures but the topic is algebraic geometry in characteristic p. 
An example of such an image in Complex Geometry would be the arrangement of contracted and blow up curves in the standard Cremona transformation of the projective plane. The one in this poster is simple but effective.
I believe this question also might be of interest for people who try to explain research to non-mathematicians or simply to mathematicians who are not geometers.

REPLY [3 votes]: This is not nearly as nice an example as the others, but I always imagined the line in characteristic five geometry as a countable set of points that glow like blue Christmas tree lights vaguely in the shape of a narrow paraboloid, with 0 at the vertex, with 1, 2,3, and 4 at the next "height", with the quadratic "irrationals" next, etc. although the makes it seem like each field of order $p^n$ contains the field of order $p^{n-1}$. Like in Carnahan's answer, I imagine the Frobenius automorphism shuffling around everything in each fixed ring. Finally, I imagine the Zariski topology as a glowy light filling in the paraboloid representing the forces that each point exerts on all others; it's constant everywhere, as in the Zariski topology, there's no real sense of distance.
More complicated varieties I imagine as a double cone with glowy lights (two Christmas trees!) or as the paraboloid distortedt to have self-intersections in some of the lights.
something like the galactic gravity simulation or spherical pendulum in this website<|endoftext|>
TITLE: Geometric interpretation of translation through the wall
QUESTION [6 upvotes]: What does translation through the wall correspond to under Beilinson Bernstein localization?
More precisely I am interested in the following:
There is a well known equivalence between the principal block of category $\mathcal O$ and 
perverse sheaves on the flag manifold, constructible along $B$ orbits:
$$\mathcal O_0 \cong \mathcal P_{(B)}(G/B)$$
Now for a singular integral weight $\lambda$ one can consider the translation through the wall functor 
$$ \theta_\lambda:\mathcal O_0 \rightarrow \mathcal O_\lambda \rightarrow \mathcal O_0$$
What does it correspond to under the above equivalence?
My naive guess/hope would be, that it is given by convolution with 
the sheaf corresponding to $\theta_\lambda (L_e)$ where $L_e$ is the antidominant simple.
Is this correct? If so, is there a geometric way to construct this sheaf?
PS: I am aware that there are descritions of the translation functors using slightly more elaborate version of localization, for example in this paper by Beilinson Ginzburg. However I would prefer to keep the above setup.

REPLY [6 votes]: I am guessing the following is well known to you/not what your are looking for, but nonetheless:
Let $s$ be a simple reflection, $P_s$ the corresponding minimal parabolic, $\pi_s\colon G/B\to G/P_s$ the projection. Translation across the $s$-wall `corresponds' to $\pi_s^*\pi_{s*}$. I use quotation marks because as stated this is clearly not true (translation across the wall is t-exact, $\pi_s^*\pi_{s*}$ is certainly not). However, $\pi_s^*\pi_{s*}$ does correspond to translation across the wall under Koszul duality. This is also the same as convolving with the $IC$-complex corresponding to $s$.
Morally (as you point out), reflection across the wall should correspond to convolving with the corresponding tilting. But there is an annoying issue here: tiltings are not $B$-equivariant. Similar problem occurs if instead of convolution using equivariant derived categories you try to use the standard Fourier-Mukai formalism and try to use an object on $G/B\times G/B$ as a kernel. However, there is a fix that comes at some technical expense. Namely, Bezrukavnikov and Yun's free monodromic sheaves http://arxiv.org/abs/1101.1253. The idea actually goes back to the paper of Beilinson and Ginzburg that you cite (look at Section 5).<|endoftext|>
TITLE: Subadditivity for Renyi entropies
QUESTION [10 upvotes]: Do the Renyi entropies satisfy the standard subadditivity of Shannon entropy? That is, 
\begin{equation}
H_\alpha(A,B) \leq H_\alpha(A) + H_\alpha(B)  ? 
\end{equation}
for $\alpha \ne 1$. If they do, for which $\alpha$? 
Here $H_\alpha(X)$ is the standard Renyi $\alpha$-entropy of the random variable X,
\begin{equation}
H_\alpha(X)=\frac{1}{1-\alpha}\log\sum_i^n p_i^\alpha,
\end{equation}
and $H_\alpha(A,B)$ is the Renyi entropy of the joint probability distribuition of the random variables A and B.

REPLY [2 votes]: In fact, to find a violation of any linear inequality you could form using the Renyi entropies of order $\alpha \in (0,1) \cup (1,\infty)$, please consult the following paper:
http://arxiv.org/abs/1212.0248<|endoftext|>
TITLE: Cohomology ring of BG
QUESTION [10 upvotes]: Let $G$ be a compact Lie group, let $T$ be a maximal torus, and let $W$ be the Weyl group.  My main question is as follows: 

How does one prove that $H^\ast(BG,\mathbb{Q})$ is isomorphic to the $W$-invariant part of $H^\ast(BT,\mathbb{Q}) \cong \mathbb{Q}[[x_1, \ldots, x_n]]$?  This is apparently basic knowledge in algebraic topology, because I keep reading "recall that..." followed by some version of this statement and no references.  But I can't find a proof in any of my textbooks.

I would ideally like a reference which also addresses the following secondary question:

When is the natural map $H^\ast(BG,\mathbb{Z}) \to H^\ast(BT,\mathbb{Z})^W$ an isomorphism, and what can one say about the integral cohomology ring of $BG$ when it is not?  Note the fact that the map above is an isomorphism for $G = U(n)$ is equivalent to the statement that the Chern classes are integral.

Thanks!

REPLY [10 votes]: Q1: Let me first note, that the statement
$$H^\ast(BG,\mathbb{Q}) \cong H^\ast(BT,\mathbb{Q})^W\tag{$\ast$}$$
made in the question requires in addition that $G$ is connected. However, the general case can be reduced to the connected case since
$$H^\ast(BG,\mathbb{Q}) \cong H^\ast(BG_0,\mathbb{Q})^{G/G_0}$$
where $G_0$ is the identity component of $G$.
A text book reference for $(\ast)$ can be found in Hsiang, Cohomology theory of topological transformation groups, Chapter III, §1, Lemma 1.1. The results of the book that are relevant for your question can also be found in the following paper: Richard Gonzales, Localization in equivariant cohomology and GKM theory (cf. Remark 9, Lemma 5). Other approaches and more information can be found in this answer.
Q2: First note that the kernel of the restriction map
$$\rho^\ast: H^\ast(BG;\mathbb{Z}) \to H^\ast(BT;\mathbb{Z})$$
is the tosion subgroup $Tors$ of $H^\ast(BG;\mathbb{Z})$. So, $\rho^\ast$ is injective iff $H^\ast(BG;\mathbb{Z})$ is torsion-free (there is a lot of literature about torsion
of $H^\ast(BG;\mathbb{Z})$ so I won't discuss it here).
A short paper of Feshbach

The image of $H^\ast(BG;\mathbb{Z})$ in $H^\ast(BT;\mathbb{Z})$ for $G$ a compact Lie group with maximal torus $T$. Topology 20(1981) 93-95 doi:10.1016/0040-9383(81)90015-X,

characterizes when the induced map
$$\bar{\rho}^\ast: H^\ast(BG;\mathbb{Z})/Tors \to H^\ast(BT;\mathbb{Z})$$
is an isomorphism:

$\bar{\rho}^\ast$ is an isomorphism iff $H^\ast(BG;\mathbb{Z})/Tors \otimes \mathbb{F}_p$ is an integral domain for each prime $p$.

There is also a counter-example, $\mathrm{Spin}(12)$, that shows that $\bar{\rho}^\ast$ is not always an isomorphism.

REPLY [6 votes]: Notice that there is a sequence of homomorphisms $T \to N \to G$, where $N$ is the maximal torus normaliser (so $W = N/T$).  $W$ acts on $BT$ (because it acts on $T$ by conjugation through group homomorphisms), and there is an equivalence from the classifying space of $N$ to the Borel construction for this action:
$$BN \simeq EW \times_W BT.$$
Consequently, we can compute the cohomology of $BN$ from the Leray-Serre spectral sequence
$$H^\ast(W; H^\ast(BT)) \implies H^\ast(BN).$$
Taking rational cohomology, this spectral sequence is concentrated in group-cohomological degree $0$, since $W$ is a finite group.  Therefore the spectral sequence collapses at $E_2$, which is $H^0(W, H^\ast BT) = H^\ast(BT)^W$.
It therefore suffices to show that the map $BN \to BG$ is an isomorphism in rational cohomology.  If we write $BN$ as $EG / N$, this map is a fibre bundle with fibre $G / N$, so it's enough to show that $G/N$ has the rational homology of a point.  
For instance, if $G = SU(2)$, $N = \mathbb{Z} / 2 \ltimes T$, and $T = S^1$.  Then $G/T = \mathbb{C} P^1$, and the action of $\mathbb{Z} / 2$ is antipodal, giving $G / N = \mathbb{R} P^2$, which is indeed rationally a point.  I don't remember the argument in general, but I think this is always true.
Hopefully this indicates how the corresponding integral statement can fail - there can be torsion contributions from the higher group cohomology of $W$, which needs to be exactly cancelled (via a differential in the second spectral sequence above) with a torsion cohomology class from $G/N$.<|endoftext|>
TITLE: Liouville's theorem with your bare hands
QUESTION [45 upvotes]: Liouville's theorem from complex analysis states that a holomorphic function $f(z)$ on the plane that is bounded in magnitude is constant. The usual proof uses the Cauchy integral formula. But this has always struck me as indirect and unilluminating. There is a proof via harmonic function theory, but this also seems to involve an unnecessarily large amount of prior buildup. So one might seek a more direct proof as below.
Assume that $f(z)$ is nonconstant. The fact that $f(z)$ is holomorphic at every point implies that at any given point, there is a direction such that moving in that direction makes $|f(z)|$ larger. But this doesn't prove that $|f(z)|$ is unbounded, because a priori its magnitude could behave like $5 - \frac{1}{|z|}$ or some such thing. 
In the case of $f(z) = \frac{1}{P(z)}$ where $P(z)$ is a polynomial, one knows that $|f(z)|$ tends toward $0$ as $|z| \to \infty$ so that there's some closed disk such that if $|f(z)|$ is bounded, then it has a maximum in the interior of the disk, which contradicts the fact that one can always make $f(z)$ larger by moving in a suitable direction. But for general $f(z)$, one doesn't have this argument.
One can try to reason based on the power series expansion of a holomorphic function $f(z)$ that is not a polynomial. Because polynomials are unbounded as $|z| \to \infty$ and grow in magnitude in a way that's proportional to their degree, one might think that a power series, which can be regarded as an infinite degree polynomial, would also be unbounded as $|z| \to \infty$. This is of course false: take $f(z) = \sin(z)$, then as $|z| \to \infty$ along the real axis, $f(z)$ remains bounded. The point is that the dominant term in the partial sums of the power series varies with $|z|$, and that the relevant coefficients change, alternating in sign and tending toward zero rapidly, so that the gain in size corresponding to moving to the next power of $z$ is counterbalanced by the change in coefficient. But there's some direction that one can move in for which $f(z)$ is unbounded: in particular, for $f(z) = \sin(z)$, $f(z)$ is unbounded along the imaginary axis.
This suggests that we write $a_n = s_{n}e^{i \theta_n}$ for the coefficient of $z^n$ in the power series expansion of $f(z)$ and write $z = re^{i \theta}$ (where $s, r > 0$) so that
$$f(z) = \sum_{n = 0}^{\infty} {a_n}z^n = \sum_{n = 0}^{\infty} sr^n e^{n\theta + \theta_n}  $$
and try to find a function $\theta = g(r)$ such that $f(z)$ is unbounded as $r \to \infty$ if one takes $\theta = g(r)$. 
But I don't know what to do next. Any ideas? Any ideas for other strategies of proving Liouville's theorem that are more direct than the ones using Cauchy's theorem?

REPLY [70 votes]: There is a truly elementary proof. Nothing but high school mathematics + the notion of limit is used.
First one proves Cauchy's inequality for polynomials:
$$|f(0)|\leq M(r),$$
where $M(r)$ is the maximum of $|f|$ on the circle $|z|=r$.
This is proved as follows: let $\epsilon_k=\exp(2\pi ik/n),$
where $n>$ the degree of the polynomial.
Then
$f(0)=\frac{1}{n}\sum_{k=1}^nf(\epsilon_kz),$
because
$\sum_k^n(\epsilon_k)^j=0$ for all $j\in[1,n-1]$, by the Geometric Progression formula,
so all other terms except the constant term, in the right hand side cancel. 
Taking absolute values, we obtain the above inequality.
Then by passing to a limit this inequality is true for all entire functions.
Applying this to $(f-f(0))/z$, we obtain
$$|f'(0)|\leq (M(r)+|f(0)|)/r,$$
for all entire functions.
If the function is bounded, we conclude that $f'(0)=0$.
Applying this to $f(z-a)$ we obtain $f'(a)=0$ for all $a$, that is $f$
is constant.

REPLY [36 votes]: I beg to differ with Andy and propose Nelson's proof as THE book proof that a bounded analytic function is constant. In fact Nelson's little gem is for harmonic functions, but the proof is incredibly beautiful and, of course, applies to analytic functions by considering their real and imaginary parts.
There are just two ingredients: 
1. The mean value theorem: the value of a harmonic function in the plane (or n-space) at some point is the average of the function over any disc centered at that point. We can even take that as definition of a harmonic function if we have a family of "discs" and a measure with which to average.
2. A geometric property of metric discs on the plane: Given any two points $x$ and $y$, for sufficiently large radii $R$, the symmetric difference of the disc of radius $R$ centered at $x$ and the disc of the same radius centered at $y$ has negligible area compared with the area of the discs. 
Now the proof goes as follows: take any two points $x$ and $y$ on the plane and choose discs of a very large radius $R$ centered at each of these points. If the harmonic function is bounded, its average over the two discs, that by (2) basically coincide, has to be almost the same. Let $R$ go to infinity and you're done.
Remark/Question Clearly the proof makes sense for a class of metric measure spaces. Has this sort of spaces (symmetric difference between large metric balls having small relative measure) been studied on its own? Normed spaces are in this category.<|endoftext|>
TITLE: For which fields does the isogeny theorem hold
QUESTION [9 upvotes]: Let $k$ be a field. We say that the isogeny theorem holds over $k$ if, for any abelian variety $A$ over $k$, there are only finitely many $k$-isomorphism classes of abelian varieties $B$ over $k$ which are $k$-isogenous to $A$.
Here are some examples of fields for which the isogeny theorem hold.

$k$ is a finite field.
$k$ is a number field (Faltings)

Does the isogeny theorem hold for $k$ a function field over an algebraically closed field of characteristic zero?
Does the isogeny theorem hold for $k$ a  function field over a finite field?
In the number field case, by results of Serre-Tate, the Shafarevich conjecture for abelian varieties implies the isogeny theorem. 
Does a similar implication hold over $\mathbf C(t)$? (Of course, the naive analogue of the Shafarevich conjecture is false, but Faltings proves a "correct" version in his paper: Arakelov's theorem for abelian varieties.)

REPLY [6 votes]: Over $\mathbb{C}(T)$ the Isogeny theorem doesn't hold, even if you throw away isotrivial components.
In his paper "Arakelov's theorem for Abelian Varieties" Faltings proves the following analogue of Shafarevich: 
Let $B$ be complete smooth complex curve and $S\subset B$ a finite subset of points, and $f:X\rightarrow B\backslash{S}$ a family of Abelian varieties. Let $G$ be the fundamental group of $B\backslash{S}$. Fix a point $p\in B\backslash S$ so that we get an action of $G$ on
$H_1(X_p,\mathbb{Z})$ by monodromy. 
Now, say $X$ satisfies (A) if 
$$End(X)=End_{G}(H_1(X_p,\mathbb{Z})).$$ Note that the LHS also injects into the RHS.
Faltings proves there are finitely many isomorphisms classes $X$ satisfying (A) of a fixed dimension, and he also gives an example of an 8-dimensional family $X$ which doesn't satisfy (A). Now we'll use this $X$ to give a negative answer to the Isogeny theorem:
By a standard argument (see Milne's Abelian Varieties online notes, Chapter IV, Theorem 2.5) if both $X$ and $X\times X$ have at most finitely many isogenous abelian varieties up to isomorphism, by an isogeny of degree a power of some prime $l$, then
$$End(X)\otimes\mathbb{Q}_l=End_G(H_1(X_p,\mathbb{Z})\otimes\mathbb{Q}_l)$$
where the second endomorphism ring is as $\mathbb{Q}_l$ vector spaces. However, the latter is false for our $X$ by assumption, since taking group invariants commutes with base change over fields, hence either $X$ or $X\times X$ don't satisfy the isogeny theorem.
$\textbf{However},$ if $X$ is a family of abelian varieties over the generic point of $B$ which extends to a family away from $S$ (in other words, it only has bad reduction at points of $S$) then every abelian varietiy isogenous to $X$ shares this property, since good reduction is an isogeny invariant. Hence the isogeny therem is true for all $X$ satisfying (A), and this includes all $X$ of dimension at most 3 by a theorem of Deligne 
(see Hodge II, 4.4.13), as long as you insist $X$ has no isotrivial subvarieties.<|endoftext|>
TITLE: Can Ext over a group ring always be expressed as group cohomology ? 
QUESTION [8 upvotes]: Given a group $G$ and $G$-modules $M,N$ with $M$ $\mathbb{Z}$-free then it's well known that 
$$Ext_{\mathbb{Z}G}^i(M,N) \cong H^i(G,Hom(M,N))$$
for all $i \ge 0$ (a reference is Brown, Cohomology of Groups, Proposition 2.2). 
But what happens if $M$ is not $\mathbb{Z}$-free ? Is it still possible to express $Ext_{\mathbb{Z}G}^i(M,N)$ by cohomology groups of $G$ (maybe in form of a spectral sequence) ?

REPLY [13 votes]: There is a long exact sequence 

$$0 \to H^1(G,Hom(M,N)) \to Ext_{\mathbb{Z}G}^1(M,N) \to \cdots $$
  $$\begin{array}{lll}
\cdots & \to & H^i(G,Hom(M,N)) \to Ext_{\mathbb{Z}G}^i(M,N) \newline 
& \to & H^{i-1}(G,Ext_{\mathbb{Z}}^1(M,N))\to H^{i+1}(G,Hom(M,N)) \to \cdots
\end{array}$$

For, as pointed out by Will and Mariano, there is a spectral sequence 
$$H^i(G, Ext^j_\mathbb Z (M,N)) \Rightarrow Ext_{\mathbb ZG}^{i+j}(M,N)$$
and since the projective dimension of $\mathbb{Z}$ is one, the spectral sequence takes the form 
$$E_2 = \quad 
\begin{array}{ccccc}
\vdots & \vdots & \vdots & \vdots &  \newline 
0 & 0 & 0 & 0 & \cdots \newline 
\bullet & \bullet & \bullet & \bullet & \cdots \newline 
\bullet & \bullet & \bullet & \bullet & \cdots 
\end{array}$$
Now the relation $E_\infty =E_3 = \ker(d_2)/\text{im}(d_2)$  yields the exact sequence 
$$0 \to \ker(d_2^{i-2,1}) \to E_2^{i-2,1} \to E_2^{i,0} \to H^i \to \ker(d_2^{i-1,1}) \to 0$$
where $H^i=Ext_{\mathbb ZG}^i(M,N)$ is the abutment. 
Remark: The statement remains true if we replace $\mathbb{Z}$ by an hereditary commutative ring $R$, $\mathbb{Z}G$ by an augmented $R$-projective $R$-algebra $A$  and $H^\ast(G,-)$ by $Ext_A^\ast(R,-)$.<|endoftext|>
TITLE: Describing the ratio of uniformizers in B_dR
QUESTION [8 upvotes]: In Conrad and Brinon's notes http://math.stanford.edu/~conrad/papers/notes.pdf, two uniformizers of $B_{dR}$ are produced: one is $\xi := [\tilde{p}]-p$ (bottom of p.58), where $\tilde{p} = (p, p^{1/p}, p^{1/p^2}, \ldots)$ is some choice of compatible $p^r$ roots of unity, and the other is $t := \log(\epsilon)$, where $\epsilon$ is some basis of the Tate module of $\mathbb{C}_K$. 
Furthermore, we know that $B_{dR}^+ / (t) \simeq \mathbb{C}_K$. Therefore, there should be some $a_1 \in \mathbb{C}_K$ such that $\xi = a_1t + ...$. My question is: how can we describe more explicitly what this $a_1$ is?

REPLY [8 votes]: What you're asking for is a description of $$a_1 = \theta(\frac{[\tilde{p}]-p}{t}).$$ It is an element of $C_p$ and you can't really "write it down explicitly". What you can do is let $G_{Qp}$ act on it and see what happens. If $g \in G_{Qp}$ then $g(\tilde{p})=\tilde{p} \cdot \epsilon^{c(g)}$ where $c(\cdot)$ is the Kummer cocycle and $g(t)=\chi(g) t$ where $\chi(\cdot)$ is the cyclotomic character. This should imply that $$g(a_1) = \theta(\frac{[\tilde{p}][\epsilon^{c(g)}]-p}{\chi(g) t})  = \frac{a_1}{\chi(g)} + p \frac{c(g)}{\chi(g)}.$$ Now that you see this formula, you should recognize it. Let $V$ be the semistable extension of $Q_p$ by $Q_p(1)$. It has Hodge-Tate weights $0$ and $1$ and $a_1$ is basically the period corresponding to weight $0$, ie the period that tells you that $(V \otimes_{Qp} C_p)^{G_{Qp}} \neq 0$.<|endoftext|>
TITLE: Center of mass from the abstract point of view, or could the ancient Greeks invent modern analysis?
QUESTION [20 upvotes]: This is a very open-ended question, which may or may not have a perfect answer, and for which I have a few ideas but nothing like a clear picture. However, I guess it won't hurt to ask to see if people think about such things at all and if they do, what their ideas are. I don't know whether to make it CW or not: on one hand, it is pure mathematics, so we are within the usual set of standards to judge what's right and what's wrong, on the other hand, it is certainly not "a question of the type MO was designed for". So, I'm hesitant to check the community wiki option myself but have absolutely nothing against someone else doing so.
I assume that the ancient Greeks had an idea of a complete normed space ($\mathbb R$ and $\mathbb R^2$ would be enough for our purposes for quite a while), a set, a linear transformation, and the center of mass. On the top of it, I assume they had as much common sense (probably more), as we have nowadays.
The task is the usual one for Archimedes: given a reasonable non-empty set $E$ in a complete linear space $V$, assign a point $C(E)$ to it that you can confidently call "the center of mass". For the purposes of this thread, let's consider bounded at most countable subsets in $V=\mathbb R$ first. If we can figure out what to do with this case to everyone's satisfaction, we can move to the next stage. It may be not a really illuminating model, but it has a few quite funny features already.
The axioms of the center of mass are just the common sense ones:
1) The center of mass is never outside the closed convex hull of the set.
2) If $A$, $B$ are disjoint, then $C(A\cup B)\in[C(A),C(B)]$.
3) If $T$ is an affine transformation, then $C(TE)=TC(E)$.
4) (this is a bit tough, so feel free to drop or to modify it if it helps) If $A,B$ are such that the sets $a+B$, $a\in A$ are disjoint, then $C(A+B)=C(A)+C(B)$
I don't know if we really need anything else (in particular, I'm not sure if the addition of the "obvious" definition of the center of mass of a finite set is needed, helpful, or hurtful), but feel free to play with this list in any reasonably way you want.
The questions are the usual ones:
A) Existence
B) Uniqueness
C) Way to find $C(E)$ given $E$.
Any ideas, constructions, counterexamples, references, etc. (not necessarily restricted to the model I described) are welcome :).
Edit: Thanks to everyone who responded! Let me clarify one thing. Yes, if we can create a meaningful notion of mass (say, if there exists a translation invariant Borel measure that is finite and positive on $E$), we can define the center of mass in the usual way and the only question will be if the definition is unambiguous. However, if I give you a symmetric set, you'll not hesitate to say that the center of symmetry is also the center of mass. Also, if I give you an infinite set and add one point to it, you'll probably (but not necessarily) agree that the center of mass won't feel this addition, the reason being that the set was infinitely times more massive than the point we added not as mush because we can measure the actual mass in some way but merely because infinitely many disjoint shifts of the point fit inside the set. In other words, quite often we can determine the relative size without being able to assign any meaning to the absolute size. This is one of the loopholes in the integration theory I'd like to exploit and see how far one can get with it. In a sense, that is a straight extension of the original Eudoxus line of thinking, which is why "ancient Greeks" entered the title of the question.

REPLY [11 votes]: If I'm not mistaken (but I often am), the physicists already have a rather simple way of defining the center of mass. But I don't think you can do it with just sets. You have to associate a mass with each set. The critical axiom is simply the one we all know:
If $A$ and $B$ are disjoint sets with masses $m(A)$ and $m(B)$ and center of masses $c(A)$ and $c(B)$ respectively, then the mass of $C$ is $m(C) = m(A) + m(B)$ and the center of mass of the set $C = A \cup B$ is given by
$$
c(C) = \frac{m(A)}{m(C)}c(A) + \frac{m(B)}{m(C)}c(B).
$$
You do need one more axiom to get started somehow. I believe physicists like to start with point masses (where the definition of the center of mass is easy) and then view a body as a limit of point masses. That's more or less what Liviu has proposed. But it also suffices to say that the center of mass of a square or cube is its geometric center. Or, more generally, the center of mass of any set with sufficient symmetry is its center.
Of course, if you really want arbitrary shapes, then you do need a countable version of the first axiom. But I think that's all you need. Note that this approach allows for bodies with different and even non-constant mass densities.
ADDED (in response to fedja's edit): It's worth noting explicitly that my answer above requires no notion of volume or choice of measure (such as Lebesgue measure) on the ambient space. It works on any length space.
But I don't see any way reduce this to just geometry (and not physics) without a notion of volume. In essence, you do it by just assuming all objects have the same constant mass density, so the mass is essentially equal to volume.
More generally, there has to be a way to measure the relative size of two sets, in order to determine the center of mass of the union of the two sets. If I understand correctly, axiom (4) in the question is an attempt to set this up.
[PREVIOUS DISCUSSION REPLACED BY THE FOLLOWING]
But for me it seems simpler to define a notion of size first and define the center of mass. And, as alvarezpaiva points out in a comment to Liviu's answer, valuations provide the appropriate setting, especially if we restrict to convex polytopes, which are objects that I believe the Greeks understood pretty well. This also allows us to avoid any issues of having to work with infinite sums or unions.
Here, a valuation $f$ is a finitely additive function on the space of convex polytopes. In other words, given polytopes $A$ and $B$,
$$f(A \cup B) + f(A \cap B) = f(A) + f(B).$$
The first observation is that the "critical axiom" stated above is equivalent to saying that $C \mapsto m(C)c(C)$ is a valuation. However, Monika Ludwig showed in her paper Moment vectors of polytopes that the only $R^n$-valued measurable valuation on convex polytopes that behaves appropriately under affine transformations is the volume of the polytope times the standard center of mass.
Ludwig also showed in her Advances article Valuations on polytopes containing the origin in their interiors that any real-valued measurable $SL(n)$-invariant valuation homogeneous of positive degree must be a constant times volume. So it is reasonable to assume $m$ is volume. This therefore implies that $c$ must be the standard center of mass.
Moreover, if you examine Ludwig's proofs, you will see that although they are quite nontrivial, the technology used was arguably within the grasp of the Greeks.<|endoftext|>
TITLE: Numerable covers from the point of view of Grothendieck topologies
QUESTION [6 upvotes]: Let $G$ be a topological group. Recall that its classifying space $BG$ is a CW-complex which  is the base of a locally trivial principal bundle of group $G$, with contractible total space $EG$. It classifies principal $G$-bundles in the sense that for any paracompact space $B$, isomorphism classes of locally trivial principal $G$-bundles are in natural bijection with homotopy classes of maps from $B$ to $BG$. 
More generally, if $B$ is any topological space, homotopy classes of maps from $B$ to $BG$ are in bijection with isomorphism classes of principal $G$-bundles for which there exists a numerable open cover $(U_i)_i$ of $B$ such that the bundle is trivial on each $U_i$. This means that there exists a partition of unity subordinate to that cover. 
I believe that this is due do A. Dold (Partitions of unity in the theory of fibrations, Annals of math., 1965).
Now observe that the family numerable covers defines a Grothendieck topology on a topological space.  Hence, this result of Dold falls within the circle of thoughts (development of étale cohomology on algebraic varieties, sites, toposes,...) that was actively developed by Grothendieck  at the time Dold wrote his paper. 
My question is whether there has been any kind of connections between these two groups of mathematicians or, in the contrary, whether this is a mere coïncidence.

REPLY [2 votes]: I think you need to check back into the history of homology and cohomology and the definition by Cech of the cohomology theory that bears his name. Then you will see the common root of both the numerable cover theory and of Grothendieck's definition of étale cohomology, sites, etc.<|endoftext|>
TITLE: Reconstructing the argument that yields Graham's number
QUESTION [31 upvotes]: Graham's number achieved a kind of cult status, thanks to Martin Gardner, as the largest finite number appearing in a mathematical proof. (It may no longer hold that record, but that is not my concern here.)  I was surprised to learn relatively recently that it is not actually the best known bound for that particular Euclidean Ramsey problem, and that the original paper by Graham and Rothschild, which predates "Graham's number," explicitly derives a better bound. I'm left to assume that Graham later found a simpler argument that gave a weaker bound, that we now know as Graham's number.
Some time ago, before I realized the above facts, I asked Graham about his "Graham's number" proof. As I recall the conversation, he no longer had the argument at his fingertips and did not seem too interested in trying to reconstruct it. This brings me to my question:

Can someone reconstruct a simple argument for the Euclidean Ramsey problem in question that naturally yields Graham's number as an upper bound?

This would not normally be that interesting a question except that Graham's number still circulates in recreational mathematics circles, so it's a bit embarrassing if nobody knows how to "derive" it.

REPLY [24 votes]: I talked to Ronald Graham last night at the Joint Mathematics Meeting in San Diego and asked him the question here.  He said he'd made up Graham's number when talking to Martin Gardner because 1) it was simpler to explain than his actual upper bound, the one that appears in his paper with Rothschild, and 2) it's bigger, so it's still an upper bound!  
So, apparently the comment on Wikipedia:

This weaker upper bound for the problem, attributed to an unpublished work of Graham [....]

is a bit misleading, though still technically true.  I'll try to fix it in a while.
Nice question!<|endoftext|>
TITLE: Understanding Specker's disproof of the axiom of choice in New Foundations
QUESTION [17 upvotes]: Hi all! I am trying to understand Specker (1953)'s proof (found here) that the axiom of choice is false in New Foundations. I am stuck on the following point. At 3.5 Specker writes:
3.5. The cardinal numbers are well ordered by the relation "there are sets $a,b$ such that $a \in n, b \in m$ and $a \subseteq b$" (axiom of choice).
I am assuming that this is a consequence of the axiom of choice, which he is using to derive a contradiction. Is that true? If so, how is it a consequence of the axiom of choice?
Another, broader question: can anybody give an intuitive explanation of why AC fails in NF?
Thank you!

REPLY [3 votes]: See lemma 2.2.1 in my dissertation of NF in 1989 (G. Wagemakers, New Foundations: A survey of Quine’s Set Theory). 
More info can be found there, for example that there is no relation between $n$ and $T(n)$ at all. The one can be larger, equal or smaller than the other.
You're asking for an intuitive explanation why AC fails in NF. Let me give that a try (terminology as in my document; note that I skip parentheses that are not essential).
For cardinals $m$ it is easy to prove the following ($\Omega = |V|$):
\begin{cases}
2^{T\Omega}=\Omega\\
m \le T\Omega \implies 2^{Tm} = T2^m\\
\Omega > T\Omega\\
2^m \ne \Lambda \implies m < 2^m
\end{cases}
This leads to a descending series of cardinals:
$$\Omega > T\Omega > T^2\Omega > ...$$
If we would be able to form the set $Clos(\{\Omega\}, T)$ containing all these cardinals, then we would contradict AC as that set would not have a smallest element (the set of all cardinals, hence every subset of it, is well-ordered by $\le$).
But, alas, this set cannot be formed as $T$ is type-raising.
So we have to resort to a real set, intuitively containing at least the cardinal numbers in this list, and hope for the best. This set is $A = \{m \in NC |\Phi m \in Fin\}$, where
\begin{cases}
\Phi m = Clos(\{m\}, 2^{.}) - \{\Lambda\}\\
Fin = \bigcap Nn\\
Nn = Clos (\{0\}, . + 1)
\end{cases}
This set exists, as $2^.$ is type-preserving, and indeed contains $\Omega$ ($|\Phi \Omega| =1$), $T\Omega$ ($|\Phi T\Omega| =2$), etcetera (note that this fact doesn't helps us much, as this series of descending cardinals could be all larger than the minimum element of the set).
Assuming AC, A is well-ordered by $\le$ (NC is, and $A \subseteq NC$). Specker proceeds to derive an inconsistency: for the smallest element $n$ of this set, it can be derived that $n=Tn$. But then $|\Phi n| = |\Phi Tn| = T|\Phi n| + (1 \text{ or } 2)$, contradicting the fact that there can be no natural number $|\Phi n|$ having this property.
Obviously, a number of lemmata is required to derive these facts.
The conclusion is that the FULL axiom of choice AC does not hold in NF. Note that for cardinals m less than that of $A$, $AC(m)$ has not been disproved by Specker's argument.<|endoftext|>
TITLE: Connection between Infinite continued fractions, elliptic integrals and AGM
QUESTION [13 upvotes]: It is known that at $x=1$, the following continued fraction represents $\frac{4}{\pi}$ and can be approximated rapidly using Gauss' Arithmetic Geometric mean.
$$C(x) = x +  \frac{1^{2}}{2x + \frac{3^{2}}{2x  + \frac{5^{2}}{2x + \frac{7^{2}}{2x + \cdots}}}}$$
Are there any other $x$ that $C(x)$ can be approximated through AGM quickly?
Is there any connection to elliptic integrals?

REPLY [5 votes]: This is not an answer.
As was noted in the comments, the function $C(x)$ can be expressed in terms of the Gamma function, see Corollary 1 on page 145 in Ramanujan's Notebook II which attributes this result to G. Bauer: Von einem Kettenbruche Euler's und einem Theorem von Wallis, Abh. Bayer. Akad. Wiss. 11(1872), 96 - 116 and to the first Ramanujan letter to Hardy.
$$
C(x) = \frac{\Gamma^2(\frac{x+1}{4})}{\Gamma^2(\frac{x+3}{4})}
$$
The AGM function can be expressed in terms of complete elliptic integral of the first kind, as is noted on the Wikipedia page, which is given by Gauss hypergeometric function  ${}_2F_1(-\frac{1}{2}, \frac{1}{2}; 1; k^2)$ whereas the function $C(x)$ is given by ${}_2F_1(-\frac{1}{2}, -\frac{1}{2};x,1)$. 
There is certainly more to the story, see related question on math.stackexchange and equation (99) at MathWorld.
The explicit relation connecting Gamma values at certain rational points and elliptic integral can be found e.g. in the article of Borwein and Zucker, book Pi and the AGM by Borwein, Borwein and a more recent article by Vidunas.<|endoftext|>
TITLE: sequences with a fractal dimension
QUESTION [14 upvotes]: This is inspired by the self-similarity of the celebrated Golay-Rudin-Shapiro sequence, more exactly, of its alternating partial sums. (This latter one is oeis 020990). The pictures show the 550 first terms, then the 9000 first terms.


It makes sense to define a certain fractal $F$ as the "limit" of the graph $\Gamma=\{(n,a_n)\}_{n\ge0}$.
More precisely:
Fix a rectangle $R\subset\mathbb R^2$, e.g. the unit square. Take the part $\Gamma_k$ of $\Gamma$ between $n=2^{2k-1}$ and $n=2^{2k+1}-1$ and rescale it to a graph $\Gamma^0_k$ that fits $R$ best.
Then because of the geometrical (almost-) similarity of the $\Gamma^0_k$,  the limit $F:=\lim\limits_{k\to\infty}\Gamma^0_k\subset\mathbb R^2$ is well-defined. Note that its Hausdorff dimension is $d=3/2$.
Other examples: 

the sequence oeis 004074 that defines likewise the Blancmange curve, dimension $d=1$
sequences linked to the Gray code, like 003188 or 006068, both with $d=1$ 
Stern's diatomic series (a.k.a. Stern-Brocot sequence or $fusc$ function) yields a fractal with dimension $d=\frac{\ln 3}{\ln 2}$
it makes sense to relate (if not to identify) the Cantor set with the sequence $1,0,1,0,0,0,1,0,1,... $ where $a_n=1$ iff the ternary representation of $n$ has only 0's and 2's (equivalently, the cellular automaton where at each step $1 \mapsto 101$ and $0 \mapsto000$), and to say this sequence has dimension $\frac{\ln2}{\ln3}$. Likewise for the "fat Cantor set" iterating 11100111 (dimension $\frac{\ln5}{\ln8}$) and all other sorts of Cantor dust.
the devil's staircase, obtained by "integrating" the Cantor set, corresponds to this sequence, and a "mirrored" version of it can be found here. 
Other sequences of Toothpick and Cellular Automata type

Like for most other sequences of this kind, the ressemblance is best seen when looking at a range from either $1$ to $2^n$, or (for some, like the Blancmange curve) from $2^{n-1}$ to $2^n$.  
Note that it is not at all straightforward or even possible to define a fractal for every self-similar integer sequence $a=(a_n)_{n\ge0}$ (self-similar meaning as usual that there is a $k\ge2$ and $\lambda$ such that $a_n=\lambda a_{kn}$). On the other hand, there are also sequences with a fractal-like appearance without being self-similar in the above sense.
Question:


Has the idea of the "fractal dimension" of certain sequences been investigated before?

REPLY [7 votes]: Interestingly, fractal dimensions of the human DNA sequence have been analyzed and different embeddings considered: http://biocomplexity.indiana.edu/jglazier/docs/papers/20_DNA_Analysis.pdf.<|endoftext|>
TITLE: The human body's random number generator
QUESTION [39 upvotes]: I remember learning in microbiology that the human body generates antibodies using a random process so that an enormous variety of antibodies can be produced with a simple genetic code.
Now that I'm trying to learn more about random processes, I find this more interesting than I used to. Has anyone developed a mathematical model for how the genetic code generates these random antibodies? Clearly, we'd like the process to be ergodic, so that as many antibody varieties as possible can be produced. The main reason that I'm interested is because there are so many ways that random and pseudo-random numbers are generated, and it would interesting to see what the body "thinks" is a good random antibody generator.
I did a quick google search, finding links such as this paper, but they seem to assume that antibodies are generated randomly without explaining the mechanism.
So, does anyone know if there is a mathematical model for the process that generates random antibodies?
And, Merry Christmas!
Edit: Steve and Tom have provided a great answer and a great comment about related topics that I think provide a good place to start reading. For now, though, I will leave the question open in case someone can provide a fuller answer in the future. End Edit.

REPLY [8 votes]: I don't know about how the human body does it, but a state-of-the art way that humans perform "complementarity-determining region sequence design" artificially is to optimize "for protein binding by utilizing a hidden Markov model that was trained on all antibody-antigen cocrystal structures in the Protein Data Bank". 
Maximum-entropy models have also been applied to understand the repertoire.
FOLLOWUP:
"Stochastic rearrangement of germline V-, D-, and J-genes to create variable coding sequence for certain cell surface receptors is at the origin of immune system diversity. This process, known as “VDJ recombination”, is implemented via a series of stochastic molecular events involving gene choices and random nucleotide insertions between, and deletions from, genes...we develop a maximum likelihood inference method to [infer the probability distribution of hidden recombination events]...The generative event statistics are consistent between individuals, suggesting a universal biochemical process."<|endoftext|>
TITLE: What is geometrically the Pontryagin class?
QUESTION [42 upvotes]: What does the Pontryagin class detects or is an obstruction to? Please avoid any answer using that it's the even Chern class of the complexified bundle or any interpretation that relies on the complexified bundle. 
As related question might be the following: when one defines the obstruction classes on a rank $4$ vector bundle (and if the first three obstruction classes do vanish) then the fourth obstruction class can be decomposed as the Euler class and the first Pontryagin class (as $\pi_3(SO_4) \simeq \mathbb{Z} \oplus \mathbb{Z}$). Is there a geometric description of a system of generators in $\pi_3(SO_4)$ which is associated to these classes?
EDIT: deleted "For example, why does the first Pontryagin class distinguishes the (tangent bundles of the) exotic $4$-spheres?" as it is wrong, see Liviu's answer below.

REPLY [7 votes]: Worth mentioning separately I believe: in "A combinatorial formula for the Pontrjagin classes", Gelfand and MacPherson construct something like analogs of Segre classes for Pontryagin classes: for a triangulation of a manifold $X$ they invoke oriented matroids to produce explicit rational simplicial cycles on its barycentric subdivision which are Poincaré duals of inverses $\bar p_i(X)$ of the Pontryagin classes of $X$.
They also describe (on half a page!) a version of the Chern-Weil theory for Pontryagin classes of a vector bundle $E$ with connection on a manifold $M$ which shows relationship between their approach and the "standard" one. It is so concise and enlightening that I decided just to reproduce it here. They consider the Grassmanian bundle $\pi:\mathscr Y\to M$ of codimension 2 planes in $E$, together with the principal bundle $\rho:\mathscr Z\to\mathscr Y$ corresponding to the tautological quotient 2-plane bundle over $\mathscr Y$. The connection on $E$ gives them a 1-form $\Theta$ on $\mathscr Z$ with coefficients in the orientation sheaf of $\mathscr Z$ and a curvature form $\Omega$ on $\mathscr Y$ determined by $\rho^*\Omega=d\Theta$. Their formula then is$$\bar p_i(E)=(-1)^i\pi_*\Omega^{\dim(E)-2(i-1)}.$$<|endoftext|>
TITLE: The origin of sets?
QUESTION [44 upvotes]: The history of set theory from Cantor to modern times is well documented. However, the origin of the idea of sets is not so clear. A few years ago, I taught a set theory course and I did some digging to find the earliest definition of sets. My notes are a little scattered but it appears that the one of the earliest definition that I found was due to Bolzano in Paradoxien des Unendlichen:

There are wholes which, although they contain the same parts $A$, $B$, $C$, $D$,..., nevertheless present themselves as different when seen from our point of view or conception (this kind of difference we call 'essential'), e.g. a complete and a broken glass viewed as a drinking vessel. [...] A whole whose basic conception renders the arrangement of its parts a matter of indifference (and whose rearrangement therefore changes nothing essential from our point of view, if only that changes), I call a set.

(The original German text is here, §4; I don't remember where I got the translation.)
According to my notes, Bolzano wrote this in 1847. Since Boole's An Investigation of the Laws of Thought was published a just few years later in 1854, it seems that the idea of sets was already well known at that time.

What was the earliest definition of 'set' in the mathematical literature?

Historical queries of this type are hopelessly vague, so let me give some more specific criteria for what I am looking for. The object doesn't have to be called "set" but it must be 
an independent container object where the arrangement of the parts doesn't matter.

It should also be fairly general in what the set can contain. A general set of points in the plane is probably not enough in terms of generality but if the same concept is also used for collections of lines then we're talking.
It shouldn't have implicit or explicit structure. Line segments, intervals, planes and such are too structured even if the arrangement of the parts technically doesn't matter.
It should be an independent object intended to be used and manipulated for its own sake. For example, the first time a collection of points in general position was considered in the literature doesn't make the cut since there was no intent to manipulate the collection for its own sake.
It should be a definition. Formal definitions as we see them today are a relatively new phenomenon but it should be fairly clear that this is the intent, such as when Bolzano says "I call a set" at the end of the quote above.
It should be mathematical concept. The strict divisions we have today are very recent but it should be clear that the sets in question are intended for mathematical purposes. Paradoxien des Unendlichen is perhaps more of a philosophical treatise than a mathematical one, but it is clear that Bolzano is considering sets in a mathematical way.

That said, any input that doesn't quite meet all of these criteria is welcome since the ultimate goal is to understand how the modern idea of set came to be.

REPLY [4 votes]: From the Wikipedia article on Euler diagrams:
"The first use of "Eulerian circles" is commonly attributed to Swiss mathematician Leonhard Euler (1707–1783)."
"Venn diagrams are a more restrictive form of Euler diagrams. A Venn diagram must contain all $2^n$ logically possible zones of overlap between its $n$ curves, representing all combinations of inclusion/exclusion of its constituent sets, but in an Euler diagram some zones might be missing if they are empty sets."<|endoftext|>
TITLE: Bipartiteness criterion
QUESTION [6 upvotes]: A graph is bipartite if and only if it does not contain odd cycles. Is there a similar criterion for hypergraphs? (A hypergraph is called bipartite if its vertices can be colored in two colors so that no hyperedge is monochromatic.)
My guess is that the answer is "no" but, maybe, there are results in this direction I am not aware of. Thanks!

REPLY [3 votes]: In "matching theory", by Lovasz and Plummer, I found the following page, which may be helpful:<|endoftext|>
TITLE: If rational points are like entire curves, then what do algebraic points correspond to
QUESTION [6 upvotes]: I read somewhere that if $X$ is a projective variety of general type over a number field $K$, then rational points are an analogue of entire curves $\mathbf{C}\to X^{an}$ (with $X^{an}$ the analytification of $X_{\mathbf{C}}$ for some $K\to \mathbf C$).
Rational points are algebraic points of degree $1$ on $X$ and they "correspond" to entire curves.
Naive question:
Let $m\geq 2$. What do algebraic points of degree $m$ on $X$ "correspond" to?

REPLY [9 votes]: Rational points are (kind of) like maps from a fixed curve (say P^1) to X.
Algebraic points of degree m are like curves endowed with a map to X and a degree-m map to P^1.<|endoftext|>
TITLE: Internal categories in simplicial sets
QUESTION [19 upvotes]: Is there a model structure (or more generally a homotopy theory) on the category of internal categories in simplicial sets, which presents the theory of $(\infty,1)$-categories?
Note that this category is closely related to other known models for $(\infty,1)$-categories.  For instance, any simplicially enriched category can be regarded as an internal simplicial category with a discrete simplicial set of objects.  And any internal simplicial category has a bisimplicial nerve which is a Segal space.  One might hope that these functors would be part of Quillen equivalences.

REPLY [17 votes]: It appears that Geoffroy Horel has solved this problem completely:

Geoffroy Horel, A model structure on internal categories in simplicial sets, Theory and Applications of Categories 30 No. 20 (2015) pp. 704–750 (journal page, arXiv:1403.6873)<|endoftext|>
TITLE: Convergence of Schwartz kernels implies convergence of operators
QUESTION [9 upvotes]: Let $K$ be a smoothing operator on $\mathbb{R}^n$, i.e., it defines a map on all Sobolev spaces $K\colon H^r(\mathbb{R}^n) \to H^s(\mathbb{R}^n)$ for all $r, s \in \mathbb{R}$. Now (a variation of) the Schwartz kernel theorem states that it is given by some smooth kernel $k \in C^\infty(\mathbb{R}^n \times \mathbb{R}^n)$, i.e., $(Kf)(x) = \int_{\mathbb{R}^n} k(x, y)f(y) dy$.
Now suppose we have kernels $k_n, k \in C^\infty(\mathbb{R}^n \times \mathbb{R}^n)$ that do define smoothing operator $K_n$ and $K$ (since not every element of $C^\infty(\mathbb{R}^n \times \mathbb{R}^n)$ defines a smoothing operator we have to assume this).

What type of convergence $k_n \to k$ is needed, so that $K_n$ does converge to $K$?

Suppose I want $K_n \to K$ only as maps $L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$. Does it suffice for this that $k_n$ converges uniformly to $k$?
Suppose I want $K_n \to K$ as maps $H^r(\mathbb{R}^n) \to H^s(\mathbb{R}^n)$ for all $r, s \in \mathbb{R}$. Is it sufficient for this that $k_n \to k$ uniformly and also all their derivatives?
I would prefer convergence in the norm topology, but any information about convergence in the strong / weak operator topology would also be nice.

REPLY [7 votes]: Here is a classical theorem. $\newcommand{\bR}{\mathbb{R}}$
Suppose that for $0< a,b<\infty$
$$\sup_x\left(\int_{\bR^n} |k(x,y)|^a dy\right)^{\frac{1}{a}}=M_1(k)<\infty, $$
$$\sup_y\left(\int_{\bR^n} |k(x,y)|^b dx\right)^{\frac{1}{b}}=M_2(k),  $$
and
$$\frac{1}{p}-\left(\frac{b}{a}\right)\frac{1}{q}=1-\frac{1}{a},\;\;1\leq p\leq q <\infty $$
then the integral operator $T=T_k$ defined by the kernel  $k$  defines a  bounded operator $L^p\to L^q$  and  its norm    satisfies the inequality
$$\Vert T_k\Vert \leq M_1(k)^{1-\frac{b}{q}}M_2(k)^{\frac{b}{q}}. $$
For  proofs and many more details, see Section 9.5 of Edwards' book,  Functional Analysis. Theory and Applications, Dover.<|endoftext|>
TITLE: Status of the 196 conjecture?
QUESTION [30 upvotes]: A palindrome is a number which remains the same when reversing it, for instance 34143. Now pick an arbitrary number, say 26: then 26+62=88 is a palindrome. If the number was 57, then 57+75=132 is not a palindrome; but 132+231=363 is.
In general, iterating $a_1\ldots a_n\to a_1\ldots a_n+a_n\ldots a_1$ always seems to lead to a palindrome. But, the point is that this doesn't work for 196! 
This problem is well-known, and there is a heavy numerical evidence for it, see http://en.wikipedia.org/wiki/Lychrel_number and http://www.p196.org/. I was wondering, is there any theoretical advance on this subject? Or at least, to which math area does this problem belong to? Many thanks.

REPLY [2 votes]: Here are some extensions to Aaron Meyerowitz's comments. (Edit: As Aaron points out in the comments, my primary claim here is actually wrong.)
As Aaron points out, it is clear that if computing the sum $s(x) = x + r(x)$ involves no carries, then $s(x)$ is a palindrome. In this case we call $x$ "special." If computing $s(x)$ does involve carries (i.e., $x$ is not special) but $s(x)$ is nonetheless a palindrome, we call $x$ "exceptional." Aaron asks how common exceptional numbers are.
I claim that exceptional numbers only occur in one very specific situation, and that this gives us a necessary and sufficient condition for $s(x)$ to be a palindrome. Specifically, I claim that a carrying computation of $s(x)$ results in a palindrome if and only if the carry happens in the first and last place of the number, and results in the first two digits and the final two digits all being one. So basically, the rule is that $s(x)$ is a palindrome iff there are no carries in its computation, except in one very specific situation. This applies in all bases.
Given a nonnegative integer $n$ and a base $b \geq 2$, we shall write $\bar{n}$ to denote the number of digits in $n$'s base $b$ representation. We write $n_i$ to denote the $i$th digit from the left, with $n_1$ being the first (least significant) digit, and $n_{\bar{n}}$ being the last (most significant) digit.
We shall write $n_{-i}$ to abbreviate $n_{\bar{n} - i + 1}$. This is the digit "corresponding" to $n_i$ in the reverse of $n$. We have $n_{-i} = r(n)_{i}$.
Given a number $n$, a "carry" is an index $1 \leq i \leq \bar{n}$ such that $n_i + n_{-i} \geq b$. It is a location where a carry happens in computing $s(n)$. If $n$ has no carries, then $s(n)$ is a palindrome.
Define an "inner carry" as a carry $i$ where $1 < i < \bar{n}$. An "outer carry" is a carry $i$ where $i = 1$ or $i = \bar{n}$. An "exceptional outer carry" is an outer carry is an outer carry where, letting $m = s(n)$, we have
$ m_{\bar{m}} = m_{\bar{m}-1} = m_{\bar{n}} = m_2 = m_1 = 1. $
That is, in an exceptional outer carry, the first two digits and the last two digits of $s(n)$ are all $1$.
Proposition. $s(n)$ is a palindrome iff every carry for $n$ is an exceptional outer carry.
(Left to right.) Let $m = s(n)$, and suppose $m$ is a palindrome. Suppose there is an outer carry. Then $m_{\bar{m}} = 1$. Then $m_1 = 1$. Then
$m_{\bar{m}-1} = m_{\bar{n}} = n_1 + n_{\bar{n}} - b = m_1 = 1.$
Then $m_2 = m_{\bar{m}-1} = 1$. So if there is an outer carry, it is exceptional. Now suppose there is an inner carry, and let $i$ be the smallest inner carry. (Observe that $i \leq \lceil \frac{\bar{n}}{2} \rceil$, since if $i$ is a carry, then $-i$ is also a carry.)
To begin, suppose there is no outer carry. Then $i$ is the smallest carry. Then
$m_{i-1} = n_{i-1} + n_{-(i+1)}.$
$-i$ is also a carry, so there is a carry into $-i+1$. But $-i$ is the largest carry, so there is no carry from $-i+1$. So
$m_{-i+1} = m_{-(i-1)} = n_{-(i-1)} + n_{--(i-1)} + 1 = n_{i-1} + n_{-(i-1)} + 1 \neq m_{i-1},$
so $m$ is not a palindrome. So in the case where there is no outer carry, there is no inner carry. Now suppose there is an outer carry. The outer carry is exceptional, and then
$m_{\bar{m}} = m_{\bar{m}-1} = m_{\bar{n}} = m_2 = m_1 = 1.$
If $i \geq 3$, then we can argue as in the case where there is no outer carry, since we have that there is no carry into the $i$th place. $i \neq 1$, since $i$ is inner. Suppose $i = 2$. Then there is a carry from $m_{\bar{n}-1}$ into $m_{\bar{n}}$. $n_1 + n_{\bar{n}} = b+1$,  (i.e., 11 in base $b$), since there is an outer carry and $m_1 = 1$. So
$m_{\bar{n}}  = n_1 + n_{\bar{n}} - b + 1 = b + 1 - b + 1 = 1 + 1 = 2,$
contradicting $m_{\bar{n}} = 1$. So there is no inner carry, and we are done with the left to right case.
(Right to left.) Suppose every carry for $n$ is an exceptional outer carry. If there are no carries, then $m = s(n)$ is a palindrome. Suppose there is an exceptional outer carry. Then $m_i = m_{-i}$ for all $i \in \{1,2,\bar{m},\bar{m}-1\}$ by definition, and $m_{i} = m_{-i}$ for all $2 < i < \bar{m}-1$ by the absence of inner carries.
Comments and criticisms are welcome; I suspect the proof could use refining, and it might actually be wrong!<|endoftext|>
TITLE: Why the sectional curvatures assume maximum on holomorphic planes for positively curved Kaehler manifold?
QUESTION [11 upvotes]: Let $M$ be a Kaehler manifold with positive holomorphic sectional curvature. then the maximum of sectional curvatures at point $p$ is assumed at the holomorphic planes. I read this claim in Klingenberg's paper "On Compact Kaehlerian Manifolds with Positive Holomorphic Curvature" He refers this result to Berger "Pincement Riemannien et pincement holomorphe"
But I don't have access to Berger's paper nor do I read French, so I ask here to see if this can be proved very quickly or has been written in somewhere else.

REPLY [14 votes]: There is a more enlightening proof of this statement than Berger's calculation and, in fact, it proves something a bit more general.  First, a definition:  Let $(M,g)$ be a Riemannian $n$-manifold with Riemann curvature tensor $R$ and let $E\subset T_xM$ be a $p$-plane with orthonormal basis $e_1,\ldots,e_p$.  Define
$$
\sigma(E) = \sum_{i,j=1}^p R(e_i,e_j,e_j,e_i),
$$ 
which is easily seen not to depend on the choice of orthonormal basis.
The quantity $\sigma(E)$ is the scalar curvature of $E$.  For $p=2$, this is twice the sectional curvature of the $2$-plane $E$, while $\sigma(T_xM)$ is simply the scalar curvature of $g$ at $x$.  In general, the scalar curvature of $E$ is $p(p{-}1)$ times the average of the sectional curvatures of the $2$-planes that lie in $E$.

Fact: If $(M,g,J)$ is a Kähler manifold, and $E\subset T_xM$ is a $4$-plane that is complex (i.e., stable under $J$), then $\sigma(E)$ is $6$ times the average sectional curvature of the complex $2$-planes that lie in $E$.  (See below for a proof.)
Proposition:  If $(M,g,J)$ is a Kahler manifold and $E\subset T_xM$ is a $4$-plane that is complex and satisfies $\sigma(E)>0$, then the only local maxima of the sectional curvature function on $\mathrm{Gr}_2(E)$ occur at $2$-planes $L$ that are complex. 
Remark:  If the holomorphic sectional curvatures of $M$ are positive, then, by the Fact, $\sigma(E)>0$ for all $4$-planes $E$ that are complex, and hence, by the Proposition, the maximum of the sectional curvature on $\mathrm{Gr}_2(T_xM)$ is attained only by complex $2$-planes (since each noncomplex tangent $2$-plane lies in some tangent $4$-plane that is complex).  Meanwhile, one can have $\sigma(E)>0$ for all $4$-planes that are complex without having positive holomorphic sectional curvature; just consider a  Kähler surface with positive scalar curvature that does not have positive holomorphic sectional curvature.  Thus, this gives a stronger result than Klingenberg's claim.
To prove the Fact and the Proposition, it clearly suffices to do it for a Kähler manifold of complex dimension $2$ with positive scalar curvature, in which case $E=T_xM$ for some $x\in M$ that one can suppose fixed for the purposes of this discussion.  This is a local argument, so one can choose an orthonormal coframing $\omega_0,\ldots,\omega_3$ such that the Kähler form is given by $\Omega=\omega_0\wedge\omega_1+\omega_2\wedge\omega_3$.  Let $\phi_{ij}=-\phi_{ji}$ be the connection forms, so that $d\omega_i=-\phi_{ij}\wedge\omega_j$ and, because of the Kähler condition, $\phi_{02}+\phi_{31}=\phi_{03}+\phi_{12}=0$.  Then, by the Bianchi identities for a Kahler metric, one has the following expression for the curvature forms $\Phi_{ij}=d\phi_{ij}+\phi_{ik}\wedge\phi_{kj}\ $:
$$
\begin{pmatrix}
\Phi_{01}+\Phi_{23}\cr
\Phi_{02}+\Phi_{31}\cr
\Phi_{03}+\Phi_{12}\cr
\Phi_{01}-\Phi_{23}\cr
\Phi_{02}-\Phi_{31}\cr
\Phi_{03}-\Phi_{12}
\end{pmatrix} =
\begin{pmatrix}
3s&0&0&r_1&r_2&r_3\cr
0&0&0&0&0&0\cr
0&0&0&0&0&0\cr
r_1&0&0&s{+}w_{11}&w_{12}&w_{13}\cr
r_2&0&0&w_{12}&s{+}w_{22}&w_{23}\cr
r_3&0&0&w_{13}&w_{23}&s{+}w_{33}
\end{pmatrix} 
\begin{pmatrix}
\omega_0\wedge\omega_1+\omega_2\wedge\omega_3\cr
\omega_0\wedge\omega_2+\omega_3\wedge\omega_1\cr
\omega_0\wedge\omega_3+\omega_1\wedge\omega_2\cr
\omega_0\wedge\omega_1-\omega_2\wedge\omega_3\cr
\omega_0\wedge\omega_2-\omega_3\wedge\omega_1\cr
\omega_0\wedge\omega_3-\omega_1\wedge\omega_2
\end{pmatrix}
$$
Here, $12s$ is the scalar curvature (which is positive by hypothesis) and $w_{ij}=w_{ji}$ satisfies $w_{11}{+}w_{22}{+}w_{33}=0$. [The reader may recognize this as the classic presentation of the curvature of a Riemannian $4$-manifold due to Singer and Thorpe; in this case, it has been adapted to the case of a Kähler metric.]
Define functions $u_i$ and $v_i$ on the Grassmann bundle $\mathrm{Gr}^+_2(M)$ of oriented tangent $2$-planes as follows:  If $L$ is an oriented $2$-plane with oriented orthonormal basis $(X,Y)$, then 
$$
\begin{pmatrix}
u_1(L)\cr
u_2(L)\cr
u_3(L)\cr
v_1(L)\cr
v_2(L)\cr
v_3(L)
\end{pmatrix} = 
\begin{pmatrix}
\omega_0\wedge\omega_1+\omega_2\wedge\omega_3\cr
\omega_0\wedge\omega_2+\omega_3\wedge\omega_1\cr
\omega_0\wedge\omega_3+\omega_1\wedge\omega_2\cr
\omega_0\wedge\omega_1-\omega_2\wedge\omega_3\cr
\omega_0\wedge\omega_2-\omega_3\wedge\omega_1\cr
\omega_0\wedge\omega_3-\omega_1\wedge\omega_2
\end{pmatrix}
(X,Y).
$$
One then has ${u_1}^2+{u_2}^2+{u_3}^2={v_1}^2+{v_2}^2+{v_3}^2=1$, and the map $(u,v):\mathrm{Gr}^+_2(M)\to S^2\times S^2$ carries each fiber of $\mathrm{Gr}^+_2(M)\to M$ diffeomorphically onto $S^2\times S^2$.  Note that a $2$-plane $L$ is complex if and only if $u_2(L)=u_3(L)=0$, i.e., if and only if $u_1(L)=\pm1$.

Now, computing from the definitions, one finds
$$
\sigma(L) = 3s\ {u_1}^2 + 2u_1(r_iv_i) + s + w_{ij}v_iv_j\ .
$$
(The proof of the Fact about the average of sectional curvatures of complex $2$-planes in $E$ follow immediately from this formula.) 

Now, $\sigma(L)$ does not depend on $u_2$ or $u_3$, and we can thus think of $\sigma$ restricted to a single fiber $\mathrm{Gr}^+_2(T_xM)$ of $\mathrm{Gr}^+_2(M)\to M$ as a function on $[-1,1]\times S^2$ instead of $S^2\times S^2$.  Because $s>0$, when one fixes a $v\in S^2$, the quadratic function of $u_1$ above cannot attain a local maximum when $-1 < u_1 <1 $.  It follows that the local maxima of $\sigma$ on $\mathrm{Gr}^+_2(T_xM)$ must occur only at those $L$ for which $u_1(L)=\pm1$, i.e., those $L$ that are complex. QED<|endoftext|>
TITLE: When is a space of measures a measurable space?
QUESTION [8 upvotes]: Let $X$ denote a measurable space, that is, a set equipped with a $\sigma$-algebra $\Sigma(X)$. Let $M(X)$ denote the space of real-valued measures over $X$. This is a vector space over the real numbers, since the sum of two measures is again a measure, as is a scalar multiple of a measure. I would like to know the most general setting for which $M(X)$ is a measurable vector space.
Does $M(X)$ admit a canonical choice of $\sigma$-algebra, turning it into a measurable space?
If the answer is "no", then what about the setting where $X$ is a localizable measurable space?
If the answer is again "no", then what is the most general setting so that $M(X)$ is admits a canonical measurable structure?
Most generally, what is the largest subcategory $\mathcal C$ of $\mathbf{Meas}$ so that $M : \mathcal C \to \mathcal C$ is an endofunctor?

REPLY [5 votes]: Let $(X,\Sigma)$ be the measurable space.  I think the sigma-algebra on $\mathcal M$ that you want is this.  The least sigma-algebra so that for all $A \in \Sigma$, the map $\mu \mapsto \mu(A)$ is measurable.<|endoftext|>
TITLE: Lower bound for exponential sums
QUESTION [11 upvotes]: Let $D$ be a subset of $\mathbb Z/n \mathbb Z$ containing $0$. For $m$ an integer, set $$\alpha(m,D)=\sum_{d \in D} e\left (\frac{m d }{n}\right ),$$
where as usual $e(x) = e^{2 i \pi x}$ This is an exponential sum (or if you like:
character sum). Obviously $|\alpha(m,D)| \leq |D|$.
Now consider
$$\sigma(D) = \frac{1}{n} \sum_{m=0}^{n-1} |\alpha(m,D)|.$$
A simple upper bound using Cauchy-Schwarz for $\sigma(D)$ is $\sigma(D) \leq \sqrt{ |D|}$,
which is better than the trivial bound $\sigma(D) \leq |D|$. But here I am interested in a lower bound:

is it true that $\sigma(D) \geq 1$, with equality only if $D$ is a subgroup of $\mathbb Z/n \mathbb Z$ ?

This seems true on some numerical tests I have done, and this seems a very elementary question , but I don't see how to prove it at this time (I might be missing something trivial).
More generally, I am interested in any information or reference on this number $\sigma(D)$.
It appears in the error term in the formula for the number of primes $p \leq x$ which residues modulo $n$ is in $D$.

REPLY [11 votes]: This seems easier than you might have expected. Up to normalization, your
quantities $\alpha(m,D)$ are Fourier coefficients of the indicator function
of $D$ (for which reason many people would rather use the notation $\hat
1_D(m)$). As such, they satisfy the Parseval identity
  $$ \sum_m |\alpha(m,D)|^2 = n|D|. $$
In view of $|\alpha(m,D)|\le|D|$, this yields
  $$ n|D| \le \sum_m |D| |\alpha(m,D)|, $$
whence
  $$ \sigma = \frac1n \sum_m |\alpha(m,D)| \ge 1. $$
Moreover, for equality to hold, one needs all $|\alpha(m,D)|$ to
be equal to either $0$ or $|D|$, which is only possible if $D$ is a coset of
a subgroup of ${\mathbb Z}/n{\mathbb Z}$.<|endoftext|>
TITLE: A novice question on Quantum Mechanics
QUESTION [19 upvotes]: I'm currently working through Dirac's book The Principles of Quantum Mechanics.  In it, he describes the nature of superpositions and at one point states:
"... if the ket vector corresponding to a state is multiplied by any complex number, not zero, the resulting ket vector will correspond to the same state."
He then goes on to state that,
"Given two states corresponding to the ket vectors $|A\rangle$ and $|B\rangle$, the general state formed by superposing them corresponds to a ket vector $|R\rangle$ which is determined by two complex numbers, namely the coefficients $c_{1}, c_{2}$ [from $c_{1}|A\rangle + c_{2}|B\rangle = |R\rangle$].  If these two coefficients are multiplied by the same factor (itself a complex number), the ket vector $|R\rangle$ will get multiplied by this factor and the corresponding state will be unaltered.  Thus only the ratio of the two coefficients is effective in determining the state $|R\rangle$.  Hence, this state is determined by one complex number, or by two real parameters.  Thus from two given states, a twofold infinity of states may be obtained by superposition."
Correct me if I am wrong, but Dirac seems to be saying that one gets an infinite number of different states of the form $c_{1}|A\rangle + c_{2}|B\rangle$ depending on $c_{1}$ and $c_{2}$ (or really determined by their ratio).  But given Dirac's earlier statement, we seem to get an equivalence class of states $\{z|A\rangle : z\in \mathbb{C}, z\neq 0\}$ and $\{z|B\rangle : z\in \mathbb{C}, z\neq 0\}$, so why wouldn't $c_{1}|A\rangle + c_{2}|B\rangle$ and $c_{3}|A\rangle + c_{4}|B\rangle$ always represent the same state no matter what the coefficients are (except 0)?

REPLY [4 votes]: I think the answers given by Konrad Waldorf and Federico Poloni are fantastic. However, if you're just starting to learn quantum mechanics—especially from an older book, like the one you're using—you may want to think only about pure states (this can be done without loss of generality). In that case, this answer may be helpful for you.

In my opinion, thinking about linear combinations of vectors as representing "superpositions" of the corresponding states is extremely misleading. It might suggest, for example, that if a photon in the state represented by $|H\rangle$ is guarnateed to pass through a horizontal polarizer, and a photon in the state represented by $|L\rangle$ is guaranteed to pass through a left circular polarizer, then a photon in the "superposition" state $|H\rangle + \sqrt{2}|L\rangle$ should have at least some chance of passing through a horizontal polarizer. However, by choosing the vectors $|H\rangle$ and $|L\rangle$ appropriately, you can set things up so that a photon in the state represented by $|H\rangle + \sqrt{2}|L\rangle$ will never pass through a horizontal polarizer.
Even when the concept of superposition isn't actively harmful, I find it totally unhelpful, and I would urge you to forget about it entirely. If you absolutely must have it, however, you may read on for a description of the only situation I know of in which "superposition" makes any kind of sense.

Suppose you have a quantum system with state space $\mathcal{H}$. You can think of each orthonormal basis for $\mathcal{H}$ as an abstract description of an experiment that could be done on the system; the basis vectors represent the possible outcomes of the experiment.
Say $|v_1\rangle, \ldots, |v_n\rangle$ is an orthonormal basis for $\mathcal{H}$, and
$$|\psi\rangle = c_1|v_1\rangle + \ldots + c_n|v_n\rangle$$
is a unit vector representing the state of the system. If you do the experiment described by the basis $|v_1\rangle, \ldots, |v_n\rangle$, you have probability $|c_k|^2$ of getting the outcome represented by $|v_k\rangle$.
Some people like to think of the state $|\psi\rangle$ as a "superposition" of the possible experimental outcomes represented by the basis vectors $|v_1\rangle, \ldots, |v_n\rangle$. Notice that if you change the coefficients $c_1, \ldots, c_n$, but keep their magnitudes the same, the state $|\psi\rangle$ will change, but the statistics of the experiment described by the basis $|v_1\rangle, \ldots, |v_n\rangle$ will stay the same. In other words, if
$$|\psi'\rangle = c'_1|v_1\rangle + \ldots + c'_n|v_n\rangle$$
is a superposition with $|c'_j| = |c_j|$ for all $j$, then the states represented by $|\psi\rangle$ and $|\psi'\rangle$ cannot be distinguished using the experiment described by $|v_1\rangle, \ldots, |v_n\rangle$.
However, consider another orthonormal basis $|w_1\rangle, \ldots, |w_n\rangle$, and write
$$|\psi\rangle = d_1|w_1\rangle + \ldots + d_n|w_n\rangle$$
$$|\psi'\rangle = d'_1|w_1\rangle + \ldots + d'_n|w_n\rangle.$$
In general, $|d_j|$ will not be equal to $|d'_j|$ for all $j$. In this case, the states represented by $|\psi\rangle$ and $|\psi'\rangle$ can be distinguished using the experiment described by $|w_1\rangle, \ldots, |w_n\rangle$.

So, if you have a "superposition" of orthonormal basis vectors, changing the coefficients without changing their magnitudes will not change the statistics of the experiment described by the basis you used, but it will generally change the statistics of the experiments described by most other bases. This is why changing the coefficients of a "superposition" generally gives you a representative of a different state.
I say "generally" because there is one exception: if you change your coefficients by multiplying them all by the same number, the statistics of all experiments will remain the same. This is why multiplying a vector by a number gives you a representative of the same state.<|endoftext|>
TITLE: Does this poset have a unique minimal element?
QUESTION [24 upvotes]: Recently I have been thinking about the following poset: the underlying set is $\mathcal{AFT}$ consisting of all (finite) automorphism-free undirected trees (with at least one edge to exclude the trivial cases pointed out by Joel) and the poset relation $\leq$ is defined by $T \leq U$ if $T$ can be obtained from $U$ by successively deleting one leaf node at a time in such a way that each intermediate tree is also an element of $\mathcal{AFT}$.
The smallest element of $\mathcal{AFT}$ is the seven node tree which is the Dynkin diagram of $E_7$ (and which I will therefore refer to simply as $E_7$ from herein):

So $E_7$ is certainly a minimal element in the above partial order.


Question: Does $(\mathcal{AFT},\leq)$ have a unique minimal element, namely $E_7$?


There are several equivalent formulations of this question which I have considered, hoping one of them might lead somewhere useful:


Question 2: Can every element of $\mathcal{AFT}$ be obtained by starting at $E_7$ and successively adjoining leaf nodes so that we remain in $\mathcal{AFT}$ at every stage?


>


Question 3: Is there an element of $\mathcal{AFT}$ besides $E_7$ such that deleting $any$ single leaf node results in a tree not in $\mathcal{AFT}$.


Question 3 in particular seems simple enough that it must have been answered somewhere before, but alas almost any search for 'automorphism-free' and 'trees' results in papers about the result that almost every tree has an automorphism or results on the fixed vertices/edges of trees under automorphisms.
Trying to construct a minimal example larger than $E_7$ for Question 3 keeps leading to near-misses where all but one leaf nodes' removal takes us outside $\mathcal{AFT}$, but after removing this one leaf node, there is still a sequence of removals remaining in $\mathcal{AFT}$ that leads back to $E_7$ which is the best evidence I have so far that Question 1 is true.
So, does anyone know where this might have been considered already, and if so, if the answer to Question 1 is affirmative?

REPLY [8 votes]: The answer is yes. 
With Ringi Kim and Paul Seymour, we proved this a few days ago, and the following is the proof. (I am not sure if this is already known or not. Please let me know if it is.)
Some definitions first. 
For a tree $T$ and $u,v \in V(T)$, $dist_T(u,v)$ is the length of the (unique) path from $u$ to $v$ in $T$.
$T \setminus u$ denotes the forest obtained from $T$ by deleting the vertex $u$ (deleting all the edges incident to $u$ as well).
$T \setminus uv$ denotes the forest obtained from $T$ by deleting the edge $uv$ (not deleting the vertices $u$ and $v$).
For each $v \in V(T)$,  $d_T(v) := \max_{u \in V(T) \setminus \{v\}} dist_T(u,v)$.
We say $v \in V(T)$ is a center of $T$ if $d_T(v)$ attains its minimum over all vertices.
The following are some easy facts about centers in a tree.
1) There are at most 2 centers in a tree.
2) If there are 2 centers $u$ and $v$, then $uv \in E(T)$. Moreover, every path of length $d_T(u)$ from $u$ passes $v$ and vise versa.
Now, here is our strategy.
We are going to look at a minimal tree $T$ in the poset AFT.
And we will choose special leaves $l_1$ and $l_2$ by certain methods, and use the fact that 
both $T \setminus l_1$ and $T \setminus l_2$ are not in AFT.
From this, we will prove various properties of $T$.
For instance, we will prove that $T$ must have two centers, and $T \setminus l_1$ must have exactly one center,
and $T \setminus l_2$ must have two centers, etc.
Eventually we will prove that $T$ must be isomorphic to $E_7$.
We first introduce the method of choosing a special leaf.
Let $T$ be a tree with $|V(T)| \geq 2$, and let $u$ be a vertex in $T$.
We are going to pick a leaf with respect to $u$ and $T$ as follows.
Consider all neighbors of $u$. Each one is in its own component $C_1,\cdots,C_k$ of $T \setminus u$.
Among those components, we take one with the least number of vertices.
(If there are more than one smallest components, just pick any one of those.)
Let $C_1$ be the component we chose and let $w$ be the neighbor of $u$ in $C_1$.
Now, look at all children of $w$ (the neighbors of $w$ except $u$). If there are no children of $w$, then we take $w$ as our special leaf.
Otherwise we consider components $D_1,\cdots,D_m$ of $C_1 \setminus w$ and again, we pick the smallest component, and we move one step ahead.
By this algorithm, we will end up with a leaf and we will take that leaf as our special leaf with respect to $u$ and $T$.
Theorem 1. Let $T$ be a minimal tree in the poset AFT. Then $T$ has exactly two centers.
Proof. For the sake of contradiction, suppose $T$ has a unique center $u$.
Let $l_1$ be the special leaf with respect to $u$ and $T$ as described above.
Now, consider the tree $T' = T \setminus l_1$. In $T'$,
$u$ is still a center because $d_{T'}(v)$ is either $d_{T}(v)$ or $d_{T}(v) - 1$ for every $v \in V(T) \setminus l_1$,
and $d_T(u)$ used to be the unique minimum in $T$. (But there might be another center of $T'$.)
1) There is another center of $T'$.
Suppose $u$ is the unique center of $T'$. Let $\phi$ be a non-trivial automorphism in $T'$.
Let $p(l_1)$ be the parent (the unique neighbor) of $l_1$ in $T$. 
Notice that $\phi$ does not fix $p(l_1)$ because otherwise we can extend $\phi$ to $T$
by assigning $\phi(l_1) = l_1$.
On the other hand, $\phi$ fixes $u$ since it is the unique center of $T'$.
Let $P$ be the path from $u$ to $p(l_1)$ in $T'$.
Then, it is clear that $\phi$ fixes a sub-path $P'$ of $P$ containing $u$,
and $\phi$ does not fix the other part of the path containing $p(l_1)$. 
Let $u'$ be the last vertex of $P'$. ($u'$ might be equal to $u$.)
Then, $T' \setminus u'$ has (at least) two components which are isomorphic. And one of them must contain $p(l_1)$
since otherwise we can extend $\phi$ to $T$.
Let $C_1$ and $C_2$ be those isomorphic components in $T' \setminus u'$ and say $p(l_1) \in C_1$.
In particular $|C_1| = |C_2|$. But in $T$, $C_1 \cup l_1$ and $C_2$ are two components of $T \setminus u'$
and $|C_1 \cup l_1| > |C_2|$.
This is a contradiction to  our choice of $l_1$. This proves (1).
Let $v$ be the other center of $T'$.
$u$ used to be the unique center of $T$, but now $u$ and $v$ are two centers in $T \setminus l_1$.
Therefore it must be the case that
$$d_{T'}(u) = d_T(u) = d_{T'}(v) = d_T(v) - 1$$
2) $l_1$ is not a neighbor of $u$.
Suppose $l_1$ is a child of $u$. Then, $d_{T'}(v) = d_T(v) - 1 = dist_T(v, l_1) - 1 = 1$. Therefore $T'$ has exactly two vertices $u$ and $v$.
This is a contradiction to the fact that $T$ is in AFT.
This proves (2).
3) $v$ is not in the component $C_1 \setminus l_1$ of $T' \setminus u$.
The path from $v$ to $l_1$ is the unique path of length $d_T(v)$ starting from $v$ in $T$.
In particular, the path from $v$ to $p(l_1)$ is a path of length $d_T(v)-1 = d_{T'}(v)$.
Recall that $u$ and $v$ are adjacent. Therefore if $v$ is in $C_1 \setminus l_1$, then the path from $v$ to $p(l_1)$ does not pass $u$.
A contradiction. This proves (3).
4) $p(l_1)$ is a leaf in $T'$.
Suppose $p(l_1)$ has a child $w$ other than $l_1$ in $T$.
Then, the path from $v$ to $w$ in $T'$ has length $d_{T'}(v)+1$ and this contradicts the definition of $d_{T'}$.
This proves (4).
5) $\phi$ switches $u$ and $v$.
Notice that either $\phi$ fixes $u$ and $v$ or switches them since they are centers.
But if $\phi$ fixes $u$, then by the same argument as in (1), we get a contradiction to our choice of $l_1$.
This proves (5).
Let $T_u$ and $T_v$ be the two components of $T' \setminus uv$.
($T_u$ contains $u$ and $T_v$ contains $v$.)
Since $\phi$ switches $u$ and $v$, $T_u$ and $T_v$ must be isomorphic.
Note that $\phi$ does not fix any vertex. Recall that $p(l_1)$ is a leaf in $T_u$ from (4).
Therefore $\phi(p(l_1))$ is also a leaf in $T_v$. Clearly it is a leaf in $T$ as well.
Let $l_2 = \phi(p(l_1))$.
(It is easy to see that this $l_2$ is actually the special leaf with respect to $v$ and $T$.)
We now consider $T'' = T \setminus l_2$.
6) $u$ is still a center of $T''$, but $v$ is not.
$u$ is still a center of $T''$ as it was in $T'$.
But, $v$ is not a center of $T''$ since $d_{T''}(v) = dist_{T''}(v,l_1) = d_{T}(v)  > d_{T}(u) \geq d_{T''}(u)$. This proves (6).
Again, there might be another center of $T''$.
And if there is one, then it must be in $T_u$ since $v$ is not a center of $T''$.
Now consider a non-trivial automorphism $\phi'$ of $T''$.
7) $\phi'$ does not fix $v$.
For the sake of contradiction, suppose $\phi'$ fixes $v$. Then $u$ is fixed as well because
$u$ is the unique center among the neighbors of $v$ (although $u$ might not be the unique center of $T''$.)
By the similar argument as before, the parent of $l_2$ is not fixed by $\phi'$
and this yields a contradiction to the fact that $l_2$ is a special leaf with respect to $v$.
This proves (7).
Clearly, $\phi'(v)$ is in $T_u$ since it is adjacent to a center of $T''$ and not equal to $v$. 
Then, there must be some component $C$ of $T'' \setminus u$ either isomorphic to $T_v \setminus l_2$ or contains it.
In any case, $C$ has size at least $|T_v \setminus l_2|$. Let $n = |T_v|$.
8) $|C| = n$ or $n-1$.
$|C| \geq n-1$ since $\phi'(V(T_v) \setminus l_2) \subseteq C$.
Recall that $|T_u| = |T_v|$ and $C$ is a subset of $V(T_u) \cup \{l_1\} \setminus \{u\}$.
Therefore $|C| \leq |V(T_u) \cup \{l_1\} \setminus \{u\}| = n + 1 - 1 = n$. This proves (8).
9) The degree of $u$ is 2. In particular, $T''\setminus u$ consists of two isomorphic components, namely $C$ and $T_v \setminus l_2$.
Note that the union of all components of $T''\setminus u$ other than $T_v\setminus l_2$ has size $|V(T_u) \cup \{l_1\} \setminus \{u\}| = n$.
Therefore if there is another component of $T'' \setminus u$ other than $C$ and $T_v\setminus l_2$, then it must be a single vertex.
Therefore $u$ has degree either 2 or 3.
If $u$ has degree 3, then it has a neighbor who has degree 1.
Then this leaf must have been our choice $l_1$. But by (2), this is impossible.
Therefore $u$ has exactly two neighbors. This proves (9).
Suppose there are some vertices of degree at least 3 in $C$.
Now let $x$ be the shortest distance from $u$ to a vertex of degree at least 3 in $C$.
And let $y$ be the shortest distance from $v$ to a vertex of degree at least 3 in $T_v$.
Since $T_u$ and $T_v$ are isomorphic, $x = y$.
On the other hand,
the shortest distance from $u$ to a vertex of degree at least 3 in $T_v$ is $y + 1$.
Since $T_v \setminus l_2$ is isomorphic to $C$,
the shortest distance from $u$ to a vertex of degree at least 3 in $C$ is $y+1$.
Therefore $x = y+1$ and this is a contradiction.
Therefore no vertex has degree at least 3 in $C$. And this implies that $T$ is a path.
And this is a contradiction to the fact that $T$ is in AFT.
This proves Theorem 1.
Theorem 2. Let $T$ be a minimal tree in the poset AFT. Then $T$ is isomorphic to $E_7$
Proof.  From Theorem 1, $T$ has two centers $u$ and $v$.
Let $T_u$ and $T_v$ be the two sub-trees in $T \setminus uv$. ($T_u$ contains $u$ and $T_v$ contains $v$.)
Let $l_1$ be the special leaf with respect to $u$ and $T_u$ and let $l_2$ be the special leaf with respect to $v$ and $T_v$.
Let $x$ be the shortest distance from $u$ to a vertex of degree at least 3 in $T_u$.
(If there aren't any vertices of degree 3 in $T_u$, then $T_u$ is a path, and set this number $x$ as the length of the path.)
Similarly, let $y$ be the shortest distance from $v$ to a vertex of degree at least 3 in $T_v$.
Without loss of generality, we may assume $|T_u| \leq |T_v|$. And further we may assume
if $|T_u| = |T_v|$ then $x \leq y$ by switching $u$ and $v$ if necessary.
We first look at $T' = T \setminus l_2$.
1) $u$ and $v$ are still two centers of $T'$.
Note that every path of length $d_T(v)$ starting from $v$ passes $u$ in $T$. Therefore this path still exists in $T'$ since $l_2 \in T_v$.
Therefore $d_{T'}(v) = d_T(v)$.
This means that $u$ is still a center of $T'$.
Suppose $u$ is the unique center of $T'$. 
Let $\phi$ be a non-trivial automorphism of $T'$.
Then, $\phi$ does not fix $v$ since otherwise we get a contradiction to our choice of $l_2$.
Then the component $T_v \setminus l_2$ of $T' \setminus u$ is isomorphic to some other component $C$ of $T' \setminus u$.
Note that $$|C| = |T_v \setminus l_2| = |T_v| - 1$$
Since $C$ is a subset of $V(T_u) \setminus \{u\}$,
$$|T_u| \geq |C| + 1 = |T_v|$$
Therefore $|T_u| = |T_v|$. And $T' \setminus u$ has exactly two components, namely $C$ and $T_v \setminus l_2$.
We may assume there is a vertex of degree at least 3 in $T_v \setminus l_2$, since otherwise $T$ is a path.
But then, $x \geq y+1$ and this is a contradiction to our assumption ($x \leq y$ if $|T_u| = |T_v|$).
Therefore $u$ is not the unique center of $T'$.
This means that $d_{T'}(u) = d_T(u)$ and $v$ is still a center as well.
This proves (1).
2) $\phi$ switches $u$ and $v$. And $|T_u| = |T_v| -1$.
Again, if $\phi$ fixes $v$, then $\phi$ fixes $u$ as well and we get a contradiction to our choice of $l_2$.
Since $\phi$ switches $u$ and $v$, $T_u$ and $T_v \setminus l_2$ are isomorphic.
In particular, $|T_u| = |T_v| - 1$.
This proves (2).
Now we consider $T'' = T \setminus l_1$.
Let $\phi'$ be a non-trivial automorphism of $T''$.
3) $\phi'$ does not fix $u$. And $v$ is the unique center of $T''$.
Again, every path of length $d_T(u)$ starting from $u$ passes $v$ in $T$. Therefore this path still exists in $T''$ since $l_1 \in T_u$.
Therefore $d_{T'}(u) = d_T(u)$.
This means that $v$ is still a center of $T''$.
Note that either $d_{T'}(v) = d_T(v)-1$ or $d_{T'}(v) = d_T(v)$.
In the former case, $v$ is the unique center of $T''$, and in the latter case, $u$ and $v$ are again two centers of $T''$.
Therefore if there is another center, then it must be $u$.
Suppose $\phi'$ fixes $u$. Then, again $v$ is fixed as well and we get a contradiction to the choice of $l_1$.
Therefore $\phi'$ does not fix $u$.
For the sake of contradiction, suppose $u$ is another center of $T''$.
Since $\phi'$ does not fix $u$, it switches $u$ and $v$.
Then, $T_u \setminus l_1$ is isomorphic to $T_v$, but $|T_u \setminus l_1| =  |T_u| - 1 = |T_v| - 2 \neq |T_v|$.
A contradiction. This proves (3).
Since $v$ is the unique center of $T''$ and $\phi'$ does not fix $u$, the component $T_u \setminus l_1$ of $T'' \setminus v$ is isomorphic to another component
$C$ of $T'' \setminus v$.
Note that the union of all components of $T''\setminus v$ other than $T_u \setminus l_1$ is exactly $T_v \setminus v$.
And $C$ has size $|T_u| - 1 = |T_v| - 2$.
This means that there are exactly three components of $T''\setminus v$, namely $T_u \setminus l_1$, $C$, and the third one with a single vertex.
Therefore $v$ has a neighbor of degree 1, and this must have been our choice $l_2$.
Now suppose there is a vertex of degree at least 3 in $T_u$. Then there is one in $T_v$ as well.
And by the usual argument, $x=y$ and $x+1 = y$ at the same time. A contradiction.
Therefore $T_u$ must be a path of length $|T_u|$ and $T_v$ must be a path of length $|T_v| = |T_u| + 1$.
Then, $T$ is a tree with a unique vertex of degree 3, namely $v$, and $T \setminus v$ 
has three components. One of them is a single vertex, namely $l_2$, and
the other two components are paths of length $k$ and $k+1$.
For every $k > 2$, $T$ is not minimal since deleting $l_1$ from $T$ yields a smaller tree $T''$ in AFT.
Therefore $k$ must be 2.
This proves that $T$ must be isomorphic to $E_7$.<|endoftext|>
TITLE: Morita equivalence for *-algebras
QUESTION [11 upvotes]: This is a reference request.  I'm looking for a definition of Morita equivalence of *-algebras, as described below.  If anyone thinks that this is not the right way to define Morita equivalence of *-algebras, I'd be interested to hear that also.

By a *-algebra, I mean an algebra equipped with an anti-involution ${}^*$, $(ab)^* = b^*a^*$.  (I'm mainly interested in finite-dimensional algebras.)
Recall that a Morita equivalence of algebras $A$ and $B$ consists of (1) a pair of bimodules $_AM_B$ and $_BN_A$, and (2) bimodule isomorphisms $f:{}_A(M\otimes N)_A \to {}_AA_A$ and $g:{}_B(N\otimes M)_B \to {}_BB_B$ such that (3) $f$ and $g$ satisfy the zig-zag identities.
If $A$ and $B$ are *-algebras, it seems to me that the definition of Morita equivalence should be modified (enhanced) as follows.  
First, note that the *-structure of $A$ allows us to convert between left and right $A$-modules, and similarly for $B$.  In particular, to any $A$-$B$ bimodule $_AX_B$ there is associated a $B$-$A$ bimodule $_BX^*_A$.  The first modification to the definition of Morita equivalence is to replace $_BN_A$ above with $_BM^*_A$.
Next, note that both $_AA_A$ and $_A(M\otimes M^*)_A$ have involutions coming from the *-structure.  The second (and final) modification to the definition is to require that the isomorphism $f:{}_A(M\otimes M^*)_A \to {}_AA_A$ intertwine with these two involutions, and similarly for $g:{}_B(M^*\otimes M)_B \to {}_BB_B$.

Recall that an (ordinary) Morita equivalence gives an isomorphism between 0-th Hochschild homologies $HH_0(A) \cong HH_0(B)$.  A *-structure on $A$ gives rise to an involution of $HH_0(A)$.  The isomorphism $HH_0(A) \cong HH_0(B)$ coming from an ordinary Morita equivalence need not commute with the involutions on either side, but the above enhanced Morita equivalences do guarantee compatibility with the involutions on $HH_0$.

So, repeating my initial question, where can i find the above definition in the literature?

REPLY [5 votes]: Usually I do not want to make to much of advertisement for my own stuff, but here it matches only too well: in a series of papers Henrique Bursztyn and myself developped the theory of Morita equivalence and inner product bimodules  for *-algebras in quite some details. This builds on one hand on the work of Ara, already mentioned in Benjamin Steinberg's answer. On the other hand, we were looking for generalizations of Rieffel's notion of strong Morita equivalence since we also wanted to incorporate positivity aspects. Thus our setting was to consider *-algebras over ordered rings like the formal power series used in deformation theory. You can find all this on my homepage together with a (not yet complete) lecture note. Therein you find also a nice categorical framework to compute and describe lots of Morita invariants and a detailed comparison of the various Morita theories (with/out *-involution with/out positivity...).<|endoftext|>
TITLE: Representations of SO(3) and vector bundles on BSO(3)
QUESTION [5 upvotes]: Let $V$ be the vector bundle over $BSO(3)$ associated to the adjoint representation of $SO(3).$  Then $V$ does not have a nonzero section.  One way to see this is that the Steifel-Whitney class $w_3(V)$ is nonzero.
Question: What about $V \oplus V$ or $V \oplus V \oplus \dots \oplus V?$

Do these bundles have a nonzero section?
If not, is there a characteristic class of some sort which obstructs the existence of such a section?

Comments:

The representation given by the direct sum of $n$ copies of the adjoint does not have a one dimensional invariant subspace.  Is that the same as (1) above??
The complexification $V \otimes {\mathbb C}$ has trivial third Chern class.

Thanks.  Jonathan

REPLY [7 votes]: $w_{3n}(nV)=w_3(V)^n\neq 0$. The (mod $2$) Euler class takes direct sums to cup products.<|endoftext|>
TITLE: Reasons for the Arnold conjecture
QUESTION [22 upvotes]: I am trying to understand the Arnold conjecture in Symplectic Geometry, which basically tells us the following: If $M$ is a compact symplectic manifold and $H_t$ be a 1-periodic Hamiltonian function, then we can consider the Hamiltonian equation of motion which defines us a family $\psi_t$ of symplectomorphisms of $M$. We then consider the fixed points of $\psi_1$ and call a fixed point $x$ non-degenerate, if $\det(1- d\psi_1(x)) \neq 0$. In the case that all the fied points are non-degenerate the Arnold conjecture then is: If every fixed point of $\psi_1$ is non-degenerate, then the number of fixed points is at least the sum off all Betti numbers of $M$.  \
I would now like to know the answer to the following questions: \
1. Why is such a result helpful for our understanding of Symplectic Geometry? Why would somebody like to know whether such a conjecture is true or not? \ß
2. Why could this result be true? Can you maybe give me an explanation or reference, why Vladimir Arnold conjectured this result?
\
Every answer to the above questions would be appreciated.

REPLY [3 votes]: The conjecture is listed as problem 1972-33 in the book Vladimir Arnold: Arnold's Problems, and in the Comments-section of this book you can you can find plenty of background information on it in a contribution by Mikhail B Sevryuk.<|endoftext|>
TITLE: Category of topological spaces with open or closed maps
QUESTION [10 upvotes]: Consider the category whose objects are topological spaces and whose morphisms are the open maps (or closed maps, open continuous maps, closed continuous maps ... that is, one whose isomorphisms are  precisely the homeomorphisms). How does such a category compare with the usual one whose objects are topological spaces and whose morphisms are continuous maps?  For example, what limits and colimits exist?
I'm probably missing something obvious, but why don't products typically exist in the category with open maps? The projections from the usual product (in the category with continuous maps) are open, yielding a canonical open map from the usual product to the putative unusual product.
Todd's observation is true enough: the product in the usual topology (contiuous maps) typically fails to realize the corresponding universal property in the unusual topology  (open maps). Nevertheless,  some other object might realize that universal property. Is it even clear that the if such a space exists its underlying set should be naturally identifiable with the underlying set of the factors? After all, while one point spaces are still terminal, maps out of such objects tend not to be open: it seems one would thereby only extract the subset of isolated points. In any event, http://christianmarks.wordpress.com/category/bagatelle
treats the special case of squares. The appropriate space is $X\times X$ with the weakest topology (stronger than the usual) which makes the diagonal embedding open. This construction is clearly not available for products of distinct spaces. My question concerns whether there isn't (as that post suggests there isn't) some devious workaround.

REPLY [11 votes]: This seems difficult to answer in any kind of uniform way, i.e., it seems we have to work on a case-by-case basis. Let's start with $\mathrm{Top}_{\mathrm{open}}$, the category of topological spaces and open continuous maps. 
A useful heuristic is to let oneself be guided by locales instead of topological spaces. (For general information on locales, see for instance Mac Lane-Moerdijk, Sheaves in Geometry and Logic, chapter IX.) Recall that the category of locales is (by definition) the category opposite to the category of frames (sup-lattices in which finite meets distribute over arbitrary joins); the frame associated with a locale (by viewing the same object in the opposite category) should be thought of as its topology. Now a frame automatically carries Heyting algebra structure, and it turns out that open continuous maps of locales correspond to frame maps that are also Heyting algebra maps (i.e., Heyting algebra maps which are sup-preserving). Thus, the category of locales and open continuous maps is opposite to the category of cocomplete Heyting algebras, which is a category of algebras of an infinitary algebraic theory. It follows that the category of locales and open continuous maps is complete. Moreover, limits of cocomplete Heyting algebras are computed just as they are in $Set$, and so the calculation of colimits in locales and open continuous maps will be similarly straightforward. 
Guided by this heuristic, one might at the very least expect that colimits in $\mathrm{Top}_{\mathrm{open}}$ are calculated just as they are in $\mathrm{Top}$. Indeed, one can easily check this directly for arbitrary coproducts (where the coproduct topology on a disjoint union $X = \sum_\alpha X_\alpha$ is the largest one such that all the canonical inclusions $X_\alpha \to X$ are continuous; notice they are open). As for coequalizers, if two maps 
$$f, g: R \stackrel{\to}{\to} X$$ 
are open and continuous, and if we consider the coequalizer $q: X \to Y$ in $\mathrm{Top}$ (where the topology on $Y$ is the usual quotient topology induced by $q$), then we claim that $q$ is also open and this is the coequalizer in the category of open continuous maps. For, if $V$ is open in $X$, then $q(V)$ is open in $Y$ iff $q^{-1}(q(V))$ is open in $X$, and clearly this holds since 
$$q^{-1}(q(V)) = V \cup \bigcup_n (g f^{-1})^n(V) \cup (f g^{-1})^n(V)$$ 
where the right side is open because $f$ and $g$ are open and continuous. 
Pausing briefly to consider the category of topological spaces and closed continuous maps, it is certainly true that finite coproducts exist and are computed just as they are in $\mathrm{Top}$. Also, it is true that quotients of equivalence relations are computed as in $\mathrm{Top}$. That is to say: if $E \subseteq X \times X$ is an equivalence relation and the two projection maps $\pi_1: E \to X$ and $\pi_2: E \to X$ are closed, then the coequalizer $q = \mathrm{coeq}(\pi_1, \pi_2)$ in $Top$ is the coequalizer in the category of closed continuous maps. Indeed, if $C$ is closed in $X$, then $q(C)$ is closed because 
$$q^{-1}(q(C)) = \pi_1(\pi_2^{-1}(C))$$ 
is closed. 
Limits are trickier, and I believe do not always exist in $\mathrm{Top}_{\mathrm{open}}$ (see below). Even when they do exist, one should not expect to calculate limits as one would in $\mathrm{Top}$ (for example, Cantor space as a countable product of copies of the discrete two-element space could not possibly be the correct product in $\mathrm{Top}_{\mathrm{open}}$ since Cantor space has no isolated points), nor even as in $\mathrm{Set}$, so this makes limits a little weird. 
Here's a partial result: if we have a diagram $F: D \to \mathrm{Top}_{\mathrm{open}}$ and all the spaces $F(d)$ are sober (i.e., of the form $\mathrm{Pt}(\mathcal{O}(X))$, where $\mathrm{Pt}: \mathrm{Frame}^{op} \to \mathrm{Top}$ is right adjoint to $\mathcal{O}: Top \to \mathrm{Frame}^{op}$), then the limit of $F$ in $\mathrm{Top}_{\mathrm{open}}$ can be constructed as $\mathrm{Pt}(\mathrm{colim} \; \mathcal{O}(F d))$ where the colimit is computed in the category of cocomplete Heyting algebras, provided that colimit exists. This uses the fact that for any topological space $X$, the set of open continuous maps $X \to \mathrm{Pt}(H)$ ($H$ a cocomplete Heyting algebra) is in natural one-one correspondence with cocomplete Heyting algebra maps $H \to \mathcal{O}(X)$, together with elementary categorical manipulations. 
I conjecture that finite colimits of cocomplete Heyting algebras exist (so that finite limits of sober spaces in $\mathrm{Top}_{\mathrm{open}}$ exist). But inasmuch as the countable coproduct of copies of the free cocomplete Heyting algebra on one generator does not exist (it would be the free cocomplete Heyting algebra on countably many generators, which is known not to exist), it seems to me that $\mathrm{Top}_{\mathrm{open}}$ does not have countable products.<|endoftext|>
TITLE: Using Quotient of Prime Numbers to Approximation Reals
QUESTION [16 upvotes]: We know a positive rational number can be uniquely written as $m/n$ where $m$ and $n$ are coprime positive integers. Particularly, we can pick out those numbers with $m$ and $n$ both prime.
Question 1: Is the collection of all such numbers dense on the positive half of the real line?
Furthermore, we can ask about the efficiency of approximation, more precisely:
Question 2: Suppose we have an inequality $1\le ps-qr\le a$. Fix some $a$, can we find infinitely many solutions where $p$,$s,$,$q$,$r$ are positive primes?

REPLY [2 votes]: I had those exact same questions today! I was so happy that I found this post that I finally decided to make an account to provide an alternate, although fundamentally equivalent, way of looking at Question 1.
Given the following generalization of Bertrand's postulate found as the most popular answer here: At what point would an elementary generalization of Bertrand's Postulate be interesting? , we can easily show $\{\frac{p}{q}:$ $p$ and $q$ are prime$\}$ is dense in $\mathbb{Q}$, hence in $\mathbb{R}$.
First, the above link tells us that for any $n$ there is a $K$ large enough so that $[nk, (n+1)k]$ contains a prime for $k>K.$
Given positive $\frac{r}{s}\in\mathbb{Q},$ for any $n>0$ there is $K>0$ and primes $p,q$ such that
$$p\in [rKn, rK(n+1)],$$
$$ q\in [sKn, sK(n+1)].$$
Hence, $$\frac{r}{s}\cdot\frac{n}{n+1}=\frac{rKn}{sK(n+1)}\leq \frac{p}{q}\leq \frac{rK(n+1)}{sKn}=\frac{r}{s}\cdot\frac{n+1}{n}.$$
We see that $n$ could be chosen arbitrarily large, so that there is a quotient $\frac{p}{q}$ of primes arbitrarily close to $\frac{r}{s}.$
Edit: small rewording.<|endoftext|>
TITLE: Poincaré line bundle
QUESTION [8 upvotes]: I am being stuck by the proof of the existence of Poincaré line bundle of complex torus in Griffiths-Harris. Here is the question: 
Let $M$ be a complex torus and $M'$ be the complex torus dual to $M$ (the one consists of all holomorphic line bundle whose chern class is zero). By Kunneth's formula and the definition of $M',$  one knows that the identity element $I$ of 
  $$ Hom(H^1(M,\mathbb{Z}),H^1(M,\mathbb{Z}))\equiv H^1(M,\mathbb{Z})\otimes H^1(M',\mathbb{Z})$$
belongs to $H^2(M\times M',\mathbb{Z})$ and is of type (1,1). Hence, there exists a line bundle $P$ over $M\times M'$ having $I$ as its chern class. For each $\xi$ in $M'$, let $P_{\xi}$ be the restriction of $P$ on  $M\times \{\xi\}$, clearly it is in $M'$. Hence we get a mapping $\Phi: M' \rightarrow M'$ that send $\xi$ to $P_{\xi}.$
Why the induced mapping of $\Phi$ on the first homology is the identity?
Thank you so much.

REPLY [2 votes]: [Not exactly an answer. Still, it may be helpful, I hope.]
I think there are two obscure points in this proof: apart from the homology 
question, why the map is holomorphic? (Is it really obvious?). 
Of course, all this can be settled. However, I prefer a more explicit 
construction of the Poincare bundle, which can be found, e.g., in 
Hindry&Silverman, "Diophantine Geometry", Exercise A.5.6. 
(I like it, but even if you won't, it is definitely worth reading).<|endoftext|>
TITLE: Applications of Floer homology
QUESTION [5 upvotes]: Can somebody tell me of other applications of Floer homology besides the proof of the Arnold conjecture. 
Every answer would be appreciated.

REPLY [8 votes]: Floer homology has, in one form or another, become ubiquitous in symplectic geometry and to give a complete list of its applications would be a mammoth task. Here are a few.
1) One early incarnation of Floer homology was in instanton gauge theory. Floer's instanton invariant is a homology group associated to a homology 3-sphere, whose dimension 'counts' flat connections. One application of this invariant is a proof (sketched in Atiyah's paper "New invariants of 3- and 4-dimensional manifolds" section 7, and further explained in Chapter 6 of Donaldson's book "Floer homology groups in Yang-Mills theory") that a 4-manifold with nonzero Donaldson invariants (like an algebraic surface) cannot be written as a connected sum of two manifolds with $b^+>0$. There is an argument which avoids Floer homology (see Donaldson and Kronheimer section 9.3.2), but the Floer homology argument is more conceptual. The idea is that by stretching the neck around a separating 3-sphere you get a pair of 4-manifolds with boundary $S^3$ and these define for you a pair of cycles in Floer homology whose intersection product is the Donaldson invariant. Since the Floer homology of $S^3$ vanishes, so too does the Donaldson invariant. The advent of more sophisticated/computable/easily-defined invariants like Heegard-Floer or Seiberg-Witten-Floer homology has led to yet more spectacular results in low-dimensional topology.
2) In Gromov's paper on pseudoholomorphic curves, he proves that there are no embedded simply-connected Lagrangian submanifolds in a symplectic vector space. To do this he constructs a holomorphic disc with boundary on the Lagrangian using the fact that it can be displaced off itself using translation. By integrating the 1-form $\sum p_idq_i$ over the boundary of this disc you get a positive number (by Stokes's theorem and holomorphicity of the disc) which means that the boundary of the disc is nontrivial in $\pi_1$ (even $H_1$). More refined topological restrictions on Lagrangians come from more refined ways of counting pseudoholomorphic discs. Lagrangian Floer homology and its algebraic structure (better still the Fukaya category) gives such a refined framework. For instance, displaceability of a Lagrangian translates into vanishing of Floer homology and there is a spectral sequence (due to Oh) converging to Floer homology whose $E_2$ page encodes the homology of the Lagrangian. The differentials involve counts of pseudoholomorphic discs with boundary on. the Lagrangian so vanishing of Floer homology means lots of discs. The algebraic structure can give very precise control over which relative homology classes are represented by holomorphic discs. This was exploited by Buhovsky and Damian to prove monotone versions of the Audin conjecture: that (monotone) Lagrangian tori (or aspherical manifolds) in $\mathbf{C}^n$ have minimal Maslov class 2.
Of course this is only one tiny tip of the iceberg of results about Lagrangian submanifolds proved using Floer theory which include the works of Fukaya-Oh-Ohta-Ono, Biran-Cornea, Seidel and many others. Let me explain two more:
3) Seidel proved that if you take a pair of Lagrangian spheres $L_1,L_2$ in a symplectic 4-manifold, intersecting transversely at a single point, then you can take a neighbourhood $X$ of the two spheres (a plumbing of cotangent bundles) and inside $X$ you can Dehn twist $L_1$ around $L_2$ twice and you obtain a Lagrangian submanifold $L_3\subset X$ which is smoothly isotopic to $L_1$ but not isotopic through Lagrangian submanifolds. This Lagrangian knotting phenomenon was detected using Floer homology.
4) There is another Arnold conjecture, asserting that compact exact Lagrangian submanifolds in cotangent bundles are Hamiltonian isotopic to the zero section. Though this is known only for $T^*S^1$ and $T^*S^2$, there are suggestive results in this direction. The further structure afforded by the Fukaya category was used by Fukaya-Seidel-Smith to prove that projection of an exact Lagrangian to the zero section induces an isomorphism on homology. This result has since been strengthened by Abouzaid.
As I said, this is just a tiny selection of the applications known.<|endoftext|>
TITLE: Etale site is useful - examples of using the small fppf site?
QUESTION [24 upvotes]: Edit: After the answers and comments, I'm hoping for a little bit of elaboration (in the comment to the answer below.) Also, question 2 was discussed here:
Points in sites (etale, fppf, ... )
There, Davidac897 gave a nice description of points in more general sites. I would be interested if there is a different description of points in the small fppf site and big etale site over an (as nice as you like) scheme (similar to the very concrete description of points that we have for the small etale site).
Thanks to everyone for all the helpful comments and answers!

The title very much sums it up. The etale site is extremely useful and the basic applications are well known. Milne also devotes time to the Flat site in his Etale Cohomology book. I am hoping that someone can give me example applications.
1.) I'm most interested in the (small) Flat site. What do you typically use this for? Let $X$ a scheme over $\mathbb{F}_P$ and $\alpha\alpha_P$ be the sheaf on the small fppf site over $X$ defined by group scheme $\mathbb{F}_P[t]/(t^p)$. The sequence of sheaves 
$$ 0 \rightarrow \alpha\alpha_P \rightarrow \mathbb{G}_a \xrightarrow{F} \mathbb{G}_a \rightarrow 0$$
($F$ is the map $z \mapsto z^p$) makes sense in the (small) Etale site, but is typically not exact there. However, it becomes exact in the (small) fppf site. This is useful, because a ses of sheaves yields a long exact sequence, and hence relations that one (at least I) cannot so easily express without the Flat site. I don't even know if this example is typical, or if there are many other examples on these lines (or many examples not along these lines).
$\textrm{Principal Homogenous Spaces}$: Cohomology in the flat site calculates the set of principal homogenous spaces over a scheme (wrt a group scheme $G/X$).
2.) What are points in the (small) flat site? I am not able to dream up a good description of these (or where to look).

REPLY [2 votes]: The cohomology of $\mu_{p^n}$ and $\mathscr{A}[p^n]$ for an Abelian scheme $\mathscr{A}$ in characteristic $p > 0$ can also be studied using syntomic cohomology.<|endoftext|>
TITLE: How to calculate zeroth crystalline cohomology
QUESTION [7 upvotes]: I am just learning crystalline cohomology, so I understand the basic set-ups. But I can't really do any calculations.
For example, let's choose the base $S=W(k)/p^n$, and let $X$ be an affine scheme over $S$, so it is represented by a $W(k)/p^n$-algebra $A$. Let's just consider the structural sheaf $\mathcal O_{X/S}$ and the sheaf of PD ideal $\mathcal J_{X/S}$ . Now, what is the zero-th cohomology:

$H^0_{cris}((X/S) ,\mathcal O_{X/S})$
$H^0_{cris}((X/S) ,\mathcal J_{X/S}) $ 
$H^0_{cris}((X/S) ,\mathcal J_{X/S}^{[i]}) $ 

I believe that there must be some books explaining this basic example, but I still can't find the reference..
Thanks!

REPLY [8 votes]: In this generality, this is unfortunately not so easy because it requires to compute universal PD-envelopes. The case where $X/S$ is embedded in a smooth scheme $Y/S$ is explained in the book by Pierre Berthelot, Cohomologie cristalline des schémas de caractéristique $p>0$, Lecture notes in mathematics 407, 1974 (see chapter V). 
You may also look at a paper by Jean-Marc Fontaine  (« Cohomologie de de Rham, cohomologie cristalline et représentations $p$-adiques », in Algebraic geometry Kyoto-Tokyo,
Lecture notes in mathematics 1016, 1983, p. 86-108). There, he proves that the 0th crystalline cohomology of $O_{\bar K}$ ($K$ local field) is the ring $B_{\rm{cris}}$ he had defined earlier.<|endoftext|>
TITLE:  a measurable cardinal  & a real-valued measurable cardinal in the same model?
QUESTION [5 upvotes]: Although I know that "ZFC & there exists a measurable cardinal" and "ZFC & there exists a real-valued measurable cardinal" are equiconsistent with one another, I am not sure whether "ZFC & there exists a measurable cardinal k & there exists a real-valued measurable cardinal b" is equiconsistent with ZFC. (Obviously k is not equal to b.)
I would be grateful for an answer.

REPLY [8 votes]: Let me interpret the question as asking for a real-valued measurable cardinal that is not measurable, plus another (two-valued) measurable cardinal, which must be above it. For example, in a more extreme form, your question would ask: can the continuum be real-valued measurable while there is also another measurable cardinal? 
The answer is yes.
Theorem. If there are two measurable cardinals, then there is a forcing extension in which the smaller one becomes the continuum and real-valued measurable, and the larger one remains measurable. 
Proof. If $\kappa\lt\lambda$ are both measurable cardinals in $V$, then the forcing to add $\kappa$ many random reals will make $\kappa$ into the continuum and still real-valued measurable (by a result of Solovay), and the larger measurable cardinal $\lambda$ remains measurable, by the Levy-Solovay theorem, because the forcing was much smaller than $\lambda$. QED
Corollary. The following are equiconsistent.

There are two measurable cardinals. 
There are two real-valued measurable cardinals. 
There is a non-measurable real-valued measurable cardinal and a measurable cardinal. 
The continuum is real-valued measurable and there is another measurable cardinal. 

Proof. Statement 1 implies 4 in a forcing extension, by the argument above. Statement 4 implies statement 3 directly. Statement 3 implies statement 2 directly. Statement 2 implies statement 1 in an inner model. QED
Of course, one can generalize the arguments to handle any number of real-valued measurable cardinals--three instead of two, or any cardinal number---with measurable cardinals above them.<|endoftext|>
TITLE: Arithmetic dynamics and dynamics on moduli spaces
QUESTION [16 upvotes]: The following question is more of a request for pointers to suitable literature on introductory material for arithmetic dynamics and dynamics on moduli spaces. 
In my dissertation, I have been working mostly with smooth dynamical systems, and a lot with a class of dynamical systems given by iteration of certain polynomial maps on smooth two-dimensional (in fact algebraic) submanifolds of $\mathbb{R}^3$. Naturally I have also been looking at holomorphic dynamics. At some point I started to ask questions (not related to my current work) about those dynamical systems, which seem to be better formulated in the algebraic context, rather than analytic, measure-theoretic or differential-geometric (all of which I am familiar with). As a result I discovered a field (which seems to be actively developing) of arithmetic dynamics and (somewhat related but not entirely, I guess) dynamics on moduli spaces. I have searched the internet for some introductory material, only to find that literature is rather scarce, and I haven't been able to find a description of major problems or conjectures which are driving the field. 
Questions:
1) What would be good references for some of the fundamental results in arithmetic dynamics? 
2) What are questions of interest in arithmetic dynamics? Are there any major actively researched conjectures? What is driving the field? Are there any strong connections with well-known problems in other fields? 
3) Is there any introductory literature of expository nature? 
4) Questions (1) - (3) applied to dynamics on moduli spaces (I really don't know much about this field, other than the phrase "dynamics on moduli spaces" that I seem to come across often lately). 
Note: Not sure whether I should make this a community wiki; please advise.

REPLY [5 votes]: In addition to the graduate textbook that Felipe already mentioned, 
The Arithmetic of Dynamical Systems, Springer-Verlag, GTM 241, 2007,
there's also the following monograph that discusses dynamical-related moduli spaces from an algebraic and arithmetic viewpoint:
Moduli Spaces and Arithmetic Dynamics, CRM Monograph Series 30, AMS, 2012.
As for the list of references http://www.math.brown.edu/~jhs/ADSBIB.pdf that Benjamin Dickman mentioned, I update it once or twice a year, so it's reasonably up-to-date, but I make no claim to its being complete. You might also try searching MathSciNet and the ArXiv for articles whose classification number is 37Pxx, which will catch most recent articles. In particular, there's
37P45 = Dynamical systems and ergodic theory / Arithmetic and non-Archimedean dynamical systems / Families and moduli spaces.<|endoftext|>
TITLE: Optimization version of the Sylvester equation
QUESTION [5 upvotes]: The Sylvester equation is a matrix equation of the form $AX-XB=C,$ where $A,B,C$ are given matrices of dimension $m\times m,n\times n$ and $m\times n$ and $X$ is an unknown matrix of dimension $m\times n.$ It is a well known fact that the equation has an unique solution if and only if the matrices $A$ and $B$ have disjoint spectrum. If they do not have disjoint spectrum, then the result in general depends on $C.$ 
While determining perturbation of eigenvalues in certain context I was naturally drawn to the the problem of determining the minimum, $min_{X}||AX-XB-C||,$ where $||.||$ is the Frobenius norm. Clearly, if the spectrum of $A$ and $B$ is disjoint then there is a choice of $X$ for which the norm is zero. Otherwise, we need to resort to certain optimization techniques. One approach could be to vectorize the matrices using Kronecker products and determine the minimum of a linear system. 
The problem is: "What is the choice of $X$ for which the norm $||AX-XB-C||$ attains minimum (if it exists) when the spectra of $A$ and $B$ are not disjoint?" 
I have not found any literature on discussion about similar problems. I would be very thankful for any references or suggestions in this direction.

REPLY [6 votes]: First recall two basic ideas.

Lemma. Let $A$, $B$, $C$ be arbitrary; then, $\text{vec}(ABC) = (C^T \otimes A)\text{vec}(B)$, where $\otimes$ denotes the Kronecker product and $\text{vec}(\cdot)$ denotes the 'vec' operator that stacks columns of a matrix to obtain a long vector.
Notation. For any matrix $Z$, I will denote by its lowercase version $z$ the vector $\text{vec}(Z)$.

Now observe that $\|X\|_F^2 = \text{trace}(X^TX) = x^Tx$. Thus, (also essentially noted by the OP) we may rewrite the optimization problem as
\begin{equation*}
  \min_x \|Hx-c\|^2,
\end{equation*}
where $H = I \otimes A - B^T \otimes I$.
This equation may or may not have a unique solution, but the unique least-norm vector $x$ that solves it is given by $x=H^+c$, where $H^+$ is the Moore-Penrose pseudoinverse of $H$. Thus, clearly, the optimization problem has a solution, which answers the question as asked.
Of course, due to the numerical expense of computing the above solution (via the truncated SVD of $H$) might be too high. The OP might be interested in consulting the numerical analysis literature on how to deal with such a case.<|endoftext|>
TITLE: Model for the (infinity,1)-category of (homotopy-)limit preserving functors
QUESTION [13 upvotes]: I've got a simplicial model category $M.$ I'd like to get my hands on the (infinity,1) category of homotopy limit preserving functors from M to Spaces in order to compare it to another simplicial model category. So it would be convenient if I could have a simplicial model category model for the functor category.
I imagine doing something like the following (sketch):
1) find a model category which models the (infinity,1) category: $\textrm{Fun}(N^{hc}_{\bullet}(M^{cf}),\textrm{SSet})$. I'll call such a model category $\textrm{Fun}(M,\textrm{SSet})$
2) use Bousfield localization on the collection of morphisms S = {for each functor F, the comparison maps  F(lim d) ---> lim Fd } in $\textrm{Fun}(M,\textrm{SSet})$ to get a model category structure which models the category of homotopy limit preserving functors.
So my questions are 

Question 1: Given a simplicial model category $M$, what model category models the functors from $M$ to Spaces?
Question 2: Given a simplicial model category $M$, what model category models the homotopy-limit-preserving functors from $M$ to Spaces?

Answers to my question don't need to address my sketch, but I am curious about whether that will work. References would also be appreciated.

REPLY [6 votes]: Even if the dual $M^{\mathrm{op}}$ of your original simplicial model category $M$ is combinatorial, so that its associated $\infty$-category $\mathcal{M}^{\mathrm{op}}$ is presentable, it is not true that the $\infty$-category of presheaves of spaces $\mathcal{P}(\mathcal{M}^{\mathrm{op}}) := \operatorname{Fun}(\mathcal{M}, \mathcal{S})$ on $\mathcal{M}^{\mathrm{op}}$ is again presentable. This only holds if $\mathcal{M}$ is small which is typically not true for a model category. Also the results of Section 5.1.1 of Lurie's Higher Topos Theory giving a model for the $\infty$-category of presheaves only speak about the small case. 
In the case you are interested in one should consider only small functors from $\mathcal{M}$ to $\mathcal{S}$ in order to get an  $\infty$-category with small mapping spaces. This is what Lurie does when considering the pro category of spaces as a full subcategory of the $\infty$-category of accessible functors from spaces to spaces. (See Definition 7.1.6.1 in Higher Topos Theory. In this case small is equivalent to accessible.) 
You should thus begin with the category of small simplicial functors from $M$ to simplicial sets, and endow it with the projective model structure that was constructed by  Chorny and Dwyer: http://arxiv.org/abs/math/0607117. Then you should left Bousfield localize this model structure in such a way that the fibrant objects in the localized model structure would be the projectively fibrant functors that preserve homotopy limits. Note that you cannot use the usual theorems on Bousfield localizations such as those of Hirschhorn or Smith-Lurie because your model category is not cellular nor combinatorial (in fact, it is not even cofibrantly generated). You will have to use techniques such as in http://arxiv.org/abs/1409.8525, which rely on the generalized small object argument due to Chorny.
The localized model structure above would indeed model the $\infty$-category $\operatorname{Fun}^R(\mathcal{M}, \mathcal{S})$ of limit-preserving functors, which is equivalent to $\mathcal{M}^{\mathrm{op}}$ itself through the Yoneda embedding by  proposition 5.5.2.2 of Higher Topos Theory, as Alberto García-Raboso mentioned.<|endoftext|>
TITLE: Fixed objects of the M endofunctor on category Meas
QUESTION [8 upvotes]: Consider the category $\operatorname{Meas}$ of measurable spaces: its objects are sets equipped with $\sigma$-algebras, and its morphisms are measurable functions between spaces. 
As Gerald Edgar & Michael Greinecker pointed out in this thread, there is the natural endofunctor $M : \operatorname{Meas} \to \operatorname{Meas}$ which sends a space $X$ to the collection $M(X)$ of (extended-real-valued) measures on $X$. This collection $M(X)$ is a measurable space, equipped with the minimal $\sigma$-algebra so that the evaluation functions $\mu \mapsto \mu(A)$ are measurable. A morphism $f : X \to Y$ is mapped to the push-forwarding map $f_* : \mu \mapsto \mu \circ f^{-1}$.
We may naturally iterate this endofunctor. Thinking of a measure on $X$ as a statistical ensemble, the space $M^2(X) = M(M(X))$ consists of ensembles-of-ensembles. Such hierarchical spaces are important in probability theory and dynamical systems. We may go further, defining $M^3(X) = M(M(M(X)))$ and so forth.
This simply generates a dynamical system on the category of measurable spaces, where the initial condition $X \in \operatorname{Meas}^{\operatorname{ob}}$ gets mapped to its successors $M(X)$, $M^2(X)$, $M^3(X)$, etc. Understanding these categorical dynamics is a hard problem, to say the least. Understanding the fixed ``points'', on the other hand, might actually be tractable. Hence the question:

What are the fixed objects of the endofunctor $M :\operatorname{Meas} \to \operatorname{Meas}$ ?

REPLY [2 votes]: Here is a candidate class of examples. I have made this community wiki so please feel free to edit it if you can answer it. Or, copy the text and make a new answer so we can give you reputation points.

Let $X_0 := X$ be any non-empty measurable space, and for each $n \in \mathbb N$, define the product $X_{n+1} := M(X_n) \times X_n$. This is a measurable space when equipped with the tensor product of $\sigma$-algebras. Define $X_{\infty}$ to be the projective limit of the sequence $X_0 \leftarrow X_1 \leftarrow X_2 \leftarrow \cdots$, where the arrows denote the projections onto the second components of the products. 
The existence of the limit gives a canonical section $X_{\infty} \to M(X_{\infty}) \times X_{\infty}$. By iterating this map with the projection onto the first component, we have defined a natural measurable function $\varphi : X_{\infty} \to M(X_{\infty})$.
Consequently, a point $x$ in $X_{\infty}$ contains the data of a measure $\varphi(x)$ on the space. It may be the case that this is all the data possessed by the point, in which case $X_{\infty} \cong M(X_{\infty})$. Is this the case? That is, 

is the function $\varphi : X_{\infty} \to M(X_{\infty})$ one-to-one and onto?

Note that the requirement that $X_0$ be non-empty is necessary. If $X_0 = \varnothing$, then $M(X_0) = \{0\}$ consists of the zero measure, but $X_1 = \{0\} \times \varnothing = \varnothing$. Consequently $X_{\infty} = \varnothing$ and $M(X_{\infty}) = \{0\} \ne X_{\infty}$.

This construction is based on the concept of an epistemic type space, which encodes the belief hierarchies of players in epistemic game theory.<|endoftext|>
TITLE: Motivation of Virasoro algebra
QUESTION [26 upvotes]: I have a question on definition/motivation of Virasoro algebra. Recall that Virasoro algebra is an infinite Lie algebra generated by elements $L_n$ $(n\in \mathbb{Z})$ and $c$ over $\mathbb{C}$ with relations
$$
[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m)\delta_{m+n,0}.
$$
A typical explanation of this definition is the following. 

Define vector fields $l_n=-z^n\frac{\partial}{\partial z}$ on $\mathbb{C}\setminus \{0\}$. They form a Lie algebra of infinitesimal conformal transformation 
  $$
[l_m,l_n]=(m-n)l_{m+n}.
$$ 
  So the Virasoro algebra is a central extension of this algebra by $c$. $c$ is called the central charge.

My questions are 

How can one see that the Lie algebra above is associate to infinitesimal conformal transformation?
What is the central charge $c$ intuitively? Why are we interested in such a central extension?

As to second question, I don't have enough physics background to check what the central charge $c$ means in physics literature. 
At this point, I don't have any intuition and have trouble in digesting the concept. I would really appreciate your help.

REPLY [3 votes]: How to see that $-z^{m+1}\frac{\partial }{\partial z}$ is related to an infinitesimal conformal transformation:
With $\omega=\frac{z^{-m}}{m}$ and $z=(m \cdot \omega)^{\frac{-1}{m}}$,  the exponential mapping gives
$exp[-t\cdot z^{m+1}\frac{\partial }{\partial z}]f(z)=exp[t\cdot \frac{\partial }{\partial \omega}]f[(m \cdot \omega)^{\frac{-1}{m}}]=f[(m \cdot (\omega+t))^{\frac{-1}{m}}]=f\left [\frac{z}{(1+\ m \cdot t \cdot z^m)^{\frac{1}{m}}}  \right ]=f(g_{m}(z,t)).$
Consistently, $g_{m}(g_{m}(z,s),t)=g_{m}(z,s+t).$
So the exponential mapping induced by the tangent vector results in composition of $f$ with $g_{m}(z,t)=z-\ t\cdot z^{m+1} +\ ....$.
Then to first order in $t$, a Taylor series expansion in $t$ about $t=0$ of the infinitesimal composition gives
$f(z-\ t\cdot z^{m+1})\approx f(z)-{f}'(z)z^{m+1}\cdot t=(1-\ t \cdot z^{m+1}\frac{\partial }{\partial z})f(z),$
and in the domain of analyticity of $g_{m}(z,t)$, the mappings are conformal. 
Some interesting associated combinatorics:
1) $g_m$ is related to the e.g.f. for planar m-ary trees and double factorials (m=1), triple factorials (m=2), quartic (m=3), etc. (Cf. OEIS A094638)
2) The compositional inverse of $h(x)= x-\ t\cdot x^{m+1}$ gives a generating function for the Fuss-Catalan numbers (e.g., OEIS A001764).
Edit (April 2018): 
More generally, let $h(h^{-1}(y))=y$ and $g(z) = 1/(dh(z)/dz)$. Then
$$ exp[t \cdot g(z) \frac{d}{dz}]f(z) = f[h^{-1}(t + h(z))] = f(W(t,z)), $$ 
which is analytic about $t=0$. Note that $W(t,W(s,z))=W(s+t,z)$ is the flow map described in OEIS A145271 related to the refined Eulerian numbers and other number arrays related to important combinatorial structures--the associahedra and noncrossing partitions among others.
About $t=0$,
$$f[W(t,z)] = f(z) + f'(z)g(z)t + \cdots = (1 + t \cdot g(z) \frac{d}{dz}) f(z) + \cdots .$$
Since the composition and product of analytic functions are analytic, conformality is locally preserved.<|endoftext|>
TITLE: twisted Poisson structures, degenerate metrics and integrability properties of (2,0)-tensors
QUESTION [9 upvotes]: Given a regular (constant rank) bi-vector $\Pi \in \Gamma(\bigwedge^2TM)$ on a smooth manifold $M$ the necessary and sufficient condition for the image of $\Pi^\sharp:T^*M\to TM$ to be an integrable distribution is that $\Pi$ is a twisted Poisson tensor with respect to a closed 3-form $\phi$, i.e. there exist a closed 3-form $\phi$ such that
$$[\Pi,\Pi]=\Pi^\sharp(\phi)$$
(see e.g. arXiv:1104.0880).
What is the corresponding condition for a degenerate regular symmetric (2,0)-tensor (a degenerate metric) $g\in \Gamma(S^2TM)$ to produce an integrable distribution as the image of $g^\sharp:T^*M\to TM$?
And what about any regular (constant rank) tensor $T\in \Gamma(TM\otimes TM)$, with no special symmetry properties?

REPLY [2 votes]: This is also only for the symmetric case.
Let $S(TM)=\bigoplus_k S^k(TM)$ and $\Gamma(S(TM))$ be the commutative graded algebra of symmetric purely contravariant tensor fields. This is the algebra of functions on $T^*M$ which are polynomial on each fiber, and it is invariant under the Poisson bracket on $T^*M$.
This is the symmetric Schouten bracket.
See the following paper for more information: 
Michel Dubois-Violette, Peter W. Michor: A common generalization of the Frölicher-Nijenhuis bracket and the Schouten bracket for symmetric multi vector fields, Indagationes Math. N. S. 6 (1995), 51--66.
(pdf)
It might be that one can just carry over the considerations from $\Gamma(\Lambda (TM))$
to $\Gamma(S(TM))$ using this bracket.
EDIT: The above did not work. So let me try again. I do not need symmetry now.
Suppose that $a\in\Gamma(TM\otimes TM)$ is of constant rank when viewed as 
$a:T^\star M\to TM$.
Choose a Riemann metric $g$ on $M$ and consider $A=a\circ g:TM\to T^\star M\to TM$ which is again of constant rank. So $\ker(A)$ and $A(TM)$ are subbundles. Choose an isomorphism $B:TM\to TM$
such that $A.B$ equals the $g$-orthogonal projection $P$ onto $A(TM)$. $P$ can be canonically constructed from $g$ and $a$, up to some free choice of an isomorphism between $Im(A)^\bot$ and $\ker(A)$. 

Now the Frölicher-Nijenhuis bracket of the vector valued 1-form $P$ is of the form
$$ [P,P] = 2R + 2\bar R,\quad R(X,Y)= P[(Id-P)X,(Id-P)Y], \quad \bar R(X,Y)=(Id-P)[PX,PY]
$$
$R$, the curvature, is exacly the obstruction against integrability of the kernel of $P$.

$\bar R$, the cocurvature, is exactly the obstruction against integrability of the image of $P$, which we are interested in. 

So the condition is $\bar R=0$ which can be written as 
$$
0=P^*[P,P]=[P,P].(P\times P):\Lambda^2TM \to TM.
$$<|endoftext|>
TITLE: Why is a ring called a "ring"?
QUESTION [23 upvotes]: Why is a ring called "ring" (or Zahlring in German)?  There seems to (naive) me nothing more ring-like to a ring than there is to a group or a field.  I am particularly interested to learn why the word "ring" seemed appropriate to the founders.  Thanks for educating me!

REPLY [12 votes]: Ring is also an outdated German word for association or coalition - there is probably a better translation, but it means something along those lines. I would guess Hilbert was thinking of this meaning of the word, not the round shiny object usually seen on fingers. 
There is also the word Verband for some algebraic structure http://de.wikipedia.org/wiki/Verband_%28Mathematik%29 for which I do not know the english name, but this notion probably predates ring. A Verband is also a word for association; nowadays, its mainly used in sports - FIFA, UEFA, IOC are Verbände (to increase the confusion, Verband is also the German word for bandage). One might speculate that Hilbert looked for a similar, but different word to describe a similar, but different algebraic structure.<|endoftext|>
TITLE: Applications of the Giry monad in probability and statistics
QUESTION [17 upvotes]: In another thread, I asked about the $M$ endofunctor on the category $\operatorname{Meas}$ of measurable spaces, which sends a space $X$ to its space of measures $M(X)$. 
Will Sawin described the monad structure of this endofunctor in an answer, and in the comments thread Tom Leinster brought up the Giry monad, saying, "it's almost inconceivable that there could be two different reasonable monad structures on this endofunctor, so I bet it's the same." 
Here's the citation to Giry's original paper:

Michèle Giry, A categorical approach to probability theory. In B. Banaschewski, editor, Categorical Aspects of Topology and Analysis, Springer LNM 915, 1982. 

It is clear that monads are very useful in general, but I am not familiar enough with the concept to see how they are useful in this context. Fortunately, as of December 2012, there are 167 citations to Giry's paper on Google Scholar, so clearly many researchers have already recognized her work. There is also a discussion on the nLab page on probability theory.
For the benefit of future researchers, I've created this community wiki thread to aggregate possible applications of the Giry monad in probability and statistics. My hope is that this thread might be a place for the structuralist and probability communities to come together and learn from each other.
If you see any interesting applications of the Giry monad, please post them here.

REPLY [6 votes]: There is paper on the Arxiv, A categorical foundation for Bayesian probability by Culbertson and Sturtz. The paper contains also a very nice discussion of related literature.<|endoftext|>
TITLE: Canonical form of a general Bilinear Form
QUESTION [6 upvotes]: There is a simple canonical form of a symmetric and antisymmetric bilibear forms. Is there a canonical form for a general bilinear form?

REPLY [3 votes]: People in matrix analysis would call this a "canonical form under congruence". Take a look at http://arxiv.org/abs/0709.2473; the solution is stated there.<|endoftext|>
TITLE: Model for the (infinity,1)-category of functors preserving certain homotopy limits
QUESTION [9 upvotes]: This question is a follow up to: Model for the (infinity,1)-category of (homotopy-)limit preserving functors.

Warm-up Question: Given a simplicial model category $M$, what model category models the $(\infty, 1)$-category of presheaves of spaces on the $(\infty,1)$-category associated to $M$? 

I'm skeptical that the projective/injective model structures on simplicial presheaves on $M$ achieve this goal because they don't seem to use the weak equivalences of $M$ at all. (Although now that I think about maybe SSet-enriched functors "see" the weak equivalences in $M$.)
I'll use $N^{hc}(M^{cf})$ to denote the homotopy-coherent nerve of the simplicial subcategory spanned by the fibrant-cofibrant objects, ie the $(\infty,1)$-category associated to $M$.

Question: Given a simplicial model category $M$ and a fixed diagram category $D$, what model category models the $(\infty,1)$-category of functors from $N^{hc}(M^{cf})$ to Spaces which preserve homotopy limits indexed by $D$?

I was hoping the answer would look something like the following. Denote by Fun(M,SSet) the model category which answers the warm-up question, and by S the collection of natural transformations {F(hlim X) ---> hlim FX } where S ranges over $F:M \to \textrm{Spaces}, X: D \to M$. Then the model category of D-shaped homotopy limit preserving functors from $M$ to Spaces is modeled by the (right?) Bousfield localization of Fun(M,SSet) by S.
If you do answer the question the way I hoped, please say something mildly conciliatory about the fact that S seems too big.

REPLY [3 votes]: You should begin with the projective model structure on the category of small simplicial functors from $M^{op}$ to simplicial sets due to Chorny and Dwyer http://arxiv.org/abs/math/0607117. This models the $\infty$-category of small functors from the underlying $\infty$-category of $M^{op}$ to spaces. Since $M$ is usually large, you must restrict to small functors in order to get small mapping spaces in the $\infty$-category of presheaves on $M$. Then you should continue as in my answer to 
Model for the (infinity,1)-category of (homotopy-)limit preserving functors.<|endoftext|>
TITLE: What do correlation functions compute in CFT? 
QUESTION [14 upvotes]: I would like to understand what correlation functions compute in Conformal Field Theory in mathematics. Let me begin with basic definitions. We define a free boson field $\phi(z)$ as a formal power series
$$
\phi(z)=q+a_0\log(z)-\sum_{n\ne0}\frac{a_n}{n}z^{-n},
$$
where $q,a_n$ for $n\in \mathbb{Z}$ are operators satisfying relations
$$
[a_n,q]=\delta_{n,0}, \ \ \ [a_m,a_n]=m\delta_{m+n,0}. 
$$
Here these operators acts on the $space$ generated by $vacuum$ vector $|0\rangle$ with properties 
$$
a_n |0\rangle=0\ (n\ge0), \ \ \langle0|q=\langle0|a_n=0\ (n<0), 
$$
where $\langle0|$ is the dual of the vacuum, i.e. $\langle0|0\rangle=1$. 
We then define the current $J(z)$ of $\phi(z)$ as a formal derivative
$$
J(z)=\sum_{n\in \mathbb{Z}}a_nz^{-n-1}. 
$$

Here is my question. In CFT one is interested in $correlation$ $functions$ such as
  $$
\langle0|J(z_1)J(z_2)|0\rangle, \ \ \langle0|J(z_1)J(z_2)J(z_3)J(z_4)|0\rangle
$$
  What do they compute in this context? 

In physics, correlation function like $\langle0|\phi(x)\phi(y)|0\rangle$ computes the possibility of the field $\phi(x)$ to become $\phi(y)$ etc where $x,y$ are coordinate in for example Minkowski space. How should one interprete the correlation function in our case?

REPLY [9 votes]: I'm not sure exactly what kind of information you want, and CFT is an enormous subject, but here is some information
on the physical interpretation of the complex coordinates and correlation functions along with an example of their
mathematical interpretation in a special CFT. 
An ordinary refrigerator magnet contains a ferromagnetic material. The atoms in such a material have electrons which act as tiny magnets and they have an interaction between them (of quantum mechanical origin) which makes the spin/magnetic moments of the
electrons have lower energy when they are aligned.  This alignment produces a macroscopic magnetic field. I am simplifying here, because in real materials this interaction only operates over short distances and one actually forms domains of aligned spins with
the domains oriented randomly and a net magnetization is produced only by subjecting the material to an external magnetic field
which aligns the domains.
If one heats up such a ferromagnet then at some temperature the random thermal motion overcomes the tendency to align and
the macroscopic magnetic field goes to zero. The transition point between the state with net magnetization and the state with zero
magnetization is known as a second order critical point. The behavior of phase transitions  between different states (magnetized vs. unmagnetized, or water vs. ice or liquid vs. gas etc. ) has been one of the central topics in condensed matter physics for many years.  Various simplified models have been invented to try to understand such behavior analytically. The simplest of these is the
Ising model consisting of spins which take the value $\pm 1$ living on a two-dimensional lattice with nearest neighbor interactions between spins. There are many more complicated models, one of these is known as the Gaussian model because of the Gaussian weight used to define the probability distribution for spins. These models at the critical point have fluctuations on all length scales and one can take a continuum limit. This continuum limit is a conformal field theory. If the model is defined in two spatial dimensions then the continuum limit is a two-dimensional conformal field theory. The continuum limit of the Gaussian model is a conformal field theory with $c=1$ equivalent to the $c=1$ conformal field theory you have defined. The continuum limit of the Ising model is a $c=1/2$ conformal field theory consisting of a free Majorna fermion. The physical interpretation of the complex coordinates is simply that they are complex coordinates in the real two-dimensional plane describing the physical space on which the model is defined. The correlation functions measure the correlations between the microscopic spins at different spatial points at the critical point.
In two-dimensional conformal field theory the dependence on the coordinates of both the two-point and three-point correlation functions
is completely determined by conformal invariance, so the only interesting information is in the numerical coefficients appearing in
these correlation functions. For $c=1$ CFT there isn't any terribly interesting information in these correlation functions. However
for other CFT's there is more interesting information, including some of purely mathematical interest. For example, Frenkel, Lepowsky and Meurman constructed a $c=24$ CFT which has the Monster sporadic group as its automorphism group. This CFT has
196884 dimension 2 fields and the three point correlation function of these dimension 2 fields can be used to compute the
structure constants of the Griess algebra which was used in the original construction of the Monster.<|endoftext|>
TITLE: Explicit homeomorphism between Thurston's compactification of Teichmuller space and the closed disc
QUESTION [24 upvotes]: Thurston's celebrated compactification of Teichmuller space was first described in his famous Bulletin paper. Teichmuller space is famously homeomorphic to an open disc of some dimension (this can be seen using Fenchel-Nielsen coordinates), which is $6g-6$ for closed surfaces of genus $g\geq 2$.
Thurston embeds Teichmuller space in some infinite-dimensional natural space (the projective space of all real functions on isotopy classes of simple closed curves) and studies its closure there. The closure is realised by adding some points that correspond geometrically to some particular objects (called projective measured foliations). The added points are homeomorphic to a sphere of dimension $6g-7$ and the resulting topological space is just a closed disc, the new points forming its boundary.
The Bulletin paper contains almost no proofs. The only complete proofs I know for this beautiful piece of mathematics is described in the book Travaux de Thurston sur les surfaces of Fathi-Laudenbach-Ponearu (an english translation written by Kim and Margalit is available here). The homeomorphism between the space of projective measured laminations and the sphere $S^{6g-7}$ as explained there is clear and natural, it's obtained by re-adapting the Fenchel-Nielsen coordinates to the context of measured foliations.
The proof that the whole compactified space is homeomorphic to $D^{6g-6}$ is however more involved and less direct. First they study some charts to prove that we get a topological manifold with boundary, and that's ok. The compactification is thus a topological manifold with boundary homeomorphic to $S^{6g-7}$, whose interior is homeomorphic to an open ball of dimension $6g-6$. Are we done to conclude that the compactification is a closed disc? Yes, but only by invoking a couple of deep results: the existence of a collar for topological manifolds, and the topological Schoenflies Theorem in high dimension. That's the argument used in the book.

Is there a more direct description of the homeomorphism between Thurston's compactification and the closed disc $D^{6g-6}$?

Is there in particular a Fenchel-Nielsen-like parametrization of the whole compactification?

REPLY [8 votes]: One natural attempt to compactify Teichmuller space is by the visual sphere of the Teichuller metric. However, Anna Lenzhen showed that there are Teichmuller geodesics which do not limit to $PMF$ (in fact, I think it was known before by Kerckhoff that the visual compactification is not Thurston's compactification).
However, it was shown by Cormac Walsh that if one takes Thurston's Lipschitz (asymmetric) metric on Teichmuller space, and take the horofunction compactification of this metric, one gets Thurston's compactification of Teichmuller space. In fact, he shows in Corollary 1.1 that every geodesic in the Lipschitz metric converges in the forward direction to a point in Thurston's boundary. I think this gives a new proof that Thurston's compactification gives a ball.
As Misha points out, it's not clear that the horofunction compactification is a ball.
Another approach was given by Mike Wolf, who gave a compactification in terms of harmonic maps, in The Teichmüller theory of harmonic maps, and showed that this is equivalent to Thurston's compactification (Theorem 4.1 of the paper). Wolf shows that given a Riemann surface $\sigma \in \mathcal{T}_g$, there is a unique harmonic map to any other Riemann surface $\rho \in \mathcal{T}_g$ which has an associated quadratic differential $\Phi(\sigma,\rho) dz^2 \in QD(\sigma)$ ($QD(\sigma)$ is naturally a linear space homeomorphic to $\mathbb{R}^{6g-6}$). Wolf shows that this is a continuous bijection between $\mathcal{T}_g$ and $QD(\sigma)$, and shows that the compactification of $QD(\sigma)$ by rays is homeomorphic to Thurston's compactification $\overline{\mathcal{T}_g}$ in Theorem 4.1. I skimmed through the proof, and as far as I can tell the proof of the homeomorphism does not appeal to the fact that Thurston's compactification is a ball, so I think this might give another proof that it is a ball.

REPLY [6 votes]: I am not aware of a single Fenchel-Nielsen type parameterization as you ask for, and I'm not sure there can be one, because even on the boundary sphere the manner in which Fenchel-Nielsen coordinates are "re-adapted" does not produce a single coordinate system. The way that proof gets coordinates for the boundary sphere does start with a pants decomposition of the surface, as do Fenchel-Nielsen coordinates. And one does get a single coordinate system for the open subset of the boundary sphere which has nontrivial intersection number with each pants curve. But then one has to patch in additional coordinate charts to cover the closed subset of measure foliations that have zero intersection number with one or more pants curve. The proof does demonstrate that these coordinates can be patched together in such an explicit way that one can see the homeomorphism to a sphere, but nonetheless one is still patching things up.
Edit: I also recall that in one of his very earliest writings on this topic, Thurston gave a different proof, certainly not explicit, that the boundary is a sphere. Namely, from the existence of a pseudo-Anosov homeomorphism $\phi$, which acts with attractor--repeller dynamics, one gets a covering by two open sets homeomorphic to Euclidean space: for any neighborhood $U_+$ of the attracting fixed point and any neighborhood $U_-$ of the repelling fixed point there exists $n>0$ such that $\phi^n(U_-)$ and $U_+$ cover the boundary. It follows that the boundary is homeomorphic to a sphere. I posted this question to verify that the same was true for manifolds with boundary, and so the same proof works for compactified Teichmuller space, as Thurston undoubtedly knew: the action of a pseudo-Anosov homeomorphism on compactified Teichmuller space also has attractor-repeller dynamics, and so it is covered by two manifold-with-boundary coordinate charts, and so it is homeomorphic to a closed ball.<|endoftext|>
TITLE: Effective vanishing of the Schwarzian Derivative
QUESTION [12 upvotes]: Recall for any complex analytic function $f:\mathbb{D}\to \mathbb{C}$
 the Schwarzian derivative of $f$ is
$$
S(f)=\frac{f'''}{f'}-\frac{3}{2} \left( \frac{f''}{f'}\right)^2.
$$
It's well known that $S(f)\equiv 0$ if and only if $f$ is a fractional linear transformation, i.e. 
$$
f(z)=\frac{a z+b}{cz+d}.$$
I was wondering if there was an effective version of this result.  In other words a result that said that if $S(f)$ was "small" in some sense then $f$ was "close" in some, possible different, sense to a fractional linear transformation.  
Of course a major part of the question is deciding what are appropriate notions of "small" and "close".  I'd be particularly interested in those that allow $S(f)$ to have poles -- i.e. weaker than $L^\infty$ bounds.
I'm (vaguely) aware of Nehari's work on univalent functions (e.g. if $|S(f)|\leq 2$ then $f$ is univalent) though am not sure how related this is to my question.

REPLY [12 votes]: Here is a revised and somewhat expanded version of my answer, with a preparatory 'toy version' to help orient the reader.
A simple warmup problem:  Before discussing a quantitative variant of the Schwarzian, let me describe the overall idea in a simpler case:  Deciding how close two nonconstant meromorphic functions on a Riemann surface are to being constant multiples of each other.

Let $D$ be a connected Riemann surface, and let $f$ and $g$ be nonconstant meromorphic functions on $D$.  They will be constant multiples of each other if the meromorphic $1$-form $\omega = f\ dg - g\ df$ vanishes identically.  However, this is an exact statement, and it is hard to see how a bound on the 'size' of $\omega$ (assuming that it doesn't vanish outright) could say anything quantitative about how 'close' $f$ and $g$ are to being constant multiples of each other, especially since $f$ and $g$ could have poles (and $\omega$ could have poles, too, for that matter).

As an alternative, consider the singular (i.e., possibly degenerate) conformal metric 
on $D$ defined by
$$
ds^2_{f,g} = \frac{4\ (f\ dg - g\ df)\circ\overline{(f\ dg - g\ df)}}
               {\bigl(|f|^2+|g|^2\bigr)^2}\ .
$$
This is the singular metric induced on $D$ by pullback under the holomorphic map $H_{f,g}:D\to\mathbb{CP}^1$ defined by $H_{f,g} = [f,g]$, where the metric on $\mathbb{CP}^1$ is the familiar Fubini-Study metric with constant Gauss curvature 1:
$$
d\sigma^2 = \frac{4\ (z\ dw - w\ dz)\circ\overline{(z\ dw - w\ dz)}}
               {\bigl(|z|^2+|w|^2\bigr)^2}\ .
$$

Note that $ds^2_{f,g}$ vanishes identically when $\omega$ does but never has poles, so it's easier to define and measure its 'size'.  Most importantly, a bound on the size of $ds^2_{f,g}$ implies a bound on the size of the image of $H_{f,g}$.  For example, if $R>0$ is the diameter of $D$ under the metric $ds^2_{f,g}$, then the image of $H_{f,g}$ fits into a disk of radius at most $R$ in $\mathbb{CP}^1$.  Obviously, the smaller $R$ is, the closer $H_{f,g}$ is to being a constant map, i.e., the closer the ratio of $f$ to $g$ is to being constant.  Meanwhile, there are many ways to estimate the diameter of $D$ under the metric $ds^2_{f,g}$; for example, one could have a pointwise bound of the ratio of this metric to another 'reference' metric whose diameter is already known.

Now, the noncanonical thing about this strategy is that there is more than one candidate for the Fubini-Study metric on $\mathbb{CP}^1$.  However, these are all equivalent under automorphisms of $\mathbb{CP}^1$, so they form a family parametrized by $\mathrm{SL}(2,\mathbb{C})/\mathrm{SU}(2)$ (which is, of course hyperbolic $3$-space).  Moreover, the ratio of any two of them is bounded, so, in a sense, switching to a different member of the family will produce a quantitative measurement that is not essentially different.

A Quantitative version of the Schwarzian:  With the above in mind, consider this geometric way to think about the problem of quantifying the Schwarzian:  To say that $f$ is a linear fractional function of $z$ is to say that there are constants $(a,b,c,d)$ with $ad-bc\not=0$ such that 
$$
c\ z f + d\ f - a\ z - b = 0.
$$
In other words, one is asking whether the mapping $H_{f,z}:\mathbb{D}\to\mathbb{P}^3$ defined by
$$
H_{f,z} = [1, f, z, fz]
$$
has image in a hyperplane $\mathbb{P}^2\subset\mathbb{P}^3$.  The condition that $f$ not be constant is what keeps this map from going into a line in $\mathbb{P}^3$.

Meanwhile, note that the image of $H_{f,z}$ always lies in the quadric $\mathbb{Q}\subset\mathbb{P}^3$ defined by $X_0X_3-X_1X_2=0$, and that this quadric is biholomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ in the obvious way: $\bigl([a_0,a_1],[b_0,b_1]\bigr)\mapsto [a_0b_0,a_1b_0,a_0b_1,a_1b_1]$.  In particular, we should be thinking of $\mathbb{P}^3$ as the projectivization of the vector space $V\simeq\mathbb{C}^4$ of $2$-by-$2$ matrices endowed with the quadratic form $X_0X_3-X_1X_2$ that is simply the determinant.  For use below, I will also fix a volume form on $V$.  That way, one can define a triple cross product of vectors $v_1,v_2,v_3\in V$ by letting $v_1\times v_2\times v_3\in V$ be the vector that satisfies
$$
(v_1\times v_2\times v_3)\cdot w = \det(v_1\wedge v_2\wedge v_3\wedge w)
$$
for all $w\in V$.

Now, let $D$ be any connected Riemann surface and let $f$ and $g$ be nonconstant meromorphic functions on $D$.  Define a holomorphic curve $H_{f,g}:D\to \mathbb{Q}\subset\mathbb{P}^3$ by
$$
H_{f,g} = [1, f, g, fg]\ .
$$
Then this curve does not have image in a line (since $f$ and $g$ are not constant).  The condition that $H_{f,g}$ have image lying in a plane is that the relative Schwarzian $\mathsf{S}(f,g)$ should vanish, where
$$
\mathsf{S}(f,g) = \left(
\frac{f'''}{f'}-\frac32\frac{(f'')^2}{(f')^2}
-\frac{g'''}{g'}+\frac32\frac{(g'')^2}{(g')^2}\right)\ dw^2
$$
and where the primes denote differentiation with respect to any local coordinate $w$.  (It is easy to verify that the meromorphic quadratic differential $\mathsf{S}(f,g)=-\mathsf{S}(g,f)$ is defined independent of the choice of local coordinate $w$.)

To get a quantitative sense of how close $H_{f,g}$ is to lying in a plane, one needs a quantitative sense of how close the 'normal' map
$$
K_{f,g} = [{\hat H}_{f,g}\times {\hat H}'_{f,g}\times {\hat H}''_{f,g}]: D\to\mathbb{P}^3
$$
is to being constant.  (Here, I am taking a (local if necessary) meromorphic lifting 
${\hat H}_{f,g}:D\to V$. This lifting is unique up to a multiple, and this multiple goes away under projectivization.  Also, the choice of local parameter $w$ used to compute the derivatives does not matter, so that $K_{f,g}$ is well-defined.)

To get a sense of how close $K_{f,g}$ is to being constant, fix some metric on $\mathbb{P}^3$; for the purposes of this argument, say we fix a Fubini-Study metric with Kähler form $\Omega$ on $\mathbb{P}^3$.  (This is where we are introducing symmetry breaking, since there is no canonical metric on $\mathbb{P}^3$, but rather a family of Fubini-Study metrics, all differing by projective transformations, so the moduli space of such is $\mathrm{SL}(4,\mathbb{C})/\mathrm{SU}(4)$.  One can do a little better than this because one can choose one that is invariant under the maximal compact in the automorphism group of the inner product, and that smaller moduli space is $\mathrm{SO}(4,\mathbb{C})/\mathrm{SO}(4)$.)

At any rate, consider the nonnegative real $(1,1)$-form on $D$ defined by 
$$
\Omega_{f,g} = (K_{f,g})^*(\Omega).
$$
This $(1,1)$-form vanishes identically if and only if $H_{f,g}$ has image lying in a plane, otherwise, it vanishes only at isolated points.  Indeed, when it doesn't vanish identically, relative to a local coordinate $w$ on $D$, it takes the form
$$
\Omega_{f,g} = \frac{i}{2}|w|^{2k}p(w,\bar w)\ dw\wedge d\bar w.
$$
where $p$ is nonvanishing and $k\ge0$.  In particular, the (possibly degenerate) metric given in local coordinates by
$$
ds^2_{f,g} = |w|^{2k}p(w,\bar w)\ dw\circ d\bar w
$$
is well-defined on $D$.  Calculation shows that $ds^2_{f,g}$ is essentially the square norm of $\mathsf{S}(f,g)$ times a factor that 'clears the denominators', so that $ds^2_{f,g}$ does not have 'poles'.  Thus, in a natural (but not canonical) sense, a bound on $ds^2_{f,g}$ is equivalent to a bound on $\mathsf{S}(f,g)$.

In particular, if you were to take the 'norm' on $\mathsf{S}(f,g)$ to be the diameter of $D$ with respect to the singular metric $ds^2_{f,g}$ (which is well-defined), then that norm would give you an upper bound on the diameter of the image of $K_{f,g}$, i.e., it would say that, for example, the distance from $K_{f,g}(p)$ to $K_{f,g}(p_0)$ is bounded, so that the image of $H_{f,g}$ would lie at most a bounded distance from the plane that $K_{f,g}(p_0)$ defines.

In geometric terms, this is exactly the kind of bound you were asking for.  The only difference is that you aren't using a norm directly on the Schwarzian, but on a closely related quantity $ds^2_{f,g}$, one that depends on an arbitrary choice (so it's not quite canonical) but that has better compactness properties (so that you can actually get estimates).<|endoftext|>
TITLE: Constructible models of New Foundations?
QUESTION [10 upvotes]: Hi all! Is there anything like Gödel's constructible universe for New Foundations?
More precisely, I would like a process for taking a model $M$ of NF, and using it to build a model $L \subseteq M$ of NF with the property that every set in $L$ is defined by a (stratified) first-order formula with quantifiers ranging over $M$. (Edited; see the comments for a discussion of some issues surrounding this definition.)
Anything not exactly that, but along those lines, would also be of interest. I would also be interested in hearing about such results for non-well-founded set theories other than NF. Has this been done? Is it possible?
I'm wondering because I am trying to build this sort of constructible model for a naïve set theory that I am studying. I haven't figured out how to apply the methods used for models of ZF to models of naïve set theory. I'm guessing that similar issues might apply in working with NF, because both theories are primarily distinguished by their use of a powerful comprehension axiom.
Thank you!

REPLY [7 votes]: You can't do it for NF, but there is a good notion of constructible model of the theory CUS of Church, that has a universal set.   But that's really just a trick of the light, since the big sets of CUS are magicked into existence by a coding trick.  None of that is in print, so there is nothing i can cite.  Sorry!<|endoftext|>
TITLE: Why is the Leibniz rule a definition for derivations?
QUESTION [26 upvotes]: In differential geometry, the tangent space is defined as a generalization of directional derivatives, which in turn are defined as functionals following Leibniz's product rule.
I understand all the proofs, but what is the intuition behind choosing the product rule to capture the notion of derivatives, rather than any other property?

REPLY [7 votes]: The derivation property
is the simplest possible property; all other are elaborations:
Let $M$ be a smooth manifold. Then we have in the category of commutative $\mathbb R$-algebras with unit (see p 296  of the reference below
for more details):

$Hom(C^\infty(M,\mathbb R),\mathbb R) = M$
$Hom(C^\infty(M,\mathbb R),\mathbb D) = TM$ where $\mathbb D$ (generated by $1,\epsilon$ with $\epsilon^2=0$) is the algebra of study numbers.
$Hom(C^\infty(M,\mathbb R),\mathbb D\otimes \mathbb D) = T(TM)$
For any finite dimensional commutative $\mathbb R$-algebra $A$ we have:
$Hom(C^\infty(M,\mathbb R),A) = Hom(C^\infty(M,\mathbb R),W(A))$ is a smooth manifold and describes a product preserving covariant endofunctor $T_{W(A)}$ on the category of smooth manifolds and smooth mappings. Here $W(A)$ is the subsalgebra of $A$ generated by all idempotent and nilpotent elements.  Up to some covering space phenomena all product preserving functors are of this form. $W(A)$ is called the Weil algebra of $A$. 
Then $T_A\circ T_B = T_{A\otimes B}$; thus composition of product preserving functors is naturally commutative.
Now consider only algebras $A$ with $A=W(A)$, and with exactly one idempotent 1 (for simplicity's sake). Then $A=\mathbb R.1\oplus N$ for a nilpotent ideal $N$.
Moreover, $T_A(M)\to M$ is a fiber bundle. 
Elements $\xi_x$ in the fiber over $x\in M$ correspond exactly to $\mathbb R$-linear mappings
$C^\infty(M,\mathbb R) \to N$ satisfying 
$\xi_x(f.g) = \xi_x(f).g(x) + f(x).\xi_x(g) + \xi_x(f).\xi_x(g)$.
Smooth section $\xi$ of this bundle correspond exactly to 
$\mathbb R$-linear mappings $C^\infty(M,\mathbb R) \to C^\infty(M,\mathbb R)\otimes N$ such that $\xi(f.g) = \xi(f).g + f.\xi(g) + \xi(f).\xi(g)$ (expansion property).
The Leibnitz rule is the simples possible expansion property.
Addition of vector fields is replaced by a non-commutative nilpotent group multiplication. 

Ivan Kolár, Jan Slovák, Peter W. Michor: Natural operations in differential geometry. Springer-Verlag, Berlin, Heidelberg, New York, (1993), (pdf)<|endoftext|>
TITLE: n-categorical description of Chern classes
QUESTION [27 upvotes]: The Chern classes of a rank $n$ vector bundle on $X$ are obtained from composing the associated classifying map in $[X, BU(n)]$ with the maps $BU(n) \to B^{2i} \mathbb{Z}$ corresponding to the cohomology of $BU(n)$, thus giving elements in the cohomology groups $H^{2i}(X,\mathbb{Z})$.
But one could instead write those maps as $BU(n) \to B^{2i-1} U(1)$; and this corresponds to associating a $(2i-1)$-circle bundle to a vector bundle. For $i = 1$, for instance, this corresponds to sending a vector bundle to its top exterior power, the determinant line bundle.
What is then the interpretation of the higher Chern classes in this geometric setting? These should associate to the original rank $n$ vector bundle some circle $(2i-1)$-bundles (e.g. principal $B^{2i-2}U(1)$ $(2i-1)$-bundles), for $1 \leqslant i \leqslant n$.
This should mirror the algebraic side with symmetric polynomials; using $e_1 = x_1 + \ldots + x_n$ we have the line bundle $L = L_1 \otimes \ldots \otimes L_n$ (corresponding to the Chern roots); for other elementary symmetric polynomials $e_i$ we should be able to build a corresponding $(2i-1)$-line bundle directly realising the $i$-th Chern class. (Perhaps it would be more natural to instead consider Schur classes, and hopefully the relationship of these $(2i-1)$-line bundles with Schur functors is elucidated.)
Is there then an obstruction-theoretic picture, when trying to build back up the original vector bundle from these successive $(2i-1)$-line bundles?

REPLY [4 votes]: The answer will depend on your realisation of a  $k$-circle bundle.  In the case of $i=2$ (second Chern class) there are results associating to any principal $G$-bundle a bundle $2$-gerbe. See: 
Bundle Gerbes for Chern-Simons and Wess-Zumino-Witten Theories.
Alan L. Carey, Stuart Johnson, Michael K. Murray, Danny Stevenson and Bai-Ling Wang.
Communications in Mathematical Physics, 159 (3) (2005), 577-613
math.DG/0410013
and the references there in to Danny Stevenson and Stuart Johnson's PhD theses and papers.  Of course you have to be happy that a $3$-circle bundle is a $2$-gerbe.
More generally you might find something useful in:
P. Gajer Geometry of Deligne cohomology, Invent. Math. 127 (1997), 155-207.
which gives a realisation of principal $B^k \mathbb{C}^*$ bundles which are another possible way of realising $(k+1)$-circle bundles or at least mathematical objects determined by a characteristic class in degree $H^{k+1}(M, \mathbb{Z})$.  There is a nice inductive classifying theory and a simplicial realisation of these spaces.<|endoftext|>
TITLE: Obstruction sheaf is a vector bundle when the moduli space is non-singular?
QUESTION [6 upvotes]: I am working on some basic of Gromov-Witten theory and stuck in understanding obstruction bundle. Recall that a perfect obstruction theory on a scheme or stack $M$ due to Behrend and Fantechi is a moprhism $\phi:\mathcal{E}\rightarrow \tau_{\ge -1}L_M$ in $D^{[-1,0]}(M)$ satisfying some conditions. Taking the first cohomology of the dual of the two-term complex $\mathcal{E}$, we get so-called obscturction sheaf $Ob=h^1(\mathcal{E}^{\vee})$. 
Assume now that $M=\overline{M}_{g,n}(X)$ for some smooth variety $X$ with the usual perfect obstruction theory on it. My question is, why the obstruction sheaf forms a vector bundle on $M$ when the moduli space $M$ is non-singular? 
I know that the modul space $M$ is intuitively obtained by cutting out the deformation space by $\dim Ob$ many equations, but I am not really convinced by this argument.

REPLY [6 votes]: The point is that if the moduli space is non-singular, then the tangent sheaf $h^0(\mathcal{E}^\vee)$ is locally free and so the map $E^{-1}\to E^0$ must be of constant rank. This implies that the cokernel is also locally free. The difference between the dimensions of fibers of the tangent sheaf and the obstruction sheaf is always constant --- it is the expected dimension of the moduli space. So if the dimension of the fibers of the tangent sheaf doesn't jump, then neither do the fibres of the obstruction sheaf.<|endoftext|>
TITLE: Polynomial bijection from ZxZ to Z?
QUESTION [26 upvotes]: It is known that the polynomial $f(n,m)=\frac{1}{2}(n+m)(n+m+1)+m$ defines bijection
$\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ (Put pairs of $\mathbb{N}$ into the semi-infinite matrix and count them by diagonals). Does there exist a polynomial bijection 
$\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$? The question is related to the open question 
about polynomial bijection $\mathbb{Q}\times\mathbb{Q}\to\mathbb{Q}$ here.

REPLY [10 votes]: It is an open problem. Maximal results about bijections from
$\mathbb N\times \mathbb N$, $\mathbb Z\times \mathbb N$, $\mathbb Z\times \mathbb Z$  to $\mathbb N$ are contained in

John S. Lew, Arnold L. Rosenberg,
  Polynomial indexing of integer lattice-points I. General concepts and quadratic polynomials, J. Number Theory 10 (1978) pp 192-214, doi:10.1016/0022-314X(78)90035-5.
Polynomial indexing of integer lattice-points II. Nonexistence results for higher-degree polynomials, J. Number Theory 10 (1978) pp 215-243,
  doi:10.1016/0022-314X(78)90036-7<|endoftext|>
TITLE: Applications of Iwasawa Theory
QUESTION [7 upvotes]: Iwasawa theory gives a formula for the power of $p$ dividing the class group of the $\mathbb{Q}(\zeta_{p^n})$ (where $\zeta_{p^n}$ is a primitive root of unity of exact order $p^n$) for sufficiently large $n$. (See, e.g., Theorem 2 of these notes.) More generally, one gets a similar result for arbitrary $\mathbb{Z}_p$ extensions of number fields. Mazur and Mazur and Rubin have studied the variation of the $p$ part of the Tate-Shafarevich group of an elliptic curve in $\mathbb{Z}_p$ extensions and the variation of the rank of an elliptic curve in $\mathbb{Z}_p$ extensions.
I find this somehow unsatisfying. The restriction to the study of the $p$ part of the ideal class group & Tate Shafarevich, and the restriction to the study of  $\mathbb{Z}_p$ extensions seem quite strong. 
Yet I've gotten the impression that Iwasawa theory is considered to be fundamental in number theory. So I feel as though I'm missing perspective on why Iwaswa theory is important.

What are some important applications
  of Iwasawa theory?

I'd also be happy with high-level philosophical comments.

REPLY [17 votes]: Aha, an excuse to quote chunks of my most recent grant proposal :-)
Iwasawa theory is heavily used in work on the BSD conjecture. For instance, the first positive result to be proved in the direction of BSD -- the Coates--Wiles theorem that analytic rank 0 implies algebraic rank 0 for elliptic curves over $\mathbf{Q}$ with complex multiplication -- was shown by using Iwasawa theory. This is a statement which has nothing obviously to do with Zp-extensions, on the face of it, although to be sure they are lurking in the proof. More generally, pretty much all the results on BSD that we now have (thanks to Kolyvagin, Rubin, Kato, Perrin-Riou, Kobayashi, etc...) use Iwasawa theory heavily.
This fits into a general philosophy which states that if $M$ is a motive over a number field $K$, the $L$-values $L(M, j)$ for integer values of $j$ encode information about the cohomology of $M$. Iwasawa theory provides a tool to attack such conjectures, by interposing a third object -- a p-adic L-function -- which one can (sometimes) relate both to the cohomology and to the L-values. This is bound up with the idea that the behaviour of $M$ over your original ground field $K$ might be quite complex, but making a tower of extensions $K = K_0 \subset K_1 \subset K_2 \subset \dots \subset K_\infty$ and taking a limit of objects defined over the $K_n$'s can serve to "smooth out" the behaviour, and then you prove things over $K_\infty$ and see what you can recover over $K$ by some kind of descent argument, splitting your problem into two hopefully easier chunks. (For more philosophy along these lines, see Colmez's Bourbaki expose on p-adic L-functions, http://www.math.jussieu.fr/~colmez/851bourbaki.pdf)<|endoftext|>
TITLE: Extreme rays in the cone of (semi)metrics
QUESTION [5 upvotes]: How many extreme rays are there on the polytopal cone formed by all semimetrics on a set with $n$ elements?
Some background. Given a set $X$ with $n$ elements, the set of all semimetrics 
$d:X \times X \rightarrow [0,\infty)$ can be seen as the cone of symmetric matrices $(d_{i,j})$
with zeroes on the diagonal and satifying the system of linear inequalities $d_{i,j} \geq 0$,
$d_{i,j} + d_{j,k} - d_{i,k} \geq 0$. The polytopal cone defined by this finite system of inequalities can also be described (at least in principle) by exhibiting its extreme rays. Some extreme semimetrics are easy to describe. Here is a class of examples: if $Y$ is a subset of $X$, define the cut semimetric $d_Y$ by setting the distance between two points to be equal to $1$ if one of the points is in $Y$ and the other in its complement, otherwise the distance between the two points is zero. Another example, given by Avis, of an extreme metric is the length metric of the graph $K_{3,2}$ on the cone of semimetrics on a set with $5$ elements.
More elaborate question. It may be hard to know exactly how many extremal rays there are on the semimetric cone, but I'm interested in a good estimate (some asymptotic estimate would also be nice). I'm also interested in knowing about classes of examples of extremal semimetrics other than the cut semimetrics and the examples given by Avis in his 1980 paper On the extreme rays of the metric cone.

REPLY [4 votes]: There has been quite a bit of work done since 1980 on this. Did you check the book by M. Deza and M. Laurent "Geometry of cuts and metrics", Springer 1997 ?
There was a quite a bit of computer search done to go beyond $n=5$ in Avis' paper. E.g. I think here you can find results for $n=7$. There are more links on a page maintained by A.Deza.
I admit I don't recall asymptotic results, although it has been a while.<|endoftext|>
TITLE: Is every functor inducing a homotopy equivalence a composition of adjoint functors?
QUESTION [8 upvotes]: It was asked here whether every functor is a composition of adjoint functors. The answer is no, because all adjoint functors induce homotopy equivalences on the nerve, and we can construct functors that do not induce homotopy equivalences.
My question is the following: can all functors inducing homotopy equivalences on the nerve be expressed as compositions of adjoint functors?

REPLY [12 votes]: The answer is no.  Let $C$ be a category such that the unique map from $C$ to the terminal category is a composition of $n$ adjoints.  Then $C$ has an object $x_0$ such that every other object of $C$ can be connected to $x_0$ by a zigzag of length at most $n$; this is easy to prove by induction on $n$.  
In particular, let $R$ be the "infinite zigzag", the unique poset such that $|BR|$ is homeomorphic to $\mathbb{R}$.  Then $BR$ is contractible, but the unique map from $R$ to the terminal category cannot be a composition of adjoints.  Note, however, that this map is a transfinite composition of adjoints (of length $\omega$).  It seems plausible to me that any functor between (small) categories which is an equivalence on nerves could be a transfinite composition of adjoints.<|endoftext|>
TITLE: Old books still used
QUESTION [90 upvotes]: It's a commonplace to state that while other sciences (like biology) may always need the newest books, we mathematicians also use to use older books. While this is a qualitative remark, I would like to get a quantitative result. So what are some "old" books that are still used?
Coming from (algebraic) topology, the first things which come to my mind are the works by Milnor. Frequently used (also as a topic for seminars) are his Characteristic Classes (1974, but based on lectures from 1957), his Morse Theory (1963) and other books and articles by him from the mid sixties. 
An older book, which is sometimes used, is Steenrod's The Topology of Fibre Bundles from 1951, but this feels a bit dated already. Books older than that in topology are usually only read for historical reasons.
As I have only very limited experience in other fields (except, perhaps, in algebraic geometry), my question is:

What are the oldest books regularly used in your field (and which don't feel "outdated")?

REPLY [2 votes]: P.A. MacMahon, Combinatory analysis, vols 1 and 2, Cambridge University Press, 1915–16.<|endoftext|>
TITLE: A Problem  about partitioning $S^2$
QUESTION [8 upvotes]: Question: Can the 2-dimensional sphere $S^2$ be partitioned into four nonempty sets such that every circle in $S^2$  passes through just three  of these four sets?
Here, "just three" means "exactly three", circle is in the ordinary sense, i.e. round circle (or say 1-sphere), not necessarily great circle. If "just three" means "at most three" as Alexandre Eremenko supposed, then the answer is yes (there are a lot of examples). For example, see the example just given by Lee Mosher.
EDIT. I prove that if such partitioning exists, then these four subsets are all dense in $S^2$. So, it seems that we cannot find a "trivial" partition.

REPLY [5 votes]: I suppose you mean circles in the literal sense, round circles.
I suppose "just three" means "at most three". The answer is "yes".
Let us identify your sphere with the extended
complex plane via stereographic projection, so that your
circles are straight lines or circles. Consider the following sets $A,B,C,D$.
Let $D=\{ \infty  \}$, $C=\{ 0 \}$, $A$ the set of points with argument
commensurable with $\pi$,
and $B$ the set of points with argument non-commensurable with $\pi$.
These sets partition the extended plane.
If a circle does not contain $D$, it intersects at most 3 sets.
If it contains $D$, it is a straight line. If this straight line
does not contain $C$, it intersects at most 3 sets.
If it contains $C$, it intersects either $C,A,D$ or $C,B,D$ but not all 4.
Sorry, do not know how to make curly braces {} with Mathjack.<|endoftext|>
TITLE: Splitting of the weight filtration
QUESTION [8 upvotes]: All varieties are over $\mathbb{C}$. Notions related to weights etc. refer to mixed Hodge structures (say rational, but I would be grateful if the experts would point out any differences in the real setting).
I am trying to get some intuition for the geometric meaning of/when to expect the weight filtration on the cohomology groups $H^i(X)$ of a variety to split. By the weight filtration splitting, I mean that the Hodge structure on each $H^i(X)$ is a direct sum of pure Hodge structures.
The simplest situation in which this happens is the classical one of smooth projective varieties. The next simplest situation I think of is the "smooth" being weakened to "mild singularities" (for instance, rationally smooth). So at least in the projective case I think of the "mixed" as encoding singularities. This also gels well with the construction of these Hodge structures using resolution of singularities. Are there other helpful perspectives?
Instead of fiddling with the "smooth" one can make the variety non-compact (but still smooth say). Examples: affine $n$-space, tori. Here I don't know how to think of or when to expect the weight filtration to split. Any intuition would be appreciated. The only rough picture I have is that this encodes information about the complement in a good compactification. But I don't find this particularly illuminating.
Generalizing affine space and tori is the situation of toric varieties for which the weight filtration always splits (thanks to a lift of Frobenius to characteristic $0$). In general when should one expect a splitting of the weight filtration to be given "geometrically" by a morphism (or say correspondence in the context of Borel-Moore homology)?
Related is the following: when should one expect the Hodge structure on each $H^i(X)$ to be pure (not necessarily of weight $i$). Here I am again more interested in weakening the "projective" rather than the "smooth".
At the risk of being even more vague, let me add some motivation from left field. There are several situations in representation theory where one expects/knows that the weight filtration on some cohomology groups splits (and is even of Hodge-Tate type). For instance, the cohomology of intersections of Schubert cells with opposite Schubert cells. However, the reasoning/heuristic has, a priori, nothing to do with geometry but more with the philosophy of "graded representation theory" (ala Soergel, see for instance his ICM94 address) I would love to have a geometric reason/heuristic for this. Pertinent to this is also the question of when should one expect the canonical Hodge structure on extensions between perverse sheaves of geometric origin to be split Tate? The only examples I know come from representation theory (see Section 4 of Beilinson-Ginzburg-Soergel's "Koszul duality patterns in representation theory").

REPLY [6 votes]: Here is a simple example to keep in mind. Let $\bar X$ be a smooth
projective curve, and let  $X=\bar X-\lbrace p_1, \ldots, p_n \rbrace$. 
The rational mixed Hodge structure on $H^1(X)$ is given by an extension
$$0\to H^1(\bar X)\to H^1(X)\to \mathbb{Q}(-1)^{n-1}\to 0$$
This  splits if and only if for each $i,j$ there exists  logarithmic  $1$-forms with
singularities only at $p_i$ & $p_j$ and rational periods.
By work of Carlson  [Extensions of mixed Hodge structures] the
obstructions are given by the   classes $p_i-p_j\in Jac(\bar X)\otimes
\mathbb{Q}$ in the Jacobian. In particular, these are usually nontrivial.
By contrast $H^1(X,\mathbb{R})$ does split because
such forms with real periods do exist.
However, for more complicated examples, the real mixed Hodge structure need
not split either. So in general, you shouldn't expect it. My intuition, however, is that
when the varieties "come from linear algebra" this is more common, but I don't have a precise statement or explanation.
I'm not sure if this answers your question. 

Addendum Here is another perspective which may or may not make the issue clearer.
The category of polarizable rational mixed Hodge structures is neutral Tannakian, so
it is the category of representations of a pro-algebraic group, what I would call the universal Mumford-Tate group $MT$. The split ones constitute  the subcategory of representations of  the quotient of $MT$ by its pro-unipotent radical.

REPLY [2 votes]: I don't have a general answer, but let me add some more examples.

For your second question, examples of smooth varieties with $H^i$ pure of the 'wrong' weight, a good example is the complement of an affine arrangement of hyperplanes in $\mathbf C^n$. In this case, the Hodge structure on $H^i$ is pure of type $(i,i)$. One way to see this is that the cohomology is generated as an algebra by $H^1$, and $H^1$ is spanned by logarithmic forms $$ \omega_H = \frac 1 {2\pi i} \mathrm d \log H,$$
where $H=0$ is the defining equation of one of the hyperplanes; the class of $\omega_H$ is of type $(1,1)$. A reference for this is Brieskorn's "Sur les groupes des tresses [d'après V.I. Arnol'd]". See also the simple and "motivic" proof in "Weights in cohomology groups arising from hyperplane arrangements" by Minhyong Kim, as well as near-simultaneous papers by Boris Shapiro and Gus Lehrer which present basically the same result.
Another example is when $X$ is an abelian variety, and $F(X,n)$ is the configuration space of $n$ points in $X$, i.e. the complement of the "big diagonal" in $X^n$. Then the mixed Hodge structure on $H^i(F(X,n))$ is always a direct sum of pure Hodge structures. A reference for this is Gorinov's preprint "Rational cohomology of the moduli spaces of pointed genus 1 curves" (on his webpage), Section 3. 
Let $Y(N)$ be the open modular curve parametrizing full level $N$ structures on elliptic curves. Then $H^1(Y(N))$ is a sum of pure Hodge structures, by the criterion in Donu Arapura's answer and the theorem of Drinfel'd and Manin.<|endoftext|>
TITLE: Manifolds with two coordinate charts
QUESTION [20 upvotes]: What is an early reference for the fact that if a compact, connected $n$-manifold $M$ is covered by two open sets homeomorphic to $\mathbb{R}^n$ then $M$ is homeomorphic to $S^n$?
And is it true that if $M$ is a compact, connected $n$-manifold with boundary, and if $M$ is covered by two open sets homeomorphic to $\lbrace(x_1,\ldots,x_n) \in \mathbb{R}^n  |  x_n \ge 0\rbrace$, then $M$ is a closed ball?

REPLY [21 votes]: I'll only discuss the first question (EDIT : Actually, I address the second question at the end).  As Agol pointed out in the comments, for $n \geq 5$ this is an easy consequence of Newman's 1966 proof of the Poincare conjecture in the topological category.
I don't know if it was explicitly stated earlier than this.  However, it can easily be derived from the main result of the paper
MR0126835 (23 #A4129)
Brown, Morton
The monotone union of open n-cells is an open n-cell. 
Proc. Amer. Math. Soc. 12 1961 812–814. 
In fact, this works in all dimensions (including $3$ and $4$).  
Brown's theorem is as follows.  Assume that $M$ is a topological $n$-manifold and that for all compact $K \subset M$, there exists some open set $U \subset M$ with $K \subset U$ and $U \cong \mathbb{R}^n$.  Then $M \cong \mathbb{R}^n$.  Brown's proof is clever, but completely elementary.
To get the desired result from this, assume that $X = U_1 \cup U_2$ with $U_i \cong \mathbb{R}^n$ and that $X$ is compact.  Let $\phi : \mathbb{R}^n \rightarrow U_1$ be a homeomorphism.  It is enough to prove that $X \setminus \{\phi(0)\} \cong \mathbb{R}^n$.  We will do this with Brown's theorem.  Consider a compact set $K \subset X \setminus \{\phi(0)\}$.  To verify Brown's criteria, it is enough to construct a homeomorphism $\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$ such that $\psi(K) \subset U_2$.     
For $r>0$, let $B(r) \subset \mathbb{R}^n$ be the ball of radius $r$.  The set $U_1 \setminus U_2$ is compact, so there exists some $R>0$ such that $U_1 \setminus \phi(B(R)) \subset U_2$.  Also, there exists some $\epsilon > 0$ such that $K \cap \phi(B(\epsilon)) = \emptyset$.  It is easy to construct a homeomorphism $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $f(B(\epsilon)) = B(2R)$ and $f(0)=0$ and $f|_{\mathbb{R}^n \setminus B(3R)} = \text{id}$.  We can therefore define a homeomorphism $\psi : X \setminus \{\phi(0)\} \rightarrow X \setminus \{\phi(0)\}$ by $\psi(p) = \phi \circ f \circ \phi^{-1}(p)$ for $p \in U_1 \setminus \{\phi(0)\}$ and $\psi(p) = p$ for $p \notin U_1$.  Clearly $\psi(K) \subset U_2$.

EDIT : Lee suggested that this might be able to address his second question too.  I thought a bit about it, and I believe that it can.  The key is the following "relative" version of Brown's theorem, which can be proven exactly like Brown's theorem.
Theorem : Let $(M,N)$ be a pair consisting of a topological $n$-manifold $M$ and a closed submanifold $N \subset M$.  Assume that for all compact $K \subset M$, there exists some open set $U \subset M$ such that $K \subset U$ and such that the pair $(U,U \cap N)$ is homeomorphic to the pair $(\mathbb{R}^n,\mathbb{R}^{n-1})$ (the second embedded in the standard way).  Then $(M,N) \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$.
To apply this, assume that $X$ is a compact manifold with boundary and that $X = U_1 \cup U_2$ with $(U_i,\partial U_i) \cong (\mathbb{R}^n_{\geq 0},\mathbb{R}^{n-1})$.  Double $X$ to get a closed manifold $Y$, and let $Y' \subset Y$ be the image of the boundary of $X$.  The open sets $U_i$ double to give an open cover $Y = V_1 \cup V_2$.  Letting $V_i' = V_i \cap Y'$, we have $(V_i,V_i') \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$.  Let $(M,M')$ be the result of deleting the image of $0$ in $(V_1,V_1')$.  It is enough to prove that $(M,M') \cong (\mathbb{R}^n,\mathbb{R}^{n-1})$, and this can be proven just like above.

Of course, Agol answered the second question first -- it follows from the topological Schonfleiss theorem applied to the double, which was proven by Brown in
MR0117695 (22 #8470b) Reviewed
Brown, Morton
A proof of the generalized Schoenflies theorem.
Bull. Amer. Math. Soc. 66 1960 74–76.
54.00 (57.00)
Mazur had earlier proven a weaker result.  This requires the sphere to be bicollared, but this holds.  Indeed, from the assumptions the sphere is locally bicollared, and Brown proved in
MR0133812 (24 #A3637)
Brown, Morton
Locally flat imbeddings of topological manifolds.
Ann. of Math. (2) 75 1962 331–341. 
that this implies that the sphere is bicollared.  See 
MR0267588 (42 #2490)
Connelly, Robert
A new proof of Brown's collaring theorem.
Proc. Amer. Math. Soc. 27 1971 180–182. 
for a super-easy proof of Brown's collaring theorem.<|endoftext|>
TITLE: Hochschild homology and change of non-ground ring
QUESTION [5 upvotes]: Let $k$ be a field, $R$ is a commutative algebra over $k$ and $A$ is an associative algebra over $R$. There is a morphism of commutative algebras $R \to T$. Is it possible to reduce calculation of Hochschild homology $HH_*(A\otimes_R T)$ (over basic field $k$) to $HH_*(A)$? I'm mostly interested in a situation when $T$ is separable over $R$, or even more specific $T=R/I$.

REPLY [4 votes]: According to the following paper:
Geller, S.; Weibel, C.: 
Étale descent for Hochschild and cyclic homology, 
Comment. Math. Helv. 66 (1991), no. 3, 368–388;
—theorem (0.1)— you need that $T$ is étale over $R$. This holds in your case only if the ideal $I$ is generated by an idempotent. Notice that "étale" = "unramified + flat". And the notion of unramifiedness is related to the condition of separable ring extension, but it depends on which of the several used definitions of "separable" you consider.<|endoftext|>
TITLE: Intuition for mean curvature
QUESTION [23 upvotes]: I would like to get some intuitive feeling for the mean curvature. The mean curvature of a hypersurface in a Riemannian manifold by definition is the trace of the second fundamental form.
Is there any good picture to have in mind, when dealing with it?

REPLY [3 votes]: Let $M$ be an oriented hypersurface in an oriented euclidean space $E$.
The normal curvature at a point $x_0\in M$ measures how much it is displaced, in the positive normal direction, with respect to the points of $M$ that are near it, in the same way that the Laplacian of a function measures how much the value of a function at a point differs from the average of the points that are near. 
In fact, if $M$ is the graph of a nearly constant function $f:\mathbb R^n\to\mathbb R$ (for example, the vibrating membrane of a drum at some instant) then the Laplacian of $f$ is a good approximation of the normal curvature of $M$.
More generally, if $M$ is possibly not nearly flat, then for any $x_0\in M$ you can find the hyperplane $\Pi$ that is tangent to $M$ at $x_0$, and find a parametrization $f:\Pi\to M$ (only defined on a neighbourhood of $x_0$) such that $p\circ f=id_\Pi$, where $p:E\to\Pi$ is the orthogonal projection. This parametrization shows that $M$ is locally the graph of a function $g=q\circ p$ (where $q$ is the projection on the orthogonal complement of $\Pi$) such that $dg(x_0)=0$. Then, the normal curvature of $M$ at $x_0$ equals $\Delta g(x_0)$.<|endoftext|>
TITLE: Direct product of rings
QUESTION [5 upvotes]: Is there an infinite family $\lbrace R_\alpha\rbrace_\alpha $ of rings (with identity $1\neq 0$) such that 
their direct product is a hereditary ring ?
I think the answer must be negative but i have no proof or counterexample yet.

REPLY [4 votes]: As @Jeremy Rickard has mentioned, the impossibility was shown by Osofsky in 1968. But it is interesting that there is another article dating back to 1968 (again!) that proves the same thing. See Cateforis, Sandomerski, The singular submodule splits off, J. Algebra, 10 (1968), 149-165, Theorem 4.1. This article was received by J. Algebra on November 20, 1967 while Osofsky's paper was received by Proc. Amer. Math. Soc. on July 10, 1967. I hope this information meets the expectations.<|endoftext|>
TITLE: Polynomial Rings 
QUESTION [14 upvotes]: Let $R$ and $S$ be non-zero rings with identity. Is it possible to have $R[x] \cong S[[x]]$ ?

REPLY [6 votes]: Thanks to Martin Brandenburg suggestion, if $R[x] \cong S[[x]]$ then their centers are isomorphic too. So without loose of generality we can assume that $R$ and $S$ are commutative. In commutative case we know $J(R[x]) = Nil(R[x])$. This means that elements in the Jacobson radical of $R[x]$ are all nilpotent. On the other hand $x \in J(S[[x]])$ and $x$ is not nilpotent. This shows that $R[x]$ and $S[[x]]$ can not be isomorphic.<|endoftext|>
TITLE: Singular values of sequence of growing matrices
QUESTION [25 upvotes]: I asked this question on math.stackexchange and haven't received an answer in two weeks, so I'm repeating it here.
Let 
$$
H=\left(\begin{array}{cccc}
0 & 1/2 & 0 & 1/2 \cr
1/2 & 0 & 1/2 & 0 \cr
1/2 & 0 & 0 & 1/2\cr
0 & 1/2 & 1/2 & 0
\end{array}\right),
$$
$K_1(\alpha)=\left(\begin{array}{c}1 \\\\ \alpha\end{array}\right)$ and consider the sequence of matrices defined by
$$
K_L(\alpha) = \left[H\otimes I_{2^{L-2}}\right]\left[I_2 \otimes K_{L-1}(\alpha)\right],
$$
where $\otimes$ denotes the Kronecker product, and $I_n$ is the $n\times n$ identity matrix.
I am interested in the limiting behaviour of the singular values of $K_L(\alpha)$ -- in particular, $K_L(0)$ -- as $L$ tends to infinity. Some calculation indicate that the $2^L\times 2^{L-1}$-matrix $K_L$ has $L$ non-zero singular values and that, for any positive integer $k$, the $k$ largest singular values converges to some limit. 
Question: Can this limit be described in terms of the matrix $H$?
I did some experiments and it seems that the limiting behaviour of the singular values of $K_L$ does not only depend on the matrix $H$, but also on the initial value $K_1(\alpha)$. This makes it unlikely for fixed-point arguments to work in this setting.
I also tried to obtain combinatorial expressions for the coefficients in the characteristic polynomial $\chi_L^\alpha(\lambda)$ of $K_L(\alpha)K_L(\alpha)^T$ but was successful only for the three highest non-trivial powers of $\lambda$.
Edit:
The analysis of $\Sigma(\alpha):=\lim_{L\to\infty}\sigma_1(K_L(\alpha))$ as a function of $\alpha$, as suggested by Suvrit, seems to be a good idea. Numerical calculations indicate that, asymptotically,
$$
\Sigma(\alpha)\sim \Sigma(0)\left(.3540+\alpha\right),\quad \alpha\to\infty,\quad \Sigma(0)\approx .8254,
$$
and that $\Sigma(\alpha)$ has a minimum at $\approx(-.2936,.7696)$.
I do not see yet, however, if this can be used to compute $\Sigma(0)$ more precisely.
Edit:
Using the improved bound $\sigma_1(K_L)\leq \frac{1}{2}\sqrt{3+2\alpha +3 \alpha ^2}$, which is sharp for $\alpha=\pm 1$, we can deduce that
$d/d\alpha \Sigma(-1)=-1/2$, and $d/d\alpha \Sigma(1)=1/\sqrt{2}$.
Edit3:
After staring at the problem a little longer I've come up with a conjecture for the characteristic polynomial $\chi_L^{\alpha}(\lambda)$ of $K_L(\alpha)K_L(\alpha)^T$. More precisely, I believe that
$$
\lambda^{-2^L+L}\chi_L^{\alpha}(\lambda) =\lambda ^L-\left(1+\alpha ^2\right)\lambda ^{L-1}+\\\\
+\frac{1}{2^L-1}\sum _{k=2}^L \left(-2\right)^{-k}\left[(1-\alpha)^{2(k-1)}\right]\left[(1-\alpha )^2+k (1+\alpha )^2 \right]\frac{(2^{-L};2)_k}{[k]_2!}\left(2^k-2+2^L\right)\lambda^{L-k},
$$
where $(a;q)_k$ denotes a q-Pochhammer symbol and $[k]_q!$ denotes a q-factorial.
It appears that this formula implies that $\lim_{L\to\infty}\sigma_1(K_L(0))$ can be characterized as $\kappa^{-1/2}$, where $\kappa$ is the smallest positive zero of $x\mapsto f(-x/2)$ and
$$
f:x\mapsto\sum_{k=0}^\infty{\frac{(k+1)x^k}{[k]_2!}}.
$$
Interestingly, this function is related to the q-exponential.
More generally, $\lim_{L\to\infty}\sigma_1(K_L(\alpha))$ can apparently be characterized as $\kappa_\alpha^{-1/2}$, where $\kappa_\alpha$ is the smallest positive solution of $x\mapsto f_\alpha(-x/2)=0$ and
$$
f_\alpha:x\mapsto\sum_{k=0}^\infty{\frac{(1-\alpha )^{2 (k-1)} \left[(1-\alpha )^2+k (1+\alpha )^2\right]x^k}{[k_2]!}}.
$$
The other singular values are similarly obtained from the remaining zeros of $x\mapsto f_\alpha(-x/2)$.
The next task will be to say something about the singular vectors.

REPLY [10 votes]: First let's change the matrix $H$ to $H=\frac{1}{2}\left(\begin{array}{cc}I_2 & I_2 \\ I_2& P_2\end{array}\right)$, where $P_2$ is a 2x2 permutation matrix.  This  swapping of the rows of $H$ won't affect the singular values  of the matrices.  Now lets consider what affect the iteration has on a matrix of the form  $\left(\begin{array}{cc}X &x \\ X &y\end{array}\right)$, where $X$ has $n$ columns.
At the next iteration we will have 
$$ \frac{1}{2}\left(\begin{array}{cccc}X &x & X &x  \\ X &y& X & y\\ X&x &X & y\\ X & y& X & x\end{array}\right)$$    To account for the linear dependence of the columns, we multiply on the right by 
$$\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}I_n & 0 &0 \\ 0 & 1& 0 \\ \frac{1}{\sqrt{2}}I_n & 0 &0 \\ 0 & 0& 1\end{array}\right)$$   The resulting matrix is 
$$ \frac{1}{2}\left(\begin{array}{ccc}\sqrt{2}X &x &x  \\ \sqrt{2}X &y& y\\ \sqrt{2}X&x & y\\ \sqrt{2}X & y& x\end{array}\right)$$
We note that if the original matrix had orthonormal columns then so will this matrix.
This accounts for why the rank of the matrix increases by one with each iteration.  Also note that if $$x=y$$ then the rank will always be one.  If we do this reduction at each step, you find that for the starting $K_1=[1;0]$ the matrix $G=K_L^TK_L$ appears to be in the limit a Hankel matrix plus a diagonal matrix with diagonal entries $G_{ii}=(\sqrt{2})^{-2i})$ and off diagonal entries $G_{ij}=(\sqrt{2})^{-i-j-2}$.  This gives you a very nice matrix but it is only for this particluar starting vector.  If you use a starting vector of $K_1=[1 ;1]/\sqrt{2}$ then $K_L$ will always have rank one and have as it's largest singular value one. 
Based on the observation above, consider the matrix $$K_1=\frac{1}{\sqrt{2}}\left(\begin{array}{rr} 1 & 1\\ 1 & -1 \end{array}\right)\left(\begin{array}{c}\alpha \\ \beta\end{array}\right)$$ such that $$K_1$$ has a 2-norm of one.  Applying the iteration to this matrix we have
$$K_2=\frac{1}{2\sqrt{2}}\left(\begin{array}{rrrr} 1 & 1 & 1 &1 \\ 1 & -1 &1 &-1\\ 1& 1 & 1&-1\\1& -1 & 1 &1  \end{array}\right)\left(\begin{array}{cc}\alpha &0\\ \beta & 0 \\ 0 & \alpha \\ 0& \beta\end{array}\right)$$
Accounting for the nullspace as outlined above requires multiplying by the transpose of the matrix used to remove the nullspace.
$$\left(\begin{array}{cccr}
\frac{1}{\sqrt{2}}I_1 & 0 &\frac{1}{\sqrt{2}}I_1&0\\
0& 1& 0&0 \\ 
0 & 0& 0 & 1\\ 
\end{array}\right) \left(\begin{array}{cc}\alpha &0\\ \beta & 0 \\ 0 & \alpha \\ 0& \beta\end{array}\right)  $$ 
We then have for $K_2$,
$$
\frac{1}{2\sqrt{2}}\left(\begin{array}{rrr} \sqrt{2} & 1  &1 \\ \sqrt{2} & -1  &-1\\ \sqrt{2}& 1 & -1\\ \sqrt{2}& -1 & 1  \end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}} \alpha & \frac{1}{\sqrt{2}}\alpha\\ \beta & 0 \\ 0 & \beta\end{array}\right)
$$
Noticing that the matrix on the left has orthogonal matrix,  we make a modification so that the matrix will have orthonormal columns which means that we only need to understand what the iteration is doing to the matrix on the right.
$$
\frac{1}{2}\left(\begin{array}{rrr} 1 & 1  &1 \\ 1 & -1  &-1\\ 1& 1 & -1\\ 1& -1 & 1  \end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}} \alpha & \frac{1}{\sqrt{2}}\alpha\\ \frac{1}{\sqrt{2}}\beta & 0 \\ 0 & \frac{1}{\sqrt{2}}\beta\end{array}\right)
$$.
All these same steps can be taken at the next iteration.  The resulting $\alpha$,$\beta$ matrix will be 
$$K_3=\left(\begin{array}{cccc}\frac{1}{2}\alpha &\frac{1}{2}\alpha &\frac{1}{2}\alpha &\frac{1}{2}\alpha \\ \frac{1}{{2}}\beta&0 &\frac{1}{{2}}\beta&0\\0&\frac{1}{{2}}\beta & 0 &0\\ 0 & 0&0&\frac{1}{{2}}\beta\end{array}\right)$$
In general the $\alpha$,$\beta$ matrix will have the form 
$$\left(\begin{array}{c}a^T\\B\\b^T\end{array}\right)$$
with the $\alpha$,$\beta$ matrix for the next iteration being
$$\frac{1}{\sqrt{2}}\left(\begin{array}{c}a^T &a^T\\B&B\\b^T& 0\\0 & b^T\end{array}\right)$$
The matrix $B$ and vector $b$ depend only on $\beta$ and at the $L$th iteration it will have entries of zero and $\beta 2^{\frac{1-L}{2}}$.  At the $L$th iteration the vector $a$ has entries $\alpha 2^{\frac{1-L}{2}}$.
The matrix $B$ will have $L-1$ rows. The $i$th row of $B$ will have $2^{L-i-1}$ nonzero entries and rows of $B$ will be orthogonal to one another as well as to the vector $b$.   The matrix will be $2^{L-1}$ wide. The singular values of the $\alpha$,$\beta$ matrix determine the singular values of $K_L$.  Due to the structure described multiplying the $\alpha$,$\beta$ matrix times its  transpose yields an arrowhead matrix
$$\left(\begin{array}{ccccc}\alpha^2 & \alpha\beta 2^{-1} & \alpha\beta 2^{-2}& \alpha\beta 2^{-3}&  \ldots \\
\alpha\beta 2^{-1} & \beta^2 2^{-1} & \ &\ & \ \\
\alpha\beta 2^{-2} & & \beta^2 2^{-2} &  &\\
\vdots & &&\ddots &\\
\end{array}\right)$$
whose eigenvalues are the roots of 
$$f(\lambda)=\alpha^2-\lambda-\sum_{i=1}^\infty\frac{\frac{\alpha^2\beta^2}{2^{2i}}}{\frac{\beta^2}{2^i}-\lambda}$$<|endoftext|>
TITLE: Authorship of Grothendieck universes
QUESTION [28 upvotes]: Universes seem to first enter Grothendieck's work in SGA 1, which is credited to Grothendieck, and a lengthy discussion is in the chapter on Prefaisceaux (presheaves) in SGA 4.  That chapter is credited to Grothendieck and Verdier.  The appendix on them there is credited to N Bourbaki.  
Is there any known evidence of who actually wrote the appendix?

REPLY [5 votes]: In the D. Monk book "Introduction to Set Theory"
i find that the (first in the bibliography dates order)   definition of Universe (in set theory) come from: 
Tarski, Alfred (1938). "Über unerreichbare Kardinalzahlen"
Fundamenta Mathematicae 30: 68–89.
And it is exatly what SGA IV.1 reports. 
See also: http://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory<|endoftext|>
TITLE: Manifolds covered by an n-dimensional torus
QUESTION [12 upvotes]: It is well-known that classification of manifolds up to homemorphism is, in general, out of question. However, this task is sometimes tractable under some additional assumptions on manifolds one would like to classify. I want to ask about one specific example of this situation.
Let $T^n$ denote the $n$-dimensional torus $(S^1)^n$. Suppose that a closed manifold is finitely covered by $T^n$. I would like to know to what extent (and whether at all) it is possible to classify such manifolds $M$. If $n=1$, then $M=S^1$; if $n=2$, then $M=S^1 \times S^1$ or $M =K$, the Klein bottle. (But then again, this is not really interesting, because in these dimensions all manifolds are classified.) What happens if $n \geq 3$, or at least if $n=3$?

REPLY [12 votes]: People have already mentioned the Bieberbach theorems, which imply that your manifold is homotopy equivalent to a Euclidean manifold.  In fact, it is homeomorphic to the Euclidean manifold, at least if the dimension is at least $5$.  This is predicted by the Borel conjecture, which asserts that homotopy equivalent aspherical manifolds are homeomorphic in high dimensions.  The necessary case of the Borel conjecture is proven in the following paper of Farrell and Jones (building on lots of earlier work of theirs): 
F. T. Farrell and L. E. Jones. Topological rigidity for compact non-positively curved manifolds.
In Differential geometry: Riemannian geometry (Los Angeles, CA, 1990), pages 229–
274. Amer. Math. Soc., Providence, RI, 1993.

EDIT : The question inspired me to do a little more reading about the history of the Borel conjecture, and I learned that for virtually abelian fundamental groups like the ones in this question, the Borel conjecture was proven in the earlier paper
MR0704219 (84k:57017)
Farrell, F. T.(1-MI); Hsiang, W. C.(1-PRIN)
Topological characterization of flat and almost flat Riemannian manifolds Mn (n≠3,4). 
Amer. J. Math. 105 (1983), no. 3, 641–672.<|endoftext|>
TITLE: Non-constructive proofs vs. efficient algorithms
QUESTION [11 upvotes]: My question concerns what is meant by "nonconstructive", and whether it has ever been defined in terms of computational complexity.
The wikipedia article on constructive proof begins, "a constructive proof is a method of proof that demonstrates the existence of a mathematical object by creating or providing a method for creating the object." On the other hand, the wiki article on the probabilistic method states, "the probabilistic method is a nonconstructive method [...] for proving the existence of a prescribed kind of mathematical object." I believe these two statements are at odds with one another.
Consider Erdős's celebrated proof of the lower bound of the Ramsey number. This proof shows that as long as $\binom{n}{r} < 2^{\binom{r}{2} - 1}$, there is some coloring of the edges of $K_n$ with $2$ colors that has no monochromatic sub-$K_r$. The proof offers no idea what such a coloring looks like; however, it does lead to a "method for creating" the object in question: try all possible colorings. The proof guarantees that this naive algorithm terminates. Of course, this algorithm quickly becomes computationally infeasible. But in principle, via exhaustive search, any proof of the existence of an object in some finite collection admits of a "method for creating" the object.
Imagine now that we had a different proof of the lower bound of the Ramsey number. This new proof constructs two possible edge-$2$-colorings of $K_n$ and shows that at least one must result in no monochromatic sub-$K_r$, although it remains silent about which of the two colorings works. I think this would also qualify as a "non-constructive" proof (based on analogy to the wiki example with $\sqrt{2}^{\sqrt{2}}$), and yet it would lead to a wonderfully efficient method for finding such colorings. For any $r$, this hypothetical proof says we have to check only two candidates to get the object we're looking for. I think this even gives us a polynomial time algorithm for finding such a coloring (but this depends on how quickly we can verify a coloring.) At any rate, I hope the distinction I am trying to draw is clear.
Does it makes sense to say that a constructive proof is a proof that leads to an efficient algorithm for creating an object with a desired set of properties? Has there been any work related to such a definition? The above is most relevant to statements in discrete math.

REPLY [13 votes]: As others have noted there are several different meanings for constructive. 
I. Constructive proof in the sense of constructive mathematics
This meaning views an object as existing if we have a description of how to construct the objects (though we don't really need to carry it out), and there are several distinct constructive views. Saying a proof is constructive or not can be ambiguous without specifying which school of constructive mathematics we are talking about. 
By the way, it can be the case that we can convert a non-constructive proof to a constructive one (Georg Kreisel's unwinding program or Ulrich Kohlenbach's proof mining program). That does not make the original proof constructive!
Note that algorithmic computability is just one of several constructive  perspectives. For example, in intuitionism there are objects which are not algorithmically computable. A way of understanding this is to remember that Churth-Turing thesis is not an axiom that is accepted by all constructivist, there can be constructions which are not algorithmic in the instuitionistic view.
II. Constructive in the sense of complexity theory
This is a more recent meaning. We mean a proof of existence of an object is constructive if it gives directly a method of efficiently computing/constructing the object. This is the more common meaning of the word in combinatorics these days, e.g. in Robin A. Moser's "A constructive proof of the Lovasz Local Lemma" paper from 2008.
Constructive is used in the sense of efficient algorithms in complexity theory, for example, constructive in Alexander Razborov and Steven Rudich's "Natural Proofs" paper means that the property used in the lower-bound proof is efficiently computable.
Note that the proof itself can be non-constructive in the sense of meaning I while remaining constructive in this sense. You can give an efficient algorithm to compute some object and the correctness and efficiency proofs can be non-constructive. We don't have many interesting examples, but a good example would be the Robertson-Seymour theorem. See also Are there non-constructive algorithm existence proofs?
III. Proof complexity perspective
This is kind of the intersection of the previous two, though I don't recall anyone refer to it as "constructive proof" (probably because they are aware of both previous meanings and don't want to confuse people further :).
Here not only the object should be computable efficiently but the correctness and efficiency proofs must use only efficient concepts. The Robertson-Seymour theorem is an example of an efficient algorithm where we don't have a proof using only efficient concepts. I can give more artificial examples to distinguish between this and the meaning in the previous section but I don't recall any other natural ones.<|endoftext|>
TITLE: Combinatorial Proof of Real Analysis Identity
QUESTION [10 upvotes]: In this question, a proof using real analysis is given of the following identity $$ \sum_{n=1}^{\infty} \frac{(n-1)!}{n \prod_{i=1}^{n} (a+i)} = \sum_{k=1}^{\infty} \frac{1}{(a+k)^{2}}$$
Is there a combinatorial proof of this identity? If so, does the proof require that $a$ be a natural number?

REPLY [7 votes]: The proof below is a WZ method proof. 
The pair of functions$G(n,k) = \dfrac{(1)_n}{(k+x+1)_n} \, \dfrac{1}{n^2}, \quad$
$F(n,k) = \dfrac{(1)_n}{(k+x+1)_n} \, \dfrac{1}{(k+x+1)n}$
is a WZ (Wilf and Zeilberger) pair. That is $\, G(n,k+1)-G(n,k) = F(n+1,k)-F(n,k)$. 
WZ-pairs are gems because of their interesting properties. For example if $$ \lim_{n \to \infty} F(n,k)=0, \quad \text{and} \quad \lim_{k \to \infty} \sum_{n=1}^{\infty} G(n,k)=0.$$ 
Wilf and Zeilberger proved that
$$\sum_{n=1}^{\infty} G(n,0) = \sum_{k=1}^{\infty} F(1, k-1).$$
The pair of functions that we are considering satisfies these conditions. Hence
$$ \sum_{n=1}^{\infty} \frac{(1)_n}{n^2(x+1)_n} = \sum_{k=1}^{\infty} \frac{1}{(k+x)^2},$$ which is the identity we wanted to prove.
NOTE: With $(a)_n$ we mean the rising factorial.<|endoftext|>
TITLE: The kissing number of a square, cube, hypercube?
QUESTION [35 upvotes]: How many nonoverlapping unit squares can (nonoverlappingly) touch one unit square?
By "nonoverlapping" I mean: not sharing an interior point.
By "touch" I mean: sharing a boundary point.
          

It seems the answer for a square in $\mathbb{R}^2$ should be $8$, 
and for a cube in $\mathbb{R}^3$, $26$.
But a 1999 paper by Larman and Zong, "On the Kissing Numbers of Some Special Convex Bodies."
Discrete Comput Geom 21:233–242 (1999). 
(Springer link) says

"In this note we determine the kissing numbers of octahedra, rhombic dodecahedra and
  elongated octahedra. In fact, besides balls and cylinders, they are the only convex bodies
  whose kissing numbers are exactly known."

In that paper, they were interested in the translative kissing number and the lattice kissing number,
whereas I want to consider arbitrary orientations of each square/cube.  Despite the quote above,
it seems this should be known...?

Update (30Dec12) The following explains (I believe) the $0.82$ in Henry Cohn's comment,
leading to his proof for $\le 9$ in $\mathbb{R}^2$:

REPLY [3 votes]: The problem for squares may have been settled first by J W T Youngs, A lemma on squares, American Math Monthly 46 (1939) 20-22. Youngs' proof can also be found in A M Gleason et al., The William Lowell Putnam Mathematical Competition – Problems and Solutions: 1938-1964, pp. 461-463, and in R Honsberger, Mathematical Morsels, pp. 82-89. A discussion of the history, and another proof, will be found in M S Klamkin et al., The kissing number of the square, Mathematics Magazine 68 (1995) 128-133. 
Likuan Zhao, The kissing number of the regular polygon, Discrete Mathematics 188 (1998) 293-296, lets $P_n$ be a regular polygon with $n$ sides, and $k(n)$ its kissing number. After citing earlier results $k(3)=12$, $k(4)=8$, and $k(6)=6$, the paper establishes $k(n)=6$ for $n>6$. Zhao and Junqin Xu, The kissing number of the regular pentagon, Discrete Math 252 (2002) 293-298, finish it off by proving $k(5)=6$.<|endoftext|>
TITLE: Points in sites (etale, fppf, ... )
QUESTION [18 upvotes]: I asked a part of this in an earlier question, but that part of my question didn't receive precedence. 
Etale site is useful - examples of using the small fppf site?
Let $X$ be a scheme (assume it is as nice as you like). There is a description of "points" in the (small) etale site $X_{et}$, and these are the geometric points of $X$. More generally, I've heard that the notion of "points" makes sense in any site (maybe "any" is a little too strong?).
1.) Can you give me a reference defining "points" in other sites. Specifically, I am interested in the small fppf site over a scheme and the big etale site. Is the notion of "points" a useless notion in sites other Zariski and small etale?
2.) What are "points" in other sites "supposed" to do? Is there an analogy that we keep in mind (as to why they are called points)? In the case of the Zariski site, the "points" have a natural structure of a locally ringed space - (the local rings being the stalks in the Zariski site) and this gives a canonically associated locally ringed space to a given site. An analogy similar to this doesn't seem to hold in the small etale site over a scheme.
3.) To whatever a "point" is, I expect one would have a naturally associated local ring. Is this the case in the small fppf site over a (nice?) scheme? This is of course the case in the etale and Zariski site. The small fppf site over a scheme seems a little strange, since limits tend not to be directed.

REPLY [18 votes]: See SGA 4, Exposé VIII, 7.8, which defines an abstract point of a site as a functor from the topos of that site (i.e. the category of sheaves of sets on that site) to the category of sets that commutes with arbitrary inductive (direct) limits and finite projective (inverse) limits.*
One should think of a functor as being the functor that takes a sheaf to its stalk at that point (the "fiber functor" of the point). Since every fiber functor preserves such limits, these are reasonable axioms to postulate.
Grothendieck then proves in this exposé that in the étale site, every functor satisfying the axiom above comes from a (unique) geometric point (up to isomorphism in a suitably-defined sense).
Furthermore, and I can only find this referenced in Brian Conrad's unavailable draft book on the Ramanujan conjecture, in a (classical) topological space where every irreducible subset has a unique generic point (also known as a sober space), such as a normal Hausdorff space or the underlying space of a scheme, every abstract point of the category of sheaves on the topological space corresponds to a unique classical point. This is Proposition 2, Section 3, Chapter IX of Sheaves in Geometry and Logic.
Why characterize points in terms of their functors? This fits in with the general Tannakian formalism, which emphasizes fiber functors, and in particular, defines the fundamental group to be the group of automorphisms of the fiber functor. I.e., we should think of a point as the functor that assigns to a sheaf its stalk at that point.
This also fits into a more general philosophy that we should ignore the site and focus on the topos entirely. One can in fact put a Grothendieck topology on the topos so that it is equal to the category of sheaves on itself.
*Edit: One can also find this in Sheaves in Geometry and Logic, by Maclane and Moerdijk, Chapter VII, Section 5.<|endoftext|>
TITLE: genus and spinor genus over a number field
QUESTION [8 upvotes]: Let $F$ be a number field with ring of integers $\mathfrak{o}$. Let $(V,Q)$ be a quadratic space of dimension $n$ over $F$, and let $L$ be a free lattice in $V$ (i.e. $L\cong\mathfrak{o}^n$). If the class number of $F$ is odd, then the genus of $L$ consists of free lattices, because the squares of their Steinitz classes are equal to the class of the volume of $L$, i.e. $1$. Also, in this case, the genus of $L$ consists of a single spinor genus.
Question 1. When is it true that the genus of $L$ consists of free lattices? 
Question 2. When is it true that the spinor genus of $L$ consists of free lattices? 
The questions above and the ones below are motivated by comparing the mass formula in its original form proved by Siegel, and in its adelic form developed by Tamagawa, Weil, etc. Assume $F$ is totally real and $Q$ is totally positive definite. The formula in its original form concerns the average number of representations of an integer $a\in\mathfrak{o}$ by the classes of quadratic forms over $\mathfrak{o}$ in the genus of $Q$. The theorems of Hasse-Minkowski and Witt allow one to reformulate this average as one over a genus of classes of representations $(x,M)$, where $x\in V$ is fixed with $Q(x)=a$, and $M$ is any free lattice in the genus of $L$ that contains $x$. In the adelic form it is no longer required that $M$ be free.
Question 3. Is there a simple reason why the classical and adelic formulations give the same answer? It seems that Siegel averages over fewer representations $(x,M)$ than Tamagawa, Weil.
Question 4. Is there a classical notion of the spinor genus of $Q$?
I would appreciate any thoughts, results, or pointers to the literature.
Added 1. It seems that the answer to Questions 1-2 are "always true" and this also gives the answer to Questions 3-4: "there is really no difference between the classical and adelic notions in the present setting". Namely, by this paper (based on a method of Kneser from 1957) each class in the genus of $L$ can be reached from the class of $L$ by successively taking $\mathfrak{p}$-neighbors for two suitable prime ideals $\mathfrak{p}$ (which can be chosen to lie outside any prescribed finite set of prime ideals), and $\mathfrak{p}$-neighbors always have the same Steinitz class. I suspect that this conclusion can be reached more directly, but I am no expert, so I am still awaiting thoughts, results, or pointers to the literature.
Added 2. Inspired by Rainer Schulze-Pillot's answer below, here is a slightly simpler way to show that the Steinitz class of a lattice $L$ is determined by the volume of $L$, hence also by the genus of $L$. We can write $L$ as
$$ L=\mathfrak{a}_1x_1+\dots+\mathfrak{a}_nx_n, $$ 
where $(x_i)$ is a suitable basis of $V$, and the $\mathfrak{a}_i$'s are fractional ideals in $F$. By definition, the volume of $L$ is the fractional ideal
$$ \mathfrak{v}L = \mathfrak{a}_1^2\dots \mathfrak{a}_n^2\cdot d(x_1,\dots,x_r), $$
where $d(x_1,\dots,x_r)$ is the discriminant of the basis $(x_i)$. The latter equals the discriminant of $V$ modulo $(F^\times)^2$, hence $\mathfrak{vL}$ really determines $\mathfrak{a}_1\dots \mathfrak{a}_n$ modulo $F^\times$ (because the multiplicative group of fractional ideals is torsion free), which equals the Steinitz class of $L$.

REPLY [10 votes]: Question 1: This is not really a question of genus theory but of elementary divisor theory. If $F$ is the quotient field of the Dedekind domain ${\mathfrak o}$ any two lattices $L_1, L_2$ of full rank on the $F$-vector space $V$ of dimension $n$ can be written as $L_1={\mathfrak a}_1 x_1+ \dots + {\mathfrak a}_n x_n, L_2={\mathfrak b}_1{\mathfrak a}_1x_1+ \ldots + {\mathfrak b}_n{\mathfrak a}_nx_n$ with a basis $x_1,\ldots,x_n$ of $V$. Then $L_1,L_2$ have the same Steinitz class if the product of the elementary divisors ${\mathfrak b}_i$ is principal. 
Now, if we have a non degenerate quadratic form on the space $V$ and $L_1, L_2$, considered as quadratic modules with this form on them, are in the same genus, it is clear that the product of the elementary divisors is ${\mathfrak o}$ (consider the local discriminants), so all lattices in the same genus have the same Steinitz class, in particular they are either all free or all non free.
Question 2: Is obsolete after Question 1.
Question 3 is nevertheless of interest: It appears not to be widely known that Siegel treated what we would now call the case of a non free lattice with a non-degenerate quadratic form. Of course, in his language of matrices this takes a different shape: In "Über  die analytische Theorie der quadratischen Formen III" (Gesammelte Abhandlungen, vol. 1., p. 469) he treats symmetric $(n+1)\times (n+1)$ matrices of rank $n$ which are not integrally equivalent to a non-singular $n \times n$-matrix. He writes:... die hierzu nötige Verallgemeinerung ... führt zu ziemlich umfangreichen Hilfsbetrachtungen" (the necessary generalizations lead to rather extensive auxiliary considerations).
Question 4: If I remember right this is in Watson's book on integral quadratic forms (which is on my bookshelf in my office, but not here at home). The equivalence of both notions has been shown in an unpublished  diploma thesis of a student of Martin Kneser in the 1970s. Since the notion of spinor genus is not much older than the adelic treatment I don't think Watson's definition has ever been used seriously.<|endoftext|>
TITLE: Math French Words
QUESTION [20 upvotes]: Hi, 
I am trying to read some papers on Algebraic Geometry in French. But I am stuck in understanding some Math.French.Words. Anybody has a good reference for it?
Thank you all.

REPLY [14 votes]: I have given a 3-hour course trying to help mathematical students with an English background to read papers in French. You can find notes here.<|endoftext|>
TITLE: units in non-commutative ring spectra
QUESTION [5 upvotes]: Let $R$ be a connective (symmetric) ring spectrum. Let $GL_1(R)$ be the space of units of $R$, i.e. the union of the components of $\Omega^{\infty}(R)$ corresponding to the units of $\pi_0(R)$. $GL_1(R)$ is a group-like monoid, therefore it makes sense to speak about $BGL_1(R)$. 
In fact, if $R$ is a commutative spectrum, then $GL_1(R)$ has the structure of an infinite loop space and defines a spectrum $gl_1(R)$. 

Are there any sensible methods to analyze $BGL_1(R)$ in the non-commutative case? 

As an example of what I mean, let me mention that $BGL_1(R)$ maps to $K(R)$ the algebraic $K$-theory of the ring spectrum $R$, which in turn maps to $THH(R)$ - the topological Hochschild homology. Thus, information about $THH(R)$ might tell you something about $BGL_1(R)$.

REPLY [6 votes]: The notation $\Omega^{\infty}$ requires care.  Definitions and comparisons for symmetric, orthogonal,
and EKMM spectra are given in Lind's thesis.  See http://front.math.ucdavis.edu/0908.1092.  For symmetric spectra, the original source is Sagave and Schlichtkrull http://front.math.ucdavis.edu/1103.2764.  In that case, one must use fibrant approximation, and it matters what model structure one uses.  When dealing with commutative ring spectra, one must use the positive stable model structure, and then one cannot just take the zeroth space.  In the noncommutative case, one can use the stable model structure and then take $\Omega^{\infty}$ to mean the zeroth space of a fibrant approximation. 
In either case, one can use Lind's work to transfer to equivalent EKMM ring spectra, which have highly structured zeroth spaces on the point set level.  Here $gl_1(R)$ is an $\mathcal{L}$-space, where $\mathcal{L}$ is the linear isometries operad.  Of course, one forgets its permutations in the noncommutative case, so that it is an $A_{\infty}$ operad.  One can apply a $1$-fold delooping machine to $gl_1(R)$ or one can construct an equivalent topological monoid and take its usual classifying space. That gives equivalent explicit constructions of $Bgl_1(R)$.  One advantage of this approach is that one uses exactly the same direct zeroth space construction in both the commutative and noncommutative cases.  This goes back to http://www.math.uchicago.edu/~may/BOOKS/e_infty.pdf; a more modern exposition is in http://www.math.uchicago.edu/~may/PAPERS/Final1.pdf.  The concrete construction may help you out.<|endoftext|>
TITLE: A question about the limit of a sequence of pointwise convergent analytic funtions
QUESTION [5 upvotes]: Question: Let $\{f_n\}$ be a sequence of analytic functions on the unit disk $\Delta$ and suppose that $f_n$ converges to a continuous function $f$ on $\Delta$ pointwisely. (1) Can we say that $f$ is analytic on $\Delta$? (2) If $f$ is analytic, is the convergence $\underline{locally}$ uniform on $\Delta$? (Note: I add the words "locally" due to obvious reason.)
If we do not assume that the limit function $f$ is continuous (of course $f$ is measurable) in advance, (3) can we say that $f$ is continuous? [I number this new question by (3)]
Note that evrey measurbale functon on $\Delta$ can be the limit of a sequence of analytic functions in the Lebesgue sense (i.e. almost everywhere).

REPLY [9 votes]: These questions were investigated by Osgood, Montel and Lavrentiev, Sur les fonctions d'une variable complexe
representable par des series de polynomes, Paris 1936. (There is a Russian translation in his
selected Works available free on Internet). If you prefer German, see Hartogs and Rosenthal,
Uber Folgen analytischer Funktionen, Math Ann., 1928, 100, 212-263, and 1932, 104, 606-610.
In general, if a sequence of polynomials converges pointwise in a region $D$, then the
limit function is analytic except for a closed nowhere dense set $E$ (Osgood).
Montel proved that $E$ is a perfect set whose union with the complement of the disc is connected.
Lavrentiev completely characterized the sets $E$ that can occur, and proved that every function
of the first Baire class
which is analytic outside $E$ is a pointwise limit of polynomials.
Thus every continuous function, analytic outside $E$ can be obtained as a limit of polynomials.
Convergence outside $E$ is locally uniform.
This answers all your questions.
By the way, similar problems for harmonic functions (characterization of their pointwise limits)
is still not solved completely.
EDIT. For example, any simple curve in the unit disc, going from $0$ to $1$, satisfies the Lavrentiev
condition. Taking this curve with positive area, we can construct a continuous function $f$
in the unit disc, which is analytic outside the curve but not analytic in the unit disc.
This function will be a pointwise limit of polynomials.

REPLY [2 votes]: Of course (1) does not imply (2) : the functions $f_n:=z^n$ converge pointwisely to $0$ on the unit disk, but the convergence is not uniform. 
In fact, assuming (1), the convergence need not be locally uniform : using Runge's Theorem, it is possible to find a sequence of polynomials $p_n$ such that $p_n \rightarrow 0$ pointwisely on the unit disk, but the convergence is not uniform in any neighborhood of $0$ : see e.g. this question
EDIT
A similar argument (using Runge's Theorem) answers your question (3) negatively : It is possible to construct a sequence of polynomials $p_n$'s with $p_n(z) \rightarrow 0$ for every $z \neq 0$ but $p_n(0) \rightarrow 1$.<|endoftext|>
TITLE: Continuum Hypothesis
QUESTION [15 upvotes]: I am new here, so forgive me if this question does not satisfy the protocols of the site. 
I know there are so many equivalents to the AC (axiom of choice) and there are books that lists this equivalent forms of AC. 
But how can I find equivalent forms of the continuum hypothesis ?
Specially I am interested in Algebraic equivalent form. 
What references do we have in the literature !? 
Thanks in Advance.

REPLY [2 votes]: A very simple algebraic statement equivalent to CH is the assertion that the Baer-Specker group $\mathbb{Z}^{\omega}$ is almost free, i.e. all its subgroups of cardinality less than $\mid \mathbb{Z}^{\omega} \mid$ are free. (This is somewhat unsatisfactory in that almost freenees refers to cardinality.)<|endoftext|>
TITLE: A question about $L^p$ integral of an entire function on $\mathbb{C}$
QUESTION [6 upvotes]: Question: Suppose that $f$ is an entire function (i.e. analytic in $\mathbb{C}$), and satisfies the condition $\iint_{\mathbb{C}}|f|^p dxdy<\infty$ for some $p\in (0,1)$. I guess that $f\equiv 0$ but I do not know how to prove it.
Note. If $p\in[1,\infty)$, it is easy to prove that $f\equiv 0$. In the settting $p\in (0,1)$, one should deal with the integral of an entire function near the essential singularity point $\infty$ carefully.
EDIT. Thank Alexandre Eremenko for his answer. I also want to know the solution to the following harmonic version of question.
Question (H): Suppose that $f$ is a harmonic function ($i.e. \Delta f=0$, $f$ may be complex-harmonic in $\mathbb{C}$), and satisfies the condition $\iint_{\mathbb{C}}|f|^p dxdy<\infty$ for some $p\in (0,1).$  I believe that $f\equiv 0$.

REPLY [5 votes]: In the case of $f$ entire a much more straightforward solution is availabe. First if $f$ is entire, then $|f|^{p}$ is subharmonic. Secondly, by subharmonicity, for any $z_0$, we have
\begin{equation}
|f(z_0)|^{p} \leq \frac{1}{\text{meas}({D(z_0,R)})} \iint_{D(z_0,R)} |f(z)|^{p} dz
\leq \frac{C}{\text{meas}(D(z_0,R))}
\end{equation}
where $D(z_0,R)$ is a disc of radius $R$ around $z_0$ and $C = \iint_{\mathbb{C}} |f(z)|^{p} dz$. Taking $R \rightarrow \infty$ in the above inequality we 
obtain $f(z_0) = 0$. Since $z_0$ was arbitrary $f \equiv 0$.
If the implications `$f$ harmonic implies $|f|^p$ subharmonic'' is correct then this would also answer the second part of the question. I haven't checked if it's true, but it seems correct at least in the case $p = 1$ (which leaves me hoping that it's correct for all $p > 0$). If anybody could check and post the answer in the comments I will be indebted.
EDIT: Here's a proof that $|f|^p$ is subharmonic if $f$ is entire and $0 < p < 1$. 
Jensen's formula implies that,
$$
\log |f(0)| \leq \frac{1}{2\pi} \int_{C} \log |f(e^{i t})| dt 
$$
Multiply by $p$, exponentiate, and then use Jensen's inequality to conclude that
$$
|f(0)|^p \leq \exp(\frac{1}{2\pi} \int_{C} p \log |f(e^{i t})| dt )
\leq \frac{1}{2\pi} \int_{C} |f(e^{i t})|^p dt
$$
Therefore $|f|^p$ is sub-harmonic.<|endoftext|>
TITLE: Are spectra really the same as cohomology theories?
QUESTION [58 upvotes]: Let $E \to F$ be a morphism of cohomology theories defined on finite CW complexes. Then by Brown representability, $E, F$ are represented by spectra, and the map $E \to F$ comes from a map of spectra. However, it is possible that the map on cohomology theories is zero while the map of spectra is not nullhomotopic. In other words, the homotopy category of spectra does not imbed faithfully into the category of cohomology theories on finite CW complexes. This is due to the existence of phantom maps: 
Let $f: X \to Y$ be a map of spectra. It is possible that $f$ is not nullhomotopic even if for every finite spectrum $F$ and map $F \to X$, the composite $F \to X \stackrel{f}{\to} Y$ is nullhomotopic. Such maps are called phantom maps. For an explicit example, let $S^0_{\mathbb{Q}} = H\mathbb{Q}$ be the rational sphere. This is obtained as a filtered (homotopy) colimit of copies of $S^0$ and multiplication by $m$ maps. The universal coefficient theorem shows that there are nontrivial maps $S^0_{\mathbb{Q}} \to H \mathbb{Z}[1]$; in fact they are parametrized by $\mathrm{Ext}^1(\mathbb{Q}, \mathbb{Z}) \neq 0$. However, these restrict to zero on any of the terms in the filtered colimit (each of which is a copy of $S^0$).
In other words, the distinction between flat and projective modules is in some sense an algebraic analog of the existence of phantom maps. Given a flat non-projective module $M$ over some ring $R$, then there is a nontrivial map in the derived category $M \to N[1]$ for some module $N$. Now $M$ is a filtered colimit of finitely generated projectives -- Lazard's theorem -- and the map $M \to N[1]$ is "phantom" in that it restricts to zero on each of these finitely generated projectives (or more generally for any compact object mapping to $M$). 
So it should not be too surprising that phantom maps of spectra exist and are interesting. 
Now spectra are analogous to the derived category of $R$-modules, but spectra also come with another adjunction: 
$$ \Sigma^\infty, \Omega^\infty: \mathcal{S}_* \leftrightarrows \mathcal{Sp}$$
between pointed spaces and spectra.  They thus come with another distinguished class of objects, the suspension spectra. (Random question: what is the analog of a suspension spectrum in algebra?) 
Definition: A map of spectra $X \to Y$ is hyperphantom if for any suspension spectrum $T$ (let's interpret that loosely to include desuspensions of suspension spectra), $T \to X \to Y$ is nullhomotopic. 
In other words, a map of spectra is hyperphantom if the induced natural transformation on cohomology theories of spaces (not necessarily finite CW ones!) is zero. 
Is it true that a hyperphantom map is nullhomotopic? Rudyak lists this as an open problem in "On Thom spectra, orientability, and cobordism."  What is the state of this problem?

REPLY [54 votes]: Consider the periodic complex $K$-theory spectrum $KU$. The integral homology group $H_i(KU)$, the direct limit of 
$$\dots \to H_{2n+i}(BU)\to H_{2n+2+i}(BU)\to\dots,$$
is a one-dimensional rational vector space if $i$ is even and trivial if $i$ is odd. It follows that $H^1(KU)$ is nontrivial. (It's $Ext(\mathbb Q,\mathbb Z)$.) But this can't be detected in the cohomology of suspension spectra, because $H^{2n+1}(BU)$ is trivial.
So that's an example of a "hyperphantom" map from $KU$ to the Eilenberg-MacLane spectrum $\Sigma H\mathbb Z$.<|endoftext|>
TITLE: 3-7 primes in base 10
QUESTION [12 upvotes]: After a quick look at the sequence (of primes) A020463,
$$
3, 7, 37, 73, 337, 373, 733, 773, 3373, \dots,
$$
the following question is straighforward:
Are there infinitely many primes compiled from digits 3 and 7 in base 10?
As any question about primes, this should be tough. Nevertheless I believe there exists some reasonable heuristics to answer it.

REPLY [11 votes]: Significant progress on this question has been made by Maynard. In this preprint on the arXiv he proves that: 

"There are infinitely many primes that do not contain the digit $9$ in their decimal expansion."

This represents the current state of knowledge on this problem, and combines various methods and techniques. From the first page of Maynard's paper:

"Our proof is based on a combination of the circle method, Harman’s sieve, the method of bilinear sums, the large sieve, the geometry of numbers and a comparison with a Markov process. In particular, we make key use of the Fourier structure of [the set of integers without that do not have a $9$ in their decimal expansion]."

See also this earlier paper by Maynard on primes not containing a certain digit modulo $q$.<|endoftext|>
TITLE: Algorithm for determining whether two polynomials have the same splitting field
QUESTION [6 upvotes]: This question asks how to tell whether two cubic polynomials with coefficients in $\mathbb{Q}$ have the same splitting field. There are several answers to the question, but they don't include proofs. Also, it's not clear how the results generalize to higher degree polynomials. Is there an algorithm for determining whether two polynomials in in $\mathbb{Q}[x]$ have the same splitting field? If so, what is it, and why does it work?
(Reposted from Math Stack Exchange.)

REPLY [10 votes]: In practice, you can reduce f and g mod the first 1000 primes (deleting those where f and g are ramified) and see if the primes all have the same splitting behavior.  If they don't, the splitting fields are not the same, and if they do, you can be pretty darn sure that they are.  If you want to replace "be pretty darn sure" with "have proved" you can use the effective Chebotarev of Lagarias and Odlyzko to figure out the value of 1000 which actually gives you a proof.<|endoftext|>
TITLE: The Practical Impact of Set-Theoretic Axioms on Measure Theory
QUESTION [14 upvotes]: The set-theoretic evidence is that we could probably safely add axioms to make many more sets measurable.  For example, we could add axioms that would make projective sets measurable.
I'm curious what would be the implications for working analysts of such a move.  I can see two potential ways in which it could potentially have an impact:

Currently, proving measurability of sets is a somewhat fussy activity.  With the additional freedom provided by extra constructions, the existing theory would become much simpler.
There are existing theories that are already straining at the limits of what can proved measurable in ZFC.  These theories could be usefully extended.

I could also see that it potentially having no real impact.  I'd be curious to hear which if any of these possibilities actually holds.

REPLY [8 votes]: Solovay's model already shows that the axiom of dependent choice (DC) is compatible with the assumption that all sets are Lebesgue measurable.  As far as I am aware, DC suffices for essentially all applications that "working analysts" care about.  If this is true, then the only practical impact of assuming that all sets are Lebesgue measurable is that you exchange slightly fussy proofs of measurability with slightly fussy proofs of results that currently invoke Hahn–Banach or other manifestations of AC, replacing AC with DC.
If I'm wrong and there are cases where DC isn't enough for "working analysts," I'd be curious to hear about it.<|endoftext|>
TITLE: Characterizing maximal powers of closed 2-forms in odd-dimensional manifolds
QUESTION [7 upvotes]: Given a nowhere-zero, closed $2n$-form $\Omega$ in a manifold of dimension $2n +1$, how do we know if there exists a closed $2$-form $\omega$ such that $\Omega = \omega^n$?
Remark. This question started off as a question on multilinear algebra because I thought that perhaps there was an algebraic point-wise condition on $\Omega$ that was non-trivial, but that is not the case: every element of $\Lambda^{2n}(\mathbb{R}^{2n+1})$ is the $n$-th power of an element of  $\Lambda^{2}(\mathbb{R}^{2n+1})$. 
Background.
The odd dimensional manifolds in which I'm really interested are spherized tangent bundles (i.e. $STM := (TM \setminus 0)/\mathbb{R}^+$. The reason for this is that this question comes up in the study of inverse problems in the calculus of variations.

REPLY [13 votes]: Thanks for explaining your motivation, because I think that the general problem as you stated it is impossibly hard, but that, fortunately, for the problem that you are really trying to tackle (the inverse problem in the calculus of variations), there is no need to solve this problem in this generality.  If you are willing to take advantage of the intrinsic geometry of the spherized tangent bundle, you only need to solve a much easier problem.  Indeed, if you don't take advantage of this geometry, solving the hard problem won't help for this purpose.
To see why the general problem is so hard, start with an orientable $2n$-manifold $M$, fix a volume form $\nu$ on $M$ (i.e., a nonvanishing $2n$-form on $M$), and let $\Omega = \pi_1^\ast(\nu)$ where $\pi_1:M\times S^1\to M$ is the projection on the first factor.  If you did have manageable necessary and sufficient conditions to solve your problem as you have stated it, then, applying it to $\bigl(M\times S^1,\Omega\bigr)$, you'd have necessary and sufficient conditions for $M$ to be a symplectic manifold.  Of course, this is known to be a very hard problem, and I don't know anyone who believes that we are going to have the solution to this general global problem anytime soon, even though, as you point out, the existence of local solutions is utterly trivial.
However, given that you are trying to solve the inverse problem in the calculus of variations, you don't really need to solve this general problem.  You can take advantage of the fact that you are working on $\mathsf{R}M = \bigl(TM\setminus O_M)/\mathbb{R}^+$, which you call the 'spherized tangent bundle', but which I like to call the 'tangent ray bundle' of a smooth manifold $M^{n+1}$ (so I use '$\mathsf{R}$').  
There are two geometric features of $\mathsf{R}M$ that are important:  First, there is the foliation by the fibers ${\mathsf R}_x M$ for $x\in M$, each of which is diffeomorphic to the $n$-sphere, and the corresponding 'vertical' $n$-plane field $V\subset T(\mathsf{R}M)$, i.e., the kernel of the differential of the projection $\pi:\mathsf{R}M\to M$.  Second, there is the $(n{+}1)$-plane field $C\subset T(\mathsf{R}M)$ that contains the vertical plane field $V$ and has the property that $\pi'(r)(C_r)= \mathbb{R}\cdot r \subset T_{\pi(r)}M$.  The plane fields $V$ and $C$ are canonical in the sense that $V$ and $C$ are preserved under the natural action on $\mathsf{R}M$ of the diffeomorphisms of $M$ (and they are the only plane fields that are preserved).   There is also an involution $\iota:\mathsf{R}M\to\mathsf{R}M$ that sends each ray to its opposite ray, but this won't be important in what follows. 
Now, an oriented path geometry on $M$ is, by one definition, a choice of a line bundle $E\subset C$ over $\mathsf{R}M$ such that $C = E\oplus V$.  The $2n$-parameter family of curves in $\mathsf{R}M$ that are tangent to $E$ (and hence foliate $\mathsf{R}M$) can be canonically oriented so that they then $\pi$-project to $M$ to be a $2n$-parameter family of oriented curves, one through each point oriented tangent to each ray based at that point.  The so-called inverse problem in the calculus of variations is to determine whether there exists a (first-order) nondegenerate Lagrangian for oriented curves in $M$ such that the extremals of that Lagrangian are exactly the oriented curves generated by $E$.  [A (classical) path geometry is simply an oriented path geometry that is invariant under $\iota$.]
In this context, a first-order Lagrangian is a section $\lambda$ of the line bundle $(C/V)^*\to \mathsf{R}M$.  The reason is that, for any immersed curve $\gamma:[0,1]\to M$, its tangential lifting to $\mathsf{R}M$ defined by $\mathsf{R}\gamma(t)= \mathbb{R}^+\cdot\gamma'(t)$ then can be used to pull back $\lambda$ to $[0,1]$, so that it can be integrated over that interval, thus defining a first-order functional on oriented, immersed curves in $M$.  (The point is that $(C_r/V_r)^\ast$ is naturally isomorphic to the dual of the tangent line $\mathbb{R}\cdot r$.)
How does one determine $E$ from a given $\lambda$?  The process is as follows:  First, one notes the classical lemma that, in this form, says that, for any given Lagrangian $\lambda$ on $\mathsf{R}M$, there exists a unique $1$-form $\delta\lambda$ on $\mathsf{R}M$ with the following properties:  First, $V$ is in the kernel of $\delta\lambda$, so that it makes sense to evaluate $\delta\lambda$, which is a section of $(T/V)^\ast$, as an element of $(C/V)^\ast$; second, this evaluation agrees with $\lambda$; and, third, $\bigl(d(\delta\lambda)\bigr)(v,w)=0$ whenever $v,w\in C$.  The $1$-form $\delta\lambda$ is known as the Poincaré-Cartan form of the Lagrangian $\lambda$.  The mapping $\delta:C^\infty((C/V)^\ast)\to C^\infty((T/V)^\ast)$ is a linear, first-order operator.  
We say that $\lambda$ is nondegenerate if $(d(\delta\lambda))^n$, which is a closed $2n$-form on $\mathsf{R}M$, is nonvanishing.  In this case, since $\mathsf{R}M$ has dimension $2n{+}1$, there is a line bundle $E_\lambda\subset T(\mathsf{R}M)$ that is the kernel of $d(\delta\lambda)$, and it is not difficult to show that $E_\lambda\oplus V = C$ and that, moreover, the oriented curves defined by the oriented path geometry $E_\lambda$ are the extremal curves (with respect to compactly supported variations) of the functional defined by the Lagrangian $\lambda$.
So, the inverse problem can be understood as, given an oriented path geometry $E$, find those nondegenerate Lagrangians $\lambda$ such that $E= E_\lambda$.
Now, one can imagine a gradual attack on this problem that tries to swim back 'upstream' from $E$ to $\lambda$.  First, note that if a nondegenerate $\lambda$ exists such that $E=E_\lambda$, then there will be a closed nonvanishing $2n$-form $\Omega = \bigl(d(\delta\lambda)\bigr)^n$ on $\mathsf{R}M$ whose kernel is $E$.  Then one must write $\Omega$ in the form $\Omega = \omega^n$ for some closed $2$-form $\omega$.  Of course, doing this in general is exactly the problem of the OP.  Supposing this could be done, though, that's far from enough because there's no guarantee that $\omega$ would vanish when restricted to each $(n{+}1)$-plane $C_r$, and this would be necessary if $\omega = d(\delta\lambda)$.  You'd have to go back and try a different $\omega$, or worse, consider all possible $\omega$s and try to select one that does do what you want.  This seems hopelessly difficult.
However, it turns out that you don't really need to do any of this.  There is a better way to proceed:  First, write the identity map of $E$ in the form $ X\otimes \xi$ where the vector field $X$ on $\mathsf{R}M$ is a nonvanishing section of $E$ (this can always be done and $X$ will be unique up to a scalar multiple) and $\xi$ is the dual section of $E^\ast$.  Define a linear, second-order differential operator 
$$
D_E:C^\infty\bigl((C/V)^\ast\bigr)\to C^\infty\bigl((T/C)^\ast\otimes E^\ast\bigr)
$$
(the vector bundle $T/C$ has rank $n$ over $\mathsf{R}M$) by the rule
$$
D_E\lambda = i_X\bigl(d(\delta\lambda)\bigr) \otimes \xi\ .
$$
(It is not hard to verify that $D_E\lambda$ is well-defined and really is a section of $(T/C)^\ast\otimes E^\ast$, i.e., an $E^\ast$-valued $1$-form on $\mathsf{R}M$ that vanishes on elements of $C$.)  Then, by construction, $E=E_\lambda$ if $\lambda$ is a nondegenerate Lagrangian that satisfies $D_E\lambda=0$.
The point is that the inverse problem is cast as a linear, second-order PDE for the unknown Lagrangian $\lambda$, with the extra condition that one is only interested in nondegenerate Lagrangians.  Locally, it is $n$ equations for $1$ unknown.
This PDE system is determined when $n=1$ and is always locally solvable.  However, when $n>1$, this is an overdetermined problem, and, for the generic $E$, there are no nondegenerate solutions to $D_E\lambda=0$.  (I believe that it was Jesse Douglas in the 1930s who first did a serious, detailed study of this overdetermined problem, and he exhibited path geometries in the case $n=2$ for which there were no nondegenerate solutions to this equation.)  
In general, the techniques of exterior differential systems provide methods for finding, or at least describing, the general solutions for a given specific $E$.  Fortunately, these methods are much easier to apply and study than the problem posed by the OP.
There is, of course, an enormous literature on this subject (not always written in the best notation, I have to say).  If there is interest, I can give a (very abbreviated) bibliography.<|endoftext|>
TITLE: extensions of IC sheaves
QUESTION [5 upvotes]: Let $X$ be a connected smooth algebraic variety (say over $\mathbb{C}$) and let $L$ be a local system on an open subvariety. Suppose that we know $H^*(X)$. How can we compute $Ext^i(IC(L),IC(L))$ for $i\ge  1$ ? Is there a general method or only ad-hoc arguments specific to examples? 
A source of examples: let $p:X'\to X$ be a small map. Then the sheaf $\mathcal{F} = Rp_*\mathbb{C}_{X'}$ is an IC sheaf of the above form. (and here suppose we also know  $H^*(X')$).
I would be interested to know if for this class of examples one can say something precise about the dimensions of the Ext groups. 
One could also replace cohomology with equivariant cohomology, or more generaly consider the question for algebraic stacks.
Any references where this is computed in specific examples would be also very useful.

REPLY [5 votes]: Not sure what to say in general, but in practice one often uses a semismall resolution. For example if your sheaf $IC(L)$ is given as $\pi_* C_Z$ (pushforward of constant sheaf - I'm assuming $Z$ smooth and oriented) for a proper $Z\to X$ you can calculate derived self-maps $$Hom(\pi_*C_Z,\pi_*C_Z)=
Hom(C_Z, \pi^!\pi_*C_Z).$$ Now write $\pi_1,\pi_2$ for the two projections from $Z\times_X Z\to Z$. Then we calculate further 
$$Hom(C_Z, \pi^!\pi_*C_Z)=R\Gamma_Z(\pi_{1,*}\pi_2^!C_Z)=R\Gamma_{Z\times_X Z}(\omega_{Z\times_X Z}),$$
i.e., Borel-Moore homology of $Z\times_X Z$ (up to a shift I've ignored). This works in the equivariant setting (i.e. for sheaves on stacks) equally well, giving equivariant Borel-Moore homology. The famous example of this is the case where $X$ is the nilpotent cone [or its quotient by the group], $Z$ the [equivariant] Springer resolution, and $Z\times_X Z$ the [equivariant] Steinberg variety, cf. the book of Chriss-Ginzburg. In this case the equivariant BM homology (=Ext algebra of the Springer sheaf) is the degenerate affine Hecke algebra. This calculation works as is in the dg setting --- for the statement on the level of derived categories you need formality, which in the Springer case is a result of Laura Rider here.<|endoftext|>
TITLE: Approximating erf by tanh
QUESTION [6 upvotes]: It appears to be well-known that $\tanh(x)\le \mathrm{erf}(x)$ on $[0,\infty)$. It's off-handedly mentioned here, for example. Where can I find a formal proof? On the one hand, it's hard to imagine that a "classic" like this wouldn't have been proven already. On the other hand, the Taylor expansions are somewhat involved (tanh involves Bernoulli numbers) and unfortunately, the inequality does not hold termwise in the expansions -- so it's certainly far from obvious.

REPLY [17 votes]: Let $f(x)={\mathrm{erf}}(x)-\tanh(x)$. It can be easily seen from Taylor series
at $0$ and from asymptotics at $\infty$ that $f(x)>0$ for small $x$ and
for large $x$.
Let us prove that $f(x)>0$ by contradiction.
Suppose that $f(x)$ is negative for some $x$, then $f'$ must have
at least $3$ positive zeros, by Rolle's theorem. This means that the equation
$$g(x):=e^{-x^2}(e^{2x}+2+e^{-2x})=2\sqrt{\pi}$$
has at least $3$ positive solutions. But this is not the case because
the LHS is monotone. Indeed, differentiating $g$, dividing by $e^{-x^2}$
and replacing $2x$ with $y$ we obtain
$$g'(x)=\sinh(y)-y\cosh(y)-y<0,$$
because $\sinh(y)  < y \cosh(y)$ as you can see from their Taylor series.

REPLY [5 votes]: First,
$$\begin{align}
1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt,
\cr
1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt .
\end{align}$$
Subtract:
$$
\mathrm{erf}(x)-\mathrm{tanh}(x) = 
\int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt
$$
So it suffices to show that this integrand is positive.  It is positive for $t>1$ (proof needed), so we establish $\mathrm{erf}(x) > \mathrm{tanh}(x)$ for $x > 1$.<|endoftext|>
TITLE: Is a conservative finite limit preserving functor of (infinity,1)-categories homotopically faithful?
QUESTION [7 upvotes]: In classical category theory we have a following criterion. If $\mathcal{C}$ and $\mathcal{D}$ are finitely complete categories and $F : \mathcal{C} \to \mathcal{D}$ is a functor which preserves finite limits and reflects isomorphisms, then $F$ is faithful. It follows easily from an observation that an equalizer of two parallel morphisms is an isomorphism if and only if those morphisms are equal.
My question is

Given two finitely complete $(\infty, 1)$-categories $\mathcal{C}$ and $\mathcal{D}$ and a functor $F : \mathcal{C} \to \mathcal{D}$ which preserves finite limits and reflects equivalences, does it follow that $\mathrm{Ho} F : \mathrm{Ho} \mathcal{C} \to \mathrm{Ho} \mathcal{D}$ is faithful?

The observation I mentioned previously doesn't work in higher category theory since for example an equalizer of the identity morphism of an object $X$ with itself is a free loop object on $X$, but I am unable to decide whether this faithfulness criterion is valid.

REPLY [8 votes]: Here's a counterexample.  Let $\mathcal{C}=\mathcal{D}$ be the stable $(\infty,1)$-category of perfect complexes over $\mathbb{Z}_{(p)}$, and let $F(X)=X\otimes \mathbb{Z}/p$ (the derived tensor product).  Then $F$ is exact, and does not send any nonzero objects to zero and hence reflects equivalences.  But $F$ is not faithful, since it kills the multiplication by $p$ map on any object.<|endoftext|>
TITLE: Fourier transform on the discrete cube
QUESTION [33 upvotes]: Notation: identify an element of $\{-1,1\}^n$ with the set $S \subseteq \{1, \ldots, n\}$ on which it takes the value $-1$.
The following is an asymptotic question. "Close to one" means "more than $r_n$" and "away from $\frac{n}{2}$" means "outside the interval $[\frac{n}{2} - k_n\sqrt{n}, \frac{n}{2} + k_n\sqrt{n}]$", for some $r_n$ and $k_n$ which respectively increase to one and infinity as $n \to \infty$.
Conjecture: for any subset of $\{-1,1\}^n$ there is a complex valued function with $l^2$ norm equal to 1, supported either on that subset or on its complement, whose Fourier transform has $l^2$ norm close to one on $\{S: |S|$ is away from $\frac{n}{2}\}$.
A positive answer would have very interesting consequences. It would mean that a single subspace of $l^2(\{-1,1\}^n)$ whose dimension is small compared to the whole space is close to every subspace spanned by standard basis vectors or its complement.
I'm not familiar with the literature on Fourier analysis on the discrete cube.  Is there anything there that would help settle this question?
Edit: I would have edited this earlier, but I assumed it had disappeared from view. Since the Kadison-Singer problem has a positive solution, the answer to the conjecture is no. Posting details in an answer.

REPLY [7 votes]: In the language of Marcus-Spielman-Srivastava, Corollary 1.5, identify $l^2(S)$ with ${\mathbb C}^d$ where $d = |S|$ and let the vectors $u_i \in {\mathbb C}^d$ be the orthogonal projections into $l^2(S)$ of the Fourier transforms of the standard basis vectors of $l^2(\{-1,1\}^n)$. By Corollary 1.5 with $r = 2$, we can find a subset $T$ of $\{-1,1\}^n$ such that
$$(.5 - O(\sqrt{\delta}))I \leq \sum_{i \in T} u_iu_i^* \leq (.5 + O(\sqrt{\delta}))I$$
where $\delta = \frac{d}{2^n}$. This means that the Fourier transform of any vector in $l^2(T)$ with $l^2$ norm equal to 1 will split into a component supported on $S$ and a component supported off of $S$, each of which has square norm in the interval $[.5 - O(\sqrt{\delta}), .5 + O(\sqrt{\delta})]$.
The conjecture fails dramatically: not only is $l^2(S)$ not close to $l^2(T)$ or $l^2(T)^\perp$, but every unit vector in either $l^2(T)$ or $l^2(T)^\perp$ is approximately $45^\circ$ from $l^2(S)$.<|endoftext|>
TITLE: Almost parallelizable 4-manifolds
QUESTION [7 upvotes]: On the first page of Milnor-Kervaire's paper "Bernoulli numbers, homotopy groups, and a theorem of Rohlin", they assert without proof or reference that if $M$ is a compact connected oriented differentiable $4$-manifold such that $w_2(M)=0$, then $M$ is almost parallelizable, that is, for all $x_0 \in M$ the tangent bundle of $M \setminus x_0$ is trivial.  Try as I might, I cannot figure out how to prove this.  Can someone help me?

REPLY [9 votes]: You want to trivialise the restriction of the tangent bundle to the 3-skeleton of $M$. Since $\pi_0 O(4) = \pi_1 O(4) = Z/2$, there are obstructions $w_1(E) \in H^1(X; Z/2)$ and $w_2(E) \in H^2(X;Z/2)$ to trivialising a rank 4 bundle over the 1- and 2-skeleta of a cell complex $X$. Because $\pi_2 O(4)$ is trivial, there is no further obstruction to extending a trivialisation from the 2-skeleton to the 3-skeleton. This is outlined in a nice way at the beginning of chaper 3 in Hatcher's book on vector bundles.<|endoftext|>
TITLE: Does 'finite + finitely presented as an algebra' equal 'finitely presented as a module'?
QUESTION [6 upvotes]: In "stack project", there is a lemma on finite locally free morphisms, saying that a finite locally free morphism of schemes is equivalent to a morphism which is finite, flat, and locally of finite presentation.
For the proof, they refer to the commutative algebra fact that a module is finite locally free iff it is flat and finitely presented. 
In order to complete the proof, I have the following gap:
Let $A \to B$ be a morphism of rings. Then $B$ can either be viewed as an $A$-algbra or an $A$-module. Is it true that the following two statements are equivalent?

$B$ is a flat $A$-module and is finitely presented as an $A$-module.
$B$ is a finitely generated flat $A$-module, and is finitely presented as an $A$-algebra.

REPLY [3 votes]: It appears that in the meantime, full proofs have been added to the Stacks Project.

Tag 0564. Let $R \to S$ be a finite and finitely presented ring map. Let $M$ be an $S$-module. Then $M$ is finitely presented as an $R$-module if and only if $M$ is finitely presented as an $S$-module.

In particular, $S$ is finitely presented as an $R$-module if $S$ is finitely generated as an $R$-module and finitely presented as an $R$-algebra.

Tag 058R. Let $M$ be an $R$-module. Then $M$ is finite projective if and only if $M$ is finitely presented and flat.<|endoftext|>
TITLE: What is a random number? (poll experiment)
QUESTION [7 upvotes]: Imagine the following experiment: you wait say at a subway exit, and ask everyone passing "please tell me a number" (positive integer, of course). You do this day after day, until you reach say 1M people. 

What is the distribution $\mu$ on the positive integers that you get?

This is a serious question, obviously some numbers are "nicer" than some other, say arithmetically speaking, so $\mu$ is probably a very interesting measure!
Of course, if you would do the poll with very small kids, $\mu$ would be more or less uniform on $2,3,4,5$ or so, perhaps with some mass at 1, and probably at $0$ too (coming from scientists's kids, proud of knowing what 0 is :) 
My question of course concerns adults, and results obtained via a real poll like the one suggested above: does anyone know, is anything written on this subject?  
(question inspired from The human body's random number generator, I mean from the title of that question.)
[Edit Jan 2: thanks very much everyone, as a partial conclusion: (1) the measure $\mu$ certainly depends on the precise location of the poll, interesting would be for instance the results of an experiment - I mean, the picture/precise formula of $\mu$ - in a "random" place, say Times Square, (2) there are lots on interesting links signaled below (by MP, Joel, JSE..), papers by cognitive scientists, plus some interesting math interpretations/speculations (by quid, Alexander, Andreas, Yuichiro..) but I'm still afraid there's no picture of some particular $\mu$ emerging from all this, (3) will keep looking etc., and of course, if I ever get a huge grant, with some freeness in spending it :) think I'll conduct such a poll experiment myself - it's probably worth it.]

REPLY [14 votes]: Cognitive sychologists study this kind of question, as you might expect.  Here's a paper (behind a paywall, sorry) where they asked people to name random digits.  You don't get uniform distribution on 0,..,9.
I learned a little about this stuff when I was writing a blog post about detecting election fraud by looking for digits which looked more like "numbers made up by humans" than "numbers arrived at randomly."
Update:  I spoke to my colleague Gary Lupyan, a cognitive psychologist here who studies such things.  There are lots of interesting results, although he hasn't done the precise experiment suggested in the question.  If you ask people to name a number between 1 and 100, the modal responses are between 1 and 10, with maybe a slight continuing dropoff afterwards.  People disprefer even numbers and multiples of 5 and 10.  He also replicated the folk belief that if you ask people to name a number between 1 and 20, the modal response is 17.
It doesn't look to me like the results he's getting are well-modeled by any particularly natural distribution, though you could certainly fit some kind of decay to it.<|endoftext|>
TITLE: What axioms are stronger than the Axiom of choice?
QUESTION [5 upvotes]: What other axioms in set theory are stronger than AC ? I mean what are those axioms that will imply AC ?

REPLY [2 votes]: The axiom of global choice. Technically this isn't really an axiom: global choice (GC) states that there is a formula $\phi(x, y)$ such that the relation $$ A\le_\phi B:= V\models \phi(A, B)$$ is a well-ordering of the universe $V$. This can't be stated as a single formula, so in some sense it's a meta-axiom. It clearly implies choice, and is implied by $V=L$: we already have a partition of the universe into $L_0$, $L_1$, . . . , $L_\alpha$, . . ., and we can get from here to a full (class-)well-ordering of the universe by fixing at the outset some well-ordering of formulas in the language of set theory (since at each stage in the construction of $L$ we are only taking definable powersets; this is why this argument doesn't work in just $ZFC$).
A couple comments on why I think global choice is interesting (even though it's not expressible in the language of set theory):

Although the relative consistency of Choice was proven rather early, it was via $L$, which satisfies $GC$ as well; the result that, assuming the consistency of $ZFC$, there is a model of $ZFC$ with no definable well-ordering of the universe came much later [NOTE: this is based on foggy memory, and I don't recall exactly when this result happened; can someone remind me?], and generally telling whether a model of $ZFC$ satisfies $GC$ is very hard.
Basically by reversing the argument that it is true in $L$, one can make an informal argument that GC implies that the universe is small (contains only "buildable" things). So there's a somewhat intuitive argument for $AC+\neg GC$: the universe should be "big enough" that for each family of sets, we have a choice function, but the universe should also be big enough that any specific definable well-order "misses something." 
$GC$ is useful in other set theories. First of all, as noted above, even stating $GC$ needs a class theory like Morse-Kelley or NBG (expansions of $ZFC$ to also talk about classes). We can also ask about the status of $GC$ in really odd set theories like New Foundations (EDIT: As Ali points out below, in NF/NFU Choice and Global Choice are essentially one and the same - but the point still stands that global choice could still be interesting in set theories different than $ZFC$.)
$GC$ makes sense even outside of set theories! It doesn't make sense to ask whether a given ring $R$ satisfies the axiom of choice, since elements of $R$ aren't (at least on the face of things) sets, but it does make sense to ask whether there is a formula in the language of rings which well-orders $R$.<|endoftext|>
TITLE: Rational Morava E-theory of cyclic groups
QUESTION [13 upvotes]: I'm confused about a seeming contradiction that is probably just a reflection of ignorance on my part.  Let's try to compute the Morava E-theory of $B \mathbb{Z}/p$ in two different ways.

First, following Ravenel-Wilson (see also Hopkins-Kuhn-Ravenel, Hunton, etc.), one computes (via knowledge of $K(n)^\ast(\mathbb{C} P^\infty)$ and a Gysin sequence) that the Morava K-theory of $B \mathbb{Z}/p$ is
$$K(n)^\ast (B \mathbb{Z}/p) = K(n)_* [x] / x^{p^n}$$
where $x$ is of degree $2$.  In particular, it is concentrated in even degrees and free of rank $p^n$ over $K(n)_\ast$.  By Bockstein arguments, the Morava E-theory is then also free of the same rank.
Second, using the knowledge that $E_n^\ast (B \mathbb{Z}/p)$ is free over $E_n^\ast$, it must embed into its rationalisation, $E_n^\ast (B \mathbb{Z}/p) \otimes \mathbb{Q}$.  Let's try to compute the rank of that rationalisation using the group cohomological Atiyah-Hirzebruch spectral sequence:
$$H^*(\mathbb{Z} / p, E_n^\ast \otimes \mathbb{Q}) \implies (E_n \otimes \mathbb{Q})^\ast (B \mathbb{Z}/p)$$
However, the $E_2$ term of the spectral sequence is rank 1 over $E_n^\ast \otimes \mathbb{Q}$ since $\mathbb{Z} / p$ is a finite group, and must collapse there.  Therefore it certainly doesn't contain a sublattice of rank $p^n$.

I've indicated my misgivings with the second argument by notating the target of the spectral sequence as $(E_n \otimes \mathbb{Q})^\ast (B \mathbb{Z}/p)$, that is, the value on $B \mathbb{Z}/p$ of the cohomology theory which is the rationalisation of $E_n$ (whose homotopy groups are $E_n^\ast \otimes \mathbb{Q}$).  My guess is that this is NOT the same as the rationalisation of the E-theory of $B \mathbb{Z}/p$.
So my real question is: is it easy to see why this is the case?  More importantly, can one compute the correct answer (i.e., coming from argument 1) through methods based upon argument 2?  I'm envisioning some sort of spectral sequence that knows how rationalisation and cohomology may fail to commute.

REPLY [15 votes]: Your guess is correct, and I think this is a confusion pretty much everyone has when they first see these computations.  What's going on here is essentially just the simple algebraic fact that $$\mathbb{Z}[p^{-1}][[x]]\neq\mathbb{Z}[[x]][p^{-1}].$$
In computation (1), you can work directly with a Gysin sequence for $E_n$, without going through $K(n)$ and Bocksteins.  The Gysin sequence tells you that $$E_n^*(B\mathbb{Z}/p)=E_n^*[[x]]/([p](x)),$$ where $[p](x)$ is the $p$-series of the formal group law.  More generally, this holds for any complex-oriented theory $E$ such that $[p](x)\in E^*[[x]]$ is a nonzero divisor.
Since the formal group law for $E_n$ has height $n$, the first coefficient of $[p](x)$ which is a unit is the $x^{p^n}$ coefficient.  It follows that $[p](x)$ `differs from a monic polynomial of degree $p^n$ by a unit in $E_n^*[[x]]$, so $E_n^*(B\mathbb{Z}/p)$ is free of rank $p^n$ over $E_n^*$.
On the other hand, if we do the same computation with $E_n\otimes \mathbb{Q}$ instead of $E_n$, $$\frac{[p](x)}{x}=p+\dots\in(E_n\otimes \mathbb{Q})^*[[x]]$$ is now a unit.  Thus $(E_n\otimes\mathbb{Q})^*(B\mathbb{Z}/p)$ is free of rank 1 over $(E_n\otimes\mathbb{Q})^*$.  However, $[p](x)/x$ is only a unit in $(E_n\otimes \mathbb{Q})^*[[x]]$, not in $E_n^*[[x]]\otimes\mathbb{Q}$.  This is because to construct a power series inverse for it, we need coefficients with arbitrarily large powers of $p$ in the denominator.  Thus we find that $(E_n\otimes \mathbb{Q})^*[[x]]/([p](x))$ and $E_n^*[[x]]\otimes\mathbb{Q}/([p](x))$ look quite different from each other, and this is exactly the discrepancy between your two computations.
As for a computation along the lines of (2) that gives the answer from (1), I don't know of anything like that.  You could make an AHSS computation for $E_n$ rather than $E_n\otimes\mathbb{Q}$ and only tensor with $\mathbb{Q}$ afterwards, which would give the right answer.  However, the AHSS $$H^*(\mathbb{Z}/p,E_n^*)\implies E_n^*(B\mathbb{Z}/p)$$ is quite subtle: I don't know how to compute the differentials in it without already knowing the final answer, and the filtration of the spectral sequence distorts almost all of the structure of $E_n^*(B\mathbb{Z}/p)$ (for instance, the associated graded you get from the spectral sequence consists mostly of $p$-torsion, whereas the actual answer is torsion-free).<|endoftext|>
TITLE: If $\kappa$ is weakly inaccessible, then is it the $\kappa$-th aleph fixed point
QUESTION [5 upvotes]: A cardinal $\kappa$ is weakly inaccessible iff $\kappa > \omega$, $\kappa$ is regular, and $\forall\lambda<\kappa(\lambda^+<\kappa)$
(here $\lambda^+$ is the successor cardinal)
A cardinal $\kappa$ is strongly inaccessible iff $\kappa > \omega$, $\kappa$ is regular, and $\forall\lambda<\kappa(2^\lambda<\kappa)$
My question is how to prove if $\kappa$ is weakly inaccessible, then it is the $\kappa$-th $\aleph$ fixed point, also if $\kappa$ is strongly inaccessible, then it is the $\kappa$-th $\beth$ fixed point?
This question is found in I.13.17 of The Foundations of Mathematics by Kenneth Kunen.

REPLY [7 votes]: If $\kappa$ is weakly inaccessible, then it is a limit cardinal and hence $\kappa=\aleph_\lambda$
for some limit ordinal $\lambda$. Since the cofinality of $\aleph_\lambda$ is
the same as the cofinality of $\lambda$, it follows by the regularity of $\kappa$ that
$\lambda=\kappa$, and so $\kappa=\aleph_\kappa$, an $\aleph$-fixed point. 
The next
$\aleph$-fixed point after any ordinal $\beta_0$ must have
cofinality $\omega$, since it is $\sup_n\beta_n$, where
$\beta_{n+1}=\aleph_{\beta_n}$. So if a weakly inaccessible
$\kappa$ is the $\delta$-th $\aleph$-fixed point, it cannot be
that $\delta$ is a successor ordinal, and so $\delta$ is a limit ordinal. Since the
$\aleph$-fixed points are closed, this implies $\kappa$ has the same cofinality as $\delta$, and so by regularity it follows that $\kappa=\delta$ and thus, $\kappa$ is the $\kappa$-th fixed point.
Essentially the same argument works with $\beth$ and strongly
inaccessible cardinals, simply by replacing $\aleph$ everywhere with $\beth$.<|endoftext|>
TITLE: Polyominoes with double contact
QUESTION [11 upvotes]: Here is a problem which arose from an earlier question.  I'll change the terminology but not the question: A polyomino is a region with a connected interior made by joining one or more unit squares edge to edge. We will allow translating, reflecting and rotating polyominos but always with corners at integer points (and sides horizontal and vertical.)

It seems obvious that any two polyominoes, neither a single square, can be positioned so that their interiors are disjoint but they share at least two edges. Find a proof, preferably an elegant one, or perhaps a counter-example.

If we only insisted on being connected at corners this would not be true.

Here is a partial argument which is perhaps not that elegant and perhaps not that easy to finish. Let us say that one is red and the other blue. Every polyomino has a minimal bounding rectangle. I won't illustrate, but clearly if  each has two consecutive filled squares on the boundary of the bounnding rectangle then those can be put in contact. So assume that the blue one does not have two consecutive colored squares on the boundary. This means that all four corners of the bounding rectangle are unoccupied. If the same is true of the red one then we can position them as shown to the left below (only a small part of each polyomino is indicated). White squares are definitely empty. The solid colored squares are definitely filled and include the rightmost blue square of the top or of that polyomino and the leftmost red square at the bottom of the red polyomino. The lightly colored squares with question marks might be empty or full. The green lines are not crossed by either polyomino in their current positions. The red polyomino can be moved down by one and there are either two edges of contact or else we can move it down another one and definitely have two edges.  To the right is a similar situation where the red poyomino has a square of the bounding rectangle filled. 

It remains to consider the case where the blue polyomino does not have two consecutive filled squares on the boundry of the bounding rectangle  but the red one does, but not at any corner.

REPLY [9 votes]: Here is a formalization of your argument. Let $A$ and $B$ be two poliominoes in question. Construct the following figure $K$ made of unit square: a square with integer coordinates $(x,y)$ is included in $K$ if and only if the translate of $B$ by the vector $(x,y)$ overlaps with $A$. ($K$ is essentially the Minkowski difference of $A$ and $B$ regarded as subsets of the integer lattice rather than figures with interior.)
Since $A$ and $B$ are connected. so is $K$. First consider the case when $K$ is not convex (i.e. not a rectangle) and find a concave corner of $K$. The cell outside $K$ near this corner is adjacent to at least 2 cells of $K$. This means there is a translate $B'$ of $B$ which does not overlap with $A$ but there are two coordinate directions (such as "up" and "left") such that if one moves $B'$ one step in any of these directions, the resulting figure overlaps with $A$. If you cannot move $B'$ up, then there is a cell of $B'$ right below a cell of $A$, so they share a boundary segment. Two such "forbidden" directions give us  two common boundary segments.
Now consider the case when $K$ is a rectangle. Clearly the upper side of $K$ corresponds to the position where the bounding box (=minimal enclosing rectangle) of $B$ touches the bounding box of $A$ from above. Similarly for the other three sides of $K$. This means that an integer cell $(x,y)$ belongs to $K$ if and only if the bounding box of $B$ translated by $(x,y)$ overlaps with the bounding box of $A$. By the definition of $K$ this means the following: translates of $A$ and $B$ overlap if and only if their bounding boxes overlap.
This implies that $A$ and $B$ contain the corners of their bounding boxes. Indeed, if e.g. $A$ does not contain the upper right corner of its bounding box, we can translate $B$ so that the intersection of the two bounding boxes is just the cell at that corner, contrary to the above.
Once we know that $A$ and $B$ contain the corners of their bounding boxes, we can find boundary segments of length 2 near these corners and position our poliominoes so that they contact by these segments (this is already pointed out in the question).
Remark. If we disallow rotations and reflections. There is a counter-example. Let $A$ be the $3\times 6$ rectangle with the following 6 cells removed: one in the middle of each short side and two in the middle of each long side. (Sorry I don't know how to make a picture here.) Let $B$ be the same figure rotated 90 degrees. Then $A$ and $B$ cannot have two common edges without rotation.<|endoftext|>
TITLE: Shortlists and job offers
QUESTION [11 upvotes]: I hope this question is appropriate for MO. I started application process this year. I've searched several online ads for a job and found a wiki page, which contains names of people that have been shortlisted. I've got listed in couple places as well and I wonder if this can harm my application process. I would like to know pros and cons of this open source.

REPLY [13 votes]: I would think that having it known that you're short-listed at some places would help rather than harm your chances at other places.  If some department ignored your application because they failed to notice anything special and it was one among $N$ (and $N\to\infty$), knowing that you're on some other place's short list might cause them to take another look.  You're probably worried that some departments will assume that you'll get an offer from better places, with which they can't compete, so they won't seriously consider you.  I doubt that this would be a real problem; hiring committees are well aware that there's a difference between being on a short list and actually getting an offer.<|endoftext|>
TITLE: Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$?
QUESTION [14 upvotes]: Let $k$ be a field. For $R=k[x_1,\ldots]$ with countably infinite number of variables, [due to the discussion in the comments] we have to make the following distinction between $k[[x_1,\ldots]]$ and the completion $\hat{R}$ of $R$ at the ideal $(x_1,\ldots)$: note that the former admits elements which can have infinitely many monomials of the same degree whereas the latter can not (e.g. $\sum_i x_i\in k[[x_1,\ldots]]$ but $\notin\hat{R}$). There are two questions
1) Is the morphism $R\to\hat{R}$ flat? If $R$ were any Noetherian ring, the map $R\to \hat{R}$ to its completion is always flat (see e.g. Atiyah-MacDonald 10.14) but my intuition breaks down for non-Noetherian rings.
2) Is the morphism $R\to k[[x_1,\ldots]]$ flat? This is answered positively by ayanta below.

REPLY [20 votes]: Let's suppose by $k[[x]]$ we mean the formal power series ring in variables $x_1, x_2, \dots$ which is literally the space of sequences of monomials that individually involve only finitely many variables per monomial (so set-theoretically a direct product of copies of $k$ indexed by such monomials, with a "cofinite" topology).  This is of course different from the $(x)$-adic completion of $k[x] := k[x_1,x_2,\dots]$ as noted by Francois, since the latter has as a cofinal system of discrete quotients the rings $k[x]/(x)^m$ of infinite $k$-dimension whereas the former has as a cofinal system of discrete quotients the artinian $k[x_1,\dots,x_r]/(x_1,\dots,x_r)^m$ of finite $k$-dimension.
I claim that $k[[x]]$ in the sense I have specified is flat over $k[x]$ (though I also think this is probably completely useless and so I don't claim this is interesting -- maybe just  amusing).  The key input is buried near the end of volume 1 of SGA3. These methods have no relevance to the $(x)$-adic completion of $k[x]$ (which is an entirely different beast than $k[[x]]$ as defined above).
First, some preliminary reductions.  We have to show that if $I$ is a finitely generated ideal of $k[x]$ then the injection $I \rightarrow k[x]$ remains injective after tensoring against $k[[x]]$ over $k[x]$.  By finite generation, $I$ "comes from" an ideal $I' \subset k[x_1,\dots,x_r]$ for some $r$, and more specifically the natural map $$I' \otimes_{k[x_1,\dots,x_r]} k[x] \rightarrow k[x]$$ is injective since $k[x]$ is certainly flat (even free) over $k[x_1,\dots,x_r]$.  So in fact 
$$I = I' \otimes_{k[x_1,\dots,x_r]} k[x],$$
and hence our problem is to show that the injection $I' \rightarrow k[x_1,\dots,x_r]$ remains injective after tensoring over $k[x_1,\dots,x_r]$ against $k[[x]]$.  More specifically, we claim this latter ring map is flat.
This final scalar extension process decomposes as a composition of two scalar extensions:
$$k[x_1,\dots,x_r] \rightarrow k[[x_1,\dots,x_r]] \rightarrow k[[x]].$$
Since the first step is known to be flat by usual commutative algebra with noetherian ring, we're reduced to proving flatness of the second map.  But this is a special case of the Gabriel-Grothendieck theory of pseudo-compact rings in SGA3, in which they systematically develop a good theory of "pseudo-compact modules" and "topological flatness" for "pseudo-compact rings", which are topological rings that are arbitrary inverse limits of artinian rings.  This theory includes as a key ingredient a relationship between topological flatness and ordinary flatness when the base ring is noetherian (analogous to completions in the noetherian setting, but logically requiring more work).
More specifically, since $A := k[[x_1,\dots,x_r]]$ is noetherian, so any finitely generated $A$-module is finitely presented (and is pseudo-compact for its max-adic topology), for any finitely generated $A$-module $M$ and pseudo-compact $A$-algebra $A'$ the natural map
$$M \otimes_A A' \rightarrow M \widehat{\otimes}_A A'$$
is bijective (ultimately because the left side is a cokernel of a map between finite free $A'$-modules and any such map automatically has closed image by a variant of Artin-Rees proved in SGA3).  Thus, the preservation of injectivity of the left as a functor in finitely generated $M$ (which is equivalent to $A$-flatness of $A'$) is reduced to topological flatness of $A'$ over $A$.  
Note that one can "distribute" formal power series over other formal power series when extracting out a finite set of variables into the coefficients over infinitely many variables (think for a minute, using our running definition of "formal power series" for a possibly infinite set of variables). Thus, in our case of interest $A' = k[[x]]$ is a formal power series ring over $A = k[[x_1,\dots,x_r]]$ in infinitely many variables. Thus, we're finally reduced to the question:  if $A$ is a pseudo-compact ring (such as $k[[x_1,\dots,x_r]]$) then is $A[[y_1,\dots]]$ topologically flat over $A$?  The answer is "yes" because such formal power series rings (in the sense of our running definition) are "topologically free", and a basic fact in the theory is that topological freeness (suitably defined...) implies topological flatness.
QED<|endoftext|>
TITLE: What are conjectures that are true for primes but then turned out to be false for some composite number?
QUESTION [10 upvotes]: Note: This is an update formulation since many people misunderstood the question before.
Of course it is easy to make a statement like "Every n is a prime or at most 1000", which is true for every prime $n$ and every small $n$ but fails for $n=1002$. What are "real" conjectures that were known to hold for primes and small values, then turned out to be false?
An excellent example about cyclotomic polynomials was given by Aaron in the comments. Here the conjecture was that the coefficients are $0, \pm 1$ for every $n$. This holds for primes and small $n$'s, but fails for $105$.
Also the existence of Carmichael numbers comes close, but here the problem itself involves primes, I would like something less "primey". I know conjuctures that are or were known only for primes. Recently solved is Colorful Tverberg, still unknown is Evasiveness or this little MO problem.

REPLY [6 votes]: An aliquot sequence is a recursive sequence in which each term is the sum of the proper divisors of the previous term. Catalan conjectured that no aliquot sequence is unbounded. The conjecture is trivially true for primes, since the aliquot sequence for $p$ is $p,1,0$ (terminating, since the sum of the proper divisors of zero is undefined). The conjecture is also true, by computation, for all $n\le275$. Strictly speaking, this doesn't qualify as an answer to the question, since there is no $n$ for which the conjecture has been proved to be false, but there are so many integers for which the aliquot sequence shows no sign of ever reaching a bound that it seems most plausible to conclude that the conjecture is false. 
Here is a website dedicated to the problem.

REPLY [4 votes]: An important question in matroid structure theory is given a matroid $M$, how many inequivalent representations does it have over a fixed finite field $\mathbb{F}$?  Two $\mathbb{F}$-matrices $A$ and $B$ are equivalent if they are related by a projective transformation, in which case they clearly represent the same matroid.  
In 1988, Kahn proved that $3$-connected $GF(4)$-representable matroids have a unique representation over $GF(4)$.  In the same paper, he conjectured that for every prime power $q$, there is an $n(q)$, such that every $3$-connected $GF(q)$-representable matroid has at most $n(q)$ inequivalent representations. Oxley, Vertigan and Whittle proved Kahn's conjecture for $GF(5)$, but also showed that it fails for all larger finite fields.  This naturally led to the question of what happens for $4$-connectivity, and the following weakening of Kahn's Conjecture. 
Conjecture 1. For every prime power $q$, there is an $n(q)$ such that every $4$-connected $GF(q)$-representable matroid has at most $n(q)$ inequivalent representations over $GF(q)$.
One would expect that Conjecture $1$ holds for all finite fields, or fails for sufficiently large finite fields. However, remarkably there is a dichotomy depending on whether $q$ is prime!  
Theorem 2. (Geelen and Whittle) For every prime $p$, there is an $n(p)$ such that every $4$-connected $GF(p)$-representable matroid has at most $n(p)$ inequivalent representations over $GF(p)$.
The proof of Theorem $2$ is not easy; the paper is $175$ pages.  
Theorem 3. (Geelen, Gerards and Whittle)
For every non-prime power $q \geq 9$, there is a $4$-connected $GF(q)$-representable matroid with arbitrarily many inequivalent representations over $GF(q)$.<|endoftext|>
TITLE: Forms satisfying the zero-energy condition on the projective plane 
QUESTION [5 upvotes]: Theorem (Michel). A $1$-form on the projective plane is exact if and only if its integral over any projective line is equal to zero.
Is there a simple proof of this result due, I think, to R. Michel ?
I'm guessing there must be a representation theoretic proof (everything is $SL(3;\mathbb{R})$ equivariant) and a complex-geometric proof (where $\mathbb{R}P^2$ is complexified to $\mathbb{C}P^2$ ) and I would appreciate reference for these, but I'm really interested in a simple proof that one could teach to seniors or first year grads. 
P.S. I have not been able to get a hold of Michel's papers. Perhaps everything is there and in that case I apologize before hand.

REPLY [11 votes]: There is a simple proof along the following lines:  Because the first deRham cohomology group of $\mathbb{RP}^2$ is trivial, a $1$-form on $\mathbb{RP}^2$ is exact if and only if it is closed.  Let $\alpha$ be a $1$-form on $\mathbb{RP}^2$ whose integral over every line vanishes, and let $\beta = \pi^*\alpha$ where $\pi:S^2\to \mathbb{RP}^2$ is the standard double cover.  Then the integral of $\beta$ over every great circle vanishes and $\beta$ is invariant under the antipodal involution $\iota:S^2\to S^2$, so $d\beta$ is a $2$-form on $S^2$ that is $\iota$-invariant and its integral over every hemisphere must vanish.  
Now $d\beta = b\ dA$ where $dA$ is the standard volume form on $S^2$ and we must have $b\circ\iota=-b$ in order for $d\beta$ to be invariant.  Now you use the representation theoretic fact that the operation $A:C^\infty(S^2)\to C^\infty(S^2)$ defined by
$$
Af(u) = \frac1{2\pi}\int_{v\cdot u\ge0} f\ dA
$$
for $u\in S^2$ is $\mathrm{SO}(3)$-equivariant and hence must, on each eigenspace of the Laplacian on $S^2$, be a multiple of the identity (since these eigenspaces are irreducible representations of $\mathrm{SO}(3)$).  To complete the proof that $Ab=0$ implies $b=0$ (and hence that $d\beta=0$), one just needs to check that $A$ is nonzero on each $\iota$-odd eigenspace of the Laplacian, and this is straightforward (see below for one proof of this).
Added comment:  Here's a simple way to see that $A$ is nonzero on the odd eigenspaces (without having to do any explicit integration). The $d$-th eigenspace $H_d$ of the Laplacian on $S^2$ is simply the restriction to $S^2$ of the harmonic polynomials on $\mathbb{R}^3$ that are homogeneous of degree $d$.  Since $A$ acts as a multiple of the identity on each of these spaces, it's enough to show that, when $d=2m{+}1$, $A$ is nonzero on at least one element of $H_d$.  To do this, let $x$, $y$, and $z$ be the standard coordinates on $\mathbb{R}^3$ and consider the polynomial
$$
f = \mathrm{Re}\bigl((x+i y)^{2m+1}\bigr) 
= \prod_{k=-m}^m \left( x-\tan\left(\frac{k\pi}{2m{+}1}\right) y\right).
$$
Clearly, $f$ belongs to $H_{2m+1}$ and $f(1,0,0)=1\not=0$.  Moreover, $f$ is odd with respect to rotation by an angle of $\pi/(2m{+}1)$ about the $z$-axis, and it vanishes on the $2m{+}1$ planes given by the above factors, which divide the sphere into $4m{+}2$ congruent 'sectors' or 'spherical wedges', on half of which $f$ is positive and on half of which $f$ is negative. Now, $Af(1,0,0)$ is the average of $f$ over the hemisphere on which $x\ge0$, and this is composed of $2m{+}1$ of these sectors, which alternate in the sign of $f$.  Since there is an odd number of them, their integrals can't cancel out, so $Af(1,0,0)$ cannot be zero.  Thus, $A$ is nonzero on the odd harmonics, as was to be shown.<|endoftext|>
TITLE: Elementary examples of the Weil conjectures
QUESTION [37 upvotes]: I'm looking for examples of the Weil conjectures---specifically rationality of the zeta function---that can be appreciated with minimal background in algebraic geometry.  Are there varieties for which one can easily calculate the numbers of points over finite fields and witness the rationality directly?  Of course, there are entirely straightforward examples coming from projective space and Grassmannians (or anything with a paving by affines).

REPLY [6 votes]: Any variety which admits a cell decomposition gives an example. Cell decomposition of $X$ is a stratification $X=\bigcup\limits_n X^n$ such that $X^n\setminus X^{n-1}=\coprod\limits_{i=1}^{k_n}\mathbb{A}^n$ is a disjoint union of affine spaces.
Then $\#X(\mathbb{F}_q)=k_0+k_1q+\dots+k_dq^d$ so zeta-function of $X$ is a product of zeta-functions of affine spaces hence rational.
Examples of such varieties are given by flag varieties of reductive groups(Bruhat decomposition).
More generally, rationality of zeta-functions of moduli spaces  of some objects can be proven elementary by counting these objects over finite field. For example, for quiver varieties it is done by V. Kac in "Root systems, representation of quivers and invariant theory".
Edit: Another, less elementary example is given by rational surfaces. For them Chow groups satisfiy the axioms of Weil cohomology theory, as explained in Manin's book "Cubic forms", so the rationality follows.<|endoftext|>
TITLE: Can one exhibit an explicit Kuratowski infinite set without invoking Replacement?
QUESTION [15 upvotes]: The customary formulation of the Axiom of Infinity within Zermelo-Fraenkel set theory asserts the existence of an inductive set: a  set  $ I$ with  $\varnothing\in  I$  such  that $x\in I$ implies $x\cup\{x\}\in I$.  Since the intersection of any nonempty set of inductive sets is itself inductive, an instance of  the Axiom Schema of Separation implies the existence of  a smallest inductive set, namely the set of von Neumann naturals $$\mathbb{N}_{\bf  vN} = \{\varnothing, \{\varnothing\},\{\varnothing,\{\varnothing\}\},\ldots\}.$$
Any inductive set is infinite (in fact, Dedekind infinite) but this formulation of the axiom  asserts more, namely the existence of a specific countably infinite set.  Given one such set, the existence of others, for example  the set of  Zermelo naturals $$\mathbb{N}_{\bf Zer}=\{\varnothing,\{\varnothing\},\{\{\varnothing\}\},\ldots\}$$  follows from appropriate instances of the Axiom Schema of Replacement. 
Consider the subsystem of  Zermelo-Fraenkel set  theory  with axioms Extensionality, Separation Schema, Union, Power Set, Pair.  Augment  this  Basic System with  an  Axiom of Infinity which asserts the existence of an infinite set, but not any particular one. Such a  formulation requires that the notion of 'finite' be defined prior to that of 'natural number',  following Kuratowski for example.  Any infinite set $ I$ determines a Dedekind-infinite  set of local naturals $$\mathbb{N}_I=\{\mbox{equinumerosity classes of finite  subsets of }  I\}$$ which (duly equipped with initial element and  successorship) yields a  Lawvere natural  number  object, as  in  the Recursion  Theorem. The  existence of  $\mathbb{N}_{\bf vN}$ and $\mathbb{N}_{\bf Zer}$ then follow from  appropriate instances of Replacement.
One might wonder  if  there is some clever way to specify an infinite  set without  recourse to Replacement. That is, does there exist (in the  language of set  theory)   a formula $\boldsymbol \phi$
with one  free variable  $x$ such  that
$$ \mbox{Basic+Infinity+Foundation } \vdash\; \exists y ( \forall  x (x\in y \leftrightarrow \boldsymbol \phi)\,\wedge \, y \mbox{ is  infinite})\>?$$
I'm inclined  to  guess no, on the following circumstantial  grounds:

For  $\mathbb{N}_{\bf vN}$ and $\mathbb{N}_{\bf Zer}$  the  use of Replacement  is  essential: Mathias has shown (Theorem 5.6 of Slim  Models of Zermelo Set Theory  that there exist transitive models ${\mathfrak  M}_{\bf  vN}$ and ${\mathfrak M}_{\bf Zer}$  of Basic+Infinity+Foundation with  ${\mathbb  N}_{\bf vN}\in {\mathfrak M}_{\bf vN}$  and ${\mathbb  N}_{\bf Zer}\in{ \mathfrak M}_{\bf Zer}$, but  such  that every  element  of ${\mathfrak M}_{\bf  vN}\cap {\mathfrak M}_{\bf Zer}$  is hereditarily finite.
The usual  definitions  of  $\mathbb{N}_{\bf vN}$ and $\mathbb{N}_{\bf Zer}$  involve unstratified formulas. Coret has shown (Corollary 9 of  Sur les cas stratifiés du schéma du replacement) that this is unavoidable:
$$ \mbox{Basic+Infinity } \vdash\; \forall y ( \forall  x (x\in y \leftrightarrow  \boldsymbol \phi)\,\rightarrow \, y \mbox{ is  hereditarily finite})$$
for any stratified  $\boldsymbol \phi$. Using  the  same technique he has shown  (Corollary 10) that  Basic+Infinity proves  every stratified instance of Replacement.

REPLY [9 votes]: The answer is no. Let ZC' be ZFC without replacement and infinity and with the assertion there is a Kuratowski infinite set. We will construct a model $M$ of ZC' such that only hereditarily finite elements of $M$ are fixed under all automorphisms of $M.$ The idea is generate a model from a $\mathbb{Z}^2$-array of objects, each of whose only element is the object below it in the array.
Define $M=\bigcup_{m<\omega} \bigcup_{n \in \mathbb{Z}} \mathcal{P}^m(\{\omega\} \times \mathbb{Z} \times [n, \infty)),$ where we adjust the $\mathcal{P}$ operator to replace each singleton of the form $\{(\omega, n_1, n_2)\}$ with $(\omega, n_1, n_2+1).$ (We use $\omega$ to differentiate the $\mathbb{Z}^2$ objects from the hereditarily finite sets).
Define a relation $E$ on $M$ by $E \restriction \mathbb{Z}^2=\{((\omega, n_1, n_2),(\omega, n_1, n_2+1)): n_1, n_2 \in \mathbb{Z}\}.$ Extending $E$ to the iterated power sets is done in the natural way. It is easy to see $(M, E)$ satisfies ZC'.
Notice that the map $(\omega, n_1, n_2) \mapsto (\omega, n_1, n_2+1)$ extends to a unique automorphism on $M,$ which only fixes hereditarily finite elements. Thus, none of the infinite sets in the model are first-order definable.
Update: The author of this question was curious whether there is such a model which also satisfies transitive containment (TC), the assertion that every set has a transitive closure. The answer is yes; here's a sketch of such a model.
Assume ZFC + "there are non-OD reals" in $V$ (or work in a ctm satisfying a sufficiently large finite fragment of this theory). We'll build an OD model $M$ of ZC'+TC such that, from every infinite element $x,$ we can define (in $V$) a non-OD real $r.$ Then $x$ is not definable in $M,$ or else we would have an ordinal definition for $r.$
For $r \in \mathbb{R} = \mathcal{P}(\omega),$ we recursively define $n_r$ by $0_r=\emptyset$ and $(n+1)_r =\{n_r\}$ if $n \not \in r,$ and $(n+1)_r = \{n_r, \emptyset\}$ if $n \in r.$ Basically, we're defining a new system of natural numbers for each real.
Let $M = \bigcup_{m=0}^{\infty} \bigcup_{S \in [\mathbb{R} \setminus OD]^{<\omega}} \mathcal{P}^m ( \{n_r: n<\omega, r \in S\} ).$
For each infinite $x,$ there are finitely many (and at least one) non-OD reals $r$ such that the transitive closure of $x$ contains every $n_r.$ The least such real is as desired. And it's routine to verify $M \models ZC'+TC.$<|endoftext|>
TITLE: faces in the discrete cube
QUESTION [6 upvotes]: This arose from a question Gil Kalai asked about a problem I posed involving the Fourier transform on the discrete cube. Maybe it is more tractable. I'm afraid I'm not sure how to do this kind of computation.
A $k$-dimensional face of the discrete cube $\{0,1\}^n$ is a set of the form: all vertices which take prescribed values (either $0$ or $1$) on some given $n-k$ coordinates and are otherwise arbitrary.
The question is: does a typical subset of $\{0,1\}^n$ approximately contain a face of dimension greater than $.6n$?
We are interested in the limit as $n\to\infty$. So "approximately contains" means "contains all but a fraction which goes to $0$ as $n\to\infty$". And "typical subset" means that as $n\to\infty$ the fraction of subsets for which this fails goes to zero. The $.6n$ can be moved a bit closer to $.5n$ but I am assuming this is not crucial.
A positive answer to this question would imply a generically positive answer to the Fourier transform question.

REPLY [9 votes]: No. The typical set doesn't contain anything like a 0.6n face. Some similar questions are considered in "The Probabilistic Method" by Alon and Spencer (which I thoroughly recommend).
Here is the calculation. Let's just deal with 0.5n faces. I want to make a crude estimate of the probability that a subset (chosen uniformly) contains 90% of a 0.5n face.
The bound I'll use is the union bound: there are $\binom{n}{0.5n} 2^{n/2}$ $0.5n$-faces (choose which indices you want to restrict, and then decide the values you want to give them). This is considerably less than $2^{2n}$.
Now what is the probability that a random subset has density at least 90% in a given 0.5$n$-face?
I'll compute the probability that a random subset has density exactly 90% in a given 0.5$n$-face, since as you increase the density, the probability decays geometrically, so the first term dominates.
Since there are $2^{n/2}$ elements in the face, we're now asking for 
$$
\binom{2^{n/2}}{0.1\times 2^{n/2}}2^{-2^{n/2}}.
$$
This is the number of ways of having exactly $0.9\times 2^{n/2}$ ones out of the $2^{n/2}$ possibilities
(assume all numbers are integers by taking floors).
A liberal dose of Stirling's formula shows that the binomial coefficient is $(0.1^{0.1}0.9^{0.9})^{-2^{n/2}}/2^{n/4}$ up to multiplicative constants, so that the number of configurations we're looking for is essentially $(0.1^{0.1}0.9^{0.9}\times 2)^{-2^{n/2}}/2^{n/4}$.
Since $0.1^{0.1}0.9^{0.9}>\frac12$, this decays fast (even when multiplied by $2^{2n}$).<|endoftext|>
TITLE: A question on the "natural metric" on the space of Bridgeland stability condition
QUESTION [7 upvotes]: I have a question on the "natural metric" on the space of Bridgeland stability condition. 
A stability condition $\sigma=(Z,\mathcal{P})$ on a triangulated category $\mathcal{D}$ consists of a linear map $Z:K(\mathcal{D})\rightarrow \mathbb{C}$ called the central charge, and full additive subcacegories $\mathcal{P}(\phi) \subset \mathcal{D}$ for each $\phi \in \mathbb{R}$, satisfying the following four conditions:

if non-zero $E \in \mathcal{P}(\phi)$, then we have $Z(E) = m_\sigma(E)exp(i\pi \phi)$ for some $m(E)\in \mathbb{R}_{+}$
forall $\phi \in \mathbb{R}$, we have $\mathcal{P}(\phi+1)=\mathcal{P}(\phi)[1]$
if $\phi_1 > \phi_2$ and $E_j \in \mathcal{P}(\phi_i)$ then $Hom_{\mathcal{D}}(E_1,E_2) = 0$
for non-zero $E \in \mathcal{D}$ there exists a finite sequence of real numbers $\phi_1 >\phi_2 >\dots>\phi_n$ and $E$ obtained as an "iterated extension" of objects $A_i \in \mathcal{P}(\phi_i)$. 

There is a "natural metric" on $Stab(\mathcal{D})$ defined on page 7 of this paper. 
A celebrated result by Bridgeland says the forgetful map
$$
\mathcal{Z}:Stab(\mathcal{D})\longrightarrow Hom_{\mathbb{Z}}(K(\mathcal{D}),\mathbb{C})
$$
induces a local homeomorphism on each connected component of $Stab(\mathcal{D})$. This seems a really nice theorem.
This generalized metric is at this point beyond my intuition and I cannot really follow the proof of the theorem above, so let me now ask

Why is the generalized metric above is "natural"? 

Of course some people may say it is the right one because the theorem holds. But I guess it is not the only reason. My problem is that I cannot really see why "distance" of two stability condition is measured by the only three quantities in $\sup_{0\ne E\in \mathcal{D}}$.
Edit
Are there any good toy example with which one can appreciate the metric or topology above? e

REPLY [4 votes]: The metric may be important for physics (and as far a I understand it comes from some physical  considerations), but mathematically the important thing is the topology. From this point of view the definition of the metric just expresses the fact that two stability conditions are close to each other if all objects have similar phases ranges in those (in particular an object stable in one stability condition is almost stable in the other), and have close masses. The particular form of the metric is chosen to satisfy the triangle inequality (I think this is an explanation of the logarithm there).<|endoftext|>
TITLE: Morse lemma with least amount of regularity.
QUESTION [7 upvotes]: I recently came across with $C^2$ Morse functions in my work and as I was reviewing some of the stuff I learned about Morse theory, I noticed that all the proofs of the Morse lemma I could come across with work only for $C^3$ Morse functions. 
A Google search was inconclusive about the existence of a Morse lemma for Morse functions $f: M \to \Bbb R$ with lesser regularity then $C^3$, where $M$ is a smooth finite dimensional manifold.
A reference is perhaps the best possible answer, but any chunk of information will be appreciated.

REPLY [4 votes]: Requiring $C^2$ is too much: you can ask only $C^1$, twice differentiable at the distinguished point with a non-degenerate Hessian matrix. More precisely the following holds true.
Theorem.
Let $\Omega$ be an open subset of $\mathbb R^n$, $x_{0}\in \Omega$ and $f:\Omega\longrightarrow \mathbb R$ be a $C^1$  function twice differentiable at $x_{0}$
such that
$$
df(x_{0})=0,\quad\det f''(x_{0})\not=0.
$$
Then there exist a neighborhood $V$ of $0$ in $\mathbb R^n$,
 a neighborhood $U$ of $x_{0}$ in $\Omega$, and a $C^1$ diffeomorphism $\kappa:V\longrightarrow U$
 such that $\kappa(0)=x_{0},\kappa'(0)=Id,$
 \begin{equation}
(f\circ \kappa)(y)=f(x_{0})+\frac12\langle{ f''(x_{0})y},{y}\rangle.
\end{equation}
For a proof, try your hand or check
Santiago Lopez de Medrano, MR 1378414.<|endoftext|>
TITLE: Relative consistency of ETCS over the theory of a well-pointed topos with NNO
QUESTION [21 upvotes]: EDIT: I'm bumping this, because I'm still curious, and because I have a relative consistency result over the theory of a well-pointed topos with NNO, and I am wondering how much baggage I save by not assuming AC in the base model (I'm not using AC in the proof either).
Gödel's well-known proof of the implication $Con(ZF) \Rightarrow Con(ZFC)$ used the construction of the inner model $L$ in $ZF$ to get a model of $ZFC$ (and in fact much more). However such a construction is not (immediately) available in a category-theoretic approach to set theory. In particular, given a well-pointed topos with NNO, which is the set theory ETCS minus the axiom of choice, I wonder whether there is any way to construct a model of ETCS. On the face of it, it doesn't seem likely, as objects of the given topos are quite amorphous. 
The only thing I can think of (admittedly I haven't tried very hard) is by passing to a model of ZF via pure sets, constructing $L$, and then taking the category of sets of $L$. But this is somewhat unsatisfactory, as it leaves the comfy realm of categories and heads out into material set theory. So:

Is there a category-theoretic construction of a model of ETCS from a well-pointed topos with NNO?

REPLY [3 votes]: I confess that I am not altogether clear what the Question requires, whilst many of the Comments are beyond me, but here is a construction of which you may not be aware and which may throw some light on the issues.
It is based on my paper Intuitionistic Sets and Ordinals, in the Journal of Symbolic Logic 61 (1996) 705-744, particularly Section 3. This in turn built on Categorical Set Theory: a characterisation of the Category of Sets by Gerhard Osius in the Journal of Pure and Applied Algebra 4 (1974) 79-119.
We are working in some given elementary topos, whose objects I will call carriers in order to avoid the word set.
A model of a fragment of set theory is given by a carrier $X$ equipped with a binary relation $\epsilon$ that it is convenient to regard as a map $\epsilon:X\to P X$ to the powerset.  That is, it is a coalgebra for the covariant powerset functor.
We say that $(X,\epsilon)$ is extensional if the map $\epsilon:X\to P X$ is mono (1-1).
The definition of when the coalgebra is well founded involves a pullback diagram and is given in the paper.
Then an ensemble is an extensional well founded coalgebra.
Applying the covariant powerset functor to one ensemble gives another.
Between any two ensembles there is at most one coalgebra homomorphism, and it is mono.  This captures inclusion in the set-theoretic sense.
The category of ensembles is therefore a preorder. It is a large category in the same sense that familiar categories such as that of groups in a topos are large.
We can rehearse the standard definitions of pairing, functions, etc from set theory using ensembles and show that this preorder provides a model of the Zermelo axioms.
Natural numbers (infinity) and the axiom of choice are inherited from the given topos if it has them.
As to the axiom-scheme of replacement, understanding this from a categorical perspective was one of my principal aims in this work.  The comments in my JSL paper are really quite naive and should be discounted in favour of those at the very end of my book, However, I got interested in other things and never pursued this to a satisfactory conclusion.<|endoftext|>
TITLE: submit the second part of a paper
QUESTION [11 upvotes]: I couldn't find similar question being asked here. The closest one I can find is When to split/merge papers?. Here is my situation: I proved a theorem. When I try to type it, I found that it's very long. Since it's long, I splitted it into two parts. I finished the first part (50+ pages) and submitted to a journal few months ago. Now I almost finished typing the second part which is also 50+ pages. My question is:  Should I submit the second part to the same journal? If I submit to the same journal, probably the editor will send the second part to the same referee. Then the referee who is familar with the first part can read the second part more easily. However, as we all know, it's hard to publish long paper, I think it would be even harder to publish two long papers in the same journal. However, if I submit to another journal, it may be even harder since the new referee may find it difficult to read the second part without reading the first part. So this also gives me the second question: should I wait until the first part is published/accepted, and then submit the second part?

REPLY [3 votes]: First of all, I wonder how you can be sure that a 100-pages-proof is correct before you even have typed it in completely -- at least unless it consists mostly of routine computations!
As to your question, I think one proof should ideally be given in one paper, or a least in two papers in the same journal. -- Well, of course there are some proofs distributed over MANY papers, like the one of the Classification of Finite Simple Groups, but that's probably something different ... .<|endoftext|>
TITLE: The symmetric monoidal category of finite sets
QUESTION [16 upvotes]: It is well-known that the (augmented) simplex category is the universal monoidal category with a monoid object. What about a commutative analogue? Consider the category $\mathsf{FinSet}$ of finite sets. It is a symmetric monoidal category with tensor product $\coprod$ and unit $\emptyset$. It contains a commutative monoid $\{\star\}$, which seems to be the universal one: For every symmetric monoidal category $\mathcal{C}$ the assignment $F \mapsto F(\{\star\})$ provides an equivalences of categories
$\mathrm{Hom}_{\otimes}(\mathsf{FinSet},\mathcal{C}) \cong \mathrm{CMon}(\mathcal{C}).$
Here, $\mathrm{Hom}_{\otimes}$ denotes the category of strong symmetric monoidal functors (not assumed to be strict).
Question 1. This should be well-known, is there a reference in the literature?
Question 2. Is there any description of the category $\mathsf{FinSet}$ using generators and relations, i.e. an elementary description of $\mathrm{Hom}_{\mathrm{Cat}}(\mathsf{FinSet},-)$? Since every map of finite sets is a bijection followed by a monotonic map, or vice versa, I expect that we need face maps, degeneracies and transpositions as generators. What are the relations? Is this written down in the literature?
Question 3. An answer to question 2 will also describe presheaves on $\mathsf{FinSet}$, which are simplicial sets with a certain extra structure. Do they have a geometric interpretation and are these geometric objects used somewhere? In some sense, this corrects the failure of the join of simplicial sets to be commutative.
Edit after Eric's comment. Ok, Q3 was already answered on MO. The objects are called symmetric simplicial sets. Relevant papers are Higher Fundamental Functors for Simplicial Sets by M. Grandis and Toposes generated by codiscrete objects in combinatorial topology and functional analysis by F. W. Lawvere and Left-determined model categories and universal
homotopy theories (Section 3) by J. Rosicky and W. Tholen.

REPLY [15 votes]: Marco Grandis has done some work on this, and you can extract answers for 1-3 from these papers

Finite Sets and Symmetric Simplicial Sets - M Grandis - TAC (pdf)
Higher Fundamental Functors for Simplicial Sets - M Grandis - CTGDC (pdf)

See also

An Alternative Presentation of the Symmetric-Simplicial Category - Eric R. Antokoletz - arxiv (link)

Question 1 and 2
The first paper by Grandis above gives a nice detailed overview of all this, including 

a description of (a skeleton of) $\mathbf{FinSet}$ as the walking symmetric strict monoidal category with a commutative monoid
in terms of generators and relations, with generators faces + degeneracies + transpositions, and relations the standard ones for faces + degeneracies, the Moore ones for transpositions, and some compatibility rels between those.

About who first proved this kind of things, in the second paper, he acknowledges that:

In November 1998, at a PSSL meeting in Trieste, Bill
  Lawvere suggested I might extend my study of the homotopy of simplicial
  complexes to symmetric simplicial sets, on the basis of his draft [19] where the
  fundamental groupoid of the latter is presented as a left adjoint. I would like to
  express my gratitude for his kind encouragement and for helpful discussions.

[19] is Lawvere's Toposes generated by codiscrete objects, in combinatorial topology and functional analysis from 1989, which I don't have access to; maybe there's some more info in there.
Question 3
You could find more about this in the second paper by Grandis. The idea is that the classical homotopy theory of simplicial complexes can be extended to symmetric simplicial sets (presheaves on $\mathbf{FinSet}$), so that the edge-path groupoid of a simplicial complex can be identified with what Grandis calls the intrinsic fundamental groupoid, which is the left adjoint of a symmetric nerve (this goes back to Lawvere notes ref above). See also 

An intrinsic homotopy theory for simplicial complexes, with applications to image analysis - M Grandis (pdf)<|endoftext|>
TITLE: Littlewood Richardson rule and seminormal basis of Specht modules
QUESTION [8 upvotes]: Background 
Seminormal Basis of Specht modules of $\mathfrak{S}_n$
Let $\lambda$ be a partition of $n$.  A $\lambda$-tableau   is a
bijection $\mathfrak{t}:\lambda \to \{1,2,...,n\}$.  We say a tableau, $\mathfrak{t}$, is standard if the entries are increasing along the rows and columns.   We let $\mathcal{T}_{\lambda}$ denote the set of standard $\lambda$-tableaux.
Given some tableau $\mathfrak{t}$ and two integers $1\leq i < j\leq n$, we
define the axial distance, $a(i,j)$, as follows: if
  $i$ occurs
in row $i_0$ and column $i_1$ and $j$ occurs in row $j_0$ and column
$j_1$, then $a(i,j)=(i_0-i_1) -(j_0-j_1)$.
If $\mathfrak{t}$ is a $\lambda$-tableau and $w \in \Sigma_n$ let
$w\mathfrak{t}$ be the tableau obtained from $\mathfrak{t}$ by replacing each entry
in $\mathfrak{t}$ by its image under $w$.
If $\mathfrak{t}$ is a standard $\lambda$-tableau, we set
$\mathfrak{t}_{i \leftrightarrow i+1}$ equal to $w \mathfrak{t}$
if this is still a standard 
 $\lambda$-tableau, and 0 otherwise. 
For a given partition $\lambda$ of $n$, the
Specht   module ${\mathbf{S}(\lambda)}$
has a basis given by the set of standard $\lambda$-tableaux.
With respect to this
basis the generators act as follows
\begin{align*}  
 {\rho_{\lambda}}(s_{i,i+1})\mathfrak{t} = \frac{1}{a(i,i+1)}  \mathfrak{t}  + \left(1 + \frac{1}{a(i,i+1)}\right)  \mathfrak{t}_{i \leftrightarrow i+1}   
\end{align*}
This basis is very compatible with induction and restriction rules (see Seminormal representations of Weyl groups and Iwahori-Hecke algebras, Arun Ram).
The Littlewood--Richardson rule
The LR rule describes the coefficients in the restriction 
$$\mathbf{S}(\nu)\downarrow_{\mathfrak{S}_{r_1}\times \mathfrak{S}_{r_2}} \cong \oplus c^{\nu}_{\lambda,\mu} \mathbf{S}(\lambda) \boxtimes \mathbf{S}(\mu)$$
There are many formulations of this rule.  For example, the Jeu de Taquin version maps standard skew-tableaux of shape $\nu/\lambda$ to those of shape $\mu$.  The LR coef, $c^{\nu}_{\lambda, \mu}$ is the cardinality of the fiber $f^{-1}(\mathfrak{t})$ for any $\mu$-tableau $\mathfrak{t}$.
So we have a map from $\nu$-tableaux to $\lambda \times \mu$-tableaux.  The fibers give the LR coefficients.  However, this map is not a homomorphism of Specht modules.
Question:
Is there a reference for an explicit construction of such a homomorphism?  I.e. a formulation of the LR rule which is compatible with the seminormal bases of Specht modules.

REPLY [5 votes]: These references solve the analogous problem for the general linear groups:

MR2166314 (2006h:20062)  Howe, Roger E. ;  Tan, Eng-Chye ;  Willenbring, Jeb F.
A basis for the $\mathrm{GL}_n$ tensor product algebra,
Adv. Math.  196  (2005),  no. 2, 531–564, doi:10.1016/j.aim.2004.09.007, arXiv:math/0407468

MR2888167  Howe, Roger ;  Lee, Soo Teck .
Why should the Littlewood-Richardson rule be true?
Bull. Amer. Math. Soc. (N.S.)  49  (2012),  no. 2, 187–236, doi:10.1090/S0273-0979-2011-01358-1

MR0955587 (89j:20046)  Tokuyama, Takeshi .
Determinantal method and the Littlewood-Richardson rule,
J. Algebra  117  (1988),  no. 1, 1–18, https://doi.org/10.1016/0021-8693(88)90237-2, (pdf)


with some work, using Schur-Weyl duality, you should be able to solve your problem.<|endoftext|>
TITLE: When two determinantal ideals together generate a power of the maximal ideal?
QUESTION [8 upvotes]: (A somewhat technical question, but maybe it is well known.)  
Consider matrices over the ring $k[[x_1,\dots,x_n]]$, whose entries vanish at the origin (i.e. belong to the maximal ideal $\mathfrak{m}$). Denote by $J(A_{k,l})$ the ideal of maximal minors of the matrix  $A_{k,l}\in Mat(k,l,\mathfrak{m})$.
Given two generic enough (and mutually generic!) matrices, consider the ideal generated by all the maximal minors: $\Big( J(A_{k_1,l_1}),J(B_{k_2,l_2})  \Big)$. (Assume $k_1\le l_1$ and $k_2\le l_2$.)
'{\bf Q.}': When does this ideal contain $\mathfrak{m}^{k_1+k_2-1}$?
Of course, an obvious necessary condition is: $(l_1-k_1+1)+(l_2-k_2+1)\ge n$, but is this condition sufficient? 
example 1. Suppose $k_1=l_1$, $k_2=l_2$, then for the generic case we have: det(A) 
 is of order $k_1$ and $det(B)$ is of order $k_2$ and the lowest order parts of the two polynomials form a regular sequence. Thus, if $n\le 2$ we get: $(det(A),det(B))\supset\mathfrak{m}^{k_1+k_2-1}$.
example 2. Suppose $k_1=1$, take $A_{1,l}$ with generic entries of first order. Again, one can show that the needed property holds.
What about the general case? How to address such questions?

REPLY [3 votes]: Let me  discuss the graded case, that is the ring is the polynomial ring and the matrices have general homogeneous entries of degree $1$. The local version should follows by taking "lowest order part" as you do in example $1$. 
Consider first of the case of the polynomial ring $S$ in variables $x_{ij}$ and $y_{ij}$ and matrices $X=(x_{ij})$ of size $a\times b$ and $Y=(y_{ij})$ of size $c\times d$ with $a\leq b$ and $c\leq d$. Then take the ideal $I$ of minors of size $a$ of $X$ and the ideal $J$ of minors of size $c$ of $Y$. 
Set $A=S/I+J$. 
The minimal free resolution of $A$ is obtained by taking the tensor product of the resolution of $S/I$ with that of $S/J$ because the ideals are generated by polynomials in different variables. We may use this fact, in combination with the fact that $I$ and $J$ are resolved by the Eagon-Northcot complex, to compute the Castelnuovo-Mumford regularity $\operatorname{reg}(A)$ of $A$, its projective dimension and its dimension. We have: 
$\operatorname{reg}(A)=\operatorname{reg}(S/I)+\operatorname{reg}(S/J)=(a-1)+(b-1)$
$\dim(A) = ab+cd-(b-a+1)-(d-c+1)$
and $A$ is Cohen-Macaulay. 
Now we specialize generically the $x_{ij}$'s and the $y_{ij}$'s  generically to linear forms in variables $z_1,..,z_n$.  The ring you want to understand gets identified with  $A/L$ where $L$ is generated  by $ab+cd-n$ general linear forms in the $x_{ij}$ and $y_{ij}$. 
Now if $n\leq (b-a+1)+(d-c+1)$ then $ab+cd-n\geq ab+cd-(b-a+1)-(d-c+1)$ and $ab+cd-(b-a+1)-(d-c+1)$ of the general linear forms generating $L$ form a maximal regular sequence in $A$. Let $U$ be the ideal generated by $ab+cd-(b-a+1)-(d-c+1)$ of the general linear forms generating $L$. 
Then $\operatorname{reg}(A/U)=\operatorname{reg}(A)=(a-1)+(b-1)$ and $\dim(A/U)=0$. The regularity for a $0$-dimensional module $M$ is the largest index $i$ such that $M_i\neq 0$. It follows that $(A/U)_i=0$ for $i>(a-1)+(b-1)$. And the same is true for $A/L$ (because it is a quotient of $A/U$). Hence we have $(A/L)_i=0$ for $i>(a-1)+(b-1)$. 
On the size of the $z$'s it says that the ideal $(z_1,..,z_n)$ to the power $(a-1)+(b-1)+1$ is contained in ideal of definition, which is exactly what you wanted to prove.<|endoftext|>
TITLE: Injectivity of the Baum-Connes assembly map for locally compact groups
QUESTION [8 upvotes]: Skandalis, Tu and Yu in "The coarse Baum-Connes conjecture and groupoids" proved that:

Let $\Gamma$ be a countable group with a proper left-invariant metric $d$. If $\Gamma$ admits a uniform embedding into Hilbert space, then Baum-Connes assembly map with coefficients is split injective. 

My question is that is it still true if we replace $\Gamma$ by a locally compact second countable Hausdorff topological group $G$? known or unknown?

REPLY [2 votes]: Property A and uniform embedding for locally compact groups by Steven Deprez and Kang Li arXiv:1309.7290
http://arxiv.org/pdf/1309.7290.pdf<|endoftext|>
TITLE: Laurent Polynomials
QUESTION [10 upvotes]: Let $R$ be a commutative ring with identity. Is there any characterization for invertible elements of $R[x,x^{-1}]$ ?

REPLY [3 votes]: Invertible elements of Laurent algebras, and more generally of algebras of torsionfree, cancellable commutative monoids, are characterised in Theorem 11.3 and Corollary 11.4 of Gilmer's Commutative Semigroup Rings (Chicago Lectures in Mathematics, 1984). The proofs given there are quite accessible.<|endoftext|>
TITLE: Linear maps preserving positive semidefiniteness
QUESTION [5 upvotes]: I know of Choi's theorem and some related problems, but not a solution to this exact problem:

Characterize the linear maps from the space $S_n$ of symmetric $n \times n $ matrices to itself that preserve positive semidefiniteness.

It looks a natural question; has a simple characterization been found? Where can I find it?

REPLY [4 votes]: Somewhat surprisingly, this seems to be still open. It is a linear preserver problem, about which there is a nice overview here. But your specific problem seems to be open, according to this recent preprint. They also say that in an earlier paper they settled the problem with some extra assumptions.<|endoftext|>
TITLE: Can we categorify the formula for the quadratic Gauss sum?
QUESTION [17 upvotes]: Background
Fix an odd prime $p$ and set $\zeta=e^{2\pi i/p}$. We define the quadratic Gauss sum as
$$g=\sum_{n=0}^{p-1} \zeta^{n^2}.$$
It's pretty easy to show that
$$g^2=
\begin{cases}
p & \textrm{if } p\equiv 1 \mod 4 \\\
-p & \textrm{if } p\equiv 3 \mod 4,
\end{cases}$$
and from this we can deduce quadratic reciprocity; it's harder to determine the modulus. We can actually find an explicit formula for $g$, namely:
$$g=
\begin{cases}
\sqrt{p} & \textrm{if } p\equiv 1 \mod 4 \\\
i\sqrt{p} & \textrm{if } p\equiv 3 \mod 4.
\end{cases}$$
This is the result I refer to for the remainder of the question.
Question

Can we categorify this result?

By categorification, I mean the opposite of decategorification, and by decategorification, I mean the process of removing structure by e.g. taking the cardinality of a set or the dimension of a vector space. (Thus an example of categorification would be interpreting some combinatorial identity of positive integers as a bijection between sets.) This is intentionally vague, because there are plenty of people who have a much better idea of what constitutes categorification than I do, so feel free to interpret "categorification" liberally.
Motivation
Gauss's original proof of our result uses q-binomial coefficients. (A modern exposition of this proof can be found in "The determination of Gauss sums" by Bruce C. Berndt and Ronald J. Evans.)
Now, $q$-binomial coefficients can be categorified by Grassmannian varieties. What I mean by that is: the $q$-binomial coefficient $\binom{n}{k}_q$ is the number of $k$-dimensional subspaces of an $n$-dimensional vector space over the finite field $\mathbb{F}_q$, i.e. the cardinality of the Grassmannian $\textrm{Gr}(n,k)$. Basically, I'm wondering if there is some way this can be connected to the formula for the quadratic Gauss sum, seeing as how the formula is clearly related to the properties of $q$-binomial coefficients.

REPLY [5 votes]: Since you are using an vague definition of categorification, the following maybe relevant. Gauss sums appear in the theory of (pre)-modular categories. The Gauss sum can be viewed as the sum of values of the quadratic form 
$$\zeta\colon \mathbb{Z}/p\mathbb{Z}\longrightarrow  \mathbb{C}^\times, \quad n\longmapsto \zeta^{n^2}.$$
According to [1], "Premodular or ribbon categories are categorical generalizations of
quadratic forms of finite abelian groups". As a reference for this analogy, see for example Example 8.13.5 and Section 8.4 of [2]. Any quadratic form $\omega\colon G\to \mathbb{C}^\times$ of an abelian group $G$ gives rise to a tensor category $\mathcal{C}(G,\omega)$ with simple objects $X_g$ corresponding to the elements $g$ of $G$. 
A pre-modular category is, a tensor category with duals, a braiding, which is a collection of functorial isomorphisms $c_{X,Y}\colon X\otimes Y\to Y\otimes X$, and other favourable properties like having a finite set of simple objects. To any pre-modular category one can associate the datum of an $S$-matrix which has a motivation from Physics which is a driving force behing the theory of modular categories. The $S$-matrix of the category $\mathcal{C}(G,\omega)$ now corresponds to $\left(b(g,h)\right)_{g,h\in G}$, where 
$$b(g,h)=\frac{\omega(gh)}{\omega(g)\omega(h)}.$$
In general, the $S$-matrix is defined by $S_{X,Y}=\operatorname{Tr}(c_{Y,X}c_{X,Y})$ for representatives of simple objects $X,Y$ in a pre-modular category $\mathcal{C}$. This used the categorical trace $\operatorname{Tr}$ of $\mathcal{C}$ which is defined using duality.
A pre-modular category is modular if the $S$-matrix is non-degenerate. For $\mathcal{C}(G,\omega)$, this condition is equivalent to non-negeneracy of $\omega$. 
Therefore, picking $G=\mathbb{Z}/p\mathbb{Z}$ with the above pairing $\omega=\zeta$ gives a modular tensor category categorifying (the group algebra of) $\mathbb{Z}/p\mathbb{Z}$, which is its Grothendieck ring, together with the Gauss sum, which is given by the sum
$$g=\sum_{g\in G} \theta_{X_g}\dim (X_g)^2,$$
where $\dim(X_g)$ is the categorical dimension of an object $X_g$, and $\theta_{X_g}$ is the scalar defining the so-called twist isomorphism of $X_g$. In the case of $\mathcal{C}(G,\omega)$, $\theta_{X_g}=b(g,g)$. Gauss sums an invariant for general pre-modular categories.
[1] S.-H. Ng, A. Schopieray, and Y. Wang: Higher Gauss Sums of Modular Categories. https://arxiv.org/pdf/1812.11234.pdf
[2] P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik: Tensor Categories, Mathematical Surveys and Monographs, AMS<|endoftext|>
TITLE: Can one recover the smooth Gauss Bonnet theorem from the combinatorial Gauss Bonnet theorem as an appropriate limit?
QUESTION [39 upvotes]: First let me state two known theorems.
Theorem 1 (for smooth manifolds): Let $(M,g)$ be a smooth compact two dimensional Riemannian manifold. Then
$$ \int \frac{K}{2 \pi} dA = \chi (M) $$
where $K$ is the Gaussian curvature, $dA$ is the area form and $\chi(M)$ is the Euler characteristic.
Theorem 2 (combinatorial version): Let $M$ be a two dimensional simplicial complex, with vertices, edges and faces (essentially a bunch of triangles glued together along the edges). Define the function $K:M \rightarrow \mathbb{R}$ to be zero at any point that is not a vertex. At any vertex sum up all the angles 
at the vertex and look at the deviation from $2 \pi$. This is the value of $K$ at a vertex. Then
$$ \sum_{p\in M} \frac{K(p)}{2 \pi} = \chi(M) . $$
Both these statements are known as Gauss Bonnet Theorem. My question is the following: Is it possible to use Theorem 2 in some way to prove Theorem 1? In other words can one recover Theorem 1 from Theorem 2 as some sort of an appropriate "limit"?
A second question is: Is there are a more general theorem from 
which both theorem 1 and theorem 2 arise as special cases? Probably 
some version of the theorem where $K$ simply has to be a measurable 
function?

REPLY [6 votes]: That should just be a comment but it's getting to long.
Also related is the theory of surfaces with bounded integral curvature developed in the 50's by A.D. Alexandrov and its followers. This class of surfaces contains both smooth and polyhedral surfaces.
To be more precise, in Alexandrov words a surface with bounded (integral) curvature is a compact topological surface $(X,d)$ with a geodesic distance $d$ such that for any finite collection of (nice) disjoint geodesic triangles $T_i$, $\sum |e(T_i)|<C$ for some $C$ indecent of the collection of triangles. Here the excess $e(T)$ of a triangle is "sum of angles of $T$ minus $2\pi$". (Actually I put quite a lot of stuff under the rug, you have to define what angles are in this context, and it's not clear they exist, so one consider upper angles instead).
From this one defines, essentially in the way Liviu did, a curvature measure $\omega$ on $M$, wich will be $K_gdv_g$ in the smooth case and $\sum K(p)\delta_p$ in the polyhedral case.
The two versions of Gauss Bonnet you gave can be rephrased as $\omega(M)=2\pi\chi(M)$, which in fact holds for all Alexandrov surfaces.
Moreover, one can deduce each one from the other by an approximation process. One can approach a smooth metric by a sequence of polyhedral metrics or a polyhedral metric by a sequence of smooth metrics in such a way such that the curvature measures with respect the approximating metrics (weakly) converge to the curvature measure of the approximated metric. 
This material can be found in 'The intrinsic geometry of surfaces with bounded curvature' by A.D. Aleksandrov and Zalgaller.<|endoftext|>
TITLE: A lost lemma about periodicity in a grid of long exact sequences?
QUESTION [7 upvotes]: This is a question about finding references and hopefully a larger
context for a lemma in homological algebra I proved recently.
The motivation is to understand properties of characteristic
classes of $T_f$, the mapping torus of a diffeomorphism $f$ of a closed
manifold, by applying the lemma to Mayer-Vietoris and a change-of-coefficients
sequence for the cohomology of $T_f$.
Let $C_{ij}, 1 \leq i,j \leq 3$ be cochain complexes, and
$$ \begin{matrix}
& & 0 & & 0 & & 0 & & \\
& & \downarrow & & \downarrow & & \downarrow & & \\
0 & \to & C_{11} & \stackrel{g}\to & C_{21} & \stackrel{h}\to & C_{31} & \to & 0 \\
& & {\scriptstyle u}\downarrow\  & & {\scriptstyle u}\downarrow\  & & {\scriptstyle u}\downarrow\  & & \\
0 & \to & C_{12} & \stackrel{g}\to & C_{22} & \stackrel{h}\to & C_{32} & \to & 0 \\
& & {\scriptstyle v}\downarrow\  & & {\scriptstyle v}\downarrow\  & & {\scriptstyle v}\downarrow\  & & \\
0 & \to & C_{13} & \stackrel{g}\to & C_{23} & \stackrel{h}\to & C_{33} & \to & 0 \\
& & \downarrow & & \downarrow & & \downarrow & & \\
& & 0 & & 0 & & 0 & &
\end{matrix}$$
a commuting diagram where the rows and columns are short exact sequences.
Let $\delta_H : H^k(C_{3j}) \to H^{k+1}(C_{1j})$ and
$\delta_V : H^k(C_{i3}) \to H^{k+1}(C_{i1})$ denote the boundary homomorphisms in the
associated long exact sequences. The long exact sequences can be arranged into
a commuting grid
$$ \begin{matrix}
H^{k-2}(C_{33}) & \stackrel{\delta_H}\to & H^{k-1}(C_{13}) & \stackrel{g}\to & H^{k-1}(C_{23}) & \stackrel{h}\to & H^{k-1}(C_{33}) & \stackrel{\delta_H}\to & H^k(C_{13}) \\
{\scriptstyle \delta_V}\downarrow\ \  & & {\scriptstyle \delta_V}\downarrow\ \  & &  {\scriptstyle \delta_V}\downarrow\ \ & &  {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \  \\
H^{k-1}(C_{31}) & \stackrel{\delta_H}\to & H^k(C_{11}) & \stackrel{g}\to & H^k(C_{21}) & \stackrel{h}\to & H^k(C_{31}) & \stackrel{\delta_H}\to & H^{k+1}(C_{11}) \\
{\scriptstyle u}\downarrow\  & & {\scriptstyle u}\downarrow\  & & {\scriptstyle u}\downarrow\  & & {\scriptstyle u}\downarrow\  & & {\scriptstyle u}\downarrow\  \\
H^{k-1}(C_{32}) & \stackrel{\delta_H}\to & H^k(C_{12}) & \stackrel{g}\to & H^k(C_{22}) & \stackrel{h}\to & H^k(C_{32}) & \stackrel{\delta_H}\to & H^{k+1}(C_{12})\\
{\scriptstyle v}\downarrow\  & & {\scriptstyle v}\downarrow\  & & {\scriptstyle v}\downarrow\  & & {\scriptstyle v}\downarrow\  & & {\scriptstyle v}\downarrow\  \\
H^{k-1}(C_{33}) & \stackrel{\delta_H}\to & H^k(C_{13}) & \stackrel{g}\to & H^k(C_{23}) & \stackrel{h}\to & H^k(C_{33}) & \stackrel{\delta_H}\to & H^{k+1}(C_{13}) \\
{\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ & & {\scriptstyle \delta_V}\downarrow\ \ \\
H^k(C_{31}) & \stackrel{\delta_H}\to & H^{k+1}(C_{11}) & \stackrel{g}\to & H^{k+1}(C_{21}) & \stackrel{h}\to & H^{k+1}(C_{31}) & \stackrel{\delta_H}\to & H^{k+2}(C_{11}) \\
\end{matrix}$$
The grid is symmetric under translation by 3 steps up and 3 to the right.
Lemma. If $[\alpha] \in H^k(C_{12})$ and $[\beta] \in H^k(C_{21})$ are classes such that $g[\alpha] = u[\beta] \in H^k(C_{22})$ then there is some $[\gamma] \in H^{k-1}(C_{33})$ such that both $\delta_H[\gamma] = v[\alpha] \in H^k(C_{13})$ and $\delta_V[\gamma] = -h[\beta] \in H^k(C_{31})$.
Proof. Take $\chi \in C^{k-1}_{22}$ such that $d\chi = g\alpha - u\beta$. By the definition of the boundary homomorphisms, $d(v\chi) = g(v\alpha)$ implies that $\delta_H([h(v\chi)]) = [v\alpha]$, and $d(h\chi) = -u(h\beta)$ implies that $\delta_V([v(h\chi)]) = -[h\beta]$. Hence we can set $\gamma = vh\chi$.

Does this lemma look familiar? Do you know some place where it's written down?

Edit: Corrected subscripts in statement of lemma.
Update: Thanks for the alternative proofs. However, what I'm after is rather a bibliography reference that I can cite when writing up my application, just to emphasise that it is an instance of something that someone somewhere has already considered (as I imagine it is).

REPLY [2 votes]: One application of your lemma is in differential cohomology.
See e.g. Ex. 3.25 in arxiv. 
I would be very interested in a generalization of this lemma to triangulated categories. So replace your grid of exact sequences by a grid of triangles in a triangulated category. Instead of cohomology you consider the group Hom(T,...)
for a fixed object T. Then you get similar long exact sequence and can state an analogous lemma. Is there a proof in this generality?<|endoftext|>
TITLE: Which metric spaces have this superposition property?
QUESTION [9 upvotes]: Let $A \subset X$ and $B \subset X$ be two isometric subsets of a metric space $X$.  So there is an isometry $f: A \to B$.
Say that a metric space $X$ has the superposition property (my terminology) if, for every pair of isometric subsets $A$, $B$, there is an isometry of $X$, $F: X \to X$, that superimposes $A$ onto $B$:  $F(A) = B$, i.e. $F$ places $A$ on top of $B$.

Which metric spaces have this superposition property?

Euclidean space $\mathbb{E}^d$ has this property.
But it seems the punctured plane does not: e.g. if $A$ is the point $(1,0)$ and $B$ is the point $(-2,0)$, then (I believe) there is not an isometry of the whole punctured plane that maps $A$ onto $B$.
Has this property been studied before?  If so, under what name?
I am (clearly) unschooled in this area. Thanks for pointers and/or examples!

REPLY [6 votes]: If you restrict sets A, B to be simply points, then you are asking for spaces which have the property that for all points A,B\in X there exists an isometry T from X onto X so that T(A)=B (am I understanding you correctly?). Such spaces are called transitive. It is known that if a finite dimensional space is transitive then it is isometric to a Euclidean space, I think that this is a result of Mazur. It is an old, still open problem, whether every separable transitive Banach space is isometric to a Euclidean space. This problem is called Banach-Mazur problem and it goes back to 1930's. There has been a lot of work on this problem and it is connected to other interesting problems.
If you require that every 2 points can be mapped by a surjective isometry onto any other 2 points with the same distance, then I think the space is called 2-transitive. Similarly one defines n-transitive. I believe, but I am not 100% certain that 2-transitive Banach spaces have to be isometric to a Euclidean space. You might check work of V. Mascioni.<|endoftext|>
TITLE: What is $A+A^T$ when $A$ is row-stochastic ?
QUESTION [16 upvotes]: This is motivated by this MO question.
If $A\in{\bf M}_n({\mathbb R})$ is row-stochastic (entrywise non-negative, and $\sum_j a_{ij}=1$ for all $i$), then $M:=A+A^T$ is 

symmetric,
entrywise non-negative. 

One finds easily the

additional property  that
$$\sum_{i\in I}\sum_{j\in J}m_{ij}\le|I|+|J|$$
for every index subsets $I$ and $J$, 

with

equality in the extremal case:
$$\sum_{i,j=1}^nm_{ij}=2n.$$


My question is whether all these four properties imply in turns that $M$ has the form $A+A^T$ for some row-stochastic $A$.

Edit. The answer is Yes when $n=2$ (obvious) or $n=3$ (more interesting).

REPLY [3 votes]: My answer builds on Brendan McKay's idea. We will show the vertices have the form he describes..
It's obvious that vertices in the polytope of row-stochastic matrices have this form, because each row is independent and the equations for each row form a simplex.
So it's enough to show that vertices in the polytope of symmetric matrices satisfying these conditions have the analogous form. First we are going to show that integer symmetric matrices (with even diagonal entries) satisfying these inequalities have the desired form. Then we are going to show that vertices are always integer matrices.
To show the first thing, use the following algorithm: Whenever any row sums to $1$, remove that row and the corresponding column, and put a $1$ in the corresponding place in the row-stochastic matrix to account for it. Whenever any row sums to $0$, derive a contradiction: that row and all previously removed rows consist of $k+1$ rows and $k+1$ columns that together contribute only $2k$ to $\sum_{i,j} m_{ij}$, so the remaining $n-k-1$ rows and $n-k-1$ columns, when intersected, contribute $2n-2k$, which is more than $(n-k-1)+(n-k-1)$. When this process is complete, every row sums to at least $2$, and the average row sums to $2$, so every row sums to $2$. then you have the adjacency matrix of a graph where every vertex has valence 2 - a union of disjoint cycles. Choose an orientation of each cycle, and complete the stochastic matrix by adding the corresponding oriented adjacency matrix.
So it's enough to show that every vertex has integer entries. Suppose not. Call a pair $I,J$ tight if $\sum_{i\in I} \sum_{j\in J} m_{i,j}=|I|+|J|$. Note that the intersection of two tight pairs is tight, by the following inequality:
$\sum_{i\in I_1} \sum_ {j\in J_1} m_{i,j} + \sum_{i\in I_2} \sum_ {j\in J_2} m_{i,j} \leq \sum_{i\in I_1 \cup I_2} \sum_ {j\in J_1 \cup J_2} m_{i,j} + \sum_{i\in I_1 \cap I_2} \sum_ {j\in J_1 \cap J_2} m_{i,j}$
Call a tight pair integral if $m_{i,j}\subset \mathbb Z$ for all $i,j \in I,J$ (and diagonal entries are even). If some $m_{i,j}$ is not integral, let $I,J$ be a minimal non-integral tight pair. It exists because the set of all indices is always a non-integral tight pair.
There must be at least one non-integral entry in $I \times J$, but the sum of all the entries is an integer, so there is another non-integral entry. Or there is an odd diagonal entry. Assume there are two entries, and that they are not just the same entry reflected around the diagonal. Then you can increase one entry and decrease the other by some small amount $\epsilon$. This will preserve all the equalities and inequalities: If $\epsilon$ is small enough it will preserve all the inequalities that are not currently tight. $m_{i,j}\geq 0$ is not tight for either of these because they are nonzero. Since $I,J$ is a minimal non-integral tight pair, any tight pair containing one contains the other, and so every tight inequality remains tight. Since one could just as well decrease one and increase the other, this shows that $m_{i,j}$ is not a vertex.
The remaining case to consider if the minimal non-integral tight pair has only $m_{i,j}\not \in \mathbb Z$ and $m_{j,i} \not\in \mathbb Z$ and is otherwise integral, or has only an odd diagonal entry and is otherwise non-integral. If $i\neq j$ then $m_{i,j}$ is clearly a half-integer, so either way the total contribution of $m_{i,j}$ and $m_{j,i}$ is odd. But since $J,I$ is also a tight pair, $I\cap J$, $I\cap J$ is also a tight pair, and contains $m_{i,j}$, so it is the same as $I,J$ and $I=J$, so $|I|+|J|$ is even. So the total contribution is even, which means there must be another non-integral or odd diagonal entry, and we are done.<|endoftext|>
TITLE: Order of elements  
QUESTION [14 upvotes]: Consider natural numbers $m,n,k > 1$. There are finite groups $G$ containing elements $x,y$ such that $o(x) = m, o(y) = n$ and $o(xy) = k$. After embedding these groups in $S_\mathbb{N}$ we drive: Given $m,n,k >1$ there are $f,g \in S_\mathbb{N}$ such that $o(f) = m, o(g) = n$ and $o(fg)=k$. This argument shows the existence of $f,g$. But how to construct such permutations $f,g$ in  $S_\mathbb{N}$ in terms of $m$, $n$ and $k$ ?

REPLY [19 votes]: It is not hard in practice, at least for reasonably small values of $m,n,k$, to find permutations with the required property, and I did once convince myself that I could find a systematic way of doing it, but it was very messy and hard to describe.
Here is one way to do it, which will not produce permutations of optimally small degree, but which is at least easy to describe and even to do on a computer using GAP or Magma if you want to.
We will produce our permutations as elements of ${\rm PSL}(2,q)$ for suitable odd $q$. We do this by finding elements $X,Y \in {\rm SL}(2,q)$ of orders $2m,2n$ with product of order $2k$, and use their images in ${\rm PSL}(2,q)$. We make use of the fact that, for elements of ${\rm SL}(2,q)$ with order greater than 2 and dividing $q^2-1$, their order is detmined by their trace.
It is easiest to do if you choose $q$ such that $q-1$ is divisible by all of $2l,2m,2k$. We can assume that $m,n,k$ are all greater than 1.
Let $ \lambda,\mu $ be primitive $ 2m^{\mathrm{th}} $, $ 2n^{\mathrm{th}} $ roots of unity in the field ${\mathbb F}_q$ of order $q$. Let
$ A = \left( \begin{array}{cc} \lambda & 0 \\ 1 & \lambda^{-1} \end{array} \right) $ and  $ B = \left( \begin{array}{cc} \mu & \alpha \\ 0 & \mu^{-1} \end{array} \right) $,
for some $ \alpha $. So $ |A| = 2m $ and $ |B| = 2n $. Now
 $ AB = \left( \begin{array}{cc} \lambda\mu & \lambda\alpha \\ \mu & \alpha + \lambda^{-1}\mu^{-1} \end{array} \right) $
 and $ {\rm Tr}(AB) = \lambda \mu + \alpha + \lambda^{-1} \mu^{-1} $. Let $ t $ be the trace of an element of $ S $ of order $ 2k $, and choose $ \alpha = t - \lambda\mu - \lambda^{-1}\mu^{-1} $. Then $ |AB| = 2k $.
As an example, I tried $m,n,k = 10,12,15$ with $q=121$, and took $\lambda=w^6$, $\mu=w^5$, $\alpha=w^4+w^{-4}-w^{11}-w^{-11}$, with $w$ a primitive field element.  Then, using Magma to project onto ${\rm PSL}(2,121)$, I got the following permutations $x,y \in S_{122}$ with
$(|x|,|y|,|xy|) = (10,12,15)$.
$x=$ (2, 105, 31, 119, 44, 29, 107, 98, 58, 120)(3, 36, 24, 18, 48, 12, 54, 30, 42, 
    60)(4, 19, 72, 87, 38, 117, 33, 102, 41, 66)(5, 73, 79, 104, 49, 56, 45, 
    118, 112, 69)(7, 57, 91, 67, 106, 109, 101, 50, 46, 114)(8, 103, 64, 81, 
    74, 21, 43, 40, 35, 77)(9, 32, 89, 34, 84, 65, 61, 47, 116, 80)(10, 26, 82,
    17, 27, 94, 100, 70, 76, 99)(11, 37, 93, 83, 39, 52, 108, 14, 86, 68)(13, 
    75, 121, 22, 78, 23, 110, 97, 62, 113)(15, 85, 63, 92, 55, 122, 115, 95, 
    59, 16)(20, 25, 96, 28, 71, 88, 90, 111, 51, 53)
$y=$ (1, 95, 18, 77, 29, 118, 22, 71, 39, 80, 38, 67)(2, 99, 65, 119, 41, 102, 89, 
    51, 75, 113, 70, 53)(4, 103, 74, 86, 106, 37, 10, 7, 88, 97, 66, 90)(5, 
    115, 109, 47, 112, 61, 31, 78, 57, 98, 121, 15)(6, 87, 91, 26, 63, 60, 43, 
    120, 73, 13, 20, 68)(8, 35, 12, 76, 114, 46, 104, 94, 32, 45, 48, 34)(9, 
    122, 72, 59, 23, 62, 101, 30, 50, 64, 42, 33)(11, 107, 108, 40, 69, 117, 
    21, 49, 27, 58, 17, 19)(14, 56, 36, 54, 116, 25, 85, 79, 84, 81, 55, 
    83)(16, 92, 93, 110, 100, 44, 52, 24, 82, 111, 96, 105).
Added later: I should perhaps add that by starting with elements $x,y \in S_8$ with orders 10,12, and replacing $y$ by random conjugates, I rapidly found a solution in $S_8$: $x=(1,2,3,4,5)(6,7)$, $y=(1,2,4,7)(3,6,8)$, so I would guess that some kind of reasonably intelligent random algorithm is the best way to solve this in practice.<|endoftext|>
TITLE: On the positive definiteness of a linear combination of matrices
QUESTION [12 upvotes]: In my work in PDE, the following problem in linear algebra came up. Any help in this direction is appreciated.
QUESTION:
Let $m,n\in\mathbb{N}$ and let $A_1,\ldots, A_m\in M_n(\mathbb{R})$ be real, symmetric, indefinite matrices. I'm interested in conditions on $A_1,\ldots,A_m$ which ensures that the set
$$P:=\big\{\sum_{i=1}^{m}\lambda_i A_i:\lambda_i\in\mathbb{R}\big\}$$
contains a positive-definite matrix. I'm aware of the following result due to Hestenes-McShane (1940) which is suffcient but not necessary.
THEOREM (Hestenes-McShane)
Let $m,n\in\mathbb{N}$ and let $A,B_i\in M_n(\mathbb{R})$ be real symmetric matrices, for all $i=1,\ldots,m$. Let us write, for each $i=1,\ldots,m$,
$$Z_{i}:=\{x\in\mathbb{R}^n:\langle B_i x;x \rangle=0\}$$
Let us suppose that

$\langle A x;x \rangle>0$, for all $x\in \cap_{i=1}^{m}Z_i$, $x\neq 0$.

$B$  is indefinite on $\mathbb{R}^n$, for all non-zero $B\in\operatorname*{span}\{B_i:i=1,\ldots,m\}$.

For every non-zero subspace $S\subseteq \mathbb{R}^n$ satisfying $$S\cap\left(\cap_{i=1}^{m}Z_i\right)=\{0\},$$ there exists $B\in\operatorname*{span}\{B_i:i=1,\ldots,m\}$ such that $B$ is positive definite on $S$.


Then, there exists $B\in\operatorname*{span}\{B_i:i=1,\ldots,m\}$ such that $A-B$ is positive definite on $\mathbb{R}^n$.
Unfortunately, in my case, condition 3 is not satisfied. Has this result been improved later?

REPLY [4 votes]: A straightforward reformulation is in terms of polynomial inequalities is by taking $n$ consecutive chief submatrices $M_{KK}(\lambda)$ for $K=(1,\dots,k)$, $1\leq k\leq n$ of the matrix $M(\lambda)=\sum_{j=1}^m \lambda_j A_j\in \mathbb{R}[\lambda_1,\dots,\lambda_m]^{n\times n}$. Then $P$ contains a positive definite matrix if and only if the basic open semialgebraic set
$\{y\in \mathbb{R}^m\mid \det M_{KK}(y) \gt 0, \  K=(1,\dots,k), 1\leq k\leq n\}$ is nonempty.
At least these kinds of conditions were used in papers by L.Khachiyan and L.Porkolab, such as "On the complexity of semidefinite programs", J. Global Optim. 10 (1997). E.g. when $m$ is fixed, one gets a strong polynomial-time algorithm for checking non-emptiness of $P$.<|endoftext|>
TITLE: An isomorphism between different Ext's coming from group cohomology
QUESTION [5 upvotes]: Let $G$ be an abelian group and $M$ a $G$-module with trivial action. It is well-known that $H^2(G,M)$ classifies extensions of $G$ by $M$, which is $\mathrm{Ext}^1_{Ab}(G,M)$.
On the other hand $H^2$ is $\mathrm{Ext}^2_{G-mod}(\mathbb Z,M)$ (by the definition of group cohomology), where $\mathbb Z$ is given the trivial $G$-action.
Is there a nice way to see this isomorphism $$ \mathrm{Ext}^1_{Ab}(G,M) \cong \mathrm{Ext}^2_{G-mod}(\mathbb Z,M) ? $$
Or is it just an accident that both happen to classify the same type of object?
Is there a high-brow reason? Is there a generalization? Why should $\mathrm{Ext}^1$ and $\mathrm{Ext}^2$ be related in such a way? Might this result from a spectral sequence? I wonder if this might be a special case of something well-known.

REPLY [14 votes]: The group $H^2(G,M)$ classifies different types of extensions than $\operatorname{Ext}^1(G,M)$. On the one hand, $H^2(G,M)$ classifies extensions
$$M\hookrightarrow H\twoheadrightarrow G$$
where $H$ may be non-abelian, and the action of $H$ on $M$ by conjugation is encoded in the $G$-module structure of $M$. On the other hand, $\operatorname{Ext}^1(G,M)$ classifies extensions
$$M\hookrightarrow A\twoheadrightarrow G$$
where $A$ is an abelian group, in particular $A$ acts trivially on $M$ by conjugation, i.e. $M$ can only be regarded as a trivial $G$-module here.
Even if you regard $M$ as a trivial $G$-module in both cases, $H^2(G,M)$ and $\operatorname{Ext}^1(G,M)$ may be different due to the existence of non-abelian but central extensions. In general, there is a universal coefficient split short exact sequence
$$\operatorname{Ext}^1(G,M)\hookrightarrow H^2(G,M)\twoheadrightarrow \operatorname{Hom}(H_2G,M).$$
You can find this in most books on group cohomology. The first morphism represents the inclusion of abelian extensions into central (but possibly non-abelian) extensions (recall that $M$ carries here the trivial $G$-module structure). The group $\operatorname{Hom}(H_2G,M)$ measures the amount of really non-abelian central extensions of $G$ by $M$.
Fortunately, $H_2G$ is very easy to compute, it is the exterior square $H_2G=\wedge^2G$, i.e. the quotient of $G\otimes G$ by the relations $g\otimes g=0$, $g\in G$. This functor is quadratic, $$\wedge^2(G_1\oplus G_2)=\wedge^2(G_1)\oplus (G_1\otimes G_2)  \oplus \wedge^2(G_2)$$
and vanishes on (finite or infinite) ciclyc groups $\wedge^2(\mathbb{Z}/n)=0$, $n\in\mathbb Z$. This gives a recipe to compute $H_2G$ for any finitely generated abelian group. In particular, if you take $G=(\mathbb{Z}/2)^2$ and $M=\mathbb{Z}/2$ you get
$$\operatorname{Ext}^1(G,M)=(\mathbb{Z}/2)^2,\qquad \operatorname{Hom}(H_2G,M)=\mathbb Z/2.$$
Hence $$H^2(G,M)=(\mathbb{Z}/2)^3.$$<|endoftext|>
TITLE: Is OEIS A007018 really a subsequence of squarefree numbers?
QUESTION [19 upvotes]: A comment in A007018 a(n) = a(n-1)^2 + a(n-1), a(0)=1 claims

Subsequence of squarefree numbers (A005117). - Reinhard Zumkeller, Nov 15 2004

Is that really so?
As far as I know, it is an open problem if a polynomial $f \in \mathbb{Z[x]}$ of degree $\ge 5$ can be
squarefree infinitely often (some sources require $f$ to be irreducible).
If the OEIS comment is correct, the sequence will give an infinite family of 
(irreducible) polynomials which are squarefree infinitely often.
Denote by $a_n$ the terms of OEIS A007018. Set $a_n = x$ and $$f(x)=a_{n+4}=x \cdot (x + 1) \cdot (x^{2} + x + 1) \cdot (x^{4} + 2 x^{3} + 2 x^{2} + x + 1) \\\\ \cdot (x^{8} + 4 x^{7} + 8 x^{6} + 10 x^{5} + 9 x^{4} + 6 x^{3} + 3 x^{2} + x + 1)$$
$f(a_n)=a_{n+4}$ will be squarefree infinitely often (including the irreducible degree 8 factor) 
and iterating $x \mapsto x^2+x$ will produce an infinite family of polynomials with this property.
Added For references of squarefree values of polynomials, the search terms are square free values of polynomials. E.g. here p.1 and here "11. Squarefree values of polynomials".

REPLY [5 votes]: Prime factors below $10^{10}$ of $a_n$ can be found in OEIS A007996, and I've tested that none of them divides $a_n$ when squared. Same was reported by Andersen earlier for primes below $2^{32}$. 
In fact, the divisibility by $p^2$ can be quickly disproved for many primes $p$ if one tries to unroll the sequence backwards starting with $a_n \equiv 0\pmod{p^2}$ for the smallest index $n$, thus giving $a_{n-1}\equiv -1\pmod{p^2}$. While this leads to solving a quadratic equation modulo $p^2$ with respect to $a_{n-2}$ and potentially can give two roots, in roughly half of the cases the equation will have no solutions. Then we solve a quadratic equation for $a_{n-3}$ and so on. On average this process stops within a constant number of steps without reaching $a_1\equiv 2\pmod{p^2}$ (meaning that $p^2$ cannot divide $a_n$).
Since solving quadratic equation (via computing square roots) modulo $p^2$ is relatively time-consuming, for "persistent" primes $p$ that survived a certain number of square-root computations it's worth to switch to direct computation of $a_n$ modulo $p^2$ until the zero residue or the period is encountered. I used the threshold of 100 square-root computations for this purpose. With this mixed strategy the search for prime divisors can be easily extended beyond $10^{10}$ if needed.<|endoftext|>
TITLE: Symbols of elliptic operators
QUESTION [15 upvotes]: First let me state the problem, then I'll  explain its origin and finally, I'll ask the main question..
Problem S.  Fix a positive integer $n$. Find all the pairs $(V, S)$, whith the following properties.
1. $V$ is a finite dimensional  complex  vector space  equipped with a Hermitian metric. $\DeclareMathOperator{\Sym}{Sym}$ We denote by $\Sym(V)$ the space of symmetric complex linear operators $V\to V$.
2. $\newcommand{\bR}{\mathbb{R}}$  $S$ is a linear map $S:\bR^n\to\Sym(V)$ such  that for any $\xi\in\bR^n\setminus 0$ the symmetric operator $S(\xi)$ is invertible.
Readers familiar with the basics of p.d.e.-s will surely recognize $S(\xi)$  as the principal symbol of an elliptic, first order  partial differential operator with constant coefficients that acts on $C^\infty(\bR^n, V)$. That explains the letter $S$ in the name of the problem.
We  denote by $\newcommand{\eS}{\mathscr{S}}$ $\eS_n$ the space of  solutions of Problem S for a given positive integer $n$.
Observe that $\eS_n$  is equipped with  two basic algebraic operations $\oplus,\otimes$ $\newcommand{\one}{\boldsymbol{1}}$
$$(V_1, S_1)\oplus (V_2, S_2):= ( V_1\oplus V_2, S_1\oplus S_2), $$
$$ (V_1, S_1)\otimes (V_2, S_2):=  ( V_1\otimes V_2, S_1\otimes\one_{V_2}+ \one_{V_1}\otimes  S_2). $$
The group $\DeclareMathOperator{\GL}{GL}$ $\GL(n,\bR)$ acts in an obvious way on $\eS_n$.  More precisely if $S:\bR^n\to\Sym(V)$  is a solution $S\in\eS_n$, and $T\in \GL(n,\bR)$, then $S\circ T\in \eS_n$.
Let us  also observe that  for each $n$, the set $\eS_n$ is not empty.    We can obtain maps $S: \bR^n\to\Sym(V)$ with the desired properties by using complex representations of the Clifford algebra generated by an    Euclidean  inner product on the   space $\bR^n$.   I will  refer to such examples as Clifford examples and I will denote by $\newcommand{\eC}{\mathscr{C}}$ $\eC_n$ the subset of  $\eS_n$ constructed  as above using representations of    Clifford algebras. Observe that $\eC_n$ is also closed  under the operations $\oplus,\otimes$ and invariant under the above action of $\GL(n,\bR)$
Main Question.  Fix $n$ Are there non Clifford solutions to Problem S? In other words, is the set $\eS_n\setminus \eC_n$ non-empty?
Addendum.   Apparently this question is related to a  classical question  discussed by Porteous  in his  book Topological Geometry.          For a given real vector space $V$ find  the largest $n$  find the maximal subspaces  $\DeclareMathOperator{\Endo}{End}$  $S\subset \Endo(V)$ such that $S\setminus 0 \subset \GL(V)$.   The answer  has to do with Radon-Hurwitz numbers, and  it basically  says that if $S$ is   such a  subspace, maximal or not,  then $V$  s a module over   the Clifford algebra generated by  an inner product  on $S$.

REPLY [12 votes]: Maybe I'm misunderstanding something, but it seems that the answer is probably 'no', at least if $d = \dim_\mathbb{C} V$ is large enough.  
What really matters is the $n$-dimensional real subspace $\mathsf{S}=\mathrm{Im}(S)\subset\mathrm{Sym}(V)$.  What you need is that this space not meet the cone of singular matrices in $\mathrm{Sym}(V)$ anywhere but at the origin.  This is an open condition on the element $\mathsf{S}\in\mathrm{Gr}_n\bigl(\mathrm{Sym}(V),\mathbb{R}\bigr)$, and this latter space has dimension $n(d^2{-}n)$.  Thus, the set of such subspaces with your property has this latter dimension, but the group of unitary transformations on $V$ (which, I gather, is the space of symmetries of the problem) is only of dimension $d^2$, so there must be at least an $(n{-}1)d^2-n^2$ parameter family of 'inequivalent' subspaces that meet your criteria.  
However, there are only a finite number of inequivalent $d$-dimensional complex representations of the Clifford algebra on $\mathbb{R}^n$ endowed with a definite inner product.<|endoftext|>
TITLE: Existence of model of ZF without AC, but with many choice function
QUESTION [6 upvotes]: Question 1: Does there exist models of the Zermelo-Fraenkel set theory without the axiom of choice, but such that every indexed family of non-void sets whose index set has a well-orderable cardinal admits a choice function ?
Question 2: The same as question 1, with "a well-orderable cardinal" replaced by "a linearly ordered cardinal".
Gérard Lang

REPLY [6 votes]: For the first question:
Yes. It is known that the axiom "Every well-orderable family of non-empty set has a choice function" implies $DC$ but not $DC_{\aleph_1}$.
You can find the proofs in Jech's "The Axiom of Choice" and in Felgner's "Models of ZF-Set Theory".
I am not sure about the second question, but I believe the answer should be "yes" as well.
In Consequences of the Axiom of Choice you can find the following principles:

Form 1 is the axiom of choice.
Form 202 is the existence of a choice function for linearly ordered families of non-empty sets.
Form 40 is the existence of a choice function for well-ordered families of non-empty sets.

Entering those three numbers in the table show that $1\implies 202\implies 40$ and neither implications is reversible. Although in the case of $202\implies 1$ this is in a weaker axioms system without regularity.

Edit: I had a bit more time now, so I went to chase after the cited source for Form 202:

Truss, J. The axiom of choice for linearly ordered families. Fund. Math. *99 (1978), no. 2, 133–139. (MR480029)

Here is an excerpt from the review:

The axiom of choice for linearly ordered families asserts that any linearly ordered family of nonempty sets has a choice function. The author shows that in ZF this statement implies the full axiom of choice, while in FM (ZF without the axiom of foundation) it does not.

So it seems that in ZF, requiring for linearly ordered families is enough to require the full axiom of choice.<|endoftext|>
TITLE: How to draw Archimedean-Galileo spiral?
QUESTION [5 upvotes]: It is known that some plane curves can be drawn with a tool. For instance, I heard at a web site that Archimedes created his spiral in the third century B.C. by fooling around with a compass and others.
Let’s however look at the spiral defined by the equation:
$r'(\theta)^2+r(\theta)^2=\theta^2$, $r(\theta=0)=0$
I am looking for a method ( a tool) which could help to plot the spiral on paper ( I named it as Archimedean-Galileo spiral.  For large $\theta$, the curve represents Archimedean spiral: $r=\theta$. When $\theta$ is small it transforms in Galileo spiral $r=\theta^2$) .
The spiral has a property that the junction point of the curve and the ray uniformly rotated in the origin coordinates when the junction point moves with uniform acceleration.
Do you think that there is a way to draw it without computer, but with other special curves (tools)?
I thought about  the spiral of Theodorus, but I am not sure how the spiral of Theodurus is connected with the equation.

REPLY [2 votes]: You have the answer: the spiral has a property that the junction point of the curve and the ray uniformly rotated in the origin coordinates when the junction point moves with uniform acceleration.
Take a ruler and fix one of its extremities to a motor, so that the ruler materialize the ray uniformly rotated.
Make a hole on the table just below the motor axis. Then pass a string trough this hole, attach a weight at the string under the table, and manage the other extremity of the string to be able to slide along the ruler.
Fix a pencil somewhere on the string along the ruler. When you drop the weight, it will fall down with a uniform acceleration, dragging the pencil along the curve you want to plot.
I wouldn't be surprised if, in facts, Archimedes was not thinking about such a device when he invented the curve. 
Note that, with the ancient Greek mentality, to draw the curve with a computer would have been an achievement much superior to build the mechanical device.
They were right after all: along the way Archimedes would have, at least invented, modern algebra, differential calculus, the planimeter and the computer.<|endoftext|>
TITLE: Non-super reflexive space
QUESTION [6 upvotes]: Suppose $X$ is a reflexive space (possibly non-separable) which is not super-reflexive. Then (by definition) there exists a non-reflexive Banach space $Y$ which is non-reflexive but is finitely representable in $X$, meaning that for each $\lambda >1$, every finite dimensional subspace of $Y$ is $\lambda$-isomorphic to a subspace of $X$. Can we always find such $Y$ (i.e. non-reflexive) which is separable? In this spirit, what are examples of reflexive but not super-reflexive spaces in which neither $\ell_1$ nor $c_0$ is finitely representable?

REPLY [6 votes]: The first question is easy: Every non reflexive space has a separable non reflexive subspace (e.g. by the Eberlein-Smulian theorem or by R. C. James' characterization of non reflexivity).  
The second question was a longstanding open problem that was solved by James in the 1970s.  Pisier and Xu gave another proof--you can find their paper by using MathSciNet.  Their approach is more conceptual and uses interpolation theory but is not easy.<|endoftext|>
TITLE: Nice Algebraic Statements Independent from ZF + V=L (constructibility)
QUESTION [19 upvotes]: Background and Motivation
I've always been fascinated about algebraic statements independent from ZFC set theory. One such fascinating example comes from considering $\rm{Ext}^1_\mathbb{Z}(A,\mathbb{Z})$. If $A$ is free then this abelian group is trivial. Is the converse true? The converse is known as the Whitehead problem.
Now the Whitehead problem was shown independent of ZFC by Shelah. This is somewhat unsatisfying, but it can be proved assuming the axiom of constructibility ($V=L$). In fact, there are other reasonable sounding and simple statements in analysis and topology that are also independent from ZFC but become theorems once $V=L$ is assumed. Another statement is that the global (a.k.a. homological) dimension of the ring $\prod_{i=1}^\infty \mathbb{C}$ is two if and only if the continuum hypothesis holds, which is implied by adding $V=L$ again. Thus I feel warmly about the axiom of constructibility.
Of course, the dissatisfaction remains, because there are other statements that are independent from $ZFC + V=L$ (well, perhaps I should write $ZF + V=L$ to save space). Question 18058 and Question 11480 are examples.
Question, Loosely Stated

Now, I am curious if there are any known algebraic (see postscript) statements, reasonably naturally sounding (use judgement), that are independent from $ZF + V=L$? Or perhaps independent from $ZF + A$ where $A$ is your favourite set-theoretic axiom independent from $ZFC$? Perhaps some easy low-hanging fruit for this search would be in the area of homological dimension theory? Has anyone done work on this type of thing?

I am sure there must be some statements of some kind. In the proof of the Whitehead problem under adding $V=L$, one can first deduce some combinatorial statement that requires little set-theoretic machinery and then use it to prove the Whitehead problem. So perhaps adding other axioms, one can also deduce various combinatorial gadgets and them use them to get new algebraic statements that are independent from the original $ZFC$? I would even like to hear about statements implied by additional axioms, but whose independence is not proved. (One of Devlin's books explains this).
(Remark: Although with enough brute force, one should be able to churn out such things no matter how many new axioms one adds, I would be interested in finding enough axioms of set theory so that the remaining independent statements would be so bizarre sounding that they would be essentially be uninteresting for all of time. Presumably as one adds more and more axioms to set theory, this would happen, no?)
Since I am not a set theorist, I would appreciate answers that are understandable to someone who knows the basics of set theory (say a typical first grad course) but knows very little about forcing.

Postscript
By "algebraic" I mean roughly some statement in the language of groups, rings, ideals, modules, fields, etc., somewhat natural sounding, that does not itself refer to the additional axiom (e.g. using some set whose cardinal is inaccessible, or something along these lines). Thus I am not looking for statements about real numbers or set theory, although those are interesting too.

REPLY [4 votes]: One can find examples of algebraic statements that are independent of ZFC + $V = L$ by considering "absolute" versions of standard algebraic propositions. This happens for example when one seeks large strongly non-isomorphic (i.e. non-$L_{\infty \omega}$-equivalent) or absolutely rigid families of structures, absolutely indecomposable abelian groups and modules, or absolute $E$-rings of arbitrarily big cardinality. Roughly, up to the first $\omega$-Erdos cardinal $\kappa(\omega)$, in general such families exist, but above $\kappa(\omega)$, they do not. 
The beauty of $\kappa(\omega)$ is that if it exists, then it exists in $L$, so that the existence of absolutely rigid systems in every cardinal is independent of ZFC + $V = L$. 
In loose terms, the absolute version of a property $P$ is obtained by requiring that any structure $M$ retains $P$ in any generic extension. This can be recast as in algebraic terms using back-and-forth systems or infinitary equivalence.
Questions about the existence of this type of strongly non-isomorphic structures were first asked by Nadel and pursued more recently by Eklof, Fuchs, Goebel, Herden and Shelah.<|endoftext|>
TITLE: Order of products of elements in symmetric groups
QUESTION [30 upvotes]: Let $n \in \mathbb{N}$. Is it true that for any $a, b, c \in \mathbb{N}$ satisfying
$1 < a, b, c \leq n-2$ the symmetric group ${\rm S}_n$ has elements of order $a$ and $b$
whose product has order $c$?
The assertion is true at least for $n \leq 10$, see
here.
Update on Sep 2, 2015: On Aug 10, 2015
Joachim König has posted
a preprint to the arXiv which gives
a positive answer to the question. Assuming that this preprint is correct,
this completely answers the question -- and thus also Problem 18.49 in the
Kourovka Notebook.
Update on Jun 18, 2014: The assertion is true at least for $n \leq 50$, see
here (4MB text file).
The list of examples in GAP-readable format can be found
here.
Added on Dec 11, 2013: This question will appear as
Problem 18.49 in:
Kourovka Notebook: Unsolved Problems in Group Theory. Editors V. D.
Mazurov, E. I. Khukhro. 18th Edition, Novosibirsk 2014.
Added on Nov 24, 2013: Is there really not enough known about, say, the
class multiplication coefficients of ${\rm S}_n$ to answer this question?
Text of the question as of Feb 12, 2013:
This question is a follow-up on Order of elements  .
Derek Holt's answer to that question is nice, but it seems that the degree of the permutations it gives is a lot larger than necessary.
So, given natural numbers $m, n, k > 1$, what is the smallest $d$ such that the symmetric group of degree $d$ has elements of order $m$ and $n$ whose product has order $k$? -
Clearly if the largest of the numbers $m$, $n$, $k$ is prime, then $d$ must be at least
$\max(m,n,k)$, and there are some cases where $d$ actually must be larger. However a quick computation suggests that $d = \max(m,n,k) + 2$ might work always. - But does this or a similar bound hold?
EDIT: Smallest-degree examples for all $m, n, k \leq 8, m \leq n$ can be found
here.

REPLY [9 votes]: The main theorem in a paper of G. A. Miller [1] is the following:

THEOREM. If $l, m, n$ are any three integers greater than unity, of which we
call the greatest $k$, it is always possible to find three substitutions $(L, M, N)$ of $k + 2$ or some smaller number of elements and of orders $l, m, n$ respectively such that $LM=N$.

Which gives a positive answer to your question.
Papers [3], [4], [5] by Brenner and Lyndon give a different proof of Miller's result, and consider the problem of finding for $l,m,n > 1$ the smallest $d$ such that $S_d$ contains permutations $x,y$ with $(|x|, |y|, |xy|) = (l,m,n)$. Other related results are e.g. in [2] and [6].
Papers [8], [9] also give a proof of the theorem by Miller. Also in [7], although it seems without assuming permutations of degree $\leq \max(l,m,n)+2$ in some cases.
The main theorem of [10] constructs for $l,m,n > 1$ elements $A,B \in PSL(2,q)$ (for suitable $q$) such that $|A| = l$, $|B| = m$, and $|AB| = n$. The construction pointed out by Derek Holt here is similar but a bit shorter. Yet another construction with matrices is in [11].
References:
[1] MR1505829 G. A. Miller, On the Product of Two Substitutions. Amer. J. Math. 22 (1900), no. 2, 185–190. JSTOR
[2] MR1505882 G. A. Miller, Groups Defined by the Orders of Two Generators and the Order of their Product. Amer. J. Math. 24 (1902), no. 1, 96–100. JSTOR
[3] MR0767585 J. L. Brenner, R. C. Lyndon, A theorem of G. A. Miller on the order of the product of two permutations. I. Jñānābha 14 (1984), 1–16. link
[4] MR0809272 J. L. Brenner, R. C. Lyndon, A theorem of G. A. Miller on the order of the product of two permutations. II. The minimal degree in case a=2. Indian J. Math. 26 (1984), no. 1-3, 105–133 (1985).
[5] MR0748120 J. L. Brenner, R. C. Lyndon, A theorem of G. A. Miller on the order of the product of two permutations. III. The minimal degree in case a>2. Pure Appl. Math. Sci. 20 (1984), no. 1-2, 37–51.
[6] MR0743150 J. L. Brenner, R. C. Lyndon, The orbits of the product of two permutations. European J. Combin. 4 (1983), no. 4, 279–293. DOI
[7] MR0053937 R. H. Fox, On Fenchel's conjecture about F-groups. Mat. Tidsskr. B 1952 (1952), 61–65. JSTOR
[8] MR3508006 J. König, A note on the product of two permutations of prescribed orders. European J. Combin. 57 (2016), 50–56. DOI
[9] MR3694453 J. Pan, On a conjecture about orders of products of elements in the symmetric group. J. Pure Appl. Algebra 222 (2018), no. 2, 291–296. DOI
[10] MR0283093 R. D. Feuer, Torsion-free subgroups of triangle groups. Proc. Amer. Math. Soc. 30 (1971), 235–240. DOI
[11] MR0207852 J. Mennicke, Eine Bemerkung über Fuchssche Gruppen. Invent. Math. 2 (1967), 301–305. DOI<|endoftext|>
TITLE: The crude monadicity theorem
QUESTION [7 upvotes]: In order to test the monadicity of a functor, there is a precise monadicity theorem (PM) as well as a crude monadicity theorem (CM), see the nlab. In CM, the forgetful functor should create reflexive coequalizers.
What is a nice and specific example which shows that CM is only sufficient, but not necessary; i.e. what is a monad which doesn't preserve reflexive coequalizers? I couldn't find such an example.
Besides, is the condition in CM satisfied in almost all practical / non-pathological examples? For example it holds for algebraic monads on $\mathsf{Set}$. What is the intuitive difference between PM and CM? The background for this rather soft question is the following: I would like to prove a certain theorem about certain monads, and in one step it would be very convenient if the monad preserved reflexive coequalizers. Now I wonder if there is any problem when I just add this assumption to the theorem.
Given a monad defined by generators and relations (on a category different from $\mathsf{Set}$), how can I test practically if the forgetful functor creates reflexive coequalizers?

REPLY [13 votes]: For $X$ an infinite set, the monad $(-)^X$ (induced from the comonoid structure on $X$ with respect to cartesian product) does not preserve reflexive coequalizers. See page 538 of this paper by Adámek, Koubek, and Velebil. Correspondingly, the forgetful functor for the category of algebras does not preserve reflexive coequalizers (hence also cannot reflect/create them since the forgetful functor reflects isomorphisms). 
Any finitary (i.e., filtered-colimit preserving) monad on $Set$ does preserve reflexive coequalizers, because finitary powers $(-)^n$ do. This implies that the forgetful functor will reflect reflexive coequalizers. So the practical difference shows up in finitary vs. infinitary monads.<|endoftext|>
TITLE: Explicit path in the unitary group of a $C^*$-algebra
QUESTION [16 upvotes]: For $G$ a discrete group, there is a canonical inclusion $g\mapsto u_g$ of $G$ into the unitary group of the reduced $C^*$-algebra $C^*_r(G)$. Denote by $[u_g]$ the class of $u_g$ in the (topological) $K$-theory group $K_1(C^*_r(G))$. It is well-known that, if $g$ is a commutator in $G$, then $[u_g]=0$ in $K_1(C^*_r(G))$: indeed the $2\times 2$ matrix $u_g\oplus 1$ is connected to the identity in the unitary group of $M_2(C^*_r(G))$. 
Now, let $G$ be the free group on two generators $a,b$, and let $g$ be the commutator of $a$ and $b$. I recently bumped into this nice paper by Haagerup, Dykema and Rordam:
http://arxiv.org/pdf/funct-an/9608001.pdf
At the top of page 3, they mention as a consequence of their main result, that $u_g$ is connected to $1$ already in the unitary group of $C^*_r(G)$. Partly out of curiosity, I was wondering whether an explicit path of unitaries between $u_g$ and $1$ had been written down somewhere.

REPLY [3 votes]: Here are some thoughts about a construction of an explicit path. Unfortunately, I could not carry it over, so take it as a long comment. I'd be glad if somebody is able to complete it, or maybe to get some useful hint!
Any skew-symmetric unitary element $J$ of a $C^*$  algebra $A$  can be connected to the identity in the unitary group of $A$ by the path $\exp(tJ)=\cos(t)I+\sin(t)J$, for $0\le t\le \pi/4$. This suggests to consider a factorization of $U:=u_g$ as a product of two  skew-symmetric unitary operators $J$ and $K$ on $\ell_2(G) $ (this can be done in many ways, indeed):
$$J\in B(\ell_2(G)),\quad K\in B(\ell_2(G))\, ,$$ 
$$J^*=J^{-1}=-J,\quad K^*=K^{-1}=-K\, ,$$ 
$$U=JK\,\quad  (\,\mathit{so\, that}\quad K=-JU)\, .$$
Then $u(t):=\exp(tJ)\exp(tK)=\cos(t)^2I+\sin(t)\cos(t)J(I-U)+\sin(t)^2U$ for $0\le t\le \pi/4$ is a path in the unitary group of $B(\ell_2(G))$ connecting $I$ to $U$. So, if we can choose the above factorization in such a way that we also have $J(I-U)\in C^*_{\mathrm {r}}(G)$, the whole path $u(t)$ would be in the reduced algebra for all $t$. Of course, $J \in C^*_{\mathrm {r}}(G)$ would do it, but this doesn't quite seem to be needed, as $I-U$ is not invertible. 
I've thus tried the simplest Ansatz for $J$, namely $Jf(x):=\alpha(x)f(\sigma(x))$, for all $f\in \ell_2(G)$ and $x\in G$, with  a convenient $\alpha\in\ell_\infty(G)$ and a  $\sigma:G\to G$ bijective. However, it seems the last requirement, $J(I-U)\in C^*_{\mathrm {r}}(G)$ can't be fulfilled in this case. A slightly more general form, $Jf(x):=\alpha(x)f(\sigma(x))+\beta(x)f(\tau(x))$ seems more promising (yet there could be some good reason why the whole approach is hopeless, though).<|endoftext|>
TITLE: Can every curve be written as $f(x)=g(y)$?
QUESTION [23 upvotes]: Does every irreducible curve admit an equation of the form $f(x)=g(y)$, where $f$ and $g$ are polynomials?  What if we allow $f$ and $g$ to be rational functions?
Actually, I'd like to understand this in the presence of an additional constraint: if we're given a finite cover of curves $\pi\colon C\to\mathbb{P}^1$, do we expect there to be a cover $\phi\colon C\to\mathbb{P}^1$ and rational functions $f(x)$ and $g(x)$ such that $C$ is isomorphic to $f(x)=g(y)$ and also $f\circ\phi=g\circ\pi$?  In other words, not only is $C$ isomorphic to $f(x)=g(y)$, but this isomorphism can be chosen so that $\pi$ is the projection onto the $y$ coordinate.
This is reminiscent of Chad Schoen's paper "Varieties dominated by product varieties", but I don't see a precise connection between the two.

REPLY [24 votes]: This would contradict the Harris-Mumford(-Eisenbud) theorem that $M_g$ is non-uniruled for $g$ at least $23$.  Let $C$ be a general curve of genus $g$.  If $C$ is in "Zieve form", then it is the normalization of the (almost certainly) singular curve in $\mathbb{CP}^1 \times \mathbb{CP}^1$,
$$D = \{ ([x_0,x_1],[y_0,y_1]) \in \mathbb{CP}^1\times \mathbb{CP}^1 \vert y_0^e f(x_0,x_1) - x_0^dg(y_0,y_1) \}, $$ where $f(x_0,x_1)$, respectively $g(y_0,y_1)$, is a homogeneous polynomial of degree $d$, resp. $e$, such that $f(0,1)$ and $g(0,1)$ are nonzero (or else the defining polynomial factors to a simpler form).  By direct computation, the singular points occur where $[x_0,x_1]$ is a multiple root of $f(x_0,x_1)$ and $[y_0,y_1]$ is a multiple root of $g(y_0,y_1)$ or the point is $([0,1],[0,1])$.  Moreover, at each point, the local analytic type of the singularity is the same as the plane curve with equation $y^n-x^m$, where $m$, resp. $n$, is the vanishing order of $f(x_0,x_1)$, resp. $g(y_0,y_1)$ at that point.  In particular, the "delta invariant" depends only on $(m,n)$.  Thus, if you "deform" $f(x_0,x_1)$ and $g(y_0,y_1)$ so that the number and type of multiple roots remains constant, then the normalizations of the corresponding curves in $\mathbb{CP}^1\times \mathbb{CP}^1$ remain of genus $g$.  However, the family of such deformations of $(f,g)$ is a rational variety.  Precisely, if you write
$$ f(x_0,x_1) = (x_1-a_1x_0)^{m_1}(x_1-a_2x_0)^{m_2}\cdots (x_1-a_rx_0)^{m_r}, $$ with $(a_1,\dots,a_r)$ pairwise distinct, 
then the deformation space for $f$ is just a Zariski open subset of the affine space with coordinates $(a_1,\dots,a_r)$, and similarly for $g(x_0,x_1)$. Since $M_g$ is non-uniruled, this is a contradiction: there is only the constant morphism from a rational variety to $M_g$ whose image contains the general point parameterizing $C$.  
Edit. Mike also asks whether this could be true if $f$ and $g$ are rational functions rather than polynomial functions.  This is equivalent to replacing the defining equation above in $\mathbb{CP}^1 \times \mathbb{CP}^1$ by the more general equation
$$
g_0(y_0,y_1)f_1(x_0,x_1) - f_0(x_0,x_1)g_1(y_0,y_1),
$$
where $f_0$, $f_1$ are homogeneous of degree $d$ with no common factor, and where $g_0$, $g_1$ are homogeneous of degree $d$ with no common factor.  The same observations apply: the number and types of singularities depend only on the number and multiplicities of the roots of $f_0$, $f_1$, $g_0$ and $g_1$.  By varying those (distinct, likely repeated) roots as in the previous paragraph, one gets a morphism from a rational, quasi-projective variety to $M_g$.  By Harris-Mumford(-Eisenbud), the only such morphism is constant if the image contains a general point of $M_g$.<|endoftext|>
TITLE: Is it necessary to use AC to solve this problem ?
QUESTION [13 upvotes]: Dear All, 
As a routine application of Zorn's Lemma, one can show that there is a subset $A$ of $\mathbb{R}$ such that $A$ contains no arithmetic progression of length 3 but for any $x\not \in A$, $A\cup \lbrace x\rbrace $ contains an arithmetic progression of length 3.
So there exists such a set (Using AC) but I failed to construct, or even prove the existence without AC, of such set which forced me to ask here if anybody can prove the existence without AC or Show that its existence is depended on AC !?

REPLY [24 votes]: I can't find the error in this argument, but everybody seems convinced (in the comments) that such a simply definable set with the desired property doesn't exist so there must be an error somewhere.
Let $A$ be the set of reals whose nonterminating base three expansion avoid the digit $1$.  As usual, there is some hassle dealing with negative reals, so a more precise definition would be that $x \in A$ iff the nonterminating base three expansion of $2\cdot 3^k + x$ avoids the digit $1$ for all sufficiently large natural $k$.  This definition makes it clear how to translate everything into the positive reals, where arithmetic is easier.
First let's check that $A$ contains no $3$-term arithmetic progression.  Suppose towards a contradiction that it contains such a sequence $x < y < z$.  For large $k$, look at the leftmost (most significant) digit where $2\cdot 3^k + x$ and $2\cdot 3^k + z$ differ; this is (eventually) independent of $k$.  Moreover, since $x < z$, this digit must be $0$ in $2\cdot 3^k + x$ and $2$ in $2\cdot 3^k + z$.  That means it must be $1$ in $2\cdot 3^k + y$ (since $y$ is the average of $x$ and $z$), contradicting the hypothesis that $y \in A$.
Now suppose that $x \not\in A$.  By considering translations as above it is enough to argue in the case that $x$ is positive.  Since $x$ is not in $A$, it has at least one digit which is $1$.  Build distinct elements $y, z \in A$ simply by "un-averaging the $1$s," i.e., $y$ and $z$ agree with those digits of $x$ which are $0$ or $2$, and on those digits of $x$ which are $1$, make the corresponding digit of exactly one of $y,z$ equal to $0$ and the other equal to $2$.  Note that if $x$ has infinitely many digits which are $1$, it's important to pick both $0$ and $2$ infinitely often for $y$ (and thus $z$) to avoid accidentally making a terminating base three expansion.  Then by construction, $y, x, z$ forms an arithmetic progression in $A \cup \{x\}$.<|endoftext|>
TITLE: Does every polynomial diophantine equation have solutions modulo p?
QUESTION [7 upvotes]: Obviously, this is not exactly true; what I am really asking is whether any diophantine polynomial equation with integer coefficients (let's call them DPEICs) who's solution does not admit contradictory results (eg, x=x+1) has a solution modulo a prime number, and more generally, whether a system of n DPEICs in n variables has a solution (again assuming non-contradictory equations).
More succinctly, if a system of PEICs has a solution in the complex numbers, does it have an integer solution modulo p?

REPLY [2 votes]: Fermat's theorem that $a^p\equiv a\pmod p$ for any inegers $a,p$ with $p$ prime 
 allows us to conclude that the equation $X^p-X+1=0$ for any prime number $p$ does not have any solution modulo that prime number $p$..
In general, as polynomial functions and polynomials are two different animals for finite fields we have many non-constant polynomials that are constant as functions. 
Actually a friend of mine used to joke that $\mathbf{Z}/2$ is an algebraically closed field:
any non-constant polynomial function over $\mathbf{Z}/2$ should assume more than one value hence it has a zero!<|endoftext|>
TITLE: Approximating Jordan curves
QUESTION [5 upvotes]: I'd like to capture the intuitive notion that a Jordan curve $\gamma_2$ “follows”  or “approximates” another Jordan curve $\gamma_1$, i.e. goes somehow “parallel” to it or “oscillates” around it.
Consider a differentiable Jordan curve $\gamma_1: [0,1] \rightarrow \mathbb{R}^2$ and its normals, seen as straight lines crossing the curve perpendicularly. Consider another Jordan curve $\gamma_2$ with the following properties:

Each normal of $\gamma_1$ crosses $\gamma_2$ at least once. I.e., for each normal $n(s_1)$ of $\gamma_1$ there is an $s_2$ such that the point $\gamma_2(s_2)$ lies on $n(s_1)$.
Furthermore when $s_1 < s_1'$ then there are unique $s_2 \le s_2'$ such that $\gamma_2(s_2)$ lies on $n(s_1)$ and $\gamma_2(s_2')$ lies on $n(s_1')$.


Do these conditions suffice to capture the notion described above? For
  which “pathological ” cases do they eventually fail? How
  then would they have to be adjusted to capture the notion?

Further questions:

Under which extra conditions does “$\gamma_2$ follows $\gamma_1$” imply that $\gamma_1$ follows $\gamma_2$? 
When $\gamma_2$ follows $\gamma_1$, (how) can the area enclosed by $\gamma_1$ and $\gamma_2$ be calculated via the distance function $d(s_1) = |\gamma_1(s_1) - \gamma_2(s_2)|$ ($s_2$ the unique parameter according to condition 2 above)?

(Let the area enclosed by $\gamma_1$ and $\gamma_2$ be the symmetric difference $X_1 \triangle X_2 = (X_1 \cup X_2) \setminus (X_1 \cap X_2)$ of the areas $X_1, X_2$ enclosed by $\gamma_1$ and $\gamma_2$.)

REPLY [7 votes]: I would use the multiscale flat norm. (I am biased of course -- see: this paper on the multiscale flat norm)
You still need the minimization over rigid motions, but the flat norm is close to what you have come up with above. It works in any ambient dimension on surfaces of any co-dimension. It also does not require the surfaces you are comparing to be boundaries. The paper I wrote with Simon Morgan (linked to above) was primarily the observation that in the codimension one boundary case, the flat norm is computed by the $L^1$TV image operator. That has lots of nice consequences, like fast algorithms to do the calculations and a very useful (though simple) generalization of the classical flat norm.
The basic idea is explained carefully (and intuitively) in the paper, but I will also explain it very briefly here. 
To find the distance between two k-dimensional currents $T_1$, $T_2$ (think of currents as surfaces with orientations), you can 

decompose $T_1 - T_2$ into two pieces, $T_1 - T_2 = E + \partial S$, where $E$ is a k-current and $S$ is a k+1-current, so $\partial S$ is again a k-current, 
charge yourself $M(E) + M(S)$ for the decomposition, where $M$ is the mass, measuring the k-volume of $E$ and the k+1-volume of $S$, and
minimize this cost over all k+1-currents $S$.
The result is the flat norm of $T_1 - T_2$, $\Bbb{F}(T_1-T_2)$.

Collecting all of this into one equation, the flat norm of a current T (we were choosing $T = T_1 - T_2$ above) is given by:
$\hspace{1in}\Bbb{F}(T) = \min_{\text{ k-currents }S} ( M(T-\partial S) + M(S)$)

Here is an image illustrating this for the case $k = 1$:
 (source)
For optimal $S$, $\Bbb{F}(T)$ is therefore the length of $T-\partial S$ plus the 2-dimensional area of $S$ (for the optimal $S$).

Finally, adding a parameter $\lambda$ permits us to controlling the tradeoff point between length and area. This gives us the $\color{blue}{\text{multiscale flat norm}}$:
$\hspace{1in}\Bbb{F}(T,\lambda) = \min_{\text{ k-currents }S} ( M(T-\partial S) + \lambda M(S)$)
If $\lambda$ is really big, we like length and try to avoid paying for area in $S$, if $\lambda$ is small then we prefer to replace cost of length with the cost of area whose boundary is used to cancel length. In the following image, for small $\lambda$'s we cancel all of the length cost in all but the largest circle:
 (source)

$\color{blue}{\text{A cool result: }}$ for minimizing decompositions, the mean curvature of $T-\partial S$ is bounded above by $\lambda$.<|endoftext|>
TITLE: Questions on how SYZ conjectures is deduced from HMS conjeture. 
QUESTION [6 upvotes]: The Strominge-Yau-Zaslow conjecture is roughly the following. Any Calabi-Yau $m$-manifold $X$ admits a special Lagrangian $T^m$ fibration (maybe at around a special point in its complex moduli space) and a mirror partner $Y$ is obtained by dualizing the tori $T^m$ with "instanton corrections" coming from singular fibers. 
If my memory serves one of heuristics of SYZ conjecture comes from Kontsevich's homological mirror symmetry conjecture; the moduli space of sky-scraper sheaves $\mathcal{O}_y$ (the easiest B-branes) on $Y$ is $Y$ itself, and there should be a corresponding moduli space of A-branes on $X$. A-brane is a pair $(L,c)$ of Lagrangian submnaifold $L\subset X$ and a flat $U(1)$ connection $c$ on $L$. By computing cohomology groups $HF^*(L,L)\cong Ext(\mathcal{O}_y,\mathcal{O}_y)=H^*(T^m,T^m)$, we expect that our $L$ is a Lagrangian $T^m$. My first question is

How do you expect that $L$ is a spcecial Lagrangian manifold?

If we assume that $L$ is special Lagrangian, then its deformation space is known and is of dimension $3$, and we expect that $L$ sweeps over $X$. My second question is

Does the flat $U(1)$ connection $c$ on $L$ play any role in this story? 

All we "deduced" from HMS conjecture is that $X$ admits a special Lagrangian fibration. Can we say anything more? Since HMS conjecture doesn't say anything about construction of mirror manifolds, I am afraid that we cannot say anything about dualizing these $L=T^n$ etc. My third question is 

What is the A-brane object on $X$ that corresponds to the obvious $6$-brane $Y$?

REPLY [6 votes]: For your first question, you can see this answer 
What is geometric intuition of special Lagrangian manifolds? 
for some idea of how the special condition maybe natural from the point of view of homological mirror symmetry.
For your second question, if we grant this, then the local system is a natural addition, because we can naively expect that $HF((L,c),(L,c))$ is also $H^*(T^3)$. Meanwhile as you have said we expect that there is only a three dimensional family of special Lagrangians, so we need more objects to correspond to the six dimensional family of skyscraper sheaves. It is therefore natural to allow the $U(1)$ flat connections on $L$ and expect that they also correspond to points in our mirror.
For your third question, the answer is that line bundles on the mirror should correspond to sections of the Lagrangian torus fibration. Normally, we fix a section to begin with and just declare that that goes to the structure sheaf $O_Y$. One motivation for this idea is a similar naive reasoning with Exts as the one you mention in the question. Namely $Ext(O_Y, O_y)$ should be isomorphic to $\mathbb{C}$, so we expect that our Lagrangian hits each fiber once. A nice case to consider is that of the elliptic curve. If you examine the functor constructed in Polishchuk and Zaslow's paper on the elliptic curve, you will see that this is in fact how mirror symmetry works in this case, namely points will correspond to local systems over (0,1) curves and line bundles will correspond to (1,n) curves.<|endoftext|>
TITLE: Minimal period of arithmetic progressions occurring in sets of positive density.
QUESTION [6 upvotes]: Let $A$ be a subset of ${\mathbb N}$ with positive upper-Banach density, and for each integer $k\geq3$, define $R_k=R_k(A)$ to be the smallest positive integer $r$ such that $A$ contains a length $k$ arithmetic progression
$$
\{a, a+r, a+2r, \dots, a+(k-1)r\}.
$$
Thus, the finiteness of $R_k$ is Szemeredi's theorem.  
What, if anything, is known about how $R_k$ grows?  More precisely, the question I am most interested in, without any luck so far, is the following: 

Does there exist an example of a set $A$ for which $R_k$ grows with $k$, with, if possible, a lower bound on $R_k$?  This lower bound may depend on $A$, but I am hoping for an explicit dependence on $k$.

If there are results in the other direction -- upper bounds on $R_k$ -- I would be interested to hear about those as well.  Such a result could be considered a quantitative strengthening of Szemeredi's theorem, so perhaps this is asking for a lot.

REPLY [2 votes]: It seems that a nice example is the set $A$ of positive integers which have an even number of 1's in their binary expansion, although I don't see a reasonable lower bound on $R_k(A)$ for now. A quick computation suggests that $R_2(A) = \dots = R_8(A) = 3$, $R_9(A) = R_{10}(A) = 9$, $R_{11}(A) = \dots = R_{20}(A) = 15$, $R_{21}(A) = \dots = R_{32}(A) = 31$, $R_{33}(A) = R_{34}(A) = 33$ and $R_{35}(A) = \dots = R_{68}(A) = 63$.<|endoftext|>
TITLE: Certain finitely presented group
QUESTION [7 upvotes]: Does the following group have a name? Is it amenable?
Fix $p$ and $q$

$\langle g,h: hg^qh^{-1}=gh^pg^{-1}\rangle$

REPLY [6 votes]: For 1-related groups, over a 2-letter alphabet, draw the relator on the plane grid: $g$'s are horizontal, $h$'s are vertical, starting at the point $O=(0,0)$. Connect $O$ with the endpoint $М$ of the resulting path $P$ by a vector $\vec{v}=\vec{OM}$. Consider the two support lines of the path $P$ that are parallel to $\vec{v}$.  If both support lines intersect the path only once, the group is free-by-cyclic. If only one of the support lines intersects the path once, the group is an ascending HNN extension of a free group, and the rank of the free group is easily computed using Magnus rewriting. It can be all found in the old paper Kenneth S. Brown, Trees, valuations, and the Bieri-Neumann-Strebel invariant. Invent. Math., 90(3):479--504, 1987.<|endoftext|>
TITLE: Roadmap to Complex Dynamics (Particularly the works of Hubbard, Douady, and Yoccoz regarding the Mandelbrot set)
QUESTION [6 upvotes]: As others have had great success with their question, I hope to ask one in a similar vein. As a student who has some background in complex analysis and dynamical systems, I am hoping to explore questions regarding connectivity (both connected and locally connected) in the Mandelbrot set. However, I have very little background in topology so I was wondering if some one could make recommendations of how to approach studying these questions. 
I suppose I am really hoping for a focused introduction to connected in topology with a focus on complex dynamics. 
Thanks!

REPLY [7 votes]: The Orsay notes have already been translated, but you want other references.
First read Milnor notes on external rays and local connectivity:

J. Milnor, "Periodic orbits, externals rays and the Mandelbrot set: an expository account," in Géométrie complexe et systèmes dynamiques: Colloque en l’honneur d’Adrien Douady, Paris: Soc. Mat. de France, 2000, pp. 277-333.
J. Milnor, "Local connectivity of Julia sets: expository lectures," in The Mandelbrot Set, Theme and Variations, Lei, T., Ed., Cambridge: Cambridge Univ. Press, 2000, pp. 67-116.

External rays are used to construct puzzle pieces. The pieces around the critical point 0 are nested (each one strictly containing the next. The goal is to compute the annuli between consecutive pieces and hopefully show their sum diverges. If so, the diameter of the pieces around 0 will shrink, and this gives you a system of neighborhoods that prove local connectivity of the Julia set at 0. A related construction gives the same result on the Mandelbrot set M. The details are in Hubbard's explanation of Yoccoz's proof of MLC for finitely renormalizable quadratic polynomials:

J. H. Hubbard, "Local connectivity of Julia sets and bifurcation loci: three theorems of J.-C. Yoccoz," in Topological Methods in Modern Mathematics, Houston, TX: Publish or Perish, 1993, pp. 467-511.

After understanding the mechanics of the puzzle construction, learn about Lyubich's principal nest. This is a subset of puzzle pieces with more complicated combinatorics:

M. Lyubich, "Dynamics of quadratic polynomials. I, II," Acta Math., vol. 178, iss. 2, pp. 185-247, 247, 1997.

This allows Lyubich to prove local connectivity of M at infinitely renormalizable pararameters of bounded type:

M. Lyubich, "Dynamics of quadratic polynomials. III: parapuzzle and SBR measures," Asterisque volume in honor of Adrien Douady's 60th birthday ``G\'eom\'etrie complexe et syst\'emes dynamiques'', v. 261 (2000), 173 - 200.

Then you are ready to read the most recent developments:

J. Kahn and M. Lyubich, "The quasi-additivity law in conformal geometry," Ann. of Math., vol. 169, iss. 2, pp. 561-593, 2009.

REPLY [5 votes]: As an introduction to complex dynamics (as well as connectivity and local connectivity) I strongly
recommend the book of J. Milnor. On my opinion, this is by far the best book on the subject.
After that, you want to know more about the Mandelbrot set, and for this I recommend
the 2-volume preprint of Douady and Hubbard in Publ. Orsay. (Now available on Internet. But exists
only in French).
Then (or simultaneously) you read McMullen's book Complex dynamics and renormalization. Annals of Mathematics Studies, 135. Princeton University Press, Princeton, NJ, 1994.
And then, papers of Yoccoz, Lyubich and others.
I have to warn you that this is a highly technical subject, and it will be hard to start your own
research in it without a proper adviser.<|endoftext|>
TITLE: What axioms are used to prove Godel's Incompleteness Theorems?
QUESTION [35 upvotes]: I understand Godel's Incompleteness Theorems to be statements about effectively generated formal systems, which basically makes them theorems about algorithms.  This is cool, because despite being very abstract, they actually constrain my expectations about how computers and human beings can behave.  But, being theorems, what formal system are they theorems in?  That is, what formal language is used to express them, how do I interpret that language as being about algorithms, what axioms are assumed, and what rules of inference are used to derive the incompleteness theorems?
I ask because I am looking for a better answer than "ZFC", which has been given to me in person a few times now.  ZFC refers to all sorts of things I don't believe exist (e.g. non recursively enumerable sets, choice functions for uncountable families...), at least not in the same way I believe in concrete things like computers and algorithms.  I can see from skimming the proofs that I could probably make up a formal system in which the theorems could be expressed and proven, which did not refer to all the monstrosities of ZFC.  I just want to know what standard, "simplest" formal system(s) can be used for this purpose.

REPLY [8 votes]: A very detailed, low-level proof of Gödel's incompleteness theorems is "Finite sets and Gödel’s incompleteness theorems". It's based on the theory of hereditarily finite sets, which is closely related to PA.<|endoftext|>
TITLE: simpler way to define modular forms
QUESTION [5 upvotes]: i'm trying to define modular forms for the full modular group for a general audience and want to make it as simple as possible. please tell me if this is correct:
definition of a modular form $f$:

holomorphic function $f$ from upper half plane to $\mathbb{C}$.
$f(z)$ bounded as Im($z$) tends to infinity.
$f$ satisfies the usual transformation law under $SL_2(Z)$

All the definitions I've seen say that, instead of 2., f should be holomorphic at infinity. I know that this is more beautiful and better for generalizations of the definition, for example to congruence subgroups... but then I have to explain what it means to be holomorphic at infinity, by using a power series representation. That's an extra step which I think is implicit in my definition because.....
By 3., $f(z)$ is periodic and so by complex analysis it must have a fourier series expansion $f(z)=\sum_{n=-\infty}^{\infty} a(n) e^{2\pi i nz}$. By 2. and complex analysis, we must have a(n)=0 for $n<0$. Then the resulting fourier series shows that $f$ is holomorphic at infinity and we arrive at the same thing.
Is all this right or am I missing something?

REPLY [11 votes]: This is not exactly an answer to the question as asked, but I was recently confronted to the same problem: how to define, for a general audience, a modular form for the full modular group
in a way which is both correct and quick? Here is a solution I used:
A modular form of weight $k$ (and level 1) is a formal series $f(q)=\sum_{n \geq 0} a_n q^n$, converging on the open unit disc, such that, in terms of the variable $z$ defined by $q=e^{2 i \pi z}$, $f$ satisfies $f(-1/z)=z^k f(z)$.
It is short and allows you to talk without further steps of the $q$-expansion of $f$, and avoid even mentioning the full modular group $Sl_2(\mathbb Z)$ or using any explicit element in it. This also avoids having to justify the existence of a $q$-expansion from $f(z+1)=f(z)$, for which I have always felt unsatisfied by the common three-word justification "by Fourier theory".  
Of course, depending of what you want to do next, this may be either an advantage or a shortcoming. If you need to discuss other congruences subgroups later, this is not the definition you want. But if what you need about modular forms is mainly their 
$q$-expansions, as in many arithmetical applications, this definition may be useful.<|endoftext|>
TITLE: Eigenvalues of the free sphere
QUESTION [8 upvotes]: Consider the usual sphere $S^{n-1}\subset\mathbb R^n$. By Stone-Weierstrass $C(S^{n-1})$ is generated by the standard coordinates $x_1,\ldots,x_n:\mathbb R^n\to\mathbb R$, and in fact we have the presentation result $C(S^{n-1})=C^*_{comm}(x_1,\ldots,x_n|x_i=x_i^*,\sum x_i^2=1)$. 
The Riemannian structure of $S^{n-1}$, or at least part of it, can be recaptured from this formula. Indeed, the eigenspaces of $D=\sqrt{d^*d}$ are $E_k=H_k\cap H_{k-1}^\perp$, where $H_k=span(x_{i_1}\ldots x_{i_r}|r\leq k)$, and the corresponding eigenvalues are $\lambda_k=k(k+n-2)$. This leads to the following question:

What is the free analogue of $\lambda_k$?

More precisely, consider the algebra $A=C^*(x_1,\ldots,x_n|x_i=x_i^*,\sum x_i^2=1)$, corresponding to the NCG-theoretic "free sphere". One can construct spaces $H_k,E_k$ as above, so this free sphere has indeed a spectral triple structure, and the question is to find the correct eigenvalues for $D=\sqrt{d^*d}$.

REPLY [10 votes]: How about
$\lambda_k = \frac{U'_k(n)}{U_k(n)}$
where the $U_k$ denote the Chebyshev polynomials of the second kind, $U_0(x)=1$, $U_1(x)=x$, and $U_k(x)=xU_{k-1}(x)-U_{k-2}(x)$ for $k\ge 2$.
In Section 10 of http://arxiv.org/abs/1210.6768 (See in particular Remark 10.4) we try to classify "Brownian motions" on $O_n^+$. The formula above follows, if you use the co-action of the free orthogonal quantum group on the free sphere to define an action of generator of "$O_n^+$-BM" on the free sphere.<|endoftext|>
TITLE: Intuitive meaning of Double Commutant Theorem
QUESTION [9 upvotes]: Is there any intuitive explanation of the Double Commutant Theorem for Von Neumann Algebras? By intuitive I mean in terms of Quantum Mechanics. For example, duality of states and observables in the case of the Gelfand-Naimark Theorem. http://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theorem

REPLY [11 votes]: Okay, here's an explanation in terms of quantum mechanics. Let ${\cal A}$ be a family of observables, modeled as self-adjoint operators on some Hilbert space, and let ${\cal U}$ be the group of all unitary transformations that leave every observable in ${\cal A}$ invariant. You can consider ${\cal U}$ to be a kind of symmetry group. Mathematically it is the set of unitaries in the first commutant ${\cal A}'$ of ${\cal A}$, and the set of all observables left invariant by ${\cal U}$ is the double commutant of ${\cal A}$. So the double commutant theorem says that the set of all observables left invariant by every transformation that leaves every observable in ${\cal A}$ invariant, is the self-adjoint part of the von Neumann algebra generated by ${\cal A}$.<|endoftext|>
TITLE: Who discovered the winding number?
QUESTION [6 upvotes]: I would guess it was well known by the time of Cauchy. But are there earlier references to it?

REPLY [13 votes]: Grünbaum and Shephard  suggest that the winding numbers (for closed polygons) have been discussed in the literature at least since 1769. 
See
A.L.F.  Meister,  Generalia de genesi figurarum  planarum et  inde pendentibus earum ajfectionibus,  Novi Comm. Soc.  Reg.  Scient.  Gotting.  1 (1769/70),  pp.  144-180.<|endoftext|>
TITLE: definition of operator valued integral with spectral measure
QUESTION [5 upvotes]: I am trying to make sense of some operators that come up on Buchholz and Summers' work on warped convolutions (two works on arxiv: 2008 and 2011). 
There, they work on a Hilbert space $H$ and on the bounded operators algebra $B(H)$ using some operator valued integrals similar to
$\int_\mathbf{R} A(x)\,dE(x)\;$ and $\;\int_\mathbf{R} dE(x)\,A(x)$,
where $E$ is the spectral resolution of a self-adjoint operator and $A$ is a $B(H)$ valued (norm-continuous) function. I don't know how one defines that.
I don't even know how that one above is defined, since both the measure and the function are operator-valued kinda. What I read about is that you can integrate vector valued functions with respect to a scalar valued measure (Bochner integral or Pettis integral), or scalar valued functions with respect to a spectral resolution (projection valued measure - spectral theorem).
If anyone knows how to define it and/or standard references for it, I would be thankful!

Some further remarks. If this helps, the authors also states that if $E$ is the spectral resolution of the self-adjoint operator $P$, then $P\,dE(p) = p\,dE(p)$, and if $B$ is a compact subset of $\mathbf{R}$ and $F$ is a finite-rank projection in $H$, then by spectral calculus $\int_B A(x)\,F\,dE(x)\;$ and $\;\int_B dE(x)\,F\,A(x)$ are well defined. I can make sense of the second type of integrals, since those will be of finite rank, but not the first type.
Thank you.

REPLY [6 votes]: Well, I would make sense of expressions like this by inserting $|e_n\rangle\langle e_n|$ between the operator and the measure and then summing over $n$.
Say we want to give a meaning to $A = \int A(x)d\mu$. If we know how to integrate scalar-valued functions against a spectral measure then we can define $$\langle v|A = \sum_n \int \langle v|A(x)|e_n\rangle\langle e_n|d\mu = \sum_n \langle e_n|\int \langle v|A(x)|e_n\rangle d\mu$$ for any $v \in H$, and if we know how to integrate vector-valued functions against scalar measures then we can define $$A|w\rangle = \sum_n \int A(x)|e_n\rangle\langle e_n|d\mu|w\rangle$$ for any $w \in H$. So take your pick, say what the operator is by saying either what happens when you apply it to a vector or what happens when you apply a covector to it.
Do these sums converge? No, not in general, even assuming $A(x)$ is a bounded operator-valued function and $\mu$ is the spectral measure coming from a compact self-adjoint operator. For example, consider the spectral measure coming from the operator of multiplication by $1/n$ on $l^2({\bf N})$. This is a spectral measure defined on $X = \{1/n: n \in {\bf N}\}$ with $\mu(\{1/n\}) = |e_n\rangle\langle e_n|$.` Define an operator-valued function on $X$ by setting $A(1/n) = |e_1\rangle\langle e_n|$. Then the integral $\int A(x)d\mu$ becomes a sum $\sum_n |e_1\rangle\langle e_n|e_n\rangle\langle e_n| = \sum_n |e_1\rangle\langle e_n|$ which does not converge.
Edit: I just looked at their 2011 paper and I found that they have some discussion about how to make their integrals rigorous in Section 2.2. So I think the full answer is that integrals of this form don't always make sense, but there is a heuristic and one can prove convergence in some special cases.<|endoftext|>
TITLE: How to show a certain determinant is non-zero
QUESTION [15 upvotes]: For any $n$ distinct points $x_1,x_2 , \ldots , x_n$ on the real line show that 
the matrix $M$ where $M(i,j) = e^{\lambda_j x_i} $ has non-zero determinant 
where $\lambda_1 \lt \lambda_2 \lt \ldots \lt \lambda_n \in \mathbb{R}$ are fixed constants. 
I am able to show this for $n=1$(duh...) and $n=2$. Is this an inductive proof?

REPLY [6 votes]: EDIT: OOPS! The solution below is FALSE. My apologies for wasting everyone's time (particularly those among you who already were aware of Schur functions) with this scrible. I'll let it stay around because maybe this thread will one day be renamed "how not to show a certain determinant..." and maybe there is something to be learned from it.
For the record, the mistake is to claim that "multiplying $\lambda_1$, $\lambda_2$, ..., $\lambda_n$ by their common denominator $p$ is easily compensated by replacing the positive reals $y_1$, $y_2$, ..., $y_n$ by the (equally positive) reals $y_1^{1/p}$, $y_2^{1/p}$, ..., $y_n^{1/p}$". This is true for the original question, but not for the more subtle (1). Sorry again!

For the sake of pushing algebraic combinatorics (specifically Schur polynomials), let me present a different proof of the nonvanishing of the determinant in question. The idea of this proof is taken from my Jan 7, 2013 comment on the original question, but it has been tweaked to work in the general case.
We want to prove that $\det\left(\left(e^{\lambda_j x_i}\right)_{1\leq i,j\leq n}\right)\neq 0$ for any $n$ distinct reals $x_1$, $x_2$, ..., $x_n$ and any $n$ distinct reals $\lambda_1$, $\lambda_2$, ..., $\lambda_n$.
Let us WLOG assume that $x_1 > x_2 > ... > x_n$ and $\lambda_1 > \lambda_2 > ... > \lambda_n$ (because interchanging the $x_i$ or the $\lambda_i$ boils down to row resp. column swaps on the matrix whose determinant we are concerned with).
Let us denote $y_i = e^{x_i}$ for every $i\in\left\lbrace 1,2,...,n\right\rbrace$. Then, $y_1$, $y_2$, ..., $y_n$ are $n$ positive reals satisfying $y_1 > y_2 > ... > y_n$, and we must prove that $\det\left(\left(y_i^{\lambda_j}\right)_{1\leq i,j\leq n}\right)\neq 0$ (since $e^{\lambda_j x_i} = \left(e^{x_i}\right)^{\lambda_j} = y_i^{\lambda_j}$ for any $i$ and $j$).
We can WLOG assume that $\lambda_1$, $\lambda_2$, ..., $\lambda_n$ are all nonnegative, since we could replace the whole $n$-tuple $\left(\lambda_1,\lambda_2,...,\lambda_n\right)$ by $\left(\lambda_1+m,\lambda_2+m,...,\lambda_n+m\right)$ for a sufficiently large real $m$ without changing much about our determinant (namely, the determinant would merely gain a multiplicative factor of $y_1^m y_2^m ... y_n^m$). So assume this.
We will actually prove that
(1) $\det\left(\left(y_i^{\lambda_j}\right)_{1\leq i,j\leq n}\right) \geq \prod\limits_{1\leq i < j \leq n} \left(y_i-y_j\right) \cdot \prod\limits_{k=1}^n y_k^{\lambda_k - n + k}$.
Once this is proven, it will follow that $\det\left(\left(y_i^{\lambda_j}\right)_{1\leq i,j\leq n}\right) > 0$ (because the right hand side of (1) is positive) and thus $\det\left(\left(y_i^{\lambda_j}\right)_{1\leq i,j\leq n}\right) \neq 0$, which is exactly what we need to prove.
So it remains to prove (1) for any $n$ positive reals $y_1$, $y_2$, ..., $y_n$ satisfying $y_1 > y_2 > ... > y_n$, and any $n$ nonnegative reals $\lambda_1$, $\lambda_2$, ..., $\lambda_n$ satisfying $\lambda_1 > \lambda_2 > ... > \lambda_n$.
Note that both sides of the inequality (1) are continuous as functions in $\lambda_1$, $\lambda_2$, ..., $\lambda_n$. Hence, we can WLOG assume that $\lambda_1$, $\lambda_2$, ..., $\lambda_n$ are nonnegative rationals (because the inequality is non-strict, and the set of strictly increasing $n$-tuples of nonnegative rationals is dense in the set of strictly increasing $n$-tuples of nonnegative reals). Assuming this, we can go further and assume WLOG that $\lambda_1$, $\lambda_2$, ..., $\lambda_n$ are nonnegative integers, because multiplying $\lambda_1$, $\lambda_2$, ..., $\lambda_n$ by their common denominator $p$ is easily compensated by replacing the positive reals $y_1$, $y_2$, ..., $y_n$ by the (equally positive) reals $y_1^{1/p}$, $y_2^{1/p}$, ..., $y_n^{1/p}$. So assume this, and let $\mu$ denote the sequence $\left(\lambda_1+n-1, \lambda_2+n-2, ..., \lambda_n+n-n\right)$. This sequence $\mu$ is a partition (in the sense of algebraic combinatorics), i. e., a finite weakly decreasing sequence of nonnegative integers.
Now, for any sequence $\kappa = \left(\kappa_1,\kappa_2,...,\kappa_n\right)$ of nonnegative integers, let $a_{\kappa}$ denote the determinant $\det\left(\left(y_i^{\kappa_j}\right)_{1\leq i,j\leq n}\right)$. Let $\rho$ be the sequence $\left(n-1,n-2,...,0\right)$. Then, $a_{\rho} = \prod\limits_{1\leq i < j \leq n} \left(y_i-y_j\right)$ (by Vandermonde's determinant) while (using the notation $\mu+\rho$ for the termwise sum of the sequences $\mu$ and $\rho$) we have $a_{\mu+\rho} = \det\left(\left(y_i^{\lambda_j}\right)_{1\leq i,j\leq n}\right)$ (since $\mu+\rho$ is the sequence $\left(\lambda_1,\lambda_2,...,\lambda_n\right)$). Hence, (1) rewrites as
$a_{\mu+\rho} \geq a_{\rho} \cdot \prod\limits_{k=1}^n y_k^{\lambda_k - n + k}$.
Since $a_{\rho} = \prod\limits_{1\leq i < j \leq n} \left(y_i-y_j\right) > 0$, this is equivalent to
(2) $\dfrac{a_{\mu+\rho}}{a_{\rho}} \geq \prod\limits_{k=1}^n y_k^{\lambda_k - n + k}$.
But it is known (e. g., Corollary 2.37 in Victor Reiner, Hopf algebras in combinatorics) that $\dfrac{a_{\mu+\rho}}{a_{\rho}}$ equals the Schur polynomial $s_{\mu}$ evaluated at $\left(y_1,y_2,...,y_n\right)$. Thus,
(3) $\dfrac{a_{\mu+\rho}}{a_{\rho}} = s_{\mu}\left(y_1,y_2,...,y_n\right) = \sum\limits_{T} \prod\limits_{k=1}^n y_k^{\text{number of }k\text{'s in }T}$,
where the $T$ on the right hand side runs over all semistandard (i. e., column-strict) Young tableaux of shape $\mu$ with entries in $\left\lbrace 1,2,...,n\right\rbrace$. One such tableau is obtained by filling each cell in row $k$ with the number $k$, for every $k \in \left\lbrace 1,2,...,n\right\rbrace$ (where the numbering of rows starts with $1$). This tableau contributes one addend to the sum on the right hand side of (3), and this addend is $\prod\limits_{k=1}^n y_k^{\lambda_k - n + k}$ (because the length of the $k$-th row is $\lambda_k - n + k$, and the $k$'s in the tableau are exactly the entries of the $k$-th row). Since all the other addends on the right hand side of (3) are nonnegative (being monomials in $y_1$, $y_2$, ..., $y_n$ with coefficient $1$), this yields
$\dfrac{a_{\mu+\rho}}{a_{\rho}} \geq \prod\limits_{k=1}^n y_k^{\lambda_k - n + k}$.
But this is exactly (2). Since we know that (2) is equivalent to (1), this completes the proof of (1), and thus solves the problem.
Remark: Why did we take the detour through (1) rather than confine ourselves to proving the (weaker but sufficient) inequality
(4) $\det\left(\left(y_i^{\lambda_j}\right)_{1\leq i,j\leq n}\right) > 0$ ?
Because (1) is a non-strict inequality, whereas (4) is strict. When proving a strict inequality, it is not enough to prove it on a dense subset of its domain, even if it is continuous; for example, Vasile Cîrtoaje's brainteaser $\left(x^2+y^2+z^2\right)^2\geq 3\left(x^3y+y^3z+z^3x\right)$ (an inequality holding for all $x,y,z\in \mathbb R$) attains its equality at $x=y=z$ but also at $x:y:z=\sin^2\dfrac{4\pi}{7}:\sin^2\dfrac{2\pi}{7}:\sin^2\dfrac{\pi}{7}$, an equality condition invisible when one restricts oneself to the dense subset of rationals.<|endoftext|>
TITLE: Seeing topological (geom.) properties of the space via corresponding C^*-algebra
QUESTION [11 upvotes]: Compact Hausdorff spaces bijectively correspond to C^*-algebras with identity. One needs to consider  the algebra of continuous functions C(X) to go in one direction and spectrum to go in the other. (See e.g. Wikipedia). The situation is similar to algebraic geometry - affine manifolds correspond to commutative algebras... Basic skill in alg.geom. is to recast algebraic properties in geometric and vice versa e.g. projective modules - vector bundles... (the dictionary is lengthy).
I wonder about similar correspondence in C^*-algebra setup. In particular:
Question 1: if  space "X" is topological manifold (i.e. locally R^n), is there some "nice" way to recognize it via C^*-algebra of continuous function ? (... is there non-commutative version ? ... )
Question 2: if "X" is smooth manifold, is there nice way to recognize it and  define sub-algebra of smooth functions entirely in terms of C^*-algebra ? (... is there non-commutative version ? ... )
Question 3 Is it possible to characterize the set of all measures on "X" in terms of C(X) ? (... is there non-commutative version ? ... ) 
If you have further comments how interesting algebraic properties can be recasted in topological or vice versa, you are welcome to post.

REPLY [6 votes]: Question 1: The topological $n$-manifold property is equivalent to every point of $X$ having a neighborhood homeomorphic to $B^n$, the closed unit ball in $\mathbb{R}^n$. The existence of such a neighborhood $x\in B^n_x \subset X$ induces the surjective algebra homomorphisms $C(X) \to C(B_x^n) \cong C(B^n) \to C(\{x\})\cong \mathbb{R}$ (actually, extremal epimorphisms, I think). The Gelfand duality between compact Hausdorff topological spaces and commutative real $C^*$ algebras ensures that the existence of surjective homomorphisms $C(X) \to C(B^n) \to \mathbb{R}$ (the first map should be an extremal epimorphism, while the second should correspond to the quotient by the maximal ideal of an interior point of $B^n$) implies the existence of continuous maps $\{x\} \to B^n \to X$, where $x$ maps to an interior point of $B^n$ and $B^n$ is embedded in $X$, and hence a neighborhood of $x$ in $X$. Having such such algebra homomorphisms for each $x\in X$ characterizes $X$ as a topological manifold.
Question 2: $C^\infty(X)$ for a compact manifold $X$ is not a $C^*$ algebra. It is at the very least a Fréchet algebra, where multiplication satisfies an extra convexity condition (though I'm fuzzy on the details). It is at least clear that one must leave the category of $C^*$ to characterize it. A point that non-commutative geometry centered discussions of this questions seem to be ignoring is that there already exists an algebraic characterization of $C^\infty(X)$ that has nothing to do with non-commutative geometric spectral triples. In my understanding, such a characterization can be found in an article by Michor and Vanžura (arXiv:math/9404228).
Question 3: As already mentioned in Vahid's answer, this is the content of the Riesz representation theorem. The topological vector space dual to $C(X)$ is the space of signed Radon measures on $X$. $C(X)$ is partially ordered by pointwise comparison, which also induces a partial order on its dual. The positive cone in the space of signed Radon measures consists of the positive Radon measures.
I cannot say anything about non-commutative versions of the above answers. But, since these correspondences are heavily based on lots of non-trivial maximal ideals, and such ideals are likely to be absent in non-commutative algebras, they probably do not translate directly.<|endoftext|>
TITLE: Characterising categories of vector spaces
QUESTION [17 upvotes]: Consider the category $FdVect_k$ of finite dimensional $k$-vector spaces, for some given field. It is abelian, semisimple, in that each object is a finite sum of simple objects (of which there is only one up to isomorphism), and also compact closed with simple tensor unit which is a progenerator.
Can we characterise $FdVect_k$ as a category of vector spaces purely by properties of the category such as above? I don't mean to demand for instance that $End(I)$ is a field, for instance, and I suspect that this may stop any such characterisation. But this I don't mind, seeing as if rings which are 'nice enough' cannot be distinguished from fields in this way, then so be it.
I should point out that if someone says 'but what about Morita equivalence?', then I'm not sure that's right answer, since I'm looking for equivalence as a compact closed semisimple abelian ... category, not just as a bare category - but I may be wrong on this point.

REPLY [12 votes]: In Schur Functors I, Todd Trimble and I proved that $FdVect_k$ is the free symmetric monoidal $k$-linear Cauchy complete category on no objects.   Here a $k$-linear category is Cauchy complete if it has direct sums (that is, biproducts) and all idempotents split.  So, we don't need to mention compact closedness, abelianness, semisimplicity....
More precisely:
Proposition.  For any field $k$, if $C$ is a symmetric monoidal $k$-linear Cauchy complete category, there exists exactly one symmetric monoidal $k$-linear functor $i:FdVect_k \to C$, up to symmetric monoidal $k$-linear isomorphism.<|endoftext|>
TITLE: Uniqueness of values in recurrence relations
QUESTION [6 upvotes]: Given an integer $k > 1$, define the sequences $X(k,n), Y(k,n)$ as follows:

$a=4k-2,$  $y_0 = 1,$  $y_1 = a + 1,y_n = ay_{n-1} - y_{n-2}$

$b   = 4k + 2,$  $ x_0 = 1,$  $x_1 = b - 1,$  $x_n = bx_{n-1} - x_{n-2}$

For example, with $k = 2$ we get  
$y_j = 7, 41, 239, 1393, \ldots$
$x_j = 9, 89, 881, 8721, \ldots$
A simple question arises, as to whether there exist $\{k, i, j\}$ such that $X(k,i) = Y(k,j)$?
This might well be an open question, and perhaps inappropriate here, but I have trawled the web for many hours and have found no evidence that anybody has even considered it.
Computational experiments suggest that in fact an even stronger result is possible, ie. that there are no $\{k_1, k_2, i>1, j>1\}$ with $X(k_1,i) = Y(k_2,j)$.
In other words, with the exception of $x_1, y_1$ which can be any odd number > 7, all values generated by these sequences appear to be unique.
Any suggestions as to a way to attack this question  would be greatly appreciated!
Update: There are explicit proofs that for $k = 2, 3$ there can be no $X(k,i) = Y(k,j)$, so we can restrict the question to $k > 3$.  Sadly these proofs are not extendable to other k

REPLY [2 votes]: You may consider the paper:
B. Ibrahimpasic, A parametric family of quartic Thue inequalities. 
Bull. Malays. Math. Sci. Soc. (2) 34 (2011), no. 2, 215–230, 
available at http://www.emis.de/journals/BMMSS/vol34_2_2.html 
It seems that Theorem 3.1, with c=2k, answers your question.<|endoftext|>
TITLE: Is the cup product of holomorphic $n$-forms with a fixed class injective?
QUESTION [5 upvotes]: Let $X$ be a compact Kahler manifold of complex dimension $n$. Fix a nonzero class $u \in H^1(X,T_X)$. This gives a linear morphism
$$
\phi_u : H^0(X,\Omega^n) \to H^{1}(X,\Omega^{n-1}),
\quad \sigma \mapsto u \cup \sigma.
$$
Is $\phi_u$ injective?
It is so for manifolds with $\Omega^n_X = \mathcal O_X$; the proof I've got is not hard but uses Ricci-flat Kahler metrics and the hard Lefschetz theorem so it cannot generalize to other situations. In the examples I know (curves, hypersurfaces in $\mathbb P^n$) we have $h^{n,0} \leq h^{n-1,1}$, so I haven't stumbled upon an obvious counterexample yet.

REPLY [12 votes]: The answer to this question is negative in dimensions $\ge 3$. For example, take a quintic in $\mathbb CP^4$  and consider its blow up $X$ in $10^{100}$ points (just to be safe). Then the space $H^1(X, T_X)$ will be huge, since it parametrises deformations of the blown up variety and you can move points as you wish. So there will be non-zero $u\in H^1(X, T_X)$  so that you map is trivial. 
Note that when you blow up the quintic $H^{2,1}$ does not change. 
Also, this trick with blow ups will not work for Kahler surfaces  as is explained for example, in appendix  1 in http://arxiv.org/pdf/1301.0478.pdf

REPLY [7 votes]: The map $\phi_u$ is not always injective.  Let $X$ be the blowing up along a large number of points inside a quartic surface in $\mathbb{P}^3$, or any surface $S$ with nonzero $h^{n,0}$. The weight $2$ Hodge structure of $X$ is a direct sum of the weight $2$ Hodge structure of $S$, the image of the pullback morphism, and a finite number of copies of the Hodge structure $\mathbb{Z}(-1)$, the Gysin images of the weight $0$ Hodge structures of the exceptional divisors.  So, as you vary the blown up points in $S$, the corresponding variation of Hodge structures is trivial.  Thus the Griffiths transversality map $\phi_u$ is zero for every $u$ coming from a variation of the points.<|endoftext|>
TITLE: Usage of set theory in undergraduate studies
QUESTION [13 upvotes]: I would like to ask my colleagues their thought on good practices concerning
set theorical framework in undergraduate studies. For example, have there been any attempt  to use another mathematical formalism, such as ETCS? (for research issues,
see this question).
While most, if not all, of our mathematics are thought, done, written using set theory, our younger students seem to struggle with these concepts. Some can well put a $4\times 6$ matrix in row reduced echelon form but plainly do not understand the meaning of a question like "If $A,B$ are two square matrices of size~$n$, prove that $\ker(AB)$ contains $\ker(B)$." The difficulties with $\varepsilon,\delta$ definition of limits may be of a similar nature.
In fact, one may argue that all set theoretical concepts presently are more or less eliminated from the lower levels of mathematical education. One may even argue that it should be so. I remember that each one of the first years of middle school (from 6th grade on, the French and US systems coincide here!) taught me one new definition in set theory; sets and mappings at the age of 11, then equivalence relations, then sets of equivalence classes (to define vectors)...  And a few years later, students are taught quotient groups like $\mathbf Z/n\mathbf Z$ as sets of equivalence classes, a definition which they of course take litteraly. 
While Set theory is very useful to formalize things, at least once you're used to it,
it is true that it allows stupid questions, requires abuses of notations (so that one does not distinguish between the $1$ of $\mathbf Z$ with the $1$ of $\mathbf R$,
not forgetting thoses of $\mathbf Q$ and $\mathbf C$). In some sense, modern mathematicians, especially algebraists, speak sets but think categories.
This may be related with the fact that the precise definition of the axioms of set theory (ZFC, say) are not so well known among mathematicians, and even not really taught (for example, no mention of the replacement axiom in my own mathematical education). In contrast, a more recent book like Terence Tao's Analysis begins with a precise exposition of these axioms, up to this replacement axiom.
I can't really make my mind between one attitude and the opposite.
So what do you think?

REPLY [6 votes]: As a set theorist, I feel some obligation to offer an answer here.  First, the difficulties students may have in proving set theoretic containments like the one you mention above or in constructing $\epsilon$-$\delta$ proofs is not a matter of them struggling with set theory but rather of them struggling with something new: constructing a proof.  In the case of $\epsilon$-$\delta$ proofs, a large part of this difficulty is in understanding quantifiers and how they work (for instance "for all ... exists ..." is not the same as "exists ... for all ...".  This is because math is not a trivial subject to learn and some difficulty is required as the students minds stretch and grow.
Surely there is no way around this.
That really has nothing to do with set theory so far.  In spite of a common misconception, set theorists do no actually care how it is that ordered pairs are defined.  Or how exactly one codes the notion of a function.  Set theory is not in competition with category theory, in spite of what category theory thinks.  A very good analogy is provided by computer science: set theory is machine language (or maybe better: a low level language like C) and category theory is object oriented programming.  While object oriented programming may provide a useful way of thinking about how to write a program, there still needs to be a machine language for the computer to run on.  Moreover, there are occasionally things for which it is just better (or even necessary) to code in a low level language.
Set theory provides an exact standard by which to discuss questions like "is there a subset of the real line which is uncountable but not of cardinality $|\mathbb{R}|$" (Hilbert's First Problem) or, maybe better,
"is there an almost free, non free group?" (Whitehead's problem) or "If $h$ is a homomorphism
a commutative Banach algebra into $C[0,1]$, is $h$ continuous?" (the negation being Kaplanski's conjecture).  With the exception of the first question, these were asked, to my knowledge at least, without any thought that there was a foundational issue involved.  Surely these are questions which could reasonably be asked regardless of how one sets up their foundations.
To my knowledge category theory has never resolved these questions; set theory has in as satisfactory a manner possible (or at least until we adopt a more complete set of axioms).  Now, one can argue at length about whether such questions are asked in poor taste or whether we should allow them to be asked at all.  Readers interested in the question of "why care about set theory" should take a look at this (which might have been titled "why care about the uncountable").<|endoftext|>
TITLE: Maximal Ideals in $R[x]$
QUESTION [8 upvotes]: Let $R$ be a non-zero ring with identity. Is it possible for $R[x]$ to have only a finite number of maximal left ideals !?

REPLY [22 votes]: Lets try something like Euclid's proof of infiniteness of prime numbers. Let $f_1 = x, f_2 = xf_1 + 1, f_3=xf_1f_2+1, f_4=xf_1f_2f_3 + 1$ and so on. The left ideal generated by each $f_i$ is proper. Thus each $f_i$ is contained in a maximal left ideal. On the other hand for any $i < j$ we have $ f_i | f_j-1$ (note that $f_i$'s are in the center of $R[x]$). This means that $f_i$ and $f_j$ can not be both in one maximal left ideal. Hence we have an infinite number of maximal left ideals.<|endoftext|>
TITLE: Origin of notion of "split Grothendieck group"?
QUESTION [5 upvotes]: In the construction of Soergel's bimodules in representtion theory , it's essential for him to work with split Grothendieck groups.   Here he starts with a certain small additive category $\mathcal{A}$ and writes $\langle \mathcal{A} \rangle$ for its split Grothendieck group: the free abelian group on objects $\langle A \rangle$ corresponding as usual to isomorphism classes, modulo sums$\langle C \rangle = \langle A \rangle + \langle B \rangle$ corresponding only to the situation $C \cong A \oplus B$.
This is a less familiar situation than the usual Grothendieck group with sums corresponding to short exact sequences which may or may not split.   

Where does the notion of split Grothendieck group originate, and why?

This is mostly asked out of curiosity, but I'm also looking for further interesting examples.

REPLY [8 votes]: These groups are mentioned in [Swan '68 - Algebraic K-Theory, p.69]. He constructs $K_0(\mathcal{A}, S)$ for a class $S$ of exact sequences in $\mathcal{A}$. Take the free abelian group mod the relations from sequences in $S$. For example the class of all exact sequences for the Grothendieck-group $K_0(\mathcal{A})$ or all exact split seqences for the group you mentioned. The name 'split-Grothendieck-group' does not appear.
He generalizes further to $K_0(\mathcal{A}, F)$ for a bifunctor $F:\mathcal{A} \times \mathcal{A} \to \mathcal{A}$ and obtains a generalized Picard-group.<|endoftext|>
TITLE: Schauder estimates for higher order linear elliptic operator on manifold
QUESTION [6 upvotes]: Hi!
Let $(M,g)$ be a smooth compact riemannian manifold without boundary. Let $L$ be a linear elliptic operator on $M$ of order $2k$ with smooth coefficients. Suppose i have $u\in W^{2k,2}(M)$ and $f\in C^{0,\alpha}(M)$ such that
$$L(u)=f$$
Do i have Schauder estimates of type
$$\left\|u\right\|_{C^{2k,\alpha}\left(M\right)}\leq C\left(L\right)\left\|f\right\|_{C^{0,\alpha}\left( M \right)}$$
I can assume also that $L$ (it would be better for $L$ of general type) is self adjoint and $u$ is $L^2$-orthogonal to $\ker\left(L \right)$.
If yes is there a reference for this kind of result?
Thank you in advance.

REPLY [4 votes]: The result is true with some caveats. Under your assumptions we have the following results.
1. If $u\in W^{2k,2}(M)$ and $Lu\in C^{j,\alpha}(M)$, then $u\in C^{2k+j,\alpha}(M)$.
2.  There exists $C>0$ depending only on $M$, $L$, $j$, and $\alpha$ such that, for any  $u\in C^{2k+j,\alpha}(M)$ we have
$$\Vert u\Vert_{C^{2k+j,\alpha}}  \leq C\Bigl(\; \Vert Lu\Vert_{C^{j,\alpha}}+ \Vert u\Vert\_{C^{0,\alpha}}\;\Bigr) $$
3. There exists $C>0$ depending only on $M$, $L$, $j$, and $\alpha$ such that, for any  $u\in C^{2k+j,\alpha}(M)\cap (\ker L)^\perp $  (the $\perp$ refers to the $L^2$-inner product)   we have
$$\Vert u\Vert_{C^{2k+j,\alpha}}  \leq C \Vert Lu\Vert_{C^{j,\alpha}} $$
For proofs and more details see Chapter 10 of these notes and the references therein.<|endoftext|>
TITLE: A (too?) simple notion of "closed multicategory"
QUESTION [5 upvotes]: Suppose I define a multicategory $M=(Ob(M),Hom_M)$ to be simply closed if 

for every sequence $S=(b_1,\ldots,b_n;x)$ of $n+1$ objects in $M$, we provide an object $Exp(S)\in Ob(M)$, and
for every sequence $A=(a_1,\ldots,a_m)$ of $m$ objects in $M$ we provide a bijection
$$\phi_S^A\colon Hom_M(a_1,\ldots,a_m,b_1,\ldots,b_n;x)\to Hom_M(a_1,\ldots,a_m;Exp(b_1,\ldots,b_n;x)),$$

subject to the condition below, for any object $x\in Ob(M)$. Fix such an object $x\in Ob(M)$ and let $X:=Exp(;x)\in Ob(M)$. (Apologies: fixing $x$ and introducing $X$ is done here only for typographical reasons: mathjax was having trouble rendering.) We name a "constant element" $c_x\in Hom_M(x;X)$ by setting $A=(x)$ and $S=(;x)$, so $\phi^A_S\colon Hom_M(x;x)\to Hom_M(x;X)$, and putting 
$$c_x:=\phi^{(x)}_{(;x)}(id_x)\in Hom_M(x;X).$$ 
We also name an "evaluation element" by setting $A=(X)$ and $S=(;x)$, so $(\phi^A_S)^{-1}\colon Hom(X;X)\to Hom(X;x)$, and putting 
$$e_x:=\left(\phi^{(X)}_{(;x)}\right)^{-1} (id_X)\in Hom_M(X;x).$$
Then our condition is that the constant element and the evaluation element for $x$ are mutually inverse in the sense that 
$$c_x\circ e_x=id_X \;\;\;\;\;\;\text{ and }\;\;\;\;\;\; e_x\circ c_x= id_x$$ in $M$.
As alluded to above, we obtain an ``evaluation map" as an element $ev\in Hom_M(Exp(A;x),A;x)$, coming as $\phi^{-1}(id)$, where $id$ is the identity element $id\in Hom_M(Exp(A;x),Exp(A;x))$. 
The multicategory of sets is simply closed in this sense.
What problems exist for, and/or what niceties are missing from, this notion of "simply closed multicategory"?

REPLY [4 votes]: In fact, if you include naturality, then your second condition becomes automatic, and moreover it suffices to consider $n=1$.  See e.g. section 3 of this paper.<|endoftext|>
TITLE: Are All Irrational Elementary Numbers Conjectured to Be Normal?
QUESTION [6 upvotes]: If one defines an elementary number to be a real root or the real part of a complex root of any finite system of elementary functions with integer coefficients, has anyone conjectured in print that every irrational elementary number is a normal number (a real number whose infinite sequence of digits in every base is distributed uniformly)?  Are any counterexamples known?
Don N. Page
Professor of Physics
University of Alberta

REPLY [3 votes]: Perhaps connected to general conjectures published here ... 
"On the Random Character of Fundamental Constant Expansions", David H. Bailey and Richard E. Crandall
Experiment. Math. Volume 10, Issue 2 (2001), 175-190.<|endoftext|>
TITLE: Trace formula for PSDOs
QUESTION [12 upvotes]: In Getzler's famous paper "Pseudodifferential Operators on Supermanifolds and the Atiyah-Singer Index Theorem", he states that for a (trace-class) pseudo-differential operator $P$ on a Riemannian manifold $M$, one has the formula
$$ \mathrm{Tr} P = \int_{T^*M} \sigma(P), $$
where $\sigma(P)$ is the full symbol of $P$, which is calculated using "the exponential map of $M$ to pull back the pseudodifferential operator near $x \in M$ to $T_xM$ and calculate its symbol on $T^*_xM$ using the Euclidean structure" (quote of p.164). 
However, as far as I know, the full symbol of a $\Psi$DO can be defined using the Levi-Civita connection, but only up to a smoothing symbol. And of course in applications, $P$ is usually a smoothing operator.Also, I could not get my hands on a full copy of the paper of Widom ("A complete symbolic calculus for pseudodifferential operators") cited there, and the preview of it on google books does not seem to help me.
So, my question is: How does one define $\sigma(P)$ to make the above formula true?

REPLY [8 votes]: Consider the modified question: How does one define the operator $P=\mathrm{Op}(p)$ so that the trace formula holds with $\sigma(P)=p$ as the full symbol? The following quantization rule answers this with the help of the exponential map of the compact Riemannian manifold $M$, $n=\dim M$:
$$Pu(x)=(2\pi)^{-n}\int_{T_x^* M} \int_{T_x M} e^{-i\eta v}\chi(x,v)p(x,\eta)u(\exp_x v)\,dv\,d\eta.$$
The quantization depends on the choice of $\chi\in C_c^\infty(TM)$ which equals unity in a neighbourhood of the zero section and localizes to an open subset mapped diffeomorphically by $\exp$ to an open neighbourhood of the diagonal in $X\times X$. Using normal geodesic coordinates one proves that this quantization rule does indeed give the standard classes $\Psi^m(M)$ of pseudo-differential operators. In a neighbourhood of the diagonal, the Schwartz kernel $K$ is given as follows:
$$K(x,y)dy=((2\pi)^{-n}\int_{T_x^* M} e^{-i\eta v} p(x,\eta)d\eta) dv, \quad y=\exp_x v.$$
In particular, a different choice of cutoff $\chi$ modifies $P$ only by adding a smoothing operator. Moreover, we get, if the order of $P$ is strictly less than $-n$, the trace formula:
$$\mathrm{Tr}P=\int_M K(x,x) dx= (2\pi)^{-n}\int_{T^*M} p(x,\xi)dx d\xi.$$
In fact, the first equality is known, and the second follows from the above.
In can now answer the question you posed. Multiply the Schwartz kernel of the operator with a fixed cutoff function which localizes to a small neighbourhood of the diagonal, pull this product back under $\exp$ to a distribution on $TM$, and (up to obvious adjustments) define the symbol by Fourier transformation in the fiber variable. 
You can find details of the full symbol calculus, but not the trace formula, for operators acting between bundles in a semiclassical setup, in the appendix of  my paper (also available at  arXiv). Full symbol calculi were developed by Bokobza (1969), Widom (1980), Pflaum (1998), Sharafutdinov (2004), and Safarov.<|endoftext|>
TITLE: What is known about arbitrary subfactors of integer index?
QUESTION [11 upvotes]: Let $N\subset M$ be an inclusion of ${\rm II}_1$ factors of finite index, $[M:N]<\infty$. I would be mostly interested in the hyperfinite case, $N\simeq M\simeq R$, but let us just take them arbitrary.
There is an evolving theory about "what can be said about $N\subset M$, in the general case'', which started with Jones' index theorem, in 1983. Well-known results here include the Pimsner-Popa basis and entropy formula, the bimodule interpretation, the planar algebra formalism.

Question: assuming that we are still in the general case, but with the extra assumption that the index is an integer, $[M,N]\in\mathbb N$, what else can be said about $N\subset M$? 

To my knowledge, at least some time ago (5-10 years), the only answer here was just that the Pimsner-Popa basis is a "clean" one, I mean as in standard linear algebra. I was wondering if any advances on this question come from the recent work on subject, in small or arbitrary index I mean, perhaps as some corollaries of the theory developed there (?) I would be interested in any comment/answer here, this is actually a question that I spent some time on, long time ago.

REPLY [2 votes]: See the issue : Non-“weakly group theoretical” integral fusion categories? 
Here is proposed some integral fusion rings, such that the prospective related fusion categories, are integral but not weakly group theoretical.
Currently, we try to build some Kac algebras $\mathbb{A}$, having these fusion rings as their Grothendieck rings. 
The related subfactors $R^{\mathbb{A}} \subset R$ would be irreducible, depth 2, integer index, but not coming from cohomology and group theory.   
You can also read my answer to the issue : Are subfactor planar algebras hard to classify at index 6?<|endoftext|>
TITLE: Smoothing L1 norm, Huber vs Conjugate
QUESTION [7 upvotes]: I'm trying to minimize a convex (not necessarily strictly convex) function involving an L1 norm (similar to lasso), which makes it non-differentiable at some points. So I'd like to smooth it and treat it as an L2 norm problem.
The two approaches I've seen ( http://www.ee.ucla.edu/~vandenbe/236C/lectures/smoothing.pdf ) are directly smoothing the L1 norm using the Huber function, and smoothing the conjugate (i.e, derive the dual norm, here it's L-infinity, which is still non-differentiable, then smooth that).
The Huber approach is much simpler, is there any advantage in the conjugate method over Huber? I can't see the point of smoothing the dual instead of just smoothing the primal.

REPLY [4 votes]: Following the suggestion of András Bátkai I post my comment as an answer:
Smoothing the dual or the primal problem are quite different things: Smoothing the dual will not give you a smooth primal. However, you get a strongly convex primal by dual smoothing (as opposed to merely a strictly convex primal by Huber smoothing). Hence, it depends on what kind of regularity you are aiming at: A smoother primal or a "more convex" primal - both can be helpful algorithmically. A smooth primal allows you to use gradients instead of subgradients and in turn allows you to apply gradient methods with appropriate stopping rules and such. A strongly convex primal leads to a proximal mapping of the primal objective which is not only non-expansive but contractive which is favorable for proximal-splitting methods.
Of course, you can also apply both primal and dual smoothing if you like.
Moreover, note that there are numerous methods to treat nonsmooth convex minimization problems efficiently<|endoftext|>
TITLE: Need reference to an assertion re: Riemann
QUESTION [5 upvotes]: I have been exploring the patterns of the greatest divisors $\leq$ to the square root $n$ for sequences of size $n$. Just for grins, I used $\textit{Mathematica}$ to produce this ListLinePlot:  

When I saw how symmetrical it was, I had to look up the sequences: OEIS A003418
The third sentence in the comments states that this sequence relates to this assertion regarding Riemann: for $n > 2$
$$|\log (\text{lcm}(n))-n |<\sqrt{n} \log ^2(n)$$
Where could I find a book or paper that describes that assertion?

REPLY [11 votes]: We have
$$ \log\text{lcm}(1,\dots,n)=\sum_p\log p\left[\frac{\log n}{\log p}\right] =\sum_p \log p\sum_{k\log p\leq \log n} 1 = \sum_{p^k\leq n}\log p.$$
The right hand side is the summatory function of the von Mangoldt function, which is known as the second Chebyshev function:
$$ \psi(n):=\sum_{m\leq n}\Lambda(m). $$
That is, the bound you are asking about is
$$ |\psi(n)-n|<\sqrt{n}\log^2 n.$$
The fact that the Riemann Hypothesis implies this bound for $n>73$ (hence probably also for $n>2$) follows from this paper. The other direction, that such a bound with any constant in front of $\sqrt{n}\log ^2 n$ implies the Riemann Hypothesis, is more classical and can be found in many textbooks (see e.g. the bottom of p.463 in Montgomery-Vaughan: Multiplicative Number Theory I).<|endoftext|>
TITLE: From the product lemma to to a result about powersets
QUESTION [6 upvotes]: Recall the product lemma from Easton's famous paper, which tells us something about when we have a forcing notion (which may be a proper class) that splits as a product with one factor $\lambda^+$-cc and the other factor $\lambda$-closed. The proof that in the resulting model the powerset of $\lambda$ can be calculated using the $\lambda^+$-cc forcing alone uses Replacement in Easton's paper, and in Jech's formalism (at the above link) he states it as being about functions from $\lambda$ to the base model $M$. 
Is there a published proof that doesn't use either Replacement or the trick of talking about functions to the universe, that is, a direct proof of the result about the powerset of $\lambda$? If not, can someone supply one?

EDIT: OK, here I am with another problem, but it is closely related, so I tack it on the end. This is separate from the issue of using Replacement.
How does Jech get Powerset for the proper class forcing from showing it just for regular cardinals $\lambda$? I gather he only considers $\lambda$ from the ground model. (Aside: Easton's proof here is totally opaque to me, I don't recognise anything in it that I can get a handle on.) Is it the case that every new set is bijective with some old set, or embeddable in some appropriate way in some old set, or otherwise bijective with something built out of old sets - and this makes it work?

REPLY [8 votes]: Let me handle just the set-forcing case, since I think the class forcing case is problematic. 
Suppose that $\mathbb{P}$
is $\lambda^+$-c.c., and $\mathbb{Q}$ is $\leq\lambda$-closed. If
$G\times H\subset\mathbb{P}\times\mathbb{Q}$ is $V$-generic, then
the claim is that $P(\lambda)^{V[G][H]}=P(\lambda)^{V[G]}$.
First, I claim that $\mathbb{P}$ remains $\lambda^+$-c.c. in
$V[H]$. To see this, if we had a $\mathbb{Q}$-name $\tau$ for an
antichain in $\mathbb{P}$ of size $\lambda^+$, then we may
undertake in $V$ the construction of what is called a
psuedo-generic, namely, we build a decreasing sequence of length
$\lambda^+$ of conditions in $\mathbb{Q}$ to decide the next
element of $\tau$. The point is that since $\mathbb{Q}$ is
$\leq\lambda$-closed, we may continue through limits of this
construction. Although the resulting filter that we build is not
actually $V$-generic, every two elements that it decides are in
$\tau$ must be incompatible in $\mathbb{P}$, and this violates the
chain condition in $\mathbb{P}$ in $V$.
(Perhaps you object here that since I am undertaking a transfinite
recursion, I am using the replacement axiom. That is true. My
answer to this objection, however, is that I don't really need a
version of replacement that is reaching arbitrarily high in the
universe, but instead I am merely undertaking a transfinite
recursion to build a descending sequence inside $\mathbb{Q}$,
which is a set. So all I need is to know that the set of all
$\leq\lambda^+$-sequences over $\mathbb{Q}$ exists, which does not
require the replacement axiom. That is, my uses of replacement are
sufficiently bounded, which seems to keep you safe. This reply doesn't work when the forcing is 
proper class forcing, since then it seems one is really using replacement.)
Continuing the argument, suppose that $A\subset\lambda$ in
$V[G][H]=V[H][G]$. Thus, $A$ has a $\mathbb{P}$-name in $V[H]$.
Since $\mathbb{P}$ is $\lambda^+$-c.c. in $V[H]$, we may find such
a nice name $\dot A$ having size $\lambda$. Since $\mathbb{Q}$ is
$\leq\lambda$-closed in $V$, it follows that $\dot A\in V$. Thus,
$A={\dot A}_G\in V[G]$, as desired.
The argument generalizes to show that any $\lambda$-sequence in $V[G][H]$ from a set in $V$ is in already $V[G]$. All that is needed for this is to know that if one has $\lambda^+$-c.c. forcing $\sigma$ is the name of a $\lambda$-sequence of elements of a set $B$, then there is a nice name, consisting essentially of a $\lambda$-sequence of $B$-labeled antichains. One argues similarly that such a name cannot have been added by the $\mathbb{Q}$-forcing, and so the name is already in $V$, putting the sequence into $V[G]$, just as with the subsets of $\lambda$. Since everything is bounded here nicely, I don't see that one is using replacement even for the general case of $\lambda$-sequences (although of course without replacement there simply are fewer such sequences that one might have in ZFC).

REPLY [6 votes]: When doing this in a set theory without replacement, the issue is really in the definition of $(\leq \lambda)$-closed. Here is a concrete example to illustrate.
Work in $V_{\omega+\omega}$, which is a model of ZC. Obvious failure of replacement in this model is that $\omega+\omega \notin V_{\omega+\omega}$ and yet there is an easy bijection between $\omega$ and $\omega+\omega$. For $\mathbb{Q}$, use the set of all finite partial functions $p:\lbrace\omega,\omega+1,\omega+2,\ldots\rbrace\to\lbrace0,1\rbrace$. Note that $\mathbb{Q}$ is isomorphic to standard Cohen forcing, except that $\mathbb{Q}$ is a proper class. Interestingly, $\mathbb{Q}$ is $(\leq\omega)$-closed for the trivial reason that $\mathbb{Q}$ has no infinite sub-set! Nevertheless, forcing with $\mathbb{Q}$ is the same as standard Cohen forcing and therefore adds a new subset of $\omega$.
For class forcing in models without replacement, the appropriate definition of $(\leq\lambda)$-closed is that every sub-class of $\mathbb{Q}$ which is a chain of size at most $\lambda$ has a lower bound in $\mathbb{Q}$. With this definition, the argument that Joel gave works perfectly.<|endoftext|>
TITLE: Conjugation of homogeneous spaces
QUESTION [5 upvotes]: Let $X$ be a smooth irreducible algebraic variety
over the field of complex numbers ${\mathbb{C}}$.
Let $x\in X({\mathbb{C}})$.
Let $\tau$ be an automorphism of ${\mathbb{C}}$ (not necessarily continuous),
and  let $\tau X$ denote the $\tau$-conjugated ${\mathbb{C}}$-variety obtained from $X$ by transport of structure
(i.e. by action of $\tau$ on the coefficients of equations defining $X$).
We consider the topological fundamental groups $\pi_1(X({\mathbb{C}}),x)$ and $\pi_1((\tau X)({\mathbb{C}}),\tau x)$.
In the papers of Serre,   Exemples de variétés projectives conjuguées non homéomorphes,
C. R. Acad. Sci. Paris 258 (1964), 4194–4196,
 and of Milne and Suh,
Nonhomeomorphic conjugates of connected Shimura varieties,
 one can find examples of $X$ and $\tau$ such that
$\pi_1((\tau X)({\mathbb{C}}),\tau x)$ and $\pi_1(X({\mathbb{C}}),x)$ are not isomorphic.
The authors conclude that in these cases the topological spaces $(\tau X)({\mathbb{C}})$ and $X({\mathbb{C}})$ are not homeomorphic.
In my very recent preprint with Cyril Demarche
(excuse me for advertising my own work!) we consider the following situation.
Let $X=G/H$, where $G$ is a connected linear algebraic group over ${\mathbb{C}}$,
and $H\subset G$ any algebraic subgroup, not necessarily connected.
Set $x:=eH\in X({\mathbb{C}})$.
We prove that in this case $\pi_1((\tau X)({\mathbb{C}}),\tau x)$ and $\pi_1(X({\mathbb{C}}),x)$ are canonically isomorphic.
I am trying to understand, what this really means.

Question. For  a homogeneous space $X=G/H$ over ${\mathbb{C}}$ as above, and for  $\tau\in {\rm Aut}({\mathbb{C}})$,
  is it  always true that
  (1) $(\tau X)({\mathbb{C}})$ and $X(\mathbb{C})$ are homotopically equivalent, or even
  (2) $(\tau X)({\mathbb{C}})$ and $X(\mathbb{C})$ are homeomorphic, or even
  (3) $\tau X$ and $X$ are isomorphic ${\mathbb{C}}$-varieties?

REPLY [2 votes]: The result in my preprint mentioned in the question was erroneous (the mistake was noticed by a referee). It is possible to construct a quotient  $X=G/H$ and and automorphism $\tau$ of $\mathbb{C}$ such that $\pi_1(\tau X(\mathbb{C}))$ is not isomorphic to $\pi_1(X(\mathbb{C}))$, see my preprint with Yves Cornulier based on his answer to this question. In our example  $G={\rm SL}(n,\mathbb{C})\times \mathbb{C}^*$ with $n\ge 5$, and $H$ is a finite nonabelian subgroup of order 55.
This time the referee noticed no mistakes... Thus the answer to the questions (1), (2), and (3) is NO.<|endoftext|>
TITLE: computational complexity
QUESTION [8 upvotes]: I am interested into computational complexity of decision problem: Does a given 2-dimensional simplicial complex contain (any)triangulation of 2-sphere? This problem trivially lies into NP, because certificate is the subset of triangles. I think that it is NP-complete.

REPLY [11 votes]: It is NP-complete. Here is a reduction from the graph 3-coloring problem.
For a given graph, consider a sphere with $n$ holes where $n$ is the number of vertices. Enumerate the holes by the vertices of the graph. Fill each hole by 3 discs (red, green and blue) by identifying each disc boundary with the hole boundary. Glue the centers of the 3 discs into one point (so that nor two of them cannot be included in one surface), Then, for every edge of the graph, pick one new point in each of the 6 discs filling the holes corresponding to the edge endpoints, and glue together each of the 3 pairs of chosen points lying on discs of the same color. The points chosen for different edges must be distinct.
The resulting topological space is homeomorphic to a simplicial complex (which is easy to construct from the graph in polynomial time). It contains a sphere if and only if the original graph is 3-colorable.
Remark. You can alter the formulation by saying that an immersed sphere which self-intersects itself in finitely many points is still a sphere. For this problem, the construction can be modified as follows. To make a pair of discs "forbidden", glue together two pairs of small discs in them rather than one pair of point. This creates a surface of positive genus if you try to pick a pair of discs connected this way.<|endoftext|>
TITLE: Meaning of a phrase from "The algebra of grand unified theories".
QUESTION [5 upvotes]: Motivated by an answer to this mathoverflow question I've been making an effort to understand Baez and Huerta's article "The algebra of grand unified theories".
As far as I can tell, mathematically, the first 30 pages or so describe one concrete finite dimensional representation of the group $U(1)\times SU(2)\times SU(3)$.  The vectors of this space are to be interpreted as the elementary particles which do not mediate forces (i.e. fermions) and their anti-particles, while composed particles (made up of several fermions and anti-fermions) are represented by vectors of tensor powers of this space which must satisfy some additional conditions (such as being fixed by the action of elements in $1 \times 1 \times SU(3)$).  Finally, the elementary force mediating particles (elementary Bosons) are represented by elements of the Lie algebra of the group.
Assuming I've got that right there are still some phrases which make absolutely no sense to me, or worse, I can make sense of them only as completely trivial statements.  The most extreme example I can find is the following (page 490 third paragraph):

"... in 1936 a paper by Cassen and Condon appeared suggesting
  that the nucleon’s Hilbert space $\mathbb{C}^2$ is acted on by the symmetry group $SU(2)$."

The literal interpretation is that Cassen and Condon suggested that it is possible to multiply a $2\times 2$ matrix with a $2\times 1$ vector.  Obviously this can't be what that authors intended to communicate.  
So my question is: What's the meaning of this phrase?
The paper refered to is "On nuclear forces" from Phys. Rev. 50 (1936) which I don't have access to.  From this page it seems they proposed some model accounting for the experimental fact that the strong nuclear force doesn't depend on the electrical charge of the particles involved.  I still don't get it.  Any help?

REPLY [10 votes]: Our sentence indeed loses its point taken out of context.  Condon and Cassen were not merely claiming that we can multiply a vector in $\mathbb{C}^2$ by an element of $\mathrm{SU}(2)$.  They were claiming that when $\mathbb{C}^2$ is used as the Hilbert space for nucleons, this transformation is a symmetry of the laws of physics.  In other words: if we have a solution of the equations of motion (whatever they are), we can apply this transformation and get a new solution.  
More physically: since $\mathbb{C}^2$ is the Hilbert space for a nucleon, with one basis vector standing for a proton and one for a neutron, it means that we can do things like replace all the protons in the universe by neutrons and all the neutrons by protons, and this will have as little effect as, say, rotating the whole universe, or moving everything in the universe two feet to the left.  
Of course this is false, taken literally.   Protons and neutrons behave differently with respect to the electromagnetic force, since protons are electrically charged and neutrons are neutral!   But of course Cassen and Condon knew this.  They (and other people, like Heisenberg) were trying to focus on the nuclear force and temporarily ignore the electromagnetic force.  They hoped that in a universe like ours, but with the electromagnetic force turned off, and only the nuclear force remaining, the laws of physics would have this $\mathrm{SU}(2)$ symmetry.
If you read on, you'll see this hope turned out to be false.  Nonetheless it was incredibly fruitful, because it got physicists thinking about Lie groups, and it got Yang and Mills to invent the Yang-Mills equations, which are very important in our current thinking about particle physics.
Summary: the difference is between a mere action of a group on a Hilbert space, and an action as symmetries of the laws of physics.  That's why we said "symmetry group", not just "group".<|endoftext|>
TITLE: Are there noncongruence subgroups (of finite index) of the modular group generated only by 2 or 3 elements?
QUESTION [6 upvotes]: By the modular group I mean either $SL(2,\mathbb{Z})$ or $PSL(2,\mathbb{Z})$.
Where can I find examples of these?
Another question: is there a good (ideally analytical, but possibly computer-aided) way to determine if a subgroup of the modular group generated by some given matrices has finite index, and possibly allowing to compute the index?

REPLY [2 votes]: $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\F{\mathbf{F}}$Let me provide a purely group-theoretic answer.
Let $H$ be a subgroup of finite index in $\mathrm{PSL}_2(\mathbf{Z})\simeq C_2\ast C_3$. Then, by Kurosh, $H$ is isomorphic to $C_2^{\ast e_2}\ast C_3^{\ast e_3}\ast \mathbf{Z}^{\ast e_0}$ for some integers $e_0,e_2,e_3\ge 0$ (clearly finite, and satisfying $e_2+e_3+e_0\ge 2$).
Since these are virtually free and since in a free group the index determines the rank, the triple $(e_2,e_3,e_0)$ determines the index $f(e_2,e_3,e_0)$ of $H$ in $\mathrm{PSL}_2(\mathbf{Z})$. Let me explicitly compute $f$.
For any groups $A,B$, we have $A\ast B\simeq A\ltimes B^{\ast A}$ (a kind of non-commutative wreath product). Hence, if $A'$ has index 2 in $A$, then $A\ast B$ has a subgroup of index 2 isomorphic to $A'\ltimes B^\ast {A'\sqcup A'}\simeq A'\ltimes (B\ast B)^{A'}\simeq A'\ast B\ast B$.
Given that $C_2^{e_2}$ has a subgroup of index 2 isomorphic to $\mathbf{Z}^{\ast e_2-1}$ for $e_2\ge 1$ (the kernel of the map onto $C_2$ that's identity on each factor), we deduce that $2f(e_2,e_3,e_0)=f(0,2e_3,2e_0+e_2-1)$ for $e_2\ge 1$.
Similarly, given that $C_3^{e_3}$, for $e_3\ge 1$ has a subgroup of index 3 isomorphic to $\mathbf{Z}^{\ast 2(e_3-1)}$, we deduce similarly that $3f(e_2,e_3,e_0)=f(3e_2,0,3e_0+2(e_3-1))$ whenever $e_3\ge 1$.
Hence $$f(e_2,e_3,e_0)=f(0,2e_3,2e_0+e_2-1)/2=f(0,0,6e_0+3(e_2-1)+2(2e_3-1))/6$$ $$=\frac16f(0,0,6e_0+3e_2+4e_3-5)$$
whenever $e_2,e_3\ge 1$; when one or both equals zero this is still valid, by direct verification and using that $f(0,0,1+k)=f(0,0,1+sk)/s$ by the values of ranks of subgroups of finite index in free groups.
Next, $\mathrm{PSL}_2(\mathbf{Z})$ has index 1 in itself, so $f(1,1,0)=1$. We just checked that $f(1,1,0)=f(0,0,2)/6$ by constructing a copy of index 6 of $F_2$. So $f(0,0,2)=6$, and, in turn, $f(0,0,n)=f(0,0,1+(n-1))=(n-1)f(0,0,2)=6(n-1)$. To conclude,
$$f(e_2,e_3,e_0)=\frac16f(0,0,6e_0+3e_2+4e_3-5)= 3e_2+4e_3+6(e_0-1).$$
When the generating rank is two, i.e., $e_0+e_2+e_3=2$ (excluding $e_2=2$ as $C_2^{\ast 2}$ can't occur), we get:
$$f(1,1,0)=1,\;f(1,0,1)=3;\;f(0,1,1)=4;f(0,2,0)=2;f(0,0,2)=6.$$
Hence subgroups of finite index of generating rank 2 in $\mathrm{PSL}_2(\mathbf{Z})$ are precisely those subgroups of index $1,2,3,4,6$. In index 1,2 there is a single such subgroup.

(Sequel: this was unfinished)
So it remains to describe subgroups of index 1,2,3,4,6 in $\PSL_2(\mathbf{Z})$.
Index 1 ($\simeq C_2\ast C_3$): the whole group (= principal congruence modulo 1).
Index 2 ($\simeq C_3\ast C_3$): there is a single one. It is congruence modulo 2, since $\PSL_2(\F_2)$ is a dihedral group of order 6.
Index 3 ($\simeq C_2\ast\mathbf{Z}$ or $\simeq C_2^{\ast 3}$): up to conjugacy there are two such subgroups. One is normal, actually $\simeq C_2^{\ast 3}$: it is congruence modulo 3, since $\PSL_2(\F_3)\simeq\mathrm{Alt}_4$ has a normal subgroup of index 3. The other ones are non-normal and congruence modulo 2, since $\PSL_2(\F_2)$ has a non-normal subgroup of index 3. [One representative consists of those matrices that are $\pm I_2$ modulo 3. If I'm correct it has a single conjugacy class of elements of order 2, so this group is isomorphic to $C_2\ast\mathbf{Z}$.]
Index 4 ($\simeq C_3\ast\mathbf{Z}$ or $\simeq C_2^{\ast 2}\ast C_3$): up to conjugacy there are two such subgroups, if I'm correct. One is congruence modulo 3, since $\PSL_2(\F_3)\simeq\mathrm{Alt}_4$ has a subgroup of index 4 (of order 3). [This one has no element of order 2, hence is isomorphic to $C_3\ast\mathbf{Z}$.] One is congruence modulo $4$, since $\PSL_2(\mathbf{Z}/4\mathbf{Z})$, which has order 48, has a subgroup of index 4 (of order 12), if I checked correctly.
Index 6 ($\simeq\mathbf{Z}^{\ast 2}$ or $\simeq C_3^{\ast 3}$ or $\simeq C_2^{\ast 4}$) is the most interesting case. I'll develop this later as it requires a number of verifications.<|endoftext|>
TITLE: Elaborating Mercer's theorem (RKHS) on Cameron-Martin space $k(x,y)=\min(x,y)$
QUESTION [5 upvotes]: Hi,
I'm trying to employ Mercer's theorem on the kernel $k(x,y)=\min(x,y)$. It is known (and easy to verify) that this is a nonnegative-definite kernel over $[0,T]$ for any $T>0$.
Fix $T>0$. Let's calculate the eigenfunctions of the transformation $ \mathscr T_kf=\intop_{0}^T k(x,y)f(y)dy$:
$$ \lambda\psi(x)=\intop_{0}^T \min(x,y)\psi(y)dy=$$
$$ \intop_{0}^x \min(x,y)\psi(y)dy - \intop_{T}^x \min(x,y)\psi(y)dy=$$
$$ \intop_{0}^x y\psi(y)dy - x\intop_{T}^x \psi(y)dy\implies$$
$$ \lambda\psi'(x)= x\psi(x)-x\psi(x)-\intop_{T}^x \psi(y)dy\implies$$
$$ -\lambda\psi''(x)= \psi(x)\implies$$
$$\psi(x)=C_1\sin\frac x {\sqrt\lambda} + C_2\cos\frac x {\sqrt\lambda}$$
it seems like we're allowed to pick $C_1=1$ and $C_2=0$. So we pick $\psi_n(x)=\sin nx $ and $\lambda_n = n^{-2}$. Then Mercer's theorem actually says that we should get:
$$ \min(x,y)=\sum_{n=1}^\infty n^{-2}\sin nx\sin ny$$
this all seem very nice, but when evaluating this numerically, it doesn't work.
I tried also to normalize $\psi$ by dividing by its norm which is $\sqrt {\frac 1 {4n}  (2nT-\sin2nT)}$, and it didn't help.
I also tried to substitute the original solution with $C_1,C_2$ in the original eigenvalue problem equation, and then to calculate $C_1,C_2$, but they turned out to depend on $x$, which is of cource unacceptable.
So what's wrong here?
I also asked the question on https://math.stackexchange.com/questions/272857/elaborating-mercers-theorem-rkhs-on-cameron-martin-space-kx-y-minx-y, but no answers there. I'll post in either forum immediately if an answer will come.

REPLY [5 votes]: What Branimir Ćaćić writes is correct. Another way to see that your $\lambda$'s where not right is as follows: 
From $$\lambda\psi(x) = \int_0^x y\psi(y) dy + x\int_x^T \psi(y)dy$$ you get that $
$$\psi(0)=0.$$
Similarly, from $$\lambda\psi'(x) = \int_x^T \psi(y)dy$$ you get $$\psi'(T)=0.$$
Hence, you have two boundary conditions for the differential equation $\lambda\psi''(x) = -\psi(x)$. The first forces $C_2=0$, the second gives $$\lambda = \frac{T^2}{\pi^2(n+\tfrac{1}{2})^2}$$
($T/\sqrt{\lambda}$ has to be a root of $\cos$) and no condition on $C_1$. Since you want an orthonormal basis, you have to normalize the functions in $L^2([0,T])$ which gives $C_1=\sqrt{2/T}$.
What you are missing in your numerics is that the series starts with $n=0$ and hence, your result differs from $\min(x,y)$ by $\psi_0(x)\psi_0(y) = \sin(\tfrac{\pi x}{2T})\sin(\tfrac{\pi y}{2T})$.<|endoftext|>
TITLE: Immersed surface with circle as a boundary
QUESTION [6 upvotes]: Is there a solution or progress of the following problem (maybe old conjecture): Is An immersed surface with constant mean curvature and with a circle as a boundary part of a sphere??. If we replace "immersed" by "embedded" I think the problem was solved by Alexandroff kind of long time ago. Is someone could enlighten about what exactly Alexandroff solves and what is to do, it would help me a lot.
Thanks
Mario

REPLY [6 votes]: Assuming your interest is in constant mean curvature surfaces with circular boundary, I found the survey, "Surfaces with constant mean curvature in Euclidean space" by R. Lopez to be a great introduction, and it contains the state of the art, and several references.
Hopf proved that the only constant mean curvature closed surfaces of genus 0 are spheres. Alexandrov's theorem says that constant mean curvature closed surfaces that are embedded are spheres. Unfortunately when we allow boundaries both analogs are conjectural:

Conjecture 1: The only constant mean curvature compact surfaces with circular boundary that are topological disks are spherical caps.
Conjecture 2: The only constant mean curvature compact surfaces with circular boundary that are embedded are spherical caps.

REPLY [4 votes]: Check out this paper of Rafael Lopez and references therein.<|endoftext|>
TITLE: How to find a sub-forcing?
QUESTION [5 upvotes]: Suppose that $M$ is a model of ZFC, $P\in M$ is a notion of forcing and $G$ is a $P$-generic filter over $M$.
It is a well-known theorem that if $M\subseteq N\subseteq M[G]$, and $N$ is a model of ZFC then $N=M[H]$ for some generic set $H$, and there is some $H'$ which is generic over $M[H]$ such that $M[G]=M[H][H']$.
But the proof I know uses Boolean valued models and is not particularly insightful about the following question:

Suppose that $M$ is a model of ZFC, $P\in M$ is a notion of forcing, and $\dot x$ is a $P$-name, such that if $G$ is a $P$-generic over $M$, and $x=\dot x^G$ then $G\notin M[x]$.
Can we give an explicit $Q\in M$ such that $x$ is $Q$-generic over $M$?

An additional question which is relevant to one particular case:

Suppose that $P$ is the Cohen forcing with $2^{<\omega}$. In such case every sub-forcing is isomorphic to $P$, what does that mean for us? Does it mean that the generic needs to be particularly chosen, or what?

REPLY [4 votes]: The intermediate model proof for $M\subset N\subset M[G]$ is not
so inexplicit. The situation is that, if $A\subset\text{Ord}$ is a
set of ordinals coding $P(\mathbb{B})^N$, the $N=M[G\cap
\mathbb{B}_0]$, where $\mathbb{B}_0$ is the Boolean algebra
generated by the $\mathbb{B}$-Boolean values of membership in
$\dot A$, where $\dot A$ is a $\mathbb{B}$-name for $A$. (This is
proved in Jech.) After all, if you know $G$ on this
$\mathbb{B}_0$, then you know $P(\mathbb{B})^N$ and hence all of
$N$ (as in Jech), and conversely, $N$ certainly knows the right
answers for $\dot A$.
In your case, you have $N=M[x]$, and so $A$ is any set of ordinals coding $P(\mathbb{B})^{M[x]}$ in $M[x]$, with $\mathbb{B}_0$ the Boolean values generated by the values $\[\check\alpha\in\dot A\]^{\mathbb{B}}$. 
Meanwhile, one cannot get too explicit about it, since the proof
in Jech uses AC and the fact is not generally true for ZF as
opposed to ZFC models. First of all, note that there can be intermediate ZF models. For example, one might start with $M=L$, say, and add $\omega_1$ many Cohen reals, to form $M[G]$. Let $N=L(\mathbb{R})^{M[G]}$, which is intermediate between $M$ and $M[G]$, but it doesn't have the form $M[G\cap\mathbb{B}_0]$ for any complete subalgebra, since it violates AC. In general, no intermediate $\text{ZF}+\neg\text{AC}$ model can be $M[G\cap\mathbb{B}_0]$, since these satisfy ZFC.<|endoftext|>
TITLE: Translations of Serre's early spectral sequences papers
QUESTION [7 upvotes]: Do there exist translations of Serre's early papers on spectral sequences?  In particular, I am interested in the following ones:
Serre, Jean-Pierre Homologie singulière des espaces fibrés. Applications. Ann. of Math. (2) 54, (1951). 425–505.
Serre, Jean-Pierre Groupes d'homotopie et classes de groupes abéliens. Ann. of Math. (2) 58, (1953). 258–294.
Serre, Jean-Pierre Cohomologie modulo 2 des complexes d'Eilenberg-MacLane. Comment. Math. Helv. 27, (1953). 198–232.
ps : I know that many people will respond by saying that I should learn French.  Well, I can read mathematical French, but it is painful and takes a long time.  Since these are classic papers, I thought that there might exist translations of them somewhere.

REPLY [6 votes]: You can look at this volume:
http://www.worldscientific.com/worldscibooks/10.1142/8444
ps: J.-P. Serre's papers are written in a very elegant style, his french is beautiful and nice to read :-)<|endoftext|>
TITLE: Fundamental group of a compact manifold
QUESTION [6 upvotes]: Why is the fundamental group of a compact manifold finitely presented?

REPLY [19 votes]: Every compact manifold has the homotopy type of a finite CW-complex, and a finite CW-complex has finitely presented fundamental group by van Kampen's theorem.<|endoftext|>
TITLE: reference for "X compact <=> C_b(X) separable" (X metric space)
QUESTION [16 upvotes]: I know (and am able to prove via Stone-Čech compactification) that the following is correct:

Theorem: A metric space is compact if and only if its space of bounded, continuous, real-valued functions is separable in the uniform topology.

I use it in a paper for readers who are presumably not familiar with this kind of topology, so I cannot call it "obvious" or "well-known".
I would be thankful for a name and/or good reference to cite this theorem!

REPLY [10 votes]: The result does appear in Dunford/Schwartz, Linear Operators Part I (page 437), but is only stated as an exercise. 
Edit after @JosephVanName' comment:
Conway's Functional Analysis has the result for completely regular spaces as Theorem 6.6 (page 140).<|endoftext|>
TITLE: Calkin Algebra and the embedding
QUESTION [13 upvotes]: Let $H$ be a separable, infinite dimensional Hilbert Space and $Calk(H):=B(H)/K(H)$ denotes
the Calkin algebra. There is obvious surjection $\pi: B(H) \to Calk(H)$ but I'm interested
in somehow opposite question: is it possible to construct embedding $j_1:Calk(H) \to B(H)$?
The same question for embedding $j_2:B(H) \to Calk(H)$ and finally is there epimorphism
$k:Calk(H) \to B(H)$?

REPLY [9 votes]: (1) One cannot embed $Q(H)$ into $B(H)$ as a Banach space either. Indeed, $Q(H)$ contains $\ell_\infty/c_0$ (as a diagonal masa). J. Bourgain gave a very clever proof of the fact that $\ell_\infty/c_0$ has no strictly convex renorming:

J. Bourgain, $\ell_\infty/c_0$ has no equivalent strictly convex norm, Proc. Amer. Math. Soc. 78 (1980), 225-226.

On the other hand, $B(H)$ is a dual space of a separable Banach space (the space of nuclear operators on $H$), so it does have a strictly convex renorming. The property of having a strictly convex renorming passes to subspaces.<|endoftext|>
TITLE: Homology of symmetric product
QUESTION [8 upvotes]: If two distinct topological spaces X and Y have the same homology groups, then their symmetric products also have the same homology?
Do you know if is it true? Do you Know any references for?

REPLY [4 votes]: A stronger statement is true.
If $X$ and $Y$ are spaces such that $H_*(X) \approx H_*(Y)$ (no need to have a map that induces this isomorphism), then $SPX$ and $SPY$ are weakly homotopy equivalent. In particular if $X$ and $Y$ are CW complexes, then $SPX$ and $SPY$ are homotopy equivalent. This can be found in Hatcher: Algebraic topology, (see Hatcher's website).
More precisely: 

By the Dold-Thom theorem $\pi_n(SPX) \approx H_n(X).$ (Thm 4.83 in Hathcher's book.)
On the other hand $SPX$ is a product of Eilenberg - Maclane spaces. (corollary 4.84 there.)

Putting these two statements together you get what I wrote.<|endoftext|>
TITLE: Rank growth of elliptic curves after cubic extensions
QUESTION [5 upvotes]: Let $E/\mathbb{Q}$ be an elliptic curve and let $N_E(3,X)$ denote the number of cyclic cubic extensions $K/\mathbb{Q}$ of conductor no more than $X$  for which $rank~E(K)> ~rank~ E(\mathbb{Q})$. Then a conjecture of David, Fearnley and Kisilevsky (stemming from considerations in random matrix theory) states that
$ \log N_E(3,X) \sim \frac{1}{2}\log X.$
My question is what the conjecture should be if we remove the condition that $K/\mathbb{Q}$ is a $\textit{cyclic}$ extension.

REPLY [2 votes]: I don't have a full answer, but if $E$ is given in Weierstrass form $y^2=f(x)$, then for most values of $c \in \mathbb{Q}$, if you look at the point with $y=c$ on $E$, you get a point in a cubic extension (usually non-cyclic) given by $f(x)-c^2=0$ which will not be in the division hull of $E(\mathbb{Q})$, i.e. the rank will grow. So the number of such cubic fields of conductor at most $X$ will be a constant times some power of $X$. The total number of cubic fields of conductor at most $X$ is a constant times some other power of $X$.<|endoftext|>
TITLE: Hopf Algebra for a physicist
QUESTION [13 upvotes]: Hello,
for my bachelor's thesis I need to understand the Hopf Algebra of Feynman Diagrams. As I have only litte knowledge in Algebra by now I wanted to ask where I could start and what preknowledge I should have in order to understand this topic. Can you perhaps recommend a textbook or some other reference regarding Hopf Algebra or Quantum Groups suitable for my need? 
Thank you in advance.

REPLY [7 votes]: I originally found the motivation for the definition of a Hopf algebra difficult to see. This article by Scott Carnahan makes the definition seem like a completely natural generalisation of the definition of a group. Think of that article as a "pre-introduction" that makes other introductions to Hopf algebras easier to grasp.<|endoftext|>
TITLE: Is every class that does not add sets necessarily added by forcing? 
QUESTION [16 upvotes]: We know there are many situations in which we can force over a model $M$ of GBC to add a class $G$ without adding any sets.  That is, the extension $M[G]$ satisfies GBC and has the same sets as $M$.  This technique is used, for example, in the proof that GBC is a conservative extension of ZFC, by forcing to add a universal choice function to a model of ZFC.  
I'd like to know whether every class that can be 'safely' added to a model (preserving GBC and adding no sets) arises through forcing:

If $M$ is a model of GBC and $G$ is a class (that is, a subcollection of the sets of $M$, not a class member of $M$) such that $M[G]$ satisfies GBC and has the same sets as $M$, then is $G$ necessarily generic for some partial order $P \in M$?

I feel 'morally certain' that the answer must be no.  Certainly the analogous question about sets, 'Can every set be added by forcing?' has a negative answer - we cannot, for example, force over $L$ to add a measure to a cardinal $\kappa$ (or to add $0^\\#$, or any other set that might increase the consistency strength).  However, I don't see how to adapt this kind of argument to classes, if we are not allowed to add sets.  I'd love to see a counterexample (a proof of a positive answer would also be welcome!).

REPLY [15 votes]: It is a fantastic question, Jonas! I've spent hours with it now, going back and forth  several times about which way it might go.
But finally, I've got a negative answer, at least for some models $M$. My idea is
that some models of ZFC admit what are called satisfaction
classes, but these can never be added by class forcing.
For any transitive model $M$ of ZFC, let $S$ be the set of pairs
$\langle{}^\ulcorner\varphi{}^\urcorner,a\rangle$, for which
$\varphi[a]$ holds in $M$. This class $S$ is the (unique) class of
pairs $\langle n,a\rangle$ satisfying the Tarskian inductive
definition of truth. Further, we may sometimes add $S$ as a class
to $M$ and still have GBC in $(M,S)$. This is true, for example,
when $M=V_\kappa$ for an inaccessible cardinal $\kappa$; but also
it is true for many other models.
So let us suppose that $M$ admits such a unique satisfaction
class.
Can $S$ be added by class forcing over $M$ without adding sets? If
we regard $M$ as a GB model by endowing it with only its definable
classes, then the answer is no.
To see this, suppose that $S$ was added in the class forcing
extension $M[H]$, where $H\subset\mathbb{P}$ is $M$-generic for
the class forcing $\mathbb{P}$, which adds no sets. So
$S=\dot{A}_H$ for some class $\mathbb{P}$-name $\dot{A}$, and in
particular, both $\dot{A}$ and $\mathbb{P}$ are definable in $M$.
Now, the key step is that although truth itself is not definable,
by Tarski's theorem, the property of $S$ that it satisfies the
inductive definition of truth has complexity merely $\Delta_1(S)$.
Thus, one of the assertions about $S$ that is true in $M[H]$ is
that it satisfies the Tarskian definition of truth. Thus, there
must be a condition forcing that $\dot{A}$ is a satisfaction
class, obeying Tarski's inductive definition.
But since the satisfaction class of $M$ is unique, this means
that there is no choice for the generic filter whether or not to
place a pair into or out of $\dot{A}$. Thus, inside $M$ by simply
looking at which pairs can be added at all to the class named by
$\dot{A}$, we can define $S$ in $M$. But this contradicts Tarski's
theorem on the non-definability of truth. QED
Let me argue next that we don't actually need the satisfaction class to be unique, and
the same argument will work whenever $M$ admits a satisfaction
class at all, with GBC in $(M,S)$. This can conceivably happen in non-standard models $M$, with non-standard Gödel codes. Nevertheless, the standard
part of the satisfaction class, that is, for standard Gödel
codes, will be unique, and so we can still get that stable part of
the satisfaction class from the name $\dot{A}$---the pairs that
are forced into $\dot{A}$ by every condition. This will still be a
satisfaction predicate, contrary to Tarski's theory of truth, even
if $M$ is not an $\omega$-model and possibly admits several
satisfaction classes.
Finally, let me point out that not every model of ZFC admits a
satisfaction class. For example, if $M$ is 
pointwise definable, as
in our joint paper on Pointwise definable models of set theory, then this property would be revealed by
the satisfaction class, and so we cannot add this class without
revealing to $M$ that it is countable.<|endoftext|>
TITLE: Orientations for pseudoholomorphic curves with totally real boundary condition
QUESTION [9 upvotes]: I am trying to understand what the obstructions are to orienting moduli spaces of pseudoholomorphic curves with totally real boundary condition. 
I believe that Fukaya-Oh-Ohta-Ono have shown that if a Lagrangian is relatively spin, the moduli spaces of disks with boundary in it can be oriented. 
My question has 3 related parts:

is there any sense in which the FOOO relative spin condition is also necessary?
if I consider curves of higher genus and/or more boundary components, do I need to impose additional conditions on the Lagrangian to guarantee orientability of the moduli spaces?
There has been a fair bit of recent work in the case in which the Lagrangian is the fixed point set of an anti-symplectic involution (Crétois, Georgieva-Zinger). Are the orientation difficulties in this case the same as in the general case, or do some special features appear here?

EDIT: Penka Georgieva pointed out that I was mistaken. Her paper (arxiv/1207.5471) deals with the general case of curves with boundary on a Lagrangian.

REPLY [4 votes]: (0) I do not know what is contained in a thesis which is not published and is not even available online (12 years after the defense).
(1) By Proposition 8.1.4 in the 1000-page FOOO book, a relatively spin structure on a (necessarily orientable) Lagrangian submanifold $L$ of a symplectic manifold $M$ determines an orientation on the moduli spaces of $J$-holomorphic disks $(D^2,S^1)\longrightarrow (M,L)$. The same reasoning applies to a family of real Cauchy-Riemann operators induced (as in Remark 1.3 of 1207.5471) by a bundle pair $(E,F)\longrightarrow(M,L)$, where $L$ is any submanifold of any manifold $M$. Proposition 8.1.7 gives an example of a non-orientable family of real Cauchy-Riemann operators, crediting it to Vin de Silva, and including a proof. I do not believe this book contains other, substantially different, statements on orientability in open GW-theory. Thus, this book discusses orienting moduli spaces of disks in some cases, but says fairly little about
their orientability in general.
(2) By Theorem 1.1 in 0606429, a relatively Pin structure on a non-orientable Langrangian induces an isomorphism between the orientation line bundle of the moduli space of open $J$-holomorphic maps from Riemann surfaces with a fixed complex structure and a product of pull-backs of the orientation line bundle of the Lagrangian by evaluation maps. I do not believe this paper contains other, substantially different, statements on orientability in open GW-theory. Thus, this paper contains a number of results on both orienting and orientability of moduli spaces.
(3) Lemma 11.7 in Seidel's book does what Tim says in (1). Unfortunately, it requires more than a quick look to understand and see that it implies Proposition 8.1.4 in FOOO and the disk case of Theorem 1.1 in 0606429.
(4) Theorem 1.1 in 1207.5471 describes the holonomy of the orientation bundle of a family of real Cauchy-Riemann operators over bordered Riemann with varying complex structures. Its statement is absolutely clear from looking at the preceding half a page, at the beginning of the introduction. In particular, it is almost immediately clear that this theorem implies Proposition 8.1.4 in FOOO and the full statement of Theorem 1.1 in 0606429. The proof, contained in Section 3, is beautifully simple and uses no K-theory or even homotopy exact sequences.
(5) In the case of anti-symplectic involutions, one often wants to orient moduli spaces of real maps, not their halves, even if they are halvable. In the case of maps from $S^2$ with the standard conjugation, if the corresponding moduli space of disk maps is orientable, the space of real maps is orientable if the flip map on the disk space is orientation-preserving. So, this becomes a problem about computing the sign of this flip map. Results on this are Proposition 5.1 in 0606429 and Theorems 1.1 and 1.3 in 0912.2646. If the involution on $S^2$ has no fixed points, the orientation problem has nothing to do with any Lagrangians. Results on orientability in this case are Theorem 1.3 and Example 2.5 in 1205.1809 and Theorem 1.1 and Corollary 1.8 in 1301.1074.  
(6) 1301.1074 says less about the Lagrangian case than 1207.5471. The point of 1301.1074 is to study the orientability problem for $J$-holomorphic maps that commute with involutions on the domain and the target. These maps need not be halvable to a map from a bordered Riemann surface with Lagrangian boundary conditions. An application is Corollary 1.8.<|endoftext|>
TITLE: The probability for a symmetric matrix to be positive definite
QUESTION [39 upvotes]: Let me give a reasonable model for the question in the title. In ${\rm Sym}_n({\mathbb R})$, the positive definite matrices form a convex cone $S_n^+$. The probability I have in mind is the ratio $p_n=\theta_n/\omega_n$, where $\theta_n$ is the solid angle of $\Lambda_n$, and $\omega_n$ is the solid angle of the whole space ${\rm Sym}_n$ (the area of the unit sphere of dimension $N-1$ where $N=\frac{n(n+1)}2$). These definitions are relative to the Euclidian norm $\|M\|=\sqrt{{\rm Tr}(M^2)}$ ; this is the most natural among Euclidian norms, because it is invariant under unitary conjugation.

Because $S_2^+$ is a circular cone, I could compute $p_2=\frac{2-\sqrt2}4\sim0.146$ . Is there a known close formula for $p_n$? If not, is there a known asymptotics?

More generally, we may define open convex cones 
$$\Lambda_n^0\subset\Lambda_n^1\subset\cdots\subset\Lambda_n^{n-1}$$
in the following way: $M\mapsto\det M$ is a homogeneous polynomial, hyperbolic in the direction of the identity matrix $I_n$. Thus its successive derivatives in this direction are hyperbolic too. The $k$th derivative defines a "future cone" $\Lambda_n^k$, these cones being nested. For instance, $\Lambda_n^0=S_n^+$. It turns out that this derivative is, up to a constant,  $\sigma_{n-k}(\vec\lambda)$, where $\sigma_j$ is the $j$th elementary symmetric polynomial and $\vec\lambda$ the spectrum of $M$. Therefore $\Lambda_n^k$ is defined by the inequalities
$$\sigma_1(\vec\lambda)\ge0,\ldots,\sigma_{n-k}(\vec\lambda)\ge0.$$
For instance, $\Lambda_n^{n-1}$ is the half-space defined by ${\rm Tr}M\ge0$.
Let us define again $p_{n,k}$ the probability for $M\in{\rm Sym}_n$ to belong to $\Lambda_n^k$. Thus $p_{n,0}=p_n$ and $p_{n,n-1}=\frac12$. 

What is the distribution of $(p_{n,0},\ldots,p_{n,n-1})$, asymptotically as $n\rightarrow+\infty$?

Edit. As Mikaël mentionned, it is equivalent, and easier for calculations, to consider the standard Gaussian measure (GOE) over ${\bf Sym}_n$.

REPLY [9 votes]: I came across this question while preparing a talk on this paper. There we gave the explicit formula for $p_n$ (in the GOE) that the other answers anticipated could be found by random matrix methods.
The formula is given as follows:
First, let $n\geq 1$ be any integer, and define $n':=2 \lceil n/2 \rceil$.
When real symmetric matrices are chosen according to the
$n$-dimensional Gaussian Orthogonal Ensemble, the probability of
positive definiteness is given by:
\begin{equation}p_n=\frac{{\mathrm{
      Pf}}(A)}{2^{n(n+3)/4}\prod_{m=1}^n\Gamma(\frac m
    2)},\label{theorem3}
\end{equation}
where $A$ is the $n'\times n'$ skew-symmetric matrix whose
$(i,j)$-entry $a_{ij}$ is given for $i<j$ by:
\begin{align}\label{pfaffianformula}
a_{ij} = \begin{cases}
\,2^{i+j-2}\Gamma(\frac{i+j}{2})\left(\beta_{\frac{1}{2}}(\frac{i}{2},\frac{j}{2})- \beta_{\frac{1}{2}}(\frac{j}{2},\frac{i}{2})\right) & \mbox{if }~ i<j\leq n, \\[.075in]
\,2^{i-1}\Gamma(\frac{i}{2})  & \mbox{if }~ i<j=n+1.\\[.075in]
\end{cases} 
\end{align}
$($Note that the second case arises only when $n$ is odd.$)$ 
Here $\mathrm{Pf}$ is the Pfaffian, $\Gamma$ denotes
the usual gamma function $\Gamma(s)=\int_{0}^\infty x^{s-1}e^{-x} \mathrm{d}
x$, and $\beta_{t}$ is the usual incomplete beta function $\beta_{t}(i,j)
= \int_0^{t} x^{i-1}(1-x)^{j-1} \mathrm{d} x.$ This formula follows from de Bruijn's identity for evaluating determinantal integrals.
As for the asymptotics, the leading term of $\log p_n$ is indeed $-\frac{\log(3)}{4} n^2$ as noted in Mikael's answer. If you are interested in the leading order asymptotic for $p_n$ it can be found from the results of Borot, Eynard, Majumdar and Nadal. Specialising to the GOE they give:
\begin{align}
p_n \sim \exp\left(-\frac{\log3}{4} n^2 - \frac{1}{2}\log\left(1+\frac{2}{\sqrt{3}}\right)n - \frac{1}{24}\log n + \alpha_1 \right)(1+o(1)).\label{asympexp}
\end{align}
The constant $\alpha_1$ is given by:
\begin{align}
\alpha_1= \frac{\zeta'(-1)}{2} - \frac{\log(2)}{12} -\frac{\log(3)}{16}+ \frac{1}{4}\log\left(1+\frac{2}{\sqrt{3}}\right)\simeq-0.0172,
\end{align}
where $\zeta(z)$ is the Riemann zeta function. 
Edit: The exact formula lends itself easily to numerics. I have added a plot of $\log(p_n)$ for $1\leq n \leq 400.$ Fitting the data for $201\leq n \leq 400$ to $a n^2 + b n + c\log n +d$ we find $$−0.27465n^2 - 0.38382n -0.041504 \log n -0.018061.$$ A numerical approximation for the asymptotic formula is $$−0.27465n^2 - 0.38382n -0.041667 \log n -0.017223,$$ so we see good agreement.<|endoftext|>
TITLE: What is a simplicial commutative ring from the point of view of homotopy theory? 
QUESTION [37 upvotes]: Let $k$ be a field. There are two natural categories to consider: 

The category of simplicial commutative $k$-algebras. 
The category of connective $E_\infty$ $k$-algebras (i.e., chain complexes of $k$-vector spaces in nonnegative dimensions with a coherently associative and commutative multiplication law). 

These categories are not the same if $k$ does not have characteristic zero. Simplicial commutative $k$-algebras are rather special, and (for example) not every commutative dga over $k$ (which determines an $E_\infty$-algebra over $k$) comes from a simplicial commutative $k$-algebra. (The homotopy groups of a simplicial commutative ring have divided powers, by an explicit construction that I don't really understand.) The category of $E_\infty$-algebras over $k$ has a nice interpretation via homotopy theory: it is the category of commutative algebra objects (in an appropriate sense) in the category of connective $k$-module spectra. (In particular, it is monadic over connective $k$-module spectra, in the $\infty$-categorical sense.)  I don't  know how to think of simplicial commutative rings in this way; all I know for motivation is that they form a nice homotopy theory (e.g. presented by a fairly concrete model category)  that allows you to extend the category of ordinary commutative rings (e.g., to resolve non-smooth objects by smooth ones). 
Is there an analog of the above discussion for $E_\infty$-algebras that works for simplicial commutative $k$-algebras? In particular, can they be described as algebras over a nice monad for $k$-module spectra?

REPLY [16 votes]: I have nothing non-trivial and non-digressive to say, but it might help to
point out in an elementary way some things that may be relevant. One way to think 
about things is that there are distinctions in algebra that lack equivalents 
in spectra. This relates to answers from the $\infty$ category point
of view of the original question, but is perhaps more explicit and concrete.
The statement of the question might be a little confusing, 
since there are perhaps six rather than two natural categories 
to consider, so let me pedantically make a fuller list, with 
everything over some commutative ring $k$.  As a joke start, notice 
that simplicial commutative algebras are the same as commutative
simplicial algebras. However, simplicial $E_{\infty}$ algebras 
make no sense (since we are not thinking about a homotopy theory
on plain algebras), whereas $E_{\infty}$ simplicial algebras might
make sense: as I understand it, that is where the problem about
symmetric powers enters.
(1) commutative DG $k$-algebras
(2) $E_{\infty}$ DG $k$-algebras
(3) commutative simplicial $k$-algebras = simplicial commutative $k$-algebras
(4) $E_{\infty}$ simplicial $k$-algebras?
(5) commutative $Hk$-algebras in any good category of spectra.
(6) $E_{\infty}$ $Hk$-algebras in any good category of spectra 
The term $k$-module spectra in the question and some answers means $Hk$-module, 
I presume, where $Hk$ is the Eilenberg-MacLane spectrum for $k$. I find it helpful
to maintain a notational distinction. Mandell proved that $E_{\infty}$ $k$-algebras 
(algebra) are equivalent to $E_{\infty}$ $Hk$-algebras (topology).  That is, 
(2) and (6) are equivalent: the algebraic and topological notions of $E_{\infty}$ 
are equivalent, of course restricting the latter to Eilenberg-MacLane algebras.
In algebra, the evident forgetful functor from commutative DG $k$-algebras to 
$E_{\infty}$ $k$-algebras is an equivalence for fields of characteristic 0 
but not in general otherwise.  That is, (1) and (2) are not equivalent.
Analogously, if sense can be made of (4), then there is a functor from commutative simplicial
$k$-algebras to $E_{\infty}$ simplicial $k$-algebras which is not an equivalence.
In topology, the world of spectra, the smash product builds in higher homotopies 
and there is no distinction between commutative $Hk$-algebras and $E_{\infty}$ 
$Hk$-algebras: (5) and (6) are equivalent. Andre's answer is an $\infty$-categorical
version of this: "in general, there is nothing more that $E_{\infty}$ to ask for".
But the real topological reason in the case of spectra is a miracle of the modern theory 
of spectra.  For good spectra in any good category of spectra, the natural map from
the homotopy symmetric power to the symmetric power
$$ (E\Sigma_n)_{+}\wedge_{\Sigma_n} X^n \longrightarrow  X^n/\Sigma_n $$
is an equivalence, where $X^n$ is the $n$-fold smash power.  (I think I've advertised
this in answer to some other question, but it is worth repeating.)  One cannot expect
anything like this in algebra, except when working over a field of characteristic $0$.
In the model categorical sense, there does not seem to be a good homotopy 
theory of commutative DG $k$-algebras (ignoring the trivial rational case), and by
analogy one does not expect such a good homotopy theory of simplicial commutative 
algebras. There is a good model theoretic homotopy theory of $E_{\infty}$ DG-algebras.
In the non-commutative case, the question has been answered, mainly by Brooke Shipley.
With the word commutative deleted and $E_{\infty}$ replaced by $A_{\infty}$, I'm pretty 
sure that all 6 homotopy categories make sense and are equivalent.
Several people have mentioned equivariant homotopy theory.  Akhil, the cartesian
power (or smash power in the based context) $X^n$ of a space $X$ is clearly a 
$\Sigma_n$-space; nothing to that. Maybe you are thinking of genuine spectra. 
Either way, Elmendorf's theorem is not particularly relevant. The earlier question 
you referred to asked for a homotopy colimit version of the infinite symmetric power 
of spaces.  The equivalence above can be jacked up to say that in modern categories 
of spectra, the naive homotopy colimit and categorical colimit construction of infinite 
symmetric powers of spectra give equivalent answers, which is bizarre from a pre-1990s 
view of homotopy theory.  There are interesting old-fashioned symmetric powers of spectra, 
and they are different.  I don't understand them in the modern world.
In the world of $G$-spectra, things are much more interesting.  There are many types of 
operadically defined commutative ring $G$-spectra.  This is new work starting with Hill and Hopkins 
and continued by Blumberg and Hill, as yet not written down (at least not yet for public consumption). The comparison between algebra and topology here is fascinating and will be relevant to representation theory as well as to algebraic topology, or so I think.  The old work by Lewis and myself is that part of the story that most directly mimics the nonequivariant case.  I think I see how to understand equivariant units in the new context, and there is a relevant paper by Santhanam in the "classical" context, but there is more to be said even there, part of a thorough treatment of equivariant infinite loop space theory now in progress at Chicago.<|endoftext|>
TITLE: D-module that is coherent as O-module
QUESTION [6 upvotes]: Suppose that $X$ is an algebraic variety over $\mathbb C$, not necessarily smooth. Is it still true that each $\mathcal D_X$-module ($\mathcal D_X$ is of course the sheaf of differential operators) that is coherent as an $\mathcal O_X$-vodule must be locally free as an $\mathcal O_X$-module?
Thank you in advance,
Serge

REPLY [4 votes]: [Edited to correct errors pointed out by David Ben-Zvi and to answer a query by serge_I. These corrections reduce this ``answer'' to the status of a comment.] 
(1) (Over $\mathbb C$) Under strong assumptions on the singularities of $X$ (namely, that $X$ should be cuspidal, Ben-Zvi and Nevins, arXiv 0212094v3), if $\mathcal E$ is coherent on $X$, then giving it a structure as a $\mathcal D_X$-module means giving an isomorphism $\phi:p_1^*\mathcal E\to p_2^*\mathcal E$, where $\mathcal X_1$ is the completion of $X\times X$ along the diagonal and $p_1,p_2:\mathcal X_1\to X$
are the projections. (There is also the cocycle condition,
$p_{31}^*\phi=p_{32}^*\phi\circ p_{21}^*\phi,$
where $p_{ij}:\mathcal X_2\to\mathcal X_1$ are the projections from the completion
$\mathcal X_2$ of $X\times X\times X$ along the diagonal, but we don't need this here.)
Since $X$ is noetherian, there is a unique flattening stratification $X=\coprod X_i$ associated to $\mathcal E$: each $X_i$ is locally closed in $X$ and for any $f:Y\to X$, $f^*\mathcal E$ is locally free if and only if $f$ factors through some $X_i$. The existence
of $\phi$ shows that $p_1^*\mathcal E$ and $p_2^*\mathcal E$ have the same flattening stratification, while $\{p_1^{-1}(X_i)\}_i$ is the flattening stratification for $p_1^*\mathcal E$
and $\{p_2^{-1}(X_i)\}_i$ is the flattening stratification for $p_2^*\mathcal E$.
Since $X$ is irreducible, there is a unique stratum $X_0$ of maximal dimension, so that
$p_1^{-1}(X_0)=p_2^{-1}(X_0)$. Now think in terms of a tubular neighborhood of the diagonal in
$X\times X$ to see that this forces $X_0=X$, so that $\mathcal E$ is locally free provided that $X$ is cuspidal.
(2) (In char. $p$, or over any base) If $X$ is smooth and $\mathcal D_X$ is taken to be the full ring of differential operators rather than the subring generated by those operators of order at most $1$, then the same argument applies.<|endoftext|>
TITLE: Is pi = log_a(b) for some integers a, b > 1?
QUESTION [30 upvotes]: Are there integers $a, b > 1$ such that $\pi = \log_a(b)$?
Or equivalently: are there integers $a,b > 1$ such that $a^\pi = b$?
Note that the transcendence of $\pi$ makes this a problem - otherwise the Gelfond-Schneider theorem would tell the answer.
When I asked this question about 10 years ago, I got an answer by Ignat Soroko  (Minsk / Belarus) arguing that the problem is likely out of reach for present methods, cf. http://www.gap-system.org/DevelopersPages/StefanKohl/problems/a%5Epi=b.dvi. Is this still the present state?

REPLY [34 votes]: There is a reason why one can say a little bit more about this question in the case of $\pi$.  Because $\pi$ is (essentially) the natural logarithm of a rational number, questions like this are easily derived from Schanuel's conjecture, which states:

If $\alpha_1, \alpha_2, \ldots, \alpha_n$ are complex numbers that are linearly independent over $\mathbb Q$, then the transcendence degree of the field $\mathbb{Q}(\alpha_1, e^{\alpha_1}, \alpha_2, e^{\alpha_2}, \ldots, \alpha_n, e^{\alpha_n})$ over $\mathbb Q$ is at least $n$.

Your proposed equation is
$$\pi = \frac{\ln b}{\ln a}\qquad (*)$$
and $(*)$ contradicts the $n=3$ case of Schanuel's conjecture as follows.  Take $\alpha_1=\ln a$, $\alpha_2 = \ln b$, and $\alpha_3 = i\pi$.  Then $(*)$ implies that $\alpha_1$ and $\alpha_2$ are linearly independent over $\mathbb Q$ (because $\pi$ is irrational), and $\alpha_3$ is trivially linearly independent of $\alpha_1$ and $\alpha_2$ because $\alpha_3$ is purely imaginary and $\alpha_1$ and $\alpha_2$ are real.  But $e^{\alpha_1}$, $e^{\alpha_2}$, and $e^{\alpha_3}$ are all rational (even integral) so Schanuel's conjecture implies that $\alpha_1$, $\alpha_2$, and $\alpha_3$ are algebraically independent.  This contradicts $(*)$.
The $n=3$ case of Schanuel's conjecture is still open. There are various partial results known (see for example the final chapter of Baker's book Transcendental Number Theory), but I very much doubt that any of these partial results suffice to disprove $(*)$.  Note that if we take $n=2$ and $\alpha_1 = \ln \beta_1$ and $\alpha_2 = \ln \beta_2$ for nonzero algebraic numbers $\beta_1$ and $\beta_2$ then we recover the Gelfond–Schneider theorem.
It's a good exercise to derive various statements of this type (e.g., that $e+\pi$ or $\pi^e$ or whatever are transcendental) from Schanuel's conjecture.  This way you will be able to figure out on your own whether the statement is likely to be known, or at least be able to approach an expert in transcendental number theory with a more targeted question.<|endoftext|>
TITLE: Belyi functions on non-compact surfaces; or: Building Riemann surfaces from equilateral triangles
QUESTION [18 upvotes]: Some background on (compact) Belyi surfaces
$\newcommand{\Ch}{\hat{\mathbb{C}}}$
A compact Riemann surface $X$ is called a Belyi surface if there exists a branched covering map $f:X\to \Ch$ such that $f$ is branched over at most three points of $\Ch$. Here $\Ch$ denotes the Riemann sphere; we can and will take the three points to be $0$, $1$ and $\infty$.
(Recall that $f$ is a branched covering map if, for every $a\in\Ch$, there is a simply-connected neighborhood $U$ of $a$ such that $f$ maps every component of $f^{-1}(U)$ like $z\mapsto z^d$ for some $d\geq 1$. The function is branched over $a$ if $d>1$ for every such $U$.)
Equivalently, $X$ is a Belyi surface if it can be created by glueing together finitely many equilateral triangles together (defining a complex structure at the vertices in the obvious manner).
Every Belyi function is uniquely determined, up to a conformal change of variable, by its line complex, also known as a dessin d'enfant. This is a finite graph that essentially tells us how to form the surface by glueing together triangles.
In particular, the set of Belyi surfaces is countable. (This also follows from Belyi's famous theorem, which states that Belyi surfaces are exactly those that are definable over a number field.) So, in a nontrivial moduli space of Riemann Surfaces, such as the space of complex tori, most surfaces are not Belyi.

Belyi functions on non-compact surfaces
It seems natural to extend this notion to noncompact surfaces.
Definition. Let $X$ be a non-compact Riemann surface. An analytic function $f:X\to\Ch$ is called a Belyi function if $f$ is a branched covering whose only branched points lie over $0$, $1$ and $\infty$, and if $f$ has no removable singularities at the punctures of $X$.
Question. On which non-compact surfaces do Belyi functions exist?
This question seems extremely natural and came up in discussions among Bishop, Epstein, Eremenko and myself. Again, a Belyi function is uniquely determined by its line complex, which is now an infinite graph. Since the space of these graphs is totally disconnected, one might expect that not every non-compact surface supports a Belyi function. However, we discovered that this initial intuition is wrong:
Theorem. For every non-compact Riemann surface $X$, there is a Belyi function $f:X\to\Ch$. In particular, every non-compact Riemann surface can be built by glueing together equilateral triangles.
(EDIT. The paper is now - finally - available: arxiv:2103.16702.)
Note that, for non-compact surfaces, as pointed out by Misha in the comments, the existence of a Belyi function is formally stronger than being built from triangles, assuming that we allow vertices to be incident to infinitely many triangles. (Such vertices would not correspond to points in the resulting surface, as we have no way of defining a complex structure near these.) To get a Belyi function, we should assume that every vertex is incident to only finitely many triangles, so that each triangle is compactly contained in the resulting surface.
(We can also prove the existence of what one might call "Shabat functions", which have two critical values and omit $\infty$.)

My question:

Have such Belyi functions, and particular the problem of their existence on arbitrary non-compact surfaces, previously appeared in the literature?

(I would also be interested to hear whether our results might be of interest outside of one-dimensional complex function theory and complex dynamics. After all, classical Belyi functions and dessins d'enfants are relevant to many areas of mathematics.)
EDIT. Since I first asked the problem, Bowers and Stephenson raised an equivalent question in their work on conformal tilings; see Appendix B of "Conformal  tilings  I:  foundations,  theory,  and  practice", Conform. Geom. Dyn. 21 (2017), 1–63.

REPLY [6 votes]: As you may possibly already be aware, there is a parallel phenomenon in circle packing riemann surfaces. 
Those compact riemann surfaces admitting full circle packings are a countable dense subset of the moduli space, where by full circle packing a riemann surface, I mean finding a circle packing $C$ such that the carrier of the nerve graph $T_C$ coincides with our surface. This is due to R. Brooks, "Circle packings and co-compact extensions of Kleinian groups", Invent. Math. 86, 1986, 461-469.
But G. Brock Williams has also proven that all noncompact riemann surfaces are packable:
"Noncompact surfaces are packable", J. D'Analyse Math, Vol 90, 2003, 243-255.
The first basic 'intuition' as to why noncompact riemann surfaces are packable is of course that any obstructions to completing a (partial) circle packing to a full packing can be `hidden' into the cusp (similar to how any noncompact surface admits lots of nonvanishing vector fields). 
I would be very interested in reading your proof that noncompact riemann surfaces admit Belyi functions. In particular, I would like to know if your functions have the same "flat euclidean" structure up within the cusps as Williams' paper. Specifically, the final corollary 6.1 in Williams states "every noncompact riemann surface of finite type supports a circle packing asymptotic to the euclidean ball-bearing packing in the cusps". I have to admit that I do not find Williams' paper to be entirely educating, and am very anxious to see other approaches.<|endoftext|>
TITLE: Deligne's 1996 note on exceptional Lie groups
QUESTION [14 upvotes]: This is about Deligne's "La série exceptionnelle de groupes de Lie, C.R. Acad. Sci. Paris Sér. I Math. 322 (1996), no. 4, 321–326". 
When this came out, that was quite something! People were often talking about it, at lunch or coffee time - even making bets if I remember well.
I totally lost contact with all this, since then, now I see that there is a lot of literature on all this, for the most very technical, and I was curious: is there any survey paper or blog post or anything, which explains (for non-specialists) the status of that conjecture of Deligne? I mean, if that mysterious category exists, and what is it.
Many thanks.

REPLY [14 votes]: The key papers to read about this are Vogel's who conjectures an even better 2-dimensional family which includes all simple Lie algebras, not just the exceptional ones.  Deligne's family would correspond to a certain line in this plane.  Personally, I find Vogel's papers difficult to read, but you have to read them if you really want to understand this stuff.  Some good expositional sources are Cvitanovic's Bird Tracks and Introduction to Vassiliev knot invariants by Chmutov-Duzhin-Mostovoy.  In particular, Cvitanovic developed some of the key ideas independently.
The basic idea is that you write down diagrammatically the universal metric Lie algebra object in a symmetric tensor category, using the category of Jacobi diagrams.  This is very closely related to Vassiliev's finite type invariants by work of Bar-Natan.  The resulting Lie algebra object won't be simple, but you can add the assumption that it is simple by assuming that certain Hom spaces are 1-dimensional.  Roughly, conjecturally this added assumption gets you down to a 2-dimensional space of possible simple metric Lie algebra objects.
To get from the universal Lie algebra in the last paragraph to Deligne's exceptional family, you need to impose one more relation which is satisfied by the exceptional Lie algebras.  Namely there is a certain invariant vector in the 4th power of the adjoint representation of any exceptional Lie algebra, and this gives an extra relation.  This relation is written down explicitly on the first page of Dylan's paper mentioned in comments.
So where are we stuck?  Well there's two problems:

Do the relations that we can write down suffice to calculate the value of all closed diagrams?
Are the relations consistent for all values of the parameters?  That is, if you evaluate the same closed diagram in two different ways do you get the same answer?

The first of these questions should be easier than the second.  But both are wide open.
What Dylan did in the paper mentioned in comments was look at some analogous situations and calculated directly that it appeared the answers to the first question in those circumstances was yes, but the answer to the second question in those situations was no, and in fact the relations imposed a polynomial relation on the parameter forcing you back to finitely many possibilities.  But for Deligne's question, the calculation would require looking at very large examples which were beyond the computer's power.

REPLY [6 votes]: My understanding is that the series corresponding to the first three rows of the magic square do not exist but the existence of the series for the last row, the original exceptional series, is still open. This follows from Dylan Thurston's (unpublished) computer calculation and from Pierre Vogel's (unpublished) papers.<|endoftext|>
TITLE: Quotients of rational surfaces
QUESTION [6 upvotes]: Let $X$ be a projective surface defined over a field $k$ of characteristic $0$, and let $G$ be a finite group acting biregularly on $X$.
Assuming that $X$ is rational over $k$, is the quotient $X/G$ always rational?
If $k=\mathbb{C}$, we can use Castelnuovo's theorem and see that $X/G$ is unirational and hence rational. If $k=\mathbb{R}$, then $X/G$ is geometrically rational and also connected for the transcendental topology, and is thus rational.
But what happens for a general $k$, in particular when $k=\mathbb{Q}$?

REPLY [4 votes]: Just to round out the picture: if the characteristic of $k$ is positive and $G$ is a finite, but non-reduced group scheme (for example, the infinitesimal group scheme $\mu_p$ of $p$.the roots of unity), then the quotient $X/G$ need not even have Kodaira dimension $-\infty$ after desingularization. Moreover, if $G$ is not linearly reductive (for example, the infinitesmial group scheme $\alpha_p$), then a resolution of singularities $f:Y\to X/G$ need not satisfy $R^if_\ast{\mathcal O}_Y=0$ for $i\geq1$. Both phenomenons occur already if $\dim X=2$ - for example, there are many examples of ``unirational surfaces of general type'' in positive characteristic.<|endoftext|>
TITLE: What are the major open problems in design theory nowaday?
QUESTION [12 upvotes]: I gather that the question whether the Bruck-Chowla-Ryser condition was sufficient used to top the list, but now that that's settled - what is considered the most interesting open question?

REPLY [5 votes]: Personally, I am most interested in design theory with an "asymptotic flavor", and I think there are (edit: were, pre-Keevash) some very interesting open questions in this direction.
To cut to the chase, I think asymptotic design theory today is effectively searching for a constructive proof of Gustavsson's Theorem:  All "admissible" graphs $G$ which are sufficiently large and dense (i.e. $\delta(G) \gtrapprox (1-\epsilon(k)) v$) can be edge-decomposed into cliques of size $k$.
There are loads of spin-off questions, such as completion of sparse partial latin squares, construction of optimal packings and coverings, or imposing extra structure on the graph decompositions.
(edit: I do agree that the Hadamard conjecture is bigger, for sure.)<|endoftext|>
TITLE: Teaching stacks to differential geometry students
QUESTION [11 upvotes]: Does anyone have any experience teaching stacks over the category of manifolds to students whose background is, say, a semester-long course on manifolds?  Does anyone know of any publicly available notes on the subject, preferably in English?  [My French is limited to the knowledge of the alphabet :).  I can read Russian.]
I am aware of a paper by Behrend and Xu, Metzler's paper in the arxiv, and notes by Heinloth.  Hepworth has a nice exposition of vector fields on stacks, but his papers are rather terse.  Vistoli's notes on descent are quite nice, but are clearly aimed at algebraic geometers.  And there differences between the categories of manifolds and schemes --- fiber products of manifolds are badly behaved, for one thing.
The challenges in teachign such a course seem many.  For one thing I don't know how to talk about stacks without getting into 2-category theory.  And most differential geometers don't know much of 1-category theory.  But I don't want to start with a crash course on category theory.

REPLY [3 votes]: I had a good experience with Heinloth's notes. I tried to explain the two-categorical stuff
in the example of the stack of principal $G$-bundles. For example, a nice way to understand 2-pull-backs is to calculate $G\cong *\times_{BG}*$ explicitly. And of course, orbifolds and gerbes, e.g.
of $Spin^{c}$-reductions of a $Spin^{c}$-principal bundle  a provide examples accessible to differential geometers.<|endoftext|>
TITLE: Seifert surfaces via Alexander duality
QUESTION [8 upvotes]: If we take a knot $K$ in $S^3$, there are several ways to construct the associated Seifert surface.  One way, which I am not familiar with, I just came across in a paper I am reading.  It goes like this:
Consider the regular neighborhood $n_K$ of $K$ in $S^3$, which is diffeomorphic to $K\times \mathbb{R}^2$. $S^3-n_K$ is the knot complement. By Alexander duality, $H^1(S^3-n_K)\cong H_1(K)\cong\mathbb{Z}$; but we also have $H^1(S^3-K)\cong [S^3-n_K,S^1]$, so we can take a (smooth) map $f:\ S^3-n_K\rightarrow S^1$ representing a generator of $H^1$. The preimage of a regular value of $f$ gives a codimension-1 submanifold of $S^3-n_K$; namely, a surface-with-boundary $S$, such that $\partial S\subset \partial (S^3-n_K)$. So far, so good.  It is the next step that throws me:
$\partial S$ is cobordant to $K$.
I don't understand how this works, and am asking how one shows this.  For example, it's not even clear that $S$ is connected, and each component could have boundary.
I should mention that this is of course trivial in this specific case, since all 1-manifolds are cobordant.  But this should work for higher-dimensional knots.  It is even used in this answer and the comments below, where $K$ is replaced by an orientable 3-manifold and $S^3$ is replaced by $S^5$ (or $S^6$).
Can someone enlighten me in how one shows this (in the general case)? Namely, without relying on the dimensions of the spaces involved, but rather their codimensions?
[It might be relevant to note that the "regular neighborhood" (normal bundle) of the two examples above - the knot and the 3-manifold - are trivial.  I don't know if this matters or not.]

REPLY [11 votes]: This is an old argument that essentially predates much modern knot theory, and goes back to Serre.   
The basic idea goes like this: let $C$ be the complement of a co-dimension two knot in $S^n$. Apply Poincare/Alexander duality to deduce that $C$ is a homology $S^1 \times D^{n-1}$.  So $H^1 C \simeq \mathbb Z$, and $H^1 \partial C$ is either $\mathbb Z^2$ or $\mathbb Z$ according to whether or not $n=3$ or $n>3$.  In either case the restriction map is an injection.  
Serre's theorem that $H^1 X = [X,S^1]$ gives you a map $C \to S^1$ which you make transverse to a point.  Because the boundary is a product $\partial C \simeq S^1 \times S^{n-2}$, you can ensure the map $C \to S^1$ is not just homotopic to but actually projection onto the first factor (when restricted to $\partial C$).  
This ensures the preimage of a regular value of $C \to S^1$ is a Seifert surface for the knot. By Seifert-surface I mean an orientable co-dimension one submanifold of $S^n$ whose boundary is the knot.  If the surface is disconnected, you throw away any components that do not touch $\partial C$.  Because $\partial C \to S^1$ is the projection onto the factor, only one path-component touches $\partial C$. 
Serre and Thom used these ideas repeatedly in their early attacks on the Steenrod realization problem.  This was the version where you're trying to realize the homology classes by embedded submanifolds. 
The above is a relative version of their arguments.   There's an old Springer Lecture Notes in Mathematics that outlines the history of this argument and when it was first brought to knot theory.  Right, SLN in Math 685.  If it's not in the essay by Cameron Gordon, it's in the other big essay, maybe it was Milgram?<|endoftext|>
TITLE: About the well ordering of finite sequences of numbers
QUESTION [5 upvotes]: We order $\mathbb{N}^{<\mathbb{N}}$ as following: if $|\sigma| < |\tau|$ then $\sigma < \tau$; if they are of same length then they are ordered lexicographically. It is provable over $\operatorname{RCA}_0 + I\Sigma_2$, that this is a well ordering (of type $\omega^\omega$).
It is said that this linear ordering could be ill founded in models of $B\Sigma_2$ (Montalban. Open questions in reverse mathematics. BSL, vol. 17, no. 3).
So, does the well foundedness of this ordering imply $B\Sigma_2$? Or is it strictly between $B\Sigma_2$ and $I\Sigma_2$?

REPLY [2 votes]: A construction of $M \models \neg B\Sigma_2 + (1)$:
Firstly, let $N$ be a countable non-standard model of $PA$. We build $M$ as $N[G]$ by forcing.
Fix some non-standard $b \in N$. The poset consists of triples $(f,c,d)$, where $f$ is an $N$-finite partial function with domain contained by $b \times N$, $0 < c < d < b$ and $d - c > \omega$. 
$(f',c',d') \leq (f,c,d)$ iff 

$f'$ extends $f$, $c \leq c' < d' \leq d$, 
$f'(x,s) = 0$ for $(x,s) \in \operatorname{dom} f' - \operatorname{dom} f$ and $c \leq x < c'$,
$f'(y,s) = 1$ for $(y,s) \in \operatorname{dom} f' - \operatorname{dom} f$ and $d' < y \leq d$.

For each $(f,c,d)$ and $e$, we can extend $(f,c,d)$ to $(f',c,d)$ forcing one of the followings:
(a) $\Phi_e(G)$ is partial;
(b) $\Phi_e(f'; n) \downarrow = \sigma$ and $\sigma$ is the $\prec$-least $\tau$ s.t. there are $g, m$ with $(g,c,d) \leq (f,c,d)$ and $\Phi_e(g; m) \downarrow = \tau$.
Of course, we can achieve (b) because $N \models PA + (1)$. For sufficiently generic $G$, $N[G]$ contains a $\Delta_2$ cut $I < b$. By (b), $N[G] \models (1)$.
However, does $B\Sigma_2 + (1)$ imply $I\Sigma_2$?<|endoftext|>
TITLE: Robin-Laplacian in unbounded domains
QUESTION [5 upvotes]: Let $\Omega\subset \mathbb R^n$ be an open domain and $\tau>0$. Consider the following boundary value problem
$-\Delta v=f $ in $\Omega$, $\partial_\nu v+\tau v=g$ on $\partial\Omega$.
If $\Omega$ is a bounded with sufficiently smooth boundary it is known that we have "maximal elliptic $L_p$-regularity", i.e. for $f\in H^k_p(\Omega)$ and $g\in W^{k+1-1/p}_p(\partial\Omega)$ there is a unique solution $v\in H^{k+2}_p(\Omega)$, where $k\in \mathbb N$ and $p\in(1,\infty)$, see e.g. Triebel, Interpolation theory, Function spaces, Differential operators.
Is anybody aware of a corresponding result for the case when $\Omega$ is unbounded, e.g. a half space or an infinite layer?

REPLY [2 votes]: For the case $p=2$, I would have a look at this paper by Wolfgang Arend and Mahamadi Warma, and its follow-up papers: Potential Analysis 19: 341–363, 2003.<|endoftext|>
TITLE: real symmetric matrix has real eigenvalues - elementary proof
QUESTION [43 upvotes]: Every real symmetric matrix has at least one real eigenvalue. Does anyone know how to prove this elementary, that is without the notion of complex numbers?

REPLY [3 votes]: Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.
Let $A\in R^{n\times n}$ symmetric of rank $\ge 1$.
By compactness let $x\in R^n$ with $\|x\|=1$ be such that
$\|Ax \| = \max_{u\in R^n: \|u\|=1} \|Au\|$
and set $y = Ax/\lambda$ for $\lambda=\|Ax\|>0$. Then
\begin{align}
  \|A(x+ y) - \lambda(x+y)\|^2
&=\| Ay - \lambda x\|^2
\\&= \|A y\|^2 + \lambda^2 - 2 \lambda y^TA x
\\&= \|A y\|^2 + \lambda^2  - 2\lambda^2
\\&\le 0.
\end{align}
Either $x+y\ne 0$ is eigenvector for $\lambda$ or $x+Ax/\lambda=0$ and $x$ is is eigenvector for $-\lambda$.<|endoftext|>
TITLE: Topological conditions of Kolmogorov Extension Theorem
QUESTION [6 upvotes]: KET is often used to construct stochastic processes in continuous time when the state space is $\Bbb R^d$. As far as I am familiar with its proof, it uses standard monotonic class-like arguments together with Caratheodory Extension Theorem. Neither of the two latter theorems requires any topological conditions. Moreover, I have not been able to find where KET uses that the space is $\Bbb R^d$ rather than just some measurable space. 
Q: is it true that KET holds for general measurable spaces, and if not - where is the topology of $\Bbb R^d$ crucial in the proof? Also, maybe any counterexample (for exsitence or uniqueness) is known?

I also asked a related question here but didn't get any answer.

REPLY [11 votes]: The KET fails for general measurable spaces, the classical example can be found in a paper by Andersen and Jessen. Topological assumptions are necessary so that the resulting measure is not only finitely additive but countably additive. There exists a quasi-topological condition of measure spaces, perfectness, that is sufficient. A probability space $(\Omega,\sigma,\mu)$ is perfect if for every random variable $f:\Omega\to\mathbb{R}$, there exists a Borel set $B\subseteq f(\Omega)$ with measure one under the distribution $\mu\circ f^{-1}$. A proof of KET under the assumption that the marginal measures are perfect due to Lamb is given here. The strategy of the proof is to employ an existence result for regular conditional probability spaces and the construct the proces for them using the Ionescu-Tulcea theorem.<|endoftext|>
TITLE: Good book on representation theory of GL(n)
QUESTION [6 upvotes]: I am interested in a recommendation for a good book which discuses representation theory of GL(n)(say over field of complex numbers).
I know only a basic representation theory. 
The question I am interested in are how looks decomposition of $GL(n)$ module $V\otimes W$, where $V$,$W$ irreps. 
I am interested in book or  chapter in book which will not require too much preliminary.

REPLY [2 votes]: It's easier to mention the keyword "Littlewood-Richardson coefficients" that gives the answer to your question than to come up with the best possible source explaining it. If you are only interested in the answer, I would suggest to read books on combinatorics, such as "Symmetric group" by Sagan, rather than trying to learn all the background from the representation theory justifying it.<|endoftext|>
TITLE: Open problems in the theory of compact quantum groups
QUESTION [13 upvotes]: What are the important open problems in the theory of compact quantum groups? Or conjectures?
Here is an example from An De Rijdt's Ph.D. thesis: Is every compact quantum group with the fusion rules of $SO(3)$ of the the form $A_{aut}(B,\varphi)$ for a finite dimensional $C^*$-algebra $B$ and a $\delta$-form $\varphi$ on $B$?
See page 83 at the end of Chapter 3 in An De Rijdt's Ph.D. thesis, available online from https://perswww.kuleuven.be/~u0018768/students/derijdt-phd-thesis.pdf

REPLY [7 votes]: I'm late to the party! Here are some I momentarily mused on. Sorry I couldn't find explicit literature stating these as open problems, but I think these are natural developments to the famous results.

Determine the von Neumann algebraic type (and the factoriality) of $A_o(F)$. Is anything known beyond the results of Vaes-Vergnioux (Duke Math. J., 2007)?
Prove the strong Baum-Connes conjecture a la Meyer-Nest for the (edit: discrete dual of) q-deformation of compact simply connected simple group $G$, beyond $SU_q(2)$ due to Voigt (Adv. Math. 2011). Even at the 'classical' limit, it seems to require the understanding of the $G_{\mathbb C}$-equivariant KK-theory of $G_{\mathbb C}/B$, which is hard as far as I understand.<|endoftext|>
TITLE: Over which fields are symmetric matrices diagonalizable ? 
QUESTION [37 upvotes]: The question is motivated by this one real symmetric matrix has real eigenvalues - elementary proof: 
Are there other fields $F$ than $\mathbb{R}$ (maybe some valued fields or real closed fields) with the property that every symmetric matrix in $M_n(F)$ is diagonalizable ?

REPLY [27 votes]: This is a generalization of the idea of Will Sawin.
The Stufe of such field should be infinite. In fact if $-1$ is a sums of
squares, i.e., $-1=a_1^2+\cdots+a_{n-1}^2$, then 
$$A:=
\begin{pmatrix}
1 & a_1&\cdots  & a_{n-1} \\
a_1 & a_1^2 &\cdots & a_1a_{n-1} \\
\vdots & \vdots&\ddots  &\vdots \\
a_{n-1} & a_{n-1}a_1&\cdots  & a_{n-1}^2\\
\end{pmatrix}
$$ would be a symmetric matrix with $A^2=0$ and is not diagonalizable.
So the base field should be a formally
real field.
A complete characterization is given in the following article (a
necessary and sufficient condition is that such field should be an
intersection of real closed fields):
MR1237224
D. Mornhinweg, D. B. Shapiro and K. G. Valente, The Principal Axis Theorem Over Arbitrary Fields
(The American Mathematical Monthly, Vol. 100, No. 8 (Oct., 1993), pp. 749-754).<|endoftext|>
TITLE: counting triangle free graphs
QUESTION [7 upvotes]: Given $n$, the number of vertices, what is the number of triangle-free simple graphs on $n$ vertices (or asymptotically)?
A more difficult problem is, given $n$, $m$, what is the number of triangle-free simple graphs on $n$ vertices with $\le m$ edges (or asymptotically)?

REPLY [11 votes]: Next terms are 581460254001, 31720840164950 (sent to OEIS).  All these numbers were found by exhaustive enumeration. As far as I know, the theoretic enumeration problem is unsolved, even for labelled graphs.  
As for asymptotics, it is an old result of Erdős, Kleitman and Rothschild that almost all triangle-free graphs are bipartite. Someone studied the case of restricted numbers of edges see http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1097-0118(199602)21:2%3C137::AID-JGT3%3E3.0.CO;2-S/abstract .<|endoftext|>
TITLE: Understanding the nature and structure of proofs; Reverse Mathematics and Proof Theory. Prerequisites? Good introductory texts?
QUESTION [13 upvotes]: I'm still studying maths at undergraduate level, but intend to continue exploring topics in pure maths after I have graduated, so am thinking already about what directions I'd like to persue now, (as I like to be several steps ahead of myself!)
One area which seems particularly interesting is 'reverse mathematics'. I'd be interested to learn what the prerequisites would be for understanding it. Presumably it is a part of logic, and is related in some way to proof theory? If say, I wanted to explore questions such as, 'are there a finite number of ways of proving something?', would this be the right direction to take? (In fact, is there already an answer to this question?)
Apologies if the above is a bit broad/vague. This is my first post, and I plan to add to it as I learn more!
Update 1: Can anyone recommend a good introductory text on reverse mathematics and/or proof theory?

REPLY [2 votes]: As for your question on prerequisites, the more logic you know the better.  Of course the basic concepts---proofs, models, Peano arithmetic, incompleteness, compactness, nonstandard models, primitive recursion---really help to understand the program of reverse math.  But also more advanced logic topics are useful.  Computability theory and proof theory have already been mentioned.  Knowing more about model theory helps, as does set theory (topics such as comprehension axioms, independence, descriptive set theory, and forcing offer insight into reverse mathematics).
Although having said that, after knowing the logical basics, it is quite possible to just jump into the subject and learn at least the main ideas of reverse mathematics.  The first chapter of Subsystems of Second Order Arithmetic is available on Steve Simpson's website.  It is a good (and long) introduction to the basics of reverse mathematics.  (It also may help one decide if they want to purchase the whole book.)
Last, reverse mathematics connects logic to other areas of mathematics.  To understand say that "$\mathsf{ACA}_0$ is equivalent over $\mathsf{RCA}_0$ to the the Bolzano Weierstrass theorem", it is helpful to know the Bolzano Weierstrass theorem and a proof of it.  In that direction, having a standard undergraduate mathematics education---real analysis, abstract algebra, topology, etc.---goes a long way.<|endoftext|>
TITLE: Elementary Embeddings and Relative Constructibility
QUESTION [7 upvotes]: Suppose $$j:M\prec N$$ is a non-trivial elementary embedding. Under what conditions on the sets (classes?) $M$ and $N$ (or even the critical point of $j$) does $j$ extend to an elementary embedding $$k:L(M)\prec L(N)?$$ 
In the case where $M=N=V_{\lambda+1}$, the existence of such a $k$ is a strictly stronger assumption (this is Woodin's $I_0$ axiom), but need this always be the case? 
A more specific question involves extendible cardinals: Recall that $\kappa$ is extendible if, for every $\eta >\kappa$ there exists a $\theta>\eta$ and an elementary embedding $j:V_{\eta+1}\prec V_{\theta+1}$ such that $crit(j)=\kappa$ and $j(\kappa)>\eta$. Does $j$ extend to a $$k:L(V_{\eta+1})\prec L(V_{\theta+1})?$$
Is the existence of such a $k$ a straight-forward construction or is it strictly stronger than the existence of an extendible cardinal?

REPLY [4 votes]: Not all such embeddings extend, as was mentioned at Bob Lubarsky's question on Extending elementary embeddings from initial segments to all of $V$. 
Specifically, suppose that $\kappa$ is the least $1$-extendible cardinal. So there is an elementary embedding $j:V_{\kappa+1}\to V_{\eta+1}$. I claim that this embedding cannot extend to $L(V_{\kappa+1})\to L(V_{\eta+1})$, since the $j\upharpoonright V_{\kappa+1}$ is determined by $j''V_{\kappa+1}$, which is a size $2^\kappa$ subset of $V_{\eta+1}$, and by means of a flat pairing function all such subsets are coded as elements of $V_{\eta+1}$. Thus, $L(V_{\eta+1})$ would see that $\kappa$ is $1$-extendible, and so by elementarity there must be a $1$-extendible cardinal in $L(V_{\kappa+1})$ below $\kappa$, contradicting the minimality of $\kappa$. 
A similar argument applies to the least $\theta+1$-extendible cardinal for any $\theta$.<|endoftext|>
TITLE: Uniform bound on the eigenfunctions of the Laplacian
QUESTION [7 upvotes]: Is it possibly to have $L_\infty$ bounds on the eigenfunctions of the Laplacian operator on bounded regular domains with Dirichlet condition?  I found several papers by Sogge but these are pretty general and apply to Laplacian-Beltrami operator on manifolds.  Their paper has bounds that are possibly too loose for my application.
Thanks,
John

REPLY [6 votes]: Moser iteration proceeds roughly like this: If the dimension $n$ is greater than $2$ and we assume homogeneous Dirichlet , we can proceed as follows: Using the Sobolev inequality on $\mathbb{R}^n$,
$$
c(n)\|u\|_{2n/(n-2)} \le \|\nabla u\|_2
$$
and integrating by parts, we get something roughly like this (you have to figure out what to do with all the absolute values) for each $p > 1$:
$$
\lambda\int |u|^p \ge \int |u|^{p-1}(-\Delta |u|)
 = \frac{4(p-1)}{p^2}\int |\nabla |u|^{p/2}|^2
\ge \frac{4(p-1)}{p^2}c(n)\|u\|_{np/(n-2)}^p.
$$
Therefore, given $p_0 > 1$, if we set $p_k = p_0(n/(n-2))^k$, we get
$$
\|u\|_{p_{k+1}} \le \left(\frac{p_k^2}{4(p_k-1)}\frac{\lambda}{c(n)}\right)^{1/p_k}\|u\|_{p_k}.
$$
Iterating this, we get
$$
\|u\|_\infty \le C(n,p_0)\lambda^{n/(2p_0)}\|u\|_{p_0}.
$$
The power of the eigenvalue in the final estimate doesn't need to be calculated explicitly. You know what it has to be by observing that the left side is invariant under rescaling space ($\mathbb{R}^n$) and therefore the right side must be, too.<|endoftext|>
TITLE: Primitive Cohomology Useful?
QUESTION [13 upvotes]: In her book, after proving the hodge decomposition, Voisin spends time discussing primitive cohomology $H^r(X, \mathbb{C})_{prim} = \ker L^{n-r+1} \subset H^r(X, \mathbb{C})$ (where $L$ is the Lefschetz operator). She proves several general theorems regarding/using primitive cohomology (Hodge index, Lefschetz decomposition, a bilinear form on $H^k(X,\mathbb{C})$ behaving in a controlled way on primitive cohomology) and establishes some technical results (if $\omega$ is a primitive form then there is a formula for $*\omega$ in terms of the Lefschetz operator and $\omega$).
$\textbf{Question: }$ I'm having a hard time understanding why one should care about primitive cohomology. Can you deduce lots of interesting facts about nonsingular complex projective varieties with say the Lefschetz decomposition as was the case with the Hodge decomposition? What are some typical applications? I'd really like some examples to illustrate if/how primitive cohomology is useful.
Specifically, I am interested in how primitive cohomology could be useful on a "daily basis". For example, let $X, Y$ be smooth complex projective varieties. Sometimes one can deduce that there are no surjective maps $X \xrightarrow{\phi} Y$ because such maps induce injective maps on cohomology (which preserve Hodge structure). Can primitive cohomology give a more refined obstruction to the existence of $\phi$ in certain cases?
$\textbf{Computing}$
1.) How about primitive cohomology? This depends on the choice of a Kahler form. Do the dimensions of the primitive cohomology groups not depend on the choice of Kahler form? It's not clear to me if primitive cohomology of abelian varieties depends only on the dimension. Does one know the dimensions of primitive cohomology groups of an abelian variety? How about other classes of varieties? For a K3 surface, it seems like one can give the dimensions of primitive cohomology groups independent of kahler form, the main point is $h^{1,1}$, where primitive cohomology has dimension 19.
$\textbf{Functoriality}$
2.) A surjective map of smooth complex projective varieties is injective on cohomology and a map of hodge structures. A finite surjective map pulls back ample divisiors to ample divisors, so if we choose kahler classes appropriately, then such a map induces a map on primitive cohomology. Does a more general class of maps induce maps on primitive cohomology (if we choose kahler classes appropriately)?

REPLY [6 votes]: One more application: the singular cohomology functor (with coefficients in a field) restricted to smooth projective complex varieties factorizes through the semi-simple category of polarizable pure Hodge structures. There is a certain (somewhat complicated) extension of this result to cohomology of all complex varieties.<|endoftext|>
TITLE: Known size invariant for Riemannian manifolds?
QUESTION [8 upvotes]: Larry Guth in his 2010 ICM address mentions the notion of a size invariant of Riemannian
metrics on a smooth manifold $M$.  These are functions $S: Metrics(M)  \to \mathbb{R}$ that are invariant under isometry and satisfy $S(g) \leq S(g')$ if $g \leq g'$, where $g \leq g'$ means $g(v,v) \leq g'(v,v)$ for all $v \in TM$. 
For an open Riemannian manifold $(M, g)$ and for a compact set $K \subset M$ define
$$
 S_M(g, K) = \inf \{||f|| : f \in C^\infty_c(M) \mbox{ such that } |\nabla f(x)|_g \geq 1 \mbox{ for all } x\in K\}
$$
where $||f|| = \max_M f - \min_M f$, and then define
$$
 S_M(g) = \sup_{K\subset M} S_M(g, K).
$$
One can prove that this is an size invariant.
This invariant appears in Kei Irie's paper 'Displacement energy of unit cotangent bundles' in Section 2, http://arxiv.org/abs/1106.2199
Irie proves that $S_M(g) \leq c_n r(M, g)$ where $c_n$ is a dimensional constant and $r(M,g)$ is the inner radius.  Irie also lower bounds $S_M(g)$ in terms of the displacement energy of the unit disk cotangent bundle $D^*_g M$ in the symplectic manifold $T^*M$.
My question is if this invariant appears elsewhere in the literature or is related to other known invariants?

REPLY [5 votes]: Remark. Now that Cadoi has disclosed the origin of the invariant (which was absent from the original formulation of the problem) my musings in trying to guess the symplectic geometry behind it seem a bit silly. I'll leave the post anyway, since it maybe useful in understanding the question. However, maybe there is something to be said for including motivation and sources (right from the start) in MO questions.
Dear Cadoi,
This is just a comment on your question, but since it is a bit long and I can see what I'm TeXing I'll post it as an answer. I myself have not seen this invariant before, however it has a symplectic look about it and may turn out to be some capacity in disguise. Note that if you define a Riemannian invariant by taking a symplectic capacity and evaluating it on unit codisc bundles, then you get a size invariant. This is because isometries lift to symplectic transformations that preserve the unit codisc bundle and because capacities are monotone.
The reason I say your invariant looks symplectic is because, seen geometrically, the differential of $f$ defines a Lagrangian submanifold of the cotangent bundle and the condition that the norm of the gradient of $f$ be at least $1$ on $K$ means this Lagrangian submanifold does not intersect the unit co-disc bundle over $K$. The quantity $\|f\|$ is a relative symplectic invariant of the pair of Lagrangians given by the zero section and the differential of $f$. I don't exactly remember Viterbo's notation for this, but it is in his paper Symplectic topology as the geometry of generating functions. 
Just for the fun of it, and without implying that this will give something non-trivial, I'll attempt a "symplectization" of your invariant, at least to the extent that it might give size invariant of Riemannian or Finsler metrics that gives the same value on two metrics that have symplectomorphic codisc bundles:
Let $D^*M$ be the unit codisc bundle of your metric and let $K \subset D^*M$ be a compact subset. Consider the set $P_K$ of pairs of Lagrangian submanifolds $(E,F)$ such that (1) $E$ is contained in  $D^*M$, (2) $F$ does not intersect $K$, and (3) both $E$ and $F$ are Hamiltonian isotopic to the zero section. 
Define $S(D^*M,K)$ as the infimum of the Viterbo capacities of all pairs of Lagrangians $(E,F) \in P_K$ and now define $S(D^*M)$ as the supremum of the quantities $S(D^*M,K)$ as $K$ ranges over all compact subsets of $D^*M$.
If you restrict $E$ to always be the zero section, $F$ to the the graph of the differential of a function, and $K$ to the the codisc bundle over some compact subset of $M$ this is exactly your invariant. 
Caveat emptor, this definition may require some fidgeting (like restricting the class of compact sets $K$) to make it non-obviously trivial.<|endoftext|>
TITLE: Rescaling positive definite matrices to force a unit eigenvector
QUESTION [9 upvotes]: Hello,
Let $X'X$ be a positive definite matrix and let $\mathbf{1}$ denote the vector of ones. 
I'm hoping to construct a positive, diagonal matrix $W$ such that
$$(W X'X W) \mathbf{1} = \mathbf{1}$$
$X$ and $W$ are all assumed to have real-valued entries, and $X'$ denotes the transpose of $X$.
I don't, yet, have a proof that such a matrix $W$ always exists, but strongly suspect it. Any ideas on algorithms, proofs, or counter-examples would be gratefully received.
The problem arises from work in statistics. 
thanks,
David.

REPLY [3 votes]: The relevant reference is 
Marshall, A. and Olkin, I. Scaling of Matrices to Achieve Specified Row and Column Sums. Numerische Mathematik  12, 83-90 (1968)
who prove the result in the affirmative for positive definite matrices (and some generalizations). The proof is elegant and construction: the diagonal matrix can be found by minimizing a particular constrained optimization problem. 
There is a good discussion of the problem and its generalizations in
Johnson, C.R. and Reams, R. Scaling of symmetric matrices by positive diagonal congruence. Linear and Multilinear Algebra, 57(2) (2009) 123-140.
-David.<|endoftext|>
TITLE: Number of Maximal Left Ideals 
QUESTION [7 upvotes]: In THIS PROBLEM it is proved that for any non-zero ring $R$ with identity, $R[x]$ has an infinite number of maximal left ideals. Is it possible for an uncountable non-zero ring $R$ with identity, $R[x]$ has only a countable number of maximal left ideals ??

REPLY [4 votes]: Generalizing the answer by wccanard: Let $R$ be a commutative ring. Then the kernel of $R[x] \twoheadrightarrow R_{red}[x]$ consists of nilpotent elements, hence this map induces a homeomorphism $\mathrm{Spec}(R_{red}[x]) \cong \mathrm{Spec}(R[x])$. It restricts to a homeomorphism between the subspaces of closed points $\mathrm{Spm}(R_{red}[x]) \cong \mathrm{Spm}(R[x])$. Now take any uncountable $R$ such that $R_{red}$ is a finite field.<|endoftext|>
TITLE: Small Implications of the Axiom of Replacement
QUESTION [7 upvotes]: The axiom of replacement implies the existence of sets larger than usual in mathematical practice, but can be used to prove theorems about sets of real numbers, such as Borel determinacy.  This is interesting because it suggests there's some sort of recursive procedure that makes sense for sets of reals, but is not provable in ZC alone.  This procedure seems like it would be of independent interest from the question of the existence of sets beyond $V_{\omega + \omega}$ in the cumulative heirarchy.
Is there a weaker axiom or recursive set of axioms that can be added to ZC that imply exactly the implications of replacement that hold for sets in $V_{\omega + \omega}$, one that explains the kind of additional constructions that replacement permits you to make?

REPLY [4 votes]: The set $M$ of all formulas $\varphi$ that are of the form $V_{\omega+\omega} \vDash \psi$ is certainly recursive.    Now the set $N:=\{ \varphi\in M:  ZFC \vdash \varphi\}$ is c.e.
A standard trick gives an equivalent set $N'$ which is recursive (decidable): replace the
$n$-th formula in $N$ (in any computable enumeration) by an equivalent formula that is much longer that all previous ones. 
Is this set $N'$ what you are looking for?   I realize that it does not have the nice form you probably wanted.<|endoftext|>
TITLE: Difference between 'generalized gradient' and 'subgradient' ?
QUESTION [9 upvotes]: Hi, I am wondering what the difference between 'generalized gradient' and 'subgradient' of a (potentially non-differentiable) convex function 'f' is. 
The generalized gradient I am interested in is meant in the sense of the paper http://www.is.tuebingen.mpg.de/fileadmin/user_upload/files/publications/ICML2010-Kim_6519[0].pdf  (see page 3, footnote 2).
many thx for any help.

REPLY [10 votes]: Generalized gradients generalize subdifferentials.  Subdifferentials are defined globally, and rely on the simple geometry of convex functions.  If in a neighborhood a linear subspace only touches the graph of a convex function without crossing it, then you know for a fact it won't cross the graph later.  So the intuition between the subdifferential is that you take each linear subspace of that is strictly below the convex function, and slide it up until it touches it.  The subdifferential tells you the relation between the slopes of linear subspaces, and the points on the graph where the linear subspace will touch without crossing.
For a general non-convex function, a linear subspace that touches the graph at one point can cross it again at another point.  So you need a definition that works locally, only in a small neighborhood of the point.  The definition of generalized gradient gives you that.  It's reasonably easy to find information online about it.  For example, the page on Clarke generalized derivative at Encyclopedia of Mathematics gives a quick introduction.  Clarke's book, Optimization and nonsmooth analysis, is nice.  A more recent book reference is Rockafeller and Wets, Variational analysis, which thoroughly covers this and related topics.
I quickly looked over the paper you link to, and I don't see why they need generalized gradients instead of just subdifferentials, though I didn't look too closely.<|endoftext|>
TITLE: A General Framework for Ramsey Theory ? 
QUESTION [11 upvotes]: There are few results in modern mathematics that I find so deep and full of philosophical implications as Ramsey's theorem. 
I am aware (at some basic level) that it has generated a plethora of further research, going well beyond graph theory, and that there is now an entire industry of Ramsey-like theorems, in many disparate domains (for instance enumerative combinatorics). 
What troubles me, though, is that I do not clearly see what the proper framework for a generalized Ramsey Theory could possibly be. 
If you browse the wiki, you find the following sentence, under the voice RAMSEY THEORY:
Problems in Ramsey theory typically ask a question of the form: "how many elements of some structure must there be to guarantee that a particular property will hold?"
That sound general enough, but a little too informal: 
can this sentence be re-formulated in a rigorous way? 
I mean: categories of structured sets are the bread-and-butter of category theory (think for instance of categories of algebras, categories of ordered sets, etc), so it looks to me as if there could be a convenient formulation of the  quoted sentence in suitable categorical form 
(something like: if .... then for every object of the category there is a large -in some suitable sense- sub-object satisfying ....., fill the dots)
Anything out there?

REPLY [7 votes]: You might be interested in the 2010 book "Introduction to Ramsey Spaces" by Stevo Todorcevic
"Ramsey theory is a fast-growing area of combinatorics with deep connections to other fields of mathematics such as topological dynamics, ergodic theory, mathematical logic, and algebra. The area of Ramsey theory dealing with Ramsey-type phenomena in higher dimensions is particularly useful. Introduction to Ramsey Spaces presents in a systematic way a method for building higher-dimensional Ramsey spaces from basic one-dimensional principles. It is the first book-length treatment of this area of Ramsey theory, and emphasizes applications for related and surrounding fields of mathematics, such as set theory, combinatorics, real and functional analysis, and topology. In order to facilitate accessibility, the book gives the method in its axiomatic form with examples that cover many important parts of Ramsey theory both finite and infinite."<|endoftext|>
TITLE: Subspaces of $l_p$ and Banach-Mazur distance
QUESTION [11 upvotes]: This is a question I posted on SE, and I have been advised to post it here. 
https://math.stackexchange.com/questions/146427/subspaces-of-l-p-and-banach-mazur-distance
It is well-known that every subspace of $l_2$ is isometric to $l_2$. When $p\neq 2$, $l_p$ has subspaces that are not even isomorphic, let alone isometric, to $l_p$. Suppose $X$ is a subspace of $l_p$ with $p\neq 2$ such that $X$ is isomorphic to $l_p$. What can one say about the Banach-Mazur distance between $X$ and $l_p$? More precisely, which one of the following mutually exclusive options holds true:
1) Given $K$, there exists a subspace $X$ of $l_p$, isomorphic to $l_p$, such that for any isomorphism $T:X\to l_p$ one has $||T||\cdot||T^{-1}||>K$.
OR
2) There exist a constant $K$ (possibly depending on $p$), such that for any subspace $X$ of $l_p$, isomorphic to $l_p$, there exist an isomorphism $T:X\to l_p$ such that  $||T||\cdot||T^{-1}||\leq K$.
Intuitively, I very strongly suspect it is 1) but I do not have an argument to exclude 2) and, if it is indeed 1), I would like to see a concrete example of a subspace having that property.

REPLY [8 votes]: (1) is correct.  It follows from the fact that there is a sequence $(E_n)$ of finite dimensional subspaces  of $\ell_p$ s.t. $\gamma_p(E_n) \to \infty$.   Here $\gamma_p(X)$ is the factorization constant of the identity on $X$ through an $L_p$ space; that is, $\gamma_p(X)= \inf \|T\|\cdot \|S\|$, where the infimum is over all $T:X\to Y$, $S:Y:\to X$, and $Y$ an $L_p$ space.  
For fixed $n$, set $X_n=\ell_p(E_n)$.  Since $E_n$ is finite dimensional, the space $X_n$ is isomorphic to $\ell_p$ (and embeds isometrically into $\ell_p$) but the Banach-Mazur distance of $X_n$ to $\ell_p$ is at least $\gamma_p(E_n)$.
There remains the sticky point of producing a sequence $(E_n)$ as above. One way is to take a subspace $X$ of $\ell_p$ that fails the approximation property and let $E_n$ be a sequence of finite dimensional subspaces of $X$ with $E_1\subset E_2 \subset \dots$ and $\cup E_n$ dense in $X$.  Or, for $p<2$, $E_n$ can be (Banach-Mazur arbitrarily close to) $\ell_r^n$ with $p<r<2$, because $\ell_r$ embeds isometrically into $L_p(0,1)$ and there are various ways of checking that $\gamma_p(\ell_r^n)\to \infty$.  Since I do not know your background (you have, unfortunately, chosen to remain anonymous), I do not know which approach of producing $(E_n)$ is best for you.<|endoftext|>
TITLE: Making CW-complexes metrizable
QUESTION [11 upvotes]: $\newcommand\met{\mathrm{met}}$It is a basic topological fact that CW-complexes aren't typically metrizable (they must satisfy a certain local finiteness condition) and the quotient topology is to blame.
Question: Suppose $X$ is a CW-complex (possibly with countably many cells and maybe even of finite dimension). Is it possible weaken the topology of $X$ to construct another space $X_{\met}$ (with the same underlying set), so that the continuous identity function $X\to X_{\met}$ is a homotopy equivalence?
Update: I will clarify (now much later) that this question has an affirmative answer for simplicial complexes. Given an arbitrary simplicial complex $K$, we have $|K|$, which has the weak topology and is not always metrizable. However, you can give the underlying set of $|K|$ a metrizable topology to form the "metric simplicial complex" $|K|_m$. The identity $|K|\to |K|_m$ is continuous and is a homotopy equivalence. A nice proof can be found in Segal and Mardesic's book Shape Theory in Appendix, $\S 1.3$, Theorem 10. As Sergey Melikhov nicely points out in his answer, the same is true for regular CW-complexes, which include simplicial complexes. Using this result, it follows that every CW-complex is homotopy equivalent to some metric space. However, my question is a bit more specific.

REPLY [3 votes]: I had trouble finding a reference online for a non-metrizable CW complex, so I figured it might be nice to record that here. Let $X = \bigvee_{n=1}^\infty S^1$ be the infinite bouquet of circles, considered as a CW complex. I claim that this space is not metrizable. An important fact about CW topology that we can exploit is that any open neighborhood of the basepoint $x_0$ intersects the interior of every 1-cell of this cell complex.
Now suppose $X$ is metrizable, i.e. has a metric $d$ which induces the same topology as that of the cell structure. Any open ball centered at $x_0$ intersects the interior of every 1-cell. So we can pick a sequence $(x_n)_{n=1}^\infty$ such that $x_n$ lies in the interior of the $n$-th 1-cell and $d(x_0,x_n)<1/n$. Then $\lim x_n = x_0$ by construction. On the other hand, the set $X-\{x_1,x_2,\dotsc\}$ is an open neighborhood of $x_0$ since it is open in each cell of the complex. Thus $x_0$ is separated from the sequence $(x_n)_{n=1}^\infty$, a contradiction. $\quad \square$<|endoftext|>
TITLE: When does the cotangent complex vanish?
QUESTION [18 upvotes]: The question is already in the title. Less succinctly, let's call a map $f:X \to Y$ of schemes $L$-trivial if its cotangent complex is quasi-isomorphic to $0$. Such maps have striking deformation-theoretic consequences; for example, any deformation of $Y$ can be followed uniquely by a deformation of $X$. 
My primary (and probably naive) question is:

Is there a classification of $L$-trivial maps? 

I am sure this question has been asked before, but I did not find any literature that deals with it. The three examples of $L$-trivial maps I am familiar with are:

Etale morphisms (and these are the only examples under finiteness constraints).
Any map between perfect $\mathbb{F}_p$-schemes.
The inclusion of the closed point in the spectrum of a valuation ring with divisible value group, or similar "divisible" constructions. For example, $\mathrm{Spec}(\mathbb{C}) \hookrightarrow \mathrm{Spec}(\mathbb{C}[ t^{\mathbb{Q}_{\geq 0}}])$ is $L$-trivial.

[ Edit: I learnt the last one in conversation after positing the first version of this question. ]
More examples can be obtained by taking filtered colimits of the above examples, but those are only slightly different. Hence, a second question is: are there other fundamentally different examples of $L$-trivial maps? 
Perhaps a classification is unreasonable to expect, so I am also happy to learn more about $L$-trivial maps in other geometric categories, like algebraic stacks, or derived/spectral schemes/stacks, or (complex/rigid) analytic spaces, etc.. In particular, I am especially curious to know if $L$-trivial maps can be better understood using derived algebraic geometry.

REPLY [9 votes]: Edit: Here is a possible characterization. As mentioned in the comments above, the vanishing of the 1-truncated cotangent complex $\tau_{\leq 1}L_{B/A}$ of a map of rings $f \colon A \to B$ is equivalent to a lifting property with respect to square-zero extensions $T' \to T$. This follows from the fact that the space $Map(L_{T/A},M[1])$ is equivalent to the groupoid of square-zero extensions of $T$ over $A$ with kernel $M$. Here $M$ is a $T$-module.
Rephrasing this, $\tau_{\leq 1} L_{B/A}$ vanishes if and only if $f$ has a lifting property with respect to morphisms of 0-truncated simplicial algebras such that the kernel is concentrated in degree 0 and squares to 0. Let's call these 0-concentrated.
Then we can go on to look at morphisms of 1-truncated simplicial algebras with kernel $K$ concentrated in degree 1. The squaring-to-zero property is vacuous here, because a product of two elements in $\pi_1(K)$ will be in $\pi_2(K)$, which is zero by assumption. Let's call these 1-concentrated Then we find that $\tau_{\leq 2} L_{B/A}$ vanishes if and only if $f$ has a lifting property with respect to all 0- and 1-concentrated maps. This again holds because the cotangent complex classifies 0- and 1-concentrated maps.
I think now it's clear how to go on: $\tau_{\leq n+1}L_{B/A}$ vanishes if and only if $f$ has a lifting property with respect to all $m$-concentrated maps with $m \leq n$. And the full cotangent complex vanishes if and only if $f$ has the lifting property with respect to $n$-concentrated maps for all $n$.
These directions one looks at if one starts to check with respect to $n$-concentrated maps for $n \geq 1$ are sometimes called the derived directions. So Avramov's theorem might be rephrased as saying that under strong finiteness assumptions, unobstructedness in the classical directions implies  unobstructedness in all derived directions.

This is an anwer to your last paragraph about L-trivial maps in other geometric categories. If you are only interested in schemes it doesn't tell you anything interesting.
One thing that the cotangent complex is good at is measuring connectivity of a morphism of simplicial rings. This also holds without any finiteness assumptions on the ring.
Recall that a morphism $f \colon A \to B$ of simplicial rings is called n-connective if it induces isomorphisms $\pi_i (A) \to \pi_i (B)$ in degrees $< n$ and a surjection $\pi_n (A) \to \pi_n (B)$ in degree $n$. There then is a result that states that if $f$ is $n$-connective, then the homology of the relative contangent complex $L_{B/A}$ vanishes in degrees $\leq n$. (I hope I got all the indices right.) So in particular, any equivalence of simplicial rings is L-trivial.
One way of intepreting your question is to ask when the converse holds. What do I know if a morphism of simiplicial rings is L-trivial? There is a partial converse to the statement above. Namely, if a morphism $f \colon A \to B$ induces an isomorphism $\pi_0(A) \to \pi_0(B)$ and is L-trivial, then it is an equivalence! I find this pretty suprprising, as L is only a linear piece of data, but still manages to detect equivalences.<|endoftext|>
TITLE: When is the determinant a Morse function?
QUESTION [9 upvotes]: This might be ridiculously obvious, but...
For each $n \in \mathbb{N}$, let $M_n$ denote the manifold of $n \times n$ matrices with real entries. It is well known that the $n$-dimensional determinant function $d_n:M_n \to \mathbb{R}$ is a Morse function if and only if $n = 2$ since the zero matrix is a degenerate critical point of $d_n$ whenever $n > 2$. (See the simple homework exercise in this pdf for a proof).
My question is as follows:

For a given $n$, is there a characterization of the sub-manifolds of $M_n$ for which $d_n$ is a Morse function?

Just to be clear, by Morse function I mean that $d_n$ must have:

Isolated critical points (not necessarily finitely many), and
Non-degenerate Hessian at each critical point.

For general subspaces of $M_n$ it seems that the critical set of $d_n$ would not be isolated (one can easily construct examples of critical subspaces), so if there is a Morse-Bott theory for submanifolds of $M_n$ that would be good to know as well.

REPLY [4 votes]: Suppose that $M$ is a manifold, $S$ is a submanifold and $\newcommand{\bR}{\mathbb{R}}$ $f: M\to\bR$ is a smooth function.  Let $T_S^*M\subset T^*M$ be the conormal bundle of $S$.  Let $\Gamma_{df}\subset T^*M$  denote  the graph of the differential of $f$, $df: M\to T^*M$, $\Gamma_{df}=df(M)$.
Then  both $T^*_SM$ and $\Gamma_{df}$ are lagrangian submanifolds of $T^*M$.  The function $f|_S$  is Morse iff the above two Lagrangian submanifolds intersect transversally. For a proof, see page 3 of these notes.<|endoftext|>
TITLE: Is the set of surfaces over Spec Z with ample  canonical sheaf empty
QUESTION [15 upvotes]: Main question. Does there exist a smooth projective morphism $X\to$ Spec $\mathbf Z$ of relative dimension two such that the canonical sheaf $\omega_{X_{\mathbf Q}}$ of the generic fibre $X_{\mathbf Q}$ is ample?
Replacing "relative dimension two" by "relative dimension one", the answer is negative by a theorem of Abrashkin-Fontaine. I highly suspect the answer to be negative in this case too. Unfortunately, it is not known yet though as confirmed by Sándor.
Question 2. Does there exist a  number field $K$ such that there are infinitely many $K$-isomorphism classes of smooth projective geometrically connected surfaces over $K$ with ample canonical sheaf and a smooth projective model over $O_K$? 
The answer is positive if we replace "surfaces" by "curves". And as Will points out the answer is positive in the higher-dimensional case.
My main question is part of the arithmetic Shafarevich conjecture. As the terminology suggests, this conjecture is the arithmetic analogue of a conjecture for geometric objects. The latter (geometric) conjecture has been resolved by Arakelov, Bedulev, Kovács, Lieblich, Möller, Parshin,  Viehweg, Zuo, et al. (Edit: Please see the references in Sándor's answer.)  Its arithmetic analogue remains widely open for relative dimension $\geq 2$ to my knowledge, and was resolved in 1983 by Faltings for relative dimension 1.
With my second question I would like to assure myself of the non-triviality of a higher-dimensional arithmetic Shafarevich conjecture. It turns out to be trivial.
Let me state the results (due to the before-mentioned) in algebraic geometry relevant to this question. The base field is  an algebraically closed field $k$ of characteristic zero.
Theorem 1. (Higher-dimensional geometric analogue of main question) There are no smooth projective (strongly?) non-isotrivial morphisms $X\to \mathbf P^1_k$ such that  the canonical sheaf of the generic fibre of $X\to \mathbf P^1_k$ is ample. 
Theorem 2. ("Folklore?" Higher-dimensional  geometric analogue of second question) Fix $d\geq 0$. There exists a smooth projective connected curve $C$ such that there are infinitely many isomorphism classes of (strongly?) non-isotrivial smooth projective morphisms $X\to C$ of relative dimension $d$ whose generic fibre has ample canonical sheaf.
Now, Theorem 2 is one of the reasons that the following grand finiteness theorem is difficult.
Theorem 3. Let $C$ be a smooth projective connected curve and let $h$ be a polynomial. Then, there are only finitely many isomorphism classes of smooth projective (strongly?) non-isotrivial morphisms $X\to C$ whose generic fibre is canonically polarized with Hilbert polynomial $h$.
Let me note that I am considering function fields over a field of characteristic zero to be analogous to Spec $\mathbf O_K$. I know some of you prefer function fields over finite fields, but regarding these questions the analogy also "works"  to a certain extent.
I might  have stated Theorems 1-3 slightly incorrectly.  In this case I apologize. (Also, I didn't state Theorem 3 in its full generality. The base curve doesn't need to be compact for instance.) Maybe, I should have only considered deformation types of families over $C$ in the statements. 
Finally, let me point out some related MO questions:
What can be the dimension of a pointless smooth proper Z-scheme?
Smooth proper scheme over Z

REPLY [9 votes]: I don't think the answer to the first question is known. 
Will has already pointed out the trivial answer to the second question. However this is not the right question. I mean this is kind of trivial. The interesting question is if you fix the genus and require that the curve over $K$ has good reduction everywhere (outside a fixed set of primes). If you ask it in that way, then the answer for curves is negative (by Faltings) and so the easy fix to do it in higher dimensions does not work.
Here are some comments and references to Theorems 1,2,3:
Theorem 1 is known in more general context. 
It does not need "strong", non-isotrivial is enough.
Relevant references are:
Kovács, Sándor J.(1-UT)
Smooth families over rational and elliptic curves. 
J. Algebraic Geom. 5 (1996), no. 2, 369–385. 
Kovács, Sándor J.(1-MIT)
On the minimal number of singular fibres in a family of surfaces of general type. 
J. Reine Angew. Math. 487 (1997), 171–177. 
Kovács, Sándor J.(1-CHI)
Algebraic hyperbolicity of fine moduli spaces. 
J. Algebraic Geom. 9 (2000), no. 1, 165–174. 
Viehweg, Eckart(D-ESSN); Zuo, Kang(PRC-CHHK)
On the isotriviality of families of projective manifolds over curves. 
J. Algebraic Geom. 10 (2001), no. 4, 781–799.
Kovács, Sándor J.(1-WA)
Logarithmic vanishing theorems and Arakelov-Parshin boundedness for singular varieties.
Compositio Math. 131 (2002), no. 3, 291–317. 
There are also generalizations for families over higher dimensional bases. See for instance:
Viehweg, Eckart(D-ESSN); Zuo, Kang(PRC-CHHK)
Base spaces of non-isotrivial families of smooth minimal models. Complex geometry (Göttingen, 2000), 279–328, Springer, Berlin, 2002. 
Kebekus, Stefan(D-KOLN); Kovács, Sándor J.(1-WA)
Families of canonically polarized varieties over surfaces. (English summary) 
Invent. Math. 172 (2008), no. 3, 657–682.
Kebekus, Stefan(D-FRBG); Kovács, Sándor J.(1-WA)
The structure of surfaces and threefolds mapping to the moduli stack of canonically polarized varieties.
Duke Math. J. 155 (2010), no. 1, 1–33. 
Patakfalvi, Zsolt(1-PRIN)
Viehweg's hyperbolicity conjecture is true over compact bases. (English summary) 
Adv. Math. 229 (2012), no. 3, 1640–1642. 
Theorem 2:
This is a triviality unless you fix some invariants. On the other hand for relative dimension $1$ and fixed genus this is not true. This is the geometric version of Shavarevich's conjecture and was first proved by Parshin:
Paršin, A. N.
Algebraic curves over function fields. I. (Russian) 
Izv. Akad. Nauk SSSR Ser. Mat. 32 1968 1191–1219,
and then in a more general case by Arakelov:
Arakelov, S. Ju.
Families of algebraic curves with fixed degeneracies. (Russian) 
Izv. Akad. Nauk SSSR Ser. Mat. 35 (1971), 1269–1293. 
In higher dimensions, the statement is true indeed by taking the product of an arbitrary family of curves and an arbitrary curve (each of the appropriate genus). The second curve can be moved in moduli which gives even a continuous family of families.
In fact, this was what led to the notion of strong isotriviality.
Some relevant references are:
Kovács, Sándor J.(1-WA)
Strong non-isotriviality and rigidity. Recent progress in arithmetic and algebraic geometry, 47–55, 
Contemp. Math., 386, Amer. Math. Soc., Providence, RI, 2005.
Kovács, Sándor J.(1-WA)
Subvarieties of moduli stacks of canonically polarized varieties: generalizations of Shafarevich's conjecture. Algebraic geometry—Seattle 2005. Part 2, 685–709, 
Proc. Sympos. Pure Math., 80, Part 2, Amer. Math. Soc., Providence, RI, 2009. 
Kovács, Sándor J.(1-WA); Lieblich, Max(1-WA)
Boundedness of families of canonically polarized manifolds: a higher dimensional analogue of Shafarevich's conjecture. (English summary) 
Ann. of Math. (2) 172 (2010), no. 3, 1719–1748. 
Zsolt Patakfalvi
Arakelov-Parshin rigidity of towers of curve fibrations, connections to the infinitesimal Torelli problem
http://arxiv.org/abs/1010.3069
Theorem 3:
as I explained above, even this is not true without the "strong" assumption.
For strongly non-isomorphic families it is proven in
Kovács, Sándor J.(1-WA); Lieblich, Max(1-WA)
Boundedness of families of canonically polarized manifolds: a higher dimensional analogue of Shafarevich's conjecture. (English summary) 
Ann. of Math. (2) 172 (2010), no. 3, 1719–1748. 
I would expect it to be true for a somewhat larger class of families, but the actual class still needs to be defined. The key modulo this paper is rigidity.
For more details see 
Kovács, Sándor J.(1-WA)
Subvarieties of moduli stacks of canonically polarized varieties: generalizations of Shafarevich's conjecture. Algebraic geometry—Seattle 2005. Part 2, 685–709, 
Proc. Sympos. Pure Math., 80, Part 2, Amer. Math. Soc., Providence, RI, 2009. 
or 
Chapter III of
Hacon, Christopher D.(1-UT); Kovács, Sándor J.(1-WA)
Classification of higher dimensional algebraic varieties. 
Oberwolfach Seminars, 41. Birkhäuser Verlag, Basel, 2010. x+208 pp. ISBN: 978-3-0346-0289-1<|endoftext|>
TITLE: Does Taranovsky's system of ordinal notations make sense?
QUESTION [28 upvotes]: Dmytro Taranovsky has a Web page on which he claims to define a system of ordinal notations strong enough to provide an ordinal analysis of full second-order arithmetic.  I think (perhaps unjustly) that this claim is suspicious, since from my passing acquaintance with the subject I seem to understand that the state of the art of ordinal analysis was around $\Pi^1_2$-comprehension (e.g., Jan-Carl Stegert's doctoral dissertation building on work by Michael Rathjen), the ordinal notation systems involved are considerably more complex (reflection instances, collapsing hierarchies), and Taranovsky mentions none of this.  On the other hand, a superficial look at his page does seem to make some kind of sense, and my interest in the subject is to choose the largest possible system of ordinal notations which isn't too fastidious to implement on a computer (i.e., I'm not concerned with the proof-theoretic aspect).
So before I decide to read it in great detail or not, I'd like an expert's opinion: what is to be thought of Taranovsky's ordinal notation systems?  (Might they define an ordinal which is not as large as claimed?  Or perhaps which could be as claimed but would be very difficult to analyse?)

REPLY [7 votes]: In this section of the paper Taranovsky gave a proof of well-foundedness of the system for the $C_0,C_1$ and $C_2$ subsystems of the version of $C$ defined in "Built-from-below with Passthrough for Lower Levels", and then, using a different approach, does so for $C_i$ for all $i$. The proof for $C_0,C_1,C_2$ is in ZFC, while the proof for $C_i$ is in ZFC + Measurable Cardinal. It's important to note that the system up to $C_\omega$ is (probably) bounded by $C(C(\Omega_2\omega,0),0)$ for the main system. People often confuse $C_i(\alpha,\beta)$ with the $n=i$ system of the main $C$. "Built from below with passthrough" C is believed to be bounded by $C(C(\Omega_2\omega,0),0)$ because for each $i$, the limit of everything constructible in $C_i(\alpha,\beta)$ for $\alpha,\beta$ are standard in the language of $\bigcup C_i$, where all ordinals $\eta$ standard in $C_n$ are built from below with passthrough by $C_{n+1}$ standard ordinals above $\eta$, seems to be $C(C(C^{i+1}(\Omega_2,\Omega_2),0),0)$ in it's standard representation where $C^0(\alpha,\beta)=\beta\land C^{n+1}(\alpha,\beta)=C(\alpha,C^n(\alpha,\beta))$. If this is true, it would imply that the limit of everything constructible in $C_i$ is $C(C(\Omega_2i+\Omega_2,0),0)$. For lower values of $i$, this seems to hold. $C_0$ is only allowed to use ordinals build from below by other ordinals within it, so in can only hold finite expressions and nothing of the form $C(\alpha,\beta)$ for uncountable ordinals $\alpha,\beta$, and is therefore limited by the least $\alpha$ such that $\alpha\mapsto C(\alpha,0)$, which happens to be $\varepsilon_0$. In the main notation, $\varepsilon_0=C(\Omega_1,0)$ which is in its standard representation in the $n=1$ system and is in the standard representation $C(C(\Omega_2,0),0)$ in the $n=2$ system. Simple enough analyses can show that the limit to everything constructible in $C_1$ is $C(C(\Omega_2 2,0),0)$ and Taranovsky showed a brief conjecture and explanation of how $C(C(\Omega_2 3,0),0)$ is possibly the limit to everything expressible in $C_2$, but a full induction for $\sup\{C_i(\alpha,\beta):\alpha,\beta\in \bigcup C_i\cap\text{Stand}\}=C(C(\Omega_2\cdot(i+1),0),0)$ implying the same sentence for $C_{i+1}$ has not been proven yet.
It's interesting that Taranovsky wasn't able to provide a proof for $C_i$ for all $i$ within ZFC and had to resort to assuming the existence of Measurable Cardinals in order to prove well-foundedness for $C_i$. Perhaps such proof doesn't exist, or perhaps it does exist but it's very long and complicated and hard to think of, possibly stretching the limits of ZFC. If a theory $T$ can prove well foundedness of a recursive ordinal notation, then the limit of that notation is an ordinal below the proof-theoretic ordinal of $T$. If ZFC can't prove $C_\omega$ recursive/well-founded and the language of $C_i$ is limited by $C(C(\Omega_2i+\Omega_2,0),0)$ are both correct, then that would imply that $C(C(\Omega_2\omega,0),0)$ is equal to or larger than the proof-theoretic ordinal of ZFC. This is highly unlikely, because a proof in ZFC for $C_i$ most likely exist and not even Taranovsky himself has given such high possible values for $C(C(\Omega_2\omega,0),0)$, which was actually proposed to be the proof-theoretic ordinal of Second Order Arithmetic by the most recent analysis.
Originally, Taranovsky believed that the limit of each $n+1$ system is within the range of $\vert\Pi^1_n-\text{CA}_0\vert_\text{Con}$ and $\vert\Pi^1_{n+1}-\text{CA}_0\vert_\text{Con}$, which is now considered wrong, and the overall system is believed to be much stronger. This is where the position of the "n-shiftedness" property comes in. This is perhaps the most important property when it comes to the strength of Taranovsky's $C$ and is one of the reasons it's so hard to analyse.
Firstly, since the ordinal notation uses a simple comparison algorithm, the strength of the notation depends almost entirely on one factor - how strong "n-built from below" works for sufficiently large ordinals. In basic terms, $\alpha$ is 0-built from below by $\beta$ iff $\alpha<\beta$ and $\alpha$ is $k+1$-built from below by $\beta$ iff the standard representation of $\alpha$ does not use ordinals above $\alpha$, except for ordinals in the scope of an ordinal that is $k$-built from below by $\beta$. We say an ordinal $\alpha$ is in the scope of an ordinal $\beta$ iff $\alpha$ appears somewhere in the standard representation of $\beta$ within the lexicographic string in the language $\{C,0,\Omega_n\}$ that represents precisely $\beta$ within the respective $n$ system.
The main system was intended to be an attempt to extend a notation Taranovsky previously made called "Degrees of Reflection", defined using a similar (but not identical) built-from-below condition, and the $n=2$ system was intended to be identical to it, namely a term in Degrees of Reflection would have roughly the same behavior as that term with all instances of $\Omega$ replaced with $\Omega_2$. But in 2014, Taranovsky discovered the $n=2$ main system is stronger, because of an irregular phenomenon happening as long as the left argument of $C$ is larger than $\Omega_2$: in many cases replacing some of Degrees of Ref.'s $\Omega$ with just $C(\Omega_2,C(\Omega_2 2,0))$ is enough. For example, $C(\Omega^\Omega,0)\, =\, C(\Omega+\Omega^\Omega,0)$ in Degrees of Ref. is expected to have similar behavior to $C(\Omega_2+C(\Omega_2,C(\Omega_2 2,0))^{C(\Omega_2,C(\Omega_2 2,0))},0)$ in the main system. Note that this fails when the left argument of the main system C is less than $\Omega_2$, for example $C(C(\Omega_2,C(\Omega_2 2,0))^{C(\Omega_2,C(\Omega_2 2,0))},0)$ (no leading "$\Omega_2+$") fails the built from below condition - this is because the built from below condition becomes sufficiently lenient enough for this to happen when $\alpha$ is large enough, since we then get subterms $<\Omega_2$ such as $C(\Omega_2,C(\Omega_2 2,0))$ "for free". Also, this phenomenon would repeat itself at higher levels in the system such as for terms like $C(\Omega_2+C(\Omega_2,C(\Omega_2 3,0),0)$. In Taranovsky's words "similar patterns repeat at different levels of the strength hierarchy." And all this is only considering the $n=2$ system!
Overall, as it turned out that n-built from below was actually stronger than expected for ordinals standard in the $n=2$ system, Taranovsky had to form a new approximation method for n-built from below, according to which, the entire notation might reach as far as the proof-theoretic ordinal of $Z_2+\text{PD}$. None of this has been confirmed, and analyses have not gone beyond $C(C(\Omega_2 2+C(\Omega_2+C(\Omega_2,C(\Omega_2 2,0))^{C(\Omega_2,C(\Omega_2 2,0))^\omega},0),0),0)$ due to exhaustion with other ordinal notations not being sufficiently strong to provide ordinals beyond that, in order to be compared further.
Unfortunately, since a couple years ago some unusual behavior that seems to be unique to the main system has been found, which makes it harder to compare with other ordinal notations. Hyp cos, the viewer mentioned a few times on Taranovsky's page, has found what they call an "erratic term" in the main system, due to how it behaves differently from not only other ordinal notations appearing in literature, but also even many of Taranovsky's other systems:

Let $a = C(\Omega_2+C(\Omega_2,\Omega_1),0)$, then $C(a+C(C(\Omega_2+C(a,\Omega_1),0),C(a,\Omega_1)),0)$ is not standard. It's standard in "Main ordinal notation system with passthrough", and "degrees of reflection", "degrees of reflection with passthrough". This is the very first erratic term.

Sadly, other than Taranovsky's notation being hard to compare with, the other reason it has not been done so much is because it hasn't yet gotten much attention from people qualified enough to do so.<|endoftext|>
TITLE: Forcing in Homotopy Type Theory
QUESTION [16 upvotes]: I apologize if this question doesn't make any sense. I'll just go ahead and delete it if that's the case.  But the question is just the title.  Is there a notion of forcing in homotopy type theory? Presumably we do homotopy type theory in some $(\infty,1)$-topos, so we can axiomatize the notions accordingly? Does anyone know of a reference for this kind of thing if it does exist or makes sense?
Thanks

REPLY [24 votes]: In as far as we regard forcing as forming internal sheaves, the question is asking how to say "internal category of sheaves" in homotopy type theory. 
It is expected that this works in directed analogy with the situation in ordinary type theory by considering internal sites, except that there is a technical problem currently not fully solved: in homotoy type theory an internal category is necessarily an internal (infinity,1)-category and in order to say this one needs to be able to say "(semi-)simplicial object in the homotopy type theory". This might seem immediate, but is a little subtle, due to the infinite tower of higher coherences involved. For truncated internal categories one might proceed "by hand" as indicated here. A general formalization has recently been proposed by Voevodsky -- see the webpage UF-IAS-2012 -- Semi-simplicial types -- but this definition does not in fact at the moment work in homotopy type theory. (Last I heard was that Voevodsky had been thinking about changing homotopy type theory itself to make this work. But this is second-hand information only, we need to wait for one of the IAS-HoTT fellows to see this here and give us first-order news on this.)
However, that all said, there is something else which one can do and which does work: if an $\infty$-topos is equipped with a notion of (formally) étale morphisms, then one can speak about internal sheaves over the canonical internal site of any object without explicitly considering the internal site itself: the "petit (infintiy,1)-topos" of sheaves on a given object $X \in \mathbf{H}$ is the full sub-(oo,1)-category of the slice over $X$ on the (formally) étale maps
It may seem surprising on first sight, but this can be saiD in homotopy type theory and it can be said naturally and elegantly:
For this we simply add the axioms of differential cohesive homotopy type theory to plain homotopy type theory. This means that we declare there to be two adjoint triples of idempotent (co)monadic modalities called
shape modality $\dashv$ flat modality $\dashv$ sharp modality
and
reduction modality $\dashv$ infinitesimal shape modality $\dashv$ infinitesimal flat modality .
Using this we say: a function is formally étale if its naturality square of the unit of the "infinitesimal shape modality" is a homotopy pullback square. Then we have available in the homotopy type theory the sub-slice over any $X$ on those maps that are formally étale. This is internally the $\infty$-topos of $\infty$-stacks over $X$, hence the "forcing of $X$" in terms of the standard interpretation of forcing as passing to sheaves.
What I just indicated is discussed in detail in section 3.10.4 and 3.10.7 of the notes
Differential cohomology in a cohesive $\infty$-topos
( http://ncatlab.org/schreiber/show/differential+cohomology+in+a+cohesive+topos )
For more details on the above axioms of cohesive homotopy type theory see the first section of my article with Mike Shulman:
Quantum gauge field theory in Cohesive homotopy type theory
( http://ncatlab.org/schreiber/show/Quantum+gauge+field+theory+in+Cohesive+homotopy+type+theory )<|endoftext|>
TITLE: Spaces parametrizing ramified covers of surfaces
QUESTION [5 upvotes]: Let $\Sigma$ be a surface (let's say oriented and of finite type). We can consider the configuration space $F(\Sigma,n)$ of $n$ ordered distinct points on $\Sigma$, i.e. $\Sigma^n\setminus \Delta$ where $\Delta$ is the "big diagonal". The cohomology of $F(\Sigma,n)$ can be computed in a very explicit way, as explained in Burt Totaro's paper "Configuration spaces of algebraic varieties": the Leray spectral sequence for $F(\Sigma,n) \hookrightarrow \Sigma^n$ degenerates after the first differential and can be written down in a concrete way, which gives us a completely explicit differential graded algebra whose cohomology is $H^\bullet(F(\Sigma,n))$. 
Now fix a finite group $G$ and consider the space $F(\Sigma,G,n)$ which parametrizes $n$ points on $\Sigma$ and a principal $G$-bundle over the complement of the $n$ points on $\Sigma$. So we have a finite sheeted covering $F(\Sigma,G,n) \to F(\Sigma,n)$ such that the fiber over the point $(s_1,\ldots,s_n) \in \Sigma^n$ is the set $\mathrm{Hom}(\Pi_1(\Sigma \setminus \{s_1,\ldots,s_n\}),G)/G$, where $G$ acts on the set of maps by conjugation. 
Q1. Is there a good way to describe or compute the cohomology of $F(\Sigma,G,n)$?
Q2. (A vaguer question.) Let's say $\Sigma$ is compact for simplicity and let $FM(\Sigma,n)$ be the Fulton-MacPherson compactification of $F(\Sigma,n)$. I think there is a natural compactification
$$F(\Sigma,G,n) \hookrightarrow FM(\Sigma,G,n)$$
covering $F(\Sigma,n) \hookrightarrow FM(\Sigma,n)$ and such that the covering $FM(\Sigma,G,n) \to FM(\Sigma,n)$ ramifies along the boundary; $FM(\Sigma,G,n)$ should parametrize principal $G$-bundles which are allowed to ramify over the nodes. Is there a natural smaller compactification of $F(\Sigma,G,n)$ which is still smooth, analogous to the compactification $$F(\Sigma,n) \hookrightarrow \Sigma^n$$ (which of course is smaller than the Fulton-MacPherson)?

Craig Westerland suggests an alternative description of the cohomology of $F(\Sigma,G,n)$. The covering $p \colon F(\Sigma,G,n) \to F(\Sigma,n)$ satisfies $R^ip_\ast\mathbf Z =0$ for $i>0$, so 
$$ H^\bullet(F(\Sigma,G,n),\mathbf Z) = H^\bullet(F(\Sigma,n),p_\ast\mathbf Z). $$
Now $F(\Sigma,n)$ is a $K(\pi,1)$ space where $\pi$ is by definition the pure braid group on $n$ strands of the surface $\Sigma$, $P_n(\Sigma)$. Hence this cohomology is given by 
$$ H^\bullet(P_n(\Sigma), \mathbf Z [ \mathrm{hom}(\pi_1(\Sigma \setminus \{s_1,\ldots,s_n\},G)/G]).$$
Also the action of $P_n(\Sigma)$ is the restriction of an action of the $n$-strand surface braid group $B_n(\Sigma)$ (i.e. where the points are unordered). 
The simplest example should be $\Sigma = \mathbf R^2$, where we get the usual pure braid group. Since the fundamental group of the punctured plane is free we find that $\mathrm{hom}(\pi_1(\Sigma \setminus \{s_1,\ldots,s_n\},G)/G = G^n/G$, where $G$ acts by elementwise conjugation on $G^n$. The action of the braid group $B_n$ can be written down in terms of Artin's generators $\sigma_i$: the element $\sigma_i$ acts by 
$$ (g_1,\ldots,g_i,g_{i+1},\ldots,g_n) \mapsto (g_1,\ldots,g_ig_{i+1}g_i^{-1},g_i,\ldots,g_n)$$
(which is well defined on equivalence classes modulo conjugation). Then I guess even these cohomology groups are hard to compute in general?

REPLY [4 votes]: As Jason Starr mentions, the set of components of this space is somewhat difficult to pin down.  I will work over the Riemann sphere $\Sigma =S^2$ for simplicity (it's already complicated enough).  If you restrict your focus to the subspace $CF(\Sigma, G,n)$ consisting of branched covers $G$-covers which are connected, then for large $n$, there is a description of the set of components in terms of a certain quotient of the $H_2(G)$ by commutators.  When $G = S_n$ (and monodromy is constrained to be given by transpositions), this case was studied by Clebsch and Hurwitz, with the result being that the space is connected.  For more general $G$, this is an unpublished result of Conway-Parker, which was expanded upon by Fried-Völklein in the appendix to their paper "The inverse Galois problem and rational points on moduli spaces."
Computing the cohomology of this space is quite difficult.  It follows from your description above that the cohomology is the same as the cohomology of the mapping class group of the surface being fibred over, with (nontrivial twisted) coefficients in the free abelian group generated by $Hom(\pi_1(\Sigma \setminus \{ s_1, \dots, s_n\}), G)$; i.e.,
$$H^\ast(F(\Sigma, G, n)) \cong H^\ast(mcg(\Sigma); \mathbb{Z}Hom(\pi_1(\Sigma \setminus \{ s_1, \dots, s_n\}), G)).$$
Note that you can identify the cohomology of individual components as the group cohomology with coefficients in the subrepresentation generated by a given orbit of homomorphisms.
Abstractly, this is nice, but it doesn't actually tell you what the groups are.  One thing that you could hope for is a form of homological stability (i.e., that the homology of a single component stabilizes as $n \to \infty$), as it does for the configuration space (here I'm using the unordered configuration space, not the ordered one).  Then, if you figure out the limiting homology, you can at least compute some of the homology groups of $F(\Sigma, G, n)$ (those to which your stability theorem applies).  
To advertise some recent work of Ellenberg, Venkatesh, and myself, in "Homological stability for Hurwitz spaces and the Cohen-Lenstra conjecture over function fields, I and II," we have proven a rational homological stability result for certain choices of $G$ (also, we often restrict the sort of monodromy around branch points that is allowed to lie in a given collection of conjugacy classes), as well as giving a method of determining the limiting homology.
It's far from an optimal answer: we can give you partial, rational information only under certain assumptions on $G$.  Also, if you're really after the ordered moduli space (we treat the unordered), our results don't quite answer your question (though I think that you can bootstrap them to get some information at least).  Any improvements in the computations of the cohomology of these spaces would be very interesting indeed.<|endoftext|>
TITLE: Connection between eigenvalues of matrix and its Laplacian.
QUESTION [9 upvotes]: Hello!
There are two definitions of graph spectrum:
1) Eigenvalues of adjacency matrix $A$.
2) Eigenvalues of Laplacian of adjacency matrix ($L$).
Different sources offer different properties based on this two definitions.
Of course it's painful to compute two different spectrums if adjacency matrix is big.
So, the question is: 
Is there a method to connect one vector of eigenvalues ($\Lambda(A)$) with another ($\Lambda(L)$)?
It is obvious that 

$L = T^{-1/2}(T-A)T^{-1/2} = E - T^{-1/2}AT^{-1}T^{+1/2}$, where 
$T$ is the diagonal matrix with $t_{v,v}=d_v$, and $t_{u,v}=0$, if $u\ne v$, 
and $t^{-1}_{v,v}=0$, if $d_v=0$, 
$d_v$ - degree of $v$. 
Also, when $G$ is $k$-regular, $L=I-\frac{1}{k}A$, so $\Lambda(L)=1-\frac{1}{k}\Lambda(A)$.
But in general case it's like I need to compute eigenvalues of $AT$, if I know eigenvalues of $A$. ($T$ is diagonal).
Thanks for any help.

REPLY [8 votes]: Essentially your question is equivalent to asking for the relation between the spectrum of $A+D$ and $A$, where are $A$ is symmetric, $D$ is diagonal and both matrices are real.
And, by change of basis, this is equivalent to asking for the relation between the spectrum of $A+B$ and $A$ when $A$ and $B$ are real symmetric. A lot of thought has been given to this question. The short summary is that eigenvalues of $A$ provide no information useful towards computing the eigenvalues of $A+D$.
It might also be worth pointing out that there are many more than two definitions of graph spectrum (normalized and unsigned Laplacian, Seidel, spectrum of complement, to mention four more that come quickly to mind). All of these provide the same information for regular graphs, but in no case does computing one provide much help in computing another.<|endoftext|>
TITLE: Structure of f.g. modules over a non-commutative ring
QUESTION [5 upvotes]: To what extent is the structure theorem for finitely generated modules over principal ideal domains true over non-commutative domains? I'm in particular interested in non-commutative euclidean domains especially the twisted polynomial ring $K\langle X \rangle$ over a field $K$ (i.e. such that $Xa = \sigma(a)X$ for some automorphism $\sigma$ of K).

REPLY [3 votes]: The question is thoroughly explored in Chapter 3 of Nathan Jacobson's Theory of Rings.  I took a quick look, and it looks like the analogous results go through in the noncommutative case.  For example, Theorem 19 in Chapter 3 states that a finitely-generated module over a noncommutative principal ideal domain is a direct sum of cyclic modules.<|endoftext|>
TITLE: Are there analogous statements for the number of zeros of a section in terms of the Euler class, even when the relevant spaces are not manifolds?  
QUESTION [8 upvotes]: Let $V \rightarrow M$ be an oriented rank $k$ vector bundle over a compact orientd manifold $M$. Let $X \subset M$ be a 
compact topological subspace of $M$ that is a smooth oriented submanifold of 
dimension $k$, except possibly at a a set of points that have 
``dimension'' less than or equal to $k-2$.  More precisely, the set of singular points is contained inside a 
submanifold of dimension $k-2$ or less. 
Let 
$s :X \rightarrow V$ be 
the restriction of a smooth section from $M$ to $V$. 
Assume that when restricted to $X$, the section vanishes only on the smooth points of $X$, 
and it vanishes transversally. Is it true that the number of zeros 
of $s$ inside $X$, counted with a sign is the Euler class of $V$ evaluated on the fundamental class of 
$X$ , ie 
$$ +-|s^{-1}(0)| = \int_{[X]} e(V)  $$ 
We need $X$ to have singularities of dimension $k-2$ or less, to 
ensure that it defines a homology class $[X]$ (ie to make sure 
that the integration actually makes sense). Note that $[X]$ is 
an element of $H_k(M, \mathbb{Z})$  and $e(V) \in H^k(M, \mathbb{Z})$. 
So the expression makes perfect sense, even though $X$ is a singular space. 
I believe this statement is true, but is there a reference for this 
fact?

REPLY [3 votes]: You make several assumptions, one being that $X$ is a stratified space, carrying an orientation class.  The answer to you question is yes,  under more restrictive assumptions.   Here they are.

The  section $s$ vanishes transversally along $M$.
The zero set  $s^{-1}(0)$ intersects the stratified space $X$ transversally. 

The complete rigorous proof  is a bit more involved, and  the clearest argument I know  is sheaf theoretic, and it involves a sheaf theoretic  version of the Poincare duality, also known as  Verdier duality.  All  the information you need you can find in B. Iversen's book Cohomology of Sheaves, especially chapters IX and X.
Addendum I realize  you do not need these stringent    conditions.      Denote by $V_X$ the restriction of $V$ to $X$ and by $\tau_X$ its  Thom class viewed as a class in the local cohomology of $V_X$ along $X$, $\tau_X\in H^k_X(V_X)$ (integer coefficients). We can then arrange that $\tau_X$ has support in a tiny neighborhood  of $X$ in $V_X$. Then $e(V_X)=s^*\tau_X\in H^k(X)$ is supported in a tiny open  neighborhood $N$ of $s^{-1}(0)\cap X$ in $X$, i.e., $e(V_X)$ is in the image of $H^k(X, X\setminus N)$ in $H^k(X)$.  Now use the technology in Iversen to conclude that 
$$ \langle e(V_X), [X]\rangle  =\sum_{s(x)=0} \epsilon(x), $$
where $\epsilon(x)\in\{\pm 1\}$ is the local Euler number of $S$ at $x$, and $\langle-,-\rangle $ denotes the pairing between  cohomology and homology.   (This is a bit long.)<|endoftext|>
TITLE: Ensuring nonempty lightface Borel sets have elements via theories of second-order arithmetic
QUESTION [9 upvotes]: This question is an outgrowth of this MathSE question: https://math.stackexchange.com/questions/276068/members-of-lightface-borel-sets. 
A Borel set $X\subseteq 2^\omega$ is a member of the smallest collection of subsets of $2^\omega$ closed under complementation and countable union. As such, Borel sets come with "codes:" well-founded, potentially infinitely-branching subtrees of $\omega^{<\omega}$ interpreted as instructions for how to combine the basic open sets corresponding to the "leaves" to form the desired Borel set. Such a subtree is codeable by a set of natural numbers, so we can talk about the Turing degree of a code for a Borel subset of $2^\omega$. A Borel set $X$ is lightface Borel if it has a computable Borel code. More usefully, the class of Borel sets is naturally stratified into $\omega_1$ many pairs of levels, with open and closed sets forming the bottom pair of levels, ${\Sigma^0_1}$ and ${\Pi^0_1}$, respectively. There are notions of codes for, and analogous lightface versions of, each of the levels ${\Sigma^0_\alpha}$ and ${\Pi^0_\alpha}$ (for $\alpha$ a computable ordinal, anyways).
According to the Low Basis Theorem, every nonempty lightface $\Pi^0_1$ subset of $2^\omega$ has a low member. Trying to push this up the Borel hierarchy leads to problems: already at the level of $\Pi^0_2$, there are computable codes for sets with no hyperarithmetic members. We still have a basis theorem, though: each nonempty lightface $\Pi^0_\alpha$ subset of $2^\omega$ has a member computable in Kleene's $\mathcal{O}$.
But in the case of lightface $\Pi^0_1$ sets, we have - in addition to the Low Basis Theorem - a "proof-theoretic" basis theorem: if $\mathcal{M}$ is an $\omega$-model of $WKL_0$ (a particular subsystem of second-order arithmetic), then every nonempty $\Pi^0_1$ subset of $2^\omega$ which has a code in $\mathcal{M}$ has a member in $\mathcal{M}$ (and the converse holds: if $\mathcal{M}$ has this property, and is an $\omega$-model of $RCA_0$, then $\mathcal{M}\models WKL_0$).
Trying to get an analogous proof-theoretic basis theorem for higher levels of the Borel hierarchy, one might want to close under hyperjump (the map $h: X\mapsto \mathcal{O}^X$); the relevant subtheory would seem to be $\Pi^1_1-CA_0$. However, it turns out that an $\omega$-model $\mathcal{N}$ of $RCA_0$ is closed under the hyperjump iff it is a model of $\Pi^1_1-CA_0$ and is also a $\beta$-model, that is, everything $\mathcal{N}$ thinks is a well-order is actually a well-order; and this latter condition cannot be phrased in a first-order manner. What I want to know is whether we can in any way remove the need to restrict attention to $\beta$-models.
My question is the following: for which $n\in\omega$ is there a first-order theory $T_n$, in the language of second-order arithmetic (and consisting only of true sentences, to avoid trivialities), such that whenever $\mathcal{M}\models T_n$, $\mathcal{M}$ is an $\omega$-model, and $X\subseteq 2^\omega$ is a nonempty $\Pi^0_n$ set with a code in $\mathcal{M}$, $X$ has a member in $\mathcal{M}$?
(Note: it's perfectly reasonable that there should be no such $T_n$ for $n$ sufficiently large - even $n=2$ - since we're already sort of cheating at the level $n=1$ by restricting attention to $\omega$-models, which is again not a first-order thing. We need to, on the face of it, in order to make the question make much sense, but there's still a sense in which non-first-order-ness has already snuck in. My suspicion is in fact that for $n>1$ there is no such $T_n$, but I don't know how to show that.)

REPLY [4 votes]: Given a tree $T \subseteq \omega^{\lt\omega}$, the statement "$B$ is an infinite path through $T$" is $\Pi^0_2$. Therefore, if $\mathcal{M}$ is such that every nonempty $\Pi^0_2$-class coded in $\mathcal{M}$ has a point in $\mathcal{M}$, then $\mathcal{M}$ is necessarily a $\beta$-model. So the models you want are precisely the $\beta$-models of $\Pi^1_1$-$\mathsf{CA}_0$.<|endoftext|>
TITLE: Commutator length modulo finite index subgroups
QUESTION [8 upvotes]: We write $cl$ for the commutator length, i.e. the least number of commutators which multiply to a given element of a group. 
Given an element $g$ in the commutator subgroup of the free group $G=F_2$ on two generators, is it true that 
$$cl_G(g) = \displaystyle \max_{\mbox{H < G finite index normal}} cl_{G/H} (g \mod H)$$ ?

REPLY [11 votes]: I think the answer is no and that actually the "finite quotient" commutator length (defined by your formula) is bounded on $[F_2,F_2]$.
Indeed by Nikolov-Segal, in the profinite completion $P$ of $F_2$, the derived subgroup $[P,P]$ is closed; since the set $C$ of commutators is compact, it follows by a Baire argument that $[P,P]$ is boundedly generated by $C$ (first get by Baire that some open neighborhood of the identity of $[P,P]$ has bounded commutator length in $P$ and then use compactness). Observe on the other hand that $[F_2,F_2]=[P,P]\cap F_2$, which is essentially trivial since the abelian group $F_2/[F_2,F_2]$ is residually finite. So in any finite quotient of $F_2$, the commutator length of $w\in [F_2,F_2]$ is bounded by the universal number that arises as upper bound of the commutator length of $[P,P]$ in $P$.
On the other hand, the commutator length of $F_2=\langle x,y\rangle$ is unbounded on $[F_2,F_2]$, as the commutator length of $[x,y]^n$ grows linearly (I think it's due to Bavard).<|endoftext|>
TITLE: What is the probability that a random subset of a finite group is generic?
QUESTION [8 upvotes]: Definition 1: Given a group $G$, a subset $X \subseteq G$, and a natural number $k$,
we say that $X$ is (left) $k$-generic in $G$ if there are $k$ many left translates of $X$ that cover $G$.
That is, if there exist $a_1, \dots, a_k \in G$ such that $G = \bigcup_i a_i \cdot X$ (where $a_i \cdot X = \{ a_i \cdot x: x \in X \}$).
Definition 2: Let $n$ be a natural number and $G$ be a group with $n$ elements.
On the family of subsets of $G$ we put the probability that gives to each subset probability $1/2^n$; equivalently, for every $g \in G$ we toss a coin to decide whether or not $g$ is in a given set.
Let $P_k(G)$ be the probability that a random subset of a group $G$ is $k$-generic in $G$.
Problem: I want to show that, for a fixed $k$, $P_k(G) \to 0$ as the cardinality of $G$ goes to infinity, independently of the choice of the group $G$ (actually, I need only the case $G = \mathbb Z / n \mathbb Z$).
I expect that the above is a known fact: I would appreciate a reference for it (my background is in logic).

REPLY [7 votes]: I think your conjecture is right. I hadn't heard of this fact before, so can't provide a reference. Here is my attempt at a proof.
The idea is that I want to reverse things: Fix a set $A=\{a_i\}$ with $|A|=k$ and show that the probability that a set $X$ is $A$-generic (i.e. $AX=G$) is exponentially small in $n$ (uniformly in $A$). On the other hand, there are only polynomially many ($\approx n^k$) $A$'s, so the probability that $X$ is $k$-generic is bounded above by $n^k\times$(exponentially small in $n$).
Let's do this in detail. Let $B=AA^{-1}$ (i.e. $\{a{a'}^{-1}\colon a,a'\in A\}$), so that $|B|\le k^2$. Now I claim that you can pick in $G$ elements $g_1,\ldots,g_m$, where $m=n/k^2$ such that $g_j\not\in\bigcup_{i < j}Bg_i$ (at the $j$th stage, there are at least $|G|-|B|(j-1)$ choices). 
The $A^{-1}g_i$ are now disjoint: if $A^{-1}g_i\cap A^{-1}g_j\ne\emptyset$ for $i < j$, then $g_j\in Bg_i$.
Now we compute for a random subset $X$: what is the probability that $AX\supset\{g_1,\ldots,g_m\}$?
Note that $AX$ contains $g_i$ if $X$ contains an element of $A^{-1}g_i$. The probability of this is $1-2^{-k}$. Since the sets $A^{-1}g_i$ are disjoint, the probability that $AX$ contains each of the $g_i$ is $(1-2^{-k})^m=(1-2^{-k})^{n/k^2}$. Hence the probability that $X$ is $k$-generic is at most $n^k(1-2^{-k})^{n/k^2}$, which tends to 0 as $n\to\infty$ for fixed $k$.<|endoftext|>
TITLE: Deformations of the punctured affine plane
QUESTION [18 upvotes]: Let $k$ be some field, algebraically closed and of characteristic $0$, if you like.
Let $U= \mathbb{A}^2_k \setminus \{ (0,0) \}$ be the punctured affine plane over $k$. Write $U$ as the union of $U_1 = \mathrm{Spec} k[x^{\pm 1},y]$ and $U_2 = \mathrm{Spec} k[x,y^{\pm 1}]$. Then deformations $U^\prime$ of $U$ to $k[t]/t^2$ are given in the following way:
We have $U^\prime = U_1^\prime\cup U_2^\prime$, where $U_1^\prime = \mathrm{Spec} (k[t]/t^2) [x^{\pm 1},y]$ and $U_2^\prime = \mathrm{Spec} (k[t]/t^2)[x^{\prime},y^{\prime \pm 1 }]$, and they are glued along an isomorphism of $\mathrm{Spec} (k[t]/t^2) [x^{\pm 1},y^{\pm 1}]$ with $\mathrm{Spec} (k[t]/t^2) [x^{\prime \pm 1},y^{\prime \pm 1}]$ which can be brought in a unique way into the form
$$
x^\prime = x + t (\sum_{i,j>0} a_{ij} x^{-i} y^{-j})\ ,\
y^\prime = y + t (\sum_{i,j>0} b_{ij} x^{-i} y^{-j})\ ,
$$
where both sums are finite. In particular, the simplest case is given by an isomorphism of the form
$$
x^\prime = x + atx^{-1}y^{-1}\ ,\
y^\prime = y + btx^{-1}y^{-1}\ ,
$$
where $a,b\in k$ are parameters.
Moreover, deforming $U$ is unobstructed, so there exists a formal lifting of $U$ to a formal scheme $\mathcal{U}$ over $\mathrm{Spf} k[[t]]$. My question is whether there exist algebraic deformations to $\mathrm{Spec} k[[t]]$. In other words:

Question: Does there exists a flat separated scheme $X$ of finite type over $\mathrm{Spec} k[[t]]$ together with an open subscheme $U^\prime\subset X\otimes_{k[[t]]} k[t]/t^2$ such that $U^\prime$ is a nontrivial deformation of $U$?

Note that up to now, I could not construct a single such example, but I couldn't figure out an obstruction either.
One attempt might be to find an affine scheme $X$. This essentially boils down to choosing a subring $R\subset k[x,y]$ which contains a power of the ideal $(x,y)$, and deforming $R$ to $k[[t]]$. One can check that for any given deformation $U^\prime$ to $k[t]/t^2$, one can find such a ring $R$ that lifts to $k[t]/t^2$, and gives rise to the deformation $U^\prime$ (upon removing the origin). However, typically $R$ will be obstructed, and it is far from clear that one can further find a lift of this $R$ to $k[[t]]$.
[Question edited to account for Angelo's comment. Sorry for a stupid second question!]

REPLY [12 votes]: I think I can prove the following

Theorem: Let $R$ be a $k$-algebra of finite type with a closed point $x\in \mathrm{Spec} R$ such that $\mathrm{Spec} R\setminus \{x\}\cong \mathbb{A}^2_k\setminus \{(0,0)\}$. Let $\tilde{R}$ be a $t$-adically complete flat lift of $R$ to $k[[t]]$. Then $\mathrm{Spf} \tilde{R}\setminus \{x\}\cong \mathbb{A}^2_{\mathrm{Spf} k[[t]]}\setminus \{(0,0)\}$.

As mentioned in the question, there do exist nontrivial deformations to $k[[t]]/t^n$ for any $n$; for that reason, I find this result surprising: Lifting to $k[[t]]$ poses a strong rigidity that I was not previously aware of.
In particular, there are no examples to the above question with $X$ affine. The proof below also shows that my intended application of algebraic deformations of the punctured affine plane cannot work, so I consider my question as answered in the negative (although strictly speaking the question posed is still open).
Let me start with something seemingly unrelated.

Proposition: Let $K$ be some field, let $X/K$ be a smooth surface, and let $C\cong \mathbb{P}^1\subset X$ be a smooth rational curve in $X$ with self-intersection $C^2 = 1$. Then there is an open subset $U\subset X$ containing $C$ and an open embedding $U\hookrightarrow \mathbb{P}^2$ (carrying $C$ into a line).

I deduced this (hopefully correctly) from the classification of surfaces (showing first that $X$ has to be rational); probably there is a more direct argument.
Now consider $Y=\mathbb{P}_k^2\setminus \{(0:0:1)\}$, with its line $C\subset Y$ at infinity. Let $R$ and $\tilde{R}$ be as above. Glue $Y$ with $\mathrm{Spec} R$ along $\mathbb{A}^2_k\setminus \{(0,0)\}$ to $\bar{Y}$ over $k$; this is projective, with $C$ an ample divisor. As $H^2(\mathbb{P}_k^2,T_{\mathbb{P}_k^2}(-C))=0$, one checks that any deformation of $R$ can be extended to a deformation of $\bar{Y}$ which also lifts $C$. In particular, $\tilde{R}$ can be extended to a proper flat formal scheme $\tilde{\bar{Y}}$ with a lift $\tilde{C}$ of $C$, over $\mathrm{Spf} k[[t]]$. By formal GAGA, this is algebraizable, to a projective surface $Z$ over $\mathrm{Spec} k[[t]]$, and $D\cong \mathbb{P}^1\subset Z$. The smooth locus of the generic fibre of $Z$, together with the generic fibre of $D$, satisfies the assumptions of the Proposition. It follows that there is a birational map from the smooth locus of the generic fibre of $Z$ to $\mathbb{P}^2_{k((t))}$. It cannot contract anything: If a curve $E$ was contracted, then this curve would meet the generic fibre of $D$ (as $D$ is ample), but in a neighborhood of $D$, the map is well-defined. Moreover, it cannot have critical points, as the target $\mathbb{P}^2_{k((t))}$ contains no exceptional curves. It follows that the smooth locus of the generic fibre of $Z$ is open in $\mathbb{P}^2_{k((t))}$. As its image contains an ample curve, it follows that the codimension of the image is at least $2$.
Now consider the reduction map $H^0(Z^{\mathrm{sm}},\mathcal{O}(nD))\to H^0(Y,\mathcal{O}(nC))$, where $Z^{\mathrm{sm}}\subset Z$ denotes the smooth locus; it induces an injection $H^0(Z^{\mathrm{sm}},\mathcal{O}(nD))/t\to H^0(Y,\mathcal{O}(nC))$. Both have the same dimension as $H^0(\mathbb{P}^2,\mathcal{O}(n))$. It follows that the map has to be surjective. It follows that $\mathrm{Proj} \bigoplus H^0(Z^{\mathrm{sm}},\mathcal{O}(nD))$ is a flat deformation of $\mathbb{P}^2_k$, thus is isomorphic to $\mathbb{P}^2_{k[[t]]}$. We get an open embedding $Z^{\mathrm{sm}}\hookrightarrow \mathbb{P}^2_{k[[t]]}$, which restricts to an isomorphism $\mathrm{Spf} \tilde{R}\setminus \{x\}\cong \mathbb{A}^2_{\mathrm{Spf} k[[t]]}\setminus \{(0,0)\}$, as desired.
Addendum: For completeness, I deduce the Proposition (in case $K$ is algebraically closed -- one can then remove this hypothesis) from the paper "Curves with high self-intersection on algebraic surfaces" of Hartshorne (I will use freely notation from there). We may assume that $X$ is proper. First, $X$ is rational: Fix a point $x\in C$. Because $C$ is very free, there is a $1$-parameter family of rational curves which contain $x$. As these curves intersect with multiplicity $1$, no two of them can have the same tangent vector at $x$. It follows that this family is defined over a rational base, thus $X$ is (uni-, and thus) rational. Now we use Theorem 3.5 in Hartshorne's paper. Case a) means that $C\subset X$ is equivalent to a section of a rational Hirzebruch surface $F_e\to \mathbb{P}^1$. The condition $C^2=1$ will then ensure that $e=1$, and that the curve $C$ does not meet the exceptional locus of $F_1\to \mathbb{P}^2$, giving the result. Case b) cannot occur. Case c) could cause trouble. Looking into the proof, only the case $m=2$ might occur. Looking at how the case $m=2$ comes about in Proposition 3.2, we see that we must have $e+n=1$. The case $e=0$, $n=1$ is impossible because of condition b) in Proposition 3.1, and the other case $e=1$, $n=0$ is impossible because of condition e) in Proposition 3.1.<|endoftext|>
TITLE: Elementary applications of linear algebra over finite fields
QUESTION [32 upvotes]: I'm teaching axiomatic linear algebra again this semester.  Although the textbooks I'm using do everything over the real or complex numbers, for various reasons I prefer to work over an arbitrary field when possible.   I always introduce at least $\mathbb{F}_2$ as an example of a finite field.  To help motivate this level of generality, I'd like to cover some application of linear algebra over finite fields.  Ideally it shouldn't make explicit reference to linear algebra or finite fields in its setup, and should require as little background as possible (the students have taken calculus, but not necessarily any other advanced math — in particular applications to group theory are out).  I've looked around a little, but haven't found anything so far that requires little enough overhead to fit into a single 50-minute lecture and wouldn't seem either too abstract or too arbitrary to motivate such students.  Any suggestions?
Alternatively, I'd be interested in elementary applications of linear algebra over any other field which isn't a subfield of $\mathbb{C}$.

REPLY [2 votes]: There is a Martin Gardner problem, reprinted in his Unexpected Hanging collection, that goes like this:

Miranda beat Rosemary in a set of tennis, winning 6–3.  There were five service breaks.  Who served first?

One solution is as follows.  The wins by the player who served first may be represented by a vector  $\mathbb{F}_2^9$ that is the sum of (1,0,1,0,1,0,1,0,1) and another vector of weight 5.  Such a vector must have even weight, so the player who served first won an even number of games.  Thus Miranda served first.
In the book, Gardner writes that his original solution was long and cumbersome, and that the shortest solution he received was by Goran Ohlin: "Whoever served first, served five games, and the other player served four.  Suppose the first server won $x$ of the games she served and $y$ of the other four games.  The total number of games lost by the player who served them is then $5-x+y$.  This equals $5$ [we were told that the non-server won five games].  Therefore $x=y$, and the first server won a total of $2x$ games.  Because only Miranda won an even number of games, she must have been the first server."  Though more elementary in some sense, this solution seems more ad hoc and less conceptual to me than the above argument using $\mathbb{F}_2$.<|endoftext|>
TITLE: Asymptotic expansion of the Schrödinger kernel?
QUESTION [7 upvotes]: My stackexchange post was somewhat unsatisfactory (also because I may not have stated clear enough what my interest was).  So here it goes!
Let $M$ be a compact Riemannian manifold and $\Delta$ be the Laplace-Beltrami operator. It is well-known that the solution operator to the heat equation $e^{t \Delta}$ is smoothing for $t>0$ and has a smooth integral kernel $k_t(x, y) \in C^\infty(M \times M)$. Furthermore, $k_t$ has an asymptotic expansion
$$ k_t(x, y) \sim \underbrace{(4 \pi t)^{-n/2} \exp \left( -\frac{1}{4t} \mathrm{dist}(x, y)^2 \right)}_{:= e_t(x, y)} \sum_{j=0}^\infty t^j \Phi_j(x, y) $$
meaning that
$$ \left| k_t(x, y) - e_t(x, y) \sum_{j=0}^N t^j \Phi_j(x, y) \right| \leq C t^{N+1}$$
uniformly in $x$ and $y$ in a neighborhood of the diagonal.
Now by by formally substituting $t \rightarrow it$, one gets the formal asymptotic series
$$ e_{it}(x, y) \sum_{j=0}^\infty (it)^j \Phi_j(x, y),$$
which has the property that it formally (i.e. termwise, as asymptotic series in $t$) solves the Schrödinger equation $ \left(i \frac{\partial}{\partial t} + \Delta\right)k_t = 0.$
Now my question is the following: Does this asymptotic series have any relation to the solution operator $e^{it\Delta}$ of the Schrödinger equation, or to its distribution kernel?

REPLY [4 votes]: No, in general, the expansion in small times of the heat kernel $k_t(x,y)$ does not tell much about the Schrodinger semigroup. 
You may think about the circle case for which the kernel of $e^{t \Delta}$ has an expansion at any order
$
k_t(x,y)=\frac{1}{\sqrt{4\pi t}} e^{-\frac{(y-x)^2}{4t}} (1 +O(t^m))$
but the Schrodinger semigroup $e^{it \Delta}$ has a rather complicated behavior. The Schrodinger kernel in that case is actually a distribution. If $t$ is a rational multiple of $2\pi$, it is a finite linear combination of delta functions  which is interestingly connected to Gauss sums.<|endoftext|>
TITLE: Algebraic definition of the Kuranishi map
QUESTION [8 upvotes]: Let $X$ be a smooth projective algebraic variety over an algebraically closed field $k$. If $k=\mathbb{C}$, we know by work of Kuranishi that the base of the versal deformation of $X$ is the germ at $0$ of the fiber over $0$ of a holomorphic map $K:H^1(X, T_X)\to H^2(X, T_X)$ (defined in the neighborhood of 0), called the Kuranishi map. 
This means that, if $S_i$ are the power series rings associated to $H^i(X, T_X)$ (i.e., the completions of $Sym^* H^i(X, T_X)^*$), there is a map $k: S_2 \to S_1$ such that $R = S_1 \otimes_{S_2} k$ pro-represents the deformation functor of $X$.
Question. Can one construct the map $k$ using algebraic methods? 
Probably one should assume that $k$ is of characteristic zero (or replace power series rings by completed divided power algebras...).

REPLY [7 votes]: In characteristic zero the answer is positive.
If $L^\bullet$ is the dgla (differential graded Lie algebra) governing your deformation problem (the Kodaira-Spencer dgla $\oplus A^{0,p}(X,T_X)$ in your example), the formal Kuranishi theorem states that
 for every splitting $\delta$ of $L^\bullet$, there exists a hull $Kur^\delta_{L^\bullet}\to Def_{L^\bullet}$,  the formal Kuranishi space. Here
$Def_{L^\bullet}:Art_{k}\to Sets$ is the deformation functor associated to $L^\bullet$.
A splitting of $L^\bullet$ is a degree $-1$ linear map, $\delta:L^\bullet\to L^\bullet[-1]$, such that $\delta^2=0$,
$d\delta d=d$, $\delta d\delta=\delta$. It plays the role of $d^\ast G$ in Hodge theory,
where $G$ is Green's operator and $d^\ast$ is the adjoint of $d$.
Manetti's article that Francesco mentions is a great introduction. Another reference that I am very fond of is Goldman and Millson's paper The Homotopy invariance of Kuranishi space,
where they compare the algebraic and analytic description of the hull.<|endoftext|>
TITLE: If M is not a free A-module, can tensoring with a bigger field make it free?
QUESTION [9 upvotes]: I need a simple commutative algebra lemma for a paper, but can't find a reference. Maybe I don't know the right keywords. Here's the setup. $F:K$ is a field extension, $A$ is an algebra over $K$, and $M$ an $A$-module. If $M \otimes_K F$ is a free $A \otimes_K F$ module, must $M$ have been a free $A$-module?
I can do the particular case I need: $K = \mathbb{F}_2$, $F$ a finite extension, $A = 1 \oplus N$ for $N$ nilpotent, $M \otimes_K F$ free of rank one. But I suspect that something more general holds, and it would be nice to quote rather than reprove.
Thanks!

REPLY [12 votes]: Yes, there are such examples even with invertible $M$ and smooth $K$-algebras of dimension 1, with $K$ any field that is not algebraically closed. 
Choose such a $K$, so we may and do also choose a nontrivial primitive finite extension $F$ of $K$.  Consider the projective line over $K$ and remove a closed point $\xi$ such that $F = K(\xi)$.  This is an affine open $U = {\rm{Spec}}(A)$ for a Dedekind $A$, and let $M$ be the maximal ideal of $A$ corresponding to an $K$-point.  This is invertible as an $A$-module (as for any Dedekind domain) but cannot be principal. Indeed, if $a \in A$ were a generator of $M$ then its divisor on $\mathbf{P}^1_K$ has restriction to $U$ that is a single point of degree 1 yet the condition ${\rm{deg}}_K({\rm{div}}(a)) = 0$ forces the "negative" part of the divisor to have degree $-1$ over $K$, contradicting that this negative part is supported at the closed point $\xi$ with $K$-degree larger than 1.
Clearly $A \otimes_K F$ is the coordinate ring of the complement in $\mathbf{P}^1_F$ of ${\rm{Spec}}(F \otimes_K F)$, which contains an $F$-point, so $A \otimes_K F$ is the coordinate ring of a non-empty affine open in the open complement $\mathbf{A}^1_F = {\rm{Spec}}(F[t])$ of an $F$-point in $\mathbf{P}^1_F$. Thus, $A \otimes_K F$ is the localization of the PID $F[t]$ at some nonzero element, so $A \otimes_K F$ is a PID and hence its nonzero ideal $M \otimes_K F$ (corresponding to a single $F$-point) is principal.

REPLY [8 votes]: The implication does not hold.
Let $K$ be the real numbers, $F$ the complex numbers, $A=K[X,Y,Z]/(X^2+Y^2+Z^2-1)$, and $M$ the kernel of the map $A^3\rightarrow A$ given by the unimodular row $(X,Y,Z)$.
Then $M$ cannot be free, by the same argument I gave in my answer to this question.
But $M$ becomes free after tensoring with the complex numbers.  To see this, write $U=(X+iY)/2$, $V=(X-iY)/2$.  Then it suffices to show that $(X,Y,Z)=(U+V,-iU+iV,Z)$ can be transformed via elementary operations to $(1,0,0)$ (so that its kernel is isomorphic to the kernel of $(1,0,0)$, which is evidently free).
To construct such a series of elementary transformations, first transform $(U+V,-iU+iV,Z)\rightarrow (2U,-iU+iV,Z)\rightarrow (2U,iV,Z)$ and then note that $2U$ is equal to 1 mod $(iV,Z)$.

REPLY [3 votes]: Example 4.7 in [Guralnick, Robert; Jaffe, David B.; Raskind, Wayne; Wiegand, Roger. On the Picard group: torsion and the kernel induced by a faithfully flat map. J. Algebra 183 (1996), no. 2, 420--455. MR1399035 (97c:14002)] is an example of an algebra $A$ —A Dedekind domain in fact— such that after an extension $K/k$ the map $Pic(A)\to Pic(A_K)$ has non-finitely generated kernel. Each element in the kernel gives a counterexample.<|endoftext|>
TITLE: Counting contracted curves
QUESTION [5 upvotes]: I'm trying to understand some aspects of Oguiso's example of a Calabi-Yau threefold of Picard number 2, described in the paper "Automorphism groups of Calabi-Yau manifolds of Picard number two".
Take $X \subset \mathbb P^3 \times \mathbb P^3$ to be the intersection of general hypersurfaces of bidegree $(1,1)$, $(1,1)$, and $(2,2)$.  The claim is that the projections $p_i : X \to \mathbb P^3$ have degree $2$ (clear enough), and contract "$(2(2+1)+2)^3 = 8^3$" rational curves.  I'm not sure how to arrive at this count.  Where does this number come from, and what can be said about the points of $\mathbb P^3$ over which there are positive-dimensional fibers?

REPLY [4 votes]: I set up the enumerative computation in a manner similar to Serge Lvovski, but I also get a different answer than the one in Oguiso's paper. Serge Lvovski's answer gives $4\times 4\times 6=96$ rather than $8^3$.  My answer gives $120$, rather than $8^3$.
The discrepancy may be due to the same issue as in my previous comment.  
Let $A$ and $B$ be the two global sections of $\mathcal{O}(1,1)$ and let $C$ be the global section of $\mathcal{O}(2,2)$.  Push forward by projection to the second $\mathbb{P}^3$ factor, i.e., consider $A$ and $B$ as global sections of the rank $4$ vector bundle $V:=H^0(\mathbb{P}^3,\mathcal{O}(1))\otimes_{\mathbb{C}} \mathcal{O}(1)$. 
 These two sections give a coherent subsheaf $S \cong \mathcal{O}\oplus \mathcal{O}$ of $V$.  The issue in my comment above is that the cokernel $V/S$ fails to be locally free at $4$ points of $\mathbb{P}^3$.
What is the image $I$ of the map $S\otimes_{\mathcal{O}} V \to \text{Sym}^2 V$, or equivalently, what is the kernel $K$?  Of course the image of $\bigwedge^2 S$ inside $S\otimes_{\mathcal{O}} S \subset S\otimes_{\mathcal{O}}V$ is in the kernel $K$.  I claim this is the entire kernel.  Since the kernel is a torsion-free subsheaf of a locally free sheaf, it can be computed at the generic point of $\mathbb{P}^3$, where the claim is basic linear algebra (in fact the same linear algebra works on the open subset that is the complement of the $4$ points where $V/S$ is not locally free).  So the image $I$ is a locally free sheaf of rank $7$, and it is a coherent subsheaf of $\text{Sym}^2V$.  As above, the cokernel $\text{Sym}^2 V/I$ fails to be locally free at the $4$ bad points.
Now form the direct sum of $I$ with the rank $1$ subbundle $T\cong \mathcal{O}$ of $\text{Sym}^2V$ 
generated by the global section $C$. 
Inclusion of each summand into $\text{Sym}^2 V$ gives a map of locally free sheaves, $$I\oplus T \to \text{Sym}^2 V$$  The rank of the domain is $7+1=8$ and the rank of the target is $10$.  By Thom-Porteous, the expected codimension of the degeneracy locus is $(10-7)(8-7) = 3$, i.e., we expect finitely many points of $\mathbb{P}^3$ where the linear subsystem of $H^0(\mathbb{P}^3,O(2))$ generated by $A$, $B$ and $C$ is not codimension $2$.  Moreover, Thom-Porteous predicts the total length of this degeneracy locus is the degree three coefficient of the power series $c_t(\text{Sym}^2 S)/c_t(I\oplus T)$.  
Of course $\text{Sym}^2 V$ is just $\mathcal{O}(2)^{\oplus 10}$, which has total Chern class $$ c_t(\mathcal{O}(2)^{\oplus 10}) = c_t(\mathcal{O}(2))^{10} = (1+2H)^{10} = 1 + 20H + 180H^2 + 960 H^3,$$ where $H$ is $c_1(\mathcal{O}(1))$ on $\mathbb{P}^3$.  Similarly, $c_t(I\oplus T) = c_t(I)c_t(T) = c_t(I)\cdot 1$, since $T\cong \mathcal{O}$.  Since also $K \cong \mathcal{O}$, $c_t(I)$ equals $c_t(S\otimes V)$, which is just $$c_t(I\oplus T) = (1+H)^8 = 1 + 8H + 28H^2 + 56 H^3.$$  Thus the power series we need is $$ c_t(\text{Sym}^2 V)/c_t(I\oplus T) = 1 + 12H + 56H^2 + 120H^3.$$  This is how I derive the number 120.
Of course we already know our $4$ bad points will be in the degeneracy locus, and they should probably have multiplicity $>1$ since the linear subsystem has codimension $5$ at each of these $4$ points.  So there should be at most 116 points (and almost certainly fewer) where the fiber is a line.  It is difficult for me to reconcile this with $8^3$.
$\textbf{Edit}$.  I just computed the length at each of the $4$ bad points is $10$: the scheme structure of the degeneracy locus at each bad point is cut out by precisely the third power of the maximal ideal of the point.  So the total contribution of the $4$ bad points to the total length is $4\times 10 = 40$.  Thus I am getting that there are $4$ fibers that are smooth plane conics, and also $80$ positive-dimensional fibers that are lines.  I do not understand how Oguiso computes $8^3$ positive-dimensional fibers. 
$\textbf{Second Edit}$.  I think I've reconstructed Oguiso's way of getting $8^3$.  The computation does have a mistake, but the method is a reasonable one.  Choose homogeneous coordinates $X_0$, $X_1$, $X_2$, $X_3$ on the first factor $\mathbb{P}^3$.  Then we can write $A$, $B$ and $C$ as $$ A = A_0X_0 + A_1X_1 + A_2X_2 + A_3X_3,$$ 
$$ B = B_0X_0 + B_1X_1 + B_2X_2 + B_3X_3,$$ 
$$ C = \sum_{0\leq i \leq j \leq 3} C_{i,j}X_iX_j, $$ where the $A_i$ and $B_j$ are global sections of $\mathcal{O}(1)$, and the $C_{i,j}$ are global sections of $\mathcal{O}(2)$.  Now we can use $A$ to "eliminate" $A_3X_3$.  Similarly, using $A_3B-B_3A$, we can use this to "eliminate" $(A_3B_2-A_2B_3)X_2$.  Thus, if we multiply $C$ by $A_3^2(A_3B_2-A_2B_3)^2$, then every instance of $X_3$ occurs as $A_3X_3$, and every instance of $X_2$ occurs as $(A_3B_2-A_2B_3)X_2$.  Thus, modulo the homogeneous ideal generated by $A$ and $A_3B-B_3A$, we have
$$
A_3^2(A_3B_2-A_2B_3)^2C \equiv F_{0,0}X_0^2 + F_{0,1}X_0X_1 + F_{1,1}X_1^2,
$$
where now $F_{i,j}$ is a linear combination of the coefficients $A_3^2(A_3B_2-A_1B_3)^2C_{i,j}$.  Of course each of these coefficients is a global section of $\mathcal{O}(8)$.  Wherever $F_{0,0}$, $F_{0,1}$ and $F_{1,1}$ are all zero, the expression above is in the homogeneous ideal generated by $A$ and $B$, thus the common zero locus of $A$, $B$ and this expression has dimension $\geq 1$ in the fiber $\mathbb{P}^3$.  If the common zero locus of $F_{0,0}$, $F_{0,1}$ and $F_{1,1}$ were finite, it would have length $8^3$ by Bezout.  Of course this common zero locus is not finite; the expression is zero on the hyperplane where $A_3$ equals $0$.<|endoftext|>
TITLE: Algebraic Morse theory
QUESTION [7 upvotes]: In 2005, prof. Emil Skoldberg developed a theory, similar to Forman's Discrete Morse Theory, but suited for arbitrary based chain complexes, in his Morse Theory from an algebraic viewpoint. I'm going through the paper and am having some difficulties. I'd be most grateful for an answer to my question 2 below.
Question 1: On p. 116, in the definition of a Morse matching, there is written:

We call a partial matching $M$ on the digraph $G_K$ a Morse matching if
for each edge $\alpha\to\beta\in M$ the corresponding component $d_{\beta,\alpha}$ is an
isomorphism, and furthermore, there is a well-founded partial order $\preceq$
on each $I_n$ such that $\alpha\succ\gamma$ whenever there is a path $\alpha^{(n)}\to\beta\to\gamma^{(n)}$ in $G^M_K$.

Is $\preceq$ defined by "exists a path $\alpha^{(n)}\to\beta\to\gamma^{(n)}$ in $G^M_K$", or is that just a necessary condition on $\preceq$? More precisely, the word "whenever" in the above quote, is that meant as $\Leftarrow$ or $\Leftrightarrow$?
Edit: Which definition is the right one (are all of them ok?): for $\alpha,\beta\in I_n$, we let:

$\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$;
$\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$
with vertices in $I_{n+1}\cup I_n$;
$\alpha\succeq\beta$ iff there exists a directed path in $G_K^M$ from $\alpha$ to $\beta$
with vertices in $I_n\cup I_{n-1}$;

Question 2: In the proof of Theorem 2 on p. 121. How do Lemmas 3 and 4 imply that for $x\in K_\alpha$ with $\alpha \in M_n^0$ there holds the equality $$\rho\pi(x)=x?$$ We have $\rho\pi(x)=\rho(x)-\rho\phi d(x)-\rho d\phi(x)$. Since $x \in C_n$ and $\rho$ is a projection, we have $\rho(x)=x$. By Lemma 3, we have $d\phi(x)= 0$. By Lemma 4, we have $\phi d(x) = \sum_{\beta\preceq\alpha}y_\beta=:(\ast)$ for some $y_\beta \in K_\beta$, but why is $(\ast)=0$ when $\alpha$ is critical?
Question 3: In Corollary 3, in the first sum, $\sigma$ ranges through $M^0_{n-1}$, right?
Question 4: If I understand correctly, the proof of Theorem 2 shows that if $\pi(K)$ has the induced boundary operator $d|_{\pi(K)}$ and $C$ has the operator $\tilde{d} := \rho(d-d\phi d) = \rho d \pi$, then the maps $\pi: C\longrightarrow \pi(C)=\pi(K)$ and $p: \pi(K)=\pi(C)\longrightarrow C$ are inverse to each other. Furthermore, $\pi\tilde{d} = \pi\rho(d-d\phi d) = d-d\phi d = d(\mathrm{id}-\phi d-d\phi) = d\phi$, so $\pi$ is a chain map. However, $\tilde{d}\rho = \rho(d-d\phi d)\rho = \rho d\rho-\rho d\phi d\rho \overset{???}{=} \rho d$.
Question 5: In general, there does not hold $\tilde{d}|_{\pi(K)}=d|_{\pi(K)}$, right?
Question 6: In the proof of Corollary 3, by Lemma 5 we have $\tilde{d}(x)$ $=$ $\rho(d-d\phi d)(x)$ $=$ $\rho(\sum_{\alpha\to\beta}d_{\beta\alpha}(x)-d\phi\sum_{\alpha\to\beta}d_{\beta\alpha}(x))$ $=$ $\rho\sum_{\alpha\to\beta}(d_{\beta\alpha}(x)-d\phi d_{\beta\alpha}(x))$ $=$ $\rho\sum_{\alpha\to\beta}(d_{\beta\alpha}(x)-d\sum_{\alpha'\in I_n,\gamma\in\Gamma_{\alpha',\beta}} m(\gamma)d_{\beta\alpha}(x))$. How do I continue to get $\rho\sum_{\sigma\in I_{n-1},\gamma\in\Gamma_{\sigma,\alpha}} m(\gamma)(x)$?
P.S. I might later add additional questions regarding p.116-122.

REPLY [2 votes]: It may be useful to note in relation to the work of Forman that his notion of discrete vector field on a complex is equivalent to the notion of marked cone complex in David W. Jones' 1984 PhD thesis on Poly T-complexes available from here, also published as "A general theory of polyhedral sets and the corresponding
  $T$-complexes" Dissertationes Math. (Rozprawy Mat.) 266 (1988) 110.
So it would be interesting if this is related also to Skoldberg's work.<|endoftext|>
TITLE: connected components of a real hyperplane arrangement
QUESTION [5 upvotes]: Let us consider the following configuration of hyperplanes in the real
vector space V with coordinates $z_1,\ldots,z_n$: the hyperplanes are
numbered by all the nonempty subsets $J\subset I=\{1,\ldots,n\}$, and the
hyperplane $H_J$ is given by $\sum_{i\in J}z_i=0$.
Question: How many connected components does the complement
$V\setminus\cup_{J}H_J$ have? That is, what are they naturally numbered by?

REPLY [3 votes]: There is a formula for the number of components which unfortunately is pretty useless,
namely, 
  $$ \sum_{k,r} \frac{(-1)^{n+k+r}}{k!}f(k,n,r), $$
where $f(k,n,r)$ is the number of real $k\times n$ $(0,1)$-matrices of rank $r$ with no zero row and no two rows equal.<|endoftext|>
TITLE: Is a manifold with flat ends of bounded geometry?
QUESTION [12 upvotes]: A Riemannian manifold $(M,g)$ is said to have flat ends if the curvature tensor of $g$ vanishes outside a compact set $K$. I was wondering if such manifolds are of bounded geometry. Recall that a manifold is of bounded geometry if

The curvature tensor and all its covariant derivatives are uniformly bounded.
The injectivity radius has a uniform positive lower bound.

It is obvious that a manifold with flat ends satisfies the first property, but it is not clear to me that a manifold with flat ends satisfies the second property. Two simple counterexamples come to mind. 

The manifold $M=\mathbb{R}^n-\{0\}$ has flat ends, but has no uniform lower bound on the injectivity radius.
A cylinder $\mathbb{R}\times S_r^1$ of radius $r$ has injectivity radius $\pi r$. Take a countable union of such cylinders with decreasing radius
$$M=\coprod_{n=1}^\infty \mathbb{R}\times S_{\frac 1n}^1.$$
This manifold is flat, but does not admit a uniform positive lower bound on the injectivity radius.

Counterexample one might be excluded by assuming that $g$ is complete, and counterexample 2 might be excluded by demanding that $M$ is connected. Therefore my question is:
 Is any connected and complete Riemannian manifold $(M,g)$ with flat ends of bounded  geometry?

REPLY [14 votes]: Here is a self-contained proof not using any classification. With some effort, it can be made to work under weaker assumptions: the curvature is nonnegative outside a compact set and is bounded from above. It is a variation of the proof of the Soul Theorem via Sharafutdinov's retraction.
Fix a reference point $o\in M$ and define a Busemann function $b:M\to\mathbb R$ by
$$
 b(x) = \limsup_{y\in M,\ d(o,y)\to\infty} ( d(o,y)-d(x,y))
$$
where $d$ is the Riemannian distance (note the non-standard sign convention). The flat ends assumption implies the following key features:

Sublevel sets of  $b$ are compact.
$b$ is (locally) convex outside a compact set.

To prove this, first observe some general properties of $b$. First, $b$ is 1-Lipschitz. Second, $b$ goes to $+\infty$ along any geodesic ray (essentially by the triangle inequality). Third, for any $x\in M$ there exists a geodesic ray starting at $x$ such that $b$ grows at unit speed along this ray (take a limit of geodesic segments $[xy_i]$ where a sequence $y_i$ realizes the $\limsup$). Let me call such rays "calibrating". Finally,  for every compact set $K$ there is a compact set $K'$ such that no calibrating ray starting outside $K'$ intersects $K$. Indeed, otherwise a sequence of such rays would converge to a geodesic line, and $b$ would  going to $-\infty$ in one of the directions along this line.
Now convexity of $b$ outside a compact set follows easily. Let $x_0\in M$ and $\gamma$ be a calibrating ray starting from $x_0$. Then $b(\gamma(t))=b(x_0)+t$ for all $t\ge 0$. Since $b$ is 1-Lipschitz, it follows that
$$
 b(x) \ge b(x_0) + (t-d(\gamma(t),x))
$$
for every $y\in M$, with equality for $x=x_0$. So $b$ is supported from below at $x_0$ by a function of the form $x\mapsto const - d(\gamma(t),x)$. If $x_0$ is sufficiently far from $o$, the ray $\gamma$ is contained in the flat part. Hence the distance to $\gamma(t)$ is bounded from above by a similar Euclidean distance function which is nearly linear near $x_0$ if $t$ is large. Existence of such lower bounds for $b$ implies that $b$ is convex.
Now let me show that sublevels of $b$ are compact. Suppose the contrary, then there is a sequence $x_i$ with $d(o,x_i)\to\infty$ but $b(x_i)\le const$. Choose a subsequence such that geodesic segments $[ox_i]$ converge to a geodesic ray $\gamma$. On this ray, mark a point $y$ where it leaves a ball containing the non-flat part of $M$. Once a segment $[ox_i]$ contains a point $y_i$ near $y$ and almost the same direction as $\gamma$ there, one observes that the derivative of $b$ along this segment at $y$ is greater than, say, 1/2. This and convexity of $b$ yield a linear lower bound for $b(x_i)$, contrary to the assumption $b(x_i)\le const$.
This proves the two key properties of $b$. Let $B_r$ denote the sublevel set $\{x:b(x)\le r\}$. Let $r_0$ be such that $b$ is convex outside $B_{r_0}$. Then, for every $R\ge r\ge r_0$, there is a distance non-increasing retraction (homotopic to identity) $f:B_R\to B_r$. Indeed, let $\varepsilon$ be smaller than the minimum injectivity radius on $B_R$. Then one retracts $B_R$ to $B_{R-\varepsilon}$ via a nearest-point projection, Locally it is just a nearest-point projection to a convex set in a Euclidean space, so it is well-defined and does not increases distances. Iterating this construction yields a retraction from $B_R$ to $B_r$. Moreover the complement of $B_r$ is mapped to the boundary of $B_r$.
Now let us turn to the injectivity radius. Let $r$ be such that $B_{r/2}$ contains the not-flat part of $M$ and $\rho_0$ is the minimum of $r/10$ and the injectivity radius on $B_{2r}$. I claim that the injectivity radius is no less than $\rho_0$ everywhere.  Suppose the contrary.  Then, somewhere outside $B_r$ there is a geodesic loop of length $2\rho<2\rho_0$. This loop is not contractible in the flat part. Apply the above retraction   $f:B_R\to B_r$ where $R$ is so large that the loop is contained in $B_R$. The image is a loop in the boundary of $B_r$ with length $\le 2\rho$ and still non-contractible in the flat part. Hence the injectivity radius at $f(x)$ is at most $\rho$, a contradiction.<|endoftext|>
TITLE: Examples of tilting objects that don't come from exceptional sequences
QUESTION [5 upvotes]: This is a question on geometric tilting theory. On smooth projective variety it is possible to define in general tilting object as perfect complex that satisfy some properties, but are there examples of tilting objects that are actually complexes? All examples I know are vector bundles obtained as sums of exceptional sequences. What would be examples of different nature?
Edit: if we denote variety $X$ then tilting object $T$ in $D(X)$ is a perfect complex (or from the point of view of abstract triangulated categories "compact object") such that $T$ generates $D(X)$ and $Ext^i(T,T)=0$ for $i \neq 0$.

REPLY [5 votes]: The answer to your more specific question is yes, there are tilting objects that involve complexes not quasi-isomorphic to any vector bundle. Consider the blowup $X$ of $\mathbb{P}^2$ at a single point $p$. Then, Orlov showed that there is a semiorthogonal decomposition $D^b(X)=\langle e,O_X,O_X(1),O_X(2)\rangle$, where $O_X(i)$ is the pullback of $O_{\mathbb{P}^2}(i)$, and where $e$ can be taken to be $i_*O_E(-1)$, where $i:E\rightarrow X$ is the inclusion of the exceptional divisor. The objects $e,O_X(i)$ are in fact all exceptional, so their direct sum is a tilting object for $D^b(X)$. However, $e$ is not quasi-isomorphic to a vector bundle, because it is supported along the exceptional divisor.<|endoftext|>
TITLE: Is equality of terms for "real" numbers with roots, logarithm, exponential, sin, cos, and other trigonometric operations decidable with a Turing-machine?
QUESTION [7 upvotes]: If yes, how? Also, I know you can't do it for arbitrary statements about real numbers, but that's not what I'm asking, and by "real" numbers, I mean the numbers constructible from 1, -, /, and the operations mentioned in the title.
Also, I don't care about numbers that can't be constructed from said operations and constant.

REPLY [13 votes]: Assuming Schanuel's conjecture, the answer seems to be yes, according to Daniel Richardson, "How to recognize zero" J. Symbolic Comput 24 (1997), 627–645 (doi:10.1006/jsco.1997.0157, available here online), in which the author defines a set of numbers he calls "elementary", which is algebraically closed and closed under exponential, logarithm and trigonometric functions, and for which equality is decidable (again, assuming Schanuel's conjecture).<|endoftext|>
TITLE: General Bruhat decomposition (with parabolic not necessarily Borel)
QUESTION [22 upvotes]: Here is the general Bruhat decomposition (which I have seen in various paper but never with a proof or a complete reference).
Let $G$ be a split reductive group, $T$ a split maximal torus and $B$ a Borel subgroup of $G$.
Let $R^+ \subset R$ be the positive roots corresponding to $B$ and $S \subset R^+$ the simple roots of $R^+$. Let $I \subset S$ and $P_I$ the standard parabolic subgroup of $G$ corresponding to $I$.
Finally let $W$ be the Weyl group of $(G,T)$ and $W_I$ the subgroup of $W$ generated by the reflections $(s_\alpha)_{\alpha \in I}$. 
Then the general Bruhat decomposition is
$$G = \coprod_{W_I \backslash W / W_I} P_I w P_I$$
and $P_I \backslash P_I w P_I$ is an affine variety of dimension $\ell(w)$ where $w$ is of minimal length in the double coset $P_I w P_I$.
My question is : is there a good choice of representatives for $P_I \backslash P_I w P_I$ ? More precisely, I am looking for an analogue of the following bijection (in the case $P=B$ Borel) :
$$B \times \lbrace w \rbrace \times U_{w^{-1}} \overset{\sim}{\longrightarrow} BwB$$
where $U$ is the unipotent radical of $B$, $U^-$ its opposite and $U_{w^{-1}}$ is the subgroup $(w^{-1}U^-w) \cap U$. What subgroup of $P_I$ would replace $U_{w^{-1}}$ ?
Also what reference exists for all this ?
Thanks in advance.
Edit : in this course of Casselman I found the following isomorphism of variety (see on top of page 12)
$$P_I \times \lbrace w \rbrace \times \prod_{\alpha \in R^+ \backslash R_I^+ ~|~ w^{-1} \alpha \notin R^+ \backslash R_I^+} N_\alpha \overset{\sim}{\longrightarrow} P_IwP_I$$
with $w \in W$ of minimal length in $W_I \backslash W / W_I$. However this seems not to work with $\mathrm{GL_3}$ : we note $S = \lbrace \alpha, \beta \rbrace$ ; if $I= \lbrace \alpha \rbrace$, $P_I = \left( \begin{smallmatrix} * & * & * \newline * & * & * \newline & & * \end{smallmatrix} \right)$ ; with $w = s_\beta$ the above product is on the set $\lbrace \beta, \alpha + \beta \rbrace$, so the isomorphism should be $P_I s_\beta P_I \cong P_I \times \lbrace s_\beta \rbrace \times \left( \begin{smallmatrix} 1 & 0 & * \newline 0 & 1 & * \newline 0 & 0 & 1 \end{smallmatrix} \right)$, which is false (the element $s_\beta \left( \begin{smallmatrix} 1 & 0 & 0 \newline 1 & 1 & 0 \newline 0 & 0 & 1 \end{smallmatrix} \right)$ is in the left side, not in the right side)...

REPLY [2 votes]: As @JimHumphreys has pointed out, [BT2] Borel and Tits - Compléments à l'article: «Groupes réductifs», specifically Proposition 3.16(i, iv), gives the decomposition $G/P_I = \bigsqcup_{w \in [W_I\backslash W/W_I]} P_I w P_I/P_I$, where $[\cdot]$ denotes the minimal-length double-coset representatives.  However, it is false that $P_I w P_I/P_I$ is an affine space.
I asked Josh Lansky about this, and he pointed me to Theorem 5.2 of his paper [La] Decomposition of double cosets in $\mathfrak p$-adic groups, which (while considering a more general situation involving parahorics instead of parabolics) suggests that, instead of trying directly to describe $P_I w P_I/P_I$, instead we should consider
$$
P_I w P_I/P_I = \bigsqcup_{w' \in [W_I/W_I \cap w W_I w^{-1}]} P_\emptyset w'w P_I/P_I.
$$
Then, as you'd expect, each $P_\emptyset w' w P_I/P_I$ is an affine space; here you can use [La, Theorem 4.6] and [BT2, Proposition 3.16(ii)], which show that $P_\emptyset w'w P_\emptyset/P_\emptyset \to P_\emptyset w'w P_I/P_I$ is an isomorphism of varieties (from an affine space).
Thus, the problematic case of $\operatorname{GL}_3$ that you mention now becomes
$$
P_\alpha s_\beta P_\alpha/P_\alpha = P_\emptyset s_\beta P_\alpha/P_\alpha \sqcup P_\emptyset s_\alpha s_\beta P_\alpha/P_\alpha,
$$
where $P_\emptyset s_\beta P_\alpha/P_\alpha$ and $P_\emptyset s_\alpha s_\beta P_\alpha/P_\alpha$ are $1$- and $2$-dimensional affine spaces, respectively; and, of course, the element $s_\beta\begin{pmatrix} 1 \\ 1 & 1 \\ && 1 \end{pmatrix}$ that you mention lies in $P_\emptyset s_\beta P_\alpha$.<|endoftext|>
TITLE: Failure of the Pointwise Ergodic Theorem
QUESTION [6 upvotes]: It is known that Birkhoff's pointwise ergodic theorem (unlike von Neumann's mean ergodic Theorem) fails to hold for general Folner sequences. 
The counter-example usually given is the Folner sequence $F_N=\{N^2,N^2+1...,N^2+N\}$, however I've only seen it referenced to the first such result by Akcoglu and del Junco in 1975, where the Folner sequence which fails is actually $G_N=\{N,N+1,...,N+\lfloor \sqrt{N}\rfloor\}$.
Since $\{F_N\}$ is just a (quite sparse) subsequence of $\{G_N\}$ it could happen that the ergodic averages converge pointwise along $\{F_N\}$ even if they don't along $\{G_N\}$.
Moreover, Lindenstrauss showed that every Folner sequence has a subsequence along which convergence holds.
So I'm just curious if there is a published proof (or if it is easy and I'm just overlooking something) that the sequence $\{F_N\}$ is not good for pointwise ergodic theorem.

REPLY [8 votes]: Bellow, Alexandra(1-NW); Jones, Roger(1-DPL); Rosenblatt, Joseph(1-OHS)
Convergence for moving averages.
Ergodic Theory Dynam. Systems 10 (1990), no. 1, 43–62.<|endoftext|>
TITLE: Errors, oversights, and misunderstandings in mathematical research
QUESTION [7 upvotes]: Possible Duplicate:
Examples of common false beliefs in mathematics. 

Hopefully this is not overly controversial, but I thought it would be instructive to compile a list of errors which are commonly (or at least not too uncommonly) made in higher level mathematical research and published mathematical works (i.e. research papers, books, etc). If nothing else this could serve as a warning to the rest of us.
Here I would like to keep the focus on mathematical errors, oversights, or misunderstandings, and not those which are related to typographical, grammatical, or purely historical issues. To start the list and to give some idea what types of errors I am thinking of, here are a few examples:

1. Confusing the finite field $\mathbb{F}_{p^k}$ for $k>1$ with the ring $\mathbb{Z}/p^k\mathbb{Z}$.


2. Assuming that every open set in $\mathbb{Q}_p$ is also closed (true for balls but not in general).


3. Assuming that the hypothesis in a conditional statement is a necessary condition, just because a weaker hypothesis does not necessarily imply the conclusion.

One remark, since this is intended solely for our own edification, it is probably better to avoid mentioning specific references, for the obvious reason.
Also, if anyone can think of better tags for this question please go ahead.

REPLY [8 votes]: Lamé's wrong proof of Fermat's Last Theorem, which led to the fact that cyclotomic fields do not have unique factorization.

REPLY [2 votes]: There have been many false proofs of the Jacobian Conjecture, and it might be interesting to have a survey of the attempts and why they ultimately failed (i.e., what holes there were in the attempted proof or what counterexamples to steps in the attempted proof were found).<|endoftext|>
TITLE: Diferent abelian varieties over local field with the same p-adic representation?
QUESTION [8 upvotes]: Let $K$ be a local field with residue field of char $p$, denote $G$ its Galois group. Is it possible that we have two Abelian varieties $A_1$ and $A_2$, defined over $K$, such that  they are  not isogeny （over $K$ or $\bar{K}$ ）, but have  isomorphic p-adic Galois representation of $G$?

REPLY [10 votes]: Yes, this can happen. Here is a counterexample (which is probably not the simplest possible, but it's the one that first came to mind). 
There are not very many 2-dimensional representations of the Galois group of $\mathbf{Q}_p$ which are "crystalline" in Fontaine's sense. Fontaine's functor $\mathbf{D}_{\operatorname{cris}}$ classifies them by linear data: 2-dimensional vector spaces over $\mathbf{Q}_p$ with a filtration and a linear operator $\varphi$ (the Frobenius) satisfying some compatibility properties. If $V$ is the $p$-adic Galois representation coming from an elliptic curve over $\mathbf{Q}_p$ with good reduction, then $\varphi$ has characteristic polynomial $X^2 - a_p(E) X + p$, where as usual $a_p(E) = p + 1 - \# \overline{E}(\mathbf{F}_p)$.
If $a_p = 0$, then this uniquely determines $\mathbf{D}_{\operatorname{cris}}(V)$ as a $\varphi$-module, and the conditions on the filtation ("weak admissiblity") mean that if $a_p(E) = 0$ then there is (up to isomorphism) a unique possibility for the filtration. So, in other words, all elliptic curves over $E$ with good supersingular reduction have isomorphic $p$-adic Galois representations [edit: if $p>3$, at least]. But they certainly aren't all isomorphic (or even isogenous) to each other, so that gives a counterexample.<|endoftext|>
TITLE: Automorphisms of $SL_n$ as a variety
QUESTION [9 upvotes]: What are the automorphisms of $SL_n$ as an algebraic variety?
In other words, let $k$ be an algebraically closed field of characteristic 0 (e.g., $k=\mathbb{C}$). Let $\tau$ be an automorphism of $SL_n$ regarded as an algebraic variety over $k$. Assume that $\tau$ takes the unit element $e$ of $G$ to itself. Is it true that $\tau$ is an automorphism of $SL_n$ as an algebraic group over $k$?

REPLY [9 votes]: The automorphism group is massive!  
Flexible varieties and automorphism groups, 
I. Arzhantsev, H. Flenner, S. Kaliman, F. Kutzschebauch, M. Zaidenberg, http://arxiv.org/abs/1011.5375.<|endoftext|>
TITLE: Moduli space of genus 1 curves with two fixed points
QUESTION [6 upvotes]: It is well-known, that the moduli space $\mathcal M_{1;1}$ of elliptic curves is isomorphic to an orbifold space $(S_3\times S_2) \backslash\backslash \mathcal M_{0;4}$, where the first factor of the group acts by permutation of the first three distinguished points on a rational curve, and the second one acts trivially. Let us try to exploit this idea to construct $\mathcal M_{1;2}$.
Note first, that Riemann-Roch implies that for any two (different) points $P$ and $Q$ on a genus 1 curve, there exists a degree 2 function having degree 1 poles in $P$ and $Q$. The set of critical points is the same for any such a function. It is not difficult to understand, that the set of critical values of any such a function is the same up to an automorphism of $P^1$ fixing the infinity.
Vise-versa, if one fixes four points on $\mathbb C\subset P^1$, it determines a genus one curve with two distinguished points coming as preimages of the infinity under a degree two function ramified in the chosen points.
That indicates, that there should exist an orbifold map $\mathcal M_{1;2} \to (S_4) \backslash\backslash \mathcal M_{0;5}$, that forgets the numbering of the distinguished points on a genus 1 curve. But from the other hand, any genus 1 curve with two fixed points admits an automorphism interchanging the fixed points.
So my question is the following: is the map $\mathcal M_{1;2} \to (S_4) \backslash\backslash \mathcal M_{0;5}$ an isomorphism?

REPLY [6 votes]: Yes, it is.
See Leila Schneps, Special loci in moduli spaces of curves (in Galois Groups and Fundamental Groups, MSRI series 41, Cambridge University Press, 2003), pages 34-35.<|endoftext|>
TITLE: Estimate on radical of $2^n \pm 1$
QUESTION [19 upvotes]: Not sure if this belongs to MO or not. 
Are there any lower bound on radical of $2^n \pm 1$?
We recall that radical of an integer $rad(k)$ is a product of primes which divide $k$.
As an example, if the abc-conjecture is true in the form $max(|a|,|b|,|c|) \leq rad(abc) ^2 $  then 
$$rad(2^n \pm 1) \geq 2^{n/2 - 1}.$$
I wonder if this estimate is proven (or perhaps conjectured) by anyone? Are there any nontrivial results here?

REPLY [5 votes]: The paper arXiv:1409.2974v1 proves a lemma regarding $2^L-1$. 
Let $n$ be a positive integer, and define $L = \text{lcm}[1, 2, \dots n]$ and $t = \lfloor \frac{\log(n)}{\log(2)} \rfloor$. Then $L*\text{rad}(2^L-1) \le 2^t(2^L-1)$.<|endoftext|>
TITLE: A question about well ordered subsets of totally ordered countable sets
QUESTION [6 upvotes]: Let us assume ZFC and let Q be the set of rational numbers ordered according to size. There is a
well known theorem which implies that if S is any totally ordered countable set containing a subset
ordinally similar to Q, then S contains well ordered subsets having arbitrarily large countable
ordinal numbers. Is there a converse to this theorem which implies that if S has the second of these
properties, then it has the first? I have been unable to find any mention of such a converse theorem
or to come up with any obvious counter-examples. (I apologize if my question is not considered
appropriate for "mathoverflow.net")

REPLY [4 votes]: The converse is a theorem of Đ. Kurepa in:
Sur les ensembles ordonnés dénombrables, Hrvatsko Prirodoslovno Društvo. Glasnik Mat.-Fiz. Astr. Ser. II. 3, (1948). 145–151.<|endoftext|>
TITLE: What, precisely, does Klein's Erlangen Program state?
QUESTION [42 upvotes]: People write that the Erlangen Program is a "program" (like the "Langlands Program"), i.e. a series of related conjectures, which in this case were all solved. There are various intuitive accounts, describing that this program is about relating algebra and geometry, about relating transformation groups of spaces (Lie groups) and different geometries, invariants, etc.
What I haven't been able to find is a precise statement in modern language (e.g. not that of his original paper) of what Klein's conjectures were. What precisely were his conjectures, or equivalently, what results constitute their resolution? Or was there never truly a precise statement?

REPLY [4 votes]: D. E. Rowe notes that one shouldn't read too much in program:

Klein wrote it to fulfill a formal requirement incumbant upon every new Ordinarius who entered the Erlangen faculty. Its name, in fact, had nothing to do with a new program for mathematical research, but rather derives from having been a “Programm zum Eintritt in die philosophische Fakultät und den Senat” at Erlangen.

(Likewise W. Killing and many others.)<|endoftext|>
TITLE: Convergence rate for product of stochastic matrices
QUESTION [5 upvotes]: Hi,
I have a system of the form $$x(t+1) = A(t + 1) x(t),$$ for $t \geq 1$, and some fixed initial condition $x(1)$. Here $A (t)$  is a time-varying $m \times m$ matrix that is  stochastic at all times $t$ (so row sums are $1$). 
In fact, I know that the matrices $A(t)$ are primitive, in that each has only one eigenvalue at $1$, and the other eigenvalues are $< 1$. (The matrices come from graph Laplacians.) 
I'm interested in number of iterations until $x(t)$ converges to a stable point where all its entries are equal to a constant. This convergence is closely related to the product of matrices $\prod_t A(t)$, for $t = 1, 2, \ldots$. 
My goal is to upper bound the convergence rate of the $x$'s as a function of the minimum second eigenvalue $\min_t \lambda_2(t)$  over all the matrices $A(t)$. This dependence seems to be a known result in the literature, but I cannot find a reference to it or a simple way of proving it.
Thanks!

REPLY [2 votes]: shIf your matrices are "time-dependent" you need additional assumptions to bound the rate of convergence and it will usually be impossible to obtain something which depends only on the minimum of the second eigenvalues. 
You may choose $a$ and $b$ at will in the example below, for example such that both matrices are doubly stochastic.
The product $\mathbf{A}\mathbf{B} \ = \ \mathbf{B}\mathbf{A}$ of the matrices
\begin{equation}
\mathbf{A}  \ := \ 
\begin{pmatrix} 
\frac{1}{3} + a^2    &    \frac{1}{3} - a^2   &   \frac{1}{3}   \\
\frac{1}{3} - a^2    &    \frac{1}{3} + a^2   &   \frac{1}{3}   \\
\frac{1}{3}          &     \frac{1}{3}        &   \frac{1}{3}   \\
\
\end{pmatrix} 
\end{equation}
and 
\begin{equation}
\mathbf{B}  \ := \ 
\begin{pmatrix} 
\frac{1}{3} + \frac{1}{4} b^2    &    \frac{1}{3} + \frac{1}{4} b^2   &   \frac{1}{3} - \frac{1}{2} b^2  \\
\frac{1}{3} + \frac{1}{4} b^2    &    \frac{1}{3} + \frac{1}{4} b^2   &   \frac{1}{3} - \frac{1}{2} b^2  \\
\frac{1}{3} - \frac{1}{2} b^2    &     \frac{1}{3} - \frac{1}{2} b^2  &   \frac{1}{3} + b^2  \\
\
\end{pmatrix} 
\end{equation}
is
\begin{equation}
\mathbf{E}  \ := \ 
\begin{pmatrix} 
\frac{1}{3}     &    \frac{1}{3} &   \frac{1}{3}   \\
\frac{1}{3}     &    \frac{1}{3} &   \frac{1}{3}   \\
\frac{1}{3}     &    \frac{1}{3} &   \frac{1}{3}   \\
\
\end{pmatrix} \, ,
\end{equation}
so whatever the values of $a$ and $b$ and independently of the initial conditions as long as all later matrices have row and column sums equal to 1, the process converges after just two iterations.
The matrices were constructed in the following way: Take an orthogonal basis of $\mathbb{R}^3$ containing the vector $(1,1,1)$, here $x = (1,1,1); y=(1,-1,0); z=(1/2,1/2,-1)$ and construct the matrices $A$ and $B$ as $k_x*x^tx + k_y*y^ty + k_z*z^tz$ by choosing the constants $k_x, k_y,k_z$ appropriately.
The eigenvalues of $A$ are $1; 2a^2; 0$, those of $B$ are $1; 1.5b^2; 0$, but while $A$ projects onto the orthogonal complement of $(1/2,1/2,-1)$, $B$ projects onto the orthogonal complement of $(1,-1,0)$, leaving only $k(1,1,1)^{(t)}$ afterwards, where $k$ is constant.  
This is clearly a more general phenomenon, so without rather specific additional assumptions it will be very difficult/essentially impossible to progress, as the above result is independent of the second eigenvalues and there is no upper bound on the convergence rate in terms of them. 
One might think about two things here: 
1) There should be a lower bound on the convergence rate in terms of the supremum of the second eigenvalues if this is bounded away from 1 as in the time independent case because projecting will not increase the norm of the image.
2) If all the matrices $A(t)$ are invertible and their eigenvalue $\lambda_{small}$ of smallest modulus is bounded away from zero, there should be an estimate of the rate of convergence from above involving $\inf (|\lambda_{small}|)$.
Concerning the literature on Markov chains I can not really help you, but you may wish to consult
"Non-negative Matrices and Markov Chains" by E. Seneta, Springer, reprint 2006.
The example I just constructed from scratch.<|endoftext|>
TITLE: A theory of bifurcation of braids ?
QUESTION [5 upvotes]: I am trying to study the braids generated by periodic orbits of diffeomorphisms of compact surfaces (for example, a punctured disk). The diffeomorphisms are generated by integrating a two-dimensional time dependent ODE on the disk, forward in time. Let us assume that the ODE system (and hence the resulting diffeomorphism) is dependent on one bifurcation parameter.
Also assume that we select some 'distinguished' periodic orbits of this system, and start varying the bifurcation parameter. For each value of this parameter, we can form a braid, where 'world-line' of each periodic orbit (in 2+1 space) is a strand. I am interested in exploring the relationship between the different braids that are formed when we vary the parameter, and especially the behavior near a bifurcation point of any one (or more) periodic points that we selected. Are there any results that shed some light on this ? I have been advised to look for connections with winding number of the periodic orbits, but so far I haven't been able to find any relevant literature.
The motivation comes from the fact that braids formed above encode quite a bit of information, including but not limited to, the topological entropy of the flow. One can obtain lower bounds on topological entropy by invoking the Thurston-Nielsen theorem, and so on.
Thanks for any insight.

REPLY [3 votes]: Here is a couple of references regarding the holomorphic world, although they do not represent an answer to your question per se (but I'm afraid this is too long a comment, and might anyhow interest people interested in your question). 
First there is a work by Cano, Moussu and Sanz where they study the way real trajectories attached to some complex ODE are entangled.
Next, in the following links, the braid structure is not studied as such, but I think you may perform the study (in this special context) by using the tools and constructions developped below.

In the context of the bifurcation of holomorphic diffeomorphisms of a disk, there is this one about the analytical classification, continued in this paper where the moduli space is completely identified.
Since you are interested in the suspension itself (what you describe as the ODE), maybe you could have a look at this one and also that one, which deal with the complex suspension of the diffeomorphisms appearing above, the first one describing more particularly the underlying geometry. You retrieve a real suspension by lifting in the complex solutions a circle included in one fo the separatrices: the subsequent holonomy is precisely the diffeomorphism you started from.
I also point out this reference for (complex) two-dimensional holomorphic dynamics, which really is also a deformation of one-dimensional diffeomorphisms. 

Hope this somehow helps you in tackling your problem.
-<|endoftext|>
TITLE: Which functions are linear combinations of irreducible characters for a given field $\Bbbk$?
QUESTION [5 upvotes]: Let $G$ be a finite group. Then it is well known that a function $f\colon G\to \mathbb C$ is a linear combination of irreducible characters iff it is constant on conjugacy classes.  What is the corresponding result for an arbitrary field $\Bbbk$, i.e., which $\Bbbk$-valued mappings on $G$ are $\Bbbk$-linear combinations of irreducible characters over $\Bbbk$?  I am assuming neither characteristic $0$, nor algebraically closed.  I expect the answer is well known and involves the characteristic and which primitive roots of unity exist. Any references are greatly appreciated. 
A more specific question:  Is it true that if $f\colon G\to \Bbbk$ is constant on conjugacy classes and is constant on cyclic subgroups, then it is a linear combination of irreducible characters over $\Bbbk$?

REPLY [4 votes]: I am interpreting your question as talking of $k$-linear combinations of traces of $k$-representations of $G.$ Note that such a function must not only be constant on conjugacy classes, but should also be constant on $p^{\prime}$-sections, where $k$ has characteristic
$p.$ Recall that every element of $G$ may be written uniquely in the form $g = ab = ba,$ where $a$ has order a power of $p$ and $b$ has order prime to $p.$ The element $b$ is called the $p^{\prime}$-part of $g.$ Two elements of $G$ are said to be in the same $p^{\prime}$-section of $G$ if and only if their $p^{\prime}$-parts are conjugate. I think this reduces us to the case where $G$ is a cyclic $p^{\prime}$-group, using Brauer's or Conlon's induction theorem.
 I think you may find the necessary analysis of that case in the 1962 book of Curtis and Reiner.
 Expanded edit: The work (of Berman I believe, if my memory is accurate), I am alluding to, 
deals with the fact that dealing with traces of $k$-valued representations forces some equality of traces at group elements at elements which are not $G$-conjugate. We can assume, as I indicated, that $|G|$ is coprime to char $k$ if char $k \neq 0$. So, for example, if $\theta$ is the trace of a $k$-representation and $k$ has $q$ elements, then we must have $\theta(g) = \theta(g^{q})$ for all $g \in G.$ Similarly, if $k = \mathbb{Q},$ and $\theta$ is the trace of a $\mathbb{Q}$-valued representation, we must have $\theta(g) = \theta(h)$ whenever $\langle g  \rangle = \langle h \rangle.$
 So, let us say that $g$ and $g^{m}$ are $k$-conjugate if $\theta(g) = \theta(g^{m})$ whenever $\theta$ is the trace of a representation over $k.$ Then a $k$-valued class function $\psi$ of $G$ is said to respect $k$ if $\psi(g) = \psi(g^{m})$ whenever $g$ and $g^{m}$ are $k$-conjugate. Then the best you can hope for is that the traces of $k$-representations span the space of $k$-valued class functions which respect $k,$ and this is indeed the case.<|endoftext|>
TITLE: Relation between TQFT and Wilson lines, boundary conditions, surface defects etc
QUESTION [8 upvotes]: I have been studying (extended) topological quantum field theories (in short TQFTs) from the mathematical point of view and I have no background of the physics point of view. Sometimes I encountered papers talking about Wilson lines, boundary conditions, surface defects and so on. I looked up these terminology but I couldn't find a good explanation how these are related to TQFTs.
I also want to know the relation of a path integral and a  TQFT.
Could you suggest me references (papars, websites, books, vidoes etc) that explains the relationship between these physical jargon and TQFTs from mathematical point of view? Or could you explain here?
Thank you in advance.
(This question was asked in mathstack exchange but there is no answer. Here

REPLY [7 votes]: Greg Moore recently gave the Felix Klein lectures and a draft of notes for his lectures is available at 
http://www.physics.rutgers.edu/~gmoore/FelixKleinLectureNotes.pdf 
You will find in the first few pages a discussion of (extended) TQFT, defects, Wilson lines and so on in a language which I imagine is more suitable to mathematicians than to (most) physicists.<|endoftext|>
TITLE: A generalization of intermediate value theorem on R^k
QUESTION [12 upvotes]: Let $f:[0,1]\to\mathbb R^k$ be a continuous function with $f(1) = \overrightarrow 0$.
Is it true that there always exist $k$ points $0 \le a_1 \le a_2 \le \ldots \le a_k \le 1$ such that $\sum_{i=1}^k (-1)^{k+1} f(a_k) = f(a_1) - f(a_2) + f(a_3) - \ldots = \frac{f(0)}{2}$?
When $k=1$ we have to find one point $a_1 \in [0,1]$ with $f(a_1) = \frac{f(0)+f(1)}{2}$, which is the intermediate value theorem.
When $k=2$ the situation is more complicated.
We believe the statement is still true.
However, we noted that if we are not finding $\frac{f(0)}{2}$ but, say, $.49 f(0)$ instead, there exist counter-examples.
A counter-example looks like a sine function $f(t) = (t, \sin 100t)$, rotated in the 2-d plane a little bit.
(This is not the example, but you can imagine that changing the constants $.49$ and $100$ a bit makes it work.)
For bigger $k$ we have no idea.
Are there any similar results known before?

REPLY [15 votes]: The statement is true. It is almost precisely Lemma 2 in the paper D.Burago, "Periodic metrics", Adv. Soviet Math. 9, (1992), 205-210. The proof is short but not easy to invent. The paper can be read on Google Books here. 
Notes on the text: the intervals in the formulation of Lemma 2 are in fact disjoint, the term "antipodal map" means "odd map" (i.e. $\varphi(-x)=-\varphi(x)$).
For $k=2$, this fact is classic and one can replace 0.5 by any inverse integer ($1/3, 1/4,\dots$) but for every other value there is a counter-example.<|endoftext|>
TITLE: Is there a characterization of groups in which at least one subgroup is not an endomorphism kernel?
QUESTION [15 upvotes]: This is a crosspost from MSE: 
Is there a characterization of groups with the property $\exists N \unlhd G : \not \exists H \leq G \;\; \text{s.t.} \;\; H\cong G/N$?

A common mistake among beginning group theory students is the belief that a quotient of a group $G$ is necessarily isomorphic to a subgroup of $G$.  Is there a characterization for groups in which this assumption fails?
If this question is too broad, I might ask if such a characterization exists for $p$-groups.
A peripheral question is how likely it is for a finite group to have this property, which I would formulate in the following way: if $g(n)$ denotes the fraction of isomorphism classes of finite groups of order $≤n$ for which the property holds, what happens to $g(n)$ as $n\rightarrow \infty$?  (Here is what $g$ looks like for small values.) If we again restrict to $p$-groups, can we say anything about the fraction of isomorphism classes of order $p^k$ with the property, e.g. a bound in terms of the exponent?
EDIT I believe Derek Holt's answer satisfies the peripheral question.  The primary question is still open - can we say anything to characterize these groups?

REPLY [2 votes]: In short, I do not think there is a reasonable characterization of such groups, beyond restating the definition you gave. Here is why. First, let's give the property of $G$ that every quotient $G/N$ embeds in $G$ "self-universality" (SU). 
First, note that there are three classes of groups which have SU:

Simple groups (and groups obtained from them via semidirect products). 
Higman-Neumann-Neumann type universal groups: They proved existence of 2-generated groups $G$ which contain every countable subgroup. In particular, such $G$ contains every quotient of $G$. 
Note that in (1), the group has too few quotients, while in (2) it has too many subgroups. One can have infinite f.g. groups which have both too few quotients and too few subgroups, like Tarski monsters (every nontrivial proper subgroup is isomorphic to $Z_p$).   

Given the diverse reason for these groups to have SU, there does not seem to be any "unifying reason" for the SU property. 
Note, however, that, at least among finitely-generated groups, all the examples 1-3 are mostly "artificial"; e.g.: most "naturally occurring" f.g. groups tend not to be simple (or universal in HNN sense, or be "monsters"). Among "natural" f.g. groups, the ones which do not have SU tend to dominate:
a. F.g. (infinite) matrix groups (over the real numbers) do not have SU, by a combination of Malcev's theorem and Selberg's lemma: If $G$ is such a group, it has finite quotients of arbitrarily high orders (Malcev), but $G$ contains only  finitely many isomorphism classes of finite subgroups (Selberg). 
b. Infinite hyperbolic groups do not have SU: For virtually cyclic groups this follows from (a), while for nonelementary groups it follows from Delzant and Olshansky who proved SQ universality for such groups. In particural, such $G$ always has a quotient containing $Z^2$ and, hence, cannot embed in $G$.<|endoftext|>
TITLE: Fixed point of $S^1$-action using roots of unity
QUESTION [8 upvotes]: Fact: For any (continuous) $S^1$-action on the closed unit disk $\mathbb{D}^n$, there is a fixed point $x_0\in\mathbb{D}^n$.
I have thought of a possible argument that re-proves this, but am not sure how to complete it:
Let $U_p\subset S^1$ be the subgroup of $p^\text{th}$-roots of unity ($p$ prime). An $S^1$-action on a compact contractible space $X$ will induce a $U_p$-action on $X$. Smith Theory then implies that $X^{U_p}$ is nonempty, i.e. there is a fixed point $x_p\in X$ under $U_p$, for any given prime $p$. Now here is where I want to say: Taking $p$ sufficiently large, we find a fixed point $x_\infty$ under $S^1$. (The intuition is that $\lim_{p\to\infty}U_p\approx S^1$, and denseness will be sufficient by continuity of the action.)
1) Is it possible to fill this gap, i.e. can this 'proof' make sense? Not sure how to make sense of this limit/sequence of $U_p$'s, and whether the fixed points hop back and forth forever.
2) Is such a sequence $\lbrace x_p\rbrace_{p=\text{prime}}$ Cauchy? Or, does there exist a prime $p_0$ where $x_p=x_{p_0}$ for all primes $p>p_0$?

REPLY [8 votes]: Not quite your question, but I'll say it anyway. 
The "fixed-points of actions on a $p$-acyclic space are $p$-acyclic" part of Smith theory easily extends to arbitrary $p$-groups. By induction on the order of the group: if $P$ acts on a $p$-acyclic $X$, choose a non-trivial proper normal subgroup $Z \leq P$ (these always exist; if $P$ is non-abelian take its centre), then $X^P = (X^Z)^{P/Z}$.
Let $U(1)$ act continuously on $D^n$, and $X(n)$ be the $\mathbb{Z}/p^n$ fixed points. This is a compact subset, and non-empty by Smith theory. Thus $X=\cap_{n=1}^\infty X(n)$ is also non-empty, by Cantor's intersection theorem. A point $x \in X$ is fixed by the subgroup of $U(1)$ of $p$-power-torsion points; this is dense, so $x$ is fixed by the whole of $U(1)$.

REPLY [6 votes]: You might have a look at Chapter VI of Borel's "Seminar on transformation groups". This is the chapter on "Isotropy groups of toral actions" by E. E. Floyd.
In particular Theorem VI.1.2 seems to be saying that the fixed point sets eventually stabilize. (I can give more details if you don't have the reference to hand.)
Added later: If you are prepared to assume that your $S^1$-action is locally smooth, then the proof that the fixed-point sets eventually stabilize is a little easier, and contains most of the ideas of the general case. A reference is Section IV.1 of Bredon's book "Introduction to compact transformation groups".
To get around the boundary problem, I suggest the following approach. The pair $(D^n,\partial D^n)=(D,\partial D)$ is of course a mod $p$ cohomology $n$-disk, so Smith theory tells you that $(D^{U_p},\partial D^{U_p})$ is a mod $p$ cohomology $r$-disk for some $0\le r\le n$. This implies in particular that $D^{U_p} \neq\partial D^{U_p}$, and so there is a $U_p$-fixed-point in the interior of the disk.
Now note that any group action on a disk must preserve the boundary and the interior. So by restriction you have an $S^1$-action on $\operatorname{int} D \approx \mathbb{R}^n$, which by the above argument has $U_p$-fixed-points for all primes $p$. Now you can apply Theorem IV.1.4 in Bredon to conclude that there is an $S^1$-fixed-point.<|endoftext|>
TITLE: What is the history of $\sqrt{}$
QUESTION [15 upvotes]: Why we use the symbol $\sqrt{}$ when we take square roots ? Anybody knows the history ?

REPLY [15 votes]: This question has already been completely answered, but here is a bit more on history.
The symbol has its origin in the Latin letter R as an abbreviation of radix (latin: root). It has been used by Leonardo de Pisa (Fibonacci) in his seminal book Liber Abaci (1202) where he treats square roots and cubic roots in chapter 14 and 15 (as well as in his later less well known Liber Quadratorum (1225)).
Fibonacci, like Euclid, did not invent all the mathematics he reports, but took much of it from the Arabian world, mainly from al-Khwarizmi and Omar Khayyam. The Arabian word for "root" had been used by al-Khwarizmi already; his word is rendered radix in translations from the Arabic to Latin by Robert of Chester, John of Seville, and Gerard of Cremona. It appears also in Alexandre de Villedieu's Carmen de Algorismo (1240) and in Sacrobosco's Algorismus (1250). By the way Fibonacci calculates approximations but considers roots as exact numbers, even if they cannot be expressed as integers or fractions.
Following the tradition of medieval writers, Nicolas Chuquet used the uppercase Latin letter R with a small stroke, looking very similar to Px when written close together. Both, R and R$^2$ indicate the square root, R$^3$ indicates the cubic root, R$^4$ the forth root and so on. Regiomontanus (1464), Luca Pacioli (1494), and Estienne de la Roche (1520) adopted this sign. 
The hook-like symbol √ that resembles a small r was introduced by Christoff Rudolff in his book "Die Coß" (1525), the first German book on algebra. He used c√ for cubic root and √√ for fourths root. (By the way he introduced also the convention $x^0$ = 1). English, French and Italien writers were slow in adopting that German sign. Even in Germany the symbol "l" for latus (side of the square) was long in use. After Michael Stifel had published the second edition of Rudolff's Coß (1553) the symbol became more and more accepted.
René Descartes (1596 bis 1650) invented (or extended) the bar above the radicand (this word being first used in 1889) in order to indicate what symbols belong to the radicand.
Moritz Cantor: "Vorlesungen über Geschichte der Mathematik", Teubner, Leipzig (1894)
http://archive.org/stream/vorlesungenber02cantuoft#page/n5/mode/2up
Florian Cajori: "A History of Mathematics" MacMillan, London (1909)
http://www.gutenberg.org/files/31061/31061-pdf.pdf
David Eugene Smith: "History of Mathematics, vol. 2", Dover Publications (1958)
http://books.google.de/books/about/History_of_Mathematics.html?id=uTytJGnTf1kC&redir_esc=y
http://jeff560.tripod.com/r.html
http://www-history.mcs.st-and.ac.uk/Biographies/Fibonacci.html
Edit As Paul Taylor said Florian Cajori favours another root of the root symbol, namely the generation of an upstroke from a row of points. In fact there may have been many sources playing together. Moritz Cantor points out that Alkasadi (or Alkasawi) an Arab living in Spain (died 1477 or 1486) wrote a book which is known under different titles like Lifting the veils of the science of the Gubar (Gubar means "dust" or "calculating with digits") where he not only abbreviated the Arab word for root dschidr by writing its first letter, but wrote it not right of the radicand (in Arabic, meaning in front of the radicand) but above. The jim can be seen in the column Initial in the table Arabic letters usage in Literary Arabic. This could also be a source for our root symbol.<|endoftext|>
TITLE: 2x2 subdeterminants of a matrix
QUESTION [17 upvotes]: If N>2, it is well known that if two invertible NxN matrices A and B have the same determinants of any 2x2 corresponding submatrices, then A=B or A=-B.
Given then all these 2x2 determinants of an invertible matrix A, is there an "explicit" way to recover/write down A?
If N=3 it is easy, as you can get the determinant of A (up to the sign) and all its cofactors, so you can obtain the inverse matrix of A or -A, but when N>3?

REPLY [21 votes]: By "corresponding submatrices" I presume you mean those $2\times2$ minors obtained by deleting $n-2$ colums and $n-2$ rows, where these columns and rows have the same $n-2$ indices. Once you've calculated the determinants of these submatrices you recover the action of $A$ on the exterior square $\Lambda^2 V$.
Now the following paper: "An algorithm for recognising the exterior square of a matrix" by Catherine Greenhill explains how to then obtain the original matrix $A$. Here's the relevant quote:

One computational problem which presents itself immediately is this: how can we
  determine whether a given matrix $Y$ is equal to the exterior square of another matrix $X$?
  In particular, if such an $X$ exists then we would like to construct one. A polynomial-time
  algorithm which solves this problem is described in Section 5.

The paper can be downloaded here.
One needs to be slightly careful here, because the exterior square does not quite determine the matrix $X$ uniquely. Here is another quote from the paper:

We prove in Section 4 that two matrices $X$, $X'$ with rank at least three
  have the same exterior square if and only $X'\in \{X, -X\}$.

So if the rank is at least three (which it is, since you are assuming invertibility), then we are pretty much done. I'm guessing that the situation where the rank is $\leq 2$ would be easy enough to resolve but in any case that's outside the scope of the question...<|endoftext|>
TITLE: Interesting thesis topic on statistical inference that is sufficiently mathematical 
QUESTION [6 upvotes]: Hello , I am a student who's gonna start honours in mathematics . Currently , I am at the stage of finding a suitable honours thesis topic . I've chosen my supervisor , who's research interest is on statistical inference and probability theory , but more on inference , I suppose. 
My undergraduate coursework consists mostly of pure mathematics , and I have not really much knowledge in statistical science , but I am really interested in the theory of statistics , that's why I chose a supervisor doing stat. 
But since my honours is in mathematics , my thesis should be sufficiently pure mathmatical. So my supervisor suggest that I could do something in shrinkage estimation , but he is even struggling on finding the right paper on it . And I guess I'd better conduct some research myself . 
I have no suffcient background in statistics , as I mentioned above , but I've done a lot of courses on mathematics( group ,rings ,field extensions , galois theory , representation theory , analysis,probability theory, general topology , algebraic topology ,moduli space etc. )  and small project also .  Now I really wanna combine theses things with theory of statists, but dont know where to start with.
I've heard that it is possible to put geometries into statistical theory , and this looks very attractive :D . I am particularly interested in geometry and topology . 
I am also thinking doing something on statistical inference and stochastic analysis. 
However , as my supervisor suggest , shrinkage estimation is a very good choice  . So maybe it would be better to follow his suggestion ? 
Anyway , I am really excited , but also struggling on choosing the right topic. Could anyone give me some advice on it ? Both reference to good papers on shrinkage estimation or perhaps suggestions on other related good topics would be both great . Thanks in advance !

REPLY [4 votes]: The intersection between computability theory and statistics is pretty interesting.  From this paper by Vovk (2009):  "It is widely accepted that advances in computing have brought about deep changes in the theory and practice of statistics. However, the use of the theory of computing, and, in particular, of its core notion of computability, has been very limited in the classical areas of statistics, such as parameter estimation and hypothesis testing."
Relatedly, Ackerman, et al. (2011) demonstrate a computable random variable $(X,Y)$ with non-computable conditional distribution $P(Y \mid X)$.
Certainly this area is pretty "mathy"; it remains to be seen if it has implications for statistical practice.<|endoftext|>
TITLE: Vertical and Horizontal Isomorphisms in 2-categories
QUESTION [5 upvotes]: Assume having a monoidal category $(\mathcal{A},I,\otimes)$ where each object has an inverse, i.e. for every object $A$ there is an object $\bar{A}$ with $A\otimes\bar{A}=\bar{A}\otimes A=I$. Then I wonder: If $f:X\rightarrow Y$ is an arrow, is $f$ invertible with respect to $\circ$ if and only if $f$ is invertible with respect to $\otimes$? The latter meaning that there is an arrow $\hat{f}:\bar{X}\rightarrow\bar{Y}$ with $f\otimes\hat{f}=\hat{f}\otimes f=I$. I'm thinking of: Define $\bar{f}:\bar{Y}\rightarrow\bar{X}$ as $\bar{X}\otimes f\otimes\bar{Y}$ (or $\bar{Y}\otimes f\otimes\bar{X}$, is it the same?). Then define $\hat{f}=\bar{f}^{-1}$. But I don't know whether that is an inverse to $f$ (wrt $\otimes$).

REPLY [10 votes]: Before answering, let me remark that it's generally considered a little weird (some would say, with tongue partly in cheek, "evil") to posit categorical axioms which enforce equalities between objects, since models of such axioms will not be invariant with respect to categorical equivalences. Such axioms ought to be replaced by enforcing corresponding isomorphisms between objects, satisfying suitable coherence axioms. 
With that in mind, let's start by putting the question a little differently: suppose that for each object $X$ we have units $\eta_X: I \to \bar{X} \otimes X$, $\eta_X': I \to X \otimes \bar{X}$, and counits $\epsilon_X: X \otimes \bar{X} \to I$, $\epsilon_X': \bar{X} \otimes X \to I$ such that the triangular equations for adjunctions $X \dashv \bar{X}$, $\bar{X} \dashv X$ hold and 
$$\eta_X' \circ \epsilon_X = 1_{X \otimes \bar{X}}, \qquad \eta_X \circ \epsilon_X' = 1_{\bar{X} \otimes X}$$ 
$$\epsilon_X \circ \eta_X' = 1_I = \epsilon_X' \circ \eta_X$$ 
(Edit: As noted by Chris and myself below, given an object $\bar{X}$ and invertible maps $\eta_X: I \to \bar{X} \otimes X$, $\phi: I \to X \otimes \bar{X}$, one can always construct an invertible map $\eta_X': I \to X \otimes \bar{X}$ such that $\epsilon_X := (\eta_X')^{-1}$ and $\epsilon_X' := (\eta_X)^{-1}$ are counits that fit into appropriate triangular equations. Details (for the more general case of 2-categories or bicategories in place of monoidal categories) are given in the nLab article mentioned in my comment.) 
Then, as you surmise, $f$ is invertible with respect to $\circ$ iff it is "invertible with respect to $\otimes$". Essentially we follow your definition of $\bar{f}$ given $f^{-1}$, defining 
$$\bar{f} = (\epsilon_X' \otimes 1_{\bar{Y}}) \circ (1_{\bar{X}} \otimes f^{-1} \otimes 1_{\bar{Y}}) \circ (1_{\bar{X}} \otimes \eta_Y)$$ 
and what we really mean by "invertible with respect to $\otimes$" is that we have equations 
$$\epsilon_Y \circ (f \otimes \bar{f}) \circ \eta_X' = 1_I = \epsilon_Y' \circ (\bar{f} \otimes f) \circ \eta_X.$$
The most perspicuous way to show these equations hold is by using string diagram calculus. Since I am not too handy with graphics, I'll just briefly sketch how to derive the first equation, leaving it to you to believe or verify that the second equation works similarly. The main steps are that 
$$\begin{array}
\epsilon_Y \circ (f \otimes \bar{f}) \circ \eta_X' & = & \epsilon_Y \circ (f \otimes 1_{\bar{Y}}) \circ (1_X \otimes \epsilon_X' \otimes 1_X \otimes 1_{\bar{Y}}) \circ (\eta_X' \otimes 1_X \otimes 1_{\bar{Y}}) \circ (f^{-1} \otimes 1_{\bar{Y}}) \circ \eta_Y' \\\ 
 & = & \epsilon_Y \circ (f \otimes 1_{\bar{Y}}) \circ (f^{-1} \otimes 1_{\bar{Y}}) \circ \eta_Y' \\\ 
 & = & \epsilon_Y \circ \eta_Y' \\\ 
 & = & 1_I
\end{array}
$$ 
where the first equation uses the definition of $\bar{f}$ and axioms of a monoidal category, the second uses a triangular equation, and the fourth uses one of the invertibility axioms for $Y$. Remaining details are similarly routine.

REPLY [3 votes]: I assume your are talking about a symmetric strict monoidial category $\mathcal{M}$. I think, you have to choose, beside $\bar{A}$, an isomorphism $\iota_A: A \otimes \bar{A} \to I$, say. I think $\iota_A = \mathrm{id}$ for all $A$ is not possible.
For any $f: X \to Y$ and $g:Y \to X$ follows, with your definition of $\bar{\square}$ and the interchange law:
$\bar{f} \circ \bar{g} := (\bar{X} \otimes f \otimes \bar{Y}) \circ (\bar{X} \otimes g \otimes \bar{Y}) = \bar{X} \otimes f \circ g \otimes \bar{Y}: \bar{X} \otimes X \otimes \bar{Y} \to \bar{X} \otimes Y \otimes \bar{Y}$
(By the symmetry, both of your definitons of $\bar{\square}$ 'agree', that is they fit in suitable commutative diagram. However for composing bars you need to insert the symmetry isomorphism $\gamma: A\otimes B \otimes C \mapsto C \otimes B \otimes A$ in suitable places, like this $ \bar{f}:= \gamma \circ (\bar{X} \otimes f \otimes \bar{Y}): \bar{X} \otimes X \otimes \bar{Y} \to \bar{Y} \otimes X\otimes \bar{X}$.)
That is, $f$ and $g$ are inverse then $\bar{f}$ and $\bar{g}$ are inverse to each other. Well, of course, this is nothing the compatability laws in your monoidial category.
As you suggested, define $\hat{f}:= \rho_{\bar{X}} \circ (\bar{X} \otimes \iota_Y)\circ (\bar{X} \otimes f \otimes \bar{Y}) \circ (\iota_{\bar{X}} \otimes \bar{Y})^{-1} \circ  \lambda_{\bar{Y}}^{-1}: \bar{Y} \to \bar{X}$
Where $\lambda_Y:I \otimes Y \to Y$ and $\rho_X: X \otimes I \to X$ are the designed natural isomorphims.
Exploiting the compatablity laws, this gives you $\hat{g} \circ \hat{f} = \mathrm{id}_X$ for $g:= f^{-1}$. Thus you obtain one implication.
I see, in the meantime Todd Trimble gave an answer which completly subsume mine. However as it took me some time to type it, i post it as a 'down to earth' version.
Edit:
I wonder if you obtain an endofunctor $\mathcal{M}^{op} \to \mathcal{M}$ via $X \mapsto \bar{X}$ and $f \mapsto \hat{f}$.
Well, probably you need some conditions like $\bar{\bar{X}} = X$ and $\iota_{\bar{X}} = {\iota_X}^{-1}$. I will think about. This may be also related to other direction.<|endoftext|>
TITLE: The set of orders of elements in a group 
QUESTION [13 upvotes]: Let $A$ be a subset of natural numbers. Consider the following problem:
Is there a group $G$ such that $\lbrace O(x) \; | \; x \in G \rbrace = A\cup\lbrace 1\rbrace$  ? (where $O(x)$ is the order of $x$) 
If you know any reference concerning this problem or any partial solution (containing a necessary or sufficient condition) , please let me know.

REPLY [16 votes]: For every fixed $n \in \mathbb{N}$, Rolf Brandl and Shi Wujie gave in Finite groups whose element orders are consecutive integers (Journal of Algebra, 143, 388-400 (1991).) a complete classification of finite groups whose spectrum is $\{1,2,\ldots,n\}$. A particularly appealing spin-off of their study is the following one:
Let $i$ be a positive integer greater than $8$. There is no finite group $G$ whose spectrum is $\{1,2,\ldots,i\}$.

REPLY [13 votes]: Obviously not for every set $A \subset \mathbb{N}$ there is a group $G$ with $A$ as
set of orders of its elements (usually called 'spectrum') -- for example if $G$ has an element of order $n$, then $G$ also has an element of order $d$ for every divisor $d$ of $n$.
For a survey of what is known on this question, you may check the following references:
H. Deng, M. S. Lucido, W. Shi: The Number of Isomorphism Classes of
Finite Groups with Given Element Orders. 
Algebra and Logic 41 (2002), Issue 1, 39-46.
Andrey Vasil'ev: On finite groups with the given set of element orders.
Talk slides, 2010.
V. D. Mazurov: Periodic groups with given element orders.
Talk slides, Mal'tsev Meeting, Novosibirsk, November 12-16, 2012.<|endoftext|>
TITLE: Permanent of a matrix of odd integers
QUESTION [13 upvotes]: It is clear that the permanent of an $n\times n$ matrix which entries are odd integers, is an even number, as it is the sum of $n!$ odd numbers. I am interested in finding the highest power of $2$ that divides the permanent of such a matrix.
Note that if $A$ is an odd $n\times n$  matrix, then $\det (A)\equiv 0 (\mod 2^{n-1})$. This can be seen by performing $n-1$ row operations to obtain an $n\times n$ matrix $B$ which $n-1$ of its rows consist of even integers (I found this nice observation in Amos Nevo Peter Sarnak's paper, "Prime and Almost Prime Points on Principal Homogeneous Spaces": http://web.math.princeton.edu/sarnak/NS-final-Oct-08.pdf).
I aim to find a similar result for the permanent. Of course, here I cannot use row operations. Running several thousands of examples on a computer, I suspect the following:
If $n=2^s -1$ for some integer $s\geq 2$ then $\text {perm}(A)\equiv 2^{n-s} (\mod 2^{n-s+1})$.
If $2^s \leq n < 2^{s+1} -1$ then $\text {perm}(A)\equiv 0 (\mod 2^{n-s})$.
I have been able to show this (by tedious case analysis) for $n=3,4,5$. Namely, I already have that: $\text {perm}(A_{3\times 3})\equiv 2(\mod 4)$,  $\text {perm}(A_{4\times 4})\equiv 0(\mod 4)$, $\text {perm}(A_{5\times 5})\equiv 0(\mod 8)$. 
Any idea how to generalize this for any $n$, or if something like this has been done before?

REPLY [10 votes]: Let us consider $s$ with $2^s\leq n<2^{s+1}$. First we prove the conjecture when all the entries of $A$ are $1$'s. Then $\mathrm{perm}(A)=n!$, hence by Legendre's formula the exponent of $2$ in it equals $n-t$, where $t$ is the number of $1$'s in the binary expansion of $n$. For $n=2^{s+1}-1$ we have $t=s+1$, hence the exponent equals $n-s-1$. For $2^s\leq n< 2^{s+1}-1$ we have $t\leq s$, hence the exponent is at least $n-s$.
For the general case we can assume that the statement holds for $n-1$. By the special case above, it suffices to show that if we increase by $2$ a single entry $a_{ij}$ of $A$, then the resulting matrix $B$ satisfies
$$ \mathrm{perm}(B) \equiv \mathrm{perm}(A) \pmod{2^{n-s}}. $$
Clearly,
$$ \mathrm{perm}(B) = \mathrm{perm}(A) + 2\mathrm{perm}(C),$$
where $C$ is the $(n-1)\times(n-1)$ matrix that results from $A$ by deleting the $i$-th row and the $j$-th column. Note that $2^s-1\leq n-1<2^{s+1}-1$, hence by the induction hypothesis we have
$$ \mathrm{perm}(C) \equiv 0 \pmod{2^{n-1-s}}. $$
The claim follows, and we are done.<|endoftext|>
TITLE: Essential uniqueness  of the real-analytic structure on $\mathbb R$
QUESTION [14 upvotes]: It is well-known that any $C^k$-smooth $1$-manifold  homeomorphic  to  $\mathbb R$ is $C^k$-diffeomorphic to $\mathbb  R$. The cases of $k\in{\mathbb N}\cup$ {$\infty$} may all be handled similarly by an elementary argument as follows: using  partitions  of unity  to construct a nowhere vanishing 1-form, integrate to  obtain  a diffeomorphism to a  connected  open submanifold of $\mathbb R$, and compose with some elementary diffeomorphism from the submanifold to $\mathbb R$.
Presented this way, the argument breaks  down in the real-analytic  case $k=\omega$  because partitions  of unity are no longer available. Even in  that case, the assertion is  still true, since Grauert-Remmert have shown that $C^1$-diffeomorphic real-analytic manifolds  are  real-analytically diffeomorphic  (assuming paracompactness, there being uncountably many inequivalent real-analytic structures  on the long ray). However, this is a very difficult general result. 
In the case at hand, it is not hard to see that the partitions of unity are merely a device for proving a cohomological vanishing theorem $$H^1({\mathbb R},{\mathcal E})=0$$ where ${\mathcal  E}$ is the sheaf  of germs of appropriately smooth real-valued functions. Indeed, consider a covering  of $\mathbb R$ by open intervals $I_n$, each intersecting only its immediate predecessor  and immediate successor, chosen small enough that each is real-analytically diffeomorphic to a standard  interval (hence also to $\mathbb R$). Note that any collection of functions  defined on the intersections $I_n\cap  I_{n+1}$ yields a 1-cocycle. Since the $I_n$ are standard intervals, there exist  everywhere positive 1-forms $\eta_n$ defined on $I_n$. Thus, there exist smooth functions $f_n$ defined on  $I_n\cap I_{n+1}$ such that  $\eta_{n+1}=(\exp f_n)\eta_n$ on  that intersection.   The vanishing theorem  implies that the 1-cocycle {$f_n: n\in{\mathbb Z}$} is a 1-coboundary, that is,  for some collection  of functions $g_n$ defined on $I_n$, we have that $f_n$  is the restriction  of  $g_{n}-g_{n+1}$ to $I_{n}\cap I_{n+1}$. By construction,  the 1-forms  $(\exp g_n)\eta_n$ on $I_n$ are the  restrictions of  a  globally defined  positive 1-form $\eta$. 
Question 1: How is the vanishing theorem established in the real-analytic  case?
I imagine this must be well-known, but I've not been able to find such a discussion in the literature.  Perhaps I am just  looking in the wrong places.  In  any event, several years ago I put together  such  an  argument. The  idea  is to consider  each consecutive pair of intervals $I_n, I_{n+1}$ each slightly thickened to a  complex neighborhood given  by a smoothly bounded Jordan domain $D_n$, the intersection  of consecutive neighborhoods being  another  Jordan  domain. If the neighborhoods are  small  enough, the given functions $f_n$ extend complex-analytically to the intersections, the real  part yielding values on $\partial(D_n\cap D_{n+1})$, which in turn  (suitably extended by 0)  yield a  function on (say) $\partial  D_n$ which we then extend harmonically, hence real-analytically, via  the Poisson  Integral formula.
Question  2: Is  such an argument been written down in the literature,  or otherwise  well-known?
Of course,  once we resort to patching suitable complex neighborhoods of the chart images, there is a quick and dirty proof via the Uniformization Theorem: it  suffices  to  glue suitable real-symmetric neighborhoods to obtain  a  simply connected  Riemann  surface with an  anticonformal involution whose fixed locus is the given 1-manfold.
Question  3: Is this surely well-known argument written  down  in the literature?
To be honest, I  started out knowing the argument via Uniformization,  and  decided  to see whether this could be reduced to more elementary considerations. The proposed argument to prove the vanishing theorem succeeds partially, but I was struck by the fact that I am still doing complex analysis, or at least  potential theory. Maybe it's unreasonable to expect to be able  to produce real-analytic functions  without sneaking a peek  into the complex plane.

REPLY [7 votes]: For Misha's comment and question 3 see the paper of David Minda Regular Analytic arcs and curves Colloq.Math 38(1977) no 1 73-82 .Regarding the vanishing theorem for real analytic manifolds this was proved by Henri Cartan Bulletin de la S.M.F tome 85 (1957) 77-99. Cartan assumed his manifolds were real analytically embedded in euclidean space . One just needs to know that the Complexification is Stein .This is due to Grauert's solution for the Levi problem.
  In case of one dimensional real analytic manifolds the complexification is Stein by the Runge type theorem of Behnke and Stein . You can also use the Behnke-Stein theorem to show that real analytic one manifolds have complete real analytic metrics and follow Milnor's proof.
   All proofs I know go through Complexifications .Existence of Complexification for real analytic manifold is due to Whitney and Bruhat<|endoftext|>
TITLE: Clifford Lie Algebras
QUESTION [5 upvotes]: I'm studying the "Clifford Lie Algebra" (see http://arxiv.org/pdf/1007.2481.pdf page 30).
It's basically a way to look at Clifford algebras and their properties in a Lie algebraic
setting (which I find appealing). I'm looking for a reference that looks at this in more
detail; one that discusses the lie subalgebras even better. Thanks.

REPLY [3 votes]: I don't know anyone else who calls this the "Clifford Lie Algebra".  It is just one of the basic applications of Clifford algebras.   Given the Clifford algebra of a quadratic form, the quadratic elements of the Clifford algebra give you the Lie algebra of the orthogonal group of that quadratic form.
There are many places to read about this, one of them would be Chapter 1.6 of "Spin Geometry" by Lawson and Michelson.  I've written up some notes for a graduate course that include this, see here:
http://www.math.columbia.edu/%7Ewoit/LieGroups-2012/cliffalgsandspingroups.pdf<|endoftext|>
TITLE: Example of fiber bundle that is not a fibration
QUESTION [25 upvotes]: It is well-known that a fiber bundle under some mild hypothesis is a fibration, but I don't know any examples of fiber bundles which aren't (Hurewicz) fibrations (they should be weird examples, I think, because if the base space  is paracompact then  the bundle is a fibration).
Does anybody know an example?
Thanks!

REPLY [22 votes]: $\newcommand{\RR}{\mathbb{R}}
\newcommand{\To}{\longrightarrow}
\newcommand{\id}{\mathrm{id}}$The example described in Tom Goodwillie's answer to a related mathoverflow question essentially solves this question. Specifically, Tom defines an orientable, non-trivial, real line bundle $L$ over a contractible space $X$. The principal $GL_1^+(\RR)$-bundle — $GL_1(\RR)$ will work as well — associated to $L$ is then a locally trivial fibre bundle $p:E\to X$. The principal bundle $p:E\to X$ does not admit a section, since the line bundle $L$ is not trivial. It follows that $p$ cannot be a Hurewicz fibration: otherwise it would admit a section, given that $X$ is contractible. For convenience, I will give below the construction of $p:E\to X$ and a few details of the proof — mostly copied from Tom's answer, and following his notation where possible. Please upvote Tom Goodwillie's answer, which is certainly shorter and more readable.
The base space $X$
Let $X$ be the space obtained by gluing two copies of $\RR$ along $\RR^+$:
$$ X = (\RR\times\{0,1\})\mathbin{/}{\sim} $$
where ${\sim}$ is generated by $(x,0)\sim(x,1)$ for $x\in\RR^+$. This space is not Hausdorff, and is closely related to the well-known line with two origins. Let $q:\RR\times\{0,1\}\to X$ be the quotient map. Define two open subspaces covering $X$ by $U=q(\RR\times\{0\})$ and $V=q(\RR\times\{1\})$. Finally, let $g:X\to\RR$ be the continuous function determined by $g(q(x,i))=x$, and define
$$f=g|_{U\cap V}: U\cap V \To \RR^+$$
Importantly, observe that $f$ does not extend to a continuous map $X\to\RR^+$.
The total space $E$
Consider the topological abelian group $G=GL_1^+(\RR)=(\RR^+,\cdot)$ given by the positive reals with multiplication. Let $E_U=U\times G$ and $E_V= V\times G$ denote the trivial $G$-bundles over $U$ and $V$, respectively. Construct the principal $G$-bundle $E$ over $X$ by gluing $E_U$ and $E_V$ along $U\cap V$ via the $G$-isomorphism
$$ \varphi_f : E_U|_{U\cap V}\overset{\simeq}{\To} E_V|_{U\cap V} $$
defined by
$$ \varphi_f(x,g)=\bigl(x,f(x)\cdot g\bigr) $$
More concretely, $E$ is obtained from $E_U \amalg E_V$ by identifying $(x,g)\in E_U$ with $\varphi_f(x,g)\in E_V$ for each $x\in U\cap V$. As in Tom Goodwillie's answer, we could just as well use any other continuous function $f:U\cap V\to\RR^+$ which does not extend to a continuous function $X\to\RR^+$.
Non-triviality of the principal bundle $E\to X$
The projection map $p:E\to X$ gives a principal $G$-bundle over $X$, which comes with canonical isomorphisms $E|_U = E_U$ and $E|_V = E_V$. We will now show $p$ does not admit a section. By the construction of $E$, a section of $p:E\to X$ determines:

a section of $E_U=U\times G\to U$, and therefore a map $s_U:U\to G=\RR^+$;
similarly, a map $s_V:V\to G=\RR^+$;
these maps verify $s_V(x)=f(x)\cdot s_U(x)$ for each $x\in U\cap V$.

In particular, $f(x)=s_V(x)/s_U(x)$ for all $x\in U\cap V$. However, this implies that $f$ extends to a continuous function $\overline{f}:X\to\RR^+$ given by
$$ \overline{f}(x)=\frac{s_V\bigl(q(g(x),1)\bigr)}{s_U\bigl(q(g(x),0)\bigr)} $$
which contradicts the known non-extension property of $f$.
Conclusion
The projection $p:E\to X$ gives a locally trivial principal $G$-bundle over $X$, since $E|_U\simeq E_U$ and $E|_V\simeq E_V$ are trivial $G$-bundles. Thus, it remains to show that $p$ is not a Hurewicz fibration. Note that $X$ is contractible. So if $p:E\to X$ were a fibration, it would necessarily admit a section. In detail, let $H:X\times I\to X$ be a null-homotopy of $\id_X$. We can obviously lift the constant map $H_1$ to $E$. Assuming $p$ is a Hurewicz fibration, the homotopy lifting property then produces a lift $\widetilde{H}: X\times I\to E$ of $H$ to $E$, and consequently a section $\widetilde{H}_0$ of $p$. But we showed in the previous paragraph that $p$ admits no sections. In conclusion, $p$ is not a Hurewicz fibration.<|endoftext|>
TITLE: Totally Geodesic Submanifolds
QUESTION [13 upvotes]: Suppose that $N$ is a totally geodesic submanifold of a complete Riemannian manifold $(M,g)$. Is it the case that a geodesic segment that minimizes length in the submanifold $N$ also minimizes length in the ambient manifold $M$?

REPLY [18 votes]: Let $M$ be the flat cylinder $\mathbb{R} \times S^1 \subset \mathbb{R} \times \mathbb{C}$ and $N = \{(t,e^{it}) | t \in \mathbb{R}\}$, which is a geodesic (hence a complete totally geodesic submanifold of $M$) minimizing between any two points of $N$ (among the geodesics of $N$). But the minimizing geodesic in $M$ between the points $(0,1)$ and $(2\pi,1)$ is the segment $\{(s,1) | s\in[0,2\pi]\}$.<|endoftext|>
TITLE: What are the general techniques for proving a variety is not toric?
QUESTION [9 upvotes]: When I read the paper "A survey on Cox rings" with the link below:
http://math.berkeley.edu/~velasco/Survey.pdf
In Section 4.2 it is mentioned that the Del Pezzo surface given by blowup of $\mathbb{P}^2$ at $s>3$ general points is not toric. Is there an easy way to see this?

REPLY [2 votes]: To be a toric variety implies many other things. For instance if $X$ is a toric variety, then

$X$ is a Mori Dream Space, i.e. $Cox(X)$ is finitely generated. Indeed $X$ is a toric variety if and only if $X$ is a Mori Dream Space and $Cox(X)$ is a polynomial ring.
$X$ is log Fano, in particular $-K_X$ is big and movable.
Any nef divisor is semiample.

Therefore to show that $X$ is not toric one could prove that $-K_X$ is not big or work out a nef divisor which is not semiample.<|endoftext|>
TITLE: What information is required for SYZ mirror symmetry? 
QUESTION [5 upvotes]: Let $X$ be a Calabi-Yau threefold. The Strominger-Yau-Zaslow conjecture suggests that $X$ should have a special Lagrangian $T^3$-fibration (when $X$ lies near a large complex structure limit) and a mirror manifold $Y$ is obtained by "dualizing the smooth fibers $T^3$". I want to know what "dualizing the smooth fibers $T^3$" rigorously means. 
To formulate my question, 

What information is required to construct the mirror manifold?

To dualize smooth fibers, we definitely need a section of the fibration to specify an origins of each smooth fiber. Does the section need to be special Lagrangian as well? 
Also, what do we need to know about the singular fibers in order to incorporate instanton corrections (and modify the gluing of the complex structure)?

REPLY [9 votes]: I'll fill in a few details here; more can be found in the references that Daniel gave.
Suppose first that $f:X\rightarrow B$ is a special Lagrangian $T^n$ fibration with only
smooth fibres. If we just want to describe the dual as a complex manifold, we do the following. Hitchin showed that $f$ induces two affine structures on $B$. The important one
for us is the one coming from the fact that $f$ is a Lagrangian fibration. Locally choose
submanifolds $\gamma_1,\ldots,\gamma_n$ of $X$ smooth over $B$ whose restriction to
each fibre provides an integral basis for $H_1$ of that fibre. We can use these to
define one-forms $\lambda_i$ on $B$ via $f_*(\omega|_{\gamma_i})$ (where $\omega$ is
the Kaehler form). Here $f_*$ denotes fibrewise integration. These forms are closed
and hence locally there exists functions $y_i$ such that $dy_i=\lambda_i$. One checks
that $y_1,\ldots,y_n$ form a coordinate system locally on $B$. Once the cycles $\gamma_i$ are chosen the $y_i$ are only well-defined up to constants, and we can always change basis, and
as a result the coordinates are well-defined up to affine linear changes of coordinates,
where the linear part of the affine linear transformation must be integral.
Now here is the dual: consider the local system $\Lambda$ of lattices contained in the tangent bundle $T_B$ of $B$, given locally by integral linear combinations
of the tangent vectors $\partial/\partial_{y_1},\ldots,\partial/\partial_{y_n}$. This
gives a lattice in each fibre of $T_B$. Define the dual torus bundle to be $T_B/\Lambda$,
i.e., divide each tangent space out by the lattice generated by the above tangent vectors.
The projection to $B$ gives the dual torus bundle.
Note that we did not need to have a section of the orginal fibration, although the dual
does have a section. It is possible to twist the dual so that it doesn't have a section;
this is discussed in my paper http://arxiv.org/pdf/math/9809072.pdf
Now this dual, which I'll write as $X(B)$, does carry the ``semi-flat'' complex structure.
This is described via complex coordinates $q_i=\exp(2\pi \sqrt{-1}(x_i+\sqrt{-1}y_i))$,
where $x_i$ is the function on the tangent bundle given by $dy_i$. 
Now we allow singular fibres, i.e., consider $f:X\rightarrow B$ whose fibres are
only smooth tori over an open subset $B_0\subseteq B$. So we get the dual $X(B_0)\rightarrow
B_0$ and the question is how we compactify this. In particular, if we want to obtain a complex manifold, we need to deform the above semi-flat complex structure via instanton
corrections. We might assume that $f$ is only a Lagrangian fibration, as that is the
only part of the structure we used to get the dual as a complex manifold, and it is easier to construct Lagrangian fibrations. In this case there has been a certain amount of success describing instanton corrections explicitly in terms of counting holomorphic disks on
$X$. This explicit program was initiated by Denis Auroux in the reference Daniel gave,
but so far can only be done for very simple singular fibres and very few singular
fibres.
The point of view taken by myself and Siebert, as well as by Kontsevich and Soibelman
in the paper Daniel referenced, is that one discards the explicit description of the dual
as a complex manifold, as this introduces too many analytic difficulties in general. Instead, Siebert and I replace this with a degenerating family of schemes over the spectrum of a formal power series ring. Most convergence issues disappear, and when the singularities
are of a relatively simple form, a purely algorithmic approach can be taken to constructing this family. A posteriori one expects that the data produced by this algorithmic approach are in fact the instant corrections, i.e., are produced by counts of holomorphic disks
on $X$. This most precise realization of this expectation is in my paper with Hacking
and Keel, http://arxiv.org/abs/1211.6367 There we explicitly construct the mirror of a surface using the Gromov-Witten invariants of the surface, with no restriction on what the expected singular fibres of a Lagrangian fibration on the surface might be.<|endoftext|>
TITLE: Irreducible divisors containing an arbitrary closed set
QUESTION [8 upvotes]: Let $X$ be a normal complex projective variety, let $V$ be a closed subset of $X$ (possibly reducible), and let $I_V$ be its ideal sheaf (consider the reduced scheme structure for example).
If $A$ is an ample divisor on $X$, then $\mathcal{O}_X(mA)\otimes I_V$ is globally generated if $m \gg 0$ by one of the definitions of ampleness.
This implies in particular that there exists a divisor $A'\in |mA|$ such that $A'$ contains $V$ in its support. In fact there is much more than one divisor with this property. 
In general there is no reason why $A'$ should be irreducible.
In particular if $V$ contains at least two prime divisors, it is clear that every divisor containing $V$ will be reducible.
My question is the following:
Suppse $\dim V\leq \dim X-2$. Can I find, for $m \gg 0$, an irreducible divisor $A'\in |mA|$ containing $V$?

REPLY [8 votes]: EDIT : Olivier Wittenberg pointed out to me that a positive answer to the question follows from Theorem 1 of [Altman-Kleiman, Bertini theorems for hypersurface sections containing a subscheme]. I keep below my previous answer (whose argument is different, but more complicated).

The answer to your question is positive. Let $X$ be an integral projective variety of dimension $n\geq 2$ (over an algebraically closed field) and $V\subset X$ a subvariety of codimension $\geq 2$. Let $A$ be ample on $X$ : up to replacing $A$ by a multiple, we may assume that $X\subset\mathbb{P}^N$ and $A=\mathcal{O}_X(1)$.
Let $\mathcal{H}_e$ be the projective space of degree $e$ hypersurfaces in $\mathbb{P}^N$,
$\mathcal{F}_e^{igr}(X)\subset\mathcal{H}_e$ the subset consisting of hypersurfaces whose intersections with $X$ are not irreducible generically reduced of codimension $1$ in $X$, and 
$\mathcal{G}_e(V)\subset\mathcal{H}_e$ the subset consisting of hypersurfaces containing $V$.
The exact sequence $0\to\mathcal{I}_V(e)\to\mathcal{O}_{\mathbb{P}^N}(e)\to\mathcal{O}_V(e)\to 0$ shows that, when $
e\gg 0$, the codimension of $\mathcal{G}_e(V)$ in $\mathcal{H}_e$ is a polynomial of degree $\leq n-2$ in $e$ (the Hilbert polynomial of $V$). On the other hand, Théorème 0.4 of arXiv:0911.1118 shows that, when $e\geq 2$, the codimension of $\mathcal{F}_e^{igr}(X)$ in $\mathcal{H}_e$ is $\geq \binom{e+n-1}{n-1}-n$, that is at least  a polynomial of degree $n-1$ in $e$. As a consequence, if $e \gg 0$, $\mathcal{G}_e(V)$ is not included in $\mathcal{F}_e^{igr}(X)$. A hypersurface in $\mathcal{G}_e(V)$ but not in $\mathcal{F}_e^{igr}(X)$ induces on $X$ the divisor you are looking for.
Note that if $X$ is moreover $S_2$ (for instance, normal), then this divisor is itself integral.<|endoftext|>
TITLE: decompositions for exceptional Lie algebras $E_6$, $E_7$ and $E_8$
QUESTION [13 upvotes]: Do anybody know , why can we write the following decompositions for exceptional Lie algebras  $E_6$, $E_7$ and $E_8$
1) $E_8=(V^{\star}\otimes \wedge^{8}V^{\star})\bigoplus \wedge^{6}V^{\star}\bigoplus \wedge^{3} V^{\star} \bigoplus \mathfrak{g}\mathfrak{l}(V)\bigoplus  \wedge^{3} V\bigoplus \wedge^{6} V \bigoplus(V\otimes \wedge^{8}V)$, here $dimV=8$
2)  $E_7= \wedge^{6}V^{\star}\bigoplus \wedge^{3} V^{\star} \bigoplus \mathfrak{g}\mathfrak{l}(V)\bigoplus  \wedge^{3} V\bigoplus \wedge^{6} V $,here $dimV=7$
3) $E_6= \wedge^{6}V^{\star}\bigoplus \wedge^{3} V^{\star} \bigoplus \mathfrak{g}\mathfrak{l}(V)\bigoplus  \wedge^{3} V\bigoplus \wedge^{6} V $,here $dimV=6$

REPLY [37 votes]: The $E_6$ and $E_7$ decompositions you list are explained in Cartan's 1894 thesis (see pages 89–92 for these formulae).  For $E_8$, Cartan instead gives a decomposition (a $\mathbb{Z}_3$-grading) of the form
$$
{\frak{e}}_8 = \Lambda^3(W^\ast)\oplus {\frak{sl}}(W)\oplus \Lambda^3(W)
$$
for a $9$-dimensional space $W$.  From this, you can get the decomposition you list for $E_8$ by writing $W = L\oplus V$ where $L$ has dimension $1$ and $V$ has dimension $8$.  Then, using the fact that $L\otimes \Lambda^8(V)\simeq \Lambda^9(W)$, which is trivial under $\mathrm{SL}(W)$, one finds that, under the action of $\mathrm{GL}(V)\subset \mathrm{SL}(W)$, these three spaces break up into the $7$ spaces (actually, $8$, since ${\frak{gl}}(V)$ has a center) you have listed for $E_8$.
There are other places in Cartan's papers where he explains this further.  For example, the $E_6$ case is explained at greater length in his paper Les groupes réels, simples, finis, et continus (Ann. Éc. Norm. 31 (1914), 263–355).  See the formulae on pp. 298 & 299.
It's probably worth adding that, in this same paper, beginning on page 313, Cartan gives a similarly nice $\mathbb{Z}_2$-graded decomposition
$$
{\frak{e}}_7 = {\frak{sl}}(W)\oplus \Lambda^4(W)
$$
where $W$ is a vector space of dimension $8$.  (The even part is the ${\frak{sl}}(W)$ subalgebra, and the odd part is the $\Lambda^4(W)$, which is isomorphic to  $\Lambda^4(W^\ast)$ as an $\mathrm{SL}(W)$-module.)  To get the decomposition you list, you use the same symmetry-breaking idea as worked for $E_8$:  Write $W = L \oplus U$ for $L$ a $1$-dimensional subspace and $U$ a $7$-dimensional subspace, then, under $\mathrm{GL}(U)$, we have $L\simeq \Lambda^7(U^\ast)$, and the above decomposition breaks into
$$
{\frak{e}}_7 = \bigl(\mathbb{R}\oplus {\frak{sl}}(U)\oplus L{\otimes}U^\ast\oplus L^\ast{\otimes}U\bigr)\oplus \bigl(\Lambda^4(U)\oplus L{\otimes}\Lambda^3(U)\bigr),
$$
so now, if you set $V = K\otimes U$, where $K^{\otimes 3}= L$, the pieces line up with what you have listed after you permute them around a bit.  (The only tricky part is seeing that you can take a cube root of the line $L$.)<|endoftext|>
TITLE: Bijective proof of Ramanujan's congruence
QUESTION [6 upvotes]: Is there a known bijective proof of Ramanujan's congruence for the partition function modulo 5?  E.g., is there a construction that for every $n$ congruent to 4 mod 5 gives a permutation of the partitions of $n$ that increases the Dyson rank of each partition by 1 mod 5?

REPLY [4 votes]: It's hard to be absolutely certain of course, but I would say no. 
The most recent reference I can find mentioning the absence of such a proof is this article by Bessenrodt and Pak (2003). I also got some negative answers from modular form specialists to whom I asked the very same question fairly recently...<|endoftext|>
TITLE: When is a smooth projective variety a fibration
QUESTION [5 upvotes]: Let $X$ be a smooth projective variety. Is there a criterion (apart from the definition) for the existence of a projective curve $C$ and a proper surjective morphism $\pi:X \to C$?

REPLY [8 votes]: Another interesting theorem in this direction is Castelnuovo-de Franchis theorem. It says that if you have two linearly independent holomorphic 1-forms $\omega_1,\omega_2$ with $\omega_1\wedge\omega_2=0$ on $X$, then there exists a morphism $f:X\to C$ with $C$ a smooth curve of genus at least 2 and forms $\omega_i'$ on $C$ such that $\omega_i=f^*\omega_i'$.

REPLY [2 votes]: By a theorem of Gromov and Schoen if the fundamental group of X is a proper amalgamated product
or HNN extension then X maps surjectively to a curve .<|endoftext|>
TITLE: Maximum distance within a subset of permutations
QUESTION [5 upvotes]: I'm modelling a scheduler that accepts a sequence of requests and outputs a sequence of responses, one response per request.  It can partially reorder requests, but only within a finite queue.  Specifically, I have a queue with maximum size q.  The numbers 1 through N are fed into the queue in order.  When the queue reaches its maximum size, the scheduler removes an item from the queue and outputs it.  When the end of input is reached, the scheduler removes and outputs items from the queue, one at a time.  The choice of which item to remove from the queue is completely up to the scheduler.  Therefore, the output is a permutation of the input, however, not all permutations are possible.  For example, if the queue size is only 2, it would be impossible to output the numbers 1 through N in reverse order.  I'll call the set of possible permutations P.
I have two permutations x and y that are elements of P.  I want to compare them by looking at the inversion count of $xy^{-1}$.  ($xy^{-1}$ is not necessarily an element of P.)  What is the maximum possible value of this inversion count?  Is there literature available on this topic?
Edit:
If it makes the problem any easier, Spearman's footrule could be used as the comparison instead of the inversion count.  The particular choice of metric is not important as long as the distance of 21345 from the identity is less than the distance of 52341 from the identity, i.e., the distance between transposed elements is important.  The triangle inequality is nice, but not absolutely necessary.  (This is an engineering problem after all, so the constraints on the math are flexible.)  Also, the most useful value for q at the moment is 32, although this will vary in the future.
It looks like P is not closed, which is a shame, but not surprising.  Using q=2, if we let x=2143 and y=1324, then $xy^{-1}$=2413, which is not in P.

REPLY [5 votes]: Let me give a very partial answer to your question, only describing what permutations lie in the set $P$: This of course depends on $q$, the size of the queue. See e.g., http://en.wikipedia.org/wiki/Permutation_pattern for the definition of avoidance used below:

If $q=1$ then clearly $P$ just contains the identity permutation, which happens to be the set of avoiders of the classical pattern $21$.
If $q=2$ then $P$ is the set of avoiders of the classical patterns $312$ and $321$. 
If $q=3$ then $P$ is the set of avoiders of the classical patterns $4123$, $4132$, $4213$, $4231$, $4312$, $4321$.

You can continue this description as one would expect:

For a general $q \geq 1$, $P$ is the set of avoiders of all patterns of the form $(q+1) \pi$ where $\pi$ is any pattern of length $q$.

The justification for this is that if the size of the queue is $q$ then you can not get a large element $x$ out of the queue before $q$ smaller ones.<|endoftext|>
TITLE: Systems of simultaneous real quadratic equations
QUESTION [5 upvotes]: Starting from a problem in spectral graph theory, I got dragged into a problem in combinatorial matrix theory about constructing $n\times n$ real orthogonal matrices with a specified pattern of zero/non-zero entries.  (I can expand on this if required.)
Anyway, I have a construction which appears to work (which I have tested in Maple for some small values of $n$), but to prove it in full generality would require showing that a certain system of simultaneous real quadratic equations has a real solution.
I know very little about algebraic geometry and, well, nothing at all about real algebraic geometry, but I was wondering if there were any existing results in the literature which may help with this---as it stands I'm not even sure where to start looking!
[edited after Alexandre's answer]
The aim (in the first instance) is to construct an orthogonal $n\times n$ matrix with zero diagonal entries and non-zero off-diagonal entries. The idea is to use the following matrix: $$ A=\left[ \begin{array}{c|ccc} 0 & 1 & \cdots & 1 \\ \hline 1 & & & \\ \vdots && B & \\ 1 &&& \end{array} \right] $$
where the rows of $B$ are formed of the $n-1$ cyclic permutations of $v=(0,x_1,\ldots,x_{n-2})$.  For this to be orthogonal (up to scaling by $\sqrt{n-1}$), the inner product of two distinct rows must be $0$, and the norm of each row must be $\sqrt{n-1}$.  This gives the equations $\sum_{i=1}^{n-2} x_i = 0$, $\sum_{i=1}^{n-2} x_i^2 = n-2$, and $\lceil (n-2)/2\rceil$ other equations arising from inner products of rows containing cyclic permutations of $v$ (from adjacent rows, rows 2 apart, 3 apart, etc.).  Also, we have the constraint that any solution for $x_1,\ldots.x_n$ may not contain a zero.
For example, when $n=5$, Maple finds the solution $x_2=1$, $x_1,x_3 = \frac{-1\pm \sqrt{3}}{2}$.

REPLY [3 votes]: This question is linked to the discrete Fourier transform. If $u=[1,\cdots,1]^T$, then $A=\begin{pmatrix}0&u^T\\u&B\end{pmatrix}$ and the equation $AA^T=(n-1)I_n$ is equivalent to $Bu=0,BB^T=(n-1)I_{n-1}-uu^T$. The spectrum of $BB^T$ is $0$ and $n-1$ ($n-2$ times). Let $F$ be the $(n-2)\times(n-2)$ matrix defined by $f_{j,k}=\exp(\dfrac{-2i\pi jk}{n-1})$ (the partial Fourier matrix associated to the non-zero eigenvalues of $B$). $B,B^T$ commute and they are simultaneously diagonalizable (using the complete Fourier matrix) and similar to $B'=diag(F[x_1,\cdots,x_{n-2}]^T)$ and to ${B'}^T=diag(F[x_{n-2},\cdots,x_1]^T)$. The eigenvalues of $BB^T$ are the $B'_{i,i}{B'}^T_{i,i}$. Finally our vector $X$ is solution of the quadratic system: for every $i\leq n-2$, $(\sum_{j=1}^{n-2}f_{i,j}x_j)(\sum_{j=1}^{n-2}f_{i,j}x_{n-1-j})=n-1$. 
EDIT: Here $X$ is real; then the equations can be rewritten: for every $i\leq n-2$, $|\sum_{j=1}^{n-2}f_{i,j}x_j|=\sqrt{n-1}$.<|endoftext|>
TITLE: 4D TQFT from a modular tensor category
QUESTION [19 upvotes]: I know the construction of a 3D topological quantum field theory (TQFT) from a modular tensor category.
I heard that we can even (mathematically) construct 4D TQFT from a modular tensor category. I would like to know how to construct it.
I'd like to study it but I don't know how to search references. Is there any good word to search this 4D TQFT? Or could you suggest references?
I also want to know if there is another mathematically constructed 4D TQFT and how it is called.
Thank you in advance.
(This question was asked in math.stackexchange but no answer was given.here)

REPLY [12 votes]: The TQFT in question should probably be called the Crane-Yetter or Crane-Yetter-Kauffman TQFT.
Crane-Yetter-Kauffman didn't work it out as a fully extended theory, and didn't notice (so far as I can tell) the relation to Witten-Reshetikhin-Turaev theories, but they definitely were the first to write down the 4d part of the theory.
The CYK TQFT contains all of the information of the WRT TQFT.  (This disagrees with David Ben Zvi's answer, but I think the difference is due to our using different axiomatic frameworks for TQFTs, not disagreement about mathematical facts.)  More specifically,
$$
  Z_{WRT}(X, \Gamma) = Z_{CYK}(\partial^{-1}(X); \Gamma).
$$
Here $X$ is a manifold of dimension 1, 2 or 3 (not necessarily closed).  $X$ is equipped with extra structure ($p_1$ structure, signature structure, null-bordism structure, ...) which makes $\partial^{-1}(X)$ sufficiently unambiguous.  (For example, if $X$ is a closed 3-manifold, then the choice of $\partial^{-1}(X)$ only matters up to bordism.)  The $\Gamma$ on the left hand side is a collection of "Wilson loops" or more generally a Wilson (labeled) graph.  The $\Gamma$ on the right hand side is a boundary condition.  (Same graph, but different interpretation.)
For more details, see Chapter 9 of these notes.
One way of looking at this is as follows.  We expect, roughly, a correspondence
$$
  \mbox{$n$-category} \;\; \leftrightarrow \;\; \mbox{$(n{+}1)$-dimensional TQFT}.
$$
The input data for for a WRT TQFT is a modular tensor category, which is a particular type of 3-category.  But the WRT TQFT is a (2+1)-dimensional theory, not a (3+1)-dimensional theory, so something weird is going on here.  The natural thing to do with a modular tensor category is to build the (3+1)-dimensional CYK TQFT, which is fully extended (a "0-1-2-3-4" theory) and anomaly-free.  One then notices that the CYK theory is almost trivial for closed manifolds (more specifically, the dimensional reduction by $S^1$ is 2-Morita trivial), so we can derive from the CYK TQFT the (2+1)-dimensional WRT TQFT via the slogan
$$
  Z_{WRT}(X, \Gamma) = Z_{CYK}(\partial^{-1}(X); \Gamma).
$$
But note that since CYK is merely almost trivial and not completely trivial, the WRT TQFT acquires an anomaly (i.e. manifolds need to be equipped with extra structure).  Also, since it's hard to make sense of $\partial^{-1}(X)$ when $X$ is a point, the WRT theory is not fully extended; it's a 1-2-3 theory rather than a 0-1-2-3 theory.
I should also note that the input data for the CYK can be a premodular category ($S$-matrix perhaps degenerate).  When the input is premodular but not modular, then the CYK TQFT is not almost trivial and we cannot construct a (2+1)-dimensional TQFT as above.<|endoftext|>
TITLE: An extension of the converse to Hall's theorem.
QUESTION [9 upvotes]: This is an extension of this MSE question, in which I asked whether there was a counterexample to the following statement,

Conjecture. If a finite group $G$ contains a $\lbrace p,q \rbrace$-Hall subgroup for every pair of primes $p$ and $q$ dividing $|G|$, then $G$ is solvable.

which is a refinement of the converse to Hall's theorem,

Theorem (Hall). Denote by $\pi(G)$ the set of prime divisors of $|G|$.  Then $G$ is solvable if and only if $G$ contains a $\pi$-Hall subgroup for every subset $\pi$ of the prime divisors of $|G|$.

Edit: As requested, we call $H\leqslant G$ a $\pi$-Hall subgroup if $|H|$ and $[G:H]$ are coprime and $p$ divides $|H|$ for every $p\in \pi$.  So, Hall subgroups are a generalization of Sylow subgroups for multiple primes.
I received a great answer from Geoff Robinson, who thinks there probably isn't counterexample, and proposed to check it case-by-case using the classification of finite simple groups.  I am still digesting his answer, however this led me to wonder whether (assuming the statement is true) there is a proof that does not rely on the classification theorem.
My original idea, before I posted the question on MSE, was to was to show that this implies the hypothesis to the to the original converse - that is, $G$ contains a $\pi$-Hall subgroup for every $\pi\subseteq \pi(G)$ iff it contains a $\{ p,q \}$-Hall subgroup for each $p,q\in \pi(G)$ - but I haven't found any way to make this work yet.  It has also been suggested to me to try to mimic the original inductive proof, which (for $|\pi(G)|\geq 3$) makes use of this lemma, but again I do not see how to put it together.
So, my question is, is this conjecture provable without using the classification of finite simple groups?
If it is false, is there a counterexample of a non-solvable group with $\lbrace p,q\rbrace$-Hall subgroups for every pair of primes (which is thus missing some other Hall subgroup, e.g. a $\lbrace p,q,r \rbrace$-Hall subgroup)?

REPLY [7 votes]: For the record, I believe that P. Hall proved that if $|G|$ has $n$ prime divisors, then $G$ is solvable if and only if $G$ has $n$ Sylow subgroups $P_{1},P_{2}, \ldots ,P_{n},$ one for each prime divisor, such that $P_{i}P_{j} = P_{j}P_{i}$ for $1 \leq i,j \leq n.$ You are asking whether the pairwise permutability condition can be dropped.
 The proof of the more difficult direction Hall's Theorem is something like the following, given Burnside's $p^{a}q^{b}$-theorem. I have forgotten Hall's proof, so have had to concoct a proof which is largely the same as Hall's except that I need to invoke Glauberman's $ZJ$-theorem, which Hall did not require. For suppose that $G$ has such a set of permutable Sylow subgroups, and we wish to prove that $G$ is solvable. Then we may suppose that $n \geq 3,$ otherwise Burnside's $p^{a}q^{b}$-theorem yields the desired result. By induction, for $1 \leq i \leq n,$ $G$ has a solvable subgroup $H_{i}$ with $G = H_{i}P_{i} = P_{i}H_{i}$ and $H_{i} \cap P_{i} = 1$ (we may take $H_{i} = \prod_{j \neq i} P_{j}$ which is a group by the permutability condition, and has order $[G:P_{i}]).$ We may also suppose that $p_{1} \geq 5,$ and we do. If $P_{1}$ normalizes a non-trivial subgroup $N_{1}$ of $H_{1},$ then we have $\cap _{g \in G} H_{1}^{g}$  = $\cap_{x \in P_{1}} H_{1}^{x} \geq N_{1},$ so $G$ has a non-trivial solvable normal subgroup $K,$ and an induction argument shows that $G/K$ is solvable. Hence for $2 \leq j \leq n,$ we have $O_{p_{j}}(P_{1}P_{j}) = 1.$ Since $P_{1}P_{j}$ is solvable, and $p_{1} \geq 5,$ we have $ZJ(P_{1}) \lhd P_{1}P_{j}$ for each such $j,$ by Glauberman's $ZJ$-theorem. But then $ZJ(P_{1}) \lhd G,$ since it is normalized by each of $P_{1},P_{2}, \ldots ,P_{n}.$ Again, and induction argument shows that $G/ZJ(P_{1})$ is solvable.
 But I emphasize that this proof requires pairwise permutable Sylow subgroups, and the hypotheses of this question do not require that.<|endoftext|>
TITLE: Irreducibility of Coxeter graphs as a function of generating sets
QUESTION [7 upvotes]: Given a Coxeter system $(W, S)$, we can form its Coxeter graph, and say that the system is irreducible if the graph is connected. Now, irreducibility is not solely a function of $W$; it depends also on $S$. E.g., for $W=D_{12}$ we could have $|S|=2$ or $3$, giving the systems $G_2$ or $A_1 \times A_2$. 
Question: For which (other) irreducible $(W,S)$ is there an $S'$ with $(W,S')$ reducible? When this phenomena occurs, does it always change $|S|$ as above? In particular, is it obvious one way or the other for $W=S_n$? 
I think this should at least be answerable for finite Coxeter groups since these are completely classified, but I'd also be happy to hear examples in the infinite case (e.g. affine Weyl groups or hyperbolic Coxeter groups). The systems for which it cannot be done would seem to be in some sense "more irreducible," or at least the irreducibility is more purely algebraic than geometric. 
Related: Coxeter group generators.

REPLY [4 votes]: For infinite Coxeter groups, there is a paper of Luis Paris, "Irreducible Coxeter groups", which proves that every infinite Coxeter group cannot be decomposed as a non-trivial direct product. In particular they must be irreducible.<|endoftext|>
TITLE: Majority vote of total orders
QUESTION [5 upvotes]: Fix an odd natural number $k$.  Suppose we have $k$ total orders on the same (finite) set $X$.  Define a tournament on the vertex set $X$ by putting a directed edge $x\rightarrow y$ if a majority of the total orders compare $x > y$.

What tournaments can be obtained this way?  Of course, if $k = 1$, only linearly ordered tournaments are possible.  I am most interested in the case of small $k$.  For example, is there an excluded-substructure characterization of these tournaments?
What if we make the problem harder and ask whether a given directed graph $G$ can be extended to a tournament $T$ such that $T$ can be obtained in this way?  Again, if $k = 1$, there are various simple characterizations, such as all digraphs that contain no directed cycles.
What can be said about the computational problem of determining the smallest $k$ that can represent a given tournament or digraph?

I assume, perhaps naively, that this problem already occurs in the literature, perhaps in the theory of voting/social choice, so I would be happy with references instead of solutions if that's easier.

REPLY [2 votes]: For $k=3$, the following paper shows an example of a non-3-majority tournament with 8 vertices. 
http://www2.isye.gatech.edu/~ctovey/publications/papers/voting__19_may_08.pdf
A few years ago, I checked that every tournament with 7 vertices (even the Paley tournament) is 3-majority by using a computer.<|endoftext|>
TITLE: "Harmonacci" recurrence and identities for $\pi$
QUESTION [25 upvotes]: While playing with something totally irrelevant I stumbled upon the recurrence:
$$a_{n+1} = \frac{1}{a_n} + a_{n-1}$$
It turns out that given $a_0 = 1, a_1 = 1$,
$$lim \frac{a_{2n}}{a_{2n-1}} = \frac{\pi}{2}$$
I have a very crude idea (or rather a hint) on proving it (the iterations sort of unfold into a sort of Viete product, which is sort of expected), but my technique is rusty at best.
With different initial conditions, things start getting really scary, for example
$ a_0 = 2, 3, 4, 5 $ yield $\frac{8}{\pi}, \frac{9\pi}{8}, \frac{128}{9\pi}, \frac{225\pi}{128}$ respectively.
So, the questions are:
Is it a known fact? If so, where can I read more on it? If not, may anybody help me to prove/disprove it? Does it mean anything?

REPLY [23 votes]: The sequence $a_n$ is closely related to the Wallis product
$$a'_n = \prod_{i = 1}^n \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$
which converges to $\pi/2$ as $n$ goes to infinity. Namely, we have
$$a'_n = a_{n + 1} \cdot \frac{2n}{2n + 1}$$.
This could be proven by induction or maybe more easily by defining $b_n = a_n a_{n - 1}$ and noticing that the recursion for $a_n$ implies the (very simple) recursion
$$b_{n + 1} = 1 + b_n$$
for $b_n$ and expressing $a_n$ in terms of the $b_n$.
For more general values of $a_0$ one gets similar formulas for $a_n$ as (up to a factor converging to 1) a Wallis product or inverse of a Wallis product where a few of the lower terms in the product are missing.<|endoftext|>
TITLE: Automorphisms of virtually abelian groups
QUESTION [5 upvotes]: Let $G$ be a finitely generated virtually abelian group (i.e., $G$ contains $\mathbb{Z}^n$  with finite index for some $n\ge 2$). Is there anything known about the outer automorphism group $Out(G)$? 
For example, is $Out(G)$ finitely presented? Is it linear? Is it residually finite?

REPLY [2 votes]: The OP has since proved a vast generalization of residual finiteness in the virtually abelian case: the outer automorphism group of a virtually compact special group is residually finite.
Antolín, Yago; Minasyan, Ashot; Sisto, Alessandro, Commensurating endomorphisms of acylindrically hyperbolic groups and applications, Groups Geom. Dyn. 10, No. 4, 1149-1210 (2016). ZBL1395.20027.<|endoftext|>
TITLE: Retracted Mathematics Papers
QUESTION [13 upvotes]: Can anyone cite an example of a mathematics paper that has been retracted?
It is said that on the order of 100,000 new theorems enter the mathematics literature every year. For a number of reasons including hyper-specialization and demands on referee resources it is, in my view, unlikely that all their proofs are correct.  Yet it seems no explicit effort is made to clean the literature.  False theorms float downstream along with the true ones, available for citation and use in constructing yet further theorems.
Thanks for any insight.
Cheers, Scott

REPLY [18 votes]: You have to notice that many of those theorems are dead-ends. They'll either not be used at all, or be superseded by a better one. Corrections sometimes happen, but it looks like pure changes of focus is also a big factor.
How many proofs have there been that given a point and a line, there is a single line parallel to the line going through the point, before non-euclidean geometry settled the matter?
Consider the vast litterature on proving that such and such type of polynomial can be solved by radicals. Galois theory made (almost) all of these obsolete -- be they right or wrong.
The notions of limits, continuity, derivability, etc... had no serious definition for very long, before people started to realize there were problems (like sequence of continuous functions converging to a non-continuous limit) and the $(\varepsilon,\delta)$ definitions were given, and people started to prove more solid results through this framework.
The italian school of algebraic geometry is another example coming to mind, where things were cleaned by a quite radical change in the paradigms of the field.
In fact, one could say that most of mathematics is about trying to get correct theorems out, and clean what is already there through obsolescence, be it gentle(correction) or cataclysmic(not interesting anymore).
As a final remark, I think it is reasonable to expect the recent works in automatic proof-checkers/proof-builders will sparkle a new revolutionary era.

REPLY [15 votes]: Here are some recent (not famous) examples of papers that have been retracted at the request of the Editor-in-Chief or the Publisher:
http://www.sciencedirect.com/science/article/pii/S089396591000265X
http://www.sciencedirect.com/science/article/pii/S0393044011002233
http://www.sciencedirect.com/science/article/pii/S0022247X12001254
http://www.sciencedirect.com/science/article/pii/S0166218X1100309X
http://www.sciencedirect.com/science/article/pii/S0898122112003008
I found them via the blog Retraction Watch, which also contains ample discussion about reasons and policies about retracting papers and many more examples, including background information.<|endoftext|>
TITLE: Formulating the calculus of varations with exterior calculus
QUESTION [7 upvotes]: I noticed that a calculus of variations problem is just an integral over a differential form. Therefore, I would think it would be possible to formulate the Euler-Lagrange equations using exterior calculus. However, I do not know of how to reconcile the notion of a functional derivative with say an exterior derivative.

PS:
I will note that I asked this question a few days ago on math stackexchange. As of the time of this writing, it remains unanswered. This led me to believe that the question is specialized enough for mathoverflow. However, I am not a mathematician, so please be pedagogical when possible.

REPLY [11 votes]: There is a large literature on this, and the roots go back more than one hundred years.  Some of the modern work along these lines can be found by looking for papers containing the term 'variational bicomplex'.  For example, look at the papers and books by Ian Anderson and his group.  
You can also look at papers and books by Phillip Griffiths and his collaborators, such as Exterior Differential Systems and the Calculus of Variations and Exterior Differential Systems and Euler-Lagrange Partial Differential Equations.<|endoftext|>
TITLE: Hyperbolic Brunnian links and rectangular cusp shapes
QUESTION [8 upvotes]: My question is as follows.

Does every hyperbolic Brunnian link have rectangular cusp shapes on all its components?

Here is what I mean:
The Borromean rings form a famous link $B$ (a smooth closed 1-manifold in the three-sphere):

Link $B$


This link has the property that removing a single component yields an unlink. This property is called the Brunnian property. Links with the Brunnian property are Brunnian links.
Here are some other Brunnian links:

Link $S$




Link $R$




Link $N$
 (Rolfsen '76, p. 67)



Link $OMG$
 (Rolfsen '76, p. 67)


The complement $S^3 - L = K_L$ of any link $L$ (Brunnian or not) may admit a complete hyperbolic structure (cf. Thurston '97, pp. 131-132, p. 147) with finite volume (cf. SnapPy). By Mostow-Prasad rigidity, this structure (if it exists) is a topological invariant. We may then write $K_L$ as the image of a quotient $D: \mathbb{H}^3 \to K_L$ of hyperbolic space by a freely acting discrete group $G \simeq \pi_1(K_L)$ of isometries. Each component $C$ of the link admits a regular neighborhood $N(C)$ that is the image under $q$ of an open horoball $\tilde{N}$. In fact it admits many such. The maximal such neighborhood is the maximal cusp neighborhood $m_C$.
The lifts of $m_C$ to the universal cover are horoballs. Pick one such lift $M_C$. The closures of all the lifts abut the (horospherical) boundary $\partial M_C$ at a discrete set $\Lambda$ of points. Let $\Gamma$ be the subgroup of $G$ that preserves $\partial M_C$, a peripheral subgroup of $G$. This subgroup acts transitively and freely on this set of points, so we may identify $\Gamma$ with $\Lambda$ by picking one point $o \in \Lambda$ to represent the identity of $\Gamma$.
As it turns out, in the induced metric, the horosphere $\partial M_C$ is isometric to the Euclidean plane, and $\Gamma$ acts on this by isometries. So $\Gamma$ is a discrete, torsion-free, freely acting group of isometries of the Euclidean plane. Therefore it is either $0,$ $\mathbb{Z}$, or $\mathbb{Z} \oplus \mathbb{Z}$.
If $\Gamma \simeq 0$ or $\mathbb{Z}$, then the quotient of the maximal cusp neighborhood by $\Gamma$ (and therefore by $G$) would have infinite volume. Conversely, assuming that $\Gamma \simeq \mathbb{Z} \oplus \mathbb{Z}$, one can show easily that the quotient of the maximal cusp neighborhood has finite volume.
Assume therefore that $\Gamma \simeq \mathbb{Z} \oplus \mathbb{Z}$. Fix an isometry $\phi$ of $\partial M_B$ with $\mathbb{C}$ sending $o$ to $0$. The elements $\gamma_m,$ $\gamma_\ell$ of $\Gamma$ corresponding to the meridian and the longitude of the link component $C$ generate $\Gamma$, so we might as well also ensure that our choice of isometry sends one of them to a positive real number.
SnapPy's convention is to send the longitude to a positive real number, apparently, so let's use that convention. Then we say the cusp shape of the link component $C$ is the ordered pair of complex numbers $(\phi(\gamma_m), \phi(\gamma_\ell))$.
Here are SnapPy's computations for the cusp shapes of the components of the above Brunnian links (I have rounded the decimals):

Link $B$: all of shape $(5\cdot 10^{-17}+0.569 i,\,1.140)$ (by symmetry)
Link $S$: all of shape $(1\cdot 10^{-16}+0.550 i,\,1.181)$ (by symmetry)
Link $N$: no hyperbolic structure
Link $OMG$: 


cusp 0: $(1\cdot 10^{-16}+0.258 i,\,2.514)$
cusp 1: $(-9\cdot 10^{-16}+0.262 i,\,2.471)$
cusp 2: $(1\cdot 10^{-16}+0.316 i,\,2.051)$
cusp 3: $(1\cdot 10^{-15}+0.431 i,\,1.506)$
cusp 4: $(2\cdot 10^{-17}+0.5848 i,\,1.111)$
cusp 5: $(-5\cdot 10^{-16}+0.232 i,\,2.796)$


That is to say, each of the above hyperbolic Brunnian links (almost certainly) has a rectangular cusp shape on every one of its components.

Does every hyperbolic Brunnian link have rectangular cusp shapes on all its components?

REPLY [11 votes]: The rectangular shapes that you find are probably due to the particular symmetries these links have. When there is a symmetry that fixes a component $C$ of a link which fixes both the meridian and the longitude but inverts the orientation to only one of the two curves, then the shape is rectangular. 
This holds because the symmetry induces an isometry on the flat cusp section, which must preserve the unsigned angle between the (geodesic representatives of the) meridian and the longitude, and the only possibility is that they stay at right angles.
Sometimes we find such symmetries when the link is very symmetric and the component $C$ is an unknot: just try to do a $\pi$-rotation that inverts $C$ and see if this gives a symmetry of the whole link. This works for instance for the Borromean rings and the second link you draw. 
Added: no, the $\pi$-rotation inverts both the meridian and the longitude... to invert only one of them, mirror the diagram and see if you get the same link up to isotopy, possibly after permuting the components other than $C$. This works for the links $B$ and $N$ drawn above, and maybe on others.
So the reason for rectangular shapes in the pictures above should be "unknotted components and many symmetries".<|endoftext|>
TITLE: Free Loops, Moore Paths and the Borel Construction
QUESTION [5 upvotes]: My question is about the relationship between the free loop space LX of a space X and the (appropriately defined) Borel construction $PX \times_{\Omega X} \Omega X$ which is a homotopy equivalent fibration. The shortest formulation of my question is:  
Is this homotopy equivalence fibre-wise deloopable? (equivalently) Is it a fibre-wise $A_\infty$ equivalence?
  
So,
The claim is that there is a model for the free loop space LX given by the Borel construction. For this we take based Moore loops in X and Moore paths in X starting at the base point. The Borel construction is then the space of equivalence classes $[p,\alpha]$ where $p$ is a Moore path in X, $\alpha$ is a Moore loop at the base point. The equivalence relation is given by $[p \beta, \alpha] = [p, \beta \alpha \beta^{-1}]$ (where $\beta$ is also a loop at base point).
What is the homotopy equivalence between this and LX? 
Is this homotopy equivalence fibre wise an $A_\infty$ equivalence?
What is the algebraic structure on the fibres of the Borel space? 
There is a natural map we can write down from the Borel space at least to the space of free Moore loops in which we map $[p,\alpha]$ to $p\alpha p^{-1}$. However the the image of this map is not even closed under composition in each fibre! (just regard the composition of $p\alpha p^{-1}$ and $p'\alpha p^{'-1}$ where $p$ and $p'$ are different.
a.

REPLY [4 votes]: When $X$ is a Riemannian manifold you can build very nice models of free loop spaces, based loop spaces and path spaces, you can have a look at 
"On "small geodesics" and free loop spaces" by Bahri and Cohen available on arxiv. This is based on Milnor's papers on classifying spaces.
Thus what you get is a topological group $GX$ which is $A_{\infty}$-equivalent to $\Omega X$ and a homotopy equivalence between the Borel construction of the adjoint action of $GX$ on itself and the free loop space.
A point in $GX$ is an equivalence class of a sequence of points $[x_0,\ldots,x_n]$ where $x_0=x_n$ is the based point, $x_i$ and $x_{i+1}$ are close to each other in a geodesic sense, modulo some cancellation rules. The group structure is concatenation of sequences of points. And the map from $GX$ to $\Omega X$ is by considering composition of geodesics segments. You can also build models of path spaces, free loop spaces in this way. With this gadget you will have maps of fibrations compatible to composition of loops and that are homotopy equivalences.
This construction can be adapted to triangulated spaces, the idea uses the fact that you have a notion of close points and a unique geodesics segment between them.<|endoftext|>
TITLE: Décomposition des nombres premiers dans des extensions non abéliennes
QUESTION [19 upvotes]: Gauß famously determined the cubic character of $2$ in his Disquisitiones : $2$ is a cube modulo a prime number $p\equiv1\mod3$ if and only if $p=x^2+27y^2$ for some $x,y\in\mathbf{Z}$.  This implies that the prime numbers which split completely in the $\mathfrak{S}_3$-extension $\mathbf{Q}(\root3\of1,\root3\of2)$ of $\mathbf{Q}$ are precisely the ones which are $\equiv1\pmod3$ and represented by the quadratic form $X^2+27Y^2$.
This was generalised by Philippe Satgé in 1977 in a paper whose title I have borrowed.  He shows for example that the prime numbers which split completely in the $\mathfrak{S}_3$-extension $\mathbf{Q}(\root3\of1,\root3\of5)$ of $\mathbf{Q}$ are precisely the ones which are $\equiv1\pmod3$ and represented by one of the quadratic forms 
$$
X^2+XY+169Y^2,\qquad 343X^2-131XY+13Y^2.
$$
I believe that similar results about all $\mathfrak{S}_3$-extensions (of $\mathbf{Q}$) can now be recovered using the known cases of Langlands reciprocity, as illustrated around the same time by Serre for the splitting fields of $T^3-T-1$ and $T^3+T-1$, which are the maximal unramified abelian extensions of 
$\mathbf{Q}(\sqrt{-23})$ and $\mathbf{Q}(\sqrt{-31})$ respectively, in his Modular forms of weight one and Galois representations, pp. 193–268 of Algebraic number fields: L-functions and Galois properties (Proc. Sympos., Univ. Durham, Durham, 1975), Academic Press, London, 1977.  
But Satgé's theorem is applicable to more general extensions : it is applicable to a $G$-extension $K$ of $\mathbf{Q}$ whenever the finite group $G$ contains a commutative normal subgroup $H\subset G$ such that 
(*) the transfer (Verlagerung) map $G/G'\to H$ is trivial, and
(**) the order of $H$ is odd if the field $K^H$ is totally real of degree $>2$ (over $\mathbf{Q}$).
Question.  Can these more general results of Satgé be recovered from known cases of Langlands reciprocity ?

REPLY [9 votes]: Far from being able to even begin to answer the question, let
me make a few remarks that perhaps shed some light on the situation.
Gauss's result on the cubic character of $2$ was generalized in 
Dedekind's highly underrated article on pure cubic fields
[Über die Anzahl der Idealklassen in reinen kubischen 
Zahlkörpern; J. Reine Angew. Math. 121 (1900), 40-123]: 
already in the early 1870s Dedekind had written

Let $k$ denote a rational integer whose cube root is irrational.
Then the equation $x^3 = k$ defines a pure cubic number field whose
discriminant has the form $D = -3g^2$, where the integer $g$ can 
easily be determined from $k$. Consider all primes $p$ of the form
$3n+1$ that do not divide $k$ and modulo which the given integer $k$
is a cubic residue. With the help of the reciprocity law we then find
the following interesting result, which essentially was known already
to Gauss (and may be extended to general cubic fields): there are three
kinds of primitive binary quadratic forms $ax^2 + bxy + cy^2$ with 
$D = b^2 - 4ac$ and representing different classes: the forms of the 
first kind are a group whose forms represent exactly those prime numbers 
modulo which $k$ is cubic residue.
He could not generalize his results, as promised, to general cubics,
although he correctly conjectured that this should be possible using
the theory of complex multiplication, i.e., class field theory for
complex quadratic number fields.
Dedekind also quotes a remark by Gauss, in which he determined the
cubic character of $5$ using binary quadratic forms (which coincides
with Satgé's example except that Gauss considered only forms
with even middle coefficient, which means that he has four forms
$(1,0,675)$, $(25,0,27)$, $(13,2,52)$, $(4,2,169)$ where  Satgé
can do with one $(1,1,169)$.
Dedekind's result was generalized by Takagi [Sur les corps résolubles 
algébriquement, C. R. Acad. Sci. Paris 171 (1920), 1202-1205]. 
Takagi presents Dedekind's theorem in the following form:

Let $D$ denote the discriminant of a cubic number field $k$;
then the class number of primitive quadratic forms with 
discriminant $D$ is a multiple of $3$. A third of the classes
forms a group which can be characterized by the property that
the prime numbers not dividing $D$ and for which $D$ is a
quadratic residue, and which split into three different factors
in $k$, and only those, are represented by a quadratic form
in this group.
Then he proves his generalization:

Let $k$ be a solvable field of prime degree and $K_0$ the 
corresponding cyclic field. If the ideal classes in $K_0$ are
defined modulo $f$, the class group contains a subgroup of 
index $\ell$, which is characterized by the property that among
the prime numbers not dividing the discriminant of $k$ and
splitting completely in $K_0$, exactly those primes that split 
completely in $k$ are the norms of ideals in $K_0$ lying in 
the subgroup above.
If $k$ is a pure cubic field, then $K_0$ is the field of cube roots
of unity, and Takagi gets back Dedekind's result. For the proof,
Takagi used his class field theory.
Special cases of this result were rediscovered e.g. by

D. Liu [Dihedral polynomial congruences and binary quadratic forms
in her Ph.D. thesis (Carleton 1992) supervised by 
Spearman & Williams 
[The cubic congruence $x^3 + Ax^2 + Bx + C \equiv 0 \pmod p$ 
 and binary quadratic forms,
 J. London Math. Soc. (2)  46  (1992), no. 3, 397-410],
[The cubic congruence $x^3 + Ax^2 + Bx + C \equiv 0 \pmod p$ 
 and binary quadratic forms II,
 J. London Math. Soc. (2)  64  (2001), no. 2, 273-274].

See also 

D. Bernardi [ Résidus de puissances,
Semin. Delange-Pisot-Poitou 1977/78, Fasc. 2, Exp. No. 28, 12 pp.

The results by 

Weinberger [The cubic character of quadratic units,
Proc. 1972 Number Theory Conf., Univ. Colorado, Boulder 1972, 241-242]

and the more recent ones by 

Sun [Cubic residues and binary quadratic forms,
J. Number Theory (2006)]

also follow this pattern (there are in fact a lot more articles 
dealing with describing the splitting of primes in ring class fields 
of quadratic number fields using binary quadratic forms).
Satgé considers the following situation: let $K$ be a normal
extension of the rationals with Galois group $G$, let $H$ be a
normal abelian subgroup, and let $k$ be the fixed field of $H$.
By applying class field theory to the abelian extension $K/k$, he
characterizes the decomposition law in $K$ by representation of
the primes in question by the norm forms attached to $k$, assuming
certain conditions on $H$. The norm forms have degree $(G:H)$;
thus the case where $H = 1$ is absolutely trivial since in this case 
the splitting of primes in $K$ (for unramified primes in normal
extensions, all we need to know is the inertia degree) is described 
by norms from $K$. 
Overall these are all "abelian" phenomena and follow readily from 
(abelian) class field theory. But of course it is legitimate to 
ask how these fit into the nonabelian Langlands conjectures, since
after all we are dealing with nonabelian extensions here.<|endoftext|>
TITLE: Understanding the "idea" behind Langlands
QUESTION [22 upvotes]: Apologies in advance if this is a bit too simple to ask here, but I think I'm probably more likely to get an answer here than at stackexchange.
I've been trying to learn the basics of the Langlands program over the last couple of months, and I've reached a funny point where I think I understand the rough idea of whats going on, but I seem to have a fairly large amount of reading to do before I can take the next step and understand everything properly.
I'm currently at the point where I'm completely happy with Tate's thesis and the basic ideas of automorphic forms and representations. My understanding of Langlands right now is roughly:

Tate's thesis shows us that by using abelian harmonic analysis on the adele ring, we can prove functional equations for L-functions attached to Hecke characters.
Hecke characters are just one-dimensional automorphic representations.
We can (or believe we can? I'm not entirely sure what the status of this part is) use nonabelian harmonic analysis to prove functional equations for the L-functions attached to more general automorphic representations. I know the point of the fundamental lemma is that it lets us use the information given by the Arthur-Selberg trace formula in a useful way - so this generalises the role of Poisson summation in Tate's thesis?
A (very) rough statement of the local Langlands conjectures is that a whole load of arithmetic L-functions that we've defined actually arise from automorphic representations.
So local Langlands would mean we can prove functional equations for a huge amount of L-functions.

I'm aware that I've missed out huge chunks of the program like functoriality, which I haven't even begun to understand yet, but I'd like to think there's enough there for me to have a pretty reasonable idea of what's going on at a basic level.
Then my question is this: is this at all accurate? The idea as I've summarised it above is one that I've pieced together myself from the various things I've read about Langlands, and it seems like a pretty simple explanation of why we should care about Langlands, yet it's not an explanation I can recall seeing anywhere before, which makes me suspect that I might have missed the point a bit.

REPLY [7 votes]: To complement the answer from GH from MO:
Your first two bullet points, about Tate's thesis and Hecke characters are completely correct.
Your third point should be made more precise as follow. Yes we can prove the functional equation for the L-function attached to automorphic representations of $GL_n$, and a bunch of other things for them: that they have a meromorphic continuation, where the poles are if any, then as you say the functional equation, and then the non-vanishing of these L-function on the line $\Re s=1$ (the analog of the Hadamard-De la Vallée Poussin theorem). Roughly we know these automorphic $L$-functions almost as well (or as badly) as the Riemann Zeta function. ("Almost"
because we don't know the Ramanujan hypothesis, that is we only know that the series defining these functions converges in a half-plane slightly smaller as what we expect, but in the grand scheme of things that's a detail. "as badly", because like for the Riemann Zeta Function, we expect them to satisfy the Riemann hypothesis, and we have no idea how to prove it.) All these results are due to Jacquet-Shalika (except for $n=1$, where they are due to Hecke and Tate), and were proved relatively early in the development of the theory of automorphic forms.And while Jacquet-Shalika's proof are hard and very clever, they don't use the Arthur-Selberg trace formula, let alone the fundamental lemma which was proved much later.
For your fourth point, as GH says, it would be correct (as a very rough statement) if you replaced "local" by "global". And the same is true for your fifth.<|endoftext|>
TITLE: The Riemann Hypothesis and the Langlands program
QUESTION [24 upvotes]: On page 263 of this book review appears the following:

Given the centrality of L-functions to the Langlands program, nothing would seem more natural (than a presentation of elementary algebraic number theory from the standpoint of L-functions and their analytic properties), but in fact the properties of L-functions traditionally of interest to analytic number theorists - for example, the location of zeroes in the critical strip (the Generalized Riemann Hypothesis) - have historically had little to do with the preoccupations of the Langlands program. Thanks largely to the efforts of a few charismatic and determined individuals, this is beginning to change and Langlands himself has in recent years turned to methods in analytic number theory in an attempt to get beyond the visible limits of the techniques developed over the last few decades.

I'd like to ask for a big picture exposition of how such questions about the location of zeroes of L-functions appear and interact with the Langlands program. My interest is mainly cultural and the answer should be tailored for the outsider to number theory (I'm viewing Langlands program algebraically as the pursuit of a nonabelian class field theory.)
A more crude question is:

Does the Langlands program say anything about the Grand Riemann Hypothesis or vice versa?

This is almost certainly too crude a question for MO, but Langlands seems to have such an amazing unifying appeal, that I feel a temptation to see how much it subsumes. I fully expect an answer like "It is impossible to coherently discuss this without years of training". Thank you for any attempt to explain things to someone who is not a number theorist, in advance!

REPLY [3 votes]: The analogue of the Riemann hypothesis for the Selberg zeta function for $\Gamma(N) \backslash \mathbb{H}$ is known as the Selberg eigenvalue conjecture. It would follow from the Langlands functoriality. However, all but finitely many of these zeros lie on $\Re s =1/2$ beforehand, so this is very different from the Riemann zeta function.<|endoftext|>
TITLE: Randomly switching street lights, in a square city
QUESTION [17 upvotes]: This is a combinatorics-probability question, best stated however in "recreational" terms. Imagine a $N\times N$ city, meaning that we have $N$ horizontal streets, and $N$ vertical streets. At each crossroads there is a street light.
When evening comes, some of the lights are switched on, namely those corresponding to a certain given subset $E\subset(1,\ldots,N)\times(1,\ldots,N)$.
Now assume that $2N$ kids come at night and start randomly playing with the switches: there is one such on/off switch at the end of each of the $2N$ streets. 
Problem: for each of the $4^N$ overall choices for the various switches, we count the number $K$ of street lights that are switched on. What is the law of this random variable $K$, as a probability measure on $(1,2,\ldots,N^2)$, depending on the initial set $E$? 
[Edit, Jan 20. As signaled by Joseph O'Rourke in his answer below, computing the upper edge of the support of the measure $\mu_E$ in my problem is known as the Gale-Berlekamp game, a difficult question (details can be found via Google search). So I realise that my problem is probably extremely difficult, adding the "open-problem" tag. I'd be interested however in the case where $E$ is an Hadamard matrix, cf. discussion with Gerhard Paseman in the comments below. Is there anything known about this measure $\mu_E$? (I mean, not only about its support.)
Btw here is the only non-trivial complete computation that I have so far: concerns the case $N=4$, where there are exactly $|E|=4$ street lights, positioned on the main diagonal of the city. Here $\mu_E=\frac{1}{32}(\delta_4+12\delta_6+6\delta_8+12\delta_{10}+\delta_{12})$. Plus an experimental remark, that I'm not able to prove abstractly: the support of $\mu_E$ seems always to be an arithmetic progression.]

REPLY [15 votes]: This paper by Fishburn and Sloane seems quite related:

Fishburn, Peter C., and N. J. A. Sloane. "The solution to Berlekamp's switching game." Discrete Mathematics 74.3 (1989): 263-290. (PDF download)
Abstract.
  Berlekamp’s game consists of a $10 \times 10$ array of light-bulbs, with $100$ switches at
  the back, one for each bulb, and $20$ switches at the front that can complement any row or
  column of bulbs. For any initial set $S$ of bulbs turned on using the back switches, let $f(S)$
  be the minimal number of lights that can be achieved by throwing any combination of
  row and column switches. The problem is to find the maximum of  $f(S)$ over all choices
  of $S$. We show that the answer is $34$. We also determine the solution for $n \times n$ arrays
  with $1 \le n \le 9$.


          


          
(Image from this link.)<|endoftext|>
TITLE: Chern-Simons for 2n-dimensional manifolds
QUESTION [9 upvotes]: In the literature I can only find Chern-Simons terms  for odd-dimensional manifolds. For example, for a $G$-bundle over a 3-dimensional manifold we have $A \wedge dA + 2/3 * A \wedge A \wedge A$ with $A$ being a $\mathfrak{g}$-valued 1-form. Why can't I write such forms for even-dimensional manifolds?

REPLY [16 votes]: It has to do with the fact that the characteristic classes (over the reals) of a principal  $G$-bundle have even degree. We can associate  Chern-Simons-like theory  to each characteristic class of  degree $2k$ together with a $G$-bundle  $P$   over  a manifold of dimension  $2k-1$.
To be a bit more technical a Chern-Simons-like form is asssociated to the following data
1. A homogeneous polynomial $\Phi$ of degree $k$ on the Lie algebra  of $G$ invariant  under the action of $G$ by conjugation.
2. A  principal $G$-bundle  $P\to M$ over $M$.
3. A pair of connections $\nabla^0, \nabla^1$ on $P\to M$.
The Chern-Weil theory  produces two closed forms
$$ \Phi(\nabla^0),\Phi(\nabla^1)\in \Omega^{2k}(M) $$
and  a form 
$$ T\Phi(\nabla^1,\nabla^0)\in \Omega^{2k-1}(M), $$
such that 
$$ d T\Phi(\nabla^1,\nabla^0)=  \Phi(\nabla^1)-\Phi(\nabla^0). $$
(For details see Chapter 8 of these  notes.)
The  transgression form $T\Phi(\nabla^1,\nabla^0)$ is the one used in Chern-Simons  theories. It depends on two connections, but usually $\nabla^0$ is some fixed  connection.<|endoftext|>
TITLE: String diagrams of special monoidal categories and higher categories
QUESTION [5 upvotes]: I have two questions related to Selinger: A survey of graphical languages for monoidal categories.
1) On page 60 there is a summary of some monoidal categories with extra structure. Among them are the well known braided and symmetric versions. Due to the periodic table, these two monoidal categories stem from 3- resp. 4-categories. Can some of the other special monoidal categories also be deduced from certain higher categories?
2) There is another type of string diagrams not present on page 60. Namely you can do the following. Assume you have a string diagram of a monoidal category with no extra structure, i.e. a string diagram on a plane square. Bend such a square to a cylinder by identifying the left and right side of that square. The diagram is now "painted" on a cylinder und you can also compose cylinders (but only "vertically"). Does this correspond also to some extra structure on a monoidal category? In fact, this extra structure should somehow reduce the dimension of the monoidal category by one, because tensor expressions are cyclically isomorphic, e.g. the objects $A_1\otimes A_2\otimes A_3$, $A_2\otimes A_3\otimes A_1$, $A_3\otimes A_1\otimes A_2$ are coherently isomorphic. This corresponds to the top and bottom of the cylindrical diagrams being a circle.

REPLY [4 votes]: 2) You may be interested in this paper.<|endoftext|>
TITLE: Unit sphere in R^\infty is contractible?
QUESTION [5 upvotes]: Hello,
We know that in Hilbert space it is, but what about these topology:
Let $\mathcal{T}_{\infty}= \{ U \subset \mathbb{R}^{\infty}: \ U \cap \mathbb{R}^n \in \mathcal{T}_n, \ for \ n=1,2,... \} $ which isn't metric space?
Of course $\mathcal{T}_{\infty}$ is topology in $\mathbb{R}^{\infty}$. How to prove or disprove that $S^{\infty} = \{ v \in \mathbb{R}^{\infty} : \ ||v||=1 \}$ is contractible? 
:)
Can we find homeomorphism without fixed point from $D^{\infty} = \{ v \in \mathbb{R}^{\infty} : \ ||v|| \le 1 \}$ onto $D^{\infty}$? I was trying to find such homeomorphism, but I failed...

REPLY [2 votes]: Don't want to be too pedantical but, to set the record straight, both topologies coincide in this situation---one of the consequences of the Banach-Dieudonne theorem.<|endoftext|>
TITLE: Even XOR Odd Infinities?
QUESTION [29 upvotes]: Modular Arithmetic (MA) has the same axioms as first order Peano Arithmetic (PA) except $\forall x (Sx \ne 0)$ is replaced with $\exists x(Sx = 0)$. 
(http://en.wikipedia.org/wiki/Peano_axioms#First-order_theory_of_arithmetic).
MA has arbitrarily large finite models based on modular arithmetic. All finite models of MA have either an even or odd number of elements. I call a model of MA "even" if it satisfies both of these two sentences:
E1) $\exists x(x \ne 0 \land x+x = 0)$
E2) $\forall x(x+x \ne S0)$
A model of MA is odd if it satisfies both of:
O1) $\forall x(x = 0 \lor x+x \ne 0)$
O2) $\exists x(x+x = S0)$
We can use compactness to prove MA has infinite "even" size models by adding the even definitions above as axioms. We can similarly prove there are infinite "odd" size models of MA. Some infinite sets, like the integers, are neither even nor odd. The integers are not the basis for a model of MA. For example, the four square theorem (every number is the sum of four squares) is a theorem of both MA and PA. The four square theorem is false in the integers. It has been conjectured the complex numbers are a basis for a model of MA. If so, the complex numbers would be an "odd" model of MA.
My question is whether every model of MA must be exclusively even or exclusively odd? Is this statement a theorem of MA?
$$\exists x(x \ne 0 \land x+x = 0) \ \overline{\vee}\  \exists x(x+x = S0)$$
I asked this question on stack exchange and got no answer.
https://math.stackexchange.com/questions/214018/even-xor-odd-infinities

[The following was merged from an answer - ed.]
Ashutosh's proof can be written as:
$\exists x\exists y( (x+x=0 \land y+y=1) \implies (x=0) )$
This answers my question when $\exists x(x+x=1)$ is true but it says nothing about when $\forall x(x+x \ne 1)$ is true.  Emil and others have stated any algebraically closed field is a model of MA. Ashutosh's proof shows any algebraically closed field is odd because $\exists x(x+x=1)$ is true.
I want to accept Ben Crowell answer, but I have some reservations. The proof starts by showing how any model of MA can be expanded into a model of PA. I have made similar arguments and always assumed it would be easy to prove. My conjecture is true of all finite models of MA so we only need consider infinite models. MA is omega inconsistent and any infinite model must have non-standard elements. Tennebaum's theorem says addition is not recursive in non-standard models of PA. Can addition actually be recursive in $A$, the model of PA he constructs? It looks like he is assuming we can add non-standard numbers from the model of MA. I also wonder if he is assuming $I$ is a standard model of PA. I don't think it makes any difference, but it might.
Obo's proof is much simpler and similar to a proof I came up with. My proof had the same error as his. I think it is fixable. In the case where we have $S(y+y)=p$ we need to also prove $y \ne p$.  $y \ne p$ can be true only in models with three or more elements.   
This isn't a discussion group so I won't go into detail why I don't think the complex numbers are a model of MA. I don't think MA has any infinite models. I will point out MA proves a lot of interesting things about odd models. In an odd model the sum of all elements is 0. This can't be stated in first order. I think if we have a successor function defined on the complex numbers we can use it to order the reals. Just ignore numbers with a non-zero imaginary component.
I want to retract my statement that the Lagrange's four square theorem is a theorem of MA. I based my claim on Andrew Boucher's paper on General Arithmetic (GA). Boucher shows GA proves the four square theorem. I thought GA was a weak sub-theory of MA because GA has much weaker successor axioms. Rereading the paper I see Boucher says he is using 2nd order induction. He also says successor is second order.

REPLY [19 votes]: The answer is no. It is enough to find a model of MA which is an integral domain of characteristic $0$ (whence O1 is true and E1 false) such that $2$ is not invertible (whence E2 is true and O2 false).
One example of such a model is the ring of $2$-adic integers $\mathbb Z_2$. This is clearly a domain, and $2$ is not a unit, hence it suffices to show
Theorem: For any prime $p$, the ring $\mathbb Z_p$ is a model of MA.
Proof: The only problem is to verify that induction holds. Assume $\mathbb Z_p\models\phi(0)\land\forall x\,(\phi(x)\to\phi(x+1))$, where $\phi$ is an arithmetic formula with parameters from $\mathbb Z_p$, and put $\phi(\mathbb Z_p):=\{a\in\mathbb Z_p:\mathbb Z_p\models\phi(a)\}$.
Since $\phi(\mathbb Z_p)$ is definable in $\mathbb Z_p$, it is also definable in the field $\mathbb Q_p$ endowed with a unary predicate for $\mathbb Z_p$. Macintyre [1] proved that such structures admit a form of quantifier elimination, and as a corollary (Thm. 2 on p. 609), every infinite definable set has a nonempty interior. Thus, there is $a_0\in\phi(\mathbb Z_p)$ and $k\ge0$ such that $a_0+p^k\mathbb Z_p\subseteq\phi(\mathbb Z_p)$. Let $a\in\mathbb Z_p$ be arbitrary, and let $b< p^k$ be a natural number such that $b\equiv a-a_0\pmod{p^k}$. Since $\phi(\mathbb Z_p)$ is closed under successor, we have $a\in b+a_0+p^k\mathbb Z_p\subseteq\phi(\mathbb Z_p)$. Thus, $\phi(\mathbb Z_p)=\mathbb Z_p$, i.e., $\mathbb Z_p\models\forall x\,\phi(x)$.   QED
I suspect the following may work as additional countermodels (they are domains where $2$ is not a unit, the issue is whether they satisfy induction):

The ring of algebraic integers $\tilde{\mathbb Z}$. A form of quantifier elimination for $\tilde{\mathbb Z}$ was proved by van den Dries [2] and Prestel and Schmid [3], but the basic formulas are somewhat messy, so it is not immediately clear to me whether this implies induction.
The localization of $\tilde{\mathbb Z}$ at a maximal ideal containing $2$. Elimination of quantifiers for this (and similar) rings is reported as Fact 3 in [2], where it is attributed to [4]. It seems it could imply induction by a similar argument as for $\mathbb Z_p$.

[1] Angus Macintyre, On definable subsets of $p$-adic fields, Journal of Symbolic Logic 41 (1976), no. 3, pp. 605–610.
[2] Lou van den Dries, Elimination theory for the ring of algebraic integers, Journal für die reine und angewandte Mathematik 388 (1988), pp. 189–205.
[3] A. Prestel and J. Schmid, Existentially closed domains with radical relations, Journal für die reine und angewandte Mathematik 407 (1990), pp. 178–201.
[4] Angus Macintyre, Kenneth McKenna, Lou van den Dries, Elimination of quantifiers in algebraic structures, Advances in Mathematics 47 (1983), no. 1, pp. 74–87.<|endoftext|>
TITLE: Top chern class under finite, unramified, dominant morphism
QUESTION [5 upvotes]: Situation: Let $\Bbbk$ be an algebraically closed field. Assume that $\pi:Y\to X$ is an finite, dominant, unramified morphism between nonsingular varieties of dimensions $n$. Let $d=\deg(\pi)$. 
What I know: For $\Bbbk=\mathbb C$, the second chern class of $X$ equals its topological Euler characteristic (i.e., the Euler characteristic with respect to the topology of a complex manifold). I know this under the name Gauss-Bonnet Formula. It then follows that $c_n(Y)=d\cdot c_n(X)$ because $\pi$ is a $d$-fold covering map of complex manifolds. 
My Question: Does $c_n(Y)=d\cdot c_n(X)$ hold under the more general assumption that $\Bbbk$ is algebraically closed? In particular, does this hold in positive characteristic? 
PS: I am mostly interested in the case $n=2$, i.e. $\pi$ is a covering map of surfaces. However, I felt that this would probably work for any $n$.

REPLY [7 votes]: Angelo's answer is complete, but I think you would be interested in the following (which is more about Euler characteristics than Chern classes).
I will assume $k= \mathbf C$, but what I will write holds for $k$ algebraically closed of characteristic zero once you replace "cohomology with compact support and  coefficients in $\mathbf Q$ on the category of para-compact Hausdorff topological spaces"  by "etale cohomology with compact support and  coefficients in $\mathbf Q_\ell$ for some prime $\ell$ on the category of finite type separated $k$-schemes". 
Let $H^\cdot_c(-,\mathbf Q)$ denote cohomology with compact support and coefficients in $\mathbf Q$ on the category of para-compact Hausdorff topological spaces. For a finite type separated $\mathbf C$-scheme, write $e_c(X)$ for the Euler characteristic of $X$, i.e., $e_c(X) = \sum_{i} (-1)^i \dim_{\mathbf Q} H^i_c(X,\mathbf Q)$. Since $X$ is separated and of finite type, this is a well-defined integer. (Of course, I'm implicitly utilizing the analytification of $X$ here.)
Theorem. Let $\pi:X\to Y$ be a finite etale morphism of finite type separated $\mathbf C$-schemes. Then $e_c(X) = \deg \pi e_c(Y)$.
Proof. We may and do assume $X$ and $Y$ are connected. Also, we may and do assume $\pi:X\to Y$ is Galois. (In fact, let $P\to Y$ be a Galois closure of $X\to Y$. Let $G$ be the Galois group of $P\to Y$. Let $H$ be the subgroup of $G$ such that $P/H = X$. Then $$e_c(Y) = \frac{e_c(P)}{\# G} = \frac{\# H}{\# G} e_c(X) = \frac{1}{\deg \pi} e_c(X)$$ and so the result follows in the general case.)
Thus, we have a finite group $G$ acting freely (without fixed points) on $Y$ such that $X=Y/G$. Note that $\deg \pi = \vert G\vert$. Apply the Lefschetz trace formula to see that $Tr(g,H^\ast_c(Y)) =0$ for any $g\neq e$ in $G$. By character theory for $\mathbf Q_\ell[G]$, we may conclude that the element $$ [H^\ast_c(Y,\mathbf Q_\ell)] := \sum (-1)^i [ H^i_c(Y,\mathbf Q_\ell)]$$ in the Grothendieck group $K_0(\mathbf Q_\ell[G])$ of finitely generated $\mathbf Q_\ell[G]$-modules is given by an integer multiple of $[\mathbf Q_\ell[G]]$; the class of the regular representation. So we may write $$[H^\ast_c(Y,\mathbf Q_\ell)] = m [\mathbf Q_\ell[G]],$$ where $m\in \mathbf Z$. Now, note that $H^i_c(X,\mathbf Q_\ell) = \left(H^i_c(Y,\mathbf Q_\ell)\right)^G$ for any $i\in \mathbf Z$. Therefore, we have that 
$$ [H^\ast_c(X,\mathbf Q_\ell)] = m$$ in $K_0(\mathbf Q_\ell[G])$. In particular, we see that $e_c(X) = \dim_{\mathbf Q_\ell} [H^\ast_c(X,\mathbf Q_\ell)] = m$. We conclude that $$e_c(Y) = \dim_{\mathbf Q_\ell} [H^\ast_c(Y,\mathbf Q_{\ell})]= m \vert G\vert  =  e_c(X) \vert G \vert = \deg \pi e_c(X). $$ QED.
For completeness, here is what you can do for "ramified covers". Not surprisingly, the same equality holds up to a "correction term" coming from the branch locus.
Lemma. Let $M$ be a finite type separated $\mathbf C$-scheme. Let $N$ be a closed subscheme of $M$. Then $e_c(M) = e_c(N) + e_c(M\backslash N)$.
Proof. Mayer-Vietoris. QED
Corollary. Let $\pi:X\to Y$ be a finite flat surjective morphism, and let $D$ be a closed subscheme of $Y$ such that $\pi$ is etale over $Y\backslash D$. Then $$e_c(X) = \deg \pi e_c(Y) + e_c(\pi^{-1}D) - \deg\pi e_c(D) .$$
Proof. Write $U=Y\backslash D$ and $V=\pi^{-1}(U)$. Then $$e_c(X) = e_c(V) + e_c(\pi^{-1}D) = \deg \pi e_c(U) + e_c(\pi^{-1}D) = \deg \pi(e_c(Y) - e_c(D)) + e_c(\pi^{-1}D).$$  The first equality follows from the Lemma, the second from the Theorem and the third from the Lemma. QED
We can use this Corollary to obtain a more precise description of the "error term" under some mild hypotheses. Recall that a strict normal crossings divisor on a smooth projective variety is a divisor whose irreducible components are smooth and intersect transversally.
Theorem. Let $D$ be a strict normal crossings divisor on a smooth projective connected variety $X$ over $k$. Let $U$ be the complement of the support of  $D$ in $X$ and let $V\to U$ be a finite etale morphism with $V$ connected. Let $\pi:Y\to X$ be the normalization of $X$ in the function field of $V$. Then

The singularities of $Y$ are quotient singularities (and thus rational singularities);
The singularities of $Y$ lie in $\pi^{-1}D^{sing}$, where $D^{sing}$ is the singular locus of $D$;
The morphism obtained by restriction $\pi^{-1}(D-D^{sing})\to D-D^{sing}$ is etale;
We have $$e_c(Y) = \deg \pi e_c(X) + e_c(\pi^{-1}(D^{sing}))-\deg \pi e_c(D^{sing}) + $$ $$e_c(\pi^{-1}(D-D^{sing})) - \deg \pi e_c(D-D^{sing}).$$

Proof. This is a long but not difficult proof. I can include the details if you'd like. For now, let me say that if you prove $Y$ has quotient singularities, it follows that $Y$ has rational singularities by a theorem of Viehweg; see "Rational singularities of higher dimensional schemes". To prove (1),  (2) and (3) you use results from SGA1 on the fundamental group. Note that (4) follows from the Corollary, the Lemma and (3). QED
Final Remark. In the last formula  $$e_c(\pi^{-1}(D-D^{sing})) - \deg \pi e_c(D-D^{sing})= e_c(D-D^{sing})(\deg \pi - d^\prime),$$ where $d^\prime$ is the degree of the finite etale morphism $\pi^{-1}(D-D^{sing})\to D-D^{sing}$. (In a previous version I thought this was always zero, because I mistakingly assumed $d^\prime = \deg \pi$.)<|endoftext|>
TITLE: Fattening of totally convex sets
QUESTION [6 upvotes]: Suppose $(M, g)$ is an open complete nonnegatively curved Riemannian manifold with $d$ its distance. 
A totally convex set $C\subset M$ has the property that for any two point $x, y \in C$ any geodesic (not only the minimal ones) joining them must lie in $C$.
If $C$ is totally convex, the fattened set $C^a=\{x \in M \mid d(x,C)\leq a\}$ is still totally convex? Is it at least for small $a$?  
This question is somehow related to question "Examples on small cut radius of totally convex set in non-negatively curved manifold", Examples on small cut radius of totally convex set in non-negatively curved manifold

REPLY [6 votes]: No. Consider a rotation-symmetric metric on $\mathbb R^2$ resembling a small spherical cap extented by a flat cone. A sufficiently short geodesic segment at the origin is totally convex in your sense. But a small neighborhood of such a  segment is not convex due to positive curvature.<|endoftext|>
TITLE: Why all irreducible representations of compact groups are finite-dimensional ? [EDIT: Subtleties: AC,etc]
QUESTION [30 upvotes]: About 20 years ago I read in textbook that
"all irreducible representations of compact groups are finite-dimensional", but
 me and the proof of this fact never met each other :) 
May I ask dear MO colleagues, is there (simple?) argument to prove it ? 
As far as I heard this result can be generalized in the realm of non-commutative geometry,
Woronowicz compact quantum groups (?). 
So the "bonus" question - what is appropriate "compactness" condition for some algebra (and/or Hopf algebra) such that it will guarantee the same property 
(i.e. all irreps are finite-dim.) ?

[EDIT] Thanks very much for excellent answers ! Let me ask about some more details, to finally clarify.
1)  What is maximal  possible relaxation of the requirement on vector space V ? Is it enough to require arbitrary
linear topological space or we need to restrict to Hausdorff (?) Banach (?) Hilbert (?), whatever spaces ?   (It seems restrictions on the space may come from the Schur lemma, it is not clear for me what is appropriate generality it holds). 
2) Do we need axiom of choice here  ?  (Probably not, we need existence of Haar measure, but Wikipedia writes that  "Henri Cartan furnished a proof of existence of Haar measure which avoided AC use.[4]"
3) Informally: what is the hardest tool one uses in the proof ? (May be existence of Haar measure ?)
[END EDIT].

[EDIT]
Let me add sketch of arguments by Aakumadula, as I understand it. It might be helpful to clarify new questions.
1) Tool: Continuous functions on the group can be mapped to operators on V. (Need measure here).  (Group algebra acts on V).
2) Fact: Continuous function will be mapped to COMPACT operators. (In R^n I know how to prove it,  in general no).
3) Observe: Conjugation invariant function are mapped to operators which commute with action of group.
4) Schur Lemma: operators commuting with group in irrep are Lamda*Id. (What do we need from the space V for this to be true ? )
5) Corollary: If we find invariant continuous function which is mapped in NON-zero in V,
then we are done, because by (2) it is compact operator and by (4) it is Lambda*Id.
So we need to find invariant function which will be non-zero in V.
6) Take arbitrary "approximate identity" i.e. sequence of continuous (non-invariant) functions f_n which
converge as functionals to delta-function in identity of the group. (It is local fact. But how to prove it ? Do we need Axiom of choice here ? )
7) Make averaging over the group of f_n - get sequence of INVARIANT continuous functions which again converge to detla(e), since delta(e) is invariant. 
8) Operators T(f_n) converge to identity operator, hence for some N they are NON-ZERO.
WE ARE DONE by (5) ! Because T(f_N) is compact and Lambda*Id and Lambda is NON-ZERO. 
[End EDIT].

REPLY [16 votes]: In comments the question is asked about semigroups. The answer is that a compact semigroup can have no non-trivial finite dimensional reps. Take the unit interval with min. Then since its connected its image under a rep is connected. But it is idempotent and commutative and hence simultaneously diagonalizable. But there are only finitely many diagonal idempotents so the image is a point. More generally you can separate points of a compact inverse semigroup iff the idempotents are totally disconnected.<|endoftext|>
TITLE: Lower bounds on the number of elements in Sylow subgroups
QUESTION [18 upvotes]: I posted this question on Math.SE (link), but it didn't get any answers so I'm going to ask here. This is an edited version of the question.

Let $p$ be a prime and $n \geq 1$ some integer. Furthermore, let $G$ be a finite group where $p$-Sylow subgroups have order $p^n$. Denote by $n_p(G)$ the number of Sylow $p$-subgroups of $G$. Denote the number of elements in the union of all Sylow $p$-subgroups of $G$ by $f_p(G)$. I am interested in finding lower bounds for $f_p(G)$ that do not depend on the group $G$, but only on $p$, $n$ and $n_p(G)$.
By Sylow's theorem, we know that $n_p(G) = kp + 1$ for some integer $k \geq 0$. What I know so far:

If $k = 0$, then $f_p(G) = p^n$.
If $k = 1$, then $f_p(G) = p^{n+1}$.
If $k \geq 2$, then $f_p(G) \geq 2p^{n+1} - p^n$.

This is a theorem due to G. A. Miller, see also this question from Math.SE. To prove the inequality in the case $k \geq 2$, you first prove that then $f_p(G) > p^{n+1}$. Then observe that $f_p(G) - 1$ is divisible by $p-1$, then the inequality follows from Frobenius theorem (*). Details are in a book of Miller, Blichfeldt and Dickson (Theory and Applications of Finite Groups) and a paper of Miller ("Some deductions from Frobenius Theorem").
My main question is the following:

What is a better lower bound for the case $k > 2$?

The case $n = 1$ is easy, because then we know the value of $f_p(G)$ precisely. If $n = 1$, then $f_p(G) = n_p(G)(p-1)+1$. What about when $n > 1$? Answers regarding particular $n$ or particular $k$ are also welcome. 
If the Sylow $p$-subgroups are cyclic, then we have $f_p(G) \geq n_p(G)\varphi(p^n) + p^{n-1}$ and this bound is okay. But most $p$-groups are not cyclic..
I think the following example shows that $f_p(G)$ gets arbitrarily large values for fixed $p$ and $n$ (not surprising). By Dirichlet's theorem, there exist arbitrarily large primes $q$ such that $q \equiv 1 \mod{p}$. Then in a direct product $G = C_{p^{n-1}} \times H$, where $H$ is a non-abelian group of order $pq$, the Sylow subgroups of $G$ have $C_{p^{n-1}}$ as their common intersection. There are exactly $q$ Sylow $p$-subgroups, because otherwise $G$ would be nilpotent but its subgroup $H$ is not. Therefore the number of elements in the $p$-Sylow subgroups is $f_p(G) = q(p^{n} - p^{n-1}) + p^{n-1}$, and this goes to infinity as $q$ goes to infinity. Thus there exist groups $G$ with Sylow $p$-subgroups of order $p^n$ such that $f_p(G)$ is arbitrarily large.
Also, $f_p(G) \rightarrow \infty$ as $k \rightarrow \infty$. This is seen by noticing that $f_p(G)^{p^n} \geq n_p(G)$, so 
$$f_p(G) \geq (kp + 1)^{p^{-n}}$$
which goes to infinity as $k \rightarrow \infty$.
One more observation: not all integers $\equiv 1 \mod{p}$ are possible amounts of Sylow $p$-subgroups. For example, there does not exist a group with exactly $22$ Sylow $3$-subgroups, although $22 \equiv 1 \mod{3}$. I don't know if this complicates things.
(*) Frobenius Theorem says that when $G$ is a finite group with order divisible by $s$, the number of solutions to $x^s = 1$ in $G$ is a multiple of $s$. We know that $f_p(G)$ is the number of solutions to $x^{p^n} = 1$ in $G$.

REPLY [2 votes]: This is just a small partial result and some comments. I think the following should settle the case $k = 2$.
Suppose that $G$ is a group with Sylow $p$-subgroups of order $p^n$ and that $n_p(G) = 2p + 1$. According to a theorem of Marshall Hall (see theorem 3.1 in [*]), this can only happen if $2p + 1 = q^t$ is a power of a prime, so let's assume that this is the case.
First of all, the lower bound $p^n(2p - 1)$ given in the question is attained. Let 
$$G = C_{p^{n-1}} \times AGL(1, q^t),$$
where $AGL(1, q^t)$ 
is the group of invertible affine transformations $x \mapsto ax + b$ of the field of order $q^t$. Here $G$ has exactly $2p + 1$ Sylow $p$-subgroups, the Sylow $p$-subgroups have order $p^n$ and $f_p(G) = p^n(2p - 1)$.
Now by Frobenius theorem $f_p(G) = tp^n$ where $t$ is an integer. Since $2p - 1 \leq t < 2p + 1$, we see that $t = 2p - 1$ or $t = 2p$.
If $p \neq 2$, then $t = 2p - 1$ because $t-1$ must be a multiple of $p - 1$ [**]. Therefore in this case $f_p(G) = p^n(2p - 1)$.
If $p = 2$ and $n \geq 2$, then $f_2(G) = 2^{n+2}$ is attained by a semidirect product $G = C_{2^n} \ltimes_\theta C_5$ (I think, I'll check this later).

I have not made much progress for the cases where $n_p(G) = kp + 1$ and $k > 2$. In the case where $n_p(G) = 3p + 1$, a theorem of Marshall Hall (see theorem 3.2 in [*]) shows that $p = 2$, $p = 3$ or $p = 5$. It seems that things get messy from now on with this approach, perhaps it's best to disregard "impossible values" like $n_3(G) = 22$. Or we could start with the case $p = 2$ where there are no impossible values. 
[*] M. Hall, On the number of Sylow subgroups in a finite group (1967) DOI link
[**] Proof: Now $f_p(G) - 1$ is the number of elements of order $p^k$, where $1 \leq k \leq n$. Since the number of elements of order $s$ is always a multiple of $\varphi(s)$, we get that $f_p(G) - 1$ must be a multiple of $p-1$. Thus $t-1$ is also a multiple of $p-1$.<|endoftext|>
TITLE: Ring structure for the motivic spectrum/complex that represents singular cohomology?
QUESTION [7 upvotes]: As the discussion here Is singular cohomology representable by a (Voevodsky's) motivic complex?
shows, the singular cohomology of (smooth) complex varieties is represented by a motivic complex (and also by a motivic spectrum). My question is: what can be proved about the ring structure for this complex/spectrum? Is it known that this a 'weak' ring spectrum? an $A_{\infty}$-spectrum? a highly structured ring spectrum? Any hints would be very welcome!

REPLY [8 votes]: Singular cohomology is represented by an $E_\infty$ motivic ring spectrum. That spectrum is $\mathbf{R}f(H\mathbb{Z})$ where $f$ is right adjoint to the stable topological realization functor and $H\mathbb{Z}$ is the topological Eilenberg-Mac Lane spectrum. Since this is a symmetric monoidal Quillen adjunction (see Theorem A.45 here: http://arxiv.org/pdf/0709.3905.pdf), $\mathbf{R}f$ preserves $E_\infty$-objects.
The same argument should show that the motivic complex is also $E_\infty$, but I don't know a reference for the required symmetric monoidal adjunction. At least if you work with $(\infty,1)$-categories this adjunction comes for free from the fact that motivic complexes are the same thing as modules over the motivic Eilenberg-Mac Lane spectrum, whose topological realization is $H\mathbb{Z}$.<|endoftext|>
TITLE: Eigenfunctions restricted on closed geodesics
QUESTION [7 upvotes]: Consider the flat torus $T^2=\frac{\mathbb{R}^2}{l_1\mathbb{Z}\oplus l_2\mathbb{Z}}$. It is easy to see that the eigenvalues of the Laplacian on torus, $-\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial y^2}$, are $\lambda_{m_1,m_2}=(2\pi)^2(\frac{m_1^2}{l_1^2}+\frac{m_2^2}{l_2^2})$ with the  associated eigenfunction 
$$f_{(m_1,m_2)}(x,y)=e^{2\pi i(\frac{m_1}{l_1}x+\frac{m_2}{l_2}y)}.$$
where $m_1,m_2\in \mathbb{Z}$. Furthermore, The closed geodesics of $T^2$ parametrized by the arc length, are
$$ \gamma_{(n_1,n_2)}(t)=\frac{1}{l}(n_1l_1t,n_2l_2t)$$
where $n_1,n_2\in \mathbb{Z}$ and $l=\sqrt{n_1^2l_1^2+n_2^2l_2^2}$. 
    A simple computation shows that an eigenfunction, say $f_{(m_1,m_2)}$, restricted on a closed geodesic,  $\gamma_{(n_1,n_2)}$, gives
 $$f_{(m_1,m_2)}\circ \gamma_{(n_1,n_2)}(t)=e^{2\pi i(\frac{m_1n_1+m_2n_2}{l})t}$$ 
Which is an eigenfunction on the circle $\mathbb{R}/l\mathbb{Z}$ with the eigenvalue $\tilde{\lambda}=\left( \frac{2\pi}{l}(m_1n_1+m_2n_2)\right)^2$.   
Now my question is: Is this true in the general cases? More precisely;
Let $\gamma:[0,l]\to M$ be a closed geodesics on the Riemannian manifold $(M,g)$ which is parametrized by the arc length. If $f\in C^\infty(M)$ is an eigenfunction for the Laplacian on $M$, i.e. $$\Delta(f)=\lambda f$$
Then 
Question 1) Is $f\circ \gamma$ an eigenfunction on the circle $S^1=\mathbb{R}/l\mathbb{Z}$? Or, Is it in the form of $$f\circ \gamma(t)=c e^{2\pi i \tilde{\lambda}t}.$$
Question 2) If so, how does $\tilde{\lambda}$ depend on $\gamma$ and $\lambda$?
Thanks.

REPLY [12 votes]: The answer is 'no', as you can see by taking the case of $M$ being the unit $2$-sphere in $\mathbb{R}^3$ and the geodesic $\gamma$ being a great circle, say, the horizontal great circle given by $z=0$.  If you consider the harmonic polynomials of degree $2$ in $x,y,z$ restricted to the $2$-sphere, these are eigenfunctions of the Laplacian on the $2$-sphere, but their restrictions to the horizontal great circle aren't usually eigenfunctions of the Laplacian on the circle.   
More generally, you take the $k$-th eigenspace of the Laplacian on the $2$-sphere for $k>1$, you'll find that the restriction of these functions to each great circle projects into a sum of a finite number of eigenspaces of the Laplacian on the great circle (I think it's about $\tfrac12(k{+}2)$ of them), but not into a single one of these eigenspaces.<|endoftext|>
TITLE: Does the proof of GAGA use the axiom of choice?
QUESTION [14 upvotes]: Serre's GAGA result roughly states the following. Let $X$ be a complex projective algebraic variety. Then the natural functor from the category of coherent sheaves over the algebraic structure sheaf of $X$ to the category of coherent sheaves over the analytic structure sheaf of $X$ is an equivalence of categories.
This theorem always seemed to have the air of magic to me. Things that are analytic must come from algebra. I want to dust away some of this magic, and get a clearer picture. With this goal in mind, I have skimmed the proof of GAGA.
The proof of GAGA is rather involved. It uses Cartan's theorem A for both the algebraic and analytic cases, the isomorphism of the completions of the stalks of the structure sheaf in the algebraic case and the analytic case, and a variety of technical results. After having done that for a few days, I still remain with a sense of amazement and a basic lack of understanding about what makes this work. This brings me to the precise phrasing of my question: (which will hopefully help me find the precise step where the magic happens)
Question
Does the proof of Serre's GAGA theorem use the axiom of choice? If so, at what step does this happen?

REPLY [9 votes]: First, let me say that the statement of GAGA you recall is true only for a complex algebraic variety that is projective. In general, it is false: take $X$ an affine space,
and consider the morphisms from the structural sheaf to itself in both the 
algebraic and analytic categories. They are just the algebraic, resp. holomorphic function on the affine spaces, which obviously are not the same.
Now, I would be very embarrassed if you asked me to justify that assertion, but I am pretty sure that GAGA doesn't use the axiom of choice. It is a very natural, very functorial 
argument that seems canonical from beginning to end.
About the air of magic of GAGA... Well, I think it is not so magic. Serre himself says
that it was a very natural thing to do at that time and place -- and he doesn't say that just out of modesty; for instance, he doesn't say that for all his articles. Instead of magic, it
was a technical tour de force, where in a prefiguration of Grothendieck's style the right notions (such as flatness) were developed and applied with exquisite precision.
But once you have observed the elementary fact that any meromorphic function on the projective
line is actually a rational function (a quotient of two polynomials), you have the prototype of all Gaga's result and it is not a tremendous stretch to imagine that everything which is analytic in the projective world is also algebraic.<|endoftext|>
TITLE: Volume-like property to upper bound lattice points in a convex body
QUESTION [12 upvotes]: The following question arises in passing in a joint paper that I am working on.  Let $K$ be a centrally symmetric convex body in an $n$-dimensional real vector space $V$ which contains a lattice $L$.  $L$ yields a natural volume scale fon $V$, and Minkowski established a well-known lower bound on the number of points in $L$ contained in $K$, namely $N \ge (\text{Vol}\ K)/2^n$.  You could view this as a primitive estimate, but actually it's often roughly correct, after you take an $n$th root for a fair comparison.
Is there a similar upper bound that (a) is invariant under the $\text{GL}(n,\mathbb{Z})$ stabilizer of $L$, (b) behaves reasonably if you dilate $K$, and (c) can be regarded as useful or simple?  Without any further restrictions on $K$, you obviously can't just use its volume, because it could be a thin rod that contains many lattice points of $L$.  An extra restriction that holds in our case is that $K$ is a lattice polytope, i.e., the convex hull of finitely many lattice points.  What we currently do is ask for basis of $L$, or a lattice that contains $L$, relative to which $K$ is contained in the $\ell^\infty$ ball of radius $c$.  Then of course you get $N \le (2c+1)^n$.  Or more generally, you could use a lattice parallelepiped. This works in the cases that we need it, but I don't know when it's a good estimate, again with the allowance of an $n$th root.  It's also only $\text{GL}(n,\mathbb{Z})$-invariant by fiat.

REPLY [4 votes]: Oded Regev and I recently published a paper that partially answers this question: https://arxiv.org/abs/1611.05979. In particular, we obtain reasonably tight bounds for the case when $K$ is a Euclidean ball in terms of the quantity $\min_{L' \subseteq L} \det(L')^{1/\mathrm{rank}(L')}$:

Section 9 of Dadush and Regev (https://arxiv.org/abs/1606.06913) has some discussion about the case of general convex bodies (as well as possible strengthenings of the above). In particular, they show that a natural generalization of the above to arbitrary convex bodies fails unless $t \geq \Omega(n^{1/4})$, but it holds for $t = O(n)$.<|endoftext|>
TITLE: Category and the axiom of choice 
QUESTION [10 upvotes]: What are (if any) equivalent forms of AC (The Axiom of Choice) in Category Theory ?

REPLY [17 votes]: Here's a somewhat trivial one, but it is one that category theorists use all the time:

Let us say that a functor $F : \mathcal{C} \to \mathcal{D}$ is a weak equivalence if it is fully faithful and essentially surjective on objects, and that it is a strong equivalence if there exists a functor $G : \mathcal{D} \to \mathcal{C}$ such that $G F \cong \textrm{id}_{\mathcal{C}}$ and $F G \cong \textrm{id}_\mathcal{D}$.
Proposition. In Zermelo set theory with only bounded separation, the following are equivalent:

Every surjection of sets splits.
Any weak equivalence between two small categories is a strong equivalence.
Any weak equivalence between two small groupoids is a strong equivalence.
Any weak equivalence between two small preorders is a strong equivalence.
Any weak equivalence between two small setoids is a strong equivalence.

Here, by "small" I mean something internal to the set-theoretic universe in question.

On the other hand, if you're asking for category-theoretic formulations of the axiom of choice inside some category of "sets", then there are several:

The usual formulation just says that every epimorphism in $\textbf{Set}$ splits. This generalises easily to any category.
In any topos $\mathcal{E}$, one can formulate the axiom schema "every surjection $X \to Y$ splits" in the internal language of $\mathcal{E}$, and this axiom schema is valid if and only if every object is internally projective, in the sense that the functor $(-)^X : \mathcal{E} \to \mathcal{E}$ preserves epimorphisms. This is called the internal axiom of choice.
The internal axiom of choice holds in $\textbf{Set}$ precisely if the usual axiom of choice holds; this is because $\textbf{Set}$ is a well-pointed topos; but in general the internal axiom of choice is weaker. For example, for any discrete group $G$, the category $\mathbf{B} G$ of all $G$-sets and $G$-equivariant maps is a topos in which the internal axiom of choice holds, but if $G$ is any non-trivial group whatsoever, then there exist epimorphisms in $\mathbf{B} G$ that do not split. (For example, $G \to 1$, where $G$ acts on itself by translation.)

REPLY [14 votes]: The following is equivalent to the axiom of choice:

A full and faithful functor which is essentially surjective on objects is an equivalence

Here by "equivalence" I mean "has an up-to-natural-isomorphism inverse".
But wondering which things are equivalent to the axiom of choice is such a set-theoretic thing to do. It is also interesting to ask whether category theory allows us for a more "algebraic" formulation of the axiom of choice. And indeed, in a topos we can express the axiom of choice in two ways:

Externally: Every epi splits.
Internally: Exponentiation by an object preserves epis.<|endoftext|>
TITLE: Visualizing polyhedra from their 1-skeletons
QUESTION [7 upvotes]: Except for a few simple cases (typically pyramids and prisms) I find it hard to visualize a polyhedron from its 1-skeleton embedded in the plane, e.g. the hexahedral graph 5, as can be seen here.
Tools that are able to take an arbitrary polyhedral graph as input and draw the corresponding polyhedron perspectively will most surely rely on an abstract representation of the graph, e.g. by its adjacency matrix. From this abstract representation - presumably - they will also draw the embedded version of the graph (without edges crossing).
I am interested in the underlying algorithms and/or heuristics of


drawing the embedded graph from the adjacency matrix
drawing the polyhedron from the adjacency matrix
drawing the polyhedron from the embedded graph 
drawing the embedded graph from the polyhedron


I am asking for references.
Computer programs will most certainly deal with (1) and (2) while humans typically have to solve problems (3) and (4). 
I guess that experts have some mental techniques to visualize a polyhedron from looking at its 1-skeleton.

Can these techniques be described, made explicit, and taught?

[Side question: If anyone could give me a visualization of the hexahedral graph 5, I would be thankful.]

REPLY [2 votes]: As has been noted one can draw diagrams of convex 3-dimensional polyhedra as graphs in the plane. Such diagrams are plane and 3-connected (Steinitz's Theorem) and are essentially unique but can be very different in visual appearance because one has the freedom to choose the infinite face. So if the polyhedron has a k-gon face one can draw a plane representation (with straight line edges and convex regions except for the infinite face) with a k-gon as the infinite face.<|endoftext|>
TITLE: Automorphism groups of general type varieties
QUESTION [7 upvotes]: The question Are automorphism groups of hypersurfaces reduced ? reminded me of the following related question that I have not seen discussed.
If $X$ is a smooth projective variety over an algebraically closed field $k$ of characteristic $p$ such that the canonical bundle $K_X$ is ample, then is the automorphism group of $X$ reduced?
Some observations: 

It is equivalent to ask if $H^0(X,T_X) = 0$. 
When $X$ is a hypersurface in $\mathbb{P}^n$, this is exactly the question answered in the link above.
If $d := \mathrm{dim}(X) < p$ and $X$ lifts flatly to $W_2(k)$, then the answer is "yes" as $H^0(X,T_X) = (H^d(X, \Omega^1_X \otimes K_X))^\vee = 0$ by Kodaira vanishing (which holds because of Deligne-Illusie-Raynaud under these hypotheses).

Arguments or references or counterexamples would be very appreciated!

REPLY [8 votes]: William Lang produced examples of surfaces of general type in positive characteristic with non-zero vector fields. Since surfaces of general type must have finite automorphism groups, this gives examples. See William Lang, "Examples of surfaces of general type with vector fields", Arithmetic and geometry, Vol. II, Progr. Math., 36, Birkhäuser, pp. 167–173<|endoftext|>
TITLE: probability of zero subset sum
QUESTION [26 upvotes]: Almost 17 years ago, I asked the following question on USENET, motivated by a method in numerology (I kid you not).
Pick integers $n \ge 2$, $k \ge 1$. Toss $n$ $k$-sided dice.  The sides of each die are numbered $0,1,\ldots,k-1$.  The dice are unbiased and the tosses are independent. 
What is the probability $P(n,k)$ that no non-empty subset of the dice adds to a multiple of $k$? 
One can get answers with inclusion-exclusion, but it becomes rapidly more difficult as $n$ increases.  Simple cases are
$$k P(1,k) = k-1,$$
$$k^2 P(2,k) = (k-1)(k-2).$$
David desJardins found that
$$ k^3 P(3,k) = k^3 - 7 k^2 + 15 k - 9 - d_2(k), $$
$$ k^4 P(4,k) = k^4 - 15 k^3 + 80 k^2 - 170 k + 104 - (10 k - 40) d_2(k) + 10 d_3(k),$$
where
$$ d_2(k) = 1 \text{ if $k$ is even, 0 otherwise},$$ 
$$ d_3(k) = 1 \text{ if $k$ is 0 mod 3, otherwise}.$$
David also found the leading terms as $k\to\infty$ for fixed $n$, starting with
$$ P(n,k) = 1 - (2^n - 1)/k
         + 1/2 (4^n - 3^n - 2^n + 1)/k^2 + \cdots .$$
However, nobody found an exact formula, recursion, or generating function, or in fact any method for rapid computation when $n$ is large.  That's my question.

REPLY [2 votes]: First off, let me prove that all $P(n,k)=0$ for $n\geq k$, which follows from a simple lemma:
Lemma. In any sequence of $k\geq 1$ integers $m_1, m_2, \dots, m_k$, there exists a subsequence summing to a multiple of $k$.
Proof. Define $s_i := (m_1+m_2+\dots+m_i)\bmod k$, including $s_0=0$. By the pigeonhole principle, among the integers $s_0, s_1, s_2, \dots, s_k\in\{0, 1,\dots,k-1\}$, there exist two equal ones, say, $s_i=s_j$ for some $i<j$. Then $m_{i+1}, m_{i+2}, \dots, m_j$ form a required subsequence. QED

Now, let me discuss computing $P(n,k)$ for fixed $k$.
For any positive integer $n$ and any $S\subset \mathbb{Z}_k$, let $Q(n,S)$ denote the number of $n$-tuples from $\mathbb{Z}_k^n$ such that sum of any nonempty subsequence is not in $S$.
Then
$$Q(n,S) = \sum_{t\in \mathbb{Z}_k\setminus S} Q(n-1,S\cup (S-t)),$$
where $S-j := \{ s-t\mid s\in S\}$.
Furthermore, by definition we have $$Q(0,S) = 1.$$
In other words, if we fix and arbitrary indexing of the subsets of $\mathbb{Z}_k$, and let $\bar{Q}(n) := [Q(n,S_1),Q(n,S_2),\dots,Q(n,S_{2^k})]^T$, then
$$\bar{Q}(n) = M\cdot \bar{Q}(n-1) = M^n\cdot \bar{Q}(0),$$
where $M$ is a $2^k\times 2^k$ matrix with $M_{ij}$ equal the number of $t\in \mathbb{Z}_k\setminus S_i$ such that $S_j = S_i \cup (S_i - t)$. Clearly, $\bar{Q}(0)=[1,1,\dots,1]^T$.
It follows that for any $S$, the sequence $(Q(n,S))_{n\geq 0}$ satisfies an order $2^k$ linear recurrence, and in particular so does $k^n P(n,k) = Q(n,\{0\})$. This, however, is not so exciting as we know that the sequence $(k^n P(n,k))_{n\geq 0}$ stabilizes to $0$ starting with $n=k$. Still, the aforementioned formulae allow to compute $P(n,k)$ routinely, at least for small $k$.

Example for $k=2$. Let us fix the indexing: $S_1 = \emptyset$, $S_2=\{0\}$, $S_3=\{1\}$, and $S_4=\{0,1\}$. Then
$$M = \begin{bmatrix}
2 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}$$
with the minimal polynomial $x^4 - 3x^3 + 2x^2$. Hence, for $n\geq 4$
$$Q(n,S) = 3Q(n-1,S) - 2Q(n-2,S).$$
In particular, $(2^n P(n,2))_{n\geq 0} = (1,1,0,0,0,\dots)$, i.e., $P(n,2)=0$ for $n\geq 2$ as we already know.
Examples for $k\leq 10$.
\begin{split}
(2^n P(n,2))_{n\geq 0} &= (1, 1, 0, 0, 0, 0, 0, 0, 0, 0, \dots), \\
(3^n P(n,3))_{n\geq 0} &= (1, 2, 2, 0, 0, 0, 0, 0, 0, 0, \dots), \\
(4^n P(n,4))_{n\geq 0} &= (1, 3, 6, 2, 0, 0, 0, 0, 0, 0, \dots), \\
(5^n P(n,5))_{n\geq 0} &= (1, 4, 12, 16, 4, 0, 0, 0, 0, 0, \dots), \\
(6^n P(n,6))_{n\geq 0} &= (1, 5, 20, 44, 10, 2, 0, 0, 0, 0, \dots), \\
(7^n P(n,7))_{n\geq 0} &= (1, 6, 30, 96, 90, 36, 6, 0, 0, 0, \dots), \\
(8^n P(n,8))_{n\geq 0} &= (1, 7, 42, 174, 240, 84, 28, 4, 0, 0, \dots), \\
(9^n P(n,9))_{n\geq 0} &= (1, 8, 56, 288, 690, 336, 168, 48, 6, 0, \dots), \\
(10^n P(n,10))_{n\geq 0} &= (1, 9, 72, 440, 1344, 984, 336, 144, 36, 4, 0, \dots).
\end{split}
P.S. I've added these values to the OEIS as the sequence A309106.<|endoftext|>
TITLE: Example: a pair of nonisomorphic parallel morphisms with isomorphic cones
QUESTION [7 upvotes]: First of all, let me fix some notation. 

Let $\mathcal D$ be a triangulated category in the sense of Verdier-Grothendieck (for example, the homotopy category $\mathbf{K}(k)$ of cochain complexes over a fixed commutative ring $k$). I call cone of a morphism $f : A \rightarrow B$ the object $C(f)$ (uniquely determined up to isomorphism) such that $A \stackrel{f}\rightarrow B \rightarrow C(f) \rightarrow A[1]$
is a distinguished triangle in $\mathcal D$. When $\mathcal D = \mathbf{K}(k)$, $C(f)$ can be identified (up to homotopy equivalence) to the mapping cone of the chain map $f$.
In any category $\mathcal C$, I say that two morphisms $f: A \to B$ and $f' : A' \to B'$ are isomorphic, if they are isomorphic in the category of morphisms $\mathrm{Mor}(\mathcal C)$, that is, there are two isomorphisms $u : A \to A'$ and $v : B \to B'$ such that $vf = f'u$.

Now, my question is the following: is there a triangulated category with a pair of parallel morphisms $f,f' : A \to B$ such that $f$ is not isomorphic to $f'$ but $C(f)$ is isomorphic to $C(f')$? I believe that an example could be found in the category $\mathbf{K}(k)$.
Of course, if we don't require $f$ and $f'$ to be parallel, then we may find examples in any reasonable triangulated category: just set $f=1_0$, the identity of a zero object, and $f' = 1_A$, the identity of a nonzero object. Then, both cones are zero objects (a general fact in triangulated categories), but clearly $f$ is not isomorphic to $f'$.

REPLY [4 votes]: Yet another example. Take $R$ any ring such that $R\cong R\oplus R$. Consider the following parallel morphisms $f,g\colon R\rightarrow R$: $f=0$ the trivial morphism, and
$$g=\left(\begin{smallmatrix}
1&0\\\0&0
\end{smallmatrix}\right)\colon R\cong R\oplus R\longrightarrow  R\oplus R\cong R.$$
Both have isomorphic mapping cone
$$C\colon \cdots\rightarrow0\rightarrow R\stackrel{0}\rightarrow R\rightarrow0\rightarrow\cdots$$
but $f\ncong g$ since $f=0\neq g$.<|endoftext|>
TITLE: Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group?
QUESTION [19 upvotes]: This question was originally posted on math.SE by myself nearly a year ago. I've been thinking again about the problem after it recently received a little attention, but little progress was made in finding a solution. So, I feel it's sufficiently difficult to post on mathoverflow.
It's an easy exercise to show that if $B_n$ is Artin's classical braid group on $n$ strings, then $B_n$ can be embedded in $B_{n+1}$ (and in a canonical way). A similar statement can be proved for the pure braid group $P_n$. This property is useful for proving various properties of the classical braid groups. One therefore asks if a similar property holds for braid groups on other surfaces such as the sphere.
Let $P\mathcal{S}_n$ be the pure $n$-string braid group on the sphere $S^2$. Fox's definition of this group is the fundamental group of the configuration space $F_{n}S^2=\prod_n S^2\setminus\{(x_1,\ldots,x_n)|\exists i\neq j, x_i=x_j\}$ and then the full braid group $\mathcal{S}_n$ is defined to be the fundamental group of the configuration space $B_nS^2=F_nS^2/\Sigma_n$ where $\Sigma_n$ is the action of the symmetric group by permuting coordinates of the elements of $F_nS^2$.
It was proven by Fadell and Van Buskirk that the braid group $\mathcal{S}_n$ has presentation given by the braid generators $\sigma_1,\ldots,\sigma_{n-1}$ and relations
$$\begin{eqnarray}\sigma_i\sigma_j\sigma_i&=&\sigma_j\sigma_i\sigma_j&(\mbox{ for }|i-j|=1)\\
\sigma_i\sigma_j&=&\sigma_j\sigma_i&(\mbox{ for }|i-j|>1)\\
\gamma&=&1&\end{eqnarray}$$
where $\gamma=(\sigma_1\sigma_1\ldots\sigma_{n-1})(\sigma_{n-1}\ldots\sigma_2\sigma_1)$.
With that framework now built up, my question is, can $\mathcal{S}_n$ be embedded in to $\mathcal{S}_{n+1}$ for $n\geq 3$ (and similarly for their pure counter parts)? The naive 'add a string on the end' map will not work because, for instance, the braid $\gamma$ becomes non-trivial when a string is added on the end.
I would think that the answer is no because of the dependence of the relation $\gamma=1$ on $n$, but a proof eludes me.

REPLY [3 votes]: An easy way to construct a section of $F_{n+1}S^2 \rightarrow F_{n}S^2$ for $n>2$ is the following: 
write $[x_1,…x_n]=A(x)\cdot[0,1,\infty,t_1,…,t_{n-3}]$ for a unique $A(x)\in PSL_2(\mathbb C)$. Then the section is given by 
$s(x)=A(x)^{-1}\cdot[0,1,\infty,t_1,…,t_{n-3},2+\sum_{i=1}^{n-3} Re(t_i)^2]$.
In fact I am interested in such sections $F_{n+1}S \rightarrow F_{n}S$ for any compact Riemann surface $S$.
If $S=\mathbb R^2/\mathbb Z^2$ we can find sections for any $n>0$ as follows.
Let $d$ be the Euclidian distance on $S$.
We can use $s(x_1,…,x_n)=(x_1,…,x_n,x_n+\tau(x))$, with $\tau(x)=\frac{1}{2}min\left(1,min_{0<i<n}\left(d(x_n,x_i)\right)\right).$
I would be very interested if someone knows if there are such sections for $genus(S)>1$ and general $n$.<|endoftext|>
TITLE: Examples of "exotic" induction
QUESTION [7 upvotes]: Next week I am going to teach two lessons on induction to very motivated students from high schools. At some point I would like to talk about ordered sets, well-ordered sets, and mention the fact that induction works on well-ordered sets. More precisely, I will prove the following formulation of induction: let $S$ be a well-ordered set with minimal element $0$, and let $P$ be a property such that $P(0)$ is true and if $P(x)$ is true for every $x < y$, then $P(y)$ is true. Then $P(x)$ is true for every 
$x\in S$. 
I would like to provide a (sufficiently elementary) application of this principle in a case when $S$ is not (isomorphic to) the set of natural numbers. For example, one could try to find an example for $S=\mathbb{N}\times \mathbb{N}$, endowed with the lexicographical order. I am looking for examples such that:

the problem is sufficiently "natural" and elementary (the audience is from high school!)
possibly, it should not reduce to an induction "in one variable" (for example, the induction on $(m,n)\in\mathbb{N} \times\mathbb{N}$ should not reduce to an induction on $n+m\in\mathbb{N}$)

Of course any induction on $\mathbb{N}\times\mathbb{N}$ may be reduced to a double induction on $\mathbb{N}$, but in order to avoid this I should look for too complicated well-ordered sets, so this does not bother me too much... Anyway, if anyone knows some reasonably simple examples with more complicated well-ordered sets, then this would be very good for me!

REPLY [8 votes]: Goodstein's Theorem 
http://en.wikipedia.org/wiki/Goodstein%27s_theorem 
is proved using an induction of length $\epsilon_0$.<|endoftext|>
TITLE: Mathematical "proof" of the stability of atoms?
QUESTION [18 upvotes]: I am trying to find proofs of the stability of an atom, says, for simplicity, the hydrogen atom. There are positive answers and negative answers in various atom models.
The naive "solar system" model of a negatively charged electron orbiting the positively charged nucleus is not stable, it radiates electro-magnetic energy and will collapse.
The Bohr-Sommerfeld atom model seems to make stability a postulate.
The Schroedinger equation seems to give a "proof" of the stability of the hydrogen atom, because we have stable solutions corresponding to bound states.
Does anybody know if the Dirac equation or Quantum Electro-Dynamics can be used to prove the stability of a hydrogen atom?
Many thanks in advance for any references where I can learn more about this.

REPLY [11 votes]: The first thing to say is that ordinary matter is actually not stable. Suppose a baseball-sized rock finds itself in the vacuum of outer space in the very distant future, isolated by the universe's accelerating expansion within its own cosmological horizon. Even within the standard model of particle physics, the rock will eventually decay by quantum-mechanical tunneling into more stable forms of matter. Over extremely long time scales, the result is believed to be that it will become a microscopic black hole, which then evaporates into other particles (mostly photons). (You will hear people say that this is the ultimate fate of all matter in the universe, which isn't actually right.) This kind of thing is discussed in Adams and Laughlin.
You asked about the stability of the hydrogen atom in various theories. There are some reasons to believe that the proton is unstable (google "proton decay"), in which case the hydrogen atom isn't actually stable. However, it is stable within specific models. Others have pointed out the Lieb paper, which in section I makes a specific technical argument about one type of stability for individual atoms according to one model. The model is the Schrodinger equation with a pointlike proton.
First off, there are really two things that are required in order to show that hydrogen is stable in this model, and Lieb only focuses on one of them, which is stability against a collapse of the electron's wavefunction so that it becomes bounded within an arbitrarily small distance from the proton.
The other type of stability that has to be demonstrated is stability against the electron's escape. Stability against escape is nontrivial. For example, the interaction between two neutrons is essentially purely attractive, and yet the two-neutron system is believed to be unbound. This is because the range of the force is so short (about $10^{-15}$ m). If the neutrons were to be confined within that distance of one another, they would have to have high kinetic energy, so they would fly apart. The reason hydrogen is bound is that the electrical force is long-range.
For hydrogen's stability against collapse, Lieb's argument is more complicated than it needs to be, because he unrealistically assumes a pointlike proton. Since protons are not really pointlike, compressing the electron to an arbitrarily small space $\epsilon$ near the center of the proton gives an electric field whose energy diverges to infinity like $1/\epsilon$. (If the proton were pointlike, then the external field would go to zero in this limit, so this argument would fail.)
Your question about quantum field theory is an interesting one. I think the nicest way to approach this is to look at the dimensionless and dimensionful quantities that you can form out of the relevant parameters. Most of the interesting physics can be understood in terms of two of these. There is the fine structure constant, $\alpha=ke^2/\hbar c\approx 1/137$, and the Bohr radius, $a_o=\hbar/mc\alpha$, where $m$ is the mass of the electron. In hydrogen, the typical velocity of the electron is $\alpha c$, and since this is small compared to c, you don't really need quantum field theory for hydrogen. The Schrodinger equation, which is nonrelativistic, is an excellent approximation. However, if you make a hydrogenlike atom consisting of a nucleus with atomic number $Z$ plus a single electron, the velocity in units of $c$ is on the order of $Z\alpha$. For large $Z$, this shows that you need relativity, and quantum field theory.
The Bohr radius is the only quantity you can form here with units of length. That suggests, without the need for explicit solution of the Schrodinger equation, that not only does hydrogen not collapse to an arbitrarily small size (as shown by Lieb's argument), but we expect it to reach a certain size which is basically the Bohr radius times some factor of order unity.
Adams and Laughlin, http://arxiv.org/abs/astro-ph/9701131
Lieb, Rev Mod Phys 48 (1976) 553, http://www.pas.rochester.edu/~rajeev/phy246/lieb.pdf<|endoftext|>
TITLE: On the Universality of the Riemann zeta-function
QUESTION [16 upvotes]: Hi,
I have a question regarding the universality property of the Riemann zeta-function. I am no expert on this, so I'd be glad for any relevant reference.
First, recall Voronin's remarkable theorem on the Universality of the Riemann zeta-function :
Let $K$ be a compact subset with connected complement lying in the strip $\{1/2 < \operatorname{Re}(z)<1\}$, and let $f : K \rightarrow \mathbb{C}$ be continuous, holomorphic on the interior of $K$, and zero-free on $K$. Then for each $\epsilon>0$, there exists $t>0$ such that
$$\max_{z \in K} |\zeta(z+it)-f(z)|<\epsilon.$$
Even more : the lower density of the set of such $t$'s is positive..!
Note that of course, the hypothesis that the complement of $K$ is connected is essential in the above theorem.
My question is the following :
Is there some sort of (modified) zeta-function universality-like result for compact sets $K$ with disconnected complements? For example, if $\mathbb{C}_\infty \setminus K$ has a finite number of components?
EDIT
Of course I know that a sequence of the form $f_n(z):=\zeta(z+it_n)$ won't work in the case when the complement of $K$ is disconnected (such a sequence cannot approximate uniformly say $1/z$ on an annulus centered at $0$). I'm asking wether there is some sequence of functions, involving the Riemann zeta-function, that could work in this case, and generalize Voronin's Theorem. Note that such functions will necessarily have poles in each component of the complement of $K$. 
2nd EDIT
Let me explain what I was looking for here. Basically, I'd like to know if there exists a result of the following form :
Let $K$ be a compact subset whose complement has finitely many components lying in the strip $\{1/2 < \operatorname{Re}(z)<1\}$, and let $f : K \rightarrow \mathbb{C}$ be continuous, holomorphic on the interior of $K$, and zero-free on $K$. Then for each $\epsilon>0$, there exists...
Here insert some uniform approximation of $f$ on $K$ by a function involving the Riemann zeta-function
Furthermore, in the case when $K$ has connected complement, I would like the above result to reduce to Voronin's Theorem.
In summary, I want to know if there exists a generalization of Voronin's Theorem to compact sets whose complement have finitely many components.
Thank you,
Malik

REPLY [2 votes]: Here is a cheaper alternative depending on what you mean by a modification. Consider $L(z)$ any Dirichlet L function different from $\zeta$.

Joint universality theorem: 
  Let $K$ be a compact set in the right half of the critical stripe $1/2< \Re s<1$ with connected complement. For any two functions $f_1$ and $f_2$ holomorphic in the interior of $K$ (vanishing or not) and every $\epsilon>0$, we have that the limit
  $$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \{ t \leq T: \sup |f_1(z) - \log \zeta(z +i t)| +  \sup |f_2(z) - \log L(z +i t)| < \epsilon\} $$
  is positive for $\lambda$ being the Lebesgue measure.

From this, we can deduce:

Corollary: Let $K_0$ be a compact set in the right half of the critical stripe $1/2< \Re s<1$.
  Let $f$ be a continuous function on $K_0$, which is holomorphic on an open set containing $K_0$. For every $\epsilon_0>0$, we have that the limit
  $$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \Big\{ t \leq T: \sup\limits_{z \in K_0} \left| f(z) - \frac{\log \zeta(z +i t)}{\log L(z+ it)}\right| < \epsilon_0\Big\} $$
  is positive for $\lambda$ being the Lebesgue measure.

Proof:
By Runge's theorem, it is sufficient to approximate rational functions, whose poles lie outside of $K_0$. Let $p(z)$ and $q(z)$ be polynomials such that $q$ does not vanish on $K_0$. Consider $\epsilon_0>0$ sufficiently small (to be made precise as we go on).
Let $K :=\mathbb{C}-O$, where $O$ is the unbounded, connected component of $\mathbb{C}-K_0$. Consider $\epsilon>0$ sufficiently small, then use the joint universality theorem for $f_1(z)=p(z)$ and $f_2(z) =q(z)$.
We want to show that 
$$\sup | f_1/f_2(z) - \frac{\log \zeta}{\log L}(z+i t) |< \epsilon_0.$$
We estimate the left-hand side:
$$ \leq \sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | +  \sup | \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)|.$$
The first summand is easy to estimate:
$$\sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | \leq \sup_{z \in K_0} \left| f_2(z)^{-1} \right| \epsilon.$$
The second one is a little bit harder:
$$  \sup \Big| \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)\Big| \leq $$
$$ \sup \Big|  \frac{\log \zeta(z+i t)}{f_2(z)\log L(z+i t)} \Big|  \sup |  \log L(z+i t) -f_2(z) | <  \sup \Big|  \frac{\log \zeta(z+i t)}{f_2(z) \log L(z+i t)} \Big|  \epsilon,$$
because we have to estimate
$$ \sup |  \frac{\log \zeta}{\log L}(z+i t) | $$
uniformly in $t$.
This is indeed possible, we have that
$$\sup | f_2(z) |  - \sup | \log  L(z + i t) | < \epsilon$$
and
$$ \sup | \log  \zeta(z + i t) | - \sup | f_1(z) |  < \epsilon$$
by the reversed triangle inequality.
So for $\epsilon \leq \sup | f_2(z) |/2$ and $\epsilon \leq \sup | f_1(z) |$ , we have that
$$  \sup | \log  L(z + i t) | > \sup | f_2(z) |/2$$
and
$$ \sup | \log  \zeta(z + i t) | < 2 \sup | f_1(z) | .$$
So 
$$\epsilon_0 := \max\{ \frac{1}{2} \sup |f_2^{-1}| \epsilon, \frac{1}{2} 4* \sup |f_1f_2^{-2}| \epsilon \}$$
will do. 
This finishes the proof of the corollary assuming the Joint universality theorem.<|endoftext|>
TITLE: Forcing to "minimally" add new reals.
QUESTION [9 upvotes]: Suppose that I want to force to add a "single" new subset of $\omega$ and not much else. For example, consider the Cohen forcing consisting of finite partial functions from $\omega$ to 2.  The forcing I am interested in is different (in fact not CCC, but Proper), but still the generic set codes a subset of $\omega$.  
Question 1:  Suppose that $G$ is $P$-generic over $M$, and there is at least one $r\in(\mathscr{P}(\omega)\cap M[G]) \setminus M$.  Is there a condition $(*)$ such that
$P$ satisfies $(*)$ if and only if $\forall t\in M[G]\cap \mathscr{P}(\omega) \exists a, b\in M\cap \mathscr{P}(\omega) [ t = (a \cap r) \cup (b\setminus r)] $ ?
Question 2 (Iterability):  Consider a model $M_\delta$ resulting from an iteration of forcings $P_\alpha : \alpha<\delta$. Say that each $P_\alpha$ adds a real $r_\alpha$. Then $ \{r_\alpha : \alpha<\delta\}  \subseteq \mathscr{P}(\omega)\cap M_\delta$.   Is there a model $M_\delta'\subseteq M_\delta$ such that the reals of $M_\delta'$ are just $(\mathscr{P}(\omega)\cap M)\cup \{ (a\cap r_\alpha)\cup (b\setminus r_\beta) : a,b\in \mathscr{P}(\omega)\cap M; \  \alpha, \beta<\delta\} $ ?

REPLY [6 votes]: There is no forcing that satisfies $(*)$. Let $r$ be a new real, $r \in M[G]\cap \mathcal{P}(\omega) \setminus M$. Take $t = \{n < \omega \mid n + 1\in r\}$, and let $a, b\in M$ such that $t = (a\cap r)\cup (b\setminus r)$. 
I will show that it is possible to reconstruct $r$ from $a, b$ and the bit $0\in r$. This implies that $r\in M$.
Let $n < \omega$ and assume that we know whether $n \in r$ or not. We want to check if $n + 1\in r$. Split into four cases: 

If $n \in a \cap b$ then $n\in t$ and therefore $n + 1\in r$. 
If $n \notin a \cup b$, $n\notin t\implies n+1\notin r$. 
If $n \in a\setminus b$ then $n\in t \iff n\in r$ and therefore $n + 1\in r \iff n\in r$.      
If $n \in b\setminus a$ then $n\in t \iff n\notin r$ and therefore $n + 1\in r \iff n\notin r$. 

So we can reconstruct $r$, bit by bit, in $M$ (up to two possibilities) from $a, b$.<|endoftext|>
TITLE: Do sparse DAGs can have large min-cuts?
QUESTION [9 upvotes]: For a graph $G$, let $e(G)$ denote the number of its edges, and $c_k(G)$ the smallest number
of edges that must be removed in order to destroy all paths of length $\geq k+1$.
Note that $c_1(G)\geq c_2(G)\geq \ldots\geq c_k(G)\geq \ldots$.
Let $K_n$ be a complete graph, and $T_n$ a complete acyclic digraph (transitive tournament) on $n$
vertices; hence $e(K_n)=e(T_n)=\tbinom{n}{2}$.
 
A classical result of Erdős and Gallai
states that 
$$
c_k(K_n)=e(K_n)-\frac{kn}{2}.
$$
In contrast, for the directed acyclic analogue $T_n$ of $K_n$, we have
$$
c_k(T_n)= k\binom{n/k}{2}=\frac{e(T_n)}{k}+\frac{n}{2}\Big(1-\frac{1}{k}\Big).
$$
Proof: To show $c_k(T_n)\leq k\tbinom{n/k}{2}$, take a topological order of vertices of $T_n$: 
vertices $i$ and $j$ are adjacent iff
$i < j$. Split the vertices into $k$ consecutive intervals of length $n/k$. If we remove all
edges whose both endpoints lie in the same interval, then we destroy all paths of length $\geq k+1$. 
 Since only $k\tbinom{n/k}{2}$ edges were removed, we are done. 

 The other direction $c_k(T_n)\geq k\tbinom{n/k}{2}$ was essentially shown
by David Eppstein in this answer:
Let $C$ be a set of edges whose removal destroys all paths of length $≥k+1$ in $T_n$. Split the vertices into $t\leq k$ layers, where the $i$-th layer
contains all vertices $u$ such that the length of a longest path to $u$ in $T_n\setminus C$
 has length $i$. Since each layer is a layer of the longest-path layering,
it is independent in $T_n\setminus C$, and therefore complete in $C$.
Thus, if $n_i$ is the number of vertices in the $i$-th layer, then
the total number of edges in $C$ is at least
$\sum_{i=1}^t\tbinom{n_i}{2}\geq t\tbinom{n/t}{2}\geq k\tbinom{n/k}{2}$, as desired. 
Q.E.D. 

Motivated by this (remarkable) difference between $c_k(K_n)$ and $c_k(T_n)$, here is my 

Question:
Does $c_k(G)$ is at most "about" $e(G)/k$ for every acyclic digraph $G$?

By "about" I mean  "times some absolute constant or times some slowly growing function in $n$".

Note that the problem is only to get rid with graphs having also short paths (shorter than $k$), 
because $c_1(G)\leq e(G)/t$ holds
for any (not necessarily acyclic) digraph $G$, where $t$ is the length of a shortest source-to-target path. This is a direct consequence of
a dual to
Menger’s theorem (attributed to Robacker):
in any directed graph, the minimum length $t$ of a path is equal
to the maximum number of edge-disjoint cuts. (The proof is elementary, see e.g.
here.)

Besides being natural in itself, an affirmative answer to my question would have some interesting consequences in boolean function complexity (see this post and references herein).

REPLY [5 votes]: I just realized that the answer
to my question is (as suspected) NO. 
Namely, in this paper, Georg Schnitger constructed a directed acyclic graph $G$ with $n$ vertices, and
$e(G)\approx n\log n$ edges such that, for every $0\leq \epsilon < 1$ and $k=n^{\epsilon}$,
we have that $c_k(G)\geq \alpha\cdot e(G)$, where $\alpha=\alpha(\epsilon)$ is a constant
depending only on $\epsilon$. This is much larger than the "desired" upper bound
$c_k(G)\leq e(G)/k$. Actually, I think that using the Kraft inequality, one can show  that
$c_k(G)=\Omega(n\log(n/k))$ holds for every $k$: show that at least $m\log m$ edges must be removed in order to disconnect any given subset of $m$ leaves, and use the argument of the proof above (haven't verified the details yet).

The graph $G$ is constructed as follows.
       (source)
Take a complete binary tree of depth $t$; hence, we have $n=2^{t+1}-1$ vertices. 
Remove all edges. Connect
each vertex with all leaves, which were previously its descendants. Direct the new
edges in the following way: the vertex receives edges from his left leaves and sends
edges to his right leaves. 

This example also shows the optimality of depth-reductions
for DAGs proved by  Erdős, Graham and Szemerédi, and generalized by
Valiant to the following important fact:

In a DAG with $m$ edges and depth (maximum length of a path) $d$, 
it is enough to take out $mr/\log d$ edges to
reduce the depth to $d/2^r$.<|endoftext|>
TITLE: Is every finite subcomplex of a contractible simplicial complex contained in a finite contractible subcomplex?
QUESTION [5 upvotes]: The question is as in the title:

Is every finite subcomplex of a contractible simplicial complex $K$ contained in a finite contractible subcomplex of $K$? What if we are allowed to take subdivisions? What if the complex is locally compact and/or finitely dimensional?

I suppose the answer is no, but I could not find any counterexamples.

REPLY [13 votes]: Let $X$ be any finite complex such that any map $X\to X$ homotopic to the identity is surjective and for which there is a surjective but nullhomotopic PL map $f:X\to X$ (for instance, $X$ could be $S^n$ for any $n>0$).  Let $K$ be the mapping telescope obtained by iterating $f$.  Then $K$ is contractible, locally compact, and finite-dimensional, but for any triangulation, the only contractible subcomplex (indeed, the only contractible closed subset) that contains the first copy of $X$ is $K$ itself.  Indeed, in order to be able to contract the first copy of $X$, you must take a subcomplex that contains the entire second copy, and to contract the second copy, you must contain the third copy, and so on.
For $X=S^1$, you can even visualize this example pretty easily.  Take a cylinder, and deform it so that one end of it looks like a crescent, and then collapse the crescent to a circle by gluing together the two "C" shapes that make up the crescent and gluing the two endpoints of the "C".  Glue that circle to one end of another cylinder, and then deform the other end of that cylinder similarly.  Repeating this infinitely gives the telescope $K$.<|endoftext|>
TITLE: Commutators in free groups
QUESTION [9 upvotes]: Let $F_m$ be the free group with $m$ generators $S:=\{x_1,\dots, x_m\}$. I am interested in the following quantity 
$$
F(n):=\frac{|\{w\in [F_m,F_m]: \|w\|_S\leq n\}|}{|B_S(n)|}
$$
where $B_S(n):=\{w\in F_m: \|w\|_S\leq n\}$ and $\|w\|_S$ is the word metric of $w$, and $[F_m,F_m]$ is the commutator subgroup of $F_m$. Do we know how $F(n)$ growth as a function of $n$? Presumably this a classic problem in group theory.  
I would be thankful if you please mention a reference that consider this function.

REPLY [11 votes]: The question is indeed very close to the one about the return probabilities on the abelian group $\mathbb Z^m$. Still, as Benjamin has noticed, there is some difference, since here one has to consider the non-backtracking paths only. 
There is a standard way to deal with it which is based on the so-called random walks with internal degrees of freedom (also known under numerous other names). In this concrete context this is the Markov chain on the product of the free group $F_m$ by the set $A$ of generators and their inverses with the transition probabilities 
$$
p( (g,a), (ga',a') ) = 1/(2m-1) \;, \qquad a'\neq a^{-1} \;.
$$
This particular chain is called non-backtracking simple random walk in the paper MR2342439 (2007) by Kaimanovich, Kapovich and Schupp. Its sample paths are precisely non-backtracking paths in the free group, and its time $n-1$ distribution for the initial distribution uniform on the set of $(a,a):a\in A$ is precisely the uniform measure on the words of length $n$ in the free group.
In the same way one can define the quotient of the above chain on $\mathbb Z^m$. Then the original question is precisely the question about an asymptotic of transition probabilities of this quotient chain. In the abelian case such chains have been extensively studied since early 80s, one of the earliest papers being the one by Krámli and Szász MR0699788 (1983). The asymptotic the OP is asking about is of course $n^{-m/2}$, which was later rediscovered and generalized in numerous other publications. However, I will refrain from going into the detailed history.<|endoftext|>
TITLE: Are $\infty$-topoi determined by their localic points ?
QUESTION [12 upvotes]: Hello !
If $T$ is an infinity topos, then you can consider the infinity category of geometric morphism from $Sh_{\infty}(\mathcal{L})$ to $T$ for any locale $\mathcal{L}$. This associate to $T$ an infinity stacks over the category of all locale (at least for the etale topology, but also probably for some stronger topology).
My question is : is there anything know about this functor ? is it fully faithful ? or does it has some kind of "conservativity" properties that could allow to give an answer to the question in the tittle ? Or in the contrary is there example of non trivial infinity topos with no (or not enough) morphism from non trivial locale ?
thank you !

REPLY [6 votes]: The functor is conservative if $T$ is hypercomplete. This follows from DAG VII, Cor. 4.14, which says that any $\infty$-topos admits a surjection from a hypercomplete locale (where $f$ is a surjection if $f^\ast$ detects $\infty$-connective morphisms).<|endoftext|>
TITLE: Mathematics with the negation of AC
QUESTION [5 upvotes]: Clearly Very important results in Math require the Axiom of choice, for example "any vector space has a base". But in the absence of AC (i.e., only in ZF) it is possible that a vector space has no basis. 
In another direction append the negation of AC to ZF.  What happens to algebra or analysis  now ? 
If you know any Theorem in this new system (other than those that can be drived in ZF alone) please let me know. A fine reference would also be helpful.

REPLY [14 votes]: The negation of the axiom of choice only allows us to prove that there is some set which cannot be well-ordered. There is some family of non-empty sets whose product is empty. There is some partially ordered set in which every chain is bounded, but there is no maximal element. And so on.
Of course, from a family without a choice function we can easily construct a set which cannot be well-ordered, and with it a partial order witnessing the failure of Zorn's lemma, and other examples. But we cannot really say much about this family. Is it a family of finite sets? Is it a family of countable sets? Is this family well-orderable? And so on.
It turns out that mathematics is not just "you either love someone or you hate them". If you have no choice, you can still have plenty of degrees of choice, and without pointing out how much choice you have, or don't have, it's very hard to say much.
Furthermore it is possible that the axiom of choice fails so very very far up the cumulative hierarchy that no set used by any mathematician (except set theorists, maybe) is a witness for this failure. In such universe it is true that some theorems will fail (e.g. there will be a commutative unital ring without a maximal ideal, and there will be a vector space without a basis), but their failure occurs so far beyond our interest that it's just as well possible to assume that it doesn't happen.
All we can say, in case we assume the negation of AC, that all those principles equivalent to the axiom of choice fail somewhere, but we cannot possible give an intelligent answer about where these failures occur.
You may also be interested in this math.SE answer of mine, which discusses a similar question.<|endoftext|>
TITLE: Convex bodies with symmetric shadows
QUESTION [8 upvotes]: Theorem. If all orthogonal projections of a convex body $K \subset \mathbb{R}^n$ onto $2$-dimensional subspaces have a center of symmetry, then $K$ has a center of symmetry. 
This is a classic result of Blaschke and Hessenberg (that I just learned thanks to Guillaume's comment.). A short simple proof of it can be found in Bonnesen and Fenchel.
I wonder if it is necessary to know what happens for every orthogonal projection or whether we can get by with less:
Question 1. Let $K \subset  \mathbb{C}^{n}$ be a convex body. Assume all orthogonal projections of $K$ onto complex lines have a center of symmetry. Does it follow that $K$ must also have a center of symmetry? 
Note. The center of symmetry of the shadows may depend on the subspace containing it. 
A similar question is:
Question 2. Let $K \subset  \mathbb{C}^{n}$ be a convex body. Assume all orthogonal projections of $K$ onto Lagrangian subspaces have a center of symmetry. Does it follow that $K$ must also have a center of symmetry?

REPLY [4 votes]: The answers are no to Question 1, and yes to Question 2 (assuming $n\ge 2$).
Let $h$ be the support function of $K$. The projection of $K$ to a linear subspace $L$ is central symmetric iff the restriction of $h$ to $L$ is a sum of an even function and a linear function. Here and below "linear" means $\mathbb R$-linear.
To construct a counter example to Question 1, begin with the unit ball and its support function $h(x)=|x|$. Then, on each complex line, pick a linear function on this line, in such a way that the coefficients of these linear functions vary smoothly from one complex line to another. Let $f$ be the union of these linear functions. It is well-defined because the complex lines are disjoint outside 0. The functions above should be chosen so that $f$ itself is not linear. Now there is $\varepsilon>0$ so small that $h_1:=h+\varepsilon f$ is still convex (just because its first and second derivatives are close to the original ones). Then $h_1$ is a support function of some convex body, which is a desired counter-example.
In Question 2, we have a function $f(x)=h(x)-h(-x)$ which is linear on every Lagrangian subspace.
Equivalently, $f$ is 1-homogeneous and $f(x+y)=f(x)+f(y)$ for all $\omega$-orthogonal pairs $x,y\in\mathbb C^n$ where $\omega$ is the symplectic form. We have to show that $f$ is globally linear, i.e., $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb C^n$. 
Fix  $x,y\in\mathbb C^n$ and pick $u,v$ from the $\omega$-orthogonal complement of $x,y$ such that $\omega(u,v)=\omega(x,y)$. 
From linearity on Lagrangian subspaces,
$$
 f(x+y+u+v) = f(x+y)+f(u+v)
$$
since $\omega(x+y,u+v)=0$, and on the other hand,
$$
 f(x+y+u+v) = f(x+v)+f(y+u) = f(x)+f(y)+f(u)+f(v)
$$
since $\omega(x+v,y+u)=\omega(x,y)+\omega(v,u)=0$ and $\omega(x,v)=\omega(y,u)=0$.
Thus
$$
 f(x+y)-f(x)-f(y) = f(u)+f(v)-f(u+v) .
$$
The same identity holds for $-u$ and $-v$ in place of $u$ and $v$,
therefore $f(x+y)-f(x)-f(y)=0$, q.e.d.<|endoftext|>
TITLE: Does Multiplicative Version of Azuma's Inequality Hold?
QUESTION [8 upvotes]: It is known that there are multiplicative version concentration inequalities for
sums of independent random variables. For example, the following
multiplicative version Chernoff bound.

Chernoff bound:
Let $X_1,\ldots,X_n$ be independent random variables and $X_i \in$
$[0,1]$. Let $Y=\sum_{i=1}^n X_i$. Then for any $\delta>0$,
$\Pr\left(Y \ge (1+\delta)EY \right) \le e^{-c\cdot(EY)\delta ^2},$
where $c$ is some absolute constant, e.g., c=1/3.

Now consider dependent random variables. A slight variant of Azuma's inequality states the following.

Azuma's Inequality:
Let $X_1,\ldots,X_n$ be (dependent) random variables and $X_i \in
[0,1]$. Assume that there exists $\mu$, such that $ \Pr \left( \sum_{i=1}^n \mathbb{E}[X_i|X_{1},\ldots,X_{i-1}] \le \mu\right) = 1$. Let $Y=\sum_{i=1}^n X_i$. Then for any $\lambda > 0$,
$\Pr\left(Y \ge n\mu+\lambda \right) \le e^{-2 \lambda^2/n}.$

Azuma's inequality is additive. My question is that does a
multiplicative version of Azuma's inequality such as the following
hold?

My question: does the following hold?
Let $X_1,\ldots,X_n$ be (dependent) random variables and $X_i \in
[0,1]$. Assume that there exists $\mu$, such that $\Pr\left( \sum_{i=1}^n \mathbb{E}[X_i|X_1,\ldots,X_{i-1}] \le \mu\right) = 1.$ Let $Y=\sum_{i=1}^n X_i$. Then for any $\delta>0$
$\Pr\left(Y \ge (1+\delta)n\mu \right) \le e^{-c\cdot n\mu \delta^2},$
where $c$ is some absolute constant.

Note: the standard Azuma's inequality does not imply the
multiplicative version when $n\mu \ll
\sqrt{n}$.

REPLY [4 votes]: $\newcommand{\de}{\delta}$
The "dependent" version of the multiplicative Chernoff bound can be proved quite similarly to the "independent" case. Indeed, let $E_{i-1}$ denote the conditional expectation given $X_1,\dots,X_{i-1}$, so that $E_{i-1}X_i\le\mu$ almost surely (a.s.) for $i=1,\dots,n$. Take any real $t\ge0$. By the convexity of $e^{tx}$ in $x$ and the conditions that $0\le X_i\le1$ and $E_{i-1}X_i\le\mu$, we have $e^{tX_i}\le1+(e^t-1)X_i$ and hence
\begin{equation*}
 E_{i-1}e^{tX_i}\le1+(e^t-1)E_{i-1}X_i\le1+(e^t-1)\mu\le\exp\{(e^t-1)\mu\} 
\end{equation*}
for $i=1,\dots,n$. So, by induction, for $j=1,\dots,n$ and $Y_j:=\sum_1^j X_i$ we have 
\begin{equation*}
 Ee^{tY_j}=EE_{j-1}e^{tY_j}=Ee^{tY_{j-1}}E_{j-1}e^{tX_j}\le Ee^{tY_{j-1}}\exp\{(e^t-1)\mu\},
\end{equation*}
whence, by induction, 
\begin{equation*}
 Ee^{tY}=Ee^{tY_n}\le\exp\{n(e^t-1)\mu\}. 
\end{equation*}
So, using Markov's inequality and then choosing $t=\ln(1+\de)$, we get 
\begin{align*}
 P(Y\ge(1+\de)n\mu)&\le e^{-(1+\de)n\mu t}Ee^{tY}
 \le\exp\{-(1+\de)n\mu t+n(e^t-1)\mu\} \\ 
 &=\exp\{-n\mu\psi(\de)\}, 
\end{align*}
where $\psi(u):=u-(1+u)\ln(1+u)$. 
Up to notation, this bound is the same as the known multiplicative Chernoff bound in the "independent" case.
Since $\psi(u)\le-u^2/3$ for $u\in[0,3/2]$, we have 
\begin{equation*}
 P(Y\ge(1+\de)n\mu)\le\exp\{-n\mu\de^2/3\} \tag{1}
\end{equation*}
if $\de\in[0,3/2]$. 
Note that (1) cannot hold for all $\de\ge0$, even in the "independent" case. Indeed, suppose that the $X_i$'s are iid with $P(X_1=1)=\mu=1-P(X_i=0)$, where $\mu:=1/n$ and $n\to\infty$. Then $Y$ will converge in distribution to  a random variable $\Pi$ with the Poisson distribution with parameter $1$, and (1) will yield 
\begin{equation*}
 P(\Pi\ge1+\de)\le\exp\{-\de^2/3\}.  
\end{equation*}
Since Poisson distributions are not subgaussian, the latter inequality cannot hold for all $\de\ge0$. So, (1) cannot hold for all $\de\ge0$.<|endoftext|>
TITLE: Computation of mod 4 Steenrod algebra [ HZ/4, HZ/4]
QUESTION [15 upvotes]: $\newcommand\Sq{\mathrm{Sq}}$I am trying to compute $ [\mathbb{HZ}/4,\mathbb{HZ}/4 ]$ the mod 4 Steenrod algebra. For some reason, I need to work it out till dimension 6 or so. My approach is to use the cofiber sequence
$\mathbb{HZ}/4 \to \mathbb{HZ}/2 \xrightarrow{\Sq^{1}}\Sigma \mathbb{HZ}/2 $
twice.
I did the computation till first five degrees. A description would be as follows,
Let $g$ be the generator of $\mathbb{Z}/4$ in $[\mathbb{HZ}/4,\mathbb{HZ}/4]_{0}$, for convenience let us denote $g' = 2g$. So whenever there is an element with $g'$ would mean that it has $2$ torsion.
Degree | elements
$ 1: \beta g$
$ 2: \Sq^{2} g' $ (this would mean that $2\Sq^{2} g' =0$)
$ 3 : \Sq^{3}g$, (but satisfies $2\Sq^{3}g =0$ ), $\Sq^{2} \beta g'$
$ 4 :  \Sq^{3} \beta g$( $2 \Sq^{3} \beta g = 0$), $\Sq^{4}g'$
$ 5 : \Sq^{5}g$( $2 \Sq^{5} g = 0$), $\Sq^{4} \beta g' = 0$
I do not know if they are right, but if somebody has done it please verify. I suspect the all the groups $\mathbb{Z}/4$ in $[\mathbb{HZ}/4,\mathbb{HZ}/4]_{n}$ are two torsion except $n=0,1$. Is this some sort of known result? Also another thing that I am worrying about is the equivalent Cartan formula in mod 4 Steenrod algebra? Is there a way to detect the Cartan formula?

REPLY [16 votes]: Let me write $H$ for $H\mathbb{Z}/2$ and $X$ for $H\mathbb{Z}/4$.  In these terms we can compute $H^*(X) = [H\mathbb{Z}/4,H\mathbb{Z}/2]$; it is isomorphic to $A/Sq^1 \oplus \Sigma A/Sq^1$, where $A$ is the Steenrod algebra.
Let's take the sledgehammer to the walnut and hit this with the Adams spectral sequence, and leave convergence for some poor other sod to clean up.  It computes the groups you're interested in and starts with
$$
Ext_A(A/Sq^1 \oplus \Sigma A/Sq^1, A/Sq^1 \oplus \Sigma A/Sq^1).
$$
This allows a change of rings to Ext over an exterior algebra:
$$
Ext_{\Lambda[Sq^1]}(\mathbb{Z}/2 \oplus \Sigma \mathbb{Z}/2, A/Sq^1 \oplus \Sigma A/Sq^1).
$$
This decomposes into a sum of four terms, all coming from $Ext_{\Lambda[Sq^1]}(\mathbb{Z}/2, A/Sq^1)$.
Now we need some structure about the Steenrod algebra: namely, the basis in terms of admissible monomials.  A sum of distinct admissible monomials is in the image of right multiplication by $Sq^1$ if and only if it the monomials all visibly end with $Sq^1$, and in the image of left multiplication by $Sq^1$ if and only if terms all end with $Sq^{odd}$.
The net result of this is that, other than $\mathbb{Z}/2$ in degree 0, the module $A/Sq^1$ is a direct sum of free modules with respect to the left action by $Sq^1$.  This means that almost all of the Ext groups vanish in positive degrees.  The only exceptions are four copies of $Ext_{\Lambda[Sq^1]}(\mathbb{Z}/2, \mathbb{Z}/2)$ in total degrees $-1, 0, 0, 1$.  These are spawning the two copies of $\mathbb{Z}/4$ that you mention in (cohomological) degrees 0 and 1.  The rest of the terms are concentrated in Adams filtration zero, and so multiplication by 2 annihilates them.
Generalization from $4$ is an exercise for the reader.  Sorry for my cryptic comments on the question of Mark Grant's, which also appear to be a description of the self-transformations of $H\mathbb{Z}_{(2)}$.  I also don't have a good answer for you about the Cartan formula.
(This could probably be phrased in a more pedestrian fashion using the cofiber sequence $H\mathbb{Z} \to H\mathbb{Z} \to H\mathbb{Z}/4$.)<|endoftext|>
TITLE: Universal property of blowing down
QUESTION [5 upvotes]: Let $X$ be a smooth algebraic surface over $\mathbb{C}$, and $Y \xrightarrow{\phi} X$ the blowup at a (reduced) point with exceptional divisor $E$. Then, the we have the following universal property: Every morphism from $Y$ to an algebraic variety $Z$ that contracts $E$ to a point factors through $X$ (Beauville, Algebraic Surfaces p.17) I'm not sure to think of this (univ. prop. of blow. down) as a property of smooth surfaces over $\mathbb{C}$ or more generally. Here are some natural questions come to mind:
Is there a more general universal property of blowing down along these lines. I don't expect there to be an answer for arbitrary blowups (say of noeth. schemes), however I would like to replace the field $\mathbb{C}$ by other algebraically closed fields like $\overline{\mathbb{F}_p}$, and if the formalism allows it, even $\mathbb{Q}$. There are a couple of cases:
1.) I can't even figure out a universal property if we stay in the context schemes smooth over $\mathbb{C}$, and blow-up at smooth, irreducible subvarieties $Z$.
2.) Let $X, Y, \phi, E$ as above (in particular, surfaces). Assume that $X$ is reduced, but possibly singular, and allow $\phi$ to be a blow-up at an arbitrary (possibly non-reduced) point. Is there a universal property that blowing-down along the exceptional divisor satisfies in this case? (The way I see it, one issue is to replace "collapsing $E$ to a point" by something else.)
3.) I'm very much interested in the most general version of the universal property one can formulate. If there is a reference where this is covered (EGA?) I would love to see it. I also don't have "good" reasons to believe that a universal property doesn't hold in complete generality (say in the context of noetherian schemes). If you're convinced that this is the case, I'd love to hear your reasoning.

REPLY [4 votes]: Artin, "Algebraization of Formal Moduli, II".<|endoftext|>
TITLE: homotopy type of embeddings versus diffeomorphisms
QUESTION [16 upvotes]: Previously, I asked a question on mathoverflow comparing smooth embeddings and diffeomorphisms, which received a very interesting and somewhat unexpected answer by Agol. I now ask a further question about the relation between the homotopy type of embedding spaces and diffeomorphism spaces.
Assume $M$ is a non-compact smooth manifold without boundary (although I would also be interested in the case where $M$ is compact with non-empty boundary). I would like to find an explicit counter-example --- or a proof --- for the following statement: The inclusion $\iota:\text{Diff}(M)\to \text{Emb}(M,M)$ of the space of diffeomorphisms of $M$ into the space of smooth self-embeddings is a weak equivalence on each component of $\text{Diff}(M)$. In other words, every homotopy fibre of the inclusion $\iota$ is either empty or weakly contractible. I am considering the compact-open $C^1$-topology on the above spaces.
Obviously, I would also be very interested in any known results relating the homotopy type of $\text{Diff}(M)$ with that of $\text{Emb}(M,M)$.
Edit: Tom Goodwillie has provided an answer for the case when $M$ is compact or the interior of a compact manifold. I placed a bounty for the case in which $M$ is open (and not necessarily the interior of a compact manifold).
Edit 2: I managed to adapt Agol's nice idea from my previous question to apparently resolve the present question. Since nobody has raised any major issues so far, and people are upvoting my answer, I will likely accept it. As a curious aside, I edited my answer so many times --- out of persnicketiness --- that it turned into community wiki. I was unaware that would happen. Oh well...

REPLY [15 votes]: Personal comment: It seems the discussion in this question finally led me to understand how to modify Agol's argument to answer the present question. In fact, my motivation when asking that question answered by Agol was mostly to resolve the present question.
Answer: For $M$ a non-compact manifold without boundary, it seems the homotopy fibre of the inclusion $\iota:\textrm{Diff}(M)\hookrightarrow\textrm{Emb}(M,M)$ at $\textrm{id}_M$ is weakly contractible. As hinted implicitly by Tom Goodwillie in his answer, it is not hard to conclude that every homotopy fibre of $\iota$ is then either empty or weakly contractible.
Before continuing, I will clarify the structure of the homotopy fibre, $F$, of $\iota:\textrm{Diff}(M)\to\textrm{Emb}(M,M)$ at $\textrm{id}_M$. Recall from the question above that $\textrm{Diff}(M)$ and $\textrm{Emb}(M,M)$ are both endowed with the compact-open (or weak) $C^1$-topology. As a set, $F$ consists of the continuous isotopies $(\varphi_t:M\to M)_{t\in I}$ through embeddings $M\to M$ for which $\varphi_0 = \textrm{id}_M$, and $\varphi_1$ is a diffeomorphism of $M$. An isotopy $(\varphi_t)_{t\in I}$ will also be written as a map $\varphi : M\times I \to M$. As a topological space, $F$ is the subspace of the space of paths in $\textrm{Emb}(M,M)$ consisting of the paths which start at $\textrm{id}_M$ and end at a diffeomorphism of $M$.
Further, let me state a strong form of the isotopy extension lemma which I will require later: when $X$ is a manifold without boundary, and $Y$ is a compact manifold, the map
$$ (\textrm{eval},\textrm{proj}):\textrm{Diff}_c(X)\times\textrm{Emb}(Y,X) \longrightarrow \textrm{Emb}(Y,X)\times\textrm{Emb}(Y,X) $$
given by $(\textrm{eval},\textrm{proj})(u,v)=(u\circ v,v)$ is a fibration. Here $\textrm{Diff}_c(X)$ denotes the space of compactly supported diffeomorphisms of $X$, equipped with the strong (or Whitney) $C^1$ topology --- the use of the strong topology is essential in the present answer, as remarked at a later point. The map $(\textrm{eval},\textrm{proj})$ is a locally trivial fibre bundle, as one can give local trivializations using tubular neighbourhoods; thus the above map is also a fibration, since the base is paracompact (even a metrizable space). The above statement implies most usual forms of the isotopy extension lemma.
I will now describe the argument that the homotopy fibre $F$ of $\iota:\textrm{Diff}(M)\to\textrm{Emb}(M,M)$ at $\textrm{id}_M$ is weakly contractible. It is a modification of Agol's ingenious answer linked above.
Fix a map $f:K\to F$ where $K$ is compact. I will describe a homotopy between $f$ and the constant map $b:K\to F$ equal to the basepoint $b=(\textrm{id}_M)_{t\in I}$ of $F$. First we construct an exhaustion of $M$ by compact submanifolds $M_i$ for $i\in {\mathbb N}$ such that:

$\bigcup_{i\in {\mathbb N}} M_i = M$;
$M_0=\emptyset$;
$M_i \subset \textrm{int}\ M_{i+1}$ for all $i\in {\mathbb N}$;
$\varphi(M_i \times I)\subset \textrm{int}\ M_{i+1}$ for all $\varphi$ in the image of $f$.

For example, we may use a proper smooth function $\rho: M\to [0,+\infty)$, and take $M_i=\rho^{-1}([0,x_i])$ for a suitable unbounded, strictly increasing sequence $(x_i)_{i\in{\mathbb N}}$ of regular values of $\rho$ (which are dense in $[0,+\infty)$ by Sard's theorem). The final condition above is where the compactness of $K$ is required, and I do not know how to avoid using compactness there.
With the compact submanifolds $M_i \subset M$ at hand, we can inductively construct a homotopy $f\simeq b$. Set $f_0=f$, and let $n\in {\mathbb N}$. Assume inductively that we have constructed, for each $k < n$, a map $f_{k+1}:K\to F$, and a homotopy $H_k : f_k \simeq f_{k+1}$. Furthermore, assume that for all $\phi$ in the image of $f_{k+1}:K\to F$, and for all $\psi$ in the image of $H_k: K\times I\to F$, the following conditions hold:

$\phi_t|_{M_k} = \textrm{id}_{M_k}$ for all $t\in I$;
$\psi_t|_{M_{k-1}} = \textrm{id}_{M_{k-1}}$ for all $t\in I$;
$\psi(M_l \times I) \subset \textrm{int}\ M_{l+1}$ for each $l\in {\mathbb N}$ with $l\geq k$.

Now we construct a homotopy $H_n : f_n \simeq f_{n+1}$ based on the above data. Let $\varphi = f_n(x)$ for a fixed $x\in K$. By condition (3) (and $M_0=\emptyset$), we have $\varphi(M_n \times I) \subset \textrm{int}\ M_{n+1}$. Therefore, the restriction $\varphi|_{M_n \times I}$ is an isotopy through embeddings $M_n \to \textrm{int}\ M_{n+1}$. Since $\varphi_0 = \textrm{id}_M$, the isotopy extension lemma implies that $\varphi|_{M_n \times I}$ extends to a compactly supported diffeotopy $\widetilde{\varphi}=g(x): (\textrm{int}\ M_{n+1})\times I \to \textrm{int}\ M_{n+1}$ with $\widetilde{\varphi}_0=\textrm{id}_{\textrm{int}\ M_{n+1}}$.
To guarantee that $\widetilde{\varphi}=g(x)$ depends continuously on $x\in K$, we now take a detour in which we apply the strong form of the isotopy extension lemma stated earlier. Consider the commutative square
$$ \begin{matrix} 
K & \xrightarrow{\ (a,b)\ } & \textrm{Diff}_c(\textrm{int}\ M_{n+1})\times\textrm{Emb}(M_n,\textrm{int}\ M_{n+1}) \\
\llap{\scriptstyle \textrm{incl}_0}\Big\downarrow  &  & \llap{\scriptstyle (\textrm{eval},\textrm{proj})}\Big\downarrow \\
K\times I & \xrightarrow{\smash{\ (c,d)\ }} & \textrm{Emb}(M_n,\textrm{int}\ M_{n+1})\times\textrm{Emb}(M_n,\textrm{int}\ M_{n+1})
\end{matrix} $$
where the components of the horizontal maps are defined by
$$ a(x) = \textrm{id}_{\textrm{int}\ M_{n+1}} \qquad b(x) = [f_n(x)]_0|_{M_n} = ( \textrm{incl}:M_n\hookrightarrow \textrm{int}\ M_{n+1} ) $$
$$ c(x,t) = [f_n(x)]_t|_{M_n} \qquad d(x,t) = b(x) = ( \textrm{incl}:M_n\hookrightarrow \textrm{int}\ M_{n+1} ) $$
(so only the map $c$ is not constant). Since the vertical map on the right of the square is a fibration, there exists a diagonal lift
$$ (e,d) : K\times I \longrightarrow \textrm{Diff}_c(\textrm{int}\ M_{n+1})\times\textrm{Emb}(M_n,\textrm{int}\ M_{n+1}) $$
and the map $g:K\to F$ (such that $\widetilde{\varphi}=g(x)$) is determined by
$$ \widetilde{\varphi}_t=[g(x)]_t = e(x,t) $$
It is easy to check the required conditions:

$\widetilde{\varphi}=g(x)$ and $\varphi=f_n(x)$ coincide on $M_n \times I$;
$\widetilde{\varphi}_0=[g(x)]_0=\textrm{id}_{\textrm{int}\ M_{n+1}}$.

as follows from the fact that $(e,d)$ is a diagonal lift for the square above.
In conclusion, we have a compactly supported diffeotopy $\widetilde{\varphi} = g(x)$  of $\textrm{int}\ M_{n+1}$ (depending continuously on $x\in K$) which extends $\varphi|_{M_n \times I}$. Extend $\widetilde{\varphi} = g(x)$ to all of $M$ by making it the identity outside $\textrm{int}\ M_{n+1}$. Fundamentally, observe that this extension operation gives a continuous map $\textrm{Diff}_c(\textrm{int}\ M_{n+1}) \to \textrm{Diff}_c(M) \to \textrm{Diff}(M)$ thanks to the strong (or Whitney) $C^1$ topology we placed on the space of compactly supported diffeomorphisms, so we are preserving continuity on $x\in K$. Then $\widetilde{\varphi} = g(x)$ becomes a compactly supported diffeotopy of the whole manifold $M$ depending continuously on $x\in K$. Moreover, $\widetilde{\varphi}_0 = \textrm{id}_M$.
It is now easy to define $f_{n+1} : K \to F$:
$$ [f_{n+1}(x)]_t = (\widetilde{\varphi}_t)^{-1} \circ \varphi_t = ([g(x)]_t)^{-1} \circ [f_n(x)]_t $$
Condition (1) above then follows from the fact that $\widetilde{\varphi}$ extends $\varphi|_{M_n \times I}$. Note that $f_{n+1}$ is continuous since mapping a diffeomorphism to its inverse is continuous in the compact-open $C^1$ topology ($\textrm{Diff}(M)$ is a topological group); alternatively, we can instead use that $g$ is actually continuous with respect to the strong $C^1$ topology on $\textrm{Diff}_c(M)$, and that inversion is also continuous in the strong $C^1$ topology.
We further define the homotopy $H_n : K\times I \to F$ between $f_n$ and $f_{n+1}$ as follows:
$$ [H_n(x,s)]_t = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t = ([g(x)]_{\min\{s,t\}})^{-1} \circ [f_n(x)]_t $$
As was done for $f_{n+1}$, we can show that $H_n$ is continuous. It is also easy to check that $H_n$ takes values in $F$: $[H_n(x,s)]_0 =  (\widetilde{\varphi}_0)^{-1} \circ \varphi_0 = \textrm{id}_M$, and $[H_n(x,s)]_1 =  (\widetilde{\varphi}_s)^{-1} \circ \varphi_1$ is a diffeomorphism of $M$. Moreover, the conditions (2) and (3) above are also straightforward:

Condition (2): Note that $\varphi_t|_{M_{n-1}} = [f_n(x)]_t|_{M_{n-1}} = \textrm{id}_{M_{n-1}}$ (which is condition (1) for $f_n$). Since $\widetilde{\varphi}$ extends $\varphi|_{M_n \times I}$, we also have that $\widetilde{\varphi}_t|_{M_{n-1}} = \textrm{id}_{M_{n-1}}$. We calculate
$$ [H_n(x,s)]_t|_{M_{n-1}} = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t|_{M_{n-1}} = (\widetilde{\varphi}_{\min\{s,t\}})^{-1}|_{M_{n-1}} = \textrm{id}_{M_{n-1}}  $$
Condition (3): Fix $l\in {\mathbb N}$ with $l\geq n$. We know that $\varphi(M_l\times I)\subset \textrm{int}\ M_{l+1}$ (either by condition (3) above if $n > 0$, or by the last condition for the submanifolds $M_i$ if $n=0$). Hence
$$ [H_n(x,s)]_t (M_{l}) = (\widetilde{\varphi}_{\min\{s,t\}})^{-1} \circ \varphi_t (M_l) \subset (\widetilde{\varphi}_{\min\{s,t\}})^{-1} (\textrm{int}\ M_{l+1}) = \textrm{int}\ M_{l+1} $$
where the last equality is a consequence of $\widetilde{\varphi}_u$ being a diffeomorphism which is the identity outside $\textrm{int}\ M_{n+1}$ (and $l\geq n$).

This completes the inductive construction of $H_n : f_n \simeq f_{n+1}$.
To finish the proof, consider the infinite concatenation of all the homotopies $H_n$:
$$ H = H_0 \ast ( H_1 \ast ( H_2 \ast ( H_3 \ast \cdots ) ) ) $$
i.e. $H$ is:

$H_0$ at twice the speed on the interval $[0,\frac 1 2]$;
$H_1$ at four times the speed on the interval $[\frac 1 2,\frac 3 4]$;
$H_2$ at eight times the speed on the interval $[\frac 3 4,\frac 7 8]$;
in general, $H_n$ at $2^{n+1}$ times the speed on the interval $[\frac{2^n-1}{2^n},\frac{2^{n+1}-1}{2^{n+1}}]$.

Finally, we declare $H(-,1)=b$. Then $H$ is the desired homotopy $H : f \simeq b$. The fact that $H$ is continuous at time $1$ is a consequence of:

most importantly, the above condition (2) for the homotopies $H_k$;
the submanifolds $M_i$ form a compact exhaustion of $M$ (their union is $M$, and $M_i \subset \textrm{int}\ M_i$);
together with the fact that we are considering the compact-open $C^1$ topology on the space of embeddings $\textrm{Emb}(M,M)$.

This concludes the proof.<|endoftext|>
TITLE: Learning through guided discovery
QUESTION [25 upvotes]: I have been working through Kenneth P. Bogart's "Combinatorics Through Guided Discovery". You can download it from this page: http://www.math.dartmouth.edu/news-resources/electronic/kpbogart/
I've found that it is a great way to learn and makes me think about the concepts as if I were discovering them. I think that a lot of people will find benefit in working through such a book.
I've looked for books in the same spirit as this one: learning through guided discovery, but my searches haven't been fruitful.
Does anyone know of any such books?

REPLY [3 votes]: Vakil's Rising Sea as a first introduction to modern algebraic geometry!
There is also quite a psychological element in his writing. Often he gives really long hints to his exercises. If one compares such hints with classical texts, then one often finds that he does not give fewer details than classical texts! Indeed, often these hints were actual proofs of theorems in a previous version of his notes which he turned to hints of an exercise.
The psychological effect is the following. While e.g. Hartshorne write "We have x, and thus y", Vakil writes "Convince yourself of x. Show that this implies y.", so in the first instance I would get frustrated as to why I can't understand a seemingly simple proof but reading Vakil I would be proud to have solved another exercise and the step would better stick to my memories.<|endoftext|>
TITLE: Does every regular Noetherian domain have finite Krull dimension?
QUESTION [6 upvotes]: Background: A Noetherian ring is said to be regular if its localizations at all prime (or maximal) ideals are regular local rings. Without this assumption, there are counter-examples.
Thanks.

REPLY [8 votes]: Here's my favorite example (is this the one nosr refers to in his comment? I'm pretty sure it's also due to Nagata).  Let $k$ be a field and $A=k[x_1, x_2, x_3, \ldots]$ be polynomial ring in countably many variables.  Let $P_1 := (x_1)$, $P_2 := (x_2, x_3)$, $P_3 := (x_4, x_5, x_6)$, and in general $P_n$ is generated by the "next" $n$ variables.  That is, $P_n := \left(x_{{n \choose 2} + 1}, x_{{n \choose 2} + 2}, \ldots, x_{{n+1} \choose 2}\right)$.  Let $W := A \setminus \bigcup_{n=1}^\infty P_n$, and let $R := W^{-1}A$.  Then every prime ideal of $R$ is in some $P_nR$, each of which is a maximal ideal of $R$, and $R_{P_n R} \cong k[y_1, \dotsc, y_n]_{(y_1, \dotsc, y_n)}$ is certainly a regular local ring.  Hence $R$ is a regular Noetherian ring.  But as it has essentially polynomial rings of every dimension as localizations, $R$ has infinite dimension.
On the other hand, every Noetherian local ring has finite Krull dimension.<|endoftext|>
TITLE: Geometric picture of scalar curvature
QUESTION [14 upvotes]: In first course differential geometry you learn, that Ricci-curvature is something like a mean-value of the curvature endomorphism, because it's a trace, and the scalar curvature is again a mean-value of the Ricci curvature, again because it's a trace.
I'm now interested, what examples of manifolds can be given, with big Ricci-curvature but small scalar curvature, i.e. how can one describe in a picture what scalar curvature forgets what Ricci curvature still sees.
The same question one can ask for ricci curvature and the curvature endomorphism.

REPLY [3 votes]: One thing the scalar curvature forgets is all the information about the coordinates you were using. The notion of having a "big Ricci curvature" is one that can only be defined in a particular coordinate system. For instance, in coordinates $(x,y)$, $R_{xx}$ and $R_{yy}$ could be big, but in some other set of coordinates $(u,v)$, $R_{uu}$ and $R_{vv}$ could be small or even zero. Because the scalar curvature is a scalar, its value is coordinate-independent.
I assume you had Riemannian spaces in mind, but there are some very well-motivated examples in relativity. For example, when Schwarzschild originally wrote down the metric for the vacuum region surrounding a spherically symmetric body, the metric had a singularity at a certain radius $r>0$, which would be external to the body if the body was very compact. In the coordinates he was using, the singularity was present in the Riemann tensor. However, decades later it was discovered that the singularity could be removed by switching to different coordinates. A hint of this nonphysical character of the singularity was that there was no singularity in any scalar measure of curvature. The scalar curvature $R^{ab}R_{ab}$ was automatically zero because the Einstein field equations require the Ricci tensor to be zero in a vacuum. However, there are other scalar measures of curvature such as Kretschmann invariant $R^{abcd}R_{abcd}$, and these also vanished.
The Kretschmann invariant does blow up at $r=0$ in the Schwarzschild solution, and this is now interpreted as the physical singularity at the center of a black hole.<|endoftext|>
TITLE: Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure?
QUESTION [12 upvotes]: This question arose a few years back when I was an assistant teacher on a course of basic (Lebesgue) measure theory, but I didn't find an answer or anyone able to solve the problem. The setting of the problem is as follows:
We say that $A$ has full outer measure in the unit interval $[0,1]$ if 
$$m^\ast(A) = m^\ast([0,1]).$$
How many disjoint subsets of the unit interval with full outer measure can there exist? By the addivity property of the Lebesgue measure these sets have to be non-measurable.
In this MO post Gerald Edgar shows that by using the Bernstein set you can compose any measurable subset $A$ of $\mathbb{R} ^n$ into two disjoint subsets, both with full outer measure in $A$. This gives a partial answer of "at least two" to my question. 

Question: Can the 'Bernstein set -construction' be modified (or is there some other method) to create $n$ disjoint subsets of $[0,1]$ with full outer measure in the unit interval? 
  Can there exist infinitely many disjoint subsets of the unit interval with full outer measure? If so, then can this family be uncountable?

REPLY [13 votes]: In 1917 Luzin and Sierpinski proved there exists continuum many pairwise disjoint subsets of the interval $[0,1]$ such that each of these subsets has outer Lebesgue measure $1.$
Nikolai N. Luzin and Waclaw Sierpinski, Sur une décomposition d'un intervalle en une infnité non dénombrable d'ensembles non mesurables [On a decomposition of an interval into nondenumerably many nonmeasurable sets], Comptes Rendus Académie des Sciences (Paris) 165 (1917), 422-424.
See Gallica site for C. R. Paris volumes or Internet Archive copy of volume 165
Although I've mentioned this paper in various internet groups several times over the past 10+ years, I don't think I've ever mentioned why I find it so fascinating. First, that such an amazingly strong result was published so early -- little more than a decade after non-measurable sets were known to exist. Second, that such a result is so rarely mentioned in texts that deal with measure theory, despite that fact that very little background is needed to state the result, which can also easily be done in a one-sentence footnote.<|endoftext|>
TITLE: Wedderburn's theorem for $\mathbb{Q}G$
QUESTION [12 upvotes]: Let $G$ be a finite group and let $\mathbb{Q}G=M_{n_1}(D_1)\times\cdots\times M_{n_k}(D_k)$ be the decomposition of $\mathbb{Q}G$ as a product of rings of matrices over divisions rings. Let $Z_i$ be the center of $D_i$. Then each $Z_i$ is an algebraic extension of $\mathbb{Q}$. The question is: how much is it known about these algebraic number fields?. More precisely, for which family of groups $G$ it is known that each $Z_i$ is a Galois extension of $\mathbb{Q}$ and in such a case what can we say about the corresponding Galois groups?.
Any comments and references will be strongly appreciated.

REPLY [2 votes]: The simplest case of your question is the case where $G$ is the cyclic group of order $n$, it is known
that $\mathbb{Q}G\simeq\mathbb{Q}[X]/(X^n-1)$.
As $X^n-1=\Pi_{d|n}\Phi_d(X)$ where $\Phi_d(X)$  is the $d$-th
cyclotomic polynomial, it follows that
$\mathbb{Q}G\simeq\Pi_{d|n}\mathbb{Q}(\zeta_d)$ where $\zeta_d\in\mathbb{C}$ is a
primitive $d$-th root of unity.
Edit: This is also a very particular but concrete case of the idea of
@Aakumadula.<|endoftext|>
TITLE: "geometric" description of the algebra of central functions on a Lie group
QUESTION [5 upvotes]: I am looking for a a description of the algebra of continuous central functions on a group, say a compact simple Lie group $G$, as the algebra of all continuous functions on a "nice" compact Hausdorff space. By central I mean of course constant on the conjugacy classes, i.e.
$f(gxg^{-1}) = f(x)$ for all $x,g\in G$.
By Gelfand's theorem such a compact Hausdorff space exists, but I am looking for a more "geometrical" construction. Something like the following two examples.
$SU(2)$: Any central function on $SU(2)$ is a function of the trace, i.e. the sum of the two eigenvalues. Since the two eigenvalues of a matrix in $SU(2)$ multiply to one, we get the interval $[-2,2]$, i.e. $C_{centr}(SU(2))\cong C([-2,2])$.
$SU(3)$: Central functions on $SU(3)$ are again functions of the trace of a matrix in $SU(3)$. Denote by $D$ the possible values of this trace, these are exactly the complex numbers that can be written as a sum of three complex numbers of modulus one that multiply to one. This set has a nice geometric description: $D$ is the closed domain in $\mathbb{C}$ bounded by the curve $2e^{i\phi}+e^{-2i\phi}$, $0\le \phi\le 2\pi$, which can be obtained by letting a disk of radius one rotate inside a disk of radius three centered at the origin, attaching a pen to a point on the boundary of the small disk and starting from the point $(3,0)$. We get $C_{centr}(SU(3))\cong C(D)$.
One can now try to continue in a similar manner for general $SU(N)$, and look for a "nice" geometrical description of the quotient of
{$\{(\lambda_1,\cdots,\lambda_n)\in \mathbb{T}^n:\lambda_1\cdots\lambda_n=1\}$}
under permutations. 
I expect that this has already been done, but haven't been able to find any references (except for some related work in the case of $SU(3)$). Who can help me?
Comment added: First of all thanks Yemon for the quick answer. I see that I have to explain better what I mean here by "nice geometric" Hausdorff space. Is it possible to find coordinates for $T/W$ such that the characters if the irreps of $G$ are polynomials in these coordinates? Or, equivalently, a compact subset $D$ in $\mathbb{R}^n$ such that $C_{centr}(G)\cong C(D)$ and such that the characters correspond to polynomials?
Note that $T/W$ (or $(D)$) inherits a measure from the Haar measure on $G$ and that characters of inequivalent irreps will be orthogonal w.r.t. this measure. (For $SU(2)$ the measure is the semi-circular distribution and the characters will be mapped to Chebyshev polynomials.)

REPLY [6 votes]: Not an answer as such, but some extended comments with explicit references (posted as community-wiki).   
1) As a non-specialist, I'm unsure whether you are asking for new information or for better references than you have.   Compact Lie groups have been studied thoroughly for more than a century, going back (as Robert Bryant points out) to foundational work by E. Cartan, followed soon by Weyl and others.   By now there are many textbook and lecture treatments of compact groups in the context of more general semisimple or reductive Lie groups, along with a few more targeted accounts of the compact groups by themselves: structure, classification, representations, topology.   So it's difficult to find much new to say, though not impossible.
2) Concerning more recent references, I'd mention the Springer GTM 98 Representations of Compact Lie Groups by Brocker and tom Dieck, which lays out the basic theory with detailed examples.   For instance, they deal explicitly with central functions in their Chapter IV, followed by more on root systems and representations.   Another source is Bourbaki's Chapter IX on compact Lie groups, where the emphasis is often quite analytic and includes a treatment in section 8.3 of Fourier transforms of central functions.   
3) Allen has provided a basic summary in terms of maximal tori and Weyl groups.  The sources I've mentioned provide a lot of details about all of this, while the topology (Stiefel, Bott-Samelson, ... ) has been exposed in lecture notes by Bott.   It's mostly a question of putting your own concerns in clear focus relative to all this literature.   The drawback of Lie theory is the multitude of approaches taken to it, but that also reflects its importance.
4) I'd mention finally that there is a remarkable algebraic parallel for all these questions about class functions in the theory of semisimple algebraic groups.    Although "tori" become algebraic tori, the Weyl groupa and root systems play much the same role.   Here the Chevalley restriction to a maximal torus leads again to study of the Weyl group orbits on a fixed maximal torus.  In this algebraic setting, the resulting geometry is that of an affine algebraic variety, which in the simply connected case is just affine space.   Moreover, the characters of irreducible representations and "characters" of maximal tori come into play in a way completely parallel to the study of compact Lie groups (even without analysis being involved).<|endoftext|>
TITLE: Models of ZFA corresponding exactly with a particular class of groups
QUESTION [6 upvotes]: I recently read [1], in which Blass exhibits a correspondence between:

Permutation models of ZFA in which the axiom of choice (AC) fails but the Boolean prime ideal theorem (BPIT) holds; and
Nontrivial extremely amenable topological groups with small open subgroups.

This correspondence is demonstrated in two theorems (Theorems 1 and 2 below) with the intermediate step of Ramsey filters. But the correspondence is not exact, in a sense I'll explain.

Background
A Ramsey filter on a group $G$ is a filter $\mathcal{F}$ of subgroups of $G$ containing a subgroup $H$ for which each $K \le H$ lying in $\mathcal{F}$ is a Ramsey subgroup of $H$, i.e. for any $2$-colouring $c : H/K \to 2$ and any finite subset $A \subseteq H/K$ there is some $h \in H$ for which $h \cdot A$ is monochromatic (where the action of $H$ on $H/K$ is the natural one).
A topological group has small open subgroups if every neighbourhood of the identity contains an open subgroup. It is extremely amenable if it is Hausdorff and whenever it acts continuously on a compact Hausdorff space, the action has a fixed point.
Theorem 1. If $G$ is a topological group with small open subgroups and $\mathcal{F}$ is the filter of all open subgroups of $G$, then $G$ is nontrivial and extremely amenable if and only if every subgroup in $\mathcal{F}$ is a Ramsey subgroup of $G$ and $\{ 1 \} \not \in \mathcal{F}$
Theorem 2. If $M$ is a model of ZFA induced by a group $G$ and a normal filter $\mathcal{F}$ of subgroups of $G$, in which all atoms are symmetric and every subgroup in $\mathcal{F}$ stabilizes some set $x \in M$, then the Boolean prime ideal theorem holds in $M$ if and only if $\mathcal{F}$ is a Ramsey filter of subgroups of $G$; and moreover the axiom of choice fails in $M$ if and only if $\{ 1 \} \not \in \mathcal{F}$.

Observation
Every subgroup in $\mathcal{F}$ being a Ramsey subgroup of $G$ is a stronger condition on $\mathcal{F}$ than $\mathcal{F}$ simply being a Ramsey filter: it's what you get if you stick $G$ in the place of $H$ in the definition of a Ramsey filter I gave above. Blass observed this fact, and demonstrated that extreme amenability cannot correspond with the Ramseyness of the filter of open subgroups.
But another question can be asked. In some sense the correspondence goes one way: any nontrivial extremely amenable topological groups with small open subgroups gives rise to a model of ZFA+BPIT+(¬AC); but we don't (automatically) have the opposite direction. Instead of weakening the condition on $\mathcal{F}$ in Theorem 1, I'd like to know what happens if we (try to) strengthen the condition on $\mathcal{F}$ in Theorem 2 and thus obtain a (more) exact correspondence.

Question
Is it known if there is some statement $\phi$ strictly intermediate in strength between AC and BPIT which forces, for a model of ZFA+$\phi$+(¬AC) generated by $G$ and $\mathcal{F}$ satisfying the hypothesis of Theorem 2, that $\mathcal{F}$ satisfy the hypothesis of Theorem 1 rather than simply being a Ramsey filter?
A slight weakening of the question: is there a statement $\phi$ as above such that, for any model of ZFA+$\phi$+(¬AC) generated by $G$ and $\mathcal{F}$ satisfying the hypotheses of Theorem 2, there exist $G'$ and $\mathcal{F}'$ which generate the same model and $\mathcal{F}'$ satisfies the hypothesis of Theorem 1.

I hope this isn't a silly question. I've given it some thought but haven't made much progress. Also, I typed this out in quite a hurry, so please forgive any mistakes I might have made.
[1] Andreas Blass, Partitions and Permutation Groups, Contemporary Mathematics (2011), no. 558, pp. 453–466.

REPLY [7 votes]: The answer to the first question is negative, because the permutation model generated by $G$ and $\mathcal F$ doesn't really "see" the group $G$.  Consider the following "shrinking" operation.  Replace $G$ by a subgroup $H$ that belongs to $\mathcal F$ and replace $\mathcal F$ by its restriction to $H$ (i.e., $\{K\in\mathcal F:K\subseteq H\}$).  Then the permutation model doesn't change, but whether $\mathcal F$ satisfies the hypotheses of Theorem 1 can change.  Specifically, if $\mathcal F$ is a Ramsey filter, as witnessed by a subgroup $H$, then using this $H$ in the shrinking construction produces the same permutation model but now with a group and filter satisfying the hypotheses of Theorem 1.  So no property of the permutation model can exactly match the hypotheses of Theorem 1.  (You could think of "Ramsey filter" as the hypothesis of Theorem 1 made invariant under (inverse) shrinking so that it can be a property of the permutation model.)
The same observations give a positive answer to your second question, with $\phi$ being BPIT itself.  The desired $G'$ and $\mathcal F'$ are obtained by suitable shrinking as above.<|endoftext|>
TITLE: Using the Yoneda embedding to talk about exactness in an additive category
QUESTION [16 upvotes]: Suppose I have an additive category $\mathcal{C}$ and a pair of composable arrows:
$$A \longrightarrow B \longrightarrow C.$$
It makes no sense to ask if this sequence is exact at $B$ since the category $\mathcal{C}$ doesn't have kernels or images.  However: the Yoneda embedding produces a sequence of functors:
$$\mbox{Hom}(A,-) \longleftarrow \mbox{Hom}(B,-) \longleftarrow \mbox{Hom}(C,-)$$
and it does make sense to ask if this sequence is exact since the functor category $[\mathcal{C},\mbox{Ab}]$ is abelian.
Similarly we may ask if:
$$\mbox{Hom}(-,A) \longrightarrow \mbox{Hom}(-,B) \longrightarrow \mbox{Hom}(-,C)$$
is an exact sequence.  So here are my questions:

What are criteria for determining if a given composable pair of arrows become exact under one of the two Yoneda embeddings?
What properties of $\mathcal{C}$ guarantee that the covariant and contravariant Yoneda embeddings agree on which sequences become exact?

REPLY [9 votes]: Here is an example: Let $A \to B \to C \to 0$ be a sequence in $\mathcal{C}$. Then it becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ (where $\widehat{\mathcal{C}} := [\mathcal{C}^{\mathrm{op}},\mathsf{Ab}]$) after applying $\mathcal{C} \hookrightarrow \widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}, X \mapsto \hom(X,-)$ if and only if $B \to C$ is a cokernel of $A \to B$ (immediate). On the other hand, it becomes exact in $\widehat{\mathcal{C}}$ after applying $\mathcal{C} \hookrightarrow \widehat{\mathcal{C}}, X \mapsto \hom(-,X)$ if and only if it is isomorphic to a sequence of the form $F \oplus E \xrightarrow{p} C \oplus E \xrightarrow{q} C \to 0$, where $E,F,C \in \mathcal{C}$, $q$ is the projection and $p$ is the projection $F \oplus E \to E$ followed by the inclusion $E \to C \oplus E$ (exercise). Thus we get a split exact complex. Therefore, exactness in $\widehat{\mathcal{C}}$ applies the one in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$, but not vice versa.
Similarily, $B \to C \to 0$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if $B \to C$ is an epimorphism, and it becomes exact in $\widehat{\mathcal{C}}$ if and only if $B \to C$ is a split epimorphism.
By duality, $0 \to A \to B \to C$ becomes exact in $\widehat{\mathcal{C}}$ if and only if $A \to B$ is a kernel of $B \to C$, and it becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if it is a split exact complex.
Combining these results, we see that $0 \to A \to B \to C \to 0$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ if and only if it becomes exact in $\widehat{\mathcal{C}}$ if and only if it is a split exact complex.
Even for abelian $\mathcal{C}$, this is not the "correct" notion of exactness. I doubt that there is anything interesting which can be said about sequences of the form $A \to B \to C$.
Edit: For abelian $\mathcal{C}$ the sequence $A \to B \to C$ becomes exact in $\widehat{\mathcal{C}^{\mathrm{op}}}^{\mathrm{op}}$ iff $A \to B \to C$ is exact and the image of $B \to C$ is a direct summand of $C$. See mathunderflow 360234.<|endoftext|>
TITLE: Automorphism group of the Turing degrees
QUESTION [16 upvotes]: It is conjectured that the automorphism group of the Turing degrees, $Aut(\mathcal{D})$, is trivial. However, to the best of my knowledge, the current state-of-the-art is that $Aut(\mathcal{D})$ is countable. 
My question is twofold: first, is my understanding correct? Is there any countable group $G$ which we know can't be isomorphic to $Aut(\mathcal{D})$?
Second, and more interesting to me: what if anything could we conclude from some group-theoretic information about $Aut(\mathcal{D})$? I.e., what new computability theory would we know if we knew that $Aut(\mathcal{D})$ is abelian, or is simple, or is not finitely generated? (This is admittedly a slightly awkward question, given that we think $ZFC$ proves "$Aut(\mathcal{D})=\lbrace e\rbrace$.")
My reason for asking is that as a rule I am interested in interaction between computability theory and other subjects (for instance, the proof via Higman that there is a universal finitely presented group is dear to my heart), but I am especially interested in examples of other mathematics being applied to computability theory, and I'd be very interested in what the group-theoretic nature of $Aut(\mathcal{D})$ (assuming it's nontrivial) has to say about other parts of computability theory.

REPLY [3 votes]: Let $p_i$ denote the $i$th prime number, and let $\oplus$ be the recursive join on $\omega$. Let $\mathcal O$ be Kleene's $\Pi^1_1$-complete set and $\mathcal O'$ its Turing jump.
For any $B$, let $G_B$ be the direct sum of $\mathbb Z/p_i\mathbb Z$ over all $i\in B\oplus\overline B$. So $G_B$ is a countably infinite abelian group.
I claim that

Aut($\mathcal D$) is not isomorphic to $G_B$ with $B=\mathcal O'$.

I'll show this by showing that Aut($\mathcal D$) has a presentation which is recursive in $\mathcal O$, hence not $\ge_T B$. This will suffice because Richter, in her famous paper,
Richter, Linda Jean, Degrees of structures, J. Symb. Log. 46, 723-731 (1981). ZBL0512.03024.
showed that for all $B$, $G_B$ has isomorphism type of degree $[B]_T$, i.e., all presentations of $G_B$ have degree $\ge_T B$.
Note that if Aut($\mathcal D$) is finite then it is not isomorphic to $G_B$ for any $B$, since the latter is countably infinite. So assume Aut($\mathcal D$) is infinite.
Slaman and Woodin showed that each automorphism $\pi$ of $\mathcal D$ is represented by an arithmetic function in the sense that there is an $n_0$ such that for all $\pi$ and all $X$, $\pi([X]_T)= [P(X)]_T$ where $P(X)=\{e\}({X^{(n_0)}})$.
Now we consider the necessary questions about numbers $e$ in order to present Aut($\mathcal D$).
We let $E=\{e_0<e_1<e_2<\dots\}$ be the set of those $e$ for which $P_e$ given by $X\mapsto \{e\}^{X^{(n_0)}}$ is an arithmetic representation of some automorphism.
We claim that the set $E$ is $\Pi^1_1$:
First, let $F$ be the $\Pi^1_1$ set of all $e$ for which
\begin{equation}
 \forall A(P_e(A)\text{ is total}),
\end{equation}
\begin{equation}
 \forall A\forall B(A\le _T B\to P_e(A)\le_T P_e(B)),\text{ and }
\end{equation}
\begin{equation}
 \forall A\forall B(P(A)\equiv_T P(B)\to A\equiv_T B).
\end{equation}
Then
$$E=\{e: e\in F\text{ and }(\exists d\in F) \forall A(P_d(P_e(A))\equiv_T A\text{ and }P_d(P_e(A))\equiv_T A)\}.$$
The multiplication is given by defining $*$ by
$$
 P_{e_1* e_2} = P_{e_1}\circ P_{e_2}
$$
which is equivalent to
$$\forall A\forall B\forall C(B=P_{e_2}(A)\text{ and }C=P_{e_1}(B)\to C=P_{e_1*e_2}(A))$$
We also have to mod out by equality of the automorphisms induced by $e_1$ and $e_2$, which we check by:
$$\forall A (P_{e_1}(A)\equiv_T P_{e_2}(A))$$
Overall, we get a subset of $\omega$ recursive in the $\Pi^1_1$-complete set Kleene's $\mathcal O$, with an $\mathcal O$-recursive group operation. This is then isomorphic to all of $\omega$ with an $\mathcal O$-recursive group operation, as desired.<|endoftext|>
TITLE: Classification of symtrivial modules over a PID
QUESTION [13 upvotes]: Let us call a module $M$ over a commutative ring $R$ symtrivial if the symmetry $M \otimes M \to M \otimes M, a \otimes b \mapsto b \otimes a$ equals the identity (the same notion applies to arbitrary symmetric monoidal categories). Equivalently, every bilinear map on $M \times M$ is symmetric. Obviously $R$, and more generally every invertible $R$-module, is symtrivial, but there are many more symtrivial modules, which follows from the following closure properties.

A directed colimit of symtrivial modules is symtrivial.
A direct sum $\oplus_{i \in I} M_i$ is symtrivial iff all $M_i$ are symtrivial and $M_i \otimes M_j = 0$ for $i \neq j$.
Every quotient of a symtrivial module is symtrivial.
Every symmetric monoidal functor, possibly non-unital, preserves symtrivial modules. For example, localizations of symtrivial modules are symtrivial.
If $A$ is a commutative $R$-algebra, then $A$, regarded as an $R$-module, is symtrivial iff $R \to A$ is an epimorphism in the category of commutative rings.
If $M_{\mathfrak{p}}$ is symtrivial for all $\mathfrak{p} \in \mathrm{Spec}(R)$, then $M$ is symtrivial.

From Nakayama one can deduce that a finitely generated $R$-module is symtrivial iff all $M_{\mathfrak{p}}$ are cyclic $R_{\mathfrak{p}}$-modules.
Question. Is there a classification of symtrivial $\mathbb{Z}$-modules?
Almost equivalent one may ask for a classification of symtrivial $R$-modules, where $R$ is a PID (or even a Dedekind domain?). There is a classification of epimorphisms of commutative rings $R \to A$, see Torsten Schöneberg's answer here. Either $A=R/rR$ for some $0 \neq r \in R$, or
$A \cong A_{n,\widehat{P}} := (\widehat{P} \setminus P)^{-1} R[(x_p)_{p \in P}] / (x_p (1-p x_p),p^{n(p)} (1-p x_p))_{p \in P},$
where $\widehat{P}$ is a set of primes in $R$, $P \subseteq \widehat{P}$ is a subset and $n : P \to \mathbb{N}^+$ is a function. These are in particular symtrivial $R$-modules.
Every locally cyclic $R$-module (:= every finitely generated submodule is cyclic) is symtrivial; these are precisely the submodules of $Q(R)$ and of $Q(R)/R$. Since $Q(R) \otimes Q(R)/R=0$, also $Q(R) \oplus Q(R)/R$ is symtrivial.
We can continue this way and use the closure properties to optain lots of examples of symtrivial modules. Nevertheless, I wonder if any classification is available (similar to the classification of epimorphisms, where in fact every epi can be optained by a canonical order of closure properties).
PS: I invented the notion of symtrivial objects for myself and also wonder if anybody else has worked with them or if this notion is already known under a different name. Any information is appreciated. The background is that for a given cocomplete symmetric monoidal category $C$ one can show that $\mathsf{gr}_{\mathbb{N}}(C)$ is the universal cocomplete symmetric monoidal category over $C$ together with a symtrivial object (namely $1_C[-1]$). Thus, in the context of graded objects symtrivial objects pop out quite naturally, and my question is equivalent to the classification of cocontinuous symmetric monoidal $R$-linear functors $\mathsf{gr}_{\mathbb{N}}(R) \to \mathsf{Mod}(R)$.

REPLY [11 votes]: Let $R$ be a Dedekind domain with field of fractions $K$. A module $M$ is symtrivial if and only if its maximal torsion-free quotient $F$ is a submodule of $K$ and its torsion submodule $T$ is a sum over each prime $p$ such that $F$ is $p$-divisible of a $p$-divisible $p$-torsion module and a $p$-power cyclic module $R/p^n$.
If $F$ and $T$ are modules as described, then we can classify extnesions $0 \to T \to M \to F \to 0$ by the group $Ext^1(F,T)$. If $F\neq 0$, this group is equal to $\prod_p R/p^n / \sum_p R/p^n$.
Proof: First we prove that such a module is symtrivial. To do this, let $a$ and $b$ be two elements and let $I$ be the submodule of $F$ generated by $a$ and $b$ modulo $T$. $I$ is a finitely generated submodule of $M$, hence it is projective, so we can find an injection $I \to M$. Hence $a$ and $b$ can be written as $x_1+t_1$, $x_2+t_2$ with $x_1,x_2 \in I$, $t_1,t_2 \in T$. We need to check:
a. $x_1 \otimes x_2 = x_2\otimes x_1$ for $x_1,x_2 \in I$.
b. $x \otimes t= t\otimes x$ for $x\in I$, $t\in T$
c. $t_1\otimes t_2 = t_2\otimes t_2$ for $t_1,t_2, \in T$.
Proof of a: $I$ is either $0$ or a fractional ideal. The case $I=0$ is trivial, so we may assume that $I$ is a fractional ideal. $I\otimes I$ is $I^2$, another fractional ideal, and the multiplication is clearly commutative.
Proof of b: $t$ is torsion, so assume $n t =0$ for some $n \in R$. For each prime $p$ dividing $n$, we know that $F$ is $p$-divisible, so $x$ is $n$-divisible up to torsion. In other words, we can write $x=ny+t'$ for $t'$ torsion. Then
$$x\otimes t= (ny+t') \otimes t = n (y\otimes t)+(t'\otimes t)= (t'\otimes t)$$
This reduces this case to case c.
Proof of c: Write $T= C+D$ for $C=\sum_p R/p^n$ and $D$ a divisible torsion module. Since $D$ is divisible, the tensor product of $D$ with any torsion module is $0$. Hence $T\otimes T = (C+D)\otimes (C+D)= C\otimes C$. But $C\otimes C = \sum_p R/p^n \otimes R/p^n = \sum_p R/p^n = C$, where the isomorphism is given by the multiplicative structure of $R/p^n$, which is clearly commutative, so $t_1\otimes t_2 = t_2 \otimes t_1$.

Next we prove that any symtrivial module has this form. Given a symtrivial $M$, we write it as an exact sequence $0 \to T \to M \to F \to 0$, with $T$ torsion and $F$ torsion-free. First we show that $F \subseteq K$. This just follows Todd Trimble's argument from the comments - tensoring with $K$ is symmetric monoidal, so preserves symtriviality, so it takes $M$ to a symtrivial vector space, which must be $0$ or $1$-dimensional. The image of the map $M \to M \otimes_R K$ is just $F$, so $F \subseteq K$.
Next we must show that $T_p=0$ if $F$ is not $p$-divisible and is a sum of a cyclic module and a $p$-divisible $p$-module otherwise. To do this we may consider the localization $M_p$, which fits into an exact sequence $0 \to T_p \to F_p \to M_p$. $M_p$ must be symtrivial. $M_p \subset R_p$. $F_p \subseteq K_p$, so $F_p=0,R_p$, or $K_p$. The middle case is the non-$p$-divisible case, and it clearly forces $T_p=0$, since $R_p$ is projective so the exact sequence splits, and we have $(R_p + T_p) \otimes (R_p + T_p) = R_p+T_p+T_p+ T_p\otimes T_p$, with the two $T_p$s switched by the symmetry, so they must both vanish.
If $F_p=0$, $T_p$ is symtrivial. We will demonstrate that if $F_p=K_p$, $T_p$ is symtrivial still. Suppose not. Find an $a,b$ in $T_p$ such that $a\otimes b \neq b \otimes a$ in $T_p \otimes T_p$, but $a \otimes b = b \otimes a$ in $M_p \otimes M_p$. The proof that $a \otimes b = b\otimes a$ must involve finitely many elements of $R_p$,. Consider the submodule $M'$ of $M_p$ generated by $T_p$ and those finitely many elements. The torsion-free quotient of $M'$ is a finitely generated submodule of $K_p$, thus is $R_p$ or $0$, thus is projective, so the submodule splits into a direct sum of torsion and torsion-free parts. But than $T_p\otimes T_p$ is a direct summand of $M'\otimes M'$, so if $a\otimes b \neq b\otimes a$ in $T_p\otimes T_p$, they do not equal each other in $M'\otimes M'$  - but a complete set of relations implying that they do are relations of $M' \otimes M'$, a contradiction.
So $T_p$ is symtrivial. We must show it is a sum of a $p$-divisible $p$-torsion module and a $p$-power cyclic module. Since tensoring with $R/p$ is a symmetric monodical functor, $T_p/pT_p$ is symtrivial, so $T_p/pT_p = 0$ or $R/p$. In the first case, $T_p$ is a $p$-divisible $p$-torsion module, so we are done. So the $p$-adic completion of $T_p$ is, by Nakayama's Lemma, generated by a single element. The $p$-adic completion has the form $\hat{R}_p/p^n$ for some $n$, and $n$ must be finite because otherwise the completion morphism would be a nontrivial morphism to a $\hat{R}_p$, a torsion-free module, which is impossible. The kernel of $T_p \to \hat{T}_p$ is $p$-divisible because its elements are exactly the elements of $T_p$ that are in $p^n T_p$ for each $n$. Since it is $p$-divisible $p$-torsion, it is injective, so the exact sequence $0 \to  \cap_n p^n T_p \to T_p \to R/p^n \to 0$ splits. (The last map is a surjection because every map to $R/p^n$ that projects nontrivially onto $R/p$ is surjective.)

Finally, we compute the Ext group. Again write $T= C+D$ where $C= \sum_p R/p^n$ adn $D$ a divisible torsion module. $Ext^1(F,T)= Ext^1(F,C)+Ext^1(F,D)$, but $D$ is injective so $Ext^1(F,D)=0$. and we just need to find $Ext^1(F,C)$. We will do so using the following lemma:
Lemma: Let $X$ be a torsion $R$-module that is the sum of finite $p$-torsion modules for different primes $p$, so $X = \sum_p X_p$. Let $Y$ be a torsion-free $R$-module that is $p$-divisible for each $p$ such that $X_p$ is nontrivial. Then
$$Ext^1(Y,X) = Hom(Y \otimes_R K, \prod_p X_p / \sum_p X_p  )$$
Proof: We use the exact sequence $0\to \sum_p X_p \to \prod_p X_p \to \prod_p X_p / \sum_p X_p$. We obtain a long exact sequence, the relevant terms are $Hom(Y, \prod_p X_p) \to Hom(Y,  \prod_p X_p / \sum_p X_p) \to Ext^1(Y,X) \to Ext^(Y,\prod_p X_p)$. We can pull out the product, replacing the first term with $\prod_p Hom( Y,X_P)$ and the second term with $Ext^1(Y, X_p)$. These modules must be $p$-divisible and must be boundedly $p$-torsion, so they vanish, and hence we get an isomorphism $Hom(Y, \prod_p X_p / \sum_p X_p) =Ext^1(Y,X)$. But $\prod_p X_p/ \sum_p X_p$ is divisible, and torsion-free, hence it is a $K$-vector space, so the Homs uniquely factor thoruhg $Y \otimes_R K$. 
In our case setting $X=C$, $Y=F$, $Y \otimes_R K$ is a one-dimensional vector space if $F$ is nontrivial, so the Homs are just $\prod_p R/p^n / \sum_p R/p^n$<|endoftext|>
TITLE: Example of collapsible complex which collapses to a non-collapsible complex 
QUESTION [5 upvotes]: In their paper http://arxiv.org/abs/0907.2954 , Barmak and Minian claim that 
"there exist collapsible complexes which collapse to nontrivial subcomplexes with no free faces"
but unfortunately do not provide a reference.  Where can I find an example of such a complex?  Are there conditions on the complex that will ensure that a collapsible complex will collapse to a point regardless of how you do the collapse? I'm not asking either of these questions "up to barycentric subdivision" but in terms of the actual complex itself.  Thank you.

REPLY [5 votes]: Check this paper by B. Benedetti and F. Lutz, which gives an explicit example (with 8 vertices) of a triangulated 3-ball that collapses to a dunce hat.

EDIT: Let me also give a conceptual proof. Notice that a simplicial complex $K$ with contractible geometric realisation is simple homotopy equivalent to a point (see M.M. Cohen's book on simple homotopy theory), which means we can transform it into a point by a sequence of elementary expansions and collapses. Moreover, all the expansions may be done first. The expanded complex is collapsible. But by reversing the sequence of expansions you can collapse it into the original contractible complex $K$. Now letting $K$ be any contractible, non-collapsible complex (e.g. the dunce hat) gives you the example you looked for.

Complexes that collapse to a point regardless of the chosen collapsing sequence are called extendably collapsible in the cited paper of Benedetti and Lutz. For example, all triangulated 3-balls with less than 8 vertices are extendably collapsible and all collapsible 1- and 2-dimensional simplicial complexes are extendably collapsible.<|endoftext|>
TITLE: When is a Baumslag-Solitar group linear? 
QUESTION [10 upvotes]: The Baumslag-Solitar group $BS(m,n)$ is given by the group presentation 
$BS(m,n)=(a,b|ba^{m}b^{-1}=a^{n})$. When does it embed into a linear group? Thanks!

REPLY [11 votes]: The metabelian groups $BS(n,1)\simeq BS(1,n)=\langle a,b\mid aba^{-1}=b^n \rangle$ are also linear (this seems not mentionened in the Wikipedia article). A faithful, linear representation $BS(1,n)\hookrightarrow GL_2(\mathbb{R})$ is given
by 
$$
a\mapsto \begin{pmatrix} n^{\frac{1}{2}} & 0 \cr 
0 & n^{-\frac{1}{2}} \end{pmatrix}, \quad
b\mapsto \begin{pmatrix} 1 & 1 \cr
0 & 1 \end{pmatrix}.
$$
This representation is not discrete, and the groups are not polycyclic (except for $n=\pm 1$).<|endoftext|>
TITLE: First-order vs second-order provability
QUESTION [6 upvotes]: Modular Arithmetic (MA) has the same axioms as first order Peano Arithmetic (PA) except ∀x(Sx≠0) is replaced with ∃x(Sx=0). Let MA2 be the second-order variation, with second-order induction.  
Answering to a question by Russell Easterly, Emil Jerabek has shown that 
Even XOR Odd Infinities?
∃x(x≠0∧x+x=0) ∨ ∃x(x+x=S0)
is unprovable in MA.  There is, however, a proof in MA2.
There are mathematical examples which distinguish first-order and second-order PA, but they are more esoteric (Paris-Harrington Theorem) or less mathematical (the consistency of first-order PA).  So the result of Jerabek seems IMHO to be of interest, by providing a simple mathematical proposition and system where the second-order system can prove the proposition but not the first-order.
Are there other simple examples of a first-order theory T and an assertion S where T cannot prove S but second-order T, with second-order induction, can prove S?  (Obviously, the interest of Emil's result increases if there are none which aren't "reasonably" equivalent.)

REPLY [7 votes]: I assume that you mean the second-order system with both second-order induction and the full second-order comprehension scheme. There are many "second order variations" of Peano Arithmetic, with different strengths, so care is required to specify which one is intended. The second-order induction axiom on its own does not allow you to prove any new sentences of first-order arithmetic, compared to Peano Arithmetic, because every model of Peano Arithmetic extends to a model of $\mathsf{ACA}_0$, and that extended model satisfies the second-order induction axiom. 
Regardless, there are not going to be any completely elementary principles of number theory that are provable in full second order arithmetic ($Z_2$) but not in PA, because of the well-known phenomenon that all elementary principles are already provable in PA. It is very difficult to find "natural" true mathematical statements that can be expressed in the language of PA but cannot be proved in PA. The Paris--Harrington principle is, in some sense, as good as it gets, which is the main reason the Paris--Harrington theorem is of interest.<|endoftext|>
TITLE: What is the characteristic property of surjective submersions?
QUESTION [11 upvotes]: In Lee's 'Introduction to smooth manifolds' he states that given smooth manifolds X,Y and a surjective submersion f:X→Y, then f is a smoothly final map, that is for any further smooth manifold Z, and any map g:Y→Z, we have g smooth iff g∘f is smooth.
He then says that problem 4.7 shows why this property is 'characteristic'. I can't see why the reverse implication should hold.
Unfortunately, google-books doesn't show that page, nor do I have access to a mathematical library, can some-one enlighten me as to what he means?
One of the answers to this question states a characteristic property, but it doesn't appear on the face of it what Lee has in mind.

REPLY [9 votes]: Here's what I had in mind:
Theorem: Suppose $M$ and $N$ are smooth manifolds and $\pi:M\to N$ is a surjective smooth submersion. Then the given topology and smooth structure on $N$ are the only ones that satisfy the characteristic property.
(That's what Problem 4-7 asks you to prove.)

REPLY [4 votes]: The reverse implication, as it is, is not true, for quite an obvious reason (though I think a local version of it should be true).
Start by any  smoothly final map $f_0:X_0\rightarrow Y$ (e.g. any surjective submersion), and a smooth map $f_1:X_1  \rightarrow Y$ which is not a submersion. Then, the disjoint union $f:=f_0\sqcup f_1: X_0\sqcup X_1 \rightarrow Y$ is not a submersion, nevertheless it is still smoothly final (indeed, for any smooth manifold $Z$ and any map $g:Y\rightarrow Z$, if $g\circ (f_0\sqcup f_1)=(g\circ f_0)\sqcup (g\circ f_1) $ is smooth, so is $g\circ f_0$, hence $g$ because $f_0$ is smoothly final).
It is true that a smoothly final map $f:X\rightarrow Y$ is necessarily surjective (note e.g. that the above construction  $f_0\sqcup f_1$ was surjective). In fact, for any $y\in Y$ there exists a map $g:Y\rightarrow\mathbb{R}$ differentiable in $Y\setminus\{y\}$ and not in $y$ (e.g., a map supported in the domain of a local chart at $y$, that in a local chart is $\|\cdot\|$ near $0$). Then, clearly, if $f:X\rightarrow Y$ is not surjective, say because there is $y\in Y\setminus f(X)$, then $g\circ f$ is smooth though $g$ is not, so $f$ is not smoothly final.<|endoftext|>
TITLE: deformation of stable curve
QUESTION [5 upvotes]: Let $R$ be a DVR, and $k$ residue field of $R$. We suppose that $X_{0}$ is a stable curve over  Spec$k$.
Dose there exist a stable model $X$ over $R$ such that the special fiber isomorphic to $X_{0}$ ?
If we assume $R=C[[t]]$, where C is complex number field, how to find a deformation which make the generic fiber is smooth?

REPLY [6 votes]: The comments above fully answer the OP's question.  I am simply collecting some of these into an answer.
First, the answer to the original question is "no" if one does not impose some additional hypothesis on $R$ such as being complete or Henselian.  As with many similar such questions, one negative answer comes from the Harris-Mumford(-Eisenbud) theorem that  $\overline{M}_g$ is non-uniruled for $g\geq 23$.  If $X_0$ is a stable, genus $g$ curve that is reducible with a single node $p$, say $(X',p') \cup (X'',p'')$ where $p$ is identified with the point $p'$ in the first irreducible component $X'$ and with $p''$ in the second irreducible component $X''$, and if $(X',p')$ and $(X'',p'')$ are sufficiently general pointed curves, then there is no deformation to a smooth curve over the (non-complete, non-Henselian) DVR $R=\mathbb{C}[t]_{\langle t \rangle}$.  If there were, this would give a rational curve in $\overline{M}_g$ that intersects a general point of a boundary divisor.  This would imply that a general point of the "interior" is also contained in a rational curve, contradicting the Harris-Mumford(-Eisenbud) theorem.
On the other hand, if $R$ is complete, or just Henselian, then there does exist a deformation.  It is clear from the comments that the OP is looking for a very explicit formulation of this result.  Here is one such formulation.  Every proper curve is projective, and for stable curves, there is even an explicit tensor power of the dualizing sheaf that is very ample.  Thus, assume that $X_0$ is given as a closed curve in some projective space $\mathbb{P}^n$.  Up to re-embedding by a $2$-uple Veronese embedding (only necessary in positive characteristic), a sufficiently general pencil of hyperplane sections is "Lefschetz".  More precisely, for a sufficiently general codimension $2$ linear subspace $L \subset \mathbb{P}^n$ that is disjoint from $X_0$, for the associated linear projection $$\pi_L:(\mathbb{P}^n\setminus L) \to \mathbb{P}^1,$$
the restriction of $\pi_L$ to $X_0$, 
$$\pi:X_0\to \mathbb{P}^1,$$ has sheaf of relative differentials $\Omega_\pi$ that is the pushforward to $X_0$ of an invertible sheaf from an effective Cartier divisor $D$ of $X_0$ with (a) no two distinct points of $D$ are contained in a common fiber of $\pi$, (b) the length of $D$ at every double point of $X_0$ equals $2$, and (c) for every smooth point of $X_0$ contained in $D$, the length of $D$ equals $1$.  
The branch divisor of $\pi$ is, by definition, $\pi_*D$: an effective Cartier divisor in $\mathbb{P}^1$ that has length $2$ at the image of every double point of $X_0$ and has length $1$ at the image of every other point of $D$.  By the analysis in Stable Maps and Branch Divisors of B. Fantechi and R. Pandharipande (the map $\pi$ is a "stable map"), for every formal deformation of the divisor $\pi_* D$ in $\mathbb{P}^1$, there exists a unique formal deformation of the stable map $(X_0,\pi:X_0\to \mathbb{P}^1)$ to $\mathbb{P}^1$ such that the associated branch divisor of the deformation equals the deformation of the branch divisor.  In particular, choosing a deformation of $\pi_*D$ to a reduced divisor in $\mathbb{P}^1$ gives a formal deformation of $X_0$ to a smooth, stable curve.<|endoftext|>
TITLE: Non-unique 2-factorization of 2k-regular graphs
QUESTION [7 upvotes]: Petersen proved that every 2k-regular graph is 2-factorable.  The factorization is in general not unique.

Given a 2k-regular graph, is it possible to move from any 2-factorization to any other 2-factorization by altering two 2-factors at a time?

If we broaden the question to multigraphs then the answer is no; consider the multigraph on three vertices with two edges between each pair of vertices and a loop on each vertex.  This suggests that the answer is no even for graphs.  In fact a colleague of mine suspects:

The answer is no even if we change "two" to any bounded number

(where by "two" here I mean only the number of factors being altered simultaneously, not the "2" in "2-factor").  However, we don't know how to prove or disprove this.
The motivation for this question comes from a conjecture in representation theory but the connection is rather convoluted so I will omit the details here.

REPLY [5 votes]: No and in fact your multigraph construction is the counterexample. Just replace each edge with an "almost 6-regular" graph, like $K_7$ minus one edge, uv, and connect u and v respectively to the endpoints of the original edge.<|endoftext|>
TITLE: Any map of a contractible complex to itself has a fixed point
QUESTION [6 upvotes]: Reading Lovasz's lecture notes on evasive graph properties, I encountered the following extension of Brouwer's fixed point theorem:
Any continues map from a contractible [finite] simplicial complex to itself has a fixed point.
Lovasz refers this to Lefschetz. Indeed, it seems Lefschetz fixed-point theorem guarantees the existence of a fixed point under some conditions. 
But I could not find a simple proof for Lefschetz theorem, while Brouwer's theorem has an elementary proof using Sperner's Lemma.
My questions are:

How should one interpret the conditions of Lefschetz fixed-point theorem,
and why do they hold for a contractible complex?
(A good reference will be appreciated.)
Is there an easy way to prove the above
statement, for the restricted case
of a collapsible complex, and a
simplicial, bijective map,
without using homology?

REPLY [12 votes]: You have to assume that your complex is finite. Then the Lefschetz Fixed Point Theorem definitely says that $f$ must have a fixed point if the (homologically defined) Lefschetz number of $f$ is not zero. If the complex is contractible, then the Lefschetz number of $f$ must be $1$.
This fact about contractible complexes reduces to the Brouwer Theorem if you use the following fact: any finite contractible complex is a retract of $D^n$ for some $n$. That is, there exist continuous maps $i:X\to D^n$ and $r: D^n\to X$ such that $r\circ i$ is the identity. For continuous $f:X\to X$, any fixed point of $i\circ f\circ r:D^n\to D^n$ gives you a fixed point of $f$.<|endoftext|>
TITLE: Sum of binomial coefficient
QUESTION [5 upvotes]: Given $\epsilon > 0$, what is the minimal $d$, such that $\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{d} \ge \epsilon 2^n$? Is there a clean answer?

REPLY [10 votes]: There's no clean answer, but you need to use the Central Limit Theorem. 
If $X_1,\ldots, X_n$ are independent random variables taking values 0 and 1 with probability $1/2$, then you're asking for a $d$ such that $\mathbb P(S_n\le d)\ge \epsilon$. By the Central limit theorem, for large $n$, you have $S_n$ is well approximated by a normal distribution with mean $n/2$ and variance $n/4$. That is $S_n$ has approximately the same distribution as $n/2+\sqrt n/2 N$ where $N$ is a Normal(0,1) random variable. 
In this language, you are now asking for $d$ such that $\mathbb P(N\le (2d-n)/\sqrt n)\approx\epsilon$, 
or $\Phi((2d-n)/\sqrt n)\approx\epsilon$.
If you solve $\Phi(x)=\epsilon$, then you can obtain a suitable value for $d$ is something like $n/2-\sqrt{-2n\log\epsilon}$.<|endoftext|>
TITLE: Relation between groups and classifying spaces
QUESTION [9 upvotes]: Let $G$ be a nonabelian group, with classifying space $BG$.
Motivation: We can compute its homology, $H_\ast(BG)=H_\ast(G)$. It would be nice to see some equivariant computations, like $H_\ast^G(BG)$ where $G$ acts by conjugation, but I need a particular model of $BG$ to work with. In any case, if $G$ acted freely it would reduce to computing the homology of $BG/G$. Although this action won't be free, I still wonder what $BG/G$ looks like.
Preferred Explicit model: Let $EG$ be the weakly contractible space (constructed simplicially using the elements of $G$), and consider the action of $G\times G$ on $G$ by $(g_1,g_2)\cdot g=g_1gg_2^{-1}$. Then we recover the classifying space $BG$ as $EG/G$ (with $G=G\times\lbrace 1\rbrace$), and we get the space $BG/G$ as $EG/(G\times G)$.
[[Edit]: As pointed out, the outcome will depend on the model. If this isn't a "good" model, then I'll settle for a better one! (Although I would want to understand this model).
So I have this space $BG/G$, dividing out the classifying space by the conjugation $G$-action. Here is where I get some big help: By the Kan-Thurston theorem, there exists a group $K$ such that $BG/G$ and $BK$ have the same homologies.
What can $K$ be?  (Note that if $G$ were abelian then we'd trivially have $K=G$).
Is there a deeper connection between these two spaces?

REPLY [2 votes]: I thought about the case when $G$ is a free group. I believe that in this case $H_2(BG/G)$ is the second exterior power of $G^{ab}$ (which is also $H_2(BG^{ab})$), and $H_3(BG/G)$ is the kernel of
$$
\oplus (C\otimes C)\to G^{ab}\otimes G^{ab},$$
where $C$ ranges over representatives of conjugacy classes of maximal cyclic subgroups of $G$ (the cokernel of this same map is yet another description of $H_2$), and $H_n(BG/G)$ is trivial for all $n>3$.
EDIT  A useful observation is that if $H$ is a subgroup of $G$ then the fixed point set $(BG)^H$ is $B(Z_GH)$, where $Z_GH$ is the centralizer of $H$ in $G$.<|endoftext|>
TITLE: How to solve this one-variable equation in a free group?
QUESTION [7 upvotes]: Let $F_n$ be a free group, $u, v \in F_n$, where $[u,v] \neq 1 $. Trying to show that the following equation does not have a solution in $F_n$ : 
$$ [v,x] = [u,v] .$$
Any ideas are appreciated. I seem to have a proof but it's ugly (combinatorial, cancellation theory etc.) and not short (so, more likely to have a silly mistake :) )
Immediate thing to notice is that if $x=w$ is a solution, then so is the "line" $x=C(v)w$, where $C(v)$ is a centralizer of $v$. 
Also, not difficult to show that there are no solution in a free group of rank two generated by $u$ and $v$. 
thanks!

REPLY [3 votes]: Hi Mark, just wanted to share the solution to this problem as well as a method used. 
I found quite a powerful tool buried at the end of this paper 
"A Classification of Fully Residually Free Groups of Rank Three or Less",
Benjamin Fine, Anthony M Gaglione, Alexei Myasnikov, Gerhard Rosenberger, Dennis Spellman.
Here is the result that follows from the proofs/discussion of that paper and stated in the Section 8, Theorem 6: 
Let $F_{u,v} = \langle u ,v \rangle $ be a free group of rank two and suppose that $a(u,v)$ and $b(u,v)$ are non-trivial and not proper powers in $F_{u,v}$. If $a(u,v)$ and $b(u,v)$ are conjugate in some overgroup $F_{u,v} < F$ then either $a(u,v)$ and $b(u,v)$ are already conjugate in $F_{u,v}$ or the element $ta(u,v)t^{-1}b(u,v)^{-1}$ is a primitive in a free group of rank three $F_{3} = \langle t,u,v \rangle $.
Let me now demonstrate that the equation $[x,u]=[u,v]$ has no solutions in $F$. 
Suppose the above equation has a solution $x=\omega$. We have then $\omega^{-1}u^{-1}\omega =[u,v]u^{-1}$, so that elements $u^{-1}$ and  $[u,v]u^{-1}$ are conjugated in $F$. First, notice that $u^{-1}$ and  $[u,v]u^{-1}$ are neither proper powers nor conjugated in $F_2= \langle u,v \rangle $. 
Hence, by the above result, element $e=t^{-1}u^{-1}tu[v,u]$ must be primitive in $F_3 = \langle t,u,v \rangle$. However element $e$ belongs to the derived subgroup $F_3 '$ and as such can not be primitive.<|endoftext|>
TITLE: When is a solution to an ODE determined by its average value?
QUESTION [5 upvotes]: Fix a smooth vector field $\vec v$ on $\mathbb R^n$.  It is well-known that any trajectory $\vec x : [0,1]\to \mathbb R^n$ solving the ODE $\dot x(t) = \vec v(\vec x(t))$ is determined by its evaluation $\vec x(t)$ for any $t\in [0,1]$.  My question is about when a solution is determined by its average value.
More precisely, let $\phi$ be a smooth function on $\mathbb R$ vanishing identically outside $[0,1]$ and satisfying $1 = \int \phi(t)\ \mathrm{d}t$.  For any $f : [0,1] \to \mathbb R$, let $\langle f\rangle = \int f(t)\ \phi(t)\ \mathrm{d}t$.  My question is:

Under what conditions does the data of the vector field $\vec v$, the averaging function $\phi$, and the value $\langle \vec x\rangle \in \mathbb R^n$ determine the classical trajectory $\vec x$?

Here are two examples.  Fix $\vec a \in \mathbb R^n$ and consider the constant vector field $\vec v(x) = \vec a$.  Then 
$$ \vec x(t) = \vec a t + \langle \vec x \rangle - \vec a \int t\ \phi(t)\ \mathrm{d}t$$
and so the answer is that any $\phi$ works.
 On the other hand, on $\mathbb R^2$ consider vector field $\vec v \bigl( \begin{smallmatrix} x_1 \\ x_2 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} 4\pi\ x_2 \\ -4\pi\ x_1 \end{smallmatrix} \bigr) $.  Then classical trajectories are of the form
$$ x_1(t) = r\cos (4\pi(t - \theta)), \quad x_2(t) = r\sin(4\pi(t-\theta))$$
for fixed $r,\theta$.  Choose a smooth function $\varphi$ which is identically $0$ on $(-\infty,0]$ and identically $1$ on $[1,\infty)$, and set $\phi(t) = 2\varphi(2t) - 2\varphi(2t - 1)$.  If I haven't made an arithmetic error, then for any $r,\theta$ we have $\langle \vec x \rangle = \bigl( \begin{smallmatrix} 0 \\ 0 \end{smallmatrix} \bigr)$.

I could imagine answers of the following forms:

For any vector field $\vec v$, there exists an averaging function $\phi$ that works: take a sufficiently good approximation of a delta-function.
If $\vec v$ is bounded in some appropriate norm (or if the trajectory $\vec x$ is known a priori to stay in a region in which $\vec v$ is bounded), then there is some $\epsilon$ depending on the bound such that any $\phi$ with domain $(0,\epsilon)$ works.  Or perhaps what works is any $\phi$ which is "within $\epsilon$ of a delta-function" in the appropriate sense.
For any $\phi$, the only vector fields that fail to have their solutions determined in the above way have such-and-such property, and hence are few and far between.
...?

REPLY [2 votes]: $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\ve}{\varepsilon}$ $\newcommand{\pa}{\partial}$ $\newcommand{\eL}{\mathscr{L}}$ 
Fix a point $p_0\in \bR^n$ and a neighborhood $U$ of $p_0$ such that  for any  $p\in U$ the solution $x_p(t)$ of the IVP
$$ \dot{x}(t)= V(x(t)),\;\;x(0)=p $$
exists for all $t\in [0,1]$.   Define
$$L_\phi: U\to \bR^n,\;\; L_\phi(p)=\int_0^1 x_p(t) \phi(t) dt. $$
Clearly $L$ is smooth, and if its differential at $p_0$ is injective, then your question has, locally, a positive answer.     If $y\in T_{p_0}\bR^n$ and $ \delta L_\phi$ denotes the differential of $L_\phi$ at $p_0$,  then
$$ \delta L_\phi(y)= \int_0^1 u_y(t) \phi(t) dt, $$
where $u_y(t)$ is the solution of the  linear Cauchy problem
$$ u_y(0)=y,\;\;\frac{d}{dt} u_y(t)= A(t) u_y(t),  $$
and $A(t)$ denotes the differential of $V$ at $x_{p_0}(t)$.   
Thus, if the linear, nonautonomous  version of your question has a  positive  answer, then the nonlinear version  has  a positive answer, locally. 
The linear  case seems quite approachable. I might  have something more to say about this a bit later.
Addendum.  (I'm back from my class.) As Pietro indicated, positivity (monotonicity) conditions on $V$ guarantee a positive answer for any nonnegative $\phi$.    Here is a condition on $\phi$.   
Observe that the above arguments work without change if we replace $\phi(t) dt$ with an arbitrary measure $\mu(dt)$ on $[0,1]$.  Denote by $\mu_0$ the Dirac measure at $0$. Note that if 
$$\phi(t) dt \to \mu_0(dt) $$
then
$$ \delta L_\phi\to \delta L_{\mu_0}. $$
Clearly $\delta L_{\mu_0}$ is  the identity operator thus invertible. In particular if $\phi(t) dt$ is sufficiently close to the  Dirac $\mu_0$ the operator $\delta L_\phi$ will also be invertible.
Thus, for any  smooth vector field $V$ the problem has a positive local answer  if $\phi(t) dt$ is sufficiently close to the Dirac measure.<|endoftext|>
TITLE: Constructible topology on schemes
QUESTION [9 upvotes]: In EGA IV Grothendieck introduced notion of constructible topology. Is it only interesting gadget or can it be use for some practical purposes in algebraic geometry?

REPLY [6 votes]: The constructible topology is used in Bhatt and Scholze - The pro-étale topology for schemes, henceforth [BS], in an essential way (as far as I understand it).
The pro-étale topology is a generalization of the étale topology where $\overline{\mathbb{Q}}_\ell$-étale cohomology is actually the sheaf cohomology (in the naive sense) of the constant sheaf $\overline{\mathbb{Q}}_l$ (rather than $\varprojlim_n H^i(X_{\text{et}}, \mathbb{Z} / \ell^n \mathbb{Z}) \otimes_{\mathbb{Z}_\ell} \mathbb{Q}_\ell$). The point is to enlarge the étale site to include "limits" of coverings, (so for example $\mathrm{Spec}(k^{\text{sep}})$ is always in the pro-étale site of $Spec(k)$, and $\mathrm{Spec}(k^{\text{sep}}) \to \mathrm{Spec}(k)$ is a pro-étale covering).
The Zariski version of this is a little more accessible. In this pro-setup, there is an "initial" "Zariski" covering $X^Z \to X$, which is the "intersection" of all Zariski coverings. The closed points $(X^Z)^c$ of $X^Z$ are in bijection with the points of $X$, and $(X^Z)^c$ equipped with the subspace topology (via the canonical inclusion $(X^Z)^c \subseteq X^Z$) is homeomorphic to $X$ equipped with the constructible topology [BS, Lemma 2.1.10].
The fact that the constructible topology is quasi-compact is used to produce a "contractible" covering of $(X^Z)^c$ in the site of pro-finite topological spaces [BS, Def.2.4.4, Thm.2.4.5, Exa.2.4.6, Exa.4.1.10], and this is one of the main steps in the construction of a "contractible" pro-étale covering of an arbitrary scheme [BS, Lemma 2.4.9, Prop.4.2.8].
The existence of "contractible" coverings is reason the pro-étale topology works [BS, Def.3.2.1, Prop.3.2.3, Def.3.1.1, Lem.3.3.2, Prop.3.3.3, Prop.5.5.4, Prop.5.6.2]. The fact that the constructible topology is quasi-compact is one of the main ingredients for this existence of "contractible" coverings.<|endoftext|>
TITLE: Algebraic characterization of the curvature operator of symmetric spaces
QUESTION [14 upvotes]: My question is the following :

Given an algebraic curvature operator $R\in S^2_B(\Lambda^2\mathbb{R}^n)$, is there an a simple criterion to know if this curvature operator can occur as the curvature operator of symmetric space ?

I would be almost equally happy if someone can point to a way to know if $R$ can be the curvature operator of some Riemannian manifold with reduced holonomy $G\subset SO(n,\mathbb{R})$.
I suspect this might be linked to the fact that $\Lambda^2\mathbb{R}^n=\mathfrak{so}(n,\mathbb{R})$ and how $R$ acts on $\mathfrak{g}\subset\mathfrak{so}(n,\mathbb{R})$. So far, the only link I found is that the image of $R$ has to be contained in $\mathfrak{g}$. I have no idea if this is sufficient. And I wonder if it is possible to write a condition NOT along the line of "There is a Lie subalgebra $\mathfrak{g}\subset\mathfrak{so}(n,\mathbb{R})$ such that..."

REPLY [22 votes]: I suppose that $S^2_B(\Lambda^2(\mathbb{R}^n))$ means the subspace of $S^2(\Lambda^2(\mathbb{R}^n))$ that satisfies the Bianchi identity, i.e., the kernel of the natural map $S^2(\Lambda^2(\mathbb{R}^n))\longrightarrow \Lambda^4(\mathbb{R}^n)$.
Certainly, you'd need that, if ${\frak{g}}\subset \Lambda^2(\mathbb{R}^n)$ is the smallest subspace such that $R$ lies in $S^2({\frak{g}})$ (and this ${\frak{g}}$ is unique and easily computable from $R$), then ${\frak{g}}$ has to be the Lie algebra of a closed, connected subgroup $G\subset \mathrm{SO}(n)$.  
This is not enough, though, as the case $n=3$ shows: The generic $R$ in this case has ${\frak{g}}=\Lambda^2(\mathbb{R}^3)={\frak{so}}(3)$ but isn't the curvature of a symmetric space. 
In addition, one needs that $R$ be invariant under the action of this $G$ on $S^2(\Lambda^2(\mathbb{R}^n))$.  Conversely, if this invariance holds, then $R$ is the curvature of an irreducible $n$-dimensional symmetric space with holonomy $G$.  
The reason is that one then can define a Lie algebra structure on ${\frak{l}} = {\frak{g}}\oplus \mathbb{R}^n$ as follows:  Let the bracket on ${\frak{g}}\subset{\frak{l}}$ be the usual bracket on ${\frak{g}}$; for $a\in{\frak{g}}\subset{\frak{so}}(n)$ and $x\in \mathbb{R}^n$, set $[a,x]=-[x,a]=ax$; and for $x$ and $y$ in $\mathbb{R}^n$, set $[x,y]= R(x{\wedge}y)$.  (Here, we are regarding $R$ as a symmetric mapping $R:\Lambda^2(\mathbb{R}^n)\to \Lambda^2(\mathbb{R}^n)$, knowing that it has image in ${\frak{g}}\subset \Lambda^2(\mathbb{R}^n)$.)  Then the assumption that $R$ lies in $S^2_B(\Lambda^2(\mathbb{R}^n))$ is just that $R(x{\wedge}y)z+R(y{\wedge}z)x+R(z{\wedge}x)y=0$ while the assumption that $R$ is invariant under $G$ implies that $a\ R(x{\wedge}y)-R(x{\wedge}y)\ a = R(ax{\wedge}y + x{\wedge}ay)$, and these are exactly the equations needed to verify that the bracket defined as above on ${\frak{l}}$ satisfies the Jacobi identity.  Then the pair $({\frak{l}},{\frak{g}})$ is a symmetric pair, so that, by Cartan's construction, there is an $n$-dimensional Riemannian symmetric space $M=L/G$ with holonomy $G$ and having $R$ as its curvature operator.
Added remark:  The OP wanted a criterion that didn't explicitly mention a Lie algebra $\frak{g}$, and one can do it this way:  If one just defines $\frak{g}$ to be the image of $R$ in $\Lambda^2(\mathbb{R}^n)$ when one regards $R$ as a symmetric map from $\Lambda^2(\mathbb{R}^n)$ to itself, then the condition that
$$
R(w{\wedge}z)\ R(x{\wedge}y)-R(x{\wedge}y)\ R(w{\wedge}z) 
= R\bigl(R(w{\wedge}z)x{\wedge}y\bigr) + R\bigl(x{\wedge}R(w{\wedge}z)y\bigr)
$$
hold for all $x,y,w,z\in\mathbb{R}^n$ implies both that $\frak{g}$ be closed under
Lie bracket (so that $\frak{g}$ is a Lie algebra and that $R$ be invariant under the action of the connected Lie subgroup $G\subset\mathrm{SO}(n)$ whose Lie algebra is $\frak{g}$.  Thus, the above system of (quadratic) equations on $R$ is exactly the algebraic condition that $R\in S^2_B\bigl(\Lambda^2(\mathbb{R}^n)\bigr)$ be the curvature operator of a symmetric space.  It's also, not surprisingly, the condition that the bracket defined above on $\frak{l}=\frak{g}\oplus\mathbb{R}^n$ satisfy the Jacobi identity.  This may be more along the lines of what the OP had in mind with his question.
Your other question about characterizing curvature operators of Riemannian manifolds with reduced holonomy is not as easy to answer.  You can, of course, tell when $R$ lies in $S^2({\frak{g}})\subset S^2(\Lambda^2(\mathbb{R}^n))$ for a given Lie algebra ${\frak{g}}\subset{\frak{so}}(n)=\Lambda^2(\mathbb{R}^n)$, but then knowing whether there is a metric with holonomy $G$ whose curvature tensor takes the value $R$ at some point is a little tricky.  
In the special case that $\frak{g}$ acts irreducibly on $\mathbb{R}^n$ and can be the holonomy of a Riemannian metric that is not locally symmetric, there is a result (I guess, due to me in several cases) that asserts that, in fact, such an $R$ does occur as the curvature operator at some point of a Riemannian metric with holonomy $G\subset\mathrm{SO}(n)$.  I proved this using Cartan-Kähler theory for the exceptional cases $\mathrm{G}_2\subset\mathrm{SO}(7)$ and $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$ in Metrics with exceptional holonomy, Ann. of Math. (2) 126 (1987).  It is obvious in the cases of $\mathrm{SO}(n)$ itself and $\mathrm{U}(n)\subset\mathrm{SO}(2n)$ and easy in the case of $\mathrm{SU}(n)\subset\mathrm{SO}(2n)$.   I gave proofs (not in all detail, I admit) for the remaining cases $\mathrm{Sp}(n)$ and $\mathrm{Sp}(n)\mathrm{Sp}(1)$ in $\mathrm{SO}(4n)$ in Classical, exceptional, and exotic holonomies: a status report, Actes de la Table Ronde de Géométrie Différentielle (Luminy, 1992), 93–165, Sémin. Congr., 1, Soc. Math. France, Paris, 1996, again using Cartan-Kähler theory.
However, it seems unlikely (though I don't know an explicit counterexample off the top of my head) that you could find a metric such that the curvature operator at every point is equivalent to $R$ up to $G$-conjugacy.  (Probably, this is not possible even for $\mathrm{SU}(2)\subset\mathrm{SO}(4)$, but I'd have to think about it to be sure.)  
Added remark:  I have now checked, and, indeed, there is no $4$-dimensional Riemannian manifold with holonomy $\mathrm{SU}(2)$ such that the curvature operators $R_p$ at all points $p\in M$ are conjugate under $\mathrm{SO(4)}$.  [Note that this condition is a priori weaker than  the condition of being homogeneous; rather it's just the assumption that the germs of the metric at all points agree up to second order, i.e., that the metric is what is sometimes called 'curvature homogeneous'.  (By comparison, there are many Riemannian $3$-manifolds such that all of the eigenvalues of the curvature are constant and so are 'curvature homogeneous', but that have no nontrivial Killing fields.)]<|endoftext|>
TITLE: combinatorial lemma (is it well-known?)
QUESTION [6 upvotes]: The following should be something well?-known, but i haven't seen it anywhere,
neither have i met any references about.
Let $M^{n}$ be a $n$-dimensional oriented closed manifold with a (sufficiently
small) triangulation $\tau$. We "colour" the vertices of $\tau$ with $n+2$
colors: $v^{o}\rightarrow w(v^{o})\in$ {$1,2,...,n+2$ } and we shall say that
the correspondence $w$ is a "coloring" of $\tau$. Take an arbitrary color
$i\in$ {$1,2,...,n+2$ } and consider the $n$-simplices whose vertices are
colored with exactly the colors {$1,2,...,n+2$ }$\backslash\{i\}$. Let
$\Delta^{n}$ be such a simplex and $v_{1},...,v_{n+1}$ be its vertices ordered
according to the positive orientation of $\Delta^{n}$ induced by the
orientation of $M^{n}$. Then we write $\sigma_{i}(\Delta^{n})=1$, if the
permutation $(w(v_{1}),...,w(v_{n+1}))$ is even, and $\sigma_{i}(\Delta
^{n})=-1$ otherwise. Set $\sigma_{i}(\Delta^{n})=0$ if some vertex of
$\Delta^{n}$ is colored $i$, or there are two identically colored vertices.
Let finally
$\sigma_{i}(w)=\sum\sigma_{i}(\Delta^{n})$,
where the sum is over all $n$-simplices.
The Claim: The number $\sigma_{i}(w)$ does not depend on $i$:
 $\sigma_{1}(w)=\sigma_{2}(w)=...=\sigma_{n+2}(w)$. So we have a global invariant
$\sigma(w)$ of the coloring $w$.
This invariant has a geometrical meaning:
Consider the dual cell complex of the triangulation $\tau$, then since each
cell corresponds to a vertex $v^{o}$ of $\tau$, we may color this cell by the
color $w(v^{o})$. Let $F_{i}$ be the union of all cells colored $i$, then we
get a covering $\lambda=\{F_{1},...,F_{n+2}\}$ of $M^{n}$. It is easy to see
that the intersection of all $F_{i}$ is empty, so the canonical map of $M^{n}$
into the nerve of $\lambda$ may be considered as a map of $M^{n}$ into the
$n$-sphere $\mathbb{S}^{n}$:
$\varphi:M^{n}\rightarrow\mathbb{S}^{n}$. Then the degree of $\varphi$ equals
$\sigma(w)$:
$\deg\varphi=\sigma(w)$.
As the proofs are not sophisticated at all and the construction seems
conceptual, maybe it is worth including this material in an elementary
topology textbook. Note also that it gives a method for calculating the degree
without smooth approximation.
Of course, i don't want to repeat well-known things without citation, so any
references are welcome.

REPLY [4 votes]: This is closely related to the Generalized Sperner's Lemma, which holds for all for simplicial manifolds with or without boundary.  See my old survey for a quick introduction (Section 8.1).  Classical references include A.B. Brown and S.S. Cairns, Strengthening of Sperner's lemma applied to homology theory, PNAS, 1960, and D.I.A Cohen, On the Sperner lemma, JCT (1967).  I don't immediately see how your result follows from the lemma, but recall that many extensions and generalizations are known.  I would start with these references and search forward to find your particular version.<|endoftext|>
TITLE: Commutativity of the Chow ring in positive characteristic 
QUESTION [10 upvotes]: I was looking in Ravi Vakil's notes on Intersection Theory, Class 20, where he introduces the bivariant intersection theory, in particular the Chow ring $A^\ast (X)$.
On p. 2, he writes the following which I had no idea about:

Alarming fact: This ring is apparently not known to be commutative in general, because the argument requires resolution of singularities. (It is known to be commutative in characteristic 0, and for smooth things in positive characteristic, and a few more things.) I think it should be possible to show that the ring is commutative in general using technology not available when this theory was first developed, using Johan de Jong’s “alteration theorem” in positive characteristic. If you would like to patch this hole, then come talk to me.

Vakil's notes for this class are from 2004, so plenty could have happened since then. Is this still an open problem?

REPLY [8 votes]: De Jong's theorem proves that the ring is commutative when tensored with $\mathbb Q$. For the torsion part, I am pretty sure that the question is still open.
[Edit:] Mikhail is absolutely right, with Gabber's theorem (http://www.math.u-psud.fr/~illusie/refined_uniformization3.pdf) one can show that the bivariant Chow ring tensored with $\mathbb Z[1/p]$ is commutative in characteristic $p > 0$.<|endoftext|>
TITLE: What is the most extreme set 4 or 5 nontransitive n-sided dice?
QUESTION [9 upvotes]: A set of nontransitive dice is a set of dice whose face numbers are such that the relation "is more likely to roll a higher number than" is not transitive. (See  wikipedia)
For some sets, the deviation from transitivity is small in the sense that A beats B beats C beats A with probabilities $p_{ij}$ only slightly greater than $0.5$ .  Efron's dice (there are 4 of them) beat each other nontransitively with probability $2/3$. 
Can we make a strictly better set of $4$ six-sided dice? That is, a set of 4 six-sided dice such that they beat each other nontransitively with all probabilities $> 2/3$ ?
Can we make a strictly better set of $4$ $n$-sided dice for some small $n$ which one can conveniently make a die out of, e.g. $n = 4, 8, 12, 20 $ ?
Can we make a strictly better set of $5$ $n$-sided dice for some small $n$ which one can conveniently make a die out of, e.g. $n = 4, 6, 8, 12, 20 $ ?
Can we make a strictly better set of $3$, $4$ or $5$ dice, each having a potentially  different number of sides ($4, 6, 8, 12$ or $20$) ?
Ideally I would like to find a fairly small set of fairly easy-to-make, preferably platonic-solid dice which beat each other nontransitively with probabilities > 80%. They would make an excellent teaching aid and magic trick. There is an answer on math.stackexchange which claims that the best you can do with 3 dice is $p = 0.58$, which is disappointingly close to $0.5$; for a teaching aid you need to be able to beat students almost every time for them to spot the pattern quickly. Efron's dice are substantially better at $2/3$, but is that really the best we can do? (Crossposted from math.stackexchange)
EDIT: I missed this answer which argues that the probability cannot be > than 0.75 irrespective of the details of the dice. Still, it would be nice to know what the "simplest" set of "simple" dice is that gets you above, say, 70%, 72%, etc.

REPLY [16 votes]: If you fix only the number of dice, but let the number of faces be arbitrary (or if you simply find a way to make arbitrary probabitities for different faces), then the answer for $n$ faces is $\displaystyle 1-\frac1{4\cos^2(\pi/(n+2))}$. This is proved (that's quite immodest, I know...) in my paper "intransitive roulettes" in Matematicheskoe prosveschenie, III series, 2010, Vol, 14, pp.~240--255, but only in Russian, sorry. Similar results (lacking, perhaps, only the explicit constants) can be found in the following papers:
S. Trybula, On the paradox of $n$ random variables, Zastos. Mat. (Appl. Math.) 8 (1965), 143--154.
Z. Usiskin, Max--min probabilities in the voting paradox, Ann. Math. Statist. 35 (1964), 857--862.
The optimal example is the following. Let $q=\frac1{4\cos^2(\pi/(n+2))}$. Define $r_n=0$, $r_{i}=q/(1-r_{i+1})$. Then one can show that $1-q=r_1>r_2>\dots>r_n=0$. Now let us make the following "dice": $i$th die ($1\leq i\leq n$) makes $i$ with probability $1-r_i$ and makes $i+n$ with probability $r_i$. Then each die wins the (cyclically) previous one with probability $1-q$. 
Consequently, $2/3$ is the optimal number for $n=4$ for any number of faces. Pitifully, the answers for $n>4$ are irrational, hence they are not achievable on regular dice.
It seems that the optimal configuration for $n>4$ on regular dice can be made by the corresponding modification of the general optimal example. E.g., for $n=5$ we have $r_4=q=0.30797\dots$, $r_3=0.445\dots$, $r_2=1-r_3$, $r_1=1-r_4$, so we cannot achieve 70%. On the other hand, these values can be approximated to make the following 5 icosahedral dice:
$$(6\times 1, 14\times 6), \quad
  (9\times 2, 11\times 7), \quad
  (11\times 3, 9\times 8), \quad
  (14\times 4, 6\times 9), \quad
  (20\times 5),
$$
where each wins the (cyclically) next one with probability at most $\frac{9\times 14}{20^2}=0.315$. 
Next, there is a bound for the answer when the number of faces is bounded (or fixed, as in our case). If the number of faces is $2k$ for each die, then consider the $k$th maximal numbers on each die. Consider the die which contains the maximal number among them; then it wins the next one with the probability at least $\frac{k+1}{4k}$. Hence for the icosahedral dice the result cannot exceed $\frac{29}{40}=0.725$. THis can be achieved on the following set:
$$
  (5\times 1,15\times 11), \quad
  (7\times 2,13\times 12), \quad
  (8\times 3,12\times 13), \quad
  (9\times 4,11\times 14), \quad
  (10\times 5,10\times 15),
$$
$$
  (11\times 6,9\times 16), \quad
  (12\times 7,8\times 17), \quad
  (13\times 8,7\times 18), \quad
  (15\times 9,5\times 19), \quad
  (20\times 10),
$$
but not on a smaller one. Some less optimal answers with smaller number of dice are
$$
  (5\times 1,15\times 9), \quad
  (7\times 2,13\times 10), \quad
  (9\times 3,11\times 11), \quad
  (10\times 4,10\times 12),
$$
$$
  (11\times 5,9\times 13), \quad
  (13\times 6,7\times 14), \quad
  (15\times 7,5\times 15), \quad
  (20\times 8)
$$
with losing probability at most $\frac{13\times 9}{20^2}=0.2925$, and
$$
  (5\times 1,15\times 7), \quad
  (8\times 2,12\times 8), \quad
  (10\times 3,10\times 9),
  (12\times 4,8\times 10), \quad
  (15\times 5,5\times 11), \quad
  (20\times 6)
$$
with losing probability at most $\frac{15\times 8}{20^2}=\frac{12\times 10}{20^2}=0.3$ --- exactly 30% on 6 dice.
Finally, for dodecahedral dice the bound is $\frac{17}{24}=0.7083\dots$, hence it is also possible to make it more than 70% (but this is impossible for octahedral dice...). The example is as follows:
$$
  (3\times 1,9\times 9), \quad
  (4\times 2,8\times 10), \quad
  (5\times 3,8\times 11), \quad
  (6\times 4,8\times 12),
$$
$$
  (7\times 5,8\times 13), \quad
  (8\times 6,8\times 14), \quad
  (9\times 7,8\times 15), \quad
  (12\times 8).
$$
I cannot claim that the numbers of dice presented above are optimal for these probabilities, but it seems so.

REPLY [5 votes]: Generalizing Efron, we can get probability $\ge 70\%$ with six 10-sided dice:

10 sides $=6$
3 sides $=11$, 7 sides $=5$
4 sides $=10$, 6 sides $=4$
5 sides $=9$, 5 sides $=3$
6 sides $=8$, 4 sides $=2$
7 sides $=7$, 3 sides $=1$

We can get probability $\ge 72\%$ (actually $.7218934911 = 122/169$) with eight 130-sided dice:

130 sides $=8$
36 sides $=15$, 94 sides $=7$
50 sides $=14$, 80 sides $=6$
58 sides $=13$, 72 sides $=5$
65 sides $=12$, 65 sides $=4$
72 sides $=11$, 58 sides $=3$
80 sides $=10$, 50 sides $=2$
94 sides $=9$, 36 sides $=1$<|endoftext|>
TITLE: Model category structures on dga's in a ringed topos
QUESTION [11 upvotes]: In the introduction to his paper "Towards a non-abelian $p$-adic Hodge theory", Olsson says that for any ringed topos $(\mathcal{T},\mathcal{O})$ with $\mathcal{O}$ a sheaf of $\mathbb{Q}$-algebras, the category $\mathrm{dga}_{\mathcal{O}}$ of $\mathcal{O}$-dga's has a model category structure, where the weak equivalences are quasi-isomorphisms, and the fibrations are surjections with level wise injective kernel (injective as objects in the category of $\mathcal{O}$-modules).
Now, it seems to me the proof of this fact goes something along the following lines. Since the category of $\mathcal{O}$-modules has enough injectives, then the category of positively graded chain complexes has a similarly defined model category structure. One takes generating sets cofibrations and acyclic cofibrations in this model category of chain complexes, and one applies the 'free algebra functor' from complexes to dga's and the small object argument to get generating sets of cofibrations and acyclic cofibrations in the category of dga's. 
My question is the following. 
Do we really need to restrict to $\mathbb{Q}$-algebras here? Or will this argument work for any ringed topos $(\mathcal{T},\mathcal{O})$? For example, will the above definitions of weak equivalences and fibrations define a model category structure on the category of sheaves of $\mathbb{Z}/\ell^n$-modules in the étale topos of some scheme?
I can't see where the argument breaks down, but I may not have understood it well enough.

REPLY [4 votes]: What you suggest should work. We're going to transport the model structure on the category of chain complexes via the adjunction using the criteria given in this answer: Transporting model structures via adjunctions
(A proof of this transport theorem for our case, where adjunction functor is monadic, can be found, for example,  as Lemma 2.3 in this paper of Schwede and Shipley, http://homepages.math.uic.edu/~bshipley/monoidal.pdf)
We have an adjunction between the functors 
$$\text{Free}: \mathbf{Ch}_{\mathcal{O}} \rightarrow \mathcal{O}\text{-}\mathbf{dga}$$ and
$$\text{Forget}: \mathcal{O}\text{-}\mathbf{dga} \rightarrow \mathbf{Ch}_{\mathcal{O}}$$
where the first is left adjoint to the second. Since $\mathcal{O}$-modules form a Grothendieck abelian category, there is a combinatorial model structure on $\mathbf{Ch}_{\mathcal{O}}$ with the fibrations and weak equivalences you described. Every object in both categories is small, since they are presentable.
Now I need to show that everything that can be obtained by sequential limits from cobase changing the arrows $\text{Free}(g)$,where $g$ is a (generating) acyclic cofibration in $\mathbf{Ch}_{\mathcal{O}}$, is a quasi-isomorphism. I think this is true, but I'm not sure so I'll include my argument in case there's something wrong with it.
First I claim that the relative tensor product $\text{Free}(D) \otimes_{\text{Free}(C)} (-)$ is exact on chain complexes. Indeed, given an exact sequence of chain complexes $0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0$ I can form the diagram

All three columns and the top two rows are exact, since free algebras are free as graded modules and the exactness diagrams of chain complexes is determined by exactness as graded objects. Therefore the bottom row is exact (diagram chase or spectral sequence argument.) (The unadorned tensors are over $\mathcal{O}$). 
Now I claim that cobase changing a morphism $\text{Free}(C) \rightarrow \text{Free}(D)$ where $C \rightarrow D$ is a monomorphism that's a quasi-isomorphism, gives a quasi-isomorphism. Indeed, we have a convergent spectral sequence
$$
\text{Tor}_{p,q}^{H^*F(C)} (H^*F(D), H^*A) \Rightarrow H(F(D) \otimes^{\mathbb{L}}A)
$$
But since $F(D)$ is a flat $F(C)$ module this converges to the cohomology of $F(D) \otimes_{F(C)} A$. On the other hand, 
$$
H^*F(C) \rightarrow H^*F(D)
$$
 is an isomorphism so the $E_2$-term collapses to an edge, and moreover the edge is just $H^*A$. The edge homomorphism is then an isomorphism, but the edge homomorphism is precisely the map induced by $A {\rightarrow} F(D) \otimes_{F(C)} A$, whence this map is a quasi-isomorphism, which was to be shown.
Since sequential colimits in the category of algebras are the same as those in the category of chain complexes, we already know that sequential colimits of quasi-isomorphisms are quasi-isomorphisms. The result follows. 
(I didn't use that the arrow $F(C) \rightarrow F(D)$ was monic... so that worries me.)<|endoftext|>
TITLE: Free subgroups of $\mathrm{GL}(2,\mathbb{Z})$
QUESTION [15 upvotes]: Is there a bound $B$ such that every 2-generator subgroup
$G = \langle a, b \rangle \le {\rm GL}(2,\mathbb{Z})$
whose generators do not satisfy a relation of length $\leq B$ is free?
If it exists, such bound must be at least 18, as the example
$$ G = \left<
\left( 
\begin{array}{rr}
 5 & 4 \\\
-1 & -1
\end{array}
\right),
\left(
\begin{array}{rr}
 6 & 1 \\\
-1 & 0
\end{array} \right) \right>
$$
shows: the shortest relation satisfied by the generators $a$ and $b$ is
$a^{-2}b(ab^{-1})^3a^2b^{-1}(a^{-1}b)^3 = 1$.
Remarks:

Obviously the question can be generalized to $m$-generator subgroups of
${\rm GL}(n,\mathbb{Z})$.
The crystallographic restriction gives a positive answer to case $m = 1$
of the above generalization, thus our case $m = n = 2$ is the minimal case
which is not covered.
By the Tits alternative, a subgroup of ${\rm GL}(n,\mathbb{Z})$ has either a free
subgroup or a solvable subgroup of finite index.

REPLY [17 votes]: In Olʹshanskiĭ, A. Yu.; Sapir, M. V. On $F_k$-like groups. (Russian) Algebra Logika 48 (2009), no. 2, 245--257, 284, 286--287; 
translation in Algebra Logic 48 (2009), no. 2, 140–146, 
we proved (Theorem 2) that every non-virtually cyclic hyperbolic group is $F_k$-like, that is for every $m$ it has a generating set consisting of $k$ elements which do not satisfy any relation of length $\le m$. The minimal $k$ that works is 1 plus the number of generators of the group. In particular, for $GL_2(Z)$, $k=3$. 
For the group $\mathbb{Z}*\mathbb{Z}_2$ (as in Yves' comment), $k=2$ also works: one can adapt the proof of Theorem 1 of the paper (we prove, in particular, that  every group with 2-generated presentation satisfying $C'(1/6)$ is $F_2$-like).<|endoftext|>
TITLE: When is $\mathbb{L}$-rank definable in inner models of $\mathbb{V} = \mathbb{L}$?
QUESTION [10 upvotes]: Suppose $\mathbb{V} = \mathbb{L}$ and there is a countable transitive model $\mathbb{M}$ of $ZFC$. 
Let $\rho$ be the $\mathbb{L}$-rank, i.e. for all $a \in \mathbb{V}$, $\rho(a) = $the least $\alpha$ with $a \in L_{\alpha+1}$. 
Define a pre-order $<'$ on $M$ by $a <' b$ iff $\rho(a) < \rho(b)$. 
Then my first question is: under what circumstances is $<'$ first-order definable over $\mathbb{M}$?
My second question is: supposing $(\mathbb{V} \not= \mathbb{L})^\mathbb{M}$, and given $a, b \in M \backslash \mathbb{L}^\mathbb{M}$, when does $\mathbb{M}$ "know" that $a <' b$? 
Formally, when is there a formula $\phi(x, y)$ (without parameters) such that $\mathbb{M} \models \phi(a, b)$ and for all $a', b'$ with $\mathbb{M} \models \phi(a', b')$, we have $a' <' b'$?

REPLY [2 votes]: The following isn't an answer to your question, as it's only one example, but I'm not able to make comments here.
It can be definable over $M$, and quite simple. (However, note that for the model I'm going to mention, things are very different if one considers $L$-order of constructibility as opposed to just $L$-rank as defined in the question.)
Let $\lambda$ be least such that $L_\lambda\models$ ZFC and let $c$ be Cohen generic over $L_\lambda$, with $c\in L_{\lambda+2}$. (This is the least level of $L$ which contains such a Cohen generic; all bounded subsets of $L_\lambda$ which are in $L_{\lambda+1}$, are already in $L_\lambda$. But $L_\lambda$ is pointwise definable, and therefore $L_{\lambda+1}$ projects to $\omega$, and in fact, it is the $\Sigma_1$-hull of the single parameter $\{\lambda\}$ in $L_{\lambda+1}$, and using this, it is easy to define a generic $c$, and in fact, there is one which is $\Sigma_1$-definable from parameter $\{\lambda\}$.)
Now rank the sets in $L_\lambda[c]$ as follows, writing $W_\alpha$ for the sets of rank $<\alpha$. Rank $<\lambda$ is just the $L$-ordering, so $W_\lambda=L_\lambda$. Then $W_{\lambda+1}$ consists
of the (bounded) subsets of $L_\lambda$ which are in $L_\lambda[c]$. More generally, given $W_\alpha$ where $\lambda\leq\alpha<\lambda+\lambda$,
then $W_{\alpha+1}$ is the set of subsets of $W_\alpha$ which are in $L_\lambda[c]$. And take unions at limits.
Then note that $W_{\lambda+\lambda}=L_\lambda[c]$, and $\lambda+\lambda$ is least such (just by rank considerations). Clearly this ranking is definable over $L_\lambda[c]$.
So it suffices to see that this ordering is exactly $L$-rank restricted to $L_\lambda[c]$.
For this, note first that every set in $W_{\lambda+1}$ is definable from parameters over $L_{\lambda+1}$.
(Given $X\in W_{\lambda+1}$, just use a name in $L_\lambda$ for $X$ and the forcing relation and the generic $c$, each of which are definable from params over $L_{\lambda+1}$.)
But no set in $Y\in L_\lambda[c]\backslash W_{\lambda+1}$ is in $L_{\lambda+2}$, because every such $Y$ contains some $X\in W_{\alpha}\backslash L_\lambda$ with $\alpha>\lambda$, for any such $X$, $X\notin L_{\lambda+1}$.
Similarly, the constructibility rank cannot be any quicker than the $W$-rank in general. So it suffices to see that the constructibility rank is quick enough. For this, let $N_\alpha$ be the $L_\lambda$-class of "rank $\alpha$ hereditarily nice names" (so every element of $L_\lambda[c]$ has such a nice name), starting with nice names for  subsets of $L_\lambda$ being those in $N_0$. (So $W_{\lambda+1+\alpha}$ is the set of all $\tau_c$ for $\tau\in N_\alpha$, where here $\tau_c$ denotes the interpretation of $\tau$ using the generic $c$; and the sequence $\left<N_\alpha\right>_{\alpha<\lambda}$ is definable over $L_\lambda$.) One observes that the name evaluation function $\tau\mapsto\tau_c$, with domain $N_\alpha$, is definable over $L_{\lambda+2+\alpha}$, and basically uniformly in $\alpha$. This is straightforward.
In some more detail: The relation of variables $(\tau,x)$ that says "$\tau\in N_0$
and $x\in L_\lambda$ and $x\in\tau_c$" is definable over $L_{\lambda+1}$, and also $W_{\lambda+1}\subseteq L_{\lambda+2}$
and the relation of variables $(\tau,x)$ that says "$\tau\in N_0$
and $\tau_c=x$" is definable over $L_{\lambda+2}$. It follows
that the relation of $(\tau,x)$ saying "$\tau\in N_1$
and $x\in W_{\lambda+1}$ and $x\in\tau_c$", is definable over $L_{\lambda+2}$, so $W_{\lambda+2}\subseteq L_{\lambda+3}$, etc. We get the evaluation functions themselves in finitely many steps later (exactly when of course depends on exactly how one codes ordered pairs etc, but this doesn't really matter), and the definitions are all uniform enough, so we can continue through limit stages.<|endoftext|>
TITLE: Which degree sequences are planar graphical?
QUESTION [16 upvotes]: The Erdős–Gallai theorem characterizes which degree sequences are graphical (i.e. realizable by a simple graph).
There has been some work on which degree sequences are planar graphical (i.e. realizable by a simple planar graph).  See, for example, On Planar Graphical Degree Sequences by Schmeichel and Hakimi (1977).
What is currently known about which degree sequences are planar graphical?

REPLY [10 votes]: It is always difficult to say what is "currently known," but at least around
2008, the paper

"A Characterization of the degree sequences of 2-trees."
  Prosenjit Bose, Vida Dujmovi, Danny Krizanc, Stefan Langerman, Pat Morin, David R. Wood, Stefanie Wuhrer.
  Journal of Graph Theory
  Volume 58, Issue 3, pages 191–209, July 2008.
  (journal link)

gives a pretty definitive summary of the state of the art:

That a sequence of $n$ positive integers is the degree sequence of a tree if and only
  if it sums to $2n − 2$ is a folklore result. Other graph
  families with known degree sequence characterizations
  include split graphs [14, 20], $C_4$-minor free graphs [21],
  unicyclic graphs [1], cacti graphs [15], Halin graphs [2],
  and edge-maximal outerplanar graphs [18]. The most
  investigated class of graphs is that of planar graphs.
  Despite the effort, no characterization of the degree
  sequences of planar graphs is known, even for edge-maximal planar graphs. 
  Partial results are obtained in
  [5, 13, 11, 16].

And, obviously, that paper characterizes 2-trees.<|endoftext|>
TITLE: Courant nodal domain Theorem for sums of eigenfunctions?
QUESTION [7 upvotes]: Courant's nodal domain theorem gives a bound on the number of nodal domains for an eigenfunction of the Laplacian.  Namely, if $M$ is a smooth compact Riemannian manifold, and $f$ is an eigenfunction for the $n$th eigenvalue, then the number of nodal domains is bounded by $n$.
Is there a bound on the number of nodal domains for a sum of eigenfunctions (with different eigenvalues)?
Note that in the case $M$ is a sphere, any linear combination of eigenfunctions (spherical harmonics) is a restriction of a homogeneous polynomial of degree $d$ (which is roughly the square root of the largest eigenvalue) and Harnack's bound says that there is a bound of order $d^2$ (same order as Courant's bound for the top eigenfunction).  Similarly, higher dimensional spheres and flat tori admit bounds with the same order as Courant's that apply for a sum of eigenfunctions.

REPLY [3 votes]: On a Riemann surface $\Sigma$, consider the space $H_\lambda$ spanned by eiigenfunctions corresponding to eigenvalues $\leq \lambda$.   By Weyl's asymptotic formula we know that
$$ \dim H_\lambda  \sim const \lambda $$
as $\lambda \to \infty$.  Denote  by $S_\lambda$ the unit sphere in $H_\lambda$ with respect to the $L^2$-norm.  Equip with with the unique rotationally invariant measure of total volume $1$ so now  you can think of $S_\lambda$ as a probability space. Thus, for any $f\in S_\lambda$, the number $N_f$ of zonal regions of $f$ is a  random   variable.  We denote by $N_\lambda$ its expectation, i.e., the average number of  zonal domains  of a function $f\in S_\lambda$.  One can show  that there exists a constant  $C>0$ such that
$$ N_\lambda \leq C\lambda $$
for $\lambda \gg 0$.     For a proof see this preprint.
I actually believe that 
$$ N_\lambda \sim C\lambda  $$
as $\lambda \to \infty$, but I have no promising idea how to approach this.<|endoftext|>
TITLE: When is a Pseudo-differential operator trace class or in Dixmier ideal?
QUESTION [7 upvotes]: Let's denote the set of all Pseudo-differential operators with symbol of “order” $d$  by $\Psi_d(M)$ and Sobolev space on $M$ by $H_s(M)$. It is known that  
If $P\in\Psi_d(M)$ Then $P$ extends to a continuous map $P:H_{s}(M)\to H_{s-d}(M)$ for all $s$. 
Moreover, since the natural inclusion $H_s\to H_t$, for $s>t$ is compact,  $P:H_{s}(M)\to H_{t}(M)$ is compact operator if $t< s-d$.
See for example Lemma 1.3.4, Gilkey's book Invariance Theory: The Heat Equation and the Atiyah-Singer Index Theorem.
In special case, when $s=0$, $L^2(M)=H_0(M)$, $P:L^2(M)\to L^2(M)$ is continuous if $d\leq 0$. and it is compact if $d<0$.
Now my question is 
when is $P:L^2(M)\to L^2(M)$  trace class? 
and when is it in Dixmier ideal $\mathcal{L}^{1,\infty}(L^2(M))$ or in general $\mathcal{L}^{(p,q)}(L^2(M))$?
Thanks

REPLY [6 votes]: For a comprehensive account of what you are looking for, see the book by Simon Scott ``Traces and determinants of pseudodifferential operators''<|endoftext|>
TITLE: Mirror symmetry for hyperkahler manifold
QUESTION [7 upvotes]: Hi there,
I have some questions about the mirror symmetry of hyperkahler manifold and K3 surface.
The well-known result said: the mirror symmetry for K3 surface is just given by its hyperkahler rotation.
1) In what sense, the rotation gives the mirror map?
2) Does this means:
if we start from $(M,\omega_I,I)$, here the k3 surface $M$ has a special lagrangian fibration structure with respect to $\omega_I$ and $I$, and also a special lagrangian section. Denote its SYZ mirror by $(M^{mirror},J_{mirror})$. Then exist a (fiberwised) diffeomorphism $\phi:  M \rightarrow M^{mirror}$, s.t. $\phi^{*} (J_{mirror}) = K$?
(Here $I,K$ are the standard base of the $S^2$-family of compatible complex structure of the hyperkahler metric on $M$.)
Thanks!

REPLY [14 votes]: Thanks, YangMills, for the references to my papers. I want to elaborate, because I disagree with the statement that mirror symmetry is given by hyperkahler rotation. It may be the case for certain choices of K3, but I think this happens by accident and that it's not a useful principle. Here is how I view mirror symmetry for K3 surfaces. Choose a rank 2 sublattice of the K3 lattice generated by $E$ and $F$ with $E^2=F^2=0, E.F=1$. Consider a K3 surface $X$ with a holomorphic $2$-form
with $E.\Omega\not=0$. We can assume after rescaling $\Omega$ that $E.\Omega=1$, and then
write $\Omega=F+\check B+i\check\omega \mod E$ for some classes $\check B,\check\omega$
in $E^{\perp}/E$. The K3 surface will be equipped also with a Kaehler form $\omega$
and a B-field $B$, which we write as $B+i\omega$. We choose this data in $E^{\perp}/E
\otimes {\mathbb C}$, although the class of $\omega$ is determined in $E^{\perp}$ by its
image in $E^{\perp}/E$ by the fact that $\omega\wedge \Omega$ must be zero. Then the mirror $\check X$ is taken to have holomorphic form
$\check\Omega=F+B+i\omega\mod E$ and complexified Kaehler class $\check B+i\check\omega$.
Note that there is no particular reason to expect this new K3 surface to be a hyperkaehler rotation, as the mirror complex structure depends on $B$, which gives far too many parameters
worth of choices: there is only a two-dimensional family of hyperkaehler rotation of $X$.
Note that we can hyperkaehler rotate $X$ so that special Lagrangians become holomorphic. The new holomorphic form is $\check\omega + i \omega \mod E$. If we multiply
this form by $i$, we get $-\omega +i\check\omega\mod E$. A change of Kaehler form
followed by another hyperkaehler rotation will give the mirror for certain choices
of $B$-field, but note this involves two hyperkaehler rotations with respect to
different metrics.<|endoftext|>
TITLE: What is the difference between a function and a morphism?
QUESTION [9 upvotes]: Say I have a function $f$. Then, if I understand correctly, $f$ can be regarded as a morphism in a suitably chosen category which has the domain of $f$ and the range of $f$ as objects.
So, the other direction. Say I have a morphism $f$ in some category from one object $A$ to another object $B$. Can I not regard $f$ as a function whose domain is $\{A\}$ and range $\{B\}$?
I'm interested in looking for contexts in which the latter move is not appropriate for some reason.

REPLY [6 votes]: There are various natural categories in topology where morphisms are not functions. For instance, you may define for every $n$ the category whose objects are all compact closed $n$-manifolds and whose morphisms are the $(n+1)$-dimensional cobordisms, i.e. compact $(n+1)$-manifolds $W$ with boundary, where each boundary component of $W$ is coloured with a "+" or "-" sign. Of course you cannot interprete a cobordism $W$ from two $n$-manifolds $M$ and $N$ as a function $M\to N$. However, you can naturally define a "composition" of morphisms by gluing the manifolds. The "identity" is just the product manifold $W = M \times [0,1]$.
You can also construct embedded versions of that. For instance: objects are compact 1-submanifolds of $\mathbb R^2$ (i.e. finitely many disjoint circles) and morphisms are proper subsurfaces of $\mathbb R^2 \times [0,1]$. You might decide to see morphisms only up to isotopy fixing their boundary.

REPLY [5 votes]: Another interesting reason why categories cannot be identified always with categories having functions for morphisms is given in this paper, by Peter Freyd in which is proven that there are some categories which aren't concrete: i.e. which don't have any faithful functor from the category in $\mathbf{Set}$ (the category of sets and functions). Having a such functor is necessary condition to have functions as morphisms.

REPLY [3 votes]: A main point is that considering objects and morphism (more generally than sets with some structure and functions that preserve it), is both a clarifying and fruitful abstraction. There are situations in which a category may be represented by a sub-category of the Set category (though it can be more complicated than what you suggested. If you are interested, you may start from the Yoneda lemma). But, in order to understand universal properties and to make categorical constructions it could even be counter-productive, like e.g. insisting in metrization with topological spaces, or coordinates with vector spaces.<|endoftext|>
TITLE: Provability in Second-Order Arithmetic without the Successor Axiom
QUESTION [5 upvotes]: Consider second-order Peano Arithmetic Z2, i.e. the two-sorted first-order theory with induction and comprehension.  Remove the assumption about the totality of the successor relationship (the Successor Axiom), i.e. the assumption that every number is successored by a number. Call this theory FPA.  FPA has as models the standard model (if it exists) and all the initial segments. FPA is "downward": if a natural number exists, it can prove all numbers less than that number exists, but none that are greater.  So it cannot prove the infinity of the primes, but then this assertion isn't even true in all its models, e.g. {0,1,2,3} has two primes, and 2 is a member of the set, so the set of primes is finite in this model.  FPA can, however, prove Bertrand's Postulate.
Are there any simple mathematical examples of assertions true in all models of FPA, provable in Z2, but not provable in FPA?
EDIT:  On François' advice, I am adding here some clarifications which appear in comments.
Full comprehension is used.
Successoring is considered to be a 2-ary relationship, addition and multiplication to be 3-ary relationships.  The usual axioms can be easily restated in these terms.
The logic is supposed to include variable n-ary relationships, for n = 1 but also for n > 1, which can be quantified over and whose existence can be proved using comprehension.  So for instance, FPA is able to define size equivalence in the straightfoward fashion: A ~ B if and only if (there exists R)(R is a 1-1 function from A onto B). (In fact, given this apparatus, addition and multiplication can be defined from successoring, so one doesn't even need axioms about addition and multiplication, although this is a detail which should not affect the question asked.)  
Induction can be considered to be:
(P)(P0 & (n)(m)(Pn & Nn & Sn,m => Pm) => (n)(Nn => Pn)),
where "N" is "is natural number" and "S" is successoring.
There are many ways to assert the infinity of primes.  One way would be to define "a < b" as
(there exists x)(x > 0 & +(a,x,b)) and 
"MP,n" (P has size n) as 
P ~ {x : x < n}.  Then 
(not there exists n)(Nn & M{p : p is prime},n) 
asserts the infinity of primes.  Or one can state in via unboundedness:  (n)(Nn => (there exists p)(p > n and p is prime)).

REPLY [9 votes]: As I already mentioned in another thread for a slightly different theory, it is possible to give a complete description of models of FPA (I mean all models, giving a complete semantics for the many-sorted first-order theory, not just proper second-order models, which abo lists in the question and which I will henceforth call “standard”) in terms of more familiar theories:

Models where successor is total are exactly the models of $Z_2$.
Models where successor is not total. If $M\models I\Delta_0+\Omega_1$ and $0< a\in M$ is such that $M\models{}$ “$2^a$ exists”, we can form the following model $A_{M,a}$: its “first-order” sort consists of the submodel $[0,a)_M$ of $M$ (where successor, addition, and multiplication are considered as relations, not functions), and for every $n$, its “second-order” universe of $n$ary relations consists of $[0,2^{a^n})_M$, where $r< 2^{a^n}$ represents the relation
$$\{\langle u_0,\dots,u_{n-1}\rangle\in[0,a)^n:M\models\mathrm{bit}(r,a^{n-1}u_{n-1}+\dots+au_1+u_0)=1\},$$
where $\mathrm{bit}(x,u)$ is the $u$th bit in the binary representation of $x$. (Note that the existence of $2^a$ implies the existence of $2^{a^n}$ by $\Omega_1$.) Then $A_{M,a}\models\mathrm{FPA}$: the main thing is that the validity of any (second-order) formula in $A_{M,a}$ translates to a formula in $M$ whose all quantifiers are bounded by some $2^{a^n}$, and $I\Delta_0+\Omega_1$ proves bit-comprehension for $\Delta_0$-definable subsets of logarithmically small intervals, which implies full comprehension in $A_{M,a}$.
Conversely, every model $A\models\mathrm{FPA}$ where successor is not total is isomorphic to $A_{M,a}$ for some $M,a$ as above. I will sketch the argument below. FPA proves that $A$ has a largest element, and satisfies full first-order induction; this first-order theory is called $\mathrm{PA^{top}}$, and it is well-known that every its model $A$ can be extended into a model $B$ of $I\Delta_0$ so that $A$ is its submodel of the form $[0,a)$, and the standard powers $\{a^n:n\in\omega\}$ are cofinal in $B$ (unless $a=1$). The construction works as follows: for every $n$, elements of the interval $[0,a^n)$ in $B$ can be represented by $n$tuples of elements of $A$; one can define in $\mathrm{PA^{top}}$ the arithmetic operations on such $n$tuples in such a way that these $[0,a^n)$ form an increasing chain of models whose union is taken as $B$. In our case, we also have the second-order universes of $n$-ary relations, and these can be used to represent exponentially larger numbers: an $n$-ary relation from $A$ (i.e., a subset of $[0,a^n)$) will represent a number below $2^{a^n}$ in binary. In this way, we can extend $B$ into a model $M$ such that $B=\{x\in M:M\models2^x\text{ exists}\}$. Since any bounded formula in $M$ translates into a second-order formula in $A$, $M$ will satisfy $\Delta_0$ induction up to logarithmically small numbers (this is called length induction), which implies $I\Delta_0$. $M\models\Omega_1$ follow from the fact that $\{2^{a^n}:n\in\omega\}$ is cofinal in $M$. By the construction, $A\simeq A_{M,a}$.
(The second part of the argument, viz. a correspondence of “second-order” models of arithmetic with bounded sets to “first-order” models with exponentially larger numbers is known as the RSUV isomorphism.)

This gives a characterization of provability in FPA: for any (second-order) sentence $\phi$, the construction above implicitly gives a first-order formula $\phi^*$ such that 
$\mathrm{FPA}\vdash\phi$ iff $Z_2\vdash\phi$ and $I\Delta_0+\Omega_1\vdash\phi^*$.
Note that $\phi^*$ is a $\Pi^0_1$-sentence; conversely, every $\Pi^0_1$-sentence is equivalent to one of this form. Note that the standard models of FPA with non-total successor are $A_{\mathbb N,n}$ for some $n\in\mathbb N$, hence the question reduces to: find sentence $\phi$ such that $Z_2\vdash\phi$, $\mathbb N\models\phi^*$, but $I\Delta_0+\Omega_1\nvdash\phi^*$.
An example of such a statement is $\mathrm{Con}_Q$ (the formal consistency of Robinson arithmetic), formulated as a $\Pi^0_1$-formula of the form $\forall x\,\theta(x)$, where $\theta(x)$ is a formula whose all quantifiers are bounded to $x$, and atomic formulas are reformulated in such a way that they do not refer to any numbers above $x$. The translation $\phi^*$ is then essentially equivalent to $\forall x\,\theta(|x|)$, where $|x|$ is the length function, that is, the statement that $Q$ has no logarithmically short proofs of contradiction. This is not provable in $I\Delta_0+\Omega_1$. Thus, $\mathrm{Con}_Q$ is not provable in FPA, but it holds in all its standard models, and it is provable in $Z_2$.
Independent $\Pi^0_1$ statements (for weak or strong arithmetic) in the literature are mostly variants of consistency statements. While this is not a precise question, it is a sort of an open problem to find natural combinatorial $\Pi^0_1$ statements independent of particular fragments of arithmetic. Let me mention two principles which are conjectured to be unprovable in $I\Delta_0+\Omega_1$, and therefore would give the wanted example for FPA:

$\Delta_0$-$\mathrm{PHP}$: the pigeonhole principle. In the language of FPA, it is the following schema: for every formula $\phi(u,X,Y)$ (possibly with other parameters not shown),
$$\begin{align}\forall u\,\neg[&\forall X\subseteq[0,u]\,\exists Y\subsetneq[0,u]\,\phi(u,X,Y)\\&{}\land\forall X_0,X_1,Y\subseteq[0,u]\,\neg(\phi(u,X_0,Y)\land\phi(u,X_1,Y))].\end{align}$$
(I.e., $\phi$ does not define an injective (multi-)function from $\mathcal P([0,u])$ into itself minus one set.)
$\mathrm{Count}_2(\Delta_0)$: the counting principle modulo $2$. In the language of FPA, it is the schema
$$\begin{align}\forall u\,\neg[&\forall X\subsetneq[0,u]\,\exists!Y\subsetneq[0,u]\,\phi(u,X,Y)\\&{}\land\forall X,Y\subsetneq[0,u]\,(\phi(u,X,Y)\to X\ne Y\land\phi(u,Y,X))]\end{align}$$
for every formula $\phi(u,X,Y)$. (I.e., $\phi$ does not define a fixpoint-free involution on $\mathcal P([0,u])$ minus one set. In general, the mod $k$ counting principle would state that some canonical class of finite cardinality not divisible by $k$ cannot be partitioned into $k$-element subclasses, but it’s easier to state it just for $k=2$.)<|endoftext|>
TITLE: Question about the Hardy-Littlewood method (quite basic)
QUESTION [7 upvotes]: Hi, I have a question about the Hardy-Littlewood method. 
Writing $R_s(n)$ for the number of ways to write $n$ as a sum of $s$ $k$-th powers and $f(\alpha )$ for the sum $\sum _{m=1}^Ne(\alpha m^k)$, we have 
$R_s(n)=\int _\mathfrak Uf(\alpha )^se(-\alpha n)d\alpha ,$
where $\mathfrak U$ is some unit interval.  The aim is to work out an asymptotic expression for $R_s(n)$ by studying this integral.  We split the domain of integration into the major and minor arcs:
$R_s(n)=\int _\mathfrak Mf(\alpha )^se(-\alpha n)d\alpha +\int _\mathfrak mf(\alpha )^se(-\alpha n)d\alpha $
and aim to work out an asymptotic expressions for the integral over the major arcs whilst making sure the minor arc contributions are "not too big".
Now, this rests on the fact that $f(\alpha )$ gets "big" near a rational number, and moreover gets "more big" the smaller the denominator of the rational number, so that the major arcs, being intervals around rationals with "small" denominators, give the biggest contribution. My problem essentially is, I think, that I don't understand why this should be true; why does $f$ get "big" near rationals, that is.
In Vaughan's book, we approximate $f(\alpha )$ near these rationals through the function $v(\beta )$ (page 14).  I think Lemma 2.7 contains what I don't understand.  For example, does this lemma say that
$f(\alpha )\sim \frac {S(q,a)}{q}v(\alpha -a/q)$, as $n\rightarrow \infty $?
And how exactly does it influence our definition of the major arcs and our choice of parameter $v$ in the definition of the major arcs?  I assume that $v$ is chosen small enough to gain a saving on the estimate $n^{s/k-1}$ in (2.13) on page 16, and that having $q\leq N^v,\alpha \in \mathfrak M(q,a)$ ensures we get an error no larger than $N^{2v}$ in the lemma, but I still feel I'm missing the big picture in the analysis somehow.
Perhaps I don't have a specific problem as such but want to clarify the situation a little bit; perhaps also it is a problem in that I'm simply not quite settled in the fact that $f(\alpha )$ is large at rationals with small denominators.
In any case, I'd appreciate any thoughts/clarifications/things to think about.  Thanks very much.

REPLY [2 votes]: @Freddie Manners gives gives a nice explanation to the above questions, especially the heuristic for (1).
Ben Green has a nice explanation for why $f$ must be big near rationals with small denominator. In particular, see his answer to another question - Does Weyl's Inequality prove equidistribution? - his answer also includes a link to his notes on the Hardy--Littlewood circle method which give a nice technical description of the answer of your question.
Best<|endoftext|>
TITLE: Tensor contraction and Covariant Derivative
QUESTION [19 upvotes]: What is the importance and intuition behind the the contraction operator on tensors (or the trace of a matrix, for that matter)?
In addition, I see that one of the requirements for a covariant derivative (in the context of connections) is to commute with contraction. Why is that a natural requirement (like it would be for, say, the product rule)?

REPLY [3 votes]: This is mainly a rehash of previous answers. In the realm of Riemannian
geometry, we have the Riemannian covariant derivative $\nabla$ and, besides
the contraction of covariant with contravariant tensors, the metric
contraction of tensors of the same type, i.e., using the isomorphism
$g:TM\rightarrow T^{\ast}M$. Let $X,Y,Z$ denote vector fields and $f$ be a
function. The basic contraction is: If $\alpha$ is a $1$-form, then
$\alpha(Y)$ is a function.
We have $\nabla_{X}f=X(f)$, i.e., $\nabla f=df$. Thinking of a contraction as
a product, commuting with contraction (CWC) is like the product rule:
$X(\alpha(Y))=(\nabla_{X}\alpha)(Y)+\alpha(\nabla_{X}Y)$. This defines the
covariant derivative of a $1$-form. Similarly, for a $2$-tensor $\beta$ we
have: $X(\beta(Y,Z))=(\nabla_{X}\beta)(Y,Z)+\beta(\nabla_{X}Y,Z)+\beta
(Y,\nabla_{X}Z)$ by CWC; think of $\beta(Y,Z)$ as $\beta\cdot Y\cdot Z$ and
apply $\nabla_{X}$ to it with the product rule in effect. The compatibility of
the $\nabla$ with $g$ is usually written as: $X(g(Y,Z))=g(\nabla
_{X}Y,Z)+g(Y,\nabla_{X}Z)$, which is equivalent to $\nabla_{X}g=0$.
At a point $x$, the trace, or metric contraction, of a $2$-tensor $\beta$ is
given by the following formula:
$$
\operatorname{Trace}{}_{g}(\beta)=\frac{1}{\omega_{n}}\int_{S^{n-1}}
\beta(V,V)d\sigma(V),
$$
where $S^{n-1}\subset T_{x}M$ is the unit $\left(  n-1\right)  $-sphere,
$n\omega_{n}$ is its volume, and $d\sigma$ is its volume form. Without loss of
generality, we may assume that we are in $\mathbb{R}^{n}$, in which case
$\beta(V,V)=\sum_{i,j=1}^{n}\beta_{ij}V_{i}V_{j}$. The formula follows from
$\int_{S^{n-1}}V_{i}V_{j}d\sigma(V)=0$ for $i\neq j$ and $n\int_{S^{n-1}}
V_{i}^{2}d\sigma(V)=\int_{S^{n-1}}|V|^{2}d\sigma(V)=\omega_{n}$ for each $i$.
In this way we see that the Ricci $2$-tensor $\operatorname{Ric}$ is an
average of the Riemann curvature $4$-tensor $\operatorname{Rm}$, since
$\operatorname{Ric}=\operatorname{tr}_{1,4}\operatorname{Rm}$. More
geometrically, the Ricci curvature of a line $L$ in $T_{x}M$ is the average of
all sectional curvatures of $2$-planes in $T_{x}M$ containing $L$. Similarly,
the scalar curvature function $R$ is the average of all Ricci curvatures of
lines in $T_{x}M$.
Another way the trace enters is: For a family of
invertible square matrices $A(t)$ we have Jacobi's formula: $\frac{d}{dt}\det
A=\det A\operatorname{tr}(A^{-1}\frac{dA}{dt})$ (using Cramer's rule). Since
the Riemannian measure is $d\mu_{g}=\sqrt{\det g_{ij}}dx^{1}\cdots dx^{n}$, if
we vary a metric by $\frac{\partial}{\partial s}g=v$, then its measure varies
by $\frac{\partial}{\partial s}d\mu_{g}=\frac{\operatorname{tr}_{g}v}{2}
d\mu_{g}$.
The covariant derivative takes a degree $r$ tensor $T$ to the degree $r+1$
tensor $\nabla T$. By tracing we have a differential operator that decreases
the degree by $1$: The divergence is $\operatorname{div}T=\operatorname{tr}
_{1,2}\nabla T$. Tracing also allows us to average the Hessian: The (rough)
Laplacian of $T$ is $\Delta T=\operatorname{tr}_{1,2}\nabla^{2}T$. Another
example is: If $v=\mathcal{L}_{X}g$, then $\frac{\operatorname{tr}_{g}v}
{2}=\operatorname{div}X$.
The trace also arises when considering the irreducible decomposition of a
tensor. For example, given a symmetric $2$-tensor $\beta$, we may write
$\beta=(\beta-\frac{1}{n}(\operatorname{tr}_{g}\beta)g)+\frac{1}
{n}(\operatorname{tr}_{g}\beta)g$. Here, the norm of the trace-free part
$|\beta-\frac{1}{n}(\operatorname{tr}_{g}\beta)g|$ is a measure of how far
$\beta$ is from a multiple of $g$. If $\beta=\operatorname{Ric}$ and $n\geq3$,
then its trace-free part is zero iff $g$ is Einstein.<|endoftext|>
TITLE: Definition of area
QUESTION [18 upvotes]: I am looking for an attractive, but rigorous definition of area;
say in Euclidean plane. Probably there is no short definition. It is OK to make it even longer, but can it be built from useful parts in a not boring way? Say in such a way that not all the students will sleep on the lecture? 
Comments. 

The real problem is to prove existence, the uniqueness is easy.
Using integral does not seem to be a good idea.
There is an approach where you write the formula for area and then proving its properties. I do not like it since it moves you to discrete geometry which is completely irrelevant and the ideas used nearly useless anywhere else (so no reason to learn this stuff). [See for example "Geometry: A Metric Approach with Models" by Millman and Parker.]
The method with measuring grid (cutting everything into small squares and counting) looks much better. One can consider this method as an introduction to integral. There is only one technical statement which has to be proved --- if you rotate square then its area does not change. The only problem is that it is not generalizable to say absolute plane or sphere...
One may define the area as a limit of $\varepsilon^2\cdot N_\varepsilon$, where $N_\varepsilon$ denotes the maximal number of points in the figure one distance $>\varepsilon$ from each other. The only hard part is to prove existence of the limit $\varepsilon^2\cdot N_\varepsilon$ for say polygons. (You can exchange the limit to ultralimit --- this way everything works smoothly, but I do not want to sell my soul just to get a def of area...)

REPLY [8 votes]: I don't know how much about areas you want to prove, and  how developed a background the audience is supposed to have,    but here is a definition of area of a bounded planar region. 
Take  such a region  $D$ in $\mathbb{R}^2$,  $\newcommand{\bZ}{\mathbb{Z}}$  $\newcommand{\ve}{{\varepsilon}}$ consider the $\varepsilon$-grid $(\varepsilon\bZ)^2$ and denote by $N_\ve(D)$ the number of $\ve$-pixels that touch $D$. (A pixel is one of the $\ve\times\ve$-tiny squarers of the grid.) We then declare $D$ to be measurable (i.e. to have area) if the limit $\newcommand{\eA}{\mathscr{A}}$
$$ \eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D) $$
exists.   If this is the case, then we  define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$.  
The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann  integrable functions  and  $D(U,L)$  is the region 
$$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace, $$
then   $D(U,L)$ is measurable
$$\eA(D(U,L))=\int_a^b  \bigl(\; U(x)-L(x)\;\bigr) dx. $$
The next thing to prove is a weak form of the inclusion-exclusion principle: if $D_1$, $D_2$   are  measurable   regions   that intersect along the graph of a $C^1$-function, then $D_1\cup D_2$ is measurable  and 
$$\eA(D_1\cup D_2)=\eA(D_1)+\eA(D_2). $$

REPLY [7 votes]: To be honest, I'm no sure if this consrtuction matches the criterion of non-boringness, but it is rather short (I promise you all this takes less than 40 pages ;) ) and works in every dimension. It is related to the answer of Andy Putman, and the subsequent comments. I should maybe have kept on commenting, but there was simply not room enough. 
For my physics classes I introduce measuring the area of bounded open and closed sets (or volume in $\mathbb{R}^n$ ), by considering countable unions of closed rectangles, the edges of all of them being parallel to a given system of cartesian coordinates (let's call them acceptable). The union must be clean, in the sense that the rectangles overlap neatly to form a rectangulation (say). Because any nonempty intersection of two acceptable rectangles is again an acceptable rectangle, any countable union of acceptable rectangles clearly admits a clean sub-rectangulation.The same argument works also pretty well for describing a common sub-rectangulation of two others having the same image.
Define the area first for finite unions, starting by assigning the usual value to the area of a single rectangle (given as axiom), and extending the area functional using the usual additivity for quasi-disjoint (intersect at most along a common edge) union of acceptable rectangles (given as axiom). You retrieve all the usual properties of an area in that case.
Now a bounded countable and clean union $\bigcup_{n\in\mathbb{N}}R_n$ of acceptable rectangles can be assigned its area as the series $\sum_{n\in\mathbb{N}}\mathtt{Area}(R_n)$ (which always converges). As pointed out above, finding a common sub-rectangulation is fairly straightforward, and the limit does not depend on the choosen sub-rectangulation (commutatively summable series).
Since any open set is rectangulable (in the sense that it breaks in a countable clean union of acceptable rectangles), we can measure any open set. Any compact $K$ is included in the interior of an acceptable rectangle $R$ and $R\setminus{K}$ is a bounded rectangulable set $O$. We assign $\mathtt{Area}(R)-\mathtt{Area}(O)$ to the area of $K$.
By adding more natural axioms you can again extend this functional to all bounded elements of the borelian tribe, but that's a classical story.<|endoftext|>
TITLE: Abelian varieties with given endomorphism algebra
QUESTION [6 upvotes]: I am confused by a statement in the very classical paper of A. A. Albert "On the construction of Riemman matrices II", Ann. Math. 1935, Thm 16. If I understand what he saying, the theorem says that given a quaternion algebra $D$ over a totally real field $F$ of degree $t$ over $\mathbb{Q}$, which is not split at any of the infinite places of $F$, there exists an abelian variety (over $\mathbb{C}$) of dimension $n$ with endomorphism ring (tensored with $\mathbb{Q}$) equal to $D$ if and only if $n = 2tr, r>1$. (I changed the notation slightly).
Now, I thought that such things existed even with $r=1$. For example, if $t=1, F = \mathbb{Q}$ one can have an abelian surface ($n=2$) with endomorphism ring $D$. Another place where these things are alluded to (for arbitrary $t$) is a paper of Y. Morita "on potential good reduction of abelian varieties" J. Fac. Sci. Univ. Tokyo, 1975, remark 3.1.
I guess my questions are, first is whether Albert is right, wrong or misunderstood by me. In case Albert is wrong, is there a place where examples with $r=1$ are constructed? Can every such $D$ occur , when $r=1$?

REPLY [8 votes]: Felipe, Albert is right: $r>1$. In particular, there are no complex abelian surfaces, whose endomorphism algebra is a definite quaternion algebra over the rationals. The same is true in any characteristic: see a survey article of Frans Oort entitled  "Endomorphism algebras of abelian varieties" (Alg. Geom. and Commut. Algebra in Honor of M. Nagata (1987, Ed. H. Hijikata et al.), Kinokuniya Cy, Tokyo 1988; Vol. II, pp. 469-502).
On the other hand, for any  quaternion algebra $D$ over the rationals, there exists a simple $2$-dimensional complex torus $T$, whose endomorphism algebra is isomorphic to $D$. ( $D$ is definite if and only if $T$ is not algebraizable.) See my paper with Frans  (Math. Ann. 303 (1995), 11--29).<|endoftext|>
TITLE: p-rank stratification in unitary Shimura variety
QUESTION [7 upvotes]: Let $K$ be a quadratic extension of $\mathbb Q$ and let $p \neq 2$ be a prime that is inert in $K$. Let $X$ be the Shimura variety associated to the unitary group $\operatorname{U}(2,1)$ over $K$ (after a choice of a suitable integral PEL datum). We have an integral model $\mathcal X$ of $X$ defined over $\mathcal O_E \otimes \mathbb Z_p$, where $E$ is the reflex field. Let $A$ be a abelian variety corresponding to a $k$-point of $\mathcal X$, where $k$ is a field $k$ of characteristic $p$. Its $p$-torsion $A[p]$ has rank $6$ and it is equipped with an action of $\mathbb F_{p^2}$ (this is true regardless the assumption on $k$ of course).

Question: What can be said about the $p$-rank of $A$? It must be $0$ or $2$ (since $A[p^\infty]$ is principally polarized), but I do not know whether both these cases really appear (I believe so) or not.

REPLY [4 votes]: Yes, both these cases appear. This follows from the Remark, page 92, of my PhD thesis.
This is in the unpublished chapter III, devoted to the study of the Shimura variety for U(2,1)
(a.k.a Picard modular surface) over $W(k)$, $k$ a field of char $p$, with level structures either spherical or Iwahori at $p$. In this chapter 
there are more precise results about the different type of abelian varieties that 
can appear, and in which dimensions.<|endoftext|>
TITLE: Are all anabelian Galois actions faithful?
QUESTION [9 upvotes]: Let $C/\mathbb Q$ be a smooth projective curve of genus $g\geq 2$ or a smooth affine curve of genus $g \geq 1$. The exact sequence
$1 \to \pi_1^{et}(C \otimes_\mathbb Q \bar{\mathbb Q}) \to \pi_1^{et}(C) \to \operatorname{Gal}(\bar{\mathbb Q}|\mathbb Q) \to 1$
gives a homomorphism from $\operatorname{Gal}(\bar{\mathbb Q}|\mathbb Q)$ to the outer automorphism group of $\pi_1^{et}(C \otimes_\mathbb Q \bar{\mathbb Q})$.
Is this homomorphism always injective?
If one instead takes $C$ to be a curve of genus $0$ with $3$ points removed, it is injective, by Belyi's Theorem.

REPLY [8 votes]: The answer is "yes" I think, even if you replace $\mathbb Q$ with a number field.
In the affine case this is a result of Matsumoto, as pointed out by Felipe Voloch, see
Matsumoto, Makoto 
Galois representations on profinite braid groups on curves. 
J. Reine Angew. Math. 474 (1996), 169–219.
In the proper case this is a more recent result of Hoshi and Mochizuki, see
Hoshi, Yuichiro; Mochizuki, Shinichi 
On the combinatorial anabelian geometry of nodally nondegenerate outer representations. 
Hiroshima Math. J. 41 (2011), no. 3, 275–342.
See also Tamás Szamuely's nice survey:
Heidelberg lectures on fundamental groups, in J. Stix (ed.) The Arithmetic of Fundamental Groups (PIA 2010), Contributions in Mathematical and Computational Sciences, Vol. 2, Springer-Verlag, 2012, 53--73
available here http://www.renyi.hu/~szamuely/pia.pdf<|endoftext|>
TITLE: Is the quantum algebra unique (up to isomorphism) in deformation quantization ?  
QUESTION [10 upvotes]: Consider a Poisson algebra A (i.e. commutative algebra with Poisson bracket).
Let $\hat A$ be a deformation quantization of the algebra A. We know that construction of deformation quantization and more generally "formality isomorphism of Kontsevich" depends on choices of certain data (in Kontsevich approach we can change "propagator" and coordinates on manifold, in Tamarakin's approach we can choose arbitrary associator.)
Question Is it known/true/expected that for different choices of "that datum" we nevertheless obtain isomorphic quantum algebras $\hat A$? 
May be one needs certain restrictions on setup (e.g. only smooth algebras A, only "generic" quantization morphism, whatever...)  to guarantee uniqueness ? 
Kontsevich also mentions that Grothendieck-Teichmuller group should act on the set of all deformation quantizations. 
Is it at least true that two quantizations living in the same orbit of that group 
give isomorphic quantum algebras ?  
Question in formally precise form
Consider a Poisson algebra. Choose two different formality isomorphisms (e.g. with different propagators or associators). 
Define two star-products $\star$ and $\star'$  with the help of these two formality isomorphisms. 
Question Are the algebras defined by these two star-products isomorphic ? 
More strongly - are these star products "equivalent"  ? (See definition of equivalence in Stefan Waldmann's answer below or in Kontsevich paper).

Some comments. 
If our manifold is R^2n  we canonical Poisson bracket { p_i q_j } = delta_ij, then
undoubtly the quantum algebra should be unique and isomorphic to Heisenberg algebra: $[ \hat p_i , \hat q_j ] = delta_{ij}$. But I am not sure that even in this case it is that much obvious - if we change coordinates we can make Poisson bracket arbitrary weird,
it would be non-obvious that we get isomorphism with Heisenberg algebra.
And moreover it might depend on category we are working with (polynomial or smooth functions).
More general example is Lie algebra $g$ - corresponding quantization should be isomorphic to universal enveloping algebra, but again it is not that much obvious. (In Kontsevich paper he devoted some special (small) arguments to prove that his quantum algebra is isomorphic to U(g)).
Concerning choices of different coordinates in classical algebra A - it already states in
Kontsevich paper that obtained algebras will be isomorphic. More strongly star-products will be "equivalent".
See the last formula on the page 3 of his paper. However nothing is said about choices of different propagators 
I had discussed this question with some experts some years ago, but there was no clear answer. 
The motivation to ask partly comes from MO-discussions here: 
Quantization of a classical system (e.g. the case of a billard)

REPLY [3 votes]: Let me give a conjectural answer. 
Point 1. $GRT$ acts non-trivially on poly-vector fields by $L_\infty$-isomorphisms (see T.Willwacher, M. Kontsevich’s graph complex and the Grothendieck-Teichm¨uller Lie algebra, arXiv:1009.1654 for an explicit description of this actions). 
Point 2. I guess that $GRT$ actually acts non-trivially on the set of equivalence classes of Maurer-Cartan elements (I haven't checked this). 
Point 3. composing Kontsevich's formality map with the action of GRT gives a negative answer to your question if Point 2 is correct. . 
Point 4. concerning the symplectic case, the main reason why the classification map doesn't depend on any choice is because the quantization is unique. Let me explain further: the choices involved in the formality isomorphism appear in the local case. By itself, the globalization procedure does not require any additional choice.<|endoftext|>
TITLE: Flat cohomology for finite infinitesimal group scheme  over a perfect field
QUESTION [5 upvotes]: Let $G$ be a finite infinitesimal group scheme (e.g.$\mu_p,\alpha_p) $ over a perfect field $k$, how much is known about $H^1_{fppf}(k,G)$?

REPLY [8 votes]: I think that $H^1(k,G)=1$ for all infinitesimal $G$.
Let us make some preliminary comments on the process of perfection. If $X=Spec(A)$ is an affine $k$-scheme then we can form its perfection $X^{perf}=Spec(A^{perf})$ where $A^{perf}=\varinjlim_{\sigma} A$ is the direct limit of the $\mathbb{N}$-indexed system formed by the Frobenius of $A$. There is a canonical map $X^{perf}\to X$. If $X$ is a $k$-group scheme, then $X^{perf}$ is a $k$-group scheme. If $X$ is a torsor under a $k$-group scheme $G$, then $X^{perf}$ is a torsor under the $k$-group scheme $G^{perf}$.
Let us come back to the question. Let $X$ be a torsor under $G$. Then $X^{perf}$ is a torsor under $G^{perf}$. Since $G$ is infinitesimal we have $G^{perf}=1$. Thus $X^{perf}=Spec(k)$. The map $X^{perf}\to X$ shows that $X$ has a $k$-point, hence is trivial.<|endoftext|>
TITLE: Why are the holomorphic line bundle sections finite dimensional?
QUESTION [16 upvotes]: I'm trying to understand the Borel--Weil theorem  at the moment (not the whole Bott--Borel--Weil theorem as has been asked elsewhere). However, I am having a little difficulty finding a direct proof, so I started to try and reconstruct it for myself. The question I can't seem to find an answer for is why should the space of holomorphic sections should be a finite dimensional space in the first place? Is there any easy way to see why this should be the case? It seems to me that if one can answer this question, then the theorem follows directly from the classification of the finite dimensional reps of $G$.
As I mention in a comment below I am more (but not exclusively!) interested in algebraic ways of proving finite dimensionality.

REPLY [10 votes]: Here is an algebraic version of Margaret's answer that gives an explicit bound on the dimension of the space of section. This reasoning can be also adapted to the case of complex (non-algebraic) manifolds, but without an explicit bound.
Claim. Suppose $X^k\subset \mathbb CP^n$ is a $k$-dimensional projective variety. Let $D$ be the zero divisor of a section of a line bundle $L$ on $X^k$. Then the dimension  of the space of sections of $L$ is at most the binomial coefficient: 
$$\binom{k+ deg(D)\cdot deg(X)}{deg(D)\cdot deg(X)}$$
The proof of this statement uses just Bezout theorem. Indeed from Bezout it follows that 
at a fixed (say smooth) point $x\in X^k$ a section of $L$ can vanish at most to order $deg(D)\cdot deg(X)$ (to see this cut $X^k$ through $x$ by a generic plane $\mathbb CP^{n-k+1}$ and intersect the obtained curve with $D$). Now, the binomial coefficient is the dimension of the space of all $deg(D)\cdot deg(X)$-jets at $x$.<|endoftext|>
TITLE: necessary and sufficient condition for existence of $SU(3)$-structure on 6-manifolds
QUESTION [10 upvotes]: Is there any necessary and sufficient condition for existence of $SU(3)$-structure on 6-manifolds $M$?

REPLY [21 votes]: Yes, it is well-known that a $6$-manifold has an $\mathrm{SU}(3)$-structure if and only if it is orientable and spinnable (i.e., it has a spin structure).  
The necessity of these two conditions is clear, since $\mathrm{SU}(3)$ is both connected and simply-connected.
For the sufficiency, first note that if $M^6$ is orientable and spinnnable (i.e., the first two Stiefel-Whitney classes of its tangent bundle vanish), then, fixing a Riemannian metric and orientation on $M$, one can choose a spin structure and construct the corresponding spinor bundle $\mathbb{S}\to M$.  Since $\mathrm{Spin}(6)\simeq\mathrm{SU}(4)$, this bundle is a complex $4$-plane bundle, which is, of course, a real $8$-plane bundle.  Since $8>6$, there is a nonvanishing section of $\mathbb{S}$ over $M$, and this will reduce the structure group of this bundle from $\mathrm{SU}(4)$ to $\mathrm{SU}(3)$.   This will, correspondingly, reduce the structure group of the tangent bundle to $\mathrm{SU}(3)$ (since this subgroup does not contain $-I_4\in\mathrm{SU}(4)$).  Thus, $M$ carries an $\mathrm{SU}(3)$-reduction of the structure group of the tangent bundle.
NB: This is a classical argument, but I don't know who first wrote it down.  Maybe, Alfred Gray did it first, but I am not sure, and I will not have any access to reference books for the next 12 days while I am traveling, so I won't be able to check or give you further information.<|endoftext|>
TITLE: The Lang isogeny
QUESTION [6 upvotes]: Let $G$ be a connected commutative algebraic group over $\mathbb{F}_q$. If $\text{Fr}_q : G \to G$ denotes the $q$-Frobenius morphism, we define the Lang isogeny $L_q$ to be the endomorphism of $G$ given by $g \mapsto \text{Fr}_q(g)g^{-1}$.
I have two questions about this important map.

It is not too hard to see that $L_q$ is etale by computing its differential, but why is $L_q$ finite?
Granted that $L_q$ is a finite Galois covering with group $G(\mathbb{F}_q)$, we get a surjection $\pi_1(G,\overline{1}) \to G(\mathbb{F}_q)$. What can we say about the kernel of this homomorphism?

Is there a good modern reference for basic results about $L_q$? If not, would someone kindly explain these two points to me?

REPLY [6 votes]: Every etale morphism is finite over some nonempty open set. (For instance, locally somewhere write it as a standard etale morphism $(A[t]/f(t))_{g(t)}$, then consider the open set where the norm of $g$, that is, the resultant of $f$ and $g$, is nonzero.) If a group homomorphism $H \to G$ is finite over some nonempty open set, it is finite over all translates of that set, so it is finite everywhere.
It is isomorphic to $\pi_1(G,\overline{1})$. In general, if $X \to Y$ is finite etale and Galois, the kernel of the map $\pi_1(Y) \to Gal(X/Y)$ is $\pi_1(X)$.

REPLY [4 votes]: Your questions are expressed in modern language, but keep in mind that Lang's original paper in Amer. J. Math. (1956) used the language of Rosenlicht and Weil.  So it's a challenge to translate the ideas.  A slightly more contemporary proof of Lang's theorem, incorporating Steinberg's useful generalization, is given by Springer in his textbook Linear Algebraic Groups (4.4.17).   In reality not very much is involved in the proof, which relies as you note on computing the differential of the Lang map.  What you see is that the map is "finite" in the older sense, via the differential criterion.   I'm not sure yet how to read your second question, but this too should be easy to answer in that older language. 
I should add that your question about commutative algebraic groups is more special than the setting of Lang's original theorem, where he considered arbitrary connected algebraic groups.   Apart from the language issue, it's worth looking at the original paper (available online from JSTOR), where a number of interrelated results on algebraic groups over finite fields are first worked out.<|endoftext|>
TITLE: DG-projective vs. K-projective complexes
QUESTION [6 upvotes]: Hello!
I'm a student learning the basics of working with the unbounded derived category $D(\mathcal{A})$. I arrived at the natural question, "is every K-projective complex formed out of projective objects?" (having learned that this isn't sufficient to guarantee K-projectivity), and a literature hunt led to the introduction of the paper "The Direct Limit Closure of Perfect Complexes" by Christensen and Holm (http://arxiv.org/pdf/1301.0731.pdf), where the authors mention K-projective and formed out of projective objects is equivalent to DG-projective (in the first paragraph).
I think I managed to prove this, but I still have two questions:
i. I gather that this last statement can be found in Avramov-Foxby-Halperin's Differential Graded Homological Algebra, but I couldn't locate a copy of this on the web. Does anyone know where I could obtain a copy of this (preferably digital)?
ii. With the result, it seems like there should be complexes which are K-projective but not DG-projective, but I have been unable to build one. Is there a reference where one is built/does anyone know of such an example (my hunch is that Avramov-Foxby-Halperin should be said reference)?
Many thanks!

REPLY [2 votes]: In response to your search for the paper by Avramov Foxby and Halperin, as of 9.01.2003 it is still a preprint entitled "differential graded homological algebra", and the second page reads:
FRIENDLY REMINDER
What follows is a partially proofread and completely preliminary version of a
paper, which has been in preparation for a number of years.
Any user of these notes is expected to adhere to the following conditions:
• No further circulation will be made without an author’s consent.
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I would recommend finding a copy however as it does have a lot of information on the kind of issue you asked about.  I'm happy to say a bit more on who you could (respectuflly!) ask for a copy, if you email me at smp12tbs@sheffield.ac.uk
Cheers,
Tom<|endoftext|>
TITLE: Line bundles and vector bundles on $\mathbb P^1 \times \mathbb P^1$
QUESTION [5 upvotes]: Is there a classification of vector bundles on $\mathbb P^1 \times \mathbb P^1$?  I know that the analogue of Grothendieck's splitting theorem is not true for $\mathbb P^1 \times \mathbb P^1$.
Is it true that any line bundle on $\mathbb P^1 \times \mathbb P^1$ is of the form $\mathcal O(m,n) = p_1^*(\mathcal O(m)) \otimes p_2^*(\mathcal O(n))$, where $p_1$ and $p_2$ denote the projections on the two factors?

REPLY [7 votes]: As Will Sawin remarks in the comments, the answer to your question (2) is "yes," either via the exponential exact sequence or by an easy application of the "theorem of the square," (which works e.g. over a general field).
The answer to your question (1) is rather more interesting--while there is no classification of vector bundles on $\mathbb{P}^1\times \mathbb{P}^1$ (that I know of), there is a splitting theorem analogous to Grothendieck's result.  First, it's worth noting that there is a result of Horrocks generalizing Grothendieck's theorem to projective spaces of arbitrary dimension:

Theorem.  (Horrocks)  Let $E$ be a vector bundle on $\mathbb{P}^n$.  Then $E$ is a direct sum of line bundles if and only if $H^i(\mathbb{P}^n, E(r))=0$ for $0<i<n$ and all $r$.

Note that if $n=1$, this is precisely Grothendieck's splitting theorem--a good place to learn about this result and its cousins is Vector Bundles on Complex Projective Spaces by Okonek et al.  It is this theorem which generalizes to $\mathbb{P}^1\times \mathbb{P}^1$.  
Indeed, a literally identical theorem to Horrocks' result above works for a quadric surface $\mathbb{P}^1\times \mathbb{P}^1$ in $\mathbb{P}^3$ (a vector bundle splits if and only if the middle cohomology of all of its twists $E(n)$ vanishes).  One reference is this paper by Buchweitz, Greuel, and Schreyer; the relevant comments are Conjecture B and Remark 2 on page 169.  Conjecture B gives a (conjectural) generalization to smooth hypersurfaces; I don't know too much about this subject, so I have no idea what the status of this conjecture is. 
Of course, this is far from a classification of vector bundles on $\mathbb{P}^1\times \mathbb{P}^1$--that said, I think that such a classification (at anywhere near the level of completeness of the classification on $\mathbb{P}^1$) is well beyond current technology.  Vector bundles on $\mathbb{P}^2$ are already quite interesting and complicated.<|endoftext|>
TITLE: Probability density that minimizes the sample range
QUESTION [7 upvotes]: Let $\mathcal{F}$ denote the set of all "concave probability distributions" on the unit interval, that is, all functions $f:[0,1]\to \mathbb{R}$ such that $f$ is concave, $f(x)\geq 0$ for all $x\in [0,1]$, and $\int_0^1 f(x) dx=1$.  Given a collection of $n$ samples $X_1 , \dots, X_n$, define $R(n) = \mathbf{E}( \max_i X_i - \min_i X_i )$ to be the expected length of the sample range of the samples $X_n$.  My question is:  what is the density $f^* \in \mathcal{F}$ that minimizes $R(n)$?  If that's too complicated, is there a useful lower bound for $R(n)$?

REPLY [10 votes]: For $n$ i.i.d. random variables $X_1,\dots,X_n$ with (cumulative) distribution function $ F(x)= \int _ 0 ^ x  f(t)dt\,  $     we have
$\mathbb{P}(\max_ {1\le i \le n} X _i \le x) =  F(x)^n \, $, $\mathbb{P}(\min _ {1\le i \le n} X _ i \le x) =  1 - \big ( 1- F(x)\big) ^n \, . $ Therefore the random variables $\max _ {1\le i \le n} X _ i$ and $\max _ {1\le i \le n} X _i$ have density functions respectively $\big( F(x)^n\big)'$ and $\big(1 - \big (1- F(x)\big )^n\big)'\, .$ Therefore 
$$ \mathbb{E} \big ( \max _ {1\le i \le n} X _ i - \min _ {1\le i \le n} X _ i\big ) =\mathbb{E} \big (\max_ {1\le i \le n} X _i \big )- \mathbb{E} \big (\min _ {1\le i \le n} X _ i\big ) = $$
$$\int_0^1 x\bigg( F ^ n + \big( 1-F \big)^ n  \bigg)'dx = 1 - \int_0^1 \bigg( F ^ n + \big( 1-F \big)^ n  \bigg)dx =\int _ 0 ^ 1 p\big(F(x)\big)dx \, ,
$$
after integration by parts; here $p(t)$ is the strictly concave polynomial function  $$p(t):= 1- t^n -(1-t)^n\, .$$
$$*$$
We have therefore to minimize the strictly concave functional 
$$J(f):=\int _ 0 ^ 1 p\Big( \int _ 0 ^ x  f dt\Big) dx \, , $$ on the convex  compact subset (see 1 below) $K\subset L^1([0,1])$ of all positive, concave, norm-one functions $f$.
A continuous strictly concave functional defined on a compact convex set $K$ attains its  infimum  on the extreme points of $K$. Here (see 2 below), the extreme points of $K$ are exactly the family of piecewise linear functions, for $\{f_c \}_ {0\le c\le 1} $, 
$$ f_c(x):=   \frac{2x}{c}\mathrm{\quad if\quad } 0\le x\le c\, , $$
$$ f_c(x):=    \frac{2(1-x)}{1-c} \mathrm{\quad if\quad } c\le x\le 1\, .$$ 
The minimization problem is therefore reduced to the  computation of the one-variable maximum: $$\min _ {f\in K}    J(f )=\min _ {0\le c \le 1}    J(f_c) =1-\max _ {0\le c \le 1} g(c)\, ,$$
where $g(c)= 1-J(f_c)$ is 
$$g(c) =\int_0^1  \bigg( c^{n+1}x^{2n} + (1-c)^{n+1}x^{2n} + c( 1- cx^2 )^n +(1-c)\big( 1- (1-c)x^2 \big)^n \bigg) dx \, ;$$
so $g$ is a polynomial in $c$ symmetrical w.r.to $c=1/2$. It should be true that  $g$ is 
concave in $c$, so that one eventually finds
$$\min _ {f\in K}    J(f )=1  -\frac{1}{2^n(2n+1)} -  \int_0^1   \bigg( 1-  \frac{x^2}{2} \bigg)^n   dx \, .$$
For $n\to +\infty$ this is (see 3 below)
$$  1- \sqrt{\frac{\pi}{2n}} + o\Big(\frac{1}{n}\Big)\, . $$

Plots of the polynomials $g(c)$, for $n=2,\dots,12$, from top to bottom.
$$*$$
More details. 
1. (Compactness of $K$). Note that by definition of $K$, each $f\in K$ satisfies 
$$ 0 \le f(x) \le 2\\ ,$$ 
for all $x\in [0,1]$ and, for any $\epsilon > 0$,
$$\mathrm{Lip}\big(f,\\ [\epsilon, 1-\epsilon]\big) \le \frac{2}{\epsilon} \, ,$$
which is sufficient to conclude that $K$ is compact in $L^1$ by the Ascoli-Arzelà and  the dominated convergence theorems.
2. (Extreme points of $K$). Note that for any $f\in K$ and $c\in [0,1]$, we have $f=f_c$  if and only if $f(c)\ge 2 $. This implies that all these $f_c$'s are extreme points of $K$ (if $f_c$ is an arithmetic mean of two elements of $K$, at least one of them has value at $c$ not less than $2$, hence it is $f_c$ itself). Conversely, if $f\in K$ has $\|f\|_\infty < 2$, then  one can  find a non-zero function $h$ with $\int_0^1 hdx =0$ and such that $f\pm h $ are concave and non-negative, hence in $K$, so that $f =\frac{1}{2}\big((f+h)+(f-h)\big)$ is not extremal.
3. (Asymptotics). Incidentally, a quick asymptotics for $\int_0^1   \big( 1-  \frac{x^2}{2} \big) ^ n   dx$  as $n\to\infty$  comes from the dominated convergence theorem, recalling that $0\le (1+t/n)^n\le e^t$ as soon as $1+t/n\ge0$. Changing variable, $x:=\xi/\sqrt n$, 
$$\int_0^1   \bigg( 1-  \frac{x^2}{2} \bigg) ^ n   dx = \sqrt{\frac{1}{n}} \int _ 0 ^{+\infty}   \bigg( 1-  \frac {\xi^2}{2n} \bigg) ^ n  \chi _ {[0,\sqrt n]}(\xi) d\xi $$
$$= \sqrt{\frac{1}{n}} \int _ 0  ^{+\infty}    e^ {  -\xi^2/2 } d\xi \, \big(1+ o(1)\big) = \sqrt{\frac{\pi}{2n}}+ o(\frac{1}{n})\, .$$<|endoftext|>
TITLE: What kinds of operations are well-defined when working with sets, classes, conglomerates, and yet higher order collections?
QUESTION [7 upvotes]: There are many foundations to set theory, ZFC, NBG, SEAR, to name a few, and while they differ in how sets, classes, and higher-order collections are represented as mathematical objects, they all attempt to provide useful tools to mathematicians without being glaringly inconsistent. 
Note that many set theories stop at the set, or only weakly partially define classes. But for those that go further, I have the following question:
What kinds of operations are well-defined when working with sets, classes, conglomerates, and yet higher order collections? Or, what kinds of operations are commonly assumed to be well-defined in fields such as category theory, etc.?
Here are some operations which I am curious about.
1) Applying the axiom of choice to the class of all sets, or to the conglomerate of all classes, and so on.
2) Given two classes $A$ and $B$, forming the set of classes $\{A,B\}$. Similarly for sets of conglomerates, etc. Of course, the set of all classes should be disallowed by Russell's paradox.
3) Given two classes $A$ and $B$ and a set $C$, make a "set-class function" mapping $C$ into $A$, a "class-class function" mapping $A$ into $B$, and a "class-set function" mapping $A$ into $C$. Similarly for any pair of higher-order collections of any type.
4) Regarding the cardinality of a set, class, conglomerate (etc) as a way to make equivalence classes of sets, classes, conglomerates (etc), respectively.
5) Regarding the cardinality of sets, classes, conglomerates, and higher-order collections as a way to define an equivalence class among all mathematical objects.
6) Forming the set of all "class-class functions" (or is it a class?)
I'm the least confident that #'s 5-6 will be acceptable, but the rest seem reasonable to my untrained eyes.

REPLY [3 votes]: This question is considered in detail in Mike Shulman's Set Theory for Category Theory.
The "big picture" way to think about it is to think that each time you do a power-set-type operation, you are constructing an object of a "higher order".  You just then need to postulate high enough orders to do whatever it is you want.  $U_0$ is the class of all sets, $U_1$ is the class of all classes of sets, etc.
Once you are forming classes of classes, etc., you are strictly outside what is provably consistent with ZFC, but set theorists routinely consider much, much stronger theories that do not seem to harbor contradiction.  So it's probably okay.<|endoftext|>
TITLE: Explicit Casselman theory: reference needed
QUESTION [6 upvotes]: Let $K$ be a nonarchimedean local field with ring of integeres $R_K$, maximal ideal $m_K$ and finite residue field $\bf k$. Let $\pi$ be an admissible irreducible complex representation of ${\rm GL}_2(K)$ with central character $\epsilon$. A fundamental result of Casselman says that there is a largest ideal $J\subseteq R_K$ such that the subspace $W_J$ of vectors in $\pi$ such that
$$
\gamma\cdot v=\epsilon(a)v\qquad
\forall\gamma=\left(\begin{array}{cc}
a & b\\
c & d 
\end{array}\right)\in{\rm GL}_2(R_K)
\ \text{with}\ c\in J
$$
is non-trivial and in fact $1$-dimensional. As every expert knows, this result is of paramount importance for the theory of modular forms.
Let $v_0$ be a generator of the $1$-dimensional space $W_J$. In some cases, it is rather easy to obtain $v_0$ explicitly. For instance if $\pi=\pi(\mu_1,\mu_2)$ is a class $1$ principal series representation with trivial central character (for which $J=R_K$) it is immediate to check that any generator of $W_{R_K}$ is of the form
$$
v_0(g)=|a|^{s_1}|d|^{s_2}|a/d|^{1/2}v_0(1)\quad
\text{where}\quad
g=\left(\begin{array}{cc}
a & *\\
 & d 
\end{array}\right)r,\quad r\in{\rm GL}_2(R_K)
$$
and $\mu_i=|\cdot|^{s_i}$, $i=1$, $2$.
My question is that if a table of generators $v_0$ has been tabulated explicitly anywhere, in particular for the supersingular representations and in other cases in which $J\subseteq m_K^2$.

REPLY [4 votes]: For the parabolically induced representation, I suggest to look at Casselman "Restriction of $GL(2, F)$ to to $GL(2,o)$"-paper.
For the Steinberg representations and the super cuspidal representation, I suggest to look at Bushnell-Henniart "Local Langlands conjectures for GL(2)". For the Steinberg, your ideal will be the maximal ideal $p$. For the supercuspidal stuff, you should try to understand the definition of a stratum.
The translation from stratum for super-cuspidals to what you are asking about is a time-consuming exercise. It suggest to argue with strata directly.
Also this article by Ralf Schmidt seems relevant: http://www.math.ou.edu/~rschmidt/papers/gl2.pdf.
Silberger has also classified representation of GL(2) in "Representations of PGL(2) over the $p$-adics" (LNM).
I am not sure if the titles of the references are all correct.<|endoftext|>
TITLE: When does continuity imply holomorphy?
QUESTION [11 upvotes]: I was studying the construction of the modular lambda function and I started thinking about the following question.  Suppose that $\Omega\subset \mathbb{C}$ is an open connected set and $f:\Omega\to \mathbb{C}$ is continuous on $\Omega$ and holomorphic except on a set of measure zero (with respect to 2-dimensional Lebesgue measure).  Is $f$ actually holomorphic on all of $\Omega$ ? 
If this statement is false, how much does one have to weaken the measure zero condition? For example if the set of measure zero is a $C^1$ curve the above follows from Morera's theorem.

REPLY [13 votes]: The following is too long for a comment. I will suppose here that your set of measure zero is compact.
In this case, Your question is closely related to so-called continuous analytic capacity. Let $K$ be a compact set in the plane, and let $\Omega$ be the complement of $K$ with respect to $\mathbb{C}_\infty$. The continuous analytic capacity $\alpha(K)$ is defined as
$$\alpha(K):= \sup \{ |f'(\infty)| : f \in A(\Omega), \|f\|_{\infty} \leq 1 \},$$
where $A(\Omega)$ is the set of all functions holomorphic in $\Omega$ that extend continuously to $\mathbb{C}_\infty$.
It is not difficult to prove that $$\alpha(K)=0$$ if and only if for every open set $U$ with $K \subseteq U$, every $f$ holomorphic on $U \setminus K$, continuous on $U$, extend analytically to the whole of $U$. 
So what you're looking for is a characterization of the compact sets $K$ with $\alpha(K)=0$. This is certainly very difficult, since it took more than a hundred years to solve a similar problem (but with analytic capacity $\gamma$ instead of $\alpha$), the so-called Painlevé's problem.
It is easy to prove however that if $\alpha(K)=0$, then the area of $K$ must be zero. You're asking if this condition is sufficient. Most likely it is not, but I don't have a counterexample right now.. Usually, for problems related to removable sets for holomorphic functions, it is more useful to consider hausdorff measure instead. 
Anyway, I suggest you look in Garnett's book "Analytic Capacity and measure". See also the recent book by Dudziak, "Vitushkin's conjecture for removable sets."
EDIT (Answer to the question)
Every compact set $K$ with Hausdorff dimension strictly between $1$ and $2$ is a counterexample. An easy way to see this is to use Frostman's lemma. Indeed, using this lemma, there exists a nonzero Borel measure $\mu$ supported on $K$ with growth $\mu(B(z,r)) \leq C r^{1+\epsilon}$. The growth of $\mu$ implies that the Cauchy Transform
$$f(z):=\int_{K} \frac{1}{z-\zeta}d\mu(\zeta)$$
is continuous on $\mathbb{C}$ and holomorphic on $\mathbb{C} \setminus K$. Furthermore, $f(z)\rightarrow 0$ as $z \rightarrow \infty$ and so in particular, $f$ is bounded on $\mathbb{C}$. However, $f$ is non-constant because $zf(z) \rightarrow \mu(K) \neq 0$. Therefore $f$ does not extend analytically to all of $\mathbb{C}$, for otherwise it would be constant by Liouville's Theorem.<|endoftext|>
TITLE: Representation theory for the exterior algebra
QUESTION [6 upvotes]: Hello everyone.
Let $V$ and $W$ be finite dimensional vector spaces over some field $K$. Consider $\rho:Sym(V)\to End(W)$ a homomorphism of algebras with unit (i.e., a representation of $Sym(V)$ on $W$). Fixing a basis $\{v_1,\ldots,v_m\}$ of $V$ there is associated an isomorphism $\Phi:K[x_1,\ldots,x_m]\to Sym(V)$ given by the extension of the mapping $\Phi(x_j)=v_j$ via the universal property of the symmetric algebra. This isomorphism makes the theory of representations of $Sym(V)$ very manageable because knowing where a basis element $v_j$ goes is tantamount to knowing where the monomial $x_j$ goes into $End(W)$ (using $\Phi^{-1}$). 
My question is: are there any analgous results for the exterior algebra $\Lambda(V)$? This algebra can be thought of as the algebra of antisymmetric polynomials in $V$. As a $\mathbb{Z}_2$-graded algebra, the representation homomorphism $\rho:\Lambda(V)\to End(W)$ would have to be zero on the odd part, but since $V$ is odd in this algebra (it is the space of generators) any analogue of the property above for the symmetric algebra would only hold for the trivial representation (1 maps to the identity mapping and everything else goes to 0). That's quite boring. Perhaps I'm utterly wrong in regarding $End(W)$ as a purely even superalgebra but that's the setting I'm mostly interested in. 
The question can be rephrased as: what are the irreducible modules over the exterior algebra?

REPLY [6 votes]: I'm not sure if this answers your question, but if $V$ is finite-dimensional and you are just asking for representations of the algebra structure, then the only irreducible representation of $\Lambda(V)$ is the trivial one-dimensional representation, in which $V$ acts as $0$.
Indeed, let $W$ be an irreducible representation of $\Lambda(V)$.  Let $0 \neq x \in V$, and consider $\mathrm{ker}(x) \subseteq W$ and $\mathrm{im}(x) \subseteq W$.  Since $x$ anticommutes with all generators of $\Lambda(V)$, both of these subspaces of $W$ are actually $\Lambda(V)$-submodules.
We are trying to show that $x$ acts as zero.  If not, then since $W$ is irreducible, we must have that $\mathrm{ker}(x) = (0)$.  Since the image is nonzero, again by irreducibility we conclude that $\mathrm{im}(x) = W$, so $x$ is an automorphism on $W$.  But then $x \wedge x = 0$ in $\Lambda(V)$, so the square of this automorphism of $W$ is zero, which is a contradiction.<|endoftext|>
TITLE: Realizing universal $C^*$-algebras as concrete $C^*$-algebras
QUESTION [18 upvotes]: How do I in general realize a universal C*-algebra generated by some generators and relation as concrete C*-algebras? For example, I know that universal C*-algebra generated by a single unitary is $C(\mathbb{T})$ by functional calculas. I am looking at the following examples to work on:

universal C*-algebra generated by single self-adjoint element whose norm is 1.
universal C*-algebra generated by single positive element whose norm is 1.
universal C*-algebra generated by single normal element whose norm is 1.
universal C*-algebra generated by single projection.

REPLY [9 votes]: Here is a further supplement: a tip for how to check if a $C^\ast$-algebra $A$ is the universal $C^\ast$-algebra for a given presentation.  (I probably learned this from Terry Loring's book "Lifting solutions to perturbing problems in $C^\ast$-algebras".)
First, check that $A$ really is generated by a set of elements satisfying the given relations.
Second, check that every irreducible representation of the universal $C^\ast$-algebra is a representation of $A$.  Say your generators are $x_1,\dots,x_n$.  Then an irreducible representation would be generated by elements $X_1,\dots,X_n$.   Since the centre of an irreducible representation is trivial, anything built out of the $X_i$'s that $*$-commutes with all the $X_i$'s is a scalar - so this approach works well if your relations entail a certain amount of commutativity, since commuting elements.
For example, to show that the universal $C^*$-algebra on a self-adjoint element of norm at most $1$ is $C_0([-1,1] \setminus \{0\})$, the second part above would go as follows.  Let $X$ be the generator in an irreducible representation.  Then $X$ is a scalar, which is self-adjoint (i.e. real) and has norm at most $1$.  So $X = t \in [-1,1]$, and this representation corresponds to evaluating at this point $t$.<|endoftext|>
TITLE: Edge-coloring of the complete graph without any rainbow paths
QUESTION [8 upvotes]: For a given $2k-1$ edge coloring of the complete graph $K_{2k}$,
say a Hamiltonian path $P$ is a rainbow path if every color appears exactly once in $P$.
My question is
"For each $2k (k \geq 2)$, is there a proper $2k-1$ edge coloring of $K_{2k}$ with no rainbow paths?"
It is easy to see that
every proper edge-coloring of $K_4$ or $K_6$ does not contain any rainbow paths.
For $K_8$, the statement is still true but some proper 7-edge-coloring contains rainbow paths.
When the number of vertices is a power of 2, then the statement is true, but I do not know if it is true for every even number (I don't even know for $2k = 10$).
Here's why the statement is true for $2^k$.
Label the vertices of $K_{2^k}$ by the elements of the group $((Z_2)^k, +)$
and label each edge by the sum(or difference) between two ends. Then the edge-labels give us a proper $2^k-1$ edge coloring, and this coloring does not have any rainbow paths.

REPLY [5 votes]: This is an open problem, see the intro of this paper for more details:
http://www.renyi.hu/~gyarfas/Cikkek/136_orthogonal.pdf<|endoftext|>
TITLE: Attack on CRT-RSA
QUESTION [9 upvotes]: The survey paper of Prof. Dan Boneh entitled 
"Twenty years of attacks on the RSA cryptosystem" mentioned that (Page 5)
one can attack CRT-RSA  in square root of decryption exponent. However no
argument is given. I know this comes from man-in-the-middle attack. 
However I cannot understand the idea clearly. 
Is there any lecture notes/paper where this attack is clearly explained?

REPLY [12 votes]: It's a good question, since it looks like Boneh's paper doesn't give a reference.  It's not actually a man-in-the-middle attack, at least not the attack I've seen.  Instead, it's reminiscent of baby-step giant-step but with an extra twist. Here's how it works.
Suppose we are given $N$ and $e$, where $N=pq$ with $p$ and $q$ distinct primes and $ed \equiv 1 \pmod{p-1}$ with $0 < d < D^2$.  We are not given $p$, $q$, or $d$, but we know the bound $D$.  We want to factor $N$ in roughly $D$ steps (up to log factors).  For what I'm about to do, we'll have to assume that the inverse of $e$ modulo $q-1$ isn't $d$, and in fact isn't anything like $d$, but this assumption holds for CRT RSA.
For a random value of $x$, $\gcd(x^{ed}-x,N)$ will be $p$ with good probability (this is where we need the assumption).  If we write $d = a + bD$ with $0 \le a,b < D$, then $\gcd(x^{ea}x^{ebD}-x,N)$ will be $p$, and in fact the gcd of $\prod_{i=0}^{D-1} (x^{ei}x^{ebD} - x)$ and $N$ will also be $p$ with good probability.  (Note that if the inverse of $e$ modulo $q-1$ were of the form $i+bD$ with $0 \le i < D$, then this would fail since we would pick up a factor of $q$ in the product.  This is what I meant by "anything like $d$" above.)
Now consider the polynomial $\prod_{i=0}^{D-1} (x^{ei}y - x)$ in the variable $y$.  In a number of steps nearly linear in $D$, we can compute this polynomial modulo $N$ and then we can evaluate it at any $D$ given points.  (This requires special algorithms, since for example multiplying the factors one by one would require about $D^2$ operations. See Chapter 10 of von zur Gathen and Gerhard's book Modern Computer Algebra for background on fast evaluation and interpolation algorithms.)
Given these fast algorithms, the final steps are easy: we compute the polynomial, compute the $D$ evaluation points $y = x^{ebD}$ with $0 \le b < D$, compute the evaluations of the polynomial at these points, and take their gcds with $N$.  All of this is nearly linear-time in $D$, and one of the gcds will give us the factor $p$.<|endoftext|>
TITLE: Sets of integers represented by degree zero rational functions
QUESTION [12 upvotes]: Suppose $f(x_1,x_2,\dots)=\frac{P}{Q}$, where $P,Q$ are polynomials in several variables with integer coefficients that have the same degree. Let's denote by $S(f)$ the set of integers $n$ for which $f(x_1,x_2,\dots)=n$ is solvable in integers.
Which sets $S\subset \mathbb Z$ can be written as $S(f)$ for some $f$ as above?
For example we have, $S(\frac{x_1^2+x_2^2}{x_1x_2+1})=\lbrace -5,0,1,4,\dots,k^2,\dots\rbrace$. 
This question is just a musing from playing around with variations to Hilbert's tenth problem. A more direct question would be: Is every Diophantine set representable as some $S(f)$?

REPLY [8 votes]: A set $T \subseteq \mathbb{Z}$ can be written as $S(f)$ if and only if $T$ is effectively enumerable.
Proof: As in zeb's comment, the restriction to degree zero doesn't matter, and we consider rational functions of arbitrary degree.
Suppose $T$ is effectively enumerable. By the MRDP theorem choose a polynomial $f(\bar{z},x)$ such that $T$ is precisely all $x$ for which the equation $f(\bar{z},x)=0$ is solvable in integers $\bar{z}$. Then the rational function \begin{equation*}
r(\bar{z},x):=x+\dfrac{f(\bar{z},x)^2}{1+f(\bar{z},x)^2}
\end{equation*}
has the value $x$ if $f(\bar{z},x)=0$, and otherwise is not an integer. Therefore $T=S(r)$.
Conversely, it is intuitively clear that every set of the form $S(r)$, with $r$ a rational function, is effectively enumerable.<|endoftext|>
TITLE: Can a model of $V\neq L$ contain a class giving the $L$-ordering on all its sets?
QUESTION [14 upvotes]: This question is inspired by the excellent question by Douglas Ulrich When is $L$-Rank definable in inner models of $V=L$?
Suppose $M \in L$ is a countable model of $ZFC$, and furthermore suppose $M \vDash V \neq L$.  These models have the funny property that, although every set in them is constructible, the model does not recognize this fact -- the "nonconstructible" sets simply arise at a level of the $L$-hierarchy that is greater than the ordinals of $M$.     Provided any countable $ZFC$ model exists, then such models abound -- for example, we can easily build forcing extensions of countable models by building an $M$-generic filter directly, and the forcing extension will not recognize that the generic is constructible.
In the answer to the question linked above it is shown that such a model $M$ cannot define the $L$-order on all of its members, at least not with the usual definition -- it can only define $<_L$ for those elements that it recognizes as constructible (that is, for members of $L^M$).  However, I am interested whether this limitation is inherent, or simply a limitation of definability over $M$.  In particular, is it consistent for us to add the $L$-order on all of $M$, as a class, without destroying $ZFC$?  I will phrase the question in terms of $GBC$ models, to make the use of classes explicit.

Is it consistent that  $M \in L$ is a countable model of $GBC$ with $M \vDash V \neq L$, and there is class $U \in M$ such that $U$ gives the $L$-order on $M$? (That is, $\langle x,y \rangle \in U$ if and only if $x <_L y$, for all sets $x, y \in M$).

Note: In the context of $GBC$ the statement $V \neq L$ may be ambiguous --  I intend it to refer only to sets, and not to classes, e.g. $V \neq L$ means "there exists a nonconstructible set."
Such a $U$ would give us a proper class well order with order type larger than $ORD^M$, but this does not seem immediately problematic - many such well orders are definable over any model of $ZFC$.  One way in which such a $U$ might be inconsistent is if, from $U$, we could show how to "construct" every set in $M$, but these constructions would be of more-than-$ORD^M$ length, and so this might not directly contradict $V \neq L$.
If such a $U$ is inconsistent, it would be nice to see why this limitation exists.  On the other hand, if we can have such a $U$, is it universally possible?

If such a $U$ is consistent, are there any restrictions on the models $M$ that can have them?

For example, it might be consistent only if every member of $M$ has $L$-rank "not too much greater" than $ORD^M$.
EDIT: changed a stray occurence of "GBC" to "ZFC" in paragraph 1, for clarity.

REPLY [5 votes]: This answer just considers the version of the question for transitive models $M$.
Under a reasonable interpretation of the order $<_L$ of constructibility, there is no such transitive model $M$. However, that interpretation is fine structurally motivated, and with a more obvious interpretation, I’m not sure how to prove it in general. There are many cases, however, where it can be ruled out with the more obvious one, and I’ll start with an example of that.
So let us start with the non-fine structurally motivated definition of $<_L$:
We order elements $x\in L$ first on stage $L_{\alpha+1}$ of construction, then on complexity $\Sigma_n$ of formula used to define $x$ over $L_\alpha$, then on finite set $s$ of ordinals used as parameters in such a definition, and finally on the specific $\Sigma_n$ formula used. Here we order $[\mathrm{OR}]^{<\omega}$ (finite sets of ordinals) top-down lexicographically, so $s<t$ iff $s\neq t$ and $\max((s\backslash t)\cup(t\backslash s))\in t$. A little more precisely, fix a recursive enumeration $\varphi_0,\varphi_1,\ldots$ of all formulas in the language of set theory. Now given $x\in L$, the rank of $x$ is the lexicographic rank of the lexicographically least tuple $(\alpha,n,s,k)$ where $x\in L_{\alpha+1}$ and $n,k<\omega$ and $\varphi_k$ is $\Sigma_n$ and $s=\{s_0,\ldots,s_{\ell-1}\}\in [\alpha]^{<\omega}$ and $x=\{y\in L_\alpha\bigm| L_\alpha\models\varphi_k(s_0,\ldots,s_{\ell-1},y)\}$.
Now let’s consider what is presumably the most obvious example of a candidate model $M$: let $\alpha$ be least such that $L_\alpha\models\mathrm{ZFC}$ and let $c$ be the $<_L$-least Cohen generic over $L_\alpha$, and $M=L_\alpha[c]$. Then $<_L\upharpoonright M$ cannot be added to $M$.
(Remark:  This should be contrasted to when we restrict attention to the prewellorder $\leq^*$ of $L$
given just by stage of construction ($x\leq^* y$ iff for all $\beta$, if $y\in L_\beta$ then $x\in L_\beta$). For this example, $\leq^*\upharpoonright M$ is definable from parameter $c$; see   When is $L$-Rank definable in inner models of $V=L$?)
Now $L_\alpha$ is pointwise definable, and this yields a surjection $\pi:\omega\to L_\alpha$ which is $\Sigma_1$-definable over $L_{\alpha+1}$ from the parameter $\alpha$. Therefore there is a Cohen generic over $L_\alpha$ in $L_{\alpha+2}$,
and in fact one that is $\Sigma_1$-definable over $L_{\alpha+1}$ from the parameter $\alpha$. This is optimal for such a Cohen generic: Because $L_\alpha\models\mathrm{ZFC}$, we have $c\notin L_{\alpha+1}$ and there is no $\Sigma_0$ definition of $c$ over $L_{\alpha+1}$ from parameters. It also implies that $L_\alpha\preceq_{\Sigma_1} L_{\alpha+1}$, and hence there is no $\Sigma_1$ definition of $c$ over $L_{\alpha+1}$ from any $s\in[\mathrm{OR}]^{<\omega}$ with $s<_{\mathrm{lex}}\{\alpha\}$.
So in terms of the 4-tuples defining $<_L$, the $<_L$-least Cohen generic $c$ has tuple $(\alpha+1,1,\{\alpha\},k)$ for some $k<\omega$.
Assuming that $<_L\upharpoonright M$ is $M$-compatible (meaning that $(M,{<_L\upharpoonright M})$ is a model of ZFC, in the expanded language with the predicate $<_L\upharpoonright M$), we want to deduce that $M$ can understand enough about $\pi$ for a contradiction.
That is, for $\beta\in\alpha\backslash\omega$ let $c_\beta=c\cup\{\beta\}$. Then note that $c_\beta$ has tuple $(\alpha+1,1,\{\alpha\},k_\beta)$ for some $k_\beta<\omega$ (using here the surjection $\pi:\omega\to L_\alpha$ mentioned above). But the class $C=\{c_\beta\bigm|\beta\in[\omega,\alpha)\}\subseteq M$ is definable over $M$, so if $<_L\upharpoonright M$ is $M$-compatible, then $<_L\upharpoonright C$ is too. But this is an ordering of $\mathrm{OR}^M\backslash\omega$ in ordertype $\omega$, which is impossible.
Now it seems one would like to generalize this to show that there is no such transitive $M$ at all. But we used some particular fine structural facts which don't easily (seem to) generalize. In particular, the use of the standard $\Sigma_n$ hierarchy becomes inconvenient when $n>1$. This is modified in fine structure theory, and replaced with a slightly different definability hierarchy (e.g. $\mathrm{r}\Sigma_n$ instead of $\Sigma_n$). One also uses the $\mathcal{J}$-hierarchy $\mathcal{J}_\alpha$, and definability over $\mathcal{J}_\alpha$, instead of the $L$-hierarchy $L_\alpha$ (though I think the more crucial thing is the definability hierarchy). If one defines $<_L$ in a natural way using these things, also incorporating the fine structural standard parameters in a natural way, then there can be no such transitive $M$ for which $<_L\upharpoonright M$ is $M$-compatible.
That is (now assuming fine structure), define $<_L$ as follows: Given $x\in L$, let the $<_L$-rank of $x$ be the lexicographic rank of the lexicographically least tuple $(\alpha,n,s,k)$  such that $x\in\mathcal{J}_{\alpha+1}$, $\varphi_k$ is $\mathrm{r}\Sigma_{n}$, $s\in[\mathrm{OR}\cap\mathcal{J}_\alpha]^{<\omega}$,
and $x=\{y\in\mathcal{J}_\alpha\bigm|\mathcal{J}_\alpha\models\varphi_k(\vec{p}_{n-1},s,y)\}$, where $\vec{p}_{n-1}=(p_1,p_2,\ldots,p_{n-1})^{\mathcal{J}_\alpha}$, the first $n-1$ standard parameters of $\mathcal{J}_\alpha$ (and $\vec{p}_{-1}=\emptyset$). Then there is no transitive $M\models\mathrm{ZFC}+V\neq L$ such that $<_L\upharpoonright M$ is $M$-compatible.
Proof: Suppose $M$ is otherwise. Let $\alpha=\mathrm{OR}^M$. Let $(\gamma,n',s,k)$
be a minimal tuple determining a set $x\in M\backslash \mathcal{J}_\alpha$. Note that $n'=n+1>0$, by rank considerations. So $x\in \mathcal{J}_{\gamma+1}\backslash \mathcal{J}_\gamma$ and $x$ is $\mathrm{r}\Sigma_{n+1}^{\mathcal{J}_\gamma}(\{\vec{p}_n^{\mathcal{J}_\alpha},s\})$. Let $\xi=\sup(x)<\alpha$. Now observe that $\rho_{n+1}^{\mathcal{J}_\gamma}\leq\xi$ and $s\backslash\xi=p_{n+1}^{\mathcal{J}_\gamma}\backslash\xi$. (If $s\backslash\xi<_{\mathrm{lex}}p_{n+1}^{\mathcal{J}_\gamma}\backslash\xi$ then use an $(n+1)$-solidity witness, which is in $\mathcal{J}_\gamma$,  to compute $x\in \mathcal{J}_\gamma$, contradicting the minimality of $\gamma$. The other direction is by $(n+1)$-soundness.) Now for $\beta\in[\xi,\alpha)$ define $x_\beta=x\cup\{\beta\}$, and note that the tuple for $x_\beta$ is $(\gamma,n+1,s_\beta,k_\beta)$, where $s_\beta\backslash\xi=s\backslash\xi$ and $k_\beta<\omega$. (If $s_\beta\backslash\xi<_{\mathrm{lex}} s\backslash\xi$, note that $x_\beta$ violates the minimality of the tuple for $x$ for giving something in $M\backslash L_\alpha$.) But then if ${<_L}\upharpoonright M$ is $M$-compatible, then $(M,{<_L\upharpoonright M})$ can define an ordering of its ordinals in ordertype $\leq$ that of the lexicographic ordering on $[\xi]^{<\omega}\times\omega$, which is impossible.
It would be very interesting to know what the answer is in general for the first (non-fine structurally motivated) definition of $<_L$.<|endoftext|>
TITLE: coincidence between minimal triangulation numbers and chromatic numbers
QUESTION [8 upvotes]: A friend and I were reading up on the classification of compact surfaces when we realized that the minimal number of triangles in a triangulation of the sphere, torus, and projective plane are 4 (routine), 7 (an exercise in several places, such as in Katok - Lectures on Surfaces), and 6 (using a similar counting trick as the torus) respectively. This was strange as we knew the chromatic numbers of the plane, torus, and projective plane to also be 4 (the Four-Color Theorem), 7 (folklore), and 6 (http://mathworld.wolfram.com/ProjectivePlane.html) respectively. 

Question: What is known about the
  relationship between these two
  invariants?

I did a little research and I think I kind of see what is going on. I'll outline what I know here, though I hope an expert will help fill in the details or give further directions: I think the relevant result is Heawood's conjecture (apparently not a conjecture), which gives a formula for the chromatic number of any surface (apparently except the Klein bottle) given the genus.  In particular, there exist triangulations of the given numbers in the above three situations, so they match. If this assessment is correct, then my question above basically becomes something like:

Question: is it obvious that when there exists a map that requires the prescribed number of colors given by Heawood's not-so-Conjecture, then there exists a triangulation with the same chromatic number? (bonus: can we force the triangulation to be a refinement of the map?)

REPLY [9 votes]: The answer to your "is it obvious" is "no". But it is a theorem in many cases. Wikipedia + one click takes to you "Solution of the Heawood map coloring problem" by Ringel and Youngs. There they completed the proof that Heawood's bound on the chromatic number was tight for surfaces
with positive genus. As they explain in their introduction, they do this by producing an embedding of a complete graph in each case. The embedding for $K_N$ is triangular if
$N$ is congruent to 0, 3, 4, or 7 mod 12.
To see this, note that Ringels and Young show that if 
$\gamma=\lceil (n-3)(n-4)/12\rceil$, then there is an
embedding of $K_n$ in the surface of genus $\gamma$, whence the chromatic number of the surface
is at least $n$. By Euler's formula $n-e+f=2-2g$ and for $K_n$ we have $e=n(n-1)/2$ and
$3f \le 2e$ (with equality if and only if each face is a triangle). Hence
$$
  -n(n-7)/6 \ge 2-2g
$$
or equivalently $g \le (n-3)(n-4)/12$. So in the mod 12 cases listed the embedding is triangular.<|endoftext|>
TITLE: Selecting $k$ integers from an interval $[0, N]$ to maximize the minimum difference between pairwise sums
QUESTION [5 upvotes]: I have an optimization problem where I need to select $k$ integers over the interval $[0, N]$ s.t. I maximize the minimum difference between any pairwise sum of the $k$ integers (where we also include the sum of one selected integer with itself).  For example, if $k = 3$, $N = 3$, and we select the set of integers $(1, 2, 3)$, we have a set of $\frac{1}{2}k(k+1) = 6$ pairwise sums:
$1 + 1 = 2$
$1 + 2 = 3$
$1 + 3 = 4$
$2 + 2 = 4$
$2 + 3 = 5$
$3 + 3 = 6$
Here, the minimum difference between any two pairwise sums is trivially: $(4 - 4) = 0$.  
As a function of $N$, how does one select the optimal set of $k \leq N$ integers?  What happens in the limit where $N \to \infty$?  I'd also be interested to understand how the number of optimal subsets of $k$ integers scales as $N$ becomes large. 

If we set $N = 100$ and do a computational experiment, we find: 

For $k = 3$: 
Optimal minimum distance between pairwise sums: $33$
Example of subset that achieves the optimal minimum pairwise difference (there are $6$ total): {{0,33,99}} 
All pairwise (and self-) sums for this example subset: {{0,33,66,99,132,198}} 

For $k = 4$: 
Optimal minimum distance between pairwise sums: $16$
Example of subset that achieves the optimal minimum pairwise difference (there are $50$ total): {{0,16,64,96}}
All pairwise (and self-) sums for this example subset: {{0,16,32,64,80,96,112,128,160,192}}

For $k = 5$: 
Optimal minimum distance between pairwise sums: $9$
Example of subset that achieves the optimal minimum pairwise difference (there are $12$ total): {{0,9,36,81,99}}
All pairwise (and self-) sums for this example subset: {{0,9,18,36,45,72,81,90,99,108,117,135,162,180,198}}


Note: From quid's comment, and also given the computational expensive of finding values for $N = 100$ and $k > 5$, it occurs to me that it would be very nice to have some kind of an upperbound, known to the achievable, for the maximum minimum difference for a $k$-sized subset as a function of $k$.  If optimal solutions are dense (I have no reason to suspect that this is the general trend), this could allow for the use of a more efficient probabilistic search procedure to find an optimal "diluted" Sidon subset.  Does anyone have any good ideas for how to construct such an upperbound?

REPLY [3 votes]: As @quiz says,  there exists $k$ integers $a_1,\dots,a_k\in [1,N]$ with $\min|a_i+a_j-(a_r+a_s)|\sim N/k^2$ when $k,\ N/k^2\to \infty$.
Indeed the asymptotic  is sharp. To see this, consider the real numbers $x_j=a_j/N$ and apply the following result of J. Cilleruelo and I. Ruzsa [1]
Theorem: Let $x_1,\dots ,x_k\in [0,1]$ and let $\delta=\min|x_i+x_j-(x_r+x_s)|$. Then
$\delta\le \frac 1{k(k-2\sqrt k)}$.
[1] J. Cilleruelo and I. Ruzsa, "Real and p-adic Sidon sets", Acta Sci. Math. (Szegez)  vol 70, nº 3-4  (2004).  http://www.uam.es/personal_pdi/ciencias/cillerue/<|endoftext|>
TITLE: Factoriality vs $\mathbf{Q}$-factoriality for threefolds hypersurfaces with isolated singularities
QUESTION [10 upvotes]: Let $X \subset \mathbf{P}^4$ be a complex threefold hypersurface with isolated singularities. 
We denote as usual by $\textrm{Cl}(X)$ the group of Weil divisors modulo linear equivalence and by $\textrm{Pic}(X)$ the group of Cartier divisors modulo linear equivalence. We also write $G(X):=\textrm{Cl}(X)/\textrm{Pic}(X)$.
Recall that 
$\bullet$ $X$ is called factorial if every Weil divisor is a Cartier divisor; by Lefschetz theorem, this is equivalent to say that $\textrm{Pic}(X)=\textrm{Cl}(X)=\mathbf{Z}$, generated by the hyperplane section;
$\bullet$ $X$ is called $\mathbf{Q}$-factorial if every Weil divisor has a multiple which is a Cartier divisor; this is equivalent to say that $G(X)$ is a torsion group.
Of course if $X$ factorial then $X$ is $\mathbf{Q}$-factorial, because if $X$ is factorial then $G(X)=0$. I'm interested in the other implication, so my first question is

Question 1. Assume that $X \subset \mathbf{P}^4$ is a $\mathbf{Q}$-factorial threefold with isolated singularities. Does this imply that $X$ is factorial? If not, what is a counterexample?

It is known that the answer to Question 1 is yes when $X$ is nodal, i.e. contains only ordinary double points. The way I see this is the following. There is an exact sequence (I think it is called Jaffe's exact sequence) $$0 \to \textrm{Pic}(X) \to \textrm{Cl}(X) \to \bigoplus_{p \in \textrm{Sing}(X)} \textrm{Cl}(\mathscr{O}_{X, p})$$
and the last group injects into $\bigoplus_{p \in \textrm{Sing}(X)} \textrm{Cl}(\widehat{\mathscr{O}}_{X, p})$. On the other hand, if $p$ is a node then $\textrm{Cl}(\widehat{\mathscr{O}}_{X, p}) \cong \mathbf{Z}$, so we have an inclusion $$G(X) \hookrightarrow \bigoplus \mathbf{Z}.$$
This implies that if $G(X)$ is a torsion group then necessarily $G(X)=0$, so factoriality and $\mathbf{Q}$-factoriality are equivalent conditions in this case. These considerations led me to the following

Question 2. Let $X \subset \mathbf{P}^4$ be a threefold hypersurface with isolated singularities and let $p \in \textrm{Sing}(X)$. Is it true that $\textrm{Cl}(\mathscr{O}_{X, p})$ is either torsion-free or zero? Or, still better, is it true that $\textrm{Cl}(\widehat{\mathscr{O}}_{X, p})$ is either torsion-free or zero?

An affirmative answer to Question 2 would imply an affirmative answer to Question 1, by the same argument used in the nodal case. 
I'm particulary interested in the case where all the singularities of $X$ are $ordinary$, i.e. the corresponding tangent cone is a cone over a smooth surface in $\mathbf{P}^3$.  
Any answer or reference to the existing literature will be greatly appreciated. Thank you!
EDIT. As pointed out by Artie in his comment, Question 1 has a positive answer when $X$ is a Gorenstein threefold with terminal singularities (here the assumption $X \subset \mathbb{P}^4$ is not necessary).

REPLY [5 votes]: The answer to Question 2 is true. If $(R,m)$ is a local complete intersection of equicharacterstic $0$ and of dimension $3$ then the Picard group of $Y  = \text{Spec} R-\{m\}$ is torsion-free. This group agrees with the class group of $R$ when $R$ has isolated singularity. The proof in this case essentially follows from the proof of the Grothendieck-Lefschetz theorem (SGA somewhere), which states that if $R$ is a  complete intersection and $\dim R\geq 4$ then $Y$ has trivial Picard group.
For reference and some extensions see the papers by Badescu and Robbiano quoted here.<|endoftext|>
TITLE: Adams-Novikov spectral sequence at p = 2
QUESTION [7 upvotes]: Does anyone know of any computer calculations of the E2-term of the Adams-Novikov spectral sequence at p=2?
I'd love to get my hands on this data.

REPLY [11 votes]: I don't think anybody knows how to compute this $E_2$-term efficiently (not just at the prime $2$). I would love to be proved wrong on this, of course. 
So far the only documented, algorithmic method that has any chance to be computationally succesful seems to be the method described by Zahler in 1969/1970.
You could use my programs to compute the $E_2$-term of the algebraic Novikov spectral sequence
$$\operatorname{Ext}_{EA}(F_2,F_2) \Rightarrow {\operatorname{Ext}}(BP_{\ast},BP_{\ast})$$
where $EA$ is an associated graded of the $2$-primary Steenrod algebra. But of course that's only a very rough approximation to the Novikov $E_2$-term.<|endoftext|>
TITLE: Question about local description of the branch locus
QUESTION [8 upvotes]: Let $\pi:Y\to X$ be a dominant, finite morphism of nonsingular varieties over an algebraically closed field $\Bbbk$. Assume furthermore that for all $Q\in Y$, with $P=\pi(Q)$, we have 
$$\mathcal O_{Y,Q}=\mathcal O_{X,P}[T_1,\ldots,T_k]/(T_1^n-x_1,\ldots,T_k^n-x_k)$$
for certain $x_i\in\mathcal O_{X,P}$. In other words, we have adjoined certain $n$-th roots. 
Now, let $R\subseteq Y$ be the ramification divisor and $H=\pi(R)$ the branch locus (both now with the reduced subscheme structure). Let $\mathcal I(H)$ be the ideal sheaf of $H$ and $\mathcal I(H)_P$ the stalk at $P$. Can I conclude that this ideal $\mathcal I(H)_P\subseteq \mathcal O_{X,P}$ is generated by the product $x_1\cdots x_k$? Or, correspondingly, can I conclude that $\mathcal I(R)_Q$ is generated by the product $y_1\cdots y_k$?
Edit. As pointed out in the answer by Dmitry Vaintrob, I want to assume $n$ not divisible by $\mathrm{char}(\Bbbk)$. Furthermore, we assume that the $x_i$ are reduced and coprime.

REPLY [4 votes]: There are two issues. You presumably mean that $x_i$ are themselves reduced and have no common divisors, or else the answer would be trivially no: just take $x_1=x_2$ and then the reduced branching ideal is generated by $x_1,$ not $x_1x_2.$ You also need to be careful what you mean in characteristic $p$, as then $T^p-x$ is everywhere ramified, no matter what $x$ is. Otherwise (in characteristic $0$ or relatively prime to $n$ and with $x_i$ reduced and having no common divisors) I don't see any reason why this would be false, just by writing out equations for the ramification and branching loci: ramification divisors under fiber products satisfy $R_{Y_1\times_X Y_2\to X}=R_{Y_1\to X}\times_X Y_2\cup Y_1\times R_{Y_2\to X}$, so it's enough to consider a single $T_1$ and you have $T_1^n-x_1=0, nT_1^{n-1}=0,$ which turns into $T_1=0, T_1^{n-1}=x_1$ (and the projection of this to $X$ is again $x_1=0$). 
Is this what you were asking or am I missing something? If this is what you need, I don't deserve the bounty as I didn't say anything deep.
 Edit: 
PS - The fact that your ring is a fiber product follows from the following general statement. Suppose $U=S[x_1,\dots, x_n]/(R_1,\dots, R_k)$ and $V = S[x_1',\dots, x_m']/(R_1',\dots, R_l')$ are commutative algebras over $S$ given by generators and relations, where $R_i$ are relations on the $x$'s and $R_j'$ are relations on the $y$'s. Then the tensor product can be written in generators and relations as follows.
$$U\otimes_S V=S[x_1,\dots, x_n, x_1',\dots, x_m']/(R_1,\dots, R_k, R_1',\dots, R_l')$$ 
You can prove this in one line by using universal properties in the category of commutative $S$-algebras. 
Namely, a map of $S$-algebras from the tensor product $U\otimes_S V$ to some $S$-algebra $W$ is equivalent to a pair of maps $f_1:U\to W, f_2:V\to W$ both over $S$, and a map from $S[x_1,\dots, x_n]/(R_1,\dots, R_k)$ to $W$ is equivalent to choosing a set of elements $w_1,\dots, w_n$ of $W$ satisfying relations $R_1,\dots, R_k$. Putting this together, $U\otimes_S V$ and the ring defined by the two sets of generators and relations as above satisfy the exact same universal property.<|endoftext|>
TITLE: Constructing expanders in Z/pZ
QUESTION [7 upvotes]: Fix a positive integer $k>0$. For $p>k$ a prime, let $A_p$ be a subset of the finite field $\mathbb{Z}/p\mathbb{Z}$ of size $k$ which contains a primitive element.
Define $G_p$ to be the (di)graph whose vertices are elements of $\mathbb{Z}/p\mathbb{Z}$, with two vertices $i,j$ joined by an edge provided $j=ia$ or $j=i+a$ for some $a\in A_p$. 
(I'm mainly interested in the situation where $A_p$ is closed under the operations of taking multiplicative and additive inverses; under these assumptions I can think of $G_p$ as a graph rather than a digraph.)
Question: Is $(G_p)_{p \textrm{ a prime}}$ a family of expanders?

Background: I'm expecting the answer to be either "possibly" or "no" (because if it were "yes" I'd hope I'd have heard about it already).
My interest comes in studying the Bourgain-Gamburd machinery for proving expansion from results about growth. For the family $(G_p)$, the relevant growth result is the Bourgain-Katz-Tao sum-product theorem for fields of prime order.
One needs more than just a growth result of course, one also needs to have some notion of `quasirandomness' (but I think I can handle this), as well as a lower bound on the girth of the graph. I've not thought much about this last aspect so I guess this is the most likely to be the sticking point.

REPLY [2 votes]: No, because solvable groups are amenable.  You're asking: Is is there a set in Z/pZ almost invariant by x->x+1 and 2x? Here's one: take the union of I, I/2, .., I/2^n, where I is an interval of length much bigger than 2^n.<|endoftext|>
TITLE: Pair correlation for the Riemann zeros and $(\zeta^\prime(s)/\zeta(s))^\prime$
QUESTION [7 upvotes]: Added Background:  The pair correlation of the zeros of the Riemann zeta function is influenced by the the derivative of the logarithmic derivative $(\zeta^\prime(s)/\zeta(s))^\prime$; see for example the answers to 
this question

I'm looking for references for computation of $\zeta^\prime(s)/\zeta(s)$, as well as the its derivative, in the critical strip $0<\text{Re}(s)<1$.  (I'm sure they're out there but google scholar/MathSciNet searches return way too many irrelevant hits.)
Of course, both $\zeta(s)$ and $\zeta^\prime(s)$ are implemented in packages like Sage, one can just take the quotient and then use this to numerically estimate the derivative via the difference quotient, but this seems computationally wasteful.  We have that 
$$
\frac{\zeta^\prime(s)}{\zeta(s)}=\log(2\pi)-1-\gamma/2-\frac{1}{s-1}-\frac12\frac{\Gamma^\prime(s/2+1)}{\Gamma(s/2+1)}+\sum_\rho\left(\frac{1}{s-\rho}+\frac{1}{\rho}\right).
$$
Similarly, one gets $(\zeta^\prime(s)/\zeta(s))^\prime$ upon differentiating term by term. 
The digamma function $\Gamma^\prime/\Gamma$ as well as the Riemann zeros $\rho$ are implemented in Mathematica.  So I think what I'm asking is a reference to answer the following:  
Given $\epsilon$, how many zeros do I need to take as a as a function of $t=\text{Im}(s)$ so the error is bounded by $\epsilon$?

Added:  In Theorem 9.6(A) in Titchmarsh's "Theory of the Riemann Zeta Function", one can compute the relevant constants to show that
$$
\left|\frac{\zeta^\prime(s)}{\zeta(s)}-\sum_{|\rho-s|\le 6}\frac{1}{s-\rho}\right|\le 4\log t.
$$
So in answer to Joro's question below, yes the sum is dominated by the zeros close to $s$.

REPLY [5 votes]: I believe the comments of joro and Carlo Benakker right under your question is to the point. Since the zeroes close to $s$ will be the ones that contributes in the sum, the zeroes close to s must be computed first and in order to do that several values of the zeta-function must certainly be calculated and the time for each such computation (and zero) will certainly not be less than computing any other particular value of $\zeta(s)$ or $\zeta'(s)$.
As for calculating a particular value $\zeta(s)$ the recent method of Ghaith Hiary  http://arxiv.org/abs/0711.5005  "Fast methods to compute the Riemann zeta function" (published in Annals of Mathematics 2011) contains the fastest method known. He shows a method that calculates the value of $\zeta(1/2+it)$ with an error term less than $|t|^{-N}$ in time $O_{\varepsilon,N}(t^{4/13+\varepsilon})$. The method of Odlyzko and Schönhage "Fast algorithms for multiple evaluations of the Riemann zeta function", doi:10.2307/2000939 is faster if sufficiently many values of $t$ needs to be calculated, i.e. it can calculate $\sqrt T$ values in the range $[T,T+\sqrt T]$ in time $O_\varepsilon(T^{1/2+\varepsilon})$. 
Now, In Hiary's paper it might look like he only considers the critical line. However in fact his method works for any value in the critical strip. Indeed in part of his argument he considers just the line Re$(s)=0$, but then he writes "It is clear that the restriction $\sigma=0$ is not important and a similar conclusions can be drawn for other values of $\sigma$" (page 6, second to last paragraph in Hiary's paper). It is true that his main interest is the critical line, but the algorithm holds for any $s$ in any vertical strip such as the critical strip.
Now to calculate $\zeta'(s)$ up to error of order $t^{-N}$ it is sufficient to calculate $\zeta(s+ih)$ and $\zeta(s-ih)$ with error less than $t^{-3N/2-1/4}$ for $h=t^{-N/2-1/4}$ and consider the difference quotient $(\zeta(s+ih)-\zeta(s-ih))/h$ by also using that $\zeta'''(s) =O(\sqrt t)$ in the critical strip and some version of the mean value theorem inequality.
It is simple to use finite difference quotients to calculate $\zeta^{(k)}(s)$ up to any desired degree of accuracy. Thus we can certainly use this method to calculate for example
$$
 \frac d {ds} \frac{\zeta'(s)}{\zeta(s)}= \frac{\zeta''(s)\zeta(s)-\zeta'(s)^2}{\zeta(s)^2}
$$
by this method. This method should not be computionally wasteful.<|endoftext|>
TITLE: Is the derived category of abelian groups a subcategory of the stable homotopy category?
QUESTION [12 upvotes]: An extension of the Dold-Kan equivalence gives an adjunction between the stable homotopy category and the (unbounded) derived category of abelian groups $SH \rightleftarrows D(Ab)$.
Question 1: Is the right adjoint $D(Ab) \to SH$ faithful?
Question 2: If not, is there a class of objects on which it is faithful (for example compact objects).

REPLY [8 votes]: I've found the following somewhat intricate way of answering Q1 in the affirmative. Any complex in $D(Ab)$ quasi-isomorphic to a graded abelian group. Hence, it is enough to consider complexes concentrated in a single degree. Given an abelian group $A$ and $n\in\mathbb Z$, let $(A,n)$ be the abelian group $A$ concentrated in degree $n$. For simplicity, I will use the same notation for the Eilenberg-MacLane spectrum $\Sigma^nHA$. In the derived category we have,
$$D(Ab)((A,n),(B,n))=\operatorname{Hom}(A,B),$$
$$D(Ab)((A,n),(B,n+1))=\operatorname{Ext}(A,B),$$
$$D(Ab)((A,n),(B,m))=0\text{ otherwise}.$$
In the stable homotopy category we have the stable Eilenberg-MacLane groups
$$SH((A,n),(B,m))=H^{m+k}(A,n+k;B),\quad k>>0.$$
It is well known, since E-ML's "On the groups..." (Annals) that
$$SH((A,n),(B,n))=\operatorname{Hom}(A,B),$$
$$SH((A,n),(B,n+1))=\operatorname{Ext}(A,B),$$
and that the functor $D(Ab)\rightarrow SH$ is the identity on the previous morphism sets. Hence we are done. The groups $SH((A,n),(B,m))$ are however non-trivial for $m>n+1$, in general.<|endoftext|>
TITLE: Good examples of random variables whose image is not a measurable set?
QUESTION [10 upvotes]: Are their simple/natural examples of real-valued Borel-measurable random variables whose image is not a Borel set? Something that occurs "naturally"?
I am teaching Doob's lemma (for two real-valued random variables $X$ and $Y$, $X$ is $\sigma(Y)$-measurable iff there exists a Borel-measurable function $f:\mathbb{R}\to\mathbb{R}$ such that $X=f(Y)$) and the main difficulty in the proof comes from the fact that $Y(\Omega)$ is in general not a Borel set. So I am wondering if there is a "natural" example that I can use to convince 4th year students that this "pathology" can naturally come up.
It is easy to construct examples, e.g., choose $A\subseteq \mathbb{R}$ any set that is not a Borel set, and equip it with the $\sigma$-algebra $\mathcal{A}=\{A\cap B; B\in \mathcal{B}(\mathbb{R})\}$, where $\mathcal{B}(\mathbb{R})$ denotes the $\sigma$-algebra of Borel sets in $\mathbb{R}$. Then the inclusion $X:(A,\mathcal{A})\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable and has $A$ as image, so its image is not a Borel set. But this feels like cheating...

REPLY [7 votes]: An analytic set that is not a Borel set...see this post* from long ago.
Such an analytic set is a continuous image of $[0,1] \setminus \mathbb Q$, and thus a Borel image of $[0,1]$.

*From: e...@math.ohio-state.edu (Gerald Edgar)
  Newsgroups: sci.math
  Subject: Re: Real Measurable, non-Borel.
  Date: 7 Oct 1993 08:13:10 -0400
  Organization: The Ohio State University, Dept. of Math.
  Message-ID: <29114m$e1b@math.mps.ohio-state.edu>
  References: 
In $\lt$CEIvIr....@undergrad.math.uwaterloo.ca> emlap...@undergrad.math.uwaterloo.ca (eli lapell) wrote:
  $\gt$What is a set of real numbers which is measurable but not Borel?  Or just not 
  Borel, period ??
An explicit example of a set of real numbers that is measurable (indeed,
  analytic) but not Borel [due to Lusin, Fundamenta Math. 10 (1927) p. 77]:
the set of all real numbers x with continued fraction expansion
x = a[0] + 1/(a[1] + 1/(...))
such that, for some positive integers r[1] < r[2] < ..., we have a[r[i]] divides 
a[r[i+1]] for all i.

Other examples of analytic sets that are not Borel can be given
  in (complete separable) metric spaces other than the line:
In the space K[0,1] of nonempty compact subsets of [0,1] with the Hausdorff
metric:  The subset consists of the uncountable compact subsets.  [Hurewicz, 1930]

${}$
In the space C[0,1] of real-valued continuous functions on [0,1] with the
unform metric:  The subset consists of the differentiable functions.
[Mazurkiewicz, 1936]<|endoftext|>
TITLE: Research trends in geometry of numbers?
QUESTION [25 upvotes]: Geometry of numbers was initiated by Hermann Minkowski roughly a hundred years ago. At its heart is the relation between lattices (the group, not the poset) and convex bodies. One of its fundamental results, for example, is Minkowski's theorem: If $L$ is a lattice in $\mathbb{R}^d$ and $C$ a centrally-symmetric convex body, then $\mbox{vol}(C) \geq 2^d \det(L)$ implies that $C$ contains another lattice point than $0$.
While there was a lot of activity in the field until at least 1960, it seems that in recent decades not so many people are working on it anymore. One of the reasons could be that the field is somewhat stuck, or in Gruber's more polite words "It seems that fundamental advance in the future will require new ideas and additional tools from other areas." (see [1]).
I would like to know more about current research trends in the geometry of numbers. What are hot topics right now? In which areas was recently considerable progress achieved? Did maybe even the "fundamental advance", that Gruber mentions, take place?
[1] P. Gruber: Convex and discrete geometry, Springer 2007, p. 353

REPLY [9 votes]: I can't say that what I'll relate is fundamental, but it does fit into the new ideas category. Since I and (my collaborator) Florent Balacheff have given talks on the subject and the paper will be in the ArXiv in a few days I feel free to comment on it. This post is an annoucement of joint work with Florent Balacheff and Kroum Tzanev.
As you comment, the basic result in the geometry of numbers is Minkowski's (first) theorem: If the volume of a $0$-symmetric convex body $K \subset \mathbb{R}^n$ is at least $2^n$, then  $K$ contains a non-zero integer point.
But what happens when the body is not $0$-symmetric? It is easy to see that Minkowski's theorem fails completely, but that's because one is not thinking symplectically. By using some Hamiltonian dynamics of the sort Balacheff and I used to study isosystolic inequalities in this paper, we guessed that the "right" result should be the following:
Conjecture. If a convex body in $\mathbb{R}^n$ contains no integer point other than the origin, then the volume of its dual body with respect to the origin is at least (n+1)/n!
In other words, one should have a sort of uncertainty principle: if the origin is localized as the unique integer point inside a convex body, the dual body cannot be too small. In fact, its volume is bounded below by $(n+1)/n!$. Another formulation  of the conjecture that seems more elementary goes as follows:
If every hyperplane $m_1x_1 + \cdots m_nx_n = 1$, where the $m_i$ are integers not all equal to zero, intersects a convex body $K \subset \mathbb{R}^n$, then the volume of $K$ is at least $(n+1)/n!$
We proved the conjecture in the case $n = 2$ and the asymptotic version:
Theorem. There exists a (universal) constant $C \leq  1$ such that if a convex body $K \subset \mathbb{R}^n$ contains no integer point other than the origin, then the volume of 
$K^*$ is at least $C^n(n+1)/n!$. 
In fact, this result is equivalent to Bourgain-Milman.  Moreover, it easily implies the asymptotic version of a conjecture of Ehrhart:
Theorem. There exists a universal constant $c \geq 1$ such that if $K \subset \mathbb{R}^n$ is a convex body with barycenter at the origin and containing no other integer point, then the volume of $K$ is at most $c^n (n+1)^n/n!$.
However, what is really interesting for us is that at least in the case $n=2$ the result trascends the geometry of numbers and is really a result in Hamiltonian dynamics. I just need a definition: 
Definition. A hypersurface in the cotangent bundle of a manifold $M$ is said to be optical
if its intersection with every cotangent space is a convex hypersurface enclosing the origin. 
To an optical hypersurface in the cotangent of a compact manifold we can associate two numbers: the symplectic volume of the region enclosed by $\Sigma$ and the least action of its periodic characteristics. 
Theorem. An optical hypersurface $\Sigma$ in the cotangent space of the two-torus carries a periodic characteristic whose action is less than or equal to the square root of two-thirds the symplectic volume enclosed by $\Sigma$. 
The inequality is sharp.
Finsler geometers will be happier if I translate: If the Holmes-Thompson volume of a (non-reversible) Finsler $2$-torus $(T^2,F)$ is $3/2\pi$, then $(T^2,F)$ carries a (non-contractible) periodic geodesic of length at most $1$. 
In other words, this is the (non-reversible) Finsler version of Loewner's systolic inequality. The reversible Finsler version (replace $3/2\pi$ by $2/\pi$) is due to Stéphane Sabourau and can be found here.<|endoftext|>
TITLE: Is an ultrafinitist Hilbert's program doomed?
QUESTION [12 upvotes]: Hilbert's program is popularly understood as an attempt to justify infinitary mathematics with a finitary consistency proof.  Godel's Second Theorem is usually considered as showing this is not possible.
Question (to be made precise below):  Can finitary mathematics be justified with a ultrafinitistic consistency proof?
Consider FPA, a multi-sorted first-order theory, with lower-case or small letters (for numbers) and upper-case or big letters for relations of n-arity (n >= 1).  (Practically, I think one can limit the theory to relations where n = 1, 2, or 3.)
Full comprehension is assumed.
FPA has a constant symbol 0, a 1-ary relationship N (natural number), and a 2-ary relationship symbol S (successoring).  
In this context the Peano Axioms can be written:
(PA1)  N0
(PA2)  $\forall$n  (Nn $\Rightarrow$ $\exists$m (Nm & Sn,m))
(PA3)  $\forall$n$\forall$m$\forall$m' (Nn & Nm & Nm' & Sn,m & Sn,m' $\Rightarrow$ m = m')
(PA4)  $\forall$n$\forall$m$\forall$n' (Nn & Nm & Nn' & Sn,m & Sn',m $\Rightarrow$ n = n')
(PA5)  $\forall$n (Nn $\Rightarrow$ $\neg$ Sn,0)
(PA6)  $\forall$P (P0 & $\forall$n$\forall$m(Pn & Sn,m $\Rightarrow$ Pm) $\Rightarrow$ 
 $\forall$n(Nn $\Rightarrow$ Pn))
FPA assumes all the Peano Axioms except (PA2), that is, everything except the totality of the successor relationship. It has as its standard models all the initial segments as well as the standard model of the natural numbers. {0} is a model. It is therefore agnostic as to whether the natural numbers go on and on. Let's call it "ultrafinitistic" even if it's not always clear to me what "ultrafinitistic" means.
Now an ultrafinitistic Hilbert's Program might be the following.  Let E(n) be the wff
     $\exists x_1 \exists x_2 ... \exists x_n$(N$x_1$ & S0,$x_1$ & N$x_2$ & S$x_1,x_2$ & ... & S$x_{n-1},x_n$)
i.e. the assertion that the number n exists.  The ultrafinitistic hope would be that FPA can prove the consistency of FPA + E(n) for any n, or even (this would be magical) prove the assertion
$\forall$n Cons(FPA + E(n)).
Now in one sense of consistency, where like Godel one uses numbers to represent sequences, one can actually get started on this.  Restricting the arities of the upper-case letters to no more than 3 (which is sufficient to develop the appropriate apparatus), it seems that FPA can prove the Godel consistency of FPA + E(1) and FPA + E(2). 
However, the proof works with a certain cheat; Godel numbering uses numbers so big that the assumption that there is a proof leading to a contradiction implies the existence of a truly big number, which provides enough space for creating a model of true-in-{0,1} or true-in-{0,1,2} for propositions whose length are <= the length of the longest proposition appearing in the proof of contradiction.  A more appropriate manner of representing consistency would surely be to use the upper-case letters, since then a sequence of length n only implies the existence of n, which is obviously much smaller than the Godel number.  Let RCons(FPA) and the like represent this notion of consistency.
According to 
Can FPA really prove its consistency?
whether FPA can prove RCons(FPA) is equivalent to a well-known open problem which is conjectured to be false.  Obviously, this does not bode well for FPA proving stronger systems to be RCons consistent. Still, "open" brings hope, so my questions are:
(1) Can it be shown, for any n, that FPA does not prove RCons(FPA + E(n))?
(2) If so what is the smallest n?
(3) If not can it be shown that FPA cannot prove $\forall$n Cons(FPA + E(n)) ?
If something significant can be said with a different formula asserting the existence of a number other than E(n) (e.g. defined using an exponential), that would obviously be of interest. (EDITED ADDITION:  The restriction of upper-case letters to arity 3 or less is welcome, especially if a "yes" answer plays on unbounded arity.)
Since this is my third question about this theory, I would also like here to refer to this question/answer, where the models (and other details) of FPA are described:
Provability in Second-Order Arithmetic without the Successor Axiom
so with one search I can later find everything.
Thanks for your time.

REPLY [6 votes]: Ulrich Kohlenbach with his coworkers in proof mining has perhaps the best modern nonfatalist position about Hilbert program. Rather than stressing over Gödel's demolition of Hilbert's ideal, we can analyze the extent to which "non-finitary" methods play a role. This has lead to some very nice results and applications of proof theory. Lecture 1 of Ulrich's notes
on "Proof mining" provides a quick but more careful explanation of the idea that proof mining is the natural outcome of Hilbert's program.<|endoftext|>
TITLE: A question on the Picard group
QUESTION [5 upvotes]: Let $X$ be a simply connected smooth projective variety, whose Picard group is generated by the classes of the irreducible codimension 1 loci $D_1, \ldots, D_k$. Let $E_1, \ldots, E_r$ be other irreducible codimension 1 loci, and suppose that $X^0$ is the complement in $X$ of the divisors $D_i$ and $E_j$.
Suppose now that $X_0$ is the complement of $n$ irreducible loci of codimension $1$ in $Y$, a smooth projective variety.
Question: Can I conclude that the Picard group of $Y$ has rank $n-r$?
I can answer the question affirmatively over $\mathbb{C}$, by using the long exact sequence with compact support associated with the inclusion $Y \setminus X^0 \to Y$, but I would like to know if there is an algebraic proof of this (valid over any algebraically closed field $k$).
EDIT: As pointed out in the answer, I am actually assuming that the Picard group of $X$ is FREELY generated by the $D_1, \ldots, D_k$.

REPLY [2 votes]: I'm going to assume that $X^0$ is the same as $X_0$ and that the Picard group of $X$ is freely generated by the $D_i$, since without that the question doesn't make much sense. In this cases, the answer is yes. This is the same as Mohan's answer, but with a little more detail.
The relevant tool is that if $Z$ is a normal variety and $D$ is an irreducible divisor in $Z$, there's an exact sequence:
$$
0 \rightarrow \mathcal O(Z)^* \rightarrow \mathcal O(Z \setminus D)^*
\rightarrow \mathbb Z \rightarrow
\operatorname{Cl}(Z) \rightarrow \operatorname{Cl}(Z \setminus D) \rightarrow 0
$$
Here $\mathcal O(Z)^*$ denotes the ring of global functions on $Z$.
I'm writing $\operatorname{Cl}$ instead of $\operatorname{Pic}$ because this applies to the Weil divisor group of an arbitrary normal variety. The right part of this exact sequence is Proposition II.6.5 in Hartshorne and it's not hard to see that the left part is also exact.
What this means is that we can define the ``virtual class rank'' of a variety $Z$ over a field $K$ to be
$$\operatorname{rank} (\operatorname{Cl}(Z)) - \operatorname{rank}(\mathcal O(Z)^* / K^*),$$
at least when these ranks are finite. Then, the exact sequence tells us that removing an irreducible divisor drops the virtual class rank by exactly 1. In your case, $X$ is projective, so it has no non-constant regular functions, so its virtual class rank is the rank of its class group, $k$. By induction, the virtual class rank of $X^0$ is $-r$ and for $Y$ it is $n-r$, which is again the rank of the class group of $Y$, because it is projective.
This only tells you about the rank of the torsion-free part of $\operatorname{Pic}(Y) = \operatorname{Cl}(Y)$. If part of your question was whether this Picard group is torsion-free, I'm not sure what you can say about that.<|endoftext|>
TITLE: Another proof of the bidisc and the ball are biholomorphically inequivalent?
QUESTION [7 upvotes]: Does this outline of a proof work?
Consider the ball and the bidisc in $\mathbb{C}^2$.  Give each space its Bergman metric.  To show that the ball and the bidisc are not holomorphic, it is enough to show that they are not isometric. 
One way to distinguish the two spaces is their sectional curvature.  I think I have shown that the sectional curvature of the Bergman metric of the ball is constant and negative, whereas the sectional curvature of the bidisc is nonpositive non constant. For example in the plane generated by the vectors $\langle 1,0 \rangle$ and $\langle 0,1 \rangle$ the section curvature is 0, but in the plane generated by $\langle 1,0\rangle$ and $\langle i,0 \rangle $ the sectional curvature is negative.
Is this true?  Is there anything subtle I might have missed?  I have seen a lot of pretty convoluted proofs of this fact, and I would think that this basic outline would be recorded in print somewhere if it is true, but I cannot seem to find it.

REPLY [6 votes]: In general, the ball $B^n$ with the Bergmann metric is isometric to the Hermitian symmetric space $SU(1,n)/S(U(1)\times U(n))$ where $SU(1,n)$ denotes the special pseudo-unitary group 
with respect to the indefinite Hermitian inner product with signature $(1,n)$ and $S(U(1)\times SU(n))$ denotes the subgroup of block matrices of sizes $1\times 1$ and $n\times n$. Note that this subgroup is the fixed point set of the involution of $SU(1,n)$
given by conjugation by the diagonal matrix with entries $(-1,1,\ldots,1)$. 
For an arbitrary Hermitian symmetric space given as $G/K$ where $G$ is connected and $K$ 
is a symmetric subgroup (i.e. open subgroup in the fixed point set of an invoutive automorphism of $G$), the identity components of the isometry group and the group of holomorphic automorphisms both coincide with $G$. Hence in the case of $B^n$ this group is $SU(1,n)$. 
Next, the $n$-polydisc is the product $B^1\times\cdots\times B^1$ ($n$-times) and it
carries the structure of a Hermitian symmetric space, product of $n$ copies of 
$SU(1,1)/S(U(1)\times U(1))\cong SL(2,\mathbb R)/SO(2)$, the unit disk in $\mathbb C$. 
As such, the identity component of its group of holomorphic automorphism is $SU(1,1)\times\cdots SU(1,1)$ ($n$ copies). It is readily seen that this group is not isomorphic to $SU(1,n)$ if $n>1$ (for instance, they have different dimensions, $3n$ and $n^2+2n$).<|endoftext|>
TITLE: What is the Weitzenböck formula for the $\bar\partial$-Laplacian
QUESTION [6 upvotes]: It is well-known that the Weitzenböck formula for the real Laplacian is
$$\frac12 Δ|∇f|2=|Hessf|2+⟨∇f,∇Δf⟩+Ricci(∇f,∇f)$$
where $Hess$ denotes the Hessian tensor of $f$. and $\nabla f$ denotes the gradient vector of $f$, $Ricci$ denotes the Ricci curvature of the manifold $M$.
If $\Delta_{\bar\partial}$ denotes the $\bar\partial$-Laplacian, it is well-known that it is half of the real Laplacian. So I am wondering is there any formula of the Weitzenböck formula in complex coordinates. (Assume the manifold is Kähler). Apparently one can devided the above formula by 2 to the one, but the expression I want should be expressed by $f_{i\bar j}$ and etc. 
ps. The Comparison Geometry of Ricci Curvature, by Shunhui Zhu, 221-262 had a very nice introduction to this formula in real case.
http://library.msri.org/books/Book30/contents.html
However I am not familiar with Kaehler case, for example, I dont know the such a formula can be derived in the same fashion as in Zhu's paper? Any book or paper with detailed calculation would be helpful.

REPLY [2 votes]: Weitzenböck Identities for Kahler manifolds is as follows
Let $\Theta $ be the Chern curvature form of $h$ for the Hermitian line bundle over Kahler manifold $(X,\omega)$
For any smooth section $\xi ∈ Γ(T^{0,1}X \otimes L)$ we have, 
$$(\bar{\partial^*}\bar\partial+\bar\partial\bar{\partial^*})\xi=\bar{\nabla^*}\bar\nabla\xi+(\Theta+Ric(\omega))(\xi,.)$$
and
$$(\bar{\partial^*}\bar\partial+\bar\partial\bar{\partial^*})\xi={\nabla^*}\nabla\xi+(\Theta(\xi,.)-(tr_\omega\Theta)\xi$$
Note that here Ricci make sense as follows
If we take a section $s\in H^0(X,L)$ and $\xi=\alpha_{\bar i}d\bar z^i\otimes s$ then Ricci can be defined as follows
$$Ric(\xi,.)=R_{i\bar j}\alpha_{\bar i}d\bar z^j\otimes s$$
I learn it from paper of Tian<|endoftext|>
TITLE: Metrization of weak convergence of signed measures
QUESTION [20 upvotes]: Edit: Changed from "Hausdorff" to "metric" spaces.
Let $\mathcal{M}(\Omega)$ denote the space of signed regular Borel measures on a compact metric space $\Omega$. By Riesz-Markov, this is the dual space of $C(\Omega)$, the space of all continuous real valued functions on $\Omega$. Denote by $$\mathcal{P}(\Omega) = \{\mu\in\mathcal{M}(\Omega)\ :\ \mu\geq 0, \mu(\Omega)=1\}$$
i.e. the set of all probability measures in $\mathcal{M}$.
The weak convergence (also called weak* convergence) in $\mathcal{M}(\Omega)$ is defined by duality and it is known that weak convergence in $\mathcal{P}(\Omega)$ can be metrizised by, e.g. the Prokhorov metric $d_P$ or the Wasserstein metrics $d_W$.
Obviously, both metrics do not metrizise weak convergence on $\mathcal{M}(\Omega)$: For the Wasserstein metric we have $d_W(\mu,\nu)=\infty$ if $\mu(\Omega)\neq\nu(\Omega)$ and for the Prokhorov metric we do not even have $d_P(\mu,\mu)=0$, as far as I see.
Googling and searching MSC did not produce any results on my question:

Are there any metrics available which metrizise weak convergence of signed regular Borel measures?

I would be surprised if there weren't (or are there any fundamental obstructions?).
I would also be happy with metrics for weak convergence of non-negative measures (but not normed ones) or for uniformly bounded measures and would also like to know the answer to the same question for vector valued Radon measures on a metric space.
2dn edit: Thanks for the great answers. I had forgotten the general procedure to define a metric for weak(*) convergence on bounded set, but in fact I had a more "geometric" metric in mind, something in the direction of R Ws and Dans answers.

REPLY [2 votes]: If you need a reference, there is a paper that covers your question.
Varadarajan (1958): If $\Omega$ is a separable metric space, then the topology of weak convergence on $\mathcal{M}(\Omega)$ is metrizable if and only if the weak convergence and norm topologies on $\mathcal{M}(\Omega)$ coincide.
This condition is obviously violated if $\Omega$ is the unit interval (or any other uncountable separable metric space).
The same paper also shows that the weak convergence topology on the subset $\mathcal{M}^+(\Omega)$ of positive measures is metrizable.<|endoftext|>
TITLE: On Weil's characters of type (A)
QUESTION [5 upvotes]: In Weil's paper 
"On a certain type of characters of the idele-class group of an algebraic number field", 
Weil introduces a class of characters on the Idele class group (of not necessarily finite order) which take algebraic values. He calls them, characters of type (A) (the (A) probably stands for algebraic, I guess). On the last 3 lines of p. 3, he says that Artin pointed out to him how to use Minkowski's theorem on units (in absolutely normal number fields) to show essentially that all such characters come from CM fields. 
Basically, this question boils down to understand the following: For which 
number fields $K$; collection of integers $f_{\iota}$; and a suitable choice of an integer $m$, is it possible to have
$$
\prod_{\iota\in S_{\infty} }\left(\frac{\iota(\epsilon)}{\overline{\iota(\epsilon)}}\right)^{mf_{\iota}}=1
$$
for all $\epsilon\in \mathcal{O}_K^{\times}$? Here $S_{\infty}$ denotes the set of infinite places of $K$.
Q1: Is there a reference in the literature where I can find a proof (or a sketch of a proof) of this result?    
Q2: If no such reference exists, then how does one prove it from Minkowski's theorem?

REPLY [7 votes]: Now that I've had the chance to look at Fargues's proof carefully, it seems incomplete for what it claims to prove, but it might answer the question under consideration. 
Suppose that you have a collection of integers $f_{\iota}$ indexed by embeddings $\iota:K\hookrightarrow\mathbb{C}$ and satisfying for some integer $m$ (this is slightly different from the OP's identity):
$$\prod_{\iota}\iota(\epsilon)^{mf_{\iota}}=1,\text{ for all $\epsilon\in\mathcal{O}_K^\times$}.$$
Then I claim that there exists a CM sub-field $L\subset K$ such that $f_{\iota}$ only depends on $\iota\vert_L$ (for the purposes of this statement, totally real fields are CM). 
The assertion that allows us to reduce this to Dirichlet's unit theorem is the following: It suffices to show that, for all $\sigma\in Aut(\mathbb{C})$, $f_{\sigma\iota}+f_{\sigma\overline{\iota}}$ is independent of $\iota$ and $\sigma$.
(EDIT: It was incorrectly stated in the original version that it was enough to require independence from $\iota$ alone). 
To see this, we reinterpret this condition as follows: After fixing an embedding $\bar{\mathbb{Q}}\hookrightarrow\mathbb{C}$, we can view the collection $(f_{\iota})$ as an element $f$ of the $G_{\mathbb{Q}}$-module of maps $Hom(K,\bar{\mathbb{Q}})\to\mathbb{Z}$. The Galois action is given by $(\sigma(f))_{\iota}=f_{\sigma^{-1}\iota}$. From this point of view, our claim amounts to: For every complex conjugation $c$ of $\bar{\mathbb{Q}}$, $f_{\iota}+f_{c(\iota)}$ is independent of $c$ and $\iota$. In particular, for all $\sigma\in G_{\mathbb{Q}}$, and all complex conjugations $c$, we have:
$$
 f_{\sigma^{-1}\iota}+f_{c(\sigma^{-1}\iota)}=f_{\iota}+f_{c\iota}.
$$
Another way to write this is, for all $\iota$:
$$ 
 \sigma(f)_{\iota}+\sigma(c(f))_{\iota}=f_{\iota}+c(f)_{\iota}.
$$
So $\sigma$ stabilizes $f$ if and only if it stabilizes $c(f)$, for all complex conjugations $c$. Let $L$ be the fixed field of the stabilizer of $f$ in $G_{\mathbb{Q}}$. We have shown that $L$ is stabilized by all complex conjugations. To show that $L$ is CM we need to now need to know that all complex conjugations have the same action on $L$. This amounts to showing that $f_{c(\iota)}$ does not depend on the choice of $c$. But, since $f_{\iota}+f_{c(\iota)}$ does not depend on $c$ by hypothesis, this is clear.
Let us return to the first identity above. Taking the logarithm of its absolute value gives us:
$$
 \sum_{\iota}f_{\iota}\vert \iota(\epsilon)\vert=0.
$$
But consider the map
$$
 \ell:\mathcal{O}_K^\times\otimes\mathbb{R}\rightarrow\oplus_{v\vert\infty}\mathbb{R}.
$$
given by $\ell(\epsilon\otimes 1)_v=\log\vert \epsilon\vert_v$ if $v$ is real and by $\ell(\epsilon\otimes 1)_v=2\log\vert \epsilon\vert_v$ if $v$ is complex. The indexing set here is the set of inequivalent archimedean norms on $K$. Then Dirichlet's unit theorem says that $\ell$ is an isomorphism onto the hyperplane $H$ where the sum of the co-ordinates is identically $0$. 
This shows that, for all $(b_v)\in H$,
$$\sum_{v}\frac{f_{\iota(v)}+f_{\overline{\iota}(v)}}{2}b_v=0.$$
Here, if $v$ is complex, $\iota(v)$ and $\overline{\iota}(v)$ are the two complex embeddings inducing $v$. If $v$ is real, they're the same embedding. If one uses the usual basis of $H$ consisting of differences of the basis vectors for the ambient space, one finds that $f_{\iota}+f_{\overline{\iota}}$ is independent of $\iota$. Applying the same argument to the character $\epsilon\mapsto\prod_{\iota}\sigma^{-1}(\iota(\epsilon))^{f_{\iota}}=\prod_\iota\iota(\epsilon)^{f_{\sigma\iota}}$, one sees that $f_{\sigma\iota}+f_{\sigma\overline{\iota}}$ is independent of $\iota$ as well. 
(EDIT: The remainder of the proof has been changed to reflect the correction in the criterion above.)
Set $w(\sigma)=f_{\sigma\iota}+f_{\sigma\bar{\iota}}$: this doesn't depend on $\iota$. To finish the proof, we must show that it is independent of $\sigma$ as well. First, let $L$ be as above. From what we have proven already, we find that $L$ is stable under all complex conjugations. Assume that $L$ is totally complex; this implies that $K$ is also totally complex. Then, if $n=[K:\mathbb{Q}]$, we have:
$$
 2nw(1)=\sum_{\iota}(f_{\iota}+f_{\bar{\iota}})=\sum_{\iota}(f_{\sigma\iota}+f_{\sigma\bar{\iota}})=2nw(\sigma).
$$
So, in this case, the independence is clear. 
To finish, it is enough to show the following: If $L$ has one real place, then $L=\mathbb{Q}$. But the hypothesis implies that there is a complex conjugation $c$ that acts trivially on $L$, and so $2f_{\iota}=f_{\iota}+f_{c(\iota)}$ is independent of $\iota$. In this case, the Hecke character must be the twist of a finite order character by a power of the norm character.<|endoftext|>
TITLE: Boolean non-hypercomplete $(\infty,1)$-toposes
QUESTION [10 upvotes]: Let's say that an $(\infty,1)$-topos is Boolean if for every object $X$, the lattice $Sub(X)$ of subobjects (i.e. $(-1)$-truncated morphisms into $X$) is a Boolean algebra.  I think this is equivalent to asking that the subobject classifier be an internal Boolean algebra, and hence to asking that the underlying 1-topos of 0-truncated objects is Boolean in the classical sense.
In particular, the $(\infty,1)$-topos of sheaves on a topological space $X$ is Boolean if and only if the lattice of open sets in $X$ is a Boolean algebra, i.e. every open set is also closed.  Thus, Booleanness is a sort of "zero-dimensionality" condition.  Lurie shows in Higher Topos Theory that other sorts of "finite-dimensionality" conditions imply that an $(\infty,1)$-topos is hypercomplete.  At the moment, however, I don't see whether Booleanness implies any of these other conditions.
Thus my question is: can a Boolean $(\infty,1)$-topos fail to be hypercomplete?

REPLY [10 votes]: If $G$ is a profinite group, then the topos of sets with a continuous $G$-action is
Boolean, but the associated $\infty$-topos is usually not hypercomplete.<|endoftext|>
TITLE: Fourier Coefficients and Hölder Continuity
QUESTION [13 upvotes]: Suppose we are given the Fourier coefficients of an $L^2$ function on the circle.  Are there necessary and sufficient conditions on the coefficients that allow us to determine that $f$ is Hölder continuous of order $\alpha$? 
Note that the necessary condition $|\hat{f}(n)| \leq C_f|n|^{-\alpha}$ is not sufficient. For example if $\hat{f}(n)=|n|^{-2/3}$ for all $n$ then $f$ is an $L^2$ function whose Fourier series does not converge absolutely.  Therefore $f$ cannot be Holder continuous of order $\alpha>1/2$.

REPLY [11 votes]: There is an excellent characterization of Hölder spaces via the Fourier transform, using Besov spaces. Let $\alpha\in (0,1)$: a function $u$
defined on $\mathbb R^n$ belongs to $L^\infty\cap C^\alpha$ if and only if it belongs to $B^\alpha_{\infty,\infty}$, i.e.
$$
\sup_{\nu\in \mathbb N}2^{\nu\alpha}\Vert\phi_\nu(D_x) u\Vert_{L^\infty}<+\infty,\quad\text{i.e.
the sequence} (2^{\nu\alpha}\Vert\phi_\nu(D_x) u\Vert_{L^\infty})_{\nu\in \mathbb N}
\in \ell^\infty.
$$
Here $\phi_\nu$ stands for a Littlewood-Paley decomposition:
$$
1=\sum_{\nu\in \mathbb N}\phi_\nu(\xi),
$$
$\phi_0$ is compactly supported and for $\nu\ge 1$,
$\phi_\nu(\xi)=\phi(2^{-\nu}\xi)$ where $\phi$ is supported in the ring 
$1/2\le \vert\eta\vert\le 2$ so that $\phi_\nu$ is supported in the ring
$2^{\nu-1}\le \vert\xi\vert\le 2^{1+\nu}$.<|endoftext|>
TITLE: General recipe for building C*-algebras out of combinatorial object
QUESTION [9 upvotes]: I want to ask what should be a nice way to build C*-algebras out of objects like groups, inverse-semigroups, semigroups, ringgs or graphs. I know there are well known construction of C*-algebras out of those objects; but I want to understand what philosophy lies under the recipe. 
Say for discrete groups I know group C*-algebras are made of as universal object of unitary elements of groups elements with relations coming from group operation. Now what I understand is they take unitary because if we fix an element $g$ in the group $G$, the map $T_g:G\rightarrow G$ defined by $T_g(h)=gh$ is one-one onto morphism. For Ring C*-algebras also this works. In that case where the map is not onty they take isometry. But my confusion starts with graph C*-algebra. I don't understand why they take projections for vertices and partial isometrys for edges (and the given relations).

REPLY [6 votes]: Benjamin's answer nicely describes where the relations for graph $C^\ast$-algebras come from.
Here is an attempt to answer the question suggested by the title: what is the general recipe for constructing $C^\ast$-algebras from other mathematical objects.  This may not work for every type of $C^\ast$-construction, but it seems to be the general formula behind at least group $C^\ast$-algebras, crossed products, graph $C^\ast$-algebras (and more generally, $C^\ast$-algebras of inverse semigroups), semigroup $C^\ast$-algebras, and ring $C^\ast$-algebras.
To the mathematical object in question is associated a natural and concrete Hilbert space and a natural collection of operators on that Hilbert space (often called the regular representation).  Take the obvious $*$-algebraic relations that hold between these operators; then these are the relations used to define the universal $C^\ast$-algebra.
Here is this formula worked out for a few examples:
Group $C^\ast$-algebras: For a discrete group $G$, the natural concrete Hilbert space is $\ell^2(G)$ (with canonical ONB $(\xi_g)_{g \in G}$), and to each group element $g \in G$ is associated the operator $\lambda_g$ defined by
$$ \lambda_g(\xi_h) = \xi_{gh}. $$
The obvious relations here are, that $\lambda_g$ is a unitary and $\lambda_g\lambda_h = \lambda_{gh}$; that is to say, that $\lambda$ is a group homomorphism between $G$ and the unitary group of $\mathcal{B}(\ell^2(G))$.
Of course, the $C^*$-algebra generated by $\{\lambda_g\}$ is called the reduced $C^\ast$-algebra of $G$, while the universal one (generated by $\{u_g: g \in G\}$ satisfying the relations that they are unitary and $u_gu_h = u_{gh}$) is called the (full) group $C^*$-algebra of $G$.
Dynamical systems: Though this can be done more generally, let's stick to a discrete group $G$ acting by homeomorphisms $\alpha_g$ on a compact metric space $X$.  The natural concrete Hilbert space is $\ell^2(G \times X)$ (with ONB $(\xi_{g,x})_{g \in G, x \in X}$).  For each $g \in G$, we may associate the operator $\lambda_g$ defined by
$$ \lambda_g(\xi_{h,x}) = \xi_{gh,x}. $$
For each $f \in C(X)$, we may also associate the operator $D_f$ defined by
$$ D_f(\xi_{g,x}) = f(g^{-1}x)\xi_{g,x}. $$
The obvious relations are: that $g \mapsto \lambda_g$ is a group homomorphism from $G$ to the unitary group; that $f \mapsto D_f$ is a $*$-homomorphism; and that
$$ \lambda_g D_f \lambda_g^* = f \circ \alpha_g. $$
Graph $C^\ast$-algebras: The details of this construction are more or less explained in Benjamin's answer.  The concrete Hilbert space has an orthonormal basis $(\xi_{\pi})$ indexed by the nonempty paths in the graph.  An edge $e$ gives rise to the operator $v_e$ given by
$$ v_e(\xi_{\pi}) = \begin{cases} \xi_{e\pi},\ &\text{if $e\pi$ is a path,} \\
 0,\ &\text{otherwise.} \end{cases} $$
A vertex $v$ gives rise to the operator $p_v$ which is the projection onto the span of 
$$ \{\xi_{\pi}: \pi \text{ is a non-empty path beginning at $v$}\}. $$<|endoftext|>
TITLE: Equivariant Cohomology for actions with finite stabilizers
QUESTION [5 upvotes]: Let $X$ be a reasonable topological space (let's say it has the homotopy type of a CW complex) and let $G$ be a topological group acting on that space. Let $E_G \rightarrow B_G$ be the universal bundle. Then the equivariant cohomology $H^*_G(X)$ is defined as $H^*(E_G \times_G X)$. We call the space $E_G \times_G X$ the homotopy quotient space. We always have a map $E_G \times_G X \rightarrow G \backslash X$, where $G \backslash X$ denotes the naive quotient space. This induces a map $H^* (G\backslash X) \rightarrow H^*_G(X)$. 
I'm curious about the following proof. From now on, we will take cohomology with rational coefficients. Suppose the action of $G$ on $X$ has finite stabilizers. Then the map $H^* (G\backslash X) \rightarrow H^*_G(X)$ is in fact an isomorphism.  This is proved in the following notes by Michel Brion (http://www-fourier.ujf-grenoble.fr/~mbrion/notesmontreal.pdf, see p. 4). The proof is as follows: the fibers of the map $E_G \times_G X \rightarrow G \backslash X$ are of the form $E_G/G_x$, where $G_x$ is the stabilizer of a point $x \in X$. Since $G_x$ is a discrete group, we have $\pi_1 (E_G/G_x) = G_x $ and vanishing of all higher homotopy groups. Since $G_x$ is finite, all the homotopy groups vanish when we tensor with $\mathbb{Q}$. I think this is what Brion means by "$\mathbb{Q}$-acyclic".
Then it is claimed that this is sufficient to prove that the map $H^* (G\backslash X) \rightarrow H^*_G(X)$ is an isomorphism. Why is this true? Could someone please explain this argument to me?
Even if the fibers were contractible, it seems to me that you would need this map to be a Serre fibration to prove a homotopy equivalence. Is it true that any surjective map with contractible fibers is a homotopy equivalence?

REPLY [5 votes]: For ease of referencing I'll prove the stated isomorphism under the hypothesis that 
$G$ is a compact Lie group and $X$ is a finite dimensional compact topological $G$-manifold. 
By the hypothesis on $G$ we can choose $EG$ to be a dimension-wise finite CW complex. The 
$(n+1)$-skeleton $E$ is compact, $n$-connected and for fixed $m$ and $n$ large enough, the inclusion $E \hookrightarrow EG$ induces an isomorphism 
$$H^m(EG\times_G X,\mathbb{Q}) \xrightarrow{\cong} H^m(E\times_G X,\mathbb{Q})$$
(that's 4) on p. 4 of the linked paper in the question). Hence it suffices to show that the map 
$E\times_G X \to X/G$ induces an isomorphism in rational cohomology in degree $m$. 
But this follows from the Vietoris-Begle theorem [Q, Corollary A.7] that applies if we have shown that $f$ is closed and the fibres of $f$ are compact, n-acyclic and relative Hausdorff in $E\times_G X$: 
Closedness of $f$ and compactness of the fibres $E/G_x$ are obvious, $n$-acyclity follows 
from $H^i(E/G_x,\mathbb{Q})=H^i(G_x,\mathbb{Q})=0$ for $0< i< n$.  Relative Hausdorff 
means different points in the fibre have disjoint neighborhoods in $E \times_G X$. This 
holds since $E \times_G X$ is Hausorff. QED
[Q] Quillen: The Spectrum of an Equivariant Cohomology ring: I 

The cited Vietoris-Begle theorem states: 

If $f: X \to Y$ is closed and its fibres are compact, relatively Hausdorff in $X$ and $n$-acyclic, i.e. $H^i(f^{-1}(y),k)=H^i(\ast,k)$ for $i< n$ ($k$ any constant coefficients), then $f^\ast:H^i(Y,k) \xrightarrow{\cong} H^i(X,k)$ for $i \lt n$.<|endoftext|>
TITLE: Analytic function avoiding elements of the modular group
QUESTION [10 upvotes]: A friend recently told me the following two facts, for which he cannot recall a proof or a reference
(but he remembers seeing them in the literature):

Let $f$ be a holomorphic function mapping the upper half-plane $H$ into itself. Let $G$ be the
group of fractional linear transformations $(az+b)/(cz+d)$ where $ad-bc=1$, $a,d$ are odd integers
and $b,c$ are even integers. Suppose that for every $g\in G$ and for every $z\in H$,
$f(z)$ is not equal to $g(z)$. Then $f$ is fractional-linear. (Or maybe such $f$ just does not exist).
Let $f$ be the same as before. Suppose that for all integers $m,n$ and all $z\in H$
we have $f(z)\neq mz+n$. Then $f$ is fractional-linear.

I will appreciate any relevant reference or any other information. 
EDIT: There is no $f\in Aut(H)$ that satisfies the condition of Problem 1. This implies that $f$ constructed by
 Aakumadula is NOT fractional-linear.
To prove this, we write $(az+b)/(cz+d)=(xz+y)/(uz+t)$, where $a,b,c,d$ are given real numbers,
and we want to find integers $x,y,u,t$, where $x,t$ are odd, and $y,u$ are even, so that this has non-real roots
$z$. This is to show that certain quadratic form in $a,b,c,d$ is indefinite. And this is performed
by an elementary calculation.

REPLY [6 votes]: Concerning question 2 see  Earle, Clifford J. On holomorphic families of pointed Riemann surfaces. Bull. Amer. Math. Soc. 79 (1973), 163–166 and Theorem 3 there.<|endoftext|>
TITLE: Is there any need to study Coxeter systems (W,S) with S infinite?
QUESTION [33 upvotes]: In their treatise Groupes et algebres de Lie, Bourbaki (no doubt heavily influenced by Tits) devoted Chapter IV (1968) to the general theory of what they dubbed "Coxeter systems" $(W,S)$ along with "Tits systems" (BN-pairs).  Here $S$ is an arbitrary set and $W$ a group generated by a subset $S$ consisting of elements of order 2, subject only to obvious relations involving pairs of generators.   This is a very large class of groups, usually infinite, which includes finite reflection groups and others of interest in Lie theory.   The axiomatic development in IV.1 doesn't require any restriction on the rank of the group: the cardinality of $S$.   
On the other hand, there seem to be almost no significant examples in which the rank is infinite.  As Bjorner and Brenti note in their book Combinatorics of Coxeter Groups, after defining Coxeter groups: "Most groups of interest will have finite rank."  Typical examples given by them and others do include the group of permutations of the positive integers which leave all but finitely many fixed; this is a direct limit of finite symmetric groups (and embeds in the much larger "infinite symmetric group").   But although the general theory applies in all ranks, it's hard for me to think of anything really new one learns about  infinite rank Coxeter groups using Coxeter theory.   Maybe I haven't looked far enough, but it's natural to ask:

Are there significant results about Coxeter groups of infinite rank which aren't obtained just as easily without Coxeter theory?

REPLY [7 votes]: I don't think this will answer your direct question either, but it is still worth mentioning. The following fact was explained to me by a combination of Lusztig and Geordie Williamson, and it motivates why one should study (completions of) simply-laced Coxeter groups $(W',S')$ with $S'$ infinite.
There is a unique 2-sided cell $C$ in any Coxeter group $(W,S)$, consisting of all non-identity elements with a unique reduced expression. Any such element has a unique simple reflection in its right (resp. left) descent set.  Within this cell, the left cells $C_s$ are parametrized by $s \in S$, and consist of all elements with a unique reduced expression and with right descent set $\{s\}$.
One can take the elements of $C_s$ and give them the structure of a graph, where $w$ is connected to $v$ if $w = tv$ for some $t \in S$. Each vertex can be labeled with the unique element of its left descent set. The resulting labeled graph does not depend on the choice of $s$.
One can view this graph as encoding a simply-laced Coxeter group $(W',S')$. Suppose that $S'$ is finite. For each $t \in S$, the reflections labeled by $t$ all mutually commute, and their product is an involution in $W'$. Together, these involutions generate a subgroup inside $W'$ which is isomorphic to the original Coxeter group $W$. In this way, any finite Coxeter group is canonically embedded inside a simply-laced one. Any Coxeter element of $W$ is sent to a Coxeter element of $W'$.
For example, this operation will produce the embedding of $H_4$ inside $E_8$. It will produce the embedding of a finite dihedral group $I_2(m)$ into $A_{m-1}$, sending each simple reflection in $I_2(m)$ to a product of every other simple reflection in $A_{m-1}$.
However, it may happen that $S'$ is infinite, and that the collection of vertices labeled by $t \in S$ is also infinite. In this case, one does not have an embedding of $W$ into $W'$, because each simple reflection would have to go to an infinite product. Though I have not seen it defined and don't know the literature, I imagine there is an embedding of $W$ into some suitable completion of $W'$.
For instance, this gives the embedding of $I_2(\infty)$ into $A_{\infty}$, sending each simple reflection in $I_2(\infty)$ to the product of every other simple reflection in $A_{\infty}$.
Anyway, the upshot of all this is that one could potentially study arbitrary Coxeter groups (with $S$ finite) using only (completions of) simply-laced ones (with $S'$ possibly infinite).<|endoftext|>
TITLE: When does $Aut(X)=Bir(X)$ hold?
QUESTION [12 upvotes]: Let $X$ be a projective complex manifold. Under what condition do we have the equality $Aut(X)=Bir(X)$? Here $Aut(X)$ denotes the group of holomorphic automorphisms of $X$ and $Bir(X)$ the group of birational morphisms of $X$.
I am interested in the case when $\dim_{\mathbb{C}}X=2,3$. Maybe there are not universal criteria, so I would appreciate your providing me with any examples for which the equality holds.

REPLY [9 votes]: A simplest example is a curve $C$, then $Bir(C)\cong Aut(C)$
A careful note that $Bir(X)$ is not a group scheme in general.Moreover if $X$ and $Y$ are birationl then $Bir(X)$ is not isometric with $Bir(Y)$
So we need to add some condition on $X$ in the sense of Minimal Model Program such that in scheme group theoretic sense $Bir(X)$ good behave since we have luck of well defined multiplication rule. In fact is $X$ be a minimal model with terminal singularities it is a theorem that $Bir(X)$ is a group scheme as an Abelian variety. In general $$Aut^o(X)\cong Bir^o(X)$$
where $Aut^o$ is the identity connected component. See this paper 
Let $X$ be a projective variety then we have the following isomorphism from MMP
$$Aut(X_{min})\cong Bir(X_{min})$$ where $X_{min}$ is the minimal model of projective variety $X$
For any surface of non-negative Kodaira dimension, we have
$$Aut(X) ⊂ Bir(X) \cong Bir(X_{min}) \cong Aut(X_{min})$$
If $X$ be a Fano variety of $dim\geq 5$ and Picard group is generated by anticanonical divisor of the variety $X$, then it is conjecture that $Aut(X)\cong Bir(X)$
Moreover for moveable log pair $(X,M_X)$ which is log Calabi-Yau pair, with at worst canonical singularities then $Aut(X)\cong Bir(X)$<|endoftext|>
TITLE: QVH characterization of virtually special groups
QUESTION [5 upvotes]: Agol's recent VHC paper gave a characterization of virtually special groups in terms of being $\mathcal{QVH}$. He remarks that this may be taken as the defining property of virtually special groups which can be used in this paper (and so presumably, in the proof of the the main theorem). 
My concern: Recall that the goal is to show that given a cube complex $X$ and hyperbolic group $G$ (satisfying the hypotheses of the theorem) there exists a finite index subgroup $G'$ such that $X/G'$ is special. Now if one was to just show that $G$ is in $\mathcal{QVH}$ (and hence virtually special), then we would know that it acts virtually specially on some cube complex $Y$. But a priori this cube complex may not be the same as $X$. 
Even though it may not be necessary for the VHC itself, it seems like we still need the full strength of the original result (i.e that we can take $X$ to be $Y$) since the proof is by induction.

REPLY [8 votes]: Being virtually special is actually a group-theoretic property, independent of the cube complex (at least in the word-hyperbolic case of interest to Agol).  More precisely, Haglund and Wise, in their seminal GAFA paper 'Special cube complexes', proved the following.
Theorem: Let $X$ be a finite cube complex with $\pi_1X$ word-hyperbolic.  Then $X$ is virtually special if and only of every quasiconvex subgroup of $\pi_1X$ is separable.
(Recall that separable means 'closed in the profinite topology' or, more simply, 'is an intersection of finite-index subgroups'.)
The 'only if' direction is Theorem 7.3, and the 'if' direction Theorem 8.13, in Haglund--Wise's paper.
The upshot of this is that you can change cube complexes without worrying.
Proof outline
The idea of the proof of is quite straightforward (though there are some technical details).  Let me quickly outline it here.
For the 'only if' direction, because $X$ is virtually special, any quasiconvex subgroup $H$ is also a quasiconvex subgroup of a right-angled Coxeter group.  This is then separable by an argument that goes back to Scott's 1979 paper 'Subgroups of surface groups are almost geometric', worked out in the general right-angled Coxeter case by Haglund in his paper 'Finite-index subgroups of graph products'.
For the 'if' direction, note that hyperplane stabilizers are quasiconvex, and therefore separable by hypothesis.  We need to eliminate certain pathologies: self-intersection, self-osculation etc.  It follows from separability that, for each hyperplane $Y$, there is a finite-sheeted cover of $X$ in which the unit-cube neighbourhood of $Y$ lifts to an embedded, trivial interval bundle over $Y$.  After doing this for all hyperplanes, there is no self-intersection and no self-osculation.
Inter-osculation is similar, but as two hyperplanes are involved, we need to use separability of double cosets of hyperplanes.  The easiest thing to do here is to quote a result of Minasyan (in 'Separable subsets of GFERF negatively curved groups'), who proved that if a word-hyperbolic group has the property that every quasiconvex subgroup is separable, then it follows that all double cosets of quasiconvex subgroups are also separable.  So we have the separability of double cosets that we require.<|endoftext|>
TITLE: Diameter of simplicial complex mirrored in property of Stanley-Reisner ring?
QUESTION [12 upvotes]: Consider a pure finite abstract simplicial complex $\Delta$. Define its diameter as the maximal distance between any two facets, i.e., between any two faces of maximal dimension $d-1$. The distance between two facets is the shortest path between them. A path of length $k$ in $\Delta$ is just a sequence of facets $(F_1, \ldots, F_{k+1})$ with $F_i \in \Delta$ such that $F_i \cap F_{i+1}$ is a ridge, i.e., a $(d-2)$-dimensional face.
For the class of simplicial complexes I am interested in, we can assume that such a path always exists.
To $\Delta$ one can associate a ring, the so-called Stanley-Reisner ring. It is defined as the quotient of a polynomial ring. Suppose $x_1, \ldots, x_n$ are the vertices of $\Delta$. Then its Stanley-Reisner ideal $I_{\Delta}$ is generated by the non-faces:
$I_{\Delta} = (x_{i_1} \cdots x_{i_s} : \lbrace x_{i_1}, \ldots, x_{i_s} \rbrace \not\in \Delta)$
Here $x_i$ denotes a vertex in $\Delta$ and at the same time also a variable in $k[x_1, \ldots, x_n]$, where $k$ is a field. The Stanley-Reisner ring is then defined as $k[\Delta] = k[x_1, \ldots, x_n]/I_{\Delta}$. These rings provide a nice bridge between combinatorics and geometry on the one hand and commutative algebra on the other.
So much for the setting. Now, what I am wondering about is if the diameter of the simplicial complex is represented by some property of $k[\Delta]$. Or is it maybe known that the diameter cannot be extracted from the ring?

REPLY [8 votes]: Interestingly enough, there have been quite a few recent attempts at answering your question. First, some relevant background. The diameter that you defined is known as the diameter of the dual graph of the Stanley-Reisner ring $R=k[\Delta]$. The dual graph $G(X)$ can be defined over any scheme $X$ as follows: the vertices are the irreducible components of $X$, and two of them are connected by an edge if the dimension of the intersection is $\dim X-1$). In the local commutative algebra community, this is also sometimes referred to as the Hochster-Huneke graph, thanks to this paper. In combinatorics, it is also called the ridge graph. 
It follows from a classical result by Hartshorne that if $R$ satisfies Serre's condition $(S_2)$ and $\dim R\geq 2$, then the dual graph of $Spec(R)$ is connected. So, if $R$ is Cohen-Macaulay, Goresntein (which all imply $(S_2)$) and $\dim R\geq 2$,  it makes sense to ask whether one can bound the diameter of $G(R)$. The elephant in the room is:

(Polynomial Hirsch Conjecture, algebraic version): If $R=k[\Delta]$ is Gorenstein, the diameter of $G(R)$ is bounded above by a polynomial $f(c)$, here $c=n-\dim R$ is the codimension. 

Note that Santos' counter-example to the original Hirsch conjecture implies that $f(c)=c$ does not work. 
Now here are some sources of recent references:
1) See this paper for some introduction on the polynomial Hirsch conjecture.
2) A lot of good introductions and recent results on bounding $G(R)$, even for general standard-graded algebras, can be found on Matteo Varbaro's website and the references in his papers on dual graphs. For example, this recent one establishes the Hirsch bound for certain class of Gorenstein ideals. 
3) Another question is: suppose we assume $R$ is just $(S_2)$, the minimal algebraic condition to guarantee that the diameter is finite (in the combinatorics community, this is sometimes known as $\Delta$ being normal). Can we bound the diameter?  
Brent Holmes, a PhD student of mine has been working on this question, and he found some exponential bounds that slightly improve the known ones in the literature, as well as compute the best bounds for small $n$ and $d$. You can find the most updated version of his work on his website (the results I mentioned can be found in the paper "On the Diameter of Dual Graphs of Stanley-Reisner Rings...").<|endoftext|>
TITLE: Monoidal model category structure on a functor category.
QUESTION [10 upvotes]: Let $A$ be a small simplicial category. The category $Fun(A,s\mathrm{Set})$ of simplicial functors from $A$ to simplicial sets can be given the projective model structure in which fibration and weak equivalences are objectwise.
Now assume further that $A$ is a symmetric monoidal category with respect to some binary operarion $\circ: A\times A\to A$. We can define the Day (or convolution) tensor product on $Fun(A,s\mathrm{Set})$ by the following coend:
$$F\otimes G(a)=A(-\circ -,a)\otimes_{A\times A}F(-)\times G(-)$$
My question is:
Is it true that the projective model structure is a monoidal model category in the sense that it satisfies the pushout-product axiom and if so is there a place where this is written down ?

REPLY [11 votes]: [This should probably be a comment, since it is so short. Nevertheless, it is an answer to the question.]
The result you ask for is a consequence of proposition 2.2.15 in Sam Isaacson's Ph.D. thesis (Harvard University, 2009). That proposition is stated for combinatorial symmetric monoidal (closed) model categories, and not just for simplicial sets. In this general case, the statement in Sam Isaacson's thesis requires the base category to have virtually cofibrant morphism objects. This condition is automatic for simplicial sets, as all simplicial sets are cofibrant in the usual model structure.
I have no idea who first proved or published the above result.<|endoftext|>
TITLE: Any abelian category as filtered colimit of categories of projective modules
QUESTION [5 upvotes]: Recently I have heard somewhere that any (edit: small) abelian category can be expressed as the colimit of categories of projective modules over some rings. The remark was that this is "basically just idempotent completion". I have been trying to figure out myself how this works but got stuck. Maybe somebody knows how to do this or where I can find a good reference.
My idea so far was to consider the rings $R=End(A,A)$ for any object $A$ in the abelian category $\mathcal A$. Then I can consider the functor $h_A:\mathcal A\to R\text{-}mod$ which maps $X$ to the $R$ module $Hom(A,X)$, which gives by Yoneda an embedding $\mathcal A\to (R\text{-}mod)^\mathcal A$. $A$ itself and its coproducts are then precisely the free $R$-modules.
If I take the idempotent completion of the category of free $R$-modules, I get the category of projective $R$-modules.
But this all doesn't really seem to go anywhere. Am I on the right way?
Also this statement seems so strong that it is odd that I can't find a reference. If I have any funtorial property for projective $R$-modules which commutes with colimits (like so many things in K-theory) I automatically have it for any abelian category. Maybe this means that the statement should be weaker?

REPLY [5 votes]: I'll denote the abelian category by $A$ and an object in it by $a$. Any $a \in A$ gives rise to a ring $\text{End}(a)$. The inclusion $a \to A$ induces a functor from the category of finitely generated projective $\text{End}(a)$-modules to $A$, since $A$ has biproducts and is idempotent complete. These module categories, over all $a \in A$, fit into a diagram of module categories of shape $A$ with $A$ as a cocone, and the claim is that $A$ is the colimit of this diagram. But it is straightforward to verify that $A$ satisfies the requisite universal property because any additive functor preserves biproducts and splitting of idempotents (these are the absolute colimits for $\text{Ab}$-enriched categories). 
Edit: If the OP wants a filtered colimit as in the title then we should look at full subcategories of $A$ on finitely many objects. Here the basic fact is that if $B$ is such a category then the category of right $B$-modules is equivalent to the category of right modules over the "category algebra" $\mathbb{Z}[B]$. This is the direct sum of all the Hom spaces in $B$ where the product is composition if that is defined and zero otherwise; equivalently, it's the endomorphism ring of the direct sum of all of the objects in $B$. We need $B$ to have finitely many objects in order for $\mathbb{Z}[B]$ as in the first definition to have a unit. 
The category of tiny objects in the category of right $B$-modules is the Cauchy completion of $B$ (its completion under direct sums and splitting idempotents) and the category of tiny objects in the category of right $\mathbb{Z}[B]$-modules is the category of finitely generated projective $\mathbb{Z}[B]$-modules, so in particular the Cauchy completion of $B$ (which, as above, naturally admits a functor into $A$) is the category of finitely generated projective modules over some ring. Now we write down the diagram of module categories whose objects are all finite collections of objects in $A$ (the module category being given by the Cauchy completion of the full subcategory on these objects) and whose morphisms are all inclusions, and $A$ is the colimit of this diagram as above.<|endoftext|>
TITLE: flatness of power series rings
QUESTION [11 upvotes]: It is known that $A[[X]]$ is flat if $A$ is noetherian (see for example Bourbaki, Algèbre commutative, Ch. III, §3, Cor. 3 p. 146). 
What happens if A is not noetherian? Is there an easy counter-example to the flatness of $A[[X]]$?

REPLY [15 votes]: As a module, $A[[X]]$ is the product of a countable family of copies of $A$. It is known that the product of flat $A$-modules is flat if and only if the ring $A$ is coherent, that is, every finitely generated ideal is finitely presented ( http://www.ams.org/journals/tran/1960-097-03/S0002-9947-1960-0120260-3/S0002-9947-1960-0120260-3.pdf ).
If you look at the proof of Theorem 2.1 in that paper, you can show that if $k$ is a field and $A$ is the quotient a polynomial ring $k[t_1, t_2, \dots]$ in countably many variables by the ideal generated by the products $t_it_j$, the product of countably many copies of $A$ is not flat over $A$.<|endoftext|>
TITLE: why are subextensions of Galois extensions also Galois?
QUESTION [8 upvotes]: Generally a Galois extension is defined to be an algebraic extension that is also normal & separable. It is then shown that in the sequence of field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This follows since the same property is valid for separable & normal extensions individually. It also follows that $L|K$ is a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$
In Robalo Delgados thesis on Galois Categories referenced in nLab-Grothendiecks Galois Theory he takes the opposite tack, and in definition 3.2.1.1 defines an algebraic extension of fields $L|K$ to be a Galois extension iff the set of elements of $L$ invariant under the action of $Aut_K L$ is $K$.
It is then shown that in the sequence of algebraic field extensions $L|M|K$ if $L|K$ is Galois then $L|M$ is. This is asserted to be an obvious deduction (and so has no details), I don't see the obviousness...can someone clarify.
In proposition 3.2.1.3 he shows that Galois extension is normal and separable.
All this appears to be in the opposite order of the standard treatments. One reason I'm interested in his formulation, if it is correct, is that one side of the Galois correspondence follows easily from this.
disclaimer: I've already asked this question on math.stackexchange but the answers there revolved around characterising Galois extensions as being normal & separable, and then showing this property follows.

REPLY [7 votes]: This proof might not count as direct. It is for finite extensions.
Lemma: $L/K$, a finite extension, is Galois if and only if $L \otimes_K L$ is a product of copies of $L$.
Proof: If $L\otimes_K L$ is a product of copies of $L$, and $x$ is fixed by every automorphism, then $s \otimes 1- 1 \otimes s$ is zero in $L \otimes_K L$, so $s \in K$ (via explicit description of tensor products of vector spaces.) 
If there are a lot of automorphisms, then each automorphism gives a different surjective map $L \otimes_K L \to L$, so we get a surjective map from $L \otimes_K L $ to the product of $[L:K]$ copies of $L$, which must be an isomorphism by dimension-counting. 
Then $L \otimes_M L$ is a quotient of $L \otimes_K L$, so is a product of finitely many copies of $L$.
I believe one can extend this to all algebraic extensions via a slightly more complicated argument. But that might just be pointless, as one could argue that my condition is just a clever way of saying "normal and separable" in different language.<|endoftext|>
TITLE: Submersions from compact flat manifold
QUESTION [5 upvotes]: Let $M=\mathbb{R}^n/G$ be a closed flat manifold, and let $F\to M \to N$ be a locally trivial submersion, where $F$ and $N$ are closed manifolds.
My question is simple: are $F$ and $N$ homeomorphic to flat manifolds?
This question seems quite natural to me, and I would expect the answer to this fact to be well known.
Any reference will be welcome.
Thank you in advance.

REPLY [4 votes]: So, the question was answered by Alexander Lytchak. I am writing down the answer for the sake of completeness.
Consider the fibration between the universal covers $F'\to\tilde{M}\to \tilde{N}$. $\tilde{M}$ is contractible and $\tilde{N}$ is simply connected, thus we can apply the Serre spectral sequence with integral coefficients, and from it we obtain that $F'$ and $\tilde{N}$ are contractible. In fact, if $H^*(F')$ has cohomological dimension $a$, and $H^*(\tilde{N})$ has cohomological dimension $b$, then $H^*(\tilde{M})$ would have cohomological dimension $a+b$ and this has to be $0$.
Then $N$ is acyclic, and from the exact sequence in homotopy so is $F$. Moreover we have and exact sequence
$$ 1\to\pi_1(F)\to \pi_1(M)\to \pi_1(N)\to 1$$
where $\pi_1(M)$ is a Bieberbach group, i.e. a torsion free group with a finite index normal abelian subgroup. As for $\pi_1(F)$, a subgroup of a Bieberbach group is again a Bieberbach group and therefore $F$ is homeomorphic to a flat manifold.
As for $N$, one can prove that there exists a normal abelian subgroup of $\pi_1(N)$ of finite index. Moreover, $\pi_1(N)$ is torsion free, since otherwise there would be a finite cyclic subgroup acting on the contractible manifold $\tilde{N}$ without fixed points. This is not possible, and therefore $\pi_1(N)$ is a Bieberbach group. Again, this implies that $N$ is a Bieberbach manifold.<|endoftext|>
TITLE: ideal operations
QUESTION [7 upvotes]: Let $R$ be a commutative Noetherian ring, $\mathfrak{a}$ an ideal and $x, y$ elements. Is it true that
$$\mathfrak{a}(\mathfrak{a}:x) \cap \mathfrak{a}(\mathfrak{a}:y) = \mathfrak{a}(\mathfrak{a}:(x,y))?$$

REPLY [8 votes]: No it is not true.
Let $R=\mathbb{K}[a,b,x,y]/\langle abx-aby\rangle$ and $\mathfrak{a}=\langle ax,by\rangle$.
Now $abx\in\mathfrak{a}(\mathfrak{a}:x)\cap\mathfrak{a}(\mathfrak{a}:y)$ but $abx\notin\mathfrak{a}(\mathfrak{a}:\langle x,y\rangle)$.<|endoftext|>
TITLE: Are rational varieties simply connected?
QUESTION [27 upvotes]: Is it true that every smooth rational variety X is simply connected? How is the proof?
Would it be still true if X has mild (for example orbifold) singularities?

REPLY [3 votes]: In characteristic zero you do not neen rational. It is enough rationally connected. 
Let $X$ be a smooth, projective, rationally connected variety over a field of characteristic zero. Then

Any finite étale cover of $X$ is trivial;
$X$ is simply connected.

You can find this, for instance, in O. Debarre "Higher-Dimensional Algebraic Geometry", Corollary 4.18.<|endoftext|>
TITLE: non-commutative finite rings
QUESTION [5 upvotes]: This is probably simple. Is there a finite non-commutative ring $R$ with identity in which all of its ideals are two-sided !?

REPLY [13 votes]: A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8. 
For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in this paper. 
Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal. 
Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done.  
Added: The comment asks for a modular representation of $Q$. Using GAP I found that $\mathbb{F}_2[Q]$ can be embedded into the matrix ring $M_4(\mathbb{F}_2)$. Write $Q=\langle x,y\mid x^4=y^4=1, yxy^{-1}=x^{-1}\rangle$. Then a faithful representation $Q\hookrightarrow GL(4,2)$ is given by 
$$x \mapsto 
\begin{pmatrix}1 & 0 & 1 & 0 \newline 0 & 1 & 0 & 0 \newline 0 & 0 & 1 & 1 \newline 0 & 0 & 0 & 1 \end{pmatrix}\qquad
y \mapsto 
\begin{pmatrix}1 & 1 & 1 & 1 \newline 0 & 1 & 0 & 1 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \end{pmatrix}$$
Since the Sylow 2-subgroup of $GL(3,2)$ is the Dihedral group $D_8$, four is the smallest degree of a faithfull representation of $Q$ over $\mathbb{F}_2$.<|endoftext|>
TITLE: Equivariant cohomology of finite group actions and invariant cohomology classes
QUESTION [7 upvotes]: Let $W$ be a finite group acting on a space $X$. In what generality is it true that $H^*_W(X) = H^* (X)^W$?  We always have a map $H^*_W(X) \rightarrow H^* (X)^W$, but it is certainly not an isomorphism with $\mathbb{Z}$-coefficients (take $X = E_W$, the total space of a universal bundle). But is it true with $\mathbb{Q}$-coefficients? Perhaps if we invert the order of the group?
This is a follow up to my prior question about equivariant cohomology. Again, I am referring to the notes http://www-fourier.ujf-grenoble.fr/~mbrion/notesmontreal.pdf on equivariant cohomology by Michel Brion.
I am interested in understanding the proof of Proposition 1 on pages 6 and 7. Let $G$ be a compact Lie group, let $T$ be a maximal torus, let $N$ be the normalizer of $T$ in $G$, and let $W = N/T$ denote the Weyl group. In part (i), we have the $W$-bundle $G/T \rightarrow G/N$, from which the author claims $H^* (G/N) = H^* (G/T)^W$ when using $\mathbb{Q}$-coefficients.
A few sentences later, a similar statement is made for a more arbitrary $W$ bundle. So it seems like the above statement about $W$-invariants of cohomology is true in some generality. Could someone explain why this is true or give a reference? Or perhaps I am mistaken in interpreting this argument.

REPLY [9 votes]: These results follow from the Cartan-Leray spectral sequence, which for a regular covering map $X\to X/W$ and a commutative ring $k$ of coefficients has 
$$
E_2^{p,q}=H^p(W,H^q(X;k))
$$
(cohomology of the group $W$ with coefficients in the $kW$-module $H^\ast(X,k)$) and converges to a graded group associated to $H^\ast(X/W)$. A reference is Ken Brown's "Cohomology of groups",  section VII.7.
In case the group $W$ is finite, if $|W|$ is invertible in $k$ then $H^p(W;H^q(X;k))=0$ for all $q$ and all $p>0$ (see Brown, Corollary III.10.2). In particular this is true if $k=\mathbb{Q}$. Thus the spectral sequence is concentrated in the $0$ column and therefore collapses, giving $H^\ast(X/W)\cong H^0(W;H^\ast(X;k))$. Since $H^0(W;M)=M^W$ for any group $W$ and any $W$-module $M$, this gives the stated results. So you were exactly right in your first paragraph!
More generally, the same collapse happens for the Serre spectral sequence of the fibration $X\to X_W\to BW$, which has
$$
E_2^{p,q}=H^p(BW;H^q(X))\cong H^p(W;H^q(X)),
$$
 giving the isomorphism $H^\ast_W(X)\cong H^\ast(X)^W$ you mentioned.<|endoftext|>
TITLE: Homogeneous Namba-like forcing
QUESTION [10 upvotes]: Let $\kappa \ge \aleph_3$ be a regular cardinal that is countably closed ($\alpha^\omega < \kappa$ for every $\alpha < \kappa$.)  I'm mostly interested in the case that $\kappa$ is strongly inaccessible.  Can there be a homogeneous notion of forcing that makes $\text{cof}(\kappa^{+V}) < \kappa$ without adding any bounded subsets of $\kappa$?
If there is a Woodin cardinal above $\kappa$ then the stationary tower forcing could do this except that it is (probably) not homogeneous.
If there is a forcing notion as desired then I believe the results of the paper "Stacking mice" would give a non-domestic mouse, so some large cardinals would be required to show that such a forcing exists.  Can we get one from, e.g. a supercompact cardinal?

REPLY [3 votes]: I think that the strongly compact Prikry forcing, for $\kappa^+$ strongly compact cardinal $\kappa$ - that forces $\text{cf }\kappa = \text{cf }(\kappa^+)^V = \omega$ without adding bounded subsets to $\kappa$, is homogeneous, but I couldn't prove it or find a reference.
Instead, I'll show something weaker that still implies that the truth value of statements of the form $\phi(a)$ in $V[G]$, where $a\in V$ doesn't depend on the generic $G$: For every $p,q\in \mathbb{P}$, we will find $p^\prime \leq p,\,q^\prime \leq q$ and an automorphism between $\mathbb{P}\restriction p^\prime = \{r \in \mathbb{P} | r \leq p^\prime\}$ and $\mathbb{P}\restriction q^\prime$.
Let $\mathbb{P}$ be the strongly compact Prikry forcing for changing both $\text{cf }\kappa$ and $\text{cf }\kappa^+$ to $\omega$, without adding bounded subsets to $\kappa$. 
Recall that a condition in $\mathbb{P}$ is a tree of finite increasing sequences in $P_\kappa (\kappa^+ )$, with finite trunk and above it every element has $U$-many successors ($U$ is fine $\kappa$-complete ultrafilter over $P_\kappa (\kappa^+)$). For the exact definitions and basic properties see section 1.4 in Gitik's chapter in the Handbook. 
Let $t$ be the trunk of $p$ and $s$ the trunk of $q$. By narrowing the trees of $p$ and $q$, if necessary, we may assume that for every $r\in p$ above $t$, $(r\setminus t)\cup s \in q$. For every $a\leq p$ define $\pi (a) = \{(r\setminus t)\cup s | r\in a\}$ - this is the required automorphism.<|endoftext|>
TITLE: Bracket of lyndon words?
QUESTION [11 upvotes]: Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free Lie algebra. Suppose further that $m < n$, so that $mn$ is a Lyndon word.
My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form 
$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_{m,n}^{(l)} B(l)$

REPLY [5 votes]: Yes it is of this form because any Lie polynomial 
"begins" by a Lyndon word, in particular, the least 
monomial of $B(l)$ is $l$. Then 
$$
B(m)=m+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}u;\ B(n)=n+\sum_{m<v\atop |v|=|n|} \alpha^{(v)}v\qquad \mathbf{(LB)}
$$
then $[B(m),B(n)]$ has only words of length $|m|+|n|$ and its least word is $mn$. 
Computationally


$$
B(m)B(n)=mn+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}mv+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}un]=mn+[sb1]
$$
all the monomials within square brackets are of same length and strictly greater than $mn$


$$
B(n)B(m)=nm+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2]
$$
all the monomials within square brackets are of same length and strictly greater than $nm$. 


But, $mn<nm$ because $mn$ is Lyndon and then  
$$
[B(m),B(n)]=mn+[sb1]-nm-[sb2]=mn+[sb3]
$$
where the square bracket $[sb3]$ is a linear combination of 
monomials that are greater than $mn$ or $nm$

Hence all (monomials of $[sb3]$) are greater 
than $mn$ which is the form you required. 
One can say a little bit more From property $\mathbf{(LB)}$, and the fact that $[B(m),B(n)]$ is a multihomogeneous Lie polynomial, one has ($\underline{w}$ being the commutative image of $w$)
$$
[B(m),B(n)]=B(mn) + \sum_{mn<l\atop l\ \mathrm{Lyndon};\ \underline{l}=\underline{mn}} \gamma_{m,n}^{(l)}B(l)
$$
the coefficients $\gamma_{m,n}^{(l)}$ are the structure constants of the free Lie algebra w.r.t. the Lyndon-Sirsov basis (they are integers, universal, i.e. characteristic free) and, up to my knowledge, their combinatorics is widely unknown.<|endoftext|>
TITLE: fourier analytic proofs
QUESTION [6 upvotes]: While searching through Mathoverflow, I found out a fourier analytic proof of the Isoperimetric Inequality.Also, by google search I found a fourier analytic proof of Quadratic Reciprocity theorem.I know of the fourier analytic approach used by combinatorialists like Ben Green. But what are the other fourier analytic proofs of some of the well known classical theorems other than what I have mentioned above specially those which admit a starkly different proofs.

REPLY [9 votes]: The sign of the quadratic Gauss sum $\tau$ can be obtained from
the spectrum of the discrete Fourier transform $\Phi$: the trace of $\Phi$
gives $\tau$, and $\det\Phi$ distinguishes $\tau$ from $-\tau$.
Recall that for an odd prime $p$ the quadratic Gauss sum
can be defined by
$$
\tau = \sum_{n=0}^{p-1} \zeta^{n^2}
$$
where $\zeta = e^{2\pi i / p}$.  It is elementary that $|\tau|^2 = p$
and that $\tau$ is real or pure imaginary according as $p \equiv 1 \bmod 4$
or $p \equiv -1 \bmod 4$.  In fact $\tau$ is always $+\sqrt p$
in the former case, and $+i\sqrt p$ in the latter, but this is
notoriously tricky to prove.
One trick is to recognize $\tau$
as the trace of the discrete Fourier transform on ${\bf C}^p$,
which has matrix
$$
\Phi = (\zeta^{mn})_{m,n=0}^{p-1}.
$$
Now $\Phi^2$ is the matrix whose $(m,n)$ entry is $p$ if $m+n \equiv 0 \bmod p$
and $0$ otherwise (this is tantamount to discrete Fourier inversion).
This matrix has eiganvalues $+1$ and $-1$ with multiplicity $(p+1)/2$
and $(p-1)/2$ respectively.  Hence $\Phi$ has eigenvalues $i^k \sqrt p$
($k=0,1,2,3$) with multiplicities $m_k$ satisfying
$m_0 + m_2 = (p+1)/2$ and $m_1 + m_3 = (p-1)/2$, and then
$\tau = \sum_{k=0}^3 m_k i^k \sqrt p$.
Since we already know $\tau$ up to sign there are only two possibilities:
if $p \equiv 1 \bmod 4$ then $m_0$ or $m_2$ is $(p+3)/4$
and the other three $m_k$ are $(p-1)/4$, while
if $p \equiv -1 \bmod 4$ then $m_1$ or $m_3$ is $(p-3)/4$
and the other three $m_k$ are $(p+1)/4$.
We are to show that the odd man out is always
$m_0$ in the former case and $m_3$ in the latter.
In each case we can decide the correct choice by computing the
sign (a.k.a. argument) of $\det \Phi = p^{p/2} \prod_{k=0}^3 i^{k m_k}$.
We can do this because $\Phi$ is a Vandermonde matrix, whence $\det\Phi$
has the product expansion $\prod_{0 \leq m < n < p} (\zeta^n - \zeta^m)$.
Each factor $\zeta^n - \zeta^m$ is a positive real multiple of
$\exp((m+n+\frac12)\pi i)$.  It soon follows that
$\det\Phi = i^{(1-p)/2} p^{p/2}$ (we already knew $\left|\det\Phi\right|$
because each eigenvalue has absolute value $\sqrt{p}$),
and conclude as desired that $\tau = \sqrt{p}$ when $p \equiv 1 \bmod 4$
while $\tau = i\sqrt{p}$ when $p \equiv -1 \bmod 4$.
[This looks like a known but not very well-known argument
that is easier to rediscover than to find in the literature.
What is the original source?]<|endoftext|>
TITLE: Why do primes dislike dividing the sum of all the preceding primes?
QUESTION [35 upvotes]: I was investigating primes with the property that the sum of the first $n$ primes is divisible by $p_n$. It turns out that these primes are extremely extremely rare. For primes less than $10^9$, I have found that there are only five primes with this property:    
$$
p_1 = 2
$$
$$
p_3 = 5
$$
$$
p_{20} = 71
$$
$$
p_{31464} = 369,119
$$
$$
p_{22096548} = 415,074,643
$$
This raises the curious and equivalent questions:
Q1. Are there infinitely many primes which divide the sum of all the preceding primes?
Q2. Even if we assume that there are infinitely many such primes, why are they so rare? In other words, why do primes dislike dividing the sum of all the preceding primes? Is there any heuristic argument to show that such primes will indeed be extremely rare?

REPLY [24 votes]: Question Q1 seems to me an extremely hard problem, but I also believe that the answer is affirmative because the heuristic argument exposed by David Speyer.
I studied with FLorian Luca [1] a related problem that could help to answer question Q2:
Let  $A$ the set of integers $n$ such that the sum of the first n primes is divisible by $n$ (instead of $p_n$). In other words, $A$ is the set of $n$ such that the arithmetic mean of the first $n$ primes is an integer:
$A=\{ n: \frac{p_1+\cdots +p_n}n \in \mathbf Z  \}=\{ 1, 23, 53, 853, 11869, 117267, 339615, 3600489,..\}   $
These numbers are not so rare because in this case the heuristic gives $ A(x)\asymp \sum_{n\le x}\frac 1n\sim \log x$.
We could not prove that $A$ has infinite elements but we proved that they are rare:
$$A(x)\ll x e^{-C(\log x)^{3/5}(\log \log x)^{-1/5}}.$$
Later Matomaki [2]  proved the stronger estimate,
$A(x)\ll x^{\frac{19}{24}+\epsilon}$.
[1] Cilleruelo, Javier; Luca, Florian, On the sum of the first n primes. Q. J. Math. 59 (2008), no. 4. http://www.uam.es/personal_pdi/ciencias/cillerue/articulos.html
[2] Matomäki, Kaisa, A note on the sum of the first n primes. Q. J. Math. 61 (2010), no. 1,<|endoftext|>
TITLE: A curious sum for integers $\equiv 7\pmod 8$.
QUESTION [8 upvotes]: For $n$ a natural integer congruent to $7$ modulo $8$, one has seemingly always
$$\sum_{k=1}^{(n-1)/2}\left(\frac{k}{n}\right)k=0$$
where $\left(\frac{k}{n}\right)$ denotes the Jacobi symbol.
First cases:
$n=7$: $1+2-3$
$n=15$: $1+2+4-7$
$n=23$: $1+2+3+4-5+6-7+8+9-10-11$
I do not see any reason for this. Did I miss something obvious?

REPLY [16 votes]: There are probably lots of ways to see this, but here's one: let $S_1$ be the sum above, and let $$S_2 = \sum_{k=(n+1)/2}^{n-1} k \left( \frac{k}{n} \right).$$ Then $$S_1 + S_2 = S = \sum_{k=1}^{n-1} k \left( \frac{k}{n} \right).$$ Now you can rewrite $S$ (since $x \to 2x$ is a bijection mod $n$) as $$S = \sum_{k=1}^{(n-1)/2} 2k \left( \frac{2k}{n} \right) + \sum_{k=(n+1)/2}^{n-1} (2k-n) \left( \frac{2k}{n} \right) = 2S - n \sum_{k=(n+1)/2}^{n-1} \left( \frac{k}{n} \right), $$ where we used $(2/n) = 1$. Finally, we have $$S = 2S + n \sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right),$$ by changing $k$ to $n-k$ in the summation and using $(-1/n) = -1$. So $S = -n\sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right)$.
On the other hand, we can switch the index in $S_2$ from $k$ to $n-k$ as well, to get 
$$S_2 = \sum_{k=1}^{(n-1)/2} (n-k) \left( \frac{n-k}{n}\right) = -n \sum_{k=1}^{(n-1)/2} \left( \frac{k}{n} \right) + \sum_{k=1}^{(n-1)/2} k \left( \frac{k}{n} \right) = S + S_1 = 2S_1 + S_2.$$
(We used $(-1/n) = -1$ again).
Therefore $S_1  = 0.$<|endoftext|>
TITLE: Basic questions on the homotopy category
QUESTION [13 upvotes]: I apologize in advance if this the answer to this question is standard or well-known. I am not in any way an algebraic topologist.
$\newcommand{\s}{\mathscr}$Let $\s T$ be the category of topological spaces, $\s T_*$ the category of pointed topological spaces. We can construct the homotopy category $\s H$ and the "pointed homotopy category" $\s H_{*}$ by letting $\hom_{\s H}(X,Y)$ be $\hom_{\s T}(X,Y)$ modulo homotopy and $\hom_{\s H_*}(X,Y)$ be $\hom_{\s T_*}(X,Y)$ modulo homotopies preserving basepoints. There are obvious functors $F:\s T\to \s H$, $F_{*}:\s T_{*}\to \s H_{*}$. 
My questions are pretty basic: do $F,F_*$ preserve products? arbitrary limits? only arbitrary finite limits? coproducts? arbitrary colimits? It not, how does one compute colimits in $\s H$ and $\s H_{*}$?
If this can be found in a standard reference, just write an answer saying "this is a stupid question - this is all worked out in...," and I'll upvote and accept.

REPLY [4 votes]: A very concrete example of a cospan having no pullback in the homotopy category, which does not require any knowledge of cohomology or Moore spaces, can be found here.  It's phrased in terms of the homotopy category of groupoids, but can easily be translated to speak about $K(\pi,1)$s in the homotopy category of spaces instead.<|endoftext|>
TITLE: The Higman group II
QUESTION [7 upvotes]: This question is related to the question  The Higman group (with a nice answer by M. Sapir). So for background, please,
see the above cited question.
The Higman group has an automorphism $h(a_j)=a_{j+1}$ ($j+1$ is mod 4). Does the Higman group have a nontrivial normal subgroup $N$, satisfying $h(N)=N$?
Motivation. It seems to be an open question if the Higman group is hyperlinear. I seem to know how to construct a nontrivial almost representation of it in the sense of hyperlinearity. I don't know if the almost representation is exact. The negative answer on the above question would imply the exactness of my almost representation...
More general groups. Consider $G_{q,r}=\langle a,b,w\;|\;a^q=b^{-1}ab,\;b=w^{-1}aw,\; w^r=1\rangle$. What is known about such a groups? For $q=2,\;r=4$ it is a semidirect product of a cyclic group of order 4 acting on the Higman group by $h$.

REPLY [11 votes]: I think that Higman's group H has plenty of such normal subgroups. Indeed, let G be the extension of H with the automorphism h. Then H has index 4 in G. By Schupp's theorem, H is SQ-universal, hence the same is true about G (that SQ-universality is stable under a passage to finite index sub/over groups was proved by Peter Neumann, I think.). Therefore G has (uncountably) many proper infinite normal subgroups M. Take one such M (of infinite index) and let N be its intersection with H. Clearly N has index at most 4 in M and is normal in G. Hence it possesses all the required properties.<|endoftext|>
TITLE: Can a model of set theory be realized as a Cohen-subset forcing extension in two different ways, with different grounds and different cardinals?
QUESTION [14 upvotes]: The question is whether, when you add a Cohen subset to a cardinal
$\kappa$, that cardinal becomes a characteristic of the resulting forcing extension $V[G]$. Or can there be strange instances in which the very
same model is realized as a Cohen subset forcing extension over
different ground models with different cardinals?
To be precise, can it happen that $M[G]=N[H]$, where $M$ and $N$
are transitive models of ZFC and $G$ is $M$-generic for the
forcing to add a Cohen subset to some cardinal $\kappa$, that is, using $\text{Add}(\kappa,1)^M$, and $H$
is similarly $N$-generic to add a Cohen subset to some other cardinal
$\delta$, using $\text{Add}(\delta,1)^N$?
For a more concrete version of the question, imagine that we have added  a Cohen real $c$ and form
the extension $M[c]$; could it be that this model might also be
realized as $N[A]$ for some other ground model $N$, where $A$ is
an $N$-generic Cohen subset of $\omega_1^N$? Note that $M\neq N$
since it must be that $c\in N$ as the higher forcing does not add
reals. For my application, I need to understand the case where the two cardinals are both inaccessible cardinals (if not much more). Also, it is not difficult to identify general situations where this kind of thing is impossible. What I really want to know is if it can ever happen at all.
I conjecture that this situation is impossible, and that indeed, when you
add a Cohen subset to a cardinal, you have in particular made that
cardinal definable, as "the cardinal for which the universe was
just obtained by adding a Cohen subset to it".
The question is really a part of the subject known as
set-theoretic geology, but it has recently arisen in another project of mine.

REPLY [7 votes]: The main part of this question is answered by theorem 10 of
my joint paper with Bagaria, Tasprounis and Usuba, below.


From J. Bagaria, J. D. Hamkins, K. Tsaprounis, T. Usuba, Superstrong
    and other large cardinals are never Laver
    indestructible.
Let $\mathcal{C}(\kappa)$ assert that $\kappa$ is a regular
    cardinal and the universe was obtained by forcing over some ground
    $W$ to add a Cohen subset to $\kappa$, that is, "$V=W[G]$ for some
    ground $W$ and some $W$-generic $G\subset\text{Add}(\kappa,1)^W$."
Theorem. If $\mathcal{C}(\gamma)$ and $\mathcal{C}(\kappa)$ hold, where $\gamma<\kappa$, then $2^\gamma=\kappa$. Consequently,

There are at most two regular cardinals satisfying property $\mathcal{C}$.
There is at most one inaccessible cardinal satisfying property $\mathcal{C}$.
If $\mathcal{C}(\kappa)$ holds, then $\kappa$ is
    $\Delta_3$-definable as
    $$\text{``the smallest regular cardinal with property $\mathcal{C}$,''}$$
    or as
    $$\text{``the second regular cardinal with property $\mathcal{C}$.''}$$
If $\mathcal{C}(\kappa)$ holds and $\kappa$ is inaccessible, then $\kappa$ is $\Pi_2$-definable as
    $$\text{``the inaccessible cardinal with property $\mathcal{C}$.''}$$

Furthermore, these definitions work also in $V_\theta$, whenever
    $\theta$ is a $\beth$-fixed point of cofinality larger than
    $2^{2^\kappa}$ or for which $V_\theta$ satisfies
    $\Sigma_2$-collection.


In other words, if the universe can be realized both as a forcing
extension by adding a Cohen subset to $\gamma$ and also as a
forcing extension (over a different ground) by adding a subset to
$\kappa$, and $\gamma\lt\kappa$, then $\kappa=2^\gamma$. It
follows that there can be at most two cardinals that arise in this
way, and furthermore, that in the case of an inaccessible
cardinal, there is at most one over which the universe was
obtained by adding a Cohen subset.
We do not know, however, whether the situation of
$2^\gamma=\kappa$ can actually arise and perhaps it is true in
general that there can be at most one cardinal to which one has
added a Cohen subset.
The main theorem of the paper is that superstrong and other large
cardinal notions (including almost huge cardinals, huge cardinals,
superhuge cardinals, rank-into-rank cardinals, extendible
cardinals, $1$-extendible cardinals, $0$-extendible cardinals,
weakly superstrong cardinals, uplifting cardinals,
pseudo-uplifting cardinals, superstrongly unfoldable cardinals,
$\Sigma_n$-reflecting cardinals, $\Sigma_n$-correct cardinals and
$\Sigma_n$-extendible cardinals, for $n\geq 3$) are never Laver
indestructible. The method of proof of the main theorem applies,
however, to answer this question, and I view this answer really as
an explanation of the main nonindestructibility result, for the
reasons that I had explained the conjecture in the question here.<|endoftext|>
TITLE: Prime number races in 2 dimensions
QUESTION [36 upvotes]: Is the mapping $$f: \ \mathbb{N} \rightarrow \mathbb{Z}[i], \ \ \ n \ \mapsto
\sum_{2 < p \leq n \ {\rm prime}} e^{\frac{p-1}{4} \pi i}$$ surjective?
In 1999, when I was an undergraduate student, I thought about writing the thesis
for my first degree on this problem. I asked Jörg Brüdern about this, and what
he said was essentially that I could do this and could probably obtain some nice
partial results, but that an answer would most likely be out of reach. I decided
then rather to specialize in group theory.
Is it nowadays possible to say more on this question?
Plots of the images of the intervals
$\{1, \dots, \lfloor e^k \rfloor\}$ for $k \in \{11, \dots, 26\}$ scaled to the same size
look as follows:

Larger plots of the images of the intervals
$\{1, \dots, 10^k\}$ for $k \in \{7,8,9\}$ are shown below:
$k = 7$:

$k = 8$:

$k = 9$:

REPLY [12 votes]: This is not an answer, but rather an explanation of why this question is so difficult.
For positive coprime integers $a,q$, let
$$\pi(x;q,a) = \# \{p \leq x : p \equiv a \pmod{q}\}.$$
For $k \in \mathbb{Z}$, let
$$A_k = \{n \in \mathbb{N} : \pi(n;8,1) - \pi(n;8,5) = k\},$$
and let
$$B_k = \{\pi(n;8,3) - \pi(n;8,7) \in \mathbb{Z} : n \in A_k\}.$$
Then your conjecture that the function
$$f(n) = \sum_{p \leq n}{e^{\pi i(p - 1)/4}}$$
is surjective on $\mathbb{Z}[i]$ is equivalent to the conjecture that $B_k = \mathbb{Z}$ for each $k \in \mathbb{Z}$.
For this to happen, the set $A_k$ must be countably infinite; that is, the equality $\pi(n;8,1) = \pi(n;8,5)$ must occur infinitely often. This is a difficult result, but it is in fact known unconditionally: it is covered by Theorem 5.1 of "Comparative prime-number theory. II" by S. Knapowski, and P. Turán. Apparently, it has now been proven unconditionally by Jason Sneed that $\pi(x;q,a) - \pi(x;q,b)$ changes sign infinitely often for all $q \leq 100$, but this is yet to appear in print (see this paper for a discussion).
If one assumes two strong conjectures, the Grand Riemann hypothesis, and the Linear Independence hypothesis (namely that the imaginary parts of the nontrivial zeroes of all Dirichlet $L$-functions are linearly independent over the rationals), then one can say a lot more. Rubinstein and Sarnak's paper on Chebyshev's bias shows that not only are there infinitely many sign changes, but the function
$$\left(\frac{\log x}{\sqrt{x}} \left(\pi(x;q,a_1) - \mathrm{Li}(x)\right), \ldots, \frac{\log x}{\sqrt{x}} \left(\pi(x;q,a_r) - \mathrm{Li}(x)\right)\right)$$
has a limiting logarithmic distribution. In particular, they can say roughly how likely $(\log x / \sqrt{x}) \pi(x;8,1)$ and $(\log x / \sqrt{x}) \pi(x;8,5)$ are to be in particular regions; unfortunately, this doesn't really tell you anything about the set $A_k$ for each integer $k$.
Once you have that $A_k$ is countably infinite, you still need to ensure that there is no "conspiracy" happening, in that the other prime number race $\pi(x;8,3) - \pi(x;8,7)$ could avoid certain configurations whenever $x$ is a zero of the prime number race $\pi(x;8,1) - \pi(x;8,5)$. This seems extremely difficult, and I don't know how one might attempt to analyse this. That being said, questions peripherally related to this were studied by Knapowski and Turán, so it is possible that there might be something in the literature that can deal with this type of problem.
As an aside, one interesting modification of this conjecture is the following. Let $\chi$ be a Dirichlet character modulo $q$, so that $\chi$ is generated by some root of unity $\zeta_Q$. Is the function
$$f_{\chi}(n) = \sum_{p \leq n}{\chi(p)}$$
surjective on $\mathbb{Z}[\zeta_Q]$?<|endoftext|>
TITLE: Should the etale cohomology of a smooth projective variety (over rationals) be semi-simple; why?
QUESTION [8 upvotes]: $\DeclareMathOperator{\char}{char}\DeclareMathOperator{\gal}{Gal}$
Let $P$ be a smooth projective variety over a field $K$ (one may certainly assume that $K$ is perfect; the case $K=\mathbb{Q}$ already seems to be interesting enough). For some $\ell\neq \char K$, $n>0$, should the $n$-th $\mathbb Q_\ell$-adic Galois cohomology of $X_{K^{sep}}$ be semi-simple as a $\gal(K)$-representation? Certainly; no proof of this fact is known, so I would rather like to know whether it is related with some 'motivic' conjectures.
Some remarks:

For a finite $K$ one can consider the 'motivic' Frobenius; thus the conjecture follows from standard (motivic) ones. Yet this argument does not seem to work already for $K=\mathbb Q$.
It is certainly tempting to apply some polarizability argument. Yet my impression is that polarizability can only be applied to Hodge structures (in general); is this true?

Upd. It seems (see the comment of Ulrich) that 'my conjecture' is wrong for $K= \mathbb Q_\ell$; this settles my question. Yet I wonder where I can find the details for this example (when is the representation corresponding to an elliptic curve with multiplicative reduction  indecomposable).

REPLY [8 votes]: For a base field finitely generated over the prime field, the Tate conjecture implies that the $l$-adic realization gives an equivalence from the category of pure motives tensor $\mathbb{Q}_l$ to the category of $l$-adic Galois representations generated by the cohomology of smooth projective algebraic varieties over the field. The standard conjectures imply that the category of pure motives  is semisimple, and hence also the category of Galois representations. (Actually, all you need for this is the Tate conjecture plus the conjecture that numerical equivalence and l-adic homological equivalence coincide, which are both in Tate's original Woods Hole article.)
For other base fields, the Galois representations need not be semisimple -- as ulrich explains in his comments.<|endoftext|>
TITLE: On the coarse moduli space of a stack
QUESTION [8 upvotes]: Consider a stack $\mathcal{X}$ over $\mathbb{C}$ as a category fibred in groupoids over the category of schemes. Let $\mathcal{X}^s$ be the $\pi_0$ of this category, i.e. objects of $\mathcal{X}^s$ are the objects in $\mathcal{X}$ and morphisms of $\mathcal{X}^s$ are the morphisms in $\mathcal{X}$ modulo automorphisms of objects. It "kills" the groupoid structure, so I think it is possible to consider $\mathcal{X}^s$ as a category fibred in sets over the category of schemes. Assume $\mathcal{X}^s$ is represented by a scheme. Should it be the coarse moduli space for $\mathcal{X}$?

REPLY [8 votes]: Yes, this would imply that $\newcommand{\X}{\mathcal X}\X^s$ is the coarse moduli space, but I don't think this is the "right" question to ask --  I believe that $\X^s$ will not even form a sheaf unless $\X$ happens to be a scheme/algebraic space to begin with. 
Anyway, any morphism from a groupoid to a set factors through $\pi_0$ of the groupoid. This implies in particular that any morphism from $\X$ to an algebraic space factors through the presheaf $\X^s$. And the map $\X \to \X^s$ is a bijection on geometric points because it's in fact a bijection on $S$-points for any scheme $S$. So if $\X^s$ is a scheme/algebraic space then it is the coarse moduli space.

Addendum. I think you are confused about some basic issues. Let us see why $BG^s$ is not the coarse moduli space of $BG$. Let $G$ be a nontrivial finite group, say.
Consider for simplicity the topological setting, so we have a topological space $X$ and an open cover $\{U_i\}$. If we have a $G$-torsor on $X$ then we can restrict to a $G$-torsor on each $U_i$, and on each overlap $U_i \cap U_j$ we have isomorphisms between the restrictions from $U_i$ and from $U_j$. These isomorphisms satisfy cocycle relation. Conversely, if we have $G$-torsors on each $U_i$ and isomorphisms satisfying the cocycle relation, we can reconstruct a $G$-torsor on the whole of $X$, unique up to canonical isomorphism. What this paragraph says is exactly that the functor $BG$ which sends a space to the groupoid of $G$-torsors over it is a sheaf of groupoids, that is, a stack. (In the usual Grothendieck topology on the category of topological spaces, where open covers are, well, open covers. And when I call $BG$ a "functor" I should say "pseudofunctor" or "fibered category".)
On the other hand we can consider $BG^s$, which is now a priori just a presheaf of sets, mapping a space to the set of isomorphism classes of $G$-torsors over it. If we have an isomorphism class of $G$-torsor on $X$ then we get well defined isomorphism classes of $G$-torsors on each $U_i$ with compatible restrictions to each $U_i \cap U_j$. But it is NOT true that if we have an isomorphism class of $G$-torsor on each $U_i$ which agree on double overlaps, then we can reconstruct a unique isomorphism class on all of $X$: consider the case when $G$ is nontrivial on $X$ and $\{U_i\}$ is a trivializing cover! What this says is that $BG^s$ is in fact only a presheaf - it is NEVER a sheaf of sets. Put simply, one can not glue together isomorphism classes.
What this shows is in fact that if we sheafify $BG^s$, then we get a point. If we only remember isomorphism classes of torsors then every $G$-torsor becomes equivalent to the trivial torsor on some open covering of your space, which means that these torsors are identified under sheafification.
The same arguments work verbatim in algebraic geometry, since every $G$-torsor is locally trivial in the étale topology.
In any case, this is why I said above that this is not the "right" question to ask: it is not natural to expect $\X^s$ to be a sheaf in the first place. I would suggest reading Heinloth or Fantechi's notes on stacks (they are somewhere online) and thinking over just what question it is you want to ask.<|endoftext|>
TITLE: Hilbert's Nullstellensatz on polynomials with integer coefficients
QUESTION [7 upvotes]: Let $f_1, f_2, \ldots, f_m \in \mathbb{Z}[x_1, \ldots, x_n]$. Assume $f_1(X) = f_2(X) = \ldots = f_m(X) = 0$ have no solutions over $\mathbb{C}^n$, then by Hilbert's Nullstellensatz, there exists polynomials $g_1, \ldots, g_m \in \mathbb{C}[x_1, \ldots, x_n]$ such that $1 = f_1 g_1 + \dots f_m g_m$.
In this case, can we always find $g_1, g_2, \ldots, g_m \in \mathbb{Q}[x_1, \ldots, x_n]$ such that $1 = f_1 g_1 + \dots f_m g_m$? In words, if the coefficients of $f_1, \ldots, f_m$ are all integers, can $g_1, g_2, \ldots, g_m$ be taken as polynomials with integer coefficients, such that $f_1 g_1 + \ldots +f_m g_m \in \mathbb{Z}^{+}$?
Added later: I checked Wikipedia on Hilbert's Nullstellensatz. Sorry, it seems to be a stupid question, and the answer is YES.

REPLY [13 votes]: Yes.  Given the degrees of the $g_i$, the equation $1 = \sum_i f_i g_i$
is tantamount to a system of linear equations in the coefficients of the $g_i$,
and those linear equations have rational coefficients.  Once such a system has
a complex solution it automatically has a rational solution.<|endoftext|>
TITLE: When must it be  sets rather  than  proper classes, or vice-versa, outside of foundational mathematics?
QUESTION [23 upvotes]: Every once  in a blue moon  it actually matters that some mathematical entity which might  a priori only be a class is in  fact  a set. For  clarification,  here are some examples of what I  do  not mean:  
A) Some colleagues of mine once  made the following disclaimer:  'The "set" of stable curves does not exist, but we leave this set theoretic difficulty to the reader.' These colleagues (names withheld to  protect the  innocent) are  of course  fully  aware of  the fact that strictly  speaking, the class of all  stable  curves, or topological  spaces, or  groups, or any the other usual suspects customarily formalized  in  terms of structured sets, cannot itself be  a set. While they recognize that  the  structure they study is transportable along arbitrary bijections between members of a proper class of equinumerous sets, they also recognize that in their setting this  same transportability could justify a  technically sufficient a  priori restriction to some fixed  but otherwise  arbitrary  underlying  set: that is, the relevant large category has a small skeletal  subcategory. (Exercise: precisely what makes this work  in the example given?) In such cases, the set versus class pecadillo is an essentially victimless  one, perhaps barring discussions  of the   admissibility of Choice, expecially  Global  Choice. 
B) There are various contexts in which seemingly unavoidable size issues are managed through the device of Grothendieck Universes. Such a move  beyond ZFC might  be regarded as cheating, sweeping the issue under the carpet for all the right reasons. Allegations of this nature regarding the use of derived functor cohomology in number theory, as in the proof  of Fermat's  Last Theorem, can  now be laid to rest, as Colin  McLarty has nicely  shown in "A finite order arithmetic foundation for cohomology" http://arxiv.org/abs/1102.1773.
C) Set  theory itself  is  replete with situations  where the set  versus class distinction is of paramount importance.  For just  one  example, my very  limited understanding  is  that forcing over  a  proper class of conditions is not  for the unwary. I'd be  interested to  hear some expert  elucidation of that, but my question  here is in  a  different spirit.   
With these nonexamples out of the way, I have a very short list of examples that do meet my criteria.  
1)  Freyd's theorem on the nonconcretizability of the homotopy category in "Homotopy  is  not concrete"  http://www.tac.mta.ca/tac/reprints/articles/6/tr6abs.html. By definition, a concretization of a  category is a faithful functor to the category of sets. The homotopy  category (of based  topological spaces) admits  no such functor.
The crux  of the  argument  is that  while  any object of a concretizable category has only a set's worth of generalized normal subobjects, there are objects  in the homotopy category - for example $S^2$ - which do not have this property (page  9). The original  closing remark (page 6)  mentions another nonconcretizability result, for the category of small categories and  natural equivalence classes of functors. A purist might try to disqualify the latter as too `metamathematical', but the homotopy example seems unassailable. 
2) A  category in  which all  (co)limits exist  is said to be (co)complete; a bicomplete category is one  which is both complete  and  cocomplete. Freyd's General Adjoint  Functor  Theorem gives necessary  and  sufficient conditions for the existence of adjoints  to  a  functor $\Phi:{\mathfrak A}\rightarrow{\mathfrak B}$ with $\mathfrak  A$ (co)complete. Let us say that  a  functor which  preserves all limits  is continuous, and that  one which preserves all colimits is cocontinuous.  A bicontinuous functor is one which is  both continuous and  cocontinuous.
Let us say that  $\Phi$ is  locally bounded  if for every $B\in  {\rm  Ob}\,{\mathfrak  B}$ there  exists a set $\Sigma$ such that for every $A\in{\rm Ob} \,{\mathfrak A}$ and  $b\in{\rm Hom}_{\mathfrak B}(B,\Phi A)$ there exist $\hat{A}\in{\rm Ob}\,{\mathfrak A}$ and $\hat{b}\in{\rm Hom}_{\mathfrak  B}(B,\Phi\hat{A})\cap\Sigma$ such that
$b=(\Phi \alpha)\hat{b}$ for some   $\alpha\in{\rm Hom}$ $_{\mathfrak A}$ $(\hat{A},A)$, and that $\Phi$  is  locally cobounded  if for every $B\in  {\rm  Ob}\,{\mathfrak  B}$ there  exists a set $\Sigma$ such that for every $A\in{\rm Ob}\,{\mathfrak A}$ and $b\in{\rm Hom}_{\mathfrak B}(\Phi A,B)$ there exist $\hat{A}\in{\rm Ob}\,{\mathfrak A}$ and $\hat{b}\in{\rm Hom}_{\mathfrak  B}(\Phi\hat{A},B)\cap\Sigma$ such that  $b=\hat{b}(\Phi \alpha)$ for some   $\alpha\in{\rm Hom}_{\mathfrak A}(A,\hat{A})$. In the literature these are known as the Solution Set Conditions.  
Theorem.
Let $\Phi:{\mathfrak   A}\rightarrow{\mathfrak B}$ be a  functor, where $\mathfrak  B$ is locally small.  
$\star$ If $\mathfrak A$ is complete then $ \Phi$ admits a left adjoint if and  only if  $\Phi$ is continuous and locally bounded.
$\star$ If $\mathfrak  A$ is cocomplete then $ \Phi$ admits a right adjoint if and  only if  $\Phi$ is cocontinuous and locally  cobounded.  
See pages  120-123 of MacLane's "Categories for the working mathematician".  
The local (co)boundedness condition has actual content. For  example:  
a) The  forgetful functor  ${\bf  CompleteBooleanAlgebra}\rightarrow{\bf Set}$ is continuous  but  admits  no left adjoint.  
b) Functors ${\bf Group}\rightarrow {\bf Set}$, continuous but  admitting no left  adjoint,  may be obtained as  follows: let $\Gamma_\alpha$   be a simple  group  of cardinality $\aleph_\alpha$ (e.g. the alternating group on a set of that cardinality,  or the projective special  linear group on a  2-dimensional vector space  over a field of that cardinality) and  take the   product (suitably construed), over  the  proper class  of all  ordinals, of the  functors  ${\rm  Hom}_{\bf Group}(\Gamma_\alpha,-)$.  
c) Freyd proposed another interesting example  (see page  -15 of the Foreword  to   "Abelian  categories" http://www.tac.mta.ca/tac/reprints/articles/3/tr3abs.html) of  a locally small bicomplete  category $\mathfrak S$ and a bicontinuous functor $\Phi:{\mathfrak S}\rightarrow {\bf Set}$ which  admits neither  adjoint: loosely speaking, the  category of  sets equipped with free  group  actions, and the evident underlying set functor. 
Does anyone know of any other examples, especially  fundamentally  different  examples?  
Finally, one could focus critical attention on the very question posed. To  what extent does the strength  and  flavor of the background  set theory  matter? Force  of habit and  comfort have me  implicitly working in some material set theory such as ZF, perhaps a bit more if I want to take  advantage of Choice, perhaps a bit less if I prefer to eschew Replacement. Indeed, I have actually checked that example b)  may be formulated in the absence of  Replacement: while the von  Neumann ordinals are no longer  available, the  same trick already used to give a kosher workaround to the illegitimate product  over all ordinals further shows that an appropriate system  of local ordinals suffices for the task. I am also quite interested in  hearing what proponents  of  structural  set theory have  to say.

REPLY [14 votes]: Terry Tao has already mentioned Zorn's Lemma in order to find maximal elements in small partial orders. More generally, colimits in categories usually only exist for small index categories. In fact, every category admitting colimits for all index categories is equivalent to a partial order.
Another typical example of this kind is the small object argument. It says that any set of morphisms in a category with certain conditions produces a functorial weak factorization system. The transfinite construction doesn't stop when we start with a class of morphisms.
Another example: A cocomplete symmetric monoidal category  is closed if and only if all functors $X \otimes -$ satisfy the solution condition. Todd Trimble has given an example where this fails. It is interesting that being closed is only a property of the data, but the property seems to depend on the size.

REPLY [11 votes]: Proper classes come up when you exhaust the means of forming sets.   You need a set when you need to know the means of set theory have not been exhausted -- for example when you want to go on and form a colimit of the structures you have formed so far.   Exactly when the means are exhausted, depends on what means of forming sets you have.  
First take an example that exhausts second order arithmetic but does not exhaust Zermelo set theory (or simple type theory): the etale fundamental group of an arithmetic scheme.  There is no universal cover like the ones for topological spaces and this is not a logical or set theoretic problem but inherent in the situation.  (The scheme has etale covers of any finite degree, so a universal cover could have no finite degree.)  So Grothendieck and others formed the colimit of all symmetries of the (non-universal, actually existing) etale covers.    Second order arithmetic suffices to give the symmetry group of any one etale cover, but because we want the colimit of all these, we need an uncountable group.  Second order arithmetic will not produce that.   Third order will.
Grothendieck and Dieudonne often found they wanted colimits sort of like this, over all cases of some structure, but not just all that exist in second order arithmetic.  Naively put, they wanted all that exist in set theory.   Maybe all algebras over some ring, or all finitely generated algebras.  They knew there is a big difference between those examples, since there is not even a set of all algebras over a ring up to isomorphism (in any set theory they considered).  Choosing one countably infinite set of generators will give you a set of all finitely generated algebras over that ring up to isomorphism.  But in either case they did not want to bother with such details.  And they were all the more eager to avoid analogous details in more complex cases. 
If you really want to talk about all sets, or all natural weak equivalences of functors from Top to Top, or all generalized normal subobjects of $S^2$ in the homotopy category then you are exhausting the means of set theory (though the last two cases are less obvious than the first).  
Grothendieck and Dieudonne appreciated the point perfectly.   They knew workarounds to fit some of their larger constructions into ordinary set theory, and they were confident other workarounds could be found.  But they were not interested in that.  They saw that when they used all sets etc., it was not "all" in any metaphysical sense.  It was all those constructed by the ordinary means of set theory, so they posited one non-ordinary means of constructing sets: each set is contained in a universe.  At any point they work inside some universe, so what would be proper classes in ordinary set theoretic accounts are sets in the next larger universe.<|endoftext|>
TITLE: Analogues of D-modules and constructible sheaves
QUESTION [7 upvotes]: For a smooth complex variety, one can consider the category of say holonomic $\mathcal D$ modules on it. It is equipped with the deRham functor, which turns a $\cal D$-module into a constructible sheaf.
My question is, is there an analogue of constructible sheaves and the deRham functor for rings which behave like $\cal D$? A typical example I would be interested in is rings of twisted differential operators.
Edit: For example I wonder whether one could complete the following scheme:
Principal block of category $\cal O$ $\rightsquigarrow$ ${\cal D}$-modules on $G/B$ $\rightsquigarrow$ Perverse sheaves on $G/B$
SIngular block of category $\cal O$ $\rightsquigarrow$ twisted $\cal D$-modules on $G/B$ $\rightsquigarrow$ ?? on $G/B$

REPLY [8 votes]: I'm not sure what you mean by "rings which look like D" but here's one point of view:
the de Rham functor is just derived Hom from the structure sheaf $O$, i.e. you're testing all D-modules against your favorite one. One can imagine an analog in any context where you have a favorite module. For D-modules with an integral twist, you can just use the corresponding line bundle instead of O, and again get a Riemann-Hilbert correspondence with constructible sheaves (this is a little silly though since all of these categories with integral twists are canonically equivalent).
More generally if your twist is a complex linear combination of line bundles (eg always in the analytic topology) you can think of twisted D-modules as monodromic D-modules: these are D-modules on the total space of a principal torus bundle, which are weakly torus equivariant, and have
some fixed monodromy along the fibers. For example in the case of the flag variety all sheaves of
twisted differential operators can be viewed this way using the "basic affine space" $G/N$.
So then you can apply the ordinary de Rham functor upstairs. This gives a Riemann-Hilbert correspondence with monodromic constructible sheaves on the total space, or if you prefer, with constructible sheaves on a gerbe over the base (I think this is addressed in another MO answer about complex powers of line bundles, I learned it from a paper of Kashiwara). The class of the gerbe in $H^2(X, C^\times)$ is just the exponential of the Chern class of the TDO, considered as a class in $H^2(X,C)$. Is that the kind of answer you're looking for?<|endoftext|>
TITLE: Trichotomies in mathematics
QUESTION [64 upvotes]: Added. Thanks to all who participated! Let me humbly apologize to those who were annoyed (quite understandably) by this thread, deeming it nothing more than an exercise in futility. If you thought the question, if legitimate at all, should have been restricted to interesting manifestations of a hyperbolic-parabolic-elliptic subdivision, then I can fully agree (although part of the idea was to interpret the question as you see fit); I left it open ended primarily because of the Weil trichotomy, which is of completely different kind and is so much more than a hierarchy, and in relation to which I was interested in hearing other people's opinions and elaborations. See, for instance, how Edward Frenkel, in a fascinating Bourbaki talk, builds upon the Weil trichotomy to introduce a parallel between Langlands and electro-magnetic dualities, which he uses as a springboard for the ideas from physics that have entered the arena of the geometric Langlands program. Or take the cherished three-sided parallel between the basic three-dimensional (from the point of view of etale cohomology) objects and their branched coverings: $\mathbb{P}_{\mathbb{F}_q}^1$, $\mathrm{Spec}(\mathbb{Z})$, and $S^3$, with primes in number fields corresponding to knots in threefolds, $\log{p}$ corresponding to hyperbolic length, etcetera.
To those who were not convinced that there was a neat trichotomy of algebraic surfaces (arguing they should instead form a tetrachotomy by Kodaira dimension), let alone in higher dimensional algebraic geometry, I refer to Sándor Kovács's answer here, which demonstrates rather eloquently the fundamental trichotomy of birational geometry: 
How "frequent" are smooth projective varieties with (anti-)ample canonical bundle?
Original post. For many purposes, notably in classification hierarchies or in Weil's "big picture" of the fundamental unity in mathematics, it seems as if mathematical reality is more accurately captured by trichotomies than by two-sided dictionaries or questions of "either/or." The most basic is of course the trichotomy negative - zero - positve embodied by the complete ordered field $(\mathbb{R},<)$ --- this is the Arrow of Time, if you will, or the conditioning of a dynamical system into states of past/present/future. As evidenced by some of the examples below, this trichotomy underlies varied, if crude, classification schemes in mathematics.
Other trichotomies arise from closer examinations of a mathematical parallel. Mathematicians have always been fond of discovery by analogy; they take very seriously the intuitions supplied by different yet loosely connected fields. In doing so, they are guided by a tacit, platonic belief in the fundamental Unity of mathematics. An example is the similarity between finite geometries and Riemann surfaces. To explain this parallel, indeed to make sense of it, it is necessary to provide a "middle column" in the dictionary: the arithmetic geometry of number fields and arithmetic surfaces. This leads to the trichotomy that Weil explained so lucidly in a letter (which he wrote in 1940 in prison for his refusal to serve in the army) to his sister, the philosopher Simone Weil. This point of view led, as we know, to an entire new field of mathematical inquiry.
Below I have listed some other cherished mathematical trichotomies. I am interested in seeing yet others, perhaps more specialized. This is my question: add more trichotomies to the list. Furthermore, I am interested in any reflections anyone might have, such as pertaining, for instance, to any of the following questions. Is 3 the most ubiquitous number in coarse classification schemes? Is it fair to say that a given trichotomy echoes the primeval trichotomy $(-,0,+)$? In a given trichotomy, is there a natural "middle column" of a corresponding three-sided dictionary? Is this "middle column" in any way the most fundamental, the most interesting, or the most elusive?
Trichotomies in mathematics: some examples.

The fabric of topology, geometry, and analysis is the real line $\mathbb{R}$. Tarski's eight axioms characterize it in terms of a complete binary total order <, a binary operation +, and a constant 1. (Multiplication comes afterwards - it is implied by Tarski's axioms - and so does the Bourbaki definition of the reals as the complete ordered field). The sign trichotomies $(<,=,>)$ and $(-,0,+)$ ensuing from those axioms have repercussions throughout all of mathematics.
For example, there are three constant curvature spaces, leading to the three maximally symmetric geometries: hyperbolic, flat (or Euclidean), and elliptic (e.g. spherical forms).
Locally symmetric spaces fall into three types: non-compact type, flat, and compact type.
In complex analysis, there are three simply connected cloths: the Riemann surfaces $\Delta$, $\mathbb{C}$, and $\hat{\mathbb{C}}$.
The connected component of the group of conformal automorphisms of a compact Riemann surface is one of the following three: trivial, $S^1 \times S^1$, $\mathrm{PGL}_2(\mathbb{C})$. 
The complexity of fundamental groups, as showcased first of all by topological surfaces: genuinely non-abelian (perhaps we could say: anabelian) - abelian (or more generally, containing a finite index nilpotent subgroup) - and trivial (or more generally, finite). This is of course related to the subject of growth of finitely generated groups, brought forward by Lee Mosher's answer.
In dynamics, a fixed point (or a periodic cycle) can be either repelling, indifferent, or attracting.
In Thurston's work on surface homeomorphisms, elements of the mapping class group are classified according to dynamics into three types: pseudo-Anosov, reducible, and finite-order.
In algebraic geometry, the positivity of the canonical bundle is central to the classification and minimal model problems. More generally, positivity is a salient feature of algebraic geometry. For a delightful discussion, see Kollar's review of Lazarsfeld's book "Positivity in algebraic geometry." (Bull. AMS, vol. 43, no. 2, pp. 279-284). The most basic example is the trichotomy of algebraic curves (rational, elliptic, general type).
In birational algebraic geometry, at a very coarse level, there are three kinds of varieties out of which a general variety is made: rational curves, Calabi-Yau manifolds, and varieties of general type (or hyperbolic type, if you prefer). For example, an algebraic surface either: 1) admits a pencil of rational curves; or 2) admits a pencil of elliptic curves or is abelian or K3 (or a double quotient of a K3); or else 3) it is of general type. Abelian and K3 are examples of Calabi-Yau manifolds.
More concretely, consider smooth hypersurfaces $X \subset \mathbb{P}^n$. They divide into three types, according to how their degree $d$ compares with the dimension. If $d \leq n$, they contain plenty of rational curves (certainly uncountably many). If $d = n+1$, they are an example of a Calabi-Yau manifold, and typically contain a countably infinite number of rational curves. (The generating function of the number of rational curves of a given degree is then a very interesting function, of significance in the physics of quantum gravity.) And if $d \geq n+2$, then $X$ is of general type, and it is conjectured to typically contain only finitely many rational curves. (More precisely, Bombieri and Lang have conjectured that a variety of general type contains only finitely many maximal subvarieties not of general type).
In diophantine geometry, rational points are supposed to come from rational curves and abelian varieties. The sporadic examples are believed to be finitely many. This leads to the following trichotomy for the growth rate of the number of rational points of bounded (big, i.e. exponential) height: polynomial growth - logarithmic growth - $O(1)$. Furthermore, even in dimension 1, it is for abelian varieties that the situation is the deepest and the most mysterious.
In topology, it seems as if the interesting dimensions fall into three qualitatively different ranges: $d = 3$, $d = 4$, and $d \geq 5$. (Although this might be stretching it a bit too much). Of these, four dimensions -- the "middle column" -- is the most mysterious, and also the most relevant for physics.
The "Weil trichotomy," of course, goes at least as far back to Kronecker and Dedekind: curves over $\mathbb{F}_q$ - number fields - Riemann surfaces. Class field theory and Iwasawa theory are particularly eloquent examples of this trichotomy. Another example is of course the zeta function and the Riemann hypothesis.
One would be tempted to extend the latter trichotomy to [non-Archimedean world ($p$-adic, profinite) - global arithmetic - Archimedean world (geometry, topology, complex variables)], if the middle column did not subsume (much of) the flanking columns. Likewise the triple [$l$-adic cohomology-motive-Hodge structure] would probably not be admissible.
Here is a variation on the theme (you may find it to be rubbish, in which case throw it away). There are two ways of completing (or taking limits of) the regular polygons $C_n$. The first is to think of $C_n$ as $\frac{1}{n}\mathbb{Z}/\mathbb{Z}$ and take the direct limit (in this case, union, or synthesis: $\rightarrow$), which is $\mathbb{Q}/\mathbb{Z}$. Completing, we get the circle $S^1 = \mathbb{R}/\mathbb{Z}$, which is the simplest manifold. The second is to think of $C_n$ as $\mathbb{Z}/n$ and take the projective limit (or deconstruction: $\leftarrow$), which is $\hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$, the profinite version of the circle. In this way, Archimedean (continuous) objects and $p$-adic objects may be seen as the two possible different limits (synthesis and deconstruction) of the same finite objects. Taking $C_n$ to be more general finite groups, we get essentially all the Lie groups, on the one hand; and all the profinite groups, on the other hand.
That we live in three perceptible spatial dimensions does not, of course, fit our bill. But in 1984, Manin published an article ("New dimensions in geometry") in which, guided by ideas from number theory (Arakelov geometry) and physics (supersymmetry), he proposed that there are three kinds of geometric dimensions, modeled on the affine superscheme $\mathrm{Spec} \mathbb{Z}[x_i;\xi_j]$, an "object of the category of topological spaces locally ringed by a sheaf of $\mathbb{Z}/2$-graded supercommutative rings." Here, $\xi_j$ are "odd," anticommuting variables, commuting with the "even" variables $x_i$. See the three coordinate axes $x, \xi$ and $\mathrm{Spec} \mathbb{Z}$ in his picture of "three-space-2000." The arithmetic axis $\mathrm{Spec} \mathbb{Z}$ is implicit in complex algebraic geometry, and is essential in problems such as the Ax-Grothendieck theorem and the construction of rational curves in Fano manifolds.
In the theory of linear groups there is, loosely speaking, a trichotomy: $\mathbb{G}_m$ (linear tori) - semisimple - $\mathbb{G}_a$ (unipotent).
Algebraic groups: reductive - abelian variety - unipotent. Especially, the classification of one-dimensional groups: $\mathbb{G}_m$ - $E$ - $\mathbb{G}_a$. (Thanks, Terry Tao!)
Variant: among commutative algebraic groups, there are: multiplicative type - abelian varieties - additive type (unipotent).
$\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ are the only finite-dimensional associative division algebras over the continuum. (Thanks Paul Reynolds, Teo B, and Sam Lewallen!)
The most basic PDEs of physics: the wave equation (hyperbolic) - the heat and Schrodinger equations (parabolic) - the Laplace equation (elliptic). (Thanks Alexandre Eremenko!)
An infinite finitely-generated group has $1,2$ or $\infty$ ends. (Thanks shane.orourke and Artie Prendergast-Smith!)
A random walk is either transient, null recurrent, or positive recurrent. (Thanks Vaughn Climenhaga!)
Zeta functions can by dynamical (Artin-Mazur); arithmetical on schemes of finite type over $\mathbb{Z}$ (Riemann and Hasse-Weil); and geometric (Selberg's zeta function of a hyperbolic surface).
In Model theory, there is an important trichotomy between super-stable theories, strict-stable (stable but not superstable) theories, and non stable theories. 
It seems fair to say that there are three kinds of three-dimensional simply connected spaces: $\mathbb{P}_{\mathbb{F}_q}^1$, $\mathbb{Spec}(\mathbb{Z})$ compactified at archimedean infinity, and $S^3$. This brings about the Mazur knotty dictionary and the fruitful analogy between primes and knots (especially hyperbolic knots).

REPLY [2 votes]: Doing Bernoulli percolation (each edge is kept with probability $p$, erased otherwise, independently) on a transitive graph yields 0, 1 or infinitely many infinite clusters, and each of these behaviours may occur.
More generally, consider a transitive (connected locally finite) graph, i.e. consider a (connected locally finite) graph where all vertices play the same role: formally, we assume that the automorphism group of the graph acts transitively on the vertices. Consider a random subgraph of this graph: you are allowed to randomly erase vertices and edges. Assume that your random process is invariant under the symmetries of your graph: for any automorphism $f$, the distribution of your random subgraph is the same as that of your $f$-translated subgraph. Finally, assume that your process is insertion-tolerant: if $x$ is a vertex or an edge, and if $E$ is an event which does not look at $x$ and which has positive probability, then there is a positive probability that $E$ occurs and $x$ is not erased. Then, almost surely, the number of infinite connected components belongs to $\{0,1,\infty\}$.
(In the case of the so-called Bernoulli percolation, this gives rise to the critical parameters $p_c$ and $p_u$. Another interesting trichotomy in statistical mechanics is subcritical/critical/supercritical; this trichotomy has already been mentioned in this thread, in the setting of PDEs.)<|endoftext|>
TITLE: Flat morphisms whose fibers are affine spaces
QUESTION [8 upvotes]: Let $f:X \to Y$ be a flat morphism, such that each fiber is isomorphic to the affine space $\mathbb{A}^n$. Then is is true that $f$ is a Zariski affine bundle? If not, is it at least an  ètale affine bundle?
If not, is there some reasonable natural ipotesis to add to make this true?
My definition of affine bundle is a map $f:X \to Y$ such that there is an open cover ${U_{\alpha}}$ of $Y$ with $f^{-1} U_{\alpha} \cong U_{\alpha} \times \mathbb{A}^n$, and $f$ restricted to $f^{-1}(U_{\alpha})$ corresponds to the first projection. 
The question is somewhat related to:
Affine bundles over varieties

REPLY [4 votes]: The question is discussed in detail in the paper "Families of group actions, generic isotriviality, and linearization" (Transform. Groups 19 (2014), no. 3, 779–792. MR 3233525) by Peter Russell and myself. There is e.g. an example, due to Nori, of a flat morphism $X \to Y$ with fibers $A^1$ and base $Y$ equal the cuspidal curve which is not locally trivial in the etale topology.<|endoftext|>
TITLE: $\kappa$-dense ideals on successor $\kappa$
QUESTION [11 upvotes]: Woodin gave a consistency proof of a normal $\omega_1$-dense ideal on $\omega_1$ from an almost-huge cardinal.  He never published this argument, but it is written up by Foreman in the Handbook of Set Theory.  In this article, and in a paper from the 90s, Foreman claims that this argument is adaptable to other cardinals to yield the consistency of an $\kappa$-dense ideal on $\kappa$ where $\kappa$ is the successor of a regular cardinal.  I have had great trouble trying to prove this claim, as one crucial part of the argument seems specific to $\omega_1$ (which I will explain if you ask).  So does anyone know how to prove Foreman's claim?

REPLY [4 votes]: This is answered in chapter 2 of my thesis.  As far as I know, this is essentially the only method for obtaining such ideals.  I am very interested in finding alternative constructions.  Please contact me if you have some ideas.<|endoftext|>
TITLE: How are Modal Logic and Graph Theory related?
QUESTION [14 upvotes]: I am currently taking a graduate logic course on Modal Logic and I can't help notice that there are a certain class of graphs characterized by the modal axioms such as (4) $\Box p \rightarrow \Box \Box p$, (5) $\Diamond p \rightarrow \Box \Diamond p$, or (B) $p \rightarrow \Box \Diamond p$ which can characterize frames as being transitive, Euclidean, and symmetric, respectively. In general, I notice many similarities between the models used in Modal Logic and the graphs in Graph Theory and I'm wondering if anyone knows if there are applications of Modal Logic to Graph Theory, or if one subject might be a special case of the other?
In any case, if anyone has studied this before or knows of any references on the interplay between Modal Logic and Graph Theory I would be very interested to read about it, and if it has not been studied before then I would be interested of any ideas regarding what open research problems could be stated to tackle the correspondence between these two topics. (A category theory perspective on this interplay would also be very interesting)

REPLY [6 votes]: Maybe look at arrow logic?<|endoftext|>
TITLE: Is the following the right definition of $L$-functions (on the Galois side)?
QUESTION [9 upvotes]: This question may be too elementary for this forum, but I have asked it on math stackexchange and didn't get an answer. I have now deleted it so there wouldn't be duplicates... Here is the question as it appeared on math stackexchange:
I know that this question is pretty elementary, but I'm still trying to grasp the basics of number theory, and I want to make sure I'm thinking about things correctly. Most likely, I am thinking about things in a way that is slightly off. Since it's so complicated, I would appreciate if you can point out for me any misconceptions that I have.
Let $X_{\mathbb{Q}}$ be a variety over $\mathbb{Q}$. I am trying to understand how to associate to $X_{\mathbb{Q}}$ an $L$-function. As I understand it (please correct any mistakes I make in my narrative) there's an $L$-function for every $i=0,1,2,...$ (for convenience call them $L_0, L_1,L_2, ...$). Starting at $2dim(X)+1$ the $L_i$'s are just $1$. The $L$-function'' associated to all of $X_{\mathbb{Q}}$, by which I mean the Hasse-Weil Zeta Function, is $\prod L_i$.
Let's fix $i$. The function $L_i$ is defined to be the product of $L$-functions $L_{i,p}$, where $p$ runs over all primes in $\mathbb{Z}$. Here things get murky again. We choose a model $X_{\mathbb{Z}}$ of $X_{\mathbb{Q}}$ over $\mathbb{Z}$. (How? It can't be that any model would work, can it?) Now we fix a Weil Cohomology theory (as I understand it, it shouldn't matter which. Is that true? Is it only a conjecture that it's true?). For example we can look at $l$-adic cohomology where $l$ is a prime coprime to $p$. Then we define $L_{i,p}(x)$ to be $P_i(x)^{(-1)^i}$ where $P_i(x)$ is the characteristic polynomial of the action of the Frobenius element on $H^i(X_{\mathbb{Z}/p},\mathbb{Q}_l)$ (or whichever Weil Cohomology theory we chose).
Questions
First of all I would like to know if the narrative above is accurate. If not, please tell me where. Particular things I am vague about are:
$1.$ How does one choose the model of $X_{\mathbb{Q}}$ over $\mathbb{Z}$?
$2.$ Is it true that we can choose any Weil Cohomology? Is that conjectural, or proven?
$3.$ As I understand it, it should be true that $\prod_{p \in Spec(\mathbb{Z})} L_{i,p}= e^{\sum_{l=1}^{\infty} |X_{\mathbb{Z}/p}(\mathbb{F}_p)|\frac{x^l}{l}}$. Is that right?
$4.$ Does this construction differ in any substantial way if we defined a Hasse-Weil Zeta function for a variety defined over a number field different from $\mathbb{Q}$?

REPLY [4 votes]: A naive way of choosing a model is to write down some equations and clear denominators. The L-functions will depend on the model but only finitely many factors coming from "bad" primes. I think you want to assume that your variety is smooth and projective, otherwise some pathologies may happen. Nailing down the factors at the bad primes is trickier and I am not sure there is a nice way to do it, independent of the model, in general.
There are theorems giving comparisons between étale (for any $\ell \ne p$) and crystalline cohomology in char $p$. If you have an abstract Weil cohomology that satisfies the Riemann hypothesis, then the local L-factors are uniquely determined (see 3. below).
Your formula is wrong. What is true is that (for a fixed prime $p$ of good reduction)
$\prod_i L_{i,p} = e^{\sum_n |X_{\mathbb{Z}/p}(\mathbb{F}_{p^n})|p^{-ns}/n}$.
From this it follows that $\prod_i L_{i,p}$ is independent of the cohomology. RH then says that the reciprocals of the zeros of $L_{i,p}$ (as a polynomial in $t=p^{-s}$) have absolute value $p^{i/2}$. From this, you can extract the individual $L_{i,p}$ from the product over $i$.
No. It's basically the same, using prime ideals instead of prime numbers.

REPLY [2 votes]: 1 One does not need a model to find the L-function. The Frobenius element is a conjugacy class, up to ramification, in the Galois group of $\mathbb Q$. The Galois group of $\mathbb Q$ acts on the $l$-adic etale cohomology for every prime $l$ not $p$, and for the $p$-adic cohomology. However there is much more ramification for $p$-adic cohomology.
One takes the characteristic polynomial of the action of Frobenius on the invariants of the action of the ramification group. This is well-defined.
2 To prove a relationship between different $l$s, one can use a model over $\mathbb Z$. In particular one wants a model that is smooth and proper at $p$. Then one uses base change theorems to relate etale cohomology of the rational fiber and the characteristic p fiber, and then relates different characteristic $p$ cohomologies via the Lefschetz Trace Formula. However this only works for $l\neq p$, so one chooses an $L$-function via $l$-adic cohomology where $l\neq p$.<|endoftext|>
TITLE: On the local structure of stacks
QUESTION [5 upvotes]: 1) Is it true that any Deligne-Mumford stack is locally a quotient stack $[X/G]$ with a finite group $G$?
2) Is it true that any Deligne-Mumford stack can be globally presented as a quotient stack $[X/G]$ with a non necessarily finite group $G$? For example, Geigle and Lenzing give such a presentation for stacky projective lines here.
3) What about Artin stacks?

REPLY [2 votes]: I don't know the answer to (2) in the algebraic setting, but the analogous statement in the topological setting is true: https://arxiv.org/abs/1906.05816<|endoftext|>
TITLE: What does "$H^*(X)$ is Hodge-Tate" mean?
QUESTION [16 upvotes]: Let $X$ be a (let us say smooth to obscure any confusions I have between $H(X)$ and $H_c(X)$) algebraic variety defined over some subfield of $\mathbb{C}$.  I have occasionally overheard the expression "$H^*(X)$ is Hodge-Tate" used to mean something which, as far as I could tell from context, resembled one of the following:
(1) $H^*(X)$ is generated by $(p,p)$ classes, i.e. those in some intersection $W_{2p} H^i(X,\mathbb{Q}) \cap F^p H^i(X,\mathbb{C})$, where $W$ and $F$ are the weight and Hodge filtrations from the mixed Hodge structure.  In particular were $X$ smooth and proper, $H^*(X) = \bigoplus H^{p,p}(X)$.
(2) Spread $X$ out as appropriate and reduce mod a good prime, then it is `polynomial count', i.e. the number of points over $\mathbb{F}_{p^n}$ is a polynomial in $p^n$.
(3) Spread $X$ out as appropriate and reduce mod a good prime, then all the eigenvalues of Frobenius are powers of $p$.
(4) The class of $X$ in the Grothendieck group of varieties is in $\mathbb{Z}[\mathbb{A}^1]$
But when I searched for "Hodge-Tate" on google, I arrived at some description of "Hodge-Tate numbers" etc which seemed to have something to do with p-adic Hodge theory and apply to any variety.  Anyway my question is as in the title, 


What does it mean for $H^*(X)$ to be Hodge-Tate?


Also I guess (4) => (3) => (2) and I vaguely recall from some appendix of N. Katz that => (1) can be tacked on the end (?)  I would also like to know


Which of the reverse implications is false, and what are some counterexamples?

REPLY [10 votes]: 2 does imply 1 (for smooth projective varieties) via $p$-adic Hodge theory and perhaps a simpler argument.
1 does not imply 2. Indeed, blow up $\mathbb P^2$ at the Galois orbit of some point that is not $\mathbb Q$-rational but is rational over some quadratic field extension, say $(1: \sqrt{-1} : 0)$ . Mod a prime $p$ where that point is not $\mathbb F_p$-rational, there are $p^2+p+1$ $\mathbb F_p$-rational points. Mod a prime $p$ where that point is rational, there are $p^2+3p+1$ points. Obviously, this cannot be explained by any polynomial.
2 does imply 3 for smooth projective varieties. Using the polynomial for the number of points, one can compute the Weil zeta function as a product of terms of the form $\left( \frac{1}{1 -p^n t} \right)$. Using the Lefschetz trace formula, this is a product of factors corresponding to the eigenvalues of Frobenius in the etale cohomology. By the Riemann hypothesis, none of these terms cancel, so all eigenvalues are powers of $p^n$.
Not sure about 3 and 4.<|endoftext|>
TITLE: Cosheafification
QUESTION [15 upvotes]: Hello all.  I have a pre-cosheaf in the category of vector spaces.  How do I cosheafify?  I've  failed to find literature on this topic.
I'll be more specific.  Let $\mathbb{X}$ be a topological space and $\mathbf{Open}(\mathbb{X})$ the open set category of $\mathbb{X}$.  Let $\mathbf{Vect}$ be the category consisting of real vector spaces as its objects and linear maps as its morphisms.  My pre-cosheaf is a functor $\mathsf{F} : \mathbf{Open}(\mathbb{X}) \to \mathbf{Vect}$.  How do I cosheafify $\mathsf{F}$?

REPLY [11 votes]: Unless I have not botched things horribly, here is a general proof that cosheafification exists following the outline J. Curry gave.
Fix a target category $\mathcal{A}$ that is complete, cocomplete and locally presentable. Examples include categories of algebraic objects like linear spaces, categories of sheaves, the category of Banach spaces and linear contractions, etc.
Let $\Omega$ be a small site and $\mathbf{PCoShv}(\Omega, \mathcal{A})$ the category of precosheaves, that is, functors $\Omega\longrightarrow \mathcal{A}$. The category $\mathbf{PCoShv}(\Omega, \mathcal{A})$ is complete and cocomplete with limits and colimits computed pointwise and locally presentable by [AR, corollary 1.54].
Cosheaves are defined dually to sheaves, that is, they send the cone induced by a sieve of the small site to a colimiting cone. Denote the full subcategory of cosheaves by $\mathbf{CoShv}(\Omega, \mathcal{A})$. By the interchange of colimits theorem, $\mathbf{CoShv}(\Omega, \mathcal{A})$ is cocomplete and the inclusion functor $\mathbf{CoShv}(\Omega, \mathcal{A})\longrightarrow \mathbf{PCoShv}(\Omega, \mathcal{A})$ is cocontinuous.
proposition: $\mathbf{CoShv}(\Omega, \mathcal{A})$ is accessible.
proof: The category of sheaves on a small site is limit-sketchable by [RP, pg. 331]. The same construction yields by duality that the category of cosheaves is the category of models of a colimit sketch, therefore it is accessible by [AR, corollary 2.61]. Q. E. D.
proposition: $\mathbf{CoShv}(\Omega, \mathcal{A})$ is complete and locally-presentable.
proof: we already know that $\mathbf{CoShv}(\Omega, \mathcal{A})$ is cocomplete and accessible, so the result follows from [AR, corollary 2.47]. Q. E. D.
theorem: the full inclusion $\mathbf{CoShv}(\Omega, \mathcal{A})\longrightarrow \mathbf{PCoShv}(\Omega, \mathcal{A})$ has a right adjoint.
proof: follows from the previous results and the fact that cocontinuous functors between locally presentable categories have right adjoints. This adjoint functor theorem can be pieced together via the concept of totality. More precisely, locally presentable categories are total by [MK, corollary 6.5 and remark 6.6] and total categories are compact, that is, every cocontinuous functor has a right adjoint -- this is [MK, theorem 5.6].
Bibliography:
[AR] J. Adamek, J. Rosicky - Locally presentable and accessible categories, Cambridge University Press (1994).
[MK] Max Kelly - A survey of totality for enriched and ordinary categories, Cahiers de Top. et Géom. Diff. Catégoriques, 27 no. 2 (1986), p. 109-132
[RP] R. Pare - Some applications of categorical model theory, in Categories in Computer Science and Logic, Contemporary Mathematics, vol. 92 (1989)
edit: cleaned up, added a couple of references and made mention of the correct adjoint functor theorem.<|endoftext|>
TITLE: Is a measurable homomorphism on a Lie group smooth?
QUESTION [7 upvotes]: Let $G$ be a Lie group, and let $\mathcal B(G)$ its Borel $\sigma$-algebra. Suppose that $f : G \to G$ is a Borel-measurable homomorphism. Is $f$ smooth?

Edit: My original question said "measurable function" instead of the more accurate "measurable homomorphism." Marc Palm and other people answered both questions very nicely: 

there are obviously non-smooth measurable functions on Lie groups, and
all measurable homomorphisms on Lie groups are smooth.

REPLY [6 votes]: Perhaps it is interesting to know that for group cocycles the answer is in general no (even up to coboundaries). For instance the extension $\mathbb{Z} \to \mathbb{R} \to S^1$ is described by a 2-cocycle $S^1 \times S^1 \to \mathbb{Z}$ that is measurable, but this cannot be chosen to be smooth or continuous.<|endoftext|>
TITLE: Disjoint sets of fixed points
QUESTION [6 upvotes]: Let $\phi$ be an acceptable programming system. For every recursive function $f$, let $(f)=\{x:\phi_x=\phi_{f(x)}\}$ the set of fixed points of $f$. Now, suppose that $f$ and $g$ are recursive functions such that $(f)$ and $(g)$ are disjoint sets. The question is: are there recursive functions $F$ and $G$ such that $(F)\subseteq(f)$, $(G)\subseteq(g)$ and $\phi_{F(x)}\neq\phi_{G(x)}$, for all $x$?

REPLY [3 votes]: This turns out to be false. I'll construct $f$ and $g$ that there are no $F$ and $G$ with the required properties.
Define $f$ such that for any $x$,

$\phi_{f(x)}(0) = 0$
If $\phi_x(0)$ has halted by $n$ steps then $\phi_{f(x)}(n) = 0$
If $\phi_x(0)$ has not halted by $n$ steps then $\phi_{f(x)}(n)$ is undefined

So the fixed points $x$ of $f$ are such that $\phi_x(0)$ is defined and equal to 0 and for large enough $n>0$, also $\phi_x(n)$ is defined and equal to 0.
Define $g$ such that for all $x$, $\phi_{g(x)}(0) = 0$ and $\phi_{g(x)}(n)$ is undefined for $n > 0$. The fixed points of $g$ are defined at 0 but then undefined thereafter, so $f$ and $g$ don't have any fixed points in common.
Now suppose for a contradiction that there are $F$ and $G$ with the required properties.
We define $x$ using the fixed point theorem so that $x$ encodes the program that feeds its own code to $F$ and $G$ to find $F(x)$ and $G(x)$ and then starts simultaneously checking for the following conditions

there exists $n$ such that $\phi_x(n) \downarrow$
$\phi_{F(x)}(0)$ is defined but not equal to 0
there exists $n > 0$ such that $\phi_{F(x)}(n) \downarrow$
$\phi_{G(x)}(0)$ is defined but not equal to 0
there exists $n > 0$ such that $\phi_{G(x)}(n) \downarrow$

If condition 1 is ever matched, then there is some $n,m$ such that $\phi_x(n) = m$. Define $x$ so that at this point it stops checking for the conditions and halts with output $m + 1$ (on any input, but in particular on input $n$).
Define $x$ so that if condition 2 is ever matched then $x$ stops checking for the conditions and switches to copying $F(x)$, ie on input $n$, $x$ outputs $\phi_F(x)(n)$.
If condition 3 is matched then for some $n$, $\phi_{F(x)} \downarrow$. "Loop round" $n$ times (to ensure that $x$ does not halt in less than $n$ steps) and then start copying $F(x)$ like before.
Define $x$ so that if condition 4 or 5 is ever matched then $x$ stops checking for the conditions and switches to copying $G(x)$, ie on input $n$, $x$ outputs $\phi_G(x)(n)$.
We now check that none of these conditions can ever actually be satisfied. For condition 1 it is clear.
For condition 2, note that if it is ever matched, then $x$ is a fixed point of $F$ and hence of $f$, but also $\phi_x(0) \neq 0$. Contradiction.
For condition 3, note that if it is satisfied, then $x$ is a fixed point of $F$ and hence of $f$ and also that for some $n$ $\phi_x(n) \downarrow$ but that $\phi_x(0)$ requires more than $n$ steps to halt. Again this is a contradiction.
If conditions 4 or 5 is matched, this implies that $x$ is a fixed point of $g$, similarly giving a contradiction.
Now we know that none of the conditions are ever matched, we can deduce the following:

for all $n$, $\phi_x(n) \uparrow$
if $\phi_{F(x)}(0) \downarrow$ then $\phi_{F(x)}(0) = 0$
for $n>0$, $\phi_{F(x)}(n) \uparrow$
if $\phi_{G(x)}(0) \downarrow$ then $\phi_{G(x)}(0) = 0$
for $n>0$, $\phi_{G(x)}(n) \uparrow$

Finally note that if $\phi_{F(x)}(0) \uparrow$, then $\phi_{F(x)}$ is undefined everywhere, and hence equal to $\phi_x$. But this would make it a fixed point of $f$ that is undefined at 0, so we get a contradiction. Similarly, $\phi_{G(x)}(0)$ must be defined.
Therefore $\phi_{F(x)}$ and $\phi_{G(x)}$ are both equal to the partial function that is 0 at 0 and undefined thereafter.<|endoftext|>
TITLE: What is a(n algebro-geometric) family of modular forms?
QUESTION [10 upvotes]: We know that a family of elliptic curves is a morphism of schemes $f:X \to Y$ such that the fiber of every point of $Y$ is an elliptic curve (and we usually require the morphism to be smooth, proper, etc).
Given such a family, we can take the $\ell$-adic representation associated to any given fiber, and in this sense we also have a "family" of Galois representations. (Alternatively, by the proper base change theorem in étale cohomology, we can take $R^1f_*(\mathbb{Z}_{\ell})$, which is a sheaf on $Y$, and the stalks of this sheaf are the duals of the above Galois representations).
Now consider different question. Can we have a "family" of cuspidal eigenforms whose associated Galois representations fit into a family in the above sense?
I'll consider the case of weight 2 (though I'm most interested in higher weight). Then such family should lead to a family of RM abelian varieties, i.e. those associated to the weight 2 cusp forms.
Let's go back to the elliptic curve (or abelian variety) side a bit, and think about what this would mean. The level of a modular form corresponds to the conductor of the associated elliptic curve, so the level of the modular forms in such a family should be just as bizarre a function of the base as is the conductor of a family of curves.
To try to engineer such a family, suppose we had a family of elliptic curves that were all known to be modular. I'm most interested in rational families, i.e. with open subsets of projective space as bases. Then if the family is defined over $\mathbb{Q}$, we at least know that the fibers of rational points are modular, and we get a "family" of modular forms over the rational points. What would this "family" look like? I have a feeling it would be pretty strange from the point of view of modular forms.
A different approach is to try to to construct a family of modular curves, then view a family of modular forms as a section of the relative cotangent sheaf or some power thereof. Maybe one could try to make it an "eigensection" of some sort of relative Hecke operators. Of course, the very idea of a family of modular curves seems strange, as there are countably-many modular curves!
In fact, this points to a general problem with this attempt: modular forms are based on discrete data, a discrete set of levels, and a discrete set of eigenforms within each level.
I have a feeling that it's impossible to make this notion work, but please let me know if you have good ideas. In particular, it's possible that experts in modular forms and curves would have more ideas.

REPLY [3 votes]: Perhaps it is easier to explain what's going on here in the context of Galois representations. I know of two almost wholly disjoint ideas that are, confusingly, both described using phrases like "families of Galois representations". 

One can consider "families of representations" of any group, which are just homomorphisms from G to GL_n(A) for some (usually commutative) ring A, or more generally GL(V) for some locally free sheaf V on a base scheme S, etc. These are "families" in the sense that the image of a group element is a matrix whose entries are functions on Spec(A) (resp. on S, etc). Note that, intuitively, the group is fixed and the coefficients are varying.
One can also consider the kind of "geometric" family you mentioned in your question and Michigan J Frog enlarged upon: given a family of geometric objects over a base S, you can do various kinds of relative cohomology to give sheaves on S whose fibres have an action of some kind of Galois group depending on the fibre, and in particular the generic fibre has an action of something like the fundamental group of S. So here the group is, so to speak, varying in the family as well.

The first kind of family, over a p-adic base and with G being a Galois group, comes up a lot in the context of modular forms (Hida, Coleman-Mazur, etc). These can be viewed as sections of a family of sheaves on a subvariety of the rigid-analytic space you get by analytifying the modular curve. Note that we are varying the coefficients and not the group, again. 
The second kind of family doesn't come up so much in modular form theory, although it makes a notable appearance in Kato's work on Iwasawa theory for modular forms.<|endoftext|>
TITLE: K-Theory and Tame Symbol
QUESTION [7 upvotes]: This might be a little bit specific but here it goes. While reading a paper (Brauer-Manin pairing...) by Yamazaki, I encountered this definition.
Let $V$ be a variety and $y$ be a one dimensional point on $V$, i.e. $dim\overline{ \{ y \} } = 1$. Then let $C(y)$ be its closure in $V$ and $\tilde{C}(y)$ be normalization and $\bar C(y)$ be smooth completion. Let $C_\infty : = \bar C(y) - \tilde{C}(y)$. Then he defined the following group:
$UK^M_{r+1}:= ker \left[K_r^M(k(y)) \to \bigoplus_{x \in C_\infty}\left(K_{r-1}^M(k(x)) \oplus K_{r}^M(k(x))\right)\right]$.
The first component is the tame symbol at $x$ and the second component is defined as $a \to \partial_x(a \cup \pi_x)$ where $\partial_x$ is the tame symbol at $x$ and $\pi_x$ is a uniformizer at $x$. 
He says that this group doesn't depend on the choice of the uniformizer, but I couldn't see why.
Is there an easy way to tell that this group doesn't depend on the choice of the uniformizer?

REPLY [4 votes]: It follows from a derivation-like property of the tame symbol (which can be found e.g. in Remark 4.6. of Bass-Tate "The Milnor ring of a global field"). 
To fix notation, let $C$ be a smooth affine curve, $\overline{C}$ the completion and $C_\infty=\overline{C}\setminus C$ the boundary divisor. For a point $x\in C_\infty$, any uniformizer is of the form $u\pi_x$ for $\pi_x$ a specific choice of uniformizer and $u\in k(C)^\times$ with $v_x(u)=0$. For a symbol $a\in K^M_r(k(C))$ and $u\pi_x\in k(C)^\times$ we get 
$$
\partial_x(a\cup u\pi_x)=\lambda(a)\partial_x(u\pi_x)-\partial_x(a)\rho(u\pi_x)\in K^M_r(k(x)).
$$
The map $\lambda:K^M_r(k(C))\to K^M_r(k(x))$ is induced from $k(C)^\times\to k(x)^\times:u\pi^n\mapsto u$, and $\rho$ is induced from mapping $k(C)^\times\to k(x)^\times:u\pi^n\mapsto (-1)^nu$. 
An element $a\in UK^M_{r+1}$ will have $\partial_x(a)=0$, so the second term of $\partial_x(a\cup u\pi_x)$ is $0$ independent of the choice of uniformizer. In the first term, $\partial_x(u\pi_x)=1\in K_0(k(x))$ independently of the generator. We get that $\partial_x(a\cup u\pi_x)=\lambda(a)$ only depends on $a$, and hence is independent of the choice of uniformizer whenever $\partial_x(a)=0$. (This, however, requires to use both the tame symbol and the second map, the second map as well as its kernel may still depend on the choice.)<|endoftext|>
TITLE: Recognizing the stablizer of a degenerate three forms in six dimension 
QUESTION [5 upvotes]: Define $Stab^{+}(\Omega )$={ $\phi \in GL^{+}(V)$ : $\phi^{*}\Omega=\Omega$ }.
we say three-form $\Omega\in\wedge^{3}V^{*}$ is non-degenerate , if $i_X\Omega\neq 0$ for all $X\in V$-{0}
Let $V\cong \mathbb{R}^{6}$ and $\Omega\in\wedge^{3}V^{*}$ be non-degenerate(by sense of Hitchin) . Then we know $Stab^{+}(\Omega )=SL(3,\mathbb{R})\times SL(3,\mathbb{R})$ OR $Stab^{+}(\Omega )=SL(3,\mathbb{C}))$
We call the three-form $\Omega\in\wedge^{3}V^{*}$ is n-fold degenerate if the annihilator of 
$\Omega$, i.e.,
$Ann(\Omega)$={ $X\in V$ : $ i_X\Omega=0$ } has $dim(Ann(\Omega)=n$
so my question is what can we say about $Stab^{+}(\Omega )$ when $\Omega$ is n-fold degenerate?( $n$ here is 0,1,3,or 6)

REPLY [15 votes]: First of all, it is not true that there are only two types of nondegenerate $3$-forms in $\Lambda^3(V^\ast)$ when $\dim_\mathbb{R}V=6$.  There are actually $3$ such nondegenerate orbits.  Here is the complete list of $\mathrm{GL}(V)$-orbit types in this space:  Let $e^1,\ldots, e^6$ be a basis of $V$.  Then the following $3$-forms are inequivalent under $\mathrm{GL}(V)$ and every element of $\Lambda^3(V^\ast)$ is $\mathrm{GL}(V)$-equivalent to exactly one of these:

$\phi_1 = e^1\wedge e^2\wedge e^3 + e^4\wedge e^5\wedge e^6$
$\phi_2 = e^1\wedge e^3\wedge e^5 - e^1\wedge e^4\wedge e^6-e^2\wedge e^3\wedge e^6- e^2\wedge e^4\wedge e^5$
$\phi_3 = e^1\wedge e^5\wedge e^6 + e^2\wedge e^6\wedge e^4+e^3\wedge e^4\wedge e^5$
$\phi_4 = e^1\wedge e^2\wedge e^5 + e^3\wedge e^4\wedge e^5$
$\phi_5 = e^1\wedge e^2\wedge e^3$
$\phi_6 = 0$

The first three types are nondegenerate, and the first two types have open $\mathrm{GL}(V)$orbits, while the orbit of the third type is a hypersurface in $\Lambda^3(V^\ast)$. For a proof of this classical fact, you can see the Appendix in my article On the geometry of almost complex $6$-manifolds (available at http://arxiv.org/abs/math/0508428). 
The dimension of the annihilators are $0$, $0$, $0$, $1$, $3$, and $6$, respectively.  The stabilizers are

$\mathrm{Stab}^+(\phi_1) = \mathrm{SL}(3,\mathbb{R})\times\mathrm{SL}(3,\mathbb{R})$
$\mathrm{Stab}^+(\phi_2) = \mathrm{SL}(3,\mathbb{C})$
$\mathrm{Stab}^+(\phi_3) = \mathrm{GL}(3,\mathbb{R})\ltimes{\frak{sl}(}3,\mathbb{R})$
$\mathrm{Stab}^+(\phi_4) = \bigl(\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^\ast\times \mathbb{R}^\ast\bigr)\ltimes\bigl(\mathbb{R}^4\oplus \mathbb{R}^5\bigr)$
$\mathrm{Stab}^+(\phi_5) = \bigl(\mathrm{SL}(3,\mathbb{R})\times\mathrm{GL}(3,\mathbb{R})\bigr)\ltimes\bigl(\mathbb{R}^3\otimes \mathbb{R}^3\bigr)$
$\mathrm{Stab}^+(\phi_6) = \mathrm{GL}(6,\mathbb{R})$

The proofs that these are stabilizers are all relatively straightforward, but, since you asked, I will put in a sketch of the arguments for Cases 4 and 5:
As I wrote in the comment below, Case 5 is very straightforward:  Assume the index ranges $1\le i,j,k\le 3 < a, b, c\le 6$. If $\bar e^1,\ldots,\bar e^6$ were any other basis of $V^\ast$ such that $\bar e^1\wedge \bar e^2\wedge \bar e^3 = e^1\wedge e^2\wedge e^3$, then we'd have to have $\bar e^i = p^i_j\ e^j$ for some $3$-by-$3$ matrix $(p^i_j)$ that satisfies $\det(p^i_j)=1$.  Then we'd also have to have $\bar e^a = q^a_j\ e^j + p^a_b\ e^b$ for some $3$-by-$3$-matrix $(p^a_b)$ such that $\det(p^a_b)\not=0$, while the $3$-by-$3$ matrix $(q^a_j)$ is arbitrary.  Thus, the changes of basis that leave $\phi_4$ fixed are the semidirect product of $\mathrm{SL}(3,\mathbb{R})\times\mathrm{GL}(3,\mathbb{R})$ with the module of $3$-by-$3$ matrices, i.e., $\mathbb{R}^3\otimes \mathbb{R}^3$, as I claimed.
Case 4 is only a little bit more tricky:  Write $\phi_5 = (e^1\wedge e^2 + e^3\wedge e^4)\wedge e^5$.  Since $e^5$ is the only linear divisor of $\phi_5$, it follows that any linear transformation $L:V\to V$ that preserves $\phi_5$ must preserve $e^5$ up to a multiple, say $\bar e^5 = L^\ast(e^5) = \lambda\ e^5$ for $\lambda\not=0$.  Moreover, such a linear transformation $L$ has to then satisfy 
$$
L^\ast(e^1\wedge e^2 + e^3\wedge e^4) 
\equiv \lambda^{-1}\bigl(e^1\wedge e^2 + e^3\wedge e^4\bigr)\quad\text{mod}\quad e^5
$$
so $L$ must be a conformal symplectic transformation in the appropriate 4-dimensional subquotient of $V^\ast$.  Up to restricting to the identity component (which means assuming that $\lambda>0$), this means that $L$ reduces on this subquotient to an element of $\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^+$.  Lifting back to the space $V$, one sees that $L$ must act on the $5$-dimensional span of $e^1,\ldots, e^5$ as an element of a group that is isomorphic to $\bigl(\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^+\bigr)\ltimes \mathbb{R}^4$.  Finally, note that $L$ can do anything to $e^6$ since it doesn't appear in the formula for $\phi_4$.  Thus, one has the identity component of the stabilizer group of $\phi_4$ in the form 
$$
\bigl(\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^+\times \mathbb{R}^+\bigr)
\ltimes \bigl(\mathbb{R}^4\oplus \mathbb{R}^4\oplus \mathbb{R} \bigr)
$$
I think that, with this sketch, you should be able to write out the matrices that define this group and supply the rest of the details yourself.<|endoftext|>
TITLE: Do torsors give a long exact sequence of cohomology?
QUESTION [7 upvotes]: Let $X$ be a finite-type scheme over a field $k$. Let $G$ be a finite-type group scheme over $k$; we write $G_X$ for the base-change of $G$ from $\operatorname{Spec}(k)$ to $X$.
Suppose $f : Y \rightarrow X$ is a $G_X$-torsor for the fppf topology (i.e. we have an $X$-group scheme action of $G_X$ on $Y$ such that the morphism $G_X \times_X Y \rightarrow Y \times_X Y$ given on points by $(g,y) \mapsto (y,gy)$ is an isomorphism). Such a $Y$ gives a class $[Y]$ in the fppf cohomology set $H^1(X,G_X)$ that classifies fppf $G_X$-torsor sheaves over $X$ (this is defined with Cech cohomology).
Consider the specialization map
$$
s : X(k) \rightarrow H^1(k,G)
$$
that sends $x : \operatorname{Spec}(k) \rightarrow X$ to the pull-back of $[Y]$ by $x$, which is an element of $H^1(k,G)$. Note: if $G$ is smooth, the fppf cohomology set $H^1(k,G)$ may be identified with the Galois cohomology set $H^1(k,G(k^{\mathrm{sep}}))$.
Supposing it exists, fix a $y \in Y(k)$. We obtain an exact sequence of pointed sets
$$
0 \rightarrow G(k) \rightarrow Y(k) \stackrel{f}{\rightarrow} X(k) \stackrel{s}{\rightarrow} H^1(k,G)
$$
where $G(k) \rightarrow Y(k)$ is just the inclusion of the fiber above $y$. From the looks of it, I'd say that this has to be the start of a long exact sequence of some kind. The question only is: what kind? I don't see an obvious way of continuing it, since $Y$ doesn't necessarily carry any group structure so as to give meaning to the expression $H^1(k,Y)$.
Question: Is this exact sequence part of a long exact sequence? For instance, are we witnessing some instantiation of homotopy theory? If not, is there any other more conceptual way of viewing the above sequence?
Example: As a motivating example, let me show you why the image of $s$ - and therefore a continuation of the exact sequence from above - is an interesting object of study. Let $E$ be an elliptic curve over $\mathbf{Q}$ given by $y^2=f(x)$. Let $Y$ be $E - E[2]$, let $X$ be $\operatorname{Spec}(\mathbf{Q}[x,f^{-1}])$ (i.e., the affine line with coordinate $x$ and with the subscheme $f=0$ deleted), and let $f:Y \rightarrow X$ be the map that sends $(x,y)$ to $x$. Let $G = \mu_2$. We endow $Y$ with the structure of a $G$-torsor by letting the non-trivial element of $G$ send $(x,y)$ to $(x,-y)$. Then the image of the map
\begin{align*}
X(\mathbf{Q}) & \rightarrow \mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2} ~~ (\cong H^1(\mathbf{Q},\mu_2)) \\\
x & \mapsto f(x) \pmod{\mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2}}
\end{align*}
consists precisely of those elements $c \in \mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2}$ such that $cy^2=f(x)$ contains rational points other than the "trivial ones", i.e. the zeros of $f$ and the point at infinity.

REPLY [8 votes]: It actually more like the ending of a long exact sequence, rather than the beginning. To see what's going on consider the analogous case in topology. For this you replace the Galois group of $k$ with a discrete group $\Gamma$ and the category of $k$-schemes with the category of $\Gamma$-spaces. Instead of an algebraic group you now have a topological group G equipped with an action of $\Gamma$ on its classifying space BG. A G-torsor is a principle fibration $Y \to X$, which in homotopy theory corresponds to a $\Gamma$-equivariant fibration sequence of the form
$$ Y \to X \to BG $$
Taking $\Gamma$-homotopy fixed points one obtains a fibration sequence
$$ Y^{h\Gamma} \to X^{h\Gamma} \to BG^{h\Gamma} $$
which leads to a long exact sequence of homotopy groups ending with
$$ ... \to \pi_1(BG^{h\Gamma}) \to \pi_0(Y^{h\Gamma}) \to \pi_0(X^{h\Gamma}) \to \pi_0(BG^{h\Gamma}) $$ 
where $\pi_1(BG^{h\Gamma}) = \pi_0(G^{h\Gamma})$. This tail is the homotopy theoretic analogue of the sequence
$$ G(k) \to Y(k) \to X(k) \to H^1(k,G) $$
and so one should consider this sequence as the ending, and not the beginning of a long exact sequence. However, as apposed to the homotopy theoretic analogue, the map $G(k) \to Y(k)$ is always injective, making it seems like the sequence is just starting.
The analogy with the homotopy theoretic case can be made more precise by considering the etale homotopy type (see http://arxiv.org/abs/1110.0164).<|endoftext|>
TITLE: A basis for Schur functors
QUESTION [8 upvotes]: Suppose $V$ is a finite-dimensional vector space (over $\mathbb{C}$) and $\lambda$ is a partition of $n$ (not necessarily the dimension). Let $S^\lambda(V)=(V^{\otimes n})_\lambda$ be the $\lambda$'th Schur functor applied to $V$ (a.k.a. the  $\lambda$-component of the $S_n$ representation $V^{\otimes n}$). Pick a basis for $V$. Is there a good basis for $S^\lambda(V)$? Even better, is there a good subset of the canonical basis of $V^{\otimes n}$ which projects to a basis of $S^\lambda(V)$? (This is the case for our favorite Schur functors coming from the trivial and alternating reps). 
Further, is there a good combinatorial description for the functoriality property, i.e. $S^\lambda(M)$ where $M:V\to W$ is a matrix?

REPLY [8 votes]: I think this exact question answered in Theorem 1 of §8.1 of William Fulton's "Young Tableaux". Lazy as I am, I am unsure whether I have ever read the proof, but the answer is the following (unless I got Fulton's notations wrong): Whenever $p$ is a filling of the Young diagram of $\lambda$ with vectors in $V$, we can define an element $x_p$ of $S^{\lambda}\left(V\right)$ by taking, for each $i=1,2,...,\lambda_1$, the wedge product $w_i$ of the elements of the $i$-th column of $p$ (from top to bottom), and then taking the tensor product $w_1\otimes w_2\otimes ...\otimes w_{\lambda_1}$ of these $w_i$, and projecting this tensor product onto $S^{\lambda}\left(V\right)$. Now, letting $e_1,e_2,...,e_n$ be a basis of $V$, we can construct, for every Young tableau $T$ of shape $\lambda$ over the alphabet $\left\lbrace 1,2,...,n\right\rbrace $, an element $e_T$ of $S^{\lambda}\left(V\right)$ by $e_T = x_{p_T}$, where $p_T$ is the filling of the shape $\lambda$ in which every cell filled with a letter $j$ in $T$ is filled with the basis vector $e_j$. Then, Fulton's Theorem 1 claims that the $e_T$ with $T$ ranging over all semistandard tableaux over the alphabet $\left\lbrace 1,2,...,n\right\rbrace $ form a vector space basis of $S^{\lambda}\left(V\right)$.
Nota bene: This is only canonical with respect to a totally ordered basis.
Related results exist for so-called "bitableaux" (and probably contain the above result for tableaux?); see Doubilet-Rota-Stein Foundations IX and the subsequent papers by Rota, Désarménien, Kung, de Concini, Procesi and others.<|endoftext|>
TITLE: Manifolds admitting CW-structure with single n-cell
QUESTION [34 upvotes]: Let $M$ be a topological $n$-manifold, closed and connected (not necessarily oriented):
When does $M$ not admit (up to homotopy-type) a CW-structure with a single $n$-cell?
By classification of surfaces we assume $n>2$. By existence of smooth structures we assume $n>3$. In particular, if $M$ is smoothable then Morse theory provides us the desired structure.
[[Edit]]: To put this question into context, we have various ways of showing that $H_{n-1}(M)$ has either $0$ or $\mathbb{Z}_2$ as its torsion subgroup depending on orientability. One way, when $M$ is such a CW-complex, is to quickly look at the chain-complex differential $d:C_n(M)\cong\mathbb{Z}\to C_{n-1}(M)$ and note that $H_n(M)\cong\mathbb{Z}$ for $M$ orientable and $H_n(M;\mathbb{Z}_2)\cong\mathbb{Z}_2$ otherwise. So I would like to see how large of a class of manifolds this argument holds for.
[[Addendum]]: After chatting with Allen Hatcher and Rob Kirby, who reaffirm the comments below, here are their resulting thoughts:
1) We should be careful with the Kirby theorem of $M$ being homotopy-equivalent to a finite complex, because this complex is obtained by first embedding $M$ into $\mathbb{R}^N$ and then wiggling the boundary of a tubular neighborhood ($M\times D^{N-n}$) of $M$ to be PL, and so the resulting complex could have $i$-cells with $i>n$.
2) When $\dim M\ne 4$ there is a handlebody-decomposition, and this can be arranged to have a single 0-handle (canceling the other 0-handles with available 1-handles -- we can do this because there are no smoothing obstructions in a neighborhood of the 3-skeleton). Taking the dual handlebody, we have a decomposition with a single n-handle. Passing from the handlebody-decomposition to the CW-decomposition (shrinking everything to their cores), we obtain the desired CW-complex with a single n-cell.
3) When $\dim M=4$ then a handlebody-decomposition exists if and only if $M$ is smoothable. So when $M$ is smoothable we can apply the argument in (2).
4) But even when $M$ is not smooth we get some positive results, in particular for the $E_8$ manifold. We build $E_8$ using Kirby calculus on an 8-link diagram, giving a decomposition of $E_8$ into a 0-handle plus eight 2-handles plus a contractible piece (without the contractible piece we get a space with boundary being a homology 3-sphere, namely the Poincare-sphere $S^3/G$ with $G=$ binary icosahedral group). In particular, flipping this structure over we see that $E_8$ is homotopy-equivalent to a CW-complex with a single 4-cell. Furthermore, Lennart Meier's remark gets us all other simply-connected 4-manifolds.
We are thus left with the scenario that $M$ (up to homotopy) is a closed connected non-simply-connected non-smoothable 4-manifold. (which the comments below assert)

REPLY [4 votes]: Let $M$ be a closed topological four-manifold and $D$ an embedded closed four-disc, then $M\setminus D^{\circ}$ is a four-manifold with boundary $\partial D = S^3$. As I learnt from this answer, the boundary has a collar neighbourhood and hence $M\setminus D^{\circ}$ is homotopy equivalent to its interior, namely $M\setminus D$. As $M\setminus D$ is an open four-manifold, it is smoothable (a fact I learnt from this answer). A smooth open $n$-manifold is homotopy equivalent to an $(n-1)$-dimensional CW complex (see this answer), so $M\setminus D$ is homotopy equivalent to a three-dimensional CW complex $X$.
As $M$ is homeomorphic to $M\setminus D^{\circ}$ with a $4$-disc attached, and $M\setminus D^{\circ}$ is homotopy equivalent to $X$, $M$ is homotopy equivalent to $X$ with a $4$-disc attached, i.e. a CW complex with one top-cell.<|endoftext|>
TITLE: When does sheaf cohomology commute with arbitrary direct sums?
QUESTION [14 upvotes]: It is well known and more or less proven in Hartshorne's 'Algebraic Geometry' (p. 209) that for every noetherian scheme $X$ and every collection of abelian sheaves $\mathcal{F}_i$ the canonical map
$$ \bigoplus_i H^*(X; \mathcal{F}_i) \to H^*(X; \bigoplus_i \mathcal{F}_i) $$
is an isomorphism.
My line of interest is whether there are other situations where these kinds of things are true. One example came to my mind:
First observe that for a finite group $G$ and a sequence of $G$-modules $M_i$ the canonical map
$$ \bigoplus_i H^*(G; M_i) \to H^*(X; \bigoplus_i M_i) $$
is an isomorphism. Indeed, the bar resolution of $\mathbb{Z}$ over $\mathbb{Z}[G]$ is finitely generated in every degree. Thus, homming out ouf this resolution commutes with direct sums, as does homology. 
Now let $X$ be a scheme with an action by a finite group $G$. Dividing by $G$, we get a map $\pi: X \to X//G$. For every quasi-coherent sheaf $\mathcal{F}$, we have a spectral sequence
$$ H^p(G; H^q(X; \pi^*\mathcal{F})) \Rightarrow H^{p+q}(X//G; \mathcal{F}) $$
(at least if $X//G$ is flat over $\mathbb{Z}$). 
It easily follow that under these conditions (and $X$ noetherian) for every collection of quasi-coherent sheaves $\mathcal{F}_i$ on $X//G$, we have an isomorphism
$$ \bigoplus_i H^*(X//G; \mathcal{F}_i) \to H^*(X//G; \bigoplus_i \mathcal{F}_i) $$
Having these examples is not totally satisfactory since there might be a more general theorem with a uniform proof. So, my question is:

What are reasonably general conditions on a Deligne--Mumford stack such that arbitrary direct sums of quasi-coherent sheaves commute with cohomology?

REPLY [11 votes]: A Grothendieck topology is called noetherian if every object is quasi-compact (which is defined as usual). On such a topology, sheaf cohomology commutes with filtered colimits (and in particular with arbitrary direct sums). A proof can be found in Tamme's Intoduction to Etale cohomology, Theorem §3.3.11.1. For the etale topology on the spectrum of a field this shows that Galois cohomology commutes with arbitrary direct sums. For the Zariski topology on a scheme we recover the well-known result cited in Hartshorne. An even more general statement can be found in the Stacks project, Lemma 19.16.2.<|endoftext|>
TITLE: ${\bar{\partial}}$-geometrically formal ?
QUESTION [9 upvotes]: A compact Riemannian manifold is called geometrically formal if the wedge product of two $d$-harmonic  forms is $d$-harmonic.  Are there any known results for when a non-Kahler compact complex manifold admits a hermitian metric such that the wedge product of two ${\bar{\partial}}$-harmonic forms is ${\bar{\partial}}$-harmonic?

REPLY [3 votes]: In a paper by S. Torelli and A. Tomassini, "On Dolbeault formality and small deformations" (to appear in Internat. J. Math.), the authors study (geometrically) Dolbeault formality. In particular, they investigate the behaviour of (geometrically) Dolbeault formality under small deformations of the complex structure.
In particular, they prove that these properties are not stable under small deformations. An example is provided on the Nakamura manifold (that is, one of the simplest non-nilpotent solvmanifolds in complex dimension $3$). It is geometrically Dolbeault formal (and so also Dolbeault formal); and it admits small deformations for which there exist non-trivial Dolbeault Massey products (and so they are non-Dolbeault formal).<|endoftext|>
TITLE: Birational Automorphisms and infinite divisibility
QUESTION [8 upvotes]: Suppose $X$ is some algebraic variety. It can be over $\mathbb{C}$, but it doesn't have to (but char $0$ preferred).
Is it possible that the additive group $\mathbb{Q}$ acts on it birationally, without this action extending to a birational action of the additive $1$-parameter group $\mathbb{G}_a$?
What if we further assume that we do have an extension of a continuous action of $\mathbb{G}_a$, but only on some subset of the variety (the field now has a topology)?
Possible variations would be:

What if we have an action of $SL_2(\mathbb Q)$?
Instead of $\mathbb{Q}$, consider the additive group $\mathbb{Z}[1/2]$.
The field I'm actually interested in is $\mathbb{R}$, but examples over $\mathbb{C}$ would be great, too.

Essentially, I would like to know if people have considered when can a very divisible group act on an algebraic variety?
Remark: What I mean by a "birational action" of a group might be vague, but one interpretation could be a birational map $\mathbb G_a \times X \to X $ with the compatibility conditions making it an action.
Added Feb 6, 2013 (Corrected, thanks to Jérémy) According to the paper linked at in this MO question by Francesco Polizzi, in the algebraically closed case, the birational automorphisms of any $X$ inject set-theoretically into those of $\mathbb P^n$, with $n>\dim X + 1$. However, the group structure need not be preserved.
I've added another possibility, say we have a birational action of $SL_2\mathbb Q$ (or $SL_2 (\mathbb Z[1/2])$). Need it extend to the full group (of $\mathbb C$ or $\mathbb R$ points)?

REPLY [7 votes]: The question is about birational actions on varieties. Here is the answer for rational surfaces.
Lemma: If an element $g$ of $Bir(X)$, where $X$ is a rational surface, satisfies that for infinitely many integers $n$ there exists $h\in Bir(X)$ with $h^n=n$, then $g$ is contained in a one-parameter group of $Bir(X)$.
Proof: We can conjugate $g$ to a birational map of $\mathbb{P}^2$ and look at the sequence $\{\mathrm{deg}(g^k)\}_{k\in \mathbb{N}}$. The fact that $g$ is "divisible" by infinitely many integers implies that the degree sequence is bounded. Indeed, it is known that the sequence, if not bounded, grows linearly, quadratically or exponentially (Diller-Favre). In the first two cases, the coefficient of the quadratic or linear polynomial is bounded by below (Theorems C and D in http://arxiv.org/abs/1109.6810 ) and in the last case the eigenvalue, or dynamical degree, is more than 1.17. This implies that we cannot divide infinitely many times an element with unbounded degree sequence.
Now, the boundedness of the degree sequence implies that $g$ has finite order or that it is conjugate to an element of $GL(2)$ (see Theorem A in loc.cit.) In the first case, the element is also conjugate to an element of $GL(2)$ (see classification of elements of finite order of Bir(X), in http://arxiv.org/abs/0809.4673. )

Now, with this lemma and the classification of centralisers of linear elements in $Bir(\mathbb{P}^2)$ made in http://arxiv.org/abs/1109.6810, you get that any action of an ininfite-divisible abelian group extends to an action of an algebraic group.<|endoftext|>
TITLE:  open problems in Seiberg-Witten Theory on 4-Manifolds
QUESTION [16 upvotes]: What are some of the open problems in Seiberg-Witten Theory on 4-Manifolds.I tried googling but couldn't any. I tried googling it, but couldn't find any resources.The places where I can a survey or review of them would be welcome.

REPLY [11 votes]: It might be useful to generalize a theorem of Donaldson and Sullivan, that the Donaldson invariants are defined for quasi-conformal 4-manifolds, to the category of Seiberg-Witten invariants. More generally, one would like to know which smooth invariants of 4-manifolds are defined for quasi-conformal 4-manifolds.<|endoftext|>
TITLE: Vanishing cycles of a locally constant sheaf for a smooth morphism in the $l = p$-case
QUESTION [6 upvotes]: $\DeclareMathOperator{\Spec}{Spec}$
My question is concerned with vanishing cycles of a locally constant sheaf for a smooth morphism in the case $l = p$. In the case $l \neq p$ this is a statement in SGA7-II. See below for the precise question. First let me start with some
Background
Let us assume that one has an affine scheme $X$ over a field $k$ of characteristic $p >0$ and a function $f: X \to \mathbb{A}^1_k$. In SGA7-II Deligne then introduces the so-called nearby and vanishing cycles of $f$ at $0$ (say).
This roughly goes as follows: Let $S$ be the strict Henselization of $k[x]_{(x)}$. Let $s$ be the closed point of $\Spec S$ and let $\bar{\eta}$ be the spectrum of a separable closure of the quotient field of $S$. Consider the cartesian diagram
$
\begin{array}{ccccccc}
  &X_s&\xrightarrow{i}&X_{S}&\xleftarrow{j}&X_{\bar{\eta}} \newline
  & \downarrow & &\downarrow && \downarrow\newline
 &s&\xrightarrow{}&S&\xleftarrow{}&\bar{\eta} 
\end{array}
$
For a constructible sheaf $\mathcal{F}$ on $X_{\acute{e}t}$ the nearby cycles of $\mathcal{F}$ are defined as $R \Psi(\mathcal{F}) = i^\ast  Rj_\ast j^\ast \mathcal{F}$ (in $D(X_s)$). There is a natural morphism $i^\ast \mathcal{F} \to R\Psi(\mathcal{F})$ whose mapping cone we denote by $R\Phi(\mathcal{F})$ -- this is called the vanishing cycles.
(Of course, Deligne's construction is much more general and there are some groups acting here which I swept under the rug. But I hope the above is sufficient to get an idea of my question).

Deligne then proves the following (SGA7-II, 2.1.5): If $\mathcal{F}$ is a locally constant sheaf of $\mathbb{Z}/l$-modules (with $l$ a prime $\neq p$) and $f: X \to S$ defines a smooth morphism, then $R\Phi(\mathcal{F}) = 0$.
  My question is the following: Does this statement also hold for $l = p$? I am content to make additional assumptions on $X$ (e.g. of finite type over $k$) or my base field (take a finite field if you want). In particular, I am looking for a reference where this is proved or a counterexample.


Edit: My previous definition of Nearby/Vanishing cycles contained an error. I am indebted to Brian Conrad for pointing this out to me.

REPLY [6 votes]: $\DeclareMathOperator{\ord}{ord}$
$\DeclareMathOperator{\Spec}{Spec}$
This negative answer to my question is a slight expansion of an email sent to me by Brian Conrad. Since I merely added some details for my own benefit this post is community wiki.
As Damian Rössler pointed out in his comment the essential ingredient is the failure of the smooth base change theorem in the $l = p$ case. This is can be done via Artin-Schreier examples (cf. Milne's book on étale cohomology VI, Remark 4.4).
Let $S$ be the spectrum of $R = k[[t]]$ where $k$ is an algebraically closed field of characteristic $p > 0$ and write $K$ for the quotient field of $R$. Let now $X = \mathbb{A}^1_{R} = \Spec k[[t]][y]$.
Let us consider the constant sheaf $\mathcal{F} = \mathbb{Z}/p\mathbb{Z}$ on $X$ and its vanishing cycles complex at the origin $0$ of the special fiber. Denote by $A$ the strict henselization of $R[y]_{(t,y)}$.
Using SGA7-I, Exposé I, Proposition 2.3 we have $$(1) \quad (i^\ast R^n j_\ast j^\ast \mathcal{F})_0 =  H^n(A_{K_s}, \mathcal{F}) = \varinjlim H^n(A_F, \mathcal{F})  ,$$ where $F$ varies over the finite extensions of $K$ in $K_s$.
Since strict henselization is compatible with finite base change each $A_F$ is the analogue of $A_K$ with $R$ replaced by its normalization in $F$ (which is again local since $R$ is henselian).
We claim that $(1)$ is non-zero (in fact infinite). Admitting this for the moment we obtain a counterexample to my question. Indeed, part of the long exact sequence associated to the exact triangle $i^\ast \mathcal{F} \to R\Psi \mathcal{F} \to R\Phi \mathcal{F}$ is $$i^\ast R^1 j_\ast j^\ast \mathcal{F} \to R^1\Phi \mathcal{F} \to H^1(i^\ast\mathcal{F}),$$ but $i^\ast$ is exact so that $H^1(i^\ast \mathcal{F})$ is zero.
In order to show that $(1)$ is nonzero we consider the Artin-Schreier sequence $$0 \to \mathcal{F} \to \mathcal{O}_{A_{K_s}} \to \mathcal{O}_{A_{K_s}} \to 0.$$ Since $\Spec A_{K_s}$ is affine the higher étale cohomology groups of $\mathcal{O}_{A_{K_s}}$ vanish. It is therefore sufficient to show that the cokernel of $f \mapsto f^p - f$ on $A_{K_s}$ is infinite.

Claim: For any non-zero elements $c_i$ in the maximal ideal of the valuation ring of $K_s$ with pairwise distinct valuation the elements $y/c_i$ represent $\mathbb{F}_p$-linearly independent Artin-Schreier classes.

It suffices to proves this in each $H^1(A_F, \mathcal{F})$ separately (assuming all $c_i$ are contained in $A_F$). Hence, by replacing $R$ with with its normalization in $F$ we may assume that $F = K$ and all $c_i$ contained in $K$.
A non-trivial $\mathbb{F}_p$-linear combination of such $1/c_i$'s is an element of $K^\times$ with negative valuation. So it suffices to show that for nonzero $c \in R$ with $\ord(c) > 0$ the element $y/c \in A_K$ is not of the form $f^p - f$ for any $f \in A_K$. Here the valuation function $\ord$ is induced by $t^n \mapsto n$. In particular, $\ord(y/c) = -\ord(c) < 0$. If such $f$ exists then we must have $$0 > \ord(f^p - f) \geq \min(p \ord(f), \ord(f)),$$ so $\ord(f) < 0$. We therefore have $f = h/t^e$ with $e > 0$ and $h \in A$ not divisible by $t$.
Thus we get $$-\ord(c) = \ord(y/c) = \ord(f^p - f) = p \ord(f) = -pe. $$ This means that $c$ is contained in $t^{ep}R^\times$ and by changing $y$ by a unit we may assume that $c = t^{ep}$. Hence, $$\frac{y}{t^{ep}} = \frac{y}{c} = f^p - f = \frac{(h^p - t^{(p-1)e} h)}{t^{ep}}.$$ Equivalently we obtain the following equation in $A$ $$y = h^p - t^{(p-1)e}h = h(h^{p-1} - t^{(p-1)e}).$$
But $A$ is a regular local ring, so in particular a UFD, in which $y$ is irreducible. It follows that precisely one of the factors on the right hand side is a unit. Since $h^p = y + t^{(p-1)e}h$ lies in the maximal ideal of $A$ so does $h$. But then, since $e > 0$, $h^{p-1} - t^{(p-1)e}$ is also contained in the maximal ideal. This is a contradiction.<|endoftext|>
TITLE: When is a Moy-Prasad filtration subgroup the stabilizer of a subset of the building (up to center)?
QUESTION [7 upvotes]: Let $G$ be a connected, simply connected, semi-simple algebraic group defined and split over a local non-arch field $k$ with integer ring $R$.  Let $B$ be the corresponding reduced building.  Fix an apartment $A$ and corresponding root system $\Phi$ and split torus $T$; then for each $x \in A$ and $r>0$, we have a corresponding Moy-Prasad filtration subgroup $G_{x,r}$.  Our hypotheses give $G_x=G_{x,0}$.
Define 
$\Omega_{A,r}$ as the set
$$ \{ y \in A : \forall  \alpha \in \Phi, \vert \alpha(x-y) \vert \leq r \}.$$

Is it true that $T(R)G_{x,r} = \cap_{y\in \Omega_{A,r}} G_y$ ?

(With $r>0$ the product gives a group which is contained in the given intersection.  Equality seems to be about whether all hyperplanes $H_{\alpha,r}$ meet $\Omega_{A,r}$.)
Let $A(x)$ be the set of all apartments of $B$ containing $x$.  Let $Z$ be the (finite) center of $G$.

Is it true that $ZG_{x,r} = \cap_{A \in A(x)}\cap_{y\in \Omega_{A,r}}G_y$ ?
  Equiv by (1): Is  $ZG_{x,r} =\cap_{g \in G_x} (gT(R)g^{-1})G_{x,r}$?  

This is part of the larger question: I would like to understand the stabilizers of certain subsets of $B$ (which are not contained within any apartment $A$).  Pointers to any literature much appreciated.

REPLY [4 votes]: Regretfully the answer to the first question is no.

For example consider a group of type $G_2$.  Normalize your roots so the short roots have length 1, for example, and let $x=0$.  Write $\Phi = \Phi^l \cup \Phi^s$ as a union of long and short roots, respectively, and define the corresponding sets $\Omega^l_{A,r}$ and $\Omega^s_{A,r}$.  Write $B_r$ for the disk of radius $r$.   
Then $\Omega^l_{A,r} \subseteq B_{2r/3}$ whereas $B_r \subseteq \Omega^s_{A,r}$.  Consequently the parenthetical comment fails. 
There was also a typo in the parenthetical comment, so let's be explicit.  (Given a root $\alpha$, write $U_\alpha$ for the corresponding root subgroup and normalize so $U_\alpha(R)= U_\alpha \cap G_0$.) 
To see that $T(R)G_{x,r} \subseteq \cap_{y\in \Omega_{A,r}} G_y$, it suffices to verify this on a set of generators.  We have $T(R) \subseteq G_y$ for all $y\in A$.  Then the rest of  $G_{x,r}$ is generated by the groups $U_\alpha(P^{\lceil r-\alpha(x)\rceil})$.
For each $y \in \Omega_{A,r}$ we have $\alpha(x)-\alpha(y)\leq r$ so $-\alpha(y) \leq r-\alpha(x)$ which implies $U_\alpha(P^{\lceil r-\alpha(x)\rceil}) \subseteq U_\alpha(P^{\lceil -\alpha(y)\rceil})$.  Thus $T(R)G_{x,r}$ is contained in each of these $G_y$.
Conversely, in the case of $G_2$ described above, the inequalities imply that for sufficiently large $r$, there is no $y \in \Omega_{A,r}$ for which the equality $\lceil -\alpha(y) \rceil = \lceil r-\alpha(x) \rceil$ ($=\lceil r \rceil$) holds; therefore the intersection will be strictly larger than $T(r)G_{x,r}$.

It follows that the answer to the first version of the second question is also no.  The second version is true for $SL(2)$, for example, so perhaps there is an argument via Chevalley generators and relations.<|endoftext|>
TITLE: Are noetherian schemes generically Jacobson?
QUESTION [7 upvotes]: Question: Let $X$ be a noetherian integral scheme. Is there a dense open subscheme $U\subset X$ such that $U$ is Jacobson?
I am happy to allow $X$ to be excellent and then the question of course immediately reduces to $X$ excellent and regular. Equivalently, we could also ask whether every noetherian/excellent scheme admits a stratification into Jacobson schemes.
Note that if $X=\mathrm{Spec}(A)$ is affine, then the Jacobson radical may be $0$ even though $X$ is not Jacobson. An example is $X=\mathrm{Spec}(D[x])$ where $D$ is a DVR (there are many closed points of $X$ sitting in the generic fiber over $\mathrm{Spec}(D)$).
What if $X$ is essentially of finite type over a field?

REPLY [10 votes]: Here is a counterexample. Let $k$ be a field, $R=k[x,y]$. Choose a set $\Sigma$ of closed points of $\mathbb{A}^2_k=\mathrm{Spec}(R)$ such that: 
(1) $\Sigma$ is Zariski-dense,
(2) for every $s\in\Sigma$ there is a curve $C$ containing $s$ such that $C\cap\Sigma$ is finite.
EDIT: (For instance, if $\mathrm{char}\,k=0$, take $\Sigma=\mathbb{Q}^2$, and for $s=(a,b)\in\Sigma$ take $C$ defined by $(x-a)^2+(y-b)^2=0$.)
Now let $R_1$ be the localization of $R$ at $\Sigma$, and $X=\mathrm{Spec}\,R_1$. Let $\emptyset\neq U\subset X_1$ be open. By (1), $U$ contains some $s\in \Sigma$, and by (2), $U$ has a one-dimensional closed subscheme with only finitely many closed points. Hence $U$ is not Jacobson.<|endoftext|>
TITLE: (Homotopy theory) When does the 2 of 3 property not imply 2 of 6?
QUESTION [11 upvotes]: A relative category is a category $C$ with a subcategory $W$ containing all the objects of $C$.  
Given a relative category $(C,W)$, $W$ is said to satisfy the ``2 implies 6'' property if, for any collection of three composable maps,
$$X\rightarrow Y\rightarrow Z\rightarrow A$$
the presence of the composites $X\rightarrow Z$ and $Y\rightarrow A$ in $W$ implies that each individual map is in $W$ (and so also the triple composition).
The property I'm more familiar with from thinking about weak equivalences is the ``2 implies 3" property, which says that, given a pair of composable maps
$$X\rightarrow Y\rightarrow Z$$
the presence of any two of the maps 
$$X\rightarrow Y$$
$$Y\rightarrow Z$$
$$X\rightarrow Z$$
in $W$ implies that the third is as well.
The "2 implies 6" property implies the "2 implies 3" property, and I've been told that "2 implies 6" is a strictly stronger property.  
QUESTION: What is the basic example of a relative category $(C,W)$ where $W$ satisfies "2 implies 3", but not "2 implies 6"?  
Edit: By "basic", I mean what is an example which comes up in applications, or better yet, what is the example to keep in mind?

REPLY [9 votes]: Here is an example, which I got from a discussion with Karol Szumilo (and maybe he got it from Cisinski?). 
Consider the notion of a cofibration category, which means essentially that you have weak equivalences (with 2 out of 3), cofibrations (closed under pushouts and the same is true for trivial cofibrations) and you can factor everything into a cofibration and a weak equivalence. According to the Radulescu-Banu article https://arxiv.org/pdf/math/0610009v4.pdf (Theorem 7.2.7) we have the following theorem:

A cofibration category is saturated iff weak equivalences fulfill 2
  out of 6 iff they are closed under retracts.

Thus, we have just to find a cofibration category where weak equivalences are not closed under retracts. Let $R$ be a ring such that its reduced $K_0$ is non-trivial (i.e. there are projective modules that are not stably free). Consider first the cofibration category $Ch_R$ of bounded chain complexes, where weak equivalences are quasi-isomorphisms and the cofibrations $\mathcal{C}$ are injections with levelwise projective cokernel. Its homotopy category $\mathrm{Ho}(Ch_R)$ is the usual bounded derived category of $R$. This is a triangulated category and its $K_0$ agrees with the $K_0$ of the ring (where we define $K_0$ of a triangulated category as the free abelian group on all isomorphism classes of objects modulo the relation that $[X]+[Z] = [Y]$ if there is a triangle 
$$X \to Y \to Z \to X[1].)$$ 
We define its reduced K-theory $\tilde{K}_0$ by taking the quotient by the subgroup generated by $R$ itself concentrated in degree $0$. 
We want to localize $Ch_R$ at the class of morphisms $f \in \mathcal{W}$ such that the cone $C(f)$ is zero in $\tilde{K}_0$. First we show that $\mathcal{W}$ satisfies $2$ out of $3$:
Let $f$ and $g$ be composable morphisms. Then we get by the octahedral axiom a triangle in $\mathrm{Ho}(Ch_R)$ of the form
$$C(f) \to C(gf) \to C(g) \to C(f)[1].$$
By the defining relation of K-theory, the third of these cones is zero in $\tilde{K}_0$ if the other two are. 
To show now that $(\mathcal{W},\mathcal{C})$ is a cofibration category, we just have to show $\mathcal{W}$ is closed under homotopy pushouts in $Ch_R$. This is clear as the homotopy pushout of a map $f$ has an equivalent cone to $C(f)$. 
Last we have to show that $\mathcal{W}$ is not closed under retracts. But this is clear: Just take a projective $R$-module $P$ that is nonzero in $\tilde{K}_0$ and write it as a retract of a free $R$-module $R^n$. Then $0 \to R^n$ is in $\mathcal{W}$, but $0\to P$ is not in $\mathcal{W}$ (where we view this maps as maps of chain complexes concentrated in degree $0$). 

Upshot: Most of the time $2$ out of $6$ is fine, but there are some natural examples where one only has $2$ out of $3$. This should be reason enough to set up the theory with $2$ out of $3$ and only later introduced $2$ out of $6$.<|endoftext|>
TITLE: Geometric interpretations of matrix inverses
QUESTION [17 upvotes]: $A$ is an invertible $n \times n$ matrix.  Interpret each row of $A$ as a point in $\mathbb{R}^n$; then these $n$ points define a unique hyperplane in $\mathbb{R}^n$ that passes through each point (this hyperplane does not intersect the origin).
Under this geometric interpretation, $A^{-1}$ has an interesting property: the normal vector to the hyperplane is given by the row sums of $A^{-1}$ (i.e. $A^{-1} * 1$, where $1 = \langle 1, \dots, 1 \rangle^T$).
Within this same geometric interpretation of $A$, what other interesting properties does $A^{-1}$ have?  Do the individual entries of $A^{-1}$ have geometric meaning?  How about the column sums (besides the obvious row sums of $A^T$ intepretation)?

REPLY [24 votes]: Here is one simple geometric relationship: if $X_1,\ldots,X_n$ are the rows of $A$, and $C_1,\ldots,C_n$ are the columns of $A^{-1}$, then the length of the column $C_i$ is equal to the reciprocal of the distance between $X_i$ and the hyperplane $V_i$ spanned by the other $n-1$ rows $X_1,\ldots,X_{i-1},X_{i+1},\ldots,X_n$ of $A$.  This is because $C_i$ is orthogonal to $V_i$ and has an inner product of $1$ with $X_i$.  If one then square sums in $i$, we obtain the negative second moment identity
$$ \sum_{i=1}^n \sigma_i(A)^{-2} = \sum_{j=1}^n \hbox{dist}(X_j,V_j)^{-2}$$
where $\sigma_1(A),\ldots,\sigma_n(A)$ are the singular values of $A$, which turns out to be a useful identity in random matrix theory (see e.g. this blog post of mine).  In particular, it highlights the importance of understanding the distance between a row and the hyperplane spanned by the other rows if one is to get some control on the small singular values.<|endoftext|>
TITLE: Can one view the Independent Product in Probability categorially?
QUESTION [21 upvotes]: One can construct a category of probability spaces, but this category has no products. Now probability theory relies strongly on the ability to build independent products, the product measure. In a sense, the notion of independence is what distinguishes probablity theory from the theory of finite measures.

Is there a categorial way to make
  sense of and enlighten the notion of independent
  products in category theory?

It is possible to formulate independence in Lawvere's category of probabilistic mappings (Borel spaces as objects and Markov kernels as morphisms) in terms of constant morphisms, but I think this is not very enlightening, conditional independence is built into the morphisms. Maybe, this is what one has to do when putting probability center stage?
I do know the rudiments of categry theory, but I would prefer an answer that does not require too much immersion in category thory, provided that is possible.

REPLY [17 votes]: Very nice question! ;) I wrote a short paper about this question about ten years ago, see http://arxiv.org/abs/math/0206017 (My apologies for advertising my own work, but this is exactly the question I asked myself at that time).
The product of probability spaces is tensor product in the sense of category, as Martin Brandeburg also pointed out in a comment. But it has an additional structure, you have natural morphisms onto the factors in the tensor product. This is because the projections onto the two factors that you have for the Cartesian product (of sets) preserve the measures, so they are also morphisms in the category of probability spaces. I called this structure a tensor product with projections: for two objects $\Omega_i=(\Omega_i,\mathcal{F}_i,P_i)$, $i=1,2$, you get $\Omega_1\otimes\Omega_2=(\Omega_1\times\Omega_2,\mathcal{F}_1\otimes\mathcal{F}_2,P_1\otimes P_2)$ and random variables $X_i:\Omega_1\otimes\Omega_2\to \Omega_i$, $i=1,2$.
You can use this "tensor product with projections" to characterise independence of random variables: two r.v. $Y_i:\Omega\to\Omega_i$, $i=1,2$, defined on the same probability space $\Omega$, are independent iff they factorise, i.e., if there exists a r.v. $Z:\Omega\to\Omega_1\otimes \Omega_2$ such that $Y_i=X_i\circ Z$, $i=1,2$.
The notion dualises to the algebras of functions on a probablity space, where it becomes a tensor product with inclusions. Generalising to not necessarily commutative algebras, it includes notions of independence used in noncommutative (or quantum) probability, like the freeness.<|endoftext|>
TITLE: Periods for 2-variable p-adic L-functions
QUESTION [7 upvotes]: Hi all,
   I am sorry to ask a stupid question but I am really confused right now. Kitagawa-Mazur constructed a $2$-variable p-adic L-function attached to Hida families of modular forms. For their construction to work they needed some nice conditions for the universal deformation rings like Gorenstein property. In recent paper by Mok using similar ideas of Panchishkin constructs a $2$-variable p-adic L-function for Hilbert modular forms. The conditions in this paper is very mild. It just uses the non-vanishing result of Rohrlich. I think I can use similar arguments to construct this p-adic L-function for modular forms. How is p-adic L-function(Mok-Panchishkin) different/similar from/to Kitagawa-Mazurs's?
 I am sorry if this is a bad question but I am really confused right now. It may be entirely possible that I made a mistake in my computation. Any help will be greatly appreciated.
 Thanks.

REPLY [4 votes]: As explained in a comment, I disagree with Olivier's answer. Let me try another one.
Let $C$ denote the eigencurve, and $x$ a point in $C$, say with integral weight $k$ (I don't require the point $x$ to be classical, and the weight $k$ can be negative if you like).
Suppose our objective is to prove the following statement:

There exists an open neighborhood $U$ of $x$ in $C$ where classical points of non-critical slope are Zariski-dense and a two variable analytic function $L_p(u,s)$, where $u$ runs in $U$, and $s$ is the natural variable of a $p$-adic $L$-function, such that for every classical point $y$ in $U$ of non-critical slope
  corresponding to a classical $p$-stabilized modular form $f_y$, one has
  $$L_p(y,s) = c(y) L_p(f_y,s)$$ where $c(y)$ is a non-zero constant (but depending on $y$ !)
  Here $L_p(f_y,s)$ is the p-adic L-function of f_y defined in AMice-Velu and Visik (and also Mazur-Tate-Teiltelbaum).

Then the question arises:
1) On what condition on $x$ can we prove such a statement ?
I claim 
1a) that the method of Mazur (I have never seen it !), used also by Kitagawa prove
such a statement for a point $x$ which satisfies an algebraic property that I will call A(x)
(roughly that the dual of the generalized eigenspace attached to x in the space of
 modular symbols is free on the Hecke algebra acting on it).
1b) That this property A(x) is implied by a geometric property G(x), which can be for example that the eigencurve is smooth at x (but contrarily to a urban legend, Gorenstein at x is not enough). 
1c) And, that no other known methods has ever proved the statement above for any $x$ that did
not satisfy A(x) ! 
My belief is that actually A(x) is a necessary condition, not a technical ones.
Let me justify 1c) by a partial review of the literature: In Mazur and Kitagawa, A(x) is explicitly assumed. In Greenberg-Stevens, Panchiskin, and (I believe) Mok,
x is a classical point of non-crtical slope (even slope 0 in the Greenberg-Stevens case)
and then A(x) follows from the Hida/Coleman control's theorem which states that the generalized eigenspace considered in A(x) is in fact an eigenspace. In my 2011
inventiones paper, x is a critical classical point but I prove G(x), which implies A(x).
So I don't think that Panchiskin (or Mok, or myself for that matter) does anything 
that really goes beyond Mazur-Kitagawa, except of course that it does that on a Coleman's family (a.k.a my open set U in the eigencurve) while Mazur-Kitagawa worked on Hida's families,
the only one that existed then.
PS: I have stated my opinions in strong and perhaps too precise terms, in order to be 
more easily corrected or refuted if needed. I may be missing something -- or even everything.<|endoftext|>
TITLE: Are epimorphisms from a division ring isomorphisms ? 
QUESTION [6 upvotes]: According to Corollary 1.2(3) of the paper  Silver: Noncommutative Localizations and Applications. J. of Alg. 7(1964), 44-67:  
If $R$ is a (commutative) field and $\alpha: R \to S$ an epimorphism in the category of rings, then $\alpha$ is an isomorphism.
Question: Is $\alpha$ also an isomorphism if $R$ is not a field but just a division ring ? 
For convenience, I repeat the short proof from Silver (which doesn't seem to work for division rings): First it is noted (Prop. 1.1) that a homomorphism $\alpha: R \to  S$ of (not necessarily commutative) rings with identity is epi iff multiplication $S \otimes_R S \to S$ is an isomorphism (of abelian groups). Then: 
"For $x\in S$, consider the subring $R[x]$ of $S$ generated by $R$ and $x$. Since $R$ is a field, one can easily see that $R \to R[x]$ is an epimorphism using 1.1. If $x$ is transcendental over $R$, then $\beta: R[x] \to R[x]$ defined by $\beta(f)=f(0)$ agrees on $R$ with the identity map of $R[x]$. So $x$ cannot be transcendental over $R$ by definition of an epimorphism. Finally, if $[R[x]:R] < \infty$, then by 1.1, $[R[x]:R]^2=[R[x]:R]$, so $[R[x]:R]=1$ and $x \in R$. Hence $\alpha$ is an isomorphism as desired."

REPLY [7 votes]: Yes, and there are more general results available. E.g. in the paper of H. H. Storrer, Epimorphic Extensions of Non-Commutative Rings, at the bottom of p. 74 there are references given for the following fact: Among those rings $R$ such that any epic monomorphism of (unital, associative) rings $\phi: R \rightarrow S$ is an isomorphism, there are the von Neumann regular rings and the self-injective rings. Compare also the first paragraphs of chapter XI in Bo Stenström's book Rings of Quotients.
And another proof in the vein of Martin's and Ralph's: If $0 \neq \alpha: R \rightarrow S$ is an epimorphism, every $R$-linear endomorphism of $S$ (as left $R$-module, say) has to be $S$-linear. (In fact, $\alpha_*: S-Mod \rightarrow R-Mod$ being full is another equivalent criterion for $\alpha$ being an epi). Now as $R$ is a skew field, $\alpha(R)$ is a direct summand in $S$. If it had a non-trivial complement, we could certainly define non-identical $R$-linear endomorphisms of $S$ whose restrictions to $\alpha(R)$ are identical; but any $S$-linear endomorphism of $S$ is determined by what it does on $1_S = \alpha(1_R)$. So $S = \alpha(R)$.<|endoftext|>
TITLE: Is the "renormalized third comultiplication" on $\mathbf{Symm}$ integral?
QUESTION [5 upvotes]: Background:
For any commutative ring $R$, let $\mathbf{Symm}_R$ be the ring of symmetric functions in countably many variables $x_1$, $x_2$, $x_3$, ... over $R$. ("Symmetric functions" really means symmetric power series of bounded degree.) It is known that $\mathbf{Symm}_R$ is generated by the elementary symmetric polynomials $e_1$, $e_2$, $e_3$, ... as an $R$-algebra. If $R$ is a $\mathbb Q$-algebra, then $\mathbf{Symm}_R$ is also generated by the power sum polynomials $p_1$, $p_2$, $p_3$, ... as an $R$-algebra. Note that $\mathbf{Symm}_{\mathbb Z} \subseteq \mathbf{Symm}_{\mathbb Q}$.
There are two ways to define a comultiplication (in the sense of coalgebras) on $R$:
The first comultiplication, called $\Delta_1$, is defined by $\Delta_1\left(e_n\right) = \sum\limits_{i+j=n} e_i\otimes e_j$ for all $n\in\mathbb N$, where the sum allows $i$ and $j$ to be zero (and $e_0$ has to be understood as $1$). This comultiplication satisfies $\Delta_1\left(f\left(x_1,x_2,x_3,...\right)\right) = f\left(x_1,x_2,x_3,...,y_1,y_2,y_3,...\right)$, where we identify the tensor product $\mathbf{Symm}_R\otimes_R\mathbf{Symm}_R$ as a ring of certain power series in "$2$ times countably many" indeterminates $x_1$, $x_2$, $x_3$, ..., $y_1$, $y_2$, $y_3$, .... In order to make sense of the term $f\left(x_1,x_2,x_3,...,y_1,y_2,y_3,...\right)$, one has to recall that $f$ is a symmetric polynomial, so that one can reorder its arguments in any way, for example as $x_1,y_1,x_2,y_2,x_3,y_3,...$.
The second comultiplication, denoted by $\Delta_2$, satisfies $\Delta_2\left(p_n\right) = p_n\otimes p_n$ for all positive integers $n$. This is not enough to define it because $p_1$, $p_2$, $p_3$, ... don't always generate the $R$-algebra $\mathbf{Symm}_R$, but at least they generate it when $R$ is a $\mathbb Q$-algebra, so one can use this definition for $R=\mathbb Q$, then show that $\Delta_2\left(\mathbf{Symm}_{\mathbb Z}\right) \subseteq \mathbf{Symm}_{\mathbb Z}\otimes_{\mathbb Z}\mathbf{Symm}_{\mathbb Z}$, so that $\Delta_2$ is also defined for $R=\mathbb Z$, and consequently (since $\mathbb Z$ is the initial object in the category of rings) also defined for any $R$. Of course, one could just as well give a more direct definition of $\Delta_2$, by setting
$\Delta_2\left(f\left(x_1,x_2,x_3,...\right)\right) = f\left(x_1y_1,x_1y_2,x_1y_3,...,x_2y_1,x_2y_2,x_2y_3,...,x_3y_1,x_3y_2,x_3y_3,...\right)$.
Again, one has to invoke symmetry of $f$ for the right hand side to make sense here. This time one also needs to check that the right hand side is well-defined at all, what with the infinitely many  terms involving the same $x_i$. Yet another way to define $\Delta_2$ is by the identity $\Delta_2\left(e_n\right) = \sum\limits_{\lambda\text{ is a partition of }n} s_{\lambda}\otimes s_{\lambda^t}$, where $s_{\mu}$ are the Schur polynomials, and $\lambda^t$ is the conjugate partition of $\lambda$. (A categorification of this identity is the isomorphism from MathOverflow question #120873 posted earlier today.)
This suggests defining a third comultiplication $\Delta_3$ by
$\Delta_3\left(f\left(x_1,x_2,x_3,...\right)\right) = f\left(x_1+y_1,x_1+y_2,x_1+y_3,...,x_2+y_1,x_2+y_2,x_2+y_3,...,x_3+y_1,x_3+y_2,x_3+y_3,...\right)$.
This, however, doesn't go well: Setting $f=p_n=x_1^n+x_2^n+x_3^n+...$, the right hand side becomes $\sum\limits_{i\geq 1,\ j\geq 1} \left(x_i+y_j\right)^n$, which involves summing infinitely many $x_i^n$'s for every $i\geq 1$, and summing infinitely many $y_i^n$'s for every $i\geq 1$.
Fortunately, these are the only undefined terms on the right hand side; all the other infinite sums do make sense. If we replace every undefined sum of infinitely many $x_i^n$'s by $rx_i^n$ for some fixed integer $r$, then we obtain
$\Delta_3\left(p_n\right) = \sum\limits_{i=1}^{n-1} \dbinom{n}{i} p_i \otimes p_{n-i} + r \otimes p_n + p_n \otimes r$.
Question:
So fix an integer $r$, and let us define a map $\Delta_3 : \mathbf{Symm}_{\mathbb Q} \to \mathbf{Symm}_{\mathbb Q} \otimes_{\mathbb Q} \mathbf{Symm}_{\mathbb Q}$ by
$\Delta_3\left(p_n\right) = \sum\limits_{i=1}^{n-1} \dbinom{n}{i} p_i \otimes p_{n-i} + r \otimes p_n + p_n \otimes r$ for every positive integer $n$.
This $\Delta_3$ is easily seen to be coassociative, and for $r=1$ even counital (with respect to the standard counity of $\mathbf{Symm}_{\mathbb Q}$).
Do we have $\Delta_3\left(\mathbf{Symm}_{\mathbb Z}\right) \subseteq \mathbf{Symm}_{\mathbb Z}\otimes_{\mathbb Z}\mathbf{Symm}_{\mathbb Z}$ for every $r$ ? In other words, can this $\Delta_3$ be defined over any commutative ring $R$ ? What is the combinatorial meaning of this $\Delta_3$ ?
Evidence:
We need to prove that $\Delta_3\left(e_n\right)$ has integer coefficients for all $n$ and $r$. Sage code and some output (please don't imitate my code) verifies my suspicion for $r=0,1$ and $n=1,2,3,...,9$. I also have some not very tangible semi-proof arguments.

REPLY [3 votes]: I have a proof of the integrality of $\Delta_3$ using Dwork's lemma (which tells when a given vector $\left(v_1,v_2,v_3,...\right)\in A^{\left\lbrace 1,2,3,...\right\rbrace}$ over some commutative ring $A$ is the vector of ghost components of a big Witt vector -- note that this is equivalent to the existence of a ring homomorphism $f:\mathbf{Symm}_{\mathbb Z} \to A$ which sends each power sum $p_n$ to $v_n$) and some binomial coefficient congruences (I am eventually going to write this up, though I cannot give a good upper bound on the "eventually").
Meanwhile this stuff doesn't seem so new. On page 14 of Andrew Baker, Birgit Richter, Quasisymmetric functions from a topological point of view, arXiv:math/0605743v4, I see a coproduct $\psi_{\otimes}$ which is exactly mine if I get it correctly that $q_0$ is an arbitrary integer. They seem to have a proof based on topology. They don't seem to notice the need for renormalization, though.
On the other hand, on the level of "algebraic rings" (commutative algebraic groups with an additional algebraic monoid structure that distributes over the algebraic group structure), the Hopf algebra $\mathbf{Symm}_{\mathbb Z}$ equipped with the comultiplication $\Delta_3$ seems to more-or-less represent the ring $\widetilde{A}$ defined in §3 of Berthelot, Grothendieck, Illusie, SGA 6, exposé 1. I am sorry for the weasel words, but I don't have the time to make more concrete assertions these days.
EDIT: Two (or three, depending on how you count) proofs of the integrality of $\Delta_3$ are now in Vic Reiner's and my notes on Hopf algebras in Combinatorics (the version with solutions). See Exercise 2.75(f) and Exercise 2.80(e). (The numbering is volatile, but the first exercise has the words "Define a $\mathbb Q$-algebra homomorphism", and the second exercise talks about "new solutions to parts (b), (c), (d), (e) and (f)".)<|endoftext|>
TITLE: Find minimum-area ellipse which encloses two ellipses
QUESTION [6 upvotes]: I need an efficient algorithm to find the ellipse with the smallest possible area which encloses two given ellipses. The given ellipses are constrained to have coincident centers at the origin but can have any orientation. The problem is limited to two dimensions. Any ideas?

REPLY [7 votes]: Below is an example to illustrate Will Jagy's solution.
(Caveat lector: I did not preserve scale from image to image.)
First, it is no loss of generality to rotate so that one ellipse $E_1$ has its major
axis along the $x$-axis:

          


Now transform by scaling the long axis down and the short axis up so that
$E_1$ becomes a circle.  The determinant of this transformation matrix is $1$, so areas are preserved. The second ellipse $E_2$ is transformed to another ellipse:

          


Compute the major axis of the transformed $E_2$:

          


Rotate $E_2$ down so its major axis is along the $x$-axes:

          


Now it is clear what the minimum area enclosing ellipse is, as its two axes are determined
by the ellipse and the circle:

          


Rotate back, and unscale:

          



Here is Will Jagy's drawing as he mentions in his comment below:<|endoftext|>
TITLE: Applications of Gauss-Bonnet theorem
QUESTION [8 upvotes]: In wikipedia,I was pretty amazed to find a proof of fundamental theorem of algebra
using Gauss Bonnet theorem.
I think given how central it is to mathematics with its far reaching generalizations like Riemann-Roch Theorem and more,I am wondering if there are more.I would also be happy to see striking applications of its generalizations.

REPLY [3 votes]: Shameless plug: in this note I showed that a two-dimensional analogue of the positive energy theorem follows essentially trivially from Gauss-Bonnet. 
(Remark: positive energy theorems are two-dimensions is not new. But previously published results assume that spatial sections are diffeomorphic to $\mathbb{R}^2$. The main contribution above is that this topological assumption can be removed as it is a consequence of the other assumptions + Gauss-Bonnet.)<|endoftext|>
TITLE: Is there a natural measurable structure on the $\sigma$-algebra of a measurable space?
QUESTION [16 upvotes]: Let $(X, \Sigma)$ denote a measurable space. Is there a non-trivial $\sigma$-algebra $\Sigma^1$ of subsets of $\Sigma$ so that $(\Sigma, \Sigma^1)$ is also a measurable space? 

Here is one natural candidate. I'm not certain, but based on answers to related questions, I think this might be the Effros Borel structure that Gerald Edgar has mentioned here and here.
The $\sigma$-algebra $\Sigma$ is an ordered set under the canonical relation given by subset inclusion $\subseteq$, and is therefore naturally equipped with a specialization topology. The closed sets are generated by downward-closed sets, and the closure of a singleton is its down-set:$$\overline{\{A\}} = \{ B \in \Sigma : B \subseteq A \}.$$ Even though this topology is highly non-Hausdorff, it's still pretty nice. For example, it's an Alexandroff space: arbitrary unions of closed sets are closed. 
Being a topological space, $\Sigma$ now has a natural measurable structure, namely, the one generated by the Borel $\sigma$-algebra $\Sigma^1 := \mathcal B_{\subseteq}(\Sigma)$. 

Is this space $(\Sigma, \Sigma^1)$ a reasonable one on which to do measure theory and probability? 

Whether it is or not, there's some non-trivial structure present. For example, we can iterate this procedure. Set $\Sigma^0 = \Sigma$, and define $\Sigma^n := \mathcal B_{\subseteq}(\Sigma^{n-1}).$ Then each one of these spaces $\Sigma^n(X) := (\Sigma^{n}, \Sigma^{n+1})$ is measurable.

Is $\Sigma : \mathrm{Meas} \to \mathrm{Meas}$ an endofunctor on the category of measurable spaces?
Under what conditions does the sequence of measurable spaces $\Sigma^n(X)$ have a limit $\Sigma^{\infty}(X)$?

REPLY [11 votes]: If $(X,\Sigma)$ is a measurable space, I think you are asking for a $\sigma$-algebra structure on $|\Sigma|$, the underlying set of $\Sigma$. We can identify this set with the set of measurable functions
$$|\Sigma|\cong \text{Hom}_{\text{Meas}}\;(X,2),$$
where $2$ is a two-point space with discrete $\sigma$-algebra. 
Thus it suffices to prove a more general result: that $\text{Meas}$ is a closed monoidal category. In other words, we would like to know that for any two measurable spaces, $X,Y$, there is a $\sigma$-algebra on the set of maps $\text{Hom}_{\text{Meas}}\;(X,Y)$, which has good formal properties (functoriality, right adjointness to $\otimes$). 
The fact that $\text{Meas}$ is a symmetric monoidal closed category was proven by Kirk Sturtz in the paper Categorical Probability Theory. See Section 2.3.<|endoftext|>
TITLE: How to understand Givental's I- and J-functions?
QUESTION [12 upvotes]: I am learning about mirror symmetry and having trouble understanding Givental's I- and J-functions. For example the J-function for the quintic threefold $X$ is defined by the formula 
$$
J:=e^{(t_0+t_1H)h}(1+h^{-2}\sum_{d\ge 0}N_dq^ddl-2h^{-3}\sum_{d \ge 0}N_dq^dpt),
$$
where $q=e^{t_1}$ and $l$ is the class of a line, $pt$ is the class of a point, and $N_d$ is the GW invariant of degree $d$. Or some consider $J_X:=i_{!}(J)$ via the inclusion $i:X\rightarrow \mathbb{P}^4$. The $I$-function is defined in a similar manner. They are cohomology valued functions (or formal power series).  
My questions are

Where do these scary looking functions come from? It looks like the Gromov-Witten potential, which is simply the generating series of GW invariants, but has a strange factor $e^{(t_0+t_1H)h}$ for example.   
What is the advantage of considering cohomology valued functions? I think this is related to Givental's proof of mirror theorem, but I cannot not follow the proof.
Are there any relation between Givental's I- and J- functions and Hosono et. al.'s cohomology valued GKZ hypergeometric series?

REPLY [2 votes]: The better source for the $J$-function now would be Dubrovin-Zhang http://arxiv.org/abs/math/9808048. The idea of that is simple - the part of the genus=1 corellators comes from the three-points corellators only. From the point of view of the Frobenius manifold - from the algebra structure of it.
The advantage of the cohomology-valued function is very simple. Everything Givental writes looks to ge thought of having certein singularity theory behind. In this case it is just natural.<|endoftext|>
TITLE: Inverse schwartz-distribution for convolution operation
QUESTION [5 upvotes]: I note here $\mathcal{D}'$ the space of all distributions and $\mathcal{S}'$ the space of tempered distributions, I am considering the following question:
Let $u \in \mathcal{D}'$ or $\mathcal{S}'$, I want to know general conditions such that we know there exists an inverse of $u$ for the convolution operation, meaning a distribution $v$ such that $u*v$ and $v*u$ can be defined and:
$$u*v = v*u = \delta$$ 
When does a solution exist? When is that solution unique and can we describe all the solutions when it is not? 
Does it change the problem to only consider right or left-inverse of $u$?
Thanks in advance.

REPLY [4 votes]: The problem of division, for $u$ having compact support, was solved by L. Ehrenpreis in the 1950's. The equation $u*v=\delta$ is solvable if and only if the Fourier transform of $u$ is slowly decreasing, which means that an estimate
$$\sup_{|\eta|< \log(e+|\xi|)} |\hat u(\xi+\eta)|\geq C(1+|\xi|)^{-N}$$
holds. See, for example Theorem 16.5.22 in volume II of Hörmander's treatise on the Analysis of LPDO.<|endoftext|>
TITLE: Base change of trace for Gorenstein or Cohen-Macaulay morphisms 
QUESTION [6 upvotes]: This is basically a question of functoriality for base change of CM morphisms.  
EDIT: $\text{ }$ Brian Conrad sent me an email explaining the that this is indeed true, and follows from his book.  I'll explain this in my case a little later.  First I will state the question.
Suppose that $f : X \to V$ is an equidimensional (dimension $d$) finite type  (reduced, if it helps) Cohen-Macaulay morphism (flat with Cohen-Macaulay fibers).  I'm also happy to assume that $V$ is integral, excellent and has a dualizing complex.  Additionally suppose that we have $f' : X' \to V$ another equidimensional (dimension $d$) finite-type (reduced)  Cohen-Macaulay morphism that factors through $f$ as below.  
$$f' : X' \xrightarrow{\phi} X \to V.$$
Further suppose that $\phi$ is finite (although the question could be asked more generally for proper $\phi$, I'll phrase it for finite $\phi$).  If it helps at any point, please feel free to assume that $f$ and $f'$ are Gorenstein morphisms.
Recall that $\omega_{f}[d] = f^! \mathcal{O}_V$ and that by Brian Conrad's book [LINK: Google books][1] we know that both $\omega_f$ and $\omega_{f'}$ are compatible with base change.   
I'd like to conclude that the following natural map is also compatible with base change:
$$\phi_* \omega_{f'} \cong R \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \cong \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \to \omega_f$$.
The map can be interpreted as evaluation at 1.  
In other words, I'd like to know that the trace map of $\phi$ is compatible with base change.  Furthermore, it would be even good enough to prove this in the $f, f'$ Gorenstein morphisms case.  
One way to do this would be as follows.  If $g : T \to V$ is any other morphism and $f_T : X_T \to T$ and $f_T': X_T' \to T$ are the base changes and $g_X : X_T \to X$ is the projection, is it true that the natural map (denoted [*] below)
$$g_X^* \phi_* \omega_{f'} \cong g_X^* \mathcal{H}om_{O_X}(\phi_* O_{X'}, \omega_f ) \to \mathcal{H}om_{O_{X_T}}(\phi_* O_{X_T'}, \omega_{f_T} ) \cong (\phi_T)_* \omega_{f_T'}$$
between abstractly isomorphic sheaves is an isomorphism?

Edit: Brian Conrad pointed out to me that this is already a special case of what is in his Theorem 3.6.1 (assuming I understand everything right).  Essentially the point is that in my case everything is finite type, which makes it all much easier.

REPLY [2 votes]: A short answer, if you are willing to upgrade your discussion to a derived category situation. The natural base-change isomorphism of duality
$$
{g'}^* f^! \cong {f'}^! g^*
$$
holds under the hypothesis of a "Tor-independent" square. In other words, you may not have $g$ flat, but if $f$ is, it holds. This is the case always if you work with varieties over a field  and the map $f$ is the structural morphism. All this is explained in Lipman's Notes on derived functors and Grothendieck duality in SLN 1960. See Theorem 4.4.1 and its corollaries.
Of course, the problem of unravelling the isomorphism and making it explicit in terms of differentials, say, might not be completely trivial.<|endoftext|>
TITLE: Rep Theory Consequences of Bott--Weil--Borel
QUESTION [28 upvotes]: I've been getting interested in the (Bott--)Borel--Weil theorem lately. As a (mainly) geometer it is very interesting to see representation appearing (from nowhere as far as I can see) in the theory of complex geometry. 
What I can't seem to find out, though, is if this fact has been used to prove any interesting results in representation theory, ie, does the geometric realization of the representation allow us to prove anything we didn't know already?

REPLY [6 votes]: Check out the book Frobenius Splitting Methods in Representation Theory by Brion and Kumar. There are lots of representation-theoretic results in positive characteristic in that book that rely crucially on the geometric realization of induced representations. For many of the representation-theoretic facts proved by Frobenius splitting methods it is true that one can also write down a non-geometric proof, but the Frobenius splitting proofs are almost always simpler and case-free (see for example chapter 4 of Brion-Kumar regarding the good filtration property for modules over a reductive group). 
And even if you're only interested in complex representations rather than positive-characteristic representations, this technique is still useful because one can base-change facts from positive characteristic to characteristic 0. There is at least one purely representation-theoretic statement over the complex numbers (regarding the so-called generalized Brylinski-Kostant filtration) that I know only a geometric proof of, using Frobenius splitting methods in positive characteristic and then base-changing.
One thing that somewhat complicates this whole picture is that often when a representation-theoretic result is first proved by geometric means it is natural to then look for an algebraic proof, so you'll often find proofs that come in both flavors. (See for example R. Brylinski's original paper on the BK-filtration, which crucially uses geometry to prove a purely representation-theoretic fact; her paper was followed by a paper of Joseph and Heckenberger which gives an alternate algebraic proof of her results. Another example is the algebraic Frobenius splitting technique devised by Kumar and Littelmann.)<|endoftext|>
TITLE: A stupid question on automorphic l-function
QUESTION [7 upvotes]: This may be a silly question for experts in this area. But I am really suffering for not being able to compute local-L function of some automorphic representation.
So, I post it hoping some benevolent one shed me a light.
Let $\pi$ be a local component of some global irr.cusp.unitary automorphic representation of $U(2)$ at split finite place.(i.e. $\pi$ is $GL(2)$ reps)
Then, there are three possible candidates for $\pi$ except for supercuspidal;
1)$B(\chi_{1} , \chi_{2})$ for $\chi_{1} \cdot \chi_{2}^{-1}=1$
2)$B(\chi_{1} , \chi_{2})$ for $\chi_{1} \cdot \chi_{2}^{-1} \ne 1$
3)irreducible quotient of $B(\chi\left\vert \cdot \right\vert^{\frac{1}{2}},\chi\left\vert \cdot \right\vert^{-\frac{1}{2}})$ for unitary character $\chi$.
(here, all $\chi, \chi_{1} , \chi_{2}$ are character of $GL(1)$ )
Then, for any character $\gamma$ of $GL(1)$, 
what is $L(s,\pi \otimes \gamma)$ for the above three each $\pi$?
(here, L-function is local L-function and we consider $\gamma$ as $GL(2)$ character through determinant map)
Since my main concern lies in computing the order of zero or pole of the above $L$-function at $s=0$, if it is hard to write explicitly in ramified case, would you just inform me the result for each cases? Then, I am very grateful for your kindness.
(For beginner in this area, getting used to L-function calculation is quite difficult.)

REPLY [4 votes]: Since I was just asking a similar case, I can give you a partial answer. In 
http://www.math.ou.edu/~kmartin/autreps/ch3.pdf   in Definition 3.1.16
you can find the $p$-adic result, which generalize to the non-archimedean, zero-characterictic cases in the obviuous manner. References are Bump or Goldfeld-Hundley.
For the tensoring by $\gamma \circ \det$ (I think that's what you are asking about), note that
$$ \gamma \circ \det \otimes B(\chi_1, \chi_2) = B(\chi_1\gamma, \chi_2 \gamma),$$
and for the Steinberg/special rep $St(\chi) \otimes \gamma \circ \det = St(\chi\gamma)$ as well. The $L$-function of a supercuspidal representation is a constant (usually chosen to be one).
The computation at the real places can be found as Lemma 5.15.1, and the complex case on page 118 ff. in Jacquet-Langlands (both with computation).<|endoftext|>
TITLE: universal property of blow up for stacks?
QUESTION [6 upvotes]: I will use as a reference Hartshorne Prop. II.7.14, the universal property of blow-up. $\tilde{X}$ is the blow up of $X$ along a sheaf of ideals. $Z\to X$ is the morphism that is to be lifted to $\tilde{X}$ (if the inverse image sheaf of ideals is invertible).
Does the same universal property hold if $Z$ is not a scheme but an algebraic (Artin) stack?

REPLY [7 votes]: Yes, this can, for example, be checked using fppf-descent.
Pick a presentation $p:U\to X$ and pull-back everything along $p$. Since blowing-up commutes with flat base change, you may then use the universal property for $Z\times_X U\to U$ and $\tilde{X}\times_X U\to U$ to deduce that if the inverse image of the ideal sheaf to $Z$ is invertible, then there is a unique lifting $Z\times_X U\to \tilde{X}\times_X U$. Similarly, there is a unique lifting $Z\times_X R\to \tilde{X}\times_X R$ where $R=U\times_X U$. Then you conclude that there is a unique lift $Z\to \tilde{X}$ by fppf descent. 
[Edit: I misread the question as asking for $X$ to be an algebraic stack. If $Z$ is an algebraic stack (and $X$ is a scheme or an algebraic stack), then you can similarly pick a presentation $q:V\to Z$ and deduce that there are unique morphisms $V\to \tilde{X}$ and $V\times_X V\to \tilde{X}$ and hence, by fppf-descent, a unique morphism $Z\to \tilde{X}$.]<|endoftext|>
TITLE: Is there any o-minimal expansion of the real field with functions of growth higher than exponential?
QUESTION [9 upvotes]: Let $\bar{\mathbb{R}}$ be the structure of the real field, that is $(\mathbb{R},0,1,+,-,*,<)$ . We say that a function $f$ is of growth higher than exponential if for all $N\in \mathbb{N}$ there $f(x)$ is ultimately greater than $\exp^N(x)$. That is there is we can find $r\in \mathbb{R}$ such that for all $x>k$ we have $\exp^N(x) < f(x)$ where  $\exp^N(x)$ stands for $\exp(\exp(\cdots \exp(x)\cdots))$ $N$ times. 
Is it possible to find an o-minimal expansion (in the model theoretic sense) of $\bar{\mathbb{R}}$ where an $f$ as above is definable?

REPLY [7 votes]: This is a well-known open problem, see e.g. http://www1.maths.leeds.ac.uk/maloa/lecturenotes/lyon.pdf . All currently known o-minimal expansions of the reals (such as the pfaffian closure of $\mathbb R$) are exponentially bounded.<|endoftext|>
TITLE: Why is Set, and not Rel, so ubiquitous in mathematics?
QUESTION [70 upvotes]: The concept of relation in the history of mathematics, either consciously or not, has always been important: think of order relations or equivalence relations.
Why was there the necessity of singling out a particular kind of relations, namely the functional ones? I guess (but I don't have data about this) historically the recognition that "operational" expressions like $x^3$ or $\sum_{i=0}^{\infty} \frac{x^n}{n!}$ could be formalized as functional relations led to devote more attention to functions understood in the modern set theoretical sense (i.e. as a special case of relations). That viewpoint permitted to consider things such as the Dirichlet function $\chi_{\mathbb{Q}}$ (which was previously not even considered to be a true "function"!) as fully legitimate objects, and to not dismiss them as pathological, with great theoretical advantage. The language and notation of functions was preferred even to deal with things that, technically, were relations: think of "multi-valued functions" in complex analysis such as $\sqrt x$ or $\log (x)$.

1) In which instances in modern mathematics are relations used as important generalizations of functions? One example that comes to mind is correspondences in the sense of algebraic geometry.


In modern Algebra the concept of homomorphism, a kind of function between algebraic structures, is central; we are used to see expressions like $f(x*y)=f(x)*f(y)$. But it would be equally possible to define a "homomorphic relation" $R$, for example on groups, by the requirement: $(xRz$ & $yRt)$ $\Rightarrow$ $(x*y)R(z*t)$, where $*$ is the group multiplication.

2) Has this kind of "homomorphic relations" been studied (on groups or other algebraic structures)? Why algebra is pervaded with homomorphisms but we never see "homomorphic relations"? Are there something more than just historical reasons?


Let Set be the usual category of sets, and Rel be the category of sets-with-relations-as-morphisms. 
There is the faithful functor Set $\to$ Rel that simply keeps sets intact and sends a function to its graph. And there is also a faithful functor Rel $\to$ Set mapping $X\to 2^X$ and $R\subseteq X\times Y$ to $R_*:2^X\to 2^Y, A\mapsto R_*(A)=\{ y\in Y\; |\; \exists x \in A : (x,y)\in R \}$.
Despite the trivial foundational fact that set theoretical functions are defined to be a special kind of relations, it seems that in category theory Set has priority on Rel. For example the Yoneda's lemma is stated for Set; and people talk of simplicial sets, not simplicial relations; and the category Rel is just retrieved as "the Kleisli category of the powerset endofunctor on Set" (I just learned this from wikipedia) and it doesn't seem to be so ubiquitous as Set (but this impression might just depend on my ignorance in category theory).

3) Are functions really more central/important than relations in category theory? If so, is it just for historical reasons or there are some more "intrinsic" reasons? E.g. is there an analogous of Yoneda's lemma for Rel?

REPLY [5 votes]: (With apologies for reviving an old question which already has many answers) here are two perspectives:

Ponder the fact that 

$\mathsf{Set}$ is complete and cocomplete. $\mathsf{Rel}$ is not.

For example, the total relation $1 \to 2$ and the graph of your favorite function $1 \to 2$ fail to have a coequalizer in $\mathsf{Rel}$; see here for a proof. See also the Milius paper linked to by Martin Brandenburg above for an example of an $\omega$-chain which doesn't have a colimit in $\mathsf{Rel}$.
Thus we "reduce" the question of why $\mathsf{Set}$ is "more fundamental" than $\mathsf{Rel}$ to the question of why limits and colimits are so fundamental to category theory. Loosely, I'd say that limits and colimits are to category theory as universal and existential quantifiers are to logic -- without them, you're simply very limited in the kinds of things it's possible to say at all.
An object $X$ of a category $\mathcal C$ is studied by looking at the set of possible "measurements" $\mathcal C(X,A)$ by objects $A$, or dually by "probing" via the set of maps in $\mathcal C(B,X)$. In category theory, we get a lot of juice out of the fact that these two ways of looking at an object $X$ are quite different, and we can play them off one another. But in a self-dual category like $\mathsf{Rel}$, you can't tell the difference between the object doing the measuring and the object being measured -- it's like there's feedback in the measurement process. Everything is simply too jumbled together to make sense of the data you're recording. To mix metaphors some more, $\mathsf{Set}$ is like a cleaner operating room to work in than $\mathsf{Rel}$, or an experimental environment where more of the environmental variables are well-controlled.<|endoftext|>
TITLE: Classical limit and Drinfelds realization of quantum groups
QUESTION [5 upvotes]: Let 
$$\hat{\mathfrak{g}}=\mathfrak{g}\otimes\mathbb{C}[t,t^{-1}]\oplus\mathbb{C}c\oplus\mathbb{C}d$$ 
be untwisted affine Lie algebra (as defined in V.G.Kac, Infinite-Dimensional Lie Algebras, 3d ed. Cambridge University Press, 1990). For $x\in\mathfrak{g}$ we can define series
$$x(z):=\sum_{k\in\mathbb{Z}}x(k)z^k:=\sum_{k\in\mathbb{Z}}x\otimes t^{k}.$$
Lusztig proved that, roughly speaking, classsical limit $q\to 1$ of quantum affine algebra $U_{q}(\hat{\mathfrak{g}})$ is universal enveloping algebra $U(\hat{\mathfrak{g}})$ and that Chevalley generators of $\hat{\mathfrak{g}}$, $e_i, f_i$, $i=0,1,...,n$, correspond to Chevalley generators of $U_{q}(\hat{\mathfrak{g}})$, $e_i, f_i$, $i=0,1,...,n$. Precise formulation (and proof) can be seen for example in J.Hong, S.-J. Kang,  Introduction to quantum groups and crystal bases, AMS, 2002.
Drinfeld found realization of quantum affine algebras $U_{q}(\hat{\mathfrak{g}})$ (here is one article about it http://arxiv.org/abs/q-alg/9610035) in terms of generators $x_{i}^{\pm}(k), a_{i}(l), K_{i}^{\pm 1}, \gamma^{\pm 1/2}, q^{\pm d}$, $i=1,2,...,n$, $k,l\in\mathbb{Z},l\neq 0$.
Classical limit $q\to 1$ of Drinfeld generators $x_{i}^{\pm}(0)\in U_{q}(\hat{\mathfrak{g}})$ are Chevalley generators $e_{i}, f_{i}\in \hat{\mathfrak{g}}\subset U(\hat{\mathfrak{g}})$ for every $i=1,2,...,n$. 
My question is following:
Is classical limit of Drinfeld generators $x_{i}^{\pm}(k)\in U_{q}(\hat{\mathfrak{g}})$, $k\in\mathbb{Z}$, $i=1,2,...,n$, equal to elements $e_{i}(k), f_{i}(k)\in \hat{\mathfrak{g}}\subset U(\hat{\mathfrak{g}})$? In other words, is subalgebra of 
$U(\hat{\mathfrak{g}})$ generated by elements $e_{i}(k)$ (or $f_{i}(k)$), $k\in\mathbb{Z}$, $i=1,2,...,n$, classical limit of subalgebra of $U_{q}(\hat{\mathfrak{g}})$ generated by elements $x_{i}^{+}(k)$ ($x_{i}^{-}(k)$), $k\in\mathbb{Z}$, $i=1,2,...,n$?
Edit
Classical limit of Drinfelds relations for 
$U_{q}(\hat{\mathfrak{sl}}_{n})$ 
should give corresponding relations in Lie algebra 
$\hat{\mathfrak{sl}}_{n}$, 
but:
In quantum algebra $U_{q}(\hat{\mathfrak{sl}}_{n})$ we have relation (Drinfelds realization)
$$(z_{1}-q^{2}z_{2})x_{i}^{+}(z_{1})x_{i}^{+}(z_{2})=(q^{2}z_{1}-z_{2})x_{i}^{+}(z_{2})x_{i}^{+}(z_{1}).$$
If classical limit of $x_{i}^{+}(z)$ is $e_{i}(z)$, then the classical limit of the above relation would give
$$(z_{1}-z_{2})[e_{i}(z_{1}),e_{i}(z_{2})]=0.$$
But in algebra $U(\hat{\mathfrak{sl}}_{n})$ should hold
$$[e_{i}(z_{1}),e_{i}(z_{2})]=0.$$

REPLY [2 votes]: It is more like comment, but seems too long.
I would say yes. I cannot give precise reference, but by all the idealogy it is yes. Or you see some "underwater stones" - problems  ?  
Ideas are like this:
If you start with RLL=LLR description, then you need to make "Gauss" or "tringular" decomposition of L = LowTriangular*D*UpperTrianular to extract Drinfeld's currents. 
In the classical limit "everything" takes the form  A = A_{cl} + O(h).
Now if you take  L = L_{cl} , Triangular = Triangular_{cla} + O(h)
The simple fact that should hint that the "yes" answer is the following.
So the decomponsition L = LowTriangular*D*UpperTrianular in classical limit
corresponds to L_{cl} = LowTriangular_{cl} + D_{cl} + UpperTrianular_{cl} 
You see multiplication in the first order corresponds to addition. 
And this means that corresponding classical currents are just currents to appropriate
upper-lower triangular parts - which corresponds to x(z).<|endoftext|>
TITLE: Hyperbolic 3-manifolds with no geometrically finite structure
QUESTION [8 upvotes]: Does there exist a compact hyperbolic 3-manifold $M$ that is not diffeomorphic to a geometrically finite hyperbolic manifold? If yes, can such $M$ have incompressible boundary?
I think the answer should be yes to both questions but I cannot find this in the literature.
Remarks: as usual, a compact hyperbolic manifold is a compact manifold whose interior carries a complete hyperbolic structure. The structure is
 geometrically finite if it is obtained as the quotient of the hyperbolic 3-space by a geometrically finite group. Thurston's hyperbolization theorem implies:

A compact 3-manifold with non-empty boundary is hyperbolizable if and only if it is irreducible and atoroidal. 
Any compact, atoroidal, pared 3-manifold is diffeomorphic to a geometrically finite one.
Any compact hyperbolic 3-manifold is homotopy equivalent to a geometrically finite one.

REPLY [8 votes]: [Edited several times] As the comments say, the answer to the first and hence to the second question is "no". Suppose that $M$ is the compact manifold and $N$ is its interior.  Let $\rho$ be the given hyperbolic structure on $N$.  If $M$ is without boundary then the volume of $\rho$ is finite and we are done.  
Suppose instead that $M$ has boundary.  Since $N$ is hyperbolic, via $\rho$, deduce $M$ is atoroidal (which includes aspherical).  Thus $M$ is Haken.  Place all tori in the boundary of $M$ into the paring locus $P$.  By Thurston's hyperbolization theorem, the interior $N$ admits a hyperbolic metric, $\rho_0$, which is geometrically finite.  (The convex core has finite volume and contains all torus boundary components.) See Theorem 1.43 in Kapovich's book.  
[A brief note - your hypotheses can be weakened.  You assumed (a) $N$ is the interior of a compact manifold and (b) N is hyperbolizable.  This can be replaced by (a') $\pi_1(N)$ is finitely generated and the same (b).  This is called the "tameness theorem", due to Agol and also Calegari-Gabai.]
In the comments below (above?) Igor asks why an atoroidal manifold with torus boundary, and admitting an essential annulus, is Seifert fibered.  This can be found as Lemma 1.16 on page 25 of Hatcher's three-manifold notes.<|endoftext|>
TITLE: From Topological to Smooth and Holomorphic Vector Bundles
QUESTION [11 upvotes]: In the last weeks I have been think of the transition from topological vector bundles to smooth and holomorphic vector bundles. This has resulted in a few questions (with a common thread) as follows: Always $\pi:E \to B$ is a topological (complex) vector bundle over a compact base,
(A) For any given smooth manifold structure on $B$, can there exist more than one differential structure on $E$ giving $\pi:E \to B$ the structure of a smooth vector bundle. If so, what is an example?
(B) Same question as above but replacing smooth by holomorphic.
(C) For a choice of smooth vector bundle structure on $\pi:E \to B$,  does the de Rham complex of $E$ have an easy relationship with the de Rham complex of $B$. A (very) naive guess would be that 
$$
\Omega^{\bullet}(E) = \Gamma^{\infty}(E) \otimes_{C^\infty(B)}\Omega^{\bullet}(B),
$$
but I can't see that there is a well-defined way to define the differential.
(D) Same question as above but for holomorphic structures and the Dolbeault complex

REPLY [4 votes]: I am under the impression that some of the answers given may be interpreting question (A) in a manner which does not seem --- at least to me --- entirely consistent with how it is stated. Interpreting question (A) quite strictly, it seems to me that there can be in fact two distinct differentiable structures on the total space of a topological vector bundle which both make it into a smooth vector bundle. In fact, for any non-empty manifold $B$ of dimension greater than zero, and any topological vector bundle $E$ over $B$ of dimension greater than zero, there exist uncountably many distinct smooth structures on $E$ for which $E$ becomes a smooth vector bundle over $B$. I will give a very simple (and detailed) example below. Also, I will work --- out of habit --- with real vector bundles, but the exact same construction with ${\mathbb R}$ replaced by ${\mathbb C}$ works equally well for complex line bundles.
Let $B$ be a manifold, and $E$ a topological vector bundle over $B$ with projection $\textrm{proj}:E\to B$ (as usual, we confuse the vector bundle with its total space). Assume the total space $E$ admits a differentiable structure which makes it into a smooth vector bundle over $B$. Denote this smooth vector bundle (and its total space seen as a smooth manifold) by $E^{(1)}$, to distinguish it from the topological vector bundle $E$. For any continuous map $f:B\to{\mathbb R}\setminus\{0\}$ we can construct the map $H_f:E\to E$ given by
$$ H_f(x)= f(\textrm{proj}(x))\cdot x $$
In other words, the map $H_f$ is the vector bundle map $E\to E$ which sits over the identity $\textrm{id}_B$ on $B$, and which is multiplication by $f(b)$ on the fibre over $b\in B$.
It is obvious that $H_f:E\to E$ is an isomorphism of topological vector bundles whose inverse is $H_{\frac 1 f}$ (since $f$ is never zero). But it does not give a map of smooth vector bundles $E^{(1)}\to E^{(1)}$ unless $f$ is itself smooth. Now transfer the smooth structure on $E^{(1)}$ via the homeomorphism $H_f$, and denote the new smooth manifold by $E^{(f)}$. More precisely, $E^{(f)}$ is the topological space $E$ equipped with the unique differentiable structure which makes $H_f:E_0\to E^{(f)}$ a diffeomorphism.
Recall that $\textrm{proj}:E^{(1)}\to B$ is a smooth vector bundle. Note also that $H_f$ is both a diffeomorphism $H_f:E^{(1)}\to E^{(f)}$ and an isomorphism of topological vector bundles $H_f:E\to E$. As a consequence, the smooth vector bundle structure on $\textrm{proj}:E^{(1)}\to B$ transfers across $H_f$ to a smooth vector bundle structure on $\textrm{proj}:E^{(f)}\to B$ whose underlying topological vector bundle is $E$. In other words, the topological vector bundle $E$ underlying $\textrm{proj}:E^{(f)}\to B$ is actually a smooth vector bundle when we consider the smooth structure $E^{(f)}$ on the total space.
To summarize, we have two smooth vector bundle structures, $E^{(1)}$ and $E^{(f)}$, on the topological vector bundle $E$ over $B$. [By the way, the notation is self-consistent: observe that for $f=1$, $E^{(f)}$ is just the original $E^{(1)}$.] To answer the question (A) affirmatively, and give an example at the same time, assume:

the vector bundle $E$ has positive dimension;
$B$ is non-empty and has positive dimension.

Then I claim:
Lemma: The identity function on $E$ is a diffeomorphism (or even just a smooth function) $E^{(f)}\to E^{(1)}$ if and only if $f$ is itself smooth. $\square$
On the one hand, it is easy to check that if $f$ is smooth then $H_f$ is a diffeomorphism $E^{(1)}\to E^{(1)}$ (both it and its inverse $H_{\frac 1 f}$ are smooth). Therefore, the smooth structure on $E^{(f)}$ is by definition the same as the smooth structure on $E^{(1)}$, i.e. the identity is a diffeomorphism $E^{(f)}\to E^{(1)}$. On the other hand, assume that the identity $\textrm{id}_E$ is a smooth function $E^{(f)} \to E^{(1)}$. I will prove that $f$ is itself smooth. Consider the composition of diffeomorphisms
$$ G : E^{(1)} \overset{H_f}{\longrightarrow} E^{(f)} \overset{\textrm{id}_E}{\longrightarrow} E^{(1)} $$
By definition of $H_f$:
$$ G(x) = f(\textrm{proj}(x))\cdot x $$
and it is fairly easy to use local trivializations for $E^{(1)}$ to conclude that $f$ is smooth. In fact, if $\varphi:U\times {\mathbb R}^n \to E$ is a (smooth) trivialization of $E^{(1)}$ over the open $U\subset B$, it follows that
$$ \varphi^{-1}\circ G\circ\varphi(u,v) = (u,f(u)\cdot v) $$
and since $E$ has positive dimension, the restriction of $f$ to $U$ is the last component of the smooth function
$$ u\longmapsto\varphi^{-1}\circ G\circ\varphi(u,(0,\ldots,0,1)) $$
Hence $f$ is smooth.
In conclusion, under condition 1 above, the smooth structure on $E^{(1)}$ is the same as the smooth structure on $E^{(f)}$ if and only if $f$ is smooth. Furthermore, under condition 2 above, we can then find a continuous map $f:B\to{\mathbb R}\setminus\{0\}$ which is not smooth. For such a choice of $f$, $E^{(1)}$ and $E^{(f)}$ are two distinct smooth vector bundle structures on $E$.
In fact, more can be said. If we start with $E^{(f)}$ in place of $E^{(1)}$ and apply the above construction, we can easily describe the result: for any continuous functions $f,g:B\to{\mathbb R}\setminus\{0\}$ it is easy to check that $H_g \circ H_f = H_{f\cdot g}$, which implies
$$ (E^{(f)})^{(g)} = E^{(f\cdot g)} $$
Applying the preceding lemma, we conclude that the smooth structure on $E^{(f)}$ coincides with the smooth structure on $E^{(g)}$ if and only if $\frac f g$ is smooth. Consequently, the construction $f\mapsto E^{(f)}$ gives a set of smooth vector bundle structures on $E$ which is in bijection with the quotient of abelian groups
$$ C^0(B,{\mathbb R}\setminus\{0\})/C^\infty(B,{\mathbb R}\setminus\{0\}) $$
where the multiplication in each of the groups is given by multiplying functions. It is fairly easy to see that this quotient is uncountable under condition 2 above: by giving for each $b\in B$ a continuous function $f_b:B\to{\mathbb R}\setminus\{0\}$ which is smooth everywhere except at the point $b$, we determine an injection of $B$ into the above quotient of abelian groups.
Essential uniqueness of the smooth vector bundle structure on $E$
In light of the above, what can be said regarding uniqueness of the smooth vector bundle structure on $E$? Well, as some of the other answers have indicated, one can use approximation of continuous functions by smooth functions to prove the following result from its topological counterpart (i.e. the usual topological classification of vector bundles).
Theorem: Let $B$ be a smooth manifold of dimension $n$. Consider the function
$$ \theta:[B,\textrm{Gr}(k,{\mathbb R}^{k+l})]^{\textrm{smooth}}\longrightarrow \textrm{Vec}^{\textrm{smooth}}_B $$
(where the domain is the set of smooth homotopy classes of smooth functions from $B$ into the Grassmannian of $k$-dimensional linear subspaces of ${\mathbb R}^{k+l}$, and the target is the set of isomorphism classes of smooth $k$-dimensional vector bundles over $B$) defined by
$$ \theta([f])=f^\ast(\gamma_{k,k+l}) $$
where $\gamma_{k,k+l}$ is the tautological smooth vector bundle over the Grassmannian. Then $\theta$ is a bijection if $l\geq n+2$. $\square$
By using this theorem, its topological counterpart (replace smooth by continuous/topological), and the approximation of continuous functions by smooth functions, one can see that the forgetful map from the set of isomorphism classes of smooth vector bundles over $B$ to the set of isomorphism classes of topological vector bundles over $B$ is a bijection.
We do not really need the above theorem to see that isomorphism classes of smooth vector bundles inject into the isomorphism classes of topological vector bundles (although it is a convenient way to prove surjectivity). Using only the approximation of continuous function by smooth functions and smooth partitions of unity, one can approximate any topological isomorphism between smooth vector bundles by a smooth isomorphism. Moreover, given isomorphisms $\varphi,\psi:E\to E'$ of smooth vector bundles over $B$, any homotopy through topological isomorphisms between $\varphi$ and $\psi$ can be approximated by a homotopy through smooth isomorphisms between $\varphi$ and $\psi$. In particular, given two smooth vector bundle structures $E_1$ and $E_2$ on a topological vector bundle $E$ over $B$, there exists a unique homotopy class of smooth vector bundle isomorphisms $E_1\to E_2$ which, as an isomorphism of topological vector bundles, is homotopic to the identity $\textrm{id}_E:E\to E$.
Continuing in the same manner, it is not too hard to conclude that the homotopy fibres of the map
$$ \textrm{Iso}^{\textrm{smooth}}_B(E_1,E_2) \longrightarrow \textrm{Iso}^{\textrm{top}}_B(E,E) $$
(between the spaces of isomorphisms of smooth/topological vector bundles over $B$) are weakly contractible. Consequently, the map is a weak equivalence. That appears to be the strongest result we can state, to the best of my current knowledge, and especially in light of my examples above.<|endoftext|>
TITLE: Does this subgroup of "even braids" have a name?
QUESTION [16 upvotes]: The full braid group on $n$ strands $B_n$ admits a surjective homomorphism $p\colon\thinspace B_n\to \Sigma_n$ onto the symmetric group on $n$ letters, which takes a braid to the induced permutation of its ends. The kernel $P_n$ is well understood; it is the pure braid group on $n$ strands.
What about $p^{-1}(A_n)$, where $A_n$ is the alternating group on $n$ letters? Let me call this group $E_n$ for now, because I think it should be called the even braid group. However an internet search using this name (and others such as "orientation preserving braids", "positive braids" and so on) came up blank.

Does this group $E_n$ have a name, and has it been studied anywhere in the literature?  

I would be particularly interested in computations of the cohomology rings of these groups.

Update: 
It occurred to me that there was one obvious name I hadn't searched for, which was "alternating subgroups of braid groups". This led me to the following preprint,
http://arxiv.org/abs/1207.3947
which has a section on finding presentations for these groups (the alternating subgroup of the braid group associated to a Coxeter system $(G,S)$ is denoted $\mathcal{B}^+(G)$ in Section 5). So it seems that they do indeed appear in the literature, although not until surprisingly recently!

REPLY [6 votes]: As well as Andy's answer above, they have another property in common with the full braid groups $B_n$: they satisfy homological stability. There are two ways (that I know of) that one can prove this. The first is in the paper:
M. A. Guest, A. Kozlowsky, K. Yamaguchi, Homological stability of oriented configuration spaces, J. Math. Kyoto Univ. 36 (1996), no. 4, 809--814.
A sketch of the argument is as follows. It's enough to consider rational coefficients and $\mathbb{F}_p$ coefficients, for each prime $p$, separately. With $\mathbb{F}_2$ coefficients $E_n$ automatically inherits homological stability from $B_n$; this is actually true for any family of index-2 subgroups, or more generally double covering spaces, by using the mod-2 Gysin sequence and considering the double covers as 0-sphere bundles. (This only works for mod-2 coefficients since in general there is no Gysin sequence for 0-sphere bundles.)
When $F$ is a field of characteristic not 2, the homology $H_k(E_n;F)$ splits as
$$
H_k(E_n;F) \cong H_k(B_n;F) \oplus H_k(B_n;F^{(-1)})
$$
where $F^{(-1)}$ is the local coefficient system where the odd braids act by multiplication by $-1$. A model of the classifying space of $B_n$ is the configuration space $C_n(\mathbb{R}^2)$ of $n$ unordered points on the plane. There is a result of Boedigheimer, Cohen, Milgram and Taylor which calculates $H_k(C_n(M);F^{(-1)})$ in the range $k<\mathrm{dim}(M)n$ for any even-dimensional manifold $M$, and the above paper uses this to compute the $H_k(B_n;F^{(-1)})$ summand in this range. It turns out to be zero in a stable range, and so homological stability for $E_n$ follows from homological stability for $B_n$.
The calculations of the Guest-Kozlowsky-Yamaguchi paper actually work more generally, for configuration spaces on any open connected surface, so homological stability is also true for ``alternating surface braid groups''.
The stable range is worse than that for the full braid groups, however: $H_k(B_n;\mathbb{Z})$ is independent of $n$ for approximately $n\geq 2k$, whereas $H_k(E_n;\mathbb{Z})$ only becomes independent of $n$ for approximately $n\geq 3k$. The obstruction to $E_n$ having the better range lies entirely in the 3-torsion.
The second way I know of proving homological stability for the alternating braid groups is kind of a shameless plug, as it's something that I wrote (apologies if this is inappropriate; this is my first MO answer). It works more generally for what could be called ``alternating configuration spaces'' (but which are actually called oriented configuration spaces since that's what they were called by the GKY paper above). It's in the preprint
Martin Palmer, Homological stability for oriented configuration spaces, arXiv:1106.4540.
and uses a method of ``taking resolutions'' adapted from that of
Oscar Randal-Williams, Resolutions of moduli spaces and homological stability, arXiv:0909.4278.
Just for historical completeness, I should also mention that homological stability for the alternating groups $A_n$ (which can be thought of as the alternating braid groups on $\mathbb{R}^\infty$ if one is so inclined) was proved much earlier, as Proposition A on page 130 of the paper:
J.-C. Hausmann, Manifolds with a given homology and fundamental group, Comment. Math. Helv. 53 (1978), no. 1, 113--134. 
using the same kind of decomposition as in the GKY paper.<|endoftext|>
TITLE: Explicit 3-cocycle of a cyclic group
QUESTION [8 upvotes]: It is well-known that $H^{3}(\mathbb{Z}/n\mathbb{Z};U(1))$ (the 3rd cohomology group of the cyclic group of $n$ in coefficient $U(1)$) is isomorphic to $\mathbb{Z}/n\mathbb{Z}$. Does there exist an explicit formula for a 3-cocycle representing a generator of $H^{3}(\mathbb{Z}/n\mathbb{Z};U(1))$? What I mean precisely is: does there exist such a formula that is uniformly expressed in terms of $n$?
In fact, I need an expression to compute the Dijkgraaf-Witten invariant invariant of a 3-manifold.
Thanks!

REPLY [6 votes]: The  corresponding 3-cocycles of $H^3(\mathbb{Z}_n,U(1))=\mathbb{Z}_n$ are very simple:
$$
\omega_{{I}}^{}(a,b,c)    = 
\exp \left( \frac{2 \pi i p^{}_{{I}}}{n^{2}} \;
a^{}(b^{} +c^{} -[b^{}+c^{}]) \right) 
$$
with $p_I \in \mathbb{Z}_n$ labels the element in $H^3(\mathbb{Z}_n,U(1))$. Also $a,b,c \in \mathbb{Z}_n$. $[b^{}+c^{}] \equiv (b^{}+c^{})$mod $n$. You can check explicitly it satisfies 3-cocycles conditions.<|endoftext|>
TITLE: Pseudo-Differentialforms
QUESTION [5 upvotes]: I'm looking for a definition of pseudo differential forms in ordinary differential 
geometry. However searching the web gave me nothing. There are definitions in supergeometry
but that is not what I'm after.
Recently I read, that pseudo-differentialforms are the natural structure to integrate, 
since integration works on any kind of submanifold (orientation not required) for them, 
but those texts don't gave a 'clean' definition of these kind of forms.
What are pseudo-differentialforms?
Can pseudo differentialforms be defined as sections of some kind of fiber bundle?
If yes that's a definition I would prefer.

REPLY [3 votes]: Pseudo-Forms:
Let $M$ be a topological manifold and $PM$ the frame bundle of $M$. If $dim(M)=n$ then $PM$
is a $Gl(n)$-principal bundle.
Let $\tau: Gl(n) \to \mathbb{R} \; ; \; A \mapsto abs(det(A))$ the map, that maps any linear
isomorphism $f \in Gl(n)$ to the absolute value of its determinant. This defines a left 
action of $Gl(n)$ on $\mathbb{R}$ by
$$\cdot: Gl(n) \times \mathbb{R} \to \mathbb{R} \; ; \; (A,x) \mapsto \tau(A)x$$
The bundle of pseudo-forms then is the associated (line) bundle 
$$PM[\mathbb{R},\cdot]$$ 
of this action and pseudo-forms are sections of this bundle. If $M$ is smooth, this is a smooth bundle,since
the action is smooth. ($\tau$ is smooth since $det(A)\neq0$ for $A\in Gl(n)$)
But this gives only pseudo-forms that behaves right in respect to integration on $dim(M)$-dimensional submanifolds. Remains the question, ow to generalize this to submanifolds of arbitrary dimension.<|endoftext|>
TITLE: Centraliser of the complex conjugation in the absolute Galois group
QUESTION [6 upvotes]: Is it known what is the  centralizer of the complex conjugation in the absolute Galois group (i.e. the Galois group of   the field of complex algebraic numbers over the rationals)? and, what would be  a good reference for this question?

REPLY [14 votes]: If some element centralizes the complex conjugation, then it must preserve the real numbers as a set. Now, since any automorphism of the real numbers preserves the set of squares, it must preserve the order; and hence be continuous. Since $\mathbb Q$ is fixed, this implies that the real numbers are fixed pointwise. It follows that any element which centralized the complex conjugation must be the identity or the complex conjugation itself.<|endoftext|>
TITLE: Nash's paper on parabolic equations
QUESTION [11 upvotes]: I am currently studying the paper "CONTINUITY OF SOLUTIONS OF PARABOLIC AND ELLIPTIC EQUATIONS" by John Nash (cf. American Journal of Mathematics, Vol. 80, 1958, https://doi.org/2372841). The author there establishes some a priori bounds on solutions of parabolic and elliptic equations.
I would be interested in the following: What was the significance and impact of Nash's results in this paper? In which way was it important for the development of the field? Also I would like to know, why this paper is special, i.e. where there any new methods used to prove a priori bounds?
Every answer will be appreciated!

REPLY [12 votes]: De Giorgi solved Hilbert's 19th problem.
Nash independently and almost simultaneously obtained the parabolic version of the same result. Nash's result implies that all quasilinear parabolic equations, under some very reasonable assumptions, have smooth solutions.
Both De Giorgi's proof and Nash's proof are very original and develop brand new methods. Pretty much everything in regularity theory for elliptic and parabolic equations that came afterwards was influenced by these two papers.
Both Nash and De Giorgi were under 40 at that time. People usually speculate that neither one got the Fields medal because they cancelled each other out. One could even argue that these papers are the most important result in the history of PDE (or at least in ellitic PDE to avoid too many complaints).<|endoftext|>
TITLE: Cohesive sets with degree below some non-high 1-generic degrees?
QUESTION [9 upvotes]: Terminology:
Cohesive sets: $A\subset \omega$, for each recursively enumerable set $W_e$, either $A\cap W_e$ is finite or $A\cap(\omega\setminus W_e)$ is finite.
Non-high degrees: Degree $a$ such that $a'\not\geq 0''$.
I'm wondering if it is possible to construct a cohesive set using some non-high 1-generic degree as an oracle? i.e. are there $A$ cohesive, and $B$ non-high 1-generic such that $A\leq_T B$? Thanks in advance!

REPLY [7 votes]: The answer to this question is in Jockusch and Stephan's 1993 paper 'A cohesive set which is not high.'
http://www.comp.nus.edu.sg/~fstephan/coh.ps
Take $A$ and $B$ such that $A \le_T B$, $A$ is Cohesive and $B' \not\ge_T 0''$. This implies that $A' \not \ge_T 0''$ and so by the paper mentioned above $A$ computes a diagonally-not-computable function. But no 1-generic can compute such a function hence $B$ cannot be 1-generic.<|endoftext|>
TITLE: Banach Manifold
QUESTION [5 upvotes]: Let $M$ and $N$ be closed manifolds. Is it true that
 $C^{k}(N,M)$, which is the space of functions  $f: N\to M$ such that $f\in C^{k}$, is a $C^{\infty}$ Banach manifold? If so, can you help me to write charts?

REPLY [10 votes]: Yes, $C^k(M,N)$ is a smooth Banach manifold, when $M$ and $N$ are smooth closed manifolds.


In fact, you can consider manifolds of maps with more general domains and regularities, this lies in the foundations of "Global Analysis". The classic work in the area is mostly due to Eells, Palais, Eliasson and others. I also particularly like this nice short note by Tausk, that considers Banach manifolds of maps from sets to manifolds, with minimum regularity requirements.
In the previous links you can find explicit descriptions of how charts are constructed; but, very roughly (in the following, I am omitting all regularities for the sake of simplicity), you can regard a neighborhood of a map $f\colon M\to N$ as consisting of maps $g\colon M\to N$ that can be written as $g(p)=\exp_{f(p)} X(p)$, where $X\colon M\to TN$ is a vector field along $f$, and the exponential map is regarded w.r.t. some choice of background metric $h$ on $N$. Of course $X$ has to be small enough, e.g., its norm (that depends of what regularity we're talking about) has to be less than the injectivity radius of $h$ along $f(M)$. The correspondence given by such chart is then $g\leftrightarrow X$, which tells you that the tangent space at $f$ to $Maps(M,N)$ consists of vector fields along $f$ (of the same regularity as the maps considered). Details can be found in any of the links above, but I hope this brief description at least gives you some intuition...

Edit: Another good reference that I forgot to mention is this book by Hirsch, see Chap 2.<|endoftext|>
TITLE: Do finitely generated groups of polynomial growth satisfy a "uniform covering property?"
QUESTION [12 upvotes]: Let $G$ be a finitely generated discrete group with a finite symmetric generating set $S=S^{-1}\subset G$. For every group element $g$, define $\|g\|_S$ to be the length with respect to $S$, i.e. the minimal length of any word in $S$ that represents $g$. For a natural number $n$, define $B_n=${$ g\in G : \|g\|_S\leq n $}.
$G$ is said to have polynomial growth, if there exist constants $C,r$ such that
$$|B_n|\leq C\cdot n^r $$ for all $n$.
Note that this is really a property of the group itself, i.e. it does not depend on the choice of the generating set.
One particular implication of this is that by going from $B_n$ to $B_{2n}$, the ratio of the cardinalities $|\frac{B_{2n}}{B_n}|$ can be controlled uniformly by a constant.

Now I would like to strengthen this a little bit. For that, I mean the following sort of uniform covering property: (just a pragmatic notion for now)

The pair $(G,S)$ has the uniform covering property, if there exists a constant $C\in\mathbb{N}$ such that for all $n$, there exist $g_1,\dots,g_C\in G$ such that $B_{2n}\subset \bigcup_{i=1}^C g_i\cdot B_n$.

This means that there exists a constant that controls the number of copies (i.e. translates) of $B_n$ that one needs to cover $B_{2n}$.
Presumably, this should not depend on the choice of $S$ either, although I have not bothered to check. So I state my question this way:

If $G$ has polynomial growth, is there a generating set $S$ such that the pair $(G,S)$ has the uniform covering property? If not, could there be an alternative description of the class of groups that have this property?

I should say that I have only limited knowledge/skills in questions concerning advanced group theoretic problems. But the questions seems elementary enough to hope that someone with more expertise in these areas could answer it.
My motivation for asking this question is that this uniform covering property came up in my research as a sufficient condition for a theorem that states something very nice about free topological dynamical systems $(X,\alpha,G)$, where $G$ acts continuously via $\alpha$ on a compact metric space $X$. On the one hand, a positive answer to this question would mean that the theorem says something nice about a very large class of amenable groups, not just standard examples like $G=\mathbb{Z}^m$. On the other hand, if the answer is no, there could still be a description of what groups I actually talk about. My intuition says that assuming this property is something very artificial/technical and should be replaced by something more natural, e.g. polynomial growth.

REPLY [9 votes]: The answer is yes.
Moreover, 
given symmetric generating set $S$, 
any sets $B_n$ has uniform covering property for all large $n$.
Indeed, let $C_n$ be the minimal number of $g\cdot B_{ n}$ needed to cover $B_{2\cdot n}$.
Let $d$ be the word metric for $S$.
Then the sequence $(G,d/n)$ is precomact in the Gromov--Hausdorff topology.
I particular, any partial limit of $(B_{2\cdot n},d/n)$ is compact.
It follows that $C_n$ is bounded.<|endoftext|>
TITLE: How submanifolds evolve under Ricci flow?
QUESTION [9 upvotes]: This may be very naive, since I just started trying to learn Ricci flow; but I couldn't really find any answer after looking for a while in all the textbooks and lecture notes I found online...


If $(M,g_t)$ is a solution of the Ricci flow (normalized or not, I don't care), and $i\colon N\hookrightarrow (M,g_0)$ is a submanifold (with the induced metric), what is known about what happens to $(N,i^*g_t)$ in terms of its intrinsic/extrinsic geometry?


This is somewhat vague, so, to be more precise: under what conditions a totally geodesic (resp. minimal) submanifold remains totally geodesic (resp. minimal)? What evolution equation is satisfied by the second fundamental form $B^t_{\xi^t}(X,Y)=g_t(\nabla^t_X Y,\xi^t)$ of $N\subset (M,g_t)$, or shape operator, in the codimension $1$ case? Note that here almost everything depends on $t$: the connection $\nabla^t$, the normal field $\xi^t$ and obviously the metric $g_t$. I tried to take the $t$ derivative using formulas for each of the objects (e.g., the ones found in Topping's notes), but it got incredibly messy very fast -- and there was nothing I could really read off the formulas. I then did some examples, but the only ones I could do all the computations for were somewhat trivial.
I would be interested in any intuition/results related to the above, it could be for hypersurfaces (instead of general submanifolds), only in low dimensions, etc...

REPLY [8 votes]: This paragraph doesn't answer the question, but discusses some more elementary
related calculations (as in Hamilton's Nonsingular Solutions paper). Consider
a closed surface $\Sigma_{t}$ in a $3$-manifold $(M,g(t))$ evolving by Ricci
flow, where $\Sigma_{t}$ evolves in the normal direction $N$ with velocity
function $V$. Then the induced metric $\operatorname{I}_{t}$ on $\Sigma_{t}$
evolves by $\frac{\partial}{\partial t}\operatorname{I}_{t}=2\operatorname{II}
_{t}V-2\operatorname{Ric}|_{T\Sigma_{t}}$, where $\operatorname{II}_{t}$ is
the second fundamental form of $\Sigma_{t}$ and $\operatorname{Ric}
|_{T\Sigma_{t}}$ is the restriction of the Ricci tensor of $g\left(  t\right)
$ to $T\Sigma_{t}$. Let $dA_{t}$ denote the area element of $\Sigma_{t}$. Then
$\frac{\partial}{\partial t}dA_{t}=\frac{1}{2}\operatorname{trace}
_{\operatorname{I}_{t}}(\frac{\partial}{\partial t}\operatorname{I}_{t}
)dA_{t}=(H_{t}V-R+\operatorname{Ric}(N,N))dA_{t}$, where $H_{t}$ is the mean
curvature of $\Sigma_{t}$. In particular, if $V=0$, then the area
$\operatorname{A}_{t}$ of $\Sigma_{t}$ evolves by $\frac{d}{dt}
\operatorname{A}_{t}=\int_{\Sigma_{t}}(-R+\operatorname{Ric}(N,N))dA_{t}
=\int_{\Sigma_{t}}(-\frac{1}{2}R-\operatorname{sect}(T\Sigma_{t}))dA_{t}$,
where $\operatorname{sect}(T\Sigma_{t})$ denotes the sectional curvature of
the plane $T\Sigma_{t}$. On the other hand, by the Gauss equations for
$\Sigma_{t}\subset M$, the intrinsic Gauss curvature of $(\Sigma
_{t},\operatorname{I}_{t})$ is $K_{t}=\operatorname{sect}(T\Sigma_{t}
)+\det(\operatorname{II}_{t})$. So $\frac{d}{dt}\operatorname{A}_{t}
=\int_{\Sigma_{t}}(-\frac{1}{2}R-K_{t}+\det(\operatorname{II}_{t}))dA_{t}$.
This formula is nice at some time, for example, if $\Sigma_{t}$ is a minimal
surface and the scalar curvature is bounded from below $R\geq-C_{t}$ (the
latter is indeed true for Ricci flow), when and where $\frac{d}{dt}
\operatorname{A}_{t}\leq\frac{C_{t}}{2}\operatorname{A}_{t}-2\pi\chi(\Sigma)$
since $\det(\operatorname{II}_{t})\leq0$ follows from $H_{t}=0$ and by the
Gauss-Bonnet formula.
The relevant computations must be somewhere in the literature, but I don't
know where; so the following is off the top of my head and needs to be
checked. About the normal $N_{t}$ in the case of a static hypersurface,
consider a family of inner products $g_{t}$ on a vector space $E$ (e.g.,
$T_{x}M$) and a fixed hyperplane $P$ (e.g., $T_{x}\Sigma$). By $g_{t}
(X,N_{t})\equiv0$ for each $X\in P$, we have $\frac{\partial g_{t}}{\partial
t}(X,N_{t})+g_{t}(X,\frac{\partial N_{t}}{\partial t})=0$. So $\frac{\partial
g_{t}}{\partial t}(N_{t})+g_{t}(\frac{\partial N_{t}}{\partial t})=cN_{t}$ for
some $c\in\mathbb{R}$, identifying $T_{x}M$ and $T_{x}^{\ast}M$ by $g_{t}$
here and below. Dotting with $N_{t}$ yields $c=\frac{1}{2}\frac{\partial
g_{t}}{\partial t}(N_{t},N_{t})$ since $g_{t}(\frac{\partial N_{t}}{\partial
t},N_{t})=-\frac{1}{2}\frac{\partial g_{t}}{\partial t}(N_{t},N_{t})$ from
$g_{t}(N_{t},N_{t})\equiv1$. We obtain $\frac{\partial N_{t}}{\partial
t}=\frac{1}{2}\frac{\partial g_{t}}{\partial t}(N_{t},N_{t})N_{t}
-\frac{\partial g_{t}}{\partial t}(N_{t})$. Combining this with some other
formulas of this nature, such as the standard $\frac{\partial}{\partial
t}\nabla_{t}$, where $\nabla_{t}$ denotes the Levi-Civita connection of
$g_{t}$, one should be able to compute the evolution of $\operatorname{II}
_{t}$, etc.
[Dec 3, 2013] In response to Chris Gerig's question:
Let $F:N\times(0,T)\rightarrow M$ be a parametrized hypersurface in a
Riemannian manifold $(M^{n},g)$. The first fundamental form (induced metric)
at time $t$ is $\operatorname{I}_{t}(X,Y)=g(dF_{t}(X),dF_{t}(Y))$, where
$F_{t}(x)=F(x,t)$. The unit normal $\nu$ and the velocity $dF_{t}
(\frac{\partial}{\partial t})=\frac{\partial F}{\partial t}=V\nu$ are vector
fields along the map $F$. We compute that
$$
\frac{\partial}{\partial t}\operatorname{I}_{t}(X,Y)=g(\frac{D}{dt}
dF_{t}(X),dF_{t}(Y))+g(dF_{t}(X),\frac{D}{dt}dF_{t}(Y)),
$$
where $\frac{D}{dt}$ is covariant differentiation along the path $\alpha
_{x}(t)=F(x,t)$. Basically, since $[\frac{\partial}{\partial t},X]=0$ in
$N^{n-1}\times(0,T)$ and by pushing this forward by $F$, we have $\frac{D}
{dt}dF_{t}(X)=\frac{D}{dX}\left(  dF_{t}(\frac{\partial}{\partial t})\right)
=\frac{D}{dX}\left(  V\nu\right)  $, where $\frac{D}{dX}$ is covariant
differentiation along $F$ restricted to a path in $N\times\{t\}$ tangent to
$X$ (heuristically, $\nabla_{dF_{t}(\frac{\partial}{\partial t})}
dF_{t}(X)-\nabla_{dF_{t}(X)}dF_{t}(\frac{\partial}{\partial t})=[dF_{t}
(\frac{\partial}{\partial t}),dF_{t}(X)]=dF_{t}([\frac{\partial}{\partial
t},X])=0$). Using $\langle X,\nu\rangle=0$ and the product rule for $\frac
{D}{dX}$, we obtain
$$
\frac{\partial}{\partial t}\operatorname{I}_{t}(X,Y)=V\left(  g(\frac{D}
{dX}\nu,dF_{t}(Y))+g(dF_{t}(X),\frac{D}{dY}\nu)\right)  =2V\operatorname{II}
{}_{t}(X,Y)
$$
by the definition of the second fundamental form in terms of the derivative of
the unit normal.<|endoftext|>
TITLE: How many proofs that $\pi_n(S^n)=\mathbb{Z}$ are there?
QUESTION [23 upvotes]: Offhand I can think of two ways in classical homotopy theory:

Show that $\pi_k(S^n)=0$ for $k\lt n$ by deforming a map $S^k\to S^n$ to be non-surjective, then contracting it away from a point not in its image.  Now use the Hurewicz theorem to show $\pi_n(S^n) = H_n(S^n) = \mathbb{Z}$, which is easy to calculate with cellular homology.
Use the Freudenthal suspension theorem to induct up from $\pi_1(S^1)=\mathbb{Z}$, which you can prove using (say) the universal covering space $\mathbb{R}\to S^1$.

What other ways are there to prove $\pi_n(S^n)=\mathbb{Z}$?

REPLY [3 votes]: This is the first theorem proven in Whitehead's book ''Elements of homotopy theory'', section I.3. The proof is by a simplicial argument, which is similar to what I believe is Hopf's original argument (if the degree is $0$, take preimages of regular values of a map and deform to reduce the number of preimages).<|endoftext|>
TITLE: Nice algebraic approximations of classifying spaces
QUESTION [7 upvotes]: Let $G=GL_k(\mathbb C)$ be the complex linear group. Then the infinite Grassmannian is a model for the classifying space $BG$. 
We can write the infinte Grassmannian as a colimit of the finite Grassmannians $Gr(k,n)$,
which are honest algebraic varieties, that have the following nice properties:
1) They admit an affine cell paving, the Bruhat stratification.
2) The intersection cohomology complexes of the strata have good vanishing properties:
They are parity sheaves, meaning that their restriction to smaller strata have cohomology only in every second degree.
3) On top of that, when "interpreting" the intersection cohomology complexes as say mixed $\mathbb Q_l$ sheaves they have good purity properties: 
Their restrictions to smaller strata are still pure and even direct sums of shifted Tate twists.
Now my question is the following: 
If we replace $GL_k(\mathbb C)$ by another connected reductive affine complex algebraic group $G$, are there still algebraic varieties approximating $BG$ such that (some of) the above properties hold?

REPLY [3 votes]: I think there is some subtlety in the case of the symplectic group that has not yet been discussed in the two other answers. 
Embedding a symplectic bundle into a trivial symplectic bundle and constructing a classifying space from that actually means we are looking at the symplectic Grassmannian as classifying space. To recall what the symplectic Grassmannian is, let $V$ be a vector space of dimension $2n$, equipped with a symplectic form. The symplectic Grassmannian  $\operatorname{HGr}(2r,2n)$ is the subset of the Grassmannian $\operatorname{Gr}(2r,2n)$ consisting of those $2r$-dimensional subspaces of $V$ on which the symplectic form has non-degenerate restriction. It can then be identified with the quotient $\operatorname{Sp}_{2n}/(\operatorname{Sp}_{2r}\times\operatorname{Sp}_{2n-2r})$. This, however, is not a flag variety, it is affine and not projective. 
The geometry of these symplectic Grassmannians is more subtle than of the ordinary Grassmannians for the general linear group. 
In particular, they do not admit a paving with affines; they admit a paving with quasi-affines which are $\mathbb{A}^1$-contractible. This failure of the existence of a paving with affine spaces can already be seen at the level of motives for $\operatorname{HGr}(2,2n+2)$. The motives of these "quaternionic projective spaces" are 
$$
\operatorname{M}(\operatorname{HGr}(2,2n+2))\cong \bigoplus_{i=0}^n\mathbb{Q}(2i)[4i].
$$
The corresponding (but too complicated to write out here) statements for the other symplectic Grassmannians exactly fit  the topological fact that Pontryagin classes live in dimensions which are multiples of $4$.
If there was a paving with affines, some motives $\mathbb{Q}(i)[2i]$ for $i$ odd would have to appear. Somehow, geometrically, what prevents the paving to be by affine spaces (instead of just contractible quasi-affine varieties) in a motive as above is that the cells have to have a boundary which is of codimension $\geq 2$, which can not happen for affine spaces.
Anyway, all of the above can be found and is beautifully discussed in the paper by Panin and Walter on "Quaternionic Grassmannians and Pontryagin classes in algebraic geometry".
The result is that (1) is not true for symplectic Grassmannians. As far as I see, this does not really cause problems with (2) and (3); only the arguments have to be made a bit more carefully, because the stratification is not by affine spaces, but by more complicated (but still contractible) quasi-affine varieties.<|endoftext|>
TITLE: Theta group representation
QUESTION [5 upvotes]: Let $(X,L)$ be a polarized abelian variety over $k=\overline{k}$, and let $K(L)$ be the kernel of the isogeny $X\to X^\vee$ that sends $x$ to $t_x^*L\otimes L^{-1}$. The theta group $\mathscr{G}(L)$ of $(X,L)$ is a central extension
$$0\to k^\times\to\mathscr{G}(L)\to K(L)\to0.$$
Now, $\mathscr{G}(L)$ acts on $H^0(X,L)$, and we get an irreducible representation 
$$\mathscr{G}(L)\to \mbox{GL}(H^0(X,L)),$$
which can be seen to descend to a projective representation $K(L)\to\mbox{PGL}(h^0(L)-1,k)$. The nice thing about this projective representation is that, if $\varphi:X\to\mathbb{P}^{h^0(L)-1}$ is the map associated to $|L|$, then for all $z\in K(L)$ and $x\in X$, we have that $z\cdot\varphi(x)=\varphi(x+z)$. Using this, it is easily verified that $\mbox{span}\{\varphi(K(L))\}$ is all $\mathbb{P}^{h^0(L)-1}$.
We have that $K(L)=K_1\oplus K_2$, where $K_2$ is the symplectic complement to $K_1$ using the Weil pairing. 
My question is the following: Is it true that the projective representation restricted to $K_1$ (or $K_2$) is also irreducible? Put into other terms, is the linear span of $\varphi(K_1)$ all of $\mathbb{P}^{h^0(L)-1}$?

REPLY [5 votes]: If I understand correctly, $K_1$ and $K_2$ are (mutually orthogonal maximal) isotropic subgroups  of $K(L)$. Therefore they both  can be lifted (non-canonically) to (finite) commutative (sic!) subgroups of the theta group. In particular, the restriction of the projective representation of $K(L)$ to $K_i$ is actually a projectivization of a certain linear representation of finite commutative $K_i$ in $H^0(X,L)$. Since a linear representation of such groups is irreducible if and only if it is one-dimensional, we get a negative answer to your question (unless $H^0(X,L)$  has dimension 1, i.e. $L$ defines a principal polarization).<|endoftext|>
TITLE: Universality of Ext functor using Yoneda extensions
QUESTION [10 upvotes]: Theses are simple and natural questions, but I could not find anything about it. If anyone has an answer or a reference this would be very much appreciated.
Let $\mathcal{C}$ be an abelian category (possibly without enough injective nor projective).
(i) Let $A,B \in \mathcal{C}$. When are the $\mathrm{Ext}^n(A,B)$ (defined using Yoneda extensions) sets ?
(ii) Let $A \in \mathcal{C}$ and suppose that $\mathrm{Ext}^n(A,B)$ is a set for all $B \in \mathcal{C}$. Is $\mathrm{Ext}^\bullet(A,-)$ a $\delta$-functor ? If yes is it universal ?
(iii) Same as (ii) in the special case where $\mathcal{C}$ has enough projective.
PS : I edited a bit the question in view of Fernando Muro's comments.

REPLY [5 votes]: In their paper entitled "Extension categories and their homotopy", Neeman and Retakh define a spectrum of extensions $\operatorname{Ext}(A,B)$ for any two objects in an exact category $\mathcal E$ such that $\pi_{-n}\operatorname{Ext}(A,B)=\operatorname{Ext}_{\mathcal E}^n(A,B)$, in the sense of Yoneda, for any $n\geq 0$. Positive-dimensional homotopy groups vanish. The spectrum $\operatorname{Ext}(A,B)$ is an $\Omega$-spectrum defined by the classifying spaces of the categories $\operatorname{Ext}^n(A,B)$ of $n$-fold Yoneda extensions. 
Given a short exact sequence $B\hookrightarrow C\twoheadrightarrow D$, 
Quillen's Theorem B shows that the homotopy fiber of $\operatorname{Ext}^n(A,C)\rightarrow \operatorname{Ext}^n(A,D)$ is $\operatorname{Ext}^n(A,B)$, $n\geq 0$. Hence, for spectra, the homotopy fiber of $\operatorname{Ext}(A,C)\rightarrow \operatorname{Ext}(A,D)$ is $\operatorname{Ext}(A,B)$. The long exact sequence on homotopy groups defines now a $\delta$-functor $\operatorname{Ext}^\bullet(A,-)$.
Universality follows from Yoneda's lemma. If $T$ is another $\delta$-functor, a natural transformation $\operatorname{Hom}(A,-)=\operatorname{Ext}^0(A,-)\rightarrow T^0$ extends uniquely to a morphism of $\delta$-functors $\operatorname{Ext}^n(A,-)\rightarrow T^n$, $n\geq 0$, as follows. An $n$-fold extension $B\hookrightarrow X_1\rightarrow\cdots\rightarrow X_n\twoheadrightarrow A$ factors as the 'composition' of short exact sequences
$$Y_{n-1}\hookrightarrow X_n\twoheadrightarrow Y_n$$
with $Y_0=B$ and $Y_n=A$. In particular we obtain morphisms
$$T^0(A)\rightarrow T^1(Y_{n-1})\rightarrow T^2(Y_{n-2})\rightarrow\cdots\rightarrow T^{n-1}(Y_1)\rightarrow T^n(B).$$
The image of the previous extension by extension by $\operatorname{Ext}^n(A,B)\rightarrow T^n(B)$ is the image by this composite of the element in $T^0(A)$ classifying the natural transformation we started with (via Yoneda's lemma). Everything is well defined by the properties defining a $\delta$-functor.
This is only a sketch of proof. If you intend to use it in a paper you should probably provide some details at some points, e.g. carefully check the hypotheses of Quillen's Theorem B.<|endoftext|>
TITLE: Nonhyperbolic groups that contain no free abelian groups or Baumslag-Solitar groups
QUESTION [9 upvotes]: I've heard it conjectured that a finitely presentable group $G$ is hyperbolic if it satisfies the following two conditions.

$G$ contains no subgroup isomorphic to a Baumslag-Solitar group $BS(n,m)$ (including $BS(1,1) \cong \mathbb{Z}^2$).
$G$ is rationally of finite type in the sense that all the groups $H_k(G;\mathbb{Q})$ are finite-dimensional and $H_k(G;\mathbb{Q})=0$ for $k \gg 0$.

Question : Can someone tell me an example of a finitely presentable group that satisfies $1$ but but not $2$?  All the examples of finitely presentable groups I know of that don't satisfy $2$ actually have plenty of copies of $\mathbb{Z}^2$ in them.

REPLY [7 votes]: Noel Brady's finitely presented non-hyperbolic group embeds in a hyperbolic group (and hence satisfies (1)), but has infinitely generated third integral homology.  
I'm guessing, but don't remember with certainty, that the rational third homology is infinitely generated.  
Noel Brady, Branched coverings of cubical complexes and subgroups of hyperbolic groups, J.
London Math. Soc. (2) 60 (1999), no. 2, 461–480. MR MR1724853<|endoftext|>
TITLE: What is the status of the PL-pseudoisotopy stability theorem?
QUESTION [14 upvotes]: Suppose that $M$ is a compact PL-manifold (possibly with boundary) and let $C^{PL}(M)$ denote the (simplicial) group of PL isomorphisms of $M \times I$ relative to $M \times \{0\} \cup \partial M \times I$, i.e. $PL(M\times I,M \times \{0\} \cup \partial M \times I)$. There is a stabilization map $\sigma: C^{PL}(M) \to C^{PL}(M \times I)$. A hypothetical PL-pseudoisotopy stability would be:

There is a function $f(k)$ so that the map $\sigma: C^{PL}(M) \to C^{PL}(M \times I)$ is $k$-connected for $n = \dim(M) \geq f(k)$.

In other words, the homotopy groups of the PL-pseudoisotopy space stabilize as one multiplies with intervals.
The history of this theorem is as follows as far as I understand. Hatcher in 'Higher Simple Homotopy Theory' gives an outline of a possible proof of this with $f(k)$ approximately $3k$. In 1988, following Hatcher's outline for the PL-case, Igusa gave a very detailed proof for the smooth case. If $M$ is a compact smooth manifold, $C^{DIFF}(M)$ is the topological (or simplicial) group $DIFF(M\times I,M \times \{0\} \cup \partial M \times I)$ and $\sigma$ again the stabilization map, then the statement of the smooth pseudoisotopy stability theorem is as follows:

Let $n = \dim(M)$. The stabilization map $\sigma: C^{DIFF}(M) \to C^{DIFF}(M \times I)$ is $k$-connected for $n\geq\max(2k+7, 3k+4)$.

This implies, by an argument of Burghulea and Goodwillie, a PL-pseudoisotopy stability theorem for smoothable PL-manifolds M with the same range as Igusa's theorem. Also a general PL-pseudoisotopy stability theorem by smoothing theory implies the smooth pseudoisotopy stability theorem. However, as far as I can find the general PL case is still open (see e.g. Waldhausen-Jahren-Rognes, page 22). My questions are thus as follows:

Is the general PL-pseudoisotopy stability theorem still open? If not, is there a reference for a detailed proof?
If the answer to the previous question is yes, do people expect the PL-pseudoisotopy stability to be true? Is the range expected to the be same as Igusa's? 
If the answer to the previous question is yes, how is a hypothetical proof expected to go? Is it a matter of making Hatcher's outline precise to get something similar to Igusa's proof or are new ideas needed?

REPLY [13 votes]: As you say, Hatcher once argued that the map $\sigma_M^{PL}:C^{PL}(M)\to C^{PL}(M\times I)$ is $k$-connected where $k$ is roughly $n/3$, but the proof was not all there.
And as you say Igusa later proved that the analogous map $\sigma_M^{DIFF}:C^{DIFF}(M)\to C^{DIFF}(M\times I)$ is $k$-connected  where again $k$ is roughly $n/3$.
Since Igusa's argument was in its broad outlines modeled on Hatcher's, it is easy to imagine that someone might be able to go back and fix Hatcher's proof. But apparently nobody ever has.
Concerning the deduction of PL stability from smooth stability, as you say, it only works in the case of smoothable PL manifolds. There's nothing tricky about it. Here's the smoothing theory ideas we need:
Smoothing theory identifies the homotopy fiber of $Diff(M)\to PL(M)$ (diffeomorphisms fixed on the boundary to PL homeomorphisms fixed on the boundary) with the space of sections (fixed on the boundary) of a bundle over $M$ with fiber $PL_n/O_n$. Likewise it identifies the homotopy fiber of $C^{DIFF}(M)\to C^{PL}(M)$ with sections of a bundle whose fiber is the homotopy fiber of $PL_n/O_n\to PL_{n+1}/O_{n+1}$. And if we write $F^{PL}(M)$ for the homotopy fiber of $\sigma^{PL}$ and likewise for $DIFF$ then it identifies the  homotopy fiber of $F^{DIFF}(M)\to F^{PL}(M)$ with sections of a bundle whose fiber is the homotopy fiber of a map $$fiber(PL_n/O_n\to PL_{n+1}/O_{n+1})\to \Omega\ fiber (PL_{n+1}/O_{n+1}\to PL_{n+2}/O_{n+2})$$
Now, by the Alexander trick the spaces $PL(D^n)$, $C^{PL}(D^n)$, and $F^{PL}(D^n)$ are contractible. Igusa's theorem tells us that $F^{DIFF}(D^n)$ is roughly $n/3$-connected. It follows that the space 
$$fiber(PL_n/O_n\to PL_{n+1}/O_{n+1})\to \Omega\ fiber (PL_{n+1}/O_{n+1}\to PL_{n+2}/O_{n+2})$$
becomes roughly $n/3$-connected after looping $n$ times. (Also this space is already known to be better than $n$-connected; in fact $fiber(PL_n/O_n\to PL_{n+1}/O_{n+1})$ is known to be $(n+1)$-connected). So this space is about $4n/3$-connected. 
It follows that that space of sections over $M$ is about $n/3$-connected, so the fiber of $F^{DIFF}(M)\to F^{PL}(M)$ is about $n/3$-connected; so $F^{PL}(M)$ is about $n/3$-connected, just like $F^{DIFF}(M)$. In other words, the $PL$ stability result for $M$ follows from the $DIFF$ stability result for $M$ and for $D^n$ as long as the $n$-manifold $M$ is smoothable. You lose just one degree of connectivity; $k$ becomes $k-1$.
For the deduction of DIFF stability from PL stability (if we knew PL stability), something more is needed, because you don't have the Alexander trick; you don't have some $n$-manifold for which $C^{DIFF}(M)$ is highly connected. This is where Burghelea and I had an idea. Use a relative connectivity argument. Suppose $M$ is obtained by attaching a handle $H$ to $M'$, where $H=D^p\times D^q$, $p+q=n$, and $M'\cap H=D^p\times S^{q-1}$. It was known, using Morlet's disjunction lemma, that the space of concordance embeddings of $H$ in $M$, $CE(H,M)$, has a $(2n-2p-4)$-connected map to the $p$th loopspace of $CE(\ast,M)$ where $\ast$ is a point, if $n-p\ge 3$, and in the case when $M$ is a disk it is also true that $CE(\ast,M)$ is $(2n-5)$-connected. So the suspension map $\sigma^{DIFF}:CE(H,M)\to CE(H\times I,M\times I)$ is a map between roughly $(2n-2p)$-connected spaces, therefore a roughly $(2n-2p)$-connected map.
That paragraph was all DIFF. There is a fibration sequence
$$
C(M')\to C(M)\to CE(H,M)$$ (DIFF or PL). If we have PL stability, so that $C^{PL}(M')\to C^{PL}(M'\times I)$ and $C^{PL}(M)\to C^{PL}(M\times I)$ are roughly $n/3$-connected maps, then $CE^{PL}(H,M)\to CE^{PL}(H\times I,M\times I)$ is also about that good. 
Now use smoothing theory to conclude that that homotopy fiber of
$$fiber(PL_n/O_n\to PL_{n+1}/O_{n+1})\to \Omega\ fiber (PL_{n+1}/O_{n+1}\to PL_{n+2}/O_{n+2})$$
must, after looping $p$ times (sections over $H=D^p\times D^q$ fixed on $S^{p-1}\times D^q$), be highly connected, the number being the smaller of, roughly, $2n-2p$ and $n/3$. Choosing $p$ so as to maximize $min(p+2n-2p,p+n/3)$, we find that the space is roughly $7n/6$-connected. We conclude that the comparison map $F^{DIFF}(M)\to F^{PL}(M)$ is about $n/6$-connected. Since we assumed $F^{PL}(M)$ to be $n/3$-connected, we learn that 
$F^{DIFF}(M)$ is about $n/6$-connected. We lost about half of the connectivity, but we got something.
Later it became clear that one could do better: Using a refinement of Morlet's disjunction lemma (proved in my thesis), I know that the Hatcher suspension for smooth concordance embeddings $\sigma:CE(H,M)\to CE(H\times I,M\times I)$ is almost $(2n-p)$-connected, much better than the connectivity of the two spaces involved, as long as $n-p\ge 3$. (I believe that my former student Guowu Meng had a proof of this, never published, using his 1992 thesis. I know a somewhat different proof.) Using this you can get from PL stability to DIFF stability without that loss of half of the connectivity.
Returning to the original question, this suggests a different approach to going from DIFF stability to PL stability. If $M$ is not smoothable, it can still be obtained from a smoothable manifold by attaching handles of index at least $3$, in fact of index at least $8$. If $M$ is $H\cup M'$ and the desired result holds for $M'$ then to get it for $M$ it would be enough to have it for $$\sigma:CE^{PL}(H,M)\to CE^{PL}(H\times I,M\times I).$$
It would be enough if the result about stability of DIFF concordance embeddings was valid also for PL. And I am sure that if that result of my 1982 thesis, a "multirelative" version of the disjunction lemma, could be replicated in the PL category then the result about suspension could be so replicated, too.
Now, that result was proved by completely different methods from those of Igusa. The one is sort of parametrized Morse theory. It's all about trying to match up the projection $M\times I\to I$ with the projection $N\times I\to I$ when you have a family of diffeomorphisms $M\times I\cong N\times I$. The other is about trying to match up the projection $H\times I\to H$ with the projection $M\times I\to M$ when you have a family of embeddings $H\times I\to M\times I$. 
One funny thing: in my long-ago student days, when I was trying to prove the multirelative smooth disjunction lemma, at one point I considered trying to work in the PL category instead. I had discovered that the method I was developing was a smooth analogue of a parametrized version of a PL technique called sunny collapsing. I even read something about parametrized PL sunny collapsing, but I couldn't understand it, so I went back to the smooth category and faced up to some singular sets and finished the project. 
In short, one potential strategy for getting PL stability would involve adapting Igusa's parametrized Morse theory to the PL category (fixing the details of Hatcher's proof). Another completely different strategy would 
involve adapting my parametrized sunny collapsing to the PL category (fixing the details of parametrized PL sunny collapsing).<|endoftext|>
TITLE: Can a Hamkins infinite time Turing Machine with infinite Super Turing jumps (from higher type oracles) get the power to decide $\Sigma_1^2$ sets?
QUESTION [8 upvotes]: Hamkins showed that his infinite time Turing machine has the power to decide some $\Delta_2^1$ sets. I wonder if some modifications of the machine could be made to reach level $\Sigma_1^2$ sets, or, if no modifications on sight, if the power of his machine plus infinitely iterated super jumps from super oracles (those consisting of uncountable sets of real numbers) could reach that level.

REPLY [17 votes]: One should think of the class $\Delta^1_2$ as truly enormous, closed under powerful set-theoretical constructions. It 
may help to keep in mind that the minimal transitive model of ZFC, if it exists, is contained inside $\Delta^1_2$, and so
one cannot jump out of $\Delta^1_2$ with a computational operation that is absolute to transitive models of 
set theory. Andy Lewis and I pointed out in our paper Infinite time Turing machines
that $\Delta^1_2$ is closed under the boldface jump: if $A$ is $\Delta^1_2$, then so is $A^\blacktriangledown$. 
In particular, let us imagine that we equip an infinite time Turing machine with a jump-operator black box, 
which whenever a real $x$ is written on a 
special tape, then the jump $x^\triangledown$ appears on another special tape. Such a machine could iteratively compute the jump 
transfinitely often, as suggested in your question. Nevertheless, these machines are 
still stuck inside $\Delta^1_2$; every function they compute and every set they decide will have complexity at most 
$\Delta^1_2$. (In 
fact, this model is simply equivalent to having the set $0^\blacktriangledown$ as a set oracle. And we can iterate this process an 
enormous number of times, so that oracle $0^{\blacktriangledown^{(\alpha)}}$ will still be in $\Delta^1_2$.)
Meanwhile, aiming to get beyond the $\Delta^1_2$ barrier, Philip Welch observed that the connection 
between infinite time Turing machines and $\Delta^1_2$ is related to the 
fact that the limit stage operation of the machine is defined by the limsup, a definition of complexity 
$\Sigma_2$ (the value is 
$0$ at the limit if there is an earlier stage, such that for all later stages, the value is $0$). With the goal of finding 
a corresponding machine-computation
model giving rise to $\Delta^1_3$ and higher levels of the projective hierarchy, Philip Welch and Sy Friedman introduced 
new machine models with more complicated limit behavior in their article 
"Hypermachines", Journal of Symbolic Logic, 76, No.2, June 2011, 620-636. As far as achieving $\Sigma^1_n$ might be concerned, this 
seems to be the most promising answer to your question.
As for $\Sigma^2_1$, I don't know of anything resembling infinite time Turing machines that approaches it. At this level of 
complexity (and even at levels of complexity within the projective hierarchy), the behavior of a computational device 
able to decide such properties would have to be sensitive to the background
set theory in which the device is operated, whereas our more ordinary conceptions of "computation" tend to be that 
they are absolute, for example, 
to forcing extensions.<|endoftext|>
TITLE: Math behind databases management and SQL ?
QUESTION [7 upvotes]: Are there some mathematical theories/theorems/... behind modern development of database management systems and in particular of SQL ?
I am refreshing my knowledge of these things which are quite down-to-earth "how to use" (create table ..., insert..., select * from ...), but I think some deeper understanding what is behind would be helpful. 
In particular in Wikipedia one may find some relations with 3-valued logic: 

Along with True and False, the Unknown
  resulting from direct comparisons with
  Null thus brings a fragment of
  three-valued logic to SQL. The truth
  tables SQL uses for AND, OR, and NOT
  correspond to a common fragment of the
  Kleene and Lukasiewicz three-valued
  logic (which differ in their
  definition of implication, however SQL
  defines no such operation).

But it is not very clear for me what it means and how deep it is ?

REPLY [3 votes]: David Spivak (who is an occasional contributor here) has done some work on categorical aspects of database management, and the creation of dictionaries between math world and the DB world that allow one to prove that certain operations are well-behaved in a robust sense.  The link gives a list of his ArXiv papers.<|endoftext|>
TITLE:  group of diffeomorphisms of a manifold 
QUESTION [11 upvotes]: How much has been the group of diffeomorphisms of a manifold  " been studied.
I got this information from wiki.
" Quite a lot is known about the group of diffeomorphisms of the circle. Its Lie algebra is (more or less) the Witt algebra, which has a central extension called the Virasoro algebra, used in string theory and conformal field theory. Very little is known about the diffeomorphism groups of manifolds of larger dimension. The diffeomorphism group of spacetime sometimes appears in attempts to quantize gravity." 
.What more is known about this?Has this group been calculated for the standard manifolds.Since this group is a big group,so what are the better ways of studying this object.

REPLY [13 votes]: One can approach the study of diffeomorphism groups from many perspectives: topology, geometry, differential equations, and dynamics. I'll mention a few results that I'm aware of, giving links to literature surveys on different topics. 
There is a short exact sequence $$Diff_0(M)\to Diff(M)\to MCG(M),$$
where $Diff_0(M)$ is the subgroup of diffeomorphisms isotopic to the identity. 
One can regard $MCG(M)=\pi_0(Diff(M))$. 
There is a huge literature studying $MCG(M)$, especially when $M$ is a surface. One question that has been answered for closed surfaces is that there is no section $Diff(M)\leftarrow MCG(M)$. I'm not sure what's known about the higher-dimensional version of this question. 
Topologists study the homotopy type of $Diff(M)$, which breaks down into computing $MCG(M)$ and the homotopy type of $Diff_0(M)$. Hatcher has a survey on the homotopy type of $Diff(M)$. 
This has more-or-less been completely resolved in dimensions $\leq 3$, but is quite complex for general higher dimensional manifolds. 
It is known that $Diff_0(M)$ is simple for closed manifolds by a result of Thurston. 
A general strategy then for understanding the group structure of $Diff_0(M)$ is to understand its subgroups. One aspect of this is the Zimmer program, to understand homomorphisms $\Lambda\to Diff_0(M)$, where $\Lambda$ is a higher rank lattice. Another aspect is to consider homomorphisms between diffeomorphism groups for different manifolds. 
There  are some results on dynamics of diffeomorphisms with relation to the diffeomorphism group. There is a huge literature on the dynamics of individual diffeomorphisms, but I think this is orthogonal to your question.<|endoftext|>
TITLE: Which finite groups are not the automorphism group of some rooted finite tree?
QUESTION [10 upvotes]: The question is as given in the title:

Which finite groups are not the automorphism group of some rooted finite tree?

A rephrasing could be: Is any finite group representable as the automorphism group of a finite tree? If not, what is typically unrepresentable?
In case of ambiguity: 
a homomorphism of finite rooted trees must preserved the root, and so does an isomorphism which is called an automorphism.
Context:
The motivation/spirit of the question is as follows. I make the isomorphism classes of finite graphs smaller by specifying a group acting on the graph's vertices, that is an isomorphism must now respect the group action (instead of the bigger $S(n)$ action).
Do I lose something by restricting myself to tree automorphisms instead of considering the group action?

REPLY [14 votes]: If I remember correctly, the automorphism groups of trees are those groups which you can make from symmetric groups by direct products and wreath products.  This is rather few groups.
An example of a group not occurring would be a nontrivial group with no subgroup of index 2. All simple groups from $Z_3$ upwards, for example. 
ADDED: An application of the constructive characterization is that a positive integer is the order of the group of a tree iff it is a product of factorials. So there are no odd numbers, only 2 and 6 as twice an odd number, etc..  Apparently these are called the Jordan-Polya numbers.
For homework, which groups have one of the allowed orders but still aren't groups of trees?<|endoftext|>
TITLE: Primes occurring as orders of elements of a finitely presented group
QUESTION [21 upvotes]: Is it true that given a finitely presented group $G$, either all primes
or only finitely many of them occur as orders of elements of $G$?

REPLY [10 votes]: Notation:
If $n\in \mathbb{N}$, then we denote by $\pi(n)$ the set of all non-trivial factors of $n$, including $n$ but excluding $1$. Call a set $X \subseteq \mathbb{N}$ factor-complete if it is closed under taking non-trivial factors. That is, $n\in X   \Rightarrow  \pi(n) \subseteq X$. 
Now we can state a strengthened answer to the main question of this post:
Complete characterisation:
Let $X\subseteq \mathbb{N}$. Then the following are equivalent:
$1$. $X$ is the set of orders of torsion elements of a finitely presented group $G$.
$2$. $X$ is factor-complete and has a $\Sigma_{2}^{0}$ description.
That $1\Rightarrow 2$ is immediate; torsion-orders are closed under taking factors, and have a $\Sigma_{2}^{0}$ description (see the proof of theorem 3.5 in arXiv:1107.1489v2). That $2 \Rightarrow 1$ can be seen from the following result, which is proved by a generalisation of an argument provided by Francois Dorais in an earlier post on this thread:
Technical result:
There is a uniform algorithm that, on input of a computable function $\phi: \mathbb{N}^3 \to \mathbb{N}$ which describes a factor-complete $\Sigma_{2}^{0}$ set $A$, outputs a finite presentation $P_{\phi}$ such that $A$ is precisely the set of torsion elements of $gp(P_{\phi})$.
As any set of primes is factor-complete, we get the following corollary.
Answer to main question:
A set of primes $A$ appears as the torsion elements of a finitely presented group if and only if $A$ has a $\Sigma_{2}^{0}$ description.
Observe that if $X\subseteq \mathbb{N}$ then the set $X_{prime}:=\{p_{i}\ |\ i \in X\}$ is factor-complete and one-one equivalent to $X$ (being a set of primes, with a computable numbering). So we conclude with the following:
Sets which can be realised up to one-one equivalence:
Given any $\Sigma_{2}^{0}$ set $A$, the set $A_{prime}$ is one-one equivalent to $A$, and can be realised as the set of orders of torsion elements of some finitely presented group $G$.
For completeness, we provide the proof of the technical result. This is the construction described by Francois Dorais, generalised and applied carefully so that none of the torsion elements `bump in to each other'. Apologies for its lenght.
Proof of technical result:
Let $\{a \in \mathbb{N}\ | \ (\exists m)(\forall n) (\phi(a,m,n)=1)\}$ be the description for the factor-complete $\Sigma_{2}^{0}$ set $A$, where $\phi: \mathbb{N}^3 \to \mathbb{N}$ is our computable function. 
Let $p_{1}, p_{2}, \ldots$ be the standard indexing of the primes ordered by size. 
Then we construct a countably generated recursive presentation as follows:
Take the infinite set of symbols $x_{1}, x_{2}, \ldots$; this is our generating set.
For any fixed $i>1$, add the relation $x_{p_{i}}^{i}=1$. Then, start successively computing $\phi(i,1,1), \phi(i,1,2), \phi(i,1,3), \ldots$ increasing the last variable by $1$ each time. If at some point $\phi(i,1,n)\neq 1$ then stop, add the relations $x_{p_{i}}=1$, $x_{p_{i}^{2}}^{i}=1$, and start successively computing $\phi(i,2,1), \phi(i,2,2), \phi(i,2,3), \ldots$. If again some $\phi(i,2,n)\neq 1$, then stop, add the relations $x_{p_{i}^{2}}=1$, $x_{p_{i}^{3}}^{i}=1$, and start computing $\phi(i,3,1),\phi(i,3,2), \ldots$. By interleaving this process for all $i \in \mathbb{N}$, we get a countably generated recursive presentation which we denote by $Q_{\phi}$. Notice that:
a) If $i \in A$, then there will be some (smallest) $m$ such that $\phi(i,m,1)=1, \phi(i,m,2)=1, \ldots$, and so the relation $x_{p_{i}^{m}}^{i}=1$ will be present, but no other relation involving $x_{p_{i}^{m}}$, and so $\langle x_{p_{i}^{m}} \rangle \cong C_{i}$ becomes a free product factor in this group.
b) If $i \notin A$ then for all $m \in \mathbb{N}$ we will have the relation $x_{p_{i}^{m}}=1$, and since $A$ is factor-complete $i$ does not divide any element of $A$. Hence no element of order $i$ occurs in the group.
So we end up with the group  $ gp(Q_{\phi}) \cong F_{\infty}$ $* _ {a \in A}  ( * _ {j \in \pi(a)}  C_{j})$ 
(With a slightly more complicated indexing, one could do away with the $F_{\infty}$ factor). As $A$ is factor-complete, we immediately see that it is the set of orders of torsion elements of $gp(Q_{\phi})$. Now apply the construction of Higman-Neumann-Neumann, and then the construction of Higman, to embed this in a finitely presented group with presentation $P_{\phi}$. Note that these embeddings strictly preserve the set of orders of torsion elements (see lemma 6.9 and theorem 6.10 of M. Chiodo `Finding non-trivial elements and splittings in groups'). Also note that this construction is completely uniform in the computable function $\phi$ used to describe the set $A$.<|endoftext|>
TITLE: Reference for rigid analytic GAGA
QUESTION [12 upvotes]: I'm looking for a reference for the following result.
Theorem. Let $K$ be a complete, non-archimedean field, and let $X/K$ be a projective scheme, with analytification $X^\mathrm{an}$. Then the analytification functor from coherent $\mathcal{O}_X$-modules to coherent ${\mathcal{O}}_{X^\mathrm{an}}$-modules is an equivalence of categories.
While I've seen this sort of statement in a lot of introductory notes on rigid analytic geometry (most attributing it to Keihl), none of them seem to give a published reference. Any help would be much appreciated.

REPLY [10 votes]: I am quite surprised by the attribution to Kiehl that you saw. Anyway, I think the result is due to Ursula Köpf (not only over a field $K$ but actually over an affinoid space): "Über eigentliche Familien algebraischer Varietäten über affinoiden Räumen", Schriftenreihe Univ. Münster, 2 Serie, Heft 7 (1974).
Brian Conrad gave another proof as an application of his results of relative ampleness in the rigid analytic setting (see "Relative ampleness in rigid geometry", Ann. Inst. Fourier (Grenoble) 56 (2006), n° 4).
I also learned a proof from Antoine Ducros in the setting of Berkovich spaces. I wrote in down in an appendix to my paper "Raccord sur les espaces de Berkovich", Algebra & Number Theory 4 (2010), n° 3). It is very close to Serre's proof in the complex analytic setting and probably very close to Köpf's proof too, but I cannot say for sure since I never saw her paper.<|endoftext|>
TITLE: The torsion point count in higher dimension
QUESTION [12 upvotes]: It is an easy consequence of the Serre open image theorem that for the torsion point count on elliptic curves, the following possibilities arise.

If $E/\bar{\mathbb{Q}}$ is an elliptic curve without CM, then the number of torsion points $x \in E$ with $[\mathbb{Q}(x):\mathbb{Q}] \leq d$ is $\asymp d^{3/2}$ as $d \to \infty$.
If $E/\bar{\mathbb{Q}}$ is an elliptic curve with CM, then this number is $\asymp d^2$ as $d \to \infty$.

This appears on page 44 of Serre's book, Lectures on Mordell-Weil. As stated there, it is easy to show that, for a $g$-dimensional abelian variety, the corresponding count is bounded by $\leq O(d^{N})$ with $N = N(g) < \infty$.
Should we expect, for $g$-dimensional abelian varieties, the count to be always asymptotic to $d^{\alpha}$ for a finite set of exponents $\alpha$ depending on $g$? Are there any clues as to the spectrum of those exponents?
Second, I wanted to ask about any results on the uniform torsion count problem giving bounds that are uniform in $E$ but still polynomial in $d$. (Thus I do not ask about Merel's uniform boundedness theorem and its relatives). For example, restricting to $g$-dimensional abelian varieties over $\bar{\mathbb{Z}}$ (that is, with integral moduli, or with everywhere potentially good reduction), or at least to the abelian schemes over the spectra of rings of integers of number fields of bounded degree, it is clear a priori that there is a uniform exponential bound in $d$. Should we expect this uniform bound to be strengthened to a polynomial bound, or even (for elliptic curves) to $Cd^{2+\epsilon}$? What about the generalization where we count the small algebraic points, say those of (canonical) height $< 1/d$?

REPLY [9 votes]: As far as I know, we expect that the image of the Galois group in $GL_{2g}(\mathbb A_\mathbb Q)$ is open in $G(\mathbb A_\mathbb Q)$ for $G$ the monodromy group of the Galois representation (which is expected to be independent of $l$. Then the asymptotic for this function clearly depends only on the connected component of the identity of this monodromy group, which is expected to be the Mumford-Tate group.
(1) For each possible Mumford-Tate group $G$, is the number of points in $(\mathbb Q/\mathbb Z)^{2g}$ fixed by an index $d$ subgroup in $G(\mathbb A_\mathbb Q)$ asymptotic to $d^\alpha$ for some $\alpha$?
(2) Are there finitely many possible Mumford-Tate groups?
The first one is not too hard to see. The index of the subgroup fixing an $n$-torsion point is the size of the orbit of that point, which is approximately the number of $\mathbb Z/n$ points of some algebraic variety, which is approximately polynomial in $n$. Then we can sum over all points $n$-torsion for $n\leq N$, another polynomial. If there are different sorts of orbits of different dimensions, then we can just sum over the different sorts of orbits and still get something of the form $d^\alpha$ up to some constant error.
The second one is clear since the Mumford-Tate group should be reductive, and thus there is a nice classification.<|endoftext|>
TITLE: Sums of two squares: What is known about the distribution of r(n)?
QUESTION [10 upvotes]: The distribution of sums of two squares has been studied by Landau. What is known about the distribution of the function $r(n)$, the number of representations of $n$ as the sum of two squares? Some specific questions of interest to us are these. Suppose $a \leq n\leq b$. What is the average value of $r(n)$? What is the standard deviation?  
Edit: Thanks for the answer and comments. I am particularly interested in the range $a\leq n\leq b$ where $b-a = \Theta(\sqrt{a})$.  Let $a,b$ be like that, and let $C(n)$ be the circle of radius $n$ around a fixed center.
Conjecture. The number of integers $n\in[a,b]$ with $r(n)>0$ is $\Omega(\sqrt{a})$. Furthermore, if $S$ is the segment of $C(b)$ cut off by a chord of length $\Omega(\sqrt{a})$ that touches $C(a)$ and if $r_S(n)$ is the number of lattice points in $C(n)\cap S$ then the number of integers $n\in[a,b]$ with $r_S(n)>0$ is $\Omega(\sqrt{a})$.
Added: Finally, with the help of Greg Martin, I got hold of relevant literature and educated myself. Thank you again for the useful answers and comments!

REPLY [8 votes]: I don't believe your new conjecture is true. Taking $b=a+\sqrt a$ for concreteness: standard sieve results show that the number of integers in $[a,b]$ that can be represented as the sum of two squares is $O(\sqrt{a/\log a})$. (This is closely related to the fact that the counting function of the sums of two squares is asymptotic to a constant times $x/\sqrt{\log x}$.)<|endoftext|>
TITLE: Skeleton category of the category of skeleton categories?
QUESTION [23 upvotes]: A category is a skeleton if, roughly speaking, no two distinct objects within the category are isomorphic. To every category is associated a skeleton, and two categories are categorically "equivalent" if and only if their skeletons are isomorphic. A fuller definition can be found here.
Consider the subcategory of $\bf{Cat}$ which takes as objects those categories which are skeletons, and morphisms the functors between them; I will call this $\bf{{Cat}_{Skel}}$. Note that this is not the skeleton of $\bf{Cat}$ itself, but a subcategory of $\bf{Cat}$ in which the objects are skeletal categories.
Within this subcategory of skeletal categories, there are a number of objects which are, themselves, isomorphic. So we can take the skeleton of this category, hence obtaining a new category, which I will call $\bf{Skel({Cat}_{Skel})}$. (I don't care which skeleton you take; pick one.)
This category is noteworthy in that it contains one object for each equivalence class of categories in $\textbf{Cat}$, making it perhaps more useful than looking at $\bf{Skel({Cat})}$ itself, which only contains one object for each isomorphism class of categories. My questions are:

Does $\bf{Skel({Cat}_{Skel})}$ have a name?
Has this category been studied in any detail, and if so, can someone please reference me towards any research that's been done on its structure?
Is there an essentially equivalent construction which might be defined more simply than the way I've laid it out here?
Are there any useful areas of study in which this category naturally arises?

Lastly, I've glossed over the usual foundational issues which arise when considering $\bf{Cat}$, mostly because I don't care whether you use Grothendieck universes, or a class-set theory, or only look at small categories, or some other way of solving the problem. Feel free to use any foundational approach that you want which makes $\bf{Skel({Cat}_{Skel})}$ to be consistent.

REPLY [9 votes]: I'm not entirely sure what you're looking for in an answer, but maybe I'll flesh out my comment.
It looks like what you're describing is equivalent to the homotopy category associated to the model structure on Cat where the weak equivalences are equivalences of categories. (I can say "the" because there is only one such, as pointed out in the comments. The cofibrations are functors injective on objects, and the fibrations are "isofibrations".)
I would say that in this context your category has been much studied. In particular, it is interesting to ask questions about homotopy limits and colimits in this category because many useful constructions arise in this way. (Homotopy (co)limits with this model structure are the same as "2-(co)limits" which is the name appearing in most of the literature, especially older literature.)
An example application of this language is the following theorem: The subcategory of presentable (resp. accessible) categories is closed under homotopy limits.
Using this one can prove that most of your favorite things are presentable (resp. accessible). For example, the category of modules over a monad arises via a homotopy limit construction, and this takes care of most things of interest.
Here's a neat application of this (which is the ordinary category version of a result that can be found, for example, in Lurie's HTT, 5.5.4.16.). 
Say you want to localize a category $\mathcal{C}$ with respect to some collection of morphisms, $S$. Usually $S$ will not be given as a set, but if $\mathcal{C}$ is presentable you're usually okay if $S$ is generated by a set. Well, it turns out that if $F: \mathcal{C} \rightarrow \mathcal{D}$ is a colimit preserving functor between presentable categories, and $S$ is a (strongly saturated) collection of morphisms in $\mathcal{D}$ that is generated by a set, then $f^{-1}S$ is a (strongly saturated) collection of morphisms generated by a set. The argument goes by way of showing that the subcategory of the category of morphisms generated by $f^{-1}S$ is presentable, using a homotopy pullback square.
Adapting this to the model category or $\infty$-category setting, one sees immediately that localizing with respect to homology theories is totally okay and follows formally from this type of argument. (Basically, after fiddling around with cells to prove the category of spectra is presentable, you don't have to fiddle any more to get localizations. This is in contrast to the usual argument found in Bousfield's paper. You've moved the cardinality bookkeeping into a general argument about homotopy limits of presentable categories.)
Anyway, apologies for the very idiosyncratic application of this language; these things have been on my mind recently. I'm sure there are much more elementary reasons why one would care about using the model category structure on Cat.<|endoftext|>
TITLE: To what extent does trajectory determine gravity sources?
QUESTION [11 upvotes]: Suppose one has in-hand an accurate time-space trajectory in $\mathbb{R}^3$ of a (small) body,
say an asteroid or satellite—effectively a point.
To what extent does this trajectory determine the point masses that could 
gravitationally determine it (according to inverse-square gravitation)?
Is this highly underdetermined, in that there are many point-mass distributions
that would lead to the (exact) same trajectory, or does the trajectory essentially uniquely determine the masses?  Perhaps
this question only has a sharp answer with some assumptions on the size of the point masses, i.e.,
planetary or star-like, as opposed to spread-out asteroid belts or dust clouds...?

        
(source: iop.org) 
        (Suggestive image from: "Spacetime symmetries and Kepler's third law,"
2012, Class. Quantum Grav.: 29. 217002 (arXiv link)).
 (Added 9Feb13).
In light of Ben Crowell's incisive analysis, and the various comments and answers
(by Joel, Abhinav, Brendan, Theo)
which point
to fundamental nonuniqueness, permit me to rephrase the question:

Given an accurate (space-time) trajectory of a point-mass (planet) in a fixed coordinate system,
  and given a number $n$, under the assumption that there are $n$ stationary/fixed point masses
  (stars) that might have caused the planet's trajectory solely due to inverse-square
  Newtonian gravitation, does the trajectory of the planet determine the fixed stars' masses and positions?

It might be useful to distinguish general positions of the stars vs. special arrangements.
If this can be answered for each $n$, then one could explore successively larger values of $n$.

REPLY [2 votes]: Although this question has been answered many times, I would like to add the following simple way to generate examples.  It is clear that in the plane any central force, no matter how complicated,  generates circular orbits as special cases.  There are many ways to distribute matter in three space to generate such a force field.  For example, any distribution of mass (which can, of course, consist of point masses) along the $z$-axis
will produce such a field in the $x,y$-plane.  As a side remark, such configurations occur
in practice---e.g., as the gravitational field generated by a wire.
I might add for those interested in the theme of trajectories that there is a considerable body of work on the question of which families of curves can occur as the trajectories of force fields, including a monograph
"Differential-geometric aspects of dynamics",  by Edward Kasner and his associates, much of which is easly accessible online. This is, however, not directly relevant to the query.<|endoftext|>
TITLE: Is this a vertex algebroid?... What is vertex algebroid?
QUESTION [15 upvotes]: A couple of day ago, I was lamenting to a friend about the fact that I have no idea what vertex algebroids are.
During our discussion, I came up with a guess of what a vertex algebroid might be.
I'm wondering whether this guess has any merit.

Question: is the definition below equivalent to what people call "vertex algebroid"?

Let me first tell you what I know about vertex algebras:
Vertex algebras
Let $\mathbb D=Spec(\mathbb C[[t]])$ be the formal disk, and
let $\mathcal M_n$ denote the moduli space of n-tuply punctured formal discs.
(While you probably have some intuition as to what it means to puncture $\mathbb D$ once,
but you might be puzzled by the idea of puncturing it more than once.
Indeed, the moduli space of twice punctured formal discs doesn't have any $\mathbb C$-points,
so it's not possible to exhibit any twice punctured formal disc!
However, given the once punctured formal disc $\mathbb D^*=Spec( \mathbb C[[t]][t ^ {-1}])$,
there's $\mathbb D^*$ many places where one can perform the second puncture.
So that produces a map $\mathbb D^*\to \mathcal M_2$, showing that $\mathcal M_2$ is at least non-trivial.)
The collection of all $\mathcal M_n$ looks a lot like the little discs operads:
But if you think about it, you'll see that it's not quite an operad: it's only a partially defined operad.
The operad multiplication
$$
\mathcal M_n\times \mathcal M_{k_1}\times\ldots\times\mathcal M_{k_n}\to \mathcal M_{k_1+\ldots+k_n}
$$
is only defined on a "subset" of $\mathcal M_n\times \mathcal M_{k_1}\times\ldots\times\mathcal M_{k_n}$.
But that doesn't matter so much: once can still define the notion of an algebra over this partially defined operad, and it turns out that an algebra over $\mathcal M_\bullet$ is exactly the same thing as a vertex algebra.
Vertex algebroids?
Fix a smooth variety $X$ and consider the moduli space of n-tuply punctured formal discs equipped with a map to $X$.
Those again form a partial operad (probably it's better to call it a colored partial operad),
and one can consider algebras over it.
Could it be that those algebras are equivalent to vertex algebroids over $X$?

REPLY [4 votes]: This was meant to be a comment to David's answer but it grew too much. At first I thought with David's intuition that your definition seemed as a sheaf of vertex algebras on X. But then I realized that I couldn't see the restriction maps. On the other hand, over each point $x \in X$ --restricting to constant maps to X -- you obtain an algebra over that partial operad of punctured formal discs. In fact,  your definition seems to agree with the definition hinted in 3.4.9 (ii)  [Beilinson and Drinfeld, Chiral Algebras] of $X$-family of factorization algebras, which in turn are defined in 3.4.6 of loc. cit.  I chose that definition instead of the standard one of 3.4.4-5 because there you see that replacing the category $C(X)$ ($X$ as in 3.4.6. of loc. cit. and not your target space that I will denote now $Y$) that fibers over affine schemes by a suitable category fibering over $Sch/Y$ (essentially replace $Z$ in that definition by maps $Z \rightarrow Y$) you obtain the notion of $Y$-family of factorization algebras. So your definition is a factorization version of $Y$-families of vertex algebras. 
Now just to give references, as David pointed out, vertex algebroids are truncated  sheaves of vertex algebras. This idea came from works of Gorbounov, Malikov, Schechtman and Vaintrob [Gerbes of chiral differential operators II]. You start with a sheaf $\mathcal{V}$ of $\mathbb{Z}_{\geq}$-graded vertex algebras on $X$ and look at the structure that you get from the filtered components $\mathcal{V}_{\leq 0}$ and $\mathcal{V}_{\leq 1}$. This definition is nicely distilled in [Bressler; Vertex algebroids I]. 
However, I thought this might give you a little intuition given that you think of vertex algebras as algebras over certain operad. It turns out [Beilinson Drinfeld - 3.3.3] that vertex algebras are Lie algebras in certain (compound) tensor category of $D$-modules on the disk $D$ (again my notation collides with yours). You can think of a Lie algebroid on $X$ as a sheaf of Lie algebras (in the category  vector spaces) with an action on a given sheaf of commutative algebras $\mathcal{O}_X$ and a compatible $\mathcal{O}_X$-module structure. A $D$-module version of that statement gives rise to the notion of a $Lie^*$-algebroids and Chiral Lie algebroids (where we replace the tensor category of $\mathcal{O}_X$-modules with some category of $D_X$-modules where the tensor product is not representable). Granted this is very sketchy, but the relevant definitions are all in Beilinson and Drinfeld's book: 1.4.11 gives the general definition of a $Lie^*$ algebroid, 2.5.16 treats them in the $D$-module setting, these ought to be thought more as "conformal algebras" or even commutative vertex algebras rather than "vertex algebras". Finally in 3.9.6 comes the definition of a Chiral Lie algebroid which corresponds to vertex algebroids. 
One last observation that perhaps someone here can comment, in the introduction to section 2.5, Beilinson and Drinfeld make a very interesting comment how how $Lie^*$ algebroids give rise to Lie algebroids on ind-schemes of sections over formal punctured discs. They refer to sections 4.6.10 and 4.6.11 for a global theory. Unfortunately those sections do not exist.<|endoftext|>
TITLE: examples of moduli functors for which coarse moduli space does not exists
QUESTION [6 upvotes]: Well, the title almost says it all. I would like to list as many examples as possible of moduli functors, for which a coarse moduli space does not exist (and maybe explain why). So, examples such as $[\mathbb{A}^1/G_m]$ are not what I mean. I would really like to see moduli functors that come from some classifying problem.

REPLY [6 votes]: Here is a classifying problem without a coarse moduli space:  Let $F$ be the stack of line bundles with section, say over an algebraically closed field k.  That is $F(X)$ is the category of pairs $(L,s)$ where $L$ is a line bundle on $X$ and $s \in \Gamma(X,L)$ is a section of $L$.  Then $F$ has no coarse moduli space:  if it did, there would be two points, corresponding to $(k,1)$ and $(k,0)$ with the former specializing to the latter.  This would be an affine scheme with a closed point and a generic point, both with residue field $k$.  That's impossible.
Of course, $F$ is just a geometric description of $[\mathbf{A}^1 / \mathbf{G}_m]$, so we already knew it didn't have a coarse moduli space.<|endoftext|>
TITLE: What kind of subset is Spec(R_P) in Spec(R)?
QUESTION [9 upvotes]: While trying to prove some properties of a subset of a prime spectrum I arrived at the following question: Let $R$ be a commutative ring and let $P \in \mathrm{Spec}(R)$. We can consider $\mathrm{Spec}(R_P)$ as a subset of $\mathrm{Spec}(R)$. My question is: What kind of subset (subspace) is this? I guess it is not true that $\mathrm{Spec}(R_P)$ is open, right? Is it true that a neighborhood of the generic point in $\mathrm{Spec}(R_P)$ is a neighborhood of $P$ in $\mathrm{Spec}(R)$?

REPLY [6 votes]: As already said, in general the image is not open and even not constructible. But it is pro-constructible (that is, locally an intersection of locally constructible subsets, see EGA IV.1.9.4, and 1.9.5(ix) because $X:=\mathrm{Spec}(R)$ is affine hence quasi-compact).  
Denote this image by $S$. This is a subset of $X$ having exactly one closed point and which is stable by generization as pointed out by Martin. I claim that these two properties characterize all possibles images $S$ when $X$ is noetherian (this hypothesis could be weakened if we notice that $S$ is always quasi-compact, but I don't know how exactly). Here $X$ needs not be affine. 
Indeed, let $s$ be the unique closed point of $S$. Let us show $S$ is the intersection of all open neighborhoods of $s$ in $X$ (then it is easy to show $S$ is the image of $\mathrm{Spec}(O_{X,s})$). Let $U$ be any open neighborhood of $s$ in $X$. Then $S\cap (X\setminus U)$ is noetherian hence admits a closed point if non-empty. But this would be a closed point of $S$ different from $s$. So $S\cap (X\setminus U)=\emptyset$ and $S\subseteq U$.  Conversely, any point $x\in X$ in the intersecion of all $U\ni s$ is a generization of $s$ (otherwise the complementary of $\overline{\{ x\}}$ is an open neighborhood of $s$ not containing $x$).<|endoftext|>
TITLE: Edge graph of the polytope of a Bruhat interval
QUESTION [5 upvotes]: Let $\Gamma$ be a Coxeter group on some generating set $S$, with reflection representation $V$. Then $\Gamma$ has two standard partial orders, the weak and strong Bruhat orders.
Moreover, if $\lambda \in V$ is chosen generically (any free orbit will do), then the covering relations in weak order are given exactly by the edge graph of the convex hull of $\Gamma\cdot \lambda$.

Let $[u,v]$ be an interval in strong Bruhat order. Has the edge graph of the polytope $hull([u,v]\cdot \lambda)$ been studied?

For example, the polytope $hull([123,321] \cdot (1,2,3))$ is a
hexagon, and the strong Bruhat cover $231 > 132$ defines an edge
through the middle of this hexagon, so not a weak cover. Whereas
the polytope $hull([132,321] \cdot (1,2,3))$ is a trapezoid, 
one edge of which connects $231$ and $132$.
EDIT: perhaps I should admit the geometry here. If $\Gamma$ is a Weyl group of a Lie group $G$ -- and I am happy to make this assumption, albeit I want $G$ Kac-Moody -- and $V$ the corresponding weight lattice, and $\lambda$ a dominant weight, then $hull(W\cdot \lambda)$ is the moment polytope for $G/B$ bearing the Borel-Weil line bundle ${\mathcal L}_\lambda$. Within $G/B$ we have the Richardson variety $\overline{BuB}/B \cap \overline{B_- vB}/B$, and $hull([u,v]\cdot \lambda)$ is the moment polytope of that.

REPLY [3 votes]: For type $A$, see Appendix A of Kodama-Williams. They prove that all the edges $Hull([u,v]) \cdot \lambda$ correspond to strong Burhat covers. They also show that $Hull([u,v]) \cdot \lambda$ is the Minkowski sum of the polytopes $c_i Hull([u,v]) \cdot \alpha_i$, where $\lambda = \sum c_i \alpha_i$ is the decomposition of $\lambda$ into simple roots.
Skimming the proofs, I think the second sentence should still be true in other types. The first sentence, however, uses the fact that weak and strong order on $S_n/(S_k \times S_{n-k})$ coincide, so I think this should not generalize.<|endoftext|>
TITLE: How are infinite-dimensional manifolds most commonly treated?
QUESTION [34 upvotes]: I originally posted this question on StackExchange, where it was suggested I post here. It was also suggested I read about Hilbert manifolds and Fréchet manifolds. Nevertheless, I am still looking for an answer to (mainly the first part of) my question.
At a summer school I recently attended, infinite-dimensional manifolds popped up. I have never worked with them before (although I'm very familiar with finite-dimensional manifolds). The lecturer at the school did not give any details about the technical realization of infinite-dimensional manifolds, mentioning that there were issues (such as picking a topology) that he would leave out for the sake of clarity, since the relevant results were true independent of the exact technical details. An internet search reveals that Banach manifolds are one way of treating infinite-dimensional manifolds, but there are others.

Are Banach manifolds the most common way of defining infinite-dimensional manifolds, or are there other notions commonly used? Is there a more or less universal consensus about when to use which treatment? What are the most important (dis)advantages of each?
Supposing I want to learn the basics of infinite-dimensional manifolds, are there any well-written introductory texts you would recommend? (on StackExchange, The Convenient Setting of Global Analysis by A. Kriegl and P. Michor was recommended)

REPLY [10 votes]: Different notions of manifolds may be useful in different approaches. Maybe more important than finding "universal consensus" on which one is suppoosed to be used where is to have a language to treat the various notions uniformly such as to be able pass between them in a useful way. 
One such more general category is that of "diffeological spaces". In 
http://ncatlab.org/nlab/show/diffeological+space
is discussed how for instance Frechet manifolds faithfully embed into these. 
Diffeological spaces form a "quasi-topos".  Following Grothendieck's lead, it is better to go one step further to an actual topos for differential geometry. The topos generalization of diffeological spaces is that of "smooth spaces" (smooth sets/smooth 0-types)
http://ncatlab.org/nlab/show/smooth+spaces
which is the sheaf topos over the category of smooth manifolds (or equivalently just over that of Euclidean spaces with smooth maps between them). Variants of this with a bit more information about the differential aspect of differential geometry include for instance the "Cahier topos"
http://ncatlab.org/nlab/show/Cahiers+topos
See there for pointers for how "convenient vector spaces" and hence the infinite-dimensional manifolds modeled on them ("convenient manifolds") are faithfully embedded into that topos.
In these toposes for instance all mapping spaces exist and can be usefully treated, while they agree with the infinite-dimensional manifold structures on mapping spaces whenever those actually exist. Similar statements hold for all other universal constructions.
Thereby topos theory transforms the question of finding "universal consensus" on which definition is best to a more relevant technical question: which concrete definition happens to constitute a presentation of a universaly existing construction in the topos. Presentations are useful, but are man-made. They may apply or not, may be useful here or there. But the smooth spaces which they present exist universally, robustly and meaniningfully irrespective of such choices.<|endoftext|>
TITLE: Plus construction considerations.
QUESTION [21 upvotes]: In order to realise the K-groups of a ring as the homotopy groups of some space associated to that ring, Quillen proposed the following (roughly-sketched) construction:
Recall that $K_1(R) = GL(R)/E(R)$, so we're, at least, looking for a space $X$ with $\pi_1(X) = K_1(R)$. The classifying space of $K_1(R)$ is obviously not a serious candidate, but we can start with the classifying space of $GL(R)$ (given the discrete topology), $BGL(R)$. Then, choosing representative loops whose classes generate $E(R) \subset \pi_1(BGL(R))$, and then attaching 2-cells using these loops on the boundaries, we end up with something that has a fundamental group of $K_1(R)$. Furthermore, now Quillen adjoins 3-cells essentially to correct the homology back to that of $BGL(R)$, which was messed up in the addition of those 2-cells. We end up with a space denoted $BGL(R)^+$, on which we define $K_i(R) := \pi_i(BGL(R)^+)$.
More specifically, Quillen sought to find a space $BGL(R)^+$ for which $(BGL(R), BGL(R)^+)$ was an acyclic pair (that is, the induced map $H_*(BGL(R), M) \to H_{*}(BGL(R)^+,M)$ is an isomorphism for all $K_1(R)$-modules $M$). My question is

In search of a space on which to define K-groups, why was it desirable to find something satisfying the above condition on homology?

My best guess is that it was observed that $K_1(R) = GL(R)/E(R) = GL(R)_{ab} = H_1(GL(R), \mathbb{Z})$, and $K_2(R) = H_2(E(R), \mathbb{Z}) = H_2([GL(R), GL(R)], \mathbb{Z})$, and so it seemed reasonable that all K-groups should be related to the homology of $GL(R)$ - and so the above is a stab at preserving that homology.
I am trying to learn about K-theory, but, as with most presentations of math, I'm sure, I'm coming across too many cleanly-shaven formulas and propositions, entirely divorced from any sorts of thought-processes, big-pictures, or mentions of what is trying to be done and to what ends. Please share with me what you think is going on; here, with the +-construction, that is. (And, if and only if it is not a completely unrelated question, what sort of K-theoretical phenomena suggested that these groups should be homotopy groups?)

REPLY [12 votes]: For convenience (at least my own) and completeness, I want to give an explanation of Tom Goodwillie's answer, as it was not obvious to me how to prove the statement he makes. I wanted to leave it as a comment to his answer, but it became way too long. In summary, the point is that applying the plus construction to the spaces defined by Tom gives the connective covers (i.e. the Whitehead tower) of $BGL(R)^+$. This follows from the fact that the plus construction preserves certain fibre sequences.
Abbreviate $X=BGL(R)$. Let $F_1=X=BGL(R)$, and denote by $F_{i+1}$ the $i$-th space constructed by Tom, so that his claim becomes $H_i(F_i) = \pi_i(X^+) = K_i R$. Explicitly, the spaces $F_k$ are defined inductively by a fibre sequence
$$ F_{i+1}\longrightarrow F_i\longrightarrow K(H_i(F_i),i) $$
for each $i\geq 1$, where the map on the right induces the "identity" on $H_i$. Applying the plus construction to this fibre sequence, we get a new sequence
$$ (FS^+): \qquad\qquad (F_{i+1})^+ \longrightarrow (F_i)^+ \longrightarrow K(H_i(F_i),i)^+ \simeq K(H_i(F_i),i) \hphantom{\qquad\qquad\qquad} $$
where we have noted that the natural map $K(H_i(F_i),i) \to K(H_i(F_i),i)^+$ is a weak equivalence. Importantly, the new sequence $(FS^+)$ is again a fibre sequence. This result is an instance of proposition 3.D.3-(2) in page 74 of Dror Farjoun's book "Cellular spaces, null spaces and homotopy localization": applying the plus construction to a fibre sequence of path connected spaces gives a fibre sequence as long as the homotopy type of the base space is unchanged by the plus construction. In fact, the book states this for any nullification functor, and the plus construction is one such functor.
The fundamental group of $(F_1)^+=X^+$ is abelian (it is $K_1 R$, after all). From the fibration sequence $(FS^+)$ for $i=1$ we then conclude that $(F_2)^+$ is simply connected. Moreover, that fibre sequence is just $(F_2)^+\to X^+\to B(\pi_1(X^+))$, and it realizes $(F_2)^+$ as the universal cover of $X^+$.
We can proceed inductively in this manner. More precisely, knowing that $\pi_1((F_1)^+)=\pi_1(X^+)$ is abelian, we can show by induction that for all $i\geq 1$:

The space $(F_i)^+$ is $(i-1)$-connected, and the Hurewicz theorem implies $\pi_i((F_i)^+)=H_i((F_i)^+)=H_i(F_i)$.
Hence, the second map in the fibre sequence $(FS^+)$ above is an isomorphism on $\pi_i$. In other words, that fibre sequence is simply killing $\pi_i$ of $(F_i)^+$.
Finally, there is a canonical equivalence $(F_i)^+ \simeq (X^+)^{\geq i}$ from $(F_i)^+$ to the $(i-1)$-connected cover of $X^+$ (also known as the $i$-th stage of the Whitehead tower of $X^+$).

Consequently, $H_i(F_i)=H_i((F_i)^+)=H_i((X^+)^{\geq i})=\pi_i((X^+)^{\geq i})=\pi_i(X^+)$ as desired.
Obviously, the above argument is not specific to $X=BGL(R)$. In fact, we only require that the pointed space $X$ is path connected, and $\pi_1(X^+)$ is abelian.<|endoftext|>
TITLE: When are the Smooth Sections of a Bundle Generated as a Module (over Smooth Functions) by the Holomorohic Sections 
QUESTION [5 upvotes]: For a holomorphic vector bundle $E$ over a complex manifold $M$, we denote its space of smooth sections by $\Gamma^{\infty}(E)$, and its space of holomorphic sections by $\Gamma^{hol}(E)$. Now I've been looking at the line bundles $L_k$ over the complex projective spaces ${\bf C} P^N$, and I have managed to show that $\Gamma^{\infty}(L_k)$ is generated as a $C^{\infty}({\bf C} P^N)$-module by $\Gamma^{hol}(E)$, which is to say that every element $\Gamma^{\infty}(E)$ is a sum of elements of the form $ef$, where $e \in \Gamma^{hol}(E)$, and $f \in C^{\infty}({\bf C} P^N)$. 
I am guessing that this result is extremely well known, and an example of a well understood general phenomenon. So I would like to ask if there is a characterization of the manifolds for which this result holds, for both the case of line bundles alone, and holomorphic vector bundles of general dimension?

REPLY [5 votes]: Swan has proved that taking global section gives an anti-equivalence between finitely generate projective $\Gamma^{\infty}(M)$-modules and $C^{\infty}$ vector bundles on $M$; this correspondence is functorial in $M$. Hence a set of section of a  $C^{\infty}$ vector bundle $E$ on $M$ generated $\Gamma^{\infty}(E)$ if and only if it generates each fiber of $E$. So if $E$ is holomorphic, the holomorphic sections generate if and only if $E$ is globally generated. In particular, this is always true if $M$ is Stein. In the case of $L_k$ on $\mathbb{CP}^N$, this is true if and only if $k ≥ 0$.<|endoftext|>
TITLE: Resembling the Levy Collapse
QUESTION [10 upvotes]: Suppose $\kappa$ is a weakly compact cardinal.  Is there a $\kappa$-c.c. forcing $\mathbb{P}$ such that $\mathbb{P} \subseteq V_\kappa$ and $\Vdash_{\mathbb{P}} \kappa = \aleph_1$, where $\mathbb{P}$ is provably NOT equivalent to the Levy collapse $Col(\omega,<\kappa)$?
I ask this because if $\kappa$ is weakly compact and $\mathbb{P}$ has the above properties, then one can prove:
a) There are stationarily many $\alpha < \kappa$ such that $\mathbb{P} \cap V_\alpha$ is a regular suborder of $\mathbb{P}$.
b) There are unboundedly many $\alpha < \kappa$ such that $\mathbb{P} \cap V_\alpha$ is equivalent to $Col(\omega,\alpha)$.
c) If $G$ is $\mathbb{P}$-generic over $V$, then there is a further forcing $\mathbb{Q}$ such that if $H \subseteq \mathbb{Q}$ is generic, then in $V[G][H]$, there is $G' \subseteq Col(\omega,<\kappa)$ which is generic over $V$, and $\mathbb{R}^{V[G]} = \mathbb{R}^{V[G']}$.
So in some sense $\mathbb{P}$ resembles $Col(\omega,<\kappa)$.

REPLY [2 votes]: Here is a different example: Let $\lambda<\kappa$ be regular uncountable. Then (perhaps surprisingly?) $\mathrm{Col}(\omega,{<}\kappa)$ and $\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\kappa)$ are not forcing equivalent. This is a consequence of the following fact:
(Let us write $\mathbb B\lessdot\mathbb C$ for $\mathbb B$ is a regular subalgebra of $\mathbb C$.)

Suppose $\mathbb B$ is a $\kappa$-cc atomless cBa that can be written as
$\mathbb B=\bigcup_{\alpha<\kappa}\mathbb B_\alpha$ so that

$\vert \mathbb B_\alpha\vert<\kappa$ for $\alpha<\kappa$.
$\mathbb B_\alpha\lessdot\mathbb B_\beta$ for $\alpha<\beta<\kappa$.
Any $\mu<\kappa$ is countable in $V^{\mathbb B_\alpha}$ for some $\alpha<\kappa$.

Then $\mathbb B\cong\mathrm{RO}(\mathrm{Col}(\omega,{<}\kappa))$ iff
$\{\alpha<\kappa\mid\bigcup_{\beta<\alpha}\mathbb B_\beta\lessdot\mathbb B\}$
contains a club.

We can write $\mathbb B=\mathrm{RO}(\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\kappa))$ as $\bigcup_{\alpha<\kappa}\mathbb B_\alpha$ with
$$\mathbb B_\alpha=\mathrm{RO}(\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\alpha))$$
and this satisfies 1.-3. above. However,
$$\{\alpha<\kappa\mid\bigcup_{\beta<\alpha}\mathbb B_\beta\lessdot\mathbb B\}\cap\mathrm{Lim}=\{\alpha<\kappa\mid\mathrm{cof}(\alpha)\geq\lambda\}$$
is costationary in $\kappa$ and hence $\mathbb B$ is not isomorphic to $\mathrm{RO}(\mathrm{Col}(\omega,{<}\kappa))$. The homogeneity of these two Boolean algebras implies that they are not forcing equivalent either.
This inequivalence here is an inconvenience that indeed comes up in practice. For example there is a difficult argument due to Woodin that shows, among other things, that one can force a dense ideal on $\omega_1$ from an almost huge cardinal (see Theorem 7.60 in Foreman's handbook article). The idea is to turn $\kappa$ into $\omega_1$ and $j(\kappa)$ (where $j$ is an embedding witnessing that $\kappa$ is almost huge) into $\omega_2$ and finally to show that a lift of $j$ exists after collapsing $\kappa$. To make that work, Woodin drops to a symmetric submodel of a forcing extension. In the inconsistent world where $\mathrm{Col}(\omega, {{<}j(\kappa)})\cong\mathrm{Col}(\omega, \kappa)\times\mathrm{Col}(\kappa, {<}j(\kappa))$ this would not be necessary and the easier naive approach would go through.<|endoftext|>
TITLE: Fixed point theorem on graphs?
QUESTION [6 upvotes]: Originally posted here: https://math.stackexchange.com/questions/276167/fixed-point-theorem-on-graphs
--
I have a graph $G=(V,E)$ where to each vertex $v$ I have associated a value, $\hat{v}$ (ie I have a "network" in the terminology here http://snap.stanford.edu/snap/index.html ).
Let $\phi : \hat{V} \rightarrow \hat{V}$ be the function which takes the value associated to each node and replaces it with median of the values of the adjacent nodes.
Empirically, iterating $\phi$ converges. Why?
edit: The graph is large and follows this model: http://en.wikipedia.org/wiki/Barab%C3%A1si%E2%80%93Albert_model . Also, for ~1% of nodes, $\phi$ is the identity. There is some related work here: http://www.cs.cmu.edu/~zhuxj/pub/CMU-CALD-02-107.pdf
edit: Looks like I can just replace the $P(j \rightarrow i)$ in the label propagation paper with $P(j \rightarrow i) = 1\; if \; \text{j is the median, }0 \text{ else}$, then I can copy their convergence result.
edit: Wait, no. If I do that then P changes each step so it's not a direct copy...

REPLY [4 votes]: Here are a few thoughts: 

It is easiest to ignore the case of unequal middle values, for example if every vertex has odd degree. Then the set of labels can only stay the same or  get smaller so eventually the states must cycle (perhaps in a cycle of length 1=fixed point).
It is enough to consider the case that the vertex values are only $0$ and $1$. If I tell you at the start only which vertices are "high" (above a threshhold $t$) and which are low (less than or equal to $t$) that information alone will be enough decide which vertices are high and which are low at each future stage. We can keep track of all the values by running a few of these thresh-hold graphs in parallel.

I can certainly see 

fixed points: Just split into two (or $k$) groups of moderate size and make sure that each vertex has the majority of its neighbors in its own group. To get fancier (with lots of label values), make the maximum and minimum groups very large, make sure everyone has at least one neighbor with its own label and then make extra edges into the maximum and minimum groups to make sure the medians stay where they should.
cycles of length $2$ Two groups each with the majority of its labels going to the other group. Or 

After I made this illustration I had the insight that there is simply alternation between a $3/7$ high/low split and $7/3$ or $4/6$ and $6/4$, but I still like the picture.
In a few experiments on the Peterson graph (starting with distinct labels) a cycle of length two happened over $95\\%$ of the time. That is probably exactly the same as saying that with $0,1$ labels a two cycle happened this often with an even split, two thirds of the time with a $4/6$ split and one sixth of the time with a $3/7$ split. In that case it would be easy to figure out the exact probabilities.
Other graphs could be explored. For a pentagonal prism the chance of a two cycle is higher. Even with a $8/2$ split it is $22\\%$ (OK, I can see why) whereas this can not happen in a Peterson Graph.
Q: Can there be cycles of length three or larger? I did not see any but I did not look all that hard. However I would guess not.
It might be worth considering weighted edges, then the examples above can have merely $k$  vertices.

Consider large  graphs (random or structured) with random labels $0$ and $1.$ Then the situation is like voting with everyone swayed by their friends (over many rounds) or some lattice with points magnetized up or down then affecting their neighbors (so this must be studied someplace). If the labels are assigned with very unequal frequencies then one would expect a sea of the majority value with islands of the minority value which shrink (or grow) for a while before stabilizing or going away (or going into a few two cycles). Big islands with lakes with little islands (containing ponds) in them would be possible by pre-planning but perhaps unlikely by chance. As the probabilities become closer to equal it would take a longer and longer time to stabilize with a phase transition at some point which might be very interesting to consider.

A final thought is that the tie breaking problem for an even number of neighbors could be avoided by saying that the value at the vertex is also counted in the median calculations when needed to break a tie. Weighted edges would also avoid this problem (if the weights were random enough to avoid ties).<|endoftext|>
TITLE: Testing simplicial complexes for shellability
QUESTION [25 upvotes]: Question

Are there efficient algorithms to check if a finite simplicial complex defined in terms of its maximal facets is shellable?

By efficient here I am willing to consider anything with smaller expected complexity than the exponential mess one gets by naively testing all possible orderings of maximal facets.
Background
Let $\Delta$ be a simplicial complex and for each simplex $\sigma \in \Delta$ let $\bar{\sigma}$ denote the subcomplex generated by $\sigma$ and all its faces. Fix an ordering of its maximal facets $F_1,\ldots,F_K$, pick some $k \in \lbrace 1,\ldots,K\rbrace$ and define $\Delta_k$ to be the subcomplex generated by $\bigcup_{1\leq j \leq k} F_j$, i.e., all facets up to and incluing the $k$-th one.
Definition: We call this ordering of maximal facets a shelling if the intersection $\overline{F_{k+1}} \cap \Delta_k$ is a simplicial complex of dimension $\dim (F_{k+1}) - 1$ for each $k \in \lbrace 1,\ldots,K-1\rbrace$.
In general, the complex $\Delta$ need not be a combinatorial manifold or have a uniform top dimension for its maximal facets. It is known that if $\Delta$ is shellable then there exists a shelling by maximal facets ordered so that the dimension is decreasing along the order. So one method to simplify the computational burden is to test only those orderings $F_1,\ldots,F_K$ of maximal facets so that $\dim F_i \geq \dim F_j$ whenever $i \leq j$, but of course in the worst case all these facets could have the same dimension.
Motivation
Shellability is an extremely useful notion in topological combinatorics: many interesting simplicial complexes and posets in this field turn out to be shellable. I refer you to the works of Anders Bjorner and others for details, see here or here or... Since every shellable complex is a wedge of spheres, establishing shellability leads to all sorts of interesting conclusions. Among other things, shellable complexes must lack torsion in homology of all dimensions.

REPLY [3 votes]: This unpublished article (collaboration welcome)
presents a systematic approach to decide shellability, which goes beyond  extending partial shellings (that can confirm but hardly disprove shellability). Furthermore, our method  (unlike Moriyama's algorithm) does not need the face-numbers. Instead of the $n!$ permutations of the $n$ facets we deal with certain admissible chains in certain posets of cardinality at most $n \cdot 2^n$ (which is $<< n!$). The shellability status of some matroid and some chessboard complexes with up to 24 facets is determined, or redetermined. Moreover the total number of shellings can be calculated. For instance the simplicial complex of all trees of the complete graph $K_4$ has exactly 722965625856 shellings. The previously known lower bound to the number of shellings was $6!=720$.<|endoftext|>
TITLE: Finite Quotients of Free Groups
QUESTION [5 upvotes]: It is interesting FACT that given $l,m,n\geq 2$, there is (are) a finite group with elements $a,b$ such that $o(a)=l, o(b)=m$, and $o(ab)=n$ (see link for a nice example by Derek Holt / B. Sury). 
Although, the group 

$G_{l,m,n}=\langle a,b\colon a^l,b^m,(ab)^n\rangle$

contains elements with above property (?), it need not be finite. But for certain values of $l,m,n$, $G_{l,m,n}$ is finite (and these are interesting groups by their geometry):
$G_{2,2,n}\cong D_{2n}$,
$G_{2,3,3}\cong A_4$,
$G_{2,3,4}\cong S_4$,
$G_{2,3,5}\cong A_5$.
A natural question I would like to ask is:

For what values of $l,m,n$, the group $G_{l,m,n}$ is finite?


I wondered by the above FACT, and very useful discussion on it in mathoverflow, with  nice answer by Derek Holt/ B. Sury. I tried to solve the question in the link, and for $(l,m,n)=(2,2,n)$, I quickly found that the group $D_{2n}$ is the best example for it.
Then I came up with the natural question above. This question may have been discussed with different point of view; but I didn't know too much about it. 
Also, if possible, one may suggest(and edit) suitable title for this question.

REPLY [7 votes]: These groups are called von Dyck groups, see e.g. here. Von Dyck group $D(l,m,n)$ is finite if and only if it is of spherical type: 
$$
\chi=-1+ l^{-1} + m^{-1} + n^{-1}>0. 
$$
A side note: Von Dyck groups are fundamental groups of 2-dimensional oriented orbifolds. The number $\chi$ above is the orbifold Euler characteristic. This is explained nicely, for instance, in Peter Scott's paper "Geometries of 3-manifolds",  Bull. London Math. Soc. 15 (1983), no. 5, 401–487.<|endoftext|>
TITLE: Interpretation of Riemann tensor antisymmetry
QUESTION [5 upvotes]: We know that the Riemann tensor is antisymmetric with respect to the first two vectors (the vectors that we parallel transport the third vector around the parallelogram made by their integral curves).
But what is the geometric interpretation?
Why do we have to expect that by changing the order of parallel transportation we will have the same vector with opposite orientation?

REPLY [7 votes]: More interpretations:
On (pseudo-) Riemannian manifolds:
the numerator of sectional curvature $-\langle R(X,Y)X,Y\rangle$ is a symmetric bilinear form
on the space of skew-symmetric bivectors. Skew symmetric bivectors describe measured 2-planes in the tangent space. Curvature in the form of  $\langle R(X,Y)Z,W\rangle$ can be recomputed by polarization from this.
Another, more general point of view: On the (orthonormal) frame bundle, curvature is a 2-form with values in the Lie algebra of the structure group: $\Omega=d\omega+\omega\wedge\omega$ for matrix valued forms. This ties in well with the fact that curvature is the obstruction against integrability of the horizontal subbundle of $TTM$.<|endoftext|>
TITLE: What is an $(\infty,1)$-topos, and why is this a good setting for doing differential geometry?
QUESTION [35 upvotes]: In this post on the n-Category Café, Urs Schreiber says that, "The theory of G-principal bundles makes sense in any $(\infty,1)$-topos." I followed the link to the nLab and tried to chase definitions, but I found too quickly my head spinning.
What is an $(\infty,1)$-topos, and why is this an appropriate setting for the study of principal bundles, i.e., doing differential geometry?

REPLY [19 votes]: If the question is not really about principal bundle theory but just about: why do we need higher differential geometry at all, then of course there are plenty of further answers:
Classical differential geometry includes orbifolds 
http://ncatlab.org/nlab/show/orbifold
as objects that handle non-free quotients of smooth manifolds. These are really the first kinds of examples of Lie groupoids
http://ncatlab.org/nlab/show/Lie+groupoid
hence of stacks on the category of smooth manifolds. All of classical foliation theory 
http://ncatlab.org/nlab/show/foliation
is secretly Lie groupoid theory. And the only sane way of understanding the collection of Lie groupoids, with their correct notion of Morita equivalence and of homotopy, is as understanding them as the objects of the $(2,1)$-topos of stacks over smooth manifolds.
Also Lie theory itself breaks out of the category of smooth manifolds. For instance where Lie's three theorems fail: not every infinite-dimensional Lie algebra integrates to a Lie group, but it instead integrates to a certain Lie 2-group, a group object in Lie groupoids/smooth stacks. This is all the more true as soon as you admit that Lie algebroids are part of differential geometry
http://ncatlab.org/nlab/show/Lie+algebroid .
Lie algebroids directly encode PDEs
http://ncatlab.org/nlab/show/exterior+differential+system
Most Lie algebroids integrate to Lie groupoids, but some want to integrate to Lie 2-groupoids, which liven in the $(3,1)$-topos over smooth manifolds. 
And so on. 
Doing differential geometry and not stopping when classical constructions fail invariably leads one to higher differential geometry, hence to working in the $(\infty,1)$-topos over smooth manifolds.
Maybe that's what the question was really asking. If in addition one feels like refining the site of smooth manifolds itself such as to include "derived smooth manifolds", then one gets something even richer, as mentioned in another reply here. Such derived and higher differential geometry is notably the home of BV-BRST formalism
http://ncatlab.org/nlab/show/BV-BRST+formalism
hence of the modern form of variational calculus, symplectic reduction and homological integration theory.<|endoftext|>
TITLE: basics of classification of trilinear forms (when is it non-discrete)
QUESTION [11 upvotes]: Consider tri-linear forms, $\{A_{ijk}\}$ where $i=1,..,n_1$, $j=1,..,n_2$, $k=1,..n_3$, over a field of zero characteristic, up to the equivalence $A\to (U_1,U_2,U_3)(A)$,  by three matrices.
What is known about the classification? (Complete) invariants?
Unlike the case of bi-linear forms, in general the classification has moduli. 
I've found only an old paper of Thrall, dealing with partial cases. Is there some general (and more modern) exposition/introduction/lecture notes?
My particular question: for which triples $(n_1,n_2,n_3)$ is the classification discrete?
 (The obvious necessary condition, $n_1n_2n_3\le n^2_1+n^2_1+n^2_3$, is certainly not sufficient.)

REPLY [6 votes]: This should really be a comment on the answer of Steven Sam, but it doesn't fit in the box. The systematic way to reduce to the cases described by Sam is via castling transformations.
It is shown in a paper of P.G. Parfenov that the group $G = GL(k_{1}, \mathbb{C})\times \dots \times GL(k_{r}, \mathbb{C})$ has finitely many orbits in $V = \mathbb{C}^{k_{1}}\otimes \dots \otimes \mathbb{C}^{k_{r}}$ if and only if $(k_{1}, \dots, k_{r})$ is one of $(n)$, $(m, n)$, $(2, 2, n)$, or $(2, 3, n)$, where $n \geq 3$. This is published in:
http://iopscience.iop.org/1064-5616/192/1/A05
(This also follows from Theorem $2$ of V.G. Kac's paper "Some remarks on nilpotent orbits")
That $G$ have finitely many orbits on $V$ means that $(G, V)$ is prehomogeneous ($G$ has a Zariski dense orbit), but the converse is not true. One can ask also when $(G, V)$ is prehomogeneous. A necessary condition was stated above by R. Bryant in a comment to the original question (for $r > 3$ the condition becomes $\sum_{i}k_{i}^{2} - \prod_{i}k_{i} \geq r- 1$). In general this can be resolved by using the Sato-Kimura classification of reduced irreducible prehomogeneous vector spaces. Here reduced means reduced with respect to castling equivalence, which is a notion of equivalence of prehomogeneous vector spaces based on the duality of Grassmannians (see Sato-Kimura and Kimura's book on prehomogeneous vector spaces). One can show that every $(G, V)$ that is prehomogeneous is castling equivalent to such a space with finitely many orbits, that is in the list due to Parfenov mentioned above (I have not been able to find this statement as such published anywhere, but it is known). For general formats $(k_{1}, \dots, k_{r})$ I don't know how to state this cleanly simply as a condition on the $k_{i}$'s, but in practice it is straightforward to check (the necessary details can be gleaned from section 3 of Sato-Kimura). For example, if $r = 3$ then the dimensions have the form $(a, b, c)$. If $ab > c$ then this format is castling equivalent to $(a, b, c - ab)$, and via this remark, one can reduce to a castling equivalent format in which (after reordering) $c \leq 2$. 
In some cases the fundamental relative invariant is a hyperdeterminant. This occurs, for example, for the formats $(2, 2, 2)$ (the classical Cayley hyperdeterminant) and $(2, 3, 3)$. See chapter 14 of the book of Gelfand, Kapranov, and Zelevinsky.<|endoftext|>
TITLE: Where in ordinary math do we need unbounded separation and replacement?
QUESTION [22 upvotes]: [I have updated the question after initial comments in the hope of clarifying it.]
I do quite a bit of reasoning, typically about topology and metric spaces, in "non-standard" foundations, such as inside of a particular topos, in type theory, or a predicative constructive setting. These typically do not have anything corresponding to unbounded separation or replacement (the constructive set theory CZF does have collection, though).
I have a pretty good feel when restricted forms of excluded middle and choice are needed, and what things powersets give us over predicative math, etc. But I never ever wish I had unbounded separation and replacement. Why is that? Is it just because of the kind of math I do, or are these two really not needed very much in ordinary math?
To make the question more specific: what are some well-known definitions and theorems in "ordinary" mathematics which require unbounded separation or replacement?
The obvious uses of replacement and unbounded separation come from set theory, so we should avoid listing those. Ideally, I am looking for theorems and definitions in algebra, topology, and analysis.
Here is a non example from order theory, which was suggested in the comments. Under the usual encoding of ordinals as hereditarily transitive transitive sets, the rank of the function $n \mapsto \omega + n$ is $\omega + \omega$ and so we need replacement to show its existence. However, even PA can speak about this sort of small countable ordinals, so we are seeing here an artifact of a particular encoding. A different encoding of countable ordinals would make this function easy to define (for example we could view the countable ordinals as orders of subsets of $\mathbb{N}$).
The only example of unbounded separation I can think of right now comes from category theory. In a large category $C$ the definition of epi is unbounded, as it requires quantification over all objects of $C$. I am looking for something that is not so directly linked to a question of size.

REPLY [4 votes]: This would be set theory  rather than `ordinary math'. Still, it's interesting to observe that without Unbounded  Separation many of the customarily equivalent formulations of finiteness diverge.
For example, it is no longer the case that the system of Zermelo naturals and the system of von  Neumann naturals are isomorphic. This is discussed in "Natural Number Arithmetic in the Theory of Finite Sets" by Mayberry-Pettigrew, http://arxiv.org/abs/0711.2922.<|endoftext|>
TITLE: Expectation of square root of binomial r.v.
QUESTION [11 upvotes]: Let $X\sim B(n,p)$ denote a binomial random variable.  Is there any approximation available for the quantity $E(\sqrt{X})$?  Clearly Jensen's inequality holds, but rudimentary tooling around with Maple hasn't turned up anything more substantial.

REPLY [18 votes]: $\newcommand{\E}{\mathbf{E}}$
$\renewcommand{\P}{\mathbf{P}}$
$\DeclareMathOperator{\var}{Var}$
If we use Taylor expansion (as Anthony suggested) for $\sqrt{x}$ around 1, we get:
$$\sqrt{x}\approx 1 + \frac{x-1}{2} - \frac{(x-1)^2}{8}  .$$
We can use this to get an approximation of
$$\E(\sqrt{X})\approx 1-\frac{\var(X)}{8} ,$$
which should be valid for any RV concentrated around an expectation of 1. Equivalently,
$$\E(\sqrt{X})\approx \sqrt{\E(X)}\bigg(1-\frac{\var(X)}{8\E(X)^2}\bigg) ,$$
for any RV concentrated around its mean.
As you noted, we can use Jensen inequality to get $\E(\sqrt{X})\le \sqrt{\E(X)}$ for any nonnegative RV. We can tweak the Taylor expansion to get a lower bound, by noticing that
$$ 1 + \frac{x-1}{2} - \frac{(x-1)^2}{2} \le \sqrt{x} \ .$$
Hence, we get
$$\sqrt{\E(X)}\bigg(1-\frac{\var(X)}{2\E(X)^2}\bigg) \le \E(\sqrt{X}) ,$$
for any nonnegative RV.
In the case of $X\sim Bin(n,p)$ we get
$$ \sqrt{np}-\frac{1-p}{2\sqrt{np}} \le \E(\sqrt{X})\le \sqrt{np} .$$<|endoftext|>
TITLE: "Classical" consequences of Bezout's theorem in dimensions $>2$
QUESTION [13 upvotes]: By Classical I mean something that could have been found before 1900 (say). 
A well known consequence of Bezout's theorem for plane curves is Pascal's theorem http://en.wikipedia.org/wiki/Pascal's_theorem .
I am curious if there are some other statements that you find pretty that can be formulated (almost) as elementarily as Pascal's theorem and proven using higher dimensional Bezout's theorem? For example, is there some statement that involves quadrics, planes and lines (cubics?...)?
Motivation. I ask this question since I want to finish to teach my (introductory) course in algebraic geometry by higher-dimensional Bezout theorem (using Hilbert polynomials, ect), and I would be extremely happy to give some pretty application :). To give you an idea of the level of the course, it is based on some bits of Harris book "Algebraic geometry first course", 
Disclaimer. I don't doubt the usefulness of Bezout theorem and am sorry if the original question sounded like I doubt it. On the contrary I based the elementary course in algebraic geometry that I teach on this theorem. Namely, the course starts with Bezout for plane curves (using resultants), intorduces projective spaces and varieties, goes through Hilbert basis theorem and Hylbert polynomials (last section of Atiyah-Macdonald) and then as an applications we get a proof of a simplest version of Bezout's theorem in high dimension.
Also, It would be difficult for me to explain what I mean by pretty in math (for myself)
but still I feel that the using of this word is justified, because we, mathematicians use this word... Sometimes we disagree on what is pretty, but personally I find pretty huge amount of facts in algebraic geometry. In other words I will be happy to see any application that can be stated in the language on the level of my course.
In the comment I put the link to the question on stackexchange

REPLY [3 votes]: I just found one more classical consequence which is quite nice. It is described in the book of Harris "First course in algebraic geometry". 
Namely, one can prove that all projective automorphisms of $\mathbb CP^n$ are projective transformations. In order to prove this one shows that any hyperplane in $\mathbb CP^n$ is sent by any automorphism to a hyperplane (otherwise it would intersect the image of the line in more than one point). From this the statement follows rather directly.<|endoftext|>
TITLE: Reference request for Plancherel measure
QUESTION [18 upvotes]: I need a good reference for the basic definitions of the dual of locally compact group (not necessarily abelian), its natural topology, $\sigma$-algebra, and the Plancherel measure on it (when they are defined). This topic seems pretty standard to me, but when I needed a basic reference on this (both to check my memories and to be able to cite it in a paper I am writing), I didn't find one. 
By the way, the wikipedia webpage "Plancherel measure" should be completely rewritten. 
There is not even  a definition, just a list of examples (and the definition given in the finite case is not compatible with the one given in the compact case). I would be happy to rewrite it, when I have a reference to check the details.

REPLY [3 votes]: In my humble opinion the best reference is Dixmier $C^*$-algebras. The first half of the book has a very complete explanation of what you need to know about $C^*$-algebras. In chapter 8 he goes over what is the decomposition of a trace for $C^*$-algebras. Then from Chapter 13 on he goes into the theory for a locally compact group. He explains necessary and sufficient conditions for the Plancherel formula to exist (has to be Type I, separable, postliminal, unimodular etc.). He also explains the topology to be given $\widehat{G}$; in fact he gives three different topologies on this set, and shows all of them agree in the case we are interested in—it is just that beautiful of a book. Chapter 18 is the statement of the Plancherel Theorem; the proof essentially is the one in Chapter 8 for $C^*$-algebras. The english version is very good, with very few typos or print mistakes that may confuse you. I have not found a typo or a mistake of any sort in the French version. I think it is a very good book, like reading a novel.<|endoftext|>
TITLE: Complexifying a real Banach space and its dual
QUESTION [11 upvotes]: A standard way to define the "complexification" $E_\mathbb{C}$ of a real Banach space $E$ is to define a complex linear structure on $E\times E$ by (1) $(x,y)+(u,v)=(x+u, y+v)$, (2) $(a+ib)(x,y)=(ax-by, bx+ay)$ and a norm by (3) $\|(x,y)\|^\mathbb{C}=\sup_{\theta\in [0,2 \pi]}\|\cos(\theta)x+\sin(\theta)y\|$. (That $\|\cdot\|^\mathbb{C}$ is a norm requires a little proving.)
This definition gives us what we want when going from real $C(K)$ or $l_p$, etc., to their complex versions.

I believe I can show that the (complex) Banach dual of $E_\mathbb{C}$ is the complexification of the real dual $E^*$, i.e., $(E_\mathbb{C})^* = (E^*)_\mathbb{c}$.  But my proof is somewhat messy.  This must surely be in the literature somewhere! Can anyone suggest a reference?
How about the converse?  What if we know that $E_\mathbb{C} = V^*$ for some complex Banach space $V$.  Is $E$ the (real) Banach dual of some Banach space?  (Again, a reference would be appreciated).

[Edit Feb. 13, 2013] Based on Bill Johnson's comment below, perhaps I should motivate the question a bit, and revise question 2.  For a compact space $K$, we know that the Banach space $C(K)$ over the real scalars is isometrically the dual of a real Banach space if and only if $C(K)$ over the complex scalars is isometrically the dual of some complex Banach space. A standard proof is particular to the situation, going through hyperstonian spaces $K$ and normal measures. I was wondering if this is just a special case of a more general fact. So here is a revision of 2. 
Q3. Suppose $E$ is a real Banach lattice and the complexification $E_\mathbb{C}$ is a Banach dual space.  Must $E$ then be the dual of some real Banach space?

REPLY [3 votes]: This is a response to the request for references, but not an answer to Q3. I came across a related paper: "Complexifications of real Banach spaces, polynomials and multilinear maps", by Munoz, Sarantopoulos, and Tonge, Studia Math. 134 (1999), 1-33.
They point out that there are many different ways to put a reasonable norm on the algebraic complexification $E\times E$.<|endoftext|>
TITLE: mapping space between classifying spaces
QUESTION [7 upvotes]: I wanted to ask a summary of known results and references about the homotopy type of the mapping space $\mathrm{Map}(BG,BK)$ (and specially the connected components) between the classifying spaces when G and K are general topological groups. 
Thank you.

REPLY [10 votes]: The Sullivan conjecture, a beautiful theorem proved by Haynes Miller
(The Sullivan conjecture on maps from classifying spaces) has striking
consequences about maps between classifying spaces when the target is
completed at a prime $p$.  Some references are Dwyer and Zabrodsky
(Maps between classifying spaces), Jackowski, McClure, and Oliver
Homotopy classification of self-maps of $BG$ via $G$-actions. I, II;
Self-homotopy equivalences of classifying spaces of compact connected 
Lie groups); Notbohm (Maps between classifying spaces).  
The classical work of J.F. Adams on this topic is also well worth
remembering: Maps between classifying spaces (I), II, III and 
Maps between $p$-completed classifying spaces.<|endoftext|>
TITLE: Epimorphisms and free submodules
QUESTION [5 upvotes]: By inspecting the accepted answer to this question 
Are epimorphisms from a division ring isomorphisms ? 
one obtains the following necessary condition for epimorphisms: 

Let $R \le S$ be rings with identity $1\neq 0$ such that $S$ is flat as right $R$-module. If the inclusion $R \hookrightarrow S$ is an epimorphism, then for each  $s \in S$ there is $r\in R, r\neq 0$ such that $rs \in R$. 

Proof: The statement says that $S/R$ has no left $R$-submodule isomorphic to $R$. If it is wrong, we have an embedding $R \hookrightarrow S/R$ and hence $0 \neq S = S \otimes_R R \hookrightarrow S\otimes_R S/R$ 
But by the quoted answer, $S\otimes_R S/R=0$ if $R \to S$ is epi. $\blacksquare$
However, without success I tried to drop the flatness condition. Whence my question: 
Question: Is the statement above still valid, if $S$ is not supposed to be flat as right $R$-module ?

REPLY [10 votes]: A simple counterexample can be constructed using free objects: 
For a set $X$ let $\mathbb{Z}\langle X\rangle$ be the free ring on $X$ and let $\mathbb{Z}F(X)$ be the group ring of the free group $F(X)$ on $X$. The inclusion $X \hookrightarrow F(X)$ induces a ring homomorphism $i: \mathbb{Z}\langle X\rangle \to \mathbb{Z}F(X)$ which is an epimorphism. If $x\neq y$ are in $X$ then $xy^{-1} \in \mathbb{Z}F(X)$ and there is obviously no "polynomial"  $f \in \mathbb{Z}\langle X\rangle$ such that $fxy^{-1} \in \mathbb{Z}\langle X\rangle$. 
To see that $i$ is an epimorphism, note that $\mathbb{Z}F(X)$ is generated (as a ring) 
by $x,x^{-1}\;(x \in X)$ and a unitary ring homomorphism $\varphi: \mathbb{Z}F(X) \to T$ satisfies 
$\varphi(x^{-1})=\varphi(x)^{-1}$. Hence $\varphi$ is determined by its values on $X$. 

Update: The property in question holds true for commutative rings. 
Proof: At first assume this has already been proved for zero-dimensional local commutative rings. Let $R \le S$ be an epimorphic extensions of comm. rings and choose a minimal prime $\mathfrak{p}$ of $R$ (exists by Zorn's lemma). $R_\mathfrak{p}$ is a local ring of dimension $\text{ht}(\mathfrak{p})=0$. Since $R\setminus \mathfrak{p}$ is a multiplicative subset of $S$ we can localize and obtain a comm. diagramm
$$\begin{array}{ccc}
R & \xrightarrow[\scriptstyle\text{epi}]{i} & S\;\;\; \newline 
{\scriptstyle\text{epi}}\downarrow &  & \downarrow\scriptstyle\text{epi} \newline 
R_\mathfrak{p} & \xrightarrow[i_\mathfrak{p}]{} & (R\setminus \mathfrak{p})^{-1}S
\end{array}$$
Hence $i_\mathfrak{p}$ is epi and it's easy to see that $i_\mathfrak{p}$ is mono as well. 
Let $s \in S$. By our assumption there are $r_i/t_i \in R_\mathfrak{p},\;r_1/t_1 \neq 0$ such that $\frac{r_1}{t_1}\frac{s}{1}=\frac{r_2}{t_2}$, i.e. there is $t \in R\setminus \mathfrak{p}$ with $(tt_2r_1)s=tt_1r_2 \in R$. Moreover $tt_2r_1 \neq 0$ (because $r_1/t_1 \neq 0$ in $R_\mathfrak{p}$ just says there is no $t \in R\setminus \mathfrak{p}$ such that $tr_1=0$ in $R)$ and we are done. 
Now suppose $R$ is zero-dimensional local comm. with max. ideal $\mathfrak{m}$. Again from the comm. diagramm 
$$\begin{array}{ccc}
R & \xrightarrow[\scriptstyle\text{epi}]{i} & S\;\;\; \newline 
{\scriptstyle\text{epi}}\downarrow\;\; &  & \downarrow\scriptstyle\text{epi} \newline 
R/\mathfrak{m} & \xrightarrow[i_\mathfrak{m}]{} & S/\mathfrak{m}S
\end{array}$$
we conclude that $i_\mathfrak{m}$ is epi and since $R/\mathfrak{m}$ is a field, $i_\mathfrak{m}$ is an isomorphism (see the link in the OP's question). Hence each $s \in S$ can be written as 
$$s=r + \sum_{i=1}^l m_is_i\qquad (r\in R, s_i \in S, m_i \in \mathfrak{m}, m_i \neq 0)$$
We want to show that there is $r_0 \in R,\;r_0 \neq 0$ such that $r_0s\in R$. 
If $l=0$ then $s=r\in R$ and we are done. Otherwise, since $\mathfrak{m}=\sqrt{0}$ there is $n_1> 0$ maximal such that $m_1^{n_1} \neq 0$ and multiplying yields 
$$m_1^{n_1}s=m_1^{n_1}r + \sum_{i=2}^l (m_1^{n_1}m_i)s_i.$$
If $m_1^{n_1}m_2=0$ ignore the corresponding summand. Otherwise, there is $n_2>0$ maximal such that $m_1^{n_1}m_2^{n_2}\neq 0$. Multiplying again yields 
$$m_1^{n_1}m_2^{n_2}s=m_1^{n_1}m_2^{n_2}r + \sum_{i=3}^l(m_1^{n_1}m_2^{n_2}m_i)s_i.$$
Proceeding this way,  we obtain the required $r_0$ in the form $r_0 = \prod_{j=1}^k m_{i_j}^{n_{i_j}}.\;\;$ qed.<|endoftext|>
TITLE: Decidability of periodic tilings of the plane
QUESTION [8 upvotes]: I'm interested in tilings of the plane by squares, with labels on the edges. It's well known that (1) the question "can one tile the plane with the following finite set of tiles?" is undecidable, and (2) there are finite sets of tiles that tile the plane, but only aperiodically. Also, (1) implies (2).
Is the question "can one tile the plane periodically with the following finite set of tiles?" decidable?

REPLY [2 votes]: Deciding whether a set of tiles admits a periodic tiling or no tiling at all is undecidable as well.
This has been shown in Y.S. Gurevich, I.I. Koryakov, Remarks on Berger's paper on the domino problem, J Sib Math J 13, 319–321 (1972).  The results can also be found in the book The Classical Decision Problem by Egon Börger, Erich Grädel, Yuri Gurevich, where Theorem 3.1.7 states

The set of domino systems that admit, respectively, no tiling and a periodic tiling of $\mathbb Z\times \mathbb Z$ or $\mathbb N\times \mathbb N$ are recursively inseparable.<|endoftext|>
TITLE: A question about definable non-empty sets containing no definable elements.
QUESTION [11 upvotes]: Can anyone provide an example of a set S which is definable in ZFC and provable in ZFC to be
denumerably infinite, while at the same time, no set definable in ZFC can be proved in ZFC to
be an element of S? Such examples are easy to find if S is allowed to be uncountable-for instance
S could be the set of all non-measurable sets of real numbers. In most of the examples where S
is uncountable the axiom of choice is needed just to prove that S is non-empty and the cardinal
number of S is greater than the cardinal number of R.

REPLY [8 votes]: Here are a few partial answers.
First, I claim that if ZFC is consistent, then for every
ZFC-definable nonempty set $S$, whether countable or uncountable, it is consistent with ZFC that $S$
contains a definable element. Further, there will be a model of ZFC in which every element of $S$ is definable. 
Suppose that we have defined a nonempty set $S$ in ZFC. What I
mean by this is that we have provided the defining formula $\phi$
of $S$ and proved in ZFC that there is a unique object, which we
call $S$, that satisfies $\phi$, and furthermore that ZFC proves
that this set is nonempty.
Now, it is known that if ZFC is consistent, then there are
pointwise definable models of ZFC, models in which every set is
definable without parameters. The main result of my paper
Pointwise definable models of set theory, joint with Jonas Reitz and David Linetsky, is that every countable
model of ZFC and indeed of GBC has an extension obtained via class
forcing that is pointwise definable, in which every set and indeed every class is definable without parameters. My point now is that if $M$ is a pointwise
definable model of ZFC, then every member of $S$ as interpreted in
$M$ will be definable, by some definition.
Although this answer seems relevant to me, it doesn't quite answer
the exact question you asked, since we don't expect that ZFC will
prove that that set is a member of $S$.
Second, I claim that there is an example if you drop the denumerability requirement. That is, there is a definable set $S$, which is provably nonempty,
but which does not provably contain any specific definable object.
(This argument therefore fills in for your proposal with
non-measurable sets, which I don't really see how to make work.)
Define $S$ to be the sets of minimal rank not in $\text{HOD}$, in the case
that $V\neq\text{HOD}$, and otherwise $S=\mathbb{N}$. So ZFC
proves that $S$ is nonempty, but there can be no definable object
$a$ such that ZFC proves $a\in S$, since in this case, in a model
of $V\neq\text{HOD}$, the object $a$ would be a definable set that
is a subset of $\text{HOD}$, by the minimality part of the definition of
$S$, and hence $a$ would be not only definable but hereditarily
ordinal-definable and hence an element of $\text{HOD}$, contrary to the
definition of $S$.
Further update. Extending the observation of Emil in the
comments, let's prove that there is a positive solution, if you
change the countability requirement, size $\aleph_0$, to require
instead $\aleph_1$.
Theorem. There is a ZFC definable set $S$, which ZFC proves
has size exactly $\aleph_1$, such that if ZFC is consistent, then
for no ZFC-provably definable object $t$ does ZFC prove $t\in S$.
Proof. Let $S$ be the set of reals not in HOD, provided that CH
holds and there are reals not in HOD, and otherwise let $S$ be
$\aleph_1$ itself. ZFC proves that this set has size $\aleph_1$, since
either $S$ is explicitly $\aleph_1$, or else there are reals not
in HOD and CH holds. But if there are reals not in HOD, then it is
not difficult to see that there must be continuum many reals not
in HOD, and so if CH also holds, this has size $\aleph_1$.
Meanwhile, if ZFC is consistent, then there is a model in which CH
holds and there are reals not in HOD. In this case, every element
of $S$ is not in HOD, and so ZFC cannot prove that any particular
definable object $t$ is in $S$. QED<|endoftext|>
TITLE: Do there exist (almost surely) $C^{\infty}$-smooth Gaussian random fields?
QUESTION [5 upvotes]: Let $d \ge 1$. Do there exist Gaussian random fields on $\mathbb R^d$ which are (almost surely) $C^{\infty}$-smooth, but which are not analytic? 
If so, what are necessary and sufficient conditions on the covariance function to assure that the field is (a.s.) $C^{\infty}$-smooth?

REPLY [2 votes]: I think, you get what you want if you convolve nonsmooth trajectories (like those of a Wiener process) with a $C^\infty$ kernel that is not analytic.<|endoftext|>
TITLE: Efficient computation of "discrete infimal convolution"
QUESTION [9 upvotes]: This question arises from an application to graphical models in probability theory, but I have abstracted that part out so only algebra remains.  Let $\mathbb{R}$ denote standard field of real numbers and $\mathcal{R} = (\mathbb{R}\cup\{\infty\},\min,+)$ the tropical semiring.  Let $\mathbb{Z}_n$ denote the cyclic group of order $n$.
Elements of the group ring $\mathbb{R}\mathbb{Z}_n$ are tuples $(x_0,\ldots,x_{n-1})$ and multiplication of these corresponds to discrete cyclic convolution.  The Fast Fourier Transform gives an embedding $\mathbb{R}\mathbb{Z}_n\to\mathbb{C}^n$ (with elementwise sum and product).  The FFT and its inverse can be computed in $O(n\log n)$ arithmetic operations, so elements of $\mathbb{R}\mathbb{Z}_n$ can be multiplied in $O(n\log n)$ arithmetic operations.
Define the group semiring $\mathcal{R}\mathbb{Z}_n$ in an analogous way.  The product of $x = (x_0,\ldots, x_{n-1})$ and $y = (y_0,\ldots,y_{n-1})$ in this semiring is given by $(x\cdot y)_k = \min _{j \in \mathbb{Z}_n} (x_j + y_{k-j})$.  This operation could perhaps be called "discrete cyclic infimal convolution" by analogy with the notion of infimal convolution in convex analysis.  I'm not sure whether there is a more standard name -- this one does not pop up in quick google searches.
The naive way of computing a discrete infimal convolution uses $O(n^2)$ operations, just as the naive method for computing a standard cyclic convolution.  My question is: is there a way to compute this "discrete cyclic infimal convolution" in $O(n\log n)$ arithmetic operations?
In convex analysis, there is an analog of the Fourier transform which turns infimal convolution into pointwise addition: the convex conjugate (or Fenchel or Legendre transformation).  However, this operation only behaves nicely for convex functions, so it is not clear to me how one would translate it to an equivalent tool for $\mathcal{R}\mathbb{Z}_n$, but perhaps there is something there.
I would be interested in answers to the question regardless of whether they go through some analog of the Fourier transform.  Also, I don't mind various restrictions such as making $n$ a power of $2$, replacing $\mathbb{R}$ with $\mathbb{Q}$, removing $\infty$ from the definition of the semiring, etc.  Really anything on this theme would be helpful.  Also any suggestions for better tags would be appreciated; perhaps this is a well-studied area I'm not aware of.

REPLY [4 votes]: You can approximate this using a fast numerical method (note this answer assumes you are performing max-convolution on an equivalent transformed problem on the ring $(\times, \max)$ rather than $(+, \min)$-- you can convert from this one to the one in the question using logarithms and negation if necessary):
If $M[m]$ is the exact result at index $m$ (i.e., $M = L ~*_\max~ R$), then consider every vector $u^{(m)}$ (for each index $m$ of the result, where $u^{(m)}[\ell] =
L[\ell] R[{m-\ell}]$). When the vectors $L$ and $R$ are nonnegative (in my case, this was true because they are full of probabilities), then you can perform the $\max_\ell u^{(m)}_\ell$ with the Chebyshev norm:
$$
M[m] = \max_\ell u^{(m)}_\ell \\
= \lim_{p \to \infty} \| u^{(m)} \|_p \\
 = \lim_{p \to \infty} {\left( \sum_\ell {\left( u^{(m)}[\ell] \right)}^{p} \right)}^{\frac{1}{p}}\\
  \approx {\left( \sum_\ell {\left( u^{(m)}[\ell] \right)}^{p^*} \right)}^{\frac{1}{p^*}}\\
$$
(where $p^*$ is a large enough constant)
$$
  = {\left( \sum_\ell {L[\ell]}^{p^*} ~ {R[{m-\ell}]}^{p^*} \right)}^{\frac{1}{p^*}}\\
  = {\left( \sum_\ell {\left(L^{p^*}\right)}[\ell] ~ {\left(R^{p^*}\right)}[{m-\ell}] \right)}^{\frac{1}{p^*}}\\
  = {\left( L^{p^*} ~*~ R^{p^*} \right)}^{\frac{1}{p^*}}[m]
$$
The standard convolution (denoted $*$) can be performed in $n \log(n)$ via FFT. A short paper illustrating the approximation for large-scale probabilistic inference problems is in press at the Journal of Computational Biology (Serang 2015 arXiv preprint). 
Afterward a Ph.D. student, Julianus Pfeuffer, and I hacked out a preliminary bound on the absolute error ($p^*$-norm approximations of the Chebyshev norm are poor when $p^*$ is small, but on indices where the normalized result $\frac{M[m]}{\max_{m'} M[m']}$ is very small, large $p^*$ can be numerically unstable). Julianus and I worked out a modified method that is numerically stable in cases when the dynamic range of the result $M$ is very large (when the dynamic range is small, then the simple method from the first paper works fine). The modified method operates piecewise over $\log(\log(n))$ different $p^*$ values (including runtime constants, this amounts to $\leq 18$ FFTs even when $n$ is on the scale of particles in the observable universe, and those 18 FFTs can be done in parallel). (Pfeuffer and Serang arXiv link which has a link to a Python demonstration of the approach) The approach (and the Python demo code linked by the second paper) generalizes to tensors (i.e., numpy.arrays) by essentially combining the element-wise Frobenius norm with multidimensional FFT. There is no other known faster-than-naive method for matrices (and tensors). 
Max-convolution is a very fun puzzle-- I'd be interested to hear about your specific application for this in comments (maybe the fast numerical method would work for you?). 
Oliver Serang<|endoftext|>
TITLE: Normal subgroup that is invariant under powering such that the quotient group is not invariant
QUESTION [5 upvotes]: I want an example of a group $G$, a normal subgroup $H$, and a prime number $p$, such that:

$G$ is powered over $p$, i.e., every element of $G$ has a unique $p^{th}$ root in $G$.
$H$ is also powered over $p$, i.e., every element of $H$ has a unique $p^{th}$ root in $H$.
The quotient group $G/H$ is not powered over $p$. Since the above conditions already guarantee the existence of $p^{th}$ roots, what I want should fail is the uniqueness condition.

While I suspect that an example exists, the example seems hard to construct, because of the following constraints I worked out for any example:

$H$ must be infinite and have infinite index in $G$ (i.e., neither $H$ nor $G/H$ can be finite).
$H$ cannot be contained in the hypercenter of $G$ (the hypercenter is the subgroup at which the upper central series stabilizes). This rules out any example involving $G$ abelian or nilpotent.
$H$ cannot have a complement (i.e., be part of a semidirect product) in $G$.

The proofs of all these assertions are straightforward, but I'll be happy to provide proofs if they are unclear to readers.
If you find a proof that no such example exists, that would be great to have too.

REPLY [8 votes]: This is a corrected answer. I apologize for not posting a complete proof here.
Recall that a group $G$ is called divisible if for every $g\in G$ and $n\in \mathbb N$, there is $x\in G$ satisfying $x^n=g$. Recall also that there exist countable (and even finitely generated) torsion free divisible groups where every element has infinitely many $n$th roots for every $n$. We fix one such a group and denote it by $D$. 
The following theorem answers the question. 
Theorem.
There exists a countable uniquely divisible group $G$ and a divisible normal subgroup $H\le G$ such that $G/H$ contains $D$. In particular, there are elements $g\in G/H$ that have infinitely many $n$th roots for every $n\in \mathbb N$.
Unfortunately, I do not know any easy proof. The only proof I know would take few pages. The main idea is to use a modification of the construction from the proof of Theorem 1.5 of my paper http://arxiv.org/abs/math/0411039. 
These groups $G$ and $H$ are very far from being finite or nilpotent; they will contain non-abelian free subgroups (this is unavoidable in my construction).<|endoftext|>
TITLE: Mathematicians whose works were criticized by contemporaries but became widely accepted later
QUESTION [67 upvotes]: Gauss famously discarded Abel's proof that an algebraic equation of degree five or more cannot have a general solution (Abel himself had rejected divergent series as the work of the devil). Cantor's theory of transfinite numbers was originally regarded as so counter-intuitive—even shocking—that it encountered resistance from mathematical contemporaries such as Leopold Kronecker and Henri Poincaré and later from Hermann Weyl and L. E. J. Brouwer, while Ludwig Wittgenstein raised philosophical objections. Ramanujan's work on divergent series was rejected by three leading English mathematicians of the time before he was discovered by Hardy.
The above stories have become mathematical folklore. I would like to know the examples of other mathematicians whose works were initially criticized or rejected by contemporaries but later became widely accepted famous. I am particularly interested in modern mathematicians or lesser known mathematicians of the classical era who stories may not be as popular as those of other mathematical giants.

REPLY [6 votes]: From Saunders Mac Lane's obituary:
"... In seeking to provide a sound conceptual framework for the subject [of homological algebra], they invented the notions of category and functor. These notions were slow to gain acceptance (the first Eilenberg-Mac Lane paper on categories was nearly rejected by the Transactions of the American Mathematical Society) on account of their seeming lack of content: for a decade or so, category theory was derided by other mathematicians as "abstract nonsense". But in time the substantial new advances made possible by the categorical way of thinking about mathematics won it acceptance: it has by now become an indispensable part of the vocabulary of the great majority of pure mathematicians (and, increasingly, of researchers in theoretical physics and computer science)."<|endoftext|>
TITLE: Connected sum of topological manifolds
QUESTION [43 upvotes]: A definition of the connected sum of two $n$-manifolds $M$ and $M'$  begins by considering two $n$-balls $B$ in $M$, $B'$ in $M'$, and glueing the varieties $M\setminus \mathring B$ and $M'\setminus \mathring B'$ along their boundary (an $(n-1)$-sphere) by an orientation-reversing homeomorphism.  The construction depends a priori on these various choices, but it is asserted at many places of the litterature (Lee's book on topological manifolds for example, as well as Wikipedia) 
 that the result does not depend on these choices.
In the differentiable case, a reference is given to a theorem of Palais (Natural operations on differential forms, Thm. 5.5) which asserts — roughly — that two embedding of $n$-balls differ by a global diffeomorphism which is isotopic to identity.
Are the details of this independence written somewhere in the litterature,
both in the continuous and the smooth case?

REPLY [12 votes]: The previous version of this post was incorrect.
EDIT:
I just taught a related fact in my undergraduate topology class (in the topological category), and in fact I usually give a version of this as a homework exercise (with hints). 
The idea is to define connected sum along the coordinate balls, i.e. the balls that are mapped via a fixed coordinate chart to standard balls in $\mathbb R^n$. For such balls the argument is easy, and no annulus theorem is needed because any two metric balls in $\mathbb R^n$ are ambiently isotopic by an isotopy that equals the identity outside a compact subset  (this can be done with bare hands). Since the homeomorphism group of manifold acts transitively on the manifold (nice exercise), any two points can be brought into the same chart. This provides some clue (not a complete proof) why the connected sum operation is independent of the choices involved.<|endoftext|>
TITLE: Disjoint Maximum Independent Sets in $\alpha$-critical graphs
QUESTION [9 upvotes]: Let $G$ be an undirected, simple graph, and let $\alpha(G)$ denote the independence number of $G$, i.e., the size of a maximum independent set (stable set) in $G$. A graph is $\alpha$-critical if for every edge $e$ of $G$, we have $\alpha(G - e) > \alpha(G)$.
Conjecture: for every $\alpha$-critical graph $G$ without isolated vertices, for every maximum independent set $S$ in $G$, there is a maximum independent set $S'$ in $G$ that is disjoint from $S$.
The conjecture is true for the simplest classes of $\alpha$-critical graphs, the graphs $K_n$ ($n \geq 2$) and the odd cycles. It also seems to be true for the list of minor-minimal obstructions to having a vertex cover of size at most five, which are also $\alpha$-critical. Is it true in general? 
Thanks in advance!

REPLY [2 votes]: Late answer, but for future reference (so that anyone reading this thread won't spend as much time as I did trying to prove the conjecture) the conjecture is false, with the following simple graph as a counterexample. Let $G$ be the following graph

One can see that $\alpha(G) = 4$. It is also easy to see that if we remove any of its edges we obtain a graph with independence number 5, whence it is $\alpha$-critical. The vertex set $S = \{1,4,7,9\}$ (or, alternatively, $S=\{2,5,8,9\}$) is a maximum independent set with the property that $H = G \setminus S$ (the graph formed by removing the vertices in $S$, and all edges incident to them, from $G$) is just the graph consisting of three $K_2$-components. Thus $\alpha(H) = 3$ and therefore the maximum size of any independent set in $G$ disjoint from $S$ is three.
Another reason for reopening this thread is that I would be interested to know if the following weakening of the conjecture could be true:
Weak conj. for every $\alpha$-critical graph $G$ without isolated vertices, there exists a maximum independent set $S$ in $G$ and a maximum independent set $S′$ in $G$ that is disjoint from $S$.
I somehow get the feeling that this is unlikely to be true, however I am unable to find a counterexample. The graph above, that provides a counterexample to the stronger conjecture, does indeed satisfy this weaker version, for example by taking the maximum independent sets $S = \{1,2,4,9\}$ and $S' = \{5,6,7,8\}$. There is one more (connected) graph with a counterexample to the stronger conjecture that I know of;
 with $S = \{1,4,7,9\}$.
However, also this graph satisfies the weaker version of the conjecture (take e.g. $S = \{1,3,4,9\}$ and $S' = \{0,6,7,8\}$).
I know of no more counterexamples to the stronger conjecture and I believe that the only graphs that provide counterexamples with less than $11$ vertices are the two graphs mentioned above. Thus I believe that any counterexample to the weaker conjecture would have to have at least $11$ vertices.<|endoftext|>
TITLE: Exponential map
QUESTION [7 upvotes]: Does there exist an infinite dimensional Lie group $G$, with Lie algebra $\mathfrak g$, such that the exponential map $exp:\mathfrak g \to G$ is not defined?
If so, can one provide an example of such a group.

REPLY [4 votes]: About Robert's example $Diff(\mathbb R)$ one can argue that it is not a Lie group, since it does not admit charts: For the compact $C^\infty$-topology it is not open in $C^\infty(\mathbb R)$. In the Whitney $C^\infty$-topology it is not locally contractible. 
Aside:
A setting where $Diff(\mathbb R)$ is a Lie group in the category of manifolds based on smooth curves etc (where the finite dimensional ones are exactly the known ones, as are the Banach ones)
is:
Peter W. Michor: A convenient setting for differential geometry and global analysis, I, II. Cahiers Topologie Geometrie Differentielle 25 (1984), 63--109, 113--178.(pdf of I)
(pdf of II)
Definition of regular Lie groups:
We consider a  smooth Lie group $G$ with Lie algebra $\mathfrak g=T_eG$ modelled on convenient 
vector spaces. 
The notion of a regular Lie group is originally due to Omori and collaborators
(see [Omori Maeda Yoshioka 1982], [Omori Maeda Yoshioka 1983]) for Frechet Lie groups, was 
weakened and made more transparent by [Milnor 1984] and carried over to convenient Lie 
groups in (here), 
see also 38.4 of (here).
A Lie group $G$
is called  regular  if the following holds:
$\bullet$ 
For each smooth curve 
$X\in C^{\infty}(\mathbb R,\mathfrak g)$ there exists a curve 
$g\in C^{\infty}(\mathbb R,G)$ whose right logarithmic derivative is $X$, i.e.,
$$
g(0) = e, \qquad
\partial_t g(t) = T_e(\mu^{g(t)})X(t) = X(t).g(t),\quad\text{where }  \mu(a,b)=\mu_a(b)=\mu^b(a) = a.b.
$$
The curve $g$ is uniquely determined by its initial value $g(0)$, if it
exists.
$\bullet$
Put $\operatorname{evol}^r_G(X)=g(1)$ where $g$ is the unique solution required above. 
Then $\operatorname{evol}^r_G: C^{\infty}(\mathbb R,\mathfrak g)\to G$ is required to be
$C^{\infty}$ also. 
Note that for $X$ constant in time, $\operatorname{evol}^r_G(X)=\exp(X)$. So each regular Lie group admits an exponential mapping.
The family of regular Lie groups is remarkably stable under constructions like extensions and quotients.
I do not know a Lie group modeled on convenient vector spaces which is not regular.
A quasi-counter-example is due to Wuestner:
Consider the space of trigonometric rational functions on $S^1$ which are everywhere positive and have no pole, with multiplication. This is not regular since $\exp(X)=e^X$ is real analytic and not trigonometric rational any more. But the modelling space is not convenient, since it is not Mackey sequentially complete for any suitable topology.<|endoftext|>
TITLE: weight monodromy conjecture for curves?
QUESTION [6 upvotes]: Hi,
Is there a simple proof of the weight monodromy conjecture in the case of a curve over a mixed characteristic local field?
Thanks!

REPLY [4 votes]: It seems to me that you can find a proof in the paper of Raynaud: 1-motifs et monodromie géométrique. Astérisque No. 223 (1994), 295–319. 
I don't know if this is "simple" but it is about curves and their Jacobian varieties. The argument involves the theory of 1-motives and the rigid uniformisation of abelian varieties.<|endoftext|>
TITLE: Two different analytic curves cannot intersect in infinitely many points
QUESTION [6 upvotes]: A curve in Euclidean space $\mathbb{R}^n$, $n \geq 2$ is $analytic$ if the coordinates of its points $x= x_{1},...,x_{n}$ can be expressed as analytic functions of a real parameter $x_{i}=x_{i}(t)$, $i=1,...,n$ and $\alpha \leq t \leq \beta$ and the derivatives $x'(t_{0})$ do not simultaneously vanish for any $t_{0} \in [\alpha, \beta]$.
I search for a proof of the following fact:
If the set of intersection points of two analytic curves is infinite, then these curves coincide.
Can we prove the same as above if we relax the condition that the coordinates are analytic to the condition that the coordinates belong to the class $C^{\infty}$?
Edit: Thanks to below remark by Ramiro, to obtain the above implication, we have to assume that any two curves $K_{1}, K_{2}$ as in the question are such that $K_{1} \cap K_{2}$ is not another analytic curve.
2nd edit: As suggested Peter, we can reformulate our question as follows:
   Consider two immersed curves which are parameterized real analytically on compact intervals. If they have an infinite number of different intersection points, then their union is again a real analytic immersed curve.

REPLY [5 votes]: Let $f\colon [0,1]\to \mathbf R^n$ and $g\colon[0,1]\to\mathbf R^n$ parametrize your curves. The set $C$ of $t\in[0,1]$ such that there exists $s\in[0,1]$ with $f(t)=g(s)$ is a closed subset of $[0,1]$. More importantly, it is definable in the sense of model theory in the language — called $\mathbf R_{\text{an}}$ — consisting of polynomial functions and so called restricted analytic functions, namely analytic functions defined over compact subsets of $\mathbf R^m$.
It is an important theorem (Gabrielov; Denef and van den Dries) that $\mathbf R_{\text{an}}$ is $o$-minimal. This means that definable subsets of the real line are finite unions of intervals. 
From there, one should be able to analyse the situation further when the coincidence set is infinite. I would conjecture that it can only be the union of one or two intervals. Ramiro gave an example with one interval, though more complicated examples are possible, e.g., of the form $s\mapsto F(u(s))$, $s\mapsto F(v(s))$, where $u,v\colon[0,1]\to\mathbf R$ are analytic and $F\colon\mathbf R\to\mathbf R^n$ is a fixed curve.  In particular, the set is a union of two intervals if $F$ is a parameterization of a circle and $u$ and $v$ are suitably chosen so as to draw two arcs which overlap twice.<|endoftext|>
TITLE: Analogue of Picard-Lefschetz formulas for more than one node?
QUESTION [6 upvotes]: It is an apparently well known result that if one has a 1-parameter family of smooth Calabi-Yau 3-folds which acquires a node at a boundary point, then there is a vanishing lagrangian 3-sphere, and the geometrical monodromy is a Dehn-twist along this sphere whose effect on the homology of our base fiber is given by the classical Picard-Lefschetz formula  $$\alpha \mapsto \alpha-(\delta\cdot \alpha) \delta,$$ where $\delta$ is the homology class of the vanishing 3-sphere.  (This formula or an analogue of it actually holds in much more general families than just CY 3-folds, but that's the case I'm interested in).  For a modern reference for the above result, see Chapter 3 of Looijenga's $\textit{Isolated Singular Points on Complete Intersections}$.  
My question is whether there is a known analogue for the case when the family in question acquires more than one $A_1$-singularity.  For example, if the smooth family acquires 4 nodes, is there a known formula for the monodromy action around such a fiber for Calabi-Yau 3-folds?
If no formula exists, is there any sense for what it's matrix might look like, roughly?  Do we know, for example, that $Rank(S-1)=1$, where $S$ is the monodromy action and $1$ is the identity matrix?

REPLY [4 votes]: In general, I think that the monodromy  should be
$$\alpha\mapsto \alpha - \sum_i (\delta_i\cdot \alpha)\delta_i $$
where $\delta_i$ are the vanishing cycles corresponding to each node. I'll see whether I can 
write down an outline of an argument later.
... Meanwhile later... Let me complement Tim's excellent comment with a more obscure explanation. The so called variation map $S-I: H^i(X_t)\to H^i(X_t)$ of the cohomology of the nearby fibre factors through a the cohomology of a (complex of) sheaf(ves) $\mathbb{R}\phi\mathbb{Z}$ supported on singular locus of the special fibre $X_0$. In the case that $X_0$ has isolated singularities, this sheaf decomposes into a sum of the corresponding sheaves supported at each of the critical points, and maps decomposes as well. The formula
I stated above would follow this together with the standard Picard-Lefschetz formula.<|endoftext|>
TITLE: Why does GL(N) have no spinor representations?
QUESTION [9 upvotes]: Does anyone know of a simple proof that the general linear group GL(N) does not admit spinor representations?
Thank you!

REPLY [17 votes]: The statement in GSW that you quote has to be interpreted properly.  When they write, "Spinors form a representation of $\mathrm{SO}(N)$ which does not arise from a representation of $\mathrm{GL}(N,\mathbb{R})$", they really mean either "Spinors form a representation of ${\frak{so}}(N)$ which does not arise from a representation of ${\frak{gl}}(N,\mathbb{R})$" (i.e., it's really a statement about representations of Lie algebras) or else they mean what some sources call "multi-valued representations", i.e., representations of the simply-connected covers of the groups $\mathrm{SO}(N)$ and $\mathrm{GL}(N,\mathbb{R})$.  Moreover, 'arise from' means that there is no (finite dimensional) representation of ${\frak{gl}}(N,\mathbb{R})$ that, when restricted to ${\frak{so}}(N)\subset{\frak{gl}}(N,\mathbb{R})$, contains a copy of a 'spinor representation' of ${\frak{so}}(N)$.
Either way, what it comes down to is this:  If $\pi:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{SL}(n,\mathbb{R})$ is the nontrivial double cover of $\mathrm{SL}(n,\mathbb{R})$, then any finite-dimensional representation $\hat\rho:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{R})$ must factor through $\pi$, i.e., $\hat\rho = \rho\circ\pi$ for some (unique) representation $\rho:\mathrm{SL}(n,\mathbb{R})\to\mathrm{GL}(k,\mathbb{R})$.
The reason for this is topological:  Although $\pi_1\bigl(\mathrm{SL}(n,\mathbb{R}),I_n\bigr)$ is nontrivial for $n\ge 2$ (being $\mathbb{Z}$ when $n=2$ and $\mathbb{Z}_2$ when $n>2$), the group $\pi_1\bigl(\mathrm{SL}(n,\mathbb{C}),I_n\bigr)$ is trivial for all $n\ge 2$.  Using this fact, the proof goes like this:
If $\hat\rho:\hat{\mathrm{SL}}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{R})$ is any representation, let $\hat\rho':{\frak{sl}}(n,\mathbb{R})\to {\frak{gl}}(k,\mathbb{R})$ denote the induced Lie algebra homomorphism.  This complexifies to a Lie algebra homomorphism $(\hat\rho')^{\mathbb{C}}:{\frak{sl}}(n,\mathbb{C})\to {\frak{gl}}(k,\mathbb{C})$ and, since $\mathrm{SL}(n,\mathbb{C})$ is simply-connected, this is induced by a Lie group homomorphism $\rho^{\mathbb{C}}:\mathrm{SL}(n,\mathbb{C})\to \mathrm{GL}(k,\mathbb{C})$, which restricts to a Lie group homomorphism $\rho:\mathrm{SL}(n,\mathbb{R})\to \mathrm{GL}(k,\mathbb{C})$ with associated Lie algebra homomorphism $\rho':{\frak{sl}}(n,\mathbb{R})\to{\frak{gl}}(k,\mathbb{C})$.  By construction, this homomorphism $\rho'$ must be the composition of $\hat\rho':{\frak{sl}}(n,\mathbb{R})\to {\frak{gl}}(k,\mathbb{R})$ with the inclusion ${\frak{gl}}(k,\mathbb{R})\hookrightarrow {\frak{gl}}(k,\mathbb{C})$.  In particular, $\rho'$ maps into ${\frak{gl}}(k,\mathbb{R})$ after all.  Now it's clear that $\hat\rho = \rho\circ\pi$.
In particular, if $\mathrm{Spin}(n)\subset \hat{\mathrm{SL}}(n,\mathbb{R})$ is the pre-image under $\pi$ of $\mathrm{SO}(n)\subset \mathrm{SL}(n,\mathbb{R})$, then any finite dimensional representation of $\hat{\mathrm{SL}}(n,\mathbb{R})$ restricts to a representation of $\mathrm{Spin}(n)$ that factors through $\mathrm{SO}(n)$ and hence cannot be (or even contain) a spinor representation, since those do not factor through $\mathrm{SO}(n)$.<|endoftext|>
TITLE: Mendelson's *Mathematical Logic* and the missing Appendix on the consistency of PA
QUESTION [20 upvotes]: A very soft question, but I hope not out of order here.
In the first edition of Elliott Mendelson's classic Introduction to Mathematical Logic (1964) there is an appendix, giving a version of Schütte's (1951) variation on Gentzen's proof of the consistency of PA. This is intriguing stuff, crisply and quite accessibly presented. The appendix is, however, suppressed in later editions (in fact, from the second onwards), even though there is plenty of room given to other materials and a new appendix
Now, a number of people have said that the appendix is one of the most interesting things about the book. I agree. I too remember being quite excited by it when I first came across it a long time ago!
So: has anyone heard a folkloric story about why Mendelson suppressed the appendix? I've never heard it suggested that there is a problem with the consistency proof as given.
Context, if you are interested: I asked this a couple of weeks ago on math.SE (without getting an answer) when starting to write up a survey of some of the Big Books on Mathematical Logic that will become part of my Teach-Yourself-Logic Guide (mostly for philosophers, though others might be interested), and I'd got to Mendelson. You can get the current version of the Guide by going to http://www.logicmatters.net/students/tyl/

REPLY [16 votes]: Before posting this question I did search around a bit (probably inefficiently and certainly quite ineffectively) to see if I could find an email for Elliott Mendelson to ask him directly! But anyway, he picked up my similar query on FOM and very kindly wrote to me:

I was intrigued by your comments about the consistency proof of PA that appeared in the First Edition of my logic book.  I omitted it in later editions because I felt that the topic needed a much more thorough treatment than what I had given, a treatment that would require more space than would be appropriate in an introduction to mathematical logic.  

I can understand that. Though I think the pointers he gave in that Appendix did spur on quite a few readers to find out more, so I still think it was a Very Good Thing, and it was perhaps a pity to drop it. 
[Prof. Mendelson has kindly allowed me to quote him.]<|endoftext|>
TITLE: Connection between complex orientations and R-orientations for a ring spectrum R?
QUESTION [5 upvotes]: We have a well defined notion of complex orientation for a spectrum (coh. theory) $E$, that is, we have a class $x_E\in \tilde{E}^2(\mathbb{C}P^\infty)$ which restricts to identity along the inclusion $\mathbb{C}P^1\hookrightarrow \mathbb{C}P^\infty$. Is there a connection between this notion of "complex orientation" and the notion of a Thom spectrum being $R$ oriented with respect to a ring spectrum $R$ in the sense of ABGHR? That is, for a Thom spectrum $Mf$ associated to a map $f:X\to BGL_1\mathbb{S}$, and a ring spectrum $R$, we say that $Mf$ is $R$-oriented if the composition $X\to BGL_1\mathbb{S}\to BGL_1R$ is null. If we consider $MU$ to be the Thom spectrum associated to some map (I'm not sure which it should be, but I suspect this is the way it's done) $BU\to BGL_1\mathbb{S}$, can we rephrase the notion of complex orientation in this language?
Thanks!

REPLY [8 votes]: When I wrote the comment above, my memory was blanking. The connection between ring maps $MU\to R$ and complex orientations that Mark describes goes back to Quillen's original work relating $MU$ to formal group laws.  (Lemma 4.6, page 52, in Adams Stable Homotopy and Generalized Cohomology characterizes ring maps $MU\to R$ in terms of complex orientations).
A discussion of $E_{\infty}$ orientations $MU\to R$ that may help with the original 
question is given in Section 5 of ``What are $E_{\infty}$ ring spaces good for?'', which relates spectrum level orientations to $E_{\infty}$ orientations on the space level.  Such orientations are equivalent to $E_{\infty}$ maps $BU \to B(U;R)$, where $B(U;R)$ classifies $R$-oriented $U$-bundles (in the classical sense of orientation).  I apologize if the relevance is unclear.  It's late.
[Continued] Here is some more background. To understand the mathematics here, you must recognize that the unit space once called $F$ and now called $GL_1(S)$ plays two very different roles, one additive and one multiplicative. This is explained in the introduction to "What are $E_{\infty}$ ring spaces good for?" The space $BF$ classifies sectioned stable spherical fibrations, and its product classifies fiberwise smash products.  You go from stable vector bundles to stable spherical fibrations by fiberwise one point compactification, and that takes Whitney sum of bundles to fiberwise smash product.  Therefore we think of this as additive structure.  The map $BU \rightarrow BF$ sees this on the represented functor level,
and it is maps like this that you are thinking of as leading to Thom spectra (as they do by
Gaunce Lewis's thesis, in "Equivariant stable homotopy theory" LMS 1213.  There is also an infinite loop space $BU_{\otimes}$, which is the identity component of the space $GL_1(KU)$.  There is no analogous $BF_{\otimes}$.  
As the unit space $GL_1(S)$, we think of the same space $F$ as multiplicative.  For any 
commutative (ie $E_{\infty}$) ring spectrum $R$, we have the unit map $S\to R$, and 
on passage to zeroth spaces it induces an infinite loop map  $GL_1(S) \rightarrow GL_1(R)$.
There is a fibration sequence $B(U,R) \to BU \to  BGL_1(R)$.  The first map says ``forget the orientation''.  The second is the obstruction to (universal) $R$-orientability of complex vector bundles.  An $E_{\infty}$ $R$-orientation of complex bundle theory is an $E_{\infty}$ map $g\colon BU \to B(U,R)$ which sections $B(U,R) \to BU$.  Such orientations $g$ correspond to maps $MU\to R$ of commutative ring spectra, as I explained in Section 5 opus cit.  Of course, everything in this general theory works equally well for other kinds of bundles. The interplay of additive and multiplicative structure is the key to the splitting of $SF$ as $J\times Coker J$ at each prime $p$ (only an $E_{\infty}$ splitting when $p$ is odd).<|endoftext|>
TITLE: Does every commutative monoid admit a translation-invariant measure?
QUESTION [7 upvotes]: Let $T$ be a commutative monoid, written additively. The set $T$ is equipped with a canonical pre-order, defined by $s \le t$ when there exists $s' \in T$ so that $s + s' = t$. Consequently, $T$ may be equipped with the specialization topology for this pre-order, where the closed sets are those which are downward-closed. Note that $T$ is typically not Hausdorff, since the closure of a singleton is its down-set: $\overline{\{t\}} = t\!\downarrow\, := \{ s : s \le t \}$. 
Let $\mathcal B(T)$ denote the Borel $\sigma$-algebra with respect to this topology. In this way, every commutative monoid is canonically a measurable space.
Equipped with the $\sigma$-algebra $\mathcal B(T)$, does every commutative monoid $T$ admit a (non-trivial) family of translation-invariant measures?

REPLY [4 votes]: I think the integers Z with max is a counterexample. First note the set $I_n$ of all integers bigger than or equal to n is open. Thus each singleton is Borel by looking at $I_n\setminus I_{n+1}$. Hence by countability of Z the measure is a weighted counting measure. But the inverse image of n under translation by n consists of all numbers less than or equal to n. Thus the weight of n is the weight of n plus the weights of all numbers less than n. Thus the weights of all numbers strictly less than n are zero. Since n is arbitrary all elements have weight 0.<|endoftext|>
TITLE: A "known" Pythagorean identity in algebra?
QUESTION [6 upvotes]: Some will recognize this as similar to a question I asked before, but
I want to ask it without the trigonometry.
Let $e_k$ be the $k$th-degree elementary symmetric polynomial in
$x_1,x_2,x_3,\ldots$.  If $k$ is more than the number of $x$s, then
$e_k$ is the sum of no terms and is $0$.  From one POV, the following
Pythagorean identities are as elementary as anything not in the
high-school curriculum:
$$
(e_0+e_2+e_4+\cdots)^2 - (e_1+e_3+e_5+\cdots)^2 =
(1-x_1^2)(1-x_2^2)(1-x_3^2)\cdots
$$
$$
(e_0-e_2+e_4-\cdots)^2 + (e_1-e_3+e_5-\cdots)^2 =
(1+x_1^2)(1+x_2^2)(1+x_3^2)\cdots
$$
So are these "known" in the sense of being in refereed publications
one could cite?  And if not, are they "known" in the sense that some
people see them mentioned or explicitly used from time to time?
(And if there should happen to be infinitely many $x$s, could this
still be considered only algebra by thinking of these as a sort of
"formal" series?)

REPLY [10 votes]: These are both simple corollaries of
$$\sum_{k\geq 0} t^ke_k(x_1,x_2,\dots)=\prod_{k\geq 0}(1+tx_k).$$
There is a typo in both your identities. They should read,
$$(e_0+e_2+\cdots)^2-(e_1+e_3+\cdots)^2=(e_0+e_1+e_2+\cdots)(e_0-e_1+e_2-\cdots)$$
$$=\prod_{k\geq 0}(1+x_k)\prod_{k\geq 0}(1-x_k)=(1-x_1^2)(1-x_2^2)\cdots,$$
and
$$(e_0-e_2+e_4-\cdots)^2+(e_1-e_3+e_5-\cdots)^2=(\mathfrak {Re}[\prod_{k\geq 0}(1+ix_k)])^2+(\mathfrak{Im}[\prod_{k\geq 0}(1+ix_k)])^2$$
$$=(1+x_1^2)(1+x_2^2)\cdots$$<|endoftext|>
TITLE: The value $\pm 1$ for the square root of Wilson's theorem, ((p-1)/2)! mod p
QUESTION [34 upvotes]: While teaching number theory this quarter, I have come across a phenomenon which was already addressed in another MO posting, but I have new questions.  Let $p$ be a prime congruent to 3 mod 4.  Then an elementary refinement of Wilson's theorem says that $\frac{p-1}2!$ is congruent to $\pm 1$ mod $p$.  In fact, a published result of Mordell says that it is congruent to $(-1)^{(h+1)/2}$, where $h$ is the class number of the imaginary quadratic number field $\mathbb{Q}(\sqrt{-p})$.  Mordell's identity is taken to mean that (1) you cannot expect $\frac{p-1}2! \in \mathbb{Z}/p$ to be governed by any very simple property of $p$ such as a congruence, and (2) we can conjecture but not prove that the density of both values is $\frac12$.  So I wonder about the following:
(1) What is fastest known algorithm to compute $\frac{p-1}2! \in \mathbb{Z}/p$?  I read somewhere that you can compute $h$ in time $O(p^{1/5})$, but for this question (actually for the interest of the class) I only care about $h$ mod 4.
(2) Never mind the density, is it known that both values occur infinitely often?
(3) If it is known that both values occur infinitely often, then is it known, for each $\text{gcd}(n,a) = 1$, that they occur infinitely often when $p = kn+a$?  (It is of course Dirichlet's theorem that there are infinitely many such primes.)

Update:  There as an subexponential time algorithm of McCurley to compute the class number $h$.  I can believe that computing $h$ mod 4 wouldn't be any faster.

REPLY [6 votes]: As suggested in the comment by Greg Martin, the determination of the parity of $(h + 1)/2$ can be performed more efficiently than by computing the factorial expression in Lagrange’s refinement of Wilson’s theorem.
Dirichlet, “Question d’analyse indéterminée,” Journal für die Reine und Angewandte Mathematik  3 (1828): 407–408; reprinted in Werke 1:107–108, noted that its parity depends upon the number of quadratic nonresidues of $p$ lying between $1$ and $p/2$ (OEIS sequence no. A178151). This number is usually designated $m$ in the literature.
Skipping over many partial results, the value of $m$ was investigated in detail by Louis C. Karpinski in his doctoral dissertation (Mathematischen und Naturwissenschaftlichen Facultät der Kaiser Wilhelms-Universität zu Strassburg, 1903), published as “Über die Verteilung der quadratischen Reste,” Journal für die Reine und Angewandte Mathematik 127 (1904): 1–19. Karpinski proved a set of formulae involving sums over Legendre symbols, and showed that the most concise sums possible contain only $\lfloor p/6 \rfloor$ terms:
\begin{equation}
m = \frac{p-1}{4} - \frac{1}{2} \sum_{k=1}^{(p-1)/2} \left(\frac{k}{p}\right) \quad (p \equiv 3 \bmod{4}; p > 3);
\end{equation}
\begin{equation}
m = \frac{p-1}{4} - \frac{2-\left( \frac{2}{p} \right)}{3-\left( \frac{3}{p} \right)} \sum_{k=1}^{\lfloor p/3 \rfloor} \left(\frac{k}{p}\right) \quad (p \equiv 3 \bmod{4}; p > 3);
\end{equation}
\begin{equation}
m = \frac{p-1}{4} - \frac{1}{2} \sum_{k=\lfloor p/4 \rfloor +1}^{(p-1)/2} \left(\frac{k}{p}\right) \quad (p \equiv 3 \bmod{8}; p > 3);
\end{equation}
\begin{equation}
m = \frac{p-1}{4} - \frac{1}{2} \quad \sum_{k=1}^{\lfloor p/4 \rfloor} \quad \left(\frac{k}{p}\right) \quad (p \equiv 7 \bmod{8}; p > 3);
\end{equation}
\begin{equation}
m = \frac{p-1}{4} - \frac{2-\left( \frac{2}{p} \right)}{1 + \left(\frac{2}{p}\right) + \left( \frac{3}{p}\right) - \left( \frac{6}{p} \right)} \sum_{k=1}^{\lfloor p/6 \rfloor} \left(\frac{k}{p}\right) \quad (p \equiv 7, 11, 23 \bmod{24});
\end{equation}
\begin{equation}
m = \frac{-5p-1}{4} + \frac{3}{2} \sum_{k=1}^{\lfloor p/6 \rfloor} \left(\frac{k}{p}\right) \quad (p \equiv 19 \bmod{24}).
\end{equation}<|endoftext|>
TITLE: Is the closed unit ball of the Hilbert space homeomorphic to the unit sphere ?
QUESTION [18 upvotes]: Is the closed unit ball of the Hilbert space (or, for that matter, of the Hilbert cube, in some metric) homeomorphic to the unit sphere (viz., its own boundary) ? This is clearly uncharacteristic of finite-dimensional cubes. This question is motivated by general considerations in dimension theory. If there is such a homeomorphism, the small inductive dimension, generalised verbatim to infinite cardinals, cannot exist for such spaces (whose "dimension" is a "strange" cardinal like w).
The initial question with 'open' ball was unwittigly typed.

REPLY [19 votes]: The answer to the question in the title is yes. 
In Bessaga and Pelczynski, Selected topics in infinite-dimensional topology, Chapter VI, §2 there is a proof of the following:

Theorem. Each of the following sets is homeomorphic to the countable product $\mathbb{R^N}$ of the real line:

The separable Hilbert space $\ell_2$.
The closed unit ball in $\ell_2$.
The unit sphere in $\ell_2$.
The "upper half space" in $\ell_2$: those vectors with non-negative first entry.<|endoftext|>
TITLE: Definition of CM modular form 
QUESTION [18 upvotes]: Dear friends,
I have some trouble finding a precise definition of what a modular form with complex multiplication. Could anyone provide such a definition and references for the study of CM modular forms and its main properties? I would be grateful

REPLY [14 votes]: There is a more down-to-earth definition. A newform $f=\sum_{n=1}^\infty a_n q^n$ of level $N$ and weight $k$ has CM
if there is a quadratic imaginary field $K$ such that $a_p=0$ as soon as $p$ is a prime which is inert in $K$. The field $K$ is then unique (if the weight $k \geq 2$), and one says that $f$ has CM by K.
A quick way to see the uniqueness of $K$, as well as other basic properties, is to consider
the $\ell$-adic ($\ell$-an auxiliary prime) Galois representation of dimension 2 attached to $f$ constructed by Deligne. If $\rho: G_{\mathbb Q} \rightarrow GL_2(\bar {\bf Q}_\ell)$ is that representation, one has tr $ \rho (Frob_p) = a_p$ (Eichler-Shimura) for all prime $p$ not dividing $N\ell$ (and $\rho$ is unramified at these primes, I should have said first). So for $p$ inert in $K$, tr $\rho (Frob_p) =0$, hence we deduce by Chebotarev and a little thought that $tr \rho=0$ on the complement on $G_K$ in $G_{\mathbb Q}$, and then by computing the hermitian  product of the character of $\rho$ with itself, that the restriction of $G_{\mathbb Q}$ to $G_K$ is reducible, hence that by Frobenius reciprocity that $\rho$ is induced from a character of $G_K$. Again some elementary 
group theory/representation theory tells you that there is a unique subgroup $G'$ of index $2$ in $G_{\mathbb Q}$ such that $\rho_{|G'}$ is reducible, except when the projective  image of $\rho$ in $K_4=(\mathbb Z/2)^2$, which is excluded because for $k \geq 2$ the projective image of $\rho$ is infinite. Hence the uniqueness of $K$, and many information gotten in the way on $\rho$. In weight $k=1$, the theory is roughly the same except from the very special modular forms whose projective image of $\rho$ is $K_4$, which have CM by two quadratic imaginary fields $K$ and $K'$, and also by a third field $K''$ with is quadratic real, in the sense that $\rho_{|G_{K''}}$ is also reducible (but then we say that $f$ has RM by $K''$, not CM).<|endoftext|>
TITLE: amalgamated product of groups and representation theory
QUESTION [6 upvotes]: Let me ask a question which could be quite stupid, but still:
let $G$ be a group which is an amalgamated product of subgroups $A$ and $B$ over $C$:$\; \;$  $G = A \ast_{C} B$ (subgroups are infinite!).
Question:  How representation theory of $G$ is connected to representation theory of $A$ and $B$ (and $C$),  in another words, how $Rep(G)$ is connected to $Rep(A)$ and $Rep(B)$?
Any comments are welcome!

REPLY [6 votes]: Let me expand on Mark Sapir's comment with a concrete example.  Burger and Mozes famously constructed a group of the form
$\Gamma = F_1*_H F_2$
where $F_1,F_2$ are finitely generated free groups, $H$ is of finite index on either side and $\Gamma$ is simple!  Because finitely generated linear groups are residually finite (by a theorem of Mal'cev, say), it follows that $\Gamma$ has no non-trivial finite-dimensional representations.  On the other hand, of course, the free groups $F_i$ have impossibly rich representation theories.
There are some positive results that hold in special cases, such as when amalgamating over cyclic subgroups.  See, for instance, this recent preprint of Jack Button and the references therein.<|endoftext|>
TITLE: Invariant polynomials for a product of algebraic groups
QUESTION [9 upvotes]: Let $G_1$ and $G_2$ be connected reductive algebraic groups defined over $\mathbb{C}$ and let $V_1$ and $V_2$ be irreducible representations of $G_1$ and $G_2$ respectively. I'm interested in general techniques for computing the (generators of the) invariant polynomial functions on the representation $V_1 \otimes V_2$ of the group $G_1 \times G_2$. 
In particular, I'm interested in computing a generating set for $\mathbb{C}[V_1 \otimes V_2]^{SO_n \otimes SO_m}$, where $V_1 = \mathbb{C}^n$ and  $V_2 = \mathbb{C}^m$ are the natural modules these special orthogonal groups ($n \geq m >2$). My first guess is to treat $V_1 \otimes V_2$ as a direct sum of $m$ copies of $V_{1}$ and use the first formulation of the Fundamental Theorem for $SO_n$ which gives me a generating set for $\mathbb{C}[V_{1}^{m}]^{SO_n}$ (which all have degree $2$, being the orthogonal contractions $(v,w)$?). But now I'm stuck trying to figure out what structure this algebra has as an $SO_m$-module, and what the degrees of the generators of the invariant polynomial algebra should thus be. 
According to a table in "Some Remarks on Nilpotent Orbits" by V. Kac, $\mathbb{C}[V_1 \otimes V_2]^{SO_n \otimes SO_m}$ should be generated by $m$ invariant polynomials of degrees $2,4,6,...$ 
This seems like it should be a fairly simple exercise, but I'm just going about it the wrong way.

REPLY [8 votes]: There is a fruitful way of thinking about the algebra $\mathbb{C}[M_{n,m}]^{{O_n}\times O_m}$ that originates from the $(GL_n, GL_m)$-duality. Namely, 

$$
\mathbb{C}[M_{n,m}]=\bigoplus_{\lambda}V_n^{\lambda}\otimes V_m^{\lambda},
$$

where the sum is over all partitions $\lambda$ with at most $\min(n,m)$ parts and $V_n^\lambda$ is the polynomial representation of $GL_n$ with highest weight $\lambda=(\lambda_1,\lambda_2,\ldots)$ with zeros appended at the end to make it length $n,$ and similarly for $m.$  Now, since $O_n$ is a spherical subgroup of $GL_n$, the space of $O_n$-invariants $V_n^{\lambda}$ is at most one-dimensional; it is non-zero precisely when $\lambda$ is even (i.e. all parts are even). A good source for these results is Roger Howe's Schur Lectures.
It follows that as a vector space, 

$$
\mathbb{C}[M_{n,m}]^{O_n\times O_m}=\bigoplus_{\lambda}(V_n^{\lambda})^{O_n}\otimes (V_m^{\lambda})^{O_m},
$$

and now the sum is over even partitions with the same restriction as before, and every summand is one-dimensional and is spanned by an explicitly described polynomial on $M_{n,m}$. Using standard techniques (described in the Schur lectures), this description can be amplified to yield the algebra structure as well. Namely, the algebra is graded by an affine semigroup of rank $\min(n,m)$ and is freely generated by explicit elements in degrees $2k$ for $1\leq k\leq \min(n,m).$ With a bit of extra work, one can also handle the more general case of $SO_n\times SO_m$ invariants.

Let me connect this description of the invariants with Abdelmalek's description following the path that Neil originally had in mind. Assume for concreteness that $n\geq m.$ The First Fundamental Theorem for $O_n$ in geometric form states that the map
$$
q:M_{n,m}\to Sym_{m,m} \qquad X\mapsto X^{T}X
$$
is the geometric quotient, i.e. it is $O_n$-invariant and gives rise to the isomorphism
$$
\mathbb{C}[M_{n,m}]^{O_n}\simeq \mathbb{C}[Sym_{m,m}].
$$
(The restriction $n\geq m$ assures that the image of $q$, which consists of symmetric matrices of rank at most $\min(n,m),$ is all of $Sym_{m,m}.$)
It is also clearly $GL_m$-equivariant, where $GL_m$ acts on the symmetric matrices of order $m$ by 
$$
Y\to g^T Y g\qquad Y\in Sym_{m,m}, g\in GL_m,
$$
and, a fortiori, $O_m$-invariant. Now pass to the $O_m$-invariants! Under this identification, 

$$
\mathbb{C}[M_{n,m}]^{O_n\times O_m} \simeq \mathbb{C}[Sym_{m,m}]^{O_m}.
$$

Thus the problem is reduced to determination of the orthogonal invariants of a symmetric matrix $Y$. It is well known that this invariant ring is freely generated by the coefficients of the characteristic polynomial of $Y.$ The coefficient of $\lambda^{m-k}$ has degree $k$ in the entries of $Y$ and, up to a sign, it is equal to the sum of the principal minors of $Y=X^T X$ of order $k, 1\leq k\leq m.$ Thus the fundamental invariants have degrees $2k, 1\leq k\leq m,$ in the entries of $X.$ Additionally, it can be verified that the $k$th fundamental invariant corresponds to the summand with $\lambda=(1^k)$ in the direct decomposition of  $\mathbb{C}[M_{n,m}]^{O_n\times O_m}$ coming from the $(GL_n,GL_m)$-duality (i.e. the second highlighted decomposition above).<|endoftext|>
TITLE: Number of idempotent $n\times n$ matrices over $\mathbb{Z}/m\mathbb{Z}$?
QUESTION [13 upvotes]: Is there any known formula for the number of idempotent $n\times n$ matrices over $\mathbb{Z}_m:=\mathbb{Z}/m\mathbb{Z}$ ? 
The number of idempotent matrices over a finite field is well-known and since we can decompose $m$ into product of prime numbers it is enough to assume that $m$ is a power of a prime number which is equivalent to assume $\mathbb{Z}_m$ is a local ring.

REPLY [18 votes]: I think it can be found the following way. If $R$ is a commutative local ring then every idempotent matrix is conjugate to a diagonal idempotent $\begin{pmatrix} I_r & 0\cr 0 &0\end{pmatrix}$ .  The point is that projective modules are free for local rings. This lets you write $R^n$ as a direct sum of the image and the kernel, both of which are free.  
So now one has to just compute the size of the stabilizer of the standard rank $r$ diagonal idempotent under the conjugation action of $\mathrm{GL}(R)$ for $R=\mathbb{Z}_{p^{k}}$. 
Added. I believe the stabilizer of the rank $r$ idempotent for a local ring is $\mathrm{GL}(R,r)\times\mathrm{GL}(R,n-r)$ like in the field case and so a formula is easily found. 
Added. I compute the answer for $n\times n$ matrices over $\mathbb{Z}_{p^k}$ to be $$\sum_{r=0}^n \frac{p^{2r(n-r)(k-1)}|\mathrm{GL}(n,p)|}{|\mathrm{GL}(r,p)|\mathrm{GL}(n-r,p)|}.$$ Here I use a matrix is invertible over a local ring iff it is over the residue field. The orders of the $\mathrm{GL}(m,p)$ are of course well known.<|endoftext|>
TITLE: Polynomials over $\mathbb F_2$ without zeros in $\mathbb F_2$ having an inverse series with support of large density.
QUESTION [5 upvotes]: Does there exist a sequence $A_n=A_n(x)\in\mathbb F_2[x]$ over the field $\mathbb F_2$ of 
two elements (represented by $0$ and $1$) such that $A_n(0)=A_n(1)=1$ and the inverse series $1/A_n=\sum_{j=0}^\infty
\alpha_{n,j}x^j\in \mathbb F_2[[x]]$
have supports with densities
$\delta_n=\lim_{k\rightarrow\infty} \frac{\alpha_{n,0}+\alpha_{n,1}+\dots+\alpha_{n,k-1}}{k}$ 
converging to $1$?
A positive answer to this question would give a positive answer to question 
[Sum of densities of support of $A$ and $A^{-1}$ for $A=1+\dots\in \mathbb F_2[[x]]$  by considering $\frac{A_n(x^2)}{1+x}$.
The highest possible density for polynomials of degree $\leq 16$ is $\frac{2}{3}$,
achieved by $1+x+x^2$.

REPLY [5 votes]: This is not an answer, rather a possible suggestion on how to deal with irreducible polynomials $A(x)$: Let $A(x)\in\mathbb F_2[x]$ be irreducible of degree $n$. Then \begin{equation}
A(x)=\prod_{i=0}^{n-1}(1+\lambda^{2^i}x)
\end{equation}
for some $\lambda\in\mathbb F_{2^n}$. The partial fraction decomposition and geometric series yield
\begin{equation}
\frac{1}{A(x)}=\sum_{i=0}^{n-1}\frac{\alpha^{2^i}}{1+\lambda^{2^i}x}
= \sum_{m=0}^\infty\sum_{i=0}^{n-1}\alpha^{2^i}(\lambda^{2^i}x)^m
= \sum_{m=0}^\infty T(\alpha\lambda^m)x^m,
\end{equation}
where $\alpha=\lambda/f'(1/\lambda)$ and $T$ is the trace map from $\mathbb F_{2^n}$ to $\mathbb F_2$.
Note that the power series is periodic with period $e$, where $e$ is the multiplicative order of $\lambda$. Thus if $U$ is the subgroup of order $e$ of $\mathbb F_{2^n}^\star$, then the density of $1$'s is the number of $u\in U$ with $T(\alpha u)=1$ divided by $\lvert U\rvert$.
An easy case is when $e=2^n-1$, so $U=\mathbb F_{2^n}^\star$. Half of the elements of $\mathbb F_{2^n}$ have trace $0$, so the density of $1$'s is $2^{n-1}/(2^n-1)$.
So when not only $n$ is prime, but even $2^n-1$ is prime, then we have this case and the density is only slightly bigger than $1/2$.
The general case seems to be more challenging. It is always difficult to relate an additive function like the trace map with subgroups of the multiplicative group of fields.<|endoftext|>
TITLE: Recovering an abelian category out of its derived category
QUESTION [9 upvotes]: I'm trying to learn more about derived category stuff and my curiosity has made me to ask these questions. Sorry if I'm being sloppy, I'm a new learner.
In Wikipedia it has been stated that since different abelian categories can give rise to equivalent derived categories, so it is impossible to reconstruct $\mathcal{A}$ (an abelian category) from its derived category $D(\mathcal{A}).$ 

To what extent it is possible to recover $\mathcal{A}?$ what would be the frame work for studying such questions?
Is there any notion of moduli space of t-structures on the derived category of a given abelian category?

REPLY [11 votes]: The derived category of an abelian category has a t-structure, so obviously that's the first thing you want.
To a t-structure corresponds a heart, which is an abelian category, whose derived category might be different than the one you started with.
To further complicate matters, as you noted, you could have a heart whose derived category is actually equivalent to the one you started with, but the heart itself is not the original abelian category.
It's not clear what you can recover from a triangulated category alone, or even from a triangulated category + t-structure.
If you like algebraic geometry and are willing to consider additional structures then you can recover the abelian category. The derived category of a scheme, considered as a monoidal category (coming from tensor products) recovers the original scheme (and thus the abelian category).
The same is true if you start with a variety with an ample canonical bundle, then the category plus the bundle do recover the variety.
Somehow this flexibility of derived categories is a nice feature, as it gives rise to interesting (and hidden?) "symmetries" and "relationships" between spaces.
As per the second question I can only think of what Sasha said, that is stability conditions.
Given a stability condition one automatically gets a heart of a t-structure (which again may not have anything to do with the original abelian category) and the slices of the stability condition may be seen as a continuous family of t-structures. 
It would indeed be really nice to have such a thing as a moduli space of t-structures!<|endoftext|>
TITLE: Free product of categories
QUESTION [14 upvotes]: Is there a general construction of the "free product" of categories? I'm not sure how to characterize such a thing precisely, but here is an analogy: the free product of categories should be to the free category on a quiver as the free product of monoids is to the free monoid on a set. This would be useful when formulating statements like "this category is generated by these subcategories subject to these relations."

REPLY [5 votes]: As Higgins pointed out in his papers and book, the useful construction for groupoids is what he calls the universal groupoid $U_\sigma(G)$ on a set $Y$ determined by a function $\sigma: Ob(G) \to Y$. Once you have set up a normal form for this, then you have normal forms for free groups and free groupoids and free products of groups and  groupoids, where the last are really determined by a pushout on object sets. 
These constructions are relevant to topology using the fundamental groupoid on a set of base points and the corresponding generalisation of the theorem of Seifert-van Kampen. 
The inclusion of the category of groups into the category of groupoids has a left adjoint which can be defined using the above universal construction. Also this inclusion of categories preserves colimits of connected diagrams. 
Similar considerations presumably apply to (small) categories. 
Jan 23, 2019: A method for constructing colimits of groupoids, and analogously for small categories, is given in Appendix B of Nonabelian Algebraic Topology, using the notions due to Grothendieck of fibred and opfibred (or cofibred) categories. They were developed initially to deal with the situation of sending a module $M$ over a ring $R$ to the ring $R$, and the lifting of ring morphisms $S \to R, R \to T$  to the module category. They  also apply to sending a crossed module of groups  $M \to P$ to the group $P$, and to the functor  $Ob$ sending a groupoid, or small category, to its set of objects.  
In the general case of a functor $\Phi: \mathsf X \to \mathsf B$ there are useful criteria involving such notions for  computing colimits of a functor $T: \mathsf C \to \mathsf X$ in terms of colimits of $\Phi T$ (Theorem B.3.4 loc.cit). 
In particular it is useful to see both small categories and groupoids in terms of the fibration and opfibration properties of the functors $Ob$ from these categories to the category of sets.<|endoftext|>
TITLE: Serre Spectral Sequence of Representations
QUESTION [10 upvotes]: Suppose that $G$ is a group acting on a fibre bundle $(F,E,B)$ by bundle automorphisms. In this case, the action automorphisms $E\to E$ give the integral homology $H_\ast(E;\mathbb{Z})$ the structure of a $G$-module. Also, the action automorphisms $F\to F$ and $B\to B$ give each module $H_p(B;H_q(F;\mathbb{Z}))$ the structure of a $G$-module. Can the Serre spectral sequence of the fibre bundle be made $G$-equivariant in the sense of being a spectral sequence of $G$-modules converging to $H_\ast(E;\mathbb{Z})$, and with second page                $E_{p,q}=H_p(B;H_q(F))$ (considered as $G$-modules in the above sense)?
Thanks!

REPLY [8 votes]: Personal comment: I feel that the two previous answers may be together creating some confusion on the subject of the question. I wish to address that with my answer, which would be more suited as a comment, were it not so long. I certainly hope my contribution will not result in even more confusion.
$\newcommand{\Ab}{\mathrm{Ab}}
\newcommand{\To}{\longrightarrow}
\newcommand{\rightset}[2]{#2\rlap{\scriptstyle #1}}
\newcommand{\leftset}[2]{\llap{\scriptstyle #1}#2}
\newcommand{\label}[1]{\qquad\qquad \text{#1}}$
Peter May's answer is quite interesting and, incisively, cautions us to pay attention to the requisite details. On the other hand, I believe his answer brings some unnecessary sophistication to a rather simple situation. More precisely, I think that Dylan Wilson's original answer, while it ignored all details, was essentially correct (if one replaces "fibre sequences" by "fibrations" in what he writes). The only subtle point is the functorial identification of the $E^2$-term of a Serre fibration $f:E\to B$, which fundamentally requires doing without a basepoint for $B$, and without the corresponding distinguished fibre of $f$.
As Peter May indicates, in the description of the Serre spectral sequence for a fibration $f:E\to B$, one usually assumes:

$B$ is path connected, and pointed;
the monodromy action of $\pi_1 B$ on $H_\ast F$ is trivial, where $F$ is the fibre of $f$ over the basepoint of $B$.

The second condition is a useful simplification which applies in most cases. Moreover, together with hypothesis 1, it allows us to write the $E^2$-term in the following neat, well-known form:
$$ E^2_{p,q}=H_p(B,H_q(F))  \label{(I)} $$
Obviously, without choosing a basepoint for $B$ that expression does not even make sense! Regardless, we can construct the Serre spectral sequence more generally: see, for example, the book More concise algebraic topology by Peter May and Kate Ponto for a generalization in which condition 2 does not necessarily hold.
For reference, I will briefly describe a common construction of the Serre spectral sequence for any Serre fibration $f:E\to B$; no assumptions are made on $B$ or $f$. We will construct it, without loss of generality, when $B$ is a CW-complex. Then, for a general topological space $B$, we first replace it with a functorial CW-approximation, and pull back the fibration $f:E\to B$ along the natural map from the approximation to $B$. Importantly, the CW-approximation gives a functor from the category of topological spaces to the category of CW-complexes and cellular maps.
For a CW-complex $B$ the spectral sequence is easy to construct. Since $B$ is a CW-complex, we can filter it by its skeleta. The inverse image by $f:E\to B$ of these skeleta gives a filtration on $E$. The Serre spectral sequence is then the spectral sequence associated with the corresponding filtration on the singular chain complex of $E$.
Observe that the functoriality of the above spectral sequence (even starting from the $E^1$-term) with respect to maps of Serre fibrations is assured by:

the naturality of the CW-approximation of a space;
the functoriality of the spectral sequence associated with a filtered chain complex.

A map of Serre fibrations $f\to \overline{f}$ is simply a commutative square
$$ \begin{matrix}
E & \To & \overline{E} \\
\leftset{f}{\Big\downarrow} & & \rightset{\overline{f}}{\Big\downarrow} \\
B & \To & \overline{B} \\
\end{matrix} $$
In particular, if a group $G$ acts on a Serre fibration by such maps, then we automatically obtain an induced action of $G$ on the corresponding Serre spectral sequence. Therefore, it becomes a spectral sequence of $G$-modules converging to the homology of the total space seen as a $G$-module. This is an immediate consequence of the exactness of the forgetful functor from $G$-modules to abelian groups. That conclusion should answer the question, apart from the identification of the $E^2$-term, which is discussed below.
Indispensably, the fibre of the Serre fibration did not actually figure into the above construction of the Serre spectral sequence. In fact, the fibres only make an appearance once one tries to compute the $E^2$-term. Nevertheless, one can also identify the $E^2$-term in this general case, and moreover, in a functorial manner. Given a Serre fibration $f:E\to B$, the $E^2$-term of the general Serre spectral sequence described above is naturally isomorphic to (which I write here as equal)
$$ E^2_{p,q}=H_p(B,H_q(F_\bullet))  \label{(II)} $$
with the naturality holding with respect to maps of Serre fibrations. Here is a description of the right hand side of (II). Let $F_x=f^{-1}(x)$ be the fibre of $f$ over $x\in B$. In the above expression, $H_q(F_\bullet)$ represents the system of local coefficients on $B$ corresponding to the functor
$$ H_q(F_\bullet):\Pi_1(B)\To\Ab $$
from the fundamental groupoid of $B$ to the category of abelian groups. This functor takes a point $x\in B$ to $H_q(F_x)$, and a path homotopy class $[\gamma]$ of paths in $B$ to the map induced by the monodromy along $\gamma$ on the homology of the fibres over the endpoints of $\gamma$. Then the right hand side of (II) is the homology of $B$ for this system of local coefficients. It is straightforward to check this homology is indeed functorial with respect to maps of Serre fibrations.
Disclaimer: I believe that the functorial identification (II) of the $E^2$-term follows the same recipe as the usual derivation of the isomorphism (I) obtained under conditions 1 and 2. To be honest, I have not carefully checked the details myself! Please let me know if I am in error. It appears to be considered folklore knowledge. Nevertheless, I cannot actually provide a complete reference to the identification (II) of the $E^2$-term. The only textbooks I have found which refer to local coefficients in the Serre spectral sequence are the aforementioned book by May and Ponto, and McCleary's A user guide to spectral sequences. The former skips the proof in the general case, but at first sight the proof in McCleary's book does seem to give (II) as a natural isomorphism, even if the result is not stated explicitly. Can anyone provide a better reference?
In particular, this functorial identification of the $E^2$-term finishes the answer to the question. The $E^2$-term, identified as the right hand side of (II), becomes a $G$-module in the canonical manner when $G$ acts on a Serre fibration. The fundamental ingredient was that (II) removed any reference to a basepoint of the base space of the fibration.
Finally, when we do have a basepoint for $B$, the naturality of the identification of the $E^2$-term as in (I) follows from the general case (II). After all, under conditions 1 and 2 above, the local coefficients over $B$ appearing in (II) are trivial, and canonically identifiable with the homology of the fibre $F$ of $f$ over the basepoint of $B$. The functoriality of the right hand side of (I) is the obvious one, at least with respect to maps of Serre fibrations which preserve the basepoint of the base spaces. For maps which do not preserve the basepoints, the functoriality makes use of the canonical identification of the homology of any two fibres of $f:E\to B$, which is a consequence of condition 2. In case condition 2 does not hold, this identification is impossible, and one is stuck working with local coefficients: it is then best to forego the basepoint of $B$ entirely.<|endoftext|>
TITLE: Inequality on Trigonometric polynomials
QUESTION [7 upvotes]: My question comes from trying to understand a technical step in this paper by Bourgain.
Let $R,L$ be positive integers and let $f(x)=\sum_{|n|\leq RL}a_ne^{2\pi inx}$ be a trigonometric polynomial. Assume $f(x)>0$.
Let $F_L(x)=\sum_{|n|\leq L}\frac{L-|n|}Le^{2\pi inx}$ be the Fejer kernel.
Define (as usual) the convolution
$$(f*F_L)(x)=\int_0^1f(t)F_L(x-t)dt=\sum_{|n|\leq L}a_n\frac{L-|n|}Le^{2\pi inx}$$
Does it follow (and how) that 
$$f(x)\leq10R(f*F_L)(x)$$?
In the paper we have a concrete $f(x)$, so maybe this is not true in general. There we have
$$a_n=1-\cos\left(2\pi\frac{RL-|n|}N\right)$$
where $N$ is a large positive integer, (at least bigger than $4RL$).

REPLY [16 votes]: That is true for all non-negative trigonometric polynomials, though not entirely obvious unless you are a Fourier analyst yourself. To see it, just note that the convolution with $K_{RL}=2F_{2RL}-F_{RL}$ recovers $f$ faithfully and $F_{RL}\le RF_L$. Of course, to Jean such things are as obvious as $2\times 2+1=5$ (he writes $10$ instead of $5$ just out of the traditional analyst's habit to have a 100% security margin in the constants), but I agree that it may be a bit perplexing for poor mortals like you and me. Joe Diestel just told me at a beer party tonight that the most common phrase in Bourgain's early papers was "By standard techniques we conclude from here that". :)<|endoftext|>
TITLE: Known decomposition of $\bigwedge^k Sym^d \mathbb C^n$ in special cases?
QUESTION [5 upvotes]: Let $V = \mathbb C^n$. Consider the plethysm $\bigwedge^k Sym^d V$ as a representation of $GL(V)$. In what special cases (e.g., for what $k$, $d$, and $n$) is this representation's decomposition into irreps known?
The only known nontrivial special case that I am aware of is when $k = 2$: in this case the decomposition is $S_{2d-1,1} \oplus S_{2d-3,3} \oplus S_{2d-5,5} \oplus \cdots$. When $n = 2$, I also know that it is equivalent to find decompositions of plethysms of the form $Sym^k Sym^i V$.
Using the Macaulay2 package SchurRings, I computed all examples with $d \leq 8$ with no obvious patterns jumping out at me.
I would be interested in any other special cases people know about (including ones which only apply to $n = 2$), conjectures along these lines, tables of computed data, or ideas about references that might be fruitful.

REPLY [8 votes]: When $d=2$, the decomposition is known for all $k$ and $n$. Given a partition $\lambda$ of $k$ with distinct parts, let $2[\lambda]$ denote the partition of $2k$ whose main-diagonal hook lengths are $2\lambda_1, \ldots, 2\lambda_k$, and whose $i$th part has length $\lambda_i + 1$. Then
$$ \bigwedge^k {\rm Sym}^2 V = \sum_\lambda S^{2[\lambda]}(V) $$
where the sum is over all partitions $\lambda$ with distinct parts such that $2[\lambda]$ has at most $n$ parts and $S^\mu$ is the Schur functor for the partition $\mu$. For a proof using the symmetric group see Lemma 7 in http://arxiv.org/abs/0903.2864.
Edit (June 2014). The constituents of $\bigwedge^3 \mathrm{Sym}^{d}(V)$ are determined on page 141 of Macdonald's book, Symmetric functions and Hall polynomials. Remark 3.6(b) in Howe, $(GL_n,GL_m)$-duality and symmetric plethysm, Proc. Indian Acad. Sci. 97 (1987) 85–109, gives a method for computing the plethysm $\bigwedge^4 \mathrm{Sym}^d(V)$. 
Apart from the case $k=2$ mentioned in this question (and the trivial cases $k=1$ or $d=1$), I think these are the only case where the complete decomposition is known.<|endoftext|>
TITLE: Semimagic Squares and Partitions
QUESTION [5 upvotes]: Say, we have a semimagic square $X$, that is, an $n\times n$ square matrix with entries from natural numbers, such that each row and column of it sums up to the same natural number $s$.
Let $M$ be a set with $s$ elements.
To each $X_{ij}$, we assign a subset $M_{ij}$ of $M$ with $X_{ij}$ elements.
 Question:  Is it possible to make the assignment in such a way, that 
    for every $i$ we have $\bigcup_j M_{ij} = M$ and
    for every $j$ we have $\bigcup_i M_{ij} = M$,
or equivalently, whenever the cells $X_{ij}$ and $X_{kl}$ lie in the same row or in the same column, then the subsets $M_{ij}$ and $M_{kl}$ are disjoint?
 PS1:  You may think of $s^2$ different balls, each labeled by a number from ${\left\lbrace 1,\dots,s  \right\rbrace}$, such that for each label $l$, there are exactly $s$ balls having label $l$. You have to distribute the balls in the cells of $X$ in such a way, that in each cell $X_{ij}$ there are $X_{ij}$ balls and in each row and column of $X$ we have a whole set of balls labeled from $1$ to $s$.
 PS2:  To me, the answer to the question seems to be positive. However, the greedy construction idea does not work.

REPLY [2 votes]: This is the same answer in different language.  Every regular bipartite multigraph of degree $s$ has a proper edge colouring with $s$ colours. Make a bipartite multigraph with $n+n$ vertices, where $X_{ij}$ is the number of edges from vertex $i$ in the first part to vertex $j$ in the second part.  This is a regular bipartite graph of degree $s$ and the edge colouring gives the necessary assignments.<|endoftext|>
TITLE: Categories where projective objects are flat 
QUESTION [15 upvotes]: In many categories arising in the theory of algebras or modules, projective objects are flat. For example each projective module over a ring with identity is flat. However, there are categories where this principle is violated: Lewis has shown that the category of Mackey functors for the orthogonal group $O(n)$ has non-flat projectives. 
Q1: Is there a classification of symmetric monoidal abelian categories $\mathcal{A}$  where all projectives are flat ? BTW: Do such categories have a particular name ?  
If a classification is too strong to establish, I would be interested in 
Q2: Are there criteria that ensure that projectives are flat ? 
An example for Q2 is, when $\mathcal{A}$ has direct sums and the unit is a progenerator of $\mathcal{A}$ (this is just the categorification of the module case mentioned above). Note that this criteria doesn't work for instance in the category of $RG$-modules ($R$  commutative ring, $G$ a group) where the tensor product is given by $M \otimes_R N$ with diagonal $G$-operation.

REPLY [4 votes]: This is the sort of question which would interest my advisor, Mark Hovey. In fact, he and Keir Lockridge wrote a paper called Semisimple Ring Spectra which addresses questions like this in triangulated categories (which cannot be abelian but make more sense to work with if you do homotopy theory). In this paper he has two results which answer your question in that context. I'll paraphrase them and highlight the bits that matter, but first some definitions.
A triangulated category come with an endofunctor $\Sigma$ which acts like suspension in topological spaces or shift in chain complexes (there are axioms for this). With this endofunctor you can define homotopy classes of maps and do cohomology theory. Call such a category $T$ a weak stable homotopy theory if every cohomology theory is representable. Suppose you have such a theory and it comes with a compact generator $S$ (i.e. the smallest localizing subcategory containing $S$ is $T$). Call such a theory a stable homotopy theory if there is also an axiom saying that $S$ plays nicely with the smash product. Call it a Brown category if all homology theories and natural transformations between them are representable. Assume below these hypotheses hold on $T$.
Proposition 2.1: The generator $S$ is semisimple iff projective objects and realizable objects coincide.
Remark 2.3: The generator $S$ is von Neumann regular iff flat objects and realizable objects coincide.
For the definitions of semisimple and von Neumann regular, see that paper. These are just the versions for triangulated categories of standard properties of rings. So semisimple implies von Neumann regular and there are classifications for when the converse holds. Furthermore, you can find a classification of semisimple objects in such theories $T$ as Theorem 4.1, which is a purely algebraic classification based on $\pi_*(S)$. So this gives some sense of a classification of theories where projectives and flats are the same. I don't know if anything like this has been done in the abelian setting. I'd be very interested to know!<|endoftext|>
TITLE: Etale homology via étale cosheaves
QUESTION [12 upvotes]: Can one develop a theory of étale homology via étale cosheaves? The hope is that this would, for example, return the Tate module (and not its dual) for an elliptic curve, and it would return group homology for $\rm Spec$ of a field.
The first step in this direction is to notice that you can define an étale cosheaf on $\rm Spec$ of a field, in which the sections are coinvariants, rather than invariants, of the associated Galois module. In this case I believe we would recover group (Galois) homology.
(See Cosheaf homology and a theorem of Beilinson (in a paper on Mixed Tate Motives) if you don't know what cosheaves are.)
I'm not expecting this to prove anything new, but it would make certain formulations nicer (for some definition of "nice"). E.g., I'm hoping we would get theorems that look similar to the relations between singular homology and cohomology, and we would get a comparison with singular homology for finite coefficients.
I feel like this should work, though I'm not entirely sure about the ability to cosheafify.

REPLY [13 votes]: Sure thing!  There's an equivalence of categories between cosheaves valued in profinite abelian groups and sheaves valued in torsion abelian groups, by Pontryagin duality one could say.  So you can directly define the first etale homology of Z_ell and it will give you the l-adic Tate module.  But it's formally the same as defining the l-adic Tate module to be Pontryagin dual to H^1 with Z/l^\infty-coefficients.  I agree with you, though, that the cosheaf perspective -- though equivalent -- seems more natural in many cases.<|endoftext|>
TITLE: Hyperbolic exceptional fillings of cusped hyperbolic 3-manifolds
QUESTION [24 upvotes]: Thurston's Hyperbolic Dehn Surgery Theorem says that all but finitely many fillings of a cusp of a hyperbolic 3-manifold result in hyperbolic manifolds that are deformations of the original manifold.  Moreover, the core curve of the filling solid torus in these fillings is a geodesic in the resulting hyperbolic manifold.
Typically the study of exceptional Dehn surgery on hyperbolic 3-manifolds is concerned with the production of non-hyperbolic 3-manifolds.  However, it is possible for a hyperbolic manifold to result without being a deformation of the original.  For example, a "random" knot in a hyperbolic 3-manifold won't be isotopic to a geodesic though is likely to have hyperbolic complement.  The meridional filling of this knot exterior would then be an exceptional hyperbolic filling.
What's the maximum number of possible exceptional hyperbolic fillings? What manifolds realize this maximum?

REPLY [11 votes]: I'll embellish on Autumn's answer. Experimentally, it seems that most one-cusped hyperbolic manifolds have at most 10 exceptional Dehn fillings in the sense you consider, i.e. points in the complement of Dehn surgery space. It is conjectured that Dehn surgery space is star-like, so that one can deform any hyperbolic metric with core geodesic by decreasing the cone angle monotonically to zero, and without affecting the singular structure. Lackenby and Meyerhoff have shown that there are at most 10 non-hyperbolic Dehn fillings, but they are not able to show that these are deformations of the complete metric in the sense of Thurston. However, their method of proof gives a bit more. The core of the Dehn filling is homotopically non-trivial, and  the kernel of the Dehn filling map on $\pi_1$ is a free group (this is true if the core is geodesic, since the fundamental group of the complement of a collection of geodesics in hyperbolic space is free). I'm not sure if they prove this in the paper, but it follows from the proof the 6-theorem. The number of negatively curved fillings is also much smaller than 60 from the $2\pi$-theorem. There's hope that the cross-curvature flow could flow these negatively curved metrics to the hyperbolic metric. If one could do this in the cone-manifold context, then this might enable one to obtain a sharpening of Hodgson-Kerckhoff.<|endoftext|>
TITLE: Is a knotted trivalent graph determined by its set of unzips?
QUESTION [10 upvotes]: A knotted trivalent graph (KTG) is a framed embedding of a trivalent graph in $\mathbb{R}^3$, modulo ambient isotopy. I believe they were first considered (in the quantum topological context, at least) in a paper by D. Thurston. There is an "unzip" move on edges, turning an "H" pattern into a pair of edges (Thanks to Kea for the hand-drawing of this image).

A knotted theta graph unzips in three ways (one unzip for each edge), giving rise to three possible 2-component framed links, related to one another by handleslide (Kirby 2). The following statement looked obvious at first to me, but now I'm beginning to doubt it's even true, and I have no idea how to prove it or to find a counterexample. It `feels' well-known.

Question: Given two KTG's (say knotted theta graphs for simplicity), all of whose unzips coincide (i.e. given an edge e in one, there exists and edge f in the other, such that unzipping along e and along f give ambient isotopic results), does it follow that the KTG's are themselves ambient isotopic? Or do there exist distinct KTG's all of whose unzips coincide?

REPLY [6 votes]: A framed theta graph (ribbon surface) is really a thrice puncture sphere embedded in $S^3$ (pants). A stronger question is whether one may have two pants whose boundaries give the same 3-component link? In fact, suppose the interiors of the two pants are disjointly embedded in the link complement. The union is a genus 2 surface. Conversely, given a genus 2 surface, one may choose a pants decomposition of it to get such a pair. I claim that if one takes a genus 2 surface bounding an incompressible surface to one side (so the other side is compressible), a generic pants decomp will give inequivalent theta graphs. One may assume the link is hyperbolic without symmetries. If they were equivalent, there would be an isometry taking one pants to the other. This isometry would be finite order, giving a symmetry of the link complement, a contradiction. It remains to find such a surface and pants decomp, but I'm sure any sufficiently complicated asymmetric example will work.<|endoftext|>
TITLE: Characterizing the Dual of $W_0^{s,p}$
QUESTION [9 upvotes]: I am interested in literature/results characterizing the dual of the fractional Sobolev space $W^{s,p}(\Omega)$, where $\Omega \subset \mathbb{R}^N$ is open, bounded, and smooth, $0< s<1$, and $1< p<\infty$.  
For example, we can characterize the dual of $W_0^{1,2}(\Omega)$ as follows.  If $f \in W^{1,2}(\Omega)$ then there exist $f_i \in L^2(\Omega)$ such that (for $v \in W^{1,2}_0(\Omega)$)
$\langle f,v \rangle = \int_\Omega f_0v+\sum_{i=1}^N f_iv_{x_i}\;dx$,
and so we write $f = f_0 - \sum_{i=1}^N (f_i)_{x_i}$.
A similar characterization holds for the dual of $W_0^{1,p}(\Omega)$, where instead the $f_i$ are in $L^p(\Omega)$ (though I do not have a reference for this - the $W_0^{1,2}$ case can be bound in Evan's PDE book).
My question is do we have such a representation $f=f_0 + (-\Delta)^\frac{s}{2} f_1$, for $f_0,f_1 \in L^p(\Omega)$, or something like this?

REPLY [2 votes]: In the case $p=2$, there is a pretty elegant formalization to be found in negative Sobolev spaces. It is mentioned briefly here, and less briefly here. 
The idea is that for any real number, the space $H^s(\mathbb R^n)$ with norm
$$
\| f \|_{H^2} = \|(1+|\xi|^2)^{s/2} \hat f(\xi) \| _{L^2}
$$
Where $\hat f$ is the Fourier transform of $f$, which may be a tempered distribution. The dual of $H^s(\mathbb R^n)$ is $H^{-s}(\mathbb R^n)$ which is the is the space of tempered distributions $f$ such that there is a an $L^2$ function $f_0$ such that $\hat  f = (1+|\xi|^2)^{-s/2}\hat f_0$. Which is at least something like $f = f_0 +(-\Delta)^{s/2} f_0$.
I don't know about the case $p\neq 2$, but would be really interested to see if anyone else does.<|endoftext|>
TITLE: Scalar curvature notion for Cartan connections
QUESTION [14 upvotes]: In Riemannian geometry, there is a well-known notion of the scalar curvature on a Riemannian manifold $M$, which is a function on $M$ given by a suitable contraction the Riemannian curvature tensor. The scalar curvature gives a rough measure how much the geometry of $M$ differs locally from the flat Euclidean case, namely how much the volume (or surface) of a small ball with a given radius differs from the usual value in flat space.
I am interested in whether (or in which cases) this notion generalises to arbitrary Cartan geometries, given by a model Klein geometry $(G, H)$, an $H$-principal bundle $P$ and a one-form $\omega$ on $P$ with values in the Lie algebra $\mathfrak g$ of $G$, the Cartan connection. For what follows, you may assume that $H$ is the trivial subgroup.
The curvature is given by the $\mathfrak g$-valued two-form $\Omega = d \omega + \frac 1 2 [\omega, \omega]$. If and only if $\Omega$ vanishes, $P$ looks locally like $G$ in that there is a local diffeomorphism between $P$ and $G$ such that $\omega$ becomes the Maurer--Cartan form of $G$. In this sense, the curvature is completely analogous to the Riemannian curvature.
Now my question is whether one can form out of $\Omega$, $\omega$ and maybe some scalar product on $\mathfrak g$ (e.g. the Killing form) a scalar valued function (or maybe density) on $P$ which gives a rougher measure for how much the geometry differs from the Lie group $G$ (just as the scalar curvature does for Riemannian curvature)? For example, does it make sense to compare the volume of a small ball around the identity of $G$ with the "same" (in terms of the connection $\omega$) ball around a point in $P$, or is there some other well-established notion?
Of course, I would like to see whether one gets back the scalar curvature of Riemannian geometry (maybe up to some term depending on the torsion) in the case that $G$ is the group of Euclidean motions and $H$ the subgroup of rotations.

REPLY [15 votes]: In the case that $H$ is trivial, $\omega$ is just a $\frak{g}$-valued coframing on the manifold $P$.  The curvature $2$-form $\Omega = d\omega + \tfrac{1}{2}[\omega,\omega]$ can then be written in the form 
$$
\Omega = R(\omega\wedge\omega)
$$
where $R:P\to {\frak{g}}\otimes\Lambda^2({\frak{g}}^\ast)$ is the curvature tensor.  Since $H$ is trivial, there is no group action to take into account, so, technically, any $\lambda\in {\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ could be used to construct a scalar-valued curvature quantity $\lambda(R):P\to\mathbb{R}$.  
Now, you seem to want to consider the scalar curvatures constructed from those $\lambda$ that are invariant under the adjoint action of $G$ on $\frak{g}$, and typically there are such elements.  For example, the Lie algebra structure itself is an element $\mu\in {\frak{g}}\otimes\Lambda^2({\frak{g}}^\ast)=\mathrm{Hom}\bigl(\Lambda^2({\frak{g}}),{\frak{g}}\bigr)$. In the semi-simple case, one can use the Killing form to dualize this to an element $\mu^\ast\in {\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$, and then the resulting 'scalar curvature' $\mu^\ast(R)$ is a very natural invariant.
More generally, though, you might want to ask for the $G$-invariant elements in ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ and see if there is anything else interesting there. However, in the case that $\frak{g}$ is simple, it is not hard to show that the induced action of $G$ on ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ only fixes multiples of $\mu^\ast$.
Remark:  One possible reason for considering the $G$-invariant elements is this:  There is a natural action of $G$ on Cartan connections for the pair $(G,\lbrace e\rbrace)$, namely, one can define $a\cdot \omega = \mathrm{Ad}(a)\bigl(\omega\bigr)$.  Then one will have $\lambda\bigl(R(\omega)\bigr)=\lambda\bigl(R(a\cdot\omega)\bigr)$ when $\lambda$ has this kind of $G$-invariance.
The more general case, when $H$ is nontrivial, can be handled similarly, and you'll want to look at the action of $H$ on ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g/h}})$ and consider the invariant elements $\lambda$.  [NB:  I have now modified this sentence to take into account the OP's comment below, as he pointed out that he wasn't restricting to torsion-free geometries.]
As to which of these invariants might be thought of as comparing some kind of local volume with some kind of 'geodesically parallel' volume (which is what the scalar curvature does in Riemannian geometry), I'd have to think about this some more.
Example: In the case that $G$ is the group of rigid motions of $\mathbb{R}^n$ and $H\simeq\mathrm{SO}(n)$ is the group of rotations about a point (in which the OP was particularly interested), one has
$$
{\frak{g}}\simeq {\frak{so}}(n)\oplus \mathbb{R}^n
\qquad\text{and}\qquad
{\frak{h}}\simeq {\frak{so}}(n),
$$
and the usual Cartan connection on the orthonormal frame bundle $P\to M$ of a Riemannian metric $g$ on an $n$-manifold $M$ is of the form 
$$
\omega = (\theta,\eta):TP\to {\frak{g}}\simeq {\frak{so}}(n)\oplus \mathbb{R}^n
$$
where $\eta$ is the 'soldering form' and $\theta$ is the Levi-Civita connection form. The curvature of $\omega$ as a Cartan connection is then
$$
\Omega = \bigl(d\theta+\theta\wedge\theta,\ d\eta + \theta\wedge\eta\bigr)
= \bigl(R(\eta\wedge\eta),0\bigr),
$$
where, now, $R:P\to \mathrm{Hom}\bigl(\Lambda^2(\mathbb{R}^n),{\frak{so}}(n)\bigr)
\simeq {\frak{so}}(n)\otimes {\frak{so}}(n)$ is the usual Riemann curvature tensor.
In this case, because the Cartan connection is torsion-free, the Cartan connection curvature takes values in 
$$
{\frak{h}}\otimes\Lambda^2\bigl(({\frak{g}/\frak{h}})^\ast\bigr)\simeq
{\frak{h}}\otimes\Lambda^2\bigl(\mathbb{R}^n\bigr) \simeq
{\frak{so}}(n)\otimes {\frak{so}}(n)
$$
and, applying the above recipe to this curvature, the unique $H$-invariant element (up to multiples) in this space does indeed give the usual scalar curvature of the underlying Riemannian metric (up to a constant multiple).  
NB: When $n=4$, the above space ${\frak{so}}(n)\otimes {\frak{so}}(n)$ actually has a $2$-dimensional space of vectors fixed under the action of $H$, but, of course, the first Bianchi identity says that the curvature takes values in a proper subspace of ${\frak{so}}(n)\otimes {\frak{so}}(n)$ that only intersects this $2$-dimensional space in a $1$-dimensional space after all.  That brings up the point that, if one does consider only Cartan connections that satisfy some given torsion restrictions, then there can be Bianchi identities that say that the curvature of the connection must take values in some proper subspace of ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$, in which case, one will have to take this into account in considering which 'scalar curvatures' can be defined.<|endoftext|>
TITLE: Partial order relation on subsets
QUESTION [9 upvotes]: I consider all $k$ element subsets of the set $\{1,\ldots,n\}$ and define a partial order relation $\prec$ as follows: $\{a_1,\ldots,a_k\}\prec\{b_1,\ldots,b_k\}$, if and only if $a_1<\cdots<a_k$, $b_1<\cdots<b_k$ and $a_i\leq b_i$ for all $i=1,\ldots,k$.
What is this partial order commonly called in the literature? Can someone point out references?
Thanks a lot,
Torsten
[edit] Here is a Hasse diagram of this partial order for $n=6$ and $k=3$:
http://www.freeimagehosting.net/7pmk7

REPLY [6 votes]: This order is sometimes denoted $L(k,n-k)$ and has many interesting properties. See for instance Chapter 6 of http://math.mit.edu/~rstan/algcomb.pdf. For the characteristic polynomial of the Hasse diagram of $L(k,n-k)$ (considered as a graph), see Remark 5.6 of http://math.mit.edu/~rstan/papers/vac.pdf.<|endoftext|>
TITLE: What sets of primes can we pick out with first-order statements?
QUESTION [15 upvotes]: For each prime $p$, we have the algebraically closed field $\bar{\mathbb F}_p$ with the Frobenius automorphism.
Given any first-order statement with no free variables using the symbols $0,1, +, \times, -, /, \sigma(),=$, we can interpret it in $\bar{\mathbb F}_p$, interpreting the field operations to mean themselves and $\sigma$ to mean Frobenius.
For each prime, it is either true of false. This gives us a set of primes.

What sets of primes can be described this way?

It is easy to pick out the primes whose Frobenius elements have a certain conjugacy class in in a Galois extension of $\mathbb Q$, and to pick out finite sets of primes.
Are there any sets of this type not generated by conjugacy classes in Galois groups and finite sets under the logical operations?

REPLY [11 votes]: You guessed the correct answer.
This is explained in the paper
of Mike Fried and  George Sacerdote,
Solving Diophantine Problems Over All Residue Class Fields of a Number Field
and All Finite Fields,
The Annals of Mathematics, 2nd Ser., Vol. 104, No. 2. (Sep., 1976), pp. 203-233.
Since the theory of fields lacks elimination of quantifiers in the language of rings
(the formula $\exists y,\ x=y^2$ which says that $x$ is a square cannot be expressed
directly as a polynomial condition on $x$), the authors introduce a richer language,
using the concept of Galois stratifications, which allows for elimination of quantifiers.
Geometrically, this basically means that one can eliminate quantifiers up to the level
of finite extensions of fields.
See also Chapters 30 and 31 on Galois stratifications in the book Field arithmetic by Mike Fried and Moshe Jarden.<|endoftext|>
TITLE: If $\mathcal{F}_t$ is separable why is $\mathcal{F}_\infty$ generated by a random variable?
QUESTION [5 upvotes]: I am reading this introduction to enlargement of filtration and at the beginning of section 2.4 there is a claim that I cannot justify but seems like it should be well known.  The author claims that if $\mathcal{F}_t$ is the natural filtration of a Brownian Motion, then since $\mathcal{F}_t$ is separable it follows that $\mathcal{F}_\infty$ is generated by a real random variable.  I assume that separable means that there is a countable collection of random variables which generates $\mathcal{F}_t$,though I have not been able to find a definition for the term 'separable' in this context.  In any case, I wonder if someone could help me with a proof (or a reference for one) of this result.

REPLY [5 votes]: Let $(\Omega,\mathcal B,\mu)$ be a probability space and $\mathcal A$ a sub-sigma-algebra of $\mathcal B$. The following statements are equivalent: 

$\mathcal A$ admits a countable set of generators. 
There is a bounded random variable $X\colon \Omega\to \Bbb R$ such that $\sigma(X)=\mathcal A$. 
There are countably many bounded random variables $X_k\colon \Omega\to \Bbb R$ such that $\sigma(X_k,k\in\Bbb N)=\mathcal A$. 

Indeed, if $1.$ is satisfied, let $\(A_k,k\in\Bbb N\)\subset 2^\Omega$ generating $X$. Then define $X:=\sum_{k=1}^{+\infty}3^{-k}\chi_{A_k}$. Then $X^{-1}(3^{-k})=A_k$. 
$2.\Rightarrow 3.$ is obvious taking $X_k=X$. 
$3.\Rightarrow 1.$ Let $S_j$ be countably many sets generating Borel sigma-algebra, for example open intervals with rational endpoints. The (countable) collection $(X_k^{-1}(S_j),(k,j)\in\Bbb N^2)$ generated $\mathcal A$.<|endoftext|>
TITLE: Analogue of cyclic homology for e_n-algebras?
QUESTION [8 upvotes]: Cyclic homology may be defined as the primitive part (with respect to a natural product) of the homology of the Lie algebra associated with the "stabilization" of an associative algebra $A$. Here the "stabilization" is the matrix algebra $Mat_n(A)$, where $n$ goes to infinity. This definition explains another name of cyclic homology - "additive K-theory".
I believe that there exists an analogous notion for $e_n$-algebras. For a $e_n$-algebra $A$ it must be the primitive part of the homology of the Lie algebra associated with a $e_n$-algebra, which is an appropriate "stabilization" of $A$. 
Whether anyone has some ideas how this thing could be defined? What is the homology of the trivial algebra? Is it an additive version of something?

REPLY [3 votes]: In the paper mentioned in the comments Francis defines a candidate for this. If $A$ is an $E_n$-algebra one has an equivalence of associative algebras $B = \int_{S^{n-1}} A \simeq (\int_{S^{n-1}} A)^{\rm op} = B^{\rm op}$. Thus, any left $B$-module has a canonical right $B$-module structure. Identifying $B$ with the $E_n$-enveloping algebra ${\rm Env}(A)$ this gives $A$ the natural structure of a left $\int_{S^{n-1}}A$-module structure. Letting $A^\tau$ be the induced right-module we can form
$$
A^\tau \otimes_{\int_{S^{n-1}} A} A
$$
which he calls the $E_n$-Hochschild homology of $A$. Using excision you can prove that $\int_{S^1} A = A \otimes_{A \otimes A^{\rm op}} A$, recovering classical Hochshild homology. Further if $A$ is commutative, or at least admits a $E_{n+1}$-refinement, then the above definition coincides with $\int_{S^n} A$.<|endoftext|>
TITLE: Best results regarding the Lang-Trotter conjecture
QUESTION [10 upvotes]: Let $E$ be an elliptic curve over $\mathbb Q$, which does not have complex multiplication over the algebraic closure of $\mathbb Q$. For $x>0$, let $P(x)$ be the number of primes $p < x$ such that $E$ has good super-singular reduction at $p$. The Lang-Trotter's conjecture states that
$$P(x) = O (x^{1/2}/\log x).$$
In Serre's paper Quelques applications du théorème de densité de Chebotarev, Serre 
proves $P(x)=O(x^{3/4})$ under the generalized Riemann hypothesis (GRH) for Artin $L$-functions. 

Has this bound been improved since? If so, what is the better known bound (under GRH)
  and where can I find it?

Thanks.

REPLY [19 votes]: A couple of years after my Ph.D. thesis (whose main result is the infinitude of
singular primes, i.e. $P(x) \rightarrow \infty$ as $x \rightarrow \infty$),
Kaneko published a result[1] that let me obtain the unconditional 
upper bound[2] $P(x) = O(x^{3/4} \log x)$ using some of the same ideas:

[1] Masanobu Kaneko: Supersingular $j$-invariants as singular moduli mod $p$, 
  Osaka J. Math. 26 (1989), 849–855.
[2] Noam D. Elkies: Distribution of supersingular primes, Astérisque 198-199-200 (1991; proceedings of Journées Arithmétiques 1989), 127–132.

As noted in [2], the factor $\log x$ can be removed with some
more care (averaging over the auxiliary discriminants $-D$ improves on
the worst-case estimate), thus exactly matching Serre's conditional bound.
As far as I know, no further improvement has been obtained since then,
even assuming GRH.
That paper [2] also reports lower bounds from my thesis, conditional on GRH
for quadratic characters: $P(x) \gg \log \log x$, and also
$P(x_n) \gg \log x_n$ for an infinite sequence $x_n \rightarrow \infty$.<|endoftext|>
TITLE: The distribution of balls in a Bean Machine that omits all the "prime pegs"?
QUESTION [5 upvotes]: The Bean Machine, also known as a quincunx or the Galton box, is a well known triangular board that contains several rows ($n$) of staggered, but equally spaced pegs. Balls are dropped from the top one by one to avoid interference. They bounce off the pegs and stack up at the bottom of the triangle in bins. The resulting stacks of balls approach the characteristic Bell curve shape for large $n$.
Consider the following altered Bean Machine that now has all the 'prime numbered pegs' (counting top to bottom, left to right) removed:
 (source)
Based on what is known about the primes, is there anything that could be predicted about the resulting distribution of balls for large $n$? Is it expected to be skewed to the left or the right? Or does it nicely balance out like the normal distribution? 
Thanks.

REPLY [8 votes]: There will be certain columns with a (much) higher than average number of missing pegs and that will cause irregularities. It turns out that quadratic residues are relevant. 
Say that we locate pegs $1,2,3,4,5,6$ at $(0,0),(-1,-1),(1,-1),(-2,-2),(0,-2),(2,-2)$ etc.
The location of the peg number $n$ will be

$(x,y)=(2n-z^2-1,1-z)$ 

where $z=\lfloor\frac{(1+\sqrt{8n-7}}{2}\rfloor.$ This means that at location $(x,y)$ (with $x,y$ of equal parity, $y \le 0$ and $|x| \le |y|$) is 

$n=\frac{1+x+(y-1)^2}{2}.$  

Imagine a ball poised just above the location of (a potential) peg $(x,y)$  If the number $n=\frac{1+x+(y-1)^2}{2}$ is prime then drop down two rows to (just above) peg $(x,y-2).$ If $n$ is not prime then the ball can go either left to $(x-1,y-1)$ or right to $(x+1,y-1).$ 
Here is a diagram showing what this setup does. A $+$ is a peg corresponding to a non-prime while a $|$ corresponds to a missing peg from a prime.  Row $i$ adds to $2^i$ as in Pascal's triangle and shows the theoretical distribution for $2^i$ balls dropped from the top. However row $i$ has about twice as many possible locations for an entry. A more detailed description of the SETUP is below.

Observe that each column contains only even pegs or only odd pegs, which is already biased. The red oval encloses part of  column $x=-3$ starting with rows rows $y=-3$ and $y=-5$ and non-pegs corresponding to the primes $7,17.$ The potential pegs below these will be for $31,49,71,97,127$ which are all prime except for $49.$ The red oval shows the drop past $71,97,107.$ the six doubling are from dropping past these three  non-pegs and, after each one, slipping through the next row. Following these there is a mix of prime and composite. Starting with row $y=-81$ the next few pegs (for $x=-3$) are  $3361,3527,3697,3871,4049,4231$ which are all prime except $3871=7^2\cdot 79.$ Below are plots I got for the first 100 stages, it reads left to right and top to bottom. The biggest spike which grows and shrinks corresponds to $x=-3.$ I'll discuss how I interpreted the setup but first let me explain the sparse columns phenomenon (which is relevant however the ambiguous details are interpreted.)
It turns out that the pegs with $x=-3$ (which we already know are odd) can have number a multiple of $q=7, 17, 23, 31, 41, 47$ but can not divide by any of $q= 3, 5, 11, 13, 19, 29, 37, 43.$ These lists  are the primes $q$ which do and do not have $2$ as a quadratic residue. This explains why a large number of pegs in that column have a prime value. From $n=\frac{1+x+(y-1)^2}{2}$ we see that some pegs in column $x=c$ will be multiples of $q$ precisely when $-c-1 \mod q$ is a square. The smallest odd primes for which $-58$ is a quadratic residue are $29, 31, 37, 47, 59.$ This means that column $x=57$ (which is an odd column) will have no multiples of any smaller primes. Of the $46$ primes less than $199$, only $16$ are potential divisors. This explains why , of the first $100$ pegs in that column, $63$ get removed for being prime. These removed pegs include a run of $8$ consecutive removed pegs and two runs of $7.$ To be fair, a column which does contain multiples of $q$ has them at twice the usual frequency, so perhaps things do balance out on a larger scale.
So here is how I interpreted the SETUP, and I did not check too hard for a programming error. In the standard setup one has $r+1$ bins if the last row has $r$ pegs. I said that when a ball gets to a missing peg it drops straight down until it hits a peg in that column and goes right or left. However, if we want to know what the balls will look like if the bins are placed below row $r$ then I decree that a ball dropping straight down will hit a bin divider and go left or right. I did not actually run a simulation, I constructed a modified Pascal Triangle where a value in a certain position doubles and distributes equally to the two neighbors below. The modification is that a number "falling past" a missing peg doubles in size and sits in a space normally empty in the usual inverted triangle. It then lands on top of a potential peg two rows down from the missing peg (having doubled again) and possible combines with values arriving from the right and/or left  and redoubles several more times if it happens to encounter several consecutive missing pegs. So the total value at each layer is a power of $2$ as with the standard setup. However some of those are quantities falling past and slipping between  two "standard" entries. When we put the bins at a certain level, the intermediate irregular quantities get split equal with the two bins on either side.
The description above of the path of a ball allows us to ivnvestigate the empty bins question without generating the whole triangle. Imagine a ball which always goes right when it can. Since the triangular number $3$ at $(-1,-1)$ is prime, the ball drops down creating an eternal empty bin on the far right. There are no more prime triangular numbers but that is irrelevant because no ball will ever get to those pegs. What is relevant is that $\frac{m(m-1)}{2}-1= \frac{(m+2)(m-1)}{2}$ is never prime for $m \gt 3.$ So coming from the right is one empty bin and then one bin which will always get 4 balls (similar to the value $1$ down the edges of the usual Pascal triangle).
On the left the number of empty bins can grow. Just following the path of a ball which falls left when it can, I come up with $1043$ empty bins on the left of row $20842.$ The number of primes less than $20842$ is $2344$ so this does not clash too badly with my rash back of the envelope calculation that the number of empty bins on the right side of row $k$ grows a bit less than half as fast as the number of primes less than $k.$ This is based on little more than that a peg at level $k$ has size about $\frac{k^2}{2}$ so has "probability" $\frac{1}{2\ln(k)+ln(2)}$ to be prime and I don't know what to do with the $\ln{2}$ so I'll ignore it.  However the growth is choppy. There is a jump of $5$ in the number of empty bins starting at row $4788$ and again at $5844$. These correspond to the "lefty" ball encountering a column with many prime divisors forbidden and dropping past 5 empty spaces in a row.
So here are the graphs.<|endoftext|>
TITLE: Strictness of the inequality relating the Iitaka dimension and algebraic dimension
QUESTION [5 upvotes]: For any (compact and connected) complex manifold $X$ and any line bundle $L$ on $X$ we have the well known inequalities $\kappa(X,L)\leq\alpha(X)\leq\dim(X)$ relating the Iitaka-dimension $\kappa(X,L)$ of $L$ over $X$, the number of algebraic independent meromorphic functions $\alpha(X)$ on $X$ and the dimension of $X$. If $\alpha(X)$ equals $\dim(X)$, i.e. if $X$ is Moishezon, there exists a big line bundle $L$ on $X$, i.e. such that all the inequalities are actually equalities.
I don't expect it to be true, but I wonder if there always exists a line bundle that "sees all the meromorphic functions". I failed in searching a manifold with strictly bigger algebraic dimension than Iitaka dimension for all line bundles (simply because I'm not able to compute $\sup_{L\to X}\{\kappa(X,L)\}$, unless there exists a big line bundle).
Consequently, I'm asking for a manifold $X$ such that for all its line bundles $L$, $\kappa(X,L)<\alpha(X)$. Do you know some? There is no word for line bundles with $\kappa(X,L)=\alpha(X)$, is it?

REPLY [3 votes]: Given a compact complex manifold $X$, there exists a meromorphic map $F: X\dashrightarrow Z$ to a projective manifold such that every meromorphic function on $X$ is the pull back under $F$ of a rational function on $Z$. The pull-back of an ample line-bundle on $Z$ realizes the equality between algebraic and Iitaka's dimensions.<|endoftext|>
TITLE: Linear groups that are nonlinear over the integers
QUESTION [22 upvotes]: What are sources of finitely generated $\mathbb C$-linear groups that are not $\mathbb Z$-linear?
Recall that a group is $R$-linear if it is isomorphic to a subgroup of $GL(n,R)$ for some $n$, where $R$ is a ring.
I know only one source: any solvable $\mathbb Z$-linear group is polycyclic. For example, the Baumslag-Solitar group $B(1,2)$ is solvable, $\mathbb C$-linear, and
it contains dyadic rationals, and therefore, is not polycyclic (abelian subgroups of polycyclic groups are finitely generated).  
My personal motivation for the question is an attempt to digest recent applications of virtual Haken conjecture implying that many $3$-manifold groups are $\mathbb Z$-linear.

REPLY [4 votes]: You can use Margulis' super-rigidity theorem on $S$-arithmetic groups to say for example, that the group $SL_n(\mathbb{Z}[\frac{1}{p}])$ is never $\mathbb{Z}$-linear.<|endoftext|>
TITLE: How to specify a finite group up to inner automorphism?
QUESTION [5 upvotes]: I want some finite set of data to which I can canoically associate a "group up to inner automorphism", and which can be constructed canoically from a "group up to inner automorphism".  I have a few answers which satisfy the first requirement, but not the second.

Give a topological space $X$.  The fundamental group of $X$ is a "group up to inner automorphism": you can get a group by picking a basepoint $p$, but given two different basepoints $p$ and $q$, there is not a canonical isomorphism $\pi_1(X,p)\to\pi_1(X,q)$, but any two automorphisms coming from a path from $p$ to $q$ differ by an inner automorphism.  This answer is unsatisfying because there are many topological spaces one can pick, even if we restrict them to be $K(\pi,1)$'s.
Give a groupoid $X$ (this generalizes the first example by taking the fundamental groupoid).  This answer is unsatisfying because again there are many (say, finite) groupoids corresponding to a given group up to inner automorphism.
Give a tensor category which is isomorphic to the category of finite dimensional representations of a finite group over an algebraically closed field of characteristic zero.  This is almost a good answer.  It looks at first that specifying such a category is a finite amount of combinatorial data: we specify the isomorphism classes of objects and specify how the tensor product of any two of them decomposes.  However we also have to specify isomorphisms $(A\otimes B)\otimes C\to A\otimes(B\otimes C)$, and this means specifying some matrix which depends on the specific bases of the isotypic components of the tensor products that we chose when specifying how they decompose.  So it is unsatisfying as well, although it is the best I have come up with.  It is almost good enough since it is easy to say what an equivalence between two such finite categories is in terms of matrices.
Give an orbifold whose coarse space is a single point.  This answer is unsatisfying because I am hoping the answer to this question will be useful for giving a nice definition of an orbifold.  Also, by unraveling the definition of an orbifold, this answer really just reduces to (1) or (2) and is thus also unsatisfying.

REPLY [2 votes]: You can give a monoidal category with a tensor functor to the category of finite sets where tensor is Cartesian product that is equivalent to the category of permutation representations of the group. By Grothendieck's Galois theory, one can recover the group from this. This improves on linear representations because your association data is finite, and should be a bit more canonical.<|endoftext|>
TITLE: When is a $*$-homomorphism between multiplier algebras strictly continuous?
QUESTION [5 upvotes]: (This question was posted on MSE here but didn't get any answers.)
The strict topology on the multiplier algebra M(A) of a C*-algebra A is that generated by the seminorms
$$ x\mapsto \|ax\|\quad x\mapsto \|xa\| \qquad (a\in A, x\in M(A)) $$
Whereas a ∗-homomorphism $\phi : M(A)\to M(B)$ between two multiplier algebras is necessarily norm-continuous, if I understand things correctly it will not always be continuous with respect to the strict topologies on either side.  Where is there a good reference for this?
On the other hand an easily-proven theorem states that $\phi$ is strictly continuous if the image of $\phi$ contains B. This is not necessary, however; take $\phi : \mathcal{B}(\ell^2)\to\mathcal{B}(\ell^2)$ to be the map $x\mapsto sxs^*$ where $s$ is the unilateral shift. This is strictly continuous even though its image doesn't contain $\mathcal{K}(\ell^2)$. Are there other conditions which guarantee $\phi$ to be strictly continuous?
I'm particularly interested in the case where $\phi$ maps A into B, and both are nonunital.  Is this enough to show that $\phi$ is strictly continuous?

REPLY [5 votes]: Non-degenerate *-homorphism from $A$ (or $M(A)$) to $M(B)$ are strict (where non-degenerate means that $\phi(A)B$ is total in $B$). An important property of strict maps $\phi:A\to M(B)$  is that they possess a unique strict extension $\tilde{\phi}:M(A)\to M(B)$.
A good reference is
E.C. Lance, Hilbert C∗-modules, London Mathematical Society Lecture Notes Series
210, Cambridge University Press, Cambridge, 1995.<|endoftext|>
TITLE: Alternate definition of vector bundle?
QUESTION [7 upvotes]: Recall the usual definition of a $k$-dimensional vector bundle (everything is assumed to be continuous/smooth/etc depending on the category):

A $k$-dimensional vector bundle is a triple $(E,B,\pi)$, where 
  $\pi\colon E \to B$, satisfying the following:
    a) The map $\pi$ is onto (I don't know if everyone requires this, but I will). 
For each $p\in B$:
    b) The fiber $E_p=\pi^{-1}(p)$ is a (real) vector space. 
    c) There's a neighborhood $U\ni p$ and a diffeomorphism $\phi\colon \pi^{-1}(U) \to U \times \mathbb{R}^k$ such that $P_1\circ \phi = \pi$, where $P_1$ is projection onto the first factor. 
    d)  for each $q\in U$, the restriction $\phi\colon E_q \to {q}\times \mathbb{R}^k$ (where $\phi$ is the diffeomorphism from c)) is a linear isomorphism.

My Question
Can d) be replaced with

    d') the restriction $\phi\colon E_p \to {p}\times \mathbb{R}^k$ is a linear isomorphism. 

In the original version, the restriction of $\phi$ to every fiber over $U$ must be a linear isomorphism. In the alternate version, this is only required for the single fiber $E_p$.
(Note: This was posted at Math.SE a few weeks ago but received no answers.)

REPLY [4 votes]: No. Take $E=[0,1] \times \mathbb{R} \to [0,1]=B$ with the usual vector space structure at all fibers except for the fiber of $1/2$ where the vector space structure is twisted by any non-linear diffeomorphism of $\mathbb{R}$, for example $x \mapsto x^3+x$.