TITLE: Cartan integral formula for a p-adic group?
QUESTION [6 upvotes]: Let $G$ denote a reductive group over a local field $F$. Suppose that $G$ is split over $F$ and fix a maximal (split) torus $A$. Let $A^+$ denote a Weyl chamber in $A$ and let $K$ be a suitable maximal compact subgroup of $G$.
Then it is known that a Cartan decomposition holds:
$$
G=KA^+K.
$$
Hence there is a weight function $w\ge 0$ such that for the Haar-integrals one has
$$
\int_G f(x)\ dx=\int_K\int_{A^+}\int_K f(kal)\ w(a)\ dk\ da\ dl.
$$
If the field is archimedean, the function $w$ can be explicitly computed as a product over positive roots, see Knapp's book on page 141.
My question is this: Is there a similar explicit formula for $w$ in the non-archimedean case?
Normalising the measures so that the compact open subgroups get measure 1 we see that
$$
w(a)=|K\backslash KaK|,
$$
but I don't know whether that's of any help.

REPLY [4 votes]: In response to Anton for a uniform proof of your formula: note that $K\backslash KaK\cong(a^{-1}Ka\cap K)\backslash K$. Let $\pi:K\rightarrow G(\mathbb{F}_q)$  be the natural quotient map, $K_1:=\ker(\pi)$ and $P:=\pi(a^{-1}Ka\cap K)$. Then $P$ is the $\mathbb{F}_q$-points of the parabolic subgroup of $G$ determined by the walls that $a$ lies on. We have
$$|(a^{-1}Ka\cap K)\backslash K|=|P\backslash G(\mathbb{F}_q)|\cdot|(a^{-1}Ka\cap K_1)\backslash K_1|.$$
Let $\mathfrak{K}$ be the Lie algebra of $K$ and likewise for $\mathfrak{K}_1$. Choose Haar measures for the group and Lie algebra so that $\mu(K_1)=\mu(\mathfrak{K}_1)$. Since $a^{-1}Ka\cap K_1$ and $K_1$ are open compact, we have $|(a^{-1}Ka\cap K_1)\backslash K_1|=\mu(a^{-1}Ka\cap K_1)\backslash\mu(K_1)=\mu(a^{-1}\mathfrak{K}a\cap\mathfrak{K}_1)\backslash\mu(\mathfrak{K}_1)=|(a^{-1}\mathfrak{K}a\cap\mathfrak{K}_1)\backslash \mathfrak{K}_1|$.
Identifying $a$ as an element in the positive Weyl chamber in the cocharacter lattice, the last term is $q$ to the power of $\sum_{\alpha\in\Phi^+}\max(\langle a,\alpha\rangle-1,0)$, while the roots of $P$ are those $\alpha\in\Phi$ with $\langle a,\alpha\rangle\le 0$.<|endoftext|>
TITLE: What does it mean for a differential equation "to be integrable"?
QUESTION [9 upvotes]: What does it mean for a differential equation "to be integrable"?
Are "integrable" and "solvable" synonyms?
The first thing that comes to my mind is to say: it's integrable if we can find the analytic solution (a function that can be written as y = f(x1, x2, ...)) that when plugged into the differential equation satisfies it.
Sometimes we cannot find such a function in closed form, so we use numerical methods to solve the differential equation. Is the equation still defined "integrable" in this case? If yes, when a differential equation is not integrable?
In the case of the Nonlinear Schrodinger equation, which is a nonlinear partial differential equation:

we say that the equation is "integrable" because we can solve it using the inverse scattering method, but also this method can provide closed form solutions only in particular cases.
we can also solve it using finite difference numerical methods. Because of this could we have said that the equation was "integrable" even before discovering the inverse scattering method?

EDIT:
Since apparently the definition of "integrability" is not straigthforward, my main concern is to understand if the inverse scattering technique to solve the equations like the Nonlinear Schrodinger equation or the KdV equation is equivalent to a Finite Difference technique to say that we "solved" the euqation. 
Moreover, "solving a differential equation" is equivalent to say "solve the initial value problem"? I guess that when numerical methods are applied (so that we cannot get an analytic solution), we are actually solving the IVP, is that correct?

REPLY [13 votes]: There is also an old-fashion meaning   attached to the attribute  integrable,    namely solvable  by quadratures, where  quadrature is  old-fashion speak for  integrals, definite or indefinite.  An  equation was also considered integrable, if you could  describe its solutions as convergent power series. So  an equation was considered integrable if you could solve it using a combination of these two methods. An implicit  description of solutions was considered acceptable.    
In the 19th century, long before  the advent  of computers, the search   for integrable  equations was a very honorable     endeavor.   Even a celebrity such as  H. Poincare  started his  mathematical career trying to produce  large classes of integrable differential equations. This is in fact the content of his  dissertation. This  search  for integrable equations lead him to ask one very natural question. 
Taking for granted that all linear differential systems are integrable (using for example the Jordan decomposition of the matrix of the system) is it possible to reduce any (real analytic) differential system to a linear one, via a clever choice of local coordinates?
Away from stationary points this is always possible so the question boils down  to understanding what happens  near a stationary point. In his dissertation   Poincare solves a special case of this problem (linearization near  stable  stationary point). He  could not  solve the general case  since he was stymied by the so called small denominators. Carl Siegel  eventually bypassed this problem in the 1930s, more that 50 years after  Poincare's dissertation.   
The problem    of small denominators     haunted the mathematicians since.  In the 1960s the Russian mathematician Brjuno  (some other spellings of this name are used)  described  very refined sufficient conditions allowing one to bypass  the small-denominator conundrum. At the end of last century Yoccoz showed that Brjuno's conditions are   also necessary.  (He even received a Fields medal for this and other work.) 
I have   to mention two other   remarkable    contributions around the concept of  integrability. The first  is  the Kolmogorov-Arnold-Moser theorem which refers to the special concept of integrability described by  Paul Siegel in his answer.  The old fashion concept of integrability is present in  Hovanskii's concept of pfaffian functions, concept  that has a very good model theoretic  incarnation (it leads to a  very useful $o$-minimal category).

REPLY [5 votes]: Integrability has a specific meaning for certain differential equations arising in geometry, but I'm not sure if it has a broader meaning for more general differential equations.
Given a collection of $1$-forms $\omega_i$ on a manifold $M$, a submanifold $N \subseteq M$ is said to be integral if the tangent space of $N$ lies in the kernel of each $\omega_i$ at every point.  $N$ is maximally integral if the tangent space of $N$ is precisely the intersection of the kernels of the $\omega_i$'s at every point.  And the collection of $\omega_i$'s is said to be integrable if $M$ admits a foliation by maximally integral submanifolds.
A typical example of an integrable system according to these definitions is the $1$ from $dx$ on $\mathbb{R}^2$: the kernel of $dx$ at each point is spanned by the tangent vector $\partial y$, and the span of $\partial y$ at a point $p$ is precisely the tangent space of the vertical line passing through $p$.  Since the set of all vertical lines forms a foliation of $\mathbb{R}^2$, we're done.
Interesting examples of non-integrable systems come from contact geometry.
So at least in this context integrability means "solutions exist and have nice geometric structure".  The actual definition that I gave makes sense only for first order systems, but probably this is the right intuition for higher order systems.  In particular, I don't think how the solutions are actually found is important for this terminology.<|endoftext|>
TITLE: Lower semi-continuity of the Hellinger-Fisher-Rao distance
QUESTION [5 upvotes]: I am currently working on unbalanced optimal transport, where the Hellinger (or sometimes Fisher-Rao) distance
$$
H^2(\rho,\mu)=\int_{\Omega}\left|\sqrt{\frac{d\rho}{d\lambda}}-\sqrt{\frac{d\mu}{d\lambda}}\right|^2 d\lambda
$$
shows up. Here $\Omega\subset R^d$ is a (possibly unbounded) smooth domain, $\rho,\mu$ are nonnegative Borel measures with finite mass on $\Omega$, $\lambda$ is any reference (nonnegative, Borel) measure on $\Omega$ such that $\rho,\mu$ are simultaneously absolutely continuous with respect to $\lambda$, and $\frac{d\rho}{d\lambda},\frac{d\mu}{d\lambda}$ denotes the corresponding Radon-Nikodym derivatives. Note that by 1-homogeneity this definition does not depend on the choice of the reference measure $\lambda$. This distance is ususally defined for probability measures (i-e with fixed unit mass $|\rho|=|\mu|=1$) but still gives a distance on the set $\mathcal M^+(\Omega)$ of nonnegative Borel measures with finite mass (thus possibly $|\rho|\neq |\mu|$).

Question: is it true that $H$ is lower semi-continuous with respect to the weak-$\ast$ convergence of measures,
  $$
\rho^n\overset{\ast}{\rightharpoonup}\rho,\mu^n\overset{\ast}{\rightharpoonup}\mu
\quad\Rightarrow\quad
H(\rho,\mu)\leq \liminf \,H(\rho^n,\mu^n)???
$$

Let me remind that the weak-$\ast$ convergence of measures is defined by duality with continuous compactly supported functions $\phi\in \mathcal C^\infty_c(\Omega)$. Also, I would be happy if this were true for the (stronger) weak $L^1(\Omega,dx)$ convergence for absoluetly continuous measures (with respect to, say, Lebesgue's measure $dx$) instead of weak-$\ast$ convergence of measures.

REPLY [3 votes]: The lower semicontinuity of the Hellinger-Rao distance follows from the following more general lower semicontinuity result (see e.g. Buttazzo, Semicontinuity, relaxation and integral representation in the calculus of variations, Thm. 3.4.1):
If $f\colon \mathbb{R}^n\to [0,\infty]$ is a proper lower semicontinuous convex positively $1$-homogeneous function, then the functional
$$
F\colon \mathcal{M}(\Omega)^n\to [0,\infty],\,\mu\mapsto \int_\Omega f\left(\frac{d\mu^1}{d\lambda},\dots,\frac{d\mu^n}{d\lambda}\right)\,d\lambda
$$
is sequentially weak* lower semicontinuous. As in the question, $\lambda$ is any scalar measure on $\Omega$ such that $\mu^1,\dots,\mu^n\ll\lambda$.
Let me sketch the proof.  First note that $f$ is the supremum of a sequence of affine-linear functions $f_i$. By monotone convergence,
$$
\int_\Omega \sup_{1\leq i\leq N}f_i^+\left(\frac{d\mu}{d\lambda}\right)\,d\lambda\to F(\mu),\;N\to \infty.
$$
For each $N\in\mathbb{N}$ we can decompose $\Omega$ into disjoint subsets $E_j$ such that $f_j^+=\sup_{1\leq i\leq N}f_i^+$. Thus
$$
F(\mu)=\sup\left\lbrace\sum_{i\in I}\int_{E_i}f_i^+\left(\frac{d\mu}{d\lambda}\right)\,d\lambda\,\Bigg|\, I\subset \mathbb{N}\text{ finite}, E_i\cap E_j=\emptyset\text{ for }i\neq j\right\rbrace.
$$
Using the regularity of the measures, one can restrict to disjoint open sets $E_i$ in the supremum above. For $U\subset \Omega$ open one has
$$
\int_U f_i^+\left(\frac{d\mu}{d\lambda}\right)\,d\lambda=\sup\left\lbrace \int_\Omega f_i\left(\frac{d\mu}{d\lambda}\right)\phi\,d\lambda\,\Bigg|\,\phi\in C_c^\infty(U),\,0\leq \phi\leq 1\right\rbrace.
$$
Since $f_i$ is affine-linear, the functional in the supremum on the right side is sequentially weak* continuous. Therefore $F$ is sequentially weak* lower semicontinuous as supremum of sequentially weak* continuous maps.
Remark: I treated the reference measure as fixed. This is justified since for a sequence $(\mu_k)$ one can always take something like
$$
\lambda=\sum_{k=1}^\infty\sum_{l=1}^n |\mu_k^l|.
$$
For this reason the argument only works for sequential semicontinuity.<|endoftext|>
TITLE: Is there any pattern to the continued fraction of $\sqrt[3]{2}$?
QUESTION [26 upvotes]: Is there any pattern to the continued fraction of $\sqrt[3]{2}$ ?  Wolfram Alpha returns for cube root of 2:

$\sqrt[3]{2}=$ [1; 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1, 3, 4, 1, 1, 2, 14,
  3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1, 121, 1, 2, 2, 4, 10, 3, 2, 2, 41, 1, 1, ...]

So the answer is likely no.  There certainly won't be any repeating pattern such as:
$$ 1+\sqrt{2} = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \dots }}}$$
But I was hoping maybe for a pattern like we might find for the number $e = 2.718\dots$:
$$ e = [2; 1, \color{blue}{2}, 1, 1, \color{blue}{4}, 1, 1, \color{blue}{6}, 1, 1, \color{blue}{8}, 1, 1, \color{blue}{10}, 1, 1, \color{blue}{12}, 1, 1, \color{blue}{14}, 1, 1, \color{blue}{16}, 1, 1, 1, ...]$$
So in fact $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ has a nicer continued fraction pattern than $\sqrt[3]{2}$.  
You can actually derive the continued fraction of $e$ (Henry Cohn) - but I haven't worked through it yet.

Proposal: Tree-Like Continued Fraction
If we try to derive a continued fraction for cube roots let's see how we get in trouble:
$$ \sqrt[3]{2} \approx 1 $$
That's really lousy guess but let's see how far we are off:
$$ \sqrt[3]{2} - 1 = \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} \tag{$\square$}$$
so let's backtrack we actually needed two pieces of information:
$$ 
\sqrt[3]{2} \approx 1 
\hspace{0.25in}
\textbf{and}
\hspace{0.25in} 
\sqrt[3]{4} \approx 1 $$
The cube root of 4 is still not quite $2 = \sqrt[3]{8}$.  Now let's try:
$$ \sqrt[3]{4} - 1 = \frac{3}{1 + \sqrt[3]{4} + \sqrt[3]{16}}
=  \frac{3}{1 + \sqrt[3]{4} + 2\sqrt[3]{2}}
\tag{$\Delta$}
$$
And then run all of this back into the equation we started from:
$$ \sqrt[3]{2} - 1 = \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}}
= \cfrac{1}{1 + \left(1 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}} \right) + \left( 1 + \frac{3}{1 + \sqrt[3]{4} + 2\sqrt[3]{2}}\right)}$$
Let's make it look a little bit cleaner but the same:
$$ \sqrt[3]{2} = 1 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}}
= 1 + \cfrac{1}{3 + \frac{1}{1 + \sqrt[3]{2} + \sqrt[3]{4}}  +  \frac{3}{1  + 2\sqrt[3]{2}+ \sqrt[3]{4}}}$$
Using the $\square$ and $\Delta$ pattern we can get a regular infinite pattern this way.

Clarification
I am being repeatedly asked to clarify the question what do I mean by "pattern" the truth is I don't know.  To this end I make two points:

the are transcedental numbers with clear patterns if their continued fraction such as $e$.  We find that every third digit is an even number starting with $2$ and increasing by $+2$.  Every other digit is $1$.
I have already proposed a "tree-like" generalized continued fraction that may have recursive properties similar to what repeated patterns havefor numberes like $\sqrt{n}$
This paper by Yann Bugeaud states continued fraction digits definitely cannot have any pattern:


Let $\alpha = [0; a_1, a_2,...]$ be an algebraic
  number of degree at least three. One of our criteria implies that the
  sequence of partial quotients $(a_n)_{n≥1}$ of $\alpha$ cannot be generated by a finite automaton, and that the complexity function of $(a_n)_{n≥1}$ cannot increase too slowly.

Personally it's hard for me to fathom a number that doesn't satisfy any pattern.  There must be some pattern.  My question is then what could we try instead?
Lastly, if the cubic case is this intractable, I might shift my attention to the quadratic case where more results are known...

REPLY [5 votes]: This is the Hermite Problem: https://en.wikipedia.org/wiki/Hermite%27s_problem
The Jacobi-Perron algorithm attempts to solve this.
Mittal and Gupta have defined Bifurcating Continued Fractions- http://arxiv.org/abs/math/0002227 http://arxiv.org/abs/math/0008060
Specifically, they say the pair $2^{1/3}$ and $2^{2/3}$ has the expansion:
$1,\overline{1,2} $ and $ \overline{1,0}$
Lehmer talks of Ternary Continued Fractions - https://oeis.org/A000962/a000962.pdf
How it all fits together I have no idea.<|endoftext|>
TITLE: Is the word problem decidable for free finitely generated self-square groups?
QUESTION [10 upvotes]: A self-square group is a group with extra structure, which encodes the fact that the group is isomorphic to its own direct square. 
To be exact, the group $G$ has a special element $1$, a unary operation of inverse $\operatorname{In}(x)$ or $x^{-1}$, and a binary operation, product,
taking the pair $(x,y)$ to $x\cdot y$, all satisfying the usual identities of groups, and in addition two unary operations $L(x)$ and $R(x)$,
and a binary operation $*$ taking the pair $(x,y)$ into $x*y$. These satisfy the identities
$L(x) * R(x) = x$ for all $x$,
$L(x * y) = x$ for all $x$, $y$,
$R(x*y) = y$ for all $x$, $y$.
So the function $*$ which maps $G \times G$ to $G$ can be inverted by the function $z$ goes to $(L(z), R(z))$,
so we have a bijection between $G \times G$ and $G$. In order to make sure that this mapping and its inverse are isomorphisms, we must require that
$L(x \cdot y) = L(x) \cdot L(y)$,
$R(x \cdot y) = R(x) \cdot R(y)$, and
$(x * y) \cdot (u * v) = (x \cdot u)*(y \cdot v)$. 
These identities insure that $L,R$ and $*$ are homomorphisms, so we have an isomorphism between $G \times G$ and $G$.
Since all these equations hold for all elements, we have an equational class or a variety in the sense of universal algebra. 
Standard theorems now guarantee the existence of free self-square groups, from here on SSG's for short. 
Now the question: is the word problem decidable for free finitely generated self-square groups, in particular for the $1$-generator SSG?
That is, given two well-formed expressions in $\cdot$, $*$, $\operatorname{In}$, $L$, $R$, $1$, and the variable $x$, can we determine if they represent the same element
in the free SSG generated by $x$?
Some comments:  
1) Notrivial SSG's exist, eg take the countable direct power of your favorite group. Less trivially, by the work of J.M. Tyrer-Jones, D. Meier, and Baumslag
there are finitely generated self-square groups, and some can be found inside fairly easy finitely presented groups. It seems it is an open question whether
there are finitely presented groups isomorphic to their direct squares.   
2) The idea of encoding a bijection between a set and its direct square using operations and identities was explored by Jónsson and Tarski, and by Higman, who used this idea to study the Thompson group.  
3) A self-square group may be finitely generated as a self-square group but not finitely generated as a group. In fact this is usually the case, and in particular it is the case for the free self-square group on one generator.

REPLY [6 votes]: You can use the identities to move the $*$s up to the top of the syntactic tree.
To see this I have to check that if I have $(a * b)$ and I apply any operation that's not $*$ to it then I can replace it with one in which the star is gone or is higher than it was.
$(a*b)^{-1} = (a^{-1} * b^{-1})$
$(a* b) \cdot c=  (a \cdot L(c)) * ( b\cdot R(c))$
$L(a * b) = a$
$R(a * b) = b$
So I can put the expressions in a form where they have a tree of $*$s at the top and the other operations down below. Using $L(x) * R(x)=x$, I can expand both trees of $*$s so that they are the same. Because $*$ is injective (being an isomorphism), it is sufficient to check equality of all the branches coming out of the tree of $*$s.
But this is just checking identity in the free group with two additional operations $L$ and $R$ that are both homomorphisms, which is the same as the free group on the elements that are a sequences of $L$s and $R$s followed by one of the variables.
In fact possibly this shows that you can put elements in canonical form.<|endoftext|>
TITLE: G-Correlation of Vectors
QUESTION [6 upvotes]: Let $\vec{a},\vec{b} \in \mathbb{R}^{n}$. Consider the function $f: S_n \to \mathbb{R}$ given by $f(\sigma):= \sum_{i=1}^{n} a_i b_{\sigma(i)}$. Let $G$ be a subgroup of $S_n$, given by $O(\log n)$ generators. 
Question: What is the most efficient algorithm for finding the maximum of $f$ restricted to $G$, and the permutation at which the maximum is attained?
Of course, for some $G$'s it should be simpler than for others. Here are several classical examples:

If $G=S_n$, one can use the rearrangement inequality to solve this problem. The inequality reduces to problem to that of sorting the entries of $a$ and $b$, so the complexity is $O(n\log n)$. The same idea also solves the case $G=A_n$.
If $G=\langle \sigma \rangle$ where $\sigma(i)=i+1$ for $ 1 \le i \le n-1$, the sums $f(G)$ are known as "circular correlations", and the best correlation can be found with $O(n \log n)$ complexity using DFT over $\mathbb{Z}/n\mathbb{Z}$.

More generally, if we identify the set $\{1,2,\cdots, n\}$ with some finite abelian group $A$, and take $G:=\{ x\mapsto x+a \mid a \in A\}$, we can use DFT over $A$ and achieve again a similar complexity.


Other examples are welcome.
It is easy to reduce to the case of transitive $G$. One can also assume that the entries of the vectors are integers.
[EDIT] I will make the question more concrete. 
More Concrete Question: Can one solve the original question for some non-commutative $G$ (apart from $S_n$ and $A_n$)?
Specifically, here are some examples of $G$'s for which I don't know what is the best algorithm:

Identify $\{1,2,\cdots, n\}$ with $\mathbb{F}_q$ (so $n$ is a power of a prime), and consider $G=\{x \mapsto ax+b \mid a \in \mathbb{F}_q^{\times}, b \in \mathbb{F}_q\}$.
Identify $\{1,2,\cdots, n\}$ with $\mathbb{F}_q \cup \{\infty\}$ and consider $G=PGL_2(\mathbb{F}_q)$ or $G=PSL_2(\mathbb{F}_q)$.

Even More Concrete: Can someone solve the original question for these two (and a half) examples of $G$?
A naive solution is to reduce to the abelian case by writing $G$ as a union of abelian groups, but I feel this is too naive.

REPLY [2 votes]: Let $n=2^m$ and let the integers from $0$ to $2^m-1$ be identified with the vector space $Z_2^m$ where operations are modulo 2 componentwise. For any $a\in Z_2^m$ the permutation $\sigma_a:x \rightarrow x+a$ can be defined on $Z_2^m$ and the correlation is then $$C_a(f)=\sum_{x \in Z_2^m} f(x) f(x+a)$$
In this case we can use the Walsh-Hadamard fast transform 
$$\hat{f}(u)=\sum_{x \in Z_2^m} (-1)^{u\cdot x} f(x)$$
(which is of course the fourier transform for this setup) to find the maximum correlation and the maximizing $\sigma_a$ in $O(n \log n)$ time complexity.
The case when $f(x)=\pm 1,$ has specific applications to coding theory and cryptography, but this is not necessary.
Note this can also be done for any prime $p,$ and the space $Z_p^m$ where complex $p^{th}$ roots of unity and a corresponding generalized Hadamard transform can be used to define the appropriate fourier transform.
A related example: If $n=2^m-1$ and the space is the multiplicative group $GF(2^m)^{\ast}$ of $GF(2^m)$ which includes the elements $1,\alpha,\ldots,\alpha^{2^m-2}$ where $\alpha$ is a primitive element of $GF(2^m)$ one can define
$$\tilde{f}(t)=f(\alpha^t),\quad t \in Z_{2^m-1}$$ taking on values in $GF(2)$ and use the transform
$$\hat{f}(a)=\sum_{x \in GF(2^m)^{\ast}}(-1)^{tr(a x)+\tilde{f}(x)},\quad a \in GF(2^m)^{\ast}$$
to compute ``cyclic'' correlations since where the shift of the function $\tilde{f}$ by $\tau$ is given by $\tilde{f}(\alpha^{t+\tau})$, if you like in discrete log order instead of integer order.
Using a self-dual basis, which always exists for a finite field, the two transforms above can be related to each other.<|endoftext|>
TITLE: "Natural" action of a semidirect product of groups
QUESTION [5 upvotes]: Suppose we have a group $N$ acting on a set $X$, and a group $Q$ acting on $Y$. Moreover, we are given $\phi: Q \to \mathrm{Aut}(N)$. 
My question is, is there a "natural" action on a set that can be constructed from $X$ and $Y$ (maybe $X \times Y$?) on which the semidirect product $N \rtimes_{\varphi} Q$ acts. 
For example, when $\phi$ is trivial (so we are considering direct product of groups), then the answer to this question is trivial (our set is $X \times Y$ and the action is coordinatewise). 
PS In general I am interested in a case when $X$ and $Y$ have more structure, but let us start with sets.

REPLY [2 votes]: Suppose you define some equivalence relation on $X$ so that the quotient $\overline{X}$ has the property that $N\rtimes Q$ acts on $\overline{X}\times Y$ coordinatewise, with the action of $n$ on $[x]$ being $[nx]$ (here, $[x]$ is the equivalence class of $x$).
Writing ${}^qn$ for $\varphi(q)(n)$, and writing the elements of $N\rtimes Q$ as pairs $(n,q)$, we would need $[nx] = [{}^qnx]$ for all $q\in Q$ (with fixed $x$ and $n$). For $({}^qn,1) = (1,q)(n,1)(1,q^{-1})$, and looking at the actions on a pair $([x],y)$ readily gives the need for $[nx]=[{}^qnx]$. 
Conversely, if $\sim$ is any equivalence relation on $x$ that satisfies this property, then the action on $\overline{X}\times Y$ given by $(n,q)([x],y) = ([nx],y)$ is readily seen to be an action:
$$\begin{align*}
(n_1,q_1)\Bigl((n_2,q_2)([x],y)\Bigr) &= (n_1,q_1)([n_2x],q_2y)\\\
&= ([n_1(n_2x)],q_1(q_2y))\\\
\Bigl((n_1,q_1)(n_2,q_2)\Bigr)([x],y) &= (n_1{}^{q_1}n_2,q_1q_2)([x],y)\\\
&= ([n_1{}^{q_1}n_2 x],(q_1q_2)y).
\end{align*}$$
The assumption on $\sim$ gives that $[{}^{q_1}n_2 x] = [n_2x]$, and hence that $[n_1{}^{q_1}n_2 x] = [n_1n_2x]$.
If you let $\sim$ be the transitive closure of the relation with $x\sim x'$ if and only if there exists $x_0\in X$, $n\in N$, and $q\in Q$ with $nx_0 = x$, ${}^qnx_0 =x'$, then any equivalence relation that is coarser than $\sim$ will do. (I may have my "coarser" and "finer" mixed up here; I mean that the equivalence classes are unions of equivalence classes under $\sim$, so that if you think of it as if it were a "topology", you would have a coarser topology, with fewer open sets).
In particular, taking $\sim$ to be the total relation ($x\sim x'$ for all $x,x'\in X$), you get the natural action on $Y$ given by mapping to the quotient (well, technically an action on $\{\bullet\}\times Y$, but that is naturally isomorphic to an action on $Y$). If the action of $Q$ on $N$ is trivial, so that ${}^qn = n$, then $\sim$ is the identity relation, so $\overline{X}=X$ and you get the natural action on $X\times Y$.<|endoftext|>
TITLE: A Graph-Theory Related Question
QUESTION [9 upvotes]: Let $n$ be a positive integer and partition a grid of $4n$ by $4n$ unit squares into $4n^2$ squares of sidelength $2$. (The squares with sidelength $2$ have all of their sides on the gridlines of the $4n$ by $4n$ grid.) What's the minimum number of unit squares that can be shaded such that every square of sidelength $2$ has at least one shaded square, and there exists a path between any two shaded squares going from shaded squares to adjacent shaded squares? (Two squares are adjacent of they share a side.)
I think the answer is $6n^2-2$. I got it by partitioning the $4n$ by $4n$ unit squares into $n^2$ squares of sidelength $4$, shading in the four center unit squares of each sidelength $4$ square (so that every square with sidelength $2$ has at least one shaded square), and connecting the shaded squares to another. There are $4n^2$ center unit squares, and a connection between any two groups of center squares requires two squares. There are $n^2$ groups of center squares, so for the shaded squares to be connected, we need at least $n^2-1$ connections. (Similar to how a connected graph on $n$ vertices needs at least $n-1$ edges.) Each connection is two squares, so we need to shade an additional $2(n^2-1)$ squares. Adding this to $4n^2$ gives $6n^2-2$.
This is by no means a proof. There are of course valid configurations in which some square with sidelength $4$ does not have the center $4$ squares shaded in. Could someone help me prove that $6n^2-2$ is the minimum? Thanks.

REPLY [2 votes]: Let $p$ be a path consisting of $m$ shaded unit squares (where every two adjacent shaded squares share a side). Define a binary string $B_p=b_0b_1b_2\dots b_m$, where $b_0b_1=10$ and for $i>1$, $b_i=1$ iff the $i$th square in $p$ belongs to a previously unseen 2x2 square. Clearly, the number of ones in $B_p$ equals the number of 2x2 squares visited by $p$.
It can be shown that every substring of length 6 of $B_p$ contains at most 4 ones. Furthermore, the maximum number of ones in $B_p$ is:

if $m\equiv 0\pmod{6}$, then $\frac{4m+6}6$ ones
if $m\equiv 1\pmod{6}$, then $\frac{4m+2}6$ ones
if $m\equiv 2\pmod{6}$, then $\frac{4m+4}6$ ones
if $m\equiv 3\pmod{6}$, then $\frac{4m+6}6$ ones
if $m\equiv 4\pmod{6}$, then $\frac{4m+8}6$ ones
if $m\equiv 5\pmod{6}$, then $\frac{4m+4}6$ ones

To visit $4n^2$ 2x2 squares, path $p$ must contain $m\geq \frac{6\cdot 4n^2 - 8}{4}=6n^2-2$ shaded squares. 
It remains to consider a case when the shaded squares form not a path but a tree, which I believe can be done similarly.<|endoftext|>
TITLE: Did Bishop, Heyting or Brouwer take partial functions seriously?
QUESTION [8 upvotes]: The partial μ-recursive functions which may or may not be provably total seem to have some direct relation to the initial motivations for intuitionistic mathematics. (Following Kronecker, one motivation might have been that mathematical statements should be reducible to statements about the natural numbers, or computable functions of the natural numbers.)
When Brouwer and Heyting did their initial work, the understanding of those partial functions was still evolving significantly. However, this excuse no longer applies to the time when Errett Bishop published his Foundations of constructive analysis. However, extramathematical observations like that "all functions in Bishop's constructive mathematics are continuous" indicate that function always meant provably total function, because otherwise a formulation like "no discontinuous function can be proved to be total in Bishop's constructive mathematics" would seem to be less misleading and better capture what is really going on.
So did Bishop, Heyting or Brouwer (or any other prominent "early" proponent of intuitionism) explicitly discussed this relationship, and clearly indicated to role (and treatment) of such partial functions in intuitionistic mathematics?

REPLY [9 votes]: The question of how Brouwer perceived partial functions is very interesting and worthy of investigation. I only have two comments about this:

Brouwer and Heyting certainly did consider partial functions such as $1/x$ and discussed its domain in great detail.
The Brouwerian Counterexamples are based on the idea of unbounded search, so both Brouwer and Heyting fully understood the idea of $\mu$-recursion.

It would be very interesting to know whether either addressed the idea of a partial function in contrast to the idea of a total function.
However, the main goal of this answer is to correct a false premise of the question: that "every provably total function is continuous" is not a good way to think about Brouwer's continuity theorem. Brouwer proved that "all total functions $[0,1]\to\mathbb{R}$ are uniformly continuous" within his intuitionistic mathematics. There is no mention of "provability" nor any other kind of restriction. Brouwer also showed that classical examples of discontinuous functions, such as the characteristic function of a singleton, are not total by giving specific examples where the functions are not defined. (Note that Brouwer's notion of real numbers allows for a great deal more possibilities than the classical notion.) So there is no room at all in Brouwer's intuitionistic mathematics for total discontinuous functions. This is very different from Bishop's constructive mathematics, which is intended to be compatible with classical mathematics and therefore must allow the possibility of total discontinuous functions.<|endoftext|>
TITLE: Intuition about Skorohod integral
QUESTION [5 upvotes]: I'm teaching myself Malliavin calculus and Skorohod integrals and with this kind of math I find myself following the logic through but lacking solid intuition about what is going on.
In particular take the Skorohod integral which generalises the Ito integral to possibly anticipating integrands (it reduces to an Ito integral when the integrand is non-anticipating)
So for example take the Skorohod integral, of the Brownian motion $B(t)$, $B(0)=0$ $$ \int_0^T B(T)\delta B(t) = B^2(T)-T$$ (see for example http://www.nhh.no/Files/Filer/institutter/for/dp/1996/wp0396.pdf)
It is emphasised that $$ \int_0^T B(T)\delta B(t) \neq B(T)\int_0^T\delta B(t)=B(T)\int_0^T dB(t)=B^2(T)$$
and I am trying to understand exactly why this is the case. I appreciate that one can't just view it as a Riemann sum which is what gives the intuition that the above is trying to ward off. So the only other kind of intuition I have is that it is because the integrand and integrator are correlated in the same way that the integrand and integrator are correlated in the Stratonovich integral which means it is not a Martingale. I.e. analogously to $$E[f(B)\circ dB]\neq E[f(B)]E[dB]=E[f(B)]\cdot 0=0\quad\left(=E[f(B)dB]\right)$$
However, in which case I don't understand why the difference between the naive and correct interpretations above is $T$ so that the difference grows independently of the correlation.
In other words I would expect that (given $T_2>T_1$) $$\lim_{(T_2-T_1)\to\infty}\int_0^{T_1}B(T_2)\delta B(t)=B(T_2)\int_0^{T_1}\delta B(t)=B(T_2)\int_0^{T_1}dB(t).$$
Or that I would expect the difference $$\int_0^T B(T)\delta B(t) - B(T)\int_0^T\delta B(t)=\int_0^T\xi(t-T)dt$$
where $\xi(t-T)\to 0$ as $T\to\infty$.
But all of this is completely at odds with the definition of the Skorohod integral above. Can someone explain this in a relatively simple with intuitive ideas?
A secondary question it raises is how to numerically simulate such an integral (say given a pre computed Wiener process)

REPLY [9 votes]: Unfortunately, calling the Skorohod integral an "integral" is a bit of a misnomer, as it doesn't really have many of the properties which you would naturally associate with integrals, except for the fact that it "magically" coincides with the Itô integral when its integrand is adapted. The best way (I think) to get some kind of intuition for the Skorohod integral is to consider the case where the unit interval is partitioned into intervals of length $\delta t_k$, the integrand $F$ is constant and equal to $F_k$ on the $k$th interval, and it depends on the Wiener process $B$ only through its increments $\delta B_k$. In this case, an elementary calculation shows that one has
$$
\int_0^1 F(t)\ \delta B(t) = \sum_k \bigl(F_k\ \delta B_k - \partial_k F_k\ \delta t_k\bigr)\;.
$$
Here, $\partial_k F_k$ is the derivative of $F_k$ (viewed as a function of the increments $\delta B_j$) with respect to $\delta B_k$. The example you give follows immediately as a special case and many properties of the Skorohod integral can be "guessed" from that formula. (See for example Section 3 in this article.)
You see that the first term looks just like what you would expect from the definition of the Itô integral, but the second term is strange: while it looks again like a Riemann sum, the "integrand" is given by the derivative of the original integrand with respect to the underlying Wiener process. In particular, this shows that there is no way whatsoever to build approximations to the Skorohod integral from a single realisation of $F$ and $B$, since this gives no information at all on the derivative of $F$ with respect to $B$... In the case where $F$ is adapted, this second term vanishes of course (since then $F_k$ is only allowed to depend on the $\delta B_j$ with $j < k$) and you get Itô's integral back.
One natural reaction might be to say "since the second term behaves strangely, let's just throw it away" and to define $\int_0^1 F(t)\ \delta B(t) = \sum_k F_k\ \delta B_k$. The problem with this is that, as an unbounded linear operator on the space of all square integrable processes (not just adapted ones), this is not closable, so there's no "reasonable" way to build a nice space of integrands for which that integral makes sense. The reason why the Skorohod integral makes sense is that the second term introduces stochastic cancellations that cause the sum to converge for far more integrands than what one might naively expect.
There are other ways of introducing such cancellations. For example, in Sections 2-5 of my book with P. Friz available to download from my homepage, we discuss how the theory of controlled rough paths developed by Lyons and Gubinelli provides a notion of stochastic integration based on a different Riemann-type approximation that has the desirable feature that $\int GF_t\ dB_t = G \int F_t\ dB_t$ for any random variable $G$. Better: if the integrand depends on a parameter $a$ and one sets $\hat F(a) = \int_0^1 F_t(a)\ dB_t$, then one has $\hat F(G) = \int_0^1 F_t(G)\ dB_t$ for any random variable $G$, which definitely fails to hold for the Skorohod integral.
This "rough path integral" isn't quite an extension of the Itô integral though since it isn't defined for all square integrable adapted processes, but it coincides with the Itô integral on the intersection of their domains of definition and it admits anticipative integrands.<|endoftext|>
TITLE: number of maximal subgroups of the symmetric group
QUESTION [7 upvotes]: What is the asymptotics of the number of the maximal subgroups of $S_n$ (as a function of $n$)? This must be written down somewhere...
EDIT I am actually more interested in the number of conjugacy classes of maximal subgroups (the difference is graphically illustrated by Derek's and Gerry's comments.)

REPLY [13 votes]: With regard to the conjugacy class question, you should refer to this:

Liebeck, Martin W.; Shalev, Aner Maximal subgroups of symmetric groups.
  J. Comb. Theory, Ser. A 75, No.2, 341-352 (1996).

The following is a quote from the ZBMath review by W. Knapp:

The purpose of this paper is to give estimates on the number of conjugacy classes of maximal subgroups of the finite symmetric groups $S_n$, on $n$ letters in terms of $n$. It is shown that this number is of the form $(\frac12+o(1))n$. The main work has to be done in establishing that $S_n$ has at most $n^{6/11+o(1)}$ conjugacy classes of primitive maximal subgroups. Of course, the O’Nan-Scott Theorem and the classification of finite simple groups are used.

The same paper is also relevant to the original question of the OP (about the number of maximal subgroups). A further quote from the review:

In the course of the proof, it is shown that any finite almost simple group has at most $n^{17/11+o(1)}$ maximal subgroups of index $n$.<|endoftext|>
TITLE: Joyal's construction of the spectrum of a commutative ring
QUESTION [21 upvotes]: I am trying to understand bits and pieces of Lawvere's article Continuously Variable Sets; Algebraic Geometry = Geometric Logic. I'm not doing very well.
I know this is a lot to ask, but basically, I would like for someone to explain in more detail the whole construction described at the end of the paper. I was told the idea is to get a "free local ring", which is free simply because of the internal language of the corresponding sheaf topos which makes any ring locally free, but I really don't know what that means and I still don't understand the idea and details of the construction in the paper. I stress that I'm looking for a detailed explanation; I realize that the idea may be encapsulated in a short elegant statement, but I won't be able to decode it..
So... What the hell is going on there? :D

REPLY [40 votes]: Since I don't know precisely which parts of Lawvere's article you have difficulties with, this answer is a bit a long and tries to give a bit of context. If you want me to be more specific at some point, just say so.
Classically, the spectrum of a ring $A$ can be defined as the set of its prime ideals equipped with the Zariski topology. This topological space then has the following universal property: For any locally ringed space $X$ we have $\mathrm{Hom}_{\mathrm{LRS}}(X, \operatorname{Spec}(A)) \cong \mathrm{Hom}_{\mathrm{Ring}}(A, \Gamma(X,\mathcal{O}_X))$.
We can also express this property using the opposite categories $\mathrm{LRS}^{\mathrm{op}}$ and $\mathrm{RS}^{\mathrm{op}}$. These can be interpreted as the category of all local rings (with local homomorphisms) respectively as the category of all rings – where "all" means that not only ordinary rings are included, but also sheaves of rings on arbitrary topological spaces. The universal property then reads: $\mathrm{Hom}_{\mathrm{LRS}^{\mathrm{op}}}((\mathrm{Spec}(A),\mathcal{O}_{\mathcal{\mathrm{Spec}(A)}}), (X,\mathcal{O}_X)) \cong \mathrm{Hom}_{\mathrm{RS}^{\mathrm{op}}}((\mathrm{pt},A), (X,\mathcal{O}_X))$.
This says that $\mathcal{O}_{\mathrm{Spec}(A)}$ is the universal localization of $A$, that is the universal (initial) way to turn $A$ into a local ring. (In contrast, if you restrict your search to ordinary rings, i.e. sheaves of rings on the one-point space, then this optimization problem only has a solution if $A$ possesses exactly one prime ideal. See this MO answer by Peter Arndt for more on this point of view.)
Starting from this result, we might want to extend it to sheaves of rings $\mathcal{A}$. That is, given a sheaf of rings $\mathcal{A}$ on a topological space $Y$, we want to find a local sheaf of rings $\mathcal{A'}$ (living on some other space $Y'$) such that $\mathrm{Hom}_{\mathrm{LRS}^{\mathrm{op}}}(\mathcal{A}', \mathcal{O}_X) \cong \mathrm{Hom}_{\mathrm{RS}^{\mathrm{op}}}(\mathcal{A}, \mathcal{O}_X)$ for any local sheaf of rings $\mathcal{O}_X$ on any topological space $X$.
By the marvelous device of the internal language of a topos, to achieve this it suffices to give a construction of the spectrum of an ordinary ring for which the universal property stated in the beginning can be verified even in constructive mathematics.
Taken literally, this is not possible. Constructively, there may be nontrivial rings with no prime ideals, so that the spectrum as classically constructed is empty. This will then fail the universal property.
However, there is a remedy. In fact, there are multiple options. Joyal's is to construct the spectrum not as a topological space, but as a topos. (A topos may be nontrivial even if it has no global points, that is geometric morphisms from $\mathrm{Set}$ into the topos.) He does this by specifying a certain site (in fact the category of a certain preorder), constructing the sheaf topos on this site, and describing a certain local sheaf of rings in this topos. This topos is then a constructive replacement for the topological space $\mathrm{Spec}(A)$, and its local ring is the replacement for the structure sheaf of $\mathrm{Spec}(A)$.
Using the wonders of the internal language, Joyal's construction applies to any ring in any topos and yields a local ring in a (new) topos. In symbols, $\mathrm{Hom}_{\mathrm{LRT}^{\mathrm{op}}}(\mathrm{JoyalSpec}(\mathcal{A}), \mathcal{O}) \cong \mathrm{Hom}_{\mathrm{RT}^{\mathrm{op}}}(\mathcal{A},\mathcal{O})$ for any local ring $\mathcal{O}$ in any topos ($\mathrm{(L)RT}$ is the category of (locally) ringed toposes).
(Note that Joyal's construction is essentially the same as Hakim's in her thesis.)
A different remedy is to not go all the way to toposes, but construct $\mathrm{Spec}(A)$ as a locale. Its frame of opens can be neatly (and without recourse to prime ideals) described, it is the frame of radical ideals of $A$. The structure sheaf is then obtained by localizing the constant sheaf $\underline{A}$ at the generic filter, a certain subsheaf of $\underline{A}$. (A filter is a subset which fulfils precisely the dual axioms of that of a prime ideal, so that in classical logic a subset is a filter if and only if its complement is a prime ideal.) The universal property enjoyed by this construction is $\mathrm{Hom}_{\mathrm{LRL}^{\mathrm{op}}}(\mathrm{LocalicSpec}(\mathcal{A}), \mathcal{O}) \cong \mathrm{Hom}_{\mathrm{RL}^{\mathrm{op}}}(\mathcal{A},\mathcal{O})$ for any local sheaf of rings $\mathcal{O}$ on any locale ($\mathrm{(L)RL}$ is the category of (locally) ringed locales). The sheaf topos over $\mathrm{LocalicSpec}(\mathrm{A})$ coincides with the topos of Joyal's description.
To go full circle, a variant of this construction is to not give the frame of opens explicitly, but construct it as the Lindenbaum algebra of a certain propositional geometric theory. For any such theory, there is a locale whose points are precisely the models of the theory in $\mathrm{Set}$. In our context, we use the geometric theory of a filter in $A$. Its Lindenbaum algebra is then a locale (verifying the right universal property for the spectrum, even constructively) whose points are the filters in $A$ – so classically, its points are in one-to-one correspondence with the prime ideals of $A$. (See this math.SE answer for more details.)
Summarizing, we obtain a constructively sensible (topos-valid) construction of the spectrum (of any ring in any topos) simply by not considering the topological space of prime ideals (or filters), but by considering the locale of filters.
[Note for the curious: The locale of prime ideals yields the spectrum equipped with the flat topology. The locale of detachable prime ideals (or detachable filters, doesn't matter) yields the spectrum equipped with the constructible topology (sometimes called "patch topology"). In constructive mathematics, a subset $U$ of a set $X$ is detachable if and only if for any $x \in X$ either $x \in U$ or $x \not\in U$.]<|endoftext|>
TITLE: Which Banach spaces are realcompact?
QUESTION [6 upvotes]: I have a question about the topological space underlying a Banach space.
A topological space $X$ is realcompact iff it is homeomorphic to a closed subset of an infinite product of the form $\mathbb R^\kappa$. Closed subsets of realcompact spaces are realcompact.
A classical result in infinite topology states that every infinite dimensional separable Banach space is hemeomorphic to $\mathbb R ^ \omega$, so in particular, a separable Banach space is realcompact.
What about non-separable Banach spaces? Are they realcompact?
Since it is consistent with ZFC that there are discrete space which are not realcompact and every discrete set $X$ is a closed discrete subset of a space $\ell^p(X)$, it seems that in general the answer will be no. But of course, this counterexample is a huge space and in particular, it is not a ZFC-counter-example.
So, one question would be: Is there a ZFC-example of a Banach space which is not realcompact?
Another question would be: Is there an easy way to determine which Banach spaces are realcompact and which are not?
Thank you!

REPLY [8 votes]: Every metric space of nonmeasurable cardinality is realcompact. [1] 15.24.  Thus, if there are no measurable cardinals, then every metric space is realcompact  
As you noted...To get a Banach space that is not realcompact: Let $X$ be a set with measurable cardinal, and then the discrete topology on it is not realcompact.  So for example Banach space $l^1(X)$ has a closed subset homeomorphic to that, so is not realcompact.
aside
There is some nice work starting with [2] on when the weak topology of a Banach space is realcompact (or normal, or Lindelof...). 
[1] L. Gillmann & M. Jerison, Rings of Continuous Functions (1960)
[2] H. H. Corson, The weak topology of a Banach space, Trans. Amer. Math. Soc. 101 (1961), 1--15.<|endoftext|>
TITLE: Definitions of negative order Sobolev spaces
QUESTION [5 upvotes]: I am having a problem with the definition of the space $W^{-k,p}$. I use Adams's definition 
$$
W^{-k,p} = \left\{T \in D'(\Omega) \ \middle| \sum \limits_{0 \leq |i| \leq k} (-1)^{|i|} \int_{\Omega} v_i D^{i} \phi \,dx \ \forall \phi \in D(\Omega), 0 \leq |i| \leq k\right\},
$$
and he says in his book Sobolev spaces that the dual space $(W_0^{k,q})'$ is isometrically isomorphic to $W^{-k,p}$ for all $1 \leq q < \infty$, where $q$ is the conjugate index of $p$. My question is if this is true also for $q = \infty$. I would really appreciate if you could tell me a source where this is proven/ claimed. This would really help me because I am trying to show that $W^{-1,1} \subset H^{-k}$ for some $k$. Do you have any other idea how I could show this embedding without using $W^{-1,1}$ as a dual space. Thank you very much in advance.

REPLY [2 votes]: $W^{-k,p}$ is the dual of $W_0^{k,q}$ if $p>1$. For $p=1$ this is not a natural definition. You should use the alternative definition that $W^{-1,1}$ is the set of all distributions of the form $f_0+\sum_m{\partial f_m\over\partial x_m}$, where $f_0,f_m\in L^1$. Using this definition for embeddings into $H^{-k}$ spaces presents no problem.<|endoftext|>
TITLE: What is higher equivariant homotopy?
QUESTION [23 upvotes]: In Lurie's "Survey of elliptic cohomology" it is claimed that there exists some mystical "2-equivariant homotopy theory" for elliptic cohomology. The classical equivariant  elliptic cohomology is described as some non-explicit functor which associates a stack $M_G $ to any compact Lie group  $G $. It is claimed that this functor can be extended from Lie groups to "extensions of Lie groups by $B\mathbb G_m $", but neither it is explained what kind of objects such extensions really are nor what is really this higher equivariant homotopy. Naturally, it is also claimed that there exists "n-equivariant homotopy theory" for spectra associated to formal groups of higher height.
While I can take some guesses at parts of this statement (e.g. a $B\mathbb G_m $ extension can be considered either as a topological group or as a category in manifolds ), most of it remains a mystery to me. Some googling didn't help me to find any references. Thus the question: what is this higher equivariant theory and where can I read about it?

REPLY [11 votes]: I would also like to know the answer to your question.  Since no one has given an answer yet, I'll speculate recklessly and irresponsibly on how this might work (by riffing off of the final paragraph of section 5.1 of the survey).
Before figuring out what "higher equivariant homotopy theory" is, we should first know what "equivariant homotopy theory" is.  In this case, we want "global equivariant homotopy theory", aka the homotopy theory of smooth stacks, as introduced by Gepner-Henriques.  
They show that the homotopy theory of smooth stacks is modelled by the homotopy theory $PSh(Orb)$ of presheaves of infinity-groupoids on $Orb$.  Here $Orb$ is a topologically enriched category (i.e., infinity category) whose objects are compact Lie groups $G$, and whose morphism spaces are given by 
$$Orb(G,H) := BFun(G,H),$$
a classifying space of the category of smooth functors (so path components of $Orb(G,H)$ correspond to conjugacy classes of homomorphisms $\phi\colon G\to H$, and the path component containing $\phi$ is the classifying space of the centralizer of $\phi$)
Remark: there is an evident map $B\colon Orb(G,H)\to Map(BG,BH)$.  When $H$ is an extension of an abelian group by a torus, it is a weak equivalence; but it is not generally a weak equivalence otherwise.
Jacob's construction apparently associates to each derived oriented elliptic curve $E\to Spec(A)$ a functor $M\colon Orb\to Sch_A$ taking values in derived schemes over $A$.  The curve $E=M(U(1))$.  Section 5.1 suggests extending this construction by incorporating a "level".
So let's define $\widetilde{Orb}$ to have objects $(G,\ell)$, where $G$ is a compact Lie group and $\ell\colon BG \to K(\Lambda, 4)$ a map, where $\Lambda$ is a free abelian group.  The morphism space $\widetilde{Orb}((G,\ell), (G',\ell'))$ is the homotopy pullback of
$$Orb(G,G') \to Map(BG,BG') \to Map(BG, K(\Lambda',4)) \leftarrow Map(K(\Lambda,4), K(\Lambda',4)).$$
I.e., a "map" $(G,\ell)\to (G',\ell')$ is a homomorphism $\phi\colon G\to G'$ together with $f\colon K(\Lambda,4)\to K(\Lambda',4)$ and a homotopy $f\circ \ell \sim \ell'\circ B\phi$.
Now we can define 2-equivariant homotopy theory to be $PSh(\widetilde{Orb})$
Ideally, at this point you would identify $\widetilde{Orb}$ with something more geometric, e.g., some suitable category of Lie 2-groups.  Maybe you would even identify $PSh(\widetilde{Orb})$ with the homotopy theory of mumble mumble smooth 2-stacks mumble (I have no actual idea here).  However, we don't need to do this if we just want to play with homotopy theory: we just need $\widetilde{Orb}$.
Note that $Orb$ can be identified as the full subcategory of $\widetilde{Orb}$ consisting of $(G,0\colon BG\to K(0,4))$.  I think Jacob is claiming that the functor $M\colon Orb\to Sch_A$ can be extended to a functor from $\widetilde{Orb}$.  This wouldn't be an arbitrary extension, but would satisfy some compatiblities, perhaps including:

Compatibility with base change: given a homomorphism $\phi\colon H\to G$, require that $M(H,\ell\circ B\phi)\to M(G,\ell)$ be the evident base change along $M(H,0)\to M(G,0)$.
Given two levels $\ell\colon G\to K(\Lambda,4)$ and $\ell'\colon G\to K(\Lambda',4)$, $M(G,\ell\oplus \ell')$ should be the pullback of $M(G,\ell)\to M(G,0)\leftarrow M(G,\ell')$.  This would imply that that the values of $M$ are determined by those on objects on $Orb$, together with $(G,\ell)$ with $\ell\colon BG\to K(Z,4)$.  
The previous implies that $M(e,Be\to K(Z,4))$ has the structure of an abelian group object in $Sch_A$.  We would require that this object be $\mathbb{G}_m$.
Require every $M(G,BG\to K(Z,4))$  be a $\mathbb{G}_m$-torsor.

Then Jacob's theorem is possibly something like: an extension of $M$ satisfying a list of compatibilities is entirely determined by specifying a value at the cup-product level $\ell\colon BU(1)\times BU(1)\to K(Z,4)$, which itself should satisfy some properties/structure ("symmetric biextension of $M(U(1))=E$ by $\mathbb{G}_m$").
One might guess that n-equivariant homotopy theory is built from pairs $(G,\ell)$, where $\ell \colon BG\to Z$ is a map to some suitable type of  $(n+2)$-truncated space $Z$.  How this ought to be specified probably depends on what you want to accomplish with it.  I don't have any idea about that.<|endoftext|>
TITLE: Deep/precise relationship between two approaches to FLT for polynomials, $n = 3$
QUESTION [11 upvotes]: David Speyer commented the following here.

I saw Brian Conrad give an excellent one hour talk to undergraduates where he proved that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$ and $c(t) \in \mathbb{C}[t]$ such that
$$a(t)^3 + b(t)^3 = c(t)^3.$$
He gave an elementary proof, then outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve.

Here is my elementary proof, which I gave here.

Suppose there are some solutions of$$a(t)^3 + b(t)^3 = c(t)^3$$in $\mathbb{C}[t]$. Choose a solution $(a(t), b(t), c(t))$ such that the maximum $m > 0$ of the degrees of $a$, $b$, $c$ is minimal possible among all solutions. Clearly, this choice ensures that $a(t)$, $b(t)$, $c(t)$ are coprime. Then we have$$a(t)^3 = c(t)^3 - b(t)^3 = (c(t) - b(t))(c(t) - \omega b(t)) (c(t) - \omega^2 b(t)),$$where $\omega$ is a third primitive root of unity. Now, we look at the factors $c(t) - b(t)$, $c(t) - \omega b(t)$. Suppose that they have a common factor $q(t)$. Considering their sum and difference, we conclude that $c(t)$ and $b(t)$ have a common factor too. Moreover, $q(t)$ is a factor of $a(t)$. Thus, $a$, $b$, $c$ are not relatively prime, which is a contradiction. Repeating the same game with other pairs of factors, we see that all three factors $c(t) - b(t)$, $c(t) - \omega b(t)$, $c(t) - \omega^2 b(t)$ are pairwise coprime. Therefore,$$c(t) - b(t) = d_1(t)^3,\text{ }c(t) - \omega b(t) = d_2(t)^3,\text{ }c(t) - \omega^2b(t) = d_3(t)^3,\text{ where }d_1,\,d_2,\,d_3 \in \mathbb{C}[t].$$Note that$$\omega^2 + \omega + 1 = 0.$$Multiplying the second equation by $\omega$ and the third equation by $\omega^2$ and adding all three, we arrive at $$d_1(t)^3 + \omega d_2(t)^3 + \omega^2d_3(t)^3 = 0.$$Choosing $\eta_1$ and $\eta_2$ as any third roots of $-\omega$ and $-\omega^2$, respectively, and letting$$a_1 = d_1^3,\text{ }b_1 = \eta_1d_2^3,\text{ }c_1 = \eta_2d_2^3,$$we get$$a_1^3 = b_1^3 + c_1^3.$$By construction, at least one of $a_1$, $b_1$, $c_1$ is a nonconstant polynomial, and the maximum of their degrees is smaller than that of $a$, $b$, $c$. This is a contradiction to the choice of $a$, $b$, $c$.

Here is my better motivated "algebraic geometric" proof, which I gave here.

Let $a(t)$, $b(t)$, $c(t)$ have degree $n_1$, $n_2$, $n_3$, respectively, and $n = \text{max}(n_1, n_2, n_3)$. Then we can define $A(u, v)$, $B(u, v)$, $C(u, v)$ as homogeneous polynomials of degree $n$ such that$$A(u, 1) = a(u),\text{ }B(u, 1) = b(u),\text{ }C(u, 1) = c(u).$$Then, by construction,$$A(u, v)^3 + B(u, v)^3 = C(u, v)^3.$$Let$$E = \{(x, y, z) \in \mathbb{P}^2 : x^3 + y^3 = z^3\}.$$$E$ is a smooth curve of genus $1$, i.e. an elliptic curve. Now, define a map$$\varphi: \mathbb{P}^1 \to E,\text{ }(u, v) \mapsto (A(u, v), B(u, v), C(u, v)).$$This map is well-defined since $A(u, v)$, $B(u, v)$, $C(u, v)$ are homogeneous polynomials of the same degree which do not vanish simultaneously. Moreover, this map is nonconstant and proper since its source is projective. As the image of a proper map is closed and it is not a point, and $E$ is an irreducible $1$-dimensional variety follows that $\varphi$ is a surjective morphism. $E$ is a topologically a torus, i..e it has genus $1$ and there is a one up to scaling differential form $\tau$ on it. This will imply that its pullback $\varphi^*\tau$ is a differential form on $\mathbb{P}^1$, which is impossible since it has genus $0$. In more formal terms, a surjective map $\mathbb{P}^1 \to E$ gives rise to the injection $H^0(E, \Omega^1) \to H^0(\mathbb{P}^1, \Omega^1)$. But this is absurd, since the former is a vector space of dimension $1$ and the latter is a vector space of dimension $0$.
Perhaps you might want to say something about why the pullback of a non-zero holomorphic differential is non-zero.
$X$ and $Y$ are Riemann surfaces and $f: X \to Y$ holomorphic differential form looks like $g(z)\,dz$, where $g$ is a holomorphic function. Then its pullback is locally given by $g(f(z))\,df$. It is obviously nonzero as long as $f$ is nonconstant.

My question is, is there anyone who knows more algebraic geometry than me who can comment on the precise/possibly deep relationship between these two different proofs, i.e. why they are fundamentally the same or different?

REPLY [8 votes]: This answer is basically a longer version of Felipe Voloch's, but maybe it will be useful. Both proofs take a class in $H^1(E)$, pull it back to $H^1(\mathbb{P}^1)$ and note that $H^1(\mathbb{P}^1)$ is zero to conclude that the map $\mathbb{P}^1 \to E$ was constant. The difference is what form of $H^1$ we work with. 
The topological version is that a continuous map from a sphere to a torus must have degree zero; then holomorphic maps of degree zero are constant. 
Proof with differentials We take the class $dy/x$ in $H^0(E, \Omega^1)$ and pull it back to $H^0(\mathbb{P}^1, \Omega^1)=0$. This gives some equations on derivatives which imply that we are pulling back by the zero map. By Hodge theory, for any curve $X$, the sheaf cohomology $H^0(X, \Omega^1)$ is half of $H^1(X, \mathbb{C})$.
The proof by descent The map 
$$ \phi(x:y:z) = (x+\omega y + \omega^2 z : x+\omega^2 y + \omega z : - 3 x y z)$$
is a degree $3$ unbranched cover of $E$, giving a class in $H^1(E, \mathbb{Z}/3)$. If we pull this cover back to $\mathbb{P}^1$, then it must become trivial. Your computations are explicitly showing that, if $a(t)^3+b(t)^3+c(t)^3=0$, then we can find some $x(t)$, $y(t)$, $z(t)$ such that $(a:b:c)=\phi(x:y:z)$. So you have shown that this $\mathbb{Z}/3$ cover has a section, and is thus trivial. Abstractly, the reason there must be a section is that $H^1(\mathbb{P}^1, \mathbb{Z}/3)=0$.
Here is an abstract proof that $H^1(\mathbb{P}^1, \mathbb{Z}/3)=0$. Look at the short exact sequence $1 \to \mu_3 \to \mathcal{O}^{\ast} \to \mathcal{O}^{\ast} \to 1$, where $\mu_3$ is the group of cube roots of unity, $\mathcal{O}^{\ast}$ is the sheaf of nonvanishing holomorphic functions and the second map is cubing. (The topology is either analytic or etale.) Looking at the long exact sequence in cohomology, we see that $H^1(\mathbb{P}^1, \mu_3)$ is the $3$-torsion in $Pic(\mathbb{P}^1)$. But $\mathbb{C}[t]$ is a PID, so $Pic(\mathbb{A}^1)=0$ and $Pic(\mathbb{P}^1) \cong \mathbb{Z}$. You wind up using the exact same fact -- that $\mathbb{C}[t]$ is a PID -- when you write down the proof in an elementary way.
I remark that the first proof using a "de Rham" description of $H^1$ (by differential forms) and the second proof uses a "Betti" description (by covering maps). Relating de Rham and Betti descriptions is always complicated.<|endoftext|>
TITLE: Information and intuition packed in the Chern character for coherent sheaves
QUESTION [6 upvotes]: even after quite some time learning it, I still get somehow puzzled by the Chern character. Let me recall some stuff to get notation and setting.
Let us consider a smooth projective algebraic variety $X$ over $\mathbb{C}$. 
Whenever I have a coherent sheaf $F$ over $X$, I can get a resolution of $F$ by a (bounded) complex $E^\bullet$ of vector bundles on $X$ and define $\mathrm{ch}(F):=\sum (-1)^i\mathrm{ch}(E^i)$. Since I (should) know how to compute Chern classes of vector bundles via splitting principle, this makes sense.
I can indeed already start with a complex of coherent sheaves $F^\bullet$ on $X$, replace by vector bundles and do the exact same thing, defining a map $\mathrm{ch}\colon D^b(CohX)\to H^\ast (X,\mathbb{Q})$, or $CH^\ast(X)$, or $HH$ something, depending on your taste (I stick to the cohomology one), which actually factors through $K(X)$ because of additivity of $\mathrm{ch}$ with respect to short exact sequences.
The other good point of vector bundles is that we have an intuition of what Chern classes are, that is some kind of objects which measure wheter certain number of generic sections of our bundle are linearly dependent.
I am now trying to reconciliate the intuition with the homological nonsense, and I would like to have some example of what can we know about some sheaf when we know its Chern character.
A precious one for me would be the possibility to characterise torsion sheaves on $X$ (e.g. $\mathcal{O_x}$ skyscrapers, $\mathcal{O_Z}$ structure sheaf of a curve/subvariety) K3 surface.
Thank you for your time!

REPLY [5 votes]: I'm expanding on my comment, not really giving a complete answer.
Say that $L$ is a polarization on your smooth projective $X$. Then HRR applied to $F\otimes L^n$ gives
$$
P_L(F)(n)=\chi(F\otimes L^{\otimes n})=\int_X ch(F\otimes L^{\otimes n})\cdot td(X)=\int_X ch(F)\cdot e^{c_1(L)}\cdot td(X)
$$
where $P_L(F)$ is the Hilbert polynomial of $F$ with respect to $L$. So the Chern character recovers the Hilbert polynomial.
The Hilbert polynomial contains some information about your sheaf. For example its degree is the dimension of the support of $F$, and if $F$ is torsion-free (so that the degree of $P_L(F)$ is $dim(X)$), then, up to a constant term that depends only on $X$, the leading coefficient of $P_L(F)$ is the (generic) rank of $F$, and the next term gives you the degree of $F$.
If you didn't know already, you can find this stuff (and more) in the book "The geometry of moduli spaces of sheaves" by Huybrechts and Lehn.
I don't know about characterizing structure sheaves of subschemes on a K3 surface, but you have good chances of finding something in that same book (a large part of it is about sheaves on surfaces), or maybe someone else will comment here.<|endoftext|>
TITLE: K theory long exact sequence
QUESTION [6 upvotes]: (1) Suppose that $Z\subset X$ is a closed embedding, $U = X\setminus Z$ is the complement. If relevant, suppose that both $X, Z$ are smooth and even (if relevant) that the normal bundle of $Z\subset X$ is trivial. Then is it true that there is an exact triangle of complexes $K^* (Z)\to K^* (X)\to K^* (U) \to$? Here by $K$ theory I mean (connective) K theory of the category of perfect complexes. For what I need (this came up in the context of trace formulae), rational coefficients are enough. I found many places where very similar results are written down, but not this one explicitly.
(2) Suppose that, further, in the example above we have a (suitably flat) sheaf of smooth, compact (DG) algebras, $\mathcal{A}$, over $X$. Then we can study $K$ theory of categories of bundles of $\mathcal{A}$-modules over the three spaces. Again, is it true that these fall into an exact triangle as above?
It seems like both of these statements should follow from some colimit-compatibility of the $K$ theory functor, but I can't find a reference.

REPLY [3 votes]: You can combine Adeel Khan's answer with Proposition 6.9 of my paper with David Gepner to prove that there are these kinds of localization sequences in a great deal of generality. (Note that our proposition is simply an analogue for stable $\infty$-categories of a dg-categorical result of Toën.) So, if $A$ is a sheaf of (quasi-coherent) dg algebras on $X$, then this proposition shows that there is a fiber sequence $$K_Z(X,A)\rightarrow K(X,A)\rightarrow K(U,A).$$ Identifying the fiber term as $K(Z,i^*A)$, where $i:Z\rightarrow X$ is the inclusion and $i^*A$ is the derived pullback, is not something I've thought about. This kind of dévissage statement is much more difficult for dg algebras than it is for ordinary algebras, and it fails in some cases. There's a discussion of this in my paper with David and Tobias Barthel.<|endoftext|>
TITLE: Okounkov-Vershik approach to representation theory of $S_n$
QUESTION [21 upvotes]: This is a rather soft question. I was wondering if someone could explain on a fundamental and intuitive level, what the Okounkov-Vershik approach to representation theory of $S_n$ is all about. It's one thing to read a few chapters of a book or a paper, but it's another to try and understand the wider picture of the theory.
For reference, here is one of the main papers:
http://www.mat.univie.ac.at/~esiprpr/esi333.pdf
Any explanations would be gratefully received.

REPLY [15 votes]: First, I'll note that they provide a pretty clear explanation of their motivation on page two of that paper.   So it would strengthen your question if you indicated you had read that, and what about it you found unsatisfying.  
My own feeling is that the key observation of their approach is that the space of elements of $\mathbb{Q}[S_n]$ that commute with $\mathbb{Q}[S_{n-1}]$ is generated (edit: over the center of $\mathbb{Q}[S_{n-1}]$) by a single element, the Jucys-Murphy element $X_n=\sum_{i<n} (i,n)$.  This adds an additional structure of the functors of induction and restriction, which is the endomorphism of multiplication by $X_n$: on a restriction, you act in the obvious way, and on an induction, you send $g\otimes v\mapsto gX_n\otimes v$.  
Thus, you can decompose these functors according to the eigenvalues of $X_n$, which gives you much more control over the situation.  Remarkably, these subfunctors corresponding to different eigenvalues actually send simple modules to simple modules, and you can naturally pull the description of simples in terms of Young diagrams out of their interactions.<|endoftext|>
TITLE: Simple question: different definitions of Bousfield localization
QUESTION [12 upvotes]: I am not an expert on model categories and I am getting lost with two different definitions I have found on Bousfield localizations. I don't see the link between them.
First definition: Let $\mathbf{C}$ be a simplicial model category and $\mathrm{A}$ be a set of morphism. A map $f\colon V\to W$ is an $A$-local equivalence if for any $A$-local object $X$ the map of simplicial sets
$$
f^*\colon \mathrm{map} (W,X)\to \mathrm{map}(V,X)
$$
is a weak equivalence. Recall that $\mathrm{map}(\ , \ )$ denotes the homotopy function complex. More concretely, a simplicial set satisfying 
$$
\pi_0(\mathrm{map} (W,X))=\mathrm{Hom}_{\mathrm{Ho}(\mathbf{C})}(W,X)
$$
The definition of an $\mathrm{A}$-local object is given also in terms of $\mathrm{map}(\ , \ )$ (cf. Hirschhorn's Model Categories and their localizations 3.1.4).
Second definition: In many contexts of Voevodsky's motivic homotopy theory people use the following definition (cf Morel-Voevodsky's $\mathbb{A}^1$-homotopy theory of schemes or Riou's Categorie homotopique...). The model category is now the category of simplicial sheaves on the big (smooth) Nisnevich site with the simplical model structure (denote $\mathbf{H}_s(S)$ the homotopy category  where $S$ is a scheme) or the category of spectra with the level structure . For example, MV definition would say that a map $f\colon V\to W$ is an $\mathbb{A}^1$-local equivalence if for every $\mathbb{A}^1$-local object $X$ the induced map
$$
f^*\colon \mathrm{Hom}_{\mathbf{H}_s(S)}(W,X)\to \mathrm{Hom} _{\mathbf{H}_s(S)} (V,X)
$$
is a bijection. The definition of local objects is also given in terms of $\mathrm{Hom} _{\mathbf{H}_s(S)} (\ ,\ )$ and not $\mathrm{map}(\ , \ )$. Everyone in motivic homotopy theory states that this is a Bousfield localization.
Does anyone know how to prove this two definitions agree in motivic homotopy theory?

Suggestion: The closest to a link between these two definitions is M-V's result 2.2.8 where they state that it is equivalent for an object $X$ to be $\mathrm{A}$-local in the category of simplicial sheaves over a site and that 
$$
f^*\colon\underline{\mathrm{Hom}} (V,X)\to \underline{\mathrm{Hom}} (W,X)
$$
is a simplicial weak equivalence for any map $f$ in $\mathrm{A}$. However, note that $\underline{\mathrm{Hom}} (V,X)(S)=\mathrm{map}(V,X)$ but simplicial weak equivalences are weak equivalences at stalks, not globally.

REPLY [5 votes]: There are many equivalent ways of defining the local equivalences. Let $\mathcal{M}$ be a simplicial model category with a cofibrant replacement functor $Q : \mathcal{M} \to \mathcal{M}$. Then, for a fibrant object $Z$, define
$$\mathbf{R}\mathrm{Hom} (-, Z) = \mathrm{Hom} (Q {-}, Z)$$
where the RHS is the simplicially enriched hom-functor. (Note that we may take $Q = \mathrm{id}$ if every object in $\mathcal{M}$ is cofibrant.) This gives a very convenient model of the homotopy function complex.
Now, fix a class $\mathcal{R}$ of fibrant objects in $\mathcal{M}$ that is closed under cotensoring, i.e. if $Z$ is a member of $\mathcal{R}$ and $K$ is a simplicial set, then there is an object $Z^K$ in $\mathcal{M}$ that is a member of $\mathcal{R}$ with an isomorphism
$$\mathrm{Hom}_\mathbf{sSet} (K, \mathrm{Hom}_\mathcal{M} (-, Z)) \cong \mathrm{Hom}_\mathcal{M} (-, Z^K)$$
of simplicially enriched functors $\mathcal{M} \to \mathbf{sSet}$. For example, the class of (fibrant) $\mathcal{S}$-local objects in $\mathcal{M}$ is such a class.

Let $X \to Y$ be a morphism in $\mathcal{M}$. The following are equivalent:

For every object $Z$ in $\mathcal{R}$,
  $$\mathbf{R}\mathrm{Hom} (Y, Z) \to \mathbf{R}\mathrm{Hom} (X, Z)$$
  is a weak homotopy equivalence of simplicial sets.
For every object $Z$ in $\mathcal{R}$ and every simplicial set $K$,
  $$\pi_0 \mathrm{Hom} (K, (\mathbf{R}\mathrm{Hom} (Y, Z)) \to \pi_0 \mathrm{Hom} (K, \mathbf{R}\mathrm{Hom} (X, Z))$$
  is a bijection of sets.
For every object $Z$ in $\mathcal{R}$,
  $$\pi_0 \mathbf{R}\mathrm{Hom} (Y, Z) \to \pi_0 \mathbf{R}\mathrm{Hom} (X, Z)$$
  is a bijection of sets.


In particular, the $\mathcal{S}$-local equivalences can be defined entirely in terms of the homotopy category of $\mathcal{M}$. The same conclusion holds for ordinary model categories as well, but the details are more complicated because we do not have such a convenient model of the homotopy function complex.

REPLY [4 votes]: Yes, they are the same. In order to prove it, we should show that they have the same new weak equivalences (this is enough, because both have the same cofibrations). Have a look at Barwick's paper On Left and Right Model Categories and Left and Right Bousfield Localization. In 4.45 he defines what "Enriched left Bousfield localization" means. In the case of a closed symmetric monoidal category (i.e. enriched over itself), it means you can test for new weak equivalences via the internal hom, i.e. your second definition is what he calls enriched localization. Gutierrez has considered a similar notion under the term "closed localization."
In 4.46 Barwick proves that enriched localization exists and is simply the usual localization (your first definition) with respect to $I \Box S$, where $S$ runs through the maps you want to invert, $I$ is the set of generating cofibrations, and the box means pushout product. 
In your situation of interest, $I$ is obtained from the generating cofibrations in sSet, i.e. built freely from maps of the form $\partial \Delta^n \to \Delta^n$ for $n\geq 0$. In particular, every map in $S$ becomes a weak equivalence in definition 2. To show that every map in $I\Box S$ is already a weak equivalence in definition 1, look at Hirschhorn section 4.2. The way in which the model structure for definition 1 is built already includes taking the pushout product with $I$. So these maps in $I\Box S$ are trivial cofibrations in definition 1, hence weak equivalences, and we're done.
Note that we are using here that for a simplicial model category (like yours) that we can avoid framings and already have a nice formula for cosimplicial resolution. That's the real reason taking the pushout product with $I$ comes in, and why those maps in $I\Box S$ are new weak equivalences for Hirschhorn.<|endoftext|>
TITLE: P.G.Goerss, J.F.Jardine, "Simplicial Homotopy Theory" prerequisites
QUESTION [8 upvotes]: I know that such questions may be better suited for math.stackexchange, but I believe that that the topic of simplicial homotopy theory is advanced enough for mathoverflow.
Besides, I know that there are a lot of people working in homotopy theory who have probably at least used the book as a reference.
I know that, obviously, the main prerequisite is category theory(and algebra).
What I'm interested in is whether any amount of algebraic topology is assumed? Or it is a self-contained introduction to the subject using simplicial sets?
Can it really be an alternative way into algebraic topology/homotopical algebra/homotopy theory?

REPLY [17 votes]: As the commenters already argued, I would not regard this book as a self-contained introduction. For instance, from a brief browse through the introductory chapters:

The reader is assumed to be familiar with CW-complexes and several of the major theorems about them already which will be generalized (e.g. the Whitehead theorem).
The reader is assumed to be familiar with homotopy in the classical sense (e.g. they point out that "homotopy" isn't an equivalence relation on maps of simplicial sets as an implicit contrast to the case of spaces).
The reader is assumed to be familiar with other important tools: e.g. they say "Recall that the integral singular homology groups $H_*(X;\Bbb Z)$ of the space $X$ are defined to be ..." (this is on page 5) and they assume structural properties of it are known.
They describe the geometric realization of a simplicial set as
$$
|X| = \varinjlim_{(\Delta^n \to X)\text{ in }\Delta \downarrow X} |\Delta^n|
$$
which is certainly concise and categorically valid, but assumes that the reader already has some familiarity with the point of this construction and some of its basic properties. For example, more introductory references would discuss how each point of the realization is in the interior of exactly one n-cell, give a proof that the result is a CW-complex, etc.

Simplicial sets are a fundamental tool used basically everywhere in modern homotopy theory. However, the reason for this is that there are concrete technical problems which they solve. I realize that it might be tempting to try to skip ahead to get to the more advanced material, but it can be very difficult for a student to "get the point" without first understanding the more basic material.<|endoftext|>
TITLE: Compute the index of the Dirac operator on $C_0(R^2)$ to obtain Bott element in $K_0$
QUESTION [10 upvotes]: I am studying the paper of Baum-Connes-Higson to understand the Connes-Kasparov conjecture. In example 4.23, they discuss the case $G=\mathbb{R}^2$. I have constructed the Dirac operator, but I’m stuck with computing its index. According to the example in the article, we should obtain the Bott element in $K_0(C_r^*G))$, but I cannot see how to get there.
For the computation, we have the following setup (in the notation of [BCH])
$G=\mathbb{R}^2$, $K= \{ e \}$, $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ with $\mathfrak{k} =0$ and $\mathfrak{p} = \mathbb{R}^2$.
We take the following representation of the Clifford algebra on the spinor space $\Delta_2 = \mathbb{C}^2$:
$$
e_1 \mapsto \begin{pmatrix} 0 & 1 \\
-1 & 0 \end{pmatrix}, \; \; e_2 \mapsto \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}.
$$
The spinor bundle is the trivial bundle $S = \mathbb{R}^2 \times \mathbb{C}^2$, and the sections are given by:
$$
C_c^\infty(\mathbb{R}^2,S) =\left\{ \begin{pmatrix} f  \\ g \end{pmatrix}: \mathbb{R}^2 \to \mathbb{C}^2 \right\}.
$$
The bundle splits as $S = S^+ \oplus S^-$, where $f: \mathbb{R}^2 \to S^+$ and $g: \mathbb{R}^2 \to S^-$.
The Dirac operator $D: \Gamma(S) \to \Gamma(S)$ is given  by:
$$
D= 2 \begin{pmatrix} 0 & \frac{\partial}{\partial \overline{z}} \\
-\frac{\partial}{\partial z} & 0 \end{pmatrix},
$$
where $z= x + i y$.
Now as in [BCH], we Fourier transform, so $C_r^*(\mathbb{R}^2)$ becomes $C_0(\mathbb{R}^2)$, and the Dirac operator becomes:
$$
D= i \begin{pmatrix} 0 & z \\
- \overline{z} & 0 \end{pmatrix}.
$$
To calculate the index we only need $D^+: C_c^\infty(\mathbb{R}^2,S^+) \to C_c^\infty(\mathbb{R}^2,S^-)$. These modules complete to the Hibert modules $C_0(\mathbb{R}^2)$ over $A= C_0(\mathbb{R}^2)$, so we end up calculating the index of the operator:
$$
D^+: C_0(\mathbb{R}^2) \to C_0(\mathbb{R}^2), \; f \mapsto i z \cdot f.
$$
According to [BCH], the index of $D^+$ should give the Bott element, so:
$$ \text{Index}(D^+) = \text{ker}[ iz ] - \text{ker}[ -i\overline{z}] = \left[ \frac{1}{1+z^2} \begin{pmatrix} 1 & \overline{z} \\ z & |z|^2 \end{pmatrix} \right] - \left[  \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right].
$$
I first tried to compute it by explicitly finding a $T$ such that $1-TD^+$ and $1-D^+T$ are compact, but I couldn't find such $T$. Then I tried to look at the graph projector $G(tD)$, see (def 4.1, Aastrup) and use theorem 4.15. This seems to give the right result, but we need here that 0 is an isolated point in the spectrum of $D^+$, which is not true in our case. 
Is there a way to fix this, or another way to calculate the index?

REPLY [4 votes]: Solved it! (Edit: typo in formula for $\tilde{S}$ and $1-\tilde{T}\tilde{S}$, added i's and signs to obtain exactly Bott element)
Denote by $E=C_0(\mathbb{R}^2)$ our Hilbert module over $A=C_0(\mathbb{R}^2)$.
First, we should make $D$ into a bounded operator, we take:
$$
T=\frac{iz}{\sqrt{1+|z|^2}}.
$$
Also define:
$$
S = \frac{-i\overline{z}}{1+|z|^2},
$$
then $1-TS=1-ST= \frac{1}{1+|z|^2}$, which is a compact operator on $E$. Hence $T$ is $A$-Fredholm, according to (def 3.1, Exel).
Now as $T$ is non-regular, we apply the procedure of (lemma 3.8, Exel) and define:
$$
\tilde{T} = \frac{1}{\sqrt{1+|z|^2}}\begin{pmatrix} i z & 0 \\ - i & 0 \end{pmatrix}, \tilde{S} = \frac{1}{\sqrt{1+|z|^2}}\begin{pmatrix} -i \overline{z} & i \\ 0 & 0 \end{pmatrix}.
$$
Then the index of $D$ is then given by (see definition 3.10):
$$
[\ker \tilde{T} ] - [ \ker \tilde{T^*}] = [1-\tilde{S}\tilde{T}] - [ 1 - \tilde{T} \tilde{S} ] \in K_0(A).
$$
We have:
$$
1-\tilde{T} \tilde {S} = 1 - \frac{1}{1+|z|^2} \begin{pmatrix} i z & 0 \\ -i & 0 \end{pmatrix} \begin{pmatrix} -i\overline{z} & i \\ 0 & 0 \end{pmatrix} = \frac{1}{1+|z|^2} \begin{pmatrix} 1 & z \\ \overline{z} & |z|^2 \end{pmatrix}
$$
and
$$
1-\tilde{S}\tilde{T} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix},
$$
so we get precisely the Bott element.<|endoftext|>
TITLE: Strongly real elements of odd order in sporadic finite simple groups
QUESTION [9 upvotes]: Recall that an element of a finite group is said to be real if it is conjugate to its inverse, and strongly real if the conjugating element can be chosen to be an involution.
Question: Is it true that for the 26 sporadic finite simple groups, all real elements of odd order are strongly real, apart from elements in the Atlas classes 3A and 5A of the McLaughlin simple group $M^cL$?
My question is motivated by this Mathoverflow question of A.Rupinski: Why are there so few quaternionic representations of simple groups ? . As noted there, $M^cL$ is the only sporadic finite simple group which has `quaternionic' representations. From the Atlas, the irreducible characters $\chi_{10}$ and $\chi_{13}$ of $M^cL$ each have Frobenius-Schur indicator $-1$.
The literature on `strongly real' finite simple groups usually looks at all conjugacy classes, not the classes of odd order elements.
I suspect that the number of (irreducible) quaternionic representations of a finite group is greater than or equal to the number of real conjugacy classes of odd order elements which are not strongly real. This might even be a known open conjecture.
PS
General discussion on relation between numbers of real/complex/quaternionic 
conjugacy classes and irreducible representations can be found here:
MO46900: Are there “real” vs. “quaternionic” conjugacy classes in finite groups?

REPLY [4 votes]: One more confirmation of OP's observation can be found in tables,
from the paper:
Jordan Journal of Mathematics and Statisticscs (JJMS) l(2), 2008, pp. 97-103
97
STRONGLY REAL ELEMENTS IN
SPORADIC GROUPS AND ALTERNATING GROUPS, 
IBRAHIM SULEIMAN
One can observe many real, but not strongly real classes, but most of them
have even order, except group McL classes 3A, 5A - exactly as OP's proposal.
Table 3 will give the real classes which are not strongly real in Sporadic Groups
M11 all real elements are strongly real.
M12 all real elements are strongly real
M22 8A.
M23 8A
HS all real elements are strongly real.
J3 all real elements are strongly real.
M24 all real elements are strongly real.
McL 3A, 5A, 6A.
He all real elements are strongly real.
Ru all real elements are strongly real. 
Suz all real elements are strongly real.
ON all real elements are strongly real.
Co3 all real elements are strongly real.
Co2 16B.
Fi22 all real elements are strongly real.
HN 8A .
Ly all real elements are strongly real.
Th 8B.
Fi23 16AB, 22BC, 23AB.
Co1 all real elements are strongly real.
J4 all real elements are strongly real.
Fi_24 all real elements are strongly real.
BM all real elements are strongly real.
M 8C, 8F, 24F, 24G, 24H, 24J, 32A, 32B, 40A, 48A. 

Table 4 will give the complete list for all non real elements in the Sporadic Groups
M11 8AB, 11AB.
M12 11AB.
J1 None
M22 7AB.
J2 None
M23 7AB, 11AB, 14AB, 15AB, 23AB.
HS 14AB.
J3 19AB.
M24 7AB, 14AB, 15AB, 21AB, 23AB.
McL 7AB, 9AB, 11AB, 14AB, 15AB, 30AB.
He 7AB, 7DE, 14AB, 14CD, 21CD, 28AB.
Ru 16AB.
ON 31AB.
Co3 11AB, 22AB, 20AB, 23AB.
Co2 14AB, 15BC, 23BC, 30BC.
Fi22 11AB, 16AB, 18AB, 22AB.
HN 19AB, 35AB, 40AB.
Ly 11AB, 22AB, 33AB. 
Fi23 16AB, 22BC, 23AB.
Co1 23AB, 39AB.
J4 7AB, 14AB, 14CD, 21AB, 28AB, 35AB, 42AB.
Fi_24 18GH, 23AB.
BM 23AB, 30GH, 31AB, 32CD, 46AB, 47AB.
 M 23AB, 31AB, 39CD, 40CD, 44AB, 46AB, 46CD, 47AB, 56BC, 59AB,
62AB, 69AB, 71AB, 78BC, 87AB, 88AB, 92AB, 93AB, 94AB, 95AB,
104AB, 119AB.<|endoftext|>
TITLE: What is known about the large cardinal strength of Shelah's categoricity conjecture?
QUESTION [7 upvotes]: Shelah's categoricity conjecture states that for every Abstract Elementary Class $\mathcal{K}$ there is a cardinal $\mu$ depending only on $\operatorname{LS}(K)$ (i.e. the Löwenheim–Skolem number of $\mathcal{K}$) such that if $\mathcal{K}$ is categorical in some $\lambda \geq \mu$ (i.e. $\mathcal{K}$ has exactly one  model of size $\lambda$ up to isomorphism), then $\mathcal{K}$ is categorical in $\theta$ for all $\theta \ge \mu$.
Due to a result of Will Boney it is known that a weaker form of this conjecture is consistent up to consistency of proper class many strongly compact cardinals. My question is about the inverse direction. 

Question: Does Shelah's categoricity conjecture have large cardinal strength? If there is no known result in this direction, what are some known notable connections or theorems that suggest possible existence of some large cardinal strength for this conjecture or an approach for proving such a statement? 

The only connection that I'm aware of and seems related to the problem of the consistency strength of Shelah's categoricity conjecture is a connection that exists between abstract elementary classes and accessible categories on one hand and accessible categories and Vopěnka's principle on the other hand. The latter appears in the following characterizations of Vopěnka's principle in terms of category theory (see here):
Theorem: The followings are equivalent:
(1) The Vopěnka's principle.
(2) Every discrete full subcategory of a locally presentable category is small.
(3) For $C$ a locally presentable category, every full subcategory $D↪C$ which is closed under colimits is a coreflective subcategory.

REPLY [3 votes]: Please note that the paper by Boney & Unger establishes (with Bone's earlier result) that tameness is a large cardinal axiom. It does not shade light on the original question.
Under the assumptions of tameness and the amalgamation property, strong versions of Shelah's categoricity conjectures are ZFC theorems. See several papers by Sebastien Vasey and others.  There is much evidence that an exciting classification theory can be developed for such AECs (tame and AP). 
While some people believe that tameness and the AP could be derived from categoricity above a certain Hanf number (I was the first to make such conjectures).  This is not clear at all.  I can imagine that a couterexample to Shelah's categoricty conjecture exists.<|endoftext|>
TITLE: Equalizing Geometric means of Graph Cycles
QUESTION [8 upvotes]: Consider a strongly connected directed graph $G$.  I have been stuck on the following question:  can you assign real numbers in $[0,1]$ to each edge of $G$ so that the geometric mean of all cycles are equal?  Also, I would like all outgoing edges to sum to 1.
For example, $p=\varphi^{-2}$ is a solution for the graph shown below, where $\varphi$ is the Golden ratio.

The geometric means for the two cycles are $(p\cdot1)^{1/2}=\varphi^{-1}$ and $(1-p)=(1-\varphi^{-2})=\varphi^{-1}$.  The appearance of the Golden ratio here can be explained by observing the relation to sequences with no consecutive 1's (the number of such sequences of length $n$ is approximately $\varphi^n$).
Even though the number of constraints is much larger than the number of variables, I have not been able to construct a counterexample and I also don't know of a simple/direct proof of existence (I am aware of a purported indirect proof that exploits the ergodicity of the Markov chain - I would like to generalize this).  I came across this question in my study of Markov chains.  I have tried many numerical examples, and such a solution always exists (and is unique in my experience).
Any ideas, hints or references would be greatly appreciated!
Question is also on math.stackexchange (with some comments, but no answers).

REPLY [6 votes]: I am not 100% sure I am not misusing the Perron-Frobenius Theorem, but I think that it justifies all the assumptions I am going to make in the following. The final construction itself is very simple.
Let $V$ denote the vertex set of the underlying graph and let $A$ be the adjacency matrix, that is $A_{v,w} = 1$ if $v \to w$ and otherwise $A_{v,w} = 0$. This matrix is non-negative and irreducible, because the graph was strongly connected. Hence there is a positive eigenvalue $\lambda$ together with a right-eigenvector $c$ all of whose entries are positive. Moreover, for every $v \in V$ we have $(Ac)_v = \lambda c_v$, that is:
$$
\sum_{w \colon v \to w} c_w = \lambda c_v.
$$
So in particular (as all entries are positive) we have that for every directed edge $v \to w$, the value $p(v \to w) := \frac{c_w}{\lambda c_v}$ is well-defined and lies in the interval $(0, 1]$.
With this definition, we have for every vertex $v \in V$
$$
\sum_{w \colon v\to w} p(v \to w) = \frac{1}{\lambda c_v}(Ac)_v = 1.
$$
Along every directed cycle, all the terms $\frac{c_v}{c_w}$ cancel out, and the accumulating power of $\lambda^{-1}$ is taken care of by definition of the geometric mean; hence every cycle has geometric mean $\lambda^{-1}$.<|endoftext|>
TITLE: Factorization of a matrix as a product of a symmetric and a skew-symmetric matrix
QUESTION [12 upvotes]: When can an $n\times n$ matrix $M$ be written as a product $M=AB$, where $A^T=A$ and $B^T=-B$?


For example, a necessary condition is that the trace of $M$ vanishes. In this case, it is easy to check that for $n=2$ all traceless matrices $M$ admit such a factorization.


Is there a way to know all pairs of matrices $A$ and $B$ as above, for a given $M$?

What conditions should satisfy $M$ so that $A$ is non-singular, and what are all factorizations with $A$ non-singular?



For example, if $n=2$ and $M$ is non-singular, $A$ and $B$ are obtained easily, and are unique up to a scalar factor and non-singular. But it is possible for $A$ to be non-singular even when $M$ is singular, for example if $M=0_n$, $M=A0_n$ is a solution for any $A$.
I expect that there is literature about this kind of factorization, but I couldn't find anything.

REPLY [9 votes]: This set of questions was considered in a recent paper by Stenzel:
https://www.evernote.com/l/ABj_1ego5jZORbVzwCyxe_EL4EJVbk-4XQA
OK, it was 1920, but it seems like yesterday.<|endoftext|>
TITLE: Can all the sporadic groups be expressed as permutation groups based on a single big cycle?
QUESTION [8 upvotes]: Working on M11, I came up with that it can be generated using the following permutations:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[[2, 0, 1, 7], [3, 4, 5, 6]]
[[4, 0, 6, 7], [2, 3, 1, 5]]
[[0, 7], [4, 6], [3, 5], [1, 2]]
The summary of that is that there's one big permutation which cycles everything, and a few more which allow GF(8) on the first 8 elements.
Do all the other sporadic groups have generators with a single cycle of everything and a few which work on the minimum number of other elements possible? If so, what are they?

REPLY [20 votes]: If a permutation group $G$ on a set of size $n$ contains an $n$-cycle then the subgroup $C$ generated by this $n$-cycle is transitive. Thus we have a factorisation $G=CG_\alpha$. Moreover, $C\cap G_\alpha=1$. All factorisations of the sporadic simple groups were determined by me in my paper 'Factorisations of sporadic simple groups' in the Journal of Algebra in 2006. This shows that the 11-cycle in the action of $M_{11}$ on 11 points and the 23-cycle in the action of $M_{23}$ on 23 points are the only examples.<|endoftext|>
TITLE: Improving Baumgartner's result?
QUESTION [8 upvotes]: Q1: Is it consistent with the failure of CH to have an $\aleph_1$-dense subset $A \subseteq \mathbb{R}$ such that for every $X \subseteq \mathbb{R}$ of size $\aleph_1$, there is a $C^{\infty}$ map $F: \mathbb{R} \to \mathbb{R}$ sending $X$ into $A$?
Q2: Is it consistent with the failure of CH to have an $\aleph_1$-dense subset $A \subseteq \mathbb{C}$ such that for every $X \subseteq \mathbb{C}$ of size $\aleph_1$, there is an analytic function $F: \mathbb{C} \to \mathbb{C}$ sending $X$ into $A$?
What I know: If we replace $C^{\infty}$ by continuous, a positive answer to Q1 follows from PFA. If we replace analytic by continuous in Q2, a positive answer follows from MA.
Thanks!

REPLY [5 votes]: The answer to your first question is no. Proposition 9.4 in Abraham, Rubin and Shelah "On the consistency of some partition theorems..." implies that there are $\aleph_1$-dense sets $A, B \subseteq \mathbb{R}$ such that there is no $C^1$ function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f[A] \subseteq B$. See also Proposition 1.2 in M. Burke, "Entire functions mapping uncountable dense sets of reals onto each other monotonically". On the other hand, if you only consider $A$ and $B$ which are everywhere of second category, you get a positive answer. This is Theorem 2.1 in the second paper.<|endoftext|>
TITLE: What characterizations of relative information are known?
QUESTION [31 upvotes]: Given two probability distributions $p,q$ on a finite set $X$, the quantity variously known as relative information, relative entropy, information gain or Kullback–Leibler divergence is defined to be 
$$          D_{KL}(p\|q) = \sum_{i \in X} p_i \ln\left(\frac{p_i}{q_i}\right)  $$
This is a natural generalization of the Shannon information of a single probability measure.   
I wrote a paper with Tobias Fritz giving a category-theoretic characterization of this quantity:

John Baez and Tobias Fritz, A Bayesian characterization of relative entropy, Th. Appl. Cat. 29 (2014), 421–456.   

Belatedly I'm wondering what other characterizations are known.  On Wikipedia I read:

Arthur Hobson proved that the Kullback–Leibler divergence is the only measure of difference between probability distributions that satisfies some desiderata, which are the canonical extension to those appearing in a commonly used characterization of entropy.

Unfortunately they don't give these desiderata, and there are lots of different characterizations of entropy, so I don't know which one is meant here.   I also don't have quick access to Hobson's book:

Arthur Hobson, Concepts in Statistical Mechanics, Gordon and Breach, New York, 1971.

Wikipedia also gives a characterization in terms of coding theory:

The Kullback–Leibler divergence can be interpreted as the expected extra message-length per datum that must be communicated if a code that is optimal for a given (wrong) distribution $q$ is used, compared to using a code based on the true distribution $p$.

This is made precise in Theorem 5.4.3 here:

T. M. Cover and J. A. Thomas, Elements of Information Theory, 2nd Edition, Wiley-Interscience, New York, 2006.

What other characterizations of relative entropy are known?

REPLY [7 votes]: I don't have Hobson's book, but I do have a paper by Hobson in which he proves the theorem that Wikipedia is presumably referring to:

Arthur Hobson, A new theorem in information theory.  Journal of Statistical Physics 1 (1969), 383-391.

(What a title.  Long gone are the days when you could get away with that!)
Here's Hobson's theorem.  Let $I(p; q)$ be defined for any pair of probability distributions $p, q$ on a finite set.  Suppose it satisfies the properties below.  Then $I$ is a constant multiple of relative entropy.
Before I list the properties, let me make a comment: the phrasing "let ... be defined" is his.  He's not precise about the codomain of the function $I$.  I see no mention of the fact that relative entropy can be infinite, and it may be tacitly assumed that $I(p; q)$ is always nonnegative.
His properties:

$I$ is a continuous function of its $2n$ variables.  (Again, I don't know how/whether he handles infinities.  As John and Tobias make clear in their paper, continuity as relative entropy tends to $\infty$ is actually a tricky point.) 
$I$ is permutation-invariant (doing the same permutation to both arguments).  Actually, he only states invariance under a single transposition, but obviously that's equivalent.
$I(p; p) = 0$ for all $p$.
$I((1/m, \ldots, 1/m, 0, \ldots, 0); (1/n, \ldots, 1/n))$ is an increasing function of $n$ and a decreasing function of $m$, for all $n \geq m \geq 1$.  (I don't know whether he means increasing and decreasing in the strict or weak sense.)
We have
\begin{align*}
&
I((p_1, \ldots, p_m, p_{m + 1}, \ldots, p_n); (q_1, \ldots, q_m, q_{m + 1}, \ldots, q_n)) \\
&
=
I((P, P'); (Q; Q')) + P\cdot I((p_1/P, \ldots, p_m/P); (q_1/Q, \ldots, q_m/Q))\\
&
\quad + P'\cdot I((p_{m + 1}/P', \ldots, p_n/P'); (q_{m + 1}/Q', \ldots, q_n/Q'))
\end{align*}
for all probability distributions $p$ and $q$ and all $m \in \{1, \ldots, n\}$, where we have put $P = p_1 + \cdots + p_m$, $P' = 1 - P$, $Q = q_1 + \cdots + q_m$, and $Q' = 1 - Q$.  (Although he divides by $P$, $P'$, $Q$ and $Q'$ here, he doesn't seem to say what to do if they're zero.)

This theorem is weaker than the result in my earlier answer.  In that other answer, his continuity axiom is replaced by measurability, the permutation-invariance and vanishing axioms are the same, and his fourth axiom (about increasing/decreasing) just isn't there.  The last axioms in both lists look different, but are actually equivalent.  This takes a bit of explanation, as follows.
As usul points out, most theorems characterizing entropies involve one axiom looking like these ones.  In fact, both Hobson's last axiom and the last axiom in my earlier answer are special cases of the following general axiom, which I'm told is known as the "chain rule".
To state it, let me introduce some notation.  Given probability distributions $\mathbf{w}$ on $n$ elements, $\mathbf{p}^1$ on $k_1$ elements, ..., $\mathbf{p}^n$ on $k_n$ elements, we get a distribution
$$
\mathbf{w} \circ (\mathbf{p}^1, \ldots, \mathbf{p}^n)
=
(w_1 p^1_1, \ldots, w_1 p^1_{k_1}, \ \ldots, \ w_n p^n_1, \ldots, w_n p^n_{k_n})
$$
on $k_1 + \cdots + k_n$.  Using that notation, the chain rule for relative entropy states that 
$$
D\bigl(\mathbf{w} \circ (\mathbf{p}^1, \ldots, \mathbf{p}^n) \,\|\, \tilde{\mathbf{w}} \circ (\tilde{\mathbf{p}}^1, \ldots, \tilde{\mathbf{p}}^n)\bigr)
=
D(\mathbf{w} \,\|\, \tilde{\mathbf{w}}) + \sum_{i = 1}^n p_i D(\mathbf{p}^i \,\|\, \tilde{\mathbf{p}}^i).
$$
It's easy to see that both Hobson's last axiom and the last axiom in my earlier answer are special cases of this general rule.  But by simple inductive arguments, either one of these special cases implies the general case.  That's why I say they're equivalent.
Whether you find the general case or one of the special cases more appealing is a matter of taste.<|endoftext|>
TITLE: Smooth manifolds as idempotent splitting completion
QUESTION [24 upvotes]: The nlab has a particularly interesting thing to say about the category of smooth manifolds: it is the idempotent-splitting completion of the category of open sets of Euclidean spaces and smooth maps.
After proving this, the following excerpt from a paper of Lawvere (cited below) is given.

“This powerful theorem justifies bypassing the complicated considerations of charts, coordinate transformations, and atlases commonly offered as a ”basic“ definition of the concept of manifold. For example the $2$-sphere, a manifold but not an open set of any Euclidean space, may be fully specified with its smooth structure by considering any open set $A$ in $3$-space E which contains it but not its center (taken to be $0$) and the smooth idempotent endomap of $A$ given by $e(x)=x/|x|$. All general constructions (i.e., functors into categories which are Cauchy complete) on manifolds now follow easily (without any need to check whether they are compatible with coverings, etc.) provided they are known on the opens of Euclidean spaces: for example, the tangent bundle on the sphere is obtained by splitting the idempotent $e'$ on the tangent bundle $A\times V$ of $A$ ($V$ being the vector space of translations of $E$) which is obtained by differentiating $e$. The same for cohomology groups, etc.” (Lawvere 1989, p.267)

Unfortunately the excerpt is not enough for me to understand neither the significance nor the idea behind the theorem, so I am looking for detailed, hand-holding explanations of as many parts of it as possible.

How does this theorem justify bypassing the considerations of charts, atlases, etc?
What are the details of the sphere example? How is the smooth structure specified by an open set containing it along with $x/|x|$?
Why are all general constructions in fact functors into cauchy complete categories?
What are some examples of general constructions and how do they follow easily? How does this approach circumvent messing with covers etc?
What is meant by "the same for cohomology groups"?

Reference: F. William Lawvere, Qualitative distinctions between some toposes of generalized graphs, Contemporary Mathematics 92 (1989), 261-299.

REPLY [11 votes]: The theorem is the following: (from 1.15 of here)

Theorem: Let $M$ be a connected manifold and
suppose that $f:M\to M$ is smooth with $f\circ f= f$. Then the
image $f(M)$ of $f$ is a submanifold of $M$.

Proof: We claim that there is an open neighborhood $U$ of
$f(M)$ in $M$ such that the rank of $T_yf$ is constant for $y\in
U$. Then by the constant rank theorem 1.13 of loc.cit. the result follows.
For $x\in f(M)$ we have $T_xf\circ T_xf = T_xf$; thus $\text{image} T_xf =
\ker (Id-T_xf)$ and $\text{rank} T_xf + \text{rank} (Id-T_xf) = \dim M$.
Since $\text{rank} T_xf$ and $\text{rank} (Id-T_xf)$ cannot fall locally,
$\text{rank} T_xf$ is locally constant for $x\in f(M)$, and since
$f(M)$ is connected, $\text{rank} T_xf = r$ for all $x\in f(M)$.
But then for each $x\in f(M)$ there is an open neighborhood
$U_x$ in $M$ with $\text{rank} T_yf\geq r$ for all $y\in U_x$. On the
other hand 
$$
\text{rank} T_yf = \text{rank} T_y(f\circ f) = \text{rank} T_{f(y)}f\circ T_yf\leq \text{rank} T_{f(y)}f =r
$$
since $f(y)\in f(M)$. So the neighborhood we need is
given by $U = \bigcup_{x\in f(M)}U_x$.
This result can also be expressed as: `smooth retracts' of
manifolds are manifolds. If we do not suppose that $M$ is
connected, then $f(M)$ will not be a pure manifold in general;
it will have different dimensions in different connected components.
Consequences: 1. The (separable) connected
smooth manifolds are exactly the smooth retracts of connected
open subsets of $\mathbb R^n$'s.
2. A smooth mapping 
$f:M\to N$ is an embedding of a submanifold if and only if
there is an open neighborhood $U$ of $f(M)$ in $N$ and a smooth
mapping $r:U\to M$ with $r\circ f=Id_M$.
Proof: Any manifold $M$ may be embedded into some $\mathbb
R^n$; see \nmb!{1.19} below. Then there exists a tubular
neighborhood of $M$ in $R^n$ , and $M$ is
clearly a retract of such a tubular neighborhood.
For the second assertion we
repeat the argument for $N$ instead of $\mathbb R^n$.
Edited and extended:
Now to your questions, as I understand them: You can reconstruct the category of smooth manifolds and smooth mappings (connect, or not connected but then not pure)
as follows: Objects are $(U,f)$ with $U$ open in some $\mathbb R^n$ and $f:U\to U$ smooth with $f\circ f= f$. Morphisms $h:(U,f)\to (V,g)$ are smooth maps $h:U\to V$ with $h\circ f = g\circ h$. A point of the manifold $(U,f)$ is any $x\in U$ with
$x=f(x)$.
Then you have to describe diffeomorphisms and get rid of the redundancy in the description of a manifold. Note that even in the classical sense, if you write a manifold $M$, you think of it either with a special atlas, or with the atlas comprised of all possible smooth charts, or $\dots$ Somehow, we mentally identify diffeomorphic manifolds. Let me try: 
The identity morphism is any $\ell:(U,f)\to (U,f)$ with $f\circ \ell = f$.
A diffeomorphism $h:(U,f)\to (V,g)$ is one which admits $\ell:(V,g)\to (U,f)$ with $f\circ \ell\circ h =f$ and $g\circ h\circ \ell = g$.   
The tangent bundle of a manifold $(U,f)$ is then just
$T(U,f) = (TU, Tf)$ as a manifold. For the vector bundle structure note that
$TU = U \times \mathbb R^n$, and for  a point $x$ in $(U,f)$, i.e., $x\in U$ with $f(x)=x$, we have $T_xf\circ T_xf = T_xf$ a projection in $\mathbb R^n$ whose image is the fiber of the tangent bundle.
I hope that I did not overlook anything.<|endoftext|>
TITLE: Definition of a normed ring
QUESTION [7 upvotes]: A normed ring "should" be a monoid object in the monoidal category of normed abelian groups. There are (at least) two choices of morphisms of normed groups, namely bounded or short homomorphisms, resulting in two definitions of a normed ring: Concretely, a normed ring is a ring $R$ (not necessarily commutative) together with a map $R_{\mathsf{Set}} \to \mathbb{R}_{\geq 0},~ x \mapsto |x|$ such that for all $x,y \in R_{\mathsf{Set}}$

$|x+y| \leq |x|+|y|$
$|{-}x|=|x|$
$|x|=0 \Leftrightarrow x=0$,

and now either (choosing short homomorphisms)

$|1| \leq 1$ and $|x y| \leq |x| |y|$ for all $x,y$,

(which easily implies $R = 0$ or $|1|=1$), equivalently

$|x_1 \cdot \dotsc \cdot x_n| \leq |x_1| \cdot \dotsc \cdot |x_n|$ for every finite sequence of elements $x_1,\dotsc,x_n$ (including the empty one!),

or (choosing bounded homomorphisms)

there is a constant $K \geq 0$ with $|xy| \leq K|x||y|$ for all $x,y$

(but no condition for the unit). In all the references that I have found so far, these two definitions are somewhat mixed. For example, $|xy| \leq |x||y|$ is assumed, but nothing about the unit. Isn't this a mistake? Also, I have almost never found the condition $|{-}x|=|x|$, but this doesn't seem to follow from the rest, right? What do you think about the stronger multiplicativity condition $|xy|=|x||y|$? This prevents zero divisors, which is certainly useful, but should one really put this into the definition? Also, which of the two definitions (short/bounded) is preferred in which contexts, and why? I would like to ask similar questions for normed modules over a normed ring, but maybe this will be a separated post.

REPLY [3 votes]: Throughout, I'll work with your "short" axioms.  I'll begin by addressing your questions about independence of the axioms.
First, the axiom $||1||\leq 1$ does not follow from the other axioms.  To see this take $R=\mathbb{Z}$, let $c>1$ be a real number, and let your norm be given by $||x||=|x|\cdot c$ (where by $|x|$, I mean the usual absolute value of $x\in \mathbb{Z}$).
Second, the axiom $|-x|=|x|$ does not follow from the others.  To see this again take $R=\mathbb{Z}$, let $c_1\neq c_2>1$ be real numbers with $c_2^2\geq c_1$, and let your norm by given by $$||x||=\begin{cases}|x|\cdot c_1 & \text{ if }x\geq 0\\ |x|\cdot c_2 & \text{ if }x<0\end{cases}.$$  A quick check should show this satisfies all of the other axioms, if I didn't make any mistake.
Third, I believe that Jérôme Poineau's comment is incorrect, in that by dividing by $||1||$ you can lose the norm property.  To see this, take $R=\mathbb{Z}[y,y^{-1}]$, let $c>1$ be a real number, and let your norm be given by $$||\sum_{\rm finite}a_m y^m|| = \sum_{m<0}|a_m|+\sum_{m\geq 0}|a_m|\cdot c.$$  Again, if I didn't make any mistakes, this should satisfy all the norm axioms except that $||1||=c>1$.  However, if you try to create a new norm $||f||'=||f||/||1||$, this loses the multiplicative property since $$||y\cdot y^{-1}||'=||1||'=1> 1\cdot c^{-1}=||y||'\cdot ||y^{-1}||'.$$
As for your question about which definition is preferred, I'm not an expert and will leave that to others.<|endoftext|>
TITLE: Hypothesis test beyond simple hypotheses (mathematical statistics)
QUESTION [6 upvotes]: In mathematical statistics, the following problem (simple hypothesis test) is considered: given a data sample, test the hypothesis $H_0$ stating that all sampled values are values of a random variable with a given distribution $P_0$ against the hypothesis $H_1$ stating that all sampled values are values of a random variable with another given distribution $P_1$. 
The question I am dealing with is a bit more general: test the same hypothesis $H_0$ against the hypothesis $H_1'$ stating that some of the sampled values are values of a random variable with a given distribution $P_0$, and some others are values of a random variable with another given distribution $P_1$. 
The original question is usually settled using Neyman-Pearson Lemma (https://en.wikipedia.org/wiki/Neyman-Pearson_lemma). What are the methods available for the more general hypothesis test? I know mathematical statistics from a postgraduate qualifying exam, but I am not a specialist. I need this for applied statistical problems I am solving, and I failed to find an answer in available textbooks and monographs. Can you direct me to some approaches?

REPLY [4 votes]: You may want to model the situation by assuming that the alternative distribution is the mixture $P_t:=(1-t)P_0+tP_1$ of the distributions $P_0$ and $P_1$, for some $t\in(0,1]$. You may then want to test for the value of the mixture parameter $t$. Let $f_t$ be the Radon--Nikodym density/derivative of $P_t$ with respect to some appropriate measure (e.g. $P_0+P_1$), so that $f_t=(1-t)f_0+tf_1$. Then for any $s$ and $t$ such that $0\le s<t\le1$ the likelihood ratio 
$$r_{t,s}:=\frac{f_t}{f_s}=\frac{tr+1-t}{sr+1-s}
$$ 
is a nondecreasing function of $r:=r_{1,0}$. (Here I am assuming for simplicity that $f_0>0$ everywhere; otherwise, rather straightforward adjustments have to be made; for instance, one may assume that $a/0=\infty$ if $a>0$, $0\cdot\infty=0$, $0/0\ge b$ and $0/0\le b$ for any $b\in[0,\infty]$.)
So, by the Karlin–Rubin theorem (see a version of it for non-randomized tests at [wikipedia]), for any $t_0\in[0,1]$ and any $t_1\in(t_0,1]$, any Neyman--Pearson test of size $\alpha\in(0,1)$ (say of the form "reject $H_{t_0}$ iff $r_{t_1,t_0}>c$ ", for some critical value $c>0$) for testing the simple hypotheses $H_{t_0}\,\colon t=t_0$ vs. $H_{t_1}\,\colon t=t_1$ will also be a uniformly most powerful test of level $\alpha$ for the null hypothesis $H_{t_0}\,\colon t=t_0$ (or $H_{\le t_0}\,\colon t\le t_0$) vs. the composite alternative hypothesis $H_{>t_0}\,\colon t>t_0$. 
In particular, for any $t_1\in(0,1]$, any Neyman--Pearson test of size $\alpha\in(0,1)$ for testing the simple hypotheses $H_0\colon t=0$ vs. $H_{t_1}\,\colon t=t_1$ will also be a uniformly most powerful test of level $\alpha$ for $H_0\colon t=0$ vs. $H_{>0}\,\colon t>0$.<|endoftext|>
TITLE: Inequality for the maximum of Gaussian variables
QUESTION [7 upvotes]: Let $X=(X_1,\dots,X_n)$ and $Y=(Y_1,\dots,Y_n)$ be centered Gaussian vectors 
with variance matrix $\Gamma_X$ and $\Gamma_Y$. We assume that the matrix 
$\Gamma_Y-\Gamma_X$ is positive definite. Is it possible to prove that 
$$ \forall x \ge 0, \mathbb{P}(\max \lbrace |Y_1|,\dots,|Y_n| \rbrace  \ge x) \ge \mathbb{P}(\max \lbrace |X_1|,\dots,|X_n|\rbrace \ge x) .$$
Thank you.

REPLY [6 votes]: I think that what you are writing is precisely Anderson's correlation inequality, see "T. W. Anderson. The integral of a symmetric unimodal function over a symmetric convex set and some probability inequalities. Proc. Amer. Math. Soc., 6(2):170–176, 1955."<|endoftext|>
TITLE: On linear integer inequalities with infinitely many solutions
QUESTION [8 upvotes]: Suppose that a linear system of inequalities $Ax \le b$, where $A\in Z^{m\times n}$ and $b\in Z^m$ have integral coefficients, has an infinite number of integral solutions $x$. 
Can one conclude that there is a ray $\{x+tp\mid t\ge 0\}$ containing infinitely many integral solutions? (If one drops the integrality condition, the answer is clearly yes.)

REPLY [2 votes]: Let the constraints be numbered 1 to $m$. Let the $i$th constraint be $a^{(i)}\cdot x\le b$. Let $S$ be the set of solutions. Inductively refine the constraints as follows: either there are $\limsup_{x\in S, x\to\infty} a^{(1)}\cdot x=-\infty$ or not. If $\limsup_{x\in S,x\to\infty} a^{(1)}\cdot x=c^{(1)}>-\infty$, then replace the system with the stronger constraint, $a^{(1)}\cdot x=c^{(1)}$. It still has infinitely many integer solutions. 
Now consider the second constraint. If the $\limsup$ is finite, replace it by an equality etc. 
In the end, you have a set of equalities, and a set of inequalities. For the inequalities, the inner products tend to $-\infty$ as the solution goes to $\infty$. In particular, you can find two solutions where for each equality, then inner product is the same, but for each inequality, the inner product of the first is larger than the inner product of the second. 
Call these $x$ and $y$. Now $x+n(y-x)$ is an infinite family of integer solutions satisfying all of the (stronger) constraints.<|endoftext|>
TITLE: non commutative polynomial which is zero for all matrix evaluation
QUESTION [8 upvotes]: I want to work on $K$ an algebraic closed (commutative) field of characteristic zero (even if it seems to be more general). 
We can define the free K-algebra of polynomials in non commutative variables $x_1, \cdots, x_n$. Il is usually denoted by $K\langle x_1, \cdots, x_n \rangle$. 
Fix a non commutative polynomial $P \in K\langle x_1, \cdots, x_n \rangle$. 
For every natural number $m$ and every choice of matrices $M_1, \cdots, M_n \in {\rm M}_m(K)$, we can evaluate $P$ at $(M_1, \cdots, M_n)$ to obtain a matrix $P(M_1, \cdots, M_n) \in {\rm M}_m(K)$. 
My question is : if the evaluation of $P$ on every matrices $(M_1, \cdots, M_n)$ for every $m$ is $0$, is necessary $P = 0 \in K\langle x_1, \cdots, x_n \rangle$ ?
I don't know if the question is totally trivial. 
1) In fact, if we restrict the condition to $m=1$, the answer is clearly no because the non commutative polynomial $x_1 x_2 - x_2 x_1$ for example is non zero. But it is yes if we consider polynomial in commutative variables.
2) Algebras ${\rm M}_m(K)$ are polynomial rings. In particular, the answer is no again if the condition $P(M_1, \cdots, M_n) =0$ only for all $M_1, \cdots, M_n \in {\rm M}_m(K)$ for a fixed $m$. Amitsur–Levitzki theorem gives an explicit counter-example.  
Extension of the question : same question where we replace $P \in K\langle x_1, \cdots, x_n \rangle$ by a non commutative formal power series $S \in K\langle\langle x_1, \cdots, x_n \rangle\rangle$ .
Thanks a lot for yours answers.

REPLY [6 votes]: This is an algebraic elaboration on Emil's answer.
Let $A = K \langle x_1,\ldots,x_n \rangle$ and let $\hat{A} = K \langle\langle x_1,\ldots, x_n \rangle \rangle$. Since $A$ is a subring of $\hat{A}$, a positive answer for $\hat{A}$ implies one for $A$.
Now for each $d \in \mathbb{N}$, let $\hat{A}_{d}$ be the $K$-linear span of all products of the generators $x_1,\ldots, x_n$ of length exactly $d$. Then $\hat{A}_{d} \cdot \hat{A}_e = \hat{A}_{d+e}$ for all $d,e\in\mathbb{N}$, and there is a vector space isomorphism
$$\prod_{d=0}^\infty \hat{A}_d \stackrel{\cong}{\longrightarrow} \hat{A}.$$
This just says that every non-commutative formal power series can be uniquely decomposed as an infinite (convergent in a natural topology on $\hat{A}$) sum of its homogeneous components.
Now let $\hat{A}_{>m} := \prod_{d>m} \hat{A}_d$ for each $m \geq 0$. These are two-sided ideals of $\hat{A}$, and for each $m \geq 0$,
$$V_m := \hat{A} / \hat{A}_{>m} \cong \prod_{d=0}^m \hat{A}_d$$
is a finite dimensional vector space over $K$, of dimension $N_m$, say. This gives us matrix representations 
$$\rho_m : \hat{A} \to End_K(V_m) \cong M_{N_m}(K).$$
Clearly $\hat{A}_{> m}$ is contained in $\ker \rho_m$, but in fact we have equality, because $\hat{A} / \hat{A}_{>m}$ (being an associative $K$-algebra in its own right) acts faithfully on itself by left multiplication.
So if $P \in \hat{A}$ is zero on every set of $n$ matrices of size $N$ for all $N \geq 0$, then we see that in particular, $P \in \ker \rho_m$ for all $m \geq 0$. Hence $P \in \bigcap_{m \geq 0} \hat{A}_{>m}$, which implies that every homogeneous component of $P$ is zero. Hence $P$ is also zero.<|endoftext|>
TITLE: Are there $n$ groups of order $n$ for some $n>1$?
QUESTION [59 upvotes]: Given a positive integer $n$, let $N(n)$ denote the number of groups of order $n$, up to isomorphism.

Question: Does $N(n)=n$ hold for some $n>1$?

I checked the OEIS-sequence https://oeis.org/A000001 as well as the squarefree numbers in the range $[2,10^6]$ and found no example. Since we have many $n$ with $N(n)<n$ and some $n$ with $N(n) \gg n$, I see no reason why $N(n)=n$ should be impossible for $n>1$.

REPLY [25 votes]: A "near-miss" is $N(19328) = 19324$, while the only $n \leq 2000$
such that $|N(n)-n| \leq 25$ are $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$,
$9$, $10$, $11$, $12$, $13$, $14$, $15$, $16$, $17$, $18$, $19$, $20$,
$21$, $22$, $23$, $24$, $25$, $26$, $27$, $28$, $32$, $36$, $48$, and $72$.<|endoftext|>
TITLE: realizing uniform boundedness of Galois representations associated to elliptic curves
QUESTION [6 upvotes]: This is less of a direct question and more of an argument that I've been worried about for a while and want to check (apologies for the length and if my writing is unclear).
Suppose I have an elliptic curve $E / \mathbb{Q}$ given by an equation of the form $y^{2} = x(x - 1)(x - \lambda)$ with $\lambda \in \mathbb{Q}$. For any rational prime $p$, let $v_{p}$ denote the $p$-adic valuation on $\mathbb{Q}$.  Let $k_{p}$ denote the fraction field of the strict Henselization of $\mathbb{Z}_{(p)}$; we fix an embedding into $\bar{\mathbb{Q}}$.
Suppose that $m := v_{p}(\lambda) > 0$ and $m' := v_{p}(\lambda') > 0$ for (distinct) rational primes $p$ and $p'$.  Now the theory of Tate curves tells us that if $p \neq 2$ or $m \geq 5$, then the $\ell$-adic image of $\mathrm{Gal}(\bar{\mathbb{Q}} / k_{p}(\mu_{\ell})$ is cyclically generated by a power of a transvection in $\mathrm{SL}(T_{\ell}(E))$.  More precisely, if $p \neq 2$ (resp. if $p = 2$ and $m \geq 5$), it is generated by the $s := \ell^{v_{\ell}(2m)}$-th power (resp. the $s := \ell^{v_{\ell}(2m - 8)}$-th power) of the transvection with respect to an element $a \in T_{\ell}(E)$.  Of course, the analogous statement for $p'$ also holds (call the analogous element $a' \in T_{\ell}(E)$ and power $s'$).  This follows from the fact that if $j$ is the $j$-invariant of $E$, then $v_{p}(j) = -2m$ (resp. $v_{p}(j) = -2m + 8$) if $p \neq 2$ (resp. if $p = 2$), and analogously for $p'$.  (See Exercise 5.13(b) of Silverman's Advanced Topics, or \SA.1.5 of Serre's Abelian \ell-adic representations.)
Let's assume (*) that $\{a, a'\}$ are always independent in $T_{\ell}(E)$.  In fact, I've proven this to be the case when $\ell = 2$ and $p, p' \neq 2$, and believe it's probably true in general.  Let's also assume (**) that if $\tilde{p} \neq p'$ is another prime such that $v_{\tilde{p}}(\lambda - 1) > 0$, then the $\ell$-adic images of $\mathrm{Gal}(\bar{\mathrm{Q}} / k_{p'}(\mu_{\ell}))$ and $\mathrm{Gal}(\bar{\mathrm{Q}} / k_{\tilde{p}}(\mu_{\ell}))$ are generated by powers of the same transvection.
Then, by considering $k_{p}, k_{p'}$ as subextensions of $\bar{\mathbb{Q}} / \mathbb{Q}$, we see that the $\ell$-adic image of the absolute Galois group of $\mathbb{Q}$ contains the powers of transvections given above.  Since $a$ and $a'$ are independent, it's not hard to show that these two powers of transvections generate a subgroup $G$ of $\mathrm{SL}(T_{\ell}(E))$ containing the principal congruence subgroup $\Gamma(\ell^{s + s'})$; it's also easy to see by reducing mod $\ell^{s + s'}$ that it can't contain any larger principal congruence subgroup.  Since $\mathbb{Q}(\mu_{\ell})$ is the intersection of the extensions $k_{p}(\mu_{\ell})$ for all primes $p$ and we only need to consider primes of bad reduction by Neron-Ogg-Shafarevich, it follows from our assumption (**) that the $\ell$-adic Galois image of $\mathrm{Gal}(\bar{\mathrm{Q}} / \mathrm{Q}(\mu_{\ell}))$ actually coincides with $G$.
Now choose primes $\ell$ and $p \neq \ell, 2$ and consider the infinite subset of fibers of the Legendre family given by $\lambda = p^{\ell^{n}}$ for $n = 1, 2, 3, ...$.  By the above arguments, for each $n$, the $\ell$-adic Galois image will contain $\Gamma(\ell^{s + s'})$ but no larger principal congruence subgroup, where $s = n$ and $s' > 0$ depends on $n$.  Thus, there is no $N$ such that the image of Galois contains $\Gamma(\ell^{N})$ for all fibers of the Legendre family.
But this contradicts well-known results (of Serre and others) on the uniform boundedness of the indices of  $\ell$-adic Galois images of non-CM elliptic curves over a fixed number field.  The only assumptions I made were (*) and (**).  In the $\ell = 2$ case, like I said, I've proven (*), and so ( **) must be false in general.  Or did I make some other unjustified assumption in my argument?  I would greatly appreciate any insight on this.
EDIT: Okay, thanks to Horace's answer below some things have become clearer to me, in particular that $\bigcap k_{p}(\mu_{\ell})$ doesn't coincide with $\mathbb{Q}(\mu_{\ell})$ as I had assumed before.  Therefore, the $\ell$-adic images of $I_{p} := \mathrm{Gal}(\bar{\mathbb{Q}} / k_{p}(\mu_{\ell}))$ do not necessarily generate the whole image of the absolute Galois group of $\mathbb{Q}(\mu_{\ell})$.  Thus, a priori the boundedness results of Serre and others aren't relevant here.
Also, Horace points out that my assumption (** ) isn't really sensible as stated, given that one has to choose a decomposition group above $p$ in order to consider $k_{p}$ as a subfield of $\bar{\mathbb{Q}}$. So suppose we amend it to claim that if $q \neq p'$ is another prime such that $v_{q}(\lambda - 1) > 0$, then the $\ell$-adic image $\rho_{\ell}(I_{q})$ is conjugate via an element of $\rho_{\ell}(I_{p})$ to a subgroup $H$ such that $H$ and $\rho_{\ell}(I_{p'})$ are each generated by powers of the same transvection. (By the way, note that we expect the images of the inertia groups to be generated by powers of transvections, not necessarily transvections themselves, so this kind of statement has nothing to do with whether $v_{q}(j)$ has higher $\ell$-adic valuation than $v_{p'}(j)$, etc.) In other words, the claim is that the subgroup of $T_{\ell}(E)$ generated by $\rho_{\ell}(I_{p})$, $\rho_{\ell}(I_{p'})$, and $\rho_{\ell}(I_{q})$ is the same as the subgroup generated by $\rho_{\ell}(I_{p})$ and $H'$ for some cyclic subgroup $H' \geq \rho_{\ell}(I_{p'})$. I believe I proved this directly for $\ell = 2$ and $p, p', q \neq 2$ by using explicit formulas for $2$-power torsion, but I'm not completely sure since I never wrote it down. Anyway, now it seems this new version of assumption (**) might be true, since it wouldn't contradict Serre's boundedness results. I'm still interested to find out when it is.
Also, thank you Horace for correcting my oversight in not including the prime $\ell$ for Neron-Ogg-Shafarevich.

REPLY [9 votes]: $\def\Q{\mathbf{Q}}$
$\def\Qbar{\overline{\Q}}$
$\def\Z{\mathbf{Z}}$
$\def\GL{\mathrm{GL}}$
There are many confusions in your post, let me point out just a few
It is dangerous to fix embeddings of $\Qbar$ into $\Qbar_p$ (equivalently, to fix decomposition groups $D_p$ inside the Galois group $G_{\Q}$ for all $p$), and leads you astray on a number of occasions. You say:
Since $\Q(\zeta_{\ell})$ is the intersection of the extensions ... we need only consider primes of bad reduction by Néron-Ogg-Shafarevich, it follows ... that the $\ell$-adic image ... coincides with $G$.
There are at least three things wrong with this paragraph. First, by Néron-Ogg-Shafarevich, one deduces that the ramification on the $\ell$-adic Tate module is unramified outside the primes of bad reduction and the prime $\ell$, which you omitted. Second, a Galois group need not be generated by the decomposition groups at primes of bad reduction; for example, this does not hold for an extension which is unramified everywhere (or which contains any such extension). You can overcome both of these issues by including the prime $\ell$ and working over $\Q$ which has no unramified extensions by Minkowski. However, you still run into trouble, because you fixed a choice of $D_p$, and for this sort of argument you need to consider all conjugates of $D_p$ as well. For example, let $L/\Q$ be the splitting field of
$$x^3 - x - 1 = 0.$$
The Galois group $G$ is $S_3$, and $G$ is ramified only at $23$. If you fix a decomposition group $D \subset G$ at a prime above $p = 23$, you will find that $D = \Z/2\Z$, which doesn't generate $G$. Correspondingly, the fixed field $K = L^{D}$ (which is isomorphic to the cubic field cut out by a root of the equation above) is unramified at one prime above $23$ but ramified at the other prime above $23$.
This leads into the second issue when fixing $D_p$. It's completely non-canonical, and yet both of your assumptions (*) and (**) depend completely on this choice. Let's assume that the $\ell$-adic representation:
$$\rho: G_{\Q} \rightarrow \GL_2(\Z_{\ell})$$
was surjective (it often is). Let's also suppose that $E$ has multiplicative
reduction at $p$, and that valuation of $j_E$ is $-1$. Then what does one know about the image of inertia $I_p \subset D_p$ inside the $\ell$-adic Tate module? Well, it will contain a transvection conjugate
to
$$\left(\begin{matrix} 1 &   u \\ 0 & 1 \end{matrix} \right)$$
for some unit $u \in \Z^{\times}_{\ell}$. However, choosing a different embedding $D_p \subset G_{\Q}$
is equivalent to conjugating this element. And every single transvection $\sigma \in \GL_2(\Z_{\ell})$
which is non-trivial modulo $\ell$ is conjugate to this element.  So for example, given another
prime $q$ with ordinary reduction and such that $j_E$ has valuation $-1$, the corresponding statement holds. But now * and ** clearly depend on the choice  you make. You can choose $D_p$ and $D_{q}$ inside $G_{\Q}$ so that you get exactly the same transvection, or  you can choose them such that together they generate $\GL_2(\Z_{\ell})$. So it doesn't make sense to ask whether * or ** holds. This is the main source of your error.
Note that (**) will be false in a stronger sense. You could ask
whether the transvection at $q$ is generated by a power
of an element in the conjugacy class in $\GL_2(\Z_{\ell})$ of the transvection at $p$.
Yet the conjugacy class will
be determined by the $\ell$-adic valuation of $v_{p}(j_E)$. So this amounts to saying that $v_q(j_E)$
has the same or higher $\ell$-adic valuation as $v_p(j_E)$. And this is unlikely to be the case, especially of one is choosing $\lambda$ specifically so that $v_p(j_E)$ is highly divisible by powers of $\ell$.
Here is a modification of your thoughts which  makes sense. Suppose $E$ is semi-stable at $N$, and we want to force the $\ell$-adic image to be small. One way is to make the $\ell$-adic image small by showing that $E[\ell^m]$ is not ramified at any of the primes dividing $N$. It's hard to eliminate ramification at $\ell$, but if $\ell$ is a prime of good reduction, then $E[\ell^m]$, although not ramified, will have some nice properties (it comes from the generic fibre of a finite flat group scheme). As a consequence of the proof of Serre's conjecture, we have:
Theorem Consider a finite flat group scheme $A$ over $\mathrm{Spec}(\Z)$ of type $(\ell^m,\ell^m)$ for some prime $\ell > 2$ such that the corresponding representation has cyclotomic determinant. Then $A \simeq \Z/\ell^m \oplus \mu_{\ell^m}$. 
So we will be able to get a contradiction to Serre's open image theorem (and Mazur's theorem) simply by ensuring that $E[\ell^m]$ is unramified at all primes above $N$. For each prime $p$ dividing $N$, to say that $D_p$ (and all its conjugates) acts trivially is equivalent to asking that $v_p(j_E)$ is divisible by $\ell^n$. So we get the desired contradiction by choosing a semi-stable elliptic curve such that $v_p(j_E) = - v_p(\Delta_E)$ is divisible by $\ell^n$ for all primes $p$. To write down such an elliptic curve, simply write
$$y^2 = x (x - a^{\ell^n})(x + b^{\ell^n}).$$
And now you get a contradiction to Serre's theorem as long as $a^{\ell^n} + b^{\ell^n} = c^{\ell^n}$
for some $c$ with $(abc,\ell) = 1$, and one can take $n$ sufficiently large. Even better, Using Mazur's theorem, one gets a contradiction in mathematics even for $n = 1$ as long as $\ell > 11$. So simply write down a solution to $a^{\ell} + b^{\ell} = c^{\ell}$ for integers $(abc,\ell) = 1$ and $\ell > 11$ and we will have found a contradiction in mathematics and can all go home and eat some pie. Good luck!

Your cubic polynomial example helped me to realize that I made one basic and crucial error in assuming that $\cap K_p = \Q$ ... That confusion seems to be the source of most of the other issues you point out. 
Hmm. I'm going to push back a little here. First, I think there are a number of different errors, and the main issue is not considering that the choice of $D_p$ is very non-canonical. The questions (*) and (**) depend completely on the choice of embeddings, and it is an error to think of them as sensible hypotheses. And also ... the statement above happens to be true! (Edit: I was assuming that $K_p$ to be the fixed field of a decomposition group $D_p$, not the inertia group $I_p$.) If you take a finite extension with Galois group $G$, then the subgroup $H$ generated by $D_p$ even for all sufficiently large primes will be $G$. This is because the subgroup $H$, by Ceboratev, will contain an element from each conjugacy class of $G$, and such subgroups have to be the entire group by a theorem of Jordan. However, it is false to say that the subgroup $H$ generated by the decomposition groups of the ramified primes generates $G$; what you can deduce is that if $N$ is the normal closure of $H$ in $G$ then the extension corresponding to $G/H$ is unramified everywhere.<|endoftext|>
TITLE: Is this theory decidable?
QUESTION [15 upvotes]: It is well-known that both Presburger arithmetic (by contrast with Peano arithmetic) and Tarski geometry are decidable. I was in the shower this morning and wondered whether there exists an elegant mutual generalisation of these theories that is also decidable. In particular, I settled on the following system:

The objects are the finite-dimensional affine subspaces of your favourite infinite-dimensional real Hilbert space (let's say $L^2$ for concreteness).
We have a binary predicate, $x \subseteq y$, which inherits its usual meaning.
We have a ternary predicate, $I(x, y; z)$, which means there is an isometry of the ambient space which fixes $z$ and maps $x$ bijectively onto $y$.

I'll show that this does indeed generalise both Presburger arithmetic and Tarski geometry.

Firstly, note that we can encode the concept of a point:
$$ x \textrm{ is a point } \iff \forall y . (y \subseteq x) \implies (y = x) $$
(It's rather cute that this is precisely how Euclid described a point, namely 'a point is that which has no part'.)
Similarly for lines:
$$ x \textrm{ is a line } \iff x \textrm{ is not a point and } \forall y . (y \subseteq x) \implies ((y = x) \textrm{ or } y \textrm{ is a point}) $$
We can continue inductively to define planes and so on.
We describe two lines $x, y$ as parallel if there is a plane which contains both $x$ and $y$, and there is no $z$ such that $z \subseteq x$ and $z \subseteq y$. This allows one to define \emph{parallelogram}, and emulate vector addition with respect to some origin $o$. That allows one to take Minkowski sums of affine subspaces with respect to $o$.

So far we haven't touched the ternary predicate $I(x, y; z)$. One rudimentary application is to equate distances between points:
$$ |a - b| = |d - c| \iff \exists e . (b + c = a + e) \textrm{ and } I(d, e; c) $$
Here we're using $(b + c = a + e)$ as shorthand for $e$ being a point and satisfying the vector addition property mentioned earlier. We can also compare distances: $|x - y| \geq |a - b|$ if and only if we can find points $c, d$ such that $b + b = c + d$ and $|x - y| = |a - c| = |a - d|$. Together with collinearity, this allows us to define Tarski's 'betweenness' predicate, so we can encode all of Tarskian geometry.

Another application of this predicate is to equate dimension:
$$ \dim(x) = \dim(y) \iff \exists z . I(x, y; z) $$
We can also add dimensions. Specifically, $\dim(x) + \dim(y) = \dim(z)$ if and only if we can find a point $o$ and spaces $a, b$ such that $\dim(a) = \dim(x)$, $\dim(b) = \dim(y)$, the intersection of $a$ and $b$ is $o$, and every point in $z$ can be expressed uniquely as a sum (as vectors relative to $o$) of a point in $a$ and a point in $b$.
This endows us with the ability to perform Presburger arithmetic on the dimensions of spaces.

Anyway, this prompts the question: is this theory (together with a suitable finite set of axioms) decidable? With 'bounded quantifiers' (i.e. bounded dimension), this reduces to $n$-dimensional Tarski geometry (and therefore is decidable). However, I feel this theory is much stronger since you can make first-order statements about arbitrary finite-dimensional vector spaces.

REPLY [2 votes]: Here are some comments and the start of a positive answer.  I can prove three things:

1) If "$x$ has square dimension" is expressible in this language, then the theory is undecidable.
2) The binary predicate is not needed for the expressive power of the theory, and the theory is equally decidable or undecidable without it.
3) For
$$\phi=(Q_1x_1)(Q_2x_2)...(Q_nx_n)P(x_1,\ldots,x_n)$$ with $P$ quantifier-free,
let
$$\phi^b = (Q_1x_1\le 2^n)(Q_2x_2\le 2^{n+1})...(Q_nx_n \le 2^{2n-1)})P(x_1,\ldots,x_n)$$
where each $Q_i$ is either $\forall$ or $\exists$, and $Qx\le n$ is restricted to $x$ of dimension $\le n$.  Then the schema $$\phi \implies \phi^b$$ would show that the theory is decidable.  I also have some ideas for why this might hold.

Proofs:
1) If the squares are definable, then we can define $x=y^2$ by "$x$ and $x+y+y+1$ are consecutive squares".  Then we can also define $x=yz$ by $(y+z)^2=y^2+z^2+x+x$, which makes the theory undecidable.
2) The predicate $x\subseteq z$ is equivalent to $\forall y\ I(x,y,z) \rightarrow x=y$.
2a) If $x$ is contained in $z$ then obviously any isometry fixing $z$ can only take $x$ to itself.
2b) Suppose that any isometry fixing $z$ can only take $x$ to itself.  Let $p$ be a point in $z$, with $x-p$ and $z-p$ being the spaces of vectors from $p$ to $x$ and $z$.  Construct an orthonormal basis $v_i$ for the ambient vector space such that $\{v_1\ldots v_n\}$ is a basis for $z-p$, and $\{v_1\ldots v_{n+k}\}$ spans both $z-p$ and $x-p$.  Then consider an isometry of the space that:

fixes $p$ and $p+v_i$ for $i\le n$
switches $p+v_i$ with $p+v_{i+k}$ for $n+1\le i\le n+k$
fixes $p+v_i$ for $i>n+2k$.

This isometry fixes $z$.  So by hypothesis, it must take $x$ to itself.  So it must not switch any basis vectors.  So $k=0$ and $x$ must be contained in $z$.
3) If the schema holds, then we can decide the truth of a sentence $\phi_0$ by putting it into prenex normal form $\phi$, and evaluating the truth of $\phi^b$ using the standard Tarski decision procedure.  The combination of $\phi \implies \phi^b$ and $\psi \implies \psi^b$, where $\psi$ is the prenex form of $\neg \phi$, is enough to show that $\phi$ and $\phi^b$ are equivalent.
Why we should expect that $\phi \implies \phi^b$?
A good test case is $$\phi = \forall w\, \exists x\, \forall y \,\exists z\, P(w,x,y,z)$$ with $P$ quantifier-free.  If $\phi$ holds then there are Skolem functions $f_2$ and $f_4$ such that
$$\forall w\, \forall y\, P( w, f_2(w), y, f_4( w, y )).$$
Then we can show $\phi^b$ by showing that other Skolem functions $g_2$ and $g_4$, whose ranges have dimensions at most $32$ and $128$, satisfy
$$\forall w \le 16\, \forall y\le 64\, P( w, g_2(w), y, g_4( w, y )).$$
The idea is to get rid of unneeded dimensions from the images of $f_2$ and $f_4$, since no configuration describable by this many quantifiers will need more than these dimensions.  This seems easy enough to prove for any particular true $\phi$, and perhaps someone will see how to articulate the argument that we can do it generally.
Which bounds should we use?
Suppose we want bounds
$$\phi^b = (Qx_1 \le a_1) \cdots (Qx_n \le a_n) P(x_1, \ldots x_n)$$ with the $a$'s depending on $n$ but independent of the $Q$'s and $P$.
 Consider the examples
$$\phi_j = \forall x_1 \cdots \forall x_{j-1} \exists x_j \cdots \exists x_n \bigwedge_{i<j} x_i \subset x_j \wedge \bigwedge_{i\ge j} x_{i+1} \subset x_i$$
Then $\phi_1$ can only be satisfied with $x_1$ of dimension at least $n-1$.
So the lowest possibility for $\phi_1^b$ is $a_1= n-1$.
The lowest simultaneous possibility for $\phi_1^b$ and $\phi_2^b$ is $a_1 = n-1,\, a_2 = n$.  
The lowest simultaneous possibility for $\phi_1^b, \ldots \phi_4^b$ is $a_1 = n-1,\, a_2 = n,\, a_3 = 2n,\, a_4 = 4n$.
Since these $a_i$ grow exponentially, the statement of claim 3 seemed easiest using only powers of 2, and that seems like a convenient form in which to try to prove the decidability.<|endoftext|>
TITLE: Distance between two knots
QUESTION [11 upvotes]: Are there some well-studied functions defining natural distance measures between two knots? One can imagine a function that counts, say, the minimum number of
moves, each of which passes one strand of a knot through a crossing strand, in order to convert one knot to another.
Or perhaps there are functions that rely on knot polynomial similarity.
Any references would be appreciated.

Update. Here is a figure from the Murakami reference kindly provided by Marco Golla:

          


          

(Murakami Fig.7, illustrating #-unknotting operations.)



Murakami, Hitoshi. "Some metrics on classical knots." Mathematische Annalen 270.1 (1985): 35-45.
  (Göttinger Digitalisierungszentrum link to PDF.)

REPLY [3 votes]: If you want to define some distance on knots, you should have a (local) transformation/move on knots or or knot diagrams such that any two knots can be connected by finitely many of them. 
As mentioned by Ryan, the most important local move is crossing change, and the corresponding distance is called the Gordian distance. 
Besides of this, $\sharp$-operation (H. Murakami, Some metrics on classical knots. Math. Ann. 270, 35-45, 1985), $\triangle$-operation (H. Murakami, Y. Nakanishi, On a certain move generating link-homology. Math. Ann. 284, 75-89, 1989) and $n$-gon move (H. Aida, Unknotting operation for Polygonal type. Tokyo J. Math. Vol. 15, No. 1, 111-121, 1992) all are unknotting operations, hence can be used to define a distance between knots. Recently, Ayaka Shimizu proved that region crossing change is also an unknotting operation for knots (A. Shimizu, Region crossing change is an unknotting operation. J. Math. Soc. Japan. Volume 66, Number 3 (2014), 693-708). While this move depends on the choice of the diagram, hence the distance defined by region crossing change are restricted on minimal diagrams.
On the other hand, there also exist some local operations that we do not know whether it can unknot every knot or not, for example the 4-move (see problem 1.59 on Kirby's list for more details and Dabkowski and Przytycki's paper: Unexpected connections between Burnside groups and knot theory. Proc Natl Acad Sci U S A. 2004 Dec 14;101(50):17357-60 for some update).<|endoftext|>
TITLE: Constructing symmetric invariants of the exceptional simple Lie algebra as restrictions
QUESTION [6 upvotes]: Let $\mathfrak{g}$ be a complex simple Lie algebra with maximal torus $\mathfrak{h}$, Weyl group $W$. The adjoint representation $\operatorname{ad} : \mathfrak{g} \rightarrow \mathfrak{gl(g)}$ extends by derivations to a representation of $\mathfrak{g}$ in $S(\mathfrak{g})$. We identify $S(\mathfrak{g})$ and $\mathbb{C}[\mathfrak{g}]$ as $\mathfrak{g}$-modules via the Killing form. The most famous description of $S(\mathfrak{g})^{\mathfrak{g}}$ goes as follows: the Chevalley restriction theorem states that the restriction map $\mathbb{C}[\mathfrak{g}]^\mathfrak{g} \rightarrow \mathbb{C}[\mathfrak{h}]^W$ is an isomorphism and the Chevalley-Sheppard-Todd theorem implies that the Weyl group invariants are a polynomial ring on $\operatorname{rank}(\mathfrak{g})$ generators. 
I am interested in another approach to constructing the invariant rings via restriction. Take an $n$-dimensional vector space $V$ and set $\mathfrak{g} = \mathfrak{gl}(V)$. The characteristic polynomial $p(t, x)$ of an element $x \in \mathfrak{g}$ can be written as $t^n + \sum_{i=0}^{n-1} p_i(x) t^i$, and the coefficients $p_i$ form a complete set of algebraically independent generators for $S(\mathfrak{g})^{\mathfrak{g}}$. Now suppose that $\mathfrak{k} \subseteq \mathfrak{g}$ is a classical Lie algebra of type $\mathfrak{so}(V)$ or $\mathfrak{sp}(V)$. We can consider the restrictions $p_{i}\vert_{\mathfrak{k}}$ and in this case we have $p_{2i}\vert_{\mathfrak{k}} = 0$ for all $i$ and the remaining invariants form a complete set of algebraically independent generators for $S(\mathfrak{k})^\mathfrak{k}$, unless of course $\mathfrak{k}$ has type ${\sf D}$, in which case $p_{n-1} = P^2$ for some $P \in S(\mathfrak{g})$ known as the Pfaffian. In this case $p_{1}\vert_{\mathfrak{k}}, p_{3}\vert_{\mathfrak{k}},...,p_{n-3}\vert_{\mathfrak{k}}, P$ form a basic set of generators for the symmetric invariants. This is all very classical and probably goes back to Weyl.
What I would like to know is whether we can play the same game with exceptional Lie algebras. This is my question: for $\mathfrak{g}$ simple and exceptional, does there always exist a simple finite dimensional representation $\mathfrak{g} \rightarrow \mathfrak{gl}(V)$ such that some of the restrictions $p_0\vert_{\mathfrak{g}},...,p_{n-1}\vert_{\mathfrak{g}}$ are zero, whilst the others form a complete set of algebraically independent generators for $S(\mathfrak{g})^{\mathfrak{g}}$, perhaps after Pfaffing around with them a bit?

REPLY [5 votes]: You can do this in GAP. To check algebraic independence you can restrict to a Cartan subalgebra.
For type $G_2$ with the minimal faithful representation we obtain $p_1(sh_1+th_2)=-2(t^2-3st+3s^2)$ and $$p_5(sh_1+th_2)=-4s^6+12s^5t-13s^4t^2+6s^3t^3-s^2t^4=-s^2(t-s)^2(t-2s)^2$$ from which you can already see the main problem - these polynomials become somewhat unwieldy as the degree increases. In this case we know that these two polynomials (in $k[s,t]$) are algebraically independent since otherwise $p_5$ has to be a multiple of $p_1^3$.
This doesn't quite answer your question, as in this case we have $p_3=(t^2-3st+3s^2)^2\neq 0$. My feeling is that for the exceptional types it is extremely unlikely that you will find a representation satisfying the condition you require (that all except $r={\rm rank}({\mathfrak g})$ of the polynomials $p_i$ are zero). On the other hand, I would expect that if you pick any faithful representation $(\rho,V)$ of a simple Lie algebra ${\mathfrak g}$ then the restrictions of $r={\rm rank}({\mathfrak g})$ of the basic invariants on $\mathfrak{gl}(V)$ to $\rho({\mathfrak g})$ are algebraically independent. (I probably should be able to see why this is true straight away.) Usually we should be able to take the same degrees as the degrees of the basic invariants on ${\mathfrak g}$. I would point out that the issue with the Pfaffian for $\mathfrak{so}_{2n}$ arises because ${\rm GL}_{2n}$ contains the full orthogonal group, which includes all of the outer automorphisms of $\mathfrak{so}_{2n}$. So in fact when we restrict ${\rm GL}_{2n}$-invariants on $\mathfrak{gl}_{2n}$ to $\mathfrak{so}_{2n}$ then we obtain elements of $k[\mathfrak{so}_{2n}]^{{\rm O}_{2n}}$, and this is a polynomial ring with homogeneous generators of degrees $2,4,\ldots, 2n$. I expect that what happens is: if $V$ is ${\mathfrak g}$-isomorphic to its twist via any outer automorphism then we get the same situation as you have for $\mathfrak{so}_{2n}\hookrightarrow\mathfrak{gl}_{2n}$; otherwise the restriction of the invariants on $\mathfrak{gl}(V)$ to ${\mathfrak g}$ gives us an exact copy of the invariants on ${\mathfrak g}$ (plus extra complications in type $D_4$). This happens, for example, for either 27-dimensional faithful representation for $E_6$.
For the higher rank cases GAP can't compute the characteristic polynomial directly as the memory required (e.g. for a 248 x 248 matrix involving 8 or 9 variables) is too great for it to handle. Magma might be able to do it, but it's not clear what value you can get out of running the computation - for example, if we take the adjoint representation in type $E_8$ and we compute ${\rm Trace}(\rho(x)^i)$ for $i=2,8,12,\ldots$ and $x=sh_1+\ldots +zh_8$ then already for $i=8$ this produces 3 pages of unreadable degree 8 polynomial in $k[s,t,\ldots ,z]$. We can then prove that this is independent of ${\rm Trace}(\rho(x)^2)$, but you still have a long way to go to prove by this route that the polynomials ${\rm Trace}(\rho(x)^i)$ are algebraically independent.<|endoftext|>
TITLE: All compact surfaces $S\subseteq \mathbb{R}^3$ are rigid?
QUESTION [20 upvotes]: Recently I've come across a lecture in differential geometry by Fernando Codá (in portuguese!) in which he stated that the following problem is (at least, up to 2014) open:

Given $S\subseteq\mathbb{R}^3$ a smooth (i.e. $C^\infty$) compact boundaryless connected surface (therefore orientable), let $f:S\times[0,1]\to \mathbb{R}^3$ be a differentiable map with $S_t:=f(S\times\{t\})$ being a surface and $f_t:S\to S_t$ an isometry (where $f_t(p):=f(p,t)$), where $f_0\equiv \mathrm{Id}_S$. Then there are rigid motions $A_t:\mathbb{R}^3\to\mathbb{R}^3$ with $f_t(S) = A_t(S)$.

More generally, in the cases of the sphere $S^2$ (Liebmann's Theorem) and "ovaloids", i.e. surfaces with Gaussian curvature $K$ strictly positive (Cohn-Vossen's rigidity theorem), it is known that isometric embeddings of it in $\mathbb{R}^3$ are exactly the restrictions of the rigid motions from $\mathbb{R}^3$. Not all compact connected surfaces $S\subseteq\mathbb{R}^3$ satisfies this, as pointed out in the book Curves and Surfaces, by S. Montiel and A. Ros:

Anyway, it is possible to show that this isometry in particular cannot be obtained by a continuous family of isometries from the identity map to the latter, so it's not a counterexample to the conjecture.
Although this seems to me to be a very studied problem, and I imagine that there must be several approachs to it, I just couldn't find enough information about it on the internet (the Montiel & Ros book was recommended by Codá later in the same lecture linked above), for "rigidity" seems to be a very comprehensive term. So my actual question is:

Actual question: Where can I find more information about this problem and the current approaches going on?

REPLY [5 votes]: Not at all an answer: I once had a conversation about this question with a friend at IHES, and Gromov was hanging around, overheard us, and said: oh, that's a stupid problem!
We said (in one voice): why do you say that?
To which Gromov responded: because the question has been around for a century, and no interesting mathematics has come out of it.
Make of this what you will.<|endoftext|>
TITLE: History of Mathematical Notation
QUESTION [28 upvotes]: I would like to see a simple example which shows how mathematical notation were evolve in time and space.
Say, consider the formula
$$(x+2)^2=x^2+4{\cdot}x+4.$$
If I understand correctly, Franciscus Vieta would write something like this. (Feel free to correct me.)
$\overline{N+2}$ quadr. æqualia $Q+N\,4+4$.
($N$ stays for unknown and $Q$ for its square.)
Can you give me the other examples?

REPLY [5 votes]: "The big picture" that can be seen in Carlo Beenakker's example is 


Rhetorical (verbal);   Syncopated  (abbreviated words);       Symbolic. 


However, this well-known picture is very algebra-oriented and does not say anything about the use of notations in say geometry as discussed here "
The Shaping of Deduction in Greek Mathematics: A Study in Cognitive History
" or as asked here Uppercase Point Labels in High-School Diagrams. 
Thus, at least for me, the interesting answers to OP questions would be those that are not related to algebra.

REPLY [3 votes]: Find a quantity such that $\varsigma\;\overline{\beta}$ squared equals
  $$\Delta^{\Upsilon}\;\varsigma\overline{\delta}\;{\stackrel{o}{M}}\overline{\delta}$$

(My translation into English from Diophantes $\alpha\rho\iota\theta\mu\eta\tau\iota\kappa\eta$.)<|endoftext|>
TITLE: What is known about Lie groups with (strictly) positive curvature?
QUESTION [12 upvotes]: If we consider $G$ a compact Lie group, there is a left invariant Riemannian metric whose the sectional curvature is nonnegative (see Milnors' paper). When can we find a left invariant metric that has positive sectional curvature?
The unique simply connected Lie group with a left-invariant metric that is positively curved is $SU(2)$. This is in Milnor's paper cited by @Igor Rivin.
Is $\mathrm{SU}(2)$ the only positively curved simply connected Lie group (not necessarily with an invariant metric)?
How could we distinguish the positively curved Lie groups?
Are there obstructions in the case of Lie groups? The fact that $G$ is a Lie group could imply the existence of some good invariant of positively curved manifolds?
Remark: I made many edits on the question because some ideas are clearer now with help of the time and the answers and comments posted here. I would like to observe that I started with a question with a pseudo-conjecture that the only positively curved Lie groups are $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ and asked for references.

REPLY [10 votes]: The following result is, for example, exercise 3 on pg. 104 of Do Carmo's Riemannian Geometry book.

Suppose $X$ is a Killing field on a compact even dimensional Riemannian manifold of positive curvature.  Then $X$ has a zero.

Using this result, it's very easy to prove the following generalization of Wallach's theorem.

Theorem:  No compact Lie group $G$ of rank $2$ or higher admits a positively curved Riemannian metric which is invariant under left multiplication by any $T^2\subseteq G$.

Proof:  Suppose there is such a group $G$.  Because the $T^2$ action on $G$ by left multiplication is free, the action fields associated to it have no zeros.  On the other hand, because the $T^2$ is isometric, the action fields are are Killing fields.  By the above result, this implies the dimension of $G$ is odd.
Because the action of $S^1\times \{1\}\subseteq T^2$ on $G$ is isometric, the even dimensional manifold $G/(S^1\times \{1\})$ inherits a Riemannian metric for which the action by $\{1\}\times S^1$ is free and isometric.  Further, by O'Neill's formulas for a Riemannian submersion, the induced metric on $G/(S^1\times \{1\})$ is positively curved.  But then, just as above, the action field is non-zero and Killing, contradicting the quoted result above.  $\square$
Incidentally, without some kind of symmetry assumption, we can say basically nothing about the existence of positively curved metrics on compact Lie groups with finite $\pi_1$.  The issue that among closed simply connected manifolds, there is no known obstruction which distinguishes those that admits metrics of non-negative curvature from those which admit metrics of positive curvature.<|endoftext|>
TITLE: Arithmetically equivalent number fields and Langlands Program
QUESTION [7 upvotes]: Two (number) fields are arithmetically equivalent if their Dedekind zeta functions are the same. It is known that any two arithmetically equivalent fields are not necessarily isomorphic; Prasad (http://www.math.tifr.res.in/~dprasad/refined-equiv.pdf) gives an example of two non-isomorphic, arithmetically equivalent fields whose idele class groups are isomorphic.
Under Langlands Philosophy, the structure of absolute Galois groups of number fields can be understood through $n$-dimensional automorphic representations of adele rings. My question is: 
Do there exist two non-isomorphic, arithmetically equivalent number fields whose $n$-dimensional automorphic representations of their adele rings are isomorphic? Here, $n>1$ is assumed.
Note: I asked this question in https://math.stackexchange.com/questions/1548060/arithmetically-equivalent-number-fields-and-langlands-program

REPLY [4 votes]: This isn't an answer, it's just too long to be a comment.
I think the question might be this. Take two number fields $K$ and $L$ such that $K$ and $L$ are not isomorphic, but $K$ and $L$ have isomorphic adele rings. This can happen! It's a fact from group theory that there's a finite group $G$ and two non-conjugate subgroups $H_1$ and $H_2$ such that $\mathbb{Z}[G/H_1]$ and $\mathbb{Z}[G/H_2]$ are isomorphic as representations of $G$, and realising everything as a Galois group over the rationals gives $K$ and $L$; for details see Dipendra Prasad's paper linked to in the question.
For such number fields, we can fix an isomorphism between the adele rings, which is completely horrible and non-canonical (e.g. we could just take two primes which split completely in all the fields in question and then just permute some of the factors in one of the adele rings but not the other, so fixing this isomorphism is an extremely un-natural and non-canonical thing to do, and there are also uncountably many ways to do it, none of which seem to me to be better than any other) and we get an induced isomorphism between $GL_n(\mathbb{A}_K)$ and $GL_n(\mathbb{A}_L)$ which is probably also completely non-canonical and random.
Having fixed this isomorphism, we can ask the following completely un-natural question: fix a ccharacter $\psi$ of $\mathbb{A}_K^\times$ and we get an induced character $\psi'$ of $\mathbb{A}_L^\times$. Does there exist some $n>1$ and an isomorphism $L^2(GL_n(K)\backslash GL_n(\mathbb{A}_K),\psi)\cong L^2(GL_n(L)\backslash GL_n(\mathbb{A}_L),\psi')$ which intertwines the actions of the isomorphic groups $GL_n(\mathbb{A}_K)$ and $GL_n(\mathbb{A}_L)$? The reason the isomorphism doesn't follow from what we have already is that $GL_n(K)$ sits in the adele group in a different way to $GL_n(L)$ so the spaces are most definitely not identified by all the choices we've made so far.
So I can't answer this question, and of course I've already made it pretty clear that I don't think it's a natural question, because although the adele rings are isomorphic, there is no canonical isomorphism between them. However, as Daniel Loughran said, if you somehow had this for all $n$ at once then one might attempt to deduce that the conjectural Tannakian category of representations of the global Langlands groups of $K$ and $L$ are isomorphic, and this would imply that the absolute Galois groups of $K$ and $L$ were isomorphic (these being the connected components of the Langlands groups) and then by Neukirch-Uchida $K$ and $L$ would be isomorphic. This is no good really though because firstly one needs the isomorphisms for all $n$, secondly one needs more than isomorphisms, one needs isomorphisms compatible with a non-existent tensor product, and thirdly one needs the Langlands program. On the other hand it's not true in general that two different discrete groups give two different spectra -- because as is well known we can't hear the shape of a drum. So this question seems theoretically hard but also perhaps very unnatural.<|endoftext|>
TITLE: Product-like structures on spheres
QUESTION [11 upvotes]: For $i=1,2$, let $j_i$ denote the inclusion of $S^n$ into the product $S^n \times S^n$ as the $i^{\text{th}}$ factor. I would very much like to know the answer to the following question, which seems to be very basic: 
For which $(a_1,a_2) \in \mathbb Z^2$ is there a map $$m \colon S^n \times S^n \to S^n,$$ such that the map $$m \circ j_i\colon S^n \to S^n$$ has degree $a_i$?
Here are some thoughts: 
(1) If we can do it for $(a_1,a_2)$ then we can also do it for $(\lambda_1a_1,\lambda_2a_2)$ for any $\lambda_1,\lambda_2$, by precomposing with a suitable map.
(2) If $n$ is even, there does not exist such a map if both $a_1$ and $a_2$ are nonzero: Indeed, if $m$ was as required, the induced map $H^{\ast}(S^n \times S^n) \leftarrow H^{\ast}(S^n)$ sends the fundamental class $[S^n] \in H^n(S^n)$ to $a_1([S^n] \times 1) + a_2 (1 \times [S^n])$. But then $[S^n] \cup [S^n] = 0$ gets send to (here we use that $n$ is even) $a_1a_2([S^n] \times [S^n]) \neq 0$, a contradiction.
(3) If $n = 1,3$ or $7$, then $S^n$ is an $H$-space, which means we can find a $m$ for $(1,1)$ and hence for all $(a_1,a_2)$.
(4) If $n$ is odd but different from $1,3,7$, it is known (by Adams) that $S^n$ is not an H-space, hence we can not find an $m$ for $(1,1)$. Can we find one for $(2,2)$ or even $(1,2)$?
(5) The question can be rephrased in terms of the homotopy group $\pi_{2n-1}(S^n \vee S^n)$ which can be computed using Hilton's theorem, but I do not see how this can be useful.

REPLY [18 votes]: Your condition determines the map $a_1 \vee a_2 : S^n \vee S^n \to S^n$ on the $n$-skeleton, so the question is when does this extend over the $2n$-cell of $S^n \times S^n$. The $2n$-cell is by definition attached along the universal Whitehead product $[\iota_1, \iota_2 ] \in \pi_{2n-1}(S^n \vee S^n)$, so the question becomes under what conditions is
$$[a_1, a_2] =0 \in \pi_{2n-1}(S^n).$$
The Whitehead product is bilinear, so this asks when $a_1 a_2 [\iota, \iota] =0 \in \pi_{2n-1}(S^n)$. 
As you point out, when $n$ is even $[\iota, \iota]$ has infinite order (as it has Hopf invariant $\pm 2$), so this happens if and only if $a_1 a_2=0$.
If $n$ is odd then the graded-commutativity of the Whitehead product means that $2[\iota,\iota]=0$. Now it follows from the EHP sequence that $[\iota, \iota]=0$ if and only if $\pi_{2n+1}(S^{n+1})$ has an element of Hopf invariant 1, which by Adams means $n=1,3,7$. Thus if $n=1,3,7$ then there are no conditions on the $a_i$, and otherwise the map exists if and only if $a_1 a_2$ is even.<|endoftext|>
TITLE: Fourier series of a Wightman field
QUESTION [8 upvotes]: From a proof that 2D Wightman CFT leads to a vertex algebra [1]:
Let
$$
Y(a,z):=\frac{1}{(1+z)^{2\Delta_a}}\Phi_a\left(i\frac{1-z}{1+z}\right),\quad\text{with}\quad |z|<1.
$$
Here $\Delta_a\ge 0$ is conformal weight of $\Phi_a$, and $\Phi_a$ is the scalar field satisfying the usual Wightman axioms (Poincare covariance, stable vacuum and positive spectrum of momentum, completeness, i.e. that polynomials of smeared fields applied on vacuum are dense in the Hilbert space, and locality (p. 6 of [1])) and in addition, satisfying conformal covariance
$$
U(q,\Lambda,b)\Phi_a(x)U(q,\Lambda,b)^{-1}=(1+2x\cdot b+|x|^2 |b|^2)^{-\Delta_a}\Phi_a((q,\Lambda,b)\cdot x)
$$
with $(q,\Lambda,b)\mapsto U(q,\Lambda,b)$ being a unitary representation of the conformal group, fixing the vacuum vector $|0\rangle$.
On p. 11 of [1] the author writes: 

We expand a chiral field $Y(a,z)$ in a Fourier series:
  $$
 Y(a,z)=\sum_n a_{(n)}z^{-n-1},\quad\quad(\star)
$$
  where $a_{(n)}\in\mathrm{End}\; \mathcal{D}$. 

Here $\mathcal{D}$ is the span of polynomials in the fields applied on the vacuum and is dense by completeness assumption.
I am not certain if this expansion is justified. Especially because it allows the author to prove that we indeed get a vertex algebra and hence we have an operator product expansion of the fields. Moreover, I found in another book [2] a similar proof on p. 190 and there the author writes

The operator product expansion (Axiom 6 on p. 168) of the primary fields allows to understand the fields $\Phi_a$ as fields
  $$
\Phi_a(z)=\sum a_{(n)} z^{-n-1} \in \mathrm{End} D [[z,z^{-1}]]
$$
  in the sense of vertex algebras.

Here $D$ is some dense domain in Hilbert space which is left invariant by fields, i.e. in [2] completeness is not assumed.
In other words, the author of [2] needs the existance of operator product expansion to write such a Fourier expansion.
My question:
$$ \text{is }(\star)\text{ true?} $$

References:
[1] V. Kac. Vertex Algebras for Beginners, 1998.
[2] M. Schottenloher. A Mathematical Introduction to Conformal Field Theory, 2008.

REPLY [2 votes]: Using Marcel Bischoff's comment:
$$
Y(a,z)=\sum_n a_{(n)} z^{-n-1} \iff a_{(m)}=\frac{1}{2\pi i} \oint\limits_{|z|=1} Y(a,z)z^{m} \,\mathrm{d}z.
$$
Since the author of [1] starts with the Schwartz space $\mathscr{S}(\mathbb{R})$, we need to show that $Y(a,z)$ is analytic in $|z|<1$.
By definition,
$$
  Y(a,z)=\frac{1}{(1+z)^{2\Delta_a}}\Phi\left(i\frac{1-z}{1+z}\right)\quad\text{with}\quad |z|<1.
$$
Clearly,
$$
\frac{1}{(1+z)^{2\Delta_a}}
$$
is analytic in the domain $|z|<1$ since it is holomorphic. Letting
$$
t:=i\frac{1-z}{1+z}
$$
we note that $|z|<1$ gets mapped to $\mathrm{Im}\, t>0$, so we need to show that $\Phi_a(t)$ can be analytically continued to $\mathrm{Im}\, t>0$. We now reintroduce the second coordinate, since the joint spectrum of momentum operators is positive. From [1]
$$
t:=x^0-x^1,\quad \bar{t}:=x^0+x^1\quad\text{and}\quad x:=(x^0,x^1)
$$
with $t,\bar{t}$ the lightcone coordinates.
Since the polynomials in fields applied to the vacuum are dense by completeness assumption, it suffices to prove that any
$$
\Phi_{a_1}(x_1)\ldots\Phi_{a_n}(x_n)|0\rangle
$$
can be analytically continued to the domain
$$
\{\mathrm{Im}\, x^0_1>0\}\times\dots\times\{\mathrm{Im}\, x^0_n>0\}\text{ such that } \mathrm{Im}\,x^0_i\ne \mathrm{Im}\,x^0_j \text{ for } i\ne j.
$$
The fact that coinciding points do not matter follows from locality [1]
$$
(z-w)^N[Y(a,z),Y(b,w)]=0 \text{ for } N\gg0.
$$
We now follow Section 2 of [3]. Let
$$
 \Psi(x_1,\ldots,x_n):=\Phi(x_1)\ldots\Phi(x_n)|0\rangle
$$
with $\Phi$ being a Hermitian scalar field. Generalizations to fields of arbitrary spin are given in Section 9 of [3] (in 4 dimensions, here 2D). Note that $\Psi$ (smeared with a test function) belongs to the Hilbert space $\mathcal{H}$ by completeness. By positive spectrum
$$
\Psi(x_1,\ldots,x_n)=\int d^2 p\; d^2 q_1\ldots d^2 q_{n-1} \tilde{\Psi}(p,q_1,\ldots,q_{n-1})e^{i\left(p x_1+\sum q_j (x_{j+1} -x_j)\right)}.
$$
Here $\tilde{\Psi}$ is non-zero only if $p^0\ge0$ and all $q_i^0\ge 0$. Thus, $\Psi$ can be analytically continued to a vector-valued analytic function $\Psi\in\mathcal{H}$, i.e. we have
\begin{align}
\Psi(z_1,\dots,z_n)\quad\text{of}\quad z_k=x_k+i y_k\quad & \text{defined and holomorphic for}\\
 &y_1\in V_+ \text{ and } y_j-y_i\in V_+ \text{ if } j>i\\
  &(V_+ \text{ is the open forward lightcone}).
\end{align}
In [4] it was shown that locality and the edge of the wedge theorem allows to extend the domain of definition and analyticity even further:
Claim. $\Psi(z_1,\ldots,z_n)$ belongs to $\mathcal{H}$ and is analytic in the $z_k$ in a connected domain which includes the Euclidean points with $z_k=(i y^0_k, x_k)$ such that $y^0_k>0$ for all $k$, and $z_i^0\ne z_j^0$ if $i\ne j$.
Proof of Claim.
    Fix $\pi$ to be any permutation of $(1,\ldots,n)$ and
$$
  \Psi^{\pi} (z_1,\ldots,z_n):=\Psi(z_{\pi(1)},\ldots,z_{\pi(n)}),\quad z_k=x_k+i y_k.
$$
    By the above, $\Psi^{\pi}$ is well-defined and holomorphic in a domain containing the Euclidean points with $0<y^0_{\pi(1)}<\ldots<y^0_{\pi(n)}$. Furthermore, by locality
$$
 \Psi^{\pi}(x_1,\ldots,x_n)=\Psi(x_1,\ldots,x_n)\quad\text{if}\quad x_k\in\mathbb{R}:(x_i-x_j)^2<0 \quad\forall i\ne j,
$$
    i.e. all $\Psi^{\pi}$'s are equal on a real neighborhood. Now the Edge of the Wedge Theorem [5] shows that they are analytic continuations of one and the same analytic function. Moreover, the domain of analyticity of this function must contain the domains of analyticity each $\Psi^{\pi}$.
Q.E.D.

Remark. I think $\Psi(x_1,\ldots,x_n)=\Phi(x_1)\ldots\Phi(x_n)|0\rangle$ generalizes to arbitrary bosonic fields trivially and for fermionic fields we would pick up some minus signs.

References:
[1] V. Kac. Vertex Algebras for Beginners, 1998.
[2] M. Schottenloher. A Mathematical Introduction to Conformal Field Theory, 2008.
[3] Lüscher, M. and Mack, G. Global conformal invariance in quantum field theory. Communications in Mathematical Physics, Springer-Verlag, 1975, 41, 203-234.
[4] Glaser, V. On the equivalence of the Euclidean and Wightman formulation of field theory. Comm. Math. Phys., 1974, 37, 257-272.
[5] Streater, R. F. and Wightman, A. S. PCT, Spin and Statistics, and All That, W. A. Benjamin, Inc., 1964.<|endoftext|>
TITLE: polynomials and symmetric functions
QUESTION [8 upvotes]: Suppose I have a polynomial function $f\in \mathbb{Z}[x_1, \dotsc, x_k],$ such that whenever $r_1, \dotsc, r_k$ are roots of a monic polynomial of degree $k$ with integer coefficients, we have $f(r_1, \dotsc, r_k) \in \mathbb{Z}.$ Is it true that $f$ is a symmetric function of its arguments?
Note: this was asked on MSE, but only resulted in an inconclusive discussion with Qiaochu...
EDIT Fedor Petrov's argument certainly answers the question quite elegantly, but I was thinking of the case where $r_1\dotsc, r_k$ are roots of an irreducible polynomial, so the reduction to $k=2$ is not quite so obvious. If there is an argument in that setting as well, that would be very much of interest.

REPLY [9 votes]: Consider the minimal polynomial $P$ of $f$ over the fixed field
$\mathbb Q(\sigma_1, \ldots, \sigma_k) = \mathbb Q(x_1, \ldots, x_k)^{S_k}$
(where the $\sigma_j$ are the elementary symmetric polynomials), so
$P$ is irreducible and $P(f) = 0$. Since $f$ is a polynomial,
the coefficients of $P$ will actually be polynomials in the $\sigma_j$.
By the Hilbert Irreducibility Theorem, $P$ will
remain irreducible when we specialize $\sigma_1, \ldots, \sigma_k$
to rational numbers, outside a thin set of such $k$-tuples.
But the set of $k$-tuples corresponding to $(x_1, \ldots, x_k)$ forming
a Galois orbit is not thin (its complement is thin: it consists of the
points $(a_1, \ldots, a_k)$ such that
$h(T) = T^k - a_1 T^{k-1} + \ldots \pm a_k$
is reducible; this set can be written as a finite union of images
of $\mathbb Q$-rational points under dominant morphisms that correspond
to the various possibilities of factoring $h$).
So there are (plenty of) integer (see this Wikipedia entry)
specializations $\boldsymbol{a}$ with irreducible $h$ such that the specialized $P_{\boldsymbol{a}}$
is irreducible. But by assumption, $P_{\boldsymbol{a}}$
has a rational root (since
$f$ evaluated at the roots $\boldsymbol{r}$ of $h$ is an integer):
$P_{\boldsymbol{a}}(f(\boldsymbol{r})) = 0$.
So $P$ must have degree 1,
and $f$ is in $\mathbb Q[\sigma_1, \ldots, \sigma_k]$.
In more detail: Write
$$P(X) = X^n + p_{n-1}(\sigma_1, \ldots, \sigma_k) X^{n-1} + \ldots + p_0(\sigma_1, \ldots, \sigma_k) .$$
Then there are integers $\boldsymbol{a} = (a_1, \ldots, a_k)$
such that $h_{\boldsymbol{a}} = T^n - a_1 T^{n-1} + \ldots \pm a_k$
is irreducible and
$$P_{\boldsymbol{a}}(X) = X^n + p_{n-1}(a_1, \ldots, a_k) X^{n-1} + \ldots + p_0(a_1, \ldots, a_k) \in {\mathbb Q}[X]$$
is also irreducible. Let $\boldsymbol{r} = (r_1, \ldots, r_k)$ be
the roots of $h_{\boldsymbol{a}}$.
Then $f(\boldsymbol{r}) = m \in \mathbb Z$. On the other hand,
$\sigma_j(\boldsymbol{r}) = a_j$, so
$$0 = P_{\boldsymbol{a}}(f(\boldsymbol{r}))
 = m^n + p_{n-1}(\boldsymbol{a}) m^{n-1} + \ldots + p_0(\boldsymbol{a}),$$
i.e., $m$ is a root of the irreducible polynomial $P_{\boldsymbol{a}}$.<|endoftext|>
TITLE: A property of Mersenne primes
QUESTION [5 upvotes]: Consider the effect of $f(x)=\frac12(x-x^{-1})$ on the residues mod $p$ (plus $\infty$) of a Mersenne prime $p$. You get the following tree (example $p=7$):
$$
\begin{array}{ccccccc}
4\\
&\searrow\\
&&1\\
&\nearrow&&\searrow\\
5\\
&&&&0&\rightarrow&\infty\\
2\\
&\searrow&&\nearrow\\
&&6\\
&\nearrow\\
3\\
\end{array}
$$
Proving that a binary tree always results is far beyond my abilities. I merely observed it. Can someone prove it?
Addendum: Primes of the form $k\cdot2^p-1$ or $k\cdot2^p+1$ with small $k$ seem to act quite similar tree-ish.

REPLY [10 votes]: The critical points of $f$ are at $x=\pm i$, so we try a projective
(a.k.a. fractional linear) change of variable  that puts these
critical points at $0$ and $\infty$, namely $x = \alpha(y)$ where
$\alpha(Y) = i(Y+1) \, / \, (Y-1)$, $\alpha^{-1}(X) = (X+i) \, / \, (X-i)$,
and find that $f(\alpha(y)) = \alpha(y^2)$.
Therefore $-$ unusually for a degree-2 rational function $-$ 
the iterates of $f$ have a closed form,
$$
f^{(n)}(x) = \alpha^{-1}(\alpha(x)^{2^n}).
$$
Now if $p=2^l-1$ is a Mersenne prime then $i$ generates
a field of $p^2$ elements, call it $F$,
and $(x+i)/(x-i)$ is an element of the norm-$1$ subgroup of
$F^*$, which is cyclic of order $p+1 = 2^l$.  Since $y \mapsto y^2$
is the doubling map on this group, its graph has the binary-tree structure
that you observed.
Likewise if $p = k 2^l - 1$ for $l \geq 2$ and $k$ odd then the graph is
the union of $k$ binary trees, and if $p = k 2^l + 1$ then
there are square roots of $-1$ in ${\bf Z} / p {\bf Z}$,
and the graph is the union of $k$ binary trees together with two
isolated points at those square roots (corresponding to $y = 0,\infty$).<|endoftext|>
TITLE: Boolean-Valued Models: Why is $\| x=y \| \cdot \| \phi(x) \| \leq \| \phi(y) \|$?
QUESTION [8 upvotes]: Let $B$ be a complete Boolean algebra. Jech defines a Boolean-valued model $\mathfrak{A}$ of the language of set theory to consist of a Boolean universe $A$ and functions of two variables with values in $B$,
$\qquad \qquad \| x=y \|, \qquad \| x \in y \|$
that safisfy the following:
$ (i)\  \ \ \| x=x \| = 1 \\
(ii) \ \ \| x=y \| = \|y = x \| \\
(iii)\  \| x=y \| \cdot \| y=z \| \leq \| x=z \| \\
(iv) \ \ \| x\in y \| \cdot \|v=x \| \cdot \|w=y \| \leq \|v \in w \|  $
For every formula $\phi(a_1,\ldots,a_n)$, we define the Boolean value $\| \phi(a_1,\ldots,a_n) \|$ of $\phi$ as follows: 
$(a)$ For atomic formulas, we have the functions $\| x=y \|, \| x \in y \| $
$(b)$ If $\phi$ is negation, disjunction, $\exists x \, \psi$,
$\qquad \| \lnot \phi(a_1,\ldots,a_n) \| = - \| \phi(a_1,\ldots,a_n) \| $
$\qquad \| (\phi \vee \psi)(a_1,\ldots,a_n) \| = \| \phi(a_1,\ldots,a_n) \| + \| \psi (a_1,\ldots,a_n) \|$
$\qquad \| \exists x \, \psi(x,a_1,\ldots,a_n) \| = \underset{a\in A}{\sum} \| \psi (a,a_1,\ldots,a_n)\|$
Next, Jech says that it's easy to prove 
$\qquad \qquad \qquad \| x=y \| \cdot \|\phi(x)\| \leq \|\phi(y)\|$
However, I'm not seeing it. It seems clear that it should be a proof by induction on the complexity of $\phi$, and the inductive step is easy in the case of disjunction and existential, but why does it hold for negation? That is, why is it that if $ \| x=y \| \cdot \|\phi(x)\| \leq \|\phi(y)\|$, then $\| x=y \| \cdot \|\lnot \phi(x)\| \leq \| \lnot \phi(y)\|$?
Thank you!

REPLY [3 votes]: One way to show this is by first noting that (i)-(iv) ensure the atomic instances of the identity axioms:
$x = y \to [\phi(x) \leftrightarrow \phi(y)]$
get value 1. Then observe, by a simple induction on the complexity of $\phi$, that the non-atomic instances follow from the atomic instances in first-order logic. So they too get value 1.<|endoftext|>
TITLE: Spectral properties of the Laplace operator and topological properties
QUESTION [17 upvotes]: Suppose that $M$ is a closed Riemannian manifold: one can construct the so called Laplace-Beltrami operator on $M$. Its spectrum contains some information of the underlying manifold: for example its dimension is encoded by the asymptotic behaviour of eigenvalues. As far as I know, there was a problem whether two isospectral closed manifolds are isometric-the negative answer was found by Milnor. So from this story I suspect that people believed that the spectrum of Laplacian will contain a lot of (geometric) information about the underlying manifold. My question is the following: 

Is it possible to extract the information about the fundamental group of $M$ from the spectrum of Laplace operator?

REPLY [3 votes]: When $M$ is negatively curved, and especially when the curvature is constant, the distribution of the eigenvalues tells something about the distribution of lengths of closed geodesics. This is because given a closed geodesic, you can construct an approximate eigenfunction. The seminal work was by A. Selberg (trace formula), improved by P. Sarnak in his PhD thesis. There have been a lot of contributors for the general theory, including Y. Colin de Verdière and more recently N. Anantharaman.<|endoftext|>
TITLE: A lift of the second Chern class
QUESTION [7 upvotes]: Let $X$ be a complex manifold (not necessarily Kahler or even compact). 
For the first Chern class $c_1(E)\in H^2(X,Z)$ of a holomorphic vector 
bundle $E\to X$ there is an obvious lift to $H^1(X, O^*)$, namely the determinant. 
Question: Is there a natural (in some sense) lift of the second Chern class to $H^3(X, O^*)$?
(I have heard that there is a lift to the Deligne cohomology group, which must be related. 
However, as I know nothing about Deligne cohomology, I have no clue if it is possible 
to lift from there to $H^3(X, O^*)$.)

REPLY [5 votes]: There are various instances of the first Chern class:

In Betti cohomology $c_1\in H^2(X, \mathbf{Z})$,
In Dolbeault cohomology $c_1\in H^1(X, \Omega^1)$,
In de Rham cohomology $c_1\in H^2_{dR}(X)$.

These can be combined to the class in Deligne cohomology $c^D_1\in H^2_D(X, \mathbf{Z}(1))\cong H^1(X, \mathcal{O}^\times)$ which is the determinant that you mention. From that class you can recover the previous classes: the morphism $\mathrm{dlog}\colon\mathcal{O}^\times\rightarrow (\Omega^1\rightarrow\Omega^2\rightarrow...)$ sends $c_1^D$ to a class in $F^1 H^2_{dR}(X)$ which is a subgroup of $H^2_{dR}(X)$ and it also projects to $H^1(X, \Omega^1)$. The connecting homomorphism for the exponential sequence gives $H^1(X, \mathcal{O}^\times)\rightarrow H^2(X, \mathbf{Z})$.
This story continues for higher Chern classes. You have a de Rham Chern class $c_n\in F^nH^{2n}_{dR}(X)$ and it has a refinement to the Deligne cohomology class $c_n^D\in H^{2n}_D(X, \mathbf{Z}(n))$. I believe Chern classes in Deligne cohomology are defined by appealing to the splitting principle by writing everything in terms of first Chern classes that you already know and then using multiplication on the Deligne cohomology.
For $n=2$ this gives the second Chern class $c_2^D\in H^3(X, \mathcal{O}^\times\rightarrow \Omega^1)$ which projects to $H^3(X, \mathcal{O}^\times)$.
If $X$ is algebraic, one instead has Chern classes in Chow groups $c_n\in \mathrm{CH}^n(X)$ which gives the same answer for $n=1$, but totally different for $n=2$.<|endoftext|>
TITLE: Confusion with practically implementing rational approximations
QUESTION [10 upvotes]: Writing a program visualizing Ford circles I've encountered a seemingly purely programmatic puzzle but then gradually realized there are some mathematical aspects of it which I don't understand. Let me start with an output (it is from a question on Mathematica SE)

This is a particular case - zooming in to a real number $x$ ($\frac1\pi$ in this case). In some detail, I do the following: at each step, I increase the zoom factor by the same fixed number $z$ ($1.04$ in this case); so, it is (roughly) $z^n$ at the $n$th step. I then consider the rectangle $(x-z^{-n},x+z^{-n})\times(0,z^{-n})$. I have certain amount of Ford circles which showed up in the previous rectangles. I then throw out those which drop out of the picture. Then, I check in succession all possible new denominators $q$ of those ratios $\frac pq$ such that their Ford circle radii $\frac1{2q^2}$ would exceed one pixel in the picture (in this case, the picture is roughly $500\times250$ pixels, so the condition is roughly $q<\sqrt{500z^n}$). Among those new $q$ I then choose those $p$ with $p$ coprime to $q$ and the Ford circle of $\frac pq$ intersecting my $n$th rectangle.
Thus at each step I throw out some $\frac pq$s with $q\leqslant\sqrt{500z^{n-1}}$ and add some new ones with $\sqrt{500z^{n-1}}<q\leqslant\sqrt{500z^n}$.

Now comes the question

From the above description, it is clear that the number of new potential denominators I have to check grows roughly by a factor of $\sqrt z$ after each step (measuring actual program runs confirms this). On the other hand, as the picture shows, the number of circles on each frame is roughly the same; again, counting them programmatically confirms this (but also it is clear that by physical reasons the number of all possible circles that can be distinguished on a drawing like this is bounded). This suggests that there might be some rational approximation algorithm which might output this fixed amount of rationals by using the same amount of resources for each step. More precisely,

Given a real number $x$ and another one $z=1+\varepsilon$, and also another one $N$, is there an algorithm which for every natural $n$ would give the list of all irreducible fractions $\frac pq$ with $\sqrt{Nz^{n-1}}<q\leqslant\sqrt{Nz^n}$ and $x-z^{-n}-\frac1{2q^2}\leqslant\frac pq\leqslant x+z^{-n}+\frac1{2q^2}$ in the same time and using the same amount of memory for all $n$?

REPLY [3 votes]: It depends on the model of computation you're using. If your memory slots can contain arbitrarily large integers (rather than single bits), and you can perform arithmetic operations in constant time, the answer is yes. Basically you use the following algorithm to step through the Farey sequence of the correct order:
https://en.wikipedia.org/wiki/Farey_sequence#Next_term
Alternatively, if you use the traditional model of computation where you can only manipulate a bounded number of bits in constant time, then clearly no such algorithm exists (since merely printing the digits of $q$ takes time proportional to $\log(q)$).<|endoftext|>
TITLE: Existence of solutions to a nonlinear algebraic equation
QUESTION [5 upvotes]: How can we prove that equation (1)  has solutions for every $p$.  I mean, is there an analytic method that can be used  to show  that there exist solutions for every $p$ for this  nonlinear equation:
\begin{equation}
\left\{
\begin{array}{ccc}
x_1^2 x_2  \cdots x_{p-1} x_p+x_1 &=& 1 \, ,\\
&&\\
 x_1 x_2^2 x_3 \cdots x_{p-1} x_p+x_1 x_2 &=& 1 \, ,\\
&&\\
x_1 x_2 x_3^2 x_4 \cdots x_{p-1} x_p+x_1 x_2 x_3&=& 1\, , \\
&&\\
\vdots & \vdots & \vdots \\
&&\\
x_1 x_2 x_3 x_4 \cdots x_{p-1} x_p^2+x_1 x_2 \cdots  x_p&=& 1\, .
\end{array}
\right.
\end{equation}
For example,  let ($p=3$). Then we have:
\begin{equation}
\left\{
\begin{array}{ccc}
x_1^2 x_2  x_3+x_1 &=& 1 \, ,\\
&&\\
 x_1 x_2^2 x_3 +x_1 x_2 &=& 1 \, ,\\
&&\\
x_1 x_2 x_3^2 +x_1 x_2 x_3&=& 1\, .
\end{array}
\right.
\end{equation}
So is there an analytic method for existence of solution for equation (2). For this example I used MAPLE software and got these solutions:
\begin{equation*}
\left\{
\begin{array}{ccc}
a &=&  0.658418845314095780 \, ,\\
&&\\
b &=& 0.849466898144101812 \, ,\\
&&\\
c&=& 0.927561975482924960  \, .
\end{array}
\right.
\end{equation*}

REPLY [10 votes]: Note that no $x_k$ can vanish. If you put $ u :=x_1x_2\dots x_p  $  the system gives inductively, for $k=1,\dots,p$:
$$\frac{1}{x_1 x_2\dots x_k}=1+u+\dots+u^k . $$
In particular $u$ solves
$$\frac{1}{u}=1+u+\dots+u^p . $$
Incidentally, $u\neq 1$, so we can express the solutions to the systems simply as
$$x_k=\frac{u^{k}-1}{u^{k+1}-1},$$
in terms of the solutions $u\neq 1$ to the equation
$$u^{p+2}-2u+1=0. $$
Also note that, by elementary arguments, if $p$ is even the latter equation has a unique real solution (besides $u=1$), which is positive, while if $p$ is odd it has a positive and a negative solution (besides $u=1$). So the system has one or two real solutions according whether $p$ is even or odd, and in any case a unique solution 
if $x_k$ are assumed to be real and positive.
To complete the analytic solution, we may solve $u^{p+2}-2u+1=0$ by
series using the Lagrange inversion formula to invert $u-u^{p+2}/2$ at $1/2$: for the solution $0<u_p<1$ one gets
$$u_p=\frac{1}{2}\sum_{k=0}^\infty {pk+2k\choose k}  \frac{2^{-pk-2k}}{pk + k+1}.$$
For instance, 
$$u_3=\frac{1}{2}\sum_{k=0}^\infty {5k\choose k}  \frac{32^{-k}}{4k+1}=0.5187900635...$$
that corresponds to the solutions $x_1=a, x_2=b, x_3=c$ you got by MAPLE.<|endoftext|>
TITLE: Does the limit of this product over primes converge for all $\Re(s) > \frac12$?
QUESTION [6 upvotes]: Numerical evidence suggests that:
$$\displaystyle F(s):= \lim_{N \to \infty}\, \ln^s\left(p_N\right)\, \prod_{n=1}^N \left(\dfrac{\left(p_n-1\right)^s}{p_n^s-1} -\frac{1}{p_n^s}\right)$$
with $p_n$ is the n-th prime number, converges for all $\Re(s) > \frac12$.
Note that for $s=1$ the function reduces to the well known limit for $e^{-\gamma}$ (see here).
Could this be proven?

REPLY [8 votes]: As you mention, $F(1)=e^{-\gamma}$, so
$$
\begin{align*}
e^{s\gamma} F(s)=\frac{F(s)}{F(1)^s}&=\frac{\lim_{x\to\infty} \log^s x \prod_{p\leq x} \frac{p^s(p-1)^s-p^s+1}{p^s(p^s-1)}}{\lim_{x\to\infty} \log^s x \prod_{p\leq x} \frac{(p-1)^s}{p^s}}\\
&= \prod_{p}\frac{p^s(p-1)^s-p^s+1}{(p^s-1)(p-1)^s}\\
&=\prod_{p}\left(1-\frac{p^s-(p-1)^s-1}{(p^s-1)(p-1)^s}\right).
\end{align*}
$$
This product converges because
$$
\frac{p^s-(p-1)^s-1}{(p^s-1)(p-1)^s}=\frac{s}{p^{s+1}}+O\left(\frac{1}{p^{2s}}\right)+O\left(\frac{1}{p^{s+2}}\right)
$$
as $p\to\infty$.

REPLY [3 votes]: If you write the general terms in terms of $x = 1/p_n,$ when $x\>1/2,$ the expansion in $x$ starts as $1-s x + O(x^{2 s})\dots,$ so your limit converges for $s>1/2.$ For $s=1/2,$ the expansion starts with $1+x/2,$ so your limit blows up (the product of the first $n$ terms, including the log fudge factor, should be of order of $\log n.$<|endoftext|>
TITLE: Examples and Counterexamples in Commutative Algebra
QUESTION [13 upvotes]: There are Counterexamples in Analysis and Counterexamples in Topology. Is there any similar book for commutative algebra? I want to see some more (counter)examples for Atiyah and MacDonald's book. Let us say "zoo" of rings and modules.

REPLY [10 votes]: The best I can think of is Harry Clayton Hutchins' "Examples of Commutative Rings" published by UChicago Press (he was a student of Kaplansky). The second part of the book is just a long list of rings satisfying strange properties ("this, this and this, but not that"). The one difference with counterexamples in top is the lack of a good index/table putting everything together, so it can be (quite) hard to find a specific example.<|endoftext|>
TITLE: Euler's constant: irrationality and proof theory
QUESTION [6 upvotes]: Let $\gamma$ represent Euler's constant. Is there a real number $x$ such that there is a proof within Zermelo-Fraenkel set theory (ZF) that $x$ is irrational and there is also a proof within ZF that $\gamma + x$ is irrational?

REPLY [8 votes]: I got the feeling that maybe what the OP wanted was: Provide an   explicit constructive example   of an irrational $x$ such that $x+\gamma$ is irrational. In any case, under this respect the question seems interesting as well, so I would like to consider it. 
While the existence is quite clear (e.g. there exists $x\in\{\sqrt{2},2\sqrt{2} \}$ with this property, and also  $ x\in\{\sqrt{2},\gamma \}$), giving an explicit example is a  bit less obvious. 
As suggested in the various answers, we may consider the problem   more in general  than for the case where $\gamma$ is the Euler-Mascheroni constant. I'd like to formalize the question in   the following Problem:

Split a given  computable  real number $\gamma$ into a sum of two irrational computable real numbers.

There is a cute construction that uses a  "balanced  factorial representation" of real numbers. Recall that any real number $\gamma$
can be written in the form 
$$\gamma=\sum_{k\ge 1}\frac{\gamma_k}{k!}$$
with coefficients $\gamma_k\in\mathbb{Z}$ verifying 
$$|\gamma_k |\le k-1 \ ,$$
for any $k\ge 2$. A number of this form is rational if and only if the sequence $\gamma_k$ is either eventually equal to  $0$, or eventually equal to $k-1$, or eventually equal to $-(k-1)$. Moreover,   $\gamma$ is a computable number  if and only if the coefficients $\gamma_k$ are given by some computable function. (Note: as remarked in comments, this is generally not true for  base systems with nonnegative coefficients: the binary expansion of a computable number may not be computable, which reflects the fact that one may not be able to decide whether $\gamma$ is larger or smaller than a given rational approximation of its, no matter how good). 
Given the  two facts above, the problem is then translated into: split computably the sequence $\gamma_k$  as a sum $ \gamma_k=\alpha_k+\beta_k$, in such a way that $$ 0<|\alpha_k|<k-1$$ and $$ 0<|\beta_k|<k-1$$
for infinitely many $k$. Again, changing sign coefficients make everything quite easy to deal with: we can e.g. eventually take  both $|\alpha_k|$ and $|\beta_k|$ in the interval $[k/4,3k/4]$.<|endoftext|>
TITLE: How to determine whether a power of eta function is a eigenform?
QUESTION [5 upvotes]: I find that it is complicated to do this from the definition. In fact, I know that $\eta^k(m z)$ is a eigenform for $k=1,2,3,4,6,8,12,24$ and $m=\frac{24}{gcd(k,24)}$. What I want to know is the cases that $k=5,7,11$. In addition, how to choose proper $N$ and find the weight $k$ as well as the $\chi(p)$. Thank you so much.

REPLY [2 votes]: The weight is $k/2$, so $\eta^{5}(24z)$, $\eta^{7}(24z)$ and $\eta^{11}(24z)$ are half-integer weight modular forms. The level $N = 576$,
because $\eta(24z) = \sum_{n=1}^{\infty} \chi_{12}(n) q^{n^{2}}$
is a single-variable theta series with level $576$ (this is stated as Corollary 1.62 in Ono's book "The Web of Modularity"). When you ask for $\chi(p)$, I presume you're asking for the character in the transformation law. This is 
$$
\chi_{12}(n) = \begin{cases} 1 & \text{ if } n \equiv 1 \text{ or } 11 \pmod{12} \\ -1 & \text{ if } n \equiv 5 \text{ or } 7 \pmod{12}. \end{cases}
$$
For clarify, the transformation law for $f(z) = \eta^{r}(24z)$ (if $r$ is odd) is
$$
  f\left(\frac{az+b}{cz+d}\right) = \chi_{12}(d) \left(\frac{c}{d}\right)^{r} \epsilon_{d}^{-r} (cz+d)^{r+1/2} f(z),
$$
where $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is any matrix in $\Gamma_{0}(576)$, $\epsilon_{d} = \begin{cases} 1 & \text{ if } d \equiv 1 \pmod{4}\\ i & \text{ if } d \equiv 3 \pmod{4} \end{cases}$. If $d > 0$,
$\left(\frac{c}{d}\right)$ is the usual Jacobi symbol, and
$$
  \left(\frac{c}{d}\right) = \begin{cases}
    \left(\frac{c}{|d|}\right) & \text{ if } d < 0 \text{ and } c > 0 \\
    -\left(\frac{c}{|d|}\right) & \text{ if } d < 0 \text{ and } c < 0.
\end{cases}
$$<|endoftext|>
TITLE: Asymptotic enumeration of magic squares
QUESTION [5 upvotes]: An order-$n$ magic square is an $n \times n$ matrix over the numbers $\{1, ... ,n^2\}$, each appearing exactly once, whose row and column sums are all equal. Sometimes the sums of the diagonals are required to be equal too.
These objects have a rich history, and they are hugely popular in recreational math. I remember being fascinated by constructions of magic squares when I was 10. It seems natural to ask how many order-$n$ magic squares are there? An exact formula is probably too much to hope for, but it is probably possible to give some asymptotic bounds.
Has anybody asked this question before? Are there known bounds on the number of order-$n$ magic squares?

REPLY [4 votes]: This has been asked before on MSE, here and here for example. According to the OEIS, not much is known. James E. Ward III wrote a paper in Mathematics Magazine where he derived an upper bound of $(n^2)!/8(2n+1)!$ (where the 8 in the denominator comes from the 8 symmetries of a magic square), but this is surely a gross overestimate.  As for lower bounds, there are numerous papers giving constructions of magic squares, but I am not aware of any that make an effort to derive an interesting lower bound on the total number of magic squares.
There has been quite a bit of research on the related question of the number of $n\times n$ squares with nonnegative integer entries whose rows and columns all sum to some given number $s$.  This question is more tractable to analyze than your problem, but is still very difficult, and for example I do not think that there is a known asymptotic formula for $s =n(n^2+1)/2$ (and in any case, this would yield an even worse upper bound than Ward's).<|endoftext|>
TITLE: Polynomials which always assume perfect power values
QUESTION [8 upvotes]: Let $f(x)$ be a non-constant polynomial with integer coefficients. It is a well-known result that if $f(n)$ is a square for all integers $n$, then $f$ must in fact be the square of a polynomial (see, for example: http://www.mast.queensu.ca/~murty/poly2.pdf).
My question is the following. Suppose that $\deg f = d$, and suppose that for every integer $n$, we have that $f(n)$ is a perfect $k$-th power for some $k_n > 1$ dividing $d$. I want to emphasize that $k_n$ is allowed to depend on $n$. For instance, it could be the case that $f$ is of degree $6$ and $f(2) = 2^3$ while $f(3) = 3^2$, and $f(n)$ is either a square or a cube (or both) for every $n$. 
Can we conclude that $f$ is a perfect $m$-th power, for some $m > 1$ dividing $d$?

REPLY [6 votes]: Here is a more elementary argument than that of nfcd23. It only assumes that $f(n)$ is a $k$th power for some $k$ depending on $n$, which need not be assumed to divide $d$. Over $\mathbf{C}$, write
$$f(x) = C \prod  (x - \alpha_i)^{r_i},$$
where the $\alpha_i$ are distinct. 
By Cebotarev, (or more simply, by Frobenius), we may find infinitely many primes $p$ such that $p$
 splits completely in $K = \mathbf{Q}(\alpha_i)$, and in addition, $p$ is prime to both $C$ and the difference $\alpha_i - \alpha_j$ of any pair of distinct roots. (The last two conditions hold automatically for all but finitely many primes.)
For each root $\alpha_i$, choose a prime $p_i$ of this form.
Then choose an integer $n$ (using the Chinese Remainder Theorem)
such that:
$$\text{$n$ is congruent to $\alpha_i$ modulo $p_i$ but not modulo $p^2_i$.}$$ 
This ensures that the $p_i$-adic valuation of $f(n)$ is $r_i$ for each $i$, because that is the valuation of $(n - \alpha_i)^{r_i}$, and by assumption, $p_i$ does not divide any of the other terms. If $f(n)$ is a perfect $k$th power, this implies that the greatest common divisor $r$ of all the $r_i$ must be divisible by $k$, and hence equal to $mk$ where $k > 1$.
 But that implies that $f(x) = A g(x)^{mk}$ for some $g(x)$. Once again letting $x = n$ and noting that $f(n)$ is a $k$th power, we deduce that $A = B^k$, and so
$$f(x) = B^k g(x)^{mk} = (B g(x)^m)^k.$$<|endoftext|>
TITLE: Convergence rate of Fagin's 0-1 law for first-order properties of random graphs
QUESTION [16 upvotes]: Fagin's 0-1 law for first-order properties of random graphs states that, for every first-order sentence in the logic of graphs, the probability that a uniformly random $n$-vertex graph models the sentence tends to either 0 or 1 as $n$ goes to infinity. It is known that testing, for a given sentence, whether the limit is 0 or whether it is 1 is PSPACE-complete (Grandjean, "Complexity of the first-order theory of almost all finite structures", Information and Control 1983). One possible approach to performing this sort of test would be to sample random graphs of sufficiently large size and test whether they model the sentence; but this would be limited both by the difficulty of testing whether a graph models a given sentence and also by the convergence rate of the 0-1 law, as that would control the size of the graphs needed in this sampling scheme.
What if anything is known and published about more explicit upper or lower bounds on the convergence rate of $P_S$, for the worst-case sentence $S$ of a given length, as a function of $S$? Or to put it another way, if one wishes to get the correct limit with probability bounded away from $1/2$, how large a graph should one sample?
One simple example of a sentence that converges slowly is given by the Ramsey property: does this graph contain either a clique of size $k$ or an independent set of size $k$? The sentence for this has length $O(k^2)$ but one needs to sample graphs of size at least exponential in $k$ to discover that the limit probability is one. But maybe there are other sentences that converge even more slowly?

REPLY [10 votes]: For definiteness, I will consider undirected graphs without self-loops, or equivalently, structures with a symmetric irreflexive binary predicate $E(x,y)$. These choices do not really matter.
The question mentions an exponential lower bound. It can be optimized as follows:
$$\phi=\forall x_0\dots\forall x_{k-1}\,\exists x_k\,\bigwedge_{i<k}(x_k\ne x_i\land E(x_k,x_i))$$
is a sentence of length $O(k)$ that holds in sufficiently large random graphs, but as it implies the existence of cliques of size $k+1$, it needs graphs of size $2^{\Omega(k)}=2^{\Omega(|\phi|)}$ to hold with probability above $1/2$.
Conversely, I will show an exponential upper bound, using a more generous measure: the number of distinct variables in the formula. Note that the number of variables is bounded by the length of the formula, but may be substantially smaller (as modal logicians know, already the $2$-variable fragment of first-order logic is quite rich, albeit in a different language).
Let me say that an extension axiom of rank $k$ is a sentence of the form
$$\forall x_0\dots\forall x_{k-1}\,\Bigl[\bigwedge_{i<j<k}x_i\ne x_j\to\exists x_k\,\Bigl(\bigwedge_{i<k}(x_k\ne x_i\land E(x_k,x_i)^{e_i})\Bigr)\Bigr],$$
where $e_0,\dots,e_{k-1}\in\{0,1\}$, and we use the notation $\phi^1=\phi$, $\phi^0=\neg\phi$. Let $E_k$ denote the conjunction of all $2^k$ extension axioms of rank $k$, together with the axiom stating that the universe has size at least $k$.

Observation: If $k\le l$, then $E_l$ implies $E_k$.

As is well known, $\def\rg{\mathit{RG}}\rg=\{E_k:k\in\mathbb N\}$ is an axiomatization of the complete theory of random graphs, which describes the sentences with limit probability $1$. Now, one easy way of showing the completeness of $\rg$ is to prove that if $A,B\models\rg$, then Duplicator has a winning strategy in the $n$-round Ehrenfeucht–Fraïssé game on $A,B$ for every $n$: indeed, whatever Spoiler plays in one of the models, Duplicator can match in the other one just by applying the relevant extension axiom.
In fact, if there are $k$ pebbles on each of the models, and Spoiler is placing a $(k+1)$th, then Duplicator can find a matching move by applying $E_k$. Thus, if we assume only that $A,B$ are models of $E_k$, then Duplicator can still win the $(k+1)$-round EF game, or even the $(k+1)$-pebble game. Consequently,

Proposition: Any two models of $E_k$ satisfy the same sentences using at most $k+1$ distinct variables.
In particular, if $\phi$ is a sentence using $k+1$ distinct variables, then $\phi$ or $\neg\phi$ follows from $E_k$.

On the other hand, if $n>k$, then a simple union bound shows that a random $n$-vertex graph fails to satisfy $E_k$ with probability at most
$$2^k\binom nk(1-2^{-k})^{n-k}\le n^k(1-2^{-k})^n\le\exp(k\log n-n2^{-k}).$$

Corollary: If $\phi$ is a sentence using $k+1$ distinct variables with limit probability $p\in\{0,1\}$, and $n>k$, then
$$|\Pr_G(G\models\phi)-p|\le\exp(k\log n-n2^{-k}),$$
where the probability is over random graphs with $n$ vertices. In particular, the probability is close to $p$ for $n$ larger than roughly $2^kk^2\log2$.<|endoftext|>
TITLE: geometric interpretation of derivation between two algebras
QUESTION [5 upvotes]: Given a smooth manifold $M,$ it is well known that any derivation of the algebra of smooth functions $C^{\infty}(M)$ can be seen (or it is associated to) a smooth vector field on $M.$ I am looking for a similar geometric meaning for a derivation $D: C^{\infty}(M)\rightarrow C^{\infty}(T^*M)$ along $\rho^*,$ where $\rho$ is the contangent bundle projection, it means: $D(fg) = \rho^*(f)D(g)+\rho^*(g)D(f),$ for every $f, g\in C^{\infty}(M).$ Thanks in advance.

REPLY [4 votes]: In general, given a map of smooth manifolds $\phi:N\to M$, derivations $C^\infty M \to C^\infty N$ are in canonical one to one correspondence with sections of the pullback bundel $\phi^* TM$ on $N$. Such a section is a smooth map that associates to each point $p\in N$ a tangent vector in $T_{\phi(p)}M$. These are sometimes called vector fields relative to the map $\phi$ and can also be pictured as infinitesimal deformations of the map $\phi$. In your case $N$ is the cotangent bundle of $M$ with its canonical projection.<|endoftext|>
TITLE: "Fractally self-similar" numbers
QUESTION [18 upvotes]: This is another question about visualization of Ford circles, the previous one being Confusion with practically implementing rational approximations. Here is an output of zooming into Ford circles at $\frac1{\sqrt2}$

Empirically I found that the pattern repeats periodically and arranged the number of frames so that one gets impression of a continuous infinite zoom (however if one looks very attentively there is a slight jump where the animation restarts).
What I want to know is whether the pattern indeed repeats rigorously for quadratic irrationalities, and whether for other irrational numbers the pattern occasionally becomes "almost" the same (clearly if the number is rational then the picture eventually starts degenerating into the horizontal line).
Since I do not actually know how exactly to formulate it, let me ask the question in form that practically occurred to me: for a real $x$ denote by $P_r(c)$ the picture of Ford circles in the rectangle $(x-c,x+c)\times(0,c)$ with resolution $r$. That is, features of size less than $rc$ cannot be observed; in particular, only the Ford circles of radius $>rc$ are visible, and moreover a circle cannot be distinguished from another one if they are both contained in an annulus of width $<rc$.
For which $x$ do there for any resolution $r$ exist two different $c$ and $c'$  such that the pictures $P_r(c)$ and $P_r(c')$ cannot be distinguished? (Well, they must be "sufficiently different" - say, there is still another $c<c''<c'$ such that $P_r(c'')$ can be distinguished from both.)
And the question about this question - what is (if any) a rigorous mathematical statement behind it?
(Later - decided to add the zoom for the golden ratio, here the jump is almost impossible to notice

Circles closest to the center alternate between left and right, ratios of their sizes must be something like consecutive Fibonacci numbers...

REPLY [7 votes]: Define the parent of a Ford circle to be the smaller of its two larger neighbors; then this parent relation defines the Stern–Brocot tree on the points of tangency of the circles. The path in this tree to a given real number is closely related to its continued fraction expansion. A path in the tree can be represented combinatorially by a binary sequence that determines whether to go left or right at each step, and this sequence is periodic exactly for the real numbers that have periodic continued fractions. That is, for the quadratic irrationals. It seems very likely to me that this periodicity in left-right steps is what is leading to your perceived periodicity in the relative sizes and positions of the circles, and that (once you determine an appropriate measure of shape similarity) this can be proven more rigorously: the shapes become approximately periodic when the left-right sequence in the tree is (after finitely many steps) exactly periodic.<|endoftext|>
TITLE: Spectrum of Ring of Smooth Functions on $\mathbb{R}^n$
QUESTION [12 upvotes]: When we define smooth manifold, we starting with topological space $M$ which localy homeomorphic to $\mathbb{R}^n$ and setting up sheaf $\mathscr{F}(M)$ of functions on it which localy isomorphic to $\mathscr{F}(\mathbb{R}^n)$.
Let's $M = \mathbb{R}^n$ and $R = \mathscr{F}(\mathbb{R}^n)$ is a ring of smooth functions on $M$.
Is it true that Spec($R$) with Zariski topology homeomorphic to $\mathbb{R}^n$?

REPLY [14 votes]: A trivial way you can tell this is false is that Spec of any ring is (quasi)compact, but $\mathbb{R}^n$ is not (if $n>0$).
A bit less trivially, this is still false if you replace $\mathbb{R}^n$ with a compact manifold (of dimension $>0$).  This follows from the following theorem.

Theorem:  Let $R$ be a commutative ring.  If $\operatorname{Spec}(R)$ is Hausdorff, it is totally disconnected.

Proof: Suppose $\operatorname{Spec}(R)$ is Hausdorff and let $p,q\in\operatorname{Spec}(R)$ be distinct.  Then there are disjoint open sets $U,V\subset\operatorname{Spec}(R)$ such that $p\in U$ and $q\in V$.  We may further choose $U$ and $V$ to be distinguished open sets, so there exist $f,g\in R$ such that $U=\operatorname{Spec}(R_f)$ and $V=\operatorname{Spec}(R_g)$.  But this implies $U$ and $V$ are compact, and hence closed since $\operatorname{Spec}(R)$ is Hausdorff.  Thus $U$ and $V$ are clopen sets separating $p$ and $q$.
However, it is true that if $M$ is a compact manifold, then $M\cong\operatorname{MaxSpec}(C^\infty(M))$.  To prove this, first note that for each $x\in M$, the ideal $m_x$ of functions vanishing at $x$ is a maximal ideal.  If $I\subseteq C^\infty(M)$ is any ideal not contained in $m_x$ for any $x$, then for each $x\in M$ we can choose an element $f_x\in I$ which does not vanish at $x$.  Replacing $f_x$ by $f_x\bar{f_x}$, we may assume that $f_x\geq 0$ everywhere.  The sets $U_x=\{y:f_x(y)\neq 0\}$ are an open cover of $M$, so there are finitely many $x_1,\dots, x_n$ such that $M=\bigcup_{i=1}^n U_{x_i}$.  The function $f=\sum_{i=1}^n f_{x_i}$ is then an element of $I$ which vanishes nowhere, and hence is a unit.  Thus $I$ is all of $C^\infty(M)$.  This shows that every proper ideal is contained in some $m_x$, and hence every maximal ideal is of the form $m_x$.  Now note that using bump functions, it is easy to show that the bijection $m_x\mapsto x$ is a continuous map $\operatorname{MaxSpec}(C^\infty(M))\to M$.  Since $\operatorname{MaxSpec}(C^\infty(M))$ is compact and $M$ is Hausdorff, this map is thus a homeomorphism.<|endoftext|>
TITLE: What are the implications of the simple loop conjecture?
QUESTION [25 upvotes]: Drew Zemke recently posted a preprint on arXiv proving the Simple Loop Conjecture for 3-manifolds modeled on Sol. 

Simple Loop Conjecture: Consider a 2-sided immersion $F\colon\, \Sigma\rightarrow M$ of a closed orientable surface $\Sigma$ into a closed 3-manifold $M$. If $F_\ast \colon\, \pi_1 \Sigma \rightarrow \pi_1 M$ is not injective then there is an essential simple closed curve in $\Sigma$ that represents an element in the kernel of $F_\ast$.

If $F$ were an embedding then this would follow from Papakyriakopoulos's Loop Theorem. "To be an embedding" is not an algebraic property, so the Simple Loop Conjecture is more of a `$ \pi_1$ to 3-manifolds' statement than the loop theorem. It allows us to replace non-$\pi_1$-injective immersions by immersions of lower genus surfaces by surgery paralleling passage to a normal subgroup; So it does translate from algebra to topology.
Joel Hass proved the conjecture for Seifert-fibered spaces using geometrical techniques in 1987, and Hyam Rubinstein and Shicheng Wang proved the conjecture in 1998 for non-trivial graph manifolds (not Sol).
In Kirby's problem list it also states that the Simple Loop Conjecture arises in trying to characterize 3-manifold groups among Poincaré duality groups, but I'm not sure what Kirby means. I also don't know what else it implies.

Question: What wonderful things would follow from the Simple Loop Conjecture if it were true? Beyond it being a "natural question" and beyond the abstract considerations brought above, what is the significance of this conjecture?

REPLY [19 votes]: The simple loop conjecture can be viewed as a statement about how to construct all surfaces in a 3-manifold.
Fix any orientable 3-manifold M. There are two well known constructions that produce oriented surfaces in M.

Start with a sphere that bounds a ball, and successively add a finite number of 1-handles (allowing self-intersections).
Start with a surface subgroup of π1(M), or equivalently with an immersed π1-injective surface, and add a finite of 1-handles, (again allowing self-intersections.)

Does this produce all immersed surfaces in M?
Yes if and only if the simple loop conjecture is true.<|endoftext|>
TITLE: Fundamental units with norm $-1$ in real quadratic fields
QUESTION [11 upvotes]: If we have distinct primes $p \equiv q \equiv 1 \pmod 4,$ with Legendre $(p|q) = (q|p) = -1,$ there is a solution to $u^2 - pq v^2 = -1$ in integers and the fundamental unit of $O_{\mathbb Q(\sqrt{pq})} $ has norm $-1.$ Stevenhagen attributes this to Dirichlet (1834).
There is no such result for $p \equiv q \equiv 1 \pmod 4,$ with Legendre $(p|q) = (q|p) = 1.$ I did a census, out of the first 300,000 such numbers, there were  99284 for which $u^2 - pq v^2 = -1$ was possible, or very close to one out of three. Here are the first 100 such, and the first 200 not:
    145    445    901   1145   1313   1745   2249   2305   2501   2545
   2605   2705   3029   3161   3341   3545   3601   3845   4045   4777
   5045   5245   5305   5545   5629   5713   5933   6145   6245   6401
   6445   6649   6757   6893   6953   6989   7045   7093   7397   7745
   7837   7897   8005   8077   8345   8545   8653   8945   9089   9305
   9673   9881   9953  10001  10081  10145  10345  10405  10445  10777
  10817  10961  11029  11141  11237  11453  11629  11729  11945  12181
  12389  12461  12629  12773  12961  13105  13169  13549  13645  13801
  14305  14933  15245  15397  15445  15509  15529  15845  15929  15949
  16153  16601  16609  16645  16801  16837  16861  16945  17305  17345

================================
    205    221    305    377    505    545    689    745    793    905
   1205   1345   1405   1469   1513   1517   1537   1717   1945   1961
   2005   2041   2045   2105   2245   2329   2353   2533   2669   2701
   2845   2993   3005   3205   3305   3497   3505   3737   3805   3893
   4069   4105   4145   4321   4369   4381   4405   4453   4633   4645
   4705   4717   4849   4981   5017   5057   5069   5105   5141   5249
   5345   5513   5645   5809   5905   5917   5989   6001   6005   6341
   6497   6505   6605   6613   6697   6773   6805   6905   7081   7145
   7157   7289   7361   7405   7445   7453   7769   7813   7957   8033
   8045   8105   8149   8333   8357   8473   8489   8605   8621   8633
   8705   8749   8801   9005   9077   9113   9445   9469   9505   9701
   9745   9797   9809   9841  10121  10237  10361  10421  10441  10517
  10573  10645  10705  10805  10841  10877  11105  11345  11357  11405
  11513  11545  11705  11905  11917  12017  12137  12205  12317  12469
  12605  12745  12869  12913  12937  13045  13073  13141  13213  13253
  13445  13637  13705  13745  13817  13837  13897  13945  13969  13973
  14005  14093  14209  14257  14309  14417  14473  14521  14545  14689
  14701  14761  14845  14893  14989  15005  15109  15205  15293  15305
  15545  15605  15677  15769  15857  15905  16021  16045  16105  16145
  16201  16237  16409  16441  16505  16769  16805  16913  17113  17197

============
Note that you can quickly check the above at http://www.numbertheory.org/php/unit.html
Question: in the limit, do exactly one third of such $n=pq,$ with  $p \equiv q \equiv 1 \pmod 4,$ and Legendre $(p|q) = (q|p) = 1,$ allow $u^2 - pq v^2 = -1?$ Or, more to the point I suppose, $x^2 + xy - \frac{pq-1}{4} y^2 = -1?$

REPLY [8 votes]: Stevenhagen "The number of real quadratic fields having units of negative norm" Exp Math 1993  makes the following analysis (last paragraph page 127):
Let $D>0$. Let $C$ be the narrow class group of $\mathbb{Q}(\sqrt{D})$ (ideals modulo principal ideals whose generator has positive norm). Then the ideal $(\sqrt{D})$ is $2$-torsion in $C$. There is a unit $\epsilon$ of norm $-1$ if and only if $(\sqrt{D})$ is trivial in $C$ (as, in that case, $\epsilon \sqrt{D}$ generates the same ideal.
Let $D=pq$ with $p \equiv q \equiv 1 \bmod 4$. Saberhagen computes that the $2$-torsion $C[2]$ is isomorphic to $\mathbb{Z}/2$. The primes $(p)$ and $(q)$ ramify, and $C[2]$ is generated by the classes of the ideals $\mathfrak{p}$ and $\mathfrak{q}$ lying over $p$ and $q$. Thus, we have a surjection $(\mathbb{Z}/2)^2 \to \mathbb{Z}/2$ sending $(a,b)$ to the class of $\mathfrak{p}^a \mathfrak{q}^b$ in $\mathbb{Z}/2$. We want the probability that $(1,1)$ is in the kernel of this surjection.
If we choose a random surjection $(\mathbb{Z}/2)^2 \to \mathbb{Z}/2$, the odds that $(1,1)$ is in the kernel are $1/3$, which recovers Will's guess. Stevenhagen reports that numerical data is consistent with the hypothesis that the surjection is randomly chosen.

I've been thinking about this, and I think that I can state Stevenhagen's conjecture in a fully elementary way. (No idea about a proof, of course.) Fix a $t \times t$ symmetric matrix $A$ over $\mathbb{F}_2$ whose rows sum to $0$. Consider $t$-tuples of primes $(p_1,p_2,\ldots,p_t)$ such that all $p_i \equiv 1 \bmod 4$ and $\left( \frac{p_i}{p_j} \right) = (-1)^{A_{ij}}$ for $i \neq j$. Let $D =p_1 \cdots p_t$ and let $(x,y)$ be the primitive solution to Pell's equation $x^2-D y^2=1$. Then $x \equiv (-1)^{b_i} \bmod p_i$ for some vector $b \in \mathbb{F}_2^t$. I will prove below that $b$ is a nonzero element in the kernel of $A$. Stevenhagen's conjecture is equivalent to saying that all nonzero elements of $\mathrm{Ker}(A)$ occur with equal probability, as $(p_1, \ldots, p_t)$ runs over primes obeying $\left( \frac{p_i}{p_j} \right) = (-1)^{A_{ij}}$. 
The negative Pell equation is solvable if and only if $b=(1,1,\ldots,1)$. (In other words, $x \equiv -1 \bmod D$.) So if $t=2$ and $\left( \frac{p_1}{p_2} \right)=-1$, then $b$ must be the unique element of $\mathrm{Ker} \left( \begin{smallmatrix} 1 & 1 \\ 1 & 1 \end{smallmatrix} \right)$ and the negative Pell equation is solvable but, if $\left( \frac{p_1}{p_2} \right)=1$, then $b$ could be any of the $3$ nonzero elements of $\mathrm{Ker} \left( \begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix} \right)$ and the negative Pell equation is solvable with probability $1/3$.
It looks like dealing with primes that are $2$ or $3 \bmod 4$ is just more bookkeeping, but you didn't ask, so I won't do it.

I will now explain some of the above claims. First, by a computation modulo $4$, note that $x$ is odd and $y$ is even. 
Put $P = \prod_{b_i=1} p_i$ and $Q = \prod_{b_i=0} p_i$, so $D=PQ$. So $x \equiv -1 \bmod P$ and $x \equiv 1 \bmod Q$. We have
$$\left( \frac{x+1}{2P} \right) \left( \frac{x-1}{2 Q} \right) = \left( \frac{y}{2} \right)^2$$
where all factors are integers.
The factors on the left are relatively prime, since
$$P \left( \frac{x+1}{2P} \right) - Q \left( \frac{x-1}{2 Q} \right)=1.$$
So $(x+1)/(2P)=u^2$ and $(x-1)/(2Q)=v^2$ for some $u$ and $v$. We deduce that
$$P u^2 - Q v^2 = 1. \quad (\ast)$$
We can now see that $b \neq 0$: If $b=0$ then $P=1$ and $Q=D$, so $(\ast)$ contradicts the minimality of $(x,y)$.
We now check that $Ab=0$, by computing each coordinate $\sum_j A_{ij} b_j$ of $Ab$. 
Case 1 $b_i=0$. Then $p_i$ divides $Q$ and, reducing $(\ast)$ modulo $p_i$, we see that $\prod_{b_j=1} p_j$ is square modulo $p_i$. This exactly says that $\sum_{j} A_{ij} b_j=0$.
Case 2 $b_i=1$. As above, we deduce that $\prod_{b_j=0} p_j$ is square modulo $p_i$, so $\sum_{j \neq i} A_{ij} (1-b_j)=0$. But this is $\sum_{j \neq i} A_{ij} + \sum_{j \neq i} A_{ij} b_j = A_{ii} + \sum_{j \neq i} A_{ij} b_j = \sum_j A_{ij} b_j$, where the first equality is because the rows of $A$ sum to $0$.
Finally, note that, if $b=(1,1,\cdots, 1)$, then $(\ast)$ says that the negative Pell's equation is solvable. Conversely, if the negative Pell's equation is solvable with solution $v^2-D u^2 = -1$, then $(x,y) = (v^2+Du^2, 2 uv)$ is the minimal solution to $x^2-Dy^2=1$ and, sure enough, $x \equiv v^2 \equiv -1 \bmod D$.<|endoftext|>
TITLE: Is it possible to make an algorithm that could predict the likelihood that a program will halt?
QUESTION [17 upvotes]: Today I began to read about computability theory. I do not even have an elementary understanding of the topic but it certainly got me thinking. I know there is there is no 'one-for-all' algorithm that solves the halting problem but I wonder if it is possible for there to be an algorithm that determines the likelihood that a program would halt. 
The reason I asked this question because I'm writing an essay about a problem I would like to solve and I thought this might be an interesting "solution" to the halting problem. 
And yes I know the halting problem is undecidable.

REPLY [25 votes]: Here is one way of interpreting your question. In my joint paper:

Joel David Hamkins and Alexei Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47 (2006), no. 4, 515--524. blog post,

the main theorem is that for some of the standard models of computation, the halting problem is decidable with probability one. Specifically, we prove that for the usual one-way-infinite Turing machine model, there is a set $A$ of Turing machine programs, such that:

Almost every program is in $A$, in the sense that the proportion of all $n$-state programs in $A$ goes to $1$ as $n$ goes to infinity; 
it is decidable whether a given program is in $A$; and
the halting problem is decidable for programs in $A$. 

So this is a sense in which the halting problem is decidable with probability one. 
The argument is sensitive, however, to the computational model, and for some other models of computation, the best currently known is that we can decide the halting problem on a set of probability $\frac 1{e^2}$, which is about 13.5%.<|endoftext|>
TITLE: References about Hasse diagrams of root systems
QUESTION [6 upvotes]: This is to ask about references of Hasse diagrams of irreducible root systems. I found here and there nice pictures of root systems of type $E$. I would like to ask for Hasse diagrams of classical root systems ($A_n, B_n, C_n, D_n$).
Thanks in advance.

REPLY [4 votes]: I put a LaTeX package up on CTAN today to draw the Hasse diagrams of all root systems, following Ringel's pictures. Code is as simple as
\documentclass{amsart}
\usepackage{lie-hasse}
\begin{document}
\hasseDiagrams{A4;B5}
\hasseDiagrams{F4;G2}
\end{document}

to display<|endoftext|>
TITLE: What does "higher monodromy" tell us about a principal bundle
QUESTION [6 upvotes]: Let $P \to X$ be a principal $G-$bundle and let $f: X \to BG$ be its classifying map. As I understand there's some way to associate a monodromy representation $\pi_1(X) \to G$ to it. I know how to construct such a presentation out of a principal connection on $P$ by taking an appropriate quotient of the holonomy representation but I failed in the past weeks to find a source that defines a purely topological construction of this representation.

Question 1: Where can I find the definition and some elementary topological treatment of the
  monodromy representation associated to a principal bundle?

Extending the classifying map $f$ to a fiber sequence gives:
$$\cdots\to \Omega G \to \Omega E_f \to \Omega X\to G \to E_f \to X \to BG$$
Where $E_f=\{(x,\gamma)\in X \times Y^I:\gamma(1)=f(x)\}$. 

Question 2: How is the map $\Omega X \to G$ above related to the usual notion of monodromy represnetation (which i'm unfamiliar with) of the
  the principal bundle $P\to X?$ How does it relate to the holonomy representation of a given connection?

My naive attempt continues by taking the fundamental group of the sequence above to get the following long exact sequence of groups:
$$\cdots \to \pi_2(G) \to \pi_2(E_f) \to \pi_2(X)\to \pi_1(G) \to \pi_1(E_f)  \to \pi_1(X) \to \pi_0(G)$$
Since the procedure was very natural I imagine the above sequence must yield some important information about the principal bundle at hand. For example ff $X$ is weakly contractible the above gives another motivation for why every fiber bundle on $X$ is trivial.

Question 3: To what extent does the above "higher monodromy sequence" determine the principal bundle $P \to X$?

REPLY [3 votes]: Question 3: To what extent does the above "higher monodromy sequence" determine the principal bundle $P \to X$? 

In general, not at all. There is no loss of generality in replacing $BG$ with an arbitrary connected space $Y$ (every such space is some $BG$, morally for $G = \Omega Y$), so your question is to what extent the homotopy class of a (pointed) map $f : X \to Y$ is determined by the long exact sequence in homotopy for the fiber sequence
$$F \to X \to Y$$
where $F$ is the homotopy fiber. What can happen in general is that $X$ and $Y$ have homotopy groups in degrees that are far apart from each other, so that this sequence tells you nothing about the homotopy class of $f$. 
For example, if $X = BG$ for some discrete group $G$ and $Y = B^3 A$ for some discrete abelian group $A$, then all you learn is that $F$ is a space with $\pi_1(F) \cong G, \pi_2(F) \cong A$, and other homotopy groups trivial, and all of the maps in the long exact sequence are either isomorphisms or zero, regardless of $f$. But of course the group cohomology $H^3(BG, A)$ is interesting in general.<|endoftext|>
TITLE: Dessins d'enfants and absolute Galois group
QUESTION [11 upvotes]: I would like to know what is the recent progress about the group homomorphism
$$ \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\rightarrow \mathrm{Out}(\hat{F_{2}})$$

$\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ is the absolute Galois group of $\mathbf{Q}$.
$\mathrm{Out}(\hat{F_{2}})$ is the group of outer automorphisms of the procompletion of the free group generated by two elements. 

It is know to be injective homomorphism, what can we say about the image ?

REPLY [7 votes]: For the sake of getting this off the unanswered stack...
Please refer to the following reference:

Guillot, P. An elementary approach to dessins d'enfants and the Grothendieck-Teichmüller group, Enseign. Math. vol 60, 2014. [arXiv version | journal version]

The article just cited develops the theory of dessins d'enfants from scratch. The final section discusses the image of the homomorphism mentioned in the OP's question. 
In particular,

Guillot includes a proof that the image must lie in the group
$\widehat{GT_0}$, a proper subgroup of ${\rm Out}(\hat{F_2})$, first
defined (I believe) by Drinfeld;
Guillot discusses (but does not prove) a stronger result due to
Ihara that the image must lie in the Grothendieck-Teichmuller group,
$\widehat{GT}$, a proper subgroup of $\widehat{GT_0}$.

Please consult the bibliography of the given reference to find the relevant works of Drinfeld and Ihara.<|endoftext|>
TITLE: Failure of universal flatification
QUESTION [7 upvotes]: Raynaud and Gruson proved a beautiful "flatification" theorem (5.2.2):  If $S$ is a quasicompact, quasiseparated scheme, and $X$ is a finitely presented $S$-scheme, $M$ is an $\mathcal O_X$-module of finite type, then there is a blowup $f : S' \rightarrow S$ such that the strict transform of $M$ along $f$ is flat over $S'$.  (The blowup can be arranged to be an isomorphism over an open subset of $S$ over which $M$ is already flat.)
They observe that when $X$ is projective over $S$, there is in fact a universal birational modification, obtained by resolving the indeterminacy of the rational map from $S$ to the Quot scheme.  (If $U$ is an open subset of $S$ over which $M$ is flat, then $M$ induces a map from $U$ to the scheme of quotients of $M$.)
What I would like to have is some intuition for, and an example of, the absence of a universal flatification in the absence of the projectivity hypothesis.

REPLY [4 votes]: I feel like there should be an example using matrix multiplication, but I tried several variations that did not work (as I commented above).  The example below is less naural than matrix multiplication, but it illustrates the key issue.
Let $k$ be a field. Let $S$ be a $3$-dimensional vector space, $V \cong \text{Spec}\ k[x,y,z]$.  Thus $\mathbb{P}(V)$ is isomorphic to $\mathbb{P}^2 = \text{Proj} \ k[r,s,t]$.  Let $\overline{X}$ be the closed subscheme of $V\times \mathbb{P}(V)$ parameterizing pairs $(v,[w])$ such that $v$ is a multiple over $w$, possibly zero.  In coordinates, $((x,y,z), [r,s,t])$, this is the zero scheme of $xs-yr, xt-zr, yt-zs$.  Denote by $\overline{\pi}:\overline{X} \to S$ the natural projection.  
Fix a line $L\subset \mathbb{P}(V)$, e.g., the zero scheme of $t$, and let $X$ be the open complement in $\overline{X}$ of $\{0\}\times L$.  Let $\pi:X\to S$ be the restriction of $\overline{\pi}$ to $X$.  Let $M$ be the structure sheaf of $X$.  This is not $S$-flat, since the fiber dimension over the origin is $2$, yet the generic fiber dimension is $0$.
The universal flatification of $\overline{\pi}$ is $\overline{\pi}$ itself,, the strict transform in $\overline{X}\times_S \overline{X}$ is the diagonal, and the exceptional set $E=\overline{\pi}^{-1}(\{ 0\})$ is $E\cong \mathbb{P}(V)$.  Certainly the strict transform of $\pi$ with respect to $\overline{\pi}$ is flat, but $\overline{\pi}$ is not a universal flatification of $M$.  
I can take any finite collection of points inside $L$ inside $E$, and I can "pinch" $\overline{X}$ along those points to form a non-normal scheme $X'$ such that $\overline{\pi}$ factors through $X'$. The morphism from $X'$ to $X$ will also be a flatification of $\pi$.  What I would "like" to do in order to make a universal (proper, birational) flatification of $\pi$ is contract / blow down $L$ inside $\overline{X}$.  But that is not possible: we cannot even contract a line $L$ in a $2$-plane $E$ (not as a scheme, not even as an algebraic space).  Thus, there is no universal (proper, birational) flatification.<|endoftext|>
TITLE: Characterizing when matrices are 'dissipative'
QUESTION [7 upvotes]: An $n$ by $n$ matrix A is said to be dissipative with respect to a norm $\|\cdot \|$ if for all $x$ and $t\geq 0$, we have $\|e^{At}x\|\leq\|x\|$. Two matrices $A$ and $B$ are said to be jointly dissipative if they are each dissipative with respect to the same norm.
Here $e^{At}$ is defined by
$$e^{At}:=\sum_{n=0}^{n=\infty}\frac{1}{n!}A^nt^n.$$ 
My question is the following: what are necessary and sufficient conditions (on $A$ and $B$) for the existence a norm for which $A$ and $B$ are jointly dissipative? 
Of course depending on what we take $n$ to be, this question can be quite hard. If anyone has any particularly clever suggestions for an approach, or solutions to special cases ($n\geq 2$), I would love to hear from you.

REPLY [2 votes]: The problem has been addressed in the infinite dimensional setting in the following paper: 
Máté Matolcsi, On the relation of closed forms and Trotter’s product formula, J. Funct. Anal. 205 (2003), no. 2, 401--413.
Theorem 2 is of special interest here. If there are any practical consequences, I do not know.<|endoftext|>
TITLE: Open questions on (finite) tensor categories
QUESTION [6 upvotes]: I would like to know about problems on (finite) tensor categories. I have read Etingof´s notes from his course at MIT. I have a question: 
There exists any reference where I can find an open problem about this topic?
Thank you very much for your attention.

REPLY [8 votes]: We made a list of open problems at an AIM conference a few years ago.<|endoftext|>
TITLE: Is Lehmer's polynomial solvable?
QUESTION [12 upvotes]: The degree 10 polynomial
$$\displaystyle x^{10} + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1$$
given by D.H. Lehmer in 1933 has the property that its largest real root, $\beta = 1.176280 \cdots$ is the smallest known Salem number. Moreover, it is a folklore conjecture that $\beta$ is in fact the smallest Salem number.
However, it is curious that one cannot find a reference for the explicit value of $\beta$. I suspect that this is because Lehmer's polynomial is not solvable. Is this the case? If so, is there a reference/relatively simple argument? Furthermore, if Lehmer's polynomial is in fact not solvable, then what is its Galois group?
Thanks for your time.

REPLY [25 votes]: Lehmer's polynomial is symmetrical,
so $x + x^{-1} =: y$ satisfies a polynomial of half the degree.
It turns out that this is the quintic $y^5 + y^4 - 5y^3 - 5y^2 + 4y + 3 = 0$, 
whose Galois group is the unsolvable $S_5$ (for instance, it's irreducible
$\bmod 2$ and decomposes as $(y^2-2y-1)(y^3-2y^2+2y+2)$ $\bmod 5$,
so the Galois group is a subgroup of $S_5$ of order divisible by $30$ that contains an odd permutation,
and the only such subgroup is $S_5$ itself).  Hence Lehmer's polynomial
is not solvable either.
[It turns out that $y$ generates the totally real quintic field
of third-smaller discriminant $36497$.
By the way, even if a polynomial is solvable, exhibiting a
solution in radicals may not be of much use; for instance,
the Salem root of $x^8 - x^5 - x^4 - x^3 + 1 = 0$ satisfies
$x + x^{-1} = y$ where $y$ is a solution of the quartic
$y^4 - 4y^2 - y + 1 = 0$ with Galois group $S_4$,
but even though this group (and thus also the octic
$x^8 - x^5 - x^4 - x^3 + 1$) is solvable I doubt that you
really want to ponder the explicit formula for $x$
involving things like the cube roots of $187/54 \pm \sqrt{-1957/108}$ . . .]<|endoftext|>
TITLE: Quotient of a vector space by a linear finite group action
QUESTION [5 upvotes]: Let the cyclic group $\mathbb{Z}_n$ act on $\mathbb{C}^n$ (or on $\mathbb{R}^n$, I'm interested in both) by permuting coordinates.  What does the topological quotient $Q$ by this group action look like?  More explicitly, I'd like to identify points under the equivalence relation $x\sim y\Leftrightarrow x=g\cdot y$ for some $g\in\mathbb{Z}_n$.  Is there some nice way to embed it as a subset of $\mathbb{C}^m$ for some $m$?
So far all I've thought of is that if $f: \mathbb{C}\to\mathbb{C}$ is any function, $e_f: \mathbb{C}^n\to\mathbb{C}^n$ is the application of $f$ elementwise, and $\mathcal{F}$ is the discrete Fourier transform, then the map $\mathcal{F}_f^n: \mathbb{C}^n\to\mathbb{C}^n$ defined by $x\mapsto \left(\mathcal{F}\left(e_f(x)\right)\right)^n$ is fixed by the group action and so defines a map out of $Q$.  It's easy to see that this map is not injective, but perhaps by concatenating $\mathcal{F}_f^n$ for several different $f$ one can get an injective map and by choosing nicely-behaved $f$ one can get a nice embedding.
But I'd bet there are more illuminating embeddings.  I'm happy to consider related questions where some bad points are removed from $\mathbb{C}^n$ (such as multiples of the all-ones vector), $n$ is assumed to be prime, etc.  I'm also interested in other (abelian, so far) group actions, like $\mathbb{Z}_n\times\mathbb{Z}_n$ acting on $\mathbb{C}^{n\times n}$, so if there is a general theory of such quotient spaces, I'd be interested to learn about it.  It seems like this sort of question must be well-studied, but I am not sure where exactly it fits, so feel free to re-tag.

REPLY [4 votes]: The action of $\mathbb{Z}_n$ on $\mathbb{C}^n$ can be diagonalized: it's conjugate to the action sending a vector $(z_0, z_1, \dots z_{n-1}) \in \mathbb{C}^n$ to
$$(z_0, \zeta_n z_1, \zeta_n^2 z_2, \dots \zeta_n^{n-1} z_{n-1}).$$
The quotient can be stratified according to which of the $z_i$ are nonzero (starting with $z_1$). On the open stratum where $z_1 \neq 0$, there's a unique representative of each orbit where $0 \le \text{arg}(z_1) < \frac{2\pi}{n}$. In general, on the stratum where $z_1 = \dots = z_{k-1} = 0$ and $z_k \neq 0$, there's a unique representative of each orbit where $0 \le \text{arg}(z_k) < \frac{2 \pi \gcd(k, n)}{n}$. Each stratum lies in the closure of the previous stratum.
There should be a similar and more complicated story for the action of $\mathbb{Z}_n$ on $\mathbb{R}^n$, coming from the decomposition of the latter into real irreducible representations.<|endoftext|>
TITLE: Banach spaces $X$ with $\ell_2(X)$ not isomorphic to $L_2([0,1],X)$
QUESTION [11 upvotes]: Let $X$ be a Banach space. I think that some time ago I read somewhere that, in general, the space $\ell_2(X)$ of all sequences $(x_n)$ in $X$ with $\sum_{n=1}^\infty \|x_n\|^2<\infty$ is not isomorphic to the space $L_2([0,1],X)$ of square integrable $X$-valued functions on $[0,1]$. 
How can I find an example of an space $X$ with $\ell_2(X)\not\simeq L_2([0,1],X)$?

REPLY [4 votes]: Non-reflexive Grothendieck spaces are examples of spaces for which $\ell_2(X)\not\cong L_2(X)$. The same holds for all $p\in (1,\infty)$.

Particular examples of such spaces include $X=\ell_\infty(\Gamma)$ for any index set $\Gamma$ (and all infinite-dimensional injective Banach spaces) as well as $X=\mathscr{B}(\ell_2)$.
Suppose that $X$ is a Grothendieck space which is not reflexive. For every $p\in [1,\infty)$ the space $L_p([0,1], X)$ is not Grothendieck. This is a result of Díaz; see Theorem 2 in:

S. Díaz, Grothendieck's property in $L_p(X)$, Glasgow Math. J., 37, 3 (1995), 379–382.

On the other hand, the Grothendieck property is preserved by $\ell_p$-sums ($p\in (1,\infty)$).<|endoftext|>
TITLE: How to prove that a monotone function is differentiable at some point?
QUESTION [12 upvotes]: This fact, which eventually belongs to Lebesgue, is usually proved with some measure theory (and we prove that the function is differentiable a.e.). Is there a significantly different approach? Let me explain how it could look like. Say, if the function is convex, we may touch its graph by a Euclidean disc (lying in the épigraphe), and in the point of touch there exists a derivative. 
Maybe, it allows to prove something about the set of points where there is no derivative, not only that it has Lebesgue measure $0$. Or is this impossible and for any set $A$ of Lebesgue measure $0$ there exists a monotone function $f$ not differentiable at any point $a\in A$? 
UPD: according to a comment by Bill Johnson, this statement is true, even for a Lipschitz function.

REPLY [10 votes]: Yes, there are a couple of alternate proofs: Here is one:
An Elementary Proof of Lebesgue's Differentiation Theorem
Michael W. Botsko
The American Mathematical Monthly
Vol. 110, No. 9 (Nov., 2003), pp. 834
And this proof by Faure (from the Real Analysis exchange, 2003), which uses Riesz's 1932 idea: http://projecteuclid.org/download/pdf_1/euclid.rae/1149698560<|endoftext|>
TITLE: notation for $(a-b)(a-qb)\dots (a-q^{n-1}b)$
QUESTION [8 upvotes]: I wonder whether there is a notation for such thing, which I denote $[a;b]_q^n$ for a moment:
$$
[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,
$$
this last equation uses $q$-Pochhammer symbol notation.
My motivation is that it is a natural analogue of $(a-b)^n$ and some identities look better with it, at least annoying (well, this is subjective) powers of $q$ disappear. Say, $q$-Vandermonde identity becomes
$$
[a;c]_q^n=\sum {\binom{n}{k}}_q[a;b]_q^k\,[b;c]_q^{n-k},
$$
compare to
$$
(a-c)^n=\sum \binom{n}{k} (a-b)^k (b-c)^{n-k}.
$$
Equivalently, for $q$-exponential generating functions $$F_{a,b}(x):=\sum \frac{[a;b]_q^k}{(k!)_q}$$we have
$F_{a,c}(x)=F_{a,b}(x)F_{b,c}(x)$.

REPLY [3 votes]: There are some different notations in the literature: 
B.A. Kupershmidt  used $ {(a\dot  - b)^n} $, 
Victor Kac and Pokman Cheung in “Quantum Calculus” used $(a-b)_q^n$. 
In my lecture notes I used ${(a\underset{\raise 0.5 em\hbox{$\smash{\scriptscriptstyle\cdot}$}}{ - } b)^n}$ (with a dot below the minus sign. ).<|endoftext|>
TITLE: The Weil numbers and modulus of an elliptic curve
QUESTION [5 upvotes]: I have an ignorant question about elliptic curves which I'll be slightly imprecise about. If I have an elliptic curve $X$ defined over $\mathbb Z$, I can base change to $\mathbb C$, and then $X(\mathbb C)$ is isomorphic to the quotient of $\mathbb C$ by the lattice generated by $1, \tau$ for some $\tau \in \mathbb C$.
I can also base change to a finite field $\mathbb F_q$, and then the Frobenius operator on $H^1 (X( \bar {\mathbb{F} }_q), \bar {\mathbb{Q} }_l)$ has eigenvalues $\sigma, \bar \sigma$.  

Q: Is there any relationship between $\sigma, \bar \sigma$, and $\tau$? 

(I may have glossed over some things, like fixing an identification embedding of $\bar {\mathbb Q}$ into $\mathbb C$, or requiring $X(\mathbb F_q)$ to be smooth, or something else.)

REPLY [13 votes]: If $E$ does not have CM, then $\tau$ will be transcendental over $\mathbb Q$, so it's hard to imagine  a relation with the eigenvalues of Frobenius, which are integers in an imaginary quadratic field. However, if $E$ does have CM, then $\text{End}(E)\otimes\mathbb Q\cong\mathbb Q(\tau)$, and Frobenius lifts to an endomorphism of $E$ that is in the ring of integers of this field, so if you choose the right value for $\tau$, then they can be very closely related. If you want more details, read any standard treatment on complex multiplication and elliptic curves.<|endoftext|>
TITLE: "Clubiness" of projective sets of ordinals
QUESTION [8 upvotes]: I'm sure this is just my google-fu failing me, but: what are sufficient, non-overkill large cardinal axioms which guarantee "Every (boldface) $\Pi^1_n$ set of (real codes for) countable ordinals contains or is disjoint from a club subset of $\omega_1$"? (I asked this question on math.stackexchange a couple weeks ago https://math.stackexchange.com/questions/1539202/clubbiness-of-pi1-n-sets, and received some attention but no answer.)
To clarify, I'm asking about the strength over ZFC.
Here's a very silly upper bound: suppose $L(\mathbb{R})$ is a model of AD, and moreover every $\Pi^1_n$-sentence with real parameters is absolute between $L(\mathbb{R})$ and $V$ (actually, I think this is already a consequence of "$L(\mathbb{R})\models AD$," but I'm not sure). Then let $A\in V$ be a $\Pi^1_n$-set of countable ordinals, via the formula (with real parameters) $\varphi$. By the absoluteness assumption, $\varphi^{L(\mathbb{R})}=A$, so $A\in L(\mathbb{R})$. And since $L(\mathbb{R})\models AD$, $L(\mathbb{R})$ thinks $A$ contains or is disjoint from a club. But inner models compute club-ness correctly, so we're done.
This seems massively overkill to me, though - what is the right bound?

REPLY [4 votes]: Meanwhile, let me provide a lower bound. Your situation is not
provable in ZFC; it is false in $L$, and it is incompatible with
having a projective well-ordering of the reals. Furthermore, it
implies that $\omega_1$ is inaccessible to reals.
Theorem. If there is a projectively definable
$\omega_1$-sequence of distinct reals (for example, if there is a
projectively definable well-ordering of the reals), then there is
a projectively definable set of countable ordinals
$S\subseteq\omega_1$, of the same complexity as the sequence, that
is both stationary and co-stationary. (Thus, it neither contains nor omits a club.)
Proof. Assume that there is a projectively definable
$\omega_1$-sequence of distinct reals $\langle
z_\alpha\mid\alpha<\omega_1\rangle$, where
$z_\alpha\subseteq\omega$. This situation occurs, for example,
under $V=L$ or indeed, if there is a projectively definable
well-ordering of the reals, since we can let $z_\alpha$ be the
$\alpha^{\rm th}$ real in that well-ordering.
Let $S_n=\{\alpha\mid n\in z_\alpha\}$. This is a projectively
definable set of ordinals, of the same complexity as the sequence.
If it is not both stationary and co-stationary, then there is a
club $C_n$ that contains or omits $S_n$. Let $C=\bigcap_n C_n$,
which is a club subset of $\omega_1$. Note that if $\alpha\in C$,
then $n\in z_\alpha$ is determined by whether $C_n$ contained or
omitted $S_n$. Thus, the $z_\alpha$ for $\alpha\in C$ must all
agree, contradicting our assumption that the reals were distinct.
So one of the sets $S_n$ must be both stationary and
co-stationary. QED
Corollary. If every projectively definable set of countable
ordinals contains or omits a club, then $\omega_1$ is inaccessible
to reals.
Proof. If $\omega_1$ is not inaccessible to reals, then
$\omega_1=\omega_1^{L[x]}$ for some real $x$. In this case, there
is an projectively $x$-definable $\omega_1$-sequence of distinct
reals, and so there will be a projectively $x$-definable set
$S\subseteq\omega_1$ that is both stationary and
co-stationary.QED<|endoftext|>
TITLE: Why is there a need for ordinal analysis?
QUESTION [9 upvotes]: Consider the Peano axioms. There exists a model for them (namely, the natural numbers with a ordering relation $<$, binary function $+$, and constant term $0$). Therefore, by the model existence theorem, shouldn't this suffice to prove the consistency of first order arithmetic? Why is Gentzen's proof necessary?

REPLY [12 votes]: In addition to the reasons Andreas gives, Gentzen's theorem gives additional information that's interesting even if you don't have any qualms about consistency.
In particular, ordinal analysis gives a fairly precise characterization of the provably total computable functions of a theory (and, along with it, a lot of information about the structure of proofs in PA).<|endoftext|>
TITLE: Parameterization of a knotted surface?
QUESTION [6 upvotes]: I'm looking for a parameterization $(x_1(u,v),x_2(u,v),x_3(u,v),x_4(u,v))$ of a knotted sphere in $\mathbb R^4$. How might one go about finding such a parameterization?

REPLY [10 votes]: If you know how to parameterize a nontrivial knot in $R^3$ (say, the trefoil knot), you can use Artin's "spinning" construction to parameterize a nontrivial knot in $R^4$ via map $S^2\to R^4$ written in spherical coordinates on $S^2$. Take, say, the trefoil $T$ in $R^3$ parameterized by some smooth periodic function $f(\theta), 0\le \theta\le a$ where $a$ is the period. Let $L$ denote the straight line in $R^3$ through the points $f(0), f(c)$, $c=a- \epsilon$ where $\epsilon>0$ is sufficiently small (if you want to know how small is small, ask me and I can explain). I will choose the coordinates $x_1, x_2, x_3$ so that $L$ is the $x_3$-axis. Let $R_\phi$ denote the rotation  by the angle $\phi$ in $R^4$ around the plane $x_1=x_2=0$ (thus, fixing $L$). Now, consider the map 
$F: [0, c]\times [0, 2\pi]\to R^4$, given by 
$$F(\theta, \phi) = R_\phi\circ f(\theta)$$ 
Its image is a knotted 2-sphere $\Sigma$ in $R^4$. It is smooth away from the points $f(0), f(c)$. If this bothers you, it can be fixed (explicitly). Now, rescale the  interval $[0, c]$ to $[0, \pi]$ via the map $s: t\mapsto \frac{\pi}{c}t$. Then the map $$G= F\circ (s^{-1} \times Id): [0,\pi]\times [0,2\pi]\to R^4$$ parameterizes the same surface $\Sigma$ as before. Lastly, use the spherical coordinates on $S^2$ to convert the map $G$ to a homeomorphism $S^2\to \Sigma$. 
This, I think, is as an explicit construction as it can be (provided you are not bothered by smallness issue for $\epsilon$). 
Edit: Upon your request, smallness of $\epsilon$ is defined as follows: There exists a round ball $B$ (or a half-space) in $R^3$ with points $p=f(a-\epsilon), q=f(a)$ on its boundary such that the arc $\alpha=f([a-\epsilon, a])$ is contained in $B$ and is unknotted in $B$, i.e., is isotopic to the line segment $pq$ in $B$ via an isotopy fixing the boundary sphere of $B$. (In practical terms, you can think of the unknottedness property of $\alpha$ as assuming that $\alpha$ is a graph of a function on the segment $pq$, with values in the plane orthogonal to $pq$.) 
See for instance here for details.<|endoftext|>
TITLE: Parametrization of positive semidefinite matrices
QUESTION [12 upvotes]: We know that a real, symmetric, positive definite matrix $A$ of size $n\times n$ can be parametrized by a vector $\theta$ of $\frac{n(n+1)}{2}$ parameters thanks to the Cholesky decomposition:
$$
A = L L^T,
$$
with $L$ a lower triangular matrix and $\theta=\mathrm{vech}(L)$. The decomposition is unique if the diagonal of $L$ is positive. The diagonal of $L$ can be expressed in log-scale so $\theta$ remains unconstrained. Hence, there is always a unique $\theta$ for a given $A$ and any $A$ can be expressed in such a way. This and other unconstrained parametrizations for positive definite matrices are discussed in this nice paper.

My question is about the existence of a similar, simple and unique parametrization when $A$ is positive semidefinite of rank $r$ (PSDr), or the best approach available. Such parametrization would rely on $\frac{(2p+1-r)r}{2}$ parameters and my intention is to use it for optimization over the set of PSDr.

So far my attempt was to work with a naive extension of the positive definite case:
$$
A = L_r L_r^T,
$$
where $L_r$ is a $n\times r$ lower triangular matrix. It is easy to see that $A$ can be always factorized in such a way:

First, by the spectral theorem, $A=V_r\Lambda_r V_r^T$, where $\Lambda_r=\mathrm{diag}(\lambda_1,\ldots,\lambda_r)$ and $V_r=(v_1,\ldots,v_r)$ is the $n\times r$ orthonormal matrix of the positive $r$ eigenvectors. We also have
$A = U_r^T U_r$, with $U_r=V_r\Lambda_r^{1/2}V_r^T$ a $n\times n$ matrix of rank $r$. The decomposition is of course unique.
Second, by the reduced rank QR decomposition,
$$
U_r=Q_rR_r=Q_rL_r^T,
$$
with $Q_r$ a $n\times r$ matrix with orthogonal columns and $R_r$ an $r\times n$ upper triangular matrix. Then, $A=L_rL_r^T$. Apparently, the decomposition is unique if $R_r$ is in row Echelon form with positive leading entries in every row (see related question here). Unfortunately, such parametrization is not-so-nice to work with as it does not display the rank deficiency explicitly.

Alternatively, but closely related, I checked the pivoted Cholesky decomposition as seen here (sorry for the .html) or in Theorem 10.9 of Higham (2009), which I quote here for completeness:

Theorem 10.9. Let $A\in\mathbb R^{n\times n}$ be positive semidefinite of rank $r$. (a) There exists at least one upper triangular $R\in\mathbb R^{n\times n}$ with nonnegative diagonal elements such that $A = R^TR$. (b) There is a permutation $\Pi$ such that $\Pi^TA\Pi$ has a unique Cholesky factorization, which takes the form
  $$
\Pi^TA\Pi=R^TR,\quad R=\left(\begin{matrix} R_{11} & R_{12} \\ 0 & 0\end{matrix}\right),
$$
  where $R_{11}$ is $r \times r$ upper triangular with positive diagonal elements.

That result seems to provide the answer to my question, but the problem is the appearance of a permutation matrix $\Pi$, which implies hidden degrees of freedom that are not captured by $R$. Or in other words, you will not know which entries have to be null and which not.
Since I want to stick to a simple parametrization and given that the previous ways lead to not-so-simple solutions, I thought just considering $A=L_rL_r^T$ with $L_r$ constrained to have positive diagonal, so ensuring $A_r$ has always rank $r$. Then I have these questions:

Question 1: Is any matrix $A$ in PSDr expressable as $L_rL_r^T$? I guess if the answer is positive, then there will not be a unique way of doing it...
Question 2: In case the answer to the previous question is negative, is  the set generated by the matrices $L_rL_r^T$ dense in PSDr (with respect to some norm, e.g. Frobenius).

Of course, any thoughts regarding the approach to the problem are much appreciated.

REPLY [7 votes]: This paper considers optimization problems on the set of low-rank PSD matrices, and in particular talks about operating in a quotient space to deal with the non-uniqueness. 
See also this work that introduces a Cholesky Manifold to parametrize low-rank PSD matrices. 
Both [1] and [2] cited above are written to deal with the parameterization, non-uniqueness, etc. problems that you are trying to address.
Finally, if you want to quickly try out a quotient-geometry parameterization, you may have a look at fixed-rank PSD matrices in the manopt solver.<|endoftext|>
TITLE: Pushing-forward morphisms
QUESTION [7 upvotes]: Let $f\colon X\to Y$ be a surjective morphism of algebraic varieties, defined over an algebraically closed field.
We take a morphism $\psi\colon X\to Z$, where $Z$ in another algebraic variety, that satisfies the following condition: for all $x,x'\in X$, $f(x)=f(x')\Rightarrow \psi(x)=\psi(x')$.
This yields the existence of a unique map $\psi'\colon Y\to Z$ such that $\psi=\psi'\circ f$. What are conditions on $f$ we can put on $f$ to be sure that $\psi'$ is a morphism of algebraic varieties?
Of course, if $f$ is an isomorphism, the result is true. But is for example enough to ask that $f$ is non-ramified? Or another local condition?
I would be happy by only considering the case where $f$ is quasi-finite, even if the general case seems also interesting.

REPLY [5 votes]: If $f$ is finite surjective and unramified, I think that $\psi'$ is always a morphism. In fact we can prove that $f$ is an effective epimorphism, which means that letting $p_1,p_2:X\times_Y X\to X$ be the projections, any morphism $\psi:X\to Z$ such that $\psi p_1=\psi p_2$ factors uniquely through a morphism $\psi':Y\to Z$. Let us prove that. This is local on $Y$ so we may assume that $Y=Spec(A)$ is affine. Since $f$ is finite, then $X=Spec(B)$ with $B$ finite over $A$. Then being an effective epimorphism is equivalent to exactness of the sequence $A\to B\rightrightarrows B\otimes_A B$, which means that $A\to B$ is injective (this is ok) and that any element $b\in B$ with $b\otimes 1=1\otimes b$ in $B\otimes_A B$ comes from $A$. Now proving exactness of the sequence is étale local on $A$. Due to the étale-local structure of unramified morphisms, we can then assume that $X$ is a disjoint sum of a finite number of closed subvarities of $Y$ covering $Y$. For simplicity let us do the case of two subvarieties. Thus we have two ideals $I,J$ in $A$ such that $B=A/I\times A/J$ and $I\cap J=0$. Let $b=(a_1,a_2)\in B=A/I\times A/J$ have equal images in $B\otimes_A B$. We have $B\otimes_A B=A/I\times A/(I+J)\times A/(I+J)\times A/J$. The condition of equality means that $a_1=a_2$ mod $I+J$, hence $a_1+i=a_2+j$ for some $i\in I$, $j\in J$ and hence $b$ comes from $A$. Thus $f$ is an effective epi.
Finally let us see why this implies what you want. Since $X\to Y$ is unramified, the fibres of the maps $X\times_Y X\rightrightarrows X$ are finite sets of reduced points and checking that $\psi p_1=\psi p_2$ is just set-theoretic i.e. it just means that $\psi$ is constant on the fibres of $f$, which is your assumption.
EDIT (Dec. 31, 2015): as so often, what is left to the reader to check turns out to be 
false. Presently, the computation above works for two subvarieties but 
fails in general for more than two. The point is that if $Y$ is a union 
of $\geq 3$ subvarieties (or better closed subschemes) $X_i$ and if $X$ 
is the disjoint union of them, then the 'glueing' condition on 
$X\times_Y X$ for a map $\psi:X\to Z$ allows only to descend $\psi$ to a map defined on the scheme $Y'$ which is obtained by topological 
glueing of the $X_i$. However $Y'\ne Y$ in general. For example let $Y$ 
be the union of three lines meeting in the origin in affine plane (e.g. 
$x^3-y^3=0$ if base field has char. $\ne 3$). Let $X$ be the 
normalization. Then the glueing condition allows to descend maps 
$\psi:X\to Z$ to $Y'=$ the union of coord. axes in affine 3-space. 
However in order to descend to $Y$ one needs an extra condition which should account for the fact that any 3 nonzero tangent vectors along the 3 
components of $Y$ lie in a common plane. This condition has the 
form of a certain linear relation between the derivatives of $\psi$ at 
the 3 origins of the lines which are the components of $X$. Concretely,
the function which is equal to the coordinate on one of the three lines
of $X$ and $0$ on the other two does not descend to a function on $Y$.
Still, something from the above can be saved to produce examples of
finite unramified effective epimorphisms :

finite unramified morphisms of fibre-degree at most 2;
normalizations of ordinary singularities. Here I call ordinary
singularity a singularity that local-analytically looks like the union
of spaces $Vect(e_i,i\in I_k)$ in affine $n$-space, for some partition
$\{1\dots n\}=I_1\cup\dots\cup I_d$.

In fact I must say that I am not well aware if there is a well-established
terminology for ordinary singularities. My understanding of this notion
is that such a singularity can be reconstructed from the normalization by
topological glueing, i.e. the normalization should be finite unramified
and an effective epimorphism. But it is likely that experts in singularities will have a more relevant viewpoint.<|endoftext|>
TITLE: Explicit construction of an element of ${\rm GL}(2, p)$ of order $p+1$
QUESTION [13 upvotes]: It is well-known that the order of $GL(2, p)$ is $(p^2-1)(p^2-p) = (p-1)^2(p+1)p.$
It is easy to construct matrices of orders $(p-1)$ and $p$ (diagonal and parabolic, respectively), but the only way that leaps to mind of constructing an element of order $p+1$ is by constructing the companion matrix of the irreducible polynomial whose root is the multiplicative generator of $GF( p^2)$ and then raising it to $p-1$st power - this is quite non-explicit. I assume there is no explicit construction which works for all $p,$ but I could be wrong: is there?

REPLY [5 votes]: Here is a slightly different way to think about this, though I admit that it is not really "explicit". Let $E$ be the field of order $p^2$, so $E$ is a two-dimensional vector space over the subfield $F$ of order $p$. The multiplicative group of $E$ is cyclic of order $p^2 - 1$, so if $x$ is a generator, then multiplication by $x$ is an $F$-linear transformation of $E$ with multiplicative order $p^2 - 1$. Now by choosing any $F$-basis for $E$, this linear transformation determines a matrix which is an element of $GL(2,p)$ having order $p^2 - 1$. A suitable power of this matrix has order $p+1$.<|endoftext|>
TITLE: Axiom of choice and a set in the plane that intersects every line in two points
QUESTION [9 upvotes]: In this question Subset of the plane that intersects every line exactly twice someone ask for a reference of a paper where they proof the result : ''There exist a subset of the plane that intersects each line exactly twice'' (called $2$-point sets [I think]).
I was talking with an students and we were wondering if, in the absence of Choice, this result is still true.
To make my question more precise, I know that we only need Choice to well order the reals and, I believe, you can make the set to be non-measureble but I'm not sure if it is non-measurable itself (I believe that using CH you can make a $2$-point set of measure zero [but, to be honest, I haven't work out the details]).
So, being more concrete, how much choice is need to create such a set? Are there models of ZF without them?

REPLY [3 votes]: I wasn't aware of A. Miller's paper http://www.math.wisc.edu/~miller/res/two-pt.pdf up until recently. In Miller's model, DC (dependent choice) is false (there is a Dedekind finite infinite set). In a paper with M. Beriashvili I show that ZF plus DC plus the existence of a 2-point set does not imply that there is a well-ordering of the reals, see https://ivv5hpp.uni-muenster.de/u/rds/mazurkiewicz_sets.pdf or https://ivv5hpp.uni-muenster.de/u/rds/mazurkiewicz.pdf .<|endoftext|>
TITLE: Generalization of a lemma of Livne
QUESTION [5 upvotes]: Let $H$ be a finite $2$-group. Let $N_{4}(H)$ be the subgroup generated by fourth powers. Let $H_{4}$ be the last term in the short exact sequence $1\rightarrow N_4(H) \rightarrow H \rightarrow H_{4} \rightarrow 1$. Suppose that every element of $H_{4}$ has a lift to an element of order $4$, $2$ or $1$ in $H$. Is it then true that $N_4(H)=1$? When $4$ is replaced by $2$ this is true, and used in Livne's generalization of Faltings-Serre.

REPLY [2 votes]: The answer is yes. First, notice that if $\phi:G\rightarrow G'$ is an epimorphism of 2-groups, then $\phi(N_4(G)) = N_4(G')$. Let now $H$ be the group in your statement. Assume that $N_4(H)$ is nontrivial. Then it contains a maximal subgroup $N$ of index 2 which is normal in $G$ (this follows from considering the action of the quotient $H_4$, which is also a 2 group, on the $\mathbb{F}_2$-vector space $N_4(H) / N_2(N_4(H))$). By considering the group $H/N$ it is enough to prove that if $|N_4(H)|=2$ and every element of $H_4$ is liftable in the way you describe we get a contradiction.
So $N_4(H)$ contains a unique nontrivial element $z$, which moreover has order 2 and is central in $H$. Since $N_4(H)$ is generated by all 4 powers in $H$, there is a $h\in H$ such that $h^4=z$. So $h$ has order 8. We know by assumption that $\bar{h}\in H_4$ has a lift of order 2 or 4. But the only other possible lift is $hz$ which also has an order 8, a contradiction.<|endoftext|>
TITLE: In a random graph which one is more probable, $k$-clique or $k$-core?
QUESTION [6 upvotes]: Recall that the $k$-core of a graph $G$ is the unique maximal subgraph of $G$ with minimum degree at least $k$.
In an Erdos-Renyi random graph, where the edge selection is independent with probability is $p$, we have the following inequalities:
$$P(\text{$G$ contains at least one $k$-clique}) \leq \dbinom{n}{k}  p^\dbinom{k}{2}$$
and
$$P(\text{$G$ has a nonempty $k$-core}) \leq \dbinom{n}{m} \prod_{i=1}^m \sqrt Q,$$
where
$$Q = \sum_{i=k}^{m-1} \dbinom{m-1}{i} p^i (1-p)^{m-1-i}$$
and $m$ is the size of the $k$-core.
I can't understand how I should compare these two probabilities to conclude which one is more probable.

REPLY [5 votes]: Every graph of average degree $2k$ contains a subgraph of minimum degree $k$, so the threshold for the appearance of a non-empty $k$-core is at most $2k/n$.  Since the threshold for the appearance of a $k$-clique is $n^{-2/(k-1)}$, $k$-cores typically appear before $k$-cliques.
The exact threshold for the appearance of a non-empty $k$-core was found by Pittel, Spencer and Wormald.<|endoftext|>
TITLE: Sum of the absolute eigenvalues of A>=B
QUESTION [7 upvotes]: Kindly help me to prove/disprove the following statement. 
Let $A$ be a symmetric matrix of order $n \times n$ with all the diagonal entry equal to $0$, and other non-diagonal entry equal to $k$ (where $k$ is a fixed positive integer). Similarly,Let $B$ be a symmetric matrix of order $n \times n$ with every diagonal entry equal to $0$, and each non-diagonal entry equal to some $\ell$ with $1\leq \ell \leq k<n$.
For example:$ A= \begin{pmatrix} \begin{array}{cccc}
 0 & 3 & 3 & 3 \\
 3 & 0 & 3 & 3 \\
 3 & 3 & 0 & 3 \\
 3 & 3 & 3 & 0 \\
\end{array}\end{pmatrix}$, 
$B=\begin{pmatrix}
\begin{array}{cccc}
 0 & 3 & 1 & 2 \\
 3 & 0 & 2 & 3 \\
 1 & 2 & 0 & 1 \\
 2 & 3 & 1 & 0 \\
\end{array}
\end{pmatrix}$.
Prove/disprove the following statement: 
Suppose $\lambda_1, \lambda_2, ..., \lambda_n$ are eigenvalues of $A$ and $\mu_1, \mu_2, ..., \mu_n$ are eigenvalues of $B$. Then  $$\sum_{i=1}^{n}{|\lambda_i|}\geq \sum_{i=1}^{n}{|\mu_i|}.$$ Note that $A$ and $B$ are the distance matrices of some vertices(diameteral) of $G_1$ and $G_2$ respectively.

REPLY [10 votes]: The claim is false.
In particular, we have $A=kee^T-kI$, so that $\lambda(A)=((n-1)k,-k,\ldots,-k)$, so that $\|A\|_* = 2(n-1)k$.
Now generate a random matrix $B$ such that $B_{ii}=0$, $B_{ij}=B_{ji}$ and $B_{ij} \le A_{ij}$ for every entry. It does not matter that the entries of $B$ are integers or not (to have a "clean" example, I round the entries below to ensure that $B$ is integral). In particular, try the following Matlab code:
n=63;k=3;B=ceil(k*rand(n));B=B-diag(diag(B));B=round((B+B')/2);sum(svd(B))-2*(n-1)*k

Quite easily one obtains $\|B\|_* > \|A\|_* = 2(n-1)k=372$.  
Other choices of $n$ and $k$ can give you smaller explicit examples if you wish to find (e.g., $n=23, k=2$ also worked after some attempts).<|endoftext|>
TITLE: Fourier transform of the critical line of zeta?
QUESTION [12 upvotes]: This was asked on MSE and got a lot of upvotes but no answers, so I'm posting it here.
Is there a known expression for the (distributional) Fourier transform of the Riemann zeta function, taken along the critical line?
I'd love to say that it's a weighted sum  of delta distributions, logarithmically spaced and decreasing in amplitude, as in
$\sum_n \frac{\delta(\omega+\log(n))}{n^{1/2}}$
but this fails to be a tempered distribution, and fails in general when the exponent in the denominator is less than 1.

REPLY [8 votes]: If $\varphi$ is in the class of Schwartz
we have
$$\int_{-\infty}^{+\infty}\varphi(t)\zeta(\frac12+it)\,dt=
\sum_{n=0}^\infty\Bigl\{
\frac{1}{\sqrt{n+1}}\widehat{\varphi}\Bigl(\frac{1}{2\pi}\log
(n+1)\Bigr)- 2\pi\int_{x_n}^{x_{n+1}}e^{\pi
y}\widehat{\varphi}(y)\,dy\Bigr\}$$
where $x_0=-\infty$ and  $x_n=\frac{1}{2\pi}\log n$.
So that $\zeta(\frac12+it)$ is the Fourier transform of the tempered
distribution defined by
$$\varphi\in{\mathcal S}\mapsto
\sum_{n=0}^\infty\Bigl\{
\frac{1}{\sqrt{n+1}}\varphi\Bigl(\frac{1}{2\pi}\log (n+1)\Bigr)-
2\pi\int_{x_n}^{x_{n+1}}e^{\pi y}\varphi(y)\,dy\Bigr\}$$
In general we can not separate the sum in two, but if $\varphi$ is
such that
$$\int_{-\infty}^{+\infty} e^{\pi
y}|\widehat{\varphi}(y)|\,dy<+\infty$$ we can simplify and put
$$\int_{-\infty}^{+\infty}\varphi(t)\zeta(\frac12+it)\,dt=\sum_{n=1}^\infty
\frac{1}{\sqrt{n}}\widehat{\varphi}\Bigl(\frac{1}{2\pi}\log
n\Bigr)-2\pi \int_{-\infty}^{+\infty} e^{\pi
y}\widehat{\varphi}(y)\,dy.$$
We can say that $\zeta(\frac12+it)$ is the Fourier transform of a
tempered distribution  that can be obtained extending the measure
$$\mu=\sum_{n=1}^\infty \frac{1}{\sqrt{n}}\delta_{\frac{1}{2\pi}\log
n}-\nu,\quad \text{where} \quad \nu(dx)=2\pi e^{\pi x}\,dx,$$
in the indicated way.
To prove this I started from the formula (2.1.5) of Titchmarsh
$$\zeta(s)=s\int_0^{+\infty}\frac{\lfloor
x\rfloor-x}{x^{s+1}}\,dx\qquad (0<\sigma<1).$$

REPLY [3 votes]: The function $\mathbb R\ni t\mapsto\zeta(\frac12+it)$
is  analytic and smaller in absolute value than $C(1+\vert t\vert)^{1/6}$
(the $1/6$ may be replaced by $9/56$ and even by a slightly smaller number). It is thus a tempered distribution. We have, with $E(x)$ standing for the floor function,
\begin{multline}
\zeta(\frac 1 2 +it)=-it\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 -it }dx
-\frac 1 2\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 -it }dx\\-\frac{1+2it}{1-2it},
\tag{$\ast$}
\end{multline}
an identity which follows from the first step of the Euler-Maclaurin formula.
With the above formula, it is easy to find an explicit expression for the Fourier transform: in fact, we need only to calculate the Fourier transform of $t\mapsto e^{-it \ln x}$, which is $\delta_0(\tau+\frac{\ln x}{2π})$
and moreover, for $\phi$ in the Schwartz space,
the integral
$$
\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 }\phi(-\frac{\ln x}{2π})dx,
$$
is absolutely converging. As a result, the Fourier transform of the second term in $(\ast)$ is given by
$$
\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 }\delta_0(\tau+\frac{\ln x}{2π})dx,
$$
which makes sense as a Radon measure. The first term has a Fourier transform which is essentially the derivative of the above Radon measure, because of the $t$ in front, whereas the Fourier transform of the last term is easy to get explicitly.<|endoftext|>
TITLE: What are some open problems regarding elliptic curves in finite fields?
QUESTION [7 upvotes]: I accept that my question seems so vague and broad, and I already looked into some similar questions in MO. But I would like to learn specifically about some open problems and conjectures regarding elliptic curves in finite fields. Also if there is something about their isogeny in particular it is highly welcomed. If you can provide also a reference to where the problem was formulated I would be glad. To clarify, both theoretical and computational open problems are welcomed.

REPLY [6 votes]: We know that, given an elliptic curve over a finite field $\mathbb{F}_q$, there exists integers $m,n$ with $m|n$ such that the group of rational points is a product of a cyclic group of order $m$ and a cyclic group of order $n$. I believe it is still open to deterministically, in polynomial time, compute $m,n$ (a big obstacle is to do it without factoring the gcd of $mn$ and $q-1$). Even if that's solved, I know it's an open problem to, again deterministically, in polynomial time, compute two points of order $m,n$ (or one point if $m=1$) that generate the group of rational points. (NB polynomial time means polynomial in $\log q$). 
Another question which just occurred to me is, given two elliptic curves over the same finite field with the same number of rational points, deterministically, in polynomial time, compute an isogeny between them. The proof of Tate's theorem seems horribly inefficient from a computational point of view.<|endoftext|>
TITLE: Topology of categories, very basic facts surrounding Quillen's Higher Algebraic K-Theory I
QUESTION [9 upvotes]: In his paper Higher Algebraic K-Theory I (see [here][1]), Quillen introduces a topological space $BC$, called the classifying space of $C$, and tries to relate its topology to the categorical structure of $C$. I am curious as to what is the importance of 1. adjoint functors inducing homotopy equivalences, and 2. computing the fundamental group of $BC$ in terms of purely categorical information. Could anyone tell me? I guess when I say "importance", I mean "having shown these facts, how might have stuff later in the paper arose naturally to Quillen?"

REPLY [11 votes]: N.B.: I have reread your question and it occured to me that you a probably asking something entirely different. However since I'm unclear what exactly is your question and since I don't want to delete this wall of text, here it is.
Firstly, both categories and (homotopy types of) topological spaces admit a description as simplicial sets which satisfy certain lifting properties. This explains why we would want to consider them on an equal footing. Specifically, a topological space gives a simplicial set for which any n-horn is fillable. Here an n-horn is the boundary of an n-simplex with one face omitted, and fillability means that any map from it can be extended to a map from the whole n-simplex. This defines an equivalence of homotopy categories. Categories can also be descibed via filling of horns, only now we 1) only consider inner horns (a combinatorial simplex has its vertices numbered; an inner horn is the one for which we drop any face other than those lying against the first or last vertex) and 2) demand that the lifting is unique. It is a nice exercise to verify that such data indeed defines a category.
In particular, there is an intermediary notion of $(\infty,1)$-categories (also called quasicategories in this specific case). A quasicategory is a simplicial set, for which any inner horn is fillable (unlike classical categories, we don't require uniqueness). It is easy to see that both categories and spaces are specific kinds of quasicategories. Morally a quasicategory is a category enriched in topological spaces, but there are complications (see e.g. introduction to J. Lurie's "Higher topos theory" for details). Quasicategories themselves form a quasicategory, just like there is a category of categories. Also there is a natural quasicategory of spaces, for which the hom-space of morphisms between two spaces is just the mapping space (especially easy to define for simplicial sets).
Now we have an inclusion of the quasicategory $Top$ of spaces into the quasicategory $Cat_\infty$ of quasicategory and this inclusion has a left adjoint. Generally it can be described as Kan fibrant replacement, but for categories this is just the functor $BC$ defined in Quillen's paper. Essentially this is the answer to your question. The classifying space of the category is the universal space in which your category maps, a truely universal way to invert all arrows of your category. Since this is a functor, it also applies to hom-categories. This means that any natural transformation between functors become invertible as natural transformation between classifying spaces, i.e. a homotopy equivalence.
This is certainly not the only way to gain topological information from a category. For example for a category $C$ we can consider its core groupoid $C^{core}$, which has the same objects but only invertible morphisms. It can also be considered as a space and thus carries much more interesting information. In particular, Quillen's Q-construction associates a K-theory space to core groupoids (with extra data). Under the equivalence stated above, the core groupoid functor is a right adjoint to the inclusion of spaces into quasicategories, so it is also functorial and universal.
Regarding your second question, any covering space $E \stackrel{f}{\to} B$ can be considered as a functor $B \to Set$ under the equivalence between spaces and certain quasicategories stated above. Geometrically you associate to any point $x: B$ the fiber $f^{-1}(x): Set$, to any path $p: \mathrm{path}(x, y)$ the map $f^{-1}(x) \to f^{-1}(y)$ obtained by the unique lifting of $p$ to a path in $E$, and more generally to any n-simplex $\Delta : B$ its lift to an n-simplex in $Set$ constructed similarly. This means that we have a chain of equivalences
$$\begin{eqnarray} \mathrm{Covers}(BC) & = & BC \to Set 
\\ & = & BC \to Set^{core}
\\ & = & C \to Set^{core}  \end{eqnarray}$$
This is precisely the theorem you stated.<|endoftext|>
TITLE: Minimize sum of $\ell_2$ norm and linear combination, on simplex
QUESTION [13 upvotes]: Let $\Delta_n := \{x \in \mathbb{R}^n | x \ge 0, \sum_{1 \le i \le n}x_i = 1\}$ be the $n$-simplex. For $a, b \in \mathbb R^n$,  with $\Delta_n \not \ni a$, consider the problem of computing the following value exactly
$$\alpha := \min_{x \in \Delta_n}\|x-a\| + \langle b,x\rangle.$$
The case when $b = 0$ corresponds to the problem of projecting the point $a$ unto $\Delta_n$. This special case is well-known to be exactly solvable in $\mathcal{O}(n)$ flops.
Now, for the general problem, it's not hard to rewrite
\begin{equation}
\begin{split}
\alpha &:= \min_{x \in \Delta_n}\max_{\|y\| \le 1} \langle y, x - a \rangle + \langle b, x\rangle = \max_{\|y\| \le 1}\left(\min_{x \in \Delta_n}\langle y + b, x\rangle\right) - \langle y, a\rangle\\
&= \max_{\|y\| \le 1}\left(\min_{1 \le i \le n}y_i + b_i\right) - \langle y, a \rangle = \max_{t \in \mathbb R}t - \min_{\|y\|^2 \le 1,\;y_i \ge t - b_i \forall i}\langle y,a\rangle.
\end{split}
\end{equation}
Question: Given $c \in \mathbb R^n$ ($c_i \equiv b_i - t$), can one compute the value of $$\min_{\|y\|^2 \le 1,\;y \le c}-\langle a,y\rangle$$
analytically ? For which values of $c$ does the latter problem have a solution ?
No polynomial time algorithm for exact solution ?
Observation: In low dimensional cases ($n = 1, 2, 3$), when non-degenerate, it's not hard to sketch that this problem has solutions which are piece-wise polynomials (or square roots of such) in the $a_i$'s and $c_i$'s, the number of pieces being in the order of $2^n$.

Update: Algorithm based on @fedja's answer + proof of Q-linear convergence in the "small perturbation" regime
For generality, let $a \in \mathbb R^n$ and $C$ be a "simple" (to be clarified) closed convex subset of
$\mathbb R^n$ not containing the point $a$. Consider the problem
\begin{equation}
\text{minimize } \|x - a\| + \langle b, x\rangle\text{ subject to }x \in C.
\end{equation}
Fedja's idea.
The idea is to introduce a radial variable $r := \|x-a\|^{-1}$. Indeed,
using the well-known elementary inequality
\begin{equation}
t + t^{-1} \ge 2\; \forall t > 0,\text{ with equality iff } t = 1,
\end{equation}
it follows that $\forall x \in \mathbb{R}^n$ and $\forall r > 0$, we
have $$\|x-a\| \le \frac{1}{2}(r\|x-a\|^2 + r^{-1}),$$
and this bound is attained at $r = \|x-a\|^{-1}.$
Thus completing the square in $x$, the optimal value $\alpha$ for the problem can be rewritten in the form
\begin{equation}
2\alpha- b^Ta = \min_{r > 0,x \in C}r\|x - (a + r^{-1}b)\|^2 +
(1-b^Tb)r^{-1}.
\end{equation}
The algorithm.
Based on Fedja's idea of introducing the radial variable $r =
\|x-a\|^{-1}$, the following alternating iterative scheme
for solving the above problem exactly, is natural
\begin{equation}
x^{(k + 1)} = \mathrm{proj}_C(a + b / r^{(k)}),\; r^{(k+1)} := \|x^{(k + 1)} -
a\|^{-1},
\end{equation}
with $r^{(0)} > 0$.
Next, we will proof some interesting things regarding the numerics for
the algorithm so-obtained.
Q-linear convergence in the "small perturbation" regime: $\|b\| < 1$.
Eliminating the $x$ variable from the above iterates, we see that the
$r$-update can be rewritten as a Picard process
\begin{equation}r^{(k+1)} = \|\mathrm{proj}_C(a + b/r^{(k)})
  - a\|^{-1}.
\end{equation}
Let's proof that this process converges after essentially $\mathcal
O(1)$ rounds.
This would mean that the cost of solving the "$b \ne 0$" problem is
essentially the cost of projecting onto $C$, i.e the cost of
solving the "$b=0$" problem.
Assumption.
Let's make the following minimal assumption: There is an oracle
  for exactly projecting onto $C$.
For example, this assumption holds for polyhedra, $\ell_p$ balls,
etc. In order to only concentrate on the "heart of the issue'', let's
take for granted that the above process has a fixed-point $r^{(*)} > 0$. Now,
\begin{eqnarray*}
\begin{split}
\left|\frac{1}{r^{(k+1)}} - \frac{1}{r^{(*)}}\right| &= \Big|\|\mathrm{proj}_{
  C}(a +
b/r^{(k)}) - a\| -\|\mathrm{proj}_{C}(a +
b/r^{(*)}) - a\|\big|\\
&\le \|\mathrm{proj}_{C}(a + b/r^{(k)}) -
\mathrm{proj}_{C}(a + b/r^{(*)})\| \\
&\le \|b/r^{(k)} - b/r^{(*)}\| =
\|b\|\Big|\frac{1}{r^{(k)}} - \frac{1}{r^{(*)}}\big|,
\end{split}
\end{eqnarray*}
where the first inequality is the "reverse triangle inequality'' and
the second follows from the nonexpansivity of the projection operator onto
a closed convex set. Thus if $\|b\| < 1$, then the residuals $\Big|\frac{1}{r^{(k)}} - \frac{1}{r^{(*)}}\Big|$ decay exponentially
fast (i.e Q-linearly) at rate $\|b\|$.

REPLY [9 votes]: I would try to approach your original problem a bit differently. Note that $|x-a|\le \frac 12(r|x-a|^2+r^{-1})$ and the equality is attained for $r=|x-a|^{-1}$. Thus, 
$$
\min_x[|x-a|+\langle b,x\rangle]=\frac 12\min_r\min_x[r|x-a+r^{-1}b|^2+r^{-1}(1-|b|^2)]+\langle a,b\rangle
$$
However, the inner minimum can be now found rather quickly and finding the outer one is a one-dimensional problem. Moreover, I suspect that even the naive algorithm of choosing $r$ arbitrarily to start with, finding the corresponding $x$, and then changing $r$ to something stupid like $(|x-a|^{-1}+r)/2$ has a decent chance to work but checking that would require some accurate analysis of the properties of the inner minimum as a function of $r$. What is true, however, is that if the minimizer $x$ for some $r$ satisfies $|x-a|=r^{-1}$, then you have the true minimum for the original expression, so, at least, you can recognize the solution when you see it this way.
Edit:
I thought it might make sense to add a few more remarks to what I wrote already.
Let's look at what we have so far and see if we can convert it into something that is guaranteed to work. Put $\rho=r^{-1}$. Then the main idea was to observe that instead of the original minimization problem, we can solve the problem
$$
\Psi(x,\rho)=\frac 12\left[\frac{|x-a|^2}\rho+\rho\right]+\langle b,x\rangle\to\min
$$ 
and that in this new problem the minimization in $x$ is equivalent to finding the closest point to $a-\rho b$. 
Since $\Psi$ is jointly convex in $x,\rho$ and the best $\rho$ has physical meaning of the distance from $a$ to $x$, the outer minimization problem is just the problem of finding the minimum of a convex function over a finite interval of length about the diameter of the convex body $C$ we project upon, so, even if the iterations fail, we can always resort to the good old bisection, which is guaranteed to give us the answer with the precision about $2^{-m/2}|b|\operatorname{diam} C$ where $m$ is the number of steps.
This is not an "exact" solution, of course. However, if you implement an exact algorithm on any real machine, you'll still never be able to get anything better than the machine precision. So, I'm not really sure how much sense the request for an exact solution makes here from the purely practical viewpoint. Still, notice that the closest point to $a-\rho b$ moves along a straight line as long as we stay in the relative interior of the same $k$-dimensional face of the simplex. Unfortunately, this gives us exponentially many lines, so just tracing all of them is impractical.  However, we can always first run the bisection for a while until we reduce the situation to a sufficiently small interval, after which we can use only the lines that correspond to the points on that interval and, in principle, we can try to trace them. The ideal scenario is that after a few iterations in the bisection, we get the values of $x$ lying in the relative interior of the same $k$-dimensional face for both endpoints of the remaining interval, in which case we have just one line left and we can finish in a single step. It is not guaranteed to always happen, of course, but I suspect that it may work more often than not.<|endoftext|>
TITLE: Thurston geometries---the geometry of the universal cover of $SL(2, \mathbb{R})$
QUESTION [6 upvotes]: In one of the eight Thurston geometries there is the geometry of the universal cover of $SL(2, \mathbb{R})$.  But from the algebraic point of view $PSL(2,\mathbb{R})$ is sufficient for building 3-manifolds i.e. we let the group act on itself and then quotient out a discrete subgroup of it.  This is what we do for the nilgeometry where we use the 2-step nilpotent Heisenberg group instead of $PSL(2,\mathbb{R})$.
Thus my question is, why passing to the universal cover?  Professor told me that it is for obtaining a simply-connected 3-manifold so as to include as many manifolds as possible for the classification i.e. there is a preference for starting with simply-connected manifolds.  But I do not really understand the reason behind it although I guess there should be a quick answer to it.

Let me try to clarify my question a bit.  My intention is to make a comparison between nilgeometry and $SL(2,\mathbb{R})$-geometry.  Their algebraic construction is the same except that one starts with the 2-step nilpotent Heisenberg group and the other with $PSL(2,\mathbb{R})$.  However for $SL(2,\mathbb{R})$-geometry there is an additional step of passing to the universal cover of $SL(2,\mathbb{R})$.  What I am asking is the motivation for this extra step.

REPLY [2 votes]: Your question "why pass to the universal cover?" is really a topological question that has little to do with the special case of solvgeometry, so it may be helpful to point out that you wouldn't want to study flat manifolds by starting with the tori, but rather you start with euclidean spaces to get more examples as quotients, including cylinders.<|endoftext|>
TITLE: How many non-isomorphic graphs of 50 vertices and 150 edges
QUESTION [6 upvotes]: Is there an way to estimate (if not calculate) the number of possible non-isomorphic graphs of 50 vertices and 150 edges?

REPLY [8 votes]: The simplest guess one could make is $\frac{1}{50!} { {50 \choose 2} \choose 150}$. That is, we first count the number of labeled such graphs, then assume that most of them have trivial automorphism group so we can approximately divide out by $50!$ when removing the labels. You can estimate how big this is using Stirling's formula. 
Edit: Actually, you can also just ask WolframAlpha to compute this estimate. You get $7.028... \times 10^{131}$, which is apparently within 1% of the true answer according to Brendan McKay.<|endoftext|>
TITLE: What categorical property of monoidal categories picks out the ones with duals?
QUESTION [18 upvotes]: Recall that a monoidal category $\mathcal C$ is rigid if every object $X\in \mathcal C$ has both left and right duals, i.e. objects $X^l$ and $X^r$ with maps $X^l \otimes X \to \mathbf 1 \to X \otimes X^l$ and $X \otimes X^r \to \mathbf 1 \to X^r \otimes X$ satisfying certain equations.  It is a fundamental fact about monoidal categories that "having a left dual" and "having a right dual" are properties of an object, not data: given $X$, the objects $X^l$ and $X^r$, if they exist, are uniquely determined up to unique isomorphism.  As such, for a monoidal category itself to be rigid is also property --- the only data is the data of being monoidal.
This should remind you of groups.  Given a monoid $G$, an element $x\in G$ is invertible if there are elements $x^l$ and $x^r$ and equalities $x^l x = 1 = x x^r$.  (Undergraduate exercise: $x^l = x^r$.)  Being invertible is property.  A monoid $G$ is a group if every element therein is invertible.  Thus this too is a property.
In some cases (e.g. algebraic geometry) you don't always want to think about the elements of a group $G$.  Fortunately, there is a very nice way to say when a monoid is a group that does not directly refer to invertibility of elements.  Let $G$ be a monoid and consider the map $G \times G \to G\times G$ (a map of underlying spaces, not of groups) that takes $(x,y)$ to $(x,xy)$.  The monoid $G$ is a group iff this map is an isomorphism (of underlying spaces).

Is there a similar characterization of when a monoidal category is rigid?  Something like "consider the map $\mathcal C \times \mathcal C \to \mathcal C \times \mathcal C$, and ask that it be a left adjoint"?

REPLY [8 votes]: The notion of rigidity can be defined for any pseudomonoid in a monoidal proarrow equipment.  See for instance Dualizations and Antipodes by Day, McCrudden, and Street (although they work only with monoidal bicategories and don't make the equipments explicit).<|endoftext|>
TITLE: Cohomology of vector bundles in families
QUESTION [9 upvotes]: Let $\pi \colon \mathfrak{X} \rightarrow B$ be a deformation of complex compact manifolds and $E$ be a holomorphic vector bundle on $\mathfrak{X}$ (or a coherent sheaf on $\mathfrak{X}$ that is flat over $B$).
The function $b \rightarrow \mathrm{h}^0(X_b, E_{|X_b})$ is known to be upper semi-continuous, but I wonder if it is constructible. 
If the deformation is algebraic, that is if there exists a relatively ample line bundle on $\mathfrak{X}$, then there exists a complex $\mathcal{L}^{\bullet}$ of locally free sheaves on ${B}$ such that for any $b$ in $B$ and any nonnegative integer $i$, 
$$
\mathrm{H}^i(\mathcal{L}^{\bullet}_{|b}) \simeq \mathrm{H}^i(X_b, E_{| X_b}).
$$
This is proved for instance in [Voisin, Complex algebraic geometry, Vol. 1, section 9.3.1], and implies constructibility. 
Is the same result known in the complex analytic case?

REPLY [5 votes]: The constructibility result you ask for is Satz 7.7(1) in Ein Kriterium für die Offenheit der Versalität (Flenner, 1981).
The appropriate generalization of the result in Voisin's book that you mention is the topic of Eine Bemerkung über relative Ext-Garben (Flenner, 1981).<|endoftext|>
TITLE: Why are monadicity and descent related?
QUESTION [26 upvotes]: This question is probably too vague for experts, but I really don't know how to avoid it.
I've read in several places that under mild conditions, a morphism is an effective descent morphism iff the base-change functor it induces is monadic.
Now, I don't really know anything about descent and I'm trying to probe around and figure out how to start learning about it. I think of monads in terms of algebraic theories. Descent on the other is supposedly of geometric nature; a formalism which generalizes familiar gluing over open subsets to a more general setting without a spatial topology. I can't imagine how or why these two notions should be related.
What lies at the core of the relationship between monadicity - a seemingly algebraic and concrete notion - and (effective) descent (morphisms)?

REPLY [24 votes]: I think of monads in terms of algebraic theories. 

Monads are substantially more general than this intuition suggests! Here is a better intuition: monads are categorified idempotents. 
The point of idempotents (acting on, say, a module) is to pick out nice subobjects: subobjects that are so nice that they are simultaneously subobjects and quotient objects (say, direct summands of modules) in a compatible way. More formally, every idempotent $m : X \to X$ wants to become a pair of a map $f : X \to Y$ and a map $g : Y \to X$ such that $g \circ f = m$ and $f \circ g = \text{id}_Y$. Similarly, the point of monads is to pick out nice categories which simultaneously map into and out of a category in a compatible way (via an adjunction). More formally, every monad $M : C \to C$ wants to become a pair of a functor $F : C \to D$ and a functor $G : D \to C$ such that $F$ and $G$ are adjoint and $G \circ F \cong M$. 
This analogy is quite robust: for example, the analogue of taking the fixed points of an idempotent is taking the category of algebras of a monad. And the analogue of an adjunction being monadic is a submodule being a direct summand. 

Descent on the other is supposedly of geometric nature; a formalism which generalizes familiar gluing over open subsets to a more general setting without a spatial topology. I can't imagine how or why these two notions should be related.

Here's a simple toy model. Let $f : X \to Y$ be a map of sets and let $\text{Sh}(X)$ be, for concreteness, the functor assigning a set $X$ the category of sheaves of sets on $X$, which just means the category of assignments, to each $x \in X$, of a set $A_x$. There is a pullback functor
$$f^{\ast} : \text{Sh}(Y) \to \text{Sh}(X).$$
It has a right adjoint $f_{\ast} : \text{Sh}(X) \to \text{Sh}(Y)$ given by taking fiberwise products: that is, if $A$ is a sheaf of sets on $X$, then 
$$f_{\ast}(A)_y = \prod_{f(x) = y} A_x.$$
(It also has a left adjoint which we'll ignore.) This adjunction induces a comonad $f^{\ast} f_{\ast} : \text{Sh}(X) \to \text{Sh}(X)$ sending a sheaf $A$ of sets on $X$ to the sheaf
$$f^{\ast} f_{\ast}(A)_x = \prod_{f(x') = f(x)} A_{x'}.$$
Now, what should descent mean in this situation? $f$ should be descent iff it is surjective, and descent should intuitively say that a sheaf on $X$ descends to a sheaf on $Y$ iff for all $y \in Y$, all of the sets $A_x, f(x) = y$ are canonically identified. This reflects the fact that $f$ is surjective iff $Y$ itself is obtained from $X$ by quotienting by the equivalence relation $x \sim x' \Leftrightarrow f(x) = f(x')$. (Which in turn says that surjections in $\text{Set}$ are effective epimorphisms.)
This is encoded by the comonad above as follows. A coalgebra for the above comonad is a sheaf $A$ on $X$ together with a map $A \to f^{\ast} f_{\ast}(A)$ satisfying some compatibilities. What does such a map look like? Stalkwise it looks like a map
$$A_x \to \prod_{f(x') = f(x)} A_{x'}$$
and this map will end up encoding a bunch of isomorphisms $A_x \cong A_{x'}$. These isomorphisms should satisfy a cocycle condition which is encoded by the coalgebra compatibilities. 
The decategorified version of this story is that a real-valued function $A : X \to \mathbb{R}$ descends to a function $Y \to \mathbb{R}$ iff $A_x = A_{x'}$ for all $x, x'$ such that $f(x) = f(x')$. Moreover, if $f$ has finite fibers, we can pick out which functions these are as the fixed points of the idempotent
$$m(A)_x = \frac{1}{|f^{-1}(f(x))|} \sum_{f(x') = f(x)} A_{x'}$$
acting on the vector space of functions $X \to \mathbb{R}$. 
A more interesting case to work out is the case that $f : X \to X/G$ is a Galois cover with Galois group $G$. In this case descent will say that $\text{Sh}(X/G)$ is the category of homotopy fixed points $\text{Sh}(X)^G$ for the action of $G$ on $\text{Sh}(X)$, and the way in which the comonad $f^{\ast} f_{\ast}$ encodes this fact is a categorification of the fact that if $G$ is a finite group acting on a vector space $V$ (over a field of suitable characteristic), the subspace $V^G$ of fixed points is a direct summand picked out by the idempotent $\frac{1}{|G|} \sum_{g \in G} g$. This is a geometric form of Galois descent.<|endoftext|>
TITLE: A proper class of formulas with every set-sized (but no proper-class-sized) subcollection satisfiable
QUESTION [5 upvotes]: What feature(s) must a (non 1st-order) language with proper-class-many formulas have in order to guarantee that:
There is a proper class P of formulas such that both
(a) every set-sized sub-collection of P is satisfiable, and
(b) no proper-class-sized sub-collection of P is satisifable?

REPLY [5 votes]: Let me give a few examples.
Example 1. Let us work in Gödel-Bernays set theory, and
assume that $T\subset {}^{<\text{Ord}}2$ is a proper class tree of
height Ord, but there is no cofinal branch.
(This theory is consistent relative to an inaccessible cardinal,
because if $\kappa$ is inaccessible and not weakly compact, then
there is a $\kappa$-Aronszajn tree $T\subset {}^{<\kappa}2$, and
then $V_\kappa$ with all subsets is a model of GBC where $T$ has
the desired property.)
In the logic $L_{\infty,\omega}$, which allows arbitrary sized
conjunctions and disjunctions, with a constant for every element
of $T$ and a unary predicate symbol $B$, consider the theory $P$
consisting of the assertions $\varphi_\alpha$ asserting first,
that there is precisely one object $u$ on level $\alpha$ of the
tree that satisfies $B$, and secondly, that in this case, every
$v<_T u$ also has $B(v)$. These assertions can be made in the
logic $L_{\infty,\omega}$ using constants for the elements of $T$.
Thus, altogether, $P$ is the theory asserting that $B$ is a
cofinal branch through the tree.
Every set-sized subtheory of $P$ mentions only a bounded number of
levels, and so we can find a model by picking any node above that
bound and using the predecessors of that node as the instantiation
of $B$.
But under our assumptions that the tree $T$ is Ord-Aronszajn,
there can be no model of all of $P$ or even of a proper class
sized subtheory of $P$, because any such subtheory will involve
the assertions concerning unboundedly many levels of $T$, and so
the model of that subtheory will pick out a cofinal branch in $T$;
but there is no such branch.
Meanwhile, there is a strong connection between your property and
(non)weak compactness, because an inaccessible cardinal $\kappa$
is weakly compact just in case we have the $\kappa$-compactness
property for $L_{\kappa,\kappa}$ theories of size $\kappa$. (And
there are diverse variations on this.)
Example 2. Here is a different kind of related example using
only first-order logic.
Theorem. There is a proper class first-order theory $P$, such
that every set-sized subtheory of $P$ has a model, but no class is
a model of the whole of $P$.
Proof. We interpret this as a theorem scheme in ZFC, where by
"class" we mean a definable class (allowing parameters). Thus, I
shall provide a definition of a theory $P$, and then prove first,
that every set-sized subtheory of $P$ is satisfiable, and second,
that no definable class is a model of $P$.
Let $P$ be the theory in the language of set theory $\in$
augmented with a predicate $\newcommand\Tr{\text{Tr}}\Tr$, meant
to serve as a truth-predicate, plus a constant for every object in
the universe. The theory $P$ asserts that $\Tr$ obeys all
instances of the recursive Tarskian truth definition:

$\Tr(a\in b)$ just in case $a\in b$ holds.
$\Tr(a=b)$ just in case $a=b$.
$\Tr(\varphi\wedge\psi)$ just in case $\Tr(\varphi)$ and
$\Tr(\psi)$.
$\Tr(\neg\varphi)$ just in case $\Tr(\varphi)$ does not hold.
$\Tr(\exists x\ \varphi)$ just in case there is $a$ such that
$\Tr(\varphi(a))$.

For any set many such assertions, we can find a model, since we
can find some $V_\theta$ large enough to contain all the
parameters mentioned in the subtheory, and then use truth in
$\langle V_\theta,\in\rangle$, which will satisfy all the
assertions made in the subtheory.
But no definable class can satisfy $P$, because this is exactly
the content of Tarski's theorem on the non-definability of truth.
QED
Meanwhile, this theory $P$ of the theorem does has proper-class
sized subtheories that are satisfiable, since we could, for
example, restrict to the quantifier-free assertions; since there
are so many constants, we can produce proper class trivially
satisfiable subtheories.<|endoftext|>
TITLE: $U_q(\mathfrak{sl}_2)$ representations of "quantum dimension" zero
QUESTION [8 upvotes]: I'm reading up on quantum groups and their applications and I've come across a question I just can't find an answer to. I know about the basic representation theory of $U_q(\mathfrak{sl}_2)$ and I know that when $q$ is a primitive root of unity (say, not equal to 1 or -1) then there are two irreducible representations in each dimension strictly smaller than the degree of $q$ and an infinite family of them in the dimension equal to this degree. Now, I've often heard people say that there is however some sort of finiteness to its category of representations because all the rep's of dimension $deg(q)$ are "of quantum dimension zero". The only notion of "quantum dimension" I know of is in pivotal tensor categories (which the aforementioned are) by taking the trace of identity morphisms (or the pivotal structure maps if you will) but I don't see why this dimension should vanish for these objects. Am I interpreting this right but just am too blind to see it? 
Edit:
Suppose $d=min(n\in \mathbb{N}:q^n=1)$ then by $deg(q)$ I mean the number $d$ if $d$ is odd and $d/2$ if $d$ is even. An example of the context I'm talking about would be the paper Topological invariants from non-restricted quantum groups as I'm trying to deal with these Turaev-Viro-type models which should be formulated over fusion categories (so in particular categories with finitely many simple objects) so that the sum over dimensions in their definition does not diverge. However, this is unfortunately all hear-say my advisors have thrown at me making it rather difficult for me to actually find reliable sources.

REPLY [7 votes]: Yes, you are interpreting it right.  For $U_q(\mathfrak{sl}_2)$, there's a very concrete interpretation of the quantum dimension: it's the trace (in the usual sense) of the element $K$.  So, on the Weyl module of highest weight $n$ (the representation generated by $v$ with the relation $Kv=q^nv$ and $F^{(n+k)}v=E^{(k)}v=0$ for all $k>0$), the quantum dimension is $\frac{q^{n+1}-q^{-n-1}}{q-q^{-1}}=q^n+q^{n-2}+\cdots+q^{-n}$.  So, if $q^{2n+2}=1$, then this representation has quantum dimension 0.<|endoftext|>
TITLE: On $XX'=I$ such that $AX=XB$ is true
QUESTION [6 upvotes]: Given list of symmetric matrices $\{A_i,B_i\}_{i=1}^r\in\Bbb R^{n\times n}$ where $r\in\Bbb N$ is arbitrary what is a good description of collection of $X\in\Bbb R^{n\times n}$ such that $$XX'=I$$ $$A_iX=XB_i$$ holds?
Note: $X'$ means the transpose of $X$.
(1) Is there a test to see if there is no such $X$?
(2) Is it easy to see if we find one such $X$ we can find ALL such $X$ from finding $U\in\Bbb R^{n\times n} $ such that $UU'=I$ and $UA_i=A_iU$ or $B_iU=UB_i$ at every $i\in\{1,\dots,r\}$?
(3) What if we have $\Bbb F_{p^r}$ instead of $\Bbb R$?

REPLY [3 votes]: A necessary condition is that, for every word $w(x,y)$ in two letters, we have
$${\rm Tr}w(A_i',A_i)={\rm Tr}(B_i',B_i).$$
If $r=1$ (one pair only), this condition is also sufficient; this is Specht's Theorem. Notice that the symmetry assumption is not important here.
In the present case, a necessary condition is that for every word $w$ in $2r$ letters, one has
$$w(A_1,A_1',\ldots,A_r,A_r')=w(B_1,B_1',\ldots,B_r,B_r').$$
Whether it is also a sufficient condition is unclear to me.<|endoftext|>
TITLE: Precise relationship between "finite" Fourier analysis and Galois theory in solving the cubic?
QUESTION [7 upvotes]: Suppose we want to solve$$x^3 - ax^2 + bx - c = 0.$$We know a priori that this can be factored as $(x - r_0)(x - r_1)(x - r_2)$; by Vieta's formulas, we know$$a = r_0 + r_1 + r_2,\quad b = r_0r_1 + r_1 r_2 + r_2r_0,\quad c = r_0r_1r_2.$$These expressions are invariant under three-cycles. Now, we make the substitution$$r_0 = u_0 + u_1 + u_2,\quad r_1 = u_0 + u_1\omega + u_2\omega^2,\quad r_2 = u_0 + u_1\omega^2 + u_2\omega^4,$$where $\omega$ is a primitive cube root of unity. Explicitly, we have$$u_0 = {1\over3}(r_0 + r_1 + r_2),\quad u_1 = {1\over3}(r_0 + r_1\omega^{-1} + r_2 \omega^{-2}),\quad u_2 = {1\over3}(r_0 + r_1 \omega^{-2} + r_2\omega^{-4}).$$Conceptually,$$\text{if }F(z) = u_0 + u_1z + u_2z^2,\text{ then }r_i = F(\omega^i).$$The first of Vieta's relations now reads$$a = F(1) + F(\omega) F(\omega^2) = 3u_0,$$which is a roots of unity filter on $F$. The second reads$$b = F(1)F(\omega) + F(\omega)F(\omega^2) + F(\omega^2)F(1),$$which is a roots of unity filter on$$F(z)F(\omega z) = (u_0 + u_1 z + u_2z^2)(u_0 + u_1z\omega + u_2z^2\omega^2).$$Since in a filter, we only care abou tthe cubic terms – here $u_0^2$ and $-u_1u_2z^3$ – we find that $b = 3u_0^2 - 3u_1u_2$. If we repeat the same thing for $c$ and compile everything together, we arrive at$$a = 3u_0,\quad b = 3u_0^2 - 3u_1u_2,\quad c = u_0^3 + u_1^3 + u_2^3 - 3u_0u_1u_2.$$The first equation gives us $u_0 = a/3$ for free, and we can then obtain $u_1u_2$ in terms of $a$, $b$, $c$. Finally, the last equation tells us what $u_1^3 + u_2^3$ are. So, we can compute the values of $u_1^3 + u_2^3$, $u_1^3 \cdot u_2^3$; this reduces to a quadratic, which we can solve.
My question is, what is the precise underlying interplay going on between the Galois theory and the "finite" Fourier analysis here, if there is any? Am I shooting in the dark for something that may possibly not exist? Or is there something deep behind all this?

REPLY [6 votes]: The idea is that you can extract the explicit Kummer extension directly from the Fourier transform of the roots.  If you have an element of the Galois group that cyclically permutes the roots, by weighting with appropriate roots of unity, i.e., taking a Fourier transform, the automorphism is diagonalized and acts on each $u_i$ via multiplication by a root of unity.  Now to determine the roots, you just have to find $u_i^3$.
Usually, one first simplifies by shifting all of the roots by $-a/3$, i.e. letting $x_{new} = x - a/3$, thereby setting $a_{new}=u_{0,new}=0$ (I'll omit the "new" from now on).  Then $b = u_1 u_2$ and $c = u_1^3 + u_2^3$.  This makes the necessary steps easier to see.<|endoftext|>
TITLE: Sophisticated treatments of topics in school mathematics
QUESTION [73 upvotes]: Sophisticated mathematical concepts typically shed light on sophisticated mathematics. But in a few cases they also apply to elementary mathematics in an interesting way. I find such examples particularly enlightening, and would love to have a list of them. To make the question well-defined, my (arbitrary) cutoff for "elementary" is USA K-12. By "sophisticated" I mean 20th-21st century research mathematics, and the applications should be nontrivial enough to have appeared in research-level publications.
I list two examples I know.  

The carrying operation in base 10 addition involves the computation of a group 2-cocycle, as explained in this MO answer and in this ncatlab page following (Daniel C. Isaksen, A cohomological viewpoint on elementary school arithmetic, Amer. Math. Monthly (109), no. 9 (2002), p. 796--805).
The passage from the poset of non-negative integers to the monoid of their differences (i.e. working with expressions of the form $x-y$ where $x$ and $y$ are numbers) is the forgetful functor from a comma category $0/\mathcal{C}$ to $\mathcal{C}$ where $\mathcal{C}$ is a monoid with unique object $0$, as described by Lawvere in Section 4 of Taking Categories Seriously.


Question: What are other examples of sophisticated mathematics elucidating elementary mathematical ideas and concepts?

Edit: One example per answer, please. Also, if you have examples which are not research-paper content but are close then that's still interesting for me, but less so as it gets further from research level. The most interesting examples for me are those with citations to research papers, as in the listed examples.

REPLY [13 votes]: Many such questions are dealt with in Klein's book "Elementary mathematics from an advanced standpoint" originally published in German in 1908 and translated into English in the 1920s as I recall. There have been many editions of this since.
To respond to Daniel's comment, one example from Klein's book that immediately comes to mind is that of small oscillations of the pendulum.  The "elementary" treatment of this is in terms of circular motion and a superposition principle (and that's the way I was taught small oscillations in a physics course in high school). Klein recommends treating this via calculus and (linear) differential equations.  That's a nice simple illustration of the power of general methods (ODE) over ad hoc "elementary" solutions (involving a somewhat mysterious superposition principle).<|endoftext|>
TITLE: Why are some solutions of these diophantine equations off the usual patterns?
QUESTION [7 upvotes]: This is inspired by a recent question about complete multipartite integral graphs. I am wondering if more can be said about tripartite integral graphs with block sizes $a<b<c$. It is easy to see that their three nonzero eigenvalues are the same as those of $$\begin{pmatrix} 0&b&c\\ a&0&c\\ a&b&0 \end{pmatrix}$$ because the corresponding eigenvectors must have the same entry inside each block. From Cardano's formula we get as a necessary condition for integer eigenvalues of this $3\times3$ matrix that $$\Delta=(ab+ac+bc)^3-27(abc)^2$$must be a square.
(Or, in terms of their reciprocals $u,v,w$, which can be wlog required to be integers,  $$\Delta'=\frac{(u+v+w)^3}{uvw}-27$$ must be a square.)
Note that this condition is not sufficient; e.g. for $c<550$ it is fulfilled by $86$ coprime triples, but only for $71$ of them the eigenvalues are integers.
Here are the triples for $c<550$.
          a,b      c             λ1    λ2   λ3
        5     8    12           -10    -6   16
    *   5    12    77           -33    -7   40
       13     4    48           -26    -6   32
       13     7    45           -26    -9   35
       13    24    28           -26   -16   42
       17     3    65           -34    -5   39
       17    12    56           -34   -14   48
       17    33    35           -34   -21   55
       25     9    91           -50   -13   63
       25    12    88           -50   -16   66
       29    36    80           -58   -32   90
       29    39    77           -58   -33   91
       37    15   133           -74   -21   95
       37    16   132           -74   -22   96
       37    63    85           -74   -45  119
       41    11   153           -82   -17   99
       41    24   140           -82   -30  112
       41    60   104           -82   -48  130
    *  45   136   396          -240   -66  306
    *  49   220   441          -315   -77  392
       53     5   207          -106    -9  115
       53    44   168          -106   -48  154
       53    95   117          -106   -65  171
       61    13   231          -122   -21  143
       61    40   204          -122   -48  170
       61    84   160          -122   -70  192
       65     8   252          -130   -14  144
       65    51   209          -130   -57  187
       65    99   161          -130   -77  207
       73    84   208          -146   -78  224
       73   105   187          -146   -85  231
       85    21   319          -170   -33  203
       85   112   228          -170   -96  266
       89    20   336          -178   -32  210
       89    57   299          -178   -69  247
       89   176   180          -178  -110  288
       97   115   273          -194  -105  299
       97   136   252          -194  -112  306
      101    63   341          -202   -77  279
      101   143   261          -202  -117  319
      101    24   380          -202   -38  240
      109    96   340          -218  -102  320
      109   189   247          -218  -133  351
      109     7   429          -218   -13  231
      113   144   308          -226  -126  352
      113    84   368          -226   -96  322
      113    17   435          -226   -29  255
      125   221   279          -250  -153  403
   *  125   357   500          -425  -175  600
      137   203   345          -274  -161  435
      137   152   396          -274  -144  418
      145   112   468          -290  -126  416
      145    99   481          -290  -117  407
      149   260   336          -298  -182  480
      149    69   527          -298   -93  391
      149    56   540          -298   -80  378
      157   276   352          -314  -192  506
      157   135   493          -314  -145  459
      169   220   456          -338  -190  528
      169   217   459          -338  -189  527
      173   285   407          -346  -209  555
      181   220   504          -362  -198  560
      185   308   432          -370  -224  594
      193   364   408          -386  -238  624
      197   255   533          -394  -221  615
  * * 200    44   525          -330   -70  400
      205   391   429          -410  -253  663
   *  225   200   252          -240  -210  450
      233   464   468          -466  -288  754
      241   451   513          -482  -297  779

Note that most often, $\lambda_1=-2a$ or $\lambda_1=-2b$ (BTW the table is sorted by this $a$ or $b$ in the first column, sometimes swapping $a$ and $b$). If we denote the entries of the first column by $a$, all these values have only divisors of the form $4k+1$, moreover each time $b+c=4a$. Further, $\lambda_2$ (and also $\lambda_3$, because $\lambda_1$ is even and the trace is $0$) has the same parity as $b$. I suppose all this can be explained easily by somebody who is knowledgeable about the properties of the related elliptical curve.
But the really intriguing thing are the exceptions where $-\lambda_1$ is not the double of $a$ or $b$. They are marked by $*$.


What about these exceptional triples? Are there infinitely many of them?
For the regular triples, does each prime $p=4k+1$ occur at least three times as one of $a$ or $b$?
  (So far, $13$ and $17$ even occur four times)  
Are there infinite families of co-prime triples which can be explicitely given (e.g. by a recursion)?


I have also done a similar exhaustive search for 4-partite graphs with block sizes $a<b<c<d$. For $d<80$, there are none, but there are some, e.g. $K_{8,15,32,40\ \ }$ or $K_{3,35,48,77\ \ }$, which have two integer nonzero eigenvalues.


Do such  4-partite integral graphs exist?

REPLY [10 votes]: I take this from my comments above and add something.
The question is about rational points on the surface $S$ given by
$$ \Delta(a,b,c) := (ab+bc+ca)^3 - 27(abc)^2 = z^2 $$
in the weighted projective space ${\mathbb P}_{1,1,1,3}$. Along the line
$b+c = 4a$ the left hand side factors as a square times $b^2 + 18bc + c^2$,
and the equation effectively reduces to $b^2 + 18bc + c^2 = z^2$, which is a
solvable conic and has lots of rational points that can be parameterized.
Also, for these points, any odd prime divisor of $a$ satisfies
$p \mid (4b)^2 + z^2$, which implies (for coprime $a,b,c$) that
$p \equiv 1 \bmod 4$.
There are other curves on $S$. For example, considering lines through
the point $(1:1:1)$ (which is a simple double point of (the rational
curve) $\Delta = 0$), we obtain curves of the form $z^2 = q(x,y)$,
where $q$ is a quartic. If we take the line $4a = b + 3c$, this
results in a curve birational to
$$ y^2 = 64 x^4 + 480 x^3 + 912 x^2 + 640 x + 117, $$
which is a curve of genus 1 with some rational points. Doing the math,
we find that this curve is isomorphic to the elliptic curve
$y^2 + y = x^3 - 1$ (243a1 in the Cremona database) of Mordell-Weil rank 1.
This implies that there are infinitely many rational points on this
curve. The smallest points with strictly positive coordinates
coming from this curve have (after re-ordering)
$$(a,b,c) = (2,3,6), \quad (6,57,74), \quad (2068,7657,9520).$$
More generally, considering the pencil of lines through $(1:1:1)$
gives us a fibration of $S$ into genus 1 curves that admits a
horiziontal curve with a degree 2 map to the base; this curve
is the conic from above. So after a base-change of degree 2, we actually
have a fibration into elliptic curves, and one would expect at least half
of them to have positive rank, resulting in infinitely many genus 1
curves on $S$ with infinitely many rational points on each.
For example, your first starred solution $(5, 12, 77)$ is on a curve
isomorphic to $y^2 = x^3 - 41659288807500$, which has even rank 2.<|endoftext|>
TITLE: Bibliographic request concerning an article by Bernstein and Robinson
QUESTION [12 upvotes]: Concerning the article "Bernstein, Allen R.; Robinson, Abraham.
Solution of an invariant subspace problem of K. T. Smith and
P. R. Halmos. Pacific J. Math.  16 1966 421-431" I am interested in
finding out the following : (1) who was the editor of the Pacific
Journal of Mathematics at the time? as well as (2) who was the
referee for the article?  The latter is obviously the harder of the
two questions but since this was nearly half a century ago perhaps
such information can be made public.  Are there books or articles that
may shed light on either of these questions? Furthermore, (3) Since Halmos published his translation of the Bernstein-Robinson proof simultaneously in the same journal, he had to know about the upcoming publication of Bernstein-Robinson; how did he find out about it?
Note 1. I looked carefully through Dauben's biography of Robinson to see if he sheds any light on the hypothesis proposed by two respondents below that Halmos may have been a referee (among others) but Dauben doesn't shed any light on this. If anybody has information about this based on other books or personal recollections they are requested to contribute a comment or answer.
Note 2.  One of the respondents below reproduced a page from Halmos' book where he claims that he first received the manuscript "early in 1966" (probably error of recollection). If that were the case, he certainly could not have been the referee since the manuscript by Bernstein and Robinson was originally submitted on 5 july 1964, and one can assume that the editor would not wait for over a year to send it out to a referee. Therefore it may be worth pointing out that Joseph Dauben documents a letter from Halmos to Robinson dated 19 june 1964 where Halmos acknowledges receipt of the manuscript.  This can be found in Dauben's book on page 328, footnote 66.
Note 3. Several NSA experts have contacted me privately with educated guesses and it seems probable that the editor was Richard F. Arens and the referee Paul R. Halmos. Anyone who has additional information about this is requested to share it if possible.
Note 4. I was just informed by the editorial office of PJM that, while they are not certain, the editor for the submission was most likely Richard Arens.

REPLY [9 votes]: I can identify one individual who scrutinized the Bernstein-Robinson manuscript and established its validity before it was published in PJM:

Paul Halmos, I Want to be a Mathematician: An Automathography (1985).
So even if the identity of Halmos as referee cannot be established, he at least did the work of a referee.

Actually, Halmos did more than the work of a referee, the preprint he received from Robinson motivated him to build on it and he wrote his own contribution, which was published in PJM (immediately following the Bernstein-Robinson paper).

REPLY [2 votes]: As a matter of pure speculation, a reasonable guess regarding a referee could be Smith and Halmos themselves (because they posed the conjecture) or perhaps
Aronszajn because he had proved the compact case.<|endoftext|>
TITLE: A conjecture based on Wilson's theorem
QUESTION [24 upvotes]: Definitions:
Lagrange's theorem implies that for each prime $p$, the factors of $(p − 1)!$ can be arranged in unequal pairs, with the exception of $±1$, where the product of each pair $≡ 1 \pmod p$. See Wiki article on Wilson's theorem.
From the example in the link above, for $p=11$ we have
$$(11-1)!=[(1\cdot10)]\cdot[(2\cdot6)(3\cdot4)(5\cdot9)(7\cdot8)]  \equiv [-1]\cdot[1\cdot1\cdot1\cdot1]  \equiv -1 \pmod{11}$$
Let the products of the pairs that $≡ 1 \pmod p$ be the multiset $A_p$, and $A_{p_n}$ the multiset for the $n$th prime.
For the above example then, $A_{p_5}=\{(2\cdot6),(3\cdot4),(5\cdot9),(7\cdot8)\}=\{12,12,45,56\}$.
Conjecture:
$$\lim\limits_{n\rightarrow\infty}\dfrac{\sum\limits_{k \in A_{p_n}}(k-1)}{(p_n)^3}\approx\frac18$$
where $p_n$ is the $n$th prime.
Examples:
For $p=11$ we have
$$\dfrac{11+11+44+55}{11^3}=\dfrac{1}{11}$$
For $p=997$ we have
$$\dfrac{123218233}{997^3}=\dfrac{123218233}{991026973}$$
Comments:
As @YCor noted below, the $-1$ in the $k-1$ can be removed, since its contribution tends to $0$. The conjecture can therefore be simplified to
$$\lim\limits_{n\rightarrow\infty}\dfrac{\sum\limits_{k \in A_{p_n}}k}{(p_n)^3}\approx\frac18$$
I have no idea whether the above statement is correct, or how to go about trying to find a proof. Any comments on the any of the above are most welcome.

REPLY [35 votes]: For an integer $n$ with $1\leq n\leq p-1$, let $n^{-1}$ be the inverse of $n$ modulo $p$. It follows from Weil's bound on Kloosterman sums that for every $\epsilon>0$ the set $\{n: xp\leq n\leq (x+\epsilon) p, yp\leq n^{-1}<(y+\epsilon) p\}$ has cardinality $\epsilon^2p+\mathcal{O}(\sqrt{p}\log^2 p)$. Hence up to a relative error tending to 0 the sum in question can be replaced by an integral, that is
$$
\sum_{n=1}^{p-1} n\cdot n^{-1} \sim p^3\int_0^1\int_0^1 xy\;dx\;dy = \frac{p^3}{4}.
$$
(Note that the $\cdot$ on the left hand side refers to the multiplication of integers, not to modular multiplication). Here each pair $(a,b)$ with $ab\equiv 1\pmod{p}$ is counted twice, with the exception of $(1,1)$ and $(-1, -1)$, which contribute less than $p^2$. Hence up to an error $\mathcal{O}(p^2)$ the left hand side of the above expression is twice $\sum_{k\in A} k$, which proves your claim.

REPLY [11 votes]: If $f(x,y)$ is a "good" function, then
$$\sum_{xy\equiv 1\mod p}f(x,y)=\frac{1}{p}\sum_{x,y=0}^{p-1}f(x,y)-R_p[f],$$
where $R_p[f]$ is a "small" error term (see Lemma 5 here). In your case $f(x,y)=xy$, so the main term is $p^3/4.$  Usually $R_p[f]=O(p^{1/2+\varepsilon}\|f\|)$ while the main term is like $p\|f\|$. In this case error term is $O(p^{5/2+\varepsilon}).$
This observation has a lot of applications in problems connected with lattices (because bases are parametrized by equation $ad-bc=n$, so $ad\equiv n\mod b$ ).<|endoftext|>
TITLE: "Let" versus "for all"
QUESTION [5 upvotes]: I have noticed that many authors tend to use "let" instead of "for all". For example, they write something like this:

Let $n$ be an even natural number. Then also $n^2$ is even.

I wonder, why they use "let" instead of "for all", also in cases where the "for all"-version sounds quite good:

For all even natural numbers $n$, $n^2$ is even.

Note that "let" has a slightly different meaning than "for all":
The statement "Let $n$ be an even natural number. Then also $n^2$ is even" translated into the logic calculus would be something like:
$\mathrm{even}(n)\vdash \mathrm{even}(n\cdot n)$ (this means that "$\mathrm{even}(n\cdot n)$" is true when we are supposing that "$\mathrm{even}(n)$" holds). On the other hand, the statement "For all even natural numbers $n$, $n^2$ ist even" can be translated into a single formula $\forall n.\ \mathrm{even}(n)\implies \mathrm{even}(n\cdot n)$.
EDIT: In the formalization of the examples the quantifier $\forall$ ranges only over natural numbers, so this is the type of "object" we are considering.
I think that in most cases the second version ("For all ...") is meant, but the authors however use "let".
Here is my question:

Why do so many authors write their statements in the form "Let [Variable] be a [Type]. Then ...", even when they actually mean "for all" and even when the version written "for all ..." sounds quite good?

Here a example where this causes confusion:

Theorem: Let $G$ be a planar graph, and let $V$ be the number of vertices, $E$ the number of edges and $F$ the number of faces. Then $V-E+F = 2$.

Proof: by induction on the number of edges $E$.
Why is this confusing? Because a proof by induction gives us a "for all"-statement.
Maybe I take the formalization of proofs too serious and exact. In this case: Sorry for the question.

REPLY [14 votes]: The authors do that for two reasons first, to give the reader a breather, second, because they want do do more with the notation than just finish this one sentence.
After "Let $n$ ne a natrual number" there is a pause. A pause in which time the notation sinks in, so that people transfer it from their ultrashort memory to their short memory so that it can then be used for various purposes. In particular, the notation then has a longer half-life than the notation in "For every even natural number $n$, the number $n^2$ ist even."
In this last sentence, the meaning of $n$ being a natural number, is erased with the period. Not so in the previous case, where it can be used on.<|endoftext|>
TITLE: Algorithm that solves every Mixed Integer Linear Program (to optimality)?
QUESTION [6 upvotes]: Given a Mixed Integer Linear Program with rational coefficients (both for the objective functions and all constraints), is it always possible to solve it algorithmically?
I know that you usually solve these problems by applying a Branch-and-X strategy and solving the subproblems as Linear Programs. But it is not clear to me whether a the branching tree has always finitely many nodes and whether you can solve any LP (e.g. given the situation that it is very hard to determine an initial feasible solution).

REPLY [3 votes]: I sometimes read the claim that Gomory cuts alone (without any branch-and-X) are sufficient for finding integer solutions to linear programs in a finite number of steps. E.g. at the bottom of page 2:

Gomory showed that alternately applying the simplex method and adding cutting planes eventually leads to a system for which the simplex method will give an integer optimum.

It gives three references to papers by Gomory, and I checked that "R.E. Gomory, An algorithm for integer solutions to linear programs, in: Recent Advances in Mathematical Programming, R.L. Graves & P. Wolfe, eds., McGraw-Hill, New York, 1963, pp. 269–302." indeed contains a mathematical proof of this statement.

This doesn't address the case of mixed-integer linear programs, but I guess my textbooks would have mentioned it explicitly, if this technique could not be adapted to mixed-integer linear programs.
You implicitly also pose the question whether branch-and-X alone would be sufficient for finding mixed-integer solutions to linear programs, even for completely stupid X. At least it seems possible to construct linear programs without valid integer solution for which such a sufficiently stupid branch-and-X algorithm would continue branching forever.

I hope it is clear to you that no known algorithm for integer programing is efficient in the sense that it runs in polynomial time, because integer programing is known to be NP-complete.<|endoftext|>
TITLE: Is a certain subset of the disc a convex set?
QUESTION [12 upvotes]: Some one asked me this question and I thought about it and I don't have any good idea to solve that. Can some one help me and give me an idea to start solve that?
Draw a Cantor set $C$ on the circle and consider the set $A$ of all chords between points of $C$. Is $A$ a convex set?

REPLY [2 votes]: Start with a 3 point set {a,b,c} on the unit circle. The union of the 3 chords forms not a convex set. Blow up the points into 3 very small Cantor sets. The union of all the new chords still fails miserably to be a convex planar set. For example the origin is not contained in any chord, despite being in the convex hull of the set in question.<|endoftext|>
TITLE: Brauer-Picard for a fusion category coming from a quantum group
QUESTION [8 upvotes]: In Fusion Categories and Homotopy Theory, ENO attatch a 3-groupoid to a fusion category. In the case of A graded vector spaces they further compute it's truncation as an orthogonal group $O(A \bigoplus A^*)$. There are also computations by Grossman and Snyder for examples that come from the Asaeda-Haagerup subfactor. But I have not seen any descriptions for quantum groups at roots of unity. If some of these are already taken care of by quantum group coincidences in other contexts, that would also be appreciated.

REPLY [8 votes]: As far as I know, no one has written this up, but I think you should be able to find the Brauer-Picard groupoid for quantum groups at roots of unity by the following techniques. Now that I've written it down, there are a lot of gaps which would need to be filled in, several of which are not straightforward.  This might well be a good project for a graduate student paper.
I'm going to assume you care about the kinds of roots of unity that appear in Reshetikhin-Turaev, the general case would be quite a bit harder because the categories wouldn't be modular, but could probably be done with a bit more thought and some time with Steve Sawin's tables.
(Also, I'm not going to compute the more detailed problem that Pinhas and I consider of classifying all other categories Morita equivalent to the given one.  Such a classification would be far too difficult because you would need the classification of "quantum subgroups" which is known only for SU(2), SU(3), and SU(4).)
Since C is modular, $Z(C) = C \boxtimes C^{op}$ where $C^{op}$ denotes the same category with the opposite braiding.  By ENO the Brauer-Picard group is the same as the group braided auto-equivalences of $Z(C)$.  We compute the braided autoequivalences of $C \boxtimes C^{op}$ in two steps: first combinatorially we want to figure out what such an autoequivalence can do on objects, and second we need to consider the case of an autoequivalence whose underlying functor is the identity.  
Let me first sketch how this would work for SU(2).  The objects in $C \boxtimes C^{op}$ are generated by $V \boxtimes 1$ and $1 \boxtimes V$ for $V$ the standard 2-dimensional representation.  We want to figure out what object $\mathscr{F}(V \boxtimes 1)$ can be.  The only objects of the same dimension as these objects are of the form $g^i V \boxtimes g^j$ or $g^i \boxtimes (g^j V)$ where $g$ is the invertible object at the far end of the Weyl alcove.  Just having the right dimensions for $\mathscr{F}(V \boxtimes 1)$ and $\mathscr{F}(1 \boxtimes V)$ isn't enough, you also need the object to be quaternionic (i.e. have the right parity), to have the right self-braiding with itself, and to "commute" (i.e. have the same over and under braidings) with each other.  Sorting this out requires some annoying futzing modulo 4, which I haven't done.  By the universal property of SU(2) (cf.) it is not hard to check that once you have the above conditions satisfied then there's a unique braided autoequivalences realizing this.  
The universal property let us skip the usual second step, which is to check that there are no nontrivial "guage automorphisms", which are tensor autoequivalences whose underlying functor is the identity.  (See Lemma 5.3 and Remark 5.6.)  This follows from the fact that the Temperley-Lieb planar algebra has no automorphisms.  It could also be done directly by a cohomological calculation following Liptrap's thesis.  In conclusion, the answer for SU(2) is trivial if there are an even number of objects and the group of order 2 otherwise.
One can calculate the BP group of SU(2) in another way, by using the classification of quantum subgroups of SU(2) to directly calculate the BP groupoid.  Invertible bimodules over C correspond to a pair of an indecomposable module category M and a tensor equivalence $C \cong M_C^*$.  The indecomposable module categories over SU(2) at roots of unity are given by ADE Dynkin diagrams with the same Coxeter number.  The dual is equivalent to $C$ only for type A and for the odd type D's.  The second piece of data is a torsor for the group of outer autoequivalences of C.  By a similar calculation as in the previous paragraph (but without braidings), you get an outer autoequivalence of C when $V$ and $gV$ have the same parity (so the number of objects is odd).  So when there is an even number of simple objects you get the trivial group, when the number is 1 mod 4 you get $\mathbb{Z}/2\mathbb{Z}$ coming from an outer autoequivalence interchanging $V$ and $g \otimes V$, when the number is 3 mod 4 you get a group of order $4$ coming from both the outer automorphism and $D_{\text{odd}}$.  It's not obvious to me which order four group you get in the last case.
In general things will be similar, but harder.  For the combinatorial step, the permutation has to preserve the dimensions and the self-braidings, which should already get you down to very few possibilities (typically they should all of which should look roughly like replacing the smallest rep $V$ by $g \otimes V$ for $g$ some invertible object in a corner of the Weyl alcove and possibly moving it to the other component).  In general this step might be a bit tricky, for example in type B and D you have to worry about where the spin representations go, which won't be determined by where the standard rep goes and so might give you some extra options.  Also in the cases where the Dynkin diagram has symmetry (A_n, D_4, E_6) you have some extra cases to consider.  You also need to check whether there's actually an autoequivalence realizing each combinatorial possibility.
Second, you need to figure out in how many ways each of these combinatorial options is realized.  Equivalently, you just need to know what the "guage automorphisms" are, that is the braided autoequivalences which act triviailly on objects.  This can be done by a kind of "cohomology"-like calculation following Jesse Liptrap's thesis, or by a "diagrammatic" argument following what Pinhas and I do.  That is, any gauge automorphism gives a very special kind of automorphism of the "planar algebra" attached to the tensor category.  In the non-exceptional types this comes down to studying automorphisms of the HOMFLY and BMW planar algebras, but in the exceptional cases I'm not sure how to do it.
I suppose I should also say a little bit about the higher structure.  $\pi_2$ is just the invertible objects $Z(C)$, which is just the product of the invertible objects in $C$ with the invertibles in $C^{\text{op}}$, so is easy to work out.  $\pi_3$ is always $\mathbb{C}^\times$.  For the full structure you need to know quite a bit more, like the action of $\pi_1$ on $\pi_2$ (which is just a calculation of where the invertible objects go), and the Postnikov k-invariants.  This could ve quite tricky.<|endoftext|>
TITLE: Question about the normal bundle of a totally geodesic submanifold of a Cartan-Hadamard manifold
QUESTION [5 upvotes]: I'm reading the exposition in Lang's Fundamentals of Differential Geometry of the following generalization of the Cartan-Hadamard theorem:
Suppose $X$ is a Cartan-Hadamard manifold (i.e. a complete, simply connected manifold with everywhere nonpositive sectional curvature) and $Y$ is a complete totally geodesic submanifold of $X$. Let $NY$ be the normal bundle of $Y$ in $X$, and let $\exp_{NY}: NY\rightarrow X$ be the restriction of the exponential map to $NY$. Then $\exp_{NY}$ is a diffeomorphism. This is Theorem 2.5 of Chapter X.
I'm confused about a detail in the proof of the preceding Theorem, 2.4. Fix $y_0\in Y$, and for each $y\in Y$, let $P_{y_0}^y$ denote parallel transport from $T_{y_0}X$ to $T_{y}X$ along the unique geodesic connecting $y_0$ to $y$. Now, define the smooth map $E:Y\times N_{y_0}Y\rightarrow X$ by $E(y,v)=\exp_y(P^y_{y_0}v)$.
At the bottom of page 273, Lang claims that $dE_{(y,v)}(z,0)$ is orthogonal to $dE_{(y,v)}(0,w)$ for any $y\in Y,v\in N_{y_0}Y,  z\in T_yY,w\in N_{y_0}Y$.
I don't understand why this is true. He says it follows from the Gauss Lemma (Lemma VIII.5.6), but I can't figure out how it follows. I'm probably missing something simple. Does anyone know why those things are orthogonal?  
Also, Lang is the only reference I can find for the fact that $\exp_{NY}:NY\rightarrow X$ is a diffeomorphism. Does anyone know if this exists elsewhere in the literature?
Thanks.

REPLY [2 votes]: I'm still not sure about that detail in the proof in Lang,
but here's a slightly different proof that $\exp_{NY}:NY\rightarrow X$ is a diffeomorphism.
Step one: $\exp_{NY}$ is surjective.
To see this, let $x\in X\setminus Y$. Let $B$ be a closed metric ball centered at $x$ containing some point of $Y$.
Since $X$ is complete, $B$ is compact, and since $Y$ is closed in $X$, $B\cap Y$ is compact.
Thus, there is $y_0\in B\cap Y$ which is closest to $x$, and so we have $d(x,y_0) = d(x, B\cap Y) = d(x, Y)$.
By differentiating $y\mapsto d(x,y)$ along any curve in $Y$ through $y_0$, we find that the geodesic connecting $y_0$
to $x$ is normal to $Y$ at $y_0$. In other words, if $\gamma:[0,1]\rightarrow X$ is the geodesic with $\gamma(0)=y_0$, $\gamma(1)=x$,
then $\gamma'(0)\in NY_{y_0}$. So $\exp_{NY}(\gamma'(0))=x$, and we have surjectivity.
Step two: $\exp_{NY}$ is injective.
Suppose there were distinct $v_0, v_1 \in NY$ such that $\exp_{NY}(v_0) = \exp_{NY}(v_1)=x$.
If $v_0,v_1$ were based at the same point of $Y$, then the exponential map based at that point would fail to be injective,
contradicting the Cartan-Hadamard theorem. So $v_1, v_2$ are based at distinct points $y_1,y_2\in Y$.
Let $\gamma:[0,1]\rightarrow Y$ be the geodesic with $\gamma(0)=y_0, \gamma(1)=y_1$. 
Then we have $\frac{d}{dt}|_{t=0}d(x,\gamma(t)) = \frac{d}{dt}|_{t=1}d(x, \gamma(t))=0$,
because the geodesic connecting $y_0$ to $x$ is normal to $\gamma'(0)$ at $y_0$ and
the geodesic connecting $y_1$ to $x$ is normal to $\gamma'(1)$ at $y_1$. This contradicts 
the convexity of the distance function of $X$.
So $\exp_NY$ is bijective.
As Deane pointed out in the comments, it now suffices to show
Step three: The differential of $\exp_{NY}$ is everywhere injective.
For this, fix $y_0\in Y$, and for each $y\in Y$, let $P_{y_0}^{y}$ denote parallel transport
from $y_0$ to $y$. Then the map $Y\times NY_{y_0} \rightarrow NY$ given by $(y, v)\mapsto P_{y_0}^y v$
is a diffeomorphism, so it suffices to show that the map $E$ defined as the composition $Y\times NY_{y_0} \rightarrow NY \stackrel{\exp_{NY}}{\rightarrow}X$
has everywhere injective differential.
To this end, choose $(y,v)\in Y\times NY_{y_0}$ and nonzero $(z,w)\in T_yY\times NY_{y_0}$.
We need to show $dE_{(y,v)}(z,w)\neq 0$. Let $\gamma(t)$ be the geodesic passing through $y$ at time 0 with velocity $z$.
We need to show that $\frac{d}{dt}|_{t=0} \exp_{\gamma(t)} P_{y_0}^{\exp(\gamma(t))} (v+tw)$ is not zero. Let 
$\alpha(s)= \exp_y P^{y}_{y_0}(sv)$.
Let $\Gamma(s,t) = \exp_{\gamma(t)} P_{y_0}^{\exp(\gamma(t))} s(v+tw)$, and 
$J(s) = \frac{\partial}{\partial t} \Gamma(s,0)$. $J$ is a Jacobi field along the geodesic $\alpha$. Set $f(s) = \langle J(s), J(s) \rangle$. We will show $f(s)>0$ for $s>0$. 
Setting $s=1$ will then yield the desired result.
Compute 
\begin{align*}
f'(0) &= 2\langle \frac{D}{\partial s}\frac{\partial}{\partial t}\Gamma(0,0), \frac{\partial}{\partial t}\Gamma(0,0) \rangle\\
       &= 2\langle \frac{D}{\partial t}\frac{\partial}{\partial s}\Gamma(0,0), \frac{\partial}{\partial t}\Gamma(0,0) \rangle\\
       &= 2\langle \frac{D}{\partial t}|_{t=0}P_{y_0}^{\exp(\gamma(t))} (v+tw), \frac{d}{dt}|_{t=0}\gamma(t) \rangle\\
       &=2 \frac{\partial}{\partial t}|_{t=0} \langle P_{y_0}^{\exp(\gamma(t))} (v+tw), \frac{d}{dt}\gamma(t) \rangle\\
       &=0,
\end{align*}
since $P_{y_0}^{\exp(\gamma(t))} (v+tw) \in NY$ and $\frac{d}{dt}\gamma(t)\in TY$ for all $t$.
Also,
\begin{align*}
f''(s) &= 2\langle J'(s), J'(s)  \rangle + 2\langle J''(s), J(s) \rangle\\ &= 2\langle J'(s), J'(s)  \rangle - 2\langle R(\alpha'(s), J(s))\alpha'(s), J(s) \rangle\\&\geq 0.
\end{align*}
Now, $J(0) = z$, so if $z\neq 0$, then $f(0)>0$, which combined with the two computations above, proves $f(s)>0$ for all $s>0$.
If $z=0$, then $w\neq 0$ and $J'(0)=w$, so $f''(0)>0$, and again we conclude $f(s)>0$ for all $s>0$.<|endoftext|>
TITLE: $C^{*}$-correspondences viewed as generalized endomorphisms
QUESTION [5 upvotes]: I've heard that $C^{*}$-correspondences (over a $C^{*}$-algebra) can be viewed as generalized endomorphisms of the algebra. I would like to understand this, and be pointed towards books or papers where this is discussed.
Thank you

REPLY [9 votes]: Expanding my comment as per suggestion above. All this is explained in more detail in the reference I gave though.
A correspondence from $A$ to $B$ is a Hilbert $B$-module on which $A$ acts non-degenerately by adjointable operators. One can think of this, roughly, as an $A$-$B$-bimodule ${}_{A}\mathcal H_{B}$. These can be composed via tensor product:
$${}_{B}\mathcal K_{C}\circ{}_{A}\mathcal H_{B}={}_{A}(\mathcal H\otimes_B\mathcal K)_{C}.$$
Obviously, this composition will not be associative (or unital), even on the underlying sets. But there are isomorphisms that identify what is supposed to agree. These isomorphisms are not only natural but also satisfy some so-called coherence laws. All this is axiomatized by the notion of a weak $2$-category (or bicategory). If you consider isomorphism classes of correspondences, you get an honest category. But then you will have forgotten about the naturality and coherence of the associator and the unitors.
How does this generalize morphisms of C$^\ast$-algebras and their composition?  A morphism from $A$ to $B$ is a non-degenerate $^\ast$-homomorphism $\varphi$ from $A$ to the multiplier algebra of $\mathcal M(B)$ of $B$, which is also the algebra of adjointable operators on the Hilbert $B$-module $B$. Thus $\varphi$ defines a correspondence $\mathcal H_\varphi$ from $A$ to $B$. Now, if $\varphi$ is a morphism from $A$ to $B$ and $\psi$ is a morphism from $B$ to $C$, then there is a natural isomorphism
$$\mathcal H_\psi\circ\mathcal H_\varphi\cong\mathcal H_{\psi\circ\varphi}.$$
Hence $\varphi\mapsto[\mathcal H_\varphi]$ defines a functor from the category of C$^\ast$-algebras and morphisms to the category of isomorphism classes of correspondences.<|endoftext|>
TITLE: Optimization of points on a plane
QUESTION [10 upvotes]: Suppose we have $n$ points on a plane. Let $D$ be the sum of the squares of all the pairwise distances between the points. Let $A$ be the area of the convex hull. What is the minimum possible value of $\frac{D}{A}$ and what arrangement achieves it for specific values of $n$?

REPLY [6 votes]: I conjecture that the minimum is $2n$, for all $n\geq 4$.
This is obtained by putting four points in a square and the remaining points in the center.
There are then $(n-4)$ vertices in the center, and thus the total sum of distances squared is $4 \cdot \frac{1}{2}(n-4) + 4 + 2\cdot 2= 2n$,
assuming that the square has area $1$.
I ran many simulations using a genetic algorithm (here is the mathematica code):
Needs["ComputationalGeometry`"];

Mutate[ptList_,f_:0.1]:=ptList + f*RandomReal[{-1,1},{Length@ptList,2}];

Fitness[ptList_]:=Fitness[ptList]=Module[{aa,dd},
    aa=ConvexHullArea[ptList];
    dd=Total[(EuclideanDistance@@@Subsets[ptList,{2}])^2];
    dd/aa
];

RunSimulation[ptsLists_]:=Module[{newLists,j=1},
    newLists=ptsLists;

    Do[
        newLists[[k]] = Mutate[newLists[[j++]],0.01];
    ,{k,Ceiling[Length[newLists]/2],Length[newLists]}];

    newLists=SortBy[newLists,Fitness];

    gg=ListPlot[ newLists[[1]], Axes->False,PlotStyle->{PointSize[0.02]},
PlotLabel->("Fitness: " <>ToString@Fitness@newLists[[1]]),Frame->True,
AspectRatio->Automatic,FrameTicks->False ];

    newLists
];

Clear[gg]
Print[Dynamic[gg]];

init=RandomReal[{0,1},{100,9,2}];
pts=SortBy[init,Fitness];

Do[
pts=RunSimulation[pts];
,{1800}];

This code above runs 1800 generations with 100 lists, each with 9 points. However, looks like a pentagon with the remaining points in the middle is a local minima, so one has to restart a few times to see the square.<|endoftext|>
TITLE: Why is the number of irreducible components upper semicontinuous in nice situations?
QUESTION [10 upvotes]: Suppose $f: X \rightarrow Y$ is a flat projective morphism of finite type schemes over an algebraically closed field so that the fibers over the closed points of Y are (geometrically) reduced. Why is it true (or do I need some additional assumptions?) that the number of irreducible components of the geometric fibers is upper semicontinuous on Y?
I have asked Joe Harris and Allen Knutson about this statement, and they both believed it was true, but I have not been able to find a reference or a proof.

REPLY [2 votes]: An expanded version of nfdc23's answer in the comments to this question can be found at http://arxiv.org/pdf/1601.05840v1.pdf, Proposition 2.9. An even more expanded version can be found at http://arxiv.org/pdf/1605.01117v1.pdf, Proposition 3.2.5.<|endoftext|>
TITLE: Geometry of Hermitian rank $\leq r$ matrices
QUESTION [7 upvotes]: Let $M_n^{sa}$ be the space of $n\times n$ complex Hermitian matrices, let $r < n$, and let $E$ and $F$ be (real) linear subspaces of $M_n$ with ${\rm codim}(E) < r^2$ and ${\rm codim}(F) = 1$. Let $V$ be the set of matrices in $E$ whose rank is at most $r$ and suppose $V$ is not contained in $F$. Is $V \cap (M_n\setminus F)$ dense in $V$?
For this to fail, there would have to be some open subset of $V$ which is contained in $F$, while not all of $V$ is contained in $F$.
I assume the answer is yes, but I think this is an algebraic geometry question and I know next to nothing about the subject $\ldots$

REPLY [6 votes]: Let $n=4$, $r=3$. Let $E = \mathbb{R} \oplus M_3^{sa}(\mathbb{C})$ (viewed as $4\times 4$ matrices which are $0$ in the first row and column except for the $(1,1)$ entry). Then the (real) codimension of $E$ in $M_4^{sa}(\mathbb C)$ is $6<9=r^2$. Let $F\subset M_4^{sa}(\mathbb{C})$ be the space of all matrices with $0$ in the $(1,1)$ entry. Clearly $V$ is not contained in $F$. 
Now consider the matrix $X= 0\oplus I_3$. We have $X\in V\cap F$, and small perturbations $Y$ of $X$ within $V$ must of course also lie in $E$, and thus have the form $Y=c\oplus (I_3+A)$ where $c\in \mathbb R$ and $A\in M_3^{sa}(\mathbb C)$, and we have $\|Y-X\|=\text{max}(|c|, \|A\|)$.  But now if $\|Y-X\|<1$, then $\|A\|<1$, so $I+A$ has rank 3 which forces $c=0$ and thus $Y\in F$.<|endoftext|>
TITLE: 3-term arithmetic progressions of terms as frequent as primes
QUESTION [5 upvotes]: Let $\ p_1\ < p_2 < \ldots\ $ be the sequence of all primes $\ (2\ 3\ 5\ \ldots)$.
Let $\ x_1 < x_2 < \ldots\ $ be an arbitrary increasing sequence of positive integers such that $\ x_n\le p_n\ $ for every $\ n=1\ 2\ldots\,$.
QUESTION: Does sequence $\ (x_1 < x_2 < \ldots)\ $ contain a 3-term arithmetic progression (of not necessary three consecutive members)? Does there exist an infinite number of such 3-term arithmetic progressions?
Acknowledgement: The simple version $\ x_n\le p_n\ $ of the assumption of this conjecture was provided by @zeb in a response to my equivalent original assumption which was clumsy and harder to read.

Reference: my MO-problem is related to a famous Klaus Roth's theorem.
Now I see from the Fedor's answer that this was indeed essentially only an MO-problem, and otherwise not essentially original.

REPLY [2 votes]: Meanwhile, this question has been answered in the positive. -- See
Thomas F. Bloom, Olof Sisask: Breaking the logarithmic barrier in Roth's theorem on arithmetic progressions.<|endoftext|>
TITLE: How to prove Liouville measure is invariant under geodesic flow?
QUESTION [7 upvotes]: Let $M$ be a complete n dimensional Riemannian manifold. $vol$ denotes the n dimensional Hausdorff measure. Let
$$
SM=\{(x,v)|x\in M, v\in T_xM, \|v\|=1\}
$$
be the unit tangent bundle of $M$. Then $SM$ will be equipped with the Liouville measure $\nu$. Given a subset $A=(U,A_x)\subset SM$, where $U\subset M$ is a subset of $M$, $A_x$ is a subset of the unit sphere of the tangent space at $x\in U$, $\nu$ is defined by
$$
v(A)=\int_U \int_{A_x} dS^{n-1} dvol(x)
$$
where $dS^{n-1}$ is the usual Lebesgue measure on the unit sphere.
Then $\nu$ is invariant under the geodesic flow on $SM$. By the comments below, I know what it means: Let $y=(x,v)\in A$,
set $\gamma_y(s)=\exp_x(sv)$, then the geodesic flow is defined by
$$
\Phi_t(y)=(\gamma_y(t),\dot{\gamma}_y(t))
$$
And $\Phi_t(A)=\{\Phi_t(y)|y \in A \}$. We have $\nu(\phi_t(A))=\nu(A)$. 
Can you give a direct proof without introducing cotangent bundle, 1-form, 2-form?

REPLY [10 votes]: A hands down proof not using the theory of Hamiltonian systems can be done by just proving that the Jacobian determinant of the transformation is zero. 
We have
$$ TSM \cong \pi^* TM \oplus VSM,$$
where $VSM$ is the vertical distribution and $\pi: SM \longrightarrow M$ is the canonical projection. The isomorphism is given by the metric, which selects a horizontal subspace. 
Now the geodesic flow is the flow generated by the vector field $X$ which is given by
$$ X(x, v) = \begin{pmatrix} v \\ 0 \end{pmatrix}$$
in this splitting. Therefore, differentiating the defining equation 
$$ \dot{\Phi}_t = X(\Phi_t), ~~~~~~ \Phi_0 = \mathrm{id}$$
with respect to some metric connection gives
$$ \frac{\nabla}{\mathrm{d} t} d \Phi_t = \nabla X|_{\Phi_t} \cdot d \Phi_t, ~~~~~ d\Phi_0 = \mathrm{id}.$$
For the determinant, we obtain
$$ \frac{\mathrm{d}}{\mathrm{d} t} \det(d\Phi_t) = \mathrm{tr}(\nabla X)\cdot \det(d\Phi_t), ~~~~~~ \det(d\Phi_0) = 1$$
Now $VSM$ carries a natural connection, and $\pi^*TM$ carries the pullback connection. The direct sum of these connections is metric (even though it is not the Levi-Civita connection of $SM$), and $\nabla X$ is given in the splitting by
$$\nabla X = \begin{pmatrix} 0 & \iota \\ 0 & 0 \end{pmatrix},$$
where $\iota$ is the inclusion of $VSM$ into $\pi^*TM$. Hence its trace is zero, and the determinant remains one for all time.<|endoftext|>
TITLE: Request for classical articles in representation theory
QUESTION [9 upvotes]: I am planning in running a Ph.D. student seminar next year on representation theory in the spirit of MIT Kan's Seminar where students give lectures on classical articles on representation theory that are typically not covered in a first course but that can none-the-less be covered in one week or so. I would like to amass a list of 12-20 articles or book chapters that should compose such a seminar. Examples would be 
Demazure Inventiones 33 (1976) 271-272 "A very simple proof of Bott's theorem", 
A. Beilinson, J. Bernstein, Localization de g-modules, C.R. Acad. Sci. Paris, 292 (1981), 15-18.
Chapter 4 of Chriss-Ginzburg "Representation Theory and Complex Geometry".
and non-examples should be the geometric Satake isomorphism or Zhu's modularity of characters of vertex algebras. 
I am not sure if this question goes here or not, but ME seemed like the wrong place to ask. Since I found similar questions here like A request for suggestions of advanced topics in representation theory and A learning roadmap for Representation Theory I figured it was alright. 
Edit: as Tobias and Jim asked for background, I would want this class to be a second year graduate class in representation theory for students with different backgrounds, however, in general I would require students to be familiar with the standard first year classes like finite dimensional Lie algebras over $\mathbb{C}$ (say Jim's GTM book), finite dimensional compact groups (as in the first chapters of Knapp's Lie groups beyond an introduction), the basics of algebraic geometry as in the first 3 chapters of Hartshorne. And hopefully some knowledge of differentiable manifolds as in Warner and/or complex geometry as in Wells. 
The focus of the seminar may/can vary from year to year, as does Kan's seminar, I myself would rather be more about geometric representations as the list supplied suggest. Finally the definition of classical is left intentionally vague as I would want it to be "those articles that most in the field have read or should have read".

REPLY [3 votes]: Bertram Kostant, Lie Algebra Cohomology and the Generalized Borel-Weil Theorem, Ann. of Math., 74, (1961), No. 2, 329-387<|endoftext|>
TITLE: Historical refererences for Castelnuovo-Mumford regularity
QUESTION [5 upvotes]: Does anyone know a good reference to understand the historical background of Castelnuovo-Mumford regularity? 
I know the backgound for the modern commutative-algebra approach (using free graded resolutions and Betti numbers) but I'd like to know the geometric motivations that led to the sheaf cohomology definition of $m$-regular sheaf. 
Mumford, in his Lectures on curves on an algebraic surface and in a later article ascribes the definition of regular sheaves to Castelnuovo. I'd like to deepen in this way.

REPLY [3 votes]: You might like looking at Eisenbud, Green and Harris "Cayley-Bacharach theorems and conjectures". They start with classical theorems of Euclidean geometry, such as Pappus and Pascal's theorem, and relate them to questions in commutative algebra regarding in what degree the saturated homogenous ideals of various sets in $\mathbb{P}^2$ are generated. This is to say, they relate them to describing the $0$-syzygies of the ideal. CM-regularity is a measure of the complexity of all syzygies.
If you more specifically want to know why people historically studied higher syzygies, I've heard good things about Eisenbud's "Geometry of syzygies", although I haven't read it.<|endoftext|>
TITLE: Complexity of linear solvers vs matrix inversion
QUESTION [18 upvotes]: Solving linear equations can be reduced to a matrix-inversion problem, implying that the time complexity of the former problem is not greater than the time complexity of the latter. Conversely, given a solver of $N$ linear equations and $N$ unknown variables with computational cost $F(N)$, there is a trivial implementation of matrix inversion using the linear solver with overall computational cost equal to $N F(N)$. 
However, the resulting algorithm is not optimal for matrix inversion. Indeed, the time complexity of linear solvers is not smaller than $N^2$, whereas the time complexity of matrix inversion is not bigger than $N^{2.375}$, as implied by the Coppersmith–Winograd algorithm. 
Thus, my question is as follows. Given any solver of linear equations, is there some algorithm for inverting matrices that uses the linear solver and with the same time cost up to some constant? In other words, does a linear-solver with time cost $N^\alpha$ induce a matrix-inversion algorithm with cost $N^\alpha$? This question comes from the observation that the most efficient known linear solvers come from matrix-inversion algorithms.

REPLY [11 votes]: A linear solver with optimal complexity $N^2$ will have to be applied $N$ times to find the entire inverse of the $N\times N$ real matrix $A$, solving $Ax=b$ for $N$ basis vectors $b$. This is a widely used technique, see for example Matrix Inversion Using Cholesky Decomposition, because it has modest storage requirements, in particular if $A$ is sparse. The Coppersmith–Winograd algorithm offers a smaller computational cost of order $N^{2.3}$, but this improvement over the $N^3$ cost by matrix inversion is only reached for values of $N$ that are prohibitively large with respect to storage requirements. An alternative to linear solvers with a $N^{2.8}$ computational cost, the Strassen algorithm, is an improvement for $N>1000$, which is also much larger than in typical applications.
So I would think the bottom line is, yes, linear solvers are computationally more expensive for matrix inversion than the best direct methods, but this is only felt for very large values of $N$, while for moderate $N\lesssim 1000$ the linear solvers are faster and have a much reduced storage requirement than direct matrix inversion.<|endoftext|>
TITLE: Automorphism group of a variety
QUESTION [7 upvotes]: Suppose $X$ is a (quasi-projective) variety over a field $k$, and let $\mathbb{P}^n(k)$ be the ambient projective space. 

When can one decide that the automorphism group of $X$ is induced by a subgroup of the automorphism group $\mathbf{P G L}_{n + 1}(k)$ of $\mathbb{P}^n(k)$ ? 

When $X$ itself is a sub-projective space over $k$, this is of course well known, but when $X$ is e.g. an affine subspace over $k$, this is in general not true (due to the existence of non-linear automorphisms).

Is it true when $X$ is projective ?

REPLY [9 votes]: I guess this should be true if and only if $f : X \to X$ fixes a polarization, i.e. there is an ample $A$ such that $f^\ast A \sim A$.
If there is such an $A$, then take the embedding of $X$ into $\mathbb P^n$ by a very ample $mA$.  The automorphism of $X$ is then induced by the linear map $f^\ast : \mathbb PH^0(X,mA) \to \mathbb PH^0(X,mA)$.  Conversely, if a map comes from a linear map on projective space, then the polarization given by restricting the hyperplane is fixed.
Typically automorphisms won't fix any polarization, so don't come from $\mathbb P^n$.  If you actually want to check it for a specific map, the first thing to do work out the induced map on $H^2(X;\mathbb R)$.  If there is an invariant ample divisor, its first chern class will be invariant under the pullback.  So you may be able to show that this doesn't happen.  If there is a fixed class, you need to know (a) whether it is represented by an ample divisor, which may or may not be easy to check depending on what $X$ is and (b) whether the divisor is linearly equivalent (rather than just numerically equivalent) to its pullback, which requires a bit more scrutiny.
As Francesco points out, $K_X$ always pulls back to itself under an automorphism, so if it's either ample or antiample you have what you want.<|endoftext|>
TITLE: On cubic reciprocity for $x^3+y^3+z^3 = 996$?
QUESTION [13 upvotes]: I. The Diophantine equation,
$$x^3+y^3+z^3 = 3w^3\tag1$$
with $x\geq y \geq z$ and $w=1$ has only two known solutions, namely $1,1,1$ and $4,4,-5$. Are there larger ones? As Noam Elkies points out in this post, Cassels in a 1985 paper showed that $w=1$ must have,
$$x\equiv y\equiv z \bmod 9\tag2$$
For general $w$, Heath-Brown in a 1992 paper showed that,
$$\text{either}\;x\equiv y\equiv z \bmod 9,\; \text{or one of}\; x,y,z\;\text{is}\;9m\tag3$$
Checking the Elsenhans-Jahnel list for $x^3+y^3+z^3=N$ with $N<1000$, one finds large solutions for $w=2,3,4$.  (The list only covers $10^{14}$, so if it can be raised higher, maybe a large one can be found for $w=1$ as well.)
II. Trying to find something similar to $(1)$, and if I did my search right, there are at least two other $N$ that obeyed $(2)$, namely $N=996$,
$$x^3+y^3+z^3 = 996\tag4$$
with,
$$x,y,z = 11,2,\,-7$$
$$x,y,z = 2169364505441,\, -631266388780,\, -2151398424325$$
and $N = 2^4\times3^3+3=435$, 
$$x^3+y^3+z^3 = 435\tag5$$
with seven known (and rather large) solutions, all of which satisfied $(2)$.

Questions: 

Is it true that all $x,y,z$ of $(4)$ and $(5)$ also obey $x\equiv y\equiv z \bmod 9$? 
In general, for what $N$ does this condition hold?

REPLY [14 votes]: Problems of this type are studied in the article
Colliot-Thélène, Wittenberg - Groupe de Brauer et points entiers de deux familles de surfaces cubiques affines.
Here, in remark 5.7, they give an explanation of Cassels' result in terms of a Brauer-Manin obstruction to strong approximation. Other examples of Brauer-Manin obstruction to strong approximation are also included, which in general imply the existence of other "non-obvious impossible congruences".
One has to work very hard to make these congruences explicit. In the Cassels example, they prove it using the fact that $3$ is the only prime of bad reduction. However in your case of $996$, there are other primes of bad reduction, namely $2$ and $83$. So the same method does not apply. Moreover, your ''evidence'' in the $996$ case is very sparse, and I would see no reason why the same Cassels congruence should hold in this case. One would need to carefully work out the Brauer-Manin obstruction in this case and see what one gets.<|endoftext|>
TITLE: Multiplicative Structure of the Atiyah-Hirzebruch/Leray-Serre spectral sequence
QUESTION [14 upvotes]: This is related to this question (edit: now answered, see below). Is there a nice explanation of the multiplicative structure on the higher pages of that spectral sequence? I want to assume that $h$ is some multiplicative extraordinary cohomology theory (satisfting the wedge axiom), and I assume
that $X\to B$ is a Serre fibration over a CW complex with typical fibre $F$. Then consider the Leray-Serre / Atiyah-Hirzebruch / Whitehead spectral sequence
$$ E^{p,q}_2=H^p(B;h^q(F))\Rightarrow h^{p+q}(X)\;.$$
Several books state that there is a cup product on each page $E_k$ such that $d_k$ satisfies a Leipniz rule, and the cup product on $E_{k+1}$ is the induced one. However, I only found a proof in G. W. Whitehead's "elements of homotopy theory", which looks rather scary. Is there a more accessible account?
Edit (again) The answers to the question mentioned above name two papers: one by Massey and one by Douady. None of these contains an actual proof, so I am still looking for a nice reference.

REPLY [5 votes]: We follow Douady's approach using Cartan-Eilenberg systems, see here.
Let $B$ be a CW complex and $\pi\colon X\to B$ a Serre fibration. Put $X^k=\pi^{-1}(B^k)$. A cellular approximation~$\Delta_B\colon B\to B\times B$ of the diagonal can be lifted to an approximation $\Delta\colon X\to X\times X$ of the diagonal such that
$$X^k\stackrel\Delta\longrightarrow\bigcup_{m+n=k}X^m\wedge X^n\;.$$
Let $(\tilde h^\bullet,\delta,\wedge)$ be a reduced multiplicative generalised cohomology theory. We define a Cartan-Eilenberg system $(H,\eta,\partial)$ by
$$H(p,q)=\tilde h^\bullet(X^{q-1}/X^{p-1})$$
for~$p\le q$ with the obvious maps $\eta\colon H(p',q')\to H(p,q)$ for $p\le p'$, $q\le q'$. The corresponding exact sequences take the form
$$\cdots\to\tilde h^\bullet(X^{r-1},X^{q-1})\to\tilde h^\bullet(X^{r-1},X^{p-1})
\to\tilde h^\bullet(X^{q-1},X^{p-1})\stackrel\delta\to
\tilde h^\bullet(X^{r-1},X^{q-1})\to\cdots$$
We ignore the grading; it is easy to fill in.
To define a spectral product $\mu\colon(H,\eta,\partial)\times(H,\eta,\partial)\to(H,\eta,\partial)$ we consider the map
\begin{multline*}
  F_{m,n,r}\colon(X\wedge X)^{m+n+r-1}/(X\wedge X)^{m+n-1}
  \cong\bigcup_{a+b=m+n+r-1}(X^a\wedge X^b)\Bigm/
  \bigcup_{c+d=m+n-1}(X^c\wedge X^d)\\
  \begin{aligned}
    \twoheadrightarrow\mathord{}&\bigcup_{a+b=m+n+r-1}(X^a\wedge X^b)\Bigm/
    \Bigl(\bigcup_{a=0}^m(X^{a-1}\wedge X^{m+n+r-a})
    \cup\bigcup_{b=0}^n(X^{m+n+r-b}\wedge X^{b-1})\\
    \cong\mathord{}&\bigcup_{a=m+1}^{m+r}(X^{a-1}\wedge X^{m+n+r-a})\Bigm/
    \bigl(X^{m+r-1}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r-1}\bigr)\\
    \hookrightarrow\mathord{}& X^{m+r-1}\wedge X^{n+r-1}\bigm/
    (X^{m+r-1}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r-1})\\
    \cong\mathord{}&(X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\;.
  \end{aligned}
\end{multline*}Together with the diagonal map $\Delta$, for $r\ge 1$, we define
\begin{multline*}
  \mu_r\colon H(m,m+r)\otimes H(n,n+r)
  \cong\tilde h(X^{m+r-1}/X^{m-1})\otimes\tilde h(X^{n+r-1}/X^{n-1})\\
  \begin{aligned}
    &\stackrel\wedge\longrightarrow\tilde h\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\
    &\stackrel{F_{m,n,r}^*}\longrightarrow\tilde h\bigl((X\wedge X)^{m+n+r-1}/(X\wedge X)^{m+n-1}\bigr)\\
    &\stackrel{\Delta_X^*}\longrightarrow\tilde h(X^{m+n+r-1}/X^{m+n-1})=H(m+n,m+n+r)\;.
  \end{aligned}
\end{multline*}
Proposition
For all $m$, $n$, $r\ge 1$, the following diagram commutes
$\require{AMScd}$
\begin{CD}
  H(m,m+1)\otimes H(n,n+1)@>\mu_1>>H(m+n,m+n+1)\\
  @A\eta\oplus A\eta A@AA\eta A\\
  H(m,m+r)\otimes H(n,n+r)@>\mu_r>>H(m+n,m+n+r)\\
  @V\partial\otimes\eta\oplus V\eta\otimes\partial V@VV\partial V\\
  {\begin{matrix}H(m+r,m+r+1)\otimes H(n,n+1)\\\oplus\\H(m,m+1)\otimes H(n+r,n+r+1)\end{matrix}}@>\mu_1\pm\mu_1>>H_{p+q-1}(m+n+r,m+n+r+1)\rlap{;,}
\end{CD}
As explained here, this Proposition allows us to define a multiplicative structure on the associated spectral sequence.
Proof.
The upper square commutes because the maps~$F_{m,n,r}$ are defined sufficiently naturally. For the lower square, we consider the boundary morphism $\delta$ of the triple
$$(X^{m+r}\wedge X^{n+r-1}\cup X^{m+r-1}\wedge X^{n+r},
X^{m+r}\wedge X^{n-1}\cup X^{m+r-1}\wedge X^{n+r-1}\cup X^{m-1}\wedge X^{n+r},\\
X^{m+r}\wedge X^{n-1}\cup X^{m-1}\wedge X^{n+r})\;.$$
The following diagram commutes:
\begin{CD}
  \tilde h^{-p}(X^{m+r-1}/X^{m-1})\otimes\tilde h^{-q}(X^{n+r-1}/X^{n-1})
  @>\wedge>>
  \tilde h^{-p-q}\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\
  @V\delta\wedge\mathrm{id}\oplus V\mathrm{id}\wedge\delta V
  @VV\delta V\\
  {\begin{matrix}
      \tilde h^{1-p}(X^{m+r}/X^{m+r-1})\otimes\tilde h^{-q}(X^{n+r-1}/X^{n-1})\\
      \oplus\\
      \tilde h^{-p}(X^{m+r-1}/X^{m-1})\otimes\tilde h^{1-q}(X^{n+r}/X^{n+r-1})
  \end{matrix}}
  @>\wedge\oplus\wedge>>
  {\begin{matrix}
      \tilde h^{1-p-q}\bigl((X^{m+r}/X^{m+r-1})\wedge(X^{n+r-1}/X^{n-1})\bigr)\\
      \oplus\\
      \tilde h^{1-p-q}\bigl((X^{m+r-1}/X^{m-1})\wedge(X^{n+r}/X^{n+r-1})\bigr)
  \end{matrix}}
\end{CD}
We extend this diagram to the right using the maps $F_{m,n,r}$ and
conclude that the lower square also commutes.<|endoftext|>
TITLE: The gonality of smooth plane curves
QUESTION [6 upvotes]: I have often seen the assertion that for a smooth plane curve $C$ of degree $d$ the gonality of $C$ is $d-1$ and each gonality pencil is obtained by projection from a point of $C$ onto a line.
(let me recall that the gonality of $C$ is by definition the minimal degree $d$ of divisors $D$ on $C$ with $r(D)=1$, those divisors which attain the minimum being called gonality pencils)
Is there a relatively simple proof for this fact?

REPLY [7 votes]: Let $X \subset \mathbb{P}^2$ be a smooth planar curve of degree $d$. It is fairly well known that $\Omega^1(X) \cong \mathcal{O}(d-3)|_{X}$, and $H^0(\mathbb{P}^2, \mathcal{O}(d-3)) \to H^0(X, \Omega^1)$ is an isomorphism. So the canonical embedding of $X$ is isomorphic to the composition $X \to \mathbb{P}^2 \stackrel{\phi}{\longrightarrow} \mathbb{P}^{\binom{d-1}{2}-1}$ where $\phi$ is the $(d-3)$-fold Veronese.
Suppose that $X$ is $k$-gonal for $k \leq d-1$. Let $(x_1, x_2, \ldots, x_k)$ be a fiber of the map to $\mathbb{P}^1$, and let $D$ be the divisor $\sum x_i$. (If our map is separable, then a generic fiber has distinct points. In general, our map is the composition of a power of Frobenius and a separable map, so we can reduce to the separable case.)  Then $\dim H^0(X, D) \geq 2$ so (by Riemann-Roch) $\dim H^0(X, K-D) > \binom{d-1}{2}-k$. This says that the points $\phi(x_1)$, $\phi(x_2)$, ..., $\phi(x_k)$ are linearly dependent in $\mathbb{P}^{\binom{d-1}{2}-1}$.
Choose a generic coordinate system $(u:v:w)$ on $\mathbb{P}^2$, and write $x_i = (u_i: v_i : w_i)$. By genericity, we can assume that the $(u_i:v_i)$ are distinct and all $u_i$ and $v_i$ are nonzero. Then $d-2$ of the coordinates of $\phi(x_i)$ are $(u_i^{d-3}: u_i^{d-4} v_i : \cdots : v_i^{d-3})$. If $k<d-1$, this is a $k \times (d-2)$ Vandermonde matrix and hence of rank $k$, a contradiction.
If $k=d-1$, our Vandermonde is $(d-1) \times (d-2)$, so it is fine for it to have rank $d-2$. But we will show that, in this case, the points $x_i$ are forced to be on a line.
In this case, we are requiring that the whole $(d-1) \times \binom{d-1}{2}$ matrix has rank $(d-2)$. The $(d-2)$ columns mentioned above already have rank $d-2$, since they are Vandermonde, so every other column is in their linear span. In particular, for every $j$, there is a degree $d-3$ polynomial $F_j(u,v)$ such that $F_j(u_i,v_i) = u_i^j v_i^{d-4-j} w_i$ for all $i$. 
Then $u_i F_j(u_i, v_i) = v_i F_{j+1} (u_i, v_i)$. But $u F_j - v F_{j+1}$ is a degree $d-2$ homogenous polynomial so, if it vanishes at $d-1$ points, it is zero. We deduce that $u F_j = v F_{j+1}$ for $1 \leq j \leq d-5$, and thus that there is a degree $1$ polynomial $G$ such that $F_j(u,v) = u^j v^{d-4-j} G(u,v)$. Thus $w_i = G(u_i, v_i)$ for all $i$, and we have shown that the points lie on a line.

REPLY [2 votes]: Check out the paper by R.~Pelikaan, 1991 - he describes one direction, and gives references to the other.<|endoftext|>
TITLE: Generalizing the Mazur-Ulam theorem to convex sets with empty interior in Banach spaces
QUESTION [16 upvotes]: The Mazur-Ulam theorem (1932) states that any isometry of a normed linear space is affine. See Nica (Expo. Math. 30 (2012), 397-398; arXiv:1306.2380) for a very elegant proof.
Question: Let $M$ be a closed convex set with empty interior in a normed linear space and $T:M\to M$ be an isometry of $M$ onto itself. Does it follow that $T$ is affine (maps linear segments to linear segments)?
This question naturally arises in connection with the work of Bader, Furman, Gelander, Monod, see p. 88 of their paper `Property (T) and rigidity for actions on Banach spaces' (Acta Math. 198 (2007), 57-105; arXiv:math/0506361). 
Related information: 1. The question has already been answered in the positive  by Mankiewicz (On extension of isometries in normed linear spaces, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 20 (1972), 367-371) for closed convex sets having nonempty interior. 

There is an active related work on the problem suggested by Tingley
(Geometriae Dedicata, 1987, p.371): Suppose that $f: S_X\to S_Y$is an (onto) isometry between spheres of normed spaces. Is $f$ necessarily the restriction to $S_X$ of a linear, or affine, transformation? But in this work a starting point is an isometry of a non-convex set.

REPLY [14 votes]: I was referred to this question by Uri Bader. Thanks Uri.
Here are two counter examples.
Consider the set $K$ of all continuous strictly increasing functions $f$ from $[0,1]$ onto $[0,1]$. Let $T$ be the map from $K$ onto $K$ that sends $f$ to its inverse. Endowing $K$ with the $L_1$ norm, $T$ is an isometry (the $L_1$ distance between two functions is the area enclosed by their graphs and, By Fubini theorem, this is the same for the functions and their inverses).
The first example, $M_1$, is the closure of $K$ in the $L_1$ norm and the extension of $T$ to $M_1$. It is easy to see it is not affine (it is also easy to describe $M_1$ and the extension of $T$ explicitly).
The second example $M_2$, is all the continuous increasing functions from $[0,1]$ onto $[0,1]$ satisfying $(y-x)/2\le f(y)-f(x)\le 2(y-x)$ for all $0\le x<y\le1$, endowed with the $L_1$ norm.
$T$ is the same as before. 
It is easy to see that both $M_1$ and $M_2$ are norm compact.<|endoftext|>
TITLE: Does the Grothendieck ring of varieties contain torsion?
QUESTION [12 upvotes]: Let $K_0(Var_k)$ be the abelian group generated by the isomorphism classes of varieties over the field $k$ with the relations
$$[X]=[U]+[X\setminus U]$$
for every variety $X$ and open subvariety $U$. 
Is this abelian group torsion free? 
(Any positive or negative result for any particular field $k$ is welcome.)

REPLY [12 votes]: As per Theo Johnson-Freyd's request, I'm converting my comment to an answer.
Larsen-Lunts show that if $k$ is algebraically closed of characteristic zero, then there is a natural isomorphisms $$K_0(\text{Var}_k)/(\mathbb{L})\overset{\sim}{\longrightarrow} \mathbb{Z}[SB],$$
where $SB$ is the monoid of stable birational equivalence classes of smooth birational varieties and $\mathbb{L}=[\mathbb{A}^1]$.  (There is a way of making sense of this if $k$ is not algebraically closed, but the target will no longer be a monoid ring, so the argument I am making will break.)
$\mathbb{Z}[SB]$ is manifestly torsion-free; thus if $nX=0$ for $n\in \mathbb{Z}$ and $X\in K_0(\text{Var}_k)$, we must have $X\in (\mathbb{L})$.  Thus either we would have an example where $\mathbb{L}$ is a zero divisor, or a proof that $$\bigcap_n (\mathbb{L}^n)\neq 0.$$  $\mathbb{L}$ was recently shown to be a zero divisor by Borisov, but this was open for a long time; as far as I know, the question of whether or not $$\bigcap_n (\mathbb{L}^n)=0$$ is open as well (I've thought about it a bit, because I used the $\mathbb{L}$-adic topology on $K_0(\text{Var}_k)$ in my paper "Symmetric Powers do not Stabilize," and I think it's quite hard).  So this question is almost certain to be open for $k$ algebraically closed of characteristic zero.<|endoftext|>
TITLE: approximate stationary distributions of a doubly stochastic matrix and its supports
QUESTION [7 upvotes]: Given a doubly stochastic matrix $M$ and a distribution $v$,let $M=\sum_{\sigma\in S_n}p_{\sigma}M_{\sigma}$ be any Birkhoff decomposition of $M$, where $M_{\sigma}$ is the permutation matrix induced by $\sigma$ and $\{p_{\sigma}\}$ is a distribution. Suppose $v$ is an $\epsilon$-approximate stationary distribution of $M$, i.e., 
$$\|v-Mv\|_{TV}\leq\epsilon,$$
where $\|\cdot \|_{TV}$ is the total variance.
Does it imply the following 
$$\sum_{\sigma}p_{\sigma}\|v-M_{\sigma} v\|_{TV}\leq \epsilon',$$
where $\epsilon'=f(\epsilon)$ only depends on $\epsilon$,  and goes to 0 if $\epsilon$ goes to 0. In other words, is $v$ also an approximate stationary distribution of $M_{\sigma}$ in expectation? It is not hard to prove when $\epsilon=0$. I wonder whether this problem has been studied.
Thanks.

REPLY [2 votes]: I think the answer is no: the rate at which $\epsilon'$ goes to 0 does depend on the size of the matrix. Here is my argument, which is not yet worked out in complete detail. 
I want to consider a circle of large radius $R$ that is discretized with a very fine mesh. Imagine an approximation to a normal distribution with variance 1 centred at a point on the circle so that it decays almost to 0 before it wraps around. This is $\nu$. Now the matrix $M$ is convolution with another approximate normal distribution with variance $\eta^2$. A Birkhoff decomposition (in general, I don't think it's unique) is just the obvious one: $M$ is a combination of rotations by different amounts. 
If you do this, I think $\|\nu-\nu M\|_\text{TV}\sim \eta^2$ while for typical $\sigma$, I think you have $\|\nu-\nu M_\sigma\|_\text{TV}\sim\eta$. 
To actually do the calculation, I would prefer to work with true normal distributions on $\mathbb R$. If $X$ is distributed as $N(0,1)$ and $\Delta$ is distributed as $N(0,\eta^2)=\eta\cdot N(0,1)$, then $X+\Delta\sim N(0,1+\eta^2)$, so that $\|\nu_{X+\Delta}-\nu_{X}\|_\text{TV}\sim\eta^2$. 
On the other hand, $\|\nu_{X+t}-\nu_X\|_\text{TV}\sim t$, so that since the typical value of $\Delta$ is of order $\eta$, the expected order here is larger.<|endoftext|>
TITLE: reference request for mod p and p-adic K-theory
QUESTION [5 upvotes]: Is there a good reference that explains mod p K-theory and p-adic or p-complete K- theory? All I know about K-theory is the topological K-theory  of "vector bundles and k-theory" in Switzer's book (and similar expositions found in a book by Hatcher, and a chapter in May's Concise course). There seems to me be a jump from that type of material to understanding statements like the " p-completion of $K(\mathbb{Z}/p)$ is $H\mathbb{Z}_p$", a result due to Quillen, according to this question. I looked at the titles of Quillen's papers and am not sure in which paper it appears. Also, perhaps there are now other good expositions introducing and computing basic things about mod p or p-complete k-theory.

REPLY [5 votes]: You have learned about topological K-theory (of topological spaces).
The Quillen result from 
"On the Cohomology and K-Theory of the General Linear Groups Over a Finite Field", Daniel Quillen, Ann. Math., Vol. 96, No. 3 (Nov., 1972), pp. 552-586,
is about algebraic K-theory (of rings).<|endoftext|>
TITLE: The Bourgain-Demeter-Guth breakthrough and the Riemann zeta function?
QUESTION [37 upvotes]: Yesterday Bourgain, Demeter and Guth released a preprint proving (up to endpoints) the so-called main conjecture of the Vinogradov's Mean Value Theorem for all degrees. This had previously been only known for degree 3 by work of Wooley. See this survey of Wooley for a more discussion.
It seems to be folklore that a proof of this conjecture should imply improved bounds on the Riemann zeta function / an improved zero free region for the zeta function / an improved error term on the prime number theorem (among other things, such as progress on Waring's problem). These, of course, are some of the most highly prized problems in analytic number theory and have been stuck for decades.
That said, to the best of my knowledge there is no blackbox reduction for these applications and one likely also needs more explicit information about the dependence of constants on various parameters in these results to reach these applications.


What is the potential of these methods? In other words, what are the implications of the most optimistic dependencies in the mean value theorem (or possible generalizations)?


(See here and here for additional discussion).

REPLY [2 votes]: And a few words about "possible generalizations". The main object here is the sum
$$S=\sum_{u\asymp a}e^{2\pi i F(u)},$$
where $F(u)=-\frac{t\log u}{2\pi}$, $t=a^{n-\theta}$, $0\le \theta<1$. It is known that $|S|\ll a^{1-\frac{c}{n^2}}$. It was mentioned by Vinogradov in his book "Trigonometrical sums in number theory" that even the much stronger estimate $|S|\ll a^{1-\frac{c}{n}}$ (unreachable by this method) will give only 
$$\pi(x)=\text{li}(x)+O\left(x\exp\left(c\log(x)^{2/3}(\log \log x)^{-1/5}\right)\right).$$<|endoftext|>
TITLE: What is the precise relationship between o-minimal theory and Grothendieck's "Esquisse d'un programme"?
QUESTION [15 upvotes]: I have seen various references in the literature to such a connection but they tend to assume that the reader is familiar with the connection, and limit themselves to providing additional detail.  So in broad terms, what is the precise connection between o-minimal models and Grothendieck's programme?

REPLY [15 votes]: To define an o-minimal geometry, one gives oneself a family of functions from $\mathbf R^n$ to $\mathbf R^m$ ($m,n$ not specified), and one considers all definable subsets of $\mathbf R^n$, ie, those which can be defined by a mathematical expression using these basic functions, addition, multiplication, constant functions, the ordering symbol, logical connectives (AND, OR, NOT) and quantifiers (FORALL, EXISTS). A function is called definable if its graph is definable.
The geometry is said to be o-minimal if the only definable subsets of $\mathbf R$ (the real line) are finite unions of intervals. 
These geometries are tame in the sense of Grothendieck's Esquisse, but what is remarkable is that the tameness axiom is on subsets of the line, not of higher dimensional $\mathbf R^n$. Here are a few tame properties of o-minimal geometries:

Every definable function $\mathbf R\to\mathbf R$ is piecewise monotone, piecewise $\mathbf C^k$ for every $k$. (No $\sin(1/x)$ curve, etc.)
The closure, the interior of a definable subset is definable (just by using the $\epsilon,\delta$ definition).
There is a cellular decomposition theorem. An open cell is a subset of $\mathbf R^n$ of the form, say, 
$$ \{(x,t)\in \mathbf R^{n-1}\times\mathbf R\, ;\,
    x\in A,  \phi_-(x) < t<\phi_+(x) \}$$
where $A$ is an open cell in $\mathbf R^{n-1}$ and $\phi_-,\phi_+$ are definable functions on $A$ such that $\phi_-(x)<\phi_+(x)$ for every $x\in A$. (Actually, there are other cells similarly defined by replacing strict inequalities by large inequalities, and taking $\phi_-=-\infty$ or $\phi_+=+\infty$.) Then for every finite family $\mathscr A$ of definable subsets of $\mathbf R^n$, there is a finite partition of $\mathbf R^n$ into cells such that each element of $\mathscr A$ is a union of some cells.
A closed definable subset of $\mathbf R^n$ is locally contractible (no hawaiian earring).
There is a nice theory of dimension (Brouwer invariance of domain is not an issue, for example).

As indicated by Todd Trimble, the book of van den Dries, Tame Topology and O-minimal Structures explains this in quite a detail. Also, a remarkable theorem of Peterzil and Starchenko states that a complex analytic subspace of $\mathbf C^n$ which is definable (when you identify $\mathbf C^n$ with $\mathbf R^{2n}$) is automatically complex algebraic.
It is also a fundamental fact that there are many interesting examples of o-minimal geometries. Let's me list some of them:

Semialgebraic geometry (where definable subsets are semialgebraic sets); by Tarki's theorem, every semialgebraic set can be defined without quantifiers, hence you see that semialgebraic subsets of the line are finite union of intervals. 
Take for a basic family of functions the restrictions to $[-1,1]^n$ of functions defined by a power series of radius of convergence $>1$. This geometry is o-minimal (Denef, van den Dries). In particular, compact analytic subsets generates an o-minimal geometry which contains real subanalytic sets. 
Take only one basic function, namely the exponential function (defined on the whole real line). This geometry is o-minimal (Wilkie).
Combine the two preceding geometries; you still get an o-minimal geometry (van den Dries, Macintyre, Marker) and this was a crucial tool in the recent solution of the André-Oort conjecture (Pila, Tsimerman, Klingler, Ullmo, Yafaev).
Speissegger shows you can add solutions of a definable Pfaff equation, etc.<|endoftext|>
TITLE: Stone-Weierstrass theorem for holomorphic functions?
QUESTION [32 upvotes]: The Stone-Weierstrass theorem has an analog for the algebras of smooth functions, called  

Naсhbin's theorem: An involutive subalgebra $A$ in the algebra ${\mathcal C}^\infty(M)$ of smooth functions on a smooth manifold $M$ is dense in ${\mathcal C}^\infty(M)$ if and only if $A$ separates the points and the tangent vectors of $M$.

See details in: "L.Nachbin. Sur les algèbres denses de fonctions diffèrentiables sur une variètè, C.R. Acad. Sci. Paris 228 (1949) 1549-1551", or in  J.G.Llavona's monograph, or here. 
This is strange, I can't find an analog for the algebras of holomorphic functions (on complex manifolds). Did anybody think about this?

Question: let $A$ be a subalgebra in the algebra ${\mathcal O}(M)$ of holomorphic functions on a complex manifold $M$ (as a first approximation, we can think that $M$ is just an open subset in ${\mathbb C}^n$). Which conditions should $A$ satisfy for being dense in ${\mathcal O}(M)$? 


Remark. By topology on ${\mathcal O}(M)$ I mean the usual topology of uniform convergence on compact sets in $M$. The algebra ${\mathcal C}^\infty(M)$ is also endowed with its usual topology, which can be described, for example, as follows. 

For each function $f\in {\mathcal C}^\infty(M)$ let us define its support as the closure of the set of the points where $f$ does not vanish:
$$
\text{supp}f=\overline{\{x\in M:\ f(x)\ne 0\}}.
$$
An equivalent definition: $\text{supp}f$ is the set of the points in $M$ where $f$ has non-zero germs:
$$
\text{supp}f=\{x\in M:\ f\not\equiv 0\ (\text{mod}\ x)\}.
$$
Let us define differential operators (see e.g. S.Helgason's book) on $M$ as linear mappings $D:{\mathcal C}^\infty(M)\to {\mathcal C}^\infty(M)$ which do not extend the support of functions:
$$
\text{supp}Df\subseteq \text{supp}f,\quad f\in{\mathcal C}^\infty(M).
$$
Equivalently, $D$ is local, i.e. the value of $Df$ in a point $x\in M$ depends only on the germ of $f$ in $x$: 
$$
\forall f,g\in{\mathcal C}^\infty(M)\quad \forall x\in M\qquad f\equiv g\ (\text{mod}\ x)\quad\Longrightarrow\quad Df(x)=Dg(x).
$$
Then we say that a sequence of functions $f_n$ converges to a function $f$ in ${\mathcal C}^\infty(M)$
$$
f_n\overset{{\mathcal C}^\infty(M)}{\underset{n\to\infty}{\longrightarrow}}f
$$
if and only if for each differential operator $D:{\mathcal C}^\infty(M)\to {\mathcal C}^\infty(M)$ the sequence of functions $Df_n$ converges to $Df$ in the space ${\mathcal C}(M)$ of continuous functions with the usual topology of uniform convergence on compact sets in $M$:
$$
Df_n\overset{{\mathcal C}(M)}{\underset{n\to\infty}{\longrightarrow}}Df
$$

Of course, this is equivalent to the convergence in ${\mathcal C}^\infty(U)$  for each smooth local chart $\varphi:U\to V$, $U\subseteq\mathbb{R}^m$, $V\subseteq M$. This is also  equivalent to what Alex M. writes about vector fields:
$$
f_n\overset{{\mathcal C}^\infty(M)}{\underset{n\to\infty}{\longrightarrow}}f
\quad\Longleftrightarrow\quad \forall k\ \forall X_1,...,X_k\in{\mathcal X}(M) \quad
X_1...X_kf_n\overset{{\mathcal C}(M)}{\underset{n\to\infty}{\longrightarrow}}X_1...X_kf.
$$

REPLY [8 votes]: $\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$
Here is a candidate counterexample for $M= \CC$: Is $e^{-z}$ in the closure of the algebra generated by $e^z$ and $e^{\sqrt{2}z}$? My current guess is "no", but I need to move on to actual work.

I will show that separating points and separating tangents is not enough for $M = \CC^2$. 
Let $A \subset \cO(\CC^2)$ be those holomorphic functions $f$ such that $f(z,z^{-1})$ extends holomorphically to $z=0$. We observe:
$A$ is a subalgebra: This is obvious.
$A$ is closed: Proof We have $f \in A$ if and only if $\oint f(z,z^{-1}) z^n dz=0$ for all $n \geq 0$, where the integral is on a circle around $0$. This fact is preserved by uniform limits on compact sets. (Specifically, by uniform limits on that circle.)
$A$ separates points: Note that the functions $f(x,y) = x$, $g(x,y) = xy$ and $h(x,y) = y (xy-1)$ are all in $A$. The functions $f$ and $g$ alone separate $(x_1, y_1)$ and $(x_2, y_2)$ unless $x_1=x_2=0$. In that case, $h$ separates them.
$A$ separates tangent vectors: Again, $df$ and $dg$ are linearly independent at all points where $x \neq 0$, and $df$ and $dh$ are linearly independent at $x=0$.
$A \neq \cO(\CC^2)$ Clearly, $y \not \in A$.
I remembered this counterexample from an old blog post of mine.
Note that we could replace $z \mapsto (z, z^{-1})$ with any map $\phi$ from the punctured disc $D^{\ast}$ to $\CC^2$. There are many such $\phi$'s, and they all appear to impose independent conditions. This makes me pessimistic about any simple criterion for equality when $M = \CC^2$.<|endoftext|>
TITLE: "Largish" cardinals
QUESTION [14 upvotes]: In what follows, $\mathsf{ZCKP}$ refers to the subset of $\mathsf{ZFC}$ consisting of the axioms of Zermelo set theory with choice and foundation ($\mathsf{ZC}$) plus those of Kripke-Platek set theory ($\mathsf{KP}$): equivalently, as these two theories have much overlap, $\mathsf{ZCKP}$ consists of $\mathsf{ZC}$ plus the axiom of replacement for $\Sigma_1$-formulæ (or equivalently, $\mathsf{KP}$ plus choice, foundation, powerset, infinity, and full separation).  A "largish" cardinal property means, informally, one whose existence is provable in $\mathsf{ZFC}$ but not in $\mathsf{ZCKP}$.
Here is the simplest example of a largish cardinal notion.  A theorem of Azriel Lévy (see, e.g., Barwise, Admissible Sets and Structures (1975), theorem II.3.5 on page 53 and theorem II.9.1 on page 76) states that for every uncountable cardinal $\kappa$, if $H(\kappa)$ is the set of sets hereditarily of cardinality $<\kappa$, then $H(\kappa)$ is a $1$-elementary submodel of the universe (meaning that every $\Sigma_1$ formula with parameters in $H(\kappa)$ is true [in $V$] iff it is true in $H(\kappa)$).  This implies that $H(\kappa)$ satisfies $\Delta_0$-collection (eqvt. $\Sigma_1$-replacement), and if $\kappa$ is a strong limit, then $H(\kappa) \models \mathsf{ZCKP}$ (and the converse is clear).  In particular, $\mathsf{ZCKP}$ does not prove the existence of strong limit cardinals.
Now I am interested in strengthenings of this condition on $\kappa$ such that the existence of these cardinals is still provable in $\mathsf{ZFC}$.  Two obvious candidates are:

$H(\kappa)$ satisfies $\mathsf{ZC}$ plus replacement for $\Sigma_n$ formulæ,
$H(\kappa)$ is an $n$-elementary submodel of the universe (i.e., every $\Sigma_n$ formula with parameters in $H(\kappa)$ is true iff it is true in $H(\kappa)$).

These should at least imply that $\kappa$ is a fixed point of the beth function, so that in fact $H(\kappa) = V_\kappa$.  (Perhaps this should be added as a precondition to be worthy of the term "largish cardinal".)
Edit (on 2015-12-11, following the answer by Joel David Hamkins): I didn't realize how very different the two notions above are: the first (call them "$\Sigma_n$-replacing" cardinals) is "local" in that it involves only sets from $V_\kappa$ and can thus be expressed as a $\Delta_1$ property of $V_\kappa$, whereas the latter ("$\Sigma_n$-correct cardinals") is "global" and involves the entire universe.  This has a consequence of size: the smallest $\Sigma_2$-correct cardinal, as explained in Joel's response, is larger than the first $\Sigma_n$-replacing cardinal for all $n$, or even the first inaccessible, etc., and there is no hope of "computing" it.  There may be some hope for the first $\Sigma_2$-replacing cardinal, however.  I really should have asked two different questions.
My question is this: Can these conditions, at least for $n=2$, or perhaps some related ones, be rephrased in purely cardinal-theoretic terms (without appealing to model theory and if possible avoiding the Lévy hierarchy)?  Even better, can the smallest cardinal satisfying such a condition be "described" or "computed" in some way?  (In the same way that $\beth_\omega$, or "the limit of the sequence defined by $\kappa_0 = \omega$ and $\kappa_{n+1} = \beth_{\kappa_n}$" are descriptions/computations of the smallest strong limit cardinal and the smallest fixed point of the beth function.)
More generally, any comments on these or related properties would be welcome (including a better term than "largish cardinal").  There is probably some connection with powerset-admissible ordinals, although the exact relation escapes me.
One reason why one might be interested in such cardinals is that the corresponding $H(\kappa)$ might serve as a drop-in replacement for Grothendieck universes in a $\mathsf{ZFC}$ formulation of category theory (they are not fully Grothendieck universes, but the point is that the use of a construction that escapes from such a "universish" set is likely to be so rare as to be very conspicuous; and unlike Grothendieck universes, their existence follows from $\mathsf{ZFC}$).

REPLY [16 votes]: Your cardinals are known as the $\Sigma_n$-correct cardinals, and they arise in diverse set-theoretic contexts. For example, we use them extensively in our paper:

J. Bagaria, J. D. Hamkins, K. Tsaprounis, T. Usuba, Superstrong
and other large cardinals are never Laver indestructible, to
appear in the Archive for Math Logic.

It is a ZFC theorem that the $\Sigma_n$-correct cardinals form a
closed unbounded proper class often denoted $C^{(n)}$.
One subtle point about the $\Sigma_n$-correct cardinals is that
although we have a concept of $\Sigma_2$-correct and
$\Sigma_3$-correct and so on, $\Sigma_n$-correct for any
particular $n$, there is no uniform-in-$n$ way to express the
concept of $\Sigma_n$-correctness in first-order set theory. The
concept is uniformly expressible in some second-order set
theories, such as Kelley-Morse set theory, which prove that there
is a truth-predicate for first-order truth.
Concerning your question, when $n\geq 2$ there can be no way to
define what it means for $\kappa$ to be $\Sigma_n$-correct by
looking only below $\kappa$, say, as a limit process, since such a
property would be too simple, as it could be verified inside
$V_\kappa$ itself, but such verifiable-in-$V_\kappa$ properties have complexity at worst $\Delta_2$. For example, the property of being $\Sigma_n$-correct cannot be
$\Sigma_n$-expressible, for then the assertion "There is a
$\Sigma_n$-correct cardinal" would reflect from $V$ to $V_\kappa$,
even when $\kappa$ is the least $\Sigma_n$-correct cardinal, which
gives a contradiction since there are none below the least one. Meanwhile, the property of being $\Sigma_n$-correct is $\Pi_n$-expressible, since one need only say that all the instances of $\Pi_n$ truth in $V_\kappa$ are actually true.
The case of $\Sigma_2$-correct cardinals is particularly attractive, and perhaps this is an example that interests you. The $\Delta_2$ properties are precisely the properties that are local, in the sense that they can be determined in any sufficiently large $H(\theta)$. You can read more on my blog post: 

Local properties in set theory

It follows that a cardinal $\kappa$ is $\Sigma_2$-correct, if whenever there is an object having a certain properties inside some possibly very large $H(\theta)$, then there is such an object inside such an $H(\theta)$ with $\theta<\kappa$. In other words, $\kappa$ is $\Sigma_2$-correct, if whenever anything verifiable happens anywhere, then it happens inside $V_\kappa$. Alternatively, everything verifiable has already happened by the time you get to $H_\kappa$. Such a way of understanding $\Sigma_2$-correctness is extremely useful, since it aligns with how set theorists often think about verifying set-theoretic facts. 
(A small matter: the distinction between $V_\kappa$ and $H_\kappa$ disappears once $n\geq 2$, since in this case the cardinals are $\beth$-fixed points and so $V_\kappa=H_\kappa$.)
Lastly, your idea of using the $\Sigma_n$-correct cardinals as a
universe replacement idea is well known. This is known as the
Feferman theory, and I also discussed it here on
MathOverflow in my answer to the question What interesting/nontrivial results in Algebraic geometry require the existence of universes?.<|endoftext|>
TITLE: Galois representations for the curve $y^{2} = x^{3} - x$
QUESTION [11 upvotes]: Let $E / \mathbb{Q}$ be the elliptic curve given by $y^{2} = x^{3} - x$.  I would like to know explicitly what the field of all $2$-power torsion looks like, as well as the image in $\mathrm{GL}(T_{2}(E))$ of the $2$-adic Galois representation (and I would be interested in analogous descriptions for $\ell$-adic Galois representations with $\ell \neq 2$ as well).  I'm sure this could be computed using Sage or Magma, but I thought I'd ask to see if it's already written down somewhere.
Here's what I do know:
1) Since $E$ has complex multiplication with $\mathrm{End}_{\mathbb{Q}(i)}(E) \cong \mathbb{Z}[i]$, and $j(E) \in \mathbb{Q}$, the dyadic torsion field must be an abelian extension of $\mathbb{Q}(i)$ ramified only above the prime $(1 + i)$.  In fact, it must be the maximal such abelian extension.
2) The dyadic torsion field must contain all $2$-power roots of unity.
3) The division field $\mathbb{Q}(E[8])$ is $\mathbb{Q}(\zeta_{16}, 2^{1/4})$, and the image of $\mathrm{Gal}(\bar{\mathbb{Q}} / \mathbb{Q}(\zeta_{8}))$ modulo $8$ is generated by the matrices $$\left[ {\begin{array}{cc} 1 & 4 \\ 4 & 1 \end{array} } \right], \left[ {\begin{array}{cc} 5 & 0 \\ 0 & 5 \end{array} } \right] \in \mathrm{SL}_{2}(\mathbb{Z} / 8\mathbb{Z})$$ with respect to a suitable basis.
[EDIT: I corrected a mistake I noticed in my claim (3) above.]

REPLY [14 votes]: I don't know that this is written down anywhere, but it's possible. It is known in general that $GL(T_{\ell}(E))$ is contained in the normalizer of $R_{\ell}^{\times}$, where $R_{\ell} = \mathbb{Z}[i] \otimes \mathbb{Z}_{\ell}$. (This follows for example from Corollary 2 on page 502 of Serre and Tate's ''Good reduction of abelian varieties'' Annals paper.) Moreover, if $\ell$ is sufficiently large, we have that the $\ell$-adic image equals the normalizer (in $GL_{2}(\mathbb{Z}_{\ell})$) of $R_{\ell}^{\times}$.
For the case of the particular curve at hand, we have that the normalizer of $R_{2}^{\times}$ is
$$
  G = \left\{ \begin{bmatrix} a & b \\ \mp b & \pm a \end{bmatrix} \in GL_{2}(\mathbb{Z}_{2}) : a, b \in \mathbb{Z}_{2} \right\}.
$$
This is not the $2$-adic image, however, because the $2$-adic representation has index $4$ in $G$. 
To see the index is at least $4$, observe that $E : y^{2} = x^{3} - x$ has full rational $2$-torsion, and also that $-I$ is not in the $2$-adic image (since the $4$-torsion field is generated by the $x$-coordinates of $4$-torsion points). 
To see the index is at most $4$, I cheat a little bit. I had already computed the $2$-adic image for $E' : y^{2} = x^{3} + 3x$ using the usual Frattini subgroup argument, this shows that the $2$-adic image for this curve is all of $G$. However, these two curves are isomorphic over $\mathbb{Q}[\sqrt[4]{3}]$, and so the $2$-adic image for $E$ has index at most $4$. More precisely, the image is 
$$
\left\{ g \in G : g \bmod 4 \in \left\langle \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}, \begin{bmatrix} 3 & 2 \\ 2 & 3 \end{bmatrix} \right\rangle \right\}.
$$<|endoftext|>
TITLE: Darboux property of non-atomic sigma-additive nonnegative measures equivalent to the AC?
QUESTION [7 upvotes]: A result commonly, and probably erroneously, attributed to W. Sierpiński is that every non-atomic, countably additive, nonnegative measure $\mu: \Sigma \to \bf R$, where $\Sigma$ is a sigma-algebra on a set $S$, has the weak, and hence the strong, Darboux property, which means that for every $X \in \Sigma$ and $a \in [0, \mu(X)]$ there exists $Y \in \Sigma$ such that $Y \subseteq X$ and $\mu(Y) = a$.
The theorem is provable in ZFC, and I'm just curious to know if it is actually equivalent to the axiom of choice.
In either case, I would very much appreciate a reference where the question is answered: Before asking, I browsed through Howard and Rubin's Consequences of the Axiom of Choice, but couldn't find anything close to what I'm looking for.

REPLY [6 votes]: This theorem follows from Dependent Choice, and thus is strictly weaker than the Axiom of Choice.  Here is a proof using only DC.  Fix $X\in\Sigma$ such that $\mu(X)>0$ and let $a\in(0,\mu(X))$.  We will use DC to inductively construct a sequence of disjoint measurable subsets $Y_n$ of $X$ such that $\mu(\bigcup Y_n)=a$.
Having defined $Y_1,\dots,Y_{n-1}$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})<a$, define $Y_n$ as follows.  Let $Z=X\setminus(Y_1\cup\dots\cup Y_{n-1})$.  Since $\mu$ is atomless, there exists a subset $W\subset Z$ such that $0<\mu(W)<\mu(Z)$.  It is also clear that there exists such $W$ with $\mu(W)$ arbitrarily small (we can get $\mu(W)\leq \mu(Z)/2$ by replacing $W$ with $Z\setminus W$ if necessary, and now iterate); in particular, there exist such $W$ with $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$.  Now choose $Y_n$ to be such a $W$ which additionally has the property that for any other such $W'$, $\mu(Y_n)\geq \mu(W')/2$ (that is, among the possible measures of such sets $W$, $Y_n$ is in the upper half).
Let $Y=\bigcup Y_n$.  By construction, the $Y_n$ are disjoint and satisfy $\sum \mu(Y_n)\leq a$, so $\mu(Y)\leq a$.  Suppose $\mu(Y)<a$.  As above, there exists a subset $W\subset X\setminus Y$ such that $\mu(W)>0$ and $\mu(W)+\mu(Y)<a$.  Now let $n$ be such that $\mu(Y_n)<\mu(W)/2$ (such $n$ exists since $\sum \mu(Y_n)$ is finite).  Then $W$ is a subset of $X\setminus(Y_1\cup\dots\cup Y_{n-1})$ such that $\mu(Y_1)+\dots+\mu(Y_{n-1})+\mu(W)<a$, so by our choice of $Y_n$, we must have $\mu(Y_n)\geq \mu(W)/2$.  This is a contradiction, so we must have $\mu(Y)=a$.<|endoftext|>
TITLE: Least-squares solution of systems of Sylvester equations
QUESTION [7 upvotes]: The Sylvester equation $AX+XB=C$ has been studied quite a lot and there are known algorithms for solving it.
But has the situation where (an over-determined) system of equations $A_{i}X+XB_{i}=C_{i}$ is to be solved has been studied?
The question is motivated by an application in robotics (see here), where $C_{i}=0$ and a non-zero solution is required (with some extra structure, actually, but never mind for now).

REPLY [3 votes]: Here's a rather obvious way to do it, in the case where all $C_i = 0$.
Let $X_1, \ldots, X_k$ be a basis of 
solutions of $A_1 X + X B_1 = 0$, so the general solution of 
$A_1 X + X B_1 = 0$ is $X = \sum_{i=1}^k t_i X_i$.  For this to satisfy
$A_2 X + X B_2 = 0$, we need $\sum_{i=1}^k t_i (A_2 X_i + X_i B_2) = 0$, which
is a set of linear equations in the $t_i$.  Solve and recurse...
Of course for this to be efficient, you'd want $k$ to be fairly small.  In the worst case (e.g. $A_1 = B_1 = I$), $k = n^2$ where these are $n \times n$ matrices, but that's rather exceptional.<|endoftext|>
TITLE: When does 'Zariski tangent space derivative' vanishes everywhere imply that a section is constant?
QUESTION [6 upvotes]: Consider an abelian algebra, $R$, over the field $K$ with the properties that every residue field of $R$ is (canonically) isomorphic to $K$ (I'm not sure but I think this is necessary, otherwise we could be talking about $\mathbb{R}[x]$ as a $\mathbb{Q}$-algebra) and that for every maximal ideal $m$, $R/m^2\cong K\oplus m/m^2$ as additive groups (and let $\phi:R\rightarrow R/m^2$ be the natural quotient map). For every $s \in R$ call the $K$ factor of $\phi(s)$ the 'value of $s$ at $m$' and call the $m/m^2$ factor 'the derivative of $s$ at $m$'. Under what conditions does 'the derivative of $s$ vanishes at each $m$' imply that 'the value of $s$ is the same at every $m$'? What are the conditions if we talk about jet spaces instead of tangent spaces (i.e. replace every instance of $m^2$ with $m^k$ for a specific or arbitrarily large $k$)?
There are obvious examples and counterexamples. The ring of polynomials over a field of characteristic 0 satisfies this condition, as does the ring of smooth functions on a connected manifold (and maybe even $p$ times differentiable functions, even though the Zariski tangent space isn't the normal tangent space). Any ring of functions on a disconnected space obviously doesn't satisfy this, but there are also connected counterexamples, such as the ring of polynomials over a field of positive characteristic, although in that case talking about arbitrarily large jet spaces gives you an analogous statement (i.e. if every jet of a polynomial is constant, then the polynomial is constant).

REPLY [4 votes]: Let me propose a more natural version of your question that can be asked about more algebras: 

What can we say about an element $r \in R$ of a commutative algebra $R$ over a field $k$ given that $D(r)$ vanishes for every $k$-linear derivation $D : R \to M$ ($M$ an $R$-module)? In particular, must we have $r \in k$? 

This condition should imply but in general should be stronger than your condition. It's equivalent to the condition that the universal $k$-linear derivation $d : R \to \Omega_{R/k}$ into Kähler differentials vanishes on $r$. I don't know what to say at this level of generality, unfortunately.<|endoftext|>
TITLE: Generalized Euler characteristics of non-motivic origin
QUESTION [6 upvotes]: By a generalized Euler characteristic $\chi$, I mean an isomorphism invariant $\chi(V)$ inside some abelian group $A$, defined for every varietiy $V$ over a field $k$, with the property that, for all varieties $X$ and open subvarieties $U$, we have $$\chi(X)=\chi(U)+\chi(X\setminus U).$$
(In other words $\chi$ is a group homomorphism $K_0(Var_k)\to A$ from the Grothendieck ring of varieties.)
There are many generalized Euler characteristics, for example the ordinary Euler characteristic (with compact support), the Hodge numbers (at least for smooth projective complex varieties), the number of rational points (over finite fields), etc.. However, all these examples are of motivic origin, i. e. they factorize over the map assigning to a smooth projective $X$ the class of its motive in the Grothendieck ring of Chow motives.
The only Euler characteristic I know of for which this is not the case is the map $K_0(Var_k)\to \Bbb Z[SB]$ where $SB$ denotes the monoid of stable birational equivalence classes, sending a smooth projective X to its equivalence class.
Are there any other known Euler characteristics factorizing neither through $K_0(Var_k)\to K_0(\mathcal Mot_k)$ nor $K_0(Var_k)\to \Bbb Z[SB]$? 
(Here, of course, $\chi$ should be somewhat sensible, in the sense that the structure of target $A$ should not be to complicated (so as to exclude the universal Euler characteristic $id:K_0(Var_k)\to K_0(Var_k)$.)

REPLY [4 votes]: See section 7 of Bondal-Larsen-Lunt's paper "Grothendieck Ring of Pretriangulated categories, where they construct a non-trivial morphism $$\mu: K_0(\text{Var}_k)\to \mathcal{PT}$$
where the latter is, as you might guess from the title of their paper, a Grothendieck ring constructed from the $2$-category of pretriangulated categories.  The idea is to send a variety to its derived category, though the paper is quite a bit more complicated than that.
This motivic measure factors through $$K_0(\text{Var}_k)/(\mathbb{L}-1);$$
hence it does not factor through $$\mathbb{Z}[SB]\simeq K_0(\text{Var}_k)/\mathbb{L}$$ (at least over an algebraically closed field of characteristic zero).
Orlov conjectures that derived equivalent varieties have isomorphic motives (which would imply this measure does factor throuh $K_0(\text{Mot}_k)$, I believe; this is very open, and I've heard some experts express skepticism about it--I'm agnostic, personally.  At the very least, this motivic measure is not known to factor through $K_0(\text{Mot}_k)$, and some people doubt it does--if you could prove this, many people would be interested.  This is the only measure I know of (off the top of my head) which is not known to factor through either $\mathbb{Z}[SB]$ or $K_0(\text{Mot}_k)$.<|endoftext|>
TITLE: Reference request: Morita bicategory
QUESTION [19 upvotes]: I have two closely related questions:

Who first recognized that Morita equivalence was equivalence in the bicategory of Rings, Bimodules, and Intertwiners?

I've heard this bicategory called the "Morita bicategory".  Who first called it this?



Here's what I've been able to figure out:
Morita introduced what now is called "Morita equivalence" in

Kiiti Morita. Duality for modules and its applications to the theory of rings with minimum condition. Sci. Rep. Tokyo Kyoiku Daigaku Sect. A, 6:83–142, 1958. MR0096700

He defined it in terms of equivalences of categories of modules, but proved that all equivalences of categories of modules are implemented by tensor product with a bimodule.  (Also by homming from a bimodule.)  It's worth also noting that Morita primarily discussed contravariant equivalences between various categories of modules satisfying some size conditions, rather than covariant equivalences between categories of all modules; his primary interest was on "dualities" $\hom_A(-,U)$ and $\hom_B(-,U)$ where $U$ is a fixed $A$-$B$-bimodule.
Eilenberg and Watts independently proved the theorem now named after them jointly, that colimit-preserving functors between categories of modules are given by tensoring with bimodules:

Charles E. Watts, Intrinsic characterizations of some additive functors, Proc. Amer. Math. Soc. 11, 1960, 5–8, MR0806.0832
Samuel Eilenberg, Abstract description of some basic functors, J. Indian Math. Soc. (N.S.) 24, 1960, 231–234, MR0125148

Bass gave an elegant survey of Morita's work and the Eilenberg–Watts theorem, focusing on applications to Wedderburn structure theory and the Brauer group in:

Hyman Bass, The Morita Theorems, Lectures given at the University of Oregon, 1962. PDF

That paper uses (and, as far as I can tell, introduces) the phrase "Morita context" for a pair of bimodules $_A P_B$ and $_B Q_A$ and bimodule homomorphisms $P \otimes_B Q \to A$ and $Q \otimes_A P \to B$, not necessarily invertible but satisfying that the two natural maps $P \otimes_B Q \otimes_A P \rightrightarrows P$ agree, as do the two natural maps $Q \otimes_A P \otimes_B Q \rightrightarrows Q$.  I feel like I don't see the phrase "Morita context" very much any more, but perhaps this is due to a biased sample on my part.
The terms "Morita equivalence" and "Morita-invariant" appear (perhaps for the first time?) in

P.M. Cohn, Morita equivalence and duality, Queen Mary College Mathematics Notes, Queen Mary College, London, 1968 (lectures written 1966, republished with additional citations 1976). MR0258885

The notion of bicategory was introduced in:

J. Bénabou. Introduction to bicategories, part I. Reports of the Midwest Category Seminar, Lecture Notes in Mathematics 47, pages 1-77. Springer, 1967. MR0220789

He discusses in some detail the bicategory of Rings, Bimodules, and Intertwiners, and observes that it receives a functor (named "modulation" by Tang, Weinstein, and Zhu, 2007) from the category of Rings and Homomorphisms.
The canonical reference for Morita theory is another work by Bass:

Hyman Bass, Algebraic K-Theory, Benjamin, 1968. MR0249491

Notably, "Chapter II: Categories of Modules and their Equivalences".  This book never uses the language of bicategories, which should not surprise: the book came out in 1968, which means it went to the publisher by 1967, which means that Bass at the time of writing didn't have Bénabou's paper.  The book does, however, make the following observation: There is a (strict 1-) category whose objects are Rings and whose morphisms are Isomorphism Classes Of Bimodules, and Morita equivalence (not a term used in the book) is isomoprhism in that category.  This almost answers Question 1, but not quite.
I then lose track of the citations, which seem to multiply quickly.  It's clear that all ingredients were available by the end of the 1960s and reasonably well known by the mid 1970s (e.g. Bunge, 1979 was doing quite sophisticated category theory).  Bounding from the other end, by Brouwer, 2003 the observation that Morita equivalence is best understood in terms of bicategories was routine.

REPLY [3 votes]: Let me venture an answer to (2), a rather recent source, but it's the oldest I have found:

The name "Morita bicategory" was introduced in January 2008 by Hellen Colman, in a talk at the Max Kelly Conference in Category Theory (Cape Town, 21–26 January, 2008): Lusternik-Schnirelmann theory for the Morita bicategory of Lie groupoids.
She subsequently abandonded the name, see this arXiv paper.<|endoftext|>
TITLE: Can Calabi-Yau manifolds have nonabelian discrete symmetry groups?
QUESTION [15 upvotes]: A particle physicist asked me the above question.  Let me try to make it more precise.  Suppose $M$ is a 3-dimensional Calabi-Yau manifold: that is, a compact Kähler manifold of complex dimension 3 whose holonomy group is contained in $\mathrm{SU}(3)$.  Let $\mathrm{Aut}(M)$ be its group of holomomorphic metric-preserving diffeomorphisms.  What can this group be like?  In particular: 
1) which nonabelian discrete groups can $\mathrm{Aut}(M)$ contain?
or if that's unmanageable:
2) which nonabelian discrete groups can appear as the group of connected components of $\mathrm{Aut}(M)$?
I believe he is particularly curious as to whether we can get $\mathrm{PSL}(2,7)$ as the answer to either of these questions.

REPLY [8 votes]: If you work one dimension down, at the level of K3 surfaces, there's a very pretty classification of finite group actions preserving the holomorphic form due to Mukai.  In that classification, the simple group of order 168 is extremal.  Oguiso and Zhang have a nice article on the properties of K3 surfaces which admit such an action.  To be specific, you can get a K3 surface with this group action (the "Klein-Mukai surface") by taking a fourfold cover of the plane branched over Klein's quartic: $x^3 y + y^3 z + z^3 x + w^4=0$.  It would be fun to look at Calabi-Yau threefolds with a geometric relationship to the Klein-Mukai surface.<|endoftext|>
TITLE: Non-existence of a prime generating polynomial recurrence relation
QUESTION [6 upvotes]: Let $f\in \mathbb{Q} [x]$ be a polynomial, and $a_0 = a$ be an arbitrary integer. Let us define a sequence $\{a_n \} $ by the recurrence relationship : $$a_n = f(a_{n-1} ). $$ I want to show that $a_n $ cannot always be a prime number, with $\{a_n \}$ being pairwise distinct. I am pretty sure that this is a very well known fact, but I cannot easily find this.
Note : I posted this question also in SE Math. 
https://math.stackexchange.com/questions/1568513/non-existence-of-a-prime-generating-polynomial-recurrence-relation

REPLY [13 votes]: Let $a_0 = 2^{2^k} + 1$ for $k$ sufficiently large, and let
$$a_n = (a_{n-1} - 1)^2 + 1.$$
Then $a_n = 2^{2^{k+n}} + 1$, so this sequence can't always be prime regardless of the value of $k$ iff there are infinitely many composite Fermat numbers, and as far as I know this is wide open.<|endoftext|>
TITLE: Self-Intersecting Geodesics in Homogeneous Spaces
QUESTION [6 upvotes]: A result of Helgason's is that any self-intersecting geodesic in a Riemannian globally symmetric space is simple and closed. To what extent does this generalise to Riemannian naturally reductive homogeneous spaces with the canonical connection?

REPLY [6 votes]: Every geodesic loop on a compact homogeneous space  is a closed geodesic. Let c be such a loop  c(L)=c(0). Take (n-1) Killing vector fields $X_i$ whose value at p are a basis of $c'(0)^\perp$.  Then $X_i$ restricted to c is a Jacobi field and hence $<X_i(c(t),c'(t)>$ is linear. Since X has bounded length, $<X_i(c(t),c'(t)>=0$
and hence $c'(L)=c'(0)$. I think this may be also true if M is not compact.<|endoftext|>
TITLE: List of known Fourier Mukai partners?
QUESTION [20 upvotes]: I'm familiar with some examples of pairs of derived equivalent varieties, for example an abelian variety and its dual, a K3 surface and certain moduli schemes on it, or the Pfaffian-Grassmannian derived equivalence.
However, when I looked for other known examples, I could only find papers constructing single examples of Fourier-Mukai partners, and no comprehensive survey.
Can anyone provide references where I can find out more about known Fourier-Mukai partners, giving me a broad overview?

REPLY [9 votes]: There are actually several known Fourier Mukai partners. 

Standard flop/Atiyah flop. See Chapter 11 of Fourier-Mukai Transforms in Algebraic Geometry by Huybrechts. 
Mukai flops (Chapter 11 of Fourier-Mukai Transforms in Algebraic Geometry by Huybrechts), stratified Mukai flops https://arxiv.org/abs/1111.0688 and Grassmannian flops https://arxiv.org/abs/1206.0219.
Abouf flops https://arxiv.org/abs/1706.04417 and Ueda flops https://arxiv.org/abs/1812.10688. 
Any birational but non-isomorphic projective Calabi-Yau 3-folds. See http://www.tom-bridgeland.staff.shef.ac.uk/publications/pub6.pdf
Examples from Homological Projective Duality (HPD), e.g. intersection of two $G(2,5)$ and the intersection of the dual Grassmannian (https://arxiv.org/abs/1707.00534), the intersection of two spinor varieties of $Spin(10)$ and the intersection of their duals(https://arxiv.org/abs/1709.07736). Both of these pairs are non-birtional Calabi-Yau's. The general theorem about the derived equivalences and the kernel functors can refer to https://arxiv.org/abs/1804.00144 and https://arxiv.org/abs/1704.01050
Pair of Calabi-Yau 3 folds in $G_2$ Grassmannian. https://arxiv.org/pdf/1611.08386.pdf It should be noted that these CY3 are non-birational.

I think there are many examples of FM partners from birational geometry, moduli spaces, and HPD theory.<|endoftext|>
TITLE: Is the modularisation of a unitary fusion category always unitary?
QUESTION [8 upvotes]: Suppose $\mathcal{C}$ is a unitary ribbon fusion category. Also assume that its symmetric centre has trivial twist and trivial pivotal structure, i.e. is tannakian. Thus, the Müger/Bruguières modularisation/deequivariantisation exists.
Is the resulting modular fusion category unitary? Is the modularisation functor unitary (dagger)?
It feels to me like it should be obviously true, coming from the philosophy that the modularisation is a kind of generalised fibre functor. But I can't find a reference, nor can I write down the dagger structure.

REPLY [5 votes]: The answer is yes to both questions (see  Müger's paper). Müger's version of modularization is done in the unitary setting. The only new information that you need for full generality is that the unitary Drinfel'd center of a unitary fusion category is equal to the usual Drinfel'd center, (see [M]) 
More generally, a de-equivariantization of a unitary fusion category is again unitary (modularization corresponds to a specific kind of de-equivariantization). In fact, a $G$-de-equivariantization of a fusion category $\mathcal{C}$ is associated to a braided inclusion of Rep$(G)$ in $\mathcal{Z}(\mathcal{C})$. The de-equivariantization is the fusion category of left $\mathcal{O}_k(G)$-modules in $\mathcal{C}$. Now, if $\mathcal{C}$ is unitary, then the Drinfel'd center is a unitary modular category. Therefore, $\mathcal{O}_k(G)$ is a $Q$-system in $\mathcal{C}$. Thus, the de-equivariantization is unitary.<|endoftext|>
TITLE: Encyclopedia of properties of nonnegative matrices
QUESTION [11 upvotes]: I'd like to buy a book that contains more or less all known properties of elementwise nonnegative nonnegative matrices, i.e. matrices $A$ such that $a_{ij} \ge 0$ for all $1 \le i,j \le n$.

Chapter 8 in Matrix Analysis of Johnson and Horn is nice but far to be exhaustive. 
In Matrix Analysis of Bhatia and Matrix Computations of Golub and Van Loan there is not even a chapter dedicated to this class of matrices.

After a bit googling I found these two books:

It seems that Nonnegative Matrices in the Mathematical Sciences of Berman and Plemmons is exactly what I want. But it is more than 20 years old (1994), so I wonder if there is something more up-to-date.
Nonnegative Matrices and Applications looks quite nice too and is a little bit more recent (1997). However, it seems that it is less oriented on the properties of nonnegative matrices themselves rather than their applications.

So, I'm a bit puzzled and my question is the following:

If I want to buy 1 book collecting the most complete list of known properties of nonnegative matrices, which one should I get?

REPLY [6 votes]: In addition to the books you list, there is also the book Nonnegative Matrices by H. Minc (1974), but it is outdated and out of print.
Also, the book Combinatorial Matrix Theory by Brualdi and Ryser does a nice treatment of the combinatorial structure of nonnegative matrices (and complex matrices in general). 
IMHO, a new textbook on nonnegative matrices is long overdue (an entire monograph could be devoted to the nonnegative inverse eigenvalue problem alone).<|endoftext|>
TITLE: Exotic line arrangements
QUESTION [14 upvotes]: I would like to discuss the following problem. Hopefully, you will suggest to me some ideas and bibliography. 
At first I will provide some basic definitions to set up the notation.
Let us consider a line arrangement $\mathcal{A}=\left\lbrace l_{1},\ldots,l_{m}\right\rbrace$ in $\mathbb{CP}^{2}$ and let $M\left(\mathcal{A}\right)$ be its complement manifold, i.e., the space 
$M\left(\mathcal{A}\right)=\mathbb{CP}^{2}\setminus\bigcup_{k=1}^{m}l_{k}$
 obtained taking the complement of the union of the lines.
Regarded as a differentiable manifold, $M\left(\mathcal{A}\right)$ is diffeomorphic to the interior of a connected  compact manifold with boundary. As a consequence, by a standard result in low dimensional topology, there is a differentiable manifold $N$ which is homeomorphic but not diffeomorphic to 
$M\left(\mathcal{A}\right).$
Here is my question:
Let $\mathcal{A}$ and $\mathcal{B}$ be line arrangements in $\mathbb{CP}^{2}$ 
with complement manifolds $M\left(\mathcal{A}\right)$ and 
$M\left(\mathcal{B}\right).$ Is that true that if $M\left(\mathcal{A}\right)$ and $M\left(\mathcal{B}\right)$ are homeomorphic, then they are diffeomorphic, too?
Otherwise, exhibit an explicit example of two line arrangements $\mathcal{A}$ and $\mathcal{B}$ in $\mathbb{CP}^{2}$ with complement manifolds $M\left(\mathcal{A}\right)$ and 
$M\left(\mathcal{B}\right)$ which are homeomorphic but not diffeomorphic.

REPLY [2 votes]: As far as I know, that's still an open problem.  The only work I'm aware of that addresses the rigidity of the topology is on the level of homotopy type: Randell's Isotopy Theorem and Rybnikov's example showing the matroid does not determine the homotopy type.  But I see you know about these from your previous question (https://mathoverflow.net/users/53064/esaini582)<|endoftext|>
TITLE: Freiman-isomorphic sets
QUESTION [5 upvotes]: Haw can we prove that an arbitrary set $A$ of $n$ positive integers is 2-Freiman isomorphic to a subset of {$ 1,2,...,4^{n}$}  and $4^{n}$ cannot be improved to $2^{n}$?

REPLY [11 votes]: It is an open conjecture from a paper of Konyagin and myself that every $n$-element set of integers is Freiman-isomorphic to a subset of $[0,2^{n-2}]$. There are some counterexamples for small values of $n$, but it is believed that the conjecture is "essentially true"; if so, your $4^n$ actually can be improved to $2^n$ (and in fact, to $2^{n-2}+1$). No further improvements are possible: the set $\{0,1,2,4,\dotsc,2^{n-2}\}$ is not isomorphic to any shorter integer set. (If it is isomorphic to a set $\{a_0,a_1,\dotsc, a_{n-1}\}$, then $a_0+a_2=2a_1$, $a_0+a_3=2a_2$, ... , $a_0+a_{n-1}=2a_{n-2}$; assuming without loss of generality $a_0=0$, this results in $a_i=2^{i-1}a_1$, so that the diameter of $\{a_0,a_1,\dotsc, a_{n-1}\}$ is $|a_{n-1}-a_0|\ge 2^{n-2}$.)
The bound $8^n$ (somewhat weaker than you indicated) can be obtained as follows. 
It suffices to show that if $A\subseteq[0,l]$ is an $n$-element integer set not isomorphic to a set of diameter smaller than $l$, then $l<8^n$. Write $A=\{a_1,\dotsc,a_n\}$ and fix a prime $2l<p\le 4l$, so that $A$ is isomorphic to its image $A_p$ under the canonical homomorphism $\Bbb Z\to\Bbb Z/p\Bbb Z$.
Consider the $n$-dimensional integer lattice $\Lambda:=(a_1,\dotsc, a_n)\Bbb Z+p\Bbb Z^n$. It is easily seen that the determinant of this lattice is $p^{n-1}$; hence, by Minkowski's First Theorem, $\Lambda$ has a non-zero point in the $n$-dimensional cube $[-p^{1-1/n},p^{1-1/n}]^n$. Consequently, there exist integers $y_1,\dotsc,y_n\in[-p^{1-1/n},p^{1-1/n}]$, not all equal to $0$, such that for yet another integer $z$ we have $y_i\equiv za_i\ (i\in[1,n])$. It is immediately seen that $z\not\equiv 0\pmod p$; hence, the set $\{y_1,\dotsc,y_n\}$ is isomorphic to $A_p$, and therefore to $A$. The assumption that $A$ is not isomorphic to a set of diameter smaller than $l$ implies now $2p^{1-1/n}\ge l$. Combining this with $p\le 4l$ yields the desired bound $l<8^n$.
For $l$ large, the prime $p$ can be chosen to satisfy $p=(2+o(1))l$, leading to $l<4^{(1+o(1))n}$. Much better estimates (something like $2^{(1+o(1))n}$, I believe) can be obtained using the method of the aforementioned paper; see also Chapter 20 of the monograph Structural Additive Theory by David Grynkiewicz.<|endoftext|>
TITLE: Fricke Klein method for isotropic ternary quadratic forms
QUESTION [7 upvotes]: Preface: the most natural way to take one isotropic vector for an indefinite quadratic form and find others is to use stereographic projection. This gives a parametrization in the same $n$ variables as the 
quadratic form. After clearing denominators, there is also not much control over the gcd of the resulting integers. So, although finding an integer multiple of every primitive solution is guaranteed, we may not be entirely sure we have found all primitive solutions with entries up to some bound in absolute value.
There is a trick for indefinite ternary forms, which leads to a parametrization by two parameters, with considerable control of the gcd's.
Most of this appears in my answers to Isotropic ternary forms
Question: is it true that the primitive integer solutions to
$$ A(x^2 + y^2 + z^2) - B (yz+zx+xy) =0  $$
can all be parametrized by a finite number  of solutions as below, in shorthand $R_j U?$ The calculations are awfully convincing, but I have proved only a few.  In case anyone gets interested, i wrote out the proof for $A=2, B=113,$ about twenty four pages pdf.
Once we have integers $B > A > 0,$ a necessary and sufficient condition that the form be isotropic in $\mathbb Q,$ and therefore $\mathbb Z,$ is that both $B-A$ and $B+2A$ have integer expressions as $s^2 + 3 t^2.$  
There is an interesting alternative, method goes back to Fricke and Klein, gives a two variable parametrization, and can be adjusted to deal with GCD's.
There is a complete answer to this, finding all primitive solutions, meaning $\gcd(x,y,z) = 1.$
We begin with finding all primitive solutions to $y^2 - z x = 0.$ If $g = \gcd(x,z) > 1,$ then $g^2 | y^2$ and $g | y,$ so $g | \gcd(x,y,z).$ However, $\gcd(x,y,z) = 1.$ So,  $\gcd(x,z) = 1.$ Since $xz = y^2,$ either $x=u^2, z=v^2,$ or $x=-u^2,z=-v^2,$ in either case with $\gcd(u,v) = 1.$ That is, possibly by changing from $(x,y,z)$ to $(-x,-y,-z)$ so as to arrange $x \geq 0,$ all primitive solutions are $$ x = u^2, y = u v, z = v^2. $$
Next, the quadratic form is $X^T G X / 2,$ where 
$$
G =
\left(
\begin{array}{rrr}
2a & d & e \\
d & 2b & f \\
e &   f &2c
\end{array}
\right)
$$
and
$$
X =
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right)
$$
The quadratic form $y^2 - z x$ is  $X^T H X / 2,$ where 
$$
H =
\left(
\begin{array}{rrr}
0 & 0 & -1 \\
0 & 2 & 0 \\
-1 &   0 & 0
\end{array}
\right)
$$
It is a theorem in Fricke and Klein (1897), pages 507-508,
that, because the quadratic form has at least one integer solution, in which $(x,y,z)$ are not all zero, there exists a square matrix of integers $R$ and a nonzero integer $n$ such that
$$ R^T G R = n H. $$
As $R$ has an inverse, we can take $S$ to have integral entries and minimize positive $k$ in $$ RS = SR = k I.  $$
We already know that we can take all solutions of $y^2 - z x = 0$ as the column vector
$$
U =
\left(
\begin{array}{r}
u^2 \\
uv \\
v^2
\end{array}
\right)
$$
for relatively prime $(u,v).$ That is, $U^T H U = 0,$ and all solutions are a scalar multiple of $U.$
What happens if $X^T G X = 0,$ the "solutions" we want, with gcd one? Well, $R^T G R = n H,$ so $S^T R^T G R S = n S^T H S,$ so
$$ G = \frac{n}{k^2} S^T H S,  $$ and  $X^T G X = 0$ says
$$ X^T S^T H S X = 0. $$ We have already shown that there is some integer $w$ with 
$$ SX = w U.  $$
This gives us $RSX = w RU$ and $kX = w RU.$ Now, as $\gcd(x,y,z) = 1,$ there is a row vector $A = (\alpha,\beta,\gamma)$ with $AX = 1.$ This tells us $k = w ARU.$ As $ARU$ is some integer, $w | k,$ and the earlier $kX = w RU$ becomes
$$  X = \frac{1}{h} R U $$ for $h = \frac{k}{w} \in \mathbb Z.$ Furthermore, as $ RS = SR = k I,  $ we know $k | \det R,$ so $h | k$ tells us $h | \det R.$ One may leave it this way: list the divisors of $\det R,$ including $\det R$ itself. For each primitive pair $(u,v),$ produce the column vector $RU,$ which will be a solution but perhaps not primitive. Divide out by the gcd of the entries of $RU.$ All integer primitive solutions are given by 
$$ X = RU/ g_1,  $$ where $g_1$ is the gcd of the three entries of $RU.$ It is worth emphasizing that $g_1$ is a divisor of $\det R.$ Also, we get some explicit bounds, as $$ |X|^2 = \frac{1}{g_1^2} U^T R^T R U, $$ since
$R$ is nonsingular integer and $R^T R$ is symmetric positive definite. So, no matter what, we have a way to find all primitive solutions $X$ with some $|X| \leq \mbox{bound}$ by taking $|u|, |v|$ up to some other bound we can figure out.
$$  \bigcirc   \bigcirc   \bigcirc   \bigcirc   \bigcirc   \bigcirc   \bigcirc  $$
A more interesting alternative: for each divisor of $\det R,$ we may rewrite the eventual primitive solution with that gcd as a new recipe, $R_1 U$ for a new integer matrix $R_1$ that also solves $R_1^T G R_1 = n H.$  
The example I like to show is solving
$$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0,  $$ four "recipes,"
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
37 u^2 + 51 uv + 8 v^2 \\
8 u^2 -35 uv -6 v^2 \\
-6 u^2 +  23 uv + 37 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
32 u^2 + 61 uv + 18 v^2 \\
18 u^2 -25 uv -11 v^2 \\
-11 u^2 +  3 uv + 32 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
38 u^2 + 45 uv + 4 v^2 \\
4 u^2 -37 uv -3 v^2 \\
-3 u^2 +  31 uv + 38 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
29 u^2 + 63 uv + 22 v^2 \\
22 u^2 -19 uv -12 v^2 \\
-12 u^2 -5 uv + 29 v^2
\end{array}
\right)
$$
For all four recipes,
$$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2  \right)^2  $$
Since $u^2 + uv + v^2 \geq 3 u^2 / 4$ and $u^2 + uv + v^2 \geq 3 v^2 / 4,$ this gives us explicit bounds on the absolute values of $u,v$ that gives us all (primitive) solutions of $ 2(x^2 + y^2 + z^2) = 113 (yz+zx+xy)  $ with the absolute values of $x,y,z$ up to a desired bound.
Indeed, we were able to choose all four coefficient matrices with this pattern:
$$
R =
\left(
\begin{array}{ccc}
\alpha & \beta & \gamma \\
\gamma & - \beta + 2 \gamma & \alpha - \beta + \gamma \\
\alpha - \beta + \gamma  &   2 \alpha - \beta & \alpha
\end{array}
\right)
$$
The rows constitute a cycle of three neighboring, but not reduced, binary quadratic forms under the action of the matrix 
 $$
P =
\left(
\begin{array}{rr}
0 & 1 \\
-1 & -1
\end{array}
\right),
$$
where $P^3 = I.$
As soon as we write $X = RU$ we get the identity
$$ x^2 + y^2 + z^2 = \left( \alpha^2 + (\alpha - \beta + \gamma)^2 + \gamma^2 \right) \cdot \left( u^2 + uv + v^2  \right)^2  $$
In all four cases we simply discard occurrences when the resulting $x,y,z$ have a common factor. With the understanding that we negate all $x,y,z$ so that the entry with largest absolute value is positive, then sort so that $$ x \geq |y| \geq |z|, $$ here are the answers with maximum up to $1200$
jagy@phobeusjunior:~$ ./isotropy_binaries_combined 2 113 1200 | sort -n
             x      y     z          first line       u   v
             29     22    -12      < 29, 63, 22 >      1  0    
             32     18    -11      < 32, 61, 18 >      1  0    
             37      8     -6      < 37, 51, 8 >      1  0    
             38      4     -3      < 38, 45, 4 >      1  0    
            188    171    -86      < 37, 51, 8 >      1  2    
            211    144    -82      < 38, 45, 4 >      1  2    
            226    123    -76      < 32, 61, 18 >      1  2    
            243     94    -64      < 29, 63, 22 >      1  2    
            246     88    -61      < 38, 45, 4 >      2  1    
            258     59    -44      < 37, 51, 8 >      2  1    
            264     38    -29      < 29, 63, 22 >      2  1    
            268     11     -6      < 32, 61, 18 >      2  1    
            396    262   -151      < 37, 51, 8 >      1  3    
            432    209   -134      < 38, 45, 4 >      1  3    
            472    129    -94      < 29, 63, 22 >      3  1    
            489     76    -58      < 32, 61, 18 >      3  1    
            516    458   -233      < 38, 45, 4 >      2  3    
            526    447   -232      < 37, 51, 8 >      2  3    
            628    311   -198      < 38, 45, 4 >      3  2    
            656    262   -177      < 32, 61, 18 >      2  3    
            671    232   -162      < 37, 51, 8 >      3  2    
            692    183   -134      < 29, 63, 22 >      2  3    
            726     47    -32      < 32, 61, 18 >      3  2    
            727     36    -22      < 29, 63, 22 >      3  2    
            804    787   -382      < 32, 61, 18 >      1  5    
            894    688   -373      < 29, 63, 22 >      1  5    
            953    946   -456      < 38, 45, 4 >      3  4    
           1034    492   -317      < 37, 51, 8 >      1  5    
           1062    443   -296      < 29, 63, 22 >      5  1    
           1102    363   -256      < 38, 45, 4 >      1  5    
           1123    314   -228      < 32, 61, 18 >      5  1    
           1159   1046   -528      < 32, 61, 18 >      1  6    
           1179    118    -88      < 38, 45, 4 >      5  1    
           1188     19      2      < 37, 51, 8 >      5  1    
           1199   1002   -524      < 29, 63, 22 >      1  6
             x      y     z          first line       u   v

I should probably point out that, while it was quite easy (after writing the C++ programs) to make a list of primitive solutions to $ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0  $ with $x \geq |y| \geq |z|$ for $x \leq 1200,$ and just as easy to identify the four square matrices $R_1,R_2,R_3,R_4$ used above, it was quite a big job to prove that these really do give all (ordered) primitive solutions. I have a pdf of the whole business in detail, about twenty pages Latex. Oh: in the above, we may always take $u,v \geq 0.$ It is a reasonable conjecture that the problem $ A(x^2 + y^2 + z^2) - B(yz + zx + xy)=0,  $ with $\gcd(A,B)=0,$ $B > A > 0,$ and both $B-A$ and $B + 2A$ expressible in integers as $s^2 + 3 t^2,$ always works out with a finite number of such $R_i.$ No proof.

REPLY [2 votes]: The same formal record.  If we look for a parameterization not in 2 and in 3 option the problem can be solved quite simply.
For the equation.
$$aX^2+bY^2+cZ^2=dXY+eXZ+fYZ$$
If you know any solution $(x,y,z)$ of this equation.  Then the formula for the solution of the equation can be written immediately.
$$X=(dy+ez-ax)p^2+(fz-2by)ps+bxs^2+cxt^2+(fy-2cz)pt-fxst$$
$$Y=ayp^2+(ez-2ax)ps+(dx+fz-by)s^2+cyt^2-eypt+(ex-2cz)st$$
$$Z=azp^2-dzps+bzs^2+(ex+fy-cz)t^2+(dy-2ax)pt+(dx-2by)st$$
$p,s,t - $  any integers.
It is seen that such formulas can be written infinitely many.  If so like to use the well-known decision, it is better to write like this.  This will allow to solve the equation.
$$A(x^2+y^2+z^2)=B(xy+xz+yz)$$
It is easy to see that there are solutions for any $A$. Ask yourself the number $c$. And place into factors.
$$c^2+A=ab$$
Then the coefficient is set to $B$ us so.
$$B=a^2+b^2-2(a+b)c+3c^2$$
Finding all the factors of $c,a,b$ you can write a formula for the parameterization of the solution of this equation.
$$x=c(ab-c^2)(k^2+s^2)+((a+b)(a^2+b^2)-(4a^2+5ab+4b^2)c+7(a+b)c^2-$$
$$-5c^3)p^2-(a^2+b^2-2(a+b)c+3c^2)cks+(b(b^2-a^2)-c(a^2+3b^2)+$$
$$+(4a+5b)c^2-5c^3)pk+(a(a^2-b^2)-c(b^2+3a^2)+(5a+4b)c^2-5c^3)ps$$
$$y=(a-c)(ab-c^2)(p^2+s^2)+(b^3-(a+2b)bc+(a+3b)c^2-c^3)k^2-$$
$$-(a-c)(a^2+b^2-2(a+b)c+3c^2)ps+(-2ab^2+c(a+b)^2-2ac^2+c^3)ks+$$
$$+(b(b^2+a^2)-(a^2+4ab+3b^2)c+(2a+5b)c^2-c^3)kp$$
$$z=(b-c)(ab-c^2)(p^2+k^2)+(a^3-(b+2a)ac+(3a+b)c^2-c^3)s^2-$$
$$-(b-c)(a^2+b^2-2(a+b)c+3c^2)pk+(a(a^2+b^2)-(3a^2+4ab+b^2)c+$$
$$+(5a+2b)c^2-c^3)ps+(-2ba^2+c(a+b)^2-2bc^2+c^3)ks$$
The transition to 3 parameters $p,s,k - $  Allows not to use too much, but to write formulas.<|endoftext|>
TITLE: How strong is Cantor-Bernstein-Schröder?
QUESTION [21 upvotes]: There are several questions here on MO about the Cantor-Bernstein-Schröder ((C)BS) theorem, but I could not find answers to what arose to me recently.
Although I don't think I need to recall it here, the theorem states that if there are embeddings $i:A\hookrightarrow B$, $j:B\hookrightarrow A$ then there is a bijection between $A$ and $B$.
First question -
As explained in several answers to one of those questions, CBS is not constructively valid. Does validity of CBS actually imply excluded middle? If not - does CBS imply any nontrivial (say, not intuitionistically valid) propositional sentence?
Some motivation comes from Diaconescu's theorem that the Axiom of Choice implies excluded middle. Again, as mentioned in several of the answers to the above questions validity of CBS does not require choice, but this does not exclude the possibility that, say, the propositional part of some form of constructive set theory + CBS is strictly stronger than "mere" intuitionism. Btw, the Wikipedia article that I linked to for Diaconescu's theorem mentions that it is in a form of a Problem (Problem 2 on page 58) in Bishop's "Foundations of constructive analysis". What is said there literally is "Construct a surjection which is not onto", and the whole set of problems has an instruction saying that the not has to be understood in the sense of discussion at the end of Section 2. What is said in that discussion seems to indicate that one has to work in intuitionistic logic although I am not sure.
I mention all this to illustrate that there might be some subtle points involved. Anyway, here is my
Second question -
Is there a constructively valid statement that becomes equivalent to CBS under excluded middle?
Again, it might be not obvious what is precise sense of this, but let me give another piece of motivation. One might replace $B$ with $j(B)\subseteq A$, then CBS becomes equivalent to the following:

Given an injective self-map $e:A\hookrightarrow A$, for any $e(A)\subseteq A'\subseteq A$ there is a bijection between $A$ and $A'$.

Now recall that Lawvere's notion of the object of natural numbers (NNO) can be equivalently given as the initial algebra for the functor $A^+:=A\sqcup 1$. That is, the NNO is characterized up to isomorphism as an object $N$ equipped with a map $N^+\to N$ and initial among objects with such structure. By Lambek's lemma it then follows that this $N^+\to N$ is an isomorphism. It can be also shown that if $A^+\to A$ is mono then the induced map $N\to A$ is mono too.
In view of this and of the above reformulation of CBS it is natural to ask what happens if one replaces 1 with some other object $T$. Similarly to the above statement in bold, if there is a mono $A\sqcup T\rightarrowtail A$ then $A$ will contain a copy of the initial $\_\sqcup T$-algebra, in particular, a subobject $I\rightarrowtail A$ with $I\sqcup T$ isomorphic to $I$.
The latter somehow suggests a form of the above equivalent of CBS with complemented "remainder".  Still the formulation escapes me although I have very strong feeling there must be something precise behind it.
 And although at the first glance this looks like an uninspiring weakening, the example of the NNO shows it might be interesting: $0:1\to N$ is a complemented element more or less by definition, but this does not make the notion of NNO too weak.
I realize all this does not add much clarity to the second question, but then if it would be clear to me I would not need to ask :D
Slightly later
Inspired by the mentions of decidability in Andrej Bauer's answer, here is a sketch of one candidate: I find it quite plausible that constructively $A\sqcup A_1\sqcup A_2\cong A$ implies $A\sqcup A_1\cong A$. Is it so? In case it is, can one do better? That is, is there a still constructively valid statement obviously implying the above but not obviously equivalent to it? Somehow I feel there must also be a "less complemented" version...
...and later - Ingo Blechschmidt commented with a link to an n-category café entry which in turn had a link to "Theme and variations: Schroeder-Bernstein" at the Secret Blogging Seminar; seems to be related, although I cannot find answers to the above questions there.

REPLY [2 votes]: The paper Cantor-Bernstein implies Excluded Middle by Pierre Pradic and Chad E. Brown shows that Excluded Middle is equivalent to CBS in constructive set theory.
I don't understand Toposes very well, but I believe the same proof goes through in a Topos with NNO.<|endoftext|>
TITLE: Which journals publish experimental results in pure maths?
QUESTION [13 upvotes]: All pure mathematicians know that the goal is to produce insight, rather than to simply obtain results. However, it might sometimes be of value to disseminate largely empirical work. In the same spirit as this question, which journals publish experimental and computational work in pure maths? Experimental Mathematics is one obvious answer, but what others exist?

REPLY [2 votes]: Computational Methods in Science and Technology is a Polish physics-oriented journal, but also lists experimental mathematics as a field of interest in its ''Aims and Scope'' section.<|endoftext|>
TITLE: On Sampling rank $r$ matrices
QUESTION [11 upvotes]: Sample $n^2$ integers $a_{11},\dots,a_{nn}$ in $\{-d,\dots,-1,0,1\dots,d\}$ uniformly. 
What is the probability that the resulting matrix $[a_{ij}]$ has rank $r$?
Is there a nice parametrization of such matrices that helps us generate such a rank $r$ matrix quickly deterministically?
In general what is a good strategy?

REPLY [4 votes]: The case $r=n$ is considered in this paper by Martin and Wong.  They prove that for every $n \geq 2$ and every $\epsilon >0$, the probability that a random $n \times n$ matrix with entries from $\{-k, \dots, 0, \dots, k\}$ is singular is $\ll \frac{1}{k^{2-\epsilon}}$. See Lemma $1$.  The discussion following Lemma $1$ shows that this is tight for $n=2$, but not for $n >2$.  By a deep theorem of Katznelson, the true probability decays as $\frac{\log (k)}{k^n}$, which is not far from $\frac{1}{k^n}$ (the probability that a random matrix contains a row of zeros).<|endoftext|>
TITLE: Is the generated subalgebra of a subset of pairwise operator-commuting element in a JB-algebra associative?
QUESTION [5 upvotes]: In a Jordan algebra elements $a$ and $b$ are said to operator-commute, whenever $a \circ (b \circ x) = b \circ (a \circ x)$ for every other element $x$.  (That is: $T_aT_b = T_bT_a$, writing $T_x(y) = x \circ y$.) In a JB-algebra elements $a$ and $b$ operator-commute if and only if they generate an associative subalgebra.  (See e.g. p.44 "Jordan operator algebras" by Hanche-Olsen and Størmer.)  Does this generalise from pairs to arbitrary subsets?

Question: Assume $A$ is a JB-algebra.  $S \subseteq A$ is a subset of pairwise operator-commuting elements.  Is the algebra generated by $S$ in $A$ associative?

REPLY [2 votes]: Yes, this is true. I couldn't find any proof of the statement you quote in the article, and even after emailing the authors I didn't get any wiser, so I decided to work out the details myself, see my paper Commutativity in Jordan Operator Algebras.
My main result is that if $a$ and $b$ operator commute then they indeed generate an associative JB-algebra of mutually operator commuting elements. In particular, $a^2$,  $b^2$ and $a*b$ also commute with each other and with $a$ and $b$.
Now suppose $S$ contains more than two elements. Consider $J(S)$ the Jordan algebra generated by $S$ which consists of "polynomials" in the elements of $S$. For such an element $p(s_1,s_2,\ldots,s_n)$ we can always decompose the operator $T_p$ into simple terms $T_{s_i}$, $T_{s_i*s_j}$ and $T_{s_i^2}$ by repeatedly applying the "normalisation equation" (Eq. 1 in my paper). By the two-element case we know that these all operator commute, hence any two elements of $J(S)$ operator commute. Now we just need to close $J(S)$ in the norm to get the JB-algebra generated by $S$. As the Jordan product is continuous, this algebra also consists of operator commuting elements.<|endoftext|>
TITLE: An apparent equivalence of the category of affine schemes over $S$ and the category of quasi-coherent $\mathcal{O}_S$-algebras
QUESTION [6 upvotes]: I had asked something very similar before on math.se (deleted now) but unfortunately it hadn't received a lot of attention. I decided to re-ask here.
Let $S$ be a fixed scheme. Is the following true?

Theorem(?): The category of affine schemes over $S$ is contra-equivalent to the category of quasi-coherent $\mathcal{O}_S$-algebras

Here the category of affine scheme over $S$ is one whose objects are arrows $\operatorname{Spec} A \to S$ where $A$ ranges over all possible rings and whose morphisms are morphisms of schemes above $S$. Here's the construction:
Let $\varphi: \mathcal{O}_S  \to \mathcal{F}$ be an $\mathcal{O}_S$-algebra.

Let $X$ be the set of all prime ideal sub-sheafs $\mathcal{P} \subset \mathcal{F}$. That is $\mathcal{P}(U)$ is either a prime or a unit on every open set $U \subset S$.
Define a topology with open sets of the form $V_{U,s} = \{\mathcal{P} \in X: s \notin \mathcal{P}(U)\}$ for open sets $U \subset S$ and $s \in \mathcal{F}(U)$. 
Define the structure sheaf by localizing $\mathcal{O}_X  (V_{U,s})=\mathcal{F}(U)_s$.
The morphism $X \to S$ that sends a prime sheaf to its pullback by $\varphi$ can be seen to satisfy the requirements. 

The inverse functor is the obvious one that sends structure sheafs to their pushforwards.
Recently I found that this equivalence supports a very slick argument that the intersection of affines are affine - which I very much suspect is not true. Here is the (probably false) argument:

Let $\operatorname{Spec} A \hookrightarrow S$ and $\operatorname{Spec} B \hookrightarrow S$ be affine subschemes. Being affine schemes over $S$ they can be considered as $\mathcal{O}_S$-algebras. Their intersection is the pullback which corresponds to their tensor product as $\mathcal{O}_S$-algebras. This gives another $\mathcal{O}_S$-algebra i.e. an affine scheme over $S$. We're done.

What are the flaws in the argument/theorem?

REPLY [15 votes]: What is true is that there is an antiequivalence between the category of schemes affine over $S$ (that is $S$-schemes for which the preimage of an open affine of $S$ is an open affine) and quasi-coherent $\mathcal{O}_S$-algebras. 
The anti-equivalence is realized by the pushforward of the structure sheaf and the relative spectrum (see Exercise 5.17 in Hartshorne's "Algebraic Geometry"). This is of course not the result you hoped for, but maybe it is still interesting.<|endoftext|>
TITLE: Mazur's Galois Deformations paper for non-residually irreducible case
QUESTION [5 upvotes]: In Barry Mazur's paper introducing Galois deformations, he hints at having a general theory for representations which are not residually Schur, but with more complicated statements. Does anyone know where/if this got written up with a similar level of detail as in Mazur's paper? I'm particularly interested in applications to Faltings-Serre.

REPLY [9 votes]: I don't know where this got written, but it's certainly well-known. Here's how it works. There are usually problems with deforming objects that have automorphisms because in many cases the corresponding functors are "obviously" not representable (Gabber produces a fine counterexample to an overoptimistic attempt to make this statement precise in Katz-Mazur, in a situation where an object and all of its deformations have precisely two automorphisms, but let's not get into this). 
So the way to fix this is to rigidify everything. You have your object (for example a mod p Galois representation) and then you add some extra structure to make it rigid (for example you don't just think of it as an abstract map to some $Aut(V)$, you choose a basis for $V$ and think of it as a map to $GL(n,k)$ and you lift the actual matrices to elements of $GL(n,A)$ -- a "framing"). Once you've killed the automorphisms, the general machinery kicks in and you get representable functors. All the dimension calculations in Mazur's paper are now off by some factor because you've changed the problem slightly, so you go through the proofs and find that your tangent spaces etc are all bigger by a factor of something like $n^2-1$. 
Because you have fixed this extra data, and you're deforming it, your universal object comes with this extra data too (in this case an action of the formal group $PGL(n)$ I guess, or maybe $GL(n)$, it all depends on how rigid you made everything) and so in particular your answer comes with an action of a group on it. The "correct" deformation ring, the deformation of the object you wanted to deform, is just the quotient of your rigidified deformation by this group -- well -- in the category of affine schemes it is, in the category of rings it's something like the invariants. But somehow that's the problem. The thing is, quotients by random group actions don't exist (or more precisely don't give the right answer) in the category of affine schemes, and that's because the actual quotient representing the answer isn't a ring, it's a stack.
So now you either go and read the stacks project, or you take a more pragmatic approach and just carry around the universal "framed" deformation ring and the group action, and you make no attempt to form the scheme-theoretic quotient. In the category of rings this corresponds to taking invariants by the group action -- but the invariants are in some sense the wrong answer -- the right answer is the framed ring with its group action up to some notion of equivalence which boils down to "gives the same stacky quotient".
If the original problem was rigid enough, e.g. irreducible mod $p$ rep, then the group acts freely on the spec of the ring and the quotient is just spec of the invariants and you recover Mazur's deformation ring.
So why didn't Mazur set things up in this generality in the first place? Because, even before Wiles, Mazur knew that his deformation rings should be related to Hecke algebras. If you deform the way Mazur does, with his conditions, then Mazur could compute the Krull dimensions (at least conjecturally) of the deformation rings, and in the 2-d case he realised that there should be a map to a Hida Hecke algebra (in the ordinary case where the det isn't fixed) and already in Mazur and Tilouine (historically before Wiles) they conjectured that this should be an isomorphism -- a result now famously known as an "$R=T$ theorem" (Mazur and Tilouine had no idea how to prove their conjecture of course, this was Wiles' insight). If Mazur had rigidified everything then his deformation rings would have had too large a Krull dimension and the conjecture would have looked far uglier -- indeed it's not immediately clear to me how to make it -- one would have to tensor the Hecke algebra over the base with the affine coordinate ring of formal $PGL(2)$ I guess, and then there would be some fiddling around. Mazur avoided all this with his approach but irreduciblility/Schur got built in. So to a certain extent it was a historical coincidence that Mazur set things up as they ended up. But I'm sure that even in the 80s Mazur would have known everything I've written here.
Nowadays people do frame things as a matter of course, but because other aspects of the theory have moved on in leaps and bounds someone who just wants to know about this part of the story is faced with the prospect of unravelling the framing stuff from all the other new ideas. It would probably make a good Masters project to unravel this stuff!<|endoftext|>
TITLE: Applications of Representation Theory in Combinatorics
QUESTION [21 upvotes]: What are the examples of interesting combinatorial identities (e.g. bijection between two sets of combinatorial objects) that can be proved using representation theory, or has some representation theoretic interpretation. Please also mention whether a purely combinatorial proof is known in each case.

REPLY [3 votes]: Here's an example that involves a group other than the symmetric group.
Let ${\mathbb F}_{q^n}$ denote the finite field with $q^n$ elements.  Since ${\mathbb F}_{q^n}$ is a vector space of dimension $n$ over ${\mathbb F}_q$, we may regard a generator of the multiplicative group ${\mathbb F}^*_{q^n}$ as an element of $GL_n({\mathbb F}_q)$; define a Singer cycle to be such a generator.  Also define a reflection to be a nontrivial element of $GL_n({\mathbb F}_q)$ whose fixed points form a hyperplane.

Theorem (Lewis–Reiner–Stanton). The number of factorizations of a Singer cycle into a product of $n$ reflections is $(q^n-1)^{n-1}$.

The only known proof uses the character theory of $GL_n({\mathbb F}_q)$ in an essential way.
Note that the above theorem is analogous to the classical result that the number of factorizations of an $n$-cycle in ${\mathfrak S}_n$ into a product of $n-1$ transpositions is $n^{n-2}$.<|endoftext|>
TITLE: Non-congruence normal subgroups of $SL_2(\mathbb{Z}[1/2])$
QUESTION [14 upvotes]: Let $G=SL_2(\mathbb{Z}[1/2])$, i.e., the modular group (if you wish) over the ring $\mathbb{Z}[1/2]$ consisting of rationals whose denominators are powers of $2$. Unlike $SL_2(\mathbb{R})$, $G$ is neither simple nor almost-simple: for example, it is possible to define congruence subgroups (for odd modulus $N$).
Is there a non-trivial (i.e., neither $\{I\}$ nor $\{\pm I\}$) normal subgroup of $G$ whose intersection with the unipotent subgroup $U = \left(\begin{matrix} 1 & * \\ 0 & 1\end{matrix}\right)$ is trivial? 
(Of course, such a subgroup would have to be of infinite index; does $G$ have non-trivial normal subgroups of infinite index?)

REPLY [6 votes]: More generally the Kazhdan-Margulis normal subgroup theorem (stating that non-central normal subgroups have finite index) applies to every irreducible lattice in a product of connected semisimple groups over real and $p$-adics of total rank $\ge 2$.
Here the ambiant group is $\mathrm{SL}_2(\mathbf{R})\times\mathrm{SL}_2(\mathbf{Q}_2)$, which has (total) rank 2.
(On the other hand this gives no information about finite index subgroups such as congruence subgroup property, but this is disjoint from the question.)<|endoftext|>
TITLE: Open Problems for Undergraduates
QUESTION [12 upvotes]: Suppose:

I am a 'problem-solver' rather than a 'theory-builder'
I am an undergraduate student
I have a passion for solving mathematical problems
The homework I get is not satisfying (in the sense that the problems are computing-problems rather than problems that require creative thinking), and I get far too little homework


Where can I find interesting problems (that require creative thinking) if I want to have fun solving mathematical problems and to practice problem-solving? Are there lists/books of such problems?

Furthermore, suppose I want to know how it is to do research.

Are there lists of the kind "open problems which can be understood by undergraduates". I guess these open problems should be in the fields of "discrete mathematics/combinatorics" and "graph theory".

I only found: http://dimacs.rutgers.edu/~hochberg/undopen/

REPLY [4 votes]: Richard Guy compiled a list of open problems in combinatorial game theory, available at http://library.msri.org/books/Book29/files/unsolved.pdf . His book "Unsolved problems in number theory" also contains parts which are more combinatorial in nature. In the realm of Davenport's constant there are many open problems, some of which are probably non-trivial but doable.

REPLY [4 votes]: There's a big list of open problems at:
Open Problem Garden
and a smaller list at:
Unsolved Problems and Rewards<|endoftext|>
TITLE: What is an Elementary "Homotopy, Model" Topos?
QUESTION [11 upvotes]: Context: 
Def (Rezk): A (Grothendieck) homotopy topos is a homotopy left exact Bousfield localization of the model category of simplicial presheaves sPsh(C) on a small simplicial category C.
Thm (Rezk): A presentable model category E is a (Grothendieck) homotopy topos iff it has descent.
Note: A (Grothendieck) topos is a presentable elementary topos.
Questions
Is there some homotopical definition of an elementary topos?
Are there some references for (approaches, attacks...) Elementary "homotopy" topos?
Motivation: Generalize Elementary "homotopy" topos to Elementary higher topos in analogy to Higher topos theory (à la Lurie, Rezk).
References:
Sheaves in geometry and logic, an intro to topos theory. Maclane and Moerdijk.
Toposes and homotopy toposes, Rezk.

REPLY [9 votes]: Since the time when Denis referred in the comments to the relevant nLab page, there has been a new proposal written up by Mike Shulman there:
An elementary $(\infty,1)$-topos is an $(\infty,1)$-category $\mathbf{E}$ such that

$\mathbf{E}$ has finite (∞,1)-limits and colimits
$\mathbf{E}$ is locally cartesian closed
There exists a subobject classifier (an object classifier that classifies the collection of all monomorphisms in an (∞,1)-category)
For any morphism $f:Y\to X$ in $\mathbf{E}$, there exists an object classifier in $\mathbf{E}$ classifying a class of morphisms that (1) includes $f$ and (2) is closed under fiberwise finite limits and colimits, composition (i.e. dependent sums), and dependent products.

For discussion see at the n-Category Café here.<|endoftext|>
TITLE: $R$ is isomorphic to $R[X,Y]$, but not to $R[X]$
QUESTION [74 upvotes]: Is there a commutative ring $R$ with $R \cong R[X,Y]$ and $R \not\cong R[X]$?
This is a ring-theoretic analog of my previous question about abelian groups: In fact, in  any algebraic category we may ask if $A \cong A + Z + Z \Longrightarrow  A \cong A+Z$ holds, where $Z$ is the free structure on one generator, and $+$ is the coproduct.
The question is similar to the Zariski Cancellation Problem from affine algebraic geometry, but not identical with it. As for the question, we may also work in the category of commutative $k$-algebras for some field $k$. Notice that $0$ is the only noetherian commutative ring with $R \cong R[X,Y]$, so that examples will be non-noetherian.

REPLY [56 votes]: The answer to this quite beautiful question is that there does exist a commutative ring $R$ with $R\cong R[X,Y]$ but $R\not\cong R[X]$.
Let $F$ be a field, and take
$$
R=F[x_i,y_i,r_i\ (i\geq 0)]
$$
subject to the relations
$$
\forall\ i\geq 0,\ r_i=x_i y_i(x_i+y_i^2)(x_i+y_i^3)(x_i+y_{i+1}^4)r_{i+1}.
$$
First, note that the relations allow us to remove $r_0$ (or more generally, any finite number of the $r_i$) from the list of generators for $R$, and hence
$$
R=(F[x_i,y_i,r_i\ (i\geq 1)])[x_0,y_0]\cong R[X,Y]
$$
where the isomorphism comes from the fact that the relations are unchanged after shifting all of the indices.
Second, by direct inspection we see that if $s$ and $t$ are irreducible factors of $r_0$ such that $s+t^2$ and $s+t^3$ are also irreducible factors of $r_0$, then $s=x_i$ and $t=y_i$ for some $i\geq 0$.  (Note that if we changed the relations by dropping the factor $x_i+y_i^3$, we would only have this fact up to associates.  By having the two irreducibles $x_i+y_i^2$ and $x_i+y_i^3$ simultaneously, we don't need to worry about unit multiples.  There might be a more clever way to deal with this minor issue.  One could also work over $\mathbb{F}_2$ to avoid unnecessary units.)
Third, the relations are homogeneous in the $r$-letters.  I claim that this will show that an element of $R$ has infinitely many non-associate irreducible factors if and only if it belongs to the ideal $I=(r_0,r_1,r_2,\ldots)$.  Let $a\in R$ be arbitrary.  After using the relations, we may write $a$ as a polynomial in $r_m$ with coefficients in $F[x_0,y_0,\ldots, x_m,y_m]$ (for $m$ sufficiently large).  Note that if we use the relations in any way (such as by increasing $m$ further), the $r$-degree doesn't change, and the constant term of the polynomial also remains unchanged.  However, it is easy to see that if we factor $a$ into irreducible factors, the number of those factors is bounded above by the $r$-degree of $a$ plus the number of irreducible factors of the constant term (which are both invariants of the relations).  This number is infinite if and only if the constant term is zero (i.e., $a\in I$), and of course conversely if $a\in I$ then it has infinitely many non-associate irrreducible factors.
Fourth, assume by way of contradiction there is an isomorphism $\varphi:R[z]\to R$.  Since the set of elements with infinitely many non-associate irreducible factors is an invariant of isomorphisms, and since the third fact above also holds for $R[z]$, we see that the preimage of $r_0$ under $\varphi$ must belong to the ideal generated by $(r_0,r_1,r_2,\ldots)$ in $R[z]$.  Thus, there is some element $b\in R[z]$ and some $k\geq 0$ so that $\varphi(r_k b)=r_0$.  Thus $\varphi(x_k),\varphi(y_k),\varphi(x_k)+\varphi(y_k)^2,\varphi(x_k)+\varphi(y_k)^3$ are all irreducible factors of $r_0$, and so by the second fact above we must have $\varphi(x_k)=x_{\ell}$ and $\varphi(y_k)=y_{\ell}$ for some $\ell\geq 0$.  The same result also holds for $\varphi(x_{k+1}),\varphi(y_{k+1})$, but since $\varphi(x_k)+\varphi(y_{k+1})^4=x_{\ell}+\varphi(y_{k+1})^{4}$ is also an irreducible factor of $r_0$, we must have $\varphi(x_{k+1})=x_{\ell+1}$ and $\varphi(y_{k+1})=y_{\ell+1}$.  Repeating the argument, we have $\varphi(x_{k+i})=x_{\ell+i}$ and $\varphi(y_{k+i})=y_{\ell+i}$ for all $i\geq 0$.
Let $J=(x_k,y_k,x_{k+1},y_{k+1},\ldots)$ be the ideal of $R[z]$ generated by the "tail" end of the $x$ and $y$ letters.  Then $\varphi(J)=(x_{\ell},y_{\ell},x_{\ell+1},y_{\ell+1},\ldots)$.  Thus, the isomorphism $\varphi$ induces an isomorphism
$$
R[z]/J\cong R/\varphi(J).
$$
The factor ring on the left is isomorphic to $F[z,x_0,y_0,x_1,y_1,\ldots, x_{k-1},y_{k-1}]$ which has odd dimension.  The factor ring on the right is isomorphic to $F[x_0,y_0,\ldots, x_{\ell-1},y_{\ell-1}]$ which has even dimension.  This gives us the necessary contradiction.
An easy modification of this construction will also show that for each modulus $m\geq 2$, there exists a ring $R_m$ such that $R_m[X_1,X_2,\ldots, X_p]\cong R_m[X_1,X_2,\ldots, X_q]$ if and only if $p\equiv q\pmod{m}$.
Edited to add: Taking the same ring but using non-commuting letters should answer the same question for noncommutative rings.<|endoftext|>
TITLE: How undecidable is the spectral gap?
QUESTION [47 upvotes]: Nature just published a paper by Cubitt, Perez-Garcia and Wolf titled Undecidability of the Spectral Gap, there is an extended version on arxiv which is 146 pages long. Here is from the abstract:"Many challenging open problems, such as the Haldane conjecture, the question of the existence of gapped topological spin liquid phases, and the Yang–Mills gap conjecture, concern spectral gaps. These and other problems are particular cases of the general spectral gap problem: given the Hamiltonian of a quantum many-body system, is it gapped or gapless? Here we prove that this is an undecidable problem. Specifically, we construct families of quantum spin systems on a two-dimensional lattice with translationally invariant, nearest-neighbour interactions, for which the spectral gap problem is undecidable".
I am curious about the undecidable part. The abstract says "our result implies that there exists no algorithm to determine whether an arbitrary model is gapped or gapless, and that there exist models for which the presence or absence of a spectral gap is independent of the axioms of mathematics". "Axioms of mathematics" is kind of vague, so in the extended version they phrase it in a Gödelian manner:"Our results imply that for any consistent, recursive axiomatisation of mathematics, there exist specific Hamiltonians for which the presence or absence of a spectral gap is independent of the axioms". But still, axiomatization of which mathematics, how much mathematics do they need to construct their Hamiltonians. Is it real analysis? ZF? ZFC? I can't figure it out even from their theorem statements.

Is this a mathematical proof or something at the "physical level of rigor"? If so, does it produce "concrete" undecidable statements, or are these Hamiltonians as obscure as "I am unprovable"? Does it represent a new way of proving independence results compared to forcing, etc.? In other words, is it an advance on Gödel sentences and the continuum hypothesis? 

EDIT: Cubitt gave an interview where he commented on the nature of the result informally:"It's possible for particular cases of a problem to be solvable even when the general problem is undecidable, so someone may yet win the coveted $1m prize... The reason this problem is impossible to solve in general is because models at this level exhibit extremely bizarre behaviour that essentially defeats any attempt to analyse them... For example, our results show that adding even a single particle to a lump of matter, however large, could in principle dramatically change its properties".

REPLY [20 votes]: I interpret your question to be asking about the transition from
computable undecidability to Gödelian or logical
undecidability, and furthermore about the extent to which this
logical undecidability might depend on which axioms of mathematics
we have adopted.
The answer is that one may quite generally deduce that there are
concrete instance of logical undecidability, whenever one has a
computably undecidable problem, no matter what theory you have
adopted as your axioms of mathematics, whether it is PA or ZFC or
ZFC plus large cardinals or what have you. The two kinds of
undecidability — computable undecidability and logical
undecidability — are tightly intertwined, and every
computable undecidable problem is saturated with instances of
logical independence with respect to every (computably
axiomatizable, sound) theory.
To see this, suppose that $A$ is a computably undecidable decision
problem, such as the halting problem or the tiling problem or,
now, the spectral gap problem, and suppose that $T$ is your
favorite foundational theory, the axioms of mathematics. We
require only that $T$ has a computably enumerable list of axioms,
that $T$ is strong enough to express the decision problem $A$, and
that $T$ is sufficiently sound.
Since $A$ is computably undecidable, there is no computable
procedure to determine on input $n$ whether or not $n\in A$.
Consider the algorithm that on input $n$ searches for a proof from
$T$ that $n\in A$ or a proof from $T$ that $n\notin A$. Since the
membership problem for $A$ was computably undecidable, this
algorithm cannot correctly settle all instances of the problem,
since otherwise it would be a computable decision procedure. Since
$T$ is sound, however, the algorithm is correct about the
instances of the problem that it does settle. So there must be a
specific instance of the problem $n$ (in fact, infinitely many
such $n$) for which $T$ does not prove $n\in A$ and $T$ does not
prove $n\notin A$. Thus, for this specific $n$ and this specific
$T$, the question of whether $n\in A$ is independent of $T$.
This argument works even as you strengthen your axioms, and so it
doesn't matter whether you use PA or ZFC or ZFC plus large
cardinals or what have you. A stronger theory may settle some of
the concrete instances that were not settled by the weaker
theories, but every theory will admit concrete instances that it
doesn't settle, for otherwise the proof-searching algorithm would
be a computational procedure deciding $A$, contrary to the
assumption that $A$ is undecidable.
See also John Pardon's question, Are the two meanings of “undecidable” related?<|endoftext|>
TITLE: Exotic smooth structures on Lie groups?
QUESTION [17 upvotes]: If a topological group $G$ is also a topological manifold, it is well-known (Hilbert's 5th Probelm) that there is a unique analytic structure making it a Lie group.
However, for a compact Lie group $G$, do we know if the underlying topological manifold supports any other exotic smooth structures (necessarily not a Lie group)?
Even a more specific example: Up to diffeomorphism, we have $SO(8)=SO(7)\times S^7$. If we replace the smooth structure on $S^7$ by an exotic one, do we get an exotic smooth structure on $SO(8)$?
Thank you!

REPLY [10 votes]: According to the introduction to the following paper of Farrell and Jones, if $n>4$ and $\Sigma^n$ is any exotic homotopy sphere, then $T^n\#\Sigma^n$ is not diffeomorphic to $T^n$. So lots of higher-dimensional tori have exotic structures. Farrell and Jones cite Section 15A of Wall's book on surgery for this result, but I couldn't pull it easily out of there. In any case, here's the paper reference: 
Farrell, F. T.; Jones, L. E.
Examples of expanding endomorphisms on exotic tori. 
Invent. Math. 45 (1978), no. 2, 175–179.<|endoftext|>
TITLE: Fast Upper Triangular Matrix Exponentiation
QUESTION [8 upvotes]: Let $Q_n$ be a $n\times n$ matrix with $Q_n=\begin{pmatrix} -\lambda_1-\mu_1 & \lambda_1 & 0 & \cdots\\ 0 & -\lambda_2-\mu_2 & \lambda_2 & \cdots\\ \vdots & \vdots & \vdots& \vdots\end{pmatrix}.$ In other words the diagonal of $Q$ is $\{-\lambda_i-\mu_i\}_{i=1}^n$ and superdiagonal is $\{\lambda_i\}_{i=1}^{n-1}$. Also, $\lambda_n=0$. 
Letting $\mu=(\mu_1,\cdots,\mu_n)^T$, I need a fast way of calculating:
$$f(t):=p\exp(Qt)\mu,$$
where $t$ is a scalar and $p=(1,0,\cdots,0)$. I know there are a lot of matrix exponentiation algorithms but I'm hoping for one that could exploit the nice structure here. I would be happy with approximations, so long as there is good control of the maximum error. I was thinking of maybe trying to find a Jordan decomposition $Q=A+N$, where $A$ is diagonalizable and $N$ is nilpotent but this seems difficult.
Generally speaking, $Q_n$ will be diagonalizable as long as the diagonal entries are all distinct. However, I'm worried that when two values are close, there's a chance of high error.
Motivation: $f(t)$ is the density of a Coxian PhaseType distribution and for doing MLE it would be really nice to have a fast way of calculating $f(t)$.

REPLY [4 votes]: You probably want the Schur-Parlett method for computing matrix functions. It is a method to compute a generic function of a triangular matrix. Essentially, you apply the function to its diagonal elements and then use a recursion (derived from the identity $f(A)A=Af(A)$) to reconstruct the elements in positions $(i,i+1)$, then $(i,i+2)$ and so on, one superdiagonal at a time.
Trouble may ensue when two diagonal elements are close, as you correctly identify, but there are techniques to work around this issue.
See Chapters 9 and 10 of Higham's Functions of Matrices for a thorough treatment.

REPLY [4 votes]: When I teach the exponential of matrices, I tell the students that the converging series is not a practical tool for calculation. It is way better to solve the differential equation. This turns out to be true here. Say $T$ is upper triangular and $M$ denotes $\exp T$. For the sake of simplicity, set $t_i=t_{ii}$Then one finds
$$m_{ii}=e^{t_{i}},\qquad m_{i-1,i}=t_{i-1,i}\frac{e^{t_{i}}-e^{t_{i-1}}}{t_{i}-t_{i-1}}$$
and
$$m_{i-2,i}=\frac{t_{i-2,i}t_{i-1,i}}{t_{i}-t_{i-2}}\left(\frac{e^{t_{i}}-e^{t_{i-2}}}{t_{i}-t_{i-2}}-\frac{e^{t_{i-1}}-e^{t_{i-2}}}{t_{i-1}-t_{i-2}}\right)$$
and so on. All the entries $m_{ij}$ are obtained by successive finite differences. As usual, when an argument is repeated, the finite difference is a derivative.

REPLY [3 votes]: The exponential $e^{Q}$ of any $n\times n$ upper triangular matrix $Q$ can be computed efficiently by solving a set of $n$ first-order differential equations, $u_{i}'(t)=\sum_{j}Q_{ij}u_j(t)$; these $n$ equations can be solved one by one: solve first for $i=n$ and then work back to $i=n-1,n-2,\ldots 1$; at each step you have a first-order linear ODE of the form $y'(t)+ay(t)+b(t)=0$, that can be readily solved with the method of integrating factors. The $j$-th column of $e^{Q}$ is the solution $u(1)$ with $u_i(0)=\delta_{ij}$.<|endoftext|>
TITLE: A conjecture of Cheeger about intersection cohomology and $L^2$- cohomology
QUESTION [7 upvotes]: Let $X$ be a projective variety and let $D$ be a simple normal
  crossings divisor on $X$
Does $$IH^*(X;\mathbb C)\cong H_{(2)}^*(X\setminus D;\mathbb C)$$ hold
  true for each Kähler metric on $X\setminus D$?  Is there any
  counterexample?
What about the Fubini-Study metric on $X\setminus D$? (This is a
  conjecture of Cheeger) What about the case when the Kähler metric on
  $X\setminus D$ has a conic model?, a cusp model(Is OK), or a
  combination of these two models?

REPLY [6 votes]: I'm not an expert, but you don't seem to be getting any answers.
First of all, I would be surprised if holds for any Kähler metric (e.g. for one with really bad singularities along $D$) but I don't have a counterexample*. Regarding the case of Fubini-Study metric, for isolated singularities, I believe it was
settled positively by Ohsawa: "Cheeger-Goreski [Goresky]-MacPherson's conjecture for the varieties with isolated singularities."  Math. Z. 206 (1991).  I don't know the status in general. I believe that there was a gap in a later paper by author on this.
There has been a lot of work with other metrics. For example, you can look at work on Zucker's conjecture, which was proved by Looijenga and Saper-Stern. Also for Poincaré type metrics, allowing coefficients in a variation of Hodge structure, there is the work by Cattani-Kaplan-Schmid and Kashiwara-Kawai. This provides the analytic basis for Saito's theory of Hodge modules.
*(added later) Here's a counterexample. Choose a diffeomorphism $f$ between $\mathbb{C}$ and the disk $D$. Pulling back the Poincaré metric gives a Kähler metric such that $\dim H^1_{(2)}(\mathbb{P}-\{\infty\})=\infty$ because $f^*(z^ndz)$ gives an infinite family of harmonic $L^2$ forms. However intersection cohomology is finite dimensional.<|endoftext|>
TITLE: Reference for t-structures on stable model categories
QUESTION [5 upvotes]: What kind of definitions of t-structures
on stable model categories have been investigated in the literature?
Of course, one can always define a t-structure on a stable model category as a t-structure on its homotopy category;
this is analogous to how Lurie defines t-structures on stable quasicategories (Definition 1.2.1.4 in Higher Algebra).
However, in the setting of model categories one normally wants a more strict presentation
that could be exploited to perform computations more easily.
In particular, one can conceive of various strictifications
of k-connective objects C_{≥k}, k-coconnective objects C_{≤k},
and their truncating functors τ_{≥k} and τ_{≤k}.
For example, in the case of symmetric simplicial spectra one can say that
a spectrum X is strictly connective if for each n the nth spectral level X_n
is a simplicial set with exactly one k-simplex for all k

Using coskeletal simplicial sets one can also conceive of a similar picture
for coconnective spectra and coconnective truncations.
Is there anything like this in the literature?
Of course, I'm not just interested in symmetric
simplicial sets, but also (say) in motivic symmetric
spectra and other stable model categories.

REPLY [6 votes]: one can always define a t-structure on a stable model category as a t-structure on its homotopy category

This is a definition, but it's extremely unsatisfying and completely unenlightening.
The paper you link from the nLab page proves that t-structures in a stable infinity-category correspond bijectively to suitable orthogonal factorization systems, called "normal torsion theories". More info are also in subsequent two papers.

Why should this address your question? 

Well, we are able to characterize t-structures as a genuinely categorical gadget, living not on the level of homotopy categories, but in the "real" higher categorical world: since every model for stable categories has its own avatar of the semantics of factorization systems, you are able to speak about t-structures in every model:

Quasicategories have quasicategorical FS, and hence "our" notion of normal torsion theory
stable model categories, the setting you are interested in, have homotopy factorization systems, and hence "homotopy normal torsion theories".
DG-categories have enriched factorization systems, and hence enriched normal torsion theories (I am not aware of anybody speaking of t-structures in enriched settings);

(I know, this is sketchy: it -especially the possibility to talk about these gadgets in stable derivators- is a work in progress with two colleagues)<|endoftext|>
TITLE: What should motives for $L(E,n)$ look like?
QUESTION [21 upvotes]: Goncharov and Manin showed in this paper that the zeta values $\zeta(n)$ can be realized as periods of framed mixed Tate motives constructed from moduli spaces $\overline{\mathcal{M}}_{0,n+3}$ of rational curves.  (See also this MO-question for a discussion.)
I would like to know if analogous statements are expected (or maybe even known?) for special values of $L$-functions of elliptic curves. 
So let $E$ be an elliptic curve over a number field $K$ and $L(E,s)$ its associated Hasse-Weil $L$-function. If $E$ is modular, then Beilinson's theorem tells us that the special values $L(E,n)$ for $n\geq 2$ are periods, see e.g. this MO-question. So it is, in principle, possible to write $L(E,n)$ as integral of a rational differential form over some cycle in an algebraic variety (roughly). 

Can the corresponding variety be made more explicit (ideally as explicit as in the work of Goncharov and Manin)? Has there been any work in this direction?  
What should framed motives realizing $L(E,n)$ look like?

(I know there is a paper by Brown and Levin on multiple elliptic polylogarithms, based on configuration spaces of marked points on the elliptic curve. However, motives realizing elliptic polylogarithms do not seem to be explicitly discussed in the paper.)  

Could we expect framed motives realizing special values $L(E,n)$ to be constructed from configuration spaces of points on $E$?

REPLY [2 votes]: Let's consider first an analogous case of a number field $F$. Then it is  conjectured that special values of the Dedekind zeta function $\zeta_{F}(n)$ is a linear combination of multiple polylogarithms, evaluated at some configurations of points of $\mathbb{P}^1_\mathbb{C}$ defined over $F$. Multiple polylogarithms are periods of the pro-unipotent completion of a punctured projective line. This is a weakened version of Zagier conjecture, known for $n\leq 4.$
I think that similar statement is expected to hold in your case, with $\mathbb{P}^1$ substituted by the elliptic curve. For $n=3$ it is known for modular elliptic curves. The relation between the generalized Eisenstein-Kronecker and pro-unipotent completion of the fundamental group of punctured elliptic curve is explained here (theorem 11.9).<|endoftext|>
TITLE: Volume form on a hyperbolic manifold with geodesic boundary
QUESTION [12 upvotes]: Let $M$ be a compact connected orientable Riemannian $n$-manifold with boundary $\partial M\ne\emptyset$. Since $H^n(M,\mathbb R)=0$, the connecting morphism $\delta: H^{n-1}(\partial M,\mathbb R)\to H^n(M,\partial M,\mathbb R)$ of the exact sequence of the pair $(M,\partial M)$ is an epimorphism, implying that there is an $(n-1)$-form $\alpha$ on $\partial M$ such that volume form of $M$ is $\delta \alpha$. Then $\operatorname{vol}(M)=\delta \alpha([M])=\alpha([\partial M])$.
Let $M$ be hyperbolic with a geodesic boundary now. Did someone figure out some nice general formula for $\alpha$ in that case? $\alpha$ cannot be the volume form on $\partial M$, at least for $n=2$, since the area of $M$ in that case is a multiple of $\pi$, while the length of $\partial M$ can be arbitrary. In fact, for $n=2$ this is the subject of Gauss-Bonnet theorem, so I guess I am asking for a generalization of it to hyperbolic manifolds with geodesic boundary for $n>2.$

REPLY [6 votes]: I don't know how to answer your question in the spirit posed, but what you're asking for is a function $f:\partial M\to \mathbb{R}$ such that $\int_{\partial M} fdA = vol(M)$, since any $(n-1)$-form on $\partial M$ is of the form $fdA$. 
Such a function must be determined globally. I can imagine many ways to write a formula for such a function. One way is to form a Voronoi region about the boundary, by taking the open neighborhood of points around the boundary which have a unique closest point on the boundary. Then this region may be written as a function $g:\partial M \to \mathbb{R}$ determining the length of the longest segment starting perpendicularly from the point and lying in the Voronoi cell, and then $f$ may be computed from $g$ by elementary calculus and Fermi coordinates.
There should be another formula determined from the orthospectrum. See this paper for a volume formula in terms of orthospectrum. This may be turned into an integral over $\partial M$ by a sum over the orthospectrum of certain functions which are rotationally symmetric and centered at the endpoints of orthogeodesics. Each orthogeodesic determines a pair of planes in the universal cover of M. A point on one plane determines a region in the unit tangent bundle, which is the set of geodesics emanating from the point which hit the other boundary plane. Then the function is the volume of this region, which is determined just by the length of the orthogeodesic and the distance from the point to the foot of this orthogeodesic. Then one sums over such functions.<|endoftext|>
TITLE: A geometric theory of Blueprints? (Algebras over the field with one element)
QUESTION [25 upvotes]: In my attempt to tackle the various approaches of defining algebraic geometry over $\mathbb F_1$, I was just reading through Lorscheid's paper The geometry of blueprints. I certainly like the idea a lot and I get the feeling of it being the 'right' approach to $\mathbb F_1$.
A blueprint is a commutative monoid $B$ with zero, together with a pre-addition, which is -the way he defines it- an  additive, multiplicative equivalence relation on $\mathbb N(B)$ respecting the $0$, i.e. ($0_\mathbb N \equiv 0_B$) and reflecting equality on simple terms $a \equiv b \implies a = b$.
A morphism of blueprints is a homomorphism of monoids respecting $0$ and the pre-addition.
The idea is that blueprints should be $\mathbb F_1$-algebras, with $\mathbb F_1$ being given by $\left\{0,1\right\}$ with standard multiplication and trivial pre-addition. $\mathbb F_1$ is then the initial object in the category of blueprints.
Blueprints generalize semirings and commutative monoids in a nice way ($\text{SemiRing}$ and $\text{CommMonoids}$ are full reflective subcategories of the category of blueprints).
They carry just enough structure to talk about ideals. He then defines blueprinted spaces and locally blueprinted spaces in much the way you'd expect, and from that also 'blue schemes' (as locally blue spaces being locally isomorphic to $\text{Spec}(B)$ for a blueprint $B$).
Now, I'm sure the answer is simple, but are 'Blueprints' a geometric theory? In particular, what would be a nice way to write down the axioms?
Let me phrase it like this

Is there a geometric theory whose $\text{Set}$ based models are blueprints? Is it coherent? Can it be extended to a theory of $\textit{local}$ blueprints?

It is a standard fact that the classifying topos of the coherent theory of local $R-$algebras is the gros zariski topos over $Spec(R)$. Given that the answer to the above question is yes, is the following true?

A blue scheme (as defined by Lorscheid) can be regarded as an object of the classifying topos of the theory of local blueprints.

In the same style of thinking, the topos of sheaves on the spectrum of a given blueprint $B$ should classify the 'theory of prime filters on $B$' in the same way as the topos of sheaves on the spectrum of a commutative ring classifies its theory of prime filters.
I'm still a novice in topos theory, which is why I'm asking: Does any of this make sense?

REPLY [14 votes]: $\newcommand{\N}{\mathbb N}\newcommand{\paren}[1]{\left(#1\right)}\newcommand{\T}{\mathbb{T}}\newcommand{\m}{\mathfrak{m}}\newcommand{\E}{\mathbf{E}}$I can answer your first set of questions:
There is a geometric theory of blueprints. It's easiest* to work from the definition of a blueprint as a pair $B = (A,R)$ consisting of a semiring $R$ and a multiplicative subset $A \subseteq R$ which contains $0$ and $1$, and which generates all of $R$. 
The theory will have a sorts $A,R$, function symbols $+, \cdot$, constant symbols $0_A,1_A, 0_R, 1_R$, and a unary function symbol $\iota$, giving the inclusion of $A$ into $R$. In addition to axioms asserting that $R$ is a semiring and that $A$ embeds monomorphically as a multiplicative submonoid of $R$ such that $\iota(0_A) = 0_R$ and $\iota(1_A) = 1_R$, we have the infinitary axiom
$$ \vdash^{x : R} (x = 0) \lor \paren{\bigvee_{n \in \N, \; \varphi \in \text{Oper}_{n+1}}\exists a_0\dotsm a_n . \varphi(\iota(a_0),\dotsc,\iota(a_n)) = x }$$
where $\text{Oper}_n$ is the set of $n$-ary semiring operations built from $0_R,1_R,+,\cdot$. This axioms states that $A$ suffices to generate all of $R$
There can be no coherent axiomatization of the theory $\T$ of blueprints. To see this, suppose that $\T$ were coherent. Then we could obtain a new coherent theory $\T\,'$ by introducing the following additional coherent axioms which require a blueprint in $\text{Set}$ to be isomorphic to $(\{0,1\} \hookrightarrow \N)$.
$$
\vdash^{a: A} \iota(a) = 0 \lor \iota(a) = 1
$$ $$
x + y = 0 \vdash^{x,y : R} x = y = 0 $$
Since any consistent finitary first-order theory with an infinite model will admit arbitrarily large models in $\text{Set}$, this is impossible.
Regarding local blueprints: The definition of local blueprints as those having a unique maximal ideal of course cannot, in its current form, be stated in geometric logic. However, we can say in $\text{Set}$ that a congruence $\sim_\m$ on a blueprint $B = (A,R)$ is the unique maximal nontrivial congruence on $B$ iff for any pair of elements $x,y \in R$, if $x \sim_\m y$ fails, then the smallest congruence $\sim$ such that $x \sim y$ is trivial. This can be stated in a conservative geometric extension of our theory of blueprints if we adjoin a binary relation symbol $\sim_m$ on $R \times R$, together with axioms stating that $\sim_m$ is a congruence, in addition to the following axiom which states that unless $x \sim_\m y$ holds, every congruence containing $(x,y)$ contains every pair of elements in $R$.
$$\vdash^{x,y,z,w:R} {x \sim_\m y \lor} \paren{\bigvee_{n \in \N} \exists a_0\dotsm a_n : A . \E(z,\iota(a_0)) \land \E(\iota(a_0), \iota(a_1)) \land \dotsi \land \E(\iota(a_n), w)}
$$ where $\E(c,d)$ denotes the sub-expression $$
\bigvee_{n \in \N, \; \varphi \in \text{Oper}_{n+2}} \exists b_0 \dotsm b_n : A. \varphi(c,b_0,\dotsc,b_n) = \varphi(d,b_0, \dotsc, b_n)
$$
*But not essential. Since the list object is a geometric construction, we could also write down a two-sorted theory which axiomatizes the behavior of an equivalence relation on the set of lists. The downside of that approach is that it would involve lots of complicated-to-state infinitary axioms.<|endoftext|>
TITLE: Finite field analogue of Chebotaryov theorem on roots of unity?
QUESTION [6 upvotes]: Chebotarev's theorem on roots of unity says that all the minors of a prime-length DFT matrix over the complex numbers are nonzero.  I was wondering if there was an analogue for finite fields.
More precisely, let $p$ be prime and $\omega=e^{2\pi i/p}$, the complex $p$th root of unity, and let $\Omega$ be the matrix given by $\Omega_{ij}=\omega^{ij}$, for $i,j \in \{0,1,\dots,p-1\}$.  Then Chebotarev's theorem on roots of unity says that every square submatrix of $\Omega$ is nonsingular; see wikipedia.
Alternatively, this is equivalent to stating that, for a complex-valued polynomial $f$ of degree less than $p$, if $f$ evaluates to zero for $t$ distinct powers of $\omega$, then either $f$ is zero, or $f$ has at least $t+1$ terms, or in other words, $$|\mathrm{supp}(f)| + |\mathrm{supp}(\hat{f})| > p.$$
Suppose then that $q$ is a prime power for which $p$ divides $q-1$, such that $\mathbf{F}_q$ contains a primitive $p$th root of unity $\omega$.  Does the analogue hold there?  I am particularly interested in the case that $q$ is prime.
The $2 \times 2$ square submatrices will be nonsingular because a polynomial $x^a + cx^b$, where $p>a>b\geq 0$, $c \neq 0$ will have a root $\omega^i$ if and only if $\omega^{(a-b)i}=c$, which will hold for at most one $i \in [p]$.  Computation shows the analogue holds for the $p=7$th roots of unity modulo $q=29$.

REPLY [3 votes]: Not for $GF(p)$ of prime cardinality. Consider a nontrivial factorization $p-1=uv$ which always exists. Let $w$ be primitive in $GF(p)^{\ast}$ The "regular" DFT submatrix which is made up of $u^{th}$ row and every $v^{th}$ column entry is an all 1 matrix and singular. More generally, when we have $GF(p^m)$ a primitive element has order $p^m-1$ and since $p^m-1=(p-1)(p^{m-1}+p^{m-2}+\cdots+1)$ is a valid factorization a similar argument holds. Thus the answer is no for any finite field. 
Edit in response to comment by OP: Thanks for clarifying the question. I believe that it would fail in that case as well, since there is an extra dependency in finite fields: All elements, including $\omega$ of order $p$ inside a $GF(q)$ with $p|(q-1)$, have a characteristic polynomial $P$ of moderate degree. In this case, the set of determinant polynomials (which is quite large) would have to avoid all multiples of $P$ which is very unlikely.
There may well be a pigeonhole type proof for this. Some quick computations confirmed this. Take $GF(23)$ and $w=2$ of order $11$. Then the indices $I\times J=(0,2,6)\times(0,1,3)$ for example lead to $$[w^{I(k)J(l)}]_{k,l=1,\ldots,3}$$ the $3\times3$ submatrix with zero determinant.<|endoftext|>
TITLE: Some Questions on the Collatz conjecture (reexpressed as "equivalence relation")
QUESTION [12 upvotes]: The set of all positive whole numbers is denoted by $\mathbb{N}_+$.
Let $f\colon\ \mathbb{N}_+\to\mathbb{N}_+:n\mapsto
\begin{cases}\frac{n}{2}&\text{$n$ even}\\3n+1&\text{$n$ odd}\end{cases}$.
Conjecture (Collatz). $\forall n\in\mathbb{N}_+.\ \exists N\in\mathbb{N}.\ f^N(n)=1$.
Let $m,n\in\mathbb{N}_+$. We define: $m\sim n:\iff\exists N_1, N_2\in\mathbb{N}.\ f^{N_1}(m)=f^{N_2}(n)$.
It is easy to see that $\sim$ is an equivalence relation on $\mathbb{N}_+$. If we suppose the truth of the Collatz conjecture, then there is only one equivalence class.


How to prove that there is only a finite number of equivalence classes of $\sim$? Has somebody ever proven this? Or is the question whether there is only a finite number of equivalence classes open?
Is there an algorithm solving the following decision problem?

INSTANCE: A pair $(m, n)\in\mathbb{N}_+\times\mathbb{N}_+$
QUESTION: Does $m\sim n$ hold?

REPLY [11 votes]: This problem is still open.  See for example Sections 2.6 and 2.7 of Lagarias's survey.<|endoftext|>
TITLE: does every vertex-cut set in a maximal planar graph contain a cycle?
QUESTION [5 upvotes]: $G = (V, E)$ is a 3-connected plane triangulation. Let $S \subset V$ such that $G(V - S)$ is disconnected.  Is it true that $G(S)$ must contains a separating cycle?
My intuition is leading me to believe it is true, but I can not prove it.

REPLY [9 votes]: Take one component $H$ of $G(V-S)$ and let $S'$ be the set of vertices in $S$ that are adjacent to a vertex in $H$.  Then $S'$ is separating: it separates $H$ from the rest of $G$. Draw $G$ in the plane. There is a simple closed (Jordan) curve $C$ in the plane which intersects the drawing exactly at $S'$ (elsewhere lying in the interiors of faces) such that there are vertices in both the exterior and the exterior of $C$.  Two vertices of $S'$ that are consecutive on $C$ must be connected by an edge of $G$, since all the faces of $G$ are triangles. So the vertices of $S'$ lie on a cycle given by the order they lie on $C$.
A shorter description: Take any component $H$ of $G(V-S)$ and contract it to a vertex.  You still have a triangulation and now it is obvious that its neighbouring vertices form a cycle.<|endoftext|>
TITLE: Which graphs are prime under the Cartesian product?
QUESTION [5 upvotes]: I'm looking for a characterization of graphs that are prime under the Cartesian product, with prime defined as in this question. Does such a characterization exist, either in general or after restricting to connected bipartite graphs?

REPLY [5 votes]: For connected graphs, they are the graphs in which every two edges are connected by a sequence of pairwise relations using one or both of the following two types of relation:

Edge $xy$ and $uv$ are related if $d(x,u)+d(y,v)\ne d(x,v)+d(y,u)$
Edge $xy$ and $yv$ are related if $y$ is the only common neighbor of $x$ and $v$.

See Imrich, Wilfried; Peterin, Iztok (2007), "Recognizing Cartesian products in linear time", Discrete Mathematics 307 (3-5): 472–483. In Theorem 3.1, they state more strongly that the transitive closure of the union of the two types of relation above is an equivalence relation whose equivalence classes give the prime factorization of the graph. They credit the theorem to T. Feder, Product graph representations, J. Graph Theory 16 (1992) 467–488.
If you need a characterization for disconnected graphs, it's going to get messier (for one thing, factorization is not unique) and involve graph isomorphism.<|endoftext|>
TITLE: How many closed measure zero sets are needed to cover the real line?
QUESTION [9 upvotes]: This question assumes familiarity with combinatorial cardinal characteristics of the continuum.
Let $\mathcal{E}$ be the $\sigma$-ideal generated by closed measure zero subsets of the real line. It is known that
$$\operatorname{cov}(\mathcal{N})\cdot\operatorname{cov}(\mathcal{M})\le 
\operatorname{cov}(\mathcal{E})\le 
\operatorname{cov}(\mathcal{N})\cdot \mathfrak{d},$$
and that the first inequality is consistently strict
(Bartoszynski-Shelah).
Is the second inequality consistently strict?
I conjecture that the answer is positive, and known, but did not find it in the cited paper (or elsewhere, but I didn't search thoroughly).
Update: This problem is solved below by Ashutosh, and the solution suggests a follow-up question.

REPLY [7 votes]: Since we can cover the reals by at most $\mathfrak{r}$ (reaping number) closed null sets, we can do a countable support iteration of rational perfect set forcing over a model of CH to get a model of $\omega_1 = \mathfrak{r} < \mathfrak{d} = \omega_2$. 
To see that $\operatorname{cov}(\mathcal{E}) \leq \mathfrak{r}$, just note that for every infinite $A \subseteq \omega$, $i \in \{0, 1\}$, the set $N_{A, i} = \{x \in 2^{\omega} : (\forall n \in A) (x(n) = i)\}$ is closed null.
$\mathfrak{d} = \omega_2$ in this model because Miller real is not dominated by any ground model real. Also the ground model p-points are preserved so $\mathfrak{r} = \mathfrak{u} = \omega_1$. The proof of this fact can be found in chapter 23 of Halbeisen's Combinatorial set theory.<|endoftext|>
TITLE: Embedding of real trees into $\ell_1(\Gamma)$
QUESTION [6 upvotes]: It seems plausible that any real tree or ${\mathbb{R}}$-tree in the sense of the definition in https://en.wikipedia.org/wiki/Real_tree admits an isometric embedding into the Banach space $\ell_1(\Gamma)$ for a set $\Gamma$ of sufficiently large cardinality (the space $\ell_1(\Gamma)$ is defined as the space of real-valued functions on $\Gamma$ with countable support and norm $||f||=\sum_{\gamma\in\hbox{supp} f}|f(\gamma)|$). My question is:
Is this known? If "Yes", I would appreciate references.
P.S. I like the proof of this fact suggested by Yves de Cornulier (see below). I also agree with Bill Johnson that the fact looks like "well known" and most probably has been published somewhere. If someone knows a reference, I would be very thankful for it (I cannot accept the second answer, but I would be happy to upvote it.)

REPLY [10 votes]: It's true: there's an isometric embedding of every real tree $T$ on some $\ell^1$-space, i.e., a set $A$ and a map $f:T\to \ell^1(A)$ that is an isometric embedding, that is, satisfies $\|f(x)-f(x')\|=d(x,x')$ for all $x\in T$.
I don't know a reference; here's a proof. Let $(x_t)_{t<\alpha}$ be an enumeration of points on the real tree $T$ by an ordinal $\alpha$, assume that the convex hull of them is $T$.  
Let $T_\beta$ be the closed convex hull of $\{x_t:t<\beta\}$ (so $T_\alpha=T$). Let us construct by induction an isometric embedding $f_\beta:T_\beta\to\ell^1(\beta)\subset\ell^1(\alpha)$, so that whenever $\beta\le\gamma$ then $f_\gamma$ extends $f_\beta$.
For $\beta=1$, $T_\beta=\{x_0\}$, and we just prescribe $f_\beta(x_0)=0$. For $\beta=\gamma+1>1$ a successor ordinal, $T_\beta$ is the convex hull of $T_\gamma\cup\{x_\gamma\}$. Let $y_\gamma$ be the projection of $x_\gamma$ on $T_\gamma$. So $$T_\beta=T_\gamma\cup [y_\gamma,x_\gamma].$$ We define $f_\beta$ so as to extend $f_\gamma$ on $T_\gamma$, and prescribe $f_\beta(y)=f_\gamma(y_\gamma)+d(y,y_\gamma)e_\beta$ for every $y\in [y_\gamma,x_\gamma]$, where $(e_t)$ is the canonical basis of $\ell^1(\alpha)$ and $d$ is the distance in the tree. This is well-defined (note that $f_\gamma(y_\gamma)$ is defined twice) and defines an isometric embedding. Finally, for $\beta$ a limit ordinal, then $T_\beta$ is the closure of $T'_\beta=\bigcup_{\gamma<\beta}T_\gamma$, we just define $f_\beta$ on $T'_\beta$ by saying its graph is the union of graphs of all $f_\gamma$ for $\gamma<\beta$; this is an isometric embedding into a complete metric space, so extends uniquely to an isometric embedding of the closure $T_\beta$ into $\ell^1(\beta)$.
So $f_\alpha$ is the required isometric embedding.<|endoftext|>
TITLE: Extension of Wigner's semicircle law?
QUESTION [7 upvotes]: It is well-known that the semicircle law holds for a wide class of matrices with independent and identically distributed (mean zero) entries.
My question is: is there any study about the more general case, in which we drop the "identically distributed" condition? That is, the matrix still have independent entries, but their distributions may be different.
I understand that the completely general case might not permit very meaningful or strong results, but I am interested in settings even many restrictions and conditions are added. Maybe the limiting distribution of the eigenvalues is no longer semicircle?
Any pointer is appreciated. Thank you!

REPLY [5 votes]: Yes, there is a central limit theorem that guarantees convergence in probability to the Wigner semicircle law, see for example Central limit theorem for linear eigenvalue statistics of random matrices with independent entries, or Chapter 2 of Tao's Topics in random matrix theory. 
The convergence to the semicircle law when the matrix dimension $n\rightarrow\infty$ requires that the variances $\sigma_{ij}^2$ of the off-diagonal matrix elements are comparable, meaning that
$$0<c_{\rm inf}\leq n\sigma_{ij}^2\leq c_{\rm sup},\;\;1\leq i<j\leq n,$$
for $n$-independent constants $c_{\rm inf}$ and $c_{\rm sup}$, see Universality of Wigner random matrices: a Survey of Recent Results.
One important class of random matrices where the universality breaks down is the class of banded random matrices, where the nonzero matrix elements are those near the diagonal, see for example Limiting eigenvalue distribution for band random matrices.<|endoftext|>
TITLE: A generalisation of $C_0$-semigroups
QUESTION [8 upvotes]: A $C_0$-semigroup is a strongly continuous family $\{T(t)\}$ of bounded linear operators on a normed space $X$, indexed in $\mathbb R_+$ and with two additional properties that make it look like an operator-valued version of the exponential function: $T(t)T(s)=T(t+s)$ for all $t,s\in \mathbb R_+$ and $T(0)=\hbox{Id}_X$. On the basis of two concrete examples I was wondering whether there is room to generalise this notions to operator families indexed on general directed sets. For this purpose, defining a sum-like operation on a directed set seems to be unavoidable and I was wondering whether this has been done in the past. 
(I admit that the only two examples I can think of are rather trivial: $\mathbb R$ viewed as a collection of two halflines directed towards 0 (which returns the definition of $C_0$-group), as well as the positive hyperoctant in $\mathbb R^n$ (which for $n=2$ reminds of how analytic $C_0$-semigroups are defined). But what I am striving for is the definition of $C_0$-semigroups indexed on some kind of tree-like sets.)

REPLY [4 votes]: This does not answer your question only adds something to your list of examples: two parameter semigroups, as investigated by 

Slocinski, M. Unitary dilation of two-parameter semi-groups of
  contractions. (English) [J] Bull. Acad. Pol. Sci., Sér. Sci. Math.
  Astron. Phys. 22, 1011-1014 (1974)

and many others later on. They are significantly different from analytic semigroups defined on a sector of the complex plane.
ADDED:
An other possible generalization is the concept of a strongly continuous cocycle, see Definition 6.1 in

Carmen Chicone,Yuri Latushkin: Evolution Semigroups in Dynamical
  Systems and Differential Equations, American Mathematical Soc., 1999.<|endoftext|>
TITLE: What do we know about the structure of $J_{0}(N)$ over $\mathbb{Q}[{\mu}_{{p}^{\infty}},{{k}}^{\frac{{1}}{{p}^{n}}}])$?
QUESTION [8 upvotes]: What is known about the structure of $J_{0}(N)$ over $\mathbb{Q}[\mu_{p^{\infty}}]$?
More generally, what do we know about $J_{0}(N)$ over 
$\mathbb{Q}[\mu_{p^{\infty}},k^{1/p^{n}}]$, where $k\in\mathbb{Z}$?
Does it have infinite rank? Does it have finite torsion?
I am especially interested in the case where $N$ is the conductor of an elliptic curve with additive reduction over $\mathbb{Q}$ at $p$. (This curve acquires, of course, a Néron model with semistable reduction st $p$ over $\mathbb{Q}[\mu_{p^{\infty}}]$.)
EDIT:
In view of the previous comments, I would like to ask whether the $rank({J}_{0}(N))$ is still finite over
$$K(k)\colon=\lim_{\stackrel{\rightarrow}{n}}(\mathbb{Q}[{\mu}_{{p}^{\infty}},{{k}}^{{\frac{{1}}{{{p}^{n}}}}}])$$

REPLY [3 votes]: The case of Kummer extensions is not yet completely understood, though partial results are obtained by Darmon-Tian in "Heegner points over towers of Kummer Extensions", Canad. J. Math. 62 (5) 2010, 1060 - 1080, following conjectures made in Dokchitser-Dokchitser "Computations in non-commutative Iwasawa theory," Proc. London Math. Soc. 94 (1) 2007, 211-272 (for instance). The results here are a bit tricky to summarize succinctly: There is a "root number dichotomy" which amounts to characterizing the forced vanishing of central values by the functional equation, and this must be considered a priori. The settings of forced vanishing typically correspond to Heegner-like growth of Mordell-Weil rank, as in the setting of anticyclotomic extensions, and this is described nicely in Darmon-Tian (who study a variation coming from varying parametrizations by Shimura curves). Anyhow, one predicts systematic growth of rank in these settings, and boundedness of rank in the others (i.e. when there is not forced vanishing coming from the functional equation). Also, results in this direction (including Kato + Rohrlich) are typically shown by combining some kind of analytic nonvanishing results for central values (such as Rohrlich's "On L-functions of elliptic curves in cyclotomic towers", Inventiones (75) 1984, 409 - 423) with an "Euler system" construction/argument (such as Kato's, as alluded to in David Loeffler's answer).<|endoftext|>
TITLE: Realization of numbers as a sum of three squares via right-angled tetrahedra
QUESTION [13 upvotes]: De Gua's theorem 
is a $3$-dimensional analog of the Pythagorean theorem:
The square of the area of the diagonal face of a right-angled tetrahedron
is the sum of the squares of the areas of the other three faces.
For certain tetrahedra, this provides a representation of
an integer $n$ as the sum of three integer squares.
Let the tetrahedron have vertices at
\begin{eqnarray}
& (0,0,0)\\
& (a,0,0)\\
& (0,b,0)\\
& (0,0,c)
\end{eqnarray}
If $a,b,c$ are integers, at least two of which are even,
then the squared areas of the three triangles incident to the origin
are each integer squares, and so "represent" $n=A^2$, the diagonal-face area squared.
Example.
Let $a,b,c$ be $2,3,4$ respectively.

                   


The diagonal face-area squared is
\begin{eqnarray}
A^2 & = & \left[ (2 \cdot 3)^2 + (3 \cdot 4)^2 + (4 \cdot 2)^2 \right] \,/\, 4\\
& = & (36 + 144 +64) \,/\, 4 \\
& = & 9 + 36 + 16\\
& = & 61\\
A & = & \sqrt{61} \;.
\end{eqnarray}
So here, $61$ is represented as the sum of three squares: $9+36+16$.
Let $N_T(n)$ be the number of integers $\le n$ that can be represented
as a sum of three squares derived from deGua's tetrahedron theorem,
as above. Call these tetra-realized.
Let $N_L(n)$ be the number of integers $\le n$ that can be represented
as a sum of three squares.
$N_L$ is determined by
Legendre's three-square theorem, which
says that $n$ is the sum of three squares except when it is
of the form $n=4^a (8 b + 7)$, $a,b \in \mathbb{N}$.
I would like to know how prevalent is tetra-realization:

Q. What is the ratio of $N_T(n)$ to $N_L(n)$ as $n \to \infty$?

I would also be interested in any characterization of the tetra-realizable
$n$.

REPLY [17 votes]: These numbers are not very prevalent and the ratio in question goes to zero.  Note first that by Legendre's theorem, a positive proportion of the numbers below $n$ may be expressed as a sum of three squares.  Now consider $N_T(n)$.   This amounts to counting (with parity restrictions on $x$, $y$, $z$, and all three positive) the number of distinct integers of the form $((xy)^2 + (yz)^2 +(xz)^2)/4$ lying below $n$. So we must have $xy$, $yz$, and $xz$ all lying below $2\sqrt{n}$, which means that 
$$ 
xyz = \sqrt{(xy) (yz)(xz)} \le 2\sqrt{2} n^{\frac{3}{4}}.
$$ 
So the total number of possibilities for $(x,y,z)$ is bounded by the number of triples with product at most $X=2\sqrt{2}n^{\frac 34}$, and this is 
$$ 
 \sum_{xyz\le X} 1 \le \sum_{x,y\le X} \frac{X}{xy} \le X(1+\log X)^2.
$$
Thus, even if these choices for $(x,y,z)$ all led to distinct integers of 
the form $(xy)^2+(yz)^2+(xz)^2$, we still would have no more than $Cn^{\frac 34}(\log n)^2$ integers up to $n$ that may be tetra-realized.<|endoftext|>
TITLE: Any PL-homology-manifold is homotopy equivalent to a manifold
QUESTION [8 upvotes]: Is it true that any compact piecewise linear homology manifold is homotopically equivalent to a (smooth?) manifold of the same dimension?

Let me say bit more since my question was wrongly understood.

Any link of homology manifold has to be a homoplogy sphere.
By double suspension every point on a simplex of dimension at least 1 is a manifold point (it has a neighborhood homeomorphic to an open set in $\mathbb R^n$. 
Therefore we have a finite discrete set of topological singularities. We can remove an $\epsilon$-neighbborhood around each, its boundary is a homological sphere so we can patch the hole by contactable manifold with the same boundary.
It seems to be an answer in the topological category. Am I right? 
I hope that starting with dimension 5 one can do the same in smooth category.

REPLY [3 votes]: On the revised question:
I am not sure what you mean by doing the same in smooth category but
there are PL -manifolds that are not homotopy equivalent to smooth ones, see e.g. [M. Davis and J-C. Hausmann, Aspherical manifolds without smooth or PL structure, Springer Lecture Notes in Math. 1370, (1989), 135--142] available here. See also example 2.1 in here. 
As for your question 4 you seem to be asking whether the identity map of a homology sphere $S$ extends to a homotopy equivalence between the cone $CS$ on $S$ and a contractible manifold $X$ with $\partial X=S$. This is an obstruction theory problem and (I think) there is no obstruction because all obstruction cocycles lie in relative cohomology groups with coefficients in $\pi_*(X)$ or $\pi_*(CS)$, which all vanish.
Perhaps by "doing it in the smooth category" you meant finding a smoothing on your manifold minus a star of the singular locus, and then replacing the remaining cones by smooth contractible manifolds. This might not be possible. The issue is that the smoothing must be such that all boundary homology spheres bound smooth contractible manifolds. For example, if a homotopy sphere smoothly bounds a contractible manifold, then removing a small ball from the interior of the contractible manifold gives an $h$-cobordism between the given homotopy sphere and the standard sphere. In higher dimensions the $h$-cobordism is trivial, so the homotopy sphere cannot be exotic.<|endoftext|>
TITLE: Motivation and potential applications of spectral algebraic geometry
QUESTION [19 upvotes]: Nowadays there is a lot of talk about derived algebraic geometry, but not so much about the related subject of spectral algebraic geometry.
Now I'm curious what future is there for spectral algebraic geometry. That is, what problems could be  possibly attacked using this formalism, rather than arithmetic algebraic geometry or derived algebraic geometry.
I know historically you have used it to study topological modular forms. 
Since it's also ''higher arithmetic geometry'' I'm curious if it can be useful in number theoretic problems. J.Lurie has a joint paper with Gaitsgory about Weil conjecture for function field. Did he use spectral algebraic geometry there? Or ''simple'' derived algebraic geometry?
I know that spectral algebraic geometry is also a type of ''higher''/''derived'' geometry, so I'm interested what advantages it has over derived algebraic geometry over simplicial commutative rings. 
I wonder if spectral algebraic geometry could be potentially a useful tool for number theory or Langlands program (geometric). And other applications are also would be appreciated compared to DAG.

REPLY [20 votes]: This is not really an answer to your question, just an attempt to address your question from the comments.
There are various flavours of homotopical or higher algebraic geometry that are commonly considered, which have different levels of connectivity, linearity, or strictness of commutativity.
These include:
1) simplicial commutative rings
2) connective $E_\infty$-algebras over $H\mathbf{Z}$
3) connective $E_\infty$-ring spectra
4) (nonconnective) $E_\infty$-algebras over $H\mathbf{Z}$
5) (nonconnective) $E_\infty$-ring spectra
In characteristic zero, one also considers (connective) commutative dg-algebras.
The flavour most suited for algebraic geometry purposes is (1): this is the minimal extension of algebraic geometry where derived tensor products and cotangent complexes live.  This was the flavour originally studied by Lurie in his thesis, and Toen-Vezzosi in HAG II.
Any of the other theories might be called "spectral algebraic geometry".  (2) is similar to (1), but is less suited for algebraic geometry purposes, because deformation theory in the $E_\infty$-world is different than in the setting of simplicial commutative rings.  In fact, the affine line is not even smooth in the $E_\infty$-world.
The difference between (2) and (3), as between (4) and (5), is linearity: in (3) and (5), you only consider objects which are linear over the sphere spectrum, so these settings are well suited to purposes of stable homotopy theory.
The main difference between the connective and nonconnective settings is the lack of converging Postnikov towers.  That is, every connective $E_\infty$-ring spectrum $R$ can be written as a homotopy limit of square zero extensions of $\pi_0(R)$.  This allows one to establish analogues of many results from classical algebraic geometry, by using induction along square zero extensions.  The nonconnective world, on the other hand, behaves much differently, and geometric intuition very often fails.
I don't know much stable homotopy theory, but I believe the main point of spectral algebraic geometry is to be able to consider $E_\infty$-ring spectra as affine schemes, and to apply algebro-geometric techniques to study them.  For example, the main application so far is Lurie's construction of tmf, the spectrum of topological modular forms, as the global sections of a sheaf of $E_\infty$-ring spectra on the moduli stack of spectral elliptic curves.<|endoftext|>
TITLE: On proof of the conditionally negative definiteness of a kernel
QUESTION [6 upvotes]: Let the kernel be $f(\mathbf{x},\mathbf{y}) = \arccos(\mathbf{x}^T \mathbf{y})$, where $\mathbf{x}$ and $\mathbf{y}$ are $\ell_2$ normalized vectors of the same dimensionality, and $\arccos(\cdot): [-1,1] \to [0,\pi]$ is the inverse cosine function.
Question: Is $f$ conditionally negative definite? If yes, how can I prove it?
I know the definition of a conditionally negative definite kernel, but I find it difficult to apply.
A kernel $f: (\mathcal{X} \times \mathcal{X}) \to \mathbb{R}$ is called (conditionally) negative definite if it it symmetric and $\sum_{i,j=1}^m c_i c_j f(x_i,x_j) \leq 0$ for all $m \in \mathbb{N}$, $\{x_1,\cdots,x_m\} \subseteq \mathcal{X}$ and $\{c_1,\cdots,c_m\} \subseteq \mathbb{R}$ with $\sum_{i=1}^m c_i = 0$.

REPLY [5 votes]: Here a direct approach. Recall the power-series
\begin{equation*}
  \arccos(z) = \frac\pi2 - \sum_{k\ge0}\binom{2k}{k}\frac{z^{2k+1}}{4^k(2k+1)}.
\end{equation*}
From this series it is clear that $\arccos(x^Ty)$ is conditionally negative definite (because it is of the form "const $-$ positive definite").

EDIT: (15/12/2015). Here are some more details. Observe that with the above powerseries representation, we have
\begin{equation*}
 f(x_i^Tx_j) = \frac\pi2 - k(x_i,x_j),
\end{equation*}
where $k(x,y)$ is a positive definite kernel (to see this observe that the power series has nonnegative coefficients, and since $(x_i^Tx_j)^{2k+1}$ is pointwise product of kernels, it is itself a kernel). 
Thus, we have in particular that the matrix $F := [f(x_i^Tx_j)] = c11^T-[k(x_i,x_j)]$, so that it immediately follows
\begin{equation*}
 z^TFz = c(z^T1)^2 - z^TKz \le 0,
\end{equation*}
because the first term is zero whenever $z^T1=0$ (as stipulated for cnd matrices), and because $z^TKz \ge 0$ as $K$ is a kernel.
Note: The above argument does not yield that $f^n$ is cnd (it may likely not be cnd, but I don't have time right now to think about it).<|endoftext|>
TITLE: Can one deform an immersion of a 3-manifold in $\mathbb R^4$ to an embedding in $\mathbb R^6$?
QUESTION [24 upvotes]: Let $M^3$ be an oriented 3-manifold, and let $f:M^3\looparrowright \mathbb R^4$ be a codimension one immersion. Is it possible to find a small deformation of the composite map
$$
M^3 \to \mathbb R^4 \to \mathbb R^6
$$
which is an embedding?
(I expect the answer to be "no", and so I'm mostly interested in the method of proof.)

REPLY [8 votes]: The existence of such an immersion is related to the existence of odd Hopf invariant in $\pi_7(S^4)$. Namely take any such element, for example the homotopy class of the Hopf map. Represent it  - by the Pontrjagin construction - as an embedded framed 3-dimensional submanifold in $R^7$. Using the so called Compression theorem (by Rourke Sanderson, it is more or less equivalent to Smale Hirsch immersion theory) you may isotope this framed immersion so, that one of the framing normal vectors becomes everywhere parallel to the 7-th coordinate direction of $R^7$. Then projecting it to $R^6$ we obtain a framed immersion in $R^6$. The (algebraic) number of double points (they have signs, since the double branches have an ordering, which was higher, which was lower) agrees with the Hopf invariant of the original map. Applying two more times the compression, you can push the immersion into $R^4$ as an immersion. When you lift back from $R^4$ to $R^6$ after arbitrary regular homotopy, the pairity of the double points remains unchanged. (This is the stable Hopf invariant of the element of the stable homotopy class in $\pi^s(3)$ represented by the obtained immersion.)<|endoftext|>
TITLE: Rational solutions of the Fermat equation $X^n+Y^n+Z^n=1$
QUESTION [20 upvotes]: As a generalisation to the equation of Fermat, one can ask for rational solutions of $X^n+Y^n+Z^n=1$ (or almost equivalently integer solutions of $X^n+Y^n+Z^n=T^n$).
Contrary to the case of Fermat, the case where $n=3$ has infinitely many solutions, because the surface is rational. For $n=4$, we get a K3 surface and for $n\ge 5$ the surface should have finitely many points or the points should be contained in finitely many curves since it is of general type (Bombieri-Lang conjecture).
But are there some non-trivial solutions for $n\ge 5$ (and $n=4$)? Do we know if there are finitely many solutions for $n\ge 5$ ?
I guess that it should be classical, but I did not find it on this site or online, after googling "Fermat, surface, rational solutions", etc

REPLY [2 votes]: I found a parameterized solution of $X^5+Y^5+Z^5=T^5$ with $X$ and $T$ rational and $Y$ and $Z$ complex rational:
$(k^2-4k+1)^5+(2k-2+(k^2-2k+3)i)^5+(2k-2-(k^2-2k+3)i)^5=(k^2-3)^5$
There is an almost Desboves form of it, as follows:
$(b-a)^5+(a+ci)^5+(a-ci)^5=(b+a)^5$, where $2a^2+b^2=c^2$<|endoftext|>
TITLE: Anything about $\prod_{n \ge 1} (1 + n^{-n})$?
QUESTION [6 upvotes]: Sophomore's dream is especially the statement that the sum, let me call it $s$, of the (convergent) real series $\sum_{n \ge 1} n^{-n}$ is equal to the (improper) integral $\int_0^1 x^{-x} dx$. A few digits of the decimal representation of $s$ are recorded on the OEIS. A couple of days ago, a user of an Italian math forum asked for information about a multiplicative analogue of $s$, namely the constant $p := \prod_{n \ge 1} (1 + n^{-n}) \approx 2.60361$. After browsing through Finch's Mathematical Constants (Cambridge University Press, 2003), I couldn't find anything there about $p$, but on the other hand, I couldn't find anything about $s$ either. So I guess my question is: 

Do you have any pointer to a piece of literature where $p$ pops up in connection to some (interesting) math?

REPLY [3 votes]: Logarithm $\log p$ of this product is also some definite integral (not a surprise, any number is a definite integral of appropriate function), but the integrand is more sophisticated than $x^{-x}$. 
Logarithm of your product equals 
$$
\log p=\sum \log(1+n^{-n})=\sum_n \sum_k(-1)^{k-1} \frac{n^{-kn}}{k}=\sum_k \frac{(-1)^{k-1}}{k} c_k, 
$$
where $$c_k:=\sum_{n\geq 1} n^{-kn}.$$
We have 
$$
\int_0^1 x^a(-\log x)^bdx=\frac{b!}{(a+1)^{b+1}}.
$$
hence $$c_k=\int_0^1 \sum_n \frac1{(kn-1)!}x^n(-\log x)^{kn} dx=\int_0^1 x^{1/k}g_k(-x\log x) dx,$$
where $$g_k(t)=\sum \frac{t^{kn-1}}{(kn-1)!}=k^{-1}\sum_{w:w^k=1} we^{wt}.$$
Thus $$c_k=\frac1k\int_0^1x^{1/k} \sum_{w:w^k=1} w x^{-wx}.$$
Therefore $\log p=\int_0^1 g(x)dx$, where
$$
g(x)=\sum_{k,w:w^k=1} \frac{(-1)^{k-1}}{k^2}x^{1/k}wx^{-wx}. 
$$
We may change order of summation, fixing $w=e^{2\pi ai/b}$ for coprime $a,b$. Then it arises for $k=b,2b,\dots$. We see that coefficient of $wx^{-wx}$ equals
$$
b^{-2}\sum_{k=1}^{\infty}(-1)^{kb-1} \frac{x^{\frac1{bk}}}{k^2}.
$$
I do not recognize this function, but maybe it may in turn be expressed via some generalized polylogarithms or whatever.<|endoftext|>
TITLE: what is the equivalent of the Euler constant for higher dimensional lattices
QUESTION [16 upvotes]: Let $\Lambda$ be a unimodular lattice in $\mathbb R^d$. Then there are constants such that
$$\sum_{\substack{\gamma\in \Lambda\\0<|\gamma|<R\\}} \frac{1}{|\gamma|^d} = c_1 \log R + c_2 + o(1).$$
My questions are: Does $c_2$ depend on the lattice ? If yes,  how ?

REPLY [10 votes]: $\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}$This is essentially the constant term in the Epstein $\zeta$-function. Given a lattice $\Lambda$ in $\RR^d$, the Epstein $\zeta$ function is
$$Z(\Lambda, s) = \sum_{g \in \Lambda \setminus \{ 0 \}} \frac{1}{(g^T g)^s}.$$
$Z$ has a simple pole at $d/2$ with residue $\tfrac{\pi^{d/2}}{\sqrt{\det \Lambda} \ \Gamma(d/2)}$, and no other poles on $\mathrm{Re}(s)\geq d/2$. Set
$$Z(s) = \frac{\pi^{d/2}}{(\det \Lambda) \ \Gamma(d/2) (s-d/2)}+c(\Lambda) + O(s-d/2)$$
There are standard tools to convert Dirichlet series estimates to partial sum estimates. If I didn't drop any constants, then 
$$c_1 = \frac{2 \pi^{d/2}}{(\det \Lambda) \ \Gamma(d/2)} \quad c_2 = c(\Lambda).$$
Theorem 4 of Terras, "Bessel Series Expansions of the Epstein Zeta Function and the Functional Equation", Trans. AMS, Vol. 183 (Sep., 1973), pp. 477-486  gives a formula for $c(\Lambda)$ in terms of other functions, but I don't feel competent to summarize it.<|endoftext|>
TITLE: What is the spin connection in 9 dimensions as opposed to 5 dimensions?
QUESTION [8 upvotes]: From Spin Connection in 5 dimensions I can define a massless fermion's covariant derivative on a curved manifold as 
$$
\nabla_\mu \psi = (\partial_\mu - {i \over 4} \omega_\mu^{ab} \sigma_{ab}) \psi
\tag{1}
$$
where $\sigma_{ab}$ are the dirac bilinears and $\omega_\mu^{ab}$ is the spin connection with three indices.
In 5 dimensions I have a $4\times 4$ spinor space, giving me three sets of irreducible matrices: $I$ as identity, $\gamma^a$ as monolinears, and $\sigma_{ab}=[\gamma_a,\gamma_b]$ as bilinears. This give me a total of $1+5+10=16$ matrices forming a complete set.
In 9 dimensions I can have $9=2(4)+1$, giving me a spinor space of $2^{(4)}\times 2^{(4)}=16\times 16$ creating additional irreducibles: $\sigma^{abc}=[\gamma^a,\gamma^b,\gamma^c]$ as trilinears and $\sigma^{abcd}=[\gamma^a,\gamma^b,\gamma^c,\gamma^d]$ as quadrilinears. This gives me a total of $1+9+36+84+126=256$. These numbers were calculated from the binomial coefficients ( binomial[d,k] ) for the total number of kth-linears in $d$ spatial dimensions.
Since there are additional irreducibles in $9$ dimensions, not found in 5 dimensions, does my covariant derivative in Eq. 1 have additional terms? For example
$$
\nabla_\mu \psi = \left(\partial_\mu - {i \over 4} \omega_\mu^{ab} \sigma_{ab}
 - {i \over 48} \omega_\mu^{abcd} \sigma_{abcd}
 \right) \psi
\tag{2}
$$
where $\omega_\mu^{abcd}$ is a new spin connection of 5 indices or is Eq. 1 still valid in $9$ dimensions?

REPLY [6 votes]: To answer this question, we have to make a few rather natural assumptions, which I give in a slightly different notation.
Let the spinor bundle $S$ be associated to a Euclidean vector bundle $E$.
Assume that both bundles are locally trivialised over some open coordinate patch $U$ by orthonormal / unitary frames $e_1,\dots, e_n$, $\psi_1,\dots,\psi_N$, such that Clifford multiplication has constant coefficients ("Pauli matrices"/"Dirac matrices") with respect to these frames, so $e_i\cdot\psi_\alpha=\sum_\beta c_{i\alpha}^\beta\psi_\beta$.
Let $\nabla^E$ be a metric connection on $E$, so the metric satisfies a Leibniz rule. Then $\nabla^E$ differs from partial derivation with respect to the given trivialisation by a one-form with values in skew-symmetric matrices,
so $\nabla^E_{\partial_i}e_j|_x=\sum_k\Gamma_{ij}^k(x)e_k$. The Christoffel symbols $\Gamma_{ij}^k\colon U\to\mathbb R$ are smooth functions with $\Gamma_{ij}^k=-\Gamma_{ik}^j$.
Assume that the spin connection $\nabla^S$ is compatible with $\nabla^E$ in the sense that Clifford multiplication satisfies a Leibniz rule.
This determines $\nabla^S$ uniquely.
Then $\nabla^S$ differs from partial derivation in the given trivialisation by
a one-form with values in the degree-2-part of the clifford algebra, more precisely $$\nabla^S_{\partial_i}\psi_\alpha=\frac14\sum_{j,k}\Gamma_{ij}^k(x)e_j\cdot e_k\cdot\psi_\alpha\;.$$
No terms of higher degree in the Clifford algebra are needed.
We check the Leibniz rule\begin{align}
\nabla^S_{\partial_i}(e_\ell\cdot \psi_\alpha)&=\frac14\sum_{j,k}\Gamma_{ij}^ke_j\cdot e_k\cdot e_\ell\cdot \psi_\alpha\\
&=-\frac12\sum_j\Gamma_{ij}^\ell e_j\cdot\psi_\alpha
+\frac12\sum_k\Gamma_{i\ell}^ke_k\cdot\psi_\alpha
+\frac14\sum_{j,k}\Gamma_{ij}^ke_\ell\cdot e_j\cdot e_k\cdot \psi_\alpha\\
&=(\nabla^E_{\partial_i}e_\ell)\cdot\psi_\alpha
+e_\ell\cdot(\nabla^S_{\partial_i}\psi_\alpha)\;.
\end{align}
You find the whole story in Lawson-Michelsohn, section II.4 or in Berline-Getzler-Vergne, Section 3.3.<|endoftext|>
TITLE: Efficiently compute the trace of a sparse matrix times the inverse of a sparse matrix?
QUESTION [8 upvotes]: How can I efficiently compute
   $\mathrm{trace}(A(B^{-1}))$
where $A$ and $B$ are both sparse symmetric PSD $n \times n$ matrices, both with $O(n)$ non-zero entries? If it helps, the pattern of non-zero entries in $A$ and $B$ can be the same.
Alternatively, is there a tight upper bound on this quantity that I can compute efficiently, e.g. in $O(n \log(n) )$ time?

REPLY [3 votes]: Simple bounds
A simple upper bound is
\begin{equation*}
 \text{tr}(AB^{-1}) \le \min\left(\lambda_{\max}(A)\text{tr}(B^{-1}), \text{tr}(A)\lambda_{\max}(B^{-1})\right).
\end{equation*}
Both these bounds are numerically "easy" to compute using Lanczos. For computing $\text{tr}(B^{-1})$ a randomized trace estimator can be used (following the more general idea outlined below).
Numerical approximation
Here is a simple approach, motivated by this nice book:

First compute $\alpha=\|B\|$ approximately using Lanczos
Now consider $B=\alpha I - C$, so that $B^{-1}=(I-\alpha^{-1}C)^{-1}/\alpha$
After that, consider
\begin{equation*}
  \text{tr}(AB^{-1}) = \frac1\alpha\text{tr}(A^{1/2}(I-\alpha^{-1}C)^{-1}A^{1/2})
\end{equation*}
Now use the von Neumann series 
\begin{equation*}
\text{tr}(A(I-\alpha^{-1}C)^{-1})=\sum_{k\ge0} (-1)^k\alpha^{-k}\text{tr}(A^{1/2}C^kA^{1/2})
\end{equation*}
Let $u \sim \mathcal{N}(0,I)$ be a mean-zero spherical Gaussian rv. Then, we approximate the above quantity by taking $m$ samples, $u_1,\ldots,u_m$ and iteratively computing $u_i^TA^{1/2}C^kA^{1/2}u_i$ for $1\le i \le m$. Observe that the key subroutine that we have is to iteratively compute $z^TC^kz = z^TC(C^{k-1}z)$.
In expectation this will be an estimator for the trace in question since $E[\text{tr}(u^TA^{1/2}C^kA^{1/2}u]=E[\text{tr}(A^{1/2}C^kA^{1/2}uu^T)]$ and $E[uu^T]=I$ by assumption.

If you do not have access to $A^{1/2}$ (or a Cholesky factorization of it) then an additional level of approximation arises by building a subroutine to compute $A^{1/2}u$. Such $f(A)b)$ family of subroutines are the subject of research interest in numerical linear algebra (see e.g., Nick Higham's webpage and his book on Functions of Matrices for further information).<|endoftext|>
TITLE: Is the sheaf of smooth functions flat?
QUESTION [8 upvotes]: Let $X$ be a smooth algebraic variety over $\mathbb{C}$. Is the sheaf of smooth functions on $X$ flat as an $\mathcal{O}_X$ module?

REPLY [14 votes]: Yes, it is. First of all, the ring of germs of holomorphic functions $\mathcal{O}^h_x$ is flat over the ring of germs regular functions $\mathcal{O}_x$ at some point $x \in X$, see for example Taylor, "Several complex variables with Connections to Algebraic Geometry and Lie Groups", Theorem 13.3.5.
Secondly, $\mathcal{O}^r_x$ is flat over the ring $\mathcal{O}^h_x$ of real analytic functions. This is so, since germs of real analytic functions in $\mathbb{C}^n$ is isomorphic to germs of holomorphic functions in $\mathbb{C}^{2n}$, and $\mathcal{O}^h_{\mathbb{C}^{2n},0}$ is flat over $\mathcal{O}^h_{\mathbb{C}^n,0}$, see for example Fischer, "Complex Analytic Geometry", Proposition 3.17.
Finally, the ring of germs of smooth functions is flat over the ring of real analytic functions, which can be found in Malgrange, "Ideals of Differentiable Functions", Corollary VI.1.12.<|endoftext|>
TITLE: Rellich-Kondrachov compacteness Theorem for the Euclidean space with Gaussian measure
QUESTION [5 upvotes]: Let $\gamma_n: \mathbb{R}^n\to\mathbb{R}$ be the Gaussian distribution function defined by 
$$
\gamma_n(x):=(2 \pi)^{-\frac{n}{2}} e^{-\frac{|x|^2}{2}}.
$$
Let $d\gamma_n$ denote the following measure (weight) $\gamma_n(x) dx$ and consequently $L^2(\mathbb{R}^n,d\gamma_n)$ and $H^1(\mathbb{R}^n,d\gamma_n)$ be the weighted versions of the Sobolev spaces with respect to the measure $d\gamma_n$.  
Does the following weighted version of the Rellich-Kondrachov Theorem hold?
$$
H^1(\mathbb{R}^n,d\gamma_n) \text{ is } \textbf{compactly} \text{ embedded in } L^2(\mathbb{R}^n,d\gamma_n).
$$

REPLY [5 votes]: I'll start with several  known facts.

Proposition 1. Suppose that  $E$ is a  real Hilbert space with norm $\Vert -\Vert$ and $K: E\to E$ is a  a compact,  selfadjoint
  positive operator.   Denote by $R(K)$ the range of $K$.  Equip the
  range with the norm 
$$ \Vert x\Vert_K:=\Vert K^{-1} x\Vert,\;\;x\in R(K).  $$
Then the inclusion $(R(K), \Vert-\Vert_K)\to (E,\Vert-\Vert)$ is
  compact.

$\DeclareMathOperator{\Tr}{TR}$

Proposition 2.  Suppose that $E$ is a real Hilbert space and $K:E\to E$  is a positive selfadjoint operator such that for a
  sufficiently large  positive integer $p$ the operator $K^p$ is trace
  class. Then the operator $K$ is compact.

Suppose now that $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bZ}{\mathbb{Z}}$ that $E=L^2(\bR^n,\gamma_n)$.   Then  $E$ admits a  natural orthogonal basis consisting of the Hermite polynomials
$$ H_\alpha(x_1,\dotsc, x_n)=H_{\alpha_1}(x_1)\dotsc H_{\alpha_n}(x_n), $$
$$\alpha=(\alpha_1,\dotsc,\alpha_n)\in\bZ^n_{\geq 0}. $$
Above, $H_k(x)$ denotes the degree $k$ Hermite polynomial in one variable $x$ defined by
$$ H_k(x) =\delta^k 1, $$
where $\delta :C^\infty(\bR)\to C^\infty(\bR)$ is the  creation operator
$$ \delta f(x)=-f'(x)+x f(x),\;\;\forall f\in C^\infty(\bR). $$
We have
$$\Vert H_\alpha\Vert^2 =\alpha!:=(\alpha_1!)\cdots (\alpha_n!), $$
where $\Vert-\Vert$ denotes the norm in $E=L^2(\bR^n,\gamma_n)$.
Denote by $E_m$ the subspace of $E$ spanned by the polynomials  $H_\alpha$such that $|\alpha|=m$, where $|\alpha|=\alpha_1+\cdots +\alpha_n$. $\DeclareMathOperator{\Proj}{Proj}$   Note that 
$$\dim E_m  =\binom{m+n-1}{n-1} = O(m^{n-1})\;\;\mbox{as $m\to \infty$}. $$
Let $\Proj_m$ denote the orthogonal projection onto $E_m$.  The norm $\Vert-\Vert_*$ on $H^1(\bR^n,\gamma_n)$   can be described in terms of the projectors $\Proj_m$. More precisely
$$\Vert f\Vert_*^2=\Vert f\Vert^2+\sum_{j=1}^n \Vert \partial_{x_j}\; f\Vert^2=\sum_{m\geq 0} (1+m)\Vert \Proj_m f\Vert^2 . $$
Now consider the operator
$$K : E\to E,\;\;K f= \sum_{m\geq 0}(1+m)^{-1/2} \Proj_m f. $$
This shows that $H^1(\bR^n,\gamma_n)$ can be identified with the range of $K$ equipped with the norm $\Vert -\Vert_K$ defined as in Proposition 1.
Clearly, for $p$ sufficiently large
$$\sum_{m\geq 0} (m+1)^{-p/2}  \dim E_m <\infty. $$
This shows that $K^p$ is trace class for large $p$ and thus, by Proposition 2,  $K$ is compact.   Now conclude using Proposition 1.<|endoftext|>
TITLE: Compact Eucledean hypersurfaces with "almost" constant H_k curvature
QUESTION [6 upvotes]: Let $M$ be an Eucledean $n$-dimensional compact hypersurface with constant $H_k$ curvature, where $k=1,...n$. A theorem by A.Ros tell us that so $M$ is an Eucledean sphere. Does anybody know if there are generalizations of such theorem?
In particular, I need something for Eucledean compact convex hypersurfaces that have $H_k$ curvature bounded and "almost" constant, in the sense that the measure of regions where $H_k$ is not constant is very small.
More precisely: let {$M_\epsilon$} be a family of convex compact Eucledean $n$-dimensional hypersurfaces depending by a parameter $\epsilon>0$. $M_\epsilon$ is $H_k$-constant except for a region whose measure is smaller than $\epsilon$. Furthermore we know that $H_k$ is bounded on $M_\epsilon$ uniformely respect to $\epsilon$.
Does exist some result that allows to say that when $\epsilon$ approaches to zero, then $M_\epsilon$ is "close" to a sphere?
A reference for A.Ros theorem is"Compact hypersurfaces with constant higher order mean curvatures." Rev. Mat. Iberoamericana 3 (1987), no. 3-4, 447–453.

REPLY [6 votes]: One can construct surfaces with mean curvature uniformly close to $1$, but look like collections of almost-touching spheres connected by perturbed catenoidal necks. The work of Ciraolo-Maggi (here) shows a certain stability result for these configurations. See the references therein for the constructions of these surfaces, and a discussion of other results (e.g. Hausdorff closeness to a single sphere with more hypotheses, like convexity or smallness of oscillation of mean curvature relative to the largest principal curvature).<|endoftext|>
TITLE: Continuity of solutions to $Av=b$
QUESTION [5 upvotes]: Let $X$ be a compact Hausdorff topological space, and $C(X)$ denote the ring of complex-valued, continuous functions on $X$. Let $A$ be a matrix with entries from $C(X)$ of size $m\times n$ and $b\in \mathbb{C}^{m\times 1}$.  Suppose that for each $x\in X$, the equation $A(x) v=b$ has a solution $v\in \mathbb{C}^n$. Then does there exists an $V\in C(X)^{n\times 1}$ such that $AV=b$?

REPLY [7 votes]: No. Take $n=m=2$, $A(x)=\pmatrix{1&0\\x&x^2}$, $x\in \mathbb{R}$, $b=\pmatrix{1\\0}$. For $x\ne 0$ the only solution of $AV=b$ is $V(x)=\pmatrix{1\\-1/x}$, it has discontinuity at 0 for any choice of $V(0)$.<|endoftext|>
TITLE: Waldhausen and Segal's delooping machinery
QUESTION [6 upvotes]: I was recently thinking about the proof of a theorem where Waldhausen compared the Segal's delooping machinery with his, in the case when the cofibration is splittable (sec.1.8 in 'Algebraic $K$-theory of space').
Assume $\mathcal{C}$ is a Waldhausen category. By forgetting some structures, one obtains a category with sum (and weak equivalence). We denote by $wN_{\cdot}$ the Segal's delooping machinery developed in `Categories and cohomology theories´
Now the comparison map is
$$wN_{\cdot}\mathcal{C}\rightarrow wS_{\cdot}\mathcal{C},$$ 
and by delooping them once, he turned the question into proving 
$$wN_{\cdot}N_{\cdot}\mathcal{C}\rightarrow wN_{\cdot}S_{\cdot}\mathcal{C}$$ 
is a homotopy equivalence. Then he purposed to prove an additive theorem for $N_{\cdot}$-construction, namely, to show 
$$wN_{\cdot}S_{n}\mathcal{C}\rightarrow (wN_{\cdot}\mathcal{C})^{n}$$
is a homotopy equivalence (Prop. 1.8.7). It is at this stage in his proof the splittabe condition is made use of. However, instead of using the approach in his proof of the additive theorem for $S_{\cdot}$- construction, he used a quite different way to show this.(see Remark below) 

My question or confusion is that: Can we still imitate the method he used in the proof of the additive theorem for $S_{\cdot}$-construction and apply it to the $N_{\cdot}$-construction, meaning, we first try to prove  
$$ wN_{\cdot}E(\mathcal{C})\rightarrow wN_{\cdot}\mathcal{C}\times wN_{\cdot}\mathcal{C}$$ 
is a homotopy equivalence, where $E(\mathcal{C})$ is the category of cofibration sequence or $S_{2}\mathcal{C}$.
  And then by the other equivalent statements of the additive theorem, one can obtain the homotopy equivalence 
  $$wN_{\cdot}S_{n}\mathcal{C}\rightarrow (wN_{\cdot}\mathcal{C})^{n}.$$ 
I saw no obvious obstruction. For example, one can replace $s_{k}\mathcal{C}\equiv Obj(S_{k}\mathcal{C})$ by  $n_{k}(\mathcal{C})\equiv Obj(N_{k}\mathcal{C})$ and $E(S_{n}\mathcal{C})$ by $E(N_{n}\mathcal{C})$, and then proceed with his argument in sec. 1.4 (the additive theorem) in the same article. But if it is the case, then we don't need splittable condition any more. I wonder where I get wrong. It seems to me very strange. They shouldn't be identical without splittable condition, right? 

Remark: Waldhausen in his article used a quite different approach to prove the statement:
$$wN_{\cdot}S_{n}\mathcal{C}\rightarrow wN_{\cdot}S_{n-1}\mathcal
{C}\times wN_{\cdot}\mathcal{C}$$ is a homotopy equivalence. He first used the homotopy fibration sequence 
$$wN_{\cdot}\mathcal{D}\rightarrow wN_{\cdot}N_{\cdot}(\mathcal{C}\rightarrow \mathcal{D})\rightarrow wN_{\cdot}N_{\cdot}\mathcal{D}$$
to turn the question into showing $$wN_{\cdot}(j_{n}:\mathcal{C}\rightarrow S_{n}\mathcal{C})\rightarrow wS_{n-1}\mathcal
{C}$$ is a homotopy equivalence. Then by studying its fiber, he concluded it is a homotopy equivalence when the cofibration is splittable.
I appreciate any comment or suggestion, and thank you in advance.

REPLY [9 votes]: There is no hope of showing that $w N. E(\mathcal C) \to w N. \mathcal C \times w N. \mathcal C$ is a homotopy equivalence without assuming that the cofibrations of $\mathcal C$ are splittable.  To see that, take a simple example and look at the direct sum Grothendieck groups which appear as $\pi_1$ of these K-theory spaces.  Let $\mathcal C$ be the category of finitely generated abelian groups with the non-split short exact sequence $M \in E(\mathcal C)$ given by $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/2 \to 0$, and consider the class $[M]$ in $K_0^\oplus \mathcal C$.  From an equation $[M]=[M']$, where $M'$ is a splittable short exact sequence, we can deduce that $M$ and $M'$ are stably isomorphic, i.e., there is another short exact sequence $N$ with $M \oplus N \cong M' \oplus N$.  The functor $E( \mathcal C ) \to \mathcal C$ that computes the kernel of the tensor product with $\mathbb Z/2$ of the first map in the short exact sequence is an additive functor that sends $M$ to $\mathbb Z/2$, $M'$ to zero, and $N$ to a finite abelian 2-group, yielding an isomorphism between two finite abelian groups of different order, giving a contradiction.<|endoftext|>
TITLE: How many closed measure zero sets are needed to cover the real line, really?
QUESTION [7 upvotes]: This is a refinement of an earlier question.
This question assumes familiarity with combinatorial cardinal characteristics of the continuum.
For the reader's convenience, I reproduce below the relevant parts.
Let $\mathcal{E}$ be the $\sigma$-ideal generated by closed measure zero subsets of the real line. It is known that
$$\operatorname{cov}(\mathcal{N})\cdot\operatorname{cov}(\mathcal{M})\le 
\operatorname{cov}(\mathcal{E})\le 
\min\{\operatorname{cov}(\mathcal{N})\cdot \mathfrak{d},\mathfrak{r}\}$$
and that the first inequality is consistently strict
(Bartoszynski-Shelah, Ashutosh).
Is the second inequality consistently strict?

REPLY [4 votes]: Mathias model (i.e., the countable support iteration of length $\omega_2$ of Mathias poset over a model of CH) satisfies $\mathrm{cov}(\mathcal{E})<\mathfrak{b}$ (recall that both $\mathfrak{d}$ and $\mathfrak{r}$ are above $\mathfrak{b}$). Refeer to Bartoszynski-Judah book Set Theory: On the structure of the real line for the citations below.
Mathias forcing adds a dominating real (Lemma 7.4.4). On the other hand, Mathias forcing satisfies the Laver property (Section 7.4A). It is enough to show that any poset $\mathbb{P}$ with the Laver property forces that the closed measure zero sets from the ground model covers the reals in the extension.
Let $\dot{x}$ be a name for a real in $\mathbb{P}$ and $p\in\mathbb{P}$. Let $\dot{h}$ be a name for a member of $\prod_{n<\omega}2^{2n}$ such that $\dot{h}(n)$ codes $\dot{x}\upharpoonright I_n$ where $\langle I_n\rangle_{n<\omega}$ is the interval partition with $|I_n|=2n$. By the Laver property, there are $q\leq p$ in $\mathbb{P}$ and $S\in\prod_{n<\omega}\mathcal{P}(2^{2n})$ such that $|S(n)|\leq 2^n$ and $q\Vdash\dot{h}(n)\in S(n)$ for all $n<\omega$. Now, $C_S:=\{z\in 2^\omega:\forall_{n<\omega}(z\upharpoonright I_n\in S(n))\}$ is a closed measure zero set (coded in the ground model) and $q\Vdash\dot{x}\in C_S$ (elements for this argument are taken from Section 2.6A).
Other example is a countable support iteration of length $\omega_2$, over a model of CH, alternating between Miller forcing and a proper poset with the Laver property that increases $\mathfrak{r}$ (e.g. Silver forcing, which has the Sacks property).
I also have some examples with finite support iterations of ccc posets (so large continuum is possible, plus having many invariants with different values), though they involve many technicalities (like preservation theorems) not published so far.<|endoftext|>
TITLE: What are the advantages of the more abstract approaches to nonstandard analysis?
QUESTION [36 upvotes]: This question does not concern the comparative merits of standard (SA) and nonstandard (NSA) analysis but rather a comparison of different approaches to NSA. What are the concrete advantages of the abstract approaches to NSA (e.g., via the compactness theorem), as compared to the more concrete approach using ultrapowers? One can name generic reasons such as naturality, functoriality, categoricity, etc., but I am hoping for a concrete illustration of why a more abstract approach may be advantageous for understanding NSA concepts and/or proving theorems.
Note 1. One of the existing answers provided a bit of information about advantages of the more abstract approach in terms of saturation. I would appreciate an elaboration of this if possible, in terms of a concrete application of saturation.
Note 2. These issues are explored in more detail in this 2017 publication in Real Analysis Exchange.

REPLY [7 votes]: You can quote me on this if you like.
This is such an old issue that I am surprised it is still up for discussion.   I know you like ultra-powers, etc,  but I have always thought they were wrong-headed;  it was Luxemburg who made them popular for NSA originally.
The point is that you actually really NEED the transfer theorems; so you essentially need the logical apparatus; in most cases the compactness theorem and some form of saturation.    Occasionally a type omitting argument could be used but that is rare.
The reason is that you are postulating that "all of mathematics" carries over to the
non-standard model and in fact, that is the underlying intuition in the applications.
I also think that the "more concreteness" of the ultraproduct construction is just an illusion.     You can not answer if the integer produced by (1,2,3,4,5, ...) is even or odd
since it depends on the ultrafilter completion of e.g. any non-principal filter.
Moreover, the basic tools in almost all the proofs is dealing with the seam between
"standard" and "non-standard" elements; which really depends on a feel for expressibility in the language being used.  Nelson's approach was to try to help people manage without the "feel".  That's a matter of taste.
Furthermore, I would differ with your point that NSA is to formalize classical mathematicians arguments.  That is "cool" and Robinson greatly enjoyed it, but really he also was sure that it is a great way to prove new theorems.    Since people sort of like to
hear about the theorems consequence in "classical" settings; then one often had to find
equivalence theorems (this was clear in the work on brownian motion etc) and in my work on inverse limits of finite groups  (by the uniqueness of inverse limits, the non-standard finite groups are essentially equivalent to inverse limits of systems of finite groups and a lot of the very early work of Lubotzky on profinite groups can be easily carried out in this setting)  but actually
a more extreme position can be seen when you just take the position that the "standard"
results are just one specific implementation and one could develop mathematics happily without them.    The argument that the non-standard models are not "unique" is just a habit and not important.   
Robinson himself put this viewpoint forward very nicely in his Brouwer medal address  "Standard and NonStandard Number Systems"
Best regards
Larry Manevitz
manevitz@cs.haifa.ac.il
Regarding the use of limit ultrapowers etc; one needs to be careful there.  I once proved there was no measureable cardinal by taking such a large ultrapower; but actually all the
standard sets dont grow in that context :).<|endoftext|>
TITLE: Decay of real continuous algebraic functions at infinity
QUESTION [9 upvotes]: Let $f$ be a real valued continuous algebraic function on $\mathbb R^n$. Suppose the zero set of $f$ is bounded, i.e., if $|x|$ is large enough, $f(x)\neq 0$. Is there any estimate of the sort $|f(x)|\ge c(1+|x|)^{-N}$ for some $c>0$ and $n\in\mathbb N$? A prototypical example of such an inequality is $\sqrt{1+x^2}-x\ge c(1+|x|)^{-1}$, but I would appreciate general estimates along this line that apply to algebraic functions of $n$ variables.

REPLY [2 votes]: Your function $f$ is, in particular, a continuous semi-algebraic function on $\mathbf R^n$, i.e., its graph is a semi-algebraic subset of $\mathbf R^{n+1}$. Such functions are known to have sub-polynomial growth. More precisely (cf. Bochnak, Coste, Roy: Real algebraic geometry, Proposition 2.6.2, p. 43):
Let $S\subseteq\mathbf R^n$ be a closed semi-algebraic subset, and let $f\colon S\rightarrow \mathbf R$ be a continuous semi-algebraic function. Then there are $c\in\mathbf R$ and $N\in\mathbf N$ such that
$$
|f(x)|\leq c(1+||x||^2)^N
$$
for all $x\in S$.
Apply this to the inverse $1/f$ of your function on the closed semi-algebraic subset $||x||\geq A$, where $A\in\mathbf R$ is such that all zeros of $f$ have norm $<A$.<|endoftext|>
TITLE: Sums of twisted products of Kloosterman Sums
QUESTION [8 upvotes]: For $m,n,c \in \mathbb{N}$, let $S(m,n;c)$ denote the Kloosterman sum
$$
S(m,n;c) := \sum_{\substack{1 \leq a < c \\ \gcd(a,c) = 1}} e \left( \frac{ma + n\overline{a}}{c} \right)
$$
where $e(n) = e^{2 \pi i n}$ and $\overline{a}$ denotes the multiplicative inverse of $a \bmod c$.
In my research, involving producing a subconvexity bound for automorphic L-functions, I've recently come across a twisted shifted sum of Kloosterman sums. Let $\chi(\cdot)$ denote a Dirichlet character mod a prime $p$. Then I'm looking at
$$ F(a, h, \chi) =  \sum_{b \bmod p} \chi(b) S(a, b; p) S(a, b + h; p) \tag{1}.$$
I've never seen sums like this appear, but it looks pretty complicated. A first thing to consider might be an upper bound. We can produce a trivial upper bound using the Weil bound for Kloosterman sums, which indicates that $(1)$ is bounded above by $p^{2 + \epsilon}$, independently of $a,h, \chi$. But I think we should expect much smaller, at the most $p^{3/2 + \epsilon}$.

So I am wondering if someone has considered sums similar to $(1)$. I would also be interested in considerations of the similar but simpler sums
  $$ \begin{align}
F(a, 1, 1) &=  \sum_{b \bmod p} S(a, b; p) S(a, b + 1; p) \tag{2} \\
F(a, 1, \chi) &=  \sum_{b \bmod p} \chi(b) S(a, b; p) S(a, b + 1; p). \tag{3}
\end{align}$$

REPLY [7 votes]: Your sum is $$\sum_{i=0}^2 (-1)^i \operatorname{tr}(\operatorname{Frob}_p, H^i_c( \mathbb A^1_{\overline{\mathbb F}_p}, \mathcal{K}\ell_2 (ab) \otimes \mathcal{K}\ell_2 (a(b+h)) \otimes \mathcal L_\chi)$$
The $H^1$ term corresponds to a square-root canncelation bound, the $H^0_c$ term vanishes by definition, and so it is sufficient to show the $H^2_c$ term vanishes.
By Poincare duality, the $H^2_c$ term is equal to the monodromy coinvariants of  $\mathcal{K}\ell_2 (ab) \otimes \mathcal{K}\ell_2 (a(b+h)) \otimes \mathcal L_\chi$. We can show this vanishes by the following arguments:
($\chi$ nontrivial) There are no local monodromy invariants at $0$, since $\mathcal{K}\ell_2(ab)$ has unipotent local monodromy at $0$, whereas $\mathcal L_\chi$ has local monodromy a nontrivial tame character.
($h$ nontrivial) One can do something similar using the irreducibility of $\mathcal{K}\ell_2$. However this is superfluous because, as noted by Alexey Ustinov in the comments, the sum simplifies in the $\chi$ trivial case.<|endoftext|>
TITLE: Is $L_q(X^*)$ complemented in $(L_p(X))^*$?
QUESTION [6 upvotes]: Let $X$ be a Banach space and let $p\in (1,\infty)$. If $q$ denotes the conjugate exponent to $p$, then $L_q(X^*)$ is easily seen to be isometric to a subspace of $(L_p(X))^*$ via the map $$f\mapsto \int\limits_{[0,1]}\langle \cdot(\omega),   f(\omega)\rangle\,{\rm d}\omega\quad (f\in L_q(X^*)).$$ One of the numerous characterisations of the Radon–Nikodym property asserts that $X^*$ has this property with respect to $[0,1]$ endowed with the Lebesgue measure if and only if this map is surjective.

Suppose that $X^*$ fails the Radon–Nikodym property with respect to the Lebesgue measure. Is the range of the above-mentioned map complemented? 

My motivation follows from this question of M. González; if the answer were positive, then we would have $\ell_2(X)\not\cong L_2(X)$ for every infinite-dimensional space $X$ with $X^*\cong L_1(\mu)$ and for every infinite-dimensional C*-algebra.
Edit (26.01.2016): This is indeed a known result. It follows from Theorem 9 in:

Z. Hu and B.-L. Lin, Extremal structure of the unit ball of $L_p(\mu, X)^*$, J. Math. Anal. Appl. 200 (1996), 567–590.

REPLY [2 votes]: Still no answers?  Perhaps we should try the following.  
Let $g : [0,1] \to X^*$ be weak* measurable, in the sense that, for every $x \in X$, the function
$$
\omega \mapsto \langle x, g(\omega)\rangle
$$
is measurable.  Then there exist weak* measurable functions $g_1, g_2 : [0,1] \to X^*$ such that 
(1) $g$ is weak* equivalent to $g_1+g_2$, in the sense that, for every $x \in X$,
$$
\langle x,g(\omega)\rangle = \langle x,g_1(\omega)+g_2(\omega)\rangle
$$
for almost all $\omega \in [0,1]$.  (But the exceptional nullset depends on $x$.)  
(2) $g_1$ is Bochner measurable in the sense that the range of $g_1$ is separable.  
(3) $g_2$ is totally non-separable in the sense that: for every separable $Y \subseteq X^*$, the outer measure of $g_2^{-1}(Y)$ is zero.  
If this works, then perhaps the elements of
$(L_p(X))^*$ can be identified as such weak* measurable functions $g$, and the projection is the map sending $g$ to its "Bochner part" $g_1$.<|endoftext|>
TITLE: Can a large transitive permutation group need many generators?
QUESTION [12 upvotes]: let $G$ be a transitive permutation group acting on $\{1, \ldots, n\}$, and let $d(G)$ be the minimal number of generators of $G$. Is it true, that for $n\rightarrow\infty$ we have $\frac{d(G)\log|G|}{n^2}\rightarrow 0$? If this is true, can you give a complete list of groups with $\frac{d(G)\log |G|}{n^2}\geq \frac{\log 2}{4}$, i.e. groups which are "worse" than $C_2$?
I believe the answer to the first question is yes, because a transitive group on $n$ letters needs $\mathcal{O}(\frac{n}{\sqrt{\log n}})$ generators, thus if $\frac{d(G)\log|G|}{n^2}$ is large, then $|G|>e^{cn\sqrt{\log n}}$. But then $G$ must involve quite large alternating or symmetric sections, which should imply that $G$ is actually quite easy to generate. I guess that the answer to the second question involves only subgroups of $S_4$, although I am not too certain about that.

REPLY [11 votes]: Let $G$ be a permutation group of degree n, and let $r>1$ be the minimal block size of $G$. Also, let $s:=n/r$, so that G may be viewed as a subgroup in the wreath product $R\wr S$, where $R\le Sym(r)$ is primitive, and $S=\pi(G)\le Sym(s)$ is transitive ($\pi$ here denotes projection $G\rightarrow Sym(s)$). Then 
$$d(G)\le \frac{a(R)bs}{\sqrt{\log_{2}{s}}}+d(S) \text{   ($\ast\ast$})$$
where b is an absolute constant, and $a(R)$ denotes the composition length of R. By a result of Pyber and Guralnick, $a(R) \le C\log_{2}(r)$ for an absolute constant $C$. Also, as you mentioned, $d(S)\le cs/\sqrt{\log_{2}{s}}$, for an absolute constant c. (For information on the constants $b,c$, and $C$, and on the bound at ($\ast\ast$), see http://arxiv.org/abs/1504.07506 .)
Now, we also have $\log{|G|}\le \log{|R\wr S|}\le rs\log{r}+s\log{s}$. This means that your first claim follows when $r>\sqrt{\log{s}}$, for example. Thus, it only remains to deal with the case $r\le \sqrt{\log{s}}$.
As you said, the key seems to be reducing the upper bounds when S is either $Alt(s)$ or $Sym(s)$. However, I think that a bound of the form $d(G)\le c'd(R)+2$ would be difficult to prove in this case, and is probably not true. For example, when $R=C_2$ and $K$ is the intersection of $G$ with the base group of the wreath product, then the number of generators for $K$ as an $S$-group can grow as $bs/\sqrt{\log{s}}$, where b is as above (see Section 3.1 of the paper I mentioned above).  
EDIT: @Jan-Christoph Schlage-Puchta The first part of your question, that is, that for a transitive permutation group of degree $n$, $\frac{d(G)\log{|G|}}{n^{2}}$ tends to $0$ as $n$ tends to $\infty$, has now been proved in http://arxiv.org/abs/1601.02561 .<|endoftext|>
TITLE: Is this generalization of eigenvalue and eigenvector studied?
QUESTION [11 upvotes]: While thinking about what it means for observables to be simultaneously measurable in quantum mechanics I came up with the following concepts, which I will call "linearly indexed" versions of standard linear algebra concepts.

A $W$-indexed linear map $L: V \to V$ is a linear map $W \otimes V \to W \otimes V$.
A $W$-indexed eigenspace for $L$ is a subspace $V'$ of $V$ such that $L = \lambda \otimes I$ on $W \otimes V'$.
$\lambda : W \to W$ is called the $W$-indexed eigenvalue of $L$ for the $W$-indexed eigenspace $V'$.

When the spaces are over $\mathbb{C}$, say, and $W = \mathbb{C}$, we just recover the normal definition of eigenspace and eigenvalue $\lambda \in \mathbb{C}$.
It seems to me that these concepts allow us to talk about multidimensional observables of a quantum mechanical system in a direct way.  An observable on a space of states $V$ is a Hermitian $W$-indexed linear map $V \to V$, and the values it can take are its $W$-indexed eigenvalues, which should to be Hermitian operators $W \to W$.
Is this something that is known and  studied, or is this a red herring?

REPLY [3 votes]: The definition of $W$-indexing requires some further clarification. As stated right now, all three properties are satisfied by the classic Kronecker product. 
To illustrate, suppose $V=\mathbb{C}^n$, $W=\mathbb{C}^m$. For argument's sake, let us assume that $L$ has simple eigenvalues, such that a diagonalization exists $L=X\Lambda X^{-1}$, and each eigenspace is dimension-1.
Then for any $w\in W$, $v\in V$, and any bijective linear map $D:W\to W$ which is also an invertible $m\times m$ matrix, we have

The $W$-indexed linear map $L$ is $\hat{L}=D\otimes L$. The domain is $w \otimes v \in W\times V$ and the range is $\hat{L}(w\otimes v)=(Dw)\otimes (Lv)\in W\times V$, so long as $L$ is nonsingular.
For any eigenpair $\lambda_i, x_i$ of $L$, we have $\hat{L}(w \otimes x_i)= (Dw) \otimes (Lx_i) = (Dw) \otimes (\lambda_i x_i) = (\lambda_i D\otimes I_n)(w \otimes x_i).$ Again, so long as $L$ is nonsingular, we have $\lambda_i\ne0$, so the range is $(\lambda_i D\otimes I_n)(w \otimes x_i) \in W \times x_i$
In the example above, the matrix $\lambda_i D$ is a $W\to W$ mapping that fits the definition of a $W$-indexed eigenvalue.

Of course if we set $D=1$, then we recover the usual eigenvalues and eigenvectors, as per stated.
Given any arbitrary $A$, $B$, the eigenvalues, eigenvectors, Jordan form, singular decomposition, etc, of $A\otimes B$ can all be expressed using properties of $A$ and $B$, and some basic linear algebra identities. See the wiki page.<|endoftext|>
TITLE: What is... A Grossone?
QUESTION [35 upvotes]: Y. Sergeyev developed a positional system for representing infinite numbers using a basic unit called a "grossone", as well as what he calls an "infinity computer".  The mathematical value of this seems dubious but numerous articles have already appeared in refereed research journals.  Thus, there are currently 23 such articles in mathscinet not to speak of numerous lectures in conferences.
In a comment accessible here Sergeyev asserts that "Levi-Civita numbers are built using a generic infinitesimal $\varepsilon$ ... whereas our numerical computations with finite quantities are concrete and not generic."  Here apparently "finite" is a misprint and should be "infinite". How is this comment on the difference between Sergeyev's grossone one the one hand, and the Levi-Civita unit on the other, to be understood?
In a 2013 article, Sergeyev compares his grossone to Levi-Civita in the following terms in footnote 5: 5 At the first glance the numerals (7) can remind numbers from the Levi-Civita field (see [20]) that is a very interesting and important precedent of algebraic manipulations with infinities and infinitesimals. However, the two mathematical objects have several crucial differences. They have been introduced for different purposes by using two mathematical languages having different accuracies and on the basis of different methodological foundations. In fact, Levi-Civita does not discuss the distinction between numbers and numerals. His numbers have neither cardinal nor ordinal properties; they are build using a generic infinitesimal and only its rational powers are allowed; he uses symbol 1 in his construction; there is no any numeral system that would allow one to assign numerical values to these numbers; it is not explained how it would be possible to pass from d a generic infinitesimal h to a concrete one (see also the discussion above on the distinction between numbers and numerals). In no way the said above should be considered as a criticism with respect to results of Levi-Civita. The above discussion has been introduced in this text just to underline that we are in front of two different mathematical tools that should be used in different mathematical contexts. It would be interesting to have a specialist in numerical analysis comment on Sergeyev's use of the term "numerical" to explain the difference between his grossone and an infinite element of the Levi-Civita field.
Sergeyev claims that his grossone has the properties of both ordinal and cardinal numbers. Does he give a definition that would ensure such properties, or is this claim merely a declarative pronouncement?
Following the publication of an article by Sergeyev in EMS Surveys in Mathematical Sciences, the editors published the following clarification:

Statement of the editorial board
We deeply regret that this article appears in this issue of the EMS Surveys in Mathematical Sciences.
It was a serious mistake to accept it for publication. Owing to an unfortunate error, the entire processing of the paper, including the decision to accept it, took place without the editorial board being aware of what was happening. The editorial board unanimously dissociates itself from this decision. It is not representative of the very high level that we expect to see in our journal, which can be assessed from all other papers that we have published.
Both editors-in-chief have assumed responsibility for these mistakes and resigned from their position. Having said that, we add that this journal would not exist without their dedication and years of hard work, and we wish to register our thanks to them.

An interesting viewpoint of a computer scientist is developed here (as well as a related discussion of legal issues in the comments).
The unanimous statement of the EMS Surveys editors is now fleshed out in the Zentralblatt review and the MathSciNet review also available here.

REPLY [8 votes]: Mathematics  is the search of truth by way of proof, as defined by Mac Lane. Mathematics is not a collection of opinions or hyphotheses about what someone has meant.  
Sergeyev defines  his grossone as follows: "We have introduced grossone as the quantity of natural numbers. Thus, it is the biggest natural number... " See, for instance, Annales UMCS Informatica AI 4 (2006) p.26 
This is not what we encounter by the way of proof.<|endoftext|>
TITLE: What is known about $\sum_{n \leq x} \mu(n) \varphi(n)$?
QUESTION [16 upvotes]: Let $\mu(n)$ denote the Möbius function and $\varphi(n)$ the Euler-phi function.  What is known about $f(x) = \sum_{n \leq x} \mu(n) \varphi(n)$?  For example:

Is it known that $f(x)$ grows without bound?
Is it known that $-f(x)$ grows without bound?
Is it known that $f(x)$ crosses the origin infinitely often?
Are there are any solutions to $f(x) = 0$ for $x \geq 3$?  (I didn't find any
for $x < 10^8$.)
What are some good upper and lower bounds for $|f(x)|$?
How many times does $f$ cross the origin for $x < X$, as a function of $X$?

REPLY [14 votes]: Fleshing out Ofir Gorodetsky's comment: if we define $G(s) = \sum_{n=1}^\infty \mu(n)\phi(n)n^{-s}$, then we have $G(s) = F(s)/\zeta(s-1)$ where
$$
F(s) = \prod_p \bigg( 1 - \frac1{p^s-p} \bigg)
$$
is absolutely convergent for $\Re s>1$. The rightmost singularities of $G(s)$ are therefore at the points $1+\rho$ where $\rho$ denotes nontrivial zeros of $\zeta(s)$. Assuming the Riemann hypothesis, we thus expect $f(x)/x^{3/2} = x^{-3/2} \sum_{n\le x} \mu(n)\phi(n)$ to have a limiting logarithmic distribution, which will be the same as the distribution of the random variable
$$
\sum_{\gamma} \bigg |\frac{F(3/2+i\gamma)}{(3/2+i\gamma)\zeta'(1/2+i\gamma)}\bigg| \Re Z_\gamma,
$$
where $\gamma$ runs over the positive imaginary parts of zeros of $\zeta(s)$, and $\{Z_\gamma\}$ is an independent set of random variables each uniformly distributed on the unit circle in $\Bbb C$. (The independence of the $\{Z_\gamma\}$ is based on our belief that the imaginary parts of the nontrivial zeros of $\zeta(s)$ are linearly independent over the rationals.)
This distribution will be roughly bell-shaped but not normal, indeed probably decaying faster than a normal distribution. The function $f(e^y)/e^{3y/2}$ will be an almost periodic function with mean $0$, but will presumably be unbounded above and unbounded below. Numerical computations up to $x=10^6$ support these claims (other than the unboundedness, which will be extremely gradually realized).
This sort of statement, for other functions, goes back to Littlewood's oscillation theorems and continues today in the subfield of comparative prime number theory. See "Limiting distributions of the classical error terms of prime number theory" by Akbary, Ng, and Shahabi for a general framework that addresses problems such as this.<|endoftext|>
TITLE: $n + 1 = 2^rm$ with $m$ odd $\implies$ do not exist $2^r$ vector fields on $\mathbb{P}^n$ that are everywhere linearly independent?
QUESTION [5 upvotes]: What is the easiest/quickest way to see the following?

If $n + 1 = 2^rm$ with $m$ odd, then there do not exist $2^r$ vector fields on the projective space $\mathbb{P}^n$ which are everywhere linearly independent.

Thanks!

REPLY [5 votes]: The standard way to treat statements like this one is to analyze the Stiefel-Whitney classes of $\mathbb P^n$. The full Stiefel-Whitney class of the tangent bundle is
$$w(T\mathbb P^n)=(1+u)^{n+1},$$
where $u\in H^1(\mathbb P^n,\mathbb Z_2)$ is the generator.
In your particualr case $(n+1)=2^rm$ and let $p=2^r(m-1)$. Then $w_p(T\mathbb P^n)=\binom{n+1}{p}u^p=u^p\neq 0$. Here we used the fact $\binom{2^rm}{2^r}=1 \pmod 2$.
On the other hand, if there are $k$ linearly independent vector fields on $\mathbb P^n$, then $T\mathbb P^n=E\oplus \mathbb R^k$ and, therefore, $w_q(T\mathbb P^n)=w_q(E)=0$ for $q>n-k$. Combining with the fact $w_p(T\mathbb P^n)\neq 0$, we conclude $k<n-(p-1)=2^r$.<|endoftext|>
TITLE: Reflex fields of Shimura varieties
QUESTION [7 upvotes]: I am currently learning the theory of Shimura varieties. Out of curiosity, is it known which number fields can occur as reflex fields? More precisely, can one find, for any number field, a positive dimensional Shimura variety which has this field as its reflex field?

REPLY [8 votes]: The answer depends on your definition of a Shimura pair $(G,X)$.
Look in Section 2.1 of Deligne's paper.
If you assume only  axioms (2.1.1.1), (2.1.1.2) and (2.1.1.3), then any number field $F$ can occur as the reflex field $E(G,X)$ with $X$ of positive dimension. 
Indeed, take $G_1=R_{F/\mathbf{Q}}\mathbb{G}_{m,F}$, then for suitable $h_1\colon\mathbf{S}\to G_{1,\mathbf{R}}$ we have $E(G_1,h_1)=F$. 
We take $X_1=\{h_1\}$, then $\mathrm{dim}_\mathbf{C}(X_1)=0$. We take $(G_2,X_2)=(\mathrm{GL}_{2,\mathbf{Q}},X_2)$ (the standard Shimura pair for $\mathrm{GL}_{2}$), then $E(G_2,X_2)=\mathbf{Q}$, $\mathrm{dim}(X_2)=1$. 
Set $G=G_1\times_\mathbf{Q} G_2$, $X=X_1\times X_2$. 
Then $\mathrm{dim}_\mathbf{C}(X)=1$, $E(G,X)=E(G_1,X_1)=F$.
However, if you  assume also axioms (2.1.1.4) and (2.1.1.5), then $E(G,X)$ must be either a totally real field or a CM-field. 
In order to show this, it suffices to consider the case when $G$ is $\mathbf{Q}$-simple adjoint and the case when $G$ is a torus. 
For the $\mathbf{Q}$-simple adjoint case see Section 2.3.4 and Proposition 2.3.6 in Deligne's paper. In the toric case, axioms (2.1.1.4) and (2.1.1.5) imply that the torus $G$ is isogenous to the product of $\mathbb{G}_{m,\mathbf{Q}}$ and a  $\mathbf{Q}$-torus which is compact over $\mathbf{R}$,  hence it splits over a CM-field, and the assertion follows.

REPLY [2 votes]: Consider the Shimura variety defined by a datum $(G,X)$, and let $T$ denote the quotient of $G$ by its
derived group, so $T$ is a torus over $\mathbb{Q}$. If $T$ splits over a CM-field then the reflex field is
either totally real or CM.  This takes care of most "naturally occurring" Shimura varieties. However, according
to Deligne's definition, you get a Shimura variety from any torus over $\mathbb{Q}$ and cocharacter, 
and then the reflex field is the field of definition of the cocharacter, which can probably be anything.<|endoftext|>
TITLE: The Picard number of the Kummer surface of an abelian surface
QUESTION [11 upvotes]: Let $A$ be an abelian surface and $\text{Km}(A)$ be the Kummer surface of $A$. If I remember correctly, the Picard number $\rho(\text{Km}(A))$ is equal to $16+\rho(A)$. 

Does anyone know any reference or proof for this fact?

REPLY [6 votes]: I have written up a self-contained proof of this fact. It is available on Thuses. The proof is entirely algebraic, works over any separably closed field of $\text{char} \neq 2$, and does not use any Hodge theory.<|endoftext|>
TITLE: closed form solution of the following iterative equation?
QUESTION [5 upvotes]: is it possible to obtain a closed-form solution w.r.t. ${P_j:\forall j}$ (or in terms of special functions) for the following equations:
$\alpha P_0=P_1$, $\alpha<1$
$\alpha P_j=P_{j+1}+P_{j+2}+\dots+P_{2j+1}$ for $j=1,2,....$
$\sum_{i=1}^\infty P_i=1$
$P_i\geq 0, \forall i$  

OR let me put the very original equations below:
$\lambda P_0=\mu P_1$
$\lambda P_{j-1} + \mu (P_{2j}+P_{2j+1})=(\lambda+\mu)P_j, \forall j>0$
$\sum_{i=1}^\infty P_i=1$
$P_i\geq 0, \forall i$

REPLY [2 votes]: There is no solution to the problem in the first version of the OP.
Proof:
We have to consider two cases:
Case a)  at least one of the $P_j$ is zero
The let $k$ be the smallest index for which $P_k = 0$.
Then from
$0 = \alpha P_k = P_{k+1} + ... + P_{2k+1}$
we have 
$P_{k+1} = P_{k+2} = .. = P_{2k+1} = 0$
and, inductively, $P_j = 0$ if $j \ge k$.
On the other hand
$\alpha P_{k-1} = P_{k} + ... = 0$
and so on downwards so that all $P_j = 0$ which contradicts the normalization condition.
Hence we can rule out case a)  
Case b) all $P_j$ are positive
I shall show that there is no solution to the recursive equations with 
(1)  $P_j \gt 0, j = 1, 2, 3, ...$ 
First from 
$\alpha P_1 = P_2 + P_3$
we conclude 
$\alpha \gt 0$
Notice also the $P_0$ appears only in the relation
$\alpha P_0 = P_1$    
which shows that 
$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization.
Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.
Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.
Define
(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$
As a sum over positive quantities we have 
$Q_i \gt 0, i = 0, 1, 2, ...$
The inversion of (2) is
(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $
Now the equations become
$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ...    $
Using (3) we get
$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$
or
(4)   $Q_{2j+1} = (1+\alpha ) Q_j - \alpha  Q_{j-1},    j = 1, 2, ...$     
This is now a recursive relation in standard form.
The inital values are
$Q_0 = P_1 + P_2 + ... = 1$
because of the normalization condition.
And
$Q_1 = 1 - P_1 = 1 - \alpha  P_0$
can be considered as a free parameter in the interval (0,1).
Before we solve (4) we observe that it defines only the elements with an odd index.
Therefore we let 
$Q_{2k} = C_k > 0, k = 1, 2, ...$
with arbitrary $C_k$ in the interval (0,1).
Performing now the first few steps of the solution to (4) the reader will find that
$P_{10} = - C_5 - \alpha  (1+\alpha ) Q_1$
But this is a negative quantity, and the contradiction proves the statement.<|endoftext|>
TITLE: Weakly compact operators between Banach spaces
QUESTION [5 upvotes]: Let $X$ and $Y$ be complex Banach spaces and $B(X,Y)$ be the Banach space
of all bounded operators. An operator $T\in B(X,Y)$ is weakly compact if
$T(\{ x\in X;\; \| x\| \leq 1\})$ is relatively compact in the weak topology
of $Y$. If $X$ or $Y$ is reflexive, then every operator in $B(X,Y)$ is weakly
compact. I guess that the converse holds as well: if every operator in $B(X,Y)$
is weakly compact then either $X$ or $Y$ has to be reflexive. But I cannot
find a satisfactory argument for this. I know about some characterizations of weakly
compact operators: factorization through a reflexive Banach space or continuity with
respect to the right topology in $X$. However it seems to me that there must be a
simple argument which forces that either $X$ or $Y$ is reflexive if every operator
in $B(X,Y)$ is weakly compact. I am asking if someone knows this simple argument
or if there is a paper with an answer to my question.

REPLY [2 votes]: I don't know if the following helps:
Let $B(X,Y) = W(X,Y)$. Then the following hold.
(i) If there exists a surjection $T\in B(X,Y)$ then $Y$ is reflexive.
(ii) If there exists an injection $T\in B(X,Y)$ with closed range then $X$ is reflexive.
Indeed, denote by $B_r$ and $K_r$ the open and closed ball around $0$, respectively. By the open mapping theorem, $TB_1$ is open. Thus, there exists $r > 0$ such that $K_r\subset TB_1\subset TK_1$, i.e., $K_r$ is weakly compact in $Y$, meaning that $Y$ is reflexive. If $T$ is an injection with closed range then a similar argument as above shows that $\operatorname{ran} T$ is reflexive. But as $T : X\to\operatorname{ran} T$ admits a bounded inverse, $X$ is reflexive.<|endoftext|>
TITLE: Relative Characteristic classes
QUESTION [6 upvotes]: A pair of vector bundles over  a  base space $X$ is  a  pair $(E,F)$ where $E$ is a vector bundle over $X$ and $F$ is  a sub-bundle of $E$. Two pairs $(E_{1},F_{1})$  and  $(E_{2}, F_{2})$ are isomorphic if there is  an isomorphism from $E_{1}$ to $E_{2}$ which send $F_{1}$ onto $F_{2}$. The direct sum of two pairs has a natural definition.
Are there some type of characteristic classes with values in the cohomology of base space with the following properties?:
It is trivial $("1")$ on the trivial pairs. Isomorphism pairs have the same characteristic classes. They commute with the pull back operation. They restrict (or have  a reasonable relation)to the usual  characteristic classes on pairs $(E,0)$  and  $(\epsilon_{n}, F)$ where $\epsilon_{n}$ is the trivial $n$ dimensional bundle.

REPLY [8 votes]: You obviously have all characteristic classes of $E$, of $F$, and of $E/F$,
with some relations because $E\cong F\oplus E/F$. To see if there are more, consider the classifying space for your pairs. If you are working over $\Bbbk$, $\mathrm{rk}(F)=k$, $\mathrm{rk}(E)=\ell$, it is the colimit over $n$ of the space of partial flags of length $(k,\ell)$ in $\Bbbk^n$. For a fixed $n$, this is homotopy equivalent to
$$G_{k,\ell,n}(\Bbbk)=U(n,\Bbbk)/(U(k,\Bbbk)\times U(\ell-k,\Bbbk)\times U(n-\ell,\Bbbk))\;.$$
This space is the total space of (at least) three fibre bundles $G_{k,\ell}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{\ell,n}(\Bbbk)$ and $G_{\ell-k,n-k}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{k,n}(\Bbbk)$ and $G_{k,n+k-\ell}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{\ell-k,n}(\Bbbk)$. Hence, you can try the Leray-Serre spectral sequence to find characteristic classes not generated by those mentioned above.
In the holomorphic setting, you see Bott-Chern forms. If $E$ and $F$ are flat, you see the forms described in the appendix of Bismut-Lott.
In both settings, you can sometimes extract cohomological information.
I am not sure there are similar classes in a purely topological setting, but I would personally go for $\Bbbk=\mathbb R$ first and look at $\mathbb Z/2$-valued cohomology.
EDIT: It seems there are no additional characteristic classes that do not come from $F$ or $E/F$. For consider the fibre bundle $G_{\ell-k,n-k}(\Bbbk)\hookrightarrow G_{k,\ell,n}(\Bbbk)\twoheadrightarrow G_{k,n}(\Bbbk)$. For sufficiently large $n$, all classes of $G_{\ell-k,n-k}(\Bbbk)$ of small degree come from the vector bundle $E/F$, and so are pulled back from $G_{k,\ell,n}(\Bbbk)$. Hence by the Leray-Hirsch theorem, in small degrees the cohomology ring of 
$G_{k,\ell,n}(\Bbbk)$ is generated by the characteristic classes of the vector bundles $F$ and $E/F$.

REPLY [6 votes]: The characteristic classes of the virtual bundle $E-F$ seem to have all the properties you asked for.
If $c$ is a characteristic class of vector bundles satisfying a sum formula
$$
c(E\oplus F) = c(E)c(F),
$$
then $c(E-F)$ can be defined via the formula
$$
c(E-F)c(F) = c(E),
$$
or if the base $X$ is compact,
$$
c(E-F) = \frac{c(E)}{c(F)}.
$$<|endoftext|>
TITLE: Does module Hom commute with tensor product in the second variable?
QUESTION [17 upvotes]: Let $A$ be a commutative ring, and $L, M, N$ be $A$-modules. Then is it true that
$$\text{Hom}_A (L, M)\otimes_A N \cong \text{Hom}_A (L, M\otimes_A N)$$
as $A$-modules? 
(Note that there is a natural morphism from the left to the right, I think it's not easy to check it is injective or surjective, but I didn't really do it; also, I think elements of the RHS are hard to decompose, so I don't hope for a (natural) arrow in the opposite direction.)
If this is not true, how about we assume that $A$ is a local ring and $N$ is a flat $A$-module or even a flat local $A$-algebra?
Could anyone give some hint or a proof, or a counterexample?
Other appropriate conditions that guarantee the isomorphism are appreciated.
$\textbf{Edit:}$ My main concern is the case when $A=\mathscr{O}_{\mathbb{C}^n,0}=M, N=\mathscr{E}_{\mathbb{C}^n,0}$, and $L$ is the stalk at $0\in \mathbb{C}^n$ of some coherent $\mathscr{O}_{\mathbb{C}^n}$-module, where $\mathscr{O}_{\mathbb{C}^n}$ and $\mathscr{E}_{\mathbb{C}^n}$ mean the sheaves of holomorphic functions and complex-valued smooth functions on $\mathbb{C}^n$ respectively. The flatness of $\mathscr{E}_{\mathbb{C}^n}$ over $\mathscr{O}_{\mathbb{C}^n}$ can be found here (Theorem 7.2.1), which cites Bernard Malgrange's book 'Ideals of differentiable functions' (page 88, Coro 1.12), and here is another discussion on MO.

REPLY [31 votes]: You can think about tensor products as a kind of colimit; you're asking the hom functor $\text{Hom}_A(L, -)$ to commute with this colimit in the second variable, but usually the hom functor only commutes with limits in the second variable. Dually, you can think about homs as a kind of limit (in the second variable); you're asking the tensor product functor $(-) \otimes_A N$ to commute with this limit, but usually tensor products only commute with colimits. 
This sort of reasoning not only suggests that your statement should be false but suggests what extra hypotheses might make it true: namely, some kind of projectivity hypothesis on $L$, or some kind of flatness hypothesis on $N$. In fact the statement is true if either $L$ or $N$ is finitely presented projective; these conditions are equivalent to requiring that $\text{Hom}_A(L, -)$ commutes with all colimits or that $(-) \otimes_A N$ commutes with all limits respectively. 
But it's also true if $L$ is finitely presented and $N$ is flat! In this case $\text{Hom}_A(L, M)$ is a finite limit (really an iterated finite limit, but this isn't an issue) in $M$, and $(-) \otimes_A N$ preserves it. Dually, it's also true if $N$ is finitely presented and $L$ is projective: in this case $M \otimes_A N$ is a finite colimit in $M$, and $\text{Hom}_A(L, -)$ preserves it. Note that in Neil Strickland's example neither $L$ nor $N$ is finitely presented.<|endoftext|>
TITLE: Ordinals which embed in surreal subfields
QUESTION [5 upvotes]: If $k$ is an ordered field, the least ordinal $s(k)$ which doesn't embed in $(k,<)$ is regular. This is because every interval of an ordered field embeds in every infinite interval so given a strictly increasing map $f: \alpha < s(k) \rightarrow s(k)$ you can define an embedding $(\sup(f(\alpha)),\in) \rightarrow (k,<)$. Moreover $s(k)$ is always uncountable since $s(\mathbb{Q}) = \omega_1$.
Note that these results make use of the field structure of $k$. They fail even for ordered rings such as the ring obtained from $({\omega}^{\omega},\underline{+},\underline{\times})$ (natural operations), by adding additive inverses and extending the operations like one usually does.
Let $\lambda$ be an $\varepsilon$-number. Let $No(\lambda)$ denote the field of surreal numbers of length $<\lambda$. I am trying to figure out what $s(No(\lambda))$ is.
Clearly, $s(No(\lambda))$ is at least equal to ${\lambda}^+$ and at most equal to ${|No(\lambda)|}^+$. Assuming GCH, this yields $s(No(\lambda)) = {\lambda}^+$ when $\lambda$ is itself a cardinal, and ${\lambda}^+ \leq s(No(\lambda)) \leq ({\lambda}^+)^+$ otherwise.
This is just a computation of $|No(\lambda)|$ using elemenraty cardinal arithmetic and the ordered-set-focused definition of $No(\lambda)$. 
Can these results (or their "equivalent" modulo GCH where $2^{\alpha}$ replaces ${\alpha}^+$) be proven in ZFC by taking advantage of the structure of $No(\lambda)$?

REPLY [4 votes]: I claim that $s(\text{No}(\lambda))=\lambda^+$.
To see this, first let me mention that it is a standard exercise
in elementary set theory to prove that there is no increasing or
decreasing $\lambda^+$-sequence in the set ${}^\lambda 2$ of all
binary $\lambda$-sequences, ordered in the lexical order, so that
$s<t$ just in case $s$ is below $t$ at the first point of
difference. I'll let you prove that on your own, or I can post
later if you want.
From this, I claim, it follows that $\lambda^+$ does not embed
into the set of surreal numbers born before $\lambda$. The reason
is that every surreal number born before $\lambda$ is
characterized uniquely by a $<\lambda$-sequence from $\{-1,1\}$,
depending on which side of the cut in the previously born surreals
the new number has landed. The surreal number order agrees exactly
with the lexical order on the these sequences. (A shorter sequence
is below a longer one, if the next digit is $1$, and above, if the
next digit is $-1$; alternatively, imagine adding a tail of $0$s
to compare sequences of difference lengths.) Since $\lambda^+$
does not embed into ${}^\lambda 2$, it cannot embed into
${}^{<\lambda}\{-1,1\}$.
The argument shows that you could even allow the surreals born on
day $\lambda$ itself, continuing up to any particular day
$\beta<\lambda^+$, and still $\lambda^+$ would not embed in.<|endoftext|>
TITLE: Is this modified bound quiver algebra necessarily representation-finite?
QUESTION [6 upvotes]: Suppose that $A = kQ/I$ is a bound quiver algebra for $k$ an algebraically closed field, $Q=(Q_0, Q_1)$ a finite connected quiver with no oriented cycles with no multiple edges or self-loops, and $I$ the ideal of commutative relations when viewing $Q$ as a commutative diagram (say of vector spaces and linear maps over the field $k$).
Suppose that I modify $Q$ by adding a single arrow $\alpha$, without introducing any oriented cycles, self-loops, or multiple edges. Call the new bound quiver algebra $A' = kQ'/I'$, where $I'$ is now the new ideal of commutative relations when viewing $Q'=(Q_0, Q_1\cup{\alpha})$ as a commutative diagram.
My question is this: If $A$ is representation-finite, is it necessarily the case that $A'$ is representation-finite?
It seems to me (inutitively) that it should be true, but doing some preliminary computations using Auslander-Reiten quivers, this type of modification alters the Auslander-Reiten quiver $\Gamma(A)$ quite a bit, and the indecomposables may be quite different. I was considering looking at using Tits quadratic form, but unfortunately there does not seem to be an if and only if statement concerning bound quiver algebras and the weak positivity of the form (the implication goes the wrong way for trying to use it here).
Any ideas from the community?

REPLY [8 votes]: The quiver
$$\begin{array}{ccccc}
&&1&\to&2\\
&&\uparrow&&\uparrow&\\
3&\to&4&&5\\
&&\uparrow&&\\
&&6&&
\end{array}$$
is a Dynkin quiver of type $D_6$, so has finite representation type. But if you add an arrow from vertex $4$ to vertex $5$, together with commutation relations, it has infinite representation type, since the representations with the zero vector space at vertex $2$ are just representations of the tame quiver $\tilde{D}_4$.<|endoftext|>
TITLE: Exceptional specializations of Galois groups in the Hilbert Irreducibility Theorem
QUESTION [5 upvotes]: Suppose $f(x,t)\in\mathbb{Q}(t)[x]$ is an irreducible polynomial with Galois group G. For any rational number $a$ we may consider the polynomial $f(x,a)\in\mathbb Q[x]$ and its corresponding Galois group $G_a$, which is a subgroup of $G$. By the Hilbert Irreducibility Theorem, the groups $G$ and $G_a$ are the same outside a thin set of values of $a$. (Here I'm using "thin" in the sense of Serre.) If I have a specific polynomial $f$, how explicitly can the set of exceptional values of $a$ (those for which $G_a$ is a proper subgroup of $G$) be described? Are there any references that discuss this question?

REPLY [7 votes]: In some sense it can be described quite explicitly. There is a finite set of curves $C_i$ and maps $\phi_i:C_i\to\mathbb P^1$ of degree at least $2$ defined over $\mathbb Q$ such that the desired thin set is contained in the union
$$ \bigcup_{i=1}^n \phi_i\bigl(C_i(\mathbb Q)\bigr).$$
And I'm pretty sure that one can effectively determine $C_1,\ldots,C_n$ and $\phi_1,\ldots,\phi_n$ from the polynomial $f(x,t)$, although in practice finding specific equations might be a rather hard commutative algebra calculation with Grobner bases, etc.
Having said that, there is still a major problem, namely we need to determine $C_i(\mathbb Q)$ for $1\le i\le n$. For those $C_i$ of genus $0$, this can be done using (1) $C_i(\mathbb Q)\ne\emptyset$ if and only if $C_i(\mathbb Q_p)\ne\emptyset$ for all completions, including $p=\infty$; (2) Hensel's lemma; (3) if $C_i(\mathbb Q)\ne\emptyset$, then one can use the known point to find a parametrization $C_i(\mathbb Q)\cong\mathbb P^1(\mathbb Q)$. But for those $C_i$ of genus $g\ge1$, we do not have effective algorithms for determining the rational points. So for a particular $f(x,t)$, you might be lucky and be able to completely describe the thin set, but in general I do not believe that there is an effective algorithm. On the other hand, there are effective upper bounds for $\#C(\mathbb Q)$ when $g\ge2$, so one might be able to write the thin set as an explicitly given infinite set plus an unknown finite set whose size is explicitly bounded. (I'd have to think a bit more about the genus 1 curves.)<|endoftext|>
TITLE: What's special about the circle problem?
QUESTION [26 upvotes]: Let $K$ be a number field, and let 
$$\zeta_{K}(s):= \sum_{0
\neq I \text{ ideal of }O_K} \frac{1}{N_{K/\mathbb{Q}}(I)^s} = \sum_{n \ge 1} \frac{a_n}{n^s}$$ 
be the Dedekind zeta function of $K$. The quantity $s_K(x):=\sum_{n \le x} a_n$ counts ideals of $O_K$ of norm up to $x$.
$\zeta_K$ is analytic in $s\ge 1$ apart from a simple pole at $s=1$, with residue given by the class number formula. The Wiener-Ikehara theorem implies: 
$$s_K(x) \sim c_{K} x$$
as $x\to \infty$, where $c_K$ is given by the class number formula.
Let $E_K(x):=s_K(x) - c_K x$.
When $K=\mathbb{Q}(i)$, the problem of studying $E_K$ is known as the Gauss circle problem. Proving $E_K(x) = O(x^{1/2})$ is easy, but it is believed that $E_K(x) = O_{\varepsilon}(x^{1/4 + \varepsilon})$.

What is known about $E_K(x)$ for general number fields (conditionally and unconditionally)? What is the heuristic for that?
What distinguishes the case $K=\mathbb{Q}(i)$ from other cases? (Apart from the elementary geometric interpretation)

REPLY [26 votes]: There is nothing special about the circle problem (except of course that it goes back to Gauss)!  The problem for number fields has been extensively investigated, and goes back to Landau.  If the number field has degree $k$, then the problem is quite analogous to the error term in the $k$-divisor problem.  One expects that the remainder term should be on the scale of $O(x^{\frac{k-1}{2k}+\epsilon})$, and an $\Omega$ result of this flavor is known.  Regarding $O$-results, less is known and for large $k$ the best may still be $1-2/(k+1)$ (the analog of the $1/3$ exponent in the usual circle and divisor problems).   You can look at Chapter XII in Titchmarsh's book (and Heath-Brown's notes) for the $k$-divisor problem, and also look at Vaughan or Nowak for more information and other references.  For example, Vaughan notes that for cubic fields the exponent $43/96$ is known in the $O$-result, and is due to Muller.<|endoftext|>
TITLE: Random walk with continuously distributed steps on [-1,1]
QUESTION [5 upvotes]: A simple random walk $S_n = X_1 +\cdots +X_n$, where $P(X_i = 1) = p \not = 0.5$ and $P(X_i=-1)= q \triangleq 1-p$, admits the following probability 
$$P(S_n \textrm{ reaches } a \textrm{ before} -b) = \left(\frac{1-\left(\frac{q}{p}\right)^{b}}{1-\left(\frac{q}{p}\right)^{a+b}}\right),$$ 
for $a$ and $b$ positive integers. The expected number of steps to hit either of these boundaries is 
$$\frac{b}{q - p} - \left(\frac{a+b}{q-p}\right)\left(\frac{1-\left(\frac{q}{p}\right)^{b}}{1-\left(\frac{q}{p}\right)^{a+b}}\right).$$
My question is whether there are analogues to these simple expressions for positive $a$ and $b$ (not necessarily integer) when the $X_i$ are continuous i.i.d. random variables on $[-1,1]$ with a given density $f$? I'm particularly interested in a case where the $X_i$ have a finite positive mean $\mu_X > 0$.

REPLY [4 votes]: As already pointed out by Anthony in a comment, you can't really expect explicit formulae. Let's write $p(x)$ for the probability that the RW starting at $x\in (a,b)$ hits $a$ before it hits $b$. Then, by the same argument as in the discrete case (condition on what happens on the very next step),
$p(x) = \int p(x+t)f(t)\, dt$; here, we must interpret $p(s)=1$ for $s\le a$ and $p(s)=0$ for $s\ge b$ on the right-hand side. So by splitting off the parts where $x+t\notin (a,b)$, we can rewrite the integral equation as
$$
p(x) = F(a-x) + \int_a^b f(t-x)p(t)\, dt ,\quad a<x<b .
$$
This is a Fredholm equation of the second type.
It won't be easy to come up with an $f$ for which this has an explicit solution. You can obtain a series representation of the solution by iterating, see here for example (notice that the $\phi_n$ increase pointwise and stay $<1$, so convergence is guaranteed).<|endoftext|>
TITLE: Optimal stacking of split logs
QUESTION [7 upvotes]: Consider firewood logs as unit-radius cylinders of the same length.
Each log is split into $k$ pieces by equiangular sectors
meeting in the circle center:
$k=2$ leads to semicircles, $180^\circ$;
$k=3$ results in $120^\circ$ sectors, etc.

          



Q. What is the optimal packing of a halfplane 
  by equiangular circular sectors, for each $k$?

I am hoping that for some $k>1$, the optimal packing is known.

          


                   


A practical application, for small $k$ (say, at most $6$, as above [gray]), is how
to optimally stack firewood for the winter.

REPLY [3 votes]: I don't know if this is the optimal packing of 1/6-circle wedges, but I do know that the following is the optimal packing of the shape formed by gluing two of those wedges along a full edge:

The two-wedge shape is convex and 2-fold symmetric, so its optimal packing is the optimal lattice packing, which can be solved for numerically. The density of this packing is $0.978769$. The horizontal edges have a small segment of length $0.0291219$ (if the radius is 1) hanging off and not touching the neighboring wedge. If the horizontal edges coincided fully, the density would only be $0.978263$. Compare to the density of circle packing, $0.906900$.<|endoftext|>
TITLE: When does $\nabla\times(\nabla\times F)=0$ imply $\nabla \times F=0$
QUESTION [10 upvotes]: On a (simply connected) domain $\Omega$ for a smooth vector field $F\colon \Omega \to \mathbb{R}^3$, when does $\nabla\times(\nabla\times F)=0$ imply $\nabla \times F=0$. I know that $n\cdot(\nabla\times F)=0$ on $\partial\Omega$ is sufficient, and also $t\cdot(\nabla\times F)=0$ is sufficient ($t$ the tangential). Is there a weaker condition?

REPLY [4 votes]: Here are some basic thoughts. Let $G$ be a vector field which is of the form $\nabla \times F$, and also obeys $\nabla \times G = 0$.
Since $\nabla \times G = 0$, the vector field $G$ is locally of the form $\nabla h$ for some scalar valued function $h$. The condition that $G = \nabla \times F$ imples that $\nabla \cdot G=0$ or, in other words, $\nabla^2(h)=0$. This says that $h$ is harmonic. So, locally, the condition is equivalent to $G$ being the gradient of a harmonic function. Globally, if $\Omega$ is not simply connected, then traveling around a loop in $\Omega$ may change $h$ by a constant.
From this, we can see that, if $G$ is compactly supported, it is zero: If $G$ is $0$ outside a ball of radius $R$, then $h$ is constant outside that ball, and thus $h$ is constant everywhere. 
We can also trot out our favorite conditions to ensure that a harmonic function is constant. For example, if $\Omega = \mathbb{R}^3$, and $G(x) \to 0$ as $|x| \to \infty$, then $|h(x)| = o(x)$ and a variant of Liouville's theorem tells us that $h$ is constant and $G=0$.
The OP seems to be interested in conditions on the flux of $G$ across $\partial \Omega$. In order to make this make sense, I am going to assume that the set up is that $\overline{\Omega}$ is a compact manifold with boundary.
Let $R$ be the flux of $G$ across $\partial \Omega$. As the OP already knows, if $R=0$ then $G=0$. And the assignment of $G \mapsto R$ is linear, so this shows that $R$ determines $G$. Let $\mathcal{V}$ be the vector space of functions on $\partial \Omega$ which can occur as such an $R$. 
The OP asks for conditions which force $G$ to be zero. In other words, he wants a vector space $\mathcal{W}$ of functions on $\partial \Omega$ which is transverse to $\mathcal{V}$. I find that an odd way to think about the question -- better to just characterize $\mathcal{V}$! 
Now, the obvious observation is that $\int_{\partial \Omega} R=0$ for any $R$ in $\mathcal{V}$, since $\nabla \cdot G=0$. It would be really cool if that were a precise characterization of $\mathcal{V}$. But I don't think it is. Let $\Omega$ be a solid cylinder, and let $R$ be zero on the sides of $\Omega$ and a hat function on top and bottom, dying out smoothly as it approaches the sides. Is $R$ the flux of some $G$? I couldn't figure out how to make it be.<|endoftext|>
TITLE: Applications of the Galois embedding problem
QUESTION [6 upvotes]: Given a finite Galois extension of number fields $L/K$ with Galois group $G$ and a surjection $E\twoheadrightarrow G$ of finite groups, the Galois embedding problem is the question of whether there exists a Galois extension $M/K$ containing $L$ such that $Gal(M/K)\cong E$.
Let's assume we can solve our Galois embedding problem. Are there common applications to such a result?
Do you know of an example where the solution of such a problem implied an interesting/seemingly unrelated result?
Thank you.

REPLY [4 votes]: Shafarevich made heavy use of embedding problems in his resolution of the inverse Galois problem for solvable groups. The (naive) viewpoint is that solving the relevant embedding problems allowed him to reduce to the case of abelian extensions, where a positive solution to the inverse Galois problem was already known.
You can find further details and precise statements in Sections IX.4 and IX.5 of:
Neukirch, Schmidt, Wingberg - Cohomology of number fields.<|endoftext|>
TITLE: Is there a Serre intersection formula in analytic geometry?
QUESTION [14 upvotes]: There is the famous Serre intersection formula in algebraic geometry using the Tor functor (see for example here). I would like to know if there is such a formula in analytic (i.e. complex) geometry. Thanks.  
In somewhat more detail: given two subvarieties of complementary codimensions in a smooth projective variety, the intersection number between them can be computed in terms of an alternating sum involving the Tor function, no matter if their intersection is transversal (and more generally, proper) or not. My question is then how far we can do so for compact complex manifolds. Is there any subtlety of using sheaf theory and so on in this general case? I would expect there is no such, the generalization should go through.
But I see no references on this topic (either on the Serre intersection formula in the compact complex manifolds setting or the more general topic of subtlety in using sheaf theory in this general setting), so I would like to know.

REPLY [6 votes]: Serre's formula works in the analytic category as well. If X is a smooth complex manifold, there is a ring structure on the Grothendieck group $K(X)$ of coherent sheaves given by the usual formula
$$ [\mathcal{F}] . [\mathcal{G}]=\sum_{i \geq 0} (-1)^i [\mathrm{Tor}^i_{\mathcal{O}_X}(\mathcal{F}, \mathcal{G})].$$ 
The Chern character $ch \colon K(X) \rightarrow H^*(X, \mathbb{Q})$ is a ring morphism. If $Z$ and $W$ are two complex subvarieties of $X$ of codimensions $p$ and $q$, write
$$ ch(\mathcal{O}_Z). ch(\mathcal{O}_W)=\sum_{i \geq 0} (-1)^i ch\,(\mathrm{Tor}^i_{\mathcal{O}_X}(\mathcal{O}_Z, \mathcal{O}_W)).$$ 
Now $ch(\mathcal{O}_Z)=[Z] + \,\mathrm{classes\,in}\, H^k(X, \mathbb{Q})$ for $k>2p$ and similarly for $W$. Hence the term with the smallest cohomological degree appearing in  $ch(\mathcal{O}_Z). ch(\mathcal{O}_W)$ is exactly the homological intersection of $Z$ and $W$ in $H^{2(p+q)}(X, \mathbb{Q})$. This proves that
$$[Z]. [W]=\sum_{i \geq 0} (-1)^i ch_{p+q}(\mathrm{Tor}^i_{\mathcal{O}_X}(\mathcal{O}_Z, \mathcal{O}_W)).$$
Next recall that if $\mathcal{F}$ is a coherent sheaf whose support consists of irreducible components $D_i$ of codimension $\geq d$, then 
$$ch_d(\mathcal{F})=\sum_{codim(D_i)=d} \ell_i(\mathcal{F})\, [D_i^{red}]$$
where $\ell_i(\mathcal{F})$ is the length of $\mathcal{F}$ at the generic point of $D_i$. This gives Serre's formula (at least over $\mathbb{Q}$).<|endoftext|>
TITLE: Relation between projective hierarchy and universally measurable sets
QUESTION [8 upvotes]: Let $X$ be Polish. It is known that every analytic and coanalytic subset of $X$ is universally measurable. The Wikipedia article about universally measurable sets notes that (assuming projective determinance) every set in the projective hierarchy is universally measurable. Are there also universally measurable sets that are not related to the projective hierarchy, i.e. can not be constructed in this way?

REPLY [10 votes]: Thanks, Joel, for mentioning our paper. 
By Martin-Steel, if there exist infinitely many Woodin cardinals, then every uncountable projective set contains a perfect set, so, by the result of Hausdorff mentioned by Joel there are universally null sets which are not projective. 
On the other hand, it is apparently an open question whether it is consistent for all universally measurable sets to be $\Delta^{1}_{2}$ (see problem CG on Fremlin's list : https://www.essex.ac.uk/maths/people/fremlin/problems.pdf). 
Finally, I should mention two improvements of Hausdorff's theorem : 
(Reclaw) Any set of reals which is wellordered by a universally measurable relation is universally null (https://projecteuclid.org/download/pdf_1/euclid.rae/1212763971)
(Grzegorek) There is a universally null set whose cardinality is the smallest cardinality of a non-Lebesgue-null set. A proof of this can be found in Bukovsky's book (http://www.springer.com/gp/book/9783034800051)<|endoftext|>
TITLE: What's so special about these $17$th deg equations?
QUESTION [15 upvotes]: While browsing the Database of Number Fields, I came across 17T8. It only had four equations, one of which is,
$$\small{x^{17} - 5x^{16} + 40x^{15} - 140x^{14} + 610x^{13} - 1622x^{12} + 4870x^{11} - 10220x^{10} + 22720x^9 - 38080x^8 + 63500x^7 - 84100x^6 + 102200x^5 - 102400x^4 + 83000x^3 \color{brown}{- 55864x^2 + 24080x - 9400}=0}$$
It may not look much, but all four examples had the same coefficients except for the  part in brown (the $x^2, x^1, x^0$ terms), so I assume there might be a parameterization. After some fiddling, I found the rather simple,
$$\frac{(x^5 - 2x^4 + 10x^3 - 10x^2 + 20x - 10)^3\,(x^2 + x + 4)}{(4x^2 - 5x + 25)} = -36m\tag1$$
The four were just the cases $m = -6, -2, -12, 9$. The discriminant of $(1)$ is,
$$D = 2^{36}\, 3^{52}\, 5^{18}\, m^{10}(16m + 81)^8$$
Update: Note that,
$$\frac{(x^5 - 2x^4 + 10x^3 - 10x^2 + 20x - 10)^\color{red}3\,(x^2 + x + 4)}{4x^2 - 5x + 25}-\frac{27^2}{2^2} = \frac{(x-1)(2x^8 - 4x^7 + 32x^6 - 40x^5 + 170x^4 - 136x^3 + 362x^2 - 166x + 185)^\color{red}2}{4x^2 - 5x + 25}\tag2$$
Note that the quintic and octic factors (both irreducible) have solvable Galois groups. Also, I've seen similar factoring behavior in formulas for the j-function like the well-known icosahedral equation,
$$j(\tau)-1728 =-\frac{(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^\color{red}3}{r^5(r^{10} + 11r^5 - 1)^5}-12^3 = -\frac{(r^{30} + 522r^{25} - 10005r^{20} - 10005r^{10} - 522r^5 + 1)^\color{red}2}{r^5(r^{10} + 11r^5 - 1)^5}\tag3$$
Makes me wonder if $(1)$ is a formula for something. 
Questions:

Does the whole family, except for special $m$, belong to 17T8?
Can one derive it from first principles, instead of a computer search? (Its simple form seem to suggest there might be others.)

REPLY [23 votes]: Many of the polynomials in the Klueners-Malle database and also in my
database with John Jones come from families in the way you correctly
describe.  So you have "reverse engineered" the source family.
This particular source family is the first of two similar families
described in Section 13 my Galois number fields with small root discriminant with Jones.  Your m and our t are related by m=-81/16 t.  The generic
Galois group of this family is 17T8 = SL_2(16).4 and so for almost
all t the specialized Galois group will be 17T8.  Section 13 discusses
some details of the specialization process.  In particular, specializing
at t=-8 keeps the Galois group at 17T8 and has the unusually 
small discriminant 2^16 3^20 5^16.<|endoftext|>
TITLE: Does $\mathfrak{N}_4$ contain at least four distinct elements?
QUESTION [5 upvotes]: How do I see that the set $\mathfrak{N}_4$ consisting of all unoriented cobordism classes of smooth closed $4$-manifolds contains at least four distinct elements?

REPLY [14 votes]: Thom proved that the unoriented bordism is a polynomial ring over $\mathbb{F}_2$ generated by elements $x_i$ with $i$ running over all numbers not of the form $2^k-1$. Thus, $\mathfrak{N}_4 = \mathbb{F}_2 \cdot \{x_2^2, x_4\}$. Thom moreover proves that the even $x_i$ can be represented by $\mathbb{RP}^i$. 
Thus, our four manifolds are the empty manifold, $\mathbb{RP}^2\times \mathbb{RP}^2$, $\mathbb{RP}^4$ and the disjoint or connected sum of the latter two. Let's give a characteristic class proof that they are all different in $\mathfrak{N}_4$:
As Tsemo Aristide mentioned, Stiefel-Whitney numbers are cobordism invariants. Denoting the generator of $H^*(\mathbb{RP}^n;\mathbb{F}_2)$ by $u$, the total Stiefel-Whitney class is $(1+u)^{n+1}$. Thus, we get that $w_1(\mathbb{RP}^4) = u$ and $w_4(\mathbb{RP}^4) = u^4$ and all other Stiefel-Whitney classes are $0$. In particular, the non-trivial Stiefel-Whitney numbers are for the indices $(4,0,0,0)$ and $(0,0,0,1)$. 
For $\mathbb{RP}^2\times \mathbb{RP}^2$ the total Stiefel-Whitney class is given by $(1+u+u^2)(1+v+v^2)$, where we denote the generator for $H^*(\mathbb{RP}^2;\mathbb{F}_2)$ of the second factor by $v$. The Stiefel-Whitney number for the index $(4,0,0,0)$ is zero because $w_1(\mathbb{RP}^2\times \mathbb{RP}^2) = u+v$ and $(u+v)^4 = 6u^2v^2 = 0$. On the other hand, $w_2 = u^2+uv+v^2$ and $w_2^2 = u^2v^2$ and thus the Stiefel-Whitney number with index $(0,2,0,0)$ is nonzero. In particular, these two manifolds have different Stiefel-Whitney numbers. 
Let $M = \mathbb{RP}^2\times \mathbb{RP}^2 \coprod \mathbb{RP}^4$. Stiefel-Whitney numbers are additive under coproducts. Thus the Stiefel-Whitney numbers for the indices $(0,2,0,0)$ and  $(4,0,0,0)$ are both $1$. Thus, they are different from the ones for the two manifolds above.<|endoftext|>
TITLE: Can a complete manifold have an uncountable number of ends?
QUESTION [8 upvotes]: Let $M$ be a complete and noncompact Riemannian manifold. Fix a point $p$ in $M$. Let $\gamma$: $[0, L]\rightarrow M$
(parametrized by its arc length) be a geodesic starting from $p$. Denote by $d(\cdot, \cdot)$ the distance function on $M$ induced by 
the Riemannian metric. $\gamma$ is minimal if $d(\gamma(s_1), \gamma(s_2))=|s_1-s_2|$. A ray is a minimal geodesic defined on $[0, +\infty)$.
Since $M$ is noncompact, there exists at least one ray from $p$.
Two rays $\gamma_1, \gamma_2$ from the same point $p$ are called cofinal if for any $r\geq0$ and all $s>r$, $\gamma_1(s)$
and $\gamma_2(s)$ lie in the same component of $M\backslash B(0, r)$, where $B(0, r)=\{x\in M| d(p, x)<r\}$. 
An equivalence class of cofinal rays is called an end of $M$.
My question is:
Can $M$ have uncountably many ends? It seems that the answer is no. But I am not very sure and I can not find a convictive proof.

REPLY [5 votes]: Let me rephrase S. Goette's example in a slightly different way. Consider the complement of the triadic Cantor set in the complex plane. This is a Riemann surface that can be uniformized: there is a complete Riemannian metric of constant minus one curvature. The ends of this surface are in one-to-one correspondence with the points of the Cantor set. 
You can see the decomposition into pairs of pants by drawing nested circles around the intervals that appear in the construction of the Cantor set. 
In some sense, uncountably many ends is the generic case. Let us restrict ourselves to planar surfaces, namely complements of compact sets in the Riemann sphere. The set of compact subsets in a compact set has a natural topology, the Vietoris topology (which coincides with the topology given by the Hausdorff metric). So we can say that two open subsets are close if their complement are close with respect to this topology. It happens that generically a compact subset of the sphere has uncountably many connected components, hence its complement has uncountably many ends.<|endoftext|>
TITLE: Always a planar-drawn cycle through $n$ points
QUESTION [8 upvotes]: Given $n$ points in the plane, can we always find a cycle through all of them that has only straight line edges and no edges intersect (planar-drawn)?

Intuitively the answer is yes, but I am struggling with a proof. In the example above, the $2^{nd}$ graph demonstrates an incorrect algorithm, whereas the last graph demonstrates a working algorithm. I have tried using induction.
For different variants there are constraints - this is not always possible if the points are colored (e.g. if two points of one color lies on non-adjacent points of the convex hull and any path between them partitions the other color).

REPLY [2 votes]: The answer is yes.
First we draw a cycle with the outmost vertices such that other vertices are within the cycle. (This is step 2 in above picture.)
Then we repeat this for vertices that are not on the first cycle and draw the second cycle.
We repeat this until all vertices lie on a unique simple closed cycle.
Now we delete an edge from each two adjacent cycles and connecting vertices such that two cycles exchange in one longer cycle. Then repeat this until we have had a cycle consists all vertices.<|endoftext|>
TITLE: elliptic curves and group cohomology
QUESTION [19 upvotes]: Recently, I've been trying to understand Jacob Lurie's 2-equivariant elliptic cohomology a bit better than I had in the past.
From what I can tell, the fragment of the story that only deals with finite groups, with characteristic zero stuff, and only talks about the degree zero part of the cohomology theories is equivalent to the following claim:

CLAIM:
  Given the following input:
  $\triangleright$ A $\mathbb Q$-algebra $R$.
  $\triangleright$ An elliptic curve $E$ over $R$
  $\triangleright$ A finite group $G$
  $\triangleright$ A cocycle $k$ representing a class in $H^4(BG,\mathbb Z)$,
  there is a way of constructing:
  $\triangleright$ An $R$-module.

.

QUESTION: What is the construction which takes $R$, $E$, $G$, $k$ as input, and produces an $R$-module as output?

.
When $k=0$, the output of the construction should be ring of functions on the moduli stack $M_G$ of $G$-bundles over $E$.
[Added later: this guess turned out to be wrong – see Jacob's answer]
When $R=\mathbb C$, I know how to define a line bundle $L_k$ over $M_G$, and I want the $R$-module to be the space of global sections of that line bundle.
When $R$ is some random $\mathbb Q$-algebra, I don't know how to define a line bundle over $M_G$, and in particular, I don't know how to construct its $R$-module of global sections.

Bonus questions: 
  $\triangleright$ What does the construction look like (in particular, does a construction exist) if one drops the assumption that $R$ is a $\mathbb Q$-algebra?
  $\triangleright$ What does the construction look like if one drops the assumption that $G$ is a finite group, and one takes it to be a compact Lie group instead?
  $\triangleright$ What does the construction look like if one drops the assumption that $E$ is an elliptic curve?
  $\triangleright$ Any combination of the above.

REPLY [17 votes]: As Charles indicates, "the moduli stack of $G$-bundles on $E$" is not quite the right thing to consider, especially if you're not working over $\mathbf{C}$. This is for two (unrelated) reasons:
1) The geometric object $M_{G}$ that you associate to a group $G$ isn't something that you can access directly (at least by the construction I know): what you can access instead is the ring of global functions
on $M_{G}$, and maybe the global sections of a few other sheaves (like the line bundles you're asking about). Consequently, it's hard to tell the difference between moduli stack of $G$-bundles and its coarse moduli space. Let me ignore this in what follows. (More precisely, instead of answering the question "what geometric object and line bundle does elliptic cohomology produce?", I'll answer the question "how can I write down a geometric object and line bundle whose global sections are related to $2$-equivariant elliptic cohomology?"
2) Saying "$M_G$ is the moduli stack of $G$-bundles" doesn't really make sense, because $G$ is a compact Lie group rather than an algebro-geometric object. A more accurate statement is "$M_G$ is the moduli stack of $G(1)$-bundles", where if $G$ is connected compact Lie group,
$G(1)$ denotes the split algebraic group having the same root datum.
(Example: if $G = U(n)$, then $G(1) = GL_n$.) 
This generally does not make sense if $G$ is not connected, for example if $G$ is a finite group. However, you can make sense of it if $G$ is a finite abelian group (for example, if $G = \mathbf{Z}/n\mathbf{Z}$, then $G(1) = \mu_{n}$; in this case, $G(1)$-bundles on $E$ correspond to $n$-torsion points on the dual abelian variety, which is just $E$ again: this recovers what Charles said. Note that we meet issue 1) here:
the scheme of $n$-torsion points on $E$ is a -coarse- moduli space for
$\mu_n$-bundles on $E$). 
For $G$ a finite nonabelian group, you can't really make sense of $G(1)$.
However, you can still make sense of a $G(1)$-bundle on $E$, at least when the order $|G|$ is invertible in $R$. To explain this, let me assume for simplicity that $R$ is an algebraically closed field of characteristic zero, and try to describe everything in way that is invariant under automorphisms of $R$. Let $\Lambda$ denote the etale fundamental group of $E$: this is a free module of rank $2$ over the ring
$\widehat{ \mathbf{Z} }$ of profinite integers. The datum of
a $G$-bundle on $E$ (up to isomorphism) is equivalent to the datum of
map $\Lambda \rightarrow G$ (up to conjugacy). So you can define
the datum of a $G(1)$-bundle on $E$ (up to isomorphism) to be the datum of a map $\Lambda(-1) \rightarrow G$ (up to conjugacy). Here
$\Lambda(-1)$ denotes the twist of $\Lambda$ by the inverse of the cyclotomic character. In the language of etale cohomology, $\Lambda(-1)$ is just $H^{1}(E; \widehat{\mathbf{Z}})$. 
So that's the sort of datum that $M_{G}$ is supposed to classify.
You get to identify $M_{G}$ with the moduli stack of $G$-bundles on $E$
if you choose a compatible system of roots of unity in $R$, for example by taking $R = \mathbf{C}$). 
Now suppose you're given a level $k$ on $G$, which we can identify with an element of $H^{4}(BG; \mathbf{Z} ) = H^3(BG; \mathbf{Q} / \mathbf{Z} )$. Let $T$ denote the profinite torus $B \Lambda(-1)$. Suitably
interpeted, we have $H_2(T) = H^2(E) = \widehat{ \mathbf{Z} }(-1)$,
and $M_{G}$ classifies maps $T \rightarrow BG$. You can now apply the construction that Qiaochu suggested: pull back the level to $T$ and integrate to get a class in $H^1( Spec R, (\mathbf{Q} / \mathbf{Z})(1) )$. Of course, that class will be trivial, but if you do everything at the level of cochains rather than cohomology classes you'll obtain the datum of a $( \mathbf{Q} / \mathbf{Z} )(1) = \mu_{\infty}(R)$-torsor,
which determines an $R$-line.
If you say all this carefully, you'll end up with something with makes sense as long as $|G|$ is invertible in $R$ (if $|G|$ is not invertible in $R$, then I don't know a construction which avoids elliptic cohomology, although in many cases you get answers which are well behaved in pure algebraic geometry; for example, when $G$ is a symmetric group and the level is zero). I don't think it makes sense for curves of higher genus: in the discussion above, it is important that $\pi_{1} E$ is abelian.<|endoftext|>
TITLE: Database of adjacency matrices on cospectral non-isomorphic graph pairs
QUESTION [5 upvotes]: Is there a repository of cospectral non-isomorphic graphs available somewhere?
I am looking for list of $0/1$ adjacency matrix pairs that can be input data in tools such as MATLAB.

REPLY [6 votes]: The simplest source of cospectral graphs is lists of strongly regular graphs, lots of which are easily available from Ted Spence's web page at http://www.maths.gla.ac.uk/~es/srgraphs.php.
Otherwise you can use Sage to generate small graphs (up to 10 or so vertices) and then filter out cospectral pairs or groups. I expect the built in Sage function for cospectral pairs just wraps this up.
I don't know what you are doing with them, but I'd probably recommend choosing the computational tool based on what you need, rather than specifying Matlab in advance. If you're working with 64 vertex graphs you'll need full symbolic computation with arbitrary length integers and you'll want to avoid, or be very very careful, in finding eigenvalues numerically.

REPLY [2 votes]: You can do this in sage for small order and then export
the adjacency matrices to say text file friendly to Matlab
and then parse in Matlab.
Tony Huynh suggests one approach. Another approach is
to enumerate with McKay's nauty in sage in keep track of cospectral.
Such database will be large:
https://oeis.org/A082104
A082104 Number of distinct characteristic polynomials among all simple undirected graphs on n nodes.
      1, 2, 4, 11, 33, 151, 988, 11453, 247357, 10608128, 901029366, 148187993520
Check the references in OIES.
From Brouwer's reference:
https://www.win.tue.nl/~aeb/graphs/cospectral/cospectralA.html
Numbers of characteristic polynomials and cospectral graphs
Consider contacting Brouwer, though the full database will take a lot of space AFAICT.<|endoftext|>
TITLE: Are the extremal points of a certain set of functions $\mathcal P(\mathbf N) \to \bf R$ weakly additive?
QUESTION [6 upvotes]: Let an upper density (on $\mathbf N$) be a (set) function $f: \mathcal P(\mathbf N) \to \mathbf R$ such that, for all $X, Y \subseteq \bf N$ and $h,k \in \mathbf N^+$, the following hold:

(F1) $f(\mathbf N) = 1$;
(F2) $f(X) \le f(Y)$ whenever $X \subseteq Y$;
(F3) $f(X \cup Y) \le f(X) + f(Y)$;
(F4) $f(k \cdot X + h) = \frac{1}{k} f(X)$, where $k \cdot X + h :=
\{kx+h: x \in X\}$.

It was noted in another thread (here) that an upper density $f$ doesn't need to be weakly additive, meaning that $f(X \cup Y) = f(X) + f(Y)$ for all disjoint $X, Y \subseteq \bf N$ such that $Y$ is an (infinite) arithmetic progression (namely, a set of the form $k \cdot \mathbf N + h$ for some $h \in \mathbf N$ and $k \in \mathbf N^+$).
There are, however, a bunch of upper densities that are weakly additive, which is notably true of the upper logarithmic, upper asymptotic, and upper Banach densities (along with at least uncountably many others).
In fact, upper densities form a convex subset, $\mathscr U$, of the real vector space, $\mathcal B(\mathcal P(\mathbf N), \mathbf R)$, of bounded functions $\mathcal P(\mathbf N) \to \bf R$, so it makes sense to ask how the extreme points of $\mathscr{U}$, relative to the linear structure of $\mathcal B(\mathcal P(\mathbf N), \mathbf R)$, look like. Hence my question:

Q. Is it true that every extreme point of $\mathscr{U}$ is a weakly additive upper density? (The answer might depend on the axiom of choice, so let us assume to work in ZFC.)

Here are some positive results: If $\preceq$ denotes the (partial) order on the set, $\hom(\mathcal P(\mathbf N), \mathbf R)$, of all functions $\mathcal P(\mathbf N) \to \bf R$ defined by $f \preceq g$ iff $f(X) \le g(X)$ for all $X \subseteq \bf N$, then it is seen that $\mathscr{U}$ is a complete subsemilattice of the join-semilattice $(\hom(\mathcal P(\mathbf N), \mathbf R), \preceq)$, meaning that, whenever $\mathscr{D}$ is a nonempty subset of $\mathscr{U}$, the set $$\{u \in \mathscr U: f \preceq u\text{ for all }f \in \mathscr{D}\}$$ has a least element, relative to the order $\preceq$, that still belongs to $\mathscr{U}$. In particular, $\mathscr{U}$ has a maximum (again, relative to the order $\preceq$), which is given by the upper Buck density (on $\bf N$), that is the function
$$
\mathfrak{b}^\ast: \mathcal P(\mathbf N) \to \mathbf R: X \mapsto \inf_{S \in \mathscr{A}: X \subseteq S} \mathsf{d}^\ast(S),
$$
where $\mathscr{A}$ is the set of all subsets of $\mathbf N$ that can be written as a finite union of arithmetic progressions, and $\mathsf d^\ast$ is the upper asymptotic density (on $\bf N$).
It is found that $\mathfrak b^\ast$ is a weakly additive upper density, and on the other hand, it can be proved, under the axiom of choice, that $\mathscr{U}$ has at least uncountably many minimal elements, relative to the order $\preceq$ (I don't know if the existence of even one minimal point can be proved in ZF).
Now, it is not difficult to prove that minimal and maximal elements of $\mathscr{U}$ are extreme points of $\mathscr{U}$; unfortunately, it is not clear to me if also the converse is true (if yes, the answer to question Q would be in the affirmative), though I don't believe so.

REPLY [2 votes]: I think that the answer is No.
The space $\mathcal B(\mathcal P(\mathbf N), \mathbf R)$ with pointwise convergence is a locally convex space. (It can be considered a subspace of $\mathbf R^{\mathcal P(\mathbf N)}$.)
The set $\mathscr U$ is compact in this space. (It can be identified with a closed subset of $[0,1]^{\mathcal P(\mathbf N)}$.)
So by Krein-Milman theorem we get that $\mathscr U$ is closed convex hull of the set of all extreme points of $\mathscr U$.
On the other hand, the set of all weakly additive upper densities is closed and convex subset of $\mathcal B(\mathcal P(\mathbf N), \mathbf R)$. (It suffices to check that limit of a convergent net of weakly additive upper densities is again a weakly additive upper density and that a convex combination of weakly upper densities is a weakly upper density.) So if all extreme points of $\mathscr U$ were weakly additive, the same would be true for the whole set $\mathscr U$.<|endoftext|>
TITLE: Path integral methods
QUESTION [5 upvotes]: Are there detailed expositions of the path integral methods in (mathematical) physics other than Feynman-Hibbs and Glimm-Jaffe?

REPLY [3 votes]: An impressive list of references to complement the answers already given can be found at the bottom of the following Scholarpedia article by Sergio Albeverio and Sonia Mazzucchi. The approach to path integrals based on "cylindrical measures", aka "promeasures", is not discussed in the body of the article but can be read about in the work by DeWitt-Morette and collaborators referenced there.<|endoftext|>
TITLE: In the $\mathbb{H}^3$ upper half space model, is a hemiellipsoid perpendicular to the plane at infinity a minimal surface?
QUESTION [7 upvotes]: Given a Jordan curve on the $\mathbb{H}^3$ boundary at infinity, there is a minimal surface (topological disk) in $\mathbb{H}^3$ with the curve as its asymptotic boundary (page.mi.fu-berlin.de/polthier/articles/diss/diss.pdf, Theorem 21 on p37).  
Consider boundary curves in the upper half space model.  If the boundary curve is a circle, the minimal surface is a geodesic plane (a hemisphere orthogonal to the boundary).  A hemiellipsoid orthogonal to the boundary (a triaxial ellipsoid with two axes in the plane at infinity, and the third axis vertical) has a boundary curve that is an ellipse, and it seems likely that a hemiellipsoid is the minimal surface for this boundary curve.  Why would it be something more complicated?
For the point on the hemiellipsoid through its vertical axis, it is easy to see that one (hyperbolic) principle curvature is positive and the other is negative, by considering the hemisphere with the same vertical axis through that point.  But it would be nice to have an argument that the hyperbolic principle curvatures are equally opposite (update: based on comments/answers, we should be able to pick the ellipsoid parameters to make this true at this specific point), and that this is true for all points on the hemiellipsoid.
The motivation for this question comes from $\mathbb{H}^3$ fibrations, specifically about lifts of geodesics from the base $\mathbb{H}^2$ surface.  For some fibrations, the lift is a hemiellipsoid in the upper half space model (see plus.google.com/+RoiceNelson/posts/1w3aoQgj61g and comments therein).

REPLY [7 votes]: On p. 21 of Polthier's thesis, equation (2.1), you'll find the explicit minimal surface equation in the upper half-space model. I plugged the formula for an ellipsoid into this equation in Mathematica (I'm too lazy to do the computation), and found that the only solutions are for hemispheres. I've included a screenshot from the Mathematica notebook: the appropriately scaled form of the differential equation is a quadratic form which must be identically zero. The only solution which makes the coefficients zero is when the ellipsoid is a sphere.<|endoftext|>
TITLE: Algebraic dynamics in finite fields
QUESTION [7 upvotes]: What is known about combinatorial structure of the rational maps of degree 2 over finite fields? From some general reasons I think it was studied. For being more specific, consider the field $\mathbb{F}_{2^n}$ and the map $x\rightarrow f(x)=x+1/x$. We may consider such a map as oriented graph with outdegrees 1 (except outdegree 0 for $x=0$) and indegrees usually 0 or 2 (except again $x=0$ with indegree 0). Each weak connected component contains exactly 1 cycle (except connected component of 0, which is a tree.) What is known about number of cycles of different lengths? About trees growing from these cycles?

REPLY [14 votes]: There is a growing body of literature on dynamics of rational maps over finite fields. The following paper would seem to be relevant. 

Ugolini, S., Graphs associated with the map $x\mapsto x+x^{-1}$ in finite fields of characteristic two, Theory and applications of finite fields, Contemp. Math. 579, 197-204, Amer. Math. Soc., 2012.

Here are a few more to get you started. 

Alberto de Faria, Joao and Hutz, Benjamin, Combinatorics of cycle lengths on Wehler K3 Surfaces over finite fields, New Zealand J. Math. 45 (2015), 19-31.
Alden Gassert, T., Chebyshev action on finite fields, Discrete Math. 315 (2014), 83–94.
Bach, Eric and Bridy, Andrew, On the number of distinct functional graphs of affine-linear transformations over finite fields, Linear Algebra Appl. 439 (2013), 1312-1320.
Burnette, Charles and Schmutz, Eric, Periods of Iterated Rational Functions over a Finite Field, 2015, arXiv:1508.04193.
Flynn, Ryan and Garton, Derek, Graph components and dynamics over finite fields, Int. J. Number Theory 10 (2014), 779-792.
Roberts, John A. G. and Vivaldi, Franco, Signature of time-reversal symmetry in polynomial automorphisms over finite fields, Nonlinearity 18 (2005), 2171-2192.
Roberts, John A. G. and Vivaldi, Franco, A combinatorial model for reversible rational maps over finite fields, Nonlinearity 22 (2009), 1965--1982.

And you can try searching for the phrase "finite field" in the arithmetic dynamics bibliography that I've compiled at http://www.math.brown.edu/~jhs/ArithDyn.bib<|endoftext|>
TITLE: Is there a short proof that the Kostka number $K_{\lambda \mu}$ is non-zero whenever $\lambda$ dominates $\mu$?
QUESTION [22 upvotes]: This is maybe a little basic for MathOverflow, but I'm hoping it will get some interesting answers.
Let $\unrhd$ be the dominance order on partitions of $n \in \mathbb{N}$.
For partitions $\lambda$ and $\mu$ of $n$, the Kostka Number $K_{\lambda\mu}$ is the number of semistandard Young tableaux of shape $\lambda$ and content $\mu$.
If $t$ is such a tableau then the $\mu_1+\cdots +\mu_r$ entries $k$ of $t$ such that $k \le r$ all lie in the first $r$ rows of $t$. Hence $\lambda_1 + \cdots + \lambda_r \ge \mu_1 + \cdots + \mu_r$ for each $r$ and so $\lambda \unrhd \mu$.

Is there a short combinatorial proof of the converse: if $\lambda \unrhd \mu$ then $K_{\lambda\mu} > 0$?

A constructive proof, maybe using the characterization of neighbours in the dominance order by single box moves on Young diagrams, would be especially welcome.
Edit. Using some representation theory it's possible to make this strategy work. The following is the symmetric functions version of Theorem 2.2.20 in the textbook by James and Kerber. Obviously $K_{\lambda\lambda} = 1$. Let $\mu$ and $\mu^\star$ be neighbours 
in the dominance order, with $\mu \rhd \mu^\star$, so $\mu^\star_i=\mu_i-1$, $\mu^\star_j = \mu_j+1$ for some $i < j$, and $\mu^\star_k = \mu_k$ if $k\not=i,j$. Since $h_{(a,b)} = h_{(a+1,b-1)} + s_{(a,b)}$ whenever $a \ge b$, we have
$$h_{\mu^\star} = \bigl(\prod_{k\not=i,j} h_{\mu_k} \bigr) h_{\mu_i-1}h_{\mu_j+1}=
\bigl(\prod_{k\not=i,j}h_{\mu_k}\bigr) \bigl( h_{\mu_i}h_{\mu_j} + s_{(\mu_i-1,\mu_j+1)} \bigr) $$
and so if $f = \prod_{k\not=i,j}h_{\mu_k}$ then
$$K_{\lambda\mu^\star} = \langle s_\lambda, h_{\mu^\star} \rangle = \langle s_\lambda, h_\mu \rangle + \langle s_\lambda, f s_{(\mu_i-1,\mu_j+1)} \rangle = K_{\lambda\mu} + \langle s_\lambda, f s_{(\mu_i-1,\mu_j+1)} \rangle \ge K_{\lambda\mu}.$$

REPLY [3 votes]: The following argument is reproduced from the proof of Lemma 3.7.3 in the book Representation Theory of the Symmetric Groups: The Okounkov–Vershik Approach, Character Formulas, and the Partition Algebras, by Tullio Ceccherini-Silberstein, Fabio Scarabotti, and Filippo Tolli (Cambridge University Press, 2010).
The proof is by induction on $n$. Suppose that $\lambda$ has $k$ parts and $\mu$ has $h$ parts, and that $\lambda\ge \mu$. Say that the positive integer $i$ is removable for the pair $(\lambda,\mu)$ if,
after setting
$$\tilde\lambda := (\lambda_1, \lambda_2, \ldots, \lambda_{i-1}, \lambda_i-1, \lambda_{i+1}, \ldots, \lambda_k)$$
and
$$\tilde\mu := (\mu_1, \mu_2, \ldots, \mu_{h-1}, \mu_h-1),$$
$\tilde\lambda$ is still a partition (i.e., $\lambda_i > \lambda_{i+1}$) and $\tilde\lambda \ge \tilde\mu$.  In particular, if the $i$th row of a SSYT of shape $\lambda$ and content $\mu$ contains the number $h$, then $i$ is removable for $(\lambda,\mu)$. Also, $k$ is always removable for the pair $(\lambda,\mu)$.
Let $i$ be the smallest removable integer for the pair $(\lambda,\mu)$, and let $\tilde\lambda$ and $\tilde\mu$ be as above. By the inductive hypothesis, we may construct a SSYT $\tilde T$ of shape $\tilde\lambda$ and content $\tilde\mu$. Moreover, by minimality of $i$, $\tilde T$ does not contain an entry $h$ in any row $j<i$. So if we add to $\tilde T$ a box at the end row $i$ with entry $h$, then we get a SSYT $T$ of shape $\lambda$ and content $\mu$.<|endoftext|>
TITLE: Characters of weight 1 cusp forms
QUESTION [12 upvotes]: Assume that the space of cusp forms of weight 1 $S_1(\Gamma_0(N),\chi)$ is
non zero. What can one say about the odd character $\chi$, for instance
concerning its order ? Serre tells me that a theorem of Tate implies that
$\chi$ is of order $2$ or $4$ if $N$ is prime, and that this can probably
be generalized to $N$ squarefree, and more difficult if not. What is known,
or conjectured ?

REPLY [11 votes]: $\def\Q{\mathbf{Q}}$
$\def\Z{\mathbf{Z}}$
$\def\Gal{\mathrm{Gal}}$
$\def\GL{\mathrm{GL}}$
$\def\Qbar{\overline{\mathbf{Q}}}$
$\def\C{\mathbf{C}}$
$\def\OL{\mathcal{O}}$
$\def\PGL{\mathrm{PGL}}$
$\def\rhobar{\overline{\rho}}$
Dear Henri, I think the case of prime conductor is a little special. The order of $\chi$ can be quite big in the dihedral case even when $N$ is a product of two primes.
First, here is a reminder of how the conductor works when $p$ exactly divides $N$. If $f$ is a classical modular cusp form in $S_1(\Gamma_0(N),\chi)$, then the representation $\rho$ associated to $f$ has the property that, if $p \| N$,
$$\rho |_{I_p} \simeq \left( \begin{matrix} \chi_p & 0 \\
0 & 1 \end{matrix} \right),$$
where $I_p \subset D_p = \Gal(\Qbar_p/\Q_p)$ is the inertia and decomposition group at $p$. This is because (when the image of inertia is tame) the conductor is given by $2 - \mathrm{dim} V^{I_p}$. As a consequence, we see that the image of $I_p$ under $\rho$ is isomorphic to the image of $I_p$ under the projective representation when $p \| N$.
Lemma: If $N$ is squarefree and $\rho$ has exceptional projective image, then $\chi^{60}$ is trivial.
Proof: The projective image of $\chi_p$ is a cyclic subgroup of $A_4$, $S_4$, or $A_5$, and hence has order $\le 5$. Hence $\chi^{60}_p$ is unramified at $p$. It follows that $\chi^{60}$ is unramified at all primes, and is hence trivial because $\Q$ has no unramified abelian extensions.
Remark: If there is only one prime, then the same argument shows that $\chi$ has order at most $5$. Since $\chi$ is odd, it has even order, and so order $2$ or $4$. 
Remark: One can do slightly better: If the exceptional subgroup is contained in $S_4$ then one gets $12$ instead of $60$, and since $A_5$ has no elements of order $4$, one gets $30$ in that case (or $6$ in the $A_4$ case). 
Example:Let $K$ be the splitting field of $x^5 - 33x^3 - 55x^2 + 803x - 1199$. The Galois group $\Gal(K/\Q)$ is $A_5$, and $K$ is odd. There is a corresponding projective representation:
$$\rhobar: G_{\Q} \rightarrow \Gal(K/\Q) = A_5 \rightarrow \PGL_2(\mathbf{C})$$
The field
 $K$ is ramified at $7$, $11$, and $197$ only (and $\infty$), and
$$I_{7} = D_{7} \simeq \Z/3\Z,$$
$$I_{11} = D_{11} \simeq \Z/5\Z,$$
$$I_{197} = \Z/2\Z, \qquad D_{197} = \Z/2\Z \oplus \Z/2\Z.$$
Each of these decomposition groups inside $\PGL_2(\C)$ admit lifts $D_p \rightarrow \GL_2(\C)$ on which inertia has non-trivial invariants. (This would be false if $D_p$ was dihedral of order $6$ or $10$ and inertia was the index two normal subgroup, or alternatively if $I_p = D_p$ was the Klein $4$-group.) Recall
the theorem of Tate: the projective representation
$$\rhobar: G_{\Q} \rightarrow \Gal(K/\Q) = A_5 \rightarrow \PGL_2(\mathbf{C})$$
admits a lift $\rho: G_{\Q} \rightarrow \GL_2(\C)$ with the following property: for each prime $\ell$,
if we choose a local lift $\rho_{\ell}: D_{\ell} \rightarrow \GL_2(\C)$ of $\rhobar_{\ell}:=\rhobar |D_{\ell} \rightarrow \PGL_2(\C)$, then we may choose $\rho$ to have the property that it agrees with all these lifts on inertia, that is, $\rho |I_{\ell} = \rho_{\ell}| I_{\ell}$. Hence, in our particular example, we may ensure that $\rho$ is
 unramified outside $7$, $11$, $197$, and moreover, for each of these primes $p$, the image of inertia has a fixed line. By the Artin conjecture (known in this case by Buzzard-Taylor), it follows that $\rho$ is modular of level $7 \cdot 11 \cdot 197$, and the Nebentypus character will have order $30$.
The Dihedral Case:
However, things can be arbitrarily bad in the dihedral case as soon as $N$ is divisible by two primes.
Fix a prime $\ell \equiv -1 \mod 4$, and let $K = \Q(\sqrt{-\ell})$.
The unit group of the ring of integers $\OL_K$ of $K$ has order $2$. Now let $p$ be an auxiliary prime which splits in $K$, and let $v$ and $w$ be the primes above $p$ in $\OL_K$. The ray class group of conductor $v$ has order $(p-1)/2$, and hence there exists a character:
$$\psi: G_K \rightarrow \C^{\times}$$
of order $(p-1)/2$ which is ramified only at $v$. Now let us consider the induction
$$\rho = \mathrm{Ind}^{\Q}_{K} \psi; \qquad \rho | G_K = \psi \oplus \psi^c$$
It is odd, because $K$ is not real. So it does correspond to a weight one modular form.  I claim it has conductor $p \ell$. It suffices to show that $\rho$, restricted to the decomposition group at either $p$ or $\ell$, is tamely ramified and has an invariant line. For $\ell$, we have $\rho | I_{\ell} = \eta \oplus 1$ for the quadratic character $\eta$ coming from $\Gal(K/\Q)$ which is tamely ramified because $\ell > 2$. For the prime $p$, we may as well restrict to the decomposition group at $w|p$. Yet $\psi$ is unramified at $w$ by construction, so $\rho | I_p = 1 \oplus \psi^c |_{I_p}$.
So the level is $N = p \ell$. Let's compute the nebentypus, which is the determinant.
Let $\chi$ be the determinant of this representation. Then
$$\chi |_{G_K} = \psi \psi^c.$$
I claim that this has order $(p-1)/2$, and that $\chi$ has order $\mathrm{lcm}(2,(p-1)/2)$. To see this, note that $\psi$ is  ramified at $v$ to this order and is unramified at $w$, whereas $\psi^c$ is ramified at $w$ to this order and is unramified at $v$. Hence $\chi$ restricted to $G_K$ is ramified at both $w$ and $v$ to order $(p-1)/2$, and so $\chi$ itself has order $p-1$. Note that $\chi^{(p-1)/2}$ is unramified at $p$ and $\chi^2$ is unramified at $\ell$, so $\chi^{\mathrm{lcm}(2,(p-1)/2)}$ is trivial.
Why is the dihedral case OK when $N = p$ is prime?
In this case, since $\Q$ has no unramified extensions, the quadratic extension $K$ must be $\Q(\sqrt{p (-1)^{(p-1)/2}})$. However, $p$ is ramified in this extension, so the only way a character can be unramified at a prime above $p$ is if it is unramified everywhere, in which case $\chi^2$ is unramified everywhere, and $\chi$ has order two.<|endoftext|>
TITLE: Repository of graph classes that are tough to test non-isomorphic pairs from isomorphic pairs
QUESTION [5 upvotes]: (1) Which graph classes are extremely tough to test for graph non-isomorphic pairs from isomorphic pairs?
(2) Is there a repository of adjacencies from such classes?

REPLY [5 votes]: There is a move towards creating a "standard" benchmark set for graph isomorphism, but it didn't happen yet.  Meanwhile, the largest collection that includes hard graphs as well as easy graphs is here. To make examples of difficult nonisomorphic pairs, take two graphs with the same parameters and randomly label them. For difficult isomorphic pairs take two random labellings of the same graph.
The graphs here that will cause trouble to most programs are in the classes ag, cfi, had, latin-sw, pg, pp, f-lex.<|endoftext|>
TITLE: The space of positive definite orthogonal matrices
QUESTION [7 upvotes]: The matrix $\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$ is orthogonal and indefinite. 
 $\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}$ is positive definite and not orthonormal.
and the Identity matrix $I$ is of course both orthogonal and positive definite. Let $S$ be the intersection of orthogonal matrices and positive definite matrices. We have seen that $S$ is non-empty. 
By a positive definite matrix, I mean either a symmetric or asymmetric matrix $M \in \mathbb{R}^{p^2}$ whose quadratic form satisfies $\forall x \in \mathbb{R}^p \setminus \{0\}: x^TMx > 0$. Let $P$ denote the set of all positive definite matrices. Then $P$ is a convex cone.
$S$ is not a convex cone, unlike $P$. Also unlike $P$, $S$ is closed under multiplication The product of any two matrices in $S$ is guaranteed to be at least PSD.[1]
I am interested in complete characterizations of this space $S$, which globally behaves more like the space of orthogonal matrices. My real motivation is that I want to know whether there are efficient procedures for testing whether a matrix $M $ is in $S$ that are faster than testing for positive definiteness which requires the calculation of eigenvalues? E.g. such procedures might take only $O(p^2)$ computations. I tried to google for resources but nothing relevant came up.
[1] Orthogonal matrices are of course closed under multiplication. The product of two PD matrices is PD PSD matrices is PSD iff their product itself is normal, which is true in $S$.
Reference: On a product of positive semidefinite matrices, A.R. Meenakshi, C. Rajian, Linear Algebra and its Applications, Volume 295, Issues 1–3, 1 July 1999, Pages 3–6.
In case someone is wondering, the real real reason, why I am interested in this question is because I want to "efficiently" find a "reasonably good" positive definite approximation of a matrix whose $QR$ decomposition is known to me. The details of this part are best left for future.

REPLY [13 votes]: You may find the Cayley transform to be useful here:
As is well-known and easy to prove, every orthogonal $n$-by-$n$ matrix $R$ that does not have $-1$ as an eigenvalue can be written uniquely in the form
$$
R = (I-A)(I+A)^{-1}
$$
for some anti-symmetric matrix $A$ for which $I+A$ is invertible, and, conversely, if $A$ is any anti-symmetric matrix such that $(I+A)$ is invertible, the matrix $R$ in the above formula is orthogonal and $I+R$ is invertible.
In fact, $A = (I-R)(I+R)^{-1}$, so $A$ is easy to find. The two matrices $R$ and $A$ are said to be Cayley transforms of each other, and this provides a 'rational parametrization' of the orthogonal group minus the hypersurface consisting of the orthogonal matrices that have $-1$ as an eigenvalue.  (Note that $R$ and $A$ commute.)
Now $R$ satisfies the stronger condition that $x^TRx>0$ for all nonzero $x\in\mathbb{R}^n$ if and only if (setting $y=(I+A)^{-1}x$ or, equivalently $x = (I+A)y$), we have
$$
0 < x^TRx = y^T(I+A)(I-A)y = y^T(I-A^2)y = |y|^2-|Ay|^2 
$$
for all $y\in \mathbb{R}^n$.  Equivalently, the matrix norm of $A$, i.e., $\|A\|$, should be strictly less than $1$.
Thus, via the Cayley transform, your set $S$ is parametrized by the open convex set in the vector space of anti-symmetric $n$-by-$n$ matrices consisting of those anti-symmetric matrices $A$ whose matrix norm is less than $1$.<|endoftext|>
TITLE: Poisson structures on non-smooth manifolds with singularities
QUESTION [8 upvotes]: It's very known how we can describe a Poisson structure on a manifold $M$, where $M$ is a smooth manifold, but what about a Non-smooth manifold with singularities? In section $(2)$ of the paper The Quantum Tetrahedron in 3 and 4 Dimensions, we found a description for a Poisson Structure on a Tetrahedron (that it's not smooth), so I was looking for some references about: Poisson structure on non-smooth manifolds with singularities, is there any progress about it?

REPLY [2 votes]: More generally, since Poisson brackets are defined on associative algebras (in fact also NC ones), the key step is usually to find the "right" associative algebra of functions (or rather seaf of algebras, as Igor Khavkine says, in some settings) to describe your space. It is in this way, for example, that orbifold singularities can also be included in Poisson geometry setting (e.g. http://arxiv.org/abs/0807.0027).<|endoftext|>
TITLE: An interesting inequality
QUESTION [12 upvotes]: Let $\mathbb{R}$ be the real field. For any homogeneous polynomial $f(X_1,\cdots,X_n)$ in $\mathbb{R}[X_1,\cdots,X_n]$, we use $S_f(X_1,\cdots,X_n)$ to denote the following homogeneous symmetric polynomial:
$$S_f(X_1,\cdots,X_n)=\sum_{\sigma=[i_1,\cdots,i_n]\in S_n}f(X_{i_1},\cdots,X_{i_n}).$$
Here the sum is computed over all permutations $\sigma=[i_1,\cdots,i_n](\sigma(1)=i_1,\cdots,\sigma(n)=i_n)$ of the set $\{1,\cdots,n\}$ and the set of all such permutations is denoted $S_n.$ We say $f(X_1,\cdots,X_n)$ is good if 
$$S_f(a_1,\cdots,a_n)\geq 0$$
for every $(a_1,\cdots,a_n)\in \mathbb{R}^n.$
For example, when $n=3$, we have $$f(X_1,X_2,X_3)=X_3^2-X_1X_2$$ is good because$$S_f(X_1,X_2,X_3)=(X_1-X_2)^2+(X_2-X_3)^2+(X_3-X_1)^2.$$
For any $n\geq 1$, define the homogeneous polynomial of degree $n^2$ in $\mathbb{R}[X_1,X_2,\cdots,X_{2n}]$ as follow:$$\varphi_n(X_1,X_2,\cdots,X_{2n})=\prod_{\substack{1\leq i\leq n\\n+1\leq j\leq 2n}}(X_i-X_j).$$
I conjecture that $\varphi_n$ is good for any $n\geq 1$, and for this conjecture I have got the following simple results:
$(1)$It is easy to proof that $S_{\varphi_n}(X_1,X_2,\cdots,X_{2n})=0$ when $n$ is odd;
$(2)$When $n$ is even,$$S_{\varphi_2}(X_1,X_2,X_3,X_4)=4[((X_1-X_2)(X_3-X_4))^2+((X_1-X_3)(X_2-X_4))^2+((X_1-X_4)(X_2-X_3))^2].$$
So the conjecture is right for $n=2$.
I can not proof it any more for $n\geq 4$, but I believe that the conjecture is right. Would you please give me some help?

REPLY [11 votes]: Alas, this is false at least for even $n\geqslant 6$, I do not know about $n=4$.
A similar question about symmetrization of $$f_{a,b}:=\prod_{1\leq i\leq a,a+1\leq j \leq a+b} (X_i-X_j)$$
may be asked (of course, it equals to 0 if $ab$ is odd). I asked this for $a=1$, $b=n-1$ here and it turned out to fail for odd $n\geq 7$. 
Now for your question. Take even $n\geq 6$ and $n-1$ variables equal taking very-very large value $M$, and some $n+1$ variables $X_1,\dots,X_{n+1}$. Then if we divide the value $$f_{n,n}(M,M,\dots,M,X_{1},X_{2},\dots,X_{n+1})$$ by $M^{n(n-1)}$, it tends to 
$$f_{1,n}(X_1,\dots,X_{n+1}),$$
therefore if symmetrization of the latter is negative, symmetrization of the former is negative for large enough $M$.
It would be interesting to describe all pairs $(a,b)$ for which inequality holds.<|endoftext|>
TITLE: A property stronger than the fixed point property
QUESTION [8 upvotes]: Assume that $X$ is  a topological space. We say that $X$ satisfies the strong  fixed point property if the graph of every surjective continuous self-map intersect the graph of every continuous self-map on $X$. For example the interval $I=[0,1]$ satisfy this property.
An equivalent definition: If two continuous  self-maps $f,g$ on $X$ have non intersecting graphs, then neither $f$ nor $g$ is  surjective. 
Are there  some examples of manifolds (with or without boundary) of higher dimension with this property? In particular, do the closed $2$_disc, or the even dimensional real or complex projective spaces satisfy this property?

REPLY [2 votes]: Three related references are the following. 
If $f:D^2\to D^2$ is such that the map $H^2(D^2,S^1)\to H^2(D^2,f^{-1}(S^1))$ induced in Cech cohomology is non-trivial, then the graph of any other map $g:D^2\to D^2$ intersects the graph of $f$. This is part of Theorem A in "W. Holsztynski, On the product and composition of universal mappings of manifolds into cubes, Proc Amer. Math. Soc, 58(1976), 311–314".
If $f,g:D^2\to D^2$ are commuting maps, do their graphs intersect? this question appears here: Two commuting mappings in the disk
If $f,g:D^2\to D^2$ are commuting homeomorphisms, do the graphs of $f,g$ and of the identity $1_{D^2}$ intersect? This question has been asked here: Do commuting homeomorphisms of the $2$-disk have a common fixed point?<|endoftext|>
TITLE: Densest Graphs with Unique Perfect Matching
QUESTION [10 upvotes]: Given a graph $G$ with $n$ vertices, that has a perfect matching $M$, what is the maximal number of edges that $G$ can have without contradicting the uniqueness of $M$? 
Are examples of such extremal graphs known?

REPLY [7 votes]: An easy construction to achieve the bound of $n^2$ edges for $2n$ vertices: embed the $n$ edges $(v_i,w_i)$ of the 1-factor as parallel segments in the plane. To these $n$ edges, add all $\binom n2 $ possible edges $(v_i,w_j)$ with $i<j$ and all $\binom n2 $ edges $(w_i,w_j)$.
For any 1-factor, each $v_i$ must belong to a different edge, from there it is clear that there is a unique one.<|endoftext|>
TITLE: Unramified extensions of quadratic fields
QUESTION [8 upvotes]: Let $K/\mathbb{Q}$ be quadratic and let $L/K$ be an (everywhere) unramified Galois extension. If $L/K$ is abelian, then one can show that $L/\mathbb{Q}$ is Galois (eg see here). Is $L/\mathbb{Q}$ necessarily Galois if $L/K$ is not abelian? What if we also assume that $L/K$ has odd degree?

REPLY [11 votes]: The following result is a special case of a sideresult in my PhD thesis (1995):
Let $k = {\mathbb Q}(\sqrt{d})$ be a quadratic extension with
discriminant $d$, let $d = d_1d_2d_3d_4$ be a factorization of $d$
into coprime discriminants, and assume that 
$$(d_1/p_2) = (d_2/p_1) = (d_3/p_4) = (d_4/p_3) = +1 $$
for all primes $p_j \mid d_j$. Then there exist
$\alpha = x_1 + y_1 \sqrt{d_1} \in {\mathbb Z}[\sqrt{d_1}]$ and 
$\beta  = x_3 + y_3 \sqrt{d_3} \in {\mathbb Z}[\sqrt{d_3}]$ satisfying 
$x_1^2 - d_1y_1^2 = z_1^2d_2$ and 
$x_3^2 - d_3y_3^2 = z_3^2d_4$ such that
$M = {\mathbb Q}(\sqrt{d_1},\sqrt{d_2},\sqrt{d_3},\sqrt{d_4},
   \sqrt{\alpha},\sqrt{\beta})$ is an unramified extension of $k$
with Galois group $D_4 \times D_4$. 
The subfield
$$ L = {\mathbb Q}(\sqrt{d_1d_2},\sqrt{d_1d_3},\sqrt{d_1d_4},
    \sqrt{\mu})$$
with $\mu = 2x_1x_3 + 2y_1y_3\sqrt{d_1d_2} + 2z_1z_2\sqrt{d_2d_4}$
has Galois group $D_4$ over $k$, but is not normal over ${\mathbb Q}$.
I have given the concrete example
$d = -3 \cdot 13 \cdot 5 \cdot 29$ with $\alpha = (-1 + \sqrt{13}\,)/2$,
$\beta = 7 + 2 \sqrt{5}$ and $\mu = -7 + 2\sqrt{65} + 2 \sqrt{-87}$.
I just verified with pari that it is indeed unramified:  
f=polcompositum(x^2+3*13,x^2+3*5)[1];
f=polcompositum(f,x^2+3*29)[1];
g=x^8 + 28*x^6 + 470*x^4 + 3836*x^2 + 380689;
The compositum $L$ of $f$ and $g$ has degree $16$ and discriminant 
$3^8 5^8 13^8 29^8$, hence is unramified over $k$.
For odd degree extensions I would try a nontrivial 3-class field tower of a quadratic number field and pick out a suitable subgroup that isn't fixed under the automorphism of order 2. I'm sure some group theorist will be able to come up with an example.<|endoftext|>
TITLE: Good references for K-theory of modular curves?
QUESTION [5 upvotes]: The title says it. I am looking for a good exposition on the K-theory of the curves $X_{i}(N)$, $Y_{i}(N)$, where $i\in\{0,1\}$.
I have some background in $K$-theory and also some background in modular curves.

REPLY [7 votes]: I wouldn't recommend Beilinson's 1985 paper as a general reference -- it's terrifyingly compressed, developing an entire new subject in a single short paper, and crashes through the necessary material on modular curves in a couple of sentences.
A much gentler reference would be Flach's 1992 Inventiones paper "A finiteness theorem for the symmetric square of an elliptic curve", which gives a much fuller introduction to K-groups, Gersten complexes, regulator maps etc. You could also try looking at parts of my paper with Lei and Zerbes "Euler systems for Rankin--Selberg convolutions" for more on this.<|endoftext|>
TITLE: A measure of how "spread out" a probability measure is
QUESTION [5 upvotes]: Consider a random variable $X$ whose variance is large. As a contrast to Markov's or Chebyshev's inequality, both of which measure the concentration of a probability distribution, is there a measure of how "spread out" a distribution is?
more specifically, I would like to have an inequality of the following sort:
$$
\Pr[|X-\mu|<r]\leq f(r)
$$
where $\mu=\mathbb{E}[X]$ and $f$ is some increasing function. This should be read: the probability of $X$ being close to its expectation is small. 
A case of particular interest to me is the following. Let $X=\sum_{i=1}^n X_i$ where $X_1,...,X_n$, are i.i.d. and non-constant. This time let $r$ be fixed, and I would like to have a bound that depends on the number of summands $n$:
$$
\Pr[|X-\mu|<r]\leq f(n)
$$
where $f$ goes to zero as $n$ goes to infinity. 
In fact, it is my intuition that for any fixed $C\in\mathbb{R}$ and $r>0$ it should be true that
$$
\Pr[|X-c|<r]\leq f(n)
$$
where $f$ goes to zero as $n$ goes to infinity, since as we have more summands, $X$ is more "smoothened out".

REPLY [11 votes]: What you need is a standard inequality for the concentration function for sums of independent random variables; see e.g. [Petrov],
Ch. III, Section 2, Theorem 3 (due to Esseen), which implies the following. 
For a random variable (r.v.) $X$ and real $c>0$, let 
$$Q(X;c):=\sup_{x\in\mathbb R} P(x\le X\le x+c). 
$$ 
Let $S_n:=X_1+\dots+X_n$, where $X,X_1,\dots,X_n$ are any independent identically distributed r.v.'s. Then for any real $c>0$ 
$$Q(S_n;c)\le \frac A{\sqrt{n D(\tilde X;c)}}, 
$$ 
where $A$ is a universal constant, $\tilde X:=X-X_1$,
$$D(\tilde X;c):=\frac1{c^2}\,E\,\tilde X^2\,I\{|\tilde X|<c\}+P(|\tilde X|\ge c)=E\Big(1\wedge\frac{\tilde X^2}{c^2}\Big), 
$$
and $I\{\cdot\}$ is the indicator function. Note that, for any real $c>0$, one has $D(\tilde X;c)=0$ iff $P(\tilde X=0)=1$ iff the r.v. $X$ is degenerate (i.e., $P(X=a)=1$ for some real $a$). Hence, if $X$ is non-degenerate, then $D(\tilde X;c)\in(0,\infty)$.  
A simpler but a bit less precise bound is due to [Rogozin]: 
$$Q(S_n;c)\le \frac A{\sqrt{n(1-Q(X;c))}}.  
$$
Addendum: If you only care about the concentration of $S_n$ around its expectation, you may want to use a Berry--Esseen type of bound (see e.g. Theorem 7 in [3]
), which implies 
$$P\Big(\Big|\frac{S_n-n\mu}{\sigma}\Big|\le c\Big)\le P\Big(|Z|\le\frac c{\sqrt n}\Big)+A\frac{\beta}{\sigma^3\sqrt n}
\le\frac C{\sqrt n}, 
$$
where $\mu:=EX_1$, $\sigma:=\sqrt{E(X_1-\mu)^2}$, $\beta:=E|X_1-\mu|^3$, $c\in[0,\infty)$, $Z$ is a standard normal r.v.,
$C:=\frac{2c}{\sqrt{2\pi}}+A\frac{\beta}{\sigma^3}$, and $A\in(0,96/100)$ is a universal constant. 
Similar but more direct and a bit more general bounds on the concentration are given e.g. in Proposition 6.1 in [4]
and Proposition 2.1 in [5].  
All these bounds are of the optimal order $O(1/\sqrt n)$ in $n$. It cannot be improved in general even for the concentration of $S_n$ around its expectation. E.g., let $P(X_i=\pm1)=1/2$. Then, by Stirling's formula, $P(S_{2n}=0)>A/\sqrt n$ for some universal constant $A>0$.<|endoftext|>
TITLE: Operator-valued measurable functions
QUESTION [5 upvotes]: Let $H$ be a non separable Hilbert space and $\Omega$ be a measurable space. 
Naturally, we say that $f:\Omega\to B(H)$ is $w$-measurable if $f^{-1}(O)$ is measurable for any open set $O$ in the weak operator topology. 
Question: Let $f$ and $g$ be two $w$-measurable functions on $\Omega$. Is the multiplication $fg$ a $w$-measurable function as well?

REPLY [3 votes]: The answer is no even for functions with values in $L_\infty(\mu)$ (for the purpose of this question embedded in $B(H)$), as long as the cardinality of the space $L_\infty(\mu)$ is larger than continuum. The following argument works for many reasonable topologies, including the w.o.t. in the question.
Let $\Omega$ be the set $L_\infty(\mu)\times L_\infty(\mu)$ with the smallest $\sigma$-algebra that makes both projections onto $L_\infty(\mu)$ measurable. The diagonal $D=\{(x,x)\mid x\in L_\infty(\mu)\}$ is not a measurable subset of $\Omega$: If it were then there would be a countable family of subsets of $L_\infty(\mu)$ separating the points of $L_\infty(\mu)$, so the cardinality of $L_\infty(\mu)$ would be at most continuum.
Define $f(x_0,x_1)=e^{x_0}$, $g(x_0,x_1)=e^{-x_1}$ for $(x_0,x_1)\in\Omega$. The function $fg=e^{x_0-x_1}$ is not measurable because the set $\{I\}$ is closed and its preimage is $D$.<|endoftext|>
TITLE: Counting refinements of partitions
QUESTION [6 upvotes]: Let $p$ and $q$ be partitions of $n$. We say $q$ refines $p$ if the parts of $p$ can be subdivided to produce the parts of $q$. For example, $(5,5,1)$ refines $(6,5)$ but not $(7,4)$. $(n)$ refines only itself, and $(1,...,1)$ refines all partitions of $n$.
For each partition of $n$, count the number of partitions refining it. Let $F(n)$ be the sum of these counts. For example, $F(3) = 3+2+1=6$, and $F(4) = 5+3+3+2+1=14$. What is known about the asymptotics of $F(n)$?
My motivation is from looking at multinomials.
Note: Refinement is not the dominance order. It is related to the refinement of set partitions.
EDIT: I found why I was looking at these numbers:
JonMarkPerryBlog:WordPress

REPLY [3 votes]: I assume you are asking for the number of pairs $(\pi,\sigma)$ of partitions of $n$ so that $\sigma$ is a refinement of $\pi$. Call this $q(n)$.
Let $p(n)$ be the number of partitions of $n$. $p(n) \sim \frac{1}{4\sqrt {3}~n} \exp \left( \pi \sqrt{\frac{2n}{3}}\right)$ by Hardy and Ramanujan, so $p(n)$ has upper and lower bounds of the form $\exp (c\sqrt{n})$.
$p(n) \le q(n) \le p(n)^2$ so $q(n)$ also has upper and lower bounds of the form $\exp (c \sqrt{n})$. It is natural to try to find the limit of $\frac{1}{\sqrt{n}}\log q(n)$. The $\liminf$ is at least $\pi \sqrt{2/3}$ and the $\limsup$ is at most $\pi\sqrt{8/3}$.
Here is a construction showing that the $\liminf$ is at least $\pi \sqrt{4/3}$: Consider pairs of partitions $(\pi^-,\sigma^-)$ of $n/2$. To $\pi^-$, add $n/2$ to the largest part to get a partition $\pi$ of $n$. To $\sigma^-$, add $n/2$ $1$s to get a partition $\sigma$ of $n$. No pair $(\pi,\sigma)$ is hit twice, and $\sigma$ is a refinement of $\pi$.<|endoftext|>
TITLE: When is Hausdorff measure a Frostman measure?
QUESTION [9 upvotes]: Let $(X,d)$ be a metric space and let $\mathcal{H}^s$ be the $s$-dimensional Hausdorff measure on $X$. 
For a measure $\mu$ on $X$, we say that $\mu$ is a Frostman measure (sometimes referred as upper Ahlfors regular measure) if for some $\alpha >0$, $\mu(B(x,r))\leq C r^\alpha$ for every $x \in X$ and $r>0$.

Under what conditions on $s$ and $d$ is $\mathcal{H}^s$ a Frostman measure?

Some (silly) partial answers. For $X=\mathbb{R}^d$, $\mathcal{H}^d$ is clearly a Frostman measure, as it is a scalar multiple of Lebesgue measure. For $s>d$, $\mathcal{H}^s$ satisfies, trivially, $0=\mathcal{H}^s(B(x,r)) \leq C r^s$, so the answer is positive also in this case. For $s<d$, we have $\mathcal{H}^s(B(x,r)) = \infty$ so the inequality can't hold.
Obviously, if the metric space equipped with the measure $\mathcal{H}^s$ is $s$-Ahlfors regular, the measure is Frostman.
I've tried to think about the more general case, but I fail to have any intuition on a metric space $X$, so apologies if the answer turns out to be easy.

REPLY [3 votes]: K. Falconer, Fractal Geometry: Mathematical Foundations and Applications (3rd ed, 2014, Wiley). page 77

Corollary 4.12 Let $F$ be a Borel subset of $\mathbb R^n$ with $0 < \mathcal H^s(F) \le \infty$.  Then there is a compact set $E \subset F$ such that $0 < \mathcal H^s(E) < \infty$ and a constant $b$ such that
  $$
\mathcal H^s\big(E \cap B(x,r)\big) \le br^s
$$
  for all $x \in \mathbb R^n$ and $r>0$.  

$ $   

Corollary 4.12, which may be regarded as a converse of the Mass distribution principle 4.2, is often called 'Frostman's lemma'.

The notes suggest for a complete proof: P. Mattila Geometry of Sets and Measures in Euclidean Space (1999, Cambridge Univ Press)<|endoftext|>
TITLE: Is it known whether this space is a suspension space?
QUESTION [7 upvotes]: For each prime $p\geq 3$ let $\alpha_p:S^{2p}\to S^3$ denote a representative of $\pi_{2p}S^3$ of order $p$. Berstein and Hilton showed that for each $p$ the homotopy cofiber $C_{\alpha_p}$ of $\alpha_p$ is a co-H-space which does not have the homotopy type of a suspension space. 
The maps $\alpha_p$ give rise to a map $\alpha:\bigvee_{p\geq 3} S^{2p}\to S^3$ whose cofiber $C_\alpha$ is a co-H-space. Is it known whether $C_\alpha$ has the homotopy type of a suspension space?

REPLY [10 votes]: Isn't the localization of a 2-connected suspension also a suspension?  Then this cannot be a suspension.  (Brayton Gray pointed this out to me.)<|endoftext|>
TITLE: An interesting integration
QUESTION [8 upvotes]: For any positive integer $n$, let
$$A_n=\idotsint\limits_{\substack{x_1+\cdots+x_n+y_1+\cdots+y_n\leq1\\x_1,\cdots,x_n,y_1,\cdots,y_n\geq0}}\prod_{i,j=1}^n(x_i-y_j)dx_1\cdots dx_ndy_1\cdots dy_n.$$
It is easy to prove that $A_n=0$ when $n$ is odd.
I conjecture that $A_n>0$ when $n$ is even. 
To prove it, I constructed the following real polynomial:
$$\varphi_n(X_1,X_2,\cdots,X_{2n})=\prod_{\substack{1\leq i\leq n\\n+1\leq j\leq 2n}}(X_i-X_j)$$
and let
$$S_{\varphi_n}(X_1,X_2,\cdots,X_{2n})=\sum_{\sigma=[i_1,i_2,\cdots,i_{2n}]\in S_{2n}}f(X_{i_1},X_{i_2},\cdots,X_{i_{2n}}).$$
Here the sum is computed over all permutations $\sigma=[i_1,i_2,\cdots,i_{2n}](\sigma(1)=i_1,\sigma(2)=i_2,\cdots,\sigma(2n)=i_{2n})$ of the set $\{1,2,\cdots,2n\}$ and the set of all such permutations is denoted $S_{2n}.$ 
Then it is easy to see
$$S_{\varphi_n}(a_1,a_2,\cdots,a_{2n})\geq0\ \text{for every}\  (a_1,a_2,\cdots,a_{2n})\in \mathbb{R}^{2n},\text{when}\ n\ \text{is even}$$
$$\Longrightarrow$$
$$A_n=\dfrac{1}{\left(2n\right)!}\idotsint\limits_{\substack{x_1+x_2+\cdots+x_{2n}\leq1\\x_1,x_2,\cdots,x_{2n}\geq0}}S_{\varphi_n}(x_1,x_2,\cdots,x_{2n})dx_1dx_2\cdots dx_{2n}>0,\text{when}\ n\ \text{is even}$$
I believed the former is right, but I can not prove it. So I asked another question here.
Unfortunately, it has been proved to be wrong:
When $n$ is even and $n\geq6$, there exists $(a_1,a_2,\cdots,a_{2n})\in \mathbb{R}^{2n}$ such that $S_{\varphi_n}(a_1,a_2,\cdots,a_{2n})<0$.
Now I have no idea how to prove or disprove my original conjecture:
$$A_n>0\quad\text{when}\ n\ \text{is even}.$$
So I ask for some help.

REPLY [19 votes]: Let's prove a bit more, namely that the integral over $x_i>0, y_j>0, \sum_i x_i+\sum_j y_j=S$ with respect to the $2n-1$-dimensional Lebesgue measure is positive. Note that due to the obvious scaling $x_i,y_j\mapsto tx_i,ty_j$, this integral depends on $S$ in a trivial way (as some pure power function of $S$). Hence, we can just as well introduce any other fast decaying positive weight $w(S)$ and prove that $\idotsint_{(0,+\infty)^{2n}}\prod\dots w(x_1+\dots+x_n+y_1+\dots+\dots y_n)dx_1\dots dx_n dy_1\dots dy_n>0$ instead. Obviously the easiest weight to deal with is just $\exp(-x_1-\dots x_n-y_1-\dots-y_n)$, in which case the integration in $x$ gives $\left(\int_0^\infty P_y(x)e^{-x}\right)^n$ where $P_y(x)=\prod_j(x-y_j)$. For even $n$ this is never negative and certainly not zero for at least some $y$ (say, all $y_j$ equal to each other). Thus the new weighted integral is positive and, thereby, the claim follows.<|endoftext|>
TITLE: Deceptively simple inequality involving expectations of products of functions of just one variable
QUESTION [18 upvotes]: For a proof to go through in a paper I am writing, I need to prove the following deceptively simple inequality:
$$(*)\qquad E(X^a) E(X^{a+1}\log X) > E(X^{a+1})E(X^a\log X) $$
where $X>e$ has a continuous distribution and $0<a<1$. The intuition, in one sentence, is that if you start from 
$$E(X^a) E(X^{a}\log X) = E(X^{a})E(X^a\log X) $$
it "pays more" (in terms of expected values) to place the added $X$ multiplying larger quantities $(X^{a}\log X)$ than smaller quantities $(X^{a})$. Simulations have confirmed the intuition, at least up to now. However, although I have tried to prove this inequality for days, using other well-known inequalities as well as relationships between expectations of products, products of expectations, and covariances, I have not been successful so far. 
Something that seems related is that we know that 
$$E(\prod_i^n f_i(X))>\prod_i^nE(f_i(X)) $$
as long as the functions $f_1...f_n$ are continuous monotonic functions of $X$, and are all, for instance, increasing and satisfy $f_i(X)>0$ (e.g, John Gurland's "Inequalities of Expectations of Random Variables Derived by Monotonicity or Convexity", $\textit{The American Statistician}$, April 1968). The inequality I am trying to prove is, in a sense, "in between" the two sides in the inequality above. 
Any suggestion would be very greatly appreciated.

REPLY [3 votes]: I am posting here an answer to my question provided by @hbp in the related site Mathematics, because I think people here may find it quite interesting (I definitely did). I am copying below @hbp's response verbatim, i.e., I have contributed absolutely nothing to this (for some reason, when I tried to use blockquote it did not work well). @hbp's answer is here.
Response by @hbp starts here
Thanks for the insightful and fun problem. Here is a proof (I think) via the Cauchy-Schwarz inequality.  Consider the function
$$
f(t) \equiv \frac{ \mathbb E[X^{a+t} \ln X] } { \mathbb E[X^{a+t}] }.
$$
So the target inequality is $f(1) > f(0)$.  We can show this by proving $f(t)$ is increasing, or $f'(t) \ge 0$.
But this is easy, because
$$
\begin{aligned}
f'(t)
&=
\frac{d}{dt} \left(
\frac{ \mathbb E[e^{(a+t)\ln X} \ln X] } { \mathbb E[e^{(a+t) \ln X}] }
\right)
\\
&=
\frac{ \mathbb E\left[ \frac{d}{dt} e^{(a+t)\ln X} \ln X \right] }
{ \mathbb E\left[e^{(a+t) \ln X} \right] }
-
\mathbb E[ e^{(a+t)\ln X} \ln X ]
\frac{ \mathbb E\left[ \frac{d}{dt} e^{(a+t) \ln X} \right] }
     { \mathbb E[e^{(a+t) \ln X}]^2 }
\\
%&=
%\frac{ \mathbb E\left[ e^{(a+t)\ln X} (\ln X)^2 \right] }
%{ \mathbb E\left[e^{(a+t) \ln X} \right] }
%-
%\mathbb E\left[ e^{(a+t) \ln X} \ln X \right]
%\frac{ \mathbb E\left[ e^{(a+t) \ln X} \ln X \right] }
%{ \mathbb E\left[e^{(a+t) \ln X}\right]^2 }
\\
&=\frac{
  \mathbb E[X^{a+t} (\ln X)^2] \, \mathbb E[X^{a+t}]
  -
  \mathbb E[X^{a+t} (\ln X)]^2
} {
  \mathbb E\left[X^{a+t}\right]^2
} \ge 0.
\qquad (1)
\end{aligned}
$$
The numerator of (1) is nonnegative by the
Cauchy-Schwarz inequality.
That is, with $U = X^{\frac{a+t}{2}} \ln X, V = X^{\frac{a+t}{2}}$, we have
$$
\mathbb E\left[U^2 \right] \mathbb E\left[V^2\right] \ge \mathbb E[U \, V]^2.
\qquad (2)
$$
It remains to argue that the equality cannot hold for all $t \in [0,1]$, which is easy.
Alternative to the Cauchy-Schwarz inequality (2)
Alternatively, we can show (1) directly by observing that
$$
\mathbb E\left[X^{a+t}(y - \ln X)^2 \right] \ge 0,
$$
holds for all $y$ (for the quantity of averaging is nonnegative), i.e., the quadratic polynomial
$$
\begin{aligned}
p(y)
&=
\mathbb E\left[X^{a+t}\right] y^2
- 2 \, \mathbb E\left[X^{a+t} \ln X\right] y
+ \mathbb E\left[X^{a+t} (\ln X)^2\right]
\\
&\equiv A \,y^2 - 2 \, B \, y + C,
\end{aligned}
$$
has no zero.
Thus the discriminant of $p(y)$, which is $4B^2 - 4AC$, must be non-positive.  This means $AC \ge B^2$, or
$$
\mathbb E\left[X^{a+t}\right] \,
\mathbb E\left[X^{a+t} (\ln X)^2\right]
\ge
\mathbb E\left[X^{a+t} \ln X\right]^2.
$$

Further discussion
There is a more intuitive interpretation of (1). We define the characteristic function of $\ln X$ as
$$
F(t) \equiv \log \left\{ \mathbb E\left[ X^{a+t} \right] \right\}.
$$
We find $f(t) = F'(t)$, and $f'(t) = F''(t) \ge 0$.  In other words, (1) is a generalized statement of that the second cumulant of $\ln X$ is non-negative at nonzero $a+t$.<|endoftext|>
TITLE: Vector bundles, finitely generated projective module?
QUESTION [5 upvotes]: Let $B$ be a Tychonoff space and let $\mathbb{R}(B)$ denote the ring of continuous real valued functions on $B$. For any vector bundle $\xi$ over $B$, let $S(\xi)$ denote the $\mathbb{R}(B)$-module consisting of all cross-sections of $\xi$. I have two questions.

If $\xi \oplus \eta$ is trivial, does it follow that $S(\xi)$ is a finitely generated projective module?
Conversely, if $Q$ is a finitely generated projective module over $\mathbb{R}(B)$, does it follow that $Q \cong S(\xi)$ for some $\xi$?

REPLY [6 votes]: Let me start by mentioning that for sure this is true when $B$ is compact (which for me includes Hausdorff): this is a celebrated theorem of Swan.  I give a fairly detailed proof in $\S$6 of these notes.
For a more general base: it's been a little while since I've thought about this carefully, but it seems to me that the proof of this part of Swan's Theorem does not use any assertions about the base.  In $\S$6.2 there is established an easy categorical equivalence between trivial vector bundles and finitely generated free modules over $\mathbb{R}(B)$.  It is then explained that every finitely generated projective module is the image of an idempotent endomorphism $P$ of a finitely generated free module, so under the categorical equivalence we have an idempotent endomorphism $P$ of a trivial vector bundle $E$ on $X$ and we need to show that $P(E)$ is a vector bundle on $X$.  This is Proposition 6.4.  (The real meat of Swan's Theorem -- and here it seems that paracompactness is needed -- is to realize every vector bundle with bounded fiber dimensions as a direct summand of a trivial bundle.) 
So...along with Qiaochu, I also say yes.
Added: I have just consulted the lovely recent text Ideals and Reality by Ischebeck and Rao.  It has a superior (e.g., to mine!) treatment of Swan's Theorem.  In fact it gives a more general result which both nails down an answer to the OP's second question (the first one is much easier) and addresses the issues raised in the comments.
For any topological space $X$, a vector bundle $E$ on $X$ is strongly of finite type if
(SFT1) There is a finite open cover $\mathcal{U} = \{U_i\}_{i=1}^n$ on which $E$ trivializes: i.e., for all $i$, the pullback of $E$ to $U_i$ is a trivial bundle, and
(SFT2) There is a (continuous) partition of unity subordinate to the cover $\mathcal{U}$.   
Some remarks:
(i) A topological space $X$ is normal (every pair of disjoint closed subsets can be enlarged to a pair of disjoint open subsets; note that we do not require $X$ to be Hausdorff) iff every finite open covering admits a subordinate partition of unity.  (By contrast, a Hausdorff space is paracompact iff every locally finite open covering admits a subordinate partition of unity.)
(ii) (SFT1) implies that the fibers have bounded dimension.  If $X$ is a direct sum of infinitely many nonempty subspaces, then there are vector bundles without this property.
And here we go:
Generalized Swan's Theorem (Theorem 5.3.15 in Ischebeck & Rao): For any topological space $X$, the global section functor gives an equivalence from the category of strongly finite type vector bundles on $X$ to the category of finitely generated projective $\mathbb{R}(X)$-modules.<|endoftext|>
TITLE: A combinatorial identity involving generalized harmonic numbers
QUESTION [20 upvotes]: The $n$-th harmonic number is defined as
$$
H_n=\sum_{k=1}^{n}\frac{1}{k},
$$
and the generalized harmonic numbers are defined by
$$
H_{n}^{(r)}=\sum_{k=1}^{n}\frac{1}{k^r}.
$$
Recently, I have found the following combinatorial identity involving the second-order harmonic numbers (I have computational evidence).
Question:
\begin{align}
\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s }{s+1}H_{s}^{(2)}=\frac{2(-1)^m}{m+1}\sum_{s=0}^m H_{s}^{(2)}. 
\end{align}
Is this a known combinatorial identity? Any proof or reference?
However, if I replace $H_s^{(2)}$ by other generalized harmonic numbers in the above identity, I can not find some similar identities.
Note:
This combinatorial identity was motivated by the following identity
\begin{align}
\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m.
\end{align}
One can refer to How to prove $\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m$?
I appreciate any hints, pointers etc.!

REPLY [14 votes]: The identity
$$
\begin{align}
\sum_{s=1}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s }{s+1}H_{s}^{(2)}=\frac{2(-1)^m}{m+1}\sum_{s=1}^m H_{s}^{(2)}. \tag{1}
\end{align}
$$
is equivalent to the following identity
$$
\sum_{s=1}^{m}{2s\choose s}\frac{H_s^{(2)}}{s+1}(x-x^2)^s=\frac{2\text{Li}_2(x)}{1-x}-\frac{\ln^2(1-x)}{x},\tag{2}
$$
where $\text{Li}_2$ is dilogarithm function. To prove this, first note that
$$
\sum_{s=1}^m H_{s}^{(2)}=\sum_{s=1}^m \sum_{k=1}^s \frac{1}{k^2}=\sum_{k=1}^m \frac{1}{k^2}\sum_{s=k}^m 1=\\ \sum_{k=1}^m \frac{1}{k^2}(m+1-k)=(m+1)H_{m}^{(2)}-H_{m}^{(1)}.
$$
Now using the generating functions (see link1, link2)
$$
\sum_{m=1}^\infty \frac{H_{m}^{(1)}}{m+1}x^m=\frac{\ln^2(1-x)}{2x},\qquad \sum_{m=1}^\infty H_{m}^{(2)}x^m=\frac{\text{Li}_2(x)}{1-x}
$$
and proceeding as in this answer one obtains
\begin{align}
\sum_{s=1}^{\infty}\binom{2s}{s}\frac{(-1)^s}{s+1}x^s (1+x)^sH_s^{(2)}=&\sum_{m=1}^{\infty}\frac{2(-x)^m}{m+1}\left[(m+1)H_{m}^{(2)}-H_{m}^{(1)}\right]=\\
&\frac{2\text{Li}_2(-x)}{1+x}-\frac{\ln^2(1+x)}{-x}
\end{align}
which is equivalent to $(2)$.
$\bf{Proof\ of\ eq.(2)}$
The generating function of Catalan numbers
$$
f(x)=\sum_{n=0}^\infty {2n\choose n}\frac{x^n}{n+1}=\frac{1-\sqrt{1-4x}}{2x}.\tag{3}
$$
The generating function of ${2s\choose s}\frac{H_s^{(2)}}{s+1}$ can be obtained from $(3)$  by integrating the identity
$$
\frac{f(x)-f(xy)}{1-y}=\sum_{n=1}^\infty {2n\choose n}\frac{x^n}{n+1}(1+y+...+y^{n-1})
$$
as follows
\begin{align}
\sum_{s=1}^{m}{2s\choose s}\frac{x^s}{s+1}H_s^{(2)}=&\int_0^1\frac{dt}{t}\int_0^t \frac{f(x)-f(xy)}{1-y}dy=\\
&\int_0^1\frac{dt}{t}\int_0^t\frac{dy}{1-y}\left(\frac{1-\sqrt{1-4x}}{2x}-\frac{1-\sqrt{1-4xy}}{2xy}\right)=\\
&-\int_0^1\frac{\ln t}{1-t}\left(\frac{1-\sqrt{1-4x}}{2x}-\frac{1-\sqrt{1-4xt}}{2xt}\right)dt
\end{align}
[the last line follows after integrating by parts with respect to $t$]. After substitution $\sqrt{1-4xt}=u$ this last integral becomes
$$
\frac{\pi^2}{6}\cdot \frac{1-\sqrt{1-4x}}{2x}+4\int\limits_{\sqrt{1-4x}}^1\frac{\ln\frac{1-y^2}{4x}}{(1+y)(y^2-1+4x)}ydy.
$$
So after replacing $x$  with $x-x^2$ one obtains
$$
\sum_{s=1}^{m}{2s\choose s}\frac{(x-x^2)^s}{s+1}H_s^{(2)}=\frac{\pi^2}{6(1-x)}+4\int\limits_{1-2x}^1\frac{\ln\frac{1-y^2}{4x(1-x)}}{(1+y)(y^2-(1-2x)^2)}ydy
$$
This last integral can be calculated by Mathematica in terms of dilogarithm functions, or manually after expanding $\frac{y}{(1+y)(y^2-(1-2x)^2)}$ into partial fractions. Then using functional equations for dilogarithm I verified that the resulting expression equals $\frac{2\text{Li}_2(x)}{1-x}-\frac{\ln^2(1-x)}{x}$.<|endoftext|>
TITLE: Do mathematical objects disappear?
QUESTION [64 upvotes]: I am asking this question starting from two orders of considerations. 
Firstly, we can witness, considering the historical development of several sciences, that certain physical entities "disappeared": it is the case of luminiferous aether with the surge of Einstein's relativity, or the case of the celestial spheres that disappeared in the passage from the pre-copernica universe to the Copernican universe.
Secondly, if we consider the evolution of mathematics, we assist to quite an opposite phenomenon: new mathematical objects are often invented and enrich already existing ontologies (let us consider the growth of the number system with negative, imaginary, etc., or the surge of non-euclidean geometries).
But can anyone think of examples of mathematical objects that have, on the contrary, disappeared (have been abandoned or dismissed by mathematicians)?

REPLY [8 votes]: There are cases like this ...
Walter Feit and John Thompson wrote a 250-page paper on nonabelian simple groups of odd order, and in the end conclude that there are none.<|endoftext|>
TITLE: What are "Artin fractions"?
QUESTION [10 upvotes]: The German Wikipedia entry for Ernst Witt https://de.wikipedia.org/wiki/Ernst_Witt has a photo of his grave in Hamburg. The bottom part has a visible text "Artin Brueche" (Artin fractions) but the numbers are obscured.
Does anyone know what the numbers are, why they are called Artin fractions, and why Witt wanted to have them on his gravestone.

REPLY [5 votes]: I have the impression that this "Artin fraction" is the Artin symbol, introduced by Helmut Hasse. Both Hasse and Emil Artin were teachers of Ernst Witt.
The inscription higher up on the grave stone shows a Witt vector, perhaps in reference to the Artin-Schreier-Witt theory. My guess would be that the inscription at the bottom explains the notation of the formula above it. Perhaps someone in the vicinity of Hamburg can make a trip to the cemetery and provide a full exposure of the grave stone?<|endoftext|>
TITLE: Prove $4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}$
QUESTION [15 upvotes]: Wolstenholme's theorem is stated as follows:
if $p>3$ is a prime, then
\begin{align*}
\sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\
\sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}.
\end{align*}
It is also not hard to prove that
$$
\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}.
$$
However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.
Question:
If $p$ is an odd prime, then
$$
4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}.
$$
I have verified this congruence for $p$ upto $7919$.
Comments: 
(1) Since 
$$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$
we need to prove
$$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$
This idea was given by Fedor Petrov.
(2) It is interesting that the congruence in the question is ture mod $p^3$ for $p\ge 7$, i.e.,
$$
4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}, \quad \text{for $p\ge 7$}.
$$
This was conjectured by tkr.
I appreciate any proofs, hints, or references!

REPLY [20 votes]: This goes way back to Emma Lehmer, see her elementary paper on Fermat quotients and Bernoulli numbers from 1938.
Assume $p\ge 7$. First, I reformulate your congruence. I replace
$$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}$$
with the equivalent
$$\sum_{k=1}^{p-1} \frac{1}{k^2} \equiv 8\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{(p-2k)^2}\pmod{p^3}$$
Lehmer has proved the following, see equation (19) in the linked paper:
$$\sum_{k=1}^{\frac{p-1}{2}} (p-2k)^{2r} \equiv p 2^{2r-1} B_{2r} \pmod {p^3}$$
(valid when $2r \not\equiv 2 \pmod {p-1}$). Plugging $2r=p^3-p^2-2$, we get
$$\sum_{k=1}^{\frac{p-1}{2}} (p-2k)^{p^3-p^2-2} \equiv p 2^{p^3-p^2-3} B_{p^3-p^2-2} \pmod {p^3}$$
Now use Euler's theorem to replace $x^{p^3-p^2-2}$ with $x^{-2}$:
$$\sum_{r=1}^{\frac{p-1}{2}} \left(\frac{1}{p-2k}\right)^2 \equiv p 2^{-3} B_{p^3-p^2-2} \pmod {p^3}$$
So your congruence becomes
$$(*)\qquad \sum_{k=1}^{p-1} \frac{1}{k^2} \equiv pB_{p^3-p^2-2} \pmod {p^3}$$
This was also proved in Lehmer's paper - in equation (15) she writes
$$\sum_{k=1}^{p-1} k^{2r} \equiv pB_{2r} \pmod {p^3}$$
(valid when $2r \neq 2 \mod {p-1}$). Plugging $2r=p^3-p^2-2$ and using Euler's theorem, we get the desired result.

It is now easy to generalize - when $2r \not\equiv 2 \pmod {p-1}$, we have
$$\sum_{k=1}^{p-1} (-1)^k k^{2r} \equiv (1-2^{2r}) \sum_{k=1}^{p-1} k^{2r} \pmod {p^3}$$
Plugging $2r=p^3-p^2-2c$ ($c \not\equiv -1 \pmod {\frac{p-1}{2}}$) we get
$$\sum_{k=1}^{p-1} \frac{(-1)^k}{k^{2c}} \equiv (1-2^{-2c}) \sum_{k=1}^{p-1} \frac{1}{k^{2c}} \pmod {p^3}$$

To get a taste of things, I include a proof of $(*)$. Via Faulhaber's formula for sum of powers, we get:
$$\sum_{k=1}^{p-1} \frac{1}{k^2} \equiv \sum_{k=1}^{p-1} k^{p^3-p^2-2} \equiv \sum_{k=1}^{p} k^{p^3-p^2-2} = $$
$$\frac{1}{p^3-p^2-1} \sum_{j=0}^{p^3-p^2-2}\binom{p^3-p^2-1}{j}B_j p^{p^3-p^2-1-j} \pmod {p^3}$$
Note that odd-indexed Bernoulli's vanish, and that by the Von Staudt–Clausen theorem, the even-indexed Bernoulli's have $p$ in the denominator with multiplicity $\le 1$. Additionally, $B_{p^3-p^2-4}$ is $p$-integral (since $p-1 \nmid p^3-p^2-4$).
Hence, for the purpose of evaluating the sum modulo $p^3$,  we can drop almost all the terms and remain only with the last one:
$$\sum_{k=1}^{p-1} \frac{1}{k^2} = \frac{1}{p^3-p^2-1} \binom{p^3-p^2-1}{p^3-p^2-2} B_{p^3-p^2-2} p  = pB_{p^3-p^2-2} \pmod {p^3}$$<|endoftext|>
TITLE: What is an ordered structure, in general?
QUESTION [8 upvotes]: This is basically a reference request, but the post is going to be relatively long (and a little bit verbose): I apologize in advance for that.
Premise. There are several examples of "ordered structures"  appearing in nature (at least for the moment, let me be vague with the actual meaning of various terms I'm using): ordered groups, ordered rings, ordered vector spaces, etc. All of these have something in common: They are algebraic structures, where each operation is compatible, in some sense and in some way, with a (partial) order. Thus, I find it reasonable to ask:

What about a fully formal definition of an ordered structure, in general?

For simplicity, I'm not seeking a definition worded in the language of categories, and I'm looking at the case where "structure" means whatever can be understood as a (classical) model of a finitary, single-sorted, first-order algebraic theory in the frame of model theory (in principle, I'm also interested in possibly non-finitary or many-sorted structures, but the general case would probably make things look more complicated than they are and overshadow some relevant details).
The most basic, non-trivial example is provided by structures with one operation. For instance, an ordered semigroup is a triple $\mathbb A = (A, {\cdot}\,, \preceq)$, where $(A, \cdot)$ is a semigroup (written multiplicatively) and $\preceq$ an order on $A$ with the property that $xz \preceq yz$ and $zx \preceq zy$ for all $x,y,z \in A$ such that $x \preceq y$.
Of course, the axiom of associativity doesn't play any role in these definitions, so from a fundamental point of view it would have been morally better to replace "semigroups" with "magmas" in the previous paragraph, but on the other hand, that doesn't really matter, insofar as I'm just using ordered semigroups as a guiding example to forge a tentative answer to my question along the following lines.
Naive (and "wrong") approach. Let $T$ be a finitary, single-sorted, first-order theory, which, to me, means a triple $(\sigma, \Xi, V)$, where:

$V$ is an infinite countable set of (logical) variables from the formal language, $\mathcal L$, underlying the axiomatic theory used to lay out the foundations (say, Tarski-Grothendieck set theory, as oversized as it may appear);
$\sigma$ is a (single-sorted) signature, namely a triple $(\Sigma_{\rm f}, \Sigma_{\rm r}, \varrho)$ consisting of a set of function symbols $\Sigma_{\rm f}$, a set of relation symbols $\Sigma_{\rm r}$, and a function $\varrho: \Sigma_{\rm f} \cup \Sigma_{\rm r} \to \mathbf N^+$ such that $\Sigma_{\rm f}$, $\Sigma_{\rm r}$, and the set of (logical and non-logical) symbols of $\mathcal L$ are pairwise disjoints (the function $\varrho$ assigns to each function or relation symbol of $\sigma$ an ariety; in particular, an $n$-ary function symbol will correspond, in any model of the theory, to a function $A^{n-1} \to A$ (for some set $A$), and an $n$-ary relation symbol to a subset of $A^n$);
$\Xi$ is a (possibly empty) subset of $\langle V; \sigma \rangle$,  the set of all (well-formed) formulas generated by combining, according to the formation rules of first-order logic, the variables in $V$, the (function and relation) symbols of $\sigma$, and those of $\mathcal L$ which are not logical variables.

(The role of $V$ is meaningless here, as we could assume that $V$ includes all the variables in $\mathcal L$; however, it starts being meaningful in the case of many-sorted theories, and that's why I beg your indulgence for being redundant in this respect.)
Now, $T$ is called an algebraic theory if $\Sigma_{\rm r}$ is empty (in which case $\sigma$ is referred to as an algebraic signature), and since we are interested in structures that are algebraic except for the fact of being endowed with an order, we may refer to $T$ as a quasi-algebraic theory if

$\Sigma_{\rm r}$ consists of one relation symbol $R$, subjected to the axioms of reflexivity, antisymmetry, and transitivity (for binary relations), which are hence comprised among the formulas in $\Xi$ (so that $R$ is interpreted as a partial order in any possible model of $T$);
$\Xi$ includes basic formulas encoding the compatibility between each function symbol of $\sigma$ and the relation $R$, which can be phrased as follows: If $\varsigma$ is a function symbol in $\Sigma_{\rm f}$ of ariety $n := \varrho(\varsigma)$, then the formula
$$
\forall \vec{x}, \vec{y} \in V^{n-1}: R^{(2n-2)}(\vec{x}, \vec{y}) \implies R(\varsigma(\vec{x}), \varsigma(\vec{y}))\qquad\qquad(\star)
$$
belongs to $\Xi$. (The notation should be self-explanatory, but let me note that the formula in the case of constant symbols is redundant, in the sense that it is implied, in any model of theory, by the very fact that $R$ is reflexive; yet, having it there makes the approach look more "uniform" than otherwise, in the same spirit of Joel David Hamkins' comment below.)

An ordered structure would then be any model of a quasi-algebraic theory.
Unfortunately, the above doesn't work even in the case of (the theory of) groups, where $\Sigma_{\rm f}$ consists, say, of the symbols $\cdot$ (binary symbol for "multiplication"), ${}^{-1}$ (unary symbol for "inversion"), and $1$ (constant symbol for "identity"). In fact, the naive approach would suggest to assume one binary relation symbol, say $\preceq$, in $\Sigma_{\rm r}$ and the following axioms in $\Xi$:
$$
\begin{split}
(1)\ & \forall x, y, z \in V: x \cdot (y \cdot z) = (x \cdot y) \cdot z \\
(2)\ & \forall x \in V: x \cdot 1 = 1 \cdot x = x \\
(3)\ & \forall x,y \in V: x \cdot x^{-1} = x^{-1} \cdot x = 1 \\
(4)\ & \forall x,y,z \in V: (x \preceq x) \land (((x \preceq y) \land (y \preceq y) \implies (x=y))) \land (((x \preceq y) \land (y \preceq z) \implies (x \preceq z))) \\
(5)\ & \forall x,y \in V: (x \preceq y) \implies (x^{-1} \preceq y^{-1}) \\
(6)\ & \forall x,y,z \in V: (x \preceq y) \implies ((x \cdot z \preceq y \cdot z) \land (z \cdot x \preceq z \cdot y)) \\
\end{split}
$$
The problem is that the 5th axiom is essentially incompatible with the 6th, as the latter yields that, in every model $(A; +, \cdot\,,1;\preceq)$ of $T$, we have $x \preceq y$ iff $y^{-1} \preceq x^{-1}$.
Remedies. One may argue that the "mistake" lies in the fact of including the function symbol ${}^{-1}$ in the signature of the theory, while an alternative could be to avoid it and replace the 3th axiom above with:
$$
\forall x \in V, \exists\, \tilde{x} \in V: x \cdot \tilde{x} = \tilde{x} \cdot x = 1.
$$
This is certainly a possibility (as long as $\mathcal L$ includes both $\forall$ and $\exists$ among the logical symbols), but I find it rather "artificial" (whatever it may mean), all the more that the same strategy fails if we try to use the naive approach outlined in the above to recover as a special case the common definition of an ordered ring, for which the compatibility between multiplication and order is "restricted to nonnegative factors", which has no hope to fit in the paradigm implied by condition $(\star)$.
So, putting it all together, my (second) question is:

Where should I look up for a sufficiently general definition of "ordered structure", which copes with the kind of issues that I've tried to point out in this post?

I've my own ideas on what to do, but there may be a much better way on how to proceed, which is what I'm looking for. In particular, one solution could be as follows:

Start with a finitary, single-sorted, first-order theory $T = (\sigma, \Xi, V)$, whose signature is of the form $((\varsigma_i)_{i \in I}, (R_i)_{i \in I \,\cup\, \{\infty\}}, \varrho)$ for some index set $I$ and $\infty \notin I$, so that each function symbol $\varsigma$ has one corresponding relation symbol of ariety $2\varrho(\varsigma)-2$, and we have an extra relation symbol $R_\infty$.
Make $\Xi$ include, for each relation symbol $R$ of $\sigma$, the axioms of reflexivity, antisymmetry, and transitivity (extended, as appropriate, from binary to $2n$-ary relations), so that, in particular, $R_\infty$ is interpreted as an
order in any possible model of the theory.
Rewrite condition ($\star$) so as to replace, for each
function symbol $\varsigma$ of $\sigma$, the symbol "$R^{(2n-2)}$" to the left of the connective "$\implies$" with the (unique) relation
corresponding to the function symbol $\varsigma$, and the symbol "$R$" to the right with "$R_\infty$".

These conditions together describe a paradigm I will refer to as (P), the last condition encoding the naive idea that the relations $R_i$ must be glued together in a "consistent way" (and can't be "completely independent" from each other, which otherwise would result into something exceedingly general, I feel).
An example. Assume the (algebraic) theory of unital rings is encoded by the (algebraic) signature whose set of function symbols is given by $\Sigma_{\rm ring} := \{+, \cdot\,, -\,, 0, 1\}$, where $+$ and $\cdot$ are, respectively, the binary symbols for "addition" and "multiplication", $-$ is the unary symbol for "additive inverse", and $0$ and $1$ are, respectively, the constant symbols for "additive identity" and "multiplicative identity". The axioms of the theory are the usual ones:
$$
\begin{split}
(1)\ & \forall x, y, z \in V: (x+(y+z) = (x+y)+z) \land (x \cdot (y \cdot z) = (x \cdot y) \cdot z) \\
(2)\ & \forall x \in V: (x+0 = 0 + x = x) \land (x \cdot 1 = 1 \cdot x = x) \\
(3)\ & \forall x,y \in V: x + (-x) = (-x) + x = 0 \\
(4)\ & \forall x,y,z \in V: (x \cdot (y+z) = (x \cdot y) + (x \cdot z)) \land ((y+z) \cdot x = (y \cdot x) + (z \cdot x)) \\
\end{split}
$$
Now, according to the paradigm (P), the theory of ordered rings would have a signature of the form $(\Sigma_{\rm ring}, \Sigma_{\rm r}, \varrho)$, where $\Sigma_{\rm r}$ is a set of relation symbols of the form $\{R_{(+)}, R_{(\cdot)}, R_{(-)}, R_{(0)}, R_{(1)}, \preceq\}$, with each relation symbol subjected to the axioms of reflexivity, antisymmetry, and transitivity, all the relation symbols glued together by the 4th condition of (P), and each function symbol $\varsigma$ in $\Sigma_{\rm ring}$ "bound" to the corresponding relation symbol $R_{(\varsigma)} \in \Sigma_{\rm r}$ by the formula:
$$
\forall (\vec{x}, \vec{y}) \in V^{n-1} \times V^{n-1}: R_{(\varsigma)}(\vec{x}, \vec{y}) \implies (\varsigma(\vec{x}) \preceq \varsigma(\vec{y})),
$$
where $n$ is the ariety of $\varsigma$.
In particular, this paradigm fits with the usual notion of an ordered ring, which is then a tuple $\mathbb A = (A; +, \cdot\,,-\,,0,1; R_{(+)}, R_{(\cdot)}, R_{(-)}, \preceq)$, where $\preceq$ is a partial order on $A$, $R_{(+)}$ the subset of $A^2 \times A^2$ consisting of those pairs $((x,y),(x,z))$ or $((y,x),(z,x))$ with $y \preceq z$, $R_{(\cdot)}$ the subset of $A^2 \times A^2$ consisting of those pairs $((x,y),(x,z))$ or $((y,x),(z,x))$ such that $0 \preceq x$ and $y \preceq z$, and $R_{(-)} = \{(x,x): x \in A\}$. (I'm intentionally omitting any explicit reference in models to the relations associated with constants, for they don't add any information, as a result of considerations already made in the above.)

REPLY [4 votes]: I agree with Gerhard about the value of a universal algebraic approach. There is one insight I can offer, that comes from a paper I wrote 25 years ago, as an undergraduate. It is simply that the defining sentences ought to have the form
(universal closure of)
$$\left(\bigwedge \Psi_i\right) \to \Phi$$
where the $\Psi_i$ are atomic inequalities and $\Phi$ is an atomic equality or inequality. E.g., $(x \geq 0 \wedge y \geq z) \to xy \geq xz$. Properties like associativity or distributivity involve an empty premise. We have a Birkhoff-type theorem which says that a class of structures is definable in this way if and only if it is closed under the formation of products, substructures, and "$*$-homomorphic" images --- an odd choice of terminology which somehow anticipated my later interest in C*-algebras. Of course, this doesn't address your main question of why your axiom (5) for ordered groups shouldn't be included, but maybe it could help in some way. The reference is Generalized varieties, Algebra Universalis 30 (1993), 27–52 if you are interested.<|endoftext|>
TITLE: Quotient of a smooth curve by a finite group and differentials
QUESTION [11 upvotes]: Let $X$ be a proper smooth connected curve over an algebraically closed field $k$ of characteristic $0$, and suppose that $X$ is equipped with a $k$-linear action of a finite group $G$. It makes sense to form the quotient curve $Y := X/G$, and $Y$ is $k$-smooth because it is normal. Is it true that the pullback of differentials gives the identification
$$H^0(Y, \Omega^1_{Y/k}) = H^0(X, \Omega^1_{X/k})^G?$$
If so, how does one prove this?

REPLY [8 votes]: Counterexample in the non-tame case.  This is not what was asked by the OP.  However, since it comes up in answers and comments above, I thought I would write down one example to illustrate the problem in the non-tame case.  
Let $k$ be a field of characteristic $p$.  Let $U$ be $\text{Spec}\ k[x][1/(x^{p-1}-1)]$, i.e., the open affine $D(x^{p-1}-1)$ in $\mathbb{A}^1_k$.  In other words, remove the closed points corresponding to elements of $\mathbb{F}_p^\times$.  Let $f:U\to U$ be the morphism of $k$-schemes with $f^*(x) = x/(1+x)$.  This is an automorphism of order $p$.  The quotient is $V = \text{Spec}\ k[y] \cong \mathbb{A}^1_k$ with quotient morphism $q:U\to V$, $q^*y = x^p/(1-x^{p-1})$.  This is a finite morphism that extends to a finite étale morphism $\mathbb{P}^1_k\setminus \{0\}\to \mathbb{P}^1_k\setminus \{0\}$ (this is one of those examples proving that $\mathbb{A}^1_k$ is not algebraically simply connected in characteristic $p$).
Consider the algebraic $1$-form on $U$, $$\alpha = \frac{x^{p-2}}{1-x^{p-1}} dx.$$  It is straightforward to compute that $f^*\alpha$ equals $\alpha$.  Yet $\alpha$ cannot be of the form $\phi^*\beta$ for any algebraic $1$-form $\beta$ on $V$.  Indeed, the module of algebraic $1$-forms on $V$ is generated by $dy$.  By direct computation, $$q^*dy = \frac{-x^{2p-2}}{(1-x^{p-1})^2}dx,$$ so that every form $q^*\beta$ vanishes to order at least $2p-2$ at $x=0$.<|endoftext|>
TITLE: Proving the irrationality of $\pi e$ and $\pi / e$
QUESTION [34 upvotes]: Rather than relying on the consequences of Schanuel's conjecture, I set about using the same ideas Apery had used to construct integer arguments converging fast enough to show $\zeta(3)$ is irrational in a form Beukers had introduced. I'm sure someone out there can crack what I have so far.
I will be using the following facts:

Theorem 1: Suppose the complex-valued function $$\begin{align}
 f{(z)} = \begin{cases} 
      -\left(\frac{1}{e}(1-z)^{1-\frac{1}{z}} \right)^q, &  z\neq 0 \\
      -1, & z = 0 
   \end{cases}
\end{align}$$ has a power series with positive radius of convergence of the form
  $$f(z) = \sum_{n=0}^\infty b_n(q) z^n$$
  Then
  $$b_n(q) = -\frac{q}{n} \sum_{k=1}^n \frac{b_{n-k}(q)}{k+1}, \quad b_0(q) = -1$$

Note that $b_n(q)$ is a polynomial of degree $n$.

Theorem 2: Let $m,q \in \mathbb{Z}$ and $m+q+1 \geq 0$; then $$\int_0^1 x^m \sin(\pi q x) \left(x^x (1-x)^{1-x}\right)^q \ dx = (-1)^{q+1} \pi e^q b_{m+q+1}(q)$$

The above can be shown by applying contour integration and residue theorem to the above function.

Theorem 3: For $n \in \mathbb{N} \cup \{0\}$, $\mathbb{P}$ be the set of primes, and let $$(n+1)!_\mathbb{P} = \prod_{p \in \mathbb{P}} p ^{\sum_{k\geq 0} \left\lfloor \frac{n}{(p-1)p^k} \right\rfloor}$$
  Then, for integer $q$, $(n+1)!_\mathbb{P} \cdot b_n(q)$ is an integer for $n \geq 0$.

This factorial like function is borrowed from Manjul Bhargava's work on the general factorial function.

Theorem 4 Let $n \in \mathbb{N} \cup \{0\}$; then $$(n+1)!_\mathbb{P}\sim e^{n(C-\gamma+o(1))}n^n$$ where $C = \sum_{p \in \mathbb{P}} \frac{\ln p }{(p-1)^2}$ and $\gamma$ is the Euler-Mascheroni constant.

If we let $P_n(x)$ be a polynomial of degree $n$ with integer coefficients and let $$I_n = \int_0^1 P_n(x) \left( \sin(\pi q x) \left( x^x (1-x)^{1-x}\right)^q -1\right) \ dx$$
We have the following inequality, in the form of Dirichlet's irrationality criterion,
$$0 < \left|C_n \pi e^q - D_n \right| = \left|(n+q+2)!_\mathbb{P} (n+1) I_n \right|$$
where $C_n, D_n \in \mathbb{Z}$. Of course, we can apply Theorem 4, and have something more familiar to work with. 

Question: Can we construct a polynomial $P_n(x)$ such that, for large $n$, $$\left|(n+q+2)!_\mathbb{P} (n+1) I_n \right| \to 0 \text{?}$$

If there does exist one, then, for $q \geq -2, q \neq 0$, the number $\pi e^q$ is irrational. Letting $q = 1, -1$, and the result follows.
I've been at this problem for some time, with no further progress. Frankly, I don't know what to do at all. If it helps, I've considered the shifted Legendre polynomials, as Beukers had done, though to no avail. 
Most of what I've seen regarding the nature of constructing a polynomial is that it belongs to the family of orthogonal polynomials.
God bless.

REPLY [3 votes]: This isn't really an answer as much as it is an "expanded" comment. 
Consider, for integer $a$, $$P_n(x) = \frac{1}{n!} \frac{d^n}{d x^n} x^n (1-ax)^n = \sum_{m=0}^n \binom{n}{m} \binom{n+m}{m} (-ax)^m$$
Given $$I_n = \int_0^1 P_n(x) \left( \sin(\pi q x) (x^x(1-x)^{1-x})^q - 1\right) \ dx$$
We have
$$I_n \leq \int_0^1 P_n(x) \left( \sin(\pi q x) a_q - 1\right) \ dx$$ where $a_q  = \max_{x\in (0,1)}\{(x^x(1-x)^{1-x})^q\}$. Furthermore, we have
$$\left|\int_0^1 P_n(x) \left( \sin(\pi q x) a_q - 1\right) \ dx \right|=  \left|\int_0^1 \frac{1}{n!} \frac{d^n}{d x^n} x^n (1-ax)^n\left( \sin(\pi q x) a_q - 1\right) \ dx\right|$$
$$=  \left|\frac{1}{a^{n+1}n!} \int_{(0,1)\cup(1,a)} \frac{d^n}{d x^n} x^n (1-x)^n\left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right) \ dx\right| $$
$$\leq  \left|\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{1}{n!a^{n+1}}\int_1^a \frac{d^n}{d x^n} x^n (1-x)^n\left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right) \ dx\right| $$
Let $$S_n = \int_1^a \left( \sin\left(\frac{\pi q x}{a}\right) a_q - 1\right)\frac{d^n}{d x^n} x^n (1-x)^n \ dx$$
So that we have
$$ = \left|\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{S_n}{n!a^{n+1}}\right|$$
Now, observing the bound in question, applying Theorem 4, and letting $A = C - \gamma + o(1)$, we have
$$\left| (n+q+2)!_\mathbb{P} (n+1) I_n \right|<\left|e^{An} e^{q+1} n^n \left(1+\frac{q+1}{n}\right)^n \left(1+\frac{q+1}{n}\right)^{q+1}n^{q+1}(n+1)\left(\left(\frac{\pi q}{4a^2}\right)^n \frac{a_q}{n! a}+\frac{S_n}{n!a^{n+1}}\right) \right| $$
If we ignore the $S_n$ term, we have that
$$\left| e^{q+1} \frac{n^n}{e^n n!} \left(1+\frac{q+1}{n}\right)^n \left(1+\frac{q+1}{n}\right)^{q+1}\frac{n^{q+1}(n+1)}{b^n}\left(\frac{e^{A+1}\pi bq}{4a^2}\right)^n \frac{a_q}{a} \right|$$
where $b > 1 $. If we consider $a$ such that $ a^2 > \frac{e^{A+1}\pi bq}{4}$, and applying Stirling's approximation to the left-most term (-ish), for large $n$, then the whole expression above tends to $0$. Now, it is left to consider the $S_n$ term, though I have a bad feeling about it. :/<|endoftext|>
TITLE: Elliptic curves and prime numbers
QUESTION [6 upvotes]: Let $p_n$ be the $n^{th}$ prime number. Suppose $E(F_{p_n})$ denotes an elliptic curve over the Galois field $GF(p_n)$ which is defined by $y^2=x^3+ax+b$. Is the below claim true?
For each integer number $n>3$, there exist integer numbers $a$ and $b$ such that $\#E(F_{p_n})=p_{n+1}$?

REPLY [14 votes]: (Sorry, I misread the question at first.)  The following result reduces your question to a problem of analytic number theory:
Theorem (Hasse-Deuring-Waterhouse): For a prime $p$ and $N \geq 1$ the following are equivalent:
(i) There is an elliptic curve $E_{/\mathbb{F}_p}$ such that $\# E(\mathbb{F}_p) = N$.
(ii) We have $|N-(p+1)| \leq 2\sqrt{p}$.
As long as $p > 3$ (i.e., $p = p_n$ for $n \geq 3$) every elliptic curve can be put in "short Weierstrass form" $y^2 = x^3 + ax +b$.
So you are reduced to asking: is it true that for all $n > 3$ we have 
$|p_{n+1} - (p_n+1)| \leq 2 \sqrt{p_n}$?
According to this esteemed source, having this kind of upper bound on the prime gap (always) is conjectured to be true but far from being proven.  In fact, if I have it right this precisely Andrica's Conjecture.  As Dror Speiser points out, it is not even known conditionally on the Riemann Hypothesis.<|endoftext|>
TITLE: When can Power Sets be Limit Cardinals?
QUESTION [13 upvotes]: My original question (posted in https://math.stackexchange.com/questions/1584430/can-all-power-sets-be-limit-cardinals) was:
Is it possible to create a model of ZFC, so that the cardinality of each power set is a limit cardinal (as opposed to GCH where they are always successor cardinals)?
Obviously, from Easton's theorem, this is possible for regular cardinals.
I also showed that in this model strong limit cardinals must also be fixed points of the $\aleph$ function (using Shelah's upper bound from PCF), and therefore such a model would mean $2^{\aleph_\delta} \geq \aleph_{\delta^+}$. But whether we can improve Shelah's upper limit from $2^{\aleph_\delta} < \aleph_{{|\delta|}^{+4}}$ to $2^{\aleph_\delta} < \aleph_{\delta^+}$ is an open question. So for strong limit cardinals, we don't know whether such a model is possible (even for one cardinal, let alone all of them).
This leaves the singular weak limit $\delta$ case, which isn't interesting unless we assume the continuum function doesn't become constant for all $\gamma$ such that $\kappa < \gamma < \delta$. Assuming this, can we create a model of where the cardinality of the power set of singular weak limit are limit cardinals?
For a concrete example, take the following rule (which satisfies Easton's theorem, and the known limitations for singular cardinals in http://www.math.tau.ac.il/~gitik/icm5.pdf):
$$2^{\aleph_\alpha} = \aleph_{\aleph_{\alpha + 1}}$$

REPLY [13 votes]: The answer to your question is yes. In the Foreman-Woodin model The generalized continuum hypothesis can fail everywhere, $2^\kappa$ is weakly inaccessible for all infinite cardinals $\kappa.$ To be more precise, Foreman and Woodin proved the following:

Theorem 1. Con(ZFC+there exists a supercompact cardinal and infinitely many inaccessibles above it) implies Con(ZFC+$\forall \kappa, 2^\kappa$ is weakly inaccessible).

In fact we can reduce their large cardinal assumption by replacing the supercompact cardinal with a strong cardinal. See my paper with Yair Hayut On Foreman's maximality principle.
Even more surprisingly, we can prove the following:

Theorem 2 Con(ZFC+there exists a strong cardinal) implies Con(ZFC+$\forall \kappa, 2^\kappa$ is a singular cardinal).

For the proof, see again my paper with Yair Hayut stated above.
Remark. I may mention that in general you can not talk about Shelah's bound in models of Theorems 1 or 2. The point is that in these models, the strong limit points are all fixed points of the $\aleph$-function, and Shelah's bound talks about non-fixed points of the $\aleph$-function.<|endoftext|>
TITLE: Maximal trivialising subspace for a vector bundle
QUESTION [8 upvotes]: Let $X$ be a locally compact Hausdorff space. Given a vector bundle $p: E\to X$, a subspace $Y$ of $X$ is called trivialising (for this bundle), if after restricting this bundle to $Y$, it is a trivial bundle. In other words, $p: p^{-1}(Y)\to Y$ is trivial. $Y$ is called maximally trivial, if it is trivial and there is no trivial subspace of $X$ which contains $Y$ as a proper subset. 

Given a point $x$ in $X$, a maximal trivialising subspace containing $x$ may
  not be unique. Does any maximal trivialising subspace must be open? Can we say
  something others about maximal trivialising subspaces?

There is no reference about this topic in the existing literatures.
A similar question has been posted on stackexchange:
https://math.stackexchange.com/questions/1581861/maximal-trivial-subspace-in-vector-bundles

REPLY [4 votes]: In addition to Mark's argument, a maximal trivialising $Y\subset X$ is also open, if we assume that $E$ is a $\Bbbk$-vector bundle with $\Bbbk=\mathbb R$ or $\mathbb C$. So every maximal trivialising subset is open and dense, but it is not (yet) clear that every trivialising subset is contained in a maximal one.
Assume $Y$ is trivialising, but not open. Then there exists $x\in Y$ such that no neighbourhood of $x$ is contained in $Y$. We choose a trivialising neighbourhood $U$ for $E$. We have maps $\varphi\colon E|_Y\to\Bbbk^r$ and $\psi\colon E|_U\to\Bbbk^r$ such that
$$p\times\varphi\colon E|_Y\to Y\times\Bbbk^r\quad\text{and}\quad
p\times\psi\colon E|_U\to U\times\Bbbk^r$$
are vector bundle isomorphisms. Hence there exists $g\colon Y\cap U\to GL_r(\Bbbk)$ such that $\psi(e)=g(p(e))\cdot\varphi(e)$ on $E|_{Y\cap U}$.
Because $g$ is continuous and $Y\cap U$ carries the subspace topology, there exists a compact neighbourhood $K\subset U$ of $x$ and a map $\xi\colon Y\cap K\to\mathfrak{gl}_r(\Bbbk)$ such that $g|_{K\cap Y}=\exp\circ\xi$. By Urysohn's lemma, there also exists a cutoff function $\rho\colon X\to[0,1]$ with $\mathrm{supp}(\rho)\subset K$ and such that $W=\rho^{-1}(1)$ is a neighbourhood of $x$. We replace the chosen trivialisation of $E|_Y$ by $$\varphi'(e)=\exp((\rho\cdot\xi)(p(e)))\cdot\varphi(e)\;.$$
Then $\varphi'$ agrees with $\varphi$ on $Y\setminus\mathring K$,
and with $\psi$ on $W\cap Y$.
Hence, we can extend $\varphi'$ by $\psi$ on $W$, contradicting the maximality of $Y$.<|endoftext|>
TITLE: What does this connection between Chebyshev, Ramanujan, Ihara and Riemann mean?
QUESTION [33 upvotes]: It all started with Chris' answer saying returning paths on cubic graphs without backtracking can be expressed by the following recursion relation:

$$p_{r+1}(a) = ap_r(a)-2p_{r-1}(a)$$

$a$ is an eigenvalue of the adjacency matrix $A$. Chris mentions Chebyshev polynomials there.
It was Will who found the generating function for the given recursion to be:

$$G(x,a)=\frac{1-x^2}{1-ax+2x^2} $$

and just recently Hamed put Chebyshev back on the table:

$$
\frac{1-x^2}{1-ax +2x^2} \xrightarrow{x=t/\sqrt{2}}\frac{1-\frac{t^2}2}{1-2\frac{a}{\sqrt 8} t+t^2}=\left[1-\frac{t^2}{2}\right]\sum_{r=0}^\infty U_r\left(\frac{a}{\sqrt{8}}\right)t^r\\
=\sum_{r=0}^\infty \left(U_r\left(\frac{a}{\sqrt{8}}\right)-\frac12 U_{r-2}\left(\frac{a}{\sqrt{8}}\right)\right)t^r\\
$$
$$
\Rightarrow p_r(a)=\begin{cases}1 & \text{if $r=0$,}\\ 2^{r/2}\left(U_r(a/\sqrt{8})-\frac12U_{r-2}(a/\sqrt{8})\right) & \text{if $r\ge1$.}\end{cases}
$$
  The final line as taken from Will's community answer...

My question how to relate Ihara's $\zeta$ function and Chebyshev seems therefore mostly settled, but...:
Is it just a funny coincidence that the scaling factor of $\sqrt 8$ coincides with $\lambda_1\leq 2\sqrt 2$, which is the definition of cubic Ramanujan graphs.
And, there is another interesting thing:

As observed by Sunada, a regular graph is a Ramanujan graph if and only if its Ihara zeta function satisfies an analogue of the Riemann hypothesis.

What does this connection between Chebyshev, Ramanujan, Ihara and Riemann mean?
EDIT
I thought maybe something like a corollary could be possible:

For Ramanujan graphs, the Ihara $\zeta$ function can be related to Chebyshev functions of the second kind, since the scaled eigenvalues of $A$ lie inside the range of convergence.
A Ramanujan graph $G$ obeys the Riemann Hypothesis.
Roots of the Ihara $\zeta$ function lie on the critical strip.

The bunch of people above have contributed to $1\leftarrow 2$.
$2 \leftrightarrow 3$ is proven here: Eigenvalues are of the form $\lambda=2\sqrt 2\cos(b\log 2)$
$3\overset{\rightarrow ?}{\leftarrow} 1$ would be nice...

REPLY [4 votes]: Let $G$ be a $d$-regular connected graph on $n$ vertices. Let
$$d=\lambda_0 \geq \lambda_1 \geq \lambda_2 \geq \dots \geq \lambda_{n-1}$$
be the $n$ eigenvalues of the adjacency matrix $A$, and let $\lambda = max( |\lambda_1|,|\lambda_{n-1}|)$.
Intuitively, the significance of the spectrum of $A$ is that many combinatorial quantities can be expressed linear-algebraically, and optimizations problems in this framework often involve the spectrum.
For instance, we saw that for two vertex subsets $S,T \subseteq V$, the cardinality of $E(S,T)$ is 
$$E(S,T)=\textbf{1}_{T}^{*} A \textbf{1}_S$$
and this in turn could be expressed as a sum over eigenvalues of $A$.\
In these cases, the summand corresponding to the eigenvalue $d$ turns out to be the "average" value that we would expect of the quantity, while the summands corresponding to the remaining eigenvalues would then be interpreted as the "error" term. In the case of $E(S,T)$, the summand corresponding to $d$ is
$$d \frac{|S|}{\sqrt{n}} \frac{|T|}{\sqrt{n}} = \frac{d}{n} |S|.|T|$$
and this is the value we expect for a $d$-regular random graph on $n$ vertices. The remaining eigenvalues and their eigenvectors, through their interaction with $S$ and $T$ determine the deviation of $|E(S,T)|$ from this average value. So when the remaining eigenvalues are small in absolute value, this gives us a bound on the error term and allows us to conclude that for all sets $S,T \subseteq V$, $|E(S,T)|$ is close to the average value.\
This idea can be used in other enumeration problems too. All graph properties, by definition, depend on the adjacency matrix, and so can be expressed in terms of $A$. The challenge is then to ensure that the expression is amenable to linear-algebraic tools.\
Another simple example is that of a random walk on the graph. Suppose we are interested in the number of walks on $G$ of length $k$ that are cycles. Clearly this is
$$Tr(A^k)$$
which is
$$d^k + \lambda_1^k + \lambda_2^k + \dots + \lambda_{n-1}^k$$
Observe that the total number of walks of length $k$ on $G$ is
$$n.d^k$$
In a random graph, we would expect a random walk to end up at its starting point with probability $1/n$. And so the "average" number of closed walks of length $k$ on a random $d$-regular graph is 
$$\frac{n.d^k}{n}=d^k$$
and this is again exactly the summand corresponding to the eigenvalue $d$ in the linear algebraic formulation of the expression for the fixed graph $G$. Furthermore the number of closed walks of length $k$ on $G$ is
$$d^k \pm (n-1)\lambda^k$$
and $(n-1)\lambda^k$ is an upper bound on the error term.\
A more non-trivial example of this is in the case of non-backtracking closed walks of length $k$ on $G$. In this case we are interested in 
$$Tr(A_k)$$
Since 
$$A_k = (d-1)^{k/2} U_k \left( \frac{A}{2 \sqrt{d-1}} \right) - (d-1)^{k/2-1} U_{k-2} \left( \frac{A}{2 \sqrt{d-1}} \right)$$
we have
$$Tr(A_k) = (d-1)^{k/2} \sum \limits_{j=0}^{n-1} U_k \left( \frac{\lambda_i}{2 \sqrt{d-1}} \right) - (d-1)^{k/2-1} \sum \limits_{i=0}^{n-1} U_{k-2} \left( \frac{\lambda_i}{2 \sqrt{d-1}} \right) $$
Since the total number of non-backtracking walks of length $k$ is
$$n d(d-1)^{k-1}$$
we would expect a $1/n$ fraction of these to be closed. The summand corresponding to the eigenvalue $d$ is
$$(d-1)^{k/2} \sum \limits_{j=0}^{n-1} U_k \left( \frac{d}{2 \sqrt{d-1}} \right) - (d-1)^{k/2-1} \sum \limits_{i=0}^{n-1} U_{k-2} \left( \frac{d}{2 \sqrt{d-1}} \right)$$
Using the closed form expression for $U_k$ given by
$$U_k(x) = \frac{(x+\sqrt{x^2-1})^{k+1}- (x-\sqrt{x^2-1})^{k+1}   }{2\sqrt{x^2-1}}$$
we get 
$$(d-1)^{k/2} \sum \limits_{j=0}^{n-1} U_k \left( \frac{d}{2 \sqrt{d-1}} \right) - (d-1)^{k/2-1} \sum \limits_{i=0}^{n-1} U_{k-2} \left( \frac{d}{2 \sqrt{d-1}} \right) = d(d-1)^{k-1}$$
as expected, and the summands corresponding to the remaining eigenvalues form the error term.\
It is here that the Ramanujan property has a natural interpretation!\
Consider the function $U_k(x)$. It is known that for $|x| \leq 1$, $U_k(x)$ has a trigonometric expression given by
$$U_k(\cos{\theta}) = \frac{ \sin{(k+1)\theta}}{\sin{\theta}}$$
So whenever $|x|\leq 1$,
$$|U_k(x)| \leq k+1$$
As for when $|x| > 1$, the expression
$$U_k(x) = \frac{(x+\sqrt{x^2-1})^{k+1}- (x-\sqrt{x^2-1})^{k+1}   }{2\sqrt{x^2-1}}$$
implies that
$$U_k(x) = O(x^k)$$
So if the graph is Ramanujan, then for every $|\lambda|\neq d$,
$$|(d-1)^{k/2}U_k\left( \frac{\lambda}{2 \sqrt{d-1}} \right) | \leq (k+1)(d-1)^{k/2} = O(k d^{k/2})$$
This means that the number of non-backtracking closed walks of length $k$ on $G$ is 
$$d(d-1)^{k-1} \pm O(nk d^{k/2})$$
On the other hand if the graph is not Ramanujan, then the error term could be significantly larger, and the most we can say is that the number of non-backtracking closed walks of length $k$ on $G$ is 
$$d(d-1)^{k-1} \pm O(n d^k)$$
which is not very useful since the error term is of the same order as the average!\
In fact this is the approach used in the original construction of Ramanujan graphs- where they forces the eigenvalues to be small by ensuring that the number of non-backtracking walks of a every given length is close to the average and the error term is small.\
Next consider the number of tailless non-backtracking cycles of length $k$ on $G$. This is given by the quantity
$$N_k=\begin{cases}
\sum \limits_{j=0}^{n-1} 2(d-1)^{k/2} T_k \left( \frac{\mu_j}{2\sqrt{d-1}} \right) & \text{ if k is odd}\\
\sum \limits_{j=0}^{n-1} 2(d-1)^{k/2} T_k \left( \frac{\mu_j}{2\sqrt{d-1}} \right) + d-2    & \text{ if k is even}\\
\end{cases}$$
https://arxiv.org/abs/1706.00851
The summand corresponding to the eigenvalue $d$ is
$$(d-1)^k + 1$$
when $k$ is odd, and
$$(d-1)^k + 1 + (d-2)$$
when $k$ is even. It is interesting to ask what this quantity can be interpreted as.\
For simplicity assume $G$ is a Cayley graph, so that the edges incident at each vertex can be represented by a symmetric generating set $S$ of size $d$. A tailless non-backtracking cycle of length $k$ is now a \emph{cyclically reduced word} of length $k$ over the alphabet $S$ that evaluates to $1$ in the group. It is known that the number of cyclically reduced words of length $k$ over a generating set of size $d$ is
$$(d-1)^k + d/2 + (d/2-1)(-1)^k$$
https://math.stackexchange.com/questions/825830/reduced-words-of-length-l
and so the total number of such special walks on the graph is
$$n \left( (d-1)^k + d/2 + (d/2-1)(-1)^k \right)$$
and so the expected number of such walks that are closed is
$$(d-1)^k + d/2 + (d/2-1)(-1)^k$$
which is precisely the expression we got earlier for the summand in $N_k$ corresponding to the eigenvalue $d$. \
For Chebyshev polynomials of the first kind, we know that
$$T_k(\cos{\theta})=\cos{k \theta}$$
and so for $|x| \leq 1$,
$$|T_k(x)| \leq 1$$
This gives us a strong expression for $N_k$ when $G$ is Ramanujan:
$$N_k = (d-1)^k + d/2 + (d/2-1)(-1)^k \pm O(n d^{k/2})$$
which is slightly better than the error we got for non-backtracking but possibly tailed cycles.<|endoftext|>
TITLE: Variant of Conceptual Completeness
QUESTION [23 upvotes]: Let $\mathcal{C}$ and $\mathcal{D}$ be pretopoi, and let $f: \mathcal{C} \rightarrow \mathcal{D}$ be a pretopos functor (that is, a functor which preserves finite coproducts, finite limits, and epimorphisms).
Let $M( \mathcal{C} )$ be the category of models of $\mathcal{C}$ (that is, pretopos functors from $\mathcal{C}$ to the category of sets) and define $M( \mathcal{D} )$-similarly. 
The conceptual completeness theorem of Makkai-Reyes asserts that if
$f$ induces an equivalence of categories $M(f): M( \mathcal{D} ) \rightarrow M( \mathcal{C} )$, then $f$ is itself an equivalence of categories.
I am wondering about the following more general situation. Suppose that the functor $M(f)$ is an op-fibration in sets (in other words, that the category $M( \mathcal{D} )$ can be obtained by applying the Grothendieck construction to a functor from $M( \mathcal{C} )$ to the category of sets). I would like to conclude that $\mathcal{D}$ can be obtained as a filtered colimit of pretopoi of the form $\mathcal{C}_{ / C}$ (in other words, that $\mathcal{D}$ is the pretopos associated to a Pro-object of $\mathcal{C}$).
Is something like this true, and/or available in the categorical logic literature?

REPLY [8 votes]: I believe the answer is:
"Yes" for pretopoi associated to classical first-order theories.
"No, but close" in general. (Close means: you cannot conclude that $\mathcal{D}$ is pro-etale over $\mathcal{C}$, but you can "cover"
$\mathcal{D}$ by objects which are pro-etale over both.)<|endoftext|>
TITLE: Is there an extrinsic-geometric viewpoint for connections?
QUESTION [14 upvotes]: As mentioned in the title of this question, I want to be able to move from an intrinsic viewpoint of vector bundles, to an extrinsic viewpoint. For manifolds, this would be described via the Nash embedding theorem:
$$\text{n-dimensional Riemannian manifold $(M,g)$} \to \text{submanifold of $\mathbb{R}^{2n}$ with canonical metric}.$$
This is incredibly useful as geometric intuition. Using the previous paragraph as context, is there an analogous "Nash embedding theorem" for vector bundles? i.e., can we describe an association
$$ \text{rank-$r$ vector bundle $(E,\nabla)$}\to \text{subbundle of $(\underline{\mathbb{R}^{2r}})$ with canonical connection}?$$
As for the definition of canonical, note that every vector bundle admits a realization as a subbundle of a trivial bundle of sufficiently-high rank. The trivial bundle has a canonical connection, and this connection induces a unique connection on the subbundle (just examine the linear algebra of horizontal distributions: a splitting of a vector space induces a splitting of a subspace). 
Note, the resulting connection of the subbundle would have to be induced by the parent bundle, analogous to the case of the Nash embedding. The extrinsic viewpoint would allow one to visualize the "Force = curvature" principle in a more concrete setting: hopefully we can make precise what it means to view the evolving connection-form of electromagnetism as the evolving geometry of some vector subbundle.

REPLY [20 votes]: The   counterpart for connections of the Nash embedding is an easier result called the Narasimhan-Ramanan theorem,  

M.S. Narasimhan, S. Ramanan: Existence of universal connections,
  Amer. J. Math., 83(1961), 563-572.

It essentially states that any  connection on a vector bundle $E$ is gauge equivalent to a  connection induced  by the trivial connection on a trivial vector bundle containing $E$ as a subbundle. 
As for the statement Force=Curvature I suggest you have a look at Th. Fraenkel's book The Geometry of Physics.<|endoftext|>
TITLE: Unusual digit sets that allow finite expansions for all (positive and negative) integers
QUESTION [12 upvotes]: Informal introduction
(If you don't like informal introductions, please skip to 'Mathematical formulation')
Whenever our 'decimal positional system' for writing numbers comes up in conversation, mathematicians are quick to denounce the prominent role played by the number 10 as a completely arbitrary choice, a relict from an ancient time when human biology played a more significant part in our abilities to do mathematics. By contrast, the use of the numbers $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ (or more generally $\{0, \ldots b-1\}$ in base $b$) as digits in our system is rarely questioned. This seems unsurprising: using digits all of which are are at least 2 (say) would give us a hard time writing down the number 1. But - newsflash! - in the current system it is equally impossible to write down all integers, at least without retorting to introducing an auxiliary non-digit symbol like '$-$'. By contrast, if we were to use digits (symbols) representing the numbers $\{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$ (perhaps with one of the two boundary cases omitted) we could easily write down any integer we want without introducing any further non-digit symbols. (E.g. 17 would be written $2[-3]$.) Now the point of this question is not to convince anybody to switch to this system. (Although I can't resist pointing out that when writing the negative digits as the mirror image of their positive counterparts, doing long multiplication manually in this system is remarkably easy.) Instead the point is to draw attention to some highly non-obvious digit sets that also have the power of representing any integer in the positional system such as (still for base 10) the set $\{-404,97,-482,49,0,471,-238,-797,64,605\}$ . The question is whether these digit sets can somehow be classified (for general base $b$ of course; I'm still a mathematician).
Mathematical Formulation
Let $b \geq 2$ be an integer (the 'base') and let $D \subset \mathbb{Z}$ (the 'digit set') be a finite set satisfying

$D$ contains exactly one element from every congruence class modulo $b$
The congruence class of $0$ modulo $b$ is represented in $D$ by the element $0$. (Shorter: $0 \in D$.)

Given $b$ and $D$, define a discrete dynamical system $T_D: \mathbb{Z} \to \mathbb{Z}$ by
$$T_D(a) = \frac{a - d_a}{b}$$
where $d_a \in D$ is the unique element of $D$ congruent to $a$ ($\mod b$).
Obviously $T_D(0) = 0$. We say that the digit set $D$ is complete if for every $a \in \mathbb{Z}$ there exist a $n \in \mathbb{N}$ such that $T_D^n(a) = 0$. (So repeated application of $T_D$ will bring every number $a$ down to zero.)
Question: Can we, for a given $b$, characterize all the complete digit sets $D$ among all sets satisfying the criteria?
(At the risk of stating the obvious: the relation between this formulation and the informal one above is that the successive numbers $d_a, d_{T(a)}, d_{T_D(T_D(a))}, \ldots$ appearing when repeatedly applying $T_D$, form, when read from right to left, the $b$-al expansion of the number $a$. Of course this expansion is only readable when it is finite, i.e. ends in an infinite string of zeroes (still reading from right to left). Thus the complete digit sets are the digit sets for which every integer has a finite $b$-expansion. (Ok, strictly speaking infinite expansions are also readable in the case that $b$ is a prime and we can interpret the sequence as a series convergent in the $b$-adic topology. But in that case I think that the set of numbers with finite expansions cut out by the digit set $D$ is an interesting subset of $\mathbb{Z}_b$ and a fortiori the question for which digit sets this interesting subset turns out to be the most interesting subset of all (i.e. all of $\mathbb{Z}$) is interesting in its own right.))
A necessary but not sufficient condition
An obvious sufficient condition for $D$ to be not complete is when all elements of $D$ are divisible by some prime $p$ (necessarily not dividing $b$) or when all elements of $D$ are either non-positive or non-negative. (Side question: is there some 'higher viewpoint' from which I can see these conditions as the same, e.g. a way in which the sets of positive and negative numbers are just 'congruence classes modulo the archimedian prime' or something?) However, avoiding these constellations (as is easily done to require $\{-1, 0, 1\}$ to be a subset of $D$) is necessary but not sufficient for completeness as illustrated by the digit set $D = \{-57, -1, 0, 1, 12\}$ for $b = 5$ which gives the cycle $(-2, 11, 2)$ under $T_D$ (thus preventing at least these numbers to ever reach $0$).
An easy but still surprising observation
When a given digit set $D$ is not complete, this can be seen in finite time by applying $T_D$ to a finite set of numbers. 
This observation allowed me to obtain the numerical results below. The reason is fairly simple. If $D$ is not complete and the dynamical system $T_D$ contains a cycle (including fixed points and infinite cycles) then there must be an integer $a \neq 0$ such that $|T(a)| \geq |a|$. If $d \in D$ is such that $a \equiv d$ then this can only happen if
$$a = d - cb \textrm{ where } d/(b+1) \leq c \leq c/(b-1) \textrm{ if } d >0$$
or
$$a = d - cb \textrm{ where }  d/(b-1) \leq c \leq d/(b+1) \textrm{ if } d <0$$
This for instance immediately shows that the digit set $\{[-b/2], \ldots, [b/2]\}$ is complete as it leaves no $a$ satisfying the criteria. (I write [] for floor since apparently \floor is not recognized.)
However, even if there are fairly many $a$ satisfying $|T_D(a)| \geq |a|$ it is still possible that all of them end up at 0 under repeated application of $T_D$ as is illustrated by the digit set $D = \{-605, -233, -309, 0, -671, 58, 31\}$ for $b = 7$. This digit set is also fairly illustrative for how little I understand about this problem.
Numerical results
Complete digit sets are rare. I wrote a program to generate random digit sets as follows: start with the set $\{[-b/2], \ldots, [b/2]\}$ and then add to each non-zero element a multiple of $b$ chosen independently uniformly at random from the interval $[-100b, 100b]$. Next the program checks whether the random digit set is complete by the above method. 
I found that for $b = 5$ in 1200 random digit sets only 4 were complete:
$$\{-262,-311,0,-134,77\}$$
$$\{-167,-71,0,21,-123\}$$
$$\{198,-346,0,-489,87\}$$
$$\{3,-21,0,381,-13\}$$
From 1200 random digit sets for $b = 7$ only 8 were complete:
$$\{382,376,-1,0,197,23,-256\}$$
$$\{242,-79,-57,0,-13,100,689\}$$
$$\{403,-79,377,0,533,142,-39\}$$
$$\{4,-597,34,0,463,-446,73\}$$
$$\{-3,-415,-526,0,449,37,675\}$$
$$\{4,593,34,0,-454,-201,-109\}$$
$$\{697,-548,13,0,-531,205,458\}$$
$$\{-605, -233, -309, 0, -671, 58, 31\}$$
From 600 random digit sets for $b = 10$ only 3 were complete:
$$\{46,77,298,339,0,-489,-748,-347,64,615\}$$
$$\{-404,97,-482,49,0,471,-238,-797,64,605\}$$
$$\{616,-33,388,-701,0,321,92,-367,-526,155\}$$
When we force our digit sets to contain the elements $1$ and $-1$ in order to satisfy the condition above, numbers increase dramatically:
200 complete digit sets in 1200 random digit sets containing $1$ and $-1$ for $b = 5$; 134 complete digit sets in 1200 random digit sets of this form for $b = 7$ and 112 out of 1200 for $b = 10$.
On the other hand demanding that the digit set is symmetric in the sense that for every $d \in D$ also $-d \in D$, seems to bring the number of complete digit sets down. Quite to my surprise I found 0 (!) complete digit sets among 1200 random symmetric digit sets containing 1 and -1 for b = 5. This does not mean it is impossible: the 'small' examples $\{-7, -1, 0, 1, 7\}, \{3, -1, 0, 1, -3\}, \{-17, -1, 0, 1, 17\}, \{23, -1, 0, 1, -23\}$ are complete. Dropping the requirement that 1 and -1 are part of the digitsets revealed one more symmetric digit set in 300 random symmetric digit sets for $b = 5$: $\{-17,-81,0,81,17\}$ so also with relatively small numbers.
Staring at these digit sets doesn't reveal very much about what makes them special. They look just as random as the others. So hence my question: do the complete digit sets share any other property by which they can be recognized, classified or constructed?

REPLY [5 votes]: This is not an answer. One way to succinctly describe both the existence and uniqueness of base-$b$ representations is as the generating function identity
$$\frac{1}{1 - x} = 1 + x + x^2 + \cdots = \prod_{i=0}^{\infty} \left( 1 + x^{b^i} + x^{2 b^i} + \dots + x^{(b-1) b^i} \right)$$
And one reason this is particularly nice is that it's easy to prove by telescoping, using the identity
$$1 + x^{b^i} + \dots + x^{(b-1) b^i} = \frac{1 - x^{b^{i+1}}}{1 - x^{b^i}}.$$
Partially telescoping gives a refinement of the existence and uniqueness of base-$b$ representations, namely that base-$b$ representations of integers between $0$ and $b^n - 1$ exist, are unique, and involve $n$ digits. 
If you want existence and uniqueness for both positive and negative integers then you are now trying to prove an identity of the form
$$\cdots x^{-1} + 1 + x^1 + \cdots = \prod_{i=0}^{\infty} \left( 1 + x^{d_1 b^i} + x^{d_2 b^i} + \dots + x^{d_{b-1} b^i} \right).$$
We need to be a bit careful about the meaning of this identity. Unlike formal power series, doubly-infinite formal power series do not form a ring. However, they do form a module over Laurent polynomials, and the RHS is a product of Laurent polynomials, so it makes sense to ask whether it converges to some doubly-infinite series with respect to the obvious topology.
Let $f(x) = 1 + x^{d_1} + \dots + x^{d_{b-1}}$ and let $\delta(x - 1)$ be the LHS above. The notation is justified due to the identity $\delta(x - 1) g(x) = \delta(x - 1) g(1)$ for any Laurent polynomial $g$. Then the identity we want has the form
$$\delta(x - 1) = \prod_{i=0}^{\infty} f(x^{b^i}).$$
It's now tempting to proceed as follows. Substituting in $x = x^b$ gives
$$\delta(x^b - 1) = \prod_{i=1}^{\infty} f(x^{b^i})$$
and hence
$$\delta(x - 1) = \delta(x^b - 1) f(x).$$
It's tempting to argue that this identity is equivalent to the desired identity, by repeatedly substituting $x = x^b$. But in fact that's not true; this identity only expresses the necessary but not sufficient condition that the $d_i$ should contain exactly one representative of each equivalence class $\bmod b$ (counting $d_0 = 0$). There's an analogous argument for formal power series which works because multiplication of formal power series is continuous in the $x$-adic topology, but the analogous statement for the action of Laurent polynomials on doubly infinite series is false. 
So prospects for a clean telescope-type argument seem poor here. In general,  the fact that we can't divide (let alone multiply) by doubly-infinite series suggests that there won't be anything extremely clean to say.<|endoftext|>
TITLE: Algorithmic complexity of formal proof verification?
QUESTION [14 upvotes]: In this question, suppose $S$ is some popular real-world automated proof system that is stronger than or equivalent to Peano Arithmetic.  I would be happy with a positive answer to the following for any such $S$, so please feel free to cherry-pick $S$ to make that easier:
Can the validity of an $S$-proof be verified in time polynomial in the
string-length of the proof?
I've been able to find some work assessing progress in formal proof verification, such as 
http://www.ams.org/journals/notices/200811/tx081101408p.pdf
but not much on the algorithmic complexity of verifying the outputs of various approaches, so I figure I must be looking in the wrong places / using the wrong search terms.  Please, help point me in the right direction!

REPLY [7 votes]: No, for languages based on Martin-Löf Type Theory
In proof systems based on Martin-Löf Type Theory, including Coq and Agda, proof-checking can involve evaluating arbitrarily complicated (proven-terminating) programs.
As a simple example, we can define a function is_positive : ℕ → Prop that evaluates to True if its argument is positive, and evaluates to False otherwise.  The size of a proof of is_positive is constant (it's just a proof of True when is_positive is given an argument that evaluates to a numeral).  However, it's relatively easy to define an exponentiation function that makes checking a proof of is_positive$2^n$ take time exponential in $n$.  Here is the Coq code:
(** Define a version of [+] which is recursive on the right argument. *)
Fixpoint plusr (n m : nat) {struct m} : nat :=
  match m with
    | 0 => n
    | S m' => S (plusr n m')
  end.
(** Define a version of [*] which is recursive on the right argument. *)
Fixpoint multr (n m : nat) {struct m} : nat :=
  match m with
    | 0 => 0
    | S m' => plusr n (multr n m')
  end.
Fixpoint pow (base exponent : nat) {struct exponent} : nat :=
  match exponent with
    | 0 => 1
    | S e' => multr base (pow base e')
  end.
(** Test [pow] *)
Compute pow 1 2. (* 1 *)
Compute pow 2 3. (* 8 *)
(** Return [True] if a number is [S _], [False] if it is [0] *)
Definition is_positive (n : nat) : Prop :=
  match n with
    | 0 => False
    | S _  => True
  end.
Time Check I : is_positive (pow 2 9). (* 0.09375 *)
Time Check I : is_positive (pow 2 10). (* 0.40625 *)
Time Check I : is_positive (pow 2 11). (* 1.875 *)

Since numbers are stored in unary, by default, in Coq, the proof term I : is_positive (pow 2 n) has $\mathcal O(n)$ nodes.
You could, hypothetically, output evidence of well-typedness.  In the degenerate case, where your well-typedness evidence is just a trace of the execution of the typechecker, you get linear time checking in the length of trace (assuming your trace-encoding isn't eliding expensive details).
I'm uncertain about languages like HOL, which are not based on dependent type theory.  It's plausible that there are some systems where proof-checking is extremely simple, and can't involve any computation.  I would look at Metamath as a likely candidate, even though it's closer to machine-checked proof than automated theorem proving (and I don't know of anyone using it for things outside of pure math).

REPLY [3 votes]: It of course depends on the proof system, but it's polynomial (actually basically linear) in any reasonable one.  The catch is that (by incompleteness) there's no effective bound on how long the proof might have to be based on the statement of the theorem.
Suppose you've got a proof system where verifying the proof takes triple-exponential time (or whatever) using some Turing machine T.  Then imagine running T on the proof, recording all the execution steps.  The sequence of recorded steps is in a sense a new proof, that can be verified in essentially linear time, though the new proof itself is much longer than the old proof.<|endoftext|>
TITLE: Division by $n$ in elliptic curves
QUESTION [5 upvotes]: Let $E/\mathbb F_{p^m}$ be an arbitrary elliptic curve over the Galois field $\mathbb F_{p^m}$, and let $$[n]^{-1}(P)\cap E(\mathbb F_{p^m})=\{Q\in E(\mathbb F_{p^m})\mid nQ=P\}.$$ Also let $N=\#E(\mathbb F_{p^m})$. Is the following claim true?
If $\gcd(n,N)=1$, then the only point in $[n]^{-1}(P)\cap E(\mathbb F_{p^m})$ is $(n^{-1} \bmod N)P$.
If this claim is evident, why are powerful programs such as the MAGMA  Computational Algebra System unable to compute this point in acceptable time?

REPLY [5 votes]: Disclaimer: I am not speaking for the Magma group.
Even though Magma tries to provide the best (i.e., most efficient)
algorithms, it is very hard to make sure that all possible cases are
taken care of. From experiments with Magma's DivisionPoint function,
it looks like the shortcut you propose is not implemented.
I would recommend that you
contact the Magma 
group and suggest to include the shortcut in the implementation.
Including a link to this MO question in your message might also be helpful.
To deal with your problem right away, write your own function, e.g.,
function myDivisionPoints(pt, n)
  N := #Parent(pt);
  g, m := XGCD(n, N);
  pt1 := m*pt;
  if g eq 1 then return [pt1]; else return DivisionPoints(pt1, g); end if;
end function;

Judging from timing "N := #E;" twice in a row, Magma caches the group order
after it was computed, so the first line in the function body will not
recompute N every time the function is called with a point on the same
curve (but just fetch the cached value).<|endoftext|>
TITLE: Eigenvalues of $X$ in the metric of $Y$
QUESTION [6 upvotes]: What does this statement describe? $X$ and $Y$ are matrices.

The eigenvalues of $X$ in the metric of $Y$.

I've not seen this language used before in this fashion and I don't really know what taking the eigenvalues of a matrix in the metric of another matrix means. 
Could someone decipher this for me?

REPLY [11 votes]: It means the $\lambda$ such that $X - \lambda Y$ is not invertible. The usual eigenvalues are those where $Y$ is the identity matrix.<|endoftext|>
TITLE: regular polygon question
QUESTION [14 upvotes]: Let $a_1,a_2,\ldots,a_n$ be distinct points on the complex plane $\mathbb{C}$ and $L$ be a circle in $\mathbb{C}$ such that 
$$f(z):=\sum_{i=1}^n|z-a_i|^{2n-2}$$
is constant on $L.$ Could somebody help me to show that $\{a_1,a_2,\ldots,a_n\}$ should be the vertices of a regular polygon, inscribed in a circle concentric to $L?$ I believe that this is true. However, I do not know how to prove it. 
Thank you.
Masih

REPLY [4 votes]: Here a proof for the case $n=3$. Consider the obvious reformulation with $\mathbb{R}^2$ instead of $\mathbb{C}$. Also let $L$ be the unit circle. Then the functional is
\begin{eqnarray*}
f(z)&=&\sum_{i=1}^3 (|z-a_i|^2)^2\\
&=&\sum_{i=1}^3 (1+|a_i|^2-2\langle a_i,z\rangle)^2\\
&=& constant-4\left\langle \sum_{i=1}^3 (1+|a_i|^2)a_i,z\right\rangle+4 z^T\left(\sum_{i=1}^3a_ia_i^T\right)z.
\end{eqnarray*}
In order for the function to be constant on the unit circle we need to have
$$
\sum_{i=1}^3 (1+|a_i|^2)a_i=0 \quad \text{and} \quad \sum_{i=1}^3a_ia_i^T=kI_2, k\geq 0, I_2 \text{ the }2\times 2 \text{ identity matrix}.
$$
Let $A$ be the $2\times 3$ matrix with $a_1,a_2$ and $a_3$ as columns. It follows that the row vectors have length $\sqrt{k}$ and are orthogonal to each other. Let $B\in \mathbb{R}^{3\times 3}$ be an orthogonal matrix for which the first two rows are the normalized first two rows of $A$. It follows that the third row of $B$ is proportional to $(1+|a_i|^2)_{i=1}^n$. If follows that $|a_i|^2/k+(1+|a_i|^2)/l=1$ for all $i\in \{1,2,3\}$ for some $l>0$. Hence we have $|a_1|=|a_2|=|a_3|$. Furthermore $\langle a_1,a_2\rangle=\langle a_1,a_3\rangle=\langle a_2,a_3\rangle$ and therefore all angles between the vectors are equal.<|endoftext|>
TITLE: Numerical invariants of symmetric products of curves
QUESTION [5 upvotes]: Let $n\ge 2$ be an integer and $C$ be a smooth projective curve of genus $g>n$. The $n$-the symmetric product $C(n)$ is a smooth variety of general type.
If $n=2$ then $C(2)$ is a minimal surface ad it is not hard to compute its numerical invariants. I would like to know whether $C(n)$ is minimal also for $n>2$ and what is the volume of the canonical bundle. 
There is famous article by MacDonald
http://www.sciencedirect.com/science/article/pii/0040938362900198
about symmetric products. In it I found the formula for $\chi(K_{C(n)})$, but I was not able to find the answer to my question above, although it may be  there.

REPLY [4 votes]: Claim:  If $C$ is a curve of genus $g \geq 2,$ the canonical class of $C^{(d)}$ is nef and big if and only if $1 \leq d \leq g-1.$
Proof:  Let $\theta \in {\rm NS}(C^{(d)})$ be the class of the pullback of the theta-divisor of ${\rm Jac}(C)$ via the Abel map, and let $x \in {\rm NS}(C^{(d)})$ be the class of the divisor $\{D + p_{0} : D \in C^{(d-1)}\},$ where $p_0 \in C$ is a given point.  (Moving $p_{0}$ keeps us in the same algebraic equivalence class, so $x$ is independent of $p_{0}.$)  These classes are linearly independent, by the formulas for $\theta^{j} \cdot x^{d-j}$ given in ACGH.  Since the theta-divisor on ${\rm Jac}(C)$ is ample, $\theta$ is nef.    
As abx points out, $C^{(d)}$ is a projective bundle over ${\rm Jac}(C)$ for $d \geq 2g-1;$ it is the subspace projectivization of a Picard bundle.  In this case $x$ is the class of the relative $\mathcal{O}(1).$  The Chern class formulas for the Picard bundle given in ACGH, together with repeated application of the adjunction formula to the embedding $C^{(d)} \hookrightarrow C^{(d+1)}$ induced by adding a point, imply that 
$$K_{C^{(d)}} = \theta + (g-d-1)x$$
A similar argument using the ampleness of the dual of the Picard bundle implies that the divisor class $x$ is ample.
The nefness of $\theta$ and the ampleness of $x$ imply that $d \leq g-1$ is sufficient for $K_{C^{(d)}}$ to be nef and big.  For necessity, note that when $d \geq g,$ the class $\theta$ spans a boundary of the nef cone since the Abel map contracts a positive-dimensional locus; if $K_{C^{(d)}}$ is nef and big, then $\theta = K_{C^{(d)}}-(g-d-1)x$ is ample, which is absurd.  This concludes the proof.
NOTE:  The previous argument shows that $K_{C^{(d)}}$ is ample if and only if $1 \leq d \leq g-2.$<|endoftext|>
TITLE: Show that the Laplacian operator on the Heisenberg group is negative
QUESTION [5 upvotes]: The Heisenberg group $H^3$ is the set $\mathbb C\times \mathbb R$ endowed with the group law
$$ (z,t)\cdot(w,s) =\left (z+w, \,t+s+\tfrac{1}{2}\Im m(z \bar{w})\right). $$
For $z=x+ i  y \in \mathbb C$ and $t\in \mathbb R$, the Laplacian operator on the Heisenberg group $H^3$  is given by 
$$ \Delta= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}  + (1+x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y}  -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} .$$ 
I want to show that the Laplacian operator $ \Delta$ is negative on $\left( L^{2}(\mathbb R^3); \left<.,.\right>_{2}\right)$, where 
$$ \left<f,g\right>_{2} =\int_{\mathbb R^3} f(x) \, \bar{g}(x) \, dx .$$
that's what I did: we have 
$\begin{align}
\Delta &= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}  + (1+x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y}  -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t}  \\
&= \Delta_{\mathbb R^2}  + (1+x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y}  -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} 
\end{align}$
we know that $\Delta_{\mathbb R^2} $ and $ (1+x^2+y^2 ) \frac{\partial^2}{\partial t^2} $ are negative operator.
My question, how to show that $A:= (x\frac{\partial}{\partial y}  -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t}$ is a negative operator. For it, let $u\in C^{\infty}_{0}(\mathbb R^3) \subset L^{2}(\mathbb R^3) $, we have 
\begin{align}
\left<A u,u\right>_{2} &= \int_{\mathbb R^3} \left[(x\frac{\partial}{\partial y}  -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t}  u(x,y,t) \right]\, u(x,y,t) \, dxdydt
\end{align}
I tried the integration by parts, but I can not show that $\left<A u,u\right>_{2}  \leq 0 $.
Thanks you in advance

REPLY [2 votes]: Using the following 
Proposition: If we consider a left invariant vector field $X$ as a
first-order differential operator on a group $G$, then it is skew-symmetric with respect to the inner product defined by a Haar measure.
So, the the left invariant vector fields $X, Y$ and $T$ are skew-symmetrics with respect to the inner product $\left<.,.\right>_{2}  \leq 0$,then $\left<X^2 u,u\right>_{2} = - \left<X u,Xu\right>_{2} \leq 0$, and hence $\Delta= X^2 + Y^2+T^2$ is a negative operator on $(L^{2}(\mathbb R^3),\left<.,.\right>_{2})$<|endoftext|>
TITLE: Is there $t\in\operatorname{Gal}(\overline{K}/K)$ s.t. $\operatorname{rank}_{\mathbf{Z}_p}((t-1)E_{p^\infty}(\overline{K}))=1$?
QUESTION [5 upvotes]: Let $E$ be an elliptic curve defined over $\mathbb{Q}$, let $$K:=\varinjlim_{k\in\mathbb{Q}[\mu_{p^\infty}]} \mathbb{Q}\left[\mu_{p^\infty},k^{1/p^\infty}\right]$$
and $G:=\operatorname{Gal}(\overline{K}/K)$. Suppose that $E_p(K)=0$.
Question: Is there always a $\tau\in G$ so that
$$\operatorname{rank}_{\mathbf{Z}_p}((t-1)E_{p^\infty}(\overline{K}))=1$$

REPLY [4 votes]: I think the answer is "yes" in the case that $E$ doesn't have complex multiplication.  In that case, Serre's Open Image Theorem says that the image of Galois in open in $\mathrm{GL}(T_{p}(E))$, and therefore, the image of $\mathrm{Gal}(\bar{\mathbb{Q}} / \mathbb{Q}(\mu_{p^\infty}))$ is open in $\mathrm{SL}(T_{p}(E))$ -- call it $H$.  Now $K$ is abelian over $\mathbb{Q}(\mu_{p^\infty})$, so the image of $\mathrm{Gal}(\bar{K} / K)$ is in turn a (normal) subgroup of $H$ with abelian quotient.  But since the Lie algebra of $H$ is all of $\mathfrak{sl}_{2}(\mathbb{Q}_{p})$, which is semisimple, any Lie subgroup of $H$ with abelian quotient must have the same Lie algebra.  So the image of $\mathrm{Gal}(\bar{K} / K)$ is open in $H$ and thus open (and of finite index) in $\mathrm{SL}(T_{p}(E))$. Thus, the image of $\mathrm{Gal}(\bar{K} / K)$ contains a (nonzero) power of a transvection, i.e. a matrix of the form $\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$ for some $a \neq 0$ with respect to a suitable basis.
In the case that $E$ has complex multiplication, I suspect I have a counterexample.  Take $E: y^{2} = x^{3} - 2$ and $p = 2$.  This has complex multiplication by $\mathbb{Q}(\zeta_{3})$, where $\zeta_{3}$ a primitive cube root of unity, and its order-$2$ points are the points $(\zeta_{3}^{i}2^{1/3}, 0)$ for $i = 1, 2, 3$.  Clearly the image of $\mathrm{Gal}(\bar{K} / K)$ is isomorphic to $\mathbb{Z} / 3\mathbb{Z}$ (since $\zeta_{3} \in K$ but $2^{1/3} \notin K$), so it doesn't fix any of the points in $E[2]$.  However, I think it shouldn't be too hard to show that the image of Galois can't contain a power of a transvection in this case.  (In fact, it should be contained in the image of the endomorphism ring, which is an order in $\mathbb{Z}[\zeta_{3}]$, and note that the endomorphism corresponding to $\zeta_{3}$ must act like $\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$ or $\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ modulo $2$.)<|endoftext|>
TITLE: Generalizing Dedekind's Factorization Theorem
QUESTION [5 upvotes]: A classical theorem due to Dedekind states the following: 

Let $O_{K}$ be the ring of integers of a number field $K$, and assume
  $K$ is generated by adjoining the algebraic integer $\alpha$ to
  $\mathbb{Q}$. Let $f(x) \in \mathbb{Z}[x]$ be the minimal polynomial
  of $\alpha$.
Then given a prime ideal $\mathfrak{p} \subseteq \mathbb{Z}$, one can deduce the
  factorization of the ideal $\mathfrak{p}O_{K}$ from the factorization of the polynomial $f(x)$ in
  $\mathbb{Z}[x]/\mathfrak{p}$ (under the assumption $\mathfrak{p} \nmid [O_K : \mathbb{Z}[\alpha]]$).

See Theorem 1 in this note by Keith Conrad for full details.
My questions are about how generalizable this theorem is. References would be highly appreciated.

Does it hold when I replace:

$\mathbb{Z}$ with $R:=\mathbb{F}_q[T]$,
$F:=\mathbb{Q}$ with the field of fractions of $R$, 
$K$ with a finite extension of $F$, 
$O_K$ with the integral closure of $R$ in $K$, 
and $\alpha$ with an element from the integral closure of $R$ in the algebraic closure of $F$? 


2
. What is the greatest generality in which the above theorem holds? For instance, is it true when $\mathbb{Z}$ is replaced with an arbitrary Dedekind domain $R$?

[EDIT]
3. It seems that the answer to the first two questions is positive when $K$ is a finite separable extension of $F$. What happens when $K/F$ is not separable?

REPLY [5 votes]: I have found two different references which answer both my questions in the affirmative. The first is Zariski and Samuel's 'Commutative Algebra I' (Chapter V, p. 317, Thm. 34) and the other is Neukirch's 'Algebraic Number Theory' (Chapter I, p. 47-48, Prop. 8.3). It seems that the only remaining interesting question is: What exactly happens when the field extension is not separable?
Zariski and Samuel describe a theorem of Kummer which applies to integrally closed domains (more general than Dedekind domains), but restricts to the case where the larger ring $R'$ is of the form $R[y]$, while in Dedekind's theorem, $R[y]$ might be a finite-index subgroup of $R'$. In Kummer's theorem, prime ideals are replaced with maximal ideals (in Dedekind domains, those two notions essentially coincide).
Neukirch describes a theorem which applies to Dedekind domains, and gives exactly Dedekind's Theorem when one works with rings of integers.
Dedekind's theorem is a slight strengthening of Kummer's theorem, specialized to the case of number fields. It doesn't follow directly from Kummer's theorem, but it does follow from the theorem described by Neukirch.
Kummer's theorem is the following:

Let $R$ be an integrally closed domain, $K$ its quotient field, $K'$ a
  finite algebraic extension of $K$, $R'$ the integral closure of $R$ in
  $K'$. We suppose that there exists an element $y$ of $R'$ such that
  $R'=R+Ry+\cdots + Ry^{n-1}$ ($n=[K':K]$) ($y$ is then a primitive
  element of $K$' over $K$). Let $F(Y)$ be the minimal polynomial of $y$
  over $K$. ($F(Y)$ necessarily has its coefficients in $R$.) Let
  $\mathfrak{p}$ be a maximal ideal in $R$; for every polynomial $G(X)$
  over $R$, we denote by $\overline{G}(X)$ the polynomial over
  $R/\mathfrak{p}$ whose coefficients are the $\mathfrak{p}$-residues of
  the corresponding coefficients of $G$. Let
  $\overline{F}(X)=\prod_{i=1}^{g} (f_i(X))^{e(i)}$ be the factorization
  of $\overline{F}(X)$ into distinct irreducible factors $f_i(X)$ over
  $R/\mathfrak{p}$; for $i=1,\cdots,g$ we denote by $F_i(X)$ a
  polynomial over $R$ such that $\overline{F_i}(X) = f_i(X)$. Then the
  ring $R'$ has exactly $g$ maximal ideals $\mathfrak{P}_i$ which lie
  over $\mathfrak{p}$, and we have $$\mathfrak{P}_i = R'\mathfrak{p}+R'F_i(y).$$ 
  Furthermore we have  $$R'\mathfrak{p} =\mathfrak{D}_1 \cap \mathfrak{D}_2 \cdots \cap \mathfrak{D}_g = \mathfrak{D}_1 \cdot\mathfrak{D}_2 \cdots \cdot \mathfrak{D}_g,$$ where
  $\mathfrak{D}_i = R'\mathfrak{p} + R'(F_i(y))^{e(i)}$.

I have found Neukirch's theorem here. $\mathcal{o}$ stands for a Dedekind domain, $K$ for its fraction field, $\mathcal{O}$ for the integral closure of $\mathcal{o}$ in $L$, a finite separable extension of $K$:<|endoftext|>
TITLE: Can an ultrapower be undone by forcing?
QUESTION [8 upvotes]: I am not 100% certain this question is appropriate for MO; I may just be missing something obvious. Also, I vaguely recall a similar question being asked here a while ago, but I can't find it; if it turns out this is a duplicate, I'll delete this question. Anyways, apologies in advance if this is too easy or is a duplicate. Note that e.g. Can measures be added by forcing? prevents the obvious nuke from working.
Also, the "descriptive-set-theory" tag is purely a guess on my part, based on the surprising ubiquity of descriptive set theory in similar-sounding questions.

Suppose I have a transitive model $M$ of $ZFC$, and - in $M$ - $U$ is a measure on $\kappa$. Then the transitive collapse of the ultrapower of $M$ along $U$ is an inner model, $N\subset M$.
My question is:

Can we ever have $M$ be a generic extension of $N$ (either by set or class forcing in $M$)?

As mentioned above, I am almost certain the answer is "no", even if $M$ has loads of large cardinals, but I don't see how to prove this.
EDIT: Douglas Ulrich answered the question for set forcing; I've asked the class forcing version as a separate question.

A small observation:
Say (inside a model $W$) a cardinal $\mu$ is

potentially measurable if $\mu$ is measurable in some forcing extension; and
reversibly measurable if $\mu$ is measurable, and $W$ is a forcing extension of the transitive collapse of the ultrapower of $W$ by a measure on $\mu$ (that is, if $\mu$ is as above).

Then suppose we had such an $M, N, U, \kappa$, with $j$ the elementary embedding. Then $N$ satisfies "There is a potential measurable below $j(\kappa)$," so - pulling back along $j$ - $M$ satisfies "There is a potential measurable below $\kappa$." This shows that - in $M$ - the least potentially measurable is strictly less than the least reversibly measurable (otherwise we get a descending chain of measurable cardinals).
Now, it feels plausible to me that there's a clever trick that can be done here to outright build a descending sequence of reversibly measurables from a single reversibly measurable; but I don't see it.

REPLY [13 votes]: For set-forcing, the answer is no, see the following article
Joel David Hamkins, Greg Kirmayer, and Norman Lewis Perlmutter, Generalizations of the Kunen inconsistency, Ann. Pure Appl. Logic 163 (2012), no. 12, 1872--1890. (see also arxiv.org/abs/1106.1951 and Hamkins's blog post)
The "Kunen Inconsistency" is the theorem that there is no nontrivial elementary embedding $j: V \to V$. The above article shows (among several other things) that if $V[G]$ is any set-forcing extension of $V$ then there is no nontrivial elementary embedding $j: V[G] \to V$. So even if you replaced ultrapowers by extenders, for example, the answer remains no.<|endoftext|>
TITLE: List of proofs where existence through probabilistic method has not been constructivised
QUESTION [33 upvotes]: The probabilistic method as first pioneered by Erdős (although others have used this before) shows the existence of a certain object. What are some of the most important objects for which we can show the existence by such a method but constructive progress (construction in polynomial time) has been very hard to come by?
An important example is the existence of good codes in coding theory. Using algebraic geometry we can show such objects not only exist but can be constructed in polynomial time.

I am thinking that in addition to graph theoretic and direct combinatorial examples, there should be plenty of natural but very difficult examples from number theory and geometry as well.

REPLY [5 votes]: There are many properties of real numbers which hold a.e. with respect to Lebesgue measure, but are hard to check for specific numbers. 
A quick example.
Take a random number $x$ in $(0,1)$. Then for a given $p>1$ we may take its $p$-expansion $x=\sum_{i>0} a_{i,p}(x)p^{-i}$, $0\leq a_{i,p}(x)\leq p-1$. Digits $a_{i,p} (x)$ are independent and distributed uniformly on $\{0,1,\dots,p-1\}$. Hence by law of large numbers a.e. $x$ satisfy $$\lim_{N\rightarrow \infty} \frac1N \left|i\leq N: a_{i,p}(x)=c\right|=\frac 1p$$
for any given $c\in \{0,1,\dots,p-1\}$.
But there is no specific $x$ for which this is proved to hold simultaneosly for $p=2$ and $p=3$.<|endoftext|>
TITLE: Grothendieck spectral sequence when one of the functors is contravariant
QUESTION [16 upvotes]: Let $f \colon X \rightarrow S$ be a morphism of schemes. I am interested in computing the cohomology groups of
$$
\mathbf{R}\mathscr{H}om(\mathbf{R}f_* \mathcal{O}_X, \mathcal{O}_S)
$$
in terms of $\mathrm{Ext}^p(R^qf_* \mathcal{O}_X, \mathcal{O}_S)$. Is there a spectral sequence that achieves this? If so, what are the entries of its pages, how do the differentials go, which way does the eventual filtration on the infinity page go, etc.? Any references to the literature where this is explained would also be highly appreciated.
I am familiar with the Grothendieck spectral sequence for a composition of functors, but I cannot see how that applies because $\mathbf{R}f_*$ is covariant and $\mathbf{R}\mathscr{H}om$ is contravariant (in the first variable).
As alluded to in the title, the question may be posed in much greater generality, but let me stick with the above for the sake of concreteness.
EDIT. I realized that I forgot that the usual Grothendieck spectral sequence has a condition that the first functor maps injectives to acyclics for the second functor. A condition of this sort does not seem to hold in my setting, so perhaps there is no spectral sequence of the type I want. Further comments are nevertheless appreciated!

REPLY [12 votes]: I think this case is actually not so obvious. The issue is that to derive $R\mathscr Hom$ in the first variable you would need to use a locally free resolution while $Rf_*$ being a covariant right derived functor would need an injective one. If you look at the usual proofs of the various forms of GSS, you'll see that the idea is to take a resolution of the first object in the first category, apply the first (derived) functor and then see that what you get in the second category computes the second derived functor. With this, as it is, you would need to switch resolutions and I am not sure I would want to get into that computation.
On the other hand, if your situation is such that $R^if_*\mathscr O_X=0$ for $i>d$ for some $d\in \mathbb Z$, for instance, $f$ is proper, $Y$ is (locally) Noetherian and $f$ has fibers of dimension at most $d$, then you can take your starting functor $R^df_*$ and consider $Rf_*\mathscr O_X$ as the left derived functor of that. (See Prop 7.4 on page 74 in [Hartshorne, R&D].) I would expect that if you know that

$R^if_*\mathscr F=0$ for $i>d$ for every coherent $\mathscr F$, and
$R^df_*\neq 0$,

then this should work. For simplicity let me assume that $f$ is a projective morphism of schemes of finite type over a field with $d$ equal the maximal fiber dimension of $f$. If you need it in more generality you can probably adjust the argument. (If it works!!).
Namely, fix a coherent sheaf $\mathscr F$ on $X$ and take a locally free resolution $\mathscr L^\bullet$ of $\mathscr F$. (Addendum: this should probably consist of locally free sheaves of the form $\oplus\mathscr O(-n)$ as we discussed in the comments, so the $R^df_*$ of this sequence would indeed be a locally free resolution of $R^df_*\mathscr F$ as claimed in the next sentence.) Then by cohomology and base change $R^df_*\mathscr L^i$ is locally free for all $i$ and hence $R^df_*\mathscr L^\bullet$ is a complex of locally free sheaves which is quasi-isomorphic to $\left(L(R^df_*\mathscr)\right)(F)$ (=the (total) left derived functor of $R^df_*$ applied to $\mathscr F$). Now if you just follow the usual proof, you should (check this!!!) get a spectral sequence something like 
$$
E_2^{p,q}= \mathscr Ext^p(\underbrace{L^q(R^df_*)}_{R^{d-q}f_*}\mathscr F, \mathscr G)\Rightarrow 
E_\infty^{p,q}= h^{p+q}\left( R\mathscr Hom(Rf_*\mathscr F, \mathscr G)\right)
$$
Actually I think that to do this right, here probably both $p$ and $q$ should be negative, since the locally free resolution goes that way. A simple first check is to see if you can write down the edge maps from obvious short exact sequences. For $E_2$ spectral sequences, the maps are usually compositions of two edge maps from two different short exact sequences. 
So, in terms of the original functors this is not a first quadrant spectral sequence, but one should expect that anyway.
Anyway, as it is probably clear from what I have written, I did not work out the details, so it is quite possible that there is some issue I didn't consider and also, clearly I didn't even give you an explicit answer, but perhaps you can work it out. And, please do work out the details carefully (and perhaps post it here if you have).

REPLY [4 votes]: If $I^*$ and $J^*$ are global sections of appropriate injective resolutions of $f_*{\cal O}_X$ and ${\cal O}_S$, then you have a double complex that looks in part like this:
$$\matrix{
Hom(I^q,J^p)&\rightarrow&Hom(I^{q},J^{p+1})\cr
&&\uparrow\cr
&&Hom(I^{q+1},J^{p+1})&\rightarrow&Hom(I^{q+1},J^{p+2})}$$
Your $E_2$ terms come from taking cohomology vertically and then horizontally,  so your
$E_2$ differentials should go from $E_2^{p,q}$ to $E_2^{p+2,q+1}$.<|endoftext|>
TITLE: Is $\ell_p$ $(1<p<\infty)$ finitely isometrically distortable?
QUESTION [12 upvotes]: Let $Y$ be a Banach space isomorphic to $\ell_p$, $1<p<\infty$. Is it true that any finite subset of $\ell_p$ is isometric to some finite subset of $Y$?
It seems to me that it is an interesting question. It can be regarded as a special case of a recent question Under which conditions is it possible to find points with same distances under bi-Lipschitz map (which was closed) and is related to my (unanswered) question Isometric embeddings of finite subsets of $\ell_2$ into infinite-dimensional Banach spaces
In connection with this question it is worth mentioning that there
is a theory developed by Krivine in Ann. Math. (2) 104, 1-29
(1976) (with important additions by Maurey and Pisier (Stud. Math.
58, 45-90 (1976)) and further simplifications by other authors,
see Chapter 12 in the book by Benyamini-Lindenstrauss on Geometric
Nonlinear Functional Analysis or Part II in the book
Milman-Schechtman on Asymptotic Theory) which implies that for any
$\varepsilon>0$ any finite-dimensional subspace of $\ell_p$ embeds
into any Banach space isomorphic to $\ell_p$ with distortion $\le
(1+\varepsilon)$.
Added on 4/3/2017: In a recent paper James Kilbane proved that the set of possible counterexamples (if they exist) is small in a certain sense.

REPLY [5 votes]: We (me and the author posing the question) have answered this question in the negative - for each $p \neq 2$ we have found a space $X$ such that $X$ is isomorphic to $\ell_p$ but there are 5 element subsets of $\ell_p$ that do not embed isometrically into $X$. Our paper is on arxiv here: https://arxiv.org/abs/1708.01570
The question about $\ell_2$ is still open.<|endoftext|>
TITLE: Sort-of converse of Kolmogorov zero-one theorem
QUESTION [10 upvotes]: Let $(\Omega, \mathscr F, \mathbb P)$ be a probability space. The Kolmogorov zero-one theorem states that

Suppose we have independent random variables $X_1, X_2, ...$. Then $\forall \ A \in \bigcap_n \sigma(X_n, X_{n+1}, ...)$, $P(A) = 0$ or $1$.

If we choose $X_k = 1_{A_k}$ for events $A_1, A_2, ...$, then we have:

Suppose we have independent events $A_1, A_2, ...$. Then $\forall \ A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$, $P(A) = 0$ or $1$.

Now, is this following conjecture true? If not, can it be modified slightly to be true?

Conjecture: Suppose we have events $A_1, A_2, ...$ s.t. $\forall \ A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$, $P(A) = 0$ or $1$. There exists an independent sequence of events $B_1, B_2, ...$ s.t.
$$\bigcap_n \sigma(A_n, A_{n+1}, ...) = \bigcap_n \sigma(B_n, B_{n+1}, ...) \tag{*}$$


I think there exists a function $f: \mathbb N \to \mathbb N$ s.t. $A_{f(n)}$'s are independent so we can choose $B_n = A_{f(n)}$. Is that true? Why/Why not? If not, how else can I prove or disprove the conjecture above? If it is true, I think it can be proven by modifying the proof of the Kolmogorov 0-1 Theorem (for events).

Perhaps one of these subsequences of sets is independent:
$$A_n$$
$$A_{2n}, A_{2n+1}$$
$$A_{3n}, A_{3n+1}, A_{3n+2}$$
$$\vdots$$
$$A_{mn}, A_{mn+1}, A_{mn+2}, ..., A_{mn+(m-1)}$$
$$\vdots$$
I think we have that
$$\bigcap_n \sigma(A_n, A_{n+1}, ...) = \bigcap_n \sigma(A_{mn+i}, A_{m(n+1)+i}, ...)$$
where $m \in \mathbb N$ and $i \in \{0, 1, 2, ..., m-1\}$.

Based on what @FedorPetrov pointed out, it seems like we need $f(n)$'s s.t.
$$\sigma(A_{f(n)}, A_{f(n+1)}...) \subseteq \sigma(A_n, A_{n+1}, ...) \tag{**}$$
which I guess is true if (and only if?) $f(n) \ge n$.

Other possible candidates for $f(n)$: (assume the variables are s.t. $f: \mathbb N \to \mathbb N$ is satisfied. If need be, $(**)$ or $f(n) \ge n$ too.)

$\sum_{i=0}^{m} a_i n^i$

$2^n, 3^n, ...$

$\sum_{i=1}^{m} b_i c_i^n$

$\lfloor{t^n}\rfloor, \lceil{t^n}\rceil$ (I guess $t > e^{1/e}$)

$\lfloor{\sum_{i=1}^{m} b_i c_i^n}\rfloor, \lceil{\sum_{i=1}^{m} b_i c_i^n}\rceil$

$\lfloor{\text{linear combination of trigonometric functions}}\rfloor, \lceil{\text{linear combination of trigonometric functions}}\rceil$

$\lfloor{\text{Some linear combination of the above}}\rfloor, \lceil{\text{Some linear combination of the above}}\rceil$



Assuming the conjecture is true, I guess it's not necessary to find $f(n)$ that works for all possible sequences of events $A_1, A_2, ...$ because such $f(n)$ may not even exist.

To disprove the conjecture: There's of course showing that any sequence that satisfies $(*)$ will not be independent, but I have a feeling it's more of showing that any independent sequence will never satisfy $(*)$.
Something that might help: we could show that $\forall \ A \in \bigcap_n \sigma(A_{f(n)}, A_{f(n+1)}, ...), P(A) = 0$ or $1$ and $\forall n \in \mathbb N, A_{f(n)}, A_{f(n+1)}, ...$ is not independent, but I'm not quite sure that the conjecture is disproved because we could construct some $B_n$'s that look like:

$$B_n = A_{n+1} \setminus A_n$$

$$B_n = A_{n} \setminus A_{n-1}, A_0 = \emptyset$$

$$B_n = \bigcap_m A_{mn}$$

$$B_n = \bigcup_m A_{mn}$$

$$B_{2n} = \bigcap_m A_{mn}, B_{2n+1} = \bigcup_m A_{mn}$$

$$B_n = \limsup_m A_{mn}$$

$$B_n = \liminf_m A_{mn}$$

$$B_{2n} = \limsup_m A_{mn}, B_{2n+1} = \liminf_m A_{mn}$$


Not to say of course that any of those $B_n$'s satisfy $(*)$ but that $B_n$ need not be in the form $A_{f(n)}$.

Borel-Cantelli:

If $\sum_n P(A_n) < \infty \to 0 = P(\limsup A_n) = P(\limsup A_{mn}) \ \forall m \in \mathbb N$. Hence $B_m = \limsup A_{mn}$ is independent.

If $\sum_n P(A_n) = \infty$, then maybe this extension of Borel-Cantelli? Not quite sure I understand it or how it would be helpful. I don't think we can conclude anything if we have $P(\limsup A_n)$.

Then there's the case of $\sum_n P(A_n) = \infty$ but the conditions earlier aren't satisfied.



Based on: https://math.stackexchange.com/questions/605301

REPLY [4 votes]: I don't think that there can be such a selection function $f$.
Define $A_i$ through independent events $C_i$ and $D_i$, each of which occurs with $50\%$ chance, as follows.
$A_1=D_1$ and for $i\ge 2$, if $C_i$, then $A_i=D_i$, otherwise $A_i=A_{i-1}$.
If $A \in \bigcap_n \sigma(A_n, A_{n+1}, ...)$, then $P(A) = 0$ or $1$ should hold, with a similar proof as for the Kolmogorov Zero-One Theorem.
On the other hand, no $A_i$ and $A_j$ are independent.<|endoftext|>
TITLE: What are some important but still unsolved problems in mathematical logic?
QUESTION [60 upvotes]: In the past, first-order logic and its completeness and whether arithmetic is complete was a major unsolved issues in logic . All of these problems were solved by  Godel. Later on, independence of main controversial axioms were established by forcing method. 
I wonder if there still exist some "natural" questions in mathematical logic that are still unsolved? Or is it the case that most of the major questions have been already answered? 
I'd love to know about some important, but still unsolved problems that puzzle logicians and why would the young logician\mathematician care about those? (that is, Whey they are important?)
I'm not an expert in logic (nor in any other mathematical field, I'm undergraduate) but I'm interested in logic so I would like to know about the current problems that logicians face and what are the trends of research in the discipline nowdays and what type of problems people are trying to solve.
I know that logic is a vast term which includes many sub-disciplines: model theory, proof theory, set theory, recursion theory,  higher-order logics , non-classical logics, modal logics, algebraic logic and many others. So feel free to tell us about problems form whichever topic you would love to.

REPLY [9 votes]: One of the very interesting open problems in Logic (and CS) I think:
Is the field of constructible numbers known to be decidable?<|endoftext|>
TITLE: Periodicity in iterated powers of sin, cos, exp
QUESTION [9 upvotes]: Given a complex number $z$, consider the sequence
\begin{align*}
a_0 & = 1\\
a_1 & = (cos(1))^z\\
a_n & = (cos(a_{n-1}))^z
\end{align*}
This question is about trying to understand periodicity in such
sequences.  For real $z < 2$, the sequence converges to a fixed point.
For larger real $z$, the sequence oscillates between two distinct
limit points.  This behavior can be explained by elementary dynamics,
and one can find that the transition point is around $2.188$.
I was surprised to find that the behavior for general complex $z$ is,
well, complex!  Define a function $P$ from the complex plane to
$\mathbb{N}$ as follows: 
$$P(z) =  \mathrm{number\ of\ limit\ points\ of\ the\ sequence}\ \{1, cos(1)^z, cos(cos(1)^z)^z, \ldots\}$$
Here is a picture of $P$: this is the region $[-8,8] \times [-8,8]$
and the pixel at $z$ is given color $n$ by some software that
estimates $P(z)$.

Black pixels are points where the software cannot detect periodicity.  The correspondence between colors and numbers is as follows, where 'Unk' means 'Unknown':

In particular, one can see the behavior along the positive real axis
matches the description above.  (Although there are no axes in this
picture, I have verified separately that the transition point in the
picture is correct.)  Unfortunately, I have no idea how to explain the
rest of the picture!  Although I have been careful writing the
software, I can't say with certainty that the picture is correct
anywhere other than the positive real axis.
This question is in danger of being too general, so here is one
specific question to answer:  Is the cardoid-shaped region colored 1
correct?
I would, of course, also be happy to hear any other verifications of features in this
picture, or other behavior of $P(z)$ not depicted.

Sin and exp
At the risk of going on too long, I think it's natural to also address
a similar question for the sine and exponential functions.  Here is
the picture for exp:

And the picture for sin is below:

This last one, for the sine function, is even more bewildering to
me.  There is something that looks very much like a Mandelbrot set
there.  Why?

Motivation
Two of my colleagues were discussing the behavior of cos along the
positive real axis on our department mailing list.  I was curious
about the complex behavior, but this is outside my field, so started
making pictures like the ones shown here.  I've been thinking about
these off and on for a little over a year, but not really made any progress or
had anyone give me a useful reference.  So I wanted to see what the
wider MO community has to say.
If you want some higher-resolution pictures for cos, see this G+ post:
https://plus.google.com/u/0/+NilesJohnson/posts/1JqWahhwGbh

Notes on the software
The software computes 500 iterates, and then looks at the next 30
iterates.  It returns the minimum $n$ such that there are two
successive subsequences of length $n$ whose corresponding terms are
within $\varepsilon = .001$.  If no such $n$ is found, it computes 500
more iterates and tries again.  This is repeated 6 times.  If no such
$n$ is found, the pixel is colored black.
Decreasing $\varepsilon$ by a factor of $10^4$ or so takes longer but does
not result in a substantially different picture.

REPLY [4 votes]: In spite of the potential issues arising from the fact that this function is not entire, there is a standard way to describe the components that you see in these types of pictures. Suppose that we are studying the iteration of a function $f_p(z)$ where $z$ is a complex variable and $p$ is a complex parameter. The cardioid-like figure that you see arises as the boundary of the set of $p$ parameters such that the corresponding function $f_p$ has an attractive fixed point. Symbolically:
\begin{align}
  f_p(z) &= z \\
  \left|\ f_p'(z)\right| &< 1
\end{align}
The equation $f_p(z)=z$ says that $z$ is a fixed point and the inequality $\left|\ f_p'(z)\right|<1$ says that the fixed point is attractive. On the boundary, we expect that $\left|\ f_p'(z)\right|=1$. Thus, the boundary may be described as 
\begin{align}
  f_p(z) &= z \\
  \ f_p'(z) &= e^{it},
\end{align}
for some $t\in[0,2\pi)$. If we can solve the equations for $z$ and (more importantly) $p$ in terms of $t$, we have an explicit description of the boundary. In the case of the Mandelbrot set, $f_p(z)=z^2+p$ and this pair of equations can be solved in closed form so we have a proof that the main cardioid is, in fact, a cardioid. This is a bit much to expect in the current case. Nonetheless, we can make some progress and finish it off numerically. To do so, write $f_p(z)=\cos^p(z)$. The equations of interest are then
\begin{align}
  \cos^p(z) &= z \\
  -p \sin (z) \cos ^{p-1}(z) &= e^{i t}
\end{align}
This second equation can be solved for $p$ in terms of $z$ and $t$ using Lambert's W function:
$$p(z,t) = \frac{W\left(-e^{i t} \cot (z) \log (\cos (z))\right)}{\log
   (\cos (z))}.$$
Note: there is certainly a potential branch cut issue here! Nonetheless, plugging that back into the first equation we get
$$\cos^{p(z,t)}(z) = z.$$
For a given $t$, this equation can be solved numerically to determine a specific $z$ value that is a neutral fixed point of $\cos^{p(z,t)}$. If we then plug that $z$ and $t$ value into $p(z,t)$ we get a point on the boundary of your cardioid-like domain.
Again, this procedure is certainly fraught with branch cut and numerical issues. proceeding undaunted, I implemented this in Mathematica together with the iteration scheme itself and came up with the following:<|endoftext|>
TITLE: Asymptotic growth rate of coefficients of generating function
QUESTION [10 upvotes]: how to calculate the asymptotic growth rate of coefficients generating function $T(z)$ satisfied this identity
$T(z)=z+\frac{T(z)^3}{6}+\frac{T(z^2)T(z)}{2}+\frac{T(z^3)}{3}$

REPLY [3 votes]: Some remarks about bounding the coefficients of $T(z)$ from above and from below. 
I assume $T_0=0$ in $T(z)=\sum_{k=0}^\infty T_k z^k$, to which it corresponds a unique formal series solution of the functional equation for $T(z)$, e.g. obtained by iteration of the composition operator on the RHS. Let's first prove that even order coefficients vanish, and odd order coefficients increase. 
The Ansatz $T(z):=zS(z^2)$ produces this equation for $S(z):=\sum_{k=0}^\infty S_k z^k$:
$$S(z)=1+\frac{1}{6}zS(z)^3 +\frac{1}{2}zS(z^2)S(z)+ \frac{1}{3}zS(z^3)\ .$$
The latter equation also have a unique formal series solution, which proves that $T$ has indeed the above form $zS(z^2)$, that is, all even order coefficients of $T$ vanish. Since $S$ can be obtained by iteration of the composition operator on the RHS, starting e.g. by $1$, it is clear that $S_k\ge0$ for all $k$, and since $S_0=1$  we also have $[z]^k S(z)^3\ge 3 [z^k]S(z)$ and $  [z^k]S(z^2)  S(z)\ge [z^k]S(z)$. Thus $$S(z)\ge 1 + z S(z) 
 \ ,$$
(coefficient-wise), proving  that $S$ has increasing coefficients. Having $S(z)$ positive, increasing coefficients also implies, again from the equation, 
$$6S(z)\le 6 + z S(z)^3 + 3zS(z)^2+2zS(z)                      \ ,$$ 
(coefficient-wise). This implies that $S$ is dominated coefficient-wise by the power series solution $y(z)$ of $6y=6+z(y^3+3y^2+y)$. By uniqueness, $y$ is the expansion at $z=0$ of the local inverse of $$h(y)=\frac{6(y-1)}{y(y+1)(y+2)}$$ at $y=1$. This proves that $S$ has a positive radius of convergence $\rho$.
A bound from above for the  coefficients of $S$ can be deduced from 
$$6S(z)\ge 6 + z S(z)^3 + 3zS(z) +2z                       \ ,$$ 
(coefficient-wise), which implies that the coefficients of $S(z)$ are larger than the coefficients of the local inverse at $y=1$ of
$$g(y)=\frac{6(y-1)}{y^3+3y+2}\ .$$
Putting $y=x+1$, by the Lagrange inversion applied to $h(x+1)$ and $g(x+1)$ one gets, for $k\ge1$
$$\frac{1}{k}[x^{k-1}](1+x+ x^2/2 +x^3/6)^k \le S_k \le \frac{1}{k}[x^{k-1}](1+ 11x/6 +x^2 + x^3/6)^k$$
etc.<|endoftext|>
TITLE: Generating function for certain partitions (with a restriction on the Durfee square)
QUESTION [12 upvotes]: First of all my apologies if this question is well known or obvious: this is not in my area of research.
Let $T(x)=\sum_{n=0}^\infty t_nx^n$, where $t_n$ is the number of partitions $\lambda$ of $n$ into $m$ parts where, if $(m-i)\times (m-i)$ is the size of the Durfee square of $\lambda$, then the partition on the right of this Durfee square has at most $m-i-1$ parts.
As usual denote by $(x)_j:=(1-x)(1-x^2)\cdots (1-x^j)$.
If I am not mistaken $$
T(x)=
\sum_{i=0}^{m-1}
\frac{(x)_{m-1} x^{(m-i)^2+i}}
{(x)_{m-i-1}^2(x)_i}.$$
In fact, given a positive integer $N$, the coefficient of degree $N$ in $$\frac{(x)_{m-1}x^i}{(x)_{m-i-1}(x)_{i}}$$
is the number of partitions of $N$ into exactly $i$ parts each of size at most $m-i$ (and this accounts for the partition at the bottom of the Durfee square because we are counting partitions with exactly $m$ parts). Similarly, the coefficient of degree $N$ of 
$$\frac{1}{(x)_{m-i-1}}$$
is the number of partitions of $N$ into at most $m-i-1$ parts (and this accounts for the partition on the right of the Durfee square because we are requiring that this piece has at most $m-i-1$ parts).
Computational evidence shows that
$$T=x^m\sum_{i=1}^m\frac{(-1)^{i-1}x^{i(i-1)/2}}{(x)_{m-i}}$$
but I have no idea why this is true.

REPLY [6 votes]: Lemma. Fix $n$ and $m$. Consider pairs of partitions $(\lambda,\mu)$ such that $\lambda$ has $m$ parts and $|\lambda|+|\mu|=n$. Let $A$ be the number of pairs for which $\max(\lambda)>\max(\mu)$ (where $\max(\emptyset)=-\infty$). Let $B$ be number of pairs for which $\lambda$ satisfies this condition with Durfee square, which we may rephrase as '$\lambda_i=i$ for some $i$, where $\lambda=(\lambda_1\geqslant \lambda_2\geqslant \dots)$'. Then $A=B$.
We construct a bijection. Take partitions $\lambda=(\lambda_1\geqslant \dots \geqslant \lambda_m>0)$ and $\mu=(\mu_1\geqslant\dots)$ such that $\lambda_1>\mu_1$ (or $\mu$ is empty). If $\lambda_i=i$ for some index $i$, take the same pair of partitions. If not, then there exists $i$ such that $\lambda_i\geqslant i+1$, $\lambda_{i+1}\leqslant i$. Take partitions $(\lambda_2-1,\dots,\lambda_{i}-1,i,\lambda_{i+1},\dots,\lambda_m)$ and $(\lambda_1-1,\mu_1,\dots)$. 
Now reduction.
At first, let's count all the other partitions onto $m$ parts. These are partitions for which $k$'s largest part never equals $k$, $k=1,2,\dots$. We want to prove that their generating function equals
$$
x^m\sum_{i=0}^m\frac{(-1)^{i}x^{i(i-1)/2}}{(x)_{m-i}}
$$
(I have used that total number of partitions onto $m$ parts have generating function $x^m/(x)_m$, that is seen by duality.) Call such partitions interesting.
Multiply by $t^m$ and sum up by $m$. We get a double generating function $f(t,x)=\sum t^{{\rm parts}(\lambda)} x^{|\lambda|}$, summation is taken over all interesting partitions $\lambda$.
We have to prove that
$$
f(t,x)=\sum_{m\geqslant i\geqslant 0} (tx)^m\frac{(-1)^{i}x^{i(i-1)/2}}{(x)_{m-i}}=
\left(\sum_{k\geqslant 0} \frac{(tx)^k}{(x)_k}\right)\left(\sum_{i\geqslant 0} (-1)^i(tx)^i x^{i(i-1)/2}\right).
$$
As for the first multiple, it is a double generating function for $t^{\max(\lambda)} x^{|\lambda|}$ taken by all partitions $\lambda$, hence by duality it is the same thing as a double generating function for $t^{{\rm parts}(\lambda)} x^{|\lambda|}$, which is $\prod_{n\geqslant 1} (1-tx^n)^{-1}$. As for the second multiple, it is a part of Jacobi triple product
$$
\prod_{n\geqslant 1} (1-tx^{n})(1-t^{-1}x^{n-1})(1-x^n)=\sum_{i=-\infty}^{\infty} (-1)^it^i x^{i(i+1)/2},$$
corresponding to non-negative $i$. Denote by $H(t,x)$ the part of double product $\prod_{n\geqslant 1} (1-tx^{n})(1-t^{-1}x^{n-1})$ with only non-negative powers of $t$. We have to prove that
$$
\prod_{n\geqslant 1} (1-tx^n)\cdot H(t,x)=f(t,x)\cdot \prod (1-x^n)^{-1}.
$$ 
Coefficient of $t^mx^n$ in RHS equals the number of pairs of partitions $(\lambda,\mu)$ for which $|\lambda|+|\mu|=n$, $\lambda$ has $m$ parts and is interesting and $\mu$ is arbitrary. By the conjecture it is the same as number of pairs of partitions $(\lambda,\mu)$ for which $|\lambda|+|\mu|=n$, $\lambda$ has $m$ parts and $\mu$ has a part which is not less then $\max(\lambda)$.
Let's obtain the same in the LHS. If $H(t,x)$ were not a part, but the whole product, then a lot would cancel when we multiple $\prod(1-tx^n)^{-1}$ and $\prod (1-tx^n)$. But something still cancels. Namely, we look at our product
$$
\prod_{n\geqslant 1} (1+tx^n+t^2x^{2n}+\dots) \prod_{n\geqslant 1} (1-tx^n) \prod_{n\geqslant 1} (1-t^{-1}x^{n-1})
$$
and see what we may take from each bracket. We are conditioned to take at least as many $t$'s from the second product than $t^{-1}$'s from the third. Consider partial involution on the set of our choices: denote by $N$ the maximal index $n$ for which we either take $-tx^{n}$ from the second product or take $t^ax^{na}$ for some $a\geqslant 1$ from the first product. We could take $1$ from the corresponding bracket in the second product and $t^{a+1}x^{n(a+1)}$ from the corresponding bracket in the first product instead. This is a sign-changing involution on the set of choices, but the set of admissible choices is not quite invariant. Namely, if total number of $t$'s from the second product equals total number of $t^{-1}$'s in the third product, we are forbidden to replace $-tx^N$ to 1. This is what remains after removing all pairs of choices formed by involution. To be more precise, what we should choose are some $k$ positive integers $0<a_1<a_2<\dots <a_k$ (second product), some $k$ non-negative integers $0\leqslant b_1<b_2<\dots b_k$ (third product) and some partition with maximal part at most $a_k$ (first product).
As for two choices from the second and from the third products, it is equivalent to the choice of a partition with maximal part $a_k$: for such a partition let $k$ be the size of Durfee square, $a_1,\dots,a_k$ correspond to the part of Young diagram on the one side of its diagonal and $b_1,\dots,b_k$ to the other side. 
So, we choose a partition with maximal part $a_k$ (this is $\mu$) and another partition onto $m$ parts with maximal part at most $a_k$. This is $\lambda$. Also, exponent $m$ of $t$ is the number of parts in $\lambda$, and exponent of $x$ is $|\lambda|+|\mu|$.<|endoftext|>
TITLE: Aren't Riemannian geodesics also geodesics of the associated Cartan geometry?
QUESTION [12 upvotes]: I was inspired by R. W. Sharpe's book on doing differential geometry through Cartan connections. Unfortunately, the book is fairly thin in terms of specific examples in Riemannian geometry, so I decided to try a few on my own. 
My first thought was to try the hyperbolic plane (modeled on $\mathbb{R}^2$ instead of just the half-plane), since the orthonormal frame bundle is still, essentially, $\mathbb{R}^2\rtimes O(2)$. This actually worked out pretty well. The metric is just $\mathrm{g}=e^{-2y}\mathrm{d}x\otimes\mathrm{d}x+\mathrm{d}y\otimes\mathrm{d}y$, and since we want to work in an orthonormal frame, I made $$\begin{bmatrix}1 \\ 0\end{bmatrix}=e^y\partial_x\text{ and }\begin{bmatrix}0 \\ 1\end{bmatrix}=\partial_y,$$ so the associated covariant derivative is given by* $${\huge\nabla}_{\begin{bmatrix}1 \\ 0\end{bmatrix}}\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}-b \\ a\end{bmatrix} \text{ and } {\huge\nabla}_{\begin{bmatrix}0 \\ 1\end{bmatrix}}\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}0 \\ 0\end{bmatrix}.$$
A few calculations give the Cartan connection as $$\omega\left(\begin{bmatrix}\cos\theta & -\sin\theta & x \\ \sin\theta & \cos\theta & y \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 & -t & v^1 \\ t & 0 & v^2 \\ 0 & 0 & 0\end{bmatrix}\right)=\begin{bmatrix}0 & -t-v^1 & v^1 \\ t+v^1 & 0 & v^2 \\ 0 & 0 & 0\end{bmatrix}.$$
The problem came when I tried to find geodesics. Clearly, $\gamma:t\mapsto (0,t)$ is a geodesic in the traditional sense, but it lifts to $$\widehat\gamma:t\mapsto\begin{bmatrix}\cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & t \\ 0 & 0 & 1\end{bmatrix},$$ which has tangent vectors $$\dot{\widehat\gamma}(t)=\begin{bmatrix}\cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & t \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}0 & 0 & \sin\theta \\ 0 & 0 & \cos\theta \\ 0 & 0 & 0\end{bmatrix},$$ so $$\omega(\dot{\widehat\gamma}(t))=\begin{bmatrix}0 & -\sin\theta & \sin\theta \\ \sin\theta & 0 & \cos\theta \\ 0 & 0 & 0\end{bmatrix}.$$ I am fairly confident that this won't develop into a curve whose image projects to a straight line.
On the other hand, the curve $$\sigma:t\mapsto\begin{bmatrix}\cos t & \sin t & \sin t \\ -\sin t & \cos t & \cos t \\ 0 & 0 & 0\end{bmatrix}$$ has tangent vectors $$\dot\sigma(t)=\begin{bmatrix}\cos t & \sin t & \sin t \\ -\sin t & \cos t & \cos t \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 1 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},$$ so $$\omega(\dot\sigma(t))=\begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$ and it develops into $t\mapsto\exp\left(t\begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\right)$, though $\sigma$ does not project to (the image of) a geodesic.

In short, as the title asks, aren't geodesics of the Riemannian geometry also geodesics of the Cartan geometry?

My ideal answer here includes:

A "yes" or "no" to the above question.
If "yes," then where did I go wrong?
In the unlikely event of "no," why aren't they?
Possibly a reference to a large collection of worked-out examples of Cartan connections for Riemannian geometry that deals with geodesics

Though, any one of the above will probably work for me.
*I apologize for the poor formatting.

REPLY [3 votes]: The geodesics of Euclidean geometry in the plane are straight lines, i.e. curves that do not change direction. Their Cartan geometry is $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $x,y,\theta$ coordinates, valued in the Lie algebra of rigid motions of the plane, where $\xi=e^{i\theta}(dx+i \, dy)$, $\gamma=i \, d\theta$.
So $\gamma=0$ says that the direction does not change: a straight line, i.e. a geodesic.
Now consider the hyperbolic plane.
(When you look at the Cartan connection in your coordinates, you spot that it must involve exponential factors $e^{\pm y}$, since these arise in coordinates for your metric. You avoid exponentials by working in the moving frame, but then your Cartan connection is not expressed in $x,y,\theta$ coordinates anymore, but in the coframing.  When you write out the velocity of $\sigma(t)$ in the coframing, it doesn't turn out to be as straightforward.)
In a complex notation, the Cartan connection is simpler. On the plane, let $\underline{\xi}=e^{-y}dx+idy$ and $\underline{\gamma}=ie^{-y}dx$. 
We see this because we want metric $\xi_1^2+\xi_2^2$.
Add a variable $\theta$ and let $\xi=e^{i\theta}\underline{\xi}$, and $\gamma=\underline{\gamma}+id\theta$. The Cartan connection is the matrix $$\omega=\begin{pmatrix}\gamma&\xi\\0&0\end{pmatrix}$$ on the oriented orthonormal frame bundle, in $x,y,\theta$ coordinates, valued in the Lie algebra of rigid motions of the plane. This differs from your expression, because there are still $e^{-y}$ factors. The 1-form $\gamma$ does not vanish on your curve $\sigma(t)$. So is not a geodesic. Your first curve $(x(t),y(t),\theta(t))=(0,t,0)$ is, since $\gamma=0$ on it.
Expanding out in these coordinates, the equation of a geodesic is $\xi=c \, ds, \gamma=0$, for a complex constant $c$. We can assume that $c\ne 0$. The 1-forms $\xi$ and $\gamma$ transform when we rotate $\theta$, say the action $r_{\theta_0} : \theta\mapsto \theta+\theta_0$, as $r_{\theta_0}^*\xi= e^{i\theta_0}\xi$, $r_{\theta_0}^*\gamma=\gamma$. So we change the equations by rotation of $c$, and we can arrange that $c>0$. Rescale the $s$ variable to get $c=1$. So our equations of geodesics expand out to
$$
\cos \theta e^{-y} \frac{dx}{ds} - \sin \theta \frac{dy}{ds}=1,
$$
$$
\sin \theta e^{-y} \frac{dx}{ds} + \cos \theta \frac{dy}{ds}=0,
$$
$$
\frac{d\theta}{ds}+e^{-y}\frac{dx}{ds}=0,
$$
if I have calculated correctly.<|endoftext|>
TITLE: A Hilbert space characterization via retractions--a conjecture
QUESTION [7 upvotes]: Given a Banach space $X$ and a functional $f:X\rightarrow \mathbb R$, let
$$ X_f := \{x\in X : f(x)\ge 0\} $$
("functional" means "non-zero linear functional"). Also, given a topological space $E$ and its topological subspace $A$, a retraction $r:E\rightarrow A$ is defined as a continuous map such that $r(x)=x$ for every $x\in A$.
CONJECTURE:
Let $\,X$ be an arbitrary Banach space such that for every functional $\ f:X\rightarrow \mathbb R\ $ there is a retraction $\ r:X\rightarrow X_f\ $ such that
$$ \forall_{x\ y\ \in\ X\setminus X_f}\ \ \ |r(x)-r(y)| = |x-y| $$
Then $X$ is isometric to a Hilbert space.

REMARK: I think that questions of this type were popular in the past in the case of finite dimensional spaces (mostly 3-dim?). I am not aware of the general case (I am not a specialist thus I have to ask :-). I still believe that my usage of the retraction language here is new (even in the finite-dimensional case).

REPLY [4 votes]: I think that the desired result can be proved on the following
lines if the dimension is at least $3$.
(1) Consider such maps for $X_f$ and $X_{-f}$. Denote them $r$ and
$r'$ respectively. One can show that $f(r(x))=-f(x)$ for $x\in
X\backslash X_f$. Similarly one can show that $f(r'(x))=-f(x)$ for
$x\in X_f$. Let $Ax=(r(x)+r'(x))/2$. Then $A$ is a $1$-Lipschitz
retraction onto $H=\{x\in X: f(x)=0\}$.
(2) Using the result of Lindenstrauss (Corollary 1 on page 270 in
Michigan Math. J. 11 (1964), 263-287), we get that there is a
linear projection of norm $1$ onto $H$.
(3) Using one of the known characterization of the Hilbert space,
see (12.8) in Amir (Characterizations of inner product spaces,
Birkhauser Verlag, Basel, 1986), one gets that the space $X$ is
Hilbert.
P.S. (1) One can replace usage of the Lindenstrauss result by the
usage of the result of Mankiewicz (Bull. Acad. Polon. Sci. Ser. Sci. Math.
Astronom. Phys. 20 (1972), 367-371) implying that the restriction
of $r$ to $X\backslash X_f$ is a restriction of a linear isometry
$\widetilde r:X\to X$ to $X\backslash X_f$. Then we observe that
$\frac12(I+\widetilde r)$ is a norm-$1$ projection onto $H$.
(2) One can use this observation to complete the $2$-dimensional
case which we divide into two subcases:
(2a) The unit ball is a polygon: this can be done by hand.
(2b) The unit ball has infinitely many strongly exposed points.
For each such point $x$ we consider the corresponding (exposing)
norm-one functional $f_x$, and the half-space $H_{f_x}$. The
corresponding $r$ maps $x$ to $-x$. This implies that
$\hbox{ker}f_x$ and $\mathbb{R}x$ satisfy the James orthogonality
relation (see Amir, page 24). Since there are infinitely many
strongly exposed points $x$, the assumption of (6.12'') (Amir,
page 53) is satisfied and the space is Euclidean.<|endoftext|>
TITLE: What is the status of Arthur's book?
QUESTION [42 upvotes]: Arthur's long-awaited book project is now published (The endoscopic classification of representations: orthogonal and symplectic groups).  However, in the book he takes some things for granted:

The stabilization of the twisted trace formula for GL($N$) and SO($2n$).
Orthogonality relations for elliptic tempered characters.
Weak spectral transfer of tempered $p$-adic characters
A twisted version of Shelstad's archimedean theory of endoscopy

Assumption 1, and implicitly 2 and 3, are in Arthur's Hypothesis 3.2.1.  Assumption 4 is made in the proof of Arthur's Proposition 2.1.1 (see also Remark 5 after Theorem 2.2.1).  (I think these are all of Arthur's working hypotheses, but let me know if I missed something.)

Q1: Recently, in a series of 10-11 preprints, Waldspurger and Moeglin
  finished the stabilization of the twisted trace formula.  This seems
  to verify Assumption 1--does it also take care of implicit Assumptions 2
  and 3?  If yes, where are 2 and 3 proved?

Arthur said that Assumption 4 should be taken care of by work in progress of Shelstad and Mezo.

Q2: Is Assumption 4 verified yet?
  Mezo and Shelstad have some recent preprints, but I don't know if they prove exactly what's needed.  If not, any explanation of what is still needed would be appreciated.  

If not all of these assumptions are proven yet, are any of the specific main results of Arthur (beyond, say, SO(3)) now unconditional?
Edit: I was recently reminded that Arthur also refers to some of his preprints ([A24]-[A27] in the quasi-split case), which are still unpublished (unfinished?) as of mid 2018, for some proofs, but based on Arthur's earlier verbal confirmation to me, I will keep my answer as accepted unless other information comes to light.

REPLY [18 votes]: Now I think the answer is yes, Arthur's work is now unconditional for quasi-split special orthogonal and symplectic groups.  Wee Teck Gan kindly directed me to the relevant papers of Waldspurger and Moeglin addressing the assumptions 2-4 above, though I was not able to verify with certainty that their stated results precisely cover what Arthur requires.  Here are the relevant papers:

The stabilization is completed in Moeglin and Waldspurger's Stabilisation X paper.
The orthogonality for elliptic tempered characters appears to be completed in Waldspurger's twisted local trace formulas paper, generalizing Arthur's 1993 Acta paper.  See Theorem 7.3.
I believe the $p$-adic weak spectral transfer that Arthur requires, specifically twisted analogues of Theorems 6.1 and 6.2 from his 1996 Selecta paper, is done by Moeglin in her Représentations elliptiques... paper (which is unofficially called "Stabilisation XI").
Twisted archimedean endoscopy seems to be done in Waldspurger's Stabilisation IV paper, which uses Shelstad's 2012 Annals paper.  This is  independent from the treatment in Mezo's Tempered spectral transfer... paper, but I cannot tell if Mezo's results are unconditionally sufficient for Arthur's needs.

Edit: Arthur confirmed that the results in his book (except the last chapter on non-quasi-split forms) are now unconditional.<|endoftext|>
TITLE: Who was the first to discover that the curvature of an embedded surface is the product of the principal curvatures?
QUESTION [10 upvotes]: The invention of intrinsic differential geometry is usually attributed to Gauss in the context of his theorema egregium but the notion of the curvature of an embedded surface existed before. Who was the first to discover that the curvature of an embedded surface is the product of the principal curvatures?
In http://arxiv.org/abs/1409.4736 the author states: "In 1763, Euler started
a thorough study of curvature of embedded surfaces. In 1767, he found an
expression of the curvature in terms of the product of principal curvatures."
This seems to suggest that the curvature Euler is talking about is what is known today as Gaussian curvature, the latter being the product of principal curvatures.  However, this comment in 1409.4736 seems to be based on a different comment in the 2008 encyclopedia of scientific biography at http://www.encyclopedia.com/topic/Johann_Tobias_Mayer.aspx#1-1G2:2830901353-full which states:
"In 1763 Euler made the first substantial advance in the study of the curvature of surfaces; in particular, he expressed the curvature of an arbitrary normal section by principal curvatures (1767)."
This of course does not imply any relation to Gaussian curvature and on the contrary suggests that Euler only dealt with curvatures of curves obtained as cross-sections of the embedded surface.  However, it is possible that 1409.4736 has a different source for this claim.
The relevant paper by Euler is apparently http://eulerarchive.maa.org/docs/originals/E333.pdf where on page 143 Euler discusses the two radii of curvature.

REPLY [6 votes]: Let me start with your first quotation:
"In 1763, Euler started a thorough study of curvature of embedded surfaces. In 1767, he found an expression of the curvature in terms of the product of principal curvatures."
Karin Reich's article in Leonhard Euler: Life, Work and Legacy (p. 482 in particular) explains what this is supposed to mean. As you know,
Euler attacked the curvature of surfaces at some point P in terms of the curvatures of curves cut out by planes normal to the tangent plane at P. He found that the planes for which the maximal and minimal curvature occurs 
are orthogonal to each other, and then he developed a formula for the curvature of the curve cut out by any plane normal to the tangent plane in terms of the
maximal and minimal curvature and the angle $\phi$ between the plane and the principal section. So "the curvature" here is definitely not Gauss's curvature.
The remark about the years 1763 and 1767 is even "curiouser": Euler presented his article E333 in 1763 to the Berlin Academy and it was published in 1767. So we are dealing with one and the same paper here.<|endoftext|>
TITLE: Proofs of main probability results from other fields
QUESTION [8 upvotes]: Making connections between different areas is very exciting and probability has already made connections with other fields (BM used in proving complex analysis and PDE results).
To keep it short, I will greatly appreciate any references to multidisciplinary proofs for any of the following results:
1)Central limit theorem (simple version, Lindeberg's or Donsker's theorem)
2)Law of large numbers
3)Recurrence in d=1,2 and transience in $d>2$ of random walk.
4)Conformal and time change: if f is conformal then $f(B_{t})\stackrel{d}{=}\widetilde{B}_{\int_{0}^{t}|f'(B_{s})|^{2}ds}$ where B,$\widetilde{B}$ are planar Brownian motions.
5)Feynman-Kac formula of Brownian motion
The more far-fetched the proofs the better. For example, I will be surprised (and glad) to see proofs using mainly topology, number theory or algebra.
For example CLT has a very interesting approach from information theory:
"An information-theoretic proof of the central limit theorem with the Lindeberg condition", Theory of Probability and its applications. 1959, Vol IV, n o 3, 288-299.

REPLY [4 votes]: A well-known example is the heat-equation proof of the central limit theorem due to [Petrovsky and Kolmogorov].<|endoftext|>
TITLE: The $\zeta$-word
QUESTION [5 upvotes]: I was wondering about classical notations in number theory. I will not ask here about special functions in general but about the more ubiquitous number theory functions. That which made me wonder originally is $\zeta$: is it really Riemann's $\zeta$? I guess so, but then why did he choose that? Thinking about it, it is interesting that number theory functions seem to have had sticky names, rather varied too. Weierstrass's $\wp$, Dirichlet's $L$, Jacobi's $\theta$,... Subsequent writers have not come up with widely-used variations. I can guess a few reasons why the original notations stuck. The inventors were aware of previous research and chose noncolliding names coherent with what came before. There were few mathematicians in the 19th century and they knew relatively well each other and each other's work. Later mathematicians were then quite rightly very respectful of the traditions set by those illustrious pioneers, especially because the whole was rather coherent and for this reason any attemp at changing  part of it would probably clash with some other piece of the "nomencla(pic)ture". Research in number theory is also probably among the most self sufficient of mathematics. Of course there are many connections to other subjects but number theorists can live "within" number theory and outside workers will "borrow" respectfully to the majestuous monument.
So the standard questions would be:
inventor? reason(s)? historical alternatives?
The notations I think about now:
$\zeta$, $\wp$ $\theta$, $\sigma$ (Weierstrass), $G$ (Eisenstein), $\tau$ (Ramanujan), $\eta$ (Dedekind), $s$ for complex numbers, $\tau$ for upper half-plane points, $q$ for $e^{2\pi i\tau}$.
For Eisenstein series I can guess he thought about $F$ but that was too common and $G$ was not used for groups then and anyway general groups are not used close to $G$. Now with general Lie groups and automorphic forms $G$ has morphed into $E$.
Feel free to add to these lists. Any comment is welcome. Thanks.
EDIT: I see my question does not satisfy moderators. If I may: I think notation reflects many deep facts about mathematical concepts and intuitions. I also think that the notations I mention are far from trivial to understand -but I probably overestimate my capacities and they are actually obvious to everybody else but me. Anyway I would really appreciate any indication, like what Carlo replied.
2ND EDIT: I consider that Carlo replied. Thanks to him I convinced myself $\zeta$ stands for "zahl...". It is quite surprising that nowhere I've read this plausible explanation for the notation. I looked at Edwards book, where he translates Riemann and could not find any comment on that. Yet reading the masters as he advises is important only to understand them best, and notation is one of the choices they make that readers should understand. Finally I would like to tell the moderators sorry that I was upset in my 1st EDIT, and also that I think they should probably be more open to questions outside the strict rules set up to now. I think, like Thurston, that mathematics is not foremost about precise statements and proofs, but about understanding. And my belief is that understanding mathematics involves some psychology, sociology, and history among other sciences. All this I would hope to find their place on mathoverflow, to efficiently contribute to the furthering of knowledge in mathematics proper.

REPLY [15 votes]: Well, Riemann himself says "I denote this function by $\zeta(s)$" ("Die Function [...] bezeichne ich durch $\zeta(s)$"), so I would think the choice of which letter to use for this function was his.

first page of Riemann's Über die Anzahl der Primzahlen unter einer gegebenen Grösse (1859).

The OP also asks: Why did Gauss choose the letter $\zeta$? For integer $s$, Euler had called this product $P$, which seems natural. 

page from Introductio in analysin infinitorum (1748)
According to this source, the use of $s$ as the variable in a series of powers of primes $p$ goes back to Dirichlet, who took $s$ to be real and positive. Riemann emphasized the importance of letting $s$ be complex. Why he chose the letter $\zeta$ for the function of $s$ is not documented, but I can guess that it would make sense to use a different character set for the function and its argument. (We still do this routinely today.) Upper case Greek letters ($\Pi$ or $\Sigma$) were already committed, so the choice for lower case Greek seems natural.<|endoftext|>
TITLE: Reduced resultant of monic polynomials
QUESTION [5 upvotes]: Let $f(x)$ and $g(x)$ be coprime monic polynomials in $\mathbf{Z}[X]$ of positive degrees $m$ and $n$ respectively. It seems that in this case their reduced resultant can be obtained from the expression $uf + vg = 1$ over $\mathbf{Q}[X]$ with $\deg u < n$ and $\deg v < m$. Namely - reduced resultant in this case is the smallest natural $D$ such that $Du$ and $Dv$ are in $\mathbf{Z}[X]$? For the notion of reduced resultant see: this MO question of Felipe Voloch. Shortly it is the generator $D>0$ of the ideal $I=(f(x),g(x))∩\mathbf{Z}$.

REPLY [8 votes]: The ideal generated by $f$ and $g$ in ${\bf Z}[x]$ is the set of all $uf+vg$ with $u,v$ in ${\bf Z}[x]$. 
Now suppose $uf+vg=d$ for some integer $d$. Assuming $f$ and $g$ are monic, we have $$v=fq_1+r_1,\qquad u=gq_2+r_2$$ with $q_i$ and $r_i$ in ${\bf Z}[x]$, $\deg r_1<\deg f$, $\deg r_2<\deg g$. Then $$fg(q_1+q_2)+r_2f+r_1g=d$$ But $\deg (fg)>\deg(r_2f+r_1g)$, so we must have $q_1+q_2=0$, and $$r_2f+r_1g=d$$ It follows that the smallest positive integer in the ideal generated by $f$ and $g$ (that is, the reduced resultant of $f$ and $g$) can be expressed as $uf+vg$ with $\deg v<\deg f$ and $\deg u<\deg g$.<|endoftext|>
TITLE: Global sections of coherent sheaves on determinantal hypersurfaces in $\mathbb{P}^n$
QUESTION [6 upvotes]: Let us consider the short exact sequence of coherent sheaves on $\mathbb{P}^n$ $$0 \to \mathcal{O}_{\mathbb P^n}(-1)^{r} \stackrel{N}{\longrightarrow} \mathcal{O}_{\mathbb P^n}^{r} \longrightarrow \mathcal F \to 0,$$
where $r \geq 2$ and $N=\{L_{ij}\}$ is a $r \times r$ matrix of linear forms in the variables $x_0, \ldots, x_n$ acting on the left (i.e., $\mathbf{x} \to N \mathbf{x}$). 
Then the sheaf $\mathcal{F}$ is supported on the determinantal hypersurface $Y:=\{\det N=0 \} \subset \mathbb{P}^n$, and it is locally free of rank $1$ (i.e., a line bundle) on the open dense locus of $Y$ given by $\textrm{rank } N = r-1$.
Passing to cohomology, we obtain an isomorphism $$\sigma \colon H^0(\mathbb{P}^n, \, \mathcal{O}_{\mathbb P^n}^{r}) \stackrel{\cong}{\longrightarrow} H^0(Y, \, \mathcal F),$$
in particular $h^0(Y, \, \mathcal F)=r.$ In other words, this is a bijective correspondence  between $n$-ples $(\alpha_1, \ldots, \alpha_r)$ of complex numbers and global sections of $\mathcal F$. 

Question. How can we describe the zero locus in $Y$ of the section 
  $\sigma(\alpha_1, \ldots, \alpha_n)$ in terms of the $\alpha_i$ and the matrix $N$?


Toy model. Consider the case $n=3$, $r=2$ and $$N = \pmatrix{x_0 & x_1 \\ x_2 & x_3}.$$ 
In this situation, $Y$ is the smooth quadric $\{x_0x_3-x_1x_2=0 \} \subset \mathbb{P}^3$, and $\mathcal{F}$ is a line bundle with $2$ global sections, hence a pencil corrisponding to one of the two rulings of $Y$. I made some (unelegant) local computation and it seems to me that in this case $\sigma(\alpha_1, \, \alpha_2)$ should be  something like $\alpha_1 \sigma_{02} + \alpha_2 \sigma_{13}$, where $\sigma_{02}$ is the section whose zero locus is the line $\{x_0=x_2=0 \}$ and $\sigma_{13}$ is the section whose zero locus is the line (in the same ruling) $\{x_1=x_3=0 \}.$ In other words, the global sections of $\mathcal{F}$ are expressed in terms of the columns of $N$.
On the other hand, dualizing our exact sequence and applying Grothendieck duality we get $$0 \to \mathcal{O}_{\mathbb P^3}(-1)^{2} \stackrel{ { }^t N}{\longrightarrow} \mathcal{O}_{\mathbb P^3}^{2} \longrightarrow \mathcal F^*(1) \to 0,$$
where $\mathcal{F}^*:=\mathscr{Hom}(\mathcal{F}, \, \mathcal{O}_{\mathbb P^3})$.
Now the sections of $\mathcal{F}^*(1)$ should correspond the other ruling of the quadric, because $\mathcal{F} \otimes \mathcal{F}^*(1)= \mathcal{O}_Y(1)$. Therefore the global sections of $\mathcal{F}^*(1)$ are given in terms of the rows of $N$, that are the columns of ${}^t N$.

Is this correct? If so, is there any intrinsec or general way to see this?

REPLY [2 votes]: This is meant to be an integration to Yusuf answer.
Consider a section 
$$
\stackrel{\rightarrow}{\alpha}=
\begin{pmatrix}
\alpha_1\\
\vdots\\
\alpha_r
\end{pmatrix} \in {\mathbb C}^r\cong H^0({\mathbb P}^n,{\mathcal O}^r_{{\mathbb P}^n}).
$$
Its image on $H^0(Y,{\mathcal F})$ vanishes in a point $p \in Y$, by definition of quotient, if and only it is in the image of the map induced by $N$ on the stalks over $p$.
Then, by the Kronecker-Rouché-Capelli's Theorem, its zero locus equals the set of points where the rank of the matrix $N$ equals the rank of the $r \times (r+1)$ matrix $(N \stackrel{\rightarrow}{\alpha})$.
In your toy model, in all points of $Y$ the rank of $N$ is $1$, so we are looking to the locus where 
$$
rank
\begin{pmatrix}
x_0&x_1&\alpha_1\\
x_2&x_3&\alpha_2\\
\end{pmatrix}=1
$$
that is $\{\alpha_1x_2-\alpha_2x_0=\alpha_1x_3-\alpha_2x_1=0\}$.
This is as in your local computation indeed one of the rulings (but the other one, if my computations are not wrong).
A similar computation in the general case, as pointed out by Yusuf, produces equations involving the $(r-1) \times (r-1)$ minors of $N$.
More precisely, we need the adjoint or cofactor matrix $N^*$ whose $(i,j)$ entry is $(-1)^{i+j}$ times the minor of $N_{i,j}$, the determinant of the $(r-1) \times (r-1)$ matrix obtained by $N$ deleting the $i^{th}$ row and the $j^{th}$ column. In your toy model
$$
N^*=
\begin{pmatrix}
x_3&-x_2\\
-x_1&x_0\\
\end{pmatrix}
$$
Then the zero locus of the image of $\stackrel{\rightarrow}{\alpha}$ is 
$$
^tN^* \stackrel{\rightarrow}{\alpha}=0
$$<|endoftext|>
TITLE: Do the analogies between metamathematics of set theory and arithmetic have some deeper meaning?
QUESTION [34 upvotes]: By "formal analogies" between the metamathematics of $\mathsf{ZFC}$/set theory and $\mathsf{PA}$(=Peano Arithmetic)/first order arithmetic, I mean facts such as the following:

We are considering a first-order theory ($\mathsf{ZFC}$ or $\mathsf{PA}$) motivated as a first-order approximation to a second-order theory (second-order $\mathsf{ZFC}^2$ or $\mathsf{PA}^2$) which is "more or less" categorical; because of this, some axioms (the separation and replacement axioms on the one hand, the induction axiom on the other) have to be stated as axiom schemes in the first-order theory.
There is an interesting hierarchy of formulæ, $\Sigma_n$ or $\Pi_n$, based on alternations of quantifiers (viz.: the arithmetic hierarchy vs. the Lévy hierarchy); at the lowest ($\Delta_0$) level of this hierarchy are formulæ with only "bounded" quantifiers.
There is a uniform truth predicate for any (concrete) given level of the hierarchy, which is built upon some kind of absoluteness of $\Delta_0$ formulæ.  As a related fact, the infinite axiom schemes are naturally stratified along the hierarchy (they can be cut off at the $\Sigma_n$ level and stated as a single formula for each concrete $n$).
There is a reflection theorem which ensures that any finite set of true statements (or one bounded in the hierarchy of formulæ) is consistent.  In particular, the full theory proves the consistency of the subtheory with the axiom schemes cut off at the $\Sigma_n$ level: that is, the theory ($\mathsf{ZFC}$ or $\mathsf{PA}$) is reflexive.  In fact, it is even essentially reflexive (every consistent extension is reflexive).
There is a conservative two-kinded extension (Gödel-Bernays on the set theoretical side, $\mathsf{ACA}_0$ on the arithmetical side) which is obtained by allowing formation of classes but only with a comprehension scheme for such classes that does not involve quantifying over classes; remarkably, this conservative two-kinded extension is finitely axiomatizable.  There is also a standard strictly stronger two-kinded extension (Morse-Kelley on the set-theoretical side, second-order arithmetic $\mathsf{Z}_2$ seen as a first-order theory on the arithmetical side).

(I hope I didn't mess things up too much, but all of these facts are standard and can be found in standard textbooks such as Jech's Set Theory for the set-theoretical side and Hájek and Pudlák's Metamathematics of First-Order Arithmetic plus Simpson's Subsystems of Second-Order Arithmetic for the arithmetical side.)
I'm sure many more examples can be found.  Maybe I should also nod to the similarity between proof theory of extensions of $\mathsf{PA}$ by analysing ordinal notations made by collapsing recursively large ordinals, and large cardinal extensions of $\mathsf{ZFC}$ — or maybe not.
Yet as striking as this analogy seems, nobody seems to comment upon it as far as I know.  (At the very least, this seems pedagogically regrettable: I'm sure all of the above statements would be more memorable to students if the analogous statements were made explicit.)
So: is there some deeper truth to be found behind this parallelism?  (Or are all my sample facts just aspects of a single phenomenon?  Or is this just a red herring?)  Might it make sense to bring $\mathsf{ZFC}$ and $\mathsf{PA}$ under an umbrella metatheory so that the above facts can be proved in a common formalism?  At the very least, is there a textbook I missed where the analogy is played out in some detail?
And perhaps more thought provokingly: can one give an example of a completely different kind of theory that is just as similar to $\mathsf{ZFC}$ and $\mathsf{PA}$ as they are to each other?
(I'm of course aware that are also huge differences between $\mathsf{ZFC}$ and $\mathsf{PA}$; but I would tend to say that they make the similarities all the more striking.)

REPLY [27 votes]: A significant amount of the parallelism can be explained by the bi-interpretabiity of $PA$ with $ZF^{-\infty}$ ("finite set theory"), which is the theory obtained from $ZF$ by replacing the axiom of infinity by its negation, and adding the sentence asserting that every set has a transitive closure. For more detail and references, see my joint paper with Schmerl and Visser entitled $\omega$-Models of Finite Set Theory here.  
But that is only part of the story since the parallelism between arithmetic and set theory is deep and mysterious. I can even say that my career as a logician has been greatly shaped by comparing and contrasting the metamathematics of arithmetic and set theory.  
My 1998-paper Analogues of MacDowell-Specker Theorem for Set Theory, available here, gives a synopsis of the similarities and differences between $PA$ (equivalently: $ZF^{-\infty}$) and $ZF$ through the lens of model theory. 
Added in the third edit: Here is a relevant quote from the above paper:

The axiom of infinity is of course only the first step in the
  progression of ever bolder large cardinal axioms. As we shall see
  below, the negation of the axiom of infinity endows $ZF^{-\infty}$
  with a model theoretic behavior that $ZF$ can only imitate with the
  help of additional axioms asserting the existence of large cardinals.
  This is partially explainable by noting that the negation of the axiom
  of infinity in finite set theory itself can be viewed as a large
  cardinal axiom, not positing the existence of a large set - indeed
  denying it - but attributing a large cardinal character to the
  universe itself. Of course the axioms of power set and replacement
  give a ”strong inaccessibility” character to the class of ordinals
  which allows models of $ZF$ to share some of the model theoretic
  properties of $ZF^{-\infty}$.

Added in the second edit: In this 2009-presentation I describe a scheme $\Lambda$ (named in honor of Azriel Levy) consisting of set-theoretic sentences of the form "there is an $n$-reflective, $n$-Mahlo cardinal" which has the surprising property that:


$ZFC + \Lambda$  is the weakest extension of $ZFC$ whose model-theoretic behavior matches that of $PA$, in several surprising respects. So $PA$ is used here as a guide to find an improvement of $ZFC$.


Most, but not all of the results in the presentation have appeared in print.<|endoftext|>
TITLE: Does every generating set of the first homology group of a Cayley graph give rise to a presentation of its group?
QUESTION [8 upvotes]: Let $G$ be a group, and fix a symmetric generating set $S$. Let $X$ be the corresponding Cayley graph. 
Let $R$ be a set of words in $S$, each corresponding to the identity of $G$, such that the set of closed walks (some authors would write `closed paths') in $X$  induced by the elements of $R$ generates $H_1(X)$, i.e. the first simplicial homology group over $\mathbb Z$. Is it true that $<S \mid R>$ is a presentation of $G$?

REPLY [4 votes]: Write $G=F/N$, where $F$ is free over $S$. Let $C_F$ and $C_G$ be the Cayley graphs of $F$ and $G$: then $C_G=C_F/N$ and in particular the fundamental group of $C_G$ is naturally $N$ (note that we need the convention that if $1\in S$ then it yields a self-loop and if $S$ has elements of order 2 then they yield double edges; if this is not fine, we can assume that $S$ has no element of order $\le 2$). Also $H_1(C_G)$ is $N/[N,N]$.
The condition that $R\subset F_S$ generates $H_1(C_G)$ means that the subgroup $L$ generated by $R$ satisfies $L[N,N]=N$. The condition that $\langle S\mid R\rangle$ is a presentation means that the subgroup of $F_S$ normally generated by $L$ is equal to $N$.
Assume that $M\subset F_S$ is a normal subgroup and is contained in $F_N$, and that $N/M$ is a perfect group. Then the subgroup normally generated by $M$ is just $M$ itself. The condition that $N/M$ is perfect means that $[N,N]M=N$. So to show that the answer is no, it is enough to pick $R=M$ and assume $N/M\neq 1$.
So the recipe is just to pick $K$ a nontrivial perfect finitely generated group, $S$ a finite symmetric generating subset (say, with no element of order $\le 2$), $G$ a cyclic group (which can be arranged to be of order 5, if $S$ has 4 elements) with an injective map $i:S\to G$ with generating image (again denoted by $S$, say with no element of order $\le 2$); let $\bar{S}\subset S\times G$ be its graph: this is a generating subset. So we pick $F_S/M$ to be the direct product $S\times G$ (mapping $S$ to $\bar{S}$ in the canonical way) and $F_S/N$ to be $G$ (mapping $S$ to $G$ using $i$).
I leave the explicit computations, which seem doable if one starts with a presentation of a nontrivial perfect group (they yield $R$ infinite but some finite subset is enough when $G$ is finite).<|endoftext|>
TITLE: On (a generalization of) the Gauss Circle Problem
QUESTION [16 upvotes]: Most (if not all) references I read about the Gauss Circle Problem that proves a bound below $O(R^{2/3})$ reduces the GCP to the Dirichlet Divisor Problem by the well known expression of $r_2(n)$, the number of ways of writing a natural number $n$ as the sum of two squares. My question is then, what happens if the circle is not centered at the origin (or any lattice point)? In this case it seems that there is no number theoretical exact formula to reduce one problem to the other, and we have to tackle the GCP directly. What are the recent results in this direction?
Addendum: I vaguely recall reading somewhere about the same problem with the circle replaced by a uniformly convex planar domain, in which case the exponent is not as good, but still better than $O(R^{2/3})$. But now I'm unable to find any reference to it, though, so I added a "reference request" tag.

REPLY [15 votes]: You can write a similar expression to the usual formula for the remainder term in the circle problem by using the Poisson summation formula.  The shift of the center of the circle simply means that one gets a variant of the usual formula for the remainder term modified by suitable exponential terms. 
Indeed Huxley has considered a more general problem and obtained analogs of what was known for the usual divisor problem.  Huxley considers a closed convex curve $C$ enclosing an area $A$, and the dilate $MC$ of $C$ by a factor $M$.  Place this dilate in any way you like (translation or rotation) on the coordinate plane, and count the lattice points inside it.  Then under suitable regularity assumptions on the boundary curve $C$, Huxley obtains estimates for the difference between the number of lattice points and the expected number $AM^2$ where $A$ is the area enclosed by $C$.  His bounds depend on the original shape $C$, but not on the embedding of $MC$ in the plane.  The results are in three papers by him Exponential Sums and Lattice points, I, 2, and 3 (developing a method of Bombieri, Iwaniec and Mozzochi) and also mentioned in a recent survey, see Huxley, which has further references.  The strongest result for the translated circle of radius $R$ has an error term of $R^{0.6298\ldots}$ (see page 593 of the third paper).<|endoftext|>
TITLE: Failure of GCH at a strongly compact cardinal
QUESTION [6 upvotes]: Does Con(ZFC+ there exists a strongly compact cardinal) imply Con(ZFC+ there exists a strongly compact cardinal $\kappa+ 2^\kappa > \kappa^+$)?

REPLY [4 votes]: The answer to the question is yes, as it is proved by Usuba in the paper
Strongly compact cardinals and the continuum function
.
In fact Usuba proves the following more general result:
Theorem. Assume $\kappa$ is a strongly compact cardinal. Then there exists a generic extension in which $\kappa$ remains strongly compact and indestructible under adding any number of new subsets to $\kappa$ (Theorem 3.1 of the above paper).<|endoftext|>
TITLE: Self-dual plane curves
QUESTION [10 upvotes]: Suppose that $C\subset \mathbb P^2$ is a plane projective curve (base field is $\mathbb C$) and $C^*\subset (\mathbb P^2)^*$ is its dual. What are the known examples in which $C$ is projectively (i.e., linearly) isomorphic to $C^*$? Besides the non-degenerate conic, I am aware of the cuspidal cubic. Anything else?
Thank you in advance,
Serge

REPLY [12 votes]: There are also the limaçons, which are rational curves of degree 4 with one node and two cusps:  There is a $1$-parameter family of these up to projective equivalence (the parameter is $a\not= \pm1,\pm2$):
$$
[x_0,x_1,x_2] 
= \bigl[2t^2{+}at^2(1{-}t^2),\ 2t^2+a(1{-}t^2),\ (1{+}a)t+(1{-}a)t^3\bigr].
$$
See L. E. Wear, Self-dual plane curves of the fourth order, American Journal of Mathematics 42 (1920), 97–118.  Up to projective equivalence, these are the only irreducible self-dual curves of degree $4$ having only traditional singularities (i.e., nodes and cusps).
Remark 1: Also, regarding the question of whether there are self-dual curves of positive genus: The quintic
$$
8\,(x^5-10x^3y^2+5xy^4)-15(x^2{+}y^2)^2+10(x^2{+}y^2)-3 = 0
$$
has genus $1$ and is self-dual.  It has 5 cusps, spaced evenly around the unit circle $x^2+y^2=1$ and including $(1,0)$, and no other singularities.  (The Plücker formulae imply that an irreducible self-dual curve of degree $5$ and genus $1$ and whose only singularities are nodes and cusps must have $0$ nodes and $5$ cusps. By Bezout, no three of the cusps can lie on a line, and hence the $5$ cusps must lie on a nonsingular conic.  If one assumes that the conic is $x^2+y^2=1$ and that the cusps are equally spaced around the unit circle, one easily finds that the above example is the only possibility.  Fortunately, it works out to be self-dual.)
Remark 2:  Here is another interesting family of self-dual rational curves: $C_{n,m}$ given by the parametrization $f_{n,m}:\mathbb{P}^1\to \mathbb{P}^2$, where
$$
f_{n,m}\bigl([1,t]\bigr) 
= \bigl[\bigl(t^n+(n{-}2m)^2\bigr),\ \bigl(t^n+(n{+}2m)(n{-}2m)\bigr)t^m,\ \bigl(t^n+(n{+}2m)^2\bigr)t^{2m}\bigr].
$$
For relatively prime positive integers $m$ and $n$, the curve $C_{n,m}$ is of degree $d=n{+}2m$ (except for the case $(n,m)=(2,1)$, when the parametrization degenerates to a conic). The curves $C_{n,1}$ (for $n\not=2$) have only traditional singularities, with $\kappa = n$ cusps (where $t^n = n^2-4 $) and $\delta = \tfrac12 n(n{-}1)$ nodes.  An interesting thing about these curves is that, for fixed $n$, the automorphism $\psi_n:\mathbb{P}^1\to\mathbb{P}^1$ of the curve $C_{n,m}$ is given by $\psi_n\bigl([1,t]\bigr) = \bigl([1,\epsilon t]\bigr)$, where $\epsilon^n=-1$ and the isomorphism $\phi_{n,m}:\mathbb{P}^2\to(\mathbb{P}^2)^*$ that defines the self-duality via $f_{n,m}^{(1)}\circ\psi_n = \phi_{n,m}\circ f_{n,m}$ depends only on the congruence class of $m$ modulo $2n$, because the tangential mapping is given by
$$
f^{(1)}_{n,m}\bigl([1,t]\bigr) 
= \left[\begin{matrix}
\bigl(-t^n+(n{+}2m)^2\bigr)t^{2m}\\ 
-2\bigl(-t^n+(n{+}2m)(n{-}2m)\bigr)t^m\\ 
\bigl(-t^n+(n{-}2m)^2\bigr)
\end{matrix}\right].
$$
Thus, this provides examples of countable families of self-dual rational curves of arbitrarily high degree that share the same automorphism $\psi_n$ and correlation mapping $\phi_{n,m}$. In particular, the curves in such a family cannot all be solutions of a nondegenerate first order differential equation.<|endoftext|>
TITLE: Axiom of choice as zero dimensionality
QUESTION [17 upvotes]: In the paper Quantifiers and Sheaves by Lawvere, at the bottom of the second page, the author writes:

"... the condition that every epi splits, which geometrically we would call 0-dimensionality and logically we would call the axiom of choice."

What is meant by this geometric interpretation of choice as zero-dimensionality? I'm guessing it has something to do with fibers but can't figure it out myself.

REPLY [23 votes]: Here are two possible motivating examples. 
First, for any topos $\mathcal{E}$, if all epimorphisms in $\mathcal{E}$ split, then $\mathcal{E}$ is a boolean topos. In particular, for a topological space $X$, if all epimorphisms in $\mathbf{Sh} (X)$ split, then every open subset of $X$ is also closed (and vice versa), so $X$ is a disjoint union of indiscrete spaces. The converse is straightforward. Thus, all epimorphisms in $\mathbf{Sh} (X)$ split if and only if $X$ is a disjoint union of indiscrete spaces.
Second, for a commutative ring $A$, if all epimorphisms in $\mathbf{Mod} (A)$ split, then every closed subset of $\operatorname{Spec} A$ is also open (and vice versa), so the Krull dimension of $A$ is zero, therefore $\operatorname{Spec} A$ is finite and discrete, and hence $A$ is a direct product of finitely many fields. Conversely, if $A$ is a direct product of finitely many fields then all epimorphisms in $\mathbf{Mod} (A)$ split.
I suppose the point is that if the category of representations of some geometric object satisfies the axiom of choice, then that object cannot have a very complicated internal structure.<|endoftext|>
TITLE: What is the relationship between Turing Machines and Gödel's Incompleteness Theorem?
QUESTION [21 upvotes]: In this article, Scott Aaronson talks about using Turing Machines for proving the Rosser Theorem.
What is the relationship between the numbering that Gödel used in his proof of incompleteness and Turing Machines?

REPLY [22 votes]: It's simple. If the halting problem is undecidable, then PA is not complete, since otherwise, you could solve the halting problem by searching for proofs in PA. And the same argument works for any sound computably axiomatizable theory $T$ able to express arithmetic. Given a Turing machine $M$ on input $i$, you formulate the assertion $\sigma$ asserting that $M$ halts on $i$, and then search for a proof in $T$ of $\sigma$ or a proof of $\neg\sigma$. If your theory were complete, then you'll find one or the other, and this would solve the halting problem. Since the halting problem is not solvable, there must be sentences of this form that are not settled by the theory. 
This argument proves the first incompleteness theorem as an elementary consequence of the halting problem. The second incompleteness theorem takes a bit more work. 
There is more discussion on Are the two meanings of undecidability related? and How undecidable is the spectral gap problem?<|endoftext|>
TITLE: Jets in synthetic differential geometry
QUESTION [7 upvotes]: As I understand it from Kostecki's notes, the $k$-jet $j^k f$ of a function $f: R^n \to R^m$ should be the map $$f^{D_k(n)} : {(R^n)}^{D_k(n)} \to (R^m)^{D_k(n)},$$
where $$D_k(n) = \{(x_1, \ldots, x_n)\in R^n \mid x_{i_1}\cdot\ldots \cdot x_{i_{k+1}}=0 ~\text{for all $k$-tuples}~ (i_1, \ldots, i_{k+1})\}$$
Thus we have that $$J^k(R^n, R^m) := \hom \left((R^n)^{D_k(n)}, (R^m)^{D_k(n)} \right),$$
where the hom here is the internal one. But can we define $J^k(M,N)$ for arbitrary objects $M$ and $N$, which are not necessarily manifolds?

REPLY [8 votes]: It's correct that a jet $u\in J^k(M,N)$ can be defined as the equivalence class of maps $f:M\to N$ whose $k$th Taylor expansions agree at a point $x\in M$. A more "synthetic" way to think of $u$ is as a map from the $k$th infinitesimal nbhd of $x\in M$ to $N$. That's the approach Kock takes in his book Synthetic Geometry of Manifolds (section 2.7). 
The theme of the book is in fact the $k$th neighbourhood relation which Kock considers mainly for smooth finite dimensional manifolds. He partly addresses the question of how to generalise the infinitesimal nbhd relation to arbitrary objects (p.41). The issue there is, that there are at least two generalisations.
A more general approach to treat infinitesimal neighbourhoods might be using Urs Schreibers differential cohesion.
Concerning your definition in the question: notice that a jet (as defined above) from $R^n$ to $R^m$ can be composed on the left with a map from $D_k(n)$ to give rise to the objects you consider. But your objects are more general since they don't have to be invariant under the action of the automorphism group of $D_k(n)$, while jets will be.<|endoftext|>
TITLE: Constructing an independent uniform random variable from two independent ones
QUESTION [10 upvotes]: Does there exist a continuous (differentiable) function $h:[0,1]\times [0,1] \to [0,1]$ such that if $\alpha,\beta\in [0,1]$ are independent and uniformly distributed on $[0,1]$, the random variable $h(\alpha,\beta)$ is uniformly distributed on $[0,1]$ independent of $\alpha,\beta$?
Clarification: By independent I mean pairwise independent, i.e.
$\mathbb{P}[h(\alpha,\beta)\leq x\mid \alpha]=x$ for all $x,\alpha\in[0,1]$ 
and
$\mathbb{P}[h(\alpha,\beta)\leq x\mid \beta]=x$ for all $x,\beta\in[0,1]$..
Thanks a lot!

REPLY [3 votes]: Inspired by Nate's answer, here is why it cannot be differentiable. First for each $y \in [0,1]$, there must be some $x \in [0,1]$ such that $h(x,y) = 0$, otherwise by compactness of the unit interval and continuity of $h$, we would have $P(h(\alpha, y) < \epsilon) = 0$ for some small $\epsilon$.
Next I claim that such $x$ cannot be in the interior of $[0,1]$. If it is, then by linear approximation near $x$ and the limit definition of derivative, $P(h(\alpha, y) < \epsilon) > \epsilon$ for sufficiently small $\epsilon$, because the derivative $\partial_1 h(x,y)$ must be zero. This is clearly enough to get a contradiction: it means that $h(0,y) = 0$ or $h(1,y)=0$ for all $y \in [0,1]$. Reversing the role of $x$ and $y$, we see that $P(h(0, \beta) < \epsilon) \geq 1/2$ or $P(h(1, \beta) < \epsilon) \geq 1/2$. Since $h$ is continuous, this argument also works in the almost sure category.<|endoftext|>
TITLE: Asymptotics of a recursion
QUESTION [7 upvotes]: suppose we have the following two sequences
$$\alpha_k = (k-1)\left(1-\frac {1}{1+(k+1)l}\right)  \quad , k \geq 2$$
$$\beta_k  = (k-1)\left(1+\frac {1}{1+(k-1)l}\right)  \quad , k \geq 2$$
where $l$ is a positive constant
and define the sequence $c_k$ recursively by:
$$c_2 = - 1/\beta_2 $$
$$c_3 = 0 $$
$$c_{k+1} = \frac{\alpha_{k-1}}{\beta_{k+1}}c_{k-1}    \quad , k \geq 3        $$
it is not hard to see that this would give 
$$c_2 = - 1/\beta_2$$
$$c_{2k} = -\frac{\alpha_2}{\beta_2  }\cdot\frac{\alpha_4}{\beta_4  }\cdot\cdot\cdot \frac{\alpha_{2k-2}}{\beta_{2k-2}  }\cdot \frac{1}
{\beta_{2k}}  \quad , k \geq 2$$
$$c_{2k+1} = 0  \quad , k \geq 1 $$
apparently we must have $d_k = c_{2k} \sim k^{-(1+1/l)}$ but I have no idea how to show this. can anyone shed some light on this?
also a similar question for the recursion
$$c_2 = - 1/\beta_2 $$
$$c_3 = \frac{\gamma_2}{\beta_3}c_2  $$
$$c_{k+1} = \frac{\alpha_{k-1}}{\beta_{k+1}}c_{k-1}  + \frac{\gamma_k}{\beta_{k+1}}c_k       \quad , k \geq 3     $$
where 
$$\gamma_k = \sigma  \frac{l(k^3-k)}{1+kl} \quad , k \geq 2$$
$\sigma$ being also a positive constant 
How would the asymptotics look like in this case?

REPLY [6 votes]: For the first sequence, $$d_k=\frac{\alpha_{2k-2}}{\beta_{2k}}d_{k-1}=\frac{k-3/2}{k+1/2-1/\ell}d_{k-1}$$
so one has 
$$d_k=C\frac{\Gamma(k-1/2)}{\Gamma(k+3/2-1/\ell)}\ ,$$
the constant $C$ being determined by the initial condition $d_1$, namely
$$C=d_1\frac{\Gamma(5/2- 1/\ell)}{\Gamma(1/2)}\ . $$ Recall that $\Gamma(x+a)=x^a\Gamma(x)(1+o(1))$ as $x\to+\infty$, so
$$d_k=Ck^{-1-1/\ell}(1+o(1)). $$
For the second sequence, $$c_{k+1}=\frac{(k-1)(k+1)}{k+2/\ell }c_k+\frac{k-2}{k+2/\ell}c_{k-1}$$
which implies $c_{k-1}/c_k=O(1/k)$; if we plug this in the recursion again we have
$$c_{k+1}=\frac{(k-1)(k+1)}{k+2/\ell }(1+O(1/k^2))\ , $$
whence $$c_k= A\frac{\Gamma(k+1)\Gamma(k-1)}{\Gamma(k+2/\ell) }(1+o(1))\ ,$$  because the infinite product of $1+O(1/k^2)$ is convergent. By the Stirling formula  $$c_k= B k^{k-1/2+2/\ell}e^{-k} (1+o(1))$$ for a certain constant $B$.<|endoftext|>
TITLE: How can you compute the maximum volume of an envelope(used to enclose a letter)?
QUESTION [22 upvotes]: It's obvious that the volume of a envelope is 0 when flat and non-0 when you open it up. However, if you were to fill it with liquid, there must be some shape where it has a maximum volume. Is there a practical way to determine that volume?
I think a starting point would be to start with a cylinder and seal one end by putting it in something like a vice. That gives you the same shape.
EDIT: The reason this problem came to mind is because Quaker's oatmeal packets have a fill line on them so that you don't need a measuring cup. Every day, as I fill up the little packet with water, I think "what a clever idea," but also I wonder about the math behind it. I expect they approximate because, after all, it is just oatmeal.

REPLY [5 votes]: As the envelope is made of paper, a mathematical model of its deformation would be an isometric coordinate transformation and thus must have zero Gauss curvature almost everywhere; that restriction is clearly violated by the cited solution of the teabag problem.  
I already pointed out in a comment, that there was an article in the German issue (Spektrum der Wissenschaft, June 1995) of Scientific American, where the topic has been discussed here (in German).
Maybe the same topic has also been treated a few month earlier in Scientific American.
The essential ressource is Edouard Baumann, especially this link (unfortunately in German only); there is also an image of such a polyhedral cushion.<|endoftext|>
TITLE: What is the first cohomology $H_{fppf}^{1}(X, \alpha_{p})$?
QUESTION [12 upvotes]: Let $X$ be a smooth projective curve of genus $g>1$ over an algebraically closed field $k$ of characteristic $p>0$. Let $\alpha_{p}$ be the group scheme of the kernel of $F: \mathbb{G}_{a} \rightarrow \mathbb{G}_{a}$, where $F$ denotes the Frobenius. It is well-know that $H_{fppf}^{1}(X, \alpha_{p})$ is the kernel of Cartier operator on $H^{0}(X,\Omega^{1}_{X,d=0})$. 
Is $H_{fppf}^{1}(X, \alpha_{p})$ a finite group? If not, what is the dim$_{k}H_{fppf}^{1}(X, \alpha_{p})$? What is the relationship between $H_{fppf}^{1}(X, \alpha_{p})$ and the $p$-rank of $X$?

REPLY [17 votes]: Lemma. Let $k$ be a perfect field of characteristic $p > 0$, and let $X$ be a reduced $k$-variety. Then $H^i(X_{\operatorname{fppf}},\alpha_p)$ is a $k$-vector space. In particular, it can only be finite if it is $0$ or if $k$ is finite.

Proof. Consider the short exact sequence
\begin{equation}
0\to \alpha_p \to \mathbb G_a \stackrel{(-)^p}\to \mathbb G_a \to 0\tag{1}
\end{equation}
on $X_{\operatorname{fppf}}$. Let $f \colon X_{\operatorname{fppf}} \to X_{\operatorname{\acute et}}$ be the natural map. Then $R^if_* \mathbb G_a = 0$ for $i > 0$, since $\mathbb G_a$ is a smooth group scheme (see for example Milne's Étale Cohomology book, Theorem III.3.9). It follows that
$$R^if_* \alpha_p = 0$$
for $i \geq 2$. On the other hand, $f_* \alpha_p = 0$, since $X$ is reduced. Hence, the Leray spectral sequence
$$E_2^{p,q} = H^p(X_{\operatorname{\acute et}},R^qf_*(\alpha_p)) \Rightarrow H^{p+q}(X_{\operatorname{fppf}},\alpha_p)$$
collapses on the $E_2$ page, and we find
$$H^i(X_{\operatorname{fppf}},\alpha_p) = H^{i-1}(X_{\operatorname{\acute et}},R^1f_*\alpha_p).$$
But $R^1f_*\alpha_p$ sits in a short exact sequence
\begin{equation}
0 \to \mathcal O_X \stackrel{(-)^p}\to \mathcal O_X \to R^1f_*\alpha_p \to 0.\tag{2}
\end{equation}
If $F \colon X \to X$ denotes the absolute Frobenius, then $F$ is a homeomorphism on the underlying spaces. Moreover, $F_*\mathcal O_X \cong \mathcal O_X$ as abelian sheaves (even on $X_{\operatorname{\acute et}}$) (but not as sheaves of $\mathcal O_X$-modules!). Thus, we can rewrite (2) as
$$0 \to \mathcal O_X \stackrel {F^\#}\to F_*\mathcal O_X \to R^1f_* \alpha_p \to 0,$$
where the map $F^\# \colon \mathcal O_X \to F_* \mathcal O_X$ is just the map of rings corresponding to $F \colon X \to X$. This is an $\mathcal O_X$-linear map, so $R^1f_* \alpha_p$ is a coherent sheaf of $\mathcal O_X$-modules, thus its cohomology is a $k$-vector space. $\square$
Remark. Some texts use the relative Frobenius instead, but since $k$ is perfect I think it is harmless to identify $X$ with $X^{[p]}$, the pullback along the geometric Frobenius on $\operatorname{Spec} k$.
Remark. If $X$ is projective, then the same argument shows that $H^1(X,\alpha_p)$ is a finite-dimensional $k$-vector space.
If $X$ is projective and geometrically integral, then $H^0(X,\mathcal O_X) = k$. Since $k$ is perfect, (1) now gives an inclusion $H^1(X_{\operatorname{fppf}},\alpha_p) \subseteq H^1(X,\mathcal O_X)$.
On the other hand, if $X = \mathbb A^1$, then we get
$$H^1(X,\alpha_p) = k[x]/k[x]^p \cong \bigoplus\limits_{\substack{m \geq 0\\p \nmid m}} k \cdot x^m,$$
which is infinite-dimensional.
Remark. On the other hand, the short exact sequence on the flat site
$$0 \to \alpha_{p^n} \to \mathcal O_X \stackrel{F^n}\to \mathcal O_X \to 0$$
realises $H^1_{\operatorname{fppf}}(X, \alpha_{p^n})$ as the kernel of the map $H^1(X, \mathcal O_X) \to H^1(X, \mathcal O_X)$ induced by the $n$-th power of Frobenius.
We can think of $H^1(X,\mathcal O_X)$ as the $p$-Lie algebra of the Picard scheme, and I believe that the Frobenius on this space corresponds to the $p$-th power operation of the $p$-Lie algebra (Mumford's Abelian Varieties proves this when $X$ is an abelian variety [Thm 15.3], and I think the case for curves should follow by embedding into the Jacobian).
Relation with the $p$-rank. In [loc. cit.], Mumford shows that the $p$-rank $r$ is the dimension of the semisimple part of $H^1(X,\mathcal O_X)$ with respect to the Frobenius, and $g - r$ is the nilpotent part. It seems that therefore $g - r$ must equal the dimension of $H^1_{\operatorname{fppf}}(X, \alpha_{p^n})$ as $n \to \infty$; note that this is bounded since all are contained in $H^1(X, \mathcal O_X$).
Remark. A somewhat weird way of looking at $\alpha_{p^n}$: as above, if $f \colon X_{\operatorname{fppf}} \to X_{\operatorname{\acute et}}$ is the natural 'inclusion', then
$$R^if_* \alpha_{p^n} = \left\{\begin{array}{ll} \operatorname{coker}(\mathcal O_X \stackrel{F^n}\to (F_X^n)_* \mathcal O_X), & i = 1,\\ 0, & i \neq 1. \end{array}\right.$$
Here, the map $F^n$ is $\mathcal O_X$-linear. Thus, $R^1f_* \alpha_{p^n}$ is a coherent sheaf $\mathscr F$ of $\mathcal O_X$-modules, which can easily be seen to be locally free of rank $p^{n \dim X} - 1$ (if $X$ is smooth). Thus, we can think of (the sheaf) $\alpha_{p^n}$ as being like $\mathscr F[-1]$, at least in the sense that
$$H^i_{\operatorname{fppf}}(X, \alpha_{p^n}) = H^{i-1}(X, \mathscr F) =  H^i(X, \mathscr F[-1]),$$
where the right hand side can be taken with respect to your favourite topology (Zariski, étale, fppf, etc.), since $\mathscr F$ is coherent.
However, $\alpha_{p^n}$ and $\mathscr F[-1]$ are not actually isomorphic in the derived category (consider sections on an infinitesimal thickening of $X$), but they become isomorphic after pushing forward to the étale site.
Remark. Happy new year!<|endoftext|>
TITLE: Ill-founded models of set theory with well-founded ordinals
QUESTION [5 upvotes]: Let $(\mathcal{M},E)$ be an internally non-well-founded model of set theory i.e of $ZFC^{\neg f}=ZFC\setminus \mathrm{foundation}+\neg \mathrm{foundation}$, then $\mathcal{M}$ includes an infinite decreasing $E$-sequence. I am interested to know about the degree of illness of internally non-well-founded models in the literature. For example we see there are many models of $ZF$ satisfying various versions of $AC$ that are really distinct. My question is:

$*$) What is the difference between internally non-well-founded models, in the sense  of axiom of foundation? I mean how the existence of different decreasing sequences in different models effect their universes. Does an especial sequence capture some interesting properties in that model, in which not satisfying necessarily by all internally non-well-founded models?

I am also interested to find the answer of the following question.

$\bigstar$) Is any of the following statements true?
$(\rm{I})~~~~$ Working in $V$, for any infinite ordinal $\beta$, there exists a model $(\mathcal{M},E)$ of $ZFC^{\neg f}$ with $Ord(\mathcal{M})=\beta$ such that $\mathcal{M}$ contains a decreasing $E$-sequence of length $\beta$.
$(\rm{II})~~~$ Working in $V$, for any infinite ordinal $\beta$, there exists a model $(\mathcal{M},E)$ of $ZFC^{\neg f}$ with $Ord({\mathcal{M}})=\beta$ such that for any $\alpha<\beta$, $\mathcal{M}$ has a decreasing $E$-sequence of length $\alpha$.

clearly $\rm{I}\longrightarrow\rm{II}$.
Edit: I have been thought that the concept of an ill-foundeded model and aninternally ill-foundeded model are the same, but Prof. Enayat and William informed me about the difference, in the following comments. The question $(\bigstar)$  answered by Prof. Enayat stems from the question $*$. I thought maybe $\bigstar$ shows me some different pictures about non-well-founded models.

REPLY [9 votes]: (I) is true (and therefore so is (II)), assuming that $ZF$ has a well-founded model $M_\beta$ of ordinal height $\beta$.  I will outline a construction that is meant to be carried out in within a model of $ZF$. It will produce the desired ill-founded model satisfying (II) when implemented within $M_\beta$. 

Given an extensional digraph $G=(X,E)$, with $X$ as the vertex set and $E$ as the edge set, define the deficiency set $D(G)$ of $G$ to be the collection of subsets $S$ of $X$ that are not "coded" in $G$, i.e., there is no element $a$ in $X$ such that $S$ = {$x \in X : xEa$}. 
In the above paragraph "digraph" stands for directed graph, and "extensional" means that if $a$ and $b$ are distinct vertices of $X$, then there is some $c\in X$ such that $cEa \leftrightarrow cEb$ does not hold. Also, $G$ is allowed to be a class.
We now can define by recursion a digraph $G_\alpha = (X_\alpha, E_\alpha)$ for each ordinal $\alpha$ as follows:
$G_0 = G$; 
$G_{\alpha+1} = (X_{\alpha+1}, E_{\alpha+1})$, where $X_{\alpha+1} = X_{\alpha} \cup D(G_{\alpha})$, and $E_{\alpha+1} = E_{\alpha}$ together with edges of the form $(x,X)$, where $x\in X_{\alpha}$, $X \in D(G_{\alpha})$, and $x\in X$.
For limit $\alpha$, $G_\alpha$ is the union of $G_\beta$ for $\beta<\alpha$.
The model/digraph $M$ we are interested in is the union of all the $G_\alpha$s, as $\alpha$ ranges over the ordinals, and $G$ is isomorphic to the linearly ordered set obtained by reversing the ordering on the class of ordinals. Here the vertex set $X$ of $G$  should be chosen so that no vertex is a set is a set of vertices, or a set of set of vertices, etc.
Intuitively speaking, the model $M$ can be thought of as the set-theoretic completion of $G$, so it should be dubbed the von Neumann completion of $G$. With the above choice of $G$, this is a class model of ZF without foundation whose class of ordinals is isomorphic to the "real class of ordinals" and which includes a decreasing chain of sets of length $Ord$.
N.B. Models of $ZF$ in which the foundation axiom fails are often constructed using the so-called Bernays-Rieger permutation method (not to be confused with the Fraenkel-Mostowski permutation method of constructing models of $ZF$ in which the axiom of choice fails). The model constructed above is based on a different idea, explored in detail for models of finite set theory in the following paper (This sort of model was also used in this answer of mine).
A. Enayat, J. Schmerl, and A. Visser, Omega Models of Finite Set Theory in Set theory, Arithmetic, and Foundations of Mathematics: Theorems, Philosophies (edited by J. Kennedy and R. Kossak), Cambridge University Press, 2011.
A preprint can be found here.<|endoftext|>
TITLE: Is this graph 3-colorable?
QUESTION [6 upvotes]: Consider the permutations of $0,1,1,2,2,3,3.$ Each permutation is corresponding to a vertex in graph $G$. So, the graph $G$ has $630$ vertices.
Each vertex has exactly 6 neighbors. $P$ is connected $Q$ if $P$ can be obtained from $Q$ by swapping 0 with another element. For example,
0112233 is connected to 1012233, 1102233, 2110233, 2112033, 3112203, 3112230.
Question: What is the chromatic number of graph $G?$ Is $G$ 3-colorable?

What we've proved: $G$ is not a perfect graph.
It has many odd holes with length $\geq 11$.

REPLY [12 votes]: If I constructed the graph correctly, according to a program the chromatic
number is $4$, so the graph is not 3 colorable.
The program is:
https://code.google.com/p/graphcol/
Got the same result after converting the problem to SAT
and ran certified UNSAT solver.
The proof for unsatisfiability was only about 11MB.
The computation took few minutes and the 4-coloring was found
very fast.
The graph was constructed with sage program:
def graphperm123():
    S=Permutations([0,1,1,2,2,3,3])
    E=[]

    for u in S:
        u=list(u)
        i=u.index(0)
        for j in xrange(len(u)):
            if j==i:  continue
            v=u[:]
            a=v[j]
            v[j]=0
            v[i]=a
            E += [(tuple(u),tuple(v))]
    G=Graph(E,multiedges=False,loops=False)
    return G<|endoftext|>
TITLE: Can an oriented closed $n(\geq 2)$-dimensional manifold be smoothly embedded in $\mathbb{R}^{2n-1}$?
QUESTION [17 upvotes]: Can anyone provide me with an example of an orientable closed manifold $M$ of dimension $n\geq 2$, which cannot be smoothly embedded in $\mathbb R^{2n-1}$?
I know these cannot exist for $n=1$, i.e. $S^1$. If we ignore orientability, then if we take $n=2^r$, $\mathbb RP^n$ cannot be embedded in $\mathbb R^{2n-1}$.
While searching on the internet, I found some sharper results here.
So can we say anything when $\dim(M)= 2^r$ and $M$ is orientable?
It would be very helpful if someone could provide me with some more references. Thanks in advance.

REPLY [3 votes]: For dim $n=4$ it has proven recently by Ghanwat and Pancholi (https://arxiv.org/pdf/2002.11299.pdf) that every closed oriented 4-manifolds smoothly embeds in $\mathbb R^7$.
The (beautiful) key idea of their proof is if we have a closed oriented smooth 4-manifold $M$ such that there exists two smoothly embedded  2-spheres $S^2_a, S^2_b$ that transversally intersect at one point and  represent non-trivial element in $H_2(M)/Tor$, then any smooth closed oriented 4-manifold $X$ smoothly embed in $M\times CP^1$. This telling us in particular $X$ embeds smoothly in $S^2\times S^2\times S^2= \partial (S^2\times S^2\times D^3)$ which embeds in $\mathbb R^7$.<|endoftext|>
TITLE: Directed homotopy in the Cayley graph of a monoid
QUESTION [5 upvotes]: There is a the notion of the Cayley graph $C(G)$ of a group $G$ (which depends on a given presentation $G \cong \mathcal F(S) / \sigma$ where $\mathcal F$ is the free group functor and $\sigma$ some congruence). This graph has several nice properties like that the fundamental group $\pi_1(C(G), e)$ ($e$ the identity of $G$) is the normal subgroup of $\mathcal F(S)$ corresponding to $\sigma$, or the subgroup of zero-sum reduced words with entries in the generators $S$ and their inverses.
One can easily imagine the analagous construction for monoids. A monoid presentation $M=\mathcal F(S) / \sigma$ defines a Cayley graph $C(M)$. The fundamental monoid $\pi_1(C(M),e)$ is the monoid of zero-sum sequences of generators $S$. Also, considering the free category generated by this graph $F(C(M))$, we have $\mathsf{Hom}_{F(C(M))}(e,x) \cong \mathsf Z_M(x)$ where $\mathsf Z_M(x)$ is the factorization set of $x$ with respect to the generators $S$. (Because these are just sets, this is just a structure-less bijection, but they both correspond to strings with entries in the generators which get evaluate to $x$ under the operation of $M$.)
Has this connection been explored? I don't think I saw it mentioned in "Directed Algebraic Topology" (but I haven't done more than skim it).
edit: I have found a variant of the basic construction I had in mind by N.D. Gilbert (Monoid presentations and associated groupoids), but this construction is nondirected and so only describes non-directed homotopy (fundamental group $\cong$ Grothendieck group of $M$, rather than fundamental monoid $\cong M$).

REPLY [5 votes]: You might want to check out this paper of Guba and Sapir on directed homotopy and Cayley complexes for monoids. http://www.math.vanderbilt.edu/~msapir/ftp/pub/2comp/dc.pdf
There are also papers by Mark Kambites studying the set of words reading a loop at some vertex of a monoid Cayley graph. 
Bob Gray and Mark Kambites have a theory of quasi-isometry of semimetric spaces that captures some of the directed geometric structure of a monoid Cayley graph. http://www.ams.org/journals/tran/2013-365-02/S0002-9947-2012-05868-5/S0002-9947-2012-05868-5.pdf<|endoftext|>
TITLE: How is Ricci flow related to computer graphics?
QUESTION [12 upvotes]: I recently came across the book Ricci Flow for Shape Analysis and Surface Registration: Theories, Algorithms and Applications by Wei Zeng and Xianfeng David Gu. Because, I just saw the book on the passing and then read a bit about it in Amazon, I still don't exactly know how Ricci flow is related to computer graphics and especially to 3D modelling in computer science. If someone can give me a comprehensive answer or point me to one, I would be glad.
On the other hand, I also would like to learn whether the Poincaré conjecture has some applications in computer graphics and 3D modelling. 
Lastly, I want to learn what are some mathematical research topics in these areas that have (or may have) direct applications in computer graphics and 3D modelling.

REPLY [11 votes]: Perhaps this—and its references both past & future
("cited by 152" subsequent papers)—will help...?

Jin, Miao, Junho Kim, Feng Luo, and Xianfeng Gu. "Discrete surface Ricci flow." IEEE Transactions on Visualization and Computer Graphics, no. 5 (2008): 1030-1043.
(DOI).

         


David (Xianfeng) Gu's work
was cited by Deane Yang
in the comments.
Here is one among the (many) later papers, coauthored by David Gu:

Wang, Yalin, Jie Shi, Xiaotian Yin, Xianfeng Gu, Tony F. Chan, Shing-Tung Yau, Arthur W. Toga, and Paul M. Thompson. "Brain surface conformal parameterization with the Ricci flow."
IEEE Transactions on Medical Imaging. 31, no. 2 (2012): 251-264.
(Journal link.)<|endoftext|>
TITLE: What sort of large cardinal can continuum be?
QUESTION [11 upvotes]: I have stumbled across a related question asking which large cardinal properties can hold for $\aleph_1$. This question is probably also related, asking in what ways $\aleph_0$ is a "large" cardinal.
To state my question:

For which large cardinal properties is it consistent with ZFC that $\frak{c}$, the cardinality of continuum, has this property? How does the answer change if we abandon choice?

Here are results I'm aware of:

$\frak{c}$ can be real-valued measurable (relative to existence of a measurable cardinal), which apparently implies it's weakly Mahlo. I believe all of these hold in ZFC.
Possibility of $\frak{c}$ being weakly inaccessible and weakly Mahlo (or, I believe, weakly 1-inaccessible or whatever intermediate condition we put on it) is also consistent relative to existence of respective cardinals, which can be established by forcing which doesn't disrupt in any way structure of ordinals, only the size of continuum.
Clearly $\frak{c}$ can't be strongly inaccessible or strongly Mahlo, because these directly require a cardinal to be strongly limit.
In ZFC, every measurable is strongly limit, so $\frak{c}$ can't be measurable. But it is also true in ZF, because we have $\aleph_0<\frak{c}$ and $2^{\aleph_0}\geq\frak{c}$, and standard proof that this can't happen for measurable goes through.
By same means, $\frak{c}$ can't be the critical point of any elementary embedding, so it can't land in any of the higher entries of large cardinals list.
Continuum can't be weakly compact since weak compactness implies strong limitness. I'm sure this is true in ZFC, but not sure about ZF.

What other properties can or can't continuum have? I believe axiom of choice disallows it to be in anywhere above weakly compact (correct me if I'm wrong), but I have hopes for ZF itself giving $\frak{c}$ more possibilites to be large.
Thank you in advance.

REPLY [7 votes]: Let me add a few examples:
(1) If we start with a supercompact cardinal $\kappa$, and force with $Add(\omega, \kappa)$, then in the extension the cardinal $\kappa=2^\omega$ becomes generically supercompact. The same holds for many other large cardinals.
(2) If we start with a weakly compact cardinal, we can find a generic extension in which $2^\omega=\kappa$ is (the least) weakly Mahlo, and tree property holds at $\kappa.$ This result is due to Boos ``Boolean extensions which efface the Mahlo property''.
(3) The consistency of the theory $ZFC$+ "there is a supercompact cardinal'' implies the consistency of the theory $ZFC$ + "there exist a uniform measure $μ$ on the cardinal $2^ω$ and a set $X⊆2^ω$ of positive $μ$-measure such that for every $y∈X$ there is a uniform measure on $y$ which is $|y|$-additive.''
(4) The consistency of the theory $ZFC$+ "there is a measure concentrating on compact cardinals'' implies the consistency of the theory $ZFC$ + "there exist a uniform measure $μ$ on the cardinal $2^ω$ and a set $X⊆2^ω$ of positive $μ$-measure such that for every $y∈X$ there is a uniform measure on $2^ω$ which is $|y|$-additive.''
For (3) and (4) see Some combinatorial properties of measures
(5) Under  $PFA, 2^\omega=\aleph_2$ has some large cardinal properties.<|endoftext|>
TITLE: Is $n=6$ the only integer satisfies ${\sigma}_x(n) \equiv 0\bmod{n}$ for every odd integer $x > 0$ and $2 (\bmod n)$ if $x$ is even integer?
QUESTION [7 upvotes]: After a few computations in wolfram alpha about the divisor function for some values of $n$ to look the behavior of $\sigma_x(n)\bmod n$ for $\,n=6,\,$ i got this result : $\sigma_x(6)=0 \bmod 6$ for $x$ odd and 2 mod 6 if $x$ is even 
Edit:01 :${\sigma}_x(n) =\sum_{d|n} d^x$ is the sum divisor function 
Note:01:I edited the question just to define the sum divisor function
My question here :
Is $n=6$ the only integer satisfies $\sigma_x(n)
\equiv 0\bmod n$ for every odd integer  $x > 0$  and $2 \bmod n$ if $x$ is even integer ? and if it is how do i show this ?
Note :02:I want to know more about periodicity of the divisor function 
Thank you for any help

REPLY [5 votes]: Let $r=\gcd(k,e+1)$, and $p$ a prime. Then $\sigma_k(p^e) \equiv r\frac{p^{e+1}-1}{p^r -1} \bmod \sigma(p^e)$. Also, $r=1$ if and only if $\sigma(p^e)$ divides $\sigma_k(p^e)$.  Thus for $k$ coprime to $\tau(n)$, we have $\sigma(n)$ divides $\sigma_k(n)$.  The relation also suggests that for a given $n$ the sequence $\sigma_k(n)\bmod \sigma(n)$ is periodic in $k$ with a period dividing $L$, the least common multiple of ($1+$ each exponent) in the prime factorization of $n$.  Edit 2016.01.04: Once can show a nonreduced representation $\sigma_k(n) = a_k\sigma(n)/b_k$ where the $b_k$ are integers not necessarily coprime to the integers $a_k$ or to $a_k\sigma(n)$, with the property that the $b_k$ are bounded and periodic with period $L$.  This is not enough to show $\sigma_k(n) \bmod \sigma(n)$ is periodic with small period, unfortunately. End Edit 2016.01.04.
If now $n$ is multiperfect (so $n$ divides $\sigma(n)$) we have $n$ divides $\sigma_k(n)$ for $k$ coprime to $\tau(n)$.  In particular if $\tau(n)$ is a power of $2$, then $n$ divides $\sigma_k(n)$ for all odd $k \gt 0$.
It is still possible that $n$ can divide $\sigma_k(n)$ for $k$ not coprime
to $\tau(n)$.  However if $L$ is not prime, it seems likely that there will be more than one nonzero value of $\sigma_k(n) \bmod \sigma(n)$.  If this is so, it would be one ingredient in a proof that 6 is the unique number having the titled properties, the other ingredient being that 6 is the only nontrivial multiperfect number with $L$ a prime.
Edit 2016.01.10: I botched an earlier edit which claimed that 6 is the only known multiperfect number $n$ which satisfies $\sigma_2(n) \bmod n = 2$.  It is true, but the analysis had some flaws.  However, one expects multiperfect numbers other than 1 and 6 to be a multiple of 4; when $n$ satisfies $\sigma(n) \bmod n = 0$ and $\sigma_2(n) \bmod n = 2$, and in addition $ n \bmod 4  = 0$, then all odd prime factors of $n$ except one must occur to an even multiplicity, and the remaining odd prime factor must occur to a multiplicity of 1 mod 4 and must be a prime that is 3 mod 4.  While simple, these observations say a lot about $n$ and suggest that any numbers satisfying the title congruences are rare indeed, perhaps more so than odd multiperfect numbers.  End Edit 2016.01.10
Gerhard "Mea Culpa, Mea Maxima Culpa" Paseman, 2016.01.03<|endoftext|>
TITLE: Law of unconsious statistician: application in characteristic function
QUESTION [6 upvotes]: Let $g(x)=(x-a)\mathbf 1_{x\ge a}$ for some $a>0$ and let $X$ be a non-negative random variable with cdf $F$ and $E[X]<+\infty$. I want to calculate $$\frac{d}{da}E[g(X)]$$ To do that I thought as a first step to use this formula here:

...Using representations as Riemann–Stieltjes integral and integration by parts the formula can be restated as
  $$\operatorname {E}[g(X)]=\int _{-\infty }^{\infty }g(x)\,\mathrm{d}P(X\le x)=g(a)+\int _{a}^{\infty }g'(x)P(X>x)\,\mathrm {d} x$$
  if $P\left(g(X)\geq g(a)\right)=1$.

So I calculated $$g'(x)=\begin{cases}0, & 0\le x< a \\ \text{undefined}, & x=a \\ 1, & a<x \end{cases}$$ and since $P(g(X)>g(a))=P(g(X)>0)=1$, I wrote: $$E[g(X)]=g(a)+\int_{a}^{+\infty}(1-F(x))\,\mathrm{d}x=\int_{a}^{+\infty}(1-F(x))\,\mathrm{d}x \tag{1}$$
My first question: Is this correct, or is $g'(x)|_{x=a}$ undefined a problem?
So, taking the derivative of $(1)$ gives me that $$\frac{d}{da}\int_{a}^{+\infty}1-F(x)\, \mathrm dx=F(a)-1 \tag{2}$$
My second question: Is this derivative correct? 
Thanks in advance!

REPLY [3 votes]: Formula (1) is correct for any random variable (r.v.) $X$ and any real $a$, even without assuming that $X$ is nonnegative and/or continuous. Indeed, by the Fubini--Tonelli theorem, 
$$\int_a^\infty (1-F(x))\,dx 
=\int_a^\infty P(X>x)\,dx =\int_a^\infty dx \int_{(x,\infty)}P(X\in du)
$$ 
$$=\int_{\mathbb R}dx\,I\{x\ge a\}\int_{\mathbb R}I\{u>x\} P(X\in du) $$ 
$$=\int_{\mathbb R}P(X\in du)\int_{\mathbb R}dx\,I\{u>x\ge a\} $$
$$=\int_{\mathbb R}P(X\in du)(u-a)_+ = E(X-a)_+; 
$$
here $I\{\cdot\}$ denotes the indicator function.  
Assume now that $E X_+<\infty$ (your conditions $X\ge0$ and $EX<\infty$ are more than enough for that). Then $E(X-a)_+<\infty$ for all real $a$. Moreover, by (1), the right and left derivatives of $E(X-t)_+$ in $t$ at $t=a$ equal $-\lim_{x\downarrow a}P(X>x)=-P(X>a)=F(a)-1$ and $-\lim_{x\uparrow a}P(X>x)=-P(X\ge a)=F(a-)-1$, respectively. If $P(X=a)=0$ then the derivative of $E(X-t)_+$ in $t$ at $t=a$ exists and equals $-P(X>a)=F(a)-1$. 
Addendum: A more direct way to compute these one-sided derivatives (again for any r.v. $X$ with $E X_+<\infty$) is as follows. 
Let  $f_{a,h}(x):=\frac1h\,[(x-a)_+-(x-a-h)_+]$. Then $0\le f_{a,h}(x)\le1$ for real $x$ and real $h>0$, and $\lim_{h\downarrow0}f_{a,h}(x)=I\{x>a\}$. 
So, the right derivative of $E(X-t)_+$ in $t$ at $t=a$ equals 
$$-\lim_{h\downarrow0}\frac{E(X-a)_+-E(X-a-h)_+}h
=-\lim_{h\downarrow0}Ef_{a,h}(X)=-EI\{X>a\}=-P(X>a),$$ 
by the dominated convergence theorem. 
Quite similarly, for the left derivative.<|endoftext|>
TITLE: Difference between the dual space of $H^1(\Omega)$ and the dual of $H^1_0(\Omega)$
QUESTION [5 upvotes]: This is cross-posted on MSE: https://math.stackexchange.com/q/1596565/9464
In the Partial Differential Equations by Evans (2nd edition p299), $H^{-1}(\Omega)$ denotes the dual space to $H^1_0(\Omega)$ where $\Omega$ is an open subset of $\mathbb{R}^n$ and $H^1(\Omega)=W^{1,2}(\Omega)$, $H^1_0(\Omega)=W^{1,2}_0(\Omega)$:
$$
W^{1,2}_0(\Omega)=\overline{C_c^\infty(\Omega)}^{\|\cdot\|_{W^{1,2}(\Omega)}}
$$
While in the Navier Stokes Equations by Constantin and Foias (p7), $H^{-1}(\Omega)$ denotes the dual space of $H^1(\Omega)$.  
Let $X$ be the (continuous) dual of $H^1(\Omega)$ and $Y$ the dual of  $H^1_0(\Omega)$. One has that $X\subset Y$. 
Could anybody give a quick example in $Y\setminus X$?

REPLY [5 votes]: In general, $X$ does not embed into $Y$. Indeed, suppose that $\partial \Omega$ is sufficiently smooth. Then there is the trace map $\gamma_0 \colon H^1(\Omega) \to H^{1/2}(\partial\Omega)$, $u\mapsto u\bigr|_{\partial\Omega}$, which fits into a short exact sequence
$$
0 \longrightarrow H_0^1(\Omega) \longrightarrow H^1(\Omega) \xrightarrow{ \ \gamma_0 \ } H^{1/2}(\partial\Omega) \longrightarrow 0.
$$ 
By duality, one gets a surjective map $X\twoheadrightarrow Y$, $\Phi\mapsto \Phi\bigr|_{H_0^1(\Omega)}$, and an embedding $H^{-1/2}(\partial\Omega)= H^{1/2}(\partial\Omega)'\to X$, $\Psi \mapsto \langle\Psi,\gamma_0 (\cdot)\rangle$. Functionals $\Psi\in H^{-1/2}(\partial\Omega)$ regarded as elements of $X$ are zero when restricted to $H_0^1(\Omega)$.<|endoftext|>
TITLE: Generator determined by finitely many translates implies zero entropy
QUESTION [6 upvotes]: Let $T$ be a measure preserving transformation of a standard probability space $(X,\mathcal{B},\mu)$. A partition $\alpha$ of $X$ is said to be a generator for $T$ if the smallest $T$ invariant $\sigma$-algebra of measurable sets containing the pieces of $\alpha$ is all of $\mathcal{B}$.
Suppose there is a generator $\alpha$ for $T$ and a finite set $F \subseteq \mathbb{Z} \setminus \{0\}$ such that $\alpha$ is measurable with respect to $\bigvee_{n \in F} T^n \alpha$. Does this imply that $T$ has zero entropy?

REPLY [3 votes]: No, this is not true. This is true if $F \subset \mathbb{N}$, but not in the general case. I am lost with the  objects of ergodic theory so I start by rephrasing in the probabilistic language. 
Let ${(Z_n)}_{n \geq 0}$ be the stationary (shift-invariant) process defined by $Z_n(x)=\alpha(T^n(x))$ where $\alpha(x)$ denotes the element of the partition $\alpha$ to which $x$ belongs. The (usual) definition of the entropy of $T$ with respect to $\alpha$ is $h(T,\alpha)=\lim \frac{H(Z_1,\ldots,Z_n)}{n}$, and it is well known that $h(T)=h(T,\alpha)$ when $\alpha$ is a generating partition. Moreover, introducing the decreasing sequence of $\sigma$-fields ${\cal F}_n=\sigma(Z_m; m \geq n)$, it is well known that $h(T,\alpha)=H({\cal F}_0 \mid {\cal F}_1)$.
Assume your $F$ is included in $\mathbb{N} \setminus \{0\}$. That means that $Z_0$ is measurable with respect to $\sigma(Z_k; k \in F) \subset {\cal F}_1$. Therefore  ${\cal F}_0 \subset {\cal F}_1$ (then ${\cal F}_0={\cal F}_1)$ because ${\cal F}_0 = {\cal F}_1 \vee \sigma(Z_0)$. Hence $H({\cal F}_0 \mid {\cal F}_1)=0$.
If your $F$ contains some negative integers, then your assumption is equivalent to $\sigma(Z_{n_0}) \subset \sigma(Z_k; k \in F)$ where $F \subset \mathbb{N}\setminus\{n_0\}$. It is easy to get a counter-example for $n_0=1$ and $F=\{0,2\}$.
Take a stationary process  ${(Z_n)}_{n \geq 0}$ with non-zero entropy, taking its values in a finite alphabet, and define a new stationary process ${(Y_n)}_{n \geq 0}$ by setting $Y_n = (Z_n, Z_{n+1})$. Then $Y_1 \subset \sigma(Y_0,Y_2)$. But it is clear from the formula defining $h(T,\alpha)$ that ${(Y_n)}_{n \geq 0}$  has the same entropy as ${(Z_n)}_{n \geq 0}$. This is also clear from the other formula because the process ${(Y_n)}_{n \geq 0}$ has the same decreasing sequence of $\sigma$-fields as ${(Z_n)}_{n \geq 0}$.
For example, if you take for ${(Z_n)}_{n \geq 0}$ a sequence of independent symmetric Bernoulli variables, then you can see ${(Y_n)}_{n \geq 0}$ as a stationary Markov chain on the vertices of a square $ABCD$, where $Y_n$ has the uniform distribution and it jumps from $A$ to $A$ or to the "next" point $B$ with equal probabilities, it jumps from $B$ to $B$ or to the "next" point $C$ with equal probabilities, etc. Its entropy is $\ln 2$ and it is easy to see that $Y_1 \subset \sigma(Y_0,Y_2)$.
So the answer is no, and it is still no with additional assumptions such as ergodicity, $K$, Bernoullicity.
I seize the opportunity to give a try to the markovchain R package:
library(markovchain)

Y <- new("markovchain", states = c("A", "B", "C", "D"),
         transitionMatrix = 0.5*matrix(c(c(1, 1, 0, 0), 
                                     c(0, 1, 1, 0), 
                                     c(0, 0, 1, 1), 
                                     c(1, 0, 0, 1)), 
                                   byrow=TRUE, nrow = 4),
         name = "Counter-example")

plot(Y)<|endoftext|>
TITLE: factorization of the regular representation of the symmetric group
QUESTION [15 upvotes]: Let $\mathbb{C}[S_n]$ be the regular representation of the symmetric group $S_n$, and let $\mathbb{C}^n$ be the vector representation.
Question:  Does there exist a representation $V$ (of dimension $(n-1)!$) such that $V\otimes\mathbb{C}^n\cong \mathbb{C}[S_n]$?  If so, does $V$ admit any particularly nice description?
For example, when $n=4$, we can take $V = V(4)\oplus V(2,1,1) \oplus V(2,2)$.
This problem has a nice solution if we only ask the isomorphism to be equivariant for the subgroup $S_{n-1}\subset S_n$.  We can realize $\mathbb{C}[S_n]$ as the cohomology of the configuration space of $n$ points in $\mathbb{R}^k$ for any odd $k$ (even $k=1$ is okay).  Then forgetting the $n$th point gives us a fiber bundle that behaves like a product on cohomology, and we obtain an isomorphism $\mathbb{C}[S_n]\cong \mathbb{C}[S_{n-1}]\otimes\mathbb{C}^n$.  But we have broken the symmetry by choosing which point to forget, so this is only an isomorphism of $S_{n-1}$ representations.  I want to do everything $S_n$-equivariantly, which is indeed possible when $n=4$, as the above example demonstrates.
Update:  As Geoff Robinson observes in the comments below, an $S_n$-representation $V$ is a solution to my problem if and only if the restriction of $V$ to $S_{n-1}$ is isomorphic to the regular representation.

REPLY [8 votes]: Let $H$ be a regular subgroup of $S_n$, for instance a transitive cyclic subgroup of order $n$. Then the permutation module $V$ of the action on the coset space $S_n/H$ has the requested property, as only the identity element of $S_n$ lies in a conjugate of $H$ and fixes a point in the natural action at the same time.
(Remark: Just noticed that in the very same minute Geoff Robinson gave the identical answer in a comment. So I marked my answer CW.)

REPLY [7 votes]: The fact that the Lie module (as proposed by Darij Grinberg) works, as well, as an explicit isomorphism of modules, follows from the theory of cyclic operads: see Corollary 6.9 in http://sites.math.northwestern.edu/~getzler/Papers/cyclic.pdf (to be precise, restrict the statement of that Corollary to $S_n\subset S_{n+1}$).<|endoftext|>
TITLE: Why is dual lattice a lattice, in the context of complex tori
QUESTION [6 upvotes]: I have a simple linear algebra question regarding the definition of dual of a lattice; it was asked by someone else here three months ago on mathstackexchange but got no answer and few views, so forgive me for asking simple minded question here.  Let $V$ be a complex vector space of dimension $n$, and let $L \subseteq V$ be a lattice (i.e a discrete free abelian subgroup) of rank $2n$. Let 
$$V^*=\{f:V \to \mathbb{C} \mid f(cv)=\overline{c}f(v) \}$$
be the space of $\mathbb{C}$-antilinear maps. This is a complex vector space of the same dimension as $V$
Let 
$$L^*=\{f \in V^* \mid Imaginary(f(L)) \subseteq \mathbb{Z}  \}$$
be the elements of $V^*$ that map $L$ to complex numbers with integer imaginary part.
My question is why is $L^*$ a lattice in $V^*$? And why does it have the same rank as $L$? 
Obviously $L^*$ is a free abelian group. The issue is showing it is discrete in $V^*$. I tried choosing a $\mathbb{Z}$ basis for $L$ and writing out the condition for an to be in $L^*$, but it turned into a mess.
(The claims I am asking about are on p. 15 of Milne's notes on Abelian Varieties, here)

REPLY [5 votes]: As explained in my comments, it is the same to treat the question where the notation $L^{\ast}$ is defined with $V^{\ast}$ taken to be the $\mathbf{C}$-dual (rather than the conjugate dual), so we do that.  That is, $L^{\ast}$ now consists of the $\mathbf{C}$-linear (rather than conjugate-linear) forms on $V$ whose imaginary part is $\mathbf{Z}$-valued on $L$. Note also that $\mathbf{R} \otimes_{\mathbf{Z}} L$ coincides with the underlying $\mathbf{R}$-vector space $V_{\mathbf{R}}$ of $V$.  We let $V_{\mathbf{R}}^{\ast}$ denote the $\mathbf{R}$-dual of $V_{\mathbf{R}}$ (not to be confused with the underlying $\mathbf{R}$-vector space of the $\mathbf{C}$-dual $V^{\ast}$!). 
The answer below is rather longer than the actual content, since we wish to explain why the relevant formulas don't just come out of thin air. If you don't care about motivation, just go immediately to the final paragraph for a quick self-contained proof.
Write $\overline{V}$ to denote the conjugate-dual space (i.e., $\mathbf{C} \otimes_{\sigma, \mathbf{C}} V$ where $\sigma$ denotes complex conjugation), so for $v \in V$ we have the associated vector $\overline{v} := 1 \otimes v \in \overline{V}$.  Thus, as $\mathbf{C}$-vector spaces we have naturally
$$\mathbf{C} \otimes_{\mathbf{Z}} L = \mathbf{C} \otimes_{\mathbf{R}} V_{\mathbf{R}} \simeq V \oplus \overline{V}$$
where the isomorphism is defined by $1 \otimes v \mapsto (v, \overline{v})$.  The inverse isomorphism is clearly 
$$(v, \overline{w}) \mapsto 1 \otimes \frac{v+w}{2} + i \otimes \frac{v-w}{2i}.$$
Now the idea is to $\mathbf{C}$-dualize both sides, exploiting that the $\mathbf{C}$-dual of $\mathbf{C} \otimes V_{\mathbf{R}}$ is naturally identified with $\mathbf{C} \otimes V^{\ast}_{\mathbf{R}}$, and that
$V_{\mathbf{R}}^{\ast}$ is $\mathbf{R}$-spanned by the $\mathbf{Z}$-dual of $L$, with this $\mathbf{Z}$-dual identified as the $\mathbf{R}$-linear functionals on $V_{\mathbf{R}}$ that are $\mathbf{Z}$-valued on $L$. 
We'll thereby eventually find that up to a harmless factor of 2, this identifies the $\mathbf{Z}$-dual of $iL$ with $L^{\ast}$ as defined in the original problem (thereby "explaining" the notation) and  from that we'll get the entire result (that $L^{\ast}$ as originally defined is both discrete in $V^{\ast}$ and spans it over $\mathbf{R}$) all at once.
To actually do the work, note that the restriction of the inverse isomorphism to the first direct summand is a $\mathbf{C}$-linear inclusion $V \hookrightarrow \mathbf{C} \otimes_{\mathbf{R}} V_{\mathbf{R}}$ given by $v \mapsto 1 \otimes (v/2) + i \otimes (v/2i)$.
Thus, passing to $\mathbf{C}$-duals gives a 
$\mathbf{C}$-linear quotient map
$\mathbf{C} \otimes_{\mathbf{R}} V_{\mathbf{R}}^{\ast} \twoheadrightarrow V^{\ast}$ defined by $$1 \otimes \lambda \mapsto (v \mapsto \lambda(v/2) + i \lambda(v/2i) = \lambda(v/2) - i \lambda(iv/2)).$$
The restriction to the $\mathbf{R}$-subspace $V_{\mathbf{R}}^{\ast}$ is an $\mathbf{R}$-linear map $V_{\mathbf{R}}^{\ast} \rightarrow V^{\ast}$ that is clearly injective and hence an isomorphism for $\mathbf{R}$-dimension reasons (or by bare hands: equivariance with $i$-scaling shows that $\mathbf{C}$-linear forms on $V$ can be uniquely written as $v \mapsto x(v) - ix(iv)$ for $\mathbf{R}$-linear forms $x$ on $V_{\mathbf{R}}$). 
The above is really just natural context for something we could have written in a single line: $V_{\mathbf{R}}^{\ast} \simeq V^{\ast}$ via the formula $\lambda \mapsto (v \mapsto \lambda(v) - i\lambda(iv))$.
Thus, for a lattice $L$ in $V$, the $\mathbf{Z}$-dual to $L$ as a lattice in the $\mathbf{R}$-dual $V_{\mathbf{R}}^{\ast}$ of $V_{\mathbf{R}} = \mathbf{R} \otimes_{\mathbf{Z}} L$ is carried over to the set of $\mathbf{C}$-linear forms $f$ on $V$ whose real part is $\mathbf{Z}$-valued on $L$, or equivalently (!!) whose imaginary part is $\mathbf{Z}$-valued on $iL$.  In particular, we conclude that for any lattice $\Lambda$ in $V$, the set of $f \in V^{\ast}$ satisfying ${\rm{Im}}(f(i\Lambda)) \subset \mathbf{Z}$ is a lattice. Now take $\Lambda$ to be $iL$ (so $i\Lambda = L$) to conclude.<|endoftext|>
TITLE: Ideals of $U(\mathrm{gl}(n,\mathbb{C}))$ and their intersection with center
QUESTION [5 upvotes]: Does every non-null two-sided ideal of $U(\mathrm{gl}(n,\mathbb{C}))$ have nonzero intersection with the center of $U$?

REPLY [7 votes]: The answer is yes. Let's say that a domain $R$ has the intersection property if every non-zero two-sided ideal $I$ of $R$ has non-zero intersection with the centre of $R$. The result follows from the Theorem and Lemma below.

Theorem Let $\mathfrak{g}$ be a semisimple Lie algebra over a field $k$ of characteristic zero. Then $U(\mathfrak{g})$ has the intersection property.

This is Proposition 4.2.2 of Dixmier's book Enveloping algebras. 
First one shows that the intersection of the annihilators in $U(\mathfrak{g})$ of all finite-dimensional representations of $\mathfrak{g}$ is zero; this is a non-trivial result due to Harish-Chandra. 
Now if $J$ is a non-zero two-sided ideal of $U(\mathfrak{g})$, then we can find some finite-dimensional representation $V$ of $\mathfrak{g}$ such that $J \nsubseteq I := \mathrm{Ann}(V)$. Since $\mathfrak{g}$ is semisimple, by Weyl's theorem on complete reducibility we can replace $V$ by one of its irreducible direct summands and thereby assume that $V$ is itself irreducible. Therefore $\mathrm{End}_k(V)$ is a simple ring, and it follows that
$$ \frac{J}{J \cap I} \cong \frac{I + J}{I} = \frac{U(\mathfrak{g})}{I} \cong \mathrm{End}_k(V) $$
because $I + J$ is a two-sided ideal in $U(\mathfrak{g})$ properly containing $I$. This is an isomorphism of $\mathfrak{g}$-bimodules.
Next, the adjoint representation of $\mathfrak{g}$ in $U(\mathfrak{g})$ is locally finite (i.e. is the union of its finite-dimensional submodules). Since $\mathfrak{g}$ is semisimple, the short exact sequence of $\mathfrak{g}$-modules
$$ 0 \to J \cap I \to J \to \mathrm{End}_k(V) \to 0 $$
splits, so we can find a $\mathfrak{g}$-module complement $W$ for $J \cap I$ inside $J$. Now the centre $Z(\mathfrak{g})$ of $U(\mathfrak{g})$ is just the set of $\mathfrak{g}$-invariants in $U(\mathfrak{g})$ under the adjoint representation, so that
$$ W^{\mathfrak{g}} = W \cap Z(\mathfrak{g}) \subseteq J \cap Z(\mathfrak{g}).$$
However $W \cong \mathrm{End}_k(V)$ as a $\mathfrak{g}$-module, and 
$$ \mathrm{End}_k(V)^\mathfrak{g} = \mathrm{End}_{\mathfrak{g}}(V) \neq 0$$
because, for example, the identity map $\mathrm{id} : V \to V$ is a non-zero $\mathfrak{g}$-invariant. Hence $J \cap Z(\mathfrak{g})$ is non-zero, so $U(\mathfrak{g})$ has the intersection property.

Lemma Let $R$ be a domain. If $R$ has the intersection property, then so does the polynomial ring $R[x]$.

Let $Z$ be the centre of $R$, let $S := Z \backslash \{0\}$, and let $Q := S^{-1}R$ be a partial localisation of $R$. Then $Q$ is a simple ring: if $J$ is a non-zero two-sided ideal of $Q$, then $J \cap R$ is a non-zero two-sided ideal of $R$, so by the intersection property $J$ contains a non-zero element of $Z$, which is a unit in $Q$ by construction. [In fact the converse is also true, but we will not need this.]
Let $I$ be a non-zero two-sided ideal of $R[x]$, and consider the two-sided ideal $S^{-1}I$ of $S^{-1} R[x] = Q[x]$. Choose a non-zero element $\alpha$ in $S^{-1} I$ of least possible degree:
$$\alpha = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in Q[x].$$
Then $a_n \neq 0$, so the ideal $Q a_n Q$ of $Q$ is equal to $Q$ by the simplicity of $Q$. Hence there exist elements $c_1,\ldots, c_m, d_1,\ldots, d_m \in Q$ such that $\sum_{i=1}^m c_i a_n d_i = 1$. The element
$$\alpha' := \sum_{i=1}^m c_i \alpha d_i = x^n + a_{n-1}' x^{n-1}  + \cdots + a_1'x + a_0'$$
lies in $S^{-1}I$, and is now monic. Let $u \in R$ and consider the commutator
$$[u,\alpha'] = [u, a_{n-1}']x^{n-1} + \cdots + [u,a_1'] x + [u,a_0'].$$
This is still an element of our ideal $S^{-1}I$ of degree stricly smaller than $n$, so the minimality of $n$ forces it to be zero. So $[u, a_i'] = 0$ for all $u \in R$ and all $i < n$. Because $Q = S^{-1}R$, it follows that each $a_i'$ is a central element of $Q$. Now if $a = rz^{-1} \in Q$ is central with $r \in R$ and $z \in Z$, then $r = az$ is central in $R$. So $a$ is actually an element of $F$, the field of fractions of $Z$.
Thus we have found a non-zero element $\alpha' \in S^{-1} I \cap F[x]$. Hence there is some $s \in S$ such that $s \alpha' \in I \cap F[x]$. This element is non-zero, and by clearing denominators, we see that $I \cap Z[x]$ is also non-zero.<|endoftext|>
TITLE: Number of real roots of an exponential polynomial
QUESTION [9 upvotes]: Let $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_n$ be real numbers, and assume that $\{a_i\} \neq \{b_i\}$. Can the equation
$$ e^{a_1 x} + e^{a_2 x} + \dots + e^{a_n x} = e^{b_1 x} + e^{b_2 x} + \dots + e^{b_n x}$$
have more than $n$ real roots including $0$ and counting multiplicities? 
There are special cases that might be familiar to some. It cannot have $(n+1)$ roots at $0$ because that would imply $\sum a_i^k = \sum b_i^k$ for all $0 \le k \le n$. Likewise, the equation cannot have all of $0, r, 2r, \dots, nr$ as roots. In fact, if $0, r, \dots, (n-1)r$ are roots, then it can be proved that there are no additional roots.

REPLY [13 votes]: Seems that it can not have more than $n$ roots. Let us use the following generalization of Descartes rule for signed measures. Namely, let $\mu$ be a Borel signed measure on a real line, with compact support (this condition may be of course weakened). We say that $\mu$ has at most $k$ changes of sign if there exist points $c_1<c_2<\dots <c_k$ such that restriction of $\mu$ to each of the intervals $(-\infty,c_1],(c_1,c_2],(c_2,c_3],\dots,(c_k,\infty)$ is either non-negative or non-positive. I claim that in this situation Laplace transform $L(t):=\int e^{tx} d\mu(x)$ has at most $k$ real roots with multiplicity counted (except the case $\mu\equiv 0$). Induction in $k$. Base $k=0$ is clear. Assume that $k\geq 1$, for $k-1$ it is established, we should prove for a measure $\mu$ having exactly $k$ sign changes. If $L(t)$ has $N$ roots, $g(t):=e^{c_1t}(L(t)e^{-c_1t})'$ has $N-1$ roots, but $g(t)=\int (x-c_1)e^{tx} d\mu(x)$, and new signed measure $d\nu(x)=(x-c_1)d\mu(x)$ has at most $k-1$ sign changes (if new measure $\nu$ is 0, then $\mu$ was proportional to the delta-measure in $c_1$, and $L(t)$ does not have roots at all.) So, by induction hypothesis $N-1\leq k-1$, as we need. 
Now we say $e^{at}-e^{bt}={\rm sign}\,(a-b)\cdot t\cdot \int_{[a,b]} e^{tx} dx$, so if we divide our equation $\sum_{i=1}^n e^{a_it}-e^{b_it}=0$ by $t$, it becomes a Laplace transform for a measure with density $\sum \pm \chi_{[a_i,b_i]}$, which has at most $n-1$ sign changes. Thus our equation has at most $n-1$ non-zero roots as desired.
(PREVIOUS VERSION, working for simple roots) It is equivalent to counting positive roots of $\sum x^{a_i}=\sum x^{b_i}$. There is always root 1, so subtract $n$ from both parts and divide by $x-1$. I hope that after this the difference has at most $n-1$ sign changes and so Descartes rule is applicable (if we assume that exponents are positive integers, as we may assume: having more than $n$ roots is open condition, so exponents may be considered rational, then by scaling change of variables integer.) Indeed, sign change appears if there were more $a$'s than $b$'s on some initial segment, but it became more $b$'s than $a$'s. It requires two points at least. 
UPD: when I say that having more than $n$ roots is an open condition, I, strictly speaking, lie. It is so for simple roots, but for multiple roots it is not: say, small perturbations of $x^2$ may have no real roots. However I believe that the statement is true for multiple roots too, but it have to be fixed somehow.<|endoftext|>
TITLE: Factorial-based constant
QUESTION [7 upvotes]: Am looking for a name for:
$$\prod\dfrac{1}{1-\dfrac{1}{n!}}$$
$$=2.529477472079152648180116154253954242$$
Wolfram|Alpha
Expanding the formula gives:
$$(1+\frac{1}{2!}+\frac{1}{2!^2}+\dots)(1+\frac{1}{3!}+\frac{1}{3!^2}+\dots)(1+\frac{1}{4!}+\frac{1}{4!^2}+\dots)\dots$$
which gives the possible denominators used in multinomials.

REPLY [15 votes]: I don't know about a name, but it does have a history. Knopfmacher, Odlyzko, Pittel, Richmond, Stark [D., not H.], Szekeres, and Wormald, The asymptotic number of set partitions with unequal block sizes, available here, find that it is the residue at $z=1$ of a generating function $$G(z)=\prod_{k=1}^{\infty}\left(1-{z^k\over k!}\right)^{-1}$$ They relate it to evaluation of sums of multinomial coefficients, and they reference page 126 of Comtet, Advanced Combinatorics.<|endoftext|>
TITLE: Which compact 3-manifolds with boundary embed in $\mathbb{S}^3.$
QUESTION [12 upvotes]: This is a more sensible (IMHO) restatement of this question: 
Which Compact $3$-manifolds with boundaries embed in $\mathbb{S}^3?$ Is there any hope of a characterization?

REPLY [11 votes]: There is a theorem of Fox that more or less deals with this.  Any such manifold is a complement of (possibly knotted) handlebodies.

Theorem: Every compact connected 3-submanifold $Y$ of the 3-sphere can be reimbedded in the 3-sphere so that the exterior of the image of $Y$ is a union of handlebodies, i.e. regular neighborhoods of embedded graphs.

R. H. Fox, On the imbedding of polyhedra in 3-space, Ann. of Math. (2) 49 (1948), 462–470.<|endoftext|>
TITLE: Adaptive version of the Azuma–Hoeffding inequality
QUESTION [9 upvotes]: The Azuma inequality states that if we have a martingale $X_1,\ldots,X_N$ that satisfies a bounded difference condition:
$$|X_k - X_{k-1}| \leq c_k$$
Then:
$$\Pr\left[X_N - X_0 \geq \sqrt{2\sum_kc_k^2 \ln(1/\delta)}\right] \leq \delta$$
My question is:

Does the same inequality hold if the constants $c_k$ are not fixed up front, but are themselves a function of the realizations of the previous terms in the martingale $X_1,\ldots,X_{k-1}$?

For concreteness, consider the following coin flipping game. An adversary has $n$ coins, which he flips in sequence (represented by random variables $B_i$ taking values uniformly in $\{-1,1\}$). After observing the outcomes of $B_1,\ldots,B_{i-1}$ (but crucially, not $B_i$), he chooses a non-negative weight $c_i$. At the end of the $n$ coin flips, we compute the quantity:
$$X_n = \sum_{i=1}^n c_i\cdot B_i$$
Can we now say that:
$$\Pr\left[X_N \geq \sqrt{2\sum_kc_k^2 \ln(1/\delta)}\right] \leq \delta$$
noting that here, the values $c_k$ are now random variables themselves?

REPLY [3 votes]: There is no such inequality even if we further restrict $c_k$ to be in $\{0,1\}$ and weaken the inequality to include a constant factor. (I think it is natural to add the condition that the $c_k$ values are uniformly bounded.) Suppose $c_k \in \{0,1\}$ and without loss of generality no $1$ follows a $0$. The choice of $c_k$ is equivalent to a stopping rule $\tau \le N$. We bet a constant amount on each toss of a coin and decide to stop after $\tau$ flips: $c_k = 0$ when $k \gt \tau$, so $c_k$ only depends on $X_1,...,X_{k-1}$.
Choose a constant $s \gt 0$. Can we bound 
$$P\left(X_N \ge s \sqrt{\sum_{k=1}^N c_k^2}\right) = P(X_\tau \ge s \sqrt{\tau})?$$
The law of the iterated logarithm implies that for an infinite sequence $\{B_i\}$, the maximum number of standard deviations of $\sum B_i$ above the mean is almost surely unbounded, hence as $N\to \infty$ the maximum exceeds $s$ with probability approaching $1$. We can let $\tau$ be the minimum of $N$ and the first time $t$ so that $\sum_{i=1}^{t} B_i > s\sqrt{t}$.
If you let the bound depend on $N$, something like 
$$P(\max \frac{X_t}{\sqrt{t}} \ge s) \le f(s,N),$$
then there are nontrivial bounds possible.<|endoftext|>
TITLE: On convergence of convex bodies
QUESTION [21 upvotes]: Let $K\subset \mathbb{R}^n$ be a compact convex set of full dimension. Assume that $0\in \partial K$. 
Question 1. Is it true that there exists $\varepsilon_0>0$ such that for any $0<\varepsilon <\varepsilon_0$ the intersection $K\cap \varepsilon S^{n-1}$ is contractible? Here $\varepsilon S^{n-1}$ is the unit sphere centered at 0 of radius $\varepsilon$.
If Question 1 has a positive answer I would like to generalize it a little bit. Under the above assumptions, assume in addition that a sequence $\{K_i\}$ of compact convex sets converges in the Hausdorff metric to $K$.
Question 2. Is it true that there exists $\varepsilon_0>0$ such that for any $0<\varepsilon <\varepsilon_0$ the intersection $K_i\cap \varepsilon S^{n-1}$ is contractible for $i>i(\varepsilon)$?
A reference would be helpful.

REPLY [7 votes]: The answer to Question 1 is yes, which is precisely Lemma 3.6 in the paper:
Boundary torsion and convex caps of locally convex surfaces, 
J. Differential Geom., 105 (2017), 427-486.
Although the lemma is stated and proved for $R^3$, the same proof works in $R^n$. The proof also indicates how to give a positive answer to Question 2.
The proof is elementary and proceeds as follows: a convex surface can be represented locally as the graph of a convex function over a convex domain in a support plane, which we may identify with $R^{n-1}$. Then the upper half of the sphere corresponds to the graph of a concave function over the same convex domain, assuming that the radius is sufficiently small and after we readjust the domain. Now the portion of the surface cut off by the sphere is the set of points where the convex graph lies below the concave graph. It is easy to check, using the standard inequalities for the convex and concave functions, that this portion projects onto a convex domain. Hence it is a disk.
In the setting of Question 2, we have a sequence of convex functions converging to the convex function mentioned in the previous paragraph. So eventually they will be defined over the same convex domain and lie below the concave function corresponding to the hemisphere, whence again we may conclude that the upper hemisphere cuts out a disk from their graphs. So the answer to Question 2 is yes as well.<|endoftext|>
TITLE: Index of a family of operators
QUESTION [11 upvotes]: In the usual setting of the Atiyah-Singer index theorem the situation is as follows: we have a closed smooth manifold $M$ without boundary and $D$ is some elliptic differential operator acting on sections of some vector bundle over $M$. In this situation $D$ turns out to be Fredholm and one can compute its index which is an integer. However I read that AS index theorem has various generalisations: one of them is about a whole family $(D_x)_{x \in X}$ of elliptic operators parametrized by some topological space. I read that in such a situation one can define the index of this family as an element of $K^0(X)$. 

Why the index of such a family is not just an integer valued function? Why it is defined as an element of $K$-theory?

REPLY [6 votes]: This is just an addendum to Sebastian Goette's excellent answer:
In fact, you can retrieve the integer-valued function that you mention from the $K$-theory class: Let $\iota_x \colon \{pt\} \to X$ be the inclusion that sends $pt$ to $x \in X$. Let $ind(D) \in K^0(X)$ be the $K$-theory class representing the family index of $(D_x)_{x \in X}$. The function you mention will be 
$$
X \to K^0(pt) \cong \mathbb{Z} \quad ; \quad 
x \mapsto \iota_x^*(ind(D))
$$
But this loses a lot of information. In fact, it will just recover the virtual dimension of the fiber over $x$ of the vector bundle Sebastian Goette is talking about in his answer. The index class $ind(D) \in K^0(X)$ also contains global information about "how non-trivial" the vector bundle is. The $K$-theory class $ind(D) \in K^0(X)$ is not a bug, it is a feature!<|endoftext|>
TITLE: How to eliminate secular terms for perturbed non-oscillatory equations?
QUESTION [10 upvotes]: Even in a linear second order equation like $x''+x'+\epsilon x=0$ the standard asymptotic expansion has a secular term already in the first order of $\epsilon$, namely 
$$x(t)=a_0+b_0e^{-t}+\epsilon(a_1+a_0t+b_1e^{-t}-b_0te^{-t})+O(\epsilon^2). 
$$
But the exact solution is obviously bounded uniformly in $t,\epsilon\geq0$, the expansion therefore is not valid for large times. The usual approach to eliminating secular terms is to use multiple time scales, but all perturbation theory texts I looked at (Holmes, Hunter, Kevorkian-Cole, Verhulst) only consider cases where the unperturbed equation is oscillatory. The Poincare-Lindstedt method or averaging that are used only make sense when there are oscillations. But the above equation is obviously non-oscillatory for small $\epsilon$. I am interested in non-linear equations that behave similarly to the linear example above, e.g. 
$$x''+x'\Big(1-\frac32 \frac{x'}{x}\Big)+\epsilon x^3=0.$$ They come up when approximately realizing mechanical constraints by viscous friction, which explains the absence of oscillations. 

The issue seems to be that bounded solutions that are not exponentially stable produce secular terms even without oscillations, so one has all the pain of secular terms with no benefit of averaging. For instance, I suspect that for small $\epsilon$ solutions to the above nonlinear equation have a limit when $t\to\infty$, which depends on the initial values, it would be nice to approximate its dependence on them and $\epsilon$. Is there something of this sort done in the literature? References are appreciated.

REPLY [4 votes]: This is probably more a comment than an answer, but too long for the former.
A different approach might be to use normal hyperbolicity theory to first find a series expansion of an invariant manifold for your system. Then, you might be able to express solutions as a combination of an outer solution converging to the invariant manifold, and an inner/boundary layer solution on the invariant manifold (I'm not sure my use of terminology is completely correct here).
In your case, the invariant manifold at $\epsilon = 0$ is $M = \{ \dot{x} = 0 \}$. This is exponentially attractive and the dynamics on $M$ is trivial. This implies that $M$ is normally hyperbolic and then allows you to expand the persistent $M_\epsilon$ in $\epsilon$ explicitly, order by order. Then, you can express the dynamics on $M_\epsilon$ using this expansion.
I replied a little later than I first saw this question, because I was just finishing an article that shows this idea, see http://arxiv.org/abs/1603.00369. I'm also studying systems with friction and show that in a limit of sending friction to infinity one obtains nonholonomic dynamics, but also one can do a (singular, not regular) expansion of the dynamics. The Chaplygin sleigh example (including section 7.1) might be instructive.
Added notes on non-compactness
First of all, even on a compact space, the solution curves of the approximate system may not converge uniformly in time to solutions of the nonholonomic system. As counterexample see e.g. the nonholonomic constraint realization example in section 3 of my preprint, but simply compactify the space away from the original circle: any approximate solution will still drift a given finite distance away from the circle, if you wait long enough. However, the solution to the approximate system is at all times an $\mathcal{O}(\epsilon)$-pseudo solution of the nonholonomic system, that is, on any finite time interval such solutions converge uniformly to a solution of the nonholonomic system. I don't know of a geometric condition for when this would improve to uniform convergence for all positive time, i.e. when the drift would not accumulate. Since this drift is essentially given by the reaction force along solutions of the nonholonomic system, you'd need to know these.
For non-compact spaces another reason that the solutions of the approximate system do not converge uniformly may be because of non-uniform exponential stability of the invariant manifold. That may lead to the perturbed invariant manifold to not be uniformly $\mathcal{O}(\epsilon)$-close to the unperturbed one. This you can check by inspecting normal attraction rate and uniform $C^1$-boundedness of the vector field near the invariant manifold. In your example the attraction rate is
$$
-\Bigl(1-\frac{3}{2}\frac{x'}{x}\Bigr),
$$
which is uniformly $1$ on $M = \{ x' = 0 \}$, but it blows up at $x = 0$, when $x' \neq 0$ but small. Hence, for solution curves passing by that point you may not even get convergence in the $\mathcal{O}(\epsilon)$-pseudo solution sense.<|endoftext|>
TITLE: Are there infinite constructions for partial circulant hadamard matrices?
QUESTION [7 upvotes]: I believe that the circulant Hadamard conjecture (that there are no circulant Hadamard matrices of size greater than $4\times4$) is still open.
I also know that examples of $(n/2) \times n$ matrices which are partial Hadamard circulant have been found experimentally for moderate values of $n.$
To clarify, a partial circulant $m(n)\times n$ Hadamard matrix $A$ has entries from $\{-1,+1\}$ and if its first row is $a,$ then its subsequent rows are $T(a),\ldots,T^{m(n)-1}(a)$ where $T$ denotes, say, the left cyclic shift operator and $T^k$ is $T$ composed with itself $k$ times.
Is there a known infinite construction for partial circulant Hadamard matrices? Equivalently does a sequence of $m_k(n_k)\times n_k$ Hadamard matrices exist, where $n_k\rightarrow \infty$ and so does $m_k(n_k)$. Note that $m_k(n_k)=o(n_k)$ is allowed, under this definition.
I am aware of references arXiv:1201.4021, Armario et al and arxiv:1003.4003, De Launey and Levin, which address generic partial Hadamard matrices, as opposed to those that are partial circulant.
Edit: To make things explicit, the best results I have found without the circulant constraint are the following: For any $\varepsilon>0$ and for $n$ large enough, $t\equiv 0~(mod~4)$ there is a $n\times t$ partial Hadamard matrix if $n\leq \frac{t}{2}-t^{\frac{113}{132}+\varepsilon}.$ Subject to the extended Riemann hypothesis the $\frac{113}{132}$ in the exponent  can be replaced by $\frac{7}{12}.$ This matches computational results, see https://codegolf.stackexchange.com/questions/55726/an-optimization-version-of-the-hadamard-problem, that have found $\frac{n}{2}\times n$ partial Hadamard matrices up to $n$ somewhere between 50 and 100.

REPLY [3 votes]: Let $r\mbox{-}H(k\times n)$ denote a $k\times n$ partial circulant Hadmard matrix in which a row (and hence all) sums to $r$. It's a known result, see Theroem 9 in [1], that $2\mbox{-}H((p+1)\times 2(p+1))$ exists for any prime power $p$. This is because negacyclic $C$-matrices of order $p+1$ exist. In [2], Paley gave a construction of $C$-matrices using the Legendre symbol $\chi$ of the Galois field GF$(p)$. A variation of this construction leads to a negacyclic form for these Paley matrices, see [3] for details.
[1] $\textit{Circulant partial Hadamard matrices}$ by Craigen, Faucher, Low, and Wares, Lin. Alg. Appl. 439 (2013)
[2] $\textit{On orthogonal matrices}$ by Paley, J. Math. Phys. 12 (1933)
[3] $\textit{Orthogonal matrices with zero diagonal II}$ by Delsarte, Goethals, and Seidel, Can. J. Math., Vol. XXIII No. 5 (1971)<|endoftext|>
TITLE: Reasons for difficulty of Graph Isomorphism and why Johnson graphs are important?
QUESTION [14 upvotes]: In http://jeremykun.com/2015/11/12/a-quasipolynomial-time-algorithm-for-graph-isomorphism-the-details/ it is mentioned:
'In discussing Johnson graphs, Babai said they were a source of “unspeakable misery” for people who want to solve GI quickly. At the same time, it is a “curse and a blessing,” as once you’ve found a Johnson graph embedded in your problem you can recurse to much smaller instances. This routine to find one of these three things is called the “split-or-Johnson” routine.'
(1) Are Isospectral Johnson graphs the hardest case to find Isomorphism?
(2) In general what makes a graph class hard to tame (Isomorphism difficult for this class)?
(3) Supposing we know that two graphs are isomorphic. Suppose our goal is to find say isomorphism between any of $\alpha\log N$ vertices of the $N$ vertices for some $\alpha >0$. How difficult is to accomplish this? If you pick a random choice of $\alpha\log N$ on an average how many spurious matches could we land up with as candidate isomorphisms on the other graph based on these fixed $\alpha\log N$ vertices?

REPLY [18 votes]: Johnson graphs do not cause difficulty to existing programs. Actually they are rather easy; nauty can handle them up to tens of millions of vertices, and so can other programs such as Traces and Bliss.
The difficulty that Babai refers to is more theoretical.  The Johnson graph $J(v,t)$, which has $n=\binom vt$ vertices, has the property that $\Theta(n^{1/t})$ vertices need to be fixed in order for partition refinement to produce a partition with all cells at most $cn$ in size ($c<1$), nor does the group have a nice recursive structure like a wreath product. Babai says that a major part of his breakthrough was to realise that only the Johnson graphs (and closely related graphs) have these undesirable properties. With that knowledge, special processing takes care of them. Previously it could have been that some other families of graphs exist with similar properties.
I think this description is reasonably close.  I recently listened to Babai speaking on his algorithm for 5 hours but some parts remain unclear to me.
Btw, the automorphism group of $J(v,t)$ for $t\lt v/2$ is $S_v$ acting on $t$-sets.  Programs find them easy because of this rich group.
About the "supplementary problem", fixing $O(\log n)$ vertices helps very little.  It is necessary to match $\Theta(n^{1/t})$ vertices in each graph before elementary methods will provide much further information about isomorphisms.  It's hard to define "elementary methods", but they include all the obvious things like classifying vertices according to their adjacencies with the fixed vertices and more generally fixed-dimensional Weissfeiler-Lehman refinement.
Babai's statement that you quote is easy to misunderstand.  It isn't the Johnson graphs themselves that were a scourge to theorists.  It's more that the Johnson graphs showed that certain difficult structural properties (mentioned above) are possible and nobody knew a general method for handling these properties. Once Babai showed that only the Johnson graphs caused that structure, the difficultly vanished.<|endoftext|>
TITLE: Frobenius complement/kernel of an infinite group
QUESTION [8 upvotes]: Happy Peaceful New Year !
In this question, I recalled that if $H$ is a proper subgroup of a finite group $G$, such that
$$({\bf A1})\qquad(g\not\in H)\Longrightarrow(g^{-1}Hg\cap H=(1)),$$
then
$$N:=(1)\bigcup\left(G\setminus\bigcup_g g^{-1}Hg\right)$$
is a normal subgroup. The normality is obvious, but the fact that $N$ is a group is proved by character theory (one doesn't know a direct proof). The terminology is that $G$ is a Frobenius group, $H$ is the Frobenius complement of $G$ and $N$ is the Frobenius kernel.

What happens when we allow $G$ to be infinite ? I suppose that the status of this question is well-known. I suspect that there be some counter-example, a pair $(G,H)$ for which $N$ is not a subgroup. Perhaps the Frobenius Theorem above adapts with some extra assumption; an additional hypothesis could have a topological flavour, in the spirit of functional analysis, where we use topology in order to extend linear algebra in the infinite dimensional context.

Edit. After the negative answers, essentially based on the use of free groups or free product, I came to the following strengthen context. Let  $K$ be the subgroup generated by the union of the conjugate subgroups $g^{-1}Hg$ (notice that $K$ is a normal subgroup). Then $G$ is the union of $K$ and $N$, where $K\ne(1)$ and $N\ne G$. If the same conclusion as in Frobenius Theorem holds true ($N$ a subgroup), we have easily that $K=G$. This leads me to add the following assumption:

Assume in addition that $$({\bf A2})\qquad\bigcup_g g^{-1}Hg\quad\hbox{generates}\quad G.$$

Then can we say that $N$ is a subgroup ?
Let me discuss a little the case where $H=\langle a\rangle$ is a cyclic subgroup of $\mathbb{L}_2$, the free group with generators $a,b$. Then $K$ is a proper subgroup, the kernel of the morphism $\phi:\mathbb{L}_2\rightarrow\mathbb{Z}$ defined by $\phi(a)=0$ and $\phi(b)=1$. This explains why $N$ cannot be a subgroup of $\mathbb{L}_2$. In a general configuration, suppose that $H$ is a malnormal subgroup of $G_0$, but that $K$ is proper. We might replace $G$ by $G_1=K$, and $H$ is still malnormal. However, the new $K$, which I denote $K(H,G_1)$ is smaller than $K$, because there are less conjugate subgroups $g^{-1}Hg$ (the constraint is now $g\in G_1$). Whence the necessity to define $G_2=K(H,G_1)$. This is the beginning of an induction. This induction is not necessatily finite or denumerable, it can be transfinite, but its length is bounded by the cardinal of $G/H$. Eventually, we reach a subgroup $G^\dagger$ in which  $H$ is malnormal and satisfies (A2). It is unclear to me whether $G^\dagger$ is bigger than $H$ or not.

What is $G^\dagger$ in the case described above, where $G=\mathbb{L}_2$ and $H=\langle a\rangle$ ? In particular, can $G^\dagger$ be equal to $H$ ?

REPLY [5 votes]: A subgroup satisfying the condition $g \not\in H \Rightarrow H \cap g^{-1}Hg=1$ is called malnormal. This class of subgroups has been much studied.
For many types of infinite groups, such as word-hyperbolic groups, there are lots of examples. As a simple example, if $G$ is free and $a$ is an element in a free generating set, then $H = \langle a \rangle$ is malnormal in $G$.
If $b$ is another element in the free basis, then neither $b$ nor $b^{-1}a$ lies in a conjugate of $H$, but their product does, so there is no equivalent of the Frobenius kernel.<|endoftext|>
TITLE: Algebraic independence of exponentials
QUESTION [10 upvotes]: First of all, a happy new year. Be it better than 2015, 
healthy, wealthy, fruitful and cross-fertilizing 
for you, familly and friends. 
In order to cope with families of solutions of evolution equations, I had to prove the following  lemma 
Lemma: Let $Z=\{z_n\}_{n\in \mathbb{N}}$ be a set of indeterminates, then 
$[e^{z_0},e^{z_1}]$ is algebraically independent on $\mathbb{C}[Z]$ 
within $\mathbb{C}[[Z]]$. In other words, if a finitely supported sum
$$
\sum_{n,m}P_{n,m}[(z_i)_{i\geq 0}]e^{nz_0}e^{mz_1}
$$
is zero, then every polynomial $P_{n,m}\in \mathbb{C}[Z]$ is zero. 
I cannot imagine it is unknown among specialists. My question is : 

Does someone have a reference for this property ?

Thanks in advance. 

On request, a proof below (I have withdrawn the - too long - previous one, using "orders of infinity".)

All relies on the following proposition which is characteristic free.  
Proposition Let $(\mathcal{A},d)$ be a commutative differential ring without zero divisor, and $R=ker(d)$ be its subring of constants. Let $z\in \mathcal{A}$ such that $d(z)=1$ and $S=\{e_\alpha\}_{\alpha\in I}$ be a set of eigenfunctions of $d$ all different ($I\subset R$) i.e. 

 $e_\alpha\not=0$
 $d(e_\alpha)=\alpha e_\alpha\ ;\ \alpha\in I$

Then the family $(e_\alpha)_{\alpha\in I}$ is linearly free over $R[z]$ (the subring generated by $R\cup \{z\}$, see remark).    
From this, one can show the 
Corollary Let $\mathcal{A}$ be a $\mathbb{Q}$-algebra (associative, commutative and unital) and $z$ an indeterminate, then $\{z,e^z\}\subset \mathcal{A}[[z]]$ are algebraically independent over $\mathcal{A}$. 
Remark 
If $\mathcal{A}$ is a $\mathbb{Q}$-algebra or only of characteristic zero, i.e. 
$$
n1_\mathcal{A}=0\Rightarrow n=0
$$
then $d(z)=1$ implies that $z$ is transcendent over $R$. This is never the case in characteristic $p$ where $z^p$ is a constant. 
End of remark
One finishes the job proving, by successive extensions, that the sets
$$
\{e^{z_0},e^{z_1},\cdots e^{z_n},z_0,z_1,\cdots ,z_n\}
$$
are algebraically independent over $\mathcal{A}$. Which is stronger than the desired result.

REPLY [6 votes]: Here is an analytic argument.  Suppose that
$$ \sum_{j,k=0}^N P_{j,k}(z_0,z_1,z_2, \dotsc,z_n)e^{jz_0}e^{k z_1}=0. $$
Fix $z_2,\dotsc, z_n$ and $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\bR}{\mathbb{R}}$ let $z_0=\ii t_0$, $z_1=\ii t_1$, $\ii=\sqrt{-1}$, $t_0,t_1\in\bR$. We obtain an equality of the form $(\vec{z}=(z_2,\dotsc, z_n)$)
$$\sum_{j,k}P_{j,k}(t_0,t_1,\vec{z}) e^{\ii (jt_0+kt_1)}=0, $$
for all $t_0,t_1\in\bR$. Fix $t_1$.  The functions $t^me^{\ii jt}$, $m,n\in\mathbb{Z}_{\geq 0}$  are linearly independent.(You can use Wronskians to prove this.) We deduce  that
$$\sum_k P_{j,k}(t_0,t_1,\vec{z})e^{\ii k t_1}=0, $$
for all $j$, $t_0,t_1\in\bR$, $\vec{z}\in\mathbb{C}^{n-2}$.  The same argument  implies 
$$ P_{j,k}(t_0,t_1,\vec{z})=0, $$
for all $j,k$, $t_0,t_1\in\bR$, $\vec{z}\in\mathbb{C}^{n-2}$.<|endoftext|>
TITLE: The scope of correspondence principle in quantum chaos
QUESTION [9 upvotes]: My understanding of the so-called correspondence principle in quantum chaos, is that it is a connection between the behaviour of a classical Hamiltonian system (chaotic/completely integrable) and the behaviour of the quantized system in the semi-classical limit $\hbar \to 0$.
For the spectrum of the quantized system, we seem to have two heuristics on the statistics of the spectrum.

When the classical system is completely integrable, one expects the spectrum to behave like a Poisson process. For example, Berry-Tabor conjectures that the nearest spacing statistics of the spectrum distributes like the waiting time between consecutive events of a Poisson process.
When the classical system is chaotic, one expects the spectrum to distribute like a suitable Gaussian orthogonal ensemble. For example, Bohigas-Giannoni-Schmit conjectures that the nearest spacing statistics to distribute like the consecutive level spacing distribution of some Gaussian orthogonal ensemble. 


Question: These heuristics only seem to hold for compact Riemannian manifolds though. What heuristics are there in more general
  cases, for example in finite-volume manifolds, as long as the spectrum is expected to have a discrete part and is large enough so that understanding its distribution makes sense? Are there any reference on this issue?

The rough heuristic above certainly seems false in that case. For example, For $\Gamma \backslash \mathbb{H}$ where $\Gamma \leq SL_2(\mathbb{R})$ is some lattice, the geodesic flow is certainly chaotic (ergodic, Anosov,..). My impression is that if $\Gamma$ is a non-arithmetic lattice (whatever that means), one expects the statistics of the spectrum of the Laplacian to be modeled by GOE as before; but for arithmetic lattice, the statistics of the spectrum for Laplacian seems to be Poissonian again. 
My vague speculative impression is that the heuristics above seems to apply even for finite-volume manifolds as long as it is "non-arithmetic", but then I still have no idea on what to expect for the spectrum of arithmetic manifolds. In the case of surfaces, Sarnak (Spectra of hyperbolic surface) seems to suggest that spectrum of Laplacian would be Poissonian/GOE-like depending on whether the lattice is arithmetic/non-arithmetic. But I have not been able to find references/speculations for higher dimensions.
Edit: The answer below is helpful, but I would like a more definitive reference/source on the current expectations/state of the art. The best reference on these issues so far remain to be Sarnak's Schur lecture notes "Arithmetic Quantum Chaos" back in 1993. There has to be a more updated survey after 20 years. Would anyone point me into more reference?

REPLY [6 votes]: A recent study of the spectral statistics of arithmetic billiards in the semiclassical limit is Arithmetic and pseudo-arithmetic billiards (2015):

The arithmetic triangular billiards are classically chaotic but have
  Poissonian energy level statistics, in ostensible violation of the
  Bohigas-Giannoni-Schmit conjecture. From the semiclassical point of
  view, the peculiar properties of arithmetic systems result from
  constructive interference of contributions of an abnormally large
  number of periodic orbits with exactly the same length and action. We
  establish the boundary conditions under which the quantum billiard is
  “genuinely arithmetic”, i.e., has Poissonian level statistics;
  otherwise the billiard is ”pseudoarithmetic” and belongs to the GOE
  universality class.

For a higher-dimensional study, see Arithmetic quantum chaos of Maass waveforms (2004):

We compute numerically the eigenvalues and eigenfunctions of the
  Laplacian that describes the quantum mechanics of a point particle
  moving freely in the non-integrable three-dimensional hyperbolic space
  of constant negative curvature generated by the Picard group. The
  Picard group is arithmetic and we find that our results are in
  accordance with the conjecture that arithmetic quantum chaos produces
  a Poisson distribution of the eigenvalue distribution.<|endoftext|>
TITLE: Probability two products are equal
QUESTION [11 upvotes]: I am interested in the following simple looking problem on which I am stuck.  Let $M$ be a fixed $m$ by $n$ matrix with $\pm1$ elements. Let $x$ and $y$  be two independently sampled  random $n$-dimensional vectors whose elements are chosen i.u.d. from $\{-1,1\}$.

Assuming that $m<n$ and both $m$ and $n$ are large, can we give a good
  estimate or bounds for $P(Mx = My)$?

Clearly $P(Mx= My) \geq P(x=y) = 2^{-n}$. 
Both $Mx$ and $My$ have elements which are distributed as a simple symmetric random walk and have covariance matrix $MM^T$. 
It feels like $\det(MM^T)$ should be part of any solution but this is just a guess and it doesn't work directly. I also attempted to solve the problem by relating the distributions of $Mx$ and $My$ to multivariate Gaussians but this didn't work out. 
Added Jan 5 2016
To narrow the question a little, let us assume that rank$(M) = m$.
Added Jan 8 2016
Is the question more tractable if all the rows of $M$ are orthogonal?

REPLY [2 votes]: I came across this problem a few days ago, and must say it was a lot of fun to think about. Thank you!
The following is, I believe, a quite complete solution of the problem for a "typical", general $M$, making no assumptions about the relative size of $n$ and $m$ (even $m>n$ is allowed), or about the rank of $M$. 
Like kodlu in his/her answer from Jan 5, I will start with slightly re-formulating the question: 
We want to find the probability $P(Ms=0)$, where the $n$ components of the vector $s=\frac{x-y}2$ are i.i.d. random variables which have values $1$ or $-1$ with probability $\frac1 4$ and value $0$ with probability $\frac1 2$. $P(Ms=0)$ can then be expressed as the expectation value of a product of Kronecker deltas for the $m$ components of $Ms$:
$$
P(Ms=0) = \left\langle \prod_{i=1}^m \delta_{0,\sum_{j=1}^nM_{ij}s_j} \right\rangle_s \ ,
\hspace{2cm} (1)
$$
where $\left\langle \cdots \right\rangle_s$ indicates the average over all $s_j$. 
For any integer $t$ with $|t|\leq n$ we can substitute $\delta_{0,t} = \frac1{n+1} \sum_{r=0}^n \omega^{rt}$ with the primitive $(n+1)^{\rm th}$ root of unity $\omega = {\rm e}^\frac{2\pi{\rm i}}{n+1}$ ("discrete Fourier transform"), so that $(1)$ becomes
$$
P(Ms=0) = \left\langle \prod_{i=1}^m \left( \frac1{n+1} \sum_{r_i=0}^n \omega^{r_i\sum_{j=1}^n M_{ij}s_j} \right) \right\rangle_s
= \left\langle \prod_{i=1}^m \prod_{j=1}^n \omega^{M_{ij} r_i s_j} \right\rangle_{r,s} \ ,
\hspace{2cm} (2)
$$
where in the second step the $r_i$ have been re-interpreted as i.u.d. random variables with values in $\{0,...,n\}$, so that $\left\langle \cdots \right\rangle_{r_i} = \frac1{n+1} \sum_{r_i=0}^n (\cdots)$, and $\left\langle \cdots \right\rangle_{r,s}$ indicates the simultaneous average over all $r_i$ and $s_j$. 
Now, obviously $P(Ms=0)$ has not the same value for all $(-1,1)$-matrices $M$ of a given size $m \times n$. As an example, if $M$ has rank $1$ (i.e., each of its column vectors is equal to either $v$ or $-v$, with some fixed $(-1,1)$-vector $v$) then the problem is equivalent to the case $m=1$, and the result is 
$$
P(Ms=0) = 2^{-n} \sum_{l=0}^{\lfloor n/2 \rfloor} 2^{-2l} \begin{pmatrix} {n} \\ {2l} \end{pmatrix} \begin{pmatrix} {2l} \\ {l} \end{pmatrix} \ ,
\hspace{2cm} (3)
$$ 
whereas usually it will be much smaller (for instance, if $M$ has rank $n$ then trivially $P(Ms=0) = 2^{-n}$). Thus $(2)$ cannot be evaluated without further assumptions on $M$. Alternatively, we can try to calculate a "typical" value of $(2)$, assuming that $M$ itself is a random sample out of a distribution of matrices with i.i.d. matrix elements $M_{ij}$, and averaging $P(Ms=0)$ over this distribution. This approach leads to
$$
\Big\langle P(Ms=0) \Big\rangle_M
= \left\langle \prod_{j=1}^n \prod_{i=1}^m \left\langle \omega^{M_{ij} r_i s_j} \right\rangle_M \right\rangle_{r,s}
= \left\langle \prod_{j=1}^n \prod_{i=1}^m \frac{ \omega^{r_i s_j} + \omega^{-r_i s_j} }2 \right\rangle_{r,s} \ .
\hspace{1cm} (4)
$$
Note that we have also changed the order of the product operations over $i$ and $j$ here. Now the average over the $s_j$ can be performed easily: for each $j$, 
$$
\left\langle \prod_{i=1}^m \frac{ \omega^{r_i s_j} + \omega^{-r_i s_j} }2 \right\rangle_{s_j}
= \frac1 2 \left( 1 + \prod_{i=1}^m \frac{\omega^{r_i} + \omega^{-r_i}}2 \right) \ ,
$$
which is independent of $j$, so that 
$$
\Big\langle P(Ms=0) \Big\rangle_M
= 2^{-n} \left\langle \left( 1 + \prod_{i=1}^m \frac{\omega^{r_i} + \omega^{-r_i}}2 \right)^n \right\rangle_{r} \ .
\hspace{2cm} (5)
$$
To evaluate the right hand side of $(5)$ we expand the $n^{\rm th}$ power, 
$$
\Big\langle P(Ms=0) \Big\rangle_M
= 2^{-n} \sum_{k=0}^n \begin{pmatrix} {n} \\ {k} \end{pmatrix} 2^{-km} \prod_{i=1}^m \left\langle \Big( \omega^{r_i} + \omega^{-r_i} \Big)^k \right\rangle_{r} \ ,
\hspace{2cm} (6)
$$
and likewise the $k^{\rm th}$ power in $(6)$: 
$$
\left\langle \Big( \omega^{r_i} + \omega^{-r_i} \Big)^k \right\rangle_{r}
= \sum_{l=0}^k \begin{pmatrix} {k} \\ {l} \end{pmatrix} \left\langle \omega^{(k-2l)r_i} \right\rangle_{r}
= \begin{cases} \begin{pmatrix} {k} \\ {k/2} \end{pmatrix} , \ k\ {\rm even} \\ \quad \ 0 \ , \quad k\ {\rm odd}  \end{cases}
\ ,
\hspace{2cm} (7)
$$
where $\left\langle \omega^{(k-2l)r_i} \right\rangle_{r} = \delta_{k,2l}$ has been used. Inserting $(7)$ into $(6)$ then leads to 
$$
\Big\langle P(Ms=0) \Big\rangle_M
= 2^{-n} \sum_{l=0}^{\lfloor n/2 \rfloor} 2^{-2lm} \begin{pmatrix} {n} \\ {2l} \end{pmatrix} \begin{pmatrix} {2l} \\ {l} \end{pmatrix} ^m \ .
\hspace{2cm} (8)
$$
Note the close similarity to Eq. $(3)$. 
The $l=0$ term in $(8)$ yields $\big\langle P(Ms=0) \big\rangle_M = 2^{-n} $, which is the lower bound mentioned in the question. For large $n$, $m$, and $l$, Stirling's approximation gives to leading order 
$$
2^{-2lm} \begin{pmatrix} {n} \\ {2l} \end{pmatrix} \begin{pmatrix} {2l} \\ {l} \end{pmatrix} ^m 
\sim \sqrt{\frac{n}{4\pi l(n-2l)}} \ {\rm e}^{n I(\frac{2l}n)} (\pi l)^{-\frac m2}
\ ,
\hspace{2cm} (9)
$$
where $I(p)$ denotes the "information entropy" of a 2-state system with probabilities $p$ and $1-p$: 
$$
I(p) = - p \ln(p) - (1-p) \ln(1-p) \ .
\hspace{2cm} (10)
$$
Note that the divergence of the right hand side of $(9)$ for $l \to 0$ is an artefact due to Stirling's approximation; the original expression on the left hand side has a  $l \to 0$ limit equal to $1$. 
To find the dominant term(s) in the sum of $(8)$, we look for the maximum of the exponential part of $(9)$ as a function of $l$, disregarding for the moment the square root. 
That is, setting $p = \frac{2l}n$ and $q = \frac{m}n$, we look for the maximum of 
$$
{\rm e}^{n \left( I(p) - \frac q 2 \ln\frac{\pi p n}2 \right)}
\hspace{2cm} (11)
$$
as a function of $p \in ]0,1]$.  The sum in $(8)$ is either dominated (for large enough $q$) by the $l=0$ term $1$, or otherwise by a term with $l \approx \frac{pn}2$, where $p$ is a solution of the equation 
$$
\frac q{2p} = I'(p) = \ln\frac{1-p}p \ .
\hspace{2cm} (12)
$$
If $q$ is larger than about $0.56$, $(11)$ has no solution $p \in ]0,1]$. For $q$ smaller than about $0.56$ there are two solutions, the smaller of the two corresponding to a local minimum of ${\rm e}^{n \left( I(p) - \frac q 2 \ln\frac{\pi p n}2 \right)}$ and the larger one to a local maximum. For large $n$, this local maximum dominates the sum in $(8)$ if and only if $I(p) - \frac q 2 \ln\frac{\pi p n}2 > 0$ for this $p$ value. For large $n$, the maximal $q$ that meets this criterion decreases logarithmically with increasing $n$, so in that case we may assume $\frac q{2p} \ll 1$ and thus obtain $p \approx \frac1 2$ from $(12)$. Inserting this back into the exponential expression $(11)$ yields an estimate of its maximum for large $n$, $m$, and $l$
$$
\max_{p \gg \frac1n} \left\{ {\rm e}^{n \left( I(p) - \frac q 2 \ln\frac{\pi p n}2 \right)} \right\} 
\approx {\rm e}^{n \left( \ln2 - \frac q 2 \ln\frac{\pi n}4 \right)} \ .
\hspace{2cm} (13)
$$
Finally, we can combine this with $(8)$,  $(9)$ and obtain, to leading order for large $n$, 
$$
2^{n} \Big\langle P(Ms=0) \Big\rangle_M
\approx 1 + \sqrt{\frac{2}{\pi n}} \ {\rm e}^{n \left( \ln2 - \frac q 2 \ln\frac{\pi n}4 \right)} 
= 1 + 2^{n-\frac1 2}  \left( \frac{\pi n}4 \right)^{-\frac{qn+1} 2}
\ .
\hspace{2cm} (14)
$$
As long as the exponent in $(13)$ is negative, i.e. $q > \frac{\ln4}{\ln\frac{\pi n}4}$ (or $m > n \frac{\ln4}{\ln\frac{\pi n}4} $), the first term $=1$ in $(14)$ dominates, which means that only the trivial solution $s=0$ of $Ms=0$ has nonnegligible probability. Otherwise, the second term dominates, quantifying how the number of solutions increases when $m$ becomes small enough compared to $n$.<|endoftext|>
TITLE: Generating functions for objects with irrational sizes
QUESTION [14 upvotes]: A problem I'm investigating concerns a combinatorial class in which the 'atoms' have irrational sizes. It seems likely that this is something that has been considered before, but I haven't been able to find it in the literature. 
As a simple example, consider the bivariate generating function $f(x,y)=(1-x-y)^{-1}$ for sequences of two different types of objects ($X$ and $Y$). Now suppose that the size of an $X$ object is $a$ and the size of a $Y$ object is $b$. If $a$ and $b$ are integers, then $f(z^a,z^b)$ enumerates sequences of a given size, and if $a/b$ is rational, we can get the enumeration by multiplying by the least common denominator.
What is the theory when $a/b$ is irrational? Consider $g(z)=f(z,z^\sqrt{2})$. This function is no longer a formal power series in $z$. However, singularity analysis of $g(z)$ seems to deliver the 'right' answer for the 'growth rate', as might be expected by continuity. On the other hand, the concept of growth rate in this case has to be modified a bit, to be something like $\lim_{t \to \infty} {S_{[t,t+1]}}^{1/t}$, where $S_I$ is the number of sequences whose sizes lie in the interval $I$.
Where can I find this kind of thing discussed?

REPLY [4 votes]: I received the following from Mireille Bousquet-Mélou:
I have in fact never seen such series in enumeration... when irrational powers are involved, even the notion of such series has to be handled carefully (I think that there could be accumulation points of exponents. This is not the case in your example though).
Such objects are called (I think) transseries. I quick google search leads for instance to this paper (for beginners...):
http://arxiv.org/pdf/0801.4877v5.pdf
There is also this book:
http://arxiv.org/pdf/1509.02588v1.pdf
I do not promise these references to be of some help, not even relevant!

REPLY [2 votes]: Some related research:
Scott Garrabrant and Igor Pak consider tilings with irrational side lengths in this article.
An open problem here is to decide if the Catalan numbers can be generated from such tilings described in the article.<|endoftext|>
TITLE: Travelling waves for nonlinear Schrödinger equation
QUESTION [6 upvotes]: Consider the following nonlinear Schrödinger equation:
$$
-\Delta \Phi - i\frac{\partial \Phi}{\partial t} = f(|\Phi|^2)\Phi,
$$
where $\Delta$ is the Laplacian on $\mathbb{R}^n$, $f$ gives the nonlinearity (I am not specifying any conditions on $f$, since I want to make this a broad reference request; $f$ can be power-type or Gross-Pitaevskii type, or otherwise). I am looking for references for existence/non-existence of (super/sub)sonic travelling wave solutions of the above equation with null condition at infinity. By travelling wave solution, I mean a solution of the form $\Phi(x, t) = \varphi (x + ct)$, where $c \in \mathbb{R}^n$.

REPLY [2 votes]: For the case of Gross--Pitaevskii type nonlinearities, you might look at the Annals paper of Maris (as well as references therein and thereof).  Here is an arXiv version: http://arxiv.org/abs/0903.0354
And, if you have access: http://annals.math.princeton.edu/wp-content/uploads/Maris.pdf<|endoftext|>
TITLE: Estimating size of greatest prime divisor of a sequence of integers
QUESTION [5 upvotes]: Consider the numbers of the form: $$A_n = \prod_{\pm}\left(\pm 1\pm \sqrt{2} \pm \cdots \pm \sqrt{n}\right)$$
where, the product in taken oven all $2^n$ terms with variations in sign. We know such numbers are integers (even perfect-squares). 

Is there a way to investigate the asymptotics of the largest prime divisor of $A_n$?

Source: Problem A-578 where it was claimed that that for all $\varepsilon>0$ there exists an $n_0$ such that for every $n>n_0$ the largest prime divisor of $A_n$ is smaller than $\displaystyle 2^{2^{\,\huge{\varepsilon} n}}$. I haven't been able to prove this claim either.

REPLY [7 votes]: The sequence grows too rapidly and I doubt that one can get meaningful asymptotics for the largest prime factor.  However, one can show that the largest prime factor is bounded by $(n\sqrt{n})^{2^{\pi(n)}}$ where $\pi(n)$ is the number of primes up to $n$, and since $\pi(n) \sim n/\log n$, this implies a stronger bound than what you wanted.  
To see this, put $\alpha =1+ \sqrt{2} + \ldots + \sqrt{n}$ and let $\epsilon=(\epsilon_1,\ldots,\epsilon_n)$ be any choice of signs.  Set $\epsilon(\alpha)= \sum_{j=1}^{n} \epsilon_j \sqrt{j}$.  Note that $\alpha$ and all the $2^n$ values $\epsilon(\alpha)$ all belong to the field $K={\Bbb Q}(\sqrt{p}, 2\le p\le n)$ which has degree $2^{\pi(n)}$ and Galois group $H \simeq ({\Bbb Z}/2)^{\pi(n)}$. Note that the choice of all possible signs $\epsilon$ forms a group $G \simeq ({\Bbb Z}/2)^n$, and $H$ may be thought of as a subgroup of $G$ in the obvious way.  Then 
$$ 
A_n = \prod_{\epsilon \in G/H} N_K(\epsilon(\alpha)), 
$$ 
where the product is over the $2^{n-\pi(n)}$ cosets $G/H$, and $N_K$ is the norm in $K$ (the product over all the elements of $H$).  Note that all the $N_K(\epsilon(\alpha))$ are non-zero integers.  Further all these norms are bounded by $(n\sqrt{n})^{2^{\pi(n)}}$ (there are $2^{\pi(n)}$ conjugates and each is bounded in size by $\sqrt{1} +\ldots + \sqrt{n} \le n\sqrt{n}$).  
Thus $A_n$ can be written as the product of $2^{n-\pi(n)}$ integers each of size at most $(n\sqrt{n})^{2^{\pi(n)}}$, and therefore its largest prime factor is no more than that.<|endoftext|>
TITLE: Forcing the negation of CH without adding Cohen reals over L
QUESTION [7 upvotes]: Suppose CH + "there are no Cohen reals over L". Can we force the negation of CH without adding any Cohen real over L?

REPLY [7 votes]: Let $\mathbb{S}$ denote Sacks forcing. Let $\mathbb{S}_{\omega_2}$ denote the $\omega_2$-length countable support iteration of Sacks forcing. 
Let $G_{\omega_2} \subseteq \mathbb{S}_{\omega_2}$ be generic over $L$ for $\mathbb{S}_{\omega_2}$. By the usual results around proper forcing, you can show that $L[G_{\omega_2}] \models 2^{\aleph_0} = \aleph_2$. 
You can show that no Cohen reals over $L$ are added by showing that every real of $L[G_{\omega_2}]$ is contained in a ground model coded closed meager set. Note that Cohen forcing is forcing equivalent to $\mathbb{P}_{I_\text{meager}}$, the forcing of nonmeager Borel sets. Hence a Cohen real is not contained in any ground model meager set. 
For simplicity, let first consider one Sacks forcing $\mathbb{S}$. Let $\tau \in V^{\mathbb{S}}$ and $\tau$ be a name for a real not in the ground model. Then do a fusion argument to produce a condition $p$ so that at every split node $s$ of $p$, $p_{s0}$ and $p_{s1}$ determines a certain finite amount of $\tau$ and what $p_{s0}$ determines about $\tau$ and what $p_{s1}$ determines about $\tau$ differs in at least two places. If you let $T$ be the set of all finite strings $t$ so that $t$ is an initial segment of what $p_s$ can determined about $\tau$ for some $s \in {}^{<\omega}2$, then you get a tree so that at each split node the two sides differ two times before splitting again. By the definability of forcing, $T \in L$. You can show that body $[T]$ is meager closed set and $p \Vdash \tau \in [\check T]$. 
If you understand how to do this for 1-Sacks forcing, now do an iterated Sacks forcing version of this. This tends to be quite heavy in notation. See Geschke and Qickert $\textit{On Sacks Forcing and the Sacks property}$ for more information. Essentially the argument above is a modified version of their proof of the 2-localization property. Also see their paper on how to do the iterated Sacks version of the 2-localization property.  

It seems that I understood the question as is it ever possible to force the continuum to be $\aleph_2$ without adding Cohen generic over $L$. It seems the far more interesting question is whether  for any universe $V$ (perhaps not equal to $L$) which does not have Cohen generic over $L$, is there a forcing extension of $V$ which does not add Cohen generics over $L$.<|endoftext|>
TITLE: Adelic and classical modular forms on quaternion algebras
QUESTION [9 upvotes]: Let $R$ be an Eichler order of an indefinite quaternion algebra $B/\mathbb{Q}$ (suppose B is not the collection of $2\times 2$ matrices) and $S$ the corresponding Shimura Curve. Modular forms of weight $2k$ on $S$ can be written as either
(1) Functions on the idele group $B^*_\mathbb{A}$ of $B$
(2) Holomorphic functions on the upper half plane that satisfy the required transformation property under the corresponding Fuchsian group.
It is well-known that these definitions are equivalent, but I can't find a reference that writes down both
(A) An isomorphism between (1) and (2) in this setting. (which I imagine is just given via pullback under $B^*_\mathbb{A} \xrightarrow{g \mapsto g_\infty(i)} \mathcal{H}$)
(B) Defines the Hecke operators.
Can you please give me a reference? An electronic reference, if available, would be very much appreciated.

REPLY [3 votes]: Many of these questions have to do with strong approximation; in (only somewhat) precise terms, it states that if $G$ is a simply connected semisimple group which is simple over $\mathbb Q$, and if $G(\mathbb{R})$ is not compact, then $G(\mathbb {Q})$ is a dense subgroup of $G(\mathbb{A}_f)$ ($\mathbb{A}_f$ is the ring of finite adeles over $\mathbb{Q}$). Consequently, if $K$ is a compact open subgroup of $G(\mathbb{A}_f)$, and $\Gamma =G(\mathbb{Q}\cap K$ (then $\Gamma $ is a congruence arithmetic subgroup of $G(\mathbb{Q})$), then $G(\mathbb{Q})K=G(\mathbb{A}_f)$ . In particular, functions on the quotient $G(\mathbb{A})/G(\mathbb{Q})$ which are invariant under $K$ may be identified with functions on $G(\mathbb{R})/\Gamma$. 
All this can be applied to $G=SL_1(B)$; to extend it to $GL_1(B)$ needs a little more book-keeping (since the latter is not simply connected) and is done in many places like the Corvallis volumes.
In particular, if $\pi$ is a discrete series representation of $SL_2(\mathbb{R})$ of lowest weight $k$ which occurs in $G(\mathbb{R})/\Gamma$ the lowest weight vector may be viewed as a modular form of weight $k$ on the upper half plane with some transformation property under the Fuchsian group $\Gamma$. By the preceding, this function on $G(\mathbb{R})/\Gamma$ may also be viewed as a function on $G(\mathbb{A}/G(\mathbb{Q})$, with prescribed behaviour on the archimedean component<|endoftext|>
TITLE: sums of zero-free entire functions and its siblings on the disk
QUESTION [5 upvotes]: Can one describe the set $\{e^f+e^g: f, g\in H(C)\}$ in some way?
For example, in unital Banach algebras, every element  has this form.
I am in particular interested in the problem whether the function $u=(1-e^L)^2$ with
$L=(1-z)^{-3}$ is a sum of two exponentials in H(D).
Here $L$ is chosen so that every value is taken by u in D (hopefully).

REPLY [2 votes]: Describing the set $\{ e^f+e^g:f,g\in H(C)\}$ is difficult, and probably it cannot be described in any reasonable form. However many conditions can be given for
a function $h$ not to be of this form. For example, $h$ cannot be a square (or any power) of an entire function. Proof: Suppose that there are entire functions
$f,g,w$ such that $e^f+e^g=w^n$, and $w$ is not constant. Then $e^{f-g}$ omits zero, and all solutions of $e^{f(z)-g(z)}+1=0$ are multiple. By the Second Main theorem of Nevanlinna, $f-g$ must be constant, so $w$ must be constant, a contradiction.
For functions in the unit disk, this argument will prove that functions
of sufficiently fast growth are not representable in the form $e^f+e^g$.
In particular, this covers your example.
Indeed, using the notation of Nevanlinna theory, we have
$$T(r,u)\geq m(r,u)\geq c(1-r)^{-2},$$
so the conditions of the Second Main Theorem hold, and the above argument shows that $u$ is not a sum of two exponentials.
Reference: Hayman, Meromorphic functions, Chap. 2.<|endoftext|>
TITLE: Ascending surfaces in the 4-ball
QUESTION [5 upvotes]: Let the standard symplectic structure on $B^4$ (viewed in $\mathbb{R}^4$ or $\mathbb{C}^2$) be given by  $\omega=(1/2) d \eta$, for 
\begin{align*}
\eta &:= 
x_1 \, dy_1 - y_1 \, dx_1 + x_2 \, dy_2 - y_2 \, dx_2 \\
&=\tfrac{i}{2}(z \, d\bar{z}-\bar{z} \, dz +w \, d\bar{w} - \bar{w}\, dw).
\end{align*}
The restriction of $\eta$ to the 3-sphere of radius $r$ defines a contact structure $\alpha_r:= \eta\mid_{S^3_r}$.

Definition. Let $F$ be a smooth oriented surface properly embedded in the unit 4-ball $B^4 \subset \mathbb{C}^2$, assumed to miss the origin and be transverse to $\partial B^4 = S^3 \subset \mathbb{C}^2$. We say that $F$ is ascending if 

the distance function $\rho(z,w)=\sqrt{|z|^2+|w|^2}$ is Morse when restricted to $F$, and
at all regular points of $\rho\mid_F$, we have $(d\rho \wedge \eta)\mid_F>0$.


The definition comes from this paper of Boileau and Orevkov. They make the following assertion, which they claim follows immediately from the definition of an ascending surface: Let $F\subset B^4$ be an ascending surface. If $p \in F$ is a critical point of $\rho\mid_F$, then $T_p F = \ker \alpha_r$, where $r=\rho(p)$; that is to say that $T_p F$ is a complex plane.

Question. Why does $T_p F = \ker \alpha_r$ at critical points $p$? Isn't the following a counterexample?

Example. Let $V$ be the affine plane $\{(\tfrac{1}{2},t,0,s) \in \mathbb{R}^4\}$ and $F=V \cap B^4$. Then $\rho\mid_F$ is Morse with a unique critical point at $(\tfrac{1}{2},0,0,0)$. For condition (2), we note that $F$ intersects $S_{1/2}$ in the above critical point and $S_{1/2+\epsilon}$ in a circle parametrized by $\gamma(t)=(\tfrac{1}{2},t,0,\sqrt{\epsilon^2+\epsilon -t^2})$. We can evaluate $\alpha_{1/2+\epsilon}$ on $F \cap S_{1/2+\epsilon}$ for $\epsilon>0$:
\begin{equation*}
(\alpha_{1/2+\epsilon})_{\gamma(t)} \gamma'(t) = \big(\tfrac{1}{2}\, dy_1-t \,dx_1+0 \, dy_2 - \sqrt{\epsilon^2+\epsilon-t^2} \, dx_2\big)\begin{bmatrix}0 \\1 \\ 0 \\ \frac{-t}{\sqrt{\epsilon^2+\epsilon-t^2}}\end{bmatrix}= \tfrac{1}{2}>0.
\end{equation*}
Therefore $F$ is an ascending surface. But at the critical point $p=(\tfrac{1}{2},0,0,0)$, we see that $T_p F$ contains $(0,1,0,0)$ and $\big(\alpha_{1/2}\big)_{(1/2,0,0,0)}(0,1,0,0)= 1/2$. Therefore $T_p F \not \subset \ker \alpha_{1/2}$.

REPLY [4 votes]: To see why $T_pF=\ker\alpha_r$ at critical points $p$ of $\rho|_F$, note that Morse-ness of $\rho|_F$ means that $d(\rho|_F)$ vanishes to precisely first order at $p$. On the other hand, we have
$$
(d\rho\wedge\eta)|_F=fd\mathrm{vol}_F
$$
for some $f\colon F\to\mathbb R_{\ge0}$. Because $f$ is nowhere negative, it can only vanish to even order, which means that $\eta|_F$ must also vanish to at least first order at $p$. But this says precisely that
$$
T_pF\subset(\ker d\rho\cap\ker\eta)=\ker\alpha_r,
$$
and equality follows from equality of the dimensions.
The reason your example doesn't contradict this is that you have only parametrized half the circle. To parametrize the other half with the same orientation, you would take
$$
\gamma_2(t)=\left(\frac12,-t,0,-\sqrt{\epsilon^2+\epsilon-t^2}\right),
$$
and there the same calculation gives $\eta_{\gamma_2(t)}(\gamma_2'(t))=-\frac12$.<|endoftext|>
TITLE: Why is $(A^\perp)^\perp = A$?
QUESTION [5 upvotes]: On page 52 of this paper, Iwasawa considered the bilinear symmetric non-degenerate pairing $\Phi_n \times \Phi_n \rightarrow \mathbb{Q}_p/\mathbb{Z}_p$ defined by
$$\langle \alpha, \beta \rangle_n := \text{the class of } T_n(\alpha\beta) \text{ in } \mathbb{Q}_p / \mathbb{Z}_p$$
where $\Phi_n = \mathbb{Q}_p(\zeta_n)$, $\zeta_n$ being $p^{n+1}$-th root of unity and $T_n$ is the trace of the field extension $\Phi_n/\mathbb{Q}_p$. He then wrote

For any closed subgroup $A$ of [the additive group] $\Phi_n$, we denote by $A^\perp$ the annihilator of $A$ in $\Phi_n$ relative to this pairing. Then $A^\perp$ is a closed subgroup of $\Phi_n$ such that $(A^\perp)^\perp = A$, ...

Unfortunately, I do not see the reasons for his statement that $(A^\perp)^\perp = A$. I know that it is true in the case of non-degenerate bilinear form and $A$ is a subspace but the pairing given is not a bilinear form. So how can I prove that?

REPLY [9 votes]: The trace form is a perfect pairing in the sense of harmonic analysis, i.e. it identifies $\Phi_n$ with its Pontryagin dual, see Weil's "Basic Number Theory" or Tate's thesis. Therefore, your statement follows from the Pontryagin duality, for example Proposition 3.6.1 in the book "Principles of Harmonic Analysis" by Deitmar/Echterhoff.<|endoftext|>
TITLE: Relations between some works by Deligne-Mostow and Thurston
QUESTION [7 upvotes]: Happy new year 2016! 
A coworker and I are interested in the relations between the works of Deligne and Mostow ([DM] and [M]) on the monodromy of Appell-Lauricella hypergeometric functions (Publ. IHES 1986) and the work of Thurston [T] on moduli spaces of flat stuctures with prescibed conical singularities on the Riemann sphere (Geom. Topol. Monogr. 1998). By different approachs, Deligne and Mostow on the one hand and Thurston on the other obtain some similar results, in particular a list of 94 complex hyperbolic orbifolds, some of which are non-arithmetic.
Thurston's paper [T] has been published more than 10 years after [DM] has been written (the latter has been received by the IHES editorial board in 1983). Hence it seems at first sight that Thurston (who refers to Picard and to [DM] and [M]) just gives a geometrical reinterpretation (in terms of flat structures with conical singularities on the Riemann Sphere) of some of the results previously obtained by Deligne and Mostow. 
However [T] is a revised version of a preprint titled `Shapes of polyhedra' dating 1987. This leads me and my coworker to wonder what were the respective motivations of Deligne and Mostow and of Thurston in the papers [DM] and [T] ?
It seems that Deligne’s motivations were not at all influenced by Thurston’s (maybe it was a bit different for Mostow but we know nothing about this). Indeed, [DM] is presented as a continuation of some classical works by Schwarz, Picard and Levavasseur and more recent ones by Terada about the monodromy groups of hypergeometric functions and there is no reference to any work by Thurston whatsoever. Moreover, Deligne wasn't aware of Thurston's paper [T] when my coworker contacted him (recently) in order to learn more about the genesis of [DM].
So, one can ask more precise forms of the above question:
(1) From where came the motivations of Thurston when he was working on the material of [T] ?
(2) When did Thurston begin to work on moduli spaces of singular flat structures on the Riemann sphere?
(3) Was he aware of or influenced by the work by Deligne and Mostow at that time?
My coworker as already received some interesting contributions about this by several people. However, none of them contains definitive answers and there are a lot of guesses in what he received.
Maybe that somebody on MO can help…
Thurston's 1987 preprint could possibly put some light on these questions but we were unable to find a version of it. We would be grateful is someone could provide us an electronic copy of this document.
Thanks in advance for any help.

References:
[DM] Deligne P., Mostow G., Monodromy of hypergeometric functions and nonlattice integral monodromy. Inst. Hautes Études Sci. Publ. Math. 63 (1986), 5–89.
[M] Mostow G., Generalized Picard lattices arising from half-integral conditions.
 Inst. Hautes Études Sci. Publ. Math. 63 (1986), 91–106.
[T] Thurston W., Shapes of polyhedra and triangulations of the sphere.
 In`The Epstein birthday schrift', 511–549, Geom. Topol. Monogr. 1, 1998.

REPLY [7 votes]: Thurston had been thinking of flat structures for at least 15 years prior to his preprint, in particular, the idea of using Teichmuller theory (via flat surfaces) to study interval exchange transformations is due to him (he told this to Bill Veech in the late seventies, Veech first thought Thurston was nuts, but quickly saw the light); in a lot of Thurston's foliations work, quadratic differentials, and thus flat structures, are central.
Troyanov's paper (on flat structures, in Ens. math.) comes from the time he was visiting Thurston (as a student) in the mid-eighties. The result, was not, shall we say, a surprise to Thurston.
As for Deligne-Mostow, yes, Thurston was familiar with it, and, in fact, this fact was one of the factors in getting the Geometry Center funded (Mostow was on the committee, and Bill, just for amusement, showed him a computer program which would generate the complex hyperbolic orbifolds - Mostow was very impressed...)<|endoftext|>
TITLE: How to cite authors from any country correctly?
QUESTION [26 upvotes]: It has always seemed to me that the Mathematical Community gives a high importance to the act of properly citing an author (Do not write Erdos! It's Erdős. Cauchy must be read as in French, not as in English...).
Hence, I thought that it might be useful to ask your opinion about how to cite correctly a foreign and/or complicated author name in the references.
It seems that Bibtex handles only standard English style names well, and requires many workarounds for foreign names (see here). So it is better not to count on it too much.
Feel free to point out any suggestion, or even the problems you have come across when citing authors.
I'll start a list:

✔ First, Middle, Last - Names. 
OK, I think there are no doubts, "John Horton Conway" should be cited as "J. H. Conway". Note the whitespaces after each period.
✔ Spanish names: See the very good answer of Leo Alonso.
✔ "Nobiliary" names (von, van der, etc.): See the answer of 
R. van Dobben de Bruyn.
Chinese names: Here I am often in trouble. I read that Chinese write their surname first and then their given name, but I think that in papers they are usually swapped. Also I heard that many Chinese have the same surname.

THE FREE LOOKUP ANSWERS

The user zeno pointed out that the tool MRLookUp http://www.ams.org/mrlookup is free an can be used to get the bibtex of any article tracked by the AMS.
Federico Poloni also suggested MRef http://www.ams.org/mref.

THE "LET'S JUST CLOSE" ANSWERS
Some users voted to close this question for different reasons. I answered this question thinking that how to cite an author correctly is often a problem for mathematicians, so address it on MO could have been useful for many users. I can agreed to close this question, probably choosing zeno answer as the best, BUT before I would like to see more comments and opinions, especially about Asian names. 
NOTES:
1) Many answered: "just look on MathSciNet". Unfortunately, MathSciNet is not free for everyone, so I think this is a quite unsatisfactory method.
2) The question is not about the transliteration of names, you can assume that the author already have a name written in a reasonable set of characters extending the Latin. The question is about the abbreviation of names in references.

REPLY [6 votes]: There is also the singular case of Yahya ould Hamidoune, the Mauritanian mathematician. In fact, ould is not a name: it just means 'the son of'. It is used in the same spirit of the Italian patronyms de or de' (as in Bruno de Finetti or Lorenzo de' Medici), so it should be treated in the very same way. However, most of Hamidoune's papers are cited (or even signed by Hamidoune himself!) as if 'Yahya' and 'ould' were two first names, which is why you will typically read 'H. O. Hamidoune' or 'Hamidoune, H. O.' in bibliographies. (In still other cases, matters are just cut short by dropping the 'ould'.)
In an ideal world, I think the mistake (if you also take it as such) should be definitely fixed: Hamidoune's papers would be indexed as 'H. ould Hamidoune' or 'ould Hamidoune, H.' (depending on the journal style), and ordered by making reference only to 'Hamidoune' (if they have to be ordered alphabetically). But I'm urged to be realistic: this won't happen for many reasons.<|endoftext|>
TITLE: Can you hear the shape of a drum by choosing where to drum it?
QUESTION [71 upvotes]: I find the problem of hearing the shape of a drum fascinating. Specifically, given two connected subsets of $\mathbb R^2$ with piecewise-smooth boundaries (or a suitable generalization to a riemannian manifold) it is in general impossible to use the spectrum of the laplacian to uniquely identify the shape of the domain.
As a simple example, reported in the Wikipedia link above, these two shapes

have identical laplacian spectra, as explained more at length by Gordon and Webb in American Scientist 84 no. 1, 46 (1996) (jstor). This can be shown by taking any eigenfunction $\varphi$ on $D_1$, cutting it up into the marked triangle regions, and taking the specified linear combinations on $D_2$ to make a laplacian eigenfunction with the same boundary conditions, correct matching at the internal boundaries, and the same eigenvalue.
What bugs me about this problem is that the spectrum is not really what you hear on a real-world drum. More specifically, the spectrum gives you the frequencies of the fundamentals and overtones that the drum is allowed to produce, but it does not tell you in what proportions it will do so. If you hit a circular drum in the centre, you will excite a large amplitude of the fundamental, resulting in a deeper sound, but if you drum it near the edge you will produce a rapping sound with a stronger proportion of higher overtones.
The set of sounds that a drum can produce is therefore better modelled by the pair of

its Laplace-Beltrami spectrum, together with
the set $T$ of timbres it can produce. 
By a timbre I mean the sequence of eigenfunction weights when the drum is hit at $x$, i.e. $$\tau(x)=\left( \sqrt{\sum_k|\varphi_{n,k}(x)|^2} \right)_{n=1}^\infty,$$ where the $\varphi_{n,k}$ are the (possibly multiple) orthonormalized eigenfunctions corresponding to the eigenvalue $\lambda_n$, and two timbres are thought of as equivalent if they only differ by a global constant (which trivially corresponds to how hard you drum). 
(This quantity is meant to model the amount of energy in each eigenspace after a point excitation to the wave equation (via e.g. $\Delta \varphi-\partial_t^2\varphi=0$, $\varphi(y,0)=0$, $\partial_t\varphi(y,0)=\delta(x,y)$), and perhaps there are cleaner or equivalent definitions for $\tau(x)$ based on that PDE. However, the definition above is plenty to satisfy the intuition of how the musical timbre depends on the drumpoint from a physicist's perspective.)
The set of all timbres is then $T=\{\tau(x):x\in D\}$: that is, you're provided all the timbres, but not which place in the geometry they come from. (As pointed out in the comments, initial knowledge of the geometry renders the whole thing moot.)

In this paradigm, two isospectral surfaces $D_1$ and $D_2$ would still be 'acoustically distinguishable' if, say, $D_1$ contains a point $x^*$ which produces a timbre $\tau^{(1)}(x^*)$ that is not matched by the timbre $\tau^{(2)}(y)$ of any point $y\in D_2$. In other words, you can't see the drum, and you can't see where the drummer is hitting it, but if drummer 1 can produce a timbre that drummer 2 cannot, then the drums must be different.
(Alternatively, one might have access to a function $u\mapsto \tau(f(u))$, where $u$ lives in some standard domain $D_0$ like the unit disk and $f:D_0\to D$ is an unknown (homeo/diffeo/etc)-morphism. That is, you get a control that specifies where the drum should be hit, but you don't know exactly what it does. You can then say two drums $D_1$ and $D_2$ are indistinguishable if there's a morphism $g:D_1\to D_2$ such that $\tau^{(2)}(g(x))=\tau^{(1)}(x)$ for all $x\in D_1$, i.e. a mapping from each point $x$ in $D_1$ to an identically-sounding drum point in $D_2$.)
So, to come to my question: has this paradigm been explored in the literature? For the known isospectral-but-not-isomorphic regions (or more complex manifolds), are any pairs known to be acoustically indistinguishable as defined above? Or must acoustically indistinguishable surfaces be isometric? Does this concept have an established name in the literature? If it has been explored, are there good introductory surveys (say, pitched at a level where a physicist might understand them)? If it hasn't, are there fundamental reasons why this is a much harder problem overall?

REPLY [36 votes]: Sometimes you can
Since there seem to be no hard results along these lines on the literature, I decided to have a look at the two shapes in the question and see if there are some relatively accessible results and happily, as it turns out, there are. In particular:

The isospectral surfaces $D_1$ and $D_2$ in the question are acoustically distinguishable: there exist points $x_1^*\in D_1$ and $x_2^*\in D_2$ such that no point $x_2\in D_2$ has the same timbre $\tau^{(2)}(x_2)$ as $\tau^{(1)}(x_1^*)$ and  no point $x_1\in D_1$ has the same timbre $\tau^{(1)}(x_1)$ as $\tau^{(2)}(x_2^*)$.

Drummers hitting those drums can therefore prove conclusively which drum they are using by an appropriate choice of the drumming point.
To explore the eigenfunctions, I used the Helmholtz solver detailed in this Mathematica.SE thread, which when implemented directly returns eigenvalues for the two surfaces within a few parts in $10^{-4}$ of each other. The first few eigenfunctions for the two domains look something like this:

What really matters here is the nodes, shown as thick black lines, though of course the timbres $\tau(x)$ as I defined them contain more information in terms of the relative weights of the eigenfunctions when they're not zero. To begin with, already by the fourth eigenfunction we're able to 'count' the shape of the drum as in Carlo Beenakker's answer (i.e. if you have access to the number of nodal domains for that eigenvalue, then you can immediately tell them apart). The nodes, however, also give quick and clear examples of the $x^*$s we seek. 
In particular, consider the nodes of $\varphi_2$ and $\varphi_4$ for the two domains, shown in black and red respectively,

and similarly the nodes of $\varphi_2$ and $\varphi_5$:

Specifically, note that the nodes of $\varphi_2$ and $\varphi_4$ cross for $D_2$ but not for $D_1$, so a drummer hitting the drum in that position will produce a sound with strong components of $\lambda_1, \lambda_3, \lambda_5, \lambda_6, \lambda_7$, and so on, but with no Fourier component along the frequencies $\lambda_2$ and $\lambda_4$; this is completely impossible for a drummer using $D_1$.
Similarly, a drummer using $D_1$ can produce a sound with no $\lambda_2$ or $\lambda_5$ component by hitting the intersection of the corresponding nodes, and such a sound is impossible to produce using $D_2$.
The Mathematica code used to produce this answer is available here.

It seems that the full question, however, is still open: whether you always can, or whether you sometimes can't. That is, it would be interesting to know whether there exist non-isometric surfaces which are not only isospectral, but also acoustically indistinguishable in the sense laid out in the question.<|endoftext|>
TITLE: Estimating the size of solutions of a diophantine equation
QUESTION [155 upvotes]: A. Is there natural numbers $a,b,c$ such that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ is equal to an odd natural number ?
(I do not know any such numbers). 
B. Suppose that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$ is equal to an even natural number ($a,b,c $ are still natural numbers) then is there any way to estimate the minimum of $a,b $ and $c$ ?
The smallest solution that I know for  $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4 $ is:

REPLY [73 votes]: This exact problem is the subject of the paper "An Unusual Cubic Representation Problem" by Andrew Bremner (ASU) and myself. It was published in Volume 43 (2014) of Annales Mathematicae et Informaticae, pages 29-41.
It is proven that strictly positive solutions never exist for $n$ odd. They sometimes do not exist for $n$ even, and, even if they do, they can be of truly enormous size - much larger than the example given.<|endoftext|>
TITLE: Perfect group of order 190080
QUESTION [10 upvotes]: I need to know some properties of the perfect group of order $190080$
which is the Schur cover of the Mathieu group ${\rm M}_{12}$, but
when using PerfectGroup(190080), GAP runs so slowly.
Is there any other method in GAP for getting this group?

REPLY [12 votes]: Just write
gap> G := PerfectGroup(IsPermGroup,190080);
M12 2^1

in order to get the desired group as a permutation group:
gap> GeneratorsOfGroup(G);
[ (3,6)(7,10)(9,12)(13,16)(15,18)(19,22)(20,23)(21,24),
  (1,2,3)(4,5,7)(6,8,9)(10,11,13)(12,14,15)(16,17,19)(18,20,21)(22,24,23),
  (1,4)(2,5)(3,7)(6,10)(8,11)(9,13)(12,16)(14,17)(15,19)(18,22)(20,24)(21,23)
 ]

Now computations should be fast.
(If you omit the first argument, you get the group as a finitely presented group, and computations with such groups are inefficient for the obvious reasons.)<|endoftext|>
TITLE: first order languages over graphs (and other discrete models)
QUESTION [5 upvotes]: A lot of research has been devoted to the study of first order language of graphs (FO). Formulas in this language are constructed using variables $x, y,\dots$ ranging over the vertices of a graph, the usual quantifiers ∀, ∃, the usual logical connectives ¬, ∨, ∧, etc., parentheses and the binary
relations =, ∼, where x ∼ y denotes that x and y are adjacent. . In FO one can for instance write “G is triangle-free” as ¬∃x, y, z : (x ∼ y) ∧ (x ∼ z) ∧ (y ∼ z).
In this context, a lot has been studied for 0-1 laws concerning sequences of graphs $\{G_i\}_{i\geq 1}$. For instance, a really strong result due to Shelah and Spencer says that that if $p = n^{−α}$ with $0 ≤ α ≤ 1$ fixed, then the FO-zero-one law holds if and only if α is an irrational number.
Possibly this is obvious for someone coming from logic (I am not from the area), but I would like to know if there are similar models not for graphs. In particular, I am interested on knowing if there (or if it has been studied or defined) FOs (or extensions of FOs) over
1.- on subsets of integers (roughly speaking, those are 1-uniform hypergraphs), here it is not clear to me how to define a binary relation. For instance, one would like to encode here that a certain set is $k$-AP free, or a sum-free set.
2.- more generally to $k$-uniform hypergraphs with $k>2$ (definitely we need more than a binary relation).
References would be greatly appreciated

REPLY [3 votes]: To complement J.-E. Pin's answer, many other structures have been considered in the literature, like words [1], discrete metric spaces [2], maps [3] or simplicial complexes [4].
In [5] the relations between some of these results are investigated as instances of a more abstract theorem. 
The case of $k$-uniform hypergraphs has been treated in [6], and fairly unsurprisingly most results on graphs convert to similar results on hypergraphs in a straightforward (yet tedious) way.
The general case of hypergraphs (non-uniform, with multiple edges allowed) has been treated in [7], using the encoding as bipartite graphs. Again, the results are fairly straightforward extensions from the graph case.
The convergence law of random graphs has three components, the structure, the language, and the probability distribution.
Most of the previously mentioned results have a general likeness to each other, which suggests rather to explore stronger langages (extensions of FO, MSO) or more complex distributions. The difficulty is to find a balance, where you can still get something; there are a number of negative results for example for fragments of Second Order logic.
If you are interested in these questions, you can start with the introduction of [8].
Two last interesting papers interesting results that are related but open up different horizons. 
In [9], the authors studies matroids in the framework of Nešetřil and de Mendez (full FO-convergence).
In [10], the author uses analogues of the Rado graph to obtain set-theoretic results.
To conclude, there is no general survey of "FO on discrete structures"; the existing ones are more specific and/or a bit old. You can start with [11], which is nice and easy to read.
This may be a bit messy since there are many things to cover, so don't hesitate to ask for details on some points!
[1] James Lynch, "Convergence laws for random words"
[2] Dhruv Mubayi, Caroline Terry, "Discrete metric spaces: structure, enumeration, and 0-1 laws"
[3] Bender, Compton, Richmond, "0-1 Laws for Maps"
[4] Andreas Blass, Frank Harary, "Properties of almost all graphs and complexes"
[5] Andreas Blass, Yuri Gurevich, "Zero-One Laws: Thesauri and Parametric Conditions"
[6] Nicolau C. Saldanha, Márcio Telles, "Some examples of asymptotic combinatorial behavior, zero-one and convergence results on random hypergraphs"
[7] Nans Lefebvre, "Convergence law for hyper-graphs with prescribed degree sequences"
[8] Peter Heinig, Tobias Muller, Marc Noy, Anusch Taraz, "Logical limit laws for minor-closed classes of graphs"
[9] Frantisek Kardos, Daniel Kral, Anita Liebenau, Lukas Mach, "First order convergence of matroids"
[10] Joel David Hamkins, "Every countable model of set theory embeds into its own constructible universe"
[11] P. Winkler, "Random structures and zero-one laws"<|endoftext|>
TITLE: Does $S^n\times H^k$ have non-isometric conformal transformations?
QUESTION [8 upvotes]: Question: Consider the product Riemannian manifold $(M,g)=(S^n\times H^k,g_{round}\oplus g_{hyp})$ of a round sphere $(S^n,g_{round})$ of curvature $1$ and hyperbolic space $(H^k,g_{hyp})$ of curvature $-1$. Is it true that the only conformal transformations of $(M,g)$ are isometries? In other words, is $\mathrm{Conf}(M,g)=\mathrm{Iso}(M,g)$?

Although I  am mainly interested in the case $n>k>1$, it would also be nice to know the answer for all $n,k\geq2$. Note that by a result of Ferrand (see Schoen), the action of $\mathrm{Conf}(M,g)$ is proper, hence there exists a metric $g_*$ conformal to $g$, such that $\mathrm{Conf}(M,g)=\mathrm{Iso}(M,g_*)$.

REPLY [4 votes]: Corollary 1 here https://projecteuclid.org/download/pdf_1/euclid.jmsj/1261148578 answer positively your question.<|endoftext|>
TITLE: Can the potential of a complete Kahler metric be bounded?
QUESTION [6 upvotes]: Let $X$ be a complex manifold and $\omega$ a Kahler form on $X$. A smooth function $\rho$ is called a potential of $\omega$ if $i\partial\bar\partial\rho=\omega$. By intuition, it seems that $\rho$ can never be bounded if $\omega$ is a complete Kahler metric. 
Is the following claim true: the potential of a complete Kahler metric can never be bounded?
We can consider an example that $X=\Delta=\{z\in \mathbb C; |z|<1\}$. Assume $\rho$ is a bounded smooth funtion on $\Delta$ such that $i\partial\bar\partial\rho$ gives a complete metric on $\Delta$. Replace $\rho$ by the averaging of it w.r.t rotations, it still induces a complete metric. So we can assume $\rho$ depends only on $r=|z|$. Then up to a constant $i\partial\bar\partial\rho=(\rho''(r)+\rho'(r)/r)dx\wedge dy$. The completeness of $i\partial\bar\partial\rho$ implies $\int^1_0\sqrt{\rho''(r)+\rho'(r)/r}dr=\infty$. 
So the above question can more or less be reduced to the following one: deos there exit a bounded smooth function $\rho$ on $[0,1)$ such that $\rho''(r)+\rho'(r)/r>0$ and $\int^1_0\sqrt{\rho''(r)+\rho'(r)/r}dr=\infty$?
I believe the answer is no, but I can not give a proof. 
Thanks a lot!

REPLY [3 votes]: The claim is false. The potential $\psi$ of the complete Kahler metric on the disk given in Theorem 1.1 of http://arxiv.org/pdf/math/0603530v7.pdf is bounded.
Indeed, such potential $\psi$ is the pullback of the potential of $P := |z_1|^2 + |z_2|^2 + |z_3|^2$ by the holomorphic map $X$. Since the image of $X$ is bounded it follows that $ \psi = X^* P$ is bounded.<|endoftext|>
TITLE: Should axiomatic set theory be translated into graph theory?
QUESTION [32 upvotes]: Recently I saw the abstract of a paper by Nash-Williams: ``Should axiomatic set theory be translated into graph theory?''. The abstract, taken from Mathscinet says the following:

The author translates the Zermelo-Fraenkel axioms for set theory ($ZF$) into graph-theoretic language by replacing "set'' by "vertex'' and the membership relation "$∈$'' by the preneighborhood relation "$→$''. A digraph $D$ is an ordered pair $(V(D),E(D))$ such that $V(D)$ is a class (of vertices) and $E(D)⊆V(D)×V(D)$ is the subclass of ordered pairs $(ξ,η)$ such that $ξ$ is a preneighbor of $η$. He describes those graph-theoretic properties a digraph must satisfy in order to be a model of $ZF$ (with or without the axiom of choice ($AC$)). The theorem that $AC$ is independent of $ZF$ can then be formulated as a theorem of graph theory asserting the existence of certain kinds of digraphs. 
The author proposes that a translation of this and other independence proofs might yield further insight into why the theorems are true. He also suggests that a textbook on axiomatic set theory as the study of digraphs satisfying certain conditions might increase its accessibility to nonspecialists. However he cautions that translations of independence theorems from one language to another must justify their existence by achieving some substantial progress in either set theory or graph theory.

Now my question is:
Question. Are there any extra works regarding the authors's suggestion ``other independence proofs might yield further insight into why the theorems are true.''?

REPLY [28 votes]: Although it may seem on the face of it that this proposal is just a question of terminology — yes, a model of set theory is a certain kind of acyclic digraph — nevertheless, my opinion is that one can indeed get some insight by thinking this way. 
In particular, the main results of my paper on the embedding phenomenon arose out of an explicitly graph-theoretic perspective on the models of set theory, viewing the models of set theory as certain special acyclic digraphs.
Joel David Hamkins, Every countable model of set theory embeds into its own constructible universe, J. Math. Log. 13 (2013), no. 2, 1350006, 27. blog post
For example, I proved that the countable models of set theory are linearly pre-ordered by embeddability: for any two such models, one of them is isomorphic to an induced subgraph of the other. Furthermore, embeddability is determined by the heights of the models, and from this it follows that there are precisely $\omega_1+1$ many bi-embeddability classes. So actually, the countable models of set theory are pre-well-ordered by embeddability! Every nonstandard model of set theory is universal for all countable acyclic digraphs. 
The proof uses universal digraph combinatorics, including an acyclic version of the countable random digraph, which I call the countable random $\mathbb{Q}$-graded digraph, and higher analogues arising as uncountable Fraisse limits, leading eventually to what I call the hypnagogic digraph, a set-homogeneous, class-universal, surreal-numbers-graded acyclic class digraph, which is closely connected with the surreal numbers.
It happens that I am just now at the JMM in Seattle, where I will speak on The hypnagogic digraph, with applications to embeddings of the set-theoretic universe on Friday afternoon at the special session on the surreal numbers. 
Finally, let me mention that it is an open question whether one can prove in ZFC that there is no graph-embedding of the set-theoretic universe $V$ to the constructible universe $L$, when $V\neq L$. In a joint project currently underway with many authors, however, we have made some significant progress, without yet settling the full question.<|endoftext|>
TITLE: Finding limit-nondecreasing sets for certain functions
QUESTION [6 upvotes]: This is a question that arose a while ago in work with Damir Dzhafarov on some pieces of reverse mathematics. As far as I know, it has no deep significance; however, it feels like the sort of thing we ought to know. The solution is probably very simple, but I don't see it.
Fix a computable total function $f$ of two variables such that $$f(x, s)\ge f(x, s+1),$$ and let $$F(x)=\lim_{s\rightarrow\infty}f(x, s).$$ Then there is an infinite set $X\subseteq\omega$ satisfying $$i, j\in X, i<j\implies F(i)\le F(j).$$ Call such an $X$ $f$-good. My question is:

How complicated must such an $X$ be?

Specifically,

Is there an $f$ as above such that any $f$-good $X$ computes $0'$?

Note that trivially, $0'$ will always compute such an $X$, so this is the upper bound. 
It's easy to show that there are not always computable such $X$ - indeed, there are $f$ such that any $f$-good $X$ computes a DNR function. However, coding $0'$ seems very difficult, because the "current guess" to whether $F(i)\le F(j)$ can only change a (bounded) finite number of times. (Note that by contrast, if we look at $f$ satisfying $f(x, s)\le f(x, s+1)$, then it is very easy to code $0'$.)

EDIT: We can ask an even weaker question:

Suppose $d$ computes an $f$-good $X$ for every computable $f$. Must $d\ge_T0'$?

This ought to be easier to show, but I am at a loss. I can't even show that if $Y$ is such that for every computable $f$ there is some $n$ such that $Y\setminus n$ is $f$-good, then $Y$ computes $0'$!

REPLY [4 votes]: You're right. This is the kind of things we ought to know.
Let $\mathsf{LNS}$ be the statement that every such function f has an f-good set.
Results
I claim that $\mathsf{LNS}$ admits 

cone avoidance (if A is non-computable, then any computable instance of $\mathsf{LNS}$ has a solution X such that A is not X-computable)
preservation of hyperimmunity (if $A_0, A_1, \dots$ are countably many hyperimmune sets, then every computable instance of $\mathsf{LNS}$ has a solution X such that the $A$'s are hyperimmune relative to $X$).

Therefore, $\mathsf{LNS}$ implies neither $\mathsf{ACA}_0$, nor $\mathsf{ADS}$.
Combinatorics
Fix a computable instance $f : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$,
let $f_0 : \mathbb{N} \to \mathbb{N}$ and $\tilde{f} : \mathbb{N} \to \mathbb{N}$
be the functions defined by $f_0(x) = f(x, 0)$ and $\tilde{f}(x) = \lim_s f(x, s)$, respectively.
Note that $f_0$ is computable, while $\tilde{f}$ is not.
If there is some $i$ such that $\tilde{f}(x) = i$ for infinitely many $x$,
then pick the least such $i$, and notice that the set $\{ x : \tilde{f}(x) = i \}$
is c.e. and every infinite computable subset of it is $f$-good.
So from now on, we will assume that for every $i$ and almost every $x$, $\tilde{f}(x) > i$.
We will build an $f$-good set by forcing, using a variant of Mathias forcing $(F, X)$, where

$F$ is a finite set over which $\tilde{f}$ is non-decreasing.
$X$ is an infinite set such that $\max F < \min X$.
for every $x \in F$ and $y \in X$, $\tilde{f}(x) \leq \tilde{f}(y)$.

In particular, $(\emptyset, \omega)$ is a valid condition,
and given a condition $(F, X)$ and a finite set $H \subseteq X$ which is non-decreasing for $\tilde{f}$,
$(F \cup H, X \setminus [0, n])$ is a valid extension for some $n \in \mathbb{N}$.
We now want to decide a $\Sigma^0_1$ formula $\varphi(G)$ given a condition $(F, X)$.
Let $\mathcal{C}$ be the $\Pi^{0,X}_1$ class of all functions $h : \mathbb{N} \to \mathbb{N}$ dominated by $f_0$,
such that $\varphi(F \cup H)$ does not hold for every finite set $H \subset X$ over which $h$ is non-decreasing.
There are two cases.

Case 1: $\mathcal{C}$ is empty. In this case, since $\tilde{f} \not \in \mathcal{C}$,
there is a finite set $H \subseteq X$ such that $\varphi(F \cup H)$ holds and over which $\tilde{f}$ is non-decreasing.
The condition $(F \cup H, X \setminus [0, n])$ for some sufficiently large $n$ is a valid extension forcing $\varphi(G)$ to hold.
Case 2: $\mathcal{C}$ is non-empty. By weak K\"onig's lemma, pick an $h \in \mathcal{C}$
and $h \oplus X$-computably thin-out the set $X$ to obtain an infinite set $Y$ over which $h$ is non-decreasing.
The condition $(F, Y)$ forces $\varphi(G)$ not to hold.

An example
Suppose for example that we want to prove that $\mathsf{LNS}$ admits cone avoidance.
Let $A$ be a non-computable set and $f$ be a computable instance of $\mathsf{LNS}$.
We work with conditions $(F, X)$ as above, with the extra property that $A \not \leq_T X$.
Given a condition $(F, X)$ and a $\{0,1\}$-valued Turing functional $\Gamma$,
I claim there is an extension forcing $\Gamma^G \neq A$. 
To see this, for each $n \in \mathsf{N}$ and $i \in \{0,1\}$, let $\mathcal{C}_{n, i}$  be the $\Pi^{0,X}_1$ class 
of all functions $h : \mathbb{N} \to \mathbb{N}$ dominated by $f_0$,
such that $\Gamma^{F \cup H}(n) \uparrow$ or $\Gamma^{F \cup H}(n) \downarrow \neq i$
for every finite set $H \subset X$ over which $h$ is non-decreasing.
Consider the $X$-c.e. set $S = \{ (n, i) : \mathcal{C}_{n,i} = \emptyset \}$.
We have three cases.

Case 1: $(n, 1-A(n)) \in S$ for some $n$. In this case, since $\mathcal{C}_{n,1-A(n)} = \emptyset$,
there is a finite set $H \subseteq X$ such that $\Gamma^{F \cup H}(n) \downarrow = 1-A(n)$ and over which $\tilde{f}$
is non-decreasing. The condition $(F \cup H, X \setminus [0, n])$ for some sufficiently large $n$
forces $\Gamma^G(n) \neq A(n)$.
Case 2: $(n, 0)$ and $(n, 1) \not \in S$ for some $n$. In this case, by the cone avoidance basis theorem,
there are two functions $h_0 \in \mathcal{C}_{n,0}$, $h_1 \in \mathcal{C}_{n,1}$ such that
$A \not \leq_T h_0 \oplus h_1 \oplus X$. By $h_0 \oplus h_1 \oplus X$-computably thinning-out
the set $X$, we obtain a set $Y$ over which both $h_0$ and $h_1$ are non-decreasing.
The condition $(F, Y)$ forces $\Gamma^G(n) \neq 0$ and $\Gamma^G(n) \neq 1$, so forces $\Gamma^G(n) \uparrow$.
Case 3: $S = \{ (n, A(n)) : n \in \mathbb{N} \}$. In this case, we can $X$-compute $A$, contradiction.

This completes the proof of cone avoidance.

EDIT: Note that one can easily use cone avoidance of $\mathsf{LNS}$ as a blackbox to build for any non-computable set $A$ a set $X$ such that $A \not \leq_T X$ and for every computable $f$, there is some $n$ such that $X \setminus [0, n]$ is $f$-good.
Indeed, fix the set $A$, and consider Mathias conditions $(F, X)$ such that $A \not \leq_T X$. For every such condition $(F, X)$ and every computable function $f$, one can apply cone avoidance of $\mathsf{LNS}$ to obtain an $f$-good set~$Y \subseteq X$ such that $A \not \leq_T Y$. The condition $(F, Y)$ forces $G$ to be $f$-good up to finite changes. Moreover, for every condition $(F, X)$ and every $e$, there is an extension $(H, Y)$ forcing $\Phi_e^G \neq A$. Every sufficiently generic for this notion of forcing yields the desired set.<|endoftext|>
TITLE: Existance of Integrating Factors, a Constructive Proof
QUESTION [11 upvotes]: Being a novice with differential equations, I recently learned that if $\mu$ is an integrating factor for $\frac{dy}{dx}f(x,y)+ g(x,y)=0$, then the corresponding 1-form, $\mu fdy+\mu g dx$, is exact.
My question is, is there some constructive proof that given a 1-form,
$fdy+gdx$, we can find a scalar function $\mu$ so that $\mu fdy+\mu gdx$ is exact, and under what conditions can we do so?  The only proofs for the existence of integrating factors I've found so far use existence theorems for solutions to ODEs and then prove the there is an integrating factor based on the existence of a solution.
To be clear, I would be very satisfied with an answer that looked like: subdivide a region into squares; linearly interpolate a scalar function to ensure the integral around each square is zero; cut each square into smaller squares; repeat.  I would also be satisfied (though less so) with a less-constructive fixed-point theorem method.  But, I'd like to stay in the context of 1-forms if possible.

REPLY [4 votes]: This seems indeed equivalent to the classical question treated by Darboux, Painleve, Poincare and others. Not surprisingly it can have several different formulations.
Classical formulation is the following: Given a polynomial equation
$f(x,y)dx+g(x,y)dy=0$, to find with finitely many operations an integrating factor, or to prove it does not exist. The next question is "what kind of integrating factor"? The simplest case is a rational integrating factor (though more general factors were considered in the classical papers). If we look for a  rational integrating factor of degree at most $n$, we can just write it with undetermined coefficients, and the question is reduced to solving a finite system of algebraic equations. There is an algorithm of doing this, if the original equation is over rational (or algebraic) numbers. So the real question is: can we bound the degree of the factor?
Then the question is "bound in terms of what"? A relatively recent breakthrough
in this problem is the counterexample: one cannot bound the degree of the factor in terms of degrees of $f,g$, and the types of singularities: there is a one-parametric family of equations, with singularities of fixed type (that is the type is independent of parameter) such that
for a countable dense set of values of parameter it has integrating factors, but their degrees tend to infinity.
MR1914932
Lins Neto, Alcides,
Some examples for the Poincaré and Painlevé problems. 
Ann. Sci. École Norm. Sup. (4) 35 (2002), no. 2, 231–266.
This paper also contains a survey of previous results.
However it remains a possibility that the degree of the integrating factor can be estimated in terms of some arithmetic properties of the coefficients. 
Here is a survey of the topic:
MR2166493 
Llibre, Jaume
Integrability of polynomial differential systems. Handbook of differential equations, 437–532, Elsevier/North-Holland, Amsterdam, 2004. 
All this was about the global aspect of the problem (as I said it has many formulations). The local aspect is usually called the Center-Focus Problem:
for the same kind of system, to determine whether an equilibrium point is a center or a focus. There is an enormous literature on this, and this problem is much better understood. Just look at the
"center-focus problem" as a keyword.<|endoftext|>
TITLE: Is this system always solvable in radicals by quartics, octics, $12$-ics, etc?
QUESTION [7 upvotes]: While considering this post, it made me wonder about its generalization in another direction and from the perspective of Galois theory. 

Question: Is it true that, given four constants ($\alpha,\beta,\gamma,\delta$), then the system,
$$\begin{aligned}
x_1^a+x_2^a+x_3^a+x_4^a &= \alpha\\
x_1^b+x_2^b+x_3^b+x_4^b &= \beta\\
x_1^c+x_2^c+x_3^c+x_4^c &= \gamma\\
x_1^d+x_2^d+x_3^d+x_4^d &= \delta
\end{aligned}\tag1$$
can be solved in radicals $x_i$ for any positive integer power ($a,b,c,d$) and $a<b<c<d$?

Example: Let's choose the year Back to the Future came out,
$$\begin{aligned}
x_1+x_2+x_3+x_4 &= \color{brown}1\\
x_1^2+x_2^2+x_3^2+x_4^2 &= \color{brown}9\\
x_1^4+x_2^4+x_3^4+x_4^4 &= \color{brown}8\\
x_1^8+x_2^8+x_3^8+x_4^8 &= \color{brown}5
\end{aligned}$$
Two solutions are given by the roots of the octic,
$$4707 + 9036 x - 7495 x^2 - 5096 x^3 + 3024 x^4 + 1848 x^5 - 896 x^6 - 256 x^7 + 128 x^8 = 0$$
Magma says this is $8T47$, has order $1152=2^7\cdot3^2$ and is the largest solvable order for deg $8$. (In fact, it factors over $\sqrt{1441}$). Using the root $r_i$ numbering system of Mathematica, we find that,
$$x_1,\, x_2,\, x_3,\, x_4 = r_1,\, r_2,\, r_5,\, r_6\quad \text{(Solution 1)}\\
\text{or}\quad\quad\\
x_1,\, x_2,\, x_3,\, x_4 = r_3,\, r_4,\, r_7,\, r_8\quad \text{(Solution 2)}$$
Remarks: Using generic ($\alpha,\beta,\gamma,\delta$)

For exponents $a,b,c,d = 1,2,c,8$ and $c=3,4,5$, one ends up with a $8T47$ octic with order $1152=2^7\cdot3^2$.
For $a,b,c,d = 1,2,6,7$, one gets a $12T294$ with order $82944 = 2^{10} \cdot 3^4$.
For $a,b,c,d = 1,2,c,8$ and $c=6,7$, one now gets a $16T1947$ with order $7962624 = 2^{15}\cdot3^5$. These three are the largest solvable orders for their respective degrees.
Using others yields deg $20,24,$ etc.

So, is the system $(1)$ always solvable in radicals?

REPLY [6 votes]: For $(a,b,c,d)=(1,2,3,20)$ and $(\alpha,\beta,\gamma,\delta)=(1,1,1,0)$ one computes that $x_1$ is a root of the irreducible polynomial $4 x^{20} - 20 x^{19} + 40 x^{18} - 40 x^{17} + 195 x^{16} - 704 x^{15} + 1050 x^{14} - 700 x^{13} + 475 x^{12} - 900 x^{11} + 900 x^{10} - 300 x^{9} + 130 x^{8} - 260 x^{7} + 130 x^{6} + 20 x^{4} - 20 x^{3} + 1$, which according to magma, has a non-solvable Galois group, namely the full wreath product $S_4^5\rtimes S_5$.
Without Magma: With the same values of $a,b,c,d,\alpha,\beta,\gamma,\delta$ as above, one computes (e.g. using Sage) that $x_1x_2x_3x_4$ is a root of $h(x)=4 x^{5} - 175 x^{4} + 300 x^{3} - 130 x^{2} + 20 x - 1$. Now $h(x-1)$ is an Eisenstein polynomial with respect to $5$, and modulo $3$ $h(x)=(x + 1) \cdot (x + 2) \cdot (x^{3} + 2 x^{2} + x + 1)$ is a factorization into irreducibles. Thus the Galois group of $h$ contains the alternating group $A_5$.<|endoftext|>
TITLE: homological 2 dimensional groups
QUESTION [5 upvotes]: In a  Commentarii Mathematici Helvetici paper  by Benno Eckman  and Heinz  Müller in  1980 (volume  50,  pages 510-520) proved  that poincaré Duality  Groups  of  dimension 2  with  positive first Betti  number  are  surface  groups. 
Is  there any  development  into  proving  that certain groups  of   (homological) or  geometric dimension  2  are surface  groups?

REPLY [4 votes]: See my paper with Bruce Kleiner "Geometry of quasiplanes" for general unital commutative rings, it also contains a reference to the paper by Eckmann  and Linnell from 1983 where they prove the theorem for PD(2) groups over Z and to the one by Bowditch which works over Q.<|endoftext|>
TITLE: Can Sacks forcing add a Cohen generic real over $L$?
QUESTION [10 upvotes]: Motivated by this question Forcing the negation of CH without adding Cohen reals over L and Todd Eisworth's comment, the question is the following:
1) Suppose $V$ has no Cohen generic reals over $L$. (i.e. every real is contained in a constructibly coded Borel meager set.) Let $\mathbb{S}$ be Sacks forcing. Let $G \subseteq \mathbb{S}$ be $\mathbb{S}$-generic over $V$. Can $V[G]$ have a Cohen generic real over $L$?

Of course, by the usual fusion argument, there are no Cohen generic reals over $V$ in $V[G]$. So in the particular case of $V = L$, the answer is no. 
Also if $V$ has stronger properties like $V$ has no unbounded reals over $L$, then Sacks forcing over $V$ can not add an unbounded real over $L$ and in particular no Cohen real over $L$. 

Also, I seem to recall it is an open question whether in countable support iterations $\langle P_\alpha : \alpha \leq \omega \rangle$ of $\langle  \dot Q_n : n \in \omega \rangle$ such that $P_n \Vdash$ "$\dot Q_n$ adds no Cohen reals", $P_\omega$ adds a Cohen real over $V$. 
This seems to suggest this question for finite length iterations is known. So a second question is
2) If $P$ does not add Cohen reals and $1_{P} \Vdash \dot Q$ does not add Cohen reals, does the two step iteration $P * Q$ add Cohen reals?
If the answer in 2 is no, then I think Question 1) has a negative answer when $V$ is a set-generic extension of $L$. 
Thanks for any information on this question.

REPLY [12 votes]: Regarding Question 2:
Zapletal's solution to the "Half-Cohen problem" gives an example of a 2-stage iteration $P*\dot Q$ such that $P$ does not add Cohen reals, and $P$ forces that $\dot Q$ does not add Cohen reals over $V^P$, yet the iteration $P*\dot Q$ does add a Cohen real over the ground model.
Reference is:
Dimension theory and forcing, Topology and Its Applications 167 (2014) 31-35
In particular, your Question 2 is mentioned at the top of page 2 of the preprint I linked to.<|endoftext|>
TITLE: embedding of quaternionic projective spaces
QUESTION [9 upvotes]: Let $\mathbb{H}P^m$ be the $m$-th quaternionic projective space. What is the smallest integer $N$ such that there exists an embedding
$$
\mathbb{H}P^2\longrightarrow \mathbb{R}^N?
$$
Are there any references?

REPLY [14 votes]: I. M. James, Lectures on algebraic and differential topology, pp. 134–174, Lecture Notes in Math., Vol. 279, Springer, Berlin, 1972,
Theorems 1.2 and 1.3 show that $$N=13.$$<|endoftext|>
TITLE: Ring of invariants for the regular representation
QUESTION [8 upvotes]: The symmetric group $S_n$ acts on $\mathbb C^n$ by permuting the coordinates. In this case the ring of invariants is generated by elementary symmetric polynomials in n-variables. Now consider the regular representation of $S_n$, the basis of this vector space is indexed by the elements of $S_n$. Then what are the generators for the ring of invariants ? Is there a reference where the generators are given explicitly ? What are the degrees of the generators ?

REPLY [11 votes]: To the best of my knowledge this is an open problem. In fact, there is strong evidence that the problem is very hard indeed: Consider the action of $S_n$ on the "two-sets," i.e., on the subsets of $\{1,\ldots,n\}$ of two elements. It is easy to see that this is a subrepresentation of the regular representation. So if generating invariants for the regular representation are known, then these map to generating invariants for the action on the two-sets, so they are also known. But such invariants have only been canclulated for $n \le 5$.
The invariants for the action on two-sets are interesting because they provide a way to decide the graph isomorphism problem. So this provides further evidence for the hardness of the problem: If there were an easy way of giving generating invariants for the regular representation, there would also be an easy way to decide the graph isomorphism problem.
You can find some background on this in Chapter 5 of the book "Computational Invariant Theory" by Harm Derksen and Gregor Kemper.<|endoftext|>
TITLE: Is there a formula for the Frobenius-Schur indicator of a rep of a Lie group?
QUESTION [18 upvotes]: Let $G$ be a simple algebraic group group over $\mathbb C$.
Let $V$ be a self-dual representation of $G$.
Let $\lambda$ be the highest weight of $V$.
Write $\lambda$ as a sum of fundamental weights: $\lambda=\sum \lambda_i\omega_i$ for $\lambda_i\in\mathbb N$.
Is there a simple formula for the Frobenius-Schur indicator of $V$ in terms of the numbers $\lambda_i$?

For the reader's convenience, I recall the definition of the Frobenius-Schur indicator. It is $1$ if the trivial rep occurs inside $Sym^2(V)$, and it is $-1$ if the trivial rep occurs inside $Alt^2(V)$.

REPLY [5 votes]: As Jeff Adams indicates, there is a detailed proof in Bourbaki's Chapter 8 (originally published in French in 1975, but later translated into English).   See especially Table I at the end of Chapter 8 for a summary of all simple types in terms of their fundamental weights, referring back to $\S13$ for classical types and to $\S7$, no. 5, for exceptional types.  While the Onishchik-Vinberg book is quite useful as a learning tool, it is not a reasonable choice as a reference book.   Anyway, it's worthwhile to take a somewhat broader look at what is really going on here.
(1)  The term "Frobenius-Schur indicator" goes back to the joint 1906 paper by Frobenius and his student Schur in which they focused on a finite group $G$ and its complex irreducible characters.   Such a character $\chi$ arises from an irreducible matrix representation $\rho$ of $G$ over $\mathbb{C}$.    By the usual averaging argument, there is certainly a unitary representation equivalent to $\rho$ which affords $\chi$.   A natural  further question is whether there is a real representation of $G$ affording $\chi$.  Clearly the answer is no if $\chi$ takes nonreal values.  Then the original Frobenius-Schur indicator takes value 0, by definition. But even if $\chi$ is $\mathbb{R}$-valued, it may come from either a real orthogonal representation (then the indicator is written 1) or from a symplectic representation (then the indicator is $-1$).   A major step in the Frobenius-Schur work was a formula for the indicator computable from the character values: sum the values $\chi(g^2)$ over all $g \in G$, then average by dividing out $|G|$.
There are accounts of all this in many textbooks, such as those by Isaacs and by Serre (but don't start with their indexes; in Serre, try 13.2).     
(2) Though I haven't gone far into all the subsequent history, it's clear that the underlying question arises whenever you start with finite dimensional representations over $\mathbb{C}$ and ask what happens over $\mathbb{R}$.  In particular, the Frobenius-Schur question makes sense for complex (semi-)simple Lie groups and their finite dimensional representations parametrized by highest weights $\lambda$.   Here it is certainly possible to have indicator $=0$, when the underlying vector space $V$ admits no nondegenerate invariant bilinear form.   This case is avoided by assuming that the representation is self-dual: in case $V \cong V^*$ as a $G$-module, we get End$(V) \cong V^* \otimes V \cong V \otimes V$, so there is always a 1-dimensonal trivial $G$-submodule (corresponding to the scalar endomorphisms).   Since an irredudible $G$-module has at most one such nondegenerate form, it remains to decide whether it is orthogonal or symplectic.   
The combinatorics of roots and weights leads to the formal recipe indicated in a couple of the answers, which at first looks opaque.   But as Bourbaki points out, writing $m:=\sigma(\lambda)$ (in the notation used here), the half-integer $m/2$ has a natural interpretation as the sum of coefficients in the expression for $\lambda$ as a $\mathbb{Q}$-linear combination of simple roots.   Note in particular that when the root lattice equals the weight lattice (in types $G_2, F_4, E_8$), this half-integer is actually an integer and $m$ is even (so all indicators are 1).
(3) Generalizations in a number of directions have also been worked out, for example in the case of finite dimensional Hopf algebras (not just finite group algebras) or compact Lie groups or complex semisimple algebraic groups.   The formula discussed in the answers depends only on the root system, so it can be potentially applied over many kinds of fields including local fields and those of prime characteristic:   involutive maps are lurking everywhere.<|endoftext|>
TITLE: Why are quantum groups so called?
QUESTION [19 upvotes]: I've recently been to a seminar on quantum matrices. In particular the speaker introduced these objects as the coordinate ring of $2$ by $2$ matrices modulo some odd looking relations (see start of Section 2 here). As a theoretical physicist, I'm struggling to understand in what sense these objects are quantum!
Has anyone got any references or knowledge which might help answer

Where do these odd looking relations come from?
In what sense are the matrices quantum (can the non-commutativity in the coordinate ring be understood as emerging from some quantisation procedure...)?
Should I think of quantum groups as a controlled mechanism for introducing non-commutativity into coordinate rings in general?

Many thanks in advance for your expertise!

REPLY [2 votes]: Let me add a few words referring in particular to the paper you cited: 
"From totally nonnegative matrices to quantum
matrices and back, via Poisson geometry" S. Launois and T.H. Lenagan
So we have a matrix 
$$( a \quad b )$$
$$ ( c \quad d ) $$
and the relations 
1) Four relations of the type XY = q YX 
 $$ ab = q ba, ac = q ca, bd = q bd , cd = q cd  $$
they cover four out of 6 relations between basic elements,
relations between diagonal terms a,d and anti-diagonal b,c are not covered.
2) relation for b,c is just commutativity: $$ bc = cb $$
3) and the last one is really "odd looking" here I will agree with you:
$$ ad - da = (q-q^{-1} ) bc $$
One way to motivate such relations is Manin's point of view - "coaction on a quantum place" - as explained by Theo Johnson-Freyd here:  Intuition behind the definition of quantum groups
Alternative point of view appeared before by Drinfeld:
Look at the Section 3.1 page 11 from the paper you cited: 
here you can see "classical limit" i.e. Poisson brackets for these relations :
1) Four relations of the type {X, Y}  = XY 
 $$ \{a, b \} =  ab, \{a, c \} =  ca, \{b, d \}=  bd , \{c, d \} =  cd  $$
they cover four out of 6 relations between basic elements,
relations between diagonal terms a,d and anti-diagonal b,c are not covered.
2) $$ \{ b, c \}  = 0$$
3) and the last one:
$$ \{a,d \} = 2bc $$ 
The claim is that quantum relations above are quantization of the Poisson bracket relations here. I.e. non-commutative algebra of observables is deformation quantization of the Poisson algebra here. 
Deformation quantization is now quite understood by efforts of Kontsevich et.al.
However without refering to big theories, as you a physicists you probably can easily see that
relations {X,Y} = XY should be quantized to XY = q YX - that are  relations of the type 1.
Why this is true: look at canonical quantization {p,q} = 1, quantized to [p,q]= h. 
Consider exponentials: {exp(p), exp(q) } = exp(p) exp(q) - we see exactly the same Poisson bracket as (1) - so it is just the exponents of canonical variables, so quantization is easy just exp(p) and exp(q) which in quantum world will give $$ exp(p) exp(q) = e^h exp(q) exp(p)$$,
so exactly the quantum relation (1) which we saw above.
{b,c} = 0, is quantized to bc = cb - is natutal
About the third relation - well I do not know easy way to quantize it from basic principles. If you know - tell me. 
So your question 
"Where do these odd looking relations come from? "
can be translated to the question: where do these Poisson relations come from ?
That is Drinfeld's insight.
He asked the following question - consider a Lie group and invariant Poisson bracket on it, we want Poisson structure would be COMPATIBLE with the group structure (in certain easy to guess sense). And from that compatibility you will get that kind of Poisson relations. Such groups are called Poisson-Lie groups as mentioned in answer by AHusain above. The group is GL(2) for examples above.
Concerning question 2:
In what sense are the matrices quantum (can the non-commutativity in the coordinate ring be understood as emerging from some quantisation procedure...)?
Yes as it was discussed above we have Poisson brackets which are quantized to that kind of noncommuative relations.
Question 3:
Should I think of quantum groups as a controlled mechanism for introducing non-commutativity into coordinate rings in general?
NO, that is not the case. 
Quantization needs only the Poisson bracket as an input.
Only in special case when Poisson bracket is somehow naturally related to some semisimple group you will get something related to quantum group.
In general you can take arbitrary Poisson bracket , quantize it and it will have nothing to do with the quantum groups.<|endoftext|>
TITLE: $\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$ implies $u\in L^{6/5}(\Omega)$
QUESTION [10 upvotes]: I posted this question first in Math.StackExchange one week ago here, but I didn't get an answer or a helpful comment so I repost it here:
Let $d=3$ and $\Omega\subset \mathbb R^d$ is a bounded Lipschitz domain and $u$ is a measurable function.
A sufficient condition for the integral $\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$ is that $u\in L^{6/5}(\Omega)$ which follows from Holder's inequality and the (continuous) embedding $H^1(\Omega)\hookrightarrow L^6(\Omega)$.
Question: Is the opposite true, i.e is it true that
$$\int\limits_{\Omega}{uvdx}<\infty,\forall v\in H_0^1(\Omega)$$ implies $u\in L^{6/5}(\Omega)$ or at least $u\in L^1(\Omega)$ ?
My thoughts: It is easy to see that $u\in L^1_{loc}(\Omega)$ by taking $v$ to be smooth cut-off functions equal to $1$ in compact subsets of $\Omega$ and $0$ in a neighborhood of the boundary $\partial \Omega$.
The motivation for this question is the "correct" weak formulation of a nonlinear problem - whether to formulate it as $(1)$ or as $(2)$:
$(1)$ Find $u\in H_0^1(\Omega)$ such that $f(u)\in L^{6/5}(\Omega)$ and 
$$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$
or
$(2)$ Find $u\in H_0^1(\Omega)$ such that $\int\limits_{\Omega}{f(u)vdx}<\infty,\forall v\in H_0^1(\Omega)$ and 
$$a(u,v)+\int\limits_{\Omega}{f(u)vdx}=0,\forall v\in H_0^1(\Omega)$$
where $a(.,.)$ is a bilinear form and $f(.)$ is in general a nonlinear function. If the answer to my question is affirmative then both formulations are equivalent. 
Note that $(2)$ is less restrictive, because the set in which we search for a solution is bigger, so it might be easier to find such.

REPLY [15 votes]: That doesn't work because $H_0^1$ functions are small near the boundary, so testing against them won't detect bad behavior of $u$ near $\partial\Omega$.
For a concrete example, take $\Omega$ as the unit ball and $u(x)=1/(1-|x|)\notin L^1$. Then
$$
\int |uv|\, dx \le \left( \int \frac{v^2\, dx}{(1-|x|)^{3/2}} \int \frac{dx}{(1-|x|)^{1/2}} \right)^{1/2} .
$$
If $v\in H_0^1$ is also smooth, then we can estimate the first integral in the same way as in this related question (by just integrating the gradient, starting from the boundary, to bound $v$). This gives $\int v^2/(1-|x|)^{3/2}\lesssim \|v\|^2_{H^1}$, so $\int |uv| \lesssim \|v\|_{H^1}$ for all such $v$, and by density of the smooth functions, this also holds for arbitrary $v\in H_0^1$.<|endoftext|>
TITLE: Uniqueness of hyperbolic rescaling
QUESTION [5 upvotes]: Let $X$ be a compact oriented surface of genus at least two, equipped with a Riemannian metric $g$.  By the uniformization theorem for Riemann surfaces, there is a conformal universal covering map $p:D\to X$, where $D$ is the unit disc.  The standard hyperbolic metric on $D$ is thus $e^{2f_0}p^*(g)$ for some function $f_0$ on $D$.  This is invariant under the group of deck transformations, and so descends to a function $f$ on $X$.  The hyperbolic metric has Gaussian curvature $-1$, and we deduce that $e^{2f}g$ also has curvature $-1$.
I think it is true that $f$ is the unique function such that $e^{2f}g$ has curvature $-1$, but this seems to be a bit harder to prove than I thought. Does anyone know a proof or reference?

REPLY [3 votes]: Here is an answer which I learned from a set of notes titled "Conformal Geometry Seminar - The Poincare Uniformization Theorem" by Gilbert Weinstein (DOI: 10.13140/RG.2.1.4644.9767).  Suppose we have found one metric $g$ in the original conformal class with curvature $K(g)=-1$.  Suppose also that $K(e^{2f}g)=-1$; we need to show that $f=0$.  A fairly standard formula tells us that $K(e^{2f}g)=(K(g)-\Delta(f))/e^{2f}$, so when $K(g)=K(e^{2f}g)=-1$ we obtain $\Delta(f)=e^{2f}-1$, so $f\Delta(f)=f(e^{2f}-1)$.  We now integrate over $X$.  It is another standard fact that $\int_Xa\Delta(b)=-\int_X\langle\nabla(a),\nabla(b)\rangle$, so we get 
$$ \int_X \|\nabla(f)\|^2 = - \int_X f(e^{2f}-1). $$
Note that the integrand $f(e^{2f}-1)$ is nonnegative, and can only be zero where $f=0$.  This is only consistent if $f$ is identically zero, as required.<|endoftext|>
TITLE: What is the "quaternionic" super Brauer group?
QUESTION [10 upvotes]: In addition to the two reasonably well-known categories $\mathrm{SuperVect}_{\mathbb R}$ and $\mathrm{SuperVect}_{\mathbb C}$ of real and complex super vector spaces, each of which is monoidally equivalent to corresponding category of $\mathbb Z/2$-graded vector spaces but with the Koszul sign rules, there is a third much less well-known symmetric monoidal category that I like to call the quaternionic super vector spaces $\mathrm{SuperVect}_{\mathbb H}$.  It appears, among other places, when studying the statistics of a certain type of pinor.
As a category,
$$ \mathrm{SuperVect}_{\mathbb H} = \mathrm{Vect}_{\mathbb R} \oplus \mathrm{Mod}_{\mathbb H} $$
The monoidal structure is a bit funny, using the Morita equivalence $\mathbb H \otimes \mathbb H \simeq \mathbb R$.  Here is a description of it.  Recall that the usual Galois correspondence between $\mathbb R$ and $\mathbb C$ identifies $\mathrm{Vect}_{\mathbb R}$ with the category of complex vector spaces $V_{\mathbb C}$ equipped with an antilinear involution, i.e. $\varphi: V_{\mathbb C} \to V_{\mathbb C}^*$, $\varphi^*\varphi = 1$.  Similarly, one can identify $\mathrm{Mod}_{\mathbb H}$ with the category of complex vector spaces $V_{\mathbb C}$ equipped with an antilinear "antiinvolution", i.e. $\varphi: V_{\mathbb C} \to V_{\mathbb C}^*$, $\varphi^*\varphi = -1$.  (By definition, $V^* = V$ as real vector spaces, but the $\mathbb C$-action is that $\lambda \in \mathbb C$ acting on $v\in V^*$ is given by the action $v\lambda^*$ in $V$.  So $\varphi^* = \varphi$ as real linear maps, but I'm thinking of them as $\mathbb C$-linear in two different ways.)  Using this, you can identify $\mathrm{SuperVect}_{\mathbb H}$ with the category of complex supervector spaces equipped with an antilinear map that squares to $1$ on the even part and to $-1$ on the odd part.  If you check carefully, you'll see that the tensor product (in $\mathrm{SuperVect}_{\mathbb C}$) of two such objects is naturally such an object, and in this way you can recover the symmetric monoidal structure on $\mathrm{SuperVect}_{\mathbb H}$.
In any sufficiently nice linear category (and $\mathrm{SuperVect}_{\mathbb H}$ is plenty nice for these purposes) you can develop a theory of associative algebras, bimodules, and Morita equivalence.  Among other things, you can define a Brauer group for your given category whose elements are Morita equivalence classes of Morita-invertible algebras.  (I.e. the group of units of the monoid whose objects are Morita equivalence classes of algebras and whose multiplication is tensor product.)
It is well known that the Brauer groups in this sense of $\mathrm{SuperVect}_{\mathbb C}$ and $\mathrm{SuperVect}_{\mathbb R}$ are respectively $\mathbb Z/2$ and $\mathbb Z/8$, and that this is closely related to periodicity in various K-theories.

Question:  What is the Brauer group of $\mathrm{SuperVect}_{\mathbb H}$?  What are the simple representatives of the elements?  What "K-theory" is it related to?

REPLY [9 votes]: The Brauer-Picard 2-category of $SuperVect_{\mathbb R}$ (let's call it $sBrPic_\mathbb R$) is the homotopy fixed points of the Brauer-Picard 2-category of $SuperVect_{\mathbb C}$ w.r.t. the involution given by complex conjugation (let's call that involution $C$).
It is plausible that the Brauer-Picard 2-category of $SuperVect_{\mathbb H}$ (call it $sBrPic_\mathbb H$) is the homotopy fixed points of  $sBrPic_\mathbb C$ w.r.t the involution $C\circ J$, where $J$ is the involution defined here: What are the "correct" conventions for defining Clifford algebras? 
As explained in the above link, $J$ acts homotopy trivially on $SuperVect_{\mathbb C}$.
The action of $C\circ J$ on the homotopy groups of $sBrPic_\mathbb C$ (which are $\mathbb Z/2,\mathbb Z/2,\mathbb C^\times$) is therefore the same as that of $C$, and so the homotopy fixed points spectral sequence for $C\circ J$ looks the same as that for $C$:
$$
\begin{matrix}
H^2(\mathbb Z/2,\pi_2(sBrPic_\mathbb C)) \\
H^1(\mathbb Z/2,\pi_1(sBrPic_\mathbb C)) & H^1(\mathbb Z/2,\pi_2(sBrPic_\mathbb C)) \\
H^0(\mathbb Z/2,\pi_0(sBrPic_\mathbb C)) & H^0(\mathbb Z/2,\pi_1(sBrPic_\mathbb C)) & H^0(\mathbb Z/2,\pi_2(sBrPic_\mathbb C))\\
\end{matrix}
$$
$$=\qquad
\begin{matrix}
\mathbb Z/2 \\
\mathbb Z/2 & 0 \\
\mathbb Z/2 & \mathbb Z/2 & \mathbb Z/2\\
\end{matrix}
$$
But the differential could be different:
there is room for a $d_2$ differential going from $(1,0)$ to $(0,2)$.
That differential is present iff
$\pi_0(sBrPic_\mathbb H)$ has order four, and also iff $\pi_1(sBrPic_\mathbb H) = 0$.
And indeed, $\pi_1 = 0$ because the ``odd line'' is not invertible in the category $SuperVect_{\mathbb H}$.
So you were right: it looks like $\pi_0(sBrPic_\mathbb H) = \mathbb Z/4$.

My answer raises the question of what is the homotopy fixed points of $J$ acting on $sBrPic_\mathbb C$, or $sBrPic_\mathbb R$?

I have no reason to believe that this is related to a version of $K$-theory.
Do you know a version of $K$-theory related to the category of (non-super) vector spaces? - probably not.
I just don't think that every linear symmetric monoidal category corresponds to a version of $K$-theory.

REPLY [2 votes]: Here I will try to record what I understand about the Brauer group of $\mathrm{SuperVect}_{\mathbb H}$.  I will suggest that the Brauer group of $\mathrm{SuperVect}_{\mathbb H}$ is a $\mathbb Z/4$, but I will not provide a complete proof.  Perhaps someone else will, or will provide a reference, or will point out something I missed.
Any Morita-invertible algebra in $\mathrm{SuperVect}_{\mathbb H}$ must complexify to a Morita-invertible algebra in $\mathrm{SuperVect}_{\mathbb C}$; the latter are either super matrix algebras or super matrix algebras tensored with (complex) Cliff(1).  Conversely, doing the descent is explained above, and it's reasonably clear that if a complex Morita-invertible superalgebra does descend to $\mathrm{SuperVect}_{\mathbb H}$, then its descendent is Morita-invertible.
So now let me try to find some Morita-invertible algebras in $\mathrm{SuperVect}_{\mathbb H}$.  A few observations:

We have, of course, the purely even algebras coming from the non-super Brauer group of $\mathbb R$, namely $\mathbb R$ and $\mathbb H$ (the latter is a real form of $\mathrm{Mat}(2)$).  But in this quaternionic world, there is a Morita equivalence $\mathbb R \simeq \mathbb H$.  Indeed, denote by $\mathbb J$ the purely-odd simple object in $\mathrm{SuperVect}_{\mathbb H}$; then $\mathrm{End}(\mathbb J) = \mathbb H$ by construction, and so $\mathbb J$ furnishes the claimed Morita equivalence.  (This is in contrast to the real and complex worlds, where the Picard group — the group of $\otimes$-invertible objects in the category — is a $\mathbb Z/2$ consisting of the even and odd simple objects; here the odd simple does not produce an auto-Morita-equivalence of $\mathbb R$ in the Brauer group, but rather a nontrivial equivalence.)
The Clifford algebras are formed by "quantizing" (graded) symmetric algebras on purely odd vector spaces.  By passing to the world of complex super vector spaces with antilinear (anti)involutions, as described above, one can check that $\mathrm{Sym}^2(\mathbb J) \cong \mathbb R^{\oplus 2} \oplus \mathbb J$.  Since $\mathbb J$ complexifies to $\mathbb C^{\oplus 2}$, one should hope to quantize this to a "quaternionic" form of $\mathrm{Cliff}(2)$.  If I did the check right, there is exactly one "quaternionic" form of $\mathrm{Cliff}(2)$, and it has the property that its even part is the algebra $\mathbb C$.  So this feels like a version of $\mathrm{Cliff}(1,-1)$, except a dimension count shows that it is Morita non-trivial.
By a dimension count, $\mathrm{Cliff}(1)$ does not have any quaternionic forms, but $\mathrm{Cliff}(3)$ might.  In fact, $\mathrm{Cliff}(3)$ has precisely two quaternionic forms; their even parts are $\mathrm{Mat}(2,\mathbb R)$ and $\mathbb H$ respectively.  (Their odd parts have dimension $\mathbb J^{\oplus 2}$.)  Again a dimension count verifies that they are not Morita-trivial — of course, this can also be seen by complexifying.

My intuition is that this is it — that these four Morita-classes are all invertible and that there aren't any others.  I don't have a strong argument in support of this intuition, and by no means do I claim a proof.
It was suggested to me that "quaternionic K-theory" usually means "symplectic K-theory" KSP.  That's an 8-periodic theory which does not have products — it is simply a shift by 4 of KO theory.  If there is a "K-theory" associated with $\mathrm{SuperVect}_{\mathbb H}$, probably it does have products, and my intuition is that it is 4-periodic.  Perhaps it is something like "4-periodicitized KO theory".<|endoftext|>
TITLE: Covering the zeros of 0/1 matrix with submatrices
QUESTION [7 upvotes]: The matrices I am dealing with are $n\times n$ of the following type (with $n=7$):
$M_7=\begin{pmatrix}1&0&0&0&0&0&1 \\ 1&1&0&0&0&0&0 \\ 0&1&1&0&0&0&0 \\ 0&0&1&1&0&0&0 \\ 0&0&0&1&1&0&0 \\ 0&0&0&0&1&1&0 \\ 0&0&0&0&0&1&1\end{pmatrix}$.
$M_n$ only has $1$'s on the main diagonal, on the diagonal just below the main diagonal and on the most upper right entry (that is, the entry (1,n)).
I define submatrices of $M_n$ with two subsets representing the rows and columns. For example the submatrix of $M_7$ defined by $r_1=\{1,2\}$ and $c_1=\{1,3,7\}$ is $\begin{pmatrix}1&0&1 \\ 1&0&0\end{pmatrix}$.

My problem is the following: I want to cover only the zeros of
  $M_n$ with as few submatrices as possible. The submatrices do not have to be disjoint.

Obviously, there is always an easy solution with $n$ submatrices. For $M_7$, this solution is:

$r_1=\{1\}$ and $c_1=\{2,3,4,5,6\}$
$r_2=\{2\}$ and $c_2=\{3,4,5,6,7\}$
$r_3=\{3\}$ and $c_3=\{1,4,5,6,7\}$
$r_4=\{4\}$ and $c_4=\{1,2,5,6,7\}$
$r_5=\{5\}$ and $c_5=\{1,2,3,6,7\}$
$r_6=\{6\}$ and $c_6=\{1,2,3,4,7\}$
$r_7=\{7\}$ and $c_7=\{1,2,3,4,5\}$

But can we cover $M_n$ with less submatrices?
I suppose the problem is difficult so I am looking for any help.
For example, if we have the cover for a matrix $M_n$, can we use the submatrices to cover $M_{n+1}$?
Thank you in advance for any advice!
EDIT: 
I found a solution with 6 submatrices for the example $M_7$:

$r_1=\{5,6,7\}$ and $c_1=\{1,2,3\}$
$r_2=\{1,2\}$ and $c_2=\{3,4,5,6\}$
$r_3=\{3,4,5\}$ and $c_3=\{1,6,7\}$
$r_4=\{1,4\}$ and $c_4=\{2,5\}$
$r_5=\{2,6\}$ and $c_5=\{4,7\}$
$r_6=\{3,7\}$ and $c_6=\{4,5\}$

EDIT II For the matrix $M_{15}$, there is a solution with 7 submatrices:

$r_1=\{1,2,7,10,15\}$ and $c_1=\{3,4,5,8,11,12,13\}$
$r_2=\{1,2,3,6,12,13\}$ and $c_2=\{4,7,8,9,10,14\}$
$r_3=\{1,4,5,11,14,15\}$ and $c_3=\{2,6,7,8,9,12\}$
$r_4=\{2,3,4,9,10,11,12\}$ and $c_4=\{5,6,7,13,14,15\}$
$r_5=\{3,4,5,6,7,8,9\}$ and $c_5=\{1,10,11,12,13,14,15\}$
$r_6=\{5,6,7,8,13,14,15\}$ and $c_6=\{1,2,3,9,10,11\}$
$r_7=\{8,9,10,11,12,13,14\}$ and $c_7=\{1,2,3,4,5,6,15\}$

REPLY [2 votes]: Here is another proof that the number of submatrices needed to cover the zeros of $M_n$ is at least $O(log(n))$.
Let $f(n)$ denotes the optimal number.
The argument uses the fact that the patterns of the columns are all different. Therefore the subsets of submatrices touching each column are also all different.
With $f(n)$ submatrices, the number of different possible subsets is obviously $2^{f(n)}-1$ (-1 because there are no columns full of ones).
Since there are $n$ columns, we have $2^{f(n)}-1\geq n$, or $f(n)\geq \log_2(n+1)$.
However, these arguments do not give the exact value of $f(n)$.
Here are the best values I obtain for$3\leq n \leq 17$. I would be interested if anyone has an idea of what the following sequence is?

$f(3)=3$
$f(4)=4$
$f(5)=5$
$f(6)=5$
$f(7)=6$
$f(8)=6$
$f(9)=6$
$f(10)=7$
$f(11)=7$
$f(12)=7$
$f(13)=7$
$f(14)=7$
$f(15)=7$
$f(16)=7$
$f(17)=7$<|endoftext|>
TITLE: Does a classical wave detect compact dimensions?
QUESTION [5 upvotes]: Please excuse if the question is too easy; I'm just not familiar enough with PDEs.
I'd like to understand a little bit classical implications of "adding compact dimensions" in Physics, that is, what would change if one considers classical Physics in $M\times \mathbb{R}^3$ instead of $\mathbb{R}^3$, where $M$ is compact and "small".
Here's a handwaving example of what I'm looking for. It's well known that a classical wave "detects" the parity of the dimension of $\mathbb{R}^n$: only for odd $n$ the wave originated at a point source would reach the observer once. Now, what would happen in $M\times \mathbb{R}^1$ when $M=S^1\times S^1$ of diameter $\epsilon$ when the distance between the source and the observer $l$ is much larger than $\epsilon$? 
One could replace a single point source in $M\times \mathbb{R}^1$ with a 2-d square grid of sources at distances $\epsilon$ from each other in $\mathbb{R}^3$, with the observer at distance $l >> \epsilon$ from the plane containing the grid. Then the waves from different sources in the grid would reach the observer at different times. This makes the point source wave in $M\times \mathbb{R}^1$ act qualitatively different from what one expect from a point source in $\mathbb{R}^1$: one would, er, "feel the disturbance in the force" long after one should, even if $M$ is "small".
My question is something like this: assuming that $\mathbb{R}^3$ is a reasonable model of a classical space, what makes it possible to consider $M\times \mathbb{R}^3$ and still arrive to reasonable conclusions?

REPLY [4 votes]: The compactification introduces periodic boundary conditions, and therefore the momentum in the extra dimension can take only discrete values. A particle that propagates in the extra compactified dimension then is associated with a ladder of momenta in our three-dimensional world. In a scattering experiment, one would observe this ladder as copies of the particle with different mass (because of the energy-momentum relation), socalled Kaluza-Klein copies. See On Kaluza-Klein States from Large Extra Dimensions.
For wave propagation in extra dimensions without compactification (where still we would only be able to observe three dimensions), see Pulse propagation in a hyper-lattice.<|endoftext|>
TITLE: Reverse-engineer forcing: am I reinventing the wheel?
QUESTION [12 upvotes]: In the course of a project I’m working on, I’ve started playing around with a sort of “reverse-engineering” forcing. It seems interesting, but
I have a sinking feeling I’m reinventing the wheel; does this construction already have a name?
(To me it feels in the same spirit as termspace forcing (https://mihahabic.wordpress.com/tag/termspace-forcing/), but not the same thing.)

Suppose $M\subset N$ are countable transitive models, $r\in N$ is a real (more generally, $r\in N$ is a subset of some $x\in M$ - but I’m only interested in the reals case for now), $\mathbb{P}$ is a forcing notion in $M$, and $\nu$ is a $\mathbb{P}$-name in $M$. Then - in $N$ - we can define a subforcing of $\mathbb{P}$, $\mathbb{P}_M[\nu=r]$, as follows:

$\mathbb{P}_M^0[\nu=r]$ is the set of $p\in\mathbb{P}$ such that there is no $n\in\omega$ such that $p\vdash$ "$\nu(n)=1-r(n)$".
$\mathbb{P}_M^\lambda[\nu=r]=\bigcap_{\alpha<\lambda}\mathbb{P}_M^\alpha[\nu=r]$.
$\mathbb{P}_M^{\alpha+1}[\nu=r]=\{p\in\mathbb{P}_M^\alpha[\nu=r]: \forall D\in\mathcal{P}_{do}^M(\mathbb{P})[\exists q\in D\cap \mathbb{P}_M^\alpha[\nu=r], q\le p]\}$. Here "$\mathcal{P}_{do}$" means "the set of dense open subsets of."
$\mathbb{P}_M[\nu=r]=\bigcap_{\alpha\in ON} \mathbb{P}_M^\alpha[\nu=r]$. Note that we can in fact just take the intersection up to $\vert\mathbb{P}\vert^+$, or $\aleph^*(\mathbb{P})^+$ in lieu of choice.

That is, we form $\mathbb{P}_M[\nu=r]$ by ``carving out” all the conditions of $\mathbb{P}$ which prevent a $\mathbb{P}$-generic filter $G$ over $M$ from satisfying $\nu[G]=r$.
We also obtain a rank function $\rho^M_{\nu=r}: \mathbb{P}\rightarrow\vert\mathbb{P}\vert^+\cup\{\infty\}$ given by $\rho^M_{\nu=r}(p)$ is the least $\alpha$ such that $p$ is not in $\mathbb{P}^\alpha_M[\nu=r]$ (and $\infty$ if $p\in\mathbb{P}_M[\nu=r]$).
It’s now easy to check a few basic properties of this construction:

If there is a $\mathbb{P}$-generic filter over $M$, $G$, such that $\nu[G]=r$, then $\mathbb{P}_M[\nu=r]\not=\emptyset$. 
Moreover, in such a case we have: if $G$ is $\mathbb{P}_M[\nu=r]$-generic over $N$, then $G$ is $\mathbb{P}$-generic over $M$ and $\nu[G]=r$.
And if $N=M[r]\subseteq M[G]$ for $r$ a real with name $\nu$, $G$ is $\mathbb{P}_M[\nu=r]$-generic over $N$.

I’m interested in questions about the ranking function $\rho$, as well as the forcing $\mathbb{P}_M[\nu=r]$. For instance:

(For “nice” = “proper,” “c.c.c.,” etc.) Suppose $M\models$``$\mathbb{P}$ is nice” and $N$ is a forcing extension of $M$ by a nice forcing. Under what conditions is $\mathbb{P}_M[\nu=r]$ nice?

and

What is $\sup\{\rho_{\nu=r}^M(p): p\not\in\mathbb{P}_M[\nu=r]\}$? That is, how hard is it to tell that a condition is bad? (I’m tentatively thinking of this as analogous to Scott rank, but that might be a bad analogy.)


Note that the monotonic nature of the construction above is a byproduct of the assumption that $r$ is a subset of a set in the ground model. If we tried to replicate this for more general sets, things get messier. In particular, a kind of injury can occur: a condition may appear bad at some stage, and then later appear safe. I have some ideas for what to do, but I’m much less certain, so I’d also be interested in extensions to such sets if this has been looked at.

REPLY [2 votes]: Turns out I was indeed reinventing the wheel! The construction is due to Solovay (page 21 of http://www.math.wisc.edu/~miller/old/m873-03/solovay.pdf) and is generalized by Kanovei (see http://arxiv.org/pdf/1403.5757.pdf).
I can't find any work on niceness properties such as c.c.c.-ness being preserved (or not) by this construction, though; if someone can point me towards such, I'd be much obliged.<|endoftext|>
TITLE: Are Sobolev spaces on non-compact manifolds separable?
QUESTION [6 upvotes]: If $M$ is a Riemannian manifold that is not compact, is it true that the Sobolev spaces on $M$, $W^{k,p}(M)$,  still be separable (for $p < \infty$)?

REPLY [6 votes]: Yes they are.
Step 1 There exists measurable sections $e_1, e_2, \dotsc, e_m$, where $m = \dim M$, of $TM$ (measurable functions mapping a point $x$ to a vector of its tangent plane $T_xM$) such that for each $x \in M$, $e_1 (x), e_2 (x), \dotsc, e_m (x)$ forms an orthonormal basis of the tangent space $T_xM$.
In a neighbourhood of a given point, it is possible to construct such continuous sections by applying a Gram−Schmidt procedure. Such continuous maps can be patched together to form measurable sections.
Step 2 
For $\alpha \in \mathbb{N}^m$, define to be $\partial^\alpha u$ to be the partial derivative in the basis $e_1, e_2, \dotsc, e_m$ 
The map
$$
  A: u \in W^{k, p} (M) \longmapsto A (u) = (\partial^\alpha u)_{\vert \alpha \vert \le k} \in L^p (M)^\nu,
$$
(where $\nu = \# \{\alpha \in \mathbb{N}^n : \vert \alpha \vert \le k\}$)
is continuous and its inverse defined on $A (W^{k, p} (M))$ is also continuous.
Step 3 The subspace $A (W^{k, p} (M))$ is separable as a subset of the separable space $L^p (M)^\nu$ (since $M$ is a separable measure space because it is a second countable topological space). Since $A (W^{k, p} (M))$ is isomorphic to the space $W^{k, p} (M)$, the space  $W^{k, p} (M)$ is separable.<|endoftext|>
TITLE: Local Langlands correspondence and Galois equivariance
QUESTION [7 upvotes]: The local Langlands correspondence $\text{rec}$ for $\text{GL}_{n}$ itself is not Galois equivariant (i.e. invariant under automorphisms of its field of definition) but rather its twist by contragradient and an unramified character is. In Carayol's article "Non-abelian Lubin-Tate theory" in Ann Arbor proceedings, this twist, which I will call $r_{\ell}$, is just referred as "Hecke correspondence," which makes me feel like there is something more about it. Is there a deeper reason to believe why not the local Langlands correspondence but its slight twist is the one that is Galois equivariant? More generally, is there an a priori reason why $r_{\ell}$ is the one that appears in geometric constructions?

REPLY [2 votes]: The reason $r_\ell$ appears in geometric constructions involving Shimura varieties is because Hecke operators act on the cohomology of a Shimura variety with coefficients in $\mathbb{Q}$; that's the source of the good behavior of $r_\ell$ with respect to automorphisms of the coefficient field.<|endoftext|>
TITLE: Adding coprimes to get a coprime
QUESTION [6 upvotes]: These questions arose in my research. Let $n$ be odd and let  $\mathcal{C}_n$ be the set of integers less than $n$ which are coprime with $n$. 
Question 1: For each integer $\ell: 0 \leq \ell < n$ can we always find a subset $\mathcal{S} \subset \mathcal{C}_n$ such that
$$ \ell+2\cdot\sum_{c \in \mathcal{S}}c \quad (\textrm{mod } n)$$
is coprime with n?
That is, can we always find some set of numbers coprime with $n$ such that twice their sum added to $\ell$ is coprime with $n$? Can we always get $|\mathcal{S}| = 1$ when $\ell$ is not coprime with $n$?
Question 2: What is the smallest set $\mathcal{B_n} \subset \mathcal{C}_n$ such that Question 1 holds with $\mathcal{C}_n$ replaced by $\mathcal{B}_n$?

REPLY [2 votes]: To make up for my  not-well-thought-out comment to Fedor Petrov's answer, here is an answer to show $B$ can have $O((\log k)^2)$ elements, where $k$ is the number of distinct prime factors of $n$.  If one pursues the literature, one can take this down to $O(\log k)$ elements.
$n$ being odd, $2$ and its powers are coprime to $n$, with $d=(n+1)/2$ being a multiplicative inverse mod $n$.  Now $\gcd(l+2b,n)=\gcd(dl+2db,n)=\gcd( c+b,n)$, so we just need to find the greatest distance between an arbitrary number $c$ and the next largest coprime integer $c+b$.  More simply put, $l+2i$ is coprime to $n$ for at least one $i$ greater than $0$ and less than a value which I will call $g(n)+1$.
This is related to Jacobsthal's function $g(n)$, of which I know a little something.  One has $g(n) \leq k^{A + C\log\log k}$, where $A,C \lt 4$, which can be found in an arxiv preprint (cf MathOverflow 37679), which lets us take $B$ to be the powers of 2 from 1 up to $\log(g(n))$, giving fewer than $O((\log k)^2)$ elements.
Gerhard "Sorry For Not Thinking Earlier" Paseman, 2016.01.13<|endoftext|>
TITLE: Intutionistic Robinson Arithmetic
QUESTION [13 upvotes]: By Friedman translation $HA$ and $PA$ prove the same $\Pi_2$ formulas.
Is it true for Intutionistic Robinson arithmetic(Robinson axioms with intutionistic logic) and classic Robinson arithmetic?
Axioms of $Q$ are:

$\neg(Sx=0)$
$Sx=Sy\rightarrow x=y$
$y=0 \lor \exists x(Sx=y)$
$x+0=x$
$x+Sy=S(x+y)$
$x\cdot 0=0$
$x\cdot Sy=(x\cdot y)+x$
$\neg(x<0)$
$0=x\lor 0<x$
$x<y \leftrightarrow (Sx<y \lor Sx=y)$
$x<Sy \leftrightarrow (x<y\lor x=y)$


Q1. Is it true that for every $\Pi_2$ formula $\phi$, $Q\vdash_c \phi$ iff $Q\vdash_i \phi$?

Let $$Q^e=Q\cup \{x=y \lor\neg(x=y),\neg(x=y)\leftrightarrow (x<y \lor y<x) \}$$
What happens to Q1 if we replace $Q$ by $Q^e$?

Q2. Is it true that for every $\Pi_2$ formula $\phi$, $Q^e\vdash_c \phi$ iff $Q^e\vdash_i \phi$?

I think the second question can be proved by strong completeness of Beth model for intutionistic logic, but I'm not sure.
Thanks.
Edit:
Definition. The set $\Delta^+_0$ formula is the smallest set such that:

$s=t\in \Delta^+_0$ for every term $s$ and $t$,
$s<t\in \Delta^+_0$ for every term $s$ and $t$,
if $\phi,\psi\in \Delta^+_0$, then $\phi\circ \psi\in\Delta^+_0$ where $\circ\in \{\lor,\land \}$,
if $\phi\in \Delta^+_0$, then $\exists x(x<s \land \phi(x))\in \Delta^+_0$ where $s$ is a term.
if $\phi\in \Delta^+_0$, then $\forall x(x<s \rightarrow \phi(x))\in \Delta^+_0$ where $s$ is a term.

By $\Pi_2$ formula $\phi$ I mean $\phi=\forall{\bf x}\exists{\bf y}\psi({\bf x},{\bf y})$ where $\psi\in \Delta^+_0$.

REPLY [7 votes]: Both are false. Consider the following Kripke model $M\vDash Q^e$ (in fact, it satisfies the intuitionistic version of $\mathrm{PA}^-$): it consists of two worlds $u,v$ such that $u$ sees $v$; the first-order structure at $v$ is the semiring $M_v$ of polynomials $f\in\mathbb Z[x]$ with positive leading coefficient (and $0$), the structure at $u$ is the substructure $M_u\subseteq M_v$ consisting of polynomials whose linear coefficient is even. Putting
$$\phi(x)=\exists y<x\,(y+y=x),$$
we see that
$$M,u\nvDash\phi(a)\lor\neg\phi(a)$$
for the element $a:=2x\in M_u$, witnessing that intuitionistic $Q^e$ does not prove the $\Pi_2$ sentence
$$\forall x\,(\phi(x)\lor\neg\phi(x))$$
provable in any classical theory. Under the restrictive definition of $\Pi_2$ as given in the question, one can take the equivalent sentence
$$\forall x\,(\exists y<x\,(y+y=x)\lor\forall y<x\,(y+y<x\lor y+y>x))$$
instead.
More generally, if the classical extension of an intuitionistic theory $T\supseteq Q^e$ is $\Pi_2$-conservative over $T$, then $T$ must prove
$$\tag{$*$}\forall x\,(\phi(x)\lor\neg\phi(x))$$
for every $\Delta_0$ formula $\phi$ (exercise: the restrictions in the question are ultimately of no consequence). A higher-level argument that this shouldn’t hold for $T=Q^e$ is that $Q^e$ is included in Cook and Urquhart’s theory IPV (which, confusingly, is the intuitionistic version of $S^1_2$ rather than PV), thus by the witnessing theorem for IPV, it cannot prove $(*)$ unless $\phi(x)$  defines a poly-time predicate (and the theory proves this). Since $\Delta_0$ formulas define arbitrary predicates in the linear-time hierarchy, this cannot happen for all $\Delta_0$ formulas unless P = NP. The argument relativizes to subtheories of intuitionistic versions of $S^i_2$ for any $i$, using non-collapse of PH as an assumption in place of P $\ne$ NP.<|endoftext|>
TITLE: Alternating binomial Dirichlet series
QUESTION [7 upvotes]: I have come across the following deceptively simple expression:
$$ H_n^s=\sum_{j=1}^n(-1)^{j-1}\left(\begin{array}{c}n\\j\end{array}\right)j^{-s} $$
We have (using eg mathematica, though probably not difficult to prove): $H_n^0=1$, $H_n^1=H_n$ (the harmonic numbers), expressions involving hypergeometric series with unit argument for integer $s<0$ and involving polygamma functions for integer $s>1$.  For fixed $n$ the sum is of course finite.  My (closely related) questions are:



Does this reduce to values of a known special function for arbitrary real (or complex) $s$?
What is its asymptotic expansion for large $n$?
Is there an efficient numerical method (avoiding cancellations) of evaluating it for large $n$?     



Edit: Using Noam's approach I found two more terms that check numerically:
$$ H_n^s=\frac{(\ln n)^s}{\Gamma(s+1)}+\frac{\gamma(\ln n)^{s-1}}{\Gamma(s)}+\frac{6\gamma^2+\pi^2}{12\Gamma(s-1)}(\ln n)^{s-2}+\ldots $$
where $\gamma=0.577\ldots$ is the Euler constant.  Further asymptotics very welcome.

REPLY [9 votes]: Does this reduce to values of a known special function for arbitrary real (or complex) $s$?


Answered by Johannes Trost in a comment: it's also known as a
"Roman harmonic number".  But this and the associated references
do not yield answers to the next two questions, so I continue:


What is its asymptotic expansion for large $n$?


I don't have a full asymptotic expansion, but for starters
$$
H_n^s = \frac{(\log n)^s}{\Gamma(s+1)} + O((\log n)^{\sigma-1})
$$
holds for each $s$ of positive real part $\sigma$.  This follows from
an integral formula that is also relevant to the final question:


Is there an efficient numerical method (avoiding cancellations) of evaluating it for large $n$?


One standard approach to alternating sums such as
$\sum_{j=0}^n (-1)^j {n \choose j} f(j)$ is to write them as
weighted averages over $X$ of
$\sum_{j=0}^n (-1)^j {n \choose j} X^j = (1-X)^n$,
which in turn requires writing $f$ as a Laplace or Mellin transform.
Here we're dealing with $\sum_{j=1}^n$, not $\sum_{j=0}^n$, but
the same method applies: taking $X = e^{-x}$, we find
$$
H_n^s = \frac1{\Gamma(s)} \int_0^\infty (1 - (1-e^{-x})^n) \, x^s \, \frac{dx}{x}.
$$
The integrand is smooth, and positive for $s \in {\bf R}$
(whence $H_n^s > 0$ for all $n$ and $s>0$),
so this formula can be used to evaluate
$H_n^s$ with numerical integration techniques,
and to estimate it asymptotically.
For large $n$ the factor $1 - (1-e^{-x})^n$
behaves like the characteristic function of the interval $[0, \log n]$,
which makes $H_n^s$ asymptotic to
$$
\frac1{\Gamma(s)} \int_0^{\log n} x^s \, \frac{dx}{x}
= \frac{(\log n)^s}{\Gamma(s+1)}.
$$
A bit more care with the difference between $1 - (1-e^{-x})^n$
and $\chi_{[0,\log n]}$ yields the error estimate $O((\log n)^{\sigma - 1})$,
and with further work it may be possible to derive more precise
asymptotic estimates.<|endoftext|>
TITLE: Why do the $2$-Selmer ranks of $y^2 = x^3 + p^3 $ and $y^2 = x^3 - p^3 $ agree?
QUESTION [24 upvotes]: I was playing around with sage, when I found that the Mordell-Weil ranks (over $\mathbb{Q}$) of the elliptic curves $y^2=x^3+p^3$ and $y^2=x^3-p^3 $ almost always agree, for $p$ prime. The first few exceptions occur at $p=37$, $p=61$, $p=157$, $p=193$, $\ldots$. This pattern struck me as odd, since the two curves are non-isogenous over the ground field, so why would their ranks be correlated?
After some reflection and further experimentation, I found out that if one looks instead at the $2$-Selmer ranks, there is even a stronger pattern: they seem to agree for all primes $p>2$.
I verified this using the following code, written in sage:

for p in primes(100):
    E1 = EllipticCurve(QQ,[0,p^3])
    E2 = EllipticCurve(QQ,[0,-p^3])
    print("p = "+QQ(p).str()+":"),
    rank1 = E1.selmer_rank()
    rank2 = E2.selmer_rank()
    print([rank1,rank2])


which gives

p = 2: [2, 1] p = 3: [1, 1] p = 5: [1, 1] p = 7: [2, 2] p = 11: [2, 2]
p = 13: [1, 1] p = 17: [1, 1] p = 19: [2, 2] p = 23: [2, 2] p = 29: [1, 1]
p = 31: [2, 2] p = 37: [3, 3] p = 41: [1, 1] p = 43: [2, 2] p = 47: [2, 2]
p = 53: [1, 1] p = 59: [2, 2] p = 61: [3, 3] p = 67: [2, 2] p = 71: [2, 2]
p = 73: [1, 1] p = 79: [2, 2] p = 83: [2, 2] p = 89: [1, 1] p = 97: [1, 1]


I have been trying to prove this by following a case distinction according to the residue class of $p$ modulo $12$, and performing a partial $2$-descent for each case, but I keep getting distracted by the thought that
there must be a neater explanation that I'm missing. Hence my question:

Is there?


Edit: It might be useful to note that similar Sage experiments suggest that also (a) the $2$-Selmer ranks of the elliptic curves $y^2=x^3 \pm p$ and $y^2=x^3 \mp p^5$ agree for all $p>2$ and (b) the $2$-Selmer ranks of the elliptic curves $y^2=x^3 \pm p^2$ and $y^2=x^3 \mp p^4$ agree tot all $p>2$. 
In fact, here's a conjecture, also born out by computer experiments, which goes even further and subsumes all cases mentioned before:

Conjecture. Let $a$ be an odd jnteger. Then the $2$-Selmer ranks of the elliptic curves $y^2=x^3 + a$ and $y^2=x^3-a^{-1}$ (which of course is isomorphic to $y^2=x^3 - a^5$) are equal. 

We get the original statement with $a=p^3$, we get (a) with $a=\pm p$, and (b) with $a=\pm p^2$.

REPLY [5 votes]: The (Selmer) ranks of the isogenous twists $y^2=x^3+D$ and $y^2=x^3-27D$ satisfy various relations that often mean the ranks are equal (or differ by 1). I realize these are not the pairs you're studying, but the theory, and especially the techniques, might well be applicable. (Actually, if you work over $\mathbb Q(\sqrt3)$, then $y^2=x^3-27D$ is isomorphic to $y^2=x^3-D$, so that would be your situation.) Here are a couple of references to get you started, they'll contain information about how the problem has been approached.

MR1451686 Chen, Yen-Mei J. The Selmer groups of elliptic curves and
the ideal class groups of quadratic fields. Comm. Algebra 25
(1997), no. 7, 2157–2167.
MR0846960 Satgé, Philippe. Groupes de
Selmer et corps cubiques. J. Number Theory 23 (1986), no. 3,
294–317.<|endoftext|>
TITLE: Spectra as functors from Spaces to Spaces
QUESTION [9 upvotes]: I will use the notation of this question. So, if $X$ is a (nice) topological space and $G$ is an abelian group, we can form its $G$-linearization $G[X]$. In McCord's article, this was denoted $B(G,X)$.
In that question it is mentioned that $\pi_*(G[X])=\tilde{H}_*(X;G)$. But more is true: the functor from Spaces to Graded Abelian Groups that maps a space $X$ to $\pi_*(G[X])$ is a homology theory, isomorphic to singular homology with coefficients in $G$. This says that the isomorphism commutes with the boundary maps in long exact sequences. And we have a good grip on what the boundary for $\pi_*(G[-])$ is, actually: if $A\to X$ is a cofibration, then McCord proves that
$$G[A]\to G[X]\to G[X/A]$$
is a fibration (actually, a principal bundle). The boundary map in homology for $A\to X$ corresponds to the boundary map in homotopy for this fibration.
With this set up, we have that $H_*(-;G)$ is naturally isomorphic to the composition $\pi_* \circ G[-]$.
If I understand correctly what I've heard, it is actually true that for any spectrum $E$ we have that $E_*$ (the associated homology functor from Spaces to Graded Abelian Groups) decomposes as a composition $\pi_* \circ E[-]$, where $E[-]$ is some functor from spaces to spaces which maps cofibrations to fibrations.

Question 1: what further hypotheses do we need on these functors to make the correspondence with the category of homology theories be a 1-1 correspondence? We certainly need $E[-]$ to map a point to a point (or to something contractible). Is this enough?

So, for $E=HG$, the Eilenberg-Mac Lane spectrum of $G$, we have that $E[-]=G[-]$: for $H\mathbb Z$, for example, $E$ is the infinite symmetric product functor.
For $E$ the sphere spectrum, we have $E[-]=Q$.

Question 2: what can be said about other spectra? Is there a clean description on how to get $E[-]$ from $E$ in general? Or maybe in other well-known cases, $KO, KU, MO, MU, K(n)$, etc...

Subsidiary question for the comments: any further references to this point of view are welcome.

REPLY [4 votes]: As a starting point you might well move backwards in time to 1960:
MR0115163
Whitehead, George W.
Homology theories and duality.
Proc. Nat. Acad. Sci. U.S.A. 46 1960 554–556. 
The details of the results announced there were written up in 
Whitehead, George W.
Generalized homology theories.
Trans. Amer. Math. Soc. 102 1962 227–283. 
The representability of homology theories in terms of spectra dates from then. I encourage you to read the Math Review of the first by Peter Hilton.  The representing spectrum is not unique, but I think the lim^1 error term measuring non-uniqueness was understood when I was a graduate student, or very soon thereafter. (I got my PhD in 1964). A clean exposition is in Section 1 of McClure's Chapter VII of $H_{\infty} ring spectra and their applications, published in 1986.<|endoftext|>
TITLE: Descent of sheaves under galois covering
QUESTION [10 upvotes]: Let $\pi: Y\rightarrow X$ be a finite Galois covering between normal projective varieties with Galois group $G$. Let $E$ be a coherent sheaf on $Y$ with a $G$-linearisation, i.e., there are isomorphisms $\lambda_g:E\cong g^*E$ for all $g\in G$ satisfying $\lambda_1=id$ and $\lambda_{gh}=h^*\lambda_g\circ\lambda_h$. Does there always exist a coherent sheaf $F$ on $X$ such that $\pi^*F\cong E$? 
I know that when $\pi$ is etale, the answer is yes by descent theory along torsors (see Vistoli: "Notes on Grothendieck topologies, fibered categories and descent theory," arXiv preprint math/0412512). But how about the general case? In the proof of Lemma 3.2.2 in the book "D Huybrechts, M Lehn: The Geometry of Moduli Spaces of Sheaves", the authors claim that the answer is yes by descent theory, but don't give us a reference. Which descent theory they used? Could someone show me a proof? Thank you very much!

REPLY [12 votes]: The answer is no. First note that you misquote the Lemma of Huybrechts-Lehn: they talk about a subsheaf of a sheaf on $Y$ which is already of the form $\pi ^*\mathcal{F}$.
For a counter-example, take for $\pi :Y\rightarrow X$ a double covering of smooth projective curves, ramified at a point $p\in Y$, and take $E=\mathcal{O}_Y(p)$. Let $\sigma $ be the nontrivial involution of $Y$ such that $\pi \circ \sigma =\pi $. Clearly $\sigma ^*E\cong E$; choose any $\lambda :\sigma ^*E\stackrel{\sim}{\longrightarrow }E$. Then $\sigma ^*\lambda \circ \lambda $ is an automorphism of $E$, hence multiplication by some $t\in\mathbb{C}^*$; dividing $\lambda $ by $\sqrt{t}$  we may assume  $t=1$, so that $\lambda $ gives a $\sigma $-linearization of $E$. But obviously $E$ does not descend since its degree is odd.<|endoftext|>
TITLE: (really) basic intuition for $\mathbb A^1$-homotopy theory
QUESTION [15 upvotes]: Apologies in advance if this question is inappropriate for MO.
I'm trying to read here and there about $\mathbb A^1$-homotopy theory in algebraic geometry. I understand some abstract machinery is needed, in particular sheaves over sites, some simplicial stuff, and some model categories. The wiki article outlines the setup and formal definition.
From what I understand, this theory is supposed to fit in with geometric intuition from topological spaces somehow. Where can I find basic examples illustrating the geometric picture? I'm looking for truly basic stuff to build on, e.g how to identify if a scheme is contractible (if that makes sense), do geometric vector bundles interact well with $\mathbb A^1$-homotopy, what's the picture of two morphisms being homotopic, etc.
If these questions are way off (or completely meaningless) and the similarity of $\mathbb A^1$-homotopy is and the usual notion in $\mathsf{Top}$ is purely formal in nature, what is the right way to think about this theory geometrically?

REPLY [18 votes]: Let me try to answer the specific subquestion "What are (intuitively) maps between schemes in the $\mathbb{A}^1$-homotopy category and what does it mean for maps to be homotopic?" 
The general-machinery answer:  computation of maps $f:X\to Y$ in the homotopy category requires a fibrant replacement $\tilde{Y}$ of $Y$, a cofibrant replacement $\tilde{X}$ of $X$ and then working out the homotopy relation on $\operatorname{Map}(\tilde{X},\tilde{Y})$. Assume we are working in a model structure where representables are cofibrant, so we only need to deal with the fibrant replacement. Morel and Voevodsky show that fibrant replacement is given by iterating infinitely often the fibrant replacement for simplicial sheaves and the explicit $\mathbb{A}^1$-singular resolution functor. We'll try to unwind this, in the special case where we only have one step instead of infinitely many, i.e., we want to compute maps $f:X\to \operatorname{Sing}_\bullet^{\mathbb{A}^1}(Y)$ (in the homotopy category of sheaves, not $\mathbb{A}^1$-localized). The Verdier hypercovering theorem theorem tells us how to do this, see e.g. the exposition in Dugger-Hollander-Isaksen: Hypercovers and simplicial presheaves. Any map has a representative of the form 
$$X\leftarrow \mathfrak{U}_\bullet\to\operatorname{Sing}_\bullet^{\mathbb{A}^1}(Y)$$ where the first map is a Nisnevich hypercovering and the second map a morphism of simplicial sheaves. Now unwind this: looking at simplicial degree $0$, we see that locally our morphism has algebraic representatives. (Ok,  not strictly true because a Nisnevich open is not a subset, but this is building in implicit functions.) However, these local algebraic representatives do not glue to a globally defined map $X\to Y$ because they do not need to agree -- in degree 1, we see that the restrictions of two local algebraic representatives $f_1:U_1\to Y$ and $f_2:U_2\to Y$ to the fiber product $U_1\times_X U_2$ are only locally naively $\mathbb{A}^1$-homotopic. More precisely, there is a Nisnevich covering of $U_1\times_X U_2$, and locally on this covering, there will be naive $\mathbb{A}^1$-homotopies $H:V_1\times\mathbb{A}^1\to Y$. The higher simplicial degrees add further locally defined coherences between the restrictions of the algebraic representatives to multiple intersections.
Well, rather terrible. And this is only maps from $X$ to the sheaf-fibrant replacement of a single application of the singular resolution. Usually, infinitely many such applications are required so that, in a way, there is no geometric intuition to be had here. But on an intuitive level, I guess one can say that 

maps in the $\mathbb{A}^1$-homotopy category are locally algebraic, and the local algebraic representatives are locally $\mathbb{A}^1$-homotopic in a compatible way. 

A similar description applies to maps being homotopic. Generally, you will want to avoid making the above really precise because it'll be a hell of a time to construct maps in the homotopy category like this. Nevertheless, it's possible to do some computations: you can see quite instructively how to work with such locally algebraic maps up to locally-defined $\mathbb{A}^1$-homotopies in the papers of Balwe-Hogdai-Sawant and Balwe-Sawant. (watch out for ghost homotopies)

C. Balwe, A. Hogadi and A. Sawant. $\mathbb{A}^1$-connected components of  schemes. Adv. Math. 282 (2015), 335--361.

A nicer, more geometric situation: If $X$ is smooth affine and $Y$ is an isotropic reductive group or a suitable homogeneous space, then it is in fact possible to prove that all maps $X\to Y$ in the $\mathbb{A}^1$-homotopy category have algebraic representatives, and homotopies are really chains of naive $\mathbb{A}^1$-homotopies $X\times\mathbb{A}^1\to Y$. Technically, the statement is that the singular resolution $\mathcal{F}=\operatorname{Sing}_\bullet^{\mathbb{A}^1}(Y)$ is almost fibrant, i.e., the fibrant replacement $\mathcal{F}\to R\mathcal{F}$ induces weak equivalences on sections over all smooth affine schemes. (A consequence of this is that the classical theory of homotopy classification of principal bundles works in these cases, answering the question on geometric vector bundles.) There is quite a bit of machinery that goes into the proof; the first result of this type (for general linear groups and Grassmannians) was proved by Morel (see the Lecture Notes on $\mathbb{A^1}$-algebraic toplogy) and there has been a recent extension in work of Aravind Asok, Marc Hoyois and myself (can be found on the arXiv here). 

Comparison to classical topology:
There is a different line of thinking to get the intuition started which you may have found already. The classical homotopy theory can be recovered by starting with the category of smooth manifolds (with Grothendieck topology given by open coverings) and then taking the Bousfield localization at $\mathbb{R}$ of the model category of simplicial sheaves on the site of manifolds. The simple reason why the result is the classical homotopy category is that every manifold has a good hypercovering, i.e., one where all spaces involved are balls in $\mathbb{R}^n$. All the homotopy information is then contained in the combinatorics of the nerve of the covering, which is just a simplicial set. In algebraic geometry, schemes look rather different even locally, so there is no way to have the structure of a scheme encoded in a single simplicial set. (Hence the need to work with a simplicial sheaf which provides a simplicial set for each scheme.) 
Of course, you can compute maps in this homotopy category by the Verdier hypercovering theorem. Why aren't people doing this? Well, continuous maps are very flexible. Assume we have topological spaces $X$ and $Y$, a covering $U_1\cup U_2=X$ and maps $f_i:U_i\to Y$ which are homotopic over the intersection $U_1\cap U_2$. We can extend this homotopy on $U_1\cap U_2$ to a continuous homotopy on $U_1$, and then subsequently change to the situation where we can actually glue the maps. So the generality of locally defined maps, locally homotopic etc., is not needed for continuous maps of topological spaces. Again, this kind of flexibility is impossible to get in algebraic geometry.
I guess, you can say generally that it is fairly difficult to relate the algebraic geometry constructions to statements about the $\mathbb{A}^1$-homotopy category. A lot more difficult than in classical topology. Nevertheless, sometimes it is possible to translate classical topological statements (like the theory of the Euler class) into $\mathbb{A}^1$-homotopy, and then it produces interesting algebraic statements.
Let me stop here before it turns into some lecture notes, I hope this gives a bit intuition what maps in $\mathbb{A}^1$-homotopy theory look like.<|endoftext|>
TITLE: How much choice does a linear or well-order on cardinals imply?
QUESTION [9 upvotes]: It is well-known that if the natural (partial) order on the class of cardinal numbers is a linear order, then it is in fact a well-order and the axiom of choice holds. I was, however, interested in how much choice we can recover given some linear ordering, or better — a well-ordering — of the class of cardinals.
I couldn't figure out by myself any results, but I would imagine it at least implies that there are no amorphous sets.
Are there any results known about this? To be more specific, let me ask the following question:

Is the axiom of countable choice implied by the existence of a linear ordering on the class of all cardinals? How about the existence of a well-order on this class?

Thanks in advance for all the feedback.

REPLY [5 votes]: No.
In Cohen's first model, as constructed over $L$, you have a uniform linear ordering of the entire model. However the axiom of countable choice fails: there is a Dedekind-finite set of reals.
To see that the first claim holds, note that the model is $L(A)$ where $A$ is the of generic Cohen reals. Therefore there is a surjection from $[A]^{<\omega}\times\sf Ord$ onto $L(A)$, and that defines a linear ordering on the entire universe. In particular on the class of cardinals.
(Remark: I am not claiming that the order extends the order of the cardinals, but it does order the class of cardinals (which is $\omega$, the $\aleph$ numbers and the Scott cardinals) in a linear order. Whether or not there is an order extension of the usual order to a linear order, I cannot say. I suspect this is the case in Cohen's model, though. Note it cannot be extended to a well-ordering since there are infinite decreasing chains.)<|endoftext|>
TITLE: Which curves have reflexive structure sheaf?
QUESTION [5 upvotes]: [This was first posted on MSE but did not get any answer. I apologize if it is not suited for MO.]
Let $C$ be a curve, i.e. a purely one-dimensional scheme, embedded in a smooth projective threefold $X$. For a coherent sheaf $E$ of codimension $c$ on $X$, let $E^D=\mathscr Ext_X^c(E,\omega_X)$ be the Grothendieck dual of $E$. It is a reflexive sheaf of codimension $c$ on $X$. Reflexive means that the natural map to the double dual is an isomorphism. Let $\mathscr O_C$ be the structure sheaf of $C$, viewed as a torsion sheaf on $X$.

Question. For which curves $C\subset X$ is $\mathscr O_C$ reflexive?

Certainly, when $C$ is smooth. Let us assume $C$ is singular. I was thinking that maybe if $C$ is a local complete intersection then the first dual $\mathscr O_C^D$ could already be isomorphic to $\mathscr O_C$. Is this true? If so, then in particular dualizing twice does not do anything. But probably someone knows a more convincing (or even more true) statement. Thank you for any help!

REPLY [4 votes]: Indeed, this condition is equivalent to $C$ being CM (Regardless of $X$ as long as it is also CM). This "dual" is called the $\omega$-dual in Kol13. You might be interested in reading section 2.5, or just the part directly dealing with this on pp.80-83. In particular, the statement you need is Cor. 2.70. Furthermore, if $X$ is not necessarily CM, then you get the same result using the dualizing complex instead of the dualizing sheaf.
As Mohan points out, the first dual will only be isomorphic to $\mathscr O$ if its own dualizing sheaf is trivial. Then again, why would you want that? That's a freak accident and doesn't give you too much mileage. Reflexivity is a (or "the") natural condition.<|endoftext|>
TITLE: Rate of Convergence of Borwein Algorithm for computing Pi
QUESTION [5 upvotes]: In a book "Pi and the AGM" in 1987, authors, Jonathan Borwein and Peter Borwein, introduced a magical algorithm to compute $\pi$. However there is a problem that I couldn't understand and couldn't find any proof or explaination. It is related to rate of convergence of the algorithm. 
Borwein brothers suggested two sequences $\{x_n\}_{ n\ge 0}$ and $\{y_n\}_{ n\ge 1}$ like: 
\begin{equation}
x_{0} := \sqrt{2} ~~~~,~~~~y_{1}:= 2^{1/4} 
\end{equation}
\begin{equation}
x_{n+1}=\frac{\sqrt{x_n} ~+~ 1/\sqrt{x_n} }{2} ~~~,~~~
y_{n+1}=\frac{y_{n} \sqrt{x_n}~+~ 1/\sqrt{x_n}}{y_{n} +1}
\end{equation}
Then my question is about a proof for the next inequality:
\begin{equation} \label{eq}
 \forall n \in \mathbb{N}~,~~~\frac{x_n -1}{y_n -1} < 2- \sqrt{3}
\end{equation}
How can I prove this inequality?

REPLY [4 votes]: Even for different starting values around $1$, both $x_n$ and $y_n$ rapidly approach $1$. 
Define $\alpha_n = x_n-1$ and $\beta_n = y_n-1$. These will be very small quantities. $\beta_5 \lt 10^{-40}$ and $\alpha_5$ is smaller. We want to estimate $\alpha_n/\beta_n$.
$$\sqrt{x_n} = \sqrt{1+\alpha_n} = 1+\frac{1}{2}\alpha_n - \frac{1}{8}\alpha_n^2 + O(\alpha_n^3).$$
$$1/\sqrt{x_n} = 1/\sqrt{1+\alpha_n} = 1-\frac{1}{2}\alpha_n +\frac{3}{8} \alpha_n^2 + O(\alpha_n^3)$$
$$x_{n+1} = \frac{1}{2}(\sqrt{x_n} + 1/\sqrt{x_n}) = 1+\frac{1}{8}\alpha_n^2 +O(\alpha_n^3) $$
So, $\alpha_{n+1} = \frac{1}{8} \alpha_n^2 + O(\alpha_n^3)$.
$$\begin{eqnarray}y_{n+1} &=& \frac{y_n\sqrt{x_n} + \sqrt{x_n} - \sqrt{x_n} + 1/\sqrt{x_n}}{y_n+1} \newline &=& \sqrt{x_n} + \frac{-\sqrt{x_n} +1/\sqrt{x_n}}{y_n+1} \newline &=&1+\frac{1}{2}\alpha_n-\frac{1}{8}\alpha_n^2 + O(\alpha_n^3) + (-\alpha_n+\frac{1}{2}\alpha_n^2+O(\alpha_n^3))(\frac{1}{2} - \frac{1}{4}\beta_n + O(\beta_n^2)) \newline &=& 1 + \frac{1}{8}\alpha_n^2 + \frac{1}{4}\alpha_n\beta_n + O(\beta_n^3)\end{eqnarray}$$
So, $\beta_{n+1} = \frac{1}{8}\alpha_n^2 + \frac{1}{4}\alpha_n\beta_n + O(\beta_n^3)$.
Instead of $\alpha_n/\beta_n$, consider the reciprocal.
$\beta_{n+1}/\alpha_{n+1} = 1 + 2 \beta_n/\alpha_n + O(\alpha_n).$ This ratio grows exponentially, and we only need to establish that it is greater than $1/(2-\sqrt{3}) = 2+\sqrt{3} \lt 4$. 
For example, $\beta_4/\alpha_4 = 99.53, \beta_5/\alpha_5 = 200.06$.
To turn this into a rigorous argument, you can use effective instead of asymptotic estimates for $\sqrt{1+\alpha_n}$, $1/\sqrt{1+\alpha_n}$, and $1/(2+\beta_n)$  when $\alpha_n$ and $\beta_n$ are small.<|endoftext|>
TITLE: Generators of pure braid groups of arbitrary Coxeter groups
QUESTION [7 upvotes]: Let $W$ be an arbitrary Coxeter group, and let $A$ be the associated Artin-Tits braid group, with standard Coxeter generators $\sigma_i\in A$.  Let $P$ be the "pure braid group", the kernel of the natural homomorphism from $A$ to $W$.  Obvious elements in $P$ include the squares $\sigma_i^2$ of the standard Coxeter generators, as well as conjugates of the $\sigma_i^2$.  Do conjugates of the $\sigma_i^2$ generate the pure braid group $P$?  If so, is there a nice presentation for $P$ in carefully chosen such conjugates?

REPLY [4 votes]: This note
http://arxiv.org/pdf/1511.08731v3.pdf
(unfortunately written in French), Corollaire 3.7, answers your question.
Writing $\bf{W}$ for the canonical positive lift of $W$ in the Artin-Tits group $B_W$, the pure braid group is generated by the elements of the form $\bf{w}\bf{s}^2 \bf{w}^{-1}$ where $\bf{w}\in\bf{W}$, $\bf{s}$ is the lift in $\bf{W}$ of a simple reflection $s\in S$ and $\bf{w}\bf{s}$ also lies in $\bf{W}$. A presentation in terms of these generators is given in Corollaire 3.7 of the above given link. 
It works for arbitrary (finitely generated) Coxeter groups.<|endoftext|>
TITLE: A strengthening of Frankl's union-closed sets conjecture?
QUESTION [9 upvotes]: A Frankl family is a nonempty finite family $\mathcal F$ of nonempty finite sets such that $A,B\in\mathcal F\implies A\cup B\in\mathcal F.$ Define $d_\mathcal F(x)=|\{A\in\mathcal F:x\in A\}|$ and $\Delta(\mathcal F)=\max_xd_\mathcal F(x).$ Frankl's union-closed sets conjecture says that $\Delta(\mathcal F)\gt\frac12|\mathcal F|,$ or in other words $|\mathcal F|\le2\Delta(F)-1,$ for any Frankl family $\mathcal F.$ Has the following stronger conjecture been considered, and is a counterexample known?

For any Frankl family $\mathcal F$ there is an element $x$ such that $\Delta(\{A\in\mathcal F:x\notin A\})\le\frac12\Delta(\mathcal F),$

or better yet,

for any Frankl family $\mathcal F$ and any element $x,$ if $d_\mathcal F(x)=\Delta(\mathcal F)$ then $\Delta(\{A\in\mathcal F:x\notin A\})\le\frac12\Delta(\mathcal F)$?

REPLY [5 votes]: It seems that here is a counterexample to the stronger statement. Perhaps, it can be modified to provide one for the weaker statement?
Our universe is $\{1,\dots,7\}$. Each set of the family contains either $6$, or $7$, or both. We asume that the elements $1,\dots,5$ are arranged into a cycle $1-2-3-4-5-1$.
The sets containing $6$ but not $7$ are obtained from $\{6\}$ by adding either (2-6) two consecutive elements from the cycle, or (3-6) three consecutive elements, or $\{1,2,4\}$, or $\{2,3,5\}$, or $\{1,3,4\}$, or (4-6) arbitrary four or five elements from the cycle. 19 sets in all.
The sets containing $7$ but not $6$ are obtained from $\{7\}$ by adding either (2-7) two non-consecutive elements from the cycle, or (3-6) three non-consecutive elements, or $\{3,4,5\}$, or $\{5,1,2\}$, or (4-6) arbitrary four or five elements from the cycle. $18$ sets in all.
The sets containing both $6$ and $7$ are obtained from $\{6,7\}$ by adding at least three elements from the cycle. $16$ sets in all.
These sets form a Frankl family. Now, $6$ is contained in $35$ sets, $7$ --- in $34$ sets, any other element --- in at most $34$ sets. Thus the unique element with the largest degree is $6$  (this was the aim for which we have added extra triples), and this degree is $35.$ But all the sets not containing $6$ do contain $7$, and there are $18>35/2$ of them.
Thus the conjectured property may fail even for every element of the largest degree.<|endoftext|>
TITLE: A Converse to Cartan–Hadamard theorem?
QUESTION [12 upvotes]: Let $M$ be a complete Riemannian  manifold, with the property that $\exp_p\colon T_pM \to M$ is a diffeomorphism for every $p \in M$.
Can we say something about it's curvature?
Is it true that its sectional curvature must be everywhere non-positive (or at least at some point)?
(If not, can we say something about the Ricci or scalar curvature?)

Note that it's clearly not true if we only assume $\exp$ is a diffeomorphism at a single point.

REPLY [12 votes]: For complete Riemannian manifolds the exponential map is a covering map at all points if 
and only if the manifold has no conjugate points. In particular, the exponential map is a diffeomorphism at all points if and only if the manifold is simply-connected and the metric has no conjugate points. 
Manifolds without conjugate points have been studied by many authors and they need not have everywhere nonpositive sectional curvature.
On the other hand it is still an open problem (as far as I know) whether every closed manifold without conjugate points admits a metric of nonpositive sectional curvature.
For one result relevant to your question is that if a complete Riemannian manifol has no conjugate points, then either it is flat (that is has zero sectional curvature everywhere) or it has negative Ricci curvature everywhere; see  Geodesics without Conjugate Points and Curvatures at Infinity
by SÉRGIO MENDONÇA and DETANG ZHOU. They also mention some related result of Leon Green involving scalar curvature.
EDIT: I misread the curvature condition in the paper linked above. Thanks to Sergei for correcting me!

REPLY [4 votes]: Assuming that there are no conjugate points, one can conclude that the curvature is zero in the following special case.  This is the case of a complete surface of nonnegative curvature.  Namely, if the curvature is positive at a point $p$, a Jacobi field whose derivative vanishes at $p$ with have a convex graph (by the curvature assumption) and therefore will be forced to vanish eventually (just by the intermediate value theorem).  Doing this in both directions along a geodesic will produce a pair of points which are conjugate to each other.
I am not sure if a similar argument will work in higher dimension. Perhaps Sergei can chip in?
Another relevant result is that of Burago and Ivanov from 1994, generalizing an older result of Hopf.  Namely, a Riemannian metric on an $n$-torus without conjugate points must be flat.<|endoftext|>
TITLE: Spectrum of unitary elements of a Banach algebra
QUESTION [6 upvotes]: Unitary elements of a Banach space have been defined in this paper as follows:
Let $A$ be a Banach space and $a\in A, \|a\|=1$. Let $S_{a}=\{f\in A':\|f\|=1=f(a)\}$. Then $a$ is said to be (geometrically) unitary if $A'=\text{ span }S_{a}$.
Here, $A'$ is the dual space of $A$.
We note that this property is true for unitary operators on a Hilbert space. Unitary operators on Hilbert spaces are invertible, and their spectra lie on the unit circle.
Now, suppose $A$ is a Banach algebra. Can we say that if $a\in A$ is geometrically unitary, then it is invertible?
I have been trying to find a counter-example, but have been unsuccessful so far. I considered the Banach algebra $l^{1}(\mathbb{Z})$, with convolution as the multiplication. The dual of $l^{1}(\mathbb{Z})$ is $l^{\infty}(\mathbb{Z})$. An element $f$ of $l^{1}(\mathbb{Z})$ is invertible iff $f(z)\neq 0 \, \forall z\in \mathbb{T}$, where $\mathbb{T}$ is the unit circle in $\mathbb{C}$. 
Let us take, for example, $f=(\cdots,0,\frac{1}{2},0,\frac{i}{2},0,\cdots)$, where the central $0$ is in the $0^{th}$ position. Then $f$ is of norm $1$ and can be shown to be not invertible. 
We now consider elements $\phi$ of $S_{f}\subseteq l^{\infty}(\mathbb{Z})$. $\phi$ must satisfy the following:
$ \|\phi\|_{\infty}=1\\$ and $\phi(-1)\frac{1}{2}+\phi(1)\frac{i}{2}=1$.
One possible solution is $\phi=(\cdots,1,x,-i,\cdots)$, where $x$ and the dots can be any scalar less than or equal to $1$.
Is there another possible solution? In that case, we would have a unitary element that is not invertible.
Alternately, is it indeed true that unitary elements described in this geometric fashion are invertible?
I'd be grateful for help with this.

REPLY [7 votes]: The answer to your immediate question is no: there is no other possible solution for $\phi$. To see this, consider that
$$\phi(-1) + i\phi(1) = 2$$
implies
$${\rm Re}(\phi(-1)) + {\rm Re}(i\phi(1)) = 2.$$
If $\|\phi\|_\infty = 1$ then both terms on the left are at most 1, and equal
1 if and only if $\phi(-1) = 1$ and $\phi(1) = -i$. So these values are forced. By a similar argument, the only unitaries in $l^1$ are the elements of the form $\alpha e_n$ where $(e_n)$ is the standard basis and $|\alpha| = 1$.
However, there is an easy counterexample. Consider $\mathbb{C}^2$ with the $l^1$ norm, i.e., $\|(a,b)\| = |a| + |b|$. Give it the product $(a,b)\cdot(c,d) = (ac, ad + bc)$. This is a Banach algebra because
$$\|(a,b)\cdot(c,d)\| = |ac| + |ad + bc| \leq (|a| + |b|)(|c| + |d|) = \|(a,b)\|\|(c,d)\|.$$
The element $(1,0)$ is the unit, and it is unitary, but so is $(0,1)$, which is not invertible.<|endoftext|>
TITLE: Morphisms of locally ringed spaces into affine schemes
QUESTION [6 upvotes]: In Görtz and Wedhorn's Algebraic Geometry I, there's the following proposition:
Proposition 3.4. Let $(X,\mathcal O_X)$ be a locally ringed space. If $Y$ is an affine scheme then the natural map below is an isomorphism $$\mathsf{Hom}(X,Y)\overset{\cong}{\longrightarrow} \mathsf{Hom}(R,\Gamma(X,\mathcal O_X))$$
A proof is given only for the case where $X$ is a scheme, and proceeds by reducing to the anti-equivalence of affine schemes with commutative rings and then gluing the morphisms.
For the general case, the authors cite Prop 1.6.3 of the 1971 edition of EGA I, to which I don't have access. Can anyone reproduce/explain the proof?

REPLY [6 votes]: Here is a sketch: Let $\alpha : R \to \Gamma(X,\mathcal{O}_X)$ be a ring homomorphism. We want to define a morphism $f:X \to \mathrm{Spec}(R)$ which is $\alpha$ on global sections. Let $x \in X$, and consider the composition of $\alpha$ with $\Gamma(X,\mathcal{O}_X) \to \mathcal{O}_{X,x}$, $s \mapsto s_x$. Pull back the unique maximal ideal $\mathfrak{m}_x$ of $\mathcal{O}_{X,x}$ to $R$, this is a prime ideal of $R$. Denote this by $f(x)$. This defines a map of sets $f : X \to \mathrm{Spec}(R)$. To show continuity, let $r \in R$ and observe that the preimage of $D(r) \subseteq \mathrm{Spec}(R)$ in $X$ is the set of points $x \in X$ such that $\alpha(r)_x$ is invertible in $\mathcal{O}_{X,x}$. This set $D(\alpha(r))$ is easily seen to be open. Also notice that by the universal property of localization, $\alpha$ induces a homomorphism $R[r^{-1}] \to \Gamma(D(\alpha(r)),\mathcal{O}_X)$, i.e. $\Gamma(D(r),\mathcal{O}_{\mathrm{Spec}(R)}) \to \Gamma(f^{-1}(D(r)),\mathcal{O}_X)$. These maps are compatible and therefore glue to a morphism of sheaves $f^{\#}:\mathcal{O}_{\mathrm{Spec}(R)} \to f_* \mathcal{O}_X$.<|endoftext|>
TITLE: What is, really, the stable homotopy category?
QUESTION [30 upvotes]: When you try to understand the fuss behind the new good categories of spectra that arose on the 90's, you read things such as the following paragraph written by Peter May (from "The Hare and the Tortoise"):

All consumers are now in agreement: Mike's stable homotopy category is 
  definitively the right one, up to equivalence. However, the really fanatical hare demands
  a good category even before passage to homotopy, with all of the modern bells
  and whistles. The ideal category of spectra should be a complete and cocomplete
  Quillen model category, tensored and cotensored over the category of based spaces
  (or simplicial sets), and closed symmetric monoidal under the smash product. Its
  homotopy category (obtained by inverting the weak equivalences) should be 
  equivalent to Mike's original stable homotopy category.

Here "Mike" is Michael Boardman. Question number zero would be, could anybody share Boardman's "Stable homotopy theory" mimeographed notes where he introduces said category? Both that and Vogt's "Boardman's stable homotopy category" are hard to find, so it's hard for me to know what they are talking about.
But suppose I know what Boardman's stable homotopy category is. Why should I agree that this is "the" right stable homotopy category, up to equivalence? 
I am guessing that the right way to formalize what I mean is: elabore a desiderata for a Stable Homotopy Category, prove that Boardman's satisfies them, and then prove that any two categories satisfying those axioms are equivalent.

Has that been done? If so, where?

I can try to answer my question. Margolis' book "Spectra and the Steenrod Algebra" from 1983 does have such a list of axioms, in section 1.2.

Is that an idiosyncratic list of axioms or is it really what homotopy theorists of the time would agree that it's exactly what they would have wanted?

I know this is perhaps argumentative. But the following is not. At the end of said chapter in Margolis' book, he conjectures that any two categories satisfying those axioms are equivalent. But at the time, it apparently wasn't established.

Has it been established since?

But maybe there is another characterization of "the" stable homotopy category, of which Boardman's (or Adams', or...) would be an example; I'd be interested by that, too.
I am mildly aware of the fact that there is a stable $\infty$-categorical universal property. That is certainly interesting, but I would be interested to see how it could be formulated in older language (model categories?) since such a formulation is, in the spirit of my question, anachronistic. (Not that I would find it uninteresting).

REPLY [19 votes]: I am mildly aware of the fact that there is a stable ∞-categorical universal property. That is certainly interesting, but I would be interested to see how it could be formulated in older language (model categories?)

Symmetric spectra have a model categorical universal property: they form the initial stable monoidal model category (theorem of Shipley). A subtlety is that this statement holds with the positive stable model structure, but at least this is Quillen equivalent to the usual one and so captures the same homotopy theory.

At the end of said chapter in Margolis' book, he conjectures that any two categories satisfying those axioms are equivalent. But at the time, it apparently wasn't established.
Has it been established since?

Shipley also shows that any category satisfying Margolis's axioms that actually comes from a stable monoidal model category, must be equivalent to the stable homotopy category (as a monoidal category). That pretty much resolves Margolis's conjecture I'd say, since by now everyone agrees that triangulated categories are just not enough for these kinds of questions.
The relevant paper of Shipley is called Monoidal Uniquness of Stable Homotopy theory.
Here are some funky bonus facts:

Given an object in a symmetric monoidal $\infty$-category, there exists a universal functor which is initial among symmetric monoidal functors which invert the given object. The symmetric monoidal $\infty$-category of spectra is obtained from pointed spaces with smash product by formally inverting $S^1$ in this way. (The symmetric monoidal $\infty$-category of pointed spaces is in turn obtained from spaces with cartesian product by freely pointing.)  See this paper of Robalo.
Given a Grothendieck derivator (another way to keep track of homotopy (co)limits in a homotopy category), there exists a universal stable derivator associated to it. Taking the derivator of spaces, this gives the stable derivator of spectra. Also, every stable derivator is canonically enriched over the derivator of spectra. See the appendix of this paper of Cisinski and Tabuada. The language of derivators is very elementary (no homotopy theory used!) and is pretty old (early 80's).<|endoftext|>
TITLE: derived categories as presentable DG-categories
QUESTION [9 upvotes]: Let $A$ be a ring. Is it true that the DG category of unbounded complexes of $A$-modules, localized by quasi-isomorphisms, is cocomplete and compactly generated? What would be a reference for that and close matters (like spelling out the compact objects, or perhaps discussing pairing with derived category of $A^{op}$-modules into $Vect$, etc.)?
Thank you,
Sasha

REPLY [6 votes]: The fact that $\mathbf{D}(A)$ is compactly generated follows from the fact that $A$ (and its suspensions) generate $\mathbf{D}(A)$ and that the smallest thick subcategory containing $A$ is $\mathrm{Perf}(A)$, in other words, compact objects are prefect complexes, i.e. bounded complex of projective $A$-modules. This goes back to Rickard and Thomason-Trobaugh.
The fact that $\mathbf{D}(A)$ is cocomplete follows from existence and exactness of coproducts in the category of $A$-modules.<|endoftext|>
TITLE: Edge chromatic number of hypergraphs
QUESTION [8 upvotes]: This is question Selection problem in a collection of non-empty sets with a simplification in criterion 3.
Is there a set $X\neq\emptyset$ and a collection ${\cal F}\subseteq {\cal P}(X)\setminus\{\emptyset\}$ of non-empty subsets of $X$ with the following properties?

$a\in {\cal F} \implies |a|\geq 2$,
$a\neq b\in {\cal F} \implies |a\cap b| \leq 1$, and
there is no function $f: {\cal F} \to X$ such that

if $a, b\in {\cal F}$ with $a\ne b$ and $a\cap b\neq \emptyset$ then $f(a)\neq f(b)$?

REPLY [5 votes]: This is equivalent reformulation of Erdös-Faber-Lovász conjecture, see Wikipedia page about it.
https://en.m.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Faber%E2%80%93Lov%C3%A1sz_conjecture<|endoftext|>
TITLE: Are all 4-manifolds $Pin^{\tilde{c}}$?
QUESTION [8 upvotes]: It's known that all oriented 4-manifolds admit a $Spin^c$ structure, ie. a spin structure on $TX\oplus\mathcal{L}$ for some complex line bundle $\mathcal{L}$. 
A usual generalization of this structure to unorientable manifolds is to ask for a spin structure on $TX\oplus\mathcal{L}\oplus\mathcal{E}$ where $\mathcal{L}$ is again a complex line and now $\mathcal{E}$ is a real line bundle (which the whole bundle being oriented will force to be the orientation line). This is called a $Pin^c$ structure. In cohomology, it amounts to an integral lift of $w_2$.
Unfortunately, not all 4-manifolds admit a $pin^c$ structure, eg. $\mathbb{RP}^2 \times \mathbb{RP}^2$. This is easy to see by a computation of $w_2$.
There is another generalization I'll call a $Pin^{\tilde c}$ structure. In this version, $\mathcal{L}$ and $\mathcal{E}$ combine into a real 2-plane bundle. In cohomology it amounts to a twisted integral lift of $w_2$.
So, do all 4-manifolds admit a twisted integral lift of $w_2$?
Here are some edits in response to Qiaochu's comment. An integral lift of $w_2$ is an element of $H^2(X,\mathbb{Z})$ mapping to $w_2$ in $H^2(X,\mathbb{Z}/2)$. A twisted integral lift lives instead in $H^2(X,\mathbb{Z}^{w_1})$, with local coefficients in the orientation line.

REPLY [8 votes]: $\newcommand{\RP}{\mathbb{RP}}\newcommand{\Z}{\mathbb Z}$No, $\RP^2\times\RP^2$ isn't pin$\tilde c$. This came up while thinking about another MathOverflow
question, but I'll rewrite the argument here.
What you call a pin$\tilde c$-structure has been studied in physics, following
Metlitski and Freed-Hopkins, where it's
also called a pin$\tilde c+$-structure.  (This is because one can also ask about pin$\tilde
c-$-structures, which are twisted integral lifts of $w_2 + w_1^2$.) These structures are discussed in
a little more detail by Shiozaki-Shapourian-Gomi-Ryu in Appendix D.
Here's the argument for $\RP^2\times\RP^2$: let $x$ denote the generator of $H^1(-;\Z/2)$ of the first $\RP^2$ and
$y$ be that for the second $\RP^2$. Using the usual CW structure on $\RP^2$ and the product CW strucure on
$\RP^2\times\RP^2$, you can check that $H_2(\RP^2\times\RP^2;\Z)\cong\Z/2$ and the reduction mod 2 map
$H_2(\RP^2\times\RP^2;\Z)\to H_2(\RP^2\times\RP^2;\Z/2)$ sends the nonzero element of $H_2(\RP^2\times\RP^2;\Z)$ to
the Poincaré dual of $x+y$.
The Poincaré duality isomorphisms $H_2(\RP^2\times\RP^2;\Z)\cong H^2(\RP^2\times\RP^2;\Z_{w_1})$ and
$H_2(\RP^2\times\RP^2;\Z/2)\cong H^2(\RP^2\times\RP^2;\Z/2)$ are natural with respect to change-of-coefficients,
which means that the reduction mod 2 map $\rho_2\colon H^2(\RP^2\times\RP^2;\Z_{w_1})\to
H^2(\RP^2\times\RP^2;\Z/2)$ sends the nonzero element of $H^2(\RP^2\times\RP^2;\Z_{w_1})\cong\Z/2$ to $x+y$.
However, $w_2(\RP^2\times\RP^2) = x^2+xy+y^2$, so it's not in the image of $\rho_2$, and therefore
$\RP^2\times\RP^2$ has no pin$\tilde c+$-structure.<|endoftext|>
TITLE: What does the axiom of replacement mean and why should I believe it?
QUESTION [8 upvotes]: Here Professor Blass describes the following cumulative hierarchy of sets:

Begin with some non-set entities called atoms ("some" could be "none" if you want a world consisting exclusively of sets), then form all sets of these, then all sets whose elements are atoms or sets of atoms, etc. This "etc." means to build more and more levels of sets, where a set at any level has elements only from earlier levels (and the atoms constitute the lowest level). This iterative construction can be continued transfinitely, through arbitrarily long well-ordered sequences of levels.
  This so-called cumulative hierarchy is what I (and most set theorists) mean when we talk about sets.

We want to agree on the following principles:

For every level there is a succeeding level.
For every sequence of levels: $l_1,l_2,l_3,\dots$ there is a level succeeding all levels $l_1,l_2,l_3,\dots$. One might call this level "limit level".

Question:
Why is the axiom of replacement true under this interpretation of the term "set" (set = anything that is formed at some level of this hierarchy)?

REPLY [4 votes]: For an argument that the iterative conception implies something weaker than unrestricted Separation (implied by unrestricted Replacement), i.e. $\Sigma_2$ Replacement, see Randall Holmes 2001 http://math.boisestate.edu/~holmes/holmes/sigma1slides.ps.  (According to Professor Holmes, “this contain[s] an error, which Kanamori pointed out to me and which I know how to fix.”)<|endoftext|>
TITLE: For a new operation on a finite group of odd order giving a loop structure, when does this also gives a group
QUESTION [7 upvotes]: For finite groups $G$ of odd order, as $x \mapsto x^2$ is bijection (but no automorphism in general) then, we can define for each $g \in G$ the element $x^{1/2}$ by requiring $(x^{1/2})^2 = x$. Then with the new operation
$$
 x\circ y := x^{1/2} y x^{1/2}
$$
we get a power-associative loop satisfiying the relations
$$
 x\circ (y \circ (x \circ z)) = (x\circ (y\circ x))\circ z
$$
and
$$
 (x\circ y)^{-1} = x^{-1} \circ y^{-1}.
$$
If the first is satisfied, such loops are also called (left) Bol loops, and if both additional equations hold they are called (left) Bruck loops. These definitions go back to ideas of G. Glauberman: On loops of odd order. Now I want to know, under what conditions for $G$ to we indeed have associativity? As $1 \in G$ is also neutral for the new operation, and also every inverse is also an inverse in the new operation, the question is under what conditions is $(G, \circ)$ also a group?

REPLY [8 votes]: As I mentioned in comments, I believe that H. Bender has used this operation in special situations(in particular for certain groups of odd order of nilpotence class 2) , and that his use of it appears in the book "Topics in Finite Groups" by T. Gagen ( published by LMS as I recall). I don't remember all details but here is a proof that when $G$ is a nilpotent group of (finite) odd order and nilpotence class 2, then we always have $x \circ y = y \circ x$.
Recall that if $G$ has nilpotence class 2, then we have $[xy,z] = [x,z][y,z]$ for all $x,y, z \in G$ as all commutators in $G$ are central. In particular, $[x,y]^{2} = [x^{2},y] = [x,y^{2}]$ for all $x,y \in G$. Also, we have always have $[x,y]^{-1} = [y,x]$.
Since when $G$ has odd order, we know that $x^{\frac{1}{2}}$ is an (integer) power of $x$, and likewise for $y$, we have  both 
$[x^{\frac{1}{2}},y]^{2} = [x,y]$ and $[x,y^{\frac{1}{2}}]^{2} = [x,y]$, so we conclude that  we always have $[x^{\frac{1}{2}},y] = [x,y^{\frac{1}{2}}] = [x,y]^{\frac{1}{2}}.$ 
Now from the definition of $x \circ y$, we may write $x \circ y = xy[y,x^{\frac{1}{2}}]$, and likewise, we may write $y \circ x = yx[x,y^{\frac{1}{2}}]$.
Now all commutators are central in $G$, so we now have 
$$(x \circ y)^{-1}(y \circ x) =  [x^{\frac{1}{2}},y]y^{-1}x^{-1}(yx)[x,y^{\frac{1}{2}}] = [x^{\frac{1}{2}},y]^{2}[y,x] = [x,y][y,x] = 1,$$
so that $x \circ y = y \circ x$, as claimed.
Later edit: Actually, I think I can reconstruct the proof of associativity in this case:
We have $x \circ (y \circ z) = x(y\circ z)[y \circ z,x]^{\frac{1}{2}}$ from above.
Again recalling that commutators are central in $G$, we may continue to obtain
$$ x \circ (y \circ z) = x(yz)[z,y]^{\frac{1}{2}}[yz,x]^{\frac{1}{2}},$$ where we make use of the fact that $[[z,y^{\frac{1}{2}}],x] = 1$.
Hence $$x \circ (y \circ z) = x(yz)[z,y]^{\frac{1}{2}}[y,x]^{\frac{1}{2}}[z,x]^{\frac{1}{2}}$$.
Similarly (and making use of commutativity), we obtain $$(x \circ y) \circ z  = z \circ (x \circ y) = z(xy)[y,x]^{\frac{1}{2}}[x,z]^{\frac{1}{2}}[y,z]^{\frac{1}{2}}$$.
Taking the product of the inverse of the first expression with the second expression ( and recalling that commutators are central), we obtain
$(yz)^{-1}x^{-1}zxy [x,z][y,z] = [z,xy][xy,z] = 1$. Hence $\circ$ is associative.
Later edit: In general, given that $(G,\circ)$ is a group, it is in fact necessary that $G$ (with its original operation) should be nilpotent, though I make no attempt to bound the nilpotency class. We have seen that if $G$ has nilpotence class $2$ (or less) then $(G,\circ)$ is a group.
Suppose then that our $G$ of odd order is such that $(G,\circ)$ is a group ( but we make no further assumptions on $G$ with respect to its original operation). Then $(G,\circ)$ is in fact an Abelian group, as noted in comments for we are told that $(x \circ y)^{-1} = x^{-1}\circ y^{-1}$ (which can also be verified by direct computation), while we have $(x \circ y)^{-1} = y^{-1} \circ x^{-1}$ if $\circ$ is associative.
Thus we have $x^{\frac{1}{2}}yx^{\frac{1}{2}} = y^{\frac{1}{2}}xy^{\frac{1}{2}}$ for all $x,y \in G$. As M. Farrokhi D.G. notes in comments below, this is equivalent to requiring that $x^{\frac{1}{2}}y^{\frac{1}{2}}$ and $y^{\frac{1}{2}}x^{\frac{1}{2}}$ commute for all $x,y \in G$. Since squaring is a bijection on $G$ this is equivalent to $[xy,yx] = 1$ for all $x,y \in G$. Thus y and $x^{-1}yx$ commute for all $x,y \in G$. Thus $[[y,x],y] = 1= [[x,y],y] = 1$ for all $x,y \in G$. Thus $G$ is an Engel group. Engel groups are well-studied and finite Engel groups are known to be nilpotent (apparently credited to Zorn), though there are examples of infinite Engel groups which are not nilpotent.
Hence $G$ is nilpotent with respect to its original group operation.
Even later edit: I will now address the question of nilpotence class, making use of a survey article by Gunnar Traustason, which is available online. The group $G$ is a $2$-Engel group, as seen above. Since we now know that $G$ is nilpotent, $G$ decomposes as a direct product of its Sylow $p$-subgroups, and clearly $(G,\circ)$ does too, in a way which respects the decomposition for $G$.
Hence it suffices to understand the situation when $G$ is a $p$-group for some odd prime $p$. It was in fact (at least implicitly) proved by W. Burnside that a $2$-Engel group of order prime to $3$ has nilpotency class $2$ ( and we have already seen that for a group $G$ of nilpotency class $2$, $(G,\circ)$ is indeed a group). It was proved somewhat later that a $3$-group which is a $2$-Engel group has nilpotency class at most $3$. Hence the only outstanding question at this point is whether there is a $3$-group $G$ of nilpotency class  $3$ such that $(G,\circ)$ is a group. For we know that a group $G$ of odd order such that $(G,\circ)$ is a group is a direct product of a $3^{\prime}$-group which is nilpotent of class at most $2$ with a $3$-group which is nilpotent of class at most $3$.
Final edit: In fact, I think it already follows from Lemma 7 of the paper of Glauberman mentioned in the statement of the question that if $(G,\circ)$ is a (necessarily Abelian as remarked before) group, then $G$ must be nilpotent of class $2$.<|endoftext|>
TITLE: Can one describe the multiplication of two Bruhat cells?
QUESTION [13 upvotes]: For $G$ a simple linear algebraic group and $B$ a fixed Borel subgroup, we have the Bruhat decomposition $G = \coprod_{w \in W} B\dot{w}B$, where $W$ is the Weyl group and $\dot{w}$ is any representative of $w$ in $N_G(T)$.
If $s_\alpha \in W$ is a simple reflection, then the following rule is known:

$(B\dot{s_\alpha} B)(B\dot{w}B) = B\dot{s_\alpha}\dot{ w} B,\hspace{5mm}$ if $\hspace{3mm}l(s_\alpha w) = l(w) + 1$,
$(B\dot{s_\alpha} B)(B\dot{w}B) = B\dot{w}B \cup B\dot{s_\alpha}\dot{ w} B,\hspace{5mm}$ if $\hspace{3mm}l(s_\alpha w) = l(w) - 1$,

where $l(w)$ is the length of $w \in W$.
This rule for simple reflections implies that for any $w_1, w_2 \in W$
$(B\dot{w}_1 B)(B\dot{w}_2B) = \coprod_{w \in U} B\dot{w}B$, 
where $U \subset W$. However, knowing what this subset $U$ is seems quite complicated (I think one can put some partial conditions on $U$ involving Bruhat order). 
My question is if anyone has seen a description of which cells appear in $(B\dot{w}_1 B)(B\dot{w}_2B)$ (i.e., what does $U$ look like)? Or, does anyone know of some nice connections between any aspects of this problem and some other theory? It seems likely to me that the decomposition of $(B\dot{w}_1 B)(B\dot{w}_2B)$ is much to complicated to have any single rule for arbitrary $w_1, w_2$. But, I thought I'd pose this question to M.O. just in case. Thanks!

REPLY [6 votes]: This is to complement Ben's answer.  The multiplication you are looking at is closely related to the 0-Hecke monoid.  This is the deformation of the Weyl group $W$ where you keep all the braid relations but you replace the involution relation $s^2=1$ with the idempotent relation $s^2=s$. This monoid $H(W)$ is in bijection with $W$ via $w\mapsto h(w)$ (where $w\in W$  and $h(w)\in H(W)$) and can be identified with the set of principal Bruhat ideals with set wise multiplication.  Richardson and Springer observed that if $w,w'\in W$, then the unique dense open $B\times B$ orbit in $BwB\cdot Bw'B$ is the $Bw''B$ where $h(w)h(w')=h(w'')$.<|endoftext|>
TITLE: Describing the crystalline extension of $\mathbb{Q}_p$ by $\mathbb{Q}_p$
QUESTION [22 upvotes]: Let $K$ be a finite extension of $\mathbb{Q_p}$. The group $\ker H^1(G_K, \mathbb{Q}_p) \rightarrow H^1(G_K, B_{crys})$ is one-dimensional, which tells us that among all extensions of Galois modules 
$$ 
0 \rightarrow \mathbb{Q}_p \rightarrow E \rightarrow \mathbb{Q}_p \rightarrow 0,
$$
which are classified by $H^1(G_K, \mathbb{Q}_p) \cong Hom_{cts}(K^*, \mathbb{Q}_p)$, there is a one-dimensional subspace corresponding to crystalline representations. 
My question is basically: What homomorphism $K^* \rightarrow \mathbb{Q}_p$ does this crystalline cocycle correspond to? 
Actually, I know the answer. It was explained to me that a theorem of Sen asserts that a crystalline representation with only $0$ as Hodge-Tate weight must have finite image of inertia, which in this case must be trivial, i.e. unramified. So the homomorphism in question is (a scalar multiple of) $x \mapsto val_p(x)$. 
I would like to come to a more elementary and concrete understanding of this puzzle. The fact that this is a coboundary in $H^1(G_K, B_{crys})$ means that I can find $b \in B_{crys}$ such that inertia acts trivially on $b$ and Frobenius translates $b$ by $1$. Is it possible to describe this period explicitly in terms of the construction of $B_{crys}$? 
As someone who has never really understood the definition of $B_{crys}$, it would be even more satisfying to me to have an explicit description in terms of elements which I believe should belong in $B_{crys}$ if my intuitive definition of the latter is merely ``the ring of periods for cohomology of varieties with good reduction''. In other words, I would love to see this period $b$ expressed in terms of periods coming from cohomology of familiar varieties, such as the period for the cyclotomic character.

REPLY [13 votes]: The representation $E$ in this case is not only crystalline, it is in fact unramified. This means we don't need much of the complicated machinery of $p$-adic Hodge theory to get a handle on the periods of $E$.
Whereas for general crystalline representations we need to use $\mathbf{B}_\text{cris}$ to find periods, for potentially unramified representations, we can work with $\mathbb{C}_p$ (potentially unramified representations are $\mathbb{C}_p$-admissible). In this case, we don't even need $\mathbb{C}_p$, as the representation is unramified (not just potentially so), and it suffices to work with $(K_0^\text{nr})^\vee$ instead (where $K_0$ is the maximal unramified intermediate extension $K / K_0 / \mathbb{Q}_p$).
To explicitly see that $E$ is $(K_0^\text{nr})^\vee$-admissible, we can start by taking $a$ to be a solution to the Artin–Schreier equation $x^q-x-1=0$ in $\mathcal{O}_K/\mathfrak{m} \cong \mathbb{F}_q$. We then have $\mathrm{Frob}([a]) = [a^q] = [a+1]$, where $[-]$ indicates Teichmüller lifts. This is nearly what we want, save for the fact that the Teichmüller lift is not additive. So you have to remedy that by hand by using the Witt addition polynomials; the upshot is that you'll obtain some element $b \in \mathrm{W}(\overline{\mathbb{F}_q})$ with $\varphi(b) = b+1$ as desired. It is necessary to go all the way up to $\mathrm{W}(\overline{\mathbb{F}_q})$; at any finite level $\mathrm{W}(\mathbb{F}_{q^n})$ there will always be the above issue of non-additivity. Indeed, only potentially trivial representations are detected at finite levels, and we need to pass to the completion $(K_0^\text{nr})^\vee$ to allow a Frobenius of infinite order.
At any rate, this means then that $\{1,b\}$ is a $(K_0^\text{nr})^\vee$-basis of $E \otimes_{\mathbb{Q}_p} (K_0^\text{nr})^\vee$. Using the inclusion $(\mathcal{O}_{K_0^\text{nr}})^\vee \subset \mathbf{A}_\text{cris}$ we can consider $b$ to be an element of $\mathbf{B}_\text{cris}$. It is a crystalline period of $E$; together with $1 \in \mathbf{B}_\text{cris}$ it provides a $\textbf{B}_\text{cris}$-basis of $E \otimes_{\mathbb{Q}_p} \mathbf{B}_\text{cris}$.   
As for the motivic question, I think no such variety is expected to exist. We are in the crystalline situation, so we'd like to be able to find $E$ inside the cohomology of some (smooth, separated, finite type) scheme over $\mathcal{O}_K/\mathfrak{m}$. By analogy with the $\ell$-adic and complex situations, I believe such extensions as $E$ shouldn't occur there, because of weight filtration considerations (e.g. there are no non-split extensions of mixed Hodge structures of $\mathbb{Q}$ by $\mathbb{Q}$).<|endoftext|>
TITLE: non-Noetherian r-Noetherian ring with Noetherian total quotient ring
QUESTION [6 upvotes]: A commutative ring is said to be r-Noetherian if every regular ideal is finitely generated, where an ideal is said to be regular if it contains a non-zerodivisor.  Does there exist a non-Noetherian r-Noetherian commutative ring whose total quotient ring is Noetherian?
EDIT: A commutative ring is Dedekind if every regular ideal is invertible. (A domain is Dedekind if and only if it is a Dedekind domain.)  Does there exist a non-Noetherian Dedekind ring whose total quotient ring is Noetherian?
The motivation for this question is that conditions like r-Noetherian and Dedekind control only the regular ideals of a ring and one can try to use the total quotient ring to control the non-regular ideals.  By this philosophy one would hope that an r-Noetherian ring is Noetherian if its total quotient ring is Noetherian.  But there is no obvious proof of this, leading to the desire for a counterexample.

REPLY [2 votes]: I believe I have answered my question in the positive.  Let $R = \mathbb{Z}+\varepsilon\mathbb{Q}[\varepsilon]$, where $\mathbb{Q}[\varepsilon]$ denotes the ring of dual numbers over $\mathbb{Q}$.  Then $R$ is the integral closure of $\mathbb{Z}$ in the total quotient ring $\mathbb{Q}[\varepsilon]$ of $R$, so in particular $R$ is integrally closed.  Every regular ideal of $R$ is principal, and every finitely generated ideal of $R$ is principal, $R$ is non-Noetherian, and yet the total quotient ring $\mathbb{Q}[\varepsilon]$ of $R$ is Noetherian.<|endoftext|>
TITLE: Does "$|{\cal P}_2(X)| = |X|$ for $X$ infinite" imply ${\sf (AC)}$?
QUESTION [6 upvotes]: This comes from a comment made by user bof in this thread. 
Let $X$ be a set, define ${\cal P}_2(X) = \big\{\{a, b\}: a\neq b\in X\big\}$.
Consider the statement


${\sf (S)}$ If $X$ is an infinite set, then there is a bijection $\varphi: {\cal P}_2(X)\to X$.


Does ${\sf (S)}$ imply ${\sf (AC)}$?

REPLY [13 votes]: Yes, the usual proof that $|X|^2=|X|$ for all $X$ implies AC works for (S) as well.  In detail, let $A$ be any infinite set, let $H$ be its Hartogs number (the least ordinal that does not inject into $A$), and let $X=A\sqcup H$.  A bijection $\mathcal{P}_2(X)\to X$ in particular restricts to an injection $i:A\times H\to X$.  For each $a\in A$, by definition of $H$ there must be some $h\in H$ such that $i(a,h)\not\in A$.  Define $j:A\to H$ by $j(a)=i(a,h)$ for the least such $h$.  Then $j$ is an injection.  Since $H$ is well-orderable, it follows that $A$ is well-orderable.<|endoftext|>
TITLE: Irreducible reps and characters of $G \rtimes A$
QUESTION [9 upvotes]: Is there a theorem which classifies irreducible representations of semi-direct product of finite groups $G \rtimes A$, where $A$ is a finite abelian group and hence write down the character table for $G \rtimes A$?
In particular, I want to write down the character table for $M_{12} \rtimes \mathbb{Z}_2$ from the character table of $M_{12}$. 
Serre's book on representation theory has a theorem on groups of the form $A \rtimes G$ for $A$ abelian, but I am studying groups of the form $G \rtimes A$.

REPLY [4 votes]: Here is a more conceptual approach to Clifford theory. Let me work with a slightly more general setup: namely, suppose we have a short exact sequence
$$1 \to N \to G \to H \to 1$$
of finite groups, which does not necessarily split. What does the representation theory of $G$ look like, in terms of $N$ and $H$? This question can be answered categorically as follows. The idea is to think of $G$ as the homotopy quotient $N//H$ of $N$ with respect to a (categorical) action of $H$. This gives an equivalence
$$\text{Rep}(G) \cong \text{Rep}(N)^H$$
where $(-)^H$ denotes taking homotopy fixed points. As a category, 
$$\text{Rep}(N) \cong \prod_{\hat{N}} \text{Vect}$$ 
where $\hat{N}$ denotes the set of isomorphism classes of irreducible representations of $N$. The action of $H$ on this category breaks up based on the orbits of the action of $H$ on $\hat{N}$, so from now on we restrict our attention to a single orbit. If $\chi$ is an irrep in this orbit then we denote this orbit by $H\chi$. 
Let $H_{\chi}$ denote the stabilizer of $H$ acting on $\chi \in \hat{N}$. Then we have an equivalence
$$\left( \prod_{H \chi} \text{Vect} \right)^H \cong \text{Vect}^{H_{\chi}}$$
where this copy of $\text{Vect}$ is spanned by $\chi$. So now it suffices to understand the homotopy fixed points of $H_{\chi}$ acting on a single copy of $\text{Vect}$. 
Actions of $H_{\chi}$ on $\text{Vect}$ are classified by classes in $H^2(H_{\chi}, \mathbb{C}^{\times})$. Given a $2$-cocycle $c$ representing such a class, the category of homotopy fixed points is the category of projective representations of $H_{\chi}$ with $2$-cocycle $c$. I don't know if there's an easy recipe for computing these $2$-cocycles; Geoff Robinson also indicates that this is the hard step.
Anyway, once the irreducible projective representations of $H_{\chi}$ with $2$-cocycle $c$ have been determined, passing through the equivalences we've been using reproduces Geoff Robinson's answer: irreps of $G$ are classified by projective irreps of the $H_{\chi}$ with appropriate cocycles. From here the nicest thing that can happen is that 
$$H^2(H_{\chi}, \mathbb{C}^{\times}) = 0$$
for all $\chi$, which in fact occurs in this example: $H_{\chi}$ is either trivial or cyclic of order $2$. So as Geoff Robinson and Derek Holt indicate, in this case it suffices to understand the orbits of the action of $\mathbb{Z}_2$ on the set of irreps of $M_{12}$. 
For some blog references see here, here, here, here, and here, but the sequence isn't done yet. For references in the literature I think Ganter and Kapranov is relevant but I haven't looked at it in detail.<|endoftext|>
TITLE: Question abouth Prokhorov metric
QUESTION [5 upvotes]: Let $X$ and $Y$ be two random variables with first order moments, i.e. $E[|X|]$, $E[|Y|]<+\infty$. Assume further that 
$$E\left[|X-Y|\right]<\varepsilon.$$
Set $Law(X)=\mu$ and $Law(Y)=\nu$, it is clear that $\mu$ and $\nu$ are  close in the Prokhorov metric, see 
https://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric
for definition. Denote by $\rho(\cdot,\cdot)$ the Prokhorov metric. My question is how to estimate $\rho(\mu,\nu)$. For example, could we show that $\rho(\mu,\nu)<\varepsilon$ or $\rho(\mu,\nu)<\sqrt{\varepsilon}$? Thanks for the reply!

REPLY [4 votes]: $\sqrt{\varepsilon}$ works. Assume that for some set $A$ we have $\mu(A)=a$ and $\nu(A^{\sqrt{\varepsilon}})<a-\sqrt{\varepsilon}$. Then with probability more then $\sqrt{\varepsilon}$ we have $X\in A$, $Y\notin A^{\sqrt{\varepsilon}}$. In this case $|X-Y|\geqslant \sqrt{\varepsilon}$ for sure. Hence expectation of $|X-Y|$ is more than $\varepsilon$.<|endoftext|>
TITLE: Conformally flat manifold with zero scalar
QUESTION [10 upvotes]: I would like to ask the following : Is there any example of a compact conformally flat Riemannian manifold $(M^n,g)$ with $n\geq 4$ which is not flat and has zero scalar curvature?

REPLY [5 votes]: Responding to the OP's comments on the duplicate question, the example that was given in dimension $4$ can be easily generalized to higher dimensions:

If $(M,g_M)$ has constant sectional curvature $+1$ and $(N,g_N)$ has constant sectional curvature $-1$, then the product manifold $(M\times N,g_M\oplus g_N)$ is conformally flat. This is proved, e.g., in Besse's book "Einstein manifolds" (see Example 1.167, p. 61).
  Furthermore, this product manifold has vanishing scalar curvature if and only if $\dim M=\dim N$.<|endoftext|>
TITLE: The Hales-Jewett Theorem for an infinite alphabet
QUESTION [8 upvotes]: Recall the Hales-Jewett Theorem:

HJT: Given a finite alphabet $A$ and some $r \in \mathbb{N}$, there is some $H \in \mathbb{N}$ such that whenever $A^H$, the set of all length-$H$ words from $A$, is $r$-colored, there is a monochromatic combinatorial line.

[A combinatorial line is defined as follows. Fix some $x \notin A$. A variable word $w$ is a length-$H$ word of members of $A \cup \{x\}$, with $x$ appearing at least once. For $a \in A$, $w(a)$ denotes the result of substituting $a$ for $x$ in $w$. A combinatorial line is any set of the form $\{w(a):a \in A\}$, where $w$ is a variable word. Notice that this definition makes sense even when $A$ or $H$ is allowed to be infinite.]
My question arises simply from wondering what happens if we let $A$ be infinite? If there is any hope of a positive result, we must be willing to let $H$ be infinite too. So I am asking about the following statement:

HJT($\kappa$): Given a set $A$ with $|A| = \kappa$ and $r \in \mathbb{N}$, there is an infinite cardinal $\lambda$ such that whenever the set $A^\lambda$ is $r$-colored, there is a monochromatic combinatorial line.

Using a result of William Weiss, I can prove that HJT($2^{\aleph_0}$) implies the existence of measurable cardinals. But that is about all I know.

Question 1: Is HJT($2^{\aleph_0}$) consistent with ZFC plus some large cardinal assumption?
Question 2: What about HJT($\kappa$) more generally?

The second question is admittedly broad, but that's because I know so little. Any results, pointers, or consistency proofs would be welcome.

REPLY [2 votes]: I returned to this problem recently, and was finally able to answer my own question (though not in the way I'd hoped):

Theorem: The principle HJT($\kappa$) is false for every infinite cardinal $\kappa$.

As a consolation prize, we can at least get:

Theorem: Suppose $A$ is countable and $A^\omega$ is colored with a Borel function. Then there is a monochromatic combinatorial line.
Theorem: If ZF is consistent, then so is ZF+DC+HJT($\aleph_0$).

In other words, the Axiom of Choice can be used to obtain a counterexample to HJT($\kappa$) for any infinite $\kappa$. On the other hand, some form of Choice is needed, because it is consistent with ZF+DC that HJT($\aleph_0$) holds.
To prove the first theorem, it suffices to show that HJT($\aleph_0$) is false, since HJT($\kappa$) implies HJT($\aleph_0$) for all infinite $\kappa$. Let $A = \omega$ and let $\lambda$ be any cardinal. We must exhibit a $2$-coloring of $\omega^\lambda$ with no monochromatic combinatorial lines.
Define an equivalence relation on elements of $\omega^\lambda$ as follows:
$$f \sim g \ \Leftrightarrow \ \{|f(\alpha) - g(\alpha)| : \alpha \in \lambda\} \text{ is bounded.}$$
If $f \sim g$, then define the distance from $f$ to $g$ to be $\max \{|f(\alpha) - g(\alpha)| : \alpha \in \lambda\}$.
Using the Axiom of Choice, pick a representative element from each $\sim$-equivalence class. Given $f \in \omega^\lambda$, determine the distance from $f$ to the chosen representative of its $\sim$-equivalence class. Color $f$ red if this distance is even, and otherwise color $f$ blue.
Now consider any combinatorial line in $\omega^\lambda$; it is determined by some variable word $w$. Notice that the combinatorial line is contained entirely within a single $\sim$-equivalence class; let $f$ be the chosen representative of this class. Let $d$ be the distance from $w(0)$ to $f$. Thus wherever $w$ has its variable, $f$ is at most $d$. It follows that, for $n \geq 2d$, the distance from $f$ to $w(n)$ is one less than the distance from $f$ to $w(n+1)$. In other words, when $n$ is sufficiently large the distance from $f$ to $w(n)$ depends only on whether $n$ is even or odd. Thus our monochromatic line contains both red and blue points.
To prove the second and third theorems, it suffices to take $A = \omega$ and to prove that if $\omega^\omega$ is partitioned into two sets, say $R$ and $B$, each of which has the property of Baire, then there is a monochromatic combinatorial line. (Note that this is HJF($\aleph_0$) for two colors only, but any finite number of colors follows by induction. For the second theorem, note that every Borel set has the property of Baire. For the third, it is known that ZF+DC+"every subset of $\omega^\omega$ has the property of Baire" is consistent relative to ZF.)
$R$ and $B$ cannot both be meagre; let us assume that $R$ is not. Since we assume that $R$ has the property of Baire, we may find a clopen subset $[s]$ of $\omega^\omega$ such that $B \cap [s]$ is meagre. In particular, $B \cap [s^\frown n]$ is meagre for every $n \in \omega$; this means that, for every $n \in \omega$, there is meagre set of $\gamma \in \omega^\omega$ such that $[s^\frown n^\frown \gamma]$ is colored blue. As a countable union of meagre sets is meagre (this is a theorem of ZF+DC), there must be some $\gamma \in \omega^\omega$ such that $[s^\frown n^\frown \gamma]$ is red for every $n \in \mathbb N$.
Let $w$ be the variable word $s^\frown x^\frown \gamma$. The conclusion of the previous paragraph states that the combinatorial line determined by $w$ is monochromatic (in red).<|endoftext|>
TITLE: Is there a locally compact, $\omega_1$-compact, not $\sigma$-countably compact space of size $\aleph_1$?
QUESTION [15 upvotes]: There are old ZFC examples due to Eric van Douwen that satisfy all the properties in the title, except that they are of cardinality $2^{\aleph_0}$, so the answer to the title question is YES if the Continuum Hypothesis (CH) is assumed, but I am asking for an example which does not use any axiom beyond the ZFC axioms. 
It would be especially nice if it were first countable, like Eric's examples, and normal, like one of his examples. A Souslin tree with the interval topology qualifies, but the existence of Souslin trees is ZFC-independent.  I have been able to weaken CH to "stick" [which says that there is a family of $\aleph_1$ countable subsets of an uncountable set, such that every uncountable subset contains a member of the family] and I also have an example if $\mathfrak b = \aleph_1$ but no ZFC example.
It would also be interesting to know whether the existence of an example implies the existence of a first countable example. There are lots of first countable examples under CH, and the "stick" example I have in mind is also first countable, as is my $\mathfrak b = \aleph_1$ example. For sure, there is a scattered example if there is one at all: the "Kunen line" qualifies under CH, while if CH is negated, then we use the fact that every crowded (= dense-in-itself) compact Hausdorff space is of cardinality at least $2^{\aleph_0}$.

REPLY [6 votes]: Lyubomyr Zdomskyy has solved this problem. He has shown:
Theorem. If P-Ideal Dichotomy (PID) holds and $\mathfrak p > \aleph_1$,
then every locally compact, $\omega_1$-compact space of cardinality $\aleph_1$
is $\sigma$-countably compact. 
I've forgotten my password, so I am using Google to "Sign up".<|endoftext|>
TITLE: quotient space of Eilenberg-MacLane space
QUESTION [8 upvotes]: Let $\pi$ be a group and $K(\pi,1)$ the Eilenberg-MacLane space. Let $G$ be a finite group acting on $K(\pi,1)$ such that the following is a covering map
$$
K(\pi,1)\longrightarrow K(\pi,1)/G.
$$
Question.  Is $K(\pi,1)/G$ an Eilenberg-MacLane space? Does
$$
K(\pi,1)/G=K(\pi \times G,1)?
$$

REPLY [16 votes]: Suppose a group $H$, not necessarily finite, acts on an Eilenberg-MacLane space $BN$. The homotopy quotient $BN/H$ (which agrees with the ordinary quotient if the action of $H$ is free) fits into a fiber sequence
$$BN \to BN/H \to BH$$
and the long exact sequence in homotopy shows that $BN/H$ has vanishing higher homotopy. Hence it is an Eilenberg-MacLane space $BG$ for a group $G$ fitting into a short exact sequence
$$1 \to N \to G \to H \to 1$$
(determined by the action). In other words, it's an extension of $H$ by $N$. 
Every such extension arises in this way. Among them, semidirect products (extensions for which the above sequence splits) correspond to pointed actions of $H$ on $BN$, or equivalently actions of $H$ on $BN$ admitting a (homotopy) fixed point. 
For example, take $BN$ to be the configuration space of $n$ ordered points in $\mathbb{R}^2$, so that $N = P_n$ is the pure braid group, and $H = S_n$ acts by permutations of the points. Then $BN/H = BG$ is the configuration space of $n$ unordered points in $\mathbb{R}^2$, so that $G = B_n$ is the usual braid group. The corresponding short exact sequence
$$1 \to P_n \to B_n \to S_n \to 1$$
does not split.

REPLY [11 votes]: $K(\pi,1)/G$ is not necessarily $K(\pi\times G,1)$ for example take $G=Z$, $K(Z,1)=S^1$. $S^1=R/(t_1=x\rightarrow x+1)$ the quotient of $S^1$ by the group $Z/n$ generated by the transformation induced by $x\rightarrow x+1/n$ is $S^1$.<|endoftext|>
TITLE: factorization of the cohomology of configuration space
QUESTION [6 upvotes]: This question is a follow-up to my previous question factorization of the regular representation of the symmetric group, which was answered in a very satisfactory way.
Let $\operatorname{Conf}(n,\mathbb{R}^3)$ be the configuration space of $n$ labeled points in $\mathbb{R}^3$, and consider the cohomology
$B_n := H^*(\operatorname{Conf}(n,\mathbb{R}^3))$, which is a graded representation of $S_n$.  Let $W_n := V[n] + q^2 V[n-1,1]$ be the graded representation that is the 1-dimensional trivial representation in degree 0 and the $(n-1)$-dimensional irreducible permutation representation in degree 2.
Question:  Does there exist a graded representation $M_n$ such that $M_n\otimes W_n\cong B_n$?
If we forget about the grading, then $B_n$ is just the regular representation and $W_n$ is the vector representation $\mathbb{C}^n$; the existence of a representation whose tensor product with the vector representation is isomorphic to the regular representation is explained in the aforementioned post.  So I am now asking a more refined version of the question in which the grading is being taken into account.
I will also note that I "know" that the answer is positive.  That is, I can define a graded representation $M_n$, I conjecture that there is an isomorphism $M_n\otimes W_n\cong B_n$, 
and I've checked this conjecture on a computer up to $n=10$.  What I really want to know is whether the graded representation $B_n$ is already known to factor in this way.  If so, then I would really like to understand why this is the case, as I believe that it will help me to understand the representation $M_n$ that I am interested in.
(In case anyone would like to know, my representation $M_n$ that conjecturally solves this problem is the intersection cohomology of the hypertoric variety associated with the braid arrangement.)

REPLY [3 votes]: Here's a geometric construction of a factorization that works for points in $\mathbb R^2$ (and not any other dimension). Given the close relationship between the cohomology rings of configuration spaces of points in $\mathbb R^d$ for varying $d$ (cf my answer to Cohomology of configuration space as a representation of the symmetric group) I'd expect that one can extract a factorization in arbitrary dimension from this construction, even though the geometric part breaks down.
Note first that the affine group $G = \mathbb C \rtimes \mathbb C^\times$ is the subgroup of Möbius transformations fixing the point $\infty \in \mathbb P^1$. It follows that the quotient $\mathrm{Conf}(\mathbb C,n)/G$ is equal to the moduli space $M_{0,n+1}$ of $(n+1)$ points on the projective line modulo symmetries. This makes $\mathrm{Conf}(\mathbb C,n)$ homotopic to a trivial circle bundle over $M_{0,n+1}$, and so
$$ H^\bullet(\mathrm{Conf}(\mathbb C,n)) \cong H^\bullet(M_{0,n+1}) \oplus H^{\bullet-1}(M_{0,n+1}).$$
So one only needs to construct such an $S_n$-equivariant factorization for the cohomology $H^\bullet(M_{0,n+1})$. Now there is an $S_n$-equivariant  fiber bundle $M_{0,n+1} \to M_{0,n}$ by forgetting the last marking. Each fiber $F$ is $\mathbb P^1$ minus $n$ points, and so $H^0(F) \cong V[n]$, $H^1(F) \cong V[n-1,1]$ as $S_n$-representations. The Leray-Serre spectral sequence degenerates, and this gives the claimed factorization.<|endoftext|>
TITLE: The positive cone of the standard representation of a Von Neumann algebra
QUESTION [8 upvotes]: Let $A$ be a von Neumann algebra, let $L^2(A)$ be the underlying Hilbert space of the standard form of $A$, and $P \subset L^2(A)$ the canonical positive cone (see for example this paper by Haagerup). We know that any normal linear positive functional $\phi$ on $A$ comes from a unique vector in $P$ and that this induces a bijection between element of $P$ and positive normal linear functional on $A$.
Is there a way to describe the addition/the scalar produt in $P$ in terms of the corresponding linear form, preferably without invoking the modular operator of Tomita's theory?
(I'm equally interested in both, and as we know that the norm corresponds to $\eta(1)$ one can go from one to the other easily...)
For example, I'm very interested in the case of the double dual algebra $C^{max}(G)^{**}$ for $G$ a discrete group: its (normal) representations are just representations of $G$ and bounded normal linear functional over it are the same as functions of positive type on $G$.

REPLY [5 votes]: The answer to your question is given in Definition 2.1.1, on page 34 of Kosaki's PhD thesis: https://dmitripavlov.org/scans/kosaki-thesis.pdf
I prefer however the description of $L^2(A)$ given by equation (6.1), on page 18 of my paper: http://arxiv.org/pdf/1110.5671v2.pdf<|endoftext|>
TITLE: Is the Steinberg representation always irreducible?
QUESTION [22 upvotes]: Let $\mathbb{F}$ be a field.  The Tits building for $\text{SL}_n(\mathbb{F})$, denoted $T_n(\mathbb{F})$, is the simplicial complex whose $k$-simplices are flags
$$0 \subsetneq V_0 \subsetneq \cdots \subsetneq V_k \subsetneq \mathbb{F}^n.$$
The space $T_n(\mathbb{F})$ is $(n-2)$-dimensional, and the Solomon-Tits theorem
says that in fact $T_n(\mathbb{F})$ is homotopy equivalent to a wedge of $(n-2)$-dimensional spheres.  The Steinberg representation of $\text{SL}_n(\mathbb{F})$, denoted $\text{St}_n(\mathbb{F})$, is $\widetilde{H}_{n-2}(T_n(\mathbb{F});\mathbb{C})$.
This is one of the most important representations of $\text{SL}_n(\mathbb{F})$; for instance, if $\mathbb{F}$ is a finite field of characteristic $p$, then $\text{St}_n(\mathbb{F})$ is the unique nontrivial irreducible representation of $\text{SL}_n(\mathbb{F})$ whose dimension is a power of $p$.
The only proof I know that $\text{St}_n(\mathbb{F})$ is an irreducible representation of $\text{SL}_n(\mathbb{F})$ when $\mathbb{F}$ is a finite field uses character theory, and thus does not work for $\mathbb{F}$ infinite (in which case $\text{St}_n(\mathbb{F})$ is an infinite-dimensional representation of the infinite group $\text{SL}_n(\mathbb{F})$).
Question: For an infinite field $\mathbb{F}$, is $\text{St}_n(\mathbb{F})$ an irreducible representation of $\text{SL}_n(\mathbb{F})$?  If not, is it at least indecomposable?

EDIT 2: In the previous edit, I said that I accepted an answer that did not answer the question as stated.  However, this has now changed since Andrew Snowden and I have written a paper giving a complete answer to this question.

EDIT: I accepted an answer, but I am particularly interested in the field $\mathbb{Q}$, which is not covered by that answer.  This case is interesting to me because it arises when studying the cohomology of $\text{SL}_n(\mathbb{Z})$; indeed, in this case the Tits building forms the boundary of the Borel–Serre bordification of the associated symmetric space and the Steinberg representation (as I defined it above) provides the "dualizing module" for $\text{SL}_n(\mathbb{Z})$.  See Section 2 of my paper

T. Church, B. Farb, A. Putman A stability conjecture for the unstable cohomology of $\text{SL}_n(\mathbb{Z})$, mapping class groups, and $\text{Aut}(F_n)$,
in "Algebraic Topology: Applications and New Directions", pp. 55–70,
Contemp. Math., 620, Amer. Math. Soc., Providence, RI. doi:10.1090/conm/620/12366, arXiv:1208.3216

for a discussion of this and references.  It is also available on my webpage here.

REPLY [9 votes]: Andrew Snowden and I managed to finally answer this question in our paper "The Steinberg representation is irreducible", available here.  As you might guess from the title, we prove that the Steinberg representation over an infinite field is always irreducible.  In fact, we prove something much more general that applies to arbitrary reductive groups over infinite fields, and also allows arbitrary coefficients for the Steinberg module.
It's worth also mentioning another recent paper by Galatius--Kupers--Randal-Williams called "$E_{\infty}$-cells and general linear groups of infinite fields", available here.  One of their results says that the Steinberg representation for $\text{GL}_n$ (as discussed in this question) is indecomposable, i.e. is not the nontrivial direct sum of two subrepresentations.  For infinite fields, the Steinberg representation is infinite-dimensional, so this is weaker than being irreducible.  However, I think their proof is quite beautiful and worth reading even if it gives a weaker result.

EDIT: My attention has been drawn to two earlier papers:
N. Xi, Some infinite dimensional representations of reductive groups with Frobenius maps, Sci. China Math. 57 (2014), no.~6, 1109--1120.
R. Yang, Irreducibility of infinite dimensional Steinberg modules of reductive groups with Frobenius maps, J. Algebra 533 (2019), 17--24.
These focus on connected reductive group $\mathbf{G}$ over the algebraic closure $k=\overline{\mathbb{F}}_q$ of a finite field $\mathbb{F}_q$.  For instance, we could have $\mathbf{G}(k) = \text{GL}(n,k)$ as in the question.  Their main theorem says that the Steinberg representation of $\mathbf{G}$ is irreducible with coefficients in any field.  Xi's paper handles the case when the coefficients have characteristic $0$ or $\text{char}(k)$, and Yang's paper handles other characteristics.<|endoftext|>
TITLE: Interpret Fourier transform as limit of Fourier series
QUESTION [11 upvotes]: Let $V=\mathbb{R}^n$, $\Lambda_r=2\pi r \mathbb{Z}^n \subset V (r>0)$ a lattice; $V^*\cong\mathbb{R}^n$ the dual vector space of $V$, and $\Lambda_r^*=\frac{1}{2\pi r} \mathbb{Z}^n =\text{Hom}(\Lambda_r, \mathbb{Z})$ the dual lattice in $V^*$.
$\Lambda_r^*$ can be thought of as the Pontryagin dual of the torus $T^n_r=V/\Lambda_r$; also, $V^*$ can be thought of as the Pontryagin dual of $V$ and can be identified with $V$ via the pairing $\left< x,\xi \right>=e^{2\pi i x\cdot\xi}$. Chapter 4 of Gerald B. Folland's book A course in abstract harmonic analysis is a nice introduction to these materials in the context of locally compact abelian groups; see also this blog of Terence Tao.
It's well known that the Fourier transform gives an isometry of Hilbert spaces
$$L^2(V)\cong L^2(V^*).$$
Also, Fourier series give an isometry of Hilbert spaces
$$L^2(T^n_r)\cong l^2(\Lambda_r^*).$$
We have the following obvious intuition: as $r>0$ becomes larger and larger, the scale of $T^n_r$ also becomes larger and larger, and finally becomes like $V=\mathbb{R}^n$; on the other hand, the dual lattice $\Lambda_r^*$ becomes more and more 'dense' in $V^*=\mathbb{R}^n$ as the distance of adjacent points is $\frac{1}{2\pi r}$, which goes to 0 as $r$ goes to $\infty$.
My question is the following:

Can we make it mathematically rigorous, both on the level of functions and on the level of spaces (e.g. $T^n_r \to V$), that the 'limit' of the isomorphisms
  $$L^2(T^n_r)\cong l^2(\Lambda_r^*)$$
  is the isomorphism
  $$L^2(V)\cong L^2(V^*)$$
  as $r$ goes to $\infty$?

The bad thing is that $V=\mathbb{R}^n$ is noncompact, while we have the notion of Bohr compactification, I hope this can be helpful.

Is there any relation between the tori $T^n_r$ and the Bohr compactification of $\mathbb{R}^n$?

Hopefully, if we can do this, then we can do similar things such as interpreting Fourier inversion as a limit. Some aspect (on the level of functions) is discussed in Exercise 40 (Fourier transform on large tori) of Tao's blog.

REPLY [2 votes]: Great question.  I've often used this heuristic but never thought about whether it had a rigorous meaning.
Let me do this in one dimension; the generalization to higher dimensions is straightforward. My first comment is that the Fourier transform between $l^2(\frac{1}{2\pi r}\mathbb{Z})$ and $L^2(r\mathbb{T})$ genuinely sits inside of the Fourier transform of distributions on $\mathbb{R}$: identify an element $(a_n)$ of $l^2(\frac{1}{2\pi r}\mathbb{Z})$ with the sum of delta functions $\sum a_n\delta_{n/2\pi r}$, and a function $f \in L^2(r\mathbb{T})$ with its periodic extension to $\mathbb{R}$. The integral of $\sum a_n\delta_{n/2\pi r}$ against $e^{-2\pi i xt}$ is $\sum a_n e^{-int/r}$. So the ordinary Fourier transform between the integers and the circle matches up with the distributional Fourier transform after making this identification.
But you want to approximate the $L^2$ Fourier transform on $\mathbb{R}$. I guess the obvious thing to do here is to convolve the embedded $l^2(\frac{1}{2\pi r}\mathbb{Z})$ with the "rectangular function" which takes the value $\sqrt{2\pi r}$ on $[-\frac{1}{4\pi r}, \frac{1}{4\pi r}]$ and is $0$ elsewhere. This isometrically embeds $l^2(\frac{1}{2\pi r}\mathbb{Z})$ into $L^2(\mathbb{R})$. In the transformed picture it corresponds to multiplying a function in the embedded $L^2(r\mathbb{T})$ by Fourier transform of the rectangular function, which is $\frac{1}{\sqrt{2\pi r}}{\rm sinc}(\frac{t}{2r})$. So now we have isometric embeddings of $l^2(\frac{1}{2\pi r}\mathbb{Z})$ and $L^2(r\mathbb{T})$
into $L^2(\mathbb{R})$ which are compatible with taking the Fourier transform before or after embedding. They converge to $L^2(\mathbb{R})$ in the sense that the orthogonal projections onto the embedded spaces converge strongly to the identity operator; this is easy to check in the untransformed picture. (It's clear that $P_nf \to f$ when $f$ is piecewise constant, and such functions are dense in $L^2(\mathbb{R})$.)<|endoftext|>
TITLE: What is modern algebraic topology(homotopy theory) about?
QUESTION [58 upvotes]: At a basic level, algebraic topology is the study of topological spaces by means of algebraic invariants. The key word here is "topological spaces". (Basic) algebraic topology is very useful in other areas of mathematics, especially, in geometry(I would say almost in all geometry).
I'm not an algebraic topologist myself, so I know only basic techniques. However, I'm intrigued by modern tool in homotopy theory. For example, we have simplicial homotopy theory, where one studies simplicial sets instead of topological spaces.
As far as I understand, simplicial techniques are indispensible in modern topology. Then we have axiomatic model-theoretic homotopy theory, stable homotopy theory, chromatic homotopy theory. Recently, we got a topological version of algebraic geometry, namely spectral algebraic geometry which is proved useful in studying topological modular forms.
But one may wonder what is it for? Those are really fancy and sometimes beautiful tools, but what are exactly the questions modern algebraic topology seeks to answer? Because It feels it's really not part of topology anymore, it's more as topology now is a small part of algebraic topology/homotopy theory.
So, I would like to hear about goals and perspectives of modern homotopy theory from those working on it. I hope this question might be useful to someone else.

REPLY [16 votes]: In response to Ryan Budney's comment, let me try to say something about topological data analysis, and other recent applications of algebraic topology outside of traditional mathematics.
Applied Algebraic Topology has been around in various forms for many years. I first learned about it in my training in computer science from Rob Ghrist's work. In fact, I wrote an MO answer back in 2011 about his work. The point seems to be efficiently computing sheaf cohomology, with applications in electrical engineering. Why sheaves? I'll illustrate with an example. All over the country a bunch of moving cell phones are trying to connect to a bunch of cell towers. The regions those towers can reach form a cover of your space. If a cell phone is in a place not covered by any tower, it's bad news, and you want to be able to detect that. Homology helps, since it finds holes. More importantly, if a cell phone is in an intersection, then it has many towers to talk to, and that can cause interference. Sheaf cohomology comes into play here, and can help you design better systems, detect interference issues, and even create coding schemes to fix the confusion interference can cause.
More recently, Gunnar Carlsson's group at Stanford (and his company) has been using algebraic topology to compute on data (my interest is that I mostly teach statistics nowadays). It's called Topological Data Analysis. If you've ever taken a basic statistics course, you know we often use linear regression, i.e. find the best fitting line and use it to make predictions for x values where we don't have any data. If the data is not linear, we transform it (via logs, square root, etc) to make it linear. But that's just because linear things were easy back in the days before computers. Nowadays you could use computational software to run much more complicated regressions. It's just as easy now to fit a curve (e.g. polynomial regression) as a line, since both involve pushing a button on any statistical software. Why stop at curves? If your data comes in the shape of a manifold, why not try to fit a manifold to the data, and use that manifold to predict values of the dependent variable for various combinations of values of the independent variables. Topological data analysis strives to give you the tools to do this. On a more basic level, persistent homology lets you detect holes in your data, by which I don't mean missing values, but rather actual regions where data is not coming to you because it's not being generated there. 
As a silly example, think of taking a picture of Lake Geneva at night. You'd probably see lots of lights ringing the lake, but none inside it. The data here are the lights, and the fact that there are no lights coming from the lake is telling you that something is not there. Similarly, you could imagine taking a picture of the sky and noticing dark spots as a way to find satellites. The examples Gunnar's group has produced are much more useful and less contrived. I believe several have to do with breast cancer data. If you google, you'll find lots of slides of talks he's given, replete with examples.
Persistent homology works by considering all possible covers of your dataset by balls of radius r drawn around the data points, as r varies. It's best to imagine 2 dimensional data where you roughly see the shape of a circle. When r is very small, the cover is entirely disconnected. When r is very large, you're probably looking at a bunch of intersecting balls, with way too many overlaps to tell you much. But for some value of r in the middle, you get a connected shape that looks roughly like $S^1$. The balls form a simplicial complex, and that's how the computations are done. When the balls form many disconnected components, $H_0$ has large dimension. Once they coalesce into a connected component, $H_0$ is $\mathbb{Z}$ and (in the circle example) $H_1$ is also $\mathbb{Z}$. It remains $\mathbb{Z}$ as $r$ gets larger and larger, till r becomes so large that the union of the covering balls forms a disc rather than a circle (up to homotopy). The word "till" in the last paragraph is why it's called "persistent" homology. One way to visualize how the homology groups change with r is to write them as barcodes, where the left-to-right axis is r and the number of bars is the dimension. When you see a long barcode, that's telling you a feature of your data that is persistent even as r varies, e.g. a hole.
There are also applications of topological data analysis (TDA) to Machine Learning, Clustering, and Classification. A simple example is barycentric clustering, which is something like a souped-up, topological version of k-means clustering. Gunnar's group has more complicated examples that have been useful in identifying previously unknown associations, that were later backed up with theory. A common problem is dividing a dataset into distinct pieces, e.g. via Support Vector Machines. Basically: if your dataset can be separated by a hyperplane then you do so. If not, you transform to a higher dimensional space where it can be and then separate it there (equivalently, you find a separating sheet or surface). I am hopeful that the methods of TDA can be used to provide improved separation algorithms.
More recently, Kathryn Hess has gotten involved with applications of algebraic topology to neuroscience. This is related to both work of Ghrist and work of Carlsson, but different from both. Now the game is to discover how information travels across the network of neurons in your brain. Working with rats, you can stimulate the brain and empirically measure how electricity moves. You can then try to uncover traits of the network based on which pathways are being used frequently, and you can try to figure out what determines the path taken and what difference the path taken makes. I know less about the work Hess is doing here, but I know it has to do with computing Betti numbers and using them as invariants. Carlsson also has work related to neural networks (I seem to recall hearing that rats have a Klein bottle in their brain, but have no idea why), but I think it has a different flavor.
In a similar vein, there was an AMS Special Session at the 2012 Joint Math Meetings entitled Generalized Cohomology Theories in Engineering Practice. I only got to one talk in that session (about K-theory as an invariant of some engineering system), but perhaps googling the speakers would lead to more useful applications. 
Incidentally, there are also algebraic topologists working in graph theory, to use algebraic topology to make new graph algorithms. Certainly computing $H_1$ is a way of detecting cycles. From what I understand, the algorithms produced so far don't do much that is new and interesting, and are much less efficient than existing algorithms. There are also people studying random simplicial complexes in the way that random graphs have been well studied. For an example, see this paper on arxiv and follow the references. Finally, there are people writing down effective algorithms to compute in simplicial sets, e.g. here. All of this may bear fruit, as we learn better how to model the world using simplicial complexes and simplicial sets, and as we find ways to wrangle data into forms where our tools can be used to attack it.<|endoftext|>
TITLE: Is there an uncountable Borel almost disjoint family?
QUESTION [5 upvotes]: Here we are considering subsets $\mathcal{F}$ of $2^\omega$, which are in correspondence with families of subsets of $\omega$ (sets of "reals").  Such a family is Borel if it is a Borel subset of $2^\omega$ under the usual topology.
Such a family is almost disjoint if, for every pair $X\not=Y$ from $\mathcal{F}$, $X\cap Y$ is finite.
Note: the question originally only required that the symmetric difference between $X$ and $Y$ be infinite, which is substantially weaker.
Countable almost disjoint families can be constructed fairly trivially.  Uncountable almost disjoint families exist, and are a standard object of study in some branches of set theory.  However, the constructions I've seen do not result in a Borel set.  Can this be done?  Can $\mathcal{F}\subset 2^\omega$ be an uncountable Borel set which is an almost disjoint family?

REPLY [3 votes]: The answers to this question: Uncountable family of infinite subsets with pairwise finite intersections describe several Borel constructions of uncountable Borel families. (Usually: perfect families)<|endoftext|>
TITLE: Complemented subspaces in the dual of James' space $J$
QUESTION [11 upvotes]: James' space $J$ is subprojective; i.e., every infinite dimensional (closed) subspace of $J$ contains an infinite dimensional subspace which is complemented in $J$. This fact can be found in Corollary 11 of [P.G. Casazza et al. Proc. Amer. Math. Soc. 26 (1977), 294-305]. 
I would like to know if the dual space $J^*$ is subprojective. It is known that a non-reflexive subspace of $J^*$ contains an infinite dimensional subspace complemented in $J^*$ (Theorem 2 in [A. Andrew. Israel J. Math. 38 (1981), 276-282]), but I do not know what happens for infinite dimensional reflexive subspaces.

REPLY [4 votes]: Yes, $J^*$ is subprojective. 
Suppose that $M$ is a closed subspace of $J^*$. By Mazur's theorem, without loss of generality we additionally suppose that $M$ has a normalised basis $(f_k)_{k=1}^\infty$. Let $\iota\colon M\to J^*$ be the inclusion map and let $Q\colon J^{**}\to M^*$ be its (surjective) adjoint. Denote by $(f^*_k)_{k=1}^\infty$ the coordinate functionals of $(f_k)_{k=1}^\infty$. 
Choose a bounded sequence $(z_n)_{n=1}^\infty$ in $J^{**}$ such that $Qz_n = f_n^*$ ($n\in \mathbb{N}$). As $J$ is quasi-reflexive, by Rosenthal's $\ell_1$-theorem, $(z_n)_{n=1}^\infty$ contains a weakly Cauchy subsequence, call it $(z^\prime_n)_{n=1}^\infty$. As $(z^\prime_n)_{n=1}^\infty$ fails to have a convergent subsequence, we may suppose that the sequence $(z^\prime_{2n+1}-z^\prime_{2n})_{n=1}^\infty$ is semi-normalised. However, by a result of Andrew [1, Theorem 2.1], every semi-normalised weakly null sequence in $J$ (so also in $J^{**}$ as $J^{**}\cong J$) contains a sequence spanning a complemented copy of $\ell_2$, so $(z^\prime_{2n+1}-z^\prime_{2n})_{n=1}^\infty$ contains such subsequence, say $(w_n)_{n=1}^\infty$. By reflexivity of $\ell_2$, $(w_n)_{n=1}^\infty$ is shrinking so the adjoint of a projection from $J^{**}$ onto $[w_n]_{n=1}^\infty$ is a projection onto a subspace of $M$. As $M\subset J^* \subset J^{***}$, the conclusion follows.
References:

A. Andrew, Spreading basic sequences and subspaces of James' quasi-reflexive space, Math. Scand. 48 (1981), 276–282.<|endoftext|>
TITLE: What is known about the algebraic variety defined by the group determinant?
QUESTION [5 upvotes]: What is known about the algebraic variety $V_G$ defined by $det(X_G) = 1$ where $X_G$ is the group matrix $(x_{g_ig_j^-1})$ of a finite group $G$? It is known that two finite groups having the same determinant are isomorphic.
Edit: Are there any results concernig the group structure on $V_G$?

REPLY [4 votes]: Per the suggestion of Neil Strickland, here is an answer.  There are some excellent notes of Keith Conrad about all of this.
If $k$ is an algebraically closed field of characteristic prime to $n$, the order of $G$, then classical results about representations of finite groups imply an isomorphism of associative, unital $k$-algebras, $$\phi:k[G]\to A_1\times \dots \times A_r,$$ between the group ring $k[G]$ is isomorphic and a product $A_1\times \dots \times A_r$ of matrix algebras $A_i \cong \text{Mat}_{n_i\times n_i}(k)$ (if $k$ is not algebraically closed, each $A_i$ may be a matrix algebra over a division ring).  Here the integers $n_1\leq \dots \leq n_r$ are the ranks of the finitely many (non-isomorphic) irreducible representations of $G$ over $k$.  In particular, because there is a trivial one-dimensional represenation, $n_1$ equals $1$.
For any associative, unital $k$-algebra $A$,
multiplication on the right in $k[G]$ defines a "regular representation", i.e., a homomorphism of associative, unital $k$-algebras, $$ \rho_A: A \to \text{Hom}_k(A,A),\ a \mapsto (b\mapsto b\cdot a).$$  If $A$ is a finite dimensional $k$-vector space, then there is the standard determinant polynomial on $\text{Hom}_k(A,A)$.  The composition with $\rho_A$ is the algebra norm, $\Theta_A$, which is some power of the reduced norm.  
For $A=k[G]$, for the standard $k$-basis of $k[G]$, $(\mathbf{b}_\gamma)_{\gamma \in G}$, and for a general element $a = \sum_{\gamma} x_\gamma \mathbf{b}_\gamma$ of $k[G]$, the matrix entry of $\rho(a)$ with respect to $\mathbf{b}_\gamma$ (in the domain of $\rho(a)$) and $\mathbf{b}_\delta$ (in the target of $\rho(a)$) is $$ \rho(a)_{\delta,\gamma} = x_{\delta\cdot \gamma^{-1}}.$$  Thus the determinant $\text{det}(x_{\delta\cdot \gamma^{-1}})_{\gamma,\delta}$ equals $\Theta_G(a)$, i.e., composition of $\rho$ with the usual determinant on $\text{Hom}_k(A,A) \cong \text{Mat}_{n\times n}(k)$.  
The algebra norm is compatible with isomorphism of associative, unital $k$-algebras.  Thus, via $\phi$, $\Theta_G$ is the same as $\Theta_{A_1\times \dots \times A_r}$.  Also the algebra norm diagonalizes, i.e., for $(a_1,\dots,a_r)\in A_1\times\dots \times A_r$, $$\Theta_{A_1\times \dots \times A_r}(a_1,\dots,a_r) = \Theta_{A_1}(a_1) \cdots \Theta_{A_r}(a_r).$$  Finally, $\Theta_{\text{Mat}_{n\times n}(k)}$ equals $\text{det}^n$, where $\text{det}$ is the usual determinant polynomial on $\text{Mat}_{n\times n}(k)$.  Now consider the multiplicative morphism,
$$ \text{det}:A_1\times \dots \times A_r \to k^r, \ (a_1,\dots,a_r) \mapsto (\text{det}(a_1),\dots,\text{det}(a_r)).$$  The algebra norm is the composite of this morphism with the morphism, $$k^r \mapsto k, \ (t_1,\dots,t_r) \mapsto t_1^{n_1}\cdots t_r^{n_r}.$$  Since $n_1$ equals $1$, for every $t\in \mathbb{G}_m$, the fiber of $\Theta_{k[G]}$ over $t$ is isomorphic to $\textbf{GL}(A_2)\times \dots \times \textbf{GL}(A_r)$ via the isomorphism, $$\textbf{GL}(A_2)\times \dots \times \textbf{GL}(A_r) \to \Theta_{k[G]}^{-1}(t), \ (a_2,\dots,a_r) \mapsto (t\cdot\text{det}(a_2)^{-n_2}\cdots \text{det}(a_r)^{-n_r}, a_2,\dots,a_r).$$  Thus, the fiber of $\Theta_{k[G]}$ over $1$ is isomorphic, as a $k$-group scheme, to the product $\textbf{GL}(A_2)\times \dots \times \text{GL}(A_r)$.  
In particular, the structure of $\text{Ker}(\Theta_{k[G]})$ as a $k$-group scheme is equivalent to the information of the sequence of integers $(n_1=1,n_2,\dots,n_r)$ of ranks of the irreducible $k$-representations of $G$.<|endoftext|>
TITLE: Is the Martin's axiom number $\mathfrak m$ regular
QUESTION [14 upvotes]: The Martin's axiom number $\mathfrak m$ is the least cardinal $\kappa$ for which $\text{MA}_\kappa(\text{ccc})$ is false, i.e. the least cardinal such that there exists a ccc poset $P$ and a family $\mathcal D$ of dense subsets of $P$ with $|\mathcal D| = \kappa$ such that there no $\mathcal D$-generic filter $G \subseteq P$.
Is $\mathfrak m$ regular?
All I know is that $\mathfrak m$ cannot possibly be singular unless $\mathfrak m < \mathfrak c$.
This is because 
$$\mathfrak m \leq \mathfrak p \leq \mathfrak c$$
where $\mathfrak p$ is the pseudo-intersection number. Hence $\mathfrak m = \mathfrak c$ implies $\mathfrak m = \mathfrak p$ and $\mathfrak p$ can be shown to be regular.

REPLY [15 votes]: Not necessarily. That $\mathfrak m$ is consistently singular is proved in 

MR0947850 (89m:03045) Kunen, Kenneth. Where $\mathsf{MA}$ first
  fails. J. Symbolic Logic 53(2), (1988), 429–433.

There, Ken shows that $\mathfrak{m}$ can be singular of cofinality $\omega_1$. 
(Both links above are behind paywalls.)<|endoftext|>
TITLE: Characters of permutation groups
QUESTION [10 upvotes]: Let $N$ be a fixed positive integer, and denote by $C(m)$ the number of
permutations on an $N$-element set that have exactly $m$ cycles (counting
$1$-cycles). Then it is in the literature that the polynomial generating
function
$$
\sum_{m=0}^N C(m)x^m = x(x+1) \dots (x+N-1),
$$
the rising factorial.
I would like a reference (or a proof) for the following.
Let $C(m,j)$ be the number of permutations (of the $N$-element set) which
have exactly $m$ cycles, $j$ of which are $1$-cycles, that is, the number
of permutations with $m$ cycles and $j$ fixed points. Then the following
should be true:
$$
\sum_{m=0}^N \sum_{j=0}^m C(m,j)(j-1)x^m = (N-1)\cdot (x-1)x(x+1) \dots
(x+N-2),
$$
that is, $N-1$ times the rising factorial of $x-1$.
I verified this for $N= 3,4,5,6$ by hand, and figured that if it works up
to $6$, it probably is true for all $N$ (since $S_4$ and $S_6$ are the
screwiest symmetric groups). But this surely must be known.
Motivation. Let $V$ be a $k$-dimensional vector space (over the
complexes or the rationals, it doesn't matter), form the $N$-fold tensor
product of $V$ with itself, $W = \otimes^N V$, and consider the obvious
permutation action of $S_N$ on $W$. This is a very heavily studied
object, going back (at least) to Schur. Let $\chi$ denote the character of
this representation of $S_N$ (this depends on $k$, of course).
It is not difficult to see that for $g \in S_N$, we have $\chi(g) =
k^{c(g)}$ where $c(g)$ is the number of cycles in $g$. Let $\chi_0$ be the
trivial character, and let $\chi_s$ be the character of the standard
$N-1$-dimensional irreducible representation of $S_N$ ($\chi_s (g) $ is the number
of fixed points less $1$, of $g$). Then the first formula entails that
$(\chi,\chi_0) = {{k+N-1}\choose N}$, and the second (if true) says that
$(\chi,\chi_s) = (N-1) {{k+N-2}\choose N}$.
If $\chi_{-}$ and $\chi_{s-}$ are respectively the irreducible characters
obtained by multiplying $\chi_0$ and $\chi_s$ by the sign character (the
other linear character, denoted $\chi_{-}$), then we also deduce from the
equations above (evaluating at $-x$ and normalizing) that $(\chi,\chi_{-})
= {k \choose N}$ and $(\chi,\chi_{s-}) = (N-1){{k+1} \choose N}$.
I was interested in the class function (on $S_N$), $\psi_k: g \mapsto
k^{c(g)}$ for $k$  positive real numbers (rather than just integers; set $x=k$); the
question was for which values of $k$ is this a nonnegative real
combination of irreducible characters. If the second displayed equation is
true, and $k < N-1$, then nonnegativity of $\psi_k$ implies that $k$ is an integer.
Presumably, it is true that for all real $k > N-1$, $\psi_k$ has only
nonnegative (and probably positive) coefficients (with respect to the
irreducible characters). I verified this for $N = 3,4$ by hand, and then
was exhausted. [A very general and easy result, using the Perron theorem
for example, is that this is true for all sufficiently large $k$.]

REPLY [4 votes]: For the reasons apparent below I shall use the notation $C_N(m,j)$, not $C(m,j)$. It is sufficient to prove
 $$
\sum_{m=0}^N\sum_{j=0}^m jC_N(m,j)x^m=Nx\cdot x(x+1)\cdots (x+N-2), 
 $$
as this formula together with your first formula implies what you want to prove. The LHS of this is the same as
 $$
\sum_{(\sigma,k)\colon \sigma\in S_N \text{ has } m \text{ cycles}, \,\sigma(k)=k }x^m, 
 $$
which, after changing the order of summation and denoting $m'=m-1$, is the same as
 $$
\sum_{k=1}^N x\sum_{m'=0}^{N-1}\sum_{j=0}^{m'}C_{N-1}(m',j)x^{m'}=\sum_{k=1}^Nx\cdot x(x+1)\cdots(x+N-2),
 $$
which is equal to $Nx\cdot x(x+1)\cdots(x+N-2)$, as required.<|endoftext|>
TITLE: Induced matrix norm less than one for matrices with spectral radius less than one
QUESTION [6 upvotes]: Let $A$ be a square matrix with elements in $\mathbb{R}$ or $\mathbb{C}$,
$\rho\left(A\right)$ stands for the spectral radius of $A$, i.e.,
the maximum absolute eigenvalue of $A$; $A^{*}$ is the conjugate
transpose of $A$; an induced matrix norm $\left\Vert *\right\Vert $ besides
the usual vector norm properties,  has the sub-multiplicative property.
Question: Given a matrix $A$, there exists a matrix norm $\left\Vert *\right\Vert $ such
that 
$\rho\left(A\right)<1\Rightarrow\left\Vert A\right\Vert <1$
and $\left\Vert A^{*}\right\Vert <1$? 
Motivation: In [Horn and Johnson 1985, Matrix Analysis, Lemma 5.6.10]
it is shown how to construct a matrix norm such that $\left\Vert A\right\Vert \leq\rho\left(A\right)+\epsilon$
for any given scalar $\epsilon>0$. With this norm it is sufficient
to choose $\epsilon$ sufficiently small to guarantee $\left\Vert A\right\Vert <1$.
However, it is possible to find examples where any of the above choice of $\epsilon$ leads to $\left\Vert A^{*}\right\Vert >1$.
The problem is to find a matrix norm with sub-multiplicative property that guarantees both, $\left\Vert A\right\Vert <1$
and $\left\Vert A^{*}\right\Vert <1.$
Someone know how to prove or a reference with a proof for that?

REPLY [9 votes]: Consider the matrix $A = \pmatrix{0 & 1\cr 0 & 0\cr}$ which has spectral radius $0$.  But $A A^*$ has spectral radius $1$, so for any sub-multiplicative norm $$1 \le \|A A^*\| \le \|A\| \|A^*\| \le \max(\|A\|,\|A^*\|)^2$$

REPLY [2 votes]: Here's another argument (when $n > 1$): all norms on the algebra of $n \times n$ matrices are equivalent since it is a finite dimensional Banach space ( this is true of all linear space norms, whether submultiplicative or not). Hence, given any algebra ( ie submultiplicative as well as subadditive) norm on the algebra of $n \times n$ matrices, we have $\rho(A) = \lim_{k \to \infty} \|A^{k}\|^{\frac{1}{k}} \leq \|A \|$ for any matrix $A$.
If there were an algebra norm of the form you ask, then if $\rho(A) <1$ we would have 
$\|A \| < 1$ and $\|A^{\ast}\| < 1$, so $\rho(AA^{\ast}) \leq \| AA^{\ast} \| \leq \|A\| \|A^{\ast}\| < 1$. But consider an upper triangular matrix $A$ with all $0$ on the main diagonal, every entry above the main diagonal $2$. Then $A$ is nilpotent, so that $\rho(A)= 0$. However $AA^{\ast}$ has trace $2n(n-1)$, so some eigenvalue of $AA^{\ast}$ has absolute value $2(n-1)$ or greater and $\rho(AA^{\ast}) \geq 2(n-1) \geq 2$. In fact we can find a nilpotent matrix $A$ 
such that $\rho(AA^{\ast})$ is as large as we like.<|endoftext|>
TITLE: Is the Fourier transform of $e^{-|x|^n}$ positive?
QUESTION [8 upvotes]: Let
$$\Phi(x) = \int_{\mathbf{R}^n} e^{-|y|^n +i (x,y)} dy.$$
Is $\Phi$ positive everywhere in $\mathbf{R}^n$?
Could someone helps me answer this question or gives a reference for it? Thanks.

REPLY [18 votes]: Here is a full characterization.
Theorem. The function $e^{-|x|^a}$ is positive definite for $0\le a \le 2$ and is not positive definite for $a>2$. Thus, its Fourier transform is positive and non positive for the said ranges.
Proof. The claim $a=0$ is trivial. Assume $0<a<2$. Then, it can be shown that
\begin{equation*}
-|x|^a = C_a\int_{-\infty}^\infty \frac{\cos (xt) - 1}{|t|^{a+1}}dt,
\end{equation*}
where $C_a$ is a constant depending on $a$. Since $\cos(xt)$ is a positive definite function, we see that $-|x|^a$ is conditionally positive definite (because of the $-1$ term). Hence, $\exp(-|x|^a)$ is positive definite, and consequently by Bochner's theorem, it's FT is positive.
For $a>2$, it is easy to construct numerical examples where the associated function is not positive definite, and hence its FT is not positive. Carlo Beenakker's answer gives an example. 
To obtain a formal proof of this, here's an outline by contradiction. In particular, suppose that for some $a > 2$, the kernel $e^{-|x-y|^a}$ is positive definite, $|x-y|^a$ is negative definite. Thus, by appealing to Schoenberg's theorem, it must be the case that $d(x,y) := |x-y|^{a/2}$ is a metric on $\mathbb{R}$. Choosing $x,y,z=(0,1,2)$ and comparing $d(x,y)=d(y,z)=1$ but $d(x,z)=2^{a/2}>2$, a contradiction to the triangle inequality. 
Reference. Chapter 5, Positive definite matrices, R. Bhatia. Princeton University Press, 2007.<|endoftext|>
TITLE: Genuine equivariant ambidexterity
QUESTION [9 upvotes]: A particular case of Lurie and Hopkins' ambidexterity theory is that if $G$ is a finite group acting on a $K(n)$-local spectrum $X$ then the norm map
$$ X_{hG} \to X^{hG} $$
is a $K(n)$-local equivalence. If we now replace $G$ with a compact lie group and $X$ with a genuine equivariant free $G$-spectrum (not necessarily $K(n)$-local), then the Adams equivalence asserts that the twisted norm map 
$$ X_{G} \to (\Sigma^{-ad(G)} X)^G $$
is an equivalence, where $ad(G)$ is the adjoint representation of $G$. One might then be led to ponder the possibility that these two phenomena are somehow part of the same thing. For example, one might consider the following (possibly overly optimistic) conjecture:
Conjecture:
Let $G$ be a compact lie group and $X$ a genuine equivariant $K(n)$-local $G$-spectrum (not necessarily free). Then the twisted norm map
$$ X_{G} \to (\Sigma^{-ad(G)} X)^G $$
is a $K(n)$-local equivalence.
Is this conjecture known to be true or false? I've had a hard time finding references which talk about genuine equivariance and ambidexterity at the same time, so any literature pointers which might be relevant to the above will be most appreciated.

REPLY [8 votes]: There's something that goes wrong even when one works strictly in the Borel-equivariant case but for compact Lie groups. 
One equivalent form of the vanishing of the Tate construction is the following. Consider the $\infty$-category $\mathrm{Fun}(BG, \mathrm{Sp})$ of spectra equipped with a $G$-action (equivalently, the $\infty$-category of $G$-spectra which are "Borel-complete" in the sense that their fixed points for every choice of subgroup is equivalent to the homotopy fixed points). 
Suppose first that $G$ is finite. Fix an object $X \in \mathrm{Fun}(BG, \mathrm{Sp})$. There is a natural norm map $X_{hG} \to X^{hG}$, whose cofiber is the Tate construction $X^{tG}$. There are certain choices of $X$ which will force this to be an equivalence: for example, if $X$ is induced from a spectrum $Y$, so that $X = G_+ \wedge Y$ (with the $G$-action on the first factor). More generally, any object of $X \in \mathrm{Fun}(BG, \mathrm{Sp})$ which belongs to the thick subcategory (i.e., smallest stable subcategory closed under retracts) will have this property, so that the Tate construction vanishes.
Let us call this the category of nilpotent objects in $\mathrm{Fun}(BG, \mathrm{Sp})$. Any nilpotent object has vanishing Tate construction. (Conversely, a ring object with vanishing Tate construction is also nilpotent.) Over the rational numbers, everything is nilpotent as any group action is a retract of an induced one via a transfer construction. There are also nilpotent objects that are more subtle: for example, the $C_2$-action on complex $K$-theory is nilpotent. (The theory of nilpotent objects in the category of genuine $G$-spectra with respect to a family of subgroups is treated in some detail in two joint papers with Naumann and Noel, see here and here, and is related to phenomena such as Quillen stratification in mod $p$ cohomology.)
The $K(n)$-local vanishing of Tate spectra (for finite groups) is equivalent to the following statement: take $K(n)$ with the trivial $G$-action. Then it is nilpotent. (This is due to Greenlees and Sadofsky.)
One can ask the same question when $G$ is now assumed to be a compact Lie group, even a circle: given a ring spectrum with trivial $S^1$-action, is it nilpotent? (That is, does it belong to the thick subcategory of the $\infty$-category $\mathrm{Fun}(BS^1, \mathrm{Sp})$ generated by the induced objects?) Any such object would have the analogous transfer maps be equivalences. However, the answer is never in this case. One way to see this is (e.g., for $R = \mathbb{Q}$ or $K(n)$) is that any $S^1$-action on a spectrum $X$ leads to a homotopy fixed point spectral sequence for computing $\pi_* X^{hS^1}$, and a consequence of nilpotence is that this must degenerate with a horizontal vanishing line at some finite stage. This doesn't happen simply because this spectral sequence (the Atiyah-Hirzebruch spectral sequences) collapses at $E_2$ but with a polynomial class in filtration two. (Note that while the Morava $K$-theory of $BG$ is a finite-dimensional graded vector space when $G$ is finite, but not if $G$ is compact Lie.)<|endoftext|>
TITLE: Evaluation of Hankel determinants for the reverse Bessel polynomials
QUESTION [5 upvotes]: Consider the sequence $(\varphi_i)$ of reverse Bessel polynomials  which begins as follows.
\begin{align*}
\varphi_0&=1\\
\varphi_1&=x\\
\varphi_2&=x^2 + x\\
\varphi_3&=x^3 + 3x^2 + 3x\\
\varphi_4&=x^4 + 6x^3 + 15x^2 + 15x
\end{align*}
In general we have 
$$\varphi_0=1;\qquad\varphi_i = \sum_{k=1}^{i} \frac{(2i-k-1)!\,x^{k}  }{(i-k)!\,(k-1)!\,2^{i-k}}   \quad \text{for}\ i>0.         
$$
This is a sequence of binomial type (I don't know if that's relevant) with exponential generating function:
$$
\sum_{i=0}^{\infty} \varphi_i(x)\frac{z^i}{i!} = \exp(x(1-\sqrt{1-2z})).
$$
I want to know if there's going to be an explicit formula for the Hankel determinants:
$$
H_n= \det\left([\varphi_{i+j}]_{i,j=0}^{n}\right).
$$
Here are the first few.
\begin{align*}
 H_0 & =
1
\\
 H_1 & =
x
\\
 H_2 & =
2 \cdot x^{2} \cdot(x + 3)
\\
 H_3 & =
 12 \cdot x^{3} \cdot(x^{3} + 12 x^{2} + 48 x + 60)
\\
 H_4 & =
288 \cdot x^{4} \cdot (x^{6} + 30 x^{5} + 375 x^{4} + 2475
x^{3} + 9000 x^{2} + 16920 x + 12600)
\end{align*}
There's clearly a factor of $\left(\prod_{k=1}^{n}k!\right) \cdot x^n$ in $H_n$.
I've had a look in the Hankel determinant literature but it seems extensive and much of it seem to be relevant to sequences of integers rather than polynomials.  So I'd be interested in any pointers.
I'm also interested in calculating the Hankel determinants of the sequence offset by two:
$$
H^{[2]}_n= \det\left([\varphi_{i+j+2}]_{i,j=0}^{n}\right).
$$
\begin{align*}
 H^{[2]}_0 & =
x \cdot (x + 1)
\\
 H^{[2]}_1 & =
x^{2} \cdot (x^{3} + 6 x^{2} + 12 x + 6)
\\
 H^{[2]}_2 & =
2 \cdot x^{3} \cdot (x^{6} + 18 x^{5} + 135 x^{4} + 525
x^{3} + 1080 x^{2} + 1080 x + 360)
\\
 H^{[2]}_3 & =
12 \cdot x^{4} \cdot (x^{10} + 40 x^{9} + 720 x^{8} + 7620
x^{7} + 52080 x^{6} + 238140 x^{5} + 730800 x^{4} + 1467900 x^{3} +
1814400 x^{2} + 1209600 x + 302400)
\end{align*}
EDIT: I hadn't realised earlier that my polynomials are the reverse Bessel polynomials.  It looks like knowing the Hankel determinants of the usual Bessel polynomials could help.

REPLY [2 votes]: There is a combinatorial formula for the Hankel determinant $H_n= \det\left([\varphi_{i+j}]_{i,j=0}^{n}\right)$ in terms of weighted sums of disjoint collections of Schröder paths (and also for the offset version).  This is given in Theorem 26 of my paper The magnitude of odd balls via Hankel determinants of reverse Bessel polynomials.  
The key observation was pointed out to me be Alan Sokal, this is that there is the following Thron-type continued fraction expansion for the generating function of the reverse Bessel polynomials.
\begin{equation*}
 \sum_{i=0}^\infty t^i \phi_i(x) 
 =
 \frac{1}{1-
   \frac{x t}{1-
     \frac{t}{1-xt -
       \frac{2t}{1-xt-
         \frac{3t}{1-xt-
           \frac{4t}{1-\dots}}}}}}
\end{equation*}
(A similar result was noted by Paul Barry in the formula section of OEIS:A001497.)
One can then use Flajolet's fundamental lemma (relating generalized continued fractions to lattice path enumeration) to give a combinatorial interpretation of the reverse Bessel polynomials as certain weighted counts of Schröder paths.
[I now know that you can prove the combinatorial interpretation of the reverse Bessel polynomials without knowing about the continued fraction interpretation: see the post on Schröder Paths and Reverse Bessel Polynomials at the $n$-Category Café.]
Using this Schröder path enumeration interpretation of the reverse Bessel polynomials, you can apply the Karlin-McGregor-Lindström-Gessel-Viennot Lemma relating determinants to counting disjoint paths to obtain a nice formula for the Hankel determinant required.  In fact, you can obtain a combinatorial (path enumeration) formula for each of the coefficients in the determinant.<|endoftext|>
TITLE: An inequality for the spectral radius of matrices used by J. Bochi
QUESTION [12 upvotes]: I am interested in the history of an inequality for the spectral radius of a $d\times d$ real or complex matrix, which occurs in Jairo Bochi's 2002 article Inequalities for numerical invariants of sets of matrices. Let $\|\cdot\|$ denote the matrix norm induced by the Euclidean norm, and let $\rho(A)$ denote the spectral radius of $A$, that is, the maximum of the absolute values of its eigenvalues. The inequality in question is
$$\left\|A^d\right\|\leq \left(2^d-1\right)\rho(A)\left\|A\right\|^{d-1}.$$
This may be proved easily using the Cayley-Hamilton formula: since
$$\sum_{k=0}^d (-1)^k\left(\mathrm{tr}\,A^{\wedge k}\right) A^{d-k}=0$$
we may write
$$\left\|A^d\right\| \leq \sum_{k=1}^{d}\left|\mathrm{tr}\,A^{\wedge k}\right|\cdot \|A^{d-k}\| \leq \sum_{k=1}^d {d \choose k} \rho(A)^k\|A^{d-k}\| \leq \left(2^d-1\right)\rho(A)\|A\|^{d-1}$$
since $A^{\wedge k}$ has ${d\choose k}$ eigenvalues all of which are bounded by $\rho(A)^k \leq \rho(A)\|A\|^{k-1}$, and since $\sum_{k=1}^d {d\choose k} =2^d-1$. The inequality yields quick proofs of Gelfand's formula for matrices and the continuity of $\rho(A)$ as a function of $A$ as corollaries. Furthermore it has powerful generalisations to numerical invariants of sets of matrices, and these generalisations can be applied to obtain nontrivial results on phenomena such as the rate of convergence in the Berger-Wang theorem and the affinity dimension of self-affine fractals. Indeed, some of these results currently admit no other known proof.
The original inequality itself is a sufficiently useful and, to my mind, interesting inequality that I am curious to know more about its history. Jairo tells me that he invented it independently of other sources and has not seen it elsewhere, but is unsure whether it has appeared elsewhere in the literature. My first question therefore is:


Has anyone seen this inequality elsewhere? What is the earliest reference for it? Is it in any textbook? Does it have a name?


My second question is whether the optimality of the constant has been investigated:


Is the constant $2^d-1$ optimal? Is it of optimal order as a function of $d$? Which matrices constitute the `worst case'?


Edited: Fedor Petrov and Mikael de la Salle have given excellent answers to the second part of the question. I've set up a bounty to see if I can get a similarly comprehensive answer to the first part.

REPLY [14 votes]: Constant is definitely not sharp. For example, let me show inequality $\|A^d\|\leqslant \frac1{\sqrt[d]{2}-1}\cdot \rho(A)\cdot \|A\|^{d-1}$. We may suppose $\|A\|=1$, denote $\rho(A)=\rho$, then we have two inequalities $\|A^d\|\leqslant \|A\|^d=1$ and $\|A^d\|\leqslant d\rho+\dots+\rho^d=(1+\rho)^d-1$, explained in the post. If $\rho\geqslant \sqrt[d]{2}-1$ use first inequality, if $\rho<\sqrt[d]{2}-1:=\rho_0$, use the second: $$\frac{(1+\rho)^d-1}{\rho}\leqslant \frac{(1+\rho_0)^d-1}{\rho_0}=\frac1{\sqrt[d]{2}-1},$$
and we are done.
Note that this constant is of linear order in $d$, it behaves like $\frac{d}{\log 2}$ for large $d$ and is less than $\frac{d}{\log 2}$ for all $d$, since $$\sqrt[d]2=e^{\frac{\log 2}d}\geqslant 1+\frac{\log 2}d.$$<|endoftext|>
TITLE: Rational cohomology of the Rosenfeld projective planes
QUESTION [6 upvotes]: The bioctonionic plane $(\mathbb{C} \otimes \mathbb{O})\mathbb{P}^2$, the quarteroctonionic plane $(\mathbb{H} \otimes \mathbb{O})\mathbb{P}^2$ and the octooctonionic plane $(\mathbb{O} \otimes \mathbb{O})\mathbb{P}^2$ are manifolds of dimension 32, 64 and 128 with isometry groups $E_6$, $E_7$ and $E_8$ respectively. They are not actual projective planes but can be defined most elegantly as quotients of $E_6$, $E_7$ and $E_8$ by a subgroup, see for an exposition John Baez's website. 
I'm interested in properties of these manifolds. Has anyone calculated their rational (or even integral) cohomology? Are they frameable?

REPLY [5 votes]: The spaces you mention are symmetric spaces. The notation from Cartan's list is:

$E\operatorname{III}$ for the quotient of $E_6$
$E\operatorname{VI}$ for the quotient of $E_7$
$E\operatorname{VIII}$ for the quotient of $E_8$. 

Computations for $E\operatorname{III}$ have been done in 

H. Toda and T. Watanabe. The integral cohomology ring of $F_4/T$ and $E_6/T$. J. Math. Kyoto Univ 14 (1974), 257-286.

Computations for $E\operatorname{VI}$ have been done in 

M. Nakagawa. The mod 2 cohomology ring of the symmetric space $E\operatorname{VI}$. J. Math. Kyoto Univ. 41 (2001), 535-556. 

I don't know of integral cohomology computations for the last one, this is more difficult due to the presence of torsion. Generally, there are a lot of computations of cohomology of Lie groups and their homogeneous spaces by the Japanese school. See e.g. Topology of Lie groups I and II by Mimura-Toda.
If you are interested in the rational cohomology, then Borel gave a presentation for the cohomology rings for homogeneous spaces $G/H$ where $H\hookrightarrow G$ where $H$ has maximal rank in $G$ (if I'm not mistaken this should apply to all the cases): 

A. Borel. Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de groupes de Lie compacts. Ann. Math. 57 (1953), pp. 115-207.<|endoftext|>
TITLE: Lifting the Hasse invariant mod $2$
QUESTION [8 upvotes]: Katz defines in Section 2.0 $p$-adic properties of modular schemes and modular forms the Hasse invariant as a mod $p$ modular form $A$ of weight $p-1$. In other words, it is a section of $\omega^{\otimes p-1}$ on the compactified moduli stack of elliptic curves $\overline{\mathcal{M}}_{ell,\mathbb{F}_p}$.
We have a morphism from integral modular forms to mod $p$ modular forms, induced by the morphism $\omega^{\otimes p-1} \to \omega^{\otimes p-1}/p$ on $\overline{\mathcal{M}}_{ell,\mathbb{Z}}$. We say that the Hasse invariant $A$ is liftable to characteristic zero if it is in the image of this morphism. By the long exact sequence in cohomology 
$$H^0(\overline{\mathcal{M}}_{ell,\mathbb{Z}}; \omega^{\otimes p-1})\to H^0(\overline{\mathcal{M}}_{ell,\mathbb{Z}}; \omega^{\otimes p-1}/p) \xrightarrow{\partial} H^1(\overline{\mathcal{M}}_{ell,\mathbb{Z}}; \omega^{\otimes p-1}) \xrightarrow{\cdot p}$$
the Hasse invariant is automatically liftable if $H^1(\overline{\mathcal{M}}_{ell,\mathbb{Z}}; \omega^{\otimes p-1})$ has no $p$-torsion. This happens iff $p\geq 5$ and for $p=2,3$ the Hasse invariant is not liftable to an integral modular form (of level $1$). 
If we consider $\Gamma_1(n)$-modular forms instead, the Hasse invariant becomes liftable to characteristic zero for $p=3$ and all levels $n\geq 2$, $(p,n)=1$, because then $H^1(\overline{\mathcal{M}}_1(n)_{\mathbb{Z}[\frac1n]}; \omega^{\otimes 2}) = 0$. 

For what $n$ does the Hasse invariant lift to a characteristic zero $\Gamma_1(n)$-modular form, $(p,n)=1$, if $p=2$?

At the time of Katz's article the problem seems to have been unsolved as he just states (with a sketch of proof) that it is liftable if $n$ is divisible by $3,5,7$ or $11$, but he does not know it in other cases. In light of the long exact sequence above, the question is equivalent to $\eta = \partial(A)$ vanishing in (the $2$-torsion of) $H^1(\overline{\mathcal{M}}_1(n)_{\mathbb{Z}[\frac1n]}; \omega)$. 
In particular, I would be interested in concrete examples (for $\Gamma_1(n)$ or $\Gamma(n)$) where the Hasse invariant is not liftable if such examples are known.

REPLY [7 votes]: $\newcommand\Q{\mathbf{Q}}$
$\newcommand\Z{\mathbf{Z}}$
$\newcommand\Zbar{\overline{\Z}}$
$\newcommand\F{\mathbf{F}}$
$\newcommand\Gal{\mathrm{Gal}}$
A lift always exists, even with coefficients in $\Z$. If suffices to consider the case when $p > 2$ is prime. (Added Note: there is a second elementary argument at the end of this post.)
Let $p$ be prime, and let $\chi$ be an odd character of $(\Z/p\Z)^{\times}$ of order $2^m$ for some $m$. This character is valued in the field $E = \Q(\zeta_{2^m})$, and is unique up to Galois conjugacy (this follows from the oddness  assumption, which implies that $2^m$ is the largest power of $2$ dividing $p-1$). The fixed field of $\chi$ considered as a Galois character is the unique field $K \subset \Q(\zeta_p)$ which has degree $2^m$; it is totally complex.
Claim: The $L$ value $L(\chi,0)$ has $2$-adic valuation at most $(2^{m-1} - 1)/2^{m-1} < 1$.
In particular, $L(\chi,0)/2$ has negative valuation, and, as in the previous answer, this shows that the Eisenstein series $E_{1,\chi}$, suitably normalized, provides a lift of Hasse to $\Z[\zeta_{2^m}]$. Hence there is a multiple of $E_{1,\chi}$ with coefficients in $\Z[\zeta_{2^m}]$ which is congruent modulo $\pi = 1 - \zeta_{2^m}$ to $1$. Write this form as follows:
$$\sum_{k=1}^{2^{m-1} - 1} \zeta^k_{2^m} f_k \in \Z[\zeta_{2^m}][[q]],$$
where $f_k \in  \Z[[q]]$. Since the Galois group preserves $M_1(\Gamma_1(p))$, all the forms $f_k$ lies in $_1(\Gamma_1(p))$, and thus so does the form
$$\sum_{k=0}^{2^{m-1} - 1} f_k \in \Z[[q]].$$
Yet this form is congruent modulo $\pi = 1 - \zeta_{2^m}$ to the multiple of $E_{1,\chi}$ above, which in turn is congruent modulo $1 - \zeta_{2^m}$ to $1$. Hence it yields an integral lift of the Hasse invariant.
Proof: It suffices to show that the norm of $L(\chi,0)$ to $\Q$ has valuation at most $2^{m-1} - 1$. We may identify this norm with the product
$$\prod_{\chi \text{ odd}} L(\chi,0)$$
Yet this is equal to the limit of $\zeta_K(s)/\zeta_{K^+}(s)$ at $s = 0$ where $K^{+}$ is the maximal totally real subfield of $K$. By the class number formula, this limit thus evaluates to:
$$\frac{w(K^{+})}{w(K)} \cdot \frac{h(K)}{h(K^+)} \cdot \frac{R(K)}{R(K^{+})}.$$
Now $w(K^{+}) = 2$, and $w(K) = 2$ or $2p$ depending on whether $p$ is a Fermat prime or not.  Hence this factor can be ignored when computing the $2$-adic valuation. I claim that the $2$-parts of $h(K^{+})$ and $h(K)$ are both $1$. To see this, let $H$ be the $2$-part of the Hilbert class field of $K^{+}$ or $K$. Then $H/\Q$ is Galois and of $2$-power order. By  class field theory (over $\Q$), $\Gal(H/\Q)^{\mathrm{ab}}$ is cyclic, because it is ramified only at $p$ and $p \ne 2$. Yet this implies that $\Gal(H/\Q)$ is cyclic, because it is a $2$-group with cyclic abelianization. But then $H$ is abelian, and so from the theory of cyclotomic fields we easily see that $H = K^{+}$ or $K$. Hence the factor from the class number is also prime to $2$. Finally, the ratio of regulators is equal to the index 
of the units $[U_K:U_{K^{+}}]$.  Let $u \in U_K$. If $c \in \Gal(K/\Q)$ is complex conjugation, then a standard argument shows that $cu/u$ is a root of unity. Now $w(K) = |\mu_K|$ has either order $2$ or $2p$. In particular, after replacing $u$ by an odd power, we deduce that $cu = \pm u$. It follows that $u^2 \in K^{+}$, and that $V_K:=U_K/U_{K^{+}} \otimes \Z_2$ is an abelian group of exponent $2$.  The power of $2$ dividing the index is $|V_K|$, so it suffices to show that the dimension of $V_K$ over $\F_2$ is at most $2^{m-1} - 1$. Since $-1 \in K^{+}$,  there is a surjection:
$$U_K/(\pm 1,U^2_K) \rightarrow V_K.$$
By Dirichlet's unit theorem, we have
$$U_K \simeq \Z^{[K^{+}:\Q] - 1} \oplus \mu_K.$$
Since $\mu_K/{\pm 1} \otimes \Z_2$ is trivial, it follows that $\dim_{\F_2} V_K \le 2^{m-1} - 1$. This proves the claim.
In fact, using circular units, one can show that $\dim_{\F_2} V_K = 2^{m-1} - 1$,  so the upper bound is actually an equality.
In the special case when $p \equiv 3 \mod 4$, then $2^{m-1} - 1 = 0$ so $L(\chi,0)$ is a unit; this is the case considered by wrigley. 
Remark:
This argument is probably more complicated than necessary. There may well be an elementary way to prove directly that 
$$\sum_{n=1}^{p-1} n \chi(n)$$
is not divisible by $2$ directly for the odd character $\chi$ of $2$-power order,  but since the argument above was immediately apparent, I didn't try to look for such an argument.
Oh, actually, now that I make this comment, I see it is actually obvious. The character $\chi$ is valued in $E = \Q(\zeta_{2^{m}})$ which is a field of degree $2^{m-1}$, and the ring of integers has a basis $\zeta^i$ for $i = 0$ to $2^{m-1} - 1$. Moreover, $\chi$ exactly takes values in this basis. So it simply suffices to sum over the integers $n$ with $\chi(n) = 1$ or $\chi(n) = -1$ and show that this sum is not divisible by $2$, because then $L(\chi,0)$ will not be divisible by $2$. Yet these integers come in pairs $(n,p-n)$  whose difference is odd (so $n \chi(n) + (p-n) \chi(p-n) = \pm n - \pm (p-n) \equiv 1 \mod 2$), so it suffices to show that the total number of integers with $\chi(n) \in \{\pm 1\}$ is divisible by $2$ and not by $4$. Yet these integers are precisely the kernel of $\chi^2$, whose order has this property by definition. Oops! Well, I guess I may as well leave the class field theory argument here as well.
Added: I guess this last argument also shows that the coefficient of $\zeta^i$ is odd for any $i$, and hence
$$L(\chi,0) \equiv 1 + \zeta + \zeta^2 + \ldots + \zeta^{2^{m-1} -1} \mod 2.$$
Yet then 
$$(1 - \zeta) L(\chi,0) \equiv 1 - \zeta^{2^{m-1}}  = 1 + 1 \equiv 2 \mod 2(1 - \zeta),$$
and hence one sees directly that the valuation of $L(\chi,0)$ is $(2^{m-1} - 1)/2^{m-1}$.
$\quad$<|endoftext|>
TITLE: Smallest length of {0,1} vectors to satisfy some orthogonality conditions
QUESTION [5 upvotes]: Let $n$ be a positive integer.
The output of the problem is another positive integer $r$ which must be as small as possible.
I want to construct $2n$ binary vectors $x_i\in\{0,1\}^r$ and $y_i\in\{0,1\}^r$, with $i\in\{1,...,n\}$, which must respect the following conditions:

$x_i^Ty_i=0$ with $i\in\{1,...,n\}$,
$x_{i+1}^Ty_i=0$ with $i\in\{1,...,n-1\}$, and $x_1^Ty_n=0$ (it is cyclic).
$x_{j}^Ty_i>0$ with $i\in\{1,...,n-1\}$, $j\in\{1,...,n\}\setminus \{i,i+1\}$.

All the vectors $x_i$ must be distinct and all the vectors $y_i$ must be distinct.

For a given number $n$ of vectors, what is the smallest dimension $r=f(n)$ for which it is possible to construct vectors $x_i,y_i$ respecting the
  conditions described above?

An obvious upper bound is $f(n)\leq n$.
Since the $x_i$ must be distinct, a lower bound is $f(n)\geq \lceil \log_2(n)\rceil$.
However, this lower bound seems rather weak.
Example for $n=7$, there is a solution with $r=6$ (but apparently, not with $r=5$):

$x_1=\begin{pmatrix}0&0&1&0&1&1\end{pmatrix}^T$ $\hspace{1cm }$ $y_1=\begin{pmatrix}0&1&0&1&0&0\end{pmatrix}^T$.
$x_2=\begin{pmatrix}1&0&0&0&1&1\end{pmatrix}^T$ $\hspace{1cm }$ $y_2=\begin{pmatrix}0&1&1&0&0&0\end{pmatrix}^T$.
$x_3=\begin{pmatrix}1&0&0&1&0&1\end{pmatrix}^T$ $\hspace{1cm }$ $y_3=\begin{pmatrix}0&1&0&0&1&0\end{pmatrix}^T$.
$x_4=\begin{pmatrix}0&0&1&1&0&1\end{pmatrix}^T$ $\hspace{1cm }$ $y_4=\begin{pmatrix}1&0&0&0&1&0\end{pmatrix}^T$.
$x_5=\begin{pmatrix}0&1&1&1&0&0\end{pmatrix}^T$ $\hspace{1cm }$ $y_5=\begin{pmatrix}1&0&0&0&0&1\end{pmatrix}^T$.
$x_6=\begin{pmatrix}0&1&0&1&1&0\end{pmatrix}^T$ $\hspace{1cm }$ $y_6=\begin{pmatrix}1&0&1&0&0&0\end{pmatrix}^T$.
$x_7=\begin{pmatrix}0&1&0&0&1&1\end{pmatrix}^T$ $\hspace{1cm }$ $y_7=\begin{pmatrix}1&0&0&1&0&0\end{pmatrix}^T$.

REPLY [3 votes]: The $\log_2 n$ lower bound is actually within a constant factor of optimal for large $n$.  
Now let $x_1, \dots, x_n$ be formed by randomly setting each coordinate of each $x_i$ to $1$ with probability $\frac{1}{3}$ (with the events $x_i(k)=1$ independent for all $1 \leq i \leq n$ and $1 \leq k \leq r$).  Let the vectors $y_1, \dots, y_n$ be defined by 
$$y_i(k)=\left\{\begin{array}{cc} 1 & \textrm{ if } x_i(k)=x_{i+1}(k)=0 \\
0 & \textrm{ otherwise } \end{array} \right.$$
The first two conditions are immediately satisfied by our definition of $y$.  What remains to check is that $x_j^T y_i>0$ for $j \notin \{i, i+1\}$.  
Note that for these $(i,j)$ the vectors $x_j$ and $y_i$ are independent.  So for any $k$, we have 
$$P(x_j(k)=y_i(k)=1)=\frac{1}{3} \left(\frac{2}{3}\right)^2 = \frac{4}{27}$$
Multiplying over all coordinates and taking the union bound over all pairs $(i,j)$, we have that the probability the third condition is violated is at most
$$n^2 \left(\frac{23}{27} \right)^r$$
So if $r=c \log n$ for $c>\log_{27/23} 2 \approx 4.323$ and $n$ is sufficiently large, then the probability some condition is violated tends to $0$, so there's a set of vectors which works.<|endoftext|>
TITLE: reverse definition for magic square
QUESTION [5 upvotes]: Recently, I saw a question in see here which is so interesting for me. This question is as follows:

Is it possible to fill the $121$ entries in an $11×11$ square with the values $0,+1,−1$, so that the row sums and column sums are $22$ distinct numbers?

This problem is interesting for me because its definition is reverse of the magic square. I think for solve this problem we must use something like design theory. Also, for $n=6$ this problem have an answer. My next question is:

For which integer number $n$, this problem has a solution and is this problem well known in math?

I will be so thankful for any helpful comments and answers.

REPLY [4 votes]: This problem is so famous. For first trivial reference, you can see:link.
$\it{R. Bodendiek}$ and $\it{G. Burosch}$ studied this problem in a paper with name:
"Solution to the Antimagic 0,1,-1 Matrix Problem."
If there is solution for integer $n$, then we have:
$1)$ $n$ is even,
$2)$ The number in $\{-n,1-n,2-n,...,n\}$ that does not appear as a line sum is either $-n$ or $n$,
$3)$ Of the $n$ largest line sums, half are column sums and half are row sums.
Also, you can search about "Alternating sum matrix", for more information.<|endoftext|>
TITLE: Complex proof of $B(a,b)=\Gamma(a)\Gamma(b)/\Gamma(a+b)$
QUESTION [11 upvotes]: It is a question in spirit of this one.
Is there a way to prove Euler's formula
$$
\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}
$$
using contour integration (and maybe something else, say, integration by parts or change of variables is ok, but double integration and Fubini theorem is not)? For $a=b=1/2$ we get the value of $\Gamma(1/2)$ which is essentially the value the cited question was about.

REPLY [3 votes]: In what follows we assume $\Re(a)>0$ and $\Re(b)>0$. 
Begin with the case $a+b=k\in\mathbb N$. Using Pochhammer contour $P$, one can relate what's going on on $[0,1]$ to what is going on on a circle $C:=x_*\mathbb S^1$, $|x_*|>1$. Indeed, looking carefully at determinations of $f(z):=z^{a-1}(1-z)^{b-1}$ one has $$ \oint_Pf(z)dz = (1-\exp 2ib\pi)\oint_C f(z)dz ~~~(*)\\=(1-\exp 2ib\pi)(1-\exp 2ia\pi)\int_0^1f(z)dz$$ the last equality being given by Pochhammer formula, so that $$ \oint_C f(z)dz = 2i(-1)^{1-a}\sin(a\pi) \int_0^1f(z)dz .$$ Since $f(z)=z^{a+b-2}(1/z-1)^{b-1}$ is holomorphic near $\infty$ we have $$\oint_Cf(z)dz = -\oint_C f(1/x)\frac{dx}{x^2}=(-1)^{b-1}\oint_C x^{-k}(1-x)^{b-1}dx , $$ which is a contour integral. It can be evaluated by looking a the expansion of $$(1-x)^{b-1} = \sum_n \frac{\Gamma(b)}{\Gamma(n+1)\Gamma(b-n)}(-x)^n .$$ The residue of $x^{-k}(1-x)^{b-1}$ at $0$ is obtained for $n+1=k$, that is $\frac{\Gamma(b)}{\Gamma(a+b)\Gamma(1-a)}$ which allows to conclude using Gamma reflection formula $\Gamma(1-a)\Gamma(a)\sin(a\pi)=\pi$.
The next step is to deal with the case $a+b=p/q\in\mathbb Q$, then conclude by analyticity and accumulation. This case is dealt with by taking a linear combination of $\oint_Cf(z)dz$ with weights $\exp (2in\pi/q)$ to obtain the same kind of relation as $(*)$. I'll write details later, but they should be straightforward.<|endoftext|>
TITLE: Is there a covering of Prym variety?
QUESTION [5 upvotes]: $\mathstrut$Hi, guys! 
Let $C$, $C^\prime$ be projective smooth irreducible algebraic curves over an algebraically closed field $k$ ($\mathrm{char}(k) \neq 2$), $\phi : C$ $\to$ $C^\prime$ a two-sheeted covering, that is a surjective morphism of degree $2$, and $\phi_* : J(C) \to J(C^\prime)$ induced to Jacobians direct image morphism.  
We know that there is the covering of $J(C)$ by symmetric power $\mathrm{sym}^{g(C)}(C)$. Consider Prym variety $\mathrm{Prym}(C/C^\prime) := \mathrm{ker}^0(\phi_*) \subseteq J(C)$ of the covering $\phi$. Is there a covering of $\mathrm{Prym}(C/C^\prime)$ by $\mathrm{sym}^{d}(C)$, where $d = \dim(\mathrm{Prym}(C/C^\prime))$?
Thank you!

REPLY [2 votes]: The answer is positive: there is a surjective, generically finite morphism $\text{sym}^d(C)\to \text{Prym}(C/C')$, at least away from small characteristics.  Fix a $k$-point $x$ of $C$, and use that to define an Abel map, $\alpha_x:C \to J(X)$.  The induced composition morphism $$C^g \xrightarrow{\alpha^g} J(C)^g \xrightarrow{\Sigma} J(C),$$ is surjective and generically étale.  Also the projection morphism, $$\pi:J(C) \to \text{Prym}(C/C')/\Gamma,$$ is a smooth morphism.  Since the Zariski tangent space of $C^g$ is generated by the Zariski tangent spaces of the $g$ fibers or the $g$ projections $C^g\to C^{g-1}$, up to a permutation of the factors, for sufficiently general $(x_{d+1},\dots,x_g)\in C^{g-d}$, the induced morphism $$C^d \times\{(x_{d+1},\dots,x_g)\} \to J(C) \to \text{Prym}(C/C')/\Gamma$$ is surjective and generically étale. Of course the morphism $$C^d \times \{(x_{d+1},\dots,x_g)\} \to J(C)$$ factors through the morphism $C^d \to \text{sym}^d(C)$ since the group law on $J(C)$ is Abelian.   Thus, in all, there is a surjective, generically étale morphism $$\psi:\text{sym}^d(C) \to \text{Prym}(C/C')/\Gamma.$$  Observe that, up to composing this morphism with a translation of $\text{Prym}(C/C')$, the morphism is independent of the choice of permutation or general point $(x_{d+1},\dots,x_g) \in C^{g-d}$.
Finally, the finite subgroup scheme $\Gamma$ is contained in the $N$-torsion subgroup scheme for some integer $N$ (that can be bounded just in terms of the topological data of $\phi$).  Thus, assuming the characteristic is larger than $N$, the multiplication by $N$ morphism, $$ \text{Prym}(C/C') \to \text{Prym}(C/C'),$$ is surjective and étale, and it factors through a surjective, étale morphism $$\chi:\text{Prym}(C/C')/\Gamma \to \text{Prym}(C/C').$$  Therefore, there is a surjective, étale morphism, $$\chi\circ \psi:\text{sym}^d(C) \to \text{Prym}(C/C').$$  
Of course that may not be the answer you want.  You still need to work out $\Gamma$ and $N$.  Also, "canonically" you only obtain a morphism from $\text{sym}^d(C)$ to a torsor for $\text{Prym}(C/C')$; trivializing the torsor depends on choosing a $k$-point of $C$ (at least as specified above).  So if you want to do this in families or over a non-algebraically closed field, you will need to do some work.  Finally, there is the question, in small characteristics, of what to do when the group scheme $\Gamma$ is not étale.<|endoftext|>
TITLE: Brownian motion in $\mathbb{R}^n$, probability of hitting a set
QUESTION [5 upvotes]: Consider a particle undergoing Brownian motion in $\mathbb{R}^n$, starting at the origin, and let $B(t)$ denote its position at time $t$. Let $X$ be an arbitrary subset of $\mathbb{R}^n$. I am trying to understand what properties $X$ needs to have so that the probability of the Brownian particle striking $X$ within time $t$ is non-zero, that is, $\mathbb{P}(B(s) \in X, s \leq t) \neq 0$. It seems to me that a sufficient condition on $X$ could be that it contains a subset $Y$ which is homeomorphic to $\mathbb{R}^{n - 1}$. Is this necessary too? Is there a more relaxed sufficient condition? Thanks in advance.

REPLY [4 votes]: Serguei Popov has the "right" answer.  Just to give a quick counterexample, consider the set $X = (\mathbb{R} \setminus \mathbb{Q})^n$ of points with all coordinates irrational.  For every $s$ we have $\mathbb{P}(B(s) \in X) = 1$, so we strike $X$ almost immediately, almost surely.  Yet $X$ is totally disconnected so it doesn't even contain a homeomorphic copy of $\mathbb{R}^1$.  So your sufficient condition is certainly not necessary.
Also, I wanted to comment on one technical point.  You say $X$ is an "arbitrary" subset, but for the question to make sense, you really need $X$ to be Borel.  Even then it is not trivial to see that your "event" $A = \{\exists s \le t : B(s) \in X\}$ is measurable, since the $\exists$ makes it an uncountable union.  But you can show it as follows: let our sample space $\Omega$ be the path space $C([0,\infty), \mathbb{R}^n)$ which is standard Borel.  Then the set $C = \{(\omega, t) : \omega(t) \in X\} \subset \Omega \times [0,\infty)$ is Borel as soon as $X$ is.  And $A$ is the projection of $C$ onto the first coordinate, so $A$ is analytic, and analytic sets are universally measurable.
You might also like to look up the debut theorem, which states that $\tau_X = \inf\{t : B(t) \in X\}$ is a stopping time for any Borel set $X$.  Again, because of measurability considerations, this is a nontrivial theorem.<|endoftext|>
TITLE: Factorization of a certain map through a CW-complex
QUESTION [5 upvotes]: Suppose that $X$ is a paracompact Hausdorff space (e.g. a metric space) with $\dim X=n$ (the Lebesgue covering dimension). I want to find a proof (or a reference) that any (continuous) map $f: X \to K(\mathbb{Z},2)=\mathbb{C}P^\infty$ can be factorized, up homotopy, through a CW-complex $Z$ of the same dimension $n$, i.e. there is a diagram

which commutes, up to homotopy.

REPLY [3 votes]: Unfortunately, I can only give a partial answer. I can show the factorization through dimension $n$ for $\mathbb{RP}^\infty$ or a weaker factorization through dimension $2n$ for $\mathbb{CP}^\infty$.
It is clear that if $X$ factors (up to homotopy) through a CW-complex of dimension $n$, then it actually factors (up to homotopy) through $\mathbb{CP}^{m}\hookrightarrow\mathbb{CP}^\infty$ with $m$ the floor of $n/2$, by cellular approximation. 
For now, I can only show that $f$ factors through $\mathbb{CP}^n$. From the classification theorem for vector bundles on paracompact Hausdorff spaces, this follows if we can prove that a complex line bundle on a space of covering dimension $n$ can be generated by $n+1$ global sections, alternatively that there is a trivializing covering by $n+1$ open sets. (The trivializing cover provides an explicit map $X\to\mathbb{CP}^n$ classifying a line bundle isomorphic to the one corresponding to the original map $X\to\mathbb{CP}^\infty$. By the  classification theorem the maps are homotopic.) 
Now we apply Ostrand's theorem, cf. Theorem 3.2.4 of Engelking's book "Dimension theory", noting that paracompact Hausdorff spaces are normal: 

A normal space $X$ has covering dimension $\leq n$ if and only if for every locally finite open cover $\{U_\alpha\}_{\alpha\in I}$ there exists an open cover $\{V_\beta\}$ of $X$ which can be represended as the union of $n+1$ families $\mathcal{V}_1,\dots,\mathcal{V}_{n+1}$ with $\mathcal{V}_i=\{V_{i,\alpha}\}_{\alpha\in I}$ such that $V_{i,\alpha}\cap V_{i,\alpha'}=\emptyset$ and $V_{i,\alpha}\subseteq U_\alpha$ for $\alpha\in I$, $i\in 1,\dots,n+1$.

So, take a trivialization of the line bundle $\mathcal{L}\to X$, and refine the trivializing cover to one satisfying the above property. Then for each $i\in\{1,\dots,n+1\}$, we can define a global section on $\bigcup_{\alpha} V_{i,\alpha}$ by taking non-vanishing sections on the open sets $V_{i,\alpha}$. Using a partition of unity precisely subordinate to our open cover, we can glue the local sections to a global section which is non-vanishing on $\bigcup_{\alpha}V_{i,\alpha}$ and zero outside. (If we assume that $X$ is metric, then we could alternatively scale the section on $\bigcup_{\alpha}V_{i,\alpha}$ with the distance to the boundary to get an extension of the section.)
Applying the above construction to the families $\mathcal{V}_i$ we get $n+1$ global sections generating the line bundle. Hence there is a classifying map $X\to\mathbb{CP}^\infty$, homotopic to the original one, which factors through $\mathbb{CP}^n$.

For what it's worth, let me remark that the proof for covering dimension zero is a lot easier. If the space has covering dimension zero, we can get a trivializing cover by disjoint open sets. Clearly, on such a cover we can define a global non-vanishing section of the line bundle, showing that the classifying map $f:X\to\mathbb{CP}^\infty$ is null-homotopic.
This actually generalizes to show that any map from a space of covering dimension zero to any CW-complex (locally contractible suffices) is null-homotopic. It is however not clear to me how to extend this to positive dimensions - the extension of homotopies from open sets to the whole space, done for sections via the partition of unity, seems to be complicated in general.

Another remark strengthening the 1-dimensional case: if $X$ is of covering dimension 1, the $\mathbb{CP}^n$ argument above shows that it factors through $\mathbb{CP}^1$. Then the composition with the projection $S^2\to\mathbb{RP}^2$ factors through $\mathbb{RP}^1$ by the $\mathbb{RP}^n$-case of the above argument. The projection $S^2\to\mathbb{RP}^2$ is in fact a Hurewicz fibration, so it has the homotopy lifting property for all spaces. We get a homotopy from $X\to\mathbb{CP}^1$ to a map which lands in the preimage of $S^1\subseteq\mathbb{RP}^2$ - but this has to be null-homotopic. 
The $\mathbb{CP}^n$-argument can be fixed if we know a variation of Ostrand's theorem, namely that for a normal space of covering dimension $\leq n$ and a locally finite open cover, there exists a refinement given by $n/2$ families of sets $\mathcal{V}_i$ such that the union of the sets in $\mathcal{V}_i$ are of covering dimension 1. But I suppose this is the point where I have give up.<|endoftext|>
TITLE: Two definitions of Lebesgue covering dimension
QUESTION [15 upvotes]: Maybe this question has already been considered here, but after a quick search I  didn't find what I was looking for.
As I see, in the literature there are two different definitions of the topological/Lebegue covering dimension.
$\textbf{Definition 1.}$ (e.g. Munkres, General topology) A topological space $X$ is said to have the Lebesgue covering dimension $d<\infty$ if $d$ is the smallest non-negative integer with the property that each open cover of $X$ has a refinement in which no point of $X$ is included in more than $d+1$ elements.
$\textbf{Definition 2.}$  (e.g. Engelking, Ryszard, "Dimension theory" or Pears,
 "Dimension theory of general spaces") A topological space $X$ is said to have the Lebesgue covering dimension $d<\infty$ if $d$ is the smallest non-negative integer with the property that each $\textbf{finite}$ open cover of $X$ has a refinement in which no point of $X$ is included in more than $d+1$ elements.
Obviously these two definitions agree for compact spaces, but I suppose the first one is stronger in general. For which other class(es) of spaces these two definitions coincide (and for which they don't?
Also, I didn't find much results in the literature about "dimension theory" with respect to Definition 1. (Reference?) For example, is the analogue of the Subset theorem (Pears, Theorem 3.6.4) true when dealing with Definition 1 (i.e. If $S$ is a subset of a totally normal space $X$, then $\dim S \leq \dim X$)?

REPLY [3 votes]: Yes, the long ray $R$ works. If $\mathcal{U}$ is a finite open cover then $\bigcap\{R\setminus U:U\in\mathcal{U}\}=\emptyset$ and at least one of these closed sets must be bounded as in $R$ any two closed ubbounded sets have closed unbounded intersection.
Fix one $V\in\mathcal{U}$ and an ordinal $\alpha$ such that $[\alpha,\omega_1)\subset V$ and refine the restriction of $\mathcal{U}$ to $[0,\alpha+1]$ to a cover of order $1$.<|endoftext|>
TITLE: Classification of knots by geometrization theorem
QUESTION [18 upvotes]: I read this interview with Ian Agol, where he says:

"...I learned that Thurston’s geometrization theorem allowed a complete and practical classification of knots."

My question is:

How does this work exactly? Is this written up somewhere?

I know that every knot in $S^3$ is either a torus knot, a satellite knot or a hyperbolic knot. But as far as I know this is not practical.
(I understand practical as follows: Given a knot diagram, there is a way to find out in which of these 3 classes the knot fits. And given two knots in the same class, there is a way to say if they are the same or not.)

REPLY [14 votes]: You have all the tools to compute the geometric decomposition of knot and link exteriors in the software Regina. I'm one of the authors, although my hands haven't been over that part of the code very much.  Most of the 3-manifold decomposition code was written by Ben Burton.  The algorithms are primarily due to people like Rubinstein and Jaco, although Ben has done quite a lot of work to find efficient implementations of these algorithms.
Regina uses SnapPea for the hyperbolic manifolds.  In a pretty strong sense it is a practical algorithm.  The only part of this software for which run-time estimates do not exist is the usage of SnapPea, but in practice it is extremely fast. The primary issue with SnapPea as an algorithm is that it's not known whether SnapPea will always find a hyperbolic structure on knot and link exteriors which are hyperbolisable -- one of the key problems is finding an appropriate triangulation on which the gluing equations have a solution.  I believe the precise mathematical question here is whether or not hyperbolisable 1-cusped manifolds admit an ideal triangulation.  It's known such manifolds admit ideal Epstein-Penner decompositions but the facets are not always tetrahedra.  One may not be able to subdivide these objects into ideal tetrahedra.
Computing the satellite decomposition tends to be slow but most of the time (as long as your diagrams aren't very complicated, say, over 50 crossings) then it performs quite well.  The algorithms here tend to be exponential run time.
One of my favourite statements of geometrization for knot and link exteriors is in terms of "splicing".  I wrote a survey paper on the topic:

Ryan Budney, JSJ-decompositions of knot and link complements in the 3-sphere, L'enseignement Mathématique (2) 52 (2006), 319--359, journal, arXiv:math/0506523

Here is an example of using geometrization of link exteriors to answer a basic question.<|endoftext|>
TITLE: Uniform partitions of a compact Riemannian manifold
QUESTION [5 upvotes]: My question is a little bit vague. I want to know if an arbitrary compact Riemannian manifold (M^d,g) admits partitions that are uniform in some sense. To be more precise, I need for every eps > 0 a partition of M into finitely many measurable subsets P_1,...,P_k such that 
(i) The diameter of each P_i is smaller than eps.
(ii) The number k goes like 1/(eps)^d w.r.t. eps (asymptotically).
(iii) The boundary of each P_i is pieced together from finitely many pieces of smooth submanifolds.
(iv) I have some control over the (d-1)-dimensional volumes of the boundaries and also over their curvatures, measured by the second fundamental form. The ideal case would be, when the boundaries are pieced together by totally geodesic submanifolds such that the second fundamental form vanishes on each piece and such that the volume of the boundary of each P_i asymptotically behaves like the volume of a (d-1)-dimensional Euclidean sphere.
Any hint or idea would be helpful.

REPLY [3 votes]: Assume first that your manifold $M$ is embedded in an Euclidean space $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$ and the metric is the induced metric. (Nash's embedding theorem shows that this is always possible.)
For each $\newcommand{\ve}{\varepsilon}$ $\ve>0$ denote by $\newcommand{\eL}{\mathscr{L}}$ $\eL_{\ve}$ the  "cubulation"  of $\bR^n$ defined by  the  infinite collection of hyperplanes $\newcommand{\bZ}{\mathbb{Z}}$
$$\frac{x_i}{\ve}\in \bZ,\;\;i=1,\dotsc n. $$
The vertices of this cubulation fill the lattice $\bigl(\,\ve^{-1}\bZ\bigr)^n$.
Consider the intersection of $\eL_\ve$ with $M$. If $M$ is in general position (meaning you may have to rotate and translate $M$ a little bit), then this intersection determines a polygonal decomposition of $M$ with the properties you desire.  For details see Whitney's  gem "Geometric Integration Theory", Chap. IV, Sec. B.<|endoftext|>
TITLE: What is the Status of Borel conjecture today?
QUESTION [12 upvotes]: Let me recall the conjecture: $M$ and $N$ two aspherical closed $n$-manifolds with isomorphic fundamental groups, then $M$ and $N$ are homeomorphic.

REPLY [4 votes]: Surely the state of the art is in this paper from a few days ago at http://arxiv.org/abs/1601.00262 by some of the leading experts in the field.<|endoftext|>
TITLE: "Partial-computably isomorphic" sets
QUESTION [6 upvotes]: For $A,B \subseteq \mathbb{N}$, define $A\sim B$ when there exist partial computable functions $f,g\colon \mathbb{N}\rightharpoonup \mathbb{N}$ such that $f$ is defined at least on all of $A$ and $g$ at least on all of $B$ and such that they restrict to inverse bijections $f|_A\colon A\to B$ and $g|_B\colon B\to A$.  (Or more concisely but also more confusingly: “there exists a bicomputable bijection $A \to B$”.)
This definition comes from an incorrect attempt of mine to remember what "computably isomorphic" meant and what the Myhill isomorphism theorem says (and then getting all confused since $A\sim B$ does not even imply that $A$ and $B$ have the same Turing degree: indeed, $\mathbb{N} \sim K$ where $K$ is the halting problem).  To remove any ambiguity, $A$ and $B$ are said to be "computably isomorphic" ($A\equiv B$) when there exists a total computable $h$ that is a bijection $\mathbb{N}\to \mathbb{N}$ and such that $h(A) = B$: I'm tempted to say that "computably isomorphic" is about being isomorphic as subsets of $\mathbb{N}$ rather than as sets of integers as the relation $\sim$ above.
But now I'd like to know more about the relation $A\sim B$ that I inadvertently defined, because it seems fairly natural and I couldn't find any reference to it anywhere (of course, it's hard to find something when one does not know a name for the thing):

Does it have a standard name?  Has it appeared in the literature previously?  Or is it completely trivial for some reason that escaped me?
What nontrivial things can be said about it?  E.g., can its equivalence classes be, if not characterized, at least related to some more classical objects?
Is there some reason to think that it's not an interesting or "good" notion to define?

Easy facts: It is easy to see that $\mathbb{N} \sim A$ iff $A$ is computably enumerable.  Also, we can bound $\sim$ by two fairly standard equivalence relations: on the one hand, $A \equiv B$ trivially implies $A\sim B$ (where $\equiv$ is defined above; the Myhill isomorphism theorem states essentially that $\equiv$ is equality of one-one degrees); and conversely, if $A\sim B$ then $\mathbf{a} \cup 0' = \mathbf{b} \cup 0'$ where $\mathbf{a},\mathbf{b}$ are the Turing degrees of $A,B$ (because using an oracles for $A$ and for $0'$ we can decide $y\in B$ by testing whether $g(y)$ is defined and belongs to $A$ and its image by $f$ is defined); so at least $\sim$ is neither insanely fine nor insanely coarse.
Addendum: If I didn't mess up too badly, $A \sim B$ is equivalent to saying that the objects of the effective topos defined by $A$ and $B$ (the source of the subobjects $A \hookrightarrow \mathcal{N}$ and $B \hookrightarrow \mathcal{N}$ of the n.n.o. $\mathcal{N}$ classified by the maps $\mathcal{N} \to \Omega_{\neg\neg} = \nabla(2)$ defined from the characteristic functions of $A$ and $B$) are isomorphic (equivalently: internally isomorphic).  (If someone can confirm this, I'd appreciate it.)  This should at least give some motivation for thinking that $\sim$ might be interesting or natural.

REPLY [4 votes]: The equivalence relation $\sim$ is referred to as recursive equivalence, and the equivalence classes as recursive equivalence types.
I don't know much about them, other than McCarty showed in his PhD thesis Realizability and Recursive Mathematics, that in the realizability model $V(\mathcal{Kl})$ this notion corresponds to the $\neg \neg$ stable subsets of $\mathbb{N}$ up to cardinality, which is basically the same as the topos theoretical definition you gave.
McCarty attributes the notion to Dekker and Myhill, who apparently wrote a book on the topic titled Recursive Equivalence Types. An internet search turns up several papers referring to them (eg Myhill, Recursive equivalence types and combinatorial functions and Nerode, Additive relations among recursive equivalence types).
By the way, the usual definition is that $A \sim B$ iff there is a partial computable $f$ which is injective on its domain and such that $f(A) = B$. To show this is implied by the other definition: note that given $f$ and $g$ as in the question, we define $f'$ by restricting $f$ to only those $n$ such that $g(f(n))$ is defined and equal to $n$, and then $f'$ is injective on its domain and is still defined on $A$.<|endoftext|>
TITLE: Does a planar triangulation always contain a Hamiltonian path?
QUESTION [6 upvotes]: What about a Hamiltonian path in a triangulation of an n-gon? If not, how long is the longest path?

REPLY [8 votes]: There are two planar triangulations on 14 vertices without hamiltonian paths. This is the smallest size. For sure these are well-known.

Those answer the question for $n=3$. For $n=4$ the first examples appear at 12 vertices. Same for $n=5$. For $n=6$ one with 10 vertices. For $n=7$ one with 11 vertices.  And so forth.<|endoftext|>
TITLE: Intuition for Picard-Lefschetz formula
QUESTION [25 upvotes]: I'm trying to develop some intuition for the (local) Picard-Lefschetz formula (which I'm encountering for the first time in Deligne's paper "La Conjecture de Weil, I").
To summarize the setup, we have a proper family of varieties $\pi : X \rightarrow D$, where $D$ is thought of as a small complex disk (or really its algebro-geometric analogue). All the fibers $\pi^{-1}(t)$ are smooth except the central fiber, which has a double point.
There is a monodromy action of the fundamental group of the punctured disk on the cohomology of the "generic fiber", which is actually only non-trivial in middle dimension, i.e. if $\dim X = n$ then the monodromy action is only non-trivial on $H^n$. See the linked page for the precise formula.
Anyway, my question is if there is some nice intuition for why the monodromy action takes this shape. Most sources that I can find online jump straight into very abstract formulations in the language of derived functors and perverse sheaves. For concreteness, here are some specific questions:

Is there some "easy" way to see that the monodromy should act as a transvection?
Is there some "easy" way to see that the action should be trivial except in middle dimension?
Is there some "easy" reason that one would expect the monodromy formula to depend on the parity of the dimension?

I thought that I could see 2 by applying the Lefschetz Hyperplane Theorem to the family. The point is that I get a family $Y \subset X$ over $D$, such that the cohomology of $Y$ maps isomorphically to that of $X$ except in middle degree. If I choose $Y$ generally, then it will miss the singular point of $X$ and thus be a smooth family, so the monodromy action should be trivial. But this argument had nothing to do with the critical point of $\pi$ being nondegenerate, so maybe I'm missing something?
I would also appreciate reference recommendations that treat this subject and how it naturally evolves into the modern derived category formalism.

REPLY [37 votes]: The fundamental observation is that the monodromy is given by twisting by vanishing cycles. A vanishing cycle is a homology class of the generic fibre that shrinks to zero in the special fibre. There is a unique (up to sign) such vanishing cycle attached to each node in the special fibre, which lies in the middle dimension, and the Picard–Lefschetz formula is precisely the statement that the effect of monodromy around the singular fibre is of twisting along vanishing cycles. If your cycle is disjoint from any vanishing cycles it isn't affected, otherwise it picks up a contribution corresponding to its intersection numbers with each vanishing cycle $\nu_i$:
$$ \gamma \mapsto \gamma + \sum_i \langle \nu_i, \gamma \rangle \nu_i. $$
(Note that one has to be careful with signs and orientations here to get the correct formula; see the references provided at the end.)
The easiest situation to visualise is of course in relative dimension 1. The following picture shows the twisting around a single ordinary quadratic singularity (where spurious self-intersections are shown in white). The vanishing cycle is depicted in red. (Click for larger version.)

It's not totally obvious to see that the blue cycle gets twisted by (minus) the red cycle, so the following topological view of the monodromy makes it more obvious (you can see it by flipping over one of the two sheets in the above picture).

The general situation is similar: the vanishing cycles appear in the middle dimension, and the same formula for the monodromy applies. There is just one subtlety is, as you noted: the behaviour is different in even relative dimension. The reason for this is that in even relative dimension, vanishing cycles have nonzero self-intersection (their self-intersection is $\pm 2$, depending on the dimension mod 4), yet are stable under the monodromy transformation. Because of this, the monodromy transformation has order 2, instead of infinite order.
A reference I recommend for a quick overview is Simon Donaldson's article Lefschetz Pencils and Mapping Class Groups, available here. The main reference for the Picard–Lefschetz formula in $\ell$-adic étale cohomology, which seems to interest you, is of course in SGA 7. See Deligne's exposés XIII, XIV and XV, in SGA 7 tome II, which use the formalism of nearby cycles to transfer computations to the special fibre.<|endoftext|>
TITLE: Book recommendation for cobordism theory
QUESTION [28 upvotes]: I am planning to organize a seminar on cobordism theory and I'm looking for a reference. Such a reference is preferably a book, but I'm open to other ideas. 
The audience is familiar with characteristic classes at the level of Milnor Stasheff. We are no experts on homotopy theory.
What would you recommend?
Edit: Thanks for the answers, I like them! I found it hard to choose one, as most seem like great sources. I chose to accept the lecture notes (even though I was looking for a book) because this seems doing what we want the seminar to be about.

REPLY [4 votes]: There is an interesting exposition on RANICKI`s book "Algebraic and Geometric Surgery", chapters 2 and 6. There is a chapter devoted to bundles and it also includes Pontrjagin Cobordism theorem...<|endoftext|>
TITLE: Problems which use S₄ → S₃
QUESTION [10 upvotes]: I need examples of problems which use, directly or indirectly, the homomorphism $S_4\to S_3$ in the solution (its kernel is $\mathbb{Z}_2\oplus\mathbb{Z}_2$).
Obvious candidates:

Lagrange resolvent (the reduction of quartic to cubic equations). 
Tait's theorem on equivalence of 4-coloring of normal map and 3-coloring of its edges.  


Do you have more interesting examples?

REPLY [13 votes]: I use this or something equivalent in teaching projective geometry to show that the cross-ratio has at most 6 distinct values (when the points are permuted) as opposed to the 24 naively expected. This involves checking that the elements in the Klein 4-group act trivially on the cross-ratio $R(A,B,C,D)$.<|endoftext|>
TITLE: Is a wild automorphism of $k[x_1,\ldots,x_n]$, $n \geq 3$, necessarily of infinite order?
QUESTION [15 upvotes]: Let $k[x_1,\ldots,x_n]$ be a polynomial ring over a field $k$ of characteristic zero.
When $n=2$, it is known that every automorphism of $k[x_1,x_2]$ is tame, namely, a finite product of elementary automorphisms.
For $n=3$, in their paper, Shestakov and Umirbaev showed that the Nagata map is wild (=non-tame, not belongs to the subgroup generated by elementary automorphisms).
My question: For $n \geq 3$, if we know that a given automorphism $g$ is of finite order (namely $g^m=1$ for some $m$),
must it be tame?
Or is it possible to have a wild automorphism of finite order?
Sorry if my question is trivial.

REPLY [12 votes]: The answer is no.
Let $K$ be a field of characteristic zero and let us take the Nagata automorphism of $\mathbb{A}^3_K$ given by 
$$N\colon (x,y,z)\mapsto (x+(x^2-yz)z,y+2(x^2-yz)x+(x^2-yz)^2z,z)$$
It is part of an action of the additive group given by 
$$f_t\colon (x,y,z)\mapsto (x+t(x^2-yz)z,y+2t(x^2-yz)x+t^2(x^2-yz)^2z,z)$$
which satisfies $f_t\circ f_s=f_{s+t}$ for each $s,t\in K$. As $N=f_1$ we get $N^{-1}=f_{-1}$. Choosing $\alpha=(x,y,z)\mapsto (-x,y,z)$ which is of order $2$, you obtain $\alpha \circ N\circ \alpha=N^{-1}$, so $\alpha\circ N$ is of order $2$. As $\alpha$ is tame and $N$ is not tame, the involution $\alpha\circ N$ is not tame.
In the article "The Nagata automorphism is shifted linearizable" of Eric Edo and Pierre-Marie Poloni, you can find that composing the Nagata automorphism with a linear map can give an element which is conjugate to a linear map even if it is not tame. So conjugation indeed changes the fact to be tame or not.<|endoftext|>
TITLE: Dimension in CW-approximation
QUESTION [11 upvotes]: The following question was something that came to my mind during my (unsuccessful) attempt at answering this MO-question.
Let $X$ be a topological space, and let $\tilde{X}\to X$ be a CW-approximation. Given that $X$ has covering dimension $n$, can anything be said about the covering dimension of the CW-approximation $\tilde{X}$?
If it's not generally true that $\dim\tilde{X}\leq\dim X$, is it possible to give explicit examples?

REPLY [22 votes]: Barratt and Milnor (An Example of Anomalous Singular Homology) proved that (for $n > 1$) the singular homology of the union of countably many $n$-spheres with one point in common and radii tending to $0$ is non-trivial in arbitrarily high dimensions. Thus any CW-replacement of this space is infinite-dimensional. On the other hand, it is a closed subspace of $\mathbb{R}^{n+1}$ so its covering dimension is at most $n+1$.<|endoftext|>
TITLE: Witt-vector vectors
QUESTION [16 upvotes]: I've never really made my way in any detail through the Witt-vector construction.  I did read all the articles that a quick Google and MSN search turned up, and none seemed to address it, but I could just be unfamiliar with the language; so please pardon me if this is a question with a well known answer.
If it makes a difference whether I'm asking my question about the 'big' Witt vectors or the $p$-typical ones, then please interpret it in the way that makes it more interesting.
With that prelude, the question itself is simple:  is there a functor $\text{$A$-Mod} \to \text{$W(A)$-Mod}$, either just for specific algebras $A$ (I have in mind the $p$-typical Witt vectors of a finite field of characteristic $p$), or, possibly, for all commutative algebras?  In my motivating case, one can do something awkward like pick a basis for an $A$-module (i.e., a vector space) and simply construct the free $W(A)$-module on that set; but, aside from the ugliness of this approach, it's not even clear to me that it's functorial.
EDIT:  As both nfdc23 and QiaochuYuan point out, I have additional assumptions in mind; namely, in both cases, that free, finite-rank $A$-modules are taken to non-$0$, free, finite-rank $W(A)$-modules, for nfdc's post that the identity morphism is taken to the identity morphism, and for Qiaochu's post that not every morphism is sent to $0$.

REPLY [9 votes]: Let me restrict my attention to the only case I understand, which is $A = \mathbb{F}_p, W(A) = \mathbb{Z}_p$ (so here I mean $p$-typical Witt vectors). You don't provide any conditions you want your functor to satisfy, but I presume you don't just want e.g. the zero functor, so here is a very simple no-go for getting extra properties: no such functor can both

be additive, and
send finite free modules to finite free modules.

This is for the very simple reason that $\mathbb{F}_p$ (the endomorphism ring of a finite free $\mathbb{F}_p$-module of rank $1$) admits no ring homomorphism to $M_n(\mathbb{Z}_p)$ for any $n$, since the latter is torsion-free. 
More generally this argument shows that the image of any additive functor from $\mathbb{F}_p$-modules to $\mathbb{Z}_p$-modules necessarily lands in modules whose underlying abelian groups are $p$-torsion. Of course there is a natural additive functor given by restriction of scalars along the quotient map $\mathbb{Z}_p \to \mathbb{F}_p$ with this property.<|endoftext|>
TITLE: What are local spaces and what are they good for?
QUESTION [6 upvotes]: Factorization structures have been popular in the past decade. Recently a variant of this structure has been suggested by Ivan Mirkovic (and possibly collaborators). This variant, which goes under the name of "local space" is supposetly useful in understanding the affine Grassmanian and geometric Langlands. 
Could a person who understands the story please explain what exactly is the difference between a factorization space (in the sense of Beilinson and Drinfeld) and a local space? Also, what do we win by considering local space instead of factorization spaces?

REPLY [12 votes]: Local spaces:finite subschemes::Factorization spaces:finite subsets.
A local space over X is a compatible collection of spaces over the Hilbert schemes of arbitrary numbers of points in X satisfying a factorization property for disjoint union. This is very close to the notion of a factorization space, in which we have a space over the Ran space Ran(X) parametrizing all finite subsets of X. The point is we're allowed (and encouraged!) to keep track of multiplicities. The idea is that in practice objects that don't care about multiplicities arise by allowing all multiplicities at once: vertex/factorization algebras/spaces naturally arise as objects living over formal discs in X, which can be filtered by objects living over increasing infinitesimal neighborhoods of points, which will form local spaces. This gives useful extra structure and allows the construction of interesting factorization spaces by "accumulation of dust" (to quote the creator, Mirkovic). 
The amazing application for which local spaces were introduced is the geometric construction of the most interesting factorization space, the affine Grassmannian (and hence of reductive groups themselves) "out of combinatorics" - starting from a torus with an invariant bilinear form and a sequence of natural operations on local spaces Mirkovic builds the affine Grassmannian. This is a beautiful reformulation of the theory of "Zastava spaces" (also a cocreation of Mirkovic), which are models for transverse slices to semiinfinite orbits in the Grassmannian --- in retrospect their construcion fits into the language of local spaces and the idea is one can assemble these slices together with the orbits themselves to reconstruct the Grassmannian. Mirkovic has many exciting related ideas, including applications in higher dimensions. A recent talk on the subject is available
here.<|endoftext|>
TITLE: Invariant ring of $S_5$
QUESTION [11 upvotes]: The irreducible representations of the Symmetric group $S_5$ are classified by the partitions of $5$. For the standard representation which corresponds to the partition (4,1) the ring of invariants is generated by the elementary symmetric polynomials and hence is a polynomial ring. For other irreducible representations except for the trivial and sign representation, $S_5$ does not represent as a pseudo-reflection group, hence for these representations the ring of invariants is not a polynomial ring any more. I also learnt that a minimal generating set for these rings is not possible (or hard) to compute using the programs available till date. Do we have any combinatorial technique to compute them explicitly ? 
For example if we take the tensor product of standard and sign representation (which corresponds to the partition 2+1+1+1) is there any way to get a minimal generating set for the ring of invariants ? 
Further it would be a great help if someone could provide me an example of a representation of $S_5$ (not necessarily irreducible and except standard, natural, sign and trivial) whose ring of invariants in terms of generators and relations is known.

REPLY [6 votes]: In fact at least in the case you seem most interested in, the (4-dimensional) standard representation tensored by the sign representation, it is possible to compute the invariants by computer, using the standard tools in MAGMA. The result is a minimal system of generating invariants of degrees 2, 4, 6, 8, 10, 13, and 15. When printerd out, the invariants look quite terrible. The invariant ring is not Gorenstein. Here's the MAGMA session:
> G:=MatrixGroup<4,Rationals() | [0,-1,0,0, -1,0,0,0, 0,0,-1,0, 0,0,0,-1],
> [0,1,0,0, 0,0,1,0, 0,0,0,1, -1,-1,-1,-1]>;
> R:=InvariantRing(G);
> time prim:=PrimaryInvariants(R);
Time: 0.030
> time fund:=FundamentalInvariants(R);
Time: 0.430
> [TotalDegree(f): f in fund];
[ 2, 4, 6, 8, 10, 13, 15 ]
> H<t>:=HilbertSeries(R);
> H * &*[1-t^TotalDegree(f): f in PrimaryInvariants(R)];
t^15 + t^13 + t^8 + 1

I don't have the matrices for the other irreducible representations (of degrees 5,6, and 6). If you give them to me, there should be a fair chance of being able to compute the invariant rings. On the other hand, in view of the ugliness of the above example, it may not be worthwhile.
Let me add that computations become much harder for sums of irreducible representations. Unlike representation theory, invariant theory does not reduce to the irreducible case. For example, invariants of direct sums of copies of the same representation, known as "vector invariants", are notoriously ugly.<|endoftext|>
TITLE: The parity of the full automorphism group order of finite non-abelian groups of prime exponent
QUESTION [8 upvotes]: Is there a finite non-abelian group $G$ of prime exponent such that the full automorphism group of $G$ is of odd order?

REPLY [13 votes]: Yes:

for every prime $p\ge 7$, there's a finite group of exponent $p$ whose automorphism group is a $p$-group.

Initial answer (Jan 18' 2016)
Start from any (finite-dimensional) complex nilpotent Lie algebra $\mathfrak{g}$ that is defined over $\mathbf{Q}$ and has a unipotent automorphism group (see e.g. Luks' Lie algebra top of p14 in (Ancochea Campoamor survey) for a 16-dimensional example), see also first edit (Jan 19' 16) below.
Fix a rational structure, so we can defined $\mathfrak{g}_p$ as the corresponding Lie algebra modulo $p$ for $p$ large enough (i.e., not dividing any of the denominators). (Formally speaking, this means writing $\mathbf{g}=\mathfrak{h}\otimes_{\mathbf{Q}}\mathbf{C}$ for some Lie algebra $\mathfrak{h}$ over $\mathbf{Q}$.)
Then for $p$ large enough, the automorphism group of $\mathfrak{g}_p$ is a $p$-group. Indeed otherwise, if for a growing subsequence $(p_i)$ this is not the case, say $\mathfrak{g}_{p_i}$ has an automorphism of prime order $\neq p_i$, passing to the algebraic closure $K_i$ of $\mathbf{F}_{p_i}$ and then to a ultralimit over $i$ we get that the ultraproduct, which is a Lie algebra over the ultraproduct $K$ of the $K_i$ has a nontrivial semisimple automorphism. But this ultraproduct is isomorphic to $\mathfrak{h}\otimes_{\mathbf{Q}}K$. This contradicts that $\mathfrak{g}$ has a unipotent automorphism group.
Note that this provides a group of order $p^{16}$ with some well-controlled nilpotency length (I haven't made the computation for Luks' Lie algebra but it's easy to do), but gives no control of $p$, although for small $p$ a computer computation is doable).

We can expect a computer computation to yield explicit examples.
Anyway no example is possible for $p$-group of nilpotency length 2 (and odd exponent $p$): indeed such a group admits an action by automorphisms of the multiplicative group modulo $p$ (of order $p-1$), because the corresponding Lie algebra is Carnot. Classification along with Malcev correspondence also shows that no group of order $p^{\le 6}$ for prime $p\ge 7$ can work, because the corresponding Lie algebras have nontrivial integral gradings according to the classification by de Graaf in here (J. Algebra 2007) [this also works in order $5^{\le 5}$, and in order $5^6$ except for nilpotency length $=5$].

Edit: (Jan 19' 2016) I have checked that the 7-dimensional (with nilpotency length 6) Lie algebra given page 17 line 8 in (Ancochea Campoamor survey), viewed with coefficients in an arbitrary field, has only unipotent automorphisms. If one takes it in the field on $p$ elements for any prime $p\ge 7$, the Baker-Campbell-Hausdorff (which realizes Malcev's correspondence) provides a group of order $p^7$ whose automorphism group is a $p$-group (thus of odd order).
According to what's above and computer check by Alireza for $p=3,5$, for $p$ odd any nontrivial $p$-group of order $\le p^6$ has an automorphism of order 2. So $p^7$ is the smallest possible order for such exceptions, for $p=3,5$ possibly one has to go a little higher since Alireza checks that every group of order $3^7$ has an automorphism of order 2. (At the moment this gives no example for $p=3,5$ although they certainly exist.)

Edit: (Jan 25' 2018) Thinking a little about the remaining cases $p=3,5$, they seem different and deserve separate additional comments:
first recall by above remark that for any odd $p$, the nilpotency length of any example should be $\ge 3$.

for $p=5$, Malcev correspondence is valid in nilpotency length $\le 4$. Therefore if one searches among nilpotent rational Lie algebras of nilpotency length in $\{3,4\}$ and unipotent automorphism groups, one can reasonably hope to find one that reduces modulo 5 and for which the reduction modulo 5 has no extra automorphisms.
$p=3$ is quite different, because exponent 3 implies nilpotency length $\le 3$ (in contrast, by Razmyslov's theorem, for all $p>3$ there are finite groups of exponent $p$ and arbitrary large solvability length). So in this case, Lie algebras are of no help. Finding a finite group of exponent 3 whose automorphism group is a 3-group (or of odd order) sounds a quite different question.<|endoftext|>
TITLE: Quasi-isometric rigidity of Nil
QUESTION [10 upvotes]: Let $Nil$ be the unique simply connected non-abelian three-dimensional nilpotent Lie group, i.e. the group of upper triangular matrices with all the eigenvalues equal to 1 (this group is also known as the 3-dimensional Heisenberg group), endowed with a left-invariant Riemannian metric. 
The following statement can be found in the literature: 
Any finitely generated group quasi-isometric to $Nil$ is in fact virtually isomorphic to a uniform lattice in $Nil$.
Recall that two groups are virtually isomorphic if they can be obtained one from the other via a finite number of steps, where each step consists in taking a finite extension or extracting a finite-index subgroup (or viceversa). The fact that virtually isomorphic groups are quasi-isometric is elementary, and any uniform lattice is quasi-isometric to its ambient Lie group as a consequence of Milnor-Svarc Lemma.
For example, the statement above is attributed to Gromov, ``Asymptotic invariants of infinite groups'' (1993), in the paper 
``Quasi-isometries preserve the geometric decomposition of Haken manifolds'', Kapovich and Leeb, Inventiones Mathematicae (1997) and it is also stated in
`` Quasi-isometric classification of graph manifolds groups'', Behrstock and Neumann, Duke Math. J. (2008).
I tried to reconstruct the proof of the quasi-isometric rigidity of $Nil$, but I am not completely sure what should be the correct attribution. By Gromov's polynomial growth Theorem, any f.g. group quasi-isometric to $Nil$ is virtually nilpotent, and it is a classical result by Malcev that every f.g. (torsion free) nilpotent group is a uniform lattice in some nilpotent Lie group. Therefore, every f.g. group quasi-isometric to $Nil$ is virtually isomorphic to a uniform lattice in a (maybe different) nilpotent Lie group. Finally, maybe using the work of Pansu in ``Croissance des boules et des géodésiques fermées dans les nilvariétés'' (1983) one could probably show that the $Nil$ is recognized among the class of all nilpotent Lie groups by its asymptotic cone. This should conclude the proof of the quasi-isometric rigidity of $Nil$.
My question is: Is there an established reference for the quasi-isometric rigidity of $Nil$? If not, did someone write somewhere the details (if correct!) for the tentative proof I outlined above? Or is there a shorter argument that I did not spot?

REPLY [6 votes]: Here's another, more topological way to see that $\mathrm{Nil}$ is not quasi-isometric to $\mathbf{R}^4$: use the fact that $\mathrm{Nil}$ is homeomorphic to $\mathbf{R}^3$, so that $\pi_2^\infty(\mathrm{Nil})$ is nontrivial, whereas $\pi_2^\infty(\mathbf{R}^4)$ is trivial.
This works because $\pi_2^\infty$ is indeed a quasi-isometry invariant within a class of metric spaces which contains $\mathbf{R}^4$ and $\mathrm{Nil}$. The key property is that they are uniformly 2-connected, i.e. spheres of dimension $\le 2$ can be filled by balls with metric control.
Concretely, to prove the result by hand you can do the following: assume that there is a quasi-isometry $f:\mathrm{Nil} \to \mathbf{R}^4$. Without loss of generality, $f$ takes the origin to the origin. Let $\phi$ be an embedding of $\mathbf{S}^2$ into $\mathrm{Nil}$ with image a metric  sphere of large radius centred at the origin. Put a sufficiently fine triangulation $\mathcal T$ on $\mathbf{S}^2$. By induction on the skeleta, construct a continuous map $g:\mathbf{S}^2\to \mathbf{R}^4$ which coincides with $f\circ\phi$ on the $0$-skeleton and is at bounded distance from $f\circ\phi$.
Since $\mathbf{R}^4$ is 2-connected at infinity, we can extend $g$ to a continuous map $\bar g$ from the 3-ball to $\mathbf{R}^4$ which stays far away from the origin. Composing with a coarse inverse of $f$ and using the same trick to make it continuous, we obtain a continous map from the $3$-ball to $\mathrm{Nil}\setminus\{O\}$ which extends $\phi$. This is a contradiction.
(Remark: we have actually only used that $\mathrm{Nil}$ is uniformly 2-connected and $\mathbf{R}^4$ is uniformly 1-connected.)<|endoftext|>
TITLE: Is a "knot knot" or "double knot" a thing in knot theory?
QUESTION [6 upvotes]: I apologize in advance for my rudimentary knowledge of knot theory, but I've been trying to find out about the significance (if any) of taking a knot (particularly a torus knot), cutting it, and forming another knot using it.  Some examples:
trefoil knot of a trefoil knot

and another "double trefoil" knot

Is there a generally accepted name for this?  Is the resulting knot prime? Are there any other generalizations one can make about these? 
Thank you for any information you have on this.

REPLY [8 votes]: A satellite knot (or link) is more general that your construction.  To make a satellite knot, you take a knot in a solid torus (not in a ball within the solid torus), and embed the solid torus so that it is knotted. 
A doubled knot (or Whitehead double of a knot) is special type of satellite knot that is different from yours. 
Your knots are cablings of a knot, where the knot in the solid torus is a torus knot following a curve on the boundary. The slope is usually parametrized $(p,q)$ and yours are $(2,n)$ cablings for some $n$. Nontrivial cablings are prime (see Burde and Zieschang, Knots, p. 93) so your knots are prime.

Thanks to Ryan Budney for pointing out that my initial definition of satellite know was more restricted than a more common one.<|endoftext|>
TITLE: Weingarten function for unitary group
QUESTION [7 upvotes]: Studying integration over unitary group I came across this function, the Weingarten function Wg, such that
$$ \int_{\mathcal{U}(N)} \prod_{k=1}^{n} U_{i_kj_k}
U^*_{m_k r_k} dU=\sum_{\tau,\sigma\in S_n} {\rm
Wg}^U(\tau^{-1}\sigma)\prod_{k=1}^n \delta_{i_k,\tau(m_k)}\delta_{j_k,\sigma(r_k)}.$$ I understand the structure of this equation, the $m$'s must be a permutation of the $i$'s, and likewise the $r$'s with the $j$'s. However, the function itself is complicated, 
$$ {\rm
Wg}(\tau^{-1}\sigma)=\frac{1}{n!^2}\sum_{\substack{\lambda\vdash n\\\ell(\lambda)\leq N}}
\frac{d_\lambda^2}{s_\lambda(1^N)}\chi_\lambda(\tau^{-1}\sigma),$$ where $s_\lambda$ are Schur functions, $d_\lambda$ is the dimension of a irreducible representation of the permutation group labelled by $\lambda$ and $\chi$ are the characters of the irreps of this group. The derivations I have seen of this result are also complicated.
Can anyone offer some insight into this function and how it is derived?

REPLY [5 votes]: I would recommend Elementary derivation of Weingarten functions of classical Lie groups by Marcel Novaes (2015).

Previous works where Weingarten functions were obtained were based
  either on representation theory and Schur-Weyl duality, the theory of
  Gelfand pairs, or Jucys-Murphy elements. In contrast, we here derive
  Weingarten functions for the classical compact groups by means of some
  elementary direct calculations (although we rely on some classical
  results that can, of course, be interpreted very naturally in the
  light of those theories).
The idea consists of five steps: 1) write the integrand as the
  derivative of a power sum function; 2) change basis from power sums to
  Schur functions; 3) perform the group integral; 4) revert back to
  power sums; 5) take the derivative to arrive at the result.<|endoftext|>
TITLE: Analogue of Tate curve for $g>1$
QUESTION [12 upvotes]: Is there any analogue of the Tate curve for (principally polarized) abelian varieties of dimension $g$ ?

REPLY [21 votes]: Yes, due to Mumford. See
Mumford, D.: An analytic construction of degenerating abelian varieties over complete rings, Comp.. Math.
24, 239-272 (1972).
Of course, there's been plenty of work done since then. See for example the survey:
W. Lutkebohmert,
From Tate's Elliptic Curve to Abeloid Varieties, Pure and Applied Mathematics Quarterly
Volume 5, Number 4
(Special Issue: In honor of
John Tate, Part 1 of 2)
1385-1427, 2009. http://intlpress.com/site/pub/files/_fulltext/journals/pamq/2009/0005/0004/PAMQ-2009-0005-0004-a007.pdf<|endoftext|>
TITLE: Is special value of Epstein zeta function in 3 variables a period?
QUESTION [16 upvotes]: Kontsevich-Zagier's article "Periods" contains the following question

Is $\displaystyle \sum_{x,y,z \in \mathbb{Z}}' \frac{1}{(x^2+y^2+z^2)^2}$ an extended period?

($\sum'$ means we do not sum over $(x,y,z) = (0,0,0)$.)
Here period is loosely interpreted as "integral of algebraic differential forms (on an smooth projective algebraic variety) over algebraic subvariety". Extended period is a period up to a factor of integer powers of $\pi$. 
(I am intentionally making this vague since I don't think even Kontsevich-Zagier tried to pin this down very precisely, but hopefully you know a period when you see one.)
My question: Is this now known? (It has been 15 years since Kontsevich-Zagier first published the article) More generally in this setup, the question would be

Let $Q(x_1,\cdots,x_n)$ be a positive definite quadratic form in an odd number ($\ge 3)$ of variables with coefficients in $\mathbb{Q}$. Is $\displaystyle \sum_{x_1,\cdots,x_n \in \mathbb{Z}} \frac{1}{Q(x_1,\cdots,x_n)^s}$ an extended period for $s > n/2$?


Some background
There is a beautiful theorem due to Beilinson and Deninger-Scholl,

Theorem Let $f$ be a modular form of weight $k \ge 2$ defined over $\overline{\mathbb{Q}}$. Then $L(f,m)$ are extended periods for all integers $m \ge k$.

(When $m < k$ this is still true; here I am merely distinguishing $m$ by whether it is a critical value or not for $L(f,s)$.)
The question above for quadratic forms with even number of variables is then solved by this theorem, since it is the $L$-function for the theta series $\sum'_{x_1,\cdots,x_n} q^{Q(x_1,\cdots,x_n)}$ of weight $n/2$ if $n$ is even.
I tried to look up generalization of this theorem to half-integral weight modular forms with no success.

REPLY [6 votes]: Nice question. I think this is what's going on:
If $Q$ is a (positive definite) quadratic form, then its Epstein zeta function $Z_Q(s)$ can be expressed as a linear combination of L-functions of cusp forms and Eisenstein series.
In the case of an even number of variables $2k$, the Eisenstein series involved are certain modular forms of weight $k$, which by the work of Hecke can be expressed in terms of Dirichlet L-functions. See for example:

Keiju Sono, Higher moments of the Epstein zeta functions (2013)

As you mention, thanks to the result of Beilinson-Deninger-Scholl, modular forms evaluated at the integers are periods. $Z_Q(n)$ is then a linear combination of periods, and therefore a period.
The problem in the odd case seems to be that the Eisenstein series are not a combination of Dirichlet L-functions, and I don't think it is known whether their values at the integers are periods or not. It might even be the case that they aren't. Remember that Epstein zeta functions are automorphic forms, and their special values might very well be trascendental.
In fact, in another paper on periods, Maxim Kontsevich, co-author of the 2001 paper you mention says:

It is not clear to me whether the analogous statement holds for
  quadratic forms in odd number of variables.


M. Kontsevich, Periods (october 2006) pp. 7<|endoftext|>
TITLE: Infinitely many $k$ such that $[a_k,a_{k+1}]>ck^2$
QUESTION [14 upvotes]: Let $a_n\in \mathbf{N}$ be an infinite sequence such that $\forall i\neq j, a_i\neq a_j$.
I have the following theorem:

For $0<c<\frac{3}{2}$, there are infinitely many $k$ for which $[a_k,a_{k+1}]>ck$, where $[\cdots]$ denotes least common multiple.

Idea of proof: By contradiction.  Suppose there is an $M$ such that for all $k>M$, $[a_k,a_{k+1}]\leq ck$.
On the one hand,
$\displaystyle \sum_{k=1}^{n}\frac{1}{[a_k,a_{k+1}]}\geq \sum_{k=M+1}^{n}\frac{1}{[a_k,a_{k+1}]}\geq \sum_{k=M+1}^{n}\frac{1}{ck}$
On the other hand,
$\displaystyle \frac{1}{[a_k,a_{k+1}]}$
$\displaystyle =\frac{(a_k,a_{k+1})}{a_ka_{k+1}}$
$\displaystyle =(a_k,a_{k+1})\frac{1}{a_k+a_{k+1}}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle =\frac{1}{\frac{a_k}{(a_k,a_{k+1})}+\frac{a_{k+1}}{(a_k,a_{k+1})}}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle \leq \frac{1}{3}(\frac{1}{a_k}+\frac{1}{a_{k+1}})$
$\displaystyle \sum_{k=1}^{n}\frac{1}{[a_k,a_{k+1}]}\leq \frac{1}{3}\sum_{k=1}^{n}(\frac{1}{a_k}+\frac{1}{a_{k+1}})\leq \frac{2}{3}\sum_{k=1}^{n}\frac{1}{k}$
And I make the following conjecture:

$\exists c>0$, there are infinitely many $k$ for which $[a_k,a_{k+1}]>ck^2$.

But I cannot prove or disprove it.

REPLY [10 votes]: In this paper
P. Erdős, R. Freud, and N. Hegyvári, Arithmetical properties of permutations of integers, Acta Mathematica Hungarica 41:1-2 (1983), pp 169-176.
http://renyi.mta.hu/~p_erdos/1983-02.pdf
the authors show that for permutations of {1,2,...,n} the answer is $\Theta(n^2/\log n)$.
See https://oeis.org/A064764 .
Also for infinite permutations they give an example where
${\rm lcm}(a_i, a_{i+1}) < i \exp (c \sqrt{\log i} \log\log i)$ for all $i$.
And I see that 
Chen, Y.-G.; Ji, C.-S.
The permutation of integers with small least common multiple of two subsequent terms. 
Acta Math. Hungar. 132 (2011), no. 4, 307–309 has improved on this still further, to  $i \exp (c' \sqrt{\log i\log\log i})$.<|endoftext|>
TITLE: Subgroups of the mapping class group of a surface generated by Dehn twists
QUESTION [8 upvotes]: Let $G$ be the mapping class group of a surface of genus $g > 1$. Is it known for which positive integer $k$ one can find a subgroup $H$ of $G$ generated by a finite number of Dehn twists and a Dehn twist $\tau \in G$ which is not in $H$ but such that $\tau^k$ is in $H$, $k$ being minimal for this property ?
Using the chain relation, one can construct examples with $k = 2$, but I cannot find any examples with higher powers.
EDIT: I realized I should put a little motivation to this question. So let us take $U_d$ the space of smooth degree $d$ planar complex curves. There is a monodromy map $\pi_1(U_d)\rightarrow G$, and the image of this map is generated by Dehn twists obtained for example by taking a generic pencil of curves. Now I can construct a loop in $U_d$ whose monodromy is a power of some Dehn twist and I am wondering if there should be a reason coming from the structure of $G$ for the Dehn twist itself to be in the image of the monodromy.

REPLY [3 votes]: Edited: 
The paper Dehn twists have roots proves that Dehn twists have roots, naturally.  The limits on that construction can be found in the paper Roots of Dehn twists; they bound the degree of the root (linearly, as I recall) in terms of the complexity of the surface.  
Now, if I understand correctly, you are asking for ways to realize $\tau_\alpha^p$, a power of a Dehn twist about $\alpha$, as a product of other twists.  If we found a root of $\tau_\alpha^p$, which was not itself a twist, then that would answer your question.  But I think that the main idea behind the above papers will work.  For: any root of $\tau_\alpha^p$ has the same canonical reduction system, namely $\alpha$.  So the root of $\tau_\alpha^p$ stabilizes $\alpha$.  You can now build a periodic map in the complement of $\alpha$ that does a fractional twist about $\alpha$: for example $p/q$ fraction of a right twist.  Take the $q$-power of this get the desired $\tau_\alpha^p$.
Note that it is far from clear that these constructions are the only way to get a subgroup $H$ of the type you desire.<|endoftext|>
TITLE: Mapping scheme from a proper variety
QUESTION [5 upvotes]: Let $X$ be a proper scheme over a field $k$. Let $T$ be a scheme over $k$. Is it true that morphisms $T \times X \to \mathbb{A}^1$ are in bijection with morphisms $T \to \Gamma (X, \mathcal{O}_X)$ (where the finite-dimensional vector space $\Gamma (X, \mathcal{O}_X)$ is interpreted as a scheme as usual)?
Or, perhaps, is something weaker (put condition on $T$, say) or stronger (more general scheme instead of $\mathbb{A}^1$, say) true?
Thank you

REPLY [6 votes]: You can just use all the universal properties one at a time. We only need that $X \to \operatorname{Spec} k$ is qcqs and that $\Gamma(X,\mathcal O_X)$ is finite-dimensional; both these assumptions are satisfied if $X$ is proper over $k$. We also need that $T \to \operatorname{Spec} k$ is flat, which is always true if $k$ is a field.
Remark. Note that $\Gamma(T \times X, \mathcal O_{T \times X}) = \Gamma(T, \mathcal O_T) \otimes \Gamma(X, \mathcal O_X)$. Indeed, this follows from flat base change (Tag 02KH), since $T \to \operatorname{Spec} k$ is flat and $X \to \operatorname{Spec} k$ is qcqs.
Remark. If $V$ is a vector space and $W$ is a finite-dimensional vector space, then the natural map
\begin{align*}
V \otimes W &\to \operatorname{Hom}_k(W^\vee, V)\\
v \otimes w &\mapsto (\phi \mapsto \phi(w) v)
\end{align*}
is a natural isomorphism. For example, one can prove the case for $W$ of dimension $1$, and use that every $W$ decomposes as a finite direct sum of $1$-dimensional vector spaces. (Note, however, that this is false for $W$ infinite-dimensional, already if $V = k$.)

Lemma. Let $X$ be a qcqs $k$-scheme such that $\Gamma(X, \mathcal O_X)$ is finite-dimensional. Let $T$ be a $k$-scheme. Then
$$\operatorname{Hom}_{\textrm{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}_{\textrm{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O_X)^\vee)).$$

Proof. By the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ given by $\Gamma$ and $\operatorname{Spec}$, we get
$$\operatorname{Hom}_{\textrm{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}_k^{\operatorname{alg}}(k[x],\Gamma(T \times X, \mathcal O_{T \times X})).$$
By the universal property of $k[x]$, the latter is just $\Gamma(T \times X, \mathcal O_{T \times X})$. By the two remarks above, this equals
$$\Gamma(T, \mathcal O_T) \otimes \Gamma(X, \mathcal O_X) = \operatorname{Hom}_k(\Gamma(X,\mathcal O_X)^\vee, \Gamma(T,\mathcal O_T)).$$
The universal property of the symmetric algebra turns this into
$$\operatorname{Hom}_k^{\operatorname{alg}}(\operatorname{S}(\Gamma(X, \mathcal O_X)^\vee), \Gamma(T,\mathcal O_T)).$$
Finally, using the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ again, this finally becomes
$$\operatorname{Hom}_{\textrm{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O_X)^\vee)),$$
which proves the claim. $\square$<|endoftext|>
TITLE: Exact determinant of a circulant matrix
QUESTION [7 upvotes]: The wikipedia gives us a formula for the determinant of a circulant matrix. That is:
$$\mathrm{det}(C) 
= \prod_{j=0}^{n-1} (c_0 + c_{n-1} \omega_j + c_{n-2} \omega_j^2 + \dots + c_1\omega_j^{n-1})= \prod_{j=0}^{n-1} f(\omega_j),$$
with $\omega_j=\exp(2\pi {\rm i}j/n)$ the $n$-th root of unity. 
Assuming a circulant matrix with only entries from $\{-1,1\}$, is there a number theoretic version of this formula which will enable you to compute the determinant exactly and efficiently using only operations on integers?
[This question was also posted to https://math.stackexchange.com/questions/1616049/exact-determinant-of-a-circulant-matrix previously.]

REPLY [2 votes]: We need to pick a prime $p$ and integer $k \ge 2$ such that $p = nk + 1$. (Alternatively you can choose a positive integer $p > 2n$ (not necessarily prime) such that $n$ divides $\phi(p)$, where $\phi$ is the Euler Totient function. However, I am not sure you can use FFT in the subsequent steps, and may have to resort to manual matrix vector multiplications.)
Next, we need to find a primitive $n$-th root of unity mod $p$. This root can be reused if you have many different circulant matrices of size $n$. This root, which we call $\omega \in \mathbb Z _p$, is needed in the next step when you compute the FFT. Various fast heuristics exist, but note that even a worst case brute force search still runs in $O(p^2 \log(p))$
Using $\omega$, we will compute the number theoretic (mod $p$) FFT of $v$, where $v = (c_0, ... c_{n-1})$. Call the resulting vector $w$. This runs in $O(n \log(n))$
The entries of $w$ are in $\mathbb Z_p$. In your notation, they represent $f(\omega_j)$ mod $p$. Note that your $f(\omega_j)$ are bounded to be in $\{-n, \ldots, n\}$. We will pick integer representatives for the entries of $w$ in $\{-n, \ldots, n\}$, and write that as $\hat w$. The product of the entries in $\hat w$ is your determinant.<|endoftext|>
TITLE: When does this interesting sum diverge?
QUESTION [5 upvotes]: For $x \gt 0,$ what is the greatest $y$ such that $$\sum_ {1\le h^x \le k^y} \frac{1}{h^x k^y}= \infty ?$$  
I don't know of any references or methods for this -- not even for $x=1$, for which the series is $$\sum_ {k=1}^{\infty} \frac{H(k)}{k^y},$$ where $H(k)$ is a harmonic number.

REPLY [11 votes]: In short, 
  $$ \begin{cases}
\text{when }1\leq x & \text{series diverges when }y\le1\\
\text{when }\frac{1}{2}<x<1 & \text{series diverges when }y\leq\frac{x}{2x-1}\\
\text{when }0<x\leq\frac{1}{2} & \text{series always diverges.}
\end{cases}
$$

When $x>1$, the inner sum of $$\sum_{k=1}^{\infty}\frac{1}{k^{y}}\sum_{h^{x}\leq k^{y}}\frac{1}{h^{x}}$$
 converges so this series diverges precisely when $y\leq 1$. When $x=1$, since $H(k^{y})\sim y\log k$, again we see that this diverges for $y\leq 1$. Lastly, when $0<x<1$
  we have that $$\sum_{h^{x}\leq T}\frac{1}{h^{x}}=\int_{1}^{T^{1/x}}\frac{1}{s^{x}}d\left[s\right]\sim\int_{1}^{T^{1/x}}\frac{1}{s^{x}}ds\sim\frac{1}{1-x}\left(T^{1/x}\right)^{1-x}=\frac{1}{1-x}T^{1/x-1},$$
 and so the convergence of the series depends on the convergence of $$\sum_{k=1}^{\infty}\frac{1}{k^{y}}k^{y\left(\frac{1}{x}-1\right)}=\sum_{k=1}^{\infty}k^{y/x-2y},$$
 and this depends on when $\left(\frac{1}{x}-2\right)y<-1.$
 Thus it diverges for every value of $y$ when $0<x\leq\frac{1}{2}$ , and for $\frac{1}{2}<x<1$
 , it diverges when $y\leq\frac{x}{2x-1}.$<|endoftext|>
TITLE: Kolmogorov complexity for matrices
QUESTION [5 upvotes]: In applications one often encounters very large matrices that barely fit in computer memory, if at all. Naturally one wishes to represent those matrices as compactly as possible. Sometimes one even sacrifices accuracy for compactness, as in L-BFGS for example.
On the other hand, often enough the linear problem naturally appears with matrices that can be presented in a relatively compact form, either with band matrices or Toeplitz matrices, etc.
Let's call an $n\times n$ matrix presentation "compact" if it requires no more than $O(n \log^{const} n)$ entries (that's the "quasilinear" rule of thumb for trackability of large problems). Is there a way to tell that a family of matrices is presentable in a "compact" form? And, if it is, what would be a computationally feasible algorithm to compute that presentation?
Or, to put it differently, has something like Kolmogorov complexity been studied for matrices? Here by "Kolmogorov complexity" one may understand a map $M_{n\times n}\to \mathbb{N}$ mapping a matrix to the length of its shortest presentation. 
This is different from sparsity of matrices, as evident from compact yet dense Toeplitz matrices.

REPLY [4 votes]: Kolmogorov’s complexity for positive definite matrices

Based on Kolmogorov’s idea, complexity of positive definite matrices
  with respect to a unit vector is defined. We show that the range of
  the complexity coincides with the logarithm of its spectrum and the
  order induced by the complexity is equivalent to the spectral one.
  This order implies the reversed one induced by the operator entropy.<|endoftext|>
TITLE: Recovering a smooth manifold from its tensor fields
QUESTION [7 upvotes]: 1) Consider the algebra $\mathcal T (M) = \bigoplus \limits _{p, q \ge 0} \mathcal T ^{p, q} (M)$, where $\mathcal T ^{p, q} (M)$ is the space of tensor fields of type $(p,q)$. I should endow it with some topology, but I do not know what to choose: the topology of pointwise convergence is one option, the compact-open topology another one, but I am open to other suggestions as well.
Is it possible to recover both the topological and the differential structure of $M$ from $\mathcal T (M)$? (This would be some sort of Gelfand-Naimark in a smooth setting; in fact, the component $\mathcal T ^{0, 0} (M)$ endowed with a suitable topology would lead to precisely some version of Gelfand-Naimark.)
A related question is: do we really need to consider all the types of tensors above? Given that covariant and contravariant are dual to each other (and we are working with manifolds of finite-dimension), shouldn't these two classes contain the same amount of information, either one being redundant for the issue under discussion? It seems that one could restrict to either $\bigoplus \limits _{q \ge 0} \mathcal T ^{0, q} (M)$, or to $\bigoplus \limits _{p \ge 0} \mathcal T ^{p, 0} (M)$.
2) I am curious about the same question but for a different algebra: if $\mathcal X (M)$ is the Lie algebra of the smooth vector fields on $M$, let $\mathcal U (M)$ be its universal enveloping algebra. Does $\mathcal U (M)$ encode all the information about $M$? Can $M$ be recovered from $\mathcal U (M)$? (This latter algebra, though, does not contain the smooth functions, so I have some doubts about it.)

REPLY [9 votes]: Look at $M=S^7$. This has 28 smooth structures, and their tangent bundles are all trivial - compare Parallelizability of the Milnor's exotic spheres in dimension 7  . Thus their tensor algebras do not carry more information about the smooth structure than $C^\infty(M)$.
On the other hand, for a compact smooth manifold, the algebra of smooth functions is sufficient to recover the smooth structure: Every maximal ideal of this algebra is of the form $\{f\in C^\infty(M): f(x) = 0\}$ for some $x\in M$, and the topology on $M$ is recovered as the weakest one on the space of maximal ideals such that $\mathfrak{m}\mapsto [f]\in C^\infty(M)/\mathfrak{m}\cong \mathbb{R}$ is continuous for all $f\in C^\infty(M)$. So this space is a compact topological manifold with a subsheaf $C^\infty$ of the sheaf of continuous functions, and we take a chart $U\to\mathbb{R}^n$ to be smooth if all of its coordinate functions are in $C^\infty(U)$.<|endoftext|>
TITLE: lower bound for Perron-Frobenius degree of a Perron number
QUESTION [7 upvotes]: A Perron number is an algebraic number which is greater than one in absolute value and is greater than all of its Galois conjugates in absolute value as well. Lind's theorem states that any Perron number is spectral radius of some Perron-frobenius matrix (a matrix $A$ is Perron-Frobenius if all of its entries are non-negative integers and there is some $n>0$ such that all entries of $A^n$ are positive).
Question: Given a Perron number $\lambda$, is there any lower bound for the size of smallest Perron-Frobenius matrix $A$ with spectral radius equal to $\lambda$ (called Perron-frobenius degree)? Besides algebraic degree off course   
PS: In particular I'm interested in Perron numbers which come from a topological construction, namely as stretch factors of pseudo-Anosov homeomorphisms of hyperbolic surfaces. There is an upper bound for size of such matrix , 6g-6 where g is the genus of closed surface, however I'm not aware of lower bounds.

REPLY [9 votes]: If a Perron number $\lambda$ has negative trace, then any Perron-Frobenius matrix must have size strictly greater than the algebraic degree of $\lambda$, for example the largest root of $x^3 + 3x^2-15x-46$. 
If $B$ denotes the $d\times d$ companion matrix of the minimal polynomial of $\lambda$ (which of course can have negative entries), then $\mathbb{R}^d$ splits into the dominant 1-dimensional eigenspace $D$ and the direct sum $E$ of all the other generalized eigenspace. Although I've not worked this out in detail, roughly speaking the smallest size of a Perron-Frobenius matrix for $\lambda$ should be at least as large as the smallest number of sides of a polyhedral cone lying on one side of $E$ (positive $D$-coordinate) and invariant (mapped into itself) under $B$.This is purely a geometrical condition, and there are likely further arithmetic constraints as well. For example, if $\lambda$ has all its other algebraic conjugates of roughly the same absolute value, then $B$ acts projectively as nearly a rotation, and this forces any invariant polyhedral cone to have many sides, so the geometric lower bound will be quite large.<|endoftext|>
TITLE: How do i show that:$\prod\frac{p^2+1}{p^2-1}=\frac{5}{2}$ without using properties of Riemann zeta function?
QUESTION [7 upvotes]: In order to know more about  product  over primes  ,I would like to know how do I show that :$$\prod\frac{p^2+1}{p^2-1}=\frac{5}{2}$$ without using properties of Riemann zeta function ?
Note01 : it is well known that $$\prod\frac{p^2+1}{p^2-1}=\frac{{\zeta}^2(2)}{\zeta(4)}=\frac{5}{2}$$ but is there other method to show that ?
Note 02 :I  wish using divisor function properties 
Thank you for any help

REPLY [4 votes]: Yes, as has been noted several times in comments, this has come up before, with a beautiful answer by David Speyer: Computing $\prod_p(\frac{p^2-1}{p^2+1})$ without the zeta function? 
It seems to me this should put the present question to rest. The only "property" of the Riemann zeta function used is the Euler expansion $\sum_{n \geq 1} \frac1{n^s} = \prod_p \frac1{1 - p^{-s}}$, but the proof of this boils down to the fundamental theorem of arithmetic, which goes back to Euclid's Elements I think, so I'd hardly call this using (analytic) properties of the zeta function. Aside from that, David's demonstration (which is elementary in the technical sense) does the rest.<|endoftext|>
TITLE: higher algebraic homotopy groups for schemes?
QUESTION [7 upvotes]: I think I understand how to define the algebraic fundamental group  $\pi^{alg}_{1}(X)$ of a scheme and I think I understand the relation between $\pi^{alg}_{1}(X)$ and $\pi_{1}(X(\mathbb{C}))$, where $X$ is a nice algebraic variety over complex numbers $\mathbb{C}$.
Is it reasonable to hope for a definition of higher algebraic homotopy groups $\pi_{n}^{alg}(X)$ of a scheme $X$ ?

REPLY [7 votes]: I am writing the comments above so that this question does not remain unanswered.  As Qiaochu Yuan writes, one good resource is the wikipedia page for etale homotopy theory.  Here are two references from MathSciNet.
MR0883959 (88a:14024) 
Artin, M.; Mazur, B. 
Etale homotopy. 
Lecture Notes in Mathematics, 100. Springer-Verlag, Berlin, 1986. iv+169 pp.  
ISBN: 3-540-04619-4 
MR0676809 (84h:55012) 
Friedlander, Eric M. 
Étale homotopy of simplicial schemes. 
Annals of Mathematics Studies, 104. Princeton University Press, Princeton, N.J.;  
University of Tokyo Press, Tokyo, 1982. vii+190 pp. 
ISBN: 0-691-08288-X; 0-691-08317-7
Here is the URL for the article of Michael McQuillan: Elementary Topology of Champs.<|endoftext|>
TITLE: Statements that Could be Forced by Ultrapowers
QUESTION [5 upvotes]: Ultrapower of a structure is a very flexible mathematical creature in comparison with the ground structure and its ordinary products. Depending on the nature of ground structure and the good properties of ultrafilter, it is often manageable to give useful properties to ultrapower.
Interestingly in some papers and theses, ultrapower of a forcing notion $\mathbb{P}$ by an ultrafilter $U$ coming from a special large cardinal $\kappa$ as the index set (e.g. $\kappa$ measurable) appeared as a useful tool for dealing with different types of consistency results (e.g. consistency of some relations between cardinal characteristics). For instance see the following references:


Diego Alejandro Mejia, Template iterations with non-definable ccc forcing notions.

Assaf Shani, Ultrapowers of forcing notions.

Saharon Shelah, Two cardinal invariants of the continuum ($\mathfrak{d} < \mathfrak{a}$) and FS linearly ordered iterated forcing. (No. 700 in Shelah's Archive)

Anda Ramona Tănasie, The splitting number and some of its neighbors.



Intuitively one can tame the bad properties of $\mathbb{P}$ in its ultrapower by throwing the bad points out of the ultrafilter and keeping the others intact. Also in ultrapower forcing usually good properties of $\mathbb{P}$ (e.g. chain condition) could be preserved under ultrapower (e.g. using a $\kappa$-complete ultrafilter over index $\kappa$) while such properties might be affected by usual forcing products.
Here I would like to ask about more examples of using ultrapower of a forcing notion as a forcing notion, for obtaining consistency results in different realms of set theory, in order to get a better idea of the type of consistency statements which could be obtained via ultrapower forcing.

Question. What are references for examples of consistency results obtained by ultrapower forcing? Any reference to lecture notes and unpublished papers are also welcome.

REPLY [8 votes]: Cichoń's maximum, arXiv:1708.03691, uses 4 consecutive ultrapowers of a finite support ccc iteration.<|endoftext|>
TITLE: concentration inequality for entropy from sample
QUESTION [7 upvotes]: Consider a measure $\mu$ on a finite set, and let $x_1, \ldots, x_n$ be i.i.d samples from $\mu$.  Then the expression $S_n = -\frac{1}{n} \sum_{i=1}^n \log \mu(x_i)$ converges by a.s. to the entropy $H(\mu)$. 
What concentration inequalities exist for finite $n$? In other words, what upper bounds are known for the expression $P(|S_n - H(\mu)|>t)$? 
Of course, we can get some upper bounds using, for instance, Bernstein's inequality for bounded variables, with bound depending on the size of the smallest atom of $\mu$. However, using the sup norm bound seems very rough in this case, since small atoms inflate the norm but also have small contributions to the mean. On the other hand, using more general versions of Bernstein's inequality with moment bounds does not seem natural. ($\sum_{i \in X} \mu(i) \log^k \mu(i)$ does not seem like a natural quantity). 
Since this is a classical situation, there probably exist sharper inequalities than sup norm, using some natural quantities. What are they?

REPLY [5 votes]: Here's a step that seems nice enough to point out. It still leaves a parameter to pick, and I'm not sure it's ever better than applying Bernstein, but it does something different.
We can get a probability bound in terms of how much $S_n$ exceeds the Renyi entropy $H_{\alpha}$ of $\mu$ (equivalently, worded in terms of the $\ell_{\alpha}$ norm of $\mu$), for any $0 < \alpha < 1$.
The unresolved question is if we can to pick $\alpha$ to get a nice closed form of some kind. Maybe someone more clever than I can speak to that.
Claim. Let $X_1,\dots,X_n$ be i.i.d. according to $\mu$ and $Y_i = \log(1/\mu(X_i))$; let $S_n = \frac{1}{n} \sum_{i=1}^n Y_i$. Then for any $0 < \alpha < 1$,
\begin{align} \Pr[ S_n \geq t ] &\leq 2^{-n (1-\alpha) \left( t - H_{\alpha}(\mu) \right) }  \\
&= 2^{-n \left( (1-\alpha)t - \alpha \log \| \mu \|_{\alpha} \right) } .
\end{align}
Here I'm writing $\mu = (\mu_1,\dots,\mu_m)$ as a vector of probabilities. Note that $H_{\alpha}$ is decreasing in $\alpha$ and $H_1 = H$, Shannon entropy. So as $n \to \infty$, we can pick $\alpha \to 1$ and get tail bounds for $t \to H(\mu)$.
Proof. Using the general Chernoff method,
\begin{align}
 \Pr[S_n \geq t] &= \Pr\left[ 2^{\lambda S_n} \geq 2^{\lambda t}\right]  & (\forall \lambda \geq 0) \\
  &\leq \frac{\mathbb{E} 2^{\lambda S_n} }{2^{\lambda t}}  & (\text{Markov's}).
\end{align}
We have
\begin{align}
\mathbb{E} 2^{\lambda S_n} &= \left( \mathbb{E} 2^{\frac{\lambda}{n} Y_1} \right)^n  \\
 &= \left( \mathbb{E} \mu(X_1)^{-\lambda/n} \right)^n  \\
 &= \left( \sum_{j=1}^m \mu_j^{1-\lambda/n} \right)^n .
\end{align}
Hence
\begin{align}
 \Pr[S_n \geq t] \leq 2^{-n \left(\frac{\lambda}{n} t - \log \sum_j \mu_j^{1-\lambda/n} \right)} .
\end{align}
Pick $\lambda$ such that $1-\lambda/n = \alpha$, for a chosen $\alpha \in [0,1]$. In other words, $\frac{\lambda}{n} = 1-\alpha$, and factoring this out and substituting,
\begin{align}
 \Pr[S_n \geq t] \leq 2^{-n (1-\alpha) \left(t - \frac{1}{1-\alpha} \log \sum_j \mu_j^{\alpha} \right)} .
\end{align}<|endoftext|>
TITLE: The principal bundle of embeddings
QUESTION [5 upvotes]: In a paper of P. Michor, it was shown that Emb(M,N) is a smooth principal diff(M)-bundle, M and N are smooth locally compact manifolds provided dim M < dim N. My question is why there is a restriction on the dimensions. Does anyone know a reference for the result when dim M = dim N ?
Thanks in advance

REPLY [8 votes]: Then an open subset of $Emb(M,N)$ is $Diff(M)$. Namely, if $M$ is compact, the image of $M$ under an embedding is open and closed, so you have a diffeomorphism onto a connected component. If $M$ is not compact, there are no smooth movements in $Emb(M,N)$ tangential to the image near infinity. If $M$ is compact with boundary, see

MR3263203 Reviewed Gay-Balmaz, François; Vizman, Cornelia Principal bundles of embeddings and nonlinear Grassmannians. Ann. Global Anal. Geom. 46 (2014), no. 3, 293–312.<|endoftext|>
TITLE: Hausdorff convergence of submanifolds in Riemannian manifolds
QUESTION [10 upvotes]: Let $(M^n,g)$ be a smooth compact Riemannian manifold. It is well known that any sequence $\{X_i\}$ of compact subsets of $M$ has a subsequence which converges in the Hausdorff metric to a compact subset $X\subset M$.
Assume now that $\{X_i\}$ are, in addition, smooth connected submanifolds with a uniform lower bound on the sectional curvature and a uniform upper bound on the diameter with respect to the induced intrinsic (!) metric.
Question. What is known about a limit space $X$?
E.g. should $X$ have an integer Hausdorff dimension?
Remark. Under the above assumptions, the Gromov compactness theorem implies that there is a subsequence converging to a compact Alexandrov space $Y$ in the Gromov-Hausdorff sense (thus $Y$ is not a subset of $M$ a priori). Is there any relation between $X$ and $Y$? (I believe there is a 1-Lipschitz map onto $Y\to X$.)

REPLY [8 votes]: You are right about 1-Lipschitz map, but that is about all you can expect.
In particular the dimension of $X$ might be not integer.
Assume your $X_i$ all isometric to to a flat torus $\mathbb{T}$ and
$M=\mathbb{E}^N$; that is your manifold is the Euclidean space of high enough dimensions $N$.
By Nash and Whitney, any short map $\mathbb{T}\to \mathbb{E}^N$ can be arbitrary well approximated by an isometric embedding.
It reamins to construct a short map $\mathbb{T}\to \mathbb{E}^N$ with non-integer Hausdorff dimension. 
An image of that type is shown on the picture.
It can be arranged that the Cantor set on the top has Hausdorff higher than 1 and lower than 2. 
,<|endoftext|>
TITLE: PDE characterisation of elementary symmetric functions?
QUESTION [7 upvotes]: For $k\leq{}n$ the elementary symmetric polynomials are defined by:
$$e_k(x_1,\ldots,x_n)=\sum_{1\leq{}i_1<...<i_k\leq{}n}x_{i_1}\cdots x_{i_k}$$
I believe I can prove (by a complex brute force method) that the function:
$$f(x_1,\ldots,x_n)=e_k(x_1,\ldots,x_n)/(x_1+\cdots+x_n)^k$$
satisfies the following PDE:
$$\left(\sum_{i=1}^nx_i\right)\left(\sum_{i=1}^nx_i\frac{\partial^2f}{\partial{}{x_i}^2}\right)+k(k-1)f=0$$
I have some questions, since both symmetric polynomials and PDEs are out of my area:

Is the above (assuming it is correct) known or easily proven, e.g. is there a reference I could use or a relatively simple proof?
Does PDE theory tell us anything about uniqueness of solutions of the given PDE(s)?

Thanks
Helmut

REPLY [7 votes]: For your first question, here's a simple proof. 

Observe that $\partial_{ii}^2 e_k = 0$ for any $i\in \{1,\ldots,n\}$. This implies 
$$ \partial^2_{ii} (\sigma^k f) = 0 \tag{1}$$
for any $i$, where $\sigma = \sum x_i$. 
Observe that 
$$ \partial_i \sigma = 1 $$
for any $i$, and hence 
$$ \partial_i \sigma^k = k \sigma^{k-1} \tag{2}$$
by the chain rule. 
The product rule applied to (1) using (2) gives
$$ 0 = k(k-1)\sigma^{k-2} f + 2 k \sigma^{k-1} \partial_i f + \sigma^k \partial^2_{ii} f$$
from which we divide by $\sigma^k$ to obtain
$$ \partial^2_{ii} f + k(k-1)\sigma^{-2} f + 2 k \sigma^{-1} \partial_i f = 0 \tag{3}$$
Recall that this holds individually for any $i$. 
Multiply now the expression (3) by $x_i$. And sum over $i$, we get finally
$$ \sum_{i = 1}^n x_i \partial^2_{ii} f + k(k-1) \sigma^{-2} f \underbrace{\sum x_i}_{= \sigma} + 2 k \sigma^{-1} \sum_{i = 1}^{n} x_i \partial_i f = 0 \tag{4} $$
Now observe that $f$ is, by definition, a homogeneous function of degree 0. So the homogeneous derivative $\sum x_i \partial_i f = 0$. This kills the last term in (4). The remainder is exactly what you wanted to show. 


Concerning uniqueness, observe that only two ingredients were used in the derivation of the PDE above:

That the original function $e_k$ satisfies $\partial^2_{ii} e_k = 0$ for any $i$. 
That the original function $e_k$ is homogeneous of degree $k$. 

There are a lot more functions that satisfy the same property. For example, when $n = 3$ and $k = 2$, we know that it works for $e_k = x_1 x_2 + x_2 x_3 + x_3 x_1$. But the same also works for $x_1 x_2$ or $x_2 x_3$ or $x_3 x_1$ or any linear combination thereof.<|endoftext|>
TITLE: a variant of the Kleene tree
QUESTION [6 upvotes]: The (a?) Kleene tree is a computable (a.k.a. decidable) sub-tree of the full binary tree with no computable path. It is well-known.
I need a variant. (For those in the know, I need a c-bar which is not a D-bar.) What I need is also based on a computable labeling of the binary tree. With the Kleene tree, once a binary sequence gets kicked off the tree, all of its descendants are kicked off too, else it wouldn’t be a tree. So you can think of a Kleene tree as a computable labeling of binary sequences by “in” and “out”, where once a sequence is “out” so are all of its descendants. In the tree I need, if a sequence is labeled “out,” a descendant is allowed to be labeled “in”. A sequence is really off the tree when it AND all of its descendants are labeled “out”. This is a $\Pi^0_1$ condition, hence not computable  (at least, not obviously computable). A sequence is in the tree when some descendant is labeled “in”. 
What I need is a tree like that, with no computable paths, not equal to any Kleene tree. Anyone know one? Or how to do it? Presumably some priority argument would suffice.

REPLY [2 votes]: Let $K_s$ be a computable monotone sequence of finite sets whose union is $K$, the halting set. Let $T$ be the tree of all $\{0,1\}$-sequences $\tau$ such that for some $s \geq |\tau|$, $\tau$ is the characteristic function of $K_s \cap \{0,\ldots,|\tau|-1\}$. The tree $T$ is of the type you want and its only infinite path is the characteristic function of $K$. $T$ cannot be a Kleene-type tree because of the Low Basis Theorem.<|endoftext|>
TITLE: Sign problem in a Calogero-Moser system: proof of integrability?
QUESTION [10 upvotes]: Everyone of us had sometimes this awful feeling that some sign is lost in a calculation and that this sign is perturbing some fundamental understanding of what is going on. I feel the same has happened for me today and I can't figure this sign problem out, so I count on you.
A Calogero-Moser system is defined as a Hamiltonian system with a Hamiltonian $$H=\sum p_i^2 + \sum_{i \neq k} \frac{1}{(x_i-x_j)^2}$$ It is widely known that this system is completely integrable.
I am trying to understand this widely known fact.
One of the proofs relies on the relation of the system with a linear flow in the space of matrices, the relationship is nicely explained in this MathOverflow entry: Is the 'massive' Calogero-Moser system still integrable?
The question that I already asked as a comment there is the following:
the standard proof of the integrability rewrites $H$ as a restriction of some other function on the space of matrices which is actually $\mathrm{Tr }Y^2$ for a matrix $Y$ defined by  $$ Y_{ii}=p_i, \; Y_{ik}=(x_i-x_k)^{-1}, \; i\neq k $$. A simple calculation will give us not $H$ but 
$$H^-=\sum p_i^2 - \sum_{i \neq k} \frac{1}{(x_i-x_j)^2}$$
This is exactly the expression Etingof obtains in his Lectures on Calogero-Moser systems,  http://www-math.mit.edu/~etingof/zlecnew.pdf.
Etingof starts from the dynamics on matrix space and defines CM system as its symplectic reduction. So no problems for him.
But for a system of the particles on the real line, I feel lost. How one can prove integrability? And also, $H^-$ is giving the trajectories that would collapse.

REPLY [4 votes]: The answer to my question (provided by BS) is the following:
We have to change the action by looking at the group $G=U(n, \mathbb{C})$ and its action by conjugacy on pairs of Hermitian matrices. The space of pairs of such matrices can be identified with $T^* (Lie U(n, \mathbb{C})^*)$ because $Lie U(n, \mathbb{C})$ consists of antihermitian matrices (so we should only divide by $i$ to obtain Hermitian matrices).
The coadjoint orbit is an orbit of a matrix with ones everywhere except for the diagonal (where it has zeroes). Its orbit is all Hermitian matrices $T$ such that $rk (T+\mathrm{Id})=1$. Moment map is $J(X,Y)=-i[X,Y]$. 
This subtle change in moment map will permit us to change the entries in $Y$ matrix.
And then a representative of each element in the orbit can be chosen in a form $(X,Y)$ where $X$ is a diagonal matrix and $Y_{j k }= \frac{i}{x_j-x_k}$.
So the idea of AHusain to multiply by $i$ was a good one -- but one has to change the action...
Note that this proof (for $H$) is quite the same as a proof for the potential $H^-$: it is related to the fact that they both come from the group $SL(n, \mathbb{C})$:this group has (among others) two real parts: $SL(n, \mathbb{R})$ and $SU(n, \mathbb{R})$. The first part corresponds to $H^-$ and the second to $H^+=H$.<|endoftext|>
TITLE: functions with orthogonal Jacobian
QUESTION [5 upvotes]: I'm working on a model that would require to use vectorial functions of $\mathbb{R}^n \rightarrow \mathbb{R}^n$, such that $\forall x, y \in \mathbb{R}^n$, $\lVert \frac{df(x)}{dx}(y) \lVert_2 = \lVert y \lVert_2$, ie with an orthogonal Jacobian.
I can only think of trivial functions (like $f(x) = Ox + c$ for $O$ orthogonal and $c \in \mathbb{R}^n$).
Are there other functions that verify this property? What would it be if we add the constraint $\forall x \in \mathbb{R}^n$, $\lVert f(x) \lVert_2 = \lVert x \lVert_2$ ?

REPLY [3 votes]: A geometric proof would be the following: First note that a diffeo with orthogonal Jacobian preserves the length of curves, in particular of straight lines, thus it preserves the Euclidean distance.
From this it follows easily that it maps lines to lines. From the fundamental theorem of affine geometry, it then follows that the diffeo is an affine map, whose linear part is clearly be orthogonal. (Or just use at once the result that a map in $R^n$ that preserves Euclidean distance is the composition of a translation with an orthogonal linear map.)<|endoftext|>
TITLE: What structure of a monoidal simplicial model category is preserved by taking the opposite category?
QUESTION [6 upvotes]: Suppose we have $(M,\otimes,1)$, a monoidal simplicial model category. Then we can consider the opposite model category $M^{op}$ with the opposite model structure (fibrations become cofibrations, etc.). This category is also still a monoidal category (the opposite of a monoidal category is canonically a monoidal category). How do all of these possible adjectives interact? Is it still closed? Does the push-out product axiom still hold? It seems like we must lose some things, for instance, if we wanted the axiom "for any cofibrant object $X$ there is an equivalence $1\otimes X\simeq X$" this can't in general be true anymore, can it, since the cofibrant objects are completely different now? 
In general it seems clear that things break down, but I'm interested in precisely what breaks down. Do we still have a simplicial model category, but we're just missing some of the monoidal model category axioms? I'm also okay with dumping the simplicial requirement.

REPLY [5 votes]: The general statement is that if $V$ is a monoidal model category (I will assume the unit object of $V$ is cofibrant, so that there are no funny extra axioms related to the unit) and $M$ is a $V$-model category, then $M^{\mathrm{op}}$ is also a $V$-model category. The enrichment comes from that of $M$, and the roles of the tensor and cotensor are interchanged.
To check that this really defines a $V$-model category structure, perhaps the clearest way is to use the cyclic structure of two-variable adjunctions. The well-known fact that a two-variable adjunction $F : C \times D \to E$ being a Quillen bifunctor can be checked using either cofibrations of $C$ and $D$, or cofibrations of $C$ and fibrations of $E$, or cofibrations of $D$ and fibrations of $E$ (Lemma 4.2.2 of Hovey's book) really says that $F$ is a Quillen bifunctor if and only if its cyclic shifts are. Specializing to the action $\otimes : M \times V \to M$, we get a Quillen bifunctor $V \times M^{\mathrm{op}} \to M^{\mathrm{op}}$ which gives the action of $V$ on $M^{\mathrm{op}}$.
In the situation of the question, this means that if $M$ is a monoidal simplicial model category, then $M^\mathrm{op}$ is a simplicial category, and also an $M$-model category. However, as Zhen Lin mentioned, there is no reason $M^{\mathrm{op}}$ should be a monoidal model category.<|endoftext|>
TITLE: Does a Riemannian manifold have a triangulation with quantitative bounds?
QUESTION [22 upvotes]: Suppose that $M$ is a closed Riemannian manifold with bounded geometry, i.e., curvature between $-1$ and $1$ and injectivity radius at least $1$.  Since $M$ is a smooth manifold, it has a triangulation.  Does it necessarily have a triangulation that is "nice" with respect to the metric?
For instance, is there an $\epsilon>0$ such that for any such $M$, there is a triangulation of $M$ whose simplices are all homeomorphic to the standard simplex by a map $f$ such that $f$ and $f^{-1}$ are both $\epsilon^{-1}$-Lipschitz?
The method I'm familiar with for constructing a triangulation is to embed $M$ in $\mathbb{R}^n$, construct a fine net of points in $M$, construct the Delaunay triangulation of those points, then project back to the manifold to get a triangulation.  But this isn't very quantitative -- it depends on the embedding, and even if the embedding is nice, an unlucky choice of points will lead to some bad simplices.  Is there a better way?

REPLY [5 votes]: A series of papers has been published (or rather is being published) on this topic, staring with Stability of Delaunay-type structures for manifolds (see http://dl.acm.org/citation.cfm?id=2261284) by Boissonnat, Dyer, and Ghosh. I think the paper in the series you might find most useful is Delaunay triangulation of manifolds (see https://arxiv.org/abs/1311.0117).<|endoftext|>
TITLE: Plugging $1-x$ into Schur polynomials
QUESTION [12 upvotes]: I have a symmetric Laurent polynomial $f$ in $k$ variables expressed as a linear combination of Schur polynomials. I'd like to know what happens when I make the substitution $p(x_1,\ldots,x_k)\mapsto p(1-x_1,\ldots,1-x_k)$ and re-expand in the Schur basis. Is there some nice combinatorial description of what happens when you do this?

That's the whole question, but I'll say more about where this comes from in case it's helpful or you're just interested. The symmetric polynomial is an element of the equivariant K-group $K_0^{GL(V)}(\mathrm{Hom}(V,W))$ where $V$ is some $k$-dimensional vector space and $W$ is some $n$-dimensional vector space (so $\mathrm{Hom}(V,W)$ is just a $kn$-dimensional affine space as a variety). This is isomorphic to the ring of symmetric Laurent polynomials in $k$ variables.
I know that my K-class is a linear combination of modules supported on the subvariety of $\mathrm{Hom}(V,W)$ consisting of non-full-rank matrices. I have an explicit but complicated description of this subspace as an ideal in the ring of symmetric polynomials, but after making the $1-x$ substitution, the ideal is exactly spanned by $s_\lambda$ where $\lambda$ doesn't fit in a $k\times(n-k)$ rectangle. So answering the question above would be enough to give a set of $\binom nk$ explicit linear equations cutting out this ideal, which is what I'm after.

REPLY [14 votes]: Let $t$ be an indeterminate. Let
    $\vartheta:\Lambda                                                                 
    \rightarrow \Lambda[t]$
be the specialization (homomorphism) defined by
  $$ \vartheta(p_k)=t+\sum_{i=1}^k {k\choose i}p_i, $$
where $p_i$ is a power sum summetric function. According to item 75 of
http://math.mit.edu/~rstan/ec/ch7supp.pdf, we have
  $$ \vartheta(s_\lambda) =\sum_{\mu\subseteq\lambda}
  \frac{f^{\lambda/\mu}}{|\lambda/\mu|!}\left( \prod_{u\in\lambda/\mu}
   (t+c(u))\right)s_\mu, $$
 where $c(u)$ denotes the content of the square $u$ of the skew shape
    $\lambda/\mu$, and $f^{\lambda/\mu}$ is the number of standard Young tableaux of shape $\lambda/\mu$.
If we restrict to $n$ variables and set $t=n$, then for any symmetric
   function $f$,
   $$ \vartheta(f)=f(x_1+1,\dots,x_n+1), $$
as may be seen by setting $f=p_k$. We need only send $x_i$ to $-x_i$
(also a homomorphism) to get an answer to the question when
   $p=s_\lambda$.
Item 75 is an improved version of Example I.3.10 on page 47 of
I. G. Macdonald, Symmetric Functions and Hall Polynomials,
second ed.
Addendum. Here is a sketch of the proof. Both sides
are polynomials in $t$, so it suffices to prove the result for
$t=n\in\{1,2,\dots\}$. Since
   $$ \vartheta(s_\lambda)(x_1,\dots,x_n) =
    s_\lambda(x_1+1,\dots,x_n+1), $$
we get (using notation from Chapter 7 of Enumerative Combinatorics,
vol. 2)
  \begin{eqnarray*}  s_\lambda(x_1+1,\dots,x_n+1) & = &
  \frac{a_{\lambda+\delta}(x_1+1,\dots,x_n+1)}
  {a_\delta(x_1+1,\dots,x_n+1)}\\
   & = & \frac{a_{\lambda+\delta}(x_1+1,
       \dots,x_n+1)}{a_\delta(x_1,\dots,x_n)}. \end{eqnarray*}
 By expanding the entries of $a_{\lambda+\delta}(x_1+1,\dots,x_n+1)$
 and using the multilinearity of the determinant we get (see
 Example I.3.10 mentioned above)
  $$ s_\lambda(x_1+1,\dots,x_n+1)=\sum_{\mu\subseteq\lambda}
      d_{\lambda\mu}s_\mu, $$
 where
  $$ d_{\lambda\mu} = \det\left( {\lambda_i+n-i\choose \mu_j+n-j}
    \right)_{1\leq i,j\leq n}. $$
 We can factor out factorials from the numerators of the row entries
 and denominators of the column entries of the above
 determinant. These factorials altogether
 yield $\prod_{u\in\lambda/\mu}(n+c(u))$. What remains is exactly the
 determinant for $f^{\lambda/\mu}/|\lambda/\mu|!$ given by
 Corollary 7.16.3 of Enumerative Combinatorics, vol. 2, and the
 proof follows. Is there a more conceptual proof that doesn't involve
 the evaluation of a determinant?<|endoftext|>
TITLE: Does anyone know the classification of fourth order surfaces?
QUESTION [6 upvotes]: Does anyone know the classification of fourth order surfaces? By "fourth order surface" I mean a surface defined by an equation of the form $$f(x, \, y, \, z)=0,$$
where $f$ is a polynomial of degree $4$. 
Edit. The two answers below (J. Silverman and F. Polizzi) concern the case of complex surfaces. Are there known results in the case of real surfaces?

REPLY [5 votes]: The classification of real non-singular quartic surfaces in $\mathbb{RP}^3$ (up to various possible equivalence relations like homeomorphism, isotopy,...) is discussed in 

A. Degtyarev, V. Kharlamov: Topological properties of real algebraic varieties: Rokhlin's way. Uspekhi Mat. Nauk 55 (2000), no. 4(334), 129-212; translation in
Russian Math. Surveys 55 (2000), no. 4, 735–814

See also Sections 3.5.3 and 3.5.4 in the arXiv-version of the paper (as well as the references given there).
Some statements concerning the structure and classification of singular real quartic surfaces are given in:

A. Degtyarev and I. Itenberg: On real determinantal quartics. Proceedings of the Gökova Geometry-Topology Conference 2010, 110–128, Int. Press, Somerville, MA, 2011.<|endoftext|>
TITLE: Why do people say DG-algebras behave badly in positive characteristic?
QUESTION [23 upvotes]: It seems to be a common wisdom in derived algebraic geometry that commutative DG-algebras are not, in general, a good model for derived rings, with the stated reason that they behave badly in positive characteristic. 
What are some examples for this bad behaivor? I know that it is unknown if the category of commutative DG-algebras has a nice model structure, but is there more to this saying?

REPLY [8 votes]: In characteristic zero, for example, it's possible to find cdgas describing the rational cohomology $H^{\bullet}(X, \mathbb{Q})$ of a (say simply connected) topological space; this is the starting point for Sullivan's approach to rational homotopy theory. In positive characteristic there are obstructions to doing this coming from cohomology operations (e.g. the Steenrod operations). It's known that one can fix this problem by using $E_{\infty}$ algebras instead; see, for example, DAG XIII.<|endoftext|>
TITLE: Is platonism regarding arithmetic consistent with the multiverse view in set theory?
QUESTION [15 upvotes]: A "truth" platonist for arithmetic believes, given a statement in the language of arithmetic, that the problem whether the statement is true has a definite answer.
Prof. Hamkins has argued for a multiverse view of set theory. Since different models of ZFC can have a different arithmetic (that is, model of the natural numbers), I wonder whether platonism regarding arithmetic is consistent with the multiverse view.

REPLY [12 votes]: I'm still uncertain of its appropriateness here, but since Joel asked, here is a quote from my book that discusses this issue:

The two kinds of independence … in geometry
  and number theory offer us strikingly different paradigms. In
  both cases there is broad agreement about the correct interpretation of
  the independence results. For instance, no one nowadays would consider it
  meaningful to ask whether the parallel postulate is "really true" in some
  universal sense; it simply holds in some two-dimensional
  geometries and fails in others.
In contrast, although the arithmetical expression of the consistency of $PA$
  is independent of $PA$, it is still widely regarded as true.
  To take a more extreme example, consider the formal system
  $MC$ = $ZFC$ + "measurable cardinals exist". Few would
  suggest that the sentence ${\rm Con}(MC)$ which arithmetically expresses
  that $MC$ is a consistent system might lack a well-defined truth value. Yet
  ${\rm Con}(MC)$ is presumably independent of $PA$, indeed, presumably
  even independent of $ZFC$.
… Should we suppose
  that the continuum hypothesis, for example, has a definite truth value in
  a well-defined canonical model? Or is there a range of models in which the
  truth value of the continuum hypothesis varies, none of which has any
  special ontological priority?
Forcing tends to push us in the latter direction. It creates the impression
  that there is a range of equally valid models of $ZFC$, and that one can always
  pass to a larger model in which the value of $2^{\aleph_0}$ changes … In an influential series of recent papers, Hamkins has vigorously argued for the
  position that there is no canonical model of $ZFC$, a position that he calls "the multiverse view".
… A picture emerges [from discussion omitted here] of a mathematical universe which is composed
  of countable structures that have absolute properties and which includes
  a range of countable models of $ZFC$ in which the truth values of questions
  like the continuum hypothesis can vary. Thus, with regard to
  independence phenomena, if we take "set theory" to be the theory of
  surveyable collections then it has an absolute meaning and behaves like
  number theory, but questions like the continuum hypothesis cannot even be
  posed; if we take it to be the theory of individuals in some model
  of $ZFC$ then it has a variable meaning and behaves like geometry.<|endoftext|>
TITLE: States in C*-algebras and their origin in physics?
QUESTION [21 upvotes]: in $C^*-$algebras with unit element, there is the definition of a state, as a functional $\omega$ with $\omega(e)=||\omega||=1.$
Now, of course there is also in classical physics and quantum mechanics the definition of a state.
In classical physics this is either a point in phase space or more generally  a probability measure on this space.
In quantum mechanics this is either a wavefunction or a density matrix.
Now there are basically two interesting examples of $C^*-$ algebras I would say: $L(H)$ the space of bounded operators on some Hilbert space $H$ (a non-comm. $C^*-$ algebra or $C_0(X)$ the space of $C_0-$ functions on some locally  compact Hausdorff-space (a commutative one).
Obviously, if $X$ is some compact subset of $\mathbb{R}^n,$ then $C(X)$ is a $C^*-$ algebra with unit element and dirac measures on $X$ and more generally probability measures are indeed states as we defined them in the functional analysis sense.
Moreover, if we work on some Hilbert space $H$ then the density matrices $\rho$ define functionals $l:L(H) \rightarrow \mathbb{R}, T \mapsto tr(\rho T).$ So these are also states in the sense of functional analysis. 
But this made me think whether
1.) Every state in the sense of functional analysis can be interpreted as a physical state?
2.) Where does the interpretation come from that the commutative $C^*-$algebra cooresponds to classical mechanics and the non-commutative one to quantum mechanics? Is there any deep interpretation of this fact? (besides the fact that non-commutativity is known to be an issue for QM?)Cause this seems to be much deeper here as this is the only distinguishing fact between the two in this setup.

REPLY [37 votes]: I'd like to try to give a more comprehensive answer.
In the elementary formulation of quantum mechanics, pure states are represented by unit vectors in a complex Hilbert space $H$ and observables are represented by unbounded self-adjoint operators on $H$. The expected value of a measurement of the observable $A$ in the state $v$ is $\langle Av,v\rangle$. We could also say that the state is represented by the linear functional $A \mapsto \langle Av,v\rangle$, and this interpretation generalizes to say that a mixed state is represented by a linear functional $A \mapsto {\rm Tr}(AB)$ where $B$ is a positive trace-class operator satisfying ${\rm Tr}(B) = 1$.
The fact that $A$ can be unbounded is forced on us by basic physical examples like position and momentum. Mathematically, it is easier to work with bounded observables, which can be obtained from unbounded observables via functional calculus: if $f: \mathbb{R} \to \mathbb{R}$ is bounded and measurable then we can give meaning to $f(A)$ as a bounded self-adjoint operator.
In the C*-algebra formulation, bounded observables are self-adjoint elements of a C*-algebra $\mathcal{A}$ and mixed states are positive linear functionals on $\mathcal{A}$ of norm one. The pure states are the extreme points of the set of mixed states.
We can always pass from the C*-algebra formulation to the elementary formulation using the GNS construction: given a state $\phi$ on a C*-algebra $\mathcal{A}$, we can find a Hilbert space $H$, a $*$-representation $\pi: \mathcal{A} \to B(H)$, and a unit vector $v \in H$ such that $\phi(x) = \langle \pi(x)v,v\rangle$ for all $x \in \mathcal{A}$.
Why bother with the C*-algebra formulation, then? Well, maybe you don't need to. But sometimes there are good reasons for looking at things this way.

Superselection sectors. A characteristic feature of quantum mechanical states is that they can be superposed. Schrodinger's cat can be alive, dead, or in some combination of the two states. But there may also be general principles which prevent certain states from being combined in this way. In elementary quantum mechanics, particle number is an unchangable quantity: an isolated system couldn't be partly in a state with one particle and partly in a state with two particles. We could tell how many particles there are in the system without measuring it, by counting how many particles are outside and subtracting this from the total number of particles in the universe. Similarly for charge. (Emphasize: I'm talking about elementary QM now.) Quantities like this produce "superselection sectors", orthogonal subspaces of $H$ which cannot be superposed. This means that certain observables are forbidden. In elementary quantum mechanics you could not have an observable with the property that the result of a measurement could be a state with an indeterminate number of particles. The mathematical condition is that every physical observable has to commute with the observable $N$ which measures the number of particles in the system. The relevant C*-algebra would not be all of $B(H)$, but only the set of operators which commute with some special family of observables (particle number, total charge, etc.) I put this example first because it applies even to the nonrelativistic, finite particle case that the OP cares about, at least if one wants to model a variable number of particles, say.
Thermodynamic states. In quantum statistical mechanics, the physical description of a system can involve macroscopic parameters like temperature which correspond to inequivalent states on the observable C*-algebra. This means that the GNS representations corresponding to one of these states does not include the other state as a unit vector or even a trace class operator (density matrix). The Hilbert space formalism simply models different temperatures with different Hilbert spaces. In some sense this formalism still works, but it does not capture the idea that the same physical system can have different temperatures. The C*-algebra formalism does capture this aspect, because it is the same C*-algebra (a CAR or CCR algebra), being represented in different ways on different Hilbert spaces.
Free fields in curved spacetime. We do not have a good mathematical theory of quantum mechanics plus general relativity, but we do have a good understanding of noninteracting quantum fields against a curved spacetime background. Phenomena like Hawking radiation can be understood rigorously in such a way, for instance. What happens here is that a single physical state may be perceived in radically different ways by different observers. In Minkowski spacetime, for instance, a ground state with no particles according to an inertial observer will be seen by a uniformly accelerating observer as a "thermal bath" of infinitely many particles --- a thermodynamic state of the type discussed in point 2. In this case there is a preferred reference frame, the inertial frame, but in general relativity there is no preferred frame. But, wonderfully, that there is a single C*-algebra (again CAR or CCR) that captures the observables for all observers and all states. The elementary Hilbert space formalism is really inadequate in this case.<|endoftext|>
TITLE: Spectral mapping theorem for polynomials in $z,\overline{z}$ and direct construction of the function calculus for a normal operator
QUESTION [7 upvotes]: Suppose that $A$ is an element in Banach algebra and $p$ is a polynomial. Then we have an equality $p(\sigma(A))=\sigma(p(A))$ where $p(A)$ has an elementary meaning. This theorem (the spectral mapping theorem) is a key point in constructing the (isometric) functional calculus in the context of $C^*$-algebras $f \mapsto f(A)$ for $f \in C(\sigma(A))$ where $A$ is self adjoint. This can be done before we prove theorems such Gelfand-Najmark: in other words even before we know that each commutative unital $C^*$-algebra is of the form $C(X)$ for some compact $X$, we can state that for a self-adjoint element $A$ in some $C^*$-algebra, the $C^*$-algebra generated by $A$ is isomorphic to $C(\sigma(A))$. I wonder if it is possible to give an analogous direct proof for a normal operator, which would involve a similar theorem as spectral mapping theorem but in the context of polynomials in $z$ and $\overline{z}$. This would be enough since every continuos complex valued function on $C(\sigma(A))$ (where $A$ is normal, therefore $\sigma(A)$ need not to be contained in the real line anymore) can be aproximated by such polynomials and if $A$ is normal then $p(A)$ makes sense for such polynomials. Of course I'm interested in a ,,elementary'' proof of such spectral mapping theorem (not using Gelfand Najmark theorem).

REPLY [3 votes]: If I interpret your question correctly, then you want an elementary proof of the fact that if $A$ is normal, then $\sigma(p(A,A^*))=f(\sigma(A))$, where $f(z):=p(z,\overline{z})$.
By looking at approximate eigenvectors, it's easy to see that $f(\sigma)\subseteq \sigma(p)$, so the question really is about the reverse inclusion. I might be wrong, but I'm not sure a completely elementary proof is possible. As I mentioned in my comment, this topic has of course been studied, see for example this paper.
Alternatively, the following argument works. It's not exactly what you are asking for; while it does avoid Gelfand-Naimark, we are using the following facts: (1) The spectrum of an operator $A\in B(H)$ agrees with its spectrum in an arbitrary $C^*$-subalgebra of $B(H)$; (2) if $A\in\mathcal A\subseteq B(H)$, with $\mathcal A$ a commutative $C^*$-algebra, then $\sigma(A)=\{\varphi(A)\}$, where $\varphi :\mathcal A\to\mathbb C$ ranges over the complex homomorphisms; (3) $\varphi(A^*)=\overline{\varphi(A)}$ (establishing this is usually the first step in proving GN)
We can now work in the $C^*$-algebra generated by $A$, which of course also contains $p(A,A^*)$. If now $z\in\sigma(p(A,A^*))$, then $z=\varphi(p(A,A^*))=p(\varphi(A),\overline{\varphi(A)})=f(\varphi(A))\in f(\sigma(A))$, as desired.<|endoftext|>
TITLE: Asymptotics of functional of i.i.d. Rademacher random variables
QUESTION [11 upvotes]: Let $X_1,\ldots, X_n$ be i.i.d. Rademacher random variables. That is, $\operatorname{Pr}(X_i = 1) = \operatorname{Pr}(X_i = -1) = 1/2$. I was wondering if the following argument is true:
$$
\mathbb{E} \exp\biggl( C\cdot \left(\sum_{i=1}^n X_i\right)^4\big/n^3 \biggr) = 1 + O(1/n),
$$
where $C \geq 0$ is a constant.
I did some numerical simulations and the results validated this argument. I would appreciate it if anyone can give a proof.
If my conjecture is correct, this problem reveals an interesting phenomenon of probability theory. In fact, if we replace $\sum_{i=1}^n X_i /\sqrt{n}$ by a standard norm random variable, the expectation is $\infty$. This means that we cannot simply plug in Gaussian approximation here. 
P.S.: As pointed out by Christian Remling, this problem has trivial lower bound $2^{-n} \exp(C n)$. So the expectation will explode when $C > \log 2$. What if $0\leq C \leq \log 2$?

REPLY [4 votes]: I believe the conjecture is true for sufficiently small $C$. Previously I was trying to disprove it using Large deviation theory. But I missed a sign at the last step. But the same argument can be turned around.
In compact form, one gets
$$ \lim \frac{1}{n} \log P(\sum_i X_i > \alpha n) = -I(\alpha),$$
where $I(\alpha) = \sup_\theta (\alpha \theta - \log \mathbb{E} e^{\theta X_1})$. One can compute 
$$I(\alpha) = (\frac{1}{2} - \alpha)\log (\frac{1}{2} - \alpha) + (\frac{1}{2} + \alpha)\log(\frac{1}{2} + \alpha) + \log 2.$$
More precisely, the lower bound (Cramer's inequality) states that for any $\epsilon > 0$, as long as $n$ is sufficient large, we get $P(\sum_i X_i \ge \alpha n) \ge e^{-n (I(\alpha) + \epsilon)}$; see this lecture notes. The upper bound is without the $\epsilon$ and can be proved easily using exponential Tchebyshev. We only need the upper bound here.
Now write the original expectation as $\int_0^1 1 + P(|\sum X_i | > \alpha n) \frac{d}{d\alpha} e^{C \alpha^4 n} d\alpha$, by the so-called integration by parts formula in probability; see page 66 lemma 1.4.30 here. The latter is bounded by
$$ \int_0^1 1 d\alpha + \int_0^1  4 \alpha^3 C n e^{(C \alpha^4 - I(\alpha))n} d\alpha.$$
After evaluating the third moment of a Gaussian of variance $n^{-1}$ near $\alpha = 0$, one sees that the integral near $0$ goes to zero as $C n^{-2}$, and the rest of the integral is clearly very small.
Note this argument doesn't work for Gaussian because the upper bound of integration is $\infty$ in that case. This is a good check of my often reckless arithmetic.<|endoftext|>
TITLE: For which divisors $a$ and $b$ of $n$ does there exist a Latin square of order $n$ that can be partitioned into $a \times b$ subrectangles?
QUESTION [6 upvotes]: There exists a Latin square of order $8$ which can be partitioned into $2 \times 4$ subrectangles:
$$
\begin{bmatrix}
\color{red} 1 & \color{red} 2 & \color{red} 3 & \color{red} 4 & \color{purple} 5 & \color{purple} 6 & \color{purple} 7 & \color{purple} 8 \\
\color{red} 2 & \color{red} 3 & \color{red} 4 & \color{red} 1 & \color{purple} 6 & \color{purple} 7 & \color{purple} 8 & \color{purple} 5 \\
7 & 8 & \color{blue} 1 & \color{blue} 2 & \color{blue} 3 & \color{blue} 4 & 5 & 6 \\
8 & 5 & \color{blue} 2 & \color{blue} 3 & \color{blue} 4 & \color{blue} 1 & 6 & 7 \\
\color{pink} 5 & \color{pink} 6 & \color{pink} 7 & \color{pink} 8 & \color{green} 1 & \color{green} 2 & \color{green} 3 & \color{green} 4 \\
\color{pink} 6 & \color{pink} 7 & \color{pink} 8 & \color{pink} 5 & \color{green} 2 & \color{green} 3 & \color{green} 4 & \color{green} 1 \\
\color{orange} 3 & \color{orange} 4 & \color{brown} 5 & \color{brown} 6 & \color{brown} 7 & \color{brown} 8 & \color{orange} 1 & \color{orange} 2 \\
\color{orange} 4 & \color{orange} 1 & \color{brown} 6 & \color{brown} 7 & \color{brown} 8 & \color{brown} 5 & \color{orange} 2 & \color{orange} 3 \\
\end{bmatrix}
$$
If we take the row-symbol parastrophe of this Latin square (i.e., replace each entry $(i,j,l_{ij})$ with $(l_{ij},j,i)$), then the entry colors define a decomposition of $K_{8,8}$ into $2$-regular spanning subgraphs of $K_{4,4}$.  I would like to generalize this.
Question: For which divisors $a$ and $b$ of $n$ does there exist a Latin square of order $n$ that can be partitioned into $a \times b$ subrectangles?
(Note: We assume $a \leq b$.  Subrectangles must have $b$ symbols.  We don't assume that the boundaries of the subrectangles align.)
Observations:

It's trivially possible when

$a$ divides $b$ (construct a Latin square with blocks that are $b \times b$ subsquares)
$a=1$ or $b=n$.

The first non-trivial case is when $a=2$, $b=3$, and $n=6$.  If my code is correct, then it is impossible (by exhaustive search).  (I'm tempted to think this is just because the parameters are too small.)  The best possible is $4$ subrectangles, which is straightforward to construct.
My code found a random Latin square which gave an $a=2$, $b=5$, and $n=10$ example:
$$
\begin{bmatrix}
\color{red} 7 & \color{red} 1 & 9 & 5 & \color{red} {10} & 4 & 8 & \color{red} 2 & 6 & \color{red} 3 \\
\color{red} 3 & \color{red} {10} & 5 & 8 & \color{red} 2 & 6 & 4 & \color{red} 1 & 9 & \color{red} 7 \\
\hline
\color{blue} 9 & \color{blue} 6 & 4 & 7 & 1 & \color{blue} 2 & \color{blue} {10} & \color{blue} 3 & 5 & 8 \\
\color{blue} 6 & \color{blue} 3 & 7 & 1 & 5 & \color{blue} 9 & \color{blue} 2 & \color{blue} {10} & 8 & 4 \\
\hline
\color{pink} 4 & 5 & 1 & \color{pink} 2 & \color{pink} 8 & 3 & 9 & \color{pink} 7 & \color{pink} {10} & 6 \\
\color{pink} 2 & 9 & 6 & \color{pink} {10} & \color{pink} 4 & 1 & 3 & \color{pink} 8 & \color{pink} 7 & 5 \\
\hline
\color{purple} 1 & 2 & \color{purple} 8 & 3 & 7 & \color{purple} {10} & 6 & \color{purple} 5 & 4 & \color{purple} 9 \\
\color{purple} 8 & 4 & \color{purple} {10} & 6 & 3 & \color{purple} 5 & 7 & \color{purple} 9 & 2 & \color{purple} 1 \\
\hline
\color{brown} 5 & 8 & \color{brown} 2 & 9 & 6 & 7 & \color{brown} 1 & 4 & \color{brown} 3 & \color{brown} {10} \\
\color{brown} {10} & 7 & \color{brown} 3 & 4 & 9 & 8 & \color{brown} 5 & 6 & \color{brown} 1 & \color{brown} 2 \\
\end{bmatrix}
$$

REPLY [5 votes]: I eventually co-authored a paper which includes this topic.  The non-trivial results are:


There exists a Latin square of order $n$ which decomposes into $2 \times (n/2)$ subrectangles for all even $n \not\in \{2,6\}$.
There exists a Latin square of order n which decomposes into $3 \times (n/3)$ subrectangles if and only if $3$ divides $n$ and $n > 9$.
For sufficiently large n, there exists a Latin square of order $n$ which partitions into $d \times (n/d)$ subrectangles if and only if $d$ divides $n$.

Akbari, Marbach, Stones, Wu, Balanced equi-n-squares Electron. J. Combin, 2020. (pdf)

The smallest cases we didn't resolve are

$20 \times 20$ Latin squares into $4 \times 5$ subrectangles;
$28 \times 28$ Latin squares into $4 \times 7$ subrectangles; and
$30 \times 30$ Latin squares into $5 \times 6$ subrectangles.<|endoftext|>
TITLE: When does an infinite model have a proper class-sized elementary extension?
QUESTION [5 upvotes]: Suppose that a set of sentences of a 1st order language has an infinite model $M$.
Under what conditions is there is a proper class-sized elementary extension of $M$?
How does the answer change if we begin with a proper class of sentences?

REPLY [8 votes]: The answer to your main question is that in ZFC there is always such a proper-class elementary-extension. 
Theorem. In ZFC, every set-sized model in a set-sized
first-order language has a proper-class elementary extension.
Proof. This is easiest to see in the case that the global
axiom of choice holds, in other words if there is a class
well-ordering of the universe. So let me first explain that case.
Fix the global well-order and consider any fixed model $M_0$ in a
set-sized first-order language. Using the upward
Löwenheim-Skolem theorem, there is a proper elementary
end-extension of $M_0$, and we may let $M_1$ be the least such
model arising in the well-order. Continuing transfinitely, picking the least elementary-extension at each stage and unions at limit stages, we may
build up an elementary chain $$M_0\prec M_1\prec\cdots\prec
M_\alpha\prec\cdots$$ by a definable procedure whose union will be a proper-class
elementary extension of each of them and in particular of $M_0$,
as desired.
But my next observation is that in ZFC you don't actually need
global choice. If we fix $M_0$, then by the axiom of choice, we
may code $M_0$ by a set of ordinals $A$. Consider the inner model
$L[A]$, which satisfies ZFC and global choice. Since $A$ codes
$M_0$, we may undertake the argument of the previous paragraph
inside $L[A]$ to get a proper-class elementary extension of $M_0$.
In the original universe $V$, then, we get an $A$-definable proper
class elementary-extension of $M_0$, as desired. QED
Your second question, however, can fail in some models. I claim that it is possible to have a
definable proper-class-sized theory $T$ in a model of ZFC, such
that every subset of $T$ has a model, and so in particular the
theory is consistent, but there is no definable (allowing
parameters) model of all of $T$. For example, assume we are
working in a model of ZFC in which there is no definable linear
order. (I explained how to construct such a model in my answer to
Asaf Karagila's question,
Does ZFC prove the universe is linearly orderable?.) Let $T$ be the
theory of a linear order $<$, with a constant symbol $\hat a$ for
every object $a$. Every restriction of $T$ to only set-many
constants will have a model, since in ZFC every set is linearly
orderable, but in this model there is no definable model of all of
$T$, since there is no definable linear ordering of the universe.<|endoftext|>
TITLE: Sample integer points of cross-polytope uniformly
QUESTION [6 upvotes]: For $r,d\in\mathbb{N}$, let 
$$C_{r,d}=\{x\in\mathbb{Z}^d: \|x\|_1\le r\}\subset\mathbb{Z}^d$$ 
be the set of integer points of the $d$-dimensional cross-polytope with radius $r$. 
What is (currently) the fastest way to sample from $C_{r,d}$ uniformly? Is there a method which is polynomial in $d$ and $\log(r)$?

REPLY [2 votes]: Look at page two of Kannan and Vempala.<|endoftext|>
TITLE: Orthonormal bases on Reproducing Kernel Hilbert Spaces
QUESTION [7 upvotes]: Recall that a Hilbert space $\mathcal{H}$ is a reproducing kernel Hilbert space (RKHS) if the elements of $\mathcal{H}$ are functions on a certain set $X$ and for any $a\in X$, the linear functional $f\mapsto f(a)$ is bounded on $\mathcal{H}$. By Riesz Representation Theorem, there exists an element $K_a\in\mathcal{H}$ such that 
$$f(a) = \langle f, K_a\rangle\ \text{ for all } \ f\in\mathcal{H}.$$
The function $K(x,y) = K_y(x) = \langle K_y, K_x\rangle$ defined on $X\times X$ is called the reproducing kernel function of $\mathcal{H}$.
It is well known and easy to show that for any orthonormal basis $\{e_m\}_{m=1}^{\infty}$ for $\mathcal{H}$, we have the formula
$$K(x,y) = \sum_{m=1}^{\infty}e_m(x)\overline{e_m(y)},\tag{Eqn 1}$$
where the convergence is pointwise on $X\times X$.
My question concerns the converse of the above statement.
Question: if $\{g_m\}_{m=1}^{\infty}$ is a sequence of functions in $\mathcal{H}$ such that $$K(x,y) = \sum_{m=1}^{\infty}g_m(x)\overline{g_m(y)}\tag{Eqn 2}$$ for all $x,y\in X$. Is the sequence $\{g_m\}_{m=1}^{\infty}$ an orthonormal basis for $\mathcal{H}$?
The answer to this question is clearly negative since equation (Eqn 1) can be re-written as
$$K(x,y) = \frac{e_1(x)}{\sqrt{2}}\overline{\frac{e_1(y)}{\sqrt{2}}}+\frac{e_1(x)}{\sqrt{2}}\overline{\frac{e_1(y)}{\sqrt{2}}}+\sum_{m=2}^{\infty}e_m(x)\overline{e_m(y)}$$
and clearly $\{e_1/\sqrt{2}, e_1/\sqrt{2}, e_2, \ldots\}$ is not an orthonormal basis for $\mathcal{H}$. So the following additional condition should be added: the sequence $\{g_m\}_{m=1}^{\infty}$ is linearly independent.
The following proof suggests that the answer is affirmative. (For those who are familiar with the proof of the Moore-Aronszajn's Theorem in the theory of RKHS, the proof here looks similar.)
Assume that we have (Eqn 2) and the sequence $\{g_m\}_{m=1}^{\infty}$ is linearly independent. Let $\mathcal M$ be the linear space spanned by the functions $\{g_m\}_{m=1}^{\infty}$. Define a sesquilinear form on $\mathcal M$ as
\begin{align*}
\left\langle\sum_{\text{finite sum}}a_jg_j, \sum_{\text{finite sum}}b_kg_k\right\rangle_{\mathcal M} = \sum_{\text{finite sum}} a_j\overline{b}_j.
\end{align*}
Since $\{g_m\}_{m=1}^{\infty}$ is a linearly independent set, the above definition is well-defined. Note that $\{g_m\}_{m=1}^{\infty}$ is an orthonormal set in $\langle,\rangle_{\mathcal M}$. For any $f\in\mathcal M$ and $x\in X$, we have
\begin{align*}
f(x) = \sum_{\text{finite sum}}\langle f,g_m\rangle_{\mathcal M}\,g_m(x).
\end{align*}
Cauchy-Schwarz's inequality gives
\begin{align*}
|f(x)| & \leq \Big(\sum_{\text{finite sum}}|\langle f,g_m\rangle_{\mathcal M}|^2\Big)^{1/2}\Big(\sum_{\text{finite sum}}|g_m(x)|^2\Big)^{1/2} \leq \|f\|_{\mathcal M}\sqrt{K(x,x)}.
\end{align*}
Let $\widetilde{\mathcal M}$ be the Hilbert space completion of $\mathcal M$. The standard argument shows that $\widetilde{\mathcal M}$ is a RKHS of functions on $X$. What is the kernel of $\widetilde{\mathcal M}$? Since $\{g_m\}_{m=1}^{\infty}$ is an orthonormal set and its span is dense in $\widetilde{\mathcal M}$, it is an orthonormal basis for $\widetilde{\mathcal M}$. The kernel of $\widetilde{\mathcal M}$ then can be computed as $$\sum_{m=1}^{\infty}g_m(x)\bar{g}_m(y),$$ which is the same as $K(x,y)$. Therefore, $\widetilde{\mathcal M}$ is the same as $\mathcal H$ (they consist of the same functions and the inner products on the two spaces are equal). Consequently, $\{g_m\}_{m=1}^{\infty}$ is an orthonormal basis for $\mathcal{H}$. This completes the proof.
Counterexample: On the other hand, there are counterexamples that provide a negative answer to the question in the infinite dimensional case. 
What part of the above proof is incorrect? I have checked but could not figure out what went wrong.
Thank you.

REPLY [7 votes]: The error is in this line:

The standard argument shows that $\widetilde{\mathcal{M}}$ is an RKHS of functions on $X$.

In fact, this is not generally true.  The completion $\widetilde{\mathcal{M}}$ may not be naturally identified with a space of functions on $X$.
The "obvious" way that one would try to prove this is as follows.  Consider an element $\phi \in \widetilde{\mathcal{M}}$.  Since $\mathcal{M}$ is dense in its completion, there is a sequence $\{f_n\} \subset \mathcal{M}$ such that $f_n \to \phi$ in $\widetilde{\mathcal{M}}$-norm, which is an extension of the $\mathcal{M}$-norm.  In particular, $\{f_n\}$ is $\mathcal{M}$-norm Cauchy.  Because of your inequality $|f(x)| \le \|f\|_{\mathcal{M}} \sqrt{K(x,x)}$ (*), we have that $\{f_n(x)\}$ is Cauchy in $\mathbb{R}$ for each $x$.  So $\{f_n(x)\}$ converges to a number which we may call $f_\phi(x)$; that is, $f_n \to f_\phi$ pointwise.  It is also easy to show that $f_{\phi}$ does not depend on the choice of sequence $f_n \to \phi$, so the linear map $\phi \mapsto f_\phi$ of $\widetilde{\mathcal{M}}$ into $\mathbb{R}^X$ is well defined.  Let's call this map $T$.
So the "obvious" thing to do is to look at the image of $T$, which is of course a function space, and make it a Hilbert space by pushing forward the $\widetilde{\mathcal{M}}$-norm.  If this works, then the same inequality (*) will show that the evaluation map is continuous in this norm, and we have ourselves an RKHS.
The problem is that $T$ might fail to be injective.   In other words, we could have a nonzero $\phi$ for which $f_\phi = 0$.  In that case, pushing forward the norm will not work; it will not be well defined on the image of $T$.
This is exactly what happens in the counterexample you discuss in the comments.  As you suggest, consider $\ell^2$ as an RKHS on $X = \mathbb{N} = \{1,2,\dots\}$, with its usual orthonormal basis $\{e_n\}$, and let $v = \sum_n \frac{1}{n} e_n$.  Let $H = \{v\}^\perp$ with the same $\ell^2$ inner product; being a closed subspace of $\ell^2$, $H$ is an RKHS.
Let $P : \ell^2 \to H$ be the orthogonal projection and let $g_n = P e_n$.  The $g_n$ are linearly independent; if $0 = a_1 g_1 + \dots + a_n g_n = P(a_1 e_1 + \dots + a_n e_n)$ then $a_1 e_1 + \dots + a_n e_n$ is a scalar multiple of $v$, which is only possible if all $a_i$ are 0.
Following your construction, let $\mathcal{M} \subset H$ be the linear span of $\{g_n\}$ and let $\langle\cdot, \cdot\rangle_{\mathcal{M}}$ be the inner product on $\mathcal{M}$ which makes the $\{g_n\}$ orthonormal.
Set $h_m = \sum_{n=1}^m \frac{1}{n} g_n$.  Clearly the sequence $\{h_m\}$ is Cauchy in $\mathcal{M}$, so it converges in the completion $\widetilde{\mathcal{M}}$ to some $\phi$.  It is also clear that $\|\phi\|_\widetilde{\mathcal{M}}^2 = \frac{\pi^2}{6}$ so in particular $\phi \ne 0$.
But on the other hand, we have $h_m = \sum_{n=1}^m \frac{1}{n} P e_n = P\left(\sum_{n=1}^m \frac{1}{n} e_n\right)$.  By continuity of $P$, we have $h_m \to Pv = 0$ in $\ell^2$, and thus also pointwise.  So $f_\phi = 0$.

This is basically an example of a somewhat paradoxical fact that's bitten me before.  Let $X,Y$ be Banach spaces, and suppose $E \subset X$ is dense.  It's well known that every bounded operator $T : E \to Y$ has a unique bounded extension $\tilde{T} : X \to Y$.  But it's possible that $T$ is injective while $\tilde{T}$ is not.  Indeed, we've just constructed an example, by letting $E = \mathcal{M}$, $X = \widetilde{\mathcal{M}}$, $Y = \ell^2$, and $T : E \to Y$ the inclusion map.<|endoftext|>
TITLE: What are the basic possibilities for a tensor product of two fields?
QUESTION [10 upvotes]: Let $k$ be a field, with $F,k'$ field extensions of $k$.  The ring $k' \otimes_k F$ is denoted by $F_{k'}$.  In Borel's Linear Algebraic Groups, it is claimed (I believe erroneously) that "each of [$F_{k'}$'s] prime ideals is minimal."  Indeed, by a result due to Grothendieck, the Krull dimension of $F_{k'}$ is the minimum of the transcendence degrees of $F/k, k'/k$.  So there may be a tacit assumption that $F/k$ is algebraic, I am not sure.  In any case, Borel claims results for three possibilities:

(a) $k'$ is separable algebraic over $k$.
Then $F_{k'}$ is reduced, but it may have more than one prime ideal.
(b) $k'$ is algebraic and purely inseparable over $k$.
Then $F_{k'}$ has a unique prime ideal (consisting of the nilpotent elements), but $F_{k'}$ need not be reduced.
(c) $k'$ is purely transcendental over $k$.
Then $F_{k'}$ is clearly an integral domain.

I was wondering if anyone might be able to provide references or proofs for any of the statements (a), (b), (c) as well as examples, e.g. an example where $F_{k'}$ has more than one prime ideal.  Unfortunately, the reference Borel gives is an old algebraic geometry conference from the 1950s which I have had a lot of difficulty understanding.

REPLY [8 votes]: Looks like nfdc23 has explanations for (a),(b), and (c). 
But: indeed, primes of $F_{k'}$ are not in general minimal if $F/k$ is not algebraic. Let $k'=k(x)$ and $F=k(y)$ be transcendental extensions of $k$. 
Then $F_{k'}$ identifies with a subalgebra of the field $k(x,y)$, hence is an integral domain [assertion (c)] - i.e. $\{0\}$ is prime.
There is a homomorphism $\phi:F_{k'} = k' \otimes_k F \to k(t)$ - where $t$ is again transcendental - which on pure tensors is given by $f(x) \otimes g(y) \mapsto f(t) \cdot g(t)$. And then $x\otimes 1 - 1\otimes y$ is a non-zero element of $P = \ker \phi$, hence $P$ is a prime ideal which is not minimal.

REPLY [5 votes]: For a), if $L$ is a finite Galois extension of $k$ with Galois group $G$, then
$$L \otimes_k L \cong \prod_{g \in G} L$$
has $|G|$ prime ideals. For b), if $k = \mathbb{F}_p(a)$ and $L = k[x]/(x^p - a)$, then
$$L \otimes_k L \cong L[x]/(x^p - a) \cong L[x]/(x - \sqrt[p]{a})^p$$
is not reduced. I haven't thought about c).<|endoftext|>
TITLE: A dictionary of Characteristic classes and obstructions
QUESTION [57 upvotes]: I apologize in advance as this is not a research level question but rather one which could benefit from expert attention but is potentially useful mainly to novice mathematicians. 
In an effort to get a hold of characteristic classes I had the idea of compiling a short dictionary relating characteristic classes to their obstructions. Unfortunately I didn't find anything of this sort on the web.
It could be a nice thing If we could compile such a list here. 

Let $E \to B$ be a real vector bundle over a compact manifold (for simplicity):

Euler class ($E$ orientable): $E \to B$ has a nowhere vanishing section $ \implies e(E)=0 $.
Stiefel-Whitney classes: 

$w_1(E)=w_1(\det E)=0 \iff E$ orientable. 
$w_1(E) =w_2(E) = 0 \iff E$ has spin structure.
$E$ has a trivial subbundle of rank $m$ $\implies$ $w_k=0$ for all $k>rank(E)-m$.
$E$ orientable $\implies$ $w_{top} (E) = e(E) \text{ mod 2}$

Pontryagin classes:

For $E$ spin vector bundle: $\frac{1}{2} p_1(E)=0 \iff E$ has string structure.
For $E$ string vector bundle: $\frac{1}{6}p_2(E)= 0 \iff E$ has 5-brane structure.
If $rank(E)$ is even: $e(E) \cup e(E) = p_{top}(E)$ 

Chern classes: Suppose $E \to B$ is now a complex vector bundle.

$E$ has a trivial complex subbundle (or is it quotient bundle here?) of rank $m$ $\implies$ $c_k=0$ for all $k>rank(E)-m$.
$c_i(E)=w_{2i}(E_{\mathbb{R}}) \text{ mod 2}$.
$c_1(E) = c_1(\wedge^{top} E) = 0 \iff E$ has reduction of structure group to $SU$. I read in several places that this has something to do with the possible number of linearly independent parallel spinors - notice $w_2( E_{ \mathbb{R}}) = c_1(E) = 0$ so $E$ is spin in particular.
$c_{top}(E)=e(E_{\mathbb{R}})$

Todd class: ?
Chern character: ?
Wu class: ?

Additions and corrections are welcome.

REPLY [12 votes]: The following classes are of a slightly different flavour because they depend on the additional choice of a connection.
Assume that $E\to B$ carries a flat connection $\nabla$. Then the Kamber-Tondeur classes are obstructions against the existence of a $\nabla$-parallel
metric on $E$. In the case of a complex line bundle, the first Kamber-Tondeur class is the only obstruction.
The Cheeger-Simons differential characters of a vector bundle $E\to B$ with connection $\nabla$ are obstructions against a parallel trivialisation. For a complex line bundle, the first Cheeger-Simons class is the only obstruction (in fact, this class classifies complex line bundles with connections).
Note that the Kamber-Tondeur classes can be interpreted as the imaginary parts of the Cheeger-Simons differential characters.<|endoftext|>
TITLE: Milnor-Witt K-theory for finite fields
QUESTION [5 upvotes]: Are $K^{MW}_*(\mathbb{F_q})$ and $K^{MW}_n(\mathbb{F_q})$ already known? Where can I read about it?

REPLY [8 votes]: I wanted to leave this as a comment, but I don't have enough reputation points yet.
There is a short exact sequence for every $n$
$$ 0 \to I^{n - 1}(k) \to K_n^{MW}(k) \to K_n^M(k) \to 0 $$
This is proven independently of Morel's article by Gille, Scully, Zhong in "Milnor-Witt $K$-groups of local rings.", see Def.3.7, Thm.3.8, Thm.5.4.
The Milnor $K$-theory of a finite field appears in Milnor's original article "Algebraic $K$-theory of quadratic forms" as Example 1.5.
For the Witt groups and powers of augmentation ideals, see page 36, 37 and Theorem 3.5 in Lam's book "Introduction to quadratic forms over fields".
The culmination of these is:
$$ K^{MW}_{\geq 0}(\mathbf{F}_q) = (\mathbb{Z} \oplus \mathbb{Z} / 2) \oplus (\mathbb{F}_q^*) \oplus 0 \oplus 0 \dots$$
and for $n < 0$ we have 
$$K^{MW}_n(\mathbf{F}_q) = \mathbb{Z} / 4$$ if $q$ is 3 mod 4, and $$K^{MW}_n(\mathbf{F}_q) = \mathbb{Z} / 2[\epsilon] / \epsilon^2$$ if $q$ is 1 mod 4.<|endoftext|>
TITLE: Polynomials with the same values set on the unit circle
QUESTION [31 upvotes]: Assume that $P(z)$, $Q(z)$ are complex polynomials such that $P(S)=Q(S)$, where $S=\{z\colon |z|=1\}$ (equality is understood in the sense of sets, but I do not know the answer even for multisets). Does it follow that there exist polynomial $f(z)$,  positive integers $m,n$ and complex number $w\in S$ such that $P(z)=f(z^n)$, $Q(z)=f(wz^m)$?
It is motivated by this question (and if above claim is true, it actually implies much more than asked therein.) I started a new question with algebraic geometry tag, since it looks reasonable and may pay attention of right people rather than comments to an old post.

REPLY [8 votes]: One can also argue algebraically.
First of all, the unit circle $S^1$ in $\mathbf C$ is a real algebraic curve in $\mathbf R^2$. Its complexification $S^1_{\mathbf C}$ in $\mathbf C^2$ is the complex projective line minus two complex conjugate points. A convenient model for $S^1_{\mathbf C}$ is $\mathbf{P}_{\mathbf C}^1\setminus\{0,\infty\}$, with complex conjugation acting as $z\mapsto1/\bar z$. The algebraic endomorphisms of the latter are the endomorphisms of the form $z\mapsto wz^m$ with $w$ a nonzero complex number and $m$ an integer. Such an endomorphism commutes with complex conjugation if and only if $|w|=1$. This will be useful below.
Another useful fact is that the set of algebraic morphisms from $S_{\mathbf C}^1$ into $\mathbf C^2$ that commute with complex conjugation can be identified with the $\mathbf C$-algebra $\mathbf C[z,1/z]$ of Laurent polynomials. Indeed, both coincide with the set of real polynomial maps from $S^1$ into $\mathbf R^2=\mathbf C$.
Now, the complex polynomial $P\in\mathbf C[z]$ is a real polynomial endomorphism of $\mathbf R^2$, and complexifies to a complex polynomial endomorphism $P_{\mathbf C}$ of $\mathbf C^2$. Assuming that $P$ is nonconstant, the image of $S_{\mathbf C}^1$ by $P_{\mathbf C}$ is an affine complex algebraic curve in $\mathbf C^2$ since $P$ is a polynomial. Moreover, it is real in the sense that it is stable for complex conjugation on $\mathbf C^2$. To put it otherwise, it is the complexification $C_{\mathbf C}$ of a real algebraic curve $C$ in $\mathbf R^2$. The curve $C$ contains the real curve $P(S^1)$, but is not necessarily equal to it. The complement of $P(S^1)$ in $C$ is a finite set of points. The normalization $\tilde C_{\mathbf C}$ of $C_{\mathbf C}$ is an open affine subset of ${\mathbf P}_{\mathbf C}^1$. Since $P$ is a polynomial, the complement of $\tilde C_{\mathbf C}$ in ${\mathbf P}_{\mathbf C}^1$ is a doubleton. We may assume that $\tilde C_{\mathbf C}$ is ${\mathbf P}_{\mathbf C}^1\setminus\{0,1\}$. The complex conjugation on $C_{\mathbf C}$ induces a complex conjugation on ${\mathbf P}_{\mathbf C}^1\setminus\{0,1\}$. Since $P(S^1)$ is compact, this complex conjugation is the same as the one above, i.e., there is an isomorphism from $\tilde C_{\mathbf C} $ to $S_{\mathbf C}^1$ that commutes with complex conjugation. Using the facts recalled above, it follows that there are a complex Laurent polynomial $f\in\mathbf C[z,1/z]$ and a nonzero integer $n$ such that $P(z)=f(z^n)$ on $S^1$, and $f_\mathbf{C}$ is a birational morphism from $S_{\mathbf C}^1$ to $C_{\mathbf C}$. Of course, one may assume that $n$ is a natural number. The Laurent polynomial $f$ is then a true polynomial. Since $f_{\mathbf C}$ is birational, one has a rational map $f_{\mathbf C}^{-1}\circ Q_{\mathbf C}$ of $S_{\mathbf C}^1$ into itself. This rational map is a true endomorphism since $S_{\mathbf C}^1$ is nonsingular, $f_{\mathbf C}$ is the normalization morphism from $S_{\mathbf C}^1$ to $C_{\mathbf C}$ and $Q$ is a polynomial. It follows that there are a nonzero integer $m$ and a complex number $w$ with $|w|=1$ such that $f^{-1}\circ Q(z)=wz^m$ on $S^1$, i.e. $Q(z)=f(wz^m)$ on $S^1$. Of course, the integer $m$ is natural and one has $Q(z)=f(wz^m)$ on $\mathbf C$.<|endoftext|>
TITLE: Atiyah-Guillemin-Sternberg convexity theorem
QUESTION [5 upvotes]: I would like to study the Atiyah-Guillemin-Sternberg convexity theorem: proof and applications. I am already familiarised with hamiltonian actions, moment maps...and with elementary Morse theory.
So my problem is to find a detailed proof of this theorem:

What are the prerequisits: Morse-Bott functions, equivariant Darboux
Theorem...?
Is the original proof by Atiyah different from the Guillemin-Sternberg's proof?
What is "the best reference" for a detailed treatment of this theorem?

Thanks for any help.

REPLY [4 votes]: Thanks Thomas, Liviu and Olga,
The Atiyah's proof is done by induction on the dimension of the torus. If $\mu$ is the moment map and

$A_m$: the level sets of $\mu$ are connected, for any $\mathbb{T}^m$ hamiltonian action.
$B_m$: the image of $\mu$ is convex, for any $\mathbb{T}^m$ hamiltonian action.

The hard part (for me) is $A_1$ which is based on the connectedness of the levels of a Morse-Bott function on a compact manifold!
The rest of the proof is very well explained in:

Ana Cannas da Silva, Lectures on Symplectic Geometry (as exercises),
Michèle Audin: Topology of torus actions on symplectic manifolds,
http://www.math.nyu.edu/~kessler/teaching/group/convexity.pdf

The book by Liviu Nicolaescu is very useful and the complete proof can be found in:
McDuff & Salamon, Introduction to Symplectic Topology.<|endoftext|>
TITLE: Wanted, dead or alive: Have you seen this curve? (circular variant of cardioid)
QUESTION [36 upvotes]: Let me start with the context.  This is definitely not a "research level" question, but I'm hoping that the research community will be able to settle for me whether or not a particular construction has been done before or not.  I'm also hoping that the wider context will mean that this question doesn't get closed instantaneously.
I do an extension class with 10/11 year olds, not selected for their interest in mathematics (ie it's just a regular class).  One of the activities I get them to do is to draw a cardioid by joining points on a circle with straight lines.  One pair decided that instead of using straight lines, they would use curves instead.  The resulting picture is similar to a cardioid, but is not the same.
My goal is to be able to say to them one of two things:
Either:

"Well done!  You took the construction and played with it, seeing what else you could make.  That's exactly what mathematicians do.  What you've discovered is called the hyperlogical antiometric cardiobloid and it was first made in 1832 by de Grundie.  It's linked to hyperlogical geometry which is a really key area of mathematics."

Or:

"Well done!  You took the construction and played with it, seeing what else you could make.  That's exactly what mathematicians do.  What you've discovered is quite probably completely new!  No-one I've asked has heard of it.  That's brilliant!  What would you like to call it?"

But to say either one, I need to know if this has been seen before.  Note that it doesn't matter what the answer is, and also please note that I'm not claiming that the curve itself has any intrinsic value.  Its value is purely extrinsic in that I can use it to show the students what being a real mathematician is like.
Without further waffle, here's a more precise construction.
On a circle, draw an arc between $\theta$ and $2 \theta$ so that the arc meets the circle at right-angles (in the manner of the Poincaré disc model of hyperbolic geometry - indeed, if this has been seen before then "curve stitching in hyperbolic space" seems a likely candidate).  This draws out an envelope, just as the straight-line version draws a cardioid.  In actual fact, one can draw the full circles - not just the arcs - and get another part of the curve outside the original circle.
Using the theory of envelopes, and sticking stuff into Sagemath, I found an equation of the curve:
$$
\begin{align}
x^8 + y^8 &+ 16x^7 + 4(x^2 + 4x + 19)y^6 + 76x^6 + 48x^5 \\
&+ 6(x^4 + 8x^3 + 38x^2 + 8x - 47)y^4 - 282x^4 + 48x^3\\
&+ 4(x^6 + 12x^5 + 57x^4 + 24x^3 - 141x^2 + 12x + 19)y^2 + 76x^2 + 16x + 1 = 0
\end{align}
$$
Using Geogebra, I drew a diagram showing the curve and its construction:

Any pointers will be gratefully received, (but please note that I no longer have access to paywalled articles (nor to MathSciNet) so if you supply references please try to find accessible ones).

REPLY [12 votes]: As per the original poster's suggestion, here is a slightly expanded version of what I posted as a comment:
If we consider the envelope of the family of lines $L_\theta$ which connect the points with arguments $\theta$ and $2\theta$ on the unit circle, we get a cardioid, as explained in the question.  (This fact is, for example, mentioned on this page where it is attributed to Cremona.)
Now consider the map $\psi$ from the closed unit disk to itself which takes the Beltrami-Klein model of the hyperbolic plane to the Poincaré disk model of the same (i.e., which takes a point $P$ of the disk to the Poincaré disk projection of the point of the hyperbolic plane whose Beltrami-Klein projection is $P$): in polar coordinates, this keeps $\theta$ unchanged and takes $r$ to $r/(1+\sqrt{1-r^2}) = (1-\sqrt{1-r^2})/r$, and it can be described geometrically in various ways.
The key facts we need are that (1) $\psi$ is a diffeomorphism on the open unit disk, a homeomorphism on the closed unit disk, and is the identity on the unit circle (i.e., ideal points of the hyperbolic plane are mapped in the same way by the Beltrami-Klein and Poincaré disk models), and (2) lines of the hyperbolic plane become line segments in the Beltrami-Klein model whereas in the Poincaré disk model they become arcs of circle perpendicular to the unit circle ("circle at infinity") at the two points where they cross it.
In particular, $\psi$ maps the segment of $L_\theta$ inside the disk to the arc of the circle $C_\theta$ that crosses the unit circle perpendicularly at the arguments $\theta$ and $2\theta$.  Since we consider the envelope of the $C_\theta$ and since $\psi$ preserves tangency in the open disk, the inner part of the curve is the image under $\psi$ of the cardioid parametrized as above.  As for the outer part, it can be described as the inversion by the unit circle of the inner part (since inversion preserves the circles $C_\theta$).
(This also suggests that maybe the nicest curve to consider is the hyperbolic one — an envelope of a family of hyperbolic lines — whose Beltrami-Klein projection is the cardioid and whose Poincaré disk projection is the one which the question was about.)
Bonus: Since I think the Beltrami-Klein model of the hyperbolic plane (and its analogy with the gnomonic projection of the sphere) deserves to be better known, I can't resist adding a couple of shameless self-plugs here for which I hope I will be forgiven: this video shows a number of azimuthal projections of the spheres and their analogous projections of the hyperbolic plane, and this JavaScript game is a maze-without-walls in a (compact quotient of) the hyperbolic plane which can be drawn in the Beltrami-Klein or Poincaré disk projections.<|endoftext|>
TITLE: Example of a non-Kähler manifold with varying plurigenera
QUESTION [7 upvotes]: Let $X \stackrel{\pi}{\to} \mathbb{D}$ be a proper holomorphic family with fibres $X_t = \pi^{-1}(t)$. Siu proved, when the $X_t$'s are projective, that the plurigenera $h^0(X_t, mK_{X_t})$ are constant. It is conjectured that the same will hold true when the fibres $X_t$ are Kähler. 
Are there examples known where this fails when the fibres are not Kähler?
My usual example of a compact non-Kähler manifold is the Hopf surface, but if the fibres of such holomorphic family are all Hopf surfaces, then the plurigenera are all zero (in particular, they are constant).

REPLY [6 votes]: The first example of this phenomenon was discovered by Iku Nakamura in his 1975 paper Complex parallelisable manifolds and their small deformations, J. Differential Geom. 10 (1975), 85-112.
The manifold $X$ is a $3$-dimensional solvmanifold, and he writes down its small deformations explicitly, see his Theorem 2 and the discussion on pages 96-99.<|endoftext|>
TITLE: Weak convergence in random measures
QUESTION [6 upvotes]: I don't understand the following as I read along a proof in a paper (Page 66, "Asymptotic Behaviour of some interacting systems", by Sylvie Meleard):

We denote by $\mathcal{P}({M})$ the space of probability measures on a metric space $M$, equipped with the weak topology.
Let $E$ be a metric space. Let $\{ \mu_n \}$ be a sequence of random measures on $E$, i.e. for each $n$, $\mu_n$ is a $\mathcal{P} (E)$-valued random variable. Also, let $\mathbb{Q}$ be a deterministic probability measure on the same probability space. We can then treat $\delta_{\mathbb{Q}}$ be a constant $\mathcal{P}(E)$-valued random variable.
Since both $\text{Law} ( \mu_n) $ and $\text{Law} ( \delta_{\mathbb{Q}})$ are measures on $\mathcal{P}(E)$, the paper defines that $\mu_n$ converges in law to $\mathbb{Q}$ if
$$\text{Law} ( \mu_n) \implies \text{Law} ( \delta_{\mathbb{Q}}). \quad \quad \quad \quad \, \, \, \, (*)$$
However, in the paper, the fact that $\mu_n$ converges in law to $\mathbb{Q}$ is concluded by establishing that
$$\mathbb{E} \bigg[ \bigg| \int_E f \,d \mu_n - \int_E f \,d \mathbb{Q} \bigg| \bigg] \rightarrow 0, \quad \quad \quad (**)$$
for all continuous bounded functions $f$ on $E$.

By definition of $(*)$, this is equivalent to saying that
$$ \int_{\mathcal{P}(E)} f \, d\text{Law} ( \mu_n) \rightarrow \int_{\mathcal{P}(E)} f \, d \text{Law} ( \delta_{\mathbb{Q}}),$$
for all continuous bounded functions $f$ on $\mathcal{P}(E)$. How does this follow from $(**)$? Any ideas?

REPLY [4 votes]: Let $F : \mathcal{P}(E) \to \mathbb{R}$ be bounded and continuous, and let $M := \sup |F|$.  In adjusted notation, we wish to show $\mathbb{E} F(\mu_n) \to \mathbb{E} F(\delta_{\mathbb{Q}}) = F(\mathbb{Q})$.  To save me some typing,  let's suppose without loss of generality that $F(\mathbb{Q}) = 0$.
Fix $\epsilon > 0$.  By continuity of $F$, there is a weakly open set $U \subset \mathcal{P}(E)$, containing $\mathbb{Q}$, such that if $\nu \in U$ then $|F(\nu)| < \epsilon$.  By definition of the weak topology, there are finitely many bounded continuous $f_1, \dots, f_k : E \to \mathbb{R}$ and $\delta > 0$ such that if $\left| \int f_i\,d\nu - \int f_i \,d\mathbb{Q}\right| < \delta$ for $i = 1,\dots,k$ then $\nu \in U$.
Now for each $i$ we have from (**) that $\mathbb{E} \left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \to 0$.  By Chebyshev's inequality we therefore have $\mathbb{P}\left(\left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) \to 0$.  So we may find a large enough $N$ such that for all $n \ge N$ and all $i = 1,\dots,k$ we have $\mathbb{P}\left(\left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) < \epsilon/k$.  For such $n$, a union bound gives $\mathbb{P}\left(\exists i : \left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) < \epsilon$.  In particular, for $n \ge N$ we have $\mathbb{P}(\mu_n \notin U) < \epsilon$.  
Now we can write
$$|\mathbb{E} F(\mu_n)| \le |\mathbb{E}[F(\mu_n) ; \mu_n \in U]| + |\mathbb{E} [F(\mu_n); \mu_n \notin U]|.$$
The first term is bounded by $\epsilon$, since we have $|F| < \epsilon$ on $U$.  And for $n \ge N$ the second term is bounded by $M \mathbb{P}(\mu_n \notin U) < M \epsilon$.  Thus we conclude
$$\limsup_{n \to \infty} |\mathbb{E} F(\mu_n)| \le \epsilon + \epsilon M.$$
Letting $\epsilon \to 0$ we have the desired conclusion.
Note this used the fact that the limiting distribution $\delta_\mathbb{Q}$ was degenerate.  I'm not sure how to do it without that.  We might need more assumptions on $E$ in that case.<|endoftext|>
TITLE: Wild half-line in a Euclidean space
QUESTION [10 upvotes]: Is there an $m$-dimensional simplicial complex $S$ with the following properties:

The cone over $S$ is homeomorphic to $\mathbb{E}^{m+1}$. Here $\mathbb{E}^{m+1}$ denoes the $(m+1)$-dimensional Euclidean space.
There is a vertex $v$ in $S$ such that the complement $S\backslash\{v\}$ is not simply connected.

Comments. 

If such example exist, the cone over $v$ has to from a wild half-line in the Euclidean space. It has to be an embedding of $[0,\infty)$ which complement is not simply connected.
The first condition is equivalent to the following: The spherical suspension over $S$ is homeomorphic to $\mathbb{S}^{m+1}$.

REPLY [4 votes]: I have got the following answer from Alexander Lytchak:
An example can be constructed the following way.
Start with a nontrivial homology sphere,
pass to its spherical suspension.
Now shrink one of the meridians of suspension to the point, which we denote by $v$.
The obtained space $S$ is the example;
it admits a natural triangulation.
The first condition is easy to check.
To check the second condition note that the spherical suspension $\Sigma(S)$ has a triangulation coming from $S$.
The space $\Sigma(S)$ is a homotopy equivalent to the sphere;
it is homological manifold 
which is also manifold everywhere except maybe the poles of suspension.
Each pole admits a simply connected punctured neighborhood.
Therefore by disjoint disc property the poles are also manifold point.
It remains to apply generalized Poincaré conjecture.<|endoftext|>
TITLE: Fiberwise criterion for a stack to be a gerbe
QUESTION [7 upvotes]: Let $f:X\to Y$ be a morphism of algebraic stacks.
If the geometric fibres of $f$ are algebraic spaces, then $f$ is representable by algebraic spaces.
I'm wondering about analogues of this fiberwise criterion in the context of gerbes and DM stacks.
For instance:
Q1. If the geometric fibres of $f$ are DM stacks, then is $f$ representable by DM stacks?
Q2. If there is a finite (abstract) group $G$ such that the geometric fibres of $f$ are $G$-gerbes, then is $f$ a $G$-gerbe?

REPLY [3 votes]: As for Q2, I don't think so. For example, consider $(BG \times (\mathbb{A}^1\setminus \{ 0 \}))\amalg BG\to \mathbb{A}^1$, everything over $\mathbb{C}$.
Added [Edit: this one doesn't work, see comments below]: for a flat, but perhaps more contrived, example, I think one can take $\mathscr{X}:=B\mu_{3,\mathbb{Q}} \amalg B(\mathbb{Z}/3)_{\mathbb{Q}}$ and $X:=\mathrm{Spec}(\mathbb{Q})\amalg\mathrm{Spec}(\mathbb{Q})$, where $\mathscr{X}\to X$ is the coarse moduli space map. Here $\mu_{3,\mathbb{Q}}$ is the nonconstant group scheme over $\mathbb{Q}$ defined as a covariant functor on $\mathbb{Q}$-algebras by $A\mapsto$ third roots of $1$ in $A$, and $(\mathbb{Z}/3)_{\mathbb{Q}}$ is the constant group scheme defined on $\mathbb{Q}$-algebras by $A\mapsto\mathbb{Z}/3$ (Edit: this is for connected $Spec(A)$, in general it's a locally constant group scheme, as by the way it has to be a locally constant sheaf of groups).
Now, I think the two basechanges along the two possible maps $\mathrm{Spec}(k)\to X$, $k$ algebraically closed, are both the gerbe $B\mu_{3,k}=B(\mathbb{Z}/3)_k$.<|endoftext|>
TITLE: Describing Levi factors and unipotent radicals of parabolic subgroups in classical groups
QUESTION [7 upvotes]: I asked this question before at Math.SE (link) but got no answer.

Let $G$ be an algebraic group over an algebraically closed field $k$ of characteristic $p \geq 0$. Then any parabolic subgroup $P$ of $G$ has factorization $P = Q \rtimes L$, where $Q$ is the unipotent radical of $P$ and $L$ is some Levi factor of $P$.
If $G = GL(V)$ with $\dim V = n$, then $$P = \{ \pmatrix{A_{n_1} &  & & * \\  & A_{n_2} & & \\  &  & \ddots & \\ 0 &  & & A_{n_t} } \}$$
for some $n = n_1 + \cdots + n_t$ and $$Q = \{ \pmatrix{I_{n_1} &  & & * \\  & I_{n_2} & & \\  &  & \ddots & \\ 0 &  & & I_{n_t} } \}$$ and for example $$L = \{ \pmatrix{A_{n_1} &  & & 0 \\  & A_{n_2} & & \\  &  & \ddots & \\ 0 &  & & A_{n_t} } \}$$
So $L \cong GL_{n_1} \times GL_{n_2} \times \cdots \times GL_{n_t}$. What about when $G = \operatorname{Sp}(V)$ or $G = \operatorname{SO}(V)$? I suppose the answer will be different for $p = 2$ and $p \neq 2$.
I know that for $\operatorname{Sp}$ and $\operatorname{SO}$ any parabolic subgroup $P$ is a stabilizer of a flag of totally singular subspaces. 
How does one describe $Q$ and $L$? Is there a good reference for this?

I suppose if $P$ is the stabilizer of the flag $0 \subset V_1 \subset V_2 \subset \cdots \subset V_t$ of totally singular subspaces of $V$, then intuitively $Q$ should consist of those maps which stabilize each $V_i$ and act trivially on each $V_{i} / V_{i-1}$. But what do the Levi factors $L$ of $P$ look like?

REPLY [9 votes]: I'll assume here that $G$ is reductive (or even semisimple), since otherwise the description of unipotent radicals is more open-ended.
There is actually quite a bit of literature over the years, often concerned especially with questions involving smaller fields of definition (such as finite fields) or with the precise action of $L$ on $Q$.   From the Borel-Tits conjugacy theorem, one can limit attention to the case of a standard parabolic subgroup arising from the choice of a set of simple roots.
In the case of classical groups a basic fact is that the semisimple derived group of $L$ itself (when nontrivial) only has as simple factors various classical groups: these are easily determined by removing nodes from the Dynkin diagram corresponding to simple roots whose negatives do not figure in the given parabolic subgroup.   Of course, these simple algebraic groups can occur with varied isogeny types ranging from simply connected to adjoint.   (On the Lie algebra level, this isn't a problem.)
Since the derived group of $L$ is only the almost-direct product of these simple factors, it's a somewhat more delicate task to examine each isogeny type of $G$ to describe the product more precisely.    Similarly, $L$ is only an almost-direct product of the derived group with a central torus.   It may help to keep in mind that when the given (reductive) group $G$ has a simply connected derived group, then the derived group of $L$ is also simply connected (a result going back to the work of Borel-Tits).   
Concerning the way $L$ acts on $Q$ (or factors in a suitable series for it), there are explicit representation-theoretic descriptions, many by Gary Seitz and some of his collaborators or students: see for example Azad-Barry-Seitz, along with  Roehrle and Anchouche.
The special case when $Q$ is commutative gets special attention.<|endoftext|>
TITLE: Transferring connection information to associated bundles and back
QUESTION [5 upvotes]: This might not be research level but I've tried more than once to ask about this in MSE and it got nowhere. So I thought It's fair to at least try.
At the risk of repeating well known stuff I tried to make the question as precise as possible. 
Let $\pi: P \to M$ be a $G$-bundle with connection form $\omega \in \Omega^1(P ;\mathfrak{g})$. Let $\rho : G \to Gl(V)$ be a linear representation and $E = P \times_{\rho} V$ the corresponding associated bundle. 
Denote by $\Omega^{\bullet}_{\rho}(P ; V) \subset \Omega^{\bullet}(P;V)$ the space of $V$-valued forms satisfying:

Vertical: $i_X \eta = 0$ for all $X \in VP =  \ker \pi_*$.
$R^*_g \eta = \rho(g^{-1}) \cdot \eta$ for all $g \in G$.

There is a one to one between forms in $\Omega^{\bullet}_{\rho}(P ; V)$ and sections of the bundle $\Omega^{\bullet}(M) \otimes E \to M$. 
We have an isomorphism $\varphi : \mathfrak{g} \to VP$ sending  $X \to (p \mapsto \frac{d}{dt}(e^{tX}\cdot p))$. Therefore, our connection $\omega \in \Omega^1(P ;\mathfrak{g})$ gives rise to a retraction $\varphi^{-1} \circ \omega :TP \to VP$ and thus to a projection onto the horizontal bundle $h= I-\varphi^{-1} \circ \omega$. This determines a derivation on the complex $\Omega^{\bullet}(P;V)$ called the exterior covariant derivative:
$$D: \eta \mapsto d\eta \circ h$$
This derivation descends to a derivation $\Omega^{\bullet}_{\rho}(P ; V)$. 
Now to the question:
In the appendix of the book "Dirac operator on riemannian geometry" there's a nice formula for how the exterior covariant derivative acts on $\Omega^{\bullet}_{\rho}(P ; V)$. Let $\eta \in \Omega^r_{\rho}(P,V)$. Here is the formula:
$$D \eta = d\eta + \rho_*(\omega)\wedge \eta$$
Where the wedge of a matrix and vector calued forms is defined as:
$$\rho_*(\omega)\wedge \eta (v_0,...,v_r) = \sum^r_{j=0} (-1)^j \rho_*(\omega(v_j))\cdot \eta(v_0,...,\hat{v_j},...,v_r))$$
First part of question:


How can this formula be neatly derived? It gets messy so quickly for me I don't manage to get very far...
Is there a deeper meaning to wedging a matrix and a vector or is it just a notational conveniance? On the face of it it seems to be a
  very awkward thing to do.


In the case where $G=GL(V)$ the representation is the identity and $P\cong F(E)$ we can get to a connection on the associated bundle $E$ by simply pulling back along a frame. Indeed if $u: U \to P$ is a local section then the covariant derivative is $\nabla = u^*(\rho_*(\omega)) \in \Omega^1(U ; End(V))$. 
This procedure doesn't seem to generalize directly.


How do I get from $\rho_*(\omega)$ to a covariant derivative $\nabla \in \Omega^{\bullet}(M;End(E))$ in the general case?


And finally:
What's a good source to read about transferring connection information from principal bundles to asscoiated bundles and back?
I can't stress enough the extent at which i feel this is glossed over in the familiar literature.

REPLY [3 votes]: Ad 1.: Since every vector can be decomposed in its horizontal and vertical part. Thus it is enough to consider the case a) where all vectors are horizontal (this is trivial) and b) where at least one is vertical and the rest is horizontal (this has to be calculated explicitly but is relatively simple since one can use the fact that the vertical vector is a Killing field).
Ad 2: As soon as one has a bilinear form $A \times B \to C$ one can form a wedge product of forms with values in $A$ and forms with values in $B$ to get forms with values in $C$ (by the above formula). For example the wedge product of Lie algebra valued forms is defined in the same way using the Lie commutator as the underlying bilinear form.
Ad 3: Use 1 and the identification of sections of $E$ with equivariant maps $P \to V$ (under which $\nabla$ corresponds to $D$)<|endoftext|>
TITLE: Totally geodesic subgroups in Lie groups
QUESTION [6 upvotes]: Let $G$ be a Lie group with a left invariant metric $g$.
Let $H$ be a (closed) Lie subgroup of $G$, and assume $g$ is right-$H$-invariant. (That is $d(R_h)_e:T_eG \to T_hG$ is an isometry for every $h \in H$). Note that this is equivalent to the statement, that for every $h \in H$ the map of right multiplication by $h$, viewed as a map $G \mapsto G$ is an isometry.
(This appears to be stronger thant requiring it will be an isometry as a map $H \mapsto H$).
Is it true that $H$ must be totally geodesic in $G$?
Since $g$ is bi-$H$-invariant, the geodesics of $H$ are the one-parameter subgroups of $H$. Hence, the question amounts to:
Are the one-parameter subgroups of $H$ geodesics in $G$?

REPLY [3 votes]: Edit. I totally misunderstood the question, maybe now its better.
Assume that $g$ is left-$G$ and right-$H$-invariant. Then one can construct a Riemannian submersion $p\colon G\to G/H$. This is now a family of Riemannian manifolds with fibre $H$ and compact structure group $H$, because you could also write it as a Borel construction $G=G\times_HH\to G/H$. Here, you regard $G$ as a principal bundle over $G/H$ with fibre $H$. $G/H$ is equipped with the induced Riemannian metric, and $T^HG\subset TG$ is defined as $\ker(dp)^\perp$. Together with the induced metric on $H$, the Borel construction gives back the original metric $g$. But then its fibres are automatically totally geodesic, so now the answer is "yes".
Old answer to a different question :-)
Consider Berger metrics on $SU(2)\cong S^3\subset\mathbb C^2$.
Here the round metric is stretched or shrunk along the fibres of the Hopf
fibration $S^3\to S^2$. These metrics are still left invariant, but
not biinvariant. For $H$, take a one parameter subgroup, so $H\cong S^1$.
The induced metric on $H$ is still left invariant, hence biinvariant, because $H$ is abelian.
If all one parameter subgroups were geodesics, then their left translates by elements of $SU(2)$ would be geodesics, too. So the Berger metrics would have the same geodesics as the round metric, but that is not the case. Hence, the answer in general is no.<|endoftext|>
TITLE: Who gave the generalized Stone-Weierstrass Theorem?
QUESTION [6 upvotes]: Let $X$ be a compact Hausdorff space and $\mathcal{A}$ be a closed self-adjoint subalgebra of $C(X)$ which contains the constants. Then $\mathcal{A}$ is the collection of continuous functions on $X$ which are constant on the sets of $\prod_\mathcal{A}$ where
  $$
\prod_\mathcal{A}=X/\sim
$$
  with $x\sim y$ iff $f(x)=f(y)$ for all $f\in\mathcal{A}$. 

Could anyone tell me with cited references that who gave the generalized Stone-Weierstrass Theorem above?

REPLY [11 votes]: This (for real-valued rather than complex-valued functions) was in Stone's original paper that proved the Stone-Weierstrass theorem, as Theorem 84.  The statement is a bit funny, since he defines the equivalence relation $x\sim y$ not in the obvious way but as "$x$ is in the intersection of all sets $f^{-1}(U)$ such that $f\in\mathcal{A}$, $U\subset\mathbb{R}$ is an open interval, and $y\in f^{-1}(U)$" (this definition appears in the statement of Theorem 81).<|endoftext|>
TITLE: Every measure on a set $X$ extends to the power set of $X$: Consistent or not with ZF?
QUESTION [12 upvotes]: Question. Is it consistent with ZF that every (countably additive, non-negative) measure $\mu: \Sigma \to \bf R$, where $\Sigma$ is a sigma-algebra on a given set $X$, extends to a (countably additive, non-negative) measure $\tilde{\mu}: \mathcal P(X) \to \bf R$? 

What I seem to know: Let us first work in the frame of ZFC. Given a set $X$ of cardinality $\ge \aleph_1$, we take $\Sigma$ to be the smallest sigma-algebra on $X$ containing all the countable subsets of $X$. It is seen that the function 
$$\mu: \Sigma \to {\bf R}: A \mapsto 
\left\{
\begin{array}{ll}
\! 0 & \text{if } |A| \le \aleph_0 \\
\! 1 & \text{otherwise}
\end{array}
\right.
$$
is a measure on $X$. Thus, if $\mu$ extends to a (countably additive, non-negative) measure $\mathcal P(X) \to \bf R$, then $|X|$ is a real-valued measurable cardinal, which implies the existence of a weakly inaccessible cardinal by an old theorem of S. Ulam, see

S. Ulam, Zur Masstheorie in der allgemeinen Mengenlehre, Fund. Math. 16 (1930), 140-150 (in German),

or Corollary 10.15 in:

T. Jech, Set Theory - The Third Millennium Edition, Revised and Expanded, Springer Monogr. Math., Springer, Berlin, 2006 (corrected 4th printing). 

However, the existence of a weakly inaccessible cardinal is unprovable from the axioms of ZFC.
On the other hand, we have from:

R. M. Solovay, ``Real-valued measurable cardinals'', pp. 397-428 in: D. Scott (ed.), Axiomatic set theory, Proc. Sympos. Pure Math. (Univ. California, Los Angeles, CA, 1967), Vol. XIII, Part I, Amer. Math. Soc.: Providence, RI, 1987 (reprinted ed.)

that the existence of measurable cardinals in ZF is equiconsistent with the existence of measurable (resp., real-valued measurable) cardinals in ZFC. However, this doesn't answer the question I'm posing, as far as I can say (in particular, it occurs to me that Solovay's result doesn't cover the case of real-valued measurable cardinals in ZF, does it?).

REPLY [6 votes]: EDIT: This answer is wrong. I am not deleting it at the request of the OP. See the counterexample at the bottom.

Here is a terrible mathematician answer that you didn't know that you didn't care for. Yes, it is consistent, at least assuming the consistency of very large cardinals.
In Gitik's model, where every limit ordinal has countable cofinality we also have the following: If we start with the class of singletons and close it under countable unions, we obtain the entire universe. And I claim that in the presence of this axiom, every countably additive measure is either atomic (and thus extends naturally) or trivial (and thus extends naturally).
To see why, if the measure gives $0$ to singletons, then by $\sigma$-additivity it has to give $0$ to every countable set, and so to every set which can be generated by a countable union of countable sets. Which is every set. If there are singletons which have positive measure, then they are necessarily the atoms and we're done.
Some relevant papers:


Arnold W. Miller, Long Borel hierarchies, MLQ Math. Log. Q. 54 (2008), no. 3, 307--322.

M. Gitik, All uncountable cardinals can be singular, Israel J. Math. 35 (1980), no. 1-2, 61--88.




Edit. So here is a counterexample, and some consequence of it.
In Gitik's model, $\Bbb R$ is a countable union of sets $A_n$, such that each $A_n$ is the countable union of countable sets. In particular, there is a countable sequence of countable sets $X_n$ whose union is uncountable.
Now, if $X_n$ is countable $R_n=\{r\in\Bbb R\mid r\text{ codes an enumeration of }X_n\}$ gives us a family without a choice function. If we could choose $r_n\in R_n$ then we could have enumerated all the $X_n$'s uniformly and their union would have been countable.
Now consider $\Sigma$ as the $\sigma$-algebra generated by the $R_n$'s, and since they are obvious pairwise disjoint (a real cannot code more than one countable set), $\Sigma$ is isomorphic to $\mathcal P(\Bbb N)$ with $R_n$ as atoms.
If we consider the atomic measure on $\mathcal P(\Bbb N)$ such that $\mu(\{n\})=1/2^{n+1}$, it induced a measure on $\Sigma$. Now I claim that $\mu$ cannot be extended to a measure on $\mathcal P(\Bbb R)$.
Suppose that it could have been extended, then it is impossible that every singleton has measure $0$, since $\sigma$-additivity would have implied that every countable set has measure $0$, thus every $A_n$ (from the partition of $\Bbb R$) has measure $0$ as a countable union of null sets; and finally $\Bbb R$ has measure $0$. But this is impossible since $\mu(\Bbb R)=1$.
So each $R_n$ has some countable subset whose elements have all positive measure summing to $1/2^{n+1}$. But if $\sum_n r_n<\infty$ it has to be the case that for every $k$, at most finitely many $r_n$'s satisfy $\frac1k < r_n$. So there is a finite set of reals with maximal weight inside each $R_n$.
So we have $R'_n\subseteq R_n$, finite sets of reals defined as above. But finite sets of reals have canonical choice functions in them: take the minimum. And this is a contradiction to the assumption that there is no choice function from $\{R_n\mid n\in\Bbb N\}$.
As a consequence we see that the "total extension principle" cannot be true if there exists an uncountable set $X$ generated by iterated countable unions from its singletons; and there is a choice function from finite subsets of $X^\omega$.<|endoftext|>
TITLE: Lattice points near a curve
QUESTION [8 upvotes]: Bombieri and Pila had a well known bound for the count of lattice points on an algebraic curve in the plane. Does it generalize to a bound for the count of lattice points near (say within a distance of $\delta=o(1)$) an algebraic curve? Is it obvious that we only need to add $O(\delta L)$ to the Bombieri-Pila count, where $L$ is the length of the curve? Or do we have to use the weaker bound of Swinnerton-Dyer?
Edit: To put things in more context, $O(\delta L)$ is the "expected" number of lattice points near the curve if the curve is situated randomly in the plane. The Bombieri-Pila bound can be thought of as arising from the arithmetic structure of the algebraic curve. My question then amounts to: do the expected count and the arithmetic count together fully account for the lattice points near an algebraic curve?

REPLY [2 votes]: See: MR2271609 (2007m:11092) Reviewed 
Huxley, Martin N.(4-CARD-SM); Sargos, Patrick(F-NANC-IE)
Points entiers au voisinage d'une courbe plane de classe Cn. II. (French. English, French summary) [Integral points near a plane curve of class Cn. II] 
Funct. Approx. Comment. Math. 35 (2006), 91–115. 
11J25 (11P21)<|endoftext|>
TITLE: Hyperimaginaries and continuous logic
QUESTION [7 upvotes]: Classical (i.e., discrete) logic is well positioned to study imaginaries in part because the $T^{eq}$ construction allows us to treat imaginary sorts as we would treat any other sort. With hyperimaginaries, on the other hand, classical  logic has no such luck. The ability to apply a $T^{eq}$-like construction to the study of hyperimaginaries motivated, in part, the study of positive model theory, a thematic precursor to the modern approach to continuous logic.
Despite that history, I've been unable to convince myself that continuous logic's $T^{eq}$ construction, when applied to a classical theory, includes sorts for the classical hyperimaginaries (even finitary ones) in the case of an uncountable theory. (The $T^{eq}$ construction I'm using is the one that adds canonical parameters for continuous logic formulas, such as in Chapter 11 of Model Theory for Metric Structures.)
Folklore as I've heard it says that the construction does include all the classical finitary hyperimaginaries, but I've been unable to find a citation for that fact, and haven't yet produced a clear proof or counterexample. Can anyone point me in the right direction?

REPLY [5 votes]: In general a $T^{eq}$ construction in continuous logic, as it's typically defined, can only possibly add hyperimaginaries corresponding to equivalence relations defined by countable partial types. If you consider the uncountable discrete theory $T$ with unary predicates $P_i$ for each $i<\omega_1$ where for each pair of disjoint finite sets $X,Y\subset \omega_1$ $T$ contains an axiom $\exists x\bigwedge_{i\in X}P_i(x)\wedge \bigwedge_{i\in Y} \neg P_i(x)$. Then if you look at the type definable equivalence relation $E(x,y)=\{P_i(x) \leftrightarrow P_i(y) : i<\omega_1\}$ the hyperimaginary where you quotient out by $E$ is not equivalent to any continuous imaginary of $T$, because any continuous formula is definable over some countable set of $P_i$'s.
EDIT3: There is a sense in which general hyperimaginaries are realized in a continuous $T^{eq}$. Specifically any type-definable equivalence relation is the intersection of a family of countably type-definable equivalence relations. It follows from this and the result below that any hyperimaginary that is just a quotient by a type-definable equivalence relation is the direct limit of some directed family of imaginaries. This is as opposed to the situation is discrete logic, where this only holds IIRC if the theory eliminates hyperimaginaries. So to put it another way continuous logic is better suited for studying hyperimaginaries in the sense that every theory, when considered as a continuous theory, eliminates hyperimaginaries.

EDIT4: It has come to my attention that this result is implicit in the proofs leading up to Theorem 2.20 in the paper 'Uncountable dense categoricity in cats' by Ben Yaacov.
EDIT3: With the help of Dap's answer to my question here I have finally  convinced myself that it does always work for countably type-definable equivalence relations. The proof (especially the proof of uniform convergence) is difficult enough that I think it's worth writing down.

Proposition. If $T$ is a countinuous first-order theory and $P(x,y)$ is some definable predicate in $T$ such that $P$'s zeroset $[P]$ is an equivalence relation, then there is a definable pseudo-metric $\rho(x,y)$ in $T$ such that $[\rho]=[P]$  (i.e. $\rho$ and $P$ have the same zeroset).

Proof. By replacing $P(x,y)$ with $P(x,y) + P(y,x)$ we may assume that $P(x,y)=P(y,x)$. Also by scaling we may assume that $P(x,y)$ takes values in $[0,1]$.
Let $\varepsilon_0 = 1$. For each $k<\omega$, given $\varepsilon_k$, find $\varepsilon_{k+1}>0$ such that for any $x,y,z,w$, if $P(x,y),P(y,z),P(z,w) \leq \varepsilon_{k+1}$ then $P(x,w) < \varepsilon_k$. Such an $\varepsilon_{k+1}$ must exist by compactness, since $[P]$ is an equivalence relation. Also choose so that  $\varepsilon_{k+1} < \frac{1}{2}\varepsilon_k$, which is clearly possible (this ensures that $\varepsilon_k \rightarrow 0$ as $k\rightarrow \infty$).
Now find a continuous, non-decreasing function $\alpha:[0,1]\rightarrow[0,1]$ such that $\alpha(\varepsilon_k)=2^{-k/2}$ for every $k<\omega$ (this is always possible). Let $Q(x,y)=\alpha(P(x,y))$.
Clearly $Q(x,x)=0$, $Q(x,y)=Q(y,x)$, and $0\leq Q(x,y) \leq 1$.
Claim: For any $x,y,z,w$, $Q(x,w) \leq 2 \max(Q(x,y),Q(y,z),Q(z,w))$.
Proof of claim: Pick $x,y,z,w$. Find $k<\omega$ maximal such that $\max(Q(x,y),Q(y,z),Q(z,w)) \leq 2^{-k/2}$. By construction this implies that $\max(P(x,y),P(y,z),P(z,w)) \leq \varepsilon_k$, so we also have $P(x,w) < \varepsilon_{k-1} $ if $k>0$ and $P(x,w)\leq 1$ if $k=0$ in any case. Therefore since $\alpha$ is non-decreasing we have that $Q(x,w) \leq 2^{-(k-1)/2}$ if $k>0$ and $Q(x,w) \leq 1$ if $k=0$.
Since $k$ was chosen to be maximal, we have that $\max(Q(x,y),Q(y,z),Q(z,w)) > 2^{-(k+1)/2}$. Therefore $$Q(x,w) \leq 2^{-(k-1)/2} = 2\cdot 2^{-(k+1)/2} < 2\max(Q(x,y),Q(y,z),Q(z,w))$$ 
if $k>0$ and 
$$Q(x,w) \leq  1 < 2\cdot 2^{-1/2} < 2\max(Q(x,y),Q(y,z),Q(z,w))$$ 
if $k=0$. 
So in any case $Q(x,w) \leq 2 \max(Q(x,y),Q(y,z),Q(z,w))$. End of proof of claim.
Let $Q_1(x,y)=Q(x,y)$ and for every $1<k<\omega$ let $$Q_k(x,y)=\inf_{z_1,\dots,z_{k-1}}Q(x,z_1)+Q(z_1,z_2)+\dots+Q(z_{k-1},y).$$ The predicate $Q(x,y)$ satisfies the conditions of my question, so by the results there the sequence $Q_k$ converges uniformly to some definable pseudo-metric $\rho(x,y)$ satisfying $\frac{1}{2} Q(x,y) \leq \rho(x,y) \leq Q(x,y)$. By this we have that $[\rho]=[Q]$ and clearly $[Q]=[P]$, so we have that $[\rho]=[P]$, as required. $\Box$
Now since any countable partial type is the zeroset of some definable predicate we get the required result for coutably definable hyperimaginaries in discrete theories.
EDIT2: There is a serious problem with this argument in the step regarding the amalgamation property, i.e. the amalgamation procedure described fails to work in all cases. If I can resolve this I'll post an edit.
EDIT: It's actually not even true for countable theories. Here is a counterexample.
 Let $\mathcal{L}=\{P_i\}_{i<\omega}$ be a countable sequence of binary predicates and let $\Sigma$ be the $\mathcal{L}$-theory consisting of the following for each $i<\omega$:

$\forall x P_i(x,x)$
$\forall xy(P_i(x,y)\leftrightarrow P_i(y,x))$
$\forall xyz(P_{i+1}(x,y)\wedge P_{i+1}(y,z) \rightarrow P_i(x,z))$

 Note that these imply $\forall xy(P_{i+1}(x,y) \rightarrow P_{i}(x,y))$. 
 Let $E(x,y)$ be the partial type given by the formulas $\{P_i(x,y)\}_{i<\omega}$. It's clear that in any model of $\Sigma$, $E$ gives a type-definable equivalence relation.
Let $\mathcal{K}$ be the class of finite models of $\Sigma$. First of all I claim that this class contains only countably many isomorphism types. To see this note that for any $\mathfrak{A} \in \mathcal{K}$ and any $a,b\in \mathfrak{A}$, the quantifier free type of $ab$ is uniquely determined by the quantity $Q(x,y)=2^{-i}$ where $\mathfrak{A} \models P_i(x,y)\wedge \neg P_{i+1}(x,y)$ or $Q(x,y)=0$ if $\mathfrak{A} \models E(x,y)$. This has countably many values and the isomorphism type of a model of $\mathcal{K}$ is uniquely determined by the values of $Q$ on its pairs of elements. 
Also since $\Sigma$ is a universal theory $\mathcal{K}$ is closed under substructures. 
I claim that $\mathcal{K}$ has the amalgamation property. To see this let $\mathfrak{A}\subset \mathfrak{B}$ and $\mathfrak{A} \subset \mathfrak{C}$ with $\mathfrak{A}$, $\mathfrak{B}$, and $\mathfrak{C}$ all models of $\Sigma$. We can give an amalgamation $\mathfrak{D}$ whose universe is $B\cup C$ and where for any $b \in B \setminus A$ and $c \in C \setminus A$ we set $Q(b,c) = \min(1,2 \min_{a \in A}\max(Q(b,a),Q(a,c)))$. In other words we introduce only the $P_i$ relationships that are absolutely necessary according to $\Sigma$. 
So we have that $\mathcal{K}$ is a Fraïssé class and we can find a Fraïssé limit $\mathfrak{F}$. Let $T=\mathrm{Th}(\mathfrak{F})$. Since the automorphism type of every finite tuple in $\mathfrak{F}$ is entirely determined by its quantifier free type and since every quantifier free type consistent with $T$ occurs in $\mathfrak{F}$, $T$ admits quantifier elimination. 
Let $\mathcal{L}^\prime$ be a continuous language with a single binary $[0,1]$-valued predicate symbol $Q$. Interpret $\mathfrak{F}$ as an $\mathcal{L}^\prime$-structure $\mathfrak{F}^\prime$ using the definition of $Q$ given above. It's clear that $\mathfrak{F}$ and $\mathfrak{F}^\prime$ are interdefinable and more than that interdefinable in a quantifier free way. In particular I claim that $T^\prime = \mathrm{Th}(\mathfrak{F}^\prime)$ eliminates quantifiers in the sense of continuous logic. In particular this implies that the $2$-type of any pair $ab$ is entirely determined by the value of $Q(a,b)$. Also note that by construction we have that $\mathfrak{F} \models E(a,b)$ if and only if $\mathfrak{F}^\prime \models Q(a,b)$ (i.e. that it is $0$). 
Let $\rho(x,y)$ be a $\varnothing$-definable $[0,1]$-valued pseudo-metric such that $\mathfrak{F} \models E(a,b)$ implies $\mathfrak{F}^\prime  \models \rho(a,b)$. By quantifier elimination there must be some continuous function $\alpha:[0,1]\rightarrow[0,1]$ such that $\rho(x,y)=\alpha(Q(x,y))$. Now for each $i<\omega$ consider the finite $\mathcal{L}$-structure $\mathfrak{A}$ with $A=\{a,b,c\}$ such that $Q(a,b)=0$, $Q(a,c)=2^{-i}$, and $Q(b,c)=2^{-i-1}$. Note that the third axiom schema for $\Sigma$ translates to $Q(x,z)\leq 2 \max(Q(x,y),Q(y,z))$, so in particular $\mathfrak{A}$ is a model of $\Sigma$. Therefore it embeds into $\mathfrak{F}$ with some map $f:\mathfrak{A}\rightarrow \mathfrak{F}$. Since $Q(f(a),f(b))=0$ we must have that $\rho(f(a),f(b))=0$. Since $\rho$ is a pseudo-metric this implies that $\rho(f(a),f(c))=\rho(f(b),f(c))$. Therefore we have that $\alpha(2^{-i})=\alpha(2^{-i-1})$. Since this is true for any $i<\omega$, we have that $\alpha$ is constant on the set $\{2^{-i}\}_{i<\omega}$. It must be the case that $\alpha(0)=0$ since $\mathfrak{F} \models E(a,b)$ implies $\mathfrak{F}^\prime  \models \rho(a,b)$, therefore by continuity $\alpha(r)=0$ for all $r\in[0,1]$ and $\rho$ is the trivial pseudo-metric. On the other hand $E$ is clearly a non-trivial equivalence relation because there is a finite model of $\Sigma$ in which there are two elements with $\neg E(a,b)$. 
So we have a theory with a countably type-definable non-trivial equivalence relation (definable over $\varnothing$) in which the only $\varnothing$-definable pseudo-metric that is at least as coarse as it is the trivial pseudo-metric. 
I'm not sure if allowing more parameters for the pseudo-metric can save it but I'm not very hopeful.<|endoftext|>
TITLE: What kind of geometric object is the Pauli spin matrix vector $\vec{\sigma} = (\sigma_1, \sigma_2, \sigma_3)$?
QUESTION [11 upvotes]: Physicists routinely wrote all 3 Pauli spin matrices as a vector.  
$$ \sigma_1 = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array} \right) \hspace{0.25in}
\sigma_2 = \left( \begin{array}{cc} 0 & -i \\ i & 0\end{array} \right)\hspace{0.25in}
\sigma_3 = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right) $$
This enables them to write shortcuts in quantum mechanics books such as:
$$ e^{ i\theta (\vec{v} \cdot \vec{\sigma})} = (\cos \theta) I + ( i \sin \theta) (\vec{v} \cdot \vec{\sigma}) $$
What kind of object is $\vec{\sigma}$ ? Is it a vector or a matrix or both?

Another example is when we write the curl as a determinant.
$$ \nabla \times F = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_x & F_y & F_z \end{array} \right| $$
Curl is not the volume of any kind of box.  Is there a rigorous geometric meaning for these "abuses of notation"?

I asked this question on Math.SE and got a very literal answer, which did not help any.  So I am asking here.  You are welcome to answer there and call it a day.
Math.SE What kind of object is the Pauli spin matrix $\vec{\sigma}$?

REPLY [5 votes]: The Pauli spin vector encodes a two-complex-dimensional unitary representation of the Lie algebra $\bf{su}_2$. A basis of the Lie algebra maps to the matrix algebra $\bf{gl}_2(\mathbb{C}),$ giving a $3\times 2\times 2$ tensor. In 3 dimensions, physicists will sometimes blur the distinction between the basis for the tangent space and the Lie algebra of the Lorenz group $SU(2)$ at the origin, by the identification of $\hat{x}_i$ with the operator $\hat{x}_i\times -$ (this is not possible in higher dimensions), but these three matrices should really be thought of as parametrized by a basis for the Lie algebra.<|endoftext|>
TITLE: Variation on the Subset Sum Problem
QUESTION [6 upvotes]: Given a nonempty set of integers, and given that there exists a subset of this set whose elements sum to zero, is finding the smallest such subset NP-complete? 

Disclaimer: The above question involves part of ongoing research for my dissertation. 
To flesh this out, this question is derived from the so-called "subset sum problem," which asks whether a given set of integers has a subset that sums to zero. For my case, however, we are assuming the existence of at least one such subset, and then wish to investigate whether finding the minimal such subset is NP-hard. 
I have a hunch (in other words, nothing resembling formal proof) that my problem above is an NP-hard problem, because we are not assuming that the subset whose elements sum to zero is explicitly given in the hypothesis, so that finding the minimal such subset leaves us with a similar search mechanism for finding the existence of such a subset in the standard Subset Sum problem (which is known to be NP-complete). 
Exploring the case if such a subset is known may also be a point of interest, as well. 
This is my first real exploration into research, and this is also the first time during the course of my research that my advisor himself is not aware of the answer, so I am feeling "stuck" for the first time in the course of researching a topic. 
The purpose of this post is not necessarily to find an explicit answer, unless someone is aware of a paper that directly or equivalently addresses this question (we have not yet been able to find one). 
The focus of my dissertation is more algebraic in nature, and the complexity discussion above relates to a curiosity worthy of including in this research, rather than ascertaining a direct solution. I would appreciate any input or advice, specific papers or even general directions to look, as I continue to develop my researching skills and run into these "ruts" for the first time.
Since this is my first post in Overflow, I apologize if this breaks any rules, ethics, or etiquette appropriate this board. I am excited to discover an interactive online place with seasoned professional mathematicians, and I look forward to developing my own researching skills from what I learn from here.

REPLY [5 votes]: This is merely a simplification of Tony Huynh's answer, but still more than a comment imho. Note that throughout, I work with multi sets (or sequences of integers) rather than sets. Reduce SUBSET SUM to SMALLEST SUBSET SUM as follows:
Given an instance $(a_1, \ldots, a_n)$ of SUBSET SUM, let $b := -(a_1 + \ldots + a_n)$ and plug the sequence $(a_1, \ldots, a_n, b)$ into your fictional SMALLEST SUBSET SUM algorithm.
(1) Clearly $a_1 + \ldots + a_n + b = 0$, so your instance has a zero-sum subset.
(2) If the original instance has a zero-sum subset, then this is still a zero-sum subset in the modified instance, hence the smallest such set will have cardinality at most $n$.
(3) If the modified instance has a zero-sum subset of cardinality at most $n$, there are two cases
(a) It does not contain $b$. Then that is a zero-sum subset of the original instance.
(b) It is $\{ b \} \cup \{ a_i : i \in I \}$ for some (possibly empty) $I \subsetneq [n]$. Then $\sum_{i \in I} a_i = \sum_{i=1}^n a_i$ and thus $\{ a_j : j \notin I \}$ is a non-empty zero-sum subset.
Hence the original instance has a zero-sum subset if and only if the modified instance has a zero-sum subset of size at most $n$.<|endoftext|>
TITLE: Fast Symbolic Linear Algebra CAS?
QUESTION [7 upvotes]: I am a regular user of Mathematica, Julia, and MATLAB but I am looking for something different. The problem I am trying to solve in Mathematica only requires (dense) linear algebra to specify but is of a size such that Mathematica takes too long (and seems to automatically halt the computation after like 18 hours, when it should finish in a week or so). I am looking for a symbolic solution and not a numeric solution (to then use a machine learning approach on the resulting equations). I was wondering if anyone knew if there are more efficient and parallel symbolic packages. Checking around it looks like Singularity or Machaly2 may be what's right for this problem?
Edit: As requested here is more detail. It's as straight-forward as it sounds. I need to solve for the symbolic solution of a system of 4x4 matrices and 1x4 vectors. I am able to succinctly write down the exact solution (attached as a photo, hard to format it right in the editor...),

but as you can see it's a little unwieldy (then it gets squared and simplified). I want to use the resulting polynomial (in terms of the matrix coefficients) in numerical computations, but I don't want to have to solve this system of equations each time the matrices change since this would increase the computational cost immensely.
So at a higher level it's really simple, and coding Mathematica to get it to try this computation is simple, but it is taking a lot of computing time (largely because Mathematica's engine is single threaded). In the simple case when the $A$ and $B$ matrices are lower triangular, I Mathematica was able to solve it after about 10 minutes. But getting rid of this assumption on even one matrix increases the complexity by a lot.
Let me lastly note that I have computing power at my disposal. I do high-performance computing, and so I have a few nodes available for a few hundred CPU cores and about a terabyte of RAM to let it churn for a month (I was hoping to change each matrix from lower triangular to full one by one to see how far I can get). However, I don't know what program/language will do this kind of "HPC Symbolic Computing". I was hoping there was some computational algebra package that has the tools to do this.

REPLY [4 votes]: I found out that the best solution is actually SymEngine.jl in Julia. Using Julia as the driver ends up increasing the speed and reducing the memory requirement, making it more efficient than using SymEngine in Python (or using SymPy). This ended up being much more efficient than Mathematica, and the few things I experimented with in Sage. Also, Julia's operator overloading lets you write a "generic" matrix function and then it will automatically compile a function which gets very good performance on symbolic expressions, and this makes it easy to extend the symbolic libraries as necessary without losing performance. This ended up being a very good solution to brute force some symbolic math to setup further analysis.<|endoftext|>
TITLE: Hodge map and the Cohomology Ring of a Riemannian Manifold
QUESTION [6 upvotes]: For a compact Riemannian manifold $M$, we know that the Hodge map $\ast$ and Laplacian $\Delta$ commute. From Hodge decomposition and its implied isomorphism between harmonic forms and cohomology classes we now that an indued on the cohomology ring $H(M)$, which is usually denoted again by $\ast$.
Now one could also naively define a map
$$
\ast:H(M) \to H(M), ~~~~~~~~~~~~~ [\omega] \mapsto [\ast(\omega)],
$$
and hope that they coincide. The thing is, as far I can tell, this map doesn't even seem to be well-defined. Can someone please prove whether or not it is well-defined. 
(In the case that it is not well-defined, any philosophical motivation for why the naive approach doesn't work would be appreciated.)

REPLY [2 votes]: You've probably seen this before, but in case you haven't:
Let $M$ be a compact oriented manifold. Let $\Omega^k$ be the smooth $k$-forms, let $Z^k$ be the closed $k$-forms and let $B^k$ be the exact $k$-forms. So $H^k=Z^k/B^k$. As Johannes's answer shows, $\ast$ does not carry $Z^k$ to $Z^{n-k}$, nor $B^k$ to $B^{n-k}$. (For an example of the latter, take $S^1$ with its usual metric, and $f:S^1 \to \mathbb{R}$ any non constant function. Then $df = f'(\theta) d \theta$ is exact, but $\ast(df) = f'(\theta)$ is not. ) So you cannot define a map $H^k \to H^{n-k}$ in your naive way.
We can put an inner product on $\Omega^k$ by $\langle \omega, \eta \rangle = \int_M \omega \wedge \ast(\eta)$. If we pretend that everything works like finite dimensional vector spaces (and the hard part of Hodge theory is justifying this), then $\Omega^k$ splits into $B^k \oplus (B^*)^k \oplus A^k$. Here $(B^*)^k$ is $(Z^k)^{\perp}$ and $A^k = (B^k \oplus (B^*)^k)^{\perp}$. The space $A^k$ is the harmonic forms. Notice that $Z^k = A^k \oplus B^k$, so $H^k = Z^k/B^k \cong A^k$.
Hodge star takes $A^k \to A^{n-k}$, $B^k \to (B^*)^{n-k}$ and $(B^*)^k \to B^{n-k}$. So $B^k$ doesn't go to $B^{n-k}$ and $Z^k = A^k \oplus B^k$ doesn't go to $Z^{n-k} = A^{n-k} \oplus B^{n-k}$, making your naive strategy not work. But we do have $Z^k/B^k \cong A^k \to A^{n-k} \cong Z^{n-k}/B^{n-k}$, giving us the Hodge map.<|endoftext|>
TITLE: Non-Cartesian Monoidal Model Structure on a Slice Category
QUESTION [14 upvotes]: Given a monoidal model category $(M,\otimes, 1)$, and a monoid therein $A$, one can take the slice model category $M_{/A}$. This category has a natural monoidal structure induced by taking fibered products over $A$. However, it should admit another monoidal structure coming from the product on $A$. In particular, given two maps $X\to A$ and $Y\to A$, there should be an object $Y\otimes X\to A\otimes A\to A$. Moreover, it seems, that monoid objects in $M_{/A}$ should be precisely the monoid morphisms $A'\to A$ for $A'$ another monoid. Is this monoidal structure described anywhere? How does it interact with the slice-category model structure? The particular case I'm thinking of is the case when $M=sSet$ with the Quillen model structure, and $A$ is a strict monoid with respect to Cartesian product, but I'd be pretty happy with this kind of statement for any nice model category of topological spaces.

REPLY [12 votes]: This construction came up in an Australian Category Seminar talk given by Ross Street last month, from which I will copy for 1. and 2. below. I'm afraid I don't know a reference. 
1. (monoidal structure) If $\mathscr{F}$ is a monoidal category and $T$ is a monoid in $\mathscr{F}$, then $\mathscr{F}/T$ becomes monoidal, with tensor product as you suggest:
$$(\theta \colon F \to T) \otimes (\varphi \colon G \to T) = \mu \circ (\theta \otimes \varphi) \colon F\otimes G\to T\otimes T \to T, $$
and unit $\eta \colon I \to T$. Note that the projection functor $\mathscr{F}/T \to \mathscr{F}$ is strict monoidal.
2. (internal hom) To your comment, if $\mathscr{F}$ is closed, then so is $\mathscr{F}/T$. The internal hom of  $\varphi \colon G \to T$ and $\psi \colon H \to T$ in $\mathscr{F}/T$ is given by the pullback in $\mathscr{F}$ of
$$[1,\psi] \colon [G,H] \to [G,T] \quad \text{along} \quad  [\varphi,1] \circ \lambda \colon T \to [T,T] \to [G,T],$$
where $\lambda$ corresponds to $\mu \colon T\otimes T \to T$ under the tensor-hom adjunction.
3. (monoidal model structure) Finally, if $\mathscr{F}$ is a monoidal model category then so is $\mathscr{F}/T$, since $\mathscr{F}/T \to \mathscr{F}$ is strict monoidal and creates colimits, cofibrations, fibrations and weak equivalences.
P.S. A general perspective on this construction which may be of interest is that the monoidal category $\mathscr{F}/T$ is the oplax limit of the arrow  $T \colon 1 \to \mathscr{F}$ in the 2-category MonCat$_l$ of monoidal categories, (lax) monoidal functors and monoidal natural transformations. The forgetful 2-functor MonCat$_l$ $\to$ Cat creates oplax limits, and the oplax limit of the underlying arrow $T \colon 1 \to \mathscr{F}$ in Cat is the slice category $\mathscr{F}/T$. See Steve Lack's papers 'Limits for lax morphisms' and the more recent 'Enhanced 2-categories and limits for lax morphisms' with Michael Shulman for the general 2-monad situation.<|endoftext|>
TITLE: (Non)-equivariant equivalence in $G$-spectra
QUESTION [8 upvotes]: In HHR, an important part is the periodicity theorem. For proving the theorem, they invert a carefully defined class $D \in \pi^{C_8}_{19\rho_8}(N^8_2MU_{\mathbb{R}})$ and they can find an element in $x \in \pi_{256}^{C_8}D^{-1}N^8_2MU_{\mathbb{R}}$, when we forget the $C_8$ equivariant structure, gives a non-equivariant equivalence from $\Sigma^{256}D^{-1}N^8_2MU_{\mathbb{R}}$ to $D^{-1}N^8_2MU_{\mathbb{R}}$. This non-equivariant equivalence is sufficient for them to show that the homotopy fixed point spectrum has the same periodicity.
However, in a following paper, they claim in several places that the periodicity theorem they proved in HHR, shall gives us an equivariant equivalence, which is stronger than the original statement I have seen in the Kervaire invariant one paper. So here is my question:

Given a finite group $G$ and a $G$-spectrum $X$, if we have an
  $G$-equivariant self-map $f:\Sigma^mX \rightarrow X$, which induces an
  equivalence of the underlying non-equivariant spectrum, will $f$
  automatically be an $G$-equivariant equivalence? If not, is there any
  condition we can apply to $G$ or $X$ to make this statement true?

If the answer is positive, the strengthened periodicity theorem will give us a computational advantage in the sense that normally slice spectral sequence in positive and negative dimension looks vastly different, and when come to resolve extension problems in $E_\infty$ page, an equivariant periodicity will induces isomorphism as Mackey functors, rather than Abelian groups. It will automatically resolve most of the extension problem by simple comparison.
I have asked all people I could reach in real life, but I do not have a satisfying answer yet.
Any answer or hint is greatly appreciated.

REPLY [6 votes]: The condition required on $X$ which makes this work is that $X$ is cofree (Definition 10.1 in the linked paper): the map $X \to F(EG_+,X)$ is an equivalence. For any equivariant map $X \to Y$ of cofree $G$-spectra which is an equivalence on the underlying spectra, the resulting maps $X^H \to Y^H$ of fixed-point sets are equivalent to the weak equivalence of homotopy fixed-point spectra $X^{hH} \to Y^{hH}$. This makes $X \to Y$ into a genuine equivalence.
The fact that $D^{-1} N_2^8 MU_{\Bbb R}$ is cofree is an important result in their paper (Theorem 10.8).<|endoftext|>
TITLE: Adding a truth-like predicate to PA
QUESTION [6 upvotes]: It is well known that adding a truth predicate to arithmetic in the most natural way leads to a contradiction.
Suppose as usual that we add a one place relation T to the language of arithmetic, and define some system of Godel numbering $\ulcorner \cdot \urcorner$ for this expanded language. Given a set of axioms A in this language (e.g., PA), we extend the usual provability relation $A \vdash \phi$ by demanding that it satisfy the following:
R1. if $A \vdash \phi$, then $A \vdash T(\ulcorner \phi \urcorner)$
R2. if $A \vdash T(\ulcorner \phi \urcorner)$, then $A \vdash \phi$
R3. if $A \vdash \neg \phi$, then $A \vdash \neg T(\ulcorner \phi \urcorner)$
R4. if $A \vdash \neg T(\ulcorner \phi \urcorner)$, then $A \vdash \neg \phi$
Notes:
• We are not adding any new axioms (in particular, we are not adding any Tarski biconditionals $\phi \leftrightarrow T(\ulcorner \phi \urcorner)$ to A), or any other rules of inference of any sort related to $T$. All we are adding to A are the above rules.
• Requirements R1-R4 are different from adding rules of inference. Obviously, adding rules of inference that tell that us we can go from $\phi$ to $T(\ulcorner \phi \urcorner)$ and back, and $\neg \phi$ to $\neg T(\ulcorner \phi \urcorner)$ and back, would be too strong and lead to contradictions in well known ways if A is sufficiently strong. Loosely speaking, with R1-R4, we can't reason hypothetically about truth, though we can talk about the truth or falsity of something when we've actually established its truth or falsity.
Question: Given the above requirements, do we nevertheless have $PA \vdash \bot$ ?

REPLY [5 votes]: Obviously, adding rules of inference that tell that us we can go from $\phi$ to $T(\ulcorner\phi\urcorner)$ and back, and $\neg\phi$ to $\neg T(\ulcorner\phi\urcorner)$ and back, would be too strong and lead to contradictions in well known ways.

Actually, that's not correct. According to a theorem of Friedman and Sheard these rules (denoted by them "$T$-intro", "$T$-elim", "$\neg T$-intro", and "$\neg T$-elim") can be consistently added to PA. (See part D of their main theorem.) [Edit: to clarify, this is in the context of Hilbert-style deduction. If these rules of inference were included in a natural deduction system then they could be converted to implications using $\to$-introduction, and then you would get the Tarski biconditionals and a liar paradox.]
I may add that I have formulated an extension $S$ of Peano arithmetic which is consistent, includes the capture scheme $\phi \to T(\ulcorner \phi\urcorner)$ for every sentence $\phi$, and for which $T(\ulcorner T(\ulcorner \cdots \ulcorner\phi\urcorner \cdots \urcorner)\urcorner)$ is not a theorem, for any false arithmetical sentence $\phi$. Thus if the release scheme "infer $\phi$ from $T(\ulcorner \phi\urcorner)$" were added it would still be consistent. Incidentally, this system has the remarkable property that it proves its own soundness and consistency in the sense that it proves the sentences $$(\forall n)({\rm Prov}_S(n) \to T(n))$$ and $$T(\ulcorner{\rm Con}(S)\urcorner).$$ See this paper or my book. (Note that my system uses intuitionistic logic, although it includes the law of excluded middle for every sentence of arithmetic.)<|endoftext|>
TITLE: Solving equations in SO(3) : an open problem by Jan Mycielski
QUESTION [20 upvotes]: I am interested in a problem closely related to a problem stated by Jan Mycielski in his paper Can One Solve Equations in Group?  (The American Mathematical Monthly, 1977, http://www.jstor.org/stable/2321255).
Let $p_1, q_1,...., p_m, q_m$ be fixed integers (both positive or negative).  I am interested in the image of the mapping $SO(3)\times SO(3) \to SO(3)$ given by $(X, Y)\mapsto X^{p_1} Y^{q_1} X^{p_2} Y^{q_2}\cdots X^{p_m}Y^{q_m}$.  By conjugacy and continuity, it is easy to see that the image is the set of all  rotations by angles in $[0,\alpha]$ for some $\alpha$ that depends on $p_1, q_1,...., p_m, q_m$.  Since $SO(3)$ contains a copy of the free group on two generators, $\alpha$ is always strictly positive.
Mycielski asked whether $\alpha$ is always equal to $\pi$.  I am intersted to know whether $\alpha$ is always at least $\pi/2$.
Is there anything new to be said about this problem, or is it still wide open?  I checked the papers citing Mycielski's paper, but none of them seem to have a solution.

REPLY [14 votes]: Let me collect a number of known results: 
i) $\alpha$ can be arbitrarily small, see my paper
Andreas Thom, Convergent sequences in discrete groups, Canad. Math. Bull. 56 (2013), no. 2, 424–433. 
ii) There is some interest in estimating how small $\alpha$ can be in terms of the word length, this has been studied in the paper above, but also in
Abdelrhman Elkasapy and Andreas Thom, On the length of the shortest non-trivial element in the derived and the lower central series. J. Group Theory 18 (2015), no. 5, 793–804.
iii) In many cases $\alpha= \pi$. This has been studied in
Abdelrhman Elkasapy and Andreas Thom, About Gotô's method showing surjectivity of word maps, Indiana Univ. Math. J. 63 (2014), no. 5, 1553–1565.
and a more recent preprint
Anton Klyachko and Andreas Thom, New topological methods to solve equations over groups, arXiv:1509.01376
iv) The shortest word that I know for which $\alpha< \pi/2$ is $w=[[[XY X,YXY],[XYX,Y]],[[XYYX,YXXY],[X,Y]]]$<|endoftext|>
TITLE: $2$-cohomology group of semi-direct products
QUESTION [5 upvotes]: Let $G=N\rtimes T$ and let $A$ be a $G$-module with a trivial $G$-action.
The action of $T$ on induce a natural action of $T$ on the second cohomology group of $N$. Denote by $H^2(N,A)^T$ the $T$-invariant cohomology classes in $H^2(N,A)$. 
For $A=\mathbb{C}^*$ and $(|N|,|A|)=1$ it is shown (e.g. in "The Schur Multiplier" by G. Karpilovsky) that 
$$H^2(G,\mathbb{C}^*)=H^2(N,\mathbb{C}^*)^T\times H^2(T,\mathbb{C}^*).$$
My question is: is this result is true when we replace $\mathbb{C}^*$ with $A$ and if not what is the generalization?
The proof given in Karpilovsky does not work.
However, in http://link.springer.com/article/10.1007%2FBF01181625#page-1, Tahara deals with similar questions but I couldn't derive the desired result from there.   
I will appreciate any help.

REPLY [2 votes]: From the spectral sequence mentioned by Anton one can derive an exact sequence
in low degrees 
$$ 0\to E_2^{1,0}\to E^1\to E_2^{0,1}\to E_2^{2,0}
    \to \mathrm{ker}\left[E^2\to E_2^{0,2}\right]\to E_2^{1,1}\to E_2^{3,0}\ , $$
where $E^n=H^n(G,V)$ for $n=1,2$. I have found this exact sequence on page 38 of Sansuc's paper   (without proof). Maybe this is what you need.
Note also that in order to construct the spectral sequence and the exact sequence in low degrees, you need only the short exact sequence
$$ 1\to N\to G\to T\to 1.$$
From the splitting of this short exact sequence you can extract additional information about the exact sequence in low degrees.<|endoftext|>
TITLE: How can any theory prove well-foundedness of ordinals above $\omega_1^{\text{CK}}$?
QUESTION [12 upvotes]: $\newcommand{\omegaoneck}{\omega_1^{\text{CK}}}$
Pardon the extremely basic question - this isn't quite my area - but I'm confused about the definition of proof theoretic ordinals.  
The proof theoretic ordinal of a theory is defined to be the smallest ordinal that the theory cannot prove is well founded.  In other words, it gives us a measure of how much transfinite induction the theory allows us to do.  There is a countable ordinal $\omegaoneck$ such that every recursive theory has a proof theoretic ordinal $\alpha\in\omegaoneck$.  
This seems fine, except that there exist recursive theories (such as ZFC) in which we can prove transfinite induction over all ordinals.  My first thought was that perhaps ZFC can't prove the existence of ordinals larger than its own proof-theoretic ordinal $\omega_{ZFC}$, so 'induction over all ordinals' is the same as 'induction up to $\omega_{ZFC}$', within ZFC.  But that can't be true: ZFC can prove Hartogs' Lemma, which gives us arbitrarily large ordinals (and they can be shown to be arbitrarily large within ZFC).  
One possibility might be that ZFC can't reason about its own proof-theoretic ordinal, and can't show that it's less than any given ordinal.  But that seems rather strange, given that ZFC can construct ordinals such as $\omega_1$ that are far far larger than even $\omegaoneck$, even in cardinality.

REPLY [12 votes]: Just slightly expanding the comment of Emil Jeřábek: on one hand, in ZFC we can define some objects we call ordinals and prove transfinite induction of each of them. This is not what we mean by the ordinal of ZFC. The latter is defined roughly as follows (although there are several nonequivalent definitions):

You take some recursive relation, such that in the actual natural numbers, this relation is well-founded and you actually fix some algorithm which describes this relation (as a set of pairs). 
Then you ask, whether the theory in question proves that the relation given by this algorithm is a well-founded one.

For instance, think of the following relation $R$: $R(m,n)$ iff $m<n$ and there is no proof of $0 \neq 0$ from the axiom of ZFC using less than  $m$ symbols or $m>n$ otherwise. This relation is easily verified to be a recursive linear order by even very weak theories but is  wellfounded (namely isomorphic to $\omega$) iff ZFC is consistent, so ZFC does not even prove that $\omega$ is well-founded under totally arbitrary presentation of $\omega$.
Notice that the above example could give an impression that no theory can have its prof-theoretic ordinal bigger than $\omega,$ therefore precise definition of this concept is a rather subtle issue and there are several nonequivalent definitions. But the rough idea, of what may go wrong is precisely this.<|endoftext|>
TITLE: Blocking sets in three dimensional finite affine spaces
QUESTION [11 upvotes]: What is the smallest possible size of a set of points in $\mathbb{F}_q^3$ which intersects (blocks) every line?
Clearly the union of three affine hyperplanes that intersect in a singleton, say $x = 0, y = 0$ and $z = 0$, forms such a set. If we denote by $s(q)$ the smallest possible size, then this gives us $$s(q) \leq 3q^2 - 3q + 1$$ for all $q$. 
Since any two points of $\mathbb{F}_2^3$ form a line, we get $s(2) = 2^3 - 1 = 7 = 12 - 6 + 1$. For $q = 3$ we can do better. In $\mathbb{F}_3^3$ the complement of such a set is a cap, i.e., a set of points no three of which are collinear. We know that the largest size of a cap in $\mathbb{F}_3^3$ is $9$ (corresponding to a quadric), and hence $s(3) = 18 < 27 - 9 + 1$. (side note: the problem of finding such a blocking set in $\mathbb{F}_3^n$ is thus equivalent to the famous cap set problem. See the survey article by Bierbrauer and Edel, large caps in projective Galois spaces and the paper by Bateman and Katz, new bounds on cap sets) 
Question 1: Can we improve the upper bound in general?
We can also give a lower bound on $s(q)$. Jamison/Brouwer-Schrijver proved using the polynomial method that the smallest possible size of a blocking set in $\mathbb{F}_q^2$ is $2q - 1$. See this, this, this and this for various proofs of their result. Now take any $q$ parallel affine planes in $\mathbb{F}_q^3$, then the intersection of a blocking set with these hyperplanes must have size at least $2q - 1$, and hence $$2q^2 - q \leq s(q).$$ 
Question 2: Can we improve this lower bound in general?
The Jamison/Brouwer-Schrijver result gives us another way of constructing a blocking set of size $3q^2 - 3q + 1$. Again take $q$ parallel hyperplanes $H_1, \dots, H_q$. Let $B_2, \dots, B_{q}$ be blocking sets of size $2q - 1$ in $H_2, \dots, H_{q}$. Then $B = H_1 \cup B_2 \cup \dots \cup B_{q}$ is a blocking set of size $(q-1)(2q-1) + q^2 = 3q^2 - 3q + 1$. 
Note that the problem of determining $s(q)$ is trivial for projective spaces. It's a classical result that a line blocking set in $PG(3,q)$ has size at least $1 + q + q^2$ with equality if and only if it is a hyperplane. See Chapter 3 of current research topics in Galois geometry for a recent survey on projective blocking sets. 
Edit 1: After Douglas Zare's answer below we have $s(q) \geq 2q^2 - 1$ for all $q$ and $s(q) \leq 3q^2 - 3q$ for $q \geq 3$. Can this be improved further?
I have also found two references that prove this lower bound of $2q^2 - 1$, Proposition 4.1 in Nuclei of pointsets in $PG(n,q)$ (1997) and Theorem 3.1 in On Nuclei and Blocking Sets in Desarguesian Spaces (1999). In fact, Sziklai has mentioned the same argument as Douglas Zare after his proof of Proposition 4.1. Their proofs are generalisations of the polynomial technique introduced by Blokhuis in On nuclei and affine blocking sets (1994).

REPLY [3 votes]: Here is an improvement of the upper bound which I found in ``The polynomial method in Galois geometries'' by Simeon Ball. See page number 4. 

The known constructions are somewhat crude. For example, let $S$ be a
  set of points of $AG(3,q)$ with the property that every line is
  incident with a point of $S$. For $q$ square, the smallest known
  example has size roughly $2q^2 + 2q\sqrt{q}$ and is constructed using
  a double blocking set of $PG(2,q)$ at infinity and forming a cone with
  a vertex point of the affine space.

Let's work out the details. A double blocking set in $PG(2,q)$ is a set of points that intersects every line in at least two points. Let $AG(3,q)$ be embedded in $PG(3,q)$ via a hyperplane at infinity $H_\infty \cong PG(2,q)$. Now let $B$ be a double blocking set in $H_\infty$. Pick any affine point $a$ and let $S$ be the affine part of the union of all lines connecting $a$ to a point of $B$ (the so-called cone). Now let $\ell$ be any line of $AG(3,q)$. The plane $H$ spanned by $a$ and $\ell$ intersects $H_\infty$ in a line $\ell_{\infty}$. The line $\ell_{\infty}$ contains at least two points $b_1, b_2$ of $B$. Since $a, b_1, b_2$ are in $H$ the lines $\ell_1 = ab_1$ and $\ell_2 = ab_2$ are contained in $H$. In the affine part of $H$ there is a unique line through $a$ and parallel to $\ell$. Thus $\ell$ intersects at least one of these lines non-trivially in their affine part, which is contained in $S$. This proves that $S$ is a blocking set in $AG(3,q)$. 
The size of $S$ is equal to $1 + (q-1)|B|$. Therefore, we can take $B$ to be as small as possible to get small blocking sets in $AG(3,q)$. Three non-concurrent lines in $PG(2,q)$ form a double blocking set of size $3q$. But this gives us the old bound of $s(q) \leq 3q^2 - 3q + 1$. When $q$ is a square we can take a union of two disjoint Baer subplanes (each one of them forms a blocking set in $PG(2,q)$, which gives us $$|S| = 1 + (q - 1)(2q + 2\sqrt{q} + 2) = 2q^2 + 2q\sqrt{q} - 2\sqrt{q} - 1.$$
Sadly, we can't do any better than this as shown by Blokhuis and Ball in ``On the size of a double blocking set in $PG(2,q)$'' for $q > 16$ . For non-square $q = p^{2d + 1} > 3$, there is lower bound of $2q + p^d \lceil(p^{d+1} + 1)/(p^d + 1)\rceil + 2$ on the size of a double blocking set, but it is not always sharp. There is further discussion for smaller planes in the paper, but that doesn't help us much in our problem and so we have $$s(q) \leq 2q^2 + 2q\sqrt{q} - q - 2\sqrt{q}$$ for $q$ square, which is strictly better than the previous bound of $3q^2 - 3q$ for all $q \geq 9$.<|endoftext|>
TITLE: Derivative of eigenvectors of a matrix with respect to its components
QUESTION [9 upvotes]: Suppose that $B$ is a real, positive-definitive symmetric ($3\times3$) matrix (more accurately, $B$ is a tensor) with distinct eigenvalues, and that we can write it as 
$$
B= \sum_{i=1}^3 \lambda_{i}(n_{i}\otimes n_{i}),
$$
where $ n_{i} $ and $ \lambda_{i} $ are the unit eigenvectors and eigenvalues of $B$, and $\otimes$ is the dyadic product.
I need to calculate the derivative of eigenvectors of $B$ with respect to its components; in other words:
 $$
\frac{\partial n_{i}}{\partial B}
$$
I read some sources such as Kato's Perturbation theory but I couldn't figure out how to solve this, so I'm looking for pointers to a closed form solution.

REPLY [16 votes]: If $B$ depends on a single parameter $t$ then derivating  with respect to $t$  the equality
$$ B n_i =\lambda_i n_i $$
we deduce
$$\dot{B} n_i +B\dot{n}_i=\dot{\lambda}_i n_i +\lambda_i\dot{n}_i. $$
Here we assume that $\Vert n_i\Vert =1$.  Hence $\dot{n}_i\perp n_i$, $\forall i$. Taking the inner product of  the above equality with $n_i$  and observing that
$$ (B\dot{n}_i, n_i)=(\dot{n}_i, Bn_i)=\lambda_i(\dot{n}_i,n_i)=0 \tag{1} $$
we deduce
$$\boxed{\dot{\lambda}_i=(\dot{B}n_i,n_i).} $$
This determines $\dot{\lambda}_i$ in terms of $\dot{B}$. 
Next, we take the inner product of (1)  with $n_j$, $j\neq i$.  Using the fact that $B$ is symmetric we deduce
$$(\dot{B}n_i, n_j)+(\dot{n}_i, Bn_j)=\lambda_i (\dot{n}_i, n_j) $$
so that
$$(\dot{B}n_i, n_j)+\lambda_j(\dot{n}_i, n_j)=\lambda_i (\dot{n}_i, n_j) $$
This shows that
$$(\dot{n}_i, n_j)=\frac{1}{\lambda_i-\lambda_j}(\dot{B}n_i, n_j), $$
that is
$$\dot{n}_i=\sum_{j\neq i} \frac{1}{\lambda_i-\lambda_j}(\dot{B}n_i, n_j) n_j.\tag{2} $$
If we let the parameter  $t$ to be  entry $b_{k\ell}$, $k\leq \ell$,  of $B$  with respect to some fixed orthonormal basis we deduce
$$\boxed{\frac{\partial n_i}{\partial b_{k\ell}}=\sum_{j\neq i} \frac{1}{\lambda_i-\lambda_j}\Biggl(\frac{\partial B}{\partial b_{k\ell}}n_i, n_j\Biggr) n_j.} $$
This is equivalent with the formula of Carlo Beenakker.<|endoftext|>
TITLE: Why do the Maynard-Tao weights work so well?
QUESTION [12 upvotes]: I am looking for an intuitive reason for why the Maynard-Tao weights work well to capture many primes of the form $n+h_1, \ldots , n+h_k$, where $(h_1, \ldots , h_k)$ is any admissible $k$-tuple.
For instance, where is the mass of the weights concentrated? Why is it good to concentrate the mass on these integers? Is there a explanation to why these work better than the GPY-weights (see this related question)?
For instance Proposition 4.2 of this polymath 8b paper asserts that the only integers the contribute are the $n$ such that $n+h_1, \ldots, n+h_k$ are $x^{\epsilon}$-rough, where $\epsilon = 1/(10k)$.

REPLY [11 votes]: In retrospect the Maynard-Tao weights are quite natural, and they can be understood as the result of the following evolution. 
1. Let us start from the naive weights
$$\nu(n):=1_\text{$n+h_1,\dots,n+h_k$ are primes}$$
If we believe in Dickson's conjecture, then these weights are perfectly fine, because they concentrate on the best $n$'s. The problem is that proving the desired inequality
$$ \sum_{x\leq n\leq 2x}\nu(n)\sum_{i=1}^k 1_\text{$n+h_i$ is prime}>\sum_{x\leq n\leq 2x}\nu(n)\tag{$\ast$}$$
is as hard as proving Dickson's conjecture. Indeed, this inequality implies that not all the $\nu(n)$'s are zero. A slightly refined analytic variant of these naive weights is
$$\nu(n):=\sum_{d\mid P(n)}\mu(d)\log^{k+\ell}\left(\frac{P(n)}{d}\right),$$
where $P(n):=(n+h_1)\dots(n+h_k)$ is as usual. The right hand side is the convolution of $\mu$ and $\log^{k+\ell}$ evaluated at $P(n)$, hence (nontrivially) these weights are nonnegative and they are nonzero if and only if $P(n)$ has at most $k+\ell$ distinct prime factors. So, if $\ell$ is not too large (e.g. $\ell<k/2$), these weights capture $n$'s such that $n+h_1,\dots,n+h_k$ contains several primes. An advantage of these weights is that they incorporate the usual inclusion-exclusion sieve technology via the Möbius values $\mu(d)$. However, they suffer from the same problem as the original naive weights: we cannot evaluate the two sides of ($\ast$) for them. At a technical level, the source of the problem is that the divisors $d$ of $P(n)$ get too large (compared to $x$), and we lose control of the error terms.
2. A key idea of sieve theory is to remedy the above mentioned difficulty by restricting to fewer divisors, e.g. by cutting them off at some artificial bound $R$. Of course, a smooth cut-off is usually better for analytic purposes than a sharp cut-off. For example, one can try to use the weights
$$\nu(n):=\left(\sum_{\substack{{d\mid P(n)}\\{d\leq R}}}\mu(d)\left(1-\frac{\log d}{\log R}\right)^{k+\ell}\right)^2,$$
which differs from the previous version (up to scaling) as follows: the restriction $d\leq R$ is in place, the fraction $P(n)/d$ became $R/d$, and the whole sum got squared. Only the last change needs an explanation: the cut-off $d\leq R$ destroys positivity of the weight, and squaring is a good analytic way to restore positivity.
3. The above squared weights are precisely the Goldston-Pintz-Yildirim weights. A key insight of Soundararajan was that it is more efficient to perform the smooth cut-off $d\leq R$ by using a more general function of $\frac{\log d}{\log R}$. That is, the Goldston-Pintz-Yildirim weights evolved to
$$\nu(n):=\left(\sum_{d\mid P(n)}\mu(d)g\left(\frac{\log d}{\log R}\right)\right)^2,$$
where $g:\mathbb{R}\to\mathbb{R}$ is a smooth function supported on $[0,1]$.
Morally, these weights try to imitate the event that $P(n)$ has few prime factors by sieving through the divisors $d\mid P(n)$. However, what we really want is that many of the individual factors $n+h_1,\dots,n+h_k$ are primes, not just that their product has few prime factors. So it is natural to further refine the above Soundararajan weights to
$$\nu(n):=\left(\sum_{\forall i: d_i\mid n+h_i}\mu(d_1)\dots\mu(d_k)f\left(\frac{\log d_1}{\log R},\dots,\frac{\log d_k}{\log R}\right)\right)^2,$$
where $f:\mathbb{R}^k\to\mathbb{R}$ is a smooth function supported on the simplex
$$\{(t_1,\dots,t_k)\in\mathbb{R}_{\geq 0}^k: t_1+\dots+t_k\leq 1\}.$$
These are precisely Maynard-Tao weights, and they really try to imitate that many of the individual factors $n+h_1,\dots,n+h_k$ are primes.
4. It is worthwhile to bear in mind that the Soundararajan weights (hence also the Goldston-Pintz-Yildirim weights) are a special case of the Maynard-Tao weights. Indeed, let us assume (without much loss of generality) that $n+h_1,\dots,n+h_k$ are pairwise coprime. Then, for
$$f(t_1,\dots,t_k):=g(t_1+\dots+t_k)$$ 
the Maynard-Tao weights reduce to the Soundararajan weights.<|endoftext|>
TITLE: Upper bound on answer for Pell equation
QUESTION [13 upvotes]: A user on MSE, @martin , asked https://math.stackexchange.com/questions/1611411/pell-equations-upper-bound about an upper bound for   $x$ in $x^2 - p y^2 = 1,$ when $p$ is prime. I checked, it appears reasonable to guess that
$$ x < p^{\sqrt p} $$
when $p > 2.$ I had the computer solve by Lagrange's method, no continued fractions, no decimal accuracy required, no memory required, but the method is still elementary.  I had the machine print out whenever $\log_p(\log_p(x))$ increased. It was necessary to take $p > 2$ because $x=3$ gives an overly large logarithm. Meanwhile, if all we do is print whenever $x$ itself increases, there are several composite numbers below $100$ that get included, after that they give way to primes $p \equiv 1 \pmod 4.$ I put in a fair amount of effort but was unable to draw any firm conclusions.
So, the questions would be, (I) what is unconditionally proved about the size of $x,$ (II) what is proved under conjectures that people mostly believe true, (III) what are the most optimistic things conjectured? 
p                                       
5        log_p(x)     1.365212388971971   log_p(log_p(x)) 0.1934277864616169   X 9  
13       log_p(x)     2.524585016802303   log_p(log_p(x)) 0.3610506760085375   X 649  
61        log_p(x)    5.17947382679923   log_p(log_p(x))  0.4000860954668999   X 1766319049  
109      log_p(x)     6.969012778576543   log_p(log_p(x)) 0.4138413148682316   X 158070671986249  
421      log_p(x)    12.79922341582056   log_p(log_p(x))  0.4218996203501611   X 3879474045914926879468217167061449  
1621     log_p(x)    23.61505725662223   log_p(log_p(x))  0.4278136548619654   X 6298101812493732343034974500091457815529942308667051412857352310169665125001  
.....................
44450701  log_p(x) 2641.408511213517     log_p(log_p(x))  0.4474228404332914   X  is rather large...

Why not, here is how it begins if we print every time $x$ increases and make no requirement about loglog, allowing $n$ composite in $x^2 - n y^2$ with $2 \leq n \leq 500$
2  log_p(x) 1.584962500721156   log_p(log_p(x))   0.6644487074538893   X 3
5  log_p(x) 1.365212388971971   log_p(log_p(x))   0.1934277864616169   X 9
10  log_p(x) 1.278753600952829   log_p(log_p(x))  0.1067868696893203   X 19
13  log_p(x) 2.524585016802303   log_p(log_p(x))  0.3610506760085375   X 649
29  log_p(x) 2.729264122987999   log_p(log_p(x))  0.298171610554983   X 9801
46  log_p(x) 2.637925539730376   log_p(log_p(x))  0.2533517055829028   X 24335
53  log_p(x) 2.79606031271967   log_p(log_p(x))   0.258976271165875   X 66249
61  log_p(x) 5.17947382679923   log_p(log_p(x))   0.4000860954668999   X 1766319049
109  log_p(x) 6.969012778576543   log_p(log_p(x)) 0.4138413148682316   X 158070671986249
181  log_p(x) 8.146702019142648   log_p(log_p(x)) 0.4035037766708247   X 2469645423824185801
277  log_p(x) 8.271023203635528   log_p(log_p(x)) 0.3756670785256742   X 159150073798980475849
397  log_p(x) 8.05129073299257   log_p(log_p(x))  0.3485719633766078   X 838721786045180184649
409  log_p(x) 8.576275777667302   log_p(log_p(x)) 0.3573497754649824   X 25052977273092427986049
421  log_p(x) 12.79922341582056   log_p(log_p(x)) 0.4218996203501611   X 3879474045914926879468217167061449

REPLY [19 votes]: Let $d$ be a positive fundamental discriminant, $\epsilon_d$ denote the fundamental unit, $h(d)$ the class number, and $\chi_d$ the primitive character associated to the discriminant $d$.  The class number formula gives 
$$ 
\log \epsilon_d = \sqrt{d} L(1,\chi_d)/h(d) \le \sqrt{d} L(1,\chi_d), 
$$ 
since the class number $h(d) \ge 1$.  Now it is known that 
$L(1,\chi_d) \le C \log d$ for a constant $C$.  This upper bound is completely effective.  The best known constant $C$ is (for large enough $d$) 
$$ 
\frac{1}{4} \Big(2-\frac{2}{\sqrt{e}} \Big). 
$$
If we knew something about how small primes split in ${\Bbb Q}(\sqrt{d})$ then this could be improved by taking those Euler factors into account (for example, we can use this if $d$ is even which would happen for primes $p\equiv 3\pmod 4$ in the question). This is a result of P.J. Stephens and uses the Burgess bound for character sums (together with an argument from multiplicative number theory along the lines of Vinogradov's $1/\sqrt{e}$ argument for the least quadratic non-residue).  For a discussion of Stephens's result and extensions, see Granville and Soundararajan.
This would be enough to give your conjecture of $p^{C\sqrt{p}}$ (one needs a little care to go from the fundamental unit to the solution to Pell's equation -- i.e. one might need to take a small power of the fundamental unit).  Also see this paper of Hua which explicitly states a bound along the lines you want, tracing it back to Schur.  Finally, Louboutin has looked at explicit upper bounds for $L(1,\chi)$ (see Theorem 5.1 there).  
The above results are unconditional.  On GRH one can do a bit better, since  $L(1,\chi_d)$ may then be bounded by $C\log \log d$, and then one would get a better bound of $(\log p)^{C\sqrt{p}}$ in your problem (see Theorem 1.5 of this paper for an explicit GRH bound), and I think that can probably happen (although this is not clear since we don't know that the class number can get down to $1$). Jacobson, Lukes and Williams report on extensive calculations on regulators, and at the end of the paper state the belief that the fundamental unit can get as large as $\exp(c \sqrt{d}\log \log d)$;  however, as also noted there, unconditionally we only know that the fundamental unit (or the solution in Pell's equation) sometimes gets as large as $\exp(c(\log d)^4)$ -- so there is a very large gap in our understanding.   (See also my answer to the related MO question Upper bound for class number of a real quadratic field .)<|endoftext|>
TITLE: Existence of a skew field with surjective inner derivations
QUESTION [7 upvotes]: In my research, I've come twice now towards a skew field $K$ that satisfies the following:
$$\text{for all non-central element $a$, the map }\quad x\mapsto ax-xa\quad\text{ is onto.}$$
I am hoping such a field does not exist but am unable to prove it.
Does this condition ring a bell to some of you ?
Any example of such a field in both zero and non-zero characteristic ?
Edit: reading the paper suggested by @TomDeMedts'comment made me realise that this is precisely Question 3 of http://www.ams.org/journals/proc/1961-012-03/S0002-9939-1961-0124355-6/ asked by G. Meisters in 1961.

REPLY [3 votes]: In The range of derivations on a skew field and the equation ax-xb=c, Journal of the Indian Math. Soc. 37 (1973), 61--69, P.M. Cohn shows the following :
Theorem (Cohn 1973). Given a field $K$ with centre $k$, there is an extension field $L$ (still with $k$ as centre) such that the equation $$ax-xa=c$$ has a solution for each $a\in L$ that is transcendental over $k$ and each $c\in L$.
In particular, taking a skew field $K$ having an algebraically closed centre (hence infinite dimentionnal over $k$ as @René notes) provides an $L$ for which every non-zero inner derivation is onto.<|endoftext|>
TITLE: finitely presented subgroup and free solvable group of class 3
QUESTION [5 upvotes]: Let $F(n)$ be free group of rank $n\geq 2$.  Denote  by $F_d(n)$ the d-th derived subgroup, that is $F_d(n)=[F_{d-1}(n),F_{d-1}(n)]$ where $F_0(n)=F(n)$. The free solvable group of rank $n$ and solvable class $d$ is $S_{n,d}=F(n)/F_d(n)$.
By Corollary 2.14. in "FINITELY PRESENTED WREATH PRODUCTS AND DOUBLE
COSET DECOMPOSITIONS",(http://arxiv.org/pdf/math/0509090.pdf), the free solvable group of class 3 and finite rank  is not a finitely presented group. Does this group contain a finitely presented subgroup of solvable class 3?

REPLY [5 votes]: No, it does not exist.
It's not too hard to check that a fp subgroup of a free metabelian group is abelian. A consequence (since any quotient of a fp solvable group is fp, by Bieri-Strebel) is that any fp subgroup of a free 3-solvable group has abelian image in the metabelianization, and in particular, has to be metabelian (I'd guess that it even has to be abelian!).
I have to justify the first claim (fp subgroup of a free metabelian group $G$ is abelian). Write, $M=[G,G]$, $Q=G/M$. Then $M$ is a torsion-free $\mathbf{Z}Q$-module [edit: justification below].
Let $H$ a fg subgroup of $G$, and $Q_H$ be the image of $H$ in $Q$.
If If $M\cap H$ or $Q_H$ is trivial then clearly $H$ is abelian.
So assume otherwise, and let us show that $H$ is not finitely presented.
Then $M\cap H$ is also, by restriction, a torsion-free $\mathbf{Z}Q_H$-module. So it is not finitely-generated over the subsemigroup $\{q\in Q_H:\ell(q)\ge 0\}$ for any choice of nonzero homomorphism $Q_H\to\mathbf{R}$. This means that the Bieri-Strebel invariant of the $Q_H$-module $M\cap H$ is the whole space of homomorphism (in conventional literature, it will be rather called empty Bieri-Strebel invariant: I always define the Bieri-Strbel invariant as the closed complement rather than the open subset, in accordance to the idea that we define the spectrum of a matrix rather than the complement of its spectrum). Since a f.g. metabelian group $G$ is f.p. iff its Bieri-Strebel invariant (a closed subset of $\mathrm{Hom}(G,\mathbf{R})$, closed under positive scalar multiplication) contains a line, we deduce that $H$ is not finitely presented.

[Edit:] That $M$ is a torsion-free $\mathbf{Z}Q$-module is a consequence of Magnus' embedding theorem. Assume, as we can, that $G$ is finitely generated free metabelian over generators $x_1,\dots,x_n$. Let $t_1,\dots,t_n,e_1,\dots,e_n$ be complex numbers, such that $(t_1,\dots,t_n)$ is algebraically independent over $\mathbf{Q}$, and $(e_1,\dots,e_n)$ are linearly independent over $\mathbf{Q}(t_1,\dots,t_n)$ (of course this holds if $(t_1,\dots,t_n,e_1,\dots,e_n)$ is algebraically independent).
Magnus says that mapping $x_i\mapsto\begin{pmatrix}t_i & e_i\\ 0 & 1\end{pmatrix}$ yields an injective homomorphism from $G$ to $\mathrm{GL}_2(\mathbf{C})$. Actually, it maps into the subgroup $H$ consisting of elements of the form $\begin{pmatrix}t & v\\ 0 & 1\end{pmatrix}$, where $t$ has the form $\prod t_i^{m_i}$, $(m_1,\dots,m_n)\in\mathbf{Z}^n$, and $v$ belongs to the $\mathbf{Z}[t_1^{\pm 1},\dots,t_n^{\pm 1}]$-submodule generated by $(e_1,\dots,e_n)$. We see $H$ is just a standard wreath product $$\mathbf{Z}^n\wr\mathbf{Z}^n=\mathbf{Z}[t_1^{\pm 1},\dots,t_n^{\pm 1}]^n\rtimes\mathbf{Z}^n=(\mathbf{Z}Q)^n\rtimes Q.$$ In particular, $M$ is isomorphic to a $\mathbf{Z}Q$-submodule of a free $\mathbf{Z}Q$-module of rank $n$, so is torsion-free.<|endoftext|>
TITLE: How do I determine a real matrix form for a group representation?
QUESTION [6 upvotes]: Hello mathoverflow community,
I am a little stucked working on my master thesis. For a representation on $\mathbb{Z}_p\ltimes\mathbb{Z}_p^*$ induced from the additive character $\chi$ of $\mathbb{Z}_p$ given by $\chi(g)=e^{\frac{2\pi i g}{p}}$, I obtain a complex matrix form. That is for $(a,b)\in\mathbb{Z}_p\ltimes\mathbb{Z}_p^*$, I find matrices $R(a,b)\in\mathbb{C}^{\mathbb{Z}_p^*\times\mathbb{Z}_p^*}$ with
$
(R(a,b))_{j,k}=\begin{cases}
\chi(aj^{-1})&\text{if }k=b^{-1}j,\\
0&\text{else.}
\end{cases}
$
In theory, there must be an equivalent representation with real matrix form (e.g. by Schur-Frobenius test). My aim is to determine a real symmetry-adapted basis and so I would like to compute this real matrix form. Is there anything known how to do that in practise or do you see any good way to do this basis transformation?
Best regards
Aron
P.S:
Notations: p is a prime number. $\mathbb{Z}_p$, resp. $\mathbb{Z}_p^*$  is the additive, resp. multiplicative, group of integers modulo $p$.

REPLY [6 votes]: Here's one way: Note that your semidirect product has a normal subgroup $\mathbb{Z}_{p} \mathbb{Z}_{2}$ which is dihedral with $2p$ elements ( I assume the semidirect product you want to work with is the holomorph of $\mathbb{Z}_{p}$).
It is easy to see a real faithful representation (of degree $2$) of that dihedral group ( using a rotation through $\frac{2 \pi }{p}$ and a reflection). Now you can induce that representation to the whole holomorph to get a real absolutely irreducible faithful representation of degree $p-1$.
In fact, the $p-1$ dimensional representation may be realized over $\mathbb{Q}$, but the above answers your immediate question.<|endoftext|>
TITLE: How many solutions does $\frac{1}{x_1}+\frac{1}{x_2}+\cdots +\frac{1}{x_n}=1$ have?
QUESTION [11 upvotes]: It is well known that $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ and this is the only solution to  $\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}=1$
with $2\leq x_1<x_2<x_3$.
My question is:  

Let $n\in \mathbb{N}$ and $x_i\in \mathbb{N} ,1\leq i\leq n$.   
How many $n$-tuples $(x_1,x_2,...,x_n)$ exist  such that    $\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}=1$ with $2\leq
x_1<x_2<...<x_n$ ?  

Any reference would be appreciated!

REPLY [9 votes]: It might be worth remarking that if $G$ is a finite group with $n$ conjugacy classes and representatives $g_{1},g_{2}, \ldots ,g_{n}$, then the class equation for $G$ (divided through by $|G|$) gives $\sum_{i=1}^{n} \frac{1}{|C_{G}(g_{i})|} = 1$, though the $|C_{G}(g_{i})|$ are not usually distinct. Long ago (around 1895, I think) , E. Landau gave a bound on the number of solutions of $\sum_{i=1}^{n} \frac{1}{x_{i}} = 1$ with the $x_{i}$ (not necessarily distinct) positive integers. This allows a crude bound on the size of a finite group with $n$ conjugacy classes though this is quite far from the presently known bounds (eg of L. Pyber) which use much more group-theoretic information.<|endoftext|>
TITLE: Taller models of ZFC
QUESTION [7 upvotes]: This question is somewhat related to a previous one, where I asked for new forms of infinite beyond the cardinal hierarchy. 
Using forcing techniques, at least the ones I know of, one starts from a ground model and then enlarge it, by adding a generic set and closing up with respect to relative constructibility.
The new model shares the same ordinals  and the same height of the original one. 
On the other hand, if one postulates some suitably large cardinal axiom, one gets as a side effect new transitive models of ZFC, shorter than the one corresponding to the large cardinal's height. 
Looks like one can fatten a model, keeping the same height, or make it shorter, but what about creating  taller ones? 
Question(s) :

1) Are there standard techniques to make a transitive model taller
  without adding new axioms to ZFC? 
2) Does an axiom like this make sense (ie is neither a theorem of ZFC, nor blatantly inconsistent):
For any transitive model of ZFC $M$ of height $\eta_0=\text{ht}(M)$,
  there is always another transitive model $N$ such that $M=V_{\eta_0}^N$ of height
  $\eta_1=\text{ht}(N)$, where $\eta_0 < \eta_1$

NOTE: in the "axiom" above I have simply asked for the second ordinal to be higher than the height of the starting model, but I am also curious as to which extent one could strengthen that, by asking for the second height to be much higher (add your own brand of -much- ) than the first one

REPLY [13 votes]: Regarding Question 1: 

It is certainly possible (i.e., consistent with ZFC) that there is only one transitive model of ZFC.  For example, if there are transitive models  $M$ and $N$ of ZFC such that $o(M)<o(N)$ (where $o(M)$ is the ordinal height of $M$), then there are ordinals $\alpha < \beta$ such that $L(\alpha)$ is the so-called Shepherdson-Cohen minimal model of set thery, which is the smallest transitive model of ZFC (where $L(\alpha)$ is the $\alpha$-th approximation to the constructible universe), and $L(\beta)$ is the next shortest transitive model of ZFC + V=L. In this situation $L(\beta)$ is a model of ZFC in which there is no  transitive model of ZFC of height greater than $\alpha$. 
So this shows that there is no universal technique within ZFC for making a transitive model of ZFC taller (where "$N$ is taller than $M$" is defined as $o(M)<o(N)$).

Regarding Question 2.

This question has already been answered by Carl Mummert's comment if "taller" is defined via: $N$ is taller than $M$ iff $o(M)<o(N)$ [since in the theory ZFC plus there are arbitrarily large inaccessibles, every transitive model of ZFC is shorter than some other transitive model].
On the other hand, if "taller'' is defined by the current version of the question which stipulates that $N$ is taller than $M$ if there is an ordinal $\alpha \in N$ such that $M = V_{\alpha}^{N}$, then the answer is in the negative since the minimal model of set theory $L(\alpha)$ (mentioned the answer above to Question 1) has no such taller extension. To see this, one needs to use the well-known fact that $L(\alpha)$ is pointwise definable [I will try to provide a reference in my next edit]. If $N$ is a model of ZFC such that $M = V_{\alpha}^{N}$, then $N$ would have to "notice" that its own $V_{\alpha}$ is countable (since $N$ can define define the satisfaction predicate for any set structure, a satisfaction predicate that will agree with the "real world" satisfaction predicate for $L(\alpha)$), which contradicts an easy theorem of ZFC that says that if $\alpha > \omega$ then $V_{\alpha}$ is uncountable.<|endoftext|>
TITLE: Does a theory of stochastic differential algebras exist?
QUESTION [20 upvotes]: My question is motivated primarily by finance, where a non-technical student will learn how to approach SDEs using the symbolic manipulation of Itô calculus and the few basic rules of Brownian motion, namely the substitutions $$
(dt)^2 = (dB_t)(dt) = (dt)(dB_t) = 0
$$ $$
(dB_t)^2 = dt
$$ and the stochastic product rule $$
d(X_tY_t) = (X_t)(dY_t) + (dX_t)(Y_t) + (dX_t)(dY_t)
$$ Differential algebra deals with rings equipped with derivations, homomorphisms on the additive structure that obey the standard Liebnitz rule. From here there is a relatively strong differential Galois theory that describes how extension of differential fields behave when you adjoin solutions of ODEs.
My question is whether there has been any work on a stochastic differential algebra with the modified Liebnitz rule that can describe when solutions to SDEs might be beyond the scope of integrating/exponentiating rational functions/Brownian motion?

REPLY [6 votes]: Yes. A systematic study of stochastic (differential) algebra could be found in 

Grenander, Ulf. Probabilities on algebraic structures. Dover Books, 1981.

Grenander studied the operation of integration on what are called "stochastic semi-groups". More specifically the Lie group representing the probability measures equipped with covariate derivatives(Lie derivative in most cases). If you want a geometric glimpse, you can have a look at some reference geometric interpretation of general stochastic processes. In fact, this idea is not new, the earliest motivation of studying the probability measures over an algebraic structure can be traced back to 1950s, the famous book

Parthasarathy, Kalyanapuram Rangachari. Probability measures on metric
  spaces. Vol. 352. American Mathematical Soc., 1967.

devoted two chapters extending the idea of studying the probability measures over locally compact groups, of which classic groups became natural candidates and hence the Lie derivative is yielded to describe the stochastic integrations. Actually this thinking is quite dominating in modern probability theory works, not only Pathasarathy's contribution but also later study carried out by Ledoux-Talagrand(they generalized the underlying space into Banch spaces instead of metric space) and Ambrosio (the study of gradient flow over the space consisting of probability measures.)
However, when you mentioned differential algebra, you are actually referring to a different object which is started by Kolchin et.al. Differential Galois theory is the correct name of the branch that studies the algebraic structure equipped with a derivation homomorphism.
The point here is that differential algebra does not provide too deep insight into the derivation homomorphism itself but focus on the D-module of derivation homomorphisms; however, the study of stochastic integration operators can be well addressed when we replaced the underlying measurable space and equipped with suitable Lie structure as shown by Grenander.<|endoftext|>
TITLE: Does every automorphism of a separably rationally connected variety have a fixed point?
QUESTION [18 upvotes]: Let $k$ be an algebraically closed field.  Let $X$ be a smooth, projective variety over $k$ that is separably rationally connected, i.e., there exists a $k$-morphism $u:\mathbb{P}^1_k \to X$ such that $u^*T_X$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1_k}(a_1)\oplus \dots \oplus \mathcal{O}_{\mathbb{P}^1_k}(a_n)$ for positive integers $a_1,\dots,a_n$.  Let $f:X\to X$ be a $k$-automorphism.  
Question. Does there exist a $k$-point of $X$ that is fixed by $f$?
This is true if $k$ is of characteristic $0$ by the Atiyah-Bott fixed point theorem.  That theorem was extended to positive characteristic by many authors; one reference is Corollaire 6.12 (Appendice), Exposé III, SGA 5.  This extension does not quite give the result above; it does give the result if $h^i(X,\mathcal{O}_X)$ vanishes for all $i>0$, but that is unknown for arbitrary separably rationally connected, smooth, projective varieties.  
By a theorem of Kollár, this is also true in positive characteristic if $f$ has finite order.  By the "spreading out" technique, this implies the result if there exists an ample invertible sheaf $\mathcal{L}$ such that $f^*\mathcal{L}$ is isomorphic to $\mathcal{L}$, i.e., if for some $n>0$, the iterate $f^n$ is in the identity component of the automorphism group scheme of $X$.  In particular, the result is true if $X$ is separably rationally connected and Fano.  However, there do exist pairs $(X,f)$ with $X$ a separably rationally connected variety and $f$ an automorphism that preserves no ample divisor class, e.g., translations on the elliptic surface obtained from $\mathbb{P}^2$ by blowing up the base locus of a pencil of plane cubics.
This question is related to the following questions: let $p$ be a prime integer, let $q$ be $p^r$, and let $Y/\mathbb{F}_q$ be a smooth, projective variety whose base change to $\overline{\mathbb{F}}_q$ is separably rationally connected.  Let $g:Y \to Y$ be an $\mathbb{F}_q$-automorphism.  What can we say about the induced permutation of the finite set $Y(\mathbb{F}_q)$ (whose cardinality is congruent to $1$ modulo $q$, by work of Esnault)?  Does there exist an integer $N$ depending only on $p$ (not $q$) and geometric properties of $Y$ and $g$ such that there exists an orbit of size $\leq N$?

REPLY [4 votes]: This is not a full answer, but a partial result. Recall that a rationally chain connected variety has rational decomposition of the diagonal: there exists $N \in \mathbb Z_{> 0}$ and a cycle $Z$ supported on $X \times D$ such that
$$N \cdot \Delta_X = N \cdot (\{x\} \times X) + Z \in \operatorname{CH}^n(X \times X).$$
This always forces $H^0(X,\Omega_X^i) = 0$ for $i > 0$ (in characteristic $p > 0$ we need $(N,p) = 1$, cf. Totaro [Tot], Lemma 2.2). (But for separably rationally connected, we always have $H^0(X,\Omega_X^i) = 0$ for $i > 0$, even in positive characteristic.)
In characteristic $0$, this forces $H^i(X,\mathcal O_X) = 0$ for $i > 0$, by Hodge symmetry. This would imply the claim, as you note, using a Woods Hole trace formula.
In analogy with Totaro's theorem, we have the following:
Theorem. Suppose $(N,p) = 1$, i.e. $X$ has decomposition of the diagonal in $\operatorname{CH} \otimes \mathbb Z/p\mathbb Z$. Then $H^i(X,\mathcal O_X) = 0$ for all $i > 0$.
Proof. Working in $\operatorname{CH} \otimes \mathbb Z/p\mathbb Z$, we may assume $N = 1$ after multiplying by its inverse. We use the cycle class map to Hodge cohomology, cf. Chatzistamatiou–Rülling [CR]. By Proposition 3.2.2 (1) of [loc cit], the map
$$\bigoplus_{i,j} H^i(X,\Omega_X^j) \to \bigoplus_{i,j} H^i(X,\Omega_X^j)$$
induced by $Z$ vanishes on the part where $j = 0$. Thus, on $H^i(X,\mathcal O_X)$, the map induced by $\{x\} \times X$ is the identity. But the map induced by $\{x\} \times X$ factors through $H^i(\{x\}, \mathcal O_{\{x\}})$, which is $0$ for $i > 0$. $\square$
Thus, any rationally connected variety with $H^i(X, \mathcal O_X) \neq 0$ needs to have $N$ (as above) divisible by $p$. We do not know if this is possible when $X$ is separably rationally connected (which is what you're interested in), but it certainly is possible for merely rationally connected:
Example. Let $X$ be a supersingular K3 surface. Then $X$ is unirational (at least if $p \geq 5$), by Liedtke [Lie]. However, since $X$ is a K3 surface, we have $h^2(X,\mathcal O_X) = h^0(X,\Omega_X^2) = 1$. But of course $X$ cannot be separably rationally connected.
To conclude, the difference between rational, mod $p$, and integral decomposition of the diagonal, as well as between rationally (chain) connected and separably rationally connected, remains very subtle. But maybe this gives some ideas for how to construct counterexamples: we need some $p$ stuff happening.
References.
[CR] Andre Chatzistamatiou and Kay Rülling. Higher direct images of the structure sheaf in positive characteristic. Algebra & Number Theory 5 (2011), no. 6, 693–775. MR2923726
[Lie] Christian Liedtke. Supersingular K3 surfaces are unirational. Invent. Math. 200 (2015), no. 3, 979--1014. MR3348142
[Tot] Burt Totaro. Hypersurfaces that are not stably rational. J. Amer. Math. Soc. 29 (2016), no. 3, 883–891. MR3486175.<|endoftext|>
TITLE: When is a general sheaf (on the projective plane) globally generated?
QUESTION [7 upvotes]: Let $v$ be a chern character on $\mathbb P^2$ so that the moduli of sheaves of chern character $v$ is non-empty of the expected dimension. When is it true that the general sheaf in moduli is globally generated?
Here is a first guess: One expects $s$ sections of a rank $r$ bundle to have rank at most $r-1$ in codimension $s-r+1$ (see e.g. the Wikipedia article on the Porteous formula). Hence, if $s \ge r+2$, we would expect the $s$ sections to generate. So:
Conjecture: If $\chi(\mathbb P^2, v) \ge r+2$ and the slope of $v$ is positive, then the general vector bundle with chern character $v$ is globally generated.
QUESTIONS
Is this true?
If not, what are some counterexamples? Is there a different theorem with this conclusion? 
If yes, does it work for other surfaces? How about for higher dimensional varieties (replacing $2$ with $\dim X$)?

REPLY [4 votes]: The result is true if  r=2 , see
Le Potier, J. Stabilité et amplitude sur P2. in Progress in Math., 7 (1980), 145–182, Birkhauser.
see also
Anghel, C., Coanda, I., Manolache, N. 
Globally Generated Vector Bundles on P^n with c_1=4
Ellia, P. 
Chern classes of rank two globally generated vector bundles on P2.
 Rend. Lincei Mat. Appl. 24 (2013), 147–163<|endoftext|>
TITLE: If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?
QUESTION [34 upvotes]: I was describing to a friend the result that a compact Hausdorff space is determined up to homeomorphism up to by its ring of continuous functions, and he asked how one could see the genus of a surface algebraically. I think it's a good question, and I don't know the answer.
Two parts to this question:

Suppose it is known that $A$ is the ring of continuous functions of a closed surface $X$ (topological manifold, real 2-dimensional) of genus $g$. Where in the algebraic structure of $A$ is the genus of $X$? (There are comments from other users about $K$-theory below... I want to emphasize that I am dreaming of a simple algebraic characterization, analogous to the relationship between idempotents and connected components.)
Does the answer to 1 say something interesting for more general classes of rings (ex, those that are rings of continuous functions on a compact Hausdorff space, or more general rings...)

Any ideas? (Analogous to the idempotents / connected components dictionary.)
Here are some thoughts of my own: 

Picking a smooth structure on $X$ amounts to passing to some subring of differentiable functions. In this setting, results about the existence of vector fields with zeros of prescribed index could be interpreted as the existence of derivations with some prescribed zeros (by choosing some arbitrary Riemannian metric and translating the results about vector fields) - though its not clear how to interpret the index of the zero algebraically, or how to describe these subrings. 
Consider all ring homomorphisms to $C(X) \to C(S^1)$ up the homotopy relation induced by factoring through $C(S^1 \times I)$ in the right way. This crudely recaptures the fundamental group, but I don't see how to turn it into an algebraic condition.
Vector bundles are the same as finite projective modules via Swan's theorem - so certain sections of the vector bundle are elements of the corresponding module? Maybe one could get at the Poincare-Hopf computation of the genus that way, but again I don't see how to compute the index. (Also as far as I know, one still needs a smooth structure to use Poincare-Hopf.)

I would be happy to just be able to distinguish $S^2$ and $T^2$ via their ring of continuous functions.

REPLY [10 votes]: The torus has two functions $f$ and $g$ which are (1) relatively prime, (2) each have two square roots, and (3) whose product has $4$ square roots. For instance take two functions which vanish on disjoint loops which are not null-homologous. There sphere does not have two such functions because (1) implies that $V(f)$ and $V(g)$ are disjoint and (2) implies that the complement of $V(f)$ and $V(g)$ are connected, and (3) implies that the complement of $V(fg)$ is not connected. These conditions are impossible to all satisfy at once on the sphere.<|endoftext|>
TITLE: Uniqueness of the fusion ring for simple finite group
QUESTION [6 upvotes]: We know that the irreducible representations $R_i$ of a group $G$ can give rise to a fusion ring: $R_i\otimes R_j = \oplus_k N^{ij}_k R_k$.
I wonder if the following statement is true or not:
If $G$ is a simple finite group, then its fusion ring is different from the fusion rings of any other groups.

REPLY [8 votes]: The fusion ring, as a ring with basis, contains the same information as the character table. So your question, phrased in language more familiar to finite group theorists, is:

Is a finite simple group determined by its character table among all finite groups? 

Here is how to see the above claim. The easy direction is that if you know the character table then you know the fusion ring, as a ring with basis: this is clear since multiplication of characters corresponds to tensor product of representations. In the other direction, for varying $i$, the matrices $N^{ij}_k$ admit simultaneous eigenvectors (namely the vectors $g \mapsto \chi_V(g)$, where $V$ runs over all irreps), and the simultaneous eigenvalues for these eigenvectors are the character values of $R_i$. 
It's known that a finite simple group is determined by its character table among all finite simple groups. And the character table determines whether or not a finite group is simple: $G$ is simple iff the kernel of every nontrivial irreducible character (the elements $g$ such that $\chi(g) = \chi(1)$) is trivial, since kernels are always normal subgroups and every proper normal subgroup is contained in a kernel. 
So the final answer is yes.<|endoftext|>
TITLE: Intuition for the Cartan connection and "rolling without slipping" in Cartan geometry
QUESTION [13 upvotes]: Consider a Cartan geometry $\pi: \mathcal{G} \to M$ with Cartan connection $\omega$ modelled on the Klein geometry $(G, H)$. 
The Cartan connection is supposed to formalize what it means to "roll without slipping" the homogeneous space $G/H$ on the manifold $M$. I am wondering if the following is one correct way to intuitively think about this.
Let $U \subseteq M$ be an open set such that we have a local section $\sigma: U \to \mathcal{G}$ of the Cartan bundle. Then we can pull down the Cartan connection to get a $\frak{g}$-valued 1-form $\sigma^* \omega$ on $U$. 
Suppose we pick a point of contact $o \in G/H$ with a point $x \in M$. Then if we pick a tangent direction $v \in T_x M$ we can imagine rolling the homogeneous space without slipping along the infinitesimal curve corresponding to the tangent vector. The Cartan form $\sigma^* \omega$ assigns to this infinitesimal curve $v$ an infinitesimal transformation $X:= (\sigma^* \omega)(v) \in \frak{g}$. Since $G$ acts on $G/H$, Then I would assume that the infinitesimal transformation $X$ tells us how the point of contact $o \in G/H$ changes as we roll along the infinitesimal curve $v$. 
Thus, if the point of contact is initially $o \in G/H$, then our new point of contact with $v(d) \in M$ after rolling for infinitesimal time $d$ along the infinitesimal curve $v$ will be $X(d).o \in G/H$, where the infinitesimal action of $\frak{g}$ on $G/H$ here is induced by the standard left action of $G$ on $G/H$. 
I am wondering whether this reasoning is a correct way to think about rolling without slipping the homogeneous space along $M$. If not, is there some other way I can think about it geometrically?
Note that while this statement in terms of infinitesimal transformations is a bit fuzzy and "handwavy" in classical differential geometry where we do not have infinitesimal objects, it is entirely rigorous in synthetic differential geometry, and we can embed the category of smooth manifolds fully and faithfully into whatever smooth topos we use to model SDG.

REPLY [14 votes]: You don't need to be 'handwavy' at all, and, in the standard modern way to understand this, one does not need one to choose local sections and talk about infinitesimal motions.  Here's the standard approach:
First, a Cartan connection of type $(G,H)$ on an $n$-manifold $M$ is, as you know, a principal right $H$- bundle $\pi:B\to M$ endowed with a $\frak{g}$-valued $1$-form $\omega:TB\to\frak{g}$ that gives a parallelism of $B$ (i.e., $\omega_b:T_bB\to\frak{g}$ is an isomorphism for each $b\in B$) subject to the following two conditions:  (1) $\omega(Y_v) = v$ for all $v\in\frak{h}$, where $Y_v$ is the vector field on $B$ induced by the flow of right action by the $1$-parameter subgroup $\exp(tv)$ in $H$, and (2) $R_h^*(\omega) = \mathrm{Ad}(h^{-1})(\omega)$ for all $h\in H$.  The curvature of $\omega$ is $\Omega = \mathrm{d}\omega + \tfrac12[\omega,\omega]$, and it is $\pi$-semibasic.  The torsion of $\omega$ is the $\frak{g}/\frak{h}$-valued $2$-form $\Omega\,\text{mod}\,\frak{h}$.
As an example, $\gamma:TG\to\frak{g}$, the canonical left-invariant $1$-form on $G$, is a Cartan connection for the natural coset quotient $q:G\to G/H$.  Of course, the curvature of $\gamma$ vanishes identically.
Now, about the rolling interpretation:  Given a Cartan connection $(B,\pi,\omega)$ over $M^n$, consider the product $B\times G$ and the modified 'difference' $1$-form
$$
\theta =\mathrm{Ad}(g)\bigl(\omega-\gamma\bigr).
$$
This $1$-form is invariant under the diagonal right $H$-action on $B\times G$
defined by $(b,g){\cdot}h = (b{\cdot}h,gh)$, and it satisfies $R^*_{a}(\theta) = \mathrm{Ad}(a^{-1})(\theta)$ where $R_a:B\times G\to B\times G$ is the right $G$-action defined by $(b,g)\cdot a = (b,a^{-1}g)$ for $a\in G$.  It also satisfies $\iota_{b,g}^*(\theta) = 0$, where $\iota_{b,g}:H\to B\times G$ is the mapping $\iota_{b,g}(h) = (b{\cdot}h, gh)$.  
It follows that $\theta$ is the pullback of a well-defined $\frak{g}$-valued $1$-form $\bar\theta$ on the quotient $B\times_HG$ of $B\times G$ by the diagonal right $H$-action. In fact, $\bar\theta$ defines a connection in the usual sense on $B\times_H G$ when this quotient is regarded as a principal right $G$-bundle over $M$, and one has the curvature identity $\bar\Theta = \mathrm{Ad}(g)(\Omega)$. (Hence, even when the torsion of $\omega$ vanishes, the curvature of $\bar\theta$ will not, generally, take values in $\frak{h}$.)
Denote by $\pi:B\times_H G\to M$ the projection to $M$, and denote by $q: B\times_H G\to G/H$ the map defined by $q\bigl([b,g]\bigr) = gH\in G/H$.  The pair $(\pi,q)$ defines a double fibration of $B\times_HG$ over the pair of bases $M$ and $G/H$.
Any element $[b,g]\in B\times_HG$ defines a pair of points $(\pi(b),gH)$, but, more than this, the horizontal space $\ker(\bar\theta_{[b,g]})\subset T_{[b,g]}B\times_H G$ maps isomorphically onto $T_{\pi(b)}M$ and onto $T_{gH}G/H$,
so it defines an isomorphism $\tau_{[b,g]}:T_{\pi(b)}M\to T_{gH}G/H$ that can be regarded as a $1$-jet of a pointed diffeomorphism between $(M,\pi(b))$ and $(G/H, gH)$, i.e., an 'infinitesimal identification' of the two spaces.  Thus, $B\times_H G$ can be regarded as providing a family of $1$-jets of identifications of $M$ with the 'model space' $G/H$.  (There can be, and, often, there is, more information in $(B\times_HG,\pi,q,\bar\theta)$ than this, but that's enough for the present purposes.)
Finally, the connection $\bar\theta$ on $B\times_H G$ defines a way uniquely to lift any differentiable curve $\alpha:[0,1]\to M$ to a $\bar\theta$-horizontal curve $\hat\alpha:[0,1]\to B\times_HG$ once one specifies $\hat\alpha(0)\in\pi^{-1}\bigl(\alpha(0)\bigr)$.  In other words, $\hat\alpha$ provides a 'rolling without slipping' interpretation of the Cartan connection $\omega$.  Note that if $\omega$ is flat, then this actually defines the developing map of the simply connected cover of $M$ to $G/H$, as you would expect.
Probably, I should also mention that the (ordinary) holonomy of the connection $\bar\theta$ is what Cartan called the holonomy of (what we now call) a Cartan connection.<|endoftext|>
TITLE: Moment matching: construction of a mixture of Gaussian distribution with lower moments identical to Gaussian
QUESTION [8 upvotes]: This is a question related to the statistical model behind independent component analysis (ICA). 
We assume that $Z \sim N(0,1)$. Our goal is to construct a random variable $X$ that follows a Gaussian mixture distribution and satisfies 
$$
E[X^j] = E[Z^j]
$$ 
for $j \leq k-1$, while 
$$
E[X^k] - E[Z^k] = \epsilon.
$$ 
Here $k \geq 2$ is an integer.
My question is: Is there a principled way to construct the Gaussian mixture random variable $X$?
For $k = 4$, one possible construction is $X\sim 1/2 \cdot N(0,1+\varepsilon^{1/2}/\sqrt{3}) + 1/2 \cdot N(0,1-\varepsilon^{1/2}/\sqrt{3})$. However, is there a construction for general $k$?

REPLY [5 votes]: If non-zero means are allowed, the answer is yes (and it is based on quite different considerations, and so, is presented separately from the previous answer). 
Indeed, let $Z$ be a standard normal random variable (r.v.). 
Let $P_k$ denote the set of all probability measures $\nu$ on $\mathbb R$ such that $\int_{\mathbb R}x^j\nu(dx)=EZ^j$ for all $j=0,\dots,k$. Let $M_k:=\sup\{\int_{\mathbb R}x^{k+1}\nu(dx)\colon\nu\in P_k\}$ and 
$m_k:=\inf\{\int_{\mathbb R}x^{k+1}\nu(dx)\colon\nu\in P_k\}$. Later it will be shown that 
$$m_k<EZ^{k+1}<M_k,\tag{-1}
$$ 
with an emphasis on the strictness of the inequalities. 
Take any 
$$c\in(m_k,M_k). \tag{0} $$ 
Let us then show that there is a mixture $\mu_c$ of normal distributions such that $\int_{\mathbb R}x^{k+1}\mu_c(dx)=c$ and $\mu_c\in P_k$, that is, $\int_{\mathbb R}x^j\mu_c(dx)=EZ^j$ for all $j=0,\dots,k$. 
For $t\in(0,1)$, let 
$$c_t:=c-(1- t^{(k+1)/2})EZ^{k+1}. 
$$
Since $c_t\to c$ as $t\uparrow1$, one can choose $t$ close enough to $1$ so that, by $(0)$, 
$$c_t\in(t^{(k+1)/2}m_k,t^{(k+1)/2}M_k).$$
So, by the convexity of the set $P_k$ and my answer to the question at 
[mathoverflow.net/questions/229646]
there is a r.v. $X_t$ taking only finitely many values such that 
$$EX_t^j=E(\sqrt tZ)^j\text{ for }j=0,\dots,k\text{ and }EX_t^{k+1}=c_t. \tag{1}
$$ Without loss of generality, $X_t$ is independent of $Z$. Let $Y_t:=X_t+\sqrt{1-t}Z$. Then the distribution of $Y_t$ is a mixture of normal distributions (each of those distributions with variance $1-t$). 
Moreover, for $j=0,1,\dots$ 
$$EY_t^j=\sum_{i=0}^j\binom ji EX_t^i\, E(\sqrt{1-t}Z)^{j-i}. \tag{2} 
$$
On the other hand, $Z$ is equal in distribution to $\sqrt t Z_1+\sqrt{1-t}Z_2$, where $Z_1$ and $Z_2$ are independent standard normal r.v.'s. So, 
$EZ^j=\sum_{i=0}^j\binom ji E(\sqrt t Z)^i\, E(\sqrt{1-t}Z)^{j-i}$ 
for all $j=0,1,\dots$. 
Comparing this with $(2)$ and recalling $(1)$, we see that $EY_t^j=EZ^j$ for $j=0,\dots,k$, whereas 
$$EY_t^{k+1}=EX_t^{k+1}+\sum_{i=0}^k\binom{k+1}i EX_t^i\, E(\sqrt{1-t}Z)^{k+1-i}$$
$$=EX_t^{k+1}+\sum_{i=0}^k\binom{k+1}i E(\sqrt t Z)^i\, E(\sqrt{1-t}Z)^{k+1-i}$$
$$
=c_t+EZ^{k+1}-E(\sqrt t Z)^{k+1}=c, 
$$
as desired.
Addendum. 
As promised, I am adding the proof of inequalities $(-1)$. 
Lemma. Let $\mu$ be any finite measure on $\mathbb R$ with finite first $k+1$ moments and with support set $S_\mu$ of cardinality at least $k+2$, where $k$ is a natural number. For any interval $J\subseteq\mathbb R$, let 
$N_k(\mu,J)$ denote the set of all measures $\nu$ on $\mathbb R$ such that $\int_J x^j\nu(dx)=\int_J x^j\mu(dx)$ for all $j=0,\dots,k$, with $N_k(\mu):=N_k(\mu,\mathbb R)$. Let $M_k(\mu):=\sup\{\int_{\mathbb R}x^{k+1}\nu(dx)\colon\nu\in N_k(\mu)\}$ and 
$m_k(\mu):=\inf\{\int_{\mathbb R}x^{k+1}\nu(dx)\colon\nu\in N_k(\mu)\}$. Then 
$$m_k(\mu)<\int_{\mathbb R}x^{k+1}\mu(dx)<M_k(\mu),\tag{-1+}
$$ 
with an emphasis on the strictness of the inequalities.
Proof. To obtain a contradiction, suppose that $\int_{\mathbb R}x^{k+1}\mu(dx)=M_k(\mu)$. 
Since $S_\mu$ is of cardinality at least $k+2$, there are pairwise disjoint closed intervals $J_0,\dots,J_{k+1}$ of nonzero $\mu$-measure. Again by 
my answer to the question at 
[mathoverflow.net/questions/229646], 
for each $i$ there is a measure $\nu_i\in N_{k+1}(\mu,J_i)$ with a finite support set contained in $J_i$, so that $p_i:=\nu_i(\{x_i\})>0$ for some $x_i\in J_i$. It also follows that the points $x_i$ are pairwise distinct. 
Let now $\nu(A):=\mu(A\setminus J_0\setminus\dots\setminus J_{k+1})+\sum_{i=0}^{k+1}\nu_i(A)$ for all Borel sets $A\subseteq\mathbb R$. 
Then $\nu\in N_{k+1}(\mu)$ and, in particular, $\int_{\mathbb R}x^{k+1}\nu(dx)=M_k(\mu)$. 
For the pairwise distinct points $x_i$ mentioned above, consider the system of equations 
$$\sum_{i=0}^{k+1}a_i x_i^j=I\{j=k+1\}\quad\text{for } j=0,\dots,k+1,
$$
for the unknowns $a_0,\dots,a_{k+1}$, where $I\{\cdot\}$ is the indicator function. The determinant of this linear system is a nonzero Vandermonde determinant, and so, the system has a solution $(a_0,\dots,a_{k+1})$. For that solution and real $t>0$, let $\nu_t:=\nu+t\sum_{i=0}^{k+1}a_i\delta_{x_i}$, where $\delta_x$ is the Dirac measure at $x$. 
If $t$ is small enough (so that $p_i+ta_i\ge0$ for all $i$), then $\nu_t\in N_k(\mu)$ and $\int_{\mathbb R}x^{k+1}\nu_t(dx)=\int_{\mathbb R}x^{k+1}\nu(dx)+t=M_k(\mu)+t>M_k(\mu)$, which is a contradiction. 
Similarly, the assumption $\int_{\mathbb R}x^{k+1}\mu(dx)=m_k(\mu)$ leads to a contradiction. QED<|endoftext|>
TITLE: Quotients of curves of genus $4$ by a free $\mathbb{Z}/ 3 \mathbb{Z}$-action
QUESTION [7 upvotes]: Let $V_2$ and $V_3$ be the two hypersurfaces of $\mathbb P^3$ defined by
\begin{equation*}
V_2:={x_2x_3 + r(x_0, \, x_1)=0}, \quad V_3:={x_2^3+x_3^3+s(x_0, \, x_1)=0},
\end{equation*}
where $r, \, s \in \mathbb{C}[x_0, \, x_1]$ are general homogeneous forms of degree $2$ and $3$, respectively. 
Then $C_4:=V_2 \cap V_3$ is a smooth, canonical curve of genus $4$.  Denoting by $\xi$ a primitive third root of unity, then $C_4$  admits a free action of the cyclic group $\langle \xi \rangle \cong \mathbb Z/3 \mathbb Z$, defined by
\begin{equation*}
\xi \cdot [x_0: x_1:x_2:x_3] = [x_0: x_1: \xi x_2: \xi^2 x_3]
\end{equation*}
and the quotient $C_2 := C_4/ \langle \xi \rangle$ is a smooth curve of genus $2$. 
A naive count of parameters shows that the number of moduli on which this construction depends is $$\dim |\mathcal{O}_{\mathbb{P}^1}(2)| + \dim |\mathcal{O}_{\mathbb{P}^1}(3)| - \dim G= 7 - \dim G,$$
where $G \subset \textrm{Aut}(\mathbb{P}^4)$ is the subgroup of projectivities sending both $V_2$ and $V_3$ to hypersurfaces of the same form. If my computations are correct, $G$ has dimension $4$ so the construction actually depends on $3$ parameters. 
On the other hand, $3$ is also the dimension of the moduli space $\mathcal{M}_2$. This shows that the general curve $C_2$ of genus $2$ can be constructed in this way.

Question.  Is it possible to obtain all smooth curves of genus $2$ by means of  this construction?

REPLY [14 votes]: Yes. Start from a genus 2 curve $C_2$, and choose a point of order 3 in $JC_2$, giving rise to  an étale $\mathbb{Z}/3$-covering $C_4\rightarrow C_2$. 
Then $C_4$ cannot be hyperelliptic: a $g^1_2$ on $C_4$ would be stable under the covering automorphism  $\sigma $, hence descend to $C_2$, which is impossible for degree reasons.   $\sigma $ acts on $H^0(C_4,K_{C_4})$ with eigenvalues $(1,1,\xi ,\xi ^2)$: the invariant part comes from $C_2$, and the trace must be 1 by the holomorphic Lefschetz formula. Writing down the  invariant quadric and cubic(s) containing $C_4$ leads to your formulas.<|endoftext|>
TITLE: Partial Orders realized by Prime Ideals on commutative rings
QUESTION [6 upvotes]: Is there a general criterion for which partial orders can be realized by the prime ideals of commutative rings (like we have for topological spaces - https://en.wikipedia.org/wiki/Spectral_space)?
And in general, what constraints on the partial order are created by additional classical properties of rings - such as being a UFD, principal domain, noetherian, etc...?
Examples for some obvious constraints -

Fields always have just one prime ideal - 0.
Local rings have just one maximal ideal.
Noetherian rings have only a finite number of minimal primes, and no infinitely ascending or descending sequence of prime ideals.
Every ring has a minimal prime and a maximal prime.

(Question originally posted in m.se https://math.stackexchange.com/questions/1628013/partial-orders-that-can-be-realized-by-prime-ideals-of-commutative-rings)

REPLY [8 votes]: The following characterization follows easily from the general theory of spectral spaces, though it isn't exactly the most explicit criterion to apply in practice.

Theorem (Hochster, Proposition 12 of this paper): Let $X$ be a poset.  Then $X$ is isomorphic to the poset of prime ideals of a commutative ring iff it is order-isomorphic to a closed subset of $\{0,1\}^V$ for some set $V$ (putting the discrete topology on $\{0,1\}$ and the product topology on $\{0,1\}^V$, and the usual order on $\{0,1\}$ and the product order on $\{0,1\}^V$).

(Given an order-isomorphism $X\cong \operatorname{Spec}(R)$, the set $V$ can be taken to be the set of compact open subsets in the Zariski topology on $X$, and the Zariski topology can be recovered as the product topology on $\{0,1\}^V$, where you topologize $\{0,1\}$ such that $\{0\}$ is open and $\{1\}$ is not.  Note that every poset embeds in some $\{0,1\}^V$, so the key point of this result is the requirement that it be a closed subset of $\{0,1\}^V$, which is a sort of compactness condition.)<|endoftext|>
TITLE: A homology theory which satisfies Milnor's additivity axiom but not the direct limit axiom?
QUESTION [21 upvotes]: Let us agree on the following: a "homology theory" means a functor $h_*$ from the category of pointed CW complexes to the category of graded abelian groups, together with natural isomorphisms $h_{*+1}(\Sigma X)\cong h_*(X)$, such that the functor $h_*$ is homotopy invariant, sends a cofiber sequence to an exact sequence, and satisfies Milnor's additivity axiom: the natural map to the homology of a wedge from the direct sum of the homologies is an isomorphism.
Adding Milnor's axiom to the list is a reasonable thing to do, partly because it guarantees uniqueness in the following sence: a morphism of homology theories (satisfying Milnor's axiom) which induces an isomorphism on the coefficients, is itself an isomorphism of homology theories.
Now enter the world of spectra. A spectrum $E$ defines a homology theory via $E_*(X)=\pi_*(E\wedge X)$. Such a homology theory satisfies an axiom a priori stronger than Milnor's additivity: it satisfies the

Direct limit axiom: The natural map to the homology of $X$ from the colimit of the homologies of the finite subcomplexes of $X$ is an isomorphism.

Reciprocally, a homology theory satisfying this axiom comes from a spectrum in the way defined above (see Adams' Stable Homotopy and Generalised Homology, pp. 199-200 and Adams "A Variant of E.H. Brown's representability theorem".)
So the homology theories given by spectra are a bit better behaved than a homology theory as is usually defined. The question is:

Could you give an example of a homology theory which does not satisfy the direct limit axiom?

I first asked this question on a comment here, and Mark Grant pointed out that by Hatcher's exercise 4.F.1, Milnor's additivity implies the direct limit axiom for countable complexes, so we'd need to go to uncountable ones.

REPLY [2 votes]: Edit: As explained in the comments, this is not an answer to the question. I 
leave it as it might at least serve as an intuition for why the direct sum axiom plus excision should be enough to imply the direct limit axiom.
Working $\infty$-categorically, we may think of a homology theory as a functor from spaces to spectra. The axioms about the suspension isomorphism and cofiber sequences are equivalent to preservation of pushouts. Now, a functor which preserves pushouts and all (small) coproducts, preserves all (small) colimits. This is proposition 4.4.2.7 in HTT by Lurie. 
Remark: Actually, we can reverse the logic by saying that we just want our the functor to preserve all (small) colimits. One can break this into (1) preservation of finite colimits and (2) preservation of filtered colimits. One can also replace (2) by (2') preservation of coproducts. Additionally, one can replace (1) by (1') preservation of pushouts. Finally, one can test that a map of spectra is an equivalence by checking the induced map on the homotopy groups and for a pushout these can be described by a long exact sequence. This allows us to reformulate everything in classical terms (i.e. the Eilenberg-Steenrod axioms).<|endoftext|>
TITLE: An example of an Azumaya algebra that isn't free over its centre
QUESTION [5 upvotes]: Azumaya originally defined an Azumaya algebra (which he called a proper maximally central algebra) to be an algebra A which is a free module of finite rank over its centre Z such that the natural map
$$A\otimes_Z A^{\mathrm{op}}\to \mathrm{End}_Z(A)$$
is an isomorphism. More modern definitions (e.g., Knus, Quadratic and Hermitian Forms over Rings) replace "free" with "faithfully projective". Is there an example of an algebra which meets this broader definition but does not meet the original one? I.e., an Azumaya algebra that is faithfully projective but not free as a module over its centre. If possible, I'd also like the centre to be a noetherian $k$-algebra, for some algebraically closed field $k$.
Even better would be some method for producing lots of examples of Azumaya algebras "to order", but that's probably asking too much!

REPLY [7 votes]: It suffices to find a vector bundle $E$ on an affine variety $V$ such that the vector bundle $\mathcal{E}nd(E)$ is nontrivial: then this vector bundle gives a non-free, projective Azumaya algebra over $\mathcal{O}(V)$. This will be the case if some Chern class of $\mathcal{E}nd(E)$, say $c_2$, is nontrivial in the Chow group $CH^2(V)$.
Take $V:=\mathbb{P}^3\smallsetminus S$, where $S$ is a smooth surface of degree $d\geq 4$, and take for $E$ the restriction of $\mathcal{O}_{\mathbb{P}}\oplus \mathcal{O}_{\mathbb{P}}(1)$. An easy computation gives $c_2(\mathcal{E}nd(E))=-h^2$, where $h$ is a hyperplane in $\mathbb{P}^3$. On the other hand we have an exact sequence
$$CH^1(S) \xrightarrow{i_*} CH^2(\mathbb{P}^3)\rightarrow CH^2(V)\rightarrow 0$$
where $i$ is the embedding of $S$ in $\mathbb{P}^3$. If $S$ is general $CH^1(S)=\mathrm{Pic}(S)$ is generated by $i^*h$; we have $i_*i^*h=dh^2$, hence $c_2(\mathcal{E}nd(E))$ is nonzero in $CH^2(V)\cong \mathbb{Z}/d$.
Edit : This is actually easier in higher dimension, where you can use Chern classes in cohomology rather than in the Chow group. For instance if $V=\mathbb{P}^n\smallsetminus H$, where $n\geq 4$ and $H$ is a smooth hypersurface of degree $d>1$, then  $c_2(\mathcal{E}nd(E))\neq 0$ in $H^4(V,\mathbb{Z})=\mathbb{Z}/d\ $ for the same $E$.<|endoftext|>
TITLE: Gaussian and the convex hull of moment curves
QUESTION [7 upvotes]: Let $c_1,\dots, c_d$ be the first $d$ moments of the standard normal distribution. Does the point $(c_1,\dots, c_d)$ lie in the convex hull of the set $\{(t,t^2,\dots,t^d)\colon t\in[-b,b]\}$, for a sufficiently large $b$? 
Here is another more involved question: Is there a way we can tell how $b$ scales with $d$ (maybe we can even get a closed form solution)?
A modified version of this problem (related to this question: Moment matching: construction of a mixture of Gaussian distribution with lower moments identical to Gaussian) is, for what kind of $\delta$, the point $(c_1,\dots, c_d+\delta)$ lie in the convex hull of the set $\{(t,t^2,\dots,t^d)\colon t\in[-b,b]\}$, for a sufficiently large $b$?

REPLY [2 votes]: For any finite sequence $c_1, \ldots , c_d$ which can be a moment sequence at all (which is characterized by a positive definiteness condition), there is a description of all probability measures $\mu$ that satisfy $\int t^j\, d\mu(t)=c_j$ for $j=1, \ldots , d$.
This is sometimes called the Nevanlinna parametrization, and it goes as follows, in outline: Form the orthogonal polynomials $p_j(t)$ with respect to your moment sequence. These satisfy a three term recurrence relation
$$
a_{n-1}p_{n-1} + a_n p_{n+1} + b_np_n = tp_n .\quad\quad\quad\quad (1)
$$
Now the solutions to the (truncated) moment problem are exactly the "spectral measures" of the associated finite Jacobi matrix (the difference operator that acts like the LHS of (1)). I put quotes because this term must be taken in a very general sense: basically, you consider arbitrary extensions of (1) from $\{ 1,2, \ldots , d\}$ to a half line (and the "extension" need not be a difference operator beyond $d$), and then take the spectral measure of this problem.
In particular, you can just impose a boundary condition at $d$, and then your measures will be spectral measures of the problem on a finite-dimensional space, so these will be finitely supported measures, as requested.
The maximum of the support of such a measure is the operator norm of the operator defined by (1), so this suggests an approach to your additional question.<|endoftext|>
TITLE: Can $C^*$-algebra of continuous functions on $R^n$ ($S^n$) be characterized alternatively?
QUESTION [6 upvotes]: Dictionary between algebra and geometry is somewhat one of the main concepts in modern mathematics. So commutative $C^*$ algebras are one-to-one with locally compact Hausdorff spaces. 
So it is natural to be curious how one can see properties of the manifold from algebra of functions.
In the realm of algebraic functions we can easily characterize algebraic functions on $C^n$  (i.e. $C[x_1,...,x_n]$ ) as free commutative algebra with $n$-generators. 
Question: So I wonder is there any (hopefully simple) characterization of $R^n$  in $C^*$ world ?
I.e. can one somehow characterize  the $C^*$-algebra of continuous functions on $R^n$ in an alternative way ? 
Or may be the question for $S^n$ would be simpler, since it is compact.

The answers given below clarify the question a lot for me.
Let me stress what I found somehow surprising for myself.
The natural description for $C^*$-algebra of $R^n$ would be just the same as in algebraic geometry setup -  commutative $C^*$-algebra freely generated by $x_1,...x_n$... But this does not work because $x_i$ are unbounded functions on R^n and hence they do not fit to $C^*$ setup. So it is probably natural to extend class of $C^*$-algebras to include some unbounded operators, rather than ask about "simple" $C^*$-description of R^n. 
I had a misunderstanding that there is NO description of $C^*$-algebras by generators and relations.
Indeed, usually we take free algebra and factorize it over relations, but in $C^*$-world  free algebra is NOT $C^*$ in any reasonable sense.
Other way to describe $C^*$-algebra by generators and relations  would be to define  appropriate completion of the algebra of polynomials in generators - but this also does not seem to work since: how to describe norms of |a+b| in terms of a,b ? how to describe what kind of power series in generators belong to completion - this seems cannot be done explicitly, it is like one to describe continuous functions in terms theirs Fourier series, which " there is no way to characterize the Fourier series of continuous functions by means of a naive "sequence space" condition on the sequence. "   
So it is somehow surprising for me that there is such a simple way to define $C^*$ by generators and relations - just say "it is universal $C^*$-algebra
with such relations"... That seems to be an example where abstract thinking does seem to have a way round... 
Well not all the relations will work, e.g. ab-ba=1 will not - see 
MO151809, but that is another story.
Okay, thanks for answers.

REPLY [10 votes]: Yes, it can be defined as the univeral commutative $C^*$-algebra with unit, generated by $n+1$ self adjoint elements $x_1,...,x_{n+1}$ subject to the relation $x_1^2+...+x_{n+1}^2=1$. Here universal mean the following: $A$ with generators $(a_j)_j$ and relations $(r_k)_k$ is called universal if whenever there is $C^*$-algebra $B$ generated by $(b_j)_j$ (the same indexing set) and satysfying relations $(r_k)_k$ then there is an epimorphim $\varphi:A \to B$ such that $\varphi(a_j)=b_j$. The problem with existence of such universal algebras is because some relations don't impose any restriction on norms of elements. For example there is no universal unital $C^*$-algebra generated by a single self adjoint element but there is universal unital $C^*$-algebra generated by a single unitary element (this algebra is in fact $C(S^1)$). However you can prove that if there is any $C^*$-algebra $B$ with generating set $S=(b_s)_s$ such that $\|b_s\| \leq 1$ and those elements satisfy some relations $(r_j)_j$ then there is a universal $C^*$-algebra $A$ with generating set $(a_s)_s$ (the same cardinality as $S$) where $(a_s)_s$ satisfy relations $r_j$. So this difficulty concerning the norms is somehow the only obstruction

REPLY [8 votes]: Podles defined quantum 2-spheres in such a universal way. In a special case they restrict to $C(S^2)$. The description in this case is: the universal unital C*-algebra generated by operators $A$ and $B$ satisfying

$A^* = A$
$AB = BA$
$BB^* = B^*B = I - A^2$.

The generalization to higher dimensions can be found in Section 2 of this paper.
In the case of $\mathbb{R}^n$, remember that $\mathbb{R}^n$ is homeomorphic to $S^n$ minus a point. So $C_0(\mathbb{R}^n)$ is realized as (any) maximal ideal of $C(S^n)$.<|endoftext|>
TITLE: What are the generating partitions of the odometer?
QUESTION [5 upvotes]: According to the countable generator theorem, every ergodic invertible measure-preserving transformation has a generating partition. 
What are the generating partitions of the dyadic odometer ? I don't find the answer in textbooks. Is there a "canonical" one among them ?

REPLY [4 votes]: Quite a nice one is the two-set partition $A,A^c$, where $A$ is the set of points with an even number of terminal 0's:
$$
A =  \bigcup_{k \geq 0} A_k
$$
where $A_k = \bigl\{(x_1, x_2, \ldots) \mid x_1=\ldots=x_{2k}=0 \text{ and } x_{2k+1}=1 \bigr\}$. 
Or, if you work with the odometer acting on the space $[0,1[$:
$$
A = \bigcup_{k \geq 0} \left[\frac{1}{2^{2k+1}}, \frac{1}{2^{2k}} \right[ 
$$
This partition is generating for the following reason. Code a trajectory $x, Tx, T^2x, \ldots$ by a sequence $(v_0, v_1, \ldots)$ of $a$'s and $b$'s according to whether it's in $A$ or not. 
Then you can get the first digit $x_1$ of $x$ by looking at the blocks of four consecutive terms of the $v$ sequence:

if the first digit of $x$ is $x_1=1$, then the block $(v_0,v_1,v_2,v_3)$ of four consecutive terms is one of $aaab$, $abaa$ or $abab$ (that is $(v_0,v_1,v_2,v_3)$ is $a*a*$, with at least one of the stars a $b$);
if $x_1=0$, then $(v_0,v_1,v_2,v_3)$ is $*a*a$ with at least one of the stars a $b$.

In particular, the sets of possible codes are disjoint, so that this information suffices to determine the first digit of $x$.
To get the second digit of $x$, look at blocks of eight consecutive terms:


if the first two digits of $x$ are 00, then the code is $*aba*aba$ with at least one of the $*$'s an $a$;

if the first two digits of $x$ are $01$, then the code is $ba*aba*a$ with at least one of the $*$'s an $a$;

if the first two digits of $x$ are 10, then the code is $aba*aba*$ with at least one of the $*$'s an $a$;

if the first two digits of $x$ are 11, then the code is $a*aba*ab$ with at least one of the $*$'s an $a$;

Again, these sets of possibilities are disjoint, so that one can recover the first and second digits of $x$ from 8 terms of the $v$ sequence.
And so on, looking at the block of $2^{n+1}$ consecutive terms of the $v$ sequence determines the first $n$ digits of $x$.<|endoftext|>
TITLE: Characterization of non-isomorphic graphs but isomorphic total graphs?
QUESTION [5 upvotes]: Given a graph $G$, the total graph of $G$, denoted $T(G)$, is the graph with vertex set $V(G) \cup E(G)$, where $a$ and $b$ are adjacent in $T(G)$ if and only if they are adjacent or incident in $G$.  

Is there any characterization of properties of two graphs $G$ and $H$ such that $T(G)$ is isomorphic to $T(H)$?

Cross-posted at MSE.

REPLY [8 votes]: It is not difficult to see that both $G$ and $L(G)$ are disjoint induced subgraphs of the graph $T(G)$. For your question, in the below paper:
[1] M. Behzad and H. Radjavi, The total group of a graph, Proc. Amer. Math. Soc.
19 (1968), 158-163. MR 36 #1358.
It is proved that, the graph $G$ is isomorphic to the graph $H$ if and only if $T(G)$ is isomorphic to $T(H)$.<|endoftext|>
TITLE: A weak kind of fixed point
QUESTION [5 upvotes]: Let $X$ be a set and let $\cal A$ be a non-empty subset of $P(X)$ with the property that whenever $A_1 \subseteq A_2 \subseteq \cdots $ is an increasing chain of elements of $\cal A$ then $\cup_i A_i \in \cal A$. Let 
$f : \cal A\longrightarrow \mathbb{R}$ be increasing and bounded from above. Is it true that there must exists $M\in\cal A$ such that for any element $C$ of $\cal A$ containing $M$, $f(M)= f(C)$ ?

REPLY [6 votes]: If not, then we can find an increasing and continuous sequence $(A_\alpha: \alpha < \omega_1)$ of elements of $\mathcal{A}$ such that $\alpha < \beta \implies f(A_\alpha) < f(A_\beta)$. This is a contradiction.

Edit. Let me add more details. By contradiction, the following holds:
$(*)$: For any $M \in \mathcal{A}$, there exists $M \subset C \in \mathcal{A}$ with $f(M) < f(C).$
Let $A_0\in \mathcal{A}$ be arbitrary. Given $A_\alpha \in \mathcal{A},$ use $(*)$ to find $A_{\alpha+1} \in \mathcal{A}$ with $A_\alpha \subset A_{\alpha+1}$ and $f(A_\alpha) < f(A_{\alpha+1}).$
Now suppose $\alpha<\omega_1$ is limit and we have defined $(A_\beta: \beta< \alpha)$ which is increasing and continuous and for all $\beta_1 < \beta_2 < \alpha, f(A_{\beta_1}) < f(A_{\beta_2}).$ Let $A_\alpha=\bigcup_{\beta<\alpha}A_\beta.$ Note that $A=\bigcup_{n<\omega}A_{\beta_n},$ where $(\beta_n: n<\omega)$ is increasing cofinal in $\alpha,$ so by assumption $A_\alpha \in \mathcal{A}.$
It is also clear that for all $\beta<\alpha, f(A_\beta) < f(A_\alpha)$ (as $f$ is increasing by assumption).
But this is impossible, as otherwise we can find $q_\alpha \in \mathbb{Q} \cap (f(A_\alpha), f(A_{\alpha+1})), \alpha < \omega_1,$ and for $\alpha \neq \beta, q_\alpha \neq q_\beta,$ so $\mathbb{Q}$ is uncountable which is impossible.<|endoftext|>
TITLE: $\mathbb{A}^1$-invariance of categories of Finite Etale Covers
QUESTION [5 upvotes]: Let $k$ be algebraically closed with characteristic $0$. For a scheme $X$, let $FEt(X)$ be the category of finite etale covers of $X$. What can be said about $FEt(X \times \mathbb{A}^1)$ and the functor $- \times_k \mathbb{A}^1: FEt(X) \rightarrow FEt(X \times \mathbb{A}^1)$? 
It seems to me that the functor induced by pullback along the zero section $0: X \rightarrow X \times_k \mathbb{A}^1$, $FEt(X \times \mathbb{A}^1) \rightarrow FEt(X)$ witnesses a fully faithful embedding since the composite of this with the one above is the identity, but that's about all I can observe immediately.
So is the functor an equivalence/final/cofinal? 
At least, the etale fundamental group is $\mathbb{A}^1$-invariant for suitably nice $X$ (the etale fundamental group preserves products for suitably nice $X$ and the etale fundamental group of $\mathbb{A}^1$ is zero) and so is the etale homotopy type after appropriate completion. Can one perhaps lift this to some equivalence between etale covers?

REPLY [6 votes]: I'm a bit surprised that this was never answered - if I'd known I'd posted something more helpful than a snippy question. Yes, $FEt(X) \to FEt(X \times \mathbb A^1)$ is an equivalence of categories in characteristic zero. This assertion is in fact equivalent to $\pi_1^{et}(X) \to \pi_1^{et}(X \times \mathbb A^1)$ being an isomorphism, a fact you seem comfortable with. An isomorphism between the respective fundamental groups necessarily induces an equivalence between the categories of finite continuous $\pi_1$-sets, which in turn are canonically equivalent to the categories of finite étale covers once we've fixed a base point. In positive characteristic you'd have to consider instead the tame fundamental group.<|endoftext|>
TITLE: The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
QUESTION [39 upvotes]: Let $(a_{n})_{n \ge 1}$ be a sequence of integers such that for all $n \ge 2$:  
$0\le a_{n-1}+\frac{1-\sqrt{5}}{2}a_{n}+a_{n+1} <1$.
Prove that the sequence $(a_{n})$ is periodic.

This question was asked at the Miklos Schweitzer Competition 2005, problem 2 (in Hungarian). 
Since $(a_{n})$ is integer it follows that $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic.  
Some discussion regarding this question can be found at Mathlinks,
and apparently we can choose $a_{1}$ and $a_{2}$ to make this period as large as we want.
Any help would be appreciated, thanks.




I would like to clarify some points about this problem:
1) There is a ceiling function ($\lceil x \rceil$) at the recursive sequence which makes it considerably harder.
The period is not 5 as claimed by some answers.
2) Miklos Schweitzer is not a conventional competition. This competition for undergraduate students is unique.
The contest lasts 10 days with 10-12 problems, which are quite challenging and basically of research level.
Moreover any literature can be used.
3) An example of Miklos Schweitzer problem can be found here at MO.
It was indeed a very nice question with an even nicer solution. I'm not sure Art of Problem Solving would be better. 
I am sorry for any inconvenience caused and I hope this question do not get closed.

REPLY [14 votes]: Define the Fibonacci numbers by $f_0 = 0$, $f_1 = 1$, $f_{n+1} = f_n + f_{n-1}$, and the golden ratio by $\phi = \frac{\sqrt{5}+1}{2}$. Then it is easy to check that $f_n = \frac{\phi^n - (\tfrac{-1}{\phi})^n}{\sqrt{5}}$ and $(\phi-1)f_n = f_{n-1}-(\tfrac{-1}{\phi})^n$.
We will show that if five terms in a row are all less than $f_{2k}$ for some $k > 1$, then all $a_n$ are less than or equal to $f_{2k}.$
Suppose that $a_{n-1}, a_n$ are both positive. As is well known, we can find sets $I,J \subset \{2,3,4,...\}$ with $I \cap (I+1) = J\cap (J+1) = \emptyset$, such that $a_n = \sum_{i\in I} f_i$ and $a_{n-1} = \sum_{j\in J} f_j$. Put $x = \sum_{i\in I} (\tfrac{-1}{\phi})^i$ and $y = \sum_{j\in J} (\tfrac{-1}{\phi})^j$. By summing a geometric series, it's easy to show that we have $\frac{-1}{\phi^2} < x,y < \frac{1}{\phi}$. Expressing $a_{n+1}, ..., a_{n+5}$ in terms of $x$ and $y$, we get
$a_{n+1} = \sum_{i\in I}f_{i-1} - \sum_{j\in J} f_j + \alpha,\ \ \alpha = \lceil -x\rceil,$
$a_{n+2} = -\sum_{i\in I}f_{i-1} - \sum_{j\in J} f_{j-1} + \beta,\ \ \beta = \lceil\phi x + y + (\phi-1)\alpha\rceil,$
$a_{n+3} = -\sum_{i\in I}f_i + \sum_{j\in J} f_{j-1} + \gamma,\ \ \gamma = \lceil-\phi x-\phi y + (\phi-1)\beta - \alpha\rceil,$
$a_{n+4} = \sum_{j\in J} f_j + \delta = a_{n-1} + \delta,\ \ \delta = \lceil x+\phi y + (\phi-1)\gamma - \beta\rceil,$
$a_{n+5} = \sum_{i\in I}f_i + \epsilon = a_n + \epsilon,\ \ \epsilon = \lceil -y + (\phi-1)\delta - \gamma\rceil.$
Note that if we instead had $a_n$ positive and $a_{n-1}$ negative, then on writing $a_{n-1} = -\sum_{j\in J} f_j$ and $y = -\sum_{j\in J} (\frac{-1}{\phi})^j$, we still get $a_{n+4} = a_{n-1} + \delta$ and $a_{n+5} = a_n + \epsilon$, with $\alpha, \beta, \gamma, \delta, \epsilon$ defined as above. Note that in this case, we have the inequalities $\frac{-1}{\phi} < y < \frac{1}{\phi^2}$.
Claim: We always have $\delta, \epsilon \in \{-1,0,1\}$. Furthermore, if $0 < x < \frac{1}{\phi^2}$ and either $\frac{-1}{\phi^2} \le y \le \frac{1}{\phi} - \frac{x}{\phi}$ or $\frac{-1}{\phi} < y \le \frac{-1}{\phi^2} - \frac{x}{\phi}$, then $\delta \ge 0$ and $\epsilon \le 0$.
Proof of Claim: For the first part, note that
$a_n - a_{n+5} = (a_n - \frac{a_{n+1}}{\phi} + a_{n+2}) + \frac{1}{\phi}(a_{n+1} - \frac{a_{n+2}}{\phi} + a_{n+3}) - \frac{1}{\phi}(a_{n+2} - \frac{a_{n+3}}{\phi} + a_{n+4}) - (a_{n+3} - \frac{a_{n+4}}{\phi} + a_{n+5}),$
which is between $-1-\frac{1}{\phi}$ and $1+\frac{1}{\phi}$ by assumption. Thus $|\epsilon| = |a_{n+5}-a_n| \le 1$ (since it is an integer).
Now suppose that $0 < x < \frac{1}{\phi^2}$ and either $\frac{-1}{\phi^2} \le y \le \frac{1}{\phi} - \frac{x}{\phi}$ or $\frac{-1}{\phi} < y \le \frac{-1}{\phi^2} - \frac{x}{\phi}$. Then we immediately see $\alpha = 0$, and from
$-1 < y \le \phi x + y \le \frac{1}{\phi} + \phi x - \frac{x}{\phi} = \frac{1}{\phi} + x < 1$,
we see that
$\beta = \lceil\phi x + y\rceil = \begin{cases} 1 & \phi x + y > 0,\\ 0 & \phi x + y \le 0. \end{cases}$
From this together with
$\gamma = \lceil -\phi x - \phi y + \frac{\beta}{\phi}\rceil = \lceil -(\phi x + y) + \frac{\beta-y}{\phi}\rceil$
and $y > \frac{-1}{\phi}$ we can easily show
$1 \ge \gamma \ge \begin{cases} 1 & y \le 0\\ 0 & y > 0.\end{cases}$
That $\delta = \lceil x + \phi y + \frac{\gamma}{\phi} - \beta\rceil$ is at least $0$ follows from
$x + \phi y + \frac{\gamma}{\phi} - \beta = (\frac{\phi x + y}{\phi} - \beta) + (y + \frac{\gamma}{\phi}) > -1 + 0.$
Finally, we must show that $\epsilon =\lceil -y + \frac{\delta}{\phi} - \gamma\rceil$ is at most $0$. We split into three cases. First case: 
$\gamma = 0\implies y>0\implies \beta = 1\implies \delta = 0\implies \epsilon = \lceil -y \rceil = 0.$
Second case:
$\gamma = 1\ \&\ y \ge \frac{-1}{\phi^2} \implies \epsilon \le \lceil \frac{1}{\phi^2} + \frac{1}{\phi} - 1\rceil = 0.$
Third case:
$\gamma = 1\ \&\ y \le \frac{-1}{\phi^2} - \frac{x}{\phi} \implies \delta \le \lceil \frac{-1}{\phi} + \frac{1}{\phi} - \beta\rceil = 0 \implies \epsilon \le \lceil \frac{1}{\phi} - 1\rceil = 0.$
Corollary: If $a_n = f_{2k}$ and $-f_{2k+2} \le a_{n-1} < f_{2k+3}-1$, then $a_{n+5} \le a_n$.
Proof: We just have to check that the claim applies. We have $x = (\tfrac{-1}{\phi})^{2k} = \frac{1}{\phi^{2k}}$, so we just need to check that either $\frac{-1}{\phi^2} \le y \le \frac{1}{\phi} - \frac{1}{\phi^{2k+1}}$ or $y \le \frac{-1}{\phi^2} - \frac{1}{\phi^{2k+1}}.$
Suppose first that $a_{n-1} \ge 0$. Then from $a_{n-1} < f_{2k+3} - 1 = \sum_{j=1}^{k+1} f_{2j}$, we see that $y \le \sum_{j=1}^k \frac{1}{\phi^{2j}} = \frac{1}{\phi} - \frac{1}{\phi^{2k+1}}.$
Now suppose that $a_{n-1} < 0$. Note that in this case we automatically have $y \le \frac{1}{\phi^2} \le \frac{1}{\phi} - \frac{1}{\phi^{2k+1}}.$ Write $a_{n-1} = -\sum_{j\in J} f_j$ for some $J \subseteq \{2,3,...\}$ with $J\cap (J+1) = \emptyset.$ If $2 \not\in J$, then $y > -\frac{1}{\phi^4}-\frac{1}{\phi^6} -\cdots = -\frac{1}{\phi^3} > -\frac{1}{\phi^2}$. If $J = \{2\}$ or if $2\in J$ and the second smallest element of $J$ is odd, then we have $y \ge -\frac{1}{\phi^2}$ as well. Now suppose that $2\in J$ and the next smallest element of $J$ is $2l$. Since $a_{n-1} \ge -f_{2k+2}$, we must have $l \le k$, so
$y < \frac{-1}{\phi^2} - \frac{1}{\phi^{2l}} + \frac{1}{\phi^{2l+3}} + \frac{1}{\phi^{2l+5}} + \cdots = \frac{-1}{\phi^2} - \frac{1}{\phi^{2l+1}} \le \frac{-1}{\phi^2} - \frac{1}{\phi^{2k+1}}.$
To finish: Note that if $a_n = f_{2k}$ and $a_{n-1}, a_{n+1} \le f_{2k}$, then $a_{n-1} + a_{n+1} = f_{2k-1}$, so $-f_{2k-2} \le a_{n-1} \le f_{2k},$ and we can apply the Corollary to see that $a_{n+5} \le f_{2k}.$ Of course, if $a_n < f_{2k}$ then we have $a_{n+5} \le a_n+1 \le f_{2k}$ (by the Claim) as well.<|endoftext|>
TITLE: Can one make high-level proofs about chess positions?
QUESTION [38 upvotes]: I realize this question is risky (as the title and the tags indicate), but hopefully I can make it acceptable. If not, and the question cannot be salvaged, I'm sorry and ready to delete it or accept closure.
Oftentimes, when one listens to chess grandmasters commentating on games, they will say things like, "The engines give a +1 advantage to White, but the postion is dead-drawn." And they proceed to provide high-level explanations of such statements. I'm a bad chess player so I cannot really attempt to verify those, but they often sound extremely convincing. Also, when they teach simple endgames, they seem to consider analyzing the crucial ideas to be enough for proof (at least in the sense of "a completely convincing explanation") and it's difficult not to agree with that. Even certain positions with a lot of pieces on the board can apparently be treated this way. "There is no way to make progress" is something that one can frequently hear. 
Sometimes these just turn out to be wrong. These statements are generally based on intuitions ("axioms") such as "you cannot give away your queen for a pawn", which are believed to be true in almost all situations and on already analyzed positions (like the Lucena position). The grandmasters' intuitions are very fine, but it can happen that they will miss a counterintuitive material sacrifice, a counterintuitive retreat or a counterintuitive something else. However, they can be extremely convincing sometimes (and never proven wrong). 
It's clear that chess can't be solved this way any time soon. A clear obstacle is the "wild" or "unclear" positions, where "anything can happen". But there are also those "tame" positions, "dead-drawn" positions and "winning advantages". (Surely, some -- or maybe a lot -- of these statements are not correct or correct with incorrect reasonings behind them.) Another indication is that the humans who make these statements get beaten by computers, which primarily use low-level tree searches.
My question is how much, if anything, of this high-level reasoning can be put to a form that would make it mathematics and mathematical proof. I think brute force is clearly not the only way of evaluating positions rigorously. In a K+Q v. K endgame, one doesn't have to analyze each possible move of the lone king to show that it's doomed. It's enough to show that a stalemate is impossible when the king is not on the side of the board, that the king can be forced to the side of the board and then that whenever the king is on the side, it can be checkmated. For a rigorous proof, one can use all kinds of symmetries and translations without having to go through every single node of the tree.
The question 

How much of that high-level thinking grandmasters employ can be made into mathematics?

is what I'm interested in, but this is vague -- it's not clear what "how much", "high-level thinking" or even "made into" mean. But I think people must have tried to pursue this and some clear mathematical statements must have been produced. If not, is it at least possible to say why it's difficult, infeasible or impossible?

Is there a chess position with a large (enough to be intractable to a human with just a pen and a piece of paper) tree that has been solved without brute-forcing the tree? Has a rigorous proof accessible to a human been produced?

REPLY [2 votes]: Although somewhat orthogonal to the thrust of the question (determining a winning strategy for a giving legal position in chess), I think the title question is in part answered by Raymond Smullyan.
Raymond Smullyan has written much to make topics in logic more accessible.  In addition, he has exemplified logical reasoning in chess with his popular books on retrograde analysis, including Chess Mysteries of Sherlock Holmes.  Here the goal often is to determine how a position was arrived at, along with certain properties like whether castling is legal.  I would recommend his titles to anyone wanting to practice mathematics, especially as related to chess.
Gerhard "Has Arabian Nights Themes Too" Paseman, 2016.02.04.<|endoftext|>
TITLE: Construction of invariants of 4-manifolds with the Kirby calculus
QUESTION [13 upvotes]: I'm an undergraduate student, interested in the low dimensional topology, in particular, the 4-manifold theory.
I have a question.
In the knot theory, the Reidemeister moves play fundamental roles.
For instance, to prove the fact that the Jones polynomial is an invariant of knots, we can use the Reidemeister moves.
On the other hand, in the 4-manifold theory, there is the Kirby calculus, which play roles similar to the Reidemeister moves.
So, are there some studies about construction of an invariant of 4-manifolds by using the Kirby calculus?
Thanks for your help.

REPLY [10 votes]: Disclaimer: Shameless self-advertising.

Yes, it can be done, and it's really beautiful! You can define the Crane-Yetter invariant with Kirby calculus, and possibly other TQFTs ("dichromatic models"). I've written this down in this article:
https://doi.org/10.1007/s00220-017-3012-9
If you want a version with more impressions and pictures, and less text, look at the talk slides:
https://www.manuelbaerenz.de/article/understanding-crane-yetter-model
The Broda invariant is a special case of the dichromatic framework, which was developed by Jerome Petit (and probably Alain Bruguières).
The dichromatic and Crane-Yetter invariants are stronger than signature and Euler characteristic. They are sensitive to the fundamental group (not just homology), but still fail to distinguish $S^2 \times S^2$ and $\mathbb{CP}^2 \# \overline{\mathbb{CP}}^2$. They are probably still homotopy invariants, but this is a conjecture.<|endoftext|>
TITLE: Automorphic representations whose local factors are tempered almost everywhere
QUESTION [10 upvotes]: Let $F$ be a global field, let $\mathbf{G}$ be a reductive algebraic group over $F$, and let $\pi$ be an irreducible discrete automorphic representation of $\mathbf{G}$.
Write $\pi$ as a restricted product of local factors $\pi=\bigotimes_\nu\pi_{\nu}$ (here, $\nu$ ranges over the places of $F$), and suppose that for almost all $\nu$, the $\mathbf{G}(F_\nu)$-representation $\pi_\nu$ is tempered. Is it true that $\pi_\nu$ is tempered for all $\nu$?
I am particularly interested in the following setting: Suppose $F$ has positive characteristic, let $D$ be a central division $F$-algebra and take $\mathbf{G}=\mathbf{GL}_1(D)$ or $\mathbf{G}=\mathbf{PGL}_1(D)$. 
Notice that the answer to the above question is yes in case $F$ has positive characteristic and $\mathbf{G}=\mathbf{GL}_n$. In fact, in this case, if just one of the local factors is tempered, then all local factors are tempered. This follows from Lafforgue's proof of the Ramanujan conjecture for $\mathbf{GL}_n$ in positive characteristic, asserting that local factors of cuspidal automorphic representations are tempered, and the description of the residual spectrum by  M\oeglin and Waldspurger (1989), which implies (after some work) that the local factors of residual representations of $\mathbf{GL}_n$ are never tempered.
Returning to the the case of $\mathbf{G}=\mathbf{GL}_1(D)$ with $\mathrm{char}(F)>0$, let  $\pi=\bigotimes_\nu\pi_\nu$ be an automorphic representation of $\mathbf{G}$ such that $\pi_n$ is tempered for almost all $\nu$. The global Jacquet-Langlads correspondence gives an automorphic representation $\rho=\bigotimes_\nu\rho_\nu$ of $\mathbf{GL}_n$ such that $\rho_\nu=\pi_\nu$ for all $\nu\notin S$. By the previous paragraph, this means $\rho_\nu$ is tempered for all $\nu$. The question therefore reduces to a question about the behavior of the global Jacquet-Langlands correspondence at ramified places, namely, does the fact that $\rho_\nu$ is tempered for all $\nu\in S$ implies that $\pi_\nu$ is tempered for all $\nu\in S$. Of course, the answer is yes if $D_\nu:=D\otimes_FF_\nu$ is a division algebra for all $\nu\in S$, so assume this is not the case.

REPLY [5 votes]: It turns out that the answer is "yes" in the case $\mathbf{G}=\mathbf{GL}_1(D)$, where $D$ is a central division $F$-algebra of degree $n$ and $\mathrm{char}(F)>0$. 
More precisely, the local Jacquet-Langlands  at a place $\nu$ takes (when defined) tempered representations of $\mathbf{GL}_n(F_\nu)$ to tempered representations of $\mathbf{GL}_1(D_\nu)$. Since the global correspondence
behaves like the local correspondence on each local factors, this gives the result as explained above.
The proof of the statement about the local Jacquet-Langlands correspondence is done by noting the following (non-trivial) facts:

The local Jacquet-Langlands correspondence is compatible with the $\times$-product.
The tempered representations of $\mathbf{GL}_r(D_\nu)$ are precisely those of the form $\pi_1\times\dots\times \pi_t$ where $\pi_i$ is an irreducible (unitary) square-integrable representation of $\mathbf{GL}_{r_i}(D_\nu)$ and $\sum r_i=r$.
The local Jacquet-Langlands (which is defined for a certain class of unitary representations of $\mathbf{GL}_{nr}(F_\nu)$) restricts to a bijection between the square integrable representations of $\mathbf{GL}_r(D_\nu)$ and $\mathbf{GL}_{nr}(F_\nu)$.

This argument is  not mine and was communicated to me independently by I.A. Badulescu and A.-M. Aubert. 
I wrote a full proof giving references to the previous three facts in Theorem 7.13 in https://arxiv.org/abs/1605.02664 .
It should be noted that this proof uses a lot of heavy machinery such as the Ramanujan conjecture for $\mathbf{GL}_n$ in positive characteristic and the global Jacquet-Langlands correspondence. I was kind of hoping that there would be a less involved proof.<|endoftext|>
TITLE: Is there a Nash-type theorem for symplectic manifolds?
QUESTION [10 upvotes]: If $(M, \omega)$ is a symplectic manifold, is it possible to embed (or injectively immerse) it symplectically into a sufficiently large $(\Bbb R ^{2N}, \Omega)$, (with the usual symplectic structure)?
The only thing that I have found is a theorem by Gromov which is conceptually very nice, but which seems to be very difficult to use. First, I should construct some embedding $i_0$ of $M$ - I guess that I could use Whitney here. Next, I should check that $i_0 ^* \Omega$ belongs to the same cohomology class as $\omega$. Finally, I should check that ${\rm d} i_0$ is homotopic through fiberwise-injective bundle maps $: TM \to T \Bbb R ^{2N}$ to some symplectic morphism. This does not look easy at all.
The thing with Gromov's theorem is that it's very general: it embeds in arbitrary symplectic manifolds (not just in $\Bbb R ^{2N}$), and it allows $\omega$ to have arbitrary non-constant rank - hypotheses that are too generous for me.
The closest thing to my needs is a corollary (corollary (a) on page 334 of Gromov's 1986 "Partial Differential Relations") which says that if $\omega$ is an exact symplectic form, then $(M, \omega)$ immerses symplectically into $(\Bbb R ^{2N}, \Omega)$ provided that $\dim M \le N$. Unfortunately, the requirement that $\omega$ be exact is too strong for me.
Does anyone, then, know of more humane conditions (but valid for arbitrary symplectic manifolds) that guarantee what I want?
ADDENDUM: It would be useful too to find symplectic embeddings into some cotangent bundle - but I expect this to be even more complicated.

REPLY [11 votes]: Not as stated, there isn't, at least for closed manifolds. For the form $\Omega$ is exact, but on a closed manifold a symplectic form cannot be exact, since a power of it is a volume form. This would apply to immersions as well as embeddings.  In particular, this explains the hypothesis of being exact in the Gromov theorem you quoted.
On the other hand, there are embedding results in $\mathbb{C}P^n$ due to Tischler (D. Tischler. Closed 2-forms and an embedding theorem for symplectic manifolds. JDG,(12):229–235, 1977) and Gromov (presumably in the book you quoted). A useful reference may be the thesis of Manuel Araujo which gives a proof; the introduction says it corrects something in Tischler's work.<|endoftext|>
TITLE: Complexity of counting MAXCUT in planar graphs -- seemingly contradicting claims
QUESTION [5 upvotes]: Confusion is likely. Appears to me two papers give contradicting claims
about the complexity of counting MAXCUT in planar graphs.
Exact Max 2-SAT: Easier and Faster p. 6

However, counting the number of Max 2-Sat or Max Cut solutions are
  $\#P$-complete even when restricted to planar graphs (results not explicitly stated but follow readily from results and reductions in [13, 20]).

[13] H. B. Hunt III, M. V. Marathe, V. Radhakrishnan, and R. E. Stearns, The complexity of planar counting problems, SIAM Journal of Computing 27 (1998), no. 4, 1142–1167.
[20] S. P. Vadhan, The complexity of counting in sparse, regular, and planar graphs, SIAM Journal of
Computing 31 (2002), no. 2, 398–427.
Exponential Time Complexity of the Permanent and the Tutte Polynomial p. 13

Proposition 4.1. If #ETH holds, the coefficients of the polynomial w → Z(G; 2, w) for a given simple graph G cannot be computed in time exp(o(m)).
  Proof. The reduction is from #MaxCut and well-known
  ...Now, the coefficient of $(1 + w)^{m-c}$ in $Z(G; 2, w)$ is the number of cuts in $G$ of size $c$.

The Tutte polynomial on the hyperbola $(x-1)(y-1)=2$ is polynomially
computable on planar graphs and (4) on the same page gives
$Z(G;2,w)$ in terms of the Tutte polynomial on the hyperbola.
The first paper doesn't define MAXCUT, but assuming the usual definition,
 the two papers imply $P=\#P$.

What is wrong with this?

REPLY [6 votes]: There was some discussion about this claim at the Dagstuhl seminar on Computational Counting in 2013. The technique you give to calculate #MAX-CUT is correct and it is not obvious how the hardness result is meant to "follow readily" from the two papers referenced, so I think it should be assumed that the claim of #P-hardness is in error.
I e-mailed Martin Fürer about this shortly after the seminar, and received no response.<|endoftext|>
TITLE: Definition of ind-schemes
QUESTION [19 upvotes]: What is the correct definition of an ind-scheme?
I ask this because there are (at least) two definitions in the literature, and they really differ.
Definition 1. An ind-scheme is a directed colimit of schemes inside the category of presheaves on the Zariski site of affines schemes, where the transition maps are closed immersions.
Definition 2. An ind-scheme is a directed colimit of schemes inside the category of sheaves on the Zariski site of affines schemes, where the transition maps are closed immersions.
This makes a difference, since the forgetful functor from sheaves to presheaves does not preserve colimits. For an ind-scheme $\varinjlim_n X_n$ according to Definition 1, we have
$$\hom(Y,\varinjlim_n X_n) = \varinjlim_n \hom(Y,X_n)$$
for affine schemes $Y$. Does this also hold for non-affine schemes? Probably not. But one certainly wants the category of schemes to be a full subcategory of the category of ind-schemes. For this, Definition 2 seems to be more adequate.
Any detailed references are appreciated.

REPLY [14 votes]: There is in fact no difference between the two definitions if you take your site to be the category of affine schemes – while it is true that the forgetful functor from sheaves to presheaves does not preserve colimits in general, for the purposes of this definition, only filtered colimits matter, and those are preserved.
Indeed, recall that the Zariski topology is a Grothendieck topology "of finite type", i.e. generated by a pretopology consisting only of finite covering families. For such Grothendieck topologies, it is straightforward to verify that the category of sheaves is closed under filtered colimits in the category of presheaves – after all, in this case, the sheaf condition amounts to saying that certain cocones under finite diagrams of affine schemes are mapped to limiting cones in the category of sets.<|endoftext|>
TITLE: Riemann zeta function: pair correlations vs. neighbor spacings
QUESTION [10 upvotes]: Montgomery's pair correlation conjecture states that the distribution of the pair correlations of the zeroes of the Riemann zeta function (normalized to have average spacing 1) is given by the function
$$
g(u) = 1 - \left(\frac{\sin(\pi u)}{\pi u}\right)^2 + \delta(u).
$$
See https://en.wikipedia.org/wiki/Pair_correlation_conjecture for some basic information. Famously, this distribution also appears in random matrix theory.
There is also a conjecture for the distribution of the nearest neighbor spacings of the normalized zeroes, which is conjectured to be given by 
$$
f(u) = \frac{32}{\pi^2} u^2 \exp\left(-\frac{4}{\pi} u^2 \right)
$$
Question: What is the conjectured distribution of the second-nearest neighbors? Can it be obtained by assuming that consecutive spacings are stochastically independent? - if "yes", then we could obtain the conjectured distribution $f_2(u)$ of the spacings of the second-nearest neighbors by convolution, that is, by the formula $f_2(u) = (f*f) (u)$. Does it work this way?
Furtheremore, writing $f^{*k}$ for $\underbrace{f* \dots * f}_{k \textrm{ times}}$ - then if all consecutive spacings were conjectured to be independent, we should have
$$
g(u) = \delta(u) + \sum_{k=1}^\infty f^{*k}(u).
$$
Is this last equality true? (I don't know how to check it.)

REPLY [10 votes]: This next-nearest-neighbor distribution of the Riemann zero's is addressed in Mehta's book on random-matrix theory. It is well reproduced by that of the Gaussian Unitary Ensemble (GUE), compare black curve and black data points:

I added the red curve in the plot, being the convolution of the nearest-spacing distribution in the GUE:
$$p(s)=\int_0^s f(u)f(s-u)du,\;\;\text{with}\;\;f(u) = \frac{32}{\pi^2} u^2 \exp\left(-\frac{4}{\pi} u^2 \right)$$ 
As you can see, it is quite different, so this "independent spacing" assumption is not reliable.
All of this is for complex matrices (GUE). For real symmetric matrices (GOE) the distribution of next-nearest-neighbor spacings is given by the distribution of the nearest-neighbor spacings of quaternion Hermitian matrices (GSE).<|endoftext|>
TITLE: Description of $\big(\ell^\infty(\mathbb N)\big)^{\!*}$ via ultrafilters
QUESTION [12 upvotes]: Let $\beta\mathbb N$ is the set of ultrafilters on $\mathbb N$ and $\mathscr F\in\beta\mathbb N$. Assume that $l_{\mathscr F}\in\big(\ell^\infty(\mathbb N)\big)^{\!*}$ is the functional which assigns to a bounded sequence $\{a_n\}_{n\in\mathbb N}\subset\mathbb C$ its limit, with respect to the ultrafilter $\mathscr F$. Set 
$$
X=\mathrm{span}\,\big\{l_{\mathscr F}: \mathscr F\in\beta\mathbb N\big\}.
$$
It is relatively straightforward to show that $X$ is dense in $\big(\ell^\infty(\mathbb N)\big)^{\!*}$, in the weak-$*$ topology.
My question is whether $X$ is also dense in the dual-norm topology.

REPLY [4 votes]: Both the answers of Eric Wofsey and Jochen Wengenroth are very elegant. Let me give a quite different, a more constructive one. (Although AC is used repeatedly.)
Definitions. A generalised limit is a bounded linear functional on $(\ell^\infty(\mathbb N))^*$, which assigns to every
convergent sequences their limit. One such limit is called positive if it assigns non-negative values to sequences with non-negative terms. A Banach limit is a positive generalised limit $\varphi$ with the property $\varphi(S\boldsymbol{a})=\varphi(\boldsymbol{a})$, where $\boldsymbol{a}=\{a_n\}$ and $S$ is the shift operator, i.e. $S\{a_n\}=\{a_{n+1}\}$.
Facts. Positive generalised limits are exactly the generalised limits of unit norm. Limits with respect to ultrafilters are positive generalised limits. Limits with respect to ultrafilters are exactly the generalised limits which assign to sequences subsequential limits. Also, if $A_1,\ldots, A_k$ is a partition of $\mathbb N$ and $\mathscr F$ an ultrafilter, then exactly one of the $A_i$'s belongs to $\mathscr F$.
For $A\subset\mathbb N$, define as $\chi_A=\{a_n\}$, with $a_n=1$ if $n\in A$ and $a_n=0$ if $n\in A^c$. If $\mathscr F$ is an ultrafilter, then $l_{\mathscr F}(\chi_A)=1$ iff $A\in\mathscr F$. Otherwise $l_{\mathscr F}(\chi_A)=0$. Moreover, if $\mathscr F_i$, $i=1,\ldots,k$, are distinct ultrafilters, there exist disjoint $A_1,\ldots,A_k\subset\mathscr N$, such that $A_i\in\mathscr F_i$, $i=1,\ldots,k$.
The fact that $X=\mathrm{span}\{l_{\mathscr F}:\mathscr F\in\beta\mathbb N\}$ is not dense in $(\ell(\mathbb N))^*$ shall be a consequence of the lemma:
Lemma. If $\varphi$ is a Banach limit, $\mathscr F_1,\ldots,\mathscr F_k$ are distinct ultrafilters on $\mathbb N$ and $\lambda_1,\ldots,\lambda_k\in\mathbb C$, then $$\Big\|\varphi-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}\Big\|_*=1+\sum_{j=1}^k\lvert\lambda_j\rvert.$$
Proof. Let $S_{m,r}=\{mj+r: j\in\mathbb N\}$. It's not hard to see that $\varphi(\chi_{S_{m,r}})=1/m$. Meanwhile, each $\mathscr F_i$ contains exactly one of the $S_{m,r}$'s, $r=0,\ldots,m-1$, and thus, for every $\varepsilon>0$, there exist $A_i\in\mathscr F_i$, $i=1,\ldots,k$, with $\varphi(A_i)<\varepsilon$. The $A_i$'s can be chosen to be disjoint, as the $\mathscr F_i$'s are distinct. Now let $B=(\bigcup_{i=1}^k A_i)^c$. Then
$$
l_{\mathscr F_i}(\chi_{A_i})=1, \quad i=1,\ldots,k\quad\text{while}\quad 1-k\varepsilon\le\varphi(\chi_B)\le 1.
$$ 
Define now the sequence
$$
a_n=\left\{\begin{array}{ccc} 1 & \text{if} & n\in B, \\
\frac{\overline{\lambda}_i}{\lvert\lambda_i\rvert} & \text{if} & n\in A_i.\end{array}\right.
$$
It's not hard to check that $\|\{a_n\}\|_\infty=1$,  $\lambda_i l_{\mathscr F_i}(\{a_n\})=\lvert\lambda_i\rvert$, while $\lvert\varphi(\{a_n\})-1\rvert\le 2k\varepsilon$. Therefore
$$
\Big\|\varphi(\{a_n\})-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}(\{a_n\})\Big\|_*\ge\Big\lvert\varphi(\{a_n\})-\sum_{j=1}^k\lambda_j\,l_{\mathscr F_j}(\{a_n\})\Big\lvert\ge 1+\sum_{j=1}^k\lvert\lambda_j\rvert-2k\varepsilon.
$$
And the Lemma follows.<|endoftext|>
TITLE: How "nondegenerate" are amalgamated free products of C*-algebras?
QUESTION [13 upvotes]: In the following, I assume all algebras are unital. Let $A$ and $B$ be C*-algebras that each contain (isomorphic copies of) a common C*-subalgebra $C$. Let $A *_C B$ denote the amalgamated free product of $A$ and $B$ over $C$, which is the pushout formed in the category of (unital) C*-algebras.  In case $C = \mathbb{C}$ is the complex field, this yields the usual "maximal" free product of C*-algebras.
It is known that the naturally induced $*$-homomorphisms from $A$ and $B$ to $A *_C B$ are injective. For instance, see Theorem 3.1 of this paper by Blackadar or Theorem 4.2 of this paper by Pedersen.
On the other hand, we may form the "purely algebraic" pushout $A \circledast_C B$ in the category of (unital) rings. By its universal property, it is equipped with a canonical algebra homomorphism $$A \circledast_C B \to A *_C B.$$ The nondegeneracy result of Blackadar implies that the natural maps from $A$ and $B$ to $A \circledast_C B$ are also injective, which is already a nontrivial piece of information since there are many examples of (purely algebraic) amalgamated free products of complex algebras that trivialize.
But how much does $A *_C B$ "know" about the purely algebraic object $A \circledast_C B$?

Question: Is the natural map $A \circledast_C B \to A *_C B$ injective for all choices of $A$, $B$, and $C$?

I believe that $A *_C B$ may be constructed from $A \circledast_C B$ in a manner similar to the method for maximal free products described in the introduction of this paper by Avitzour, by defining $\|x\|$ for $x \in A \circledast_C B$ to be the supremum of the operator norm of $x$ over all Hilbert space $*$-representations and completing with respect to $\|\bullet\|$. If this defines an honest norm, then I can see that the answer to the question would be affirmative because the algebra will embed in its completion.
However, do we know that each $x \in A \circledast_C B$ truly acts nontrivially in some Hilbert space $*$-representation? It is not clear to me how to deduce this from the nondegeneracy proofs of Blackadar or Pedersen. I don't even see that this is the case for the free product when $C = \mathbb{C}$ as in Avitzour's paper, where he actually refers to $\|\bullet\|$ as a norm.

REPLY [4 votes]: This is a partial solution:
If $\varphi$ and $\psi$ are faithful states on $A$ and $B$, respectively, then the proposition in subsection 2.3 in Avitzour proves that $A \circledast_\mathbb C B$ injects into the reduced free product of $(A,\varphi)$ and $(B,\psi)$. This is obtained by showing that the cyclic vector belonging to the state $\varphi * \psi$ is separating.
I can imagine that one can obtain a similar result when there are faithful expectations onto $C$ in the amalgamated case.<|endoftext|>
TITLE: Weighted projective spaces as stacks
QUESTION [7 upvotes]: As stacks are the weighted projective line $\mathbb{P}$(1,n-1) and $\mathbb{P}$(k,n-k) isomorphic? Is there any reference for this?

REPLY [13 votes]: Let $X$ be a weighted projective (stacky) line. At least if we work over a field $K$, the Picard group of $X$ is $\mathbb{Z}$. Only powers of one of the two generators have global sections. Call this generator $\mathcal{O}(1)$ and its powers $\mathcal{O}(l)$. Thus, we can attach to $X$ the graded ring $R_*(X) =\bigoplus_{l\in\mathbb{Z}}H^0(X; \mathcal{O}(l))$. 
We have $R_*(\mathbb{P}(n,k)) = K[x,y]$ with $|x| = n$ and $|y| = k$. Thus, we can recover the degrees $n$ and $k$ from the graded ring $R_*(X)$. In particular, the weights $k$ and $n$ are uniquely determined.

REPLY [10 votes]: This is not true, and you can verify that it is not so by checking the group of automorphisms of the $k$-points for $k$ a field. In the case of $\Bbb P(1,n-1)$, the inclusion of graded rings $k[x,y^{n-1}] \to k[x,y]$ (where $|y| = 1$ and $|x| = n-1$) gives a projection down to a coarse moduli space $\Bbb P^1$. The stabilizers of the points where $y \neq 0$ are all trivial (there is an open embedding $\Bbb A^1 \to \Bbb P(1,n-1)$ with this image), but the stabilizer of the point where $y=0$ is the cyclic group of $(n-1)$'st roots of unity.
By contrast, $\Bbb P(2,2)$ also has coarse moduli space $\Bbb P^1$ but the automorphism group of any point is $\{\pm 1\}$.<|endoftext|>
TITLE: What is the right adjoint of the tensor product in a closed monoidal functor category?
QUESTION [7 upvotes]: The nLab says the following about closed monoidal functor categories:

Let $C$ be a complete closed monoidal category and $I$ any small category. Then the functor category $[I, C]$ is closed monoidal with the pointwise tensor product, $(F \otimes G)(x) = F(x) \otimes G(x)$.

Now I wonder what the right adjoint of $F \otimes {-}$ is. I suppose, that it fulfills the following equation (which is a generalization of the equation for exponentials in functor categories):
$$(F \multimap G)(x) = \int_{y : I} \prod_{I(x, y)} F(y) \multimap G(y)$$
Is this correct? And if yes, is there a more standard way of representing the right adjoint?

REPLY [8 votes]: The formula you give is correct. However, I prefer to use a power/cotensor rather than an indexed product, for while they are equivalent for ordinary categories, only the former gives the correct formula for the general enriched case. So, if $\mathscr{A}$ is a $\mathscr{V}$-category and $\mathscr{X}$ is a closed monoidal $\mathscr{V}$-category, the internal hom in $[\mathscr{A},\mathscr{X}]$ for the pointwise monoidal structure is given by any of the following equivalent formulas:
\begin{equation}
\begin{split}
\left[F,G\right]A & \cong \int_B \mathscr{A}(A,B) \pitchfork [FB,GB] \\
& \cong \int_B [\mathscr{A}(A,B) \otimes FB,GB] \\
& \cong \int_B [ \mathscr{A}(A,B) \otimes I,[FB,GB]]
\end{split}
\end{equation}
when the indicated limits and colimits exist in $\mathscr{X}$ (here $\pitchfork$ and $\otimes$ denote cotensoring and tensoring with objects of $\mathscr{V}$ and $I$ is the unit object of $\mathscr{X}$).<|endoftext|>
TITLE: Is hyperelliptic cryptography "practical"?
QUESTION [22 upvotes]: Previosly my impression on this subject was that hyperelliptic cryptography systems (as well as other possible cryptosystems based on abelian varieties of dimension $>1$) have no advantages over elliptic curve ones and are more difficult to implement. Yet Google search has demonstrated me that people are working on  improving hyperelliptic cryptosystems and also implement them as computer programs. So my question is: is anybody currently using hyperelliptic cryptography  "in practice", and what is the chance that anyone will prefer hyperelliptic cryptography to the elliptic one "soon enough".
Sorry if this question is not appropriate for MO.

REPLY [18 votes]: I am not aware of anybody seriously considering hyperelliptic curves for actual real-world usage, beyond toys, and I would be rather surprised to hear differently from anyone. 
As you say, hyperelliptic provide comparatively few (if any!) advantages over elliptic curves but have the huge disadvantage that virtually nobody has well-tested, battle-hardened code for them. As a matter of fact, even elliptic curves are difficult to implement right, which is one of the reasons Dan Bernstein developed Curve25519 -- this is an elliptic curve (and more) which is designed to make it hard to screw up, compared to previously used ones.
Given all that, why should a practical cryptographer then bother with hyperelliptic curves?
To convince people who implement crypto to use a new cryptosystem is hard. Heck, some standard committees and vendors still are unsure about these new-fangled elliptic curves ;-).
And if they switch to something new, they usually want it to give some serious advantage, to justify the cost and the incompatibility. Such as (a high chance for) being safe against quantum computer attacks -- and hyperelliptic curves are no better in that regard than RSA or elliptic curves (i.e. they are completely broken if we ever get "real" quantum computers).
My best guesses as to why people still research crypto based on hyperelliptic curves are:

habit,
they are genuinely curious, and do the research for its own sake, not caring about applications,
they suffer from the very disturbing, but also sadly widespread disconnection between crypto researchers who work purely theoretically, and people who actually implement it, and know how to do it properly. The intersection between these two sets is rather small, and a lot of crypto systems that look good on paper are completely useless in reality for that reason.

Being an algebraist myself, whose most applied results at best have application in theoretical physics (string theory for that), I am very reluctant to judge any research based on how applicable it is to real-world scenarios. But in this case, I am tempted... ;-)<|endoftext|>
TITLE: A sum over characters of the symmetric group
QUESTION [12 upvotes]: Let $C_\mu$ be the size of the conjugacy class in $S_n$ of permutations whose cycletype is the partition $\mu\vdash n$. Let $\chi$ be the characters of the irreducible representations of $S_n$.
Let $\omega\vdash m$ and let $\theta\vdash(n+m)$. I am interested in the sum 
$$ \frac{1}{n!}\sum_{\mu\vdash n} C_\mu\chi_\lambda(\mu)\chi_\theta(\mu\cup\omega),$$ where $\mu\cup\omega$ is a partition of $n+m$ containing the parts of $\mu$ and of $\omega$. Numerics suggest that for most pairs $(\lambda,\theta)$ this sum is zero. In the simplest case of $\omega=0$, for example, the only non-vanishing pair is $\theta=\lambda$ (and the result is 1).
To illustrate, these are the values of the sum when $\omega=(3)$, with $n=4$ and $n+m=7$ ($\lambda$ is labelling the rows and $\theta$ the columns, both in lexicographic order)
$$\left[ \begin {array}{ccccccccccccccc} 1&0&0&1&0&-1&0&0&1&0&0&0&0&0&0
\\ 0&1&0&0&0&0&0&-1&0&0&0&1&0&0&0
\\ 0&0&1&-1&0&0&0&0&0&0&-1&0&1&0&0
\\ 0&0&0&0&1&0&-1&0&0&0&0&0&0&1&0
\\ 0&0&0&0&0&0&0&0&1&-1&1&0&0&0&1\end {array}
 \right] $$
A possible solution would be to write $\chi_\theta(\mu\cup\omega)=\sum_{\rho\vdash n} a_\rho(\omega) \chi_\rho(\mu)$. Is there a known way to accomplish this decomposition? (I'm thinking something like the Murnaghan-Nakayama rule)

REPLY [12 votes]: Since $\chi_\theta(\mu\cup\omega)=\langle s_\theta,p_\mu p_\omega\rangle$, your sum is given by
  $$ \frac{1}{n!}\sum_{\mu\vdash n}
     C_\mu\chi_\lambda(\mu)\langle s_\theta,p_\mu p_\omega\rangle =
     \left\langle s_\theta,p_\omega\cdot \frac{1}{n!}\sum_{\mu\vdash
     n} C_\mu \chi_\lambda(\mu)p_\mu\right\rangle $$
  $$ \qquad\qquad = \langle s_\theta,p_\omega s_\lambda\rangle. $$
We can then expand $p_\omega s_\lambda$ in terms of Schur functions by
Theorem 7.17.1 (the basis for the Murnaghan-Nakayama rule) of
Enumerative Combinatorics, vol. 2. Note also that $\langle s_\theta,p_\omega s_\lambda\rangle = \langle s_{\theta/\lambda},p_\omega\rangle =
\chi_{\theta/\lambda}(\omega)$, a value of the skew character $\chi_{\theta/\lambda}$.<|endoftext|>
TITLE: stabilization of Legendrian knots
QUESTION [6 upvotes]: There are two ways to stabilize a Legendrian knot $k$ in standard contact sphere $(S^3,\xi_{st})$ i.e. adding right cusps or left cusps, let's call these two stabilized Legendrian knots $k_R$ and $k_{L}$ respectively. If the knot $k$ is embedded on the page of an open book $(\Sigma,\phi)$ of $(S^3,\xi_{st})$, then $k_R$ and $k_L$ can also be embedded on page of an open book obtained by positive stabilization of $(\Sigma,\phi)$. I want to know what are the monodromies of the two new open books relative to $(\Sigma,\phi)$.
I know they cannot be equivalent, since $(-1)$-contact surgery on each of the stabilized knots gives two non-homotopic contact structures (their first chern class is different), so the contact structures are not isotopic and therefore the open books cannot be equivalent.

REPLY [4 votes]: As you mentioned, $k_L$ and $k_R$ both live in a stabilisation of the open book $(\Sigma,\phi)$.
Namely, suppose you have the triple $(\Sigma,\phi,k)$, where now I think of $k$ as an embedded, nonseparating curve in $\Sigma$. Recall that $k$ is oriented (in order to make sense of "left" and "right" stabilisations), therefore it makes sense to consider a path $\gamma$ that goes from $k$ to $\partial \Sigma$ to the left of $k$ or to the right of $k$, to a given point $p\in\partial\Sigma$.
Once you've chosen such a path, stabilise $(\Sigma,\phi)$ near $p$, choosing a stabilisation arc that is boundary-parallel. There is a natural way to make $k$ go through the new handle, running along $\gamma$ and than back along $\gamma^{-1}$.
The choice of a left/right path $\gamma$ determines whether the stabilisation is a left or a right stabilisation (although the standard terminology is positive vs negative).
I'm not 100% sure that a left path corresponds to a left stabilisation in your language, but there's a 50% chance that I'm right. In any event, I have learnt this from Lisca-Ozsváth-Stipsicz-Szabó (the definition of the LOSS invariant in Heegaard Floer homology), and they credit this to John Etnyre, in his (extremely valuable) Lectures on open book decompositions and contact structures.<|endoftext|>
TITLE: Definition of E-infinity operad
QUESTION [6 upvotes]: What is the definition of $E_\infty$-operad in the category of chain complexes over $\mathbb{Z}/p\mathbb{Z}$? J. Smith in http://arxiv.org/abs/math/0004003 define it for complexes over $\mathbb{Z}$ (Definition 2.24). Sorry but I'm confused with how does the version in
$\mathbb{Z}/p\mathbb{Z}$ must looks like. Thanks!

REPLY [11 votes]: Operads $\mathcal{C}$ can be defined in any symmetric monoidal category, and then
$E_{\infty}$ operads are specified in accordance with the (or a) notion of equivalence 
relevant to that category.  In any category, I prefer to insist that they be $\Sigma$-free, 
in the sense that the symmetric group $\Sigma_n$ acts freely on $\mathcal{C(n)}$.   
In spaces, it is required that each $\mathcal{C}(n)$ be contractible, so that it is a 
universal cover of the orbit space $\mathcal{C}(n)/\Sigma_n$, which is a $K(\Sigma_n,1)$.
(Some might prefer weakly contractible, which is the same when the spaces of 
the operad are CW homotopy types, as is true in all of the examples I know).
There is not and should not be a canonical example: that would lose the whole
force of the examples, where very different $E_{\infty}$ operads act on different
naturally occurring spaces.
In algebra, we can take our symmetric monoidal category to be
the category of chain complexes of modules over any commutative 
ring $R$.  In that case, it is reasonable to require $\mathcal{C}(n)$ to be free 
(or at least projective) over the group ring $R[\Sigma_n]$.  While 
grading is somewhat negotiable, in homological grading I would
insist that $\mathcal{C}(n)$ is zero in negative degrees and that it be an
$R[\Sigma_n]$-projective resolution of the trivial $\Sigma_n$-module $R$.
Again there are many examples, none thought of as canonical.
This gives the pretty picture that the chain complex with coefficients
in $R$ of an $E_{\infty}$ operad of spaces is an $E_{\infty}$ operad
of $R$-chain complexes, automatically giving many different examples.
There are many other examples that do not arise from spaces.
The $\Sigma$-freeness forces one not to confuse things with $\mathcal{Com}$,
which is not homologically correct or interesting; see
Are $E_n$-operads not formal in characteristic not equal to zero?
which is especially relevant to the case $\mathbf{Z}/p\mathbf{Z}$ in the question.<|endoftext|>
TITLE: Nonsolvable finite quotients of matrix groups
QUESTION [10 upvotes]: Suppose that $\Gamma$ is a finitely generated nonsolvable subgroup of $GL(n, R)$. Is it in the literature that $\Gamma$ has a nonsolvable finite quotient? I know how to prove it (the hardest ingredient is due to Nori), but would prefer to give a reference instead of a proof.

REPLY [7 votes]: It is a straightforward consequence of the combination of two old results of Malcev and most likely Malcev was aware of this consequence.
1) The first is the well-known Malcev's residual finiteness result (every finitely generated linear group over, say, a field, is residually finite), or rather the stronger result that follows from the proof. Indeed, the main ingredient is the lemma saying that every finitely generated domain is residually a finite field. As a corollary, we have that for every $n$, every field $K$, every finitely generated subgroup of $\mathrm{GL}_n(K)$ is residually a subgroup of $\mathrm{GL}_n(F)$ where $F$ ranges over finite fields (with the same $n$).
2) The second is Malcev's result (1951, English translation 1956) that solvable subgroups of $\mathrm{GL}_n(K)$ have solvability length $\le r_n$ for some $r_n$, independently of the field $K$. This can be found (Theorem 3.21) in the book of D. Robinson Finiteness conditions and generalized soluble groups I, Theorem 3.21. Actually, Malcev proved the stronger result that for some $s_n$ independent of the field, every solvable subgroup of $\mathrm{GL}_n(K)$ has a subgroup of index $\le s_n$ that is conjugate, in $\mathrm{GL}_n(\hat{K})$, to the upper triangular group ($\hat{K}$ denoting an algebraic closure).

3) The combination of these two facts entails the result for linear groups over fields. Now suppose we have a non-solvable subgroup $\Gamma$ of $\mathrm{GL}_n(R)$ where $R$ is an arbitrary (associative unital) commutative ring. Let $N$ be the nilradical of $R$. It is no restriction to assume that $R$ is finitely generated. So the kernel of $\mathrm{GL}_n(R)\to\mathrm{GL}_n(R/N)$ is nilpotent; thus the image of $\Gamma$ in $\mathrm{GL}_n(R/N)$ is not solvable. Thus, we can suppose that $R$ is reduced (i.e., $N=0$). So $\{0\}$ is the intersection of the finitely many minimal prime ideals $P_i$ of $R$. It follows that the image of $\Gamma$ in at least one of the $\mathrm{GL}_n(R/P_i)$ is non-solvable. This reduces to the case of a field.<|endoftext|>
TITLE: How often can subsets of a universe intersect exactly once?
QUESTION [5 upvotes]: My question is inspired by the following observation:
Claim: It is not possible to choose $n$ subsets of the universe $[n]$, each of size $\Omega(n)$, such that for each subset $S$ and each element $s \in S$, there is another subset $S'$ such that $S \cap S' = \{s\}$.
Proof: Suppose, towards a contradiction, that this system of subsets exists.  First, build a bipartite graph, in which nodes on the left side of the partition correspond to subsets, nodes on the right side of the partition correspond to members of the universe, and edges correspond to set membership. Next, add an edge between each pair of subsets that intersects on exactly one element.  Note that this graph contains $\Omega(n^2)$ triangles, and each edge that participates in any of these triangles participates in exactly one triangle.
We now obtain a contradiction by applying the Triangle Removal Lemma, which states: for all $\epsilon > 0$, there is a $\delta > 0$ such that any graph with at most $\delta n^3$ triangles can be made triangle-free by removing at most $\epsilon n^2$ edges.  Our graph has $\Theta(n^2)$ triangles, so it satisfies the premise regardless of the value of $\delta$.  Thus, we can make the graph triangle-free by removing $\epsilon n^2$ edges, for any $\epsilon > 0$.  By choosing $\epsilon$ so small that the graph has more than $\epsilon n^2$ edge-disjoint triangles, we obtain a contradiction.

My Question: Suppose that we are now choosing $n$ subsets, each of size $\Omega(n)$, out of a universe $[u]$.  How large does $u$ have to be such that it is possible that for each $s \in S$, there exists $S'$ with $S \cap S' = \{s\}$?

The above proof shows that $u = \omega(n)$.  The trivial upper bound -- which I have not been able to beat -- is $u = O(n^2)$.  I am interested in improving either of these bounds.  I am especially interested in whether $u = \Omega(n^{1 + c})$ for some absolute $c > 0$ (i.e. a polynomial increase in $u$ is required).

REPLY [3 votes]: It seems that for every $k\geq 2$, there exists a desired cpllection of $6k$ subsets of $[3k]$ with size $k$ each. 
Let $a_1$, $a_2$, $a_3$ be three rays with a common origin $O$, and let $s_1,\dots,s_k$ be $k$ different circles centered at $O$. Denote $x_{ij}=a_i\cap s_j$; we may identify $[3k]$ with the set of all the $x_{ij}$.
Now take $6k$ sets as follows. Each set consists of all the points on one of the rays $a_i$ except for some element $x_{ij}$, this element being replaced with some $x_{i'j}$, $i'\neq i$. It seems that this collection satisfied all the requirements.
One may also restrict to the sets with $i'=i+1\mod 3$.
So, if I'm right, there is something wrong with the proof of the claim.<|endoftext|>
TITLE: A nice subcategory of the category of measurable spaces
QUESTION [21 upvotes]: Is there some notion of "nice" measurable spaces and "nice" maps between them which satisfies the following properties?

The real line equipped with the Lebesgue $\sigma$-algebra is nice.
Any translation $\mathbb R \to \mathbb R$ is nice. (Ideally, any continuous map $\mathbb R \to \mathbb R$ is nice, but right now I'm not picky.)
Any finite or countable set equipped with the discrete $\sigma$-algebra is nice.
Nice measurable spaces and nice maps form an elementary topos.

Observations:

If such a topos exists, it is Boolean but does not satisfy the axiom of choice. To prove the latter: Construct the unit circle as a quotient of $\mathbb R$, and then consider the quotient of the unit circle by an irrational rotation. This quotient map cannot split, because if it did the image of its section would be a Vitali set.
If such a topos does not exist, any proof of its nonexistence must rely on the axiom of choice, since in the Solovay model the category of discrete measurable spaces satisfies the conditions.

Edited to add: I would also be interested in the answer to the same question with the Lebesgue $\sigma$-algebra replaced by the Borel $\sigma$-algebra.

REPLY [3 votes]: Take the category of measurable locales, equip it with its natural Grothendieck topology,
and take the topos of sheaves of sets on the resulting site.
(Apply standard disclaimers about universes, coaccessibility, or cosmall sheaves to avoid size issues.)
The resulting Grothendieck topos contains the category of measurable locales
(or, equivalently, the category of measurable spaces,
see https://mathoverflow.net/a/20820 for a detailed description of the latter, and https://mathoverflow.net/a/49542 for more information).
Therefore it satisfies all of your properties, except for the
part about arbitrary continuous maps: the preimage of a measure 0
set under a continuous map R→R is not necessarily of measure 0,
so you cannot hope to get such maps.
But you do get all continuous maps whose preimage preserves measure 0 sets.<|endoftext|>
TITLE: Height function on 2-torus with only 3 critical points
QUESTION [39 upvotes]: It is well-known that a Morse function on $T^2$ has at least $4$ critical points, but also that there exist functions $f\colon T^2\to\mathbb R$ with only 3 critical points (the least possible number by Lusternik-Schnirelmann theory): a minimum, a maximum, and a degenerate saddle.
It is not hard to describe these functions by means of their levelsets, but it seems difficult to produce immersions of $T^2$ into $\mathbb R^3$ with a height function that does the job. According to Banchoff and Takens, there are no smooth embeddings with such height functions, only immersions. I was trying to look for pictures of such immersions and came across the following beautiful image by Cassidy Curtis:

I expected that these would be somehow easier to produce -- but probably I am wrong... still:

Is there a "simpler" immersion of $T^2$ into $\mathbb R^3$ whose image has a height function with exactly 3 critical points? Or is the above example "optimal" in some sense?

The only necessary conditions I see are that the saddle point $p$ must be degenerate and that there are 3 arcs with both endpoints at $p$ that lie in the same levelset as $p$. I find it hard to believe that these conditions are not sufficient, and that there are no easier immersions; but I have not been able to prove this or find any such immersions.

REPLY [10 votes]: Warning: this is not an immersion (it has twelve Whitney-umbrella-like pinch points)
Here is a relatively simple explicit realization: the $z$ coordinate for the parametric surface$$(x,y,z)=(\sin(2u),\sin(2v),\sin(u)\sin(v)\sin(u-v));$$after an affine shift leaving $z$ unchanged the graph looks like this:

It is thus similar to the Steiner's Roman surface except that the latter has three double lines and this one has six.
I've tried to cut it to make the central monkey saddle more visible:

The implicit equation, in slightly different coordinates, is$$(X+Y+Z)(X+Y-Z)(X-Y+Z)(-X+Y+Z)=(XYZ)^2,$$
 
with function $X+Y+Z$:

Mathematica codes:
ParametricPlot3D[
 {3Sin[2u]+4Sin[u]Sin[v]Sin[u-v],3Sin[2v]-4Sin[u]Sin[v]Sin[u-v],Sin[u]Sin[v]Sin[u-v]},
 {u,0,\[Pi]},{v,0,\[Pi]},
 BoxRatios->{1,1,1},
 Mesh->None,
 PlotPoints->150,
 PlotStyle->FaceForm[Red,Cyan],
 Boxed->False,Axes->False,
 SphericalRegion->True
]

ParametricPlot3D[
 {3Sin[2u]+4Sin[u]Sin[v]Sin[u-v],3Sin[2v]-4Sin[u]Sin[v]Sin[u-v],Sin[u]Sin[v]Sin[u-v]},
 {u, 0, \[Pi]}, {v, 0, \[Pi]},
 BoxRatios -> {1, 1, 1}, Mesh -> None,
 RegionFunction -> (Abs[#3]<.05 \[Or] .2<Abs[#3]<.4 \[Or] .6<Abs[#3]<.8 \[Or] Abs[#3]>1 &),
 PlotPoints -> 250,
 PlotStyle -> FaceForm[Red, Cyan],
 BoundaryStyle -> Black,
 Boxed -> False,
 Axes -> False,
 SphericalRegion -> True
]

With[{d=2},
 ContourPlot3D[
  (U-V-W) (U+V-W) (U-V+W) (U+V+W)+(U V W)^2==0,
  {U,-d,d},{V,-d,d},{W,-d,d},
  BoxRatios->{1,1,1},
  PlotPoints->150,
  MeshFunctions->{#1+#2+#3&},
  ImageSize->Full,
  Mesh->40
 ]
]<|endoftext|>
TITLE: Does this notion related to species/operads/FI-modules have a name?
QUESTION [9 upvotes]: Let $B$ be the symmetric monoidal category of finite sets and bijections with disjoint union. Let $C$ be a symmetric monoidal category. Is there a standard name for a lax monoidal functor $F:B \to C$? In other words, we are considering a sequence $F(n)$ of $S_n$-representations in $C$ (i.e. a species in $C$) together with $S_n\times S_m$-equivariant maps $F(n) \otimes F(m) \to F(n+m)$ satisfying certain associativity conditions.
I could invent a name for this notion (like "a $B^\otimes$-module in $C$") but if there is already an established terminology I'd prefer to use that.

REPLY [8 votes]: Depending on whether you want it to agree with the symmetric structure or only with monoidal structure, this would be usually referred to, respectively, as twisted commutative algebras or twisted associative algebras. See, for example, http://arxiv.org/pdf/0710.3392.pdf, and a more classical reference http://www.sciencedirect.com/science/article/pii/0022404993901064 . (Or Chapter 4 in http://www.maths.tcd.ie/~vdots/AlgebraicOperadsAnAlgorithmicCompanion.pdf :) )<|endoftext|>
TITLE: What did Euler do with multiple zeta values?
QUESTION [15 upvotes]: When reading about multiple zeta values, I often find the claim that the case of length two 
$$
\zeta(s_1, s_2)=\sum_{n>m \geq 1} \frac{1}{n^{s_1}m^{s_2}}, \qquad s_1 \geq 2, \quad s_2 \geq 1
$$
was first considered by Euler. Does anybody know what precisely he proved about these numbers?

REPLY [19 votes]: Euler proved in 1775 in Meditationes circa singulare serierum genus ("Meditations about a singular type of series") that
$$\sum_{i+j=n,\,i\geq 2,\;j\geq 1}\zeta(i,j)=\zeta(n),$$
as a special case of a more general sum rule.
Notice that Euler's definition of the multiple-zeta-value is slightly different than in the OP ($n\geq m$ instead of $n>m$)<|endoftext|>
TITLE: Why the reflection rule trivializes higher paths in Martin-Löf Extensional Type theory?
QUESTION [14 upvotes]: Martin-Löf Extensional Type theory differs from its intensional counterpart in that it contains the so-called reflection rule that says that if $p : x = y$, then actually $x \equiv y$ (i.e. $x$ and $y$ are definitionally or judgementally equivalent). Its known that this causes strong normalization to fail and type-checking to be undecidable.
More to the point, it is known that it also trivializes higher paths in a type in the sense that for any $p, q: x = y$, $p \equiv q$. I wonder why is this so?
I heard this is because it implies that for any $p : x = x$, $p \equiv refl_x$. But I can't see why this is true. As far I can see, given any such $p$ the reflection rule only implies that $x \equiv x$, which is true anyway, since each term is definitionally equal to itself - with or without the rule. So how come we conclude that $p \equiv refl_x$? What am I missing?
Thanks!

REPLY [16 votes]: The point is that the reflection rule makes $p = \mathsf{refl}_x$ a well-formed expression. This turns out to be incredibly dangerous: now we can prove it by induction on equality.
More precisely:

In a context where $x$ and $y$ are variables of the same type $T$ and $p$ is a variable of type $x = y$, we have $x \equiv y$, so $p$ is also a variable of type $x = x$, and therefore $p = \mathsf{refl}_x$ is well formed.
Thus $\prod_{x : T} \prod_{y : T} \prod_{p : x = y} p = \mathsf{refl}_x$ is also well formed.
By induction on equality, we have:
$$\left( \prod_{x : T} \mathsf{refl}_x = \mathsf{refl}_x \right) \to \left( \prod_{x : T} \prod_{y : T} \prod_{p : x = y} p = \mathsf{refl}_x \right)$$

Of course, we have $\mathsf{refl}_{\mathsf{refl}_x} : \mathsf{refl}_x = \mathsf{refl}_x$, so the claim follows.<|endoftext|>
TITLE: Do we care about multiple zeta functions?
QUESTION [18 upvotes]: Coming from a number-theoretic background, I certainly care about $L$-functions and in particular automorphic ones. For automorphic forms on $SL_2(\mathbb{Z}) \backslash SL_2(\mathbb{R})$, $L$-function can be interpreted as the Mellin transform of Fourier expansion at the cusp.
If we look at automorphic forms on $SL_(n,\mathbb{Z}) \backslash SL(n,\mathbb{R})$, we still have a Fourier expansion in $n-1$ variables since it is periodic with respect to the super-diagonal unipotent group. This gives us Fourier coefficients $A(m_1,\cdots,m_{n-1})$. In forming the $L$-functions for $SL(n,\mathbb{R})$, we just look at all the Fourier coefficients $A(m,1,\cdots,1)$ and define
$$L(s) = \sum_{m=1}^{\infty} A(m,1,\cdots,1)m^{-s}$$
But the most natural thing to do (comparing to the $SL_2(\mathbb{R})$ case) would be to form the multiple Dirichlet series instead
$$\sum_{m_1,\cdots,m_{n-1}} \frac{A(m_1,\cdots,m_{n-1})}{m_1^{s_1} \cdots m_{n-1}^{s_{n-1}}}$$
It certainly seems that people care "less" about this multiple Dirichlet series: in the whole Langlands business we always take about L-parameters and stuff, which seems to imply that we track only the information of $L$-function but not the whole multiple Dirichlet series.
So here are my questions,

Is there any conceptual reason why we care more about $L$-functions rather than multiple Dirichlet series?
Of course, Fourier expansion is available when there is a cusp. For cocompact arithmetic quotients of $GL_n(\mathbb{R})$ (or other reductive groups in general), can one similarly define a multiple Dirichlet series that should incorporate the data of $L$-function?

Thank you.

REPLY [5 votes]: To address the conceptual question, the $L$-function essentially characterizes the automorphic representation and can be studied locally (associating local $L$-functions to local components of the global representation), so it seems to me there is little need (a priori) for using a more complicated Dirichlet series to study these representations.  
On the other hand, multiple Dirichlet series typically do not have Euler products, so do not admit a local-global study, at least in a naive way (though see Bump's survey article).  Of course, as Matt indicates, they are useful for studying $L$-functions.  
Moreover, from an arithmetic point of view, $L$-functions are naturally related to varieties such as elliptic curves.  As far as I know (though I am not an expert on multiple Dirichlet series), there is no direct connection between MDS and counting points on varieties.<|endoftext|>
TITLE: First Galois cohomology of Weil restriction of $\mathbb{G}_m$
QUESTION [6 upvotes]: Let $L/K$ be a finite Galois extension, write $G:= Gal(L/K)$. Denote by $R = Res(\mathbb{G}_m)$ the Weil restriction of $\mathbb{G}_m$, from $L$ to $K$. I want to show that its first Galois cohomology vanishes: $H^1(G, R(L)) = 0$.
Question: Is there a simple way to do this, without calculation (something formal; Any torsor must be trivial because...)?
I think that I checked that this is true, but I just calculate: The Galois module in question is $(L \otimes_K L)^{\times}$, with action on the first coordinate only. Using Galois theory, I rewrite it as $\prod_{\sigma\in G}L^{\times}$ with some action. Writing $M:=L^{\times}$, this is just $Hom_{\mathbb{Z}}(\mathbb{Z}[G] , M)$ (inner $Hom$ in $G$-modules). Now I check that such modules are acyclic: As a functor of $M$, this has an exact left adjoint, and hence sends injectives to injectives. Thus, $$R^1Hom_{\mathbb{Z}[G]} (\mathbb{Z} , Hom_{\mathbb{Z}}(\mathbb{Z}[G],M)$$ equals to the first derived functor of $Hom_{\mathbb{Z}[G]} (\mathbb{Z} , Hom_{\mathbb{Z}}(\mathbb{Z}[G],M))$, which equals to the identity functor.

REPLY [13 votes]: One can do much better: it is not necessary to assume $L/K$ is Galois (merely separable is sufficient). And in fact one can formulate the result in a manner which works beyond that of fields, working over more general rings.  Namely, let $f:S' \rightarrow S$ be a finite etale map of schemes, and let $G'$ be a commutative finitely presented and relatively affine $S'$-group scheme.  Then the finitely presented and relatively affine $S$-group scheme $G = {\rm{R}}_{S'/S}(G')$ defines a sheaf for the etale topology on $S$, and one can contemplate the etale cohomology group ${\rm{H}}^i(S, G)$; when $S$ and $S'$ are spectra of fields and $G' = {\rm{GL}}_1$ and $i=1$ then this recovers the setup of the question with the Galois hypothesis relaxed to separability of the field extension.
So what?  Well, I claim that naturally ${\rm{H}}^i(S, G) = {\rm{H}}^i(S', G')$.  In fact, this is a special case of something more general, and which in greater generality is clearer to prove:  if $\mathscr{F}'$ is any abelian etale sheaf on $S'$ then we get the pushforward $f_{\ast}(\mathscr{F}')$ as an abelian etale sheaf on $S$, and when $\mathscr{F}'$ is the functor of points on $G'$ on the category of etale $S'$-schemes then $f_{\ast}(\mathscr{F}')$ is the functor of points of $G$ on the category of etale $S$-schemes (since by definition Weil restriction is a "functorial pushforward").  Consequently, this is all a special case of the general claim that ${\rm{H}}^{\bullet}(S, f_{\ast}(\cdot)) \simeq
{\rm{H}}^{\bullet}(S', \cdot)$ on the category of abelian etale sheaves on $S'$.  But $f_{\ast}$ is exact between categories of abelian etale sheaves since $f$ is finite (i.e., higher direct images of $f$ for the etale topology vanish), so this is just the degenerate Leray spectral sequence.  
[Of course, in the special case of a finite separable extension of fields this recovers the Shapiro Lemma identification by relating everything to edge maps in spectral sequences.]
It's overkill for the specific question posed, but since you asked for a way to do it "without calculation" (which I'll interpret to mean "by more conceptual means") it seemed worth mentioning. One virtue of this more general approach is that it illuminates the importance of considering the etale topology rather than the fppf topology (e.g., over fields perhaps one might want to consider coefficients in a non-smooth group schemes, and then fppf cohomology is more appropriate to consider than etale cohomology, though the two coincide in all degrees with coefficients in a smooth commutative relatively affine group scheme by section 11 of Grothendieck's Brauer III paper).  The point is that the preceding argument breaks down for the fppf topology because pushforward of finite flat maps are not exact for abelian sheaves relative to the fppf topology, even for something as concrete as a non-separable finite extension of fields (and correspondingly degree-1 higher direct image sheaves on $\alpha_p$ and $\mu_p$ are generally nonzero when considering imperfect fields of characteristic $p$).  So there is no version of Shapiro's Lemma for the fppf topology.
Addendum: Although there is no version of Shapiro's Lemma for the fppf topology, when the coefficients upstairs are a smooth relatively affine commutative group scheme then (by Brauer III, as noted above) fppf cohomology coincides with etale cohomology and so by exactness of pushforward for the etale topology relative to any finite morphism we do have a Shapiro Lemma with $f$ above merely finite locally free rather than just finite etale. In the special case of a finite (possibly non-separable) extension of fields and coefficients in a smooth commutative group scheme of finite type this recovers (by a different viewpoint of proof) the (unique) Corollary in section 2.3 of Chapter IV of Oesterle's beautiful paper "Nombres de Tamagawa et groupes unipotents en caracteristique $p$" in Inventiones Math 78 pp. 13--88 (1984) (which in sections 2.3 and 2.4 gives related useful results in the setting of local and global fields).<|endoftext|>
TITLE: What is this distance about?
QUESTION [5 upvotes]: For points $a,b\in \mathbb{R}^n\setminus \{0\}$ denote $$d(a,b)=\frac{\|a-b\|}{\|a\|+\|b\|}.$$
This question by Ritesh Ahuja (positive answered by Iosif Pinelis) says that $d$ is a metric. My questions are:
1) was it studied, does it have a name?
2) may it be viewed as a mix of Eucledian metrics in the sense 
$$
d(a,b)=\int \|F(a,t)-F(b,t)\|d\mu(t)
$$
for some function $F(a,t)$ on $a\in \mathbb{R^n}\setminus \{0\}$, $t\in (\Omega,\mu)$ is some measure space? For example, metric
$$
\frac{\|a-b\|}{\|a\|\cdot \|b\|},
$$
corresponding to Ptolemy inequality, is nothing but $\|F(a)-F(b)\|$, where $F(a)=a/\|a\|^2$ is inversion. Here $\Omega$ is a trivial 1-point measure space.

REPLY [5 votes]: Later generalizations notwithstanding (as noted already), this particular one...
$$
d(a, b) = \frac{\lVert a - b \rVert}{\lVert a \rVert + \lVert b \rVert}
$$
...was originally called "a multiplicative metric" in the namesake article published in 1976 by Schattschneider, Doris J.. "A Multiplicative Metric". Mathematics Magazine 49.4 (1976): 203–205 where she proved that $d(a, b)$ is in fact a metric on $\mathbb R$.<|endoftext|>
TITLE: Is there an alternate name for the symplectic convolution?
QUESTION [6 upvotes]: Looking into the Wigner-Weyl transformation mapping Hilbert space operators to functions on phase-space, I've run up against the need for a symplectic convolution
$$[F\star G](x,p) = \int \!dy\,dk\, F(y,k)G(x-y,p-k)e^{i (xk-yp)}$$
or more compactly
$$[F\star G](\alpha) = \int \!d\beta\, F(\beta)G(\alpha-\beta)e^{i \alpha \wedge \beta}$$
where $\alpha = (x,p)$ and $\wedge$ is the symplectic form.  Unlike the symplectic Fourier transform,
$$\hat{F}(\xi) = \int \!d\alpha\, F(\alpha)e^{i\alpha\wedge\xi}$$
which can be completely understood as normal Fourier transform followed by a change of variables $(\xi_x,\xi_p) \to (-\xi_p,\xi_x)$, the symplectic convolution appears to have nontrivial properties.  However, I have not been able to find almost any references that discuss it.  The most clear reference to it I could find is in "Toeplitz and Hankel operators and Dixmier traces on the unit ball of $\mathbb{C}^n$" by Englis et al., but they seem to use it without much discussion.
Does the symplectic convolution go by another name?  What introductory references discuss its basic properties?

REPLY [5 votes]: This operation (generalised slightly by replacing $e^{i (xk-yp)}$ by $e^{i\lambda (xk-yp)}$ for a parameter $\lambda$) is known as "twisted convolution" in the harmonic analysis literature, see e.g. Chapter XII.3.3 of Stein's "Harmonic analysis".  (Side note: the citation tool does not seem to cover books, which is very strange.)  As noted in that text, it arises naturally in studying convolution on the Heisenberg group (with group law $(x,p,t) (y,k,s) = (x+y,p+k,t+s+ xk-yp)$), after reducing to an isotypic component of the action of the centre (i.e. to functions of the form $f(x,p,t) = F(x,p) e^{i \lambda t}$ for some fixed $\lambda$).  It also shows up in the composition law for pseudodifferential operators under Weyl quantisation: if $Op(a) Op(b) = Op(c)$, then the Fourier transform of $c$ is the twisted convolution of the Fourier transforms of $a$ and $b$ for a suitable choice of parameter $\lambda$. Equivalently, as noted above by Igor, $c$ is the Moyal product of $a$ and $b$. 
I'm partial to Folland's "Harmonic analysis in phase space" for a treatment of all of these topics (e.g. twisted convolution is introduced on page 25).<|endoftext|>
TITLE: Minimal "subset" of a set of homogeneous polynomials with same solution space
QUESTION [5 upvotes]: Suppose $A:=\{f_1,\dots,f_m\}\subset \mathbb{C}[x_1,\dots,x_n]$ with $m>n$ is a set of homogeneous polynomials of equal degree $d>0$. Suppose further that the variety they define consists of a single point (i.e. the zero vector).

Is it true that there exists a subset $g_1,\dots,g_n$ of the $\mathbb{C}$-linear span of $f_1,\dots,f_m$ with the same property? I.e. they have exactly one common zero?

That this is true for $d=1$ is basic linear algebra. For general $d$, my geometric intution (which, I admit, is probably rather bad given my inexperience in algebraic geometry), tells me that any subset $g_1,\dots,g_n$ "in general position" should have the desired property, which is why I believe this statement to be true. 
Thus, two "bonus questions":

Is it true that any "sufficiently generic" subset $g_1,\dots,g_n$ does it, and that "almost all" subsets of size $n$ are "sufficiently generic" in some suitable sense?

Finally (this holds for $d=1$, I think):

Can one always choose $g_1,\dots,g_n\in A$?

I wouldn't be surprised if for somebody who is familiar with algebraic geometry, this is a basic exercise, and so I apologize if this is deemed unsuitable for MO. I'd still be grateful for any references, proofs or hints.

REPLY [2 votes]: While subsets of $A$ will not work as Tony pointed out, the other questions have a positive answer. There are more general statements, but you situation is particularly vivid in the geometric sense (of course you need an infinite field, but I will assume algbraically closed for simplicity). Your condition implies that the set $A$  gives a morphism $\mathbb{P}^{n-1}\to\mathbb{P}^{m-1}$. If $m>n$, then the image is not all of $\mathbb{P}^{m-1}$ and so choose a point not in the image and project. We get a morphism $\mathbb{P}^{m-2}$. Instead of a point, you could choose a linear subspace of maximal dimension not intersecting the image and then you will end up with a map $\mathbb{P}^{n-1}$ to itself, giving you  what you need.<|endoftext|>
TITLE: Examples of NIP fields of characteristic $p$
QUESTION [7 upvotes]: Definition. According to Shelah, a field $K$ does not have the independence property (i.e. is NIP) if for every first order formula $\varphi(x, \bar y)$ in the language of fields $(+,\times,0,1)$, the Vapnik–Chervonenkis dimension of the family of subsets$\{\varphi(K, \bar k) : \bar k \in K^n\}$ is finite, i.e. there do not exist arbitrarily large finite sets $A \subset K$ whose subsets are all of the form $A \cap \varphi(K, \bar k)$ for some $\bar k$ from $K^n$.
Examples of NIP fields include, in characteristic $p$ finite fields (easy), $\mathbf F^{alg}$, and in zero characteristic, $\mathbf C$, $\mathbf R$ and $\mathbf Q_p$.
A Theorem of I. Kaplan asserts that a NIP field has no Artin-Schreier extension. I found Theorem (I. Kaplan, F. Wagner and T. Scanlon). A valued field of characteristic $p$ with perfect infinite NIP residue field, with $p$-divisible value group and which is algebraically maximal ($i.e.$ with no proper algebraic valued extension having both same residue field and same valued group) is NIP.
I am not familiar with valuation theory, so the above is of little help to me, but for those who know,
Question: would you have any concrete examples of NIP fields of characteristic $p>0$ ? (non separably closed)
I am looking for one which has a cyclic (Galois) extension.

REPLY [4 votes]: Understanding concretely which fields (in the "pure" language of rings, $\mathcal{L} = \{0, 1, +, \cdot\}$) are NIP is a topic of current interest in model theory. The 2015 paper "Dp-minimal valued fields" by Jahnke, Simon, and Walsberg begins:
"Very little is known about NIP fields. It is widely believed that an NIP field is either real closed, separably closed or admits a definable henselian valuation. Note that even the stable case of this conjecture is open."
(http://arxiv.org/pdf/1507.03911v1.pdf)
So conjecturally, you may have to become familiar with a little valuation theory even to understand some interesting concrete examples.
To complement the 2/6/16 answer by Drike, we do have a fairly good idea of which fields are dp-minimal, where dp-minimal theories are a subclass of NIP theories in which one-variable definable sets are "indecomposable" in a certain sense (they generalize strongly minimal stable theories). A remarkable characterization of dp-minimal theories was given in 2015 by Will Johnson, and the answer involves a nontrivial amount of valued field theory:
"On dp-minimal fields," Will Johnson, http://arxiv.org/pdf/1507.02745v1.pdf
From that paper: "[the main theorem] almost says that all dp-minimal fields are elementarily equivalent to ones of the form $K((t^\Gamma))$ [AKA Malc'ev-Neumann fields] where $K$ is $\mathbb{F}^{alg}_p$ or a characteristic $0$ local field, and $\Gamma$ satisfies some divisibility conditions. The one exceptional case is the mixed characteristic case, which includes fields such as the spherical completion of $\mathbb{Z}^{un}_p(p^{1/{p^\infty}})$.''<|endoftext|>
TITLE: Relation between eigenvalues of $A$ and $A^TA$?
QUESTION [7 upvotes]: For an $n\times n$ diagonizable matrix $A$, is there a relation between the eigenvalues of $A$ and the eigenvalues of $A^TA$?
I ask this because I am looking into the relation between $A$ and $A+cI$, specifically their condition numbers:
$$\kappa(A)=\sqrt{\frac{\lambda_{\mathrm{max}}\left(A^TA\right)}{\lambda_{\mathrm{min}}\left(A^TA\right)}}\quad \mbox{ and }\quad \kappa(A+cI)=\sqrt{\frac{\lambda_{\mathrm{max}}\Big((A+cI)^T(A+cI)\Big)}{\lambda_{\mathrm{min}}\Big((A+cI)^T(A+cI)\Big)}}$$
I know for SPD matrices this reduces to
$$\kappa(A)=\frac{\lambda_{\mathrm{max}}(A)}{\lambda_{\mathrm{min}}(A)}\quad \mbox{ and }\quad \kappa(A+cI)=\frac{\lambda_{\mathrm{max}}(A)+c}{\lambda_{\mathrm{min}}(A)+c},$$
such that I can derive stuff like 
$$\kappa(A)<\kappa(A+cI)\quad \mathrm{ or } \quad \kappa(A)>\kappa(A+cI) $$
depending on $c$.
However, for general $A$ when I try rewriting $\kappa(A)$ I don't know what to do with the $\lambda(A^TA)$ part. 
Can anyone help me? Or provide another trick to relate $\kappa(A)$ and $\kappa(A+cI)$ for general $A$, like I did for SPD matrices.

REPLY [8 votes]: Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$ and $s_1,\ldots,s_n$ be those of $\sqrt{A^*A}$, ordered by
$$|\lambda_1|\ge\cdots\ge|\lambda_n|,\qquad s_1\ge\cdots\ge s_n.$$
Then it holds
$$\prod_{j=1}^k|\lambda_j|\le\prod_{j=1}^ks_j,\qquad \forall\,k=1,\ldots,n.$$
For instance, if $k=1$, this is $\rho(A)\le\|A\|$ where we use the operator norm. For $k\ge1$, this can be viewed as the same inequality applied to he $k$-th exterior power of $A$.

REPLY [6 votes]: A simple relation between singular values and eigenvalues does not exist in general, as far as I know.
This is an old question, e.g. 
A. Horn, On the eigenvalues of a matrix with prescribed singular values Proc. Am. Math. Soc 5 4-7 (1954)
H. Weyl H, Inequalities between the two kinds of eigenvalues of a linear transformation Proc. Natl. Acad. Sci. USA 35 408-11 (1949)
For a more recent paper, that treats this problem from a statistical point of view, you can try this
On the mean density of complex eigenvalues for an ensemble of random matrices with prescribed singular values, Yi Wei and Yan V Fyodorov, J. Phys. A: Math. Theor. 41 502001 (2008)<|endoftext|>
TITLE: What's in the genus of the cubic lattice?
QUESTION [8 upvotes]: I'll write $\mathbf{Z}^n$ for the integral quadratic form $x_1^2 + \cdots + x_n^2$.  For which values of $n$ is $\mathbf{Z}^n$ unique in its genus, i.e. isolated in Kneser's graph?  In particular can someone verify or shoot down the following guesses:

For $n \leq 8$, any lattice in the same genus as $\mathbf{Z}^n$ is isomorphic to $\mathbf{Z}^n$.
Any lattice in the same genus as $\mathbf{Z}^9$ is isomorphic to either $\mathbf{Z}^9$ or to $\mathbf{Z} \times \mathrm{E}_8$.

REPLY [3 votes]: Both statements are well-known in the arithmetic theory of quadratic forms (aka integral lattices).  
The first statement for $n \leq 5$ is a consequence of Hermite's bound on the nonzero minimum of an integral lattice: the minimum of a unimodular lattice of rank $\leq 5$ is 1.  
Both statements are  consequences of Kneser's theory of neighbors.  You can find it in the last theorem in O'Meara's book.  It also mentioned that Kneser completed the analysis of the genus of $I_n$ for $n \leq 13$.<|endoftext|>
TITLE: Almgren's mimeographed lectures notes on varifolds
QUESTION [7 upvotes]: I am trying to get some insights for the combinatorial argument of Pitts (in his PhD thesis 'Existence and regularity of minimal surfaces in Riemannian manifolds', Princeton University Press, 1981) to prove the existence of suitable class of varifolds in the so-called Almgren-Pitts theory. Pitts quotes the reference 'The theory of varifolds' (Almgren, mimeographed notes, Princeton, 1965) as the origin of his argument. 
Unfortunately, I have no idea where to find this text, so any help would be very appreciated.

REPLY [15 votes]: I just bumped into your post after doing some google search to find a bibtex entry for Almgren's notes. 
I have a copy of them: I could copy it and send it to you (or maybe scan it and share it via dropbox). However before getting to the combinatorial arguments you would have to go through quite a lot of material. I am writing a paper with a PhD student where we modify Pitts' theory for the case of surfaces with boundaries and we have a separate section with the relevant combinatorial argument, which is probably easier to understand. If you are interested please contact me by email. My name is Camillo De Lellis and I work at the math department of the University of Zuerich: you find my email address in my web page there<|endoftext|>
TITLE: Is there a finite test for isomorphisms of rigid monoidal abelian categories?
QUESTION [12 upvotes]: Let $G$ be a semisimple algebraic group. (I'm already interested in the case $G=SL_2$.)
Let $\mathcal C$ be a semisimple rigid monoidal abelian category endowed with pair of exact tensor functors $Rep_G \to \mathcal C \to Vect$ that whose composition is the forgetful functor $Rep_g \to Vect$. Assume every object of $\mathcal C$ is a summand of an object in the image of this functor $Rep_G \to \mathcal C$.
To check that this functor $Rep_G \to \mathcal C$ is an equivalence, it is sufficient to check that it is full. (Faithfulness follows from the composition being the forgetful functor, and essential surjectivity follows from fullness and the summand condition.)
Is it sufficient to check the fullness condition on finitely many objects of $Rep_G$?
I know the answer is "yes" for $G=SL_n, Sp_{2n},$ or $SO_n$ if $\mathcal C$ is symmetric, at least under the mild assumption that both functors are symmetric tensor functors. Then $\mathcal C$ corresponds by the Tannakian correspondence to a subgroup $H$ of $G$. Then as long as the space of $H$-homomorphisms from the fourth tensor power of the standard representation of $G$ to itself has the correct dimension then $H$ is all of $G$. This is "Larsen's alternative", and you can see why this might be true by considering that as long as the group of $H$-homomorphisms from the adjoint representation of $G$ to itself is one-dimensional, the adjoint representation of $G$ is $H$-irreducible, so the Lie algebra of $H$ is either all the Lie algebra of $G$ or none of it. The second case can be eliminated by considering a slightly more complicated representation of $G$.
Is it still true without using the Tannakian correspondence?

REPLY [4 votes]: In the situation you describe here, the category $\mathcal{C}$ will automatically be symmetric, and the functors will automatically be symmetric functors. The reason is the following: If you have two objects $X$ and $Y$ of $\mathcal{C}$ which are direct summands of $F(A)$ and $F(B)$ respectively (where $F:Rep-G\rightarrow \mathcal{C}$), then we have the following composition of morphisms in $\mathcal{C}$: $X\otimes Y\rightarrow F(A)\otimes F(B)\rightarrow F(A\otimes B)\rightarrow F(B\otimes A)\rightarrow F(B)\otimes F(A)$. By composing with the map $F(B)\otimes F(A)\rightarrow F(B)/Y\otimes F(A)\oplus F(B)\otimes F(A)/X$ and apply the second functor $F':\mathcal{C}\rightarrow Vect$ which is faithful, we see that the first map must split via $X\otimes Y\mapsto Y\otimes X$. One can then show that we get in this way a well defined symmetric structure on $\mathcal{C}$, for which both functors $F$ and $F'$ are symmetric. So your argument about assuming that $\mathcal{C}$ is symmetric and using Tannaka reconstruction is always valid.
About the general question: I only know to say that semisimplicity of $G$
is necessary here: if for example $G=\mathbb{G}_m$ then the collection of subgroups $H_n = n$th roots of unity in $G$ (for different $n$'s) gives a counterexample.<|endoftext|>
TITLE: Hard maths on viXra?
QUESTION [8 upvotes]: A few years ago a nice paper surveyed the differences in quality between papers submitted to arXiv and those submitted to arXiv's rough cousin, viXra. However, that paper was about generic contributions to natural sciences, whereas this post on Quora is somewhat focussing on physics. Whence my question: what's the shape of maths on viXra? Should one ever bother to take a glance there, too? (I've just checked the contributions in analysis and I must admit I was a bit scared.)
Are you aware of any mathematical article later quoted by, say, more than ten different peers and originally published on viXra?

REPLY [14 votes]: From a quick search I found Adjugates of Diophantine Quadruples by Philip Gibbs, which was originally posted on the viXra and has since appeared in INTEGERS.  I do not think it meets the quota of being cited more than ten times, but according to Google Scholar, it has been cited by these two articles. 
Aguirre, J., Dujella, A. and Peral, J.C., 2012. On the rank of elliptic curves coming from rational Diophantine triples. The Rocky Mountain Journal of Mathematics, 42(6), pp.1759-1776.
Dujella, A. and Jurasić, A., 2011. Some Diophantine triples and quadruples for quadratic polynomials. Journal of Combinatorics and Number Theory, 3(2), pp.123-141.<|endoftext|>
TITLE: The Weyl group of E8 versus $O_8^+(2)$
QUESTION [15 upvotes]: Right now Wikipedia says:

The Weyl group of $\mathrm{E}_8$ is of order 696729600, and can be described as $\mathrm{O}^+_8(2)$.

The second part feels wrong to me.  $\mathrm{O}^+_8(2)$ is the group of linear transformations of an 8-dimensonal vector space over $\mathbb{F}_2$ preserving a quadratic form of plus type, meaning a nondegenerate quadratic form that vanishes on some 4-dimensional subspaces.   The Weyl group of $\mathrm{E}_8$ has center $\mathbb{Z}/2$, consisting of the transformations $\pm 1$.   I believe $\mathrm{O}^+_8(2)$ has trivial center. I suspect that $\mathrm{O}^+_8(2)$ is the quotient of the Weyl group by its center.
Is this correct?

REPLY [4 votes]: Let me describe this in my language. $E_8$ lattice has 120 axes (240 vectors). Weyl group of $E_8$ Lie group (let's call it $W(E_8)$) is generated by reflections in these vectors. This is conjugacy class of size 120 in $W(E_8)$. It contains elements of determinant $-1$. If we take just "rotations" in Weyl group i.e. elements of determinant $1$ then we obtain group of size 348 364 800 which is $2.O_8^+(2)$ according to Atlas notation.
The center of $W(E_8)$ contains two elements $\pm I$. The simple group $O_8^+(2)$ of size 174 182 400 is obtained by dividing "rotation" part of $W(E_8)$ by two elements center.
The proof that obtained factor group is really $O_8^+(2)$ may go via octonions over field $\mathbb F_2$. Taking $E_8$ lattice modulo 2 (which is not easy for my intuition) we obtain 120 vectors of length 1 and 135 vectors of length $\sqrt 2$ (sums of perpendicular pairs of length 1). Multiplication of integral octonions is then mapped to multiplication in octonions over $\mathbb F_2$ (let's call it $\mathbb O_{F_{2}}$). There are 120 invertible elements and 135 zero divisors in $\mathbb O_{F_{2}}$. 
Each $W(E_8)$ element is mapped to norm preserving automorphism i.e. it permutes 120 points of norm 1 and 135 points of norm 0. In order to obtain representation of this group as matrices 8x8 over $F_2$ we need to fix some basis in 255 points of $\mathbb O_{F_{2}}$. Actually it will be $O_8^+(2).2$, because $-I$ does not change points of $\mathbb O_{F_{2}}$.
The size of the group can be calculated as 240*126*60*2 * 24*2*2*2 = 696 729 600, because for standard basis $e_0,...,e_7$ first vector $e_0$ can be mapped to any of the 240 vectors.
The second vector $e_1$ can be mapped any of the 126 perpendicular. Third one $e_2$ can be mapped any of the 60 vectors perpendicular to given perpendicular pair of vectors.
The axis of fourth vector is already defined - for given perpendicular triple there is just one fourth vector in $E_8$ lattice such that resulting 4-space contain $D_4$ sublattice of $E_8$.
We may also see that if first vector is octonion $1$ then fourth vector is $\pm$ product of second and third one.
Remaining four vectors are determined by the fact that they belong to perpendicular $D_4$ lattice to the one generated by first four vectors - there are just 24 vectors there to use.
I see now that wikipedia article about $E_8$ is corrected.<|endoftext|>
TITLE: Does the sum $\sum_{n=1}^{\infty}\frac{1}{p_n(p_{n+1}-p_n)}$ converge?
QUESTION [15 upvotes]: Prove, if possible in an elementary way, that $\sum_{n=1}^{\infty}\frac{1}{p_n(p_{n+1}-p_n)}$ converges/diverges, where $p_n$ denotes the $n^{\textrm{th}}$ prime.

REPLY [23 votes]: By popular demand, I am converting my comment above to an answer. Yes, the series converges. In the paper
Erdös, Paul(H-AOS); Nathanson, Melvyn B.(1-CUNY7) On the sum of the reciprocals of the differences between consecutive primes. Number theory (New York, 1991–1995), 97–101, Springer, New York, 1996
the authors show that $$\sum_{n=3}^{\infty} \frac{1}{n(\log\log{n})^c (p_{n+1}-p_n)}$$ converges for every choice of $c>2$. (I start the sum at $n=3$ rather than $n=2$ as as is done in the paper to avoid the annoyance that $\log\log{2} < 0$.) They also present a heuristic argument that the series diverges when $c=2$. The main tool in the proof is Brun's classical upper bound sieve estimate for the number of prime pairs $p, p+N$ with $p$ below a given bound.
Since $p_n$ is bounded below by a positive constant multiple of $n\log{n}$, the convergence of your series follows by comparison.<|endoftext|>
TITLE: How to simplify the proof of right-properness?
QUESTION [5 upvotes]: Question. Is it true that to check that a model category is right proper, it suffices to check the property for weak equivalences with fibrant codomain ? (if the domain is also fibrant, the pullback is always a weak equivalence). Or is there a close statement that I can't remember (browsing nLab did not help me) ?
Comments. Consider the diagram $\mathbf{D}=X\rightarrow Y \leftarrow Z$ where the left-hand map is a fibration and the right-hand map a weak equivalence. Let $T=\projlim \mathbf{D}$. Choose a fibrant replacement $Y^{fib}$ for $Y$ and a trivial cofibration $Y\rightarrow Y^{fib}$. Factor the composite map $X \rightarrow Y \rightarrow Y^{fib}$ as a composite trivial cofibration-fibration. We obtain a diagram $\mathbf{E}=X^{fib}\rightarrow Y^{fib}\leftarrow Z$ such that the left-hand map is a fibration between fibrant objects and the right-hand map is a weak equivalence. Let $U=\projlim \mathbf{E}$. By hypothesis, the map $U\rightarrow X^{fib}$ is a weak equivalence. By construction, the map of diagrams $\mathbf{D} \rightarrow \mathbf{E}$ is a weak equivalence of diagrams. The map $T\rightarrow X$ is a weak equivalence iff the map $T\rightarrow U$ is a weak equivalence. What next ?
Why. I found this cryptic remark in my notebook, and I can't remember where it comes from. The reason why I want to simplify the proof of right properness is that I have to deal with model categories where a set of generating trivial cofibrations is not known. I only know what I call a set of generating anodyne cofibrations. And the trivial fibrations which are the anodyne fibrations (i.e. having the RLP with respect to the set of generating anodyne cofibrations) which are a dual strong deformation retract. And the reason why I am interested in right properness is that I want to study right Bousfield localizations.

REPLY [5 votes]: To complete the argument you need to apply K. Brown's Lemma. Call your model category $\mathcal{M}$, then the map $Z \to Y$ induces a pullback functor $\mathcal{M} \downarrow Y \to \mathcal{M} \downarrow Z$ and the lemma implies that it preserves weak equivalences between fibrations over $Y$. If you define $V \to Y$ as the pullback of $X^{\mathrm{fib}} \to Y^{\mathrm{fib}}$, then $X \to V$ is a weak equivalence by the assumption and 2-out-of-3. Then apply the previous observation to $X \to V$ to see that $T \to U$ is a also weak equivalence and hence so is $T \to X$.
A reference for this is Lemma 9.4 in Bousfield's On the Telescopic Homotopy Theory of Spaces, but I imagine it was known well before that.<|endoftext|>
TITLE: Applications of Jordan algebras
QUESTION [15 upvotes]: Jordan algebras are non-associative algebras satisfying a somewhat strange (to me) list of axioms, see wikipedia. Basic examples are real symmetric and complex hermitian matrices with the product $A\circ B=\frac{1}{2}(AB+BA)$. According to the above article in wikipedia P. Jordan introduced this notion in 1933 to  formalize the notion of an algebra of observables in quantum mechanics.
Question 1. Did this operation become really useful in quantum mechanics in any non-trivial way? 
Question 2. I would be happy to see some explanations why the notion of Jordan algebra is useful/ natural. Are there applications of it to other parts of mathematics?
Here are very few interesting facts I was able to find so far. 
1) There is a really beautiful classification of the finite dimensional formally real Jordan algebras due to  Jordan, von Neumann & Wigner (1934). 
2) The above wikipedia paper mentions a classification of some special class of infinite dimensional Jordan algebras (Zelmanov, 1979). 
3) I have recently heard about the Koecher-Vinberg classification of symmetric cones: such cones are precisely cones of squares in Euclidean Jordan algebras.
Thus the only way I heard Jordan algebras are related to other parts of mathematics is via the classification lists of various subclasses of them.

REPLY [5 votes]: Jordan algebras were originally introduced by Pascual Jordan in a hope to generalize the orthodox formulation of quantum mechanics, but this program was not successful as far as the generalization of quantum mechanics is concerned. In this respect, Jordan algebras are neither essential nor terribly useful in understanding quantum mechanics. 
On the other hand the importance of Jordan algebras in mathematical physics is undisputed and related to the fact that they are tightly interconnected with another nonassociative structures: Lie algebras. As remarked by Kevin
McCrimmon, "if you open up a Lie algebra and look inside, 9 times out of 10 there is a Jordan algebra (of pair) which makes it work”.
In addition to McCrimmon's book "A taste of Jordan algebras", mentioned in comments, the following works might be useful in understanding the role of Jordan algebras in mathematical physics and mathematics:
http://projecteuclid.org/euclid.bams/1183540925 (Jordan algebras and their applications, by Kevin Mccrimmon).
http://arxiv.org/abs/1106.4415 (Jordan structures in mathematics and physics, by Radu Iordanescu).
http://arxiv.org/abs/0809.4685 (Black Holes, Qubits and Octonions, by L. Borsten, D. Dahanayake, M.J. Duff, H. Ebrahim and W. Rubens).
http://www.worldscientific.com/worldscibooks/10.1142/3282 (On the Role of Division, Jordan and Related Algebras in Particle Physics, by F. Gursey and C.-H. Tze). 
The biography of Pascual Jordan can be found here: http://arxiv.org/abs/hep-th/0303241 (Pascual Jordan, his contributions to quantum mechanics and his legacy in contemporary local quantum physics, by Bert Schroer). Jordan was a very important figure in developing quantum mechanics and quantum field theory but his reputation was greatly damaged by his relations to the Nazi regime, although, according to Schroer, "he never received benefits for his pro-NS convictions and the sympathy remained one-sided. Unlike the
mathematician Teichmueller, whose rabid anti-semitism led to the emptying of the Gottingen mathematics department, Jordan inflicted the damage mainly on himself".<|endoftext|>
TITLE: Elementary prime-generating sequences
QUESTION [13 upvotes]: A student of mine keeps coming again and again and telling "I've found a formula $n\mapsto f(n)$ giving all primes" or sometimes "infinitely many primes", where $f$ is a classical function (I mean made of exponentials and polynomials, just like Mersenne primes). Then follows a long discussion, me saying "nice try, but it doesn't work because ...", and him argumenting ... and trying to restrict $f$ to the primes ...
Today I told him something like: "you know, many great mathematicians tried to find such formulas, but never succeeded" and he answered: "maybe they didn't see something elementary, and I can find it".
Now the questions: let $\mathcal S$ be the algebra of sequences of integers (under $+$, $\times$ and $\circ$). Let $\mathcal S_0$ be the smallest subalgebra containing all polynomial sequences $n\mapsto P(n)$ for $P\in \mathbf Z[X]$, and all exponentials $n\mapsto a^n$ for $a\in\mathbf Z$. Let $\mathcal S_1$ be the subset of $\mathcal S_0$ made of sequences $n\mapsto f(n)$ satisfying $f(n)>n$ for any integer $n$.
Q1) Is there any result saying that no element of $\mathcal S_1$ can have an infinite subset of the set of primes as its image?
Or even better :
Q2) Is there any result saying that the image by an element $f$ of $\mathcal S_1$ of the set of prime numbers must contain infinitely many composite numbers?
And finally :
Q3) Is there any result saying that if $f$ is an element of $\mathcal S_0$ satisfying $f(n)>n$ for all $n$, and $u_n$ is the sequence obtained by choosing an initial value $u_0$ and by setting $u_{n+1}=f(u_n)$, then $(u_n)_n$ must contain at least one composite numbers?
(Ok, this is not exactly a research question, but answers would help research since it would enable me to reject his proposals directly, without the need of finding counter-examples). From this point of view, negative answers wouldn't be welcome !)

REPLY [7 votes]: Do a web search for Diophantine Representation of prime numbers.  You will find an article of that title containing a polynomial of degree 25 in 26 variables whose range intersected with the positive integers is exactly the prime numbers.
There are general results for polynomials of one variable which extend to several variables which show that the range (both positive and negative) cannot be exclusively primes.  Also the Prime Pages online, and books by Ribenboim, contain essays on the subject addressing your question, references on such are easily found by web searches.
In short, he is right, there could be something elementary that others might have overlooked.  If he can derive the degree 25 polynomial above on his own, I say have him keep looking, AFTER he reads and understands the essays.
Gerhard "Don't Spoil His Prime Fun" Paseman, 2016.02.04.<|endoftext|>
TITLE: Intuition/idea behind a proof of the splitting principle?
QUESTION [13 upvotes]: The splitting principle is as follows.

Given a vector bundle $E \to X$ with $X$ compact Hausdorff, there is a compact Hausdorff space $F(E)$ and a map $p: F(E) \to X$ such that the induced map $p^*: K^*(X)  \to K^*(F(E))$ is injective and $p^*(E)$ splits as the sum of line bundles.

My question is, what is the idea/intuition behind the proof of the splitting principle?

REPLY [17 votes]: Perhaps my very short (4 pages plus bibliography) paper ``A note on the splitting principle'' http://www.math.uchicago.edu/~may/PAPERS/Split.pdf
may be illuminating. It shows that the splitting principle can be viewed as a statement about the reduction of the structural group of a $G$-bundle $\xi$ from $G$ to a maximal torus $T$, where $G$ is a compact Lie group. It applies more generally than in just the usual examples. One starts with the bundle $BT\to BG$ with fiber $G/T$. For a $G$-bundle over $X$ classified by $f\colon X\to BG$, one has a pullback bundle $q\colon Y\to X$ with fiber $G/T$ together with a reduction of the structure group of $q^*\xi$ to $T$. When $H^*(BG;R)$ is concentrated in even degrees, $q^*\colon H^*(X;R)\to H^*(Y;R)$ is a monomorphism. That is easily seen to imply the splitting theorem as usually stated, and many variants thereof.
As stated and explained briefly in the paper, the argument adapts to $K$-theory.<|endoftext|>
TITLE: Invertibility of group Laplacian in $\ell^1$
QUESTION [7 upvotes]: Let $G$ be a discrete group and let $S$ be a generating set for $G$; assume that $S$ is symmetric (i.e., $g\in S$ iff $g^{-1}\in S$).  Let $L=L_S=\frac{1}{|S|}(\sum_{g\in S} g-1)$ be an element of the group algebra.  For $F\in \ell^p(G)$ and $g,h\in G$ let $\lambda_p(g) F$ be defined by $\lambda_p(g) F (h) = F(hg^{-1})$; this is the regular representation of $G$ on $\ell^p(G)$.  Finally let $L_p = \pi_p(L)$.
It is a classical result of Kesten that $G$ is non-amenable iff $L_2$ has a bounded inverse (i.e., $0$ is not in the spectrum of $L_2$).
What is known in the case of $p=1$? Clearly if $G$ is amenable, then there are functions $f_n\in \ell^1(G)$ with the property that $\Vert L_1 f_n \Vert_1 / \Vert f_n \Vert_1 \to 0$ showing that $0$ is in the spectrum of $L_1$ and so $L_1$ is not invertible.  Also, if $G$ is not amenable, then such $f_n$ cannot exist under the additional assumption that $f_n\geq 0$. 
Given $G$ non-amenable, can one find a generating set $S$ with the property that $L_1$ is invertible?

REPLY [2 votes]: The Laplacian is never invertible on $\ell^1(G)$.
There is already a correct answer by ARG. 
I am writing this answer as there is a demand for more information. 
I will thus try to provide a more conceptual point of view. But I should remark that my answer is essentially the same as ARG's answer.

Let $\mu$ be a probability measure on $G$ and assume the Laplacian $\Delta=1-\mu$ is an automorphism of $\ell^1(G)$.
By taking the double dual, $\Delta$ extends to an automorphism of $\ell^\infty(G)^*$.
Recall that a mean on $G$ is a norm 1 positive element of $\ell^\infty(G)^*$ and let $M$ be the space of all means on $G$. It is easy to see that $M$ is convex and compact for the w*-topology. Note that $G$ acts on $M$ preserving its structure. In particular, $M$ is $\mu$-invariant. By Markov–Kakutani fixed-point theorem there is a $\mu$-invariant mean $m\in M$. Thus $\ker \Delta\neq 0$ as 
$$ \Delta m=m-\mu m =0. $$
This is a contradiction.

Note that, by the same argument, $\Delta$ is never an automorphism of $\ell^\infty(G)$.
For completeness let me add that for $1<p<\infty$ and $\mu$ with a generating support, $\Delta$ is an automorphism of $\ell^p(G)$ iff $G$ is not amenable.
For this recall that the Mazur map $f \mapsto \text{sgn}(f)|f|^{2/p}$ is a $G$-equivariant uniformly continuous homeomorphism from the unit sphere of $\ell^2(G)$ to the unit sphere of $\ell^p(G)$. It follows that there exist almost invariant vectors in $\ell^p(G)$ iff there exist almost invariant vectors in $\ell^2(G)$, and this happens iff $G$ is amenable (by Kesten). Thus, if $G$ is amenable $\Delta$ is not an automorphism (and this works also for $p=1$). If $G$ is not amenable, by the uniform convexity of the unit ball of $\ell^p(G)$ (which fails for $p=1$), we get that $\|\mu\|<1$, thus the sum $\sum_{n=0}^\infty \mu^n$ exists in the space of bounded operators, and since $(1-\mu)\sum_{n=0}^\infty \mu^n=1$, $\Delta$ is indeed an automorphism.<|endoftext|>
TITLE: Symplectic orthogonality and projective duality: how do they work together?
QUESTION [6 upvotes]: Let $(V,\omega)$ be a $2n$-dimensional linear symplectic space, and $(\mathbb{P}V,\theta_\omega)$ the corresponding $(2n-1)$-dimensional contact manifold.
Given a smooth $(n-1)$-dimensional smooth projective variety $X\subset \mathbb{P}V$, I can define its orthogonal
$$
X^\perp:=\{ \pi\in \mathbb{P}V^*\mid\pi=v^\perp\textrm{ for some }[v]\in X  \}\, ,
$$
as well as its dual
$$
X^\ast:=\{ \pi\in \mathbb{P}V^*\mid\pi\supset T_v\widehat{X}\textrm{ for some }[v]\in X  \}\, ,
$$
where $\widehat{X}$ is the cone over $X$.
Here's the point: I'm trying to define a "companion" $X_\bullet$ of $X$ via the equation
$$
X_\bullet^\ast=X^\perp\, .\quad\quad (1)
$$
Example. If $D<V$ is a linear $n$-dimensional subspace, and $X=\mathbb{P}D$, then $(1)$ unambiguously defines $X_\bullet$, which turns out to be $\mathbb{P}D^\perp$. In particular, if $D$ is $\omega$-Lagrangian, then $X_\bullet=X$ is also $\theta_\omega$-Legendrian.

QUESTION: how to characterise the $X$'s possessing a well-defined (by means of $(1)$) "companion" $X_\bullet$? In particular, does this property hold for arbitrary (i.e., not necessarily linear) $\theta_\omega$-Legendrian $X$'s? Is it true that in this case $X=X_\bullet$?

I'm sure that this problem has been already addressed somewhere: any reference/keyword will be most welcome.
EDITED after Libli's and Robert Bryant's comments
Of course, by using the reflexivity theorem (see, e.g., this paper by Tevlev, Theorem 1.7a), one can solve $(1)$ and obtain
$$
X_\bullet=(X^\perp)^*\, .
$$
So, the answer to my question is "for all $X$". Nevertheless (recall that I'm assuming $X$ to be smooth), I can still ask, how to characterise those $X$ whose $X_\bullet$ enjoys some extra properties, like being smooth and (which this is what really I'm interested in) having the same dimension as $X$ - or even being $X$ itself (the only purpose of the symplectic orthogonality is to have $X$ and its dual sitting in the same space, and compare them).

REPLY [3 votes]: Varieties having the same dimension as their duals have been classified by Ein (see http://link.springer.com/article/10.1007%2FBF01391495). They are:
_hypersurfaces,
_$\mathbb{P}^1 \times \mathbb{P}^{n-1} \subset \mathbb{P}^{2n-1}$;
_$G(2,5) \subset \mathbb{P}^9$,
_$\mathbb{S}_{10} \subset \mathbb{P}^{15}$. 
EDIT : this results holds if one assumes the variety non-degenerate (that is non-included in a hyperplane).<|endoftext|>
TITLE: Algebraic Geometry needed for Kähler-Einstein metric
QUESTION [9 upvotes]: I am a Master's student interested in Differential Geometry / Geometric Analysis. Currently active research is going on in Kähler-Einstein / Extremal Kähler metric. I was wondering how much Algebraic geometry (by Algebraic Geometry I DON'T mean Complex Analytic Geometry) I need to know if I want to study Kähler-Einstein / Extremal Kähler metrics in my PhD. Is the role of Algebraic Geometry in Kähler-Einstein metric something like black-box or do I need to have a thorough understanding of Algebraic Geometry? Frankly speaking, I dislike Algebraic Geometry and I don't understand it.

REPLY [11 votes]: Some references 
Finding Kahler-Einstein metric on a Kahler Variety is related to MMP
  (Minimal model program) in algebraic geometry

Ricci flow and birational surgery
Jian Song http://arxiv.org/abs/1304.2607
Riemannian geometry of Kahler-Einstein currents II: an analytic proof of Kawamata's base point free theorem
Jian Song http://arxiv.org/abs/1409.8374
Canonical measures and Kahler-Ricci flow, with G. Tian, J. Amer. Math. Soc. 25 (2012), no. 2, 303-353, arXiv:0802.2570
The Kahler-Ricci flow through singularities
Jian Song, Gang Tian http://lanl.arxiv.org/abs/0909.4898
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This question has long history, In fact the idea for connecting Algebraic geometry to PDE comes from your vision to meaning of "surgery"

In fact philosophically
$$Algebraic \; surgery<===>PDE \; surgery<===> Geometric \; Surgery$$

Here algebraic surgery I mean flips and flops(If it exists). In fact PDE surgery here means surgery by flow, like Kahler Ricci flow, because this type of flows smooth out singularities and resolve the singularities
Perelman used PDE surgery<===> Geometric surgery for solving Poincare conjecture.
But for finding Canonical Kahler metric huristicly you need 
$$Algebraic \; surgery<===>PDE \; surgery$$
Now I come back to your question

One of fundamental questions in Kahler geometry is about finding
  canonical metrics(Kahler-Einstein metrics, twisted Kahler Einstein
  metrics, constant scalar curvature,...)

You need to know birational geometry in algebraic geometry for finding canonical metrics.

One of connections of Algebraic geometry to canonical metric theory is
  the Minimal Model Program and a nice program introduced by Jian Song
  and Gang Tian, refes as Song-Tian program.

Now let you are facing with a variety which first Chern class is not definite, we prefer to find twisted Kahler-Einstein metric which is canonical metric(in fact finding generic Kahler Einstein make no sense here).
If $X$ be a projective variety(which first Chern class is not definite ) and by definition the canonical metric $g_{can}$ is the  metric which is attached to  canonical model $X_{can}=\text{Proj}\bigoplus_{m\geq 0}H^0(X,K_X^{m})
$.
Let the canonical Ring $$\bigoplus_{m\geq 0}H^0(X,K_X^{m})$$ is finitely generated and let the Kodaira dimension is positive then 
$$\pi:X\to X_{can}=\text{Proj}(\bigoplus_{m\geq 0}H^0(X,K_X^{m}))$$
gives a canonical metric which is twisted by Weil-Petersson metric on $X$ via
$$Ric(g_{can})=-g_{can}+g_{WP}$$
where $g_{WP}$ is the Weil-Petersson metric and corresponds to moduli space of Calabi-Yau fibers.
So
If the Kodaira dimension 
$$\kappa(X)=\limsup_{m\to \infty}\frac{\log dim H^0(X,K_X^m)}{\log m}$$
is negative then you can find canonical metric by applying Mori fiber space (by assuming base is of general type)and some condition on Chow-Mumford line bundle and extending the result of Ross-Fine we have
$$Ric(g_{can})=-g_{can}+g_{WP}$$
where $g_{WP}$ is the Weil-Petersson metric on moduli-space of Fano fibers of positive first Chern class which can be introduced by using Deligne pairing.
Note that if base is Fano K-stable then along Mori fibre space we have 
$$Ric(g_{can})=g_{can}+g_{WP}$$
If base is Calabi-Yau variety and along Mori fibre space we have
$$Ric(g_{can})=g_{WP}$$
where $g_{WP}$ is the Weil-Petersson metric on moduli-space of K-stable Fano fibers 
If the first Chern class be positive then we need the notion of K-stability of Tian-Donaldson for finding canonical metric(Kahler-Einstein metric)
If the first chern class be zero or negative then we can apply the result of Yau and Aubin for finding canonical metric(Kahler-Einstein metric in this case)
See papers of Jian Song about your question  https://www.math.rutgers.edu/~jiansong/<|endoftext|>
TITLE: Algebraic spaces as locally ringed spaces
QUESTION [18 upvotes]: Let $S$ be a scheme (although I am more than happy to have $S=\text{Spec}(k)$ for a field $k$) and $\mathsf{AlgSp}/S$ the category of algebraic spaces over $S$. 
Does there exist an embedding $\varphi:\mathsf{AlgSp}/S\hookrightarrow \mathsf{LRS}$ (where $\mathsf{LRS}$ is the category of locally ringed spaces) such that we have a natural equivalence
$$X\cong \text{Hom}_\text{LRS}(-,\varphi(X))$$
in the category of presheaves on $\mathsf{Sch}/S$?
It would also be desirable to have that if $X$ is finite type over $\mathbb{C}$, that there is a map $X^\text{an}\to \varphi(X)$ which is initial amongst maps from $X^\text{an}$ to elements of $\varphi(X)$.

REPLY [17 votes]: Probably you want your functor $\varphi$ to restrict to the "identity functor" on the full subcategory of algebraic spaces that are schemes (which doesn't seem to be a purely formal consequence of your hypotheses). And to be "reasonable" you likely want such a hypothetical functor $\varphi$ to carry open immersions to open immersions and etale maps to flat maps.  
Under these very mild assumptions (without which $\varphi$ would be rather weird and hard to use) the answer is no even in the quasi-separated case, where one has a good theory of "associated topological space". 
To discover examples showing this cannot be done (as you must surely be expecting anyway), let's first address what must be the actual motivation: to give a characterization of analytification of algebraic spaces locally of finite type over $\mathbf{C}$ (and non-archimedean fields)  in terms of an "initial object" property akin to that for such schemes mapping to locally ringed spaces. (At the end of the question, I think you meant to write "maps from analytic spaces to $\varphi(X)$".) That would be a "replacement" for the usual definition of $X^{\rm{an}}$, perhaps felt to be a bit ad hoc, as $U^{\rm{an}}/R^{\rm{an}}$ for an etale chart $R \rightrightarrows U$ in schemes for $X$. 
As you likely know, the existence or not of the quotient $U^{\rm{an}}/R^{\rm{an}}$ is independent of the etale chart $R \rightrightarrows U$ for $X$ and assuming existence that quotient is (uniquely up to unique isomorphism) independent of the chart and uniquely functorial in such an $X$ (in a sense made precise in section 2.2 of http://arxiv.org/abs/0706.3441, for example). 
To give an "initial object" characterization of $X^{\rm{an}}$ it is more natural (as pbelmans has noted) to use locally ringed topoi (or a broader notion that replaces local rings with henselian local rings) and not to try to force onself into the setting of locally ringed spaces, but that would then kill the motivation for the question posed. 
Note that for such algebraic spaces $X$ over $\mathbf{C}$ (or non-archimedean fields $k$), $X^{\rm{an}}$ exists only if the quasi-compact diagonal morphism $\Delta_{X/\mathbf{C}}$ (or $\Delta_{X/k}$) is an immersion; this is remarked without proof in Ch. I, 5.17ff. in D. Knutson's book on algebraic spaces. The necessity of this diagonal condition is proved in Theorem 2.3.4 of the above arxiv link, whereas sufficiency is shown to fail badly over non-archimedean fields beyond the separated case in section 3 of loc. cit. The sufficiency of the diagonal condition over $\mathbf{C}$ (proved as Ch. I, Prop. 5.18 in D. Knutson's book on algebraic spaces) is a bit more interesting in view of the necessity (i.e., that one needs some non-trivial diagonal hypothesis for the existence).  
So to speak of $X^{\rm{an}}$ as an analytic space you need to assume the diagonal of $X$ is an immersion (this property is usually called "locally separated", somewhat unfortunate terminology); as you know, this is stronger than quasi-separatedness. Of course, one could get around that by enlarging the scope of what is meant by a (complex) analytic space, to go a bit beyond the framework of locally ringed spaces, but that would again somewhat kill the motivation for the question.
This finally brings us to examples showing that no reasonable $\varphi$ satisfying the additional mild properties mentioned above, and even if we limit attention to quasi-separated algebraic spaces of finite type over a field $k$.  Consider a "locally separated" (in the sense of diagonal being an immersion) smooth geometrically connected algebraic space of dimension 2 built by modifying a smooth surface along a geometrically integral curve by "replacing" the curve with a $2:1$ finite etale cover from another geometrically integral curve.  See Example 3.1.1 of the arxiv link above for a discussion of why such examples made over any non-archimedean field cannot be analytified in the sense of non-archimedean geometry (a la Tate, Berkovich, or even Huber, all by the same obstruction), whereas when built over $\mathbf{C}$ they can be analytified (so it is most fun with such examples over $\mathbf{Q}$, which communicates to both $\mathbf{Q}_p$ and $\mathbf{C}$).
By using an etale chart presentation for such a surface $S$ as in Example 3.1.1 mentioned above, the hypotheses that $\varphi$ carries open immersions to open immersions and etale maps to flat maps enables one to deduce (by some elementary considerations that I prefer not to write out here) that $\varphi(S)$ must have as its underlying topological space that of the algebraic space $S$ (not just as a set!) and its structure sheaf must be the sheaf of rings $A$ whose value on an open set $\Omega \subset |S|$ is $O_S(U)$ where $U \subset S$ is the unique Zariski-open subset such that $\Omega = |U|$.  That is, $\varphi(S)$ has to be the naive ringed space (in fact locally ringed) attached to the algebraic space $S$ and its associated topological space. Concretely, if 
$R \rightrightarrows U$ is an etale scheme chart for $S$ then $\varphi(S)$ is the quotient $U/R$ in the category of locally ringed spaces. 
The assumed main property of $\varphi$ (and the argument I omitted which identifies exactly what $\varphi(S)$ must be) implies that the "obvious" morphism of locally ringed spaces $f:S \rightarrow \varphi(S)$ has the property that for any scheme $T$ a morphism of locally ringed spaces $T \rightarrow \varphi(S)$ uniquely factors through $f$. But consider the generic point $\eta$ of the distinguished geometrically integral curve $C$ in the construction of $S$; that curve is a 2:1 cover of another curve $C_0$ such that the local stalk of $O_{\varphi(S)}$ at $\eta \in |S| = |\varphi(S)|$ has residue field $k(C_0)$ whereas $O_S$ has local henselian stalk at $\eta$ with residue field $k(C)$. So we have a natural map $T := {\rm{Spec}}(k(C_0)) \rightarrow \varphi(S)$ of locally ringed spaces, and this does not factors through $f:S \rightarrow \varphi(S)$.  Indeed, if it were to factor then the resulting map $T \rightarrow S$ would have to hit exactly $\eta \in |S|$, and so would factor through the closed subscheme $C \subset S$, sandwiching $k(C) = O_{C, \eta}$ between $k(C_0)$ and $k(C_0)$, forcing $[k(C):k(C_0)] = 1$, a contradiction.<|endoftext|>
TITLE: What is the probability that two random permutations have the same order?
QUESTION [18 upvotes]: I am interested in the orders of random permutations. Since the law of the logarithm of the order of a permutation converges to a normal law (for instance Erdös-Turan Statistical group theory III), one expects that the probability for two permutations of $\frak S_n$ to have the same order goes to 0 as n goes to infinity. Indeed experimentally this seems to happen with speed $O(1/n^2)$
I know that Wilf proved an asymptotic for a permutation in $\frak S_n$ to be of order $d$ (https://www.math.upenn.edu/~wilf/website/Asymptotics%20of%20exp%28P%28z%29%29.pdf) but I don't think it can be used directly.  
On the other hand it is clear that the probability that two permutations have same order is more than probability that two permutations are conjugate. This is $K/n^2$ according to Flajolet et al. (http://arxiv.org/abs/math/0606370), but here again I failed to generalize the method for the order.

REPLY [6 votes]: Nice problem! I claim that $\limsup n^2 p(n) = \infty$.
Suppose $k < n/2$ is such that $n-k$ is divisible by $L_k = \text{lcm}(1,2,\dots,k)$. Then if $\pi \in S_n$ has a cycle of length $n-k$ (this happens with probability $1/(n-k)$) then $\text{ord}(\pi) = n-k$, so the probability gets a contribution of $1/(n-k)^2 \geq 1/n^2$ from such $\pi$. Let $K_n$ be the set of all such $k$. Then $p(n) \geq |K_n|/n^2$.
Now the great thing about $n \mapsto K_n$ is that it is "lower semicontinuous on $\widehat{\mathbf{Z}}$". What I mean is this: Suppose $K = K_n$, and let $k = \max K$. Then the condition that $K_n \supset K$ is $L_k$-periodic, so provided we alter $n$ only by multiples of $L_k$, the set $K_n$ can only get bigger. Moreover, note that we would have $n \in K_n$, except for our stipulation that $k < n/2$. It follows that
$$
  K_{n + L_n} \supset K_n \cup \{n\}.
$$
This shows that $|K_n|$ gets arbitrarily large, which proves the claim.
I have no idea about the $\liminf$! The above argument constructs very particular $n$ (of the shape $n_1 + \dots + n_k$, where each $n_i$ is large and highly divisible in comparison with $n_{i-1}$), and the lower bound is very weak besides (roughly $\log^*(n)/n^2$), so it's tempting to conjecture that the $\liminf$ is finite. But I'm not sure this is supported by the numerics. From the numerics it just looks like we're forgetting something.<|endoftext|>
TITLE: How many uniquely colored degree two vertices in 3-coloring of subcubic graph?
QUESTION [8 upvotes]: Is there a graph with maximum degree three that has 3 degree two vertices that must get the same (resp. different) color in every 3-coloring of the graph?
I'm interested in any similar results as well.
ps. Note that if we change the maximum degree to four, then it is easy to construct such graphs. Just take a long path and double every other vertex, connecting it to its old neighbors and the two copies with each other. Then you can add a degree two vertex connected to both copies for each doubled vertex.

REPLY [8 votes]: No such graph exists (that is, you cannot have a subcubic graph with three degree-$2$ vertices all forced to the same color). Suppose that such a graph exists; we may assume the graph is connected. The vertices of degree $2$ forced to the same color must be pairwise nonadjacent (otherwise, your graph has no $3$-coloring at all, contradicting Brooks' Theorem). Add a new edge among any two of these vertices to form a new graph $G'$; now $G'$ is subcubic and has no proper $3$-coloring. Since there are still missing edges from the other degree-$2$ vertex to the endpoints of the new edge, $G'$ is not a complete graph, but by Brooks' Theorem, any connected subcubic graph that is not a complete graph has a proper $3$-coloring.
Similarly, you cannot have a connected subcubic graph with three degree-$2$ vertices all forced to different colors in every $3$-coloring, unless that graph is a triangle: add a new vertex adjacent to those three vertices to form a new subcubic graph $G'$ with no proper $3$-coloring. If your original graph is not a triangle, then $G'$ is not a complete graph, so Brooks' Theorem again gives a contradiction.<|endoftext|>
TITLE: Obstructed automorphisms of schemes
QUESTION [12 upvotes]: Let $X$ be a smooth projective scheme over a field $\mathbf{k}$ of characteristic zero such that $\mathrm{H}^0(X, \mathrm{T}X)$ vanishes, and let $f$ be an automorphism of $X$. I would like to have an explicit example of such a pair $(X, f)$ such that:
1) The automorphism $f$ lifts to some non trivial first order deformation $\mathfrak{X}$ of the scheme $X$, that is some nonzero element of $\mathrm{H}^1(X, \mathrm{T}X)$ is fixed by the action of $f$.
2) Given any nontrivial deformation $\widetilde{\mathfrak{X}}$ of $X$ over a small extension $A$ of $\mathbf{k}[t]/t^2$ such that the pullback of $\widetilde{\mathfrak{X}}$ is $\mathfrak{X}$, then $f$ doesn't lift to $\widetilde{\mathfrak{X}}$.
edit: by nontrivial, I mean that $\widetilde{\mathfrak{X}}$ is not obtained by pull back from $\mathfrak{X}$ via a section of the map $A \rightarrow \mathbf{k}[t]/t^2$.

REPLY [9 votes]: Choose a homogeneous  polynomial $f(x_0,x_1, \ldots, x_n) = x_0^d+ f_1(x_1,\ldots, x_n)$  such that $V(f)$ is smooth.  Then multiplication by a $d$th root of unity on $x_0$ gives an automorphism $\sigma$ of $X= V(f)$.
By choosing $d$ large, we can assume that there are no vector fields. Now choose a second polynomial $g= x_0^d + g_1(x_1,\ldots)$ satisfying the same conditions. Let $h(x_0,\ldots)$ be a generic homogeneous degree $d$ polynomial. Then
$$ Proj\, k[t]/(t^3)[x_0,\ldots, x_n]/(f + tg+ t^2h)$$
provides an example where the automorphism $\sigma$ extends to first order but not to second. I realize, after I answered it, that I missed the "any" in your condition 2. Your question is either very hard or trivially false (the automorphism always extends to 
$$\widetilde{\mathfrak{X}}= \mathfrak{X}\times_{Spec k[t]/t^2} Spec A$$ 
whenever $A\to k[t]/(t^2)$ has a section, but I suspect you meant to disallow this sort of family).<|endoftext|>
TITLE: No normal coordinates on general Finsler manifolds
QUESTION [5 upvotes]: I recently read a footnote in Chern's article stating that a non-Riemmanian Finsler manifold does not possess normal coordinates.  
As I'm still new to non-Riemmanian Finsler geometry I don't see why this is the case.  Can someone provide me with a simple example in conjunction to an intuitive explanation why this fails in general?
Many thanks.

REPLY [14 votes]: I think it's important to keep two things separate here:  
First, if $(M,F)$ is a smooth Finsler manifold (which means that $F^2:TM\to [0,\infty)$ is smooth and strongly convex away from the zero section of $TM$), then the unit sphere bundle $\Sigma\subset TM$ (aka the tangent indicatrix) is a smooth hypersurface in $TM$ and there is an open neighborhood $U$ of $\Sigma\times\{0\}$ in $\Sigma\times\mathbb{R}$ on which there is defined a smooth mapping $\exp:U\to M$ such that, for each fixed $u\in \Sigma$, the curve $\gamma_u(t) = \exp(u,t)$ (defined for all $t\in\mathbb{R}$ such that $(u,t)$ lies in $U$) is the maximally extended unit speed $F$-geodesic with initial velocity $u\in\Sigma$.
Second, for any given $p\in M$, there is a $\delta>0$ such that, if $B_\delta(p)\subset T_pM$ is the set of vectors $v\in T_pM$ satisfying $F(v)<\delta$, then there is a well-defined mapping $\exp_p:B_\delta(p)\to M$ such that $\exp_p(0_p) = p$ and $\exp_p(tu) = \exp(u,t)$ for all $u\in \Sigma_p$ and all $t\in(0,\delta)$.  By taking $\delta$ sufficiently small, one can ensure that $\exp_p:B_\delta(p)\to M$ is a homeomorphism onto its (open) image, one that is a smooth diffeomorphism away from $0_p$.  
Now, unless $F$ is reversible (i.e., $F(-v) = F(v)$), the map $\exp_p:B_\delta(p)\to M$ need not be $C^2$ at $0_p$, because it can easily happen that $\exp(-u,-t)\not=\exp(u,t)$, so the $\exp_p$-image of the line segment $\{tu\ |\ |t|<\delta\}$ need not be a $C^2$ curve in $M$ (it will be $C^1$, though).
Even when $F$ is reversible (or, more generally, geodesically reversible), so that the exponential map $\exp_p$ is smooth on lines through $0_p\in B_\delta(p)$, it may still not be smooth at $0_p$.
Unfortunately, writing down a simple, explicit example for which smoothness clearly fails is not easy because, usually, it is not possible to integrate the geodesic equations and compute the map $\exp:U\to M$ explicitly.  In one case where it is possible, namely the case of a Minkowski Finsler metric (i.e., $M=\mathbb{R}^n$ and $F$ is invariant under translations in space), the exponential map 
$$\exp:\Sigma\times \mathbb{R} \bigl(= (\mathbb{R}^n\times\Sigma_0)\times\mathbb{R}\bigr)\to\mathbb{R}^n$$ 
is $\exp((p,u),t) = p+tu$ for $p\in \mathbb{R}^n$ and $u\in \Sigma_0\subset T_0\mathbb{R}^n = \mathbb{R}^n$, so the map $\exp_p:T_p\mathbb{R}^n\to\mathbb{R}^n$ is a smooth diffeomorphism after all, regardless of how non-Riemannian $F$ is.  (This shows that it's not just a matter of looking at the shapes of the unit spheres of $F$. The failure of smoothness is more subtle than that.)
However, in an answer to this question, I gave an example of a (homogeneous!) Finsler metric $F$ on a surface $M$ that is reversible and has the property that $F^4$ is a smooth function on $TM$, but the fourth power of the Finsler distance function from any given point of $M$ is not smooth.  Thus, this is an example in which the map $\exp_p$ cannot be smooth at the origin $0_p$ (for any $p\in M$), because, if $s_p:M\to[0,\infty)$ is the function that gives the $F$-distance from $p\in M$, then, for $\delta>0$ sufficiently small, we have $F(v) = s_p\bigl(\exp_p(v)\bigr)$ for all $v\in B_\delta(p)\subset T_pM$.
Remark (28 Feb 2016): I was curious as to what are the necessary and sufficient conditions for a Finsler surface to have $\exp_p$ be at least $C^2$ at every point.  It turns out that this is very restrictive:  Either the Finsler structure must be Minkowskian (i.e., invariant under a $2$-dimensional abelian group of translations as discussed above) or else the Cartan scalar (usually denoted $I$) must be constant.  The latter case includes the Riemannian case ($I$=0) as a special case; the cases where $I$ is a non-zero constant are actually 'generalized' Finsler structures (at each $p\in M$, the unit tangent vectors in $T_pM$ form a logarithmic spiral rather than a closed convex curve).   In all these cases, the map $\exp_p$ is actually $C^\infty$.<|endoftext|>
TITLE: Asymptotics of product of Euler's totient function (A001088)?
QUESTION [9 upvotes]: Conjecture:
\begin{align}
\lim_{n\to \infty } \, \frac{\left(\prod _{k=1}^n \phi (k)\right){}^{1/n}}{n}\sim 0.2059\text{...} 
\end{align}
The numerical result from 100000 terms is:


My questions are:
1) exists this limit ?
2) if yes, what is closed form of this constant ?
I am sure that this limit is NOT equal to 6/Pi^2 * exp(-1) = 0.2236438825...

REPLY [12 votes]: We may just write down $\varphi(k)=k\cdot \prod_{p|k} (1-1/p)$, $p$ runs over the primes which divide $k$, then 
$$
\frac{\left(\prod_{k\leqslant n}\varphi(k)\right)^{1/n}}{n}=\frac{\sqrt[n]{n!}}n\prod_{p} (1-1/p)^{\frac{[n/p]}{n}}.
$$
First multiple tends to $1/e$ by weak version of Stirling's formula, and in the product we may replace each exponent $\frac{[n/p]}n$ to $1/p$ for $p\leqslant n$ with multiplicative error between 1 and
$$
\prod_{p\leqslant n} (1-1/p)^{1/n}\geqslant \prod_{2\leqslant k\leqslant n}(1-1/k)^{1/n}=n^{-1/n}\rightarrow 1.
$$
Hence the limit exists and equals $e^{-1}\prod_p (1-1/p)^{1/p}$ as Lucia and 
so-called friend Don have already said.

REPLY [8 votes]: Yes, the limit exists, and is equal to $\exp(-1 + \sum_{p} \frac{1}{p}\log(1-\frac{1}{p}))$ (as Lucia said), where $p$ runs over primes. Hint: Reduce the problem to computing $\lim_{N\to \infty} \frac{1}{N} \sum_{n \le N} \log \frac{\phi(n)}{n}$. The arithmetic function $\log\frac{\phi(N)}{N}$ is strongly additive and (under favorable conditions which hold here) the mean value of a strongly additive function $g$ is given by $\sum_{p} g(p)/p$.<|endoftext|>
TITLE: What (fun) results in graph theory should undergraduates learn?
QUESTION [27 upvotes]: I have the task of creating a 3rd year undergraduate course in graph theory (in the UK). Essentially the students will have seen minimal discrete math/combinatorics before this course. Since graph theory is not my specialty and I did not take a graph theory course until grad school, I am seeking advice on the content.
I can of course consult excellent books like Diestel's, and numerous online lecture notes, course outlines on university webpages, etc. (although if any of these are particularly great that would be useful to know).
I am also aware of most of the standard results that should go in such a course.
But, what I'm really looking for is, slightly less well known, interesting/fun results that are at the right level to make suitable content (or could be adapted to assignment questions).
So, what is your favourite, unusual fact, that would be suitable for such a course?
Apologies if this is not a suitable question for MO. Let me know and I will delete. I'm aware of the related question Interesting and Accessible Topics in Graph Theory which gave me some good ideas, but was generally aimed at topics for high school students rather than final year undergraduates.

REPLY [2 votes]: A concrete and fun problem: Draw $n$ circles (can be intersecting). This divides the plane into regions. Show that this map is two-colorable.<|endoftext|>
TITLE: When does the radius of convergence of the product of two $p$-adic power series increase?
QUESTION [5 upvotes]: Let $p$ be a prime number and denote by $R(f)$ the radius of convergence of a power series $f(x) \in \mathbb{C}_p[[x]]$, where $\mathbb{C}_p$ is the completion of the algebraic closure of $\mathbb{Q}_p$, the field of $p$-adic numbers. Given two power series $f(x), g(x) \in \mathbb{C}_p[[x]]$, it is known that the radius of convergence of the product $h(x) = f(x)g(x)$ is at least the minimum of the radius of convergence of the two series $f(x)$ and $g(x)$. In other words, we have
$$ R(h) \geq \min\{R(f), R(g)\}.$$
Keep in mind $(1-x)(1+x+x^2+\dots) = 1$ as an example for the strict inequality. Is there a way to easily predict when $R(h) > \min\{R(f), R(g)\}$ and find $R(h)$ explicitly? More specifically, I'm interested in computing the radius of convergence of power series of the form $\exp(f(x))$ for $f(x) \in x\mathbb{C}_p[[x]]$.
For example, let $f(x) = \exp(x)$ and $g(x) = \exp(x^p/p)$. Then $R(f) = R(g) = (1/p)^{1/(p-1)}$ and using the fact that the Artin-Hasse exponential series $$\text{AH}(x) = \exp(x + x^p/p + x^{p^2}/p^2 + \cdots)$$ 
lies in $\mathbb{Z}_p[[x]]$ (which implies $R(\text{AH}) \geq 1$), $h(x) = \exp(x + x^p/p)$ has radius of convergence 
$$ R(h) = R\left(\exp\left(\frac{x^{p^2}}{p^2}\right)\right) = \left(\frac{1}{p}\right)^{\frac{(2p-1)}{p^2(p-1)}}> \left(\frac{1}{p}\right)^{\frac{1}{p-1}}. $$
The importance of this example comes from the fact that if we set $\pi$ to be a root of $x+x^p/p = 0$, then $h(\pi)$ is a non-trivial $p$-th root of unity in $\mathbb{C}_p$. This provides an analytic representation of $p$-th roots of unity, exploited in particular in Dwork's proof of the rationality of zeta functions over finite fields. 
More generally, using this method one can show that for any $n \geq 1$, we have
$$ R(\exp(x+x^p/p+\cdots + x^{p^n}/p^n)) = R(\exp(x^{p^{n+1}}/p^{n+1})). $$
Even though I understand the details involved in this calculation, I don't know if there is some more general theory underlying these examples. It may be helpful to share similar examples that you know. For instance, is it always the case that 
$$ R(\exp(f(x))) > R(\exp(x)),$$
 given $f(x) \in x\mathbb{C}_p[[x]]$ has a nonzero root $\alpha \in \mathbb{C}_p$ of absolute value $R(\exp(x)) = (1/p)^{1/(p-1)}$ and no non-zero roots of smaller absolute value?

REPLY [8 votes]: Here is a counterexample to your question at the end, for each $p$. Let $f_u(x) = x + ux^p/p$ for $u \in \mathbf C_p$ with $|u|_p = 1$ and $|u-1|_p = 1$. (Such $u$ can be taken in $\mathbf Z_p^\times$ if $p > 2$, but you need to go outside $\mathbf Q_p$ if $p = 2$ to an extension with residue field of size greater than $2$.) Since $|u|_p = 1$, all the nonzero roots of $f_u(x)$ in $\mathbf C_p$ have absolute value $(1/p)^{1/(p-1)} = R(\exp)$.
To find the $p$-adic radius of convergence of $\exp(f_u(x))$, write
$$
\exp(f_u(x)) = \exp\left(x+\frac{x^p}{p}\right)\exp\left((u-1)\frac{x^p}{p}\right)
$$
as formal power series. On the right side, the first factor has radius of convergence greater than $(1/p)^{1/(p-1)}$, as you noted. 
Since $|u-1|_p = 1$, the second factor has radius of convergence equal to that of $\exp(x^p/p)$, which is $(1/p)^{1/(p-1)}$. The reciprocal $(\exp(x + x^p/p))^{-1}$ has the same radius of convergence as $\exp(x+x^p/p)$, even for $p=2$, so $\exp(f_u(x))$ has radius of convergence equal to $(1/p)^{1/(p-1)}$. This is the counterexample to your question.

REPLY [6 votes]: I can only give a very partial answer, and that only from my very parochial point of view.
I will use the additive valuation $v$ rather than absolute value, normalized so that $v(p)=1$, and in terms of which $R(\sum a_nx^n)=-\liminf\bigl(v(a_n)/n\bigr)$, so that when $v(z)>R(f)$, $f(z)$ is a convergent series. Examples are: if $\exp(x)=\sum_{n\ge1}x^n/n!$ and $\log(x)=\sum_{n\ge1}(-1)^{n-1}x^n/n$, then $R(\exp)=1/(p-1)$ and $R(\log)=0$. It’s for this reason that I prefer the logarithm to its inverse.
The log that I’ve named above is the logarithm of the multiplicative formal group $\hat{\mathbf G}_{\mathrm m}(x,y)=x+y+xy$, that is a formal-group homomorphism from $\hat{\mathbf G}_{\mathrm m}$ to the additive formal group $\hat{\mathbf G}_{\mathrm a}(x,y)=x+y$ with $\log'(0)=1$.
The $p$-typical logarithm $\log_{\mathrm{AH}}=x+x^p/p+x^{p^2}/p^2+\cdots$ is the logarithm of another formal group $\mathscr M$. which we might call the $p$-typical recoordinatization of $\hat{\mathbf G}_{\mathrm m}$, and the $\Bbb Z_p$-formal-group isomorphism $u:\mathscr M\to\hat{\mathbf G}_{\mathrm m}$ is exactly what’s called the Artin-Hasse Exponential. It satisfies $\log\circ u=\log_{\mathrm{AH}}$.
Now here’s the moral of my story. These two logarithms, $\log$ and $\log_{\mathrm{AH}}$, being convergent throughout the open unit disc of $\Bbb C_p$, have all sorts of interesting behavior that is not seen at all by the exponential series $\exp(x)$, except at a very far remove. In particular, they have zeros. The closest such to the origin has $v(\zeta)=1/(p-1)$, which explains immediately the radius of convergence of the exponential function.
And so I am pretty sure that your last conjecture has no chance of being correct. I tried $p=3$, $\zeta=\omega-1$, where $\omega^2+\omega+1=0$, and found that $\exp(x-x^2/\zeta)$ seemed to have as bad convergence properties as the exponential itself.
EDIT:
You’ve asked me to explain further why the existence of zeros of the logarithm prevents wider convergence of the exponential, and indeed, it is not quite so obvious as I was pretending.
Let $\lambda$ be a root of the above-described $\log$ with $v(\lambda)=1/(p-1)$, in fact $\lambda+1$ will be a primitive $p$-th root of unity. If $R(\exp)\ge1/(p-1)$, then the Newton polygon of $f(x)=\exp(x)-\lambda$ will have a vertex $(n,v(b_n))$ in addition to the vertex $(1,0)$. (You don’t need to know that $b_n=1/n!$.) To see this, you may look at $f(\lambda x)$ and note that its coefficients must go to zero for convergence. In particular, There will be a segment of the polygon whose negative slope is $\ge1/(p-1)$, and thus a root $\mu$ of $f$ in an algebraic extension such that $\exp(\mu)=\lambda$, impossible if $\log(\exp(x))=x$.<|endoftext|>
TITLE: Numerical invariants for a graph or its complement that are bounded by some constant
QUESTION [6 upvotes]: I'm looking for numerical graph invariants that are bounded by a constant either for a graph $G$ or its complement $\bar{G}$.  (The complement graph $\bar{G}$ has the same set of vertices as $G$ but the edges are complemented.)  More specifically I’m looking for what numerical “Invariant $X$” is out there for which the following statement is true:
“Invariant $X$ is bounded by a constant $c$ either for $G$ or $\bar{G}$”.
One example is diameter $d(G)$ of a graph $G$ where it is known that:
"Either $d(G) \leq 3$ or $d(\bar{G}) \leq 3$"  
(Reference: F. Harary, R. W. Robinson: The diameter of a graph and its complement , The American Mathematical Monthly, Vol. 92, No. 3. (Mar., 1985), pp. 211-212”)
My question  is what other such numerical graph invariants are out there? I looked at the list of invariants on wikipedia but there was no mention of such bound on their respective pages.
 Again, all I’m interested is that the condition holds for either $G$ or its complement $\bar{G}$ and not necessarily for both.

REPLY [4 votes]: Another one: girth. By Ramsey's theorem, for every graph $G$ on six or more vertices, either it or its complement has girth at most three.

REPLY [3 votes]: There are a lot of properties which $G$ and $\overline{G}$ both have them: number of vertices (edges), automorphism group, and etc. So, naturally this question is interesting for me. I want to mention a spectral property. Let $G$ be a graph with $n$ vertices, and $\rho(G)$ denotes the largest eigenvalue of the adjacency matrix of $G$. Then we have:
$$\rho(G)\geq \frac{n-1}{2}\; \; \text{or}\;\; \rho(\overline{G})\geq \frac{n-1}{2}.$$
I will give the simplified proof for the above fact. 
For any $n\times n$ Hermitian matrices $A$ and $B$, we have:
$$\lambda_i(A+B)\leq \lambda_j(A)+\lambda_{i-j+1}(B),\;\; n\geq i\geq j\geq 1,$$
where, $\lambda_i(A)$ denotes the $i$th greatest eigenvalue of the matrix $A$. Now, for graph $G$, let $A$ be the adjacency matrix of $G$ and $B$ be the adjacency matrix of $\overline{G}$. Since we have $A+B=J-I$, which is the adjacency matrix of complete graph, the result is clear.<|endoftext|>
TITLE: If the natural density (relative to the primes) exists, then the Dirichlet density also exists, and the two are equal
QUESTION [6 upvotes]: On p. 76 of the 1996 edition of Serre's A Course in Arithmetic, one reads the following (inline) remark:

One can prove that, if $A$ has natural density $k$, the analytic density of $A$ exists and is equal to $k$.

Here, $A$ is a subset of $\bf P$ (the set of all positive rational primes), and the natural density of $A$ is actually the natural density of $A$ relative to $\bf P$, viz. the limit $$\lim_{n \to \infty} \frac{|A \cap [1,n]|}{|\mathbf P \cap [1,n]|}$$ (if it exists), while the analytic density of $A$ is actually the analytic density relative to $\bf P$, viz. the limit $$\lim_{s \to 1^+} \frac{\sum_{p \in A} p^{-s}}{\sum_{p \in \mathbf P} p^{-s}}$$ (again, if it exists). Here are then my questions:

Q1. Was it Serre the first who made this observation explicit? Q2. Do you know of a paper or book where a proof is provided? Serre doesn't even give a hint about it.

Notes (added later). 
On Q1: In the light of Lucia's comment below, let me make it clear that I myself find it very unreasonable that the result hadn't been known before Serre's remark in the 1970 French edition of his book (p. 126). I'd just like to find out if it was Serre the first who made it explicit. 
On Q2: I have my own proof, but would appreciate a reference. The reason is that something sensibly stronger is true, and I'm hoping to understand from the inspection of the proof he may have had in mind if this is intentional (e.g., it is evident from the proof he may have had in mind that something sensibly stronger is true, but he just didn't care), or not.
Edit (Feb 09, 2016). For future reference, I think it can be useful to make order and summarize, here in the OP, what has emerged from the answers and comments of those who have so far contributed to this discussion:
1) As expected, it wasn't Serre the one who first made explicit the relation between the analytic and natural densities relative to the primes. The result is already stated on p. 118 of:

E. Landau, Handbuch der Lehre von der Verteilung der Primzahlen, Erster Band, Teubner: Leipzig, 1909,

where a detailed proof is also presented. This answers both Q1 and Q2.
2) Franz Lemmermeyer, in a comment to the OP, had suggested since the outset that the result should have appeared almost surely in some of Landau's books. This was confirmed by so-called friend Don in his answer (here), where it's also reported that the result was mentioned on p. 225 of the 1st edition of:

H. Hasse, Vorlesungen über Zahlentheorie, Die Grundlehren der mathematischen Wissenschaften 59, Springer-Verlag: Berlin, 1950.

Interestingly enough, Hasse made a mistake here, by stating that not only the existence of the natural density (relative to the primes) implies that the analytic density (always relative to the primes) also exists, and the two are then equal: He went on asserting that also the converse is true! As still noted by so-called friend Don, the mistake was fixed in the 2nd (1964) edition of the book (p. 236), and it was mentioned in a comment to his answer that we know by now that Hasse was really wrong, for an example attributed by Serre to a private communication from E. Bombieri (p. 126 in the 1970 French edition of A Course in Arithmetic, or p. 76 in the 1996 English edition) proves the existence of a set of primes that has analytic (relative) density, but not natural (relative) density.
3) Comparison results in the same spirit of those considered in this question, but involving densities on $\mathbf N^+$, are not so rare in the literature. Most notably, it is known (and easy to prove by Abel's summation formula) that the upper analytic density (on $\mathbf N^+$) is not greater than the upper logarithmic density, which is in turn not greater than the upper asymptotic density, see, e.g., Theorem 2 in Section III.1.3 of:

G. Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, Cambridge Stud. Adv. Math. 46, Cambridge Univ. Press: Cambridge, 1995.

It follows at once that the existence of the natural density (on $\mathbf N^+$) implies the existence of the logarithmic density, and the existence of the logarithmic density implies the existence of the analytic density.
4) On the other hand, it is known that upper and lower asymptotic and natural densities are pretty much independent from each other, in a sense that was first made precise by L. Mišík in:

L. Mišík, Sets of positive integers with prescribed values of densities, Math. Slovaca 52 (2002), No. 3, pp. 289-296.
  see here for further reading on the subject. 

You may want to read the comments to Question 103111: Prescribed values for the uniform density for a more accurate account of Mišík's results and generalizations theoreof.
5) Furthermore, it is known that the existence of the analytic density (on $\mathbf N^+$) implies that also the logarithmic density exists, and the two are then equal. This is a non-trivial result, which goes back at least to H. Davenport and P. Erdős, who make an implicit reference to it in the proof of Theorem 1 from:

H. Davenport and P. Erdős, On sequences of positive integers, Acta Arith. 2 (1936), No. 1, 147-151.

The proof is based on the Hardy-Littlewood tauberian theorem. All of this was pointed out by so-called friend Don in a comment to GH from MO's answer (here). An alternative proof, that rather uses Karamata's tauberian theorem, is given by Tenenbaum in his book (Theorem 3 in Section III.1.3). The same Tenenbaum mentioned in a private communication that the special case of Karamata's theorem needed here goes back to:

O. Szász, Münchner Sitzungsberichte (1929), 325-340.

6) Last but not least, Christian Elsholtz added some further elements to the story (here).

REPLY [3 votes]: Scholz (in Jahresbericht der Deutschen Mathematiker-Vereinigung 45, 1935, p. 110, (Aufgabe 207) 
https://www.digizeitschriften.de/dms/toc/?PID=PPN37721857X_0045
https://www.digizeitschriften.de/download/PPN37721857X_0045/PPN37721857X_0045___log39.pdf
poses the problem below.
Given a set $M$ of positive integers $m_1<m_2< m_3 <...$ with natural density
$\lim_{n \rightarrow \infty} \frac{n}{m_n}=h$.
Take the set $P=\{ p,...\}$ of those integers (e.g. in base 10) with $m_1, m_2, ...$ digits,
which obviously have no natural density. Prove that it has a Dirichlet density $\lim_{s \rightarrow 1} (s-1) \sum_P p^{-s}=h$.
...
This shows that Scholz was certainly aware of the distinction between natural and Dirichlet density, by oscillating examples, similar to those
alluded to (by Salvo) in the more modern literature.
For completeness, the Solution by Stöhr is in the 1938 volume 
https://www.digizeitschriften.de/dms/toc/?PID=PPN37721857X_0048
https://www.digizeitschriften.de/download/PPN37721857X_0048/PPN37721857X_0048___log26.pdf<|endoftext|>
TITLE: What is a foliation and why should I care?
QUESTION [62 upvotes]: The title says everything but while it is a little bit provocative let me elaborate a bit about my question. First time when I met the foliation it was just an isolated example in the differential geometry course (I was the Reeb foliation) and I didin't pay many attention to it. In the meanwhile I get interested in the noncommutative theory in particular in $C^*$-algebras. While reading about Noncommutative Geometry I came across foliations as the one of the main motivating examples of the theory. I learned that in general the space of leaves of the foliation is badly behaved as a topological space and I believe that it is more worthwile to deal with these spaces using algebraic methods. But I don't have something like a mental picture of what foliations is and why should I even care about those objects?

REPLY [7 votes]: Here's why I care about foliations. It is always interesting when a structure can be expressed in terms of simpler structures. For instance a torus is the union of circles making it into a cartesian product. Or a Klein bottle is the union of circles making it into a non-trivial fibre bundle.
A foliation is a description of an n-manifold M as the union of submanifolds of fixed lower dimension k, whose tangent planes fit together cleanly. (I.e., such that every point of M has a neighborhood that is decomposed topologically just as n-dimensional Euclidean space is the cartesian product of k-dimensional Euclidean space and (n-k)-dimensional Euclidean space.) This is a natural generalization of two things: a) the concept of a fibre bundle, and b) the decomposition of a manifold into the trajectories of a non-singular vector field.
A foliation can be defined in terms of the reduction of a manifold's atlas to a certain simple pseudogroup.
The quintessential example of a foliation is the Reeb foliation of the 3-sphere. Although the 3-sphere possesses infinitely differentiable codimension-1 foliations such as the Reeb foliation, it is a beautifully subtle fact, proved by André Haefliger, that it has no real-analytic one. Another fascinating fact, proved by Sergei Novikov, is that every codimension-1 foliation of the 3-sphere must possess a compact leaf (which is in fact a torus). Yet another fascinating fact, proved by Krystyna Kuperberg, is that the 3-sphere possesses a real-analytic 1-foliation that has no closed trajectory. (This counterexample was discovered only 43 years after the original conjecture, by Herbert Seifert, suggested that every 1-foliation of the 3-sphere must have a closed trajectory.)<|endoftext|>
TITLE: An inequality for the minimal number of generators of a finite group
QUESTION [12 upvotes]: Let $G$ be a finite group, $n(G)$ the minimal number of generators and $m(G)$ the minimal number of irreducible complex representations generating (with $\otimes$ and $\oplus$) the left regular representation.    
Notation: the word "generating" does not mean "generating exactly", but as a direct factor.
Question: Is it true that $n(G) \ge m(G)$ ?
Remark: a group $G$ is linearly primitive iff $m(G) = 1$, so it's obviouly true in this case.

REPLY [4 votes]: Let $P$ be a $p$-group. Then the smallest number of irreducible characters of
$P$ whose kernels intersect trivially is at least $n(Z(P))$, and thus
$m(P) \ge n(Z(P))$. So to find an example where $n(P) < m(P)$, it suffices to find $P$ such that $n(P) < n(Z(P))$. An example in the Magma or GAP data
base of small groups is SmallGroup($3^6$,9). Here, $n(P) = 2$ but $n(Z(P)) = 3$.<|endoftext|>
TITLE: Generalization of the rigidity lemma in birational geometry
QUESTION [13 upvotes]: Let $X,Y,Z$ be projective varieties, and let $f:X\rightarrow Y$, $g:X\rightarrow Z$ be dominant morphisms. Assume that all the fibers of $g$ have the same dimension and are connected.
If there exists a point $z_0\in Z$ such that $f(g^{-1}(z_0))\subseteq Y$ has dimension $k$ is it true that $f(g^{-1}(z))\subseteq Y$ has dimension $k$ for any $z\in Z$ ? 
When $k=0$, that is $f(g^{-1}(z_0))$ is a point, this is true. It is known as the rigidity lemma.

REPLY [11 votes]: EDIT: I've just realized that this holds under somewhat weaker assumptions. It is not necessary that the fibers of $g$ are connected.
EDIT#2: Apparently, in my previous edit I weakened the conditions too far... properness of $g$ is back as it is needed, but connectivity of fibers is not as it is not. I think it is OK now. :) 
Cool question. Actually, I think this is true! 

Generalized Rigidity Lemma     Let $X,Y,Z$ be (irreducible) varieties, $f:X\rightarrow Y$ a morphism and $g:X\rightarrow Z$ a surjective proper morphism whose fibers are of the same dimension $n$. Then 
  $f(g^{-1}(z))\subseteq Y$ has the same dimension for any $z\in Z$.

Proof [This proof is inspired by the proof of the original Rigidity Lemma given in [Kollár-Mori, 1.6]]
Let $z_0\in Z$ and  $k=\dim f(g^{-1}(z_0))$. Further let $W=\mathrm{im}(f\times g)\subseteq Y\times Z$ with projection $p:W\to Z$. Let $h=f\times g: X\to W$.
Notice that since $g$ is proper, and $g=p\circ h$, so $h$ is also proper and hence $W=\mathrm{im}\, h\subseteq Y\times Z$ is closed and so it is a variety. 
 Then for all $z\in Z$, $p^{-1}(z)=h(g^{-1}(z))$ and hence $\dim p^{-1}(z_0)=k$. By semi-continuity of fiber dimension there exists a non-empty open set $z_0\in U\subseteq Z$ such that for all $z\in U$ $\dim p^{-1}(z_0)\leq k$. 
This implies that $p^{-1}U\subseteq W$ is a non-empty open set (in fact, since $g$ is surjective, this is dense) such that for all $w\in p^{-1}U$, $\dim h^{-1}(w)\geq n- k$.  However, then by semi-continuity of fiber dimension again the same is true for all $w\in W$. 
Now fix a $z\in Z$ and consider the morphism $h:g^{-1}(z)\to p^{-1}(z)$. By assumption $g^{-1}(z)$ is $n$-dimensional and we have just observed that the fibers of this map have dimension at least $n-k$, so the image has dimension at most $k$.
In other words, we proved that for any $z\in Z$, $\dim h(g^{-1}(z))\leq k$. Finally, observe that if there was a $z_1\in Z$ for which $\dim h(g^{-1}(z_1))< k$, then by repeating the same proof with $z_0$ and $z_1$ exchanged we would get that for any $z\in Z$, $\dim h(g^{-1}(z))< k$, but this would contradict the assumption that $\dim h(g^{-1}(z_0))= k$. Therefore we must have equality for every $z\in Z$. $\square$<|endoftext|>
TITLE: When do powers and ends in functor categories act pointwise?
QUESTION [7 upvotes]: $\newcommand{\C}{\mathcal C}\newcommand{\I}{\mathcal I}\newcommand{\D}{\mathcal D}\newcommand{\J}{\mathcal J}$Let $(\C, \otimes, I, \multimap)$ be a complete closed monoidal category and $\I$ a small category. Then the functor category $[\I, \C]$ is closed monoidal with the pointwise tensor product and its right adjoint defined as follows:\begin{align}(F \otimes G)X & = FX \otimes GX\\(F \multimap G)X & = \int_Y \I(X, Y) \pitchfork (FY \multimap GY)\end{align}Now assume that $\C$ is a functor category $[\J, \D]$ where $\D$ is complete and $\J$ is small. We can write the definition of ${\multimap} : [\I, \C]^{\mathrm{op}} \times [\I, \C] \to [\I, \C]$ as follows:\begin{equation}(F \multimap G)XZ = \left(\int_Y \I(X, Y) \pitchfork (FY \multimap GY)\right)Z\end{equation}I was thinking about pushing the application to $Z$ under the end and the power, which would result in the following equation:\begin{equation}(F \multimap G)XZ = \int_Y \I(X, Y) \pitchfork (FY \multimap GY)Z\end{equation}Is this actually correct? What are the conditions for ends and powers acting pointwise?

REPLY [6 votes]: Yes, this is correct. It is generally true that the evaluation functors $\text{ev}_Z \colon [\mathcal{J},\mathcal{D}] \to \mathcal{D}$ for $Z \in \mathcal{J}$ jointly create limits (and colimits). So a limit in $\mathcal{C} = [\mathcal{J},\mathcal{D}]$ can be computed pointwise if and only if the corresponding pointwise limits exist in $\mathcal{D}$. In particular when $\mathcal{D}$ is complete, the limit defining $(F \multimap G)X$ in $\mathcal{C} = [\mathcal{J},\mathcal{D}]$ exists and is preserved by evaluation at $Z$.<|endoftext|>
TITLE: Classifying Low Dimensional Solutions of the Yang--Baxter Equation
QUESTION [7 upvotes]: What is the present situation with classifying solutions of the Yang--Baxter equation in low dimensions?
To make my question more specific, have all solutions for dimension $2$ and $3$ been classified?

REPLY [6 votes]: Let $(V,c)$ be a braided vector space, that is: $V$ is a vector space and  $c\colon V\otimes V\to V\otimes V$ is an invertible linear map that satisfies $c_{12}c_{23}c_{12}=c_{23}c_{12}c_{23}$, where $c_{12}=(c\otimes\mathrm{id})$ and $c_{23}=(\mathrm{id}\otimes c)$. 
As far as I know, the classification of  braided vector spaces is completed in the case where $\dim V=2$:

Hietarinta, Jarmo. All solutions to the constant quantum Yang-Baxter equation in two dimensions. Phys. Lett. A 165 (1992), no. 3, 245--251. MR1169634 (93d:16050). doi. 

Other related interesting results:

Dye, H. A. Unitary solutions to the Yang-Baxter equation in dimension four. Quantum Inf. Process. 2 (2002), no. 1-2, 117--151 (2003). MR2032002 (2004k:81168). doi
Galindo, César; Rowell, Eric C. Braid representations from unitary braided vector spaces. J. Math. Phys. 55 (2014), no. 6, 061702, 13 pp. MR3390645. doi

Edit: In general, producing solutions of the Yang-Baxter equation is a very hard problem. In MO Question 201901, you will find some information on the so-called set-theoretic solutions.<|endoftext|>
TITLE: Does $|A+A|$ concentrate near its mean?
QUESTION [8 upvotes]: Fix $N$ to be a large prime. Let $A \subset \mathbb{Z}/N\mathbb{Z}$ be a random subset defined by $\mathbb{P}(a \in A) = p$, where $p = N^{-2/3 + \epsilon}$ for some fixed $\epsilon > 0$. My question is what kind of concentration inequalities do we have for the random variable $|A+A|$?

REPLY [8 votes]: If $\epsilon>1/6$ then $|A+A|=N$ with high probability. Suppose that $\epsilon<1/6$.  Call a pair $(x,y) \in {\mathbb Z}/N{\mathbb Z} \times {\mathbb Z}/N{\mathbb Z} $ "bad" if $(x,y) \in A \times A $ and there is another pair $(z,w) \in A \times A $ (also different from $(y,x)$), such that $z+w=x+y$. The probability that a particular pair $(x,y) \in {\mathbb Z}/N{\mathbb Z} \times {\mathbb Z}/N{\mathbb Z} $ will be "bad" is at most $Np^4$ so the expected number of "bad" pairs is at most $N^3 p^4$. Every good pair $(x,y)$ contributes one element to $|A+A|$, with $(y,x)$ yielding the same eklement. By Markov's inequality, 
$|A+A| \ge (|A|^2+|A|)/2-O(N^3 p^4)$ with high probability. Under  the assumption 
 $\epsilon<1/6$, we have that $N^3p^4$ is negligible compared to $N^2p^2$
so $|A+A|$ will be well approximated by $|A|^2/2$, which is half the square of a Binomial variable.<|endoftext|>
TITLE: Direct axiomatization of ordinal and cardinal numbers
QUESTION [8 upvotes]: Again, this question is  related  (**) to a previous one:
in standard books on basic set theory, after stating the axioms of ZFC, ordinal numbers are introduced early on. Afterwards cardinals appear: they are special ordinals which are minimal with respect to equinumerosity. 
Ordinals and cardinals live happily  within standard set theory.
    Nothing wrong with that, but what about axiomatizing them directly, ie without the underlying set theory? 
What I mean is: develop a first-order theory of some number system (the intended ordinals), such that they are totally ordered, there is an initial limit ordinal, and satisfy induction with respect to formulas in the language of ordinal arithmetic (here, of course, one will have to adjust the induction schema to accommodate the limit case).
The cardinals could be introduced by adding an order-preserving  operator $K$ on the ordinal numbers mimicking their definition in  ZFC: cardinals would then be the fixed points of $K$.

has some direct axiomatization along these lines been fully developed? I would suspect that the answer is in the affirmative, but I have no refs. 
the induction schema would be limited to first order formulae, so, assuming that the answer to 1 is yes, is there a theory of non-standard models of Ordinal Arithmetic? 
Assuming 1 AND 2, what about weaker induction schemas for ordinals (*)? 

(*)  I am thinking again of formal arithmetics and the various sub-systems of Peano
(**) it is not the same, though: here I am asking for a direct axiomatization of ordinals, and indirectly of cardinals, via the operator $k$

REPLY [3 votes]: You might consider taking a look at a paper by Athanassios Tzouvaras titled "Cardinality without enumeration" (look under title on the Web), especially at Definition 3.1.  Since it is short, I will quote it verbatim:
"Definition 3.1  Let $M$ be a model of $ZF$.  A notion of cardinality for $M$ is a mapping $C$$\subset$$M$ such that:
(1) dom($C$)=$M$ and  r ng($C$)=Card
(2) $C$($\kappa$)=$\kappa$ for every $\kappa$$\in$_Card_
(3) For any disjoint sets $x$, $y$ $C$($x$$\cup$$y$)=$C$($x$)$+$$C$($y$)
(4) For any $x$, $y$ $C$($x$$\times$$y$)=$C$($x$) $\cdot$ $C$($y$)
(5) If $f$ : $x$$\rightarrow$$y$ is an injective mapping, then $C$($x$)$\le$$C$($y$)
A cardinality notion $C$ is said to be standard if in addition the converse of (5) holds, i.e. if $C$($x$)$\le$$C$($y$) implies that there is an injective $f$  from $x$ to $y$."
An especially interesting consequence of the cardinal notion 
$C$ being standard is Remark 3.2(ii):
"If $C$ is standard, then $C$($x$)=$C$($y$) implies that there is an injective $f$ from $x$ onto $y$.  So the existence of a standard notion of cardinality implies $AC$ [and the Kunen inconsistency as well--my comment].  Moreover in the presence of $AC$ there is a unique notion of cardinality, the standard one.  Thus in order to depart from the standard notion of cardinality, we must drop $AC$." 
Is Definition 3.1 more along the lines of what you are looking for, at least regarding cardinality?<|endoftext|>
TITLE: Representation theorem for modular lattices?
QUESTION [18 upvotes]: Birkhoff's representation theorem implies that every distributive lattice embeds into the lattice of subsets of a set. Is there also some representation theorem for modular lattices?
For example, I wonder: Does every modular lattice embed into the lattice of submodules of some module? After all, I cannot think of any relation which holds in the lattice of submodules of some module, except for modularity.

REPLY [7 votes]: By adding a couple of conditions we can indeed obtain a representation theorem as the OP suggests.  Specifically, von Neumann's coordinatization theorem says that every complemented modular lattice of order at least 4 is isomorphic to the lattice of principal right ideals of a von Neumann regular ring.  See the wikipedia article on continuous geometry or any of the books on continuous geometry mentioned in this mathoverflow post.
If we replace modularity with the stronger Arguesian law mentioned in Todd and Pedro's answers, we get the more general coordinatization theorem of Jónsson, which says that every complemented Arguesian lattice with a large partial 3-frame is again isomorphic to the lattice of principal right ideals of a von Neumann regular ring.  See Proposition 10.1 of this paper by Wehrung for a first order characterization of "having a large partial 3-frame".<|endoftext|>
TITLE: Collection of dense subsets as a "fingerprint" for Hausdorff topologies?
QUESTION [10 upvotes]: Let $(X,\tau)$ be a Hausdorff space and let ${\cal D}$ denote the collection of dense subsets of $(X,\tau)$. Is it possible that there is another Hausdorff topology $\tau_1 \neq \tau$ on $X$ such that the collection of dense subsets of the space $(X,\tau_1)$ also equals ${\cal D}$?

REPLY [10 votes]: By the following proposition, the only Hausdorff spaces which are completely determined by their dense subsets are the discrete spaces.
Proposition. Let $(X,\mathcal{T})$ be a $T_{0}$ topological space such that there does not exist an $\mathcal{S}$ where

$(X,\mathcal{S})$ is a topology
$\mathcal{T}\subseteq\mathcal{S}$ and
$\mathcal{S}\neq\mathcal{T}$
$(X,\mathcal{S})$ and $(X,\mathcal{T})$ have the same dense sets.

Then $(X,\mathcal{T})$ is discrete.
Proof. Suppose to the contrary that $(X,\mathcal{T})$ is not discrete. Then there exists an open set $U$ which is not clopen since the only $T_{0}$ space where every open set is clopen is the discrete space.
Therefore, let $U$ be an open set which is not clopen. Now let $C=\overline{U}$. Let $\mathcal{S}$ be the topology generated by $\mathcal{T}\cup\{C\}$. Then $\mathcal{S}$ is generated by the basis consisting of sets in $\mathcal{T}$ along with the sets of the from $O\cap C$ where $O\in\mathcal{T}$. I claim that $\mathcal{T}$ and $\mathcal{S}$ have the same dense sets. If $D$ is dense in $(X,\mathcal{S})$, then $D$ is clearly dense in $(X,\mathcal{T})$. Conversely, suppose that $D$ is dense in $(X,\mathcal{S})$. Suppose that $O\in\mathcal{T}$ and $O\cap C=O\cap\overline{U}$ is non-empty. Then $O\cap U$ is also non-empty. Therefore, $D\cap O\cap U$ is non-empty, hence $D\cap O\cap C=D\cap O\cap\overline{U}$ is non-empty as well. We therefore conclude that $(X,\mathcal{S})$ and $(X,\mathcal{T})$ both have the same dense sets.<|endoftext|>
TITLE: A seemingly simple combinatorial object that must have an easy generating function
QUESTION [14 upvotes]: One more question related to my earlier "Special" meanders.
I am trying to isolate simplest problems related to it. Here is one.
For a composition (i. e. a tuple of natural numbers) $\boldsymbol a=(a_1,...,a_k)$, define the set of its midpoints by
$$
\operatorname{mid}(\boldsymbol a):=\{a_1+...+a_{i-1}+\frac{a_i+1}2\mid1\leqslant i\leqslant k\}.
$$
For example, $\operatorname{mid}(4,1,3)=\{\frac52,5,7\}$.
Call compositions $\boldsymbol a$ and $\boldsymbol b$ unmatchable if $\operatorname{mid}(\boldsymbol a)\cap\operatorname{mid}(\boldsymbol b)=\varnothing$.
I need any explicit information (formula, generating function, ...) for the numbers
$F(n):=$ number of unmatchable pairs $\langle\boldsymbol a,\boldsymbol b\rangle$ with $\sum a_i=\sum b_j=n$.
The sequence starts $0,2,6,24,78,284,960,3402,11710,41020,...$
My attempts so far have led to increasingly absurdly complicated approaches (like inverting infinite matrices with power series coefficients) and gave nothing in the end.
I believe a specialist can either give an answer immediately or relate it to some known difficult problem.

REPLY [5 votes]: I've now got an explicit formula for the generating function; after several simplifications it is still quite messy. Still decided to show it (in a separate answer, adding this to the old one would make it too long I think), maybe somebody can use it for a better answer. The proof is equally messy and uninspiring, it consists in making the functional equation from the recursion which I described in my previous answer.
So let $f(q)=\sum F(n)q^n$. Then,
$$
f(q)=\frac AB,
$$
where
$$
A=\sum_{n=0}^\infty
(-1)^{n-1}\prod_{k=1}^n
\frac{(1-2q^k)^2}{(1-q^k)(1-4q^{k+1})}
\frac{1+q-4q^{n+1}}{1-2q^n}\left(\frac{q^2}{2(1-q)}\right)^n
$$
and
$$
B=\sum_{n=0}^\infty
(-1)^n\prod_{k=1}^n
\frac{(1-2q^k)^2}{(1-q^k)(1-4q^{k+1})}
\left(\frac{1-q}{1-2q^n}+2\frac{1-2q}{(1-2q^n)^2}\right)\left(\frac{q^2}{2(1-q)}\right)^n.
$$
Or, if you prefer,
\begin{multline*}
A=1-3 q
+\frac{(1-2 q) \left(1-4 q^2+q\right) }{(1-q) \left(1-4 q^2\right)}\frac{q^2}{2 (1-q)}
-\frac{(1-2 q)^2 \left(1-2 q^2\right) \left(1-4
   q^3+q\right) }{(1-q) \left(1-4 q^2\right) \left(1-q^2\right) \left(1-4 q^3\right)}\left(\frac{q^2}{2 (1-q)}\right)^2
+\frac{(1-2 q)^2 \left(1-2
   q^2\right)^2 \left(1-2 q^3\right) \left(1-4 q^4+q\right) }{(1-q) \left(1-4 q^2\right) \left(1-q^2\right) \left(1-4
   q^3\right) \left(1-q^3\right) \left(1-4 q^4\right)}\left(\frac{q^2}{2 (1-q)}\right)^3\mp...
\end{multline*}
and
\begin{multline*}
B=1-3 q-\frac{\left(3-7 q+2 q^2\right)}{(1-q) \left(1-4 q^2\right)}\frac{q^2}{2 (1-q)}
+\frac{(1-2 q)^2 \left(3-5 q-2 q^2+2 q^3\right)
   }{(1-q) \left(1-4 q^2\right) \left(1-q^2\right) \left(1-4 q^3\right)}\left(\frac{q^2}{2 (1-q)}\right)^2
-\frac{(1-2 q)^2 \left(1-2 q^2\right)^2 \left(3-5
   q-2 q^3+2 q^4\right) }{(1-q) \left(1-4 q^2\right) \left(1-q^2\right) \left(1-4 q^3\right) \left(1-q^3\right) \left(1-4
   q^4\right)}\left(\frac{q^2}{2 (1-q)}\right)^3\pm...
\end{multline*}
(Later)
Let me add one more version:
$$
f(q)=\frac{(1-4 q) \, _2\phi _1\left(\begin{smallmatrix}2,\ 2 q\\4 q\end{smallmatrix};q,\frac{-q^2}{2 (1-q)}\right)+q \, _2\phi _1\left(\begin{smallmatrix}2,\ 2 q\\4 q^2\end{smallmatrix};q,\frac{-q^2}{2 (1-q)}\right)}{2 (1-2 q) \,
   _2\phi _1\left(\begin{smallmatrix}2,\ 2\\4 q^2\end{smallmatrix};q,\frac{-q^2}{2 (1-q)}\right)-(1-q) \, _2\phi _1\left(\begin{smallmatrix}2,\ 2 q\\4 q^2\end{smallmatrix};q,\frac{-q^2}{2 (1-q)}\right)}
$$
where $_2\phi _1$ is the basic hypergeometric series<|endoftext|>
TITLE: Random links and $3$-manifolds
QUESTION [11 upvotes]: In Jeffrey Weeks book  "The Shape of Space" he explaines at the end of Chapter 18 (on page 255) the following about the geometrization conjecture:

A non-trivial connected sum $M_1\# M_2$ admits a geometric structure if and only if $M_1=\mathbb{R}P^3=M_2$.
"Most $3$-manifolds admit a hyperbolic structure.
All other $3$-manifolds admit one of the other geometric structures or can be cut along $2$-spheres and $2$-tori into pices that admit geometric structures.

From this it follows that a "randomly chosen" $3$-manifold is not a connected sum.
I thought one way to make this precise is to use the hyperbolic Dehn surgery theorem:

Every $3$-manifold can be obtained by Dehn surgery along a link in $S^3$.

If the link is a hyperbolic link, then the resulting manifold is hyperbolic except for only finitly many surgery coefficients.

A knot in $S^3$ is hyperbolic if and only if it is not a torus or a satellite knot.

On Wikipedia it is written: "Every non-split, prime, alternating link that is not a torus link is hyperbolic by a result of William Menasco."


I have the following questions:

In which ways one can make claims like "a random knot is hyperbolic" precise?
Are there hyperbolic split links?
Is it true that a random link is hyperbolic? (Again how to make this precise?)
Is a random link a split link?

REPLY [6 votes]: As others have indicated, there are many different notions of random 3-manifold or random link. Here are two other types of models of random linking:

The Petaluma Model (http://front.math.ucdavis.edu/1411.3308). This model is based on link diagrams with a single multicrossing, with the randomness given by a choice of random permutation, which determines the heights of the arcs relative to one another. They're actually able to compute the distribution of (appropriately scaled) linking numbers on the nose with this model. I don't think that this has been done with any other model.  They do some computations of some of the moments of other low order finite type invariants too.
The random projection model (http://front.math.ucdavis.edu/1602.01484). This model starts with a fixed embedding of some circles in some high-dimensional Hilbert space, and randomly projects these onto a 3-dimensional subspace. In principal, the moments of the linking numbers ought to be computable. This is an intriguing model because there are continuously many parameters. It's possible that by varying the initial embeddings, these models can be made to limit to other types of models. Maybe that could explain some of the universality observed experimentally and discussed in the Petaluma paper.

As far as I'm aware, no one has examined hyperbolicity in either of these models.<|endoftext|>
TITLE: Solutions of equations characterizing a complex structure
QUESTION [11 upvotes]: Let $(S^n,g)$ denote the unit $n$-sphere endowed with its induced metric $g$ from its embedding into $\mathbb{R}^{n+1}$. The Levi-Civita connection of $g$ induces a splitting of the tangent bundle of $\pi:TS^n\to S^n$ into horizontal and vertical parts as $TTS^n = V\oplus H$, where each summand is canonically isomorphic to $\pi^*(TS^n)$.  Thus, every vector field $X$ on $S^n$ can be uniquely lifted to $TS^n$ as either a horizontal vector field $X^h$ or a vertical vector field $X^v$.
Given a triple of (smooth) real-valued functions $(\alpha,\delta,\beta)$ on $TS^n$ satisfying $\alpha \delta - \beta ^2 = 1$, one can define an almost-complex structure $J_{\delta, \beta}$ on $TS^n$ by the conditions
\begin{equation}
J_{\delta , \beta}(X^h)=\beta X^h + \alpha X^v,\\
J_{\delta , \beta}(X^v)=-\beta X^v - \delta X^h,
\end{equation}
where $X^h$ and $X^v$ are the horizontal and vertical lifts of any vector field $X$ on $S^n$. 
Question: What are the possibilities for $(\alpha,\beta,\delta)$ if we require that $J_{\delta,\beta}$ be integrable?  
Remarks: 
(1) I am particularly interested in integrable $J_{\delta,\beta}$ for which $\beta$ is not constant. I am also interested in understanding the integrable $J_{\delta,\beta}$ in which $(\alpha,\beta,\delta)$ are only defined on some open subset of $TS^n$.
(2) I already know some local solutions that take a special form with respect to conformal coordinates on $(S^n,g)$:  If $x = (x_1,\ldots,x_n)$ are local conformal coordinates on $U\subset S^n$, so that $g = \lambda^2(x) \sum _{i=1}^n dx_i ‎\otimes dx_i$, let $y_i:TU\to\mathbb{R}$ defined by $y_i(v) = \mathrm{d}x_i(v)$ be the associated tangential coordinates.  Then one can compute that a basis for the $(1,0)$-forms for $J_{\delta,\beta}$ on $TU$ are given by 
$$
\zeta_k = \mathrm{d}y_k + y_j(\delta_{jk}\,\mu_l\,\mathrm{d}x_l +\mu_j\,\mathrm{d}x_k-\mu_k\,\mathrm{d}x_j) - z\, \mathrm{d}x_k
$$
where the summation convention is assumed and 
$$
\mu_j = \frac{\partial (\log\lambda)}{\partial x_j}\qquad\text{and}\qquad z = \frac{i+\beta}{\delta}.
$$
Then $J_{\delta,\beta}$ is integrable if and only if
$$
\mathrm{d}\zeta_k\equiv0\mod \zeta_1,\ldots,\zeta_n\,.
$$
When $n>1$, this works out to be $2n$ first-order partial differential equations for $z$, so the system is overdetermined (and nonlinear).
I already know the solutions when one supposes that the function $z$ is assumed to be a function of $E(x,y)=\lambda ^2(x)\sum _{i=1}^n (y_i)^2$. I want to know whether there are any solutions that are not of this form. 
Addendum: I have some questions on Robert Bryant's answer:


Are our leaves of foliation $‎\lbrace(u,v)‎\rbrace \times H_+$?


What is the proof of the proposition in the answer?


Why the functions $\delta, \beta$ of the associated complex structure $J_{\delta, \beta}$ (as the associated complex structure  for the equation $Z.Z+1=0$) are not functions of $E(x,y)=\lambda^2 \sum_{i=1}^n (y^i)^2$? And are there other types of solutions?

REPLY [11 votes]: The answer is 'Yes, there are many other solutions, even global ones, and, when $n>1$, the local solutions depend on one holomorphic function of n complex variables.'  (Of course, when $n=1$, $J_{\delta,\beta}$ is always integrable.)
For a geometric context of this question, see Remark 3 below.
Briefly, here is a description of how they can be understood:  Let $\mathbb{R}^{n+1}$ be given its standard inner product (and extend it complex linearly to a complex inner product on $\mathbb{C}^{n+1}$, which will be used below).  Then
$$
S^n = \bigl\{ u\in\mathbb{R}^{n+1}\ \bigl|\ u{\cdot}u = 1\ \bigr\}
$$
and 
$$
TS^n = \bigl\{ (u,v)\in\mathbb{R}^{n+1}{\times}\mathbb{R}^{n+1}
\ \bigl|\ u{\cdot}u = 1,\ u{\cdot}v = 0\ \bigr\}.
$$
Let $H_+ = \{ a+ib\ |\ b>0\ \}\subset\mathbb{C}$ be the upper half-line in $\mathbb{C}$.  Define a mapping
$$
\Phi:TS^n\times H_+\to\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}
$$
by $\Phi\bigl((u,v),z\bigr) = v - z\,u$.  The mapping $\Phi$ is easily seen to be a diffeomorphism (one-to-one, onto, and smoothly invertible).  One can think of $\Phi$ as establishing a smooth (though not holomorphic) foliation of $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ by half-lines, one for each point of $TS^n$.  
Now, given real-valued functions $(\alpha,\beta,\delta)$ on an open set $U\subset TS^n$ defining an almost complex structure $J_{\delta,\beta}$ on $U$ as per the OP's description, one has $\alpha\delta-\beta^2 = 1$, so without loss of generality, one can replace $J_{\delta,\beta}$ by $-J_{\delta,\beta}$, so one can assume that $\alpha$ and $\delta$ are positive. Set $z = (\beta{+}i)/\delta = \alpha/(\beta{-}i):U\to H_+$.  (One way to interpret this is that $TS^n\times H_+$ can be thought of as pairs $\bigl((u,v),z\bigr)$ consisting of a point $(u,v)$ of $TS^n$ and a complex structure (parametrized by $z\in H_+$) on the vector space $T_{(u,v)}TS^n$.  The map $\Phi$ is just an identification of this $H_+$-bundle over $TS^n$ with the complex manifold $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$.)
Then one has the following result, easily proved by a moving frames calculation (see Remark 2 below):
Proposition: When $n>1$, the almost complex structure $J_{\delta,\beta}$ 
on $U\subset TS^n$ is integrable if and only if the image of the mapping 
$$
\Phi_z = \Phi\bigl((u,v),z(u,v)\bigr):U\to \mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}
$$
is a holomorphic hypersurface in $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$.
Moreover, in this case, the map $\Phi_z$ is a holomorphic embedding of
$(U,J_{\delta,\beta})$ into $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$.
Note that this Proposition implies that, if $X\subset \mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ is a nonsingular holomorphic hypersurface that is transverse to the half-line foliation determined by $\Phi$ and intersects each such half-line in at most one point, then $X$ is the image of $\Phi_z$ for some $z:U\to H_+$ where $U\subset TS^n$ is an open subset, and thus, writing $z=(\beta{+}i)/\delta$ for real-valued functions $(\delta,\beta)$ on $U$ (with $\delta>0$), the almost complex structure $J_{\delta,\beta}$ is integrable on $U\subset TS^n$.
Thus, when $n>1$, the space of local solutions is identified with the space of holomorphic hypersurfaces in $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ satisfying some open conditions, which, locally, are 'parametrized' by one holomorphic function of $n$ complex variables.  (As already remarked, when $n=1$, $J_{\delta,\beta}$ is always integrable.)
Now, there are many holomorphic hypersurfaces in $\mathbb{C}^{n+1}\setminus \mathbb{R}^{n+1}$ that satisfy these conditions.  For example, the hyperquadric in $\mathbb{C}^{n+1}$ defined by $Z{\cdot}Z +1 = 0$ has no real points and meets each half-line of the $\Phi$-foliation transversely in a unique point, and this gives a solution of the kind that the OP has already found.
More generally, consider a hyperquadric of the form
$$
Z{\cdot} AZ + B{\cdot} Z + 1 = 0,
$$
where $A$ is a complex symmetric $(n{+}1)$-by-$(n{+}1)$ matrix and $B\in \mathbb{C}^{n+1}$ is a complex vector.  The conditions needed to guarantee that this hyperquadric has no real points and meets each half-line of $\Phi$ transversely in a single point are open conditions on the pair $(A,B)$, and this set is not empty because it contains the above example. 
This already works out to provide a space of globally defined complex structures on $TS^n$ that has (real) dimension $(n{+}1)(n{+}4)$, which is far more than the dimension of the space of solutions that the OP mentioned as already known.  The generic one is not of the kind mentioned by the OP.
Remark 1:  It turns out that this problem has a much larger symmetry group than one might, at first, suspect.  It is well-known that the tangent bundle of $S^n$ can be regarded as the space of oriented lines in $\mathbb{R}^{n+1}$; simply let $L(u,v)\subset\mathbb{R}^{n+1}$ be the oriented line through $v$ with direction $u$.  Then the complexification of $L(u,v)$ as a line in $\mathbb{C}^{n+1}$ is divided into two (complex) half-lines, a positive half and and a negative half.  The above map $\Phi$ simply gives this foliation of $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ explicitly. All of this is equivariant under the group $G$ of affine transformations of $\mathbb{R}^{n+1}$, a Lie group of dimension $(n{+}1)(n{+}2)$.  It turns out that the entire problem is invariant under this group.  However, even under this larger group, the space of hyperquadric solutions listed above is not homogeneous (i.e., they are not all equivalent) since the dimension of this space is $(n{+}1)(n{+}4)$.
Remark 2: The following is the moving frames calculation 
I used to prove the Proposition.  While it can
be done as a bundle calculation without taking a local
moving frame, choosing a local moving frame reduces
the number of 'auxiliary variables' and so may be 
easier to understand for the novice.  The usual
summation convention for repeated indices in a term
will be assumed in what follows.
Let $U\subset S^n\subset\mathbb{R}^{n+1}$ be any proper
open subset and let $e_0:U\to \mathbb{R}^{n+1}$ 
denote the (vector-valued) inclusion mapping.  Let $e_1,\ldots,e_n:U\to\mathbb{R}^{n+1}$ be any (smooth) orthonormal
tangential frame field extending $e_0$, 
i.e., $e_a\cdot e_b = \delta_{ab}$ for $0\le a,b\le n$.
There exist smooth $1$-forms $\omega_i$ and $\omega_{ij}=-\omega_{ji}$
(where $1\le i,j\le n$) on $U$ satisfying the structure equations
$$
\mathrm{d}e_0 = e_i\,\omega_i
\qquad \text{and}\qquad
\mathrm{d}e_i = -e_0\,\omega_i + e_j\,\omega_{ji}\,. 
$$
The $n$ $1$-forms $\omega_i$ are linearly independent on $U$
and, in fact, form a basis for the $1$-forms on $U$.
These forms satisfy the structure equations
$$
\mathrm{d}\omega_i = -\omega_{ij}\wedge\omega_j
\qquad \text{and}\qquad
\mathrm{d}\omega_{ij} = -\omega_{ik}\wedge\omega_{kj} + \omega_i\wedge\omega_j\,.
$$
 Define functions $v_i:TU\to\mathbb{R}$ 
by $v_i(u,v) = e_i(u){\cdot}v$ for $1\le i\le n$,
so that $v = e_i(u)\,v_i$ for all $(u,v)\in TU$ and
define $1$-forms $\eta_i$ on $TU$ 
by $\eta_i = \mathrm{d}v_i + v_j\,\omega_{ij}$.
Then the $2n$ $1$-forms $\omega_i$, $\eta_i$ 
are a basis of the $1$-forms on $TU$. 
Moreover, they have the property that a vector $w\in T(TU)$ 
is tangent to the fibers of the basepoint projection $TU\to U$
if and only if $\omega_i(w) = 0$ while $w$ is tangent to the
horizontal plane field on $TU$ if and only if $\eta_i(w) = 0$.
 Now, the OP defines, for any (say, smooth) triple of functions 
$(\alpha,\beta,\delta)$ on $TU$ satisfying $\alpha\delta-\beta^2 = 1$, 
an almost-complex structure $J_{\delta,\beta}:T(TU)\to T(TU)$ 
by the rules
$$
\omega_k\circ J_{\delta,\beta} = \beta\,\omega_k - \delta\,\eta_k
\quad\text{and}\quad
\eta_k\circ J_{\delta,\beta} = -\beta\,\eta_k + \alpha\,\omega_k\,.
$$
It is easy to check that this does define an almost-complex structure
and that a basis for the $(1,0)$-forms on $TU$ with respect
to this almost-complex structure is given by
$$
\zeta_k = \eta_k - z\,\omega_k
$$
where $z = (\beta{+}i)/\delta = \alpha/(\beta{-}i)$ 
has nonzero imaginary part (in particular, $z$ is nowhere vanishing).
Note that the $\mathbb{C}$-valued $1$-forms $\zeta_k$ and $\overline{\zeta_k}$
for $1\le k\le n$ provide a basis for the $1$-forms on $TU$.
Since $J_{-\delta,-\beta} = - J_{\delta,\beta}$, 
one can always reduce to a case in which $\delta>0$ 
(and, of course, $\alpha>0$) when investigating integrability.  For simplicity, I will assume this from now on, which is equivalent to the condition that $z$ have positive imaginary part.
 Now, the condition that $J_{\delta,\beta}$ 
be integrable is the condition that the ideal generated
by $\zeta_1,\ldots,\zeta_n$ be closed under exterior
differentiation, i.e., that
$$
\mathrm{d}\zeta_k \equiv 0\  \text{modulo}\ \zeta_1,\ldots,\zeta_n\,.
$$
Using the structure equations, one calculates
$$
\mathrm{d}\zeta_k = \zeta_j\wedge(\omega_{kj} + (v_j/z)\,\omega_k)
  - \tfrac{1}{2z}\,\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2) \wedge \omega_k\,.
$$
Since $\omega_k = (\zeta_k-\overline{\zeta_k})/(\bar z - z)$, this yields
$$
\mathrm{d}\zeta_k \equiv \frac{\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2)}{2z(\bar z - z)}\wedge \overline{\zeta_k}\ \  \text{modulo}\ \zeta_1,\ldots,\zeta_n\,.
$$
Thus (since $n>1$), $J_{\delta,\beta}$ is integrable if and only if
$$
\mathrm{d}(z^2{+}{v_1}^2{+}\cdots {+}{v_n}^2) 
\wedge\zeta_1\wedge\cdots\wedge\zeta_n = 0.
\tag1
$$
 On the other hand, 
consider the mapping $\Phi_z:U\to\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$
defined by 
$$
\Phi_z = v - z u = v_1\,e_1 + v_2\,e_2+ \cdots +v_n\,e_n - z\,e_0.
$$
Again, using the structure equations, one computes
$$
\mathrm{d}\Phi_z 
= -e_0\,\tfrac{1}{2z}\,\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2)
     + (e_k{+}(v_k/z)\,e_0)\,\zeta_k\,. 
$$
Write $\mathrm{d}(z^2{+}{v_1}^2{+}\cdots{+}{v_n}^2) = p_k\,\zeta_k + q_k\,\overline{\zeta_k}$ for some functions $p_k$ and $q_k$. 
It follows that the map $\Phi_z:U\to\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$ 
has a complex linear differential (when $U$ is endowed 
with the almost-complex structure $J_{\delta,\beta}$)
if and only if $q_k = 0$ for all $k$, which, by (1), holds if and only if $J_{\delta,\beta}$ is integrable.  
Furthermore, the image of the tangent space at a point of $U$ is the set of vectors of the form
$$
-e_0\,\tfrac{1}{2z}\,(p_k\,a_k + q_k\,\overline{a_k})
     + (e_k{+}(v_k/z)\,e_0)\,a_k
$$
where the $a_k$ are complex numbers.  But this is a complex subspace of $\mathbb{C}^{n+1}$ if and only if $q_k=0$ for all $1\le k\le n$, i.e., if and only if (1) holds.  Thus, the image $\Phi_z(U)$ is a complex submanifold of $\mathbb{C}^{n+1}$ (of complex dimension $n$) if and only if (1) holds.  Thus, the Proposition is proved.
Remark 3: This question can be placed in a larger 
context that partly explains its interest.
The underlying geometry is that of a $2n$-manifold $M^{2n}$ 
endowed with a splitting of its tangent bundle $TM = S_1\oplus S_2$
into two $n$-dimensional subbundles $S_1$ and $S_2$ 
together with a bundle isomorphism $\iota:S_1\to S_2$.
Given such a structure on $M$ a family of almost-complex structures 
can be defined on $M$ as in the OP's question:  For each mapping $h = (\alpha,\beta,\delta):M\to\mathbb{R}^3$
satisfying $\alpha\delta-\beta^2 = 1$, one has an almost-complex
structure $J_h$ that satisfies
$$
J_h(s) = \beta\,s + \alpha\,\iota(s)
\quad\text{and}\quad
J_h\bigl(\iota(s)\bigl) = -\beta\,\iota(s) + \delta\,s
$$
for any section $s$ of $S_1$.
This can be understood more geometrically in terms of
a somewhat weaker structure, namely, a pair of bundles
$P\to M$ and $S\to M$ where $P\to M$ is an orientable $2$-plane
bundle and $S\to M$ is an $n$-plane bundle, together
with a vector bundle isomorphism $\iota:TM\to P\otimes S$.
(Such a structure is sometimes called a(n) (oriented) Segre structure
of type $(2,n)$.)  Then to each reduction of structure
of $P$ from an oriented real $2$-plane bundle to a complex line
bundle $L$, one associates a corresponding almost-complex
structure on $M$ using the isomorphism 
$\iota:TM\to L\otimes_{\mathbb{R}}S$.  Such reductions of structure of $P$ are given by sections
of an associated bundle $J(P)\to M$ whose fiber 
is $\mathrm{GL}(2,\mathbb{R})/\mathrm{GL}(1,\mathbb{C})$,
a homogeneous space that is topologically and geometrically
the hyperboloid of $2$-sheets in $\mathbb{R}^3$.
Then one finds that the data $(M,P,S,\iota)$ can be
used to define a natural almost-complex structure on $J(P)$
such that a section of $J(P)$ represents an integrable 
almost-complex structure if and only if the image of the
section is an almost-complex submanifold of $J(P)$.
Usually, this almost-complex structure on $J(P)$ is
not, itself, integrable, but this is determined by 
the torsion and curvature of the natural $G$-structure
associated to the data $(M,P,S,\iota)$.  
In the particular case of the OP's problem, 
the underlying data $(M,P,S,\iota)$ determine a flat $G$-structure, 
and the almost-complex structure on $J(P)$ is integrable.
Indeed, $J(P)$ is the disjoint union of two copies of $\mathbb{C}^{n+1}\setminus\mathbb{R}^{n+1}$
as complex manifolds.<|endoftext|>
TITLE: Blinking graphs
QUESTION [14 upvotes]: For any simple graph $G$, assign its nodes a weight/bit of $0$ or $1$.
Call this a bit assignment for $G$.
Now, generate a new bit assignment as follows:
Each node $x$'s bit is replaced by $1$ if the sum of the bits of $x$'s
neighbors is odd, and $0$ if that sum is even.
For example, the $3$-cycle $K_3$ with bits $(1,0,0)$ goes to $(0,1,1)$ and then is stable:

          


The $3$-path starting with $(1,0,0)$ goes in three steps to $(0,0,0)$ and then
is stable:

          


Say that $G$ has a blinking assignment if there is a bit assignment that
flips to its complement in one step, and then returns to the original assigment,
forming a cycle of length two:

          


(If animated, the graph would "blink.")

Q. Which graphs have blinking assignments?

For example, $K_3$ has no blinking assignment,
but the star $S_k$ for $k$ odd does.
Many other questions could be asked (e.g., concerning longer cycles),
but I'll focus on the above for now.
Because the update rule is to replace a node's weight with the sum
of its neighbors' weights $\bmod 2$, it seems possible this process has been
studied for some $\mathbb{Z}_n$. If so, I would appreciate a pointer.

Added animation just for fun:

          


          

(Bit assignment thanks to Tony Huynh.)

REPLY [12 votes]: This question seems quite similar to the problem of parity domination. Let $S$ be the set of vertices with label $1$. The condition that $S$ flips to its complement after updating is equivalent to the condition that $|N[v] \cap S|$ is odd for all $v \in V(G)$, where $N[v] = N(v) \cup \{v\}$. Such a set is called an odd dominating set, and it was proved by Sutner that every graph has such a set. In fact, Sutner's proof involved considering a similar CA. See, e.g., this paper for more about parity domination, or see here for Sutner's article.
Asking whether $G$ has a blinking assignment, then, is equivalent to asking whether $G$ has an odd dominating set whose complement is also an odd dominating set. Writing $d_S[v]$ and $d_{G-S}[v]$ for the size of $N[v] \cap S$ and $N[v] \cap (V(G)-S)$ respectively, we need that for every $v \in V(G)$, the quantities $d_S[v]$ and $d_{G-S}[v]$ are both odd. Since $d_S[v] + d_{G-S}[v] = d(v) + 1$, this means every vertex of $G$ needs to have odd degree. On the other hand, if every vertex of $G$ has odd degree, then any odd dominating set should do the trick, and we're guaranteed that one exists.
So, I believe that $G$ should have a blinking assignment if and only if all its vertices have odd degree.

REPLY [8 votes]: Here is an exact characterization of the graphs with blinking assignments.  Let $H$ be an arbitrary bipartite graph with all vertices having odd degree.  Let $(X,Y)$ be a bipartition of $H$.  Now add edges to $H$ to form a new graph $G$ such that all vertices in $G[X]$ and $G[Y]$ have even degree.  A graph constructed in this way has a blinking assignment; namely all vertices in $X$ are assigned $0$ and all vertices in $Y$ are assigned $1$.  
Conversely, every graph with a blinking assignment is constructed in this way. Let $G$ be a graph with a blinking assignment and let $X$ be the vertices of weight $0$ and $Y$ be the vertices with weight $1$. Since zeros become ones, each vertex in $X$ has an odd number of neighbours in $Y$.  Since ones become zeros, each vertex in $Y$ has an even number of neighbours in $Y$.  Repeating the argument for the second iteration, we have that each vertex in $Y$ has an odd number of neighbours in $X$, and each vertex in $X$ has an even number of neighbours in $X$.<|endoftext|>
TITLE: Does the truncated Hausdorff moment problem admit absolutely continuous solutions?
QUESTION [5 upvotes]: Let $\mu$ be a (Borel) probability measure on $[0,1]$ and define $m_j(\mu) = \int x^j\,\mu(dx)$. Let $k$ be a positive integer and consider the set $\mathcal C_{\mu,k}$ of probability measures $\nu$ on $[0,1]$ such that $m_j(\nu) = m_j(\mu)$ for $j = 1,\dotsc,k$.
We are interested in whether $\mathcal C = \mathcal C_{\mu,k}$ contains an absolutely continuous probability measure.
Some restrictions are obviously necessary: When $k=1$, evidently we must rule out $\mu \{0\} = 1$ or $\mu \{1\} = 0$, as either statement implies that $\mathcal C$ contains one point.  More generally, if we define the set 
$$
\mathcal M_k = \{ (m_1(\nu),\dotsc,m_k(\nu) ) \}
$$
of achievable moments,
where $\nu$ ranges over the probability measures on $[0,1]$, then it would seem (though we have not shown) that the boundary $\partial \mathcal M_k$ corresponds to discrete distributions supported by at most $k$ points.
We are aware of some literature on truncated Hausdorff moment problems, and have looked through it, but our particular question does not seem to be addressed.  When explicit representations (say, by densities that are sums of Bernstein polynomials) are used, the emphasis is usually on showing that the moments can be arbitrarily well approximated, but this does not rule out the possibility that the limit is no longer absolutely continuous.

REPLY [2 votes]: Here is a positive answer, for the interior points of the set $M_k:=\mathcal M_k$. Let indeed $c=(c_1,\dots,c_k)$ be any point in the interior of $M_k$. 
Let $P$ stand for the set of all probability measures on $[0,1]$. 
Let us show that then there is an absolutely continuous measure $\nu\in P$ such that $m_j(\nu)=c_j$ for $j=1,\dots,k$. 
For each $\mu\in P$ and each real $h\in(0,1/2)$, define $\mu_h\in P$ by the convolution-like condition that 
$$\int_{[0,1]} f\,d\mu_h=\int_{[0,1/2]}\mu(dx)\int_0^h\frac{du}h\,f(x+u)
+\int_{(1/2,1]}\mu(dx)\int_0^h\frac{du}h\,f(x-u) \tag{1}
$$
for all nonnegative Borel functions $f$ on $[0,1]$. Then $\mu_h$ is clearly absolutely continuous, with density 
$$[0,1]\ni y\mapsto\frac1h\,\mu\big((y-h,y)\cap[0,1/2]\big)+\frac1h\,\mu\big((y,y+h)\cap(1/2,1]\big).$$ 
Since $\mu\mapsto\mu_h$ is a linear operator, the set 
$$M_{k,h}:=\{(m_1(\mu_h),\dots,m_1(\mu_h))\colon\mu\in P\}
$$
is convex.  
It is enough to show that $c=(c_1,\dots,c_k)\in M_{k,h}$ for small enough $h$. Suppose the contrary. Then there is a hyperplane $H$ in $\mathbb R^k$ through the point $c$ such that the set $M_{k,h}$ is contained in a closed half-space $H_+$ whose boundary is $H$. Since $c$ is in the interior of $M_k$, there is a sphere $S_c(r)$ of positive radius $r$ centered at $c$ such that $S_c(r)\subseteq M_k$. Take the point $b=(b_1,\dots,b_k)\in S_c(r)$ that is at distance $r$ from $H_+$ and hence at distance $\ge r$ from the set $M_{k,h}$. 
Since $b\in S_c(r)\subseteq M_k$, there is some measure $\mu\in P$ such that $m_j(\mu)=b_j$ for all $j=1,\dots,k$. 
Since $|(x+u)^j-x^j|\le j|u|\le k|u|$ for $j=1,\dots,k$ and $x,x+u$ in $[0,1]$, by $(1)$ we have $|m_j(\mu_h)-b_k|=|m_j(\mu_h)-m_j(\mu)|\le kh/2$ for $j=1,\dots,k$, so that the distance from the point $(m_1(\mu_h),\dots,m_k(\mu_h))$ to $b$ is $\le k^{3/2}h/2<r$ if $h<2r/k^{3/2}$, which contradicts the condition that $b$ is at distance $\ge r$ from the set $M_{k,h}$. QED 
Addendum: For a point $c=(c_1,\dots,c_k)$ to be in the interior of $M_k$, it is enough that $c_j=m_j(\mu)$ for some measure $\mu\in P$ with at least $k+1$ points of support and all $j=1,\dots,k$; cf. e.g. the lemma in my answer to the question at [moment-matching]. 
This condition of $\mu$ having at least $k+1$ points of support is however not necessary; indeed, for large $k$, the measure $\mu$ representing the point $c$ may have only about $k/2$ points of support -- but those points need to be specially chosen; namely, the points need then to be the so-called roots of a principal representation of $c$; see e.g. Definitions 3.1 and 3.2 and Corollary 3.1 in Ch. II in [Karlin--Studden].
Addendum 2: In particular, it follows from the first sentence of the above Addendum that the condition that the point $c$ lie in the interior of $M_k$ is, not only sufficient, but also necessary for $c$ to be representable by an absolutely continuous probability measure. That is, $c$ is in the interior of $M_k$ if and only if $c_j=m_j(\mu)$ for some absolutely continuous $\mu\in P$ and all $j=1,\dots,k$.<|endoftext|>
TITLE: Simply connected noncompact surfaces
QUESTION [11 upvotes]: Is there a theorem saying that any noncompact, simply connected topological surface is homeomorphic to the plane ? There seems to be many well-known results about the classification of compact surfaces bu I can't find the same kind of results in the noncompact case. I am sorry if it is well-known, I am no geometer.

REPLY [7 votes]: There is in fact a classification of non-compact surfaces, though it is more involved; see https://en.wikipedia.org/wiki/Surface#Non-compact_surfaces for a decent explanation. This classification does indeed imply that a non-compact simply-connected surface is homeomorphic to the plane.<|endoftext|>
TITLE: Admissibility in Heegaard Floer, especially with torsion Spin^c structures
QUESTION [7 upvotes]: I'm confused about the relationship between strong admissibility and weak admissibility for pointed diagrams in Heegaard Floer theory. For reference, here are Ozsváth-Szàbo's original definitions:

A pointed Heegaard diagram is called strongly admissible for the ${Spin}^c$ structure $\mathfrak{s}$ if for every nontrivial periodic domain $\mathcal{D}$ with $$\langle c_1(\mathfrak{s}),H(\mathcal{D})\rangle=2n \geq 0,$$
  $\mathcal{D}$ has some coefficient $>n$. A pointed Heegaard diagram is called weakly admissible for $\mathfrak{s}$ if for each nontrivial periodic domain $\mathcal{D}$ with $$\langle c_1(\mathfrak{s}),H(\mathcal{D})\rangle =0,$$
  $\mathcal{D}$ has both positive and negative coefficients. 

Here's a concrete case that puzzles me: If $c_1(\mathfrak{s})$ is torsion, then it evaluates to zero on every homology class. In that case, it seems that a diagram is 

strongly admissible for $\mathfrak{s}$ if every nontrivial periodic domain $\mathcal{D}$ has some positive coefficient, and
weakly admissible for $\mathfrak{s}$ if every nontrivial periodic domain $\mathcal{D}$ has both positive and negative coefficients.

If these conditions are to coincide for $c_1(\mathfrak{s})$ torsion (or at least for "strong" to be stronger), then it seems like any nontrivial periodic domain with positive coefficients must also have negative coefficients. Is this true?

REPLY [4 votes]: Yes, this is correct. It is spelled out explicitly on page 20 of András Juhász' A survey of Heegaard Floer homology.

Edit: it was not clear to me that you were also asking why, for torsion spin$^c$ structures, strong admissibility implies weak admissibility. Here's a possible proof: if $\mathcal D$ is periodic, then so is $-\mathcal D$. Hence, both $\mathcal D$ and $-\mathcal D$ have some positive coefficient, i.e. $\mathcal D$ has both positive and negative coefficients.<|endoftext|>
TITLE: Degree formalism for line bundles on Deligne-Mumford stacks
QUESTION [7 upvotes]: Let $k$ be an algebraically closed field and let $\mathcal{C}$ be proper, Cohen-Macaulay, purely $1$-dimensional Deligne-Mumford stack over $k$. From looking at section 4.3 on page 135 of the paper "Les schemas de modules des courbes elliptiques" by P. Deligne and M. Rapoport, I understand that there is a  notion of degree for line bundles $\mathscr{L}$ on $\mathcal{C}$. My question is about some of the compatibility claims left to the reader in that reference and has several subquestions listed below.
The degree of $\mathscr{L}$ is defined starting from a "rational section" $f$ of $\mathscr{L}$. My first question is:

What precisely is $f$? How is it defined?

The only reasonable candidate in my mind is a rational section of the pushforward to $\mathscr{L}$ to the coarse space of $\mathcal{C}$, but does this really work? For instance, why can't it happen that this pushforward is zero?
Once we have $f$, we look at strict Henselizations of $\mathcal{C}$ at closed points $x$ and, letting $R_x$ be the coordinate algebra of such a Henselization, we look at the "order" $\mathrm{deg}_x(f)$ with respect to $R_x$ of the pullback of $f$ to the total ring of fractions of $R_x$. In the text it is implicitly claimed that either the pullback of $f$ or its inverse is actually in $R_x$; I cannot see why this is so, but the order makes good sense nevertheless by expressing the pullback as a fraction (presumably at this stage one sees that the Cohen-Macaulayness condition cannot actually be dropped?). One finally defines
$$ \mathrm{deg}(\mathscr{L}) = \sum_x \frac{1}{|\mathrm{Aut}(x)|} \mathrm{deg}_x(f).$$
My other subquestions are:

Why is $\mathrm{\deg}(\mathscr{L})$ independent of $f$?
Why is $\mathrm{\deg}(\mathscr{L})$ invariant under base change to an algebraically closed overfield (I bet this one is easy, but I cannot quite see it)?
Why is $\mathrm{\deg}(\mathscr{L})$ "invariant under specialization", as claimed in $\gamma)$ of the reference? In other words, if $S$ is a spectrum of a DVR and $\widetilde{\mathcal{C}}$ and $\widetilde{\mathscr{L}}$ are over $S$, why is $\mathrm{deg}(\mathscr{\widetilde{L}}_\eta) = \mathrm{deg}(\mathscr{\widetilde{L}}_s)$?

I apologize for so many multiple parts, but comments on any one of them would help.

REPLY [2 votes]: Let us do the example where $\mathscr{M}$ is a proper Deligne-Mumford moduli stack of  varieties. Note that if $\mathscr{M}$ is one-dimensional, then the Cohen-Macauleyness of $\mathscr{M}$ is equivalent to the absence of embedded points in $M$; this assumption together with the properness is to insure that the degree map $\mathrm{Pic}(M) \to \mathbf{Z}$ is well defined.
A line bundle $\mathscr{L}$ over $\mathscr{M}$ in this case is an assignment of each family $f : \mathscr{X} \to S$ of varieties parameterized by $\mathscr{M}$  to a line bundle $\mathscr{L}_f$ over $S$ such that whenever we have a fibered product
\begin{array}{c}
\mathscr{X'} & \to & \mathscr{X} &  \\
 \downarrow& & \downarrow \\
S' & \xrightarrow{p} & S
\end{array}
the line bundles $\mathscr{L}_{f'}$ and $p^*\mathscr{L}_f$ coincide. The first Chern class of $\mathscr{L}$ is defined by picking a rational section $s_f$ of $\mathscr{L}_{f}$ for each $f$ such that $\mathrm{div}(s_{f'}) = p^*\mathrm{div}(s_{f})$ in $\mathrm{Pic}(S')$ for all $p$ as above. 
One can also define a rational section $s$ of $\mathscr{L}$ by picking a rational section $s_f$ of $\mathscr{L}_{f}$ for each $f$ such that the induced map $S \to \mathscr{M}$ is étale, compatible with $\mathscr{L}_{f'} = p^*\mathscr{L}_f$ for all $p$  as above which are necessarily étale. For such an $f$, let $q : S \to M$ be the map induced by the family $f: \mathscr{X} \to S$. The $\mathbf{Q}$-divisor class
$$\frac{1}{\deg(q)} q_*\mathrm{div}(s_f) \in \mathrm{Pic}(M)_\mathbf{Q}$$
is independent of the choices of $s$ and $f$, which is precisely the image of $\mathscr{L}$ under the pushforward map
$$\pi_* : \mathrm{Pic}(\mathscr{M})_\mathbf{Q} \to \mathrm{Pic}(M)_\mathbf{Q}$$
as Niels already mentioned in his comment.
When $\mathscr{M}$ is one-dimensional, the degree of $\mathscr{L} \in \mathrm{Pic}(\mathscr{M})_\mathbf{Q}$ that Deligne and Rapoport define is the degree of $\pi_* \mathscr{L}$, and your questions 2), 3) and 4) follow from the same statements for line bundles over schemes.<|endoftext|>
TITLE: Numerical and topological density
QUESTION [7 upvotes]: Let $\mathbb{N}$ denote the set of positive integers, and let's say that $A\subseteq \mathbb{N}$ is numerically dense if $$\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 1.$$
Is there a topology $\tau$ on $\mathbb{N}$ such that the collection of numerically dense subsets of $\mathbb{N}$ equals the collection of topologically dense subsets of $(\mathbb{N},\tau)$?

REPLY [4 votes]: Suppose we have such a topology on $\mathbb{N}$. Let $\{A_\alpha : \alpha < \mathfrak{c}\}$ be an uncountable almost disjoint family of subsets of $\mathbb{N}$. For each $\alpha$, let
$$B_\alpha = \bigcup\{[2^n,2^{n+1}) : n \in A_\alpha\}.$$
Notice that $\{B_\alpha : \alpha < \mathfrak{c}\}$ is still an almost disjoint family, but it has the additional property that each $B_\alpha$ meets every dense set of our alleged topology. In other words, each $B_\alpha$ has nonempty interior. These interiors must be disjoint, since if $\alpha \neq \beta$ then the complement of $B_\alpha \cap B_\beta$ is dense. Thus, taking interiors, we get an uncountable family of disjoint nonempty subsets of $\mathbb{N}$, a contradiction.<|endoftext|>
TITLE: Is deciding if one planar graph is dual to another really NP-hard (Wikipedia claim)?
QUESTION [15 upvotes]: Wikipedia claims (permanent link) without reference:

Testing whether one planar graph is dual to another is NP-complete.

Another claim with reference:

For any plane graph G, the medial graph of G and the medial graph of the dual graph of G are isomorphic. Conversely, for any 4-regular plane graph H, the only two plane graphs with medial graph H are dual to each other.

One can decide if a planar graph is dual to another by
checking if the medial graphs are isomorphic.
The two Wikipedia claims mean graph isomorphism is NP-complete,
which is unlikely collapse.

Q1 What is wrong with this?

REPLY [23 votes]: The dual graph and medial graph depend on the choice of an embedding in the plane. The Wikipedia claim seems to be that testing whether there are choices of embeddings for which two graphs are dual is NP-complete.
It looks as though Wikipedia makes a distinction between a planar graph (can be embedded in the plane) and plane graph (is embedded in the plane).
Edit: Actually, later on in the Wikipedia article, there is a reference, to
Angelini, Patrizio; Bläsius, Thomas; Rutter, Ignaz (2014), "Testing mutual duality of planar graphs", International Journal of Computational Geometry and Applications 24 (4): 325–346, arXiv:1303.1640
and the precise statement is indeed as I understood.<|endoftext|>
TITLE: $K$ theory and singular cohomology
QUESTION [9 upvotes]: For cell complexes${}^1$ $X$ we have an isomorphism
$$
K^*(X)\otimes \mathbb{Q}\cong H^{*}(X;\mathbb{Q}),
$$
which is induced by the Chern character. 
What is the analogous statement for $KO(X)$?
${}^1$:Hatcher states finite, but I've seen arbitrary CW-complexes stated as well.
edit: The footnote seems wrong, as by the comments.

REPLY [12 votes]: In the following, $X$ is a finite complex. The Adams operator $\psi^{-1}$ (complex conjugation) acts on $K^0(X)$. After inverting $2$, the group $KO^0(X) \otimes \Bbb Z[1/2]$ maps isomorphically to the subset of $K^0(X) \otimes \Bbb Z[1/2]$ fixed by $\psi^{-1}$.
We can then tensor this with $\Bbb Q$. Then $K^0(X) \otimes \Bbb Q \cong \prod_{n \geq 0} H^{2n}(X;\Bbb Q)$. The operator $\psi^{-1}$ acts on this by fixing the factors with $n$ even and negating the factors with $n$ odd. The fixed set is $KO^0(X) \otimes \Bbb Q \cong \prod_{m \geq 0} H^{4m}(X; \Bbb Q)$.<|endoftext|>
TITLE: Lefschetz on étale fundamental group for quasi-projective varieties
QUESTION [7 upvotes]: If $X$ is a smooth projective variety of dimension at least $3$ over $\mathbb{C}$, Lefschetz's Hyperplane theorem says that for every hyperplane section $H$
$$\pi^1(H)\to\pi^1(X)$$
is an isomorphism, where $\pi^1$ can be taken to be the topological fundamental group or the étale one (as it is the profinite completion of the topological).
The same holds for the étale fundamental group over any field in any characteristic (using Leff conditions, e.g. SGA 2, XII Cor. 3.5), when $X$ is a smooth projective variety and $H$ is a regular hyperplane section.
Over the complex numbers, though, more is true: if $X$ is any smooth quasi-projective variety (of dimension at least $3$) and $H$ is a general hyperplane section, then 
$$\pi^1(H)\to\pi^1(X)$$
is again an isomorphism, as far as I know this is proven using Morse Theory (see e.g. Sec. 5.1 in Goresky and MacPherson "Stratified Morse Theory").
My question is:

Could one expect a similar statement in positive characteristic? Is there a counterexample? 

Note that the tame part of the fundamental group should not pose any problem, as there is a Lefschetz theorem for it (once we have a good compactification).
EDIT: I always thought that generic is commonly used for "at the generic point", general for "every rational point of something containing an open" and very general for "every rational point of something dense" (in this case, in the projective space parametrizing hyperplanes). Hence by general I meant for every hyperplane in some open of the parametrizing space.

REPLY [5 votes]: I am posting my comments above as an answer, together with a reference (per the OP's request).  The difference between Will's answer and my answer results from the ambiguous term "general".  Only the OP can specify what is meant by "general" in the question.  In this answer, I interpret "general" as meaning "the property is true for the geometric generic point".
Let $k$ be an algebraically closed field.  Let $X$ be an integral $k$-scheme of dimension $\geq 2$, and let $f:X\to \mathbb{P}^n_k$ be a finite type morphism that is unramified on a dense, open subscheme of $X$.  Denote by $(\mathbb{P}^n_k)^\vee$ the projective space parameterizing hyperplanes in $\mathbb{P}^n_k$.  In particular, for every algebraically closed field extension $\kappa/k$, there is a natural bijection between the hyperplanes $H\subset \mathbb{P}^n_\kappa$ and the $k$-morphisms $[H]:\text{Spec}(\kappa) \to (\mathbb{P}^n_k)^\vee$.  
Let me restate Corollary 2.2 of the following (which, in turn, depends on Théorème 4.10 and 6.10 of Jouanolou's "Théorèmes de Bertini et applications").
MR3114946 
Graber, Tom; Starr, Jason Michael 
Restriction of sections for families of abelian varieties. 
A celebration of algebraic geometry, 311–327, 
Clay Math. Proc., 18, Amer. Math. Soc., Providence, RI, 2013. 
For every generically finite morphism $g:Y\to X$ that admits no rational section, there exists a dense, Zariski open subscheme $U\subset (\mathbb{P}^n_k)^\vee$ such that for every algebraically closed field $\kappa$ and for every geometric point $[H]:\text{Spec}(\kappa) \to U$, also $Y\times_{\mathbb{P}^n_k} H \to X\times_{\mathbb{P}^n_k} H$ admits no rational section.
Now let $K$ denote the algebraic closure of the function field of $(\mathbb{P}^n_k)^\vee$ as an extension of $k$, and denote by $[H]:\text{Spec}(K)\to (\mathbb{P}^n_k)^\vee$ the corresponding $k$-morphism.  Then for every connected, finite, etale morphism $g:Y\to X$, also the morphism $Y\times_{\mathbb{P}^n_k} H \to X\times_{\mathbb{P}^n_k} H$ is connected, finite and etale.  Therefore, the induced homomorphism, $$\pi^1(X\times_{\mathbb{P}^n_k} H) \to \pi^1(X),$$ is surjective.<|endoftext|>
TITLE: Topology in space of test functions $\mathcal{D}(\Omega)$ and space of distributions $\mathcal{D}'(\Omega)$
QUESTION [6 upvotes]: We can concluded that $\mathcal{D}(\Omega):=\bigcup_{K \in \mathcal{K}(\Omega)} \mathcal{D}_K(\Omega)$ (where $\mathcal{K}(\Omega)$ denotes the union of all compacts set content in a open subset $\Omega \subset \mathbb{R}^n$) it's a countable union of Fréchet spaces, and $\mathcal{D}(\Omega)$ is a locally convex space not metrizable.
 We can also conclude that space of distributions $\mathcal{D}'(\Omega)$ is a locally convex space.
More specifically, for the second question, it's correct to say that $(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ is a dual pair, since $\langle \varphi , u \rangle := u(\varphi)=0$ $\forall u \in \mathcal{D}'(\Omega)$ implies $\varphi=0$ in the sense that since $\mathcal{D}(\Omega) \subset L^1_{loc}(\Omega) \subset \mathcal{D}'(\Omega)$ we have $\int_{\Omega} |\varphi|^2 dx = \langle \varphi , \overline{\varphi} \rangle =0$. 
Therefore, $\mathcal{D}'(\Omega)$ is equipped with the weak* topology $\sigma(\mathcal{D}'(\Omega), \mathcal{D}(\Omega))$ making it a locally convex space, with topology defined by sufficient (separable) family of seminorm $\mathcal{F}:=\lbrace p_{\varphi}(u):=|u(\varphi)| : \varphi \in \mathcal{D}(\Omega) \rbrace $ which it is the restriction ad $\mathcal{D}'(\Omega) \subset \mathbb{K}^{\mathcal{D}(\Omega)}:=\prod_{x \in \mathcal{D}(\Omega)} \mathbb{K}$  of the product topology or the topology of pointwise convergence. In particular, it is the minimal topology that makes continuous distributions: okay, $\mathcal{D}'(\Omega)$ is a locally convex space, it's correct?
Therefore we say that $u_k \rightarrow u$ if $\langle \varphi , u_k \rangle \rightarrow \langle \varphi , u \rangle$ $\forall \varphi \in \mathcal{D}(\Omega)$. Many authors, call this property "sequential continuity" of a distributions sequence $\lbrace u_k \rbrace \subset \mathcal{D}'(\Omega)$, but it would be more appropriate to call it weak* convergence. I do not know if you agree.
I'm getting another doubt, that in the space of Fréchet $\mathcal{D}_K(\Omega)$  is defined topology $\mathcal{T}_K$, through seminorm
$\displaystyle p_N(\varphi)=\sup_{|\alpha| \leq N} \left \| D^\alpha \varphi \right \|_{K_{N}}$ , $\displaystyle \left \| D^\alpha \varphi \right \|_{K_{N}}:=\sup_{x \in K} |D^\alpha \varphi|$ 
and $\varphi_k \rightarrow \varphi$ in $\mathcal{D}(\Omega)$ if and only if
(1) $\exists K \in \mathcal{K}(\Omega): \mathrm{supp}(\varphi_k), \mathrm{supp}(\varphi) \subset K$ $\forall k \in \mathbb{N}$. 
(2) $D^\alpha \varphi_k \rightarrow D^\alpha \varphi$ uniformly on $K$ $\forall \alpha \in \mathbb{N}^n$
By definition, a distribution is a continuous linear functional than the previous convergence.
In particular, this convergence has compared a locally convex topology $\mathcal{T}$ on $\mathcal{D}(\Omega)$, where it's local base $\mathcal{U}$ is formed by the family of all convex balanced subsets $U$ such that $U \cap \mathcal{D}_K(\Omega)$ is an open subset in $\mathcal{D}_K(\Omega)$. 
Now, on $\mathcal{D}(\Omega)$, there is also the weak topology $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega))$ defined by sufficient (separable) family of seminorm  $\mathcal{F}=\lbrace p_u(\varphi)=|u(\varphi)| : u \in \mathcal{D}'(\Omega)\rbrace$ and $\varphi_k \rightharpoonup \varphi$ if $u(\varphi_k) \rightarrow u(\varphi)$ $\forall u \in \mathcal{D}'(\Omega)$.
then we can say that $\sigma(\mathcal{D}(\Omega), \mathcal{D}'(\Omega)) \subset \mathcal{T}$ ?

REPLY [5 votes]: In Schwartz's Théorie des distributions, chapter III, Théorème VII : $\mathcal D$ is a Montel space, where bounded sets are relatively compact. Then the weak and strong topologies, restricted to bounded sets, coincide, and convergent sequences are the same in these two topologies (and also in weaker Hausdorff topologies). In more concrete terms: whenever a sequence $\varphi_k$ is such that $u(\varphi_k)\to u(\varphi)$ $\forall u\in \mathcal D'$, there does exist a compact $K\subset\Omega$ such that $\varphi_k\to\varphi$ in the Fréchet space $\mathcal D_K(\Omega)$.<|endoftext|>
TITLE: Schur multiplier of $Sp(2g, \mathbb{Z}/2)$ for $g \geq 3$
QUESTION [14 upvotes]: This question is about the computation of $H_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z})$, where $Sp(2g, \mathbb{Z}/2)$ is the group of symplectic $2g \times 2g$ matrices over $\mathbb{Z}/2$. 
With respect to this computation I have seen quotations to the following two papers:

Stein "Surjective stability in dimension $0$ for $K_2$ and related functors" (https://www.dropbox.com/s/jcy7sep2hpxy7gt/Stein.pdf?dl=0) 
Steinberg "Générateurs, relations et revêtements de groupes algébriques"
(https://www.dropbox.com/s/jidchwwjssnmmlm/Steinberg1.pdf?dl=0)

According to the quotations of Stein's and Steinberg's papers the result is that
            $H_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z}) = 0$ for $g \geq 3$. 
My two concerns are:
1) The papers by Stein and Steinberg do not directly give the computation of  $H_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z})$. So my first question is if it is really possible to extract this result from these papers?
2) There is a discrepancy in the case $g=3$ between the quotations of Stein's and Steinberg's papers - which say that $H_2(Sp(6, \mathbb{Z}/2), \mathbb{Z}) =0$ - and the computation of this group done with GAP (the computational system for discrete algebra). GAP gives $H_2(Sp(6, \mathbb{Z}/2), \mathbb{Z}) = \mathbb{Z}/2$.  Checking on GAP, the process that the computer goes through to get the computation looks correct. Is there an error in Stein's and/or Steinberg's paper or is it a misinterpretation of the case $g=3$ in the quotation?

REPLY [8 votes]: I've been corresponding via email with the OP about this (it is a paper of mine that she got these citations from), and she asked me to post an answer summarizing what I told her.  I apologize for the length of this answer -- this is really quite a long story.  I also apologize for sometimes butchering people's names.  I am copying this answer from a long document I sent the OP, and unfortunately autocorrect screws things up without telling me (eg it wants to call Steinberg "Sternberg", though I hope caught most of those occurrences).
$\DeclareMathOperator{\Sp}{Sp} \DeclareMathOperator{\ESp}{ESp} \DeclareMathOperator{\SL}{SL} \newcommand\Z{\mathbb{Z}} \DeclareMathOperator{\HH}{H} \newcommand\tG{\widetilde{G}} \newcommand\tD{\widetilde{D}} \newcommand\Field{\mathbb{F}} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\St}{St}$ 
For $\Sp_{2g}(\Z/2)$, the correct theorem is that $\HH_2(\Sp_{2g}(\Z/2)) = 0$ for $g \geq 4$.  This should be attributed to Steinberg and is contained in the paper cited by the OP.  More generally, Steinberg showed that a similar theorem holds for $\Sp_{2g}(\Field_q)$.  What Stein did in the cited paper was show how to extend what Steinberg did to $\Sp_{2g}(\Z/k)$ where $k$ is not prime.  They both in fact dealt not just with the symplectic group, but also with more general finite Chevalley groups.
Another good reference for Steinberg's work is sections 6 and 7 of Steinberg's Yale lecture notes, which were never published but which are available here.
It is not easy to extract the above homological statement from Steinberg and Stein's papers since they are written in the language of algebraic K-theory.  What they are imitating is the calculation of $\HH_2(\SL_n(\Field_q))$ that is described in Milnor's book on algebraic K-theory, which I highly recommend reading.
To help you understand these papers, below I have written a guide to the calculation of $\HH_2(\SL_n(\Field_q))$ from Milnor's book.

I will begin by recalling the theory of universal central extensions.  Let $G$ be a group.  A central extension of $G$ is a group $\tG$ together with a short exact sequence
$$1 \longrightarrow C \longrightarrow \tG \longrightarrow G \longrightarrow 1$$
such that $C$ is contained in the center of $\tG$.  This central extension is a universal central extension
if for any other central extension
$$1 \longrightarrow C' \longrightarrow \tG' \longrightarrow G \longrightarrow 1,$$
there exists a unique homomorphism $\tG \rightarrow \tG'$ such that the diagram
$\require{AMScd}$
$$\begin{CD}
1 @>>> C @>>> \tG @>>> G @>>> 1 \\
@.     @VVV   @VVV     @VV{=}V   @. \\
1 @>>> C' @>>> \tG' @>>> G @>>> 1
\end{CD}$$
commutes.  The usual argument shows that universal central extensions are unique if they exist, but they might
not exist.  The following theorem summarizes their properties.  A proof of it can be found in Theorem 5.7 and Corollary 5.8 of Milnor's book
Theorem 1: Let $G$ be a group.  Then $G$ has a universal central extension
$$1 \longrightarrow C \longrightarrow \tG \longrightarrow G \longrightarrow 1$$
if and only if $\HH_1(G;\Z) = 0$, in which case we have $C \cong \HH_2(G;\Z)$.
For perfect groups, this reduces the computation of $\HH_2(G;\Z)$ to the construction of the universal central
extension of $G$.

I now describe what happens in "infinite rank".  Let $R$ be a ring.  Define
$$\GL(R) = \bigcup_{n=1}^{\infty} \GL_n(R)$$
and let $E(R)$ be the subgroup of $\GL(R)$ generated by elementary matrices.
The low-dimensional homology groups of $\GL(R)$ and $E(R)$
are closely connected to the algebraic K-theory of $R$.  In particular,
$K_1(R)$ is by definition equal to $\HH_1(\GL(R);\Z)$.
It also turns out that $K_2(R) = \HH_2(E(R);\Z)$, but this
is a theorem rather than a definition.  To understand what happens in finite rank, we'll have to understand
where this comes from.
There a group $\St(R)$ called the Steinberg group.
It is defined via generators and relations, and informally can be described as the group generated by
elementary matrices in $\GL(R)$ with the ``obvious relations'' between elementary matrices.  Its precise
definition is as follows.

The generators are symbols $e_{ij}(r)$ with $i,j \geq 1$ distinct
and $r \in R$.  This corresponds to the elementary matrix obtained from the identity matrix by putting an $r$ in
position $(i,j)$.
The relations are as follows.

For distinct $i,j \geq 1$ and $r,r' \in R$, we have $e_{ij}(r)
   \cdot e_{ij}(r') = e_{ij}(r+r')$.
For distinct $i,j,k \geq 1$ and $r,r' \in R$, we have
$[e_{ij}(r),e_{jk}(r')] = e_{ik}(r r')$.
For distinct $i,j \geq 1$ and distinct $k,l \geq 1$ such that $j
    \neq k$ and $i \neq l$, we have $[e_{ij}(r),e_{kl}(r')] = 1$ for all
$r,r' \in R$.


Since all these relations hold in $\GL(R)$, there is a group homomorphism $\St(R) \rightarrow \GL(R)$ whose
image is $E(R)$.  By definition, $K_2(R)$ is the kernel of this homomorphism, so we have a short
exact sequence
$$1 \longrightarrow K_2(R) \longrightarrow \St(R) \longrightarrow E(R) \longrightarrow 1.$$
The main theorem concerning the Steinberg group is as follows (see Theorem 5.10 of Milnor's book).
Theorem 2: Let $R$ be a ring.  Then the extension
$$1 \longrightarrow K_2(R) \longrightarrow \St(R) \longrightarrow E(R) \longrightarrow 1$$
is the universal central extension of $E(R)$.  In particular, $K_2(R)$ is an abelian group
and $\HH_2(E(R);\Z) = K_2(R)$.
This allows many concrete calculations.  One very simple one is as follows.
Example: Since $\SL(\Z)$ is generated by elementary matrices, we have $E(\Z) = \SL(\Z)$.  The group
$\HH_2(\SL(\Z);\Z)$ is then cyclic of order $2$.  Identifying it with $K_2(\Z)$, the generator
is $(e_{12}(1) e_{21}(1)^{-1} e_{12}(1))^4$.  Here the matrix
$$e_{12}(1) e_{21}(1) e_{12}(1) = \left(\begin{matrix}
0 & 1 \\
-1 & 0 
\end{matrix}\right)$$
is the one that rotates the plane by $90$ degrees, and hence has order $4$.
Assume now that $R$ is a commutative ring and let $u,v \in R^{\ast}$ be units.  Define
$$D_u = \left(\begin{matrix}
u & 0 & 0 \\
0 & u^{-1} & 0 \\
0 & 0 & 1
\end{matrix}\right)$$
and
$$D_v' = \left(\begin{matrix}
v & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & v^{-1}
\end{matrix}\right).$$
It is not hard to see that diagonal matrices like these can be written as products of elementary matrices, so
$D_u, D_v' \in E(R)$.  What is more, the matrices $D_u$ and $D_v'$ commute.  Letting $\tD_u$ and $\tD_v'$ be
lifts of $D_u$ and $D_v'$ to $\St(R)$, we obtain an element
$[\tD_u,\tD_v'] \in K_2(R)$, which will be called a symbol and denoted $\{u,v\}$.  It is not hard to see
that this does not depend on the choice of lifts.
The following theorem says that for fields, the symbols generate $K_2(R)$; see Corollary 9.13 of Milnor's book.
Theorem 3: Let $\Field$ be a field.  Then $K_2(\Field)$ is generated by the set of symbols $\{u,v\}$ as $u$ and $v$
range over $\Field^{\ast}$.
For finite fields, the final piece of the puzzle is as follows; see Corollary 9.9 of Milnor's book.
Theorem 4: Let $\Field$ be a finite field.  Then $\{u,v\} = 0$ for all $u,v \in \Field^{\ast}$.
Combining everything above with the fact that $\SL(\Field)$ is generated by elementary matrices for a field $\Field$,
we deduce the following theorem.
Theorem 5: Let $\Field$ be a finite field.  Then $\HH_2(\SL(\Field);\Z) = 0$.

Of course, what we are really interested in is $\GL_n(R)$, not $\GL(R)$.  Define $E_n(R)$ and $\St_n(R)$ in the obvious
way.  There is still a surjection $\St_n(R) \rightarrow E_n(R)$; denote its kernel by $C_n(R)$, so we have a short
exact sequence
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1.$$
Associated to the natural inclusion $\GL_n(R) \hookrightarrow \GL_{n+1}(R)$ used to define $\GL(R)$ are an inclusion
$E_n(R) \hookrightarrow E_{n+1}(R)$ and homomorphisms $\St_n(R) \rightarrow \St_{n+1}(R)$ and $C_n(R) \rightarrow C_{n+1}(R)$
that fit into a commutative diagram
$$\begin{CD}
1 @>>> C_n(R) @>>> \St_n(R) @>>> E_n(R) @>>> 1 \\
@. @VVV @VVV @VVV @. \\
1 @>>> C_{n+1}(R) @>>> \St_{n+1}(R) @>>> E_{n+1}(R) @>>> 1.
\end{CD}$$
It is clear that $\St_n(R)$ is the limit of
$$\St_1(R) \rightarrow \St_2(R) \rightarrow \St_3(R) \rightarrow \cdots$$
and that $K_2(R)$ is the limit of
$$C_1(R) \rightarrow C_2(R) \rightarrow C_3(R) \rightarrow \cdots.$$
In the ``ideal'' situation, we would have theorems of the following form.

For $n$ sufficiently large, the exact sequence
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1$$
is the universal central extension of $E_n(R)$; and hence $C_n(R)$ is an abelian group and $\HH_2(E_n(R);\Z) = C_n(R)$.
For $n$ sufficiently large, the map $C_n(R) \rightarrow C_{n+1}(R)$ is surjective (this is called surjective stability).
For $n$ sufficiently large, the map $C_n(R) \rightarrow C_{n+1}(R)$ is injective (this is called injective stability).

If these conditions are satisfied, then we would have $\HH_2(E_n(R);\Z) = K_2(R)$ for $n$ sufficiently large.
Unfortunately, one can give examples where these fail.  However, they do hold for many rings; in particular, they
hold for fields.  Our goal is
to talk about finite fields, so we will not try to give particularly general statements.  We begin with the
bit about being a universal central extension.  The proof of Theorem 2 can
be followed to deduce the following.
Theorem 6: Let $R$ be a ring and let $n \geq 5$.  Assume that $C_n(R)$ is contained in the center of $\St_n(R)$.  Then the extension
$$1 \longrightarrow C_n(R) \longrightarrow \St_n(R) \longrightarrow E_n(R) \longrightarrow 1$$
is the universal central extension of $E_n(R)$.  In particular,
$\HH_2(E_n(R);\Z) = C_n(R)$.
As the following theorem shows, the condition in this theorem is satisfied for fields; see Theorem 9.12 of Milnor's book.
Theorem 7: Let $\Field$ be a field.  Then $C_n(\Field)$ is contained in the center of $\St_n(\Field)$ for $n \geq 3$, and hence
$\HH_2(\SL_n(\Field);\Z) = C_n(\Field)$ for $n \geq 5$
Remark: In fact, it turns out that
$$1 \longrightarrow C_n(\Field) \longrightarrow \St_n(\Field) \longrightarrow \SL_n(\Field) \longrightarrow 1$$
is the universal central extension of $\SL_n(\Field)$ for $n \geq 3$ except for the following exceptions:

$n=3$ and $\Field = \Field_2$, and
$n=3$ and $\Field = \Field_4$, and
$n=4$ and $\Field = \Field_2$.

This was proved by Steinberg.
We now turn to injective and surjective stability.  The key is the notion of symbol.  If $R$ is a commutative ring
and $u,v \in R^{\ast}$ are units, then we can define a symbol $\{u,v\}_n \in C_n(R)$ for any $n \geq 3$ in the obvious
way.  We then have the following; see Theorem 9.11 of Milnor's book.
Theorem 8: Let $R$ be a commutative ring and let $n \geq 3$.  Assume that $C_n(R)$ is contained in the center of $\St_n(R)$.
Then $C_n(R)$ is generated by the set of symbols $\{u,v\}_n$ as $u$ and $v$ range over the units of $R$.
Remark: Theorem 7 and Theorem 8 combine to show that if $\Field$
is a field, then $K_2(\Field)$ is generated by symbols $\{u,v\}$ for $u,v \in \Field^{\ast}$.  This is
precisely Theorem 3 above, and in fact this is how Theorem
3 is proved.
The proof of Theorem 4 above also works for the symbols $\{u,v\}_n$ and
gives the following.
Theorem 9: Let $\Field$ be a finite field and $n \geq 3$.  Then $\{u,v\}_n = 0$ for all $u,v \in \Field^{\ast}$.
Combining everything above, we deduce the following.
Theorem 10: Let $\Field$ be a finite field.  Then $\HH_2(\SL_n(\Field);\Z) = 0$ for $n \geq 5$.
Remark: In fact, as in the remark after Theorem 7, one can show that
$\HH_2(\SL_n(\Field);\Z) = 0$ for $\Field$ a finite field and $n \geq 3$ except for the following exceptions:

$n=3$ and $\Field = \Field_2$, and
$n=3$ and $\Field = \Field_4$, and
$n=4$ and $\Field = \Field_2$.<|endoftext|>
TITLE: Why should intersection cohomology and quantum cohomology be related for a symplectic resolution?
QUESTION [14 upvotes]: In http://arxiv.org/pdf/1410.6240.pdf M. McBreen and N. Proudfoot conjectured a precise relationship between the quantum cohomology of a symplectic resolution and the intersection cohomology of the cone which it is resolving. However, I could not find the motivation for this conjecture in the above article. Is there a heuristic reason why the two sides should be related?

REPLY [8 votes]: I'll try to explain the genesis of the paper, then some intuition we developed later.
The original motivation was a coincidence. For $X$ hypertoric, Tom Braden and Nick Proudfoot had defined a ring structure on $IH^*(X^{aff})$ whereas Daniel Shenfeld and I found a generators & relations presentation of $QH_{\mathbb{C}^*}(X)$. When the quantum parameter $q$ is specialized to the 'most singular value' $q_0$, our two rings coincide.
We ended up finding something much stronger. Namely, the decomposition theorem tells us $H^*(X)= IH^*(X^{aff}) \oplus K$ for some perverse sheaf $K$, and hence gives an inclusion $IH^*(X^{aff}) \to H^*(X)$. The latter maps to a `polynomial subbring' $QH^*_{\mathbb{C}^*}(X)_{pol} \subset QH^*_{\mathbb{C}^*}(X)$, which admits a specialization $QH^*_{\mathbb{C}^*}(X)_{pol} \to R'$ at $q=q_0$.
In our paper, we show that for $X$ hypertoric, the composition
$$ IH^*(X^{aff}) \to H^*(X) \to QH^*_{\mathbb{C}^*}(X)_{pol} \to R' $$
is an isomorphism. This is (to me) still rather surprising! Neither surjectivity nor injectivity are given to you for free, and the fact that they hold for all hypertorics is very suggestive.
There are nonetheless some general heuristics. For $X$ a symplectic resolution, the Steinberg algebra $A = H^{top}(X \times_{X^{aff}} X)$ acts on $H^*(X)$, and the trivial isotypical component of this module is exactly $IH^*(X^{aff})$. In fact, the Steinberg algebra decomposes as $\mathbb{C} id \oplus A^+$, where $id$ is the diagonal correspondence, and in the decomposition
$$ H^*(X) = IH^*(X^{aff}) \oplus K $$
we have $K = image(A^+)$.
The conjecture can be roughly reformulated as `specializing $q=q_0$ kills the image of $A^+$'. 
Now, let $u \in QH^2_{\mathbb{C}^*}(X)$. Then [BravermanMaulikOkounkov] tells us that the operator of quantum product by $u$, viewed as a function of the quantum parameter $q \in H^2(X,\mathbb{C}^*)$, has poles along certain divisors $D_{\alpha} \subset H^2(X,\mathbb{C}^*)$. The residues are Steinberg operators $Z_{\alpha} \in A^+$.
For many symplectic resolutions, the $D_{\alpha}$ intersect in a 'maximally singular point $q_0$' (when this is not the case, our conjecture runs into some trouble and needs to be corrected). 
One would like to say: specializing $q$ to $D_{\alpha}$ kills the image of $Z_{\alpha}$ in $H^*(X)$; specializing to $q_0$ kills the image of all $Z_{\alpha}$. In fact, modulo some technical difficulties it seems likely one can prove such a statement.
Hence, very roughly, our conjecture ressembles the claim that the $Z_{\alpha}$ generate $A^+ \subset A$, or at least the set of idempotents in $A^+$. I do not know of a general reason why the $Z_{\alpha}$ should generate, but perhaps you prefer this mystery to the one we started with.
There are also some motivations coming from representation theory, which I won't go into here as they require a lot of set-up.
More speculatively, the limit $q \to q_0$ should correspond to a singular degeneration $Y \to Y_0$ of the mirror variety of $X$. Since symplectic resolutions are almost 'self-mirror', one may hope that the cone singularity of $X^{aff}$ appears in $Y_0$. Perhaps one can directly identify $R'$ with the intersection cohomology of this singularity. Of course, for now there are too many difficulties to make this precise.<|endoftext|>
TITLE: Upper bound on the number of convex connected components of the complement of the zero set of a polynomial
QUESTION [7 upvotes]: The classic Milnor-Thom upper bound on the sum of the Betti numbers of real algebraic sets (for a nice exposition and references, see e.g. N.R. Wallach, On a Theorem of Milnor and Thom, in S. Gindikin (ed.), Topics in Geometry: In Memory of Joseph D'Atri, Progress in Nonlinear Differential Equations and their Applications 20 (Birkhäuser, 1996), pp. 331-348) allows one to give an upper bound on the number of connected components of the complement 
$$
U_P=\{\xi\in\mathbb{R}^n\ |\ P(\xi)\neq 0\}
$$
of the zero set of a real polynomial function $P:\mathbb{R}^n\rightarrow\mathbb{R}$, since $U_P$ is homeomorphic to the real algebraic subset $\tilde{U}_P=\{(\xi,\lambda)\in\mathbb{R}^{n+1}\ |\ \lambda P(\xi)=1\}$ of $\mathbb{R}^{n+1}$ through the map 
$$
U_P\ni\xi\mapsto\left(\xi,\frac{1}{P(\xi)}\right)\in\tilde{U}_P\ ,
$$
whose inverse is simply the (restriction of the) projection $$\mathbb{R}^{n+1}\ni(\xi_1,\ldots,\xi_n,\xi_{n+1})\mapsto(\xi_1,\ldots,\xi_n)\in\mathbb{R}^n$$ (to $\tilde{U}_P$).
The aforementioned bound tells us that if the degree of $P$ is $r$, then the sum of the Betti numbers of $U_P$ (and, therefore, its number of connected components) is bounded above by $(r+1)(2r+1)^n$.

Question: is there an upper bound on the number of convex connected components of $U_P$ which is sharper than the Milnor-Thom bound but also only depends on $n$ and $r$?

My intuition is that the Milnor-Thom bound is too crude to this end, even in the exceptional case when all connected components of $U_P$ are convex - take, for instance, $P(\xi)=\prod^n_{j=1}\xi_j$, in which case $r=n$ but the connected components of $U_P$ are precisely the $2^n$ orthants 
$$
\mathbb{R}^n_I\doteq\{\xi\in\mathbb{R}^n\ |\ \xi_j>0\text{ if }j\in I\ ,\,\xi_j<0\text{ if }j\not\in I\}\ ,\quad I\subset\{1,\ldots,n\}
$$
of $\mathbb{R}^n$, which are clearly convex. More generally, if $P(\xi)$ is the elementary symmetric polynomial of order $1\leq r\leq n$ $$P(\xi)=\sum_{1\leq j_1<\cdots<j_r\leq n}\xi_{j_1}\cdots\xi_{j_r}\ ,$$ then $U_P$ has $2^r$ connected components, all of them convex.
My motivation for the above question is related to Lars Gårding's theory of hyperbolic polynomials (see e.g. L. Gårding, An Inequality for Hyperbolic Polynomials, J. Math. Mech. 8 (1959) 957-965). Recall that an homogeneous polynomial $P:\mathbb{R}^n\rightarrow\mathbb{R}$ is said to be hyperbolic with respect to $0\neq\tau\in\mathbb{R}^n$ if $P(\tau)\neq 0$ and the roots of the one-variable polynomial $P_{\xi,\tau}(\lambda)\doteq P(\xi-\lambda\tau)$ are all real. The hyperbolicity cone $C(P,\tau)$ to which $\tau$ belongs is the connected component of $U_P$ to which $\tau$ belongs. Gårding has shown that 
$$
C(P,\tau)=\{\xi\in\mathbb{R}^n\ |\ \text{all roots of }P_{\xi,\tau}\text{ are positive}\}
$$
and, as a consequence, that $C(P,\tau)$ is an open convex cone. Roughly speaking, dual cones of hyperbolicity cones provide upper bounds to the propagation speed of the support of solutions of hyperbolic partial differential equations with principal symbol $i^rP$.
Due to the homogeneity of $P$, hyperbolicity cones always come in opposite pairs - more precisely, if $P$ is hyperbolic with respect to $\tau$, then it also is with respect to $-\tau$, and $C(P,-\tau)=-C(P,\tau)$. Since $P$ in the above examples is hyperbolic with respect to any $\tau$ in $U_P$, we conclude that the number of hyperbolicity cones of an hyperbolic homogeneous polynomial $P:\mathbb{R}^n\rightarrow\mathbb{R}$ is an even number bounded above by the number of convex connected components of $U_P$, and such an upper bound is sharp. Therefore, a positive answer to the above question would provide an upper bound to the number of hyperbolicity cones of $P$ which is sharper than the Milnor-Thom bound.
Update: This is not exactly an answer to my question (hence it does not appear as a separate answer) but rather to the original question that motivated it - to wit, a sharp upper bound on the number of hyperbolicity cones of an (homogeneous) hyperbolic polynomial. Since the result is connected to Aaron Meyerowitz's bounty-awarded answer, I feel it is worthwhile to bring it up here.
Thorsten Jörgens and Thorsten Theobald have just posted (March 15th, 2017) the arXiv preprint Hyperbolicity Cones and Imaginary Projections, arXiv:1703.04988 [math.AG].
The main result of the paper (Theorem 1.1) states (I have retained only the part of the statement that interests us here and added a short explanation on the notation):

Let $P\in\mathbb{R}[\mathbf{z}]$ be an homogeneous $n$-variable polynomial 
  of degree $r$ with real coefficients. Then the number of hyperbolicity cones 
  of $P$ is at most 
  $$\mathcal{H}(n,r)=\begin{cases} 2^r & \text{for }r\leq n\ ,\\ 2\sum^{n-1}_{k=0}\binom{r-1}{k} & \text{for }r>n\ .\end{cases}$$ 
  The maximum is attained if and only if $P$ is a product of linear 
  polynomials.

To get a feeling for the saturation of the above bound, we point out that $\mathcal{H}(n,r)$ is exactly the number of connected components of the complement of a central arrangement of hyperplanes in general position (here "central" means that all hyperplanes in the arrangement contain the origin) - the centrality comes from the fact that we are requiring here that $P$ is homogeneous. Notice that the saturation of the bound occurs precisely as conjectured in Boris Bukh's comment below (in the homogeneous case).
Of course, the conjecture of whether the above bound also applies to the total number of connected components of the complement of the zero set of an homogeneous polynomial $P$ remains. If so, the bound will also be sharp.

REPLY [4 votes]: It may be that the bound is only fairly good for $r \ll n.$ This is supported by the following which is an elaboration on the comment by Boris Bukh. 
We can get a lower bound of 
$$ f(r,n)=\sum_{j=0}^n \binom{r}{j}.$$ This is the number of regions (all of which happen to be convex) in the complement of the union of $r$ hyperplanes in general position. This would be $$\{\xi\in\mathbb{R}^n\ |\ P(\xi)\neq 0\}$$ for $P$ the product of $r$ linear polynomials.

Interesting, but not obviously relevant to this question is that for such a $P$, when $n \ge 2$,
  the surface   $$\{\xi\in\mathbb{R}^n\ | \ |P(\xi)| = \epsilon\}$$ where
  $\epsilon \gt 0$ is small enough,  has $ {f(n,r)} $
  components.  This is the union of two polynomial surfaces $P(\xi) = \pm \epsilon$ each with about half that many components.

I had the vague impression of hearing that both bounds are optimal, at least in the case $n=2.$ But I have nothing to go on there. Are any examples known where the number of components exceeds $\mathbf{ f(n,r)}?$
As you note, $f(n,r)=2^r$ for $1 \le r \le n$ and this is quite a bit less than $(r+1)(2r+1)^n.$ Again, it may be that the bound is only fairly good for $r \ll n.$ Then (as always) the upper bound is $O(r^{n+1})$ and the lower bound is $O(r^n)$.
Certainly for $n=1,$ The zero set of a degree $r$ polynomial can divide $\mathbb{R}^1$ into $f(1,r)=r+1$ intervals but no more.
In the case that $2n \le r$ we have $$\binom{r}n \lt f(n,r) \lt \binom{r}n \frac{r-n+1}{r-2n+1}. $$ The upper bound is found in Sum of 'the first $k$' binomial coefficients for fixed $n$ (which we would prefer to call Sum of 'the first $n+1$' binomial coefficients for fixed $r$) and is quite good for $n \lt \frac{r}2+\sqrt{r}.$<|endoftext|>
TITLE: Example to $2^\kappa\nrightarrow (3)^2_\kappa$, plus closed walks of odd length?
QUESTION [6 upvotes]: Let $\kappa$ be an infinite cardinal. Consider the following example to $2^\kappa\nrightarrow (3)^2_\kappa$.
$V$ is a set of vertices, each of which is an element of $2^\kappa$. Color the edge between two vertices $f,g\in 2^\kappa$ by the least ordinal on which $f,g$ disagree. It follows that there is no homogeneous set of size $3$, i.e. there is no triangle with all three edges the same color. 
More is true: There is no closed walk of length 3,5,7,... which is monochromatic, i.e. all edges are the same color.
Definition:A closed walk consists of a sequence of vertices starting and ending at the same vertex.
Question: Can you find an example to $2^\kappa\nrightarrow (3)^2_\kappa$, therefore no monochromatic triangles are allowed, but which has closed walks of length $n$, for any odd number $n>3$?

REPLY [3 votes]: The answer is yes. In fact, we can get an example by slightly modifying your example.
Let $V = 2^\kappa \cup \{a,b,c,d,e\}$. Color the edges $ab$, $bc$, $cd$, $de$, and $ea$ with some ordinal $< \kappa$, say $0$. Color all other edges from $a$ with a different color, say $A$. Color all other edges from $b$ with a different color, say $B$. Similarly for $c$, $d$, and $e$. The remaining edges are all between elements of $2^\kappa$, and we color them just as you did in your example, with the least ordinal on which they disagree. This gives a coloring of $V$ with $\kappa$ colors, namely $\kappa \cup \{A,B,C,D,E\}$. We have a $5$-cycle in color $0$. We obviously have no triangles in colors $A,B,C,D,E$, and we have no triangles in our other colors $> 0$ because there are no triangles in your example. We have no triangles in color $0$ because in that color we only added a $5$-cycle that is not connected to anything else.
This shows that we can extend your example to one including a $5$-cycle in color $0$ by using five new colors and five new vertices. But we didn't use any specific properties of your example -- any example would have done just fine -- and the number $5$ wasn't special either. So after adding a $5$-cycle in color $0$, we can then add a $7$-cycle some other color, a $9$-cycle in some other color, etc. In fact, we can add monochromatic $n$-cycles for any set of odd $n$ we like, using any colors we like. You can even iterate the construction transfinitely ($\kappa$ steps) to get a graph of size $2^\kappa$ colored in $\kappa$ colors with no monochromatic triangles, but $\kappa$ $n$-cycles in every color for every odd $n > 3$.<|endoftext|>
TITLE: Number of Plücker relations for a Grassmannian
QUESTION [6 upvotes]: Is it true that the number of Plücker relations for a Grassmannian $Gr(k,n)$ is equal to the dimension $k(n-k)$ of said Grassmannian? So far, for $Gr(2,5)$, I get exactly five Plücker relations: $$p_{12}p_{34}+ p_{23}p_{14}- p_{13}p_{24}=0,$$ $$p_{12}p_{35}+p_{23}p_{15}- p_{13}p_{25}=0,$$ $$p_{12}p_{45}+p_{24}p_{15}- p_{14}p_{25}=0,$$ $$p_{13}p_{45}+p_{34}p_{15}- p_{14}p_{35}=0,$$ $$p_{23}p_{45}+p_{34}p_{25}- p_{24}p_{35}=0.$$ I'm not sure, but I think that three of the above relations are algebraically independent, no? Are there supposed to be six Plücker relations for $Gr(2,5)$?

REPLY [6 votes]: "The number of Plucker relations" is a little ambiguous, but there is no sense in which it is $k(n-k)$.
The number of Plucker coordinates is $\binom{n}{k}$, so the number of degree $2$ monomials in Plucker coordinates is $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right)$. The vector space they span inside the homogenous coordinate ring of the Grassmannian has dimension 
$$\frac{1}{k+1} \binom{n}{k} \binom{n+1}{k}.$$
(Derivation available on request.) So a minimal list of relations between them would be of size $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right) - \tfrac{1}{k+1} \binom{n}{k} \binom{n+1}{k}$. Note that, if we fix $k$ and let $n$ grow, then $\tfrac{1}{2} \left( \binom{n}{k}^2 + \binom{n}{k} \right) \approx \tfrac{n^{2k}}{2 (k!)^2}$ and $\tfrac{1}{k+1} \binom{n}{k} \binom{n+1}{k} \approx \tfrac{n^{2k}}{(k+1) (k!)^2}$. So the number of relations is growing like $\tfrac{k-1}{k+1} \tfrac{n^{2k}}{(k!)^2}$, not $kn$.
Now, many people mean specific lists of relations when they say "the Plucker relations". They don't always agree on which relations they mean, and they don't always mean an irredundant list. But,  adding redundant relations would just make the list longer.

In your particular case of $G(2,5)$, the $5$ relations you gave form a basis for the space of relations. In general, there is an $\binom{n}{4}$-dimensional space of relations for $G(2,n)$, and I think everyone would agree that the best basis is the relations of the form $p_{ab} p_{cd} - p_{ac} p_{bd} + p_{ad} p_{bc}$, for $1 \leq a < b < c < d \leq n$.<|endoftext|>
TITLE: Uniform bounds on the number of integer points on a family of elliptic curves
QUESTION [6 upvotes]: Let $P(x,y)$ be a binary cubic polynomial with integer coefficient. Let $n$ be an integer. Suppose the (complex) curve $P(x,y)=n$ is nonsingular, so is an elliptic curve. Is there any bound on the number of the integer points on it that is uniform in $n$? I'm particularly interested in the case when $P(x,y)=xy(x+y)$ (and $n$ is a nonzero integer).

REPLY [6 votes]: As the other answers have made clear, there is a big difference if $n$ is arbitrary, or if $n$ is cube-free. But actually, the right formulation in the latter case is to instead ask for solutions $(x,y)$ satisfying $\gcd(x,y)=1$. So for any $n\in\mathbb Z\setminus\{0\}$, let
$$ N^*(F,n) 
  = \#\bigl\{(x,y)\in\mathbb Z^2 : F(x,y)=n~\text{and}~\gcd(x,y)=1\bigr\}.
$$
Also let $E_{F,n}:F(x,y)=nz^3$ be the associated elliptic curve. Then there is a bound of the form [1]
$$ N^*(F,n) \le C(F)^{1+\text{rank}\,E_{F,n}(\mathbb Q)}, $$
where $C(F)$ is independent of $n$. In particular, if you could find any one $F$, for example $F(x,y)=xy(x+y)$ as you specified, for which you could prove that 
$$ \sup_{n\in\mathbb Z\setminus\{0\}} N^*(F,n) = \infty, $$
then you would also have proven that 
$$ \sup_{n\in\mathbb Z\setminus\{0\}} \text{rank}\,E_{F,n}(\mathbb Q) = \infty, $$
which would be a spectacular result. I've listed a few references that deal with uniform bounds for integer points on (minimal) affine models of elliptic curves. The following MO questions/answers also seem related/relevant:
unboundedness of number of integral points on elliptic curves?
Can the number of solutions $xy(x-y-1)=n$ for $x,y,n \in Z[t]$ be unbounded as n varies?
[1] J.H. Silverman, Integer points and the rank of Thue elliptic
curves Invent. Math. 66 (1982), 395-404.
[2] D. Abramovich, Uniformity of stably integral points on elliptic
curves, Invent. Math. 127 (1997), 307-317.
[3] J.H. Silverman, A quantitative version of Siegel's theorem: Integral
points on elliptic curves and Catalan curves, J. Reine Angew. Math.
378 (1987), 60-100.
[4] M. Hindry, J.H. Silverman, The canonical height and integral
points on elliptic curves, Invent. Math. 93 (1988), 419--450.<|endoftext|>
TITLE: Counting isomorphism classes in open subsets of Bun_G
QUESTION [10 upvotes]: Let $G$ be a split semisimple algebraic group and let $C$ be a curve of genus $g$ over $\mathbb F_q$. Assume $g \geq 2$.
The number of $\mathbb F_q$-points of $\# \operatorname{Bun}_G(C)$, where each point is weighted by the inverse of the order of its automorphism group, is $(1+o(1))q^{ (g-1) \dim G}$. This follows from the Tamagawa number 1 theorem, though I think this basic estimate is easier to prove.
Is this still true if we do not divide out by automorphisms? Obviously not. If $G = SL_2$ we can take $L + L^{-1}$ for infinitely many line bundles $L$ to get infinitely many different isomorphism classes of vector bundles with determinant $1$. But it is true if we restrict to stable vector bundles, because the number of automorphisms is bounded.
Let $U$ be an open subset of $\operatorname{Bun}_G(C)$ that is of finite type. For instance, if $G=SL_n$ we can take the set of determinant $1$ vector bundles $V$ such that $V \otimes L^{-1}$ has no sections for a line bundle $L$ of large degree.
Is the number of isomorphism classes of $G$-bundles on $C$ contained in $U(\mathbb F_q)$, not weighted by automorphisms, still $(1+o(1)) q^{(g-1)\dim G}$? Or does the number of points on $\operatorname{Bun}_G$ jump when you add even midly unstable bundles?
I promise I have a really good reason to care about this "evil" count.
I think I can prove this estimate in the case $G=SL_2$. As I already mentioned stable vector bundles are fine. The remainder are split sums $L + L^{-1}$, the number of which with $L$ of degree $n$ is $(1+o(1)) q^g$, and nontrivial extensions of $L^{-1}$ by $L$, which are classified by $H^1(L^2)$, so they are cardinality of them is at most the cardinality of $H^0 ( K L^{-2})$. Summed over all line bundles $L$ of degree $n$, this is at most a constant times the total number of nonzero sections of all line bundles of degree $2g-2-2n$ up to scaling, which is equal to the number of effective divisors of that degree, or $q^{2g-2-2n}$. As $n \geq 0$, either one is less than $q^{3g-3}$.

REPLY [6 votes]: I believe that the claim is true, and that this is related to the so-called 'very good' property of the stack $\mathrm{Bun}_G$. Following [Beilinson, Drinfeld, Quantization of Hitchin's integrable system and Hecke eigensheaves], we call a smooth stack $X$ very good if for every $k>0$, the locus of points $$\{x\in X:\dim \mathrm{Aut}(x)= k\}$$ has codimenion that is strictly larger than $k$. Beilinson and Drinfeld show (Proposition 2.1.2) that $\mathrm{Bun}_G$ is very good if $g>1$. Thus, codimension of the locus $$X_k:=\{x\in\mathrm{Bun}_G:\dim \mathrm{Aut}(x)= k\}$$ is at least $k+1$. Keep in mind that this is the codimension in the sense of stacks; its `evil' counterpart is the claim that the dimension of the space corresponding to $X_k$ is less than $\dim(\mathrm{Bun}_G)=(g-1)\dim G$. This implies that $|X_k(\mathbb{F}_q)|=o(q^{(g-1)\dim G})$, which is what you need, right?
Disclaimer. Beilinson and Drinfeld work in characteristic zero. It seems likely that the result is valid in characteristic $p$, but I didn't look at it carefully.<|endoftext|>
TITLE: For a sufficiently large $a$, are there distinct (mod $a$) integers such that all powers up to the $n$-th are "close" modulo $a$?
QUESTION [5 upvotes]: Given $n\in\Bbb N$ is there an $a_n\in\Bbb N$ such that for every $a>a_n$ there are two distinct integers $0<b<c<a$ such that $b^i\bmod a,c^i\bmod a\in(\sqrt a,\sqrt a\log a)$ for every $i\in\{1,\dots,n\}$? 
If so what is minimum such $a$?

REPLY [2 votes]: This is only a heuristic, but perhaps it can be made rigorous with some
effort (using Weil's bound etc).
I write $m$ instead of your $a$. Let $I:=(\sqrt m,\sqrt m\log m)\pmod m$. You
want to show that there are at least two distinct values of $b\in{\mathbb
Z}_m$ such that $b,\dotsc,b^n\in I$. The number of such $b$ is
  $$ \frac1{m^n} \sum_{x_1,\dotsc,x_n\in I} \sum_{u_1,\dotsc,u_n\in{\mathbb Z}_m}
      \sum_{b\in{\mathbb Z}_m} e^{ \frac{(b-x_1)u_1+\dotsb+(b^n-x_n)u_n}{m} }. $$
The main term obtained for $u_1=\dotsb=u_n=0$ is $|I|^n/m^{n-1}$, which is of
the order of magnitude $m^{1-n/2}(\log m)^n$. This is a small number for
$n\ge 3$ and $m$ large, and it is thus plausible to expect that in this case you cannot (in general) find even one single $b\in{\mathbb Z}_m$ with the property in question.<|endoftext|>
TITLE: Up to $2000$, $A145722(n-1) \equiv \sigma(4n-3) \pmod{5}$
QUESTION [5 upvotes]: A145722 is
Expansion of f(q) * f(q^5) / phi(-q^2)^2 in powers of q where f(), phi() are Ramanujan theta functions.
Using the pari program and offset 0, up to $2000$, 
$$A145722(n-1)  \equiv \sigma(4n-3) \pmod{5}$$

Q1 Is this congruence true?

@Gjergji Zaimi proved similar congruence in another answer.
Since sigma is multiplicative, the offsets bug me a bit.
I would expect if $a(f(n)) \equiv \sigma(g(n))$ and $a$
is defined in a "natural" way, then either $f(n) \mid g(n)$ or
$g(n) \mid f(n)$.

Q2 In case the congruence is true, where $n-1$ and $4n-3$ come from?

Certain sequences (like modular forms, sigma, totient) can be computed significantly
faster if the index is factored.

Q3 Does factoring $n-1$ helps in computing $A145722(n-1)$? (I expect no).

REPLY [9 votes]: Yes, the congruence is true. Here's a modular forms proof. Let $\eta(z) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})$, where $q = e^{2 \pi i z}$. Define
$$
g(z) = \frac{\eta(16z) \eta(40z)^{3}}{\eta(4z) \eta(8z) \eta(20z) \eta(80z)} = \sum_{n=0}^{\infty} {\rm A145722}(n) q^{4n+1}.
$$
This is a modular form of weight zero for $\Gamma_{0}(80)$ and character $\chi_{5}$. Fermat's little theorem shows that $\eta^{5}(z) \equiv \eta(5z) \pmod{5}$ (in the sense that their Fourier coefficients are congruent modulo $5$). Applying $\eta(40z) \equiv \eta(8z)^{5}$, $\eta(20z) \equiv \eta(4z)^{5}$ and $\eta(80z) \equiv \eta(16z)^{5}$ shows that $g(z) \equiv h(z) \pmod{5}$ where
$$
h(z) = \frac{\eta(8z)^{14}}{\eta(4z)^{6} \eta(16z)^{4}}.
$$
This is a holomorphic modular form of weight $2$ for $\Gamma_{0}(16)$.
It is reasonably well-known that $F(z) = \frac{\eta(4z)^{8}}{\eta(2z)^{4}} = \sum_{n=0}^{\infty} \sigma(2n+1) q^{2n+1}$ is a modular form of weight $2$ for $\Gamma_{0}(4)$. Thus $F \otimes \chi_{-1} = \sum_{n=1}^{\infty} \chi_{-1}(2n+1) \sigma(2n+1) q^{2n+1}$ is a modular form of weight $2$ and level $16$.
The space of modular forms of weight $2$ and level $16$ is $5$-dimensional and by checking the coefficients of $q^{0}$ through $q^{4}$, we can see that
$$
h(z) = \frac{F + F \otimes \chi_{-1}}{2} = \sum_{n=0}^{\infty} \sigma(4n+1) q^{4n+1}.
$$
This proves the congruence. Hopefully this explains naturally why the shift in offset occurs.
The most efficient way to compute ${\rm A145722}(n)$ is probably using some analogue of the Rademacher formula for $p(n)$, and in no instance can this be computed more quickly by factoring $n$. In fact, since $\log({\rm A145722}(n)) \sim \frac{2 \pi \sqrt{n}}{\sqrt{5}}$ computing ${\rm A145722}(n)$ must take $\Omega(n^{1/2})$ time. Factoring $n$ is much faster than that.<|endoftext|>
TITLE: When is a map from a logarithmic tangent bundle to a normal bundle surjective?
QUESTION [8 upvotes]: Suppose that $X$ is an algebraic variety with divisor $D$ such that the logarithmic tangent bundle $\mathcal{T}_X(-\log D)$ is locally free. Suppose moreover that $\iota\colon Y\to X$ is a regular embedding of varieties with $Y$ smooth so that the normal bundle $\mathcal{N}_{Y/X}$ is also locally free and the natural map $\iota^\ast\mathcal{T}_X\to \mathcal{N}_{Y/X}$ is surjective. 
Question: Are there natural geometric conditions on $D$ and $Y$ that guarantee that the restriction of the map $\iota^\ast \mathcal{T}_X\to \mathcal{N}_{Y/X}$ to a map $\iota^\ast\mathcal{T}_X(-\log D)\to \mathcal{N}_{Y/X}$ remains surjective? Moreover if the answer is yes, does it appear in the literature?
I suspect that the answer is along the lines that the map is surjective when $D$ and $Y$ intersect transversely for some suitable notion of transverse intersection in this context but I'd be grateful for something precise.
My motivation comes from the analogous question where 'algebraic variety' is replaced in my first sentence by 'rigid analytic variety' but I'm guessing I'm more likely to get an answer as stated and that it will translate easily. I don't mind assuming that $X$ is smooth as well as $Y$.

REPLY [4 votes]: I guess my answer will be in the category of complex manifolds, therefore we definitely need to assume that $X$ is smooth. I am not that familiar with this language, so I apologize for any stupid mistakes coming from this. 
Assume that $D$ is simple normal crossing and that $Y$ is smooth. Denote by $X_0 = X \setminus D$, $X_1 = D \setminus \text{Sing}(D)$, and inductively $X_{k+1}$ is the non-singular part of $\text{Sing}(\overline{X_k})$. Then the sections of $\mathcal{T}_X(- \log D)$ are those holomorphic vector fields on $X$, which are tangent to all strata $X_k$. I think a sufficient criterion for surjectivity is:
Y intersects all the strata $X_k$ transversally (in the sense $T_p Y + T_p X_k = T_p X$ for all $p \in Y \cap X_k$)
Indeed, to check whether the morphism $i^* \mathcal{T}_X(- \log D) \to \mathcal{N}_{Y/X}$ of locally free sheaves on $Y$ is surjective, it suffices to check in the fibres. At a point $p \in Y \cap X_k$ the divisor $D$ has equation $z_1 z_2 \ldots z_k =0$ in local coordinates and a local basis of $\mathcal{T}_X(- \log D)$ is given by 
$$z_1 \frac{\partial}{\partial z_1}, \ldots, z_k \frac{\partial}{\partial z_k}, \frac{\partial}{\partial z_{k+1}} \ldots, \frac{\partial}{\partial z_{\dim X}}.$$ 
The assumption, that $Y$ intersects $X_k$ transversally means that $T_p Y$ together with $\frac{\partial}{\partial z_{k+1}}|_p \ldots, \frac{\partial}{\partial z_{\dim X}}|_p$ span the vector space $T_p X$, thus the sections $\frac{\partial}{\partial z_{k+1}} \ldots, \frac{\partial}{\partial z_{\dim X}}$ restrict to generators of the fibre of $\mathcal{N}_{Y/X}$ at $p$. 
If we drop the assumption that $Y$ is smooth, we might be able to phrase the transversality condition in terms of the matrix of derivatives of a regular sequence $x_1, \ldots, x_l$ defining $\iota$ along tangent directions to $X_k$ having rank $l$. [For this see also the discussion in the comments.]<|endoftext|>
TITLE: Splitting subspaces and finite fields
QUESTION [6 upvotes]: Hellow. I'm sure that the following is truth, but I can't prove it.
Let $R<S<K, R=\mathrm{GF}(q),\  S= \mathrm{GF}(q^n), \ K= \mathrm{GF}(q^{mn})$ be a chain of finite fields and
$A = \{\theta\in K: \mathrm{ord}\theta = q^{mn}-1\}$ be the set of primitive elements. I want to prove the following statement. If $_RW$ be a subspace of $_RK$, $\mathrm{dim}_RW = n$  such that for every $\theta \in A$
$$
W\oplus W\theta\oplus\ldots\oplus W\theta^{m-1} = K,
$$
then $W = aS$ for some $a\in K$.
I will be grateful for the help.

REPLY [2 votes]: Clearly, the statement is invariant under multiplication by $a\in K$, so we may assume that $W\ni 1$. This implies that $W\supseteq R$, and we want to show that $W=S$. 
Suppose that $t\in W$. I claim that $t$ can not be written as a value of a  non constant rational function $f(x)/g(x)$ with coefficients in $R$ and degrees of numerator and denominator $<m$ at a primitive root $x=\theta$.
Indeed, if we have $tg(\theta)-f(\theta)=0$, then this translates into
linear dependence on $W,W\theta,\ldots,W\theta^{m-1}$, since $1,t\in W$.
It remains to argue that the complement in $K$ to the set of non constant "degree $\leq m-1$" rational functions in primitive roots is contained in (and is equal to) $S$. This sounds plausible, but I don't have a proof.<|endoftext|>
TITLE: Circle Action on Quaternionic Projective Space
QUESTION [8 upvotes]: Quoting from Wikipedia article on quaternionic projective space:

Therefore the quotient manifold
  $$
\mathbb{HP}^{2}/\mathrm{U}(1)
$$
  may be taken, writing $U(1)$ for the circle group. It has been shown that this quotient is the $7$-sphere, a result of Vladimir Arnold from 1996, later rediscovered by Edward Witten and Michael Atiyah.

Does such a situation hold for higher quaternionic projective spaces? More specifically, does there exist a $U(1)$ action on $\mathbb{HP}^n$, and is the quotient a sphere?

REPLY [9 votes]: $\mathbb{HP}^n\cong \mathrm{Sp}(n+1)/(\mathrm{Sp}(n)\times \mathrm{Sp}(1))$ is a symmetric space, so every one-parameter subgroup of $\mathrm{Sp}(n+1)$ acts on it. As was noted in the comments, such an action always has at least $n+1$ fixed points. Let me restrict to the action $z\cdot [x_0:\cdots: x_n] = [x_0z:\cdots:x_nz]$, since this is what was used in the case $n=2$.
For $i\in\{0,\cdots,n\}$, there is a coordinate chart $\phi_i:\mathbb{C}^{2n}\cong U_i\subset \mathbb{HP}^n$ defined by
$$
\phi_i(x_1,\cdots,x_n,y_1,\cdots,y_n) = [x_1+jy_1:\cdots:x_{i-1}+jy_{i-1}:1:x_i+jy_i:\cdots:x_n+jy_n]
$$
where the $1$ is in the $i$-th homogeneous coordinate. Then
\begin{align*}
\hspace{5pt}&\hspace{-5pt}z\cdot \phi_i(x_1,\cdots,x_n,y_1,\cdots,y_n)\\
 &= [(x_1+jy_1)z:\cdots:(x_{i-1}+jy_{i-1})z:z:(x_i+jy_i)z:\cdots:(x_n+jy_n)z]\\
&= [z^{-1}(x_1+jy_1)z:\cdots:z^{-1}(x_{i-1}+jy_{i-1})z:1:z^{-1}(x_i+jy_i)z:\cdots:z^{-1}(x_n+jy_n)z]\\
&= [x_1+jy_1z^2:\cdots:x_{i-1}+jy_{i-1}z^2:1:x_i+jy_iz^2:\cdots:x_n+jy_nz^2]\\
&= \phi_i(x_1,\cdots,x_n,z^2y_1,\cdots,z^2y_n)
\end{align*}
Of course, $-1$ commutes with all quaternions and thus acts by the identity. We can replace $U(1)$ by $U(1)/(\mathbb{Z}/2)$ which amounts to replacing $z^2$ by $z$. (In fact, this action is just that of the automorphisms of $\mathbb{H}$ restricting to the identity on $\mathbb{C}$, which explains the relation to complex conjugation.)
Thus $U_i$ is stable under the $U(1)$-action, and $U_i/U(1)\cong \mathbb{C}^n\times (\mathbb{C}^n/U(1))$, where $U(1)$ acts on $\mathbb{C}^n$ by scalar multiplication. Using polar coordinates, one easily sees that $\mathbb{C}^n/U(1)\cong C\mathbb{CP}^{n-1}$ is homeomorphic to the cone on $S^{2n-1}/U(1)\cong \mathbb{CP}^{n-1}$. Hence the quotient $\mathbb{HP}^n/U(1)$ is covered by $n+1$ open sets, each of which is homeomorphic to $\mathbb{C}^n\times C\mathbb{CP}^{n-1}$. When we remove the tip of the cone from any neighbourhood of it, the resulting open set retracts onto $\mathbb{CP}^{n-1}$. Looking at $H_2$, we see that if the tip of the cone (with some element of $\mathbb{C}^n$ in the first component) has a euclidean neighbourhood, we must have $n = 2$. Conversely, in this case $C\mathbb{CP}^{n-1}\cong CS^2 \cong \mathbb{R}^3$, so $U_i/U(1)\cong \mathbb{R}^7$ and the quotient is a manifold. In higher dimensions, the quotient is not even an orbifold, but one still understands its singularities.
Here is a proof that for $n=2$ the quotient is homeomorphic to $S^7$: Note that $U_n = \mathbb{HP}^{n}\setminus \mathbb{HP}^{n-1}$. For $n=0$, the quotient is a point; for $n=1$ it is compact and the union of a point and $\mathbb{C}\times C\mathbb{CP}^0\cong \mathbb{R}^2\times\mathbb{R}_{\ge 0}$, so it is the one-point compactification of the half-space, which is contractible. Consequently, $\mathbb{HP}^{2}/U(1)\to (\mathbb{HP}^{2}/U(1))/(\mathbb{HP}^{1}/U(1))$ is a homotopy equivalence, and the target is the one-point compactification of $\mathbb{C}^2\times C\mathbb{CP}^1\cong \mathbb{R}^7$ which is $S^7$. So $\mathbb{HP}^2/U(1)$ is a manifold and a homotopy sphere; by the Poincare conjecture, it is homeomorphic to $S^7$.<|endoftext|>
TITLE: Stable homotopy groups of $RP^{\infty}$
QUESTION [12 upvotes]: Are the stable homotopy groups $\pi^s_i(\mathbb R P^{\infty})$ known for small $i$? In particular, I would be interested in the values for $i = 5,6$. A quick Internet search did not lead to anything.

REPLY [7 votes]: The middle column of table IV on page 82 of George W. Whitehead's "Recent Advances in homotopy theory" Regional Conference series in mathematics Number 5
lists the groups in dimensions up to 30 (including the 2 quoted by Matthias Wendt).<|endoftext|>
TITLE: Discrete spectrum and almost periodicity
QUESTION [5 upvotes]: According to Vershik, an ergodic invertible measure-preserving transformation $T$ on a Lebesgue space $X$ has discrete spectrum if and only if for every bounded measurable function $f\colon X \to \mathbb{C}$ and  almost all $x_0 \in X$, the function from $\mathbb{Z}$ to $\mathbb{C}$ defined by $n \mapsto f(T^nx_0)$ is Besicovitch almost periodic (for short, BAP). I am mainly interested in the only if part of this result: the proof of BAP under the discrete spectrum assumption. 
When $T$ has discrete spectrum, it is isomorphic to an ergodic isometry, and using this version of $T$,  it is known that $n \mapsto f(T^nx_0)$ is uniformly almost periodic when $f$ is continuous. This is proved in Petersen's book, but I have failed to adapt the proof to the  Besicovitch almost periodicity (that does not mean the proof is hard, I am not an expert in spectral theory, or even in ergodic theory in general). 
Here is the definition of BAP. On the space of bounded functions $g\colon \mathbb{Z}\to\mathbb{C}$, define the Besicovitch semi-norm:
$$
{\Vert g \Vert} = \limsup_{K \to \infty} \frac{1}{2K+1}\sum_{|k|\leq K} \bigl|g(k)\bigr|.
$$
For a bounded function $g\colon \mathbb{Z}\to\mathbb{C}$ define its shift $g_n(k)=g(n+k)$. Say that $g$ is BAP if the set $\{g_n, n \in \mathbb{Z}\}$ is relatively compact for ${\Vert \cdot \Vert}$.
Note that ${\Vert \cdot \Vert} \leq {\Vert \cdot \Vert}_\infty$ (the sup norm), therefore a function is BAP whenever it is uniformly almost periodic (the definition of uniform almost periodicty is obtained by replacing ${\Vert \cdot \Vert}$ with ${\Vert \cdot \Vert}_\infty$).
The two following conditions are equivalent to BAP:

$g$ is the ${\Vert \cdot \Vert}$-limit of a sequence of trigonometric polynomials $p_N$ having form $p_N(k) = \sum_{j=1}^N a^{(N)}_je^{2i\pi\alpha_j^{(N)}k}$ with $\bigl|\alpha_j^{(N)}\bigr|=1$. 
The continuation of $g$ on the Bohr compactification of $\mathbb{Z}$ exists and belongs to $L^1$ (when the Bohr compactification is endowed with the Haar probability measure). 

I don't know whether there is a characterization of BAP looking like the syndeticness of $\epsilon$-periods, which is known to be equivalent to uniform almost periodicity (see the child's garden of almost periodic functions where the uniform almost periodicity is called Bochner almost periodicity, and the syndeticness property is called the Bohr almost periodicity). 
One of my attempts was to prove that the set $\bigl\{f \text{ bounded} \mid k \mapsto f(T^kx_0) \text{is BAP}\bigr\}$ is a vector space containing the uniformly continuous functions and stable by pointwise limit. Then one could conclude by a Stone-Weierstrass result. 
I would prefer a proof that deduces the result directly from the definition, not using a characterization such as 1. and 2., but this is secondary.
Another claim by Vershik is for the case when $T$ is given as a shift on $A^{\mathbb{Z}}$, where $A$ is a finite alphabet. For two sequences $x, y \in A^{\mathbb{Z}}$, define the Besicovitch-Hamming distance
$$
\rho(x,y) = \liminf_{K \to \infty} \frac{1}{2K+1} \#\left\{k \;\big\vert\; |k| \leq K, x_k \neq y_k\right\}. 
$$
Note: I obtained my definition of ${\Vert g \Vert}$ by adapting the Wikipedia page to $\mathbb{Z}$. It is strange that it is defined with a limit superior whereas there is a limit inferior in Vershik's definition of the BH distance. 
Then say that $x$ is BH almost periodic if the set consisting of $x$ and all its shifted sequences is relatively compact for $\rho$. Then it is claimed that an ergodic $T$ has discrete spectrum if and only if almost every orbit of $T$ is BH almost periodic. I don't know whether this result is easier than the previous one, but I am interested in one or the other. 
Vershik gives an argument for this second claim: $T$ is isomorphic to an ergodic isometry by Von Neumann's theorem, and "observe that the restriction of any bounded measurable function on a compact Abelian group to a countable $\mathbb{Z}$-subgroup, regarded as a function on $\mathbb{Z}$, is BH almost periodic" (sic). 
UPDATE
Ten minutes after posting this question, I found this document which provides a proof of the result (the result is stated with "$\mathbb{N}$-orbits" but this should be equivalent). It makes use of the characterization 1. Instead of deleting this post, I ask for a proof that more directly deduces the result from the definition.

REPLY [5 votes]: It follows from ergodic theorem that for a.e. $x \in X$ and any  $f \in L^1(X)$ the limit $\lim\limits_{N\to \infty}\frac{1}{2N+1}\sum\limits_{k =-N}^N |f(T^kx)|$ exists and coincides with $\|f\|_{L^1(X)}$. For $k \in \mathbb{Z}$ we can use this for the function $f(T^k\cdot)-f(\cdot)$ and obtain that for a.e. $x \in X$ the following identity holds
$$\lim\limits_{N\to \infty}\frac{1}{2N+1}\sum\limits_{k =-N}^N |f(T^kx)-f(x)|=\|f(T^k\cdot)-f(\cdot)\|_{L^1(X)}.$$
Therefore, for a.e. $x \in X$ BAP-property of the function $n\mapsto f(T^nx)$ is equivalent to the relative compactness of the set $\{f(T^k\cdot)\}_{k \in \mathbb{Z}}$ in $L^1(X)$. But this relative compactness is equivalent to the discrete spectrum assumption.<|endoftext|>
TITLE: Sufficient conditions for $\sum_{n \ge 1} a_n e^{-(a_1+\cdots+a_n) s} \sim \frac{1}{s}$ as $s \to 0^+$
QUESTION [5 upvotes]: Let $(a_n)_{n \ge 1}$ be a sequence of non-negative real numbers such that $\sum_{n \ge 1} a_n = \infty$, and set $\lambda_n := a_1 + \cdots + a_n$ for each $n$. Then the (generalized Dirichlet) series $\sum_{n \ge 1} a_n e^{-\lambda_n s}$ converges for all $s > 0$, as its abscissa of convergence is $0$ by the main result of Chapter II, Section 6 from:

G. H. Hardy and M. Riesz, The General Theory of Dirichlet's Series, Cambridge Univ. Press: Cambridge, 1915.

So my question is:

Q. What about sufficient conditions for having that $\sum_{n \ge 1} a_n e^{-\lambda_n s} \sim \frac{1}{s}$ as $s \to 0^+$? 

To be clear: I'm just looking for references. 
For the record: I'm especially, though not uniquely, interested in the case where $a_n = n^{\alpha}$ for all $n$, with $\alpha$ being a given exponent $\ge -1$.

REPLY [3 votes]: Set $\lambda_0:=0$. We have 
$$
s^{-1}=\sum_{n=1}^{\infty} s^{-1}\left(e^{-\lambda_{n-1}s}-e^{-\lambda_n s}\right)=\sum e^{-\lambda_ns}\frac{e^{a_ns}-1}{s}\geqslant \sum a_ne^{-\lambda_ns}.
$$
On the other hand,
$$
s^{-1}=\sum_{n=1}^{\infty} s^{-1}\left(e^{-\lambda_{n-1}s}-e^{-\lambda_n s}\right)=\sum e^{-\lambda_{n-1}s}\frac{1-e^{-a_ns}}{s}\leqslant \sum a_ne^{-\lambda_{n-1}s}=a_1+\sum a_{n+1}e^{-\lambda_ns}.
$$
Thus if $a_n\sim a_{n+1}$, we have desired relation $\sum a_ne^{-\lambda_ns}\sim s^{-1}$ for sure. Of course, this may be weakened.<|endoftext|>
TITLE: Chern-Simons forms, characteristic numbers, and boundary terms?
QUESTION [8 upvotes]: For any principal $G$-bundle $P \to M$ with principal connection $\omega$, given a $G$-invariant polynomial $p: \mathfrak{g} \to \mathbb{R}$ we can construct a form $p(F_\omega)$ on $P$ which descends to a characteristic form on the base $M$ from the curvature $F_\omega$. This form is exact when considered as a form on $P$, and the Chern-Simons form $\text{cs}(\omega)$ on $P$ satisfies $p(F_\omega) = d \text{cs}(\omega)$. 
I have seen it mentioned that Chern-Simons forms are related to boundary terms when computing characteristic numbers, but I have not been able to find a reference explaining this. In particular, I have heard that the Chern-Simons forms are somehow involved in the generalization of the Chern-Gauss-Bonnet theorem to manifolds with boundary.
Could anyone shed some light on this?

REPLY [4 votes]: You might have heard the following. I am actually referring to the second meaning of Chern-Simons classes $\tilde p(\nabla^0,\nabla^1)\in\Omega^\bullet(M)$ satisfying $d\tilde p(\nabla^0,\nabla^1)=p((\nabla^1)^2)-p((\nabla^0)^2)$. They can of course be recovered from the classes you described.
Given a vector bundle $(V,\nabla^V)$ with connection over $M$, you get the Chern-Weil form $p(F)\in\Omega^\bullet(M)$ that you described. A lift to relative forms would be a pair $(p(F),\alpha)$ with $\alpha\in\Omega^\bullet(U)$ satisfying $d\alpha=p(F)|_U$ on a collar neighbourhood $U$ of $\partial M$. With $\alpha$, you could easily replace $p(F)$ by a cohomologous form with compact support.
To construct $\alpha$, you need more information than just $(V,\nabla^V)$ - you need an explicit reason for $p(V)$ to vanish on $U$. In this context, a plausible reason is the existence of another connection $\nabla'$ on $V|_U$ such that $F'=(\nabla')^2=0$. This could come from a trivialisation of $V|_U$, but there might be other  possibilities. Now you get the relative term $\alpha=\tilde p(\nabla',\nabla^V|_U)$, and you can compute a characteristic number
$$\langle(p(F),\tilde p(\nabla',\nabla^V|_U)),[M,\partial M]\rangle
=\int_Mp(F)-\int_{\partial M}\tilde p(\nabla',\nabla^V|_U)\;.$$
For Gauss-Bonnet-Chern, you need a "reason" for the Euler class of $TM$ to be trivial in a neighbourhood of $\partial M$. This is given by the fact that $TM|_{\partial M}\cong T\partial M\oplus\underline{\mathbb R}$. But the Levi-Civita connection does not respect this decomposition unless $M$ has totally geodesic boundary. If it does not, then you use a Chern-Simons form for the Euler form as a correction term. In fact, this is also related to Mathai-Quillen currents, which can be used to reduce Gauss-Bonnet-Chern to Poincaré-Hopf.<|endoftext|>
TITLE: Hahn-Banach theorem for arbitrary locally compact fields?
QUESTION [7 upvotes]: Does anyone know if the Hahn-Banach theorem is true for every locally
compact field?  Specifically, let $F$ be a finite algebraic extension of
either $Q_p$, the $p$-adic completion of $Q$, or of
$S_p$, the $t$-adic completion of $F_p\{t\}$, the field of
formal Laurent series over the field of $p$ elements.  Then is $F$
injective with respect to topological embeddings in the category of
normed linear $F$-spaces?

REPLY [12 votes]: There exists a $p$-adic Hahn Banach theorem.
Reference Alain M. Robert. A course in p-adic analysis. Graduate text in Math.
Theorem (Ingleton)
Let $K$ be a spherically complete ultrametric field. $E$ a $K$-normed space and $V$ a subspace of $E$. For every bounded linear functional $\phi$ defined on $V$, there exists a bounded linear functional $\psi$ defined on $E$ whose restriction to $V$ is $\phi$ and such that $\|\phi\|=\|\psi\|$.<|endoftext|>
TITLE: Have Grothendieck's notes in Montpellier already been investigated?
QUESTION [26 upvotes]: Grothendieck, who passed away on November 13, 2014, left a huge amount (around 20.000 sheets) of personal notes in the University of Montpellier that he thought he was the only one to be able to decipher. I heard that this intellectual treasure has been stored even though Grothendieck did not want anyone to publish any of his works. 
My question is therefore: has someone begun to examine these notes?

REPLY [24 votes]: Apparently 18,000 pages of Grothendieck's notes are going to be released on May 10, 2017 at 4:30 pm on the site of the University of Montpellier. I get this information from the following article: http://www.liberation.fr/sciences/2017/05/05/les-notes-du-mathematicien-alexandre-grothendieck-arrivent-sur-le-net_1567517

...and here is a link to the Université de Montpellier site where 18,000 of Grothendieck's notes are now posted!<|endoftext|>
TITLE: Looking for Severi varieties
QUESTION [5 upvotes]: Let $K$ be an algebraically closed field of characteristic $0$, and let $\mathbb{O}$ be the Cayley algebra over $K$. Let 
$$
\mathfrak{J}_{3}=\{A\in\mathcal{M}_{3}(\mathbb{O}):A\text{ is Hermitian}\},
$$
that may be considered as a $K$-vector space of dimension $27$. We are going to denote $\mathbb{P}^{26}=\mathbb{P}(\mathfrak{J}_{3})$.
Given $A,B\in\mathfrak{J}_{3}$ we may define:
$$
A\circ B:=\frac{1}{2}(AB+BA),
$$
$$
A*B:=A\circ B-\frac{1}{2}(A\cdot tr(B)+B\cdot tr(A))+\frac{1}{2}(tr(A)tr(B)-tr(A\circ B)),
$$
$$
A\times A:=A\circ A-tr(A)\cdot A+\frac {1}{2}(tr(A\circ A)-tr(A)^{2})\cdot Id,
$$
$$
\det A:=\frac{1}{3}tr((A*A)\circ A).
$$
We say that $rk(A)=1$ if $A\times A= 0$. Then, I want to prove that
$$
X=\{[A]\in\mathbb{P}^{26}:rk (A)=1\},
$$
is an algebraic variety of dimension $16$ such that its secant variety is 
$$
SX=\{[A]\in\mathbb{P}^{26}:\det A=0\},
$$
and that $\dim SX=25$ (i.e. I want to prove that $X$ is the fourth Severi variety).
It is stated in the last parragraph of p.18 of this text, but I haven't been able to find a sketch of the proof of this fact. 
There is an alternative way via representation theory that is explained in Lazarsfeld's book on Zak's work or in Zak's 'Tangents and secants of algebraic varieties', but I would like to know if it is not very hard to see it this way.

REPLY [4 votes]: Partial answer:
Let $A =  \left( \begin{array}{ccc}
a & b & c \\
\overline{b} & e & d \\
\overline{c} & \overline{d} & f \end{array} \right) $ be the generic hermitian matrix with octonionic coefficients. This means that $b,c,d$ can be written as $X_1.1 +X_2.i_1 + \cdots + X_8.i_7$, where the $X_p$ are abstract variables (which will take value in $K$) and $1,i_1, \cdots, i_7$ are basis of $\mathbb{O}$ over $K$.
The symbols $\overline{b}, \overline{c}, \overline{d}$ are the conjugate of $b,c,d$ and $a,e,f$ are self conjugate, that is they are of the form $X_1.1$, with $X_1$ a variable which will take value in the field $K$. Overall, you have $27$ abstract variables $X_1,\cdots X_{27}$ (which will take value in $K$) coming into the picture.
Now, if you compute de determinant of $A$ (using Sarrus rule for instance), you will find a cubic equation in $X_1, \cdots, X_{27}$ which involves only integer numbers (the purely octonionic numbers disappear, which is kind of extraordinary). This computation makes sense because of the pseudo-associativity of $\mathbb{O}$, you don't need it to be commutative or even associatif.
This equation is the equation of a cubic hypersurface in $\mathbb{P}^{26}$ and it is irreducible. Now let $Z$ be the scheme cut out by the $2 \times 2$-minors of $A$. Again one finds that the minors are equations in $X_1, \cdots, X_{27}$ with only integers coefficients (again the purely octonionic coefficients vanish).
Each minor is a quadratic equation in $X_1, \cdots, X_{27}$ and it is easy to see that they correspond to the partial derivatives of the equation of $det A$ with respect to each variables $X_1, \cdots, X_{27}$. Hence, the scheme $Z$ is the singular locus of $det A = 0$. 
The scheme $Z$ is smooth. Indeed, if it had a singular point, this point woud be a triple point of $det A =0$. Being a cubic hypersurface, $det A =0$ would then be a cone, which isn't the case (this is a bit more complicated to check).
Now, $Z$ being the singular locus of $det A=0$ and this hypersurface being cubic, Bezout's theorem insures that every line meeting two points of $Z$ will be included in $det A=0$. This proves that $S(Z) \subset \{ det A = 0 \}$, where $S(Z)$ is the secant variety of $Z$.
Now two things are left to prove :
_$\dim Z = 16$,
_ the secant variety $S(Z)$ is actually the whole hypersurface $det A =0$.
I don't remember a simple proof of the first fact. Let's admit it. Since $Z$ is smooth of dimension $16$ and included in $\mathbb{P}^{26}$, ZAK's theorem on tangency gurantees that the dimension of $S(Z)$ must be at least $25$. The hypersurface $det A = 0$ being irreducible, we have $S(Z) = \{det A = 0 \}$.
EDIT : There is a mistake in the description of the $2 \times 2$ minors. Three of them will indeed be quadartic equations over $K$ with integer coefficients. The other $3$ will have fully octonionic equations, each of them giving $8$ equations over $K$ with integer coeffcients. At the end, wet $27$ equations over $K$ with integer coefficients, corresponding indeed to the $27$ partial derivatives of the equation of $det A$.<|endoftext|>
TITLE: A non locally compact group of finite topological dimension?
QUESTION [7 upvotes]: Is there a topological group which is Hausdorff, first countable,
locally connected and has finite topological dimension, yet fails
to be locally compact?

REPLY [4 votes]: The Gleason-Montgomery theorem on the local compactness of locally path-connected finite-dimensional topological groups was generalized by Banakh and Zdomskyy (http://topology.auburn.edu/tp/reprints/v36/tp36027.pdf) who proved that a topological group $G$ is locally compact if it is compactly finite-dimensional and locally continuum-connected.
On the other hand, for every $n\ge 2$ the Euclidean space $\mathbb R^n$ contains an additive subgroup $G$, which is connected, locally connected, and not Borel (so, not locally compact). 
To construct such a group $G$, consider the family $\mathcal B$ of all  uncountable Borel subsets of $\mathbb R^n$ and observe that it has cardinality of continuum. So, $\mathcal B$ can be enumerated by ordinals $<\mathfrak c$ as $(B_\alpha)_{\alpha<\mathfrak c}$. Fix any non-zero point $p$ in $\mathbb R^n$ and by transfinite induction, for every ordinal $\alpha<\mathfrak c$ choose a point $x_\alpha\in B_\alpha$ so that the subgroup $G_\alpha$ generated by the set $\{x_\beta\}_{\beta\le\alpha}$ does not contain the point $p$. Such a choice is always possible since $B_\alpha$ has cardinality of continuus and $x_\alpha$ should be chosen to avoid the subset $\{\frac1n(p-g):n\in\mathbb Z\setminus\{0\},\;g\in \bigcup_{\beta<\alpha}G_{\beta}\}$ of cardinality $<\mathfrak c$.
After completing the inductive construction, consider the additive subgroup $G$ of $\mathbb R^n$, generated by the set $\{x_\alpha\}_{\alpha<\mathfrak c}$. By construction, this subgroup $G$ does not contain the point $p$ and meets every uncountable Borel subset $B$ of $\mathbb R^n$.  In particular, it meets the uncountable Borel set $B-p$, which implies that $p+G\subset \mathbb R^n\setminus G$ meets $B$. This means that $G$ is a Bernstein set in $\mathbb R^n$.
It remains to prove that every Bernstein set $G$ in $\mathbb R^n$, $n\ge 2$, is connected and locally connected. This will follow as soon as we check that for every open subset $U\subset\mathbb R^n$, homeomorphic to $\mathbb R^n$ the intersection $U\cap G$ is connected. To derive a contradiction, assume that $U\cap G$ is disconnected and find two disjoint non-empty open sets $U_1,U_2\subset U$ such that $U\cap G=(U_1\cap G)\cup(U_2\cap G)$. It follows that $F=U\setminus (U_1\cup U_2)$ is a closed subset of $U$ separating $U$, which is disjoint with the Bernstein set $G$. So, $F$ is at most countable and hence $\dim(F)=0$.
But this contradicts Mazurkiewicz Theorem (saying that $\mathbb R^n$ cannot be separated by a closed subset of dimension $<n-1$).<|endoftext|>
TITLE: Is the restriction map for embeddings of manifolds with boundary a fibration?
QUESTION [17 upvotes]: Let $M$ and $W$ be smooth manifolds (possibly with boundary) and $V\subseteq W$ a submanifold. We have a map between embedding spaces
$$Emb(W,M)\rightarrow Emb(V,M)$$ given by restriction.
Richard Palais proved around 1960 that if all manifolds have no boundary, $V$ is compact and the embedding spaces are equipped with the weak Whitney $C^{\infty}$-topology, then this is a fiber bundle, in particular a Serre fibration. (The path-lifting property of the fibration reflects the isotopy extension theorem.)
In conclusive remarks (section 6 of the cited paper), he mentions that the result is also true allowing $W$ and $V$ to have boundary but still insisting on $V$ being compact and $M$ having no boundary. It is said that these results will appear, among others, in a joint paper with Morris Hirsch.
It seems to me that this paper was never published.

Has a proof of the announced result been published elsewhere in the meanwhile?
Does it fail for $M$ having boundary or $V$ being noncompact? If so, is the map still a fibration? I am especially interested in the case of $V,M,W$ all being compact and with boundary.

REPLY [9 votes]: In certain cases it's true, in others its not.  The proof for manifolds with boundary follows very much in the same spirit as the proof for ones without, but there are a few extra complicating details. 
It helps to think about the case where it fails.  For example, consider embeddings of a compact manifold with boundary $W$ in another compact manifold with boundary $M$.  Any example works, but take $W=[0,1]$ and $M=[0,1]$.  For such embeddings the isotopy extension theorem does not hold.  This fibration theorem you are interested in is just the isotopy extension theorem "with parameters" so it clearly fails. 
The reason for the failure: In the proof you need to extend a vector field on the image of $W$ to a vector field with prescribed behaviour outside of a small neighbourhood of the image of $W$.   This is generally impossible when both manifolds have boundary -- in this case because the complement can be empty. 
So the theorem is true if both $W$ and $M$ have boundary but you require the embeddings to be proper meaning they send boundary to boundary (and being transverse to $\partial M$).  
The theorem is also true if $W$ has boundary but $M$ does not.  The theorem is also true if $W$ has no boundary, and $M$ has boundary, but you require the embedding to be disjoint from $\partial M$.  etc. 
I suggest reading the proof of the isotopy extension theorem in Hirsch's Differential Topology text.  That proof gives you the basic idea, even for the "with parameters" versions of the theorem that you are interested in.
(edit)
Regarding (1), for example, Cerf's dissertation made heavy usage of these facts.  It is a large IHES publication and I believe it has all these results in it, with fairly detailed proofs.  Most publications that use these results have sketches of them. The Goodwillie-Weiss embedding calculus, for example, makes heavy usage of these theorems. 
Regarding (2), yes it fails for $M$ having boundary precisely in the instances I described above.   
When $V$ is non-compact it can fail as well. For example, $Emb(\mathbb R, \mathbb R^3)$ is pathwise connected, so if the the map $Emb(\mathbb R^3 , \mathbb R^3) \to Emb(\mathbb R, \mathbb R^3)$ were a fibration there would be an ambient isotopy between a "long trefoil" and an unknot.<|endoftext|>
TITLE: Topological Derivation of Leray Spectral Sequence
QUESTION [9 upvotes]: I'm interested in computing - to the extent possible - the Leray spectral sequence for a particular map which is almost, but not quite, a fiber bundle (e.g. a Seifert fiber space). The hardest step currently is writing down the edge maps on the $E_2$ page.
Can someone point me to a reference that derives the Leray sequence from a topological/geometric point of view (i.e. not by appealing to the Grothendieck spectral sequence)? I recall reading somewhere that a CW-structure on the base space can induce the needed filtration, but the reference I found (which I can't remember now) only did this for the case of an actual fiber bundle.

REPLY [5 votes]: Bott and Tu do this in their book Differential forms in algebraic topology, see Section 14, ``Leray's construction" (starting on page 179).<|endoftext|>
TITLE: Generic Smoothness Type of Results in Positive Characteristic
QUESTION [6 upvotes]: Let $f:X\to Y$ be a surjective morphism between two projective varieties over a  field of characteristic $p>0$. Also assume that $f_*\mathcal{O}_X=\mathcal{O}_Y$, and $X$ is smooth. 
We know that the general fiber of $f$ is not smooth in general. But can we say that the general fiber is an integral scheme?
Maybe this is too much to ask for an arbitrary morphism $f$; are there any known classes of morphsims for which this property holds? What if $f$ is given by an Iitaka fibration of a nef line bundle, does it make any difference?   
$\textbf{Note}:$ It is known that if the $\textit{generic}$ fiber of $f$ is $\textit{geometrically integral}$, then the general fiber is an integral scheme. So my question basically reduces to asking what are some known classes of morphisms with geometrically integral generic fibers?

REPLY [7 votes]: Correction.  I just realized that there are examples where the geometric generic fiber is NOT generically reduced.  In all of my comments and the answer below, I was assuming that the geometric generic fiber is generically reduced.  When that is true, then the geometric generic fiber is integral.  However, without the hypothesis (or some other hypothesis that implies this hypothesis), the argument below only proves that the geometric generic fiber is an irreducible, LCI scheme.  It may be everywhere nonreduced.  There is a numerical condition that will guarantee this.  For any ample divisor class $A$ on $X$, and for any ample divisor class $B$ on $Y$, if the intersection number $A^{\text{dim}(X)-\text{dim}(Y)}.(f^*B)^{\text{dim}(Y)}$ is prime to $p$, then the geometric generic fiber of $f$ is generically reduced.  I noticed the counterexamples below by looking for examples where $p$ divides this integer.
Original post.
For a normal, integral scheme $X$, for an integral scheme $Y$, for a proper, locally finitely presented morphism $f: X\to Y$, if the natural map $f^\# : \mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism, then the geometric generic fiber of $f$ is integral and LCI (thus Cohen-Macaulay), but quite possibly not $R1$, i.e., not normal.
Denote by $K(Y)$ the function field of $Y$.  Denote by $X_\eta$ the fiber product of $f$ and the natural morphism $i:\text{Spec}\ K(Y) \to Y$.  Since $X$ is a normal scheme and $X_\eta$ is obtained by localization, also $X_\eta$ is a normal $K(Y)$-scheme.  Also the natural map $K(Y)\to H^0(X_\eta,\mathcal{O}_{X_\eta})$ is an isomorphism since it is the generic fiber of the isomorphism $f^\#$.  
Denote by $K(Y)^{\text{sep}}/K(Y)$ the separable closure of $K(Y)$.  Denote by $X_{\eta^s}$ the base change of $X_\eta$ to $K(Y)^{\text{sep}}$.  Since $K(Y)^{\text{sep}}/K(Y)$ is a limit of etale extensions, the same holds for $X_{\eta^s}\to X_\eta$.  In particular, $X_{\eta^s}$ is a normal scheme.  Since the base change is flat, also $K(Y)^{\text{sep}} \to H^0(X_{\eta^s},\mathcal{O}_{X_{\eta^s}})$ is still an isomorphism.  Thus, $X_{\eta^s}$ is still a normal, integral $K(Y)^{\text{sep}}$-scheme.
Finally, the field extension to the algebraic closure $\overline{K(Y)}/K(Y)^{\text{sep}}$ is purely inseparable.  Thus for the corresponding base change $X_{\overline{\eta}}$ of $X_{\eta^s}$, the morphism $X_{\overline{\eta}} \to X_{\eta^s}$ is also purely inseparable.  In particular, it is a homeomorphism (with respect to the Zariski topologies).  Since $X_{\eta^s}$ is irreducible, also $X_{\overline{\eta}}$ is irreducible.  As explained in my comments above, $X_{\overline{\eta}}$ is LCI and generically reduced, hence (everywhere) reduced.  Thus, the geometric generic fiber $X_{\overline{\eta}}$ is integral and LCI.  However, there are examples (like quasi-elliptic fibrations) where $X_{\overline{\eta}}$ is not $R1$.      
Edit. The OP did not ask this, but I (and my advisees) have occasionally found it useful.  In the setting above, since the divisorial part of the closed subscheme $X_{\eta}^\text{sing} \subset X_{\eta}$ (cut out by the Fitting ideal of the sheaf of relative differentials, for example) must be inseparable over $K(Y)$, this forces certain numerical invariants to be divisible by the characteristic $p$.  So if you study varieties with those numerical invariants equal to a specified (nonzero) integer, if you exclude "small characteristics" that divide that integer, then $X_{\overline{\eta}}$ will be normal. 
Counterexample to Generic Reducedness. Let $k$ be a field of characteristic $p$.  Let $Y$ be $\mathbb{P}^2_k = \text{Proj}\ k[s_0,s_1,s_2]$.  Let $T$ be $\mathbb{P}^2_k = \text{Proj}\ k[t_0,t_1,t_2]$.  Let $X\subset T\times_{\text{Spec}\ k} Y$ be the closed subscheme $\text{Zero}(t_0^ps_0 + t_1^ps_1 + t_2^ps_2)$.  The projection $\text{pr}_T:X\to T$ is a smooth, projective morphism; it is Zariski locally a $\mathbb{P}^1$-bundle.  Define $f:X\to Y$ to be the other projection.  This morphism is nowhere smooth.  The geometric generic fiber is everywhere nonreduced.    
The argument above does prove that the geometric generic fiber is irreducible.  Moreover, if the geometric generic fiber is generically reduced, then it is everywhere reduced.  However, there are examples where the geometric generic fiber is everywhere nonreduced.<|endoftext|>
TITLE: Calculation of the integral related to the gravitational shock wave
QUESTION [11 upvotes]: The following integral
$$\int\limits_0^\infty \frac{\cos{\left(\frac{1}{2}\sqrt{3}s\right)}}{\sqrt{\cosh{s}-\cos{\theta}}}\,ds$$
can be found in the paper 

Tevian Dray and Gerard 't Hooft, The gravitational shock wave of a massless particle, Nuclear Physics B 253 (1985) 173--188, doi:10.1016/0550-3213(85)90525-5. 

They write that they "have not attempted to perform the integration explicitly". Was this integral ever calculated explicitly?

REPLY [6 votes]: A closed form exists in Gradshteyn and Ryzhik, 8.842.1.

As a comment to this answer:
Referring to the Gradshteyn and Ryzhik, the following result is given in http://arxiv.org/abs/hep-th/9408169 (On Gravitational Shock Waves in Curved Spacetimes, by K. Sfetsos):
  $$\int \limits_0^\infty \frac{\cos{(\sqrt{c-1/4}\,s)}}{\sqrt{\cosh{s}-\cos{\theta}}}\,ds=\frac{\pi}{\sqrt{2}\cosh{(\sqrt{c-1/4}\,\pi)}}F\left(\frac{1}{2}-i\sqrt{c-1/4},\frac{1}{2}+i\sqrt{c-1/4},1,\cos^2{\frac{\theta}{2}}\right)=\frac{\pi}{\sqrt{2}\cosh{(\sqrt{c-1/4}\,\pi)}}P_{-1/2+i\sqrt{c-1/4}}(-\cos{\theta}).
$$<|endoftext|>
TITLE: Bounding and dominating numbers ${\frak b}, {\frak d}$ via ultrafilters
QUESTION [7 upvotes]: Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$ and suppose that ${\cal U}$ is a free ultrafilter on $\omega$. We write $f \leq_{\cal U} g$ if $$\{n\in\omega: f(n) \leq g(n)\}\in{\cal U}.$$ 
Similar to the usual bounding and dominating numbers, we define $${\frak b}_{\cal U} = \min\{|B|: B\subseteq \omega^\omega \land \forall f\in\omega^\omega \exists g\in B(g\not \leq_{\cal U} f)\},$$
and 
$${\frak d}_{\cal U} = \min\{|D|: D\subseteq \omega^\omega \land \forall f\in\omega^\omega \exists g\in D(f \leq_{\cal U} g)\}.$$
Do we have ${\frak b} = {\frak b}_{\cal U}$ and ${\frak d} = {\frak d}_{\cal U}$?

REPLY [10 votes]: The relation $\leq_{\cal U}$ is very well-studied: it is the order relation on the model of the hyperreals obtained by taking an ultrapower of $\mathbb{R}$ with respect to $\cal U$.
Letting $\mathbb{R}^*_{\cal U}$ denote this ultrapower, you are asking for the smallest size of an unbounded set in $\mathbb{R}^*_{\cal U}$ (which you call $\frak b_{\cal U}$) and the smallest size of a cofinal set in $\mathbb{R}^*_{\cal U}$ (which you call $\frak d_{\cal U}$).
As Wojowu points out, these cardinals are the same because $\leq_{\cal U}$ is a linear order on $\mathbb{R}^*_{\cal U}$. Also, this cardinal is bounded below by $\frak b$ and above by $\frak d$.
But one can say a bit more. Andreas Blass has proved (in this paper) that $\frak g \leq \frak b_{\cal U}$. Mike Canjar proved that, if you add lots of Cohen reals to a model of CH, then any uncountable regular cardinal $\leq \frak c$ is equal to $\frak b_{\cal U}$ for some free ultrafilter $\cal U$. In other words, he showed that the value of $\frak b_{\cal U}$ can depend not only on your model of set theory, but within a single model it can also depend on $\cal U$.
If you want to learn more, I suggest you look at Andreas's paper, and also his answer to this related question of Joel David Hamkins.<|endoftext|>
TITLE: Weak*-closure of finite rank operators on dual space
QUESTION [14 upvotes]: Given a Banach space $X$, we consider the space $B(X^*)$ of bounded, linear operators on $X^*$ with the weak*-topology from its canonical predual $B(X^*)_*=X^*\hat{\otimes}X$. What is $\overline{F(X^*)}^{wk*}$, the weak*-closure of the finite rank operators on $X^*$? Since this is rather vague, here are some concrete questions:

Q1: Do we always have $K(X^*)\subseteq\overline{F(X^*)}^{wk*}$, i.e., is every compact operator in the weak*-closure of finite-rank operators?
Q2: Is there a characterization when $B(X^*)=\overline{F(X^*)}^{wk*}$, i.e., when the finite-rank operators are weak*-dense?
Q3: Is there a characterization when $B(X^*)=\overline{K(X^*)}^{wk*}$, i.e., when the compact operators are weak*-dense?

Considering $B(X)$ as a subalgebra of $B(X^*)$ in the usual way, we may ask related questions in connection with $F(X)$ and $K(X)$. The principle of local reflexivity implies $\overline{F(X)}^{wk*}=\overline{F(X^*)}^{wk*}$ in $B(X^*)$. However, it is not clear to me if we always have $\overline{K(X)}^{wk*}=\overline{K(X^*)}^{wk*}$ (I guess not). Therefore, we may also ask:

Q4: Do we always have $K(X)\subseteq\overline{F(X^*)}^{wk*}$?

REPLY [8 votes]: Regarding Q2: If and only if $X$ has the approximation property.  I'll use Ryan's book "Introduction to tensor products of Banach spaces" as a reference, see Prop 4.6 (but this is all standard stuff).
Theorem: $X$ has the approximation property if and only if, whenever $u=\sum_{n\geq 1} \mu_n\otimes x_n \in X^*\hat\otimes X$ is such that $ \sum_n \mu_n(x) x_n = 0$ for all $x\in X$, then $u=0$.
As we have $\sum_n \|\mu_n\| \|x_n\|<\infty$, the condition is equivalent to $\sum_n \mu_n(x) \mu(x_n)=0$ for all $x\in X,\mu\in X^*$ (Hahn-Banach) and hence also equivalent to $\sum_n \mu(x_n)\mu_n = 0$ for all $\mu\in X^*$, and so finally also equivalent to $\sum_n \mu(x_n) f(\mu_n) = 0$ for all $\mu\in X^*, f\in X^{**}$.
This in turn is equivalent to $\langle u, F \rangle=0$ for all finite rank operators $F$ on $X^*$, under your dual pairing between $X^*\hat\otimes X$ and $F(X^*)$.
Finally, observe that $F(X^*)$ is weak$^*$ dense in $B(X^*)$ if and only if the only element of $X^*\hat\otimes X$ which annihilates all of $F(X^*)$ is $0$.
There is a related definition of the "compact approximation property".  If I recall it correctly, then you can adapt the proof, and get
Thm: $X$ has the compact approximation property if and only if, whenever $u=\sum_{n\geq 1} \mu_n\otimes x_n \in X^*\hat\otimes X$ is such that $ \sum_n \langle\mu_n,T(x_n)\rangle = 0$ for all $T\in K(X)$, then $u=0$.
Thus, if $X$ has the compact approximation property, but not the approximation property (I think there is an example due to Willis) then we can find $u\in X^*\hat\otimes X$ which annihilates all of $F(X^*)$ but is not zero.  There is then $T\in K(X)\subseteq K(X^*)$ with $\langle T,u\rangle \not=0$, and as
$$ \overline{F(X^*)}^{wk^*}  = \{ T\in B(X^*) : \langle T,u\rangle=0
\text{ for all }u\in X^*\hat\otimes X\text{ with } \langle S,u\rangle=0
\text{ for all } S\in F(X^*) \} $$
we conclude that $K(X)$ is not contained in the weak$^*$-closure of $F(X^*)$.  So Q4 is a negative.<|endoftext|>
TITLE: Cells in affine Weyl groups
QUESTION [5 upvotes]: This may sound like a very general question, which pretty much reflects my ignorance on the subject.
In the case of Weyl groups $W$, there is a notion of left/right/double cells, which is roughly some partitions on the elements of $W$. It turns out that these cells afford representations of $W$ (or $W \times W$ in the case of double cells).
The structure of these cells are well-known, for example in the case of $W = S_n$, the partition of cells is related to the Robinson-Schensted algorithm. 
There are applications of cells in other aspects of mathematics. Due to my ignorance of the subject, I will simply refer what I mean by 'other aspects' to the notes here. 
On the other hand, Lusztig also defined 'cells' in affine Weyl groups in a series of papers under the title (Cells in affine Weyl groups), which I have literally zero knowledge about. Here is a couple of questions I am particularly interested in:
1) I read from Sommers-Gunnells (here p.7) that the double cells in affine Weyl group are parametrized by nilpotent orbits in the Langlands dual. This, to my knowledge, is quite different from the case of double cells in Weyl group, whose double cells are parametrized by special orbits in the Langlands dual (correct me if I am wrong). Can anyone give us some intuition on how the parameterization in the affine Weyl group case works?
2) There is a notion of canonical left cell in each double cell of affine Weyl group (here). Are there any results on how these canonical left cells are computed?

REPLY [7 votes]: You are asking a number of related questions here, most of which require more reading of Lusztig's papers.   See the reference list in my conference paper here, for example.  But note first that the notions about cells you mention are defined quite generally for Coxeter groups in the landmark 1979 paper by Kazhdan and Lusztig here.    What is special about affine or finite Coxeter groups is the precise detail Lusztig later worked out.   For example, in the fourth part of his series of papers on cells in affine Weyl groups, he arrived at a proof of his conjectured bijection between 2-sided cells and nilpotent orbits in the Lie algebra of a Langlands dual algebraic group of the relevant type.  [But it is very hard to get an intuitive feeling for this bijection.   As far as I know, there is not yet an alternative to Lusztig's complicated proof, which uses virtually all possible tools.  So this part of the subject remains challenging.]
There is still no proof of an older 1983 conjecture by Lusztig on the precise number of left cells in a 2-sided cell.    This and other such combinatorial matters are related to unsolved problems in representation theory.   Anyway, the precise relationship between cells in affine Weyl groups and finite Weyl groups has been worked out in Lusztig's papers and is nicely illustrated in the rank 2 cases in his Japanese conference paper. (See also the photo of Lusztig in his specially made $G_2$ t-shirt in the paper by Gunnells here.)   
Concerning canonical left cells, there has been further computational work by Chmutova and Ostrik in a 2002 paper here.  Gunnells and others have gone on to study further the difficult problem of characterizing the location of "distinguished involutions" in all left cells, which I suspect is closely related to modular representations of the corresponding Lie algebras. 
ADDED: To comment a little further, the occurrence here of the Langlands dual group is perhaps mysterioous, but for the reflection groups and cells it shows up mainly in the interchange of Lie types $B_\ell$ and $C_\ell$.   The embedding of a Weyl group into an affine Weyl group is less affeccted, since Weyl groups of these types are isomorphic.   But the dual version of an affine Weyl group, with its translation lattice expanded by a prime factor $p$, is crucial in characteristic $p$ representation theory of the algebraic groups involved (as Verma first noticed).
From Lusztig's work (especially his definition of "special" nilpotent orbits), one sees that a two-sided cell of the Langlands dual affine Weyl group meets a (unique) two-sided cell of the finite Weyl group precisely when the corresponding nilpotent orbit is special.    In type $A_\ell$ where $W = S_{\ell+1}$, all orbits are special, but otherwise not.  This adds a further layer of interest to the kind of parametrization you ask for.   Special orbits have a Lusztig-Spaltenstein duality, for type $A_\ell$ just the transpose involution on partitions of $\ell+1$.   This leads further, to Lusztig's special pieces of the nilpotent variety, which people now understand better in terms of an "exotic" version of the variety: see for example here. 
Some short unpublished notes (in a more general context, where the finite Weyl group appears as one of the parabolic subgroups of an affine Weyl group) are here, but don't deal with the geometric side.<|endoftext|>
TITLE: Maximal TB number and slice genus relation of a knot in any 3-manifold
QUESTION [6 upvotes]: Lisca-Stipsicz say that if a knot $K\subset{S^{3}}$ satisfies $g_{s}(K)>{0}$ and                  $TB(K)=2g_{s}(K)-1$ then $S^{3}_r(K)$ carries positive, tight contact structures
for every $r\neq{TB(K)}$ where $S^{3}_r(K)$ is the new manifold obtained by performing $r$-surgery to $K$ in $S^{3}$. Can we say this result is true for any 3-manifold not only for $S^{3}$?

REPLY [3 votes]: If $K \subset (M,\xi)$ is null-homologous, then it makes sense to define $TB(K)$ as in $S^3$.  The right definition of $g_s(K)$ is as a minimum over the genera of smooth surfaces in $M \times [0,1]$ with boundary $K \times \{1\}$.
With a null-homologous knot satisfying $TB(K) = 2g_s(K) -1$, and when the Heegaard Floer contact invariant $c(\xi)$ is non-vanishing, then $M_r(K)$ has a tight contact structure for all $r \neq TB(K)$.  For $r < TB(K)$, this follows from the fact that Legendrian surgery on any link preserves the non-vanishing of the Heegaard Floer contact invariant (in particular, choose the link as in Ding and Geiges's Surgery diagrams for contact 3-manifolds); for $r > TB(K)$, this result can be found as Theorem 1.7 in my pre-print Transverse Surgery on Knots in Contact 3-Manifolds.
When $c(\xi) = 0$ but $\xi$ is still tight, I believe that we know nothing except in the case where $K$ is fibred, whereupon $M_r(K)$ supports a tight contact structure for all $r \neq TB(K)$ (this is Theorem 1.6 in the above pre-print).<|endoftext|>
TITLE: Asymptotics for the number of abelian groups of order at most $x.$
QUESTION [7 upvotes]: The number of abelian groups of order $n$ (call it $a(n)$ is a studied subject (see http://oeis.org/A000688), but I can't seem to find any asymptotic results. Obviously, there is no asymptotic for $a(n)$ itself (it is much too irregular), but there should an asymptotic for $\sum_{n\leq x} a(n),$ but I can't seem to find a reference.
Roberto answered the above, but another question is whether one has any distributional results (how high are the maxima, is there a limiting distribution, etc).

REPLY [14 votes]: The problem was studied quite a bit; a complete summary will be complicated to give (in any case I cannot). A standard reference for classical results on this is A. Ivić "The Riemann Zeta-function: Theory and Applications" (1985); it seems there is a recent Dover edition. Ivić has various papers on this problem, too. The book has a chapter on this subject (14.5 Non-isomorphic abelian groups of a given order). At the moment I cannot recall what exactly is in there.
Some information related to this type of problem:
The correct asymptotics for $\sum_{n\in \mathbb{N}} a(n)$ were obtained by 
Erdős and Szekeres (1934), they proved: 
$$\sum_{n\le x} a(n) \sim A x + O(\sqrt{x})$$
with $A= \prod_{n \ge 2}\zeta(n)$. 
By now there are more precise results known. For example  the paper by Kendall and Rankin (1947), mentioned in another answer, gives $Ax + B x^{1/2} + O(x^{1/3} \log x )$, and developments continued, like the result mentioned by Roberto Pignatelli.
The to my knowledge latest improvement there is by Sargos and Wu (2000)
$$Ax + Bx^{1/2} + Cx^{1/3} + O(x^{55/219} (\log x)^7 )$$
The first to get the third term of the main term was Richert (1952). 
For the historical development of the error see the introduction of a paper by  Calderón (2003).
Related questions were also studied. For example,  Kendall and Rankin (1947) showed that the "local densities" of $a(n)$ exist, that is   $\sum_{n \le x, \, a(n)=k}1 \sim d_k x$ (good error terms are known). 
There are also estimates "in short intervals" so on $\sum_{ x \le n \le y, \, a(n)=k}1$.
For two recent papers on this see: 
Emre Alkan, On the enumeration of finite abelian and solvable groups, J. Number Theory 101 (2003), no. 2, 404--423.
Ekkehard Krätzel, The distribution of values of the enumerating function of finite, non-isomorphic abelian groups in short intervals, Arch. Math. (Basel) 91 (2008), no. 6, 518--525.
The introductions and references there will lead to various additional paper. (The first is freely accessible.) 
References
C. Calderón:  Asymptotic estimates on finite abelian groups. Publications De L’institut Mathematique, Nouvelle série, 74, 57-70 (2003).
P. Erdős, G. Szekeres: Über die Anzahl der Abelschen Gruppen gegebener Ordnung und über ein verwandtes zahlentheoretisches Problem (in German), Acta Litt. Sci. Szeged 7 (1934), 95--102; Zentralblatt 10,294. Free PDF
D. G. Kendall and R. A. Rankin, On the number of Abelian groups of a given order, Quart. J. Math., Oxford Ser. 18 (1947), 197--208.
Hans-Egon Richert, Über die Anzahl Abelscher Gruppen gegebener Ordnung. I, Math. Z. 56 (1952), 21--32.
P. Sargos and J. Wu, Multiple exponential sums with monomials and their applications in number theory, Acta Math. Hungar. 87 (2000), no. 4, 333--354.<|endoftext|>
TITLE: Mixed Hodge structure on sheaf cohomology of a variation of Hodge structures
QUESTION [15 upvotes]: I'm new here. I hope to do it right!
I am interested in studying mixed Hodge structures over complex algebraic surfaces and their generalizations.
Let us take a smooth complex variety $X$ and a variation of Hodge structures $\mathbb{V}$ over $X$ of weight $k$. That is, $\mathbb{V}$ is a local system of complex vector spaces with a finite decreasing filtration $\left\{\mathcal{F}^p\right\}$ of the holomorphic vector bundle $\mathbb{V} \otimes_\mathbb{C} \mathcal{O}_X$ defining a pure Hodge structure of weight $k$ on each stalk and such that, for the associated flat connection $\nabla$ over $\mathbb{V} \otimes \mathcal{O}_X$ whose horizontal sections are $\mathbb{V}$ we have $\nabla(\mathcal{F}^p) \subseteq \mathcal{F}^{p-1} \otimes \Omega_X^1$.
My question is: Is there a natural (i.e. functorial) way of obtaining a (mixed?) Hodge structure in the sheaf cohomology $H^p(X, \mathbb{V})$?
Related to this problem, Griffiths proved that, given a proper morphism $f: X \to S$ of maximal rank between complex varieties, with $X$ birrationally equivalent to a compact Kähler manifold, then, the relative de Rham cohomology sheaf
$$
H^p(X/S) :=R^pf_*\underline{\mathbb{C}} \otimes \mathcal{O}_S
$$
underlies a variation of Hodge structures compatible with the pure Hodge structure in the stalk $\left(H^p(X/S)\right)_s \cong H^p(f^{-1}(s), \mathbb{C})$ obtained as subvariety of a variety birrationally equivalent to a compact Kähler manifold.
Another try was to mimic Deligne's proof of existence of mixed Hodge structures on the cohomology of a smooth complex variety via the complex of logarithmic forms. However, it is not not clear what auxiliar resultion take when substituying the de Rham resolution $\underline{\mathbb{C}} \to \Omega^*_X$ by a resolution of $\mathcal{F}$.
Thank you so much in advance!

REPLY [20 votes]: The answer to your question is yes (provided the VHS is polarized). This is due to Morihiko Saito. It is implicitly contained his two long papers on (mixed) Hodge modules, and there is an explicit statement in his note
"Mixed Hodge modules and admissible variations" Compte Rendus 1989. 
When the base is a curve, it goes back to Zucker, "Hodge theory with degenerating coefficients" Annals 1979; and even earlier due to Deligne when the base is projective (unpublished, but see Zucker).
For a VHS of geometric origin there is a different construction due to me, "The Leray spectral sequence is motivic" Invent. 2005, but the MHS is the same as Saito's.<|endoftext|>
TITLE: Elementary congruences and L-functions
QUESTION [24 upvotes]: In a recent article, Emmanuel Lecouturier proves a generalization of the following surprising result: for a Mersenne prime $N = 2^p - 1 \ge 31$, the element 
$$ S = \prod_{k=1}^{\frac{N-1}2} k^k $$ 
is a $p$-th power modulo $N$, and observed that he did not know an elementary proof. Neither do I.
Numerical experiments suggest that $s_p$ is actually a $6p$-th power
modulo $N$. I can't even see why it is a quadratic residue, i.e., why
the following result (not proved in the article cited) should hold:
$$ T = \prod_{k=1}^{\frac{N+1}4} (2k-1) $$ 
is a square mod N.
For arbitrary primes $N \equiv 3 \bmod 4$, the following seems to hold: 
$$ \Big(\frac{T}{N} \Big) = \begin{cases}
    - (-1)^{(h-1)/2} & \text{ if } N \equiv  3 \bmod 16, \\
    - 1      & \text{ if } N \equiv  7 \bmod 16, \\
      (-1)^{(h-1)/2} & \text{ if } N \equiv 11 \bmod 16, \\
    + 1      & \text{ if } N \equiv 15 \bmod 16, 
\end{cases} $$
where $h$ is the class number of the complex quadratic number field with discriminant $-N$. This suggests a possible proof using L-functions (i.e. using methods in (Congruences for L-functions, Urbanowicz, K.S. Williams) and explains the difficulty of finding an elementary proof. 
My questions:

Is this congruence for $(T/N)$ known?
How would I start attacking this result using known results on L-functions?

REPLY [22 votes]: 1. First we show that
$$\left(\frac{T}{N}\right) = \begin{cases}
    - (-1)^{(h-1)/2} & \text{ if } N \equiv  3 \bmod 16, \\
      (-1)^{(h-1)/2} & \text{ if } N \equiv 11 \bmod 16. \\
\end{cases}$$
Consider the sets
$$A_0:=\{1,2,\dots,\tfrac{N-1}{2}\},\quad A_1:=\{1,3,\dots,\tfrac{N-1}{2}\},\quad A_2:=\{2,4,\dots,\tfrac{N-3}{2}\},$$
$$ A_3:=\{1,2,\dots,\tfrac{N-3}{4}\},\quad A_4:=\{\tfrac{N+1}{4},\tfrac{N+5}{4},\dots,\tfrac{N-1}{2}\}.$$
Consider also the corresponding character sums
$$S_i:=\sum_{a\in A_i}\left(\frac{a}{N}\right),\qquad i=0,1,2,3,4.$$
Modulo $N$, we have $2A_3=A_2$ and $-2A_4=A_1$. Therefore, $S_3=-S_2$ and $S_4=S_1$, and hence $S_1-S_2=S_3+S_4=S_0$. But also $S_1+S_2=S_0$, hence $S_1=S_0$ and $S_2=0$. On the other hand, it is known that $S_0=3h$, see e.g. Theorem 70 in Fröhlich-Taylor: Algebraic number theory. This means that $S_1=3h$, i.e. the number of quadratic nonresidues in $A_1$ equals
$$ n=\frac{N+1}{8}-\frac{3h}{2}=\frac{N+5}{8}-2h+\frac{h-1}{2}.$$
We conclude
$$ \left(\frac{T}{N}\right)=(-1)^n=(-1)^{(N+5)/8}(-1)^{(h-1)/2}. $$
This is the claimed formula (with Jeremy Rouse's correction).
2. Now we show that
$$\left(\frac{T}{N}\right) = \begin{cases}
    - 1      & \text{ if } N \equiv  7 \bmod 16, \\
    + 1      & \text{ if } N \equiv 15 \bmod 16.
\end{cases}$$
We proceed as before, but this time we get $S_3=S_2$ and $S_4=-S_1$, so that $S_2-S_1=S_3+S_4=S_0$. Together with $S_2+S_1=S_0$, this yields $S_1=0$ and $S_2=S_0$. Hence the formula for $n$ simplifies to $n=(N+1)/8$, and we conclude
$$\left(\frac{T}{N}\right)=(-1)^n=(-1)^{(N+1)/8}.$$<|endoftext|>
TITLE: Poincaré lemma for distributions
QUESTION [10 upvotes]: Let us consider a current on $\mathbb R^n$, that is a differential form whose coefficients are distributions. For simplicity, let us check the case of a $1$-form
$$
u=\sum_{1\le j\le n} u_j dx_j,\quad u_j\in \mathscr D'(\mathbb R^n),
$$
and assume that $du=0$, i.e. $\partial u_j/\partial x_k=\partial u_k/\partial x_j$. I want to prove that there exists $a\in\mathscr D'(\mathbb R^n)$
such that 
$da=u$.
The same question can be raised for tempered distributions and also for higher degrees.

REPLY [6 votes]: The case of a 1-form is also given in Proposition 4.3.9 on page 334 in John Horváth's Topological Vector Spaces and Distributions.<|endoftext|>
TITLE: "Set theory" founded on lists rather than sets
QUESTION [10 upvotes]: On a computer, sets are often represented rather "indirectly / implicitly", e.g. in terms of some properties that they or their members satisfy. But some sets can be represented more "directly / explicitly",  in the sense that the representation either directly contains, or gives an algorithm for enumerating, a representation of every individual element of the set. I realize that might still be quite a vague distinction, but hopefully it'll be clear enough for what follows.
Any "directly represented" set must of course be countable, and hence well-orderable regardless of one's stance on the axiom of choice. But more than that, the representation of the set must actually come with a particular well-ordering, since the elements must be represented or enumerated in some order regardless of whether that order is actually used for anything else. It's possible to some extent to work with such direct representations of sets without caring about the ordering, but one must first define an appropriate notion of equivalence up to reordering.
What would happen if we made such (implicitly) well-ordered countable sets the basis for a form of set theory, insisting on some sort of ω-rule to avoid ω-inconsistency, and then treated uncountable sets e.g. as proper classes, or as sets whose only properties we can discuss are those pertaining to their well-ordered countable subsets [edit: or quotients]? Has such a theory been characterized?

What would we still be able to prove? What statements would definitely be independent of such a theory?
What about statements in-between, where the definiteness depended on the precise formulation of the theory?
What kinds of nonstandard models could such a theory have?


Edit: I realize that one possible ambiguity is which countable well-orders are allowed in this theory. My intended meaning was that only well-ordered sets of type ≤ω would be given a priori, but at least some ordinals larger than this could be modelled in the sense that if X has given order-type ω then we can construct X×X, also with given order-type ω (via e.g. Cantor's pairing function) and then construct a well-ordering relation on X×X of type ω2, say.

REPLY [6 votes]: [O. Deiser, An axiomatic theory of well-orderings, Rev. Symb. Logic 4, No. 2, 186-204 (2011)] discusses an approach to the foundations of mathematics with lists as the primary objects.  It turns out that the expressive power of the new theory is the same as ordinary set theory, i.e., ZFC, if I remember correctly.  Note, however, that this theory allows for uncountable wellorderings.<|endoftext|>
TITLE: Is there a consistent arithmetically definable extension of PA that proves its own consistency?
QUESTION [15 upvotes]: I asked this on stackexchange with no answer.
The negation would be the obvious generalization of Gödel's second incompleteness from r.e. extensions of PA to any arithmetically definable extension of PA.
I see that there is a complete consistent $\Sigma^0_2$ extension of PA. Said theory therefore contains either the sentence "I am consistent" or "I am inconsistent". However, just because it's consistent and complete doesn't mean it contains the sentence "I am consistent" — it could contain the sentence "I am inconsistent", and remain consistent by being such that any model is a nonstandard model in which there exists a strictly nonstandard proof of its inconsistency

REPLY [8 votes]: I add some history to the Joel's answer and also some generalizations.
A similar example can be found in this article by R. Jeroslow. It is also possible to construct a complete and $\Sigma_2$(in fact $\Delta_2$) definable theory which proves its own consistency, see this aricle of M. Kasa. I have presented another method for constructing such a theory 
in this paper which also proves this generalization :
For every $n>1$ there is a complete $\Delta_n$-definable
theory $T$ which contains PA, "T contains PA" is PA-provable and $T$ proves its own $\Sigma_{n-2}$-soundness.
(note that for $\Sigma_1$-complete theories, $\Sigma_0$-soundness is equivalent to simple consistency). 
It is also proved that if the theory $T$ is $\Sigma_n$-definable, $\Sigma_{n-1}$-sound and "T contains PA" is PA-provable, then $T$ can not prove its own $\Sigma_{n-1}$-soundness and it can be seen as a generalization of the second incompleteness theorem.<|endoftext|>
TITLE: Rational homology sphere that is not Seifert manifold
QUESTION [10 upvotes]: I wonder if there is an example of rational homology sphere that is not a Seifert manifold. If there is, how can one construct such a rational homology sphere from a surgery of a knot in $S^3$?

REPLY [13 votes]: By Thurston, all but finitely many $(p,q)$-surgeries on a hyperbolic knot in $S^3$ result in hyperbolic rational homology spheres for $p\neq 0$. In particular there are infinitely many integral homology spheres among them.

REPLY [5 votes]: If you glue opposite faces of a dedecahedron with a twist of $\frac\pi5$, you obtain the classical Poincaré sphere. It is an integral homology sphere, and a Seifert manifold, see BS' comment below.
But if you do the gluing with a twist of $\frac{3\pi}5$, then you obtain the Seifert-Weber manifold, which is a rational homology sphere with a hyperbolic structure, so definitely not a Seifert manifold. It has $H_1\cong(\mathbb Z/5)^3$, see Neil Hoffman's comment. I do not know what knot would give this manifold. But you can deduce many of its properties by drawing a picture of a dodecahedron with the appropriate corners, edges and so on identified.<|endoftext|>
TITLE: Why do we want $p$-permutation modules in splendid equivalences?
QUESTION [9 upvotes]: First Rickard (in Splendid Equivalences: Derived Categories and Permutation Modules ) and then Rouquier (Block theory via stable and Rickard equivalences, Appendix A.1) define splendid equivalences between (principal) blocks of algebras $A$ of $\mathbb{K}G$ and $B$ of $\mathbb{K}H$ with an isomorphic defect group $P$ this way:
$1)$ There exists a complex $X$ of finitely generated $(A,B)$-bimodules such that $\operatorname{Hom}_A (X,X) \simeq B$ and $\operatorname{Hom}_B(X,X) \simeq A$ in the homotopy category of complexes of $B$-modules (resp. $A$-modules), and all terms of $X$ are projective as left and right modules (this is called a split-endomorphism two sided tilting complex)

$2)$ All the terms of $X$, considered as modules of the group algebra of $G \times H$, are relatively projective with respect to the diagonal embedding of $P$ and are $p$-permutation modules (direct summands of permutation modules)
Now, reading both papers, I haven't really been able to understand the requirement that these modules have to be $p$-permutation modules, meaning I don't understand what we lose if we have a split-endomorphism two sided tilting complex made of $\operatorname{diag}(P)$-relatively projective modules that are not $p$-permutation modules.
I am kind of new to this theory, so I'm probably missing something huge. Thanks to anyone who will help me.

REPLY [14 votes]: The motivation for the definition was an attempt to explain structurally the phenomenon of an "isotypy". This makes sense for arbitrary blocks, but let's stick to principal blocks for simplicity.
Suppose $G$ is a finite group with abelian Sylow $p$-subgroup $P$, and $H=N_G(P)$ is the normalizer of $P$. Then Broué's Abelian Defect Group Conjecture predicts that the derived categories of the principal blocks of $kG$ and $kH$ (where $k$ is a sufficiently large field of characteristic $p$) should be equivalent. For any subgroup $Q\leq P$, the principal blocks of $kC_G(Q)$ and $kC_H(Q)$ should also have equivalent derived categories, and it seems reasonable to hope for some kind of compatibility between these equivalences for varying $Q$. This is what, at the level of character theory, Broué's notion of "isotypy" gave.
If you have a splendid equivalence between the principal blocks of $kG$ and $kH$, then you can obtain one between each pair $kC_G(Q)$ and $kC_H(Q)$ by applying a certain functor (the "Brauer construction") to the complex $X$. The main reason that it's important that the terms of $X$ are $p$-permutation modules is that, although it's possible to define the Brauer construction for general modules, it behaves much better for $p$-permutation modules, and this is needed to prove that it gives a tilting complex for $kC_G(Q)$ and $kC_H(Q)$.
There are also a couple of secondary reasons.
One is that it's useful to be able to pass between characteristic $p$ and characteristic zero by considering representation theory over a complete discrete valuation ring $\mathcal{O}$ of characteristic zero with residue field $k$. The fact that $p$-permutation modules over $k$ lift to ($p$-permutation) modules over $\mathcal{O}G$, which is not true for modules in general, is needed to prove that a splendid equivalence lifts to characteristic zero.
The other, rather vaguer, is that if there's some geometric origin for the tilting complexes then one might expect that the complexes should have terms related to permutation modules.<|endoftext|>
TITLE: What are fun elementary subjects in probability?
QUESTION [13 upvotes]: I have to read several lectures on probability or applications of probability for high school students (of high level). There is no necessary part I must lecture, that is, my aim is just advertisement. What are possible subjects? 
There are many nice examples in Alon-Spencer book, with applications to combinatorics and number theory. Is there something in this spirit, but rather on probability itself?

REPLY [3 votes]: A quincunx (balls dropping through what is typically a triangle of nails, creating a roughly binomial distribution) is fun to watch. Real ones are extremely sensitive to errors in the locations of the nails; there are simulations on the internet, somewhere.<|endoftext|>
TITLE: Weight filtration on certain Galois representations
QUESTION [9 upvotes]: Let $G$ be the absolute Galois group of a number field $K$. Let $\ell$ be a prime number. There are representations $\mathbb{Z}_\ell(n)$ of $G$ on the group of $\ell$-adic integers given by the formula $g.x=\chi(g)^nx$ where $\chi:G\to \mathbb{Z}_\ell^{\times}$ is the cyclotomic character. 
Question: Is it true that the ext groups $\mathrm{Ext}^*(\mathbb{Z}_\ell(0),\mathbb{Z}_\ell(n))$ vanish for $n$ negative ?
The reason I am asking this is that this would imply that the triangulated subcategory of the derived category of Galois representations spanned by the objects $\mathbb{Z}_\ell(n)$ has a weight filtration, as constructed for instance in Lemma 1.2. of this paper of Marc Levine:
https://www.uni-due.de/~bm0032/publ/TateMotives.pdf

REPLY [8 votes]: No, life is not so easy I'm afraid. For instance, the group $\mathrm{Ext}^1_{G_{\mathbf{Q}}}(\mathbf{Z}_\ell, \mathbf{Z}_\ell(n)) = H^1(\mathbf{Q}, \mathbf{Z}_\ell(n))$ has positive rank for all odd integers $n$, whatever the sign; this is easy to see from Tate's global Euler characteristic formula. 
The point is that if you have two irreducible Galois representations $V_1, V_2$, and you know that $V_1$ and $V_2$ arise in geometry (as the realisations of motives $M_1, M_2$), then there are in general many more extensions of $V_1$ by $V_2$ in the category of Galois reps than there are extensions of $M_1$ by $M_2$ in the category of mixed motives. 
But all is not lost: there is a beautiful and deep theory that seeks to characterise in terms of local properties at $\ell$ those extensions which come from geometry. You might like to read Bloch and Kato's article in the Grothendieck Festschrift. The upshot is that for a geometric Galois representation $V$, one defines a group $H^1_\mathrm{f}(K, V) \subseteq H^1(K, V)$, which parametrises those extensions of the trivial rep by $V$ which are expected to arise in geometry. It is expected that $H^1_\mathrm{f}(\mathbf{Q}, V)$ is zero if the Hodge--Tate weights of $V$ are $\le -1$, and this is known for the representations $V = \mathbf{Q}_\ell(n)$ by a theorem of Soule.<|endoftext|>
TITLE: What is a Futaki invariant, what is the intuition behind it, and why is it important?
QUESTION [16 upvotes]: As the question title suggests, what is a Futaki invariant, what is the intuition behind it, and why is it important?

REPLY [9 votes]: The Futaki invariant $F(X,[\omega])$ is a quantity that needs two pieces of information on a compact complex manifold $M$. 
1) A Kahler class $[\omega]$
2) A holomorphic vector field $X$.
It is an obstruction for the existence of a constant scalar curvature Kahler (cscK) metric in the Kahler class $[\omega]$. If it does not vanish then there cannot exist a cscK metric in that class. (The converse is not true in general.)
Its definition is as follows. Let $\hat{S}$ be the average scalar curvature with respect to any Kahler metric. This is a topological quantity. Let $S-\hat{S} = \Delta f$. Then $F(X,[\omega])=-\displaystyle \int _M X(f) \omega^n$. It is a miraculous fact that this quantity depends solely on the Kahler class (as opposed to the specific Kahler metric chosen).
Now, one question could be "How the heck could one have come up with such an invariant by oneself?" The deeper idea is the following simple one, originally due to Bourgougnon. If $G$ is a lie group acting on a manifold $M$, and $\alpha$ is a $G$-invariant closed $1$-form, then $\alpha(X)$ is a constant for every $X$ arising from the Lie algebra of $G$. When one applies this philosophy for appropriately chosen $G, M,$ and $\alpha$ (where infinite dimensional things are allowed) then one gets a wealth of invariants including the Futaki invariant.<|endoftext|>
TITLE: Are these two quotients of $\omega^\omega$ isomorphic?
QUESTION [14 upvotes]: Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$. For $f,g\in\omega^\omega$ we say $f\simeq_{\text{fin}} g$ if there is $n\in \omega$ such that $f(k) = g(k)$ for all $k\geq n$.
We say that $A\subseteq \omega$ has measure 1 if $$\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n+1\}|}{n+1} = 1.$$We define another  equivalence relation on $\omega^\omega$ by setting $f\simeq_1 g$ if $$\{n\in\omega:f(n)=g(n)\} \text{ has measure } 1.$$
When I look at the posets $(\omega^\omega/\simeq_{\text{fin}}, \leq_{\text{fin}})$ and $(\omega^\omega/\simeq_{1}, \leq_{1})$ (where the ordering relations are defined "pointwise modulo the equivalence relation"), I get the feeling that they are not isomorphic and am quite certain this is correct. However a proof has eluded me. Any help is appreciated!

REPLY [18 votes]: Very nice question!
They are not isomorphic.
What I claim is that when we take the quotient with respect to density, there is a countably infinite antichain above $0$ having a minimal upper bound, but when we take the quotient modulo finite, there is no such antichain. This is a property that distinguishes the isomorphism types.
To see this, I shall use the characteristic of the size of the
smallest infinite maximal almost disjoint family of sets in two
quotients. It turns out that these cardinals are different for the
two quotients.
Namely, on the one hand, the size of the smallest infinite maximal
almost-disjoint family modulo the finite sets is well-known to be
uncountable. This is the cardinal characteristic known as $\frak
a$.
On the other hand, when we take the quotient of $P(\omega)$ modulo
the ideal of asymptotic density zero sets, usually denoted $Z_0$, then it turns out that there is a countably infinite maximal almost disjoint family.
Simply let $A_0$ be half the numbers (the evens, say); let $A_1$
be half of what is left; let $A_2$ be half of what is left after
that, and so on. These sets are disjoint, union to the whole space and $A_n$ has density $1/2^{n+1}$; so for any set $A$ with positive density, there is large enough $N$ such that it does not concentrate on the tail sets $\bigcup_{k\geq N}A_k$, since this union is too sparse, and so it must have $A\cap A_n$ with positive density for some $n<N$. (See additional detail in lemma 2.5 in
Barnabás Farkas, Lajos Soukup, The zero density
ideal, cardinal invariants and related forcing problems.)
If $f_n$ is the characteristic function of $A_n$, then it follows
that $0<f_n<1$ for all $n$, modulo density, and the $f_n$ are incomparable and
have meet $0$ and the constant-$1$ function is a minimal upper
bound of the $f_n$, since if $h<f$ modulo density, then the set
where it dropped would have positive density on some $A_n$, and in
this case, we wouldn't have $f_n\leq h$ modulo density. So in the
density quotient $\omega^\omega/Z_0$, we have the countable
antichain above $0$ with a minimal upper bound.
But this situation cannot arise in the other quotient
$\omega^\omega/\text{Fin}$. Namely, if we have $0<g_n<h$ modulo finite and
$g_n\wedge g_m=0$ for all $n\neq m$, then the support of the
$g_n$'s must form an almost-disjoint family. But since it is
countable, there is another set almost disjoint from all these
supports, and contained in the support of $h$. So we can reduce
$h$ on that set to find a strictly smaller $h'<h$, which is still
an upper bound of the $g_n$. So $h$ is not a minimal upper bound
of the $g_n$ modulo finite.
So the two quotient order structures are not isomorphic.<|endoftext|>
TITLE: Recent observation of gravitational waves
QUESTION [54 upvotes]: It was exciting to hear that LIGO detected the merging of two black
holes one billion light-years away. One of the black holes had 36
times the mass of the sun, and the other 29. After the merging the
mass was that of 62 suns, with the rest converted to gravitational
radiation. Could someone explain, without assuming a deep knowledge of
general relativity, how these conclusions were reached?

REPLY [23 votes]: The numbers 36,29,62 were obtained as the best match between the received signals and the output of computer simulations. The 90% confidence intervals on these numbers are about $\pm 4$.  The details (largely beyond my understanding) are in the technical paper here.<|endoftext|>
TITLE: Basic question about polytope duals
QUESTION [6 upvotes]: The following must be well known.  Is there a beginning or midlevel
text where the answer is discussed?  Thanks.
Along with a polytope one has the notion of its dual which is officially
defined via the inner product.  However, in three dimensions at least,
the dual is often pictured simply by placing a point in each face and then 
taking the convex hull.  Will this same method work in general?
Question: Let P be an n-dimensional polytope.  Place a point at the barycenter
of each facet of P and designate by Q the convex hull of these points.  Is the
resulting polytope Q combinatorially equivalent to the dual of P ?

REPLY [3 votes]: While the answer to the question of OP is "no", I would like to mention a question raised by Grünbaum and Shephard in "Some problems on polyhedra.", J. Geom. 29 (1987), no. 2, 182–190, for the information of those who may be interested.
They say that a polytope $P$ is of DV-type if it has a combinatorially equivalent realization such that every vertex $v$ of $P$ is an interior point of the face of a dual $P^*$ that corresponds to $v$ in the duality.  The questions
1. Is every polyhedron of DV-type?
2. Does every polyhedorn of DV-type have a dual of DV-type?
3. Which polyhedra of DV-type are also of SV-type? (inscribable w.r.t. a sphere)

I'm not sure about the current status of the question.<|endoftext|>
TITLE: Reference Request for Hilbert Schemes
QUESTION [16 upvotes]: I'm a physicist working on Fractional Quantum Hall effect. The mathematical subjects of study are symmetric, translational invariant, homogeneous polynomials on $\mathbb{C}$. Very early in my study I realized that there is some beautiful geometry underneath all of the physics. As I dug deeper, turns out ($\to$Nakajima), other than symmetric polynomials, the Heisenberg algebra has also a representation via Hilbert schemes. So I decided to learn about Hilbert schemes from ZERO knowledge about algebraic geometry.
It has been months that I've been trying to learn more and more algebraic geometry. As fascinating as it is, I realized it is about time I ask for some serious guidance. Assuming I know nothing about algebraic geometry (which is not true), what is the most direct (maybe a 20 step program) to learn at least what Nakajima is doing and how to modify his method to my less well-behaving (translational invariant) polynomials?
The problem is algebraic geometry, as I've seen it, is gigantic. Any text explores lots of different avenues. It is kind of impossible to learn it all in one year; it takes time and I'm prepared to give it time. But in the meantime my research is hurting! 
So if someone could be so kind to introduce to me a series of concepts (+maybe good references for those concepts) to get step by step closer to understand specifically what is going on in Hilbert scheme, I would be very grateful. Forgive me if maybe this is not the best place to ask such a thing.

REPLY [4 votes]: I am not an expert on Hilbert schemes but I would recommend the following references:

Eisenbud, Harris, "the geometry of schemes". A beautiful, relatively short introduction to scheme theory. Specifically section II.3 will be relevant, giving a concrete feeling for the notion of multiple points by means of concrete examples. 
Nakajima, "Lectures on Hilbert schemes of points on surfaces". As you said, difficult, but some sections are easier than others. For example section 1.4 should be fairly accessible, especially after understanding II.3 of Eisenbud and Harris.    
Stromme, "Elementary introduction to representable functors and hilbert schemes". A more difficult reference than the previous ones but nicely written. Contains a proof of the existence of Hilbert schemes in general.
Bolognese, Losev, "A general introduction to the Hilbert scheme of points on the plane". In the spirit of Nakajima. 
de Cataldo, Migliorini, "The Douady space of a complex surface". A good reference for the analogue of the Hilbert scheme in complex analytic geometry. 
Voisin, "on the Hilbert scheme of an almost complex four-fould"
Technical, but written in more differential geometric terms. 

In general I would say it's important to get a good feeling for the notion of multiple points, flat morphisms (which are relatively easy to understand in the case the fibers are all 0-dimensional, see also Fischer, "complex analytic geometry" section 3.13) and representable functors.<|endoftext|>
TITLE: Who said "the naive counting numbers don't exhaust $\Bbb N$"?
QUESTION [8 upvotes]: In the context of Robinson's framework, or more precisely its reformulation by Ed Nelson, one of the practitioners in the field expressed the sentiment something like "the naive counting numbers don't exhaust $\mathbb{N}$."  Who was this mathematician and what was the precise wording?
For a follow-up question see What's Reeb's take on naive integers?

REPLY [3 votes]: Georges Reeb, I believe. At least that was his slogan, and C. Lobry made it the title of an uproarious book: Et pourtant... ils ne remplissent pas N! Reeb would go around conferences and confront random attendees with the quote, insisting that even Bourbaki said that.<|endoftext|>
TITLE: Real rank zero of group $C^*$-algebras
QUESTION [9 upvotes]: The concept of real rank zero of a $C^*$-algebra is introduced as non-commutative analogue of dimension ( topological dimension ). For example, it shown (by Brown-Pedersen) such that, if $X$ is a compact Hausdorff space, then $RR$ $C(X)=dim(X)$.  
Let $A$ be a $C^*$-algebra. we define $RR(A)=0$ iff every self-adjoint element in $A$ can be approximate by invertible, self-adjoint element in $A$.
 for example every Von-Neumman algebra has real rank zero.
So, on the other hand since $C^*(\mathbb T)=C(\mathbb Z)$ and $C^*(\mathbb Z)=C(\mathbb T)$, then $RR$ $C^*(\mathbb T)=0$ but
$RR$ $C^*(\mathbb Z)$$\not=$$0$ .
Question: Which locally compact groups $G$ satisfy $RR$ $C^*(G)=0$?
NOTE: If $G$ is locally compact group with fixed  Harr measure, let  full group $C^*$-algebra $C^*(G)$ be the  completion of the convolution algebra $L^1(G)$ with respect to the norm $\lVert f\rVert_{C^*(G)}$=$\sup $$\{\lVert\phi(f)\rVert\}$, where the supremum is taken over all $*$-representation $\phi$ of $L^1(G)$ as an algebra of bounded linear operators in a  Hilbert space.

REPLY [8 votes]: Here is a partial answer, and a pointer towards some literature that should be relevant.$\newcommand{\Cst}{{\rm C}^*}$
Doing some searching online brings up an old result of Kaniuth (Proc. AMS, 1993), which implies that if $G$ is a connected non-compact group then its full $\Cst$-algebra has real rank strictly greater than zero. The proof relies heavily on structure theory for Lie groups, but the structure theory used seems to be fairly standard material from that area.
Theorem 2 of Kaniuth's paper gives some examples of nilpotent locally compact groups whose $\Cst$-algebras have real rank zero.
Note that the $c_0$-direct sum of finite-dimensional matrix algebras always has real rank zero (this is a nice exercise in understanding the definition) and hence $\Cst(G)$ has real rank zero for every compact group $G$.<|endoftext|>
TITLE: Has anything ever been done with the set $\{1,2,3,4,\ldots\}$ equipped with the operation $a \oplus b = a+b-1$ and the usual notion of multiplication?
QUESTION [18 upvotes]: Definition. $$\mathbb{J} = \{1,2,3,\ldots\}.$$
We can refer to the elements of $\mathbb{J}$ as "joiners."

The product of joiners is inherited from $\mathbb{Z}$.

The sum of joiners will be defined by $$a \oplus b = a+b-1.$$



The motivation is that if we're thinking of the elements of $\mathbb{J}$ as non-empty finite totally ordered sets, then we can think of $a \oplus b$ as the result of "identifying" or joining the greatest element of $a$ to the least element of $b.$
This makes the $\mathbb{J}$ into a commutative monoid in two different ways. The good thing is that the identity elements of both these monoids are the same (namely $1$). The bad thing is that the distributivity law doesn't hold. Nonetheless, we do have the "variant distributivity" law:
$$(a(b \oplus c)) \oplus a = ab \oplus ac$$
Proof.
$$\mathrm{LHS} = a(b+c-1) \oplus a = (ab+ac-a) \oplus a = ab+ac-a+a-1 = ab+ac-1= \mathrm{RHS}$$
From this, we can deduce further "variant distributivity" identities, one for each natural number. For example, the $n=3$ case is $$a(b \oplus c \oplus d) \oplus a \oplus a = ab \oplus ac \oplus ad$$
This motivates the following definition.

Definition. Define that a flask is a pair of monoids on a common set, one denoted concatenatively, the other denoted with $\oplus$, such that:

$\oplus$ is commutative
their identity elements are equal and denoted $1$, and
the variant distributivity law holds (on both sides).


It makes sense to speak of "commutative flasks" of course, the prototypical example being $\mathbb{J}$. There's lots of other flasks, though; for example, every ring $R$ can be viewed as a flask in the obvious way, and if $R$ is commutative, the corresponding flask will obviously be commutative. Note that we can view $\mathbb{J}$ as a subflask of $\mathbb{Z}$. This implies that subrings can have subflasks that aren't themselves subrings (because they don't contain $0$ and aren't closed under subtraction.)

Question. Has anything interesting ever been done with the algebraic structure I'm denoting $\mathbb{J}$ and/or the class of algebraic structures I'm referring to as flasks?
If so, is there a book or article that provides a good reference?

REPLY [19 votes]: I believe this could be related to an algebraic structure useful for studying set-theoretical solutions of the Yang-Baxter equation, the so-called braces. At least the funny distributive property appears in the theory of set-theoretical solutions of the Yang-Baxter equation.
Definition. An abelian group $(A,+)$ is a left brace if there is a multiplication $m\colon A\times A\to A$, $(a,b)\mapsto ab$, turning $A$ into a group such that 
$$a(b+c)+a=ab+ac$$ 
holds for all $a,b,c\in A$.
Similarly one defines right braces.
The connection to the Yang-Baxter equation is as follows. First, for each left brace $A$ 
one has a canonical solution $(A,r_A)$ of the Yang-Baxter equation: 
$$r_A\colon A\times A\to A\times A,\quad r_A(a,b)=(ab-a,(ab-a)^{-1}ab).$$
One also has the following.
Theorem. Let $X$ be a set, $$r\colon X\times X\to X\times X,\quad r(x,y)=(\sigma_x(y),\tau_y(x))$$ be a 
non-degenerate involutive solution of the Yang–Baxter equation. (Non-degenerate means that $\sigma_x$ and $\tau_x$ are permutations for each $x\in X$, and involutive means that $r^2=\mathrm{id}_{X\times X}$.) Then there exists a unique left brace structure over the group 
$$G=G(X,r)=\langle X:xy=\sigma_x(y)\tau_y(x)\rangle$$
such that its associated solution $r_G$ satisfies
$r_G(\iota\times \iota) = (\iota\times \iota)r$, where $\iota\colon X\to G(X,r)$ is the canonical map.
Furthermore, if $B$ is a left brace and $f\colon X\to B$ is a map such that $(f\times f)r=r_B(f\times f)$,
then there exists a unique brace homomorphism $\varphi\colon G(X,r)\to B$ such that $f = \varphi\iota$.
and $(\varphi\times\varphi)r_G = r_B(\varphi\times\varphi)$. 
The theorem is implicit in the work of Rump where braces were discovered:

Rump, Wolfgang. Braces, radical rings, and the quantum Yang-Baxter equation. J. Algebra 307 (2007), no. 1, 153--170. MR2278047

You can find some interesting examples of braces here:

Bachiller, David. Classification of braces of order $p^3$. J. Pure Appl. Algebra 219 (2015), no. 8, 3568--3603. MR3320237, arXiv: 1407.5224

See also this survey:

Rump, Wolfgang. The brace of a classical group. Note Mat. 34 (2014), no. 1, 115--144. MR3291816

There is a non-commutative version of braces useful to study 
non-involutive solutions, see this paper.
Added. Rump proved that braces on both sides are in bijective correspondence with radical rings. The correspondence is as follows. If $R$ is a non-zero radical ring (for all $x\in R$ there is $y\in R$ such that $x+y+xy=0$) then $R$ with $a\circ b=ab+a+b$ is a two-sided brace. Conversely, if $A$ is a two-sided brace, then $A$ with $a*b=ab-a-b$ is a radical ring.<|endoftext|>
TITLE: Generalized Plateau problem with non-Jordan boundary
QUESTION [5 upvotes]: Let $C_\pm$ be the two circles obtained by intersecting the cylinder $x^2+y^2=R^2$ with the planes $z=\pm 1$, on which we mark four points $A_\pm:(R,0,\pm 1)$ and $B_\pm:(-R,0,\pm 1)$.  Assume that $R$ is big enough such that minimal surface between $C_\pm$ is a catenoid.
I want to find a minimal surface of disk type $f:D \to \mathbb{R}^3$ such that $\partial D$ is mapped to the curve $\gamma$ as follows: $\gamma$ starts from $A_+$, then around $C_+$ by $+3\pi$ (clockwise) to $B_+$, then straight down to $B_-$, around $C_-$ by $-3\pi$ (anticlockwise) to $A_-$, finally straight back to $A_+$.
This curve is not a Jordan curve.  I wonder if there are any results on "Plateau problem" with this kind of boundary.

REPLY [7 votes]: The paper below shows that a least area can be found spanning any rectifiable curve.  Sometimes the disk is not immersed however, as with a disk bounding a figure 8 in the plane.
J. Hass, Singular curves and the Plateau problem, International J. of Math. 2, (1991) 1-16.
For your example, there is a more straightforward construction of a minimal disk with boundary $\gamma$ that gives a disk that is immersed. You assume that a catenoid spans $C_+$ and $C_-$.  The union of this catenoid and the cylinder between $C_+$ and $C_-$ is a torus that bounds a solid torus.  The boundary of this piecewise smooth solid torus has mean curvature that points into the solid torus (or at least never out of it) and the angles at the curves where the two surfaces intersect are less than $\pi$. This "mean convex" boundary condition suffices to allow the classic solution of Plateau's problem to be solved for simple curves, as shown in the Meeks Yau papers- see "Topology of Three Dimensional Manifolds and the Embedding Problems in Minimal Surface Theory" William H. Meeks and Shing-Tung Yau, Annals of Mathematics, Vol. 112, No. 3 (Nov., 1980), pp. 441-484.
Now your curve is not embedded on the boundary of the solid torus, but the curve lifts to an embedded curve in a double cover of the solid torus.  Solve the Plateau problem there (using the induced metric, also mean convex) and project to get a minimal disk.
For large $R$ this will be the least area disk spanning $\gamma$. It stays close to the cylinder. For small $R$ I think there might be a smaller disk that cuts across the axis.<|endoftext|>
TITLE: Cap product on Leray-Serre spectral sequences
QUESTION [10 upvotes]: Let  $\ p:X\to B\ $ be a fibration of spaces with fiber $F$. Then there are homological and cohomological Leray-Serre spectral sequences $E^{r}_{pq}$ and $E_r^{pq}$ that converge to $H_*(X)$ and $H^*(X)$ respectively. It is well known that the cohomological spectral sequence $E^{pq}_r$ is multiplicative (spectral sequence of algebras) with the product induced by the cup product $$\smile\: :E_r^{pq}\times E_r^{p'q'}\longrightarrow E_r^{p+p',q+q'}.$$
Moreover, the product on $E_\infty^{**}$ is induced by the cup product on $H^*(X)$ and the product on $E_2^{**}$ given by the 'double' cup product on $H^*(B,H^*(F)).$
Question: Is it possible to define the cap product on the spectral sequences $$\frown\: :E_{pq}^r\times E_r^{p'q'} \longrightarrow E^{r}_{p-p',q-q'} $$
with the similar properties? I believe it is possible but I need a reference.

REPLY [8 votes]: I am sorry that I don't know a reference, so I will try to give a proof instead.
We follow Douady's approach using Cartan-Eilenberg systems, see here.
Let $B$ be a CW complex and $\pi\colon X\to B$ a Serre fibration. Put $X^k=\pi^{-1}(B^k)$. A cellular approximation~$\Delta_B\colon B\to B\times B$ of the diagonal can be lifted to an approximation $\Delta\colon X\to X\times X$ of the diagonal such that
$$X^k\stackrel\Delta\longrightarrow\bigcup_{m+n=k}X^m\wedge X^n\;.$$
Let $(\tilde h_\bullet,\partial,\wedge)$ be a reduced multiplicative generalised homology theory, and let $(\tilde h^\bullet,\delta,\wedge)$ be the corresponding cohomology theory. We define Cartan-Eilenberg systems $(H,\eta,\partial)$,  $(H',\eta',\partial')$ by
$$H(p,q)=\tilde h_\bullet(X^{-p},X^{-q})\qquad\text{and}\qquad
H'(p,q)=\tilde h^\bullet(X^{q-1}/X^{p-1})$$
for~$p\le q$ with the obvious maps $\eta\colon H(p',q')\to H(p,q)$, $\eta'\colon H'(p',q')\to H'(p,q)$ for $p\le p'$, $q\le q'$.
The maps $\partial\colon H(p,q)\to H(q,r)$, $\partial'\colon H'(p,q)\to H'(q,r)$ come from exact sequences of triples.
Sorry for the weird indexing,
but this makes it the easiest to use Douady's result.
The spectral sequence will not end up in the first quadrant, but you can surely bring it back into its rightful place. 
We ignore the grading; it is easy to fill in.
To define a spectral product $\mu\colon(H,\eta,\partial)\times(H',\eta',\partial')\to(H,\eta,\partial)$ we consider the map
\begin{multline*}
  F_{m,n,r}\colon(X\wedge X)^{m+n}/(X\wedge X)^{m+n-r}
  \cong\bigcup_{a+b=m+n}(X^a\wedge X^b)\Bigm/
  \bigcup_{c+d=m+n-r}(X^c\wedge X^d)\\
  \begin{aligned}
    \twoheadrightarrow\mathord{}&\bigcup_{a+b=m+n}(X^a\wedge X^b)\Bigm/
    \Bigl(\bigcup_{a=0}^{m-1}(X^a\wedge X^{m+n-a})
    \cup\bigcup_{b=0}^{n-r}(X^{m+n-b}\wedge X^b)\\
    \cong\mathord{}&\bigcup_{a=0}^{r-1}(X^{m+a}\wedge X^{n-a})\Bigm/
    \bigl(X^{m-1}\wedge X^n\cup X^{m+r-1}\wedge X^{n-r}\bigr)\\
    \hookrightarrow\mathord{}& X^{m+r-1}\wedge X^n\bigm/
    (X^{m-1}\wedge X^n\cup X^{m+r-1}\wedge X^{n-r})\\
    \cong\mathord{}&(X^{m+r-1}/X^{m-1})\wedge(X^{n}/X^{n-r})\;.
  \end{aligned}
\end{multline*}
Together with the diagonal map $\Delta$, for $r\ge 1$, we define
\begin{multline*}
  \mu_r\colon H(-n-m,r-n-m)\otimes H'(m,m+r)\\
  \begin{aligned}
    &\cong\tilde h_\bullet(X^{m+n}/X^{m+n-r})\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\
    &\stackrel{\Delta_X^*\otimes\mathrm{id}}\longrightarrow\tilde h_\bullet\bigl((X\wedge X)^{m+n}/(X\wedge X)^{m+n-r}\bigr)\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\
    &\stackrel{F_{m,n,r,*}\otimes\mathrm{id}}\longrightarrow\tilde h_\bullet((X^{m+r-1}/X^{m-1})\wedge(X^{n}/X^{n-r}))\otimes\tilde h^\bullet(X^{m+r-1}/X^{m-1})\\
    &\stackrel{/}\longrightarrow\tilde h_\bullet(X^{n}/X^{n-r})=H(-n,r-n)\;.
  \end{aligned}
\end{multline*}
Proposition
For all $m$, $n$, $r\ge 1$, the following diagram commutes
$\require{AMScd}$
\begin{CD}
  H(-m-n,1-m-n)\otimes H'(m,m+1)@>\mu_1>>H(-n,1-n)\\
  @A\eta'\oplus A\eta''A@AA\eta A\\
  H(-m-n,r-m-n)\otimes H'(m,m+r)@>\mu_r>>H(-n,r-n)\\
  @V\partial\otimes\eta'\oplus V\eta\otimes\partial'V@VV\partial V\\
  {\begin{matrix}H(r-m-n,r+1-m-n)\otimes H'(m,m+1)\\\oplus\\H(-m-n,1-m-n)\otimes H'(m+r,m+r+1)\end{matrix}}@>\mu_1\pm\mu_1>>H_{p+q-1}(r-n,r+1-n)\rlap{\;.}
\end{CD}
As explained here, this Proposition allows us to define a multiplicative structure on the associated spectral sequence. In this setting, it will be the cap product you ask for:
$$\frown\colon E^r_{m+n}\otimes E_r^m\to E^r_n\;.$$
Proof.
The upper square commutes because the maps $F_{m,n,r}$ are defined sufficiently naturally. For the lower square, we use the Leibniz rule for the slant product and continue as here<|endoftext|>
TITLE: Tensor product of fields over integers
QUESTION [8 upvotes]: Inspired by this question we ask: Is there a name for each of the following properties about fields? what are some examples other than $\mathbb{Q}$? 

A field $K$ with the property that $K\otimes_{\mathbb{Z}} K$ is a field.
A field $K$ is isomorphic to $K\otimes_{\mathbb{Z}} K$ (as field, or equivalently as rings)

The tensor products are taken over $\mathbb{Z}$ (the tensor product of $\mathbb{Z}$-modules, i.e., Abelian  groups).

REPLY [22 votes]: Here is a self-contained argument. First, as Jeremy Rickard observes, $K \otimes K \cong K \otimes_k K$, where $k$ is the prime subfield of $K$ (so $\mathbb{Q}$ if $K$ has characteristic zero and $\mathbb{F}_p$ if $K$ has characteristic $p$). If $K \otimes_k K$ is a field, then as Denis Nardin observes, the multiplication map
$$K \otimes_k K \xrightarrow{m} K$$
must be injective, and since it's surjective it must be an isomorphism. But $K$ has dimension $1$ as a $K$-vector space, while $K \otimes_k K$ has dimension $\dim_k K$. It follows that $\dim_k K = 1$, hence that $K = k$.<|endoftext|>
TITLE: Open problems in Hopf algebras
QUESTION [21 upvotes]: I couldn't find a list of open problems in Hopf algebras. So my question is the following:

In the theory of Hopf algebras, what are the (big) open problems?

Any kind of problem/question will be welcome. I am interested in any kind of problem involving Hopf algebras, say  going from combinatorics or topology to abstract algebra.

REPLY [8 votes]: An open problem in the theory of Hopf algebras is the classification of pointed Hopf algebras.
One method to classify finite-dimensional pointed Hopf
algebras is the Lifting Method of Andruskiewitsch and Schneider. The method was
proved to be successful in the case of abelian coradical, see for example

Andruskiewitsch, Nicolás; Schneider, Hans-Jürgen. On the classification of finite-dimensional pointed Hopf algebras. Ann. of Math. (2) 171 (2010), no. 1, 375--417. MR2630042, doi

The problem in the case where the coradical is a non-abelian group is still open.
I will be more precise. The heart of the method is the understanding of the structure of certain
finite-dimensional braided Hopf algebras
known as Nichols algebras. Nichols algebras are constructed from braided vector spaces. The braided vector spaces
interesting for the classification mentioned are Yetter-Drinfeld modules over groups.
In the survey

Andruskiewitsch, Nicolás. About finite dimensional Hopf algebras. Quantum symmetries in theoretical physics and mathematics (Bariloche, 2000), 1--57, Contemp. Math., 294, Amer. Math. Soc., Providence, RI, 2002. MR1907185, link

one finds the following problems:

Problem 1. Classify finite-dimensional Nichols algebras.
Problem 2. Obtain a "nice" presentation by generators and relations of
finite-dimensional Nichols algebras.

These problems are in general open; they were solved in the case of braided vector spaces of diagonal
type, i.e. Yetter-Drinfeld modules over abelian groups.  The first one was
solved by Heckenberger; the second one, by Angiono.  Both solutions deeply use the so-called
Weyl groupoid. References:

Heckenberger, I. The Weyl groupoid of a Nichols algebra of diagonal type. Invent. Math. 164 (2006), no. 1, 175--188. MR2207786, link
Heckenberger, I. Classification of arithmetic root systems. Adv. Math. 220 (2009), no. 1, 59--124. MR2462836, link
Angiono, Iván Ezequiel. A presentation by generators and relations of Nichols algebras of diagonal type and convex orders on root systems. J. Eur. Math. Soc. (JEMS) 17 (2015), no. 10, 2643--2671. MR3420518, link

So I would add the following as an problem in the theory of Hopf algebras:

Classify finite-dimensional Nichols algebras over non-abelian groups.

Partial results are known. However, several questions are still open. An interesting particular case is related to symmetric groups. This particular problem is connected to
some quadratic algebras known as Fomin-Kirillov algebras.
Small comment. The Weyl groupoid is an analogue of the usual Weyl group. It also works for Lie super algebras, see this MO Question.
Update. Another interesting open problem related to pointed Hopf algebra is a conjecture of Andruskiewitsch and Schneider related to generation in degree one. For some information and a categorical generalization see page 109 of:

Etingof, Pavel; Gelaki, Shlomo; Nikshych, Dmitri; Ostrik, Victor. Tensor categories. Mathematical Surveys and Monographs, 205. American Mathematical Society, Providence, RI, 2015. xvi+343 pp. ISBN: 978-1-4704-2024-6 MR3242743

The conjecture is known to be true in several cases. For example, it is true for finite-dimensional pointed Hopf algebras with abelian coradical:

Angiono, Iván. On Nichols algebras of diagonal type. J. Reine Angew. Math. 683 (2013), 189--251. MR3181554, link<|endoftext|>
TITLE: Does there exist some $C$ independent of $n$ and $f$ such that $ \|f''\|_p \geq Cn^2 \| f \|_p$, where $1 \leq p\leq \infty$?
QUESTION [14 upvotes]: Let $f$ be a trigonometric polynomial on the circle $\mathbb{T}$ with $\hat{f}(j) = 0$ for all $j \in \mathbb{Z}$ with $\lvert j \rvert < n$. Does there exist some $C$ independent of $n$ and $f$ such that
$$
\|f''\|_p \geq Cn^2 \| f \|_p,
$$
where $1 \leq p\leq \infty$?

REPLY [3 votes]: Here is more pedestrian argument. If $T_{n}$ is a trigonometric polynomial of degree at most $n$ with total mass $\frac{1}{2\pi}\int_{-\pi}^{\pi}T_{n}=1$ then $f*T_{n}=0$, and in particular, 
$$
f(x) =\frac{1}{2\pi}\int_{-\pi}^{\pi}(f(x)-f(x-s))T_{n}(s)ds.
$$
Therefore, by the triangle inequality 
$$
\|f\|_{p} \leq \frac{1}{2\pi}\int_{-\pi}^{\pi}\|f(x)-f(x-s)\|_{L^{p}(dx)}|T_{n}(t)|dt \leq \|f'\|_{p}\frac{1}{2\pi}\int_{-\pi}^{\pi}|s||T_{n}(s)|ds,
$$
where the inequality $\|f(x)-f(x-s)\|_{L^{p}(dx)} \leq |s| \|f'\|_{p}$ follows, for instance from Schur test applied to the operator $(Af')(x)=\int f'(t) 1_{[x-s,x]}(t)dt$. 
Now how can we make $\frac{1}{2\pi}\int_{-\pi}^{\pi}|s||T_{n}(s)|ds$ of order $\frac{1}{n}$? Let us be not too demanding and seek for $T_{n}$ among even nonnegative  trigonometric polynomials to reduce the matters to $\int_{0}^{\pi}sT_{n}(s)ds$. 
One immediate choice is Fejer kernel 
$$
k_{n}(s) = \frac{1}{n}\left(\frac{\sin(\frac{ns}{2})}{\sin(\frac{s}{2})}\right)^{2}.
$$
Now $k_{n}(s) \asymp  n$ on $[0,\frac{1}{n}]$, and $k_{n}(s)<C \frac{1}{n}\frac{1}{s^{2}}$ on $[\frac{1}{n}, \pi]$, therefore $\int_{0}^{\pi}sk_{n}(s)ds \leq C' \frac{\log(n)}{n}$. Well, not too bad but not exactly what was requested. 
What else can we do? Let us look at $k_{n}^{2}(s)$. It is even nonnegative trigonometric polynomial of degree $2n$. A small jump in degree is okay (we can just start from $f$ of degree $\geq 2n$). Since $k_{n}^{2}(s) \asymp  n^{2}$ on $[0, \frac{1}{n}]$ its total mass is at least $\geq n$. Then 
$$
\frac{k_{n}^{2}}{n} \asymp n \quad \text{on}\quad [0, \frac{1}{n}], \quad \text{and} \quad \frac{k_{n}^{2}}{n} \leq C\frac{1}{n^{3}} \frac{1}{s^{4}} \quad \text{on} \quad [\frac{1}{n}, \pi]
$$
therefore $\int_{0}^{\pi} s \frac{k^{2}_{n}}{n}\leq C' \frac{1}{n}$ voila!<|endoftext|>
TITLE: Free open-access peer-reviewed math journals
QUESTION [75 upvotes]: Is there any free (as in free beer, i.e., no publication fees or other fees whatsoever), open-access (free and open access to everyone) and peer-reviewed mathematics journal?
I am interested in a list of journals. I am particularly interested in my field which is low dimensional topology and geometry and geometric group theory.
The observation how well a site like MathOverflow or Math Stack Exchange works makes me wonder if a similar system does exist for the peer-reviewing and publication process itself.

REPLY [2 votes]: The Bulletin of the Irish Mathematical Society<|endoftext|>
TITLE: Is there a closed form of $\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$?
QUESTION [12 upvotes]: For naturals $n\ge m$, define
$$I(n,m):=\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$$
with $\text{arcsinh}\ x=\ln(x+\sqrt{1+x^2} )$, so $\text{arcsinh} \frac12=\ln \frac{\sqrt{5}+1}2 $.
Is it possible to find closed form expressions of $I(n,m)$? I mean closed form in a broad sense, i.e. involving any other "known" constants.  
Motivation: it is known that
$$I(2,1)=\int_0^\frac12\dfrac{\text{arcsinh}^2x}xdx=\dfrac{\zeta(3)}{10},$$
further $$I(1,0)= \int_0^\frac12 \text{arcsinh}\  x\ dx=\frac12\left(2-\sqrt{5}+\ln \frac{\sqrt{5}+1}2 \right)$$
$$I(1,1)= \int_0^\frac12\dfrac{\text{arcsinh}\ x}xdx=\frac{\pi^2}{20}$$ and $$I(2,2)=\int_0^\frac12\dfrac{\text{arcsinh}^2x}{x^2}dx=\frac{\pi^2}6-5\ln^2\frac{\sqrt{5}+1}2 $$
so there might be some hope that at least some others of the $I(n,m)$ have closed forms involving values of $\zeta(k)$, ideally odd zeta values.

REPLY [3 votes]: Letting $y=\text{arcsinh}\, x$ and integrating by parts, we have 
$$I(n,m)=-\frac{2^{m-1}a^n}{m-1}+\frac n{m-1}\,[J(n-1,m-1;a)-J(n-1,m-1;0)]
$$
if $n\ge m\ge2$, where $a:=\text{arcsinh}\,\frac12$ and 
$$J(p,q;y):=\int\frac{y^p}{\sinh^qy}\,dy.$$
Formula 1.4.24.1 in Prudnikov--Brychkov--Marichev (PBM, Vol. 1, ISBN 5-9221-0323-7) tells us that
$$J(p,q;y)=
-\frac{py^{p-1}}{(q-1)(q-2)\sinh^{q-2}y}
-\frac{y^p\cosh y}{(q-1)\sinh^{q-1}y}$$
$$
+\frac{p(p-1)}{(q-1)(q-2)}\,J(p-2,q-2;y)
-\frac{q-2}{q-1}\,J(p,q-2;y).  
$$
Also, formulas 1.4.24.2 and 1.4.24.4 in PBM tell us that
$$J(p,1;y)=\sum_{k=0}^\infty\frac{(2-2^{2k})B_{2k}}{(2k)!(p+2k)}\,y^{p+2k}\quad (|y|<\pi,\ p>0)$$
and 
$$J(p,2;y)=-y^p\coth y+p\sum_{k=0}^\infty\frac{2^{2k}B_{2k}}{(2k)!(p+2k-1)}\,y^{p+2k-1}\quad (|y|<\pi,\ p>1).$$
These formulas in PBM should be easy to obtain/check. 
Since $|a|=a=\text{arcsinh}\,\frac12<1/2<\pi$, the above formulas provide a recursion to compute the values of $I(n,m)$ in terms of the Bernoulli numbers $B_{2k}$ or, alternatively, in terms of $J(p,1;a)-J(p,1;0)$ and $J(p,2;a)-J(p,2;0)$, that is, in terms of $I(n,2)$ and $I(n,3)$.<|endoftext|>
TITLE: Growth of an integer vector under the action of a matrix in $GL_n(\mathbb{Z})$
QUESTION [10 upvotes]: I have some questions regarding the dynamics of elements of $GL_n(\mathbb{Z})$ acting on $\mathbb{Z}^n$. In particular, given an invertible integer matrix $M \in GL_n(\mathbb{Z})$, and given an integer column vector $v \in \mathbb{Z}^n$ which is not equal to the zero vector, I want to know about the asymptotics of the sequence of vector norms $| M^j v|$ for positive integers $j$, with particular emphasis on boundedness and polynomial growth.
One can deduce some things about this using the Jordan decomposition for the action of $M$ on $\mathbb{C}^n$. For example, here is a nice, precise characterization of boundedness. The following are equivalent:

$| M^j v|$ is bounded
The vector $v$, regarded as a complex vector, is contained in the direct sum of the eigenspaces corresponding to eigenvalues of $M$ that are roots of unity. 

What one uses to deduce $1 \implies 2$ is a discreteness argument: $\mathbb{Z}^n$ has only finitely many elements with a given bound, and hence boundedness of $|M^j v|$ implies the existence of $j$ such that $M^j v = v$.
I would like a similarly precise characterization of polynomial growth. The best I know at the moment is that the following are equivalent:

$| M^j v|$ is bounded above by a polynomial function of $j$, for positive integers $j$.
$v$ is contained in the direct sum of generalized eigenspaces corresponding to eigenvalues $\lambda \in \mathbb{C}$ such that $\lambda \le 1$, i.e. to those $\lambda$ that lie on or inside the unit circle of $\mathbb{C}$.

But statement 4 is somewhat loose, in that it allows some seeming possibilities that turn out to be impossible using a discreteness argument. For example, it is impossible that $v$ be contained in the direct sum of the generalized eigenspaces corresponding to eigenvalues that lie inside the unit circle, for in that case one can show that the sequence $M^j v$ converges to the zero vector, an impossibility for $M \in GL_n(\mathbb{Z})$ and $v \in \mathbb{Z}^n$.
What stronger statements are there which characterize polynomial growth of $|M^j v|$? For example, given $M \in GL_n(\mathbb{Z})$ and $v \in \mathbb{Z}^n$, are Statements 3 and 4 above equivalent to the following nice, strong statement?

$v$ is contained in the direct sum of generalized eigenspaces corresponding to eigenvalues of $M$ that are roots of unity.

REPLY [5 votes]: The linear algebra needed to answer this question is spelled out in my paper Dynamical Properties of Quasihyperbolic Toral Automorphisms (Ergodic Th. & Dynam. Sys. 2 (1982), 49-68), in particular Section 2.
The observation that a monic polynomial with integer coefficients all of whose roots lie on the unit circle must be cyclotomic goes back to Kronecker: Zwei Sätze über Gleichungen mit ganzzahligen Coefficienten, J. reine angew. Math. 53 (1857), 173-175. David Boyd proved a generalization of this for polynomials of several variables, which says that a polynomial whose logarithmic Mahler measure vanishes (for one variable this is equivalent to having all roots on the unity circle) must be a product of generalized cyclotomic polynomials (Kronecker's Theorem and Lehmer's Problem for Polynomials in Several Variables, J. Number Theory 13 (1981), 116-121.
There are some much deeper results along the following lines. Let $E$ be the direct sum of the generalized eigenspaces of $M$ corresponding to eigenvalues on or inside the unit circle, and assume there are no eigenvalues that are roots of unity. Katznelson showed that there is a constant $C>0$ such that if $v$ is a nonzero integer vector then its Euclidean distance to $E$ is greater than $C\|v\|^{-\dim E}$ (Ergodic Automophisms of $T^n$ are Bernoulli, Israel J. Math. 10 (1971), 186-195). This is the key diophantine component in his proof that all ergodic toral automorphisms are measurably isomorphic to Bernoulli shifts. Subsequently this was extended by me and others to show that all ergodic group automorphisms are Bernoulli.<|endoftext|>
TITLE: Moduli spaces in applied mathematics and condensed matter physics?
QUESTION [10 upvotes]: In this MO question it is stated that there is a relation between some aspects of condensed matter physics (namely the fractional quantum Hall effect) and the algebraic geometry of Hilbert schemes.
Having made a quick google search without immediate results, I'm curious to know:


How does this interaction between the two topics happen?



Let's have a more general look. I'm aware of the relation between Hilbert schemes -and also other kinds of moduli spaces of sheaves- and instantons. Anyway, here I would be interested to know of other examples more in the vein of the MO question linked above.


Do moduli spaces in the sense of algebraic geometry -of which Hilbert schemes are a special case- have any link with physics (except string theory, high energy and elementary particle physics)?



Last but not least:


What about moduli spaces and applied mathematics?

REPLY [5 votes]: Regarding question 2: FQHEs are examples of topological phases of matter, whose effective field theories are topological quantum field theories, whose quasi-particle excitations are (essentially) described by modular tensor categories (MTCs). Since what we're usually interested in for constructing models of topological quantum computing (and related things) is the algebraic data of these quasi-particles, the problem of classifying topological phases is very closely related to the problem of classifying modular tensor categories. 
Since we're talking about categories what we really want to do is classify them up to something, which in this case is (braided) monoidal equivalence. It turns out though that, for every equivalence class of MTCs, representatives of that equivalence class can be constructed from solutions to certain polynomial equations called the pentagon, hexagon, and pivotal equations. The collection of these solutions define an algebraic set $X$. 
For a given $X$ there exists an algebraic group $G$ which acts on $X$ such that for points $F\text{ and }F' \in X$, $F$ and $F'$ give rise to monoidally equivalent categories if and only if there exists $g \in G$ such that $g \cdot F = F'$. Thus the orbits of $G$ in $X$ are in 1-1 correspondence with equivalence classes of categories, and so now we can consider the problem of classifying orbits of $G$.
It turns out that in doing this we have almost the nicest possible situation imaginable - $G$ is reductive, all orbits have the same dimension and are in fact the irreducible components of $X$. This then implies that we can construct another algebraic set $Y$ which is an orbit space for $X$ - that is to say that the points of $Y$ are in 1-1 correspondence with orbits of $G$ in $X$ and the regular functions on $Y$ are those regular functions on $X$ which are invariant under the action of $G$. All of this allows us to classify orbits (i.e. MTCs) by looking at the evaluations of $G$-invariant functions on $X$. 
Picking these functions is a really hard problem in general, but for MTCs we have a set of generic candidates: Every MTC gives you a pair of matrices $(S,T)$ which are the so called modular data of the category. They are called this because they specify a representation of the modular group $SL(2,\mathbb Z)$. It is conjectured that MTCs are classified by their modular data. 
Bringing things back to physics, the $(S,T)$ matrices have physical meaning - the entries of the $S$-matrix encodes the mutual statistics between particle types and the $T$-matrix encodes the self-statistics. Additionally, the entries of $S$ and $T$ are given by the evaluations of regular functions on $Y$ which is to say that they are given by the evaluation of $G$-invariant regular functions on $X$. 
That this has physical meaning can be seen by noting that $G$ is (essentially) the gauge group for our TQFT. Given two quasi-particles $a$ and $b$, the state space $V_{a b}$ for their composite system is finite dimensional and decomposes in to subspaces $V_{a b}^c$, where $c$ is another quasi-particle type (including the vacuum) and whose dimension is the number of fusion channels from $a\otimes b$ to $c$. $G$ is the direct product of the groups of basis transformations on the $V_{ab}^c$ spaces.
The information in paragraphs 1,6, and 7 is pretty standard and can basically be found in these lecture notes. The details for paragraph 2 can be found in arxiv:1305:2229 and for paragraphs 3 and 4 in arxiv:1509.03275.

REPLY [3 votes]: You can consider Nekrasov-Shatashvili for application to Toda chain or Calagero-Moser type systems. That nominally looks like a string or high energy particle link, but those systems are integrable chain systems. Looking for any experiments on these sorts of systems, the setup might be some optical trap like http://www.nature.com/nature/journal/v440/n7086/full/nature04693.html
I don't know if you would call this sort of Atomic, Molecular, Optical system as a cheating answer because it is designed to have that integrable structure.
Same principal for any other $\mathcal{N}=2$ way of getting any other integrable chain that could be engineered with laser traps.<|endoftext|>
TITLE: Gabber's original proof of his purity theorem
QUESTION [14 upvotes]: Gabber's purity theorem is the statement that if $\mathscr{F}$ is a pure perverse sheaf on an open subvariety $j : U \hookrightarrow X$ then so is $j_{!*} \mathscr{F}$.
It is remarkable because it gives many objects besides (pure) local systems on smooth varieties for which purity and the Weil conjectures hold.
It is proved in $\S$5 of Faisceaux pervers by BBD(G). Here the proof is quite natural once one accepts deep theorems of Deligne: $j_{!*} \mathscr{F}$ occurs as the image of a map between an object with weights $\le m$ and an object with weights $\ge m$ and hence has weights $\{ \ge m \} \cap \{ \le m \} = \{ m \}$.
However I understand that Gabber found another proof earlier. Is this the case? Or is it similar to the one given by [BBD(G)]?
Gabber's work is often cited as:
[Ga1] O. Gabber : Pureté de la cohomologie de MacPherson-Goresky rédigé par P. Deligne, prépublication I.H.E.S., 1981

REPLY [29 votes]: It's different, but it also uses Weil II. See
 Purity for intersection cohomology after Deligne-Gabber
for my translation of the original.<|endoftext|>
TITLE: Eigenspace of a specific operator
QUESTION [6 upvotes]: Consider the operator $T:\ell^\infty({\mathbb N})\to\ell^\infty({\mathbb N})$ defined by
$$
(Tx)_m=\sum_{k=m+1}^\infty p_{k,m} \ \ x_k,
$$
where
$$
p_{k,m}=\frac k{(k-1)(k-m)(k-m+1)}.
$$
Then $T$ is a bounded operator of norm $\zeta(2)=\frac{\pi^2}6$ as an easy calculation shows.
I need to know the dimension of the eigenspace to the eigenvalue $1$, i.e.
$$
E=\{x\in \ell^\infty: Tx=x\}.
$$
Ideally, I would like to have $\dim(E)=1$, however, I already don't know whether this space is finite-dimensional.
So my question is, what is the dimension of the space $E$?

REPLY [3 votes]: Theorem.
Let $T:\ell^\infty\to\ell^\infty$ be defined by
$$
(Tx)_m=\sum_{k=m+1}^\infty p_{k,m}x_k,
$$
where $p_{k,m}=\frac k{(k-1)(k-m)(k-m+1)}$.
Let $L=\{x\in\ell^\infty: Tx=x\}$.
Then $\dim L= 1$.
We first show $\dim L\ge 1$.
Lemma.
The element $\alpha$ with $\alpha_1 = 1$ and
\begin{align*}
\alpha_k &= (1-k)\int_0^1 \frac{t^{k-1}}{\log(1-t)} dt\quad\text{for } k = 2,3,\dots
\end{align*}
lies in $L$.
For $m=1$ we compute
\begin{align*}
(T\alpha)_1&=\sum_{k=2}^\infty \frac 1{(k-1)^2}\alpha_k\\
&=-\sum_{k=2}^\infty \frac1{k-1}\int_0^1\frac{t^{k-1}}{\log(1-t)}\,dt\\
&=-\int_0^1\sum_{k=2}^\infty \frac{t^{k-1}}{k-1}\frac{1}{\log(1-t)}\,dt=\int_0^11\,dt=1=\alpha_1.
\end{align*}
For $N\in{\mathbb N}$, $N\ge 2$ and $x\in\ell^\infty$ we get
\begin{align*}
\sum_{m=1}^N(Tx)_m&=\sum_{m=1}^N\sum_{k={m+1}}^\infty\frac k{(k-1)(k-m)(k+1-m)}x_k\\
&=\sum_{k=2}^\infty\frac{kx_k}{k-1}\sum_{m=1}^{\min(N,k-1)}\frac1{(k-m)(k+1-m)}\\
&=\sum_{k=2}^\infty\frac{kx_k}{k-1}\sum_{m=1}^{\min(N,k-1)}\left(\frac1{k-m}-\frac1{k+1-m}\right)\\
&=\sum_{k=2}^\infty\frac{kx_k}{k-1}\left(\frac1{k-\min(N,k-1)}-\frac1k\right)\\
&=\sum_{k=2}^N\frac{kx_k}{k-1}\left(1-\frac1k\right)+\sum_{k=N+1}^\infty\frac{kx_k}{k-1}\left(\frac1{k-N}-\frac1k\right)\\
&=\sum_{k=2}^Nx_k+N\sum_{k=N+1}^\infty\frac{x_k}{(k-1)(k-N)}.
\end{align*}
Now assume we have shown $(T\alpha)_j=\alpha_j$ for $1\le j\le N-1$.
Then we get
$$
1+(T\alpha)_N=\alpha_N+N\sum_{k=N+1}^\infty\frac{\alpha_k}{(k-1)(k-N)}.
$$
We compute
\begin{align*}
N\sum_{k=N+1}^\infty\frac{\alpha_k}{(k-1)(k-N)}
&=-N\sum_{k=N+1}^\infty\frac{1}{(k-N)}
\int_0^1 \frac{t^{k-1}}{\log(1-t)} dt\\
&=-N\int_0^1\sum_{k=N+1}^\infty\frac{t^{k-N}}{(k-N)}
 \frac{t^{N-1}}{\log(1-t)} dt\\
&= N\int_0^1t^{N-1}\,dt=1.
\end{align*}
The lemma follows.
$\square$
In order to show $\dim L\le 1$, let now $x\in L$ with $x_1=0$. We need to show that $x=0$. 
Let $y_k=\frac {k+1}{k}x_{k+1}$.
Then $y\in\ell^\infty$ and $Tx=x$ as well as $x_1=0$ lead to
\begin{align*}
0&=\sum_{k=1}^\infty\frac{y_{k}}{k(k+1)},\\
\frac m{m+1}y_m&=\sum_{k=m+1}^\infty\frac{y_k}{(k-m)(k-m+1)},\quad m\ge 1.
\end{align*}
We have to show $y=0$.
For $m\in{\mathbb N}_0$ let $z^{(m)}\in\ell^1$ be defined by
\begin{align*}
z^{(0)}_k&=\frac1{k(k+1)},\\
z^{(m)}_k&=\begin{cases}0&k<m,\\
\frac m{m+1}&k=m,\\
\frac{-1}{(k-m)(k-m+1)}&k>m.
\end{cases}
\end{align*}
Now $\ell^\infty$ is the dual space of $\ell^1$.
Let $<.,.>$ denote the duality pairing.
The formulas above say that
$$
<z^{(k)},y>=0\qquad\text{for }k=0,1,\dots
$$
Therefore the theorem follows if we can show that the sequence $z^{(0)},z^{(1)},\dots$ spans  a dense subspace of $\ell^1$.
Let $Z$ denote the span of $z^{(0)},z^{(1)},\dots$. 
For $m\in{\mathbb N}$ let
$$
h^{(m)}_k=\begin{cases}0&1\le k\le m-1,\\
\frac m{(k-m+1)(k+1)}&k\ge m. \end{cases}
$$
Then $h^{(1)}=z^{(0)}$ and a quick computation shows 
$h^{(m)}=h^{(m-1)}-z^{(m-1)}$ for $m\ge 2$.
Therefore, by induction we have $h^{(m)}\in Z$ for every $m$.
Next, let
$$
v^{(m)}=\frac1m h^{(m)}+z^{(m)}.
$$
then $v^{(m)}\in Z$ for every $m\in{\mathbb N}$ and
$$
v^{(m)}_k=\begin{cases}0&1\le k\le m-1,\\
1& k=m,\\
\frac{-(m+1)}{(k+1)(k-m)(k-m+1)}&k\ge m+1.\end{cases}
$$
Let $n\in{\mathbb N}$ and let $e_n$ be the $n$-th standard basis element  of $\ell^1$, i.e., $e_n=(0,\dots,0,1,0,\dots)$ with the $1$ at the $n$-th position.
We show that $e_n$ lies in the closure of the span of $v^{(n)},v^{(n+1)},\dots$.
For this purpose let $a^{(n)}=v^{(n)}$ and for $m>n$ let $a^{(m)}=a^{(m-1)}-a^{(m-1)}_mv^{(m)}$.
Then $a^{(m)}\in Z$ and 
$$
a^{(m)}_k=\begin{cases}0&1\le k\le m,\ k\ne n,\\
1 &k=n,\\
a^{(m-1)}_k-a_m^{(m-1)}v^{(m)}_k&k\ge m+1.\end{cases}
$$
Iterating the recursive relation yields for $k>m$,
\begin{align*}
a^{(m)}_k&=a^{(n)}_k-\sum_{j=n}^{m-1}a^{(j)}_{j+1}v_k^{(j+1)}\\
&=\frac{-(n+1)}{(k+1)(k-n)(k-n+1)}
+\sum_{j=n}^{m-1}a_{j+1}^{(j)}\frac{j+2}{(k+1)(k-j-1)(k-j)}.
\end{align*}
For $m\ge n$ set $b_m=a^{(m)}_{m+1}$, then in particular,
$$
b_m=\frac{-(n+1)}{(m+2)(m-n+1)(m-n+2)}+\sum_{j=n}^{m-1}\frac{b_j(j+2)}{(m+2)(m-j)(m-j+1)}.
$$
 Lemma.
For every $m\ge n$ we have $|b_m|\le \frac{n+1}{m+2}$.
Proof of the lemma.
For $m=n$ we have
\begin{align*}
|b_m|=\frac{n+1}{n+2}\frac12\le \frac{n+1}{n+2}=\frac{n+1}{m+2}.
\end{align*}
For the induction step we assume $m>n$ and the claim proven for smaller indices. Then
\begin{align*}
|b_m|&\le \frac{n+1}{(m+2)(m-n+1)(m-n+2)}+\sum_{j=n}^{m-1}\frac{|b_j|(j+2)}{(m+2)(m-j)(m-j+1)}\\
&\le \frac{n+1}{m+2}\left(\frac1{m-n+1}+\sum_{j=n}^{m-1}\frac1{m-j)(m-j+1)}\right)\\
&= \frac{n+1}{m+2}\left(\frac1{m-n+1}+1-\frac1{m-n+1}\right)=\frac{n+1}{m+2}.
\end{align*}
$\square$
One has
\begin{align*}
\parallel{e_n-a^{(m)}}\parallel_1&=\sum_{k=m+1}^\infty |a_k^{(m)}|\\
&\le \sum_{k=m+1}^\infty\frac{n+1}{(k+1)(k-n)(k+1-n)}\\
&\ \ \ +\sum_{k=m+1}^\infty \sum_{j=n}^{m-1}|b_j| \frac{j+2}{(k+1)(k-j-1)(k-j)}.
\end{align*}
We estimate the first sum
\begin{align*}
\sum_{k=m+1}^\infty\frac{n+1}{(k+1)(k-n)(k+1-n)}
&\le \frac{n+1}{m+2}\sum_{k=m+1}^\infty \frac1{(k-n)(k+1-n)}\\
&= \frac{n+1}{m+2}\frac1{m+1-n}=A(m,n)\to 0,\quad m\to\infty.
\end{align*}
For the second sum we have
\begin{align*}
\sum_{k=m+1}^\infty \sum_{j=n}^{m-1}|b_j| \frac{j+2}{(k+1)(k-j-1)(k-j)}
&\le \sum_{k=m+1}^\infty \sum_{j=n}^{m-1} \frac{n+1}{(k+1)(k-j-1)(k-j)}\\
&=\sum_{k=m+1}^\infty \frac{n+1}{k+1}\left(\frac1{k-m}-\frac1{k-n}\right),
\end{align*}
which tends to zero as $m\to\infty$.
We have shown that
$$ \parallel{e_n-a^{(m)}}\parallel_1\to 0
$$ as $m\to\infty$. The theorem follows.
$\square$<|endoftext|>
TITLE: Is the automorphism group of a Calabi-Yau variety an arithmetic group
QUESTION [5 upvotes]: Let $X$ be a smooth projective variety over the complex numbers with trivial canonical bundle. Suppose that $X$ is Calabi-Yau.
Is the automorphism group of $X$ an arithmetic group?
What if $X$ is a K3 surface?

REPLY [10 votes]: The answer for $K3$ surfaces is no. A counterexample, where the group is not even commensurable with an arithmetic group, was given by Totaro in Example 6.3 of this paper.<|endoftext|>
TITLE: Are Picard stacks group objects in the category of algebraic stacks
QUESTION [6 upvotes]: I've been wondering about what a "group algebraic stack" should be, and ran into the notion of a Picard stack.
I'm slightly confused by the terminology here.
Given an algebraic stack $\mathcal X$ over $S$, one can define the Picard stack $\mathcal{Pic}_{\mathcal X/S}$. Is this a Picard stack as defined in "Smooth Toric Deligne-Mumford stacks" by Fantechi, Mann and Nironi?
Are Picard stacks group objects in the category (or 2-category?) of algebraic stacks?

REPLY [6 votes]: Converting my comment into an answer: stacks form a 2-category, not a category. If a stack takes values in groupoids, then a "group stack" takes values in 2-groups, or equivalently in monoidal groupoids where every object is invertible. But it's natural to go further and ask for "abelian group stacks," which take values in symmetric monoidal groupoids where every object is invertible (e.g. the symmetric monoidal groupoid of line bundles on a scheme). These groupoids are sometimes called Picard groupoids, and this notion of stack is sometimes called a Picard stack.<|endoftext|>
TITLE: Is there an explicit description of a cobordism between $\mathbb{CP}^n$ and $\mathbb{RP}^n\times\mathbb{RP}^n$?
QUESTION [29 upvotes]: With a little bit of work, one can show that $\mathbb{CP}^n$ and $\mathbb{RP}^n\times\mathbb{RP}^n$ have the same Stiefel-Whitney numbers, so by a theorem of Thom, they are (unorientedly) cobordant. 

Is there an explicit description of a cobordism between them? 

I know the answer for $n = 1$: it reduces to finding a cobordism between $S^2$ and $S^1\times S^1$ (take a solid ball and remove a solid torus from the interior, or take a solid torus and remove a solid ball from the interior).

REPLY [46 votes]: An explicit cobordism is given by Stong:
R. E. Stong, A Cobordism, Proceedings of the American Mathematical Society
Vol. 35, No. 2 (Oct. 1972), pp. 584-586
I do like the short title "A Cobordism".
The construction is as follows:
Write $\mathbb{CP}^n$ with homogeneous coordinates $[z_0,z_1,\dots ,z_n]$ and consider the space $$W=(\mathbb{CP}^n\times [0,1])/\sim,$$
where $([z_0,z_1,\dots ,z_n],1)\sim([\bar{z_0},\bar{z_1},\dots ,\bar{z_n}],1)$ if $|z_0^2+z_1^2\dots +z_n^2|\leq\frac{3}{5}$.
Then Stong takes a page to prove that $W$ is the desired cobordism. 
But one should also read the note:<|endoftext|>
TITLE: Hausdorff dimension of Apollonian circle packing, 1.305686729, 1.305688 or yet something else?
QUESTION [15 upvotes]: I am interested in the Hausdorff dimension of the Apollonian circle packing.
There seem to be two numerical calculations of the value:
1.305686729(10)

from P.B Thomas and D.Dhar, The Hausdorf[sic!] dimension of the Apollonian packing of circles Journal of Physics A: Mathematical and General, Volume 27, Number 7, April 1994
and later from Curt McMullen (he does not cite Thomas and Dhar):
1.305688

McMullen, Curtis T. Hausdorff dimension and conformal dynamics. III. Computation of dimension. Amer. J. Math. 120, no. 4, 691–721. 1998
His algorithm is also explained in Section 5.5 on page 156 in "Indra's Pearls" by Mumford and Series (beware of some typos on that page!).
It is somewhat unsatisfying to have two contradicting numerical approximations. The OEIS gives 1.3056867 and does not cite McMullen, see A052483 (as of today).
With modern computers it might be worthwhile trying to settle this question. Curt McMullen has a C-program on his website that can calculate the Hausdorff dimension, that I am interested in:
hdim.tar on
http://abel.math.harvard.edu/~ctm/programs/index.html
Running
./hdim -a -e .00005

gives as output
Apollonian gasket


Epsilon   Dimension   Cover  Matrix Steps
5.00e-05 1.305687542911558287346746 76

So I guess after 76 steps (and about a week of computation time) we get
1.30568754291
as an numerical approximation. 

Does this mean that actually the first few digits are
1.305687?
What are the correct first few decimal digits of this number? 
What are the best proven exact bounds and what are the most promising numerical experiments to get a good approximation?

REPLY [3 votes]: It seems in the meantime, there are new results:
Bai, Zai-Qiao; Finch, Steven R. Precise calculation of Hausdorff dimension of Apollonian gasket. Fractals 26 (2018), no. 4, 9 pp.
claims a better approximation is
$$1.3056867280498771846459862068510\dots$$<|endoftext|>
TITLE: The saturation of Murray von Neumann relation
QUESTION [7 upvotes]: Edit: According to comment of  Pace Nielsen, I remove question 2 of the previous version:
Let $R$ be a unital ring. We define  Murray Von Neumann  relation $M$ on $R$ as follows:
We say $a M b$ iff $a=xy,\;b=yx$ for some $x,y\in R$. (This is inspired by the usual Murray Von Neumann equivalent relation in K theory, which is defined on the set of idempotents of  a  ring). The relation $M$ is a reflexive  and symmetric relation but is not a transitive relation. So we consider its  saturation. The saturation of this relation is an equivalent  relation denoted by $\simeq$. In fact we say $a\simeq b$ if there are $p_{i}\in R\;$  with $p_{0} M p_{1},\;\;\;p_{1}  M p_{2},\ldots p_{n-1} M p_{n}$  where $p_{0}=a,\;p_{n}=b$.
Put $R=M_{n}(\mathbb{C})$. One can show that the equivalent class containing $0$ is $$[0]=\{A\in M_{n}(\mathbb{C})\mid A^{n}=0\}$$
(In fact one can prove the following: If $A\in B(H)$ satisfy $A^{k}=0$ then there are $X,Y\in B(H)$  with $A=XY$  and $(YX)^{k-1}=0$. Here $B(H)$ is the space of bounded operators on a  Hilbert space. The same is true by replacing $B(H)$   with an  arbitrary  Von Neumann  algebra. The  same  also is true   without any topological consideration, that is by replacing $B(H)$  with $L(V)$, the space of linear endomorphisms of  a vector space $V$.)
So for  $R=M_{n}(\mathbb{C}),\;\;[0]$  is  an algebraic variety,i.e: the variety of nilpotent  matrices  $A^{n}=0$ 

1.Is every equivalent class  of  $M_{n}(\mathbb{C})$ an  algebraic variety?(the zero set of  polynomials on $M_{n}(\mathbb{C}) \simeq \mathbb{C}^{n^{2}}$ or the zero set of  polynomials in the form $f(A)=0$ where $f\in \mathbb{C}[x]$?What is the precise description of equivalent classes?

Assume that  $A$ is  a  $C^{*}$  algebra  and $a\in A$  satisfies $a^{k}=0$  for  some  $k>1$. Are there two elements $x,y \in A$  with $a=xy$  and $(yx)^{k-1}=0$?


**Note:**Inspired by methods from K theory, I tried to construct a functor $NK$ based on the constructions above. please see  A functor on the category of rings, algebras or compact Hausdorff topological space
Perhaps, it would be interesting to ask "Is there a kind of periodicity property for this functor?"

REPLY [3 votes]: In answer to the second question, yes this is true. Say $x^k=0$.   Let $x=v|x|$ be the polar decomposition of $x$ in $A^{**}$ (the bidual of $A$). Let $a=v|x|^{\frac 1 2}$ and $b=|x|^{\frac 1 2}$. Then clearly $x=ab$.
Both $a$ and $b$ belong to $A$. In $b$'s case, by functional calculus. It is a well-known property of polar decompositions that $v|x|^{\frac 1 2}$ is also in $A$. To see this, write $p_n(|x|)\to |x|^{\frac 1 2}$, where each $p_n$ is a polynomial such that $p_n(0)=0$. Then $vp_n(|x|)\to v|x|^{\frac 1 2}$ in norm and $vp_n(|x|)\in A$ for all $n$ because we can factor out $|x|$ from $p_n(|x|)$. 
Now consider $ba=|x|^{\frac 1 2}v|x|^{\frac 1 2}$ (the Aluthge transform of $x$). Then
    $$
(ba)^{k-1}(ba)^*=
(|x|^{\frac 1 2}v|x|^{\frac 1 2}\cdots |x|^{\frac 1 2}v|x|^{\frac 1 2})\cdot 
|x|^{\frac 1 2}v^*|x|^{\frac 1 2}= 
|x|^{\frac 1 2}x^{k-1} v^*|x|^{\frac 1 2}=0,
 $$ 
    where we have used that $|x|^{\frac 1 2}x^{k-1}=0$ (since  $|x|^{\frac 1 2}\in C^*(x^*x)$ and $(x^*x)x^{k-1}=0$). It follows that $(ba)^{k-1}((ba)^{k-1})^*=0$ which implies that $(ba)^{k-1}=0$.<|endoftext|>
TITLE: Simple lie algebras, (almost-)simple groups of Lie type
QUESTION [6 upvotes]: Take an algebraic group $G$ defined over a finite field $K$. Suppose its Lie algebra $\mathfrak{g}$ is simple. It should follow that $G$ is almost-simple. (By this I mean not that $G(K)$ is simple -- I've heard there is a tricky though in some sense inessential counterexample -- but rather that $G$ has no normal algebraic subgroups of dimension greater than $0$ and less than $
\text{dim}\, G(K)$.)
Does anybody know of a concise and straightforward proof of this fact? One can't proceed as for $K=\mathbb{R}$, since $\exp$ is not defined (except on nilpotent subalgebras of $\mathfrak{g}$). I feel this should be simpler than studying the simplicity of $G(K)$. Is this so?
IMPORTANT EDIT: I just realized that I meant to ask the converse of this question. If $G$ is almost-simple in the sense of algebraic groups, how do you prove that $\mathfrak{g}$ is simple? I agree that the direction stated above is very easy.
Yet another important edit: there are exceptions to the converse! See the answers below. They all seem to come from non-trivial centers due to small characteristic. What I really need is the following: 
Let $G$ be an almost-simple linear algebraic group. Let $V$ be a subvariety of $G$ going through the origin. Assume $0<
dim(V)<
dim(G)$. Let $\mathfrak{v}$ be the tangent space to $V$ at the origin. Prove that there is no ideal $\mathfrak{w}$ of $\mathfrak{g}$ containing $\mathfrak{v}$. (Or, what is the same: prove that the conjugates of $g \mathfrak{v} g^{-1}$ of $\mathfrak{v}$ span $\mathfrak{g}$.)
Yet another edit: Aha - not even that is true.
Well, can we at least give a quick proof, without case-work, of the following? Given a classical almost-simple group $G$ (defined as a subgroup of $\SL_n$ by means of equations with integer coefficients), show that there is a constant $c$ such that, for every field $K$ of characteristic $>c$, the Lie algebra $\mathfrak{g}$ of $G(K)$ is simple.
Surely there should be a reduction to the case over $\mathbb{C}$, or a proof from scratch?

REPLY [2 votes]: Your final request, even if augmented to require the locally closed subvariety $V$ to be smooth, is false over every field $k$ of characteristic 2 and 3, with counterexamples of any rank $n \ge 1$ over any $k$ of characteristic 2, with $V \subset G$ of very large dimension (unbounded as $n$ grows).  So avoiding finite fields of bounded size is insufficient in characteristics 2 and 3.  A source of counterexamples is certain "well-known" exceptional (non-central) isogenies through which Frobenius non-trivially factors, and these isogenies only exist in such characteristics. 
I will describe counterexamples with $G = {\rm{Spin}}_{2n+1}$ for any $n \ge 1$, and a variant of the same ideas (but carried out by more conceptual methods based on the structure theory of semisimple groups rather than based on manipulations in linear algebra) gives counterexamples for ${\rm{Sp}}_{2n}$  ($n \ge 1$) and ${\rm{F}}_4$ over any field of characteristic 2 and for G$_2$ over any field of characteristic 3; see section 7.1 of the book "Pseudo-reductive groups" for a unified development of the relevant general principles
(but again, the phenomenon at hand has been widely known for many decades). 
Let $V = k^{2n+1}$ and
$$q = x_0^2 + x_1 x_2 + \dots + x_{2n-1} x_{2n}.$$
The associated symmetric bilinear form $B_q(v,v') = q(v+v') - q(v)-q(v')$ has defect space $V^{\perp} = \{v \in V\,|\,B_q(v,\cdot) = 0\}$
that is a line since ${\rm{char}}(k)=2$ (this line is $k e_0$). On $V/V^{\perp}$, the induced bilinear form $\overline{B}_q$ is symplectic since ${\rm{char}}(k)=2$. Thus, we get a natural composite homomorphism
$$f: G = {\rm{Spin}}_{2n+1} \rightarrow {\rm{SO}}_{2n+1} \rightarrow
{\rm{Sp}}(\overline{B}_q) = {\rm{Sp}}_{2n}.$$
Let $T$ be a split maximal $k$-torus in $G$, and let $\Delta$ be the base  of $\Phi(G, T)$ associated to a choice of positive system of
roots $\Phi^+$. The multiplication map
$\prod_{a \in \Delta} \mathbf{G}_m \rightarrow T$ defined by
$(\lambda_a) \mapsto \prod a^{\vee}(\lambda_a)$ is an isomorphism since $G$ is simply connected. In particular, ${\rm{Lie}}(T)$ is the direct sum of the "coroot lines" ${\rm{Lie}}(a^{\vee}(\mathbf{G}_m)) = {\rm{Lie}}(a^{\vee})(\partial_t|_{t=1})$.
The kernel $\mathfrak{n} := \ker {\rm{Lie}}(f) = {\rm{Lie}}(\ker f)$ is an ${\rm{Ad}}_G$-stable
subspace of $\mathfrak{g}$, and it is nonzero and proper.  Explicitly, it is the direct sum of the root lines for the short roots and the coroot lines 
${\rm{Lie}}(a^{\vee}(\mathbf{G}_m))$ for the coroots associated to the 
short roots in $\Delta$. (In fact, $\mathfrak{n}$ is the unique
minimal non-central ${\rm{Ad}}_G$-stable subspace of $\mathfrak{g}$.) Hence, $\mathfrak{n} = {\rm{T}}_e(V)$ for the smooth locally closed subvariety $V \subset G$ through the identity $e$ given as the closed subvariety of the open cell associated to $\Phi^+$ by forming the direct product (embedding via multiplication into $G$) of the $T$-root groups for the short roots and the direct factors $a^{\vee}(\mathbf{G}_m)$ for short $a \in \Delta$.<|endoftext|>
TITLE: Is Faltings' $p$-adic Eichler-Shimura isomorphism the $p$-adic comparison isomorphism?
QUESTION [13 upvotes]: This is a question about Faltings' $p$-adic Eichler-Shimura isomorphism from his 1987 article "Hodge-Tate structures and Modular Forms".
Let $N\ge5$, $k\ge2$ be integers. Denote by $X(N)$ the proper modular curve of full level $N$ (over $\mathbb Q$, say), $f\colon\overline E(N)\rightarrow X(N)$ the universal generalized elliptic curve over it and by $e\colon X(N)\rightarrow E(N)$ the unit section. Further let $\omega=e^*\Omega^1_{E(N)/X(N)}$ and $\Omega^1=\Omega^1_{X(N)}$. Let $Y(N)$ be the open modular curve and denote the universal elliptic curve over it still by $f$. Put $V=H^1_{\mathrm{p}}(Y(N)\times\overline{\mathbb Q},\operatorname{Sym}^{k-2}R^1f_*\mathbb Z_p)$, $W_0 = H^0(X(N),\omega^{k-2}\otimes\Omega^1)$ and $W_{k-1}=H^1(X(N),\omega^{2-k})$.
Faltings' $p$-adic Eichler-Shimura isomorphism, which is Thm. 6 (iii) in his paper, states that canonically
$$ \mathbb C_p\otimes V(1)\cong \mathbb C_p\otimes W_0 \oplus \mathbb C_p(k-1)\otimes W_{k-1}. $$
It seems to be well-known that this Eichler-Shimura isomorphism "is" the $p$-adic comparison isomorphism for the modular motive ${}^N_kW$ introduced by Scholl in his 1990 paper "Motives for Modular Forms".
In trying to make this statement precise, I obtained the following.
The motive introduced by Scholl which he calls ${}^N_kW$ has $V$ as its $p$-adic étale realization and $W:=W_0\oplus W_{k-1}$ as its Hodge realization, with the $W_i$ sitting in degree $i$, respectively.
The comparison isomorphism from $p$-adic Hodge theory gives
$$ B_{\mathrm{HT}}\otimes V \cong B_{\mathrm{HT}}\otimes W. $$
Taking the degree $0$ part of this one obtains an isomorphism
$$ \mathbb C_p\otimes V \cong \mathbb C_p\otimes W_0 \oplus \mathbb C_p(1-k)\otimes W_{k-1}, $$
which differs from Faltings' isomorphism above.
What is the precise relation between Faltings' Eichler-Shimura isomorphism and the general comparison isomorphism? Since I am not familiar with the techniques used by Faltings, I was not able to prove anything about this by myself.

REPLY [7 votes]: The issue seems to be about notation.  Your $k$ is what Faltings calls $k + 2$.  And what Faltings calls $\underline{V}_k$ is what you would call $V(k-1)$.  (Possibly you confused Faltings' $\underline{V}_k$ with his $V_k$?)  So when I write out Faltings' comparison isomorphism and then twist, I get (using your notation)
\begin{equation}
\mathbb{C}_p \otimes V \simeq \mathbb{C}_p \otimes W_{k-1} \oplus \mathbb{C}_p(1-k) \otimes W_0, 
\end{equation}
as one wants. 
Faltings' modular forms paper seems to be a prelude (a special case even) of his proof of the general comparison isomorphism which you refer to.  So it would be strange if the two comparison maps were not literally the same map.<|endoftext|>
TITLE: A search for theorems which appear to have very few, if any hypotheses
QUESTION [5 upvotes]: I'm interested in theorems which appear to have very few, if any hypotheses. Essentially a search for unexpected regularity or pattern in a relatively unstructured situation.
By "few hypotheses" I mean theorems which start "take any triangle", or "take any three circles". Similarly, the conclusion of the theorem ought to be really surprising. I know this is a little vague, but I've deliberately left it that way.
Perhaps my favourite here is Morley's theorem. This applies to any triangle, but has a very surprising conclusion. Contrast this with Pythagoras' theorem (needs a right angled triangle: too special!) or Viviani's Theorem (needs an equilateral triangle: too special!).
Can you help me gather a collection? I've made a very preliminary start here: http://tube.geogebra.org/book/title/id/2673817
Part of my underlying interest is in the aesthetic, and what professional mathematicians think is "significant", "surprising" or when exceptional cases mean the "take any .... except ..." means the theorem isn't so general after all.
Please don't be shy. I'd love to know what your favourites are. They don't have to be in geometry either.....
Chris Sangwin

REPLY [3 votes]: Every vector space (or module over a division ring) admits a basis (this includes the empty set as a basis for the zero module). Implicit is the axiom of choice, which I don't consider an extra hypothesis. 
All bases of a vector space have the same cardinality.<|endoftext|>
TITLE: boundary of semihyperbolic groups
QUESTION [7 upvotes]: There are various definitions of boundary of a hyperbolic group. Which of those generalize to semi-hyperbolic groups (in the sense of Alonso and Bridson)? 
The example I have in mind is a semisimple group over a nonArchemedian local field.The answer I'd like to get is something like a nonHausdorff space which is a union of partial flag manifolds $G/P$ (which can be described as the quotient of the complement to $G$ in the De Concini-Procesi compactification by the right action of $G$).

REPLY [5 votes]: I don't think any notion of boundary has been developed specifically for semi-hyperbolic groups, but there have been some recent generalizations of boundaries of hyperbolic groups that may be of interest to you.

Charney and Sultan defined the contracting boundary of a CAT(0) group (arXiv:1308.6615). (I don't know enough about your setting to know whether your groups act nicely on some sort of CAT(0) building, which might help.)
Charney's student Cordes generalized the contracting boundary to define the Morse boundary of any finitely generated group (arXiv:1502.04376).

Note that both of these boundaries can be, and frequently are, empty.  I think they're also both always Hausdorff, so may not give exactly the answer you want.  But hopefully it's somewhere to start.<|endoftext|>
TITLE: Example request: seriously deficient homogeneous spaces
QUESTION [5 upvotes]: In a previous post, I cite a dimension condition commonly satisfied by homogeneous spaces and claim that a counterexample must have deficiency at least $3$. For convenience, I reproduce the definition of deficiency in the next paragraph.
Let $G/K$ be a homogeneous space with $K$ connected. The quotient map $G \to G/K$ induces a ring map $H^*(G/K) \to H^*(G)$ in rational cohomology whose image is an exterior algebra $\Lambda \hat P$. The dimension of $\hat P$ is bounded above by $\mathrm{rk \,}G - \mathrm{rk \,}K$, and the deficiency is the difference from this upper bound:
$$\mathrm{df}(G/K) := \mathrm{rk \,}G- \mathrm{rk \,}K - \dim_{\mathbb Q} \hat P.$$
Now one thing that is clear about deficiency is that it is additive in the sense that given two homogeneous spaces $G_1/K_1$ and $G_2/K_2$, one has
$$\mathrm{df}\Big(\frac{G_1 \times G_2}{K_1 \times K_2}\Big) = \mathrm{df}(G_1/K_1) + \mathrm{df}(G_2/K_2).$$ 
Morally, this is an uninteresting way of picking up deficiency, and particularly, it is not useful in constructing counterexamples to the dimension condition of the previous post. Call a homogeneous space $G/K$ reducible if it admits a direct product factorization in terms of other homogeneous spaces. 
All examples of irreducible homogeneous spaces that I know of from the literature are of deficiency $0$ or $1$. I have been able to construct high-deficiency irreducible examples by modifying reducible ones, but these examples have a feel of cheapness to them and wind up satisfying the dimension condition anyway.
Is anyone aware of any examples in the literature of irreducible homogeneous spaces of deficiency at least $2$?

REPLY [2 votes]: def=2: see investigation of the case def=2 in [On coset-spaces of compact Lie groups by subgroups of corank 2. A.N.Shchetinin. Russian Mathematics, volume 61, pages 60–68(2017)]
and other papers of this author.<|endoftext|>
TITLE: $8$-ary operation $(\mathbb{P}^2)^8 \text{ }-\to \mathbb{P}^2$, can we say anything about what this formula would look like?
QUESTION [20 upvotes]: My friend, who is currently taking an algebraic geometry course from an unnamed prolific poster on MO, told me about the following bonus question on one of his problem sets a few weeks ago.

Preliminary discussion. By Bézout's Theorem (which we will prove later in the course), two plane cubics in general position with regard to one another, and hence any two memebrs of the pencil that they span, intersect in $9$ points. The previous question shows that in fact this pencil is already determined by $8$ of the $9$ intersection points.
From this you can deduce two things:
(a) $9$ points in general position cannot arise as the intersection of two cubic projective curves.
(b) $8$ points in general position determine a $9$th point (namely the $9$th common point of intersection of the pencil of cubics determined by the $9$ give points).
We can regard this process as defining a kind of $8$-ary operation$$(\mathbb{P}^2)^8\text{ }-\to \mathbb{P}^2.$$(We use a broken arrow because this operation is not really defined on all $8$-tuples of points in $\mathbb{P}^2$, but only on those in general position. It is an example of what is called a rational map in algebraic geometry.)
Actual bonus question. Can you say anything about what this formula would look like (e.g. its degree)?

This begs the natural question, does there exist a natural moduli space $X$ of $9$ points so that our map $X \to$ $9$th point is actually a map and not just a rational thing? We could start with $\text{Hilb}^8 \,\mathbb{P}^2$ and figure out how we need to modify it.
At this point, we could remark the following. Presumably, it would be an open subscheme of $\text{Hilb}^8\,\mathbb{P}^2$? We can not get general position already as an open subscheme of $(\mathbb{P}^2)^8$ because we would have to account for infinitely many closed things to throw out. but perhaps we could get this to work on passing to the Hilbert scheme. Actually intuitively, probably not – we would probably have to add extra parameters or something to parameterize all the possible things we would need to throw out? But it definitely seems possible.
Or wait, linear dependence is definitely finitely encodable, so we can get finitely many things to throw out already in in the $(\mathbb{P}^2)^8$ case. Obviously, since it is a rational map, it is defined on an open subset. Basic degree questions like "if we fix $7$ points, and a target point, what is the locus of the possible $8$th point" can be addressed in that context geometrically too. So does the Hilbert scheme buy us anything? Obviously, the symmetry of the point arrangements and stuff – so we guess that is a goal in itself. And it would be useful, for example, if we wanted to consider "families" of points over something, the Hilbert scheme would allow us to see the possible behaviors more cleanly. We do not know if we could convince someone else that the Hilbert scheme adds something here, though.
Also, but wait, the weird blowup behavior of the Hilbert scheme of points only happens at diagonals, so in this case, the open subscheme of the Hilbert scheme of points would literally just be the symmetric quotient of $(\mathbb{P}^2)^8$, so in particular, any behavior of it is recoverable in an extremely simple way from the latter.
With regards to those remarks, the entire point of using the Hilbert scheme is we want to understand what happens at nongeneric points to do a lot of things (e.g. intersection theory) because we need properness, which we are going to get from modifying the Hilbert scheme in some, hopefully small, way.
At this point, we could remark that we do not see an obvious way to extend the rational map to all of $\text{Hilb}$. Conceivably, $\text{Hilb}$ might help, since the blowups give us extra information about how points approach the degenerate positions, but we do not know how to actually do it.
With regard to that remark, right, the rational map does not extend to $\text{Hilb}$, and that is why we need to modify $\text{Hilb}$ first.
At this point, we could remark yeah true, but we do not see why a priori modifying $\text{Hilb}$ to have this behavior is easier than just modifying $(\mathbb{P}^2)^8$.
With regard to that remark, we think $\text{Hilb}$ is the correct starting point because the intersection of $2$ different cubic curves is a length $9$ subscheme.
Question. Hopefully building on the blath I have above, is there anything more we can say about what this formula looks like?

REPLY [18 votes]: This is very much a question of recent research, solved with varying degrees of generality in the papers listed below.  Here is a summary.
The rational map $\mathbb{P}^{2[8]}\dashrightarrow \mathbb{P}^2$ from the Hilbert scheme of $8$ points in $\mathbb{P}^2$ is the map corresponding to an extremal effective divisor on the Hilbert scheme.  This suggests that you should run the minimal model program for the Hilbert scheme, studying the birational modifications that arise on the way from the standard model of the Hilbert scheme until you reach the model where the map to $\mathbb{P}^2$ becomes a morphism.  (Note that the Hilbert scheme is a Mori dream space, so this goal is actually realistic.)
Along the way, you'll find that the models you obtain can be realized as moduli spaces of certain Bridgeland semistable objects.  In the general case of $n$ points, the Hilbert scheme will be replaced by a Grassmannian bundle over a moduli space of representations of a Kronecker quiver; in your $n=8$ point case this simplifies to a $G(2,9)$-bundle over $\mathbb{P}^2$, as follows.
Consider the space consisting of pairs $(p,\Lambda)$ where $p\in\mathbb{P}^2$ is a point and $$\Lambda\in G(2, H^0(\mathcal{O}_{\mathbb{P}^2}(3) \otimes I_p))$$ is a two-plane in the space of cubics passing through $p$.  Thus, this space is a $G(2,9)$-bundle over $\mathbb{P}^2$.  It is evidently birational to the Hilbert scheme, since a general $(p,\Lambda)$ determines a length $8$ scheme residual to $p$ in the intersection of the cubics in $\Lambda$.  And, the forgetful map to $\mathbb{P}^2$ is your Cayley-Bacharach map.
The references for more general questions of this type are:
Daniele Arcara, Aaron Bertram, Izzet Coskun, and Jack Huizenga, The minimal model program for the Hilbert scheme of points on $\Bbb{P}^2$ and Bridgeland stability, Adv. Math. 235 (2013), 580--626. (Explicitly runs the MMP for $n\leq 9$ points, so your question is a special case.)
Coskun, H., Woolf, "The effective cone of the moduli space of sheaves on the plane."
H., "Effective divisors on the Hilbert scheme of points in the plane and interpolation for stable bundles."
Also there is a survey which might help get you started:
Izzet Coskun and Jack Huizenga, The birational geometry of the moduli spaces of sheaves on $\Bbb P^2$,  Proceedings of the Gökova Geometry-Topology Conference 2014 (2015), 114--155.<|endoftext|>
TITLE: An explicit description of neighborhoods of the rank 2 boundary in the Satake Compactification of $\mathbf{A}_2$
QUESTION [7 upvotes]: My Motivation: I'm having a hard time following the description of the topology in the Satake Compactification of locally symmetric spaces. The group theory is something I'm finding a bit tricky to follow, so I'd prefer answers in terms of explicit matrix representations, though a concrete descriptions of cylindrical sets or something might be helpful too.
Background: Let $\mathbf{A}_2$ be the moduli space of dimension 2 principally polarized abelian varieties. If you prefer, just think of it as the Siegel upper half space of $2\times 2$ symmetric matrices $Z=X+iY$ with $Y$ positive definite, quotiented out by $\mathbf{SP}_4(\mathbf{Z})$. If one compactifies $\mathbf{A}_2$ according to the Satake compactification, there is a rank 1 boundary component and a rank 2 boundary component. 
Main Question: What is the preimage- in Siegel upper half space - of a simple system of fundamental neighborhoods for the rank 2 boundary component of $\mathbf{A}_2$, in terms of an explicit description for the matrix co-ordinates.
Form of Answer that I'd like: Something about $Y$ as a quadratic form, involving its determinant, minimal vector length, etc...
I'd strongly prefer an answer entirely specific to this case, if possible, and not something for general locally symmetric spaces. 
Additional question I'd be grateful to get the answer to: Same question but for the rank 1 boundary component.
Thanks!!

REPLY [3 votes]: This answer is incomplete but I'll try to finish it later.
First note that you can't express the answer only using $Y$, even in the $g=1$ case. If the imaginary part of $\tau$ goes to $0$, you do not know anytihng about what $j(\tau)$ is doing.
I think you want to work with the induced quadratic form on pairs $(v,w) \in \mathbb Z^2 \times \mathbb Z^2 = \mathbb Z^4$ given by 
$$ (v_1,w_1) \cdot (v_2,w_2) = \operatorname{Re}\left( \overline{(v_1 + Z w_1)}^T Y^{-1}(v_2 + Z w_2) \right)$$
If I calculated correctly, this is the actual Riemann form of your torus, because it comes from a Hermitian form $Y^{-1}$ on $\mathbb C^2$ whose imaginary part resticts to the standard symplectic from on $\mathbb Z^4$.
The action of $SP_{4}(\mathbb Z)$ on this is the natural action on quadratic forms in four variables. Indeed, $v^T Y^{-1} v$ on $\mathbb C^2$ is the unique Hermitian form whose imaginary part restricts to the standard symplectic form on the lattice $\mathbb Z^4$ embedded by the map $(v_1, v_2) \mapsto v_1 + Z v_2$. The action of $SP_{4}$ on $Z$ comes from the action of $SP_4$ on this lattice plus some weird action on $\mathbb C^2$, but since the action preserves the standard symplectic form it will preserve the uniquely determined Hermitian form.
Now I think a basis of neighborhoods of the rank $k$ component of the boundary of this lattice consists of points where there are $k$ vectors $v_1,\dots,v_k$ in $\mathbb Z^4$ that are linearly independent, whose pairing under the symplectic forms with each other all vanish, and whose lengths under the quadratic form are all $<\epsilon$. 
I think this because if you have a sequence of such quadratic forms with vectors $v_1,\dots,v_k$ so that the length of those vectors goes to $0$, you obtain a lattice of rank $4-2k$ with a symplectic form by modding out by $v_1,\dots,v_k$ and taking the kernel of the map to $v_1,\dots, v_k$, and you obtain a corresponding quadratic form by taking the limit of the original quadratic form, which will be well-defined as the lengths of $v_1,\dots,v_k$ goes to $0$. So a sequence of points that are in all the rank $k$ neighborhoods but not all the rank $k+1$ neighborhoods will have an accumulation point in $\mathcal A^{2-k}$. That is precisely what the Satake compactification is supposed to look like. I will try to rigorously justify this...
In fact the condition that $v_1,\dots,v_k$ have zero pairing under the symplectic form is irrelevant. As the symplectic form and quadratic form come from the same Hermitian forrm, they are related by a Cauchy-Schwartz inequality, so if $v_1,v_2$ have length at most $\epsilon$, their pairing is at most $\epsilon^2$, so because the symplectic form is integral it is $0$. So we can express the condition only in terms of tuples of short vectors in the quadratic form with matrix
$$\begin{pmatrix} Y^{-1} & Y^{-1} X \\ X Y^{-1} & XY^{-1}X + Y\end{pmatrix}$$
That's totally explicit but might not be too fun. An alternate desciption of the system of neighborhoods of the rank $2$ part which may or may not be easier is the $SP_4(\mathbb Z)$ orbit of the set of pairs $X,Y$ where the shortest vector of $Y$ has length at least $1/\epsilon$. This is the same because you can always transform your two short vectors to be the first two basis vectors.<|endoftext|>
TITLE: Levelt-Turrittin Theorem over p-adics (or the monodromy theorem)
QUESTION [6 upvotes]: Let $V$ be a finite dimensional vector space over $\mathbb{C}((t))$. Let $D:V\rightarrow V$ be a differential operator; i.e., an additive $\mathbb{C}$-linear map satisfying 
$$
D(a.v)=(t\frac{d}{dt}a).v,\quad \quad a\in \mathbb{C}((t)), \quad v\in V. 
$$
The fundamental theorem of Turrittin and Levelt states that $D$ has a Jordan decomposition. In more details, after possibly extending the field by adding a root of $t$, there exists a basis of $V$ in which $D$ is represented by a Jordan matrix. A good reference here is Levelt's original paper on the subject. 
In the past decade or so, several people have developed a "$p$-adic analogue" of this theorem. Some of the people involved are Mebkhout, Andre, and Kedlaya. Christol has an unfinished book on this topic. Sometimes the result is known as $p$-adic monodromy theorem. 
I would like to understand the statement of this result which shows its resemblance to the usual Levelt-Turrittin theorem mentioned above. Presumably, one has to replace $\mathbb{C}((t))$ by another field, and then the result states that differential operators over this field also have Jordan decompositions. 
Unfortunately, most of the references on the subject are full of technical jargon and seem impossible to penetrate for a novice. So I appreciate any helpful thoughts are references on this question.

REPLY [4 votes]: Kiran Kedlaya has written a nice book called "$p$-adic differential equations" (Cambridge University Press), where the theorem you're looking for is stated and proved
http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521768795
I wrote a short review of the book for the Bulletin of the AMS, which contains a very brief explanation of how this theorem is a $p$-adic generalization of Turrittin's theorem, see
http://www.ams.org/journals/bull/2012-49-03/S0273-0979-2012-01371-X/<|endoftext|>
TITLE: Is there a polynomial-time algorithm to check if a signed graph contains an odd-K5 minor?
QUESTION [5 upvotes]: I suspect this exists, if anyone has a reference please that would be very helpful.
By signed graph, I mean each edge is designated either odd or even (e.g. as in Guenin's result for weakly bipartite graphs).

REPLY [11 votes]: Yes.  This result is contained in my PhD thesis, which is available here (see Theorem 1.1.10).  We prove that for any finite abelian group $\Gamma$ and fixed $\Gamma$-labeled graph $H$, there is a polynomial time algorithm to determine if an input $\Gamma$-labelled graph $G$ contains an $H$-minor.  The case you are interested in is $\Gamma=\mathbb{Z} / 2 \mathbb{Z}$ and $H$ equals odd-$K_5$.  For signed graphs, this result was also obtained independently by Kawarabayashi, Reed, and Wollan (although I am not sure that a journal version is available yet).<|endoftext|>
TITLE: Zero-sum sets in union-closed families
QUESTION [6 upvotes]: The Davenport constant $D(G)$ of a finite abelian group $G$ is the minimum integer $n$ such that whenever $a_1, \ldots, a_n \in G$ (not necessarily distinct), there is a non-empty $I \subseteq [n]$ such that $\sum_{i \in I} a_i = 0$. 
I wonder what happens when we cannot just add the entries up:
Given a finite ground-set, say $[n]$, and a union-closed family $\mathcal{A}$ of (non-empty) subsets of $[n]$, let's say $\mathcal{A}$ has the zero-sum property with respect to $G$ if for every map $\nu : [n] \to G$, there is a $A \in \mathcal{A}$ with $\sum_{x \in A} \nu(x) = 0$.
If $\mathcal{A}$ contains $D(G)$ disjoint sets $A_1, \ldots, A_{D(G)}$, then under any $\nu : [n] \to G$ there will be some $I \subseteq [D(G)]$ such that $\sum_{x \in \bigcup_{i \in I} A_i} v(x) = 0$. Thus $\mathcal{A}$ has the zero-sum property.
If $\mathcal{A}$ can be covered by some $X \subseteq [n]$ of size $|X| < D(G)$, meaning that $X$ meets every element of $\mathcal{A}$, then $\mathcal{A}$ does not have the zero-sum property: By definition, there exists a map $\nu : X \to G$ without zero-sum subsequence and we can extend this map by setting $\nu(y) = 0$ for every $y \notin X$.
The question I ultimately want to get at is the following: Does the converse of the last statement also hold, namely: If $\mathcal{A}$ does not have the zero-sum property, does it necessarily admit a cover of size less than $D(G)$? Or, possibly weaker, is there some constant $q = q(G)$ such that if $\mathcal{A}$ does not have the zero-sum property, then it admits a cover of size at most $q$?
Using the inclusion-exclusion principle, this is easy to show for $G = \mathbb{Z}_2$, but it is not clear to me how to extend this to other groups.
[The question is related to, but different from this one that I asked on this site a while ago.]
Edit: Trimmed the question to make it more accessible.

REPLY [3 votes]: Let me prove that for $G=\mathbb{Z}_{p^s}$, $p$ is prime, we indeed have $q(G)=p^s-1=D(G)-1$. It is again inclusion-exclusion, as for $p^s=2$.
Let $\mathcal{A}$ be a union-closed family, assume that a function $\nu:[n]\rightarrow \mathbb{Z}$ satisfies the following condition: $p^s$ does not divide $S(A):=\sum_{x\in A} \nu(x)$ for any set $A\in {\mathcal A}$. 
We use the following polynomial
$$
\varphi(x)=\binom{x-1}{p^s-1}+(-1)^{p^s}.
$$
We have $\varphi(0)=0$ and $\varphi(x)\equiv (-1)^{p^s} \pmod p$ for integer $x$ not divisible by $p^s$.
Let $\mathcal A$ be union-generated by sets $A_1,\dots,A_m$ (we may even assume that these are all sets of our family $\mathcal A$). Consider the following sum
$$
\sum \varphi(S(A_i))-\sum_{i<j} \varphi(S(A_i\cup A_j))+\sum_{i<j<k}\varphi(S(A_i\cup A_j\cup A_k))-\dots
$$
Modulo $p$ it equals 
$$
(-1)^{p^s}\left(m-\binom{m}2+\binom{m}3-\dots\right)=(-1)^{p^s}.
$$ 
On the other hand, expand this as a polynomial in variables $\nu(1),\nu(2),\dots,\nu(n)$. There is no free term, since $\varphi(0)=0$.
Consider any specific term, say $X=\nu(1)^3 \nu(2)\nu(5)\nu(8)^{p^s-6}$. Assume that elements 1,2,5,8 do not cover our family $\mathcal A$, that is, there exists index $t$ such that $\{1,2,5,8\}\cap A_t=\emptyset$. Then we may partition summands with the term $X$ onto pairs $\{\cup_{i\in I} A_i,\cup_{i\in I\cup\{t\}} A_i\}$, where $I$ runs over non-empty subsets of $[m]\setminus t$ such that $\{1,2,5,8\}\subset \cup_{i\in I} A_i$. For each such pair term $X$ cancels. So, total coefficient of $X$ is 0. But some coefficient is non-zero, since tital sum is not divisible by $p$. Thus there exists a cover of $\mathcal A$ of size at most $\deg \varphi=p^s-1$.<|endoftext|>
TITLE: Asymptotics of a special function
QUESTION [8 upvotes]: In my research, I came up with a special function which I denote by $B(q)$ and is defined by the integral 
$$B(q)\equiv \int_{-\pi/2}^{\pi/2} \frac{\sin\left(\frac{q}{2}\tan\theta\right)}{\sin\theta}d\theta\ .$$
I only care about real $q$'s, but that is probably not important for my question. Using a simple change of variables $w=\frac{q}{2}\tan\theta$ this can also be written as 
$$B(q)=\int_{-\infty}^{\infty} \frac{\sin\left(w\right)}{w}\frac{1}{\sqrt{1+4 w^2/q^2}}dw
$$
(if you're interested, this came up as one of the components of the response function of an elastic half-space).
I am interested in the asymptotics of $B(q)$. For large $q$ it is easy to show that $B(q)$ approaches $\pi$. My problem is with the small $q$ limit, $q\to0$. Numerically I have found that the leading order seems to behave as $B(q)=(\pi \gamma-\log(q))q+\mathcal{O}(q^2)$, where $\gamma$ is the Euler-Mascharoni constant. You can see in these plots that this is a very good guess (blue line is the approximation, red points are calculated numerically):

What I need is a way to prove that this is indeed the asymptotic behavior of $B$. Any suggestions (or references to relevant techniques) will be much appreciated.

REPLY [12 votes]: Gradshteyn & Ryzhik equation 3.773.1 gives (for $q>0$)
$$\frac{1}{q}B(q)=\int_0^\infty \frac{\sin x}{x(q^2/4+x^2)^{1/2}}\,dx=\frac{1}{q}G^{21}_{13}\left(\frac{q^2}{16}\biggl|^1_{1/2,1/2,0}\right)$$
$$\qquad\qquad=1-\gamma+\log 4-\log q+{\cal O}(q^2)=1.80908-\log q+{\cal O}(q^2).$$
The asymptotic result desired by the OP is $B(q)/q=\pi\gamma-\log q=1.81338-\log q$. This is sufficiently close to the above answer that I am tempted to conclude that the correct order 1 term is actually $1-\gamma+\log 4$ instead of the conjectured $\pi\gamma$.
In support of this, the plot shows the numerical evaluation of the integral (green line) and the two asymptotes (blue and red lines). The upper red line is the asymptote guessed by the OP and the lower blue line is the asymptote obtained here, which seems to be the correct one.

and this is j.c. 's plot, showing the difference between $B(q)$ and the two asymptotes (gold from the OP, and blue from this answer):<|endoftext|>
TITLE: Do vanishing characteristic classes of the tangent bundle imply a manifold is stably frameable?
QUESTION [16 upvotes]: Suppose we know that all stable characteristic classes of the tangent bundle of a manifold $M$ vanish, i.e. the map $f:M\rightarrow BO$ stably classifying the tangent bundle is trivial on cohomology, i.e. $0=f^*: H^*(BO, A)\rightarrow H^*(M, A)$, for all coefficients $A$. Is it then possible to find a stable framing of $M$? In other words, can we lift $f$ to a map $\tilde{f}: M\rightarrow B\{1\}\simeq *$? 
I know this is not the case for a general vector bundle (as discussed in Hatcher's "Vector bundles and K-theory", p.75-76), but I was wondering whether it might be true when restricting to the case where the bundle is the tangent bundle of a manifold. And how about if it is a normal bundle?

REPLY [9 votes]: Suppose that $X^n$ is a finite complex and $V \to X$ a vector bundle which has trivial characteristic classes but is not stably trivial. Examples for such bundles were given in the answers to 
Non-stably trivial bundle with trivial characteristic classes
The stable normal bundle of $S^{2n}$ admits a bundle map to $V$ (because $S^{2n}$ is stably parallelizable). This gives a map $f: S^{2n} \to X$, covered by a bundle map. By surgeries below the middle dimension, we can change $S^{2n}$ and $f$ to a map $f':M \to X$ which is covered by a bundle map $\nu_M \to V$ and such that $f$ is $n$-connected.
To see that $\nu_M$ is not stably trivial, note that since $X$ is $n$-dimensional and $f: M \to X$ $n$-connected, we can find a map $g: X \to M$ with $f \circ g \simeq id_X$, by basic obstruction theory. 
But  $g^* \nu_M \cong g^* f^* V  \cong V$ is not stably trivial, and so $\nu_M$ isn't, either.<|endoftext|>
TITLE: Is there a primitive recursive real number which cannot have a primitive recursive expansion?
QUESTION [5 upvotes]: Using the definitions given below, my question can be restated as

Does there exist a primitive recursive (PR) real $\{s_n\}$ such that for every scale $r \geq 2$ and every PR sequence $a_n$ with $0 \leq a_{n+1} \leq r-1$ it is NOT the case that 
  $$
\{s_n\} = \{ \sum_{p \leq n} a_p r^{-p} \}
$$

If $\{s_n\}$ is a PR real number and if $a_n$ is a PR sequence such that
$$
0 \leq a_{n+1} \leq r-1, r \geq 2
$$
and
$$
\{s_n\} = \{ \sum_{p \leq n} a_p r^{-p} \}
$$
then $\{s_n\}$ is said to have the PR expansion $\sum_{p \leq n} a_p r^{-p}$ in the scale of $r$.
A rational PR function which is PR convergent is called a PR real number.
PR real numbers $\{s_n\}$ and $\{t_n\}$ are equal, denoted $\{s_n\} = \{t_n\}$,  if $\vert s_n - t_n \vert < 1/10^k$ for majorant $n$.
A rational PR function $f(x)$ is PR convergent if there is a PR $N(k)$ such that $N(k+1) \geq N(k) \geq k$ and
$$
i,j \geq N(k) \rightarrow
\vert f(i)-f(j) \vert < 1/10^k
$$

REPLY [4 votes]: I think it's useful to first consider an analogous situation in constructive mathematics. It is not possible to show constructively that every real number has a base $b$ expansion. This is because if the expansion starts $0.xxxx$, then we know the number is less than or equal to $1$, and if the expansion starts $1.xxxx$ we know the number is greater than or equal to $1$. But to be able to decide whether a number is $\leq 1$ or $\geq 1$ is equivalent to the principle LLPO, which is not constructive. Similarly it is impossible to find a base $n$ expansion of all computable reals in a uniformly computable way. We'll use this basic idea in the proof below.
There is one curious difference. For general computable functions, we can't find a expansion uniformly, but each computable real individually does have a computable expansion, because both rational and computable irriational numbers have computable expansions. For p.r. reals we can do something much better, which is to find one particular p.r. real with no p.r. expansion.
We basically do this by a diagonal argument. Assume that we are given a primitive recursive surjective pairing operation on $\mathbb{N}$, denoted $(,)$. Also say that we are given a p.r. enumeration of codes for all p.r. functions, represented as codes for Turing machines. We will define a p.r. real $s = (s_i)_{i \in \mathbb{N}}$ in a sequence of "blocks" where in the $n$th block we will ensure that if $n$ codes the pair $(e, r)$ then the $e$th p.r. function is not a p.r. expansion of $s$ in the scale $r$. Along the way we will ensure that for $j, k > i$, $|s_j - s_k| < \frac{1}{10^i}$. Together with defining $s_i$ in a p.r. manner, this will ensure that $s$ is in fact a p.r. real.
Within each block $(e, r)$, we define $s_i$ as follows. Let $i_0$ be the first $i \in \mathbb{N}$ that lies in the block $(e, r)$. Define $q \in \mathbb{Q}$ to be $s_{i_0 - 1}$ if $i_0 > 0$ and $1$ otherwise. Now let $N$ be such that $r^{N} > 2 \times 10^{i_0}$. We can then find in a p.r. way, $k \in \mathbb{N}$ so that $|\frac{k}{r^N} - q| < \frac{1}{2\times 10^{i_0}}$. Let $s_{i_0} := \frac{k}{r^N}$. Now suppose that $(a_j)_{j \in \mathbb{N}}$ is some sequence such that $0 \leq a_j < r$ for all $j$. Then we have that for some $k'$, $\sum_{j = 0}^{N} a_j r^{-j} = \frac{k'}{r^N}$. Note that if $k' < k$, then $\sum_{j = 0}^\infty a_j r^{-j} \leq \frac{k}{r^N}$ and if $k' \geq k$ then $\sum_{j = 0}^\infty a_j r^{-j} \geq \frac{k}{r^N}$. We now ensure that the $e$th p.r. function is not an expansion of $s$ in scale $r$. We can't evaluate $\{e\}(j)$ for $j\leq N$ in a p.r. way, but what we can do is ask whether $\{e\}(j)$ has halted in less than $i$ steps for $j \leq N$. If $\{e\}(j)$ has not yet halted at stage $i$ for some $j \leq N$, we set $s_i := s_{i_0}$. If $\{e\}(j)$ has halted at stage $i$ for every $j \leq N$, we calculate $k'$ as above. If $k' < k$, we define $s_i := s_{i_0} + \frac{1}{3 \times 10^i}$ and from now on ensure that for future $i' > i$, we have $s_{i'} > s_{i_0} + \frac{1}{4 \times 10^i}$. This makes sure that $s > s_{i_0}$. If $k' \geq k$, then we similarly define $s_i := s_{i_0} - \frac{1}{3 \times 10^i}$ and ensure that for $i' > i$, we have $s_{i'} < s_{i_0} - \frac{1}{4 \times 10^i}$, to get $s < s_{i_0}$. However, by the above reasoning this has made sure that $e$ is not a p.r. expansion of $s$ of scale $r$ as required. Finally, note that

Since $e$ is primitive recursive, we know that eventually $\{e\}(j)$ will halt for every $j \leq N$, so we will at some point be able to diagonalise and move on to the next block. This is the crucial difference to general recursive functions.
We can't find the length of each block in a primitive recursive way, but also we don't need to for the argument to work.<|endoftext|>
TITLE: "Lagrangian" subalgebra of cohomology, with respect to Poincare duality?
QUESTION [9 upvotes]: Let $M$ be a compact oriented $n$-manifold, and let $H^*(M)$ denote
its cohomology ring with coefficients in $\mathbb{R}$.
Let's say that a graded subalgebra $K^\bullet \subset H^\bullet(M)$ is a
Lagrangian subalgebra if we have an isomorphism of graded vector
spaces with bilinear forms
$$ H^\bullet(M) \simeq K^\bullet \oplus (K^{n-\bullet})^*$$
where the left hand side has the bilinear form coming from Poincare
duality, and the right hand side has the sum of the tautological bilinear forms
$$ \tau_i : K^i \oplus (K^{n-i})^* \times K^{n-i} \oplus (K^i)^* \to
\mathbb{R}$$
given by
$$ \tau_i((x,\phi), (y,\psi)) = \psi(x) + (-1)^{i(n-i)}\phi(y) $$
This terminology is meant to make an analogy with the case of a
symplectic vector space $(V,\omega)$, where a subspace $L \subset V$
is Lagrangian if $(V,\omega) \simeq (L \oplus L^*, \tau)$ with $\tau$ the
tautological alternating form.
The question is simply

When does $H^*(M)$ admit a Lagrangian subalgebra?  Is there
a simple criterion?

More generally I wonder if this notion has been studied under some
other name, and if there is a structure theory classifying such
subalgebras in cases when they do exist.
Examples
If $M=S^n$ then the trivial subalgebra $K = \mathbb{R}$ is the unique
Lagrangian subalgebra.
If $M$ is an oriented surface of genus $g$, then using standard
generators $\{ a_i, b_i \}_{i=1\ldots g}$ for $H^1(M)$, the algebra
generated by $a_1, \ldots, a_g$ is a Lagrangian subalgebra.  More generally
one can take the algebra generated by any Lagrangian subspace of the
symplectic vector space $H^1(M)$.
If $M = \mathbb{CP}^2$, or more generally if $M$ has dimension $2k$
and $H^k(M)$ is odd-dimensional, then there is no Lagrangian
subalgebra.  This is because a Lagrangian subalgebra would give an
isomorphism $H^k(M)$ with the even-dimensional space $K^k \oplus
(K^k)^*$.
Motivation
Lagrangian subalgebras seem to correspond to subcomplexes with
interesting topological properties.
For example, suppose $\iota : X \hookrightarrow M$ is a subcomplex "representing" a Lagrangian subalgebra of $H^\bullet(M)$ in the sense that there is an
exact sequence $$ 0 \to K^\bullet \hookrightarrow H^\bullet(M) \xrightarrow{\iota^*}
H^\bullet(X) \to 0 $$ with $K^\bullet$ a Lagrangian subalgebra. 
Suppose also that $X$ can be thickened to an embedded trivial $S^k$-bundle $Y
\subset M$, for some $k \geq 0$. Then the cohomology ring of the
complement $M \setminus Y$ satisfies Poincare duality in dimension
$(n-k-1)$.  (In this sense the complement of $Y$ "looks" like a bundle over the $(k+1)$-disk with compact fiber, at least cohomologically.)
In the case where $M = S^n$ and $X$ is a point, representing the trivial Lagrangian subalgebra, one recovers here the familiar homotopy equivalence between $S^n \setminus S^k$ and $S^{n-k-1}$.

REPLY [3 votes]: By definition, a closed oriented $n$-dimensional manifold $M$ has a twisted double structure if $M=N\cup_h-N$ for an orientation-preserving self-homeomorphism $h:\partial N \to \partial N$ of the boundary of an oriented $n$-dimensional manifold $N$. The graded subalgebra $K^{\bullet}=ker(H^{\bullet}(M)\to H^{\bullet}(N)) \subset H^{\bullet}(M)$ is a Lagrangian subalgebra of the asymmetric Poincare duality structure on $H^{\bullet}(M)$. In fact, using ${\mathbb Z}[\pi_1(M)]$-coefficients the converse is true: a closed oriented $n$-dimensional manifold $M$ has a twisted double structure if and only if there exists such an asymmetric Lagrangian subalgebra. The obstruction to the existence of a twisted double structure is the Quinn obstruction to the existence of an open book structure on $M$, which takes value in the asymmetric Witt group for $n$ even, and is 0 for $n$ odd. (As usual, one has to treat the cases $n>4$ and $n \leq 4$ separately and be particularly careful with the low-dimensional cases). See Chapters 29 and 30 of my 1998 Springer book "High-dimensional knot theory" http://www.maths.ed.ac.uk/~aar/books/knot.pdf for the algebraic surgery treatment of this obstruction, including the references to the relevant work of Milnor, Smale, Barden, Wall, Winkelnkemper, Quinn and  T.Lawson. The book also includes an appendix by Winkelnkemper on the history and applications of open books.<|endoftext|>
TITLE: Largeness and arithmetic progression properties of generic reals
QUESTION [8 upvotes]: Consider the following properties for a subset $A$ of $\mathbb{N}$:
(1) $A$ is large: $\sum_{n \in A}$$ 1\over n$$=\infty,$
(2) $A^\infty=\limsup \frac{|A \cap \{ 1, \dots, n\}|}{n} >0$,
(3) $A_\infty=\liminf \frac{|A \cap \{ 1, \dots, n\}|}{n} >0.$
By a conjecture of Erdős-Turán, if $A$ is large, then it contains arithmetic progressions of any given (finite) length.
By a result of Szemerédi, if $A^\infty >0,$ then $A$ contains infinite arithmetic progressions of length $k$ for all positive integers $k$.
Let's consider these properties for generic subsets of $\omega$ added by forcing (I assume they do not contain $0$) and let me say a few examples:
(a) If $r$ is Cohen, then $r$ is large, $r^\infty=1, r_\infty=0$ and for any $K,L$, we can find $M$ such that $M, M+L, M+2L, \dots M+KL$ are in $r$.
(b) If $r$ is Random, then $r$ is large, and $r^\infty=r_\infty=$$1\over 2.$ So
by Szemerédi's result, it contains arbitrary large arithmetic progressions.

Question 1. Is there a direct proof, without using of Szemerédi's result, that $r$ contains arbitrary large arithmetic progressions? 

(c) If $r$ is Mathias, related to an ultrafilter $U$, then the properties of $r$ depends on $U$.
It is possible to say the same results for some other generic reals, but as there are many generic reals that I do not know, I would like to ask a more general question: 

Question 2. Suppose that $r$ is a generic real added by forcing (we assume it does not contain $0$). Disuss if $r$ is large, and what are $r^\infty$ and $r_\infty$. Also say if $r$ contains arbitrary large arithmetic progressions or not (I would rather direct proofs instead of referring to some known results (say for example, as $r^\infty>0,$ it contains arbitrary large arithmetic progressions)).

REPLY [12 votes]: Nice question! I happened to be thinking about some similar things a few weeks ago. Here is what I found:
Cohen and random reals:
Cohen and random reals have just about any Ramsey-theoretic property you could want. In fact, they will satisfy the conclusion of the central sets theorem, which implies automatically that they will also satisfy the conclusions of van der Waerden's Theorem, Schur's Theorem, and Hindman's Theorem (and more). Let me briefly explain why.
If $r$ is random, then $r$ must be thick, which means that it contains arbitrarily long intervals (already, this is much stronger than the conclusion of van der Waerden's Theorem). This is not hard to prove, and you certainly don't need anything as powerful as Szemeredi's Theorem to do it. Informally: if you flip a coin over and over and over again, you expect that (for any fixed $n$) you will eventually get $n$ heads in a row. Formally: you may check that, for each $n$, the set
$$D_n = \{r : r \text{ contains a sequence of }n \text{ consecutive } 1s\}$$
has measure $1$ in $2^\omega$. $D_n$ is also a dense $G_\delta$, which shows that Cohen reals are also thick.
If $r$ is thick, it automatically satisfies the conclusion of van der Waerden's Theorem. It is known that if a set $A \subseteq \mathbb{N}$ is thick, then $\overline{A}$ contains a minimal idempotent in the semigroup $(\beta \mathbb{N},+)$, which is equivalent to $A$ being a central set. If you're not familiar with central sets, then you still might be familiar with the ultrafilter-based proof of Hindman's Theorem, and you'll know that any set in a (not necessarily minimal) idempotent ultrafilter will satisfy the conclusion of Hindman's Theorem (and Schur's Theorem is implied by Hindman's).
Other kinds of reals:
For Mathias reals arising from an ultrafilter $U$, you're right to say that the Ramsey-theoretic properties of the real will depend on $U$. More specifically, it seems that the generic real will satisfy the conclusion of van der Waerden's Theorem if and only if every member of $U$ does. This is the case, for example, when $U$ is a minimal ultrafilter. However, Grigorieff reals (when $U$ is a $P$-point) will not satisfy the conclusion of van der Waerden's Theorem, because it is known that every $P$-point contains reasonably sparse sets not containing arbitrarily long progressions.
It seems impossible to say anything positive about Sacks reals. Given a set $A \subseteq \mathbb{N}$, it's not hard to find a perfect tree $T$ that forces your Sacks real to be contained in $A$. Thus a Sack's real can be contained in any very sparse set you like, and it will not have nice Ramsey-theoretic properties, it will not be large, and it will have $A^\infty$ and $A_\infty$ both zero. It is also not too hard to find a perfect tree $T$ that forces your Sacks real to be thick. So Sacks reals can also have very nice Ramsey-theoretic properties -- it just depends on the particular Sacks real you are looking at. A similar analysis applies to Prikry-Silver reals.<|endoftext|>
TITLE: Heyting algebras originating from directed graphs
QUESTION [6 upvotes]: The category RefGph of reflexive directed graphs is the functor
category $\hat{∆}_1=\mbox{Fun}(∆^◦_1,$Set), where $∆_1$ is 
the simplex category truncated at level 1. 
Hence the poset Sub(X) of subobjects of such a graph X is a Heyting algebra.
I am interested in the category $K_1$, whose objects are the Heyting algebras, which are isomorphic to those of the form Sub(X) for some reflexive directed graph X and in the corresponding superintuitionistic propositional logic for which these Heyting algebras are algebraic models i.e. in the set $L(K_1)$, constisting of those propositional formulas $\varphi$ in the variables $p_i$, which are valid in exactly the objects of $K_1$. (graph logic)
In particular I would like answers or explanations to the following questions: 

How does one construct (finite) coproducts in $K_1$ explicitly?
How does one describe free objects of $K_1$ on finite generating sets explicitly?
Can the objects of the category $K_1$ be characterized by a finite set of explicit identities? 
Can the propositional logic $L(K_1)$ be described by a finite set of explicit axioms?

Can you give some references to relevant papers?
May be you even have ideas concerning the following question:

Can you think of an interpretation of the logic $L(K_1)$ as a special type of constructive reasoning?

REPLY [5 votes]: Your category of graphs is dually equivalent to an equationally defined category 
of algebras. The algebras don't form a variety in the technical sense that François intends, since infinitary operations are required, although there is a variety whose category of finite algebras is dually equivalent to the category of finite graphs. If we remain open about what the algebras are, that is, don't require that they appear as $Sub(G),$ then there is a simple answer to what they can be. The category of sets is dually equivalent to the category of complete atomic Boolean algebras, which are equationally definable using equations for joins and meets, and equations for complete distributivity. The action of the two element monoid, whose Karoubi envelope is $\Delta_1$, used to describe the graph category can be transferred to the Boolean algebras as an extra unary operation.
But there's another answer that is closer to your question. What if the subobject classifier Omega is used directly to induce a duality, and we try to describe the Heyting algebras that arise from the poset of subgraphs of the graphs ? If we use the axioms for Heyting algebras plus the two equations
(mentioned already in my second post)
   $p \vee (p \Rightarrow (q \vee \neg q))=1$ and
   $\neg (p  \wedge q \wedge r) = \neg (p \wedge q)\vee   (q \wedge  r)\vee   (r \wedge p)  $
 as well as infinitary join and meet rule, will we be able to recover, up to an isomorphism, a single graph ? Well no, because the opposite of a graph, obtained by reversing the direction of arrows, will have a subgraph lattice that is isomorphic to the subgraph lattice of the original graph, and of course a directed graph is not generally isomorphic to its opposite. There is something that can be done to correct the situation however. If we add a single unary "orientation" operation
  $ \tau $
 satisfying
 $\tau p \vee \tau \neg p =1$ and $\tau p \wedge \tau \neg p = p \vee \neg p,$
 then these "graphic" algebras, with the two extra Heyting algebra rules and the infinitary operations and equations, form a category dually equivalent to the category of graphs.The unary operation $\tau,$ has two possible interpretations on the algebra of subgraphs of all graphs. One interpretation assigns to a subgraph $H$ of each graph $G$ the subgraph $\tau H$ of $G$ that has the same vertices as $G$ and whose arrows are all arrows that start in $H$ and end outside $H.$ For the other interpretation the arrows start outside and end inside the subgraph. For the first, $\tau$ on $\Omega$ is the morphism that classifies the subgraph of $\Omega$ generated 
by the arrow that starts at 1 and ends at 0. For the second, the arrow starts at 0 and ends at 1. 
If you are interested in more details, here's a url where I posted a similar
duality (it was published in Jan 2005 in "Mathematical Reports of the Academy of Sciences of the Royal Society of Canada".)
https://www.researchgate.net/publication/266304536_A_duality_for_the_category_of_directed_multigraphs
Lawvere calls the graphs you are looking at reflexive and the graphs in my article irreflexive. The generator, in the sense of category theory, of the category of reflexive graphic algebras, is the free reflexive graphic algebra on one generator, $x$; it  has 12 elements
 $(0,x,\neg x, \neg \neg x, x \vee \neg x, \neg x  \vee   \neg \neg x,  \neg \neg x \Rightarrow x, \tau x, \tau \neg x, \tau x \vee  \neg \neg x, \tau \neg x \wedge (\neg \neg x \Rightarrow x),1)$,
 whereas for the irreflexive algebra its more complicated, there are 39 elements
(all 39 polynomials are identified at the url).
The irreflexive graphs have an interesting graph that is internally isomorphic to but not isomorphic to the subgraph classifier, which is not the case for reflexive graphs ( this is related to my first post (version: 2011-03-02): https://mathoverflow.net/q/57032).
I don't have an answer to question 5, but the other four questions have answers that follow from the duality. My article has the references I used,
but in addition there is an earlier papers by Marta Bunge that shows that in principle there should be such a duality for Grothendiek toposes( sorry I don't have the exact
 reference). I was strongly influenced by a paper of Andrew Pitts : On an Interpretation of Second Order Quantification in First Order Intuitionistic Propositional Logic, Jour. Symbolic Logic 57(1992) 33-52. In fact I started writing the duality paper shortly after reading his paper. 
I have seen essentially this duality in a suprisingly different context:
a paper by Katarzyna Idziak  , Undecidability of Brouwerian Semilattices,
Algebra Universalis 1986, that establishes the undecidability of the first order theory of algebras that are similar to the Heyting algebras that satisfy the two extra equations I gave above. The author constructs undirected graphs out of this type of algebra (in the same way that I construct directed graphs out of graphic algebras)
and uses the fact that the first order theory of the undirected graphs is undecidable.
There is no category theory in the paper.<|endoftext|>
TITLE: $A_{\infty}$-structure on closed manifold
QUESTION [11 upvotes]: Is there an exmaple of a closed smooth connected manifold $M$ having a structure of $A_{\infty}$-space (with unit) but $M$ is not homeomorphic to a compact connectd Lie group as space ? 
Edit: First, I would like to thank Fernando, Vladimir and Jesper for the excellent answers!
My original motivation was the following question:
Suppose that $\mathcal{A}_{\infty}$ is a cofibrant replacement of the associative operad $\mathcal{A}$ in the category $\mathsf{Op}$ of topological operads (non-symmetric) and suppose that $X$ is a finite connected $CW$-complex. Is it true that $$Hom_{\mathsf{Op}}(\mathcal{A}_{\infty},End(X))\neq \emptyset \Rightarrow Hom_{\mathsf{Op}}(\mathcal{A},End(X))\neq \emptyset ?$$
Here is a counterexample: Suppose that $M$ is a smooth closed connected manifold with a structure of $\mathcal{A}_{\infty}$-space (i.e, Loop space) such that $M$ is not homotopy equivalent to a Lie group. By definition $Hom_{\mathsf{Op}}(\mathcal{A}_{\infty},End(M))\neq \emptyset$. On another hand if $Hom_{\mathsf{Op}}(\mathcal{A},End(M))\neq \emptyset$, this will implie that $M$ has at least one structure of topological monoid. By Wallace Theorem any closed manifold with a structure of topological monoid is a Lie group, hence $M$ is a Lie group. Contradiction.

REPLY [13 votes]: Yes:
As has been pointed out, admitting the structure of a connected $A_\infty$-space is the same thing as being homotopy equivalent to a connected loop space.
The Hilton-Roitberg criminal, mentioned by Vladimir Dotsenko, was the first example of a closed manifold homotopy equivalent to a loop space, but not to a Lie group. The space has the rational homotopy type of the compact Lie group Sp(2).
In fact there exists "rational criminals", i.e., finite loop spaces that are homotopy equivalent to closed smooth manifolds (as always, by the Broto-Kitchloo-Notbohm-Pedersen theorem), but not even rationally homotopy equivalent to a Lie group. However, they only start occurring in rank 66 and onwards.
See this paper that I wrote with Andersen-Bauer-Pedersen<|endoftext|>
TITLE: Examples of two-dimensional Riemannian manifolds that can't be isometrically embedded into $\mathbb{R}^4$
QUESTION [6 upvotes]: Can anyone give some examples of two-dimensional Riemannian manifolds $(M,g)$ that can't be isometrically embedded into $\mathbb{R}^4$? (Further more Globally)
What if it is smooth?

REPLY [9 votes]: I found the following in the notes to chapter 2 of the book of Han and Hong: "Isometric embedding of Riemannian manifolds in Euclidean spaces" which is a great source for questions about isometric immersions and embeddings.  
The real projective plane $\mathbb{RP}^2$ (with the standard (constant Gauss curvature 1) metric from $S^2$) has no $C^2$ isometric embedding in $\mathbb{R}^4$ (in particular this rules out smooth embeddings as in your question).  Han and Hong cite the 1970 survey "Embeddings and immersions in Riemannian geometry" by Gromov and Rokhlin, in particular see Appendix 4.  
This result deserves a little more comment.  Note that there do exist embeddings of $\mathbb{RP}^2$ into $\mathbb{R}^4$, see e.g. wikipedia.  What Gromov and Rokhlin prove is that there are none with strictly positive curvature.
Gromov and Rokhlin state that it is open whether isometric embeddings for $\mathbb{RP}^2$ in $\mathbb{R}^4$ exist for $C^r$ with $1<r<2$ (recall that for $r\leq1$, Nash-Kuiper applies).

REPLY [8 votes]: This is more of a remark than an answer and addresses the weaker condition of asking for an immersion rather than an embedding.
I assume that you want the immersion to be smooth (or at least $C^2$) since, otherwise, by Nash-Kuiper, there is no obstruction.
On the other hand, according to Gromov's Partial Differential Relations (published in 1986, so it may be out of date on this), "There is no single known obstruction for isometric $C^\infty$-immersions of surfaces into Riemannian $4$-manifolds". (See 3.2.4, pp. 289-294.)
More specifically, the last time I checked, it was unknown whether $\mathbb{RP}^2$ endowed with any metric of positive curvature could be isometrically immersed into $\mathbb{R}^4$ and it was still unknown whether the hyperbolic plane (with Gauss curvature $-1$) could be isometrically immersed in $\mathbb{R^4}$.  The latter can be isometrically embedded in $\mathbb{R}^5$, as can $\mathbb{RP}^2$ with a metric of constant positive Gauss curvature.  See Gromov's PDR for details.<|endoftext|>
TITLE: The mean of points on a unit n-sphere $S^n$
QUESTION [7 upvotes]: A unit n-sphere is defined as $$\mathcal{S}^n = \{\mathbf{p} \in \mathbb{R}^{n+1}: \|\mathbf{p}\| = 1\}$$
The distance between two points $\mathbf{p}$, $\mathbf{q}$ on $\mathcal{S}^n$ is the great-circle distance:
$$\rho(\mathbf{p},\mathbf{q}) = \arccos(\mathbf{p}^T \mathbf{q})$$
where $\arccos(\cdot): [-1,1] \to [0,\pi]$ is the inverse cosine function.
Given a set of points on $\mathcal{S}^n$, I would like to compute the mean (centroid) of these points. Since the surface of a unit n-sphere is not a Euclidean space, I guess we cannot use arithmetic mean in this case.
Question: how do we compute this mean point $\boldsymbol{\mu} \in \mathcal{S}^n$, taking into account the spherical geometry?

REPLY [3 votes]: I think what you're looking for is the field of statistics known as directional statistics. Even for the circle $S^1$ it's not obvious how things should be defined, but it is possible, depending on context.<|endoftext|>
TITLE: Inequality on Kähler classes
QUESTION [8 upvotes]: Let $X$ be a compact Kähler manifold of complex dimension $n$, and let
$\omega_1, \omega_2$ be Kähler classes on $X$. Denote the Lefschetz
operator of a Kähler class $\omega$ by $\Lambda_{\omega}$. Then
$$
\Lambda_{\omega_1} \omega_2 + \Lambda_{\omega_2} \omega_1 \geq 2n
$$
with equality if and only if $\omega_1 = \omega_2$.
Has anyone seen this inequality? I checked in the usual places like
Lazarfeld's Positivity, but didn't find it. It seems simple enough that
someone must have come across it before.
(Aside one: We can rewrite this as an inequality on intersection numbers if
we want, since $\Lambda_{\omega_1} \omega_2 = n \int_X \omega_2 \cup
\omega_1^{n-1} / \int_X \omega_1^n$.)
(Aside two: On a complex torus, everything reduces to linear algebra. There
one can whittle the statement of the inequality down to that
$\operatorname{tr}(H) + \operatorname{tr}(H^{-1}) \geq 2n$ for any
Hermitian positive-definite matrix $H$, which follows from the
arithmetric-geometric inequality.)
The proof I found is a little amusing. The Kähler cone of $X$ has a
natural Riemannian metric $g$, which at a point $\omega$ is given by the
inner product that $\omega$ induces on real $(1,1)$-classes. If $u,v$
are such classes, then
$$
g(u,v)_{\omega} = \Lambda(u) \Lambda(v) - \tfrac12 \Lambda^2(u \cup v).
$$
Given two points $\omega_1, \omega_2$ in the Kähler cone, we can
estimate the Riemannian distance between them by calculating the length
of any path between them. Doing so for the segment of the Euclidean line
that runs through the two gives the estimate
$$
d(\omega_1,\omega_2)^2 
\leq \Lambda_{\omega_1} \omega_2 + \Lambda_{\omega_2} \omega_1 - 2n,
$$
from which the inequality follows.

REPLY [4 votes]: It is a consequence of the Khovanskii-Teissier inequality for Kähler classes (which was proved by Gromov and Demailly on Kähler manifolds, the algebraic case is also in Lazarsfeld's book):
$$\int_X \omega_1^{n-1}\wedge\omega_2 \geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{1}{n}},$$
which implies
$$\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)\geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n+1}{n}}.$$
Take this inequality and the same inequality with the roles of $\omega_1$ and $\omega_2$ exchanged, and sum them up:
$$\begin{split}&\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)+\left(\int_X \omega_2^{n-1}\wedge\omega_1 \right)\left(\int_X\omega_1^n\right)\\
&\geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n+1}{n}}+\left(\int_X \omega_1^n\right)^{\frac{n+1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n-1}{n}}.\end{split}$$
Now use the Young inequality
$$2xy \leq x^{\frac{n-1}{n}}y^{\frac{n+1}{n}}+x^{\frac{n+1}{n}}y^{\frac{n-1}{n}},$$
and get
$$\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)+\left(\int_X \omega_2^{n-1}\wedge\omega_1 \right)\left(\int_X\omega_1^n\right)\geq 2\left(\int_X\omega_1^n\right)\left(\int_X\omega_2^n\right),$$
which is equivalent to your inequality. If equality holds, then equality must hold in Khovanskii-Teissier, hence $\omega_1$ and $\omega_2$ are proportional by Boucksom-Favre-Jonsson, but equality also holds in Young, hence $\omega_1=\omega_2$.

REPLY [3 votes]: Here is a simple proof using Theorem 1.6.1 in Lazarsfeld book, which is the following:

Theorem (Demailly)
If $H_1,\ldots,H_n$ are Kähler classes in a compact Kähler manifold of dimension $n$, then the following inequality holds:
$$(H_1 \cdots H_n)^n \ge (H_1^n)\cdots(H_n^n).$$

Let $H_1$ and $H_2$ be Kähler classes. By the AM–GM inequality,
$$\left[\frac{1}{2}\left( \frac{{H_1^{n-1}H_2}}{H_1^{n}} + \frac{H_2^{n-1}H_1}{H_2^{n}} \right)\right]^2 \ge\frac{{H_1^{n-1}H_2}}{H_1^{n}} \cdot \frac{H_2^{n-1}H_1}{H_2^{n}}.  $$
Using the above theorem, it is easy to see that
$$({H_1^{n-1}H_2}) \cdot ({H_2^{n-1}H_1}) \ge {H_1^{n}}{H_2^{n}},$$
which finishes the proof.<|endoftext|>
TITLE: If $G$ is absolutely simple simply connected, why is G(F_v) quasisimple for almost every valuation v?
QUESTION [5 upvotes]: Let $G$ be an absolutely simple simply connected and connected algebraic group defined over a global field $k$ with ring of integers $\mathcal{O}$. Fix an embedding of $G$ into $GL_n$. Given $v$ a non-archimedean valuation of $k$, let $\mathfrak{p}_v\subset \mathcal{O}$ the associated maximal ideal, $\mathcal{O}_v$ its completion, $\mathfrak{m}_v$ the corresponding maximal ideal , $\mathbb{F}_v=\mathcal{O}_v/\mathfrak{m}_v$ the residue field and $G_v$ the reduction of $G$ modulo $\mathfrak{p_v}$.  Let $G(\mathbb{F}_v)$ be the image of $G(\mathcal{O}_v)$ in $GL_n(\mathcal{O}_v/\mathfrak{m}_v)$. I am trying to understand why the following statement holds:

For all but a finite number of valuations $v$, $G(\mathbb{F}_v)$ is perfect and is a central extension of a finite simple group of Lie type $H(\mathbb{F}_v)$, where $H$ is of the Lie type of $G$ (either twisted or untwisted).

In trying to understand why this holds I have thought there are several steps:

The reduction of $G$ modulo $\mathfrak{p_v}$ is absolutely simple simply connected and of the same Lie type as $G$ for almost every $v$.
$G(\mathbb{F}_v)=G_v(\mathbb{F}_v)$ for almost every $v$.
If $H$ is an absolutely simple simply connected algebraic group defined over a finite field $\mathbb{F}_q$, then for almost every $q$, $H(\mathbb{F_q})$ is perfect and the quotient by its center is a finite simple group.

If this is a right way to prove the statement, I would appreciate help concerning (1.). Any other proof or idea is also very welcome.

REPLY [4 votes]: This is basically an instance of "spreading out" principles in algebraic geometry, combined with arguments specific to simply connected semisimple groups over general fields away from a few low-rank examples over fields of size 2 and 3. The formulation of the question is a bit imprecise, since the notation $G_v$ isn't really defined for all $v$, and making precise what it means for all but finitely many $v$ implicitly entails many of the steps below. We'll work directly with group schemes over rings since that is the natural framework to systematically discuss and analyze all questions of this type. 
First, the spreading out. Let $R$ be a domain with fraction field $K$ (for example, $K$ a global field and $R$ a ring of $S$-integers for some non-empty finite set $S$ of places of $K$ containing the archimedean places) and $G$ a connected reductive $K$-group. By general spreading-out results in EGA IV, there exists a nonzero $r \in R$ such that $G$ is the generic fiber of a smooth affine $R[1/r]$-group scheme $\mathscr{G}$ whose geometric fibers are connected; we may and do rename $R[1/r]$ as $R$. 
By Corollary 3.1.11 of the article "Reductive group schemes" in the Proceedings of the 2011 Luminy summer school on SGA3 (applied to $K$ viewed as the direct limit of $R[1/r]$'s) we can replace $R$ with $R[1/r]$ for some nonzero $r \in R$ to arrange that all geometric fibers of $\mathscr{G}$ over ${\rm{Spec}}(R)$ are reductive, and even semisimple when $G_{\overline{K}}$ is semisimple. 
There is a finite separable extension $K'/K$ such that $G_{K'}$ is split (as $G_{K'_s}$ is split). Fix an $R$-finite domain $R' \subset K'$ with fraction field $K'$, so $K'$ is the direct limit of subrings $R'[1/r]$ for nonzero $r \in R$ (as $R' \otimes_R K = K'$). Hence, by inverting some such $r$ (again using the EGA IV spreading-out principles, for properties such as being an open immersion) we can arrange that $\mathscr{G}_{R'}$ admits an "open cell" with fiberwise maximal split $R'$-torus and root groups over $R'$ recovering the usual ones on the generic fiber.  That already tells us the root datum on every fiber, and the root datum detects being simply connected.
Thus, now assuming $G_{\overline{K}}$ is semisimple, absolutely simple, and simply connected (all detected by the root datum: irreducible root system and coroots spanning the character lattice of a maximal torus), it follows that the same holds for all fibers of $\mathscr{G}$ over ${\rm{Spec}}(R)$.  
In the special case that $K$ is a global field and we did spreading-out relative to some ring of $S$-integers, by enlarging $S$ we see that for each $v \not\in S$ the pullback $\mathscr{G}_{O_v}$ over the $v$-adic completion of $O_{K,S}$ is a smooth affine $O_v$-group. Hence,
$\mathscr{G}(O_v) \rightarrow \mathscr{G}(\mathbf{F}_v)$ is surjective.  The role of GL$_n$ in the setup is subsumed in the theory of group schemes over rings (namely, again by enlarging $S$ depending on some initial inclusion of $G$ into some ${\rm{GL}}_n$ we ensure that the schematic closure of $G$ inside ${\rm{GL}}_n$ over the $S$-integers coincides with $\mathscr{G}$ as an $O_{K,S}$-group).
In other words, the crutch of GL$_n$-embeddings is just a device to discuss flat affine integral model group schemes (of finite type) without speaking in terms of schemes.  But it is precisely the relative smoothness of such $\mathscr{G}$ (or more concretely of its Zariski closure in such a GL$_n$ over $R$ at the cost of passing to some $R[1/r]$) which guarantees that the "ad hoc" definition of $G(\mathbf{F}_v)$ in the question actually agrees with the more conceptual notion $\mathscr{G}(\mathbf{F}_v)$ (of $\mathbf{F}_v$-points of the $O_{K,S}$-group scheme $\mathscr{G}$). 
Consequently, we can now forget about the entire setup with group schemes over rings of integers (that has already served its purpose, bringing us to a suitable $R = O_{K,S}$ over which we've performed enough nice spreading-out)
and are reduced to problem entirely about connected semisimple groups over fields:  if $H$ is a connected semisimple group over a field $F$ such that $H$ is absolutely simple and simply connected, then away from a few low-rank cases over $F$ of size 2 or 3, prove $H(F)$ is perfect and that $H(F)/Z_H(F)$ is a simple finite group (with $Z_H$ denoting the center of the $F$-group $H$). This is an application of the role of Tits systems in the Borel-Tits structure theory for connected reductive groups over fields, and is discussed by a variety of methods in  a variety of references (depending on your taste).<|endoftext|>
TITLE: An extension of Hadamard maximum determinant problem
QUESTION [5 upvotes]: Consider the Vandermonde product $\prod_{1\le j < k \le n} |z_j - z_k|$. It is well-known that under the constraint $|z_j| \le 1$ for all $j$, the product is maximized at a picket fence configuration of the form $z_j = e^{2 \pi i j / n}$, with a maximum value of $n^{n/2}$; see here.
Now instead of the simple constraint above, I impose $|z_j| \le r_j$, with the convention that $0 < r_1 \le r_2 \le \ldots \le r_n$. The maximizing configuration is obviously going to be more complicated, but I conjecture the following upper bound on the maximum value:
$$ \prod_{1 \le j < k \le n}|z_j - z_k| \le n^{n/2}\prod_{j=1}^n r_j^{j-1}.$$
Is this true? Any partial result (including weaker bound) is welcome. 
Update: the case $n=2$ is trivially verified since the left hand side is $\le r_1 + r_2 \le 2 r_2$. For $n=3$, consider projecting the optimizing $z_j$'s radially to the outermost circle of radius $r_3$, and call them $\zeta_j$'s, so that $\zeta_3 = z_3$. One must have $|\zeta_j - \zeta_k| > r_3$. This ensures that the radial projection can only increase the product. But for higher $n$, this last condition does not necessarily hold, so one has to account for the tradeoff between the losses and gains from outward radial projection.

REPLY [2 votes]: Due to homogeneity, we assume that $r_n=1$.
Set $B=\{z\in \colon |z|\leq 1\}$ and $T=\{z\colon |z|=1\}$. Consider the function 
$$
  f(z_1,z_2,\dots,z_n)=\frac
    {\displaystyle \prod_{1\leq i<j\leq n}(z_i-z_j)}
    {\displaystyle \prod_{i=1}^n z_i^{i-1}}
$$
on the set $S=\{(z_1,\dots,z_n)\in B^n\colon 0<|z_1|\leq\dots\leq |z_n|\}$ (yes, this set is not closed). We claim that the maximum of $|f|$ is attained at a point with $|z_i|=1$ for all $i$ (that is, at a point in $T^n$); the required result will follow, since this reduces the problem to the case of equal radii.
For any point $\mathbf z=(z_i)\in S$, set $\mu(\mathbf z)=\max\{m\leq n\colon |z_1|=|z_2|=\dots=|z_m|\}$. To prove our claim, we will show that any point $\mathbf z=(z_i)\in S\setminus T^n$ can be modified into a point $\mathbf w=(w_i)\in S$ such that $|f(\mathbf w)|\geq |f(\mathbf z)|$ and $\mu(\mathbf w)>\mu(\mathbf z)$. After a chain of such modifications, we will arrive at a point with $\mu(\mathbf w)=n$, which is what we need (by homogeneity again).
So, assume that $\mu(\mathbf z)=m<n$. Set $t_i=z_i/z_1$ for all $i\leq m$ (then $|t_i|=1$), and consider the function $g(z)=f(zt_1,zt_2,\dots,zt_m,z_{m+1},\dots,z_n)$ (so $g(z_1)=f(\mathbf z)$). Notice that the whole power of $z$ in the denominator cancels out, so this function is a polynomial in $z$. Therefore, the maximum of absolute value of $g$ on $\{z\colon |z|\leq |z_{m+1}|\}$ is attained at some point $z_0$ with $|z_0|=|z_{m+1}|$, and we may set $\mathbf w=(z_0t_1,\dots,z_0t_m,z_{m+1},\dots,z_n)$. This finishes the proof.<|endoftext|>
TITLE: Elementary linear algebra over a (possibly skew) field $K$
QUESTION [5 upvotes]: I have a number of questions which seem linked to me, about basic  (?) linear algebra:
Given a field (possibly skew) $K$, and an superfield $L$, one can do linear matrix algebra with coefficients in $L$ (any reference for basic facts about that ?)
So given a square matrix $M\in M_{n\times n}(L)$, it is left invertible if and only if it is right invertible. What are the known relations between the left kernel of $M$ in $L^n$ and its right kernel ? 
By left kernel, I mean the space (left-vector space) of rows $\lambda\in L_{1\times n}$ (one row $n$ columns) such that $\lambda M=(0)$ and by right kernel, I mean the set of $\lambda \in L_{n\times 1}$ (one column n rows) such that $M\lambda=(0)$.
Question 1. For instance, given $M\in M_{n\times n}(L)$, if there is a non-trivial point  of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?
If $K$ is commutative, I am trying to understand why the notion of dimension  of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).
Question 2. What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew determinant ?). Does the notion of $K$-dimension of the $K$-vector $L$ space depends on possible enlargements of $K$ ?
Finally:
Question 3. Why does the Zariski dimension of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?

REPLY [3 votes]: If I understand correctly what Question 1 is asking, then there are easy counterexamples even using commutative fields.
Let $K=\mathbb{R}$ and $L=\mathbb{C}$. Then $\begin{pmatrix}1&i\\1&i\end{pmatrix}$ is a matrix over $L$ that has no nonzero elements of $K^2$ in its right kernel, but the element $\begin{pmatrix}1&-1\end{pmatrix}$ of $K^2$ is in its left kernel.<|endoftext|>
TITLE: Primary obstruction to the existence of a cross-section of $V_{n - q}(\omega)$ is a cohomology class in $H^{2q+2}(B, \pi_{2q+1} V_{n - q}(F))$?
QUESTION [11 upvotes]: Let $V_{n -  q}(\mathbb{C}^n)$ denote the complex Stiefel manifold consisting of all complex $(n - q)$-frames in $\mathbb{C}^n$, where $0 \le q < n$. This manifold is $2q$-connected, and$$\pi_{2q + 1} V_{n-q}(\mathbb{C}^n) \cong \mathbb{Z}.$$Given a complex $n$-bundle $\omega$ over a CW-complex $B$ with typical fiber $F$, we can construct an associated bundle $V_{n-q}(\omega)$ over $B$ with typical fiber $V_{n-q}(F)$; consider the vector bundle $\text{Hom}(B \times \mathbb{C}^{n-q}, \omega)$ over $B$, and take the open subvariety of homomorphisms $u$ such that $u_b$ is injective for each $b \in B$. How do I see that the primary obstruction to the existence of a cross-section $V_{n-q}(\omega)$ is a cohomology class in$$H^{2q+2}(B, \pi_{2q+1}V_{n-q}(F))$$which can be identified with the Chern class $c_{q+1}(\omega)$?
Edit. To clarify what definition of the Chern classes I am using, I am using the one in Milnor-Stasheff where the top Chern class is the Euler class of the realification and the lower ones are defined by induction.

REPLY [8 votes]: The fiber $V_{n - q}(F)$ is $2q$-connected, so it is not hard to construct a equivalence over the $(2q + 1)$-skeleton. We clearly can take sections over each vertex in the $0$-skeleton in the same connected component, then we can connect them on the 1-skeleton via paths because it is connected, then we can fill in with disks on the 2-skeleton because it is simply connected, and so forth. We then construct the "primary obstruction" as usual in $H^{2q + 2}(B, \pi_{2q + 1} V_{n - q}(F))$. Trying to extend to the $(2q + 1)$-skeleton gets us, through the attaching maps, an element$$\mathfrak{o}_{q + 1} \in H^{2q + 2}\left(B, \pi_{2q + 1}(V_{n - q}(F))\right) = H^{2q + 2}\left(B, \mathbb{Z}\right)$$in exactly the same way we discussed in class from traditional obstruction theory. According to Chapter 12 of Milnor, this construction is natural.
So in order to show that $\mathfrak{o}_{q + 1} = c_{q + 1}$, we can simply show it for the tautological bundle $\gamma^n$ on the infinite complex Grassmannian of $n$-planes, because any relation will pull back. By the results on the structure of the cohomology ring, we have$$\mathfrak{o}_{q + 1} = p(c_1, c_2, \ldots, c_q) + \lambda c_{q + 1}$$for some unique polynomial $p$ and some unique constant $\lambda$. This relation we are taking at first to be for the tautological bundle $\gamma$, but by pullback, it has to hold for any bundle.
Let us look at the bundle $\gamma^q \oplus \epsilon^{n - q}$, over the infinite complex Grassmanninan of $q$-planes, where $\epsilon$ denotes the trivial line bundle. Evidently, this has $n - q$ independent linear sections, so $\mathfrak{o}_q$ should vanish. However, clearly $c_{q + 1} = 0$, because this is stably equivalent to a $q$-dimensional bundle, hence we get$$p\left(c_1\left(\gamma^q\right), c_2\left(\gamma^q\right), \ldots, c_q\left(\gamma^q\right)\right) = 0,$$so $p = 0$ because we know from structure results that there is no nontrivial polynoimal relation on these classes.
Lastly, we find that $\lambda = 1$. Let $\gamma_1$ be the tautological line bundle  over $\text{Gr}_\mathbb{C}(q + 1, q + 2)$, identified through orthogonal complements of the moduli as $\mathbb{C}P^{q + 1}$. Then let us look at$$\mathfrak{o}_{q + 1}\left(\gamma_1^{q + 1} \oplus \epsilon^{n - q - 1}\right) = \lambda c_{q + 1}\left(\gamma_1^{q + 1} \oplus \epsilon^{n - q - 1}\right).$$We want them to actually be equal to conclude that $\lambda = 1$. However, when $q + 1$ is the dimensional of the bundle, i.e. $q + 1 = n$, which is clear because the Chern class and the Euler class are both the obstruction class by Theorem 12.5 in Milnor, because a complex nonzero global section exists evidently if and only if a section of the bundle considered as a real bundle does. Otherwise, it follows because obstruction classes are clearly stable from their primary obstruction interpretation, i.e. we can add, delete trivial factors without modifying anything. Hence, we can reduce the case to $\mathfrak{o}_{q + 1}(\gamma_1^{q + 1}) = w_{q + 1}(\gamma_1^{q + 1})$, which follows again from the Euler class argument.<|endoftext|>
TITLE: Is there research on Machine Learning techniques to discover conjectures (theorems) in a wide range of mathematics beyond mathematical logic?
QUESTION [13 upvotes]: Although there already exists active research area, so-called, automated theorem proving, mostly work on logic and elementary geometry. 
Rather than only logic and elementary geometry, are there existing research results which by using machine learning techniques(possibly generative learning models) to discover new mathematical conjectures (if not yet proved), in wider branches of mathematics such as differential geometry, harmonic analysis etc...?
If this type of intellectual task is too difficult to study by far, then for a boiled-down version, can a machine learning system process a compact, well-formatted notes (to largely reduce natural language processing part) about for instance real analysis/algebraic topology, and then to ask it to solve some exercises ? Note that the focus and interest here are more about "exploring" new possible conjectures via current(or possible near future) state-of-the-art machine learning techniques, instead of proof techniques and logic-based knowledge representation which actually already are concerned and many works done in classical AI and automated theorem proving.   
So, are there some knowing published research out there, particularly by generative model in machine learning techniques. 
If there are not even known papers yet, then fruitful and interesting proposals and discussions are also highly welcomed. 
As a probably not so good example I would like to propose, can a machine learning system "re-discover" Cauchy-Schwarz inequality if we "teach" it some basic operations, axioms and certain level of definitions, lemmas etc, with either provided or generated examples(numerical/theoretical) ? e.g. if artificial neural networks are used as training tools, what might be useful features in order eventually output a Cauchy-Schwarz inequality in final layer.

REPLY [7 votes]: A pre-print by Li-An Yang, Jui-Bin Liu, Chao-Hong Chen and Ying-ping Chen was submitted to arXiv last month (24 Feb) giving some preliminary results of an ad-hoc evolutionary algorithm used to prove some simple theorems within the Coq proof assistant:
Automatically Proving Mathematical Theorems with Evolutionary Algorithms and Proof Assistants
The paper also discuss some possible future work within this context and the code used was released as open-source at GitHub.
Update (04/26/2019):
Found today in a NewScientist article about the Deepmath project, a Google project seeking to improve automated theorem proving using deep learning and other machine learning techniques. In particular, they made a software named DeepHOL that is a deep reinforcement learning driven automated theorem prover (1).
(1): Kshitij Bansal, Sarah M. Loos, Markus N. Rabe, Christian Szegedy, Stewart Wilcox. HOList: An Environment for Machine Learning of Higher-Order Theorem Proving (extended version)<|endoftext|>
TITLE: Cyclic vectors for the shift operator
QUESTION [9 upvotes]: Let $S:\ell^2\to\ell^2$ be the shift operator
$$
S(x_1,x_2,\dots)=(0,x_1,x_2,\dots).
$$
Let $x\in \ell^2$ with $x_1=1$. Is $x$ cyclic for $S$? In other words, is the span of the vectors $x,Sx,S^2x,\dots$ dense in $\ell^2$?
If this does not hold for every $x$, is there a handy criterion to decide whether a given $x$ is cyclic or not?

REPLY [8 votes]: This is part of what the theory of Hardy spaces is for. There is a complete characterization of the cyclic vectors for the shift, though it applies to the Fourier transform of $x$ rather than the sequence $x=(x_0, x_1, \dots)$ itself. (For convenience I am starting the indexing at 0 rather than 1.) As in Fedor's answer, we can view the $x$ sequence as the sequence of Fourier coefficients of an $L^2$ function $h$ on the unit circle, which can be extended analytically to the disk $|z|<1$ by 
$$
h(z) = \sum_{n=0}^\infty x_n z^n.
$$
The function $h$ is said to belong to the Hardy space $H^2$, which sits as a closed subspace of $L^2$ on the circle. In this model, the right shift becomes the map $h(z)\to zh(z)$. It is then a fact that $\log|h|$ belongs to $L^1$ of the unit circle, and the inequality
\begin{equation}
\log|h(0)|\leq \frac{1}{2\pi} \int_0^{2\pi} \log|h(e^{i\theta})|
\,d\theta\end{equation}
always holds. Such an $h$ is called an outer function if equality holds (so in particular $x_0=h(0)\neq 0$ is necessary but not sufficient). Finally, it is a theorem that $h$ is cyclic for the shift if and only if $h$ is outer. All of these facts may be found e.g. in the book Banach Spaces of Analytic Functions by Kenneth Hoffman.<|endoftext|>
TITLE: Connectivity of complements of Stein opens
QUESTION [8 upvotes]: Let $Y$ be an affine open subset of a locally noetherian scheme $X$. Then, $X \setminus Y$ has pure codimension one [EGAIV$_4$, Cor. 21.12.7]. Moreover, if $X$ is proper and of finite type over a field $k$, and $\dim X \ge 2$, then $X \setminus Y$ is connected [Hartshorne, Cor. II.6.2].
My question is the following:

If $X$ is a complex projective variety of dimension $\ge 2$, and $Y$ is a Zariski open
  subset that is Stein, then is it still true that $X \setminus Y$ is
  connected?

The fact that $X \setminus Y$ is of pure codimension one is [Neeman, Prop. 3.4], and is a consequence of Hartog's theorem. Jing Zhang cites this proposition for the statement that $X \setminus Y$ is connected in Algebraic Stein Varieties, specifically in the proof of Thm. 2.8, but it is not clear to me how Neeman's proposition implies this result.
It could also be true that the algebraic proof when $Y$ is affine can be modified in the case when $Y$ is Stein instead. But a key step in the proof by Hartshorne (and Goodman in his original article) is that we can embed $Y$ into an affine space $\mathbf{A}^n$, then take its projective closure $\overline{Y}$ to get a compactification of $Y$, and compare this with $X$ using theorems about blow-ups from Nagata's paper on compactifications. I can't make this work in the Stein case.

REPLY [3 votes]: If you look at corollary 4.10 page 45 of the book of Banica and Stanasila titled Algebraic methods in the global theory of complex spaces,you will find a proof of the following .Any irreducible Stein space of dimension at least two has one end.This answers your question when X is irreducible .<|endoftext|>
TITLE: Possible limits of $(1/n) \sum_{k=0}^{n-1} e^{i2\pi \cdot 2^k\alpha}$
QUESTION [22 upvotes]: I made a throwaway comment on math stackexchange the other day that got me thinking about the following question. Let
$$ f_n (\alpha) = \frac1n \sum_{k=0}^{n-1} e(2^k\alpha),$$
where $e(x) = \exp(i2\pi x)$.

Can you determine the set of possible limits $$ L = \{z : f_n(\alpha)\to z~\text{for some}~\alpha\}?$$

Of course by the ergodic theorem $f_n(\alpha)\to 0$ for almost every $\alpha$. Also $f_n(\alpha)\to 1$ for every dyadic rational, so for a comeagre set of $\alpha$ the sequence $f_n(\alpha)$ has a subsequence converging to $1$. So we're interested in a particular meagre null set of $\alpha$. :)
Here is everything I know about $L$:

$L$ is disjoint from a neighbourhood of $S^1\setminus\{1\}$. Indeed suppose $f_n(\alpha)\approx z$, where $|z|=1$. Then $e(2^k\alpha) \approx z$ for almost all $k\leq n$, so also $e(2^{k+1}\alpha)\approx z$ for almost all $k\leq n$, so $z^2 \approx z$, which implies $z\approx 1$.
$L$ is a closed, convex subset of $\{z: |z|\leq 1\}$. Hand-waving argument for convexity: Given $f_n(\alpha)\to z$ and $f_n(\beta)\to w$, with $\alpha,\beta\in[0,1]$, find $N$ such that $f_N(\alpha)\approx z$ and $f_N(\beta) \approx w$, write $\alpha_N$ and $\beta_N$ for $\alpha$ and $\beta$ rounded to the nearest multiple of $2^{-N}$, and then consider $$\gamma_N = \alpha_N + \beta_N/2^N + \alpha_N/2^{2N} + \beta_N/2^{3N} + \cdots.$$ Then the sequence $f_n(\gamma_N)$ hovers around the point $(z+w)/2$ for all sufficiently large $n$. Now patch the sequence $(\gamma_N)$ together in a similar way. Other convex combinations $pz + (1-p)w$ can be obtained similarly. Closedness is also similar.
Obviously $f_n(\alpha)$ converges for every rational $\alpha$, and the limits tend to be interesting. For example, the points $1,-1/2,(1\pm i\sqrt{7})/6$ are in $L$, and so also by 2 their convex hull is contained in $L$. Thus at least $L$ contains a neighbourhood of $0$. By continuing in this way we begin to see the following picture.


Most, but not all, of the points on the boundary appear to be obtained from $\alpha$ of the form $1/(2^n-1)$.

In fact $L = \{\int z\,d\mu : \mu~\text{a}~\times 2\text{-invariant measure on}~S^1\}.$ To see the inclusion $\subset$, take $\mu$ to be any weak* limit point of the sequence of measures $\frac1n \sum_{k=0}^{n-1} \delta_{2^k\alpha}$. To see the inclusion $\supset$, from the ergodic decomposition and the convex of $L$ observe that we may assume $\mu$ is ergodic for $\times 2$, and in this take $\alpha$ to be a $\mu$-random point and apply the ergodic theorem. This observation suggests the points $\int z \,d\mu_p$, where $\mu_p$ is the $\times 2$-invariant measure in which binary digits are independent and come up $1$ with probability $p$. However, this appears to constitute only a strict subset of the above picture, as shown below.


To me these observations strongly suggest the following question.

Is $L$ the convex hull of $\{\lim_{n\to\infty} f_n(\alpha) : \alpha\in\mathbf{Q}\}$?

REPLY [12 votes]: See the paper "Le poisson n'a pas d'arêtes" by Thierry Bousch. This is a joke that was explained to me much later. The set is considered to resemble a fish. The French word arête means both bone and edge (of a polygon); so the English title of the paper would be "The fish has no bones/The fish has no edges". The paper proves that every point on the boundary is an extreme point. (so that the answer to your question is No). The set was also studied in works of Oliver Jenkinson.<|endoftext|>
TITLE: Coefficient problem for univalent harmonic functions on unit disk
QUESTION [6 upvotes]: The Clunie Sheil Small conjecture for the second coefficient of a univalent harmonic function on the unit disk is as follows:
Suppose, $h(z)+\overline{g(z)}$ is a one-to-one harmonic function on the unit disk where $h(z)$ and $g(z)$ are analytic. Assume that $h(0)=g(0)=g'(0)=0$ and $h'(0)=1$. Then the conjecture is:
$|h''(0)|\le 5$.
This is a open problem as far as I know, but I think I heard somewhere that some bound for $|h''(0)|$ has been proved. I could find some references online where they proved a bound for this for certain class of univalent harmonic functions. But I can't find any reference where they prove some bound for $|h''(0)|$ in the general case.
Can anyone please give a reference where I can find this.

REPLY [4 votes]: In page 86 of "Harmonic mappings in the plane" you can find references for these bounds.
In section 6.3 they prove the bound $|a_2| \leq 49$. They also reference to an earlier bound proved by Clunie and Sheil-Small.
The conjecture is related to the now proved Bieberbach conjecture, so I wouldn't be surprised if more progress has been done in recent years using SLE.<|endoftext|>
TITLE: Non semi-simple monodromy in an algebraic family
QUESTION [14 upvotes]: I am looking for an example of a (edit: projective) family
$f : X \to Y$
of complex algebraic varieties which is a topologically locally trivial fibration in (singular) varieties and such that there exists an $q$ such that the monodromy representation
$\pi_1(Y,y) \to GL(H^q(F))$
is not semi-simple, where $F := f^{-1}(Y)$.
I guess that such examples should be plentiful but I don't know any.
By Deligne's theorem (the decomposition theorem for smooth families) such behaviour is impossible if $f$ is smooth. Hence the requirement that the fibres should be singular.
I am actually interested in the following: A variety $X$ and a semi-simple local system $\mathcal{L}$ on a Zariski open subset such that some cohomology sheaf of $IC(X, \mathcal{L})$ has non-semi-simple monodromy.
Bonus points: Is there an example with $\mathcal{L}$ trivial?
EDIT (following Piotr's relevant comment below): Note that it is important that $Y$ be an algebraic variety. Indeed, if $Y = \mathbb{A}^1 \setminus \{ x_1, \dots, x_n \}$ and $f$ is smooth then the monodromy around any $x_i$ will be quasi-unipotent. However the representation of the free group will still be semi-simple.

REPLY [4 votes]: So it seems passerby's example can be modified to give a projective example.
(Thanks for de Cataldo and Migliorini for some of the following. All mistakes are mine.)

Fix $E$ an elliptic curve and consider a family over $B = E - \{ id \}$ where the fibre over $s$ is $E$ with $id$ joined to $s$. Let us call this singular elliptic curve E_s. (I have not checked that such a family exists, but I guess it isn't difficult.)
We have an exact sequence
$0 \to H_1(E) \to H_1(E_s) \to Z[c] \to 0$
where c is any cycle that passes through the singular point of E_s. In Deligne's theory $H_1(E)$ is of weight -1 and $Z[c]$ is of weight 0.

Now $\pi_1(B)$ is a free group on 2 generators. Let $p : \pi_1(B) -> H_1(E)$ denote the canonical map (the abelianization). Then I think that if $\gamma \in \pi_1(B)$ then $\gamma$ acts on $[c]$ by
$\gamma(c) = c + p(\gamma).$
This picture might help...

Anyway, this means that the representation of $\pi_1(B)$ is certainly not-semi-simple. (It "mixes weight 0 with weight -1".)
In this context BBD, Proposition 6.2.3 is useful: the weight filtration for a topologically locally trivial family is by locally constant subsystems.
Now suppose that $f : X \to Y$ is some family of stable curves of genus 2 such that the fibres are generically smooth and such there exists some subvariety $E - \{ id \} \in Y$ such that over this subvariety the family is the above example. Then applying the decomposition for $f_* \mathbb{Q}_X$ one gets the non-semi-simple local system above occurring.
I am not sure if such a family exists. But in any case the above seems to suggest that considerations of stable curves should give many such examples of non-semi-simplicity.
(It is nice in this example to imagine the genus 2 curve degenerating and deducing the mixed Hodge structure on the IC from the limit mixed Hodge structure.)<|endoftext|>
TITLE: Are quadric hypersurfaces toric varieties?
QUESTION [8 upvotes]: A quadric hypersurface (over an algebraically closed field of characteristic zero) in $\mathbb{P}^n$ for $1\leq n\leq 3$ is a toric variety.  (Namely, it's isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$, $\mathbb{P}^1$, or two points, depending on $n$.)
Is it true for general $n$ that quadric hypersurfaces are toric varieties?  Again, over an algebraically closed field of characteristic zero.
According to Harris' book (Algebraic Geometry A First Course), on page 288, the blowup of a smooth quadric in $\mathbb{P}^n$ at a smooth point is isomorphic to the blowup of $\mathbb{P}^{n-1}$ along a smooth quadric $C$ of dimension $n-3$, but I'm not sure if that helps.  And I'm assuming that the smooth case is the only interesting one, since singular quadrics are just cones over smooth ones, so I'd be content with an answer just about the smooth quadrics.

REPLY [13 votes]: (Rewriting comment for the sake of having an answer)
The quadric is not toric for $n \geq 4$.  If $n \geq 4$, the Picard rank of a quadric in $\mathbb P^n$ is $1$ by the Lefschetz hyperplane theorem.  The only smooth $(n-1)$-dimensional toric variety of Picard rank $1$ is $\mathbb P^{n-1}$ (see e.g. here).  But a quadric in $\mathbb P^n$ isn't isomorphic to $\mathbb P^{n-1}$, for example because the canonical divisor of an $(n-1)$-dimensional quadric is divisible by $n-1$, while that of $\mathbb P^{n-1}$ isn't.<|endoftext|>
TITLE: Does the functor Sch to Top have a right adjoint?
QUESTION [17 upvotes]: Let $S$ be a scheme, let $T$ be an $S$-scheme, and let $M$ be a set. Let $M_{S}$ be the disjoint union of $M$ copies of $S$, considered as an $S$-scheme. (Notation from [SGA 3, Exp. I, 1.8].) Then $S$-scheme morphisms $T \to M_{S}$ correspond to locally constant functions $T \to M$, i.e. continuous functions $T \to M$ where $M$ is given the discrete topology. The functor $G_{0} : \operatorname{Set} \to \operatorname{Sch}/S$ sending $M \mapsto M_{S}$ is a sort of "partial right adjoint" to the functor $F : \operatorname{Sch}/S \to \operatorname{Top}$ sending $(T,\mathscr{O}_{T}) \mapsto T$, i.e. taking the underlying topological space of the $S$-scheme.

Can the functor $G_{0}$ be extended to a right adjoint $G : \operatorname{Top} \to \operatorname{Sch}/S$ of $F$?

My naive guess is to take a topological space $X$, give $X_{S} := S \times X$ the product topology and set $\mathscr{O}_{X_{S}} := \pi^{-1}(\mathscr{O}_{S})$ where $\pi : X_{S} \to S$ is the projection. Then $(X_{S},\mathscr{O}_{X_{S}})$ is indeed a locally ringed space and gives the usual construction when $X$ is a discrete space, but in general it is not a scheme. Consider $S = \operatorname{Spec} k$ and $X = \{x_{1},x_{2}\}$ the two-point set with the trivial topology; then the only open subsets of $X_{S}$ as defined above are $\emptyset$ and $X_{S}$ itself, so that $X_{S}$ is not even a sober space.

What if I restrict the target category of $F$ to the category of sober spaces?

The product of sober spaces is sober, so it's no longer immediately clear to me whether the above construction fails.

REPLY [24 votes]: Another way to see that the functor $\mathrm{Sch} \to \mathrm{Top}$ is not a left adjoint is to see that it does not preserve colimits. In this MO answer, Laurent Moret-Bailley gives an example of a pair of arrows $Z \rightrightarrows X$ in $\mathrm{Sch}$, such that the canonical map from $X$ to the coequalizer $Y$ is not surjective (as a function between the sets of points of the underlying spaces). Since in $\mathrm{Top}$ those canonical maps to the coequalizer are always surjective, this coequalizer cannot be preserved by the forgetful functor.<|endoftext|>
TITLE: $\omega$-colorings of $\kappa^2$
QUESTION [6 upvotes]: Let $\kappa\le 2^{\aleph_0}$ be an infinite cardinal. We have a collection of functions $\{f_i|i<\kappa\}$ such that $f_i:i\rightarrow \omega$ and the collection is "triangle-free", i.e. there are not $i<j<k<\kappa$ such that $$f_j(i)=f_k(i)=f_k(j).$$
Is it always possible to extend this collection by adding one more function $f_\kappa:\kappa\rightarrow\omega$ so that the collection $\{f_i|i\le\kappa\}$ remains triangle-free?

REPLY [4 votes]: No, it is not always possible.
I find it helpful to translate the problem slightly. Note that, if $\alpha$ is an ordinal (not necessarily $\leq 2^{\aleph_0}$), a sequence of functions $\langle f_i \mid i < \alpha \rangle$ as specified is the same as a single function $f:[\alpha]^2 \rightarrow \omega$ (where $[\alpha]^2$ is the set of 2-element subsets of $\alpha$), and the requirement that $\langle f_i \mid i < \alpha \rangle$ is triangle-free precisely the same as requiring $f$ to be triangle-free (i.e. there is no $i < j < k < \alpha$ such that $f(\{i,j\}) = f(\{i,k\}) = f(\{j,k\})$).
Suppose that, for every $\alpha < (2^{\aleph_0})^+$ and every triangle-free coloring $f:[\alpha]^2 \rightarrow \omega$, $f$ can be extended to a triangle-free coloring $g:[\alpha+1]^2 \rightarrow \omega$ with $g \restriction [\alpha]^2 = f$. Then we can recursively build a sequence $\langle g_\alpha \mid \alpha < (2^{\aleph_0})^+ \rangle$ such that $g_\alpha:[\alpha]^2 \rightarrow \omega$ and, for $\alpha < \beta$, $g_\beta$ extends $g_\alpha$. Then, letting $g = \bigcup_{\alpha < (2^{\aleph_0})^+}g_\alpha$, we have that $g:[(2^{\aleph_0})^+]^2 \rightarrow \omega$ is a triangle-free coloring, contradicting the Erdos-Rado theorem. There thus must be some $\alpha < (2^{\aleph_0})^+$ and $f:[\alpha]^2 \rightarrow \omega$ such that $f$ is triangle-free and cannot be extended to a triangle-free coloring with domain $[\alpha+1]^2$. Taking a bijection between $\alpha$ and its cardinality, we can assume $f$ is actually a triangle-free coloring $f:[\kappa]^2 \rightarrow \omega$ that cannot be extended to a triangle-free coloring on $[\kappa+1]^2$, where $\kappa \leq 2^{\aleph_0}$ is a cardinal. Translating back to your formulation, we get $\langle f_i \mid i < \kappa \rangle$ by letting, for $i < j < \kappa$, $f_j(i) = f(\{i,j\})$. This sequence cannot be extended by a function $f_\kappa$.
As a side note, there is a very natural triangle-free coloring of size continuum that cannot be extended. Let ${^\omega}2$ denote the set of functions $F:\omega \rightarrow 2$, and define $f:[{^\omega}2]^2 \rightarrow \omega$ by letting $f(F,G)$ be the least $n$ such that $F(n) \neq G(n)$. $f$ is obviously triangle-free, and there's a nice argument showing that it cannot be extended.<|endoftext|>
TITLE: Is the hom-simplicial set in the hammock localization a nerve?
QUESTION [11 upvotes]: Let $(C,w)$ be a relative category. Then associated to it we have its hammock localization, $L^H(C,w)$, which is a simplicially enriched category.
If $X,Y\in C$, the description of the simplicial set $L^H(C,w)(X,Y)$ can be found in Dwyer-Kan, "Calculating Simplicial Localizations", 2.1. The 0-simplices are zig-zags where the reversed arrows are in $w$. The $k$-simplices are defined by taking "natural transformations" of such zig-zags that fix the endpoints, all go in the same direction, and are in $w$. (See the article for more details).
In the nLab, it is claimed that this is the nerve of a certain category (groupoid, actually), whose objects are some equivalence classes of zig-zags (under an equivalence relation whose formulation I don't really understand), and whose morphisms are similar to the 1-simplices of the simplicial set above.
Is this formulation correct? I'm failing to see whether they're equal (or equivalent). In particular, I'm troubled by the fact that the $\pi_0$ of the nerve of a groupoid gives you the set of isomorphism classes of the groupoid, so I'm nervous about whether $\pi_0$ of the nerve defined in the nLab will actually really give $C[w^{-1}](X,Y)$.
If the formulation is not correct, the next question would be: is the simplicial set $L^H(C,w)(X,Y)$ defined by Dwyer-Kan the nerve of some category?

REPLY [10 votes]: The nLab description is not correct.  
For each "shape" of zig-zag, there is a "hammock category" for it (not a groupoid, and the nLab page I am looking at never mentions groupoids here), whose objects are functors $f\colon Z\to C$ ($Z$ is an abstract zig-zag of a particular shape) such that the backwards arrows of $Z$ are sent into $W$.  The morphisms are natural transformations $f\to f'$ which are identities at the ends (and which in the original formulation of Dwyer and Kan are such that the vertical arrows of the transformation must also be in $W$, though this condition turns out not to really be necessary, so it is nowadays often dropped).  
The full hammock mapping space $L^H(X,Y)$ is a quotient of all the nerves of these hammock categories by an equivalence relation which is not compatible with the category structure (though it is compatible with the simplicial structure).  Thus, $L^H(X,Y)$ is not a nerve of a category.
Go look at the original Dwyer-Kan paper, or at the book by Dwyer-Hirschhorn-Kan-Smith.<|endoftext|>
TITLE: When a journal doesn't give your work a fair chance
QUESTION [27 upvotes]: There is no way around the fact that determinations about the relative contributions of papers and journal acceptances will always be highly subjective. In addition, editors and referees are busy people and contribute their time on a near volunteer basis. However, even keeping this in mind, I can't help but feel that occasional my work doesn't get a fair shake. To help frame my question, let me describe some situations I have witnessed firsthand (often as the author):
1) The paper receives strong acceptance recommendations from every referee report. The editor rejects the paper and declines to give any additional feedback beyond a boilerplate sentence that the journal gets many more good papers than it can accept.
2) The paper is rejected without a referee report or any comments regarding the content of the paper by the editor. The rejection note solely cites that due to the backlog the journal isn't even able to send the paper out for an opinion or refereeing. However, the journal is processing papers for colleagues.
3) The editor removes positive comments or quashes altogether a positive referee report and then rejects the paper. This was discovered when the referee approached the author after the paper was rejected against the referee's opinion. There is a similar notorious case involving Duke that has been discussed on the internet.
4) The paper is rejected with a referee report that makes significant objectively inaccurate comments about the results and how they fit into the related literature. The editor declines to discuss or respond to these issues.
5) The referee report compliments the work but doesn't find the problem compelling. The report applies verbatim to other papers on the same problem in the same journal and no distinction or comparison to those papers is made. The editor rejects the paper based on this report.
6) I've also seen this from the other side. A graduate student across the world who I had never heard of solved a problem that I and others had worked on for several years, using some ingenious new ideas. The paper was submitted to a ~15th ranked journal. I was a referee and I wrote the most positive review I have so far in my career. Despite this the editor rejected the paper with some boilerplate language. I was never given a good explanation as to why (and I highly doubt there was any secret/political dimension to the story I am unaware of). This leaves me questioning if an unqualified glowing review by me wasn't sufficient, why even bother asking me to spend the time refereeing the paper in the first place?
I've heard other appalling stories from colleagues about their experiences with certain journals and editors, but I have been personally close enough to all of the instances described above to know the details of the particular cases. 
My question is the following:


Is it reasonable to demand some minimal level of due diligence / feedback / justification / consistency / logic from a journal? Do authors frequently push back on editors? Has anyone had success changing an editor's decision based on a discussion following a rejection?


For the sake of this question, assume you are advising a younger established researcher at a well rated research university who has previously published papers in similarly selective journals. Also assume that this person is aggressive and aims high but is not completely delusional.

REPLY [13 votes]: As a data point: I have changed an editor's mind after a negative decision that was based on a referee's report that was demonstrably wrong. When I was an editor, whenever an author raised a potentially valid objection to a referee's report based on more than a difference of opinion, I made it a practice to seek other expert opinions, and change my decision if warranted. To misquote Orwell, one has the right to expect ordinary decency even of an editor.<|endoftext|>
TITLE: Why do the adjoint representations of three exceptional groups have the same first eight moments?
QUESTION [27 upvotes]: For a representation of a compact Lie group, the $n$th moment of the trace of that representation against the Haar measure is the dimension of the invariant subspace of the $n$th tensor power. The sequence of moments determines the distribution of the trace.
For the adjoint representations of $G_2,F_4$, and $E_7$, the $0$th through $7$th moments are all 1,0,1,1,5,16,80,436.
Is there a conceptual reason for this, or is it just a numerical coincidence? 
Nick Katz asked this question in his graduate class.
One possible way to make this more conceptual is to note that there is more structure on the vector spaces of invariants than just their dimensions. There is an action of $S_n$, and maps front the $n$th invariants tensor the $m$th invariants to the $n+m$th invariants. Is this structure the same for these three groups? Is there a single structure with a conceptual definition that contains all three?
Some potentially related sequences, where the first $6$ elements agree with this one, are the moment sequences of the adjoint representations of $SP_6$ and $E_8$, and the sequence with exponential generating function  $e^{-\int_{0}^x \log(1-y)dy}$
Is there some kind of stabilization phenomenon occurring where high-dimensional exceptional groups, if they existed, would also agree with this moment sequence? If so it doesn't seem to agree with the first few stable moments of the adjoint representation of any sequence of classical groups. For $A_n$ the moment sequence is $1,0,1,2,9, 44, 265, 1854$ (counting derangments) and $C_n$ and possibly the others have moment sequence $1,0,1,1,6,22,130, 822$ (counting graphs where each vertex has two edges, with multiple edges but without loops).

REPLY [14 votes]: Yes, look for "Deligne's exceptional series". There are no theorems, but several beautiful conjectures.
The basic idea is that there should be a symmetric pivotal category generated by a trivalent vertex, with just a few local relations, depending on a parameter. At special values of the parameter, the category becomes degenerate, and the quotient by the negligible ideal recovers the representation category of one of the exceptional Lie algebras. (More or less; in some cases you get an equivariantization or subcategory.)
Working over rational functions in the parameter instead, it is expected that the category is semisimple, and its moments should agree with the sequence you describe. The exceptional algebra $F_4$ is the `least degenerate' point, in that its moments fall short the least.
Here are some pointers to the literature.
Pierre Deligne, La série exceptionnelle de groupes de Lie, C. R. Acad. Sci. Paris Sér. I Math. 322 (1996), no. 4, 321--326.
Pierre Deligne and Ronald de Man, La série exceptionnelle de groupes de Lie. II, C. R. Acad. Sci. Paris Sér. I Math. 323 (1996), no. 6, 577--582.
Arjeh M. Cohen and Ronald de Man, Computational evidence for Deligne’s conjecture regarding exceptional Lie groups, C. R. Acad. Sci. Paris Sér. I Math. 322 (1996), no. 5, 427--432.
Pierre Deligne and Benedict H. Gross, On the exceptional series, and its descendants, C. R. Math. Acad. Sci. Paris 335 (2002), no. 11, 877--881.
J. M. Landsberg and L. Manivel, Series of Lie groups, Michigan Math. J. 52 (2004), no. 2, 453--479.
J. M. Landsberg and L. Manivel, Triality, exceptional Lie algebras and Deligne dimension formulas, Adv. Math. 171 (2002), no. 1, 59--85.<|endoftext|>
TITLE: A converse of the abc conjecture?
QUESTION [17 upvotes]: Let ${\rm rad}(n)$ denote the radical of a positive
integer $n$, i.e. the product of its distinct prime divisors.
Given positive integers $a$ and $b$, the triple $(a,b,a+b)$ is
called an abc triple if $a$ and $b$ are coprime and
${\rm rad}(ab(a+b)) < a+b$. The quality of an abc triple
$(a,b,a+b)$ is defined as the quantity
$$
  q(a,b,a+b) \ := \ \frac{\log{(a+b)}}{\log{({\rm rad}(ab(a+b)))}},
$$
and the abc conjecture asserts that for any $\epsilon > 0$
there are only finitely many abc triples of quality greater than $1+\epsilon$. 
Now let the smoothness of an abc triple $(a,b,a+b)$ be
$$
  s(a,b,a+b) \ := \ \frac{\log{(\log{(\frac{a+b}{{\rm rad}(ab(a+b))})})}}{\log{(\log{(a+b)})}}.
$$
Clearly the smoothness of any abc triple is strictly smaller than $1$.

Question: Is it true that for any $\epsilon > 0$ there are
  abc triples with smoothness $\geq 1-\epsilon$?

Examples:

The triple $(1, 8, 9)$ has quality $1.22629$
and smoothness $-1.14676$.  
The triple $(3, 125, 128)$ has quality $1.42657$
and smoothness $0.23562$.
The triple $(10, 2187, 2197)$ has quality $1.28975$
and smoothness $0.26825$.
The triple $(1, 2400, 2401)$ has quality $1.45567$
and smoothness $0.43400$.
The triple $(1, 4374, 4375)$ has quality $1.56789$
and smoothness $0.52238$.
The triple $(343, 59049, 59392)$ has quality $1.54708$ 
and smoothness $0.56635$.
The triple $(3200, 4823609, 4826809)$ has quality $1.46192$
and smoothness $0.57855$.
The triple $(2, 6436341, 6436343)$ has quality $1.62991$
and smoothness $0.65457$.
The triple $(283, 8251953125, 8251953408)$ has quality $1.58076$
and smoothness $0.67991$.
The triple $(125, 11174240024064, 11174240024189)$ has quality
$1.53671$ and smoothness $0.690851$.
The triple $(24833, 5020969537415167, 5020969537440000)$ has
quality $1.62349$ and smoothness $0.73326$.
The triple $(8654525279998779296875, 229727166528260169448321941,
    229735821053540168227618816)$
has quality $1.48053$ and smoothness $0.72594$.

REPLY [14 votes]: Robert, Stewart and Tenenbaum have put forward a refined version of the abc conjecture (other variants are due to Granville, Baker, van Frankenhuijsen, ...) which states that if $a+b=c$ with $a$, $b$, $c$ positive, and if $k$ denotes the radical of $abc$ then (for an $abc$-triple with $k\le c$)
$$ 
\frac{c}{k} \le \exp \Big(A \sqrt{\frac{\log k}{\log \log k}}\Big) 
\le \exp\Big( A \sqrt{\frac{\log c}{\log \log c}}\Big), 
$$ 
for some constant $A$ and all $c$ large enough.  In fact they even conjecture that $A> 4\sqrt{3}$ is enough, and that this constant is best possible.  This conjecture implies that 
$$ 
s(a,b,c) = \frac{\log \log (c/k)}{\log \log c} \le \frac 12, 
$$ 
if $c$ is large enough.  The conjecture of Robert, Stewart and Tenenbaum is based on the work of Robert and Tenenbaum which gives a detailed analysis of the number of integers up to $x$ with radical at most $y$.<|endoftext|>
TITLE: Dimension in Whitney's theorem
QUESTION [9 upvotes]: There are some interesting examples of (classes of) manifolds for which I suspect that the Whitney theorem can be strenghtened. For example it is known that every (smooth) $n$-dimensional manifold can be embedded into $\mathbb{R}^{2n-1}$ for example if $M$ is orientable or is not of dimension of the form $2^k$. In particular I'm interested what is the minimal number $emb(n)$ with the property that each $n$-dimensional manifold can be embedded into $R^{emb(n)}$ if we assume additionally that $M$ is:
a) simultaneously orientable and not of dimension of the form $2^k$,
b) a Lie group, and
c) the boundary of some other manifold?

REPLY [8 votes]: The following paper of Elmer Rees addresses precisely your question for Lie groups:
Rees, Elmer
Some embeddings of Lie groups in Euclidean space. 
Mathematika 18 (1971), 152–156. 
The main result is that if a compact connected Lie group $G$ of rank $\ell$ admits a faithful representation on $\mathbb{R}^k$, then $G$ admits an embedding into Euclidean space whose codimension is $k-\ell$. This improves on Whitney's Theorem when $k-\ell<\operatorname{dim}G$, which is the case for many of the classical families of Lie groups.<|endoftext|>
TITLE: Add a multiple of $I$ to a matrix to minimize its operator norm
QUESTION [7 upvotes]: Given $A\in\mathbb{C}^{n\times n}$, what is $s_* = \arg\min \|A-sI\|$?
Here $\|A\|$ is the operator norm, $\|A\|=\rho(A^*A)^{1/2}$, and $I$ is the identity.
The corresponding problem for the Frobenius norm $\|A\|_F = \left(\sum A_{ij}^2\right)^{1/2}$ has a simple solution: reduce to the triangular case with a Schur factorization $A=QTQ^*$ (which does not alter $\|A\|_F=\|T\|_F$), then the problem becomes the one of minimizing $\sum_i |\lambda_i-s|^2$, with $\lambda_i=T_{ii}$ the eigenvalues of $A$, and then the solution is the arithmetic mean $s_{*,F}=\frac{1}{n}\sum \lambda_i$.
I do not see a simple solution for the case of the operator norm, though; I have tried applying the Schur factorization or the SVD, but the resulting problem is not immediate. Do you have any ideas? Or is this problem a known one?

REPLY [4 votes]: Not a closed form answer, but this can be solved as a semidefinite program. In particular, we can rewrite the task as
\begin{equation*}
\min_{t\ge 0, s}\ t\quad \text{s.t.}\quad (A-sI)^*(A-sI) \preceq t^2I,
\end{equation*}
which results in the SDP
\begin{equation*}
\min_{s,t}\ t\quad \text{s.t.}\quad \begin{bmatrix} tI & A-sI\\ (A-sI)^* & tI\end{bmatrix} \succeq 0.
\end{equation*}<|endoftext|>
TITLE: Kalman filters and stock price prediction
QUESTION [5 upvotes]: Could someone be so kind as to direct me to a good source that would explain time series (more specifically) stock price prediction using Kalman filters, Extended kalman filters or particle filters. Many thanks

REPLY [6 votes]: This web site provides a good entry point on Kalman filtering. It has a listing of books, software and more. The applications are biased towards navigation, but the applications to economic time series are also covered.
For an older introduction, specifically to the use of Kalman filters for stock price prediction, see this thesis on Kalman filtering approach to market price forecasting.<|endoftext|>
TITLE: Integral quaternary forms and theta functions
QUESTION [6 upvotes]: The following question arises when I attempt to understand the modular parameterization of the elliptic curve $$E:y^2-y=x^3-x$$
In Mazur-Swinnerton-Dyer and Zagier's construction, a theta function associated with a positive definite quadratic form is induced:
$$\theta(q)=\sum_{x\in\mathbb{Z}^4}q^{\frac{1}{2}x^{T}Ax}$$
where $$A=\left(\begin{matrix}2 & 0 & 1 & 1\\
0 & 4 & 1 &2\\
1 & 1 & 10 & 1\\
1 & 2 & 1 & 20
\end{matrix}\right)$$
$A$ is a positive definite matrix of determinant $37^2$, and we have $37A^{-1}=K^TAK$ where $K$ is an integral matrix of determinant $\pm 1$.
Question: Suppose $A$ is a positive definite $4\times 4$ matrix with integral entries. All diagonal entries are even numbers. The determinant of $A$ is a square number $N^2$. Is it true that for every $N=p$ ($p>2$ is a prime number), there is at least one $A$ that $NA^{-1}=K^TAK$, where $K$ is an integral matrix of determinant $\pm 1$?

REPLY [4 votes]: The answer is true, using the following construction.
Let $B$ be the quaternion algebra of discriminant $p$ and let $O$ be a maximal order with an element $x$ satisfying $x^2 = -p$. The reduced norm is a quadratic form on $O$, with positive definite Gram matrix $A$ of determinant $p^2$. The matrix $A^{-1}$ then represents the reduced norm on the dual lattice $O^{\sharp}$. Recall that $\text{nrd}(\text{diff}(O)) = \text{discrd}(O) = p$ and since $O$ is maximal, $\text{diff}(O)$ is invertible and $O = \text{diff}(O) O^{\sharp}$ (see e.g.
Voight, John, Quaternion algebras,  ZBL07261776. Section 16.8), hence
$$
p O^{\sharp} \subseteq \text{diff}(O) O^{\sharp} = O
$$
which implies that $p \cdot (O^{\sharp} / O) = 0$ (therefore $O^{\sharp} / O$ is an abelian group of exponent $p$ and size $p^2$, so isomorphic to $(\mathbb{Z} / p \mathbb{Z})^2$). Next, we note that the matrix $A^{-1}$ is also the change of basis matrix between the chosen basis of $O$ and its dual, therefore $pA^{-1}$ is integral.
This is not enough, but so far we have not used the element $x$. The matrix $pA^{-1}$ is the matrix representing the norm form on the ideal $xO^{\sharp}$, since $x^2 = -p$. But $nrd(xO^{\sharp})=p$ is a maximal order with an element $x$ such that $x^2 = -p$. By a theorem of Ibukiyama (see reference below), it is isomorphic (hence isometric as lattices) to $O$.
This could also be seen directly (referring to some of the answers above - it is possible to do it using a finite number of cases) by explicitly writing down the maximal orders. I will list below explicit constructions, which are based on
Ibukiyama, Tomoyoshi, On maximal orders of division quaternion algebras over the rational number field with certain optimal embeddings, Nagoya Math. J. 88, 181-195 (1982). ZBL0473.12012. -
If $p \equiv 3 \bmod 4$, then $B = (-p,-1)$, and $O = \mathbb{Z}<(1+i)/2,j>$. In this case, we have
$$
A = \left( \begin{array}{cccc} 
2 & 1 & 0 & 0 \\ 
1 & \frac{p+1}{2} & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 1 & \frac{p+1}{2}
\end{array} \right)
,
pA^{-1} = \left( \begin{array}{cccc} 
\frac{p+1}{2} & -1 & 0 & 0 \\ 
-1 & 2 & 0 & 0 \\
0 & 0 & \frac{p+1}{2} & -1 \\
0 & 0 & -1 & 2
\end{array} \right),
K = \left( \begin{array}{cccc} 
-1 & 1 & 0 & 0 \\ 
1 & 0 & 0 & 0 \\
0 & 0 & -1 & 1 \\
0 & 0 & 1 & 0
\end{array} \right)
$$
If $p \equiv 1 \bmod 4$, then choosing $q \equiv 3 \bmod 4$ such that
$\left( \frac{q}{p} \right) = -1$, we can find $c$ such that
$c^2 \equiv -p \bmod q$, and then $B = (-p,-q)$ and
$$
O = \mathbb{Z} \oplus \mathbb{Z} \frac{1+j}{2} \oplus
\mathbb{Z} \frac{i(1+j)}{2} \oplus \mathbb{Z} \frac{(c+i)j}{q}
$$
In this case, we compute that
$$
A = \left( \begin{array}{cccc} 
2 & 1 & 0 & 0 \\ 
1 & \frac{q+1}{2} & 0 & c \\
0 & 0 & \frac{p(q+1)}{2} & p \\
0 & c & p & \frac{2(p+c^2)}{q}
\end{array} \right)
,
pA^{-1} = \left( \begin{array}{cccc} 
\frac{(q+1)(c^2+p)}{2q} & -c-\frac{c^2+p}{q} & -c & \frac{c(q+1)}{2} \\ 
-c-\frac{c^2+p}{q} & 2(c^2 + \frac{c^2+p}{q}) & 2c & -c(q+1) \\
-c & 2c & 2 & -q \\
\frac{c(q+1)}{2} & -c(q+1) & -q & \frac{q(q+1)}{2}
\end{array} \right)
$$
and if $\{e_1, e_2, e_3, e_4 \}$ is the above basis for $O$, then we see that $ \{e_2 e_4, ce_1 - e_4, e_1, j e_2 \} $ is a basis for the resulting module, which written in terms of the original basis yields the matrix
$$
K = \left( \begin{array}{cccc}
0 & c & 1 & -\frac{q+1}{2} \\
-c & 0 & 0 & 1 \\
-1 & 0 & 0 & 0 \\
\frac{q+1}{2} & -1 & 0 & 0
\end{array}
 \right).
$$
Referring to Will Jagy's wondering in the first answer, the reason that this does not work for most lattices is that they do not correspond to a maximal order (as in the answer by few_reps, the quotient of the lattices is actually cyclic) and there are only one or two (depending on $p \bmod 4$ isomorphism classes of maximal orders which contain a root of $-p$. For example, for $p = 37$, there are only two isomorphism classes of maximal orders in the quaternion algebra, and only one of them contains a root of $-p$.<|endoftext|>
TITLE: Convergence of zeta functions for schemes of finite type over the integers
QUESTION [10 upvotes]: In his lecture "Zeta functions and $L$-functions", Serre presents a very elegant proof of the convergence of the zeta function 
$ \zeta (X,s) = \prod_{x \in |X|} (1- N(x)^{-s})^{-1}$ in the half plane $R(s) > dim(X)$, where $X$ is a scheme of finite type over $\mathbb{Z}$, $|X|$ the set of closed points of $X$ and $N(x)$ the number of elements in the residue field $k(x)$.
He reduces the claim to the case where $X = Spec \, A[x_1, \ldots x_n]$ and $A$ is either $\mathbb{Z}$ or $\mathbb{F}_p$.
The decisive input is the following lemma:
a) If $X$ is the finite union of the schemes $X_i$, and the claim holds for all $X_i$, then it holds for $X$. 
b) If $f: X \to Y$ is finite and the claim holds for $Y$, then it holds for $X$ as well.
I've been trying to prove b) but I seem to be missing something. Here's what I've tried so far:
I was considering $\zeta(X,s) = \prod_{y \in |Y|} \zeta(X_y \, ,s)$, where $X_y$  is the fiber of $f$ at $y$. I now the fibers are finite but I don't know how to connect this with the fact that $\zeta(Y,s)$ converges. Is it true that the residue field $k(y)$ is a finite extension of $k(x)$ for all $x \in X_y$ (of degree $\deg f$)? I know this is the case for the function fields.
Any help is very appreciated!

REPLY [7 votes]: I think part of your confusion lies in the definition of a finite morphism (see e.g. wikipedia). By definition, there is a finite affine open cover $U_i = \mathrm{Spec}(B_i)$ of $Y$ such that $f^{-1}(U_i) = \mathrm{Spec}(A_i)$ is affine and $A_i$ is a finite $B_i$-module.
Therefore there exists some $d$ such that for all closed points $y \in |Y|$, the fibre $f^{-1}(y)$ is a finite scheme of degree at most $d$ over the residue field of $y$. In particular, there are at most $d$ points above $y$, and each such point $x$ satisfies $N(x) \geq N(y)$.
One therefore obtains
\begin{align*}
  |\zeta (X,s)| & = \prod_{x \in |X|} \left|1- N(x)^{-s}\right|^{-1} \\
  & = \prod_{y \in |Y|} \prod_{\substack{x \in |X| \\ f(x) = y}} 
  \left|1- N(x)^{-s}\right|^{-1} \\
  & \leq \prod_{y \in |Y|} \left|1- N(x)^{-s}\right|^{-d} \\
  & = |\zeta (Y,s)|^d.
\end{align*}
Hence the absolute convergence of $\zeta (X,s)$ for $\mathrm{re}(s) > \mathrm{dim}(X)$ follows from the absolute convergence of $\zeta (Y,s)$ for $\mathrm{re}(s) > \mathrm{dim}(X)=\mathrm{dim}(Y)$.<|endoftext|>
TITLE: ω-categorical, ω-stable structure with trivial geometry not definable in the pure set
QUESTION [13 upvotes]: Briefly, my question is the following.

does every countable ω-categorical, ω-stable structure with 
disintegrated strongly minimal sets interpret in the countable pure set?

By countable pure set I mean a structure with countable universe and equality relation only.

This is a repetition of this question A totally categorical structure with trivial geometry which is not interpretable in the trivial structure. However, I do not understand why the answer provided there is marked correct. (I agree with Dima Sustretov in the comments that the structure does interpret in the pure set).
The background to my question is the following.
It is shown in the paper of Cherlin, Harrington and Lachlan that
every ω-categorical, ω-stable structure is coordinatized by a collection of projective spaces, affine spaces and pure sets (which appear as strictly minimal sets in the expansion of the original structure by imaginaries). I'm interested in those structures in which only the pure sets appear. These were studied in the paper of Lachlan titled "Structures coordinatized by indiscernible sets". In this paper, it is shown that every such structure interpret in an arbitrary countable linear order,
and also, that such structures correspond precisely to reducts of totally categorical structures with trivial geometry (of the strongly minimal sets).
It is easy to see that every structure which interprets in the pure set is 
ω-categorical and ω-stable, and it follows from the paper of Lachlan that it is coordinatized by indiscernible sets (this even shown for structures which interpret in a dense linear order). 
Therefore, we have the following implications:

interprets in countable pure set → ω-categorical, ω-stable, with disintegrated strongly minimal sets → interprets in countable dense linear order.

The second implication cannot be reversed (the dense linear order itself is not ω-stable). My question is whether the first implication can be reversed. In other words, this is the same question as asking about the existence of A totally categorical structure with trivial geometry which is not interpretable in the trivial structure.

REPLY [4 votes]: This is a follow-up to Szymon Toruńczyk's answer in which I will prove the claims in it.
Proposition.
$\mathrm{Th}(M)$ is totally categorical with trivial geometry.
Proof. First, note that $M = \mathrm{acl}(D)$. This implies that $D$ is stably embedded in $M$. Furthermore, we have that every permutation of $D$ extends (non-uniquely) to an automorphism of $M$. This implies that the induced structure on $D$ is that of an infinite pure set, so, in particular, $D$ is strongly minimal.
For any $N \equiv M$, we will also have that $N = \mathrm{acl}(D^N)$, so we get a unique model of each infinite cardinality. $\square$
Proposition. $\mathrm{Th}(M)$ is not interpretable in the theory of an infinite pure set.
Proof. Let $N$ be a model of $\mathrm{Th}(M)$, and let $X$ be an infinite pure set. Suppose that $N$ is interpretable in $X$. So in particular, we have imaginary sorts $D'$ and $C'$ as well as a definable binary relation $E' \subseteq C'\times C'$ and a definable function $f':C' \to D'$ such that $(D',C',E',f')$ is isomorphic to $N$.
Since $\mathrm{Th}(X)$ is uncountably categorical, there is a strongly minimal set $S\subseteq D'$. Furthermore, there must be a definable finite-to-finite correspondence between $X$ and $S$. Since $X$ is a pure set, this must actually be a bijection on some cofinite set.
Let $Y\succ X$ be an elementary extension such that $|Y \setminus X| = 2$. Let the two new elements be $a_0$ and $a_1$. Let $S^Y$ and $C'^Y$ be the corresponding sets in the elementary extension, and let $b_0$ and $b_1$ be the new elements of $S^Y$ corresponding to $a_0$ and $a_1$, respectively. Finally let $c_0,$ $c_1,$ $c_2,$ and $c_3$ be elements of $C'^Y$ satisfying that $f'(c_0) =f'(c_2)=b_0$, $f'(c_1)=f'(c_3)=b_1$, and $c_0E'c_1E'c_2E'c_3E'c_0$.
Let $\sigma$ be the automorphism of $Y$ that fixes $X$ and has $\sigma a_0=a_1$ and $\sigma a_1 =a_0$. We necessarily have that $\sigma b_0 = b_1$ and $\sigma b_1 = b_0$.
There are only two possibilities for the value of $\sigma c_0$. Either $\sigma c_0 = c_1$ or $\sigma c_0 = c_3$. In both cases $\sigma^2 c_0 = c_2$, but $\sigma^2$ is the identity automorphism, so this is absurd.
Therefore no such interpretation can exist. $\square$<|endoftext|>
TITLE: Epsilon regularity for minimal surfaces in arbitrary Riemannian manifolds
QUESTION [6 upvotes]: For experts in the analysis of minimal surfaces I will state the question first; then I will follow up with details.
Question:  Does the $\varepsilon$-regularity theorem of Choi and Schoen (http://link.springer.com/article/10.1007%2FBF01388577)
for minimal surfaces in three-dimensional Riemannian manifolds extend to Riemannian manifold targets of arbitrary dimension?  
For non-experts, let me state the $\varepsilon$-regularity result of Choi and Schoen.  Let $(M,h)$ be a three dimensional Riemannian manifold and suppose
\begin{align}
f: \Sigma \rightarrow M
\end{align}
is an immersed minimal surface; here the topology of $\Sigma$ is arbitrary.  Let $(g,B)$ denote the induced metric and second fundamental form via the immersion $f.$  The following $\varepsilon$-regularity theorem posits that sufficient control on the $L^{2}$-norm of $B$ on balls of controlled size implies $L^{\infty}$-bounds on $B$ on balls of controlled size.  
Theorem: There exists constants $\varepsilon, \rho>0,$ (which depend only on the curvature of $M$ and it's covariant derivatives), such that if $r_{0}<\rho, \ x\in M$ and
\begin{align}
f(\Sigma)\cap \partial B_{r_{0}}(x)=\partial(f(\Sigma)\cap B_{r_{0}}(x)),
\end{align}
the following property holds.  If there exists $0<\delta\leq 1$ such that
\begin{align}
\int_{B_{r_{0}}(x)\cap f(\Sigma)} \lVert B\rVert^{2} dV_{g}< \delta\varepsilon,
\end{align}
then for all $0<\sigma\leq r_{0}$ and all $y\in B_{r_{0}-\sigma}(x)\cap f(\Sigma)$,
\begin{align}
\sigma^{2}\lVert B\rVert^{2}(y)<\delta.
\end{align}
In the above theorem, $B_{r}(x)$ is the Riemannian ball of radius $r>0$ in the manifold $M$ and $dV_{g}$ is the Riemannian volume element of the metric $g$ on $\Sigma.$  Lastly, the symbol $\partial$ indicates the topological boundary of a set.
My question is if the manifold $M$ can be replaced by a Riemannian manifold of arbitrary dimension.
These types of regularity estimates date back to the work of Sacks-Uhlenbeck on minimal surfaces, and are an indispensable tool for the study of asymptotics to solutions of many systems of partial differential equations that have a geometric origin: examples include the Yang-Mills equations, the self-duality equations, the study of minimal surfaces, and more generally, harmonic maps.  
Unfortunately, for the most part I'm a casual observer to these developments and some basic google searches haven't pulled up any papers claiming this can be extended to arbitrary co-dimension.  The obvious thing to do is to dive into the proof of Choi and Schoen, but before I undertake such a task, I was hoping someone here might save me the time.  I appreciate any suggestions and/or references.  Thank you.

REPLY [3 votes]: The desired bound is correct (and in fact you get a fairly explicit value for $\epsilon$ of anything below $4\pi$) .  A proof can be found in these beautiful notes  of a course by Brian White (it's Theorem 8.12).<|endoftext|>
TITLE: Galois representations along eigenvarieties
QUESTION [5 upvotes]: This question is about the status of the following. 
Meta-hypothesis. Let $X$ be an irreducible component of an eigenvariety. Then there exist: (a) a pseudo-representation/character $\psi$ along $X$,  specializing to what it should at classical points; (b) a "cover"  $f\colon X' \rightarrow X$, a locally free sheaf $V$ on $X'$, and an ${\scr O}_{X'}$-linear Galois action on $V$, such that that $f^*\psi$ is the pseudocharacter associated with $V$. 
Questions: 

What instances of this statement are known, and with what constructions?
In what generality do we expect the hypothesis to hold, and under what meaning of "cover" (and "eigenvariety")?
What about the Coleman-Mazur-Buzzard eigencurve of tame level $N$? Can one take $X'$ to be the normalization of $X$, and if so, are the details written anywhere?

Some remarks. There are quite general results on (a) by Chenevier, Bellaiche-Chenevier,...; the less clear part (to me) is (b). Two constructions which come to mind and work "on the nose" ($X'=X$) in special cases are the following: (1) for the eigencurve of level 1, take (the Jacquet module of) completed cohomology; (2) if the mod $p$ reduction $\overline{\psi}$ of the universal $\psi$ on $X$ corresponds to an irreducible mod $p$ Galois representation  $\overline{\rho}$, then the generic fibre of the Spf of the universal deformation ring $T_{\overline{\psi}}=T_{\overline{\rho}}$ admits a map from $X$ and carries a universal Galois-sheaf $W$, which can then be pulled back to a Galois-sheaf $V$ on  $X$.

REPLY [2 votes]: The answer depends what kind of "cover" you need down the road. 
For a strong definition of cover, like a "Zariski cover", the answer is no in general. I believe it is still no for an "étale cover" or "fpqc cover".
Now if you ready to consider as covers not only Zariski covers but any proper and birational map $X' \rightarrow X$, then the answer is yes in full generality, and for general reasons which have not much to do with eigenvarieties. See Lemma 3.4.2 in my book with Chenevier (Astérisque 324) and how it is used in the proof of Theorem 3.4.1. (I can give more details if needed).<|endoftext|>
TITLE: Switching left and right adjoints in recollement situations
QUESTION [6 upvotes]: In Fascieaux pervers, Beilinson, Bernstein and Deligne define a recollement situation as a triple of triangulated categories $\mathcal{D}_U,\mathcal{D}_F$ and $\mathcal{D}$, together with functors
$$
i_*\colon \mathcal{D}_F\to \mathcal{D},\qquad j_*\colon \mathcal{D}\to \mathcal{D}_U
$$
with suitable properties. In particular, $i_*$ has both a left adjoint $i^*$ and a right adjoint $i^!$. Beilinson, Bernstein and Deligne show that, in this situation, given
$t$-structures $(\mathcal{D}^{\leq0}_U,\mathcal{D}^{\geq 0}_U)$ on $\mathcal{D}_U$ and $(\mathcal{D}^{\leq0}_F,\mathcal{D}^{\geq 0}_F)$ on $\mathcal{D}_F$ one obtains a $t$-structure $(\mathcal{D}^{\leq0},\mathcal{D}^{\geq 0})$ on $\mathcal{D}$ by setting
$$
\mathcal{D}^{\leq0}=\{K\in \mathcal{D}\,|\, j^*K\in\mathcal{D}_U^{\leq 0} \text{ and } i^*K\in \mathcal{D}^{\leq0}_F\}
$$
$$
\mathcal{D}^{\geq0}=\{K\in \mathcal{D}\,|\, j^*K\in\mathcal{D}_U^{\geq 0} \text{ and } i^!K\in \mathcal{D}^{\geq0}_F\}
$$
It is not hard to check that this is indeed a $t$-structure on $\mathcal{D}$. One may wonder what happens by switching the role of the left and right adjoint of $i_*$ in the above definition, i.e., if one tries to define a $t$-structure by setting
$$
\tilde{\mathcal{D}}^{\leq0}=\{K\in \mathcal{D}\,|\, j^*K\in\mathcal{D}_U^{\leq 0} \text{ and } i^!K\in \mathcal{D}^{\leq0}_F\}
$$
$$
\tilde{\mathcal{D}}^{\geq0}=\{K\in \mathcal{D}\,|\, j^*K\in\mathcal{D}_U^{\geq 0} \text{ and } i^*K\in \mathcal{D}^{\geq0}_F\}
$$
The argument that shows that $(\mathcal{D}^{\leq0},\mathcal{D}^{\geq 0})$ is a $t$-structure breaks down if one tries to adapt it to $(\tilde{\mathcal{D}}^{\leq0},\tilde{\mathcal{D}}^{\geq 0})$, precisely since the adjoints are not in the right place. However this is not a proof that $(\tilde{\mathcal{D}}^{\leq0},\tilde{\mathcal{D}}^{\geq 0})$ is not a $t$-structure, but only that it is not obviously so. And indeed, if one tries to set up a recollement situation in the much more symmetric setting of stable $\infty$-categories (where all constructions show a very nice symmetric behaviour due to the fact that every pullback is a pushout and vice versa), then it seems that not only also $(\tilde{\mathcal{D}}^{\leq0},\tilde{\mathcal{D}}^{\geq 0})$ is a $t$-structure, but that this actually coincides with $(\mathcal{D}^{\leq0},\mathcal{D}^{\geq 0})$ (this however does not imply that $i^*=i^!$). Or at least, this is what I believe we have shown with Fosco Loregian in http://arxiv.org/abs/1507.03913 Assuming our result is correct (which may well not be the case), the natural questions are:

is this actually true also for triangulated categories? (I have not been able to locate a statement like this in the literature)
it is this maybe false for a general triangulated category but true for a triangulated category which is the homotopy category of a stable $\infty$-category?
is this always manifestly false for triangulated categories? (something that would make me suspect of some mistake in my argument for stable $\infty$-categories, or a hint of the fact that giving a recollement situation in the stable setting is such a strong requirement that there are actually no natural examples of stable recollements)

REPLY [4 votes]: The unexpected (and actually undue) symmetric behavior of stable recollements noticed in http://arxiv.org/abs/1507.03913 (Lemma 4.3 of version 1, therein called the Rorschach lemma) turned out to be the far reaching consequence of a typo in one of the commutative diagrams on page 9. This has now been corrected (i.e., Lemma 4.3, together with all its corollaries, has been removed).
Luckily, this was only minimally affecting the remaining part of the article, which has now been revised accordingly.<|endoftext|>
TITLE: Rigorous scaling limit for Navier-Stokes and Boltzmann equation
QUESTION [9 upvotes]: In the now 35 years old survey paper ''Kinetic equations from Hamiltonian dynamics'', Herbert Spohn mentions two important unsolved problems in mathematical physics: On p.571 the hydrodynamic limit, and on p.603 the derivation of the nonlinear Boltzmann equation from quantum mechanics.
Where can I find the current state of affairs? In particular, are there now rigorous derivations of the Navier-Stokes equations and/or the Boltzmann equations from either nonrelativistic many-particle quantum mechanics or quantum field theory? If yes, under which assumptions?

REPLY [7 votes]: The Boltzmann Equation from Quantum Field Theory

We show from first principles the emergence of classical Boltzmann
  equations from relativistic nonequilibrium quantum field theory as
  described by the Kadanoff-Baym equations. Our method applies to a
  generic quantum field, coupled to a collection of background fields
  and sources, in a homogeneous and isotropic spacetime. We show that the system follows a generalized Boltzmann equation whenever the WKB approximation holds. The generalized Boltzmann equation, which includes off-shell transport, is valid far from equilibrium and in a time dependent background, such as the expanding universe.<|endoftext|>
TITLE: More general than semidefinite program?
QUESTION [12 upvotes]: I was TAing my convex optimization class and explaining that Linear Programs are a special case of Second Order Cone Programs, which are themselves special cases of Semidefinite Programs.  My question is, is there any well-established class of optimization problems that is more general than semidefinite programs?  Conic optimization problems would be an example of this, but I'm hoping for something a little more algebraic, if it exists.

REPLY [5 votes]: If you like, you might look at cones of sums of squares of polynomials (cones of PSD matrices are the same thing as cones of sums of squares of linear polynomials). This is the starting point of the modern technique of solving optimisation problems on semi-algebraic sets, due to J.Lasserre and others.
More generally, you might look at cones of nonnegative polynomials, and this opens up the whole Hilbert 17th problem business.<|endoftext|>
TITLE: Deciding isomorphism between graphs which interpret in the pure set
QUESTION [6 upvotes]: I am interested in the following decision problem:

Given descriptions of two graphs $G,H$ which interpret in the pure set $\mathbb N=(\{0,1,2,\ldots\},=)$, decide whether $G$ and $H$ are isomorphic.

The question is: is this problem decidable at all?
(A graph $G$ is represented by a triple of formulas $(\phi_{\text{dom}},\phi_{\sim},\phi_E)$ describing the interpretation in the usual way (namely, $\phi_{\text{dom}}$ is a formula with $d$ free variables for some $d\in\mathbb N$, defining a set  $V\subset \mathbb N^d$, $\phi_{E}$ has $2d$ free variables and defines a symmetric binary relation $E\subset V\times V\subset \mathbb N^{2d}$, and $\phi_\sim$ is a formula with $2d$ free variables defining a binary relation $\sim\subset V\times V$ which is a congruence of the graph $(V,E)$ (i.e. the edge relation is invariant under $\sim$). The graph $G$ is defined as the quotient graph $(V,E)/\sim$ with vertices $V/\sim$ and edges $\{[v],[w]\}$ such that $(v,w)\in E$.)
The graphs $G,H$ are ω-categorical and therefore, whenever they are non-isomorphic, there is a sentence $\phi$ which distinguishes $G$ from $H$. Since it can be effectively tested whether $\phi$ holds in $G$ and in $H$,
it follows that non-isomorphicity is recursively enumerable. 
Therefore, the question which remains is whether there is an effective witness of isomorphicity, which would probably require some form of structure theorem for these graphs.
These graphs are ω-categorical, ω-stable, and moreover they are coordinatized by indiscernible sets, as in the eponymous paper by Lachlan,
so perhaps the structure theory developed in this paper could be of use.
(In this question I ask if those two classes coincide:ω-categorical, ω-stable structure with trivial geometry not definable in the pure set).
My question could be generalized to graphs which interpret in other structures with decidable first order theory, e.g., the dense linear order.

Edit 1: I modified the question so that it talks about graphs rather than arbitrary structures. Those two questions are easily seen to be equivalent. 

Edit 2: I wrote down some preliminary observations concerning this problem here: http://atoms.mimuw.edu.pl/?p=1063

REPLY [2 votes]: Update. As noted in the comments, this answer applies only to definable quotients of $\mathbb{N}$, rather than $\mathbb{N}^d$, and so it doesn't answer the question.

The answer is yes, your relation is computably decidable.
To see this, observe first that the theory of the structure
$\langle\mathbb{N},=\rangle$ admits elimination of quantifiers.
This can be easily proved by induction on formulas in the usual
elimination-of-quantifiers manner.
If $A$ is a definable structure in $\langle\mathbb{N},=\rangle$,
then the domain, quotient relation and fundamental relations of
$A$ are each defined by a quantifier-free formula with parameters.
Furthermore, given the defining relations of $A$, then since the
elimination of quantifiers argument is effective, we can
computably find those quantifier-free formulas. You didn't seem to
allow parameters in the definitions, but I claim that we can even
allow parameters in the definition and it will still be
computable.
The quantifier-free definable sets, without parameters, are
exactly the empty set and the full set. With parameters
$b_0,\dots,b_n$, the quantifier-free definable sets will be the
subsets of the parameter list, plus possibly the complement of the
parameter set. So, the finite and co-finite sets.
Furthermore, we can tell from the quantifier-free definition which
case we are in. So if we are given the definitions of $A$ and $B$,
we can tell whether the pre-quotient domains have the same size or
not.
The equivalence relation used in the quotient will be similarly
trivial, since outside the parameter set, it must either identify
all points or none, and we can computably tell exactly which. And
on the parameter set, we will be able to read off from the
quantifier-free definition exactly what it does on the parameter
list. So given the definitions of $A$ and $B$, we can computably
decide whether the quotients have the same size or not. Indeed,
the quotient is essentially trivial outside the parameter set,
since it must either collapse the entire complement of the
parameter set or none of it.
Similarly, I claim that we can computably decide the nature of the
other relations. Basically, the quantifier-free definition of any
relation using parameters is specified in a computably decidable
manner on the elements of the parameters themselves, and all
distinct elements outside the parameter list are indiscernible for
that relation. So there are finitely many types of elements, which
can simply be read directly from the definition of the relation.
I claim that this implies that the isomorphism relation for your
structures is computably decidable. Each $n$-ary relation has
essentially finitely many types of instantiations and
non-instantiations, by considering the points of the parameters
and then the indiscernibles of the non-parameters. Given pairs of
finitely many such relations, we can determine if there is an
isomorphism of the quotient structures: the domains should have
the same size, and then the relations should be determined by a
rearrangement of the parameter set and a matching of the
indiscernibility nature of the relations outside the parameter
set.<|endoftext|>
TITLE: Uniqueness of $\infty$-adjoints
QUESTION [5 upvotes]: Adjoints in a 2-category are essentially unique, in the following strong sense.  If $\mathbf{2}$ denotes the "walking arrow" category $(\cdot \to \cdot)$, then there is a 2-category $\mathrm{Adj}_1$ with two objects, and a map $\mathbf{2}\to \mathrm{Adj}_1$, such that for any 2-category $K$ the precomposition functor $\mathrm{Cat}(\mathrm{Adj}_1,K) \to \mathrm{Cat}(\mathbf{2},K)$ is fully faithful and its image consists of the 1-morphisms in $K$ with right adjoints.
Is an analogous fact true for adjoints in an $(\infty,2)$-category?  If so, what is the $(\infty,2)$-category $\mathrm{Adj}_{\infty}$?  A natural conjecture would be that $\mathrm{Adj}_{\infty} = \mathrm{Adj}_1$; I'm not quite sure whether to hope for that.
(I'm assuming that an "adjoint" in an $(\infty,2)$-category $K$ means a morphism whose image in the homotopy 2-category of $K$ has an adjoint in the usual sense.  So in a sense, this question is about "coherentification" of adjoints.)

REPLY [9 votes]: Yes, and $\mathrm{Adj}_\infty = \mathrm{Adj}_1$.  This is Theorem 4.4.18 of Riehl-Verity.  Thanks David, I should have remembered that.<|endoftext|>
TITLE: The action of an S-arithmetic group on the hyperbolic plane
QUESTION [5 upvotes]: I have a really quick question. I am interested in $G=SL_2(\mathbb{Z}[1/p_1,...,1/p_n])$, where $p_1$,..., $p_n$ are prime numbers. Since $G$ is a subgroup of $SL_2(\mathbb{R})$, it acts in the hyperbolic plane $\mathbb{H}$. My question is: are all the isotropy groups of this action finite?
Thank you

REPLY [7 votes]: This answer addresses the question posed in the comments. 
It follows from Bass-Serre theory that every finite subgroup of $G$ is conjugate into $SL_2(\mathbb{Z})$ (see Section II.1.4 of Serre's book Trees). So maximal finite subgroups will be order $4$ or $6$, generated by an element of trace $0$ or $1$, and thus property (M) holds. More specifically, to every $p_i$ is associated an action on a tree. A finite subgroup of $SL_2(\mathbb{Q})$ has to fix a vertex of this tree, and by Bass-Serre theory is conjugate into $SL_2(\mathbb{Z}_{(p_i)})$. Applying this to each $i$, one sees that a finite group is conjugate into $SL_2(\mathbb{Z})$, for which the finite subgroups are well-known to be cyclic. 
The second property (NM) is false in general. Consider the matrix $\left[\begin{array}{cc}a & b \\-b & a\end{array}\right]$, $a^2+b^2=1$, then it normalizes the maximal finite subgroup generated by $\left[\begin{array}{cc}0 & 1 \\-1 & 0\end{array}\right]$. Such matrices are plentiful coming from Pythagorean triples. 
On the other hand, for certain denominators $p_i$, I think property (NM) will hold. If $-1$ and $-3$ are not quadratic residues $(\mod p_i)$ for all $i$, then I think that the stabilizers of maximal finite subgroups will be trivial. Equivalently, $p_i\equiv -1 (\mod 12)$ for all $i$.<|endoftext|>
TITLE: Berry-Esseen bound for martingale sequence with varying and dependent variances
QUESTION [9 upvotes]: Let $(X_{1},\ldots,X_{k},\ldots)$ be a martingale difference sequence, i.e. 
$$
E[X_{k}|\mathcal{F}_{k-1}] = 0
$$
where $\mathcal{F}_{k-1}$ is the $\sigma$-algebra filtration at $k-1$. 
Let $\sigma_{k}^2 = E[X_{k}^2|\mathcal{F}_{k-1}]$. Here note that $\sigma_{k}^2$ is a random variable measurable w.r.t. $\mathcal{F}_{k-1}$. If we know that $\sigma_{k}^2 \in [\underline{\sigma}^2, \overline{\sigma}^2]$, where both $\underline{\sigma}$ and $\overline{\sigma}$ are positive constants, then do we have, as $\beta\to 0^+$,
$$
\frac{\sum_{k=1}^\infty (1-\beta)^k X_{k}}{\sqrt{\sum_{k=1}^\infty (1-\beta)^{2k} \sigma_{k}^2}} \rightarrow N(0,1)
$$
in distribution, and what will be the correct order of Berry-Esseen bound?

REPLY [5 votes]: In the setting you describe, there's generally no CLT. Let me describe a counterexample showing that $\sum_{i=1}^{k-1} X_i / \sqrt{\sum_{i=1}^{k-1} \sigma_i}$ doesn't tend to normal. I'm pretty sure the same holds for the discounted averages as well.
The example is a random walk which is lazy when positive. More precisely, conditioned on $\mathcal{F_{k-1}}$, if $\sum_{i=1}^{k-1} X_i <0$ then $X_k$ is uniformly distributed in $\{-1,1\}$ and if $\sum_{i=1}^{k-1} X_i \ge 0$ then $X_k$ is uniformly distributed in $\{-1,0,1\}$. The ratio between $P( \sum_{i=1}^{k-1} X_i <0)$ and $P(\sum_{i=1}^{k-1} X_i >0)$ tends to the ratio between the speed of the random walk when negative and when positive, which is 2:3, so the probability that the sum is positive tends to $\frac35$. I expect the same limit also for the discounted average.<|endoftext|>
TITLE: real and complex vector spaces as topological categories
QUESTION [11 upvotes]: Let $Vect_{\mathbb{R}}$ be the category of (say, finite dimensional) vector spaces over $\mathbb{R}$. The automorphism group of the object $\mathbb{R}^n\in Vect_{\mathbb{R}}$, is $GL_n(\mathbb{R})$. We usually like to think of it as a topological group. For example $BGL_n(\mathbb{R})$ classifies real $n$-dimensional vector bundles. This suggests that perhaps we can make $Vect_{\mathbb{R}}$ into a topological category in such a way that the topological group $BGL_n(\mathbb{R})$ will be the space of automorphisms of $\mathbb{R}^n$. Namely, we will get an $\infty$-category such that the $BGL_n(\mathbb{R})$-s are the connected components of its space of objects.
The obvious idea is just to take the hom-sets of linear maps with their natural topology, but note that this makes all the mapping spaces contractible. Of course one can make an artificial definition, like taking only isomorphisms or chopping the mapping space into connected components according to the rank, but this has other disadvantages. For example, if we let $Vect_{\mathbb{C}}$ be the category of (finite dimensional) vector spaces over $\mathbb{C}$, we have the extension/restriction of scalars adjunction 
$$
Vect_{\mathbb{R}}\leftrightarrows Vect_{\mathbb{C}}
$$
and we would like to make it into an $\infty$-adjunction. The isomorphism of hom-sets that comes from the adjunction does not respect ranks. So my question is basically, 

can one enrich $Vect_{\mathbb{R}}$ and $Vect_{\mathbb{C}}$ over topological spaces in such a way that the restriction/extension of scalars will induce an $\infty$-adjunction and such that the automorphisms of $\mathbb{R}^n$ (resp. $\mathbb{C}^n$) will be equivalent to $GL_n(\mathbb{R})$ (resp. $GL_n(\mathbb{C})$)?

REPLY [5 votes]: I think the answer is no. Suppose there exists an enrichment satisfying your requirements, and let $U: Vect_{\mathbb{C}} \to Vect_{\mathbb{R}}$ be the forgetful functor. Let $C \subseteq Map(\mathbb{R}^n,U(\mathbb{C}^n))$ be the subspace consisting of those maps which are adjoint to equivalences $\mathbb{R}^n \otimes \mathbb{C} \to \mathbb{C}^n$. Then $C$ is a connected component of the mapping space, and since the connected group $GL(U(\mathbb{C}^n))$ acts on this mapping space by post-composition it must preserve this component. On the other hand, if one restricts this action to $GL(\mathbb{C}^n) \hookrightarrow GL(U(\mathbb{C}^n))$ then $C$ becomes a principal homogeneous space. This implies that for every $n$ the subspace $GL_n(\mathbb{C}) \cong GL(\mathbb{C}^n) \hookrightarrow GL(U(\mathbb{C}^n)) \cong GL_{2n}(\mathbb{R})$ is a retract up to homotopy. At $n=1$ this happens to be ok, as $GL_1(\mathbb{C}) \simeq S^1$ is indeed a retract of $GL_2(\mathbb{R}) \simeq O(2) \simeq S^1 \coprod S^1$, but I would bet you would be able to find an $n$ where this inclusion is not a homotopy retract.
-- Edit -- 
As pointed out in the comments below:
1) Already when $n=2$ the map $GL(\mathbb{C}^n) \to GL(U(\mathbb{C}^n))$ is not a homotopy retract.
2) $GL(U(\mathbb{C}^n))$ is not connected, but one can replace it with the connected component of the identify $GL^0(U(\mathbb{C}^n)) \subseteq GL(U(\mathbb{C}^n))$, replace $C$ with the corresponding $C^0 \subseteq C$, and continue the argument as before (since $GL(\mathbb{C}^n)$ is connected its image in $GL(U(\mathbb{C}^n))$ lies in $GL^0(U(\mathbb{C}^n))$, and the inclusion $GL(\mathbb{C}^n) \hookrightarrow GL^0(U(\mathbb{C}^n))$ is again not a retract already for $n=2$).<|endoftext|>
TITLE: What can we say about the differences between roots of a polynomial with large Galois group?
QUESTION [5 upvotes]: Suppose that $K$ is a number field and $L$ is the splitting field of a monic polynomial in $\mathcal{O}_{K}[x]$ of degree $d \geq 5$ with roots $\alpha_{1}, ... , \alpha_{d}$.  Assume that the $\mathrm{Gal}(L / K)$ is the full symmetric group $S_{d}$ (there is an obvious variant of this question where we assume that the Galois group is $A_{d}$ instead).
Vaguely stated, my question is, given the relative lack of algebraic relations among the roots, what can one conclude about whether there exist primes $\mathfrak{P}$ of $L$ such that certain subsets of the roots become equal modulo $\mathfrak{P}$?  More specifically, could one conclude, for instance, that there exists a prime $\mathfrak{P}$ of $L$ such that $\alpha_{i} - \alpha_{j} \in \mathfrak{P}$ for exactly one choice of $\{i,j\} \subset \{1, ... , d\}$?
Here is a (possibly) closely related question: what can we say about the powers of primes of $K$ which contain the discriminant $D$ of this polynomial (again assuming its Galois group is $S_{d}$)?  Can one conclude that there exists a prime $\mathfrak{p}$ of $K$ such that $D \in \mathfrak{p} \setminus \mathfrak{p}^{2}$?  Is it possible for $D \in (K^{\times})^{n}$ for some $n \geq 3$?
I'm sorry that I'm asking several vaguely related questions here rather than narrowing things down to be more concrete.  But I've been trying but failing to get small results on this theme using elementary algebraic number theory for a while now, and I'm curious as to whether any statements of this kind are already known.  (Perhaps unsolvability and $2$-transitivity are the only properties of $S_{n}$ which we really need to use here.)

REPLY [7 votes]: Kedlaya proved that for every $n$ there is a monic polynomial $f\in\mathbb Z[X]$ with square-free discriminant and Galois group $S_n$. See here (published version) or here (preprint).<|endoftext|>
TITLE: Torsion-freeness of two groups with 2 generators and 3 relators and Kaplansky Zero Divisor Conjecture
QUESTION [13 upvotes]: Let $G_1$ and $G_2$ be the groups with the following presentations:
$$G_1=\langle a,b \;|\; (ab)^2=a^{-1}ba^{-1}, (a^{-1}ba^{-1})^2=b^{-2}a, (ba^{-1})^2=a^{-2}b^2 \rangle,$$
$$G_2=\langle a,b \;|\; ab=(a^{-1}ba^{-1})^2, (b^{-1}ab^{-1})^2=a^{-2}b, (ba^{-1})^2=a^{-2}b^2 \rangle,$$
Are these groups torsion-free?
Motivation: In both of these groups $1+a+b$ as an element of the group algebra $\mathbb{F}_2[G_i]$  over the field with two elements is a zero divisor. Thus one has a counterexample for the Kaplansky zero divisor conjecture if one of $G_i$s is torsion-free! 
$$(1+a+b)(b^{-1}a^{-2}ba^{-1}+a^{-1}ba^{-2}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+a^{-1}b^2a^{-1}ba^{-1}+aba^{-1}ba^{-1}+1+a^{-2}ba^{-1}+a^{-1}b^{-1}a^{-1}+b+baba^{-1}ba^{-1}+ba^{-1}ba^{-1}+a^{-1}+b^{-1}a)=0$$
$$(1+a+b)(aba^{-1}ba^{-1}+a^{-1}b^2a^{-1}ba^{-1}+a^{-1}ba^{-1}+b^{-1}a^{-1}+b^{-1}a^{-2}ba^{-1}+ab+1+ba^{-1}ba^{-1}+baba^{-1}ba^{-1}+b+bab+a^{-2
}ba^{-1}+a^{-1}+b^{-1}a)=0$$

REPLY [8 votes]: Similar to the answer of Mark Sapir, for the second group let $x=ba$ and $y=b^{-1}ab^{-1}$. From the second relation we have $y^2=x^{-1}y^{-1}$. So, from the first relation we have $x=y^4$. These relations implies that $y^4=y^{-3}$ or $y^7=1$.<|endoftext|>
TITLE: Applications of the Central Limit Theorem in dynamical systems
QUESTION [12 upvotes]: There are very many papers in the area of (possibly non-uniformly) hyperbolic dynamical systems whose aim is to prove the Central Limit Theorem. In a dynamical context, this means that one:

has a metric space $\Omega$, a map (or flow, but let's stick to maps) $T:\Omega\to\Omega$ and a (Borel) invariant probability measure $\mu$,
and aims at finding a suitable space $\mathcal{H}$ of functions $f:\Omega\to \mathbb{R}$ such that for all $f\in \mathcal{H}$, denoting by $X$ a random variable of distribution $\mu$, the sequence of random variables $f(X), f(T X), f(T^2 X),\dots$ satisfies the conclusion of the CLT for some variance $\sigma^2$.

I can see the beauty of this, first as it yields for a sequence of completely dependent random variables the conclusion of a Theorem most usually proved with a high degree of independence (thus enforcing the principle that deterministic chaos is tamed by time averaging); and second as it gives a strengthening of ergodicity, giving insight on the rate of equidistribution of orbits.

My question is whether there are application of this kind of CLT? By application I don't (necessarily) mean real-life or applied science application, but rather mathematical statements who do not look like they involve the CLT, but whose proof use it in an essential way.

REPLY [10 votes]: These limit theorems can be useful when studying systems preserving an infinite invariant measure. 
For example, Jean Pierre Conze has used it in his paper "Sur un critere de recurrence en dimension 2 pour les marches stationnaires" to get a criteria for recurrence for  $\mathbb{Z}^2$ skew product extensions. See also Klaus Schmidt's "Recurrence of cocycles and stationary random walks". In general, a local central limit theorem can be used to establish multiple recurrence in systems with infinite measure. 
Another nice use is in Ledrappier and Sarig's paper where they use Ratner's central limit theorem to prove rational ergodicity (a stronger form of ergodicity for an infinite measure preserving transformation) for $\mathbb{Z}^d$ covers of horocycle flows.  
Another place where it is used is in homogenisation of fast-slow systems, see for example the papers of Dolgopyat and Kelly-Melbourne.<|endoftext|>
TITLE: How to prove that $\int _0^\infty\frac{\text{arcsinh}^nx}{x^m}dx$ is a rational combination of zeta values?
QUESTION [12 upvotes]: For $n\ge m\ge 2$, define $$I(n,m):= \int _0^\infty\dfrac{\text{arcsinh}^nx}{x^m}dx$$ Computer algebra systems say that the indefinite integral can be expressed in terms of polylog functions (of rapidly increasing complexity), but I see no way of doing the limit $\to\infty$. On the other hand, I have found numerically that $I(n,m)$ can (conjecturally always) be written as a rational combination of zeta values, moreover this happens in a rather beautiful way.
Here are the values for small $m$
and then what seems to be the general formula.
Defining $\bar\zeta(n):=(2^n-1) \zeta(n)$, it seems like
$  I(n,2)=\dfrac{n!}{2^{n-1}}  \bar\zeta(n)$
$  I(n,4)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1 {6}[  4\bar\zeta(n-2)-\bar\zeta(n)]$
$  I(n,6)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1{120}[ 16\bar\zeta(n-4)-40\bar\zeta(n-2)+9\bar\zeta(n)]$
$  I(n,8)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1{5040}[ 64\bar\zeta(n-6)-560\bar\zeta(n-4) +1036\bar\zeta(n-2)-225\bar\zeta(n)]$
(you may want to stop here for a moment before reading on and look if the coefficients tell you something already)
and
$  I(n,3)=\dfrac{n!}{2^{n-1}}  \zeta(n-1)$
$  I(n,5)=\dfrac{n!}{2^{n-1}} \cdot\dfrac13[ \zeta(n-3)-\zeta(n-1)]$
$  I(n,7)=\dfrac{n!}{2^{n-1}} \cdot\dfrac2{45}[\zeta(n-5)- 5 \zeta(n-3)+4\zeta(n-1)]$
Well, in umbral notation introducing a formal variable $\bar Z$, for even $m=2k$ $$  I(n,2k)=\frac{n!}{2^{n-1}} \cdot\frac1{(2k-1)!}\bar Z^{n}\prod_{j=-k+1}^{k-2} \left(\frac2{\bar Z}-(2j+1)\right)$$ where each power $\bar Z^r$ is to be replaced by $\bar\zeta (r)$. E.g. for $m=6$, the "zeta polynomial" is related to $16z^4-40z^2+9=(2z-3)(2z-1)(2z+1)(2z+3)$.
Likewise for odd $m=2k+3$ (sic to make notation more elegant), $$  I(n,2k+3)=\frac{n!}{2^{n-1}} \cdot\frac{2^{2k}}{(2k+1)!}Z^{n}\prod_{j=-k}^{k}\left (\frac1{Z}-j\right)$$ where each power $Z^r$ is to be replaced by $\zeta (r)$. E.g. for $m=7$, the "zeta polynomial" is related to $z^5-5z^3+4z=(z-2)(z-1)z(z+1)(z+2)$.
We can rewrite the expression for even $m$ as follows to look amazingly similar to the one for odd $m$:
$$  I(n,2k+2)=\frac{n!}{2^{n-1}} \cdot\frac{2^{2k}}{(2k+1)!}\bar Z^{n}\prod_{j=-k+1/2}^{k-1/2} \left(\frac1{\bar Z}-j\right)$$ where in the product $j$ runs over the half-integers.

So far, this is all speculation, but once these patterns found, it seems obvious that there must be some deeper connection (which might even yield an elegant proof). Any insights? Has anybody encountered similar families of integrals?

EDIT for the record:
A similar pattern occurs for the integral
$$J(n,m):=\int _0^\infty\dfrac{x^n}{\cosh^{m}x}dx,$$ which features for even $m$ the Dirichlet eta function $\eta(s)=  \sum\limits_{n=1}^{\infty}\dfrac {(-1)^{n-1} }{ n^s}= \left(1-2^{1-s}\right) \zeta(s)$,
and for odd $m$ the Dirichlet beta function $\beta(s) = \sum\limits_{n=0}^\infty \dfrac{(-1)^n} {(2n+1)^s}$.
Using the same umbral notation as above for $\eta$ and $\beta$ respectively, we have for even $m=2k$
$$\bbox[10px,#cFF]{ \int _0^\infty\dfrac{x^n}{\cosh^{2k}x}dx=\frac{2^{2k-n}~n!}{(2k-1)!} \eta^{n} \prod_{j=1}^{k-1}\left ( j^2-\frac {  1}{ \eta^2}\right) } $$ and for odd $m=2k+1$
$$\bbox[10px,#cFF]{\int _0^\infty\dfrac{x^n}{\cosh^{2k+1}x}dx=2\frac{n!}{(2k)!}\beta^n \prod_{j=1}^{k} \left ( (2j-1)^2-\frac {  1}{ \beta^2}\right).}  $$
For example, if $m=8$, we get from $(x^2-1)(4x^2-1)(9x^2-1)=36x^6-49x^4+14x^2-1$ (note that these polynomials for even $m$ are essentially the same as the ones in the case of odd $m$ for the original integrals $I(n,m)$ above)
$$J(n,8)=\int _0^\infty\dfrac{x^n}{\cosh^{8}x}dx=\frac{n!}{7!~2^{n-8}}\Bigl[36~\eta(n)-49~\eta(n-2)+14~\eta(n-4)-\eta(n-6)\Bigr]$$
and from $(x^2-1)(9x^2-1)(25x^2-1)=225x^6-259x^4+35x^2-1$, we get for $m=7$
$$J(n,7)=\int _0^\infty\dfrac{x^n}{\cosh^{7}x}dx=2\frac{n!}{6!} \Bigl[225~\beta(n)-259~\beta(n-2)+35~\beta(n-4)-\beta(n-6)\Bigr]$$

REPLY [5 votes]: Letting $y=\text{arcsinh}\, x$ and integrating by parts, we have 
$$I(n,m)=\frac n{m-1}\,J(n-1,m-1)
$$
if $n\ge m\ge2$, where 
$$J(p,q):=\int_0^\infty\frac{y^p}{\sinh^qy}\,dy.$$
Formula 1.4.24.1 in Prudnikov--Brychkov--Marichev (PBM, Vol. 1, ISBN 5-9221-0323-7) implies
$$J(p,q)=
\frac{p(p-1)}{(q-1)(q-2)}\,J(p-2,q-2)
-\frac{q-2}{q-1}\,J(p,q-2)    
$$
if $p\ge q>2$.
That formula in PBM (which is reproduced in my answer at Is there a closed form of $\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$?) should be easy to obtain/check. 
Thus, we have a recursion to reduce the value of $I(n,m)$ to those of $J(p,1)$ and $J(p,2)$, that is, to those of $I(n,2)$ and $I(n,3)$.<|endoftext|>
TITLE: Regular CW complex arising from a Morse decomposition
QUESTION [9 upvotes]: Suppose $(M,g)$ is a Riemannian manifold equipped with a Morse function $f: M \rightarrow \mathbb R$. It's been shown that $f$ gives rise to a CW decomposition homeomorphic to $M$ under the generic assumption that $(f,g)$ satisfies Morse-Smale transversality. What additional conditions does one need to impose in this situation to ensure that the CW complex is regular? (Regular means the attaching maps are homeomorphisms.)
It is true that every CW complex is homotopy equivalent to a simplicial complex, and those are all regular, but I'm asking for regularity on the nose.
(Note: I originally asked the question on MSE.)
EDIT: I'm particularly interested in the case when Morse functions give rise to pseudo-regular CW complexes, by which I mean the incidence numbers of the cells all lie in $\{0,\pm 1\}$. By this definition, regular implies pseudo-regular, and pseudo-regular allows for a trivial boundary operator on the Morse-Smale complex.

REPLY [4 votes]: As I mentioned earlier today, a manifold like $S^2 \times S^2$ with a standard Morse function (giving one 0-cell, two 2-cells and one 4-cell) can never be regular for silly reasons -- the attaching map from the 4-cell is a map $S^3 \to S^2 \vee S^2$ which can't be an embedding for dimension reasons. 
This example shows there's no hope for anything generic satisfying your embedding condition.  A minimal necessary condition would be for the k-skeleton to be locally k-dimensional, i.e. every point is in a closed k-cell.   I'm not certain if that is a strong enough condition, but its a start.<|endoftext|>
TITLE: Finiteness Conjecture (New Doomsday conjecture)
QUESTION [17 upvotes]: This is completely out of curiosity. 
I wonder if there has been any recent progress reported on the Finiteness or New Doomsday conjecture, in the form of a talk, preprint or possibly a paper? 
Just to recall that the conjecture says that any $Sq^0$-family
$$\{x,Sq^0(x),Sq^0Sq^0(x),...,(Sq^0)^n(x),...\}$$
in the Adams spectral sequence $\mathrm{Ext}_A(\mathbb{F}_2,\mathbb{F}_2)$ converging to ${_2\pi_*^s}$ detects only finitely many elements. Here $A$ is the mod $2$ Steenrod algebra.
I also wonder if the conjecture has any equivalent formulations other than the one proposed by Minami? and if there is any written work on the relation between this conjecture and other important problems related to the Steenrod algebra (I am interested in the case of the prime $p=2$)?
I will appreciate any advise on this.
EDIT: I have recently looked at Bruner's Edinburgh talk, also cited in the below answer, and therein he mentions relations to some other problems - but the details of these relations I have not seen. Also, Minami in his paper in which he formulates the New Doomsday Conjecture, he mentions Wood's work on the Hit problem for $B\mathbb{F}_2^{\times n}$ as a new ingredient in his proof. In this case, what I like to know is that if the hit problem and the Finiteness Conjecture are related in a systematic way and whether or not if the details of this relation exists somewhere in the literature?!

REPLY [11 votes]: Two talks from 2011:

The
New Doomsday Conjecture and the motivic homotopy theory (Norihiko Minami) 
The
Finiteness Conjecture (Robert Bruner)<|endoftext|>
TITLE: A sum over characters of $S_{2n}$ and zonal spherical functions of $(S_{2n},H_n)$
QUESTION [6 upvotes]: The hyperoctahedral group $H_n$ can be seen as the centralizer of the permutation $(12)(34)\cdots (2n-1\,2n)$ in $S_{2n}$.  It has $2^nn!$ elements.
The quantities $$ \omega_\lambda(\pi)=\frac{1}{2^nn!}\sum_{h\in H_n}\chi_{2\lambda}(h\pi)$$ are called the zonal spherical functions of the Gelfand pair $(S_{2n},H_n)$. Here $2\lambda=(2\lambda_1,2\lambda_2,...)$ and $\chi$ are irreducible characters of $S$.
I have observed the following very nice result:
$$ \frac{2^nn!}{(2n)!}\sum_{\lambda\vdash n}\chi_{2\lambda}(1^{2n})\omega_\lambda(\pi)=\begin{cases} 1, &\pi\in H_n\\0, &\pi\notin H_n\end{cases}$$
Does anyone know how to prove this? (Orthogonality of characters is not straightforwardly useful) If so, can it be generalized to $\sum_{\lambda\vdash n}\chi_{2\lambda}(\mu)\omega_\lambda(\pi)$?

REPLY [8 votes]: The essential thing here is that the characters $\chi_{2\lambda}$ are exactly the irreducible constituents of the induced character $(1_{H_n})^{S_n}$. The result generalizes to an arbitrary subgroup $H\leq G$ of a finite group $G$ as follows: For $x$, $y\in G$, we have $\DeclareMathOperator{\Irr}{Irr}$
$$ \frac{ |H| }{ |G| } 
   \sum_{ \chi\in \Irr(G \mid 1_H) }
   \chi(x^{-1} ) \omega_{\chi}(y) 
   = \frac{ |x^G \cap Hy| }{ |x^G| }.  \tag{*}
$$
Here $\Irr(G \mid 1_H)$ denotes the set of irreducible constituents of $(1_H)^G$, which by Frobenius reciprocity is the set of $\chi\in \Irr(G)$ such that $1_H$ is a constituent of the restriction $\chi_H$.
As above, 
$$ \omega_{\chi}(y) = \frac{1}{|H|}
                      \sum_{h\in H} \chi(hy),
$$ 
but we do not have to assume that $(G,H)$ is a Gelfand pair. ($x=1$ and $y=\pi$ is your result.) 
Proof: Let
$$ e_H = \frac{1}{|H|} \sum_{h\in H} h \in \mathbb{C} H,  
$$
the central primitive idempotent of the group algebra belonging to $1_H$. Then $\omega_{\chi}(g) = \chi(e_H g)$ by defintion. 
If $1_H$ is not a constituent of the restriction $\chi_H$ for $\chi\in \Irr(G)$, then $\chi(e_Hg) = 0$ for all $g\in G$. 
Thus we can let run $\chi$ over all of $\Irr(G)$ in the sum in (*). 
We get
$$ \frac{ |H| }{ |G| } 
     \sum_{ \chi\in \Irr G } \chi( x^{-1} ) \omega_{\chi}(y)
   = \frac{ 1 }{ |G| } 
     \sum_{ h\in H } \sum_{ \chi\in \Irr G } 
                     \chi(x^{-1})\chi(hy), 
$$
and the second orthogonality relation for characters yields the result.<|endoftext|>
TITLE: When does the projective model structure on functors exist?
QUESTION [5 upvotes]: What this boils down to is: if $\mathcal{K}$ is cofibrantly-generated model category which permits the small object argument and $\mathcal{D}$ is a small category, then when does $\mathcal{K}^\mathcal{D}$ permit the small object argument? This might be trivial depending on what exactly it means to "permit the small object argument".
Since this is really a question about Quillen's small object argument, allow me to frame it in a minimal way. Let $\mathcal{K}$ be a cocomplete category and let $I$ be a set of morphisms in $\mathcal K$. Recall that the small object argument says:


If $(\mathcal K,I)$ permits the small object argument, then $(\mathrm{cof}(I),I^\square)$ forms a weak factorization system on $\mathcal K$.


Here $\mathrm{cof}(I)$ is the closure of $I$ under coproducts, retracts, cobase change, and transfinite composition, and $I^\square$ is the class of morphisms which have the right lifting property with respect to the maps of $I$.
Depending on the author, "permitting the small object argument" can mean various things. Some authors use a strong sense which I am not so interested in:

$(\mathcal{K},I)$ permits the small object argument if: Every object of $\mathcal{K}$ which is the domain of an arrow in $I$ is small.

(1) encompasses many important examples, including the weak factorization systems found in all combinatorial model categories. But (1) does not encompass the weak factorzation systems found in such important model cateogories as $\mathsf{Top}$ (in fact, the only small objects of $\mathsf{Top}$ are the discrete spaces), so it is sometimes necessary to use the more liberal sense which I am more interested in:

$(\mathcal{K},I)$ permits the small object if: Every object of $\mathcal{K}$ which is the domain of an arrow in $I$ is $\mathrm{cof}(I)$-small.

Recall that an object $A$ of a category $\mathcal{K}$ is called small (or presentable) if there exists a regular cardinal $\lambda$ such that the covariant hom-functor $\mathcal{K}(A,-)$ preserves $\lambda$-filtered colimits. Here, if $S$ is a collection of morphisms in $\mathcal{K}$, I say that $A$ is $S$-small if $\mathcal{K}(A,-)$ preserves $\lambda$-filtered colimits over functors that land in $S$.
Now, if $(\mathcal{K},I)$ permits the small object argument and $\mathcal{D}$ is a small category, I'm interested in knowing 

When does $(\mathcal{K}^\mathcal{D}, I^\mathcal{D})$ permit the small object argument?

Here $I^\mathcal{D} = \{ \mathcal{D}(D,-) \cdot i \mid D \in \mathcal{D}, i \in I\}$ where "$\cdot$" is a tensor, a.k. copower. Under sense (1), it's not hard to see that the answer is: always. But under sense (2), it's not so clear. I don't even see why the domains of $I^\mathcal{D}$ should be small with respect to the maps of $I^\mathcal{D}$ themselves, nevermind $\mathrm{cof}(I^\mathcal{D})$! (If this can be established, I think I can see that the class of maps with respect to which $A$ is $\lambda$-small is closed under retract and transfinite composition of length $< \lambda$, but I don't see why it should be closed under cobase change.) So if anyone can help me out with an argument, or at least a pointer to the literature, I'd appreciate it!
(The connection to the projective model structure is that if $(\mathcal{K},I)$ is a cofibrantly-generated model category permitting the small object argument and $\mathcal{D}$ is a small category, then I thought that $(\mathcal{K}^\mathcal{D},I^\mathcal{D})$ was supposed to form a cofibrantly generated model category. But the proof requires $(\mathcal{K}^\mathcal{D},I^\mathcal{D})$ to permit the small object argument. The nlab article on the projective model structure actually currently neglects to mention the hypothesis of permitting the small object argument entirely. The nlab's main reference is the 2nd appendix of Higher Topos Theory, where Lurie simply requires requires that $\mathcal{K}$ be combinatorial, so that (1) holds. Emily Riehl also uses sense (1) in her book. She mentions that Hovey uses sense (1) in his book, but I can't find a discussion of the projective model structure in Hovey. In fact, I was surprised at how difficult it was to find an account of the projective model structure! I also looked in Hirschhorn, Goerss-Jardine, and Quillen's original monograph.)

REPLY [5 votes]: I found it in Hirschhorn's book. It's in 11.6 Diagrams in a cofibrantly generated model category. The answer is always, provided K
is cofibrantly generated (Theorem 11.6.1). For him (as well as for Hovey), the small object argument is (2), actually replacing cof with cell (Definitions 10.5.15 and 10.5.12).<|endoftext|>
TITLE: Principal bundle approach to general relativity
QUESTION [7 upvotes]: I am curious if there is any literature (texbooks, mainly, but articles will do too, though I don't have easy access to any paid journal) that deals with general relativity by using Ehresmann connections on the (orthonormal) frame bundle in a rigorous manner, rather than Koszul connections on the tangent bundle, and develops calculus directly on the frame bundle, rather than on spacetime itself.
Basically, I am interested in this, for the sake of being interested in it, however I have hopes that I might be able to use this formalism to attack some problems in my research.
To clarify a bit more:

I am not looking for local tetrad formalism, however I am hoping that if what I am asking for exists, it will resemble local tetrad formalism a lot, but with globally defined quantities, as opposed to just local ones.
Although I'll take any materials on the subject gladly, I really would prefer if the resource treated lagrangian formalism. I am mostly aware of how Ehresmann connections on the frame bundle work, but I have absolutely no clue how to do langrangian formalism with it, and this is an absolute necessity for my work.
I have Kobayashi & Nomizu for necessary extra mathematical details, but I'd prefer this resource to be generally self-contained.

REPLY [2 votes]: You might enjoy Bleeker's book "Gauge Theory and Variational Principles" (http://www.amazon.com/Gauge-Theory-Variational-Principles-Physics/dp/0486445461). His focus is definitely more on particle theory than relativity, though there is a good section on it towards the end of the book. He spends a great deal of time dealing with Lagrangians in the abstract with the occasional example (Electromag and QED I think).
It's what I read after Kobiyashi and Nomizu to pick up the physical side of principal fiber bundles. And it is one of my favorites. It is very dense, but (almost) everything is spelled out is gratuitous detail (including a lot of mappings that are left implied by others).
You might also like to take a look at Kaluza-Klein theory (https://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory), if you haven't already.<|endoftext|>
TITLE: Positivity of the alternating sum of indices for boolean interval of finite groups
QUESTION [6 upvotes]: Let $G$ be a finite group and $H$ a subgroup such that the interval $[H,G]$ is a boolean lattice.     
Let $L_1, \dots , L_n$ be the maximal subgroups of $G$ containing $H$.
Let the alternative sum of the interval $[H,G]$ defined as follows:
$$\chi([H,G]):= \sum_{r=0}^n (-1)^{r} \sum_{  \ i_1 < i_2 < \cdots < i_r } [L_{i_1} \wedge \cdots \wedge L_{i_r}: H] $$ 
Notation: $L_{i_1} \wedge \cdots \wedge L_{i_r} = G$ for $r=0$.    
Theorem: $\chi([H,G]) > 0$.
Proof: Observe that
$$\chi([H,G]) = \frac{\vert G \vert - \vert \bigcup_i L_i \vert}{\vert H \vert} $$  but a boolean lattice is distributive so by a result of Oystein Ore (see here) $\exists g \in G$ with $\langle H,g \rangle = G$, which precisely means that $g \not \in L_i \  \forall i$, and so $\chi([H,G])> 0$ $\square$    
Let $K_1, \dots , K_n$ be the minimal overgroups of $H$.
Let the dual alternative sum of the interval $[H,G]$ defined as follows:
$$\hat{\chi}([H,G]):= \sum_{r=0}^n (-1)^{r} \sum_{  \ i_1 < i_2 < \cdots < i_r } [G: K_{i_1} \vee \cdots \vee K_{i_r}] $$
Notation: $K_{i_1} \vee \cdots \vee K_{i_r} = H$ for $r=0$.    
Question: Is $\hat{\chi}([H,G]) > 0$ ?
Remark: after GAP checking, it is true for $[G:H]<32$ (recall that $[H,G]$ is assumed boolean).

REPLY [4 votes]: UPDATE:  The original poster of the question, together with Mamta Balodi, have shown that the labeling I suggest below is an EL-labeling if and only if group (product) complements coincide with lattice complements on the given Boolean interval in the subgroup lattice.  The latter condition does not always hold; in the same paper they present examples where it does not.  The Cohen-Macaulay and/or shellability questions still seem to be open in general.
Unless I'm mistaken, this is (up to sign) the Möbius number of the poset $C(H,G)$ formed by all cosets of all groups on $[H,G]$, together with an artificial $\hat{0}$ element.  This is because the Möbius number of a Boolean interval is $\pm 1$.  Thus, your sum is really
$$ (-1)^n \cdot \sum_{H \leq K \leq G} \mu(K,G) \cdot [G:K].$$ 
The Möbius number observation I made now follows from definition.
I believe that $C(H,G)$ is Cohen-Macaulay.  Since the Möbius number of a Cohen-Macaulay poset of height $n$ is positive or negative depending on whether $n$ is even or odd, this would give you the desired positivity.  
There are lots of ways to prove a poset to be Cohen-Macaulay.  One way would be to construct a dual EL-labeling.  I think that you can do this as follows, but have not written down a proof.  (Edit:  the following works some of the time, but not all of the time; see UPDATE above.)  Label the bottom edges $\hat{0} \lessdot Hh$ with 0.  Let $M_i$ be the join of all $K_j$ except for $K_i$, so that $M_i$ is maximal in $[H,G]$ for any $i$.  If $Y = X \vee K_i$, then label $Xh \lessdot Yh$ with $-i$ if $Xh = Yh \wedge M_i$, and label $Xh \lessdot Yh$ with $+i$ otherwise.
Now verify (assuming it is actually true) that every interval has a unique increasing chain that is lexicographically first, as required for an EL-labeling.
The labeling that I suggest here is extremely similar to that constructed in Section 4.2 of my paper "Cubical convex ear decompositions".  The observation that the quantity you're interested in is a Möbius number is attributed to Bouc, but is in a paper of Ken Brown, "The coset poset and probabilistic zeta function of a finite group".  The latter is rather well-written, and also discusses some other ways of computing the Möbius function of the full coset lattice.<|endoftext|>
TITLE: What's the probability distribution of a deterministic signal or how to marginalize dynamical systems? (functional integrals in probability theory)
QUESTION [6 upvotes]: Because I still have no idea how it is possible for me to write down seemingly important equations ... that don't make any sense (at least for me) and because I haven't got any helpful comment so far, I'll be happy to offer a +100 bounty (that is, almost all my reputation!) not for a definitive anwser to this weird question but only for any serious, relevant feedback, thought, opinion or advice. 
In many signal processing calculations, the prior probability distribution of the theoretical signal of interest (not the noisy experimental signal) is required. Here is the concrete problem from which my question has arisen:
Is there a Bayesian theory of deterministic signal? Prequel and motivation for my previous question
In random signal theory (à la Shannon), this probability distribution typically 
is a stochastic process, e.g. an i.i.d. stochastic process if you are a frequentist or an exchangeable stochastic process if you are a Bayesian, in the most basic cases.
What do such probability distributions become in deterministic signal theory/dynamical system theory?, that is the question.
To make it simple, consider a discrete-time real deterministic signal 
$ s\left( {1} \right),s\left( {2} \right),...,s\left( {M} \right) $
For instance, it can be obtained by sampling a continuous-time real deterministic signal.
By the standard definition of a discrete-time deterministic dynamical system, there exists:

a phase space $\Gamma$, e.g. $\Gamma  \subset  \mathbb{R} {^d}$, $\Gamma  = \left[ {0,1} \right]$, etc.
a state-space equation $f:\Gamma  \to \Gamma $  such as $z\left( {m + 1} \right) = f\left[ {z\left( m \right)} \right]$;
an output or observation equation $g:\Gamma  \to \mathbb{R}$ such as $s\left( m \right) = g\left[ {z\left( m \right)} \right]$;
an initial condition $ z\left( 1 \right)\in \Gamma $ in the domain of definition of $f$.

Hence, by definition we have
$\left[ {s\left( 1 \right),s\left( 2 \right),...,s\left( M \right)} \right] = \left\{ {g\left[ {z\left( 1 \right)} \right],g\left[ {f\left( {z\left( 1 \right)} \right)} \right],...,g\left[ {{f^{M - 1}}\left( {z\left( 1 \right)} \right)} \right]} \right\}$
or, in probabilistic notations
$p\left[ {\left. {s\left( 1 \right),s\left( 2 \right),...,s\left( M \right)} \right|z\left( 1 \right),f,g,\Gamma ,d} \right] = \prod\limits_{m = 1}^M {\delta \left\{ {g\left[ {{f^{m - 1}}\left( {z\left( 1 \right)} \right)} \right] - s\left( m \right)} \right\}} $
Therefore, by "total probability and the product rule", the "marginal joint prior probability distribution" for a discrete-time deterministic signal conditional on phase space $\Gamma$ and its dimension $d$ formally/symbolically writes
$p\left[ {\left. {s\left( 1 \right),s\left( 2 \right),...,s\left( M \right)} \right|\Gamma ,d} \right] = \int\limits_{{\mathbb{R}^\Gamma }} {{\text{D}}g\int\limits_{{\Gamma ^\Gamma }} {{\text{D}}f\int\limits_\Gamma  {{{\text{d}}^d}z\left( 1 \right)\prod\limits_{m = 1}^M {\delta \left\{ {g\left[ {{f^{m - 1}}\left( {z\left( 1 \right)} \right)} \right] - s\left( m \right)} \right\}p\left( {z\left( 1 \right),f,g} \right)} } } } $
Should phase space $\Gamma$ and its dimension $d$ be also unknown a priori, they should be marginalized as well so that the most general "marginal prior probability distribution" for a discrete-time deterministic signal I'm considering formally/symbolically writes
$p\left[ {s\left( 1 \right),s\left( 2 \right),...,s\left( M \right)} \right] = \sum\limits_{d = 1}^{ + \infty } {\int\limits_{\wp \left( {{\mathbb{R}^d}} \right)} {{\text{D}}\Gamma \int\limits_{{\mathbb{R}^\Gamma }} {{\text{D}}g\int\limits_{{\Gamma ^\Gamma }} {{\text{D}}f\int\limits_\Gamma  {{{\text{d}}^d}z\left( 1 \right)\prod\limits_{m = 1}^M {\delta \left\{ {g\left[ {{f^{m - 1}}\left( {z\left( 1 \right)} \right)} \right] - s\left( m \right)} \right\}p\left( {z\left( 1 \right),f,g,\Gamma ,d} \right)} } } } } } $
where ${\wp \left( {{\mathbb{R}^d}} \right)}$ stands for the powerset of ${{\mathbb{R}^d}}$.
Dirac's $\delta$ distributions are certainly welcome to "digest" those very high dimensional "integrals". However, we may also be interested in "probability distributions" like
$p\left[ {s\left( 1 \right),s\left( 2 \right),...,s\left( M \right)} \right] \propto \sum\limits_{d = 1}^{ + \infty } {\int\limits_{\wp \left( {{\mathbb{R}^d}} \right)} {{\text{D}}\Gamma \int\limits_{{\mathbb{R}^\Gamma }} {{\text{D}}g\int\limits_{{\Gamma ^\Gamma }} {{\text{D}}f\int\limits_\Gamma  {{{\text{d}}^d}z\left( 1 \right)\int\limits_{{\mathbb{R}^ + }} {{\text{d}}\sigma {\sigma ^{ - M}}{e^{ - \sum\limits_{m = 1}^M {\frac{{{{\left\{ {g\left[ {{f^{m - 1}}\left( {z\left( 1 \right)} \right)} \right] - s\left( m \right)} \right\}}^2}}}{{2{\sigma ^2}}}} }}p\left( {\sigma ,z\left( 1 \right),f,g,\Gamma ,d} \right)} } } } } } $
Please, what can you say about those important "probability distributions" beyond the fact that they should better not be invariant by permutation of the time points, i.e. not finitely De Finetti-exchangeable, otherwise the chronological order, that is the time would be lost (conjecture)?
What can you say about such strange looking "functional integrals" (for the state-space and output equations $f$ and $g$) and even "set-theoretic integrals" (for phase space $\Gamma$) over sets having cardinal at least ${\beth_2}$? Are they already well-known in some branch of mathematics I do not know yet or are they only abstract nonsense?
Clearly, the noninformative case is the most important one. Hence, a definitive answer to my question could be something like this:
If $p\left( {z\left( 1 \right),f,g} \right)$ is the "improper non-informative prior probability distributions" over $\Gamma  \times {\Gamma ^\Gamma } \times {\mathbb{R}^\Gamma }$ $p\left( {z\left( 1 \right),f,g} \right) \propto 1$ 
then the "marginal probability distribution" 
$p\left[ {\left. {s\left( 1 \right),s\left( 2 \right),...,s\left( M \right)} \right|\Gamma ,d} \right] = \int\limits_{{\mathbb{R}^\Gamma }} {{\text{D}}g\int\limits_{{\Gamma ^\Gamma }} {{\text{D}}f\int\limits_\Gamma  {{{\text{d}}^d}z\left( 1 \right)\prod\limits_{m = 1}^M {\delta \left\{ {g\left[ {{f^{m - 1}}\left( {z\left( 1 \right)} \right)} \right] - s\left( m \right)} \right\}} } } } $ 
is the improper uniform probability distribution over ${\mathbb{R}^M}$.
Your are free to restrict the sets of the state-space equations and output equations if necessary. However, contrary to what is suggested below, for a discrete-time dynamical system the state space equation $f$ needs not be continuous (the classical counterexample for phase space $\Gamma  = \left[ {0,1} \right]$ is the Bernoulli shift that is discontinuous at $1/2$). Hence, a priori ${\beth_2}$ of them must be marginalized out. 
Those beasts look very important to me because there are many problems of interest where we know a priori that the signal is deterministic but we don't known the underlying dynamical system (and output equation) and there is little hope to ever know it (generally speaking, dynamical systems identification is very delicate and difficult). Hence the deterministic model is unknown and not operationnal and by definition it is not legitimate to introduce a (objectively)  stochastic one. So how to handle and process those signals properly? In theory, we perfectly know how to do it: just marginalize over all possible dynamical systems. But that does not make sense mathematically, so that it seems that we don't know how to model and process deterministic signals of unknown origin.
Thanks.

REPLY [2 votes]: Classical Bayesian analysis rests on first chosing a prior measure $m$, either finite (proper) or infinite (improper), then deriving the posterior probability of an event $A$ conditional on observed $B$ as $Pr(A|B)=m(A\cap B)/m(B)$.
As I said in my first answer, this is possible for $[0,1]^{[0,1]}$ with the product Lebesgue measure, but not for $\mathbb R^{\mathbb R}$ because there is no such thing as "product Lebesgue measure" there, albeit an improper prior.
But we might consider a wider concept of prior, that of conditional probability space axiomatised by Alfred Renyi (1955): a family of ordinary probability spaces $(\Omega,\mathcal A,P_B)_{B\in \mathcal B}$ (where $P_B(A) $ represents $Pr(A|B)$), satisfying some compatibility condition.
In the case of $\Omega=\mathbb R^{\mathbb R}$, the family $\mathcal B$ of conditioning events could be that of products $\prod_{z\in\mathbb R}E_z$ where all $E_z$s have finite Lebesgue measure. For such a $B$ the proba $P_B$ will be defined as the product of normalised Lebesgue measures $\mathcal L(\cdot\cap E_z)/\mathcal L(E_z)$. The $\sigma$-algebra $\mathcal A$ is that of sets $\prod_{z\in\mathbb R}E_z$ where (contrary to members of $\mathcal B$) all but uncountably many $E_z=\mathbb R$.
While maybe (I don't know) conceptually new for Bayesian theory, I think this doesn't change much for what is searched here, a foundation for Bayesian processing of signals arising from a dynamical system: you still need restrictions to define probabilities...<|endoftext|>
TITLE: Unexpected applications of transcendental number theory?
QUESTION [35 upvotes]: In the last pages of "Equations Différentielles à points singuliers réguliers", Deligne provides a proof, attributed to Brieskorn, of the so-called local monodromy theorem (on the quasi-unipotence of the monodromy operator acting on the cohomology of a degenerating family of complex algebraic varieties).
The argument uses the base change compatibility and regularity of the Gauss-Manin connection to reduce the problem to the following statement: if $M$ is a complex square matrix such that, for every field automorphism $\sigma$ of $\mathbb{C}$, $\exp(2\pi i M^{\sigma})$ is conjugated to a matrix with integer coefficients, then $\exp(2\pi i M)$ is quasi-unipotent. This in turn is a simple consequence of Gelfond-Schneider theorem!
I always found this proof quite surprising and I was wondering if there aren't other unexpected applications of transcendental numbers out there.

REPLY [11 votes]: Here is an application of transcendental number theory to differential geometry that I think would count as unexpected to all but a small group of experts in the area (who would probably view the application as being a very natural one).
Let $G$ be a connected absolutely simple real algebraic group and $\mathcal G=G(\mathbb R)$ the corresponding real Lie group. We'll call $\mathcal G$ absolutely simple. Let $\mathcal K$ be a maximal compact subgroup of $\mathcal G$ and $\mathfrak X=\mathcal K\backslash \mathcal G$ the symmetric space of $\mathcal G$.
If $\Gamma_1$ and $\Gamma_2$ are discrete subgroups of $\mathcal G$ then denote by $\mathfrak X_{\Gamma_1}=\mathfrak X/\Gamma_1$ and $\mathfrak X_{\Gamma_2}=\mathfrak X/\Gamma_2$ the associated locally symmetric spaces. We say that such a locally symmetric space is arithmetically defined if the corresponding discrete subgroup of $\mathcal G$ is arithmetic.
In their paper Weakly commensurable arithmetic groups and isospectral locally symmetric spaces, Gopal Prasad and Andrei Rapinchuk prove a number of interesting results about the spectral theory of such locally symmetric spaces, for instance:
Theorem (Prasad and Rapinchuk) Let $\mathfrak X_{\Gamma_1}$ and $\mathfrak X_{\Gamma_2}$ be two arithmetically defined locally symmetric spaces of the same absolutely simple real Lie group $\mathcal G$. If they are isospectral, then the compactness of one of them implies the compactness of the other.
Theorem (Prasad and Rapinchuk) Any two arithmetically defined compact isospectral locally symmetric spaces of an absolutely simple real Lie group of type other than $A_n (n > 1)$, $D_{2n+1} (n\geq 1)$, $D_4$ and $E_6$, are commensurable to each other.
While these results are unconditional for rank one locally symmetric spaces, for spaces of higher rank the results depend on Schanuel's conjecture in transcendental number theory. 
As I mentioned above, I think that this geometric application would probably come as a complete surprise to non-experts, whereas to people working the field it is extremely natural, the idea being that the Laplace spectrum of such a space is related to the geodesic length spectrum (by results of Duistermaat and Guillemin), and the lengths of geodesics are in turn given by logarithms of algebraic numbers.<|endoftext|>
TITLE: Symmetry Group of a Polynomial
QUESTION [6 upvotes]: Given a polynomial $P \in \mathbb{Z}[X_1,\ldots,X_n]$, is there a poly-time algorithm which computes the group of permutations of variables that leaves $P$ unchanged? (Clearly, the trivial $O(n!)$-time algorithm is not poly-time.)

REPLY [14 votes]: If the input has fully-expanded polynomials, then this is equivalent to graph isomorphism.
In one direction, given a graph, create a variable for each vertex and consider the polynomial $\prod_{vw\in E(G)} (1 + x_vx_w)$. Unique factorization assures that this representation is reversible.
In the other direction, given a polynomial there are multiple ways of encoding it as a graph. I'll use a coloured bipartite graph, which can be converted to a uncoloured simple graph by standard means. Vertex colors are numbers that appear as coefficients and edge colours are numbers that appear as powers of variables. Make one vertex for each variable. For each monomial, make a new variable coloured by the coefficient and join it to each variable it contains by an edge coloured by the degree of that variable in the monomial. (Eg., monomial $6x^2y$ is a vertex coloured "6" joined to vertex "$x$" by an edge of colour "2" and to vertex "$y$" by an edge of colour "1".) This representation is also reversible.
I'll explain how the second example works, assuming you have an oracle that provides one (colour-preserving) isomorphism between two coloured graphs or says that there is none. Let $P_1,P_2$ be two polynomials and $G(P_1),G(P_2)$ their corresponding graphs. The construction ensures that any isomorphism between $P_1$ and $P_2$ can be converted to an isomorphism between $G(P_1)$ and $G(P_2)$, and vice versa. Now consider one polynomial $P$ and two of its variables $x,y$. Say $n$ is the total degree of $P$.  Then $P$ has an automorphism (symmetry as in the original question) that maps $x$ onto $y$ iff $G(P+x^{n+1})$ is isomorphic to $G(P+y^{n+1})$, and the oracle will give you such an automorphism if there is one. By continuing this process by marking more variables, you can build up a strong generating set for the automorphism group of $P$. For example, comparing $G(P+x^{n+1}+u^{n+2})$ to $G(P+y^{n+1}+v^{n+2})$ will get you an automorphism, if any, that maps $x$ onto $y$ and $u$ onto $v$.<|endoftext|>
TITLE: Essential dimension and the moduli space of abelian varieties
QUESTION [5 upvotes]: The following problem is listed here: http://www-personal.umich.edu/~erman/Papers/Questions2.pdf and attributed to Vistoli:
Let $\mathcal A_g$ denote the moduli stack of principally polarized abelian varieties over a field $k$, let $A_g$ denote the associated coarse moduli space, and $K(A_g)$ its function field.  What is the minimal degree of a finite extension $L/K(A_g)$ so that there exists a commutative diagram:
$$\begin{matrix}\operatorname{Spec}L&\to&\mathcal A_g\cr\downarrow&&\downarrow&&?\cr\operatorname{Spec}K(A_g)&\to&A_g\end{matrix}$$
The paper "Essential dimension of moduli of curves and other algebraic stacks" by Brosnan--Reichstein--Vistoli--Fakhruddin is listed as a reference.  The last section of this paper (by Fakhruddin) calculates the essential dimension of the stack $\mathcal A_g$.  My question is: how is the notion of essential dimension of a stack related to the problem posed by Vistoli?
For reference, the essential dimension of a functor $F:\mathfrak F\mathfrak i\mathfrak e\mathfrak l\mathfrak d\mathfrak s_{/k}\to\mathfrak S\mathfrak e\mathfrak t\mathfrak s$ is the minimal integer $n$ such that every element of  $F(K/k)$ comes from an element of $F(L/k)$ for some $L/k$ of transcendence degree at most $n$ over $k$ (if no such $n$ exists, then the essential dimension of $F$ is infinite).  The essential dimension of a functor $F:\mathfrak F\mathfrak i\mathfrak e\mathfrak l\mathfrak d\mathfrak s_{/k}\to\mathfrak G\mathfrak r\mathfrak o\mathfrak u\mathfrak p\mathfrak o\mathfrak i\mathfrak d\mathfrak s$ is just the essential dimension of $\left|F\right|$ (where $\left|\cdot\right|:\mathfrak G\mathfrak r\mathfrak o\mathfrak u\mathfrak p\mathfrak o\mathfrak i\mathfrak d\mathfrak s\to\mathfrak S\mathfrak e\mathfrak t\mathfrak s$ denotes taking isomorphism classes).

REPLY [4 votes]: The two notions are related using Theorems 4.1 and 6.1 of the paper of Brosnan, Reichstein and Vistoli: 
Theorem 6.1 reduces the computation of the essential dimension of the stack to that of the generic gerbe $\mathcal{X}_g$. Theorem 4.1 says that the essential dimension of $\mathcal{X}_g$ is $\mathrm{cd}(\mathcal{Y}_g) - 1$, where $\mathrm{cd}(\mathcal{Y}_g)$ is the canonical dimension of the Brauer--Severi variety $\mathcal{Y}_g$ associated to $\mathcal{X}_g$. Since the index of $\mathcal{Y}_g$ is a power of $2$ (as the order is $2$), the last line of Theorem 4.1 implies that $\mathrm{cd}(\mathcal{Y}_g) + 1$ is equal to the index, which is by definition the degree of the field $L$.<|endoftext|>
TITLE: Have there been any updates on Mochizuki's proposed proof of the abc conjecture?
QUESTION [67 upvotes]: In August 2012, a proof of the abc conjecture was proposed by Shinichi Mochizuki. However, the proof was based on a "Inter-universal Teichmüller theory" which Mochizuki himself pioneered. It was known from the beginning that it would take experts months to understand his work enough to be able to verify the proof. Are there any updates on the validity of this proof?

REPLY [29 votes]: Today (3 April 2020) his papers have been accepted for publication on RIMS journal.
https://www.nature.com/articles/d41586-020-00998-2<|endoftext|>
TITLE: Which groups are Galois over some p-adic field?
QUESTION [13 upvotes]: Suppose I have some finite $p$-group $G$, or a little extension of it. 
How do I know if there exists a prime $l$ and a finite extension $K$ of $\mathbb{Q}_l$ such that $G$ is the Galois group of some finite Galois extension  $M/K$?
For instance, I am interested in the case when $G$ is the semidirect product of a cyclic group of order $8$ by its automorphism group, a direct product of two groups of order $2$.

REPLY [8 votes]: When $p \neq \ell$, if $N/K$ has Galois group $G$ then $N/K$ is tamely ramified. It follows that $N = K(\sqrt[e]{\pi}, \zeta)$ where $e$ is the ramification degree of $N/K$, $\pi$ is some uniformizer of $K$, $\zeta$ is a primitive $(p^f-1)$st root of unity and $f$ is the residue degree. It is not hard to work out $G$ given $e$, $f$ and $\pi$. From this one can work out the list of possible $G$. I'll leave the details to the reader.
The more interesting case is when $p=\ell$, which I now assume. Suppose $K/\mathbb{Q}_p$ is finite, and let $K^p$ be the maximal pro-$p$ extension of $K$ (i.e. the compositum of all finite normal $N/K$ with $(N:K)$ a $p$-power). Then $G_{K,p} := \operatorname{Gal}(K^p/K)$ is known in the literature, and can be written down in terms of generators and relations. Given this, then there is an extension of $K$ with Galois group $G$ (a $p$-group) if and only if there is a surjective homomorphism $G_{K,p} \to G$. For specific $G$, one could use a computer algebra system (like MAGMA) to find all such homomorphisms, and deduce the number of extensions with Galois group $G$.
Note that the theory behind $G_{K,p}$ uses Galois cohomology, and as such is somewhat nonconstructive. In particular, given a homomorphism $\varphi: G_{K,p} \to G$ I don't know of an easy method to produce the corresponding extension $N/K$ (i.e. the fixed field of $\ker \varphi$).
When $K$ does not contain a primitive $p$th root of unity, then $G_{K,p}$ is a free pro-$p$ group with $n+1$ generators where $n=(K:\mathbb{Q}_p)$.
When $K$ does contain a primitive $p$th root of unity, then $G_{K,p}$ is a pro-$p$ group with $n+1$ or $n+2$ generators and one relation. It is known as a Demushkin group and these have been fully classified, although the details are too much for a MO answer. The result of this is that given $K$, one can easily compute $G_{K,p}$. For example, $$D_{\mathbb{Q}_2,2} = \langle a, b, c : a^2 b^4 [b,c] \rangle.$$
The following articles will tell you more:

H Koch. Galois theory of $p$-extensions. (On the general theory of $p$-extensions, including the full answer in the $p \neq \ell$ case and much of the $p = \ell$ case. See chapter X in particular.)
J. Labute. Classification of Demushkin groups. PhD thesis, 1965. (Fully describes Demushkin groups in terms of generators and relations.)
M. Yamagishi. On the number of Galois $p$-extensions of a local field. Proc. Amer. Math. Soc., 1995. (A neat summary of the previous article, and one algorithm for computing the number of extensions with a given Galois $p$-group.)<|endoftext|>
TITLE: open subgroup scheme closed
QUESTION [6 upvotes]: Let $G/S$ be a group scheme and $H \leq G$ an open subgroup scheme. Is $H \subseteq G$ closed? I want to apply this to $G^0 \leq G$ (see SGA 3, VI_B, Théorème 3.10) for $G$ commutative.
(*) If $S = \mathrm{Spec}(K)$, this is proven in http://jmilne.org/math/CourseNotes/iAG200.pdf Proposition 1.27 by the usual argument: the complement is the disjoint open union of the cosets of $H$.
If $G/H$ is representable by a group scheme and everything is separated, apply Exercise 1(ii) of http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf to $\pi: G \to G/H$. But I want to prove this without having to assume $G/H$ being representable. It seems that if $H \subseteq G$ is closed, $G/H$ is representable (under certain conditions) ...
(This is a cross-post from https://math.stackexchange.com/questions/1670288/open-subgroup-scheme-closed)
Edit: Assume $S$ the spectrum of a DVR. Then, by (*), $G^0_\eta \subseteq G_\eta$ and $G^0_s \subseteq G_s$ are closed. Now, if everything is separated and flat, does it follow that $G^0 \subseteq G$ is closed?

REPLY [3 votes]: For $k=\mathbb{C}$ (or any field of characteristic $\ne2$) let $G$ be the subgroup scheme of $SL(2)\times\mathbb{A}^1$ consisting of all matrices of the form $\begin{pmatrix}x&sy\\y&x\end{pmatrix}$. The defining equation is $x^2-sy^2=1$ which shows that $G$ is an irreducible variety. For $s\ne0$ the fiber is a torus, hence connected. The fiber over $s=0$ is $\mathbf{G}_a\times\{\pm1\}$, hence disconnected. Thus $G^0$ is $G$ minus $\{s=0,x=-1\}$, hence open but not closed.<|endoftext|>
TITLE: Every self-adjoint trace class operator on $L^2$ has integral kernel
QUESTION [7 upvotes]: I have asked this question on MSE but did not receive an answer. I thought I could try it here.
Let $T$ be a self-adjoint trace-class operator on $L^2(\mathbb{R})$. Is is true that it can be represented as an integral operator. 
I thought the kernel would be $$k_T(x,y) =\sum_{i=1}^\infty \lambda_i \phi_i(x) \bar\phi_i(y).$$
Here $\{\phi_i\}$ is an eigenbasis of $T$, i.e. $T=\sum_i \lambda_i |\phi_i\rangle\langle\phi_i|$. Then, we have $$\int k_T(\cdot,y) f(y) = \int\sum_i \lambda_i \phi_i(\cdot) \bar\phi_i(y) f(y) dy = \sum_i \lambda_i \phi_i \langle \phi_i, f\rangle=\sum_i \lambda_i |\phi_i\rangle\langle\phi_i|f\rangle = Tf.$$
Is this correct?

REPLY [5 votes]: Yes, this is correct. Actually, "self-adjoint trace-class" is more than you need; any Hilbert-Schmidt operator can be represented as an integral operator. The Hilbert-Schmidt operators from $L^2(X)$ to $L^2(Y)$ are precisely the integral operators with kernel in $L^2(X\times Y)$ (at least for $\sigma$-finite $X$ and $Y$). This should be in a standard reference, probably Dunford-Schwartz, for example.<|endoftext|>
TITLE: sum of squares of Schur polynomials indexed over partition valued functions on a set
QUESTION [5 upvotes]: Fix a finite set $X$ and two natural numbers $d$ and $n$.
For a partition $\lambda$ and a number $d$ denote by $s_\lambda^d(x_1,\dots,x_d)$ the Schur polynomial in $d$-many variables $x_1,\dots,x_d$. Denote by $\mathcal P_n(X)$ the set of partition-valued functions on $X$ of total size $n$, i.e. the set of functions $\lambda\colon X\to\{\text{partitions}\}$ such that $\|\lambda\|:=\sum_{x\in X}|\lambda(x)|=n$.
Let $a\colon X\to \mathbb{N}\backslash\{0\}$ be a function on $X$. EDIT: I am mostly interested in the case where $a$ is the constant function with value $1$.
I am trying to evaluate the expression
$$F_X(d,a):=\sum_{\lambda\in \mathcal P_n(X)}\left(\prod_{x\in X}s^d_{\lambda(x)}(a(x),\dots,a(x))\right)^2.  $$  

EDIT: 

I have now good reason to believe that if we take for $a$ the constant function with value $1$ then we have $$F_X(d,a)=\left(\binom{d^2|X|}{n}\right)$$
Here $\left(\binom{m}{n}\right)$ denotes the multiset number
More generally I am hoping that if $G$ is a finite group, $X$ is the set of irreducible characters (over $\mathbb C$) of $G$ and $a$ maps each character to its dimension then we have $$F_X(d,a)=\left(\binom{d^2|G^{\mathrm{ab}}|}{n}\right)$$ Here $G^\mathrm{ab}$ is the abelianisation of $G$. This reduces to case 1 if $G$ is taken to be an abelian group.

EDIT2: my conjecture 2 above is totally wrong (as can be seen by hand in small examples like $G=S_3$) and I have no good substitute hypothesis. I am now mostly interested in a proof of point 1 (where $a$ is the constant function $1$)

The special case where $X=\{\star\}$ is a one-element set can be done using the identity
$$\prod_{i,j}(1-x_iy_j)^{-1}=\sum_n\sum_{|\lambda|=n} s_\lambda(x_1,\dots)s_\lambda(y_1,\dots)$$
by putting $x_i=y_j=t$ for $i,j=1,\dots,d$ (and zero otherwise) and picking out the coefficient of $t^{2n}$ obtaining
$$c^d_nt^{2n}=\sum_{|\lambda|=n}s^d_\lambda(t,\dots,t)^2,$$
where $c^d_n$ is the coefficient of $t^n$ in $(1-t)^{-d^2}$ and can be calculated to be the multiset number $\left(\binom{d^2}{n}\right)$.
All in all we obtain $F_{\{\star\}}(d,a)=\left(\binom{d^2}{n}\right)\cdot a(\star)^{2n}$

REPLY [2 votes]: assume $a$ is the constant function with value $1$.
as in the case of $X=\{\star\}$ we have the equality
$$(1-t^2)^{-d^2|X|}=\left(\sum_{\lambda\in \{\mathrm{partitions}\}}s^d_\lambda(t,\dots,t)^2\right)^{|X|}=\sum_{\lambda\colon X\to\{\mathrm{partitions}\}}\prod_{x\in X}s_{\lambda(x)}^d(t,\dots,t)^2$$
and picking out the monomial $t^2$ gives
$$\left(\binom{d^2|X|}{n}\right)t^{2n}=\sum_{\lambda\in P_n(X)}\prod_{x\in X}s_{\lambda(x)}^d(t,\dots,t)^2$$
The result follows.<|endoftext|>
TITLE: comparing homology of a space and homology of the classifying space of its fundamental group
QUESTION [6 upvotes]: Let $X$ be a (connected) closed $n$-manifold and $G=\pi_1(X)$ be the fundamental group of $X$. There is a classifying map $f: X \rightarrow K(G, 1)$ which induces an isomorphism on $\pi_1$. I would like know when the map $f_*: H_n(X, \mathbb{Z}) \rightarrow H_n(K(G,1), \mathbb{Z})$ is injective, or even when $f_*: H_n(X, \mathbb{Q}) \rightarrow H_n(K(G,1), \mathbb{Q})$ is injective. (Here $K(G, 1)$ is not assumed to be a manifold.)
For example, if $X$ is simply connected, the induced map $f_*$ is the zero map on $H_n$ and when $X$ is the n-torus $T^n$, the induced map $f_*$ is an isomorphism. Is there any condition on $\pi_1(X)$ that will imply injectivity of $f_*$ on $H_n$?

REPLY [6 votes]: There can not be such a condition.
For any finitely presented group $G$, we can find a closed 4-manifold $N$ with fundamental group $G$. In dimension $n \geq 6$, we can now take $M = N \times S^{n-4}$, a $n$-manifold with fundamental group $G$. Then the classifying map $f\colon M \to BG$ of the universal cover $\tilde{M} \to M$ factors through $N$, hence
$$f_{\ast}\colon \ H_n(M;\mathbb Z) \to H_n(BG;\mathbb Z)$$ is zero.

REPLY [2 votes]: Here is another source of counterexamples. Take $N$ as in Jens Reinhold's answer. Then let $M$ be the connected sum $M=N\#\mathbb C P^2$. It has the same fundamental group $G$ as $N$, and the classifying map $M\to BG$ simply collapses $\mathbb C P^2$ to a point. But $H_2(M)\cong H_2(N)\oplus H_2(\mathbb C P^2)\cong H_2(N)\oplus\mathbb Z$, and the extra $\mathbb Z$ maps to $0\in H_2(BG)$. Examples like this exist in all dimensions $\ge 4$.<|endoftext|>
TITLE: Applications of the Cayley-Hamilton theorem
QUESTION [46 upvotes]: The Cayley-Hamilton theorem is usually presented in standard undergraduate courses in linear algebra as an important result. Recall that it says that any square matrix is a "root" of its own characteristic polynomial.
Question. Does this theorem have important applications?
Being not an algebraist, I am aware of only one application of this result which I would call important; it is very basic for commutative algebra, algebraic geometry, and number theory. It is as follows. Let a commutative unital ring $A$ be imbedded into a field $K$. Consider the set of elements of $K$ which are integral over $A$, i.e. are roots of a polynomial with coefficients in $A$ with the leading coefficient equal to 1. Then this set is a subring of $K$.

REPLY [4 votes]: There is a very important theorem by Procesi that derives all polynomial identities of the ring of matrices from the  Cayley-Hamilton theorem. For a review of rings with polynomial identities see this paper by Procesi.<|endoftext|>
TITLE: Lower Bounds for the Roots of Polynomials
QUESTION [5 upvotes]: I'm interested in the "size" of the roots of a sequence of Taylor Polynomials of an entire function. 
For example, consider $\mathrm f(z) = \mathrm e^z$. The Taylor Polynomials, or $k$-jets, are 
$$\mathrm P_k(z) = 1 + z + \frac{1}{2!}z^2 + \cdots + \frac{1}{k!}z^k$$
I want to find the best possible lower bound for the modulus of the roots for any given $\mathrm P_k$. Let
$$ L_k := \min\{ |w| : \mathrm P_k(w)=0 \}$$
I would like to know the limit of $L_k$ as $k \to \infty$. In this case, I think that $L_k \to \infty$ and $k \to \infty$.
I've used the standard bound based on the Rouché theorem, but this does not give a useful lower bound; it gives a lower bound independent of $k$. In the case of $\mathrm e^z$, the lower bound is $\frac{1}{2}$. 
I was hoping for a bound dependent on $k$, so that I could then find the limit of the bound as $k \to \infty$ and show that $L_k \to \infty$.
In general: Given a sequence of polynomials, how could I find the greatest lower bound of the modulus of the roots for each of these polynomials? Specifically, I am interested in the limit of these lower bounds as the polynomials tend to an infinite power series.

REPLY [2 votes]: Your question is related to the Szego curve, which has been much studied (just google it). A lower bound for the roots is given by $W(1/e)n=0.278..n$ where $W$ is the Lambert function.
Theorem 4 of http://www.math.kent.edu/~varga/pub/paper_184.pdf should give you an expansion of a lower bound, up to order $O(1/n)$.<|endoftext|>
TITLE: Finite union of affinoid is affinoid in proper smooth rigid curves (unless it is everything)
QUESTION [6 upvotes]: In several papers I have found the surprising statement that finite unions of affinoid subspaces of a proper smooth and connected rigid curve are either the whole curve or again affinoid.
Could you give me a reference for this fact or help me to sketch a proof?
Thank you in advance.

REPLY [6 votes]: I think that the original reference is "Zariski's Main Theorem für affinoide Kurven" by K.-H. Fieseler (Mat. Ann. 251, 1980). He proves that a finite union of affinoid domains of a one-dimensional affinoid space is affinoid, but this is probably not enough to answer your question.
More generally, though, J. Fresnel and M. Matignon prove that a quasi-compact irreducible one-dimensional rigid space is either affinoid or projective ("Sur les espaces analytiques quasi-compacts de dimension 1 sur un corps valué complet ultramétrique", Annali di Matematica Pura ed Applicata 145, 1986). You can also find a proof in chapter 6 of A. Ducros's book (see https://webusers.imj-prg.fr/~antoine.ducros/livre.html) in the language of Berkovich spaces.<|endoftext|>
TITLE: Complexifying a real-analytic singularity
QUESTION [6 upvotes]: This is probably a well-known issue, but I could not find a clear discussion in the literature, and I think others could find it useful.
Consider a real-analytic function germ $f:(\mathbb R^2,0) \rightarrow \mathbb R$, it is represented by a convergent power series $\sum_{i,j}a_{ij}x^iy^j \in \mathbb R\{x,y\}$. Suppose $f$ has a singularity at $0$, i.e. $df(0)=0$. Define its Jacobian ideal as the ideal generated by its partial derivatives: $I_{df} = \mathbb R\{x,y\}\langle f_x,f_y\rangle$, and its real local algebra as:
$$Q_f := \frac{\mathbb R\{x,y\}}{I_{df}}.$$
Let $\mu_{\mathbb R} := \dim_{\mathbb R} Q_f$. The singularity is called algebraically isolated if $\mu_{\mathbb R}< \infty$. It is called isolated if, in some small neighborhood of $0$, $df(x,y)=0$ only at $(x,y)=0$ (i.e., the usual meaning).
Now take the complexification of $f$ by simply considering the series as a complex one: $\sum_{i,j}a_{ij}z^iw^j$. It is convergent in some polydisc and defines an holomorphic germ $f^{\mathbb C}:(\mathbb C^2,0) \rightarrow \mathbb C$. Take its (complex) Jacobian ideal, that is: $I_{df^{\mathbb C}} = \mathbb C\{x,y\}\langle f^{\mathbb C}_z,f^{\mathbb C}_w\rangle$, and the complex local algebra
$$Q^{\mathbb C}_f := \frac{\mathbb C\{x,y\}}{I_{df^{\mathbb C}}}.$$
The Milnor number of the complexified singularity is defined by: $\mu_{\mathbb C} := \dim_{\mathbb C} Q_f^{\mathbb C}.$
A complex singularity is called isolated if $\mu_{\mathbb C} < \infty$. (In the holomorphic case it can be proven that a singularity is isolated if and only if, in some small neighborhood of $0$, $df(z,w)=0$ only at $(z,w)=0$, so we don't need to specify ''algebraically'' here.)
Questions:  

What is the relationship between the real-analytic singularity and its complexification? In particular: between the two local algebras, and between $\mu_{\mathbb R}$ and $\mu_{\mathbb C}$? Is there a class of functions where things work better?
Is true (or false) that a real-analytic singularity is algebraically isolated if and only if its complexification is isolated?
In the real setting: is ''algebraically isolated'' stronger then simply ''isolated''?
Any reference!

Thank you.

REPLY [7 votes]: As David Speyer has already suggested, we have $Q_f^\mathbb{C}\simeq Q_f\otimes\mathbb{C}$, so that answers the first question (and the second question).
For the third question, consider $f = (x^2+y^2)^2$.  The singularity at $(x,y)=0$ is clearly isolated, but it is not algebraically isolated, since $\mu_\mathbb{R} = \infty$.  Of course, over the complex numbers, $(z^2+w^2)^2$ has a non-isolated singularity at $(z,w)= (0,0)$; the two lines of singularities $z = \pm i\,w$ intersect at $(0,0)$.
The various confusions will be cleared up by noting that the ring of germs of analytic functions in two variables is a UFD, and, the singularity will be algebraically isolated as long as $f$ does not have any repeated irreducible factors.  'Isolated' is weaker over the reals simply because all the factors of $f$ might have no real zeros other than at the origin.<|endoftext|>
TITLE: On what kind of condition of a compact set $K$ in the plane, $C(K)$ has a generator?
QUESTION [7 upvotes]: Let $K\subset \Bbb{C}$ be a compact subset of the complex plane, and let $C(K)$ be the space of all complex continuous functions on $K$.
We say that $f\in C(K)$ is a generator of $C(K)$ when the set $\{p(f) \mid p\ \style{font-family:inherit;}{\text{is a polynomial}}\}$ is dense in $C(K)$.
If $K$ is any set consisting of finite points, it is easy to check that $C(K)$ has a generator.
If $K=[0,1]$, then we know that $f(x)=x$ is a generator of $C[0,1]$ by Stone-Weierstrass theorem. It follows that $C(\gamma)$ has a generator whenever $\gamma$ is a not a closed simple curve.
Now if $K=S^1$, the unit circle on the plane, then it has been proved that $C(K)$ does not have a generator. It follows that $C(\gamma)$ does not have a generator for any $\gamma$ that is a simple connected curve.
If $K=\overline{\mathbb{D}}$, the unit disk, then by using invariance of domain, a similar method as in the case $K=S^1$ can be given to prove that $C(\overline{\Bbb{D}})$ does NOT have a generator. Similarly, It can be proved that $C(K)$ does NOT have a generator if $K$ has an interior point in $\Bbb{C}$.
How to solve this problem if $K$ does not have an interior point?

REPLY [3 votes]: I believe Yemon's conjecture is correct: $C(K)$ has a generator in the sense of the question if and only if $K$ has empty interior and $\mathbb{C}\setminus K$ is connected. As he points out, the reverse direction follows from Lavrentiev's theorem. For the forward direction, if $K$ has nonempty interior then Mr. Li has already pointed out that we know $C(K)$ has no generator: this easily follows from David Ullrich's answer to a previous question which says that $C(S^1)$ has no generator, and the fact that $K$ contains a homeomorphic copy of $S^1$.
Now suppose $\mathbb{C}\setminus K$ is disconnected and assume $C(K)$ has a generator $f$. I claim that $K$ has a component $K_0$ such that $\mathbb{C}\setminus K_0$ is disconnected. Granting the claim, $C(K_0)$ also has a generator $f$, namely the restriction to $K_0$ of any generator of $C(K)$. It is obvious that $f$ must be 1-1 on $K_0$, so by a standard fact it is a homeomorphism between $K_0$ and $f(K_0)$. According to the answer to this question, $\mathbb{C}\setminus f(K_0)$ is also disconnected. Then $f(K_0)$ contains the boundary of a bounded open set, namely any bounded component of its complement. But now David Ullrich's argument for $S^1$ carries over verbatim to this setting to yield a contradiction. End of proof, modulo the claim.
To verify the claim, let $U$ be a bounded component of $\mathbb{C}\setminus K$, let $V$ be the unbounded component of $\mathbb{C}\setminus\overline{U}$, and let $K_0 = \partial V$.<|endoftext|>
TITLE: Integer sets with forbidden differences
QUESTION [9 upvotes]: Given a finite set $S$ of positive integers, and a positive integer $n$, let $F(n,S)$ be the largest possible cardinality of a subset of {$1,2,\dots,n$} no two of whose elements differ by a number in $S$.
E.g., if $S=${$2,3$} and $n=10$, we have $F(n,S)=4$ corresponding for instance to the set {$1,2,6,7$}.
What is known about the computational complexity of $F$?
A purely greedy approach doesn't work. E.g., if $S=${$3,5$} and $n=20$, a greedy approach gives the set {$1,2,3,9,10,11,17,18,19$} which has smaller cardinality than {$1,3,5,7,9,11,13,15,17,19$}.
Dynamic programming works, but takes computational resources that are exponential in max($S$). Can one do better?
I don't like the name of this question; if anyone can think of a better way to describe the problem, please feel free to revise it.

REPLY [3 votes]: Here is one way to view this problem:
Form the graph $G$ with vertices $\{1,\dots,n\}$ and edges $(i,j)$ for all $i,j$ such that $|i-j|\in S$. Then $F(n,S)$ is the size of a maximal independent set in $G$. 
From this it is clear $F$ has at most the computational complexity of finding the size of a maximal independent set. It is not quite clear to me how to use the special properties of this graph to get something better.
For special sets $S$, one immediately gets the value $F(n,S)$:

$F(n,\{k\})=\sum_{i=1}^{k+1}\lfloor\frac{n+k+i-1}{2k}\rfloor$
$F(n,\{1,\dots,k\})=\lceil\frac{n}{k+1}\rceil$

For your two examples the graphs look like this:

.
All instances for $F(10,\{2,3\})=4$ are:

$\{1, 2, 6, 7\}, \{1, 2, 6, 10\}, \{1, 2, 7, 8\}, \{1, 2, 8, 9\}, \{1, 2, 9, 10\}, \{1, 5, 6, 10\}, \{1, 5, 9, 10\}, \{2, 3, 7, 8\}, \{2, 3, 8, 9\}, \{2, 3, 9, 10\}, \{3, 4, 8, 9\}, \{3, 4, 9, 10\}, \{4, 5, 9, 10\}$

All instances for $F(20,\{3,5\})=10$ are:

$\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}, \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$<|endoftext|>
TITLE: The origin of the Ramanujan's $\pi^4\approx 2143/22$ identity
QUESTION [20 upvotes]: What is the origin of the Ramanujan's approximate identity $$\pi^4\approx 2143/22,\;\;\tag 1$$ which is valid with $10^{-9}$ relative accuracy? For comparison, the relative accuracy of the well known $\pi\approx 22/7$ is only $4\cdot10^{-4}$ and in this case we have the identity
$$\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2}, \tag{2}$$
which explains why the difference is small (concerning this identity, see Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$?).
Of course, (1) can be rewritten in the form $$\zeta(4)\approx 2143/1980,$$
so maybe some fast convergent series for $\zeta(4)$ can be used to get this approximate identity (in the case of $\frac{22}{7}-\pi$,  a series counterpart of (2) is $$\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)}=\frac{22}{7}-\pi$$ - see Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$?).
P.S. I just discovered that this question was discussed in  https://math.stackexchange.com/questions/1359015/is-there-an-integral-for-pi4-frac214322 and in https://math.stackexchange.com/questions/1649890/is-there-a-series-to-show-22-pi42143 is anything to add to the answers given there?

REPLY [4 votes]: Perhaps the most intuitive way of explaining this identity is with the help of a continued fraction
$$
97+\frac{1}{2+\frac{1}{2+\frac{1}{3+\frac{1}{1+\frac{1}{16539+\frac{1}{\ddots}}}}}}
$$
The unexpectedly large partial quotient of 16539 means that using all partial quotients up to (but not including) this value gives an incredibly close approximation of $\pi^4$ given the number of terms used. What this large number means essentially is that it takes a denominator of that many times more than the previous in order to obtain a better approximation, implying that the previous denominator is a really close approximation. The same method can be used to show that 355/113 is a very close approximation of $\pi$ for the number of partial quotients used.
Evaluating the truncation
$$
97+\frac{1}{2+\frac{1}{2+\frac{1}{3+\frac{1}{1}}}}
$$
gives the famous approximation of $\frac{2143}{22}$ for $\pi^4$.<|endoftext|>
TITLE: Is the action of $G$ on $H_1(T^n, \mathbb{Z}) = \mathbb{Z}^n$ faithful?
QUESTION [10 upvotes]: Let $G$ be a finite group of diffeomorphisms of the torus $T^n$ fixing some point $p$, i.e. $p$ is fixed by every element of $G$. I have two questions.

Is the action of $G$ on $H_1(T^n, \mathbb{Z}) = \mathbb{Z}^n$ faithful?
What if $G$ is not assumed to fix a point $p$?

REPLY [5 votes]: This follows from a 1931 theorem of M. H. A. Newman. Theorem 1 of the paper says that for a connected manifold $M$ with a metric and an integer $m>1$, there is a constant $d$ so that any (uniformly continuous) periodic transformation of $M$ of order $m$ must move a point of $M$ at least a distance $d$. 
Given a homeomorphism $f$ of a torus $T^n$ which acts trivially on $\pi_1(T^n)$, has order $m$, and fixes a point, there is a lift $\tilde{f}$ of $f$ to the universal cover $\mathbb{R}^n$ which fixes a point and has order $m$ and is uniformly continuous. Moreover, $f$ is homotopic to the identity, since $T^n$ is a $K(\mathbb{Z}^n,1)$. Then in some Euclidean metric on $T^n$, every point will be homotopic to its image under $f$ by a path of length at most $D$ (the tracks of the homotopy have bounded length). 
The homotopy of $f$ to the identity lifts to a homotopy of $\tilde{f}$ to the identity, so $\tilde{f}$ also has the property that it moves points at most $D$ and has order $m$. Let $d$ be the constant from Newman's theorem for the manifold $\mathbb{R}^n$ with Euclidean metric and $m$, then rescale the metric by $d/D$, we see that $\tilde{f}$ is periodic of order $m$ and moves points at most distance $d$, hence is the identity. But then $f$ is also the identity. 
Hence for a finite group $G$ acting (faithfully) on a torus $T^n$, every non-trivial element $f$ of $G$ must act non-trivially on $\pi_1(T^n)$.<|endoftext|>
TITLE: Are finitely generated amenable groups positively finitely generated?
QUESTION [12 upvotes]: Let $G$ be a finitely generated amenable group.
Is there a positive integer $n$ such that $n$ random elements of $G$ generate it with positive probability?
Being more formal, note that $G^n$ is amenable, so it carries a translation invariant finitely additive probability measure $\mu$. I am asking whether there is a positive integer $n$ such that:
$$\mu (\{(g_1, \dots, g_n) \in G^n \ : \  \langle g_1, \dots g_n \rangle = G\}) > 0$$
Alternatively, we can take a Folner sequence $\{F_k\}_{k \in \mathbb{N}}$ in $G^n$ and ask whether 
$$ \liminf_{k \to \infty} \frac{|\{ \text{Generating n-tuples} \}\cap F_k|}{|F_k|} > 0$$

REPLY [10 votes]: The answer is already no for $\mathbb{Z}$, assuming the question is whether this holds for every meaure.  Let $n\in\mathbb{N}$, and let
$$
S \,=\, \{(a_1,\ldots,a_n)\in\mathbb{Z}^n \mid \gcd(a_1,\ldots,a_n)=1\}.
$$
I will demonstrate a Følner sequence in $\mathbb{Z}^n$ that does not intersect $S$, which leads to a translation-invariant, finitely additive probability measure $\mu$ on $\mathbb{Z}^n$ for which $\mu(S) = 0$.
The idea is that there are large cubes in $\mathbb{Z}^n$ that do not intersect $S$.  Specifically, define a $\boldsymbol{k}$-cube in $\mathbb{Z}^n$ to be any product of the form
$$
A_1 \times \cdots \times A_n
$$
where each $A_i$ is a set of $k$ consecutive integers. I claim that for each $k\in\mathbb{N}$ there exists a $k$-cube $C_k$ in $\mathbb{Z}^n$ so that $C_k\cap S = \emptyset$.  Then $\{C_k\}$ is a Følner sequence in $\mathbb{Z}^n$, and
$$
\liminf_{k\to\infty} \frac{|C_k\cap S|}{|C_k|} \,=\, 0.
$$
To prove that the $k$-cubes $C_k$ exist, let $k\in\mathbb{N}$, let $\mathcal{I} = \{1,\ldots,k\}^n$, and choose a family
$$
\{p_{(i_1,\ldots,i_n)}\}_{(i_1,\ldots,i_n)\in\mathcal{I}}
$$
of $k^n$ distinct primes.  By the Chinese remainder theorem, there exists a tuple $(c_1,\ldots,c_n)$ of integers so that
$$
c_j \equiv -i_j\pmod{p_{(i_1,\ldots,i_n)}} 
$$
for each $(i_1,\ldots,i_n) \in\mathcal{I}$ and each $j\in\{1,\ldots,n\}$.  Then
$$
C_k \,=\, (c_1,\ldots,c_n) + \mathcal{I}
$$
is a $k$-cube in $\mathbb{Z}^n$ that does not intersect $S$, since
$$
p_{(i_1,\ldots,i_n)}\;\bigl|\; \gcd(c_1+i_1,\ldots,c_n+i_n)
$$
for each $(i_1,\ldots,i_n) \in \mathcal{I}$.
Edit: This technique can also be used to show that there exists a Følner sequence $\{F_k\}$ in $\mathbb{Z}$ itself so that for all $n\in\mathbb{N}$
$$
\liminf_{k\to\infty} \frac{|\{\text{generating $n$-tuples}\}\cap F_k^n|}{|F_k^n|} = 0.
$$
Specifically, we can let each $F_k$ be a disjoint union
$$
F_k \,=\, A_{k,1} \uplus A_{k,2} \uplus \cdots \uplus A_{k,k}
$$
where each $A_{j,k}$ is a set of $k$ consecutive integers, and these sets are chosen to have the property that $\gcd(a_1,\ldots,a_k)\ne 1$ for all $(a_1,\ldots,a_k)\in A_{k,1}\times\cdots\times A_{k,k}$.  If we fix $n$, then the probability that a randomly chosen element of $F_k^n$ has two coordinates from the same $A_{j,k}$ goes to $0$ as $k\to\infty$, and thus a randomly chosen element of $F_k^n$ has a high probability of not generating $\mathbb{Z}^n$.<|endoftext|>
TITLE: Independence between the number of prime factors of $n$ and $n+2$
QUESTION [9 upvotes]: I am interested in having an upper bound for the cardinality of  
$\#\left\{n\leq x\,:\quad\omega(n)=k, \omega(n+2)=\ell\right\}$
for $k,\ell\geq 1$,  
where $\omega(n)=\sum_{p\vert n}1$ counts the number of (distinct) prime factors of the integer $n$.
I would actually like a sharp upper bound (up to a constant) for $k=\ell=\log\log x$. 
I thought this was known, but I can't find anything in the literature. For $k\ll\sqrt{\log\log x}$ and $\ell\ll\log\log x$ I can have a good upper bound, but I can't reach $k=\ell=\log\log x$.
Does anyone has a reference or an idea please ? 
Thank you very much !

REPLY [3 votes]: The problem as been solved and generalized. Thread can be closed.<|endoftext|>
TITLE: Relationship between first and second incompleteness theorems
QUESTION [7 upvotes]: By my understanding, Gödel's first incompleteness theorem says that any theory with sufficient1 interpretability strength is essentially incomplete, that is, any consistent recursively enumerable extension of the theory must be incomplete.
Meanwhile, the second theorem says that any theory with sufficient2 interpretability strength can formulate, but not prove, a statement of its own consistency (and this is also true for any consistent, recursively enumerable extension of the theory).
Obviously, any theory satisfying the second theorem must also satisfy the first. But what about the converse: are there (consistent, recursively enumerable) essentially incomplete theories which can prove a statement equivalent to their own consistency, or which cannot formulate such a statement?

REPLY [5 votes]: It is possible to construct first-order theories which prove their own consistency (and which are incomplete).  Here is one example.
Consider a first-order theory with equality with a constant symbol 0 (zero), a unary predicate symbol N (the natural numbers), and a binary predicate symbol σ (the sequential relationship). Make the usual Peano Axiom assumptions, with the exception that one does not assume that σ is a total function, i.e. one does not assume that, for every natural number, there exists a natural number which is its successor.  That is, it assumes every axiom in the following list except PA3:
(PA1) Ν0
(PA2) ∀n∀m ( Νn & σn,m ⇒ Νm )
(PA3) ∀n ( Νn ⇒ ∃m σn,m )
(PA4) ∀n∀m∀m' ( Νn & σn,m & σn,m' ⇒ m = m' )
(PA5) ∀n∀m∀n' ( Νn & Νn' & σn,m & σn',m ⇒ n = n' )
(PA6) ∀n ( Νn ⇒ ¬ σn,0 )
(PA7) Induction schema. Let φ be a well-formed formula where m does not occur
and n is not a bound variable. Suppose φ[0\n] and
∀n∀m ( Νn & σn,m & φ ⇒ φ[m\n] ). Then ∀n ( Νn ⇒ φ ).
Now add a 3-ary symbol ∂.  ∂f,x,n can be thought of as meaning that f is a sequence whose nth term is x.  Make four assumptions about ∂:
(SEQ1) ∀x ( Nx ⇒ ∃f ∀i∀y (∂f,i,y ⇔ i = 0 & y = x) )
(SEQ2) ∀f ∀n∀m∀x ( ∂f,m,x & Nn & σn,m ⇒ ∃y ∂f,n,y )
(SEQ3) ∀f ∀n∀m∀x∀y ( ∂f,n,x & Ny & Nm & σn,m
⇒ ∃g ( ∂g,m,y & ∀i(¬ i = m ⇒ ∀z(∂f,i,z⇔∂g,i,z)) ) )
(SEQ4) ∀f∀i∀x∀y ( ∂f,i,x & ∂f,i,y ⇒ x = y )
Call this theory fpa.  The most important thing to take away from fpa is that its ontological assumptions about the naturals are downward.  That is, if a natural number n exists, then one can prove that all natural numbers less than n exist and have all the normal properties, but not that any number greater than n exists.  As to sequences, if a natural number n exists, one can prove that there is a sequence from 0 to n but not any longer, and all properties of sequences and subsequences up to length n hold.
It is easy to define addition and multiplication formulas which have all the usual properties, with the exception of totality. E.g. one can prove commutativity of addition
+(x,y,z) ⇒ +(y,x,z) 
even though one can't prove, given x and y, that there exists such a z.
Continue in this way.  Define a predicate one(x) to abbreviate Nx & σ0,x. Again, one can prove all the usual properties of 1 except that it exists. Define two(x) and so forth.
Prove simple facts about sequences, in particular what it means for h to be the concatenation of f and g.  Again, remark that given f and g, one cannot prove that h exists. 
Now provide some Godel numbering for your system. Say use one(x) to represent left-parenthesis and two(x) to represent right-parenthesis and so forth. variable(x) can be defined by ∃y (ten(y) & x > y).  (I haven't said how to define > but it's defined in the obvious way.)
A string, a formula, and a proof are just sequences of symbols. You can define what it means to be a term, an atomic wff (it's a particular kind of sequence), and then a wff (another particular kind of sequence).  Eventually you can define the particular sequences which are proofs.  That is, Proof(s) will state that sequence s represents a proof (as usual one will not be able to prove that there exists a sequence which is a proof).  Finally, define the consistency of fpa as the wff Cons asserting the non-existence of a proof of the wff ¬ 0 = 0.
One proves Cons as follows. Suppose there exists a sequence-proof of the contradiction ¬ 0 = 0. But fpa can be modeled by a model of a single element, 0.  For atomic wffs Nt, ∂f,t,u, and t = u assign the value of true.  For atomic wffs St,u assign the value false.  Continue in the obvious way.  For instance a wff equivalent to ∀x(W) gets assigned the value that W has. Then one can prove that every wff gets assigned one and only one value (true or false), all the axioms will be assigned true, and that Modus Ponens and Generalization preserve truth.  Hence, one can prove that every step in any proof in fpa must be true.  But ¬ 0 = 0 is assigned the value false.  This is a contradiction.  Hence there does not exist a sequence-proof of a contradiction, i.e. fpa is consistent. But one has proved this in fpa, i.e. fpa has proved its own consistency.<|endoftext|>
TITLE: Tangent bundle of smooth closed simply-connected $4$-manifold $w_1 = w_2 = 0$ can be trivialized in complement of point?
QUESTION [6 upvotes]: Let $M$ be a smooth closed simply-connected $4$-manifold with $w_1 = w_2 = 0$. Can $TM$ be trivialized in the complement of a point?

REPLY [6 votes]: Yes. For manifolds of dimension $\le 7$ the obstructions to trivializing any real vector bundle are $w_1, w_2$, and a class $\frac{p_1}{2} \in H^4(BSpin(n), \mathbb{Z})$ ($n \ge 3$), the fractional first Pontryagin class, defined in the presence of a spin structure. On an open $4$-manifold $H^4(-, \mathbb{Z})$, and hence $\frac{p_1}{2}$, vanishes.

REPLY [4 votes]: $M$ is smooth, so there is a cellular decomposition with a single 4-cell. Thus we want $TM$ trivializable over the 3-skeleton, so we look to analyze sections of the frame bundle which has Lie group fiber $SO(4)$ (here $M$ is orientable because $w_1=0$). Obstruction theory says we can extend over the 2-skeleton, because the class is $w_2\in H^2(M; \pi_1SO(4)=\mathbb{Z}_2)$ and you assumed that to be zero. The next obstruction (extending over the desired 3-skeleton) is automatically vacuous, because $\pi_2SO(4)=0$.<|endoftext|>
TITLE: Covering the unit sphere by open hemispheres
QUESTION [12 upvotes]: Suppose $H_1,\ldots,H_{2n}$ are open hemispheres which cover $S^{n-1}$ with the property that removing any one of them leaves $S^{n-1}$ uncovered. Is it necessarily the case that the hemispheres can be grouped into antipodal pairs? That is, possibly after reordering we have $H_i=-H_{i+1}$ for $i=1,3,5,\ldots,2n-1$?
Another way to phrase this question is as follows: Suppose there are $2n$ points in $S^{n-1}$ with the property that the origin lies in the interior of their convex hull, but the origin does not lie in the interior of the convex hull of any proper subset of the points. Is it necessarily the case that the points can be grouped into antipodal pairs?

REPLY [8 votes]: It looks so. We start with inductive proof of
Steinitz Theorem. Let $A$ be a finite set of rays starting from the origin in $\mathbb{R}^d$. Assume that positive span of these rays is the whole $\mathbb{R}^d$. Then there exists a subset of at most $2d$ rays from $A$ with the same property.
Proof. Induction in $d$. Base $d=1$ is clear. Induction step. Choose minimal $k$ such that some $k$ vectors $a_1,\dots,a_k$ which generate rays from $A$ are linearly dependent with positive coefficients: $\sum t_i a_i=0$, $t_i>0$. Then $0$ lies in interior of a $(k-1)$-dimensional simplex with vertices $a_1,\dots,a_k$. Let $X={\rm span}\,(a_1,\dots,a_k)$, $\dim X=k-1$. Factor everything modulo $X$, we get a space of dimension $d-k+1$, 0 lies in an interior of a convex hull of the image of $A\setminus {\mathbb R}_+\cdot \{a_1,\dots,a_k\}$, and it suffices to use $2(d-k+1)$ rays by induction proposition, add to them rays generated by $a_1,\dots,a_k$ to get totally at most $2(d-k+1)+k=2d+(2-k)\leqslant 2d$ rays. 
If on the first step it was $k>2$, we get improved bound $2d-1$ on the number of used rays. If $k=2$, then we find a pair of opposite rays.  If we proceed by induction proving your statement, we may think that two rays have generators $\pm e_d$ and generators of others are partitioned onto pairs $(x_k,\alpha_d)$, $(-x_k,\beta_d)$ for some $x_1,\dots,x_{2d-2}\in \mathbb{R}^{d-1}$. If $\alpha_d\ne \beta_d$, then considering above 4 rays, which lie in a 2-plane, it is easy to see that the whole 2-plane is generated by some three of them, so safely remove one ray.<|endoftext|>
TITLE: Is every triangulation of a Euclidean ball by convex tetrahedra shellable?
QUESTION [5 upvotes]: Suppose you are given a 3-ball $B$ in $\mathbb{R}^3$ that is bounded by a PL sphere, a triangulation $T$  of $B$ by Euclidean tetrahedra. Is that triangulation necessarily shellable?
I know that if $T$ can be lifted to a convex hypersurface in $\mathbb{R}^4$, then it is shellable; this applies if $T$ comes from a Delaunay triangulation. But I believe most triangulations are not liftable.
I suspect the answer is "no", that $T$ is not necessarily shellable. The first obstruction I found was the existence of a knotted arc made out of two edges of the triangulation; this obviously cannot happen for triangulations of the kind that I described.

REPLY [8 votes]: I haven't had time to check through the details of the construction, but the example B_3_9_18 found in the proof of Theorem 2 here by Frank Lutz appears to be embeddable in 3-space.
Frank specializes in creating wickedly tiny polytopes with unexpected properties, his website may also be useful for those studying similar questions: http://page.math.tu-berlin.de/~lutz/ (see in particular "the manifold page").<|endoftext|>
TITLE: What part is left unsolved in the Unknotting problem? (after results of Bar-Natan, Khovanov, Kronheimer and Mrowka)
QUESTION [30 upvotes]: Kronheimer and Mrowka showed that the Khovanov homology detects the unknot.
Bar-Natan showed a program to compute the Khovanov homology fast: there was no rigorous complexity analysis of the algorithm, but it is estimated by Bar-Natan that the algorithm runs in time proportional to the square root of the number of crossings, so it is even less than linear in the number of crossings.

What I might understand from this, is that if we have:

a proof for the correctness of Bar-Natan's algorithm
a rigorous algorithm analysis showing the run time of the algorithm is polynomial or less

then we have a proof that the unknotting problem is in P.
I guess this is not really the case.
Is what I assume here true? if not, why? (maybe Bar-Natan's algorithm is not deterministic?)

REPLY [24 votes]: This isn't directly what you ask, but it's also worth noting that unknot detection is in $\text{NP} \cap \text{co-NP}$, that is, there are polynomial-checkable certificates that will show that either a knot is the unknot or that the knot is not the unknot. The $\text{NP}$ certificate uses normal surface theory:

Ian Agol, Joel Hass, William Thurston, The computational complexity of knot genus and spanning area. Trans. Amer. Math. Soc. 358 (2006), no. 9, 3821–3850. 

The $\text{co-NP}$ certificate is currently conditional on the Generalized Riemann Hypothesis, and is based on finding representations of the knot group:

Greg Kuperberg, Knottedness is in NP, modulo GRH. Adv. Math. 256 (2014), 493–506.

(There ought to be another proof of the $\text{co-NP}$ result using normal surface theory and the notion of hierarchies. Ian Agol sketched that some time ago, but the details are difficult.)
The class $\text{NP} \cap \text{co-NP}$ is relatively small, and this gives reasons to believe that detecting the unknot is in $\text{P}$.
(I would have made this a comment, but I think it's too long for that.)<|endoftext|>
TITLE: Evaluating an infinite sum related to $\sinh$
QUESTION [17 upvotes]: How can we show the following equation
$$\sum_{n\text{ odd}}\frac1{n\sinh(n\pi)}=\frac{\mathrm{ln}2}8\;?$$
I found it in a physics book(David J. Griffiths,'Introduction to electrodynamics',in Chapter 3,problem 3.48), which provided no proof.

REPLY [15 votes]: In addition to GH's answer: this result can most easily be proved using Thompson–Lampard theorem of electrostatics: http://scitation.aip.org/content/aapt/journal/ajp/67/2/10.1119/1.19203 (A curious and useful theorem in two-dimensional electrostatics, by J. D. Jackson). From the concluding remarks of this paper:

For the mathematical physicist, the theorem elicits wonder
and amusement. ‘‘Unbelievable’’ results such as
$$\small \sum_{n \,even} \frac{8\sin^2{(n\pi a/2)}\sinh^2{(n\pi a/2)}}{n\sinh{(n\pi)}}+\sum_{n \,odd}\frac{8\cos^2{(n\pi a/2)}\cosh^2{(n\pi a/2)}}{n\sinh{(n\pi)}}=\ln{2} \tag{1}$$
valid for any $a$ ($0\le a<1$), and its special case ($a=0$),
$$\sum_{n \,odd}\frac{8}{n\sinh{(n\pi)}}=\ln{2}, \tag{2}$$
are causes for wonder, as is the equivalent product,
$$\prod\limits_{k=0}^\infty \left[\frac{1+e^{-(2k+1)\pi}}{1-e^{-(2k+1)\pi}}\right]^8=2.$$
The physicist, if not the mathematician, may view these relations as true curiosities, most easily proved by means of the very practical Thompson–Lampard theorem of electrostatics.

Completely mathematical proof of (1) (without using Thompson–Lampard theorem, but using Jacobi’s theta functions, as in the GH's proof) was given in http://scitation.aip.org/content/aip/journal/jmp/42/2/10.1063/1.1332123 (A mathematical proof of a result derived from the application of the Thompson–Lampard theorem of electrostatics, by D.A. Morales).
The special case (2) was evaluated in the paper
Link
(A new theorem in electrostatics with applications to calculable standards of capacitance, by D. G. Lampard) and the evaluation is based on a more general result $$\sum_{n \,odd}\frac{1}{n\sinh{(n\pi\beta)}}=\frac{1}{2}\ln{\frac{\theta_3(0;e^{-\pi\beta})}{\theta_4(0;e^{-\pi\beta})}},$$
proved in the appendix of the paper.<|endoftext|>
TITLE: Elliptic Operators on Vector Bundles
QUESTION [5 upvotes]: I know the kernel of an elliptic operator on a compact manifold has finite dimension. Is the kernel of an elliptic operator on sections of a vector bundle a finite dimensional space?

REPLY [4 votes]: In general, no. As an example, the spin Dirac operator on even-dimensional complex hyperbolic space has nontrivial $L^2$-sections in its kernel. Because the space is symmetric, in particular homogeneous, one can conclude that the $L^2$-kernel must be infinite-dimensional. The proof uses some results of Harish-Chandra on representations of semi-simple noncompact Lie groups. For more details, see this article or arXiv:math/9905089.<|endoftext|>
TITLE: Cluster algebras and cluster varieties
QUESTION [9 upvotes]: I have a really basic question about cluster algebras and cluster varieties. According to the definition of Fomin-Zelevinsky a cluster algebra is generated by a bunch of polynomial rings inside the ring of Laurent polynomials. It means that the corresponding variety is covered by a bunch of affine spaces (all which have some $\mathbb G_m^n$ in common). Let's call this variety $X$. 
Question: Is it true that all the above maps from $\mathbb A^n\to X$ are open embeddings (a priori it is only a birational map)? A related question: is it true that $X$ is always a smooth variety?
Edit: Of course, the question is wrongly asked as the maps are from $X$ to $\mathbb A^n$.

REPLY [13 votes]: A bunch of points:
$\def\Spec{\mathrm{Spec}\ }$
• Let $A$ be a cluster algebra over a field $k$, let $(x_1, \ldots, x_n)$ be a cluster and let $L$ be the Laurent polynomial ring $k[x_1^{\pm}, \ldots, x_n^{\pm}]$. I imaging your intended question is whether the map $\Spec L \to \Spec A$ is an open immersion. (You ask about a map $\mathbb{A}^n \to X$, but there isn't a natural such map; $\Spec L$ is a torus, not affine space.) 
The answer is yes. The question is equivalent to asking whether $L$ is a localization of $A$. I claim that $L = A[x_1^{-1}, \ldots, x_n^{-1}]$. Proof: On the one hand, $A \subset L$ (by the Laurent phenomenon) and $x_1^{-1}$, ..., $x_n^{-1} \in L$ (obviously), so $A[x_1^{-1}, \ldots, x_n^{-1}] \subseteq L$. On the other hand, $L$ is generated by the $x_j$ and their reciprocals, and these are all in $A[x_1^{-1}, \ldots, x_n^{-1}]$, so $L \subseteq A[x_1^{-1}, \ldots, x_n^{-1}]$. $\square$.
• $\Spec A$ is not generally the union of cluster tori. This is true even in the simplest case: The extended exchange matrix $\left( \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right)$ gives the cluster algebra
$$\mathbb{C}[x,x', y^{\pm 1}]/x x'-y-1.$$
The two tori are $xy \neq 0$ and $x' y \neq 0$. The point $(x,x',y) = (0,0,-1)$ is in neither torus.
• Some sources define "the cluster variety" as the (quasi-affine) union of cluster tori. If you take that as the definition, it is obviously smooth.
• But I imagine what you care about is whether $\Spec A$ is smooth (or possibly $\Spec U$, where $U$ is the upper cluster algebra. No, that doesn't have to be smooth. (The singular points are not in any of the cluster tori, of course.) I think the simplest example is the Markov clsuter algebra, with $B$-matrix $\left( \begin{smallmatrix} 0 & 2 & -2 \\ -2 & 0 & 2 \\ 2 & -2 & 0 \end{smallmatrix} \right)$. This ring is not finitely generated, and there is a maximal ideal generated by all cluster variables where the Zariski tangent space is infinite dimensional. If you like at the upper cluster algebra instead, it is $k[\lambda, x_1, x_2, x_3]/x_1 x_2 x_3 \lambda - x_1^2- x_2^2 - x_3^3$, which is singular along the line $x_1 = x_2 = x_3 = 0$.
For another example, which doesn't have the $A$ versus $U$ issue, look at the $A_3$ cluster algebra with no frozen variables. From Corollary 1.17 in Cluster Algebras III, this is generated by $(x_1, x_2, x_3, x'_1, x'_2, x'_3)$ module the relations
$$x_1 x'_1 = x_2+1,\ x_2 x'_2 = x_1 + x_3,\ x_3 x'_3 = x_2+1.$$
Look at the point $(x_1, x_2, x_3 , x'_1, x'_2, x'_3) = (0, -1, 0, 0,0,0)$. The Jacobian matrix of these $3$ equations with respect to the $6$ variables is
$$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 \\  0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix}$$
which has rank two, so this is a singular point.
• We do have Theorem 7.7 of Greg Muller's Locally Acyclic Cluster Algebras -- if the cluster algebra is locally acyclic, the $B$-matrix has full rank and our ground field has characteristic zero, then $\Spec A$ is smooth. 
See Muller 1 and Benito-Muller-Rajchgot-Smith 2 for more on singularities of cluster varieties.<|endoftext|>
TITLE: question about fiber functors and fundamental groups
QUESTION [6 upvotes]: I apologize for the vague question title, but I can't think of a better subject line.
Let $f : X\rightarrow S$ be a morphism of connected schemes. Let $x,y\in X$, $s\in S$ be geometric points with $f(x) = f(y) = s$. 
Let $FEt_S$ be the category of finite etale schemes over $S$, and similarly with $X$. We have fiber functors $F_s : FEt_S\rightarrow Sets$, and $F_x,F_y : FEt_X\rightarrow Sets$, and $\pi_1(X,x),\pi_1(X,y),\pi_1(S,s)$ are just the automorphism groups of the associated fiber functors.
There are canonical isomorphisms
$$\beta_x : F_x\circ f^*\stackrel{\sim}{\longrightarrow}F_s\qquad\text{and}\qquad \beta_y : F_y\circ f^* \stackrel{\sim}{\longrightarrow}F_s$$
Any isomorphism $\alpha : F_x\stackrel{\sim}{\longrightarrow} F_y$ induces an automorphism $$\overline{\alpha} := \beta_y\circ\alpha\circ\beta_x^{-1}\in Aut(F_s) = \pi_1(S,s)$$
Suppose the induced map $f_* : \pi_1(X,x)\rightarrow \pi_1(S,s)$ is trivial (sends everything to the identity), and similarly $f_* : \pi_1(X,y)\rightarrow\pi_1(S,s)$, then I believe I can prove that $\overline{\alpha}$ does not depend the choice of $\alpha$. However, I'm having a lot of trouble arguing that $\overline{\alpha}$ should always be $id\in Aut(F_s)$.
Is this even true? (I mean, surely it is right. "What else could it be?") How would one prove something like this?
EDIT 1: The specific case I'm interested in is where $X\rightarrow S$ actually factors through the geometric point $s$.
EDIT 2: Okay it seems that this additional assumption makes the statement pretty easy to prove, though I would be interested in a counterexample to the original question.

REPLY [6 votes]: I don't think that $\bar\alpha$ is always the identity. You should think of $\alpha$ as a ``path'' from $x$ to $y$, so in principle it is possible that the loop obtained as the image of $\alpha$ is not nullhomotopic, even if the image of every loop is. In fact, if the map $f$ is the universal covering space you can get every possible element in $\pi_1(S,s)$ by choosing appropriately $x$ and $y$. To support this let me try to get you an example where $\bar \alpha$ is not in the image of $f_*$ (although $f_*$ is not trivial). 
Fix the ground field $\mathbb{C}$. Take $f:\mathbb{A}^1\smallsetminus \{0\}\to \mathbb{A}^1\smallsetminus \{0\}$ be the square map, and take $x=1$ and $y=-1$. Let the isomorphism $\alpha: F_x\xrightarrow{\sim} F_y$ to be obtained by "sliding" the points of the fiber along the half unit circle in the positive semiplane. Then $\bar\alpha$ should be the generator of $\pi_1(\mathbb{A}^1\smallsetminus\{0\})=\hat{\mathbb{Z}}$, while the image of $f_*$ is the index 2 subgroup.<|endoftext|>
TITLE: An elementary inequality for a recursive double sequence
QUESTION [6 upvotes]: Here is what looks like (but is not) an Olympiad problem. Is it really that tough, or am I overlooking a simple solution?
I have a system of sequences $\sigma_0(m),\sigma_1(m),\ldots\,$ defined for all integer $m\ge 0$ as follows. If $\min\{m,n\}=0$, then $\sigma_n(m)=0$. If
both $m$ and $n$ are positive, then find the (uniquely defined) integers
$K\ge 1$ and $j\in[0,n-1]$ such that $K^n\le m<(K+1)^n$ and indeed,
$K^{n-j}(K+1)^j\le m<K^{n-j-1}(K+1)^{j+1}$, and writing $m=K^{n-j}(K+1)^j+R$,
let
  $$ \sigma_n(m) := (Kn+n-j)K^{n-j-1}(K+1)^{j-1} + \sigma_{n-1}(R). $$
(The expression $(Kn+n-j)K^{n-j-1}(K+1)^{j-1}$ can be understood as the "formal
derivative" $\frac{d}{dK}K^{n-j}(K+1)^j$.) Thus, for instance,
$\sigma_1(m)=1$ for all $m\ge 1$, and
  $$ \sigma_2(m) = \begin{cases}
                     2K+1\ &\text{if}\ K^2<m\le K^2+K, \\
                     2K+2\ &\text{if}\ K^2+K<m\le (K+1)^2.
                   \end{cases} $$

Is it true that for all $k,n,m_1,\ldots,m_k\ge 1$, we have
    $$ \sigma_n(m_1+\dotsb+m_k)
       \le \sigma_{n-1}(m_1)+\dotsb+\sigma_{n-1}(m_k)
                                            + \max\{ m_1,\ldots,m_k \} ? $$

The cases where $n=2$ and $m_1=\dotsb=m_k=1$ follow without much effort, but already the general case where $k=1$ seems difficult:

Is it true that $\sigma_n(m)\le\sigma_{n-1}(m)+m$ for all integer $m,n\ge 1$?

Computations seem to confirm the inequality in question.

REPLY [2 votes]: So far I was not able to prove the inequality $\sigma_n(m)\leq\sigma_{n-1}(m)+m$ and have not even convinced myself that it is true. Still, I have deduced a number of properties that one may find helpful. I state them as lemmas below with only brief proof ideas, but can provide complete proof(s) upon request.
Everywhere below I assume $m,n\geq 1$ be integer and $K,R,j$ be defined as in the question.
Lemma 1. 
$$\sigma_{n}(m) = \frac{Kn+n-j}{K(K+1)}(m-R) + \sigma_{n-1}(R).$$
Proof. Trivial.
Lemma 2. 
$$R < \frac{m}{K+1} < m^{(n-1)/n} \leq \frac{m}{K}.$$
Lemma 3. For any fixed $n\geq 0$, $\sigma_n(m)$ is a non-decreasing function of $m$.
Proof idea. Use Lemmas 1,2 and induction on $n$.
Lemma 4. 
$$\frac{mn}{K+1} \leq \sigma_n(m) \leq \frac{mn}{K-1},$$
where the upper bound holds for $K>1$. For $K=1$ (i.e., $m<2^n$), we have
$$\sigma_n(m) \leq mn.$$
Proof idea. Use Lemmas 1,2,3 and induction on $n$.
Lemma 5.
$$\sigma_n(m) \geq \frac{(K+2)n-j-1}{(K+1)^2}m.$$
Furthermore, for $1<K\leq n$, we have
$$\sigma_n(m) \leq \frac{K}{K^2-1}mn - \frac{j+1}{(K+1)^2}m \leq \frac{K}{K^2-1}mn,$$
while for $K=1$ (i.e., $m<2^n$), we have
$$\sigma_n(m) \leq mn-\frac{j+2}{4}m \qquad (m>1).$$<|endoftext|>
TITLE: Partitioning a convex object without harming existing convex subsets
QUESTION [7 upvotes]: $C$ is a convex planar figure and $C_1,\dots,C_n$ are pairwise-disjoint convex subsets of $C$, like this:

A convex-preserving partition of $C$ is a partition $C=E_1\cup\dots\cup E_N$, , such that $N\geq n$, the $E_i$ are pairwise-disjoint convex figures, and for every $i=1,\dots,n$: $C_i\subseteq E_i$, i.e, each existing figure is contained in a unique new figure, like this:

For every $C,C_1,\dots,C_n$, let $F(C,C_1,\dots,C_n)$ be the smallest cardinality $N$ of a convex-preserving partition.
For every $n$, let $G(n)$ be the largest value of $F(C,C_1,\dots,C_n)$, for all combinations of $C,C_1,\dots,C_n$. 
What is $G(n)$?

Obviously $G(1)=1$.
By the half-plane separation theorem, $G(2)=2$.
By the above figure, $G(3)\geq 4$; there is apparently no convex-preserving partition with $N=3$.

What more can be said about $G(n)$?
Remark: I asked a similar question in cstheory.SE. There, $C$ and all its subsets were axis-parallel rectangles. In that case, I found an algorithm that proves $G(n)\leq 3n+1$.

REPLY [6 votes]: I think the number of holes is not greater than two times the number of sets $C_i$.
Note that the sets $C_i$ can be extended to $E_i$ in a such way that all holes will be convex (see page 8 in paper of Rom Pinchasi, On the perimeter of $k$ pairwise disjoint convex bodies contained in a convex set in the plane. http://www2.math.technion.ac.il/~room/ps_files/perim_k_convex.pdf). Lets think about $C_i$ as about already extended sets.
Each side of a hole is formed by one of the $C_i$-s. Lets call two sets $C_i$ and $C_j$ neighbors, if they form two adjacent sides of some hole. Note that:

To each hole correspond at least 3 neighbor-pairs - since each hole has at least 3 sides.
To each neighbor-pair $C_i,C_j$ correspond at most two holes - since all such holes must have a side co-linear with the segment in which the boundaries of $C_i$ and $C_j$ intersect.

Therefore, the number of holes is at most 2/3 the number of neighbor-pairs.
The "neighbor" relation defines a planar graph with $V=n$ vertexes. Euler's formula implies that in a planar graph (with at least 3 vertexes) the number of edges is bounded by: $E\leq 3V-6$. Hence, the number of holes is at most $2n-4$. Hence, $G(n)\leq 3n-4$.
The lower bound gives the tiling shown on the figure:

In the infinite tiling, each hexagon touches 6 holes and each holes touches 3 hexagons, so the number of holes is exactly $2n$. In the finite tiling, the number of holes is smaller since the holes near the boundary can be attached to their neighboring hexagons. So the number of holes is $2n-o(n)$ and $G(n)\geq 3n-o(n)$.<|endoftext|>
TITLE: Can the Dedekind zeta function distinguish between real and imaginary quadratic number fields?
QUESTION [18 upvotes]: Suppose I am given a machine that gives me the coefficients $a_1$, $a_2$, $a_3$, ... of a Dirichlet series
$$\sum_1^{\infty} \frac{a_n}{n^s} $$
and assume that I know that this Dirichlet series is the Dedekind zeta function of a quadratic number field. Is there any kind of algorithm which allows me to determine whether the number field is real or imaginary?

REPLY [11 votes]: Asking in terms of $B$ how many $a_n$ are needed is equivalent to asking the following question: 
What is the largest $N$ such that there exists two quadratic characters  $\chi_1, \chi_2$ of conductor $<B$ with $\chi_1(n)=\chi_2(n)$ for $n<N$  but $\chi_1(-1)\neq \chi_2(-1)$?
An obvious approach is to note that then $\chi= \chi_1 \chi_2^{-1}$ is a character of conductor $<B^2$  with $\chi(n)=1$ for $n<m$, which by the bound for the least quadratic nonresidue problem can only happen for $n< \left(B^2\right)^{1/4\sqrt{e}+o(1)}=B^{1/2\sqrt{e}+o(1)}$
Of course any improvement on this problem would also represent improvement on the least quadratic nonresidue problem, at least for residues modulo primes congruent to $3$ modulo $4$.
Given this many you could perform the algorithm of enumerating all the characters of conductor $<B$ and seeing which agree with your sequence.
The only efficient algorithm I see requires  $B$ coefficients - you simply look to see for which primes $\chi(p)$ is $0$ (or $a_p$ is $0$ for the Dirichlet $L$-function). These are precisely the ramified primes. If you check up to $B$ you find all the ramified primes. Knowing the ramified primes determines the Dirichlet character up to multiplication by a Dirichlet character modulo $8$, since $2$ is the only prime that can be ramified in multiple different ways. Simply check the four possibilities to see which one matches the first few coefficients of your sequence.<|endoftext|>
TITLE: Term for a metric space for which the triangle inequality is strict?
QUESTION [12 upvotes]: Is there a standard term for a metric space in which $\rho(p,r) < \rho(p,q) + \rho(q,r)$ for any distinct $p$, $q$, $r$? Sort of the opposite of metric convexity.
For instance, a subset of euclidean space with the inherited distance function has this property if and only if no three points are colinear. Another example: $\rho^\alpha$ has this property for any metric $\rho$ and any $\alpha \in (0,1)$.
(Context: I'm working on the second edition of my book Lipschitz Algebras, and I realized that a property I called "concavity" in the first edition is equivalent to this simple condition.)

REPLY [3 votes]: The only thing which I recall in this connection is: Blumenthal in his "Theory and Applications of Distance Geometry" (page 56) calls three points a linear triple if the triple is congruent with a triple in $\mathbb{R}^1$. See also page 242 in Blumenthal-Menger, "Studies in Geometry". (Possibly one can find in these books something more relevant, but I do not recall such things now.) So one can call such spaces without linear triples.<|endoftext|>
TITLE: A cubic and six conics problem
QUESTION [15 upvotes]: I am an electrical engineer system. I live in Viet Nam. I am not a Mathematician. I construct and found a problem as follows:
Let a cubic, and five conics $(C_1)$, $(C_2)$, $(C_3)$, $(C_4)$, $(C_5)$. The Cubic and a conic $(C_i)$ have six common points. Conic $(C_i)$ and conic $(C_{i+1}$ and the cubic have three common points for $i=1,2,3,4$. Let $P_1, P_2, P_3$ lie on $(C_1)$ and the cubic but don't lie on $(C_2)$ and $P_4, P_5, P_6$ lie on $(C_5)$ and cubic but don't lie on $(C_4)$. Then $P_1, P_2, P_3, P_4, P_5, P_6$ lie on a conic. I need a proof? Could you help me give your proof?

REPLY [23 votes]: Recall the quartic version of the Cayley-Bacharach theorem:
Theorem: Consider two quartics in general position, which intersect in $16$ points (by Bezout's theorem). Then if a third quartic passes through $13$ of those points, it also passes through the other $3$.
This can be used to establish a simpler result:
Lemma: Let $\Gamma$ be a cubic, and let $T_1, T_2, T_3, T_4$ be triples of points on $\Gamma$. Suppose that for each $i \in \{1, 2, 3\}$, the six points $T_i \sqcup T_{i+1}$ lie on a conic $C_i$. Then the six points $T_1 \sqcup T_4$ also lie on a conic.
Proof: We have an obvious set of $12$ points, and wish to extend it to a set of $16$ points so that Cayley-Bacharach is applicable. We let $P$ be the fourth intersection point of $C_1$ and $C_2$ (i.e. the one that isn't in $T_2$), and $Q$ be the fourth intersection point of $C_2$ and $C_3$. Now draw the line $l$ through $P$ and $Q$; it intersects $C_1$ again at a point $R$, and $C_3$ again at a point $S$.
Then $T_1 \sqcup T_2 \sqcup T_3 \sqcup T_4 \sqcup \{P, Q, R, S \}$ is a set of $16$ points. There are two quartics passing through it, namely $C_1 \cup C_3$ and $\Gamma \cup \ell$. Note that $C_2$ passes through $8$ of these $16$ points; let $C_4$ be the conic passing through any $5$ of the other $8$ points. Then $C_2 \cup C_4$ is a quartic passing through $13$ of the points, so passes through the other $3$ by Cayley-Bacharach. The result follows.
Corollary: Let $\Gamma$ be a cubic, and let $T_1, T_2, \dots, T_{2k}$ be triples of points on $\Gamma$. Suppose that for each $i \in \{1, 2, \dots, 2k - 1\}$, the six points $T_i \sqcup T_{i+1}$ lie on a conic $C_i$. Then the six points $T_1 \sqcup T_{2k}$ also lie on a conic.
Proof: Repeated application of previous result.
Your theorem is the case $k = 3$.<|endoftext|>
TITLE: "Anticanonical sections" on singular varieties
QUESTION [9 upvotes]: Let $X$ be a reduced scheme, so, generically regular; you may assume extra conditions like equidimensional and seminormal (though normal is stronger than I'd like, as is Gorenstein). 

Is there a reasonable definition of "section of the anticanonical bundle" over $X$? At the very least I'd like to be able to restrict such a "section" to the regular locus and get an actual section. If $X$ is normal, then I'd like any (honest) anticanonical section over $X_{reg}$ to uniquely extend to one of these objects on $X$.

Note: a divisor and a section are not quite the same thing, in that the divisor of zeroes only determines the section up to a global invertible function (which need not even be constant). I really want a section, not just its divisor of zeroes.
The motivation comes from Frobenius splitting. If $X$ is defined over a perfect field of characteristic $p$, and $\sigma$ is a section of the anticanonical over $X_{reg}$ (not vanishing on any component), then away from $\sigma=0$ and the singularities we can define a morphism of sheaves $\varphi:\ F_* \mathcal O_X \to  \mathcal O_X$ by $g \mapsto \mathcal C(g/\sigma) \sigma$ (here $F$ is the Frobenius and $\mathcal C$ is the Cartier operator on (top) forms). This extends over $X_{reg}$, and over all $X$ if $X$ is normal. Such maps $\varphi$ make sense even when $X$ is more singular, but I'd rather talk about an anticanonical section than about $\varphi$. 
Feel free to add tags.
EDIT: since this hasn't produced an answer "of course you want this standard object well-known to those who know it well", here's more detail about exactly what I want.
Over $X$, we have the sheaf $\mathcal{Hom}(F_* \mathcal O_X, \mathcal O_X)$. If we restrict to $X_{reg}$, where $F$ becomes a (finite) flat morphism, then using Serre duality for such morphisms we can identify this sheaf with $F_* \mathcal{Hom}(\omega^p,\omega) = F_* \omega^{1-p}$ (as in $\S$1.3 of the book [Brion-Kumar]). Now, I'd rather work with $\omega^{-1}$, and take the $p-1$ power (which gets me only very special sections, but that's okay). But I'd like to do that on $X$, where $F$ isn't flat.

REPLY [5 votes]: Allen, it seems to me that you are asking what generality you can go to to define canonical/anticanonical (divisorial) sheaves (not bundles!). I think that divisorial sheaves are actually easier to define on a non-normal scheme than divisors and I am guessing this is essentially what Jason is saying. As long as you do not care about actual divisors of zeros than you probably need very little. In fact, it seems to me that you can probably do this without assuming anything additional.
Dualizing complexes exist under fairly general assumptions. If you are dealing with schemes that are finite type over a field, then you are good. Since you are OK with equidimensional schemes, let's assume that $X$ is a reduced scheme of finite type over a field of pure dimension $d$. Then it has a dualizing complex $\omega_X^\bullet$ and define the canonical sheaf of $X$ as the unique cohomology sheaf of $\omega_X^\bullet$ which is not supported on the singular locus of $X$. Usually $\omega_X^\bullet$ is normalized so that this cohomology sheaf is $h^{-d}$. Let us denote this by $\omega_X$ and assume that $\omega_X^\bullet$ is normalized so that $h^{-d}(\omega_X^\bullet)=\omega_X$. In fact, let's assume that $\omega_X^\bullet$ is normalized so that on the regular locus it restricts to $\omega_{X_{\rm reg}}[d]$.
Reduced implies $S_1$, so your $X$ is actually Cohen-Macaulay in codimension $1$. This implies that the support of $h^{i}(\omega_X^\bullet)$ has codimension $2$ for all $i\neq -d$. In other words, there is a closed subset $Z\subseteq X$ of codimension at least $2$ such that $X\setminus Z$ is Cohen-Macaulay and $\omega_{X\setminus Z}^\bullet\simeq \omega_{X\setminus Z}[d]$. 
Now here is where you get lucky. In general you'd want to define Weil divisorial sheaves (which is probably what you are looking for) as reflexive sheaves of rank $1$. If $X$ is $S_2$ and $G_1$, then a coherent sheaf is reflexive if and only if it is $S_2$, so you get a relatively easy way to decide what being reflexive means and this is the usual context in which Weil divisorial sheaves (a replacement of divisors as cycles) are considered.
On the other hand, I've just realized that you might not need this much actually. The point is that $\omega_X$ is always $S_2$ in this context even if $X$ might not be. And then you are actually better off with the dual $\omega_X^\vee=\mathscr Hom_X(\omega_X, \mathscr O_X)$, because that is reflexive already if $X$ is $S_1$ and $G_0$ and with $X$ reduced you have more than that (reduced $\Leftrightarrow$ $S_1$ and $R_0$). So, you get that $\omega_X^\vee$ is reflexive and its global sections are the same as those you get on the Cohen-Macaulay locus. 
Restricting to the regular locus will give you actual sections. I think that's automatic, no? If $\mathscr F$ is a sheaf and $U\subseteq X$ is an open set, then being a sheaf means that you have a restriction map
$$
H^0(X,\mathscr F)\to H^0(U,\mathscr F|_U).
$$
So I guess what you are really asking is that the anticanonical sheaf of $X$ would restrict to the usual anticanonical line bundle on the regular locus. This is true for the above definition.
Of course without assuming that $X$ is $R_1$, you cannot expect all sections from the regular locus to extend, but you are wisely only asking that for $X$ normal. If $X$ is normal, then pushing forward the anticanonical sheaf of the regular locus to $X$ will coincide with $\omega_X^\vee$ because it is reflexive, so in particular, sections
extend uniquely.<|endoftext|>
TITLE: A possible mistake in Walter Rudin, "Fourier analysis on groups"
QUESTION [16 upvotes]: I have the following lemma 4.2.4 on page 80 in the book (we have locally compact abelian topological groups $G_1, G_2$ and their duals $\Gamma_1, \Gamma_2$):

Suppose $E$ is a coset in $\Gamma_2$ and $\alpha$ is an affine map of $E$ into $\Gamma_1$. THen $\alpha$ can be extended to an affine map of the closure $\overline{E}$ of $E$, and $\alpha(\overline{E})$ is a closed coset in $\Gamma_1$

I think $\alpha(\overline{E})$ is not necessary a closed coset in $\Gamma_1$. The book just states that it is a corollary of the uniform continuity of $\alpha$ but consider $E = \Gamma_2 = \mathbb{Q}$ with discrete topology, $\Gamma_1 = \mathbb{R}$ and $\alpha$ is identity map. In that case, $\overline{E} = E$ and the image is not closed in $\mathbb{R}$.
Edit: Since nobody proved the statement or pointed out where my example is wrong I'm going to assume the lemma is indeed incorrect.

REPLY [6 votes]: [This is just moving my comment to an answer so that the question does not keep showing up as unanswered.]
The discussions above seem to demonstrate that the OP is correct, and that this part of the lemma is incorrect. According to my Canadian sources, it seems that this error has been noticed tacitly by other people: see the remarks before and after Lemma 1.3 in:

M. Ilie, N. Spronk. Completely bounded homomorphisms of the Fourier algebras, J. Funct. Anal. 225 (2005) no. 2, 480–499<|endoftext|>
TITLE: Counting lattice points inside a three-dimensional ellipsoid
QUESTION [15 upvotes]: I want to answer the following simple question:

Given a three-dimensional ellipsoid defined by $Q(x, y, z) \leq Z$ for a positive definite quadratic form $Q$, how many lattice points in $\mathbb{Z}^3$ are inside the ellipsoid?

The answer is given by the volume of the ellipsoid (of order $Z^{3/2}$) plus an error term, and I am interested in the strongest possible error term, which I hope would be $O(Z^{3/4})$ or better. The error term should also depend on $Q$ in some fashion which is explicitly described.
It seems like this question must have been extensively studied by now. But I was unable to find a suitable reference. Two references that approach what I'm looking for are:

Schmidt (Lemma 1, Northcott's theorem on Heights II. The quadratic case, Acta Arith., 1995) proves that if $S$ is (for example) a convex body in $n$ dimensions, lying in a ball of radius $r$, and $\Lambda$ is any lattice of rank $r$, the number of lattice points of $\Lambda$ inside $S$ is $$\frac{\rm{Vol}(S)}{\det(\Lambda)} + O\left(\frac{\lambda_n r^{n - 1}}{\det(\Lambda)}\right),$$ where $\lambda_n$ is the largest successive minimum of $\Lambda$. After a change of variable this is equivalent to my question, and the answer is of the form I am looking for -- but I believe a better error term should be possible when $S$ is an ellipsoid.
Bentkus and Götze (main theorem, On the lattice point problem for ellipsoids, Acta Arith., 1997) formulate the question in the same way that I did, and obtain a power savings of $Z$ in the error term (as opposed to $Z^{1/2}$, which is what can be deduced from Schmidt's paper or any similar geometry of numbers method). Writing the quadratic form as $\langle Qx, x \rangle$, the theorem specificies the dependence of the error term on the eigenvalues of $Q$. This is of the shape that I'm interested in, but the paper requires that the dimension be at least 9.

I skimmed through the likelier looking references in the latter paper, as well as the books on counting lattice points of Krätzel and Fricker, and I found nothing. The problem of counting lattice points in three-dimensional ellipsoids is addressed, but if the dependence of the error term on the ellipsoid is made explicit, then I missed it.
Finally I should mention that I know how to solve my own problem: write down the Epstein zeta function associated to $Q$, and estimate its partial sums using Perron's formula and the method of Landau and Chandrasekharan-Narasimhan. (The methods in the books above don't use Epstein zeta functions, but after a brief reading they seem like equivalent arguments that don't go through the usual zeta function machinery.) The dependence of the error term on $Q$ can be described in terms of the functional equation for the Epstein zeta function.
But I would prefer to avoid inventing the wheel if I can help it. Does anyone know if such a theorem has already been proved? Thank you very much.

REPLY [4 votes]: As far as I can tell, the best result (too painful to typeset) is in:
Jingwei Guo, On lattice points in large convex bodies, Acta Arith. 151 (2012), no. 1, 83--108.<|endoftext|>
TITLE: Automorphisms of cartesian products of curves
QUESTION [10 upvotes]: Let $C$ be a smooth projective curve. Is it true that
$$\textrm{Aut}(C\times C)\cong S_2 \ltimes (\textrm{Aut}(C)\times \textrm{Aut}(C))$$
and in case, what would be a reference for this? Thanks.

REPLY [9 votes]: This is a particular case of a more general rigidity result, whose proof (similar to the one given in abx's answer) can be found in Lemma 3.8 of 
F. Catanese, Fibred surfaces, varieties isogenous to a product and related moduli spaces, Amer. J. Math. 122 (2000), no. 1, 1-44.

Proposition. Let $f \colon C_1 \times C_2 \longrightarrow B_1 \times B_2$ be a surjective holomorphic map between products of curves. Assume that both $B_1$ and $B_2$ have genus $\geq 2$. Then, after possibly exchanging $B_1$ with $B_2$, there are holomorphic maps $f_i \colon C_i \to B_i$ such that $$f(x, \, y) = (f_1(x), \, f_2(y)).$$<|endoftext|>
TITLE: Reference request: seminar report of Serre from late 60s on possibility of Galois representations attached to modular forms?
QUESTION [8 upvotes]: See here for a comment of Matt Emerton.

There are also various seminar reports of Serre, e.g. his report on mod p modular forms, but also his report from the late 60s on the possibility of Galois reps. attached to modular forms.

My question is, with regards to "his report from the late 60s on the possibility of Galois reps. attached to modular forms," what is the specific report he is referring to here?

REPLY [5 votes]: Googling reveals these lectures slides of Ken Ribet, which say the following.

After Serre's article on elliptic curves was written in the early 1970s, his techniques were generalized and extended in different directions. In particular, Serre and Swinnerton-Dyer began to study the mod $\ell$ representations of $\text{Gal}(\overline{\textbf{Q}}/\textbf{Q})$ attached to the cusp form $\Delta$ of weight $12$ on $\textbf{SL}(2, \textbf{Z})$.
These representations weren't always with us: It was only in his 1967-1968 DPP seminar on modular forms ("Une interprétation des congruences relatives à la fonction $\tau$ de Ramanujan") that Serre proposed the possibility of linking Galois representations to holomorphic modular forms that are eigenforms for Hecke operators. Almost immediately afterwards, P. Deligne constructed the representations whose existence was conjectured by Serre.

This suggests the specific report is the following, see here.

Serre, Jean-Pierre. "Une interprétation des congruences relatives à la fonction $\tau$ de Ramanujan." Séminaire Delange-Pisot-Poitou. Théorie des nombres 9.1 (1967-1968): 1-17.

A pithy one sentence review of the report by B. Stolt on MathSciNet says the following.

The author makes a survey of results on Ramanujan's $\tau$-function.<|endoftext|>
TITLE: Evolution of partial sum of a sequence of induced Dirichlet characters
QUESTION [5 upvotes]: Let's consider the Dirichlet Character $\chi_3(n)$ modulo 3 given by $\chi_3(1)=1$, $\chi_3(2)=-1$ and $\chi_3(3)=\chi_3(0)=0$.
Lets consider the sequence of induced characters $\chi^{P_N} $ obtained from $\chi_3$ having modulo $3.P_1...P_N$ where $P_N$ is the $N$-th prime number.
So at each step to define the new Character we replace by zero the value of multiple of $P_N$, we have :
$\chi_3 = (0,1,-1,0,1,-1,0,1,-1,...)$
$\chi^{P_1} =(0,1,0,0,0,-1,0,1,0,0,...)$ (all multiple of 2 have been put ot zero)
$\chi^{P_3} =(0,1,0,0,0,0,0,1,0,0,...)$ (all multiple of 5 have been put to zero)
So we obtain a sequence of induced characters with increasing modulo. My question is on the behavior of the maximum of the partial sum of the $\chi^{P_N}$ as $N$ increases :
$$S(\chi^{P_N},x) = |\sum_{n=1}^{x} \chi^{P_N}(n)|$$
$$Max_x(S(\chi^{P_N},x)$$
How this maximum will evoluate ?
My feeling is that this maximum will go up and down when $N$ goes to infinity and so that for a fixed constant $M$ (big enough) we can find an infinity of induced characters with this maximum lower than $M$. But is it true ? How to prove this is right or wrong ? Any reference on similar problem or idea ? Sieve theory can solve this kind of problem ?
(I already made a post on the subject in a more general way here : On partial sum of non-primitive Dirichlet characters)

REPLY [2 votes]: The problem you are looking at can be stated in terms of the sum
$$
S(x,y):= \sum_{\substack{n\le x \\ P^-(n)>y}} \chi_3(n) ,
$$
where $P^-(n)$ denotes the smallest prime divisor of $n$, with the convention that $P^-(1)=+\infty$. If $y\ge3$, then this sum is $Q$-periodic, where
$$
Q=\prod_{p\le y} p = e^{(1+o(1))y}.
$$
Because the modulus is very smooth, the Polya-Vinogradov bound can be improved quite a bit using the Eratosthenes-Legendre sieve:
$$
S(x,y)= \sum_{n\le x} \chi_3(n) \sum_{d|(n,Q)} \mu(d)
 = \sum_{d|Q} \mu(d)\chi_3(d) \sum_{m\le x/d}\chi_3(m) .
$$
The inner sum is $O(1)$ by the 3-periodicity of $\chi_3$, so that
$$
S(x,y) \ll \sum_{d|Q} 1 = 2^{\pi(y)} = Q^{(\log 2+o(1))/\log\log Q} ,
$$
which is $Q^{o(1)}$ as $y\to\infty$.
It is even possible to go further: note that
\begin{align}
\sum_{n\le N} \chi_3(n) 
 =\sum_{\substack{k\ge0 \\ 1+3k\le N}} 1 - \sum_{\substack{k\ge0 \\ 2+3k\le N}} 1
 &= \left\lfloor \frac{N-1}{3} \right\rfloor -  \left\lfloor \frac{N-2}{3}
 \right\rfloor \\
 &= \begin{cases} 1 &\mbox{if $N$ is in $[1,2)$ mod 3},\\
0 &\text{otherwise}
\end{cases},
\end{align}
so that
$$
S(x,y) = \sum_{d|Q,\, d\in D} \mu(d)\chi_3(d),
$$
where
$$
D := \bigcup_{n\in\mathbb{Z}_{\ge0}} \left( \frac{x}{3n+2}, \frac{x}{3n+1}\right].
$$
This can be now estimated using smooth number technology. In particular, the interesting case is when $x$ is of size $Q$, in which case lattice point count heuristics start becoming accurate. What you could try and use now is either the saddle point method (see Tenenbaum's book "Introduction to Analytic and Probabilistic Number Theory"), or to try and adapt recent work of R. de la Breteche and G. Tenenbaum (see http://iecl.univ-lorraine.fr/~Gerald.Tenenbaum/PUBLIC/Prepublications_et_publications/Psi-resM.pdf.)<|endoftext|>
TITLE: Reference for Nori motives
QUESTION [11 upvotes]: I would like to study Nori motives and I am a complete outsider of the subject. I do, however, have background on Chow motives, Voevodsky motives $\mathrm{DM}$ and his stable homotopy category $\mathrm{SH}$.
My question is:
Do you know an introductory reference and the main papers about Nori motives?
Thank you very much

REPLY [4 votes]: The PhD thesis of D. Harrer is now available on Arxiv. 
https://arxiv.org/ftp/arxiv/papers/1609/1609.05516.pdf<|endoftext|>
TITLE: Which closed 3-manifolds can be embedded in $R^4$?
QUESTION [14 upvotes]: I wonder which closed orientable 3-manifolds can be embedded in $\mathbb R^4$ and which in $\mathbb R^5$. Is there a way to determine whether given closed 3-manifold, obtained, say by Dehn surgery on knot, can be embedded into $R^4$ ?
Is the answer known for spherical 3-manifolds (finite fundamental group) ?
I am mainly interested in topological properties of manifolds. The known answers for PL or smooth embedding are of value as well for me. I appreciate any kind of answer. I am not interested in such peculiarities as exotic $R^4$.

REPLY [12 votes]: There is an analogy to surfaces in a sense.  For 3-manifolds that fibre over surfaces there is a complete answer.  For a variety of Seifert-fibred manifolds there are complete answers -- but not all.  For example, Seifert-fibred homology spheres are still problematic.   The preprint that Ian linked to in his comments has much more results of this kind in it. 
At present, in summary:
1) We likely do not have a complete set of invariants that obstruct embedding into $\mathbb R^4$.  
2) We appear to be far from knowing all the "natural" constructions of embeddings of 3-manifolds into $\mathbb R^4$ for the manifolds that are known to embed.  
It is quite possible there are elements of formal logic obstructing both 1 and 2.  
For example, if a compact boundaryless connected 3-manifold embeds in $S^4$ it separates it into two components.  It is possible that one or even both of these components has a fundamental group with an unsolvable word problem.  This would restrict the kinds of techniques one could use for creating obstructions in (1). 
edit: I see Agol and Freedman's paper on this topic as connected to this last concern.  2-manifolds in $S^3$ have the Fox re-embedding theorem.  So you could hope for some nice re-embedding theorems for $3$-manifolds in $S^4$.  You shouldn't expect too nice a re-embedding theorem in $S^4$, since the tool that makes Fox's theorem work is Dehn's lemma, and the analogies to Dehn's lemma in 4-manifold theory are generally not true.<|endoftext|>
TITLE: A Hom-Tensor identity - $\text{Hom}_{R}(P,B)\otimes _SC \cong \text{Hom}_{R}(P,B \otimes_S C) $
QUESTION [6 upvotes]: let $R,S$ be associative algebras over $\mathbb{C}$. Let $\mathcal{C} \subseteq$ $R$-Mod be a full abelain subcategory of $R$-Mod which is the category of $R$-modules. Let $B$ and $C$ be, a $(R,S)$-bimodule, a left $S$-module, respectively.
In addition, let $P\in \mathcal{C}$ be a projective object in $\mathcal{C}$.  Assume that $P$ is finitely-generated $R$-module.
Note that $\text{Hom}_{R}(P,B)$ and $B\otimes_S C$ are right $S$-modules and left $R$-modules in a natural way.
$\bf{My \ \ Question:}$ 

Prove that $\text{Hom}_{R}(P,B)\otimes _SC  \cong \begin{align} \text{Hom}_{R}(P,B \otimes_S C) \end{align}$ provided $\text{dim}C< \infty$, $\text{dim}B\otimes_S C <\infty$ and $B\in \mathcal{C}$.

In fact, I am not sure whether the condition on finite-dimensional spaces are suitable. It is just to make this equality correct. I would be appreciated if anyone can help me improve or correct this conditions. 
My idea is that the map $\Phi:\text{Hom}_{R}(P,B)\otimes _SC  \rightarrow \begin{align} \text{Hom}_{R}(P,B \otimes_S C) \end{align}$ given by  
$\Phi(f \otimes v)(a) : = f(a)\otimes v$, for all $a\in P, f\in \text{Hom}_{R},(A,B) $ and $v\in C$, is an isomorphism. Thank you in advance!

REPLY [5 votes]: Fix a finitely generated left $R$-module $_RP$ that is in $\mathcal{C}$ and projective as an object of $\mathcal{C}$, and a finite dimensional left $S$-module $_SC$.
As noted in the question, for any $R$-$S$-bimodule $_RM_S$ there is an obvious map
$$\Phi_M:\operatorname{Hom}_R(P,M)\otimes_SC\to\operatorname{Hom}_R(P,M\otimes_SC),$$ 
and it is natural in $M$.
If $M$ is of the form $X\otimes_\mathbb{C}S$ for $_RX$ a left $R$-module, then $\Phi_M$ is an isomorphism, using the fact that $_RP$ is finitely generated as a left $R$-module, so that the natural map
$$\operatorname{Hom}_R(P,X)\otimes_\mathbb{C}V\to
\operatorname{Hom}_R(P,X\otimes_\mathbb{C}V)$$ is an isomorphism for any vector space $V$.
The standard projective bimodule presentation
$$S\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to S\otimes_\mathbb{C}S\to S\to0$$
of $S$ induces an exact sequence
$$\tag{*}
B\otimes_\mathbb{C}S\otimes_\mathbb{C}S\to
B\otimes_\mathbb{C}S\to B\to0,$$
which is split as an exact sequence of left $R$-modules, so that applying the functor $M\mapsto\operatorname{Hom}_R(P,M)\otimes_S C$ to it preserves exactness.
Also, applying $-\otimes_SC$ to the exact sequence $(*)$ gives an exact sequence
$$\tag{**}
B\otimes_\mathbb{C}S\otimes_\mathbb{C}C\to
B\otimes_\mathbb{C}C\to B\otimes_SC\to0.
$$
As a sequence of left $R$-modules, the first term is a (possibly infinite) direct sum $\bigoplus_{i\in I}B$ of copies of $B$ and the second is (since $C$ is finite dimensional) a finite direct sum of copies of $B$.
This is the filtered colimit, over finite subsets $J\subseteq I$, of exact sequences of the form
$$\tag{***}
\bigoplus_{j\in J}B\to B\otimes_\mathbb{C}C\to U_J\to 0,$$
where the first two terms are in $\mathbb{C}$ since $_RB$ is in $\mathcal{C}$, and since $\mathcal{C}$ is an abelian subcategory, the third term is also in $\mathcal{C}$. Therefore $\operatorname{Hom}_R(P,-)$ preserves exactness of the sequences $(***)$ since $P$ is projective in $\mathcal{C}$. Since $P$ is finitely generated as an $R$-module, 
$\operatorname{Hom}_R(P,-)$ also preserves filtered colimits, and so applying $\operatorname{Hom}_R(P,-)$ to $(**)$ also gives an exact sequence.
So applying
$$\Phi:\operatorname{Hom}_R(P,-)\otimes_SC\to\operatorname{Hom}_R(P,-\otimes_SC)$$
to the sequence $(*)$ gives a map of exact sequences which is an isomorphism on the first two terms and therefore also on the third term.
So $\Phi_B$ is an isomorphism.
This doesn't seem to require $B\otimes_SC$ to be finite dimensional.<|endoftext|>
TITLE: A general version of the 5 lemma
QUESTION [10 upvotes]: Suppose you have an abelian category $\bf A$, and $A\to B\to C$, $A'\to B'\to C'$ two exact sequences, in a diagram
$$
\begin{array}{cccccccc}
0 &\to & A &\to& B &\to& C &\to &  0\\
&&\downarrow && \downarrow  && \downarrow \\
0 &\to &  A' &\to& B' &\to& C' &\to &  0
\end{array}
$$
(i.e., suppose you have a morphism of exact sequences $(f,g,h)$); let $(\cal E,M)$ be a factorization system on $\bf A$, and suppose $f,h$ lie in $\cal E$. 
Under which conditions (on the category, or on the factorization systems) $g\in\cal E$? Same question for $f,h\in\cal M$ $\Rightarrow g\in\cal M$.

I'm not assuming that $(\cal E,M)$ is proper (so there can be no relation between $(\cal E,M)$ and $(Epi, Mono)$);
For the motivating example I have in mind with this question, assuming something more on the factorization system is more natural than assuming it for the category.

REPLY [3 votes]: One condition on the factorization system that should work is that "factorizations preserve exact sequences", i.e. if you have a map between exact sequences and you $(E,M)$-factor it componentwise, then the intermediate objects also form an exact sequence.  If this is true, then you can factor your given map of exact sequences and apply the ordinary 5-lemma, using the fact that $f\in E$ iff the $M$-part of its $(E,M)$-factorization is an isomorphism.
However, this assumption may not be much easier to check in an example than to check your desired conclusion directly!  (-:<|endoftext|>
TITLE: Affinely flat structures. How many different ones on the same manifold?
QUESTION [5 upvotes]: Let $M$ be a smooth manifold endowed with $\nabla$ a flat torsion-free connection. Let $\operatorname{Diff}(M)$ be the group of smooth diffeomorphisms of $M.$ Obviously if $ \phi \in \operatorname{Diff}(M)$, then $\phi^* \nabla$ is torsion-free and flat, but equivalent (through $\phi$) to $\nabla.$ 
Is it known whether the set of nonequivalent affinely flat connections is infinite (if it is not empty)?. In the case of the $2$-dimensional torus it is known that the set of flat, torsion free connections quotient out by the group of diffeomorphisms is a $4$-dimensional manifold.

REPLY [4 votes]: The set of affinely flat structures has been studied by many authors. Among them, we can quote William Goldman who has studied the deformations of $(X,G)$-structures. Let $X$ be a manifold, $G$ a Lie group which acts on $X$, we suppose that if two elements of $G$ coincide on an open subset of $X$, they coincide on $X$. A manifold $M$ is an $(X,G)$-manifold if it is endowed with an atlas $(U_i,\phi_i)$, where $\phi_i\colon U_i\subset M\rightarrow X$ is a diffeomorphism between $U_i$ and its image. We suppose that $\phi\circ {\phi_j}^{-1}$ is the restriction of an element of $G$. 
A flat torsionless connection defined on the $n$-dimensional manifold $M$ defines on it a $(\mathbb{R}^n,\operatorname{Aff}(\mathbb{R}^n))$-structure.
The $(X,G)$-structure of $M$ can be lifted to its universal cover $\tilde M$, and there exists a developing map: $D\colon\tilde M\rightarrow X$ such that for every element $\gamma\in\pi_1(M), D\circ \gamma =h(\gamma)\circ D$. We thus have a representation $h\colon\pi_1(M)\rightarrow G$ called the holonomy representation of the $(X,G)$-structure on $M$. It is a result of Thurston that if $M$ is closed the set of holonomy of $(X,G)$-structures defined on $M$ is an open subset of $\operatorname{Rep}(\pi_1(M),G)$, the set of representations $h:\pi_1(M)\rightarrow G$.
Goldman, William M. "Varieties of representations." Geometry of Group Representations: Proceedings of the AMS-IMS-SIAM Joint Summer Research Conference Held July 5-11, 1987 with Support from the National Science Foundation. Vol. 74. American Mathematical Soc., 1988.
Available here:
http://www-users.math.umd.edu/~wmg/geost.pdf
For the dimension 2, a theorem of Benzecri shows that an oriented surface is endowed with an affine structure if and only if its genus is one.<|endoftext|>
TITLE: a question about Bockstein spectral sequence
QUESTION [6 upvotes]: I find the following theorem for Bockstein spectral sequence at http://pages.vassar.edu/mccleary/files/2011/04/MC10.fin_.pdf, page 459:

Question. for a fixed $k$, if $\beta$ does not hit $H_k(X;\mathbb{F}_p)$ hence $B^1_k$ survives at the second page, can we conclude that $B^1_k$ survives at the infinity page thus $B^1_k$ is a summand of $(H_*(X;\mathbb{Z})/Torsion)\otimes \mathbb{F}_p$?

REPLY [8 votes]: No, you can't necessarily make this conclusion. Your spectral sequence is obtained from another doubly-graded Bockstein spectral sequence of the form
$$
E^1_{r,s} = \begin{cases}
H_{r+s}(X;\Bbb F_p) &\text{if }s+t \geq 0, s \geq 0\\
0 &\text{otherwise}
\end{cases}
$$
with $d_1$ differential the Bockstein and $E^\infty$-page the associated graded of $H_*(X;\Bbb Z_p)$ by the multiples-of-$p$ filtration. (You can think of the $E^1$-page as $H_*(X;\Bbb F_p) \otimes \Bbb F_p[v]$, where $v$ is a polynomial generator in bidegree $(r,s) = (-1,1)$ representing multiplication-by-$p$.)
In this spectral sequence (with the assumption that $X$ has finite type), you get a $d^k$-differential if and only if there was a summand of the form $\Bbb Z/p^k$ in $H_*(X;\Bbb Z)$; this becomes a summand $\Bbb F_p[v]/v^k$ in the $E_\infty$ page. You basically get your spectral sequence by inverting $v$, and so the same holds there: there is a $d^k$-differential if and only if the original space had a $\Bbb Z/p^k$-summand in its homology.<|endoftext|>
TITLE: Completion of spaces of measures w.r.t. weak norms
QUESTION [8 upvotes]: For the sake of concreteness denote by $M_0(X)$ the linear space of all signed Borel measures $\sigma$ with $\sigma(X)=0$ on some metric space $(X,d)$ and fix some base point $x_0\in X$. On this space define the norm
$$
\newcommand{\norm}[1]{\|#1\|}
\newcommand{\Lip}{\mathrm{Lip}}
\norm{\sigma}_0^* = \sup\{\int_X f\,d\sigma\ :\ \Lip f\leq 1,\ f(x_0)=0\}
$$
where the supremum is taken over all Lipschitz functions on $X$ and $\Lip f$ denotes the Lipschitz constant of $f$. In Bogachev's "Measure theory" (§8.10) this norm is called Kantorovich-Rubinshtein norm and it is shown that convergence in the Kantorovich-Rubinshtein metric implies weak convergence. Moreover, it is stated that $M_0(X)$ is not complete with this norm provided that $X$ is not complete. This can be seen by as follows: Assume that there exist sequences $x_k,y_k\in X$ which do not converge, but $d(x_k,y_k)\to 0$. Then define the sequence $\sigma_k = \delta_{x_k}-\delta_{y_k}$ and observe that
$$
\norm{\sigma_k}_0^* \leq d(x_k,y_k)\to 0
$$
in other words $\sigma_k\to 0$ w.r.t $\norm{\cdot}_0^*$. However, $\sigma_k$ does not convergence weakly to zero.
My question is:
What is the completion of $M_0(X)$ w.r.t. $\norm{\cdot}_0^*$?
I would like some description as a dual space or derived from some space of measures.
It feels like it should be something like a dual space of differentiable functions, or derivatives of some measure…

REPLY [11 votes]: This is known as the Arens-Eells space $AE(X)$. In the nonlinear Banach space literature it's also called the Lipschitz-free space $\mathcal{F}(X)$. It is not a dual space in general, but rather the predual of the space ${\rm Lip}_0(X)$ of Lipschitz functions which vanish at $x_0$.
In some cases it is a dual space. For instance, if $X$ is a compact metric space and $0 < \alpha < 1$, let $X^\alpha$ be the set $X$ equipped with the "Holder" or "snowflake" metric $\rho^\alpha$. In this case $AE(X^\alpha)$ is the dual of the space of "little" Lipschitz functions which are locally flat.
The space $AE(X)$ is characterized abstractly by its universal property: there is a natural isometric embedding of $X$ in $AE(X)$ (in your formulation, take the point $p$ to the Dirac measure at $p$ minus the Dirac measure at $x_0$), and for any Banach space $E$ and any Lipschitz function $f: X \to E$ which takes $x_0$ to $0$ there is a unique bounded linear extension $T: AE(X) \to E$. Moreover, $\|T\| = Lip(f)$.
You can find out all you want about this space in Chapter 3 of my book Lipschitz Algebras (second edition).<|endoftext|>
TITLE: Weak Mordell-Weil over number fields
QUESTION [11 upvotes]: I have a question regarding the Mordell Weil theorem a number field $K$.
I read the proof of the Mordell Weil theorem in "rational points on elliptic curves" by Tate and Silverman.
They presented a proof for the case where $E[2] \in E (\mathbb{Q}) $ 
where $E : Y^2 = X(X^2 + AX + B)$ and mentioned before hand that it is possible to prove the case over number fields in the same fashion (thus without the use of group cohomology) with a little help of algebraic number theory.
After a bit of investigating, I concluded that every reasoning also holds for number fields except for the proof that the image of the map $\alpha : E(K) \rightarrow K^* / K^{*2}$ is finite (proposition 3.8(c)).
In the proof they claimed that every squarefree integer representing the corresponging quadratic residue class divides $B$.
My reasoning for number fields was as follows:
Let $P = (x,y) \in E$, then $\alpha(P) = x \pmod {K^{*2}}$.
Since $K$ is the field of fractions of $O_K$ (ring of integers) 
we have that $x = \frac{a}{b}$ for $a,b \in O_K$.
Now consider $S_P := \{ \rho \in \text{Max}(O_K) \: : \: v_{\rho}(\alpha(P)) \neq 0 \pmod 2 \}$ where $v$ is just the valuation of the prime ideal factorization.
My claim is that every prime ideal in $S_P$ must be a divisor of the ideal generated by $B$.
My question: is this correct? And if so, how do I prove this?
Thanks in advance.

REPLY [11 votes]: That's more or less the right way to do it. The proof of your claim (or something similar) will require (1) finiteness of the ideal class group (or at least, the 2-part of the class group) and (2) finite generation of the group of units in $O_K$, i.e., Dirichlet's unit theorem, although again what's really needed is that $O_K^*/(O_K^*)^2$ is finite. Anyway, you'll find all this in Chapter VIII Section 1 of The Arithmetic of Elliptic Curves. The claim is more-or-less the content of Propositions VIII.1.5 and VIII.1.6, although the specific statement is a bit buried in the proofs of those propositions. There's no group cohomology involved, but in some sense one is looking at 1-cocycles for the step where one goes to an extension field where the 2-torsion is rational; cf. the reduction Lemma VIII.1.1.1.<|endoftext|>
TITLE: Isometry group of low dimensional Alexandrov space
QUESTION [8 upvotes]: It is known by the work of Galaz-García and Guijarro, that the dimension of the isometry group of an $n$-dimensional Alexandrov space (of curvature bounded below) is bounded above by $\frac{n(n+1)}{2}$, however little is said about the groups or the isometries themselves.  
Is it known what the isometry groups of low dimensional ($n=2,3$) Alexandrov spaces are? Is it known for example for spaces with $\mathrm{curv}\geq0$?

REPLY [2 votes]: I don't think much is known about this. As far as I know, in dimension three a complete answer is only known in the case of spherical space forms: McCullough, Darryl. Isometries of elliptic 3-manifolds. J. London Math. Soc. (2) 65 (2002), no. 1, 167--182. MR1875143.<|endoftext|>
TITLE: Dimension of a homotopy type
QUESTION [11 upvotes]: What is the state of knowledge about the dimension of homotopy types? By the latter I mean the minimal number which is the dimension of a topological space representing the homotopy type. The open Eilenberg–Ganea conjecture concerns 1-types. What is known for n-types?
By the dimension of a CW complex I mean the dimension of its highest cell. For general spaces one can take Lebesgue covering dimension, however the primary interest is in CW complexes.

REPLY [12 votes]: If you are willing to make "niceness" assumptions, assuming that the n-type is, say, nilpotent with finite fundamental group, then the dimension, in any reasonable sense, will always be infinite, since it will have homology in infinitely many dimensions. 
The first result along these lines is due to Serre:
J.-P. Serre, Cohomologie modulo 2 des complexes d’Eilenberg–Mac Lane, Comment. Math. Helv. 27 (1953), 198–232.
I wrote a couple of papers about this back in the days:
http://www.math.ku.dk/~jg/papers/postnikov.pdf
http://www.math.ku.dk/~jg/papers/serre.pdf
where I prove that the cohomology, also as a ring, will be huge.
If you don't want to make assumptions on the fundamental group, then it is an open problem.
There are no known examples of n-types that are not $K(G,1)$s but that are homotopy equivalent to finite dimensional CW-complexes. And that there are no such examples has been stated as a conjecture by Casacuberta-Castellet in "Mislin's book of conjectures":
https://atlas.mat.ub.edu/personals/casac/articles/Mislin.pdf<|endoftext|>
TITLE: Inverse of a polynomial map
QUESTION [5 upvotes]: Ax-Grothendieck Theorem states that if $\mathbf K$ is an algebraically closed field, then any injective polynomial map $P:\mathbf K^n\longrightarrow \mathbf K^n$ is bijective.
Question 1. What does the inverse map of $P$ look like ? What kind of map is that ?
$P^{-1}$ need not be polynomial, as the example $x^p$ in $\mathbf F_p^{alg}$ shows.
Question 2. Are there conditions under which $P^{-1}$ is polynomial ?

REPLY [4 votes]: In light of abx's comment, I came up with the following argument. I'm sure some version of this must be in the literature somewhere.
Recall that a dominant morphism $Y \to X$ of varieties is separable if $K(X) \to K(Y)$ is a separable field extension. In characteristic $0$, this is automatic.

Theorem. Let $k$ be an algebraically closed field, and let $f \colon Y \to X$ be a morphism of $k$-varieties, with $X$ normal. If $f$ is bijective (in particular, dominant) and separable, then $f$ is an isomorphism.

Proof. By a suitable version of Zariski's main theorem (see e.g. Tag 05K0), there exists an open immersion $Y \subseteq Z$ and a finite morphism $g \colon Z \to X$ extending $f$. By restricting to the underlying reduced scheme of the unique irreducible component of $Z$ containing $Y$, we may assume $Z$ is integral. In particular $K(Z) = K(Y)$, and $g$ is separable since $f$ is.
Since $g$ is separable, it is generically étale, hence there exists an open $U \subseteq X$ such that
$$g \colon g^{-1}(U) \to U$$
is finite étale. In particular, $g$ is finite flat over $U$ of some rank $r \geq 1$. If $x \in U$ is a closed point, then $f^{-1}(x) \to x$ is finite étale of rank $r$, thus a disjoint union of $r$ copies of $\operatorname{Spec} k$ (since $k$ is algebraically closed). Since $f$ is bijective, we have $r = 1$, so $f$ is birational, i.e.
$$K(X) \stackrel \sim \to K(Z).$$
Since $X$ is normal and $g$ finite, this forces $g$ to be an isomorphism. Since $f$ is bijective, we conclude that $Y = Z$ and $f = g$. $\square$
Remark. I do not know how to get rid of the algebraically closed condition without weakening the hypotheses to $f$ being universally bijective (in a restricted sense: we only need it for base change along field extensions) and $X$ geometrically normal (when $k$ is perfect, this is equivalent to normal).<|endoftext|>
TITLE: Sixteen points circle - A conjecture on Möbius plane
QUESTION [5 upvotes]: The conjecture refer the reader about the Bundle's theorem configuration. (This conjecture from a note)
Consider the Bundle theorem configuration :
Points $A_1, A_2, A_3, A_4$ lie on a circle,
points $B_1, B_2,B_3, B_4$ lie on a circle, 
points $A_1, A_2, B_1, B_2$ lie on a circle, 
points $B_1, B_2, A_3, A_4$ lie on a circle, 
points $B_3, B_4, A_3, A_4$ lie on a circle, 
and points $A_3, A_4, A_1, A_2$ lie on a circle.
Let the pair of circles $(P_1P_3Q_i)$ and $(P_2P_4Q_j)$ is such that {if $P=A$ then $Q=B$} or {if $P=B$ then $Q=A$} and if {$i=1$ then $j=2$} or {if $i=2$ then $j=1$} or {if $i=3$ then $j=4$} or {if $i=4$ then $j=3$}. Hence, there are eight pair of circles with this definition.
With one pair of circles $(P_1P_3Q_i)$ and $(P_2P_4Q_j)$ we have two common points. Hence, we have 16 points of intersection of $8$ pairs of circles.

The conjecture: The sixteen points of intersection of the eight pairs of circles lie on a circle.

The problem is true for Euclidean plane geomtry, and it be constructed on Bundle theorem configuration, the bundle theorem true for Möbius plane. I don't know the conjecture is also true for the Möbius plane? I didn't find what I'm looking for.

REPLY [2 votes]: There is a circle $\omega$ perpendicular to all yellow circles (the center of it is intersection of  $A_1A_2$ and $A_3A_4$). If you make inversion in $\omega$ points $A_1$ and $A_2$ go to each other, the same holds for pairs $A_3$ and $A_4$, $B_1$ and $B_2$, $B_3$ and $B_4$. That means that your circles $(P_1P_3Q_i)$ and $(P_2P_4Q_j)$ symmetric with respect to $\omega$, therefore the points of intersection of them lie on $\omega$, which is the green circle on your figure.<|endoftext|>
TITLE: Is the number of representations as the sum of two elements of a polynomial sequence always small?
QUESTION [15 upvotes]: Let $f(x) \in \mathbb{Z}[x]$ be a degree $d>1$ polynomial with integer coefficients. Define 
$$r(n) := | \{x,y \in \mathbb{Z} : f(x)+f(y) = n  \}|. $$
My question is:


Is it true that $r(n) \ll_{\epsilon} n^{\epsilon} $ for $\epsilon >0$? 


In certain cases (such as $ f(x)=x^{2k}$), one can "factor" the problem and deduce the desired result from the divisor bound. However, I do not see how to approach the general case in this manner. I am aware that there is a weaker but more general result of Bombieri and Pila which states that 
$$r'(n,M) := | \{x,y \in \mathbb{Z} : f(x,y) = n, |x|,|y| < M  \}| $$
satisfies $r'(n,M) \ll_{\epsilon} M^{1/d + \epsilon} $when $f(x,y)$ is an absolutely irreducible polynomial of degree $d$. In this greater level of generality this is nearly best possible as can be seen by taking $f(x,y) = x^d -y$.

REPLY [2 votes]: I don't have the rep to comment so I will just mention something here, though I would guess you might have recognized it already. 
You can certainly bound the number of solutions to $f(x)-f(y)=n$ by writing $$f(x)-f(y)=(x-y)H(x,y)$$ for some polynomial $H$ that depends only on $f$. Then $x-y=d$ divides $n$ and $H(x,x-d)=n/d$ so there are $O_f(1)$ choices for $x$. Going from differences to sums is probably hard.<|endoftext|>
TITLE: Bousfield Localization and Quillen Equivalence
QUESTION [6 upvotes]: The notion of a (left, say) Bousfield localization of a model category doesn't seem to be invariant under Quillen equivalence. There are a lot of things that could go wrong. But I don't know any examples. So if anyone could help me out with even one of the below questions, I'd appreciate it.
Let $M$ be a model category and $\mathrm{Ho} M$ its homotopy category. 

At the most basic level, what is an example where there exists a localization of $\mathrm{Ho} M$ which isn't the homotopy category of a left Bousfield localization of $M$? This isn't usually how left Bousfield localizations are presented (although it seems the most invariant way I can think of), so, as a bonus: what is an example where there is a localization of $\mathrm{Ho} M$ where the inverted objects aren't of the form $\{S-\text{local objects}\}$ for some class $S$?
More specifically, what is an example of (1) where the implied acyclic cofibrations aren't generated by a small set (actually I'd be happy to know just this, but of course what I'm driving at is that they shouldn't admit factorizations)?
Alternatively, what is an example of (1) where the implied acyclic cofibrations are perhaps generated by a small set, but not one which admits the small object argument (and they don't have factorizations)?
Pressing forward, if $F: M \overset{\to}{\to} N: U$ is a Quillen equivalence and $M'$ is a left Bousfield localization of $M$, then if the model structure $N$' induced by $M'$ along $F$ exists, it is a left Bousfield localization of $N$ and $F: M' \overset{\to}{\to} N': U$ is a Quillen equivalence. What's an example where $N'$ fails to be a model structure? Again it's interesting to ask about the various ways this could go wrong.
In the other direction, it seems dicier to try to induce a left Bousfield localization along a right Quillen equivalence. I guess that dualizing the argument from (4) will show that if the induced model structure exists, it is a left Bousfield localization and the Quillen equivalence restricts to a Quillen equivalence between the localizations. But it must be not unusual that this induced model structure fails to exist, right?
What is an example of a Bousfield localization that can't be induced from a Bousfield localization along some particular Quillen equivalence? How about two different Bousfield localizations that induce the same Bousfield localization along a Quillen equivalence?
Are there any model categories where all Bousfield localizations are known to exist, not just those generated by a small set?

REPLY [7 votes]: For (1)-(2), look at work of Carles Casacuberta. He has lots of good examples. His paper with Chorny on the orthogonal subcategory problem has an example for your (2), on the last page. This paper of Casacuberta-Chorny goes into great depth about (3) as well, and it led to Chorny's work on class combinatorial model categories and localizations (when the model structure is generated by a class, and you have a generalized small object argument). This paper may also include an example for (1). If it doesn't, I can dig out an example Casacuberta showed me last summer and send you a sketch by email (it's unpublished and not mine, so I don't want to post it here).
Part of my thesis can answer (4). You could have a Quillen equivalence of monoidal model categories $M_1$ and $M_2$, and suppose both satisfy the commutative monoid axiom so that commutative monoids inherit transferred model structures. There was a condition in my thesis that if the free commutative monoid functor Sym(-) preserves $C$-local equivalences then a localization exists at the level of commutative monoids. Since this condition need not be preserved by a Quillen equivalence, you can have a left Bousfield localization of $M = CMon(M_1)$ that does not transfer to a left Bousfield localization of $N = CMon(M_2)$.
For (5), I don't have an example off the top of my head, but I agree with you that I would not expect this to come for free. This would be a strange example, because on the infinity category level there's no difference between left Quillen equivalence and right Quillen equivalence, and there I do expect localizations to be induced (assuming everything in sight is presentable and accessible). So maybe look at some very simple model structures with a lot going on with the cofibrations and fibrations.
For the first part of (6), I see two ways to interpret this. Every Bousfield localization is induced by the identity functor, so I assume you didn't mean that. For the other interpretation, if someone hands me a Quillen equivalence and a Bousfield localization of one of the two model categories I see no reason at all to expect it to be induced by a Bousfield localization on the other piece. For the other part of (6), what do you mean by "two different Bousfield localizations"? If they provide Quillen equivalent model structures most people would not say they are different.
(7) was answered in the comments, by Phillipe.
Another paper you might enjoy is Casacuberta-Neeman "Brown Representability does not come for free." It's about a localization failing to exist at the homotopy category level, when it was expected to exist.<|endoftext|>
TITLE: Flat versus etale cohomology
QUESTION [21 upvotes]: Although the definition of etale ($\ell$-adic) cohomology is scary, I have at least some intuition for how it should behave: for instance, when it makes sense, I expect that it should be ``similar'' to the classical (singular) cohomology. 
I'm in a situation in which I'm wondering if I can generalize a result in the realm of etale cohomology by using flat cohomology instead. (For instance, I want to be able to consider varieties over characteristic 2 and take cohomology with coefficients in the sheaf $\mu_2$, as an analogue of the case where one of the 2's is replaced by $p \neq 2$.) 
Unfortunately, I have no intuition whatsoever for how flat cohomology should behave, so I am looking for mental models, slogans, etc. An answer of "don't use flat cohomology because it is impossible to work with" would also be helpful (although a little disappointing). For the sake of concreteness, here are some specific questions (but my interest is not limited to these)

Can I expect analogues of familiar nice things such as Poincare duality and cycle class maps? 
Should this behave like (or even coincide with) etale cohomology in ``nice'' cases? 
What are non-trivial examples where flat cohomology can be effectively computed? Especially examples where the computation is not by formally showing that it must be the same as etale cohomology, and then computing the latter. 
Should I expect to have a good theory of characteristic classes?

REPLY [12 votes]: Here is a good example of some pathological behaviour that will shatter all your hopes and dreams.

Claim. Let $k$ be a perfect field, and let $X$ be a smooth, proper, integral $k$-scheme. Then the cohomology $H^1(X,\alpha_p)$ is a $k^p$-vector space. In particular, it is only finite if it is $0$ or $k$ is finite.

For a proof, see this MO post.
The reason this is disappointing is that $\alpha_p$ should be a constructible sheaf on the flat site, and from étale cohomology we expect the cohomology of a constructible sheaf to be finite. But it isn't!
On the other hand, for $\mu_p$-coefficients, we do have a fairly nice theory, thanks to Milne's arithmetic duality theorems. The first paper in this direction was his Duality in the flat cohomology of a surface, and this has been extended and rewritten many times. Some references were already given in the comments; I am not aware of any others (but they might well exist).
There are many other interesting sheaves to consider other than just sheaves represented by finite group schemes over the ground field. But I don't think there is a nice and unified theory in the same that there is for étale cohomology.
Remark. A nice conceptual remark is the fact that flat cohomology can be computed by the quasi-finite flat site. This is discussed in Milne's étale cohomology book, Example III.3.4. Thus, even though the flat site seems infinitely larger than the étale site (flat morphisms can have arbitrary relative dimension), it somehow isn't as big as you would think.<|endoftext|>
TITLE: Examples of calculating perverse sheaves on algebraic varieties with easy stratification
QUESTION [11 upvotes]: I have been learning intersection homology and perverse sheaves in the following way. I started by reading the first $7$ chapters of Kirwan and Woolf's book 
http://www.amazon.com/Introduction-Intersection-Homology-Theory-Edition/dp/1584881844.
Then, I read chapter $8$ of the book http://www.math.columbia.edu/~scautis/dmodules/hottaetal.pdf which introduced the theory of perverse sheaves using the language of $t-$ structure. After reading the abstract construction of the category of perverse sheaves for an algebraic variety or analytic space, I hope to see examples of calculating the perverse sheaves for spaces with easy stratification, such as $\mathbb{CP}^n$. Could anyone please provide some interesting examples? Thanks!

REPLY [3 votes]: Things are easiest when the automorphism group of $M$ (with its stratification) acts with finitely many orbits on $T^* M$: see Perverse sheaves on  Grassmannians, by Tom Braden.<|endoftext|>
TITLE: Projective resolutions for commutative monoids
QUESTION [5 upvotes]: What is the right notion of a projective resolution of a commutative monoid?

The category Mon of commutative monoids has plenty of projective (and even free) objects. Indeed, for every set $X$ one can consider the commutative monoid $\mathbb{N}[X]$, with generators $\delta_x$ for $x\in X$. Then, for every commutative monoid $M$ one can consider the free monoid $P_0:=\mathbb{N}[M]$ together with the canonical surjective homomorphism $\varphi_0\colon P_0\to M$, which sends the (formal) sum $\sum_m c_m \delta_m$ in $P_0$ (with $c_m\in\mathbb{N}$, and all but finitely many $c_m$ zero) to the sum $\sum_m c_m m$ in $M$.
One can consider this as the first step in building a projective resolution for $M$. In the category of groups, one would next consider the kernel of $\varphi_0$ and find a free group surjecting onto that kernel.
However, the notion of kernel in Mon is more subtle. If one simply considers $L:=\{x\in P_0 : \varphi_0(x)=0\}$, then it is in general not true that $M$ is isomorphic to $P_0 / L$. One needs to consider kernel pairs, which categorically are the pullbacks of the two maps $(\varphi_0,\varphi_0)$.
Thus, a "better" definition of kernel for the map $\varphi_0$ is 
$$ K_0 := \{(x,y)\in P_0\oplus P_0 : \varphi_0(x)=\varphi_1(y) \}, $$
together with the two natural maps $p_0,p_1\colon K_0\to P_0$ satisfying $p_0(x,y)=x$ and $p_1(x,y)=y$. Then, it is true that $M$ is isomorphic to the quotient of $P_0$ by the congruence relation generated by $p_0$ and $p_1$. Categorically, $\varphi_0$ is the co-equalizer of $p_0$ and $p_1$. 
Next, I can set $P_1:=\mathbb{N}[K_1]$ and consider the natural surjection $\varphi_1\colon P_1\to K_1$. Doing this, I obtain two maps from $P_1$ to $P_0$, namely $p_0\varphi_1$ and $p_1\varphi_1$. Thus, we have obtained the following situation:
$$P_1 \rightrightarrows P_0 \to M.$$
What is the next step? I could consider the kernels of the two maps from $P_1$ to $P_0$. This would mean to construct two free monoids with two morphisms each. Thus, at each next level it seems that this (naive) approach doubles the number of arrows. Is this really how it should be done? Has this been considered? Are there better (simpler) notions of projective resolutions for commutative monoids? In either case, the following seems like an interesting question to aks as well:

Do commutative monoids have projective resolutions of finite length?

The motivation for this question is to define derived functors in the category Mon.
Just to avoid some misunderstanding:
I am not interested in homological algebra over the field with one element.
I am aware of some other questions that have been asked about homological algebra of commuative monoids, such as:
Homological Algebra for Commutative Monoids?
Structure Theorem for finitely generated commutative cancellative monoids?

REPLY [5 votes]: Homological algebra for monoids have been done by a lot of people in theoretical computer science. What is done for associative algebras over a field in a brilliant paper of David Anick (http://www.jstor.org/stable/2000383) of 1986, was discovered roughly at the same time by Craig Squier (http://link.springer.com/chapter/10.1007%2F3-540-17220-3_7) and then generalised and conceptualised by many other people, see e.g. the following links:
http://www.sciencedirect.com/science/article/pii/S0022404908002089
http://www.emis.de/journals/TAC/volumes/11/7/11-07abs.html
http://www.sciencedirect.com/science/article/pii/S1571066104808363
http://iml.univ-mrs.fr/~lafont/pub/agr.pdf<|endoftext|>
TITLE: Is $[729,57]$ a Sylow $3$-subgroup of some well-known group?
QUESTION [7 upvotes]: Let $G$ be the group $[729,57]$, using GAP's notation. I have so far two descriptions of the group:

a presentation
an embedding (not surjective!) of the group into a Sylow $3$-subgroup of the unit group of a finite ring

Is $G$ isomorphic to a Sylow $3$-subgroup of some well-known group?

REPLY [10 votes]: Yes (to my surprise) it appears to be isomorphic to the Sylow $3$-subgroup of $3.J_3$, the $3$-fold cover of the Janko sporadic simple group $J_3$. Here is some Magma code:
> C:=MatrixGroup("3J3",1);
> P:=Sylow(C,3);   
> IdentifyGroup(P);  
<729, 57>

And here is GAP code for the same computation (thanks to  Stefan Kohl and Frieder Ladisch)
LoadPackage("AtlasRep");;
G:=AtlasGroup("3.J3");;
v:= Z(4)^0*[1,0,1,0,0,0,0,0,0];;
Gperm:= Action( G, Orbit( G, v, OnRight), OnRight );;
P := SylowSubgroup(Gperm,3);;
IdGroup(P);

[ 729, 57 ]<|endoftext|>
TITLE: $\mathcal{O}_{\infty}$ and $\mathcal{Z}$ stable isomorphism as equivalence
QUESTION [8 upvotes]: If $O$ is a (strongly ?) self absorbing $C^*$-algebra, one has an equivalence relation on (separable) $C^*$-algebras: "being $O$-stably isomorphic" i.e. $A$ and $B$ are $O$-stably isomorphic if and only if there is an isomorphism $A \otimes O \simeq B \otimes O$.
In the case where $O$ is the algebra of compact operator, (and if we restrict to separable algebra) this relation is better understood as Morita equivalence, i.e. equivalence of the category of Hilbert modules, or isomorphisms in the category of (proper) correspondences.
My question is: for other self absorbing algebra, and more precisely for the Cuntz algebra $\mathcal{O}_{\infty}$ and for the Jiang-su algebra $\mathcal{Z}$, is there something comparable to Morita equivalences to have a better (more categorical if you prefer) understanding of stable isomorphisms.
For example, is there something that plays the role of the category of Hilbert modules that one can attach to any $C^*$-algebra and that classifies them up to $\mathcal{Z}$-stable isomorphism. Or maybe some sort of category of correspondences ?

REPLY [5 votes]: Maybe the following can count if you are willing to restrict to separable, simple, stable, nuclear $C^*$-algebras $A$ and $B$. Then the Kirchberg-Phillips Classification Theorem says that $A\otimes\mathcal O_\infty\cong B\otimes\mathcal O_\infty$ if and only if $A$ and $B$ are KK-equivalent, that is, isomorphic in the KK-theory category. KK-theory is defined in terms of certain bimodules, so there is some analogy to correspondences.
(One can get rid of the simplicity assumption above by asking $A$ and $B$ to have isomorphic ideal lattices and being equivalent in a version of KK-theory that keeps track of the ideals, see here.)<|endoftext|>
TITLE: What does the unique mean on weakly almost periodic functions look like?
QUESTION [7 upvotes]: There is a unique invariant mean $m$ on WAP functions on any discrete group (see definitions below, theorem of ?). However, the proofs I found are fairly non-explicit on how to obtain this invariant mean.
By "explicit", I would like to know (preferably, for the free group on two generators, but other non-amenable groups are welcome), if there is a sequence of functions $f_n \in \ell^1\Gamma$ so that $f_n$ tends weak$^*$ to the mean $m$?
There is of course bound to be a big amount of non-explicitness due to the fact that, ultimately, elements of $(\ell^\infty \Gamma)^* \setminus \ell^1\Gamma$ require [a weak form of] the axiom of choice (the "Hahn-Banach axiom").
Here is an example of what I mean by explicit. When the group is amenable, uniqueness implies that this means coincides with any other invariant mean on $\ell^\infty\Gamma$. So it would suffice to pick a sequence $F_n$ of Folner sets, and let $f_n = \chi_{F_n} /|F_n|$ be the normalised characteristic functions of those sets.
As for the definitions... 

$\Gamma$ acts on $\ell^\infty \Gamma$ by translation by the left: $(\gamma \cdot f)(x) = f(\gamma^{-1}x)$. (Likewise, it acts on the right.) A function $f$ is WAP if the weak closure of its orbit under this action is weakly compact. (WAP is a subspace of $\ell^\infty$ which contains $c_0$.)
An invariant mean on a subspace $X$ of $\ell^\infty \Gamma$ is a linear map $m\colon X \to \mathbb{R}$ so that $m(\chi_\Gamma) = 1$ (where $\chi_\Gamma$ is the constant function taking value $1$) $m(f) \geq 0$ if $f \geq 0$ and $m( \gamma \cdot f) = m(f)$ (likewise on the right). Linearity, positivity and $m(\chi_\Gamma) =1$ automatically implies that $m$ is bounded, i.e. $m \in (\ell^\infty\Gamma)^*$.
A [non-explicit] characterisation of the invariant mean on WAP functions goes as follows. Let $f$ be WAP. Then there is a unique constant function in the weak closure of its orbit under translations. The value that this constant function takes is the value of the mean.

REPLY [7 votes]: I don't think you will ever get what you want.
I will explain here the way I like to think of this invariant mean. I hope you will find it nice, but I doubt you will find it explicit.
There are two natural classes of compactifications which we associate with a group $\Gamma$ that I would like to recall.
A group compactification is just a group homomorphism from $\Gamma$ into a compact topological group.
A semi-group compactification is a semi-group homomorphism from $\Gamma$ into a compact semi-topological semi-group (a semi-group endowed with a topology for which both left and right multiplication are (separately) continuous).
Both types of compactifications could be seen as objects of corresponding categories in which morphisms are continuous homomorphisms between the target (semi-)groups making the obvious diagram commute. It is not hard to see that both categories have initial objects. The universal group compactification is called the Bohr compactification and the universal semi-group compactification is called the WAP compactification. Not surprisingly, when you pull back to $\Gamma$ the continuous functions on these compact objects you get correspondingly the algebras of Almost Periodic and Weakly Almost Periodic functions.
Let me remark here that AP / WAP functions are the matrix coefficients of finite dimensional / reflexive Banach representations. I will not elaborate on that here, though it is crucial for a better understanding.
Since every group is a semi-group there is a canonical homomorphism $\text{WAP}(\Gamma)\to \text{Bohr}(\Gamma)$. It turns out that there is a section to this map, a homomorphism $\text{Bohr}(\Gamma)\to \text{WAP}(\Gamma)$. Its image is called the Sushkevich kernel of $\text{WAP}(\Gamma)$. It is the unique minimal two sided ideal in $\text{WAP}(\Gamma)$ and it is usually denoted by $K(\Gamma)$.
So $K(\Gamma)$ is an isomorphic copy of $\text{Bohr}(\Gamma)$ which happens to live inside $\text{WAP}(\Gamma)$.
As I mentioned before, every WAP functions on $\Gamma$ extends canonically to a continuous function on the compact semi-group $\text{WAP}(\Gamma)$. You can now take this extended function and restrict it to the subgroup $K(\Gamma)$. The resulting function on $K(\Gamma)$ you can average wrt the Haar measure of $K(\Gamma)$. This process takes a WAP function on $\Gamma$ and gives back a scalar. This is the invariant mean you wanted to understand.
Of course, since $\Gamma$ is dense in $\text{WAP}(\Gamma)$, the finitely supported probability measures on $\Gamma$ are w*-dense in $\text{Prob}(\text{WAP}(\Gamma))$, so in particular you could find a net of finitely supported functions in $\ell^1(\Gamma)$ that will converge to the Haar measure on $K(\Gamma)$. But I seriously doubt that you can find a sequence with this property. In general these objects are not metrizable.<|endoftext|>
TITLE: Parametrizing all cyclic extensions of the rational numbers of degree 5
QUESTION [11 upvotes]: Is there a polynomial $f(T,X) \in \mathbb{Q}(T)[X]$ in the indeterminate $X$ over the field $\mathbb{Q}(T)$ with $\mathrm{Gal}(f/\mathbb{Q}(T)) \cong \mathbb{Z}/5\mathbb{Z}$ such that for every Galois extension $K/\mathbb{Q}$ with $\mathrm{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/5\mathbb{Z}$ there exists some $t \in \mathbb{Q}$ such that the splitting field of $f(t,X)$ over $\mathbb{Q}$ is isomorphic to $K$.

REPLY [11 votes]: Evidently no.  In "Generic Polynomials Constructive Aspects of the Inverse Galois Problem" by Jensen, Ledet, Yui (2002): generic dimension (i.e. minimum number of parameters in a generic polynomial) of $\mathbb{Z}/5$ over $\mathbb{Q} = 2$ (p.203) and moreover essential dimension (i.e. minimum transcendence degree of parameters) of $\mathbb{Z}/5$ over $\mathbb{Q} = 2$ (p.190).  They conjecture (p.202) that if generic dimension is finite then essential dimension = generic dimension (over any field).  One story for $\mathbb{Z}/3$ that breaks down for $\mathbb{Z}/5$ uses an action of $\mathbb{Z}/3$ on $\mathbb{P}^1_{\mathbb{Q}}$. 
Edit: As Pablo noted, this answer is about a generic polynomial over $\mathbb{Q}$ but it doesn't answer the original question about existence of a 1-parameter parametric polynomial over $\mathbb{Q}$.  I didn't appreciate this nuance in the original question when I wrote the above answer.<|endoftext|>
TITLE: Growth of $\zeta_{\mathbf Q[\cos(\frac{\pi}{2^{n+1}})]}(2)$
QUESTION [14 upvotes]: Let $K_n$ be the field $\mathbf Q[\cos(\frac{\pi}{2^{n+1}})]$ (the real subfield of the cyclotomic field $\mathbf Q[e^{\frac{i\pi}{2^{n+1}}}]$). 
Is there anything known about the growth of the special values of the Dirichlet $\zeta$ function 
$$\zeta_{K_n}(2)=\prod_{\mathfrak p}\frac{1}{1-\frac{1}{N(\mathfrak p)^2}}$$ when $n\to \infty$ ?
The best I could do was to use the inequalities
$$\left(1-\frac{1}{p^2}\right)^{ab}\leq \left(1-\frac{1}{p^{2a}}\right)^{b}\leq \left(1-\frac{1}{p^{2ab}}\right)$$
(that are verified in our range) to obtain the inequalities
$$2\ \left(\frac{\pi^2}{3}\right)^{2^n}\ =\ 2^{1-2^n}\ \zeta(2)^{2^n}\geq \zeta_{K_n}(2)\geq 2\ \left(1-\frac{1}{2^{n+1}}\right)\ \zeta(2^{n+1})$$
that are quite unilluminating since the left hand side tends to infinity exponentially fast, while the right hand side converges to 2. The only thing we use here is that $2$ is totally ramified and that the other primes are not ramified. I guess it is possible to use more of our knowledge of the arithmetic in $K_n$ to obtain a precise idea of the behaviour of the sequence $(\zeta_{K_n}(2))_n$ ... (maybe by a standard use of the Chebotarev theorem - I'm quite ignorant in this kind of questions). 
I'm almost certain that $\zeta_{K_n}(2)$ tends to infinity exponentially fast (and that's what I'd like to check).

REPLY [18 votes]: Actually $\zeta_{K_n}(\sigma)$ is bounded for any fixed $\sigma > 1$.
Let $N = 2^n = [K_n : {\bf Q}]$.  Then all the local factors of 
$\zeta(\sigma)$, other than the factor $(1-2^{-\sigma})^{-1}$ for
the prime above $2$, are of the form $(1 - q^{-\sigma})^{-g}$,
where $q$ is a prime power congruent to $\pm 1 \bmod 4N$, and $g \mid N$.
Thus 
$$
\zeta_{K_n}(\sigma) < \frac1{1-2^{-\sigma}} 
  \prod_{m=1}^\infty \frac1{(1 - (2mN)^{-\sigma})^N}
$$
and the product approaches $1$ as $n \to \infty$ because
its logarithm behaves like
$\sum_{m=1}^\infty N/(2mN)^\sigma = \zeta(\sigma) \, / \, 2^\sigma N^{\sigma-1} \to 0$.
Since each local factor exceeds $1$, it follows that in fact
$\zeta_{K_n}(\sigma) \to (1-2^{-\sigma})^{-1}$ as $n \to \infty$.<|endoftext|>
TITLE: The Fourier transform of a function supported on $B_1$ is essentially constant on $B_1$?
QUESTION [8 upvotes]: I'm going through the last steps of Bourgain and Demeter's proof of the $l^2$ decoupling conjecture, but I'm unable to see how the first inequality in (43) goes through. I'll water down the question a little to make it more transparent.
Let $f_1, \dots, f_n$ be $n$ functions whose Fourier transform is supported on $B(1)$. Bourgain claimed that by the heuristics he and Guth used in their paper we can think that $f_i$ is "essentially constant" on $B_1$. This seems to allow the following "reverse Holder" inequality
$$\prod_{i=1}^n \|f_i\|_{L^p(B_1)}^{1/n}\ll_{n,p} \|\prod_{i=1}^n f_i^{1/n}\|_{L^p(B_1)}. $$
But I skimmed that paper and only found an estimate morally like the following:
$$ f(x)=f*\eta(x)\le \|f\|_{L^1}\|\eta\|_{L^\infty}, $$
where $\hat\eta$ is compactly supported and equals 1 on $B_1$. How do I deduce the reverse Holder inequality from that?
Added: Just for reference, the inequality (43) that I referred to is
$$ \prod_{i=1}^n \|\widehat{g_id\sigma}\|_{p,\delta,B_{\delta^{-1/2}}}^{1/n}\ll_{n,p} \|\prod_{i=1}^n (\sum_{\theta\subset\tau_i:\delta^{1/2}\text{ cap}} |\widehat{g_{i,\theta}d\sigma}|^2)^{1/2n}\|_{L^p(w_{B_{\delta^{-1/2}}})}. $$
In the above, the $f_i$ are supposed to be $\widehat{g_{i,\theta}d\sigma}$, and I'm assuming $\tau_i=\text{supp }g_i$ contains only one $\delta^{1/2}$ cap, which is a slab of size $\delta^{1/2}\times\cdots\times\delta^{1/2}\times\delta$. $\tau_i$ are assumed to be $\nu-$transverse, but I feel this is unimportant in this inequality. $w_{B_{\delta^{-1/2}}}$ is just a smooth cutoff weight that equals 1 on $B_{\delta^{-1/2}}$. I have normalized $\delta$ to 1 in the above.

REPLY [12 votes]: I checked with Jean and Ciprian about this, and there is indeed a small issue here; the bound indicated is "morally" correct, and may possibly even be true (using the weights $w_B$ rather than a sharp truncation $1_B$), but it does not quite follow from the standard device of representing a Fourier-localised function by a convolution with a Schwartz function.  In later papers, most notably in the Vinogradov conjecture paper with Guth, the issue was addressed by inserting an additional $L^p$ norm at small scales, replacing expressions such as $\| \widehat{g d\sigma} \|_{p,\delta,\Delta}$ with somewhat more complicated looking expressions $A_p(u,B^n,u)$.  From what I understand, it is possible to adapt the arguments in these later papers (which are more streamlined than the original arguments in some respects) to reprove the hypersurface decoupling estimate rigorously.  The authors plan to update the arXiv version of the original decoupling paper with some additional remarks addressing these issues.<|endoftext|>
TITLE: Examples of étale covers of arithmetic surfaces
QUESTION [14 upvotes]: Define an arithmetic scheme $X$ to be a separated, integral scheme, flat and finite type over $\mathbb{Z}$. I am interested in obtaining examples of finite étale covers of arithmetic schemes. I am mainly interested in the case of arithmetic surfaces, but I would enjoy examples of finite étale covers of arithmetic schemes of any dimension $>1$.
Let us restrict to schemes proper over $\mathbb{Z}$ to rule out the possibility of forming a finite integral extension and then restricting to an open subscheme on which the morphism is étale. Let us also restrict to geometrically connected (generic fiber?) to rule out the possibility of base ring extensions.
In the case where $X\to Spec(\mathbb{Z})$ is also smooth, the Katz-Lang finiteness theorem tells us that there will be at most finitely many abelian étale covers of $X$.
Abelian varieties can be a convenient way to create covers: if $A$ is an abelian variety, and $H$ a finite subgroup, then the dual isogeny to $A\to A/H$ is an étale cover of $A$. However, there are no abelian varieties over $Z$. I'm not sure whether this approach could be modified to produce an example.
It would be preferable to stick to regular or at least normal schemes, but that's not strictly necessary.
So, can anyone provide some nice examples of étale covers satisfying these conditions? If you would like to weaken some of the conditions to provide an interesting example (particularly, allowing nonproper or working over another $\mathcal{O}_K$), I guess I'll allow it.
To summarize, the optimal conditions on $X$ are regular, geometrically connected, proper over $\mathbb{Z}$.
Thanks.

REPLY [6 votes]: You can, of course, use Bertini's theorem to make examples of a finite, flat, Galois extension with arbitrary finite Galois group $\Gamma$.  Let $M$ be $\mathbb{Z}[\Gamma]$, the group ring of $\Gamma$ with coefficients in $\mathbb{Z}$.  This is a finite, free $\mathbb{Z}$-module.  Let $r>0$ be a positive integer.  Then $M^{\oplus r}$ is also a finite, free $\mathbb{Z}$-module with a natural action of $\Gamma$.  Form $S=\mathbb{Z}[M^{\oplus r}]$, the $\mathbb{Z}_{\geq 0}$-graded polynomial ring over $\mathbb{Z}$ such that the first graded piece $S_1$ equals $M^{\oplus r}$.  Then $\mathbb{P}:=\text{Proj}(S)$ is a projective space over $\mathbb{Z}$ that has a natural action of $\Gamma$.  
Denote by $U\subset\mathbb{P}$ the maximal open subscheme on which $\Gamma$  acts freely.  Denote by $F$ the complement of $U$.  The fibers of $F$ (over $\text{Spec}\ \mathbb{Z}$) have codimension at least $r$ inside the fibers of $\mathbb{P}$.  Hence, assume $r$ is sufficiently large (say $r\geq 2$) so that the fibers of $F$ have codimension $\geq 2$.   Denote by $q:\mathbb{P} \to Q$ the quotient of the action of $\Gamma$ on $\mathbb{P}$.  This morphism is finite, and it is flat when restricted to $U$.  The quotient $Q$ is a projective scheme, say $Q\subset \mathbb{P}^N_{\mathbb{Z}}$.  By Bertini's Theorem, for sufficiently large integers $d$, there exist degree $d$ hypersurfaces in $\mathbb{P}^N_{\mathbb{Z}}$ whose intersection with $Q$ is a regular, $2$-dimensional scheme $X$ that is disjoint from the image of $F$.  Define $\widetilde{X} = \mathbb{P}\times_Q X$.  Then $\widetilde{X}\to X$ is a finite, flat, étale morphism that is Galois with Galois group $\Gamma$.<|endoftext|>
TITLE: About the logarithmic derivative of the Riemann zeta function
QUESTION [6 upvotes]: Let $\rho=\beta+i\gamma$ a non-trivial zeros of the Riemann zeta function and $s=\sigma+it$ a complex number. It is possible to prove that $$\frac{\zeta'}{\zeta}\left(s\right)=\sum_{\left|t-\gamma\right|\leq1}\frac{1}{s-\rho}+O\left(\log\left(t\right)\right) \tag{1}$$ uniformly for $-1\leq\sigma\leq2$ (see for example Titchmarsh, “The theory of the Riemann zeta function”, second ed., page $217$). So in particular if we take $\sigma=0$ holds $$\frac{\zeta'}{\zeta}\left(it\right)=\sum_{\left|t-\gamma\right|\leq1}\frac{1}{it-\beta-i\gamma}+O\left(\log\left(t\right)\right).
 $$ Question: is it possible to prove that $$ \sum_{\left|t-\gamma\right|\leq1}\frac{1}{it-\beta-i\gamma}\ll\log\left(t\right)?
 $$ The problem is that it could be some zeros with real part very close to $0$ and so for $t=\gamma$ the sum is very "big". Thank you.
Addendum: My final goal it's to prove that $$\frac{\zeta'}{\zeta}\left(it\right)=O\left(\log\left(t\right)\right)$$ so also another proof of it (if exists) without the use of $(1)$ is welcome.

REPLY [12 votes]: I think your final goal follows by taking the logarithmic derivative of the functional equation:
$$\frac{\zeta'}{\zeta}(s)+\frac{\zeta'}{\zeta}(1-s)=\log\pi-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s}{2}\right)-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{1-s}{2}\right).$$
Applying this with $s=it$ and using the familiar asymptotic expansion of $\Gamma'/\Gamma$, we get
$$\frac{\zeta'}{\zeta}(it)+\frac{\zeta'}{\zeta}(1-it)=O(\log t),\qquad t>2.$$
In fact we get 
$$\frac{\zeta'}{\zeta}(it)+\frac{\zeta'}{\zeta}(1-it)=-\log t+\log(2\pi)-\frac{i}{2t}+O(t^{-2}),\qquad t>2,$$
but observing $O(\log t)$ is sufficient on the right hand side. Indeed, the second term on the left hand side is $O(\log t)$, hence the same holds for the first term.
P.S. Of course this means that your sum is indeed $O(\log t)$. See also my comment below the original post.<|endoftext|>
TITLE: Can we count the number of simple modules for a reduced enveloping algebra?
QUESTION [8 upvotes]: Let $G$ be a reductive algebraic group over a field of positive characteristic $p$, which I'll assume to be very good for $G$. Then the Lie algebra $\mathfrak{g}$ is restricted and each simple $\mathfrak{g}$-modules belongs to precisely one of the reduced enveloping algebras $U_\chi(\mathfrak{g})$ with $\chi \in \mathfrak{g}^*$. Thanks to a theorem of Kac–Weisfeiler the simple $\mathfrak{g}$-modules may be entirely understood by focusing on the case where $\chi$ is nilpotent, i.e. $\chi$ vanishes on some Borel subalgebra.  Assume $\chi$ is nilpotent, let $\mathcal{B}$ denote the flag variety of $G$, and $\mathcal{B}_\chi$ the Springer fibre over $\chi$. As a set this is just the collection of all Borel subalgebras of $\mathfrak{g}$ where $\chi$ vanishes.
Now let $p > 2h - 2$ where $h$ is the Coxeter number. In a famous paper by Bezrukavnikov, Mirković and Rumynin the authors proved (amongst other things) a certain conjecture of Lusztig, which asserts that the number of simple $U_\chi(\mathfrak{g})$-modules with trivial central character is precisely the rank of the Grothendieck group of the coherent sheaves on $\mathcal{B}_\chi$. By a later theorem in that same paper this number is actually equal to the sum of the dimensions of the $\ell$-adic cohomology of $\mathcal{B}_\chi$. It is well known that the blocks of $U_\chi(\mathfrak{g})$ are determined by their central character, and after Jantzen's results on translation functors (see Proposition B5 in Jantzen's paper listed below) it is also known that each block contains the same number of simple modules.
My question is the following: is there a known formula for the dimensions of the cohomology of the Springer fibre over a nilpotent element? Has somebody used this to write down a formula for the number of simple $U_\chi(\mathfrak{g})$-modules? Using the results listed above is there anything interesting/unexpected/useful which we can learn about the category of $U_\chi(\mathfrak{g})$-modules?
When $\chi$ has standard Levi type the simple $U_\chi(\mathfrak{g})$-modules are classified by a theorem of Friedlander and Parshall. The classification shows that simple modules correspond to the orbits of a certain subgroup of the Weyl group (depending on $\chi$) on an $\mathbb{F}_p$-lattice of the chosen torus, and this leads to a formula for the number of simple modules. It would be wonderful if there existed a formula outside standard Levi type which generalises this, but it would also be interesting to see a list of numbers corresponding to Bala–Carter labels for nilpotent orbits in exceptional Lie algebras, for instance.
References:
R. Bezrukavnikov, I. Mirkovic & D. Rumynin, “Localization of modules for a semisimple Lie algebra in prime characteristic”, Annals Math., Vol. 167 (2008), 945–991.
E. Friedlander & B. Parshall, “Modular representation theory of Lie algebras”, Amer. J. Math., Vol. 110 (1988), 1055–1093.
J. C. Jantzen, “Subregular nilpotent representations of Lie algebras in prime characteristic”, Rep. Theory, Vol. 3, (1999), 153–222.

REPLY [3 votes]: In the case when $\mathfrak{g}$ is a finite-dimensional restricted Lie algebra over an algebraically closed field of characteristic $p>0$, the number of isomorphism classes of simple $U_{\chi}(L)$-module is bounded above by $p^{MT(L)}$, where $MT(L)$ denotes the maximal dimension of a torus in $L$. See Section 3 in the following joint paper with J. Feldvoss and Th. Weigel:
http://link.springer.com/article/10.1007%2Fs00031-015-9362-5<|endoftext|>
TITLE: Generalized ordering on simplicial complex
QUESTION [5 upvotes]: The vertices of simplicial complexes are usually totally ordered so that face maps of each simplex can be defined easily for the purposes of homology. That gives an "oriented" simplicial complex.
But that's not always necessary, since a partial order that restricts to a total order on each simplex would also suffice. In fact, that is called an "ordered" simplicial complex.
One could generalize this even more to consider irreflexive binary relations that restrict to a total order on each simplex; as a consequence these relations will also be antisymmetric. Is there a name for such relations?
(Cross posted from Math Stackexchange)

REPLY [2 votes]: You may look into what is called $\Delta$-complexes, and may be the notion of a $\Delta$-complex will suit you. There you give an orientation to each simplex and then glue them by using face maps, which become orientation-reversing or preserving. An orientation-reversing glueing gives you an orientable complex. Beyond that, I have no knowledge of relevant terminology.<|endoftext|>
TITLE: Can any path in the diffeomorphism group of a smooth compact manifold be approximated by a smooth path?
QUESTION [8 upvotes]: Given a smooth compact $n$-dimensional manifold $M^{n}$, let $\operatorname{Diff}(M)$ denote the group of smooth diffeomorphisms $M \rightarrow M$ equipped with the Whitney $C^{\infty}$-topology. Let $h_{\phi} \colon [0, 1] \times M \rightarrow M$ denote the homotopy associated to a path $\phi \colon [0, 1] \rightarrow \operatorname{Diff}(M)$. My question is: Given such $\phi$, how can one prove that there exists a path $\psi \colon [0, 1] \rightarrow \operatorname{Diff}(M)$ with the same endpoints as $\phi$ and such that $h_{\psi}$ is smooth?
So far, I have neither succeeded in finding a formal proof nor a detailed reference in the literature. I am particularly interested in the case where $M$ has non-empty boundary. I am looking forward to helpful comments!

REPLY [3 votes]: Here is an example of a very hands-on (and standard) construction of the path that you're looking for. The goal will be take a sequence of diffeomorphisms in the path that are located sufficiently close to each other in the $C^\infty$ topology, and then to join these diffeomorphisms by smooth paths.
Choose an auxiliary Riemannian metric on $M$, and consider the "exponential map" $E \colon TM \to M \times M$ given by $v \mapsto (p(v),\exp_{p(v)}(v))$ where $p \colon TM \to M$ is the tangent bundle.
For a diffeomorphism $\varphi$ of $M$ we write $\Gamma_\varphi:=\{(x,\varphi(x))\} \subset M \times M$ for its graph (this is a smooth submanifold). Given that a diffeomorphism $\varphi$ is sufficiently $C^\infty$-close to the diagonal $\Gamma_{\mathrm{id}_M} :=\{(x,x); x \in M \} \subset M \times M$, it follows that $E^{-1}(\Gamma_\varphi) \subset TM$ consists of a smooth section $\zeta$ of $TM \to M$ together with possibly additional loci contained outside of some a priori fixed tubular neighbourhood of the zero-section. (Here I am omitting some applications of standard transversality results as well as properties of the exponential map.)
In other words $x \mapsto \exp_{x}(t\zeta)$ is the sought smooth path parametrised by $t$ which connects $\mathrm{id}_M$ (at $t=0$) and $\varphi$ (at $t=1$).<|endoftext|>
TITLE: Resolution of the $E_8$ singularity with a weighted blowup
QUESTION [6 upvotes]: I am reading Miles Reid's notes on weighted projective spaces, and I'm a little confused about a particular paragraph (notes here, page 8):

A famous case is the $E_8$ singularity $X: (x^2+y^3+z^5=0)$, which is
  naturally weighted homogeneous with weights 15,10,6. The
  $\mathbb{G}_m$ quotient morphism $X \to \mathbb{P}^1$ defined by the
  ratio $x^2:y^3:z^5$ has stabiliser of order 2, 3, and 5. The weighted
  blowup $Y \to X$ (the graph of the quotient morphism $X \to
\mathbb{P}^1$) is a surface having cyclic quotient singularities of
  order 2,3,5 at the 3 points, giving rise to the Dynkin diagram of
  $E_8$.

I'd like to see this very explicitly. I agree that $\mathbb{P}(15,10,6) \cong \mathbb{P}^2$, and I can see that the equation $x^2 + y^3 + z^5$ becomes $u+v+w$ in $\mathbb{P}^2$, with coordinates $(u,v,w)$, so I agree that $X \to \mathbb{P}^1$. However, I am having trouble writing the equations for the graph and observing the singularities Reid describes.
Here is what I can do:
The map $\mathbb{A}^3 \setminus \{0\} \to \mathbb{P}(15,10,6)$ is given by $(x,y,z) \mapsto [x:y:z]$, and the isomorphism $\mathbb{P}(15,10,6) \to \mathbb{P}^2$ is $[x:y:z] \mapsto [x^2:y^3:z^5]$. The graph of this map is
$$
\Gamma = \{(x,y,z) \times [u:v:w] \,|\, uy^3=vx^2, uz^5=wx^2,wy^3=vz^5\}
$$
Restricting to $w=-u-v$ I get the equations $(uy^3=vx^2, uz^5=(-u-v)x^2,(-u-v)y^3=vz^5)$. Taking partial derivatives, this appears to be singular everywhere. What have I done wrong?

REPLY [4 votes]: Edit: Your computation is correct. The weighted blowup $Y \to X$ as defined in Example 3.7 of Reid's notes (i.e. the graph of the quotient morphism $X \to \mathbb{P^1}$) is singular at all the points of the exceptional line. However, the statement about $Y$ having 3 cyclic singularities becomes true when you replace $Y$ by its normalization $Y'$.
As you write, $\Gamma$ is defined in $X \times \mathbb{P}^2$ by the equations $$uy^3 = vx^2,\ uz^5 = wx^2,\ wy^3 = vz^5,\ w+u+v = 0$$
Now $w+u+v = 0$ defines a hypersurface in $X \times \mathbb{P}^2$ isomorphic to $ X \times \mathbb{P}^1$, with coordinates $((x,y,z), [u:v])$. Now $\Gamma$ is defined in this $ X \times \mathbb{P}^1$ by your equations 
$$uy^3 = vx^2,\ uz^5 = -(u+v)x^2,\ -(u+v)y^3 = vz^5$$
It follows that on $ X \times \mathbb{A}^1$ where $v \neq 0$, $\Gamma$ is defined by 
$$u'y^3 = x^2$$
where $u' := u/v$. Let $U := \Gamma \cap \{v \neq 0\}$. As you noted, $U$ is singular at all the points on $u'$-axis. In particular, $U$ is  not  normal. Indeed, it is straightforward to note that $x/y, z^2/y$ and $xz/y^2$ are integral over the coordinate ring of $U$. It turns out that you need to only adjoin these elements to get the normalization $U'$ of $U$; i.e. $U'$ is the closure in $U \times \mathbb{A}^3$ of the graph of the map $U\setminus\{y=0\} \to \mathbb{A}^3$ given by $(u',x,y,z) \mapsto (x/y, z^2/y, xz/y^2)$. $U'$ has $2$ singular points: $(0,\ldots,0) $ and $(-1,0,\ldots, 0) $ (where the first coordinate corresponds to $u'$) - corresponding respectively to the stabilizers of $x=0$ and $z=0$ on $X$. The other singular point (corresponding to the stabilizer of $y=0$ on $X$) on $Y'$ is on (the normalization of) the chart $\Gamma \cap \{u \neq 0\}$.<|endoftext|>
TITLE: Property of bundles with connections on abelian variety doesn't hold for additive or multiplicative group?
QUESTION [5 upvotes]: This question is a followup to two of my previous questions, see here and here.


Let $A$ be an abelian variety over a field $k$ of characteristic $0$. How do I prove, without using transcendental methods, that if $\nabla$ is an integrable connection on a vector bundle $L$ on $A$ then $(L, \nabla)$ comes from a $G$-bundle with connection for some abelian algebraic subgroup $G \subset GL(n)$? And the corollary there exists a $\nabla$-stable flag$$L = L_n \supset L_{n-1} \supset \dots \supset L_1 \supset 0$$with $\text{rank }L_i = i$? It is easy to prove for $k = \mathbb{C}$ by looking at the monodromy of $(L, \nabla)$, and the general case follows by the Lefschetz principle, but I want to find a different proof which does not use $\mathbb{C}$ at all.

I am curious as to whether or not we can, in a similar way, show that if $k$ is an algebraically closed field (of any characteristic) then the algebraic fundamental group of $A$ is abelian (given a connected étale covering $E \to A$ define an algebraic group structure on the group $G$ of pairs $(a, \tau)$, where $a \in A$ and $\tau$ is a lift of the translation automorphism $T_a: A \to A$ be an automorphism of $E$, etc.).



Does the above property of bundles with connections on an abelian variety hold for the additive or multiplicative group?
Thoughts. I suspect it does not, and it would suffice to construct a rank $2$ vector bundle on the affine line over $\mathbb{C}$ with a connection $\nabla$ such that $L$ has no $\nabla$-stable subbundles of rank $1$ to show this. But I am not quite sure on how to do this. Could anybody help? Thanks in advance!

REPLY [2 votes]: No, it fails for the additive and multiplicative group as they are not compact. Consider the differential equation of the Airy function $d^2 f/dx^2 = x f$. We can write this in first order form as $df/dx=u$, $du/dx = x f$. This becomes a vector bundle with connection by taking the vector bundle to be a rank $2$ free bundle and the connection to be $\nabla (f,u) = (\frac{df}{dx}-u, \frac{du}{dx}-xf)$ so that flat sections are the same as solutions to the differential equation.
We will show that the vector bundle has no invariant flag. If it did, it would have some sub-bundle of the form $\nabla (g) = \frac{dg}{dx} - p(x) f$ for a polynomial $p(x)=x$, which would have an invariant section of the form $g(x)= e^{ \int p(x)}$ Since that vector bundle with connection is a sub-bundle of this one by some polynomial map $(f(x),u(x))=(a(x)g(x), b(x)g(x))$ for polynomials $a$ and $b$, it follows that there is a solution to the Airy differential equation $d^2 f/dx^2 = x f$ of the form $f(x)=a(x) e^{\int p(x)}$. But there is no such solution, as we can see from the asymptotic formulas for the two solutions of the Airy equation, which do not match.
This gives a rank two vector bundle with flat connection generated by $f$ and $df/dx$, and it can be seen that it is irreducible (because the growth rate of a solution is $e^{x^{3/2}}$ but any solution to a rank one differential equation grows like an exponential of a polynomial function). Of course this also pulls back to the multiplicative group.
Your second question, on the etale fundamental group, fails for the additive and multiplicative groups in characteristic $p$ for the same reason. Indeed, there is a perfectly analogous example - the Airy sheaf.<|endoftext|>
TITLE: Automorphisms of singular varieties
QUESTION [5 upvotes]: If $f : X \to X$ is an automorphism of a singular variety over $\mathbb C$, does there exist a resolution of singularities $g : Y \to X$ such that the induced birational map on $Y$ is also an automorphism?

REPLY [3 votes]: Edited after Mark's comment. Also after ACL's comment. Cheers.
Take a resolution $g:Y\to X$ and 
let $Y'=X\times_{f,g} Y$ with projection maps $f':Y'\to Y$ and $g':Y'\to X$. Next let $Y''=X\times_{f^{-1},g'} Y'$ with projection maps $f'':Y''\to Y'$ and $g'':Y''\to X$ and $Y'''=X\times_{f,g''} Y''$ with projection maps $f''':Y'''\to Y''$ and $g''':Y'''\to X$.
Observe that $Y''=X\times_{f^{-1},g'} Y'= X\times_{f^{-1},g'}(X\times_{f,g} Y)\simeq X\times_{\mathrm{id}_X ,g}Y\simeq Y$ where the last isomorphism is given by $f'\circ f''$ and also, clearly, $g''=g$.
Similarly $Y'''=X\times_{f,g''} Y''\simeq X\times_{f,g'}(X\times_{f^{-1},g} Y')\simeq X\times_{\mathrm{id}_X ,g'}Y'\simeq Y'$  where the last isomorphism is given by $f''\circ f'''$ and also, clearly, $g'''=g'$ and then $f'''=f'$.
So we get that for an arbitrary resolution $g:Y\to X$ there is another resolution $g':Y'\to X$ such that $f':Y'\to Y$ is an isomorphism with inverse $f'':Y\to Y''$.
Now if we choose $g:Y\to X$ to be a functorial resolution of $X$ (as in 3.4 of Kollár's book), then it follows that $g'=g$ and so $f'$ is the required lift. Functorial resolutions exist by  3.36 of Kollár's book.<|endoftext|>
TITLE: Monomorphisms in operad algebras
QUESTION [7 upvotes]: Setup: Let $\mathcal{O}$ be an operad in the category of sets, and let $\mathcal{O}\text{-Alg}$ denote the category of algebras on it (i.e., operad functors $\mathcal{O}\to\mathbf{Set}$. This category is cocomplete; let $\kappa$ denote an initial object. Call an algebra $X$ initially monic if the unique map $!_X\colon\kappa\to X$ is monic. 
Question 1: If $X$ and $Y$ are initially monic, is their coproduct $X\sqcup Y$ initially monic as well?
Remark: If, in the setup, we replace "operad" by "algebraic theory", then the answer to the question becomes no, as shown here by Zhen Lin in the case of $R$-algebras. I'm wondering if the issue might somehow come down to the diagonal maps.
Question 2: Can one characterize algebraic theories $\mathcal{T}$ for which the coproduct of initially monic algebras is initially monic?

REPLY [4 votes]: Even for single-sorted operads, the coproduct of initially monic algebras need not be initially monic. 
First a general construction. Let $A$ be a commutative monoid. Then the comma category $A \downarrow \mathrm{CMon}$ (aka the undercategory or co-slice under $A$) is the category of algebras of an operad $\mathcal{O}_A$. I'll indicate a proof of that later, but it allows us to relativize the question to consideration of pushouts of monos in $\mathrm{CMon}$, where we ask: if $i_B: A \to B$ and $i_C: A \to C$ are monos in $\mathrm{CMon}$ and we take the pushout $P = B \oplus_A C$, is it true that the canonical map $A \to P$ is monic? This is equivalent to the question of initial monicity for the operad $\mathcal{O}_A$. 
The answer is 'no'. The following is a modification of an example that appears in 

Kimura, On semigroups, PhD dissertation, Tulane University 1957. 

Take $A = \{u, v, 1, 0\}$ to be a commutative monoid with identity element $1$ and null element $0$, with all products $x y = 0$ for $x \neq 1, y \neq 1$. Define $B = A \sqcup \{b\}$, adjoining a new element $b$ to $A$, and extending the multiplication on $A$ by $b u = u b = v$ and $b \cdot 1 = 1 \cdot b = b$ and $b x = x b = 0$ for any $x \notin \{u, 1\}$. Similarly define $C = A \sqcup \{c\}$ with $c v = v c = u$ and $c \cdot 1 = 1 \cdot c = c$ and $c x = x c = 0$ for $x \notin \{v, 1\}$. Let $i_B: A \to B$ and $i_C: A \to C$ be the inclusion maps, and let $P$ be the pushout of $i_B$ and $i_A$ with coproduct coprojections $j: B \to P$, $k: C \to P$. I claim $k \circ i_C: A \to P$ is not monic. The short version of the calculation is that in $P$ we have 
$$u = v c = b u c = b \cdot 0 = 0$$ 
and the long version (if anyone really needs it) is that $(k \circ i_C)(u) = (k \circ i_C)(0)$ according to the string of equations 
$$k(i_C u) = k(i_C v \cdot c) = k(i_C v) k(c) = j(i_B v) k(c) = j(b i_B u) k(c) = j(b)j(i_B u)k(c) = j(b)k(i_C u)k(c) = j(b)k(i_C u \cdot c) = j(b)k(i_C 0) = j(b)j(i_B 0) = j(b \cdot i_B 0) = j(i_B 0) = k(i_C 0).$$ 
Now for the claim of an operad $\mathcal{O}_A$ whose algebras are $A \downarrow \mathrm{CMon}$. We have pairs of adjoint functors 
$$F_A \dashv U_A: A\downarrow \mathrm{CMon} \to \mathrm{CMon}, \qquad F \dashv U: \mathrm{CMon} \to \mathrm{Set}$$ 
where $U F$ is the monad attached to the commutative monoid operad and the forgetful functor $U_A: A \downarrow \mathrm{CMon} \to \mathrm{CMon}$ is monadic. The composite $U \circ U_A$ is also monadic; this follows for example from the crude monadicity theorem (both $U$ and $U_A$ reflect isomorphisms and preserve reflexive coequalizers). What remains to be seen is that the monad attached to $U \circ U_A$ is an analytic monad in the sense of the theory of Joyal species. This has two parts: that the underlying functor of the monad $U U_A F_A F$ is an analytic functor, and that the monad unit and multiplication are cartesian natural transformations. 
One way to characterize analytic functors $\mathrm{Set} \to \mathrm{Set}$ is that they are functors which preserve filtered colimits and weak wide pullbacks (see for example this paper), so we check that the three functors $U, F,$ and $U_A F_A = A \oplus -: \mathrm{CMon} \to \mathrm{CMon}$ have these properties. That $U$ and $F$ do follows from the fact that $U F$ does (being an analytic functor) plus the fact that $U$ preserves and reflects limits and even weak limits, plus the fact that $U$ preserves filtered colimits. Meanwhile, the underlying functor of the monad $A \oplus -: \mathrm{CMon} \to \mathrm{CMon}$ preserves filtered colimits (this is true for the functor $A \oplus - \cong A + -$, seen as taking the coproduct with $A$) and preserves wide pullbacks (this is true for the functor $A \oplus - \cong A \times -$, seen as taking the product with $A$). 
That the monad data for the monadic functor $A \downarrow \mathrm{CMon} \to \mathrm{Set}$ are cartesian follows from (1) the observation that the unit and multiplication of the monad $A \oplus -: \mathrm{CMon} \to \mathrm{CMon}$ are cartesian; this is most easily seen by considering $A \oplus -$ in the guise of $A \times -$, where the cartesianness boils down to consideration of products of pullback squares, and (2) the fact that cartesian natural transformations are closed under pasting, e.g., given that the units $u: 1_{\mathrm{Set}} \to U F$ and $\eta: 1_{\mathrm{CMon}} \to A \oplus -$ are cartesian, so is the unit of the monad formed by pasting 
$$1 \stackrel{u}{\to} U F \stackrel{1 \eta 1}{\to} U(A \oplus -)F$$ 
(using the fact that $U$ preserves pullbacks).<|endoftext|>
TITLE: The number of Hamiltonian paths in a tournament
QUESTION [11 upvotes]: If $h(T)$ denotes the number of (directed) Hamiltonian paths in the tournament $T,$ what is the range of $h(T)$ as $T$ ranges over all (finite) tournaments $T$?

By a classical theorem of Rédei (László Rédei, Ein kombinatorischer Satz, Acta Litteraria Szeged 7 (1934), 39–43), every tournament has an odd number of Hamiltonian paths. Moreover, one can verify that there is no tournament with exactly $7$ Hamiltonian paths.

Is there an odd number $p\gt7$ such that no tournament has exactly $p$ Hamiltonian paths?

REPLY [6 votes]: This sounds bizarre, but my computations suggest that 21 might be a possibility..
I checked the tournaments on up to 9 vertices (10 vertices is under way) and got the following results:

by 6 vertices, every odd number (except 7 and 21) less than 33 had occurred at least once
by 7 vertices, every odd number (except 7 and 21) less than 63 had occurred at least once
by 8 vertices, every odd number (except 7 and 21) less than 611 had occurred at least once
by 9 vertices, every odd number (except 7 and 21) less than 2883 had occurred at least once

So there is an obvious conjecture based on those numbers...
Edit (Additional data): By 10 vertices, every odd number (not 7, 21) less than 14685 has occurred. So now I am convinced that there is no tournament with exactly 21 hamilton paths
Edit (Even more additional data): By 11 vertices, every odd number (not 7, 21) less than 80557 has occurred as the number of hamiltonian paths of a tournament on at most 11 vertices.<|endoftext|>
TITLE: Uniform elimination of imaginaries
QUESTION [9 upvotes]: Does the following principle follow from uniform elimination of imaginaries?
For every formula $\varphi(x;y)$ there is a formula $\vartheta(x;z)$ such that 
$$\forall y\;\exists^{=1}z\;\forall x\;\Big[\varphi(x;y)\leftrightarrow\vartheta(x;z)\Big]$$
The answer is affirmative if we restrict the question to formulas such that $\forall y\;\exists x\;\varphi(x;y)$. Is this limitation necessary?
Edit (to answer a request in the comments). Uniform elimination of imaginaries says that every definable equivalence relation is the kernel of a function (that is, $aEb\leftrightarrow fa=fb$).
Elimination of imaginaries says that for every $\varphi(x;y)$ and every parameter $a$ there is a formula $\vartheta_a(x;z)$ such that 
$$\exists^{=1}z\;\forall x\;\Big[\varphi(x;a)\leftrightarrow\vartheta_a(x;z)\Big]$$
Elimination of imaginaries is equivalent to uniform elimination under very weak hypotheses (there are two definable elements).

REPLY [2 votes]: Yes, the statement follows from uniform elimination of imaginaries, with a slight warning: you need to allow the range of the variable $z$ to be a definable set (of tuples), rather than all tuples of a given length. More on that caveat later.
Let $\varphi(x,y)$ be a formula in tuples $x$ and $y$. Define an equivalence
$$y_1 E y_2 := \forall x : (\varphi(x,y_1) \leftrightarrow \varphi(x,y_2)).$$ That is, $a E b$ iff $\varphi(x,a)$ and $\varphi(x,b)$ define the same set. By uniform elimination of imaginaries, we have a definable $f$ such that $E$ is the kernel of $f$. Moreover, we can take $f$ to be surjective by considering its codomain to be its range, which is definable.
Now, set $$\vartheta(x,z) := \exists y : f(y) = z \wedge \varphi(x,y).$$ Fix $y$. We need to show existence and uniqueness of $z$ such that $\varphi(x,y)$ and $\vartheta(x,z)$ define the same set in $x$. Existence is easy: $z := f(y)$ works. For uniqueness, assume we have $z_1,z_2$ such that $\vartheta(x,z_1)$ and $\vartheta(x,z_2)$ both define the same set (in particular, that they both define $\varphi(x,y)$). By the surjectivity of $f$, take $y_1,y_2$ such that $f(y_1) = z_1, f(y_2) = z_2$. We now need only show that $y_1 E y_2$, which we already know from the definitions of $E$ and $\vartheta$.

About the caveat. The caveat appears necessary (though I don't have a counterexample) in order to consider $f$ surjective. I don't consider the caveat a problem, though, since there's no real reason to consider definable sets of tuples as "second class citizens." And even if you do, just make the definable set explicit by modifing the statement to say there is $\vartheta(x,z)$ and a definable set $Z$ such that $$(\forall y) (\exists^{=1} z \in Z) (\forall x) \varphi(x,y) \leftrightarrow \vartheta(x,z).$$<|endoftext|>
TITLE: Is Lemma D4.5.3 in the Elephant correct? (“In a topos, weakly projective implies internally projective.”)
QUESTION [23 upvotes]: Lemma D4.5.3 of Johnstone’s Sketches of an Elephant states:

Lemma. For an object $A$
   of a topos $\newcommand{\E}{\mathcal{E}}\E$, the following are equivalent:

$A$ is internally projective [i.e. $\Pi_A : \E/A \to \E$
  preserves epimorphisms];
$(−)^A : \E \to \E$ preserves epimorphisms;
for every epimorphism $\newcommand{\epito}{\twoheadrightarrow} e : B \epito A$ in $\E$, there exists $C \epito 1$ such that $C^∗(e)$ is split epic.


(These conditions should be (i), (ii), (iii), but lettered lists don’t seem to be available in this Markdown dialect.)
The implications (i) $\newcommand{\Iff}{\Leftrightarrow} \Iff$ (ii) and (ii) $\newcommand{\Imp}{\Rightarrow}\Imp$ (iii) are fine, but (iii) $\Imp$ (i) is rather murky.   The argument given in the Elephant is a very brief sketch; I and a couple of colleagues spent some time today trying to figure out the details, couldn’t, and ended up doubting that this implication is correct.  Searching around, this nforum thread shows that Mike Shulman and Jonas Frey have previously come to essentially the same conclusion.
However, we couldn’t find a counterexample, and the nforum thread doesn’t give one either, so the question is still a bit unsettled.  Does anyone know either a counterexample to the implication (iii) $\Imp$ (i), or else an argument that it holds?
To keep any discussion clear, I suggest the term weakly projective for condition (iii) (unless someone knows a more established term for it).  As Mike says in the nforum thread, if you strengthen this to its stable version, by quantifying over epis $e : B \to U \times A$ for all objects $U$, then it does imply internal projectivity.

REPLY [8 votes]: There is a counterexample, due to Todd Trimble, in another nForum thread; cf. the nLab entry on internally projective objects. Theorem 2 of loc. cit. tells that for an object $A$ to be internally projective, it is necessary and sufficient that for all objects $B$ and $U$ and every epimorphism $e\colon B\to A\times U$ there is an epimorphism $f\colon C\to U$ such that the pull back morphism $(\operatorname{id}_A\times f)^*(e)\colon (\operatorname{id}_A\times f)^*B\to A\times C$ is split. Right after the proof it is noted that in Lemma 4.5.3 of the Elephant, the third statement is just the the special case $U=1$ and that this is insufficient to prove that $A$ is internally projective, 

[...] since if so, then every projective object would be internally projective, which as we show below is not the case.

In fact, this is not "shown below", but a link is provided to this counterexample (which is just the same as the one in the aforementioned nForum entry by Todd Trimble). Note that by proposition 1 of the nLab entry, if the topos has enough projectives and projectives are closed under binary products, then every projective object is internally projective.<|endoftext|>
TITLE: Combinatorial aspects of continued fractions
QUESTION [8 upvotes]: Recently, I got interested in the study of the combinatorial aspects of continued fractions. Precisely, I read of the following lemma of Flajolet (see here):
Lemma. It holds
$$\sum_{\omega} \nu(\omega) \, z^{|\omega|} = \frac1{1 - c_0 z - \displaystyle\frac{a_0b_0z^2}{1 - c_1 z - \displaystyle\frac{a_1 b_1 z^2}{1 - c_2 z - \ldots}}} ,$$
where $\omega$ runs over all the Motzkin paths, $|\omega|$ is the length of the Motzkin path $\omega$, and $\nu(\omega)$ is its weight, assigning the weights $a_i$, $b_i$, and $c_i$ to up, down, and horizontal steps, respectively (see the linked PDF for more details).
I would like to know more about this kind of connections between combinatorics and continued fractions and I am looking for a book about this subject, or at least with a detailed chapter about.
Until know I just found some articles, all pointing to the 1980 article of Flajolet (1). I also see that the book (2) has a chapter on "Combinatorial interpretations of continued fractions", but it regards tilings and continued fractions with integers numerators and denominators, so it is about other things. The lecture notes of Viennot (3) might be good, but I cannot read French and they are only 3 years after Flajolet paper, it would be better something more updated.
Thank you in advance for your help.
(1) P. Flajolet, Combinatorial aspects of continued fractions, Discrete Mathematics 32 (1980) 125--161.
(2) A. T. Benjamin, J. J. Quinn, Proofs that Really Count: The Art of Combinatorial Proof, Mathematical Association of America, 2003.
(3) Une theorie combinatoire des polynomes orthogonaux generaux http://www.xavierviennot.org/xavier/polynomes_orthogonaux.html

REPLY [5 votes]: One book that I know which covers this topic is (available online):
Analytic Combinatorics, P. Flajolet and R. Sedgewick. In particular, (as the OP now also notes in a comment), the relevant section is V.4.
The following slides of Flajolet are also relevant.

Remarks

The OP may also find this paper on restricted permutations and continued fractions of interest.
Another interesting reference (not on the original topic, but I thought it worthwhile to mention here) is: Analytic Combinatorics in Several Variables, by R. Pemantle and M. Wilson.

REPLY [3 votes]: Hint: No book recommendation, but in case you are not aware of it you might appreciate the nice survey Combinatorial aspects of continued fractions and applications by Xavier G. Viennot in honor of P. Flajolet.

REPLY [3 votes]: See Chapter 5 of the book Combinatorial Enumeration by Ian P. Goulden and David Jackson, Wiley, 1983, reprinted by Dover in 2004.<|endoftext|>
TITLE: Is there a geometric construction of hyperbolic Kac-Moody groups?
QUESTION [33 upvotes]: Just as the theory of finite-dimensional simple Lie algebras is connected to differential geometry and physics via the theory of simple Lie groups, the theory of affine Lie algebras was connected to differential geometry and physics via the realization that these are the Lie algebras of central extensions of loop groups:

Andrew Pressley and Graeme Segal, Loop Groups, Oxford U. Press, Oxford, 1988.
Graeme Segal, Loop groups.

Indeed it's not much of an exaggeration to say that central extensions of loop groups are to strings as simple Lie groups are to particles.
The finite-dimensional simple Lie algebras and affine Lie algebras are both special cases of Kac--Moody algebras.  The next class of Kac--Moody algebras are the hyperbolic Kac--Moody algebras, and these have been completely classified:

Lisa Carbone, Sjuvon Chung, Leigh Cobbs, Robert McRae, Debajyoti Nandi, Yusra Naqvi and Diego Penta, Classification of hyperbolic Dynkin diagrams, root lengths and Weyl group orbits.

My question is whether a geometrical construction of any of the corresponding hyperbolic Kac--Moody groups is known.  Since after 'particle' and 'string' one naturally says '2-brane', one might naively hope that these are connected to 2-brane theories, or perhaps 2+1-dimensional field theories.  But maybe that's the wrong idea.
Jacques Tits famously gave a way to construct Kac--Moody groups, not only over the real and complex numbers but over arbitrary commutative rings:

Jacques Tits, Uniqueness and presentation of Kac–Moody groups over fields, J.
Algebra 105 (1987), 542–573.

This has been simplified for a certain class of hyperbolic Kac--Moody groups, namely the simply-laced ones:

Daniel Alcock, Lisa Carbone, Presentation of hyperbolic Kac-Moody groups over rings.

This construction does not feel 'geometric' to me: it's in terms of generators and relations.  
Just for fun, here are the Dynkin diagrams of the finite-dimensional simple Lie algebras:

Here are the Dynkin diagrams of the untwisted affine Lie algebras:

and the twisted ones:

A Dynkin diagram describes a hyperbolic Kac--Moody algebra if it's not among those shown above, but every proper connected subdiagram is. There are infinitely many hyperbolic Kac--Moody algebras whose Dynkin diagrams have $2$ nodes, but only 238 with $\ge 3$ nodes.  The simply-laced ones were nicely drawn by Allcock and Carbone:

The last of these diagrams is called $\mathrm{E}_{10}$, and there are lots of interesting conjectures about its role in physics --- see Allcock and Carbone's paper for references.  These could be clues to a geometric construction of the corresponding group.

REPLY [27 votes]: Thanks for the questions. The initial set-up for the Chevalley type construction (exponentiating from the Kac-Moody algebra) for all symmetrizable Kac-Moody groups (including the hyperbolic case) over arbitrary fields was worked out with Howard Garland (Existence of lattices in Kac-Moody groups over finite fields, Comm. Cont. Math. 5 no 6 (2003) pp 1-55). For Kac-Moody groups over Z, it is summarized in my paper with Frank Wagner (Uniqueness of representation--theoretic hyperbolic Kac--Moody groups over $\mathbb{Z}$, arXiv:1512.04623v2 [math.GR]) where we compared the Chevalley construction with the finite presentation for Tits' group obtained in the paper with Daniel Allcock. The full Chevalley theory for Kac-Moody groups is currently being worked out. In joint work with Alex Feingold and Walter Freyn, we have constructed an action of hyperbolic Kac-Moody groups over C on a simplicial structure in the Lie algebra of the compact form. This simplicial complex has the structure of a Tits building, but is constructed geometrically in terms of Cartan subalgebras, rather than in terms of group cosets like the usual Tits building. This paper will also appear soon. There is an algebro-geometric construction of Kac-Moody groups by Olivier Mathieu, Construction du groupe de Kac-Moody et applications, C. R. Acad.
Sci. Paris Sér. I Math. 306 (1988), no. 5, 227–230. The book of Kumar Shrawan Kumar, Kac–Moody groups, their flag varieties and representation theory,
Progress in Mathematics, vol. 204, Birkhauser Boston Inc., Boston, MA, 2002 has a lot of useful information. In spite of hard work by many people, we don't have a uniform construction of the "overextended" hyperbolic Kac-Moody groups starting from the corresponding finite-dimensional simple Lie groups. As you must have seen, only certain of the hyperbolic Dynkin diagrams are "overextensions" of finite dimensional ones. The constructions described above are "uniform" (or rather formal) for all hyperbolic Kac-Moody groups. Individual "zoo-like" constructions could be very revealing, but there are no examples currently.<|endoftext|>
TITLE: Shape whose translated and scaled copies are closed under intersection
QUESTION [8 upvotes]: The translated and scaled copies of an equilateral triangle with fixed orientation are closed under intersection - the intersection is again an equilateral triangle with the same orientation.

What other convex shapes in 2D have this property?

REPLY [7 votes]: I think triangles (degenerated ones included) are the only such convex shapes. The idea is: If $C$ is such a convex shape, let  $p_i\in\partial C$ for $i=1,2,3$ be three smooth points. One  can obtain an approximate triangle as an intersection of three large copies $C$, say $m(C-p_i)+p_i$ for large $m$. Since these intersections are similar to $C$, and converge to a triangle as $m\to\infty$, $C$ itself is  a triangle.

REPLY [3 votes]: Since the shape $A$ is convex its boundary is differentiable almost everywhere. Take a point $p$ where it is differentiable. A line $l$ is tangent to A in $p$. Take two more points $q_1$ and $q_2$ with tangent lines $l_1$ and $l_2$, translate the shape by vectors $q-q_1$ and $q-q_2$ and intersect results. Suppose we get a shape $B$. The only point which can possibly have a tangent line parallel to $l$ will be $p$, but $\partial B$ is not differentiable there (unless all lines are parallel). Therefore, $B$ is not a homothetic image of $A$ and that means $B$ is at most one-dimensional. 

$B$ being one-dimensional means $\angle (l_2, l) + \angle (l, l_1) \le \pi$ (we take angles which "look" at $A$). If there is at least 5 different directions for tangent lines we get a contradiction since in a convex pentagon the sum of angles is $3\pi$ but from our inequality it must be no more than $5\pi/2$. Therefore our shape is either a triangle (for which it is true) or parallelogram (false).<|endoftext|>
TITLE: Two elementary inequalities for real-valued polynomials
QUESTION [9 upvotes]: I am looking for references discussing two inequalities that come up in the study of the dynamics of Newton's method on real-valued polynomials (in one variable). The inequalities are fairly different, but it seems to make sense to ask about both of them in the same post.
Most of the details below are fairly elementary, they are mostly included for completeness.

1. The first inequality is quite easy to verify: Suppose that $f$ is a real-valued polynomial in one variable, all of whose roots are real. Then 
 $$ f f''\le (f')^2, $$
with equality holding only at the double roots of $f$. I've seen this referred to as Turan inequality [a], but do not know of any places where it is discussed, or what a primary reference would be. It is pretty enough that I would imagine it well-known.
Also, I would be interested in knowing about generalizations, by which I mean either more relaxed conditions on the roots of $f$ that still suffice for the inequality to hold, or a discussion of intervals where the inequality is ensured in general, or a combination of both. Some assumption is needed, as $f(x)=x^2+1$ verifies.
I would also be curious to hear of applications outside of the context of Newton's dynamics.

[a]. MR0729188 (85a:58060). Saari, Donald G.(1-NW); Urenko, John B.
  Newton's method, circle maps, and chaotic motion.
  Amer. Math. Monthly 91 (1984), no. 1, 3–17. 


2. The second inequality is more specialized. It is due to Barna [b] and the proof is straightforward but I would not call it trivial. Since [b] may not be easily accessible, I wrote a short note with the details, which can be found here. 
Some notation is needed before I can state the result: Given a differentiable function $f$, the Newton function for $f$ is $N=N_f$ given by $$ N_f(x)=x-\frac{f(x)}{f'(x)}. $$ 
This is the function that results from applying the familiar Newton's method for approximating the zeros of $f$. (Note that $x^*$ is a zero of $f$ iff it is a fixed-point of $N$.)
Under reasonable assumptions on $f$, if $x^*$ is a zero of $f$, then there is an interval $I$ about $x^*$ such that for any $x_0\in I$, the Newton sequence $x_0,x_1,\dots,x_{n+1}=N(x_n),\dots$ is well defined and converges to $x^*$. Perhaps the simplest proof of this goes by observing that 
 $$ N'=\frac{f f''}{(f')^2}, $$ 
so $N'(x^*)=0$, and there is a small interval $J$ centered at $x^*$ and a positive $\rho<1$ such that $|N'(t)|<\rho$ for all $t\in J$. But then we see that $$ |N(t)-x^*|=|N(t)-N(x^*)|\le\rho|t-x^*| $$ by the mean-value theorem, from which it readily follows that the Newton sequence starting at $t$ indeed converges to $x^*$. 
The largest interval $I$ as above is the immediate basin of attraction of $x^*$. One easily checks that it is open and that, if bounded, say $I=(a,b)$, then $N(a)=b$ and $N(b)=a$. 
Barna's inequality is as follows: Suppose that $f$ is a real-valued polynomial, all of whose roots are real. If $r$ is a root of $f$ and its immediate basin of attraction $I=(a,b)$ is bounded, then $|N'(a)|>1$ and $|N'(b)|>1$.
I would like to know of references other than [b] presenting details of a proof or some discussion of it ([a] refers to [b] in a brief remark, without details). I would also like to hear of generalizations or strengthenings. 

[b]. MR0135224 (24 #B1274). Barna, Béla. Über die Divergenzpunkte des Newtonschen Verfahrens zur Bestimmung von Wurzeln algebraischer Gleichungen. III. (German) Publ. Math. Debrecen 8 1961, 193–207.

REPLY [2 votes]: Regarding Newton's method for polynomials with all zeros real, there is some discussion in the paper by J. H. Hubbard, D. Schleicher and S. Sutherland:http://www.math.cornell.edu/~hubbard/NewtonInventiones.pdf,
especially in Section 7.
There is a reference to another paper by Barna there, and lots of other references, which may be useful.<|endoftext|>
TITLE: Counting nonzero hyperdeterminants over $\mathbb{F}_q$
QUESTION [11 upvotes]: The hyperdeterminant  $D(A)$ is a multidimensional generalization of the
determinant. It is a polynomial in the entries of a $(k_1+1)\times
(k_2+1)\times\cdots \times (k_n+1)$ array $A$. The hyperdeterminant is defined
when $2\max k_i\leq \sum k_i$. For further information see for
instance http://arxiv.org/pdf/1301.0472v1.pdf.
Assuming that $(k_1,\dots,k_n)$ satisfies the above condition, let
$N(q)$ be the number of nonzero $(k_1+1)\times\cdots\times(k_n+1)$
hyperdeterminants over the finite field $\mathbb{F}_q$. For instance,
if $n=2$ then $k_1=k_2$, and the number of $m\times m$ matrices over
$\mathbb{F}_q$ with nonzero determinant is well-known and easily seen
to equal $(q^m-1)(q^m-q)\cdots (q^m-q^{m-1})$. A few years ago some
M.I.T. graduate students computed that for $2\times 2\times 2$
hyperdeterminants we get a polynomial in $q$, though I seem to have
misplaced the formula. Can anything be said about the general case? Do
we always get a polynomial in $q$? (This seems unlikely to me.) Can
someone compute the $2\times 2\times 3$ case?

REPLY [3 votes]: I saw this question years ago and thought it was interesting. I never got around to doing anything with the question, but today I happened to come across some slides of Steven Sam with some data on the problem. I imagine the slides refer to the same M.I.T. students alluded to in the question. The slides can be found on Steven Sam's website and information on hyperdeterminants is on pages 16, 17, and 18.
The count for $2 \times 2 \times 2$ is reported to be $(q^4 - 1)(q^4 - q^3)$ and is attributed to Musiker--Yu. Attributed to Lewis--Sam are counts for $2 \times 2 \times 3$, $2 \times 3 \times 3$, and $2 \times 2 \times 4$ which are
$$q^4(q-1)^4 [2]^2 [3]\\ q^{10}(q-1)^3[2]^2[3] \\ q^4(q-1)^2[3][4](q^3+q^2-1).$$
It says "Caveat: these need to be double-checked..." which I have not done. But now maybe I (or someone else) will be inclined to check since we have an expression to compare to.<|endoftext|>
TITLE: Modular Tensor Categories: Reasoning behind the axioms
QUESTION [16 upvotes]: (Sorry for the length of the question, I'm trying to communicate what is bothering me as thoroughly as possible)
In the construction of modular tensor categories (MTC) from ground zero, we put structures one by one:

Tensor Product Structure $\to$ Monoidal Categories
Dual Objects $\to$ Rigid Monoidal Categories
Representation of Braid Group $\to$ Rigid Braided Monoidal Categories
Abelian and $k$-Linear Category
Semisimplicty
Finiteness conditions $\to$ Locally Finiteness and Finite Number of Isomorphism Class of Simple Objects
$\mathrm{End}(\mathbf{1})=k$: The unit object is simple.
$X\simeq X^{**}$ $\to$ Pivotal Categories

Demanding the above we get a Ribbon Fusion Category. If we look back at these structures there is a very good and mathematically natural reason why we demand them. The most non-trivial reasons are probably related to $\mathrm{End}(\mathbf{1})=k$ and finiteness conditions. But even then there are good reasons, for example $\mathrm{End}(\mathbf{1})=k$ guarantees not only that unit is simple, but also that quantum traces are a $k$-number. The finiteness conditions on the other hand gives us access to Jordan-Holder and Krull-Schmidt theorems, and isomorphisms like $(X\otimes Y)^*\simeq Y^*\otimes X^*$. So every last demand is quite reasonable and natural for a mathematician to study!
Then comes the last structure of an MTC:

The following structures on a ribbon fusion category are equivalent and this last structure defines a Modular Tensor Category

(Turaev) The matrix $\tilde{s}$ with components $\tilde{s}_{ij}=\mathrm{tr}(\sigma_{{X_i}X_j}\circ \sigma_{{X_j}X_i})$
  is invertible (trace is quantum trace, $\sigma$ is braiding
  isomorphism and $X_i$ are simple)
(Bruguières, Müger) Multiples of unit $\mathbf{1}$ are the only transparent objects of the category. (An object $X$ is called
  transparent if $\sigma_{XY}\circ\sigma_{YX}=\mathrm{id}_{Y\otimes X}$
  for all objects $Y$.)
Our premodular category $\mathscr{C}$ is factorizable, i.e. the functor $\mathscr{C}\boxtimes \mathscr{C}^\text{rev}\to
 \mathcal{Z}(\mathscr{C})$ is an equivalence of categories.


I've been desperately searching for a good reason on why this last demand is also mathematically natural to ask. If I may be so bold, seems a bit random to me! The best I could came up with is that

An MTC is the polar opposite to a symmetric ribbon fusion category. In other words in a symmetric category every object is transparent, so it is, in a way, maximally transparent. While an MTC is minimally transparent (maximally opaque maybe!).

But this still doesn't satisfy my irk! So what? What about the categories in between? Then there are indirect reasons why one should demand this structure:

By demanding this last structure a lot of `nice' things happen.

We have a projective representation of $\mathrm{SL}(2,\mathbb{Z})$ in our MTC: Namely matrices
  $s=\tilde{s}/\sqrt{D}$ with $D$ being the global dimension, $t$ with
  $t_{ij}=\theta_i\delta_{ij}$ and $c$ (charge conjugation)
  $c_{ij}=\delta_{ij^*}$ such that $$(st)^3=\zeta^3s^2, \quad s^2=c,
\quad ct=tc, \quad c^2=1$$
Modular Functors and their $\mathscr{C}$-extensions and they are very closely related to MTCs.
Verlinde Formula: The data $s,t,\zeta$ determines an MTC uniquely.
Vafa Theorem and it's generalizations: The scalars $\theta_i$ and $\zeta$ are roots of unity.

and even more nice things...

But still... Unless there is some good reason why we demand this last conditions naturally, all of this seems like a happy coincidence. So basically my question is

Is there a deep fundamental mathematical (or physical, physical is also fine) reason behind this last structure, which makes this condition a priori?

REPLY [16 votes]: Interesting question ! As far as I know, there are at least two secretly equivalent answers. 
You somehow already gave the first one: a modular tensor category is the same as a modular functor (though the precise statement is quite subtle, see the beautiful introduction to this paper : https://arxiv.org/abs/1509.06811). A modular functor is, roughly, a collection of compatible (projective) representations of mapping class groups, so in particular you need a representation of $SL_2(\mathbb{Z})$. It turns out this is also sufficient, in the sense that a premodular tensor category gives representations of mapping class groups in genus 0, and modularity is exactly what you need to extends this to higher genus.
A somehow more conceptual reason is related to the fact that those theories have an anomaly, i.e. you get only projective representations of MCG's, and invariant of 3-fold with some extra structure (I learned this point of view from Walker and Freed-Teleman). The origin of the story is the 3d Chern-Simons TFT introduced by Witten using Feynman integrals. It turns out that there is a 4d TFT around, a very simple topological version of Yang-Mills theory. This theory is simple because it involves integrating 4-forms on 4-folds, and those forms are actually exact. So if Feynman integrals really were integrals, by Stokes theorem this would be trivial on closed manifold, and for a 4-fold W with boundary we could define
$$Z_{CS}(\partial W):=Z_{YM}(W).$$
SInce any oriented 3-fold bound a 4-fold this would indeed be enough to define your theory.
It turns out $Z_{YM}$ is not trivial, but it is close to be: it is an invertible theory. It means in particular that it attaches 1-dimensional vector spaces to 3-fold, and that every 4-fold $W$ gives an isomorphism
$$Z_{YM}(\partial W) \longrightarrow Z_{YM}(\emptyset)=\mathbb{C}$$
which depends on the choice of $W$ only up to bordism. Hence you get a number defined up to this choice, and the bordism class of $W$ is precisely the extra structure occurring in this anomaly I was talking about.
How does it relates to your question ? Well, it is expected that every premodular category gives rise to a 4-dimensional TFT. Roughly speaking, to know whether this theory is invertible, it's enough to check what you get for the 2-sphere, which is a category, and it turns out this category is equivalent to the Müger center, i.e. the category of transparent objects. At the categorical level, invertibility means being equivalent to the category of vector spaces. 
Therefore, the modularity condition is exactly what you need to go from a 4d TFT to a 3d TFT with anomaly.<|endoftext|>
TITLE: Why is the Frankl conjecture hard?
QUESTION [47 upvotes]: This is a naive question that could justifiably be quickly closed.
Nevertheless:
Q. Why is 
Péter Frankl's
conjecture so difficult?

If any two sets in some family of sets have a union that also belongs to the family, must some element belong to at least half of the sets in the family?

This has remained unsolved for ~$37$ years.
It seems that, unlike other conjectures (say, concerning prime conjectures), it has not been confirmed for vastly huge sets (just: families of at most $50$ sets).
More specifically, can anyone indicate why this conjecture seems so
difficult to prove or disprove? Why it has withstood assaults so long?

REPLY [31 votes]: (Migrated by request from the comments.)
Bruhn and Schaud's (2013) The journey of the union-closed sets conjecture provides a rather readable write-up. Particularly relevant is the section Obstacles to a proof; for example, you may check just after Conjecture 15 in which the authors ask (essentially) your question here:

"So, why then has the conjecture withstood more than twenty years of proof attempts?" (p. 14)

Bruhn and Schaud then list three possible techniques of proof, and go into a bit of detail around why they do not seem to work out; these techniques are: injections, local configurations, and averaging.
The paper also provides a few relevant re-formulations using, e.g., lattices, (maximal stable sets of bipartite) graphs, and the "Salzborn" formulation (p. 12). In each case, a re-formulation of the Frankl (or union-closed sets) conjecture brings corresponding ideas and techniques with varying potential; the authors of this particular survey do well by their promise early on:

"The focus of this survey is on the methods employed to attack the conjecture. Our treatment of the literature is therefore somewhat uneven. Whenever we can identify a technique that, to our eyes, seems interesting and potentially powerful we discuss it in greater detail" (p. 3).<|endoftext|>
TITLE: Need a good name for an algorithmic problem in groups that generalizes the conjugacy problem
QUESTION [8 upvotes]: I am looking for a good name for the following problem:

Given elements $g_1,\dotsc,g_n$ in a (finitely generated) group $G$, determine if the product of their conjugacy classes $g_1^G\dotsb g_n^G$ contains the identity element $1$.

In some situations it might be more natural to pose this problem slightly differently:

Given elements $g_1,\dotsc,g_n$ and $g$ in $G$, determine if $g_1^G\dotsb g_n^G$ contains $g$.

This problem generalizes the conjugacy problem.
Geometrically the problem can be stated as follows: given a path-connected space $X$ (with fundamental group $G$) and a sphere with holes $S$, determine for every mapping $\partial S\to X$ (a mapping of the boundary circles of $S$ to $X$) if this mapping can be extended to a mapping $S\to X$.
It would be nice to also have consistent names for other related problems.  In particular, for this one:

Given elements $g_1,\dotsc,g_n$ in $G$, determine if the product of their conjugacy classes $g_1^G\dotsb g_n^G$ contains a commutator.

Geometrically this problem can be stated as follows: given a path-connected space $X$ and a torus with holes $S$, determine for every mapping $\partial S\to X$ if this mapping can be extended to a mapping $S\to X$.

Suggestions
Would generalized conjugacy problem be an acceptable name for this problem?  I've seen that this term is already used for other problems, but it does not seem to have a generally accepted meaning.  For example i've seen it used for the simultaneous conjugacy problem, but we do not need two names for the same problem.

REPLY [3 votes]: There is a name for this problem: (the solvability of) a genus 0 quadratic equation over G. Check out Sections 2.1 and 3.3 in http://arxiv.org/abs/0802.3839.
This terminology is also consistent with your topological interpretation. Also, whether the set $$
g_1^G\cdots g_n^G
$$ contains a commutator is equivalent to the solvability of the genus 1 quadratic equation $$
[x,y]\prod_{i=1}^n t_i^{-1}g_it_i=1.
$$ in variables $x,y,t_1,\ldots,t_n$. This is terrible terminology.<|endoftext|>
TITLE: A matrix norm inequality
QUESTION [13 upvotes]: Suppose that $A, B$ are Hermitian positive definite matrices of the same order and $0\le p\le 1$. Using a standard approach in matrix analysis, one can show that 
$\|A^{1-p}B^p\|\ge \|A\sharp_p B\|$, where $A\sharp_p B:=A^{1/2}(A^{-1/2}BA^{-1/2})^pA^{1/2}$ which is sometimes called the weighted geometric mean and has a geometric interpretation. The matrix norm here is spectral norm (i.e., largest singular value).  
I tried to play with $\|A^{1-p}B^p\|\ge \|A\sharp_p B\|$ a little bit, a question suddenly occured to me: Is it true $$\|AB\|\ge \|(A\sharp_p B)(A\sharp_{1-p}B)\|?$$ 
I ran some simulations yet no counterexample showed up... the standard approach I know seems not work, so I am looking for some new ingredients.

REPLY [4 votes]: Her is an elementary proof of the case $p=\frac12$. As noted by Suvrit, it is enough to prove  $\|A^{1/2}B^{1/2}\|^2 \le \|AB\|$, or equivalently $\|HK\|^2\le\|H^2K^2\|$. This is true because of
$$\|HK\|^2=\rho((HK)^*HK)=\rho(KH^2K)=\rho(H^2K^2)\le\|H^2K^2\|.$$
Hereabove, $\rho$ is th spectral radius and the last inequality is true for every subordinated norm.<|endoftext|>
TITLE: Contact distributions on $(G_2,P)$-type Cartan geometries in dimension 5
QUESTION [8 upvotes]: Up to topology, the 5D homogeneous space
$$
G_2/P
$$
of the (real form of the) 14D exceptional Lie group $G_2$ is the 5D jet space
$$
M:=J^1(2,1)=\{(x,y,u,p,q)\}
$$
of scalar functions in two independent variables.

WHY? Because $G_2/P=S^2\times S^3\approx\mathbb{P}T^*S^3\approx\mathbb{P}T^*\mathbb{R}^3=\mathbb{R}^2\times(\mathbb{R}\times\mathbb{R}^{2\ast})=J^2(2,1)$, where $\mathbb{R}^2$ is the plane of independent variables, $\mathbb{R}$ the line of the dependent one, and $\mathbb{R}^{2\ast}$ the plane of the derivatives of the latter w.r.t. the former.

Now $M$ comes equipped with a flag of distributions, viz
$$
(M,V,C)\, ,
$$
where $V=\langle\partial_p,\partial_q\rangle$ is the 2D vertical and $C=V\oplus\langle \partial_x+p\partial_u, \partial_y+q\partial_u\rangle$ the 4D contact one.
Now there is a result (I've learned it from P. Nurowski) saying that

(THEOREM) There is a one-to-one correspondence between $(G_2,P)$-type Cartan geometries and generic 2D distributions in dimension 5.

Since - as textbooks recite - a $(G_2,P)$-type Cartan geometry is obtained by rolling without slipping the homogeneous space $G_2/P\approx M$, I guess that a generic 2D distribution in dimension five is obtained by "rolling without slipping" the vertical distribution $V$ of the homogeneous model $M$. This should be the essence of the theorem.
My point is simple: if I can "roll" the vertical distribution $V$, why I cannot "roll" the contact distribution $C$ as well?

QUESTION: Is it true that a $(G_2,P)$-type Cartan geometry in dimension five is the same as a manifold equipped with a $(2,4)$-type flag of distributions $(\widetilde{V},\widetilde{C})$? Under which circumstances $\widetilde{C}$ is contact? (This makes sense, being so at least for the flat case $M$.)

Maybe the question is ill-posed, due to my still poor grasp on Cartan geometries, but I hope that its essence will be clear: the "flat model", i.e., $M$, is a contact manifold equipped with a distinguished 2D contact-subdistribution (no doubts about that), and I'm just wondering how much of this structure descends to the "curved cases". 
"I'm just a simple man trying to make my way in the universe of Cartan geometries" - J. Fett.

REPLY [5 votes]: As Ben wrote, the question appears to conflate two different parabolic geometries of type $\newcommand{bfD}{{\bf D}}\newcommand{bfE}{{\bf E}}\newcommand{bfH}{{\bf H}}G_2$:
Let $\Bbb V$ be the standard (i.e., $7$-dimensional irreducible) representation of $\mathfrak{g}_2$ (either the split real or the complex form); recall that $G_2$ is the stabilizer of cross product map $\times : \Bbb V \times \Bbb V \to \Bbb V$. The inclusion $G_2 \hookrightarrow SO(3, 4)$ ($G_2 \hookrightarrow SO(7, \Bbb C)$) determines an indefinite, nondegenerate, symmetric bilinear form $H$ on $\Bbb V$.
The "first" parabolic subgroup $P_1$ (corresponding to a cross on the first node of the Dynkin diagram of $G_2$ in the usual Bourbaki ordering) is the stabilizer of an isotropic $1$-dimensional subspace of $\Bbb V$. The cone $\mathcal C$ of nonzero isotropic vectors inherits an invariant filtration of tangent distributions, whose fibers at $Y \in \mathcal C$ are $$\ker (Z \mapsto Y \times Z) \subset \operatorname{im} (Z \mapsto Y \times Z) \subset T_Y \mathcal C$$ of dimensions $3, 4, 6$ (to apply the cross product, we implicitly use here the identifications determined by the canonical isomorphism $T_Y \Bbb V \leftrightarrow \Bbb V$). By linearity, this descends to a filtration $$\bfD \subset \bfD' \subset T\Bbb Q_5,$$ of dimensions $2, 3, 5$, on the null quadric $\Bbb{Q}_5 := \Bbb P(\mathcal C) \subset \Bbb P(\Bbb V)$, which is diffeomorphic to $(\Bbb S^2 \times \Bbb S^3) / \Bbb Z_2$, where $\Bbb Z_2$ acts by the antipodal map on both factors. Since the ingredients are $G_2$-invariant, so is $\bfD$ under the induced action on $T\Bbb Q_5$, and as one expects, it turns out that $[\bfD, \bfD] = \bfD'$ and $[\bfD', \bfD] = T\Bbb Q_5$. (NB there are no $G_2$-invariant linear or hyperplane distributions on $T\Bbb Q_5$.)
On the other hand, consider differential equations of the form $z' = F(x, y, y', y'', z)$. Any function $F(x, y, p, q, z)$ determines a total derivative $D_x := \partial_x + p \partial_y + q \partial_p + F \partial_z$ on the corresponding partial jet space $J^{2, 0}(\Bbb R, \Bbb R) \cong \Bbb R^5_{xypqz}$. Suitably regarded, the vertical fibers of the jet truncation map $J^{2, 0}(\Bbb R, \Bbb R) \to J^{1, 0}(\Bbb R, \Bbb R)$ are spanned by $\partial_q$. Computing directly shows that the distribution $$\bfD_F := \langle D_x, \partial_q \rangle$$ is generic iff $F_{qq}$ vanishes nowhere; conversely, a theorem of (I believe) Monge states than any generic $2$-plane distribution on a $5$-manifold is locally equivalent to $\bfD_F$ for some function $F$. If $F(x, y, p, q, z) := q^2$, then the resulting distribution has infinitesimal symmetry algebra isomorphic to $\mathfrak{g}_2$, so by a general fact about parabolic geometries the distribution $(J^{2, 0}(\Bbb R, \Bbb R), \bfD_F)$ corresponding to the differential equation $z' = (y'')^2$ is everywhere locally diffeomorphic to the homogeneous model distribution $(\Bbb Q_5, \bfD)$ above. In particular, it follows from this that neither of the distributions $V$ and $\widetilde{C}$ are invariant under the action of the infinitesimal symmetry algebra $\mathfrak{g}_2$ of $(J^{2, 0}(\Bbb R, \Bbb R), \bfD_F)$.
The correct statement of the theorem you mention is that there is an equivalence of categories between generic $2$-plane distributions on $5$-manifolds and normal, regular parabolic geometries of type $(G_2, P_1)$. (See the end of Subsubsection 4.3.2 in Cap & Slovak's book, Parabolic Geometries.)
On the other hand, we can consider the action of $G_2$ on the space of isotropic $2$-planes in $\Bbb V$. This action has two orbits, according to whether the cross product $\times$ restricts to the zero map on each $2$-plane. (Bryant calls the $2$-planes on which the restriction is zero special in his highly enjoyable lecture notes Elie Cartan and Geometric Duality [pdf], which treats the correspondence space construction for $G_2 / P_1 \leftarrow G_2 / (P_1 \cap P_2) \to G_2 / P_2$, as well as the analogous construction for $A_2$ and $B_2 \cong C_2$. NB that this article seems contains a few typos, replacing $\Bbb N_5$, introduced in a moment, with $\Bbb Q_5$, which is the essential apparent confusion in the question here.) The isotropy subgroup of a point in the $5$-dimensional space $\Bbb N_5$ of special $2$-planes is the "second" parabolic $P_2 \subset G_2$. Analogously to the situation for the first parabolic, we can view $\Bbb N_5$ as a subset of $\Bbb P^{13} = \Bbb P(\mathfrak{g}_2)$, but its geometry is apparently much more complicated than that of $\Bbb Q_5$: In the complex case, $\Bbb N_5$ is a variety of degree $18$, and its complete intersection with three hyperplanes in a general configuration is a K3 surface of genus $10$, but NB other geometric descriptions of this space (which look less daunting to non-algebraic geometers like myself) are available, too. Apparently this is worked out in the paper of Borcea cited below, but I can't find an ungated copy. See also the accessible historical survey paper of Agricola, also cited below.
Now, $\Bbb N_5$ inherits an invariant contact distribution $\bfH$. (Surely this can be written down with some much effort in terms of the cross product on $\Bbb V$, but to my knowledge this hasn't been done anywhere.) Moreover, the representation $P_2$ induces on each fiber of $\bfH$ turns out to be a trivial extension of a representation of $GL(2, \Bbb F) \subset P_2$, and this representation is isomorphic to $S^3 \Bbb F^2$ (this representation is conformally symplectic and so determines equivalently a nondegenerate cone in each fiber of $\bfH$). All of the $G_2$-invariant structure on $\Bbb N_5$ can be recovered from these objects, corresponding to the fact that a (normal, regular) parabolic geometry of type $(G_2, P_2)$ is a $5$-manifold $M$ equipped with a "$G_2$ contact structure", which is a contact structure $\bfH \subset TM$ together with an auxiliary rank-$2$ vector bundle $\bfE \to M$ and a vector bundle isomorphism $S^3 \bfE \stackrel{\cong}{\to} \bfH$ such that the Levi bracket $\bfH \times \bfH \to TM / \bfH$ (the tensorial map induced by the Lie bracket) is invariant under the induced action of $\mathfrak{sl}(\bfE)$. See Subsubsection 4.2.8 of Cap & Slovak's book. I don't know of any sensible analog of the Monge (quasi-)normal form $z' = F(\cdots)$ for $G_2$ contact structures, but I would be pleased to hear about one.
There are connections between these two types of parabolic geometry beyond the mentioned correspondence space construction. See this preprint of Leistner, Nurowski, & Sagerschnig.

I. Agricola, Old and New on the Exceptional Group $G_2$ [pdf], Notices Amer. Math. Soc. 55(8) (2008), 922-929.
C. Borcea, Smooth global complete intersections in certain compact homogeneous complex manifolds, J. Reine Angew. Math. 344 (1983), 65–70.
A. Cap, J. Slovak, Parabolic geometries I: Background and general theory. Math. Surveys Monogr. 154,  Amer. Math. Soc., Providence, RI, 628pp.
T. Leistner, P. Nurowski, K. Sagerschnig, New relations between $G_2$-geometries in dimensions $5$ and $7$, Internat. J. Math. 28(13) (2017).  arXiv:1601.03979<|endoftext|>
TITLE: non-triviality of the underlying real vector bundle of the complexification of a real vector bundle
QUESTION [9 upvotes]: Let $M$ be a given manifold and $\xi$ be a given $k$-dimensional vector bundle over $M$. How to determine whether the underlying real vector bundle of $\xi\otimes\mathbb{C}$, i.e. the Whitney sum $\xi\oplus\xi$, is trivial or not? I want to try different methods as many as possible.
My attempt: To prove $\xi\oplus\xi$ is not a trivial bundle, we only need to prove the Stiefel-Whitney class
$$
1\neq (w(\xi))^2=(1+\sum_{i=1}^k w_i(\xi))^2=1+\sum_{i=1}^k (w_i(\xi))^2,
$$
i.e. there exists $1\leq i\leq k$ such that 
$$
(w_i(\xi))^2\neq 0.
$$
But even the cup-product of
$$
H^*(M;\mathbb{Z}_2)
$$
is known, I still do not know how to determine the element $w_i(\xi)$?
Are there any other methods to try?

REPLY [5 votes]: It is a bit unclear what the question is asking exactly, so let me try to give some examples what can happen to a vector bundle $E$ when taking  $E\oplus E$ (for simplicity, let me call this Whitney doubling). 
Certainly, as already mentioned in the question, the non-vanishing of the Stiefel-Whitney classes is a sufficient condition for non-triviality of $E\oplus E$, and the Stiefel-Whitney classes of $E\oplus E$ are of the form $w_{2i}(E\oplus E)=w_i(E)^{\cup 2}$. 
Another possibility to show non-triviality of  $E\oplus E$ would be using the Pontryagin classes which are classes in $H^{4k}(M,\mathbb{Z})$. We have $2p(E\oplus E)=2p(E)^{\cup 2}$, and non-vanishing of the Pontryagin classes of $E\oplus E$ will imply non-triviality. In favourable cases, this will be possible by computations in rational cohomology.

However, real vector bundles are not determined by their characteristic classes in general. Generally, classification results can be obtained using obstruction theory (to the extent that homotopy groups of $O(n)$ can be computed). The general formalism is explained e.g. in Hatcher's book on algebraic topology. For $M$ a manifold, the obstruction to lifting a map $M\to BSO(n)^{(k)}$ along $BSO(n)^{(k+1)}\to BSO(n)^{(k)}$ lies in  $H^{k+1}(M,\pi_{k-1}BSO(n))$, and the possible lifts are classified by $H^k(M,\pi_{k-1}BSO(n))$. So the vector bundle classification hinges on knowledge of cohomology with coefficients in homotopy groups. Moreover, knowing the induced morphism on homotopy groups $\pi_k(O(n))\to\pi_k(O(2n))$ allows to analyze what happens to vector bundles under Whitney doubling. The following examples can be obtained by obstruction theory: 

As a very simple example, the morphism $O(1)\to O(2)$ is null-homotopic, hence every line bundle $L$ will have $L\oplus L$ trivial (generalizing Ben McKay's answer). 
For $O(2)\to O(4)$, the only nontrivial map induced on homotopy groups is the reduction mod 2 $\mathbb{Z}\cong\pi_1(O(2))\to\pi_1(O(4))\cong\mathbb{Z}/2\mathbb{Z}$. In particular, the Whitney doubling of a rank 2 bundle will be trivial iff the Euler class is even. 
To get some more interesting examples, recall the paper of Kervaire ( Non-parallelizability of the $n$-sphere for $n>7$. Proc.N.A.S. 44 (1958), 280-283.) 
Lemma 1 states that the induced morphism $\pi_k(O(n))\to\pi_k(O(2n))$ is twice the stabilization morphism for $k<n-1$ (where the stabilization morphism happens to be an isomorphism). In particular, the stable range lifting classes  in the obstruction classification will just be multiplied by $2$ under the Whitney doubling.
Now let's use this to get examples of bundles $E$ with trivial Stiefel-Whitney classes such that $E\oplus E$ is stably non-trivial. Consider a manifold $M=S^5/(\mathbb{Z}/8\mathbb{Z})$ which has $H^4(M,\mathbb{Z})=\mathbb{Z}/8\mathbb{Z}$. Any such class $\sigma$ gives a rank $5$ bundle on $M$ whose Stiefel-Whitney $w_4$ class will be the reduction mod $2$, and whose Pontryagin class is $2\sigma$. If we choose $\sigma$ to be twice a generator, we get a bundle with trivial Stiefel-Whitney classes and Pontryagin class $2$-torsion. Now employ the result of Kervaire, which states that the morphism $\pi_3(SO(5))\to\pi_3(SO(10))$ induced by Whitney doubling is multiplication by two. This means, that the Whitney doubling will have lifting class four times a generator of $H^4(M,\mathbb{Z})$. Hence the Whitney doubling is stably non-trivial, but has all characteristic classes trivial. 
Better examples exist in higher dimensions. The homomorphism $\pi_{4k}(BO)\to\mathbb{Z}$ defining the $k$-th Pontryagin class is multiplication with $(2k-1)!$ if $k$ is even and multiplication by $2(2k-1)!$ if $k$ is odd (see the paper of Bott and Milnor on parallelizability of spheres). In particular, we can construct bundles over $S^{13}/(\mathbb{Z}/8\mathbb{Z})$ whose only lifting class is twice a generator of $H^{12}$. These bundles will have all characteristic classes trivial, but will have non-trivial Whitney doubling.<|endoftext|>
TITLE: Functional limit theorem under random change of time
QUESTION [6 upvotes]: FINAL EDIT: There is one main question left: According to the answer, we have choosen $\theta=1$ , where we could choose $0<\theta<\infty$ as we like. His this sufficient, if we regarde the convergence to $\theta$ as a finite positive random variable?
This post seems long, but its almost everything proofed except the last step. The unknown part is marked especially.
Assumptions
Given a Levy-Process $U_{t}$ with  with $E(U_t)=0$ (then $U_t$ is a martingale). Let $U_t$ have finite variance and $Var(X_1)=\sigma^{2}$ and the limit theorem  holds:
\begin{align}
F_t:=\sqrt{t}\left(\frac{U_t}{t}-E(U_1) \right)=\frac{U_t}{\sqrt{t}}\xrightarrow{d}\mathcal{N}(0,\sigma^{2})\quad as \,\,t\rightarrow \infty.\tag1
\end{align}
Let $K_t$ a non-decreasing positive ($K_{t}>0$ a.s.) process with cadlag-paths with the property that $K_{t}\rightarrow \infty$ almost sureley, as $t\rightarrow \infty$.
I want to show that 
\begin{align}
F_{K_t}:=\frac{U_{K_t}}{\sqrt{K_{t}}} \xrightarrow{d}\mathcal{N}(0,\sigma^{2})\quad as \,\,t\rightarrow \infty. \tag2
\end{align}
For this one requires a positive non-random cadlag-function $a(t)$ with $a(t)\rightarrow \infty$ as $t\rightarrow \infty$ such that
\begin{align}
\frac{K_{t}}{a(t)}\rightarrow \theta\quad P\, a.s. \tag3
\end{align}
holds. Where $\theta$ is a positive finite random-variable.
Then the convergence in distribution of $F_{t}\xrightarrow{d} \mathcal{N}(0,\sigma^{2})$ implies the convergence in distribution of $F_{K_t}\xrightarrow{d} \mathcal{N}(0,\sigma^{2})$.
The suggestion how it has to be proofed is given in the post below:
Here he takes $\theta=1$. So that we have $K_{t}\in ((1-\epsilon)a(t),(1+\epsilon)a(t))$ a.s. as $t\rightarrow \infty$. 
Is this legit, concerning that we have generally $\theta$ a positive ($\theta>0$) finite random variable?
For small $m$ we have
$$
P(U_{K_t}<x\sqrt{K_t})\leq P\left(K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))\right)+P\left(U_{a_t}<x\sqrt{(1+\epsilon)a(t)}+m\cdot \sqrt{\epsilon a(t))}\right)+ P\left(\sup_{s\in ((1-\epsilon)a(t),(1+\epsilon)a(t))}|U_{s}-U_{a(t)}|>m\cdot \sqrt{\epsilon a(t))}\right)
$$
The first term converges to 0 due to (3). The second term converges to $\Phi(x+m)$ (Why?) by the central limit theorem (1).
Due to the martingale maximale inequality the third term is bounded by 
$$\frac{1}{(m\cdot \sqrt{\epsilon a(t))})^{2}}$$ and tends to zero as $a(t)\rightarrow \infty$.
Why should this proof (2)? So far we have only that the distribution $P(U_{K_{t}}/\sqrt{K_t}\leq x)$ is bounded by $\Phi(x+m)$ then. A lower bound converging to $\Phi(x)$ is necessary i guess?
Idea:
$$
P(U_{K_t}<x\sqrt{K_t})\geq P(U_{K_t}<x\sqrt{K_t},|U_{a(t)}-U_{K_t}|<m\sqrt{\epsilon a(t)},K_{t}\in ((1-\epsilon)a(t),(1+\epsilon)a(t))\\
\geq P(U_{a(t)}<x\sqrt{(1-\epsilon)a(t)}-m\sqrt{\epsilon a(t)})
\\ \rightarrow \Phi(x-m)
$$
And we have it sandwiched, converging to $\Phi(x)$ right?
Btw: How to come to this inequality is due to
$$
P(U_{k_t}<x \sqrt{K_t})\\
\leq P[U_{k_t}<x \sqrt{K_t},K_{t}\in((1-\epsilon) a(t),(1+\epsilon) a(t))]+P[U_{K_t}<x\sqrt{K_t},K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))] \\ \leq P[K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))]+ P[U_{k_t}<x \sqrt{K_t},K_{t}\in((1-\epsilon) a(t),(1+\epsilon) a(t))] \\
\leq P[K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))]  \\
+P(U_{K_{t}}<x\sqrt{(1+ \epsilon)a(t)},|U_{K_t}-U_{a(t)}|\leq m\sqrt{\epsilon a(t)},|U_{K_t}-U_{a(t)}|> m\sqrt{\epsilon a(t)}]
\\ \leq P[U_{a(t)}<x\sqrt{(1+\epsilon)a(t)}+m \sqrt{\epsilon a(t)}] + P\left(\sup_{s\in ((1-\epsilon)a(t),(1+\epsilon)a(t))}|U_{s}-U_{a(t)}|>m\cdot \sqrt{\epsilon a(t))}\right)+P\left(K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))\right)
$$

REPLY [2 votes]: For square integrable Levy processes this should be easy because you can control the difference between $U_{K_t}$ and $U_{a(t)} $ with the martingale maximal inequality. Subtract out the mean and so suppose one has a square integrable independent increments process.  Then e.g. $$P(U_{K_t} < x \sqrt{K_t }) \le I_t+J_t+L_t$$ with $$I_t=P\left(K_t \notin ((1-\epsilon)a(t) , (1+\epsilon a(t))\right)$$ $$J_t= P\left(U_{a(t)} < x\sqrt{(1 + \epsilon) a(t)} + m \sqrt{\epsilon a(t)} \right)$$ and $$L_t= P\left(\sup_{s \in ((1-\epsilon)a(t) , (1+\epsilon a(t))} |U_s - U_{a(t)}| > m\sqrt{\epsilon a(t)}  \right) $$  $m$ should be small.  Use convergence is probability on $I_t$, CLT on $J_t$, and maximal inequality on $L_t$ (which will tell you that the sup can't be worse than $\frac 1 {(m \sqrt{\epsilon a(t)})^2}$).  Don't take every term literally as I am tex-ing on the fly, but that is the idea.
The assumption that $\frac{K_t}{a(t)}$ converges to a constant shows that $I_t$ is small. $J_t$ converges to $\Phi(x+m)$  by the CLT applied to $U_t$ or $U_{a(t)}$ as you prefer, and $L_t$ is  used as mentioned above.  Btw, I call this Anscombe's theorem.<|endoftext|>
TITLE: Profinite groups as absolute Galois groups
QUESTION [18 upvotes]: It is a well-known result that all profinite groups arise as the Galois group of some field extension.

What profinite groups are the absolute Galois group
  $\mathrm{Gal}(\overline{K}|K)$ of some extension $K$ over
  $\mathbb{Q}$?

The answer is simple enough in the finite case:

(Artin-Schreier) Only the trivial one and $C_2$.

This might tempt us to think that absolute Galois groups are not that diverse. But 0-dimensional anabelian geometry shows us differently:

(Neukirch-Uchida) There's as many different absolute Galois groups as non-isomorphic number fields.

The answer might still turn out to be boring, but I haven't seen this discussed anywhere. The closest is Szamuely in his book "Galois Groups and Fundamental Groups", where he seems delighted by the fact that the absolute Galois group of $\mathbb{Q}(\sqrt{p})$ and $\mathbb{Q}(\sqrt{q})$ are non-isomorphic for primes $p\neq q$.

REPLY [4 votes]: In order to study absolute Galois groups of fields one may also use another consequence of the aforementioned Rost-Voevodsky theorem: in fact, for any field $K$, the cohomology algebra $H^*(G_K,\mathbb{F}_p)$ is a quadratic algebra over the field $\mathbb{F}_p$ also for $p$ odd; and this is still true for the algebra $H^*(G_K(p),\mathbb{F}_p)$, where $G_K(p)$ denotes the maximal pro-$p$ quotient of $G_K$, if $K$ contains a primitive $p$-th root of 1 - i.e., $G_K(p)$ is the Galois group of the compositum of all Galois $p$-extensions $L/K$, and it is called the maximal pro-$p$ Galois group of $K$.
Note that the class of such Galois pro-$p$ groups includes all absolute Galois groups which are pro-$p$.
One reduces to pro-$p$ groups as in general they are easier to deal with than profinite groups.
Pro-$p$ groups whose $\mathbb{F}_p$-cohomology is a quadratic algebra - and thus they are «good candidates» for being realized as maximal pro-$p$ Galois groups - are studied by S. Chebolu, I. Efrat and J. Minàc in the paper Quotients of absolute Galois groups which determine the entire Galois cohomology (2012), and in my paper Bloch-Kato pro-$p$ groups and locally powerful groups (2014). Both papers are available also on the arXiv.
Also, recently I. Efrat, E. Matzri, J. Minàc and N.D. Tan proved that the $\mathbb{F}_p$-cohomology algebra of maximal pro-$p$ Galois groups of fields (containing a primitive $p$-th root of 1) satisfies another property, called the «vanishing of triple Massey products»: in the papers Triple Massey products and Galois theory and Triple Massey products vanish over all fields Minàc and Tan produce some examples of pro-$p$ groups which are not realizable as maximal pro-$p$ Galois groups (and thus as absolute Galois groups).
All the papers I mentioned are available in the arXiv: if you wish I may provide the links.<|endoftext|>
TITLE: Is the intersection of all p-adic fields equal to Q?
QUESTION [8 upvotes]: Fix an algebraic closure $\overline{\mathbb Q}$ of the field $\mathbb Q$ of rational numbers. For a prime $p$ let $K_p$ the field of all algebraic elements in ${\mathbb Q}_p$.

Question: Is $K_p$ normal over $\mathbb Q$?

If so, it defines a unique subfield of $\overline{\mathbb Q}$ which we might want to write as ${\mathbb Q}_p\cap \overline{\mathbb Q}$. Then the second question is:

Is $\bigcap_p{\mathbb Q}_p\cap \overline{\mathbb Q}=\mathbb Q?$

REPLY [13 votes]: No: the field $\mathbb{Q}_5$ has a cube root of $2$, but does not contain a square root of $-3$. (It's easy to form such examples for all primes.)
An alternate version of the question is as follows: Given an algebraic number $\alpha$ such that there is an inclusion $\mathbb{Q}(\alpha) \rightarrow \mathbb{Q}_p$ for all $p$, is $\alpha$ necessarily rational? The answer is yes. Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial with $\alpha$ as a root, and assume the degree of $f(x)$ is $> 1$. Let $G$ be the Galois group of the splitting field $F$ of $f(x)$. The group $G$ acts transitively on the roots of $f(x)$. By a theorem of Jordan, there exists an element $\sigma \in G$ which has no fixed points. If $p$ is a prime such that the corresponding Frobenius element of $p$ in $F$ is (the conjugacy class of) $\sigma$, then there will be no inclusion $\mathbb{Q}(\alpha) \rightarrow \mathbb{Q}_p$. 

A replacement concept would be to take the elements in $\overline{\mathbb{Q}}$ which land in $\mathbb{Q}_p$ for any embedding of $\overline{\mathbb{Q}} \rightarrow \overline{\mathbb{Q}}_p$. The corresponding elements $K_p$ do form a field which is Galois over $\mathbb{Q}$.

REPLY [7 votes]: No:  $x^3-2$ factors linear $\times$ quadratic over ${\mathbb Q}_5$.
Yes: any algebraic extension of ${\mathbb Q}$ is ramified at some prime.<|endoftext|>
TITLE: Local differential geometry and invariant theory
QUESTION [6 upvotes]: Can someone please give me pointers to the literature for local differential differential geometry according to invariant theory in the following sense, provided such a literature exists?
Start with the observation that any given diffeomorphism $\rho:M\rightarrow M$ of a Riemannian manifold $M$ allows for pulling the metric on $M$ back along the map $\rho$.  
Now consider just those Riemannian metrics on a neighborhoods of $0$ in ${\Bbb R}^n$ given by analytic functions $m$ on ${\Bbb R}^n$. $m$ takes values, of course, in real, symmetric, positive-definite $n\times n$ matrices. Assume also (without serious loss of generality) that $m(0)$ equals the identity matrix. 
Locally, one can get a new (still analytic) metric by pulling such an $m$ back along any analytic injection $\iota$ from a neighborhood of $0$ into ${\Bbb R}^n$ that fixes $0$.  If, furthermore, $\iota$ has its Jacobian at $0$ sitting in the orthogonal group, the new metric at $0$ will still get represented by the identity matrix. 
Viewed only up to order $j+1$, these analytic injections $\iota$ will form a finite-dimensional Lie group.  This group acts on metrics $m$ viewed up to order $j$ (as one takes derivatives to form the Jacobians needed to pull back metrics).  
Now metrics themselves don't form a linear space (taking differences can kill positive definiteness).  But (germs of) functions $m - m(0)$ do comprise a linear space, and when viewed only up to order $j$, a finite-dimensional linear space.
So if I said everything right, I have a finite-dimensional Lie group acting on a finite dimensional linear space, and invariant theory should have something to say about classifying the orbits.  Metrics in a common orbit will then share all local geometric invariants up to order $k$.

REPLY [12 votes]: This question was answered decisively by Weyl and Cartan.  The essential point is that there is a canonical way to reduce this nonlinear action to the linear action of $\mathrm{O}(n)$ on a finite dimensional vector space:  Use normal coordinates.
The point is that a metric $g$ can be written in $p$-centered coordinates $x=(x^i)$ in the form
$$
g = g_{ij}(x)\,\mathrm{d}x^i\,\mathrm{d}x^j
$$
where $g_{ij}(x)$ satisfies the system of $n$ (affine) linear equations
$$
g_{ij}(x)x^j = x^i.
$$
Such coordinates are normal coordinates for $g$ about $p$ and are unique up to a rotation of the form $\bar x^i = A^i_j\,x^j$ where $A = (A^i_j)$ is orthogonal.  The $k$-jet of $g$ in these coordinates is determined by the $k$-jets of the $g_{ij}(x)$ satisfying the above relation. (Of course, the $0$-jet is fixed, since that is $g_{ij}(0)=\delta_{ij}$.)  The group $\mathrm{O}(n)$ acts on the rest of the polynomial solutions of the above equation of degree $k$ or less in the obvious linear fashion.  
Now one can apply standard invariant theory to that sum of representations, as Weyl does, to write down a generating set of all of the $\mathrm{O}(n)$-invariant polynomials on this vector space.  What Weyl does is show that these invariants are all generated by taking the Riemann curvature tensor and its first $k{-}2$ covariant derivatives with respect to the Levi-Civita connection at $p$, and then  forming all possible contractions on the direct sum using the metric tensor.  Consequently, all of the pointwise invariants can be expressed in terms of the Riemann curvature tensor and its covariant derivatives.<|endoftext|>
TITLE: Godel's proof of Completeness
QUESTION [6 upvotes]: Where could I find a detailed exposition in English of Godel's proof (not Henkin's) of Completeness Theorem for first order logic? The wikipedia article omits certain details that I am not clear about, and Godel's original dissertation is not in English.

REPLY [11 votes]: In addition to an English translation of the original version contained in the Collected Works referred to by Andrés, there is an English translation of a rewritten version of the paper contained in From Frege to Gödel edited by Jean van Heijenoort. The latter is also contained in the Collected Works.<|endoftext|>
TITLE: Is there a rank for higher degree homogeneous forms analogous to that of quadratic forms?
QUESTION [8 upvotes]: Given a quadratic form $Q(x_1, ..., x_n)$, there is a natural notion of rank defined by looking at the rank of the unique symmetric matrix associated to the quadratic form, i.e. we consider the symmetric matrix $A$ such that
$\mathbf{x}^T A \mathbf{x} = Q(\mathbf{x})$. 
Suppose we have a $F(\mathbf{x})$, a degree $d$ homogeneous form. I naively thought that perhaps the rank of $F$ should be defined to be the rank of a hypermatrix $A$ attached to $F(\mathbf{x})$ (Assuming such notions makes sense). But I have never seen anything like that in the literature. 
So perhaps there are issues if one wants to define a rank of forms this way. 
I was wondering if someone could explain me why considering this may not make sense or not considered? Thank you very much!

REPLY [8 votes]: When $F$ is a degree $d$ homogeneous form on $n$ variables in a finite field ${\bf F}_q$, in the regime when $n$ is large and the order of the field and the degree is small, there are two related notions of rank that are useful.  One is an "algebraic" notion of rank - $F$ has rank at most $r$ if $F(x)$ can be expressed as $F(x) = f( Q_1(x),\dots,Q_r(x))$ for some forms $Q_1,\dots,Q_r$ of degree strictly less than $d$, and some function $f$.  (For comparison, note that a rank $r$ quadratic form $F(x)$ can be written in terms of $r$ linear functions of $x$.)  See for instance this paper of Ben Green and myself where this notion is used to control the equidistribution of $F$.
A related notion is the "analytic" notion of rank introduced by Gowers and Wolf - a form $F$ has analytic rank $r$ if the exponential sum $\frac{1}{q^n} \sum_{x \in {\bf F}_q^n} \chi( F(x) )$ has magnitude $q^{-r/2}$, where $\chi$ is some fixed character on ${\bf F}_q$.  For quadratic forms in odd characteristic this matches the usual notion of rank by the familiar properties of Gauss sums.<|endoftext|>
TITLE: Counting problems where unlabeled is easier than labeled
QUESTION [19 upvotes]: I was encouraged to post this question by Jim Propp during a meeting of the Cambridge Combinatorics and Coffee Club. It is a counterpoint to the MathOverflow question "Counting Problems where Labeled is Known but Unlabeled is Not" which asks for examples where it is known how to count labeled objects but not known how to count unlabeled objects. 
I think the general expectation in enumerative combinatorics is that it should be easier to count labeled objects as opposed to unlabeled objects. For instance, we have much nicer formulas for the number of labeled trees, labeled graphs, labeled connected graphs on $n$ vertices than for the corresponding unlabeled objects. (Here labeled means the vertices are labeled.)
Nevertheless I am asking for counterexamples to this general trend. That is, I am looking for examples of counting problems where the unlabeled objects have a nicer formula than the labeled objects. I know of two such examples.

Semiorders. A semiorder, also known as a unit interval order, is a poset that avoids $2+2$ and $3+1$ as induced subposets. The number of semiorders on $n$ unlabeled elements is the $n$th Catalan number $C_n := \frac{1}{n+1}\binom{2n}{n}$. The number of labeled semiorders is sequence A006531 in the OEIS. Labeled semiorders on $n$ elements have an exponential generating function of $C(1-e^{-x})$ where $C(x) = \frac{1-\sqrt{1-4x}}{2x}$ is the ordinary generating function for the Catalan numbers.
Threshold graphs. A threshold graph is a graph $G=(V,E)$ for which there is some threshold function $\omega\colon V\to\mathbb{R}$ on the vertices such that $\{i,j\}\in E$ iff $\omega(i)+\omega(j) > 0$. The number of threshold graphs on $n$ unlabeled vertices is $2^{n-1}$ (because there is a simple recursive construction of these graphs). The number of labeled threshold graphs is sequence A005840 in the OEIS. Labeled threshold graphs on $n$ vertices have an exponential generating function $\frac{e^x(1-x)}{(2-e^x)}$.

Does anyone know other examples like these? Interestingly, for both semiorders and threshold graphs we have an associated hyperplane arrangement; see Stanley's notes.

REPLY [3 votes]: A generalization of Per Alexandersson's comment is to take the isomorphism class of any graph (or digraph) $G$.  There is exactly one such unlabelled graph, so it is obviously harder to count the labelled objects.  
For Per Alexandersson's comment, $G$ is a directed path with $n$ vertices, in which case there are $n!$ labelled versions of $G$.  If $G$ is a cycle with $n$ vertices, there are $\frac{(n-1)!}{2}$ labelled versions of $G$.  If $G$ is $K_{n,n}$, there are $\frac{1}{2}\binom{2n}{n}$ labelled versions of $G$. These examples are quite structured, making it fairly easy to count the labelled versions, but for a random graph $G$ on $n$ vertices, it will likely be difficult to count the number of labeled versions of $G$.<|endoftext|>
TITLE: Is ordinal arithmetic more complicated than classical arithmetic?
QUESTION [12 upvotes]: Consider the first-order language $\mathcal{L}_{\text{OA}}:=(+,\cdot,0,1)$; in this language, we can formulate statements of ordinal arithmetic. Clearly, the theory $T_{\text{OA}}$ of $(\text{On},+,\cdot,0,1)$ is not recursive, as $\omega$ is definable in $\mathcal{L}_{\text{OA}}$ (as being the only ordinal $\gamma$ different from $0$, not having a predecessor and having no $\alpha$ and $\beta$ different from $0$ such that $\alpha+\beta=\gamma$ with $\alpha$ not having a predecessor) and hence the theory TA (true arithmetic) of $\mathbb{N}$ is computable from $T_{\text{OA}}$. Note that $T_{\text{OA}}$ is absolute between transitive class models of ZFC.
My question is whether the reverse reduction holds: Can we decide $T_{\text{OA}}$, given TA? More precisely, what is the Turing degree of $T_{\text{OA}}$?

REPLY [11 votes]: Using what are now called Ehrenfeucht–Fraïssé Games and extensions thereof, Ehrenfeucht showed that
$$(\mathrm{Ord},{<}) \sim (\omega^\omega,{<})$$
$$(\mathrm{Ord},{<},{+}) \sim (\omega^{\omega^\omega},{<},{+})$$
$$(\mathrm{Ord},{<},{+},{\cdot}) \sim (\omega^{\omega^{\omega^\omega}},{<},{+},{\cdot})$$
where $\sim$ denotes elementary equivalence. Since $(\omega^{\omega^{\omega^\omega}},{<},{+},{\cdot})$ is a computable structure, its first-order theory is computable from $0^{(\omega)}$ ($\equiv_T TA$).
The question omits the relation $<$ but that doesn't matter for the last two results since $x < y$ is definable by $\exists z(x + 1 + z = y)$. [Thanks to Andrés Caicedo for pointing out the correct definition.]
A. Ehrenfeucht, An application of games to the completeness problem for formalized theories, Fund. Math. 49 (1960-1961), 129–141.<|endoftext|>
TITLE: Sum of skew characters over hooks and "odd" partitions
QUESTION [5 upvotes]: Let us call a partition odd if all its parts are odd, and let $Odd(n)$ be the set of all odd partitions of $n$, e.g. $Odd(6)=\{(5\,1),(3\, 3),(3\,1^3),(1^6)\}$. 
Let $H(n)$ denote the set of all hook partitions of $n$. I have made the following surprising observation (to me, at least): For every $\mu\in H(m)$,
$$\frac{1}{2^{\ell(\nu)}}\sum_{\lambda\in H(n+m)} \chi_{\lambda\backslash\mu}(\nu)=\begin{cases} 0, \text{ if }\nu\notin Odd(n)\\
1, \text{ if }\nu\in Odd(n),\end{cases}$$ where $\chi_{\lambda\backslash\mu}$ are skew characters of the symmetric group.
I think a proof would come from the combinatorial description of skew characters in terms of tableaux (although I didn't really develop this all the way through). My question is: is this result already in the literature somewhere? 
EDIT: Upon a moment of reflection it becomes clear that my sum is indeed independent of $\mu$. So we could use any $\mu$ to compute it, even the trivial one $\mu=(0)\vdash0=m$.
However, user61318 mentioned a paper in his comment which proves precisely that
$$ \frac{1}{2^{\ell(\nu)-1}}\sum_{\lambda\in H(n)} \chi_{\lambda}(\nu)=\begin{cases} 0, \text{ if }\nu\notin Odd(n)\\
1, \text{ if }\nu\in Odd(n),\end{cases}$$ 
which differs from the $m=0$ case of my sum by a factor of $2$. I have checked that both formulas are right as they stand, so this difference is a puzzle.

REPLY [5 votes]: I know you were asking for a reference, and there may be better approaches, but just to offer one proof of your statement based on the Murnaghan–Nakayama formula. Assume $m>0$ then any skew tableaux $\lambda/\mu$ has shape $(a,1^b)$ with $a+b=n$. This means $a$ boxes in the first row and $b$ boxes in the first column but the row and column are disconnected. We want to evaluate the sum
$$
\Lambda_n = \sum_{a=0}^n \chi_{(a,1^{n-a})}
$$
If $1 \leqslant k \leqslant n$ then the skew partition $(a,1^b)$ has at most two $k$-hooks, one of leg length 0 if $k \leqslant a$ and one of leg length $k-1$ if $k \leqslant b$. Now assume $\nu\vdash n$ is a partition with part equal to $k$ and let $\hat{\nu} \vdash n-k$ be the partition obtained by removing this part. By the Murnaghan–Nakayama formula for skew partitions we have
\begin{align*}
\Lambda_n(\nu) &= \sum_{a=0}^{k-1}\chi_{(a,1^{n-a})}(\nu) + \sum_{a=k}^{n-k}\chi_{(a,1^{n-a})}(\nu) + \sum_{a=n-k+1}^n\chi_{(a,1^{n-a})}(\nu)\\
&= \sum_{a=0}^{k-1}(-1)^{k-1}\chi_{(a,1^{n-a-k})}(\hat{\nu}) + \sum_{a=k}^{n-k}(\chi_{(a-k,1^{n-a})}(\hat{\nu}) + (-1)^{k-1}\chi_{(a,1^{n-a-k})}(\hat{\nu})) + \sum_{a=n-k+1}^n\chi_{(a-k,1^{n-a})}(\hat{\nu})\\
&= \sum_{a=0}^{n-k}\chi_{(a,1^{n-k-a})}(\hat{\nu}) + (-1)^{k-1}\sum_{a=0}^{n-k}\chi_{(a,1^{n-k-a})}(\hat{\nu})\\
&= (1+(-1)^{k-1})\Lambda_{n-k}(\hat{\nu}).
\end{align*}
Hence if $k$ is even then $\Lambda_n(\nu) = 0$ and an easy induction shows that $\Lambda_n(\nu) = 2^{\ell(\nu)}$ if $\nu \in \mathrm{Odd}(n)$.
Maybe I would just add that the same argument essentially works when $m=0$. For this case let $\tilde{\chi}_{(a,1^{n-a})}$ be the irreducible character labelled by a hook partition and let $\tilde{\Lambda}_n = \sum_{a=1}^n \tilde{\chi}_{(a,1^{n-a})}$. The calculation above goes through in this case except when $\nu = (n)$ in which case we have
\begin{equation*}
\tilde{\Lambda}_n(\nu) = \sum_{a=1}^n \chi_{(a,1^{n-a})}(\nu) = \sum_{a=1}^n (-1)^{n-a}
\end{equation*}
which is 0 if $n$ is even and 1 if $n$ is odd. This explains why one gets $\tilde{\Lambda}_n(\nu) = 2^{\ell(\nu)-1}$ when $\nu \in \mathrm{Odd}(n)$.
EDIT: It should also be possible to show that $\Lambda_n = 2\tilde{\Lambda}_n$ directly using the fact that $\chi_{\lambda/\mu} = \sum_{\tau \vdash n} c^{\lambda}_{\mu\tau}\tilde{\chi}_{\tau}$, where $c^{\lambda}_{\mu\tau}$ is the Littlewood–Richardson coefficient.<|endoftext|>
TITLE: Majority coloring for directed graphs
QUESTION [9 upvotes]: I came up with the following coloring concept when studying neural networks (which are often modelled using directed graphs). No idea whether there is already an established name for it.
If $X$ is a non-empty set, we say that $M\subseteq X$ is a majority if $|M| > |X\setminus M|$.
Let $G=(V,E)$ be a finite directed graph. For $v\in V$ we set $\text{In}(v)=\{x \in V: (x,v) \in E\}$.
Let $n$ be a positive integer. We say that a map $c:V(G) \to \{1,\ldots,n\}$ is a majority coloring if the following condition is satisfied:

For every $v\in V(G)$ with $\text{In}(v) \neq \emptyset$, if for some $k \in \{1,\ldots, n\}$ we have that $c^{-1}(\{k\}) \cap \text{In}(v)$ is a majority of $\text{In}(v)$, then $c(v) \neq k$.

We set the majority coloring number $\chi_m(G)$ to be the least positive integer $j$ such that there is a majority coloring $c:V(G) \to \{1,\ldots,j\}$.
Many directed graphs I've looked at have majority coloring number $2$, but for instance $K_3$ with the orientation $1\to 2\to 3\to 1$ has majority coloring number $3$. 
Questions: 

For $n\in\mathbb{N}$ is there a directed graph $G$ such that $\chi_m(G) = n$?
For $n\in\mathbb{N}$ is there even a tournament $T$ such taht $\chi_m(T) = n$?

REPLY [5 votes]: The paper mentioned above has been published:
Stephan Kreutzer, Sang-il Oum, Paul Seymour, Dominic van der Zypen, David R. Wood. 
"Majority Colourings of Digraphs"
Electronic Journal of Combinatorics, 24(2):#P2.25, 2017. 
http://www.combinatorics.org/ojs/index.php/eljc/article/view/v24i2p25
Other papers have appeared on this topic: 
Marcin Anholcer, Bartłomiej Bosek, Jarosław Grytczuk. 
"Every digraph is majority 4-choosable"
http://arxiv.org/abs/1608.06912
Fiachra Knox, Robert Šámal. 
"Linear Bound for Majority Colourings of Digraphs"
https://arxiv.org/abs/1701.05715
A. Girao, T. Kittipassorn, and K. Popielarz. 
"Generalised majority colourings of digraphs"
https://arxiv.org/abs/1701.03780<|endoftext|>
TITLE: The coefficient of a specific monomial in the expansion of the following polynomial
QUESTION [6 upvotes]: Let $a_{n,k}$ be the coefficient of $$X_1^{\frac{k(n-1)}{2}}X_2^{\frac{k(n-1)}{2}}\cdots X_n^{\frac{k(n-1)}{2}}$$ in the expansion of the real polynomial $$\left(\prod\limits_{1\leq i<j\leq n}(X_j-X_i)\right)^k,$$where $n,k$ are positive integers such that $n>1$ and $k(n-1)\equiv0 \pmod 2$.
Since $$\prod\limits_{1\leq i<j\leq n}(X_j-X_i)=\sum\limits_{\sigma\in S_n}sgn(\sigma)\prod\limits_{i=1}^nX_{\sigma_i}^{i-1},$$
here the sum is computed over all permutations $\sigma$ of the set $\{1,2,\cdots,n\}$, we can get the following simple results:
$(1)a_{n,1}=0$ for every odd positive integer $n>1$;
$(2)a_{n,2}\neq 0$ for every positive integer $n>1$.
I want to ask whether $a_{n,k}$ is equal to 0 or not when $k>2$.

REPLY [10 votes]: This is not an answer but a conjecture based on some trials.
$$a_{n,k} = \begin{cases} 1, & n=1 \\
                          0, & n\gt 1, k\text{ odd} \\
                          (-1)^{t\binom n2}\displaystyle\frac{(tn)!}{(t!)^n}, & k=2t.
            \end{cases}
$$
Note that $\frac{(tn)!}{(t!)^n}$ is the number of ways of partitioning a set of size $tn$ into $n$ sets of size $t$.
Oh, now I see that this is a special case of Macdonald's constant-term conjecture, see this article, which has references.<|endoftext|>
TITLE: When is the generalized Cantor space $\kappa$-compact?
QUESTION [5 upvotes]: My M.Sc. student has the following question, that I assume has an answer in the literature, and we are looking for references.
The generalized Cantor space is the space $2^\kappa$, with basic open sets
$$
[\sigma] := \{f\in 2^\kappa : \sigma\subseteq f\},
$$
for $\sigma\in 2^{<\kappa}$.
A space is $\kappa$-compact if every open cover has a subcover of cardinality (strictly) smaller than $\kappa$.
For which cardinals $\kappa$ is the generalized Cantor space $2^\kappa$ $\kappa$-compact? We are interested in strong limit cardinals $\kappa$, mainly strongly inaccessible $\kappa$.
We would especially appreciate references.
Update: A related question.

REPLY [10 votes]: A cardinal $\kappa$ is weakly compact if and only if $2^{\kappa}$ is $\kappa$-compact and $\kappa$ is strongly compact if and only if $2^{I}$ is $\kappa$-compact for all sets $I$ where $2^{I}$ is given the topology with basis of open sets of the form $[\sigma]$ where $\sigma:J\rightarrow 2$ and $|J|<\kappa$.
To prove the tree property from the Tychonoff theorem characterization of weak compactness, one uses the following argument. Suppose that $T$ is a $\kappa$-tree. Then for each $\alpha<\kappa$, let $T_{\alpha}$ be the $\alpha$-th level of this tree. Then by a compactness argument, one can show that the inverse limit $\varprojlim_{\alpha<\kappa}T_{\alpha}\subseteq\prod_{\alpha<\kappa}T_{\alpha}$ is a non-empty closed subset (the proof of this fact is the same in the ordinary compactness case). However, $\varprojlim_{\alpha<\kappa}T_{\alpha}$ is the set of all $\kappa$-branches of $T$. Therefore $\kappa$ satisfies the tree property.
To prove that every weakly compact cardinal satisfies the $\kappa$-Tychonoff theorem, one could use the tree property and inaccessibility to show that every $\kappa$-filter on a $\kappa$-algebra of sets can be extended to a $\kappa$-ultrafilter and from this fact one could show that weakly compact cardinals satisfy a $\kappa$-version of Alexander's subbase theorem. From Alexander's subbase theorem, one then establishes that weakly compact cardinals satisfy the $\kappa$-Tychonoff theorem.
This result was originally proven in the paper Additions to some Results of Erdos and Tarski by Donald Monk and Dana Scott.
Furthermore, a long list of characterizations of weak compactness including several topological characterizations is given in the book The Theory of Ultrafilters (1974).<|endoftext|>
TITLE: What are parabolic bundles good for?
QUESTION [18 upvotes]: The question speaks for itself, but here is more details: Vector bundles are easy to motivate for students; they come up because one is trying to do "linear algebra on spaces". How does one motivate parabolic bundles (i.e., vector bundles with flags at finitely many points)?  Said differently, how do parabolic bundles arise in nature?

REPLY [18 votes]: Parabolic bundles were introduced in the 70's by Mehta and Seshadri in the set
up of a Riemann surface with cusps. They were trying to generalize the
Narasimhan-Seshadri correspondence on a compact Riemann surface (between
polystable bundles of degree $0$ and unitary representations of the
fundamental group). In the non-compact case, they were able to determine the
missing piece of data - partial flags and weights at each cusp. They
established what is now called the Mehta-Seshadri correspondence. Then they
proceeded to study the moduli space.
Mehta, V. B.; Seshadri, C. S.
Moduli of vector bundles on curves with parabolic structures.
Math. Ann. 248 (1980), no. 3, 205–239.
https://link.springer.com/article/10.1007/BF01420526
Since then, the definition of a parabolic bundle has been clarified (tensor
product with the initial definition is not really computable for instance) and
generalized. This is a long story starting with C.Simpson, I.Biswas, and many
authors. The upshot is that given a scheme $X$, a Cartier divisor $D$, and an
integer $r$, there is a one to one tensor (and Fourier-like) equivalence
between parabolic vector bundles on $(X,D)$ with weights in
$\frac{1}{r}\mathbb Z$ and standard vector bundles on a certain orbifold
$\sqrt[r]{D/X}$, the stack of $r$-th roots of $D$ on $X$. So one can turn your
question in: why are these orbifolds natural ? They were first introduced by
A.Vistoli in relation with Gromov-Witten theory. They also turned out to be
related to the section conjecture (rational points of stack of roots are
Grothendieck's packets in his anabelian letter to Faltings). So parabolic
sheaves - and stack of roots - are ubiquitous. They are also very strongly
related to logarithmic geometry.<|endoftext|>
TITLE: What is an example of an orbifold which is not a topological manifold?
QUESTION [22 upvotes]: In Thurston's book The Geometry and Topology of Three-Manifolds it is proven that the underlying space of a two-dimensional orbifold is always a topological surface. 
Are there any easy examples of higher dimensional orbifolds whose underlying spaces are not topological manifolds?

REPLY [3 votes]: To give a compact version of Igor's example, consider the antipodal map in the tangent space of a round 3-sphere at a point.  This extends to an isometry of the 3-sphere with a pair of fixed points.  The quotient by the isometry is a manifold with two conical singularities which is not homeomorphic to a topological 3-manifold.<|endoftext|>
TITLE: A spectral sequence for computing cohomology of a space from that of its strata
QUESTION [19 upvotes]: Let $X$ be a smooth complex variety (not necessarily compact) and let $D$ be a normal crossings divisors with components $D_1$, $D_2$, ..., $D_N$. For a set of indices $I$, let $D_I = \bigcap_{i \in I} D_i$ and let $D^{\circ}_I = D_I \setminus \bigcup_{J \supsetneq I} D_J$. I would like to compute $H^{\ast}(X)$ in the case that I know all of the $H^{\ast}(D_I^{\circ})$.
It seems to me that there should be a spectral sequence whose first page is $\bigoplus_{\#(I)=p} H^{q-p}(D^{\circ}_I)$. If $I' \supset I$ with $\#(I') = \#(I)+1=p+1$, then the map $H^{q-p}(D^{\circ}_{I}) \to H^{q-(p+1)}(D^{\circ}_{I'})$ would given by the Gysin map (up to standard sign issues).
Does this sequence have a name, or is it a special case of something which has a name? Where can I read about it?

REPLY [17 votes]: Let $X = T_n \supset T_{n-1} \supset \cdots \supset T_{-1} = \varnothing$ be a topological space filtered by closed subspaces, where for simplicity I assumed the filtration bounded. Then there is a spectral sequence
$$ E_1^{pq} = H^{p+q}_c(T_p \setminus T_{p-1}) \implies H^{p+q}_c(X).$$
(Some mild point-set assumptions are required for this and what follows to be true, but let me disregard that.) As you noted in a comment, this gives the spectral sequence you want by Poincaré duality.
Usually, the spectral sequence of a filtered space is written in terms of ordinary cohomology rather than cohomology with compact support. One reason is that the compactly suppported spectral sequence is a special case of the usual one. Recall that if $U \subset \overline U$ is any compactification of a space $U$, then $H^\bullet_c(U) = H^\bullet(\overline U, \partial U)$, where $\partial U = \overline U \setminus U$. So if we choose an arbitrary compactification $X \subset \overline X$ and let $\overline T_p$ be the closure of $T_p$ in $\overline X$, then we get a filtration 
$$\overline X = \overline T_n \cup \partial X \supset \overline T_{n-1} \cup \partial X \supset \overline T_{n-2} \cup \partial X \supset \cdots \supset \overline T_{-1} \cup \partial X = \partial X.$$
Now the usual spectral sequence of a filtration  reads
$$ E_1^{pq} = H^{p+q}(\overline T_p \cup \partial X,\overline T_{p-1}\cup \partial X) = H^{p+q}(\overline T_p,\overline T_{p-1})  \implies H^{p+q}(\overline X, \partial X),$$
and this is the spectral sequence we wanted. 
(Often one chooses $\overline X$ and $\overline T_p$ to be the one-point compactification of $X$ resp. $T_p$, and then the above is just the spectral sequence for the reduced cohomology a filtered based space. But allowing yourself to use any compactification is useful in the algebraic setting, which you are interested in.)
Here's an alternative "sheafy" derivation of the spectral sequence. The constant sheaf $\mathbf Z_X$ is filtered:
$$\mathbf Z_X = \mathbf Z_{T_n} \supset \mathbf Z_{T_{n-1}} \supset \cdots \supset \mathbf Z_{T_{-1}} = 0$$ where I denote by $\mathbf Z_A$ the pushforward of the constant sheaf on the subspace $A$. The successive quotients in the associated graded for this filtration are of the form ${j_n}_!\mathbf Z$, where $j_n$ is the locally closed inclusion of $T_n \setminus T_{n-1}$. Taking compactly supported cohomology of this filtered object gives rise to a spectral sequence which is exactly the one we want. A useful reference for making sense of this is the section about filtered objects and spectral sequences in Lurie's "Stable $\infty$-categories" (Chapter 1 of Higher Algebra). 
To answer a question you made in the comments, the spectral sequence of a filtered space is indeed compatible with mixed Hodge structure, when $X$ is an algebraic variety filtered by closed subvarieties. So the one you consider (by Poincaré duality) is compatible up to Tate twist. I don't know what is the canonical reference for this fact, but Sections 3 and 4 of Arapura's "The Leray spectral sequence is motivic" proves rather generally that the spectral sequence of a filtered algebraic variety is compatible with all kinds of extra "motivic" structure.<|endoftext|>
TITLE: Is the set of subsequences of branches through a tree Borel?
QUESTION [5 upvotes]: Let $T$ be pruned subtree of $\omega^{<\omega}$. For my cases of interest, we may assume that $T$ is infinitely branching at every node, and consists of increasing sequences.
Let $A=\{x\in\omega^{\omega}:\exists y\in[T](x \text{ is a subsequence of } y)\}$, where $[T]$ denotes the (closed) set of infinite branches through $T$. Is $A$ necessarily a Borel set?
Clearly, $A$ is analytic. It might be useful to note the following: Given an infinite increasing $x\in\omega^\omega$, let $T'(x)=\{s\in T: x\upharpoonright m(s) \text{ is a subsequence of } s\}$, where $m(s)$ is the least integer $m$ such that $x(m)$ is greater than the last (greatest) entry of $s$. Then, $T'(x)$ is a tree and $x\in A$ if and only if $T'(x)$ is ill-founded. This doesn't improve the complexity, but might suggest how to show that $A$ is proper analytic (for certain choices of $T$).

REPLY [6 votes]: With your assumption that the tree consists of increasing
sequences only, then the answer is yes, this is Borel. The reason
is that we can identify whether or not $x$ is a subsequence of a
branch through the tree simply by checking the arithmetic-in-$T$
condition that every finite initial segment of $x$ is a
subsequence of a node on the tree. We don't need any real
quantifiers. To see this, suppose that for every $n$, we have that
$x\upharpoonright n$ is a subsequence of some node of the tree
$t_n\in T$. We may assume that the final node of $x\upharpoonright
n$ is the last element of $t_n$. Let $T'$ be the tree of initial
segments of the $t_n$'s, which is a subtree of $T$. Note that
because the sequences are all increasing, it follows that $T'$ is
finitely branching, because no initial segment of any $t_n$ can be extended in $T'$ so as to 
skip over the next higher element of $x$. So we have essentially reduced to
the case where the tree is finitely branching. In particular, it
follows that $[T']$ is sequentially compact. For each $n$, pick
$y_n\in[T']$ with $x\upharpoonright n\subset y_n$. Because $T'$ is
sequentially compact, there is a convergent subsequence, and so we
may find a single $y\in [T']$ with unboundedly many
$x\upharpoonright n$ contained in $y$. Thus, $x$ is a subsequence
of $y$, which is a branch through $T'$ and hence through $T$, as
desired. So the set is arithmetically definable from the tree and
hence Borel.
If you don't insist on increasing sequences, then I believe the
set of subsequences can be complete analytic, and I can explain
this if you want.<|endoftext|>
TITLE: Optimization of a function of a positive definite matrix and its inverse
QUESTION [5 upvotes]: This question is a little ill-posed, but I've been playing with some equations and am just wondering if this resembles any known problems that have been solved.
Suppose I have two real, positive definite (square) matrices $\mathbf{A}$ and $\mathbf{C}$, and I wish to find another real, positive definite matrix $\mathbf{B}$ such that
$\mathbf{A B} + \mathbf{B}^{-1}\mathbf{C}$ is as close as possible to identity.
I'll entertain any reasonable definition of "close" that makes the problem tractable.  Maybe minimization of $\| \mathbf{A B} + \mathbf{B}^{-1}\mathbf{C} - \mathbb{I} \|$ for some choice of norm.  
Does anyone have any insight or experience with such a problem?
Thanks,
Nick

REPLY [2 votes]: You are actually looking to solve the continuous algebraic Riccati equation. For convenience, I will write your $B$ as $X=X^T$. Then the equation you're trying to solve is simply
$$ X - XAX + (-C) = 0$$
Or even more explicitly, writing the Cholesky factorization of $A=BB^T$
$$ \left( \frac{1}{2}I \right)^TX + X\left( \frac{1}{2}I \right) - XBB^TX + (-C) = 0$$
The solution of an algebraic Riccati equation using Hamiltonian matrices is a standard topic in control theory. But I would instead refer you to the "care" command in MATLAB.<|endoftext|>
TITLE: non-orientability of vector bundles induced from a symmetric group action
QUESTION [6 upvotes]: Let $\Sigma_k$ be the symmetric group on $k$-letters. Let $M$ be a manifold with a free $\Sigma_k$-action. Then we can form a $k$-dimensional vector bundle
$$
\xi:\mathbb{R}^k\longrightarrow M\times_{\Sigma_k}\mathbb{R}^k\longrightarrow M/\Sigma_k.
$$
We notice that the transition functions of $\xi$ are given in the form
$$
(a_{i,j}=\delta_{j,\sigma(i)})_{k\times k}
$$
where $\sigma\in \Sigma_k$ and $\delta_{j,\sigma(i)}=1$ if $j=\sigma(i)$ and $0$ otherwise. These matrices have determinants both $1$ and $-1$. 
Question. For any $k\geq 2$, can we conclude that $\xi$ is always non-orientable? 

Thanks for the answer given by Will Sawin! In his solution, I do not understand the following part:
Suppose $M$ is connected and the covering map from $M$ to $M/\Sigma_k$ induces a  surjective homomorphism 
\begin{eqnarray*}
h: \pi_1(M/\Sigma_k)\longrightarrow \Sigma_k. 
\end{eqnarray*}
(I obtained this surjective homomorphism by Prop. 1.40 (c), Algebraic Topology, A. Hatcher). 
Let $r: \Sigma_k\longrightarrow O(k)$ be the regular representation of $\Sigma_k$ given by permuting the coordinates of $\mathbb{R}^k$.  
Question: Does a $k$-dimensional vector bundle over $M/\Sigma_k$ with structure group $\Sigma_k$ is uniquely determined by a map from $\pi_1(M/\Sigma_k)$ to $\Sigma_k$?  Why the bundle $\xi $ comes from the map 
 $r\circ h$?

REPLY [2 votes]: Yes if $M$ is connected, no otherwise.
If $M$ is connected, the covering gives a surjective homomorphism $\pi_1(M/\Sigma_k)\to \Sigma_k$. The vector bundle comes from a representation of $\pi_1$ defined by composing this with the permutation representation of $\Sigma_k$. Taking determinants, the determinant line bundle is the same homomorphism composed with the sign representation of $\Sigma_k$. Because the homomorphism is surjective and the sign representation is nontrivial, the determinant line bundle is nontrivial. (Here I am using the fact that the map from homomorphisms $\pi_1(X) \to \pm 1$ to isomorphism classes of line bundles is a bijection.)
If $M$ is disconnected then the vector bundle could be trivial. For instance, $M$ might itself equal the symmetric group.<|endoftext|>
TITLE: What was a "cusp" to Hurwitz in 1892?
QUESTION [20 upvotes]: Let $d\in\mathbb{N}$
be squarefree.
Let $\mathcal{O}_d$
be the ring of integers of $\mathbb{Q}(\sqrt{-d})$.
Let $\Gamma_d=\mathrm{PSL}_2(\mathcal{O}_d)$.
Let $\mathcal{H}^3$
be the upper half-space model for hyperbolic 3-space.
Let $X_d=\mathcal{H}^3/\Gamma_d$
be the orbifold obtained by the usual action of $\Gamma_d$
on $\mathcal{H}^3$
(do Mobius transformations on the boundary, and extend them upward isometrically).
$X_d$ is called a Bianchi orbifold.
Bianchi was an Italian mathematician who died in 1928.
An orbifold is like a manifold but with cone points,
and was defined by Thurston in the 1970's.
First question: to what extent did Bianchi understand the geometry of $X_d$,
and what were his main tools in studying these objects?
Theorem: The number of cusps of $X_d$
is equal to the class number of $\mathbb{Q}(\sqrt{-d})$.
In the literature, this is attributed to Hurwitz in 1892.
I have that reference but it is in German (which would take me a long time to translate).
Class numbers have been around since Gauss, but my understanding of a cusp (like orbifolds) is from a rather contemporary topological standpoint. We could define a cusp of $X_d$ as a conjugacy class of maximal parabolic subgroups of $\Gamma_d$ but when you say it like that it doesn't sound nearly as interesting. What I love about this theorem is that it's pure number theory on one end and pure topology on the other. So, second question: what was a "cusp" as far as Hurwitz was concerned, and why would he have found them interesting?

REPLY [20 votes]: It is very difficult to find a paper of Hurwitz dealing in any way with the geometry or topology of $\mathbb{H}^3/PSL_2(\mathcal{O}_{\mathbb{Q}(\sqrt{-D})})$. (And I am not sure I like what this implies about our ways of keeping knowledge alive, attribution and such.) The closest two things I could find were the following:


A paper of Bianchi on computations of fundamental domains for $PSL_2(\mathcal{O}_{\mathbb{Q}(\sqrt{-D})})$: 

L. Bianchi. Geometrische Darstellung der Gruppen linearer Substitutionen mit ganzen complexen Coefficienten nebst Anwendungen auf die Zahlentheorie. Math. Ann. 38 (1891), 313-333.
The main purpose of the investigation seems to have been reduction theory (and classification) of quadratic forms. The description of the fundamental domain is based on explicit generators for the group. The discussion of the topology is based on Poincaré's hyperbolic space model, the explicit transformation formulas for the group action, and the metric of hyperbolic space. 
On page 2 of Bianchi's paper there is a reference to 
A. Hurwitz: Über die Entwicklung complexer Grössen in Kettenbrüche. Acta Math. 11 (1888), 187-200. 
That paper is mostly about continued fractions and discusses what could possibly be interpreted as related to the fundamental domains for $PSL_2(\mathcal{O}_{\mathbb{Q}(\sqrt{-D})})$ for $D=1$ and $D=3$. It seems, however, to be a rather indirect link of Hurwitz to cusps. 


Next, there is a 1892 paper of Bianchi: 

L. Bianchi. Sui gruppi di sostituzioni lineari con coefficienti appartenenti a corpi quadratici immaginarî.  Math. Ann. 40 (1892), 332-412.
In this, he states the relevant theorem: "Il numero dei vertici singolari eguaglia il numero delle classi degli ideali nel corpo quadratico corrispondente." Singular vertices for him are orbits of boundary points (or boundary points in each fundamental domain), and it seems to me that in §§2 and 3 he essentially proves this result by computing the orbits of $PSL_2(\mathcal{O}_{\mathbb{Q}(\sqrt{-D})})$ on the boundary projective line (as already suggested in various comments).  
There is a reference to Hurwitz's work here as well: 
A. Hurwitz. Grundlagen einer independenten Theorie der Modulfunctionen. Math. Ann 18 (1881), pp. 528-592. 
For some reason this paper does not seem to be in MathSciNet. Anyway, Bianchi says
"Ma senza riferirci al teorema generale di Poincaré daremo qui
una dimostrazione diretta di questa proprietà affatto analoga a quella
che il Sig. Hurwitz ha fatto conoscere pel gruppo modulare." 
Meaning that Bianchi's computation of the fundamental domain does not make use of Poincaré's general theorem, but is proved directly analogous to Hurwitz's arguments for the modular group. Hurwitz's treatment of fundamental domains for the modular group and its congruence subgroups  on the upper half plane (in the abovementioned paper) seems to consider only the upper half plane, and not discuss boundary points.

Conclusion: At least from these two findings, it seems that Hurwitz was not directly involved in the proof that cusps for the Bianchi groups are in bijection with ideal classes. Maybe additional information could be inferred from Hurwitz's mathematical diaries which are available from the ETH library. 
It is not clear to me if Bianchi really did consider the topology of the orbifold. He considered the group $PSL_2(\mathcal{O}_{\mathbb{Q}(\sqrt{-D})})$ (or $PGL_2$) as generated by an explicit set of transformations, and he viewed the group action on $\mathbb{H}^3$ in terms of explicit formulas. From the papers above it seems he did not think about an orbit space, but just about the construction of some fundamental polyhedron which contained representatives for all orbits. From this point of view, of course, cusps and the quotient topology are not an issue at all since everything takes place in $\mathbb{H}^3$ or $\overline{\mathbb{H}^3}$. (The same applies to Hurwitz's treatment of the action of the modular group on the upper half plane.)
In any case, in view of the above references it seems to me more appropriate to credit Bianchi (in 1892) with the identification of cusps and ideal classes.<|endoftext|>
TITLE: Bounded Arithmetic vs Complexity Theory
QUESTION [10 upvotes]: In this post, when I talk about bounded arithmetic theories, 
I mean the theories of arithmetic according to "Logical Foundations of Proof Complexity", which capture the complexity classes between $AC^0$ 
and $PH$, and the theories capturing $PSPACE$ and $EXPTIME$ (see Sam 
Buss's PhD thesis (Bounded Arithmetic, Chapter 9) or Alan Skelley's PhD thesis, Theories and proof systems for $PSPACE$ and the $EXP\text{-}Time$ Hierarchy).  
According to some belief, it ought to be easier to separate bounded arithmetic theories than complexity classes. Is there any evidence which kind of supports this belief?

REPLY [7 votes]: If $T_1$ and $T_2$ are theories corresponding to complexity classes $C_1$ and $C_2$ (resp.), then separation of $C_1$ from $C_2$ from $C_2$ implies separation of $T_1$ from $T_2$, but not necessarily vice versa. (This is already mentioned in T. Chow’s answer.) In fact, with details somewhat dependent on the pair of theories, generally separation of $T_1$ from $T_2$ tends to be equivalent to separation of $C_1$ from $C_2$ in some model of the weaker of the two theories, as opposed to separation of $C_1$ from $C_2$ in the standard model $\mathbb N$. Thus, in principle, separation of the theories is a weaker statement than separation of the corresponding complexity classes.
In practice, separation of theories appears essentially as hard as separation of complexity classes.
Most of the known unconditional separation results for theories of bounded arithmetic actually go the opposite way, i.e., they are based on separations of complexity classes: the two typical cases are separations of relativized theories, such as $S^i_2(\alpha)\ne S^{i+1}_2(\alpha)$ (which follow from separation of complexity classes with oracles), and separations of theories corresponding to very small complexity classes, such as $V^0\ne V^0[p]$ (which follows from $\mathrm{AC}^0\ne\mathrm{AC}^0[p]$).
The exceptions are separations of various “pathologically weak” theories. Primarily, this concerns variants of $\Sigma^b_0$ induction, about which you can read more in https://mathoverflow.net/a/228102, including a list of references. Some of these results also apply to “very short” $\Sigma^b_1$ induction (such as $\Sigma^b_1$-LLLIND); for this see [6], and the paper S. Boughattas, J.-P. Ressayre: Bootstrapping, part I, Annals of Pure and Applied Logic 161 (2010), no. 4, pp. 511–533. Now, what makes these unconditional separations work is precisely the fact that these theories do not correspond to reasonable complexity classes, so this does not provide any evidence for the thesis in the question.<|endoftext|>
TITLE: Atiyah-Singer index theorem, pairing between K-homology and K-theory and Chern character
QUESTION [12 upvotes]: There is a general (abstract) index theorem in noncommutative geometry: you take a 
K-theory class and K-homology class (which is represented by a triple $(A,H,F)$) and 
you pair them together. This pairing is computed as an index of certain operator. 
There is a notion of (noncommutaive analog of) Chern character in this context 
which takes values in cyclic cohomology and cyclic homology. Therefore you can
apply this Chern character to both: K-theory class and K-homology class obtaining
two classes, in cyclic cohomology and homology. In this context you have a natural 
pairing between cohomology and homology. The remarkable result is that this equal to 
the previous pairing between K-theory and K-homology (I would like to omit all
technical details involving precise definitions: detailed discussion can be found in
the book "Basic Noncommutative Geometry" by Masoud Khalkhali). My question is:


Can one deduce the 'usual' Atiyah Singer theorem from this abstract index theorem?


If so, how to proceed? For example, one problem is that in the data defining K-homology
cycle the operator $F$ is bounded which is not the case for order $>0$ differential 
operators.

REPLY [3 votes]: No.
What you describe is purely analytic (the definitions of the groups, of the pairings, and of the Chern-Connes characters), but the Atiyah-Singer index theorem has also a topological part.
By the way, I wouldn't call what you describe an "index theorem" (because, as I said, you are missing completely the topological part). What you have is just the compatibility of the Chern-Connes characters with the pairings.<|endoftext|>
TITLE: Can a smooth diffeomorphism of a Riemannian manifold have only positive Lyapunov exponents on a large set?
QUESTION [8 upvotes]: Let $M$ be a compact Riemannian manifold, $f: M \to M$ a diffeomorphism, and $\mu$ an ergodic measure for $M$. Suppose that the support of $\mu$ is not a finite set. Is it possible that all the Lyapunov exponents of $\mu$ will be positive?
This is a version of a previous question with one added condition, to remove a specific example.
The answer is no if $\mu$ is absolutely continuous with respect to Lebesgue measure. Then $\mu$ has a density function $d$, and there is some interval $I$, not containing $0$ or $\infty$, such that $\mu(d^{-1}(I))>0$. Hence for $x$ with $d(x)<I$, there are infinitely many $n$ such that $f^n(x)\in I$. By the change of variables formula $\left|\det \left(\frac{d f^n}{dx}\right)\right| (x) d(f^n(x))=d(x)$, there is a bounded interval that $\left|\det \left(\frac{d f^n}{dx}\right)\right|$ stays in infintely often and hence the sum of the Lyapunov exponents is $0$ so one is nonpositive.
The same thing is true if $\mu$ is merely absolutely continuous with respect to Lebesgue measure on a submanifold.
In fact, there is a much weaker condition that is sufficient to rule this out. Define $r_x(\delta)$ be the infimum of $r$ such that a ball of radius $r$ around $x$ has measure at most $\delta.$ Suppose there is a function $s(\delta)$ such that for $x$ in a set of positive measure, $r_x(\delta)$ is bounded above and below by a constant times $s(\delta)$. Then one of the Lyapunov exponents is nonpositive - we can take universal constants $C_1,C_2$ such that $c_1 s(\delta) < r_x(\delta)< C_2 s(\delta)$ on a set of positive measure. Then as long as $x$ and $f^n(x)$ are both in that set, the lowest singular value of $\frac{d f^n}{dx}(x)$ is at most $C_2/C_1$ and so one of the Lyapunov exponents is nonpositive.
So any counterexample must at the very least have quite nontrivial local structure.

REPLY [6 votes]: The answer is no. In fact, the following result holds: if $M$ is a compact manifold, $f\in\mathrm{Diff}^{1+\alpha}(M)$ and $\mu$ is an ergodic $f$-invariant probability measure such that all its Lyapunov exponents are positive, then $\mu$ is a periodic measure, i.e. there exists a periodic source $\{p,f(p),\ldots,f^{n-1}(p)\}$ such that $\mu=1/n\sum_{i=0}^{n-1} \delta_{f^i(p)}$. 
A reference for this result (stated for negative Lyapunov exponents and a periodic sink) is the book of Katok and Hasselblatt, "Introduction to the Modern Theory of Dynamical Systems", Corollary S.5.2, page 694.<|endoftext|>
TITLE: Non-isomorphic graphs with isomorphic edge vectors
QUESTION [8 upvotes]: Let $G$ and $H$ be graphs on the vertex set $\{1, \ldots, n\}$ and let $(e_i)$ be the standard basis of $\mathbb{R}^n$. For each edge $\{i,j\}$ define edge vectors $e_i - e_j$ and $e_j - e_i$ in $\mathbb{R}^n$.

Question 1: If there is a linear isomorphism of $\mathbb{R}^n$ with itself that takes the edge vectors of $G$ bijectively onto the edge vectors of $H$, must $G$ and $H$ be isomorphic?

The answer to this question is no: if $G$ and $H$ are both trees then there is such a linear isomorphism. Aside from $e_i - e_j$ being the negation of $e_j - e_i$, there are no linear dependences among the edge vectors, and the edge vectors of $G$ can be mapped to the edge vectors of $H$ in any manner.

Question 2: Same as Question 1, but now assuming that $G$ and $H$ both have central vertices, i.e., each of them has a vertex which is adjacent to every other vertex.

I assumed a counterexample to Question 1 would easily yield a counterexample to Question 2, but I don't see this. A counterexample to Question 2 is what I need.

REPLY [2 votes]: Attempt to show that there is no example for Question 2. 
Let $G$ be a graph with central vertex $v_0$, $H$ be a graph with central vertex $u_0$ and cycle structures (cyclic matroids) of $G$ and $H$ are isomorphic. I claim that $H$ and $G$ themselves are isomorphic as graphs. Let $T$ be a spanning tree in $G$ formed by edges incident to $v_0$. It corresponds to some spanning tree $f(T)$ in $H$, here $f$ is an isomorphism of matroids (so, $f$ is defined on edges of $G$). Note that in $G$ any edge $e\notin T$ belongs to a triangle with two edges from $T$. Thus the same holds in $H$. Apply this to edges in $H$ incident to $u_0$ but not coming from $u_0$. We see that maximal path in $f(T)$ going from $u_0$ consists at most two edges. Let $u_0u_1,\dots,u_0u_k$ be edges incident to $u_0$ and belonging to $f(T)$, $v_0v_i$ be their $f$-preimages. Next, if $u_i$, $1\leqslant i\leqslant k$, is incident to some edge $u_iu_m\in f(T)$, $m>k$, then denote by $v_0v_m$ $f$-preimage of the edge $u_0u_m\in H$. Then $v_i\rightarrow u_i,i=0,1,\dots$ is isomorphism of $G$ and $H$.<|endoftext|>
TITLE: Atiyah-Singer theorem-a big picture
QUESTION [61 upvotes]: So far I made several attempts to really learn Atiyah-Singer theorem. In order 
to really understand this result a rather broad background is required: you need
to know analysis (pseudodifferential operators), algebra (Clifford algebras, 
spin groups) and algebraic topology (characteristic classes, K-theory, Chern 
character). So every time I have stuck in some place and have too many doubts to 
continue my journey into the realm of index theory. Finally I came to conclusion that what I'm lacking is some big picture about various approaches to index theorem: there are many (excellent) books that treat the subject but before you arrive at the proof of index theorem itself you have to go through about 150 pages of background material (which usually you partially know). 
Some issues which I already figured
out:

There are at least two approaches: the one which uses K-theory and the second
so called heat kernel approach. As I understood for the first one good knowledge of algebraic topology is required while for the second you have to know differential  geometry stuff (connections, curvatures etc.)  
As far as I know these approaches have different range of applicability: the
heat kernel approach is more restrictive and do not include all differential elliptic operators.    
There are some other proofs (using K homology - see the book of Higson and Roe, via tangent groupoid, there is also quite recent proof due to Paul Baum)

So what answer(s) I would be very pleased to see? 


I would like to gather in one place 
    various possible approaches to the proof of index theorem together with discussion of
    the main ingredients needed to complete such proof (for example: "you need good
    understanding of K-theory not only for compact spaces, but also for locally compact. 
    You also have to understand Bott periodicity and Thom isomorphism theorem), the 
    formulation of the index formula together with comparison and relation to other 
    index formula (for example: you obtain topological index as a K-theory class- 
    in order to obtain the famous cohomological formula (involving Todd class) you apply 
    Chern character) and finally the level of generality of formulation of index theorem
    (for example: this approach works for all elliptic differential operators).   


Maybe this question is broad but I have an impression that I'm not the only person who
will find such a discussion useful.

REPLY [16 votes]: Just for completeness let me add a few words on the heat kernel proof. It is true that some analysis is needed. But one should not forget that even the $K$-theoretic proof depends on the notion of pseudo-differential operators and therefore cannot avoid analysis completely.
It is also true that the proof really only works well for Dirac operators. 
But then, Dirac operators generate $K$-homology, and most applications of the index theorem are formulated in terms of (0-order perturbations of) Dirac operators. Contradicting comments are welcome! The most prominent examples are mentioned in Coudy's answer, but one should add the signature operator (which is the Euler operator with respect to a different grading), and the so-called untwisted Dirac operator.
The only real drawback is that one only sees the image of the index in rational cohomology. But then, using family index techniques or $\eta$-invariants, one can sometimes recover a bit more of the topological content of the index.
The proof consists of the following main ideas. 

The McKean-Singer trick. This still works for all elliptic differential operators. If $P\colon\Gamma(E)\to\Gamma(F)$ is an elliptic differential operator on a manifold $M$, consider its adjoint $P^*\colon\Gamma(F)\to\Gamma(E)$. Then it makes sense to consider
$$\operatorname{ind}(P)=\operatorname{tr}\bigl(e^{-tP^*P}\bigr)-
\operatorname{tr}\bigl(e^{-tPP^*}\bigr)\;.$$
In the limit $t\to 0$, the traces above become local, that is, they can be represented by an integral over $M$ that depends only on the coefficients of $P$ in local coordinates. 

From here on, one could try to prove that only finitely many geometric terms can contribute to the index. This was the idea of Gilkey and Patodi. One difficulty comes from the fact that the relevant contribution to the index sits in the $t^0$-term of the asymptotic expansion, which is by far not the leading term. However,
in the special case of Dirac operators, one gets around this problem.

Getzler rescaling. If $P=D$ is a Dirac operator, then $\Delta=D^*D\oplus DD^*$ is a Laplace operator, and $e^{-t\Delta}$ is the standard heat kernel. One performs a standard parabolic rescaling of space and time together with Getzler's rescaling of the Clifford algebra to show that as $t\to 0$, $\Delta$ converges against a model operator that looks like a harmonic oscillator with coefficients in $\Lambda^{\mathrm{even}}T^*M$. The relevant contribution to the index now appears in the leading order term thanks to Getzler's trick. This only works if $D$ is compatible with a Riemannian metric (here one makes use of the homotopy invariance of the index to achieve this), and one has to chose nice local coordinates.
Mehler's formula. The heat kernel for the model operator has an easy explicit formula in terms of the Riemannian curvature and the coefficients of the Dirac operator.
Chern-Weil theory. The output of step 3 is an integral over $M$ of an invariant polynomial, applied to the Riemannian curvature and the so-called twisting curvature of the Dirac bundle on which $D$ acts. Hence it can be interpreted as the evaluation on the fundamental cycle of $M$ of an expression in characteristic classes.

Indeed, as pointed out in the comments to Paul's answer, one does not need any advanced analysis. Some notions from a first course in Riemannian geometry are needed. The most tricky part seems to be Getzler's rescaling, which is mostly algebraic.<|endoftext|>
TITLE: Restriction of the Picard group of a surface to a curve
QUESTION [10 upvotes]: In a paper by Griffiths and Harris on the Noether-Lefschetz theorem, they use the following fact which they don't comment as if it is obvious: 
For a general (smooth) surface $S$ in $\mathbb{P}^3$ over the complex numbers which is not ruled, and for a general curve $C$ on this surface which is proportional to some multiple of a plane section $H$, the restriction map $$\mathrm{Pic}\ S \rightarrow \mathrm{Pic}\ C$$ is injective. 
I can only see the following: we always can assume that the surface has irregularity $0$, and we can check injectivity only for divisors $L$ such that $L.H = 0$, and hence $L^2 < 0$ by Hodge index theorem. But I cannot see how to use that the surface is non-ruled. Thanks in advance for any suggestions.

REPLY [14 votes]: Edit. I edited the argument below to make it work in all characteristics.  By SGA $7_{II}$ Exposé XVII, this requires working with a sufficiently general pencil of divisors in $\mathcal{O}_{\mathbb{P}^3}(e)|_S$ for $e\geq 2$ (as before, $e=1$ suffices if $d\geq 3$ and the characteristic is sufficiently large).
These kinds of arguments used to be "common knowledge", but times change, we use different methods now, etc.  The statement is true for all smooth surfaces $S$ in $\mathbb{P}^3$ and every "very general" divisor $C\subset S$ in the linear system $\mathcal{O}_{\mathbb{P}^3}(e)|_S$ such that either $e\geq 2$ (all characteristics) or $e=1$ and $d\geq 3$ (sufficiently positive characteristics).  In this case, there exists a Lefschetz pencil $\{C_p\}$ of divisors in the linear system $\mathcal{O}_{\mathbb{P}^3}(e)|_S$.  A pencil of divisors is a Lefschetz pencil if (a) for all but finitely many $p$, $C_p$ is smooth, and (b) the finitely many singular $C_p$ are (geometrically) irreducible and reduced with a single ordinary double point.  The existence of a Lefschetz pencil follows by SGA $7_{II}$, Exposé XVII by Nicholas Katz.
Let $\nu:\widetilde{S} \to S$ be the blowing up of the base locus of this pencil of divisors. Let $\pi:\widetilde{S}\to \Pi$ be the projection of $\widetilde{S}$ to $\Pi\cong \mathbb{P}^1$ such that the fiber of $\pi$ over $p\in \Pi$ equals the strict transform of the divisor $C_p \subset S$ in the pencil.  
Denote by $\text{Pic}^\perp(S)$ the subgroup of $\text{Pic}(S)$ of classes $[\mathcal{L}]$ of invertible sheaves $\mathcal{L}$ on $S$ such that $(c_1(\mathcal{L}).c_1(\mathcal{O}_{\mathbb{P}^3}(1)|_S))_S$ equals $0$.
For every such $\mathcal{L}$ on $S$, denote by $F(\mathcal{L})$ the associated coherent sheaf on $\Pi$, $$F(\mathcal{L}) = R^1\pi_*(\omega_\pi\otimes_{\mathcal{O}_S} \nu^*\mathcal{L}^\vee).$$  Denote by $\eta$ the generic point of $\Pi$, and denote by $C_\eta \subset \widetilde{S}$ the generic fiber of $\pi$.  Then, by cohomology and base change, $F(\mathcal{L})$ is a torsion sheaf unless $\nu^*\mathcal{L}|_{C_\eta}$ is isomorphic to $\mathcal{O}_{C_\eta}$, in which case $F(\mathcal{L})$ is a rank $1$ sheaf (a priori, not necessarily pure). 
First consider the case that $F(\mathcal{L})$ is a rank $1$ sheaf. By cohomology and base change, there exists a dense open subscheme $U\subset \Pi$ such that $\nu^*\mathcal{L}|_{\pi^{-1}(U)}$ is isomorphic to the structure sheaf on $\pi^{-1}(U)$.  Thus, there exists a rational section of $\nu^*\mathcal{L}$ whose support is contained in finitely many fibers of $\pi$.  Since every fiber of $\pi$ is integral, it follows that $\nu^*\mathcal{L}$ is isomorphic to $\pi^*\mathcal{M}$ for some invertible sheaf $\mathcal{M}$ on $\Pi$.  Since $\Pi$ is isomorphic to $\mathbb{P}^1$, $\nu^*\mathcal{L}\cong \pi^*\mathcal{O}(m)$ for some integer $m$.  On the dense open subscheme $V\subset \widetilde{S}$ on which $\nu$ is an isomorphism, $\pi^*\mathcal{O}(1)$ is isomorphic to $\nu^* \mathcal{O}_{\mathbb{P}^3}(e)|_S$.  Thus, by S2 extension, $\mathcal{L}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^3}(me)|_S$.  Since $[\mathcal{L}]$ is in $\text{Pic}^\perp(S)$, $me$ equals $0$.  Thus $\mathcal{L}$ is isomorphic to $\mathcal{O}_S$.  
Thus, for every nonzero element $[\mathcal{L}]$ in $\text{Pic}^\perp(S)$, the coherent sheaf $F(\mathcal{L})$ is torsion.  Of course $\text{Pic}^\perp(S)\setminus\{0\}$ is a countable set by the theorem of Néron-Severi-Lang, i.e., the "Theorem of the Base", and the computation that $h^1(S,\mathcal{O}_S)$ is zero. Thus also the subset of $\Pi$, $$\Pi^\perp = \bigcup\{ \text{Support}(F(\mathcal{L})) : [\mathcal{L}] \in \text{Pic}^\perp(S)\setminus\{0\} \},  $$ is a countable set of closed points of $\Pi$.  Assuming that the ground field is uncountable (or using the geometric generic point), for a "very general" point $p$ of $\Pi$, for every $[\mathcal{L}]$ in $\text{Pic}^\perp_(S)\setminus\{0\}$, $\mathcal{L}|_{C_p}$ is not isomorphic to $\mathcal{O}_{C_p}$.
Edit. It may be instructive to consider what goes wrong if $d$ equals $2$ and $e$ equals $1$. Then every pencil of hyperplane sections has two reducible fibers, $C_0 = A_0 \cup B_0$ and $C_\infty = A_\infty \cup B_\infty$.  So in the argument above, we conclude that if $F(\mathcal{L})$ has rank $1$, then $\mathcal{L}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^3}(m)|_S(a_0\underline{A}_0 + b_0\underline{B}_0+a_\infty\underline{A}_\infty+b_\infty\underline{B}_\infty)$ for integers $a_0$, $b_0$, $a_\infty$ and $b_\infty$.  We can rewrite with $b_0=a_\infty=0$.  Since $A_0\cup B_\infty$ is a hyperplane section, we can also rewrite with $b_\infty = 0$.  Finally, since $\mathcal{L}$ is in $\text{Pic}^\perp(S)$, $a_0$ equals $-2m$.  Thus, the kernel of the restriction map is generated by the class of $\mathcal{O}_{\mathbb{P}^3}(1)|_S(-2\underline{A}_0) \cong \mathcal{O}_S(\underline{B}_0-\underline{A}_0)$.<|endoftext|>
TITLE: Concrete Applications of knowing $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$
QUESTION [21 upvotes]: I have very little experience with Galois representations, mostly as they relate to class field theory, elliptic curves, and modular forms, but they seem to have quite a reputation in number theory as one of the most important objects of study, in particular because Tannakian philosophy states that they allow us to recover $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. So my question is rather simple: what are some "basic" things (if any) we can do it we really have a strong understanding of the structure of this group?
The question is meant to be flexible, meaning you can assume as much additional information as you want to consider "understanding" the group to mean (i.e. understanding the decomposition groups at every prime, etc.). However, I'm looking for a collection concrete applications of this sort of thing, in the spirit of number theory. Anything from solving diophantine equations, to statements about the distribution of primes/representations of primes, to reciprocity laws. Things a high school student could understand even if the machinery is rather complicated, or something you could tell another mathematician very far removed from number theory about what motivates this research.
One thing that comes to mind immediately is resolving the inverse Galois problem, but this is something that doesn't seem to yield a concrete application immediately (unless someone has an example, which I'd be interested to see). I'm curious to see if there's any sort of conjectured techniques to solving classical number theory problems that might become easier if we knew more about this group. Of course some speculation is allowed, as the actual (currently unknown) structure of the group might determine which applications are admissible.

REPLY [23 votes]: Of course, Galois theory intervenes as a basic tool in the study of some diophantine equation. But deeper aspects, in the form of Galois representations, were crucial for the proof of at least 4 fantastic theorems in the last century.

The Mordell-Weil theorem concerning rational points of elliptic curves over the field of rational numbers (Mordell, 1922) or an abelian variety over a number field (Weil, 1929). It says that this group of rational points is finitely generated.
Galois representations also play an important role (via the Tate conjecture) in Faltings's proof (1983) of Mordell's conjecture. 
The proof of Fermat's Last Theorem wouldn't have been possible without a deep understanding of properties representations of the group $\mathop{\rm Gal}(\bar{\bf Q}/{\bf Q})$. In fact, the core of the proof by Wiles/Taylor-Wiles (1995) consists in establishing the link between representations associated with elliptic curves and representations associated with modular forms. 
Finally, Mihailescu's proof (2002) of Catalan conjecture, although more elementary than the previous ones, makes great use of cyclotomic fields and their class groups (hence, by class field theory, looks closely at the abelian quotients $\mathop{\rm Gal}(\bar{\bf Q}/{\bf Q})$).<|endoftext|>
TITLE: Are the abelian absolute Galois groups of these local fields isomorphic?
QUESTION [6 upvotes]: For a field $F$ we denote by $F^{\mathrm{ab}}$ the compositum of all finite Galois abelian extensions of $F$.

Is $\mathrm{Gal}(\mathbb{Q}_2(\sqrt[8]{3})^{\mathrm{ab}}/\mathbb{Q}_2(\sqrt[8]{3})) \cong \mathrm{Gal}(\mathbb{Q}_2(\sqrt[8]{48})^{\mathrm{ab}}/\mathbb{Q}_2(\sqrt[8]{48})) ? $

Equivalently, 


Is every finite abelian group which is a Galois group of some finite Galois extension of $\mathbb{Q}_2(\sqrt[8]{3})$, is also a Galois group of some finite Galois extension of $\mathbb{Q}_2(\sqrt[8]{48})$ (and vice versa) ?

REPLY [8 votes]: Although @znt has gotten the answer through pari, I think it may be instructive to outline my argument.
It all depends on the transition function of Higher Ramification Theory: for a finite extension $K\supset k$ of local fields, $\varphi^K_k$ is a function from $\Bbb R^{\ge0}$ to itself, polygonal and concave, in which, if the (finitely many) vertices are $\{(x_i,y_i)\}$, each $x_i$ is a lower number of a ramification break, and $y_i$ is the corresponding upper number. Perhaps the most useful property of the transition function is that it is functorial: if $L\supset K\supset k$, then $\varphi^L_k=\varphi^K_k\circ\varphi^L_K$.
Although you’ll see below that there are other methods of calculating $\varphi^K_k$, here’s a relatively painless way for totally ramified extensions, if you know a prime element $\pi$ of the integers of the big field. Let $F(X)\in k[X]$ be the minimal $k$-polynomial for $\pi$, and draw the Newton copolygon of $F(X+\pi)$.
The copolygon of $g(X)=\sum_na_nx^n$ is the intersection of all the lower halfplanes $\eta\le n\xi+v(a_n)$, where you may take $v$ to be the (additive) valuation for which $v(k^\times)=\Bbb Z$. One sees that the segments of the copolygon are in one-to-one correspondence with the vertices of the Newton polygon: there’s a segment along the line $\eta=n\xi+v(a_n)$ if and only if $(n,v(a_n))$ is a vertex of the polygon.
And you get the transition function from this process by stretching the boundary of the copolygon horizontally by a factor of $e^K_k$, the ramification index (equal to the degree since our extension is presumed totally ramified). That is, every point $(\xi,\eta)$ on the boundary of the copolygon gets moved to the point $(e^K_k\xi,\eta)$ on the graph of the transition function. The function achieved in this way is shifted one unit up and one to the right from that defined in most standard texts, like Corps Locaux of Serre. In all cases, the infinite segment in the graph has slope $1/e^K_k$, and in the case of wildly ramified extensions, all slopes are powers of $1/p$.
Now calculate the transition functions of $\Bbb Q_2(3^{1/8})$, $\Bbb Q_2(\sqrt3\,)$, and $\Bbb Q_2(\sqrt2\,)$, all as extensions of $\Bbb Q_2$.
I won’t go through the details, except to say that $(X+1)^8-3$ and $(X+1)^2-3$ are the Eisenstein polynomials for the first two of the fields above. You find that the transition functions are: $\varphi^{\Bbb Q_2(3^{1/8})}_{\Bbb Q_2}$ has vertices $(2,2)$, $(4,3)$, and $(8,4)$;
$\varphi^{\Bbb Q_2(\sqrt3\,)}_{\Bbb Q_2}$ has the single vertex at $(2,2)$, and $\varphi^{\Bbb Q_2(\sqrt3\,)}_{\Bbb Q_2}$ has the single vertex at $(3,3)$.
Even though the transition function of $\Bbb Q_2(\sqrt2,\sqrt3\,)$ over $\Bbb Q_2$ can be calculated directly knowing that a prime element is $\pi=1-(\sqrt3-1)\big/\sqrt2$, you can also get it from the transition functions of the subfields, using functoriality. The upper numbers ($\eta$-values) of the subfields must appear among the upper numbers of the whole extension. Since these numbers are $2$ and $3$, $\varphi^{\Bbb Q_2(\sqrt2,\sqrt3\,)}_{\Bbb Q_2}$ must have the vertices $(2,2)$ and $(4,3)$.
All the above is straightforward and easy. But I wanted the transition function of the extension $\Bbb Q_2(\sqrt2,3^{1/4})$ over $\Bbb Q_2(\sqrt2,\sqrt3\,)$. For that I needed a prime element, and I found
$$
\beta=\frac{\frac{3^{1/4}-1}{\pi}-1}{\pi^2}-1\,,
$$
for which the minimal polynomial over $\Bbb Q_2(\sqrt2,\sqrt3\,)$ is of form $X^2+u_1\pi X+u_2\pi$, for units $u_i$. The upshot is that the transition function has the single vertex $(2,2)$. Compose this function first, then the transition function $\varphi^{\Bbb Q(\sqrt2,\sqrt3\,)}_{\Bbb Q_2}$, and you see that the transition function for the degree-eight extension $\Bbb Q_2(\sqrt2,3^{1/4})$ over $\Bbb Q_2$ has two vertices $(2,2)$ and $(6,3)$. The slope ratio at the lefthand vertex is $4$ rather than $2$.
And that’s enough to get our conclusion that $\Bbb Q_2(\sqrt2,3^{1/8})$ is totally ramified of degree $16$ over $\Bbb Q_2$, since its transition function must  have a vertex with height $4$, coming from the subfield $\Bbb Q_2(3^{1/8})$. So the full transition function has vertices $(2,2)$, $(6,3)$, and $(14,4)$.<|endoftext|>
TITLE: What are the different ways of defining 3-manifolds?
QUESTION [11 upvotes]: I wonder what are the different ways of defining the 3-manifold. Obviously for average human being it is difficult to imagine the 3-manifold. Therefore the presentation or visualisation of such object is important to understand what we are talking about.
Here are the possible ways which I know:
1) Dehn surgery on framed link in $S^3$.
2) Homeomorphism of the surface generate 3-manifold by gluing solid handlebodies along this homeomorphism - Heegaard splitting.
3) Gluing faces of polyhedron (is it somehow formalized ?).
4) Define real smooth function $f$ on $\mathbb R^4$, then $f^{-1}(p)$  is 3-manifold for not critical value $p$ (see Morse theory). More general try smooth function $\mathbb R^5 \to \mathbb R^2$.
5) Spherical 3-manifolds can be determined by finite subgroup of $SO(4)$.
6) Brieskorn manifolds looks as a nice way to define 3-manifold. Is there any generalization to obtain 3-manifolds by intersecting complex manifold of complex dimension 2 with hyperplane or sphere.
Such methods are preferred which could lead to step-by-step "manufacturing" of any 3-manifold. Methods 1) and 2) above satisfy this criteria. Method 1) is the best since we can visualize the link easily in dimension 3 and even 2 as link diagram with numbers.
Note: I have not mentioned hyperbolic manifolds, since I know very little about it.

REPLY [10 votes]: This question ought to be community-wiki, since there is no correct answer. 
As indicated in the comments, one may represent a 3-manifold as a triangulation (simplicial complex), special spine, branched cover, real algebraic variety,  mechanical linkage,.... In fact, Thurston showed the existence of ``universal links", such that any 3-manifold is a branched cover over $S^3$ with branch locus the link (such as the figure 8 knot). It is known that any closed 3-manifold is a 3-fold branched cover over $S^3$. Also, the Borromean rings with orbifold locus of order 4 along each component were shown to be a universal orbifold. 
There are various combinatorial encodings of triangulations called gems and crystallizations.
Turaev introduced a notion of shadows to describe 3-manifolds. For relations with other presentations of 3-manifolds, see the paper of Costantino and Thurston.
A quite powerful way related to 5) to represent closed 3-manifolds is via the geometrization theorem. One considers orientable 3-manifolds (with torus
boundary) admitting geometric structures (on the interiors), and then
glue together by homemorphisms of tori (or to Klein bottles), and then connect sums. 
 Although it is somewhat trickier to describe geometric 3-manifolds this way, 
the advantage is that one may reproduce 3-manifolds without duplication (if done with care),
unlike the other techniques. 
2) is called a Heegaard splitting. 
For 3), there are notions of polyhedral representations 
of 3-manifolds (even allowing self-gluing of faces). 
For example, Cannon, Floyd and Parry showed that any
3-manifold admits a "bi-twist" description. 
Not every 3-manifold may be obtained by 4). 
6) is an interesting question: links of singularities are
restrictive, but I'm not sure what's known about intersecting
with a general hyperplane or sphere.<|endoftext|>
TITLE: Simplicity of alternating group $A_n$
QUESTION [22 upvotes]: I am teaching an introductory group theory course, and it has come to the inevitable proof that $A_n$ is simple for $n\geq 5$. Now, there seem to be a number of proofs that I can find – one the "standard" one with $3$-cycles, and the others using primitivity or conjugacy class size estimation. Does anyone have a list of proofs of simplicity of $A_n$ with comments? I am sure there are a number of different ideas which could be used... In particular, any method that generalizes to some of the other finite simple groups would be of particular interest...

REPLY [6 votes]: I prefer the proof going via 3-cycles. It is probably the least elegant proof, but it is the one which you probably would have found when considering the problem without any prior knowledge. Also the general strategy generalizes to classical groups as well as Lie groups: Look for elements $g_1$, which cannot be contained in a normal subgroup, then look for elements $g_2$, such that the normal subgroup generated by $g_2$ contains $g_1$, and continue until you run out of elements to check. The disadvantage is that for each new class of groups you have to come up with the right elements, e.g. 3-cycles for $A_n$, transvections for $Sl_n$, but if you choose the element to start with "close to the identity", you will succeed quite often.<|endoftext|>
TITLE: Descent of Higher categorical structures along geometric morphisms
QUESTION [15 upvotes]: Let $f: \mathcal{E} \rightarrow \mathcal{T}$ be a geometric morphism between two (Grothendieck) toposes (or maybe more generally a bounded geometric morphism between elementary toposes).
It is well known that if $f$ is an open or proper surjection then objects (and even locales) descend along $f$.
What I want to known is whether one also have descent for Higher categorical structures in $\mathcal{E}$ ?
To be precise, I am still speaking about ordinary $1$-toposes, but I want to consider for example the $2$-category of groupoid object, category object or toposes inside $\mathcal{E}$. 
I would like to know if for example groupoid objects/category objects in $\mathcal{E}$ satisfies a 2-categorical version of descent.
Or, even more optimistically do Grothendieck toposes in $\mathcal{E}$ satisfies a $2$-categorical version of descent (i.e. does open or proper surjections are effective descent morphism in the $(2,1)$-category of toposes.)
Maybe one need to add some hypothesis on $f$, for example being locally connected instead of just open , or tidy instead of proper ?
any results even for a restricted class of morphism will be of interest (for example, hyperconnected morphisms)

REPLY [5 votes]: I have been able to gather all the elements for the case of $2$-categorical descent along open surjections, so I will write it as an answer.
It is worth noting that the proof below follows the exact same path as the proof of descent for locales and objects along open surjections given in the Elephant.
So what follow is probably optimal and obtaining statement for Higher categorical structures probably involves working with Higher toposes (with additional difficulties related to hypercompletness in the case of infnity categorical structures). Their is also a High chance that everything works exactly the same for proper surjections (and it suffices to proove the first theorem for proper surjections and everything will follow)
Theorem (Moerdijk) : Open surjections are effective descent morphism in the $2$-category of toposes.
Moerdijk's proof is in a paper called "Descent for toposes" in 1989 in Bulletin of the BMS, which is unfortunatly not very easy to find.
It is a very nice proof that roughly goes along the same line as the modern proof that open/proper surjections are of effective descent for locales (the one in the elephant): it follow formally from:
1) the stability properties of open surjections
2) The fact that open surjection are (stably) the coequalizer of their kernel pair.
3) The fact that "equivalence relation" (groupoids) whose legs are open surjection have stable co-equalizer.
With the exception that for (3) he was only able to prove it for groupoids object whose action map $G_1 \rightarrow G_0 \times G_0$ is localic. But as toposes have localic diagonal it is enough for the proof of the descent theorem.
Proposition: Constructively, any topos $\mathcal{T}$ such that $p:\mathcal{T} \rightarrow *$ is open and 
 $\Delta: \mathcal{T} \rightarrow \mathcal{T} \times \mathcal{T}$ is etale is isomorphic to the category of functor from a groupoid $G$ to sets.
Proof: Let $\mathcal{T}$ be such a topos, in particular $\mathcal{T}$ and its diagonal are open, hence $\mathcal{T}$ is atomic. 
As $\Delta$ is etale, there is an object $X$ in $\mathcal{T} \times \mathcal{T}$ such that $\mathcal{T} \simeq (\mathcal{T} \times \mathcal{T}) / X $. $X$ can be covered by objects
 of the form $a \otimes b$ where $a$ and $b$ are atoms of $\mathcal{T}$ (and $a \otimes b$ denotes the object $\pi_1^*(a) \times \pi_2^*(b)$). We fixe such an object $a \otimes b \rightarrow X$.
As $a \otimes b$ is in $\mathcal{T} \times \mathcal{T}/X$ is corresponds to an object $c$ in $\mathcal{T}$. Moreover one has: 
$(\mathcal{T} \times \mathcal{T} )/( a \otimes b) \simeq \mathcal{T}/a \times \mathcal{T}/b \simeq \mathcal{T}/c$ and all those isomorphisms are compatible to the maps to $\mathcal{T} \times \mathcal{T}$.
In particular, the map $\mathcal{T}/c \rightarrow \mathcal{T}/a \times \mathcal{T}/b$ is induced by morphisms from $c$ to $a$ and $b$, hence can be factored as the diagonal map $\mathcal{T}/c \rightarrow \mathcal{T}/c \times \mathcal{T}/c$ followed by an etale map $\mathcal{T}/c \times \mathcal{T}/c \rightarrow \mathcal{T}/a \times \mathcal{T}/b$.
So we have an isomorphism which is factored as $f \circ g$ with $f$ etale, this implies that $g$ is an inclusion (indeed, internally in the target of $f$ it is a map from a point to a set). In our case, the diagonal map of $\mathcal{T}/c$ is an inclusion, by C2.4.14 this implies that $\mathcal{T}/c$ is localic and atomic hence discrete. As $\mathcal{T}$ can be covered by such toposes this impies the result.
From this, one easily deduces:
Corollary: Let $\mathcal{T}$ be a topos, then the $2$-category of groupoid object of $\mathcal{T}$ is equivalent to the full subcategory of the $2$-category of toposes over $\mathcal{T}$ whose map to $\mathcal{T}$ is open with an etale diagonal.
Corollary: Open surjections are of effective descent for the $2$-categories of groupoids objets.
Indeed, topos descend along open surjection, and the condition of the above corollary are detected by open surjection (C5.1.7 of the elephant) hence one has descent for toposes which are open with étale diagonal and those are the same as groupoids.
Proposition: Open surjections are of effective descent for categories.
Roughly, if you have an open surjection $f:\mathcal{E} \rightarrow \mathcal{T}$ and $\mathcal{C}$ a category object in $\mathcal{E}$ endowed with a descent data for $f$, then the descent data restrict to a descent data on the core groupoid of $\mathcal{C}$ hence one can apply descent to it and we obtain a groupoid $G_0$ in $\mathcal{T}$.
Then for any pair of object $x,y \in G$ internally in $\mathcal{T}$ one has a pair of object $f^*x, f^* y$ in $f^* G$ and $f^* G$ is equivalent to the core of $C$, one easily see that the set of morphisms between $f^* x$ and $f^* y$ in $C$ comes with a descent data and the rest is just a boring manipulation of some $2$-categorical diagram.
One can of course gives some more abstract proof of this relying in the end on the fact that the $2$-category of categories can be written explcitely as the category of model in groupoids of some "cartesian $2$-theory" (whatever that precisely means).<|endoftext|>
TITLE: What does "game theory" cover and how should it be called?
QUESTION [15 upvotes]: There seems to be a huge discrepancy in what people refer to when they speak of "game theory".  I tend to think of it as including, among other things:

Combinatorial game theory dealing with certain games of perfect information, i.e., things like the Sprague-Grundy theory of (terminating, perfect information) impartial games, but also partizan games à la Berlekamp, Conway and Guy, and more (e.g., not necessarily terminating games where non-termination is counted as a draw).
Questions of determinacy: Martin's proof of Borel determinacy, for instance, is probably more commonly classified as being part of set theory, but it seems odd to me not to (also) consider it part of game theory.
Games with application to model theory, like semantic games, Ehrenfeucht-Fraïssé games, etc.
Differential game theory.
Various questions of algorithmic computability or complexity linked to some of the above.
...and probably much more that I've never even heard about (quantum game theory?).

On the other hand, if we look at this online course (by Jackson, Leyton-Brown and Shoham) that is simply called "game theory", it is obvious from the syllabus that their focus is much narrower:

Week 1. Introduction: Introduction, overview, uses of game theory, some applications and examples, and formal definitions of: the normal form, payoffs, strategies, pure strategy Nash equilibrium, dominated strategies.
Week 2. Mixed-strategy Nash equilibria: Definitions, examples, real-world evidence.
Week 3. Alternate solution concepts: iterative removal of strictly dominated strategies, minimax strategies and the minimax theorem for zero-sum game, correlated equilibria.
Week 4. Extensive-form games: Perfect information games: trees, players assigned to nodes, payoffs, backward Induction, subgame perfect equilibrium, introduction to imperfect-information games, mixed versus behavioral strategies.
Week 5. Repeated games: Repeated prisoners dilemma, finite and infinite repeated games, limited-average versus future-discounted reward, folk theorems, stochastic games and learning.
Week 6. Coalitional games: Transferable utility cooperative games, Shapley value, Core, applications.
Week 7. Bayesian games: General definitions, ex ante/interim Bayesian Nash equilibrium.

Evidently they don't intend to teach about the various things I mentioned above.
Now this presents the unfortunate problem that I don't know how to refer to this particular sub-branch of game-theory-in-the-wide-sense, or, alternatively, if we reserve the term "game theory" to this game-theory-in-the-narrow-sense, how to rename game-theory-in-the-wide-sense.  I've thought of calling it "Nashian game theory", but I don't want to be caught making up names for branches of mathematics that I'm in no way a specialist of, at least not without some kind of asking about.
Sadly, neither the Wikipedia article (which sort-of favors using "game theory" in the wide sense) nor the AMS classification (which is really unclear as to what goes into 91Axx — do we seriously think that "Games involving topology or set theory" should be part of "Game theory, economics, social and behavioral sciences"?) seem to have an idea as to what game-theory-in-the-narrow-sense might be called.
So, can someone suggest less ambiguous terminology, and possibly a map of the layout of game theory (how the different sub-domains stand in relation to each other)?  I'm also curious to know how wide the intersections are (some descriptions seem to suggest that combinatorial game theory and Nashian(?) game theory are almost a partition, but the picture is probably much more complex).

REPLY [11 votes]: If we include the larger research community -- economics, computer science, social sciences, business schools, operations research, etc -- I think there really is a partition between combinatorial game theory and what I would propose to call "equilibrium" game theory. Most in econ and related fields who study/use game theory extensively have probably never even heard of the combinatorial kind! And even those who have usually don't encounter it in their research. (For instance the course you link is by two computer scientists and an economist.)
And I think this divide illustrates the nature of the partition, namely, "equilibrium" game theory is at home in microeconomics: Its purpose is understanding the behavior of groups of strategic / self-interested agents. I think the following motivation helps illustrate. A choice facing a single agent is simply an optimization problem about maximizing utility and hence, once written down, has a well-defined solution. With multiple agents making decisions, however, one needs to propose a "solution concept" describing or predicting how groups of agents might behave. There is no necessarily right or best answer. The insight developed by von Neumann, Nash, etc was to propose as solution concepts equilibria, where the key point of equilibrium is that all agents are simultaneously optimizing. Essentially all game theory of the second kind, in my experience, follows this motivation and solution approach, hence my proposal for "equilibrium" as the descriptive term.
(By the way, for this reason I would argue that "cooperative game theory" is a misnomer. Although it is also taught in the same economics classes, it has little to do with "game theory proper". It fits better within social choice.)
On the other hand, while I have almost no experience with combinatorial GT and probably shouldn't risk putting my foot in my mouth, my impression is that it is not generally motivated by modeling strategic agents. Instead, it tends to use a "game" as an analogy or mental picture for describing a well-defined mathematical problem in which issues surrounding strategic behavior, and especially the problem of solution concepts, do not tend to play a role. The question, although described as involving multiple agents, is more about understanding the (well-defined and uncontroversial) optimization problem.
To highlight this, in my (limited) experience, even in artificial intelligence where problems related to combinatorial game theory come up (planning, alpha-beta pruning, solving perfect-info zero-sum games like checkers/chess/go), the problem is not really described as falling under game theory (which to that crowd tends to mean equilibrium game theory) but rather simply algorithm design or optimization.
So in summary, I'm hoping to put forward two points. The first is that "equilibrium game theory" may be a good disambiguation name for the second kind of game theory you mention. I think the notion of equilibrium actually quite closely capture both necessity and sufficiency for falling into that category. The second point is that, if you look at the broader research community than in mathematics (not to mention popular culture), a large majority equate "game theory" exclusively to equilibrium game theory (and generally out of ignorance rather than conscious choice). I'm not sure what the point of that point is, but maybe it's useful.<|endoftext|>
TITLE: analog of Lusztig nilpotent scheme
QUESTION [5 upvotes]: Fix a quiver $Q$ without loop. Denote the set of vectices of $Q$ by $I$.
Let $\Lambda_V$ be the Lusztig nilpotent scheme with associated vector space $V$ over $I$. Briefly speaking, when $Q$ is a $ADE$ quiver, $\Lambda_V$ is just the scheme of $\Lambda$-module of dimension $|V|$ where $\Lambda$ is the preprojective algebra of $Q$.
Lusztig's theorem:
Let $\frak{g}$ be the Kac-Moody algebra constructed from $Q$. Let $\frak{n_+}$ be the postive part of $\frak{g}$. Then there is an algebra embedding $$Un_+\to \sum_{V} M(\Lambda_V)^{G_V}$$,where $V$ runs through all isomorphic type vector spaces over $I$.
$M(\Lambda_V)^{G_V}$ denotes the set of constructible function over $\Lambda_V$ which are $G_V$ invariant.
For more details (like the algebra structure on $\sum_V M(\Lambda_V)^{G_V}$), you can have a look on [1] , section 5.2 of [2] and section 12 of [3].
Lusztig constructed nilpotent scheme and considered algebra of construcible function on them so that we get an embedding from $U\frak{n}_+$ to $\sum_VM_V(\Lambda_V)^{G_V}$, a convolution algebra of construcible function.
My question is: Is there an analog of nilpotent scheme such that there is an embedding from $U\frak{g}$ to the convolution algebra of construcible function on that analogous space? In this case, we have a geometric interpretation of the whole universal enveloping algebra but not just its positive part.
EDIT: I am aware of the construction of representation of modified enveloping algebra via Nakajima variety. I am more interested in a construction via Euler characteristic measure (like the construction of $U\frak{n}_+$ by construcutible function the convolution of which involves Euler characteristic measure) .
I am mainly interested in Dynkin case and more specifically type A. Any references on this aspect is appreciated. It would be great to know if people have considered this before or not. Thank you so much!
Reference:
1.Note from Quantum groups, combinatorics and geometry seminar Spring 2011
2.Christof Geiss, Bernard Leclerc, Jan Schröer's paper on semicanonical basis
3.Lusztig's original paper

REPLY [2 votes]: One thing that fits the role you're looking for is the Nakajima quiver varieties. These are discussed later in the seminar you linked to. There's a natural map of $U(\mathfrak g)$ to functions on these spaces (defined by Nakajima), but rather than being injective, the spaces depend on a highest weight and the intersection of the kernels over all weights is trivial.  See also the notes of Ginzburg.
EDIT: Lusztig wrote to me in an email that there is a (partial) answer to that
question in his paper "Constructible functions on varieties attached to
quivers",  in "Studies in memory of I. Schur."  I haven't had a chance to look at that paper myself, but it's another place to look.<|endoftext|>
TITLE: Characteristic classes for odd $K$-theory
QUESTION [14 upvotes]: There are different models of odd $K$-theory. In one case,
one takes the group $U=\lim\limits_{\longrightarrow}U(n)$ as classifying space. Similarly, if $\mathcal U$ denotes the unitary group  of a separable Hilbert space, one can consider the subspace
$$\mathcal U_0=\bigl\{\,g\in\mathcal U\bigm|g+\mathrm{id}\text{ is Fredholm}\,\bigr\}\;.$$
In another model, one considers instead the space $\mathcal F$ of selfadjoint Fredholm operators on a Hilbert space that have an infinite number of positive as well as of negative eigenvalues.
Both spaces are related by a Cayley transformation
$$\mathcal F\owns D\mapsto \frac{D-i}{D+i}\in\mathcal U_0\subset\mathcal U_0\;.$$
Characteristic classes are cohomology classes on these classifying spaces. It is well-known that the cohomology ring is the exterior algebra over $\mathbb Z$ with one generator $\gamma_k$ in each odd degree $2k-1$. There is a de Rham type description for maps $u\colon M\to U$ of these generators of the form
$$\gamma_k(u)=c_k\,\mathrm{tr}\bigl((u^{-1}\,du)^{2k-1}\bigr)$$
with appropriate constants $c_k\in\mathbb C$.
On the other hand, for a map $D\colon X\to\mathcal F$, the characteristic classes of $[D]\in K^1(X)$ should be determined entirely by the way that $\mathrm{spec}(D)\cap(-\varepsilon,\varepsilon)$ (and maybe the corresponding eigenspaces) behave for some small $\varepsilon>0$. 
My question is, how are these descriptions related? In there some book/article that makes characteristic classes of selfadjoint Fredholm operators more explicit and relates them to those of maps to $U$?

REPLY [9 votes]: I know something about it when $X$ is compact.
Since $U$ is the classifying space of $K^1$, for $x\in K^1(X)$, there exists $u\in C(X, U)$, such that $[u]=x$. In fact, we could choose  $N$ large enough, such that $u\in C(X, U(N))$. 
For $(b,t,v)\in X\times [0,1]\times \mathbb{C}^N$, the relation $(b,0,v)\sim (b,1, u(b)v)$ makes a vector bundle $W$ over $X\times S^1$. Let $U=X\times S^1\times \mathbb{C}^N$ be the trivial bundle. Then $[W]-[U]\in K^0(X\times S^1)$ corresponds to $[u]\in K^1(X)$.
Let $\nabla^u=d+t\cdot u^{-1}du$. Then $\nabla^u$ is a connection on $W$ over $X\times S^1$ under the paste above.
$$\mathrm{ch}([u]):=\int_{S^1}\mathrm{ch}(W)=\left[\int_{S^1}\mathrm{Tr}\left[\exp\left(\frac{(\nabla^{u})^2}{2\pi i}\right)\right]\right].$$ 
(There is a sign notation here for the integral on odd dimensional manifold.)
Calculate it carefully, 
$$\mathrm{ch}([u])= \mathrm{Tr}\left[\sum_{n≥0}\frac{(−1)^n} {(2πi)^{n+1}}\frac{n!}{ (2n + 1)!}(u^{−1}du)^{2n+1}\right].$$
This is the familiar form in the paper of Getzler: The odd chern character in cyclic homology and spectral flow
For Fredholm type interpretation, we could only consider the Dirac type case. Let $\pi:Z\rightarrow X$ be a submersion with odd-dimensional compact Spin fiber $Y$. Let $E$ be a complex vector bundle over $Z$. We have the fiberwise Dirac operator $D_Y^E$ and $\mathrm{ind}(D_Y^E)\in K^1(X)$. 
For any $x\in K^1(X)$, there exists such submersion with appropriate geometric structure such that $\mathrm{ind}(D_Y^E)=x$. See Section 4.2.3 in the book 
Index theory, eta forms and Deligne cohomology.
In fact, choose submersion $X\times S^1\rightarrow X$ above, we have $\mathrm{ind}(D_{S^1}^W)=[u]\in K^1(X)$. 
For the Caylay transform, maybe you could find something in Section 5 of the paper of Melrose-Piazza: An index theorem for families of Dirac operators on odd-dimensional manifolds with boundary.<|endoftext|>
TITLE: Are there 'finitistic' nonrecursive functions (assuming Church's Thesis is false)?
QUESTION [8 upvotes]: [Note: In what follows, I will be using the same type of argument Laszlo Kalmar did in his paper "An Argument Against the Plausibility of Church's Thesis" found in Constructivity in Mathematics, (Amsterdam, 1959, pp. 72-80).  Kalmar considered the following nonrecursive function derived from the nonrecursive function $\mu_y[T_1(x,x,y)]$ found in Kleene's paper, "General Recursive Functions of Natural Numbers" (pp. 237-253 of Martin Davis' book The Undecidable, in particular Thm. XIV):
$$\psi(x)=\begin{cases} \mu y(\phi(x,y)=0) & \text{if } (\exists y)(\phi(x,y)=0) \\
0 & \text{if }(\forall y)(\phi(x,y)\ne0)\end{cases}$$
where $\phi$ is an appropriate general recursive function of two arguments (since Elliot Mendelson argues in his paper,  "On some recent criticism of Church's Thesis", Notre Dame Journal of Formal Logic Vol. IV, No. 3, July 1963, that, "if [Kalmar's--my comment] arguments were correct, then Church's Thesis would be false", I will not concern myself with the unplausible consequences).
Kalmar argues as follows:   "...for any natural number $p$ for which a natural number $y$ with $\phi(p,y)=0$ exists, an obvious method for the calculation of the least such $y$, i.e. of $\psi(p)$  can be given:  calculate in succession the values of $\phi(p,0), \phi(p,1),\phi(p,2),\ldots$ (each of which can be calculated, on acount of the general recursivity of of $\phi$, in a finite number of steps) until we obtain a natural number $q$ for which we have $\phi(p,q)=0$ and take this $q$.  On the other hand for any natural number $p$ for which we can prove, not in the frame of some fixed postulate system but by means of arbitrary--of course, correct--arguments that no natural number $y$ with $\phi(p,y)$ exists, we also have a method to calculate the value of $\psi(p)$ in a finite number of steps:  prove that no natural number $y$ with $\phi(p,y)$ exists, which requires in any case but a finite number of steps, and gives immediately the the value $\psi(p)=0$."  The reader should note that in Mendelson's paper, no account is given as to why this argument is false.
Question 1. Is Kalmar's argument as stated above false, and if so, why?] 
Now to my argument.  It is based on Kleene's theorem XV in the same paper.  In it Kleene makes the following statement: 
"...nonrecursive functions can be defined by the schema
$$\tau(x)= \begin{cases}0 & \text{if } (\exists y)R(x,y)\\
1 & \text{if } (\forall y)\lnot R(x,y)\end{cases}$$
where $R(x,y)$ is primitive recursive [pg. 251 of Davis--my comment]  ."  Recall that $R(x,y)$ is primitive recursive iff there is a primitive recursive function $\pi(x,y)$ such that 
$$R(x,y)\Longleftrightarrow \pi(x,y)= \begin{cases}
0 & \text{if } R(x,y)\text{ true} \\
1 & \text{if } R(x,y)\text{ false} \end{cases}$$
By the Law of Excluded Middle if $R(x,y)$ is primitive recursive then $\lnot R(x,y)$ is primitive recursive also.          
Now consider Kalmar's argument as applied to Kleene's schema.  For any natural number $p$ for which a natural number $y$ with $R(p,y)$ ($\pi(p,y)=0$) exists, one can calculate in succession $\pi(p,0), \pi(p,1), \pi(p,2),\ldots$, until one obtains a natural number $q$ such that $\pi(p,q)=0$ and for which $R(p,q)$ holds.   On the other hand, consider,  for any natural number $p$ for which one can prove, not in the frame of some fixed postulate system, but by means of arbitrary--of course, correct--arguments one also has a method to calculate the value $\pi(p)$ in a finite number of steps:  prove that no natural number $y$ with $\pi(p,y)=0$ (i.e. $\lnot R(p,y)$ is true) exists (that this can be done is immediate by the fact that $\lnot R(x,y)$ is primitive recursive if $R(x,y)$ is for any substitution of numerals $p,y$ for the variables $x,y$), which requires in any case but a finite number of steps.  Therefore, it would seem to be reasonable to conclude that, given Church's Thesis to be false, my aforementioned example is an example of a 'finitistic', nonrecursive function (since Mendelson claims that "if his [Kalmar's] arguments were correct, then Church's Thesis would be false.")  The following question now presents itself:
Question 2:  Have I misinterpreted or misapplied Kleene's Schema? 
Addendum:  I want to respond to two assertions made by Noah Schweber--one positive and one negative.  The assertion I wish to respond positively to is the following:

Kalmar's argument is a complicated way of making the following claim:  There is an effective way for a person to recognize "correct" $\Pi^0_1$ sentences.

The assertion that I want to respond negatively to is the folowing:

Kalmar's argument can't survive the choice of a fixed background theory...

In fact it can, and the background theory in question is $QF$--$IA$ which is $PRA$ transformed into a first-order theory by adding first-order logic.  This is a conservative extension of $PRA$ and has the virtue of using $\Pi^0_1$-sentences instead of open formulas in $PRA$ (since there is a finitistic procedure for transforming a proof in $QF$--$IA$ of a $\Pi^0_1$-sentence into a proof in $PRA$ of the corresponding open formula, making provability, and the consistency of these theories, finitistically equivalent--all this found on pg. 2 of Ignjatovic's paper, "Hilbert's Program and the omega-rule").  Since it is known that R(x,y) and $\lnot$R(x,y) are primitive recursive and therefore decidable for any substitution of numerals $p$,$q$, then for each $p$,$q$ instance R($p$,$q$) ($\lnot$R($p$,$q$ )) is finitistically provable.  Now Ignjatovic (pg.5) quotes Detlefsen paraphrasing Herbrand:

And, again, he says that a universal claim is merely a description or manual of operations which are to be executed in each particular case

Ignjatovic now quotes Detlefsen stating his $\omega$-rule (which I relate to Kleene's schema using brackets):

This view of the universal quantifier would seem to sponsor the following restricted $\omega$-rule:  if I have an effective procedure $\mathbf P$ (a manual of operations $\mathbf P$) for showing of each individual [$p$,$q$] that [R($p$,$q$), $\lnot$R($p$,$q$)] is finitistically provable, then [$\forall$yR(x,y), $\forall$y$\lnot$R(x,y)] is also finitistically provable.  

Now that I have Detlefsen's $\omega$-rule, I would like to be able to use Buldt's Lemma 3.6 to show that $QF$--$IA$ under the $\omega$-rule given by Buldt in his paper "The Scope Of Goedel's First Incompleteness Theorem" (an instance of Detlefsen's $\omega$-rule) is complete in Hilbert's sense for $\Sigma^0_1$ and $\Pi^0_1$ sentences of $QF$--$IA$:

Lemma 3.6:  Let $\mathcal F^{\omega}_{\alpha}$ denote a semi-formal system of arithmetic that admits $\alpha$ applications of the $\omega$-rule [$\vdash_{\mathcal F}$$\varphi$(n), for all n$\in$$\mathbb N$ $\Rightarrow$ $\vdash_{\mathcal F}$($\forall$x)$\varphi$(x)--my comment] ($\alpha$ an ordinal number; $\mathcal F$ extending, say, $PA$), then for all n$\in$$\mathbb N$ $\mathcal F^{\omega}_{n}$ is $\Sigma^0_{2n}$- and $\Pi^0_{2n}$-complete.

It is easily seen that for 1 application of the $\omega$-rule one has that $\mathcal F^{\omega}_1$ is $\Sigma^0_2$- and $\Pi^0_2$-complete and if $\mathcal F$ is an extension of, say $PA$ then Lemma 3.6 will hold for fragments thereof, in particular for $QF$--$IA$ and also for the fragment of $QF$--$IA$ dealing with $\Sigma^0_{2}$ and $\Pi^0_{2}$ sentences. Note that 'complete in Hilbert's sense" means, for Buldt, the following:

$\mathcal F$ is Hilbert-complete iff $\sigma$$\vdash$$\varphi$ or $\Sigma$$\vdash$$\lnot$$\varphi$ for all $\varphi$$\in$$\mathscr L_0$, where $\Sigma$ is the set of axioms of $\mathcal F$ and $\mathscr L_0$ ={$\varphi$|$\varphi$ closed and $\mathfrak U$$\vDash$$\varphi$}

Observe that for $\mathfrak U$$\vDash$$\forall$x$\varphi$(x), $\vDash$$( \forall$x)$\varphi$(x) $\Longleftrightarrow$ $\vDash$$\varphi_x$($\mathbf i$) for all $\mathbf i$ $\in$ $\mathscr L$($\mathfrak U$) where $\mathscr L$($\mathfrak U$) is the first-order language obtained from $\mathscr L$ 
by adding all the names of individuals of $\mathfrak U$ (as new constants) to $\mathscr L$ and adding $\mathbf i$, $\mathbf j$ as syntactical variables which vary through names (Shoenfield, Mathematical Logic, pp. 18-19).  Observe also that the implication in the $\Leftarrow$ direction is identical in form to Buldt's $\omega$-rule, 

$\vdash_{\mathcal F}$$\varphi$(n), for all n$\in$$\mathbb N$ $\Rightarrow$ $\vdash_{\mathcal F}$$\forall$(x)$\varphi$(x) 

and indeed is just the syntactical form of the $\Leftarrow$ direction of the semantical definition of $\forall$x$\varphi$(x).  This being the case, it is clear that since R(x,y) is primitive recursive iff $\lnot$R(x,y) is primitive recursive for all substitutions of numerals formed from 0 by the application of the successor operation (these being the syntactical 'names' for numbers in $\mathbb N$), a single application of this $\omega$-rule to the variable y of the p.r. relations R(x,y), $\lnot$R(x,y) will produce only the true $\Pi^0_1$ sentences of $QF$--$IA$ for R(x,y), $\lnot$R(x,y) (and applying a single instance of the $\omega$-rule according to Buldt's proof of Lemma 3.6 will produce only the true $\Sigma^0_2$ and $\Pi^0_2$ sentences of $QF$--$IA$, so the term 'complete' in Lemma 3.6 can be construed to be 'Hilbert-complete'), one (through the use of Buldt's $\omega$-rule) will have an effective means to 'recognize' 'correct' (true) $\Pi^0_1$ sentences of $QF$--$IA$ and in such manner, Kalmar's argument against Church's Thesis is (at least to me) correct.

REPLY [17 votes]: Kalmar's argument is indeed wrong. The problem, of course, lies in his justification of our ability to compute $\psi(x)=0$, where he writes 

"We can prove, not in the frame of some fixed postulate system but by means of arbitrary - of course, correct -arguments that no natural number $y$ with $\varphi(p,y)$ exists, we also have a method to calculate the value of $\psi(p)$ in a finite number of steps: prove that no natural number $y$ with $\varphi(p,y)$ exists, which requires in any case but a finite number of steps, and gives immediately the the value $\psi(p)=0$." $\quad$ [Emphasis mine.]

By not relying on any fixed axiom system, Kalmar has indeed dodged Godel - but not in a good way. There is a gigantic unspoken assumption here: how do we identify a correct proof? Kalmar is implicitly assuming that "correct"ness can be recognized. Even defining correctness is no easy task if we're not Platonists; and even assuming we're happy with a "true" $\mathbb{N}$, how on earth do we argue that 

(1) Every true $\Pi^0_1$ sentence has an identifiably correct proof

and

(2) We have some procedure to never mistake an incorrect proof for a correct proof?

Kalmar's argument is a complicated way of making the following claim:

There is an effective way for a person to recognize "correct" $\Pi^0_1$ sentences.

This flies in the face of experience: we've all (I suspect) made the embarrassing mistake of thinking that some statement was true - a fortiori consistent with the axiom system we're using - only to find out later that it is disprovable from that axiom system. Rephrased appropriately, this is a case of thinking some Turing machine doesn't halt, and then later finding out that it does. Kalmar claims implicitly that there is some process we can perform to avoid ever doing this, while at the same time eventually deducing each true $\Pi^0_1$ sentence. 
EDIT: Let me clarify what I mean when I say that Kalmar's argument is incorrect. I am not saying (although I do believe) that Kalmar's conclusion is incorrect; rather (and ironically given his claim about correctness) he is assuming a principle which is not at all obviously true, and in fact is basically what he wants to prove in disguise. So his argument doesn't really do anything. If we accept Kalmar's mathematical optimism, then Church's Thesis is clearly false; but Kalmar gives us no reason to do this.
As to your main question, I'm not really sure what it means - what does "finitistic" mean here?<|endoftext|>
TITLE: p-groups such that the center is contained in many cyclic subgroups
QUESTION [9 upvotes]: I'm looking for examples of $p$-groups $G$ with the following three properties:

the center of $G$ is $\mathbb{Z}/p$, and
$G^{\text{ab}} = (\mathbb{Z}/p)^n$ for some $n$, and
for every $g \in G$ whose image in $G^{\text{ab}}$ is nonzero, the cyclic group generated by $g$ contains the center.

Easy examples include cyclic groups of order $p$ and the quaternion group.
I'm particularly interested in examples with large $n$.

EDIT: Stefan Kohl's answer is helpful, but not quite what I'm looking for.  Let me ask a slightly more focused question: do there exist examples with $n$ arbitrarily large?

REPLY [6 votes]: We can use the following GAP functions to search examples of groups
with the desired properties:
IsExample := function ( G )

  local  p;

  if not IsPGroup(G) then return false; fi;
  p := PrimePGroup(G);
  if Size(Centre(G)) <> p then return false; fi;
  if Set(AbelianInvariants(G)) <> [p] then return false; fi;
  return ForAll(Difference(AsList(G),AsList(DerivedSubgroup(G))),
                g->IsSubgroup(Group(g),Centre(G)));
end;

AllExamplesOfOrder_n := n -> Filtered(AllGroups(n),IsExample);

What one finds is that all examples whose order is less than $10000$
and not equal to any of $2^9$, $2^{10}$, $2^{11}$, $2^{12}$, $2^{13}$
or $3^8$ and which are neither cyclic nor generalized quaternion groups
are the groups with the following GAP catalogue numbers:
gap> smallexamples := List([ [ 81, 10 ], [ 128, 802 ], [ 729, 96 ],
>                            [ 729, 101 ], [ 2187, 268 ], 
>                            [ 2187, 272 ], [ 2187, 4486 ] ],SmallGroup);;
gap> 
gap> List(smallexamples,AbelianInvariants);
[ [ 3, 3 ], [ 2, 2, 2 ], [ 3, 3 ], [ 3, 3 ], [ 3, 3 ], [ 3, 3 ],
  [ 3, 3, 3 ] ]
gap> List(smallexamples,StructureDescription);
[ "C3 . ((C3 x C3) : C3) = (C3 x C3) . (C3 x C3)", 
  "C2 . ((C2 x (C4 : C4)) : C2) = (C4 x C2 x C2) . (C2 x C2 x C2)", 
  "C3 . ((C9 x C9) : C3) = (C9 x C9) . (C3 x C3)", 
  "C3 . ((C9 x C9) : C3) = (C9 x C9) . (C3 x C3)", 
  "C3 . (((C9 : C9) : C3) : C3) = ((C9 x C9) : C3) . (C3 x C3)", 
  "C3 . (((C9 : C9) : C3) : C3) = ((C9 x C9) : C3) . (C3 x C3)", 
  "C3 . ((C3 x C3 x ((C3 x C3) : C3)) : C3) = (C3 x C3 x C3 x C3) . \
(C3 x C3 x C3)" ]

Thus in particular the maximum $n$ which you can obtain for a group of
order less than $10000$ and not equal to any of $2^9$, $2^{10}$, $2^{11}$, $2^{12}$, $2^{13}$ or $3^8$ is $3$.<|endoftext|>
TITLE: Why is there a Parity Problem in Sieve Theory and not a Mod p problem for any other p?
QUESTION [25 upvotes]: The "parity problem" in sieve theory, so far as I understand it, is the fact that sieves can't distinguish between primes and $2$-almost primes, numbers with exactly two prime factors, and will always be off by a factor of two. Of course as stated this is false: there are way more $2$-almost primes ($\frac{N\log \log N}{\log N}$) than primes ($\frac{N}{\log N}$) - what we really mean is either (a) $2$-almost primes weighted by the second von Mangoldt function, or (b) $2$-almost primes with no small prime factors.
Terry Tao gives an explanation for this phenomenon: look at the sequence  $a_n = 1+\lambda(n)$ where $\lambda(n)$ is the parity of the number of prime factors of $n$, counting multiplicity. This sequence is the indicator (x2) of the numbers with an even number of prime factors. Assuming Riemann Hypothesis this sequence looks almost exactly like the constant function along arithmetic progressions, so sieves should not be able to distinguish $\sum_{prime} a_p= 0$ and the prime counting function.
My question is why this problem doesn't happen for any other moduli. If we picked $a_n = 1+\zeta_3 ^{\Omega(n)}$ instead, where $\Omega(n)$ is the number of prime factors and $\zeta_3$ is a cube root of unity, why doesn't this force a "mod 3" problem? Why can sieves distinguish between numbers with $0$, $1$, or $2$ mod $3$ prime factors?

REPLY [3 votes]: I found the answer from Zeb Brady. The relevant method is the Selberg-Delange method, see for example this paper. In particular Selberg proved that the average order of $z^{\Omega(n)}$, where $|z| < 2$ is not $-1$ (when it would be a pole of $\Gamma(z)$) is estimated by
$$ \sum_{n\le x} z^{\Omega(n)} = (\frac{f(1,z)}{\Gamma(z)}+o(1)) x (\log x)^{z-1} $$
where $f$ is some convergent Euler product.
So for $z$ any root of unity other than $-1$ we get a comparatively huge main term (as Terry's comments predict). Maybe other people can comment on where the $1/\Gamma(z)$ comes from and how it allows Mobius to cancel unlike everything else.<|endoftext|>
TITLE: Meaning of topological tensor products in Frenkel-Gaitsgory
QUESTION [10 upvotes]: The appendix to http://arxiv.org/abs/math/0508382 by Frenkel & Gaitsgory (following an earlier work of Beilinson) describes three different monoidal structures, denoted by $\otimes^!,\otimes^*,$ and $\otimes^\to,$ on the category of topological vector spaces, which are used to define various kinds of structures and actions in this category. They are defined as follows:

If we write $V=\lim_i V^i$ and $W=\lim_i W^i$ for $V^i,W^i$ discrete, then $$V\otimes^!W:=\lim_{i,j} (V^i\otimes W^j)$$
If $W$ is discrete and $W=\bigcup_kW_k$ with $W_k$ finite-dimensional, then $$V\otimes^\to W:=\text{colim}_k (V\otimes W_k).$$ If we write $W=\lim_j W^j$ for $W^j$ discrete, then $$V\otimes^\to W:=\lim_j (V\otimes^\to W^j)$$
Finally, $V\otimes^* W$ is defined so that $\text{Hom}(V\otimes^*W,U)$, for $U$ discrete, is the set of bilinear continuous maps $V\times W\to U.$

What follows is a series of definitions that combine these operations in ways which I have never been able to keep straight in my head. For instance, topological algebras are defined with respect to the $\otimes^\to$ structure, their modules are defined with respect to the $\otimes^!$ structure, and topological Lie algebras are defined with respect to the $\otimes^*$ structure. Then various combinations of these are defined with respect to some combination of the monoidal structures.
I'm sure that there's some much better ways to think about these tensor products that makes clearer what each of them is good for and why certain kinds of structure ought to be defined with respect to certain of them. This is my question. In other words,

What is the right way to understand these three monoidal structures? How can I keep straight what each of them is good for?

REPLY [5 votes]: This isn't a complete answer, but here are two things they're good for. It may be helpful to observe that linearly topologized vector spaces (Hausdorff, complete, with countable base) embed fully faithfully into the category of pro-vector spaces, with the essential image being countable (cofiltered) diagrams with surjective transition maps. These monoidal structures make sense on the pro-category.
Thing one is that if $\mathscr{C}$ is a category enriched over vector spaces, then the pro-category $\text{Pro}(\mathscr{C})$ is enriched over pro-vector spaces with the $\stackrel{\to}{\otimes}$ monoidal structure.
Thing two is a remark from Beilinson's paper. A topological algebra $A$ is an algebra with respect to $\stackrel{*}{\otimes}$. If the topology on $A$ has a base consisting of left, resp. two-sided ideals then $A$ is an algebra with respect to $\stackrel{\to}{\otimes}$, resp. $\stackrel{!}{\otimes}$. So e.g. the ring of functions on an affine formal scheme is a commutative algebra for $\stackrel{!}{\otimes}$.<|endoftext|>
TITLE: Can every trace preserving isomorphism of unital self-adjoint matrix algebras be realized as conjugation by a unitary?
QUESTION [7 upvotes]: In this paper, Friedland shows (in Lemma 3.4) that if $\phi$ is an isomorphism of coherent algebras, then there exists a unitary $U$ such that
$$ \phi(M) = UMU^\dagger$$
for all $M$. I am wondering if the same is true of any trace-preserving isomorphism $\phi'$ between two self-adjoint unital matrix algebras as long as $\phi'(M^\dagger) = \phi'(M)^\dagger$.
Further comments:
A coherent algebra is a unital (contains identity) matrix algebra containing the all ones matrix which is additionally closed under conjugate transpose (i.e. is self-adjoint) and closed under Schur (entrywise) multiplication. An isomorphism of coherent algebras is an algebra isomorphism that also preserves conjugate transposition and Schur products.
In his proof, Friedland says that since a coherent algebra is semisimple and has a representation of the form given in equation (2.2) of the paper, it suffices to show that any such isomorphism $\phi$ preserves the trace map.
If I understand correctly, then the fact that a coherent algebra is semisimple and has a representation of the form given in (2.2) is a consequence of the fact that it is self-adjoint (closed under conjugate transpose).
Am I correct in my understanding? In other words, is it true that if $\phi$ is a trace-preserving isomorphism of unital self-adjoint matrix algebras such that $\phi(M^\dagger) = \phi(M)^\dagger$, then there is a unitary matrix $U$ such that $\phi(M) = UMU^\dagger$ for all $M$?

REPLY [4 votes]: It is true that every unital self-adjoint algebra $A$ is semisimple. For $A$ contains no non-zero nilpotent right ideal (given any such ideal $I$ and any $M \in I$, we have trace($MM^{\ast}) =0$ since $MM^{\ast} \in I$ is nilpotent, and this forces $M = 0$). Thus $A$ is semisimple, and is a direct sum of full matrix algebras.
If $A$ and $B$ are both unital self adjoint subalgebras of a full matrix algebra $C$, and $\phi: A \to B$ is a trace preserving isomorphism which respects $\ast$, then $A$ and $B$ have the same size full matrix algebra summands with the same multiplicities, since the sizes of the full matrix algebra summands of $A$ are the traces of the primitive idempotents of $Z(A)$ and likewise for $B$.
The primitive idempotents of $Z(A)$ are self-adjoint, for if $e$ is one such, then so is $e^{\ast}$, and $ee^{\ast}$ has positive real trace so certainly $ee^{\ast} \neq 0$. Hence the (commuting) primitive idempotents of $Z(A)$ may be simultaneously diagonalized by a single unitary matrix $V$, and likewise for $B$ ( with another unitary matrix $W$).
Hence it suffices to consider the case (possibly replacing $A$ and $B$ by unitary conjugates) that $\phi(e) = e$ for each primitive idempotent $e$ of $Z(A)$. But then $eA = \phi(eA)$ for each such $A$, and $\phi$ induces an automorphism of the full matrix algebra $eA$ which respects the Hermitian adjoint. Any such automorphism is induced by conjugation of a unitary matrix in $eA$, so the desired conclusion holds.<|endoftext|>
TITLE: The properness of the special singular simplicial spaces
QUESTION [6 upvotes]: This is a question related to another one in MO
Background : A special simplicial space $X_{\cdot}$ is a simplicial space with $X_{0}=\ast$ and $X_{n}\simeq X_{1}^{n}$ via the simplicial map $v_{i}:[1]\rightarrow [n]$ where 
$v_{i}(1)=i$ and $v_{i}(0)=i-1$.
Denote the category of special simplicial spaces by $SsTop$ and the category of  pointed spaces $Top_{\ast}$ (i,e.the pointed compactly generated weak Hausdorff spaces). 
Now $\Omega_{\cdot}$, being defined via $$\Omega_{k}X\equiv hom((\triangle^{k},(\triangle^{k})^{0}),(X,x_{0})),$$ 
  is a right adjoint of the realization functor $\vert-\vert$, where $(\triangle^{k})^{0}$ means the vertices of $\triangle^{k}$ (face and degeneracy defined as in the singular simplicial space). We call $\Omega_{\ast}X$ the special singular simplicial space of $X$, for $X\in Top_{\ast}$.  
We thus obtain
$$\vert-\vert: SsTop\leftrightarrows Top_{\ast}: \Omega$$
Now my first question is: Do we have $\vert\Omega_{\cdot}X\vert\rightarrow X$ is a (weak) homotopy equivalence? Or under what condition of $X$, one can have this?
This has been actually proved in [May: Classifying spaces and fibrations, Porp. 15.5], under the assumption that $X$ is well-pointed. However there some properness assumption of $\Omega_{\cdot}X$ is made use of, but which is not quite clear to me. 
More precisely, he first indicated there is a simplicial map from $B_{\cdot}\Lambda X$, the bar construction of the Moore loop space $\Lambda X$ which is a topological monoid, to $\Omega_{\cdot}X$ and it is a level-wise homotopy equivalence as both simplicial spaces are special and the simplicial map induces a homotopy equivalence on the $1$-simplices. 
Then by applying the lemma:
If a simplicial map between two proper simplicial spaces is level-wise homotopy equivalence, then it induces a homotopy equivalence after applying realization functor. 
and using the homotopy equivalence $\vert B_{\cdot}\Lambda X\vert \rightarrow X$ as well as the decomposition $$\vert B_{\cdot}\Lambda X\vert\rightarrow \vert\Omega_{\cdot}X\vert\rightarrow X,$$ the assertion follows.  
But in applying this lemma, we need to have $\Omega_{\cdot}X$ and $B_{\cdot}\Lambda X$ are proper simplicial spaces. For the latter it is rather fine (see below), but for the former it is then not so clear.
So my second question is: How to see $\Omega_{\cdot}X$ is proper when X is well-pointed? Or it is actually not the case?
For the bar construction of the Moore loop space, $B_{\cdot}\Lambda X$, I think I can see it is proper as the Moore loops space is well-pointed if $X$ is. For $\Omega_{\cdot}X$, though, it is not so clear to me. Beside, as far as I can see, one also cannot apply the approach in the solution of the MO question cited above as well as the construction of the counterexample there. 
Any comment or solution will be very much appreciated.
Thank you in advance.

REPLY [3 votes]: On page 21 of the cited paper of May, $\mathcal{T}$ is defined to be the category of nondegenerately based spaces, but he inconsiderately fails to say that that convention remains in place in Section 15.  With that convention, as he also fails to say but is true, the right adjoint ($\Omega_*$ in the question, $S$ in May) takes values in May's category  $\mathcal{S}^+\mathcal{T}$ of special proper simplicial based spaces and all is well.<|endoftext|>
TITLE: Explicit formulas for certain elements in $Z(U(\mathfrak{gl_n}))$
QUESTION [6 upvotes]: Let $\lambda$ be a partition with $\leq n$ rows and let $L_{\lambda}$ be the corresponding irreducible representation of ${\rm GL}_n(\mathbb{C})$. Let $e_m(X_1,\dots,X_n)$ be the $m$th elementary symmetric polynomial. Is there a known formula for the element $c_m \in Z(U(\mathfrak{gl}_n))$ which acts on $L_{\lambda}$ by the scalar $e_m(\lambda_1,\dots,\lambda_n)$?
It is easy to check that $c_1 = E_{11} + \dots + E_{nn}$, so I suspect that $c_m$ should be related to the coefficients of the characteristic polynomial. Also, I am aware of the Harish-Chandra isomorphism, but have not been able to use it to produce a formula.
I would be particularly interested in references where this is worked out. I couldn't find it in any of the standard representation theory text books.

REPLY [4 votes]: I believe the elements of $\mathcal{Z}(\mathfrak{g})$ you are looking for are (suitably renormalised) the Capelli elements; these come from the (renormalised) Capelli determinant
\[C(u) = \det\left(E_{jk} + \left(u - \frac{n - 2j + 1}{2}\right) \delta_{jk}\right),\]
where by the determinant of a matrix $A = (A_{jk})$ we mean the column determinant
\[\det A = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) A_{\sigma(1),1} \cdots A_{\sigma(n),n}. \]
For all $u$, $C(u)$ lies in $\mathcal{Z}(\mathfrak{g})$.
One can show that given an irreducible representation $\pi_{\lambda}$ of $\mathrm{GL}_n(\mathbb{C})$ of highest weight $\lambda = (\lambda_1,\ldots,\lambda_n)$, the Harish-Chandra isomorphism $\chi_{\pi_{\lambda}}$ acts on $\mathcal{Z}(\mathfrak{g})$ by sending $E_{jj}$ to $\lambda_j + \frac{n - 2j + 1}{2}$ and $E_{jk}$ to $0$ whenever $j \neq k$. So from the definition of $C(u)$, $\pi_{\lambda}(C(u))$ acts by the scalar
\[\prod_{\ell = 1}^{n} (u + \lambda_{\ell}).\]
Expanding $C(u)$ as a polynomial in $u$, we see that the coefficients of each power of $u$ are elements of $\mathcal{Z}(\mathfrak{g})$ that act by the scalar $e_m(\lambda_1, \lambda_2, \ldots, \lambda_n)$.<|endoftext|>
TITLE: Convex body with affine-equivalent cross-sections
QUESTION [7 upvotes]: I recently discovered the following fact: Let $K\subset\mathbb R^3$ be an origin-symmetric convex body with smooth and strictly convex boundary. Suppose that all central cross-sections of $K$ (that is, its intersections with planes through the origin) are affine equivalent planar bodies. Then $K$ is an ellipsoid.
Equivalently, if we consider the norm in $\mathbb R^3$ defined by $K$, this fact can be restated as follows: If a 3-dimensional normed space (with a smooth strictly convex norm) is such that all 2-dimensional linear subspaces are isometric to one another, then the normed space is Euclidean.
My motivations and my proof are from differential geometry, but the fact itself looks like something that might be studied in convex geometry. I know there are many ellipsoid characterization theorems but I could not find this particular one.
So the question is: Is the above fact known, and what are appropriate references? If it is known, does it generalize to non-smooth bodies and to higher dimensions?

REPLY [6 votes]: For given integers $n>2$ and $k\in \{2,3,\dots,n-1\}$ the question of Banach asks whether any $n$-dimensional real Banach space with isometric $k$-dimensional sections is a Hilbert space. For $k=2$ this is proved by 
H. Auerbach, S. Mazur, S. Ulam (here I am not completely sure: Gromov attributes this result to Mazur without direct reference). For given $k$ and infinite-dimensional space this follows from Dvoretzky almost spherical section theorem. Gromov (Izvestiya 1967 31(5)) proves this for any even $k$ and also for odd $k$ and $n\geqslant k+2$ (in complex case, for even $k$ and also for odd $k$ and $n\geqslant 2k+2$). So, in real case even-dimensional spaces with isometric hyperplanes remain (as far as I know) open. Gromov's proof uses algebraic topology of Grassmanians.<|endoftext|>
TITLE: Example where Calabi invariant is nontrivial?
QUESTION [7 upvotes]: Let $D^2$ denote the closed unit disk in $\mathbb{R}^2$. Let $\omega := dx \wedge dy$ denote the standard area form on $\mathbb{R}^2$ (and on $D^2$ by restriction). Let $\phi$ be a diffeomorphism of $D^2$ which is equal to the identity in a neighborhood of $\partial D^2$, and which preserves area; i.e. $\phi^* \omega = \omega$. There is a $1$-form $\alpha$ with $d\alpha = \omega$. We have $\phi^*\alpha - \alpha$ is exact, and equal to $df$ for some smooth function $f$ which vanishes on $\partial D$. The Calabi invariant of $\phi$ is the integral$$C(\phi) = \int_{D^2} f\omega.$$What is an example where the Calabi invariant is nontrivial?

REPLY [3 votes]: There are equivalent definitions of the Calabi invariant that allow to build easily such examples.
1) Given a Hamiltonian $H:[0,1]\times \mathbb{R^2}\to \mathbb{R}$, the Calabi invariant of its time-one map is given by
$$\int_0^1\int H(t,x)\,\omega dt.$$
(See the book "Introduction to symplectic topology" by McDuff and Salamon.)
So any non negative and non identically zero Hamiltonian yields a positive Calabi invariant.
2) There is a dynamical interpretation of the Calabi invariant: Intuitively, it measures how pair of points wind around each other along an isotopy. This is due to Fathi, and two different proofs can be found in the paper of Gambaudo and Ghys "Enlacements asymptotiques". To be more precise, given a Hamiltonian isotopy $f^t$, consider for every pair $x\neq y$, the map 
$t\to Ang(\overrightarrow{f^t(x)f^t(y)},\overrightarrow{xy})$
from $[0,1]$ to $\mathbb{S}^1$, and lift it to a map $A^t(x,y):[0,1]\to\mathbb R$. Then the Calabi invariant of $f^1$ is 
$$\int_{\mathbb{R}^2}A^1(x,y)\,dx\,dy.$$
So to build an example, take an isotopy that preserves all the circles with center in 0 and with non negative rotation on each of these circles. If your map is not the identity, you'll get a positive Calabi invariant.<|endoftext|>
TITLE: Classify $K(\pi,n)$ that are manifolds
QUESTION [6 upvotes]: Inspired by `Naturally occuring'  $K(\pi, n)$ spaces, for $n \geq 2$. and When is a classifying space a topological manifold?, I'd like to formulate a precise question:
For which $n \in \mathbb{Z}_{\ge 2}$ and (isomorphism classes of) groups $\pi$ can the homotopy type $K(\pi,n)$ be represented by a topological manifold?
By a smooth manifold?
In case it's not clear, I'm looking to see if one can prove that one has a complete list.

REPLY [17 votes]: The answer is that this never happens for manifolds which are of finite type in the sense that they are homotopy equivalent to finite CW complexes. Serre showed that a simply connected finite CW complex has infinitely many nonzero homotopy groups. 
Loosely, there's a kind of uncertainty principle relating homotopy and cohomology: it's hard for a space to simultaneously have few homotopy groups and few cohomology groups. So on the one hand it's hard for classifying spaces $B^n A$ to have bounded cohomology (I think they never have bounded cohomology if $n$ is even?), and on the other hand it's hard for finite CW complexes to have bounded homotopy. 
The cohomology of classifying spaces $B^n A$ is extensively studied because they describe cohomology operations, so presumably someone who's more familiar with these can tell you more about no-go results from this direction. It's not hard to show that $B^n \mathbb{Z}$ has nonzero cohomology in arbitrarily high degrees when $n$ is even; you can take the cohomology operations to be cup powers. Similarly it's not hard to show that $B^n \mathbb{Z}_2$ has nonzero cohomology in arbitrarily high degrees for all $n$; here you can also take cup powers, but Steenrod operations are also available.<|endoftext|>
TITLE: Identities of commutators
QUESTION [39 upvotes]: Let $G$ be a group and set $[x,y]:= x^{-1}y^{-1}xy$ as usual and consider it as a binary operation.

Question: Is there a description of the identities that the operation $[.,.]$ satisfies for all groups? 

Just to clarify, those identities should not involve ordinary group multiplication, conjugation or inversion (such as the Hall-Witt identity and various other identities) but only commutators and the neutral element.
Of course one has identities of the form $[x,x]=1$ and $[[x,y],[y,x]]=1$ (as pointed out in a comment), but there are also more complicated ones. As an example, one can check the following three-variable identity 
$$[ [[x,y], z],[z,[y,x]]] = [ [[x,y], z],[[x,y],[z,[y,x]]]]$$
and derive one other of similar type. Are all other identities derived from this?

REPLY [2 votes]: The paper Commutators of flows and fields
contains the following result:

Let $M$ be a manifold, let
$\phi^i:\Bbb R\times M\supset U_{\phi^i}\to M$ be smooth mappings for
$i=1,\dots,k$ where each $U_{\phi^i}$ is an open neighborhood of
$\{0\}\times M$ in $\Bbb R\times M$, such that each $\phi^i_t$ is a
diffeomorphism on its domain, $\phi^i_0=Id_M$, and
$\frac{\partial}{\partial t}|_0\, \phi^i_t=X_i\in\mathfrak X(M)$. We put
$[\phi^i_t,\phi^j_t]
    :=(\phi^j_t)^{-1}\circ(\phi^i_t)^{-1}\circ\phi^j_t\circ\phi^i_t.$
Then for each formal bracket expression $B$ of length $k$ we have
\begin{align}
0&= \tfrac{\partial^\ell}{\partial t^\ell}|_0
    B(\phi^1_t,\dots,\phi^k_t)\quad\text{ for }1\le\ell<k,\\
B(X_1,\dots,X_k)&=\tfrac1{k!} \tfrac{\partial^k}{\partial t^k}|_0
    B(\phi^1_t,\dots,\phi^k_t)\in \mathfrak X(M)
\end{align}

This suggests that any relation holding for Lie brackets holds also for the group commutator.
#Edit:
I am not sure about the suggestion above. Cor.12 of the paper points in this direction.
A more precise conclusion is: The algebraic structure for the group commutator (alone) has as quotient the algebraic structure of Lie brackets (involving brackets alone). What is the kernel? Is there a kernel?<|endoftext|>
TITLE: Tangent bundle of a homogeneous space and the euler exact sequence
QUESTION [7 upvotes]: Let $H \subset G$ be a closed subgroup of a lie group and $G/H$ the homogeneous coset space. There's an exact sequence of adjoint representations of $H$:
$$0 \to \mathfrak{h} \to \mathfrak{g} \to \mathfrak{g/h}\to 0  $$
The canonical principal $H$-bundle $G \to G/H$ gives an exact functor from representations of $H$ to vector bundles over $G/H$. The corresponding sequence of vector bundles is:
$$0 \to G \times_H \mathfrak{h} \to G \times_H \mathfrak{g} \to G \times_H \mathfrak{g/h}\to 0  $$
I'm pretty convinced that $G\times_H \mathfrak{g/h}$ is canonically isomorphic to $T(G/H)$ but a simple proof alludes me so help here would be welcome.
Taking the projective space as an example we have $\mathbb{CP}^n \cong \frac{U(n+1)}{U(n)\times U(1)}$ and the corresponding exact sequence:
$$0 \to U(n+1) \times_{U(n)\times U(1)} \mathfrak{u(n)\times u(1)} \to U(n+1) \times_{U(n)\times U(1)} \mathfrak{u(n+1)}  $$
$$\to U(n+1) \times_{U(n)\times U(1)} \mathfrak{\frac{u(n+1)}{u(n)\times u(1)}}\to 0  $$
Where the last vector bundle is the tangent bundle $T\mathbb{CP}^n$. How does this relate to the euler sequence if at all?
$$0 \to \mathcal{O}_{\mathbb{P}^n} \to \mathcal{O}_{\mathbb{P}^n}(1)^{n+1} \to \mathcal{T}_{\mathbb{P}^n} \to 0$$

REPLY [8 votes]: The isomorphism $G\times_H(\mathfrak g/\mathfrak h)\to T(G/H)$ is induced by the map $G\times (\mathfrak g/\mathfrak h)$ mapping $(g,X+\mathfrak h)$ to $T_gp\cdot L_X(g)\in T_{gH}(G/H)$. 
The Euler sequence corresponds to an exact sequence for the restriction of the standard representation of $G$ to $H$. This is better seen when viewing $\mathbb CP^n$ as a homogeneous space of $G=SL(n+1,\mathbb C)$. Then $H\subset G$ is the group of block-upper-triangular matrices with blocks of sizes $1$ and $n$, so this is a semi-direct product of $S(GL(1,\mathbb C)\times GL(n,\mathbb C))\cong GL(n,\mathbb C)$ and $\mathbb C^{n*}$. In particular, there are natural completely reducible representations for $H$ on $\mathbb C$ and $\mathbb C^n$, say $V$ and $W$. Now consider the standard representation $\mathbb C^{n+1}$ of $G$ and restrict it to $H$. The result is indecomposable but not irreducible, since is contains an $H$-invaraint line but no invariant complement. It fits into an exact sequence $0\to V\to\mathbb C^{n+1}\to W\to 0$ and the Euler-sequence is the short exact sequence of homogeneous vector bundles over $\mathbb CP^n$ corresponding to the tensor product of this sequence with $V^*$. In the picture of unitary groups, the picture is not as clear, since $H$ itself is semisimple, so the restriction of $\mathbb C^{n+1}$ is completely reducible, and it is not obvious, which "direction" of the resulting exact sequence to use. This corresponds to the fact that the Euler sequence admits a split which is  $U(n+1)$-equivariant, but not one, which is $SL(n+1,\mathbb C)$-equivariant.<|endoftext|>
TITLE: Does the Fano plane "embed" in the complex projective plane?
QUESTION [7 upvotes]: $PSL(2,7)$ acts on the projective plane over $\mathbb{F}_2$ (the Fano plane) through its identification with $GL(3,2)$. It also acts on the projective plane over $\mathbb{C}$ through either of its pair of 3-dimensional complex representations. Does the Fano plane embed in $\mathbb{P}_\mathbb{C}^2$ so that the action restricts? To make the question precise:

Does there exist a 7-point orbit in $\mathbb{P}_\mathbb{C}^2$ so that the permutation representation of $PSL(2,7)$ obtained from the action on this orbit is conjugate to the permutation representation of $PSL(2,7)$ on the 7 points of the Fano plane?

REPLY [8 votes]: (As suggested, comment turned into answer) : The answer is "no". ${\rm PSL}(2,7)$ has two non-conjugate subgroups of index $7$, each isomorphic to the symmetric group $S_{4}$. Either of the three dimensional irreducible representations of ${\rm PSL}(2,7)$ restrict irreducibly to each subgroup of ${\rm PSL}(2,7)$ isomorphic to ${\rm S}_{4}$ (a direct and easy calculation), so no such subgroup stabilizes any one-dimensional subspace.<|endoftext|>
TITLE: Do strong embeddings always provide all the ultrafilters that exist?
QUESTION [5 upvotes]: Let $\kappa$ be a strong cardinal. Then for each $\lambda\geq\kappa$ does there exist a $\mu>\lambda$ such that if $U$ is a $\kappa$-complete ultrafilter on $\lambda$ and $j:V\rightarrow M,V_{\mu}\subseteq M,j(crit(j))>\mu$ then there is some $\alpha$ and $x\in V_{\alpha}$ so that the ultrafilter $\{R\subseteq V_{\alpha}|x\in j(R)\}$ is Rudin-Keisler equivalent to $U$?

REPLY [7 votes]: If all we want is that for every measure $U$ there is a strongness
embedding with a seed for $U$, then the answer is yes. To see
this, suppose $\kappa$ is a strong cardinal, and let $U$ be any
$\kappa$-complete ultrafilter on some $\lambda$. Let $j_0:V\to
M_0$ be any $\mu$-strong embedding, so that the critical point of
$j_0$ is $\kappa$ and $V_\mu\subset M_0$ and $j_0(\kappa)>\mu$.
Now consider $j_0(U)$ inside $M_0$, which is a
$j_0(\kappa)$-complete ultrafilter on $j_0(\lambda)$. Let
$h:M_0\to M$ be the ultrafilter by $j_0(U)$ as computed inside
$M_0$, and let $j:V\to M$ be the composition $j=h\circ j_0$. Since
the critical point of $h$ is $j_0(\kappa)$, it follows that
$V_\mu\subset M$, and so the composition embedding is a strongness
embedding for $\kappa$. Let $x=[\text{id}]_{j_0(U)}$ be the
canonical seed for $j_0(U)$ via $h$, so that $X\in j_0(U)\iff x\in
h(X)$. It follows now easily that $$X\in U\iff j_0(X)\in
j_0(X)\iff x\in h(j_0(X))\iff x\in j(X),$$ which is what you
wanted.
Update. But your question asks for something much stronger:
you want every $\mu$-strongness embedding to have seeds for
every ultrafilter $U$.
Here is one easy way to see that this can fail. Suppose that
$\kappa$ is strong and also $\kappa^+$-supercompact. Let
$\lambda=\kappa^+$ and let $U$ be isomorphic to a
$\kappa^+$-supercompactness measure, that is, a normal fine
measure on $P_\kappa(\kappa^+)$.
Now suppose that $j:V\to M$ be a $\mu$-strongness embedding with
critical point $\kappa$, for some $\mu>\kappa$, and furthermore,
such that $j$ is an extender embedding (a direct limit of
ultrapowers on $\kappa$). It follows that $j$ is continuous at
$\kappa^+$, in the sense that $\sup(j''\kappa^+)=j(\kappa^+)$,
since every ordinal below $j(\kappa^+)$ is represented by a
function from $\kappa$ to $\kappa$.
I claim that $U$ is not generated by a seed via $j$ for such an
embedding. If it is, then by basic seed theory we get a factor
diagram $j_U:V\to M_U$ and $k:M_U\cong X\prec M$, with $j=k\circ
j_U$, where $j_U$ is the ultrapower by $U$, and
$X=\{j(f)(\alpha)\mid f:\kappa\to V\}$, where $\alpha$ is the seed
for $U$, meaning that $A\in U\iff
\alpha\in j(A)$.
Since $U$ is a $\kappa^+$-supercompactness embedding, it follows
that $j_U$ is not continuous at $\kappa^+$, in the sense that
$\sup j_U''\kappa^+<j_U(\kappa^+)$. From this, it follows that
$k\circ j_U$ also is not continuous at $\kappa^+$. That is, $j$ is
not continuous at $\kappa^+$, contradicting our earlier assertion
that is was. QED
Here is another counterexample with lower consistency strength.
Suppose that $\kappa$ is strong and there is a measurable cardinal
$\lambda>\kappa$ that is measurable. Let $U$ be a measure on
$\lambda$, that is, a $\lambda$-complete nonprincipal ultrafilter
on $\lambda$. In particular, it is also $\kappa$-complete. Suppose
$\mu>\lambda$ and $j:V\to M$ is a $\mu$-strongness extender
embedding for $\kappa$, so that $V_\mu\subset M$, the critical
point of $j$ is $\kappa$ and $j(\kappa)>\mu$. Since it is an
extender embedding, it follows that $j$ is discontinuous only at
ordinals of cofinality $\kappa$, and continuous at all other
ordinals. If $j$ had a seed generating $U$, then $j_U:V\to M_U$
would be a factor embedding, with $j=k\circ j_U$ for some
$k:M_U\to M$, and since $j_U$ is discontinuous at $\lambda$, this
means that $j$ also would be discontinuous at $\lambda$, contrary
to our earlier observation. So $U$ does not arise in your desired
manner via $j$. QED<|endoftext|>
TITLE: Lifting of Frobenius on torsors over abelian varieties
QUESTION [6 upvotes]: This is related to my previous question Assume that $A$ is an abelian variety over a field $k$ of characteristic $p$, $\mathcal{L}$ is a line bundle on $A$. Assume that $A$ is ordinary and $\mathcal{L}$ lifts on the canonical lifting of $A$ on $W(k)$.
Consider $X=(\mathrm{Spec}_A\mathcal{L})\setminus A$ -- the total space of $\mathcal{L}$ without the zero section. $X$ lifts to $W(k)$. When Frobenius morphism of $X$ lifts to $W(k)$? In the previous question I asked about the case $\deg \mathcal{L}=0$.
Clearly, if a lifting of Frobenius maps fibers to fibers then reduction of $\mathcal{L}$ to $k$ has to be trivial which is impossible if $\deg\mathcal{L}\neq 0$. But maybe it is possible to construct a lifting which does not respect projection on $A$?

REPLY [4 votes]: The canonical lifting $\mathcal{A}$ of $A$ has a canonical lift of the relative  Frobenius $F_{\mathcal{A}/W}:\mathcal{A}\to \mathcal{A}'$, where $\mathcal{A}'=F_W^* \mathcal{A}$ (=the canonical lift of the Frobenius twist $A'=F_k^*(A)$). Moreover, any line bundle $L$ on $A$ has a canonical `Teichmueller' lift $\mathcal{L}$ to $\mathcal{A}$ such that $F_{\mathcal{A}/W}^* (\mathcal{L}) \cong \mathcal{L}'^p$,  $\mathcal{L}' = w^* \mathcal{L}$ where $w$ is the ($W$-Frobenius-linear) projection $\mathcal{A}'=F_W^*\mathcal{A}\cong \mathcal{A}$. 
Let $T(L)$ be the total space of $L$ (treated as a $\mathbb{G}_m$-torsor), in other words, $T(L) = {\rm Spec}_A \bigoplus_{n\in \mathbb{Z}} L^n$. Then $F_{A/k}^* (T(L)) = T(L^p)$, and the relative Frobenius $F_{T(L)/A} : T(L)\to T(L^p)$ is simply ${\rm Spec}_A$ of the inclusion $\bigoplus_{n\in p\mathbb{Z}} L^{n}\subseteq \bigoplus_{n\in\mathbb{Z}} L^n$.
The upshot is that the relative Frobenius $F_{T(L)/A}$ naturally lifts to an $F_{T(\mathcal{L})/\mathcal{A}}: T(\mathcal{L})\to T((\mathcal{L}')^p) = T(F_{\mathcal{A}/W}^* \mathcal{L})$. Then we get a lift of the relative Frobenius $F_{T(L)/k}$ by composition with the natural projection $T(F_{\mathcal{A}/W}^* \mathcal{L})\to T(\mathcal{L}')$.
The above argument works similarly for any torsor under a torus over $A$, or a family of toric varieties.
EDIT. The following commutative diagram with Cartesian squares might be helpful:
$\require{AMScd}$
\begin{CD}
    T(\mathcal{L}) @>F_{T(\mathcal{L})/\mathcal{A}}>> T(\mathcal{L}^p) @>>> T(\mathcal{L}') @>>> 
T(\mathcal{L})  \\
     @. @VVV \square @VVV \square @VVV \\
    @. \mathcal{A} @>F_{\mathcal{A}/W}>> \mathcal{A}' @>w>> \mathcal{A} \\
      @. @. @VVV \square @VVV\\
    @. @. {\rm Spec}(W) @>F_W>> {\rm Spec}(W) \\
    \end{CD}
The composition of the first two arrows in the top row is the desired lift $F_{T(\mathcal{L})/W}$ of $F_{T(L)/k}$ to $W$.<|endoftext|>
TITLE: What is the complexity of finding a third Hamilton Cycle in cubic graph?
QUESTION [9 upvotes]: According to Smith Theorem:  if a cubic graph has a hamilton circuit then it must have a second one. SMITH : Given a Hamilton circuit in a 3-regular graph, find a second Hamilton circuit. It is known that SMITH is in PPA, but it is unknown whether is it PPA-complete. 
More details of this problem can be found here: https://kintali.wordpress.com/tag/ppa-completeness/
According to Tutte[1]: Every edge of a cubic graph lies on an even number of Hamilton cycles.Consequently a cubic hamiltonian graph has at least three Hamilton cycles.
My question: Given one Hamilton Cycle, what is the complexity of finding a third Hamilton Cycle in cubic graph? 
For the problem that given one Hamilton Cycle in a cubic graph, to find the second one, Thomason[2] gave an exponential time algorithm, which can be converted into a PPA-membership proof of SMITH. How about if given one Hamilton Cycle in a cubic graph, to find the THIRD one? If we still use the algorithm of Thomason, it is no better than exponential time, and it is at least in PPA. But can we do better? 
[1] W.T. Tutte, On Hamiltonian circuits, J. London Math. Soc., 21 (1946), 98–101.
[2] A. G. Thomason, Hamiltonian cycles and uniquely edge colourable graphs, Ann. Discrete. Math. 3 (1978), 259-268.

REPLY [5 votes]: Thomason's algorithm surely is superpolynomial, and shows that the problem is in PPA. In [3] I described another algorithm, also exponential and shows PPA, which is just as simple and has the added feature that it is easier to show that it is superpolynomial. Towards your question, given the way in which each of those algorithms behaves, I would very much suspect that finding a third Hamilton cycle given the first one is no easier than to find the second one. But I am not sure whether that helps to solve the related question: if you are given two distinct Hamilton cycles already, can you find a third one? Which I suppose is what you had in mind here.
[3] T.R. Jensen, Simple algorithm for finding a second Hamilton cycle, Siberian Electronic Math. Reports 9 (2012) 151–155.<|endoftext|>
TITLE: What is the motivation behind inner model theory?
QUESTION [16 upvotes]: Inner model theory aims to construct canonical inner models which captures as much of V as possible, which now is formulated more concretely as to build (fine structural) mice that contain many large cardinals. I understand that supercompact is the next step.
But, what is the motivation for achieving this goal? I have some ideas, but I'm wondering if there's a motivation I'm missing completely:

It provides a tool for showing lower consistency bounds of axioms of interest by constructing inner models with certain large cardinals. Most notably, finding such a lower bound for PFA.
It supplies us with more knowledge about V, especially if it turns out that, say, no mouse with a supercompact can be found.

REPLY [6 votes]: Grigor Sargsyan, towards the end of this talk, discusses the "motivational problem" in some detail:

One of the main open problems in set theory is the conjecture that
  "PFA is equiconsistent with a supercompact cardinal". As it is already
  known that one can force PFA from supercompact cardinals, the
  direction that is open is whether one can produce a model of
  supercompactness from a model of PFA. We know essentially one method
  of doing such things and that is via solving the inner model problem
  for large cardinals. The approach is via developing two things at the
  same time.

Develop tools for proving determinacy from hypothesis such as PFA.
Develop tools for proving equiconsistencies between determinacy hypothesis and large cardinals.<|endoftext|>
TITLE: What advantage humans have over computers in mathematics?
QUESTION [56 upvotes]: Now that AlphaGo has just beaten Lee Sedol in Go and Deep Blue has beaten Garry Kasparov in chess in 1997, I wonder what advantage humans have over computers in mathematics?
More specifically, are there any fundamental reasons why a machine learning algorithm trained on a large database of formal proofs couldn't reach a level of skill that is comparable to humans?
What this question is not about
We know that automated theorem proving is in general impossible (finding proofs is semi-decidable). However, humans are still reasonably good at this task. I'm not asking for a general procedure for finding proofs but merely for an algorithm that could mimic human capability at this task.
Another caveat is that most written mathematics at the moment is in a form that is not comprehensible to computers. There do exist databases of formal proofs (such as Metamath, Mizar, AFP) and, even though they are quite small at the moment, it is conceivable that in future we could have a reasonably sized database. I'm not asking whether you believe that a substantial amount of mathematics will be formalized one day -- I'm willing to make this assumption.
Finally, there is the issue of the sheer machine power required to run this. Again, I'm willing to assume that we have a large enough computer to train an AlphaGo-style algorithm and then use reinforcement learning for "practice runs".

REPLY [2 votes]: What advantage humans have over computers in mathematics?

There is an active research in machine learning world, more specifically in NLP (natural language processing) that tries to answer contrary, and even more general question:

What computers can do, that we previously thought that only humans can do?

In my opinion the most interesting approach is an ongoing collaborative benchmark for NLP models called BIG-bench; it is just a bunch of tasks which you can test your model on, or test yourself (or your friends, family or foes)! What is important, you can also add your own tasks.
Regarding the OP question, you can browse the tasks by the keywords here. For example you can test if your model is capable to "understand" Mathematical Induction. You may ask it if the following is true:

1 is an odd integer. Adding 4 to any odd integer creates another odd integer. Therefore, 9 is an odd integer.

You can even "check" if you model is self-aware. Moreover, regarding @logicute answer, you can "measure" your models creativity in various of tasks.
So, can computers do math already or not?... It hasn't been tested carefully yet, but not yet perhaps. Will they?... To answer this question I believe we should keep track of the projects like BIG-bench.<|endoftext|>
TITLE: Is the biproduct of dualizable objects itself dualizable
QUESTION [5 upvotes]: In a monoidal category with biproducts, let $A$ and $B$ be objects with right duals. Then does $A \oplus B$ have a right dual?
The question is a bit subtle. Suppose I already know that $A \oplus B$ has a dual. Then certainly, given the data of right duals for $A$ and $B$, I can construct a right dual for $A \oplus B$, because knowing that $A \oplus B$ has a dual implies certain nice properties. But this does not seem useful in the more restricted case described above.

REPLY [3 votes]: This is not an answer, just an explanation of the mistake I made in my original (now deleted) answer.
It is very tempting to conjecture that an object $X$ has a right dual $X^{\ast}$ iff there is an adjunction of the form
$$\text{Hom}(X \otimes A, B) \cong \text{Hom}(A, X^{\ast} \otimes B)$$
since among other things there are natural choices of a candidate for the unit and the counit. If this were true, then the desired conclusion would follow assuming the monoidal structure distributes over biproducts. But you can't get the zigzag identities this way; as Todd Trimble informed me, there was a long discussion on the n-category cafe about this in 2008. 
Abstractly, the problem is the following. The above condition says that the image of $X$ under the canonical monoidal functor
$$C \ni X \mapsto (A \mapsto X \otimes A) \in \text{End}(C)$$
has a right dual, also in the image. We would be able to conclude that $X$ itself has a right dual if this functor were fully faithful. But it's just not, in general! (In other words, there is no "monoidal Yoneda lemma" in general.)
Here is an explicit counterexample. Let $G$ be a finite group and let $C = \text{Rep}(G)$ be the monoidal category of complex $G$-representations. When we talk about endofunctors of $C$ let's restrict our attention to cocontinuous $\mathbb{C}$-linear endofunctors. Then $\text{End}(C)$ is a "matrix algebra": it's equivalent to the category of matrices $M_{ij}$ of vector spaces, where $i, j \in \hat{G}$ runs over the complex irreps of $G$, acting via
$$\bigoplus_i n_i V_i \mapsto \bigoplus_{i, j} n_j M_{ij} \otimes V_i.$$
Natural transformations between two such matrices correspond to tuples of linear maps $M_{ij} \to N_{ij}$, and composition is given by matrix multiplication. 
In particular, the tensor product functor $V \mapsto X \otimes V$ always lives in $\text{End}(C)$. As a matrix of vector spaces we have $X_{ij} \cong (X \otimes V_j)_i$ where the subscript $i$ denotes taking the $i$-isotypic component. It follows that there is a nontrivial natural transformation $X_{ij} \to Y_{ij}$ iff there are some $i, j$ such that $V_i$ occurs in both $X \otimes V_j$ and $Y \otimes V_j$. But there is a nontrivial morphism $X \to Y$ iff there is some $i$ such that $V_i$ occurs in both $X$ and $Y$. If $G$ is nonabelian then this second condition is strictly stronger.
Probably you can use this to get a counterexample to the original claim about right duals vs. right adjoints but I haven't worked out what the dualizable objects in "matrix algebras" are yet. 
In the positive direction, I believe the "monoidal Yoneda lemma" holds if $C$ is generated under iterated colimits by invertible objects.<|endoftext|>
TITLE: For an arithmetic hyperbolic 3-manifold group, when is its trace field not its invariant trace field?
QUESTION [6 upvotes]: Edit: In my original post I failed to require the group to be a manifold group. The answer below from @BenLinowitz works in that case. I am really interested though in when the group is torsion-free, so have edited this to reflect that.
For a non-elementary finite-covolume Kleinian group $\Gamma$,
let $\mathbb{Q}(\mathrm{tr}\Gamma)$
and $k\Gamma$
be the trace field and the invariant trace field of $\Gamma$,
in the sense of Maclachlan-Reid and Neumann-Reid.
If $\Gamma$
is derived from a quaternion algebra,
then $\mathbb{Q}(\mathrm{tr}\Gamma)=k\Gamma$.
What I want to know is,
what is an example of a case where $\Gamma$
is arithmetic and torsion-free (but not derived from a quaternion algebra, of course),
and $\mathbb{Q}(\mathrm{tr}\Gamma)\neq k\Gamma$?
Definitions:
The trace field of $\Gamma$
is $\mathbb{Q}(\mathrm{tr}\Gamma):=\mathbb{Q}\big(\{\mathrm{tr}\gamma\mid\gamma\in\Gamma\}\big)$.
The invariant trace field of $\Gamma$
is $k\Gamma:=\mathbb{Q}(\mathrm{tr}\Gamma^{(2)})$,
where $\Gamma^{(2)}:=\langle\gamma^2\mid\gamma\in\Gamma\rangle$ (the group generated by squares in $\Gamma$).
$\Gamma$ is derived from a quaternion algebra
if there exists a quaternion algebra $B$
that has a unique complex place,
and is ramified at all real places,
and an order $\mathcal{O}\subset B$
so that $\Gamma$
is a finite-index subgroup of $\mathrm{P}\mathcal{O}^1:=\{x\in\mathcal{O}\mid\mathrm{nrd}(x)=1\}/\{\pm1\}$.
$\Gamma$ is arithmetic if it is commensurable to a group derived from a quaternion algebra.
Properties:
$\mathbb{Q}(\mathrm{tr}\Gamma)$ and $k\Gamma$ are number fields.
If $\Gamma$ and $\Gamma'$ are commensurable (up to conjugation),
then $k\Gamma=k\Gamma'$.
The definition given of arithmeticity is equivalent to $\Gamma$ being Kleinian and and an arithmetic group in the sense of Borel (the ramification conditions guarantee discreteness).

REPLY [3 votes]: For cusped 3-manifolds that are link complements in $\mathbb{Z}/2\mathbb{Z}$ homology spheres, the trace field and invariant trace field are equal (see Neumann and Reid Arithmetic of Hyperbolic Manifolds Corollary 2.3), so it makes sense to look at manifolds with non-peripheral $\mathbb{Z}/2\mathbb{Z}$ homology or closed manifolds. 
In fact, $m007(3,1)$ aka Vol3 (the third smallest volume manifold in the Hodgson-Weeks' census of closed orientable manifolds) is an example of a arithmetic hyperbolic manifold of this form. Its trace field is degree 4 ($\mathbb{Q}(\sqrt{1 + i \sqrt{3}})$) and its invariant trace field is $\mathbb{Q}(i\sqrt{3})$. In some sense, this is the 'first' such example, but there are many more of this form in the Hodgson-Weeks' census and the early manifolds in this census have a high probability of being arithmetic (compare to the Chapter 13.4 of Maclachlan and Reid's "The Arithmetic of Hyperbolic 3-Manifolds", with the caveat that the surgery descriptions of closed manifolds are non-unique), so if you wanted more examples of this form the following code will help: 
sage: import snappy
sage: CC = snappy.OrientableClosedCensus()
sage: distinct_field_manifolds = []
sage: CUTOFF = 10
sage: for c in CC:  
    citfg = c.invariant_trace_field_gens()
    ctfg = c.trace_field_gens()
    kI = citfg.find_field(degree=30,prec=200)
    k = ctfg.find_field(degree=30, prec=200)
    try:
       if kI[0].degree() < k[0].degree():
          print c, "invariant trace field", kI, "trace field:", k, "\n"
          distinct_field_manifolds.append(c)
          if len(distinct_field_manifolds) >= CUTOFF:
              break;
    except:
          print "field computation failed for c=",c, "kI= ", kI, "k=",k

(Note there are a lot of manifolds in the census and computations in exact arithmetic are expensive so the code cuts out after the first 10, of course that can be adjusted by CUTOFF.)
UPDATE: There are cusped arithmetic manifolds with this property as well. For example, 'm009' which has homology $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, trace field $\mathbb{Q}(\sqrt{\frac{5}{2} - \frac{i \sqrt{7}}{2}})$ and invariant trace field $\mathbb{Q}(i\sqrt{7})$.<|endoftext|>
TITLE: $C(X)$ as finitely generated $C^*$-algebra
QUESTION [5 upvotes]: Let $X$ be a Hausdorff space. Suppose that $C(X)$ (or $C_0(X)$) is a finitely generated $C^*$-algebra. What we can say about $X$ ? For example can we characterize its inductive dimension, axioms of countability etc. ?

REPLY [7 votes]: The maximal ideal space $\Delta$ of a finitely generated Banach algebra is homeomorphic to a compact subset of $\mathbb{C}^n$. On the other hand, evaluation at each point of $X$ is clearly a complex homomorphism of $C(X)$. We conclude that $X$ is homeomorphic to a subset of a compact subset of $\mathbb{C}^n$. 
Edit: actually, if $X$ is not compact $C(X)$ is not a Banach algebra at all (locally compact for $C_0(X)$). In locally compact case, as Nik Weaver showed in comments, $\Delta \cup \{0\}$ is compact in $\mathbb{C}^n$, so $\Delta$ is homeomorphic to a closed subset of some Euclidean space. 
Also, Stone-Weierstrass theorem shows that all such $X$ satisfy the condition.<|endoftext|>
TITLE: Embedding a cancellative monoid into another in such a way that a prescribed element becomes left-invertible
QUESTION [6 upvotes]: Let $\mathbb A = (A, +_A)$ be a cancellative, but possibly non-commutative, monoid with identity $0$, and fix an element $x \in A$. Does there always exist a cancellative monoid $\mathbb B = (B, +)$ such that $\mathbb A$ is a submonoid of $\mathbb B$ and $x$ is left-invertible invertible in $\mathbb B$ (see Benjamin Steinberg's comment below)?
It is known from a paper of A. I. Mal'cev [Math. Ann. 113 (1937), No. 1, 686-691] that there exist (finitely generated) cancellative monoids that do not embed into a group, but the question here is about an embeddability condition that looks very much weaker.

REPLY [6 votes]: The answer is no. 
In a cancellative monoid left invertible elements are right invertible. If yx =1 then xyx=x and so xy =1 by cancellation. So you are asking to invert a given element in some cancellative monoid. 
Take a fg cancellative monoid M
 not embeddable in a group.  Invert the generators one by one to embed M in the group of units of a monoid and get a contradiction.<|endoftext|>
TITLE: Extending rational Diophantine triples to sextuples
QUESTION [7 upvotes]: (This is a follow-up to a previous post.) A rational Diophantine $m$-tuple is a set of rationals {$a_1,a_2,\dots a_m$} such that (with $i\neq j$), all $a_i a_j+1$ is a square. Problem: Find a class of triples that can be extended to sextuples.

I. Elliptic curve

One solution is described in the paper by Dujella et al. Find rational $u,v$ such that,
$$(-27 - 9 u^2 - 27 u v + u^3 v - 54 v^2) (1 + v^2) (-1 + u v + 2 v^2)=y^2\tag1$$
This is a quartic in $u$ to be made a square. Since it has a rational point, it is birationally equivalent to an elliptic curve. Define,
$$\alpha_2 = \frac{-3-6uv-12v^2+u^2v^2}{4+4v^2}\tag2$$
Let $a,b,c$ be the roots of,
$$z^3-uz^2+\alpha_2 z-v =0\tag3$$
Then $a,b,c$ are special rational triples such that $ab+1,\;ac+1,\;bc+1$, and,
$$\begin{aligned}
&a^2b^2c^2+1 = p^2\\
&2(a^2+b^2+c^2)-(a+b+c)^2-3 = q^2\\
&2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcbd+(abcd)^2=(pd+q)^2
\end{aligned}\tag4$$
are all rational squares (for any $d$ for the last), implying they obey,
$$\big(2(a^2+b^2+c^2)-(a+b+c)^2-3\big)(1+a^2b^2c^2)=(a + b + c + 3 a b c)^2$$
A second elliptic curve can then yield infinitely many three rational $x_i$ such that $a,b,c,x_1,x_2,x_3$ is a sextuple, but we need not go to that step.

II. Solutions $a,b,c$

Using $v = \frac{t^2-1}{2t}$ and $u = \frac{1 + 130 t^2 - 390 t^4 + 130 t^6 + t^8}{-3 t + 105 t^3 - 105 t^5 + 3 t^7}$, Dujella et al found,
$$\begin{aligned}
a &= \frac{18t\,(t^2-1)}{(t^2-6t+1)(t^2+6t+1)}\\
b &= \frac{(t - 1)(t^2 + 6t + 1)^2}{6t\,(t + 1)(t^2 - 6t + 1)}\\
c &= \frac{(t + 1)(t^2 - 6t + 1)^2}{6t\,(t - 1)(t^2 + 6t + 1)}
\end{aligned}\tag5$$
Using $v = \frac{t^2-1}{2t}$ and $u = \frac{1 + 1996 t^2 + 102 t^4 + 1996 t^6 + t^8}{-8 t + 2584 t^3 - 2584 t^5 + 8 t^7}$, I found,
$$\begin{aligned}
a &= \frac{128t\,(t^2+1)^2}{(t^2-1)(t^2-18t+1)(t^2+18t+1)}\\
b &= \frac{(t^2 - 1)(t^2 + 18 t + 1)^2}{16t\,(t^2 + 1)(t^2 - 18t + 1)}\\
c &= \frac{(t^2 - 1)(t^2 - 18 t + 1)^2}{16t\,(t^2 + 1)(t^2 + 18t + 1)}
\end{aligned}\tag6$$
Notice that the triples $(5)$ and $(6)$ have an aesthetically pleasant similarity. And since $(1)$ is an elliptic curve, there should be infinitely many families.

Q: Using $(1)$, can you find another triple $a,b,c$ that has polynomial numerator/denominator of small degree?

REPLY [2 votes]: Alternative constructions of rational Diophantine sextuples can be found in 
"Rational Diophantine sextuples containing two regular quadruples and one regular quintuple" https://arxiv.org/abs/1904.00348
"There are infinitely many rational Diophantine sextuples with square denominators" https://arxiv.org/abs/1903.02805<|endoftext|>
TITLE: A log inequality for positive definite trace-one matrices
QUESTION [13 upvotes]: Let $\{v_i\}_{i=1}^N$ be a set of $n$-dimensional real vectors and let $X=X^\top\in\mathbb{R}^{n\times n}$ be a positive definite trace-one matrix. I would like to prove (or disprove) the following inequality:
$$
\sum_{i=1}^N \log \left(\frac{1}{N}\sum_{j=1}^N \frac{v_i^\top X^{1/2}v_j v_j^\top X^{1/2}v_i}{v_j^\top X v_j}\right) - \sum_{i=1}^N \log v_i^\top X v_i \geq 0,\qquad (*)
$$
where $X^{1/2}=(X^{1/2})^\top$ denotes the principal square root of $X$. Note that the previous inequality holds for $N=1$. Indeed,
$$
\log \frac{\left(v_1^\top X^{1/2}v_1\right)^2}{v_1^\top X v_1} - \log v_1^\top X v_1 = 2 \log\frac{v_1^\top X^{1/2}v_1}{v_1^\top X v_1}\geq 0,
$$
where the last step follows from the fact that, since $X$ is a trace-one matrix, $v^\top(X^{1/2}-X)v\geq 0$ for all $v\in\mathbb{R}^n$. I carried out a large series of numerical simulations for $N>1$ in order to find a counterexample to $(*)$, but I couldn't find it. Hence my guess is that $(*)$ is true.
Thanks for your help!
Some comments. Notice that $(*)$ can be written as
$$
\sum_{i=1}^N \log \left(\frac{1}{N}\sum_{j=1}^N \frac{v_i^\top X^{1/2}v_j v_j^\top X^{1/2}v_i}{v_j^\top X v_jv_i^\top X v_i }\right) \geq 0.
$$
A lower bound to the LHS of previous inequality can be found by using the inequality $\log(x)\geq 1-1/x$, which yields,
$$
\sum_{i=1}^N \log \left(\frac{1}{N}\sum_{j=1}^N \frac{v_i^\top X^{1/2}v_j v_j^\top X^{1/2}v_i}{v_j^\top X v_jv_i^\top X v_i }\right) \geq N -N\sum_{i=1}^N  \left(\sum_{j=1}^N \frac{v_i^\top X^{1/2}v_j v_j^\top X^{1/2}v_i}{v_j^\top X v_jv_i^\top X v_i }\right)^{-1}.
$$
Hence if we prove that
$$
\sum_{i=1}^N\left(\sum_{j=1}^N \frac{v_i^\top X^{1/2}v_j v_j^\top X^{1/2}v_i}{v_j^\top X v_jv_i^\top X v_i }\right)^{-1}\leq 1
$$
we are done. Numerical simulations suggest that the latter inequality holds.

EDIT. A follow-up of this question has been posted here.

REPLY [9 votes]: The proof of the general case, in a strong form suggested in the end of OP.
Denote $X^{1/4} v_i=u_i$, $X^{1/2}=S$, then we have ${\rm tr}\,S^2=1$ and need to prove that 
$$
{\rm tr}\,S^2\geqslant 
\sum_i \frac{(Su_i,u_i) (Su_j,u_j)}{\sum_j (u_i,u_j)^2},
$$
then the very original inequality follows by applying AM-GM to $N$ summands in RHS.
By homogeneity in $u$'s we may suppose that $(Su_i,u_i)=1$ for all $i$. On the set of symmetric operators on $\mathbb{R}^n$ we have an inner product $(Y,Z)={\rm tr}\, YZ$, so ${\rm tr}\,S^2=\|S\|^2$. Our restriction on $S$  may be rewritten as $(S,u_i\otimes u_i)=1$, where we naturally identify $u\otimes u$ with the (rank at most 1) operator $x\rightarrow (u,x)u$. For operators $T_i=u_i\otimes u_i$, $i=1,\dots,N$ we have $(u_i,u_j)^2=(T_i,T_j)$. That is, we get the following problem: given that $(T_i,T_j)\geqslant 0$ for all $i,j$ and $(S,T_i)=1$ prove that 
$$
\|S\|^2 \geqslant \sum_{i=1}^N (T_i,T_1+\dots+T_N)^{-1}
$$
For $c_i=(T_i,\sum T_j)$ we choose numbers $\mu_1,\dots,\mu_N$ such that $\sum 1/c_i=(\sum \mu_i)^2/\sum \mu_i^2c_i$, for example take $\mu_i=1/c_i$. Then 
$$
\|S\|^2\sum \mu_i^2c_i=\|S\|^2 \sum_i \sum_j \mu_i^2 (T_i,T_j)\geqslant \|S\|^2\|\sum \mu_iT_i\|^2\geqslant (S,\sum \mu_iT_i)^2=(\sum \mu_i)^2
$$
as desired (we have used that $\mu_i^2 (T_i,T_j)+\mu_j^2 (T_j,T_i)\geqslant 2\mu_i\mu_j (T_i,T_j)$ for all pairs $i\ne j$.)<|endoftext|>
TITLE: Identifying the canonical principal polarization of a Jacobian
QUESTION [5 upvotes]: Let $X$ be a curve over an algebraically closed field $k$ (even over $k = \mathbb{C}$ if you want), let $J = Pic^0_{X/k}$ be its Jacobian, let $P \in X(k)$ be a point, and let $i \colon X \hookrightarrow J$ be the closed immersion that sends $Q$ to the divisor class of $[Q] - [P]$. How does the map 
$$Pic^0(i): Pic^0_{J/k} \rightarrow J$$
compare with the canonical principal polarization 
$$p \colon J \rightarrow Pic^0_{J/k}$$
induced by the $\Theta$-divisor? Are $Pic^0(i)$ and $p$ inverse to each other?
I came across Lemma 6.9 in the following notes http://www.jmilne.org/math/xnotes/JVs.pdf by James Milne, and, if I understand correctly, this lemma claims that $Pic^0(i) = -p^{-1}$. I am very confused about the minus sign though: if $X = J$ is an elliptic curve and $P = 0$ is the origin, aren't both $Pic^0(i)$ and $p$ ``the identity''?

REPLY [9 votes]: The answer depends very much on how you define the homomorphism $p$ associated to the polarization. My personal choice is $p(a)=\mathcal{O}(\Theta _a-\Theta )$, where $\Theta $ is a theta divisor and $\Theta _a:=\Theta +a$. Realizing $\Theta $ as $i(\mathrm{Sym}^{g-1}C)$, you find that $i^*\Theta _a$ is the divisor $Q_1+\ldots +Q_g$, where the $Q_i$ are the solutions of 
$h^0(Q_i+(g-2)P-a)\geq 1$, or equivalently $h^0(K-(g-2)P+a-Q_i)\geq 1$; this implies
$i^*\Theta _a=K-(g-2)P+a\ $ for general $a$, hence for all $a$, and therefore
 $i^*p(a)=i^*\Theta _a-i^*\Theta=a$, so that $i^*\circ p$ is the identity.
Now Mumford in Abelian varieties, and others following him, take the opposite sign for $p\,$, which leads to the formula (unfortunate in my view) $\ i^*\circ p=-\mathrm{Id}$.<|endoftext|>
TITLE: Do all non-degenerate quadratic forms come from positive even lattices?
QUESTION [6 upvotes]: Let $(G,+)$ be a finite Abelian group. We say $q\colon G\to \mathbb{T}$ is a non-degenerated quadratic form, if $q(-a)=q(a)$ and the symmetric function
$$
b(g,h) =q(g+h)q(g)^{-1}q(h)^{-1}
$$
is a non-degenenerate bicharacter on $G$, i.e. $b(g+h,k)=b(g,k)b(h,k)$ and 
$b(k,g)=1$ for all $k\in G$ implies $g=0$.
Let $(\Gamma,\langle\,\cdot\,,\,\cdot\,\rangle)$ be a positive even lattice, i.e. a free abelian group with an symmetric bilinear form $\langle\,\cdot\,,\,\cdot\,\rangle\colon \Gamma \times \Gamma \to \mathbb{Z}$,
such that
$\langle a,a\rangle \in 2\mathbb N$ for all $a \in \Gamma$ and $\langle a,a\rangle=0$ if and only if $a=0$. Let $\Gamma^\ast =\mathrm{Hom}(\Gamma,\mathbb Z)$ be the dual lattice. 
Then $G_\Gamma=\Gamma^\ast/\Gamma$ is a finite group
and $q_\Gamma([a])=\exp(\pi i \langle a,a\rangle)$ is a non-degenerate quadratic form on $G_\Gamma$.
Question: Is the map from positive even lattices to finite Abelian groups with non-degenerate quadratic forms
$$
   (\Gamma,\langle\,\cdot\,,\,\cdot\,\rangle)\longmapsto (G_\Gamma,q_\Gamma)
$$
 surjective (up to the obvious equivalences)?
It is for sure not injective, because all self-dual lattices map to the trivial group.
If one drops the assumption of positiveness the answer is, apparently, yes.
Motivation: A pointed unitary modular tensor category is completely characterized by the fusion rules which give a finite Abelian group $G$ and a non-degenerate quadratic form on $G$ given by the twists (basically type I Reidmeister move). Every even lattice $\Gamma$ gives a unitary modular tensor category using, for example, the lattice Vertex Operator Algebra or Conformal Net associated with $\Gamma$ and consider its representation category, which turns out to be a modular tensor category which is characterized by $(G_\Gamma,q_\Gamma)$.
So the question can be reformulated to: Do all unitary pointed modular tensor categories come from positive even lattice CFTs?

REPLY [4 votes]: Edited: I have missed your "positive". The signature of the lattice modulo 8 depends on the form only (some people call this Brown invariant and van der Blij theorem; Nikulin below calls this just the signature of the form).
Otherwise (given the right signature), I would suggest that the map is surjective, and the easiest proof would be the known classification of forms and an explicit construction for each class. For more details (e.g., the classification) see Nikulin, V. V. Integer symmetric bilinear forms and some of their geometric applications. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 43 (1979), no. 1, 111–177, 238 [MR0525944]
Edit The existence of some positive form is given by Theorem 1.10.1 (which does not assume that the form is indefinite). For the construction, the two forms on $\mathbb{Z}/2$ are $[2]$ and $E_7$. Then, do induction on $p$ to construct the two forms on $\mathrm{Z}/p$. First, take $[2pq]$ with appropriate $q<p$ to get the correct $p$-primary part; then, use Nikulin's gluing and induction to kill the extra $2$- and $q$-primary parts. For the two remaining series of $(2\times2)$-forms, one can probably start from $D_4(2^k)$ or $E_8(2^k)$ and pass to appropriate finite index extensions. (Though, I did not work out the details.)<|endoftext|>
TITLE: Hilbert symbol averages
QUESTION [7 upvotes]: Let me call a pair of integers $a, b$ acceptable if the equation $ax^2 + by^2 = z^2$ has a non-trivial rational solution. Theorem 4.5.4 of Cojocaru-Murty's book on Sieves says that the number of acceptable pairs of integers $a, b$ with $1 \leq a, b \leq H$ is $\ll H^2 / \log \log H$. This is proved using the Turan sieve. Serre had previously proved in a short paper in 1990 that the estimate can be improved to $H/(\log H)^\delta$ for some $\delta >0$. This proof of the latter result uses the large sieve. 
Question.  Is it possible to give an asymptotic formula for the number of acceptable pairs of integers $a, b$ with $1 \leq a, b \leq H$ as $H \to \infty$?

REPLY [3 votes]: This problem has been considered by a few authors. Versions for $a$ and $b$ rational were considered by Hooley and Guo (independently). See 
Hooley - On ternary quadratic forms that represent zero.
Guo - On solvability of ternary quadratic forms.
Hooley obtained the (sharp) lower bound of the form $H^3/(\log H)^{3/2}$ for the corresponding counting problem. Guo obtained an asymptotic formula for the slightly different problem where $a$ and $b$ are assumed to be square-free (i.e. the numerators and denominators are square-free).
An asymptotic formula without the square-free assumption does not seem to be known.
More recent work:
Friedlander, Iwaniec - Ternary quadratic forms with rational zeros
considers the case where $a$ and $b$ are integers, and obtained an asymptotic formula of the order $H^2/(\log H)$ under the additional assumption that $a$ and $b$ are odd, coprime and square-free. An asymptotic formula for general $a$ and $b$ might be possible to obtain from their methods, but probably quite messy.<|endoftext|>
TITLE: Singular in $V$ regular in $HOD$
QUESTION [10 upvotes]: Prikry forcing can be used to produce a model $V$ of $ZFC$ such that fo rsome cardinal $\kappa$ we have:
(1) $\kappa$ is singular in $V$ of cofinality $\omega,$
(2) $\kappa$ is regular (and in fact measurable) in $HOD$.
Now my question is can this happen with $\kappa$ having uncountable cofinality in $V$. So

Question Can we find a model $V$ of $ZFC$ which contains a cardinal $\kappa$ so that
(1) $\kappa$ is singular of uncountable cofinality in $V$,
(2) $\kappa$ is regular in $HOD$.

Let me explain why Magidor or Radin forcing do not work in general. Let's start with core model $K$ in which $\kappa$ is large enough, and let $V$ be the generic extension obtained by Magidor or Radin forcing to change the cofinality of $\kappa$ to, say, $\omega_1.$ Let $C$ be the resulting club. We can assume all elements of $C$ were regular in $K$.
Claim. $Lim(C) \in HOD,$ where $Lim(C)$ is the set of limit points of $C$.
Proof. We have $Lim(C)=\{\alpha \leq \kappa: \alpha$ is singular, but regular in $K     \} \in HOD.$
In particular $\kappa$ is  singular in $HOD$.

REPLY [4 votes]: The answer to the question is yes. In fact, it is possible to prove something stronger:

Theorem (Omer Ben-Neria, Spencer Unger) Assuming the existence of suitable large cardinals, there exists a model $V$ of $ZFC$ which contains a club of cardinals all of them are regular in $HOD$ of $V$.

Their work is under preparation.
Remark. Their result is optimal in the sense that we can not hope to build a model in which all infinite cardinals are regular in $HOD$, as for example $\aleph_\omega$ is always singular in $HOD$.

Update: The paper by Omer Ben-Neria and Spencer Unger is now available:

 Homogeneous changes in cofinalities with applications to HOD<|endoftext|>
TITLE: The quantum group SUq(n) as von Neumann algebra
QUESTION [7 upvotes]: i have a question about a "presentation" of the quantum $SU(n)$. Here presentation means the following. Let $(M,\Delta)$ be a quantum group in the sense of Kustermanns and Vaes. One can show that the von Neumann algebraic quantum group $SU_q(2)$, which will be denotes by $\mathscr{L}^{\infty}(SU_q(2))$ is $B(\ell^2(\mathbb{N}))\overline{\otimes}\mathscr{L}(\mathbb{Z})$ (where the tensor product denotes the von Neumann algebraic tensor product). Can one generalize this to $SU_q(n)$. Can one find an explicit expression for $\mathscr{L}^{\infty}(SU_q(n))$ as above?
Thank you very much

REPLY [4 votes]: Theorem 3.1 of [2] is basically what you need. It implies $L^\infty(G_q) \simeq B(\ell^2(\mathbf{N})) \otimes L^\infty(T)$ for the maximal torus $T$ in $G$ (connected semisimple compact Lie group), and up to isomorphism the translation action of $T$ is just the standard one on the second factor.
As suggested there, Soibelman's work [1] on the classification of irreducible representations of $C(G_q)$ brings you most of the way for this. It implies that $C(G_q)$ is of type I and does not have finite dimensional irreducible representations of dimension $>1$, so any von Neumann algebraic closure has to be of the form $A_1 \oplus B(\ell^2(\mathbf{N})) \otimes A_2$ for some commutative von Neumann algebras $A_1, A_2$. Then you take into account of the torus action to show $A_1 = 0$ (edit: and that $A_2$ is diffuse) in the regular representation.

Ya. S. Soibelman, Algebra of functions on a compact quantum group and its representations, Algebra i Analiz 2 (1990), no. 1, 190--212.
Reiji Tomatsu, Product type actions of $G_q$, Adv. Math. 269 (2015), 162--196.<|endoftext|>
TITLE: Homotopy type of diffeomorphism which are the identity on and near the boundary
QUESTION [7 upvotes]: Let $M$ be a compact manifold with boundary. Denote by $Diff(M), Diff_\partial(M)$ and $Diff_{U\partial}(M)$ the groups of diffeomorphisms of $M$ and the subgroups of the ones that are the identity on the boundary and the ones that are the identity on some (non-fixed) neighborhood of the boundary. Equip them with the usual smooth Whitney topology. We have inclusions
$$Diff_{U\partial}(M)\subseteq Diff_\partial(M) \subseteq Diff(M)$$ and I am interested in the effect of those in homotopy groups.
The second inclusion is usually far away from being a (weak) homotopy equivalence. In the case $M=D^2$, $Diff_\partial(M)$ is contractible and $Diff(M)$ has the homotopy type of $O(2)$.
The situation of the first inclusion seems to be more subtle. In the case of low dimensions, I believe that it should be a weak homotopy equivalence in general and I expect that this is no longer true in higher dimensions. This fits to the literature in the sense that people are usually very vague between the difference of $Diff_{U\partial}(M)$ and $Diff_\partial(M)$ in low dimensions and mostly emphasize the use of $Diff_{U\partial}(M)$ opposite of $Diff_\partial(M)$ when working in higher dimensions.

Is my guess correct?
Is there anything else useful to say about the relation of the homotopy type between those three groups?

REPLY [3 votes]: The first inclusion $Diff_{U\partial} M \to Diff_{\partial} M$ is a homotopy-equivalence provided you do not let the neighbourhood get "too big".  If you fix the neighbourhood there is a fibre sequence
$$Diff_{U\partial} M \to Diff_{\partial} M \to Emb(U, M)$$
where the embeddings of the collar neighbourhood $U$ of $\partial M$ in $M$ are required to be the identity on the boundary.  The fact that this embedding space is contractible boils down to the uniqueness of tubular neighbourhoods theorem + the convexity of small linear collars. 
I almost never see people use the space where you allow the neighbourhood $U$ to vary.  What literature are you reading?   But you can adapt this argument to describe the homotopy-type of that space, as well.  As yours is basically the union of these spaces.  As these spaces all intersect over a homotopy-equivalent subspace, you get the result at the weak homotopy-type level. 
Regarding the last inclusion $Diff_{\partial} M \to Diff M$ this also sits in a fibre sequence
$$Diff_\partial M \to Diff M \to Diff(\partial M)$$
this space is more closely related to the idea of pseudoisotopy.  In particular, this fibre-sequence is not always onto as not every diffeomorphism of the boundary extends to the interior of the manifold.<|endoftext|>
TITLE: combinatorics on cyclic sequences
QUESTION [9 upvotes]: Given $m\geq 1$, let $I=(a_1,\ldots,a_{3m})$ be a sequence such that $I$ contains exactly $m$ zeros, $m$ ones, and $m$ twos.
Given $i=1,2$ and $j\leq 3m,k\leq m$ we can define $$U_{i,j}(k)=\text{number of $i$'s before finding $k$ zeros, starting from position $j$}.$$
(moving to the right, in a cyclic way)
For instance, for the sequence $(0,2,1,1,0,2)$, we would have $U_{1,2}(1)=2,U_{1,6}(2)=2$ (here is necessary to move to the beginning to continue counting) and, in total (using a matrix notation), 
$$U(1)=\begin{bmatrix}U_{i,j}(1)\end{bmatrix}=\begin{bmatrix}0 & 2 & 2 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 1\end{bmatrix}$$
$$U(2)=\begin{bmatrix}U_{i,j}(2)\end{bmatrix}=\begin{bmatrix}2 & 2 & 2 & 1 & 0 & 2\\ 1 & 2 & 1 & 1 & 1 & 2\end{bmatrix}$$

Question: Is it true that for all such sequence there is $k\leq m$ such that at least $3m$ of the coefficients $U_{i,j}(k)$ satisfy
  $U_{i,j}(k)\geq k$?

At first, I thought that it was enough to take either $k=1$ or $k=m$, which are cases that I understand very well, but the sequence $(0,0,1,2,2,0,2,1,1)$ only works when $k=2$.
My first atempt was that for a fixed $k$, the position $j$ was $2g$ (too good) if both $U_{1,j}(k)\geq k$ and $U_{2,j}(k)\geq k$, $2b$ (too bad) if both where $<k$, and $1g1b$ otherwise. Then, if we define an "interval" to be a subsequence starting immediately after a zero and ending with the next zero, one can measure the "goodness" of the sequence by measuring how good or bad is each interval.
It turns out that for $k=1$ we have:

Bad intervals have the form: $(1g1b,1g1b,\ldots,1g1b,2b)$ and the "sum" will be $-2$
Good intervals have the form: $(2g,\ldots,2g,1g1b,\ldots,1g1b,2b)$ and the sum is at least $+2$ and depends on the number of elements that are $2g$ at the beginning.
neutral intervals: $(2g,1g1b,\ldots,1g1b,2b)$ and the sum is zero.

However, I do not know how to extend this idea, or how to solve the general question.

REPLY [6 votes]: Answer: No this is not true.

For $m=5$, a counter example is: $(0, 2, 0, 2, 0, 2, 1, 2, 1, 1, 1, 0, 1, 2, 0)$. In this case we have:
$$\begin{align}
U(1) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
0 & 0 & 0 & 0 & 0 & 4 & 4 & 3 & 3 & 2 & 1 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0
\end{array}\right) \\
U(2) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
0 & 0 & 0 & 4 & 4 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\
1 & 2 & 1 & 3 & 2 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & 0
\end{array}\right) \\
U(3) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
0 & 4 & 4 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\
2 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 1 & 1 & 1 & 1 & 2 & 2 & 1
\end{array}\right) \\
U(4) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
4 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 1 & 0 & 0 \\
4 & 5 & 4 & 4 & 3 & 4 & 3 & 3 & 2 & 2 & 2 & 2 & 3 & 3 & 2
\end{array}\right) \\
U(5) &= \left(\begin{array}{rrrrrrrrrrrrrrr}
5 & 5 & 5 & 5 & 5 & 5 & 5 & 4 & 4 & 3 & 2 & 1 & 5 & 4 & 4 \\
5 & 5 & 4 & 5 & 4 & 5 & 4 & 4 & 3 & 3 & 3 & 3 & 5 & 5 & 4
\end{array}\right) 
\end{align}$$
Each $U(k)$ has only $14$ entries $\geq k$. 
To elaborate a bit: 

For $m\leq4$ the statement in your question is true, as can be checked by complete enumeration of all cases.
For $m=5$ if you mod out cyclic permutations there are precisely 20 counter examples:
$$\{(1, 1, 1, 1, 0, 2, 0, 2, 0, 0, 2, 0, 2, 1, 2), (1, 1, 1, 1, 0, 2, 0, 2,
0, 0, 0, 2, 2, 1, 2), (1, 1, 1, 1, 0, 2, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1,
1, 1, 0, 1, 2, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 2, 1, 0, 2, 0,
0, 2, 0, 2, 1, 2), (1, 1, 1, 0, 2, 1, 0, 0, 2, 0, 2, 0, 2, 1, 2), (1, 1,
1, 0, 0, 2, 0, 2, 0, 2, 1, 2, 0, 1, 2), (1, 1, 1, 0, 0, 2, 0, 2, 0, 2,
1, 0, 2, 1, 2), (1, 1, 1, 0, 0, 2, 0, 2, 0, 2, 0, 2, 1, 1, 2), (1, 1, 1,
0, 0, 2, 0, 0, 2, 2, 0, 2, 1, 1, 2), (1, 1, 0, 1, 2, 2, 1, 2, 2, 2, 0,
0, 1, 0, 0), (1, 1, 2, 1, 2, 2, 2, 2, 0, 1, 0, 1, 0, 0, 0), (1, 0, 1, 0,
1, 2, 1, 0, 2, 1, 2, 2, 2, 0, 0), (1, 0, 1, 0, 1, 0, 1, 2, 2, 1, 2, 2,
2, 0, 0), (1, 0, 1, 0, 1, 2, 1, 2, 2, 2, 0, 1, 2, 0, 0), (1, 0, 1, 0, 1,
2, 1, 2, 2, 2, 2, 0, 1, 0, 0), (1, 0, 1, 0, 1, 2, 1, 2, 2, 2, 0, 2, 1,
0, 0), (1, 0, 1, 0, 1, 2, 0, 1, 2, 1, 2, 2, 2, 0, 0), (1, 0, 1, 2, 1, 2,
2, 2, 0, 1, 2, 0, 1, 0, 0), (1, 0, 1, 2, 1, 2, 2, 2, 2, 0, 1, 0, 1, 0,
0)\}$$
I would expect that there counter examples for all $m\geq5$.<|endoftext|>
TITLE: On the ordered set of real numbers, does sheaf+cosheaf imply constant?
QUESTION [6 upvotes]: I have a technical question about unbounded chain complexes. I couldn't think of a descriptive title for it.
Let $P$ be a chain complex of contravariant functors on $\mathbf{R}$ (the real numbers regarded as an ordered set), valued in the category of abelian groups.
Suppose that for every real number $s$, both of the maps
$$
P(s) \to \varprojlim_{t < s} P(t), \qquad P(s) \leftarrow \varinjlim_{t > s} P(t)
$$
are quasi-isomorphisms, where the lims indicate homotopy limit and colimit.  Does it follow that all of the maps $P(s_1) \to P(s_2)$ are quasi-isomorphisms?
When the $P(s)$ have homology groups that are uniformly bounded above, this is a result of Kashiwara.  It plays a key role in the foundations of microlocal sheaf theory, especially for sheaves that are not necessarily constructible.
Dima Tamarkin showed me how to remove the boundedness assumption, if you replace abelian groups by $k$-vector spaces for a field $k$. In that case the abelian category of presheaves of vector spaces on $\mathbf{R}$ has homological dimension $1$, so the derived category has a simple structure.
I'm interested in the case where $P$ takes values in the $\infty$-category of spectra, where you can formulate the same question, but already I don't know what to expect for abelian groups.

REPLY [2 votes]: Here is one possible way to proceed.
(Caveat lector: I haven't checked all the details carefully.)
Denote by S the sphere spectrum and by N the (contractible) spectrum that implements a nullhomotopy for S.
A map A→B of spectra is a weak equivalence if and only if
for any map Σ^k(S)→A with a nullhomotopy Σ^k(N)→B for the composition Σ^k(S)→A→B (i.e., Σ^k(S)→Σ^k(N)→B is Σ^k(S)→A→B),
the map Σ^k(S)→A is itself nullhomotopic via Σ^k(N)→A, with the composition Σ^k(N)→A→B being homotopic to the original map Σ^k(N)→B relative boundary Σ^k(S).
I refer to such data as a lift for a nullhomotopy.
(Depending on the (co)fibrancy of the spectra involved this criterion can be either made more strict or less strict.)
Both S and N are homotopy compact objects in spectra, hence their corepresentable functors commute with homotopy filtered colimits.
Suppose we want to prove P_s→P_t is a weak equivalence for s>t.
By the limit condition, a map Σ^k(S)→P_s defines maps Σ^k(S)→P_{s'} for all s>s'
and a nullhomotopy Σ^k(N)→P_t defines nullhomotopies Σ^k(N)→P_{t'} for all t>t'.
Denote by r the supremum of all r such that s≥r and there is a compatible system of lifts Σ^k(N)→P_q of the above nullhomotopy for all r>q≥t.
By the limit condition, the compatible system of lifts for all r>q yields a lift Σ^k(N)→P_r.
(Initially, we start with r=t and the lift is already given to us.)
Suppose s>r, then by the colimit condition the nullhomotopy lifts through some P_{r'}, s>r'>r,
which gives us a compatible system of lifts for r' by the limit condition, which contradicts the definition of r.
Thus s=r and we have constructed a lift Σ^k(N)→P_s as desired.
Addendum:
Here is how one can use model categories to simplify the above
discussion by eliminating homotopies.
Start with the model category of simplicial spectra or simplicial symmetric spectra.
Replace P fibrantly in the injective model structure on functors
R→Spectra; this guarantees that X_p is fibrant and X_p→X_q is a fibration for all p>q.
The weak equivalence criterion in the first paragraph can be formulated for fibrations between fibrant spectra by requiring a strict lift (in the ordinary sense) with respect to S→N.
The (cofiltered) homotopy limits can be computed as strict limits using fibrancy,
and the (filtered) homotopy colimits can be computed as strict colimits
by the compact generatedness of the model category of simplicial (symmetric) spectra.
With these conventions, the argument in the second paragraph can now be interpreted strictly.<|endoftext|>
TITLE: Branching behavior in string diagrams/monoidal categories?
QUESTION [7 upvotes]: I am currently working through Peter Selinger's paper "Towards a Quantum Programming Language", and trying to connect it with what I already know about monoidal categories and string diagrams.
However, his flow charts involve branching behavior for measurements, and I don't know how to tie that to the categorical machinery.
More specifically, how could I take the following specification and represent it in a monoidal category? What additional structure do I need?
f: Int -> Int
f(x) = if {x > 10}
       then x+3
       else 0

My best guess would be to require coproducts, as that is how I would approach the problem in a Cartesian context.

REPLY [3 votes]: You can represent such a branching behaviour in bimonoidal categories (also known as rig categories).
In addition to your multiplicative monoidal structure $\otimes$, which is used to represent compound systems, you need another monoidal structure $\oplus$ to model the branching histories. A category is bimonoidal when it has these two structures and $\otimes$ distributes over $\oplus$.
Let us consider a slight variation on your example specification to illustrate that:
f(x,y) = if (x > 10)
         then x + 3
         else y + 3

You can model it like this:

$a : \text{Int} \rightarrow \text{Int} \oplus \text{Int}$ is the map that sends its input to the first component if it is greater than 10, and to the second component otherwise
$b : \text{Int} \rightarrow \text{Int}$ is the morphism that adds 3 to its argument
$c : \text{Int} \rightarrow I$ is the discarding map, where $I$ is the monoidal unit for $\otimes$.
$d : \text{Int} \oplus \text{Int} \rightarrow \text{Int}$ is the codiagonal map (mapping both sides of the sum to the same object).

Then your algorithm can be represented as $d \circ ((b \otimes c) \oplus (c \otimes b)) \circ (a \otimes 1_{\text{Int}})$. You can represent it graphically using sheet diagrams for bimonoidal categories, which generalize string diagrams for monoidal categories:

Manipulate diagram interactively
(The reason why I tweaked your example a bit is to demonstrate how the multiplicative monoidal structure can be used in these diagrams. Your example only had a single variable so it could be represented with the additive monoidal structure only, so with regular string diagrams for that monoidal structure.)<|endoftext|>
TITLE: How to stop worrying about enriched categories?
QUESTION [21 upvotes]: Recently I realized that ordinary category theory is not a suitable language for a big portion of the math I'm having a hard time with these days. One thing in common to all my examples is that they all naturally fit into the enriched categorical context.

2-Categories - Enriched in categories. Examples: Stacks ($BG$, $QCoh$) are 2-sheaves, 2-category of rings and bi-modules. 
DG categories - Enriched in chain complexes. Prime example: The dg category of chain complexes of $\mathcal{O}_X$-modules over $X$.
Topological/Enriched categories - Enriched in topological spaces/simplicial sets. Prime example: $\mathsf{Top}$.

I now have the impression that many of the difficulties I face in trying to learn about math that involves the three above originate in the gap between the ordinary categorical language and the enriched one. In particular, the natural constructions from ordinary category theory (limit, adjunctions etc.) are no longer meaningful and I'm practically blindfolded.
Is there a friendly introduction to enriched category theory somewhere where I can get comfortable with this general framework? Is it a bad idea to pursue this direction?

REPLY [5 votes]: Have a look around on my n-Lab 'home page': 
https://ncatlab.org/timporter/show/HomePage
and go down to the `resources'.  There are various  quite old sets of notes that look at simplicially enriched categories, homotopy coherence etc. and that may help you with homotopy limits, homotopy coherent / $\infty$-category ends and coends, etc.
With Cordier, I wrote a paper: Homotopy Coherent Category Theory, Trans. Amer. Math. Soc. 349 (1997) 1-54, which aimed to give the necessary tools to allow homotopy coherent ends and coends (and their applications) to be pushed through to the $\mathcal{S}$-enriched setting and so to be used `without fear' by specialists in alg. geometry, non-abelian cohomology, etc. 
You can also find stuff in my Menagerie notes, mentioned on that Home Page.<|endoftext|>
TITLE: Could this unexpected bias in the distribution of consecutive primes have any impact on the security of encryption algorithms?
QUESTION [16 upvotes]: In a recent paper a quite unexpected result about a new pattern in prime numbers emerged:
Unexpected biases in the distribution of consecutive primesby Oliver, R. J. L.; Soundararajan, K. (Submitted on 11 Mar 2016)

While the sequence of primes is very well distributed in the reduced
  residue classes (mod $q$), the distribution of pairs of consecutive
  primes among the permissible $ϕ(q)^2$ pairs of reduced residue classes
  (mod $q$) is surprisingly erratic. This paper proposes a conjectural
  explanation for this phenomenon, based on the Hardy-Littlewood
  conjectures. The conjectures are then compared to numerical data, and
  the observed fit is very good.

My question
Could this result have any impact on the security of encryption algorithms which are based on prime numbers?

REPLY [3 votes]: There's another discussion of the bias on math.SE.  The phenomenon is not limited to consecutive primes, but its repulsion effect seems to be attenuated (or masked by noise) too quickly to have powerful consequences.  Oliver and Soundararajan note there may be implications for semiprimes, which would be of cryptographic interest.
There are hints that the prime numbers are leaking information to nearby composites.  It's conceivable that one might be able to factorize an arbitrary composite quickly by collecting enough information from the primes near it.<|endoftext|>
TITLE: invariants that can be measured by Local Cohomology
QUESTION [7 upvotes]: What invariants can be measured by Local Cohomology (and what application it has)?

As an example of what I mean:   
Local Cohomology can measure invariants like depth and dim. So in some cases Local Cohomology help us detect Cohen-Macaulay-ness: Let $(R,m)$ and $(S,n)$ be local rings and $S$ is an $R$-Algebra via homomorphism $f:R\to S.$   Assume that for all $i$ we have following isomorphism $H^i_n(S)\cong H^i_m(R)$ of $R$-modules. Then knowing Cohen-Macaulay-ness of $R$ we know if $S$ is Cohen-Macaulay.  
Thank you.

REPLY [2 votes]: There are many invariants can be measured by local cohomology. Here are a few examples.
I assume $(R, \mathfrak{m})$ is a complete local ring of dimension $d$. The non Cohen-Macaulay locus of $R$ is defined as follows
$nCM(R) = \{\mathfrak{p} \in \mathrm{Spec}(R) \ | \ R_{\mathfrak{p}} \text{ is not Cohen-Macaulay}\}.$
We can ask some questions about the non Cohen-Macaulay locus as: Is $nCM(R)$ a closed subset of $\mathrm{Spec}(R)$? What is the dimension of $nCM(R)$? Local cohomology give us the answers of these questions.
Indeed, let $\mathfrak{a}_i = \mathrm{Ann}(H^i_{\mathfrak{m}}(R))$, $i \ge 0$, and $\mathfrak{a} = \mathfrak{a}_0 \ldots \mathfrak{a}_{d-1}$. It is easy to see that if $R$ is Cohen-Macaulay i.e. $nCM(R) = \emptyset$, then $\mathfrak{a} = R$ since $H^i_{\mathfrak{m}}(R) = 0$ for all $i = 0, \ldots, d-1$. The Faltings annihilator theorem give us the answer in general case (see Brodmann-Sharp: local cohomology)
Theorem 1. Suppose $R$ is equidimensional. Then we have 
(i) $nCM(R) = Var(\mathfrak{a})$ is a closed subset of $\mathrm{Spec}(R)$.
(ii) The dimension of $nCM(R)$ is $\dim R/\mathfrak{a}$.
We can define a generalization of the class of Cohen-Macaulay ring as follows.
Definition 2. The ring $R$ is called generalized Cohen-Macaulay if $\dim R/\mathfrak{a} \le 0$, that is $H^i_{\mathfrak{m}}(R)$ has finite length for all $i = 0, \ldots, d-1$.
It should be noted that the affine cone of any nonsingular projective variaty is generalized Cohen-Macaulay. Working with generalized Cohen-Macaulay rings are much more difficult than Cohen-Macaulay rings but we have some chances to do so with $\ell(H^i_{\mathfrak{m}}(R))$. For example, let $\mathfrak{q}$ be a parameter ideal of $R$. Then we always have $\ell(R/\mathfrak{q}) \ge e(\mathfrak{q})$ the multiplicity of $q$. Moreover $R$ is Cohen-Macaulay if and only if $\ell(R/\mathfrak{q}) = e(\mathfrak{q})$ for some (and any) $\mathfrak{q}$. When $R$ is generalized Cohen-Macaulay, the different $\ell(R/\mathfrak{q}) - e(\mathfrak{q})$ are bounded above by an invariant in terms of local cohomology. More precisely we have
$$\ell(R/\mathfrak{q}) - e(\mathfrak{q}) \le \binom{d-1}{i} \ell(H^i_{\mathfrak{m}}(R))$$
(see Trung's paper: towards a theory of generalized Cohen-Macaulay modules, Nagoya, 1986)<|endoftext|>
TITLE: Does infinitesimal variance imply continuity?
QUESTION [8 upvotes]: Let $u:[0,1]\to\mathbb{R}^n$ be a bounded Borel function.
It is well-known that if, for any compact interval $I\subseteq [0,1]$,
$$ \int_I|u-u_I|^2\le C|I|^{1+\alpha} $$
for some $C,\alpha>0$ (here $u_I:=\frac{1}{|I|}\int_I u$), then $u$ is in fact $\frac{\alpha}{2}$-Holder continuous: this was first proved by Campanato in 1963.
Q: Is it true that if
$$ \int_I|u-u_I|^2\le |I|\omega(|I|) $$
for any $I$ then $u$ is continuous? Here $\omega$ denotes an arbitrary modulus of continuity. If this is false in general, can one characterize the $\omega$'s for which this is true?

REPLY [8 votes]: Since $u$ is assumed bounded, your condition is equivalent to
$$
\frac{1}{|I|} \int_I |u-u_I|\, dx =o(1)
$$
as $|I|\to 0$, uniformly in $I$, and this is the condition that defines VMO.
So you are asking if functions in $VMO\cap L^{\infty}$ are continuous, and this is known to be false. This classical paper by Sarason introduced VMO; Theorem 1(iv) there answers your question, modulo facts about the Hilbert transform. Here's a more direct reference.<|endoftext|>
TITLE: Which Heyting algbras arise out of some elementary topos which satisfies the ultrafilter principle?
QUESTION [6 upvotes]: It is known that AC implies LEM constructively, and also that AC implies the ultrafilter principle. Is there a similar relationship between the ultrafilter principle and classical logic? In other words, are there any inference rules of propositional logic which are classically-but-not-constructively admissible, but are constructively admissible assuming the ultrafilter principle?
Edit: François G. Dorais points out that defining ultrafilters in constructive settings can be somewhat subtle. I'd naively guess that the "right" way to state it would be that every consistent boolean algebra $B$ admits a homomorphism $B \to 2$. This would be "right" in the sense of implying the completeness theorem of classical FOL. I think this works in François' example, since we can take the consistent boolean algebra $B$ of decidable subsets of $\mathbb{N}$, quotient by the Fréchet filter $F$, and then take the preimage of $\top$ under $B \to B/F \to 2$ to be our desired ultrafilter.
To clarify the question, consider this: we can associate to every elementary topos $E$ the subobject lattice of the terminal object $1$, which is equivalently the Heyting algebra of global elements of the subobject classifier $\Omega$. My question would then be to ask which Heyting algebras can arise this way from an elementary topos $E$ whose internal logic satisfies the ultrafilter principle.

REPLY [5 votes]: I think your formulation of the ultrafilter principle implies the de Morgan law.
Let $U$ be any proposition, consider the boolean algebra $A$ freely generated by an element $v$.
so $A = \{ 0,1,v,\neg v \}$.
Consider the following equivalence relation on $A$:
$$ \{(0,0),(1,1),(v,v),(\neg v, \neg v) \} \cup U \times \{ (0,v),(1,\neg v) \} \cup (\neg U) \times \{ (1,v),(0,\neg v) \} $$
(I didn't write it, but of course, as soon as you have an element $(x,y)$ in the relation you also add the element $(y,x)$ in the relation)
A long and uninteresting case by case treatment shows that it is an equivalence relation compatible to the boolean algebra structure and hence the quotient is again a boolean algebra $B$.
$0$ and $1$ are different in $B$, so by the ultrafilter principle there should exist a morphism $B \rightarrow 2$. the image of $v$ is either $0$ or $1$.
If it is zero, then as $\neg U \Rightarrow v=1$ one has $\neg \neg U$ and if it is one then $\neg U$ for the same reason hence $\neg U $ or $\neg \neg U$ which is an equivalent form of De Morgan's law.
If you want to avoid this kind of problem you need to restrict to decidable boolean algebra, that is those which satisfies $\forall v \in B,v=0$ or $v \neq 0$ (this also prevent the example in François G dorsais's answer)<|endoftext|>
TITLE: Homological contractibility of a prestack
QUESTION [7 upvotes]: This question is in reference to Gaitsgory's preprint Contractibility of the space of rational maps. On p. 5 of the preprint, Gaitsgory defines a prestack $\mathscr{Y}$ (say over affine $\mathbb{C}$-schemes) to be homologically contractible if the functor $\text{Vect}\longrightarrow\mathfrak{D}(\mathscr{Y})$ taking $V\mapsto V\otimes\omega_{\mathscr{Y}}$ is fully faithful. Here $\mathfrak{D}(\mathscr{Y})$ denotes the DG category of D-modules on $\mathscr{Y}$. Further down the same page, Gaitsgory mentions that homological contractibility of $\mathscr{Y}$ is equivalent to the condition that $H_{\bullet}(\mathscr{Y}(\mathbb{C})^{\text{top}},\mathbb{Q})\cong\mathbb{Q}$. What is the argument for the equivalence of these two formulations of homological contractibility? I would be happy to see the argument even in a simple case, like the case where $\mathscr{Y}$ is a smooth complex variety.

REPLY [2 votes]: This is proven in some detail in section 3 Gaitsgory's writeup of his the Atiyah-Bott formula. He starts with the fully faithfulness definition, then proves the equivalance with homological statement at the very end of the section.<|endoftext|>
TITLE: Separable von Neumann algebra
QUESTION [11 upvotes]: What is the simplest argument which shows that each infinite dimensional von Neumann algebra is not separable (in the norm topology)? It seems that this is a kind of folklore: at least I never saw the proof of this fact.

REPLY [4 votes]: Suppose that we have an infinite-dimensional von Neumann algebra. In particular, this is a C*-algebra so it contains an infinite-dimensional abelian C*-algebra $A$. Consequently, $\overline{A}^{{\rm WOT}}$ is a non-separable abelian von Neumann algebra. No need to invoke the fact that it is isomorphic to $L_\infty(\mu)$ for some measure $\mu$, simply use the Borel functional calculus to build an uncountable, closed discrete subset in $\overline{A}^{{\rm WOT}}$ just like you would do in $L_\infty$.
The fact that every infinite-dimensional C*-algebra contains an infinite-dimensional abelian subalgebra is fairly elementary and can be found in Kadison-Ringrose.<|endoftext|>
TITLE: Do smooth manifolds admit linear atlases?
QUESTION [9 upvotes]: There is a theorem of Whitney showing that a smooth manifold can be endowed with a compatible real-analytic atlas (later, it was proven that this analytic structure is essentially unique).
I am curious, how much stronger structures can be put on a smooth manifold in a compatible way? Most importantly, is it possible to find an atlas such that its transition maps be elements of $GL(n)$ ($n$ being the dimension of the manifold)?
The only related thing that I have found is the concept of piecewise-linear manifold, which seems not to be what I am looking for.

REPLY [3 votes]: In dimension 3, radiant affine manifolds have been classified by Choi. To quote from the AMS Math Review: 

The main result is a decomposition theorem, which says that such a manifold admits a decomposition along finitely many disjoint totally geodesic tori or Klein bottles into two kinds of pieces. The first kind are convex radiant affine 3-manifolds, i.e. those for which the universal cover is real projectively homeomorphic to a convex domain in affine space. The second kind of piece is the suspension of a real projective surface of a special kind by a real projective automorphism. If Σ is a real projective (n−1)-manifold and ϕ is a real projective automorphism of Σ, there is a natural radial affine structure on the mapping torus of the automorphism ϕ, which is a quotient by a cyclic group of a component of the complement of the zero section of the tautological line bundle over Σ coming from its projective structure. 
     If M is a closed manifold with a radiant affine structure, there is a natural developing map from the universal cover M˜ toRn−{0}, unique up to composition with an element of GL(n,R). The radial vector field X=∑ixi∂/∂xi on Rn−{0} pulls back by the developing map to a vector field on M˜ which by naturality covers a vector field on M. This vector field is known as the radial vector field on M. Carrière asked whether every compact radiant affine 3-manifold admits a total cross section to the radial flow. Together with some results of Barbot and the author, contained in an appendix, the decomposition theorem answers the Carrière conjecture in the positive. Together with some basic results in 3-manifold topology, this implies that every compact radiant affine 3-manifold with empty or totally geodesic boundary is homeomorphic to a Seifert space with Euler number zero, or is finitely covered by a surface bundle over a circle with fiber homeomorphic to a compact surface of Euler characteristic zero.<|endoftext|>
TITLE: Counting the number of permutations of $(1,\ldots,i,\ldots,j,\ldots,m)$, where $i < j$ and number of inversions is $k$
QUESTION [5 upvotes]: How can I prove the following:
$d^{ij}(m,k) > d^{ji}(m,k)$ for all $k < \frac{1}{2}\binom{m}{2},$
where $d^{ij}(m,k)$ denotes the number of permutations of $(1,\ldots,i,\ldots,j,\ldots,m)$ with $k$ inversions and where $i$ is on the left of $j.$ Similarly $d^{ji}(m,k)$ denotes the number of permutations where $i$ is on the right to $j.$ 
Example: $m=4$, $k=1.$ We have the following permutations $2134,1243,1324.$ So, $d^{12}(4,1)=2$ and $d^{21}(4,1)=1.$

REPLY [4 votes]: Denote $\Delta^{ij}(m,k)=d^{ij}(m,k)-d^{ji}(m,k)$. By reverting the permutation we observe that $d^{ij}(m,k)=d^{ji}(m,{m\choose 2}-k)$, so 
$$
  \Delta^{ij}(m,k)=-\Delta^{ij}(m,{m\choose 2}-k).
  \qquad (*)
$$
Now we prove the statement by the induction on $m$. Firstly, we present the step, and then we establish the base case (which is $i=1$, $j=m$).
For the step from $m$ to $m+1$, assume that $j\leq m$ and $k<{m+1\choose 2}$. Take any permutation counted in $d^{ij}(m+1,k)$ and remove the number $m+1$ (say, from the $(m+1-t)$th position); the number of inversions decreases by $t$. Thus,
$$
  d^{ij}(m+1,k)=d^{ij}(m,k)+d^{ij}(m,k-1)+\dots+d^{ij}(m,k-m).
$$
Similarly, we write the formula for $d^{ji}(m+1,k)$; this gives us
$$
  \Delta^{ij}(m+1,k)=\Delta^{ij}(m,k)+\Delta^{ij}(m,k-1)+\dots+\Delta^{ij}(m,k-m).
$$
If $k\leq {m\choose 2}$, then the induction hypothesis immediately yields $\Delta^{ij}(m+1,k)>0$. Otherwise, several first terms cancel due to $(*)$, but there are still some remaining terms, and they are positive in view of the induction hypothesis again.
If $j=m+1$ but $i>1$, then we may similarly remove $1$ from the permutation. So we are left with the base case only.
Denote by $n(m,k)$ the total number of permutations on $m$ elements with $k$ inversions. Now, if we remove $i=1$ and $j=m$ from the permutations counted in $d^{1m}(m,k)$ and $d^{m1}(m,k)$, the number of inversions reduces controllably, provided that the positions of $i$ and $j$ are known. Indeed, if $I$ and $J$ are their positions, then the number of inversions reduces by $(I-1)+(m-J)$ (if $I<J$) and by $(I-1)+(m-J)+1$ otherwise. This way, we obtain
$$
  d^{1m}(m,k)=n(m-2,k)+2n(m-2,k-1)+\dots+(m-1)n(m-2,k-(m-2))
$$
and
$$
  d^{m1}(m,k)=n(m-2,k-(2m-3))+2n(m-2,k-(2m-2))+\dots+(m-1)n(m-2,k-(m-1)).
$$
Now the inequality $\Delta^{1m}(m,k)>0$ follows from the unimodularity (and symmetry) of the sequence $n(m-2,0), n(m-2,1),\dots,n(m-2,{m-2\choose 2})$. This unimodularity can be proved similarly to our step above (and, as I assume, is well-known).<|endoftext|>
TITLE: On the tensor product of presentable categories
QUESTION [5 upvotes]: I am trying to understand how the tensor product of presentable categories works: let $\otimes\colon {\cal A}\times {\cal B}\to {\cal A}\otimes{\cal B}$ the universal bilinear functor corresponding to $\text{id}_{{\cal A}\otimes{\cal B}}$ under the correspondence defining ${\cal A}\otimes{\cal B}$,
$$
\text{Bilin}({\cal A}\times{\cal B}, {\cal E})\cong \text{Func}({\cal A}\otimes{\cal B},{\cal E}).
$$
What is the answer to these questions?

Does the essential image of $\otimes$ generate ${\cal A}\otimes{\cal B}$ under colimits? This seems the analogue of $V\otimes W$ being made by formal sums of monomials. Does $\otimes$ have some other remarkable properties (fullness, faithfulness, commutes/reflects/creates co-limits?)
We have a fairly explicit (albeit constructed by absolute nonsense) model for ${\cal A}\otimes{\cal B}$, that is $\text{Func}({\cal A}^\text{op}, {\cal B})_R$ (functors ${\cal A}^\text{op}\to \cal B$ which commute with limits (and filtered colimits? I see different definitions here). This means that to $\otimes$ correspond a canonical functor ${\cal A}\times {\cal B} \to \text{Func}({\cal A}^\text{op}, {\cal B})_R$, sending $(A,B)$ to... who?
The adjoint functor theorem gives $A\otimes -$ and $-\otimes B$ right adjoints $A/-$ and $-\backslash B$; how are these functors ${\cal A}\otimes{\cal B}\to {\cal A,B}$ defined? There are Kan extensions one can write, but...

REPLY [7 votes]: I think in 2 the correct definition is that $\text{Func}({\cal A}^\text{op}, {\cal B})_R$ denotes the category of functors which are right adjoints or equivalently in this case, functors that preserve limits. Indeed, $\text{Func}({\cal A}^\text{op}, {\cal B})_R \cong \left(\text{Func}({\cal A}, {\cal B}^\text{op})_L\right)^\text{op}$, and because $\cal A$ is presentable a functor ${\cal A} \to {\cal B}^\text{op}$ is a left adjoint if and only if it preserves colimits.
(I think I know why you mentioned filtered colimits though: if $\cal C$ and $\cal D$ are presentable, then a functor $\cal C \to \cal D$ is a right adjoint if and only if it preserves limits and $\kappa$-filtered colimits for some sufficiently large regular cardinals $\kappa$; but notice that if $\cal A$ is presentable then $\cal A^\mathrm{op}$ is never presentable, so this does not apply here!)
OK, so what is $(A,B)$ sent to in $\text{Func}({\cal A}^\text{op}, {\cal B})_R$? Let $\text{ev}_A : \text{Func}({\cal A}^\text{op}, {\cal B})_R \to {\cal B}$ be the functor of evaluation at $A \in \cal A$ and let $L_A : {\cal B} \to \text{Func}({\cal A}^\text{op}, {\cal B})_R$ be its left adjoint, then the pair $(A,B)$ is sent to $L_A(B) \in \text{Func}({\cal A}^\text{op}, {\cal B})_R$.
I tried to make this more explicit but failed: it tempting to ignore the $_R$ and just look at the left adjoint to $ev_A : \text{Func}({\cal A}^\text{op}, {\cal B}) \to {\cal B}$, which is $\hat{L}_A(B) = \hom(-,A) \times B$, but since $\hom(-,A) \times B$ (almost?) never preserves limits, this $\hat{L}_A$ can't be the $L_A$ we're looking for!
You can find a proof of my $L_A(B)$ claim by reading between the lines in the proof of Proposition 4.8.1.16 of Lurie's Higher Algebra.
EDIT: I finally understood what Mike Shulman is saying, and it's the same as what I just wrote! Oh well, I'll leave this here in case anyone finds it easier to read than Mike's version.<|endoftext|>
TITLE: A question on the commutativity degree of the monoid of subsets of a finite group
QUESTION [5 upvotes]: The commutativity degree $d(G)$ of a finite group $G$ is defined as the ratio
$$\frac{|\{(x,y)\in G^2 | xy=yx\}|}{|G|^2}.$$It is well known that $d(G)\leq5/8$ for any finite non-abelian group $G$. If $P(G)$ is the monoid of subsets of $G$ with respect to the usual product of group subsets, then the commutativity degree $d(P(G))$ of $P(G)$ can be defined similarly: $$d(P(G))=\frac{|\{(A,B)\in P(G)^2 | AB=BA\}|}{|P(G)|^2}.$$Which are the connections between $d(G)$ and $d(P(G))$? Is there a constant $c\in (0,1)$ such that $d(P(G))\leq c$ for any finite non-abelian group $G$?
Additional Question: Let $P_k(G)$ be the subset of $P(G)$ consisting of all $k$-subsets of $G$ and $$d(P_k(G))=\frac{|\{(A,B)\in P_k(G)^2 | AB=BA\}|}{|P_k(G)|^2}.$$Clearly, $d(P_1(G))=d(G)$ and, for every $k\in\{1,2,3\}$, $G$ is abelian iff $d(P_k(G))=1$. A similar question can be asked for $d(P_k(G))$, $k=2,3$: is there a constant $c_k\in (0,1)$ such that $d(P_k(G))\leq c_k$ for any finite non-abelian group $G$?

REPLY [5 votes]: @Derek Holt is completely right: as $|G|\to \infty$, the fraction of subsets $(A,B)$ with $AB=G$ tends to $1$, so there is no such $c$.
Indeed, assume that $|G|=n$. Let us choose the subsets $A$ and $B$ uniformly and independently. Fix any $g\in G$; there are $n$ pairs $(a,b)$ with $ab=g$, each pair belongs to $A\times B$ with probability $1/4$, and these events are independent for distinct pairs. Thus the probability that $g\notin AB$ is $(3/4)^n$, so the probability that $AB\neq G$ is at most $n(3/4)^n$ which tends to $0$ as $n\to\infty$. 
To conclude: if $n$ is large, almost all pairs $(A,B)\in P(G)\times P(G)$ satisfy $AB=BA=G$.<|endoftext|>
TITLE: Super-plethysm?
QUESTION [10 upvotes]: Let $U$ be a representation of $S_m$ and $V$ a representation of $S_n$.  Then the representation $\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}(U^{\otimes{n}}\otimes V)$ has a nice interpretation in terms of symmetric functions.  If $\operatorname{ch}(U)$ and $\operatorname{ch}(V)$ are the Frobenius characteristics of $U$ and $V$ (symmetric functions of degree $m$ and $n$), then
$$\operatorname{ch}\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}(U^{\otimes{n}}\otimes V)= \operatorname{ch}(V)[\operatorname{ch}(U)],$$
where the square brackets denote plethysm.  I am interested in a graded version of this result.
More specifically, let $X$ be a space with an action of $S_m$ and $Y$ a space with an action of $S_n$.  I am interested in the representation
$$\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}\left(H^*(X^n)\otimes H^*(Y)\right)$$ of $S_{mn}$.  The naive statement would be that
$$\operatorname{ch}\operatorname{Ind}_{S_m\wr S_n}^{S_{mn}}\left(H^*(X^n)\otimes H^*(Y)\right) = \operatorname{ch}(H^*(Y))[\operatorname{ch}(H^*(X))],$$
where $\operatorname{ch}(H^*(X))$ and $\operatorname{ch}(H^*(Y))$ are symmetric functions of degree $m$ and $n$ with coefficients in the polynomial ring $\mathbb{Z}[t]$.  
I'm pretty sure that this statement is correct when the cohomology of $X$ is concentrated in even degree, but otherwise it is wrong.  The issue is that, at the level of $S_n$-representations, the Kunneth isomorphism $H^*(X^n)\cong H^*(X)^{\otimes n}$ needs to be interpreted in terms of super-vector spaces: the action of the simple transposition $(i,i+1)$ picks up a sign when the two cohomology classes being swapped have odd degree.  For example, if $X=S^1$ and $n=2$, then $H^2(S^1\times S^1)\cong H^1(S^1)\otimes H^1(S^1)$ is the sign representation of $S_2$ rather than the trivial representation.
So my question is:  Is there a "super version" of plethysm for symmetric functions with coefficients in $\mathbb{Z}[t]$ for which the last displayed equation is correct?  
I know that the answer is tautologically "yes"--one can just translate from symmetric functions to representations, do the induction, and translate that, and take this as the definition of super-plethysm.  But I'm looking for something explicit enough to allow me to do calculations in SAGE.

REPLY [6 votes]: This amounts to study composition of "linear species" in the category of complexes.
The correct way to handle these computations using plethysm is
to introduce an auxiliary variable $t$ and to weight the cohomology $H^i$ with the weight $(-t)^i$. The plethysm must act on $t$ by $p_n(t)=t^n$.
I do not know a written reference.<|endoftext|>
TITLE: Does X(13) have potentially good reduction at 13?
QUESTION [21 upvotes]: The complete level modular curve $X(p)$ does not have potentially good reduction at $p$ for any $p \neq 2,3,5,7,13$ because then there are cusp forms on $X_0(p)$ showing up in the cohomology of $X(p)$, whose associated Galois representations have unipotent local monodromy at $p$, which contradicts potentially good reduction.
$X(p)$ clearly does have good reduction at $p$ for $p=2,3,5$ because it has genus $0$ and a rational point.
$X(7)$ nonobviously has potential good reduction at $p=7$ where $X(7)$, the Klein quartic curve in $\mathbb P^2$, degenerates to a smooth hyperelliptic curve. See, for instance, Elkies's paper on the Klein quartic.
So $p=13$ is the only remaining case. What's the answer there?
A good first step might be to try to answer the question for $X_1(13)$, which only has genus $2$.

REPLY [12 votes]: Regarding Will's question on $X(13)$, it follows from Michael Stoll's answer and the following lemma that $X(13)$ does not have potentially good reduction.
Lemma. Let $X\to Y$ be a finite morphism of smooth projective geometrically connected curves over a number field $K$. Suppose that $Y$ has non-zero genus. If $X$ has good reduction over $O_K$, then $Y$ has good reduction over $O_K$.
Proof. This follows from the existence of Neron models for hyperbolic curves (not only abelian varieties); see Corollary 4.7 in Liu-Tong http://arxiv.org/abs/1312.4822 . QED
One now concludes as follows. Suppose that $X(13)$ has good reduction everywhere over some number field $K$. Replacing $K$ by a finite field extension if necessary, there  is a (natural) finite morphism $X(13)\to X_1(13)$. Since  $X_1(13)$ has genus two, it follows from the above lemma  that $X_1(13)$ has good reduction over $O_K$. This contradicts Stoll's answer.<|endoftext|>
TITLE: Twisted Day convolution
QUESTION [5 upvotes]: Has anyone studied a version of Day convolution for an enriched presheaf category $V^{A^{\mathrm{op}}}$ where the monoidal structure of $V$ is "twisted" on one side by an action of $A$?  I'm thinking of modifying the standard formula
$$(F \otimes G)(a) = \int^{b,c} F(b)\otimes G(c) \otimes A(b\otimes c,a)$$
to something like
$$(F \otimes G)(a) = \int^{b,c} F(b)\otimes G(c)^b \otimes A(b\otimes c,a)$$
where $G(c)^b$ denotes an action of $b\in A$ on $G(c)\in V$.

REPLY [3 votes]: The following is a proof that the twisted monoidal structure $\otimes^\rho$ is a convolution on $[A, V]$, where $A$ is a monoidal category, $V$ the cosmos on which $A$ is enriched, and $\rho\colon A\times V\to V$ a monoidal action (in the sense that $(v^a)^b \cong v^{a\cdot b}$, $v^i \cong v$ where $i$ is the monoidal identity of $A$, and $(v\otimes w)^a \cong v^a \otimes w^a$).
To make things work, I must assume that

Every object $a$ of the monoidal structure $(A, \cdot, i)$ has a monoidal inverse $a^{-1}$;
The functor $(-)^a \colon V\to V$ is cocontinuous.

This is somehow restricive, but since we know (coends, Prop. A.3) that (strong) promonoidal structure in $C$ correspond bijectively with convolution structures on $[A, V]$, there must exist a convolution product such that
$$
F\otimes^\rho G \cong \int^{xy} Fx \otimes Gy \otimes P_\rho(x,y;-)
$$
so that the action $(x,y)\mapsto (Gy)^x$ is "absorbed" into the promonoidal functor $P\colon A^\text{op}\times A^\text{op} \times A \to V$. 
I'm not able to find the $P_\rho$, but I'm able to find sufficient assumptions (those above) to ensure associativity. As I mentioned in the comment, I come up with a "twisted ninja Yoneda lemma" (coends, Prop. 2.1), and I learned something more about coend-nonsense, so thank you (-:
Let's start the coend-juggling:
\begin{align}
[(F \otimes^\rho G)\otimes^\rho H](a) &= \int^{bc} (F\otimes^\rho G)(b)\otimes (Hc)^b \otimes A(b\cdot c, a) \\
 & \cong \int^{bcxy} Fx \otimes (Gy)^x \otimes A(x\cdot y, b)\otimes (Hc)^b \otimes A(b\cdot c, a) \\
 & \cong \int^{bcxy} Fx \otimes (Gy)^x \otimes (Hc)^b \otimes A(x\cdot y, b)\otimes A(b\cdot c, a) \\
 & \cong \int^{cxy}Fx \otimes (Gy)^x \otimes \left( \int^b (Hc)^b \otimes A(x\cdot y, b)\otimes A(b\cdot c, a) \right) \\
 & \cong \int^{cxy}Fx \otimes (Gy)^x \otimes (Hc)^{x \cdot y}\otimes A((x\cdot y)\cdot c, a)
\end{align}
where I applied the ninja Yoneda lemma to the functor $S(b) = (Hc)^b \otimes A(b\cdot c, a)$ ($S$ results as the composition $b\overset{\Delta}\mapsto (b,b) \overset{\hat S}\mapsto (Hc)^b \otimes A(b\cdot c, a)$; this type of dependence is essential, and implicit in the definition). We now have to manipulate the other parenthesization of $F,G,H$: to obtain the same object, we need the assumptions of cocompleteness of the action and invertibility:
\begin{align}
[F \otimes^\rho (G\otimes^\rho H)](a) & \cong \int^{mn} Fm \otimes ((G\otimes^\rho H)(n))^m \otimes A(m\cdot n, a) \\
& \cong \int^{mnzt} Fm \otimes \Big( Gz\otimes (Ht)^z \otimes A(z\cdot t, n)\Big)^m \otimes A(m\cdot n, a) \\
& \cong \int^{mnzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes A(z\cdot t, n)^m \otimes A(m\cdot n, a) \\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \left( \int^n A(z\cdot t, n)^m \otimes A(m\cdot n, a) \right)\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \left( \int^n A(z\cdot t, n) \otimes A(m\cdot n, a)^{m^{-1}} \right)^m\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \Big( A(m\cdot (z\cdot t), a)^{m^{-1}} \Big)^m\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes A(m\cdot (z\cdot t), a)
\end{align}
where I applied the assumptions, and the ninja Yoneda lemma.
The same manipulations show that the presheaf $\hom(i,-)$ is the monoidal unit.
\begin{align}
F \otimes^\rho I & = \int^{bc} Fb \otimes A(i,c)^b \otimes A(b\cdot c ,a ) \\
& \cong \int^b Fb \otimes \left( \int^c A(i,c) \otimes A(b\cdot c, a)^{b^{-1}} \right)^b \\
&\cong \cdots
\end{align}<|endoftext|>
TITLE: Symmetric polynomial separating points
QUESTION [5 upvotes]: I've been looking for references/answers to this problem for several days and I couldn't find anything.
If we consider the closed unit ball $B$ in $\mathbb C^2$ then for any point $(z_1,z_2)\notin B$ we can find a polynomial $P$ of two variables with $$|P(z_1,z_2)|>\sup_{(w_1,w_2)\in B}|P(w_1,w_2)|.$$ 
In fact $P$ can be chosen to be just a line, but this statement written with polynomials is related to a more general theory of polynomially convex sets. See for instance https://www.encyclopediaofmath.org/index.php/Polynomial_convexity.
My question is, can we find $P$ being symmetric i.e. $P(w_1,w_2)=P(w_2,w_1)$ for all $w_1,w_2\in\mathbb C$?
I found a non-costructive proof of the result but it uses several strong results. However, the question can be easily solved if we consider the polydisc of center zero and radius one instead of $B$. We can consider the polynomial $P_n(w_1,w_2)=w_1^n+w_2^n$ with $n$ big enough such that the argument of $z_1$ and $z_2$ is almost the same. However, this argument does not work for $B$ since we can have that $z_1$ and $z_2$ both have modulus smaller than one, and as a consequence when $p$ is big the values of the polynomials $P_n$ go to zero at $(z_1,z_2)$. 
Any reference would be appreciated. Thanks in advance.

REPLY [6 votes]: I think combining the two tricks in your post answers the question. Let $(x_1, x_2)$ be a point not in $B$. Set
$$\ell(z_1, z_2) = \frac{\overline{x_1}}{\sqrt{|x_1|^2+|x_2|^2}} z_1 + \frac{\overline{x_2}}{\sqrt{|x_1|^2+|x_2|^2}} z_2.$$
Then $\ell$ is a linear polynomial with $\ell(x_1,x_2)>1$ but $|\ell(y_1, y_2)| \leq 1$ on $B$.
Set $f_n(z_1, z_2) = \ell(z_1, z_2)^n + \ell(z_2, z_1)^n$. Then $f$ is a symmetric polynomial and $|f| \leq 2$ on $B$, but $\lim \sup_{n \to \infty} |f_n(x_1, x_2)|$ is infinite.<|endoftext|>
TITLE: Steenrod operations in etale cohomology?
QUESTION [25 upvotes]: For $X$ a topological space, from the short exact sequence 
$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0 $$
we get a Bockstein homomorphism 
$$H^i(X, \mathbb{Z}/2) \rightarrow H^{i+1}(X, \mathbb{Z}/2)$$
This is also known as the Steenrod square $Sq^1$. 
Now suppose instead that $X$ is a variety over a (not algebraically closed) field. We still get a sequence
$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0 $$
inducing a Bockstein homomorphism in etale cohomology
$$H^i_{et}(X, \mathbb{Z}/2) \rightarrow H^{i+1}_{et}(X, \mathbb{Z}/2).$$
However, there is also a short exact sequence 
$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4(1) \rightarrow \mathbb{Z}/2 \rightarrow 0 $$
where $\mathbb{Z}/4(1)$ denotes the Tate twist, probably less confusingly written as $\mu_4$. This also induces a [presumably different] Bockstein map in etale cohomology 
$$H^i_{et}(X, \mathbb{Z}/2) \rightarrow H^{i+1}_{et}(X, \mathbb{Z}/2).$$
Question: Which of these is the "right" Bockstein homomorphism? 
This question is a little open-ended. My main criterion for "right" is that the map be "$Sq^1$ on etale cohomology", meaning it fits into an action of the Steenrod algebra. 
There are other possible criteria. For instance, a literature search revealed that people have defined notions of "Bockstein homomorphism" and "Steenrod operations" on Chow rings, motivic cohomology, ... so "right" could also mean "compatible with these other things". (Hopefully the answer is the same.) 
Relevant literature:

Jardine's paper https://link.springer.com/chapter/10.1007%2F978-94-009-2399-7_5 gives a version with $\mathbb{Z}/4$ 
Guillou and Weibeil (http://arxiv.org/pdf/1301.0872.pdf) describe a $Sq^1 : H^k(\mu_2^i) \rightarrow H^{k+1}(\mu_2^{2i})$, so the weight has doubled. I don't know if it fits into a Bockstein story. 

Unfortunately, I can't really parse what's happening in these papers.  
Some motivation/example: For $X \subset Y$ a closed embedding of smooth varieties of codimension $r$, I have a cycle class $[X] \in H^{2r}_X(Y; \mathbb{Z}/2)$. I would like to show that $Sq^1 [X]=0$ (it may not be true). Since the cycle class lifts to $H^{2r}_X(Y;\mathbb{Z}_{2}(r))$, this depends on which sequence is related to $Sq^1$.

REPLY [26 votes]: You maybe want to have a look at 

P. Brosnan and R. Joshua. Comparison of motivic and simplicial operations in mod-$\ell$ motivic and étale cohomology. In: Feynman amplitudes, periods and motives, Contemporary Math. 648, 2015, 29-55. 

There are two sequences of cohomology operations in motivic and étale cohomology which can rightfully be called Steenrod operations. One comes from a "geometric" model of the classifying space of the symmetric groups, the other one from a "simplicial" model. The first one is related to Voevodsky's motivic Steenrod algebra, the second one to the more classical Steenrod algebra (as in Denis Nardin's answer). 
The paper mentioned above provides a comparison between these sequences of cohomology operations (see Theorem 1.1 part iii for the étale cohomology) in terms of cup products with powers of a motivic Bott element in $H^0(\operatorname{Spec} k,\mathbb{Z}/\ell\mathbb{Z}(1))$. 

Addendum: Actually, the Bockstein operations are the same in both sequences of cohomology operations, and they are the ones defined by the $\mathbb{Z}/4\mathbb{Z}$ extension. (So probably the first one should be the "right".) This follows from the paper of Brosnan and Joshua as well as the paper by Guillou and Weibel in the question. Note that $\mu_2^{\otimes i}\cong\mu_2^{\otimes 2 i}\cong\mathbb{Z}/2\mathbb{Z}$ and the $Sq^1$ in the paper of Guillou-Weibel actually fits the Bockstein story.
Another note: If the motivation/example is defined over a field with $4$-th root of unity then $\mathbb{Z}/4\cong\mu_4$ (compatible with the extensions) and so both operations you defined agree.<|endoftext|>
TITLE: Can the groupoid completion of a topological category be recovered from its classifying space?
QUESTION [10 upvotes]: Let $C$ be a category. The groupoid completion of $C$ is the free groupoid on $C$, i.e. the category $C[C^{-1}]$ obtained by localizing at everything. Recall that the classifying space $\mathbf{B}C$ of $C$ is the geometric realization of the nerve of $C$. It's well-known that there is an equivalence of groupoids $C[C^{-1}] \simeq \Pi_1(\mathbf{B}C)$ between the groupoid completion of $C$ and the fundamental groupoid of the classifying space of $C$.
Now let $\mathcal{C}$ be a topologically-enriched category. Now the groupoid completion $\mathcal{C}[C^{-1}]$ is a topological groupoid. There is also a classifying space $\mathbb{B}\mathcal{C}$, which is the geometric realization of the nerve of $\mathcal{C}$ (where the nerve of $\mathcal{C}$ is a simplicial space). But the fundamental groupoid $\Pi_1(\mathbb{B}\mathcal{C})$ is discrete, so it is generally not equivalent to $\mathcal{C}[\mathcal{C}^{-1}]$.
Question.
Can we recover $\mathcal{C}[C^{-1}]$ up to equivalence of topological groupoids from $\mathbb{B}\mathcal{C}$ in some other way? More generally, what kind of homotopical significance does $\mathcal{C}[\mathcal{C}^{-1}]$ have? Does it fit into a sequence of topological groups analogous to the usual homotopy groups, and is there perhaps a natural model structure on topologically-enriched categories or simplicial spaces where the weak equivalences are the morphisms inverted by these functors?

REPLY [5 votes]: I'm not sure how well-posed this question is, what exactly is the topological structure of $\mathcal{C}[\mathcal{C}^{-1}]$?
However, the question certainly makes good sense for simplicial categories and the answer is affirmative. Dwyer and Kan proved that if $\mathcal{C}$ is a cofibrant simplicial category, then $\mathcal{C}[\mathcal{C}^{-1}](X, Y)$ is the loop space of $N \mathcal{C}$ (at $X$). See the proof of Proposition 9.3 in Simplicial Localizations of Categories (this part of their results really goes back to Segal and his delooping machine).<|endoftext|>
TITLE: A decomposition of $w_0$ which is similar to the reduced decomposition
QUESTION [5 upvotes]: Some basic definitions about reduced decomposition:
In the symmetric group $S_n$, let $s_i$ denote the adjacent transposition $(i,i+1),i\in \{1,2,\cdots,n-1\}.$ Since $S_n$ is generated by adjacent transpositions, any permutation $w\in S_n$ can be written as $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ and an expression $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ of minimal possible length $l$ is called a reduced decomposition, $l=\ell(w)$ is the length of $w$. In fact, $\ell(w)=\#\{(i,j):1\leq i<j\leq n\ \text{and}\ w(i)>w(j)\}$. So $w_0$, given by $w(i)=n+1-i$, is the longest element in $S_n$ whose length is $\frac{n(n-1)}{2}$.
For any $1\leq i<j\leq n$, let $t_{ij}$ be the transposition that interchanges $i$ and $j$. My question is how to find all the following decompositions:
$$w_0=t_{i_1j_1}t_{i_2j_2}\cdots t_{i_mj_m}$$
such that $\ell(t_{i_1j_1}\cdots t_{i_kj_k})=k,\ k=1,2,\cdots,m$. Is there any method or function to characterize such decompositions?

REPLY [6 votes]: Consider the (strong) Bruhat order on the symmetric group: This order is defined for $a,b \in \mathcal{S}_n$ as $a \leq b$ if $\ell(a) < \ell(b)$ and $at=b$ for a transposition $t$.
Therefore, your factorizations are exactly the maximal chains in this poset.
You find some further information about such maximal chains in A Weighted Enumeration of Maximal Chains in the Bruhat Order by John Stembridge. You indeed see the $\mathcal{S}_3$ example with 4 factorizations at the end of Section 1.<|endoftext|>
TITLE: Is there any real quadratic ring for which the Euclidean algorithm is polynomial?
QUESTION [8 upvotes]: We know from Rolletschek's work that the Euclidean algorithm of $\mathbb{Z}[i]$ is polynomial. Indeed, let $n$ be the maximum number of steps in the Euclidean algorithm applied to $u,v \in\mathbb{Z}[i]$ such that $\mid v\mid \leq\mid u\mid \leq N, N\in\mathbb{N}$. We have $n=1.0526\log_{2}(N)+M$, where $-1\leq M\leq 2$.
My question is if there is any Euclidean real quadratic ring for which the Euclidean algorithm is polynomial?

REPLY [7 votes]: $\mathbb{Z}[\sqrt{2}]$ is such a ring.
First, let us recall a simple algorithm that suffices to give a smaller remainder in this ring: 
If $a,b,c,d \in \mathbb{Z}$, then $\frac{a+b\sqrt{2}}{c+d\sqrt{2}} \in \mathbb{Q}(\sqrt{2})$. Write $\frac{a+b\sqrt{2}}{c+d\sqrt{2}} = (n_{1}+r_{1}) + (n_{2}+r_{2})\sqrt{2}$, where $n_{1}, n_{2} \in \mathbb{Z}$, $r_{1}, r_{2} \in \mathbb{Q}$, and $|r_{1}|, |r_{2}| \leq \frac{1}{2}$. Then $|r_{1}^{2}-2r_{2}^{2}| \leq |r_{1}|^{2} + 2|r_{2}|^{2} \leq \frac{3}{4}$, so it is always possible to divide in $\mathbb{Z}(\sqrt{2})$ and get a remainder whose norm is at most $\frac{3}{4}$ times the norm of the element you divided by.
Then such a bound for the number of divisions (starting with dividing $u$ by $v$) is given by $\frac{\log_{2} |N(v)|}{2-\log_{2}3}$, where $N$ denotes the $\mathbb{Z}[\sqrt{2}]$ norm. If you want a bound in terms of another norm, it's probably not hard to relate it to this bound.<|endoftext|>
TITLE: Changing cofinalities above supercompact cardinals
QUESTION [14 upvotes]: Question. Suppose $\kappa$ is a supercompact cardinal and $\lambda > \kappa$
is measurable (or even larger large cardinal if necessary).
Is there a set generic extension of the universe in which $\kappa$
remains supercompact, $\lambda$ is preserved and $cf(\lambda)=\omega?$

Remark 1. By Gitik-Shelah indestructibility result, if supercompact cardinal is replaced with strong cardinal, then the answer is yes.
Remark 2. If we require that the forcing preserves $\lambda^+,$
then the answer is no, as it is shown by Yair.
Remark 3. In A note on sequences witnessing singularity - following Magidor-Sinapova, Gitik has conjectured the following:

Conjecture. Suppose that

$V ⊆ W$ models of ZFC with same ordinals,

$κ$ is a regular cardinal in $V$,

$cof(κ) = ω$ in $W$,

$\aleph_1^V=\aleph_1^W,$

$V, W$ agree about a final segment of cardinals.


Then there is a subclass $V′$ of $V$ which is a model of $ZFC$, agree with $V$ about a final segment
of cardinals, and there is a sequence witnessing singularity of $κ$ (in $W$) which is generic over
$V′$
for either Namba, Woodin tower or Prikry type forcing.

Assuming this conjecture, it seems quite plausible that the answer to the question might be no in general.

Edition. I realized that the question has connection with recent work of Woodin:
Theorem. Assume $\kappa$ is an extendible cardinal. If Woodin's $HOD$-conjecture holds, then we can not change the cofinality of some large cardinal $\lambda > \kappa$, preserving the supercompactness of $\kappa,$ by set forcing without collapsing $\lambda.$

REPLY [11 votes]: There is no such forcing that preserves $\lambda^+$. Since $\lambda$ is measurable, $2^{<\lambda} = \lambda$ and therefore $\square_{\lambda,\lambda}$ holds in $V$. Since $\lambda^{+}$ is preserved, the same sequence will witness that there is still a weak square at $\lambda$ in the generic extension. But weak square fails at singular cardinals of small cofinality above a supercompact cardinal, so $\kappa$ cannot be supercompact in the generic extension. 
In fact, a theorem of Dzamonja and Shelah shows that if you change the cofinality of an inaccessible cardinal $\lambda$ while preserving $\lambda^+$, there is even a $\square_{\lambda,\omega}$ sequence in the generic extension.<|endoftext|>
TITLE: $L^\infty-L^2$ smoothing for heat equation on manifold using Nash-Moser-De-Giorgi technique
QUESTION [6 upvotes]: Let $M$ be a compact and closed smooth Riemannian manifold, and consider weak solution $u$ of the equation
$$u_t - \Delta u = f$$
given $f \in L^2(Q)$ and $u(0)=u_0 \in L^\infty(M)$.
I'm looking for a reference or sketch proof using Nash-Moser-De-Giorgi iterations (and not heat kernels or semigroup stuff, since I wish to adapt it to a different setting) of an $L^\infty-L^2$ smoothing effect for $u$, i.e., 
$$\lVert u \rVert_{L^\infty(Q)} \leq C(\lVert f \rVert_{L^2(Q)}, \lVert u_0 \rVert_{L^\infty(M)})$$
where $Q:=[0,T]\times M$.
Note that $f$ is only in $L^2$. 
I tried books by Davies etc. but for the proofs involving bounded  domains it seems the proofs really depend on the Sobolev inequalities, which varies on manifolds.

REPLY [8 votes]: To do De Giorgi-Nash-Moser it is important that $f$ is in $L^q$ for $q$ larger than half the dimension. A heuristic is scaling: if $v$ solves $-\Delta v = f$, then the right side for the rescaling $v(\epsilon x)$ is $\epsilon^2 f(\epsilon x)$, whose $L^q$ norm is like $\epsilon^{2-n/q}$, so zooming in helps when $q > n/2$.
When in dimension $4$ or higher, one can take e.g. $v = \sum h_k\varphi(2^k x)$ for some smooth $\varphi$ that is $1$ in $B_1$ and vanishes outside $B_2$. Taking $h_k = 1/k$ makes $v$ unbounded ($\log\log$ growth near the origin) while $f := -\Delta v$ is $L^2$ (the integral of $f^2$ is like $\sum h_k^2 2^{(4-n)k}$).
It seems to me that for such a choice of $f$, the solution $u$ will immediately become unbounded since $u - v$ is unbounded initially and solves the heat equation. 
Note also that indeed, in Deane's computation, the (reciprocal) exponent on $f$ looks like $n/2$ for large $p$. It is important to use that $f \in L^q$ for $q > n/2$, for instance as follows (in the elliptic case for simplicity): Let $A_p = \left(\int u^{p\chi}\right)^{\frac{1}{p \chi}}$, with $\chi = n/(n-2)$. Applying Holder to the last term in the first line of Deane's computation gives an inequality like
$$A_p \leq (pC(f,S))^{1/p} A_{\gamma p},$$
where $\gamma < 1$ depends on $q/(q-1) < \chi$, and $C(f,S)$ depends on $f$ and the Sobolev constant. Iteration gives
$$A_{2\gamma^{-k}} \leq C(f,S)^{\sum j\gamma^j} \|u\|_{L^2(Q)}.$$
Taking $k \rightarrow \infty$ gives an $L^{\infty}$ bound in terms of the desired quantities and the Sobolev constant. I'm not sure how to remove dependence on the Sobolev constant.<|endoftext|>
TITLE: Quantum Grassmannians?
QUESTION [7 upvotes]: In noncommutative algebraic geometry a commonly studied family of objects are  quantum projective spaces. Theses are certain deformations of the homogeneous coordinate ring of $\mathbb{CP}^n$. For example, see this mathoverflow post. The obvious question I would like to ask is whether or not people consider a Grassmannian generalisation of such objects, and if so, what are some well-known references.

REPLY [2 votes]: There is a deformation quantization approach to the quantization of the Grassmannians taking their Kähler symplectic form as the starting point. You can find this in the preprint by Schirmer arXiv:q-alg/9709021 from the nineties. He wrote a PhD on that. There are many earlier works on the quantization of $\mathbb{CP}^n$ in the same spirit, starting perhaps with works of Moreno and collaborators in the early eighties. You find many references in Schirmer's preprint.
However, this is not directly a $C^*$-algebraic approach, if you are interested in things like that. Nevertheless, the (a priori formal) star product is algebraic on many nice functions (the representative functions).<|endoftext|>
TITLE: Grothendieck says: points are not mere points, but carry Galois group actions
QUESTION [45 upvotes]: Apologies in advance if this question is too elementary for MO. I didn't find an explanation of these ideas in any algebraic geometry books (I don't know French).
The following is an excerpt from this archive:

Thierry Coquand recently asked me
"In your "Comments on the Development of Topos Theory" you refer
  to a simpler alternative definition of "scheme" due to Grothendieck.
  Is this definition available at some place?? Otherwise, it it possible
  to describe shortly the main idea of this alternative definition??"
Since several people have asked the same question over the years, I
  prepared the following summary which, I hope, will be of general interest:
The 1973 Buffalo Colloquium talk by Alexander Grothendieck had as
  its main theme that the 1960 definition of scheme (which had required as a
  prerequisite the baggage of prime ideals and the spectral space, sheaves
  of local rings, coverings and patchings, etc.), should be abandoned AS the
  FUNDAMENTAL one and replaced by the simple idea of a good functor from
  rings to sets. The needed restrictions could be more intuitively and more
  geometrically stated directly in terms of the topos of such functors, and
  of course the ingredients from the "baggage" could be extracted when
  needed as auxiliary explanations of already existing objects, rather than
  being carried always as core elements of the very definition.
Thus his definition is essentially well-known, and indeed is
  mentioned in such texts as Demazure-Gabriel, Waterhouse, and Eisenbud;
  but it is not carried through to the end, resulting in more
  complication, rather than less. I myself had learned the functorial point
  of view from Gabriel in 1966 at the Strasbourg-Heidelberg-Oberwolfach
  seminar and therefore I was particularly gratified when I heard
  Grothendieck so emphatically urging that it should replace the one
  previously expounded by Dieudonne' and himself.
He repeated several times that points are not mere points, but
  carry Galois group actions. I regard this as a part of the content of his
  opinion (expressed to me in 1989) that the notion of
  topos was among his most important contributions. A more general
  expression of that content, I believe, is that a generalized "gros" topos
  can be a better approximation to geometric intuition than a category
  of topological spaces, so that the latter should be relegated to an
  auxiliary position rather than being routinely considered as "the" default
  notion of cohesive space. (This is independent of the use of localic
  toposes, a special kind of petit which represents only a minor
  modification of the traditional view and not even any modification in the
  algebraic geometry context due to coherence). It is perhaps a reluctance
  to accept this overthrow that explains the situation 30 years later, when
  Grothendieck's simplification is still not widely considered to be
  elementary and "basic".

I'm trying to slowly digest the last paragraph. As a novice in algebraic geometry I'm always looking for geometric and "philosophical" intuition, so I very much want to understand why Grothendieck was insistent on points having Galois group actions.
Why, geometrically (or philosophically?) is it essential and important that points have Galois group actions?

REPLY [5 votes]: The points of a Topos have natural transformations between them; restricting to natural isomorphisms you get a groupoid. You can also represent the points of a bounded Topos as principal bundles; I.e. Something with a G -action. Not sure which is being referred to, but you can see this aspect of points without having any rings around.<|endoftext|>
TITLE: Does the Cayley-Dickson construction preserve isomorphism of quaternion algebras?
QUESTION [6 upvotes]: I posted this on math.stackexchange to no avail, so I hope it's appropriate to post here despite that it might not be research-level. I expect the answer to this is well-known to people studying non-associative algebras, but I cannot find it in my references and the more thorough literature on the topic is expensive! On a similar note, I would appreciate a recommendation for a reference covering octonion algebras over number fields and the Cayley-Dickson construction, in more generality. I don't mind expensive if it gives a good thorough treatment.
Let $K$
be a number field and let $\mathcal{B}=\Big(\frac{a,b}{K}\Big)$
be a quaternion $K$-algebra.
Then its norm is the Pfister form $\langle\langle a,b\rangle\rangle$
over $K$.
Apply the Cayley Dickson construction to $\mathcal{B}$,
yielding an octonion $K$-algebra $\mathcal{C}$.
What is the Pfister form of the norm of $\mathcal{C}$?
Is it $\langle\langle a,b,ab\rangle\rangle$?
Does the isomorphism class of $\mathcal{B}$
(i.e. different choices for $a,b$ preserving the ramification set of $\mathcal{B}$)
determine the isomorphism class of $\mathcal{C}$?
Equivalently, does the isometry class of the quadratic form $\langle\langle a,b\rangle\rangle$
determine the isometry class of the Pfister form of $\mathcal{C}$ as a quadratic form?
If it does, it makes me wonder about the octonion $K$-algebras with norm the Pfister form of $\langle\langle a,b,c\rangle\rangle$, non-isomorphic to $\mathcal{C}$. It would seem that these also contain $\mathcal{B}$ as a quaternion subalgebra,
and that some variations of the Cayley-Dickson construction would take you from $\mathcal{B}$ to these. How does this work?

REPLY [6 votes]: Suppose $K$ is a field, and $B$ your quaternion algebra $K$.  The octonion algebra $C$ made from $B$ using the Cayley-Dickson construction depends on an auxiliary choice of an element $c \in K^{\times}$.  Namely, $C$ is the set of pairs $(u,v)$ with $u,v \in B$ with addition 
$(u_1,v_1) + (u_2, v_2) = (u_1 + u_2, v_1 + v_2)$
and multiplication 
$(u_1,v_1)(u_2,v_2) = (u_1u_2 + c\overline{v_2}v_1, v_2u_1 +v_1\overline{u_2}).$
Here $x \mapsto \overline{x}$ is the involution on the quaternion aglebra $B$.  The involution on $C$ is then $(u,v) \mapsto (\overline{u},-v)$.  With this multiplication and involution, one computes that the norm on $C$ is
$(u,v)(\overline{u},-v) = u\overline{u} - cv \overline{v}$.
It follows that if the norm form on $B$ is the Pfister form $<<a,b>>$,  then the norm form on $C$ is the Pfister form $<<a,b,c>>$.
The resulting octonion aglebra $C$ depends not just $B$ but also on $c$.  For example, suppose $B$ is the quaternion division algebra over $\mathbf{R}$ (i.e., Hamilton's quaternions).  If one chooses $c=1$, then the resulting octonion algebra $C$ contains nonzero elements with norm $0$, while if one chooses $c = -1$ then the norm form on $C$ is anisotropic.<|endoftext|>
TITLE: Independence of $\ell$ of Betti numbers
QUESTION [17 upvotes]: When $X$ is a smooth proper variety over $\mathbb F_q$, we know by Deligne's theory of weights that the dimension of $H^i_{\operatorname{\acute et}}(\bar X, \mathbb Q_\ell)$ does not depend on $\ell$. In fact, we even get that Frobenius acts with the same characteristic polynomial.
Do we have a similar theorem when $X$ is no longer smooth and proper (e.g. $X$ smooth affine, or proper but singular)? How about if we assume resolution of singularities? (We could also assume other things like the Tate conjecture or a suitable subset of the standard conjectures, but that feels a bit like cheating.)
A related question: in the smooth proper case, do we know that there is some sort of canonical isomorphism
$$H^i(\bar X,\mathbb Q_\ell) \otimes_{\mathbb Q_\ell} A \stackrel \sim \to H^i(\bar X,\mathbb Q_{\ell'}) \otimes_{\mathbb Q_{\ell'}} A$$
for some big 'period' ring $A$ containing $\mathbb Q_\ell$ and $\mathbb Q_{\ell'}$?
Remark. (Also remarked by Ben Webster) In characteristic $0$, the result is known by comparison to the singular cohomology of the associated analytic space. However, in positive characteristic there cannot exist a $\mathbb Q$-valued Weil cohomology theory to compare with. In fact, there does not even exist an $\mathbb R$-valued or a $\mathbb Q_p$-valued Weil cohomology theory. The argument is outlined in 2.2 of these notes, and is attributed to Serre.
An obvious workaround would be to lift everything to characteristic $0$ and use comparison there, but again no luck: Serre gave an example of a smooth projective variety that does not lift to characteristic $0$.

REPLY [3 votes]: If you looks at Milne's notes, Chapter 21, you'll see that the comparison theorem with the singular cohomology of the $\mathbb{C}$-points has no dependence on properness; there is a smoothness hypothesis there, but Milne says it can be removed.  Since independence of $\ell$ is true for singular cohomology, it must hold for etale cohomology as well.<|endoftext|>
TITLE: Whitehead products and Framed Manifolds
QUESTION [16 upvotes]: The attaching map for the top cell of the torus $S^n \times S^n$ is a map
$$
[x,y]: S^{2n-1} \to S^n \vee S^n
$$
where the notation is such that 
$x,y : S^n \to S^n \vee S^n$ are the two inclusions–––the map $[x,y]$ is the generalized Whitehead product of $x$ and $y$. (More generally, if $f: \Sigma X \to \Sigma Z$ and $g: \Sigma Y \to \Sigma Z$ are maps, we have a map $[f,g]: \Sigma (X\wedge Y) \to \Sigma Z$.)
It is not difficult to show that this map has a framed manifold description via the Pontryagin construction: Consider the standard inclusion
$$
S^{n-1} \times D^n \subset \partial (D^n \times D^n) = S^{2n-1}
$$
Then $P = S^{n-1}\times 0$ and $Q:=  \ast \times S^{n-1}$ (where $\ast$ is the basepoint of $S^{n-1}$) are a pair of disjoint framed manifolds in $S^{2n-1}$ (with trivial framings in each case) having linking number one. Then a version of Pontryagin construction applied to $P \amalg Q \subset S^{2n-1}$ defines the map $[x,y]$. (The map is given by sending a point in a tubular neighborhood of $P$ to the first sphere and a point in a tubular neighborhood of $Q$ to the second sphere in each case using the Pontryagin construction.)
I am really interested in finding an analogous description for iterated Whitehead products: for example there is a map
$$
[[x,y],z]: S^{3n-2} \to S^n \vee S^n \vee S^n
$$
where $x,y,z$ are the three inclusions of $S^n$ in the three fold wedge.
I'd like to have a framed manifold description of this map. Presumably, there should be a "link" 
$$
S^{2n-2} \amalg S^{2n-2} \amalg S^{2n-2} \subset S^{3n-2}
$$
where each component is representing the trivial framed bordism class. 
This link  should represent the map
$[[x,y],z]$ via the Pontryagin construction.
How does this work?

REPLY [6 votes]: By Pontryagin's Theorem, you are asking for the preimages of any chosen points
 in each wedge factor of $S^p \vee S^q \vee S^r$ for the iterated Whitehead product map $p : S^{p+q+r-2} \to S^p \vee S^q \vee S^r$.  This can be obtained by chasing through the factorization of this map as 
$$ S^{p+q+r-2} \to S^{p+q-1} \vee S^r \to S^p \vee S^q \vee S^r.$$
Let's call the three points we want to take preimage of $0_p$, $0_q$ and $0_r$.  Then as John said, in the standard model for this product map the preimage of $0_p$ in $S^{p+q-1}$ is a copy of $S^{q-1}$, which we can view as $0 \times S^{q-1} \subset \partial(D^p \times D^q) \subset D^p \times D^q$.  Then the preimage of $0_p$ in $S^{p+q+r-2}$ will be the preimage of this $S^{q-1} 
\subset S^{p+q+1} \vee S^r$, which will be homeomorphic to $S^{q-1} \times S^{r-1}$.  Note that this isn't what John anticipated; if he is looking for
a (framed) $S^{q+r-2}$, then my answer falls short of that, and in fact I'm not sure if there whether or not there is a model for the Whitehead product where this preimage is spherical.  Note as well that I haven't given an embedding of this in $S^{p+q+r-2}$, but that is "straightforward" once one identifies $S^{p+q+r-2}$ as a codimension one subspace of $\partial(D^p \times D^q \times D^r)$.  (That is, I'm being lazy.)
Similarly, the preimage of $0_q$ is a copy of $S^{p-1} \times S^{r-1}$.  Finally, the preimage of $0_r$ requires just one step and is a $S^{p+q-2}$.  Again, embeddings can be "explicitly" worked out, though one runs into repeated use of the homeomorphism $\partial(D^n \times D^m) \cong S^{n+m-1}$, which clouds things a bit.
The reason I've thought of this is the following question: in what sense are these $S^{p-1} \times S^{r-1}$, $S^{q-1} \times S^{r-1}$ and $S^{p+q+r-2}$ linked, and in particular what topological invariants are there of this linking?  Here's an answer: Call these three sub manifolds $K, M$ and $N$.  Because they are disjoint and disjoint from $\infty \in S^{p+q+r-2}$, taking one point on each defines a map $K \times M \times N \to {\rm Conf}_3({\mathbb R}^{p+q+r-2})$.  Taking the image of the fundamental class defines a homology class in the configuration space, which in this example is non-trivial, and in general is a complete framed cobordism invariant.  Choosing some cohomology class to evaluate on this gives a numerical linking invariant, and one such would "count over crossings" of $K$ over $L$ over $M$, etc.
This is all worked out in my paper with Ben Walter on Hopf invariants, where we show that such linking invariants "with correction terms" faithfully measure all rational homotopy groups of simply connected spaces.<|endoftext|>
TITLE: Temporal semantics for string diagrams
QUESTION [7 upvotes]: Suppose I have a string diagram $D$ which involves a set of strings $S$ and atomic processes $A$. Formally, we should think of this as a canonically chosen map in the free symmetric monoidal category (SMC) $Free(D)$ generated by objects $S$ and arrows $A\rightrightarrows List(S)$.
Now suppose that I assign a duration to each atomic process $d:A\to\mathbb{R}^+$. I would like to extend this to all the maps in $Free(D)$ and, ideally, this assignment would define a functor on $Free(D)$.
Intuitively, the extension (also denoted $d$) would be defined by $d(f\circ g)=d(f)+d(g)$ and $d(f\otimes g)=max(d(f),d(g))$.
A problem arises for diagrams like this:
---(f)---(g)---
---(h)---(k)---
On one hand, the SMC laws tell us that 
$$(g\circ f)\otimes (k\circ h)=(g\otimes k)\circ(f\otimes h),$$
but in general we only know that
$$\max(d(f)+d(g),d(k)+d(h)) \leq \max(d(f),d(h))+\max(d(g),d(k)).$$
Of course, the assignment of durations to diagrams is not difficult to describe algorithmically: for each path through the diagram, sum up the durations along that path, and take the maximum over those sums (though there are certainly better algorithms).
Ultimately, I am less interested in how to calculate the duration of a diagram, and more interested in whether/how we can express that assignment functorially?

REPLY [2 votes]: As you correctly note, the overall duration alone is not compositional data on your processes/diagrams.
However, the algorithm you describe gives a clue as to how one could obtain duration data on processes which IS compositional, given the same data on the atomic processes.
Instead of keeping track of the overall duration of a process, you can keep track of the minimum duration from any input of the process to any output of the process.
When two processes compose (in sequence or in parallel) you can then use the data for each process to compute data for the composite process, as sketched in your algorithm.
For atomic processes, you can set the minimum time to be the same number from any input to any output (or you can use a more sophisticated assignment, if you wish).
The overall duration for a process is obtained (non-compositionally) by taking the maximum of the minimum times between any input and any output.
To make it formal, let $(R, \max, -\infty, +, 0)$ be any semiring (e.g. the max-plus semiring $\mathbb{R}^+ \sqcup\{-\infty\}$) and consider the symmetric monoidal category $\text{Mat}(R)$ of $R$-valued matrices, with matrix multiplication as composition and direct sum as monoidal product.
The duration data described above for a process is easily encoded in a matrix $M$: columns indexed by inputs, rows indexed by outputs, element $M_{i, j}$ encoding the "minimum duration" from the $i$-th input to the $j$-th output.
The overall process duration is then the maximum $\max_{i,j}M_{i,j}$ of the matrix elements.
Any assignment of $R$-valued matrices to the atomic processes automatically extends to a monoidal functor, by sending sequential composition of processes to the multiplication of the corresponding matrices and parallel composition of processes to the direct sum of the corresponding matrices.
The only thing to check is that the matrices thus associated to sequential and parallel compositions actually encode the data described informally above.
If two processes with matrices $M$ and $N$ are composed in sequence, the matrix for the sequential composition is the matrix product $NM$, encoding the following minimum process duration from input $k$ to output $i$:
$$
(NM)_{ik} = \max_j\left(N_{ij}+M_{jk}\right)
$$
This is the maximum over all paths from input $k$ to output $i$, as expected.
If two processes with matrices $M$ and $N$ are composed in parallel, the matrix for the parallel composition is the matrix direct sum $M \oplus N$, encoding the following minimum process duration from input $k$ to output $i$:

if $k$ and $i$ are an input and an output of the first process, $(M \oplus N) = M_{ik}$;
if $k$ and $i$ are an input and an output of the second process, $(M \oplus N) = N_{ik}$;
otherwise, $(M \oplus N) = -\infty$.

This is also as expected: the $-\infty$ indicates that there are no paths between inputs of one process and outputs of the other, and hence there are no minimum duration requirements from one to the other.<|endoftext|>
TITLE: How many papers are posted a year?
QUESTION [6 upvotes]: How many pure math papers are published a year? I vaguely remember seeing a figure of 10,000 but that might be old, and I may be wrong.

REPLY [24 votes]: At SCImago you can find pretty much the entire statistics:
The first graph gives the total number of math papers per year.
The second graph breaks it down per subject area (so you can distinguish "pure" from less pure). I would say that about 20% is "pure math", so some 30.000 papers per year. The number 10.000 you mention was the annual count twenty years ago.<|endoftext|>
TITLE: Why is multiplication on the space of smooth functions with compact support continuous?
QUESTION [16 upvotes]: I asked the question 
Why is multiplication on the space of smooth functions with compact support continuous? on M.SE 
sometime ago but I didn't receive a satisfactory answer. 
I was reading this post of Terence Tao
and I'm not able to prove the last item of exercise 4. 
I have a map $F:C_c^{\infty}(\mathbb R^d)\times C_c^{\infty}(\mathbb R^d)\to C_c^{\infty}(\mathbb R^d)$ given by $F(f,g) = fg$. 
The question is: Why is $F$ continuous?
I proved that if a sequence $(f_n,g_n)$ converges to $(f,g)$ then $F(f_n,g_n) \to F(f,g)$, that is, $F$ is sequentially continuous. But, as far as i know, this does not implies that $F$ is continuous because $C_c^\infty (\mathbb R^d)$ is not first countable. 
The topology of $C_c^{\infty}(\mathbb R^d)$ is given by seminorms $p:C_c^{\infty}(\mathbb R^d) \to \mathbb R_{\geq 0}$ such that $p\big|_{C_c^{\infty}( K)}:{C_c^{\infty}( K)} \to \mathbb R_{\geq 0}$ is continuous for every $K\subset \mathbb R^d$ compact; the topology of ${C_c^{\infty}( K)}$ is given by the seminorms $ f\mapsto \sup_{x\in K} |\partial^{\alpha} f(x)|$, $\alpha \in \mathbb N^d,$ and $C_c^{\infty}( K)$ is a Fréchet space.

REPLY [15 votes]: You can spare yourself the functional analytic abstract nonsense by using an explicit set of seminorms on $\mathcal{D}(\mathbb{R}^d)=C_{c}^{\infty}(\mathbb{R}^d)$
which, unfortunately, are not well-known but can be found in the excellent book "Topological Vector Spaces and Distributions" by Horváth on p.171. 
Let $\mathbb{N}=\{0,1,\ldots\}$, and denote the set of multiindices by $\mathbb{N}^d$. A locally finite family $\theta=(\theta_{\alpha})_{\alpha\in\mathbb{N}^d}$ of continuous functions $\mathbb{R}^d\rightarrow \mathbb{R}$ is one such that for all $x\in\mathbb{R}^d$ there is a neighborhood $V$ such that $V\cap {\rm Supp}\ \theta_{\alpha}=\varnothing$ for all but finitely many $\alpha$'s.
Let
$$
\|f\|_{\theta}=\sup_{\alpha\in\mathbb{N}^d}\sup_{x\in\mathbb{R}^d}
|\theta_{\alpha}(x)D^{\alpha}f(x)|\ ,
$$
then the seminorms $\|\cdot\|_{\theta}$ where $\theta$ runs over all such locally finite families define the topology of $\mathcal{D}(\mathbb{R}^d)$.
Continuity of the pointwise product follows once you show that for every $\theta$, there exists $\theta'$ and $\theta''$ such that
$$
\|fg\|_{\theta}\le \|f\|_{\theta'}\|g\|_{\theta''}
$$
for all test functions $f$ and $g$, which one can do by hand.
For instance, you can use the Leibniz or product rule
$$
D^{\alpha}(fg)=\sum_{\beta+\gamma=\alpha}\frac{\alpha!}{\beta!\gamma!} D^{\beta}f D^{\gamma}g\ ,
$$
and the brutal $l^1$-$l^{\infty}$ estimate
$$
|D^{\alpha}(fg)|\le \prod_{i=1}^{d}(\alpha_i+1)
\times\max_{\beta+\gamma=\alpha} \frac{\alpha!}{\beta!\gamma!} |D^{\beta}f| |D^{\gamma}g|\ ,
$$
in order to see that $\theta'=\theta''$ works if it is defined by
$$
\theta'_{\beta}(x):=\frac{1}{\beta!} \sup_{\alpha\ge \beta}
\sqrt{\prod_{i=1}^{d}(\alpha_i+1)!}\times\sqrt{|\theta_{\alpha}(x)|}\ .
$$

Brief Feb 2020 addendum:
@Martin Sleziak: Thank you for the edit. I didn't know one could link to a specific page as you did for the reference to Horváth. That's great!
Request for references: I attribute these explicit seminorms to Horváth because I only saw them in the book I mentioned. If you are aware of an earlier reference where these seminorms appeared, please let me know.

REPLY [3 votes]: The problem is solved here: https://math.stackexchange.com/a/1710723/41494
The solution given by the user Vobo is the following:
Let $B_n$ be the ball with radius $n$, $K_n=C_c^\infty(B_n)$ with its metrizable topology, $\varphi_n\in K_n$ a function with support contained in $B_{n}$ and $\varphi_n(x)=1$ for $x\in B_{n-1}$. First observe that $$
F_n\colon K_n\times K_n \to K_n $$ is a continuous map, which can be easily seen by the defining seminorms for these metric spaces. 
Now let $U$ be a convex neighbourhood of $0$, i.e. $U\cap K_n$ is a convex neighbourhood of $0$ in $K_n$ for each $n$. Inductively for each $n$, you can find a $0$-neighbourhood $V_n$ of $K_n$ such that $$
F[V_n,V_n] \subseteq U\cap K_n $$
(by the continuity of $F_n$) and $$ \varphi_k V_n \subseteq V_k\,\,\,\,\, (1\leq k < n).$$
Set $W_n:=V_n\cap K_{n-1}$ and $W$ as the convex hull of $\bigcup_n W_n$. Observe that for each $n$, $W_n$ is neigbourhood of $0$ in $K_{n-1}$, so $W\cap K_{n-1}\supseteq W_n$ is one too, hence $W$ is a neighbourhood of $0$ in $C_c^\infty(\mathbb{R}^d)$. Now $F[W,W]\subseteq U$ would establish the continuity of $F$.
Let $\psi, \chi\in W$, i.e. $\psi=\alpha_1\psi_1+\cdots + \alpha_m\psi_m$ and $\chi=\beta_1 \chi_1 + \cdots + \beta_m \chi_m$ with $\alpha_i, \beta_i\geq 0$, $\sum \alpha_i = \sum \beta_i =1$ and $\psi_i,\chi_i\in V_i$. As $$
F(\psi,\chi)=\psi\cdot \chi = \sum_{i,j} \alpha_i\beta_j \cdot \psi_i\chi_j $$ and $\sum_{i,j} \alpha_i\beta_j = 1$, it it sufficient to verify $\psi_i\chi_j\in U$. Now if $i=j$,
$$ \psi_i\chi_i = F(\psi_i,\chi_i)\in F[V_i,V_i]\subseteq U\cap K_i \subseteq U.$$
If $i\neq j$, e.g. $i<j$, then $\psi_i\in V_i$ and $\chi_j\in V_j$ and so $$
\psi_i\chi_j = (\psi_i \varphi_i) \chi_j =\psi_i (\varphi_i\chi_j)\in  V_i\cdot V_i \subseteq U\cap K_i \subseteq U.$$<|endoftext|>
TITLE: The minimum number of Hamiltonian paths in a strongly connected tournament of order $n$
QUESTION [5 upvotes]: For $n\ge3$ let $a(n)$ be the minimum number of Hamiltonian paths in a strong (i.e., strongly connected) tournament of order $n.$

Where is $a(n)$ discussed in the literature? Is the exact value known? If not, what nontrivial bounds are known?

My attempt: (It is assumed throughout that $n\ge3.$)
(I) $a(n)$ is always odd.
This is because of Rédei's theorem, that a tournament has an odd number of Hamiltonian paths.
(II) $a(n+1)\ge a(n)+n-1.$
This follows easily from the fact that every strong tournament of order $n+1$ contains a cycle of length $n.$
(III) $\lfloor(n-1)^2/2\rfloor+1\le a(n)\le3\cdot2^{n-4}+3.$
These are the trivial bounds. The lower bound (which happens to be OEIS sequence A099392) follows from (II) by induction. For $n\gt4,$ the upper bound (OEIS sequence A060013) is achieved by taking the transitive tournament with vertices $v_1,\dots,v_n$ and edges $v_iv_j\ (i\lt j),$ and reversing the edges $v_1v_3$ and $v_2v_n.$
(IV) $a(3)=3,\ a(4)=5,\ a(5)=9,\ a(6)\in\{13,15\}.$
That's all I know. I already asked this question on math.stackexchange.com without result.

REPLY [4 votes]: Arthur H. Busch proved that the answer is about $5^{n/3}$.<|endoftext|>
TITLE: Topology on the space of constructible sheaves
QUESTION [8 upvotes]: Let $X$ be a nice compact topological space with a fixed finite stratification by locally closed topological manifolds. At the beginning one may assume that $X$ is a complex algebraic manifold with algebraic strata, though this is not my case. Consider the set of isomorphism classes of sheaves on $X$ constructible with respect to this stratification (say, with complex coefficients).
Question. Is there any natural topology on this set? Is there a modular interpretation of this set (in the algebraic case)?
For example if there is just a single stratum, the set of such sheaves is the set of isomorphism classes of representations of $\pi_1(X)$. The topology can be defined choosing generators and relations in $\pi_1(X)$.
Another version of this question is to consider instead of sheaves the set of isomorphism classes of objects of the bounded derived category of sheaves with cohomology sheaves constructible with respect to the given stratification.
Sorry if my question is too vague. I am not an expert in the field, and I am trying to figure out what is known, in order to understand what approximately I can hope for.

REPLY [8 votes]: If you triangulate your space refining the stratification, a constructible sheaf is given by the data of a vector space $V_{\sigma}$ (a stalk at the barycenter, say) on each simplex $\sigma$ and a restriction map $V_{\sigma} \to V_{\tau}$ for any pair of incident simplices $\sigma \subset \tau$.  If you fix all those vector spaces ("frame those stalks"), that defines a point in $\prod_{\sigma,\tau \mid \sigma \subset \tau} \mathrm{Hom}(V_{\sigma},V_{\tau})$.
The restriction maps are subject to some conditions: the triangle of restriction maps associated to a triple $\sigma \subset \tau \subset \upsilon$ should commute (cutting out a closed subset of $\prod$), and $V_{\sigma} \to V_{\tau}$ should be an isomorphism if the relative interiors of $\sigma$ and $\tau$ belong to the same stratum (cutting out an open subset of that closed subset).
That space depends on the triangulation, but the quotient (as a set or as a stack) by the action of $\prod \mathrm{GL}(V_{\sigma})$ doesn't.
The situation for sheaves of chain complexes is worse, even of chain complexes themselves (sheaves on a point).  If you fix the chain groups and vary the differential, a sequence of acyclic complexes can converge to a non-acyclic complex of any size.  If you want to consider the "moduli of quasi-isomorphism classes of sheaves" you have to decide what to make of this (use a stability condition, or make a derived moduli space, or there could be other options); maybe it depends on taste and circumstances.<|endoftext|>
TITLE: Rigid analytic geometry in characterstic 0 vs positive characteristic
QUESTION [5 upvotes]: This question is motivated purely by curiosity. In algebraic geometry there is a major distinction between the world of characteristic $0$ and that of characteristic $p > 0$ with different methods, different results available etc. 
From reading a number of books and papers I got the idea that in the case of rigid analytic geometry the distinction between the two worlds is not that important. So my question is:
Are there results in rigid geometry that are only known to be true in characteristic $0$ (like Hironaka desingularisation in algebraic geometry), or vice-versa, results that are only known to be true in positive characteristic?

REPLY [5 votes]: Resolution of singularities for rigid analytic varieties of equal characteristic zero follows from resolution of singularities for schemes of characteristic zero (Nicaise, A trace formula for rigid analytic varieties etc., 2009, Proposition 2.43).
There are more examples where the characteristic plays a role, e.g. in Van der Put, Cohomology on affinoid spaces, 1982. Here the reason is the radius of convergence of the logarithm.<|endoftext|>
TITLE: Applications of algebraic geometry to machine learning
QUESTION [41 upvotes]: I am interested in applications of algebraic geometry to machine learning. I have found some papers and books, mainly by Bernd Sturmfels on algebraic statistics and machine learning. However, all this seems to be only applicable to rather low dimensional toy problems. Is this impression correct? Is there something like computational algebraic machine learning that has practical value for real world problems, may be even very high dimensional problems, like computer vision?

REPLY [26 votes]: One useful remark is that dimension reduction is a critical problem in data science for which there are a variety of useful approaches.  It is important because a great many good machine learning algorithms have complexity which depends on the number of parameters used to describe the data (sometimes exponentially!), so reducing the dimension can turn an impractical algorithm into a practical one.
This has two implications for your question.  First, if you invent a cool new algorithm then don't worry too much about the dimension of the data at the outset - practitioners already have a bag of tricks for dealing with it (e.g. Johnson-Lindenstrauss embeddings, principal component analysis, various sorts of regularization).  Second, it seems to me that dimension reduction is itself an area where more sophisticated geometric techniques could be brought to bear - many of the existing algorithms already have a geometric flavor.
That said, there are a lot of barriers to entry for new machine learning algorithms at this point.  One problem is that the marginal benefit of a great answer over a good answer is not very high (and sometimes even negative!) unless the data has been studied to death.  The marginal gains are much higher for feature extraction, which is really more of a domain-specific problem than a math problem.  Another problem is that many new algorithms which draw upon sophisticated mathematics wind up answering questions that nobody was really asking, often because they were developed by mathematicians who tend to focus more on techniques than applications.  Topological data analysis is a typical example of this: it has generated a lot of excitement among mathematicians but most data scientists have never heard of it, and the reason is simply that higher order topological structures have not so far been found to be relevant to practical inference / classification problems.<|endoftext|>
TITLE: Dao's theorem on six circumcenters associated with a cyclic hexagon
QUESTION [14 upvotes]: This questions from Ngo Quang Duong's paper
In 2013, O. T. Dao published without proof a theorem with title Another seven circles theorem in Cut the Knot, a free site for popular expositionsof many topics in mathematics.
The calculation of barycentric coordinate for concurrence given by N. Dergiades takes more than 72 pages A4. In 09-2014, N. Dergiades gave an elegant proof of this theorem and renamed this theorem: Dao’s theorem on six circumcenters associated with a cyclic hexagon. In 10-2014, T. Cohl, a Taiwan student, gave a synthetic proof for this theorem. Two proofs were published in the Forum Geometricorum journal.

We consider the following configuration: Let $L_1, L_2, L_3, L_4, L_5, L_6$ be six lines and let $P_{ij}= L_i \cap L_j$, such that $P_{12},P_{23}, P_{34}, P_{45}, P_{56}, P_{61}$ lie on a circle. Let
$(O_{ijk})$ is circle $(P_{ij}, P_{jk}, P_{ik})$ with center
$O_{ijk}$. Let $(O_{ijk})$ meets $(O_{jkh})$ again at $P'_{jk}$. (We
taking subscripts modulo 6.)

I am looking a solution for the problem as follows:

Problem 1 Ngo Quang Duong: Let $P_{12}P_{45}, P_{34}P_{61}, P_{56}P_{23}$ are concurrent at $P$. Then six points $P'_{12}$, $P'_{23}$, $P'_{34}$, $P'_{45}$, $P'_{56}$, $P'_{61}$ lie on a circle.


PS: I posted this topic because the problem 1 look like Miquel's Pentagram Theorem

REPLY [2 votes]: The solution of problem in our paper On an Extension of Miquel's Theorem to a Cyclic Hexagon;

Relative configuration in here On the eight circles theorem and its dual<|endoftext|>
TITLE: Self-complementary block designs
QUESTION [7 upvotes]: For what $n$ does there exist a self-complementary 
$(2n,4n-2,2n-1,n,n-1)$ balanced incomplete block design?
(All I know is that a self-complementary design with these parameters does exist for all $n$ of the form $2^k$ but doesn't exist for $n=3$.)
Motivating problem: One wants to divide $2n$ players into 2 equal-sized teams in $2n-1$ different ways in such a way that each player is on the same team as each other player exactly $n-1$ times.
For a follow-up question, see More about self-complementary block designs .

REPLY [4 votes]: These are the parameters for Hadamard 2-designs. Let $H$ be an $n\times n$ Hadamard matrix, where $n=4m$, normalized so that all entries in the first row are equal to 1. Each row apart from the first gives a partition of 
$\{1,\ldots,n\}$ into two sets of size $n/2$, hence we get
$2n-2$ blocks of size $n/2$. The resulting incidence structure is a 
2-design: because the columns of $H$ are pairwise orthogonal, any two columns differ in exactly $n/2$ positions and so any two points lie in exactly $(n-2)/2$ blocks. 
Also the designs constructed from Hadamard matrices as above are 3-designs, and it
can be show that any 3-design with these parameters arises in this way.<|endoftext|>
TITLE: Non-commutative Galois theory
QUESTION [15 upvotes]: Recall that an finite-dimensional algebra $A$ over a field $k$ is central simple iff there is an iso
$A \otimes_k A^{op} \cong M_n(k)$
where $A^{op}$ is the opposite ring and $M_n(k)$ is the matrix ring.
On the other hand, a finite field extension $K / k$ is Galois iff there is an iso
$K \otimes_k K \cong K^{Aut_k(K)}.$
These two statements strike me as similar and I am wondering:

Is there a Galois theory for non-commutative extensions in which the central simple algebras take the role of the Galois extensions in the usual Galois theory? Or, is there a theory which subsumes these two facts as some special cases?
Is there some sort of absolute "non-commutative" Galois group which features automorphisms of all non-commutative extensions of $k$ at a time? I could imagine this group $Gal_{nc}(k)$ to map (on?)to the usual group $Gal(k)$. 

Thank you!

REPLY [15 votes]: Let $k$ be a field. Say that a $k$-algebra $A$ is separable if any of the following equivalent conditions holds (it is not obvious that they are equivalent):

$A$ is projective as an $(A, A)$-bimodule.
$A$ is geometrically semisimple in the sense that $A \otimes_k L$ is semisimple for any field extension $L$ of $k$.
$A \otimes_k A^{op}$ is semisimple.


Theorem: The separable $k$-algebras are precisely the finite products of matrix algebras over finite-dimensional division algebras $D$ over $k$ whose centers $Z(D)$ are separable (in the usual sense) field extensions of $k$.

Hence the classification of separable algebras up to Morita equivalence is a mix of usual Galois theory and the classification of central simple algebras: the ones which are indecomposable with respect to product are classified by pairs of a finite separable extension $k \to L$ and a class in the Brauer group $\text{Br}(L)$ of $L$. 
Over an arbitrary base $k$ the correct generalization requires an additional finiteness hypothesis: an algebra $A$ is 2-dualizable if $A$ is finitely generated projective both as an $(A, A)$-bimodule and as a $k$-module. I think the classification of these is a mix of the classification of finite etale $k$-algebras and the classification of Azumaya algebras. The terminology comes from topological field theory: a 2-dualizable $k$-algebra defines a 2-dimensional topological field theory taking values in the Morita 2-category of $k$-algebras, bimodules, and bimodule homomorphisms.<|endoftext|>
TITLE: When two vertex (operator) algebras can be patched-up to a full CFT on a genus 0 surface?
QUESTION [6 upvotes]: Theorem 3 of the nLab article "Full field algebra" states that

Theorem 3. Two vertex operator algebras $V$ may appear as the left and right chiral halfs of a full conformal field theory precisely if their modular tensor categories of representations have the same Witt class.

I checked the references therein and searched more, but could not find neither the theorem stated anywhere else, nor the proof. I managed to find  out that one can take two copies of the same vertex operator algebra to get a full field algebra in the sense of Huang and Kong, or that there exist unphysical diagonal modular invariants (Davydov2015). Since I am interested in CFTs on the plane, modular invariants are not interesting to me because they are associated with tori?
I am aware of (as in, I know that it exists) the FRS construction, but it only works for rational CFTs and Theorem 3 seems to be more general, unless it's just sloppiness since 
"The purpose of the nLab is to provide a public place where people can make notes about stuff. The purpose is not to make polished expositions of material; that is a happy by-product."
Any help will be appreciated. In particular, a little summary of the state of the art would suffice, if Theorem 3 turns out to be very mysterious.

REPLY [5 votes]: If your VOA $V$ is not rational, then it is quite unlikely that its category of representations is a modular tensor category.  That is, you can safely conclude that Theorem 3 contains an unstated assumption that $V$ is rational.  At this point, we only know the modular tensor property when $V$ is rational, $C_2$-cofinite, of CFT type, and self-dual as a $V$-module (due to Huang).<|endoftext|>
TITLE: Reconstructibility of topological spaces
QUESTION [7 upvotes]: Let $(X,\tau), (Y,\sigma)$ be topological spaces with $|X|$ infinite and suppose $\varphi:X\to Y$ is a bijection such that for all $x\in X$ we have that $(X\setminus\{x\}) \cong (Y\setminus\{\varphi(x)\})$.
Does this imply that $(X,\tau) \cong (Y,\sigma)$?

REPLY [4 votes]: Edit. As pointed out by anonymous, the following argument assumes that the homeomorphisms $X\setminus\{x\}\cong Y\setminus\{\varphi(x)\}$ are all induced by $\varphi$. I will leave it for a while and maybe delete it later.
Recall that a subset $A\subset X\setminus\{x\}$ is open in the subspace topology if and only if there exists an open $V\subset X$ such that $A=V\cap(X\setminus\{x\})$. The subsets $V\subset X$ with $A=V\cap(X\setminus\{x\})$ are $A$ and $A\cup\{x\}$.
Consider $U\subset X$. If $U$ is open in $X$, then $U\setminus\{x\}$ is open in $X\setminus\{x\}$ for all $x\in X$.
On the other hand, let $U\setminus\{x\}$ be open in $X\setminus\{x\}$ for all $x\in X$. Assume that $U$ is not open in $X$, then $U\setminus\{x\}$ must be open in $X$ for all $x\in U$, and $U\cup\{y\}$ must be open in $X$ for all $y\notin U$.
Assume $X$ has at least three distinct elements. Then there are two cases.
If $U$ has at least two elements $x_1\ne x_2$ then $U\setminus\{x_i\}$ must be open in $X$, so $U=(U\setminus\{x_1\})\cup(U\setminus\{x_2\})$ is open, too.
If $X\setminus U$ has at least two elements $y_1\ne y_2$ then $U\cup\{y_1\}$ and $U\cup\{y_2\}$ must be open in $X$, so $U=(U\cup\{y_1\})\cap(U\cup\{y_2\})$ is open, too.<|endoftext|>
TITLE: Why such an interest in studying prime gaps?
QUESTION [44 upvotes]: Prime gaps studies seems to be one of the most fertile topics in analytic number theory, for long and in lots of directions : 

lower bounds (recent works by Maynard, Tao et al. [1])
upper bounds (recent works by Zhang and the whole Polymath 8 project [2])
statistics on most frequent gaps ("jumping champions" [3])
mean gaps (prime number theorem)
median gaps (Erdös-Kac and related conjectures)

I keep wondering about why so many efforts ? indeed it can be for pure knowledge of prime number distributions and properties for themselves, and that would already be a sufficient motivation, but is there any hope for further applications and consequences ? 
What I am thinking about is the following. Since Weil's works on explicit formulae, prime distribution knowledge is useful to deduce properties on operator's spectrum or zeroes of L-functions. For instance, all the works since Montgomery around pair correlation of zeroes and $n$-densities estimates, and their relations with primes properties (underlined by Montgomery and Goldston [4]). 
So my question is, mainly related to the jumping champions problem [3] : could we, by the mean of explicit formulae or whatever else, deduce from prime gaps properties some properties out of this apparently very specific field (zeroes of zeta functions, spectral informations, families statistics, etc?) ? 
Hoping the question will not be an affront to those for who the answers will be obvious and trivial, I keep impatiently waiting for possible motivations and external relations for this prime gap world ;)
Best regards

== References ==
[1] James Maynard, Small gaps between primes, Ann. of Math. (2) 181 (2015), no. 1, 383--413.
[2] Yitang Zhang, Bounded gaps between primes, Ann. of Math. (2) 179 (2014), no. 3, 1121--1174.
[3] Andrew Odlyzko, Michael Rubinstein, and Marek Wolf, Jumping champions, Experiment. Math. 8 (1999), no. 2, 107--118.
[4] D. A. Goldston, S. M. Gonek, A. E. Özlük, and C. Snyder, On the pair correlation of zeros of the Riemann zeta-function, Proc. London Math. Soc. (3) 80 (2000), no. 1, 31--49.

REPLY [9 votes]: There is also a recent paper "Unexpected biases in the distribution of consecutive primes" by Robert J. Lemke Oliver and Kannan Soundararajan, which resulted in some sensational headlines.  The relation to prime gaps is currently under discussion on math.SE.

REPLY [2 votes]: A new (strong) result may affect proving Legendre's conjecture
The first thing you can read there is "prime gaps".<|endoftext|>
TITLE: Massey products in the Steenrod algebra
QUESTION [17 upvotes]: When building $kU/2$ via its Postnikov tower, there are some interesting Massey products that show up in the Steenrod algebra, and I'd like to understand them. I bet these appear somewhere in the litterature, but I have not been able to find a reference. I will call $X = kU/2$, $H = H\mathbb{F}_2$ and I'll use $X_{\leq n}$ for the truncated guy, with homotopy in $\pi_0, \ldots, \pi_n$.
Start with $X_{\leq 0} = H$. The $k_0$-invariant to attach $\pi_2$ is the map $X_{\leq 0} =H \stackrel{Q_1}{\to} \Sigma^3 H$. We take its homotopy fiber to get the stage $X_{\leq 2}$. 
Now we can rotate and we have $\Sigma^{-1} X_{\leq 0} \stackrel{Q_1}{\to} \Sigma^2 H \to X_{\leq 2}$. The $k_1$-invariant should be given by the extension of $\Sigma^2 H \stackrel{Q_1}{\to} \Sigma^5 H$ to $X_{\leq 2}$. Since $Q_1 Q_1 \simeq 0$ this map does extend to the Postnikov truncation $X_{\leq 2} \stackrel{k_1}{\to} \Sigma^5 H$ giving us the $k$-invariant. (here I am not sure about uniqueness, maybe it is easier to put an $HZ$ at the bottom $\pi_0$, because then $k_1$ would be unique as $[HZ, H] = A/Q_0$). 
Anyways, we take the fiber again and get $X_{\leq 4}$. The $k_2$-invariant here will exists if and only if $Q_1 k_1 \simeq 0$. By drawing the diagram defining the Massey product $\langle Q_1, Q_1, Q_1 \rangle$ we see that the product $Q_1 k_1 = 0$ if and only if the product $\langle Q_1, Q_1, Q_1 \rangle$ is $Q_1$-divisible.
How to show this ? I am of course not interested in arguing that $kU$ exists and so this has to happen. Thank you for any help, or any reference.

REPLY [8 votes]: As near as I've been able the find, the primary reference for a proof is probably Kristensen and Madsen's "On the structure of the operation algebra for certain cohomology theories." This result (in fact, its generalization to all the Milnor primitives) occurs just after Proposition 3.1 in this document; it appears to be a consequence of a general result for things in the kernel of the cap-product with $\xi_1$.
An alternative reference is in Baues' book "The algebra of secondary cohomology operations": he works out an algebraic method for computing triple Massey products in the Steenrod algebra, and in Table 1 (starting on page 456) computes all the (non-matric) Massey products of three homogeneous elements up through degree 22. In particular, your bracket $\langle Q_1, Q_1, Q_1\rangle$ is listed as $\langle 3+2.1, 3+2.1, 3+2.1 \rangle$ in the degree-9 portion of their table, and this bracket contains zero. (In fact, they only found one triple product that doesn't contain zero: $\langle Sq(0,2),Sq(0,2),Sq(0,2) \rangle$.) Their method, so far as I understand it, consists of using their machinery of "track algebra" to take a presentation of the Steenrod algebra and lift it up to a more enriched algebraic structure that also remembers homotopy data that enforces the Adem relations. To do this he needs to do some real work with Eilenberg-Mac Lane spaces. I would be misrepresenting things if I claimed that I understood how this works.
(A quick comment: You're correct that there's not uniqueness of the next k-invariant or the triple product. The indeterminacy amounts to the indeterminacy in the Massey product $\langle Q_1, Q_1, Q_1\rangle$, which in this case consists of multiples of $Q_1$, so you could equally well say that the next stage in the tower exists if and only of the Massey product contains zero.)<|endoftext|>
TITLE: Trivial algebras given by generators and relations
QUESTION [6 upvotes]: Let $V$ be a finite dimensional vector space over a field $K$ of characteristic zero. Assume that we are given a set of (not necessarily homogeneous) elements $f_1,\ldots f_n$ in the tensor algebra $T(V)$. How can we decide if the ideal $I$ generated by the elements $f_i$ is the entire ring $T(V)$ or not? Is there any particular algorithm which is suitable for this problem? I am not interested in the particular structure of the ring $T(V)/I$, but only in the question of whether or not it is trivial.
Another related question: how can we decide if $T(V)/I$ is finite dimensional or not?

REPLY [5 votes]: It is well known that the analogous question is undecideable for Groups. (By Encoding a Turing machine into the Generators and relations and using undecideablity theorems like the undecideabilty of the halting Problem there.) 
You can expect that the question for algebras is undecideable as well, as you can for example encode Groups (or Turing machines) in your Algebra.<|endoftext|>
TITLE: How to compute $\sum_{x \in \mathbb{Z}^n} e^{-x^TMx}$ efficiently
QUESTION [10 upvotes]: Let $M$ be a real symmetric integer valued positive definite matrix with $\det(M) \geq 1$. I would like write code to compute
$$S_M= \sum_{x \in \mathbb{Z}^n} e^{-x^TMx}.$$
One option is to simply iterate over the vectors $x$ starting with ones with small coefficients and stop if things seem to be converging. Apart from the obvious problem of determining when to stop, this method is tremendously slow if $M$ is even $10$ by $10$.  
It is tempting to look at the integral $\int_{x \in \mathbb{R}^n} e^{-x^TMx}\;dx$ instead but this equals $\sqrt{\frac{\pi^n}{\det(M)}}$ which is potentially a terrible approximation (it can for example be much less than $1$ where $S_M \geq 1$).

How can one compute a good approximation for $S_M$?  Even an algorithm
  that runs $2^n$ time would be a huge improvement over what I have
  currently.

REPLY [14 votes]: You are trying to compute a multi-dimensional theta function, and this question is studied in depth in this 2003 Math. Comp. article by Deconinck, Heil, Bobenko, van Hoeij,and Schmies.<|endoftext|>
TITLE: Does every (generalized?) Markov chain admit transition probabilities?
QUESTION [7 upvotes]: To pose the question let us start by recalling the following notions:

Transition Probabilities.  A  transition probability matrix between two measurable spaces $(S,\mathcal{S})$ and $(V,\mathcal{V})$ is a function $P:S\times\mathcal{V}\to [0,1]$ such that for every $s\in S$, $P(s,\cdot)$ is a probability measure in $\mathcal{V}$ and for every $A\in\mathcal{V}$, $P(\cdot,A)$ is $\mathcal{S}-$measurable. If $(S,\mathcal{S})=(V,\mathcal{V})$ we will speak of a transition probability matrix in $(S,\mathcal{S})$.
Random Elements.  A random element of a measurable space $(S,\mathcal{S})$ is a $\mathcal{F}/\mathcal{S}$ measurable function $\xi:\Omega\to S$ defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$ (we can identify two random elements if they coincide at $\mathbb{P}-$a.e. $\omega$, but this definition will be enough for our purposes). Note that, in this definition, the word random comes from the introduction of a probability measure in $(\Omega,\mathcal{F})$, but this measure has no role in the property defining $\xi$.
Markov Chains (with fixed state space). A sequence $(\xi_{k})_{k\in\mathbb{Z}}$  of random elements of $(S,\mathcal{S})$ defined on the same probability space $(\Omega,\mathcal{F},\mathbb{P})$ is a Markov chain if for every $k\in\mathbb{Z}$ there exists a transition probability matrix $P_{k}$ in $(S,\mathcal{S})$ such that for every $A\in\mathcal{S}$,
$$\omega\mapsto {P}_{k}(\xi_{k}(\omega),A)$$
defines a version of $\mathbb{P}[\xi_{k+1}\in A|\sigma(\xi_{j})_{j\leq k}]:= E[I_{A}\circ \xi_{k+1}| \sigma(\xi_{j})_{j\leq k}]$ (the conditional expectation is taken with respect to $\mathbb{P}$), where $I_{A}$ is the indicator (or "characteristic'') function of $A$. This implies in particular that for every $A\in\mathcal{S}$ and every $k\in\mathbb{Z}$
$$\mathbb{P}[\xi_{k+1}\in A|\sigma(\xi_{j})_{j\leq k}]=\mathbb{P}[\xi_{k+1}\in A|\sigma(\xi_{k})].$$ 
Generalized (?) Markov Chains. If we can verify the last equation (regardless of whether the family of transitions matrices $(P_{k})_{k\in\mathbb{Z}}$ as before exists), we will say that $(\xi_{k})_{k\in\mathbb{Z}}$ is a generalized Markov chain.

Question: Is every generalized Markov chain a Markov chain? I.e. can we always find, for a generalized Markov chain, a family of transition probability matrices $(P_{k})_{k\in\mathbb{Z}}$ such that $\omega\mapsto P_{k}(\xi_{k}(\omega),A)$ is a version of $\mathbb{P}[\xi_{k+1}\in A|\sigma(\xi_{j})_{j\leq k}]$?
Again, I don't know if this question is elementary or its answer is well-known. If such is the case, I would appreciate a reference/explanation of the answer before it gets closed. Note also that the question can be generalized easily to the case in which the state spaces vary with $k$, but an answer to this version is sufficient for my purposes.
Thanks for your attention!

REPLY [5 votes]: Under mild conditions on the state space $(S, \mathcal{S})$, it is true.   For instance, it is sufficient that it be standard Borel.  In that case, each $\xi_k$ admits a regular conditional probability, which is (in your language) a transition probability matrix $Q_k$ such that for each event $B \in \mathcal{F}$, we have $\mathcal{P}(B \mid \xi_k) = Q_k(\xi_k, B)$ almost surely.  Then we can simply let $P_k(x, A) = Q_k(x, \xi_{k+1}^{-1}(A))$.
The existence of regular conditional probabilities is discussed in many advanced probability texts.  For instance, you can find it in Durrett's Probability: Theory and Examples, 4e in Section 5.1.3 (for standard Borel spaces).  If you want a more general version, the standard place to look is Dellacherie and Meyer, or Bogachev's Measure Theory section 10.4.  For instance, I believe regular conditional probabilities still exist if $(S, \mathcal{S})$ is merely Radon; i.e. $\mathcal{S}$ is the Borel $\sigma$-algebra of some topology on $S$ for which every Borel probability measure is Radon.
The result is also a special case of the disintegration theorem, which you can also find treated in depth in the aforementioned references.
The case where the state spaces $S_k$ are distinct also follows, simply by letting $S$ be the disjoint union of all the $S_k$.
But I don't think it's true in full generality.  I think the counterexample from Stoyanov's Counterexamples in Probability can be adapted into a generalized Markov chain. If I ever have some spare time I may try to work it out, or anyone else is welcome to do so.<|endoftext|>
TITLE: $G$-action on the integral homology of a compact surface
QUESTION [16 upvotes]: Let $S$ be a compact connected orientable surface, and let $G$ be a nontrivial finite group acting freely on $S$ and preserving orientation (note the the action being free is a strong condition, since automorphisms usually have fixed points). Then $H^1(S)$ also has an action of $G$. I know how to prove, using Riemann-Hurwitz and Artin's induction theorem, that
$$
  H^1(S,\mathbb{Q}) \cong \mathbb{Q}^2 \oplus \mathbb{Q}[G]^{2k}
$$
for some integer $k\ge 0$, where $\mathbb{Q}$ denotes the trivial $\mathbb{Q}[G]$-module.

What is the $\mathbb{Z}[G]$-module $H^1(S,\mathbb{Z})$ ?

I could not figure this out, although it feels like it should be well-known. In particular, when do we have $H^1(S,\mathbb{Z}) \cong \mathbb{Z}^2 \oplus \mathbb{Z}[G]^{2k}$ ?

EDIT: In case it helps, here is a proof over $\mathbb{Q}$:
Let $k = g(M/G)-1$ and $V = \mathbb{Q}^2\oplus\mathbb{Q}[G]^{2k}$, where $g(\cdot)$ denotes the genus.
For every subgroup $H\le G$, the covering $M\to M/H$ is unramified, so we have $g(M)-1 = |H|\cdot(g(M/H)-1)$, so that
$$
\dim H^1(M,\mathbb{Q})^H = \dim H^1(M/H,\mathbb{Q}) = 2+2\frac{\dim H^1(M,\mathbb{Q})-2}{|H|}\cdot
$$
On the other hand, we have
$$
\dim V^H = 2+2k[G:H] = 2+2\frac{\dim V-2}{|H|}\cdot
$$
By Artin's induction theorem, $\mathbb{Q}[G]$-modules are characterized by the dimensions of the spaces fixed by the subgroups of $G$, so $H^1(M,\mathbb{Q})\cong V$.

REPLY [2 votes]: This answer complements Oscar Randal-Williams's. First of all, if $G=\mathbb{Z}/p$ acts freely and orientation-preservingly on a closed orientable surface $S$ then $H_{1}(S) =\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$. One way to see this is to put the action into a sort of 
"normal form". The covering $S\to S/G$ is determined by a classifying map $c:S/G\to BG$, or, equivalently, the corresponding surjective homomorphism $c_{\#}:\pi_{1}(S/G)\to G$. Such a homomorphism can be completely described by listing its values on a standard set of  simple closed curves representing a symplectic basis for $H_{1}(S/G)$. One can argue that a system of such curves can be chosen so that all but one curve maps trivially. This result goes back to P.A. Smith, if not further. For details see my old paper 
Allan L. Edmonds, MR 654478 Surface symmetry. I, Michigan Math. J. 29 (1982), no. 2, 171--183. 
From this normal form, the main claim is now immediate, since the covering is trivial over all except a torus.
On the other hand if $G=(\mathbb{Z}/p)^{2}$ it can happen that the representation has the desired form and also happen that the representation is different. Again one can understand the covering $S\to S/G$ by putting the classifying homomorphism $c_{\#}:\pi_{1}(S/G)\to G$ into standard form. It turns out in this case that up to equivariant homeomorphism and automorphisms of $G$ there are exactly two such forms with respect to a suitable system of simple closed curves representing a symplectic basis for $H_{1}(S/G)$. If $G$ has generators $x,y$, then the normal forms are
(1)
$(x,y;1,1;\dots ;1,1)$
(2)
$(x,1; y,1; 1,1;\dots ;1,1)$
Again see my old paper for more details. For Case (1) the corresponding representation on $H_{1}(S)$  again has the form $\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$, since the covering $S\to S/G$ is trivial over all but the core torus.
In Case (2), however, things are different. The two cases are distinguished by $c_{*}[S/G]$ in $H_{2}(BG)=\mathbb{Z}/p$. In Case (1) $c_{*}[S/G]\neq 0$. But in Case (2) $c_{*}[S/G]= 0$, as one can visibly see, by surgering curves to create a cobordism from the map $c$ to a map $S^{2}\to BG$, which is necessarily null-homotopic. In this situation we can adapt Oscar's argument to see that $H_{1}(S)$ cannot be of the form $\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$.
The spectral sequence of the fibration $S\to S/G \to BG$ leads to a five-term exact sequence of integral homology groups
$$
H_{2}(S/G)\to H_{2}(BG)\to H_{0}(BG; H_{1}(S))\to H_{1}(S/G)\to H_{1}(BG)\to 0.
$$
Because we are in Case (2), the left hand homomorphism is trivial, and the sequence becomes
$$
0\to \mathbb{Z}/p \to H_{0}(BG; H_{1}(S))\to H_{1}(S/G)\to (\mathbb{Z}/p)^{2}\to 0.
$$
Now if $H_{1}(S)=\mathbb{Z}^{2}\oplus (\mathbb{Z}[G])^{n}$ we must have $H_{1}(S/G)=\mathbb{Z}^{n+2}$ by consideration of Euler characteristics. Then the exact sequence becomes
$$
0\to \mathbb{Z}/p \to \mathbb{Z}^{n+2}\to  \mathbb{Z}^{n+2}\to (\mathbb{Z}/p)^{2}\to 0,
$$
which is impossible. One could also work with coefficients $\mathbb{F}_{p}$ and reach a similar contradiction. One can describe the $\mathbb{Z}[G]$ module $H_{1}(S)$ more or less explicitly in this case, presumably involving  augmentation ideals. Note also that in Case (1) this sequence becomes
$$
0\to \mathbb{Z}^{n+2}\to  \mathbb{Z}^{n+2}\to (\mathbb{Z}/p)^{2}\to 0,
$$
which does occur.<|endoftext|>
TITLE: Bounding $p$-adic characters and Jacquet-Langlands transfert
QUESTION [7 upvotes]: I would like to bound uniformly in $\pi$ the $p$-adic Harisch-Chandra characters $\Theta_\pi$ for division quaternion algebras. By the Jacquet-Langlands correspondence, it is sufficient to bound it on $GL_2(\mathbf{Q}_p)$. 
An idea, for instance appearing in Kin, Shin and Templier [1], is to use the Sally-Shalika formula giving explicit calculations for the characters of $SL_2$, providing a bound for all supercuspidal representations :
$$|\Theta_\pi(\gamma_p)| \leqslant 1 + 2D(\gamma_p)^{-1/2} \ll 1$$
I would like to do the same for division quaternion algebras. By Jacquet-Langlands, we can tranfer to the $GL_2$ setting. By Labesse and Langlands and the bound above we can bound any $\Theta_{\tilde{\pi}}$ where $\tilde{\pi}$ is the restriction of $\pi$ to $SL_2$. My question follows: could we lift this bound obtained for restrictions to $SL_2$ to a bound on the characters for the whole representations on $GL_2$?
Perhaps my question only betray my deep ununderstanding of the relations between representations of $GL_2$ and those of $SL_2$. Anyway, every enlightening comment or answer will be warmly welcome.
Best regards

[1] Kim, Shin and Templier, Asymptotics and Local Constancy of Characters of $p$-adic Groups, 2015

REPLY [9 votes]: First some remarks on the Jacquet-Langlands correspondence. The image of the local Jacquet-Langlands transfer $\mathrm{JL}$ from $G$, the group of units in a non-split quaternion algebra, to $\mathrm{GL}_2(\mathbb{Q}_p)$ is the set of discrete series representations.
This set contains the supercuspidal representations as well as the Steinberg representation $St$ and all its twists by characters.
Facts:  1. The image $\mathrm{JL}(\mathbf{1}_G)$ of the trivial 1-dimensional representation $\mathbf{1}_G$ of $G$ is given by $St$.

For any character $\chi:\mathbb{Q}_p^*\rightarrow \mathbb{C}^*$ and any irreducible smooth representation $\pi$ of $G$, 
$$\mathrm{JL}(\pi \otimes (\chi\circ \mathrm{Nrd})) = \mathrm{JL}(\pi)\otimes (\chi \circ \det).$$

Together, 1 and 2 give that $\mathrm{JL}(\pi)$ is supercuspidal, iff $\dim(\pi)>1$.
Regarding the relation between representations of $\mathrm{GL}_2(\mathbb{Q}_p)$ and $\mathrm{SL}_2(\mathbb{Q}_p)$:
Given an irreducible smooth representation $\pi$ of $\mathrm{GL}_2(\mathbb{Q}_p)$, the restriction to $\mathrm{SL}_2(\mathbb{Q}_p)$ decomposes into a finite direct sum of irreducible smooth representations $\pi_i$ of $\mathrm{SL}_2(\mathbb{Q}_p)$
$$\pi|_{\mathrm{SL}_2(\mathbb{Q}_p)}\cong \bigoplus_{i=1}^n\pi_i,$$ 
each $\pi_i$ occurring with multiplicity one. The number $n$ is either 1,2 or 4. 
If $\pi$ is supercuspidal, then so are all the $\pi_i$. 
The restriction of the Steinberg representation is the Steinberg representation of $\mathrm{SL}_2(\mathbb{Q}_p)$. 
For more details I think Section 2.3 in 
http://msp.org/pjm/2007/231-1/pjm-v231-n1-p08-s.pdf 
and the references mentioned there should be helpful.<|endoftext|>
TITLE: Solutions-set first order ODE's without uniqueness
QUESTION [12 upvotes]: In short: What can we say about the collection of all solutions of an ODE when we don't have uniqueness?
When we teach a first course in ODE's, we look at the equation
$f:D\to \mathbb{R}, \quad D\subseteq \mathbb{R}^2,$
$y'(x) = f(x,y),\quad y(x_0 ) = y_0, \quad (x_0,y_0 )\in D $
and prove two theorems

If $f\in C(D)$, then there exists a neighbourhood of $x_0$ for which there is a solution $y(x) $ Peano Theorem.
If $f$ is also Lipschitz in $y$, then there exists a neighbourhood of $x_0$ in which $y(x)$ exist and is a unique solution.Picard Lindelöf.

The natural question, which I tried to "google" and did not find an answer to, is

What can be generally said about the set of all solutions when there is no uniqueness, i.e. $f$ is continuous but not Lipschitz?

REPLY [4 votes]: Check the exercises in chapter existence theorem in  Dieudonne Foundations on Modern Analysis you will get a nice overview for an answer to your question. 
best jorge<|endoftext|>
TITLE: Relation of some Euclidean geometry theorems and more conjecture generalizations
QUESTION [6 upvotes]: In this topic I want to share relation of the Pythagorean theorem, the Stewart theorem and the British Flag theorem, the Apollonius' theorem and the Feuerbach-Luchterhand. Since that I posed two conjectures of generalizations of these theorems. My question: I am looking for get a solution of my conjecture.
I. Relation of some Euclidean geometry theorems 

Pythagorean theorem: Let $ABC$ For a right triangle with legs $AB$ and $AC$ and hypotenuse $BC$ then:

\begin{equation}AB^2+AC^2=BC^2 \end{equation}


Apollonius' theorem in any triangle $ABC$, if $AD$ is a median, then

\begin{equation}AB^2 + AC^2 = 2(AD^2+BD^2)\end{equation}


Let $ABC$ be Isosceles triangle with $AB=AC$. Apply Apollonius' theorem we have $AB^2=AD^2+BD^2$, in this case $AD \perp BC$. This is the Pythagorean theorem with the right triangle $ABD$. So the Apollonius' theorem is a generalization of the Pythagorean theorem.

Stewart's theorem Let $A$, $B$, $C$ be points on a directed line $l$ in the Euclidean plane, and $P$ be a point anywhere in the plane. Then


\begin{equation}PA^2.\overline{BC} + PB^2.\overline{CA} + PC^2.\overline{AB} + \overline{BC}.\overline{CA}.\overline{AB} = 0\end{equation}


Let $B$ is midpoint of $AC$ we have: $\overline{BC}=\overline{AB}$, $\overline{CA}=-2\overline{AB}$. Since Stewart's theorem we have:

\begin{equation}PA^2.\overline{AB} - 2PB^2.\overline{AB} + PC^2.\overline{AB} - 2.\overline{AB}.\overline{AB}.\overline{AB} = 0\end{equation}
$\Leftrightarrow$ 
\begin{equation}PA^2 + PC^2 = 2(PB^2+AB^2) = 0\end{equation}
This is the Apollonius' theorem  with the triangle $PAC$ with median $PB$. So the Stewart's theorem is a generalization of the Apollonius' theorem.

British flag theorem if a point $P$ on the plane of rectangle $ABCD$ then:

\begin{equation}PA^2+PC^2=PD^2+PB^2\end{equation}


Let $P \equiv D$ we have \begin{equation}DA^2+DC^2=DB^2\end{equation}

$\Leftrightarrow$
\begin{equation}DA^2+DC^2=AC^2\end{equation}
This is the Pythagorean theorem with the right triangle $DAC$. So the British flag theorem is a generalization of the Pythagorean theorem.
5. Feuerbach-Luchterhand Let $ABCD$ be a cyclic quadrilateral, $P$ be a point on the plane then:
\begin{equation}PA^2.DB.BC.CD-PB^2.AC.CD.DA+PC^2.BD.DA.AB-PD^2.CA.AB.BC = 0\end{equation}


Let circles through $A, B, C, D$ is a line, and $D$ at infinity. Then $DB.CD=CD.DA=BD.DA=PD^2$. From the Feuerbach-Luchterhand we have:

\begin{equation}PA^2.BC-PB^2.AC+PC^2.AB-CA.AB.BC = 0\end{equation}
This is the Stewart theorem with three collinear points $A, B,C$ and $P$ on the plane. So the Feuerbach-Luchterhand is a generalization of the Stewart theorem.

Let cyclic quarilateral $ABCD$ is a rectangle. We have $AB=CD$ and $AD=BC$ and $AC=BD$. From the Feuerbach-Luchterhand we have:

\begin{equation}PA^2.AC.AD.AB-PB^2.AC.AB.DA+PC^2.AC.DA.AB-PD^2.CA.AB.AD = 0\end{equation}
$\Leftrightarrow$
\begin{equation}PA^2-PB^2+PC^2-PD^2 = 0\end{equation}
This is the British flag theorem with rectangle $ABCD$ and $P$ on the plane. So the Feuerbach-Luchterhand is a generalization of the British flag theorem.
II. More conjecture generalization-I am looking for get a solution
We write Feuerbach-Luchterhand with a form following:
Let $A_1A_2A_3A_4$ be a cyclic quadrilateral, $P$ be a point on the plane then:
\begin{equation}PA_1^2.\frac{A_{4}A_{2}}{A_{1}A_{4}.A_{1}A_{2}}-PA_2^2.\frac{A_{1}A_{3}}{A_{2}A_{1}.A_{2}A_{3}}+PA_3^2.\frac{A_{2}A_{4}}{A_{3}A_{2}.A_{3}A_{4}}-PA_4^2.\frac{A_{3}A_{1}}{A_{4}A_{3}.A_{4}A_{1}}=0 \end{equation}
Since the new form, I posed two generalization of Feuerbach-Luchterhand as follows(I check it is true with geogebra sofware):


Conjecture 1:(First generalization) Let 2n-convex cyclic polygon $A_1A_2A_3...A_{2n}$, let $P$ be a point on the plane, then:
  \begin{equation}
\sum_{i=1}^{2n} (-1)^{i+1}.PA_i^2.\frac{A_{i-1}A_{i+1}}{A_{i}A_{i-1}.A_{i}A_{i+1}}=0
\end{equation}
Conjecture 2:(Second generalization)  Let two direct similar 2n-convex cyclic polygon $A_1A_2A_3...A_{2n}$ and $B_1B_2B_3...B_{2n}$, then:
  \begin{equation}
\sum_{i=1}^{2n} (-1)^{i+1}.B_iA_i^2.\frac{A_{i-1}A_{i+1}}{A_{i}A_{i-1}.A_{i}A_{i+1}}=0
\end{equation}
  Where $A_0=A_{2n}$ and $A_{2n+1}=A_1$

My question: I check the two conjectures by Geogebra and see that it is true, but I don't have a solution. I am looking for a solution.


See case hexagon
See case octagon

REPLY [5 votes]: First problem may be solved, for example, as follows. The locus of $P$ satisfying this relation is either a circle or a line or a point or the whole plane or the empty set. Thus it suffices to check this when $P$ is on the same circle or when $P$ is the center $O$ of this circle. For $P=O$, we have alternating sum of $A_{i-1}A_{i+1}/A_{i-1}A_i\cdot A_{i}A_{i+1}$. Denote by $2\varphi_i$ the arc $A_{i-1}A_i$, then this equals (assuming that radius is 1) $\cot \varphi_{i}+\cot \varphi_{i+1}$, and everything cancels telescopically. 
If $P$ lies on a circle, make inversion in $P$. Your points now lie on a line, after substituting the formulae for change of distance under inversion you see that everything containing point $P$ cancels, and you have again an alternating sum of $A_{i-1}A_{i+1}/A_{i-1}A_i\cdot A_{i}A_{i+1}$, but on the line, not on a circle. Of course it is 0 again. 
As for the second problem, it asks for complex numbers. Denote $a_i,b_i$ complex coordinates of $A_i,B_i$, assume that 0 is circumcenter of $B_1\dots B_{2n}$. Then $B_iA_i^2=|b_i-a_i|^2=|b_i|^2+|a_i|^2-b_i \bar{a_i}-a_i\bar{b_i}$. The sums for $|b_i|^2$ and for $|a_i|^2$ are equal to 0, it is already proved. Since polygons are similar, we have $b_i=ua_i+v$ for some fixed complex $u,v$. Then $b_i \bar{a_i}=u|a_i|^2+v\bar{a_i}$. So, everything containing $u$ also is equal to 0. Thus we may suppose that $u=0$, i.e. all $B_i$ are the same, it is also already proved.<|endoftext|>
TITLE: Fibered knots vs Heegaard genus
QUESTION [5 upvotes]: We call a knot $K$ in a 3-manifold $M$ fibered if $M\backslash K$ fibers over $S^1$ with fibers $\Sigma$ and such that $K$ is ambient isotopic to the boundary of the compactified fiber $\overline{\Sigma}$.  Given a fibered knot $K$ of genus $g$, by gluing two copies of $\overline{\Sigma}$ together we get a Heegaard surface for $M$ of genus $2g$.
Is it known whether the difference between the Heegaard genus of $M$ and the minimum of twice the genus of a fibered knot in $M$ can be arbitrarily large?
In general, are there techniques besides Heegaard genus to bound from below the minimal genus of a fibered knot in a 3-manifold?  I know that Ken Baker has worked on counting genus one fibered knots in lens spaces, but I don't know of any other work in this area.

REPLY [3 votes]: This difference can be made arbitrarily large, so that there is a manifold of Heegaard genus $g$ such that the minimum genus of a fibered knot is at least $g$. This follows from examples of Hass-Thompson-Thurston of closed orientable hyperbolic 3-manifolds of Heegaard genus $g$ which have a genus $g$ oriented Heegaard splitting which requires $g$ stabilizations to become isotopic to the stabilization of the Heegaard splitting with the opposite orientation. 
Now, suppose one has such a manifold of Heegaard genus $g$ which has curve complex distance $\geq 2g$ (which is essentially a corollory of the manner in which the manifolds are constructed). Suppose one has a fibered knot of genus $<g$. The corresponding Heegaard splitting of genus $<2g$ has the property that it is a stabilization of the Heegaard splitting of genus $g$ by a Corollary 4.5 of Scharlemann and Tomova. However, the Heegaard splitting of a fibered knot is equivalent to its orientation reversal, a contradiction.<|endoftext|>
TITLE: Understanding "Decategorified" symplectic Khovanov homology
QUESTION [6 upvotes]: In http://arxiv.org/abs/math/0405089 Seidel and Smith constructed a link invariant using Lagrangian Floer theory that was conjectured to be equivalent to Khovanov homology. The equivalence was recently proved by Abouzaid and Smith in http://arxiv.org/abs/1504.01230. Their construction involves associating a Hamiltonian diffeomorphism $\phi_B$ with a braiding between two crossingless matchings C and C', which are in turn associated with Lagrangian submanifolds $L_C,L_{C'}$ of a particular symplectic manifold. Then the invariant of the resulting link is defined to be $HF^*(L_C,\phi_B(L_{C'}))$, the Floer cohomology of the pair $L_C,L_{C'}$. Modulo a few additional details this is the extent to which I understand the theory. Given that there are a number of starting points for understanding the Jones polynomial (e.g. state sums, quantum group intertwiners etc.), is there a natural symplectic geometry-based explanation that may first help me to understand a "decategorified" version of symplectic Khovanov homology? I have no idea why all the different ingredients make a sensible starting point for defining something that eventually turns out to give the correct answer.

REPLY [8 votes]: One thing you didn't mention is the manifold in which this calculation happens: the Slodowy slice to a nilpotent of type $(n,n)$. This manifold is the key to everything, since it is a geometric avatar of the invariants inside the representation $(\mathbb{C}^{2})^{\otimes 2n}$.
In what sense is this true?  First, we have an embedding of this tensor product into $\bigwedge{}^{\! 2n}(\mathbb{C}^{2}\otimes \mathbb C^{2n})$ as the elements of weight 0 under $\mathfrak{sl}(2n)$.  By skew Howe duality,  the invariants are the intersection of this weight space with the unique copy of the simple representation of $\mathfrak{sl}(2n)$  with highest weight $(2,\dots, 2, 0, \dots, 0)$.  The Slodowy slice under discussion is actually the Nakajima quiver variety for this weight space, so that's one level of geometric avatarage; the Hamiltonian diffeomorphisms correspond to the action on invariants of the braiding for quantum groups, and the Lagrangians for crossingless matchings to the invariant vector for that matching.
This actually fits into a much more general picture: Lauda, Queffelec and Rose have proven that you always get Khovanov homology out of a category when you have a categorical action of $\mathfrak{sl}(2n)$ categorifying the simple with highest weight $(2,\dots, 2, 0, \dots, 0)$. Nakajima's construction of the action of $\mathfrak{sl}(2n)$ on the cohomology of the quiver varieties for this representation has an obvious candidate for a lift to the Fukaya category: he pushes and pulls on some Lagrangian correspondences, now just think of them as Lagrangian correspondences and look at the induced functor on the Fukaya category.  There's a deformation quantization version of this that really works (http://front.math.ucdavis.edu/1208.5957), but the Fukaya category is a much spookier place.
Seidel-Smith is in all likelihood just implementing the LQR construction there; of course, good luck to whoever wants to check that those correspondences really give a categorical action.<|endoftext|>
TITLE: Rational inscribed realization of the regular dodecahedron
QUESTION [20 upvotes]: While it is clear that the regular dodecahedron $D$ cannot be realized with all integer coordinates, it is easy to find a polytope, which is combinatorially equivalent (face lattice isomorphic) to $D$ with all coordinates rational. Since $D$ is simple you can just take the hyperplanes that define $D$ and perturb them slightly to make them rational; this does not change the combinatorial type and the resulting vertices will all be rational (and almost lie on a sphere).
There is an especially nice embedding of a dodecahedron combinatorially equivalent to the regular dodecahedron with small integer coordinates, given by André Schulz, see this link:


Is there a polytope combinatorially equivalent
  to the regular dodecahedron with all rational coordinates, inscribed
  into the unit sphere?

REPLY [16 votes]: An example

Yes, here is a list of rational coordinates lying on the unit sphere, the convex hull of which is combinatorially equivalent to a regular dodecahedron. This polyhedron is invariant under reflections in three orthogonal hyperplanes (having a symmetry group $C_2 \times C_2 \times C_2$ of order 8, much smaller than the order-120 symmetry group of a regular dodecahedron):
verts = {
{304/425, 297/425, 0},
{1, 0, 0},
{52/173, 132/173, -99/173},
{52/173, 132/173, 99/173},
{54/175, 22/175, -33/35},
{54/175, 22/175, 33/35},
{44/125, -108/125, -9/25},
{44/125, -108/125, 9/25},
{3236/4325, -1452/4325, -99/173},
{3236/4325, -1452/4325, 99/173},
{-304/425, -297/425, 0},
{-1, 0, 0},
{-52/173, -132/173, 99/173},
{-52/173, -132/173, -99/173},
{-54/175, -22/175, 33/35},
{-54/175, -22/175, -33/35},
{-44/125, 108/125, 9/25},
{-44/125, 108/125, -9/25},
{-3236/4325, 1452/4325, 99/173},
{-3236/4325, 1452/4325, -99/173}
}

Mathematica confirms that the convex hull has twelve pentagonal faces:
In[4]:= InputForm@MeshCells[Region`Mesh`MergeCells[ConvexHullMesh[verts]], 2]                                                                                                                 

Out[4]//InputForm= 
{Polygon[{4, 17, 19, 15, 6}], Polygon[{1, 4, 6, 10, 2}],
 Polygon[{1, 2, 9, 5, 3}], Polygon[{1, 3, 18, 17, 4}],
 Polygon[{2, 10, 8, 7, 9}], Polygon[{6, 15, 13, 8, 10}],
 Polygon[{3, 5, 16, 20, 18}], Polygon[{11, 12, 20, 16, 14}],
 Polygon[{5, 9, 7, 14, 16}], Polygon[{7, 8, 13, 11, 14}],
 Polygon[{12, 19, 17, 18, 20}], Polygon[{11, 13, 15, 19, 12}]}

and that the vertices lie on the unit sphere:
In[3]:= Map[Norm, verts]                                                                                                                                                                      

Out[3]= {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}

How I found the solution
The main idea was to use stereographic projection to transform the unit sphere onto the plane. This reduces the question to the following equivalent problem:

Does there exist an embedding of the dodecahedral graph into $\mathbb{Q}^2$ such that the five vertices of each face are all concyclic?

Now, twenty points are a lot to contend with, so I decided to instead search for a symmetrical solution which has the following three symmetries:

reflectional symmetry about the x-axis;
reflectional symmetry about the y-axis;
inversive symmetry in an appropriate-sized circle centred on the origin.

This reduced the problem to finding six positive rationals $a,b,c,d,x,y > 0$ such that:

$y < b < d$ and $ad > bc$;
$a < x < \sqrt{c^2 + d^2}$;
$(0, y)$ lies on the circle through $(c, d), (-c, d), (a, b), (-a, b)$;
$(x, 0)$ lies on the circle through $(a, b)$ and $(c, d)$ which intersects the origin-centred circle of radius $\sqrt{c^2 + d^2}$ at right-angles;
$(x, 0)$ lies on the circle through $(a, b), (0, y), (a, -b), (0, -y)$;

I wrote an integer-arithmetic C program to search through all possibilities of positive integers $a,b,c,d < 1000$ satisfying the inequality constraints. For each of these 4-tuples, it computes $x$ and $y$ (from the penultimate and antepenultimate constraints, respectively), checks whether they're rational (using an analytic formula), and then verifies the last condition.
All of the solutions found by my program were integer scaled-up versions of the same primitive solution:
$ ./dodeca 
22, 21, 22, 54, 40/1, 10/1
44, 42, 44, 108, 80/1, 20/1
66, 63, 66, 162, 120/1, 30/1
88, 84, 88, 216, 160/1, 40/1
110, 105, 110, 270, 200/1, 50/1
132, 126, 132, 324, 240/1, 60/1
154, 147, 154, 378, 280/1, 70/1
176, 168, 176, 432, 320/1, 80/1
198, 189, 198, 486, 360/1, 90/1
220, 210, 220, 540, 400/1, 100/1
242, 231, 242, 594, 440/1, 110/1
264, 252, 264, 648, 480/1, 120/1
286, 273, 286, 702, 520/1, 130/1
308, 294, 308, 756, 560/1, 140/1
330, 315, 330, 810, 600/1, 150/1
352, 336, 352, 864, 640/1, 160/1
374, 357, 374, 918, 680/1, 170/1
396, 378, 396, 972, 720/1, 180/1

(The numbers in order are the values of $a, b, c, d, x, y$ in order; the latter two are rendered as rationals because they're not a priori integers.)
I took this primitive solution and computed the vertices as points in $\mathbb{Q}[i]$ (identifying the plane with the complex numbers). I divided all points by the Gaussian integer $c + di$ to eliminate the arbitrary scale factor, resulting in a set of rational points invariant under inversion in the unit circle. After stereographically projecting back to the unit sphere, this inversion invariance corresponds to a reflectional symmetry through the equator.
EDIT: After porting the code to the GPU and expanding the search range to 10000, the number of known primitive solutions has increased to 3.<|endoftext|>
TITLE: Cohomology of central extensions of groups
QUESTION [16 upvotes]: Let G be a central extension of a finite group H by $Z/2$. I need an explicit description of the differentials $d_2$ and $d_3$ in the Lyndon-Hochschild-Serre spectral sequence which converges to the cohomology of G with coefficients in either $Z/2$ or $U(1)$. It appears that $d_2$ is related to the cup product with the extension class, while $d_3$ is related to the Bockstein of the extension class, but I could not find it in any standard books on cohomology of groups.

REPLY [7 votes]: Ok, maybe not in a standard book, but a classic anyway: the description of the differentials $d_2$ and $d_3$ can be found in the paper 

J.-P. Serre. Cohomologie modulo 2 des complexes d'Eilenberg–MacLane. Comment. Math. Helv. 27 (1953), 192-232. (MR)

As expected in the question, $d_2$ is the cup product with the extension class and $d_3$ is related to the Bockstein of the extension class via the product structure on the $E_3$-page.
I actually found this in the introduction of the following paper

I. Leary. A differential in the Lyndon–Hochschild–Serre spectral sequence. JPAA 88 (1993), 155-168. (MR)

which gives some more discussion and identifies the $d_4$-differential.<|endoftext|>
TITLE: Weyl Groups as Galois groups
QUESTION [14 upvotes]: I am looking for explicit examples (for  all positive integers $n \ge 5$) of  degree $2n$ even polynomials $f(x)=h(x^2)$ over the field $\mathbb{Q}$ of rational numbers such that the Galois groups of $f(x)$ over $\mathbb{Q}$ are the Weyl groups $W(B_n), W(D_n)$ or their  normal subgroups of small index 2 or 4. (In this case the Galois group of $h(x)$ should be the full symmetric group $S_n$ or the alternating group $A_n$.)

REPLY [4 votes]: It's possible to produce examples using congruence conditions.  If there is a prime modulo which $h(x)$ is irreducible, a prime modulo which it splits into the product of a linear polynomial and an irreducible polynomial, and a prime modulo which it splits into the product of a linear polynomial and $n-2$ irreducible polynomials, then the Galois group of $h(x)$ is $S_n$ (because every subgroup of $S_n$ containing an $n$-cycle, an $n-1$-cycle, and a transposition is all of $S_n$).
Under these conditions, the kernel of the map from the Galois group to $S_n$ is an $S_n$-invariant subgroup of $(\mathbb Z/2)^n$. As long as there is a prime modulo which $h(x)$ splits completely and some of the roots, but not all, are squares, this kernel will contain the index $2$ normal subgroup of elements that sum to $0$, and if the number of non-squares is odd, it will contain the whole group. So if you want $W(B_n)$ you want the number of non-squares to be odd, and if you want $W(D_n)$ you want it to be even.
To construct explicitly a polynomial with $W(B_n)$ Galois group, take 4 primes and modulo each prime find a polynomial satisfying one of the four conditions, then use the Chinese remainder theorem to find an integer polynomial that reduces mod $p$ to the desired polynomials.
This applies immediately for $W(B_n)$, because we have shown the kernel contains all of $(\mathbb Z/2)^n$.
For $W(D_n)$, the congruence conditions force the kernel to contain an index $2$ subgroup, but to get exactly $W(D_n)$ you also need the constant term to be a perfect square. This will still work with the Chinese remainder theorems as long as the constant terms of your polynomials mod $p$ are squares mod $p$. It should be possible to find polynomials satisfying the first three conditions with the desired reduction mod $p$ (it is trivial if $n$ is odd as you can just scale the polynomial appropriately) and a polynomial that splits into linear factors will ave a perfect square constant term as long as the number of non-squares is even.<|endoftext|>
TITLE: what is the injective hull of indecomposable module of preprojective algebra
QUESTION [5 upvotes]: Let $Q$ be a ADE type quiver and $s_i$ ($i$ runs through the vertices of $Q$) be the simple $\Lambda$-module with 1-dimensional vector space at vertex $i$ and zero-dim at other vertices. Here $\Lambda$ is the preprojective algebra of $Q$. 
In Geiss, Leclerc, and Schröer's paper, they gave a geometric realization of irreducible $\lambda$ highest weight module by considering constructible function on the variety of nilpotent $\Lambda$-module which are isormorphic to submodule of $q_\lambda$. Here $q_\lambda$ is a direct sum of some $q_i$, where $q_i$ is the injective hull of $s_i$.
I want to calculate some small examples which are mainly about ADE type quiver. Are there any reference finding the injective hull $q_i$ of simple module $s_i$? 
Thank you so much.
EDIT: I have found a reference. In Alistair Savage, Peter Tingley's paper:Quiver grassmannians, quiver varieties and the preprojective algebra definition 2.7. 
Assume $Q$ is a quiver of finite type. For any vertex $i$, let $$q^i=\text{Hom}_{\mathbb{C}}(e_i\mathcal{P},\mathbb{C}).$$The left $\mathcal{P}$-module structure on $q^i$ is by setting $a\cdot f(x)=f(xa)$.
EDIT 2: 
I have found a way to draw $q^i$ in type A. The next thing I want to know is about indecomposable module.
For ADE type quiver, what is the injective hull of an indecomposable module $M$ over the preprojective algebra $\mathcal{P}$? Is its injective hull equal to its socle?

REPLY [3 votes]: The preprojective algebra of an ADE graph is non-symmetric Frobenius, so the injective hull of $s_i$ is isomorphic to the indecomposable projective module $$\mathcal P e_{\phi(i)},$$ where $\phi$ is the inverse of the Nakayama permutation.
For preprojective algebras of ADE graphs the Nakayama permutation is either the identity or it has order two, and it is given as follows. 
In type $\mathsf {A_n}$: the unique graph automorphism of order two.
In type $\mathsf {D_n}$: the identity if $n$ is even and the unique graph automorphism of order two if $n$ is odd.
In type $\mathsf {E_6}$: the unique graph automorphism of order two.
In type $\mathsf {E_7}$ and $\mathsf {E_8}$: the identity.
Sources: 
Karin Erdmann and Nicole Snashall, On Hochschild cohomology of preprojective algebras. I, II, J. Algebra 205 (1998), no. 2, 391--412, 413--434.
Erdmann, Karin; Snashall, Nicole. Preprojective algebras of Dynkin type, periodicity and the second Hochschild cohomology. Algebras and modules, II (Geiranger, 1996), 183--193, CMS Conf. Proc., 24, Amer. Math. Soc., Providence, RI, 1998<|endoftext|>
TITLE: A question about homogenous polynomials of degree $\frac{n(n-1)}{2}$
QUESTION [6 upvotes]: Let $n$ be a positive integer and $S_n$ be the symmetric group on $\{1,2,\ldots,n\}$.
For any $w\in S_n$ and polynomial $f\in \mathbb{R}[x_1,x_2,\ldots,x_n]$, denote $w(f)=f(x_{w(1)},x_{w(2)},\ldots,x_{w(n)})$. Furthermore, define
$$\alpha(f)=\sum\limits_{w\in S_n}\epsilon(w)w(f),$$
where the sum is over all the $w\in S_n$ and $\epsilon(w)$ denotes the sign of the permutation $w$. 
It is easy to see that for any $f\in \mathbb{R}[x_1,x_2,\ldots,x_n]$, $\alpha(f)$ is divisible by $\prod\limits_{1\leq i<j\leq n}(x_i-x_j).$
So when $f$ is a homogeneous polynomial of degree $\frac{n(n-1)}{2}$ , we have $$\alpha(f)=a_f\prod\limits_{1\leq i<j\leq n}(x_i-x_j)$$
for some $a_f\in \mathbb{R}$ .
Suppose $f$ is a homogeneous polynomial of degree $\frac{n(n-1)}{2}$, how to know whether $a_f$ is $0$ or not when the expansion $$f(x_1,x_2,\ldots,x_n)=\sum\limits_{i_1+i_2+\cdots+i_n=\frac{n(n-1)}{2}}a_{i_1,i_2,\ldots,i_n}x_1^{i_1}x_2^{i_2}\cdots x_n^{i_n}$$
is not easy to get?
For example, let $n=5$ and 
$$f(x_1,x_2,x_3,x_4,x_5)=(x_1+x_2)^2(x_2+x_3)^2(x_3+x_4)^2(x_4+x_5)^2(x_5+x_1)^2,$$
then $f$ is homogeneous polynomial of degree $10$.
With the expansion of $f$ expanded by the computer, I can calculate that $a_f$ is nonzero. But how to get this without the expansion of $f$?
Maybe it is hard to answer my question in general, so I hope someone can answer my question for the specific polynomial I gave above.

REPLY [4 votes]: As for your polynomial $f=\prod_{i=1}^5 (x_i+x_{i+1})^2$, you may do the following trick. At first, you replace your polynomial to $g=\prod_{i=1}^5 (x_i+x_{i+1}-4)(x_i+x_{i+1}-3)$. Antisymmetrizations of $f$ and $g$ are the same, since their difference has degree less than 10. Next, we use points 0,1,2,3,4 as $x$'s. I think, there is unique (up to dihedral group symmetry) permutation 10234 for which $g$ is non-zero. Hence all non-zero summands in the sum $\sum {\rm\,sign}\,(\pi) g(x_{\pi_1},x_{\pi_2},\dots,x_{\pi_5})$ are equal, so this sum is non-zero.<|endoftext|>
TITLE: Paradoxical spherical caps
QUESTION [8 upvotes]: All spherical caps (i.e. sets $C_L:=\{(x,y,z)|x^2+y^2+z^2=1, z\geq L\}$) admit a paradoxical decomposition in the sense of Banach-Tarski, meaning $C_L \tilde{} 2C_L$; here $\tilde{}$ stands for the relation of equivalence by finite equidecomposition.  (See http://www.maths.usyd.edu.au/u/athomas/amenability/Lecture5_Spheres.pdf)
Since flat disks do not admit paradoxical decompositions, a compactness argument suggests that no universal upper bound exists for the number of pieces necessary for a paradoxical decomposition of a spherical cap.
So what upper and lower bounds are known for the minimum necessary number of pieces, as a function of $L$?

REPLY [4 votes]: There are absolute lower and upper (we assume here that we do not make the pieces smaller by crushing) bounds on the number of pieces in the paradoxical decompositions in question. Namely:
if $L=-1$, the sphere is paradoxical using 4 pieces;
if  $L \in (-1,0\rangle$, the cap $C_L$ is paradoxical using 5 pieces;
if $L \in (0,1)$, the cap $C_L$ is paradoxical using 6 pieces.
This follows from the papers of R. M. Robinson "On the decomposition of spheres", Fund. Math. 34 (1947), 246-260 and of G. A. Sherman "Minimal paradoxical decomposition for Mycielski's square", Fund. Math. 139 (1991), 151-165, who proved also that the paradoxical sets contained in the caps $C_L$ use the same number of the pieces in the decompositions as $C_L$, where $L$ runs through $\langle-1, 1\rangle $.<|endoftext|>
TITLE: $T_2$ topologies that are "as disjoint as possible"
QUESTION [14 upvotes]: Let $X$ be an infinite set. Are there Hausdorff topologies $\tau_1, \tau_2$ on $X$ such that $\tau_1\cap\tau_2 = \{\emptyset\} \cup \{U\subseteq X: X\setminus U\text{ is finite}\}$? 
(That is, the intersection is as small as it can get.) 
And what about the special case $X=\mathbb{R}$ and $\tau_1$ being the topology coming from the Euclidean metric?

REPLY [9 votes]: Let $\varepsilon$ denote the Euclidean topology on $\mathbb R$.
Proposition: There is a 0-dimensional $T_2$ topology $\tau$ on $\mathbb R$
such that $\tau$ and $\varepsilon$ intersect in only the co-finite sets.
Proof: 
We need the following lemma:
Lemma:
Let $Y=(Y,\nu)$ be an infinite  topological space such that
$$|\overline{E}^\nu|\ge 2^{\omega}$$ for all $E\in [Y]^\omega$.
Then there is an injective map $f:\mathbb R\to Y$
such that 
$$
\forall D\in [\mathbb R]^\omega \text{ if }\overline D\ne \mathbb R
\text{ then }
\exists x_D\in (\mathbb R\setminus \overline D) \ f(x_D)\in \overline{f[D]}^\nu.
$$
Proof of the lemma:
Enumerate $[\mathbb R]^\omega$
as $\{D_{\alpha}:{\alpha}<2^{\omega}\}$.
By transfinite induction define 
an increasing continuous sequence $(f_{\alpha}:{\alpha}\le 2^{\omega})$
of injective functions from subsets of $\mathbb R$ into $Y$ such that 
$|f_{\alpha}|\le {\alpha}+{\omega}$,  $D_{{\beta}}\subset dom(f_{\alpha})$
for ${\beta}<{\alpha}$, and 
$$\text{ if ${\beta}<{\alpha}$ and    }\overline D_{\beta}\ne \mathbb R
\text{ then }
\exists x_{\beta}\in (dom(f_{\alpha})\setminus \overline D_{\beta}) \ 
f_{\alpha}(x_{\beta})\in \overline{f_{\alpha}[D_{\beta}]}^\nu.
$$
Assume that ${\alpha}={\beta}+1$, and we have constructed $f_{\beta}$.
Let $g\supset f_{\beta}$ be an injective function from 
$dom(f_{\beta})\cup D_{\beta}$ into $Y$.
If $\overline{D_{\beta}}=\mathbb R$, then$f_{\alpha}=g$ works.
Assume that $U=\mathbb R\setminus \overline{D_{\beta}}\ne \emptyset$.
Since $|U|=2^{\omega}$ and $|\overline{g[D_{\beta}]}^{\nu}|\ge 2^{\omega}$   we can pick $x_{\beta}\in U\setminus  dom(g)$ 
and $y_{\beta}\in \overline{g[D_{\beta}]}^{\nu}\setminus ran(g)$.
Let 
$$f_{\alpha}=g\cup\{(x_{\beta},y_{\beta})\}$$
This completes the inductive construction.
QED.
Pick a 0-dimensional $T_2$  space $(Y,\nu)$ which meets the requirements of the lemma. 
(For example, $Y=\omega^*$ works)
Apply the lemma  to obtain an injective 
$f:\mathbb R\to Y$.
Define the topology $\tau$ by declaring that $f$
 is a homeomorphism between $(\mathbb R,\tau)$ and $(f[\mathbb R],{\nu})$.
To show that $\tau $ is as required assume that $\emptyset\ne U\in \varepsilon $ such that $F=\mathbb R\setminus U$
is infinite. Let $D$ be a countable $\varepsilon$-dense subset of $F$.
Then there is  $x_D\in \mathbb R\setminus \overline D$ such that
$f(x_D)\in\overline{f[D]}^{\nu}$.
Since $f$ is a homeomorphism, $x_D\in\overline{D}^{\tau}$.
Thus $D\subset F$ implies that $x_D\in \overline{F}^{\tau}\setminus F$,
and so $F$ is not $\tau$-closed. Thus $U\notin\tau$.
Thus we proved the proposition.<|endoftext|>
TITLE: Algebraic cobordism (of a point) outside the geometric diagonal
QUESTION [6 upvotes]: This question is about the state of current knowledge regarding Voevodsky's algebraic cobordism of a point $\mathrm{MGL}^{*,*}(\mathrm{Spec}\,k)$. That the geometric diagonal $\mathrm{MGL}^{2*,*}(\mathrm{Spec}\,k)$ is isomorphic to the Lazard ring $\mathbb{L}$ if $char(k)=0$ has been shown by Levine using the Hopkins-Morel isomorphism. For positive characteristic fields, Hoyois showed the same isomorphism holds after inverting the characteristic exponential. It is also well-known that $\mathrm{MGL}^{n,n}(\mathrm{Spec}\,k)$ is identified with the Milnor $K$-theory of a field. 
Spitzweck, in Algebraic Cobordism in mixed characteristic, section 7, computed some of the homotopy groups $\pi_{p,q}\mathrm{MGL}_S$ of the algebraic cobordism spectrum in the stable homotopy category $SH(S)$ over $S$, $S$ being the spectrum of a Dedekind domain of mixed characteristic. For instance, he described the cases $(p,q)=(2n+1,n)$, $(n+1,n)$. My question is:


Are similar descriptions already known in the case $S=\mathrm{Spec}\,k$?
For $S=\mathrm{Spec}\,k$, is it known that $\mathrm{MGL}^{2n+i,n}(\mathrm{Spec}\,k)$ vanishes for $i\geq 1$.

REPLY [4 votes]: The answer to both question is yes, provided you invert $p$ in characteristic $p$ (though conjecturally this is not necessary). In fact, as far as I can see, all of Spitzweck's computations apply to fields as well. He only assumes mixed characteristic because some of the arguments are more difficult in that case (e.g. Corollary 5.9). So for example $\mathrm{MGL}^{2n-1,n}(\operatorname{Spec} k) = k^\times \otimes L_{1-n}$ and $\mathrm{MGL}^{n-1,n}(\operatorname{Spec} k)$ is an extension of $H^{n-1,n}(\operatorname{Spec} k)$ by a quotient of $K_{n+1}^M(k)$.
For the vanishing see also this question: The vanishing of $MGL^{2n+i,n}(X)$; do spectra of smooth projective varieties generate $SH_{l}$?
It's worth noting that the vanishing of $\mathrm{MGL}^{p,q}(\operatorname{Spec}k)$ for $p>q$ and the computation $\mathrm{MGL}^{n,n}(\operatorname{Spec}k)=K_n^M(k)$ hold without inverting the characteristic (they come from Morel's results about the sphere spectrum).<|endoftext|>
TITLE: What does it tell us, if we know a unital C*-algebra has approximately inner (half-)flip?
QUESTION [13 upvotes]: This is a somewhat vague question, but I think it is not too open-ended and should admit well-circumscribed answers by specialists in operator algebras.$\newcommand{\Cst}{{\rm C}^*}$ It arises from some background reading I was doing while working on a different area/problem.
For sake of simplicity I'll restrict to unital separable $\Cst$-algebras, although I think everything can be discussed more generally with appropriate modifications. $\otimes$ will denote the completed minimal tensor product of $\Cst$-algebras.
Definition. A (unital, separable) $\Cst$-algebra $A$ is said to have approximately inner flip if there is a sequence of unitaries $(u_n)\subset A\otimes A$ such that $u_n(x\otimes y)u_n^* \to y\otimes x$ for every $x,y\in A$. It is said to have approximately inner half-flip if $u_n(x\otimes 1_A)u_n^* \to 1_A\otimes x$ for all $x\in A$.
It's known that having a.i. half-flip already imposes some fairly strong restrictions on $A$, such as being simple and nuclear. We also know that ${\mathcal O}_2$ and ${\mathcal O}_\infty$ have a.i. flips.
From a quick look at the 1978 paper of Effros and Rosenberg, I gather that these properties are analogues of von Neumann algebraic properties of the hyperfinite ${\rm II}_1$ factor. What I would like to know is: why is it interesting/important to characterize those $\Cst$-algebras with a.i. (half-)flip? and if we can show that a given simple, nuclear $\Cst$-algebra has a.i. (half-)flip, what further structural consequences does one typically hope to deduce?
I guess part of what I would like to find out is whether we can think of these properties in any of the following ways:
1) "goes unseen by many of the usual invariants"
2) "has some kind of homogeneity not shared by general unital nuclear simple $\Cst$-algebras"
3) "looks like one of several standard examples on a short list".
Of course, if any or all of these three claims are wide of the mark, I'd welcome any clarifications or corrections.

REPLY [4 votes]: If $A$ is a simple AF algebra, then having an approximately inner flip entails more than that it has unique trace—because $K_0(A \otimes A)$ is order isomorphic to $K_0(A) \otimes_{Z} K_0(A)$, the latter as partially ordered abelian groups—it also says that the infinitesimals are trivial, so in fact, it forces $K_0(A)$ to be a totally ordered subgroup of the rationals, i.e., $A$ is UHF. (The key and perhaps only point is that approximately inner automorphisms induce the identity on taking $K$-groups.) This is drastic, and the result presumably extends to much larger classes of C*-algebras. This was surely in Effros-Rosenberg? (I am too tired to check, and it's fairly late for me.) So maybe (2) and (3) apply?
More generally, if $K_1(A)$ is trivial and UCT applies (whatever that is), then $K_0(A \otimes A)$ is isomorphic to the group tensor product, but I don't know whether anyone has checked that it is a pre-order isomorphism; however, if traces behave well (meaning, are determined by their effects on projections), ai flip should imply at most one trace. 
What's an example of an ai half-flip that is not an ai flip?
Edit: It's clear (without using K-theory) that only C*-algebras with zero or one trace can have an ai flip ... .<|endoftext|>
TITLE: Tube of a mod p point on a smooth Z_(p)-scheme
QUESTION [9 upvotes]: Let $R$ be a smooth, integral, finite-type $\mathbb{Z}_{(p)}$-algebra of relative dimension $n$ and $\overline{f} \colon R \to \mathbb{F}_p$. Then Hensel's lemma tells us that this lifts to a map $R \to \mathbb{Z}_p$. My understanding is that the space of lifts looks like an affine space, but I would like to understand this more explicitly.
I'm in particular hoping that it's always possible to choose an injective lift. In geometric terms, this is asking for a $\mathbb{Z}_p$-point such that the associated $\mathbb{Q}_p$-point maps to the generic point of $X:=\mathrm{Spec}(R)$.
I'm hoping this has something to do with the tangent space - like we want a tangent vector whose coordinates in $\mathbb{Z}_p$ are algebraically independent over $\mathbb{Q}$. But I don't understand this deformation space.
I'm listed this as "reference-request" because it might be fairly standard from deformation theory, but I couldn't find a reference and don't know where to look.

REPLY [7 votes]: Take $x_1,\dots,x_n$ in $R$ that lie in the kernel of $f$ and generate the tangent space a the point $f$. By Hensel's lemma, the map $(\frac{x_1}{p},\dots,\frac{x_n}{p})$ from the space of $\mathbb Z_p$-points of $R$ to $\mathbb Z_p^n$ is a bijection.
In fact the inverse function can be seen to be analytic. So each nonzero element of $R$ may be seen as a power series in $x_1,\dots,x_n$, hence a convergent power series in $\frac{x_1}{p},\dots,\frac{x_n}{p}$. By the integrality assumption none of these power series are identically zero. There are countably many power series.
Hence there is a point that is not in the vanishing locus of any of them by the standard cardinality / induction on the dimension argument. View each power series as a power series in $x_n$ with coefficients in $x_1, \dots,x_{n-1}$. There are countably many of these, so by induction there must be a tuple of values of $x_1,\dots,x_{n-1}$ where each power series has some nonvanishing coefficient of a power of $x_n$. Hence each has finitely many zeros as a function of $x_n$, so some point is a zero of none.<|endoftext|>
TITLE: Explicit free subgroup in Thompson's group $V$
QUESTION [9 upvotes]: R. Thompson introduced three groups $F\subset T\subset V$. The question concerning amenability of $F$ is still unanswered and has attracted much attention. I have read that Thompson group $V$ contains a copy of the free group $F_2$ (with two generators), in particular it is not amenable. Does anyone know an explicit embedding or can give me a reference for it? Thank you very much for the help.

REPLY [2 votes]: As already mentioned, a standard strategy to construct (non-abelian) free subgroups in $T$ or $V$ is to play ping-pong on the Cantor space. Just for fun, I would like to describe another point of view.
A topological interpretation of $T$ I really like is the following. Let $T_3$ denote the $3$-regular tree $T_3$, thought of as drawned on the plane (such that its vertex-set is discrete), and let $\mathscr{S}$ denote the ribbon surface obtained by thickening it. For each edge of $T_3$, fix a transverse arc in $\mathscr{S}$ with its endpoints on the boundary of $\mathscr{S}$. This collection of arcs decomposes $\mathscr{S}$ as a union of hexagons. We refer to a homeomorphism of $\mathscr{S}$ which sends all but finitely many hexagons to hexagons as an asymptotically rigid homeomorphism.

Theorem. The asymptotically rigid mapping class group $$\{ \text{orientation-preserving asymptotically rigid homeomorphisms of $\mathscr{S}$} \}/ \text{isotopy}$$ is isomorphic to the Ptolemy-Thompson group $T$.
See for instance Section 1.3 in Kapoudjian and Funar's article The braided Ptolemy-Thompson group is finitely presented (and references therein).
Some isometries of $T_3$ induce asymptotically rigid homeomorphisms of $\mathscr{S}$. For instance, the rotation $\alpha$ of order $3$ around a vertex (say $v$) and the rotation $\beta$ of order $2$ around the middle point of an edge (having $v$ as one of its endpoints). Clearly, the subgroup $\langle \alpha, \beta \rangle$ acts on the subdivision of $T_3$ with a single orbit of edges and with trivial edge-stabilisers. It follows that
$$\langle \alpha, \beta \rangle = \langle \alpha \rangle \ast \langle \beta \rangle \simeq \mathbb{Z}/2 \mathbb{Z} \ast \mathbb{Z} /3 \mathbb{Z}.$$
Consequently, $T$ contains a non-elementary free product (and a fortiori a non-abelian free subgroup).<|endoftext|>
TITLE: Is this a characterization of commutative $C^{*}$ algebras?
QUESTION [7 upvotes]: Assume that $A$ is  a  $C^{*}$ algebra with self adjoint elements $A_{sa}$. Assume that for all  $a,b\in A$  we  have $$ab\in A_{sa} \iff  ba \in A_{sa}$$

Is  $A$ necessarily a  commutative  algebra?

This question is in line of this post

REPLY [17 votes]: Yes. I will show that any two positive elements of $A$ commute. Since every element is a linear combination of positive elements, this suffices.
Say $a$ and $b$ are positive. Then $a^{1/2}ba^{1/2} \in A_{sa}$, so by hypothesis $ba^{1/2}a^{1/2} = ba \in A_{sa}$. That is, $ba = (ba)^* = a^*b^* = ab$. QED<|endoftext|>
TITLE: Construction of a graph
QUESTION [5 upvotes]: To construct a specific kind of undirected graph $G=(V,E)$, which $|V|=n>2$. For convenience, label the vertices with $v_1,v_2,\dots ,v_n\in V$, and $(v_i,v_j)\in E$ means there is a edge between vertices $v_i,v_j$.
And the graph has the following property:


$(v_1,v_2)\in E$
For $i\geq 3$, $(v_i,v_1)\in E\Rightarrow (v_i,v_2)\notin E$,and $(v_i,v_2)\in E\Rightarrow (v_i,v_1)\notin E$
For $i\neq j,(v_i,v_j) \notin E \Rightarrow \exists v_{k_1},v_{k_2}(k_1\neq k_2)$ which $(v_{k_1},v_i),(v_{k_2},v_i),(v_{k_1},v_j),(v_{k_2},v_j)\in E$ and $\forall l\neq k_1,k_2, (v_l,v_i) \notin E$ or $(v_l,v_j)\notin E$


If $n$ is even, for instance, $n=12$, we can construct the graph like this:

But I couldn't construct one when $n$ is odd, for instance, when $n=5$, etc.

If $n\geq 3$ is an odd number, is it possible to construct a graph meet the above-mentioned property?

REPLY [3 votes]: Notice that for every $n>3$ there should be a $C_4$ with no diagonal. For $n=5$, it is easy to see then that no such graphs exist. 
For $n=7$, the Mycielski construction over $C_3$ fits. In other words, mark the vertices of an equilateral triangle with side length 2 and the midpoints of its sides; conect each two vertices at distance 1. Now add one more vertex which is connected to the three vertices of the initial triangle.<|endoftext|>
TITLE: No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
QUESTION [33 upvotes]: A while ago a professor of mine said something along the lines of

No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants

Where algebraic invariants are functors from $Top$ to $Grp$ I guess. I've not been able to find this result anywhere online (partly because I don't know what to google), so I'm looking for a source where this result is discussed.

REPLY [3 votes]: There are interesting jumps in the responses from the question to other questions, e.g. from homeomorphisms to homotopy equivalences, from spaces to pointed spaces, .... One reason for the homotopy  study is that invariants up to homeomorphy might be uncountable, except for restricted classes of spaces, while for classes of "nice" spaces we can hope for a countable number of homotopy invariants. 
In my own work I have been forced into considering "structured spaces" to get higher homotopy groupoids,  and so colimit theorems, in higher homotopy; this pdf file on "A philosophy of modelling and computing homotopy types" is a relevant 2015 presentation. These notions of structured spaces, i.e. of topological data,  extend to higher dimensions the idea of using many base points for the fundamental groupoid, and do lead to some not previously available precise nonabelian colimit calculations of homotopy types, and so of homotopy groups, of pointed spaces. It should not be forgotten that homotopy groups are but a pale shadow of pointed homotopy types. 
Part of the philosophy is that in order to specify a space you need some kind of data, and that data will have some kind of structure. It could be wise to keep that structure in mind when constructing "invariants". It is for example standard to consider the cellular chain complex of a cell complex, and this is extended to crossed complexes and filtered spaces in the book Nonabelian Algebraic Topology. Grothendieck in Section 5 of his "Esquisse d'un Programme" argues that the notion of topological space is too far from the geometry, and wants more structure, such as a stratification.<|endoftext|>
TITLE: Are Hilbert-Schmidt operators on separable Hilbert spaces "Hilbert Schmidt" on the space of Hilbert Schmidt Operators?
QUESTION [5 upvotes]: Let's consider a separable Hilbert space $(\mathcal H, \langle\cdot, \cdot\rangle_{\mathcal H})$ with Norm $||\cdot||_{\mathcal H} := \langle\cdot, \cdot\rangle^{1/2}_{\mathcal H},$ orthonomal basis $(e_j)$ of $\mathcal H$ and let $s\colon \mathcal H \rightarrow \mathcal H$ be a Hilbert Schmidt operator, denoted by $s \in \mathcal{S_H}.$ It's known that $(\mathcal{S_H}, \langle\cdot, \cdot\rangle_{\mathcal{S_H}})$ is a (separable ??) Hilbert space and for $s_1, s_2 \in \mathcal{S_H}\colon$ $\langle s_1, s_2\rangle_{\mathcal{S_H}}=\sum_{j=1}^\infty\langle s_1(e_j), s_2(e_j)\rangle_{\mathcal H},$ such that $||s_1||^2_{\mathcal{S_H}} = \sum_{j=1}^\infty||s_1(e_j)||^2_{\mathcal H}.$ Now, assuming that $\mathcal{ S_H}$ indeed is separable, let $(\phi_j)$ be an orthonormal basis of $\mathcal S_{\mathcal H}.$ It easily can be shown, since $||\cdot||_{\mathcal{S_H}}$ is sub-multiplicative, that for $s \in \mathcal{S_H}\colon s\colon \mathcal{S_H} \rightarrow \mathcal{S_H}$ is a bounded operator as well, denoted by $s \in \mathcal{L_{S_H}}.$
Is $s\colon\mathcal{S_H} \rightarrow \mathcal{S_H}$ "Hilbert-Schmidt" as well ($s \in \mathcal{S_{S_H}}$)? 
Thank you in advance!

REPLY [11 votes]: No it's not.
If $s:H\to H$ is a rank one projection, then
$s\otimes\mathrm{id}:H\otimes \overline H\to H\otimes \overline H$
is a projection of rank $\dim(H)$.
In particular, if $H$ is infinite dimensional, then
$s\otimes\mathrm{id}:H\otimes \overline H\to H\otimes \overline H$
is a projection of infinite rank, and thus not Hilbert-Schmidt.
Here, I've identified the space of Hilbert-Schmidt operators with the Hilbert space tensor product $H\otimes \overline H$.

REPLY [9 votes]: The answer is No, assuming of course that for you $s$ acts on $\mathcal{S}_\mathcal{H}$ by left-multiplication. There is no need to assume that $\mathcal{S}_\mathcal{H}$ is separable, since it actually is, with an explicit countable orthonormal basis given by the elementary operators $e_{ij}$ that act as $e_{ij}(e_k) = e_i \delta_{jk}$ on the orthonormal basis of $\mathcal{H}$.
The Hilbert-Schmidt norm of $s$ on $\mathcal{S}_\mathcal{H}$ is $\operatorname{tr}(s' s)$, where $s'$ is the adjoint of $s$ on $\mathcal{S}_\mathcal{H}$. It is not hard to tell that $s' = s^*$, that is, left-multiplication by $s^*$, where $s^*$ is the adjoint of $s$ on $\mathcal{H}$. Let $r = s^* s$ on $\mathcal{H}$ and denote by $r$ also the left-multiplication by $r$ on $\mathcal{S}_\mathcal{H}$. Its trace is $\operatorname{tr}(r) = \sum_{ij} (e_{ij}, r e_{ij})_{\mathcal{S}_\mathcal{H}}$. A quick calculation gives $(e_{ij}, r e_{ij})_{\mathcal{S}_\mathcal{H}} = (e_i, r e_i)_\mathcal{H}$. So, each of the diagonal elements $(e_i, r e_i)_\mathcal{H}$ occurs infinitely many times in the summation for $\operatorname{tr}(r)$, which hence cannot be finite unless all diagonal elements of $r$ vanish. But we know that that's impossible from $r = s^* s$.
This answer is essentially the same as the one by André Henriques, which appeared as I was typing, but goes into more explicit detail.<|endoftext|>
TITLE: Are definable sets in an o-minimal expansion of the real field locally analytic?
QUESTION [6 upvotes]: I have a strong suspicion that yes, but as I am not a specialist in o-minimal structures, I thought that I might have overlooked some corner case. 
The precise statement is as follows: let $X \subset \mathbb{R}^n$ be a set definable in an o-minimal structure. Then there exists a point $P \in X$ and an open neighbourhood $O \subset X$ of $P$ such that $O$ is a closed analytic subset of a real analytic manifold embedded into some open $U \subset \mathbb{R}^n$

REPLY [7 votes]: No.
In their paper Quasianalytic Denjoy-Carleman classes and o-minimality, J. AMS, vol. 16 (4), 2003, p. 751—777, Rolin, Speissegger and Wilkie show that there exists a function on $[-1,1]$ which belongs to an o-minimal structure but is nowhere analytic.
Their theorem 1 (p. 752) asserts that for appropriate choices of a sequence $M=(M_n)$, the class $\mathcal C_M$ of $\mathcal C^\infty$ functions $f$ such that $\mathopen|f^{(n)}\mathclose|\leq A^{n+1} M_n$ (for some $A$) generates an o-minimal structure. The precise condition is that $(M_n)$ satisfies the two conditions:

$\sum_{n=0}^\infty\frac{ M_n}{M_{n+1}}=+\infty$;
$(M_n/n!)^2 \leq (M_{n-1}/(n-1)!) (M_{n+1}/(n+1)!)$.

By theorem 2 (2), p. 752, there exists such a sequence $(M_n)$ and a function $f$ in the corresponding class which is nowhere  analytic. This furnishes an o-minimal structure without a real analytic cell-decomposition (Corollary, (2), p. 752).<|endoftext|>
TITLE: Regularity properties of convolution
QUESTION [11 upvotes]: Let $f$ be a compactly supported $C^{\alpha}$ function (that is Holder continuous with exponent $\alpha$) and let $g$ be a compactly supported $C^\beta$ function. What can we say about Holder continuity of their convolution
$$
h(x):=\int f(z-x) g (z) dz?
$$
It is quite clear that $h$ is in class $C^{\max(\alpha,\beta)}$, but is it possible to say something better? Is it true that $h$ is in class $C^{\alpha+\beta}$? How one can prove something like this?

REPLY [4 votes]: While the proof given by Kore-N is very nice and easy to extend to more general scales of function spaces, I thought someone may find it useful to have a more direct argument. So here it is.

Write $\Delta_h f(x) = f(x + h) - f(x)$. Suppose that $f \in C^\alpha$ and $g \in C^\beta$, and one of them — say, $g$ — is compactly supported. Since
$$ \Delta_h^2 (f * g)(x) = (\Delta_h f) * (\Delta_h g) ,$$
we have
$$ |\Delta_h^2 (f * g)(x)| \leqslant \|\Delta_h f\|_\infty \|\Delta_h g\|_1 \leqslant C |h|^\alpha |h|^\beta ,$$
where $C$ depends on the Hölder constants of $f$ and $g$ and the size of the support of $g$.
It is now a well-known fact that the above condition implies that $f * g$ is in the Hölder–Zygmund space $\Lambda^{\alpha + \beta}$, which:

coincides with the space of Hölder continuous functions when $\alpha + \beta < 1$;
coincides with the space traditionally denoted by $C^{1,\alpha+\beta-1}$ if $1 < \alpha + \beta < 2$;
is strictly larger than the class of Lipschitz-continuous functions, but it is reasonably close when $\alpha + \beta = 1$.

I am not aware of any simple proof of this fact. A very nice and reasonably elementary argument is given by Stein in his Singular Integrals and Differentiability Properties of Functions book (see Proposition V.8 there).

Remarks:

More generally, we may assume that $p, q, r \in [1, \infty]$ are such that $\frac{1}{p} + \frac{1}{q} = 1 + \frac{1}{r}$ (as in Young's inequality), and $\|\Delta_h f\|_p \leqslant C |h|^\alpha$ and $\|\Delta_h g\|_q \leqslant C |h|^\beta$. Then $\|\Delta_h^2 (f * g)\|_r \leqslant C' |h|^{\alpha + \beta}$.
The above argument clearly carries over to arbitrary locally compact Abelian groups.
Yet another approach uses the fact that Hölder–Zygmund spaces are interpolation spaces between the usual classes $C^k$, and proceeds as in the standard proof of Young's inequality by real interpolation.<|endoftext|>
TITLE: Can $b^4+1$ be a pseudoprime to base 2 (except for Fermat numbers)?
QUESTION [6 upvotes]: Cross-post: This very elementary question was first posted to Mathematics Stack Exchange but the response I got there (even after offering a bounty) was not useful.
For the purpose of this question, a pseudoprime is a composite number $n$ satisfying $2^{n-1} \equiv 1 \ (\text{mod}\,n)$, also known as a (composite) odd, weak Fermat pseudoprime to base two.
If we look at $n$ of the form $n=b^2+1$ we find many primes ($b$ in OEIS A005574) and pseudoprimes (A135590).
However, if we move to $n=b^4+1$ we still find many primes (A000068), but the only pseudoprimes we have been able to find are of the form $b=2^{2^k}$ which makes $n=b^4+1$ a Fermat number (which is obviously either prime or pseudoprime).

Question: If $b^4+1$ is composite but we still have
  $2^{b^4} \equiv 1 \ (\text{mod}\ b^4+1)$,
  will $b$ necessarily be of the form $2^{2^k}$?

If there is no obvious reason why this should be true, can someone provide some heuristics on the "expected" asymptotic behavior
of such numbers $b$? A computer search seems to demonstrate that there are no examples $b \le 2\cdot 10^{10}$. Maybe there is a smarter way to locate an example?
Or maybe this has been asked/answered before in the literature?

REPLY [10 votes]: Carl Pomerance conjectured in
On the Distribution of Pseudoprimes, Math. Comput. 37, 587-593 (1981)
that for large $x$, the number of pseudoprimes $\leq x$ is
$$
  \frac{x}{e^{(1+o(1))\log{x}\frac{\log{\log{\log{x}}}}{\log{\log{x}}}}}
$$
If Pomerance's conjecture holds, for sufficiently large $x$ there are more
than $x^{\frac{3}{4}}$ pseudoprimes $\leq x$. Now there are also about
$x^{\frac{1}{4}}$ integers of the form $n^4+1$ in this range which are not
Fermat numbers. Since $\frac{3}{4} + \frac{1}{4} = 1$, common heuristics
suggest that for large enough $x$ there are coincidences, i.e. numbers
which are both of the form $n^4+1$ (and not Fermat numbers) and
pseudoprimes. Here we make the plausible assumption that the properties
"$n$ is a pseudoprime" and "$n$ is of the form $b^4+1$, but not a Fermat 
number" are independent of one another.
The lack of examples for small numbers is easily explained by the $o(1)$
term converging to $0$ only slowly -- in fact there are only $118968378$
pseudoprimes below $x = 2^{64}$, which is just about $x^{0.419} < x^{0.75}$,
cf. this table.
The probability that two random subsets of $\{1, \dots, 2^{64}\}$ of 
cardinalities $118968378$ and $2^{16}$ intersect nontrivially is pretty low.
Exact counts of pseudoprimes up to the fourth power of your search limit
$2 \cdot 10^{10}$ are not yet known, but it seems likely that $x$ must be
considerably larger to push the count of pseudoprimes $\leq x$ above
$x^{\frac{3}{4}}$. Thus finding an example of a pseudoprime of the desired
form by means of computation may be very hard.<|endoftext|>
TITLE: Chern classes of PU(n)-bundles
QUESTION [7 upvotes]: Let $PU(n) = U(n)/U(1)$ be the projective unitary group and denote by $BPU(n)$ its classifying space. Consider the algebra $M_n(\mathbb{C})$ as an $n^2$-dimensional Hilbert space equipped with the inner product $\langle T, S \rangle = \mathrm{tr}(T^*S)$. Then we have a group homomorphism
$$
PU(n) \to U(n^2)\ ,
$$
which sends $[u] \in PU(n)$ to the linear map $\mathrm{Ad}_u \colon M_n(\mathbb{C}) \to M_n(\mathbb{C}) \ ;\ T \mapsto uTu^*$. The corresponding map $BPU(n) \to BU(n^2)$ on classifying spaces induces 
$$
\mathrm{Ad}^* \colon \mathbb{Z}[c_1, \dots, c_{n^2}] \cong H^*(BU(n^2), \mathbb{Z}) \to H^*(BPU(n), \mathbb{Z})\ .
$$
The generators of the polynomial ring are the Chern classes of the universal bundle. Since every bundle of matrix algebras has the unit section as a non-vanishing section, we have $\mathrm{Ad}^*(c_{n^2}) = 0$. 


For which $n \in \mathbb{N}$ do we have $\mathrm{Ad}^*(c_{n^2-1}) \neq 0$ ? 


(I just realized that the map could rightfully be called $\mathrm{BAd}^*$, but decided against it :-)

REPLY [3 votes]: You may as well look at what happens on maximal tori, since Chern classes are seen there. Let $T$ be the maximal torus of $PU(n)$ and $D$ the maximal torus of the unitary group. Let's write the cohomology rings of their classifying spaces as $\mathbb{Z}[t_i-t_j]/~$ and $Z[x_{i,j}]$ resp. Here the quotient is by the evident linear relations on the generators (you get a polynomial ring in $n-1$ generators I just wanted to avoid choosing a basis).
It's not hard to check that the induced map on cohomology sends $x_{i,j}$ to $t_i-t_j$. (In particular the diagonal guys go to zero). So the effect on Chern classes is the effect on symmetric polynomials- you're asking when those go to zero in the target.
You've observed that the top polynomial vanishes, which here follows from the fact that the top symmetric polynomial has a factor of, say, $x_{1,1}$ in it.
Unless I'm mistaken- none of the other terms vanish. You can see this by noting that the total Chern class is given by coefficients of the polynomial in $z$, $\prod (z-t_i - t_j)$. Substitute $t_1=1$ and $t_i=0$ otherwise. The resulting polynomial is like $z(z-1)^k$ which has vanishing constant term but non vanishing coefficients otherwise (binomial coefficients).<|endoftext|>
TITLE: Exact eigenvalues of a specific tridiagonal matrix
QUESTION [11 upvotes]: I'm studying the following tri-diagonal matrix
$$
  X = \begin{pmatrix}
  0 & x_0 & 0 & 0 &\cdots & 0 & 0 & 0 \\\
  x_0 & 0 & x_1 & 0 &\cdots & 0 & 0 & 0 \\\
  0 & x_1 & 0 & x_2 &\cdots & 0 & 0 & 0 \\\
  \vdots  & \vdots & \ddots & \ddots & \ddots & \vdots & \vdots& \vdots \\\
  \vdots  & \vdots & \vdots & \ddots & \ddots & \ddots & \vdots& \vdots \\\
  \vdots  & \vdots & \vdots & \vdots & \ddots & \ddots & \ddots& \vdots \\\
  0 & 0 & 0 & 0 &\cdots & x_{N-1} & 0 & x_N \\\
  0 & 0 & 0 & 0 & \cdots & 0 & x_N & 0
 \end{pmatrix} 
$$
where the entries are given by the sequence $x_n = \frac{\sqrt{(N+1-n)(n+1)}}{N+1}$
Numerically I find that the eigenvalues are given by $\lambda_k = 1-\frac{2k}{N+1}$ for $k\in[0,N+1]$.
Also I find numerically that the orthogonal transformation $V$ diagonalizing the above matrix $D = V^T X V$ (such that the columns of $V$ are the eigenvectors of $X$) can be chosen to be symmetric: $V = V^T$. 
I have been trying to find/come up with a proof for the above two statements, however, so far unsuccessful. I do have an exact expression for the eigenvector $\vec{v}_0 = c_i\vec{e}_i$ corresponding to $\lambda_0=1$, being
$$
c_i = \frac{1}{\sqrt{2^{N+1}}}\sqrt{\begin{pmatrix} N+1 \\ i \end{pmatrix} 
  }
$$
Also $X = V^TDV$ implies a recurrence relation on the eigenvectors $\vec{v}_i$ of $X$, being $\lambda\vec{v}_k = x_{k-1}\vec{v}_{k-1} +x_k\vec{v}_{k+1}$ (this is  equivalent the statement $V=V^T$). Unfortunately this has not helped me so far.
Any help on proving either of the above statements  would be greatly appreciated!
Thanks.

REPLY [4 votes]: The Matrix $X$ I was looking at was nothing more than the angular momentum operator $L_x$ in the spin $l=(N+1)/2$ representation, see angular momentum on wikipedia.
It can be written as $L_x = 1/2(L_+ + L_-)$ where $L_+$ and $L_-$ obey $[L_z,L_\pm] = L_\pm$ and $L_z$ is the diagonal matrix with integer spaced entries ranging from $-l$ tot $l$. From the knowledge of spin representations (or $SL(2,\mathbb{C}$) if you which) one can easily find the spectrum of $L_x$.
Also, the matrix diagonalizing $X$ is just $V = e^{\pi i/2 L_y}$ which in itself is not symmetric but from it a matrix $V' = VD$ can be constructed which also diagonalizes $X$ and which is symmetric. Here $L_y = i/2(L_+ - L_-)$.
I would like to thank all who have put thought in my question!<|endoftext|>
TITLE: Artin reciprocity $\implies $ Cubic reciprocity
QUESTION [35 upvotes]: I asked this on math.SE a few days ago with no reply, so I'm reposting it here. Hope this is not considered too elementary for MO (feel free to close if so).
I'm trying to understand the proof of cubic reciprocity from Artin reciprocity as outlined in this well-known previous math.SE question and the link KCd mentions there. However, there's one final step that I can't get to work. I suspect that the proofs linked above are in fact incomplete but I'd like to confirm that I'm not missing something. (Reason: if Artin alone worked, I think Cox would have done it in his book $x^2+ny^2$, which he conspicuously does not.)
So let me write out what I've figured out so far.
Let $K = \mathbb Q(\sqrt{-3})$, and let $\pi$ be a primary prime in $K$.
(For me, primary means $\pi \equiv 1 \pmod 3$.) We want to prove
Cubic Reciprocity: If $\theta$ is a primary prime in $K$ distinct from $\pi$ then
$$\left( \frac{\pi}{\theta} \right)_3 = \left( \frac{\theta}{\pi} \right)_3.$$
Now the proof proceeds as follows.
The main idea is to consider the diagram

Here, the top row is the Artin symbol,
followed by the ``evaluation'' map
$$ \text{ev }_\pi : \sigma \mapsto \frac{1}{\sqrt[3]{\pi}} \sigma(\sqrt[3]{\pi}) $$
so that the composition yields the cubic Legendre symbol (top row).
Then Artin reciprocity implies that the Artin map is surjective and
factors through $I_K(3\pi)/P_K(3\pi)$,
which is isomorphic to $\left( \mathcal O_K/\pi \right)^\times$
by taking any prime ideal and sending it to its primary generator
(this is the map in the bottom row).
Finally, the rightmost arrow
$\left( \mathcal O_K/\pi \right)^\times \to \{1, \omega, \omega^2\}$
is surjective from the rest of the diagram.
Now the claim I don't believe is that this implies the rightmost arrow
is $\left( \frac{\bullet}{\pi} \right)_3$, which would imply cubic reciprocity.
The argument is that the kernel of the rightmost arrow is an index three
subgroup of $(\mathcal O_K/\pi)^\times$,
hence it consists of the cubes in $(\mathcal O_K/\pi)^\times$.
This means that
$$
\left( \frac{\theta}{\pi} \right) = 1
\implies \left( \frac{\pi}{\theta} \right) = 1.
$$
However, it doesn't seem to work for the other values,
for the reason that there are in fact two different
nontrivial homomorphisms $(\mathcal O_K/\pi)^\times \to \{1, \omega, \omega^2\}$,
namely
$\left( \frac{\bullet}{\pi} \right)_3$
and $\left( \frac{\bullet}{\pi} \right)_3^{-1}$.
(This is different from the quadratic case, in which there was only one.)
In other words, from this we can only conclude that for a fixed $\pi$, either
$$
\left( \frac{\theta}{\pi} \right)_3 = \left( \frac{\pi}{\theta} \right)_3
\quad\text{ or }\quad
\left( \frac{\theta}{\pi} \right)_3 = \left( \frac{\pi}{\theta} \right)_3^{-1}
\qquad \forall \theta \equiv 1 \pmod 3.
$$
Question: how do you prove that we are in the former case and not the latter?
One idea I had was to pick a convenient value of $\theta$ and just check it directly,
but I haven't been able to find a way to make this work.

REPLY [19 votes]: I was just thinking about this the other day! Here is one solution.
As in the original post, let $K = \mathbb{Q}(\sqrt{-3})$ and let $\theta$ and $\pi$ be distinct primary primes. Also, set $L = K(\sqrt[3]{\theta \pi})$ and $M = K(\sqrt[3]{\theta}, \sqrt[3]{\pi})$.
We note that $M/L$ is unramified. The only primes which might ramify are those lying over $(\sqrt{-3})$, $(\pi)$ and $(\theta)$ in $K$. We can check that there is no ramification at these places using, respectively, a $3$-adic computation and the fact that $\theta$ and $\pi$ are primary, the presentation $M = L[T]/(T^3-\theta)$ and the presentation $M = L[U]/(U^3-\pi)$.
Let $\pi'$ be the prime of $L$ lying over $\pi$, and let $\theta'$ be the prime of $L$ lying over $\theta$. Now, $\theta' \pi'$ is the principal ideal generated by $\sqrt[3]{\theta \pi}$. So Artin reciprocity gives the equation 
$$\mathrm{Frob}_{M/L}(\theta') \mathrm{Frob}_{M/L}(\pi') = \mathrm{Id} \qquad (\ast)$$
in $\mathrm{Gal}(M/L)$.
Now, $(\sqrt[3]{\pi})^{N(\pi')} = (\sqrt[3]{\pi})^{N(\pi)} \equiv \left[ \frac{\pi}{\theta} \right]_3 \sqrt[3]{\pi} \bmod \theta$. So 
$$\mathrm{Frob}_{M/L}(\theta')(\sqrt[3]{\pi}) = \left[ \frac{\pi}{\theta} \right]_3 \sqrt[3]{\pi}.$$
Since $\sqrt[3]{\pi} \sqrt[3]{\theta} \in L$, it is fixed by $\mathrm{Frob}_{M/L}(\theta')$, so we compute
$$\mathrm{Frob}_{M/L}(\theta')(\sqrt[3]{\theta}) = \left[ \frac{\pi}{\theta} \right]_3^{-1} \sqrt[3]{\theta}.$$
Similarly,
$$\mathrm{Frob}_{M/L}(\pi')(\sqrt[3]{\theta}) = \left[ \frac{\theta}{\pi} \right]_3 \sqrt[3]{\theta} \quad 
\mathrm{Frob}_{M/L}(\pi')(\sqrt[3]{\pi}) = \left[ \frac{\theta}{\pi} \right]_3^{-1} \sqrt[3]{\pi} .$$
Plugging into $(\ast)$ and acting on $\sqrt[3]{\pi}$ we get
$$ \left[ \frac{\pi}{\theta} \right]_3  \left[ \frac{\theta}{\pi} \right]_3^{-1} = 1$$
as desired.

On a linguistic note, I recently realized that "reciprocal" is a word, like "moot" or "cleave", which can denote two opposite things: 

Life cannot subsist in society but by reciprocal concessions. (Samuel Johnson)

involves the exact opposite meaning from

...the pressures and expansions [volumes] to be in reciprocal relation. (Boyle, p. 60)

The higher reciprocity laws use reciprocal in the former sense.<|endoftext|>
TITLE: Equivalence of the Banach–Tarski paradox
QUESTION [7 upvotes]: I am working on the Banach–Tarski paradox and the fact that the Hahn–Banach theorem implies that paradox. The proof involves the equivalence of the Hahn–Banach theorem and the fact that for every Boolean algebra $\mathcal A$ there is a finitely additive "measure" $\mathrm m:\mathcal A \to [0,+\infty]$  such that some element of $\mathcal A$ has finite measure.
$\mathsf{ZF}$ is not enough to derive the Banach-Tarski paradox, nor is $\mathsf{ZF}+\mathsf{DC}$.
However, I am interested in reversing the roles, in order to characterize the Banach–Tarski paradox in terms of choice-like axioms. Hahn–Banach is strictly weaker than the axiom of choice or the ultrafilter lemma/boolean prime ideal theorem.

Can we find a weaker axiom to derive the Banach–Tarski paradox? Do you know any reference or have any ideas on this? Are there weaker axioms that should suffice in order to get the Banach–Tarski paradox? Is it possible to reformulate the Banach–Tarski paradox in a way that this assumption allows us in $\mathsf{ZF}$ to recover the Hahn–Banach extension theorem?

These questions do not seem clear to me, so I would be very pleased if they are somehow absurd. Thanks in advance.

REPLY [7 votes]: The Banach-Tarski paradox is about real numbers, so it is limited to special kind of sets. The Hahn-Banach theorem is much more general. Therefore, trying to recover the Hahn-Banach theorem, we have to reformulate the Banach-Tarski paradox in a reasonable way. Recall that the core of the paradox is the existence of a free group of rotations acting on the sphere $\mathbb{S}^2$ in the way that the set of fixed points can be eliminated (we can absorb the points and recover two spheres from the one). Clearly, we can generalize the paradoxical construction to arbitrary group actions on some abstract spaces. The point is that we must care about fixed points. So consider a group $G$ acting on a space $X$, we say that the set $F$ of fixed points of the $G$-action is absorbable if $X$ is $G$-equidecomposable to $X \setminus F$, that is $X$ can be partitioned into finitely many sets $A_1,...A_k$ and $X \setminus F$ can be partitioned into finitely many sets $B_1,...,B_k$ such that $g_i(A_i) = B_i$ for $i=1,...,k$ and $g_i \in G$. Thus we have the following generalization of the Banach-Tarski paradox:
If a free non-abelian group of rank two $F_2$ acts on a set $X$ in the way that the set of fixed points is absorbable, then $X$ is $F_2$-paradoxical (there is a Banach-Tarski decomposition of X via elements of the group $F_2$).
I think that this general setting should correspond with the existence/non-existence of measures that is used in the implication Hahn-Banach theorem $\Rightarrow$ Banach-Tarski paradox.<|endoftext|>
TITLE: Orientability of orbit type strata of Lie group actions
QUESTION [5 upvotes]: Let $G$ be a compact Lie group that acts on a smooth, finite dimensional, oriented manifold $M$, and suppose that such action preserves orientation, i.e., for each $g\in G$, the diffeomorphism $\mu_g$ of $M$ induced by the action preserves orientation. Consider the stratification
$$M=\bigcup M_j$$
by orbit types, i.e., $M_j$ is the set of points that have the same isotropy, up to conjugation.
My question is, are there "simple" conditions under which the quotient manifold $M_j/G$ is orientable? Is it automatic true with the orientation hypothesis on $M$ and the action?

REPLY [3 votes]: In your question, you may as well take $M=M_j=$ a ($G$-invariant) tubular neighborhood of a single orbit $G/S$. Near the point $S/S$, the space looks like $\mathfrak g/\mathfrak s \times N$, where $N$ is the normal space. Indeed, using a $G$-invariant metric, we can make $S$ act on $N$ and make this identification $S$-equivariant. By your condition that the stabilizers are all conjugate, the $S$-action on $N$ should be trivial.
You say that $\mathfrak g/\mathfrak s \times N$ is oriented, and ask whether $N$ (the tangent space in the quotient) is naturally oriented. This is therefore equivalent to $G/S$ being oriented. (Which is easy to violate, e.g. $SO(3)$ acting on $\mathbb{RP}^2$.) If we assume $G$ to be simply connected (which we don't get for free, since taking the universal cover might spoil its compactness), then for $G/S$ to be nonorientable you'd need $S$ to be disconnected. (Here's it's $S(O(2)\times O(1))$.)
For a counterexample, we want a $G/S$-bundle over a non-orientable manifold $M$. These are easy to cook up but I haven't yet seen how to guarantee that the total space is oriented.<|endoftext|>
TITLE: CM $j$-invariants in $p$-adic fields
QUESTION [21 upvotes]: I'm trying to understand the $p$-adic distribution of $j$-invariants for elliptic curves with complex multiplication.
Specifically, suppose $\sigma$ is some embedding $\sigma:\overline{\mathbb Q}\to \overline{\mathbb Q_p}$ from the algebraic numbers to the algebraic closure of the $p$-adic rationals. Let's consider the set $$J_{p}=\{\sigma(j) \ : \  j \ \text{ is the } j\text{-invariant of some CM elliptic curve over } \overline{\mathbb Q}\}.$$
Is $J_p$ dense in any neighborhood in $\overline{\mathbb Q_p}$? Does it help if we restrict to a finite extension of $\mathbb Q_p$, or even just to $\mathbb Q_p$ itself?

REPLY [7 votes]: All accumulation points of $J_p$ in $\mathbb{C}_p$ are roots of degree two monic equations over $\mathbb{Z}_p$, and their approximants are necessarily supersingular at $p$. Moreover, there exist accumulation points. (There is then a further restriction: the reduction of the point has to be one of the $\approx p/12$ supersingular residues in $\mathbb{F}_{p^2}$. Is this perhaps the only restriction?)
First, extending Pete Clark's remarks, the ordinary CM points have no accumulation point in $\mathbb{C}_p$. (So no, the CM invariants aren't dense in any of the ordinary disks.) This is similar to the corresponding fact about the $p$-adic roots of unity. The analogy here is substantiated by the Serre-Tate theory; cf. Prop. 3.5 in de Jong and Noot's paper Jacobians with complex multiplication. Building upon this, P. Habegger (The Tate-Voloch conjecture in a power of a modular curve, Int. Math. Res. Notices 2014) established much more generally that no algebraic subvariety $V/\mathbb{C}_p$ in a power of the modular curve is $p$-adically approximated by ordinary CM points not lying in $V$. The prototypical $\mathbb{G}_m^r$ case, where the special points are the torsion ones, had been established by Tate and Voloch in the same journal (Linear forms in $p$-adic roots of unity, 1996).
So the question reduces to approximating with supersingular points. These belong to the valuation ring of a quadratic extension of $\mathbb{Q}_p$, hence the claim in my opening paragraph. Habegger's Proposition 2 proves that $0$ is an accumulation point of supersingular CM points, in order to demonstrate that the restriction to ordinary points in his main result is essential. This should work on other examples, though I am unsure exactly which quadratic integral elements over $\mathbb{Z}_p$ may be approximated with Habegger's method. At least this shows the existence of accumulation points.<|endoftext|>
TITLE: Help in understanding result from publication on operator theory
QUESTION [5 upvotes]: in my research on dilations of contractions on Hilbert spaces and manifolds I have come across this nice publication concerning the classic Sz-Nagy theorem on the Arxiv by Levy and Shalit which states the classical Sz-Nagy theorem on the existence of unitary and isometric dilations for contractions and their minimality:
http://arxiv.org/pdf/1012.4514.pdf
Right below quoting the famous theorem they mentioned that if the operator is not a unitary contraction then we must have that the larger Hilbert space K is infinite dimensional. Please forgive my ignorance but I cannot seem to understand why and it seems like a nice result I wish to understand. If the contraction to be dilated is not a unitary operator then why is the larger space embedding H is necessarily infinite dimensional? I am sorry if it was trivial and I missed it but I think I need to understand this. I thank all helpers

REPLY [3 votes]: A unitary dilation $U\in B(K)$ of $T\in B(H)$ is said to be minimal if
$$\overline{span}\{U^nH : n\in \mathbb Z\} = K.$$
It is a well known result that the minimal unitary dilation is unique up to unitary equivalence.
For any contraction $T\in B(H)$, Schaeffer found a constructive version of Sz.-Nagy's dilation theorem. Specifically,
$$
U = \left[\begin{array}{cccccc} \ddots \\ \ddots & 0\\
& 1& 0 \\ && D_{T^*} & T \\ && -T^* & D_T & 0 \\ &&&&1 & 0 \\ &&&&&\ddots&\ddots \end{array}\right]
$$
is a unitary dilation of $T$, where $D_X = (1-X^*X)^{1/2}$. 
Hence, let $K' = \overline{span}\{U^nH : n\in \mathbb Z\} \subset K$. $K'$ is a reducing subspace of $U$ and $U|_{K'}$ is the minimal unitary dilation of $T$. If $T$ is not a unitary already then either it isn't an isometry ($D_T \neq 0$), a co-isometry ($D_{T^*} \neq 0$) or both. Either way it is direct to see that $K'$ is infinite dimensional from the form of the Schaeffer dilation. Therefore, by minimality, any unitary dilation of $T$ must be on an infinite dimensional Hilbert space.<|endoftext|>
TITLE: Deformation long exact sequence of GW theory in the analytical setting
QUESTION [5 upvotes]: Let $f\!=\!(u\colon (\Sigma,p_1,\ldots,p_k) \to X)$ be an element of the moduli space of genus $g$ $k$-marked degree $A$ $J$-holomorphic maps $\mathcal{M}_{g,k}(X,A,J)$. For simplicity assume $C=(\Sigma,p_1,\ldots,p_k)$ is stable.
In the algebraic (or holomorphic) setting, deformation long exact sequence of $\mathcal{M}_{g,k}(X,A,J)$ around $f$ has the form
\begin{align}
0\to \operatorname{Def}(u) &\to \operatorname{Def}(f) \to \operatorname{Def}(C)  \\ 
\to \operatorname{Obs}(u) &\to \operatorname{Obs}(f) \to 0\;;
\end{align}
see Section 24 of "Mirror Symmetry" book by Hori, Katz, Klemm, etc.
In the analytical (symplectic) setting, the first column corresponds to kernel and cokernel of linearization of Cauchy Riemann operator 
$$
D_u\bar\partial\;\colon \Gamma(u^*TX)\to \Gamma(\Omega^{0,1}_\Sigma\otimes u^*TX);
$$
i.e. $\operatorname{Def}(u)=\ker(D_u\bar\partial)$ and $\operatorname{Obs}(u)=\operatorname{coker}(D_u\bar\partial)$.
$\operatorname{Def}(C)$ is as in the algebraic case:
$$
\operatorname{Def}(C)=H^1(T\Sigma(-p_1\cdots -p_k))
$$
Question 0: Does such long exact sequence even always exist in the analytical setting? 
Question 1: What is the analytical description of the map $\operatorname{Def}(C)\to \operatorname{Obs}(u)$? Whether the answer to Q0 is Yes or No, this map should be naturally definable.
Question 2: What are the analytical descriptions of $\operatorname{Obs}(f)$ and $\operatorname{Def}(f)$?
Question 3: Do you know of any reference where this sequence is explained for the analytical setup of GW theory?
Comments: In the case of no-marked points, $\operatorname{Def}(C)=H^1_{\bar\partial}(T\Sigma)$ and we can get the map via $du:T\Sigma \to TX$. In the case there are marked points, the map should be similar, I just have hard time visualizing it. In question 2, if the map is immersion and no-marked points the spaces are similar to that of $u$ with $N_{u(\Sigma)}X$ instead of $TX$.

REPLY [2 votes]: In addition to the nice description of Jason in the comments, 
there is a fairly detailed description of the deformation long exact sequence in Section 3.2 of the article of Siebert-Tian in "Symplectic 4-manifolds and algebraic surfaces".<|endoftext|>
TITLE: Holomorphy of a function with values in a Hilbert space
QUESTION [6 upvotes]: Denote by $\mathbb C^\infty $ the Hilbert space $\ell^2 (\mathbb C)$. Fix $1\leq N,M \leq \infty$, and let $U$ be an open subset of $\mathbb C^N $. Following Mujica's book "complex analysis in Banach spaces", a function $f:U\to \mathbb C ^M $ is called $holomorphic$ if for every $p \in U$ there is a bounded linear map $A:\mathbb C^N \to \mathbb C^M$ such that
$\displaystyle \lim_{h\to 0} \frac{||f(p+h)-f(p)-Ah||}{||h||}=0$.
In Mujica's book, theorem 8.12 says that a function $f:U\to \mathbb C^M $ is holomorphic if and only if, for every bounded linear map $\psi : \mathbb C^M \to \mathbb C$, the function $\psi \circ f$ is holomorphic.
Denote $\pi_j:\mathbb C^M \to \mathbb C$ the map that to every $z\in \mathbb C^M$ associates its $j$-th component. Then $\pi_j $ is a bounded linear map, so $f^j := \pi_j \circ f$ is holomorphic for every $j$.
My question is: is the converse true? I.e. is it true that if every $f^j$ is holomorphic then $f$ is holomorphic?
The answer is obviously yes if $M$ is finite. I think that, if $f$ is continuous, then the answer is yes also in the infinite-dimensional case. Can we drop the continuity hypotesis?
Thank you.
EDIT2: I add my proof attempt of the fact that it suffices to assume $f$ continuous. So, let $f:U\to \mathbb C^\infty$ (with $U\subseteq \mathbb C^N$ and $N\leq \infty$) continuous such that, for every $j$, $f^j$ is holomorphic. Let $\psi:\mathbb C^\infty \to \mathbb C$ be a bounded linear map: then $\psi$ acts like $z\mapsto <z,u>$ for some $u\in \mathbb C^\infty$, so $f$ is holomorphic if and only if the map $z\mapsto <f(z),u> $ from $U$ to $\mathbb C$ is holomorphic for every $u\in \mathbb C^\infty$. Now, select $u\in \mathbb c^\infty$ and denote $g_n (z) = \sum _{j=1}^{n} f^j (z) \overline u^j $ and $g=\lim_n g_n $: since every $g_n $ is holomorphic because every $f^j$ is, it suffices to prove that $g_n \to g$ uniformly on compact subsets. But, if $K$ is a compact of $U$, then $\|g_n (z) -g(z)\|\leq \|<f(z),u^{>n}>\|\leq \|f(z)\|\|u^{>n}\|\leq R_K \|u^{>n}\|$ for $z\in K$, where $u^{>n}$ is the vector of $\mathbb C^\infty$ with $j$-th component equal to $0$ if $j\leq n$, and equal to $u^j$ if $j>n$, and $R_K$ is a positive constant that bounds $\|f\|$ on $K$. Since $\|u^{>n}\|\to 0$ for $n\to \infty$, by generality of $u$, we have the thesis.
EDIT3: following this question, I found a counterexample in the case $N=\infty$ and $M=\infty$. Define $f:\mathbb C^\infty \to \mathbb C^\infty$ by setting
$f^1 (z) = z^1$;
$f^2 (z) = f^3 (z) = \frac{1}{\sqrt 2} (z^2 + z^3)$;
$f^4 (z) = f^5 (z) = f^6 (z) = \frac{1}{\sqrt 3} (z^4 + z^5 + z^6)$;
and so on. Then the Jacobian matrix of $f$ is the one given in this example, and does not represent a bounded linear operator.
However, the question still remains unanswered for finite $N$ and infinite $M$.

REPLY [7 votes]: No. There are non-holomorphic functions $f:\mathbb D\to \ell^2$ such that all components $f_n=\pi_n\circ f$ are holomorphic. This follows from a general result of Arendt and Nikolski [Vector-valued holomorphic functions revisited. Math. Z. 234 (2000), no. 4, 777–805]: 
Theorem 1.5 Let $X$ be a Banach space and $W$ a subspace of $X'$ which
does not determine boundedness. Then there exists a function $f : \mathbb D \to X$
which is not holomorphic such that $\varphi \circ f$ is holomorphic for all $\varphi\in W$.
Here, a subspace $W\subseteq X'$ is said to determine boundedness if, for all subsets $B\subseteq X$, the condition $\sup\lbrace |\varphi(x)|: x\in B\rbrace <\infty$ for all $\varphi\in W$ implies that $B$ is bounded. This theorem is applied to $X=\ell^2$ and the linear span $W$ of all projections $\pi_n$. It does not determine boundedness by considering $B=\lbrace ne_n:n\in\mathbb N\rbrace$ with the unit vectors $e_n\in\ell^2$.<|endoftext|>
TITLE: "Common-neighbor-regular" graphs
QUESTION [5 upvotes]: Let $G=(V,E)$ be a finite, simple, undirected graph. For $v\in V$ we set $N(v)=\{w\in V:\{v,w\}\in E\}$.
We say that $G$ is $k$-common-neighbor-regular if for all $v\neq w\in V$ we have $|N(v)\cap N(w)|=k$. (I haven't been able to find out whether there is a canonical term for this concept.)
Clearly the complete graph on $k+2$ points is $k$-common-neighbor-regular, but for $k>0$ I haven't been able to find non-complete examples.
Question: Given $k>0$, for which $n\in\mathbb{N}$ does there exist a $k$-common-neighbor-regular graph $G=(V,E)$ with $|V|=n$?

REPLY [2 votes]: This problem was (partially?) solved by Kelmans in 1973 and by Ralucca Gera and Jian Shen in 2008.  See this paper including the comment. They consider the generalization that friends have $\lambda$ common friends and non-friends have $\mu$ common friends. If the graph is regular, it is strongly-regular by definition. If it is irregular, it is either a disjoint union of complete graphs of some specified sizes, or a disjoint union of complete graphs of the same size plus an extra vertex joined to everything.<|endoftext|>
TITLE: When does a $C^*$-algebra have no nonzero projection?
QUESTION [7 upvotes]: Let $A$ be a $C^*$-algebra and $\hat{A}$ its spectrum of $A$,the set of classes of non-zero irreducible representation of $A$ endowed with hull-kernel topology. suppose $\hat{A}$ is a non-compact connected Hausdorff space.
Why  $A$ cannot contain a nonzero projection?

REPLY [6 votes]: Given $x\in A$ and $\alpha>0$ the set of $\pi\in \hat A$
such that $\|\pi(x)\|\geq \alpha$ is compact (Proposition 3.3.7 in Dixmier's ``C*-algebras"). Applied to a non-zero projection $p$, the set
of $\pi\in \hat A$ such that $\pi(p)\neq 0$ is compact (since then $\|\pi(p)\|=1$)
and non-empty. This set is also open (because $\pi\mapsto \|\pi(x)\| $ is l.s.c.; Proposition 3.3.2 in Dixmier's). In the question the spectrum is Hausdorff and connected. So we get a non-empty clopen set which must be the whole space. But the spectrum has also been assumed noncompact.<|endoftext|>
TITLE: Multiplicative infinitesimals in q-analogs?
QUESTION [12 upvotes]: Risking to be downvoted, here is a very lightweight question.
In various fields - say, algebraic geometry, nonstandard analysis, synthetic differential geometry - infinitely small quantities, i. e. those very-very close to zero, are represented by nilpotents, sometimes just square zero elements suffice to do quite a big portion of analysis. To give just one example, a generic $\varepsilon$ with $\varepsilon^2=0$ is used to represent things like tangent vectors, etc.
Now thinking about $q$-analogs, like $q$-derivative and similar gadgets, I am wondering what is the multiplicative analog of the above? That is, is there a way to capture algebraically quantities very-very close to 1? Or concisely -

"$\varepsilon$ very-very close to zero" is to "$\varepsilon^2=0$"
as
"$q$ very-very close to 1" is to what?

REPLY [2 votes]: There is more than one question that is being asked here so I will leave aside the one about $q$-analogues for the simple reason that one can take any question, say $X$, in mathematics, and ask for its $q$-analog, $X_q$, so things can get pretty monotonous.
As far as the multiplicative version of being "very small" is concerned, this can be expressed in Smooth Infinitesimal Analysis and algebraic geometry in terms of nilsquare infinitesimals, but it also has a straightforward meaning in the hyperreals as "being smaller than any positive real", so that one defines the additively invariant relation $x\approx y$ by requiring $x-y$ to be infinitesimal. Meanwhile, Fermat and Leibniz had a (multiplicatively invariant) relation that had more of a multiplicative character, which we will denote $x\;{}_{\ulcorner\!\urcorner}\; y$, which holds if and only if $\frac{x}{y}\approx 1$. The reason this is closer to Fermat's ideas is because Fermat would divide his adequalities at will by his $E$ which is of course impossible for the additively invariant relation.<|endoftext|>
TITLE: Useful, non-trivial general theorems about morphisms of schemes
QUESTION [43 upvotes]: I apologize in advance as this is not a research level question but rather one which could benefit from expert attention but is potentially useful mainly to novice mathematicians.
I'm trying to compile a list of non-obvious theorems about morphisms of schemes which are useful for general intuition but whose proofs are not easy/technical. Here are some examples:

Zariski's Main Theorem: Let $Y$ be quasi-compact, separated and $f:X \to Y$ be separated, quasi-finite, finitely presented. Then there is a factorization $X \to Z \to Y$ where the first map is an open immersion and the second is finite. Mnemonic: (quasi-finite) $\sim$ (finite) $\circ$ (open immersion)



Nagata's compactification theorem: Let $S$ be qcqs and $f:X \to S$ be separated, finite type. Then $X$ densely embeds into a proper $S$-scheme.Mnemonic: non-horrible schemes have compactifications 



Temkin's factorization theorem: Let $Y$ be qcqs and $f: X \to Y$ be separated, quasi-compact. Then there's a factorization $X \to Z \to Y$ with the first being affine and the second proper. Mnemonic: (separated + quasicompact) = (proper) $\circ$ (affine).  



Chow's lemma: Let $S$ be noetherian and $f: X \to S$ separated finite type. Then there exists a projective, surjective $S$-morphism $\bar{X} \to X$ which is an isomorphism on a dense subset and where $\bar{X} \to S$ is quasi-projective. Moreover $X$ is proper iff $\bar{X}$ is projective, and if $X$ is reduced $\bar{X}$ can be chosen to be so as well. Mnemonic: reasonable schemes have quasi-projective "replacements" and proper schemes have projective "replacements" 


Hopefully it's clear now what I'm looking for. All theorems above have very weak assumptions and very satisfying conclusions. These are what I'm after.

REPLY [14 votes]: Here's a theorem I find useful:


Theorem. Let $\phi \colon X \to Y$ be a smooth morphism of schemes of relative dimension $d$. Then there exists an open cover $X = \bigcup U_i$ of $X$ such that each $U_i \to Y$ factors as
    $$U_i \stackrel \pi \to \mathbb A^d_Y \to Y,$$
    with $\pi$ étale. Mnemonic: smooth morphisms have étale coordinates.


See Tag 054L.<|endoftext|>
TITLE: Etale cohomology of $\mathrm{Spec}(k\{X,Y\})\backslash\langle0,0\rangle$
QUESTION [10 upvotes]: Illusie in "Grothendieck et la cohomologie étale" says Artin's Harvard notes on Grothendieck Topologies prove: The étale cohomology with coefficients in $Z/nZ$ of the variety $\mathrm{Spec}(k[X,Y])\backslash\langle0,0\rangle$ for any algebraically closed field $k$ agrees with the cohomology of the 3-sphere.  
This seems to refer to 3.5 on Artin's p. 110.  But that result is actually stated a little differently.  It describes the étale cohomology with those coefficients of  $\mathrm{Spec}(k\{X,Y\})\backslash\langle0,0\rangle$ where $k\{X,Y\}$ is the the Henselization  of $k[X,Y]$ localized at the origin.  Artin does suggest you should think of this scheme as a 4 ball minus a point. (Also Artin notes $k$ need only be separably algebraically closed.)
Does Artin's 3.5 somehow easily yield the result for  $\mathrm{Spec}(k[X,Y])\backslash\langle0,0\rangle$?
Besides that Will Sawin's answer is easy enough for me to see, it is easy in a historic sense: These are largely considerations Artin would have known in the fall of 1961.  He did not have the affine dimension theorem in general -- but he did for curves. If he did not yet have a rigorous étale Kunneth formula, he expected something like it, and might have worked around it in the case of constant sheaves on $\mathbb A^2$ over a separably closed field. So this supports the idea that this result on $\mathrm{H}^3$ of $\mathbb A^2 \backslash \langle 0,0\rangle$ is what Artin later called the first theorem in higher dimensional étale cohomology (in an interview, in Joel Segel, ed., Recountings: Conversations with MIT Mathematicians, A K Peters, 2009, pp. 351–74).  And it seems this could be close to how he first proved it.

REPLY [14 votes]: I'm not sure what "easy" means in the context of etale cohomology but there is a way of passing from Artin's result to the stated one.
Let $j$ from $\mathbb A^2 \backslash \langle 0,0\rangle$ to to $\mathbb A^2$ be the open immersion. Then there is a Leray spectral sequence relating the etale cohomology of $\mathbb A^2 \backslash \langle 0,0\rangle$ to the etale cohomology of $\mathbb A^2$ with coefficients in $R^i j_* \mathbb Z/n\mathbb Z$. 
The key point is that the stalk of $R^i j_* \mathbb Z/n\mathbb Z$ at $\langle 0,0\rangle $ is precisely the $i$th etale cohomology group of $\operatorname{Spec} k\{X,Y\} \langle 0,0\rangle $ - because by definition it is the limit of the etale cohomology groups of punctured neighborhoods of $ \langle 0,0\rangle $, and these converge to the etale cohomology group of $\operatorname{Spec} k\{X,Y\} \langle 0,0\rangle $.
Furthermore, note that $R^0 j_* \mathbb Z/n\mathbb Z$ is the constant sheaf $\mathbb Z/n\mathbb Z$ by direct computation and $R^i j_* \mathbb Z/n\mathbb Z$ vanishes away from $\langle 0,0\rangle $ for $i>0$.
So $R^i j_* \mathbb Z/n\mathbb Z$ is zero for $i \not\in\{ 0,3\}$, is the constant sheaf $\mathbb Z/n\mathbb Z$ for $i=0$, and is the skyscraper sheaf for $i=3$. Taking cohomology, $H^p(\mathbb A^2, R^q j_* \mathbb Z/n\mathbb Z)$ vanishes for $p \neq 0$ or $q \not\in\{ 0,3\}$ and is $\mathbb Z/n$ for $p=0, q\in\{0,3\}$ so the spectral sequence degenerates and we get the desired result.<|endoftext|>
TITLE: Positivity of power of positive PSD matrices
QUESTION [10 upvotes]: Background: Let $M$ be an $n\times n$ matrix with nonnegative entries. It is immediate that for any integer $k$, $M^k$ has nonnegative entries.
Suppose now that, on top of having nonnegative entries, $M$ is a positive semi-definite matrix (i.e., it is symmetric and has nonnegative eigenvalues). Now one may ask whether for any real $p\ge 1$, $M^p$ has nonnegative entries. Surprisingly, the answer is a no. The matrix
$$
M=
  \begin{bmatrix}
0.96523 & 2.6398 & 0.012905 & 0.059013\\
2.6398 & 10.053 & 3.0808 & 0.029887\\ 
0.012905 & 3.0808 & 26.252 & 3.2929\\
0.059013 & 0.029887 & 3.2929 & 0.52308
  \end{bmatrix}
$$
has positive eigenvalues, however $M^{1.5}$ has negative entries. This example is due to Koenraad Audenaert.  
A nicer looking matrix of the same kind is $$
M=
  \begin{bmatrix}
2 &6 &0.1 &0.1\\
6 &30 &6 &0.1\\
0.1 &6 &30 &6\\
0.1 &0.1 &6 &2
  \end{bmatrix}
$$
Question: Are there such (i.e., $M^p$ has negative values) positive PSD matrices whose eigenbasis is the discrete Fourier transform?
To be more explicit, let us fix the Fourier transform on $\mathbb F_2^n$ as
$$H_{x,y} = (-1)^{\langle x,y\rangle}\quad x,y\in\mathbb F_2^n$$
Is there a nonnegative function $f:\mathbb F_2^n\to\mathbb R_+$ such that $Hf\geq 0$ however, $Hf^p$ has negative values for some $p\geq 1$? Here, $f^p$ denotes the coordinate-wise powering of $f$.
Recall that $H/2^{n/2}$ diagonalizes matrices of the form $M(x,y)=f(x+y)$, so the existence of such an $f$ is equivalent to existence of positive PSD matrices diagonalized in the Fourier basis whose $p$th power contains negative entries.
Thank you

REPLY [3 votes]: There is an $f$ such that $f\ge 0$, $Hf\ge 0$ however $Hf^{p}$ contains negative values for some $p\ge 1$. Take
$$
f = \begin{bmatrix} 727 & 200 & 163 & 234 & 429 & 448 & 437 & 6 \end{bmatrix}^T
$$
The vector $Hf$ is positive, however $(Hf^{8/7})_{100}\approx -3.35$. Here the subscript $100\in \mathbb F_2^3$ is the 5th index in lexicographic order.<|endoftext|>
TITLE: Density of polynomials in $C^k(\overline\Omega)$
QUESTION [19 upvotes]: Let $\Omega$ be an open and bounded subset of $\mathbb{R}^2$ and let $C^k(\Omega)$, $1\leq k<\infty$, be the space of functions $f$ with continuous derivatives of order $\leq k$ in $\Omega$, endowed with the usual topology of semi-norms
$$|f|_{K}=\sup_{|p|\leq k}~\sup_{x\in K}|(\partial/\partial x)^p f(x)|,$$
where $K$ runs over the compact subsets of $\Omega$. Then, the polynomials are dense in $C^k(\Omega)$, see e.g. Treves, Topological vector spaces, distributions and kernels.
Let us now consider the subset $C^k(\overline\Omega)\subset C^k(\Omega)$ of functions whose derivatives extend continuously up to the boundary of $\Omega$, endowed with the topology of the norm
$$\|f\|_{\overline\Omega}=\sup_{|p|\leq k}~\sup_{x\in\overline\Omega}|(\partial/\partial x)^p f(x)|.$$
Is it true that the polynomials are dense in $C^k(\overline\Omega)$ ?
By Whitney extension theorem, if $\Omega$ has some smoothness, like quasi-convexity, functions of $C^k(\overline\Omega)$ can be extended, which implies density of polynomials, so my question is about general $\Omega$.
I would appreciate any reference, counter-example,...
I am also interested in the case of $C^\infty(\overline\Omega)$ defined as the intersection of all $C^k(\overline\Omega)$.

To answer the comment by ACL, here is an exemple of a domain $\Omega\subset\mathbb{R}^{2}$ and a function $f\in C^{1}(\overline\Omega)$ which cannot be extended to any neighborhood of $\overline\Omega$ :
Define the domain $\Omega$ to be the square centered at the origin, of length 2, from which a spine delimited by
$y=\pm e^{-1/x}$, $x\geq 0$, has been removed.
The function $f$ is the zero function except on the first quadrant where
 $f(x,y)=x^{2}$. Then $f$ is $C^{1}$, and $f$ and its derivatives of order 1 admit continuous limits on the boundary of $\Omega$, hence, by definition,
$f\in C^{1}(\overline\Omega)$. If $f$ could be continued to a 
$C^{1}$ function on a neighborhood of
$\overline\Omega$, it would be Lipschitz at 0, 
which is not the case since there is no constant $C$ such that
$x^{2}\leq 2Ce^{-1/x}$ when $x\to 0$.

REPLY [9 votes]: No, the polynomials will not be dense in general.
The following example is essentially one-dimensional. Let $C\subset[0,1]$ be the usual ternary Cantor set and $g\colon[0,1]\to[0,1]$ the Cantor function (a.k.a. Devil's staircase). Then $g$ is continuous and locally constant on the open set $U := (0,1)\setminus C$.
Set $\Omega := U\times(0,1)$ and let $f\colon\Omega\to\mathbb{R}$ be given by $f(x,y)=g(x)$. Obviously $f\in C^\infty(\Omega)$ and $f$ and all of its partial derivatives have continuous extensions to $\overline{\Omega}=[0,1]\times[0,1]$. So $f\in C^\infty(\overline{\Omega})$ by definition.
But $f$ cannot be approximated in $C^1(\overline{\Omega})$ by a sequence of polynomials $(p_k)$ as $\partial_x p_k$ would need to converge uniformly to $0$ on $\overline{\Omega}$, which is impossible since for any fixed $y\in(0,1)$ one then has both
$$p_k(1,y)-p_k(0,y)=\int_0^1 \partial_x p_k(t,y)dt\to 0$$
and
$$p_k(1,y)-p_k(0,y)\to g(1)-g(0)=1$$
as $k\to\infty$.
By a straightforward modification (add a horizontal box to the bottom of $\Omega$ and multipy $f$ with a suitable smooth cut-off function) one can obtain an example where $\Omega\subset\mathbb{R}^2$ is connected.

It is also possible to make $\Omega$ topologically regular, i.e. make it satisfy $\operatorname{Int}(\overline{\Omega})=\Omega$. This can be accomplished by punching tiny holes $B_k$ into $\Omega$ (so the $B_k$ are disjoint closed balls contained in the original $\Omega$) that accumulate exactly at $C\times[0,1]$. Note that the holes can made arbitrarily tiny to ensure that $\sum_k\mathcal{H}^1(\partial B_k)<1/4$, where $\mathcal{H}^1$ denotes the one-dimensional Hausdorff measure. In particular, one can still apply the above argument based on the fundamental theorem of calculus on some horizontal line that does not hit the holes.<|endoftext|>
TITLE: K-theory of non-compact spaces
QUESTION [17 upvotes]: This is a question on nomenclature of $K$-theory in the topological category.
The $K$-theory of a compact space $X$ is defined as the Grothendieck group of the vectorbundles on $X$. The Atiyah-Jänich Theorem states that this is the same thing as the homotopy classes of maps $X\rightarrow \Phi(\mathbb{H})$, where $\Phi(\mathbb{H})$ is the space of Fredholm operators on some separable infinite-dimensional Hilbert space.
Now, for a non-compact space $Y$ the $K$ theory $K(Y)$ is not defined as the Grothendieck group of vector bundles on $Y$. One can do a couple of things:

Define it as the $K$-theory of its one point compactification
$K(Y):=K(Y_+)$. One needs to assume $Y$ is locally compact for this
to make sense.
Another option is to assume that $Y$ is nice, for example an
infinite $CW$ complex such as $CP^\infty$, and to define the $K$-theory of $Y$ as the limit of its finite subcomplexes. Note that $CP^\infty$ is not locally compact.
Yet another option is to define the $K$-theory of $Y$ as maps
$Y\rightarrow \Phi(\mathbb{H})$. I believe this is called representable
$K$-theory.

I have a couple of questions.

Why is 1. a good definition? I like my cohomology theories to be
      functorial under maps. Theory 1. clearly is not. I believe it is functorial
      with respect to proper maps. This reminds me of compactly supported
      cohomology. But why is this then not called compactly supported $K$-theory? 

And:

How are 2. and 3. related? 

More specifically:

What exactly does 3. describe? Are these virtual vector bundles that admit numerable trivializations?

finally:

Is there a reference where all these definitions are discussed?

REPLY [4 votes]: What exactly does 3. describe? Are these virtual vector bundles that admit numerable trivializations?

Virtual vector bundles, when defined as formal differences (i.e.,
elements in the homotopy group completion) of vector bundles,
do not form a sheaf when considered on noncompact manifolds.
The reason for this is that a compatible system of such differences
considered on some open cover might use differences
of higher and higher dimensional vector bundles as one approaches
the infinity, whereas any virtual vector bundle is a difference
of two bundles of fixed dimension.
Put differently, the homotopy group completion functor, being
a left adjoint, does not preserve the sheaf condition,
which is defined using a homotopy limit.
However, the problem can be easily rectified by replacing
the above presheaf with its sheafification.
With such an extended notion of virtual vector bundles
one then proves that their isomorphism classes are represented
by the connective K-theory spectrum.
(The above is literally true for smooth manifolds with the standard
Grothendieck topology; for topological spaces one must
take the numerable Grothendieck topology, which produces numerable bundles.)<|endoftext|>
TITLE: What is the mistake in the proof of the Homotopy hypothesis by Kapranov and Voevodsky?
QUESTION [95 upvotes]: In 1991, Kapranov and Voevodsky published a proof of a now famously false result, roughly saying that the homotopy category of spaces is equivalent to the homotopy category of strict infinity categories that are weak infinity groupoid.
In 1998 Carlos Simpson showed that their main result could not be true, but did not explain what was precisely wrong in the paper of Kapranov and Voevodsky.
In fact, as explained by Voevodsky here, for a long time after that, Voevodsky apparently thought his proof was correct and that Carlos Simpson made a mistake, until he finally found a mistake in his paper in 2013 !
Despite being false, the paper by Kapranov and Voevodsky contains a lot of very interesting things, moreover, the general strategy of the proof to use Johnson's Higher categorical pasting diagram as generalized Moore path to strictify an infinity groupoid sound like a very reasonable idea and it is a bit of a  surprise, at least to me, that it does not work.
In fact when Carlos Simpson proved that the main theorem of Kapranov and Voevodsky's paper was false he conjectured that their proof could allow one to obtain that the homotopy category of spaces is equivalent to the homotopy category of strict non unital infinity category that are weak (unital) infinity groupoid (this is now known as Simpson's conjecture).
So:
Can someone explain what precisely goes wrong in this paper ?

REPLY [35 votes]: It's been more than a year and a half since I asked this question and I had a lot of thought about it so I decided I will post my own answer.
First I entirely agree with Yonatan that the main problem is with lemma 3.4. The specific problem that he mentions appears exactly because of the "degeneracy maps" that are added by Voevodsky and Kapranov to their category of diagrams. More precisely, one can, using the degeneracies, construct a diagrammatic set whose realization will "collapse" because of the Eckman-Hilton argument, and quite interestingly if one modifies their definition so that the category has no degeneracies then this no longer happens (the free $\infty$-category is just obtained by "freely adding arrows" gradually as they assume it behaves in the paper). So if one thinks that degeneracies correspond exactly to units, this is very encouraging for the Simpson conjecture. I haven't been able to make this into a clear counterexample of the lemma, but only because the lemma actually has other problems that appear before that.
In the end, I believe the main obstruction to their proof is the following: the initial idea to use "generalized Moore homotopy" parametrized by some class of diagrams seems (at least intuitively) to need the following two properties of the class of diagrams: 
1) One should be able to formally "compose" the diagrams (and that it corresponds to pushout on the level of geometric realization) so that when you look at all the continuous functions $|D| \rightarrow X$ for all diagrams $D$ you indeed get an $\infty$-category.
2) That given two "parallel" $n$-diagrams, you can construct an $(n+1)$-diagram whose source and target are the two given $n$-diagrams, so that if two diagram shapes are used to represent homotopically equivalent $n$-arrows then one can actually have an $(n+1)$-arrow that represents this homotopy.
It appears that both these properties fail for the kind of diagrams (Johnson diagrams) they are using! Unfortunately, due to the fact that they actually use a slightly different construction than the one they explain in the introduction, these do not immediately translate into mistakes in their paper.
This being said, they actually seem to use that Johnson diagrams can be composed within the proof of Lemma 3.4 mentioned above, so that it is probably a second reason for which this lemma fails.
It is not clear to me if (and where) they use the second property somewhere, but I expect some property of this kind should be important in order to prove that the geometric realization of diagrammatic sets indeed induces an equivalence with the category of spaces (and they are extremely imprecise about how this equivalence is obtained, they just say that "one does exactly as for simplicial and cubical sets"!).
For more details (and a third reason why Lemma 3.4 fails) I have a very recent preprint (https://arxiv.org/abs/1711.00744) which constructs a category of diagrams that has the two properties mentioned above as soon as you work in a 'non-unital' framework, unfortunately this category of diagrams is a lot more complicated than the category of Johnson diagrams (and it is unique so this complication is unavoidable) and this prevents from using the exact same strategy as they do. I discus in details in the appendix of the paper the proof of Kapranov and Voevodsky (this will expand a lot on this answer) and explain some ideas on how to make it into a proof of the Simpson conjecture using the category of diagrams that I constructed. This new version also has the advantage to make the two ways of explaining the construction (in terms of generalized Moore homotopies and using two adjunctions with a presheaf category of diagrams in the middle) actually equivalent.
Update : In fact, in a subsequent preprint, I did proved a version of the Simpson conjecture using essentially the strategy of Kapranov and Voevodsky with a modified category of diagrams.
Note that there are still some difficulties appearing (due to the increased complexity of the category of diagram) and at the moment I'm still not capable of proving the most general version of the Simpson conjecture. To be precise, at this point I'm only able to strictify a certain set of composition operations, which I call the "regular composition operations", (informally they are those whose pasting diagram is "topologically regular") which are such that any kind of composition operation that you have in an $\infty$-category can be obtained as a regular composition of identities and non identities arrow. So it does gives a notion where you have a bunch of operations that are strictly compatible, and only weak identities on top of that, but one can still hope to find stronger statement with more strict operations.<|endoftext|>
TITLE: Is there a short expression for height and width of product and coproduct of posets?
QUESTION [7 upvotes]: I am trying to derive some basic relations for the height and width of the direct product and the coproduct of posets. I feel that these are very basic and should be written somewhere, however, I cannot find a reference.
Short question is: is there a short expression for the following quantities, representing height and width of product and coproduct of posets? And do they hold also in the case of infinite cardinality? 
Edit: current status (with help from Harry Altman, David Spivak) of this question:
[resolved] $\color{green}{ w(P\coprod Q) = w(P)+w(Q) }$  
[resolved] $\color{green}{ h(P\coprod Q) = \max\{h(P), h(Q)\}}$ 
[resolved] Assuming $P$ and $Q$ not empty, then $\color{green} {h(P \times Q) = h(P)+h(Q)−1 }$.  (For empty posets, then $h(P \times Q) = 0 \neq h(P)+h(Q)-1$.)
[resolved] From a theorem in Berzukov, Roberts, "On antichains in product posets", it follows that the width can be bounded as follows, with both bounds attainable:
$\color{green}{ w(P)w(Q)\leq w(P\times Q) \leq \min\{|P|\ w(Q), |Q|\ w(P)\}}  $ 
Original question below.

Preliminaries Define:

$C_n$ to be a chain of size $n$. For example take $C_n = \langle\{1, \dots, n\}, \leq\rangle$.
$A_n$ to be an antichain of size $n$, that is, a set with $n$ incomparable elements.
$P \times Q$ the direct product of two posets.
$G_{m,n}$  is a grid; for example $G_{m,n} = C_n \times C_m$.

The height and width of a poset are defined as:

the height $h(P)$ is the cardinality of the longest chain in $P$.
the width $w(P)$ is the cardinality of  the longest antichain in $P$.

Some simple examples:
Width of a chain: $w(C_n) = 1$.
Height of a chain: $h(C_n) = n$.
Width of an antichain: $w(A_n) = n$.
Height of a antichain: $w(A_n) = 1$.
Width of an $m\times n$ grid: $w(G_{m\times n}) = \min\{m,n\}$
Height of an $m\times n$ grid: $w(G_{m\times n}) = m + n -1$
Questions
Is there a simple expression for the height and width of a product and a coproduct of a poset?
This is what I got so far.
For a co-product: 
The height must be the maximum of the two heights, because chains belonging to different factors are uncomparable:
$ h( P \coprod Q) = \max\{ h(P), h(Q) \}$
For the width, the widths of the factors sum together:
$ w( P \coprod Q) =  h(P) + h(Q) $
This is because I can take an antichain $S_1$ in $P$ and one antichain $S_2$ in Q, and then $S_1\cup S_2$ is an antichain in $P \coprod Q$.
For a product, I am not sure.
For the height of a product I can certainly say that 
$h(P\times Q) \geq  h(P) + h(Q) - 1$
because I can construct a chain of that size. If $C=\{1,2,\dots,h(P)\}$ is the longest chain in $P$ and $D = \{a,b,\dots\}$ the longest chain in $Q$ then I can construct the chain $E = \{(1,a), (2,a), \dots, (h(P), a), (h(P), b), \dots\}$ that has height $ h(P) + h(Q) - 1$.
I am also not sure if any of the above fails for posets of infinite cardinality.

REPLY [6 votes]: Sticking first to finite sets, for the question of $h(P\times Q)$, one does in general have $h(P \times Q)=h(P)+h(Q)-1$.  You've already proven the lower bound.  For the upper bound, take a chain $(a_1,b_1),\ldots,(a_n,b_n)$ in $P\times Q$; let's assume this is written in increasing order.  (Note this is strictly increasing.)  Then each time we go from $(a_i,b_i)$ to $(a_{i+1},b_{i+1})$, at least one of the coordinates must increase (the other is allowed to stay the same).  But the first coordinate can only increase $h(P)-1$ times, and the second only $h(Q)-1$ times.  So the total number of elements in the chain is at most $(h(P)-1)+(h(Q)-1)+1=h(P)+h(Q)-1$.
If you want to generalize with infinite posets, you should make sure you know exactly what definitions you want to work with -- is it really cardinality that you want to look at?  I suppose for $w(P)$ you'd have to just use cardinality, as I don't think there's really any other good way to measure the "size" of an antichain in a general poset.  But for height you can possibly do more.  For instance, if you are working with well-founded partial orders, you might want to look at the largest embedded ordinal to get more information.
Note also that in this case your maxima might have to be replaced by suprema; for instance, consider the disjoint union of a chain of length $k$ for every $k$.  This has no longest chain!
For well-founded partial orders, the supremum of all embedded ordinals is (I'm pretty sure?) the same as what's generally known as the height of the order in that context.  There is actually a similar formula for $\ell(P\times Q)$ in that case but I will come back and edit in later if nobody else has already stated it, I am typing this in a bit of a hurry, sorry.
Edit: Let me also add briefly -- your grid example already shows that there can be no formula for $w(P\times Q)$ in terms of $w(P)$ and $w(Q)$, since there you have $w(P)=w(Q)=1$ but $w(P\times Q)$ arbitrarily large.  This can be extended to the infinite realm as well; if you have some given cardinal, take a totally ordered set $X$ of that cardinality, and consider $X\times X'$, where $X'$ is the same set as $X$ but with the reverse order.  Then the "diagonal" is an antichain.  But maybe you were intending to allow for other quantities in the formula?
(Also, obviously all the coproduct stuff will work for anything infinite, and should continue to unless you are using some very strange definitions.)
Edit: More on the width -- here's an example that shows that $w(P)$, $w(Q)$, $h(P)$, and $h(Q)$ are not enough to determine $w(P\times Q)$.  Say $P$ is a poset on three elements, with two elements forming an antichain and the third on top; and say $Q$ is the reverse.  Then $w(P)=w(Q)=h(P)=h(Q)=2$, but (if I've done this correctly) $w(P\times P)=4$, while $w(P\times Q)=5$.
As for if one only wants a bound -- well, for a finite set $P$, Dilworth's theorem implies that $|P|\le h(P)w(P)$, and certainly $w(P)\le P$, so one thereby gets the trivial bound $w(P\times Q)\le w(P)w(Q)h(P)h(Q)$.  But I rather doubt that's what you wanted... (there is of course also the easy lower bound $w(P\times Q)\ge w(P)w(Q)$).
Also, to handle $h(P\times Q)$ in the infinite case, if we use the definitions you've give above where we just care about cardinality -- if either $h(P)$ or $h(Q)$ is infinite (and neither is zero, see below), then $h(P\times Q)=\max\{h(P),h(Q)\}$.  Certainly it is at least this; and it's easy to see that $h(P\times Q)\le h(P)h(Q)$ (since its projection onto either coordinate is a chain).  But the product of two nonzero cardinals, at least one of which is infinite, is simply their maximum, answering the question.  Of course, you could ask for a solution that works without axiom of choice, and that I do not have at the moment!
Actually, if we want to nitpick, the formula for $h(P\times Q)$, here and in the finite case, has an exception -- if either $h(P)$ or $h(Q)$ is zero, then of course so is $h(P\times Q)$.
Edit: OK, one last edit -- about the well-founded case I mentioned above: We can define the height of an element of a well-founded partial order, it's the least ordinal greater than the heights of all elements less than it; the height of the partial order is then the least ordinal greater than the heights of all the elements.  The height is usally denoted $\ell$ in this context so that's what I'll do.  Then if you have two WFPOs $X$ and $Y$, and you have $(x,y)\in X\times Y$, then $\ell(x,y)=\ell(x)\oplus\ell(y)$, where $\oplus$ is natural addition.
This means that $\ell(X\times Y)$ is the smallest ordinal greater than any $\alpha\oplus\beta$ for any $\alpha<\ell(X)$ and $\beta<\ell(Y)$.  How can we compute this?  I'll use the "order" of an ordinal to mean the smallest term that appears in its Cantor normal form.  Take the natural sum of $\ell(X)$ and $\ell(Y)$.  If $\ell(X)$ has the higher order, drop everything below the lowest term in $\ell(X)$.  If $\ell(Y)$ has the higher order, same but with $\ell(Y)$.  If the orders are equal, just drop one copy of the lowest term.
So you can see that this genralizes the $h(P\times Q)=h(P)+ h(Q)-1$ that occurs when $P$ and $Q$ are finite and all the terms are $1$.
(And of course again if either of them is zero you get zero.)<|endoftext|>
TITLE: A different equivalence relation on partizan combinatorial games
QUESTION [6 upvotes]: The following definitions are fairly standard, but reworded in a way that will be more appropriate for my question (so what follows is fairly long, but should be easy to read for the experts and might even serve as an introduction to the subject):

A partizan combinatorial game is a set $G$ (of positions or states), together with an element $x_0 \in G$ (the starting position or initial state) and two relations, $L, R \subseteq G^2$ which we call the blue and red edges of the graph $(G, L\cup R)$ (an edge which is both blue and red can be called green), such that the graph in question is well-founded (=progressively finite, =there is no infinite sequence $z_0,z_1,z_2,\ldots$ of elements of $G$ with $(z_i,z_{i+1}) \in (L\cup R)$ for each $i$).  The game is played as follows: starting with $x = x_0$, each player (called Left and Right) selects an edge $(x,y)$ in his respective set ($L$ for Left and $R$ for Right), replacing the position $x$ by its out-neighbor $y$; the player who cannot play loses (something which always happens in finite time by the well-foundedness assumption).
Define the extensional collapse of a game $G$ as follows: first, remove all vertices unreachable from the starting position; then, repeatedly (transfinitely) identify two vertices $x,x'$ which have the same sets of blue-neighbors and the same sets of red-neighbors (i.e., identify $x,x'$ if $\{y : (x,y) \in L\} = \{y : (x',y) \in L\}$ and the same is true for $R$).  Call two games $G,G'$ extensionally equivalent when they have isomorphic extensional collapses and write $G \cong G'$.  For example, any two games in which no player has any legal play ($L=R=\varnothing$) are extensionally equivalent to the trivial game $0$ with only one position.  Essentially everything will be studied up to extensional equivalence.
The sum $G'' := G \oplus G'$ of two games $G,G'$ is the Cartesian product $G\times G'$ with starting position $x''_0 := (x_0,x'_0)$ and edge relations $L'' := (L\times\Delta_{G'}) \cup (\Delta_G\times L')$ and similarly for $R''$, where $\Delta$ stands for the diagonal: i.e., playing a move in $G\oplus G'$ consists of playing a move in $G$ xor playing one in $G'$.  Clearly this is well-founded, and clearly this operation is compatible with extensional equivalence, and it is commutative and associative (up to extensional equivalence, in fact, even isomorphism).  The opposite $-G$ of a game $G$ is the same set $G$ but with $L$ and $R$ exchanged; we write $G \ominus G'$ for $G \oplus (-G')$.
A game $G$ is Conway-positive when Left has a winning strategy no matter who plays first, Conway-negative when Right has (i.e., $-G$ is positive), and Conway-zero when the second player has a winning strategy.  Two games $G,G'$ are Conway-equal when $G-G'$ is Conway-zero.  Let me write $G \doteq G'$ for this.  It is not hard to see that $\doteq$ is an equivalence relation, which is weaker (=larger) than extensional equivalence.
The product $G'' := G \otimes G'$ of two games $G,G'$ is defined informally as follows.  Its positions will be finite combinations of "plus" and "minus" tokens on the vertices of $G \times G'$.  The starting position has a single "plus" token on $(x_0,x'_0)$ (the starting positions of $G$ and $G'$).  At each turn, each player chooses a single token ("plus" or "minus"), at $(x,x')$ say, and chooses edges $(x,y)$ from $G$ and $(x',y')$ from $G'$: for a "plus" token, Left must choose two blue edges or two red edges (i.e., $(x,y) \in L$ and $(x',y') \in L'$, or else $(x,y) \in R$ and $(x',y') \in R'$) and Right must choose a blue edge and a red edge; for a "minus" token, it's the reverse (Left chooses a blue edge and a red edge, and Right chooses two blue edges or two red edges); in either case, the token that was at $(x,x')$ is removed and three new tokens are created: two of the same sign as the original token at $(x,y')$ and $(y,x')$, and one of opposite sign at $(y,y')$.  In brief, we move from $x\otimes x'$ to $(x\otimes y') \oplus (y\otimes x') \ominus (y\otimes y')$.

The product has some nice properties: it is commutative and associative (up to extensional equivalence), it is distributive over the sum (ditto), has $1$ (defined below) as unit element (again, up to extensional equivalence) and $-G \cong (-1) \otimes G$.
For impartial games (those with $L=R=:E$), Conway-equality is exactly the equality of the Grundy ordinals (the Grundy ordinal of an impartial game $(G,x)$ being recursively defined as the least ordinal not equal to the Grundy ordinal of $(G,y)$ for any out-neighbor $y$ of $x$, i.e., $(x,y) \in E$), and the sum and product operations are compatible with this equivalence, and define over the ordinals two operations called "nim sum" and "nim product" which have plenty of nice properties (e.g., each infinite cardinal is an algebraically closed field of characteristic $2$; Conway calls these "nimbers", a portmanteau of "nim" and "number").
Sadly, whereas the sum is always compatible with Conway equality (if $G_1 \doteq G_2$ then $G_1 \oplus G' \doteq G_2 \oplus G'$), product is not in general.  For a counterexample, leg $1$ be the game with two positions $0,1$ and a blue edge between $1$ and $0$, let $2$ be the game with three positions $0,1,2$ and blue edges $L = \{(1,0),(2,0),(2,1)\}$, and let $1\oplus 1$ be the same game as $2$ but without the $(2,0)$ edge (this is, indeed, extensionally equivalent to the sum $1\oplus 1$ as defined above): then $1\oplus 1 \doteq 2$ yet $(1\oplus 1)\otimes * \cong * \oplus * \doteq 0$ is not Conway-equal to $2 \otimes * =: * 2$, where here $*$ and $* 2$ are the same games as $1$ and $2$ but with green (=both blue and red) edges instead of just blue (multiplying by $*$ is extensionally the same as making the game impartial by making all edges green): the Grundy ordinals of $*$ and $*2$ are $1$ and $2$ respectively, so they are not Conway-equal.
So there are two obvious ways to try to solve this problem:

Restrict the definition of the product $G \otimes N$ for certain games $N$ for which $G \doteq G'$ implies $(G\otimes N) \doteq (G'\otimes N)$.  This is Conway's approach in On Numbers and Games, the games $N$ in question being called "numbers" (I'm not sure whether all the games for which $G \doteq G'$ implies $(G\otimes N) \doteq (G'\otimes N)$ are numbers, but numbers satisfy this property).  This gives a nice theory where the numbers are a field and the games a partially ordered group, and we can multiply games and numbers, but we can't multiply two games.
But the other would be to change the equivalence relation $\doteq$, and define a stronger (=smaller, =finer) equivalence relation, say $\triangleq$, by $G\triangleq G'$ iff $(G\otimes H) \doteq (G'\otimes H)$ for all games $H$ (i.e. not only does the second player win $G' \ominus G$, but in fact $(G'\ominus G) \otimes H$ for all games $H$; in particular, this is required for $H = *$, which demands that the impartial games $G \otimes *$ and $G' \otimes *$ have the same Grundy ordinals).  It seems that this should still give the games (up to a certain size…) a ring structure, of which the nimbers are a factor or something like that.

My question (at last) is whether this second approach has been studied, and whether it leads to an interesting theory or whether on the contrary there is something obviously wrong with it.

REPLY [2 votes]: As soon as we have 0, 1 , 1/2 (assumed to satisfy the usual properties)
and distributivity, we cannot have an element of additive order 2. Since if 
* has order 2, then
$ * = 1 \otimes * = (1/2 + 1/2) \otimes * =  
1/2  \otimes * +  1/2 \otimes * = 1/2 \otimes (* + *) = 1/2 \otimes 0 = 0$.
[EDIT: The next sentence goes the other way wrt the question: in the question $\triangleq$ is finer than $\doteq$; here on the contrary $\triangleq$ is assumed to be coarser than $\doteq$] Hence it seems that $\triangleq$ collapses all first-winner games (which 
have order 2). So I guess that when we take the quotient modulo 
$\triangleq$ we get a Ring isomorphic to the Surreals.
Hence we should renounce to something. I guess everyone wants to keep 
0 and 1, at least. Hence we can
(A) Do as Conway, restrict ourselves to numbers.
(B) Renounce to 1/2. I do not know whether this could work and how 
could  be done exactly. We have to
consider some class of games in which there are no "fractions". This 
should be given a precise meaning, surely we cannot have "divisors of 1", 
but probably many other games should be taken out.
(C) As above, but discarding only "dyadic" fractions. In this case some 
identifications are necessary, for example, arguing as above, $1/3 \cdot * $
should be identified with $*$. Again, I do not know whether this could 
work
(D) Renounce to distributivity. This does not change a lot with the present 
situation, but leaves open the problem of giving a good definition of the 
product which does not depend on the representatives.
(E)  Consider instead a coarser [EDIT: I meant finer] equivalence relation (classes are smaller) 
than Conway's. Then we can find a way to make * have infinite order. 
Alas, in this case it is probably hard to mantain the group structure, that is, to have inverses.
EDIT Not exactly, it seems that we can actually have inverses even
with an equivalence relation finer than $\doteq$.
Let $\mathbf{PG}$ be the class of  all
(extensional equivalence classes of)
partizan combinatorial games. Set-theoretical complications can be 
dealt with in the usual way.
$\mathbf{PG}$ is a commutative Monoid wrt $\oplus$ and $0$.
Moreover, the operation $-$ satisfies $-(G \oplus H) =( -G )\oplus (-H)$. 
It is then obvious that if we define $\sim$ by
$G \sim H$ if and only if there are games $X$ and $Y$ such that
$G \oplus X \ominus X \cong  H \oplus Y \ominus Y $, then
$\sim$ is an equivalence relation on    $\mathbf{PG}$  and
$(\mathbf{PG}/ {\sim}, \oplus/{\sim})$ becomes a Group
with $(-G)/{\sim}$ the additive inverse of $G/{\sim}$. 
I do not know whether  $ \sim$ and $\triangleq$  are the same relation.
However, it is  interesting to notice that
one arrives at $ \sim$ just by group theoretical considerations,
with no need of considering products. 
The relation $\sim$ is the finest relation which makes the quotient
 of $\mathbf{PG}$ a Group, at least if $-$ is supposed to be the inverse.
Were actually 
$ \sim$ equal to $\triangleq$, this would give additional arguments to the 
supposition  that  $\triangleq$
is an interesting relation.
Now for the product.
We need not have 
$(G\oplus H) \otimes K \cong (G\otimes K) \oplus (H\otimes K)$ in
$\mathbf{PG}$, 
for example
as a consequence of the counterexample presented in the question.
However, the classical (and standard)
proof that
$(G\oplus H) \otimes K \doteq (G\otimes K) \oplus (H\otimes K)$
actually shows that
$(G\oplus H) \otimes K \sim (G\otimes K) \oplus (H\otimes K)$,
since in the proof two opposite games annihilate, hence we do not
actually need the full strength of  $\doteq$. Given distributivity, there is no problem in proving associativity (modulo $\sim$).
By the definition of $\sim$ and the above formula,
 we easily get that
$ G \sim L$ implies
$ G  \otimes K \sim L \otimes K$,
for every game $K$. Just use
$(-X) \otimes K = - (X \otimes K )$. 
Thus $\otimes$ passes to the quotient with respect to 
$\sim$.
Hence 
  $(\mathbf{PG}/ {\sim}, \oplus/{\sim},  \otimes/{\sim})$
is a Ring.
This implies that 
 $ \sim$ is finer than $\triangleq$.
Arguing in the same way, we get that
$(\mathbf{PG}/ {\triangleq}, \oplus/{\triangleq},  \otimes/{\triangleq})$
is a Ring. Otherwise, just check that
the $\triangleq$-class of $0$
is an ideal of   $\mathbf{PG}/ {\triangleq}$.
Strictly formally, we should work with 
the $(\triangleq/{\sim})$-class of $0 /{\sim}$.
To check whether  $ \sim$ is the same as $\triangleq$
means to check if the following is true. 
Given any game $G$, if
$G \otimes H \sim 0$, for every game $H$,
then  $G \sim 0$.
All the above structure definitely deserves further study, so your question
is a really great question!<|endoftext|>
TITLE: von Neumann algebras as C*-algebras with multiplicative conditional expectation $A^{**}\to A$
QUESTION [8 upvotes]: Let $A$ be a C*-algebra. We identify $A$ with its canonical image in the bidual $A^{**}$. Consider the following conditions:
(1) $A$ is a von Neumann algebra.
(2) There is a multiplicative conditional expectation from $A^{**}$ onto $A$, that is, a map $\pi\colon A^{**}\to A$ that is a *-homomorphism and such that $\pi(a)=a$ for all $a\in A$.
Then (1) implies (2): Consider the predual $A_*$ of $A$, and the canonical embedding $\kappa\colon A_*\to (A_*)^{**}$. Then the dual of $\kappa$ has the desired properties.
Note that $A^{**}$ is always a von Neumann algebra.
One can show that (2) implies that $A$ is a monotone complete AW*-algebra.
However, is it also a von Neumann algebra?

Q1: Does (2) imply (1) ?

Assuming (2), let $J$ denote the kernel of $\pi$. Then $A$ is (*-isomorphic to) the quotient $A^{**}/J$. 
However, a quotient of a von Neumann algebra by a closed, two-sided ideal need not be a von Neumann algebra. (For example, the Calkin algebra is such a quotient and not a von Neumann algebra.)
In the commutative case, we might also consider the following more general question:

Q2: Is the quotient of a commutative von Neumann algebra by a closed, two-sided ideal again a von Neumann algebra?

Equivalently, is every closed subset of a hyperstonean space again hyperstonean?

REPLY [11 votes]: The answer to Q2 is no, and $l^\infty/c_0$ is already a counterexample. Its lattice of projections is $\mathcal{P}(\omega)/fin$, which is not complete.<|endoftext|>
TITLE: Classification of $O(2)$-bundles in terms of characteristic classes
QUESTION [16 upvotes]: I had asked this question in stackexchange but there seems to be no consensus in the answer
It is well-known that $SO(2)$-principal bundles over a manifold $M$ are topologically characterized by their first Chern class. I was wondering what was the characterization of $O(2)$-bundles in terms of characteristic classes. I guess the first and second Setiefel-Whitney classes are necessary for the topological characterization of $O(2)$-bundles, but they can't be enough, because if $w_{1} = 0$ then one should recover the classification of $SO(2)$-bundles, which is given by the first Chern class and not by the second Stiefel-Whitney class.
Thanks.

REPLY [9 votes]: To complement Mark Grant's excellent answer, I'll say something more about the general case. This topic goes under the name of obstruction theory.
The first observation is that a $G$-bundle on $X$ is the same thing as an homotopy class of maps $X\to BG$. To study them we will use the Postnikov tower of $BG$. This is a tower assembled by spaces $P_n(BG)$ together with a map $BG\to P_n(BG)$ such that

The  map $\pi_i(BG)\to \pi_i(P_n(BG))$ is an isomorphism for $i\le n$;
$\pi_i(P_n(BG))=0$ for $i>n$.

We can assemble this spaces together so to form a tower as follows:
$\require{AMScd}$
\begin{CD}
    @. \vdots\\
    @. @VVV \\
    @. P_2(BG)\\
    @. @VVV \\
    X @>>d> P_1(BG)
\end{CD}
and moreover the limit of the tower is $BG$. So we can study the homotopy classes $[X,BG]$ by studying the collections of arrows $[X,P_i(BG)]$ making the diagram commute.
Now let's start at the bottom of the diagram. By definition we have that $P_1(BG)$ is a $K(\pi_1BG,1)=K(\pi_0G,1)$, so we have
$[X,P_1(BG)] = [X,K(\pi_0G,1)] = H^1(X;\pi_0G)$
This is our first cohomology class, corresponding to $w_1$ in the case of $BO(n)$.
Now let us suppose that we have lifted our map all the way to $P_n(BG)$ and we want to see what algebraic information corresponds to a lift to $P_{n+1}(BG)$. It turns out that there is a cartesian diagram
$\require{AMScd}$
\begin{CD}
    @.  P_{n+1}(BG) @>>> K(\pi_0G,1)\\
    @.  @VVV @VVV\\
    X @>>> P_n(BG) @>>> K(\pi_{n+1}G,n+2)_{h\pi_0G}
\end{CD}
(don't be scared by all those homotopy quotients you see: they're just the homotopy theorist's way of saying that we're dealing with twisted cohomology classes). So lifting a map from $P_n(BG)$ to $P_{n+1}(BG)$ is the same thing as lifting a map from $K(\pi_{n+1}G,n+2)_{h\pi_0G}$ to $K(\pi_0G,1)$. This is saying that the lift exists if and only if some class in $H^{n+2}(X,\pi_{n+1}G)$ vanishes (not all choices of characteristic classes will correspond to a $G$-bundle!) but, more importantly for us, this is exactly the same situation as in Mark Grant's answer and so the possible choices are parametrized by a class in $H^{n+3}(X,\pi_{n+1}G)$.
So, to sum up we will have

A class $\alpha$ in $H^1(X;\pi_0G)$
An infinite sequence of classes in $H^{n+1}(X;\pi_nG)$ for $n\ge1$ where the coefficients are twisted by $\alpha$.<|endoftext|>
TITLE: Has it been proved that odd perfect numbers cannot be triangular?
QUESTION [14 upvotes]: (Note:  This question has been cross-posted from MSE.)
Euclid and Euler proved that every even perfect number is of the form $m = \frac{{M_p}\left(M_p + 1\right)}{2}$ where $M_p = 2^p - 1$ is a prime number, called a Mersenne prime.  Thus, an even perfect number is triangular.
On the other hand, Euler showed that an odd perfect number, if one exists, takes the form $N = q^k n^2$, where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n) = 1$.  (Descartes, Frenicle and subsequently Sorli conjectured that $k = 1$ always holds.)
Here is my question:

Has it been proved that odd perfect numbers cannot be triangular?

Added March 26 2016
If $\sigma(q) = 2n^2$, then it would follow that $n < q$, which implies that $k = 1$.  The odd perfect number $N = q^k n^2$ then takes the form $N = \frac{q(q + 1)}{2}$.  Unfortunately, it is known that $\sigma(q^k) \leq \frac{2n^2}{3}$.
Any pointers to the existing literature containing such a proof would be most appreciated.

REPLY [3 votes]: Not an answer, but I just want to point out some thoughts that recently occurred to me, which are related to this problem.
By this answer, we know that every odd perfect number $N = q^k n^2$ can be written in the form
$$N = \dfrac{q(q+1)}{2} \cdot d$$
where $d > 1$.  (That is, an odd perfect number is a nontrivial multiple of the triangular number
$$T(q) = \dfrac{q(q+1)}{2},$$
where $q$ is the Euler prime of $N$.)
If $k=1$, then it is easy to show that
$$d = D(n^2)$$
where $D(n^2) = 2n^2 - \sigma(n^2)$ is the deficiency of the non-Euler part $n^2$.
(I have asked a separate question for the form of $d'$ here, if $N = q^k n^2$ is an odd perfect number with $k > 1$, and $N = \dfrac{q(q+1)}{2} \cdot d'$ with $d' > 1$.)<|endoftext|>
TITLE: Cardinality of factors of infinite non-abelian groups
QUESTION [6 upvotes]: Let $A$ and $B$ be arbitrary nonempty subsets of a group $G$. Then
the product $AB$  is called  direct, and we denot it by $A \cdot B$,
if the representation of every its element by $x=ab$  with $a\in A$, $b\in B$ is unique.
It is obvious that if $AB=A \cdot B$ then $|A B|=|A||B|$, and the converse is true if both are finite.
Now, let $A \cdot B_1=A \cdot B_2$. If $A$ is finite then $|B_1|=|B_2|$. Also, if $G$ is abelian then we can define an injective map between $B_1$ and $B_2$
(and also between $B_2$ and $B_1$). Hence, the question is:

Let $G$ be an infinite non-abelian group, and $A,B_1,B_2$ their nonempty subsets. Is it true that if $A \cdot B_1=A \cdot B_2$ then $|B_1|=|B_2|$?
(what about $A \cdot B_1=A \cdot B_2=G$, if the answer is negative?)

REPLY [2 votes]: Here's an example with $|B_1|=1, |B_2|=2$:  Let $G=\mathbb Z*\mathbb Z$ (free product). Write non-identity elements of $G$ multiplicatively as reduced words with $X^{\mathbb Z-\{0\}}, Y^{\mathbb Z-\{0\}} $(with symbols $X,Y$).  Let $A_0=X^{\mathbb Z - 2\mathbb Z}, A_1 = \{$reduced words with $X^{2\mathbb N},Y^{2\mathbb N}\}, A=A_0A_1, B_1=\{1\}, B_2=\{X^2,Y^2\}$.  Note we can't have $A \cdot B_1=A \cdot B_2=G$ with $|B_1|=1, |B_2|>1$.<|endoftext|>
TITLE: How fine an invariant of a representation is its quotient singularity?
QUESTION [8 upvotes]: This is a refinement of a question asked on MSE.
Let $G$ be a finite group and let $V$ be a finite-dimensional faithful complex representation of $G$. Consider $V$ as an affine complex variety. In general (though not always), the image of the origin in the quotient variety $V/G$ is singular. By the "singularity type" of the pair $G,V$ I refer to the isomorphism class of the ring $\widehat{\mathcal{O}_{V/G,0}}$, the completion of the local ring at this point. 
At broadest (soft-question) level, what I want to know is, "how fine an invariant of the pair $G,V$ is the singularity type?" But let me narrow the scope in order to be able to ask something precise:

Suppose $G$ is nonabelian, $V$ has no one-dimensional subrepresentations, and $V/G$ is singular at the image of the origin. Further suppose $A$ is a finite abelian group and $W$ is a faithful representation of $A$. Is it possible for $G,V$ to have the same singularity type as $A,W$?

Comments: (1) The answer is certainly "yes" without the stipulation that $V/G$ be singular at the origin; both abelian and nonabelian groups can have smooth quotients, by Chevalley-Shephard-Todd since both abelian and nonabelian groups can be reflection groups. (2) The stipulation "$V$ has no one-dimensional subrepresentations" is there to rule out a match resulting from $G,V$'s singularity "actually" coming from $G$'s abelianization. This can happen for example if $V$ decomposes as a sum $W\oplus L$ with $W$ such that $W/G$ is smooth and $L$ is a sum of one-dimensional representations.

REPLY [6 votes]: To cherry-pick the answer from Jason Starr's comment above:  the condition you want is that $G$ acts on $V$ without any element giving a pseudo-reflection.  Chevalley-Shephard-Todd tells you that you must lose all information about pseudo-reflections, but you don't lose anything else.  When there are no pseudo-reflections, the singular set of $V/G$ is exactly the image of the elements in $V$ with non-trivial stabilizer, so the smooth locus of $V/G$ is a quotient of a simply connected space by $G$, and thus its $\pi_1$ is $G$.  This should be reconstructable from the singularity type using the Galois theory of the fraction field of the completion at the origin (looking at the Galois group of the largest Galois extension with no ramification at smooth points, I think).<|endoftext|>
TITLE: The coefficient of a specific monomial of the following polynomial
QUESTION [7 upvotes]: Let the real polynomial 
$$f_{a,b,c}(x_1,x_2,x_3)=(x_1-x_2)^{2a+1}(x_2-x_3)^{2b+1}(x_3-x_1)^{2c+1},$$
where $a,b,c$ are nonnegative integers.
Let $m_{a,b,c}$ be the coefficient of the monomial $x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}$ in the expansion of $f_{a,b,c}(x_1,x_2,x_3)$.
It is easy to see $m_{a,b,c}=0$ when two of $a,b,c$ are equal. I want to ask whether $m_{a,b,c}=0$ or not when $a,b,c$ are pairwise unequal.

REPLY [17 votes]: This coefficient $L$ is a constant term of the Laurent polynomial $g(x_1,x_2,x_3)=f(x_1,x_2,x_3)/x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}$, this guy $g$ satisfies $g(x_1,x_2,x_3)=-g(1/x_1,1/x_2,1/x_3)$, thus $L=-L$.

REPLY [4 votes]: This answer is really just a convoluted edition of Fedor's answer.  By Cauchy's theorem,
$$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi
   \frac{(e^{i\theta_1}-e^{i\theta_2})^{2a+1}(e^{i\theta_2}-e^{i\theta_3})^{2b+1}(e^{i\theta_3}-e^{i\theta_1})^{2c+1}}{e^{i(a+c)\theta_1+i(a+b)\theta_2+i(b+c)\theta_3}}
\,d\theta_1\,d\theta_2\,d\theta_3.$$
Advancing each variable by $\pi$ changes the sign of the integrand but not the value of the integral, so its value must be 0.
To obtain the above integral start with the contour integral
$$\frac{1}{(2\pi i)^3} \oint\oint\oint \frac{(x_1-x_2)^{2a+1} (x_2-x_3)^{2b+1} (x_3-x_1)^{2c+1}}{x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}} \,dx_1dx_2dx_3 $$
that is just Cauchy's formula applied three times. Now take the contours to be unit circles $x_j=e^{i\theta_j}$ for $j=1,2,3$.<|endoftext|>
TITLE: A prime ideal $\mathfrak{p}$ decomposes in $\mathbb{Q}(\zeta_{24})/\mathbb{Q}(\sqrt{-6})$ iff it is generated by $\alpha\in1+2\Bbb{Z}[\sqrt{-6}]$
QUESTION [7 upvotes]: For a nonzero prime ideal $\mathfrak{p}$ of $\mathbb{Z}[\sqrt{-6}]$ which does not divide $2$, does $\mathfrak{p}$ decompose completely in the extension $\mathbb{Q}(\zeta_{24})/\mathbb{Q}(\sqrt{-6})$ of degree $3$ if and only if $\mathfrak{p} = (\alpha)$ for some $\alpha \in \mathbb{Z}[\sqrt{-6}]$ such that $a \equiv 1 \text{ mod }2\mathbb{Z}[\sqrt{-6}]$?
I suspect this is a true based off working out the following example. $5 + 6\sqrt{-6}$ is a prime divisor of the prime number $241$ in $\mathbb{Z}[\sqrt{-6}]$ and $5 + 6\sqrt{-6} \equiv 1 \text{ mod }2\mathbb{Z}[\sqrt{-6}]$. We have$$241 = \prod_{a = 1, 5, 7, , 11, 13, 17, 19, 23} (2 - \zeta_{24}^a), \quad 5 + 6\sqrt{-6} = -\prod_{a = 1, 5, 7, 11}(2 - \zeta_{24}^a).$$

REPLY [3 votes]: OK, so here are the details. The decomposition law in cyclotomic extensions tells you that a prime $p$ coprime to $m$ splits completely in ${\mathbb Q}(\zeta_{m})$ if and only if $p \equiv 1 \bmod m$. Since ${\mathbb Q}(\zeta_{24})  = {\mathbb Q}(\sqrt{-1},\sqrt{2},\sqrt{-3})$ this is equivalent to the condition $(-1/p) = (2/p) = (-3/p) = +1$.
Prime ideals in the subfield $k = {\mathbb Q}(\sqrt{-6}\,)$ split completely in $K$ if and only if they split in $k/{\mathbb Q}$ and in $K/k$. Primes $p$ split in $k$ iff $(-6/p) = +1$, i.e. iff $(2/p) = (-3/p)$. By genus theory, the prime ideals above $p$ are principal if both Legendre symbols are $+1$, and nonprincipal otherwise. 
For finding out which principal primes split completely in $K/k$ we look at their norms: clearly $p = x^2 + 6y^2 \equiv 1 \bmod 4$ is equivalent to $2 \mid y$, which means that exactly those principal primes split completely whose norm is a prime $p \equiv 1 \bmod 4$. 
On a more advanced level, the extension $K/k$ consists of the Hilbert class field and genus field $F = {\mathbb Q}(\sqrt{2},\sqrt{-3})$ and a quadratic  extension $K/F$ unramified outside $2$ that happens to live inside the ray class field modulo $(2)$ over $k$. The Hilbert class field step explains why only principal primes split, the ray class part explains the congruence condition modulo $2$.<|endoftext|>
TITLE: Smoothness of the "Archimedean special fiber" in Arakelov geometry
QUESTION [10 upvotes]: If $X$ is a scheme over, let's say, $\mathbb{Z}_p$, one can consider its special fiber obtained by reduction modulo $p$ ans it certainly makes sense to ask if this special fiber is smooth or not. 
The ring of p-adic integers $\mathbb{Z}_p$ gives a local (formal) picture near $p$ in $Spec(\mathbb{Z})$. One basic intuition at the source of Arakelov theory is that a similar picture should exist near the "point at infinity" of an hypothetical compactification of $Spec(\mathbb{Z})$ obtained in some sense by adding the Archimedean place. In particular, a variety over $\mathbb{C}$ should be an analogue of a variety over $\mathbb{Q}_p$ (or maybe $\mathbb{C}_p$, anyway) and there should be an Archimedean analogue of a variety over $\mathbb{Z}_p$. Arakelov theory gives a proposal for such an analogue: a variety over $\mathbb{C}$ endowed with a Hermitian metric. There should also exist an analogue of the special fiber but I don't think that such thing exists yet (it should be something over something of characteristic one). 
Without knowing what is the "Archimedean special fiber", my question is: is there an expectation for the notion of smoothness of this special fiber? Given a variety over $\mathbb{C}$ endowed with an Hermitian metric, is there a well-defined yes/no criterion for the smoothness of the hypothetical Archimedean special fiber?
This question is motivated by the fact that the notions of good/bad reductions play a key role in p-adic theory and I would like to understand what is the analogue story at the Archimedean place.

REPLY [13 votes]: In Arakelov geometry, the conventional wisdom is that the ``closed fibre at $\infty$'' should be viewed as totally degenerate. This is the extreme opposite of smoothness. A visualization in the case of a curve is proposed in:
Yu. Manin: Three-dimensional hyperbolic geometry as $\infty$-adic Arakelov geometry, Inventiones, 1991.
This was inspired by Mumford's work on $p$-adic Schottky groups, extending to higher genus the rigid analytic picture of the Tate curve.
This work of Consani and Marcolli could be also helpful:
C. Consani, M. Marcolli: Noncommutative geometry, dynamics, and $\infty$-adic Arakelov geometry, Selecta Math. 2004.
The Tate curve may be enough to convince you of such a point of view. Analytically, over $\mathbb{C}_p$ or $\mathbb{C}$, a $\mathbb{G}_m$-reduction elliptic curve or a complex elliptic curve are, respectively, both a quotient of $\mathbb{G}_m$ by a one-dimensional group of periods. There is little similarity with an elliptic scheme over $\mathbb{Z}_p$. In the Tate curve, making increasingly ramified extension of the base field, the closed fibre of the minimal desingularization is a Neron polygon of projective lines, whose group of irreducible components converges to a circle $\mathbb{R}/\mathbb{Z}$: a tropical elliptic curve, and the skeleton of the Berkovich analytification. In contrast, for an elliptic curve over $\mathbb{Z}_p$  the analytification is contractible, giving the wrong Betti number in comparison to what happens Archimedeanly. This consideration extends to curves of higher genus $g$: the first Betti number of the analytification is $\leq g$, with equality if and only if the curve is totally degenerating.
The dynamics of rational maps illuminate another, though related, point. If the map has a good reduction at $p$ then its dynamics is completely predictable: the Julia set has no $\mathbb{C}_p$-points. But the Julia sets of complex dynamics are always non-empty, and they are typically quite intricate. 
Finally, we may compare the geometry or dynamics over $\mathbb{Q}$ with the corresponding situation over the rational function field $k(t)$. If in the function field model we have a rational map over $k(t)$ with everywhere potential good reduction, then the map is isotrivial. The irrelevance of isotriviality for algebraic varieties or rational maps over number fields (as in Bogomolov's problem, Lehmer's problem, the $abc$ conjecture) leads us to admit the degeneration of at least some of the fibres: the Archimedean ones. If, for instance, we took the map $z \mapsto z^2$ over $\mathbb{Q}$ to be modeled by the squaring map over $k(t)$, which is isotrivial, the Lehmer problem (on the spectral gap of the dynamical Mahler measure) has a trivial, affirmative answer. Not so if we admit a  non-isotrivial map: conceivably, this will have the same answer and comparable difficulty as the classical Lehmer problem. In the case of $z \mapsto z^2$, a more faithful model for the Lehmer question is not the squaring map of $\mathbb{F}_q(t)$, but the map $c(z) = tz + z^q$, which is the isogeny of multiplication by $t$ on the Carlitz module over $\mathbb{F}_q(t)$. In either case we have exactly the same progress on Lehmer's question, namely, Dobrowolski's bound; and the map $c$ has a bad reduction at the ``infinite place.''
Added. Since the question concerned an arbitrary hermitian metric, I felt I should add a pointer to Chambert-Loir and Ducros's work on Chern forms and their associated measures: Formes differentielles reelles et courants sur les espaces de Berkovich. Arakelov geometry considers line bundles $\bar{L} = (L,\|
\cdot\|)$ with a smooth hermitian metric. The Chern (curvature) form $c_1(
\bar{L})$ is then a smooth $(1,1)$-form, and its determinant (or self intersection) is either zero or a full support signed measure on $X(\mathbb{C}) = X_{/\mathbb{C}}^{\mathrm{an}}$. For a $p$-adic metric $\|\cdot\|_p$ in $L$ there are a corresponding form and measure on $X_{/\mathbb{C}_p}^{\mathrm{an}}$. But if this metric arises from an integral model $(\mathfrak{X},\mathfrak{L})$ over $\mathbb{Z}_p$, where $\mathfrak{X}_{/\mathbb{F}_p}$ is regular, then the obtained (signed) measure charges a single point of the Berkovich space: the unique point reducing to the generic point of the scheme $\mathfrak{X}_{/\mathbb{F}_p}$.<|endoftext|>
TITLE: Is there a class of mathematical structures with non-isomorphic natural representations as a standard Borel space?
QUESTION [14 upvotes]: Background. The field of Borel equivalence relation theory
provides a robust, unifying theory that organizes most of the
classification problems of classical mathematics into a hierarchy,
allowing us precisely to compare their relative difficulty.
Namely, every such classification problem amounts to an
equivalence relation on a class of mathematical structures, and
one can generally present the class of structures as a standard
Borel space. An equivalence relation $E$ on such a space reduces
to another $F$, if there is a Borel function $f$ for which
$$x\mathrel{E}y\quad\iff\quad f(x)\mathrel{F} f(y).$$ Thus, the classification problem of $E$ is reduced to that of $F$, and a rich hierarchy has emerged. Anyone with a classification problem in any part of
mathematics should seek to situate it into this hierarchy. The
subject is a pleasing mix of ideas from many parts of mathematics.
The issue. Since the hierarchy works essentially on coded versions of the classification problems as standard Borel spaces, it is important for the subject that these encodings are authentic. Su Gao discussed the importance of this issue in his
book,
Invariant descriptive set theory (p. 328),
where he proposed the following principle:

Gao's thesis. For any collection $H$ of mathematical
  structures and natural standard Borel structures $B_1$ and $B_2$
  on $H$, there is a Borel isomorphism $\psi:\langle
H,B_1\rangle\cong\langle H,B_2\rangle$ for which $\psi(x)$ is
  isomorphic to $x$ for every $x\in H$.

In other words, all natural presentations of a given class of
mathematical structures as a standard Borel space are the same up
to Borel isomorphism. Many instances of such isomorphisms have been observed, and empirical evidence is accumulating in support of the thesis. 
In his dissertation, Burak Kaya formulates the principle as asserting: for any class $H$ of mathematical structures, if $\langle X,B_1\rangle$ and $\langle Y,B_2\rangle$ are two standard Borel spaces naturally encoding the structures of $H$, then there is a Borel isomorphism $\psi:X\to Y$ such that the structures in $H$ coded by $x$ and by $\psi(x)$ are isomorphic. It is this formulation of the thesis that seems to arise more often in practice, as researchers give different encodings of their class of structures.
Gao discusses the philosophical nature of his thesis, citing
specifically the difficulty of the issue of what counts as
natural. He describes the thesis as an analogue of the
Church/Turing thesis in computability theory.
The questions. My view is that Gao's thesis is critically
important for the subject, because if as we desire we are to view
the results of Borel equivalence relation theory as being about
the actual classification problems arising in mathematics, we need
to know that we have successfully captured those problems in our
presentations of them as standard Borel spaces.
In the case of computability theory, Turing in his famous paper
gave a forceful philosophical argument that in principle any
effective means of computation can be simulated by Turing
machines. In the case of Gao's thesis, however, we seem to lack
comparable forceful philosophical grounds. So are the two theses analogous? 
I'd like to get a grasp on the nature of possible counterexamples
to Gao's thesis, if there might be any. Presumably, if we weaken
the naturality requirement, we can find counterexamples to the
thesis. How unnatural do the counterexamples have to be? Can anyone provide me with unnatural counterexamples?
Question 1. What are examples of collections $H$ of
mathematical structures that have almost-natural standard Borel
structures $B_1$ and $B_2$ on $H$, with no Borel isomorphism
respecting isomorphism of structures in $H$?
In other words, how close to natural can we get while violating the conclusion of Gao's thesis? It seems important for us to be aware of the range of possibility when discussing this thesis.
Question 2. What are the philosophical grounds that we have in
support of Gao's thesis?
I became interested in the issue of Gao's thesis because it arose
in the recent dissertation and dissertation defense of Burak Kaya,
where we had an interesting discussion about it during the
question session.

REPLY [4 votes]: Perhaps I am misunderstanding the question, but here is what seems to be a counterexample (Lemma 9.2.2 from Su Gao Invariant Descriptive Set Theory):
Let $H$ be the class of countable well-orders. The first Borel representation will be $(\mathbb{R}, F_{\omega_1})$ where $a F_{\omega_1} b$ iff $\omega_1^{CK(a)} = \omega_1^{CK(b)}$. The second is $(LO, E_{\omega_1})$, where $LO$ is the subset of $\mathcal{P}(\omega \times \omega)$ that encode linear orders of $\omega$, and $\phi E_{\omega_1} \psi$ iff either they are both non-wellordered, or they are both well-ordered of the same rank.
Both of these have $\omega_1$-many classes, so can be used to represent classes of $H$ (for $E_{\omega_1}$, we have an extra class, which we let represent $0$, and we let the linear orders $\phi \in LO$ of rank $n+1$ represent well-orders of rank $n$.)
But there is no Borel reduction at all from $E_{\omega_1}$ to $F_{\omega_1}$, or conversely. (Addendum: they are $\Delta^1_2$-bireducible, as described in Section 9.2.)
Discussion Su Gao gives this as an example of the fact that when the equivalence classes involved are not Borel, then Borel reducibility often seems inadequate to capture the notion of "definable reduction." Some weaker notions of reduction are "provably $\Delta^1_2$," or "$\Delta^1_2"$. See Section 9.2 of Su Gao.
Note that in general, when given two Borel representations $(\mathbb{R}, F)$ and $(\mathbb{R}, E)$ of the class of structures $H$, then presuming the maps $H / \cong \to \mathbb{R}/F$ and $H/ \cong \to \mathbb{R}/E$ are sufficiently definable in some sense, then there is a sufficiently definable bijection $\mathbb{R}/F \to \mathbb{R}/E$. So if there is no Borel such map, then no worries, we just figure out in what sense the representation is definable.
One can ask: what is the least restrictive notion of reduction that still allows a good theory? I'll put forward $HC$-reduction, defined in the paper A New Notion of Cardinality for Countable First Order Theories, by myself, Richard Rast, and Chris Laskowski. (Here $HC$ is the set of hereditarily countable sets.)<|endoftext|>
TITLE: Cross between the nil-Hecke ring and the group ring of a Coxeter group
QUESTION [10 upvotes]: A Coxeter system $(W,S)$ has a set of generators $S=\{s_1,s_2,\ldots\}$ and the Coxeter group $W$ is determined by relations of the form $(s_is_j)^{m_{ij}}=1$ for some integers $m_{ij}$, where $m_{ii}=1$ for all $i$ and if $i\neq j$ then either $m_{ij}=\infty$ or $m_{ij}\geq 2$. 
Let $V$ be a vector space containing a root system for $(W,S)$ with simple roots $x_1,x_2,\ldots$ and a corresponding action of $W$. The skew group ring is the left $F(V)$-vector space (where $F(V)$ is field of fractions of the symmetric algebra of $V$) with basis $W$ (it is also a right vector space over $F(V)$ but the action is different) where elements of $W$ are multiplied in the usual way and if $p\in F(V)$ and $w\in W$ then
$$wp=(w\cdot p)w$$
The nil Hecke ring is the subring generated over the symmetric algebra of $V$ by the divided difference operators
$$\partial^i=\frac{1}{x_i}(1-s_i)$$
We have $(\partial^i)^2=0$ for all $i$ and if $i\neq j$ then 
$$\underbrace{\partial^i\partial^j\partial^i\cdots}_{m_{ij}\text{ times}}=\underbrace{\partial^j\partial^i\partial^j\cdots}_{m_{ij}\text{ times}}$$
The nil-Hecke ring contains the group ring of $W$ as a subring because
$$s_i=1-x_i\partial^i$$
 I decided to study what happens when you remove the polynomials and just consider the subring generated by the $s_i$ and $\partial^i$ over the integers because this subring is the smallest subring in which you can express the Leibniz formula for divided difference operators (from my dissertation).
I'm trying to find sufficient relations to characterize this ring, but I'm having a hard time. Let me show examples from type $A$, where $s_i$ is the transposition $(i,i+1)$. We have the relations
$$\partial^i=s_i\partial^i=-\partial^is_i$$
$$\partial^is_j=s_j\partial^i$$
if $|i-j|\geq 2$,
$$\partial^is_{i+1}s_i=s_{i+1}s_i\partial^{i+1}$$
$$s_i\partial^{i+1}s_i=s_{i+1}\partial^is_{i+1}$$
$$\partial^is_{i+1}\partial^i=s_{i+1}\partial^i\partial^{i+1}+\partial^{i+1}\partial^{i}s_{i+1}$$
It'd be hopeful to expect this to be enough, but I can give you tons of identities that don't come from these relations. Has anyone found sufficient relations to characterize the ring, or at least studied it before?

REPLY [3 votes]: Too long for a comment, some references. 
The question in the case "A"  has been considered by 
Florent Hivert et Nicolas Thiéry
https://www.lri.fr/~hivert/PAPER/HeckeGroupJAlg.pdf
(Journal of algebra, 321 (2009), no. 8, 2230-2258. )
An analogous question (action of $s_i+\partial_i$) has been 
treated by B. Leclerc, A. Lascoux and Jean-Yves Thibon
http://www-igm.univ-mlv.fr/~jyt/ARTICLES/schubaf6.ps
( Banach Center Publications vol. 36 (1996), 111-124. )<|endoftext|>
TITLE: A symmetric embedding of manifolds
QUESTION [10 upvotes]: Assume that $M$ is  a manifold.
Is there an embedding of $M$ in  some $\mathbb{R}^{n}$ such that the image of  $M$ in $\mathbb{R}^{n}$ is invariant under each reflection $(x_{1},x_{2},\ldots x_{i},\ldots,x_{n}) \mapsto (x_{1},x_{2},\ldots , -x_{i},\ldots,x_{n})$, for all $i\in \{1,2,\ldots ,n\}$?

REPLY [20 votes]: No. There exist closed manifolds which do not admit any compact group actions, in particular, no $Z_2$ actions. For example, Shultz showed in "Group actions on hypertoral manifolds. II." that in  dimensions $\ge 4$ every oriented cobordism class contains such a manifold.<|endoftext|>
TITLE: On the complexification of a Riemannian manifold
QUESTION [5 upvotes]: Let $(M,g)$ be a Riemannian manifold and $TM$ be its tangent bundle. If we suppose $TM\otimes\mathbb{C}$ is the complexification of $TM$ then how can we define a natural metric on the complex bundle $TM \otimes\mathbb{C}$ as what there exists and how can we do the differentiation from the sections of this bundle by using the Levi-Civita connection of the metric $g$?
I mean If $X, Y$ are two sections of $TM \otimes\mathbb{C}$ then the inner product of $X, Y$ how can be defined by using the metric $g$?
Moreover, how can we differentiate from $X$ along $Y$ in a natural way by using the Levi-Civita connection of $g$?
I can not find a definition which describes such a metric and differentiation.

REPLY [7 votes]: Each section of $TM \otimes \mathbb{C}$ has a unique decomposition $Z=X+iY$ as a sum with $X$ and $Y$ sections of $TM$. Define your metric using this, for example as $\left<Z_1,Z_2\right>=\left<X_1,X_2\right>+\left<Y_1,Y_2\right>$. Use an affine connection as $\nabla_{X+iY} U+iV=\nabla_X U - \nabla_Y V + i \left(\nabla_Y U + \nabla_X V\right)$ to get complex linearity.
Edit: the natural Hermitian metric on $TM \otimes \mathbb{C}$ is $$\left<Z_1,Z_2\right>=\left<X_1,X_2\right>+\left<Y_1,Y_2\right>+i\left<Y_1,X_2\right>-i\left<X_1,Y_2\right>$$.<|endoftext|>
TITLE: Intersection between integral closures is algebraically closed field
QUESTION [5 upvotes]: Consider an algebraically closed field $k$, a finite field extension $K$ of $k(T)$, the integral closure $A$ of $k[T]$ in $K$, and the integral closure $A'$ of $k[T^{-1}]$ in $K$. Does it follow that $A \cap A' = k$?

REPLY [9 votes]: Yes. Each element of $K$ satisfies a unique irreducible monic polynomial over $k(T)$. It is integral over $k[T]$ if and only if the coefficients lie in $k[T]$ and integral over $k[T^{-1}]$ if and only if the coefficients lie in $k[T^{-1}]$. If it is integral over both, the coefficients lie in $k[T] \cap k[T^{-1}]=k$. Because $k$ is algebraically closed, an irreducible polynomial with coefficients in $k$ is linear.<|endoftext|>
TITLE: Why do people study representations of 3-manifold groups into $SL(n,\mathbb{C})$?
QUESTION [31 upvotes]: Varieties of representations and characters of $3$-manifold groups in $SL(2,\mathbb{C})$ have been intensively studied. They have provided tools to identify geometric structures on manifolds, and are related to $A$-polynomials and thus to various conjectures concerning asymptotics of quantum invariants such as the AJ-conjecture.
Varieties of representations and characters of surface groups in algebraic groups have also been intensively studied, this being a survey. 
In recent years, there have been a large number of papers on varieties of representations and characters of $3$-manifold groups in $SL(n,\mathbb{C})$ for $n>2$, including the development of various tools to compute invariants associated to such varieties, e.g. work of Garoufalidis, Thurston, Zickert, and Goerner.
I have leafed through various of these papers on various occasions and heard some talks, but although understanding any sorts of representations of $3$--manifold groups is surely a worthy goal, as is generalizing results known for $n=2$ or for surface groups, I haven't really understood what motivates this direction of research or what people are aiming to achieve by studying such representations.

Question: What is the motivation for the study of representations of 3-manifold groups into $SL(n,\mathbb{C})$ for $n>2$? 

For example, is it expected that such representations should provide tools to identify geometric structures on manifolds? Are they expected to offer insight into conjectures concerning asymptotics of quantum invariants? Or maybe something else?

REPLY [14 votes]: Three manifold groups are quite special and their representation varieties have more structure than that of a random group. Casson's view point is helpful to see the point. The Heegaard decompoistion of a three manifold $Y= H_1\cup_\Sigma H_2$ means that the fundamental group of a three manifold has a quite special presentation. There is pushout diagram,
\begin{eqnarray}
\pi_1(\Sigma) & \to & \pi_1(H_1)\\
\downarrow & & \downarrow \\
\pi_1(H_2) & \to & \pi_1(Y)
\end{eqnarray} 
and hence maps in the opposite direction on representation varieties.
\begin{eqnarray}
R_G(\Sigma) & \leftarrow & R_G(H_1)\\
\uparrow & & \uparrow \\
R_G(H_2) & \leftarrow & R_G(Y)
\end{eqnarray} 
The representation variety $R_G(\Sigma)$ is a symplectic manifold when $G$ is compact Lie group (thanks to Goldman and Atiyah-Bott) and gets a Kahler structure
once metric is chosen on $\Sigma$.  If $G$ is the complexification of a compact Lie group and a metric on $\Sigma$ is choose $R_G(\Sigma)$ is Hyperkahler manifold.
(It is a slight lie (small l ;-) that these are manifolds.)
The maps $R_G(H_i) \to R_G(\Sigma)$ are injective and images (to the extent they are manifolds) are Lagrangian in case $G$ is compact and complex Lagrangian when $G$ is the complexification of a compact group.  The intersection 
$$
R_G(Y)=R_G(H_1)\cap R_G(H_2) \subset R_G(\Sigma)
$$
is then much more special, so in the compact case lies in the setting of Lagrangian Floer homology and the in the complex case in the setting is for example discussed recently by Witten and Haydys (see for example arXiv: 1010.2353 )
Related to this point of view is the fact that representation varieties of three manifolds are the set of critical points of a Chern-Simons functional on a suitable space of gauge equivalence classes of connections.
Apologies for the inaccuracies above due to haste and lazy typesetting.<|endoftext|>
TITLE: Application of Factorization Theory to Oscillatory Integral Estimates
QUESTION [7 upvotes]: In the article "Some New Estimates on Oscillatory Integrals" by Bourgain in the book Essays in Honor of Elias M. Stein, Bourgain considers operators of the form
$$S_{N}g(x):=\int_{\mathbb{R}^{n}}g(y)e^{iN|x-y|}b(x-y)dy, \quad x\in\mathbb{R}^{n}$$
where $g\in\mathcal{S}(\mathbb{R}^{n})$, $N>0$, and $b$ is a standard bump function adapted to a neighborhood of the origin in $\mathbb{R}^{n}$. The goal is to prove an estimate of the form
$$\|S_{N}g\|_{L^{p}}\lesssim_{\epsilon}N^{-\frac{n}{p}+\epsilon}\|g\|_{L^{p}}$$
for all $0<\epsilon\ll 1$, where $\frac{n}{p'}<\lambda+\frac{n+1}{2}$, where $\lambda>0$ is fixed. The context for this operator is proving $L^{p}$ bounds for the Bochner-Riesz multiplier.
Assume $p>2$. Bourgain claims that "by general factorization theory and the rotational invariance, it suffices to get the $L^{\infty}\rightarrow L^{p}$ estimate
$$\|S_{N}f\|_{L^{p}}\lesssim_{n,p} N^{-n/p}\|f\|_{L^{\infty}}$$
Since he refers earlier to the article, in the context of Fourier restriction/extension estimates, to Nikishin-Maurey theory, I am assuming he means something along the lines of the following theorem of Nikishin, which I have taken by Garcia-Cuerva, Rubio De Francia's Weighted Inequalities

Theorem (Nikishin) Let $(Y,\mu)$ be an arbitrary measure space, $(X,m)$ be a $\sigma$-finite measure space, and let $T: L^{p}(\mu)\rightarrow L^{0}(m)$ be a continuous sublinear operator, with $0<p<\infty$. Then there exists a weight $w(x)>0$ a.e. such that $T: L^{p}(\mu)\rightarrow L^{q,\infty}(wdm)$, where $q=p\wedge 2$; i.e.
  $$\int_{\{|Tf|>\lambda\}}w(x)dm(x)\leq (\|f\|_{L^{p}}/\lambda)^{q}, \quad f\in L^{p}(\mu); \lambda >0$$

One can take the weight $w$ to be bounded above. 
If we apply Nikishin's theorem to the adjoint operator $S_{N}^{*}$, we obtain that there is such a weight $w$ such that $S_{N}^{*}: L^{p'}(\mathbb{R}^{n})\rightarrow L^{p',\infty}(\mathbb{R}^{n}, w(x)dx)$, for $p>2$. By translation invariance, we can also take $w$ to be continuous and everywhere positive. By rotation invariance and averaging over the orthogonal group, we can take $w$ to be radial. 
How does one eliminate the weight completely to obtain that $S_{N}^{*}: L^{p'}(\mathbb{R}^{n})\rightarrow L^{p',\infty}(\mathbb{R}^{n})$, which I presume is the goal (it's quite possible I am completely offbase here)? If my understanding is correct, in setting where the operator maps to functions on a homogeneous space, such as the sphere, one can use the associated group action to average out the weight; however, this doesn't seem to work in the current setting. I was unable to find a reference where this argument was sketched. Other papers seem to refer to the aforementioned Bourgain paper.

REPLY [2 votes]: So here's a stab at how to obtain the desired estimate; I still don't know if it's what Bourgain intended. I would appreciate if someone could comment if my understanding of a result from the theory of absolutely summing operators is correct.
Observe that the operator $S_{N}$ is local in the sense that if $\text{supp}(f)\subset B_{r}(c)$, then $\text{supp}(S_{N}f)\subset B_{Cr}(c)$, where $C>0$ is some constant which only depends on the dimension $n$ and the function $b$. So since $S_{N}$ is translation invariant, it suffices to show that
$$\|S_{N}f\|_{L^{p}}\lesssim_{n,p,\epsilon}N^{-\frac{n}{p}+\epsilon}\|f\|_{L^{p}}, \quad \text{supp}(f)\subset Q:=[0,1]^{n}$$
Fix $q>p>2$ (in particular, $q$ can be outside the admissible range for Bochner-Riesz). Since $L^{p'}(Q)$ has type $p'$ and $\|S_{N}\|_{(\infty,p)}\lesssim N^{-n/p}$, the theory of absolutely summing operators (See [Pisier 86]) tells us that there exists a probability measure $\mu$ on $Q$, such that
$$\|S_{N}f\|_{L^{p}(dx)}\lesssim_{n,p,q}N^{-n/p}\|f\|_{L^{q}(\mu)}$$
By translation invariance, we have that for any $h\in\mathbb{R}^{n}$,
$$\|S_{N}f\|_{L^{p}(dx)}\lesssim_{n,p,q}N^{-n/p}\|f(\cdot-h)\|_{L^{q}(\mu)}$$
Now integrating both sides of the inequality over the cube $2\sqrt{n}Q$ with respect to $dh$ and using Fubini together with the fact that $\mu$ is a probability measure, we obtain
$$C_{n}\|S_{N}f\|_{L^{p}(dx)}^{q}\lesssim_{n,p,q} N^{-n/p}\|f\|_{L^{q}(dx)}^{q}$$
To obtain the desired $(p,p)$ estimate, we use complex interpolation with the trivial $(2,2)$ estimate. Specifically, writing $\frac{1}{p}=\frac{\theta}{q}+\frac{1-\theta}{2}$, we obtain
$$\|S_{N}f\|_{L^{p}}\lesssim_{n,p,q}N^{-\theta\frac{n}{p}}\|f\|_{L^{p}}$$
from which the desired conclusion follows.<|endoftext|>
TITLE: How long does the slow inefficient algorithm for computing the product in classical Laver tables take?
QUESTION [8 upvotes]: Let $(A_{n},*)$ denote the $n$-th classical Laver table. Let
$X_{n}$ be the set of all finite sequences of elements from $A_{n}$.
Define a function $E_{n}:X_{n}\rightarrow X_{n}$ by letting

$E_{n}((x))=(x)$.
$E_{n}((2^{n},x_{1},...,x_{k}))=(x_{1},...,x_{k})$
$E_{n}((x,1,x_{1},...,x_{k}))=(x+1,x_{1},...,x_{k})$ whenever 
$x_{1},...,x_{k}\in A_{n}$ and $x<2^{n}$
$E_{n}((x,y,x_{1},...,x_{k}))=(x,y-1,x+1,x_{1},...,x_{k})$ whenever $x<2^{n}$ and $y>1$.

The motivation behind the function $E_{n}$ is that if
$E_{n}(x_{1},...,x_{r})=(y_{1},....,y_{s})$, then $(...(x_{1}*x_{2})...)*x_{r}=(...(y_{1}*y_{2})...)*y_{s}$. Furthermore, for all $x_{1},...,x_{r}$, there is some $m$ with $E_{n}^{m}(x_{1},...,x_{r})=((...(x_{1}*x_{2})*...)*x_{r})$. Therefore, the function $E_{n}$ represents going to a next step in an algorithm that computes the classical Laver tables (and this algorithm is simply an application of the double recursive definition of the classical Laver tables). Although this algorithm has no practical uses, a modification of this algorithm can be used to calculate Laver tables.
Let $t_{n}$ be the least natural number such that $E_{n}^{t_{n}}((x,y))=(x*y)$ whenever $x,y\in A_{n}$. The motivation behind the number $t_{n}$ is that $t_{n}$ measures the number of steps it takes to compute the product $x*y$ in $A_{n}$.
We have $t_{0}=1,t_{1}=3,t_{2}=145$ and $t_{3}=599$ and $t_{4}>15000000$.
Can anyone give any good bounds on the function $n\mapsto t_{n}$? Is the function $n\mapsto t_{n}$ primitive recursive?

REPLY [3 votes]: Let $m=2^n$ for simplicity. Then the worst input pair seems to be $(1,m-1)$, in which case the number of steps is roughly on the order of $m!/2^m$. Meaning, although the algorithm is quite inefficient, still the number of steps is boringly small compared to non-primitive recursive functions.
Analysis:
If we start with the pair $(m-1,k)$, then after $k-1$ steps we get to $(m-1,1,m^{(k-1)})$ (where $j^{(i)}$ denotes $j,j,\ldots,j$, with $i$ repetitions), then one more step gives $(m^{(k)})$, and then $k-1$ more steps give $(m)$. Total: $2k-1$ steps. Had we started with the tuple $(m-1,k, \overline x)$ for some tuple $\overline x$, then after $2k$ steps we would have eliminated the first two entries and reached $(\overline x)$.
Similarly, if we start with $(m-2, k)$, then after $k-1$ steps we get to $((m-1)^{(k)})$. Then by the previous paragraph, each pair of $(m-1)$'s takes $2(m-1)$ steps to eliminate. (The final result of the computation depends on whether $k$ is even or odd, but we only care about the running time so this is a minor issue.) Total: Roughly $2(m-1)\cdot k/2 \approx (m-1)k$ steps.
If we start with $(m-3, k)$, then after $k-1$ steps we get to $((m-2)^{(k)})$. By the preceding paragraph, each pair of $(m-2)$-s takes roughly $(m-1)(m-2)$ steps to eliminate. Total: Roughly $(m-1)(m-2)\cdot k/2$.
And so on. At stage $i$ of the argument, we replace $k$ by $m-i$ and we multiply by $k/2$. Therefore, at the end we get roughly $(m-1)!/2^{m-2}$, as claimed.
By the way, the technique that would turn your inefficient algorithm into an efficient one is called dynamic programming or "memoizing". It consists of storing the previously computed results so we don't have to compute them again. For example, with dynamic programming we can compute $F_n$, the $n$-th Fibonacci number, in roughly $n$ steps, whereas if we blindly recompute all intermediate results over and over again, the number of steps is roughly $F_n$. (!)<|endoftext|>
TITLE: Computing algebraic properties of trace fields, as given by SnapPy
QUESTION [6 upvotes]: SnapPy can tell you the trace field of a hyperbolic $3$-manifold (which is awesome), but it specifies the field by outputting:

the minimal polynomial of the field over $\mathbb{Q}$, and
a decimal approximation of the field's primitive element.

I am interested in algebraic properties of the trace field, such as the lattice of field extensions between the trace field and $\mathbb{Q}$,
and algebraic expressions for a set of generators of the trace field over $\mathbb{Q}$. This is in general difficult to find based on the data given (and after degree five, as we know ever since Galois, it is often literally impossible).
But presumably, SnapPy "knows" more than the output it gives.
It just gives the trace field in this way because it's useful for lots of other applications.
So, a general inquiry is:
Can I get SnapPy to answer algebraic questions about the trace field? For instance: What is the maximal real subfield of the trace field? Does its real subfield admit complex embeddings? What is its Galois group? (Forgive me if it is well-known how to do these things, I am pretty new to SnapPy.)
Below is a more specific question about what I need to do at the moment,
with a more precise explanation of the SnapPy output.
Let $K$
be some number field, and suppose all we know about it is

an irreducible polynomial $m(t)∈\mathbb{Z}[t]$ such that $K≅Q(t)/m(t)$, and
a decimal approximation $z∈\mathbb{Q}[i]$ of a root $s∈\overline{\mathbb{Q}}$ of $m(t)$ satisfying $K=\mathbb{Q}(s)$, where we assume that $z$ closer to $s$ than it is to any of the other roots of $m(t)$.

I want to know whether $K$
can be written in the form $F(\sqrt{−d})$, where $F\subset\mathbb{R},d∈F^+$.
Preferably I'd like some very easy way to check this just by looking at $m(t)$.
There may be something from Galois theory that has rusted away in my brain,
in which case I apologize for asking about a sub-research-level topic.
Otherwise, perhaps there is some computer implementation that can check this?
I can show that when $[K:\mathbb{Q}]=4$,
my condition holds if and only if $m(t)$ is of the form $t^4+2at^2+b$ where $a,b\in\mathbb{Z}$.
This is done casewise according to whether or not $d\in\mathbb{N}$,
then forming the primitive element and solving for $m(t)$.

Addendum:
Here's a sub-question that might be more fun (following a suggestion from @DimaPasechnik).
What are similar forms that $m(t)$
must take if and only if $[K:\mathbb{Q}]=2n$
for other $n$-values?
Perhaps there is a recognizable pattern that can be written for general $n$.

REPLY [3 votes]: Although there has been extensive discussion of this question in the comments, I thought I might contribute something that one might consider an answer to this question. This treatment will focus on a narrow bit of the code and programs that are available and not the rich and wonderful underlying theory. 
The kernel code for SnapPy (based on Jeff Weeks' SnapPea) focuses on a narrow set of things and seems to do them very efficiently. Later, Marc Culler and Nathan Dunfield provided an updated interface for the SnapPy kernel and in that process has added functionality to the original version. One facet of this interface is that SnapPy can be run inside of sage. 
First let's discuss the original SnapPea kernel (for simplicity I will ignore some of the technical details involving incomplete geometric structures and non-peripheral homology, trusting that the experts will know how to deal with these issues). Essentially, this kernel can take as input an ideal triangulation of a cusped 3-manifold. From this triangulation, combinatorial invariants can be computed, like a homology, a presentation of the fundamental group, etc. However, the true utility of the SnapPea kernel is that it builds and provides an approximate solution to the Thurston gluing equations (see Thurston's notes or Chapter E of Benedetti and Petronio). This associates approximate geometric information to each ideal tetrahedron. As part of the standard SnapPy interface one can uses quadruple-double (128 bit) precision to find an approximate solution.
To promote this to an exact solution, Coulson, Goodman, Hodgson and Neumann collaborated on snap (Goodman's PhD thesis work represents a substantial contribution to the coding):
David Coulson, Oliver A. Goodman, Craig D. Hodgson, and Walter D. Neumann, MR 1758805 Computing arithmetic invariants of 3-manifolds, Experiment. Math. 9 (2000), no. 1, 127--152.
which uses pari and it's LLL algorithm to find number field that contains all of the exact values of the shapes of the ideal tetrahedra (one should consider the DNA of the manifold, because the geometric invariants of the manifold can be computed from these shapes). 
Sadly, snap is no longer directly supported (although with luck and will power some have got a version of this running on a current system) however, SnapPy has functionality in sage which mimics some of snap's more important features. In addition, sage has several methods to manipulate number fields. 
Perhaps this is a long way of saying no, "SnapPy" will not do the desired field computations but sage can (assuming SnapPy can describe the tetrahedral shapes of one's example with enough precision).<|endoftext|>
TITLE: Minimum length of a convex lattice polygon containing k lattice points?
QUESTION [5 upvotes]: Let $f(k)$ denote the minimum length of a convex lattice polygon containing exactly $k$ lattice points (including lattice points on the boundary).
It is not too hard to show that $k = \frac{1}{4\pi} f(k)^2+o(f(k)^2)$ as $k \to \infty$.  
Is it known whether in fact $k = \frac{1}{4\pi} f(k)^2+o(f(k))$?  I naively expect this to be the case, but can not prove it [nor am I sure of it's validity].  Any help would be appreciated!

REPLY [5 votes]: This may be a delicate problem. I have a few comments but no real progress to a solution.
It might be worth trying to figure out what polygon(s) give $f(k).$ 
I think it is not the convex hull of the lattice points in circles centered at the origin (the Gauss circle problem.) More promising is circles centered at $(1/2,1/2).$ The number of points if the diameter is ${\sqrt{4n^2-4\sqrt{2}n+2}}$ is the series $4,12,24,44,68,\cdots$ obtained by quadrupling the   Number of nonnegative solutions to $x^2 + y^2 \le n^2$. 
For example a circle of radius $5$ centered at the origin seems promising since it has $12$ points on the boundary. It has $49$ points in all and the boundary has length $8\sqrt{10}+4\sqrt{2} \approx30.955$ But a $7 \times 7$ square gives length $28.$ The optimum may be $13+5\sqrt{2}+3\sqrt{5}\approx 23.779$ coming from columns of lengths $4,6,8,8,8,6,6,3.$ In other words, a $6 \times 6$ square with $4,3,3$ and $3$ additional points parallel to the sides.
It seem plausible that the optimum is the convex hull of the points in some circle but getting the right one may not be easy. Perhaps for larger numbers the jumps are not as extreme. 
I do suspect that $\log_{f(k)}\left(k - \frac{1}{4\pi} f(k)^2\right)$ may oscillate. But that is just a guess.<|endoftext|>
TITLE: Criterion for deciding the conformal class of a metric on a complete surface
QUESTION [6 upvotes]: For orientable closed Riemannian surfaces $(S,g)$, there is a constant curvature metric $\overline{g}$ on $S$ that is conformal to $g$ in the sense that $\overline{g} = e^ug$ for some smooth function $u$. Moreover, by the Gauss-Bonnet theorem the sign of the curvature of $\overline{g}$ is uniquely determined by the topology of $S$.
However, when $S$ is non-compact, it may admit different kinds of complete metrics of constant curvature. For example, on $\mathbb{R}^2$ you have many flat metrics but you also have the hyperbolic metric coming from some identification of the plane with the unit disk. On the cylinder $C = \mathbb{R}^2 \backslash \mathbb{Z}$, there is the natural flat metric coming from the Euclidean metric but there is also the hyperbolic metric arising from the quotient of the upper half-plane by a horizontal translation.
I am interested in formulating some (geometric) criterion for a metric $g$ on $\mathbb{R}^2$ or on the cylinder $C$ to be conformal to a flat metric. For example, if $(C,g)$ is asymptotic to the product metric $dx^2 + d\theta^2$ with suitable decay assumptions, I was shown how to solve the PDE for a conformal deformation to a flat metric using weighted Sobolev spaces. But for weaker notions of "asymptotic flatness" of $g$, this method wouldn't work although I feel like such a metric should be conformally flat, even for very weak notions of asymptotic flatness...
So, given a metric $g$ on the plane or on the cylinder, how can I differentiate between $g$ being conformally flat or conformally hyperbolic?

REPLY [4 votes]: This is a classical problem which is called the Type Problem of a simply connected Riemann surface: If you have a metric on an open simply connected surface, to determine whether it is conformally equivalent to the plane or to the disk.
The general situation is the following: there are necessary and sufficient conditions, but they are usually difficult to verify for specific metrics.
And there are very many separate necessary or sufficient conditions which are
easier to verify, especially for some special classes of metrics.
There was an intensive research on this in 1930-1950, and a few modern papers. 
Some references are:
MR2019938 
Benjamini, Itai; Merenkov, Sergei; Schramm, Oded,
A negative answer to Nevanlinna's type question and a parabolic surface with a lot of negative curvature. 
Proc. Amer. Math. Soc. 132 (2004), no. 3, 641–647,
MR0954627 Doyle, Peter G. On deciding whether a surface is parabolic or hyperbolic. Geometry of random motion (Ithaca, N.Y., 1987), 41–48, Contemp. Math., 73, Amer. Math. Soc., Providence, RI, 1988. 
MR0428232 Milnor, John On deciding whether a surface is parabolic or hyperbolic. Amer. Math. Monthly 84 (1977), no. 1, 43–46. 
MR0279280 Nevanlinna, Rolf Analytic functions. New York-Berlin 1970 viii+373 pp.
MR0049330 Volkovyskiĭ, L. I. Investigation of the type problem for a simply connected Riemann surface. (Russian) Trudy Mat. Inst. Steklov. 34, (1950). 171 pp. This is the most comprehensive exposition. Not much was added to this theory since 1950.<|endoftext|>
TITLE: What does the notation $[b_1,b_2]$ in M. Hochster's "Prime Ideal Structure in Commutative Rings" mean?
QUESTION [7 upvotes]: I'm reading the article 
M. Hochster, Prime ideal structure in commutative rings, Trans. Amer. Math. Soc. 142 (1969), 43--60. Freely available here on the journal's website.
But, I can not find the definition of the symbol "$[b_1 , b_2]$" in the proof of Theorem 4 (page 49, which is 7th page of the pdf). 
Could someone explain what this notation means there?
Possibly it is explained somewhere in the paper, but I could not locate it.

REPLY [4 votes]: My best guess for what $[b_1,b_2]$ is supposed to mean is the subring of $A$ generated by $b_1$ and $b_2$.  However, even with this interpretation, there are some statements that aren't quite right (though as I recall, this particular paper has a lot of minor errors of this sort, so this shouldn't be too surprising).  Here is how I would rewrite the final three sentences of the proof of Theorem 4:
Let $f:\mathbb{Z}[x_1,x_2]\to A$ be the unique homomorphism sending $x_i$ to $b_i$ and let $I=(x_1,x_2)\subset\mathbb{Z}[x_1,x_2]$.  For any $z\in I$, $f(z)$ vanishes off of $d(b_1)\cup d(b_2)$, and its image restricted to $d(b_1)\cup d(b_2)$ is equal to its image restricted to the finite set $Y$.  Thus we need only find $z\in I$ such that $z\not\in f^{-1}(\phi(y))$ for each $y\in Y$, and then $c=f(z)$ will be an element of $(b_1,b_2)$ such that $d(c)=d(b_1)\cup d(b_2)$.  But each $f^{-1}(\phi(y))$ is a prime ideal in $\mathbb{Z}[x_1,x_2]$ which does not contain $I$ (since every point of $Y$ is in $d(b_1)\cup d(b_2)$), and there are only finitely many of them, so by prime avoidance such a $z$ must exist.<|endoftext|>
TITLE: Primitive log-divergent graphs and convergence of Feynman amplitudes
QUESTION [5 upvotes]: To a connected graph $G$, quantum field theory attaches the integral
$$
I_G=\int_{\sigma} \frac{\Omega_G}{\Psi_G^2}
$$ where $N_G$ is the number of edges of the graph, $\sigma$ is the simplex of points $(x_1: \ldots : x_{N_G}) \in \mathbb{P}^{N_G-1}(\mathbb{R})$ such that $x_i \geq 0$ for all $i$, $\Omega_G$ is the volume form
$$
\Omega_G=\sum (-1)^i x_i dx_1 \wedge \cdots \widehat{dx_i} \cdots \wedge x_{N_G}
$$ and $\Psi_G$ is the first Symanzik polynomial, defined by
$$
\Psi_G=\sum_{T \subseteq G} \prod_{e \notin T} x_e, 
$$ where the sum runs over spanning trees. 
The integral is in general divergent, but converges for a class of graphs called primitive log-divergent. These are those for which $N_G=2b_1(G)$ (first Betti number) and $N_\gamma>2b_1(\gamma)$ for any proper subgraph $\gamma$. I have found this statement in several places but no proof. 
Does anybody know how to prove the convergence of $I_G$ in the primitive log-divergent case?
Is it an if and only if?
Any intuition about what the condition means?

REPLY [4 votes]: A convergence proof ("if and only if") follows proposition 5.2 of On Motives Associated to Graph Polynomials (2005).<|endoftext|>
TITLE: What are the special parahoric subgroups in unitary groups?
QUESTION [8 upvotes]: Let $L$ be a $p$-adic field and let $L'/L$ be a quadratic extension.  Let $U_{L'/L}(n)$ be a quasi-split unitary group of $n\times n$ matrices with entries in $L'$.  I'm curious about what the special maximal subgroups of such a group are.
More specifically, let $\mathbf{K}$ be the subgroup consisting of unitary matrices whose entries lie in the ring of integers $\mathfrak{o}_{L'}$.  If the residue characteristic of $L'/L$ is not $2$, and $L'/L$ is unramified, then $\mathbf{K}$ is hyperspecial.  However, when the residue characteristic is $2$, then $\mathbf{K}$ is not even maximal when $n = 3$.
My question is:

When $L'/L$ is ramified and has residue characteristic not $2$, is $\mathbf{K}$ still maximal special (even though it is not hyperspecial)? If not, what are some of the special subgroups?
If the residue characteristic is $2$, when is $\mathbf{K}$ special? When it isn't, can we find a special subgroup containing it?

REPLY [2 votes]: If you can read french, explicit models of Bruhat-Tits buildings of classical groups over local fields (of any characteristic) are given in:
Bruhat, F.; Tits, J. Schémas en groupes et immeubles des groupes classiques sur un corps local. II : groupes unitaires. Bulletin de la Société Mathématique de France, 115 (1987), p. 141-195<|endoftext|>
TITLE: Further directions of index theory
QUESTION [8 upvotes]: The Atiyah-Singer theorem is a major achievement of twentieth century mathematics. It has inspired a lot of work and people started to develop generalizations of this theorem. I would like to know the current state of the theory and current central problems in index theory. Let me mention some possible generalisations:  

More than one operator This approach was initiated by Atiyah himself in the index theorem for families of operators. Possible generalizations are index theorems for foliations (for easy foliations which are fibrations it implies the index theorem for families).  
Higher index theory This line of research concerns manifolds which are $G$-spaces and equivariant operators or more generally, noncompact (complete, Riemannian) manifolds (as in the work of John Roe). The prominent examples are universal covers of non-simply connected manifolds (where the fundamental groups acts).
Non-smooth manifolds Examples include combinatorial manifolds, Lipschitz manifolds and quasiconformal manifolds. This direction was developed mostly by Nicolae Teleman.   
Non-commutative geometry This whole discipline is heavily inspired by index theory: in fact the points mentioned above use the machinery of noncommutative geometry. There is an abstract local index theorem for spectral triples of Connes-Moscovici.   


What are most important current directions and trends in index theory? What are the most important problems in index theory?

REPLY [5 votes]: Let me add some more current trends to the list.

Local and smooth index theory. Starting from a geometric index problem, one regards the cohomological index as a form on the underlying manifold, rather than just a cohomology class. One regards the index itself as an element of a smooth $K$-group, which includes differential form data obtained from the local index. As an example, the smooth family index of a family of Dirac operators typically involves an $\eta$-form.
Indices of non-elliptic operators. An important example is Bismut's hypoelliptic Dirac operator.<|endoftext|>
TITLE: Just how close can two manifolds be in the Gromov-Hausdorff distance?
QUESTION [22 upvotes]: Suppose that we have two compact Riemannian manifolds $(M,g)$ and $(N,h)$. Define the Gromov-Hausdorff distance between them in your favorite way, I'll use the infimum of all $\epsilon$ such that there are $\epsilon$-Gromov-Hausdorff approximations between $(M,g)$ and $(N,h)$.
Then define $d(M,N)$ to be the infimum of the Gromov-Hausdorff distance between $(M,g)$ and $(N,h)$ taken over all Riemannian metrics $g$ and $h$ with sectional curvatures bounded in absolute value by 1. 
If you know $d(M,N)$ for various choices of $N$, what can you conclude about $M$?
I know that Cheeger-Fukaya-Gromov theory on collapsed manifolds exactly covers the case when $d(M,N)=0$; for example $d(M,pt)=0$ if and only if $M$ is almost flat. I'm interested in situations where these numbers do not vanish. For example:
If $\Sigma$ is some compact orientable surface how should $d(\Sigma,pt)$ depend on the genus?
If $S^n$ is the standard sphere of dimension $n$, to what extent do the numbers $d(M,S^n)$ determine $M$? Are there manifolds $M$ and $N$ for which $d(M,N)\neq 0$ but $d(M,S^n)=d(N,S^n)$ for all $n$?
Is there an $\epsilon$ depending only on dimensions such that $d(M,N)<\epsilon$ implies $d(M,N)=0$?
If anyone can point me toward a reference/paper it would be very appreciated.

REPLY [8 votes]: This is only an answer to one point of your question: for surfaces of large genus $g$ the distance should be 
$$
d(S, \mathrm{point}) \asymp \log(g). 
$$
The lower bound should follow from volume estimates (by Gauss-Bonnet the volume is at least $\gg g$ and the volume growth of balls in spaces with bounded curvature is at most exponential). 
The upper bound can be proven at least in two ways, explicit construction and random. First it follows from the existence of expanding sequences of covers of arithmetic surfaces, which have diameter $\ll \log(g)$ by general facts about expanders (see eg. Lubotzky's book Discrete groups, expanding graphs and invariant measures, in particular 7.3.11(ii)). I think that probably one can get any genus this way but I am unsure whether somebody wrote it down. On the other hand, by results of Mirzakhani (see II) in the intro of Growth of Weil--Petersson volumes and random hyperbolic surfaces of large genus, 1012.2167) a typical surface of genus $g$ has diameter $\ll \log(g)$.<|endoftext|>
TITLE: Complexity of graph isomorphism
QUESTION [7 upvotes]: Last year, Laszlo Babai proved that the graph isomorphism problem can be solved in time:
$$ \exp(O(\log^c n)) $$
where $n$ is the number of vertices.
What is the best bound we have for $c$? (The case $c = 1$ would correspond to a polynomial-time algorithm for graph isomorphism.)

REPLY [4 votes]: Babai apparently retracted some parts of his proof, now he claims that he can do $O(\exp(n^c))$ for some small $c$ (say, $c=0.01$), but not all $c>0$. See http://people.cs.uchicago.edu/~laci/update.html<|endoftext|>
TITLE: Existence of a particular kind of polygonal subdivisions of surfaces
QUESTION [7 upvotes]: Let $\Sigma$ be a closed, compact, connected, oriented smooth $2$-manifold (in other words, a sphere or a torus with $g$ handles). We may draw smooth arcs on the surface in such a way that they cut it up into polygons. Call this configuration a polygonal subdivision of $\Sigma$. We say that such a subdivision is $n$-regular if it consists only of $n$-gons and is such that exactly $n$ polygonal faces meet at every vertex of the subdivision.
For instance, the tetrahedron is a $3$-regular polygonal subdivision of the sphere. As a second example, the following figures induce $4$-regular subdivisions of the torus after identifying opposing sides of the square:
$4$-regular subdivisions of the torus" />
One may obviously produce an infinite family of this kind by refining the mesh.

Q: Is it possible to construct $n$-regular subdivisions of surfaces of arbitrarily high genus?

By an Euler characteristic argument we have some fairly obvious conditions over the possible genus of a surface admitting an $n$-regular subdivision. Indeed, if $v,e,f$ denote the number of vertices, edges and faces respectively, then $v=f$ and $nf/2=e$, so the Euler formula $v-e+f=2-2g$ implies:
$$4g-4=(n-4)f$$
This imposes some restrictions that an eventual example must satisfy. Are there finer invariants that present obstructions to this kind of construction? If there are no obstructions, is there an easy way to construct them?

REPLY [6 votes]: There are no further obstructions beyond the Euler obstruction you point out and there is a simple construction of examples in each class.
You have already given constructions for all possibilities for $g<2$ except $g=0$ and $n=2$, which is easy.
Douglas Zare, in a deleted answer, described how constructing the surface of genus $2$ by identifying sides of an octagon in pairs provides an $8$-regular subdivision of the surface of genus $2$ (with $f=1$). More generally, a similar standard construction works for all $g\ge 2$, giving a $4g$-regular subdivision of the surface of genus $g$ (with $f=1$). Below, I will outline two other similar constructions, one for $g=2k$ and $n=2k+3$ with $f=4$ and the other for $g=2k$ and $n=4k+2$ with $f=2$. This gives a base construction for every $n>4$.
Suppose we have these base constructions for all $n>4$. Let $(g,n)$ be an arbitrary pair (with implicit $f$) satisfying your Euler equation with $g>1$. Then $n>4$ and by Euler characteristic the surface of genus $g$ covers the surface with the base construction (this can be seen in three cases which I will omit for now but can add if desired). Then as Douglas pointed out in the deleted answer, the base construction can be lifted to the genus $g$ surface.
Thus the problem is reduced to giving the other two base constructions. The following two constructions are probably known classically but were described to me by Seonhwa Kim.
Consider first the case with even genus $g=2k$ and $n=4k+2$. Cut the surface of genus $g$ into two surfaces with boundary, each of genus $k$ with one boundary component. Using a variation of Douglas' construction with two extra edges for the boundary component, a surface of genus $k$ with one boundary component can be constructed from a $4k+2$-gon identifying $4k$ of the faces in pairs. The resulting surface has two vertices each of valence $2k+2$. Gluing the two of these together along their boundary gives a $4k+2$-regular decomposition of the genus $g$ surface.
To get the case with even genus $g=2k$ and $n=2k+3$, take the $4k+2$-gon described in the previous paragraph and cut it with an edge between the midpoints of the non-identified edges making up the boundary component. This yields a decomposition of the surface of genus $g$ with one boundary component into two $2k+3$-gons with  four vertices, two of valence $3$ and two of valence $2k+2$. Gluing two such decompositions together with a quarter-twist so that the vertices of valence $3$ on one side are paired with the vertices of valence $2k+2$ on the other side yields a $2k+3$-regular decomposition of the genus $g$ surface.<|endoftext|>
TITLE: A chain of six circles associated with a conic
QUESTION [9 upvotes]: I found this problems three years ago. But I never have been a proof. Recently I posted in math.stackexchange.com.  I am looking for a solution of the following problems:
A chain of six circles associated with a conic.

Let $A_1, ..., A_6$ and $B_1, ..., B_6$ be 12 points lying on a conic, and suppose that for $i=1, ..., 5$ through $A_i, A_{i+1}, B_{i+1}, B_i$ passes a circle $(O_i)$. Then through $A_6, B_6, A_1, B_1$ as well passes a circle $(O_6)$. Let $P_1, P_4$ be intersection points of $(O_1)$ and $(O_4)$; the same for $P_2, P_5$ and $P_3, P_6$. Show that:

Three lines $O_1O_4$, $O_2O_5$, and $O_3O_6$ have a common point $O$.

Six points $P_1, ..., P_6$ lie on a circle with center in $O$.




My remark: With six points and six lines we get the Pascal theorem. With 12 points and six circles we have this problem

I checked it by Geogebra, But I can not calculate.

See also: Sequences of Concyclic Points on a Conic and Geogebra A chain of six circles associated with a conic

There are two other chains of six circles

REPLY [3 votes]: Since $С_1$, $C_2$, and your conic $\alpha$ pass through $A_2$ and $B_2$ we get that $A_1B_1$ and $A_3B_3$ are parallel (it calls three conic theorem http://mathworld.wolfram.com/ThreeConicsTheorem.html). Applying it several times to your construction you get that $C_6$ exists.
Intersection of $O_1O_4$, $O_2O_5$ and $O_3O_6$ follows from the Pappus theorem applied to the triple parallel lines $O_1O_2$, $O_3O_4$, and $O_5O_6$ and the triple $O_2O_3$, $O_4O_5$, and $O_6O_1$.
Upd. Regarding the second question:
@zeb was almost right. We need lemma with little bit different combinatorics:
$(abcd)$, $(aba_1b_1)$, $(abc_1d_1)$, $(cda_1b_1)$, $(cdc_1d_1)$ => $(a_1b_1c_1d_1)$.
Lets show that $P_1$, $P_4$, $P_3$, and $P_6$ lie on a circle. For that we note that the following quadruples circumscribed
$(A_2B_2A_5B_5)$, $(A_2B_2P_1P_4)$, $(A_2B_2P_3P_6)$, $(A_5B_5P_1P_4)$, $(A_5B_5P_3P_6)$ and then apply the Lemma.
It is clear that center of this circle is $O$ because it is lie on the corresponded perpendicular bisectors of $P_iP_{i+3}$.
Therefore all $P_i$ lie on the fixed circle with center at $O$.<|endoftext|>
TITLE: Examples of varieties with many automorphisms acting trivially on co-homology
QUESTION [12 upvotes]: Let $X$ be a smooth projective variety over the complex numbers. Denote by $Aut(X)$ its automorphism group, by $Aut(X)_0$ the connected component of the identity, and by $G$ the quotient $Aut(X)/Aut(X)_0$. 
The pull-back gives a representation
$$
\rho \colon G \to GL(H^{1,1}(X,\mathbb{C}))
$$
I look for examples of varieties $X$ such that the kernel of $\rho$ is not a finite group.

REPLY [11 votes]: The group of connected components of the subgroup of $Aut(X)$ fixing $H^{11}(X)$ is finite. This follows from the main result of 
Polarized Varieties, Fields of Moduli and Generalized Kummer Varieties of Polarized Abelian Varieties,
T. Matsusaka,
American Journal of Mathematics,
Vol. 80, No. 1 (Jan., 1958), pp. 45-82 
I quote from the mathscinet review:

The main result (section 3) is as follows. Let $U$ be a projectively embeddable non-singular variety; then there is a maximal connected algebraic group $G$ of automorphisms of $U$. Moreover, $G$ is of finite index in the group $G′$ of those automorphisms of $U$ which, for a given projective embedding of $U$, carry a hyperplane section $H$ into a cycle numerically equivalent to $H$.

Since the group of numerical equivalence classes of divisors embeds in $H^{11}$, the group of automorphisms preserving $H^{11}$ is a subgroup of the group $G'$ above. It is clearly a closed algebraic subgroup, so it likewise has finitely many components.
I must admit, on a quick read, I also endorse the last sentences of the review:

Unfortunately, the author's style, added to the intrinsic difficulties of the subject, makes it exceedingly hard to check the accuracy of many technical details which seem essential for the validity of the proofs; this is all the more to be regretted, since his results are so valuable and important. It is to be hoped that he, or someone else, will some day give a completely lucid and completely convincing exposition of these topics.<|endoftext|>
TITLE: Tightness and Functional Analysis
QUESTION [7 upvotes]: Let $(\Omega , \mathbb{P})$ be a probability space and $X$ be a real-valued random variable. Then we immediately have the push-forward measure $\mu$ on $\mathbb{R}$ and one can think of $\mu$ as an element of the unit ball of $C_0(\mathbb{R})^*$. By Banach-Alaoglu, the unit ball is weak-$^*$ compact and since $C_0(\mathbb{R})$ is separable, we have that every sequence of probability measures has a weak-$^*$ limit. This limit might not be a probability measure, but is if the sequence is tight. My first question is there a nice functional analytic viewpoint of tightness, in this scenario?
In applications, it is important to extend the action of $\mu$ to all of $C_B(\mathbb{R})$ (since, for instance, that's where the characters are). Now $\mu$ no longer lives in the separable dual space of $C_0(\mathbb{R})$, but in the dual of $C_b(\mathbb{R})$ which is not separable. Thus Banach-Alaoglu no longer implies anything about sequences. In probability texts, they use very specific Lebesgue theory arguments to assert that if $\mu_n$ is a sequence of tight measures that there is a convergence sub sequence. My second question is in this more general setting, can we still find a functional analytic view point that will allow us to see this?

REPLY [5 votes]: The functional analysis setting for this was established by R.C. Buck in the $1950$'s.  He introduced a natural complete locally convex topology on $C^b(S)$, the space of bounded, continuous functionss on a locally compact space S$, the so- called strict topology with the properties
1) the dual space is the space of tight (Radon) measures on $S$;
2)  if $ S$ is $\sigma$  compact, then a set of measures is weakly compact (for this duality) if and only if it is tight.
This was later extended to the case of general completely regular spaces and the question of which $S$ display the second property  (in particular, with respect to families of probability measures) attracted much attention.<|endoftext|>
TITLE: How many ways can I factor a matrix (over $\mathbb{Z}$)?
QUESTION [21 upvotes]: Let $A$ be a fixed matrix in $M_2\mathbb{Z}$ with determinant $n \neq 0$.
Question 1 How many ways can I write $A = XY$ for $X, Y \in M_2\mathbb{Z}$?
The answer to this question is pretty clearly infinite, since for any $\gamma \in GL_2\mathbb{Z}$ and any such pair $(X, Y)$, then $(X \gamma^{-1}, \gamma Y)$ is another such factorization. So let's get rid of that.
Question 2 Consider the set $S_A = \{(X, Y) \in (M_2\mathbb{Z})^2 \mid A = XY\}$. The group $G = GL_2\mathbb{Z}$ acts naturally on this set via
$$
\gamma \cdot (X, Y) = (X \gamma^{-1}, \gamma Y)
$$
What is the cardinality of $S_A/G$?
Unfortunately, this depends on our choice of $A$ (fixing $n$). If we compare
$$
A_1 = \begin{pmatrix}4 & 0 \\ 0 & 1 \end{pmatrix} \qquad \qquad 
A_2 = \begin{pmatrix}2 & 0 \\ 0 & 2 \end{pmatrix}
$$
then in the first case, we find $|S_{A_1}/G| = 3$ while in the second, $|S_{A_2}/G| = 4$. This is due to the fact that the second matrix is not primitive. We could exclude such matrices, or we could define an equivalence relation on factorizations via
$$
(X, kY) \sim (kX, Y)
$$
which is compatible with the action of $G$. For a fixed $A$, define $T_A = S_A/_\sim$.
The Real Question Does the cardinality of $T_A/G$ depend on $A$? If not, what is it?
By computation, it seems that this does not depend on $A$ and moreover, 
$$
|T_A/G| = \sum_{d \mid n} 1 = \sigma_0(n)
$$
This seems like it should be a pretty obvious question about arithmetic/algebraic groups, but it's not really my area of expertise.

REPLY [15 votes]: $\def\ZZ{\mathbb{Z}}$Your conjecture is right when you require $A$ to be primitive. The version where you set $(X,kY) \sim (kX,Y)$ doesn't work even in your chosen example.
$A$ gives a map $\ZZ^2 \to \ZZ^2$. Set $K = \ZZ^2/A \ZZ^2$; this is an abelian group of order $n$ generated by $2$ elements and the condition that $A$ is primitive implies that $G \cong \ZZ/n \ZZ$.
You want to know how many ways you can factor this as $\ZZ^2 \to L \to \ZZ^2$, where $L \cong \ZZ^2$ and we work up to isomorphisms on the middle factor. Such a factorization is uniquely determined by the subgroup $L/ZZ^2$ of $K$. So we are counting subgroups of $\ZZ/n \ZZ$, which there are $\sigma_0(n)$ of.
If we take $A = \left( \begin{smallmatrix} 2 & 0 \\ 0 & 2 \end{smallmatrix} \right)$, then $G \cong (\ZZ/2 \ZZ)^2$, with five subgroups. Representative factorizations are
$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}, \quad \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},$$
$$\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad
\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}, \quad \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1& 1 \\ 1 & -1 \end{pmatrix}.$$
Your equivalence relation collapses these $5$ cases to $4$, not $3$.
It is clear that $\#(T_A/G)$ will only depend on the isomorphism type of the abelian group $\ZZ^2/A \ZZ^2$ (in other words, on the Smith normal form of $A$). If you really need it, I could work it out; I suspect you could too.<|endoftext|>
TITLE: Frankl's conjecture and Oeis sequence A188163
QUESTION [7 upvotes]: For every natural number $c \geq 2$, let $f(c)$ denote the least natural number $f$ with the following property : every union-closed family of sets with at least $f$ members has $c$ members whose intersection is nonempty. If I'm not wrong, it is easy to prove that $f(c)$ always exists. Frankl's conjecture amounts to say that $f(c) \leq 2c-1$.
If $c-1$ is a power of $2$, then $f(c) \geq 2c-1$ (look at the family of all subsets of a set with cardinality $k$, where $c-1 = 2^{k-1}$). Thus, if $c-1$ is a power of $2$, then Frankl's conjecture (if true) is optimal.
But if $c-1$ is not a power of $2$, it can happen that $f(c) < 2c-1$, so that Frankl's conjecture is not optimal in these cases.
For example, if I'm not wrong, $f(4) = 6$ and $f(6) = 10$. So, I searched the Oeis site for a sequence $A$ with the following properties :
$A(1+2^{k-1}) = 1+2^{k}$;  $A(4) = 6$; $A(6) = 10$.
A188163 ( http://oeis.org/search?q=3%2C5%2C6%2C9%2C10&language=english&go=Search) :
1, 3, 5, 6, 9, 10, 11, 13, 17, 18, 19, 20, 22, 23, 25, 28, 33, 34, 35, 36, 37, 39, 40, 41, 43, 44, 46, 49, 50, 52, 55, 59, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 79, 81, 82, 84, 87, 88, 89, 91, 92, 94, 97, 98, 100, 103, 107, 108, 110, 113, 117, 122
seems a good candidate, since (numbering $A(n)$from $A(1) = 1$)
$A(3) = 5$, $A(5) = 9$, $A(9) = 17$ and so on, thus $A(1+2^{k-1}) = 1+2^{k}$ in the limits of the table (I don't know if this is always true, I discover this sequence today)
and also  $A(4) = 6$; $A(6) = 10$.
(For $c = 1$, we have $f(c) = 2 \neq A(c) = 1$, but it doesn't seem very important to me.)
Is there literature about a possible relation between $f(c)$ and $A188163(c)$ ?

REPLY [7 votes]: Let me elaborate a bit on what I said over at Polymath11.
There is a nice survey on Frankl's conjecture, where the following similar question is discussed in Section 8. Let $\phi:\mathbb{N}\to\mathbb{N}$ be the function where $\phi(n)$ is the smallest maximal abundance of an element of the ground-set of a union-closed family with $n$ members. Frankl's conjecture asserts that $\phi(n)\geq \tfrac{n}{2}$, and computing $\phi(n)$ for small $n$ suggests that $\phi$ may coincide with A004001. The survey notes that this sequence always provides an upper bound for $\phi$, but this upper bound is not always tight and there remains a gap. This has been shown by Renaud.
Concerning your $f$, note that it is by definition the Galois adjoint of $\phi$,
$$f(c)\leq n \quad\Leftrightarrow\quad c\leq\phi(n),$$
meaning that $f(c)$ equals the smallest $n$ for which $c\leq \phi(n)$ holds. Since $\phi$ is monotone and surjective, this implies that $f(c)$ is the smallest $n$ for which $\phi(n) = c$ holds; this reflects exactly the relation between A188163 and A004001. Therefore your question is equivalent to asking about the relation between $\phi$ and A004001, which has been answered by Renaud as above.<|endoftext|>
TITLE: Nuclear operators/spaces and transfer operators
QUESTION [12 upvotes]: While studying for my thesis (in dynamical systems) I've encountered multiple times with the concept of nuclear operators and nuclear spaces, often linked with the works of Grothendieck. For example, when studying the generalized transfer operator (or Ruelle operator) for the Gauss Map, Dieter Mayer points out that this operator is in fact nuclear (On the thermodynamic formalism for the Gauss map). While I can understand the definition of a nuclear operator, I still cannot get the real importance of being nuclear of order zero. Usually I'm interested in spectral gap properties for transfer operators, but is there any implication of the nuclear property? Also, any reference for nuclear operators and Fredholm kernels would be appreciated, since trying to learn directly from Grothendieck's works has been really difficult for me. I have already read these notes by Paul Garrett introducing nuclear Frechet spaces in order to get a categorically genuine tensor product in the enlarged (with countable limits) category of Hilbert spaces.
Thanks in advance.

REPLY [4 votes]: I believe your question is: What is the importance of being nuclear of order zero for the transfer (=Perron-Frobenius) operator in Ergodic theory?
The importance of being nuclear is a theorem by Grothendieck: If $L$ is nuclear of order less or equal than $2/3,$ then $det(I-zL)$ is entire in $z$ and we have that $det(I-zL)=\prod_n(1-z\lambda_n)$ where $\lambda_n$ are the eigenvalues of $L.$ From this and the trace $tr(L)=\sum_n\lambda_n$ we can write $\log det(I-zL)=-\sum_n \frac{z^n}{n}Tr(L^n).$ This is quite helpfull because of many reasons. The main reason is that Ruelle succeeded to prove the nuclearity (of order zero) of the Perron-Frobenius operator associated to Axiom A diffeomorphisms in the seventies, so by Grothendieck's theorem he related the eigenvalues of the Perron-Frobenius operator with the inverses of the zeros of $\exp(-\sum_n \frac{z^n}{n}Tr(L^n)).$ This motivates many accurate approximation results in Ergodic theory by Pollicott, as eigenvalues of the Perron-Frobenius operator are known to be related to mixing properties of the associated dynamical system, and on the other hand, the inverses of the zeros of $\exp(-\sum_n \frac{z^n}{n}Tr(L^n))$ can be approximated using the Taylor expansion of holomorphic functions. A few examples of these mixing properties are the following:

Invariant densities are fixed points of the Perron-Frobenius operator.
If there exist an ergodic invariant density, then the first eigenvalue of the Perron-Frobenius operator is equal to 1 and it is simple.
If there exist a mixing invariant density, then the first eigenvalue of  the Perron-Frobenius operator is equal to 1 and moreover, every other eigenvalue has modulus strictly less than 1.

I suggest you to read Pollicott, Baladi and Ruelle, in this order. They have some lecture notes available on the web. On the other hand, I do not think you will find anything useful directly from Grothendieck's 1956 notes, Ruelle did it!
Another reason of its importance (of $L$ being nuclear) it is that isolated eigenvalues are stable under perturbations (in Kato's sense). Therefore, as a consequence, one can usually use that $L$ is nuclear in order to prove stability results. For example, of the invariant density.<|endoftext|>
TITLE: Is there a modern account of Veblen functions of *several* variables?
QUESTION [11 upvotes]: Veblen $\phi$ functions extend the $\xi \mapsto \phi(\xi) := \omega^\xi$ and the $\xi \mapsto \phi(1,\xi) := \varepsilon_\xi$ functions on the ordinals by repeatedly taking fixed points (I won't repeat the definitions, which are on Wikipedia).  For a reason that isn't entirely clear to me, Veblen functions of just two variables seem to have gotten all the attention (to the point that the range $\phi(1,0,0)$ of the notation system they provide has a special name: the "Feferman-Schütte ordinal").
Essentially the only text I was able to find which discusses the Veblen functions of more than two variables is Veblen's original paper (published in 1908, "Continuous Increasing Functions of Finite and Transfinite Ordinals"), which is difficult to read: he starts his ordinals at $1$ and uses slightly different conventions on the $\phi$ functions than are now more common (e.g., on the order of the variables), and so on; more importantly, he does not discuss normal forms for ordinals produced by these functions.
So, is there a more modern account that I missed?  Ideally one that would describe in some detail the system of ordinal notations that can be extracted from Veblen functions of $\omega$ variables (the "small" Veblen ordinal) or of the smallest fixed point of $\xi \mapsto \phi(1^{(\xi)},\ldots,0)$ (the "large" Veblen ordinal).  But even a modern "translation" of Veblen's paper would be of some help.
Edit: There seems to be some kind of ordinal notation system related to Veblen functions of three variables (I think) in Ackermann's 1951 paper "Konstruktiver Aufbau eines Abschnitts der zweiten Cantorschen Zahlenklasse", though the notations are also very old (and the ordinals also start at $1$, apparently zero still hadn't been invented in 1951).  I'll try to read it and see whether it's interesting.

REPLY [5 votes]: I think the following paper by Buchholz should provide very useful information. http://www.mathematik.uni-muenchen.de/~buchholz/articles/jaegerfestschr_buchholz3.pdf<|endoftext|>
TITLE: Gauss proof of fundamental theorem of algebra
QUESTION [17 upvotes]: My question concerns the argument given by Gauss in his "geometric proof" of the fundamental theorem of Algebra. At one point he says (I am reformulating) :
A branch (a component) of any algebraic curve either comes back on itself (I suppose that means : it is a closed curve) or  it goes to infinity on both sides.
I have a geometric intuition of what it means, but I am not sure 
where or how to get a "modern" formulation of such result, and a proof.

REPLY [4 votes]: $\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}$Not sure this is useful, but a few years ago I put some thought into what it would take to rigorously give this proof in a Honors Multivariable calculus course (the sort where everything is proved). Here is what I came up with. Let $f:\CC \to \CC$ be a polynomial of degree $n$. It turns out to be convenient to make the following simplifying assumption -- letting $w_1$, $w_2$, ..., $w_k$ be the zeroes of $f'$, none of $f(w_i)$ are either purely real or purely imaginary. This is possible because, if $f(w_i)=0$ we are done, and otherwise we can replace $f$ by $e^{i \theta} f$ for some generic $\theta$.
Let $R = f^{-1}(\RR)$ and $I = f^{-1}(i \RR)$. The argument which we are trying to make rigorous is "near infinity, $R$ and $I$ look like $4n$ interleaved spokes, so they must cross somewhere in the interior." The implicit function theorem shows that $I$ and $R$ are closed one dimensional submanifolds of $\CC$. (This is why we required that the zeroes of $f'$ be disjoint from $I \cup R$.)
From this perspective, we can see that it would be bad if one of the components of $R$, for example, stopped at a point, or spiralled infinitely into a point -- we need them to disconnect $\mathbb{C}$.
I have heard people say that fixing this proof comes down to proving the Jordan curve theorem in the form "If $\phi: \RR \to \CC$ and $\psi: \RR \to \CC$ are smooth maps which are interleaved at infinity, then they $\phi(\RR)$ and $\psi(\RR)$ cross. In fact, I claim the hard thing is to show that the unbounded components of $R$ may be parametrized by $\RR$ in the first place, and that each component has two ends!
First, let's see why things are easy if we assume such a parametrization exists. Let $\Gamma$ be a connected component of $R$ touching one of the unbounded spokes. Suppose we could show there is a parametrization $\phi: \RR \to \Gamma$. Note that the composite $f \circ \phi: \RR \to \RR$ has nowhere vanishing derivative, so it is monotone and without loss of generality we can assume it is increasing. If we know that $\Gamma$ is unbounded in both directions (which is what is implicitly assumed when you draw a picture of $R$ as a bunch of strands connecting the spokes at infinity), then there is no need to use the Jordan curve theorem -- just apply the implicit function theorem to $f \circ \phi$! We have $\lim_{t \to \pm \infty} f(\phi(t))=\pm \infty$, so somewhere in the middle $f(\phi(t))=0$ and we win.
So the real challenge is to show that $\Gamma$, a connected unbounded $1$-dimensional submanifold of $\CC$, can be parametrized by $\RR$ and goes to $\infty$ in both directions. I looked up various proofs of the classification of $1$-dimensional manifolds, but they all seemed a little messier than I wanted to do in class.
Then I came up with the idea of just trying to invert the map $f: \Gamma \to \RR$, which is how I came up with this argument.  I must admit, though, that the geometric origins are no longer visible.<|endoftext|>
TITLE: Tensor Product of Convex Sets?
QUESTION [15 upvotes]: I was wondering if such a concept was used anywhere. What I was thinking of is this. Consider two vectors spaces $V,W$ and convex sets $C_1 \subseteq V$ and $C_2 \subseteq W$ if we define $C_1 \otimes C_2 := \text{Convex Hull}(\{c_1 \otimes c_2 \in C_1 \otimes C_2 : c_1 \in C_1,c_2 \in C_2 \}) $. If $C_1$ and $C_2$ are bounded convex sets then so is $C_1 \otimes C_2$. Also, I believe that it is the case that if $a$ is a vertex of $C_1$ and $b$ is a vertex of $C_2$ then $a \otimes b$ is a vertex of $C_1 \otimes C_2$ and all vertices of $C_1 \otimes C_2$ can be constructed in this way.
Does anybody know of a construction like this actually being used anywhere? In particularly interested in the case where $C_1$ is given by as the solution to a list of inequalities $\alpha_i(v) \leq 1$, $\alpha_i \in V^*$, and $C_2$ is likewise given by a list $\beta_j(w) \leq 1$, $\beta_j \in W^*$. Then I can see that $C_1 \otimes C_2$ is contained in the convex set given by the inequalities $\alpha_i \otimes \beta_j \leq 1$. I've been try and failing at showing that $C_1 \otimes C_2$ is in fact exactly equal to the solution of these inequalities. Any advice that may be helpful would be appreciated.

REPLY [4 votes]: This is well-documented in the case where $C_1$, $C_2$ are unit balls for some norm, and in that case your $C_1 \otimes C_2$ is the unit ball for the so-called projective norm. The set you compare to in the last paragraph is the unit ball for the so-called injective norm, which indeed is a different norm. Googling these keywords should give many results.
Also, you may look at Chapter 4.1.4 in the book "G. Aubrun & S. Szarek, Alice and Bob Meet Banach: The Interface of Asymptotic Geometric Analysis and Quantum Information Theory", where we consider such projective tensor products for general convex sets.<|endoftext|>
TITLE: On the parity of $[x^n]$
QUESTION [7 upvotes]: I am trying to find a problem which appeared years ago in the American Mathematical Monthly.  It went something like this:  There was a Putnam Competition question which asked to show that there is a number $x$ with the property that $[x^n]$ has the same parity as $n$ for all positive integers $n$. The square brackets indicate the floor function.  Demonstrate such a number. 
Does anyone know when this problem and solution appeared?

REPLY [9 votes]: Google turns up this "Mock Putnam Exam" from the U[niversity] of I[llinois], whose unattributed second problem asks to show that $[(\sqrt2+1)^n]$ has parity opposite to that of $n$ for each $n=1,2,3,\ldots\,$.  Perhaps that's what you remember?  (It should also be possible to make any infinite binary sequence arise as  $\{[x^n] \bmod 2\}_{n=1}^\infty$ for some $x$ by starting from $x$ large enough and using the sequence of conditions to define a sequence of approximations converging to $x$.)
[added later: an explicit example of $[x^n] \equiv n \bmod 2$ is the largest root $3.214319743377535\!\ldots$ of the cubic $x^3 - 3x^2 - x + 1$.]<|endoftext|>
TITLE: Dimension of preprojective algebra of Dynkin type
QUESTION [5 upvotes]: Fix a field $\Bbbk$. Let $Q$ be a Dynkin quiver and let $\Pi(Q)$ be its preprojective algebra. It is well-known that in this case $\Pi(Q)$ is finite-dimensional, but I've been unable to find a reference for what the dimension actually is for each type. I'm also interested in knowing the dimension of $e_i \Pi(Q)$, where $e_i$ is the idempotent corresponding to vertex $i$ (i.e. the dimension of the space of all paths starting at vertex $i$).
I've already calculated by hand that the dimension of  $\Pi(\mathbb{A}_n)$ is given by ${n+2 \choose 3}$, and that the dimension of $e_i \Pi(\mathbb{A}_n)$ is given by the $i$th entry of the $n$th diagonal in the multiplication table for positive integers. However, the other Dynkin types seem to be more difficult.

REPLY [3 votes]: As a module over $kQ$, a finite-type preprojective algebra is a direct sum of each of the indecomposable $kQ$-modules once.  Thus, the total dimension is the sum over all positive roots of the height of the root (where the height is the sum of the coefficients in the simple root expansion).  If we write $\lambda_i$ for the number of roots of height $i$, the partition conjugate to $\lambda$ consists of the exponents of the root system.  (This fact was first observed empirically by Shapiro and Steinberg, then proved by Kostant.)
Thus, we can think of an exponent $e$ as accounting for $e$ roots, one of height 1, one of height 2, etc., up to one of height $i$.  Thus, each exponent $e$ contributes $1+2+\dots+e={e+1\choose 2}$ to the total dimension.  
In type $A_n$, the exponents are 1 to $n$, so the answer is $\sum_{e=1}^n {e+1 \choose 2} = {n+2\choose 3}$, as you found.<|endoftext|>
TITLE: Convex Hull of univalent functions and Bieberbach Conjecture
QUESTION [7 upvotes]: Consider the class $S$ of univalent functions on the unit disk $D$ normalized so that $f(0)=0$ and $f'(0)=1$. 
Each function in $S$ satisfy the Bieberbach conjecture, that is the $n$-th coefficient in the power series  expansion is less or equal than $n$.  I am looking at the convex hull of class $S$. Any function there satisfies the Bieberbach conjecture by an easy computation.
Are there functions which do not belong to the convex hull of $S$ but satisfy the Bieberbach conjecture? 
More generally, how do you prove that a function is not in the convex hull of $S$ without using the Bieberbach conjecture? There are characterizations of convex hulls of subfamilies of $S$, for example starlike maps, in terms of integral representations.

REPLY [4 votes]: By extremality, a function of the form $f(z)= z+\sum_{n\geq 2} n a_n z^n$ with $|a_n|=1$ belongs to the convex hull of $S$ if and only if it belongs to $S$. 
According to wikipedia, the only univalent such functions are those for which $a_n= \omega^{n-1}$ for some $|\omega|=1$, so you get many examples of functions satisfying Bieberbach's conjecture but not in the convex hull of $S$. For example, $f(z) = 2z - \frac{z}{(1-z)^2}$.<|endoftext|>
TITLE: The class of the diagonal in the symmetric product of a smooth curve
QUESTION [9 upvotes]: Let $C$ be a smooth curve of genus $g$, and let us consider its $d$-th symmetric product $\textrm{Sym}^d(C)$ and its Jacobian $J(C)$. Fixing a point $p_0 \in C,$ there are two maps $$u_d\colon C_d \to J(C), \quad u_d(D):=\mathcal{O}_C(D-dp_0),$$ $$i_{d-1} \colon \textrm{Sym}^{d-1}(C) \to \textrm{Sym}^d(C), \quad  i_{d-1}(D):=D+p_0.$$ Then the following two divisor classes in the Néron-Severi group of  $\textrm{Sym}^d(C)$ are independent on $p_0$: $$\theta:=u_d^*(\Theta), \quad x:=i_{d-1} (\textrm{Sym}^{d-1}(C)),$$
where $\Theta$ is the theta divisor in the Néron-Severi group of $J(C)$. It is well-known that, if $C$ is with general moduli, (i.e., it defines a very general point in the moduli space $\mathcal{M}_g$), then the following holds:

(1) the Néron-Severi group of $\textrm{Sym}^d(C)$ is generated by $\theta$ and $x$;
(2) the class $\delta$ of the diagonal of $\textrm{Sym}^d(C)$ is given by $$\delta = 2((d+g-1)x - \theta).$$

Now, it is not difficult to produce counterexamples (for instances, curves with non-trivial correspondences) showing that (1) does not hold if $C$ is not with general moduli. So let me ask the following

Question. Is (2) true for every smooth curve of genus $g$?

A. Kouvidakis, in Lemma 7 of his paper Divisors on symmetric product of curves (Trans. Amer. Math. Soc. 337), claims that (2) is due to MacDonald, and refers to Arbarello-Cornalba-Griffiths-Harris'book Geometry of Algebraic Curves I, Proposition 5.1 p. 358  for the proof.
Actually, the (much more general) result quoted in the book is apparently stated without assumptions on $C$, but the proof is based on a Lemma whose statement begins with "Let $C$ be a curve with general moduli", and whose content is essentially (1). So I am a bit confused.
My impression is that the answer to the question should be yes and that some kind of specialization argument should do the job. Any reference to the relevant literature will be particularly appreciated. I am primarily interested in the case $d=g=2$, but I would like to know the answer to the general question, too.

REPLY [5 votes]: Let me try to give a slightly different approach, which hopefully complements Jason's answer. Basically, let us compute as much as we can about the natural line bundles on the symmetric power.
First of all, the line bundle $O(-\Theta)$ on $\mathrm{Pic}^{g-1}(C)$ coincides with the line bundle $\det \mathrm{R\Gamma}$ (whose fiber over $\ell\in\mathrm{Pic}^{g-1}(C)$ is $\det\mathrm{R\Gamma}(C,\ell)$). I prefer to work with degree $g-1$ bundles on $C$ here, so that the $\Theta$ divisor is defined canonically, not just up to a shift. Your formula involves the pullback of $\Theta$ (or, equivalently, of this line bundle) under the map $$u_d:\mathrm{Sym}^dC\to\mathrm{Pic}^{g-1}(C):D\mapsto O(D+(-d+g-1)p_0),$$ but I would prefer to work with the map $$v_d:\mathrm{Sym}^dC\to\mathrm{Pic}^{g-1}(C):D\mapsto O(-D+(d+g-1)p_0),$$ which is algebraically equivalent to $-u_d$ (and $\Theta$ is invariant under the inversion anyway).
Now in these terms, we can state the following identity:
$$v_d^*(\det\mathrm{R\Gamma}^{\otimes 2})\simeq O(\Delta-2(d+g-1)H),$$ where following Jason's answer, $\Delta$ is the diagonal in $\mathrm{Sym}^d C$, and $H$ is the reduced divisor whose complement is $\mathrm{Sym}^d(C-{p_0})$. (Note that this is an isomorphism of line bundles, not just an algebraic equivalence.) Indeed, the left-hand side is the line bundle whose fiber over $D\in\mathrm{Sym}^dC$ is $$\det\mathrm{R\Gamma}(C,O(-D+(d+g-1)p_0))^{\otimes 2}\simeq\det(O/O(-D))^{\otimes-2}\otimes (O(-D))_{p_0}^{\otimes 2(d+g-1)}.$$ It now remains to notice that the line bundle whose fiber over $D$ is $\det(O/O(-D))^{\otimes 2}$ is isomorphic to $O(-\Delta)$, while the line bundle whose fiber is $O(-D)_{p_0}$ is isomorphic to $O(-H)$.<|endoftext|>
TITLE: Signs in Chevalley's commutator formula
QUESTION [8 upvotes]: I am trying to understand presentations of twisted groups of Lie type (specifically $^2D_5$) over finite fields using Steinberg's presentations (for instance from Gorenstein, Lyons and Solomon, Number 3, Chapter 2.4). Such presentations (for instance from Theorem 2.4.8 in the above book) typically involve the Chevalley commutator formula, which involves terms often of the form (in the most straightforward case)
$[x_{\tilde{\alpha}}(t),x_{\tilde{\beta}}(u)] = x_{\tilde{\alpha} + \tilde{\beta}}(\epsilon_{\alpha, \beta} t u)$
where $\epsilon_{\alpha,\beta} = \pm 1$ is a parameter dependent only on the roots $\alpha$ and $\beta$. This is the extent of the information I can find about $\epsilon_{\alpha, \beta}$. I suspect that I am not free to choose these completely arbitrarily, but I'm not sure on what restrictions I need to place on these choices; in other words, how to tell whether a given set of choices of values for $\epsilon_{\alpha,\beta}$ would result in a valid presentation.
If anyone has any information or references on how to understand these signs in more detail, I would be very grateful.

REPLY [5 votes]: Here are some supplementary comments, in community-wiki format:
For either the original Chevalley groups or the twisted variants, the concrete, detailed treatment in Roger Carter's 1972 book here is also a good resource: see Chapters 4-5 (especially 5.2) for the split groups and their complicated Chevalley commutation relations, as well as Chapter 13 for the twisted groups.  It's useful to note that there are two kinds of integer constants involved: those occurring with signs in a Chevalley basis of the simple Lie algebra over $\mathbb{C}$ and those coming from the root strings when two linearly independent roots are given.    Both are relatively simple for the simply-laced types, especially classical types such as $D_\ell$ where root vectors can be specified in natural ways.  But the twisted groups add some complications, and I haven't seen examples of rank as large as 5 treated explicitly.    (Though people have by now done a lot of computation.)
Any way it's approached, the basic idea is to start with a Chevalley basis and then work with exponential power series in characteristic 0; after that one reduces modulo a prime, etc.    Another version, dealing with more than the adjoint groups, is found in Steinberg's 1967-68 Yale
lectures (in the early sections).   These notes used to be available online, but since his death the homepage at UCLA has been taken down as noted elsewhere on MO.    But there are copies including LaTeX versions circulating, apart from libraries.
One other comment is that some previous questions on MO have focused on the sign choices in a Chevalley basis, and they too provide some theoretical references which may (or may not) be helpful.    For practical advice, you might try Frank Luebeck here.

REPLY [4 votes]: There are lots of ways to fix the signs of the structure constants, so I will just provide a bunch of references I use when needed:

Section 2 of A third look at weight diagrams gives an easy and unified treatment for the ADE-type groups;
Section 9 of Chevalley Groups Over Commutative Rings I. Elementary Calculations works out groups of all normal types in a bit harder form, but has some tables of signes inside;
Can one see the signs of structure constants? is mostly about the microweight representations for $\mathsf{E}_6$ and $\mathsf{E}_7$, but contains a lot of reference on the methods of determining the sighs.

Now all of the above is for the groups of normal types and you indicated you are interested in twisted groups such as ${}^2\mathsf{D}_\ell$. For those you can always fix the signs in the corresponding non-twisted groups and use the fact that the elementary generators of the twisted group are the products of the elementary generators over the orbits of roots — see the Steinberg lectures on the Chevalley groups (original mimeographic notes (PDF), TeXified version (PDF)) or Carters's Simple Groups of Lie Type.<|endoftext|>
TITLE: Étale morphisms in SDG
QUESTION [9 upvotes]: On page 70 here, Kock defines a formally étale morphism $f:M\rightarrow N$ as one for which the following square is a pullback for each $d:\mathbf 1\rightarrow D$
$$\require{AMScd} \begin{CD} M^D @>{f^D}>> N^D\\ @V{M^d}VV @VV{N^d}V\\ M^\mathbf{1} @>>{f^\mathbf{1}}> N^\mathbf{1}. \end{CD}$$
Concretely this means the tangent space $M^D=TM$ is isomorphic to set of pairs $(t,x)\in N^D\times M$ satisfying $t(d)=f(x)$. This is not the same as just taking tangents whose basepoint $t(0)=f(x)$, so how is this definition equivalent to the one saying the induced map on tangent spaces is an iso?

REPLY [4 votes]: You have omitted a crucial part of the sentence:

... in the sense of §16.

The point is small objects admit canonical basepoints via the augmentation of the Weil algebras defining them. See this MSE question for the definition of $\operatorname{Spec}_R(\pi)$ for the augmentation $\pi:W\rightarrow R$. The author's definition only asks for pullbacks involving these canonical base points.
For the dual numbers, this notion does pick zero as a basepoint, since dual numbers can be identified with $R^2$ with-dual-number-multiplication, making the projection of $\bar x=(0,x)$ clearly zero.<|endoftext|>
TITLE: Can two rational rotations $F_2 = \langle A, B \rangle \to SO(3)$ efficiently approximate the $3 \times 3$ identity matrix?
QUESTION [11 upvotes]: Let $A,B$ be two rational rotations:
$$ A = \left[\begin{array}{rcc} \frac{3}{5} & \frac{4}{5} & 0 \\
-\frac{4}{5} & \frac{3}{5} & 0 \\
0 & 0 & 1 \end{array}\right] 
\quad\text{ and }\quad
B = \left[\begin{array}{crc} 
1 & 0 & 0  \\ 
0 & \frac{3}{5} & \frac{4}{5}  \\
0 & -\frac{4}{5} & \frac{3}{5}   \end{array}\right]
$$
Then we build the free group $\mathbb{F}_2 = \langle A, B \rangle$ of all possible products of $A, B, A^{-1}, B^{-1}$.  Or I think since we chose $A,B$ to be rotations, we have a homomorphism
$$ \phi:  \mathbb{F}_2 \to SO(3)$$
just by evaluating the product.  The image $\langle A,B \rangle$ is obviously dense in $SO(3)$, but how much work is it to find products of $A$ and $B$ that are very close to the identity as a function of the word length $n$?
$SO(3)$ only has a few discrete subgroups.  Proving that $\overline{\langle A,B \rangle} = SO(3)$ seems to be a matter of the pigeonhole principle or maybe showing this group is dense Zariski topology.
Kaloshin and Rodnianski suggest this is always the case for pretty much any collecton of rotations in Diophantine Properties of $SO(3)$.  I am wondering if the argument simplifies for any specific choice of group elements.
On my computer this morning I was able to find:
$$ AB^{-1}A^{-3}B^{-1}A^2 =
\left[\begin{array}{rrr}
 0.98189568 & -0.12261376 &  0.144384  \\
        0.15538176 &  0.95731968 & -0.243712  \\
       -0.10833920 &  0.26173440 &  0.959040    \end{array}\right]$$
which is not that close to the identity matrix, but it's the best one I can find with 8 letters.
Maybe there's a "cheap trick"  that works here.  I would especially love that.  This seemed like such an easy problem, but all the solutions I found are quite hard.

REPLY [5 votes]: Have you tried b^{-34}*a^{34}
It gets pretty close...
  $b^{-34}*a^{34} = \left[ \begin{matrix} 0.9937  & 0.1119  &  0  \\ -0.1112 & 0.9875 & -0.1119 \\ -0.0125 & 0.1112 & 0.9937 \end{matrix} \right]$
in fact chaining b^-17*a^17*...*b^-17*a^17 an even number of times will get you closer while maintaining unit determinant...
$b^{-17}a^{17}b^{-17}a^{17}b^{-17}a^{17}b^{-17}a^{17}=$
$\left[ \begin{matrix} 1.0000  &  0.0002  & -0.0063  \\ -0.0002 &   1.0000   &-0.0002 \\    0.0063 &   0.0002   & 1.0000 \end{matrix} \right]$
Are these results product of rounding errors?
No!
The geometric meaning is that $17 \cdot \arccos(3/5) \approx 5\pi$, so $a^{17}$ is like rotating two and a half times around the $\left[ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix}\right]$ axis, and $a^{34}$ is 5 way round, so you obtain the identity (approximately).
You could chase down the decimals after $\arccos(3/5)$ and get so close to identity as you wish. And the same goes with b. The question about combinations of powers of multiples other than 17 is far more difficult.<|endoftext|>
TITLE: Factor matrix ${\bf A}$ into the product ${\bf B}{\bf C}$ where ${\bf C}$ has no negative entries and ${\bf B}$ has few non-zero entries
QUESTION [8 upvotes]: This is a more carefully worded version of this question, here tailored to professional mathematicians.
Consider a matrix ${\bf A}\in{\bf M}_{n\times n}({\mathbb R})$ with possibly positive, negative and zero-valued entries and ${\rm Det}[{\bf A}] \neq 0$.  

Is there an algorithm to write ${\bf A}$ as a product of two matrices
  ${\bf B}{\bf C}$ where ${\bf B} \in{\bf M}_{n\times n}({\mathbb R})$ and ${\bf C} \in{\bf M}_{n\times n}({\mathbb R}_\ge)$ in which ${\bf B}$ has the maximum number of $0$
  entries (i.e., is sparse) and all the entries of ${\bf C}$ are
  non-negative?  Again, the cost metric is the number of non-zero
  entries in ${\bf B}$.

Example
Suppose ${\bf A} = \left( { \ \ 1\ \ \ \ 2 \atop -6\ -8} \right)$.  
Here are three factorizations, ${\bf B}{\bf C}$, with their associated costs.

$\left( { \ \ 1\ \ \ \ \ 2 \atop -6\ -8} \right)\left( {1\ \ \ \ 0 \atop 0\ \ \ \ 1} \right)$,  Cost = $4$
$\left( {0\ \ \ \ \ 2 \atop 1\ \ \ \ -14} \right)\left( {1\ \ \ \ \ \ 6 \atop 1/2\ \ \ \ 1} \right)$,  Cost = $3$
$\left( {1\ \ \ \ \ 0 \atop 0\ \ \ \ -2} \right)\left( {1\ \ \ \ \ 2 \atop 3\ \ \ \ \ 4} \right)$,  Cost = $2$

I do not need an algorithm to find a unique decomposition, just a principled method for finding at least one having minimum cost.  
As far as I know, despite immense work on matrix factorization, this precise problem has never been solved.  (Polar decomposition, Cholesky decomposition, LUD decomposition, Gram-Schmidt orthogonalization and Sparse matrix approximation are not quite appropriate.)
Motivation
The general computational task is to perform the linear operation ${\bf A}{\bf x}$, where ${\bf A}$ has the conditions listed above and ${\bf x}$ is an $n$-dimensional real-valued vector of non-negative entries.  The overall computational task can be split into two linear systems.  The first system can perform ${\bf C}{\bf x}$ at extremely low computational cost (assume zero cost), but the entries of ${\bf C}$ must be non-negative.  The second system can perform ${\bf B}{\bf y}$ (where ${\bf y} = {\bf C}{\bf x}$) and the entries of ${\bf B}$ can be positive or negative or zero but there is a unit cost for each non-zero entry of ${\bf B}$.  
We seek to split the overall computation of ${\bf A}{\bf x}$ into the two systems to minimize the overall computational cost.

REPLY [8 votes]: Let $v_1, \dots, v_n$ denote the rows of $A$. 
Let $u$ be a vector with all positive entries that is not a linear combination of $v_2, \dots, v_n$. Then we may write $v_1 = c u + a_2 v_2 + \dots a_n v_n$ for some scalars $c, a_2, \dots, a_n$.
Now choose $\epsilon$ small enough that for all $j$ from $2$ to $n$, $u + \epsilon \sum_{i=2}^j a_i v_i$ has all positive entries.
Let $C$ be the matrix with rows $u + \epsilon \sum_{i=2}^j a_i v_i$ for $j$ from $1$ to $n$.  Then by construction $C$ has all positive entries.
Each vector $v_i$ can be written as a linear combination of two rows of $C$. For $v_i$ this is by subtracting two adjacent rows and dividing by $\epsilon$, and for $v_1$ it is $c-1/\epsilon$ times the first row plus $1/\epsilon$ times the last row.
We conclude that there is a sparse matrix $B$ with $2n$ entries such that $BC = A$.
This is optimal assuming each of the $v_i$ has both positive and negative entries, as the corresponding row of $B$ must have both positive and negative entries and hence at least $2$ entries.  If $A$ has any rows that are either non-negative or non-positive, then this algorithm might not be optimal, as the example in the original question shows.  However, this algorithm produces matrices where $BC$ is very sensitive to slight changes in $C$ and thus may be unsuitable for practical applications.<|endoftext|>
TITLE: Does $\mathbb C\mathbb P^\infty$ have a group structure?
QUESTION [42 upvotes]: Does $\mathbb C\mathbb P^\infty$ have a (commutative) group structure? More specifically, is it homeomorphic to $FS^2$, (the connected component of) the free commutative group on $S^2$?
$\mathbb C\mathbb P^\infty$ is well known to have the homotopy type of the classifying space of a commutative group, $\mathbb C^\times$. One model for the classifying space is the bar construction, which is a functor from groups to spaces and is product preserving, so the classifying space of a commutative group is again a commutative group (since the group operation is a homomorphism iff the group is a commutative). The Dold-Thom theorem says that the component of the free commutative group on $S^n$ is a $K(\mathbb Z,n)$. So those are two fairly nice group models. But is either actually homeomorphic to the space we started with, $\mathbb C\mathbb P^\infty$?
By $\mathbb C\mathbb P^\infty$ I mean the colimit of the $\mathbb C\mathbb P^n$ under the closed inclusions. Similarly, I topologize the free commutative group on a space $X$ as the colimit of the inclusions of the subset of words of length $n$, which is topologized as a quotient of $(X\times\{\pm1\}\cup\{*\})^n$, that is, sequences of $n$ letters, each of which may be a generator, the inverse of a generator, or the identity element. So $\mathbb C\mathbb P^\infty$, $BS^1$, and $FS^2$ are CW complexes with finite skeleta. Are they all locally modeled on $\mathbb R^\infty$? Do such infinite dimensional manifolds have homeomorphism types determined by their homotopy types?
Appendix: Motivation
In the question, I invoked the Dold-Thom theorem to motivate the particular model $FS^2$ under consideration. But this case is easy and should motivate the general theorem, rather than vice versa. So I will suggest a proof. Moreover, I will motivate the choice of space, how we get from $\mathbb C\mathbb P^\infty$ to $FS^2=F\mathbb C\mathbb P^1$. We want to make a model for $B\mathbb C^\times$ that is as nice as possible. If we identify $\mathbb C^\infty=\mathbb C[x]$, it is a ring, so its multiplication makes $\mathbb C\mathbb P^\infty=\mathbb P(\mathbb C[x])$ into an topological monoid. That’s a pretty nice structure; it just lacks inverses. So we should consider $\mathbb P(\mathbb C(x))$. For any reasonable topology on $\mathbb C(x)$, the vector space is contractible, as is the complement of the origin; and the multiplication on the ring is continuous. Thus $\mathbb P(\mathbb C(x))$ is a topological monoid homotopy equivalent to $\mathbb C\mathbb P^\infty$. This is better than the previous model because every element has an inverse. However, the inversion map is continuous only for some topologies. For counterexample, the finest possible topology is given by making the topological vector space the colimit of its finite dimensional subspaces (a colimit indexed by an uncountable filtering poset). But then the inverse is not continuous: the copy of $\mathbb C$ given by $y\mapsto x-y$ has inverses a set of linearly independent vectors, thus discrete. 
Turning away from topological vector spaces, how else can we put a topology on $P(\mathbb C(x))$? A polynomial is determined (up to scale) by its zeros and a rational function by its zeros and poles. Then we can think of $\mathbb P(\mathbb C[x])$ as something like the free commutative monoid on $S^2=\mathbb C\mathbb P^1$ (specifically, the free monoid modulo the point at infinity). Similarly, $\mathbb P(\mathbb C(x))$ is the component of the free commutative group on $\mathbb C\mathbb P^1$, that is, $F\mathbb C\mathbb P^1$. To put the zeros and poles on the same footing, we need the topology described from the beginning.

REPLY [19 votes]: I noticed that this question still has no accepted answer and all existing answers are rather long. It seems that the answer can be easily obtained using some results of infinite-dimensional topology, namely, the theory of manifolds modeled on the direct limit $\mathbb R^\infty$ of Euclidean spaces (see Chapter 5 of Sakai's book).
Two results of this theory will be important for our purposes:
Characterization Theorem 5.4.1. A topological space $X$ is is homeomorphic to an open subspace of $\mathbb R^\infty$ if and only if any embedding $f:B\to X$ of a closed subspace $B$ of a finite-dimensional compact metrizable space $A$ can be extended to an embedding $\bar f:A\to X$. 
Classification Theorem 5.5.1. Two $\mathbb R^\infty$-manifolds are homeomorphic if and only if they are homotopically equivalent.
Now using the Characterization Theorem, it can be shown that $\mathbb{CP^\infty}$ is an $\mathbb R^\infty$-manifold. 
Next, in his question Ben Wieland writes that $\mathbb{CP}^\infty$ is homotopically equivalent to the connected component $G_0$ of the free topological Abelian group $G$ over the sphere. Using the Characterization Theorem of Sakai once more, one can show (and this was done by Zarichnyi in 1982) that $G_0$ is an $\mathbb R^\infty$-manifold. Since $\mathbb{CP}^\infty$ and $G_0$ are two homotopically equivalent $\mathbb R^\infty$-manifolds, the Classification Theorem ensures that $\mathbb{CP}^\infty$ is homeomorphic to $G_0$ and hence has a compatible topogical group structure.<|endoftext|>
TITLE: Problems in some parts of Monique Hakim thesis?
QUESTION [6 upvotes]: In "Reminiscences of Grothendieck and his school" Luc Illusie says:
"I heard from Deligne that there were problems in some parts. (of Monique Hakim thesis).
Topos annelés et schémas relatifs, Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 64, Springer, Berlin, New York (1972).
My question is for a reference for these "problems". What are the problems? Where?

REPLY [2 votes]: I have no idea if it's what Illusie and Deligne are referring to, but W. D. Gillam points out some strange behavior of Hakim's Spec functor in Remark 11 of this paper. Gillam says that "there is no meaningful situation in which Hakim's Spec functor agrees with the 'usual' one," and that "if $X$ is a locally ringed space at least one of whose local rings has positive Krull dimension, Hakim's sequence of spectra yields an infinite strictly descending sequence of [morphisms of ringed spaces] $\cdots\to\text{Spec}(\text{Spec}_{\ \!}X)\to\text{Spec}_{\ \!}X\to X$."<|endoftext|>
TITLE: Shortest path through $n^{1/3}$ points out of $n$
QUESTION [14 upvotes]: Say I sample $n$ points uniformly at random in the unit cube in $\mathbb{R}^3$, and then I look for the shortest path through $n^{1/3}$ of those points (rounding up, say). What happens to the length of this path as $n\to\infty$? Does it increase, decrease, or converge (or "none of the above")?
This question is the three-dimensional version of the following earlier question:
Shortest path through $\sqrt{n}$ points out of $n$
I have tried to apply the lower bounds proposed in that question, but it does not seem to me that they scale in the necessary way when we go from two to three dimensions.

REPLY [3 votes]: The lower bound argument I gave for $\sqrt{n}$ points in a square works here, too. I have tried to simplify it. The idea is to use the union bound: The probability that a random path with $m=\lfloor \sqrt[3]{n} \rfloor$ steps has length less than some constant is small. It is so small that the expected number of ways to choose a short path is less than $1$, so the probability is less than $1$, in fact with probability going to $1$ there is no such short path using $m$ vertices out of $n$. 
It helps to suppose the points are chosen from a fine lattice $\{1/\ell,2/\ell,...\}^3$ with $\ell \gg m$, and to estimate the probability that the sum of the $L^1$ distances along the path is small rather than the $L^2$ distance. Pushing the points to be on a fine lattice does not change the length much, less than $2\sqrt{3}m/\ell$.
The count of sequences of $m$ nonnegative steps of total $L^1$ length up to $d$ is the number of ways of distributing $d$ objects among $3m+1$ categories, $d+3m \choose 3m$. There are at most $2^{3m}$ choices of signs. So, the probability that a random path with $m$ steps has $L^1$ length at most $c \ell$ is at most $2^{3m}{c \ell +3m \choose 3m} /\ell^{3m} \le \frac{2^{3m}(c \ell + 3m)^{3m}}{(3m)!\ell^{3m}} = \frac{(2c+6m/\ell)^{3m}}{(3m)!}$. If $c \ell \gt 3m$ we can estimate this as less than $\frac{(4c)^{3m}}{(3m)!}$. (Actually, we don't need to accept this factor of $2$.)  Using $x! \gt (x/3)^x$, this is less than $\left(\frac{4c}{m}\right)^{3m}$.
The number of ways to choose an $m$ step path from $n$ points is at most $n\times(n-1)\times(n-2)...\times(n-m) \le n^{m+1} \approx m^{3m+3}.$
The expected number of paths with $m$ steps of $L^1$ length less than $c$ is at most $m^{3m+3}\left(\frac{4c}{m} \right)^{3m} =m^3 (4c)^{3m}$. So, if we choose $c \lt 1/4$ then as $m,n \to \infty$, the probability that there is a path with $L^1$ length smaller than $c$ goes to $0$. 
The Euclidean length is up to $\sqrt{3}$ times smaller than the $L^1$ length, so the probability that there is a path through $\sqrt[3]{n}$ points with Euclidean length less than $1/(4\sqrt{3})$ goes to $0$ as $n \to \infty$. 
The constant can be improved easily by a factor of $2$, and with a little work one can estimate the probability that the $L^2$ distance is small directly instead of the $L^1$ distance, and this should improve the constant significantly, too. The convenient estimate $x! \gt (x/3)^x$ can be improved to a lower bound of roughly $(x/e)^x$ from Stirling's formula which improves the constant a bit more.<|endoftext|>
TITLE: When does the generalized Cantor space embed in a $\kappa$-compact space
QUESTION [6 upvotes]: The generalized Cantor space is the space $2^\kappa$, with basic open sets
$$
[\sigma] := \{f\in 2^\kappa : \sigma\subseteq f\},
$$
for $\sigma\in 2^{<\kappa}$.
A space is $\kappa$-compact if every open cover has a subcover of cardinality (strictly) smaller than $\kappa$.
Problem. For which cardinals $\kappa$ does the generalized Cantor space $2^\kappa$ 
embed as a subspace of every $\kappa$-compact set $C\subseteq 2^\kappa$ with $|C|>\kappa$?
It is a classic result that this holds for $\kappa=\omega$. 
In the original formulation of this problem, I mentioned that,
in light of an earlier answer, the case where $\kappa$ is weakly compact
is particularly interesting, and the expected answer is "for all of these".
According to Yair Hayut's answer below, this is wrong. The full problem,
as stated above, remains open (we need cardinals for which the
answer is positive).

REPLY [7 votes]: Let $\kappa$ be an uncountable regular cardinal that is not weakly compact and let $C$ be a $\kappa$-compact subset of ${}^\kappa 2$. Assume, towards a contradiction, that there is a continuous injection of ${}^\kappa 2$ into $C$. Since $C$ is closed in ${}^\kappa\kappa$, Lemma 2.9 of this paper shows that $C$ contains a closed set homeomorphic to ${}^\kappa 2$. By our assumption, $\kappa$ is not weakly compact and a theorem of Hung and Negrepontis (see Corollary 2.3 of the above paper) implies that the spaces ${}^\kappa 2$ and ${}^\kappa\kappa$ are homeomorphic. This is a contradiction, because an easy argument shows that $\kappa$-compact subsets of ${}^\kappa\kappa$ do not contain closed subsets homeomorphic to ${}^\kappa\kappa$ (see Fact 1.5 of the above paper). 
Let $\kappa$ be indestructible supercompact, let $\lambda>\kappa$ be inaccessible and let $G$ be $Col(\kappa,{<}\lambda)$-generic over $V$. 
In this situation, standard arguments (see, for example, Lemma 7.6 of this paper) show that, in $V[G]$, all subtrees of ${}^{{<}\kappa}\kappa$ with more than $\kappa$-many $\kappa$-branches contain a subtree isomorphic to ${}^{{<}\kappa}2$. In particular, in $V[G]$, $\kappa$ is weakly compact and every $\kappa$-compact subset of ${}^\kappa 2$ of cardinality greater than $\kappa$ contains a continuous injective image of ${}^\kappa 2$. 
Let $\kappa$ be a weakly compact cardinal with the property that there is a subset $A\subseteq\kappa$ such that $\kappa^+=(\kappa^+)^{L[A]}$ and the set $\{\alpha\in S^\kappa_\omega \mid cof^{L[A\cap\alpha]}(\alpha)=\omega\}$ is stationary in $\kappa$. In this situation, a modification of a classical argument of Solovay shows that there is a weak $\kappa$-Kurepa tree, i.e. a subtree $T$ of ${}^{{<}\kappa}2$ with $\kappa^+$-many $\kappa$-branches and $\vert T(\alpha)\vert=\vert\alpha\vert$ for stationary many $\alpha<\kappa$. Since $\kappa$ is weakly compact, the tree $T$ contains no $\kappa$-Aronszajn subtrees and a theorem of Juhász and Weiss (see Lemma 2.2 of the first paper) shows that the corresponding closed subset $[T]$ of ${}^\kappa 2$ is $\kappa$-compact. By results of Mekler and Väänänen, there is no continuous embedding of ${}^\kappa 2$ into $[T]$. 
The existence of weak $\kappa$-Kurepa trees at every inaccessible cardinal $\kappa$ is consistent with the existence of very large large cardinals (including supercompact cardinals). This is discussed on page 33 of this paper by S. Friedman, Hyttinen and Kulikov.<|endoftext|>
TITLE: References for higher descriptive set theory surveys
QUESTION [6 upvotes]: A student of Adi Jarden and mine attempts at generalizing results on selection principles from the Baire space $\omega^\omega$ to the higher Baire space $\kappa^\kappa$ ($\kappa$ uncountable), and similarly for the higher Cantor space $2^\kappa$, with the initial segment topology, as defined here and here. This leads (see linked questions) to many elementary questions whose answers are likely to be known.
We would appreciate references to surveys of this type of "higher descriptive set theory" and its basic results. In particular, it would be nice if these deal with results like Cantor-Bendixon Theorem, and combinatorial cardinals of the (higher) continuum.

REPLY [5 votes]: I would like suggest the following three course notes given by Sy Friedman:

1) Invariant descriptive set theory (on classical and generalised Baire space),
2) Cardinal characteristics of the uncountable,
3) The higher descriptive set theory of isomorphism.

Also the monograph

Generalised descriptive set theory and classification theory

by Sy Friedman, Tapani Hyttinen and Vadim Kulikov must be very useful, in particular its introduction gives a survey of previously works on the subject.
All these papers are available at Friedman's home page.<|endoftext|>
TITLE: Bott Chern cohomology via currents
QUESTION [11 upvotes]: Let $X$ be a compact complex manifold. Is the space of $(p,p)$ $d$-closed currents modulo $\partial\bar{\partial}$-exact ones naturally isomorphic to the Bott-Chern cohomology (made in the same way with smooth forms)? I can prove this statement for $p=1$ or for any $p$ on Kähler manifolds (via the $\partial\bar{\partial}$-lemma for currents), but I can not prove it for $p > 1$ on a general manifold. Does anyone know how to prove it?

REPLY [10 votes]: When $X$ is a compact complex manifold, its Bott-Chern cohomology groups can be computed either by smooth forms or by currents. The proof of this fact can be found for instance in Demailly's book (link here), page 326, considerations after the proof of Lemma 12.2. 
Demailly derives there this result from hypercohomological considerations.

REPLY [5 votes]: This is a theorem which was first proved by Bruno Bigolin in this paper:
Osservazioni sulla coomologia del $\partial \overline{\partial}$, Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Sér. 3, 24 no. 3 (1970), 571-583,
stated there as Proposition 2.2. The argument is pretty much the same as the one in Demailly's book that diverietti mentions.<|endoftext|>
TITLE: Finding an "optimal" quotient in a free group
QUESTION [6 upvotes]: Consider the abelian free group $G = \mathbf{Z}^n$ of rank $n$ and a finite subset $A \subset G \setminus \{0\}$. Since $G$ is residually finite, there is a subgroup $H \subset G$ such that $A \cap H = \emptyset$. Call such a subgroup optimal if it has minimal index in $G$.
Is there an efficient algorithm which computes an optimal subgroup $H$? By efficient I mean polynomial complexity in $n$ and $\mathrm{Card}\,A$.
What happens in the nonabelian case ($G$ is a nonabelian free group)? I think it should be more difficult. Is there at least an upper bound for the index of a normal subgroup $H$ such that $A \cap H = \emptyset$?

REPLY [2 votes]: For abelian groups, you need to find a number $k$ with the property that at least one entry of each $a \in A$ is not divisible by $k$. In the simplest case, when $n=1$ and $A=\{m\}$ a prime of size roughly $\log|m|$ can be found -- this is a consequence of the prime number theorem. If $A=\{1,\dots,m\}$, something like $m+1$ is best possible, so it depends on the structure of the set $A$.
In general, the optimal $k$ (for a worst case $A$) will be of size 
$$\log({\rm lcm}(\{{\rm gcd}(a_1,\dots,a_n) \mid (a_1,\dots,a_n) \in A \})).$$
In the non-abelian case, the case $A=\{w\}$ has been studied and it is known that sometimes a quotient of size at least $n \log(n)^2$ is needed to detect $w$ of length $n$. This can be found in http://arxiv.org/abs/1508.07730. If you want to detect $A=\{w_1,w_2\}$, then this reduces to detecting $[w_1,w_2]$ (or some variation of this in case the commutator is trivial in the free group), so that one reduce (iterating this idea) the general case to the case of a single word. 
An upper bound (in the non-abelian case) is $n^3$, since $SL(2,p)$ has size roughly $p^3$ and does not satisfy a law of length shorter than $p$. I believe that something like $n \log(n)^2$ should also be an upper bound, but this is an open problem.<|endoftext|>
TITLE: Do non-stable Banach spaces exist?
QUESTION [11 upvotes]: Let $K$ be $\mathbb{R}$ or $\mathbb{C}$. A Banach space $X$ over $K$ is stable if $X\cong X\times K$. I encountered the following question in some papers in the sixties:

Is every infinite dimensional Banach space stable?

Is this question still open?

REPLY [13 votes]: No, see the following paper of Gowers
https://blms.oxfordjournals.org/content/26/6/523.full.pdf

REPLY [12 votes]: This is the famous Banach's hyperplane problem that was solved in the negative by W. T. Gowers. There is a whole industry in Banach space theory concerning spaces which have even more peculiar properties (google for hereditrarily indecomposable Banach spaces).<|endoftext|>
TITLE: Hopfian modules
QUESTION [7 upvotes]: My question is slightly motivated by basic results in linear algebra, for example, that if $F$ is a field then a surjective linear map $F^n \rightarrow F^n$ is injective. More generally, any surjective endomorphism of a finitely generated module for a Noetherian ring is injective.
Let $R$ be a unital ring, not necessarily commutative. An $R$-module $M$ is Hopfian if any surjective $R$-module endomorphism $f : M \rightarrow M$ is injective. 

Is there a ring $R$ and an $R$-module $M$ such that $M$ is Hopfian but $M \oplus M$ is not?

As a follow-up, in the event the answer is `yes', is there an example where $M=R$?

REPLY [7 votes]: Let $R$ be Shepherdson's ring, that is, a domain for which the two by two matrix ring, $M_2 R$, contains
elements $x$ and $y$ such that $xy = 1$, but $yx \neq 1$. With right module $M =
R$, every homomorphism $M \to M$ is given by left multiplication by an
element of $R$, so every nonzero endomorphism of $M$ is one to one (as $R$
has no zero divisors).
On the other hand, $x$ is an endomorphism of $ M \oplus M$ that is
clearly onto, but is not one to one, as $x(1-yx) = 0$.
To clarify, a finitely generated module free on $n$ generators is Hopfian iff the $n \times n$ matrix ring is directly finite (latter means one-sided inverses are two-sided). The argument is simple: if $x$ is onto, then the map splits.<|endoftext|>
TITLE: Relative version of Quillen's theorem A
QUESTION [8 upvotes]: Quillen's Theorem A is formulated as follows:
Let $F:X\to Y$ be a functor between small categories. Suppose for each $y\in Y$ the category
$F/y$ is contractible. Then $F$ induces a weak equivalence between the nerves $N(X)\to N(Y)$.
I am not a topologist, but it seems can prove the following statement, which I believe is much more powerful:
Relative Theorem A.
Let $F:X\to Y$ and $G:Y\to Z$ be functors between small categories. Suppose for each $z\in Z$ the induced functor $(G\circ F)/z\to G/z$ induces a weak equivalence on the nerves. Then $F$ induces a weak equivalence on the nerves $N(X)\to N(Y)$.
When $G$ is the identity functor we recover the usual Theorem A.
This seems to be so basic that it must be in textbooks. Is it well-known?

REPLY [10 votes]: The condition that appears in the assumption of what you call "Relative Theorem A" was introduced by Grothendieck in Pursuing Stacks. It is a part of the definition of a basic localizer, i.e. a class of functors between small categories that behaves like the class of weak homotopy equivalences.
Grothendieck posed a conjecture that weak homotopy equivalences form the smallest basic localizer which was eventually proven by Cisinski here. In particular, Théorème 2.1.13 shows (already a well-known) fact that weak homotopy equivalences indeed form a basic localizer, i.e. that the "Relative Theorem A" holds.<|endoftext|>
TITLE: Special topics to include in course in algebraic number theory
QUESTION [17 upvotes]: I'll be teaching an introductory course in algebraic number theory this fall (stopping before class field theory).  I'm looking for a good list of "special topics" I can include to illustrate the general theory.  In other words, attractive theorems (preferably off the beaten path) that can be proved using the basic results and that illustrate their power.  For example, one standard one is the proof of quadratic reciprocity using cyclotomic fields.  Can people suggest other ones?  Especially ones that connect to other branches of mathematics (e.g. combinatorics, geometry/topology, group theory, etc)?

REPLY [7 votes]: Here's something within algebraic number theory, but is sort of a "special topic" in the sense that it's not treated in most algebraic number theory courses, but could easily be.  It also has some combinatorial aspects. Everyone knows that the class group of a number field $K$ measures the failure of unique factorization into irreducibles in its ring of integers.  
Question: what does this really mean?  That is, in what way does it quantitatively measure this?  
A sample result is Carlitz's theorem, which says all irreducible factorizations of a given element have the same length if and only if the class number is at most 2.  You can actually count the number of factorizations of $x$ based on the prime ideal factorization of $(x)$ in $\mathcal O_K$.  I discussed this in my course a few years ago, based on this note.  The note make more clear connections with combinatorics.  Narkiewicz's book also discusses this problem from a different perspective.<|endoftext|>
TITLE: Cartan subspaces for general algebraic representations
QUESTION [5 upvotes]: So I feel like asking the following likely open-ended question: What good generalizations of the notion of Cartan subspace do we have?
To be precise, let $G\curvearrowright V$ be an algebraic representation over a field $k$ satisfying some conditions you like (e.g. $\mathrm{char}(k)=0$, $G$ is connected reductive). When can we call a subspace $S\subset V$ a Cartan subspace with a good reason? (hopefully generalizing the notion of Vinberg as duplicated below, at least that of Cartan subalgebras for adjoint representations for sure)
My motivation is that I am trying to generalize some work of myself about affine Springer fibers and possibly a bit Springer theory. I end up deciding that the notion of Cartan subspace is essential.
-
When $\mathrm{char}(k)=0$, $H$ is a reductive group, $\theta$ is an automorphism on $H$ of finite order $m$, $\zeta_m\in k$ is a primitive $m$-th root of unity, $G:=(H^{\theta})^o$, $V:=\mathfrak{h}^{\theta=\zeta_m}$ and $G\curvearrowright V$ comes from the original adjoint representation $H\curvearrowright\mathfrak{h}$, Vinberg defined a Cartan subspace to be a maximal subspace consisting of commuting semisimple elements in $V$. Here commuting means commuting in $\mathfrak{h}$, and semisimple means its $G$-orbit is closed, or equivalently it is semisimple in $\mathfrak{h}$. Vinberg proved that all Cartan subspaces are conjugate, generalizing previous results for symmetric spaces when $m=2$.

REPLY [4 votes]: Assume $\text{char}\,k=0$ for simplicity. Then there are two further generalizations of Cartan subspaces which come to my mind: 

A linear subspace $S\subseteq V$ such that the restricted quotient morphism $S\to V/\!\!/G$ is finite and surjective. Such a thing exists if and only if the quotient morphism $\pi:V\to V/\!\!/G$ is equidimensional or, equivalently, the nullcone $\pi^{-1}(0)$ has the minimal possible dimension, namely $\dim V-\dim V/\!\!/G$. There are classifications for $G$ simple (Schwarz) or $V$ irreducible (Littelmann).
A generalization of Chevalley's restriction theorem due to Luna-Richardson. For that let $H\subseteq G$ be the principal stabilizer, i.e., $H$ is conjugate to the stabilizer of a generic closed orbit. Let $S:=V^H$ be the fixed point set. Then $S/\!\!/N_G(H)\to V/\!\!/G$ is an isomorphism. For the adjoint representation, $H$ is a maximal torus, $S$ is a Cartan subspace and the action of $N_G(H)$ on $S$ factors through the Weyl group. So one gets exactly Chevalley's theorem.

There is also a nice discussion of so-called sections in a survey on Invariant Theory by Popov-Vinberg (Encyclopedia of Math. Sci. Vol 55).<|endoftext|>
TITLE: Examples of math hoaxes/interesting jokes published on April Fool's day?
QUESTION [111 upvotes]: What are examples of math hoaxes/interesting jokes published on April Fool's day?
For a start P=NP.
Added 2022-04-01 Anything new in 2022?

REPLY [4 votes]: A genuinely natural information measure [arXiv:2103.16662]

The theoretical measuring of information was famously initiated by
Shannon in his mathematical theory of communication, in which he
proposed a now widely used quantity, the entropy, measured in bits.
Yet, in the same paper, Shannon also chose to measure the information
in continuous systems in nats, which differ from bits by the use of
the natural rather than the binary logarithm.      We point out that
there is nothing natural about the choice of logarithm basis, rather
it is arbitrary. We remedy this problematic state of affairs by
proposing a genuinely natural measure of information, which we dub
gnats. We show that gnats have many advantages in information theory,
and propose to adopt the underlying methodology throughout science,
arts and everyday life.<|endoftext|>
TITLE: Source request for $H^*(B\mathrm{TOP},\mathbb{Q}) \cong H^*(BO,\mathbb{Q})$
QUESTION [7 upvotes]: Let $B\mathrm{TOP}$ denote the classifying space for microbundles, i.e. $B\operatorname{Homeo}(\mathbb{R}^n,0)$. Now we get a map from $BO$ to $B\mathrm{TOP}$ via the inclusion. Let $f$ denote the corresponding map in the rational cohomology rings. 
Andrew Ranicki claims in his paper "On the construction and topological invariance
of the Pontryagin classes" that the surjectivity of $f$ is equivalent to the topological invariance of the Pontryagin numbers, which is fine. But then he also writes without a source, that it is now known, that these two rings are actually isomorphic.
So on the one hand I would like to get a source for this "fact" on the other hand I would like to know if there is some nice meaning behind the injectivity of this map, like for surjectivity. I mean there is the obvious translation, that whenever to rational characteristic classes for micro bundles agree on vector bundles then they have to be the same, but maybe there is a stronger statement.
Link to the paper: http://www.maths.ed.ac.uk/~aar/papers/invtop.pdf p.311

REPLY [9 votes]: The canonical map $BPL\rightarrow BTOP$ is a rational homotopy equivalence by works of Thom, Novikov, Kirby-Siebenmann and others. In fact the homotopy fiber $TOP/PL$ is a $K(\mathbb{Z}/2,3)$. A nice reference is the survey "Piecewise Linear Structures on Topological Manifolds" by Rudyak.
The canonical map $BO\rightarrow BPL$ is also a rational homotopy equivalence, this follows from smoothing theory. And the $i$-th homotopy groups of $PL/O$ are equal to zero when $i\leq 5$ isomorphic to $\Theta_i$ when $i\geq 6$. Where $\Theta_i$ is the group of exotic spheres:  the  equivalence  classes  of  smoothing  on
$S^i$ under  orientation-preserving diffeomorphism. As $\Theta_i$ are finite abelian groups you get your answer. 

3.The situation is completely different in the unstable range. A nice survey is "Dalian notes on rational Pontryagin classes" by Michael Weiss.
Refs: 

The Hauptvermutung book (Kluwer 1996). 
M. Hirsch and B. Mazur "Smoothings of piecewise linear manifolds", Annals of Mathematics Studies 80. (Princeton University Press, Princeton, 1974).<|endoftext|>
TITLE: Is $MGL$ an $H\mathbb{Z}$-algebra?
QUESTION [7 upvotes]: Let $\mathrm{MGL}$ be the $\mathbb{P}^1$-ring spectrum over a field $k$ representing algebraic cobordism. Suppose, for simplicity, that $k$ is of characteristic 0. Let $H\mathbb{Z}$ be the motivic Eilenberg-Maclane spectrum. My question is:

Can $\mathrm{MGL}$ be given the structure of an $H\mathbb{Z}$-module?

One knows that $$ \mathrm{MGL}\wedge H\mathbb{Z}=H\mathbb{Z}[b_1,b_2,...]$$ where $b_i$ is of bidegree $(2i,i)$ in $\mathrm{MGL}_{*,*}(\mathrm{Spec}\,k)$.
Thus, in view of the Morel-Hopkins isomorphism
$$ \mathrm{MGL}/(b_1,b_2,...)\simeq H\mathbb{Z}$$
shown by Hoyois, is there a map $H\mathbb{Z}[b_1,b_2,...]\to\mathrm{MGL}$ in $SH(k)$. It seems to me that such a map would be an analog of  Levine-Morel's generalized degree formula for $\mathrm{MGL}$.

REPLY [9 votes]: $MGL$ does not admit a structure of $H\mathbb Z$-module. There are many ways to prove this. As Sean said in the comments, if it were true over $\mathbb C$, topological realization would imply that $MU$ is an $H\mathbb Z^{top}$-module, which is false. By rigidity this takes care of all characteristic zero fields. The same argument can presumably be made in positive characteristic using etale realization over a separable closure, but that would require computing the $l$-completed etale homotopy type of Thom spaces over Grassmannians which could be tricky.
Another argument is the following. If $MGL$ were an $H\mathbb Z$-algebra, $KGL$ would be too. Taking mapping spectra in $SH(k)$, this would imply that the $K$-theory spectrum $K(k)$ is an $H\mathbb Z^{top}$-module, which is false, as the first $k$-invariant of $K(k)$ is nontrivial (if the characteristic of $k$ is not $2$).
Here's an expansion of my comment about the $b_i$'s.
Let's use $x_i$ for the elements in the Hopkins-Morel isomorphism. These are homogeneous generators of the Lazard ring $L$, and there is no canonical choice for them. The Hurewicz map $L \to \mathbb Z[b_1,b_2,...]$ is injective but not surjective, though it becomes surjective after tensoring with $\mathbb Q$.
A complete formula for $x_i$ in terms of the $b_i$ depends of course on a specific choice of the generators $x_i$. You can see one such choice and the resulting formula in Hazewinkel Constructing formal groups II, equation (7.5.1), where $U_n$ is $x_{n-1}$ and $m_n(U)$ has degree $n-1$ and is determined by $\sum_{n\geq 1}m_n(U)(\sum_{i\geq 0}b_i)^n=1$.
Usually one only cares about the image of $x_i$ modulo decomposable, which is $\pm pb_i$ if $i+1$ is a power of a prime $p$ and $\pm b_i$ otherwise (with Hazewinkel's choice of generators the signs are minus signs).<|endoftext|>
TITLE: Tricky two-dimensional recurrence relation
QUESTION [6 upvotes]: I would like to obtain a closed form for the recurrence relation
$$a_{0,0} = 1,~~~~a_{0,m+1} = 0\\a_{n+1,0} = 2 + \frac 1 2 \cdot(a_{n,0} + a_{n,1})\\a_{n+1,m+1} = \frac 1 2 \cdot (a_{n,m} + a_{n,m+2}).$$
Even obtaining a generating function for that seems challenging. Is there a closed form for the recurrence relation or at least for the generating function? Alternatively, is there a closed form for $a_{n,0}$?

REPLY [5 votes]: Set $b_{n,m}=2^na_{n,m}$ if $m\geq 0$, and set $b_{n,m}=b_{n,-m-1}$ if $m<0$. Then the relation simplify to
$$
  b_{0,m}=f(m), \quad b_{n+1,m}=b_{n,m-1}+b_{n,m+1}+2^{n+2}f(m),
$$
where $f(0)=f(-1)=1$, $f(m)=0$ otherwise.
Now define $c_{n,m}$ in the same way, but without a constant term:
$$
  c_{0,m}=f(m), \quad c_{n+1,m}=c_{n,m-1}+c_{n,m+1}.
$$
Then it is clear that $c_{n,m}={n\choose \lfloor \frac{n-m}2\rfloor}$, where ${a\choose b}=0$ if $b\notin[0,a]$. On the other hand, the constant terms in the relations for $b_{n,m}$ merely add some more multiples of the $c_{i,m}$, so that
$$
  b_{n,m}=c_{n,m}+\sum_{0<k\leq n} 2^{k+1}c_{n-k,m}=-c_{n,m}+\sum_{0\leq k\leq n} 2^{k+1}c_{n-k,m}\\
  =-{n\choose \lfloor \frac{n-m}2\rfloor}+2\sum_{0\leq k\leq n-m} 2^k{n-k\choose \lfloor \frac{n-k-m}2\rfloor}.
$$

If $m=0$, ths simplifies as
$$
  b_{n,0}=-{n\choose \lfloor \frac{n}2\rfloor}
  +2\sum_{0\leq k\leq n/2}2^{2k}{n-2k\choose \lfloor \frac{n-2k}2\rfloor}
  +2\sum_{0\leq k\leq (n-1)/2}2^{2k+1}{n-2k-1\choose \lfloor \frac{n-2k-1}2\rfloor}.
$$
Recall that ${2a\choose a}=2{2a-1\choose a-1}$, so these sums are almost the same.
Say, if $n$ is even, then
$$
  b_{n,0}=-{n\choose \frac{n}2}-2^{n+1}
    +4\sum_{0\leq k\leq n/2}2^{2k}{n-2k\choose \frac{n-2k}2}
  =-{n\choose \frac{n}2}-2^{n+1}+4\frac{(n+1)!}{\frac{n}2!^2}
$$
(for the last sum, see Philippe Deléham's comment at http://oeis.org/A002457). Simplifying a bit, we get
$$
  b_{n,0}=(4n+3){n\choose \frac n2}-2^{n+1},
$$
which agrees with Per Alexandersson's formula.
The computation for odd $n$ is similar (one needs to separate the first term in the first sum of the general formula for $b_{n,0}$).<|endoftext|>
TITLE: Moduli space of (all) vector bundles on $\mathbb{P}^1$
QUESTION [12 upvotes]: It is well known that, by a theorem of Grothendieck, every vector bundle (always assumed coherent in this question; and everything is over the complex numbers) splits as a direct sum of line bundles.
This is not true in families, though, not even locally on the base of the family, as it is possible to construct flat $1$-parameter families of VB that have generically the same splitting type and "jump" to a different splitting type at the special fiber. This prevents the moduli stack $\mathfrak{M}$ of all VB on $\mathbb{P}^1$ (or the corresponding functor $F$ of isomorphism classes) to be separated (in the sense of valuative criteria). Here we took the notion of isomorphism of families to mean, as usual: the families $\mathscr{E}$ and $\mathscr{E}'$ of VB on $X$ parametrized by $T$ are isomorphic if there is a line bundle $L$ on $T$ such that $\mathscr{E}\simeq\mathscr{E}'\otimes L$.
For a smooth projective curve $X$, the moduli functor $F^s$ of stable bundles is representable by a smooth separated scheme $M^s$, and the stack $\mathfrak{M}^s$ of stable bundles is a gerbe over the latter. The moduli functor $F^{ss}$ of isomorphism classes of semistable bundles has a non separated coarse moduli space with proper connected components, and collapsing $S$-equivalence classes in the sense of Seshadri fixes this: now the functor $F'^{ss}$ of such $S$-equivalence classes of semistable bundles has a coarse moduli space (each connected component of) which is a projective variety (in particular separated).
By Grothendieck's theorem, the relation of S-equivalence on $\mathbb{P}^1$ coincides with the relation of isomorphism, so we can avoid talking of $S$-equivalence in this case. Also, there are no stable VB on $\mathbb{P}^1$ (of rank $>1$).
In what follows the curve $X$ is $\mathbb{P}^1$.
Since every VB is uniquely determined by its splitting type, and there is a countable set of choices for the possible line bundles occurring as summands, each connected component of $M^{ss}$, which is a variety, has to be a reduced point $\mathrm{Spec}(\mathbb{C})$.
Recall we let $\mathfrak{M}$ be the stack in groupoids of all VB on $\mathbb{P}^1$, with isomorphisms of families as $1$-arrows, and $F$ the corresponding functor of isomorphism classes.

Q.1: How does $\mathfrak{M}$ look like?
Q.2: Does $F$ have a coarse moduli space $M$? If yes, how does it look like?

It seems possible to me that an answer to question 1 or 2 has something to do with the partial order structure on the set $\boldsymbol{\mathrm{HNP}}$ of "Harder-Narasimhan polygons" (see Shatz, The decomposition and specialization of algebraic families of vector bundles). Maybe the specialization order on (field-valued) points of $M$ (or of $F$, or of $\mathfrak{M}$) reflects somehow the structure of the lattice $\boldsymbol{\mathrm{HNP}}$?

REPLY [3 votes]: The stack $\mathfrak{M}$ was discussed in section 7 of On the motivic class of the stack of bundles by Behrend and Dhillon, where the same stratification as discussed by Alexander Braverman was described.  Moreover, the various strata are identified with classifying stacks (for the automorphism groups of the corresponding bundles).<|endoftext|>
TITLE: How to construct particular De Bruijn sequences
QUESTION [5 upvotes]: For $n \ge 2$, there is at least one binary DeBruijn sequence beginning with $n$ zeros followed by $n$ ones.  Is there a straightforward way to construct such a sequence for each $n \ge 2$?  Examples:
$n=2: 0011$
$n=3: 00011101$
$n=4: 0000111101100101$
$n=5: 00000111110111001101011000101001$

REPLY [7 votes]: One of the simple ways to generate De Bruijn sequenceis the following rule. You write $n$ zeros, and after that "$1$ is better than $0$" (left hand rule):
$$0000111101100101000$$
For example 9th digit is $0$ because we already have $1111$. This algorithm was taken from the article N. M. Korobov, "Concerning some questions of uniform distribution". In the article Normal periodic systems and their applications to the estimation of sums of fractional parts he proposed a more general algorithm which allows to generate all possible De Bruijn sequenceis.<|endoftext|>
TITLE: A button for individual reals
QUESTION [5 upvotes]: Hamkins introduced the notion of a "button" in forcing.  This is a set-theoretic statement that can be forced, and can never be made false by further forcing.  An example is $V \not= L$. Another example (with parameters) is "$S$ is a nonstationary subset of $\kappa$."
Question: Is there an analogue of the second example for reals?  I mean a property $\varphi(v,a)$, where $a$ is a parameter, such that there is some forcing $\mathbb P$ and some class of reals $C$ such that for all $r \in C$, $\Vdash_{\mathbb P} \varphi(\check r, \check a)$, and for all $\mathbb P$-names $\dot{\mathbb Q}$ for partial orders, $\Vdash_{\mathbb P * \dot{\mathbb Q}} \varphi(\check r, \check a)$?

REPLY [7 votes]: The way you've stated it, you haven't said that $\varphi$ isn't already true, and so technically any tautological statement would work. But I assume that you want $\varphi(r,a)$ to start out false in the original model $V$. This would be an unpushed button, which hasn't yet been pushed. 
It is consistent with ZFC that there are no such statements (with real parameters). This is exactly equivalent to the axiom $\text{MP}(\mathbb{R})$, which is discussed in my paper 

J. D. Hamkins, A simple maximality principle, J. Symbolic Logic, vol. 68, iss. 2, pp. 527-550, 2003. 

Namely, the axiom $\text{MP}(\mathbb{R})$ asserts that all buttons with real parameters have already been pushed. 
Meanwhile, it is also consistent that there are many such statements. For example, let $C$ be the set of reals coding ordinals, and let $\varphi(r)$ assert that $\aleph_\alpha^L$ is collapsed, where $r$ is a code for $\alpha$. This is not true in $L$, but it is possibly necessary by forcing. More generally, if $W$ is any definable inner model that is sufficiently absolute, then you can refer to collapsing cardinals of $W$ in a similar way.
You might be interested in having large independent families of buttons, which means that they can each be pushed without pushing any others, as desired, in any forcing extension. These were important in the work I did with Benedikt Löwe on the modal logic of forcing.
For example, if you have a stationary partition $\omega_1=\bigsqcup_\alpha S_\alpha$ of $\omega_1$ in $L$, then you can make the assertion $\varphi(r)$ that $S_\alpha$ is no longer stationary, using the least such partition, where $r$ codes $\alpha$. If $C$ has one code for each ordinal, then these buttons can be controlled independently, since the usual way of killing stationarity will preserve all stationary subsets of the complement. 
You can obviously modify this trick in many other ways. For example, add a large family $C$ of mutually generic Cohen reals, and then let $\varphi(r)$ for $r\in C$ assert that the least Suslin tree in $L[r]$ is no longer Suslin.<|endoftext|>
TITLE: Liouville property - a very basic question
QUESTION [6 upvotes]: Let $\mathbb{F}_2$ be the free group on two generators. By a result of Kaimanovich and Vershik, for each measure $\mu$ on $\mathbb{F}_2$ such that the support of $\mu$ generates $\mathbb{F}_2$, we have that the random walk is not $\mu$-Liouville, i.e. there is a bounded $\mu$-harmonic function on $\mathbb{F}_2$ which is non-constant. Can one construct this function geometrically without involving KV result? I see this for finitely supported measures. Likely this has been clarified somewhere, I would like to have a citation in the latter case.

REPLY [3 votes]: If I understand correctly the question is a request for a formula expressing a specific non-constant bounded $\mu$-harmonic function on $\mathbb{F}_2$, where $\mu$ is a fixed generating probability measure. 
I assume below that $\mathbb{F}_2$ is freely generated by $\{a,b\}$. Let $A$ be the set consisting of all words starting with $a$ in $\mathbb{F}_2$. I claim that
$$ h(x)=\lim_{n\to\infty}\frac{1}{n+1} \sum_{k=0}^n\mu^k(xA), \quad x\in \mathbb{F}_2 $$
is such a $\mu$-harmonic, where $\mu^k$ stands for the $k$-th convolution power of $\mu$ ($\mu^0=\delta_e$).
In fact, it is not hard to see that the formula for $h$ above converges pointwise to a $[0,1]$-valued $\mu$-harmonic. With some work you can see that this function is non-constant.
A slightly more sophisticated way to see this, which is in line with my comments above, is as follows.
Consider the compact space $\bar{\mathbb{F}}_2=\mathbb{F}_2\cup \partial\mathbb{F}_2$. Then $\nu=\lim_{n\to\infty}\frac{1}{n+1} \sum_{k=0}^n\mu^n*\delta_e$ is a (in fact, the unique) stationary measure on $\bar{\mathbb{F}}_2$. It is supported on $\partial\mathbb{F}_2$, because $\mathbb{F}_2$ supports no stationary measure. It is fully supported there, by minimality. It is not hard to see that $(\partial \mathbb{F}_2,\nu)$ is a $\mu$-boundary in the sense of Furstenberg, thus the Poisson transform of any non constant $L^\infty$ function is non-constant. The expression for $h$ above is the Poisson transform for $\chi_\bar{A}$ wrt $\nu$ on $\bar{\mathbb{F}}_2$, which is the same as the Poisson transform for $\chi_{\partial A}$ wrt $\nu$ on $\partial \mathbb{F}_2$. The latter is non-constant by the fact that $\nu$ is fully supported.
Of course, the choice of $A$ in the construction above was quite arbitrary.<|endoftext|>
TITLE: Difference between the Laplacian and the sub-Laplacian of a Lie group
QUESTION [6 upvotes]: Given a Lie group $G$, what is the difference between the Laplacian $\Delta$ and the sub-Laplacian $\Delta_{sub}$ of $G$. And what are the properties that we lose when going from sub-Laplace to Laplace and vice versa.
For example, what I know, for $G$ being the Heisenberg group $H^3= \mathbb C \times \mathbb R$, the difference between the Laplacian $\Delta$ and the sub-Laplacian $\Delta_{sub}$ of $H^3$ is the standard Laplacian $\Delta_{\mathbb R} = \frac{\partial^2}{\partial t^2} $ of $\mathbb R$, because 
$$ \Delta= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}  + (x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y}  -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} + \frac{\partial^2}{\partial t^2},$$
which can be rewritten in terms of the sub-Laplacian $\Delta_{sub}$ as 
\begin{align}
\Delta &= \Delta_{sub} + \Delta_{\mathbb R},
\end{align}
and for the properties that we lose when going from sub-Laplace to Laplace are for example the ellipticity, because $\Delta_{sub}$ is sub-elliptic but not elliptic, however $\Delta$ is elliptic.
Thank you in advance

REPLY [8 votes]: As Sebastian Goette explained in his comment, the sub-Laplacian $\Delta_{sub}$ depends in general from an additional structure. And so does the Laplace-Beltrami $\Delta$ that you use to compute the difference. Let me elaborate.
SUB-LAPLACIANS
On a given smooth manifold $M$, we consider a sub-Riemannian structure $(\mathcal{D},g)$, where $\mathcal{D} \subseteq TM$ is a vector distribution (a sub-bundle of the tangent bundle) and $g$ is a smooth metric defined on it. Furthermore, let $\mu$ be a smooth measure on $M$ (i.e. given by a smooth density).
We define the (horizontal) gradient of a smooth function $f$ as the unique vector field $\nabla f \in \Gamma(\mathcal{D})$ such that
$$ g(\nabla f, X) = df(X), \qquad \forall X \in \Gamma(\mathcal{D})$$
(here $\Gamma(\mathcal{D})$ denotes the space of smooth sections of the distribution, that is horizontal vector fields). Clearly this depends on the distribution and the metric.
Furthermore, we define the divergence of a smooth vector field $X \in TM$ as the smooth function $\mathrm{div}_\mu(X)$ such that
$$ \mathcal{L}_X \mu = \mathrm{div}_\mu(X) \mu$$
where $\mathcal{L}_X$ denotes the Lie derivative. Then the sub-Laplacian is
$$\Delta_\mu f := \mathrm{div}_\mu(\nabla f), \qquad f \in C^\infty(M)$$
Such an operator depends on the sub-Riemannian structure $(\mathcal{D},g)$ but also on the measure $\mu$.
Example: In the Riemannian case $\mathcal{D} = TM$ and $g$ is defined on the whole tangent space at any point. Moreover it is customary to choose the standard Riemannian measure in place of $\mu$ (that is $\mu = \mathrm{vol}_g = \sqrt{|g|}|dx^1\wedge\ldots \wedge dx^n|$ in local coordinates). In this case we obtain the standard Laplace-Beltrami.
Properties: The sub-Laplacian $\Delta_\mu$ is always symmetric on the space of smooth and compactly supported functions $C^\infty_c(M)$, with respect to the product of $L^2(M,\mu)$. If the distribution $\mathcal{D}$ is Lie-bracket generating (a standard assumption in this field, dating back to Hormander work on hypoelliptic operators), then $\Delta_\mu$ is hypoelliptic (and indeed subelliptic) for any choice of $\mu$. Moreover it is well known that if $M$ equipped with its sub-Riemannian distance is a complete metric space, then $\Delta_\mu$ is essentially self-adjoint on $C^\infty_c(M)$.
Lie groups: On Lie groups one can choose $\mu$ to be any left-invariant measure (any such a measure differs up to a constant rescaling, which does not change the divergence and thus the sub-Laplacian). Moreover it is natural to choose a left-invariant distribution $\mathcal{D}$. This gives you a left-invariant sub-Laplacian.
Local formula: In terms of a local (possibly left-invariant if you are on a Lie group) orthonormal frame $X_1,\ldots,X_k$ of $\mathcal{D}$ we have:
$$\Delta_\mu = \sum_{i=1}^k X_i^2 + \mathrm{div}_\mu(X_i) X_i $$
where $X_i^2$, when applied to functions, means that we apply it twice, that is $X_i^2(f) = X_i(X_i(f))$.
DIFFERENCE OPERATOR
Riemannian extensions: The question you raised is well posed if you choose a Riemannian complement of the sub-Riemannian structure, that is a Riemannian metric $\hat{g}$ such that $\hat{g}|_{\mathcal{D}} = g$. In this case we define a "vertical distribution" $\mathcal{V}$ as the orthogonal complement to $\mathcal{D} w.r.t. $\hat{g}$, in such a way that
$$TM = \mathcal{D} \oplus \mathcal{V}$$
and $\hat{g}(\mathcal{D},\mathcal{V}) = 0$. Now you also have a well defined Laplace-Beltrami, the one of the Riemannian structure $\hat{g}$.
Difference operator: It is then a simple exercise to compute the difference between the sub-Laplacian $\Delta_\mu$ and the Laplace-Beltrami pf the Riemannian structure. On Lie groups, where all left-invariant measures are proportional, then the difference between the two operators is precisely the sub-Laplacian associated with the (possibly non-bracket generating) sub-Riemannian structure $(\mathcal{V},\hat{g}|_{\mathcal{V}})$. More explicitly, let $Z_1,\ldots,Z_{n-k}$ be a (left-invariant) local orthonormal frame for $\mathcal{V}$, in such a way that $X_1,\ldots,X_k,Z_1,\ldots,Z_{n-k}$ is a frame for the Riemannian metric $\hat{g}$. Then your difference operator is
$$ \Delta - \Delta_{sub} = \sum_{i=1}^{n-k} Z_i^2 + \mathrm{div}_\mu(Z_i)Z_i $$
Example: In the Heisenberg group $M = \mathbb{R}^3$, and following your notation, $(\mathcal{D},g)$ is generated by the left-invariant vector fields:
$$X_1 = \partial_x - y \partial_t, \qquad X_2 = \partial_y +x\partial_t$$
The Lebesgue measure $\mu=dx dy dz$ is left-invariant (and also right-invariant) and the divergence term vanishes (but this is just a coincidence on unimodular groups, where the divergence of left-invariant fields vanishes). Denoting with $\Delta_{sub}$ the sub-Laplacian associated with the standard sub-Riemannian structure and left-invariant measure $\mu = dxdydz$, we have:
$$\Delta_{sub} = X_1^2 + X_2^2. $$
You recover your computation by choosing the "trivial" Riemannian extension $\hat{g}$ obtained by promoting $\partial_t$ to a global unit vector orthogonal to $\mathcal{D} = \mathrm{span}\{X_1,X_2\}$.
Remark: In any case, the difference operator depends on the choice of a complementary Riemannian structure.<|endoftext|>
TITLE: Very stable vector bundles
QUESTION [8 upvotes]: Let $X$ be a smooth curve, and $E$ a rank $r$ vector bundle over $X$, $E$ is said very stable if every nilpotent map $$u:E\rightarrow E\otimes K_X$$ is zero (nilpotent means that the composition $E\rightarrow EK_X\rightarrow\cdots\rightarrow EK_X^r$ is zero) 

Is there any sufficient condition for vector bundle to be very stable?

Thanks

REPLY [2 votes]: This is not a complete answer but only deals with the rank 2 case with trivial determinant:
Let $u=\Phi\neq0$ be a nilpotent Higgs field, then $\Phi^2=0,$ so the kernel
line bundle $L\subset E$ (assuming $\Phi$ is not identically zero) contains the image of $\Phi.$ 
Hence, $\Phi$ is uniquely 
determined by a holomorphic section $\phi\in H^0(X;L^2K).$
This shows that a sufficient condition of a holomorphic rank 2 bundle with trivial determinant is that it does not contain any holomorphic line bundle
$L$ such that $H^0(X;L^2K)\neq\{0\}.$ 
In particular, in the case of a surface of genus 2, stable holomorphic bundles of rank 2 with trivial determinant are uniquely determined by the divisor of holomorphic line subbundles in $Pic_{-1}(X)$ by Narasimhan-Ramanan. Therefore, in this situation the aforementioned condition is very natural. In particular it also follows from Narasimhan-Ramanan that the set of very-stable rank 2 bundles with trivial determinant over a genus 2 surface is $\mathbb CP^3$ without the Kummer surface (of $X$, corresponding to strictly semi-stable bundles) and without several copies of $\mathbb CP^2$ corresponding to the space of non-trivial extensions
$$0\to S^*\to E\to S\to 0$$
of the 16 spin bundles $S$ on $X.$<|endoftext|>
TITLE: Braided Hopf algebras and Quantum Field Theories
QUESTION [15 upvotes]: It is well-known, that there are a lot of applications of classical Hopf algebras in QFT, e.g. Connes-Kreimer renormalization, Birkhoff decomposition, Zimmermann formula, properties of Rota-Baxter algebras, Hochschild cohomology, Cartier-Quillen cohomology, motivic Galois theory etc. 
But all these structures are based on Hopf algebras in monoidal categories with trivial braiding, e.g. given by $\tau(a\otimes b)=b\otimes a$.
Are there some applications of braided Hopf algebras (i.e. an object in some braided monoidal category) in QFT ? I'm looking for some examples which are both nontrivial and "calculable".

REPLY [3 votes]: I think that -apart from the applications in CFT and TFT already mentioned in previous answers- one of the most fundamental applications of braided Hopf algebras (with both non-trivial and "calculable" braiding), which underlies a significant part of various quantum field theories, is the mathematical foundations of the idea of supersymmetry itself: the notions of  super vector space ($\mathbb{Z}_2$-graded vector space) and superalgebra ($\mathbb{Z}_2$-graded algebra) and their "super" tensor products can be conceptually understood in the framework of braided monoidal categories as applications of the (unique) non-trivial braiding of the category ${}_{\mathbb{CZ}_{2}}\mathcal{M}$ of representations of the quasitriangular group Hopf algebra $\mathbb{CZ}_2$.  
This can be understood better if you take into account that the following statements: 

$V$ is a $\mathbb{Z}_{2}$-graded vector space (equivalently: a super vector space).
$V$ is a $\mathbb{CZ}_{2}$-module, through the $\mathbb{Z}_{2}$-action 
\begin{equation}
\begin{array}{cccc}
  1 \cdot v = v &  &  & g \cdot v = (-1)^{|v|}v=\left\{
                                                \begin{array}{r}
                                                v, \ \ v\in V_0 \\
                                                -v, \ \ v\in V_1
                                                \end{array}
                                                \right. \\
\end{array}
\end{equation}
for any homogeneous element $v\in V$, where $\ |v| \ $ stands for the degree of  $av$. (In other words $|v|=0$, if $v\in V_{0}$ ($v$ is an even element) and $|v|=1$, if $v\in V_{1}$ ($v$ is an odd element)). 
$V$ is a vector space in the braided monoidal Category ${}_{\mathbb{CZ}_{2}}\mathcal{M}$ of representations of the group Hopf algebra $\mathbb{CZ}_{2}$. 

are equivalent. 
Furthermore, the above correspondence generalizes to superalgebras: The statements:  

$A$ is a $\mathbb{Z}_{2}$-graded algebra (equivalently: a super algebra).
$A$ is a $\mathbb{CZ}_{2}$-module algebra through the $\mathbb{Z}_{2}$-action 
\begin{equation}
\begin{array}{cccc}
  1 \cdot a = a &  &  & g \cdot a = (-1)^{|a|}a=\left\{
                                                \begin{array}{r}
                                                a, \ \ a\in A_0 \\
                                                -a, \ \ a\in A_1
                                                \end{array}
                                                \right. \\
\end{array}
\end{equation}
for any homogeneous element $a\in A$, where $\ |a| \ $ stands for the degree of  $a$. (In other words $|a|=0$, if $a\in A_{0}$ ($a$ is an even element) and $|a|=1$, if $a\in A_{1}$ ($a$ is an odd element)). 
$A$ is an algebra in the braided monoidal Category ${}_{\mathbb{CZ}_{2}}\mathcal{M}$ of representations of the group Hopf algebra $\mathbb{CZ}_{2}$

are equivalent. 
In the above, the braiding $\Psi$, of the braided monoidal Category
${}_{\mathbb{CZ}_{2}}\mathcal{M}$ (i.e. the Category of $\mathbb{CZ}_{2}$-modules), is given by the family of natural isomorphisms $\psi_{V,W}: V\otimes W \cong W\otimes V$ explicitly written:
$$
\psi_{V,W}(v\otimes w)=(-1)^{|v| \cdot |w|} w \otimes v
$$
It can furthermore been shown, that the above (non-trivial) braiding is induced by the non-trivial quasitriangular structure of the group Hopf algebra $\mathbb{CZ}_{2}$, given by the $R$-matrix:
\begin{equation} 
R_{\mathbb{Z}_{2}} =\sum R_{\mathbb{Z}_{2}}^{(1)} \otimes R_{\mathbb{Z}_{2}}^{(2)}= \frac{1}{2}(1 \otimes 1 + 1 \otimes g + g \otimes 1 - g \otimes g)
\end{equation}
To be more specific, this $R$-matrix, induces the -the above mentioned- braiding through
$$
\psi_{V,W}(v \otimes w) = \sum (R_{\mathbb{Z}_{2}}^{(2)} \cdot w)
\otimes (R_{\mathbb{Z}_{2}}^{(1)} \cdot v)=(-1)^{|v| \cdot |w|} w \otimes v
$$
In the above, $v,w$ are any elements of the $\mathbb{CZ}_{2}$-modules (super vector spaces according to the above) $V,W$.  
Now, the so-called super tensor product algebra or $\mathbb{Z}_2$-graded tensor product algebra, of superalgebras, is the superalgebra $A\underline{\otimes} B$, whose multiplication $m_{A\underline{\otimes} B}$ given by 
$$
m_{A\underline{\otimes} B}=(m_{A} \otimes m_{B})(Id \otimes \psi_{B,A} \otimes Id): A \otimes
B \otimes A \otimes B \longrightarrow A \otimes B
$$
or equivalently:
$$
(a \otimes b)(c \otimes d) = (-1)^{|b| \cdot |c|}ac \otimes bd
$$
where $A,B$ are superalgebras, $m_A, m_B$ are their multiplications, $b,c$ are homogeneous elements of $B$ and $A$ respectively and $a,d$ any elements of $A$ and $B$ respectively.
(for further details and the generalization of the above for any finite abelian group, you can see this article, sect. $3$, pages 78-81).
Remark: It is interesting to note that most of the formalism of super vector spaces, superalgebras and their super tensor products was known to mathematicians since the late $'40$'s and the idea of supersymmetry in physics dates back to the $'70$'s. However it was not until the advent of quasitriangular Hopf algebras (in the late $'80$'s) and the investigation of their relation to the braided monoidal categories that the above connections were realized. Since, it is usual to speak (mainly in the math. physics literature) rather of "braided" than "graded" tensor products. 
Notice also, that the above application involves the simplest non-trivial braiding. This stems from the non-trivial $R$-matrix $R_{\mathbb{Z}_{2}}$ of the group Hopf algebra $\mathbb{CZ}_{2}$ rather than from its trivial quasitriangular structure $R=1\otimes 1$ (i.e. its cocommutativity). 
Finally, if you are further interested on the mathematical basis of supersymmetry and its importance in physics, among others I would recommend:

Supersymmetry for Mathematicians: an Introduction
these notes<|endoftext|>
TITLE: Is the regularization of a Fourier transform unique?
QUESTION [11 upvotes]: The Fourier transform of the Coulomb potential $1/\vert \mathbf r \vert$   of an electric charge doesn't converge because one obtains
$$F(k)=\frac {4\pi}{k} \int_0^\infty \sin(kr) dr.$$
The standard way to obtain a sensible value is to multiple the integrand by $f(\alpha,r)=e^{-\alpha r}$ and after doing the integral, taking the limit $\alpha\to 0$ (which has a nice physical reason). So one gets 
$$F(k)=\frac{4\pi}{k^2}.$$
Would any other function $f(\alpha,r)$ that makes the integral converge and that satisfies $\lim_{\alpha\to\alpha_0}f(\alpha,r)=1$ give the same result?  For example
$$F(k)=\lim_{\alpha\to 0}\frac {4\pi}{k} \int_0^\infty \frac{\sin(kr)}{\Gamma(\alpha r)} dr\stackrel{?}{=}\frac{4\pi}{k^2}.$$
In this case, Cesàro integration gives the same result. What would be the sufficient condition for uniqueness of regularization (maybe the theory of tempered distributions can answer this).
This question was asked in Math StackExchange.

REPLY [15 votes]: Yes, the answer is unique.  What this "regularization" is doing is computing the Fourier Transform in the sense of distributions.
Editing to add (see comment of Christian Remling below): For your precise question, the hypotheses you propose for the regularization are too weak: it's not enough that $f(\alpha,r)$ converge to the constant function pointwise.  What is true that most "natural" choices will give the same answer, because reasonable regularizations would converge weakly (i.e. in the sense of tempered distributions).<|endoftext|>
TITLE: Differential geometric interpretation of cohomology
QUESTION [12 upvotes]: I'm not sure whether this is an appropriate question for this forum. I'm afraid that this is not a research level question however:
1. It's about reference request therefore the answer does not require any tedious computation.
2. As it is reference request, I think that it is better to ask it here instead of mathstackexchange since here there are more experts familiar with good literature.
3. I believe that this discussion may be useful not only for me.
So what I would like to ask is some good reference for differential geometric interpretation of cohomological stuff like for example: representatives for characteristic classes in geometric terms, cross products, cap product, cup product for deRham forms, pairing with the fundamental class of $M$ as integration over $M$, and Poincare duality.

REPLY [11 votes]: For geometric interpretations of cup product and Poincaré duality let us assume that in the following (dual) homology classes are representable by smooth submanifolds (and everything is orientable). Then we can look at intersection theory.
For an integral class $x\in H^i(M;\mathbb R)$ we have that the Poincaré dual $x^* \in H_{m-i}(M,\partial M;\mathbb R)$ is represented by any submanifold which gives, by counting intersections with $i$-manifolds, the same homomorphism on $H_i(M;\mathbb R)$ as $x$. This is what allows you to understand cohomology classes geometrically.
This geometric situation behaves natural in some situations, e.g. consider the inclusion of a submanifold $i:N\to M$. Then we can pull back homology classes in the above sense, but also with the shriek map:
$$ H_i(M ,\partial M)\to H^{m-i}M \stackrel {i^*}\to H^{m-i}N \to H_{n-m+i}(N,\partial N) \to H_{n-m+i}(N,\partial N)/tors.$$
Claim: This map coincides with the above sense, i.e. the map 
$$[(X,\partial X)\hookrightarrow (M,\partial M)] \mapsto [(X\pitchfork N,X\pitchfork \partial N) \hookrightarrow(N,\partial N)],$$ after representing (if possible) $x\in H_i(M,\partial M)$ by an $i$-submanifold $X\hookrightarrow M$ which is transverse to $(N,\partial N)$.
Proof: In terms of dimensions and transverse intersections this makes sense. Note that the codomain is canonically isomorphic to $Hom(H_{m-i}N,\mathbb Z)$, as the map $H_{n-m+i}(N,\partial N) \to H^{m-i}N \to Hom(H_{m-i}N,\mathbb Z)$ factors. In other words the image of $x$ in $H_{n-m+i}(N,\partial N)$ is determined by what it does as a homomorphism on $H_{m-i}N$ up to torsion. To check that the above map does precisely the same as the pullback, defined on the $Hom$ quotient, we just need the above intersection counting. Let $Y^{m-i}\subset \mathring N$ be a submanifold transverse to $(X\cap N)$, then $Y\hookrightarrow N \hookrightarrow M$ is transverse to $X$, and everything evaluates already on $N$.
Okay, now that we got familiar with Poincaré duality we can now look at a Cup product interpretation on homology (which will actually just be a generalization of the above). As above we want $x \in H^iM, y\in H^jM$ and $x^*,y^*$ denote their duals, represented by transverse submanifolds $X,Y$. Then we have:
$$ x \smile y = [X]^* \smile [Y]^* = [X \pitchfork Y]^*,$$
i.e. Theorem: Cup product is dual to transverse intersection.
You should have a good idea in your head now about this interpretation. You should do some own calculation to get more familiar (or excited) with this interpretation.
To combine this with characteristic classes, you will get some insights by noting that the Thom class of the normal bundle of $N \hookrightarrow M$ in $M$ is Poincaré dual to the class represented by $N$. By looking at $M$ the total space of a vector bundle over $N$ (or rather the disk bundle for compactness), you see e.g. why the Euler class of the tangent bundle evaluates to the self intersection of the manifold. This also allows further playing around and applying to other cases.
For references see e.g. Topology books of Bott & Tu, Guillemin & Pollack, Bredon, stratifold theory (Kreck), and you should definitely check out Hutchings' nice notes on this.
I really hope this monologue also helps you and not only my desire to procrastinate.<|endoftext|>
TITLE: $RO(G)$-graded homotopy groups vs. Mackey functors
QUESTION [15 upvotes]: Everything here is model-independent: either take co/fibrant replacements wherever appropriate, or work $\infty$-categorically.
Also, I've looked through other similar MO questions, but I didn't find answers to the questions below.  Of course, please let me know if I missed anything!

In motivic homotopy theory, the category of motivic spectra comes with a set of "bigraded spheres", $S^{i,j} = \Sigma^i (\mathbb{G}_m)^{\wedge j}$.  It is well-known that homotopy classes of maps out of these don't detect equivalences.  This gives rise to a notion of "cellular" motivic spectra (as studied by Dugger--Isaksen): these sit as a right localization -- that is, a full subcategory whose inclusion admits a right adjoint -- which is by definition the largest subcategory in which bigraded homotopy groups do detect equivalences.
Back in equivariant homotopy theory, the analogous "tautological Picard elements" are the virtual representation spheres $S^V$, i.e. $V$ is a formal difference of finite-dimensional $G$-representations.  These corepresent what are therefore known as "$RO(G)$-graded homotopy groups".

A. If a map $X \to Y$ of $G$-spectra induces an isomorphism $[S^V,X] \xrightarrow{\cong} [S^V,Y]$ for all $V \in RO(G)$, is it necessarily an equivalence?

I would of course be interested to hear partial answers, e.g. if this is only known when $G$ is finite/discrete but is open for $G$ a (compact) Lie group -- similarly if this depends on what $G$-universe I'm working over (though the answer is pretty obviously "no" e.g. for the trivial universe).
Moreover, if the answer to the above is "no", then I have a follow-up question.

A'. What is known about the subcategory of cellular $G$-spectra?  Is it known to contain or not contain any particular $G$-spectra of interest?

Another question I have is the following.

B. Are there any particular $G$-spectra of interest with known $RO(G)$-graded homotopy groups?

In my limited understanding, the more common notion of "homotopy groups" in the equivariant world are given by homotopy classes of maps out of the stabilized orbits $\Sigma^n \Sigma^\infty_+ (G/H)$ for $n \in \mathbb{Z}$, which altogether assemble into a Mackey functor.  By definition, these detect equivalences.  (The motivic analog would be mapping out of $\Sigma^{\infty+n} X_+$ for all $X$ in the Nisnevich site.)

C. Besides the obvious ones, are there any known relationships between $RO(G)$-graded homotopy groups and the "homotopy" Mackey functor?

REPLY [4 votes]: A.  The brackets are the same computed in any model, as you say, and for most that entails 
fibrant approximation.   For genuine $G$-spectra (complete universe), $G$ a compact 
Lie group, it goes back to LMS http://www.math.uchicago.edu/~may/BOOKS/equi.pdf
that a map is a weak $G$-equivalence if it induces an isomorphism on $Z$--graded homotopy
groups, so on the $\pi_n^H$ for all integers $n$ and (closed) subgroups $H$.   That is the right 
definition of weak $G$-equivalence by Theorem I.4.6 there, which says that a map is a weak
$G$-equivalence if and only if it is a spacewise weak $G$-equivalence.   Taking $H=G$, one
is looking at trivial representations, and so the question amounts to asking when looking at all spheres $S^V$ for $V\in RO(G)$  is equivalent to looking at all $G/H_+\wedge S^n$.  Justin's
answer says that this is true for $G = Z/2$ and false for $G=Z/p$ for an odd prime $p$.
A'.  For G-CW spectra, as defined in LMS where all $G$-spectra are fibrant, a map is a weak 
$G$-equivalence if and only if it is a $G$-homotopy equivalence. For other models, one must 
first fibrant approximate and then construct $G$-CW spectra, giving the right model independent
notion.  Such $G$-CW spectra are not generally well related to model theoretic cell $G$-spectra.      See Chapter 24 of 
http://www.math.uchicago.edu/~may/EXTHEORY/MaySig.pdf
for discussion.
B.  Very few, as Justin says.  Don't have time to think this through right now.
C.  Question is cryptic.  Spell it out and I can try to answer.  Not clear what you have in mind.<|endoftext|>
TITLE: Coherence and rewriting
QUESTION [9 upvotes]: In category theory there are numerous coherence theorems (https://ncatlab.org/nlab/show/coherence+theorem). One example is the Mac Lane's coherence theorem for monoidal categories. This and probably others as well, I believe, can be formulated in the language of term rewriting (https://en.wikipedia.org/wiki/Rewriting). 
What work has been done in the term rewriting theory related to coherence in category theory? Are there general results in the term rewriting theory from which various categorical coherence theorems follow? 
There may be some advantages to the term rewriting formulation in practice too. For example, analogous to a monoidal category, one can define a pseudomonoid (also called a monoidale) internal to a monoidal 2-category. This should satisfy coherence similar to Mac Lanes's coherence, which will involve 2-cells rather than natural transformations. Some other similar coherences also can occur (see Pseudomodules, "general coherence theorem"). I think that all these should follow from one term rewriting result.
I am also interested in term rewriting for higher categorical coherence.

REPLY [3 votes]: There are two kinds of "coherence theorem". One of them states for example "bicategories are equivalent to 2-categories", the other kind are results that look like "every diagram of a certain form commute".
As far as I know, rewriting has only ever be applied to the second kind of problem. The basic result for proving this kind of thing is Squier's homotopical Theorem, and its extension by Guiraud and Malbos: http://arxiv.org/abs/0810.1442

Guiraud and Malbos then used their theorem to prove coherence for monoidal, symmetric monoidal and braided monoidal categories: http://arxiv.org/abs/1004.1055
I just recently uploaded this paper to the arXiv. Using the same technique as Guiraud and Malbos I am able to prove coherence for bicategories and pseudofunctors. I also prove coherence for pseudonatural transformations, but this requires significantly more work http://arxiv.org/abs/1508.07807<|endoftext|>
TITLE: Cayley graphs of $A_n.$
QUESTION [17 upvotes]: Consider the Cayley graphs of $A_n,$ with respect to the generating set of all $3$-cycles. Their properties must be quite well-known, but sadly not to me. For example: what is its diameter? Is it an expander family? (I assume not) Is there a nice description of the eigenvalues? The tree number?

REPLY [13 votes]: For $n\geq 5$, all the eigenvalues of the Cayley graph you describe are integers.  The eigenvalues correspond to character values on $3$-cycles multiplied by the number of $3$-cycles (because the generating set of $3$-cycles is a conjugacy class). The multiplicity is the dimension of the irreducible representation.
Since the character values are algebraic integers, it is enough to observe that on a $3$-cycle all character values are rational. This follows from the well-known fact (see Noam Elkies answer to this question) that if $g$ is an element of a finite group $G$, then $\chi(g)$ is rational for all characters $\chi$ of $G$ if and and only if $g^m$ is conjugate to $g$ for all powers $m$ coprime to the order of $g$.  
Since every power of a $3$-cycle that is not the identity is a $3$-cycle and $3$-cycles are all conjugate in $A_n$ for $n\geq 5$, this gives the integrality of character values on $3$-cycles.<|endoftext|>
TITLE: Fine structure question: when do levels of $L$ look "a lot" like each other?
QUESTION [6 upvotes]: (Everything is assuming $V=L$.)
Fix an uncountable regular cardinal $\kappa$, and let $$E_\kappa=\{\mu<\kappa: \mbox{there is an elementary substructure of $L_\kappa$ isomorphic to $L_\mu$}\},$$ and let $$E_\kappa^+=\{\mu<\kappa: \mbox{$\exists$ an elementary substructure of $(L_{\kappa^+}, L_\kappa)$ which is isomorphic to $(L_\kappa, L_\mu)$}\}.$$ (In each definition $\mu$ ranges over ordinals.) Here "$(L_\alpha, L_\beta)$" denotes the structure $(L_\alpha; \in)$ augmented by a predicate for $L_\beta$.
Clearly $E_\kappa^+\subseteq E_\kappa$. In the course of a problem I'm working on, I assumed that $E_\kappa$ and $E_\kappa^+$ are quite different, but thinking about it in more detail this seems unjustified. My question is:

Is $E_\kappa^+$ necessarily a proper subset of $E_\kappa$?

I would also be interested in specific cases, e.g. $\kappa=\omega_1$.

EDIT: Oh dear, this is what I get for doing math late at night: the question above has a terrible typo - $E_\kappa^+$ is supposed to be the set of $\mu$ such that for some $\alpha<\kappa$ we have $(L_\alpha,L_\mu)$ is isomorphic to an elementary substructure of $(L_{\kappa^+}, L_\kappa)$. This completely changes the meaning of the question, of course. 
I'm leaving the question as is and accepting Joel's answer, since it is a good answer; but I've asked a corrected version of the question at Levels of L resembling each other, take 2.

REPLY [6 votes]: Under your assumptions, the set $E_\kappa^+$ is empty. Indeed, we don't even need the predicates that you mention. (In any case, since $\kappa$ is definable in $L_{\kappa^+}$, the predicate for $L_\kappa$ in $\langle L_{\kappa^+},\in,L_\kappa\rangle$ would be definable and therefore offer no additional expressive power.) 
What I claim is that if $\kappa$ is a cardinal, then there is no elementary substructure of $\langle L_{\kappa^+},\in\rangle$ in $L$ that is isomorphic to $\langle L_\kappa,\in\rangle$. 
To see this, suppose that $\kappa$ is a cardinal and there is an elementary substructure $X\prec L_{\kappa^+}$ in $L$ that is isomorphic to $L_\kappa$. This amounts to an elementary embedding $j:L_\kappa\to L_{\kappa^+}$. Since $\kappa$ is definable in the target structure, we must have $j(\mu)=\kappa$ for some $\mu<\kappa$, and so $j$ has a critical point $\delta\leq\mu$. Since all the subsets of $\delta$ in $L$ appear in $L_\kappa$, we may define in $L$ a measure on $\delta$ via $X\in U\iff \delta\in j(X)$. So $\delta$ is a measurable cardinal in $L$, contradicting Scott's theorem that there are no measurable cardinals in $L$.<|endoftext|>
TITLE: is every element in a C* algebra a product of normal elements?
QUESTION [8 upvotes]: I have the following question and since I am not an expert on C*-algebras, I thought I ask it here:
I know that in general the sum and product of normal elements need not be normal. It is even true that every element in a C*-algebra is the sum of two normal elements. What I do not know is if that is true for multiplication, too, i.e. if every element in a C*-algebra is a product of normal elements.
I think I know that every invertible element can be written as a product of a unitary and a positive element but this decomposition does not work for non-invertible elements.
My standard example is the right shift operator on the l^2(N)-space. This is a non-normal operator, which is not invertible and I have no idea how to write it as a product of two normal operators.
I would be very grateful if anyone could help.
Greetings, Tom.

REPLY [10 votes]: Since the question was asked wrt C$*$-algebras, I guess there is room for a general remark. Suppose that $xy = 1$ where $x$ is a product of normal elements, say
$v_1v_2\cdots v_n$. Then $v_1$ is normal and has a right inverse;
therefore it is (two-sided) invertible. Thus $v_2\dots v_n y $ is
invertible, so $v_2$ has a right inverse, and thus is invertible. By
induction, we obtain that all the $v_i$ are two-sided invertible, so that
$x$ is too.
The upshot of this argument is that a product of normal elements that is
one-sided invertible, is actually invertible. In particular, the right
shift is not a product of normal elements---although of course, the index argument above is decisive in this case.
If every element of a C$*$-algebra were a product of normals,
then it would be directly finite (all one-sided inverses are two-sided).
Whether this is sufficient is unclear (as is the question as to whether the product of normals property extends to matrix rings over the original). But if the C*-algebra has one in
the stable range, then every element is an invertible times a positive, so
is a product of (three) normals. 
And for von Neumann algebras, if of finite type, then every element is a product of two normals, as the isometry in the polar decomposition argument (below) can be modified to be a unitary in that case. If not of finite type, the property fails.<|endoftext|>
TITLE: An equality involving roots of unity which holds most of the times, but not always
QUESTION [16 upvotes]: Let $m$ and $n$ be distinct odd positive integers. The equality
$$
  \prod_{k=0}^{mn-1} \left( e^{\frac{2\pi i k}{m}}
                          + e^{\frac{2\pi i k}{n}} \right)
  \ = \ 2^{\gcd(m,n)}
$$
holds for all pairs of such $m$ and $n$ less than $50$ except for
$(3,21)$, $(3,39)$, $(15,33)$ and $(21,39)$, where the left-hand side
takes the value $512 = 2^9$.
Question: For which values of $m$ and $n$ does the equation
not hold, and which values takes the left-hand side in these cases?

REPLY [30 votes]: It always equals $2^{\gcd(m-n,mn)}$. Proof:
denote $x=e^{2\pi i/mn}$. Denote by $A$ the set of residues $k$ modulo $mn$ for which $mn$ divides $k(m-n)$, $|A|=\gcd(m-n,mn)$, $B$ is the set of other residues. We have
$$
\prod_k (x^{km}+x^{kn})=\prod_{k\in A} 2x^{km}\cdot \prod_{k\in B} \frac{x^{2km}-x^{2kn}}{x^{km}-x^{kn}}.
$$
The second product equals 1, since the multiples in the numerator and denominator are the same. The first product equals $2^{|A|}$, since $k$ and $-k$ belong to $A$ simultaneously, and we may couple multiples corresponding to them (with exception $k=0$).<|endoftext|>
TITLE: Most natural equivalence between $C^*$-algebras in NCG
QUESTION [7 upvotes]: I have listen or read that, in the context of noncommutative geometry,  Morita equivalence is a more natural equivalence for $C^*$-algebras than $*$-isomorphism. 
Can someone explain this sentence or know some text that could be helpful? 
Does anybody know some comparisons of different $C^*$-algebras categories? 
NOTE: I asked this same question en SE last week, it's still unanswered.

REPLY [11 votes]: Here I list some facts that may be useful for building your intuition:
1. Two commutative Morita equivalent $C^*$-algebra are in fact $*$-isomorphic.
2 If $A$ is $C^*$-algebra and you take $B=M_n(A)$ then $A$ and $B$ are Morita equivalent.
3 Many invariants for $C^*$-algebras such as $K$-theory or Hochschild or cyclic homology are the same for Morita equivalent $C^*$-algebras.
4 Two Morita equivalent $C^*$-algebras have the same representation theory so from this point of view they should represent the same "noncommutative space".<|endoftext|>
TITLE: Sphere packings : what next after the recent breakthrough of Viazovska (et al.)?
QUESTION [21 upvotes]: Given the march 2016 breakthrough concerning sphere packings by Viazovska for the case of dimension 8, and by Cohn, Kumar, Miller, Radchenko and Viazovska for the case of dimension 24, it follows that the known cases $\Delta_1=1$, $\Delta_2=\frac{\pi}{\sqrt{12}}$, $\Delta_3=\frac{\pi}{\sqrt{18}}$, $\Delta_8=\frac{\pi^4}{384}$ and $\Delta_{24}=\frac{\pi^{12}}{12!}$ all have a lattice solution with neat number-theoretic properties, yet the densities obtained do not seem to follow any obvious pattern, except for the presence of $\pi$.  

Question A: Is there a consensus among experts that one should not hope for a synthetic formula for densities, or are there still reasonable candidate formulas standing ?  

and

Question B: are there workshops planned on the recent papers soon (say, this fall, maybe even this summer) ?

REPLY [18 votes]: Here's an attempt at answering your Question A:
Currently, one of the most powerful methods for proving upper bounds on sphere packing densities is the linear programming bound of Cohn and Elkies (which is what Viazovska used).
Exact answers
According to the numerical computations of Cohn and Elkies, we do not know any dimensions other than 1, 2, 8, and 24 where their linear program has a chance of proving the exact bound (see their Figure 1). There are exceptionally good lattices in those four dimensions that match the bounds given by the Cohn-Elkies linear program. It could be the case (likely?) that these are the only dimensions where the bounds match, although we don't know how to rule out other possibilities. There could be some deep reason (perhaps related to finite simple groups) why these four dimensions are special.
Hales' proof of Kepler's conjecture (sphere packing in dimension 3) uses a completely different method.
Asymptotic bounds
An important open question, for which there is still an exponential amount of room for improvement, is the problem of sphere packing bounds in very high dimensions.
The current best asymptotic upper bound to the packing density $\Delta_n$ in $\mathbb{R}^n$ is by Kabatiansky and Levenshtein (1977)
$$
\Delta_n \le 2^{-(0.5990\dots + o(1))n}
$$
(see, e.g., my paper with Henry Cohn on the matter; in particular, this asymptotic bound lies within the scope of the Cohn-Elkies linear program). The current best lower bound is due to Venkatesh. For all $n$,
$$
\Delta_n \ge c n 2^{-n}
$$
for some explicit $c$ (that has been improved over time), and for a specific infinite set of $n$,
$$
\Delta_n \ge c n (\log\log n) 2^{-n}.
$$
Closing the gap seems to be a difficult problem.
Further reading

Henry Cohn's slides
Henry Cohn's PCMI 2014 lecture notes<|endoftext|>
TITLE: Iteration of random reals
QUESTION [7 upvotes]: Consider two random reals $x, y$ over a transitive model $V$ of ZFC. More specifically, if $\mathcal C^V={}^\omega2$ is the Cantor space, composing the canonical homeomorphism with the projections $\mathcal C^V\stackrel\cong\longrightarrow \mathcal C^V\times \mathcal C^V\longrightarrow \mathcal C^V$ we obtain two continuous maps inducing complete embeddings $h_i:\mathcal B^V/\mathcal N\longrightarrow \mathcal B^V/\mathcal N$, where $\mathcal B^V$ is the Borel $\sigma$-algebra of $\mathcal C$ and $\mathbb N$ is the ideal of null sets with regard to the usual product measure ($i=1,2$). If $G$ is a generic ultrafilter, then $h_i^{-1}[G]$ are also generic ultrafilters determining two random reals $x,y$ (over $V$ in principle).
I would like to know a proof that $y$ is in fact random over $V[x]$. All the texts I have seen pass over this fact very quickly.
Let me indicate what I have tried to do:
I tried to show that $y$ does not belong to any null Borel subset $B$ of $\mathcal C^{V[x,y]}$ with code in $V[x]$. On the contrary, if $y\in B$, I use that $B$ can be represented as $B=C^{V[x]}_x$, where $C$ is a Borel subset of $\mathcal C^V\times \mathcal C^V$ with code in $V$ (Lemma 3.1.6 in Bartoszynsky and Judah Set Theory. On the Structure of the Real Line). Here $C^{V[x]}$ represents the Borel set with the same code as $C$ and $C^{V[x]}_x=\{y\in\mathcal C^{V[x]}:(x,y)\in C^{V[x]}\}$.
If $C$ is null, then $(x,y)\in C^{V[x,y]}$ and a contradiction follows, but, unfortunately, the fact that $C_x^{V[x]}$ is null does not imply that $C$ is necessarily null (see here). I do not know if the Borel set $C$ can be taken null anyway.

I edit the question after the Andreas Blass' answer:
I thought I have completed the proof, but I have realized that my reasoning was circular. Namely:
According to Blass' indication, we must prove that $(P'\times {}^\omega2)\cap C$ has measure zero. On the contrary, there exists a Borel set $E\subset P'$ of positive measure in $V[x]$ such that $m(((P'\times {}^\omega2)\cap C)_u)>0$ for each $u\in E$.
We choose a random real $x'\in E$ and then we obtain a contradiction, since $C_{x'}$ should be null and non-null.
The problem is why there exists a random real in $E$. I had used that the set $R(V)$ of random reals in $V[x]$ over $V$ has outer measure $1$. But now I have realized that the proof I knew of this fact uses the fact we are trying to prove, namely, that null Borel sets of $V[x]$ can be represented as sections of null Borel sets of $V$.
The proof I know is that appearing in Bartoszynsky and Judah's book (Lemma 3.2.16). The argument is as follows: $R(V)$ is a tail set, so its outer measure must  be $0$ or $1$. Hence, it suffices to show that it is not $0$. Since $x+\mathcal C^V\subset R(V)$, it is enough to show that $\mathcal C^V$ is not null (in $\mathcal C^{V[x]}$). This is shown in Lemma 3.2.39, which starts by representing a null set of the random extension as a section of a null set of the base model $V$.
This is stated without any comment, so I wonder if it is obvious for some reason.

I have found a different argument for proving that $\mathcal C^B$ is not null avoiding circularity. Mi doubt is definitely solved. Thank you again.

REPLY [8 votes]: Since $x$ is random over $V$, the fact that $C_x^{V[x]}$ is null is (like any fact about $x$ in $V[x]$) forced by some condition in $V$.  This condition is the equivalence class, modulo the null ideal, of some positive-measure Borel set $P$ in $V$.  Let $P'$ be the Borel set in $V[x]$ with the same code.  Then $x\in P'$, so $(x,y)\in (P'\times{}^\omega2)\cap C$, and this has measure zero because $P$ forces $C_x$ to have measure zero.
In other words, although $C$ need not have measure zero, intersecting it with a suitable cylinder containing $(x,y)$ (namely the cylinder over $P'$) does, and that's all you need.<|endoftext|>
TITLE: If $U,D$ are $\kappa$-complete nonprincipal ultrafilters on $\kappa$ and $j_U(U) = j_D(D)$, is $U=D$?
QUESTION [17 upvotes]: Here, $\, j_U, \, j_D$ are the canonical elementary embeddings induced by $U,D$ respectively.
I note that it is consistent with the existence of a measurable that the answer be yes: it is true in the model $L[D]$ for $D$ a measure on $\kappa$.

REPLY [11 votes]: It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, not only for $\kappa$-complete ultrafilters on $\kappa$, but also for arbitrary countably complete ultrafilters.
First I'll show that in the Kunen-Paris model, there exist distinct normal ultrafilters $U_0$ and $U_1$ such that $j_{U_0}(U_0) = j_{U_1}(U_1)$. Moreover, $M_{U_0} = M_{U_1}$.
Let $U$ be a normal ultrafilter on a measurable cardinal $\kappa$ and let $j : V\to M$ denote its ultrapower. Assume $2^\kappa = \kappa^+$. Let $\mathbb P$ be the Easton product $\prod_{\delta\in I}\text{Add}(\delta,1)$ where $I\subseteq\kappa$ is a $U$-null set of cardinals.
Let $\mathbb Q = j(\mathbb P)$ and let $\mathbb Q/\mathbb P$ denote the product  $\prod_{\delta\in j(I)\setminus \kappa}\text{Add}(\delta,1)$ as computed in $M$. Thus $\mathbb Q \cong \mathbb P\times (\mathbb Q/\mathbb P)$. Since $\kappa\notin j(I)$, $\mathbb Q/\mathbb P$ is ${\leq}\kappa$-closed and $j(\kappa)$-cc in $M$, so by standard arguments, one can construct an $M$-generic filter $G\subseteq \mathbb Q/\mathbb P$ in $V$.
Let $H\subseteq \mathbb P$ be a $V$-generic filter. Let $j_0:V[H]\to M[H\times G]$ be the unique lift of $j$ such that $j_0(H) = H\times G$. Let $\sigma_{\alpha,\kappa}$ denote the automorphism of $\mathbb P$ given by $$\sigma_{\alpha,\kappa}((p_\delta)_{\delta\in I}) = (p_\delta)_{\delta\in I\cap \alpha}{}^\frown(p^*_\delta)_{\delta\in I\setminus \alpha}$$
where for $q\in \text{Add}(\delta,1)$, $q^*$ denotes the result of flipping the bits in $q$. Denote the similar automorphism of $\mathbb Q$ by $\sigma_{\alpha,j(\kappa)}$. Let $j_1 : V[H]\to M[H\times G]$ be the lift of $j$ such that $j_1(H) = \sigma_{\kappa,j(\kappa)}(H\times G)$.
Now it's time to show $j_0(j_0) = j_1(j_1)$. Since $j_0(j_0)\restriction M = j_1(j_1)\restriction M$, it suffices to show that $j_0(j_0)(H\times G) = j_1(j_1)(H\times G).$ This follows from a long fun computation:
\begin{align*} 
   j_0(j_0)(H\times G) 
        &= j_0(j_0)(j_0(H))\\
        &= j_0(j_0(H))\\
        &= j_0(H\times G)\\
        &= j_0(H)\times j(G)\\  
        &= H\times G\times j(G)\\
        &= \sigma_{\kappa,j(j(\kappa))} \circ \sigma_{\kappa,j(j(\kappa))}(H\times G\times j(G))\\
        &= \sigma_{\kappa,j(j(\kappa))}\circ \sigma_{j(\kappa),j(j(\kappa))} (\sigma_{\kappa,j(\kappa)}(H\times G)\times j(G))\\
        &= \sigma_{\kappa,j(j(\kappa))}\circ \sigma_{j(\kappa),j(j(\kappa))} (j_1(H\times G))\\
        &= \sigma_{\kappa,j(j(\kappa))} (j_1(\sigma_{\kappa,j(\kappa)}(H\times G)))\\
        &= \sigma_{\kappa,j(j(\kappa))}(j_1(j_1(H)))\\ 
        &= \sigma_{\kappa,j(j(\kappa))}(j_1(j_1)(j_1(H)))\\
        &= j_1(j_1)(\sigma_{\kappa,j(\kappa)}(j_1(H)))\\
        &= j_1(j_1)(j_0(H))\\
        &= j_1(j_1)(H\times G)
\end{align*}
Finally, let $U_0$ and $U_1$ be the normal ultrafilters derived from $j_0$ and $j_1$. Since $j_0(j_0) = j_1(j_1)$, $j_0(U_0) = j_1(U_1)$, as desired.
Second I'll sketch a proof that under the Ultrapower Axiom, the answer to your question is yes for arbitrary countably complete ultrafilters.
I'll do this by answering Trevor's question from the comments:
Fact (UA). Suppose $U_0$ and $U_1$ are countably complete ultrafilters on ordinals $\delta_0$ and $\delta_1$. Let $j_0 :V\to M_0$ and $j_1:V\to M_1$ denote their ultrapower embeddings, and assume $j_{0}(P(\delta_0)) = j_{1}(P(\delta_1))$. Then $j_0 = j_1$.
The fact suffices, since if $j_0(U_0) = j_1(U_1)$, the hypotheses of the fact hold, and hence $j_0 = j_1$, which means $j_0(U_0) = j_0(U_1)$, so $U_0 = U_1$. I'll sketch a direct proof of the fact assuming $2^{{<}\delta_0} = \delta_0$, although with significantly more work, one can do without.
Proof of fact. Apply UA to obtain internal ultrapower embeddings $i_0 : M_0\to N$ and $i_1:M_1\to N$ such that $i_0\circ j_0 = i_1\circ j_1$. Let $\alpha_0 = [\text{id}]_{U_0}$, $\alpha_1 = [\text{id}]_{U_1}$, and $\delta_* = j_0(\delta_0) = j_1(\delta_1)$.
Note that $U_0$ is the ultrafilter derived from $i_0\circ j_0$ using $i_0(\alpha_0)$ and likewise for $U_1$, so if $i_0(\alpha_0) = i_1(\alpha_1)$, then $U_0 = U_1$, and we're done. So assume without loss of generality that $i_0(\alpha_0) < i_1(\alpha_1)$. Let $D$ be the $M_1$-ultrafilter on $\alpha_1$ derived from $i_1$ using $i_0(\alpha_0)$, let $k_1 : M_1\to P$ be its ultrapower, and let $\ell:P \to N$ be the factor map. One can define $k_0 : M_0\to P$ by $k_0([f]_{U_0}) = [j_1(f)]_D$. Since $\ell\circ k_0 = i_0$ is an internal ultrapower embedding, one can conclude that $k_0$ is too. Since $D$ is an ultrafilter on an ordinal less than $\delta_*$, $D$ is coded in $M_1$ by a subset of $\delta_*$ (using that $2^{<\delta_*} = \delta^*$ in $M_1$), so $D\in M_0$. One can show $j_D^{M_0}\restriction \text{Ord} = k_1\restriction \text{Ord}$, and as a consequence $M_1\subseteq M_0$: if $A$ is a set of ordinals in $M_1$, then $k_1(A)\in P\subseteq M_0$, so $k_1(A)\in M_0$, so $A = (k_1\restriction \text{Ord})^{-1}[k_1(A)]\in M_0$. Under UA, $M_1\subseteq M_0$ implies that there is an internal ultrapower embedding $h : M_0\to M_1$ such that $h\circ j_0 = j_1$. (See Corollary 5.4.21 here.) From the perspective of $M_0$, the embedding $h$ preserves the powerset of $\delta_*$. But $h(\delta_*) = h(j_0(\delta_0)) = j_1(\delta_0)\leq j_1(\delta_1) = \delta_*$, so by the Kunen inconsistency theorem, $j\restriction\delta_*$ is the identity. Therefore $h$ is surjective: if $a\in M_1$, $a = j_1(f)(\alpha_0) = h(j_0(f)(\alpha_1))$. So $h$ is the identity, and since $h\circ j_0 = j_1$, we finally conclude that $j_0 = j_1$.<|endoftext|>
TITLE: Examples of high level math that can be motivated to laypeople
QUESTION [6 upvotes]: One of the difficulties of mathematics over other sciences is that our problems are harder to motivate to a general audience.  A biologist studying a particular pathway in the body can say that he's looking to understand or cure some disease, a physicist at CERN can say he's trying to understand why particles have mass, a string theorist can say he's trying to find a single coherent picture that explains both the physics of the very small and very large.  However, a lot of mathematics seems far removed from motivating problems that can be appreciated by non-mathematicians.
So what are some examples of research level mathematics that can be motivated to a non-mathematician in a paragraph or two?  Ideally, I would like examples of motivation for things you have personally worked on.
A caveats.  I don't want examples where the motivation is "the problem is old and unsolved, and the challenge of it captured my imagination."
To make this easier, I am happy with a motivation that would have to be fleshed out.  Perhaps a paragraph aimed at a mathematician that would allow them to motivate the problem to a layperson in a 10 minute conversation?
Edit: In response to a comment, I would like to contrast what I am looking for with Not especially famous, long-open problems which anyone can understand.  
First, I do not care about whether a problem has been long opens. In fact, as I explicitly stated above, I am NOT looking for motivation based on how hard a problem is (or the contrast between how hard a problem is and how hard it initially seems).  If you are working to explore the properties of a newly defined algebraic object, for example, that would be perfectly fine, so long as you can explain why someone who doesn't know math might view that as a worthwhile pursuit.  You don't have to be able to perfectly explain what the object is, just hint at a context of where it lives and what you hope would happen if it were better understood.  This could be applications of the understanding, in a traditional sense, or something more nebulous, like a belief that there is a connection between two seemingly unconnected ideas and that the research might give "insight" into that connection.
Second, I am not necessarily for specific problems.  What do you tell your mother when she asks why you study cluster algebras or string topology or loop erased random walks?  You wouldn't drive down into the specifics of your research, you would pull back towards a general description of the landscape and where you are hoping to travel within it.  
I would also like to contrast what I am looking for with How does one justify funding for mathematics research?.  While I am looking for answers in that vein, I want things that are more pointed.  Not "How does one justify funding for pure research at large" but "How does one justify funding for your particular field of research."  What motivation do you jump to when intrinsic motivation wanes, when existential doubt begins to set in, or when you realize that your cousin isn't going to pass the mashed potatoes until he feels that he appreciates why you do what you do?

REPLY [2 votes]: A concrete example would be the application of categories to software development as described e.g. in this presentation ; even layman can anticipate the importance of correct software and even more being sure about its correctness and that is, where mathematics hooks in: what has been proven correct is so forever.  
So essentially, mathematics provides safe grounds for the progress of human knowledge.<|endoftext|>
TITLE: Can the graph removal lemma be proved directly from the triangle removal lemma?
QUESTION [8 upvotes]: The Triangle Removal Lemma states that any graph with $o(n^3)$ triangles can be made triangle-free by removing only $o(n^2)$ edges.  More generally, the Graph Removal Lemma states that for any graph $H$ on a constant $|V(H)| = k$ number of nodes, any graph with $o(n^k)$ copies of $H$ can be made $H$-free by removing at most $o(n^2)$ edges.
These theorems are proved using essentially the same set of techniques.  However, I wonder if there is a direct proof of the Graph Removal Lemma, assuming the Triangle Removal Lemma, that does not need to pass through any of the usual regularity lemmas used to prove these things.

REPLY [8 votes]: A proof of the Graph Removal Lemma that avoids using the regularity lemma can be found in A new proof of the graph removal lemma (2010). For an explanation why a direct proof of the Graph Removal Lemma from the Triangle Removal Lemma is not viable, see this discussion:

The proof of the Graph Removal Lemma is more intricate than that of
  the Triangle Removal Lemma. Actually, it depends on the structure of
  the graph $H$. If, for example, $H$ is a four-cycle, then the argument
  applied in the proof of the triangle removal lemma does not work,
  mainly because, once the "impure" edges are discarded, the copy of $H$
  that remains may have two vertices in a same cluster. In other words,
  the connectivity properties of $H$ influence the distribution of the
  vertices along the clusters.<|endoftext|>
TITLE: What is the smallest $x$ such that $\lfloor x^n\rfloor$ has the same parity as n?
QUESTION [18 upvotes]: A previous MO question asked for information about a number $x$ such that $\lfloor x^n\rfloor$ has the same parity as $n$ for all positive $n$.  Answers to this post included two values of x which meet the requirements.  From Noam Elkies: the largest root of $x^3-3x^2-x+1$, giving $x=3.2143...$.  From Max Alekseyev: $(3+\sqrt{17})/2$, giving $x= 3.56155...$.
What is the smallest such positive number?  Failing this, what are the first few digits of the smallest such positive number?
This question was suggested to me by Moshe Newman

REPLY [4 votes]: In a sense, there is no smallest: the smaller root of $x^2+k x+2$ has the desired property, and this goes to $-\infty$ with $k$. I assume the problem is meant to ask for a positive $x$.
Here's some code (at the bottom) for Mathematica (infinite precision!) that computes the half-open set $[a_0,b_0) \cup [a_1,b_1) \cup \cdots$ of real numbers $x$ that are less than the root of $\alpha^3-3\alpha^2-\alpha+1$ and for which $\lfloor x^k \rfloor\equiv k \bmod 2$ for all integers $k$ between 1 and $n$, inclusive.
For $n=16$, the smallest $x$ is $\sqrt[16]{100801956}$, a number that appears in the answer by user35593. This lower bound on $x$ is improved, but only by a minuscule amount by taking $n=45$: we learn that $x$ is at least $\sqrt[45]{32341223721862945369971}$. This is still not large enough to provably dismiss user35593's claim.
To enrich the examples we have, I offer the following. The larger root $\alpha$ of $x^2 -(2k+1)x-2\ell$, where $1\leq \ell \leq k$, is slightly above $2k+1$, while the smaller root $\bar \alpha$ is between $-1$ and 0. Because of the recurrence satisfied by $\alpha^n+\bar\alpha^n$, the real number $\alpha$ has the desired property $\alpha^n \equiv n \bmod 2$ for all $n\geq 1$.  
alpha = Root[1 - #1 - 3 #1^2 + #1^3 &, 3];
NewUpInt[1] = {{1, 2}, {3, alpha}};
NewUpInt[n_] := NewUpInt[n] =
  Module[{int},
    If[EvenQ[n],
      int[{a_, b_}] := 
         Table[{Max[a, (2 k + 2)^(1/n)], Min[b, (2 k + 3)^(1/n)]},
           {k, Floor[(a^n - 3)/2] + 1, Ceiling[(b^n - 2)/2] - 1}],
      int[{a_, b_}] := 
         Table[{Max[a, (2 k + 1)^(1/n)], Min[b, (2 k + 2)^(1/n)]},
           {k, Floor[(a^n - 2)/2] + 1, Ceiling[(b^n - 1)/2] - 1}]];
    Flatten[Map[int, NewUpInt[n - 1]], 1]];<|endoftext|>
TITLE: $\mathfrak{ufo}$: An unidentified combinatorial cardinal characteristic of the continuum?
QUESTION [15 upvotes]: An ultrafilter ornament is a chain of free filters on $\mathbb{N}$ that are not ultrafilters, whose union is an ultrafilter.
Let $\mathfrak{ufo}$ be the minimal cardinality of 
an ultrafilter ornament. 
I arrived at this definition back in 2008, while teaching Ramsey theory at the Weizmann Institute of Science, based on a nice book by I. Protasov.
Topologically, the cardinal number $\mathfrak{ufo}$ is the minimal length of a transfinite convergent sequence in $\beta(\mathbb{N})$, but you may proceed with the combinatorial definition if you prefer.
Basic facts. 

$\aleph_1\le\mathfrak{ufo}\le\mathfrak{c}$. 
$\mathfrak{ufo}$ is a regular cardinal number.
$\mathfrak{ufo}\le\mathfrak{u}$, indeed $\mathfrak{ufo}\le\operatorname{cof}(\kappa)$ for each cardinality $\kappa$ of a basis for an ultrafilter.

Item 1 follows from the topological interpretation, but can also be proved combinatorially. Blass supplied me with one such proof. 
This is a cute exercise (or see below). Items 2-3 are immediate.
Item 3 was pointed out to me by Blass. 
Problem. Can the cardinal number $\mathfrak{ufo}$ be identified as a classic combinatorial cardinal characteristic of the continuum?
(For me, $\aleph_1$ is definitely a (potential) positive answer.)
Here is Blass's proof. I thank him for the permission to include it here.
I made tiny changes. If you find errors, this must be my fault. :)
Suppose, toward a contradiction, that some ultrafilter $U$ is the union of a countable, strictly increasing sequence of filters $F_n$.  For each $n$, pick a set $A_n\in F_{n+1}\setminus F_n$. By intersecting each $A_n$ with all the earlier $A_m$'s, we may assume that the $A_n$ sequence is decreasing (with respect to set-inclusion).  Let $D_n$ be the set-difference $A_n\setminus A_{n+1}$; so the sets $D_n$ are pairwise disjoint.  Let $X$ be the union of the sets $D_n$ for the even $n$, and let Y be the union the remaining sets $D_n$.  Then 
$$
\mathbb{N}=X\cup Y\cup (\mathbb{N}\setminus A_1)\cup \bigcap_n A_n.
$$
The last of these isn't in $U$, because, being a subset of $A_n$, it can't be in $F_n$ for any $n$.  And $\mathbb{N}\setminus A_1$ isn't in $U$, because $A_1$ is.  So either $X$ or $Y$ is in $U$; without loss of generality, suppose $X$ is in $U$.  Then $X$ is in $F_n$ for some $n$, and we may assume that $n$ is odd (just add 1 to $n$ if necessary).  So $F_n$ contains $X$ and $A_n$ but not $A_{n+1}$.  But $An\setminus A_{n+1}=D_n$ is disjoint from $X$ (because $n$ is odd), which means that $A_{n+1}$ is a superset of the intersection of $X$ and $A_n$.  The preceding two sentences contradict the fact that $F_n$ is a filter.

REPLY [17 votes]: This invariant is known for Boolean algebras in general as pseudo-altitude. That it is $\omega_1$ is proved in van Douwen's chapter in the Boolean algebra handbook 11.1 and 12.7, in the more general form that this holds for any weakly countably complete BA.<|endoftext|>
TITLE: Generating function of a sequence is not algebraic
QUESTION [11 upvotes]: Let we have a sequence $\{a_{n}\}$, such that $\forall n \,\, a_{n}>0$ and $a_{n} \rightarrow\infty, n\rightarrow\infty$. Also let's suppose that we have a subsequence $\{a_{n_{k}}\}$ such that $\exists C>0$ $\forall k:$ $a_{n_{k}}<Clog\, n_{k} $. How can we prove that the generating function of $\{a_{n}\}$ can't be an algebraic function?

REPLY [13 votes]: Yes, we can. Asymtotics of the coefficients of an algebraic function is determined
by the finitely many singularities on the circle of convergence. At every singularity the algebraic function has a Puiseux exansion, from which follows
that the function has a power asymptotics (no logs). This paper is a convenient reference
for the relation between the asymptotics of the function and asymptotics of its power series:
MR1039294
Flajolet, Ph., Odlyzko, A.,
Singularity analysis of generating functions. 
SIAM J. Discrete Math. 3 (1990), no. 2, 216–240. 
Inspection of all cases shows that no subsequence of coefficients of an algebraic function can have the intermediate growth between the constant and the logarithm.
More precisely, constant can happen but log cannot. So your conditions that
the coefficients tend to infinity but no faster than the log exclude algebraic functions.<|endoftext|>
TITLE: When may "summand of" be dropped from the definition of perfect dg module?
QUESTION [6 upvotes]: Let $\mathcal{A}$ be a small dg category. In Section 1 of Lunts-Orlov http://arxiv.org/pdf/0908.4187v5.pdf, $Perf(\mathcal{A})$ is defined to be the full DG subcategory of $\mathcal{S}\mathcal{F}(\mathcal{A})$ consisting of DG modules homotopy equivalent to a direct summand of a finitely generated semi-free DG module over $\mathcal{A}$. Here, $\mathcal{S}\mathcal{F}(\mathcal{A})$ is the DG category of semi-free DG modules over $\mathcal{A}$, and a module is considered ``free'' if it is a direct sum of shifts of representable modules.
Example 1.9 of Lunts-Orlov is the assertion that the bounded derived category of $\mathcal{A}$ is (the homotopy category of) $Perf(\mathcal{A})$.
Under what conditions on $\mathcal{A}$ are all objects of $Perf(\mathcal{A})$ actually homotopy equivalent (isomorphic in $H_0$) to finitely generated semi-free DG modules?
The motivation of this question is the case when $\mathcal{A}$ is a strands algebra from bordered Heegaard Floer homology (the structure as a small dg category comes from the distinguished idempotents of $\mathcal{A}$).The papers that decategorify such $\mathcal{A}$ typically identify the unbounded derived category of $\mathcal{A}$ with $H_0$ of the category of Type D structures homotopy equivalent to  operationally bounded Type D structures. The bounded derived category is identified with the full subcategory on Type D structures homotopy equivalent to finitely generated operationally bounded Type D structures.
An operationally bounded Type D structure is (more or less) the same as a semi-free dg module $X$ with a choice of decomposition of $X$ as $\mathcal{A} \otimes_{\mathcal{I}} N$ with respect to the $\mathcal{A}$-action, where $\mathcal{I}$ is the idempotent ring of $\mathcal{A}$ and $N$ is a left $\mathcal{I}$-module. Type D structures are isomorphic or homotopy equivalent iff the corresponding dg modules are isomorphic or homotopy equivalent. The same is true for finite generation.
So, based on the usual definition of perfect dg modules, I'd expect at first to have to use``Type D structures which split inject in $H_0$ into a finitely generated operationally bounded Type D structure'' as the bounded derived category. But this is less convenient for computing $K_0$, and I'd rather be able to replace an object of the bounded derived category with an isomorphic object which literally is a finitely generated bounded Type D structure.
In other words, my question is: are there any sufficient conditions one can place on dg algebras $\mathcal{A}$ for this to work, which are general enough to cover some interesting examples of $\mathcal{A}$ with nonzero differential?

REPLY [4 votes]: For a DGA $A$, all perfect modules are equivalent to semi-free modules if and only if $K(A)$ is generated by the class of the rank 1 free module. Furthermore, a particular perfect module is equivalent to a semi-free module if it is in the subgroup generated by the rank 1 free module. $K$-theory is exactly the gap between semi-free and perfect modules, so this does not seem to me useful for computing $K$-theory. You may as well ask: which DGAs have trivial reduced $K$-theory; maybe you should, but I think that is a different question.
It is easy to see that this $K$-theory condition is necessary. The definition of semi-free can be rephrased as that the module is generated from shifts of $A$ by a finite sequence of cones; so its class in $K(A)$ is in the subgroup generated by $[A]$. For example, for the product of two fields $A=E\times F$, an $A$-module is just a pair of $E$- and $F$-vector spaces. They could have different dimensions, so $K(A)=\mathbb Z^2$. But a semi-free module must have equal Euler characteristics. For example, the cone of $(1,0)\colon A\to A$ has homology zero on the $E$ side and nontrivial on the $F$ side, but its Euler characteristics are both zero. So $(E,0)$ is a perfect module not equivalent to a semi-free one.
Thomason proved that this necessary condition is actually sufficient. Given a triangulated category $C$, a triangulated subcategory $D$ (ie, a full subcategory closed under cones (and isomorphisms)) is called dense if every object of $C$ is a summand of an object of $D$. Thomason proved that dense subcategories of $C$ are determined by their $K$-groups, which necessarily inject into $K(C)$. The dense subcategory $D\subset C$ corresponding to a subgroup $H\subset K(C)$ consists of the objects whose class is in $H$. The semi-free modules are dense in the perfect modules by definition of the perfect modules, so a perfect module is equivalent to a semi-free module if and only if its class is in the subgroup generated by the ring $\langle [A]\rangle$. In particular, for every perfect module $M$, the module $M\oplus M[1]$ has trivial class and thus has a semi-free model.<|endoftext|>
TITLE: Are infinite simplicial complexes all manifolds?
QUESTION [14 upvotes]: Are infinite dimensional simplicial complexes manifolds locally modeled on $\mathbb R^\infty=\operatorname{colim}\mathbb R^n$? If they are homotopy equivalent, are they homeomorphic?
Of course not. Part of such a complex might not be infinite dimensional: glue an interval to $\mathbb R^\infty$ by one endpoint and it stops being homogeneous. Or take two copies of $\mathbb R^\infty$ and glue them at a point and that point is special. Also, for $\alpha$ an uncountable cardinal, $\mathbb R^\alpha$, topologized as the colimit of the finite dimensional subspaces, is not locally isomorphic to $\mathbb R^\infty$ because every open subset contains an uncountable discrete subset, namely an appropriate choice of basis. But are those three obstacles the only ones?

If a simplicial complex is countable, every link* is contractible, and every simplex is the face of another, does it follow that it is locally homeomorphic to $\mathbb R^\infty$? 

Is it PL isomorphic? What if we generalize to CW complexes? This leads to a second class of question. If those hypotheses are sufficient for local homeomorphism with $\mathbb R^\infty$, then apply them to the open cone on a contractible space satisfying the hypotheses. The space is now homeomorphic to $\mathbb R^\infty$. Was it already homeomorphic before applying the cone? If two such $\mathbb R^\infty$ manifolds are homotopy equivalent, are they, in fact, homeomorphic?
Is this topic in the literature? What about the special case of the product of a finite simplicial complex with $\mathbb R^\infty$? What about the product with a finite CW complex?
* The homotopy link of a point $x$ in a space $X$ is the (pro?) homotopy type of the homotopy inverse limit of the deleted neighborhoods $U-\{x\}$. This is a homeomorphism invariant of a space. Asking for the local homology $H_*(X,X-\{x\})$ to vanish is not quite sufficient for the link to be contractible because of the possibility that it is not simply connected. In a simplicial complex, the link is a specific simplicial complex, namely the union of all simplicies that are opposite $x$ in some simplex; and its PL isomorphism type is a PL invariant of the space.
Here is a potential counterexample to the PL statement. Take a closed manifold with the homology of a sphere but with nontrivial fundamental group, such as the Poincaré homology 3-sphere. The Cannon-Edwards theorem says that the double suspension is homeomorphic but not PL isomorphic to a sphere. Equivalently, the double cone is homeomorphic but not PL isomorphic to a disk. It is not PL isomorphic because a link of the link of the cone point is the homology sphere, not a real sphere. Now cross the double cone with $\mathbb R^\infty$. The result is homeomorphic to $\mathbb R^\infty$, but is it PL isomorphic? Is the cone point special? Is there some way of extracting the original fundamental group from the space, or has it been pushed off "to infinity"?
Motivation:
Is there an abelian group structure on $\mathbb C\mathbb P^\infty$? There are a couple of other models for the classifying space of $S^1$ that do have abelian group structures, such as the bar construction $BS^1$ and $FS^2$ the connected component of the free abelian group on $S^2$; we’d like to transfer them over to our favorite model. I think that we managed to cobble together a proof that $BS^1$ is PL isomorphic to $\mathbb C\mathbb P^\infty$, but what about $FS^2$? For many groups there is a model for the classifying space as the union of manifolds, but the filtration of the bar construction is only by manifolds if the group is a sphere, as fails for, eg, $\mathbb Z/3$ and $SO(3)$. But even though each $B_n\mathbb Z/3$ fails to be homogeneous, the full $B\mathbb Z/3$ is a group, thus homogeneous. So, in some sense it is a manifold. How many types of manifolds are there among infinite dimensional simplicial complexes? Perhaps they are all locally $\mathbb R^\infty$? And what about $BSO(3)$? It is neither a group nor filtered by manifolds, so it has no reason to be homogeneous. But I think it is.

REPLY [5 votes]: This question concerns manifolds modeled on the direct limit $\mathbb R^\infty$ of Euclidean spaces. The theory of such manifolds is well-developed. The most important results of this theory were obtained by Katsuro Sakai and can be found in Chapter 5 of his book. Let us mention here two results of this theory implying affirmative answers to both (topological) questions posed by Ben Wieland.
Characterization Theorem 5.4.1. A topological space $X$ is is homeomorphic to an open subspace of $\mathbb R^\infty$ if and only if any embedding $f:B\to X$ of a closed subspace $B$ of a finite-dimensional compact metrizable space $A$ can be extended to an embedding $\bar f:A\to X$.
Classification Theorem 5.5.1. Two $\mathbb R^\infty$-manifolds are homeomorphic if and only if they are homotopically equivalent.
In Sections 5.6 and 5.7 of Sakai's book one can find some results on PL-theory of $\mathbb R^\infty$-manifolds.<|endoftext|>
TITLE: Limits of rearranged sequences along ultrafilters
QUESTION [10 upvotes]: Suppose that a bounded sequence of real numbers $s_i$ ($i\in\omega$) has a limit $\alpha$ along some ultrafilter $\mu_1\in \beta{\Bbb N}\setminus{\Bbb N}$. Then given another ultrafilter $\mu_2\in \beta{\Bbb N}\setminus{\Bbb N}$, surely there exists some rearrangement $s_{r(i)}$ of $s_i$ that has the same limit $\alpha$.
One can easily extend this simple observation to a countable family of sequences. 
Now given $s_{i;j}$ ( $i,j\in \omega$; values bounded for each fixed $j$) with limits $\alpha_j$ along a fixed $\mu_1\in \beta{\Bbb N}\setminus{\Bbb N}$, and given another ultrafilter $\mu_2\in \beta{\Bbb N}\setminus{\Bbb N}$, there exists a simultaneous rearrangement $s_{r(i);j}$ having the same limits $\alpha_j$ along $\mu_2$.  
All this fails if we pass to size $c=2^\omega$ families of sequences.  Indeed $s_{i;j}$ could then enumerate all bounded sequences. But all the limits $\alpha_j$ together would determine $\mu_1$.  Taking limits of a simultaneous rearrangement of all the sequences amounts, equivalently, to taking limits of the original sequences along an ultrafilter $\mu_2'$ in the orbit of $\mu_2$ under the action of the symmetric group of $\Bbb N$ extended to $\beta\Bbb N$.  Equality of all those limits thus forces $\mu_1=\mu_2'$, and that places $\mu_1$ and $\mu_2$ in the same orbit of the symmetric group action, a severe restriction on $\mu_2$.
Question:  If CH fails, what happens for a size $\omega_1$ family of sequences?

REPLY [7 votes]: As Joel David Hamkins commented, this is likely to be a cardinal characteristic.  Let me define (I hope temporarily) the relevant characteristic $\mathfrak{dfpmk}$ (named after the OP and the author of the accepted answer) to be the smallest cardinal $\kappa$ such that, for some non-principal ultrafilters $\mathcal U$ and $\mathcal V$ on $\omega$ and some family of $\kappa$ sequences $(s_n^\xi)_{n\in\omega}$ (indexed by $\xi\in\kappa$), the sequences all converge to limits $\lambda^\xi$ with respect to $\mathcal U$ but no simultaneous rearrangements $(s_{r(n)}^\xi)_{n\in\omega}$ all converge to the corresponding $\lambda^\xi$ with respect to $\mathcal V$.  So what Paul McKenney proved is that $\mathfrak p\leq\mathfrak{dfpmk}\leq\mathfrak u$.
Of course, what Joel probably intended is that $\mathfrak{dfpmk}$ might be a known cardinal characteristic.  I can't prove anything like that, but I can improve in some models the upper bound $\mathfrak u$ in Paul's answer.  Namely, if there exists a P-point, then $\mathfrak{dfpmk}\leq\mathfrak d$.  (This is an improvement only if there is a P-point and $\mathfrak d<\mathfrak u$; there are models where that happens, for example, the random real model.)
Before giving the proof, let me indicate an alternative way to view $\mathfrak{dfpmk}$, which will be useful (at least for me) in the proof.  To avoid excess verbiage, assume that all filters in the following extend the cofinite filter on $\omega$, so in particular all ultrafilters are non-principal.)  I claim that $\mathfrak{dfpmk}$ is the smallest number of generators needed for a filter $\mathcal F$ on $\omega$ such that, for some ultrafilter $\mathcal U$, no isomorphic copy of $\mathcal U$ extends $\mathcal F$.  
To prove the equivalence, assume first that $\kappa<\mathfrak{dfpmk}$, let $\mathcal F$ be any filter with a basis $\mathcal B$ of size $\kappa$, and let $\mathcal U$ be any ultrafilter.  Consider the characteristic functions of the sets in $\mathcal B$ and consider any ultrafilter $\mathcal V$ extending $\mathcal F$.  These characteristic functions all converge to 1 with respect to $\mathcal V$, and there are only $\kappa<\mathfrak{dfpmk}$ of them, so some simultaneous rearrangements of them, say by a permutation $r$, all converge to 1 with respect to $\mathcal U$. But that means that the original (not rearranged) characteristic functions converge to 1 with respect to $r^{-1}(\mathcal U)$. That means $r^{-1}(\mathcal U)$ contains all the sets in $\mathcal B$ and therefore extends $\mathcal F$.
For the converse, suppose $\kappa$ is smaller than my proposed alternative view of $\mathfrak{dfpmk}$, and let $\kappa$ sequences $(s_n^\xi)$ converge to limits $\lambda^\xi$ with respect to an ultrafilter $\mathcal U$.This convergence means that $\mathcal U$ contains all the sets $A_{\xi,k}=\{n\in\omega: |s_n^\xi-\lambda^\xi|<2^{-k}\}$ for $\xi<\kappa$ and $k\in\omega$.  These $\kappa$ sets $A_{\xi,k}$ generate a filter $\mathcal F$, and, by choice of $\kappa$, every ultrafilter $\mathcal V$ has an isomorphic copy $r(\mathcal V)$ extending $\mathcal F$.  Then the sequences $(s_n^\xi)$ converge to $\lambda^\xi$ with respect to $r(\mathcal V)$, and therefore the simultaneous rearrangements $(s_{r^{-1}(n)}^\xi)$ converge to $\lambda^\xi$ with respect to $\mathcal V$.
This completes the proof of the equivalence between the two views of $\mathfrak{dfpmk}$.  Now to prove my claim about $\mathfrak d$, consider the filter $\mathcal F$ on the "plane" $\omega\times\omega$ consisting of those sets $X$ such that, for all but finitely many $x\in\omega$, the "column" at abscissa $x$ has all but finitely many of its points in $X$.  That is, for all but finitely many $x$, for all but finitely many $y$, $(x,y)\in X$.  (This filter is often called the tensor square or the Fubini square of the cofinite filter on $\omega$.) This filter is generated by $\mathfrak d$ sets, namely the sets $\{(x,y):x>n\}$ for each $n\in\omega$ and the sets $\{(x,y):y>f(x)\}$ for each $f$ in some dominating family of cardinality $\mathfrak d$.
If an ultrafilter $\mathcal U$ extends $\mathcal F$, then $\mathcal U$ cannot be a P-point, because any set on which the projection to the first factor, $(x,y)\mapsto x$, is finite-to-one or constant is the complement of a set if $\mathcal F$ and therefore cannot be in $\mathcal U$.  Furthermore, since the property of being a P-point is preserved by isomorphism, no P-point can be isomorphic to an extension of $\mathcal F$.  So, as long as there is a P-point, the filter $\mathcal F$ witnesses, in the alternative view of $\mathfrak{dfpmk}$ above, that $\mathfrak{dfpmk}\leq\mathfrak d$.
A very similar argument shows, under the weaker hypothesis that there exists a nowhere dense ultrafilter, that $\mathfrak{dfpmk}\leq\mathfrak{cof}(B)$.  (Here "nowhere dense" is in the sense of Baumgartner's $I$-ultrafilters; it means that the image of $\mathcal U$ under any map $\omega\to\mathbb Q$ contains a nowhere dense set. $\mathcal{cof}(B)$ is the cofinality number for Baire category, the minimum number of sets needed to generate the ideal of meager subsets of $\mathbb R$.)<|endoftext|>
TITLE: Applications of functional analysis beyond analysis(towards algebra, geometry, number theory...)
QUESTION [10 upvotes]: So far, We have seen the applications of functional analysis in PDE, probability and many areas in applied mathematics. On the other hand, methods of algebraic topology are introduced to functional analysis via operator K-theory.
From the mathematical research point of view, (as a crazy fan of pure math) I wonder if functional analysis(theory of Banach and Hilbert spaces and operator algebras), as a well-built theory in mathematics, has any applications to the areas in pure math like algebraic topology, algebraic geometry and number theory, solving problems for which purely algebraic or geometric methods seem to be powerless.

REPLY [5 votes]: Recently a new application of functional analysis in geometry appeared, the study of envelopes of topological algebras. It allows to look at "big" geometric disciplines -- complex geometry, differential geometry, topology -- from the point of view of category theory, so that these disciplines become "purely categorical constructions". This can be considered as a developement of Klein's Erlangen program.
According to this view, different geometric disciplines are just pictures that appear in the imagination of an outlooker after applying different "observation tools" for studying a given category of topological algebras. Formally, this "projection of functional analysis to geometry" is established by a categorical  construction, called envelope (and there are many different envelopes that give different geometries as disciplines).      
This activity allows to build different generalizations of Pontryagin duality to classes of non-commutative groups (including some quantum groups).<|endoftext|>
TITLE: Missing citations of "to appear" papers on MathSciNet
QUESTION [41 upvotes]: Recently, looking at my author profile on MathSciNet (Am I a narcissist?), I saw that my citations counter is lower than what I expected. After a while, I realized what the problem is:
Many of my papers are cited (both by myself alone, by me plus some coauthors, or by other authors) as
"Authors names, Article title, Journal, (to appear)"
because they were accepted but not yet published at the time.
Unfortunately MathSciNet does not count those citations.
Do you know how it is possible to fix this problem?
Have any of you had the same problem?
Is there any official email address/website for suggesting corrections to MR citations? (quoting Sam Hopkins)
I guess it is a matter of contact in some way the MathSciNet's reviewers. I tried MathSciNet "Support Mail", but I guess it is only for browser/connection related problems, in fact I got no answers.
Thank you very much for any suggestion.
EDITs:

I found that when the "to appear" papers are cited as

"Authors names, Article title, Journal (year) (in press)"
then they are correctly handled by MathSciNet citations database. Thus I suggest to do so to everyone who has to cite a "to appear" paper.

This question is "on hold" since somebody thinks it is off-topic. I can agree that it is quite different from the typical MO questions. However, it seems to me that this kind of problem can interest many researchers, so I will wait at least until some satisfactory answer before close it.

REPLY [37 votes]: For citations from a reference list at the end of an article, we try to match items in the list to items already in the Math Reviews database.  If a paper is "to appear" at the time it is cited, there is probably nothing in the database to match it to.  We are currently developing a new matching program, one that should be better at matching papers that were cited as “to appear” but are now in the database.  Once that is finished (and tested), it will replace the current method.  We will also be running it regularly, so that we can find more matches of this type.  
More information about citations in MathSciNet is available online at http://www.ams.org/mathscinet/help/citation_database_help_full.html#matched
If you have questions about the content of MathSciNet, including missed citations, you should write to mathrev@ams.org.  Questions about the functioning of the site, such as problems with connectivity or math not rendering properly, you should write to msn-support@ams.org.  If you are not sure which address to use, just pick one.  The people who read the messages are good at triage.  Or you can write to me: egd@ams.org. 
There is another type of citation in MathSciNet: from the body of a review of another paper.  These can be particularly interesting, because a third party has decided that the two papers are related.<|endoftext|>
TITLE: Generic set that is a proper subgroup
QUESTION [7 upvotes]: For a group $G$ generated by a finite set $S$ we denote by $B_{G,S}(n)$ the ball of radius $n$, that is the set of all elements in $G$ which are expressible as products $x_1x_2\ldots x_n$ where $x_i\in S\cup S^{-1}\cup\{1\}$. One calls the set $Q$ generic in $G$ with respect to $S$ if $$\lim_{n\to\infty}\sup \frac{|Q\cap B_{G,S}(n)|}{|B_{G,S}(n)|}=1.$$
My question is whether there exist a group $G$ and a proper subgroup $H<G$ such that $H$ is a generic subset in $G$ with respect to some finite generating set $S$ of the group $G$.

REPLY [9 votes]: No, it's not possible. For notational convenience assume $S$ is symmetric and $1\in S$, so that $B_{G,S}(n) = S^n$. Suppose
$$
  |H\cap S^{n_i}|/|S^{n_i}|\to 1
$$
for some subsequence $(n_i)$. Let $x$ be an element of $S$ not in $H$. Since $H$ and $Hx$ are disjoint we have
$$
  |H\cap S^{n_i}x^{-1}|/|S^{n_i}| = |Hx\cap S^{n_i}|/|S^{n_i}| \to 0.
$$
Thus in particular
$$
  |S^{n_i} \cap S^{n_i} x^{-1}|/|S^{n_i}| \to 0.
$$
But
$$
  S^{n_i-1} x \subset S^{n_i},
$$
so
$$
  S^{n_i-1} \subset S^{n_i} \cap S^{n_i} x^{-1}.
$$
Also $S^{n_i} \subset S^{n_i-1} S$. Thus
$$
  |S^{n_i} \cap S^{n_i} x^{-1}| / |S^{n_i}| \geq |S^{n_i-1}|/|S^{n_i}| \geq 1/|S| \not\to 0,
$$
a contradiction.
But I suppose you can have subgroups $H$ with $\limsup |H\cap S^n|/|S^n|$ arbitrarily close to $1$? However, we would need to have (1) lots of generators, (2) exponential growth. (If $|S^n|$ has subexponential growth then you have an invariant measure so you can't get bigger than upper density $1/2$.) Something in the free group $F_n$?<|endoftext|>
TITLE: Shafarevich conjecture for abelian varieties
QUESTION [6 upvotes]: In the paper "Arakelov's theorem for abelian varieties" Faltings proves the Shafarevich conjecture for abelian varieties.
The statement is the following: 

Let B be smooth projective a curve, S a finite set in B. There exist
  only finitely many families of principally polarized abelian varieties
  of dimension $g$ over $B$, with good reduction outside S and
  satisfying $(*)$

where $(\ast)$ is some technical condition not important in this post.
My concern is about the ground field $K$ on which the abelian varieties of the theorem are defined. It seems that $K$ should be a number field, but in his paper Faltings says that $K$ is any field (but on the other hand at the beginning he fixes the curve $B$ on an algebraically closed field $k$ of char, $0$). I need a clarification about the ground fields because I'm confused.
What happens when $K=\mathbb C$? Here there is no notion of good reduction since we don't have discrete valuations. Is still the theorem true? Does in this case "good reduction outside S" mean that the maps $f:A\to B$ attain non-singular values outside $S$, like in the classical one dimensional Shafarevich conjecture?

REPLY [11 votes]: Let $B$ be a smooth projective curve over an algebraically closed field of characteristic zero. Let $K$ be the function field of $B$. Let $S$ be a finite set of closed points of $B$. 
You might find the following reformulation of Faltings's theorem less confusing.
Theorem 1. (Faltings, geometric Shafarevich conjecture)
Let $g$ be an integer. Then the set of $K$-isomorphism classes of $g$-dimensional principally polarized abelian varieties $A$ over $K$ with good reduction over $B-S$  and satisfying condition $(*)$ is finite.
If you're used to thinking about number fields, then you might be interested in reading Faltings's papers on the arithmetic analogue of the aforementioned finiteness result:
Theorem 2. (Faltings, arithmetic Shafarevich conjecture) Let $K$ be a number field and let $S$ be a finite set of closed points of $B = $ Spec $O_{K}[S^{-1}]$.
Let $g$ be an  integer. Then the set of $K$-isomorphism classes of $g$-dimensional principally polarized abelian varieties over $K$ with good reduction over $B-S$ is finite.
Remark 1. You can get rid of the words "principally polarized"   via Zarhin's trick. See Szpiro's 1985 asterisque on the Mordell conjecture.
Remark 2. The reason you need condition *
 in Theorem 1 (i.e., the geometric Shafarevich conjecture) is because of the existence of non-rigid families of abelian varieties. This is an artifact of the function field realm which does not appear in the number field realm. If you don't want to use assumption * the best you can hope for is finiteness of the set of $K$-deformation classes (and not $K$-isomorphism classes).
Remark 3. Shafarevich didn't actually conjecture the above finiteness statements, as far as I know. He conjectured the analogous statements for curves of genus at least two (proven by Arakelov-Parshin in the function field case, and Faltings in the number field case).<|endoftext|>
TITLE: How many primes can there be in a short interval?
QUESTION [19 upvotes]: Given $n \in \mathbb{N}$, let $\pi(n)$ denote the number of prime numbers $\leq n$.
What is
$$
  \limsup_{m \rightarrow \infty} \left( \limsup_{n \rightarrow \infty} \frac{\pi(n+m) - \pi(n)}{\pi(m)} \right)?
$$
If needed, answers may be conditional under the assumption that a suitable generalization
of the Bunyakovsky conjecture holds.

REPLY [36 votes]: As observed by Hensley and Richards in
Douglas Hensley and Ian Richards, Primes in intervals, Acta Arith. 25 (1973-74), 375--391, 
if the prime tuples conjecture is true, then $\limsup_{n \to \infty} \pi(n+m) - \pi(n) = \rho^*(m)$, where $\rho^*(m)$ is the length $k$ of the largest admissible $k$-tuple of diameter less than $m$ (a $k$-tuple is admissible if it avoids a residue class modulo $p$ for every prime $p$). So your quantity would then be equal to $\limsup_{m \to \infty} \rho^*(m)/\pi(m)$, which is also easily seen to be equal to $\limsup_{k \to\infty} k \log k / H(k)$, where $H(k)$ is the minimal diameter of an admissible $k$-tuple.
The best known upper and lower bounds for $H(k)$ can be found in Section 3 of the (unabridged version of) the Polymath8a paper.  The best asymptotic upper bound known is $k \log k + k \log\log k - (1+2 \log 2) k + o(k)$ (due to Schinzel), and the best asymptotic lower bound is a bit tricky to state (coming from the sharp Montgomery-Vaughan version of the large sieve) but is roughly of the form $\frac{1}{2} k \log k - o(k \log k )$.  So the best that is known about your quantity is (as was pointed out in comments) it is somewhere between $1$ and $2$.  (One does not need the best results on $H(k)$ for these bounds: the lower bound comes from observing that the first $k$ primes larger than $k$ are automatically admissible, while the upper bound follows from (the proof of) the Brun-Titchmarsh inequality.)  Based on the known numerical values and bounds for $H(k)$ (see this online database, as well as Figures 2 and 3 of the Polymath8a paper referenced above) one can tentatively conjecture that your quantity is equal to $1$, but the evidence is not terribly strong for this conjecture yet.  Improving the upper bound of $2$ would be a major breakthrough in sieve theory (closely related to the parity problem), and is one of the motivations of the nascent field of "inverse sieve theory", discussed in
Ben Green and Adam J. Harper, Inverse questions for the large sieve, Geom. Funct. Anal. 24 (2014), no. 4, 1167--1203.
The upper bound of $2$ is unconditional (following directly from Brun-Titchmarsh).  No non-trivial unconditional lower bound (beyond the trivial bound of $0$) is known; the best result is by Maynard in
James Maynard, Small gaps between primes, Ann. of Math. (2) 181 (2015), no. 1, 383--413 
with slightly better implied constants obtained by the Polymath8b project
D. H. J. Polymath, Variants of the Selberg sieve, and bounded intervals containing many primes, Res. Math. Sci. 1 (2014), Art. 12, 83,
basically showing that $\limsup_{n \to \infty} \pi(n+m) - \pi(n) \gg \log m$.  This is known to be the limit of the existing Selberg sieve method to unconditionally produce many primes in short intervals; getting beyond the $\log m$ bound here would be a significant breakthrough (though I think breaking the parity upper bound of $2$ mentioned earlier would be even more revolutionary).
Note that until the famous recent breakthrough of Zhang in
Yitang Zhang, Bounded gaps between primes, Ann. of Math. (2) 179 (2014), no. 3, 1121--1174.
it was not even known unconditionally that $\limsup_{n \to \infty} \pi(n+m) - \pi(n)$ was larger than $1$ for any $m$.
EDIT: Ben Green pointed out to me an observation of Selberg in Vol. I of his "Collected works" (page 610 in the new edition) that one cannot obtain effective improvements to the $1/2$ constant in the lower bound $H(k) \geq \frac{1}{2} k \log k - o(k \log k)$ without also getting effective bounds in Siegel's theorem, since a Siegel zero implies the existence of a medium-sized arithmetic progression with about twice as many primes as one would expect without the zero.  This leads to an admissible k-tuple that is also about twice as dense as expected.  So the upper bound of 2 in the original question is unlikely to be improved without some sort of breakthrough on the Siegel zero problem (or else some massive extension of the repulsion phenomenon).<|endoftext|>
TITLE: Intuition behind the definition of finite correspondences
QUESTION [7 upvotes]: Finite correspondences were introduced by Suslin-Voevodsky (if I am not wrong) to define motivic complexes that compute motivic cohomology. Let $X$ and be smooth separated schemes of finite type over a field $k$.  An elementary finite correspondence from $X$ to $Y$ is defined to be an irreducible closed subset $W$ of $X\times Y$ whose associated integral subscheme $\widetilde{W}$ is finite and surjective over $X$. The group $Cor_k(X,Y)$ of finite correspondences is the free abelian group on the elementary finite correspondences. My question is:

Why does one puts the conditions of being finite and surjective on the morphism $\widetilde{W}\to X$? 

Given elementary finite correspondences $V$ and $W$ from $X$ to $Y$ and $Y$ to $Z$ respectively, it is shown in Mazza-Voevodsky-Weibel Lemma 1.7 that $V\times Z$ and $X\times W$ intersect properly and also $\widetilde{V}\times_Y\widetilde{W}$ is finite and surjective over $X$, so that composition of correspondences can be defined. 

Is this statement true in the reverse direction? Is this the only reason fro the choice of the definition of finite correspondences?

REPLY [12 votes]: Traditionally correspondences were defined simply as cycles on the product, but then you need a moving lemma just to define composition. This limits you to working on smooth varieties. The beauty of finite correspondences is that composition can be defined much more directly, essentially as a composition of multivalued functions (cf Denis Nardin's comment). So by limiting the kind of correspondences you allow, you increase the range applicability.<|endoftext|>
TITLE: A question regarding lines on a cubic surface
QUESTION [5 upvotes]: Let $X$ be a smooth cubic surface in $\mathbb{P}^3$. It is a classical theorem of Cayley and Salmon that $X$ contains exactly 27 lines over an algebraically closed field. 
In 2002, Heath-Brown proved in his paper "The density of rational points on curves and surfaces" that if $F$ is a binary form of degree $d \geq 3$ which is not a perfect $d$th power, then the only lines on $X: F(x_1, x_2) = F(x_3, x_4)$ correspond to $F(x_1, x_3) = F(x_3, x_4) = 0$, i.e., the line corresponds to a pair of roots of $F$, or there exists $a_1, a_2, a_3, a_4$ such that $F(x_1, x_2) = F(a_1 x_1 + a_2 x_2, a_3 x_1 + a_4 x_2)$, i.e., the line corresponds to an automorphism of the binary form $F$.
Now let $F$ be a smooth cubic form (i.e., its discriminant is non-zero). Then the surface $X : F(x_1, x_2) - F(x_3, x_4) = 0$ is a smooth cubic surface, hence it should have 27 lines on it. However, I count 9 lines coming from roots ($3^2$ many) and $6$ coming from automorphisms of $F$ (the automorphism group of $F$ in $\operatorname{GL}_2(\mathbb{C})$ is always isomorphic to $D_3$, the dihedral group on 3 letters). This only gives 15 lines, so there are 12 missing lines. By Heath-Brown's theorem, the only way to account for the discrepancy seems to be to count certain lines with multiplicity. I am not sure which are the ones that need to be counted with multiplicity. 
Any help would be much appreciated.

REPLY [2 votes]: To each automorphism of the binary form there correspond actually $d$ lines, not only one. In case $d = 3$, since there are always 6 automorphisms, we get $6\cdot 3 = 18$ lines, which summed to the 9 lines coming from roots give a total of 27.
This is very well explained in this article (Theorem 3.1):

S. Boissière and A. Sarti, Counting lines on surfaces, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (5) Vol. VI(2007), 39-52.

Curious fact: the 27 lines on a smooth cubic are indeed always distinct. If one allows the surface to have isolated rational double points, then the number is always strictly smaller, but there is a "natural" way to assign a multiplicity to each line such that the sum of the multiplicities is still equal to 27, see the following article:

J. W. Bruce and C. T. C. Wall, On the classification of cubic surfaces, J. London Math. Soc. (1979) s2-19 (2): 245-256.<|endoftext|>
TITLE: Finite morphism from a smooth projective curve.
QUESTION [8 upvotes]: Let $k$ be an algebraically closed field and $C$ be a smooth projective curve over $k$. Let $p$ be a prime number. Does there exists a finite morphism $f : C \to \mathbb{P}^1$ such that the degree of the Galois closure of the field extension $k(\mathbb{P}^1) \to k(C)$ is coprime to $p$ ?

REPLY [8 votes]: Following Jason Starr's suggestion, let $C$ be a very general genus $2$ curve, forming a covering of degree $n$ of $\mathbb P^1$. Without loss of generality there is no intermediate curve between $C$ or $\mathbb P^1$ (it would have to have genus $0$ or $1$. In the first case, we may simplify by replacing $\mathbb P^1$ with that curve, and in the second case we get an elliptic factor in the Jacobian, contradicting "very general".)
Let $G \subseteq S_n$ be the Galois group of the covering. As Jason pointed out, it is solvable. It is also transitive and primitive. Because $G$ is solvable, it has "soluble socle", hence $n=p^d$ is a prime power and $G$ acts as a subgroup of the group of affine transformations of $\mathbb F_{p^d}$ by a fragment of the O'Nan-Scott theorem (The argument goes that because $G$ is solvable, it has some normal subgroup $J$ isomorphic to $\mathbb F_p^d$ for some $d$. $J$ must act transitively, or else the action would not be primitive - the cover would factor through the cover defined y the orbits of $J$. Hence the $n$ points acted on are isomorphic to $\mathbb F_p^d$, and $G$ acts by transformations that normalize the group of translations, i.e. affine transformations).
Now the local monodromy around each ramification point is an affine transformation, hence has at most $p^{d-1}$ fixed points. Because the ramification at the non-fixed points has order at least $3$, the local contribution to Riemann-Hurwitz is at least $(2/3) (p^{d} - p^{d-1})$. Because the curve is very general, the number of ramification points is at least $6$, so that after moving the first three to $0,1,\infty$ we still have three parameters. So the total contribution to Riemann-Hurwitz is at least $4 (p^{d}- p^{d-1})$ and we obtain the inequality
$$-2 = 2-2g  \leq 2 p^d - 4 (p^{-d}  - p^{d-1}) = 4 p^{d-1} -2 p^d $$
$$\left(1-\frac{2}{p} \right) p^d \leq 1 $$
$p$ is at least $3$ so $1-\frac{2}{p}$ is at least $1/3$ so $p^d$ is at most $3$ which is an obvious contradiction as for $3$-coverings every point has full ramification and then we obtain $2-2g \leq 6 - 6 \cdot 2 = - 6$.
Because this is a contradiction, a very general genus $2$ curve cannot be so represented.<|endoftext|>
TITLE: Homotopy fiber of a map between classifying spaces
QUESTION [16 upvotes]: I'm looking for a reference (and precise hypothesis if more are needed) for the following facts (or a correction, if I'm just plain wrong):
Let $G$ and $H$ be topological groups and $f : G \to H$ be a continuous homomorphism. Applying the classifying space functor gives a map $Bf: BG \to BH$.

The homotopy fiber of $Bf$ is the homotopy orbit space $H_{hG}$
 where $G$ acts on $H$ via $g \cdot h = f(g)h$.
The canonical map $H_{hG} \to H/G = H/\mathrm{im} f$ has homotopy fiber given by $BK$ where $K = \ker f$.

Maybe more well-known that those are the corollaries for when $f$ is either injective or surjective:

If $G$ is a subgroup of $H$, the homotopy fiber of the map induced by the inclusion $BG \to BH$ is the coset space $H/G$.
If $1 \to K \to G \to H \to 1$ is a short exact sequence of topological groups, applying the classifying space functor gives a fiber sequence $BK \to BG \to BH$.

EDIT: I sketched a proof here. I'm mostly looking for references, but would also appreciate alternate proofs (specially a proof that avoids using the generalized Mather cube property on a diagram whose shape is not a 1-category).
EDIT 2: Tyler Lawson's nice example  shows that more hypothesis are needed for part 2 of the "fact" and for corollary 1. My current guess is that for corollary 1, it is enough that $H \to H/G$ locally have a section.

REPLY [4 votes]: One source for some of this is section 8 of May's "Classifying spaces and fibrations"
He has G and H interchanged, unfortunately uses a coset notation for what turns out to be the homotopy fibre, and doesn't talk about the special cases you are after (his interest was more in the topological monoid case).
May is using the two-sided bar construction.
Your sketch doesn't say very much about the topologies involved, and I did wonder if, for example, you need $im f$ closed for your coset space identification.<|endoftext|>
TITLE: Famous results about the value of a given limit assuming it exists
QUESTION [8 upvotes]: Chebyshev got famous showing that if the limit $l:=\lim_{x\to\infty}\frac{\pi(x)}{x/\log x}$ exists, then necessarily $l=1$, constituting a major breakthrough towards a proof of the famous prime number theorem conjectured by Gauss and Legendre. What I would like to know is whether other famous similar results are known both inside and outside the realm of number theory, and if general probability theorems can be used to obtain such results (like mimicking a deterministic quantity by a random variable that follows a given distribution law, the sum of expected values of which converges almost surely to the desired value of the considered limit, or the like).
Many thanks in advance.

REPLY [4 votes]: The story with sharp thresholds for random constraint satisfaction problems somewhat fits into this picture.  In the random k-SAT problem with $n$ Boolean variables, one includes each of the $2^k \binom{n}{k}$ potential clauses independently with probability $p$, where $p$ is chosen so that the expected clause density (number of clauses over number of variables) is some fixed constant $\alpha$.  
It is very widely believed that for each $k$ there is a critical density $\alpha_k$ such that for all $\epsilon > 0$ the following holds:

if $\alpha \leq (1-\epsilon)\alpha_k$ then the resulting formula is satisfiable with probability $1-o(1)$;
if $\alpha \geq (1+\epsilon)\alpha_k$ then the resulting formula is unsatisfiable with probability $1-o(1)$.

However in general the existence of this $\alpha_k$ is unknown.  Friedgut's Theorem [Fri99] comes extremely close to showing that $\alpha_k$ exists. His result implies that for each $k$ the above statement is true for a sequence of real numbers $\alpha_k(n)$ depending on the number of variables. (It is also easy to show that $\alpha_k(n)$ is bounded in the range $[1,2^k \ln 2]$ for all $n$.)  It's utterly inconceivable that this sequence could oscillate, rather than tend to a limit, but in general this hasn't been proven.
For each $k$, the value of $\alpha_k$ is "known", via sophisticated heuristics from statistical physics [MPZ02, MMZ06].  In this sense, we sort of have an example answering Sylvain's question.
In a recent major breakthrough, Ding, Sly, and Sun rigorously established the physics prediction for $\alpha_k$ for all sufficiently large $k$.  It is worth mentioning that this does not obviate the need for Friedgut's Theorem; by virtue of that theorem, it was enough for Ding--Sly--Sun to show that 

if $\alpha < \alpha_k$, the random k-SAT formula is satisfiable with probability bounded away from 0.

[Fri99] Ehud Friedgut, Sharp thresholds of graph properties, and the $k$-sat problem, J. Amer. Math. Soc. 12 (1999), no. 4, 1017--1054.
[MMZ06] Stephan Mertens, Marc Mézard, and Riccardo Zecchina, Threshold values of random $K$-SAT from the cavity method, Random Structures Algorithms 28 (2006), no. 3, 340--373.<|endoftext|>
TITLE: How do you mentor undergraduate research?
QUESTION [48 upvotes]: Lets say you had an undergraduate who wanted to do some advanced work and some research, possibly for a thesis, or things like that. 
There are two slightly more specific groups of questions I have about this process:

How would you go about choosing a problem? Specifically, should the student work on open problems or work through existing proofs? Are there lists of problems at that might be fruitful and approachable? 
What sort of guidance would you provide them? With what frequency would you meet? Would your meetings be closer to teaching or guiding them along on their own?

REPLY [11 votes]: To give a viewpoint from the other side, I am an undergraduate B.S. student in Engineering, and spent the last year on a research in the Biomedical Engineering area, in Brazil.
It was based on a hot-topic of the area, and my part on the research was to verify the usability of a method invented by my mentor. So, to answer your first question, working on open-problems is very interesting to a student, as it gives you the opportunity to do something no one has ever done before, even if it gives no relevant results.
As it was a research in which I could do most work at home (programming, reading articles, etc.), my mentor and I would encounter only once a week. Each week my mentor would give me some guidance as "what to do this week", and see what I have done in the past week. We would also discuss through e-mail, when necessary. This would make sure that I was progressing in the research, and if I was not, we would discuss what should we do as an alternative.
So, answering your second question with my own experience, it was very important a regular meeting, with a goal between two of them, so that it would progress on a regular basis. More than once in a week probably would result in many goals not achieved, (which is demotivating) and less than once a week would be a very slow-paced research, with less guidance (also demotivating). So once a week worked really fine for us.
Also, you should always ask the student what he thinks could be done to solve the next step of the problem. If you don't agree with him, try to make him explain why it is a good idea. If you're still not convinced, say to them why you think that won't work, and maybe give another way to proceed the research (stating things like "I strongly suggest you do this next, instead, because of (...)"), but let him choose what should be done next, as it's his research.
Also, leave all the learning to the student. He is learning how to make research, not how to do something. You could recommend some textbooks and/or papers to read of the related theme, but it is not your job to tell him what the books/papers states. 
Also, it is a good idea to ask the student to give an presentation of what the papers are stating, on some of the meetings. This way, he would also be learning how to learn, instead of just learning a tool to make progress with his research. This will be very important when he starts his Master's.
This was how I was mentored. This made the research both interesting and fulfilling to me, as an scholar, and resulted in a published paper on the biggest international conference in Biomedical Engineering.
Just to be clear, again, I'm only stating my own experience as a mentored undergraduate student, that worked well for both me and my mentor. I hope that helps!<|endoftext|>
TITLE: Maximum size of minimal sequence of transpositions whose product is a given permutation
QUESTION [5 upvotes]: Consider the sequence $S = 1,2,3,\ldots n$ of elements, along with a sequence $T = t_1, t_2, \ldots, t_m$ of transpositions. Each transposition $t_i$ is a tuple $(a_i, b_i) \in [n]^2$. When applying a transposition $t_i$ to (a permutation of) $S$, it exchanges the elements at positions $a_i$ and $b_i$ in the sequence.
Applying all transpositions $t_1, \ldots, t_m$ one by one on $S$ yields a final permutation $R$. Call a transposition sequence $T$ minimal if there is no proper subsequence $T'$ of $T$ whose application to $S$ would result in the same permutation $R$. My question is the following: what is the maximum length of a minimal transposition sequence, in terms of $n$? It is clear that this maximum is at most $n!$: if we have a sequence with more than $n!$ transpositions, then when applying the transpositions in $T$ there will be 2 points at which we have obtained the same intermediate permutation, and the transpositions in between can be removed from the sequence without changing the final outcome. What I am wondering is the following: can there be a polynomial upper bound on the length of a minimal transposition sequence? If not, then what does a minimal transposition sequence of superpolynomial length look like?
Edit: Apologies for my elementary exposition of the question; I'm a computer scientist and relatively unfamiliar with the literature in group theory. Let me clarify some points: my use of subsequence follows that of Wikipedia. The sequence of transpositions $T$ will, in general, contain repeated transpositions, otherwise a polynomial ($n^2$) bound on the length of any such sequence would be trivial. I do not think my question can be rephrased in terms of properties of the Cayley graph with a given set of transpositions as generators of the symmetric group, because I care for the order in which the transpositions are applied: in making a given sequence $T$ minimal you are allowed to remove transpositions from the sequence, but not to re-order them.

REPLY [2 votes]: Thank you for clarifying the question.  (I would edit further to indicate subsequence is different from substring or subword, so not necessarily a contiguous portion of the given sequence.)  The short answer is I don't know, but I suspect the length is not superquadratic for reasons that will appear.
Pick a sequence T of transpositions on the n letters, set S to the singleton set of the identity permutation and i to 1. Now compute for each  uncolored (non-green) permutation s in S the permutation s' = st_i.  If s' is already in S, mark it green, otherwise add uncolored s' to S.  Then increment i. Check if all members of S are green.  If so stop, else loop back to 'Now'.Edit 2016.04.07: Thanks to Ilya Bogdanov, I should also compute s'=st_i for green s as well, and then color those s' instantly as I put them in S; I might get a smaller maximal value of i.  To address the concerns of the poster, this algorithm should be run for enough sequences T to determine the largest value of i given n.  Also, Ilya's suggestion of looking at inversion count using sequences of adjacent transpositions has merit, as it might lead to a quadratic in n lower bound on maximal i.  However, more rigor is needed to ensure minimality of a long enough sequence of adjacent transpositions.  End Edit 2016.04.07.
This will give you an algorithm to determine the longest initial substring of T which is minimal.  Once the algorithm stops, i is too big.  As S has the potential to be 2^i in size, and we stop after duplicating at most n! permutations in S, my gut says i is bounded by cnlog n and definitely by n^2.  However, this is not a proof, as there may be some holdouts.  Note though that if the initial segment duplicates a subset of permutations, then any sequence including that initial segment as a substring is not minimal.
For small values of n, it should be only mildly onerous to compute the maximal length of such a minimal sequence.  I predict that 4n^2 is a weak upper bound on i for all n.  If you choose substring instead of subsequence, I think the growth might be superpolynomial, and you can write a similar algorithm above to test it.
Gerhard "Needs Practice In Formatting Pseudocode" Paseman, 2016.04.06.<|endoftext|>
TITLE: Moment problem on [-1,1]: necessary and sufficient conditions
QUESTION [6 upvotes]: Consider a sequence of real numbers $s=(s_0,s_1,\ldots)$. When is there a Borel measure $\mu$ supported on $[-1,1]$ so that
$$
s_k = \int_{[-1,1]} x^k\,\mathrm{d}\mu,\quad \forall k\in\mathbb N\;?
$$
The well known Hausdorff moment problem asks the same question on the interval $[0,1]$. In that case, $s$ is a moment sequence of a measure supported on $[0,1]$ if and only if $s$ is completely monotone. It follows that, for our problem, the even terms of $s$ must be completely monotone. What is known about the odd terms?

Question: Are the necessary and sufficient conditions for $s$ to be a moment sequence on $[-1,1]$ known? What are the simplest known necessary conditions?

The following question maybe easier to solve.

Assume now that $s$ is a sequence of nonnegative reals. When is there measure supported on $[-1,1]$ with the moment sequence $s$?

There is a well known way of obtaining necessary conditions: Let $P(x) = \sum a_i x^i$ be a polynomial that is nonnegative on $[-1,1]$. Form the Hankel matrix $H_P$
$$
H_P(j,k):= \sum_i a_i s_{i+j+k},\quad j,k\in \mathbb N.
$$
Then we observe
Claim: Suppose $P(x)\ge 0$ on $[-1,1]$. Then $H_P$ is positive semi-definite.
Pf. Let $c_0,c_1,\ldots$ be a complex sequence with finitely many nonzero elements. We have
\begin{align*}
cH_Pc^* &= \sum_{j,k}c_j\bar{c_k}\int \sum_i a_i x^{i+j+k}\,\mathrm{d}\mu\\
&=\int_{[-1,1]} P(x)\left|\sum_j c_j x^j\right|^2\,\mathrm{d}\mu \ge 0.
\end{align*}
A result due to Riesz implies that if $H_P$ is PSD for all polynomials $P$ that are nonnegative on $[-1,1]$ then $s$ is a moment sequence. However if the interval is $[0,1]$ this PSD criterion is equivalent to a much much simpler condition: that the sequence is completely monotone. Is there a simpler characterization for $[-1,1]$?

REPLY [3 votes]: The following paper answers a more general version of this question, namely the Hausdorff moment problem for compact sets in $\mathbb{R}^n$.
This is a fairly broad topic, and you may enjoy learning more by looking at these notes or this really nice book.<|endoftext|>
TITLE: Cofinality of countable ordinals in ZF, and in toposes
QUESTION [5 upvotes]: A countable limit ordinal $\kappa$ has cofinality $\omega$.  One proves this in ZF, say, using the usual trick for representing $\kappa$ as a countable set of reals having closed convex span $[0,1]$ (with the usual order) and then comparing with any increasing sequence in $[0,1]$ converging to 1.
Nevertheless, I suspect independence from ZF for the following uniform version of this claim: 
There exists a function $f:\omega \times \omega_1 \rightarrow \omega_1$ such that
1) $f(\alpha,\beta) < \beta$;
2) ${\rm sup}_\alpha f(\alpha,\beta) =\beta $ for $\beta$ a limit ordinal.
(For fixed $\beta$ assume that $f(\alpha,\beta)$ increases with $\alpha$, if you like.)
Briefly, such an $f$ would support, by induction and coding tricks, the construction of an injection from $\omega_1$ to ${\Bbb R}$; that in turn would mean that CH implies the existence of a well-ordering of the reals.  (Details on demand.)
Questions:
1) Does this independence come up in the literature?
2) Can someone point me to a model of ZF+CH where the reals have no well-ordering?
3) Is it easy to get directly a model of ZF having no such $f$ by forcing?
4) Is the existence of $f$ equivalent to any well-known consequences of AC?
5) Are there toposes where even the original cofinality statement (on some reasonable interpretation) fails (for lack, say, of a global bijection between $\kappa$ and the natural number object)?

REPLY [9 votes]: Yes, this comes up. For example, Raisonnier showed that the existence of an injection $\omega_1 \to \mathbb{R}$ implies the existence of a subset of $\mathbb{R}$ which is not Lebesgue measurable over ZF + DC.
Jean Raisonnier, A mathematical proof of S. Shelah’s theorem on the measure problem and related results, Israel J. Math. 48 (1984), no. 1, 48--56.
Important examples of models where this property fails are:

The Feferman-Levy model where $\mathbb{R}$ is a countable union of countable sets. [See Theorem 10.6 in Thomas J. Jech, The Axiom of Choice, Studies in Logic and the Foundations of Mathematics, Vol. 75. North-Holland Publishing Co., Amsterdam-London; Amercan Elsevier Publishing Co., Inc., New York, 1973. xi+202 pp.]
The Solovay model where every subset of $\mathbb{R}$ is Lebesgue measurable. [Robert M. Solovay, A model of set-theory in which every set of reals is Lebesgue measurable, Ann. of Math. (2) 92 (1970), 1--56.]

Both of these model constructions use an inaccessible cardinal. This is necessary for if $\omega_1$ is not inaccessible in $L$ then there is a real $x$ such that $\omega_1^{L[x]} = \omega_1$ and then the required function can be constructed in $L[x]$ (which always satisfies AC).
Raisonnier's Theorem in a topos-theoretic context was discussed my answer this MathOverflow question:
Alex Simpson, How strong is “all sets are Lebesgue Measurable” in weaker contexts than ZF?<|endoftext|>
TITLE: Applications of Level Lowering
QUESTION [5 upvotes]: What are some applications/consequences of level lowering of Galois representations? I understand the application of Ribet's theorem in the proof of Fermat's last theorem but I am wondering what other kinds of results can follow from knowing a given modular representation, of level $Np$, also arises as a modular representation of level $N$. Are there other well known results whose proof uses such a result?

REPLY [4 votes]: First, a small clarification: level-lowering tells you that a modular representation in level $Np$ occurs in level $N$ only if by occurs you mean "is congruent to modulo $p$".
That said, my answer to your question is essentially trivial: level-lowering is useful every time you have a property $(P)$ which is known to satisfy
$$P(f)\textrm{ true}\Rightarrow P(g)\textrm{ true}$$ 
provided $f\equiv g\mod p$ and its usefulness is in reducing the general case to the minimally ramified case (that is to the case where $\rho_f$ is no more ramified than $\bar{\rho}_f$).
For a concrete example (for the property $P$ "The Iwasawa Main Conjecture is true") see (among many other) Variation of Iwasawa invariants in Hida families (Invent. math. 163) by M.Emerton, R.Pollack and T.Weston.<|endoftext|>
TITLE: Identity types: What makes Intuitionistic Type Theory *intuitionistic*?
QUESTION [7 upvotes]: In the opening passage of Martin-Löf's (1975) he famously says that

"the theory of types with which we shall be concerned is intended to be a full scale system for formalizing intuitionistic mathematics".

Although we are dealing with type theory and not logic, the proposition-as-types paradigm teaches us how the introduction and elimination rules for, say, product types or function types, can be justified by the BHK informal semantics - in this case, conjuncion and implication, respectively.
However, what about the identity type - the sugar of Martin-Löf Type Theory (MTT)?
Specifically, I am concerned about the intuitionistic justification of the elimination rule of the identity type in the intensional version of MTT. The point is that, as far as I am concern, the BHK interpretation is limited to explain the meaning of conjunction, disjunction, implication, contradiction, existential and universal quantifiers as logical constants, but it has absolutely nothing to say about identity. (I wonder if Brouwer et al have held any position in respect to identity?)
Question:

Can we intuitionistically justify the elimination rule of the identity type? 

My initial guess is that the rule can be substantiated via the meaning explanations, but I am not sure in which extent it characterizes the intuitionistic position: it is undoubtedly a constructive position though, more like in Bishop's sense.

REPLY [6 votes]: You are right to note that Martin-Löf's treatment of identity as a type is an extension of the BHK-interpretation as originally presented by H(eyting) and K(olmogorov/reisel). 
Your guess is also right that Id-elimination can be justified on the basis of Martin-Löf's meaning explanations for his type theory. First it must be clarified what it means to justify this rule. 
$$\begin{array}{l}
p:\mathrm{Id}(A,a,b) \\
z:A\vdash c:C[z,z,\mathrm{refl}(A,z)]\\
\hline
J(p,z.c):C[a,b,p]
\end{array}$$
In order to justify the rule one must make the conclusion evident on the assumption that one knows the premisses. To make the conclusion evident requires, according to the meaning-explanations, to make it evident that $J(p, z.c)$ evaluates to a canonical object in $C[a,b,p]$. 
Assuming that we know $p:\mathrm{Id}(A,a,b)$, we know, by the explanation of the form of judgement $a:A$ that $p$ evaluates to a canonical element of $\mathrm{Id}(A,a,b)$. We stipulate, in general, that a canonical element of type $\mathrm{Id}(A,a,b)$ has the form
$$\mathrm{refl}(A', a')$$
where 
$$A=A':type\\
a = a' : A \\
b = a' : A $$
Hence we know 
$$p=\mathrm{refl}(A', a')=\mathrm{refl}(A,a):\mathrm{Id}(A,a,b)$$
Therefore
$$J(p, z.c)=J(\mathrm{refl}(A,a), z.c)=c[a]:C[a,a,\mathrm{refl}(A,a)]$$
where the final equality follows from the (definitional) rule of Id-equality. 
By the (judgemental) equalities already stated we have
$$C[a,a,\mathrm{refl}(A,a)]=C[a,b,p]:type$$
By the premiss $z:A\vdash c:C[z,z,\mathrm{refl}(A,z)]$ and the meaning explanation for hypothetical judgements, we therefore know that $c[a]$ evaluates to a canonical object in $C[a,b,p]$.
It follows then from the meaning explanation for the form of judgement $a=b:A$ that $J(p, z.c)$ also evaluates to such a canonical object.
For more details perhaps I can refer to a paper of mine, forthcoming (at the time of writing) in Topoi, The justification of identity elimination in Martin-Löf's type theory.<|endoftext|>
TITLE: Is the analytification functor part of a geometric morphism of topoi?
QUESTION [9 upvotes]: Let $Sh(\mathsf{\mathbb{C}-fAlg}^{op})$ be the topos of zariski sheaves on finitely genertaed $\mathbb{C}$-algebras. 
A complex analytic space for our purpose is a locally ringed space locally isomorphic to an analytic subset of $\mathbb{C}^{n}$ (with the sheaf of holomorphic functions). Denote the category of these by $\mathsf{An}$ (for analytic). There's a grothendiek topology on the category given by jointly sujective open immersions (where open immersion are topological embeddings with locally isomorphic structure sheaves). Let $Sh(An)$ be the topos of analytic sheaves on analytic spaces. We have an affine analytification functor:
$$f: \mathsf{\mathbb{C}-fAlg}^{op} \to \mathsf{An}$$
Which associates to an affine $\mathbb{C}$ scheme the corresponding analytic subspace of $\mathbb{C}^n$. This morphism induces a pushforward for presheaves:
$$f_*: Psh(\mathsf{An}) \to Psh(\mathsf{\mathbb{C}-fAlg}^{op})$$
I'm afraid to say anything about sheaves at this point since I'm not so comfortable with toposes yet really. What I'd like to be able say is the following: 

(Extremely conjectural!) The pushforward $f_*$ extends to a morphism of topoi:
$$F_*: Sh(\mathsf{An}) \to Sh(\mathsf{\mathbb{C}-fAlg}^{op})$$
This morphism is geometric with left adjoint the analytification
  functor:
$$F^*:  Sh(\mathsf{\mathbb{C}-fAlg}^{op}) \to Sh(\mathsf{An})$$
In partuicular this functor sends schemes in
  $Sh(\mathsf{\mathbb{C}-fAlg}^{op})$ to analytic spaces.

Is there a way to make this precise?
I'm trying to see the analytification functor come out of general nonsense because it seems like the existence of it should be natural.
I'm obviously not talking about any powerful statements about properties of this functor like GAGA etc. I'm only looking for a natural way to define this functor.

REPLY [7 votes]: I am not very familiar with the analytic side of the pictures or with the analytification functor but here is what I can claim, it seems from your comment that this answer your question:
If you have two subcanonical site $C$ and $D$ (it means that representable presheaves are sheaves, so it is the case with your example).
If $C$ has all finite limits.
If you have a functor $f :C \rightarrow D$ that preserves finite limits and send covering family in $C$ to covering family in $D$ then:
the precompostion with $f : Prsh(D) \rightarrow Prsh(C)$ send sheaves over $D$ to sheaves over $C$ and the restriction of this functor to sheaves defines the $f_*$ part of a geometric morphism:
$f_*: sh(D) \rightarrow sh(C)$
Moreover (easy to check from the universal property) $f^*$ send representable sheaves in $sh(C)$ to representable sheaves in $sh(D)$ in a way that extend $f: C \rightarrow D$.
You have more general claim of this sort on the nLab page morphism of sites
Or in more standard litterature, have a look to MacLane&Moerdijk "sheaves in geometry and logic" chapter VII, more specifically setion 9 and 10.<|endoftext|>
TITLE: Integer solutions of (x+1)(xy+1)=z^3
QUESTION [6 upvotes]: Consider the equation
$$(x+1)(xy+1)=z^3,$$
where $x,y$ and $z$ are positive integers with $x$ and $y$ both at least $2$ (and so $z$ is necessarily at least $3$). For every $z\geq 3$, there exists the solution
$$x=z-1 \quad \text{and} \quad y=z+1.$$
My question is, if one imposes the constraint that
$$x^{1/2} \leq y \leq x^2 \leq y^4,$$
can there be any other integer solutions (with $x,y \geq 2$)? 
Moreover, $z$ may be assumed for my purposes to be even with at least three prime divisors.
Thank you in advance for any advice.
Edit: there was a typo in my bounds relating $x$ and $y$.

REPLY [4 votes]: There are other solutions. Try $x+1=a^3$, $xy+1=b^3$, where $b$ is chosen so that $b^2+b+1$ is divisible by $a^3-1$. Such $b$ is always possible to choose provided that $a-1$ is not divisible by 3 and by primes of the form $3k-1$. Moreover, $b$ may be replaced to its remainder modulo $a^3-1=x$, in this case $xy<xy+1=b^3<x^3$, thus $y<x^2$. If $b<a^3/2-1$, replace $b$ to $a^3-b-2$, it makes $y$ of order at least $x^2/2+O(x)$.<|endoftext|>
TITLE: Can we just use effective descent morphisms (pure morphisms) as covers?
QUESTION [23 upvotes]: There are a number of notions of "cover" for a scheme: etale, faithfully flat, fpqc, fppf, Zariski, Nisnevich, etc. Most of these have a nice property, which is that a cover of that type satisfies effective descent for modules or quasicoherent sheaves. For me it will be easiest to state this just for rings (hence for affine objects): A morphism $f:R\to S$ of commutative rings satisfies effective descent for modules if the category of $R$-modules can be recovered from the category of $S$-modules with descent data. There are many interpretations of descent data, but it should basically be thought of as gluing data for $Spec(S)\times_{Spec(R)} Spec(S)$. 
Grothendieck showed that faithfully flat morphisms always satisfy this condition, and Joyal and Tierney showed that a more general class of morphisms, called pure morphisms, completely classifies the morphisms which satisfy this condition (i.e. a morphism satisfies effective descent for modules if and only if it is "pure"). Does anyone know what the "pure" site looks like? Is it subcanonical? How come nobody ever uses it for anything?
More generally, suppose I have some other stack (i.e. not the stack of modules, but something else). I can associate to it a topology where the covers are exactly the morphisms along which this stack descends. What does this topology look like? Has anyone studied it?

REPLY [4 votes]: Every faithfully flat morphism is of effective descent. However, the topology consisting of all faithfully flat morphisms is not subcanonical (i.e. it is not the case that every representable functor is a sheaf with respect to this topology). Therefore, the "pure" topology, if it is even a topology, also cannot be subcanonical. So even if it's a topology, it's not really of much use.<|endoftext|>
TITLE: What is meant by a Lie group acting by affine transformations?
QUESTION [5 upvotes]: This question is a continuation of this one, where I did not receive a complete answer so am moving to MO. I am trying to understand a paper by JP Szaro that was referred to there, and in particular, what is meant by the action of a Lie group on an affine manifold by affine transformations. 
I'm assuming Szaro's definition of $n$-dimensional affine manifold $M$ coincides with the Wikipedia definition, meaning that $M$ can be covered by an atlas of coordinate charts whose transition functions are in $Aff(\Bbb R^n)$. The general statement I'm having trouble with is "assume that a Lie group $G$ acts on $M$ (complete in sense of Wikipedia) by affine transformations", e.g., see Corollary 3.7 of Szaro's paper. 
Obviously $Aff(\Bbb R^n)$ acts on the universal cover of any complete affine manifold (just $\Bbb R^n$) and so $G$ may act by affine transformations after factoring through this one. However, I have no feel whatsoever for the conditions under which $G$ descends to an action on $M$—nor precisely what is meant by "acts by affine transformations". Is the statement a local one saying that in each affine coordinate chart $\{x_i\}$ the components of the generating vector field are given by $V= (a_{ij}x_i+b_j) \partial/\partial x_j$, or is it simply saying that the $G$ action factors through an action of the affine group on $M$ (to what extent are these the same thing)? 
If the latter is correct, then does this necessarily imply $M$ admits an action of the affine group and we therefore must restrict our attention to a select set of complete affine manifolds? 
An explanation with explicit examples of affine manifolds other than $\Bbb R^n$ would be very useful. Hopefully the lack of answers to my previous question warrants the transfer to MO. Thanks.

REPLY [2 votes]: If the definition of Szaro corresponds to the definition of Wikipedia, then the affine structure of $M$ is defined by a connection $\nabla$ whose curvature and torsion forms vanish identically. An affine transformation of $(M,\nabla)$ is a diffeomorphism which preserves the connection $\nabla$. When $M$ is complete, it is the quotient of $\mathbb{R}^n$ by a subgroup $H$ of affine transformation which acts properly and freely on $R^n$. In this case the group of affine transformations of $(M,\nabla)$ is the quotient of the normalizer of $H$ in $\operatorname{Aff}(\mathbb{R}^n)$ by $H$. When $(M,\nabla)$ is compact and complete, I have shown in my thesis that the connected component of the group of affine transformations of $(M,\nabla)$ is nilpotent. This implies that the connected component of $G$ is nilpotent in this case.
To show that the connected component of $\operatorname{Aff}(M,\nabla)$ (the group of affine transformations of $(M,\nabla)$) is nilpotent when $(M,\nabla)$ is compact and complete, one remarks firstly that if $M$ is the quotient of $\mathbb{R}^n$ by the group of affine transformations $H$, then the action of $H$ is irreducible, that is, does not preserve a proper affine subset. To see this, suppose that $H$ preserves an affine subset $V$, then $V/H$ and $M$ are $K(\pi,1)$-Eilenberg McLane space. Thus the cohomology of $M$ and $V/H$ is the cohomology of $H$, since $M$ is compact. (We can assume $M$ is oriented up to a 2-cover.) Thus $H^n(M,\mathbb{R})=H^n(\pi,\mathbb{R})\neq 0$. This implies that $H^n(V/H,\mathbb{R})\neq 0$, thus the dimension of $V/H$ is at least $n$.
Next, you remark that $N(H)_0$ the connected component of the normalizer of $H$ in $\operatorname{Aff}(\mathbb{R}^n)$ commutes with $H$. You deduce that this implies that it acts freely on $\mathbb{R}^n$ since the action of $H$ on $\mathbb{R}^n$ is irreducible.
Now, let $\operatorname{aff}(M,\nabla)$ be the Lie algebra of $\operatorname{Aff}(M,\nabla)$. For every vector $X,Y\in \operatorname{aff}(M,\nabla)$, $\nabla_XY$ is again in $\operatorname{aff}(M,\nabla)$. Since $\operatorname{Aff}(M,\nabla)$ is the quotient of $N(H)$ by the discrete group $H$, their Lie algebras are isomorphic. Remark that the product induced by $\nabla$ on $n(H)$ the Lie algebra of $N(H)$ is the associative product of $\operatorname{aff}(\mathbb{R}^n)$ the Lie algebra of $\operatorname{Aff}(R^n)$ defined by $(A,a).(B,b)=(AB,A(b))$. The associative algebra $n(H)$ does not have an idempotent $(C,c)$ because if $(C,c)$ is such an idempotent, $(C,c).(C,c)=(C^2,C(c))=(C,c)$. This implies that the 1-parameter group generated by $(C,c)$ fixes $-c$. This is in contradiction with the fact that the action of $N(H)_0$ on $\mathbb{R}^n$ is free. An associative algebra which doesn't have a nilpotent element is nilpotent so $n(H)$ is nilpotent, hence $\operatorname{aff}(M,\nabla)$ and $\operatorname{Aff}(M,\nabla)_0$ are nilpotent.
When $(M,\nabla)$ is compact and complete, I have also shown that the natural action of $\operatorname{Aff}(M,\nabla)_0$ on $M$ is locally free, thus its orbits define on $M$ a foliation.
The typical example is the torus $T^n$ which is the quotient of $\mathbb{R}^n$ by the group generated by $n$-translations $t_{e_1},...,t_{e_n}$ such that $(e_1,...,e_n)$ is a basis of $\mathbb{R}^n$. The connected component of $\operatorname{Aff}(T^n)$ is $T^n$. Remark that $\operatorname{Aff}(T^n)$ is not nilpotent although its connected component is nilpotent since the action of $\operatorname{GL}(n,\mathbb{Z})$ on $\mathbb{R}^n$ induces an action of $\operatorname{GL}(n,\mathbb{Z})$ on $T^n$.
Tsemo Aristide
Dynamique des variétés affines. Journal of the London Mathematical Society 63.2 (2001): 469-486.<|endoftext|>
TITLE: Fat stationary sets
QUESTION [6 upvotes]: Recall a stationary subset $S$ of a regular cardinal $\kappa$ is fat when for every $\alpha < \kappa$, and every club $C$, there is a closed set of order type $\alpha$ contained in $S \cap C$.  It is a result of Stavi, proved here, that:
(1) For every regular cardinal $\kappa$ and every stationary $S \subseteq \kappa^+ \cap \mathrm{cf}(\kappa)$, $S \cup \mathrm{cf}(<\kappa)$ is fat.
(2) If $S \subseteq \kappa$ is fat, and $2^{<\alpha} < \kappa$ for all $\alpha< \kappa$, then there is a $<\kappa$-distributive forcing of size $2^{<\kappa}$ which forces a club $C \subseteq S$.  Furthermore, the forcing preserves every stationary subset of $S$.
Questions: Suppose $\kappa$ is either (a) inaccessible or (b) the successor of singular cardinal. Is it true that there is a sequence of disjoint stationary sets $\langle S_\alpha : \alpha < \kappa \rangle$ and some set $T$ disjoint from all $S_\alpha$ with the following property?  For all clubs $C$, and all $\alpha,\beta < \kappa$, there is a closed subset $p$ of $C \cap (T \cup S_\alpha)$ with order type $\geq \beta$, and $\max p \in S_\alpha$.  An answer under additional combinatorial assumptions (known to be consistent) would be welcome.

REPLY [8 votes]: Suppose that $\lambda$ is a singular cardinal, $\square_\lambda$ holds, and $2^\lambda=\lambda^+$. Then:

There exists a partition of $\lambda^+$ into $\lambda^+$ many pairwise disjoint fat stationary sets.
There exists a family of $2^{(\lambda^+)}$ pairwise almost-disjoint fat stationary sets.

Both clauses follow from Theorem D of http://www.assafrinot.com/paper/11 . 
An application of Lemma 2.3 of the same paper shows that Clause (1) follows already from $\square_\lambda$ (without the arithmetic hypothesis). 
For $\lambda$ regular, one obtains Clause (1) from $\square(\lambda^+)$ by feeding $\Gamma:=E^{\lambda^+}_\lambda$ to Lemma 3.2 of http://www.assafrinot.com/paper/18 . 
Update Jan/2017: http://settheory.mathtalks.org/?p=7240<|endoftext|>
TITLE: Symmetries of the flag variety
QUESTION [8 upvotes]: Let $\mathfrak g$ be a finite dimensional simple Lie algebra over $\mathbb C$, and let $\mathcal B=G/B$ be the associated Flag variety.
Is it true that the obvious map
$$
\mathfrak g\to \Gamma (T\mathcal B)
$$
from $\mathfrak g$ to the Lie algebra of globally defined algebraic vector fields on $\mathcal B$ is an isomorphism?
Remark:There are examples where the corresponding statement for $G/P$ is false.

REPLY [7 votes]: Yes, this is true. In fact, the homogeneous spaces $G/P$ such that $\mathfrak{g}\rightarrow \Gamma (T_{G/P})$ is not an isomorphism have been classified (see e.g. M. Demazure, Inventiones math. 39, 179-186 (1977)): they are the odd-dimensional projective spaces, the Grassmannian of linear subspaces of maximal dimension in a smooth odd-dimensional quadric, and the 5-dimensional quadric. None of them is of the form $G/B$ (in fact they all have $b_2=1$).<|endoftext|>
TITLE: Embed a bordered Riemann surface into punctured Riemann surfaces?
QUESTION [7 upvotes]: Let $U$ be a bordered Riemann surface of genus $g$ with $n -1$ punctures and one hole (i.e., the border has one connected component). For any punctured Riemann surface $\Sigma$ of genus $g$ with $n$ punctures, does there necessarily exist an embedding (i.e. an injective holomorphic map) $\Phi: U \to \Sigma$?

REPLY [15 votes]: The answer is no. For $g=0$, the problem is equivalent to the following: is there a univalent function in the unit disk which takes given values at finitely many given points. It is well known that the answer is no: there are many inequalities
relating the points and the values. For example, Theorem 1 on p. 119 of Goluzin's book Geometric theory of functions of a complex variable, Engl. tranls. AMS, 1969
(available on the Internet). Certainly, similar situation prevails for other genera.
The intuitive explanation is this: suppose all your points (punctures) are far away from the hole. And two of them are very close together, while the third is on certain distance from these two. Then this situation (two close together and one
relatively far away must persist no matter how you embed it.<|endoftext|>
TITLE: Structure sheaf of affine variety consists of noetherian rings (again)
QUESTION [5 upvotes]: Let $X\subseteq \mathbb{A}^n$ be an affine variety.
The local ring of $X$ at $p\in X$, given by $\mathcal{O}_{X,p}=\{f\in k(X):f \text{ regular at } p\}$ is noetherian because it is a localization of $k[X]$.
If $U\subseteq X$ is open, let $\mathcal{O}_X(U)=\bigcap_{p\in U}\mathcal{O}_{X,p}$. Is this ring noetherian as well?
Note This question has already been asked here before, but was migrated to stackexchange. After the migration, we seem to realize that it is not as trivial as we have thought. Hence I take the risk to crosspost it back here.
Arguments against the claim comes from Section 19.11.13 of Ravi Vakil's notes, where he produced a variety whose ring of global section is not Noetherian. His example, though, is not affine, and it remains a question if we can produce an affine version of that example. This argument is suggested by Marco Flores et al on stackexchenge.
Added later: here is the link to the questin on stackexchange.

REPLY [4 votes]: I am just reposting my comment as an answer.
Vakil shows in this note, §3 that you can modify the construction you refer to to get an affine example. For completeness, I'll write down the example here.
Let $E$ be an elliptic curve over a field $k$, let $N$ be a degree $0$ non-torsion invertible sheaf, and $P$ an invertible sheaf of degree $\ge3$. Then, the total space $Y$ of the bundle $N \oplus P^*$ (where $P^*$ is the $k^*$ bundle associated to $P$) is quasi-affine, and has a ring of global sections that is non-noetherian.<|endoftext|>
TITLE: Closed formulas for topological K-theory?
QUESTION [8 upvotes]: Let $X$ be a compact manifold. I'm interested in whether any of the following cases admits a general closed formula for (complex)-$K$-theory. Let $E$ be a complex vector bundle with a given line bundle decomposition $E = \bigoplus_{\alpha}{L_\alpha}$.

Let $S(E)$ the pointed sphere bundle obtained by fiberwise compactification. $K(S(E))$ is a $K(X)$ algebra. Is there a formual in terms of $[E]\in K(X)$? In terms of all the $[L_\alpha]\in K(X)$?
Let $\mathbb{P}(E)$ the associated projective bundle obtained fiberwise projectivization. $K(\mathbb{P}(E))$ is a $K(X)$ algebra. Is there a formual in terms of $[E]\in K(X)$? In terms of all the $[L_\alpha]\in K(X)$?
Let $Fl(E)$ the flag bundle of $E$ obtained by associated bundle construction from the frame bundle via the representation on the flag $GL_n/B_n$. $K(Fl(E))$ is a $K(X)$ algebra. Is there a formula in terms of $[E] \in K(X)$? In terms of all the $[L_\alpha]\in K(X)$?
Let $Fr(E)$ be the frame bundle of $E$. $K(Fr(E))$ is a $K(X)$ algebra. Is there a formula in terms of $[E] \in K(X)$?

Even if all none of the above admit closed formulas references for computations of these will help me a lot too.
A last tiny question:
Is this correct $K(X \times \mathbb{P}^n)=K(X)[T]/(T)^n$?

REPLY [10 votes]: Put $u_\alpha=[\mathbb{C}]-[L_\alpha]$.
It is a standard fact, known as the projective bundle theorem, that
$$K(\mathbb{P}(E))=K(X)[t]/\prod_{\alpha}(t-u_\alpha)$$
One can express $Fl(E)$ as the top of a tower in which each level is the projective bundle associated to a vector bundle over the level below.  Using this one can obtain
$$ K(Fl(E)) = K(X)[t_1,\dotsc,t_d]/(r_1,\dotsc,r_d), $$
where $r_i$ is the difference between the $i$'th elementary function in the variables $t_j$, and the $i$'th elementary function in the variables $u_\alpha$.   Next, there is a natural cofibration sequence $S(E)\xrightarrow{}X\xrightarrow{}X^E$, where $X^E$ is the Thom space.  In combination with the Thom isomorphism, this gives a long exact sequence 
$$ K^*(X) \xrightarrow{f} K^*(X) \xrightarrow{} K^*(S(E)) \xrightarrow{d} K^{*+1}(X).  $$
Here $f$ is just multiplication by the $K$-theory Euler class, which is $\prod_\alpha u_\alpha$.<|endoftext|>
TITLE: Are knots determined by their complements within a homotopy class?
QUESTION [8 upvotes]: Suppose $M$ is a closed 3-manifold and $K_1,K_2$ are two homotopic knots in $M$.  That is, they are two embeddings $f_1,f_2\colon S^1 \to M$ such that there exists a homotopy $h\colon S^1 \times [0,1] \to M$ with $h(x,0) = f_1(x)$ and $h(x,1) = f_2(x)$.  Suppose also that there is an orientation preserving homeomorphism $M - K_1 \to M - K_2$.  Must $K_1$ and $K_2$ be isotopic in $M$?
A positive result seems like it would have fairly large implications, not to mention being necessarily more difficult than Gordon and Luecke's result (and is therefore unreasonable to expect here) so I am mainly wondering if there are any known or simple counterexamples.

REPLY [7 votes]: The answer is no. 
It stems from the fact that links are not determined by their complements.  If you take the Borromean rings (for example), and think of one component as being a knot in the exterior of the other two components, i.e. a knot in the connect sum of two copies of $S^1 \times D^2$, then this knot is homotopic to another knot, which when put together with the other two components of the Borromean rings, is not isotopic to the Borromean rings, but the complement is diffeomorphic, preserving orientation. 
You get these homotopic but not isotopic knots by doing a Dehn twist along a spanning disc for one of the other two components -- or you could do it on both simultaneously. 
I suppose the Whitehead link would also work. 
Perhaps instead of describing the second link via the Dehn twist it would be more appropriate to describe the homotopy as a "finger move" pushing one crossing over the other.<|endoftext|>
TITLE: How much do I need to learn algebraic geometry to understand arithmetics over number fields
QUESTION [11 upvotes]: I am at the stage of learning. Mostly, I am attracted by algebraic number theory. Roughly speaking, I am interested in the rational points of algebraic varieties. I am little bit afraid to start to learn algebraic geometry, since I find the modern language used there is category theory (topos, sheaves, schemes,...). Here is my question:
Is it necessary to know the modern algebraic geometry (Grothendieck formalism) to explore the theory of rational points of algebraic varieties?
Concrete examples are welcome.

REPLY [4 votes]: I'm going to say no: it's not necessary to know the modern algebraic geometry to explore the theory of rational points of algebraic varieties.  Obviously, as pointed out in some of the answers, it's helpful, because modern algebraic geometry is used in many of the main results proved in the last century.
But Arithmetic geometry is a HUGE subject, and the techniques people use to prove new theorems, improve existing ones, etc. very often don't involve modern algebraic geometry.  Even more often, one can get away with just black-boxing some basic ideas from modern algebraic geometry, especially if you're trying to prove results about certain varieties that are well-understood geometrically.  (For example, you can read a ton of papers about elliptic curves without knowing more than classical algebraic geometry.)  In addition to scheme theory, étale cohomology, stacks, etc., there are a wealth of other concepts that are super important, such as (some of these have been mentioned by others) 

the circle method, exponential sums,...
zeta- and $L$-functions
heights
modular forms
geometry of numbers (counting lattice points)
group theory & representation theory (e.g. for studying Galois representations)
probability theory (e.g. many asymptotic problems in arithmetic geometry and number theory in general are modeled by random phenomena, which leads to powerful intuition.)
computation (like on a computer)
classical algebraic geometry (a good understanding of which can be disjoint from a good understanding of scheme theory -- you have to stay grounded!)
transcendence theory (big in unlikely intersections lately, in combination with tools from model theory...)
diophantine approximation (later proofs of Faltings' Theorem)
$p$-adic analysis, Berkovich spaces,...
dynamics (not just in "arithmetic dynamics," but also big in diophantine approximation)
ergodic theory
geometric group theory

Understanding everything is a tall order, but that's why we collaborate!  Progress in arithmetic geometry is made by combining ideas from all of these areas and more.  Don't be afraid of what you don't know.<|endoftext|>
TITLE: Does the category of (algebraically closed) fields of characteristic $p$ change when $p$ changes?
QUESTION [37 upvotes]: Let $\mathrm{ACF}_p$ denote the category of algebraically closed fields of characteristic $p$, with all homomorphisms as morphisms. The question is: when is there an equivalence of categories between $\mathrm{ACF}_p$ and $\mathrm{ACF}_l$ (with the expected answer being: only when $p=l$)?
I'd also be interested in the answer with the words "algebraically closed" deleted -- I'm not sure if this is easier or harder. Also, there is a natural topological enrichment of $\mathrm{ACF}_p$ where one gives the homset the topology of pointwise convergence (which yields the usual topology on the absolute Galois group). I'd be interested to hear of a way to distinguish these topologically-enriched categories, which in principle might be easier.
Here are some easy observations (with the "algebraically closed" condition):

First, any equivalence of categories must do the obvious thing on objects, preserving transcendence degree because $K$ has a smaller transcendence degree than $L$ if and only if there is a morphism $K \to L$ but not $L \to K$ (and using the fact that the category of ordinals has no nonidentity automorphisms, as pointed out by Eric Wofsey).
We can distinguish the case $p=0$ from the case $p \neq 0$ because $\mathrm{Gal}(\mathbb{Q}) \not \cong \mathrm{Gal}(\mathbb{F}_p)$. But $\mathrm{Gal}(\mathbb{F}_p) \cong \hat{\mathbb{Z}}$ for any prime $p$. So we can't distinguish $\mathrm{ACF}_p$ from $\mathrm{ACF}_l$ for different primes $p \neq l$ in such a simple-minded way. So for the rest of this post, let $p,l$ be distinct primes.
The next guess is that maybe we can distinguish $\mathrm{ACF}_p$ from $\mathrm{ACF}_l$ by seeing that $\mathrm{Aut}(\overline{\mathbb{F}_p(t)}) \not \cong \mathrm{Aut}(\overline{\mathbb{F}_l(t)})$. To this end, note that there is a tower $\mathbb{F}_p \subset \overline{\mathbb{F}_p} \subset \overline{\mathbb{F}_p}(t) \subset \overline{\mathbb{F}_p(t)}$. The automorphism groups of these intermediate extensions are respectively $\hat{\mathbb{Z}}$, $\mathrm{PGL}_2(\overline{\mathbb{F}_p})$, and a free profinite group (the last one is according to wikipedia).

From (3), there is at least a subquotient $\mathrm{PGL}_2(\overline{\mathbb{F}_p}) \subset \mathrm{Aut}(\overline{\mathbb{F}_p(t)})$ which looks different for different primes. But I don't see how to turn this observation into a proof that $\mathrm{Aut}(\overline{\mathbb{F}_p(t)}) \not \cong \mathrm{Aut}(\overline{\mathbb{F}_l(t)})$ specifically, or that $\mathrm{ACF}_p \not \simeq \mathrm{ACF}_l$ more generally.
I initially posted this on math.SE, thinking that there must be some easy way to resolve it lying just beyond the reaches of my algebraic competency, but after the question has languished there for a week I think maybe I should try it out here.

REPLY [8 votes]: The answer is "only when $p=l$" if you work with the category of algebraically closed fields enriched over the category of topological spaces. That means I want to put a topology on the set of homomorphisms from one field to another. I will do this with the compact-open topology for the discrete topology for each fields - i.e. a basis for the open sets consists of the sets $B_{x,y} =\{ f \mid f(x)=y\}$. This is the standard way of topologing Galois groups, for instance.
We will construct $PGL_2(\overline{\mathbb F}_p)$ from this category in several steps. (This is sufficient, because of what you noted about maximal $p$-subgroups.)
First, as you noted, we can find in this category the unique algebraically closed field of transcendence degree $0$, namely $\overline{\mathbb F_p}$ and of transcendence degree $1$, namely $\overline{\mathbb F_p(t)}$. Take any map from the first to the second. Consider the group of automorphisms of the second that form a commutative triangle with this map. This is the topological group $\operatorname{Aut}(\overline{\mathbb F_p(t)}/\overline{\mathbb F}_p)$.
Next we classify the compact open subsets of this group. Any open subgroup contains some nonempty open set of the form: $\{ f \mid f(x_1)=y_1,\dots, f(x_n)=y_n\}$. By composing with element of that set, it contains $\{f \mid f(x_1)=x_1,\dots, f(x_n)=x_n\}$.  This is an open subgroup, so it is finite-index. It follows that this subgroup contains the Galois group of the curve with function field $\mathbb F_p(x_1,\dots,x_n)$ as a finite index subgroup, so it contains the Galois group of some cover of that curve as a finite index normal subgroup, so it is the Galois group of a quotient of that cover by a finite subgroup of its automorphism group. So every compact open subgroup is the Galois group of some curve.
Let $H$ be a compact open subgroup fixing some field $F$. The normalizer $N(H)$ of $H$ acts naturally on $F$, and the kernel of that action is $H$, so $N(H)/H$ is a subgroup of the automorphisms of $F$. In fact it is all the automorphisms of $F$, as these clearly normalize $H$. So we may construct the set of all automorphism groups of curves over $\overline{\mathbb F}_p$.
Among these, the only infinite nonsolvable groups are the automorphism groups of curves of genus $0$, which are all isomorphic to $PGL_2(\overline{\mathbb F}_p)$. So we may construct $PGL_2(\overline{\mathbb F}_p)$.<|endoftext|>
TITLE: Generalized Dirac operators
QUESTION [9 upvotes]: So far I met three definitions of the so called generalized Dirac operator(or Dirac type operators. Everything takes place over Riemannian manifols $M$ and we have smooth hermitian vector bundle $S \to M$ over $M$.
First definition The first order differential operator $D$ is called Dirac type if its symbol $\sigma$ has the property: 
$$\sigma(x,\xi)^2u=-\|\xi\|^2u$$ for $x \in M, \xi \in T^*_xM, u \in S_x$.
Second definition $D$ is called Dirac type operator if $D^2$ is of the form $$\sum_{i,j}g^{ij}\partial_i\partial_j$$ modulo the lower order terms (here $g^{ij}$ are components of Riemannian metric on cotangent bundle).
Third definition $D$ is called Dirac type if there is a Clifford action $c$ and a Clifford connection $\nabla$ (i.e. metric connection compatible with this Clifford action) such that $D=c \circ \nabla$ (some authors put $i$ in front of this operator).   


Are these definitions equivalent? If so, why is it true?  


The one implication which is evident for me is that the second definition implies the first

REPLY [7 votes]: A very good place to read about this is the 3. chapter of the book "Heat kernels and Dirac operators" by Nicole, Getzler & Vergne.
Up to the wrong sign in your second definition, 1. and 2. are equivalent, as the symbol just collects the highest order terms. Also you obtain from the basic properties of the symbol map that the bundle which admits a Dirac operator of the 1. definition admits a Clifford representation of the cotangent (and via the Riemannian metric by the tangent) bundle. The action is just given by the symbol of $D$. A slightly tricky computation (carried out in the aforementioned book) gives you a metric connection (if $D$ is self-adjoint) on your bundle, such that the corresponding Dirac operator differs from $D$ only by a endomorphism. As you cannot get rid of this zeroth order term in general, you see that the 3. definition is a little bit more special than 1. & 2.<|endoftext|>
TITLE: Philosophy behind cohomological representations
QUESTION [13 upvotes]: For a given real reductive Lie group $G$, we have the notion of a representation being cohomological using the Lie algebra cohomology. In particular we know that the discrete series representations of $G$(whenever it exists) is cohomological. I am trying to understand what information do we exactly obtain when we know that these representations are cohomological. Also, why are cohomological representations important and what role do they play in the global setting, i.e in the setting of automorphic representations?
It would be great even if you could direct me to an appropriate reference.
Thanks in advance.

REPLY [5 votes]: I'm far from an expert, but here is a comment. In the case of a Shimura variety, the Matsushima-Murakami formula and the (proof of the) Zucker conjecture shows that cohomological representations are precisely the ones that contribute to the intersection cohomology of the Shimura variety with coefficients in some local system. This "piece" of the intersection cohomology is a candidate for a compatible system of $\ell$-adic Galois representations attached to the representation (or a "motive"). Thus cohomological representations are precisely the ones for which Deligne's original construction of Galois representations attached to Hecke eigenforms of weight $\geq 2$ could be optimistically hoped to generalize. For instance, Weissauer has constructed 4-dimensional Galois representations attached to Siegel cusp forms of genus 2 in the cohomological case, using this strategy. But e.g. for the interesting case of Siegel cusp forms of weight $(2,2)$, which are expected to have Galois representations given by $H^1$ of an abelian surface, such a construction is wide open.<|endoftext|>
TITLE: Does there exist finite dimensional irreducible representation of Euclidean or Poincare group in which translation and rotation both act nontrivially?
QUESTION [6 upvotes]: Does there exist any finite dimensional irreducible rep. of Euclidean or Poincare group in which translation and rotation both act nontrivially? 
Let me firstly clarify my question. For example, we obviously have a faithful rep. of Poincare group, $\begin{pmatrix} \Lambda & x \\ 0 & 1 \end{pmatrix}$, where $\Lambda$ is Lorentz transformation and $x$ is translation. But this is a reducible and indecomposable representation. We can always define an irreducible rep. of Poincare group by
$$f:\begin{pmatrix} \Lambda & x \\ 0 & 1 \end{pmatrix}\rightarrow D_{(i,j)}(\Lambda)$$
where $D_{(i,j)}(\Lambda)$ is some irreducible rep. of Lorentz group. But in this case, the translation acts trivially.
Because Euclidean and Poincare group are noncompact, their unitary and irreducible rep. must be infinite dimensional. However it doesn't say we don't have finite dimensional and irreducible but not unitary rep. of these groups. The above is an example. So I want to know whether there is other finite dimensional irreducible rep. of Euclidean and Poincare group in which rotation and translation both act nontrivially? If it doesn't exist, how to prove or tell me the name of theorem.

REPLY [12 votes]: The answer is "No".
You don't specify which kind of representation you have in mind. I assume these are finite dimensional complex representations. Thus you ask about (continuous) homomorphisms $G\to \text{GL}_n(\mathbb{C})$. In the groups $G$ you consider you have a non trivial normal nilpotent subgroup $N$, the group of translations (aka the unipotent radical).
Let $Z$ be the center of $N$. It is still normal in $G$. In your case $Z=N$.
Decompose $V=\mathbb{C}^n$ to a direct sum of $Z$-eigenspaces. Each of these is $G$-invariant. Thus, if $V$ is $G$-irreducible then each element of $Z$ must act by a scalar multiplication on $V$. In particular, in the image of $G$, $Z$ must be central.
In your specific groups the only normal subgroups are $\{e\},G$ and $N$ ($=Z$ ), and $N$ is not central in $G$. So $N$ must be in the kernel of the representation.<|endoftext|>
TITLE: Resolution of singularities in étale cohomology
QUESTION [15 upvotes]: The lack of a suitable resolution of singularities comes up often in work on étale cohomology from the 1960s and 70s,  And I think even the latest version of Milne's lecture notes says "It is likely that de Jong’s resolution theorem (Smoothness, semi-stability and alterations. Inst. Hautes Etudes Sci. Publ. Math. No. 83 (1996), 51–93) will allow many improvements" to both theorems and proofs.  Is there currently some comprehensive treatment of étale cohomology that uses this?
In other words has anyone worked on realizing the hope that Milne expressed, that basic questions already prominent in the SGA could be usefully "re-thought" by greater use of Artin's algebraic spaces, de Jong's removal of singularities, or other new ideas?
Ten days after posting the question, the community wiki answer makes me think the situation is as follows, though I would love to learn otherwise:  On one hand, étale cohomology finds so many concrete applications treating the general theory as a "black box," that there is not much pressure to rethink the general theorems.  Proofs can cite theorems from the books by Freitag and Kiehl or Milne.  Those books in turn give many general theorems without proof, citing SGA for the proofs. 
On the other hand, new fundamental ideas keep coming in.  Notably Artin's algebraic spaces, plus de Jong's theorem, made Milne say "the structure of the subject needs to be re-thought" (http://www.jmilne.org/math/CourseNotes/LEC.pdf page 6).  More recently Ofer Gabber's work suggests the same but up to now that work has been presented on the base of the SGA and not as a way to reorganize the SGA.  
Because such results are still coming in, and raise many possibilities, comprehensive treatments of étale cohomology even through Lei Fu, Etale Cohomology Theory (Nankai Tracts in Mathematics) are still substantially based on the SGA, including SGA 4.5.  SGA 4.5 did revise several major proofs from SGA 4.  But no one has yet attempted a thoroughgoing synthesis with the latest ideas.
I will mention de Jong's http://stacks.math.columbia.edu/download/etale-cohomology.pdf does not seem to reach the issues where resolution of singularities would matter.

REPLY [6 votes]: I am turning Giulia's comment into a CW answer.  At over 400 pages, I think that "Travaux de Gabber", published in Astérisque, ought to count as a  "comprehensive treatment of [some aspects of ] étale cohomology" using the method of alterations. The style is very SGA, and indeed there is a nonempty intersection among the set of authors.<|endoftext|>
TITLE: commutators in upper triangular matrices
QUESTION [10 upvotes]: Consider the group $T_p(n)$ of all non-singular upper triangular matrices with entries in $\mathbb{F}_p.$ Its commutator subgroup is $U_p(n)$ (all elements in $T_p(n)$ with $1$s on the main diagonal). The question is: is every element in $U_p$ a commutator of matrices in $T_p?$

REPLY [8 votes]: Too long for a comment and only a partial answer for $n>2$. A note on commutators in the group of infinite triangular matrices over a ring, by Agnieszka Bier and Waldemar Holubowski, Linear and Multilinear Algebra 63  (2015) no. 11, 2301-2310; seems relevant. It includes results on finite matrices, despite the title.
Let $R$ be an associative ring with $1$, and let ${U}(R)$ be its group of invertible elements. $T(n,R)$ denotes the group of upper triangular $n\times n$ matrices; $UT(n,R)$ the set of upper unitriangular matrices ($1$s in the diagonal), $UT(n,m,R)$ the subgroup containing exactly all those matrices which have zero entries on the first $m$ super diagonals; and the basic commutators $c_k(x_1,\ldots,x_k)$ are defined inductively by $c_1(x_1)=x_1$, $c_{i+1}(x_1,\ldots,x_{i+1}) = [c_i(x_1,\ldots,x_i),x_{i+1}]$; let $\gamma_k(G)=G$ be the $k$th term of the lower central series of $G$. Finally, the Engel words $e_k(x,y)$ are given by $e_2(x,y)=[x,y]$, $e_{m+1}(x,y) = [e_m(x,y),y]$. 
The authors prove:

Theorem 1.5 Let $R$ be an associative ring with $1$ such that $U(R)$ is commutative. Then:

$\gamma_k(UT(n,R))=UT(n,k,R)$ and every element of $\gamma_k(UT(n,R))$ is a value of the basic commutator $c_k$.
Every element of $\gamma_k(UT(n,R))$ is a value of the Engel word $e_k$.

Moreover, if $1$ is a sum of two invertible elements, then

$[T(n,R),T(n,R)] = UT(n,R)$ whenever $n\geq 2$.
Every element of $\gamma_2(T(2,R)) = UT(2,R)$ is a commutator.
Every element of $\gamma_2(T(n,R)) = UT(n,R)$ with $n>2$ is a product of at most two commutators.


In particular, for $R=\mathbb{F}_p$ with $p>2$, then the answer is "yes" for $n=2$ by point 4; but for $n>2$ they only show every element is a product of at most two commutators. 
Perhaps this was already known for the special case of matrices over commutative rings/fields?<|endoftext|>
TITLE: Vector fields, diffeomorphism subgroups and lie group actions
QUESTION [6 upvotes]: Let $M$ be a compact smooth manifold. Since any vector field is complete we get a $1$-parameter subgroup for each vector field. Consider the following generalization:
Let $\{X_j\} \in Vect(M)$ be a finite "basis" of some integrable subbundle of $TM$ (meaning that their locally linearly independent and closed under lie bracket). We have $Span\{X_j\} \cong \mathfrak{g}$ for some finite dimensional lie algebra. The exponential map $\varphi : \mathfrak{g} \to G$ gives a group equipped with a natural action $\rho: G \to Diff(M)$. Denoting the flow of $X$ by $\varphi_t^X$ the action looks like:
$$\rho : g=e^{X_1}e^{X_2}\dots e^{X_n} \mapsto \varphi_1^{X_1}\circ e_1^{X_2}\circ\dots \circ\varphi_1^{X_n}(-)$$

Question 1: Is this map well defined?

In any case if $\mathfrak{g}$ is abelian the action is well define and we get a product of circles and lines inside $Diff(M)$. For every point $x \in M$ the action of the torus part will carve an embedded submanifold of $M$ and the action of the euclidean part will carve an immersed submanifold. 

Question 2: Is there anything more substantial to say here? When will the
  action of the torus part yield an actual torus?
Question 3: Does this construction give a $G/\ker\rho$-fiber bundle? (does it help if $G$ is compact?).

REPLY [6 votes]: Every orbit of a torus is a torus, since every orbit of a Lie group action is a homogeneous space of the Lie group.<|endoftext|>
TITLE: Index of Modified Dirac Operator
QUESTION [6 upvotes]: Let's say we have an oriented compact 4-d Riemannian spin manifold $(M,g)$. Everybody who's anybody has heard about the index of the Dirac operator $D: S^+\rightarrow S-$; it's the $\hat{A}$-genus, which is $\displaystyle\frac{-1}{8}\tau(M)$ ($\tau$ is the signature). I don't know too much about the index theorem and it's inner workings, but I'm wondering what happens if you perturb the Dirac operator and consider $$D_{f,s}=D+s\textrm{grad}(f)\cdot$$
where s is a (let's say small) real parameter and $f\in C^{\infty}(M)$; "$\cdot$" indicates Clifford multiplication, and maps $S^+\rightarrow S^-$ since elements of $TM$ anticommute with the volume element in dimension 4. 
Does the index stay the same? Change in a predictable way related to $f$? What if $f$ is special somehow? What about in the special case where $\tau=0$? Or the even-more-special case where $\tau=0$ because the dimension of the harmonic spinors is $0$, e.g. when $R>0$? How does the dimension of the kernel jump as $s$ changes?
Sorry for the avalanche of questions. I'm interested in any information people have about any subset of them.

REPLY [5 votes]: For this particular case the two operators are conjugate (though note the conjugacy is not unitary unless $s$ is imaginary)  
$$
D_{f,s}=e^{-sf}D e^{sf}
$$
so that the dimension of both the kernel and cokernel are independent of $s$.
Note that Chris Gerig's comments on the other hand apply more generally. The index of 
$$
D_{s,\theta}=D+s\theta⋅
$$
for any one-form $\theta$ is independent of $s$.<|endoftext|>
TITLE: From relative categories to marked simplicial sets
QUESTION [16 upvotes]: Both relative categories and marked simplicial sets (over Δ^0) present the ∞-category of ∞-categories.
Naturally, one could ask whether there is a reasonably direct way to pass between these two models.
There are some obvious candidates, e.g., the nerve of a relative category is a marked simplicial set.
Is the functor that sends a relative category (C,W) to the marked simplicial set (NC,NW)
a weak equivalence from the relative category of relative categories to the relative category of marked simplicial sets?
The closest reference I could find is Remark 1.3.4.2 in Higher Algebra, which states it on the level of individual objects, not relative categories.
Note that although the right adjoint functor (C,W)↦(NC,NW) does compute the correct answer,
its left adjoint is badly behaved because it only depends on the marked 2-skeleton.
Thus there is no chance that this right adjoint functor is a right Quillen equivalence of model categories.
However, Barwick and Kan treat the case of nonmarked simplicial sets in §6.6 of their paper on relative categories,
and they recover Thomason’s theorem (Theorem 6.7 in their paper),
which can be seen as the analog of the above question in the nonmarked setting.
Their result is stated as a Quillen equivalence,
which is achieved by replacing the naive cosimplicial object (n↦(0→⋯→n)) in relative categories with its double subdivision as a relative poset.
The resulting nerve functor N_ξ is weakly equivalent to the naive nerve functor N,
but their left adjoints are not weakly equivalent,
and it is the nonnaive left adjoint that computes the correct category that corresponds to a simplicial set.
Thus one can refine the above question from relative categories to model categories as follows.
Is the subdivided relative nerve (N_ξ C,N_ξ W) weakly equivalent to the naive relative nerve (NC,NW) as a marked simplicial set?
Is (C,W)↦(N_ξ C,N_ξ W) a right Quillen equivalence?

REPLY [7 votes]: Concerning the first question: the simplicial localization functor $L^H$ induces an equivalence from the relative category of small relative categories to the relative category of small simplicial categories (see, e.g., Barwick and Kan http://arxiv.org/pdf/1012.1540.pdf). The right derived coherent nerve functor $\mathbb{R}N$ induces an equivalence from the relative category of small simplicial categories to the relative category of quasi-categories. The functor $X \mapsto X^{\natural}$ induces an equivalence from relative category of quasi-categories to the relative category of marked simplicial sets over $\Delta^0$ (and Cartesian equivalences). According to Hinich (http://arxiv.org/pdf/1311.4128v4.pdf, Proposition 1.2.1), there is a natural equivalence of marked simplicial sets
$$ (N(C),N(W)) \stackrel{\simeq}{\to} \mathbb{R}N(L^H(C,W))^{\natural}. $$
It follows that the association $(C,W) \mapsto (N(C),N(W))$ induces an equivalence from the relative category of small relative categories to the relative category of marked simplicial sets.
Concerning the second question: I think it's highly likely that the answer is yes (on both counts), but as far as I know it has not been written up anywhere yet (though someone will probably write a paper about it soon).<|endoftext|>
TITLE: Current status of computable spectral theorem and interpretation of quantum mechanics
QUESTION [6 upvotes]: The spectral theorem states if $A$ is a Hermitian operator acting on an $n-$dimensional Hilbert space space $H$, and $\lambda_1, ... \lambda_m$ are $m \leq n$ distinct eigenvalues of $A$, then 
$$ H=\oplus_{i=1}^{m} H_i,$$
where $H_i$ are the corresponding eigenspaces, and in turn
$$ A = \sum_{i=1}^{m}\lambda_i P_i,$$
where $P_i$-s are the corresponding projections. It is known that the spectrum of a Hermitian operator acting on a finite-dimensional Hilbert-space can be computed provided that its cardinality is known beforehand, or, alternatively, that for each pair of eigenvalues $\lambda_i, \lambda_j$, it is known whether $\lambda_i = \lambda_j$ or $\lambda_i \neq \lambda_j$. It is a crucial condition when we try to build a computable foundation of quantum mechanics since deciding which eigenvalue is actually measured essentially defines what eigenspace the system falls into after the measurement. The problem of degeneracy is described in Bridges and Svozil, 2000 in the following example:

Under this link, I came across this book of Ye where he explains the degeneracy problem as follows:

The representation in [when $\forall i,j. \lambda_i \neq \lambda_j \lor \lambda_i = \lambda_j$] above is exactly the same as the classical case. It seems
  that indistinguishable $\lambda_i$-s arise only in artificial constructions. In other words, we
  expect that the operators on finite-dimensional spaces appearing in natural physics
  contexts all satisfy the condition $\forall i,j. \lambda_i \neq \lambda_j \lor \lambda_i = \lambda_j$. Because, in quantum mechanics, the spectrum of an operator A is supposed to consist of the possible values
  of the observable corresponding to A. Any realistic observation can be performed
  only up to a finite precision. So, in a finite dimensional case, we can expect that the
  eigenvalues are all mutually distinguishable.

What is actually behind such a justification and what does it have to do with the fact that each measurement can be done up to a finite precision?
I also have a side question regarding the following theorem of Ziegler and Brattka (2001):

where a matrix is called computable iff the corresponding linear map is computable. How does this fact enable us to reconstruct exactly the classical theorem? I can't see how we'd obtain the cardinality of the spectrum just from the condition that $A$ be computable (normal) matrix.
WARNING: there is no such a corollary in the preprint!

REPLY [5 votes]: I think the point is this.  It is impossible in general to decide computationally whether two computable real numbers $\alpha$ and $\beta$ are equal.   If in fact they are not equal, by computing sufficiently good approximations you can determine that they are not equal, but no finite computation can determine that they are equal.
For a polynomial $P(\lambda)$ (in particular the characteristic polynomial of a matrix), the roots of $P$ are distinct iff the discriminant of $P$ is nonzero.  This discriminant is a polynomial in the coefficients, so it is computable.  But again, if the discriminant happens to be $0$ we can't determine that by a finite computation.
However, in "real-life" or "natural" situations, we should expect eigenvalues to be distinct unless they have a good reason (typically involving a symmetry) to be equal.
The situation in Corollary 15 is different: there is no requirement here for the $\lambda_j$ to be distinct.
EDIT: In fact, Corollary 15 is false.  Consider the family of $2 \times 2$ matrices
$$ A(p) = \pmatrix{\min(p,0) & \max(p,0)\cr
          \max(p,0) & \max(-p,0)\cr} $$
where $p$ is a computable real number.  Then $A(p)$ is a computable real symmetric matrix.  If $p < 0$ it has orthonormal eigenvectors $$ \pmatrix{1/\sqrt{2}\cr 1/\sqrt{2}\cr}, \ \pmatrix{1/\sqrt{2}\cr -1/\sqrt{2}\cr}$$
while if $p > 0$ it has orthonormal eigenvectors $$ \pmatrix{1\cr 0\cr},\ \pmatrix{0\cr 1\cr}$$
The 
 orthonormal eigenvectors $x_1,\; x_2$ mentioned in Corollary 15 can't be computable functions, because they can't be continuous at $p=0$.<|endoftext|>
TITLE: Is locally freeness of a sheaf (of fixed rank) around a divisor detectable from a first order neighbourhood?
QUESTION [5 upvotes]: Assume you have a smooth projective variety $X$ over the complex numbers, a smooth prime divisor $D$ on it, and a torsion free coherent sheaf $E$ on $X$ of rank $r>0$. Let $E|_{2D}:=E\otimes_{\mathcal{O}_X}\mathcal{O}_{2D}$, where $\mathcal{O}_{2D}:=\mathcal{O}_{X}/\mathcal{I}_{D}^2$, be the restriction of $E$ to the first order neighbourhood of $D$ in $X$.
If $E|_{2D}$ is a locally free $\mathcal{O}_{2D}$-module of rank equal to $r$, can we conclude that $E$ is locally free in a (Zariski) neighbourhood of $D$ (necessarily of rank $r$)?
Does it change anything if we assume $X$ is a surface and/or that $E$ is semistable?

REPLY [6 votes]: Let $X$ be a connected reduced noetherian scheme and $\mathscr F$ a coherent sheaf on $X$. Let $\varrho(x)=\dim_{\kappa(x)}\mathscr F_x\otimes \kappa(x)$ where $x\in X$ is a point and $\kappa(x)$ is the residue field at $x$. 
Using Nakayama's lemma you can prove the following:
The function $\varrho$ is upper semi-continuous on $X$ in the sense that the set
$$
Z_m:=\{x\in X \ \vert \ \varrho(x)\geq m \} 
$$
is closed for any $m\in \mathbb Z$. 
It follows that if $\mathscr E$ is a torsion-free sheaf of rank $r$, then $Z_m=\emptyset$ for $m<r$ and $Z_r=X$. 
Furthermore, $\mathscr E\left|_U\right.$ is locally free where 
$$
U:=\{x\in X \ \vert \ \varrho(x)=r \}. 
$$
It looks like that in your case $D\subseteq U$ and hence $U$ gives you the open set you are looking for. In fact, it seems that a lot less is enough for this than what you have. 
I don't think $X$ being a surface or $\mathscr E$ being semistable makes any difference. After all how could this be any better? :)<|endoftext|>
TITLE: Summation of series involving $\sinh$ of a square root
QUESTION [13 upvotes]: Consider the following series:
$$
S = \sum_{\text{odd } n} \sum_{\text{odd } m} \frac{(-1)^{(n+m)/2}}{nm} \frac{\sinh( \pi \sqrt{n^2 + m^2}/2)}{\sinh( \pi \sqrt{n^2 + m^2})}
$$
From the physical context, one can argue that the series should converge to
$$
S = - \frac{\pi^2}{96},
$$
and numerically calculating the first 10–20 terms of this series seems to show rapid convergence to this value.  
Is it possible to prove this statement without appealing to a physical argument?  I honestly have no idea how to begin tackling a problem like this.

REPLY [13 votes]: Here is a solution that I have found while working on other lattice sums.
It utilizes a very simple result:

Define $f$ by 
  $$f(x)=\sum_{n=0}^{\infty} (-1)^n (2n+1) e^{-\pi x (n+\frac12)^2}.$$
  Then 
  $$\int_0^{\infty} e^{-y x} f(x)dx=\operatorname{sech}\sqrt{\pi y}.\tag{$\star$}$$

The proof is simple - just integrate term by term, and the result follows from the series 
$$\sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)}{(2n+1)^2+(2x)^2}=\frac{\pi}{4} \operatorname{sech} \pi x.$$
Therefore, 
$$\large \operatorname{sech} \frac{\pi}{2} \sqrt{n^2+m^2} = \int_0^{\infty} e^{-\frac{\pi}{4} x(n^2+m^2)} f(x)dx.$$
Now put $(2n+1)$ and $(2m+1)$ instead of $n$ and $m$, multiply by $\displaystyle \, \frac{(-1)^{n+m}}{(2n+1)(2m+1)},$ and
sum both $n$ and $m$ over the non-negative integers:
$$ 2 S=\sum_{n,m=0}^{\infty} \frac{(-1)^{n+m}}{(2n+1)(2m+1)} \operatorname{sech}\left(\frac{\pi}{2} \sqrt{(2n+1)^2+(2m+1)^2}\right)
= \int_0^{\infty} f(x) \left(\sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} \right)^2 dx$$
Here comes the neat part: we notice that
$$\frac{d}{dx}  \sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} = -\frac{\pi}{4} f(x),$$
and conclude that 
$$ 2 S = \frac13 \left(-\frac{4}{\pi}\right)  \left(\sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} \right)^3\Biggr{|}_0^\infty
\\= \frac13 \left(\frac{4}{\pi}\right) \left(\frac{\pi}{4}\right)^3=\frac{\pi^2}{48}.$$
Can you see how nicely it generalizes to higher dimension versions of this sum? 
Note: 
The function $f(x)$ is actually $\eta^3(i x)$ where $\eta$ is the Dedekind eta function (This is a consequence of Jacobi's triple product identity). The result $(\star)$
was obtained by Glasser, along with other beautiful formulas for integrals involving $\eta(i x)$, in his article 
Some Integrals of the Dedekind Eta-function (2008) (arxiv link). I omitted this detail simply becuase it is not important for 
this purpose.<|endoftext|>
TITLE: Tableaux with limited rows and complementary skew shapes
QUESTION [11 upvotes]: Given a partition $\mu=(\mu_1,\mu_2...,\mu_d)$, define $\bar\mu=(\mu_1-\mu_d,\mu_1-\mu_{d-1},...,\mu_1-\mu_2,0)$, the complementary shape in the $d\times \mu_1$ rectangle. Then the number of skew standard Young tableaux with at most $d$ rows and skew part $\mu$ are (I believe) in bijection with those with skew part $\bar\mu$ (where they each have $n$ labels).
I have a proof for the case $\mu=(i+j,i,0)$, $\bar\mu=(i+j,j,0)$ but I heard the general case may have appeared in the literature. I have been unable to find the reference. It might work with jeu de taquin and I'd love to see the details.
Is anyone familiar with this paper?
EDIT: Equivalently, given a fixed lattice path from $(-\infty,0)$ to $(+\infty,d)$, we want to fill in $n$ labels of a standard Young tableau below the path and above the line $y=0$. The claim is that there's a bijection between these objects and the objects we get if we flip the lattice path over (180 degrees).
For example, if $n=6$, $d=3$, $\mu=(3,1,0)$, and $\bar\mu=(3,2,0)$, these two are paired up by my bijection:

REPLY [4 votes]: Here is another solution, based on the paper "Robinson-schensted algorithms for skew tableaux" by Sagan and Stanley (Darij Grinberg was the one who suggested that this algorithm might work, I'm just filling in the details in a pretty straightforward fashion).
Fix integers $k$ and $n$, and denote $[n]=\{1,2,\dots,n\}$. We say that 
$$\pi=\begin{pmatrix}i_1 & \dots & i_m\\ j_1 & \dots & j_m\end{pmatrix}$$ is a partial two-line array 
if the integers $i_1,\dots,i_m$ are distinct elements of $[n]$, the same is true for $j_1,\dots,j_m$, and $i_1<\dots<i_m$. 
Also, fix some partition $\mu$ with at most $k$ rows, and assume all parts of $\mu$ are much bigger than $n$ (the correct way is to think of $\mu$ as a lattice path inside an infinite horizontal strip of height $k$). 
We say that an SYT $P$ with at most $k$ rows has right boundary $\mu$ if it has shape $\mu/\nu$ for some $\nu$. A similar definition goes for SYT with left boundary $\mu$, and the original conjecture becomes $$\#\mathrm{SYT}_k^n(\mu/*)=\#\mathrm{SYT}_k^n(*/\mu)$$
where the LHS denotes the number of SYTs with $n$ boxes with at most $k$ rows and right boundary $\mu$ while the RHS denotes the number of SYTs with $n$ boxes with at most $k$ rows and left boundary $\mu$. 
One more definition: a partial SYT with entry set $S$ is just a Young tableau (i.e. the entries increase along rows and columns) with distinct entries such that the set of its entries is $S$. For example, an SYT with $n$ boxes is a partial SYT with entry set $[n]$.
We are going to describe a bijection $(\pi,T,U)\leftrightarrow (\tau,P,Q)$ where

$\pi=\begin{pmatrix}i_1 & \dots & i_s\\ j_1 & \dots & j_s\end{pmatrix},\quad \tau=\begin{pmatrix}i'_1 & \dots & i'_t\\ j'_1 & \dots & j'_t\end{pmatrix}$ are partial two-line arrays;
$T,U$ are partial SYT's of the same shape $\mu/\lambda$ with entry sets $[n]\setminus \{j_1,\dots,j_s\}$ and $[n]\setminus \{i_1,\dots,i_s\}$;
$P,Q$ are partial SYT's of the same shape $\nu/\mu$ with entry sets $[n]\setminus \{j'_1,\dots,j'_t\}$ and $[n]\setminus \{i'_1,\dots,i'_t\}$.

This bijection is going to work in exactly the same way as in Sagan-Stanley's paper, so we're going to redefine their internal and external insertions. 
Let $R$ be a partial SYT with entry set $S$, and let $q$ be an integer not belonging to $S$. Then define the external insertion of $q$ into $R$ to be the new partial SYT denoted $R\leftarrow q$ and defined by the following algorithm: we first row-insert $q$ into the first row of $R$, then it bumps out some element $q_1$ which we then insert into the second row of $R$, and so on until either $q_i$ is just appended on the right to the row $i+1$ of $R$ (i.e. nothing is bumped) or we reach $k$-th row of $R$ in which case we get one extra entry $q_k$ of $R$ which we memorize. 
Similarly, let $(i,j)$ be an inner (that is, left) corner of $R$. Define the internal insertion of $(i,j)$ into $R$ to be the new partial SYT denoted $(i,j)\to R$: remove the entry $q$ from $(i,j)$ and insert it into the next row $i+1$ of $R$. It bumps out some entry $q_{i+1}$ which we then insert into row $i+2$ of $R$ and so on until either nothing is bumped or we reach row $k$ in which case we memorize the extra entry $q_k$ that we get. 
Now, here is the RSK algorithm that is going to solve the problem: 

initially, set $P=T$, $Q=\mu/\mu$ an empty tableau and $\tau:=\begin{pmatrix} \\ \end{pmatrix}$ an empty partial two-line array;
for each $q=1,2,\dots,n$, do the following:
if $q$ belongs to the entry set of $U$, then $q$ lies in an inner corner $(i,j)$ of $U$, so remove it from $Q$ and replace $P$ by $(i,j)\to P$; 
otherwise, if $q$ is equal to $i_m$ for some $m$, replace $P$ by $P\leftarrow j_m$;
if we memorize an extra entry $p$ of $P$ as a result of our insertion, append $\begin{pmatrix}q\\ p \end{pmatrix}$ to $\tau$;
otherwise, append $q$ to $Q$ in the box that was added to $P$ at the final step of the insertion.

It is easy to see that this is indeed a bijection. Another property that can be proved analogously to Theorem 3.3 in Sagan-Stanley's paper is the following symmetry of the algorithm:
if $(\pi,T,U)\leftrightarrow (\tau,P,Q)$ then $(\pi^{-1},U,T)\leftrightarrow (\tau^{-1},Q,P)$.
Here $\pi^{-1}$ is obtained from $\pi$ by first switching $i_m$ with $j_m$ for all $m$ and then sorting the columns so that the first row would be increasing. As a corollary of that, we see that the triples $(\pi,T,T)$ where $\pi=\pi^{-1}$ is an involution correspond to the triples $(\tau,P,P)$ where $\tau$ is an involution (Sagan-Stanley, Corollary 3.4). Therefore the following formula is true:
$$\sum_{s=0}^n \mathrm{Inv}(n,s) \#\mathrm{SYT}_k^{n-s}(\mu/*)=\sum_{t=0}^n \mathrm{Inv}(n,t) \#\mathrm{SYT}_k^{n-t}(*/\mu),$$
where $\mathrm{Inv}(n,s)$ is the number of involutions $\pi$ with $s$ columns and with entries from $[n]$. This recurrence immediately implies the result by induction on $n$.
One can consider the semistandard version (the Knuth version) of this RSK algorithm which shows the following identity:
$$\sum_{\lambda\subset \mu} s_{\mu/\lambda}=\sum_{\mu\subset \nu} s_{\nu/\mu},$$
where the sums are infinite and taken over all partitions $\lambda,\nu$ with at most $k$ parts (that are respectively to the left and right of $\mu$ in this infinite horizontal strip of height $k$). Of course, this is a stronger statement since the number of SYT's is recovered by taking the coefficient of $x_1\dots x_n$ on both parts. The stronger identity was conjectured by Alex Postnikov.<|endoftext|>
TITLE: A curious eigenvalue inequality
QUESTION [18 upvotes]: Suppose $A, B$ are positive definite Hermitian matrices, $U$ is a unitary matrix such that $AUB$ is Hermitian. The spectral radius of a square matrix $X$ is denoted by $\rho(X)$. In my study, I guessed an inequality
$$\rho(U^*AU+B)\le \rho(A+B).$$
That is, the largest eigenvalue of $U^*AU+B$ is no larger than the largest eigenvalue of $A+B$.  How to prove it?
The condition $AUB$ being Hermitian is indispensible, but I have no clue how to use it.

REPLY [2 votes]: I still don't have enough reputations points to comment but here is another related iteration.  Geometrically the inequality says that using the matrix $U$ to transform $A$ moves the hyperellipse associated with $A$ further away from the hyperellipse associated with $B$. This has something to do with the angles between the eigenvectors of $A$, $U^*AU$ and $B$.
If we diagonalize $A=V_A\Lambda_AV_A^*$ and $B=V_B\Lambda_BV_B^*,$
then $$AUB=V_A\Lambda_A V_A^*UV_B\Lambda_B V_B^*$$.  Taking a similarity transformation with $V_A\Lambda_A$ and letting  $\Theta_2=V_A^*UV_B$ and $\Theta_1=V_A^*V_B$ denote the matrices whose entries have the information regarding the angles between the eigenvectors of $A$ , $U^*AU$ and $B$, then we have 
$$H_1=\Theta_2\Lambda_B \Theta_1^*\Lambda_A^{-1} $$
which is a Hermitian matrix.  We have 
$$ H_1\Lambda_A\Theta_1=\Theta_2\Lambda_B$$
At the next iteration we would have 
$$ H_2\Lambda_A\Theta_2=\Theta_3\Lambda_B$$
Eliminating $\Theta_3$, we have
$$ H_2^2=\Lambda_A^{-1}\Theta_2\Lambda_B^2\Theta_2^*\Lambda_A^{-1},$$
and now eliminating $\Lambda_B$ and $\Theta_2$ and taking a square root,
$$H_2=(\Lambda_A^{-1}H_1\Lambda_A^2H_1\Lambda_A^{-1})^{1/2}.$$
$\Lambda_B$ is absent from this iteration though it can be recovered.  It seems that the 2-norm of the $H_i$ is increasing and that $H_i$ must converge to $\Lambda_B\Lambda_A^{-1}$ where the eigenvalues of $A$ are in ascending order and the eigenvalues of $B$ are in descending order.<|endoftext|>
TITLE: Two questions about higher Souslin trees
QUESTION [9 upvotes]: Assume $V=L$ and let $\kappa$ be a Mahlo cardinal. Let $L[G]$ be the generic extension obatined by Mitchell forcing to make $2^{\aleph_0}=\aleph_2=\kappa.$
It is known that in the extension there are no special $\aleph_2$-Aronszajn trees but there are $\aleph_2$-Aronszajn trees. 

Question 1. Is there any $\aleph_2$-Souslin tree in $L[G]?$

I assume the believe is that there are, but I don't know how to prove it. Surprisingly, if, instead of $\aleph_2$, we consider a cardinal $\lambda^+ > \beth_\omega,$ with $\lambda$ regular and apply the Mitchell forcing to get 
$2^\lambda=\lambda^{++}=\kappa,$ then the results of Assaf Rinot show that there are $\lambda^{++}$-souslin trees in the extension.

My second question is motivated by the work of Laver-Shelah. Assume $V=L$ and $\kappa$ is weakly compact. In order to produce a model of $CH+$there are no 
 $\aleph_2$-Souslin trees, Laver and Shelah, first force with Levy collapse $Col(\aleph_1, < \kappa),$ and over it do an iteration to kill all possible $\aleph_2$-Souslin trees.

Question 2. Are there any $\aleph_2$-Souslin trees just after doing the Levy collapse $Col(\aleph_1, < \kappa)$?

REPLY [6 votes]: About question 1: If $\kappa$ is not weakly compact in $L$, then there is an $\aleph_2$-Suslin tree in $L[G]$. 
In $L$ there is a $\kappa$-Suslin tree, $T$, and since Mitchell's forcing is $\kappa$-Knaster, it cannot add an antichain of cardinality $\kappa$ to $T$: If $\dot{\mathcal{A}}$ is a name for unbounded antichain in $T$, we can pick for every $\alpha < \kappa$, a condition $p_\alpha$ in Mitcell's forcing, such that $p_\alpha$ forces that the $\alpha$-th member of $\dot{\mathcal{A}}$ is some specific $t_\alpha\in T$. Let $I\subseteq \kappa$ be an unbounded subset such that for all $\alpha, \beta\in I$, $p_\alpha$ is compatible with $p_\beta$ then $\{t_\alpha \mid \alpha \in I\}$ form an antichain in $L$.

About Question 2:
As it is stated in the comments by Assaf, the answer to question 2 is also affirmative.<|endoftext|>
TITLE: Discrete Morse theory: how do zig-zag paths determine homotopy type?
QUESTION [14 upvotes]: Let $\Delta$ be a simplicial complex (or more generally, a regular CW complex). Let $\mathcal{M}$ be a Morse matching (or equivalently, a discrete Morse function) on $\Delta$. 
By Forman's theorems, $\Delta$ is homotopy equivalent to a CW-complex whose cells are $\mathcal{M}$-critical (=unmatched) simplices. Its homology is computed from the chain complex, in which an entry of the $k$-th boundary matrix equals the sum of signs of all zig-zag paths (=directed paths in the Hasse diagram of $\Delta$ with the edges from $\mathcal{M}$ reversed) between a critical $k$-cell and critical $k\!-\!1$-cell.


Do zig-zag paths determine how the critical cells are glued onto each
  other?

A CW-complex $X$ is determined, up to homotopy, by the number of cells in each dimension, and the homotopy class of each gluing map $S^k \!\longrightarrow\! X^{(k)}$. 

Q1: If all critical cells are in dimensions $0,k,k\!+\!1$, then gluing
  maps are determined by equivalence classes in
  $\pi_k(S^k)\cong\mathbb{Z}$. Does the sum of the signs of all zig-zag
  paths from a $k\!+\!1$-cell to a $k$-cell equal the degree of the
  gluing map?

I suspect this to be true. More intriguingly:

Q2: If all critical cells are in dimensions $0,k,k\!+\!2$ with $k\!\geq\!3$, then gluing
  maps are determined by equivalence classes in
  $\pi_{k+1}(S^k)\cong\mathbb{Z}_2$. Does the sum of the signs of all paths (of a new type) from a $k\!+\!2$-cell to a $k$-cell determine the gluing map?

In general, the homotopy groups of spheres are not all of the form $\mathbb{Z}_m$, so probably summing the signs of zig-zag paths is not sufficient, i.e. some information is lost when applying the matching. For instance, if critical cells are of dimension $1,4,8$, then gluing maps are determined by equivalence classes in $\pi_7(S^4)\cong\mathbb{Z}\!\oplus\!\mathbb{Z}_{12}$. Furthermore, the gluing maps of $k\!+\!1$-cells go into the $k$-skeleton, so they are determined by their representatives in $\pi_k(X^{(k)})$, but according to Mihai Damian, On the higher homotopy groups of a finite CW-complex, Topology Appl. 149 (2005), no. 1-3, 273--284., a homotopy group of a finite CW-complex may be infinitely-generated!
I have an example of $\Delta$ and $\mathcal{M}$ that produce critical  $1$ $0$-simplex, $1$ $5$-simplex, $10$ $7$-simplices. Can I conclude that $\Delta\simeq S^5\vee\bigvee_{\!10}S^7$ by inspecting certain (???) paths in the Hasse diagram?

REPLY [6 votes]: Thanks to Cosheaf Overlord Justin Curry for bringing this question to my attention. I'm only going to address the first question here, and I think with some computations (whose complexity depends on your individual complexes) you can extract answers to the other two questions yourself.
There is a very precise relation between zigzags of an acyclic partial matching $\mathcal{M}$ and the homotopy type of $\Delta$, namely:

There exists a poset-enriched category whose objects are the $\mathcal{M}$-critical cells, whose morphisms are equivalence classes of zigzags where only arrows in $\mathcal{M}$ can point backwards, and whose classifying space is homotopy-equivalent to $\Delta$.

All details are in the preprint here; the category mentioned above is a discrete analogue of the flow category described by Cohen, Jones and Segal in this paper.

Here is a brief summary of how to prove the result: given a regular CW complex $\Delta$ one has the "entrance path category" $E_\Delta$ whose objects are the cells and morphisms $E_\Delta(x,y)$ are descending sequences of faces $(x > z_0 > \cdots > z_n > y)$. You compose by concatenation, and note that these descending sequences have a poset structure (by inclusion of subsequences) where $(x > y)$ is minimal in $E_\Delta(x,y)$ if $x \neq y$ and $(x)$, which serves as the identity, happens to be unique and therefore minimal in $E_\Delta(x,x)$. Regularity guarantees that the classifying space of $E_\Delta$ lies in the homotopy class of $\Delta$.
Now, every acyclic partial matching picks out a class of minimal morphisms $\{(x_\bullet > y_\bullet)\}$ which I'll call $\mathcal{M}$. And the zigzags you see are precisely the sorts of morphisms which arise when you localize about (i.e., formally invert) elements of $\mathcal{M}$ in the poset-enriched category $E_\Delta$. Call this localized poset-enriched category $E_\Delta[\mathcal{M}^{-1}]$ and consider its full subcategory $C_\Delta(\mathcal{M})$ whose objects are the critical cells. Here's the main result of that preprint:

The localization functor $E_\Delta \to E_\Delta[\mathcal{M}^{-1}]$ and the inclusion functor $C_\Delta(\mathcal{M}) \hookrightarrow E_\Delta[\mathcal{M}^{-1}]$ both induce a homotopy-equivalence of classifying spaces.

The proofs of both equivalences require investigating fiber 2-categories and the appropriate use of Quillen's Theorem A. Things get slightly technical, but you can find an explicit computation in Sec 7.1 of the preprint. In any case, one immediate corollary is the quoted theorem above.<|endoftext|>
TITLE: Fundamental difference between Poisson Point Process and Binomial Point Process
QUESTION [6 upvotes]: What is the fundamental difference between Poisson Point Process and Binomial Point Process?
I am evaluating a solution in a Binomial Point Process setup. If I want to evaluate that in a Poisson Point Process setup, what all issues need to be considered?

REPLY [5 votes]: This figure (copied from these notes) serves to illustrate the difference between a binomial and a Poisson point process. Shown are $n=100$ points randomly and independently placed in the unit square. This is a binomial point process, because the distribution of the number of points in an area $W$ inside the unit square is the binomial distribution with $n$ attempts and success probability $p=W$.
The same figure can also be seen as a realization of Poisson point process, given that there are $n=100$ points in the unit square. If this is not given, different realizations of the Poisson process will have different $n$'s, and the distribution of $n$ is the Poisson distribution. This is the only difference between the two processes, the binomial point process has a fixed number of points, while in the Poisson point process this number varies between different realizations.
If $n$ is large enough, the number of points in the unit square for the Poisson process will be peaked around the average of the Poisson distribution, so you might neglect the fluctuations around the average and just fix $n$ at the average value. In that approximation the Poisson and binomial point processes amount to the same thing.<|endoftext|>
TITLE: Is there a relation between the Hessian matrix and the structure Tensor?
QUESTION [6 upvotes]: for a 2 dimensional image, i am interested to find a relation between the Hessian
$H = \begin{pmatrix} \frac{\partial^2 I}{\partial x^2}&\frac{\partial^2 I}{\partial x \partial y}\\ \frac{\partial^2 I}{\partial x \partial y}&\frac{\partial^2 I}{\partial y^2} \end{pmatrix}$ and the structure tensor $ST = \begin{pmatrix} w*\frac{\partial I}{\partial x}\frac{\partial I}{\partial x}&w*\frac{ \partial I}{\partial x}\frac{ \partial I}{\partial y}\\ w*\frac{\partial I}{\partial x}\frac{ \partial I}{\partial y}&w*\frac{\partial I}{\partial y}\frac{ \partial I}{\partial y}, \end{pmatrix}$, where $w$  is a weight function. I am using a Gaussiankernel. As far as i know, the structure tensor is also called "autocorrelation matrix" or "second moment matrix". 
I am using these two matrices to find so called "interesting points" in an image. For the $ST$ i can find the eigenvectors $v_1$ und $v_2$ with the corresponding eigenvalues $\lambda_1>\lambda_2$.
$v_1$ should have a orientation with the highest grayvalues fluctuation, so i think you can say it is a normalvector. $v_2$ stands orthogonal to $v_1$ and should have a orientation with the lowest grayvalues fluctuation.
I found out that the eigenvalues of the Hessianmatrix can be seen as the principal curvatures $\kappa_1$ $\kappa_2$. Because for both matrices the determinant of the matrix gives us $det(M)=\lambda_1 \cdot \lambda_2$, i can calculate for the hessian $det(H)=\lambda_1 \cdot \lambda_2=\kappa_1 \cdot \kappa_2$. The eigenvectors of the Hessian gives us the principal directions.
If i want to find a interesting point, i calculate for all points in the image, the determinante of both matrices. If i say that the determinant should be max i find with both matrices some interesting point. This means that both eigenvectors have a large magnitude and the point should be a corner in the image or any other singularity. Generally i observed that the ST gives me fewer points than the Hessian and that most of the points detected with ST is also found with the Hessian (the interesting points dont have exactly same position, but mostly the are only one pixel away. so i take it as the same points). 
So what i want to know is: If i take the ST can i transform it somehow and get the Hessianmatrix? Because the eigenvalues of both matrices are important to me,i want to know if i can find the eigenvalues or an approximation of the Hessian using the ST. I am also interested to find a reverse way for this problem. I found an approach which i think could help but i cant make anything out of this. The Weingarten mapping matrix is $W = M \cdot T$ with.
$M = \begin{pmatrix} 1+\frac{\partial I}{\partial x}\frac{\partial I}{\partial x}&\frac{ \partial I}{\partial x}\frac{ \partial I}{\partial y}\\ \frac{\partial I}{\partial x}\frac{ \partial I}{\partial y}&1+\frac{\partial I}{\partial y}\frac{ \partial I}{\partial y}, \end{pmatrix}$ and 
$T = (1+\frac{ \partial I}{\partial x}\frac{ \partial I}{\partial x}+\frac{ \partial I}{\partial x}\frac{ \partial I}{\partial x})^{-1/2} \cdot H $ 
thank you.

REPLY [2 votes]: The ST can be related to the hessian of the first order taylor expansion of a non-linear function you evaluate the hessian on.
There is a lot of information on this subject from people performing gradient descent algorithm as the outer product of the jacobian (ST for an image) is a one of the quasi-Newton methods known as Gauss-Newton method.
While the gradient contains information on the orientation, the structure tensor is a way to have an average of the orientation that respect orthogonality as it is needed to compute eigen{values|vectors}. This helps to find corner by analyzing invariant of this tensor, which contain only orientation information. So if you have two high eigenvalues you are at a corner, one you are at an edge.
The hessian matrix would tend to be infinite on the edge and corner as the curvature is infinite ... So indeed it could be interesting for corner detection. You have to keep in mind that the impact of the noise on derivation (derivation ca be seen as a high pass filter) is very important so when you have a filter with sufficient characteristic length for the structure tensor, it won't be necessarily the case for the hessian matrix, so you will have to increase your characteristic length ...
It is not obvious to state if it is better to increase your characteristic length or to approximate your hessian matrix for detection purposes.<|endoftext|>
TITLE: why use Crofton's formula in order to prove log-analyticity of volume?
QUESTION [6 upvotes]: In their interesting paper "Integration des fonctions sous-analytiques et volumes des sous-ensembles sous-analytiques" Lion and Rollin show that the volume of a fibre $Y_x$ of a globally subanalytic set $Y \subset \mathbb{R}^{m+n}$ is a function of $x \in \mathbb{R}^n$ which is of the form
$$
P(t_1, \ldots, t_d, \log(t_1), \ldots, \log(t_d))
$$
where $P$ is a polynomial, and $t_1, \ldots, t_d$ are globally subanalytic functions of $x$.
I will only speak about the situation where all $Y_x$ are uniformly bounded (the general case can be reduced to it).
The first step in the proof is to apply Crofton's formula in order to write the volume of $Y_x$ as the integral
$$
\int_{e \in \mathrm{Gr(k,m)}} \delta(x,e) d\mu
$$
where $\mu$ is some measure (with analytic density) on the Grassmanian of affine $k$-subspaces of $\mathbb{R}^m$, and $\delta(x,e)$ is the number of intersections of $Y_x$ with $k$-dimenional affine subspace $e$ (if it is finite; otherwise $\delta=0$).
The authors then say that, by application of Gabrielov's complement theorem, and uniform finiteness of subanalytic sets, the above integral is just an integral of a subanalytic function on the Grassmanian, which in turn can be reduced to an integral over a unit box of appropriate dimension by an analytic base-change. The proof then proceeds to use integration and elemination theorems to conclude. 
My question is: why does one have to resort to Cauchy-Crofton's formula in order to reduce calculation of the volume of $Y_x$ (as a function of $x$) to an integral of this type? It seems that if one takes just the integral of the indicator function of $Y_x$, then such integral can be reduced to the integral as in the previous paragraph after an appropriate base change.

REPLY [2 votes]: This is an embarrasing confusion, really, the answer to this question seems to be less surprising than I expected. 
Of course, the integral of the indicator function mentioned in the question has nothing to do with the volume of $Y_x$ which is a $k$-dimensional submanifold of $\mathbb{R}^n$, so one has to integrate a $k$-dimensional volume form inhereted from $\mathbb{R}^n$ over it. One can of course proceed formally by cutting $Y_x$ according to some atlas, and taking care that one ends up with an integral over an analytic domain of an analytic volume form, but Crofton's formula just does this in one sweep.<|endoftext|>
TITLE: A long-lasting conjecture of Pólya & Szegő
QUESTION [31 upvotes]: There is a conjecture by Pólya & Szegő (~1950, stated in p. 159 of their book Isoperimetric Inequalties in Mathematical Physics) which is as follows:
"Of all $n$-gons of a fixed area, the regular $n$-gon minimizes the first Dirichlet eigenvalue."
Surprisingly, this is still open (to my knowledge) for the general case. The only settled cases are the triangles and the quadrilaterals (see Henrot's survey). 
 Is there any progress on the general case?

REPLY [12 votes]: I'm pretty sure that this is still open for $n$-gons (with $n\geq 5$). As far as I know, basically no progress has been made since the original proofs for triangles/quadrilaterals. 
There have been some numerics as well as some refined inequalities for triangles. This article might point you towards some of these results.
Interestingly, a related conjecture of Pólya & Szegő is resolved "the regular $n$-gon has least logarithmic capacity among $n$-gons of a fixed area" http://www.ams.org/mathscinet-getitem?mr=2052355.<|endoftext|>
TITLE: Square root of dirac delta function
QUESTION [58 upvotes]: Is there a measurable function $ f:\mathbb{R}\to \mathbb{R}^+ $ so that $ f*f(x)=1 $ for all $ x\in \mathbb{R} $, i.e  $$\int\limits_{-\infty}^{\infty} f(t)f(x-t) dt=1 $$ for all $ x\in \mathbb{R} $.

REPLY [48 votes]: So I guess my initial intuition was wrong; there is enough "room at infinity" to concoct such a function $f$.  The key lemma is

Lemma.   Let $m_1,m_2,m_3,\dots$ be an enumeration of the integers.  Then there exists an increasing sequence $0 = f_0 \leq f_1 \leq f_2 \leq \dots$ of finitely supported functions $f_n: {\bf Z} \to {\bf R}^+$ such that for any $n \geq 0$, the function $f_n*f_n$ equals one on $m_1,\dots,m_n$, and is bounded above by $1-2^{-n}$ for all other integers.

Indeed, with such a sequence $f_n$, if one sets $F := \sup_n f_n$, then by  monotone convergence $F: {\bf Z} \to {\bf R}^+$ is such that $F*F = 1$ on the integers, and if one then sets $f: {\bf R} \to {\bf R}^+$ to be $f(x) := F(\lfloor x \rfloor)$ (thus, $f$ is the convolution of $\sum_n F(n) \delta_n$ with $1_{[0,1]}$, with $\delta_n$ the Dirac delta at $n$) then $f*f=1$ on the reals also.
One proves the proposition by a recursive construction on $n$.  Clearly $f_0=0$ obeys the required properties.  Now suppose that $n \geq 1$ and that $f_0,\dots,f_{n-1}$ have already been constructed.  Set $M$ to be a sufficiently large natural number (depending on $n$ and $f_{n-1}$), and let $a_1,\dots,a_M$ be very large natural numbers that are generic in the sense that $(a_1,\dots,a_M)$ avoids a finite number of affine hyperplanes in ${\bf R}^M$.  Set
$$ f_n := f_{n-1} + \sqrt{\frac{1-f_{n-1}*f_{n-1}(m_n)}{2M}} \sum_{i=1}^M (\delta_{a_i} + \delta_{m_n - a_i})$$
where $\delta$ now denotes the Kronecker delta.
Clearly $f_n$ is finitely supported with $f_n \geq f_{n-1}$.  If $M$ is large enough, and $a_1,\dots,a_M$ are in generic, one can verify that $f_n*f_n - f_{n-1} * f_{n-1}$ vanishes on $m_1,\dots,m_{n-1}$, equals $1 - f_{n-1}*f_{n-1}(m_n)$ on $m_n$, and is bounded by $2^{-n}$ elsewhere, giving the required claim.
EDIT: It seems the basic reason why such a construction works is because convolution is not bounded on $\ell^\infty({\bf Z})$ (or on $L^\infty({\bf R})$); this unboundedness makes the problem significantly more underdetermined, and thus easier to solve.  For instance, since convolution is bounded from $L^2({\bf R}) \times L^2({\bf R})$ to $L^\infty({\bf R})$, there is no solution to $f*f=1$ with $f \in L^2({\bf R})$; indeed, a dense subclass argument or the Riemann Lebesgue lemma plus Plancherel shows that $f*f \in C_0({\bf R})$ whenever $f \in L^2({\bf R})$.  In constrast, the function $f$ constructed above can be seen to lie in $L^{2,\infty}({\bf R})$ but no better, and convolution is not bounded from $L^{2,\infty}({\bf R}) \times L^{2,\infty}({\bf R})$ to $L^\infty({\bf R})$.<|endoftext|>
TITLE: H-space structures on non-sphere suspensions?
QUESTION [15 upvotes]: It is well known that $S^n$ admits an H-space structure if and only if $n=0,1,3,7$. I'm interested in whether there are other suspensions $\Sigma X$ that admit H-space structures:
Question 1 For which $X$ (not a sphere) is $\Sigma X$ an H-space? And what about $\Sigma X$ that are associative H-spaces?
My motivation is that we have a construction (in the framework of homotopy type theory, and presumably portable to a wide range of model categories) that gives an H-space structure on the join $\Sigma X * \Sigma X$ whenever $\Sigma X$ has a homotopy-associative H-space structure (that is compatible with an involution on $X$ – for details, see these slides). Thus, it would be interesting to know some more examples where this construction applies. 
This also leads to a follow-up question (mostly in case the answer to Q1 is “none”):
Question 2 If we go to a localization, do we get more answers to Q1? What about in other (non-stable) model categories?
Any references would be appreciated.
Finally, let me share a little scratch-work in trying to answer Q1 (feel free to ignore!): Any H-space structure $\mu : \Sigma X \times \Sigma X \to \Sigma X$ gives rise a Hopf map $H(\mu) : \Sigma X * \Sigma X \to \Sigma^2 X$. Assuming $X$ is pointed, we get a map $\Sigma^3(X \wedge X) \to \Sigma^2 X$. Applying $K$-cohomology to the cofiber sequence should yield information restricting $X$, but I failed to get much milage out of it without a notion of “bidegree” for $\mu$ (generalizing the case of $X$ a sphere).

REPLY [3 votes]: Here are some comments about the case where $X$ is not assumed to have finite type.  Put $Y=\Sigma X$.  For any field $K$, the groups $H_*(Y;K)$ form a Hopf algebra in which all elements of the augmentation ideal are primitive.  If $u$ and $v$ lie in the augmentation ideal, then $u$, $v$ and $uv$ are all primitive, which gives $u\otimes v+(-1)^{|u||v|}v\otimes u=0$.  If $u$ and $v$ are nonzero, it follows that $|u|=|v|$ and $Ku=Kv$, and $|u|$ is odd unless $K$ has characteristic $2$.  Thus, $H_*(Y;K)$ is either $K$ or $K\oplus Ku$ for some element $u$, usually of odd degree.  In particular, we see that $Y$ is rationally trivial or an odd-dimensional sphere.
Now consider the groups 
\begin{align*}
 A(p)_k &= \widetilde{H}_k(Y;\mathbb{Z}_{(p)}) \\
 B(p)_k &= A(p)_k/p \\
 C(p)_k &= \widetilde{H}_k(Y;\mathbb{Z}/p) \\
 D(p)_k &= \text{ann}(p,A(p)_{k-1})
\end{align*}
so that 

$A(p)\otimes\mathbb{Q}$ has dimension $0$ or $1$ over $\mathbb{Q}$
$C(p)$ has dimension $0$ or $1$ over $\mathbb{Z}/p$
There is a short exact sequence $B(p)\to C(p)\to D(p)$, so $B(p)$ and $D(p)$ have dimension $0$ or $1$, and at least one of them is zero.

There are a number of different possibilities here.

If $D(p)=0$ then $A(p)$ is torsion-free and so injects in $A(p)\otimes\mathbb{Q}$.  This means that $A(p)=0$ or $A(p)\simeq\mathbb{Z}_{(p)}$ or $A(p)\simeq\mathbb{Q}$.
If $B(p)=0$ then $A(p)$ is divisible, and so is injective as a $\mathbb{Z}_{(p)}$-module.  If we let $T(p)$ denote the torsion part of $A(p)$ then we find that $T(p)$ is also divisible and therefore injective and therefore a summand in $A(p)$.  We therefore have $A(p)=T(p)\oplus Q(p)$, where $Q(p)$ is a $\mathbb{Q}$-module.  It follows that $A(p)\otimes\mathbb{Q}\simeq Q(p)$, so $Q(p)$ is $0$ or $\mathbb{Q}$.  If $D(p)=0$ then $T(p)=0$.  If $D(p)=\mathbb{Z}/p$ then I think it follows that $T(p)=\mathbb{Z}/p^\infty$.

The most interesting question arising from this analysis is as follows.  Let $Y$ be the Moore space with $H_2(Y)=\mathbb{Z}/p^\infty$ for some prime $p$.  To avoid trouble from the low-dimensional homotopy groups of spheres, we may want to take $p\geq 5$.  Note that $\widetilde{H}_*(Y;K)=0$ unless $K$ has characteristic $p$, and that $\widetilde{H}_*(Y;\mathbb{Z}/p)$ is a copy of $\mathbb{Z}/p$ in dimension $3$.  Does $Y$ have an $H$-space structure?  I do not see an easy way to answer that.  Note that $Y\wedge Y$ is a Moore space with $H_5(Y\wedge Y)=\mathbb{Z}/p^\infty$, but that does not immediately give a good hold on $[Y\times Y,Y]$.<|endoftext|>
TITLE: Is $\max_{\|x\|_p=\|y\|_p=1} |\langle x, Ay\rangle|$ equivalent to $\max_{\|x\|_p=|} |\langle x, Ax\rangle|$ for symmetric $A$ & $p\geq 2$?
QUESTION [7 upvotes]: Let $A\in \mathbb{R}^{n\times n}$ be a symmetric matrix, and consider the $l_p$ norm ($p\geq 2$).
Can we prove that the following problems are equivalent:
$$\max_{\|x\|_p=\|y\|_p=1} \left| \langle x, Ay\rangle \right|$$
and 
$$\max_{\|x\|_p=1} \left| \langle x, Ax\rangle \right| $$
Can the result be generalized to symmetric tensor and symmetric multilinear form? In particular, if $F: R^n\times\cdots\times R^n$ is a $m$-order symmetric multilinear form, can we prove the problem
$$\max_{\|x_1\|_p=\cdots=\|x_m\|_p=1} \left| F(x_1,\ldots,x_m) \right|$$
is equivalent to
$$\max_{\|x\|_p=1} \left| F(x,\ldots,x) \right| $$
If not, can we prove if for any special case? (the case $p\neq 2$ would be more interesting.)
Thank you very much!
PS: I would even highly appreciate if someone could give me an example where the above statement fails.

REPLY [7 votes]: This is false for every $p>2$. Take $A=\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$. Then the maximum of $|\vec{x}^T A \vec{x}|$ on $|x|_p=1$ is $1$, achieved on the coordinate axes. (The proof is an easy computation with Lagrange multipliers.) But the maximum of $\vec{x}^T A \vec{y}$ is $2^{1-2/p}$ (achieved at $(\sqrt[p]{1/2}, \sqrt[p]{1/2})$, $(\sqrt[p]{1/2}, - \sqrt[p]{1/2})$), which is greater for $p>2$. 
(Of course, the claim is true if $A$ is positive definite. By the Cauchy-Schwartz inequality, if $A$ is positive definite, $|\vec{x}^T A \vec{y}| \leq \sqrt{(\vec{x}^T A \vec{x}) (\vec{y}^T A \vec{y})} \leq \max(\vec{x}^T A \vec{x}, \vec{y}^T A \vec{y})$, so $\max_{\vec{x}, \vec{y} \in S} |\vec{x}^T A \vec{y} | = \max_{\vec{z} \in S} \vec{z}^T A \vec{z}$ for any $S$ at all.)
For $p=2$ and $F$ of any degree, this is true and is a theorem of Banach: "Über homogene Polynome in $(L^2)$" Studia Mathematica (1938) Volume: 7, Issue: 1, page 36-44 . A proof in English can be found as Proposition $1.2^{\circ}$ in  Polynomials and multilinear mappings in topological vector-spaces Jacek Bochnak; Józef Siciak Studia Mathematica (1971) Volume: 39, Issue: 1, page 59-76
I learned all of this from "Estimates for polynomial norms on $L_p(\mu)$ spaces", I. Sarantopoulos, Mathematical Proceedings of the Cambridge Philosophical Society, Volume 99, Issue 02, March 1986, pp 263-271, which has much more to say about the problem.<|endoftext|>
TITLE: Elements of infinite order in the topological mapping class group
QUESTION [10 upvotes]: Let $M$ be a closed topological manifold, and let $\operatorname{MCG}(M):=\operatorname{Homeo}(M)/\operatorname{Homeo}_0(M)$ denote the topological mapping class group of $M$ ($\operatorname{Homeo}_0(M)$ denotes the identity component of $\operatorname{Homeo}(M)$).  More generally, we could let $M$ be compact with boundary and consider homeomorphisms fixing the boundary.

What known methods/invariants can be used to certify that a given element $\varphi\in\operatorname{MCG}(M)$ has infinite order?

Of course, $\varphi$ might have infinite order for homotopical reasons, but I am primarily interested in finer invariants.  So, for the sake of this question, let's assume $\varphi$ is homotopic to the identity as maps $M\to M$.  Of course, if $M$ has boundary, then this means homotopic through maps fixing the boundary.
For instance, the kernels in this answer are not detected by their action on homotopy, however unfortunately they are purely torsion (though I would still like to know how they are detected).
The focus of this question is on the high-dimensional case, though an answer in any dimension $\geq 4$ would be interesting.

REPLY [4 votes]: Let me suppose that $M$ is a closed manifold of dimension $d$, and let $\varphi : M \to M$ be a diffeomorphism / homeomorphism which is homotopic to the identity, and choose such a homotopy $h_t$. The mapping torus $X_\varphi$ of $\varphi$ is a smooth / topological manifold fibering over $S^1$, and our choice of homotopy $h_t$ yields a preferred homotopy equivalence
$$ H : X_\varphi \longrightarrow S^1 \times M.$$
For each cohomology class $v \in H^{d-4i+1}(M;\mathbb{Q})$ we may form
$$\int_{X_\varphi} p_i(TX_\varphi - H^*TM) \smile H^*(1 \otimes v) \in \mathbb{Q}$$
where $p_i$ is the (rational) Pontrjagin class. This defines a linear functional $H^{d-4i+1}(M;\mathbb{Q}) \to \mathbb{Q}$ which is Poincare dual to a class $\xi_i \in H^{4i-1}(M;\mathbb{Q})$.
Morally this class measures the following: on the one hand $\varphi$, by virtue of being a homeomorphism, preserves the rational Pontrjagin class $p_i(M)$, so preserves a given cocycle representative $c$ up to a "preferred" coboundary. On the other hand, the homotopy $h_t$ gives another coboundary for $c-\varphi^*c$; the difference between these cobounding cochains is thus a cycle, and it represents the cohomology class $\xi_i$.
In principle $\xi_i$ is an invariant of the pair $(\varphi, h_t)$. However, a different choice of homotopy $h_t'$ yields a homotopy equivalence $H'$ which differs from $H$ by postcomposition by a homotopy automorphism of $S^1 \times M$ which i) is over $S^1$ and, ii) fixes a fibre $M \times \{*\}$. This means that for each $v$ we have
$$(H')^*(1 \otimes v) = H^*(1 \otimes v) + H^*(u \otimes \bar{v})$$
for some $\bar{v}$, where $u \in H^1(S^1;\mathbb{Q})$ is the canonical class. But
$$\int_{X_\varphi} p_i(TX_\varphi - H^*TM) \smile H^*(u \otimes \bar{v})$$
can be written as
$$\int_{S^1 \times M} (H^{-1})^*(p_i(TX_\varphi-H^*TM)) \smile (u \otimes \bar{v}) = \int_M p_i(TM-TM) \cup \bar{v}=0$$
so in fact $\xi_i$ only depends on $\varphi$.
As $X_{\varphi \circ \psi}$ is cobordant to $X_{\varphi} \sqcup X_{\psi}$, one sees that $\xi_i(\varphi \circ \psi) = \xi_i(\varphi) + \xi_i(\psi)$. In particular if $\xi_i(\varphi) \neq 0$ for some $i$ then $\varphi$ has infinite order.
There is a theorem due to Sullivan that a simply-connected high-dimensional manifold is determined "up to finite ambiguity" by its homotopy type and rational Pontrjagin classes. The analogous statement for automorphisms of simply-connected (or perhaps 2-connected?) manifolds says that elements in
$$\mathrm{Ker}(\pi_0(Top(M)) \to \pi_0(G(M)))$$
(or the smooth analogue) are determined up to finite ambiguity by their associated $\xi_i$'s. This can (surely?) be proved using the surgery exact sequence.<|endoftext|>
TITLE: Easier Girard's paradox in a circular pure type system (PTS)
QUESTION [7 upvotes]: System U is an inconsistent PTS in that one has a term of type $\bot = \forall p\colon \ast \ldotp p$, and such a term is explicitly constructed in Hurkens' A Simplification of Girard's Paradox.
One-sorted circular PTS $\lambda\ast$ ($S = \{\ast \}, A=\{\ast\colon \ast\}, R=\{(\ast,\ast)\}$) (Geuvers' Logics and Type Systems, p.78) is also inconsistent since a term of type $\bot$ in System U can be translated into $\lambda\ast$ by mapping $\ast, \square$ and $\triangle$ to $\ast$. 
Is there any easier way to construct a term of type $\bot$ in $\lambda\ast$? Smaller terms, or constructions easier to understand, are appreciated.

REPLY [7 votes]: I feel like most of my posts on mathoverflow and cstheory.stackexchange consist of this answer, but the most perspicuous (in my opinion) proof of inconsistency of U and $*:*$ is a construction by Alexandre Miquel, given in his phd dissertation. Tragically, it is in French, so I'll summarize the idea below, and maybe you'll be able to fill in the remainder using his other paper lamda-Z: Zermelo's Set Theory as a PTS with 4 Sorts..
The idea is to build a model of naive set theory in $*:*$. The naive theory allows for unrestricted comprehension, and in particular the paradox falls out quite easily. The proof term of $\bot$ itself is longer than in Hurkens' presentation, but I think it's safe to say that the process is at least more straightforward.
The crucial idea is to model a set as a directed pointed graph. The point represents the set, the graph contains all the members of the set, and the membership relation is represented by the edges. We rely on the encoding of $\Sigma$ types in the impredicative theories.
A set $S$ is thus an element of
$$\mathrm{Set}=\Sigma (A:*)(x :A)(R:A\rightarrow A\rightarrow *) $$
where the carrier is $A$, the base is $x$ and the edge relation is $R$. What's a bit counter-intuitive is that sets are not necessarily well-founded. To get sane notions of equality and membership, we need to define equality as bisimilarity $\simeq$ of pointed graphs.
Then it's pretty easy to define membership: $S\in T$ if

$T = (A, x, R)$
There is some $y$ with $R\ y\ x$ ($y$ is an element of $x$)
$S\simeq (A, y, R)$ ($S$ is the "same set" as $y$)

Then we can show:

For each predicate $P:\mathrm{Set}\rightarrow *$ that respects bisimilarity there is a set
  $$ \{\ x\ |\ P(x)\ \}$$

It's then easy to show that $P(x)=x\notin x$ respects bisimilarity.<|endoftext|>
TITLE: Does there exist a bijection of $\mathbb{R}^n$ to itself such that the forward map is connected but the inverse is not?
QUESTION [146 upvotes]: Let $(X,\tau), (Y,\sigma)$ be two topological spaces. We say that a map $f: \mathcal{P}(X)\to \mathcal{P}(Y)$ between their power sets is connected if for every $S\subset X$ connected, $f(S)\subset Y$ is connected. 
Question: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$? 
Full disclosure: I originally asked this on Math.SE a year and a half ago; there were some discussion in the comments as well as a few answer attempts that were unfortunately flawed. Remarks 3 and 5 below capture the essences of many of those attempts. 
Various remarks

If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = \sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not. 
With the bijection, it holds true in $n = 1$. But this is using the order structure of $\mathbb{R}$: a bijection that preserves connectedness on $\mathbb{R}$ must be monotone. 
As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See https://math.stackexchange.com/q/949168/1543 which inspired this question for more about this.)
Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:

Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open? 

In fact, it is a Theorem of Tanaka's (see my answer here) that if $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection such that both $f$ and $f^{-1}$ are connected, then $f$ is a homeomorphism. So an equivalent formulation of the question is
Equivalent Question: Does there exist a bijection $f:\mathbb{R}^n\to\mathbb{R}^n$ such that $f$ is connected but discontinuous?
Some properties of $\mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $\mathbb{R}^n$ with the standard topology to itself, by self-maps of some arbitrary topological space, it is easy to make the answer go either way. 

For example, if the topological space $(X,\tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected. 
On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = \mathbb{Z}$ equipped with the topology generated by 
$$ \{\mathbb{N}\} \cup \{ \{k\} \mid k \in \mathbb{Z} \setminus \mathbb{N} \} $$
then the map $k \mapsto k+\ell$ for any $\ell > 0$ maps connected sets to connected sets, but its inverse $k\mapsto k-\ell$ can map connected sets to disconnected ones.

REPLY [4 votes]: I have a partial answer which should at least help to restrict our attention to a certain case. 
As noted, if either $f$ of $f^{-1}$ is continuous, then we are done by Invariance of Domain. So suppose $f$ is discontinuous at $x\in \mathbb{R}^n$. Then there exists $\epsilon_0>0$ such that for all $\delta>0$, there exists $y_\delta\in\mathbb{R}^n$ with $|x-y_\delta|<\delta$ and $|f(x)-f(y_\delta)|\geq \epsilon_0$. Note that
$$f(B(x,\delta))=\Big[f(B(x,\delta))\cap (B(f(x),\epsilon_0))\Big]\coprod\Big[f(B(x,\delta))\cap (B(f(x),\epsilon_0))^c\Big]=:U_\delta\coprod V_\delta$$
We know that each of these disjoint sets is non-empty since $f(x)\in U_\delta$ and $y_\delta\in V_\delta$. Furthermore, $U_\delta\subseteq f(B(x,\delta))$ is clearly open in the subspace topology. If only $V_\delta$ were open in $f(B(x,\delta))$ then $(U_\delta,V_\delta)$ would be a separation of $f(B(x,\delta))$, hence $f$ would not be connected!
But the problem happens at the boundary where $|x-y|<\delta$ and $|f(x)-f(y)|$ is precisely equal to $\epsilon_0$. If there are no such $y$, then $V_\delta=f(B(x,\delta))\cap \mathrm{Int}\big(B(f(x),\epsilon_0)\big)$ which is open in $f(B(x,\delta))$ and we are done. 
Otherwise, $\forall \delta>0, \exists y_\delta\in \mathbb{R}^n$ such that $|x-y_\delta|<\delta$ and $|f(x)-f(y_\delta)|=\epsilon_0$. Now, define 
$$E:=\{\epsilon>0:\hspace{.3cm}\forall\delta>0, \exists y_\delta\in\mathbb{R}^n\text{ such that }|x-y_\delta|<\delta\text{ and }|f(x)-f(y_\delta)|=\epsilon \}$$
Note that $E\neq\varnothing$ since $\epsilon_0\in E$. Moreover, $E$ is bounded below by zero so (by completeness of $\mathbb{R}^n$),  $\hat{\epsilon}:=\inf(E)$ exists and $0\leq \hat{\epsilon}<\infty$. 
Case 1: Suppose $\hat{\epsilon}>0$. 
Then $\frac{\hat{\epsilon}}{2}<\hat{\epsilon}$, so $\frac{\hat{\epsilon}}{2}\notin E$. Thus, $\exists \hat{\delta}>0$ such that  $|x-y|<\hat{\delta}\Longrightarrow |f(x)-f(y)|\neq \frac{\hat{\epsilon}}{2}$. 
But by the discontinuity of $f$, we know that $\exists y_{\hat{\delta}}\in\mathbb{R}^n$ such that $|x-y_{\hat{\delta}}|<\hat{\delta}$ and $|f(x)-f(y_{\hat{\delta}})|\geq  \epsilon_0\geq \hat{\epsilon}>\frac{\hat{\epsilon}}{2}$. 
Then we have
\begin{align*}f(B(x,\hat{\delta}))&=\Big[f(B(x,\hat{\delta}))\cap B(f(x),\frac{\hat{\epsilon}}{2})\Big]\coprod \Big[f(B(x,\hat{\delta}))\cap B(f(x),\frac{\hat{\epsilon}}{2})^c\Big]\\&=\Big[f(B(x,\hat{\delta}))\cap B(f(x),\frac{\hat{\epsilon}}{2})\Big]\coprod\Big[f(B(x,\hat{\delta}))\cap \big\{y\in \mathbb{R}^n: |f(x)-y|>\frac{\hat{\epsilon}}{2}\big\}\Big]\end{align*}
Thus, $f(B(x,\hat{\delta})$ is disconnected. 
Case 2: Suppose $\hat{\epsilon}=0$. 
This is where the trouble lies. If $f$ is not assumed to be bijective, then this can actually happen. The classic example of  $f(x)=\begin{cases}\sin(\frac{1}{x}),&x\neq 0\\0,&x=0\end{cases}\hspace{.3cm}$ satisfies $\hat{\epsilon}=0$. For any $0<\epsilon< 1$, any neighborhood of the origin contains a point $x$ for which $|f(x)|=\epsilon$ (in fact infinitely many). 
Therefore, if there exists a bijective discontinuous connected function, it must be of this form. That is, the $\epsilon$ which witnesses the discontinuity of $f$ can be taken arbitrarily close to zero. 
I hope this helps the search for a proof/counter-example!<|endoftext|>
TITLE: Fast Fourier Transforms for non-trigonometric bases
QUESTION [8 upvotes]: The fast Fourier transform allows decomposition into a sin/cos basis in $N \log(N)$ complexity. Can one generalize the algorithm (or the ideas used) to other bases, e.g. orthogonal polynomial bases such as Hermite, Legendre, Chebyshev,...? I have a special interest in Zernike polynomials as well.
If so, how?
If not, what is special about the sin/cos that separates it from the other cases?
Also asked previously on: https://math.stackexchange.com/questions/647613/fast-fourier-transform-for-non-trigoniometric-bases/1730645#1730645
not having quite a satisfying answer yet. 
Also, in a related question here on MO (here), I see pointing to this document on why the FFT would be fast. How could I use this principle on general orthogonal bases?

REPLY [2 votes]: Related questions have been considered in some depth by Dan Rockmore and collaborators. For more, check out Rockmore's web page.<|endoftext|>
TITLE: Interpreting Robinson arithmetic in a very weak set theory
QUESTION [9 upvotes]: It is known that adjunctive set theory interprets Robinson arithmetic, and that extensionality is not needed for that. (Montagna and Mancini, "A minimal predicative set theory", Notre Dame Journal of Formal Logic, 1994). 
I wonder whether it is known if the following weak set theory (without extensionality) also interprets Robinson arithmetic. Axioms:

$\exists y\forall x(x\notin y)$
For every n, $\forall x_1\ldots\forall x_n\exists y\forall z(z\in y\leftrightarrow z=x_1 \vee \ldots\vee z=x_n)$

REPLY [9 votes]: This theory was introduced by Vaught, and it does not interpret Robinson’s arithmetic. See Visser [1] for a thorough discussion of related theories; Vaught’s theory is denoted VS in the paper. (Note that the axioms are stated more concisely there: axiom 1 is a special case of axiom 2 for $n=0$.)
That VS does not interpret Robinson’s arithmetic Q follows by the following argument:

Q is finitely axiomatized. Thus, if it were interpretable in VS, it would be interpretable in its finite fragment.
VS is locally interpretable in the theory of nonsurjective pairing (using the obvious construction of $n$-tuples from pairs).
However, said theory of pairing is known to have decidable extensions, hence it does not interpret Q (or indeed, all of VS, which is also an essentially undecidable theory). Cf. Theorem 2 in [1,§3.2].

Reference:
[1] Albert Visser, Pairs, sets and sequences in first-order theories, Archive for Mathematical Logic 47 (2008), no. 4, pp. 299–326.<|endoftext|>
TITLE: Are quadratic units cyclotomic norms?
QUESTION [18 upvotes]: Consider the fundamental unit $\varepsilon$ of a real quadratic number field $k = {\mathbb Q}(\sqrt{p})$ for primes $p \equiv 1 \bmod 4$, and let $h$ denote its class number. By Dirichlet's work on class number formulas, $\varepsilon^h$ is a norm of a cyclotomic unit in the maximal real subfield $K^+$ of the field $K = {\mathbb Q}(\zeta_p)$ of $p$-th roots of unity. In particular, $\varepsilon$ is the norm of a cyclotomic unit if $h = 1$.
If $h > 1$ (in our case, $h$ is odd), we know that $h$ divides the class number $h^+$ of $K^+$, and that the cyclotomic units form a subgroup of index $h^+$ in the group of all units of $K^+$. Thus it might well be possible that $\varepsilon$ is the norm of a cyclotomic unit even in this case, and my question is: Is the fundamental unit of $k$ (for prime values of $p$) always the norm of a unit from $K^+$?

REPLY [7 votes]: Edit:Let $K=\mathbb Q(\zeta_p)^+$
Since it's easy to show $-1 \in N(\mathcal O_K ^{\times})$ and $\mathrm{Gal}(K/k)$ is cyclic, your question is equivalent to ask whether $\mathrm{H}^2(K/k,\mathcal O_K^\times)=0$. By a fact about  Herbrand quotient ([1]Proposition1.2.4) , this is equivalent to ask whether  $|\mathrm{H}^1(K/k,\mathcal O_K^\times)|=n$, where $n=[K:k]$.
Apply the exact sequence([1]Proposition1.2.3) to our case $K/k$. Since $K/k$ is totally ramified, and the raimified primes are principal. We have the following exact sequence
$$0\longrightarrow \mathrm{Ker}(J) \longrightarrow \mathrm{H}^1(K/k,\mathcal O_K^\times) \longrightarrow \mathbb{Z}/n\mathbb{Z}\longrightarrow 0,$$
where $J$ is the natural map from $Cl(k)$ to $Cl(K)$. 
Then we know $|\mathrm{H}^2(K/k,\mathcal O_K^\times)|=|\mathrm{Ker}(J)|$. So your question is to ask whether there is a nonprincipal ideal of $k$ becomes principal in $K$.  If $h_k$ is coprime to $n$, then $J$ is injective, so the norm of units is surjective as Pound Sterling said. If $h_k$ is not coprime to $n$, $J$ may be injective or not. For example, $p=229$,$|\mathrm{Ker}J|=3$, as Pound Sterling says the norm index is $3$.  $p=2089,h_k=3$, and $\mathrm{gcd}(h_k,n)=3$, but $J$ is injective, see[2,Page 2728], so the norm map between units is surjective. Numerically, most $p$ such that $\mathrm{gcd}(h_k,n)>1$ adimt a nontrival $\mathrm{Ker} J$, hence the norm map between units is not surjective. See the discussion in [2]Page 2727.
References:[1] Topics in Iwasawa theory. Greenberg.
[2] Visibility of ideal classes. Schoof and Washington.<|endoftext|>
TITLE: Roadmap to Geometric Representation Theory (leading to Langlands)?
QUESTION [37 upvotes]: I believe there has been at least one question similar to this one and yet I still think this particular question deserves to have a thread of its own.
I'm becoming increasingly fascinated by stuff related to geometric representation theory ($D$-Modules, geometric quantization, Langlands & CFT).
It's fair to say I'm mainly drawn to the geometric stuff but recently after reading a bit about the classification of semisimple lie algebras, I discovered the beauty in the non-geometric aspects as well. Still I feel like my knowledge of the representation side is a lot more basic than the geometric side.
Thus, my question is three-fold:

What are the most important/must-know results in the field so far? (even just buzzwords like "BB-lcalization" can be helpful). 
How much beyond the finite representations of finite groups should one know before diving into this subject? (Example: Is knowledge of the classification of real semisimple lie algebras a prerequiste for "D-Modules, Perverse Sheaves, and Representation Theory"?).
What would be a good roadmap to geometric representation theory with an eye toward geometric Langlands? In particular one aimed at a geometer with a minimal representation theory background?

By a "geometer" I mean someone with enough background in Differential/Complex/Modern-Algebraic Geometry to find himself around the geometric side of the story while having only knowledge of representation of finite groups on the representation side.

REPLY [12 votes]: Since this has many answers, I'll just put down what comes to mind.
It seems like you would be content to not worry about real semisimple or other fields and just grant yourself use of $\mathbb{C}$ and of $\mathbb{C}((t))$. That would be the safest option to specialize everything to this.
So you've already read some classification of semisimple so I presume you have some Weyl character computations done.
Some ordinary representation theory:

Why you are going to restrict yourself to category $\mathcal{O}$. How it is broken into blocks.
Translation functors on above
Classical Satake isomorphism using $\mathbb{C}((t))$. ( Note that wikipedia uses a stricter definition ). Pre-Geometric Satake.

Computations to do:

Do some Birkhoff factorizations
Trivialize some small rank vector bundle V on complex curve over the complement of finitely many points.
D-modules on $\mathbb{P}^1$, get an example of BB. Write these differential operators and pick some $\mathcal{D}$ modules to see which reps they give.
Some Bruhat Decompositions. Maybe do a non-type-A example. Parameterize a double bruhat cell as nicely as you can while here.

Buzz-People:

Mirkovic-Vilonen
Gan-Ginzburg
Algebraic Peter-Weyl
Higgs Bundle and Hitchin System
Riemann-Hilbert
Fourier-Mukai
Deligne - Geometric Abelian Class Field Theory
Kazhdan-Lusztig polynomials
Braden-Licata-Proudfoot-Webster

The lists are in the order that I thought of them; the order is not significant.<|endoftext|>
TITLE: Characterizing surface area
QUESTION [8 upvotes]: (This question is a variant of an unanswered question at math.stackexchange.)
The Definition section of Wikipedia's article on surface area currently starts as follows:

While the areas of many simple surfaces have been known since antiquity, a rigorous mathematical definition of area requires a great deal of care. 
  This should provide a function
  $$S \mapsto A(S)$$
  which assigns a positive real number to a certain class of surfaces that satisfies several natural requirements. The most fundamental property of the surface area is its additivity: the area of the whole is the sum of the areas of the parts. More rigorously, if a surface $S$ is a union of finitely many pieces $S_1, \dots, S_r$ which do not overlap except at their boundaries, then 
  $$A(S) = A(S_1) + \cdots + A(S_r).$$
  Surface areas of flat polygonal shapes must agree with their geometrically defined area. Since surface area is a geometric notion, areas of congruent surfaces must be the same and the area must depend only on the shape of the surface, but not on its position and orientation in space. This means that surface area is invariant under the group of Euclidean motions. These properties uniquely characterize surface area for a wide class of geometric surfaces called ''piecewise smooth''.

It seems that not all the properties being hinted at are mentioned explicitly. 
Indeed, I can define the "constant-curvature surface area" of $S$ to be
the supremum of the sums of surface areas of finitely many disjoint subsets of $S$, all of which have constant curvature,
and it will satisfy all the mentioned properties. But the constant-curvature surface area of a general ellipsoid, say, will be zero. So:
Which properties are missing in the Wiki article?

REPLY [4 votes]: I would add some kind of monotonicity. 
Say if there is a distance nonexpanding map between surfaces  $f\colon S\to S'$ then 
$$\mathop{\rm area} S\ge \mathop{\rm area} S'.$$
P.S. Instead of all distance nonexpanding maps one can take only orthogonal projections to planes.<|endoftext|>
TITLE: Reference for Manin's idea on algebraic geometry over the symmetric monoidal model category of Motives
QUESTION [11 upvotes]: Reference for Y. Manin's idea of "algebraic geometry over the symmetric monoidal model category of motives."
Has been sugested to me that this was made in a Manin's letter. There is an escaned copy?
Some work in this direction has been made in the thesis of Spitzweck, but I refer to first references of Manin.

REPLY [5 votes]: Not really an answer, but as a complement to Myshkin's, let me just mention a predecessor : Deligne's very famous (and magnificent) article
Deligne, P. Le groupe fondamental de la droite projective moins trois points.
available here
especially the two paragraphs 
§5  Géométrie algébrique dans une catégorie tannakienne
§7  Géométrie algébrique dans la catégorie tannakienne des systèmes de réalisations: interprétations<|endoftext|>
TITLE: What is the optimal size in the finite axiom of symmetry?
QUESTION [10 upvotes]: Freiling's axiom of symmetry states that if you assign to each real number $x$ a countable set $A_x\subset\mathbb{R}$, then there should be two reals $x,y$ for which $x\notin A_y$ and $y\notin A_x$. 
This principle turns out to be equivalent to the failure of the continuum hypothesis, a fact on which I spoke today at the CUNY math graduate student colloquium, Freiling's axiom of symmetry or, throwing darts at the real line.
Informally, the axiom asserts that if you have attached to each element $x$ of a large set $\mathbb{R}$ a certain comparatively small subset $A_x$, then there should be two independent points $x,y$, meaning that neither is in the set attached to the other. 
At the conclusion of the talk I had assigned an exercise to the audience to prove the finite analogue of the axiom: 
The finite axiom of symmetry. For each finite number $k$ there is a sufficiently large  finite number $n$ such that for any set $X$ of size $n$ and any assignment $x\mapsto A_x$ of elements $x\in X$ to subsets $A_x\subset X$ of size $k$, there are elements $x,y\in X$ such that $x\notin A_y$ and $y\notin A_x$. (See my blog post.)
That is, for any concept of smallness $k$, there is a large size $n$ such that whenever we attach a small subset $A_x$ to each element $x$ of a large set $X$, then there are two independent elements, neither of which are in the set attached to the other.
Here is one way to see it. Suppose we are given a finite number $k$. Let $n$ be any number larger than $k^2+k$. Consider any set $X$ of size $n$ and any assignment $x\mapsto A_x$ of elements $x\in X$ to subsets $A_x\subset X$ of size at most $k$. Let $x_0,x_1,x_2,\dots,x_k$ be any $k+1$ many elements of $X$. The union $\bigcup_{i\leq k} A_{x_i}$ has size at most $(k+1)k=k^2+k$, and so by the choice of $n$ there is some element $y\in X$ not in any $A_{x_i}$. Since $A_y$ has size at most $k$, there must be some $x_i$ not in $A_y$. So $x_i\notin  A_y$ and $y\notin A_{x_i}$, and we have fulfilled the desired conclusion.
It seems unlikely to me that this argument gives the optimal bound on the size of $n$, since first of all it is simply the first argument I thought about, and secondly, the argument gives more than is necessary, since it shows with $n>k^2+k$ that we can find at least one of the pair of independent elements inside any $k+1$ size subset of $X$ and the other outside. So perhaps we might hope for a smaller bound.  MathOverflow is overrun with talented finite combinatoricists, and I am hoping they can help me find the optimal size:
Question. What is the optimal size of $n$ as a function of $k$? 
That is, what is the smallest size $n$ for which the conclusion of the finite axiom of symmetry holds for assignments in a set of size $n$ to subsets of size $k$? 
Please feel free to post better bounds than I have given, even if you don't know it is optimal. Or perhaps there is an easy argument for the optimal bound.

REPLY [3 votes]: This seems really easy? Let $A \subseteq X^2$ be the set of all $(x,y)$ such that $y \in A_x$, and let $A^r$ be the reflection of this across the main diagonal. You are asking whether $A\cup A^r = X^2$. Suppose, for a contradiction, that $n \ge 2k$ and that $A\cup A^r = X^2$. Then $A$ must certainly contain the diagonal of $X^2$, so by the Principle of Inclusion and Exclusion, we have
$|X^2| = |A\cup A^r| = |A| + |A^r| - |A \cap A^r| \le 2nk - n < n^2 = |X^2|,$
a contradiction. On the other hand, if $n \le 2k-1$ then we can take $X = \mathbb{Z}/n$ and $A_x = \{x, x+1, ..., x+k-1\}$, so that $A = \{(x,y) \mid y-x \in \{0, ..., k-1\}\}$.<|endoftext|>
TITLE: Extension of functions from geodesically convex compact sets in a Riemannian manifold
QUESTION [5 upvotes]: In the paper Extension operators for spaces of infinite differentiable Whitney jets (J. reine angew. Math. 602 (2007), 123—154, DOI:10.1515/crelle.2007.005) by Leonhard Frerick, a convenient condition for the existence of an extension operator (for Whitney jets on compact $K\subset \mathbb{R}^n$ to $\mathbb{R}^n$; if $K$ is the closure of its interior I gather this is the same as smooth functions on $K$) is given:

Assume there exist $\varepsilon_0,\ \rho\gt 0$, $r\geq 1$ such that for all $z$ in the boundary of $K$ and $0\lt \varepsilon \lt\varepsilon_0$ there is $x\in K$ with $|x-z|\lt\varepsilon$ and $\{y: |y-x| \leq \rho\varepsilon^r\} \subset K$. Then $\mathcal{E}(K)$ satisfies (DN) and therefore it admits an extension operator.

This implies the version proved much earlier by Stein that covers the case of $Lip_1$ or $C^1$ boundary, or convex sets with non-empty interior.
I'm interested in the case where I don't just take compact sets in Euclidean space, but some more general (in fact compact) manifold $M$, possibly with boundary (or even corners). I certainly am only interested in sets $K$ that are closures of open sets, so (as far as I can tell) I'm really working with smooth functions here.
For our general compact manifold-with-boundary $M$ pick a metric $g$, giving $M$ bounded geometry. (Aside: I believe one could even find a real-analytic structure on $M$, see e.g. K. Shiga, Some aspects of real-analytic manifolds and differentiable manifolds J. Math. Soc. Japan Volume 16, Number 2 (1964), 128-142. However I'm suspicious about the existence of a real analytic metric. Once one exists, then certainly one exists giving $M$ bounded geometry by ibid.)
So I am interested to know whether given some $K\subset M$ contained in a single chart, hence diffeomorphic to $K' \subset \mathbb{R}^n$, the two ways of defining a topology on what one might call $\mathcal{E}(K)$ coincide. That is, consider $\mathcal{E}(K')$ as defined in eg Frerick's paper linked above, or consider a variant of the topology defined directly from $K$ using the chosen metric $g$ for estimates. Clearly this reduces to the case where we have transported $g$ across the chart map, so that we are reduced to considering the case of $K'$, where now $\mathbb{R}^n$ is given some metric $g'$ (and one can choose it to be of bounded geometry, as the closure of the chart in $M$ is compact). Call this space $\mathcal{E}_{g'}(K')$.
Now I'm interested in compact sets that are closures of finite intersections of geodesically convex balls in general position, hence closures of sets that are geodesically convex. If I can believe that $\mathcal{E}(K') \simeq \mathcal{E}_{g'}(K')$ as Fréchet spaces, then I would like to apply an analogue of Frerick's result, using the metric induced by $g'$ instead of the usual norm. Then, relying on geodesic convexity of my set of interest, I hope to conclude that an extension operator exists.
Does this argument pan out?

REPLY [4 votes]: A solution to this problem has now been accepted for publication:

David Michael Roberts, Alexander Schmeding, Extending Whitney's extension theorem: nonlinear function spaces, to appear, Annales de l'Institut Fourier, arXiv:1801.04126.


Original answer:
OK, here is a fairly general answer based on the sledgehammer theorem of Frerick mentioned in the question, going beyond what I asked above. I will state his result in the less general case of closed sets that are closures of their interior, and which satisfy the interior corkscrew condition (this is a nonlinear version of an interior cone condition). Call these closed interior corkscrew domains, for arguments' sake, or corkscrew domains for short (though this name has been used for sets with stronger conditions).

Theorem (Frerick) Let $K \subset \mathbb{R}^n$ be a closed interior corkscrew domain. Then the surjective map of Fréchet spaces $C^\infty(\mathbb{R}^n) \to C^\infty(K)$ has a continuous linear section.

Closed interior corkscrew domains make sense in any metric space, and we shall be considering flat $\mathbb{R}^n$, and a compact Riemannian manifold $(M,g)$ with the geodesic metric.
Note that for a closed Lipschitz map $M \to N$ between metric spaces, corkscrew domains (in my sense above) are sent to corkscrew domains. Since a compact Riemannian manifold is locally uniformly bi-Lipschitz to flat $\mathbb{R}^n$, for any corkscrew domain $K$ (in the geodesic metric) in $M$ that is contained in a single chart $M \supset U \xrightarrow{\phi} \mathbb{R}^n$, we know that $\phi(K)$ is a corkscrew domain. (Note that we also know that $C^\infty(K) \simeq C^\infty(\phi(K))$, where the Fréchet seminorms are defined using the metric $g$ and the flat metric respectively; the same holds for a ball surrounding K$ in the ambient chart and in its image).
Thus there is an extension operator $C^\infty(\phi(K)) \to C^\infty(\mathbb{R}^n)$ (or, if you like, to a large ball containing $\phi(K)$), and hence an extension operator for $K$ to a neighbourhood of $K$, and hence, by a smooth partition of unity, to $M$.
Now the case above, that of geodesically convex closed sets, follows since we can show that such sets satisfying an interior (geodesic) cone condition, hence the interior corkscrew condition.

We can do this all in greater generality, for instance by using the actual condition Frerick was using (weaker than the usual corkscrew condition, and in fact quite close to uniformly polynomially cuspidal), by considering non-compact manifolds with bounded geometry, and by considering closed subsets $K$ that are larger than any one chart (the case of compact boundary is easy, but I believe that is also not necessary, given enough uniform bounds in the geometry of $M$ in the neighbourhood of he boundary). A natural setting is that of homogeneously regular manifolds (i.e. complete, of bounded geometry and positive injectivity radius). Details will be forthcoming, but I would be interested to know of analogous results, or if this is all too obvious to bother writing down.<|endoftext|>
TITLE: An inequality on the simplex involving $x^x$
QUESTION [9 upvotes]: Is anything known about the behavior of the function $$f(x)=\prod_{i=1}^n x_i^{x_i}$$ on the standard simplex, i.e. the set $\{x\in\mathbb{R}^n:\sum_{i=1}^n x_i=1, x_i\geq0\}$?    I ask because I have done a lot of numerical experiments that suggest that $$1\leq\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\left(\sum_{i=1}^{n}x_{i}^{1/2}\right)^{2}\leq 2$$ for all $n$ and all $x$ on the standard simplex, and I have no clue how to prove this.  Of course, one could also re-write this more compactly as $$1\leq\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\|x\|_{1/2}\leq 2$$but this still doesn't seem particularly helpful to me.

REPLY [12 votes]: The upper bound is not correct. E.g., let $n=101$, $x_1=\dots=x_{100}=1/1000$, $x_{101}=9/10$. Then 
$$\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\left(\sum_{i=1}^{n}x_{i}^{1/2}\right)^{2}>7>2.$$
More generally, take any $a\in(0,1)$ and any natural $k$, and then let $n=k+1$, $x_1=\dots=x_k=a/k$, $x_{k+1}=1-a$. Then
$$\ln\left[\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\left(\sum_{i=1}^{n}x_{i}^{1/2}\right)^{2}\right]=a\ln\frac ak+(1-a)\ln(1-a)+2\ln(\sqrt{ak}+\sqrt{1-a})\to\infty$$
as $k\to\infty$.<|endoftext|>
TITLE: Doubt regarding the definition of slice filtration
QUESTION [6 upvotes]: Voevodsky defined the slice filtration on the motivic stable homotopy category $SH(S)$ over a Noetherian scheme $S$. In the article Open Problems in the Motivic Stable Homotopy Theory, I, Section 2, he defined $SH^{eff}(S)$ to be the smallest triangulated subcategory in $SH(S)$ which is closed under direct sums and contains suspension spectra of spaces. It follows from a result of Neeman that the inclusion functor 
$i_n:\Sigma^n_TSH^{eff}(S)\to SH^{eff}(S)$ has a right adjoint $r_n$. Since $i_n$ is full, we have a natural isomorphism $\mathrm{id}\simeq r_n\circ i_n$. Denote $f_n=i_n\circ r_n$. Applying the counit $f_{n+1}\to\mathrm{id}$ to $f_n$, we get a natural transformation $f_{n+1}\circ f_n\to f_n$. Voevodsky claims that $f_{n+1}=f_{n+1}\circ f_n$ so that we get a natural transformation $f_{n+1}\to f_n$. The slice functor $s_n$ is defined to be the cofiber of this map. Denoting the inclusion functor $\Sigma^n_TSH^{eff}(S)\to \Sigma^{n-1}_TSH^{eff}(S)$ by $j_n$, (so that $i_{n+1}=i_n\circ j_{n+1}$), I calculated that 
$$f_{n+1}=i_{n+1}\circ r_{n+1}=i_n\circ j_{n+1}\circ r_{n+1}=i_n\circ\mathrm{id}\circ j_{n+1}\circ r_{n+1}$$
$$\simeq i_n\circ r_n\circ i_n\circ j_{n+1}\circ r_{n+1}=f_n\circ f_{n+1}$$
but don't see why  $f_{n+1}=f_{n+1}\circ f_n$. Note that $f_{n+1}=f_n\circ f_{n+1}$ also induces a natural transformation $f_{n+1}\to f_n$.
Am I missing something here? Or is there a typo in that paper?

REPLY [5 votes]: The key here is that $SH^{eff}(S)$ is closed under suspensions, so there's an inclusion $j_{n+1}:\Sigma^{n+1}_T SH^{eff}(S)\subseteq \Sigma^n_T SH^{eff}(S)$. Hence you can write $i_{n+1}=i_n \circ j_{n+1}$ and $r_{n+1} = l_{n+1}\circ r_n$ where $l_{n+1}$ is the right adjoint of $j_{n+1}$ (which exists because $j_{n+1}$ commutes with colimits). So
$f_{n+1}f_n = i_{n+1}r_{n+1}i_nr_n = i_n j_{n+1} l_{n+1} r_n i_n r_n = i_n j_{n+1}l_{n+1}r_n = f_{n+1}\,.$<|endoftext|>
TITLE: Is the following integral nonzero?
QUESTION [6 upvotes]: Recently I met an integral as follow:
$$\int_0^{2\pi}\cdots\int_0^{2\pi}\left(\prod\limits_{1\leq i<j\leq9}\sin\frac{\theta_i-\theta_j}{2}\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+1}))\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+2}))\right)d\theta_1\cdots d\theta_9,$$
where $\theta_{10}=\theta_1,\theta_{11}=\theta_2$.
I want to know whether the above integral is $0$ or not. I do not know how to connect this integral with some known integral. So I ask it here for some ideas.
Well, my trick was seen through by Fedor Petrov.
Originally I wanted to prove the antisymmetrization of 
$$f(x_1,x_2,\cdots,x_9)=\prod\limits_{i=1}^9(x_i+x_{i+1})^2\prod\limits_{i=1}^9(x_i+x_{i+2})^2,\ \text{where the indices are module}\ 9,$$
is nonzero, just like what I asked before, see my previous question.
Then I did some transformations as follow:
$$\text{the antisymmetrization of}\ \prod\limits_{i=1}^9(x_i+x_{i+1})^2\prod\limits_{i=1}^9(x_i+x_{i+2})^2\ \text{is nonzero}$$
$$\Longleftrightarrow$$
$$\text{the constant term of this Laurent polynomial}\\\prod\limits_{1\leq i<j\leq 9}(\frac{1}{z_i}-\frac{1}{z_j})\prod\limits_{i=1}^9(z_i+z_{i+1})^2\prod\limits_{i=1}^9(z_i+z_{i+2})^2\ \text{is nonzero}$$
$$\Longleftrightarrow$$
$$\int_0^{2\pi}\cdots\int_0^{2\pi}\prod\limits_{1\leq i<j\leq9}(e^{-i\theta_i}-e^{-i\theta_j})\prod\limits_{i=1}^9(e^{i\theta_i}+e^{i\theta_{i+1}})^2\prod\limits_{i=1}^9(e^{i\theta_i}+e^{i\theta_{i+2}})^2d\theta_1\cdots d\theta_9\neq0$$
$$\Longleftrightarrow$$
$$\int_0^{2\pi}\cdots\int_0^{2\pi}\left(\prod\limits_{1\leq i<j\leq9}\sin\frac{\theta_i-\theta_j}{2}\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+1}))\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+2}))\right)d\theta_1\cdots d\theta_9\neq 0,$$
where the indices above are all module $9$. That is how my above question came from.
I thought maybe it is easy to turn into an integral, but now it seems that it is not easy to see whether the above integral is zero or not. So come back to my original question:
Is the antisymmetrization of 
$$f(x_1,x_2,\cdots,x_9)=\prod\limits_{i=1}^9(x_i+x_{i+1})^2\prod\limits_{i=1}^9(x_i+x_{i+2})^2$$ nonzero?

REPLY [9 votes]: Denote $z_k=e^{i\theta_k}$, where $i$ is imaginary unit. Up to some non-zero factor like power of 2, this is the integral of $\prod_j z_j^{-8}\cdot \prod_{j<k}(z_j-z_k)\cdot \prod_j (z_j+z_{j+1})^2 (z_j+z_{j+2})^2$, indices are modulo 9, integral with respect to $\prod d\theta_j$. Only constant term survives after integration, that is, your question is whether constant term of this Laurent polynomial, or, equivalently, coefficient of $\prod_j z_j^8$ in $\prod_{j<k}(z_j-z_k)\cdot\prod_j (z_j+z_{j+1})^2 (z_j+z_{j+2})^2$, equals 0 or not. At first, you may calculate it on the computer (this is much easier task for them than integrating). At second, this looks to be less or more the same question as your previous question.<|endoftext|>
TITLE: Gromov-Hausdorff limits of 2-dimensional Riemannian surfaces
QUESTION [21 upvotes]: Let $\{M_i\}$ be a sequence of 2-dimensional orientable closed surfaces of genus $g$ with smooth Riemannian metrics with the Gauss curvature at least $-1$ and diameter at most $D$. By the Gromov compactness theorem, one can choose a subsequence converging in the Gromov-Hausdorff sense to a compact Alexandrov space with curvature at least $-1$ and Hausdorff dimension 0,1,or 2. One can show (see below) that if $g\geq 2$ then the limit space cannot be a point, thus the dimension of the limit space is at least 1 (while for $g=0,1$ it can be 0). 
Let us assume that the limit space has dimension 1. Then it is either circle or segment. 
Whether these both possibilities (circle and segment) can be obtained in the limit? 
ADDED: It is not hard to see that one can get segment for $g=0$ and circle for $g=1$. I suspect (but cannot prove) that for $g\geq 2$ and $g=0$ one cannot get circle in the limit. In fact I do not even know whether in the case $g\geq 2$ a 2-dimensional limit is the only possibility.
UPDATE: Based on the answer by Igor Belegradek, let me summarize the situation. Let $\{M_i\}$ be a sequence of genus $g$ orientable closed surfaces with Riemannian metrics with Gauss curvature at least -1 which converges in the Gromov-Hausdorff sense to an Alexandrov space $X$. 
1) If $g=0$ then $X$ is either a point, or a segment, or $X$ is homeomorphic to $S^2$ (by Perelman stability theorem), and all the three cases are possible.
2) If $g=1$ then $X$ is either a point, or a circle $S^1$, or homeomorphic to the 2-torus, and all the three cases are possible.
3) If $g\geq 2$ then $\dim X=2$ and hence $X$ is homeomorphic to an orientable genus $g$ closed surface.
ADDED: Let me add a proof that if $g\geq 2$ then a point cannot be the limiting space. Indeed otherwise we would have $d_i:=diam(M_i)\to 0$. Let us divide the metric of $M_i$ by $d_i$ and denote the new metric space by $N_i$. Then the sectional curvature of $N_i$ is at least $-d_i^2$ and diameter 1.
By the Gauss-Bonnet $$4\pi(1-g)=\int_{N_i}K\geq -d_i^2vol(N_i).$$
By the Bishop inequality $vol(N_i)$ is bounded from above. Hence the right hand side in the above inequality tends to 0. Hence $1-g\geq 0$ which is a contradiction.

REPLY [8 votes]: I remembered another reason why closed surfaces of negative Euler characteristic cannot collapse under a lower bound on sectional curvature. 
Much more is true: if a sequence of $n$-dimensional closed manifolds $M_i$ of Ricci curvature $\ge -k^2$ Gromov-Hausdorff converges to a compact space of (Hausdorff) dimension $<n$, then the simplical volume of $M_i$ is zero for large $i$. 
Indeed, on the bottom of p.244 of of Gromov's Volume and bounded cohomology he shows that a lower Ricci curvature bound implies that the simplicial volume is bounded above in terms of volume and the dimension. If a sequence of manifolds converges to a space of dimension $<n$ under the lower Ricci bound, then
their volume goes to zero; this is due to Colding, as mentioned e.g. on p.91 of 
Aspects of Ricci Curvature.
The simplicial volume of a closed surface of negative Euler charactersistic $\chi$ is $2|\chi|$, see p.217 in Gromov's paper, so it cannot collapse under a lower curvature bound. There are of course many high-dimensional manifolds of nonzero simplicial volume.<|endoftext|>
TITLE: Why are quasi-isomorphisms of homotopy algebras only defined for arity 1?
QUESTION [10 upvotes]: When reading about homotopy algebras (e.g. $L_\infty$-algebras, $A_\infty$-algebras), an $\infty$-morphism $f$ is called an $\infty$-quasi-isomorphism if $f_1$ is a quasi-isomorphism.
Recall/Example ($A_\infty$-algebras):
An $A_\infty$-morphism between two $A_\infty$-algebras $(A,\mathfrak{m})$ and $(A', \mathfrak{m}')$ (here $\mathfrak{m}$ and $\mathfrak{m}'$ are the structure maps) is a collection $\{f_k\}_{k\geq1}:(A,\mathfrak{m}) \rightarrow (A',\mathfrak{m}') $ of degree zero (degree preserving) multilinear maps
\begin{equation*}
f_k: A^{\otimes k}\rightarrow A', \hspace{1cm}k\geq 1
\end{equation*}
that satisfy the following relation for $n\geq1$:
\begin{equation*}
\sum_{k+l=n+1}\sum^k_{i=0} (-1)^{a_1+\dots+a_n}f_k(a_1, \dots, a_i, m_l(a_{i+1}, \dots, a_{i+l}), a_{i+l+1}, \dots, a_n).
\end{equation*}
\begin{equation*}
=\sum_{\substack{1\leq k_1\leq \dots \leq k_j \\ k_1+\cdots+k_j= n}} m'_j(f_{k_1}(a_1, \dots, a_{k_1}), f_{k_2}(a_{k_1+1}, \dots, a_{k_1+k_2}), \dots, f_{k_j}(a_{k_{j-1}+1}, \dots, a_n))
\end{equation*}
Furthermore, we call such morphisms $A_\infty$-quasi-isomorphisms if $f_1$ induces isomorphism in cohomology.
Q1: Why do we normally omit higher arity maps when talking about quasi-isomorphisms? 
Q2: Would it be possible to have a weak equivalence that only appears in higher arity maps?
Q3: In case we only care about $f_1$, wouldn't that imply an equivalence at the level of homotopy categories between, for example, Ho(DGLA) and Ho(L$_\infty$), as all the higher arity maps between $A$ and $B$ in L$_\infty$ with the same $f_1$ give isomorphism in Ho(L$_\infty$)?

REPLY [3 votes]: $\text{}$Hi Marcel, I don't believe you're confused about this anymore, but I stumbled upon this question and wanted to add a remark that I think is missing from the other answers.
I claim that before understanding what a quasi-isomorphism of $A_\infty$-algebras is, one should understand what an isomorphism of $A_\infty$-algebras is.
But there is absolutely no ambiguity in what it means for a morphism in a category to be an isomorphism. An isomorphism always means a morphism with a two-sided inverse. And one can easily prove the following result:
Proposition: If $f \colon A \to B$ is a morphism of $A_\infty$-algebras, given by components $f_n \colon A^{\otimes n} \to B$, then $f$ is an isomorphism in the category of $A_\infty$-algebras if and only if $f_1$ is an isomorphism in the category of chain complexes.
Given this result it is very natural that quasi-isomorphisms are defined the way they are.<|endoftext|>
TITLE: Reference for affine Grassmanian
QUESTION [5 upvotes]: Could someone please provide a precise reference in the literature where the following well-known fact is proved. Also if someone could write out the proof that would be great. 
$G(\mathbb{C}((t)))/G(\mathbb{C}[[t]])$ is in bijection with the set of $G$-bundles on the disk together with a trivialisation away from the center.

REPLY [5 votes]: This is correct as long as you say "up to isomorphism over the disk." An element of $G(\mathbb{C}((t)))$ is the same as a trivialization away from the center of the trivial $G$-bundle on the disk. Two such trivializations differ by an automorphism of the trivial $G$-bundle over the disk if and only if the elements of $G(\mathbb{C}((t)))$ differ by an element of $G(\mathbb{C}[[t]])$. This shows that the quotient injects into the set of isomorphism classes we're considering.
Surjectivity is just the statement that any $G$-bundle on the disk is trivial. To do this choose a trivialization over the center, then use the fact that $G$ is smooth, hence formally smooth, to extend to a trivialization over the disk (basically Hensel's lemma).<|endoftext|>
TITLE: Is the Hofer topology second countable?
QUESTION [5 upvotes]: Let $(M,\omega)$ be a symplectic manifold and let $\operatorname{Ham}^c(M,\omega)$ denote the group of compactly supported Hamiltonian diffeomorphisms of $(M,\omega)$. Is the Hofer topology on $\operatorname{Ham}^c(M,\omega)$ second countable?
By the Hofer topology I mean the topology induced by the Hofer metric.

REPLY [3 votes]: Since it is a metric space, your question is equivalent to asking whether it is separable (i.e. has a dense countable subset). Now, under some assumption on the manifold (compact is ok for sure, but second-countable is probably enough too), $\mathrm{Ham}_c(M,\omega)$ is second-countable (hence separable) for the $C^1$-topology, as a subspace of the second-countable space $C^1(M,M)$. Since $C^1$ convergence implies Hofer convergence, $\mathrm{Ham}_c(M,\omega)$ is separable for the Hofer topology.<|endoftext|>
TITLE: Why is it true that if two 4-manifolds are homeomorphic then their squares are diffeomorphic?
QUESTION [23 upvotes]: Near the top of the second page of this paper, it is claimed that if two 4-manifolds $X$ and $Y$ are homeomorphic, then their squares $X \times X$ and $Y \times Y$ are diffeomorphic. Why is this true? It is trivially correct for 3-manifolds and lower dimensions. What about higher dimensions?
References would be greatly appreciated as well.

REPLY [6 votes]: The original paper of Ivan Smith assumes that the 4-manifolds are simply-connected.  So this controls $H^3(X; \mathbb{Z}/2) \cong H_1(X; \mathbb{Z}/2) = 0$.  
Possibly this gives an interesting invariant for non-simply-connected 4-manifolds in general?<|endoftext|>
TITLE: Are there non-smoothable homotopy/homology spheres?
QUESTION [14 upvotes]: A homotopy sphere is a topological $n$-manifold $M$ which is homotopy equivalent to $S^n$.
A homology sphere is a topological $n$-manifold $M$ such that $H_i(M) \cong H_i(S^n)$ for all $i$.
Note, by (one version of) Whitehead's Theorem, every simply connected homology sphere is a homotopy sphere. However, there are homology spheres which are not homotopy spheres, for example, the Poincaré homology sphere.
I've seen in some references that homotopy/homology spheres are assumed to be smooth, but of course the smooth structure is not needed to define them. My question is whether or not this assumption is restrictive. That is:

Are there homotopy/homology spheres which admit no smooth structure?

REPLY [22 votes]: Since the Poincaré conjecture is known in all dimensions any homotopy $n$-sphere is homeomorphic to $S^n$ and hence admits a smooth structure.
Any manifold of dimension $\le 3$ admits a smooth structure. 
Kirby-Siebenmann famously showed that a closed manifold $M$ of dimension $>4$ admits a PL structure if and only if a certain element of $H^4(M;\mathbb Z_2)$ vanishes. The element is called the Kirby-Siebenmann class. Thus any homology sphere of dimension $\neq 4$ admits a PL structure.
Kervaire noted on p.71 of Smooth homology spheres and their fundamental groups that any PL homology sphere of dimension $\neq 3$ admits a smooth structure (because it bounds a contractible PL manifold and all such manifolds are smoothable). 
Thus the only remaining case is homology $4$-spheres. Some of them have large fundamental groups for which $4$-dimensional surgery is not yet known to work. It seems smoothability of such spheres is open. Note that any closed PL $4$-manifold admits a smooth structure.<|endoftext|>
TITLE: Must $L_\alpha$ be correct about well-foundedness?
QUESTION [10 upvotes]: If $R \in L_\alpha$ is a binary relation so that $L_\alpha$ thinks $R$ is well-founded, must $R$ truly be well-founded? (Edit) That is, if $L_\alpha$ thinks that every nonempty subset of the domain of $R$ has a least element, is the same true in $V$?
(Edit) If $L_\alpha$ satisfies a sufficiently strong fragment of $\mathsf{ZFC}$, then the answer is yes because then $L_\alpha$ can build a rank function from $R$ into its ordinals and thus an ill-foundedness in $R$ would give rise to an ill-foundedness in $\mathrm{Ord}$, which is impossible. But if $L_\alpha$ does not have rank functions for all well-founded relations then it only has access to the $\Pi_1$ characterization of well-foundedness. It is conceivable in this case that $L_\alpha$ could be wrong about well-foundedness, that it has a relation which it wrongly believes to be well-founded. 
For comparison, Zermelo set theory (= $\mathsf{ZFC}$ minus Replacement and Foundation) has transitive models which are wrong about well-foundedness. The problem with these models is that they fail to have enough ordinals to capture the ordertype of every well-order.
That these bad models exist is an easy consequence of a theorem by Harvey Friedman.

Theorem (H. Friedman, 1973): Fix a countable admissible set $A$. Consider $T$, a theory extending $\mathsf{KP}$ in the infinitary logic $L_A$ which has a model containing $A$ and is $\Sigma_1$-definable over $A$. Then there is an ill-founded $M \models T$ so that the ordinals of the well-founded part of $M$ are exactly the ordinals of $A$ and the well-founded part of $M$ contains $A$.

To build these bad models of Zermelo set theory, consider the theory consisting of $\mathsf{KP}$ plus the assertion that $V_{\omega + \omega}$ exists. This theory has models containing your favorite countable admissible set, which let's say is $L_{\omega_1^{CK}}$. Let $M$ be the model Friedman's theorem produces when applied to this theory and admissible set and consider $N = V_{\omega+\omega}^M$. Then $N$ is transitive, since everything in $N$ has rank far less than $\omega_1^{CK}$, and $N$ agrees with $M$ about well-foundedness. Take any countable "ordinal" from the ill-founded part of $M$ and you can find an isomorphic copy, call it $R$, which is a subset of $\omega^2$ and hence in $N$. Then, $N$ wrongly thinks $R$ is well-founded.
This argument won't answer the question for non-admissible $L_\alpha$s. Any well-order in $L_\alpha$ must have ordertype less than the least admissible $\beta > \alpha$ as otherwise $L_\beta$ would see an isomorphic copy of its ordinals in an initial segment of itself. If we tried to run a variation of the above argument to produce an ill-founded $M$ with $L_\alpha$ in its well-founded part, we would have that the well-orders in $L_\alpha$ have ordertype in the well-founded part of $M$. As such, we cannot by this means produce an $L_\alpha$ which is wrong about well-foundedness.
(Edit) As pointed out by François and Noah, admissibility isn't sufficient to make $L_\alpha$ correct about well-foundedness. The particular case I'm interested in is when $\alpha$ is the successor of an ordinal whose corresponding fragment of $L$ is correct about well-foundedness.

Question: (Edit) Is there $\xi$ so that $L_\xi$ satisfies enough of $\mathsf{ZFC}$ to be correct about well-foundedness but $L_{\xi+1}$ is wrong about well-foundedness?

REPLY [7 votes]: Coming back to this years later I've just noticed that my original answer was massively flawed; I've corrected it now. The issue is essentially that I took it for granted that all admissible sets look like $L_{\omega_1^{CK}}$ more than they actually do - see e.g. here.
The exact relationship between correctness about well-foundedness and closure properties of ordinals depends what is meant by "thinks is well-founded." (I'll restrict attention to linear orders as opposed to general partial orders for simplicity here.) I'll consider two notions: isomorphisms with ordinals and non-existence of descending sequences.

Isomorphisms with ordinals
The most well-behaved interpretation is via ordinals. Classical a linear order is a well-order iff it is isomorphic to an ordinal; so, for what levels of $L$ does this hold as well?
The nice thing about this correctness property - which I'll call weak wf-correctness - is that it can't yield false positives: if $L_\alpha\models R\cong\gamma$, then $R$ really is a well-ordering. The interesting issue is false negatives.
It turns out that every admissible ordinal is weakly wf-correct. This is a neat application of external induction. Suppose $R\in L_\alpha$ is a well-ordering and every proper initial segment of $R$ is isomorphic to some ordinal via an isomorphism in $L_\alpha$. Then for each $x$ in the domain of $R$, there is a unique ordinal $\gamma_x<\alpha$ such that there is an isomorphism between $R_{<x}$ and $\gamma$ in $L_\alpha$; this means that the map $x\mapsto \gamma_x$ is $\Sigma_1$. Since $R\in L_\alpha$ we may apply $\Sigma_1$-replacement to get $\sup\{\gamma_x:x\in dom(R)\}\in L_\alpha$ and so the ordertype of $R$ is some ordinal $\lambda<\alpha$. The set $g=\{\langle x,\gamma_x\rangle: x\in dom(R)\}$ yields an isomorphism $R\cong\lambda$ and is in $L_\alpha$ since it is a $\Delta_1$-over-$L_\alpha$ subset of $dom(R)\times\lambda$.
The converse, interestingly, fails badly: we can have an $\alpha$ such that $L_\alpha$ is weakly wf-correct but is not admissible. The simplest way to see this is to note that there are non-admissible limits of admissibles (e.g. $\sup_{i\in\omega}\omega_i^{CK}$, which is annoyingly sometimes called "$\omega_\omega^{CK}$"). However, there are even stronger examples: e.g. letting $\theta$ be the next admissible above $\omega_1$, we have that $(i)$ for every limit ordinal $\beta<\theta$ the supremum of the ordertypes of the $\beta$-recursive well-orderings is $<\theta$ and $(ii)$ $cf(\theta)=\omega_1$, so combining these facts we can get a weakly wf-correct ordinal strictly between $\omega_1$ and $\theta$ - in particular, this ordinal is not a limit of admissibles either.

Non-existence of descending sequences
The other obvious correctness notion here is "sees a descending sequence in:" say that $L_\alpha$ is strongly wf-correct if for each ill-founded linear order $R\in L_\alpha$ there is a descending sequence through $R$ in $L_\alpha$.
Here we get the opposite situation: even admissible ordinals need not be strongly wf-correct! The classic example of this is $\omega_1^{CK}$: there are computable linear order (the Harrison orders) which are ill-founded but have no hyperarithmetic descending sequence. (FWIW, each of these is of the form $\omega_1^{CK}(1+\eta)+\mu$ where $\eta$ is the ordertype of the rationals and $\mu<\omega_1^{CK}$.)
On the other hand, it turns out that any ill-founded linear order in an admissible set $L_\alpha$ does have a descending sequence in the next admissible set since quantifying over initial segment isomorphisms in $L_\alpha$ is bounded at that point. This means that every limit of admissibles is strongly wf-correct, so there are non-admissible strongly wf-correct ordinals.

Now what about your final question?
The construction of Harrison orders does relativize to some levels of $L$. Specifically, suppose $L_{\zeta+1}\models$ "$L_\zeta$ is countable." Then there is an ill-founded linear order in $L_{\zeta+1}$ with no descending sequence in $L_\xi$, where $\xi$ is the next admissible above $\zeta$. Extracting an appropriate theory (e.g. a small fragment of KP + "Every set is contained in an admissible set) gives an affirmative answer to your question.
If you're willing to go one level higher, though, things get much cooler. Suppose $L_\alpha$ is pointwise-definable. Then there is a bijection between $L_\alpha$ and $\omega$ in $L_{\alpha+2}$, and so again a la Harrison there is an ill-founded linear order in $L_{\alpha+2}$ with no descending sequence in the next admissible above $L_\alpha$. The point is that pointwise-definable levels of $L$ exist which have very strong theories - e.g. the least level of $L$ satisfying ZFC is pointwise-definable!
I don't know if the above works with merely "$+1$," however.<|endoftext|>
TITLE: Passing from T-equivariant to G-equivariant cohomology
QUESTION [12 upvotes]: Let G=GLn(ℂ) and let T be a maximal torus. Let X be a topological space with a G-action. My question is: when is the canonical map $$H^*_G(X;\mathbb{Z})\to H^*_T(X;\mathbb{Z})$$ injective?
Some remarks: I am trying to understand Torsten's answer here, which claims injectivity for the case where X is the space of length m quotients of $\mathcal{O}_C^m$ for a smooth proper curve C over ℂ. (there is a remark "G being special" that is especially cryptic to me and is probably where the answer lies). I am familiar with some arguments to prove injectivity when the ring of coefficients has n! invertible, but here I specifically want to consider integral coefficients.

REPLY [12 votes]: This isn't a complete answer, but for example, if you know that $H_T^*(X;\mathbb{Z})$ injects into the cohomology of the fixed point set $X^T$, then for $G=GL_n(\mathbb{C})$, the canonical map $H_G^*(X;\mathbb{Z})\to H_T^*(X;\mathbb{X})$ is injective.  See Theorem 2.10 and Corollary 2.11 in T. Holm and R. Sjamaar, "Torsion and abelianization in equivariant cohomology." Transform. Groups 13 (2008), no. 3-4, 585–615.<|endoftext|>
TITLE: Canonical scheme structure on the singular locus of a variety
QUESTION [19 upvotes]: I first asked this question on Math StackExchange but no answers were given. 
Let $X$ be subvariety of affine space $\mathbb{A}_{k}^n$, where $k$ is a field, and suppose $X$ is given by equations 
$$X:(F_1=\cdots = F_m=0)\subset \mathbb{A}^n.$$
Then $X$ is singular at $p\in X$ if 
$$\text{rank}(a_{ij}(p))<d,$$
where $d=n-\text{dim}(X)$ (i.e., the codimension of $X$ in $\mathbb{A}^n$), and $(a_{ij})$ is the matrix valued function 
$$(a_{ij})=\frac{\partial F_i}{\partial x_j}.$$ Thus the ideal generated by the defining equations for $X$ along with the $d\times d$ minors of the matrix $(a_{ij})$ define a natural scheme structure on the singular locus of $X$ (as a subscheme of $\mathbb{A}^n$). Let's call this scheme $S_X$. My question then is the following:
Is $S_X$ intrinsic to $X$?

REPLY [10 votes]: The answer is yes. Let $X$ be a scheme of finite type over the field $k$, of pure dimension $r$; then $S_X$ can be defined as the closed subscheme of $X$ defined by the $r$th Fitting ideal of the sheaf $\Omega^1_{X/k}$.
(See Chapter 20 of the commutative algebra book by Eisenbud for the definition of Fitting ideals of finite modules. It is easy to show that the construction globalises to coherent sheaves on locally Noetherian schemes.)
This construction of $S_X$ coincides with your definition when $X$ is a closed subscheme of $\mathbb{A}^n_k$. Indeed, consider the exact sequence
$$
I/I^2 \longrightarrow \Omega^1_{\mathbb{A}^n_k/k} \vert_X \longrightarrow \Omega^1_{X/k} \longrightarrow 0
$$
where $I$ is the ideal sheaf of $X$ in $\mathbb{A}^n_k$. If $I$ is generated by the polynomials $f_1, \dots, f_m$
one can obtain another exact sequence of $\mathcal{O}_X$-modules
$$
\mathcal{O}_X^{\oplus m} \overset{J}\longrightarrow \mathcal{O}_X^{\oplus n} \longrightarrow \Omega^1_{X/k} \longrightarrow 0
$$
where $J$ is the Jacobian matrix of the polynomials $f_i$. Then the $r$th Fitting ideal of $\Omega^1_{X/k}$ is the ideal in $\mathcal{O}_X$ generated by the $(n-r) \times (n-r)$ minors of the matrix $J$.<|endoftext|>
TITLE: Harmonic function with injective boundary conditions is an immersion?
QUESTION [6 upvotes]: Let $(M,g)$ be an $n$-dimensional, connected, compact Riemannian manifold with boundary. Assume we are given an immersion $f:M \to \mathbb{R}^n$. (i.e $df$ is invertible at every point $p \in M$, note that I assume $n$ is the dimension of $M$). 
Let $\omega:(M,g) \to (\mathbb{R}^n,e)$ be the harmonic function corresponding to the Dirichlet problem imposed by $f$, i.e $\omega|_{\partial M}=f|_{\partial M}$.
Is it true that $\omega$ must be an immersion? Does anything change if we assume $f$ is (globally) injective?
Remark. Note that since $f$ is an immersion, it is locally injective, hence $f|_{\partial M}$ is locally injective. (In particular naive "counter-examples" like taking $\,f|_{\partial M}$ to be constant so $\omega$ is constant do not work). In fact we have $\text{rank}(d\omega_p)\ge \text{rank}\big(d(\omega|_{\partial M})_p\big)= \text{rank}\big(d(f|_{\partial M})_p\big)=n-1$ for every $p \in \partial M$.

REPLY [10 votes]: Let $M$ be the closed unit disk in the plane. Let $f$ be a diffeomorphism of $M$  onto a smooth Jordan region $D$ in the plane. If $D$ is not convex, we can easily arrange that the average
$$\int_{\partial{M}}f(z)ds$$
does not belong to $\overline{D}$. This implies that the harmonic extension $F$ of $f$ maps
some interior points of the unit disk $M$ to points outside $D$. Let $z_0$ be some point
in the interior of $M$ such that $F(z_0)\in\partial F(M)\backslash \overline{D}$.
Evidently, $F$ is not an immersion
at this point. 
Remark. It is a classical theorem of Rado, Kneser and Choquet that if $D$ is convex then $F$ is a diffeomorphism inside $M$. I have not looked in their papers, but I am sure this example must be there.<|endoftext|>
TITLE: Foundations of topology
QUESTION [15 upvotes]: I recently went to a talk of Oleg Viro where he expressed his dissatisfaction with current foundations of differential topology parallel to what has been discussed here. 
Also some time ago I read about Grothendieck's "Denunciation of so-called “general” topology" with interesting comments also made here:

According to Winfried Scharlau's book, Grothendieck described his work
  in a letter to Jun-Ichi Yamashita as: "some altogether different
  foundations of 'topology', starting with the 'geometrical objects' or
  'figures', rather than starting with a set of 'points' and some kind
  of notion of 'limit' or equivalently) 'neighbourhoods'. Like the
  language of topoi (and unlike 'tame topology'), it is a kind of
  topology 'without points' - a direct approach to 'shape'. ...
  appropriate for dealing with finite spaces...

So I am wondering what progress has been made here and in what directions. Does there currently exist an approach to the foundations of general topology that is not based on a notion of "points", in the spirit of Grothendieck's denunciation?

REPLY [12 votes]: Reading section 5 in Grothendieck's essay Esquisse d'un programme it becomes clear that with regard to topology Grothendieck was bothered by some artificial foundational problems introduced by the fact that the foundations of topology were created by analysts rather than by geometers and topologists. Specifically he refers to phenomena such as space-filling curves which he thinks should be ruled out at the foundational level by a more careful choice of definitions of the basic objects we work with.
The basic model is Hironaka's semianalytic sets (or what Grothendieck proposes to call piecewise analytic sets) where such phenomena do not occur, and which on the other hand is sufficiently rich to accomodate various constructions in geometry and topology, such as coning, stratification, etc.  What Grothendieck seeks to do is provide an axiomatisation that would be more or less satisfied by Hironaka's proposal, but that would be realizable in other models as well.  Notes Grothendieck:
This situation, like so often already in the history of our science,
simply reveals the almost insurmountable inertia of the mind, burdened by
a heavy weight of conditioning, which makes it difficult to take a real look
at a foundational question, thus at the context in which we live, breathe,
work – accepting it, rather, as immutable data.
My conclusion is that Grothendieck's proposal in this context does not necessarily amount to a search for a foundation not based on points.  Rather the idea is to get away from the continuous category with its odd phenomena that are viewed by Grothendieck as being a function of inadequate foundations rather than intrinsic mathematical merit. Not an uncommon phenomenon I must say.<|endoftext|>
TITLE: Relation between unipotent cuspidal representations and cuspidal local systems
QUESTION [7 upvotes]: This could well be a question for reading suggestion. Hope it's not too bad and thanks a lot.
So the question is as in the title. What are the relations between the notion of unipotent cuspidal representations of $G(\mathbb{F}_q)$, and that of cuspidal local systems as in generalized Springer theory, if any? For example, I notice that for classical groups of type $\mathbf{C}$, there exists a cuspidal local system for $\mathbf{C}_n$ iff there exists a unipotent cuspidal representations for $\mathbf{C}_{2n}$. Same for type $\mathbf{D}$. (For type $\mathbf{B}$, while there exist cuspidal local systems for $\mathrm{SO}_{n^2}$ and $\mathrm{SO}_{(n+1)^2}$, there exists a unipotent cuspidal representations for $SO_{n^2+(n+1)^2}$.)
This doesn't look like coincidence to me. Though by the time of the 1977 paper of Lusztig on representations of finite classical groups, he hasn't invented generalized Springer theory yet...

REPLY [5 votes]: Have you looked at Lusztig's 1995 IMRN paper "Classification of Unipotent Representations of Simple p-adic Groups"? He discusses this relationship in general, and it is indeed not a coincidence. To the question in Aswin's answer: see Sections 6.6. and 6.7. of the paper.<|endoftext|>
TITLE: Spin structures on schemes
QUESTION [8 upvotes]: This is a very naive question, but I have been wondering about the role of spin geometry and spinor structures in the context of algebraic geometry. I know the definition of spin structures and associated spinor bundles on manifolds and CW-complexes. The question is: is there analog notions of the following classical objects:

Spin structure
Bundle of Clifford algebras
Spinor bundle/bundle of Clifford modules

in the context of schemes (over the complex numbers) and which can be developed in a purely algebraic language? For example, is there a notion of "spin structure" or sheaf of Clifford algebras on a scheme?
Thanks.

REPLY [4 votes]: You might consider holomorphic spin structures on complex algebraic manifolds. If you try to reduce the frame bundle of a complex manifold to a holomorphic spin structure, you will induce a reduction to a holomorphic Riemannian metric, giving a holomorphic affine connection, forcing the vanishing of the Atiyah class, and therefore the vanishing of all Chern classes. Therefore the manifold admits a finite unramified covering by a complex torus, and the holomorphic Riemannian metric is translation invariant. So you don't want a holomorphic spin structure, unless you want to allow singularities.<|endoftext|>
TITLE: Theorem of Bukovsky characterizing ground models
QUESTION [8 upvotes]: It was mentioned in a talk that Bukovsky proved the following are equivalent for inner models $M \subseteq V$:
(1) There is a partial order $\mathbb P \in M$ and a $\mathbb P$-generic filter $G \in V$ over $M$ such that $V = M[G]$.
(2) There is a cardinal $\kappa$ such that for every ordinal $\alpha$ and every function $f : \alpha \to \mathrm{Ord}$ in $V$, there is $F : \alpha \to [\mathrm{Ord}]^{<\kappa}$ in $M$ such that $(\forall \beta < \alpha) f(\beta) \in F(\beta)$.
Is this written up somewhere?  If not, can you prove it?  (Obviously (2) $\rightarrow$ (1) is the hard part.)

REPLY [4 votes]: The original paper is ``Characterization of generic extensions of models of set theory''
A proof can be found in the master thesis of Giorgio Audrito,  "Characterizations of set generic extensions", which is available at Viale's home page here.
Also the paper ''On the set-generic multivers'' by Friedman-Fuchino-Sakai gives a proof. See Friedman's home page.
I may mention that $\kappa$ can be chosen to be the cardinal for which the forcing notion is $\kappa$-c.c.<|endoftext|>
TITLE: Expected number of changes in the sign of a rolling sum of independent normal variables
QUESTION [5 upvotes]: Imagine we define $Y(t+n)=
X(t+1)+.....+X(t+n)$ where $X(i)$ is an independent normal (i.e. everyday we remove the starting observation and we add a new one). We have $n$ consecutive observations of $Ys$. I am trying to find the expected number of changes in the sign in the sequence of $Ys$ as a function of $n$. 
Thanks

REPLY [3 votes]: I'm assuming that $n$ is fixed, so that $Y(t)$ is the sum of the previous $n$ observations. You didn't say what the mean and variance of the random variables $X$ was, so for now, I will assume that they are general: $\mu$ and $\sigma^2$. Now your question boils down (by the strong law of large numbers, or by the ergodic theorem) to computing the probability that $Y(t)$ and $Y(t-1)$ have opposite signs. In fact, there are (in the long run) the same number of switches from negative to positive as from positive to negative, so it suffices to compute $\mathbb P(Y(t)>0, Y(t-1)<0)$. Now $Y(t)$ and $Y(t-1)$ are themselves normal random variables with means $n\mu$ and variance $n\sigma^2$. The covariance is $(n-1)\sigma^2$. Now $(Y(t-1),Y(t))$ has a multivariate normal distribution. If $U\sim N(n\mu,(n-\frac 12)\sigma^2)$ and $V\sim N(0,\frac 12\sigma^2)$ are independent normal random variables, then it is easy to see that $(U-V,U+V)$ has the same mean and covariance matrix as $(Y(t-1),Y(t))$. Hence, since they are both multivariate normal distributions, they have the same distribution.
Now it suffices to compute $\mathbb P(U-V<0<U+V)$. That is $\mathbb P(V>|U|)=2\mathbb P(V>U>0)$.
At this point, I'll make the assumption that you probably intended $\mu=0$. By scaling the random variables, you can also assume that $\sigma=\sqrt 2$. Now $V\sim N(0,1)$ and $U\sim N(0,2n-1)$. Let's write $U=\sqrt{2n-1}N'$, so that we are asking for $2\mathbb P(0<N'<V/\sqrt{2n-1})$. Since $(V,N')$ is radially symmetric, $\mathbb P(0<N'<V/\sqrt{2n-1}) = \arctan(1/\sqrt{2n-1})/(2\pi) $. The probability of a sign change in each location is $\frac{1}{\pi} \arctan(1/\sqrt{2n-1}) = \frac{1}{\pi}\left((2n-1)^{-1/2} - \frac{1}{3}(2n-1)^{-3/2} + \frac{1}{5}(2n-1)^{-5/2} - \cdots \right) $. If you consider $n$ consecutive values for $Y$, there are $n-1$ possible sign changes, and the expected number of sign changes is $\frac{n-1}{\pi}\arctan(1/\sqrt{2n-1}) \sim \frac{\sqrt{n}}{\pi\sqrt{2}}$.<|endoftext|>
TITLE: Can a surface group act on a finite-valence simplicial tree?
QUESTION [7 upvotes]: Question. Let $S$ be a closed surface of genus $> 1$. Can $\pi_1(S)$ act faithfully and minimally on a simplicial tree of finite valence? Here "minimal" means that there is no invariant sub-tree.
Things I know related to this: a minimal action on an $\mathbb{R}$-tree $T$ (like a simplicial tree) gives a canonical measured foliation $F$ on $S$, and a measured foliation in turn gives an action on an $\mathbb{R}$-tree $T_F$, with a surjective equivariant map $T_F \to T$. For simplicial trees, the measured foliation will come by taking a finite collection of simple closed curves and expanding them to make a measured foliation. The resulting tree $T_F$ is then an infinite-valence simplial tree, on which $\pi_1(S)$ acts faithfully. So the question is basically whether you can fold that infinite-valence tree to get a finite-valence tree on which the action is still faithful.
Another point of view is to look at the resulting graph of groups (from Bass-Serre theory). But the edge groups cannot be cyclic (Skora proved this), and it's not clear to me how to proceed from this point.
Richard K. Skora, Splittings of surfaces, J. Amer. Math. Soc. 9 (1996), no. 2, 605--616.

REPLY [6 votes]: Theorem 1.2 of Breuillard-Gelander-Souto-Storm in "Dense embeddings of surface groups" (http://arxiv.org/abs/math/0602635) says the following:
Let $G$ be a locally compact group. Suppose that $G$ contains a nondiscrete
free subgroup $F$ of finite rank $r > 1$. Then $G$ has a subgroup $\Gamma$ containing $F$ such that
$\Gamma$ is isomorphic to a surface group (of genus $2r$). In particular, if $G$ has a dense free subgroup of finite rank, then it has a dense surface group.
Apply this in the case $G$ being the automorphism group of a locally finite regular tree, which clearly contains a free group. Of course, you can always pass to a minimal sub-tree, or alternatively start with a dense free group and get a dense surface which thus will act minimally on the original tree.<|endoftext|>
TITLE: Finite graph colorings without symmetries
QUESTION [5 upvotes]: Let $G$ be a connected finite simple graph with vertex set $V$, $F$ a finite set and let $\Delta(G)$ denote the degree of $G$, i.e. $\Delta(G)= \max_{v\in V} \deg(v)$. We say that a coloring $\phi\colon V \to F$ is asymmetrical if $\text{Iso}(G,\phi) = \{Id_V\}$ , where $\text{Iso}(G,\phi)$ denotes the set of bijections $h\colon V\to V$ preserving both the coloring and the graph structure (note that two adjacent vertices may share the same color). Then let $\Gamma(k)\in \mathbb{N}$ be the infimum of the numbers such that every finite graph $G$ with $\Delta(G) = k$ admits an assymmetrical coloring by $\Gamma(k)$ colors. 
For any $k\in \mathbb{N}$, $\Gamma(k)\leq k^2+2$, to see this consider the graph $G'$ whose set of vertices is the same as that of $G$, and where two vertices $x,x'\in G'$ are adjacent if and only if $d_G(x,x')\leq 2$. This graph admits a proper coloring (any two adjacent vertices have different colors) $\phi$ by $k^2+1$ colors, which we can think of as a coloring on $G$. It is easy to check that the set $\text{Iso}(G,\phi)$ acts freely on $G$. Define now $\tilde{\phi}$ by choosing an arbitrary point $x_0\in V$ and mapping it to a color different from those in $\phi$. Then $\text{Iso}(G,\tilde{\phi})$ acts freely on $G$ and must map $x_0$ to $x_0$.
However, this bound seems too large. For example, for $k=2$, it seems that $\Gamma(2)=3$.
Question: Has the function $\Gamma$ been studied? If so, what values of $\Gamma(k)$ are known? Is there a better bound for $\Gamma(k)$ than $k^2+2$?

REPLY [2 votes]: I think $k+1$ always suffice. Here is an algorithm that produces such a coloring. 
Suppose the set of colors is $\{0,\ldots,k\}$. First, pick a vertex, say $v$, and color it $0$. We will never use that color again. Now, repeat the following procedure, iteratively: choose a vertex $u$ which is colored but has at least one uncolored neighbour, and color the uncolored neighbours of $u$, in such a way that all neighbours of $u$ have different colors (using only the colors from $\{1,\ldots,k\}$). This is clearly possible, since we have $k$ colors available, and $u$ has at most $k$ neighbours. Keep doing this until everything is colored. (Since the graph is finite and connected, this will eventually happen.)
Now, $v$ is clearly fixed by any color-preserving automorphism, since it is the unique vertex of that color. Moreover, any neighbour of a fixed vertex is clearly also fixed, since it is the unique neighbour of that vertex with that color. By induction (and connectedness), a color-preserving automorphism must fix every vertex.
As I commented earlier, this is sharp, as can be seen with the complete and complete bipartite graph. There are also other "sharp" graphs, such as the cycle of order $5$. I wonder if it might be possible to classify them.<|endoftext|>
TITLE: Conjugacy classes of $\mathrm{SL}_2(\mathbb{Z})$
QUESTION [31 upvotes]: I was wondering if there is some description known for the conjugacy classes of $$\mathrm{SL}_2(\mathbb{Z})=\{A\in \mathrm{GL}_2(\mathbb{Z})|\;\;|\det(A)|=1\}.$$ I was not able to find anything about this. Most references only give solutions for $\mathrm{SL}_2(\mathbb{R})$.
Thank you for your help.

REPLY [40 votes]: One can proceed as follows for $SL_2(\mathbb{Z})$.

First, the trace is a conjugacy invariant.
For trace $0$ there are two conjugacy classes represented by $\pmatrix{0 & 1 \\ -1 & 0}$ and $\pmatrix{0 & -1 \\ 1 & 0}$. These representatives can be thought of as $90^\circ$ and $270^{\circ}$ degree rotations of a lattice generated by the corners of a square centered on the origin.
For trace $1$ and $-1$ there are two conjugacy classes each, represented by the matrices
$$M=\pmatrix{1 & -1 \\ 1 & 0}, M^2=\pmatrix{0 & -1 \\ 1 & -1}, M^4=\pmatrix{-1 & 1 \\ -1 & 0}, M^5 = \pmatrix{0 & 1 \\ -1 & 1}
$$
These representatives can be thought of as $60^\circ$, $120^\circ$, $240^\circ$, and $300^\circ$ degree rotations of a lattice generated by the vertices of a regular hexagon centered at the origin.
For trace $2$ there is a $\mathbb{Z}$-indexed family of conjugacy classes, represented by $\pmatrix{1 & n \\ 0 & 1}$; these are all "shear" transformations except for the identity. For trace $-2$ there is a similar $\mathbb{Z}$-indexed family of conjugacy classes represented by $\pmatrix{-1 & n \\ 0 & -1}$.
In general, for nonzero trace the conjugacy classes come in opposite pairs, represented by a matrix $M$ with trace $t>0$ and an opposite representative $-M$ with trace $-t<0$.
For trace of absolute value $> 2$, there is one conjugacy class for each word of the form
$$\pm R^{j_1} L^{k_1} R^{j_2} L^{k_2} \cdots R^{j_I} L^{k_I}
$$
up to cyclic conjugacy, where $I \ge 1$ and all the exponents are positive integers. A matrix representing this form is obtained from the above word by making the replacements
$$R=\pmatrix{1 & 1 \\ 0 & 1}, \quad L=\pmatrix{1 & 0 \\ 1 & 1}
$$
The transformations represented by such words are all "hyperbolic" transformations, having an independent pair of real eigenvectors. The slope of the expanding eigenvector is a quadratic irrational, and hence has eventually repeating continued fraction expansion. The cyclic sequence $(j_1,k_1,j_2,k_2,\ldots,j_I,k_I)$ can be thought of as the fundamental repeating portion of the continued fraction expansion of the slope of the expanding eigenvector, or, better, as an appropriate power of the fundamental repeating portion where the power is equal to the exponent of the given matrix.

Number theorists will tell you that the number of conjugacy classes of each trace $t>2$ is closely related to the class number of the number field generated by $\sqrt{t^2-4}$.

REPLY [13 votes]: The conjugacy classes of elements of ${\rm SL}(2,\mathbb{Z})$
with given trace are counted in:
S. Chowla, J. Cowles and M. Cowles:
  On the number of conjugacy classes in SL(2,Z).
  Journal of Number Theory 12(1980), Issue 3, Pages 372-377.<|endoftext|>
TITLE: how to define the injectivity radius of manifolds with boundary?
QUESTION [11 upvotes]: For manifolds without boundary one defines the injectivity radius as the maximal radius where the exponential map is a diffeomorphism. One can then show that the injectivity radius is the maximum number that such any two points with distance less than that number have a unique geodesic length minimizer between them.
If one looks at a (smooth complete compact) manifold with boundary, using the exponential map definition, points close to the boundary have small injectivity radius and the injectivity radius of the entire manifold would be zero. Using the interpretation about unique length minimizer instead would give a nontrivial notion of injectivity radius for manifolds with boundary.
My question is mostly a reference request, is it written down somewhere how to define the injectivity radius of a manifold with boundary?
My question came from papers that have general statements of the form: Let $(M,g)$ be a manifold with boundary, such that some curvature bounds hold and the injectivity radius is bounded from below. Then bla bla bla. I assume these are not theorems about the empty set.
Edit: 
Maybe this is more complicated than I thought. It seems we have three alternative proposals:
1. my original idea: there exists a unique length minimizer
2. Thomas Rot's idea: use the injectivity radius of the double of the manifold
3. Schick's paper, referenced by user44172: using a collar near the boundary
I think 1. and 2. are equivalent. Whether these two are equivalent to 3. looks nontrivial. Additionally, all these definitions are usually only used in combination with some curvature bounds, so the definitions might be distinct in full generality but equivalent under some suitable curvature bounds.

REPLY [8 votes]: http://arxiv.org/abs/math/0001108 
This paper by Schick is nicely written and contains coordinate-wise and coordinate-free definitions of bounded geometry for manifolds with boundary. 
In particular, it says that in this case the condition of having injectivity radius bounded from below translates to the following conditions:

Normal collar: there is $r_c>0$ such that the geodesic collar $\partial M \times [0,r_c] \to M$ sending $(x,t)$ to $K(\exp_x(tv_x))$ (where $v_x$ is the unit inward normal vector) is a diffeomorphism into its image.
Positive injectivity radius in $\partial M$.
Positive injectivity radius in $M\setminus K(\partial M \times [0,r_c])$.

Most theorems that assume that $M$ is a borderless manifold with positive injectivity radius should have an analogue for the case of manifolds with border satisfying the above conditions.<|endoftext|>
TITLE: Proof of Lefschetz-Hopf Fixpoint Theorem with de Rham cohomology?
QUESTION [6 upvotes]: Looking for a proof of the Lefschetz-Hopf Fixpoint Theorem with the de Rham Cohomology. 
(I´m more interestet in the Formula then just the simple statement that if the Lefschetz number is not zero then the function has a fixed point. But the weak version would be helpful too.)
Does anyone know a book or article? Is this even possible?

REPLY [3 votes]: Mark Stern, Fixed point theorems from a de Rham perspective, http://arxiv.org/abs/math/0606564

REPLY [3 votes]: This is exercise 11.26 in Bott and Tu.<|endoftext|>
TITLE: Expressing $SO_8$ element as product of $L_u$ and $R_u$ for unit octonions $u$
QUESTION [8 upvotes]: Welcome octonions friends !
Long time ago when I travelled through octonion land, I conjectured that every $SO_8$ element can be expressed as product $L_a L_b R_c R_d$ for unit octonions $a$, $b$, $c$, $d$. Is this true ?
It can be seen as generalization of the fact that $SO_4$=$S^3 \otimes S^3$ i.e. every element in $SO_4$ can be seen as product $L_h R_\bar{k}$ for unit quaternions $h$, $k$.
Next step is to see what elements in $SO_{16}$ we obtain by multiplying $L_u$ and $R_v$ for unit sedenions $u, v \in \mathbb S$. We need at least 8 such elements sincce 15*8=120. Sedenions are not so nice as octonions since we have zero divisors there. Therefore we need to check the rank of matrix $L_u$ for unit sedenion $u$. There is work of Moreno showing that set $\{(a,b): ab=0; a,b \in S^{15} \subset \mathbb S\}$ is homeomorphic to $G_2$. Indeed, the subset of zero divisors on sphere $S^{15}$ is 11-dimensional set $\{a+b \iota: a,b \in S^6 \subset \mathbb O; a \perp b; a \perp 1; b \perp 1 \}$ (a,b are perpendicular imaginary unit octonions; $\iota$ is extra element in Cayley-Dickson formula defining sedenion multiplication). (The incorrect statement in wikipedia article on sedenions was corrected on June 28th, 2016 by John Baez). The space of zero-divisors on unit sphere $S^{15}$ is 11-dimensional.
EDIT 0 2018-04-27
I am trying to write article about octonions to magazine for high school students in Poland. While doing this I discovered that Moreno was right. There is no such thing as "zero divisor" in sedenions. Zero divisor should have norm equal to zero. In sedenions there is norm defined as square root of $x\bar x$ where conjugation is defined as in octonions from Cayley-Dickson formula $\bar x=\bar a-b\iota$. In this case sedenion norm is just length of the vector. So Moreno correctly considered pairs of elements whose product is zero.
EDIT 1: Since some people have justified doubts whether element $L_u$ can be in $SO_{16}$ for unit sedenion $u$ I present following.
EDIT 2: I extended 7-dimensional to 9-dimensional set of unitary sedenions u such that $L_u \in SO_{16}$.
$\underline{Fact}$: $L_x$ is in $SO_{16}$ for $x=(u+v\iota)/\sqrt 2$ such that $u,v \in \mathbb C \subset \mathbb O$.
$\underline {Proof}:$
From Cayley-Dickson formula:
$$L_u=\pmatrix{ L_u & \\ & R_u}$$
$$L_{v\iota}=\pmatrix{ 0 & -R_v S \\L_v S & 0}$$.
Calculate the transposed matrix:
$$L_x^T=\pmatrix{ L_{\bar u} & R_v S \\ -L_v S & R_\bar u }$$
Now $$L_x L_x^T=\pmatrix{2 & 0 \\ 0 & 2 }$$
because position 1,1 of the matrix is $L_u L_\bar u+R_v R_\bar v$ and position 1,2 of the matrix is $(L_u R_v-R_v L_u)S$. Elements $L_u$ and $R_v$ commutes only when $u,v$ are in the same complex subalgebra of octonions.
I am using following facts in calculations:
$L_u S=S R_\bar u$
$L_u^T=L_\bar u$ and similar. $\square$
The set defined in the Fact is roughly bundle $S^6 \times \times S^3$, because for each imaginary octonion $a$ there is sphere $S^3$ spanned by $<1,a,\iota,a\iota>$ belonging to this set. I said "roughly", becuase points {1,-1} belong to all fibers $S^3$.
Next, we can continue to trentaduonions $\mathbb T$. I claim that set of zero divisors is 27-dimensional subset of $S^{31}$. We need 16 elements of shape $L_u$ or $R_u$, since 16*31=496.
EDIT 3: Regarding trentaduonions I don't have ready theory yet. I tested in GAP, so I present following experimental results. For all 32 base elements $e_k$ the matrices $L_{e_k}$ and $R_{e_k}$ are in $SO_{32}$. Among 496 pairs of 32 base elements there 202 such pairs $k$, $l$ that $L_{e_k+e_l}$/length($e_k+e_l$) is in SO(32). 550 out 4960 triples satisfy it; 1002 out of 35960 fours; 1274 out of 201376 fives; 1218 out of 906192 sixs; 970 out of 3 365 856 sevens; 740 out of 10 518 300 eights.
If I try to prove it then I would see whether similar formulas for conjugation $S$ multiplied by $L_u$ are valid in sedenions.
... and so on ...:)
I have also some generalization proposal for any Lie group, which I post in separate question.

REPLY [9 votes]: Your conjecture on octonions is false.  In Conway and Smith's book On Quaternions and Octonions (A.K. Peters 2003), §8.5 theorem 9 on page 94, it is shown that the set of $L_a L_b R_c$ for $a,b,c$ unit octonions is $20$-dimensional.  So the set of $L_a L_b R_c R_d$ for $a,b,c,d$ unit octonions is at most $27$-dimensional and cannot be $SO_8$.  Actually, based on the dimensions given in the above-cited theorem, I would conjecture that it is $26$-dimensional, but it seems, based on the comments the authors make near the end of the section (just before the conjecture they state), that they do not know the dimension.
On the other hand, they make a conjecture about products of five multiplications, namely that if $V,W,X,Y,Z$ are any five letters from the set $\{L,R,B\}$, not all equal, then $V_a W_b X_c Y_d Z_e$ gives all of $SO_8$.<|endoftext|>
TITLE: How to calculate the PSD of a stochastic process
QUESTION [5 upvotes]: This question was asked on math.stackexchange about 2 months ago, but it hasn't been very successful in attracting answers yet, so I'm posting it here.
Say we have a stochastic process described by a stochastic differential equation (in the Itô sense), and maybe we are able to find an explicit solution of it in terms of deterministic and Itô integrals (and maybe not). 
In any of these cases, how can we compute the power spectral density of the process?
For instance, for the Ornstein-Uhlenbeck process, the steps would be

Find autocorrelation function. This is easily done with the Itô isometry and the properties of the Wiener process. 
Compute Fourier transform 

And that's it, because the Wiener-Khinchin theorem assures us that the process autocorrelation and the PSD are a Fourier transform pair as long as the process is wide sense stationary (and the Ornstein-Uhlenbeck process satisfies this). However, The Ornstein-Uhlenbeck process is too nice in reality. 
What can be done to obtain the PSD in a more general, non stationary case? What can be then said about the autocorrelation function relation to the PSD? Could you provide any references with theory/worked examples?
For instance, I would really appreciate your help in the not-so-nice (but still nice) second order harmonic oscillator with additive noise term
$$
\ddot{x} + \gamma\dot{x} + \Omega^2x = \sigma \frac{\mathrm{d}W_t}{\mathrm{d}t}
$$
(forgive my abuse of notation). I have seen physicists doing awful things with this stochastic process, and I'm in need of some (mathematical) cleanliness and mental peace.

REPLY [6 votes]: You ask for the spectral analysis of a nonstationary stochastic process. Because the autocorrelation function $C(s,t)$ now depends on the two times $s$ and $t$ separately, and not only on their difference, the power spectral density $P(\omega,\omega')$ will depend on two frequencies, and not just on a single frequency. Alternatively, one can define a power spectrum $P_t(\omega)$ that depends on one frequency and one time variable, and has the interpretation of a time dependent power spectral density. This makes sense if the nonstationary characteristics change sufficiently slowly in time (in a way that can be made precise).
This approach was developed by M.B. Priestley, formally in Evolutionary Spectra and Non-Stationary Processes, with applications in Power spectral analysis of non-stationary random processes.<|endoftext|>
TITLE: Morita equivalence base equivalence relation for discrete groups
QUESTION [7 upvotes]: In the context of "discrete groups", is there an equivalence relation that implies the Morita equivalence of their reduced group C*-algebras?
We define $G \sim H$ for discrete groups $G$ and $H$, when $C^*_r (G)$ and $C^*_r (H)$ (or even their full group C*-algebras) are Morita equivalent as the two C*-algebras.
Does $\sim$ have an explicit description?
thanks
PS: By Alain Valette's first comment in below, the $C^*_r (-)$ case shouldn't be considered so the problem is about $C^*(-)$.

REPLY [6 votes]: Not really an answer, but too long for a comment: let $G,H$ be countable, abelian, torsion-free groups. Claim: $C^*G\sim C^*H$ if and only if $G$ is isomorphic to $H$. To prove the non-trivial implication: denote by $\hat{G}$ the Pontryagin dual of $G$, a compact connected abelian group. Then $C^*G\simeq C(\hat{G})$ by Fourier transform. Since $C^*G\sim C^*H$, the groups $\hat{G}$ and $\hat{H}$ are homeomorphic, hence the first $\check{C}$ech cohomology groups $H^1(\hat{G},\mathbb{Z})$ and $H^1(\hat{H},\mathbb{Z})$ are the same. It remains to see that $H^1(\hat{G},\mathbb{Z})\simeq G$. Write $G=ind-lim_k G_k$, an inductive limit of finitely generated free abelian groups, i.e $G_k\simeq \mathbb{Z}^{n_k}$. Dualizing, $\hat{G}$ is a projective limit of tori, i.e. $\hat{G}=proj-lim_k \mathbb{T}^{n_k}$.Since $\check{C}$ech cohomology converts projective limits into inductive limits, we get $H^1(\hat{G},\mathbb{Z})=ind-lim_k H^1(\mathbb{T}^{n_k},\mathbb{Z})=ind-lim_k \mathbb{Z}^{n_k}=G$.<|endoftext|>
TITLE: minimum number of bases of a matroid, that comes from a convex polytope
QUESTION [7 upvotes]: Given a d-dimensional polytope P with n points, then what is the minimum number of simplices that are spanned by vertices of P? This question led my research to matroids and so my question is: what is the minimum number of bases af a matroid, that comes from a fulldimensional convex set with n vertices in dimension d? Here you have to think of affine independence or linear independence in the projective plane in dimension d+.
The upper bound is obviously $n \choose d+1$, but the lower bound seems to be quite hard.
My theory is, that it should be bounded by $n-d+2 \choose 3$. This refers to a (n-d+2)-gon in a 2-dimensional subspace and the other vertices spanning the other d-2 dimensions. But until now i couldn't find a proof. As I do not know a lot about matroids i hope that in terms of matroids this turns out to be easier, than in terms of polytopes.

REPLY [4 votes]: It looks that David Speyer strengthening holds. Namely, if a matroid $M$ on a set $E$, $|E|=n$, has rank $k\leqslant n$ and minimal crcuit size at least $p$, $k\geqslant p-1$, than $M$ has at least $\binom{n-k+p-1}{p-1}$ bases. (In our situation $k=d+1$, $p=4$.) Prove by induction. Cases $n=k$ and $k=p-1$ are clear. So assume that $n>k>(p-1)$ and the claim holds for less $n$.
If there exists a bridge $a\in E$, we simply remove it, $n-k$ does not change. If there are no bridges, then for any $a\in E$ induction proposition gets as at least $\binom{n-k+p-2}{p-1}$ bases without $a$, this have to be multiplied by $n$ (choices of $a$) and divided by $n-k$ (number of times we counted each base). We get even more than we need: $\frac{n}{n-k}\binom{n-k+p-2}{p-1}\geqslant \binom{n-k+p-1}{p-1}$.<|endoftext|>
TITLE: What is this Lie algebra?
QUESTION [6 upvotes]: Consider two matrices $A,B \in \mathfrak{su}(N)$ which are both diagonal in the standard basis and non-zero.
If we consider the new matrix $\tilde{B} := FBF^{\dagger}$ where $F$ is the `quantum' fourier transform matrix:
$$F_N = \frac{1}{\sqrt{N}} \begin{bmatrix}
1&1&1&1&\cdots &1 \\
1&\omega&\omega^2&\omega^3&\cdots&\omega^{N-1} \\
1&\omega^2&\omega^4&\omega^6&\cdots&\omega^{2(N-1)}\\ 1&\omega^3&\omega^6&\omega^9&\cdots&\omega^{3(N-1)}\\
\vdots&\vdots&\vdots&\vdots&&\vdots\\
1&\omega^{N-1}&\omega^{2(N-1)}&\omega^{3(N-1)}&\cdots&\omega^{(N-1)(N-1)}
\end{bmatrix}$$
(where $\omega=\exp(2i\pi/N)$, see the wikipedia page on Quantum Fourier Transform)
what are the possibilities for the Lie algebra generated (i.e. all linear combinations of all bracket expressions) as $\left<A, \tilde{B}\right>_{Lie}$? I.e. what algebras can be generated this way by varying $B$ while keeping it diagonal.
Specifically, will the Lie group associated to whatever algebra is generated contain a subgroup isomorphic to any non-trivial Clifford group?

REPLY [2 votes]: I can't see your question having a completely general answer. In the 'generic' case, I think the subalgebra generated by $A$ and $\tilde{B}$ is just $\mathfrak{su}(N)$.
To see this, first of all note that $\tilde{B}$ is of the form $$\begin{bmatrix} 0 & b_1 & b_2 & \dots & b_{N-1} \\ b_{N-1} & 0 & b_1 & \dots & b_{N-2} \\
b_{N-2} & b_{N-1} & 0 & \dots & b_{N-3} \\
\vdots & \vdots & \vdots & \dots & \vdots \\
b_{1} & b_{2} & b_{3} & \dots & 0 \end{bmatrix}$$
with $b_{N-i}=-\overline{b_i}$ for $1\leq i\leq N-1$, and all such matrices are possible.
Let $A$ have diagonal entries $a_1{\rm i},\ldots ,a_N{\rm i}$ (where the $a_i$ are real and sum to zero). In the generic case, the square differences $(a_i-a_j)^2$ for $i<j$ are all distinct.
Then the subspace of elements $x$ of $\mathfrak{su}(N)$ satisfying $[A,[A,x]]=-(a_i-a_j)^2 x$ is two-dimensional, spanned by $e_{ij}-e_{ji}$ and ${\rm i}(e_{ij}+e_{ji})$, where $e_{ij}$ denotes the matrix with $1$ in the $(i,j)$ position, and $0$ elsewhere.
Generically, the $b_i$ are all non-zero, and hence $\tilde{B}$ is supported on this subspace of elements satisfying $[A,[A,x]]=-(a_i-a_j)^2 x$, specifically the component of $x$ is $$b_{j-i}e_{ij}-\overline{b_{j-i}}e_{ji}=\beta(e_{ij}-e_{ji})+\gamma{\rm i}(e_{ij}+e_{ji})$$ where $\beta,\gamma$ are real (and at least one is non-zero). Since the $(a_i-a_j)^2$ are distinct, it follows that $\beta(e_{ij}-e{ji})+\gamma{\rm i}(e_{ij}+e_{ji})$ belongs to $\langle A,\tilde{B}\rangle$.
Now since $$[A,\beta(e_{ij}-e_{ji})+\gamma{\rm i}(e_{ij}+e_{ji})]=(a_i-a_j)(-\gamma(e_{ij}-e_{ji})+\beta{\rm i}(e_{ij}+e_{ji}))$$ and since we cannot have $\gamma/\beta=-\beta/\gamma$, then $\langle A,\tilde{B}\rangle$ contains both $e_{ij}-e_{ji}$ and ${\rm i}(e_{ij}+e_{ji})$.
Finally, $[e_{ij}-e_{ji},{\rm i}(e_{ij}+e_{ji})]=2{\rm i}(e_{ii}-e_{jj})$ is also in $\langle A,\tilde{B}\rangle$.<|endoftext|>
TITLE: Transitive permutation groups which all of their proper subgroups are intransitive
QUESTION [7 upvotes]: Let $G$ be a transitive permutation group on a finite set $\Omega$. It is clear
that if $G$ is regular, then every proper subgroup of $G$ is intransitive. Is
there any other class of groups with this property? I mean that which transitive groups has no proper transitive subgroup?

REPLY [3 votes]: A very repeatable construction to yield such permutation groups is to let a group act on the cosets of a small subgroup.
The idea is that if $H$ is a subgroup of $G$, a subgroup $T \leq G$ acts transitively on the cosets of $H$ if and only if $HT = G$. If $H$ is small, $T$ is very restricted (beginning, for example, with $[G:T] \leq |H|$) and it's easy, in many cases, to prove $T$ must be the whole $G$. If $G$ is regular, this corresponds to choosing $H = 1$, which forces $T=G$ without further discussion.<|endoftext|>
TITLE: Equivalence between a derived subcategory and a subcategory of the derived category
QUESTION [7 upvotes]: Let $\mathcal{A}$ be a subcategory of $\mathcal{C}$. Let $D(\mathcal{A})$ and $D(\mathcal{C})$ be the associated derived categories. We can define $D_\mathcal{A}(\mathcal{C}) = \{X \in \mathcal{C}\hspace{2pt} |\hspace{2pt} H^i(X) \in \mathcal{A}\}$. We have natural inclusions $$D(\mathcal{A}) \to D_\mathcal{A}(\mathcal{C}) \to D(\mathcal{C})$$
My question is this: When is $D(\mathcal{A}) \to D_\mathcal{A}(\mathcal{C})$ an equivalence of categories? 
I have read that it is not always an equivalence, but that it often is. I have examples of when it is: Is there an easy example for when it isn't?

REPLY [4 votes]: This is more comment than answer, but for bounded derived categories we have the following: let $\mathcal{C}$ be an abelian category and $\mathcal{A} \subset \mathcal{C}$ be a Serre subcategory. Then, a sufficient condition for the natural functor $\mathrm{D^b}(\mathcal{A}) \to \mathrm{D}^{\mathrm{b}}_{\mathcal{A}}(\mathcal{C})$ to be an equivalence is that for every exact sequence $S_1:0 \to A \to B \to C \to 0$ in $\mathcal{C}$ with $A \in \mathcal{A}$, there is an exact sequence $S_2: 0 \to A \to B' \to C' \to 0$ in $\mathcal{A}$ and a map of exact sequences $S_1 \to S_2$ which is the identity on $A$.
This result can be found in Keller's ``On the cyclic homology of exact categories'', who in turn refers to SGA5.<|endoftext|>
TITLE: How do the roots of a polynomial change when another polynomial is added?
QUESTION [5 upvotes]: I need to obtain an analytical solution to an equation of the following form:
$$
(x-a)(x-b)(x-c)=d(x-e)(x-f),
$$
where $a$, $b$, $c$, $d$, $e$, and $f$ are known numbers and $x$ is the variable.
Of course, the equation can be reduced to a "simple" equation for the roots of a 3rd-order polynomial, and its solution is provided by the Cardano formula (and other similar methods), but the result is too complex...
So the question arises: does any simpler method exist that makes use of the known roots of the "constituent polynomials" (left-hand and right-hand side of my equation)?
P.S. It is possible to try the Cardano method with the polynomial coefficients expressed through the roots $a$...$c$,$e$,$f$ and then to simplify the resulting gargantuan formulae using a CAS (computer algebra system) hoping that the CAS will cope with it... But I am not so sure that the CAS is powerful enough in the dark art of formulae simplification.

REPLY [9 votes]: As you point out, this is no harder than solving a general cubic. It turns out that it is also no easier.
For any choice of the $6$ parameters $a,b,c,d,e,f$ your desired equation can be multiplied out to a cubic $x^3+Ax^2+Bx+C$ which can be solved by Cardano's formula. You hope for a nicer solution perhaps based on the form. However one might expect that with $6$ parameters one can easily get $A,B,C$ to be anything desired. This turns out to be the case.
Here is one way assuming all parameters constrained to be real. The complex case is as easy.
Pick $a=b=0$ and $c=-(A+d)$ where $d$ is $+1$ or $-1$ according as $C \ge 0$ or  $C \lt 0.$  Then $e$ and $f$ are the (real) roots $$\frac{Bd \pm \sqrt{B^2+4|C|}}{2}$$ of the quadratic equation.
$$x^2-\frac{B}{d}x-\frac{C}{d}.$$<|endoftext|>
TITLE: (Geometric) Proof for the projective bundle formula in K-theory
QUESTION [5 upvotes]: I'm trying to piece together a proof of the projective bundle formula from several incomplete sources. Here's the statement I'd like to prove:

Projective bundle formula: Let $\pi: E \to X$ be a vector bundle of rank $r$ over a compact space $X$. Let $\mathbb{P}(E)$ be the projective bundle of $E$ with distinguished line bundle $H$. Then we have the following formula:
   $K(\mathbb{P}(E))=K(X)[H]/( \sum^r_{k=0}(-1)^k [H]^k[\bigwedge^k E] )$.

At this point I feel like I have all the ingredients for the proof but I can't seem to piece them together. Here's what I have:
There's a canonical Kozul complex over $E$:
$$0 \to \mathbb{C} \to E \to \bigwedge^2 E \to \dots \to \bigwedge^r E \to 0$$
Over each point $e \in E$ the maps are given by wedge multiplication $(-) \wedge e$. Therefore this complex is exact outside of the zero section and gives us the thom class in $\lambda_E \in K(E,E-0)=\tilde{K}(Th(E))$. What's a neat way of proving that $K(E,E-0)$ is a free rank 1 module generated by $\lambda_E$? 
Suppose wer'e past that. Observe that there's a cofiber sequence
$$\mathbb{P}(E) \to \mathbb{P}(E \oplus \mathbb{C}) \to Th(E)$$
This suggests that the pullback of $\lambda_E$ to $\mathbb{P}(E \oplus \mathbb{C})$ might tell me the relavant information. Two things are not clear to me here. 

Why is the pullback of $\lambda_E$ to $\mathbb{P}(E \oplus \mathbb{C})$ equal to $\sum^r_{k=0}(-1)^k [H]^k[\bigwedge^k E]$? (Twisting the kozul complex by powers of $[H]$ must have some geometric interpratation... no?)
It seems rather implausible that for every cofiber sequence $X \to Y \to C$ one has $K(X) = K(Y)/Im(K(C)\to K(Y))$. So can we really justify and prove the formula using these ingredients? Maybe there's an inductive step here that might help simplify the statement?

I apologize if there's a well known source that discusses this in detail. The very reason I'm here is I didn't find one.

REPLY [5 votes]: First, for any bundle $V$ of dimension $d$ over $Y$ put 
$$ \lambda(V)(t) = \sum (-1)^k[\Lambda^k(V)]t^{d-k} \in K^0(Y)[t]. $$
This is a monic polynomial of degree $d$ over $K^0(Y)$.  It satisfies $\lambda(L)(t)=t-[L]$ if $L$ is a line bundle, and $\lambda(A\oplus B)(t)=\lambda(A)(t)\lambda(B)(t)$.  Thus, if $L$ is isomorphic to a subbundle of $V$ then $V\simeq L\oplus W$ for some $W$ and we find that $\lambda(V)([L])=0$.
Let $p\colon \mathbb{P}(E)\to X$ be the obvious projection.  For any $U\subseteq X$, put 
\begin{align*}
 A^*(U) &= K^*(U)[t]/\lambda(E|_U)(t) \\
 B^*(U) &= K^*(p^{-1}(U))
\end{align*}
The bundle $H$ over $\mathbb{P}(E)$ is tautologically a subbundle of $p^*E$, and using this we obtain a ring map $A^*(X)\to B^*(X)$ sending $t$ to $[H]$.  Essentially the same construction gives maps $\phi_U\colon A^*(U)\to B^*(U)$ for all $U$.  Recall also that $\lambda(E)(t)$ is a monic polynomial of degree $d$.  It follows that the set $T=\{1,t,\dotsc,t^{d-1}\}$ is a basis for $A^*(U)$ over $K^0(U)$.
Now suppose we have open subsets $U$ and $V$.  There is a Mayer-Vietoris sequence relating the $K^*$ groups of $U$, $V$, $U\cup V$ and $U\cap V$.  Using the basis $T$, we obtain a long exact sequence relating the $A^*$ groups of $U$, $V$, $U\cup V$ and $U\cap V$.  The Mayer-Vietoris sequence for $p^{-1}(U)$ and $p^{-1}(V)$ gives another long exact sequence relating the $B^*$ groups of $U$, $V$, $U\cup V$ and $U\cap V$.  The maps $\phi$ link these two long exact sequences.  They are obviously compatible with ther restriction maps $A^*(U)\to A^*(U\cap V)$ and so on, but a little work is needed to check that they are also compatible with the connecting morphisms.  
Now put $\mathcal{U}=\{U\;|\;\phi_U \text{ is iso } \}$.  If $U$ is contractible then $E|_U\simeq U\times\mathbb{C}^d$ and the standard calculation of $K^*(\mathbb{C}P^{d-1})$ shows that $U\in\mathcal{U}$.  If $U$, $V$ and $U\cap V$ lie in $\mathcal{U}$ then the Mayer-Vietoris sequences together with the five lemma show that $U\cup V\in\mathcal{U}$.  Now suppose that $X$ can be covered by open sets $U_1,\dotsc,U_n$ such that all intersections $U_{i_1}\cap\dotsb\cap U_{i_r}$ are empty or contractible.   Then one can check by induction on $n$ that $\phi_X$ is iso.  If $X$ is a finite simplicial complex then the open stars of vertices provide a cover of the required type, so $\phi_X$ is iso.  The case of a completely general $X$ follows by homotopy invariance and a limit argument.<|endoftext|>
TITLE: What is the relation between the sphere spectrum and supersymmetry?
QUESTION [32 upvotes]: In this this google+ post of Urs Schreiber, he says: "Grading over the sphere spectrum is supersymmetry" and then he redirect us to the abstract idea of superalgebra (in nLab).
Are there some references (other than Kapranov and nLab) about these ideas?

REPLY [8 votes]: Like Schreiber does in his post, I would advertise the point of view developed by Sagave and Schlichtkrull in their Adv. Math 2012 paper, and used by us to study topological logarithmic geometry. Each symmetric spectrum $X$ has a graded underlying space that is really a $J$-shaped diagram $Y$. Here $J$ is a category with nerve $QS^0$, so $hocolim_J Y$ maps to $QS^0$. If $X$ is a commutative symmetric ring spectrum then $hocolim_J Y$ is an $E_{\infty}$ space over $QS^0$, i.e., it is graded over the sphere spectrum. Sagave started developing this while a postdoc in Oslo. I noted that the nerve of $J$ was not $Z$ but something with $\pi_1 = Z/2$.<|endoftext|>
TITLE: Connection between solution for Schrödinger equation and solution for heat equation
QUESTION [12 upvotes]: It's known, that if you write imaginary unit into a heat equation you'll get time-dependent Schrödinger equation. Recently one guy discovered a connection between solutions for these two equations (see this paper). Specifically, he found a way to obtain a solution for a Schrödinger equation from a solution for a heat equation (and even from computationally simpler objects related to а heat equation). The method is rather general and is for instance applicable to equations with arbitrary configurational space. 
My question is who and in what papers suggested to study connections between Schrödinger and heat equations? Are there any results? Are these problem included in any sort of list of problems? I know about Wick rotation and Doss trick (see refernces in here). But what else exists in this field?

REPLY [2 votes]: Let me add some more comment on Mugnolo's answer. It is rather standard to study the heat group with complex time $t$ provided $\Re t>0$. The question of what happens as $t$ tends to the imaginary axis is a natural one and is in the mind of most people working on the subject I guess. It all depends on what you want to prove, of course. If you are interested in proving some 'typically Schroedinger' properties, like e.g. Strichartz estimates, using only information from the heat kernel, then there are not many results. A recent and rather interesting result is this paper. Let me also mention that the connection with the wave equation (more precisely, with the finite speed of propagation property for the wave flow) is a well known and useful one.<|endoftext|>
TITLE: Is the domain of an operator valued weight closed under Hahn-Jordan decomposition?
QUESTION [13 upvotes]: Let $N\subseteq M$ be an inclusion of semi-finite factors with normal faithful semi-finite traces $\operatorname{Tr}_N$ and $\operatorname{Tr}_M$ respectively. Let $T: M^+\to \widehat{N^+}$ be the unique trace-preserving normal faithful semi-finite operator valued weight.
As for normal weights, we define
\begin{align*}
\mathfrak{n}_T 
&= 
\left\{x\in M \,\middle|\, T(x^*x)\in N^+\right\}
\\
\mathfrak{m}_T 
&= 
\mathfrak{n}_T^*\mathfrak{n}_T
\\
\mathfrak{p}_T 
&=
\left\{x\in M^+\,\middle|\,T(x)\in N^+\right\}
\end{align*}
It is straightforward to show that $\mathfrak{n}_T$ is a left ideal, $\mathfrak{m}_T\subset \mathfrak{n}_T\cap \mathfrak{n}_T^*$ is a hereditary $*$-algebra, both $\mathfrak{n}_T$ and $\mathfrak{m}_T$ are $N-N$ bimodules, and $\mathfrak{m}_T$ is spanned by its positive part $\mathfrak{m}_T^+$, which in turn is equal to $\mathfrak{p}_T$.
Moreover, $T$ has a canonical extension to a map $\mathfrak{m}_T\to N$.
Thus if $x\in \mathfrak{m}_T$ is self-adjoint, we can write $x=x_1-x_2$ with $x_1,x_2\in \mathfrak{p}_T$.
But it is not clear to me if $\mathfrak{m}_T$ is closed under taking the Hahn-Jordan decomposition of $x$. That is:

Suppose $x\in \mathfrak{m}_T$ is self-adjoint and $x=x_+-x_-$ is the Hahn-Jordan decomposition of $x$, where $x_+$ and $x_-$ are positive and $x_+x_-=0$. Does it follow that $x_\pm \in \mathfrak{p}_T$?

REPLY [3 votes]: Here is a counter-example.  Which is sadly rather long.  I use Haagerup's original paper for the definition of $T$ below.
Let $M$ be the $\ell^\infty$ direct sum of the matrix algebras $\mathbb M_n$ with $\tau_M$ being the sum of the un-normalised traces on $\mathbb M_n$.  Let $N$ be the subalgebra of all diagonal matrices, and give $N$ the ``weighted'' trace
$$ \tau_N(x) = \sum_n \frac{1}{n} \sum_i x^{(n)}_i. $$
where $x = (x_n)\in N$ and each $x_n$ has diagonal entries $(x^{(n)}_i)$.
Below I'll regard $N$ as the direct sum of $\ell^\infty_n$.
Let $T:M_+\rightarrow \widehat{N_+}$ be the operator valued weight, so
$$ \tau_M(y^{1/2}xy^{1/2}) = \tau_N(y^{1/2} T(x) y^{1/2}) \qquad (x\in M, y\in N). $$
In general, $T(x)$ is not bounded, so a little care is needed to define the right-hand side.  However, it follows from this that $T$ must respect the direct-sum decomposition, and so we obtain maps $T_n:\mathbb M_n \rightarrow \ell^\infty_n$ which satisfy
$$ \sum_i y_i x_{i,i} = \frac{1}{n} \sum_i y_i T_n(x)_i
\qquad (x=(x_{i,j})\in \mathbb M_n^+, y=(y_i)\in(\ell^\infty_n)^+) $$
Thus $T_n(x)_i = n x_{i,i}$, that is, project onto the diagonal and multiply by $n$.
Suppose we can find positive matrices $a_n, b_n$ in $\mathbb M_n$ so that $\|a_n\|\leq 1, \|b_n\|\leq 1$, with the diagonal entries bounded by $1/n$, but with $n |a_n - b_n|$ having diagonal entries which are unbounded, as $n\rightarrow\infty$.  Set $a=(a_n) \in M$ and $b=(b_n)$.  By the assumptions on $(a_n)$ and $(b_n)$ we see that $a,b \in \mathfrak{m}_T^+$ but that $|a-b|\not\in\mathfrak{m}_T^+$.  This is equivalent to $x=a-b \in \mathfrak{m}_T$ but $x_+\not\in\mathfrak{m}_T^+$.
How do we find $a_n,b_n$?  Here's one construction.  Let $(\delta_i)_{i=1}^n$ be the usual orthonormal basis of $\mathbb C^n$ and let
$$ \xi = \frac{1}{\sqrt n} \sum_{i=1}^n \delta_i, \qquad
\eta = \frac{1}{\sqrt n} \sum_{i=1}^{n-1} \delta_i. $$
Let $a_n = a = \theta_{\xi,\xi}$ the rank-one positive operator, and let $b_n = b = \theta_{\eta,\eta}$.  As matrices, $a$ has all entries $1/n$ and $b$ has all entries $1/n$ except for the last row and column which are 0.  We claim that $|a-b|$ has diagonal entries of size order $\sqrt n$, to be precise, $|a-b|_{n,n}$ is about $\sqrt n$.
I don't see an easy way to show this.  I spent ages working out a formula for $|\theta_{\xi,\xi} - \theta_{\eta,\eta}|$ for arbitrary vectors $\xi,\eta$ in a Hilbert space, and then performed the particular calculation here.  If anyone has a nice argument, I'd like to see it.<|endoftext|>
TITLE: Chern-Einstein metrics on complex Hermitian manifolds
QUESTION [6 upvotes]: Metric on a Riemannian manifold $(M,g)$ is Einstein, if for some function $\lambda\colon M\to \mathbb R$
$$
Ric(g)=\lambda g.
$$
It is well know, that such $\lambda$ is, in fact, a constant.
The notion of Einstein metric fits perfectly into the world of K\"ahler manifolds, in this case such a metric is called K\"ahler-Einstein. Existence of K\"ahler-Einstein metrics is related to many deep results in differential and algebraic geometry.
I wonder whether there is something interesting going on for an arbitrary complex Hermitian manifold $(M, g, J)$. More specifically, let $\nabla$ be  the Chern connection on a Hermitian manifold $(M, g, J)$ and let $\Omega\in \Lambda^{1,1}T^*M\otimes\Lambda^{1,1}T^*M$ be the curvature form of this connection
$$
\Omega(\xi,\bar\eta,\zeta,\bar\nu):=g\Bigl(
(
\nabla_\xi\nabla_{\bar\eta}-\nabla_{\bar\eta}\nabla_\xi-
\nabla_{[\xi,\bar\eta]}
)\zeta, \bar\nu
\Bigr).
$$
Note that unlike the K\"ahler situation, $\nabla$ has torsion and $\Omega$ has less symmetries, then usually. In particular, we can define two different Chern-Ricci forms:
$$
\Theta^{(1)}_{i\bar j}=g^{k\bar l}\Omega_{i\bar jk\bar l},\quad
\Theta^{(2)}_{k\bar l}=g^{i\bar j}\Omega_{i\bar jk\bar l}.
$$
One might play the same game as in Riemannian case, and introduce two versions of Chern-Einstein metrics (is there a common name for these phenomena?)
$$
g_{i\bar j}=\lambda_1 \Theta^{(1)}_{i\bar j}, \mbox{ and } g_{i\bar j}=\lambda_2 \Theta^{(2)}_{i\bar j}.
$$
Remark. In the second case the metric $g$ on $T^{1,0}M$ is Hermitian-Einstein. Hermitian-Einstein is not necessarily Chern-Einstein for $\Theta^{(2)}$.
Very vaguely, I would like to know
Q1 What is know about the existence of Chern-Einstein metrics on Hermitian manifolds (I am mostly interested in dimensions >2)?
Q2 What are topological/geometrical obstructions to the existence of such metrics?
A more direct question is
Q3 Is it true that $\lambda_1$ and $\lambda_2$ (whic are apriori functions on $M$) are necessarily constant?

REPLY [6 votes]: Let me write your equations instead as
$$\Theta^{(1)}_{i\bar{j}}=\lambda_1 g_{i\bar{j}},$$
$$\Theta^{(2)}_{i\bar{j}}=\lambda_2 g_{i\bar{j}},$$
where $\lambda_1,\lambda_2$ are real-valued functions. The Chern-Ricci form $\Theta^{(1)}=\sqrt{-1}\Theta^{(1)}_{i\bar{j}}dz^i\wedge d\bar{z}^j$ is closed.
The case $\lambda_1\equiv 0$ is easy to deal with. Such a Hermitian metric exists if and only if the Chern-Ricci form defines the zero class in the Bott-Chern cohomology $H^{1,1}_{\rm BC}(X,\mathbb{R})$, i.e. if and only if for some (and hence any) Hermitian metric $g$ we have $\Theta^{(1)}_g=\sqrt{-1}\partial\bar{\partial}F$ for some smooth function $F$ on $X$. In this case the conformal metric $e^{F/n}g$ has vanishing Chern-Ricci form. The vanishing of this cohomology class implies the vanishing of $c_1(X)$ in $H^2(X,\mathbb{R})$, but not conversely.
When $\lambda_1$ is not identically zero, then $\lambda_1$ is constant if and only if the torsion $1$-form $\tau$ of $g$ vanishes identically, where 
$\tau=T_{ij}^j dz^i,$ and $T_{ij}^k$ is the torsion of $g$. This is a result of Goldberg. In this case, we must have that $g$ is in fact Kähler, since $d\Theta^{(1)}=0$.
In general, the equation $\Theta^{(2)}_{i\bar{j}}=\lambda_2 g_{i\bar{j}}$ does not imply that $\lambda_2$ is constant. Also, even if $\lambda_2$ is constant and nonzero this does not imply that $g$ is Kähler. A simple example is the standard Hermitian metric
$$g_{i\bar{j}}=\frac{\delta_{ij}}{|z|^2+|w|^2},$$
on the Hopf surface $\mathbb{C}^2/((z,w)\sim (2z,2w))$, which satisfies $\Theta^{(2)}_{i\bar{j}}=g_{i\bar{j}}$.
In general since such metrics are Hermitian-Einstein, you get topological obstructions from the Chern number inequality.
Lastly, you could also define a third Ricci curvature
$$\Theta^{(3)}_{i\bar{j}}=g^{k\bar{\ell}}\Omega_{i\bar{\ell}k\bar{j}}.$$
Then the result of Goldberg also holds in this case.<|endoftext|>
TITLE: $\mathsf{AD}_\mathbb{R}$ and Elementary Embeddings
QUESTION [8 upvotes]: Suppose $\mathsf{AD}_\mathbb{R} + V = L(\mathscr{P}(\mathbb{R})) + \mathsf{DC}$ holds. (We can use more if it is helpful.)
I believe under $\mathsf{AD}_\mathbb{R}$, every $A \subseteq \mathbb{R}$ is homogeneous Suslin. Let $T$ be some tree so that $A = p[T]$. Suppose $\mathbb{P}$ is a small forcing in $V_{\omega + \omega}$. (I am most interested by forcing whose conditions are reals like Sacks forcing, Cohen forcing, or Random forcing.) Let $G \subseteq \mathbb{P}$ be $\mathbb{P}$-generic for $V$. Let $A_G = p[T]^{V[G]}$. The question is
Is there an elementary embedding $j : L(A,\mathbb{R}) \rightarrow L(A_G,\mathbb{R}^{V[G]})$?

Ultimately, I would like, under $\mathsf{AD}_\mathbb{R}$, to have some generic absoluteness about statements involving $A$ (as defined as $p[T]$). 
This statement also resemble results that are proved in choice setting assuming large cardinals. 

The only ideas I have are: In Solovay's paper about independence of DC, he showed that under $\mathsf{AD}_\mathbb{R}$, for all $A \subseteq \mathbb{R}$, $(A,R)^\sharp$ exists. In $\mathsf{AD}_\mathbb{R}$, $(A,\mathbb{R})^\sharp$ is homogeneously Suslin. So $(A,\mathbb{R})^\sharp = p[U]$ for some homogeneously Suslin tree $U$. I am hoping perhaps one can show that $p[U]^{V[G]} = (A_G,\mathbb{R}^{V[G]})^\sharp$. Therefore, $(A,\mathbb{R})^\sharp \subseteq (A_G, \mathbb{R}^{V[G]})^\sharp$. I believe this will imply that there is an elementary embedding from $L(A,\mathbb{R})$ into $L(A_G,\mathbb{R}^{V[G]})$. 
I would like to know if the embedding above can exists under $\mathsf{AD}_\mathbb{R}$. Can my vague argument possibly work? Are there other ways to obtain this embedding or the desired generic absoluteness. Thanks for any information and references.

REPLY [4 votes]: Yes, you can show this using your assumption that every set of reals in $L(A,\mathbb{R})$ is $\delta$-weakly homogeneously Suslin (Woodin's Pmax book, theorem 2.30). In that case $(A,\mathbb{R})^{\#}$ exists by closure of pointclass under countable unions. Alternatively you can assume that all sets of reals in $L(A,\mathbb{R})$ are $\delta$-universally Baire and $(A,\mathbb{R})^{\#}$ exists, see theorem 6.3 in http://www.math.yorku.ca/~ifarah/Ftp/2005l14-extender.pdf<|endoftext|>
TITLE: Are the Gessel sequence integers composite for all $n\ge 3$?
QUESTION [5 upvotes]: The Gessel sequence is known for Ira Gessel's Lattice Path Conjecture of $2001$, which has been proved by Kauers, Koutschan and Zeilberger in $2009$ with the aid of a computer. Later, other proofs were found  ("human proofs"), e.g., by using Weierstrass elliptic functions (see here).
I was wondering whether or not it is true that the integers $a_n$ of the Gessel sequence are never prime numbers for $n\ge 3$. There is an easy recursion for this sequence, namely $a(0):=1$ and 
$$
a_{n+1}=\frac{4(6n+5)(2n+1)}{(3n+5)(n+2)}a_n
$$ 
for all $n$. Is there some (obvious) reason that all $a_n$ are composite ?

REPLY [4 votes]: It is easy to see from the recurrence that every prime factor of $a_{n}$ is less than $6n$.  But the sequence grows much faster than that.<|endoftext|>
TITLE: Equivariant maps from simplicial complexes to spheres
QUESTION [5 upvotes]: Given a topological space $X$ with  involution $\nu$, the $\mathbb Z_2$-index $\text{ind}(X)$ is the minimum integer $n$ such that there exists a map $f:X \to S^n$ which is equivariant with respect to the antipodal map on the sphere $S^n$.
Let $K$ be a (finite) simplicial complex with a fixed-point-free involution such that $\text{ind}(|K|)=d$ and $|K|$ is not homotopy equivalent to $S^d$. Can we always find a (maximal) simplex $\sigma$ such that deleting $\sigma$ and $\nu(\sigma)$ does not decrease the $\mathbb Z_2$-index?

REPLY [3 votes]: I believe the following should work as a counter-example. Let $K$ be a simplicial torus $T^2$, obtained as the orientation double cover of a triangulated Klein bottle $\overline{K}$. Then $K$ comes equipped with an (orientation-reversing) involution $\nu$. I believe $\operatorname{ind}(|K|)=2$, and that removing any simplex $\sigma$ and its involute $\nu(\sigma)$ decreases the index.
The key observation (for me at least) is this: if $X$ is a (reasonably nice) free $\mathbb{Z}_2$-space and $p:X\to \overline{X}$ is the resulting quotient double cover, then
$$
\operatorname{ind}(X) = \operatorname{secat}(p:X\to \overline{X}).
$$
Here $\operatorname{secat}(p)$ is the sectional category (or normalized Schwarz genus) of the double cover, which by definition is one less than the smallest number of open sets needed to cover $\overline{X}$, on each of which $p$ admits a section. It follows that
$$
\operatorname{cup-length}\ker(p^*:H^*(\overline{X})\to H^*(X))\le \operatorname{ind}(X)\le \dim(\overline{X}).
$$   
Returning to the example of the torus covering the Klein bottle, the calculation of the index follows from mod 2 cohomology calculations, and the fact that the index decreases on removing any top-dimensional cell follows from homotopy invariance of the sectional category together with the fact that the Klein bottle minus a cell deformation retracts onto its 1-skeleton.  
In fact, this should generalize: whenever $X$ is a closed $d$-dimensional free $\mathbb{Z}_2$-manifold with $\operatorname{ind}(X)=d$ we should see this behaviour.<|endoftext|>
TITLE: Are square tiled surfaces dense in the moduli space of translation surfaces?
QUESTION [7 upvotes]: I'm reading the survey "An introduction to Veech surfaces" by Pascal Hubert and Thomas Schmidt. 
At page 19 they state "In any fixed stratum, the set of square-tiled surfaces of that stratum is dense.". The reason should be that in the coordinates for the moduli space of translation surfaces given by the period map (integration of the 1-form on relative periods) the coordinates of the translation surfaces are exactly $\mathbb{Q}+i\mathbb{Q}$. So the assertion follows because of the density of $\mathbb{Q}$ in $\mathbb{R}$.
Well, it's hard for me to believe in Hubert and Schmidt's assertion.
Isn't it true that the square tiled surfaces are the "integer points" of the moduli space of translation surfaces? If I have a translation surface tiled by squares with the side of length one, isn't it true that the relative periods are contained in $\mathbb{Z}+i\mathbb{Z}$?
So my questions are:
Is Hubert and Schmidt's assertion wrong?
If so, which are the translation surfaces corresponding to rational points?
Thank you

REPLY [3 votes]: Is Hubert and Schmidt's assertion wrong?

No, they are correct.  They allow squares where the sidelength is not equal to one. 
If you like, we can say that the two translation surfaces $(X, \omega)$ and $(X, r\omega)$ (for $r$ positive and real) are "scalar multiples" of each other.  Then under any definitions, the scalar multiples of the square-tiled surfaces are dense in the space of translation surfaces.<|endoftext|>
TITLE: Are all formal schemes *really* Ind-schemes?
QUESTION [19 upvotes]: $\newcommand\LRS{\mathsf{LRS}}\newcommand\FormalSch{\mathsf{FormalSch}}\DeclareMathOperator\Spf{Spf}\newcommand\IndSch{\mathsf{IndSch}}\newcommand\ALRS{\mathsf{ALRS}}\newcommand\FSch{\mathsf{FSch}}$I'm trying to understand whether there's a fully faithful functor $\LRS \supset \FormalSch \to \IndSch$ and in what sense. Here's my progress so far:
Let $\mathsf{A}$ be the category of adic rings. The objects are topological rings whose topology is generated by a descending filtration of ideals whose intersection is $\{0\}$. Morphisms are continuous homomorphism of rings.
There's a functor $\Spf: \mathsf{A} \to \IndSch$ which takes an adic ring to the formal spectrum which is naturally a filtered colimit of (affine) schemes). The target of the functor could be that of adic locally ringed spaces (topological spaces with sheaves of adic rings and morphisms between for which the comorphism of sheaves is continuous). Denote this category $\ALRS$.
In $\ALRS$ we have an adjunction with the "continuous" global section functor $\Gamma_{\text{cont}} \dashv \Spf $. Continuous here just means it remembers the topology (i.e. the filtration).
Now the definition of formal schemes feels inevitable:

Definition: A formal scheme is an adic locally ringed space locally isomorphic to
a formal spectrum of an adic ring. Denote the subcategory of formal schemes by $\FSch\subset \ALRS$.

This raises a problem though. There's no obvious way to turn a "formal scheme" in this sense into an ind-schemes (which are much more convenient for certain purposes). We could try to define the ind-scheme as the formal colimit over the Čech nerve of a chosen covering by formal spectra (which are themselves filtered colimits of affine schemes). However, this is probably a very bad idea since it will most likely depend on the choice of covering.

Question: Can we construct a functor $\FSch \to \IndSch$ with some good properties? (Hopefully fully faithful but if not maybe at least full.) If not is there a better definition of a formal scheme which enables you to play in both worlds (ind schemes and locally ringed spaces)?

REPLY [4 votes]: In Yasuda's article Non-adic formal schemes, https://arxiv.org/abs/0711.0434, he redefines formal schemes as certain topological spaces equipped with a sheaf of pro-rings (as opposed to a sheaf of topological rings).
Complete linearly topologized rings embed fully faithfully into the category of pro-rings (Paragraph 5.2 of loc.cit) so maybe this is in the direction you are looking for? Admissibility is discussed in Paragraph 5.3.<|endoftext|>
TITLE: Confusion surrounding the Koszul-Malgrange theorem
QUESTION [6 upvotes]: I recently had the need to appeal to some complex geometry in my research and have been trying to unravel the various relationships surrounding the Koszul-Malgrange theorem.
According to nlab, the theorem goes as follows.

Theorem 1. (Koszul-Malgrange theorem)
Holomorphic vector bundles over a complex manifold are equivalently complex vector bundles which are equipped with a (hermitean) holomorphic flat connection. Under this identification the Dolbeault operator $\overline{\partial} $ acting on the sections of the holomorphic vector bundle is identified with the holomorphic component of the covariant derivative of the given connection.

Since the original reference is in french, which I have trouble reading, I haven't been able to go though the proof of the theorem. From the bit of complex geometry I know though, I am a bit confused.
I know that, given a complex vector bundle $E\to X$ and a flat connection, the $(0,1)$-part of the connection defines a holomorphic structure, while the $(1,0)$-component defines a holomorphic connection on the corresponding holomorphic bundle. By Chern-Weil theory, the rational chern classes of $E$ vanish, as do the rational Chern classes of a holomorphic vector bundle equipped with holomorphic connection.
On the other hand, I know that in general, a holomorphic bundle may have nonvanishing rational Chern classes. In fact, over compact, complex projective manifolds, these classes generate certain Hodge classes (or possibly all according to the Hodge conjecture).
This makes me think the theorem should be a correspondence between flat bundles and holomorphic vector bundles with holomorphic connection.
Question
What is the exact relationship between smooth complex vector bundles equipped with connection and holomorphic structures? I realize this question is probably a bit elementary, but complex geometry is not my specialty. References are also welcome.

REPLY [10 votes]: Since I was  rather surprised that those authors would make such a claim, I looked up the original reference [Koszul, Malgrange, Sur certaines structures fibrées complexes.] Here is a rough translation of the relevant theorem:

Theorem 2. Let $G$ be a complex Lie group, $V$ a complex manifold, $P$ a principal bundle with group $G$ over $V$. Let $\gamma$ be a connection form on $P$. The following are equivalent

The almost complex structure defined by $\gamma$ is a complex structure.
...
The $(0,2)$ part of the curvature $R_\gamma$ is $0$.
...

[Footnote in orignal: ...For $R_\gamma$ to be of type $(2,0)$ it is necessary and sufficient that $\gamma$ be holomorphic...]

So that they are not saying that the curvature is zero, or that the connection is holomorphic. Clearly the version of the theorem you quoted is needlessly ambiguous, at best. Perhaps you should contact the author of that nlab page.<|endoftext|>
TITLE: New separation axiom?
QUESTION [9 upvotes]: I am looking for the name and notation of the following separation axiom , temporarily denoted by $T_i$ (where $i=\sqrt{-1}$ is the imaginary unit):
Axiom $T_i$: For any point $x$ of a topological space $X$ and any neighborhood $O_x$ of $x$ there is a closed subset $F$ in $X$ that contains $x$ and is contained in $O_x$.
It is easy to see that a topological space $X$ satisfies the axiom $T_i$ if and only if each open set in $X$ is a union of closed sets.
It is easy to check that the separation axiom $T_1$ is equivalent to $T_0+T_i$.
The connected doubleton is an example of a $T_0$-space which is not $T_i$.
Any anti-discrete space satisfies $T_i$ but not $T_0$. So, the axioms $T_0$
and $T_i$ are incomparable.
Question: Is the axiom $T_i$ known? If yes, where is it introduced and how is it denoted and called?

REPLY [12 votes]: According to the Wikipedia article about ${\mathrm T}_1$ spaces your ${\mathrm T}_i$-spaces are called $\it symmetric$ or ${\mathrm R}_0$-spaces. There are several equivalent conditions, my personal favorite being that point closures are antidiscrete.
Unfortunately I was not able to pin down the initial place where this axiom has been introduced or used. The article refers to two books, but I could not find anything about ${\mathrm R}_0$ there.<|endoftext|>
TITLE: Has there been a computer search for a 5-chromatic unit distance graph?
QUESTION [29 upvotes]: The existence of a 4-chromatic unit distance graph (e.g., the Moser spindle) establishes a lower bound of 4 for the chromatic number of the plane (see the Nelson-Hadwiger problem).
Obviously, it would be nice to have an example of a 5-chromatic unit distance graph. To the best of my knowledge, the existence of such a graph is open. Has there been any (documented) attempt to find such a graph through a computer search? For instance, has every $n$-vertex possibility been checked up to some $n$?

REPLY [43 votes]: As of this morning there is a paper on the ArXiv claiming to show that there exists a 5-chromatic unit distance graph with $1567$ vertices.  The paper is written by non-mathematician Aubrey De Grey (of anti-aging fame), but it appears to be a serious paper.  Time will tell if it holds up to scrutiny.
EDIT: in fact, it must be the one with 1585 vertices, according to checkers, see https://dustingmixon.wordpress.com/2018/04/10/the-chromatic-number-of-the-plane-is-at-least-5/<|endoftext|>
TITLE: structure of maximal tori in semisimple algebraic groups
QUESTION [5 upvotes]: I feel experts might be able to  answer this question immediately.
Let $G$ be a connected $\mathbb Q$-simple and $\mathbb Q$-isotropic  algebraic group. 
Let $S$ be a maximal $\mathbb Q$-split torus of $G$ and
let $T\supset S$ be a maximal torus defined over $\mathbb Q$. Let $T_a$ be the maximal anisotropic subtorus of $T$. Is $T$ a direct product of $S$ and $T_a$?

REPLY [5 votes]: As the example of nfdc23 shows, the answer is generally no.  But maybe it helps to think about this question in a somewhat wider context, where the notions of split and anisotropic tori arise: the study of a connected reductive algebraic group defined over an arbitrary field $k$ (as in the 1965 paper by Borel and Tits).  In the structure theory of such groups, it quickly becomes clear that the nature of $k$-anisotropic groups depends heavily on $k$ (and is not understood for many familiar fields).    Leaving that aside, Borel and Tits got a lot of unified information about the structure of a $k$-isotropic group.    Modulo the knowledge of $k$-anisotropic groups, this leads ultimately to the Tits classification method.  Of course, the special case $k=\mathbb{Q}$ is part of this story, but the main ideas are developed for all $k$.
Note especially that the question raised here never gets answered explicitly in the structure theory.  Indeed, the maximal $k$-anisotropic subtorus $T_a$ here is mentioned but does not play an important role.   The key players include: a (nontrivial!) maximal $k$-split torus $S$ (unique up to $k$-conjugacy), along with its (reductive) centralizer in $G$ (which of course contains $T_a$), a minimal $k$-parabolic subgroup containing $S$, and various data about the associated root systems and Weyl groups. 
What the general theory reveals is the existence of an almost-direct product $ T=T_a\, S$: see for example Borel's 8.15 in GTM 126.   But toward the end of their respective textbooks, Borel (in his expanded second edition) and Springer (in his later framework of $F$-groups) develop a lot of finer detail about classical groups somewhat in the spirit of the answer by nfdc23.   
One extreme, however, is the case of a quasi-split group $G$, in which a minimal $k$-parabolic subgroup is a Borel subgroup (and which is the only type possible for finite or some other special fields).<|endoftext|>
TITLE: Does the consistency strength hierarchy coincide with the "arithmetic consequence" hierarchy at ZF + Reinhardt?
QUESTION [7 upvotes]: In these slides (see especially slide 26), Steel emphasizes the phenomenon that for all known "natural" extensions of ZFC, the ordering by consistency strength agrees with the ordering by containment of arithmetic consequences. That is, if $T$ and $T'$ are natural extensions of ZFC, then $T \leq T'$ with respect to consistency strength if and only if $T'$ proves all the arithmetic consequences of $T$. Indeed, Steel points out that as long as $T$ and $T'$ have high enough consistency strength, the same could be said of their analytical consequences, not just their arithmetical consequences.
But I find it remarkable that the first-order theories of the strongest consistency strength known (which are not known to be inconsistent) are most naturally formulated in ZF without the axiom of choice. Namely, if you look at the theories listed at Cantor's attic, the highest consistency strength theories are ZF + Reinhardt cardinals or ZF + Berkeley cardinals (I'm being vague about how many of these cardinals are stated to exist). Of course you can get stronger theories by taking Con(ZF + Reinhardt) etc. in some language, but these are the basis of things.
My question is, basically: does the phenomenon Steel discusses continue to hold for natural extensions of ZF, rather than ZFC? In particular, does ZF + Reinhardt cardinals imply all the arithmetical consequences of ZFC + $I_0$ (the strongest large cardinal principle not known to be inconsistent with choice)? How about ZF + Berkeley cardinals?
EDIT
Steel says a little more about "naturalness" on p. 6, footnote 10 of the related paper. I'll state it using some notation:

If $A$ is a set of statements in the language of set theory, let $\overline{A}$ denote the deductive closure of $A$
If $A$ is a set of statements in the language of set theory, let $\mathrm{Con}^\mathrm{Fin}(A)$ denote the set of statements $\{\mathrm{Con}(\phi) \mid \phi \in A\}$ where $\mathrm{Con}(\phi)$ is the statement that $\phi$ is consistent.
If $S,T$ are extensions of ZFC, write $S \leq_{\mathrm{Con}}^{\ast} T$ if $\mathrm{Con}^\mathrm{Fin}(S) \subseteq T$. Note by compactness that $S\leq_\mathrm{Con}^\ast T$ implies $S \leq_\mathrm{Con} T$, where $\leq_\mathrm{Con}$ is the usual consistency strength preorder.
If $T$ is an extension of ZFC, let $(\Pi^0_1)_T$ denote the $\Pi^0_1$ consequences of $T$.

According to Steel, the reflection principle implies for any $T$ extending ZFC that $(\Pi^0_1)_T = \overline{\mathrm{Con}^\mathrm{Fin}(T)}$. This implies that

$(\Pi^0_1)_S \subseteq (\Pi^0_1)_T \Leftrightarrow S \leq_\mathrm{Con}^\ast T$

That is, the ordering on extensions of ZFC by $\Pi^0_1$ consequences can be recast in terms of consistency statements. Steel says that for the "natural" extensions he's talking about, it turns out that $S \leq_\mathrm{Con} T \Leftrightarrow S \leq_\mathrm{Con}^\ast T$. In some sense this is just a re-statement of another version of the phenomenon, but I think it sheds some light.
I'm not sure whether all of this should go through with ZF in place of ZFC, but if it does, it would be interesting to know whether $ZFC +I_0 \leq_\mathrm{Con}^\ast ZF + \mathrm{Reinhardt}$ for example. Of course, this is still quite a ways from talking about containment of all arithmetic statements. I would be happy to say something about the $\Sigma^0_1$ statements -- I'm particularly keen (for kind of frivolous reasons) to know in $ZF + \mathrm{Reinhardt}$ that a given Turing machine will halt so long as $ZFC + I_0$ thinks it will halt.

REPLY [11 votes]: Steel makes his assertions only for "natural" theories extending
$\newcommand\ZFC{\text{ZFC}}\ZFC$, and so in order to focus
attention on the extent to which his phenomenon relies on this
idea of naturality, let me describe a few counterexamples to the
phenomenon, even in the $\ZFC$ case. The fact of the matter is
that there are some natural-seeming theories for which the
phenomenon can fail.
First example. Imagine a large cardinal set theorist who
asserts that inconsistency is no more likely at the level of Mahlo
cardinals than it is at the level of inaccessible cardinals. I
believe that many large cardinal set theorists actually hold this
view, and I have heard some prominent set theorists assert similar
claims explicitly. Indeed, Woodin has asserted that if $\ZFC$ is
found inconsistent, then he would look to $\text{PA}$ to fall
next. So in this sense, the view is not unreasonable.
Let $T$ be the theory that expresses the view explicitly, so that
$T$ asserts $\newcommand\Con{\mathop{Con}}\ZFC$ plus the
implication $\Con(\ZFC+I)\to\Con(\ZFC+M)$, where $I$ asserts that
there is an inaccessible cardinals and $M$ asserts that there is a
Mahlo cardinal. So $T$ asserts over $\ZFC$ that if an inaccessible
cardinal is consistent, then so is a Mahlo cardinal.
This theory is actually equiconsistent with $\ZFC$, because if
$\ZFC$ is consistent, then there is a model of $\ZFC$ where $\ZFC$
is inconsistent, which would make $T$ vacuously true, by denying
the antecedent. Thus, $T\leq_{\Con}\ZFC$ in the consistency
strength hierarchy. But meanwhile, $T$ has additional arithmetic
consequences, such as the implication
$\Con(\ZFC+I)\to\Con(\ZFC+M)$ itself, which is not provable in
$\ZFC$ unless inaccessible cardinals are inconsistent, since
$\Con(\ZFC+M)$ implies $\Con(\ZFC+\Con(\ZFC+I))$, which would
violate the incompleteness theorem for $\ZFC+\Con(\ZFC+I)$ if this
theory were consistent. So this is a counterexample to Steel's
phenomenon.
One can construct many similar counterexamples to the phenomenon,
by using theories of the form
$\ZFC+\Con(\ZFC+LC_1)\to\Con(\ZFC+LC_2)$, whenever $LC_1$ is a
weaker large cardinal than $LC_2$. Steel will have to claim that
these theories are not natural in order to preserve his
phenomenon. But are they so unnatural? Most large cardinal set
theorists would find them to be true.
A Better example. Let me now explain a much better
counterexample to the phenomenon (thanks to Emil for explaining
how to make it work). If $T$ is a natural theory extending $\ZFC$,
then most people would find $\ZFC+\Con(T)$ also to be a natural
theory. Indeed, the literature is full of theorems proved under
the assumption $\Con(T)$, where $T$ is any of the usual large
cardinal theories. The theory is strictly stronger than $T$ in
consistency strength. But meanwhile, I claim that if $T$ is $\ZFC$
plus any of the usual large cardinal axioms, then $T$ will have
arithmetic consequences not provable in $\ZFC+\Con(T)$, and so
this is a counterexample to the phenomenon.
The reason is that all of the usual large cardinal axioms prove
that there is a transitive model of $\ZFC$, and this is enough to
establish the implication
$$\newcommand\Prov{\text{Prov}}\Prov_{\ZFC}(\varphi)\to\varphi$$
for arithmetic assertions $\varphi$. Note that this implication is
itself an arithmetic assertion. For example, if $\kappa$ is
inaccessible, then since $V_\kappa\models\ZFC$, it follows that
$\Prov_{\ZFC}(\varphi)$ implies that $V_\kappa$ thinks the
arithmetic assertion $\varphi$ is true, and this is absolute to
$V$, since $V_\kappa$ has the same arithmetic as $V$. So $\varphi$
is true.
But meanwhile, $\ZFC+\Con(T)$, if consistent, will not be able to
prove all instances of this implication. The reason is that by the incompleteness
theorem, there is a model of
$\ZFC+\Con(T)+\neg\Con(\ZFC+\Con(T))$. Because of the negated
consistency assertion, the model will satisfy
$\Prov_{\ZFC}(\neg\Con(T))$, while also satisfying $\Con(T)$, and
this violates the desired implication
$\Prov_{\ZFC}(\varphi)\to\varphi$ in the instance where $\varphi$
is $\neg\Con(T)$.
So the theory $\ZFC+\Con(T)$ is a counterexample to the phenomenon
discussed by Steel, where $T$ is any of the usual large cardinal
axioms, because $T$ proves $\Prov_{\ZFC}(\varphi)\to\varphi$ for
arithmetic assertions $\varphi$, but these implications are not
all provable in $\ZFC+\Con(T)$, if this theory is consistent.<|endoftext|>
TITLE: Weil conjectures for higher dimensional cycles?
QUESTION [18 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{F}_{q}$. For each pair of positive integers $n$ and $d$, let $\text{Chow}_{n,d}(X)$ denote the (coarse) moduli space of $n$-cycles of degree $d$ on $X$, as defined in §3 of Kollár's book Rational Curves on Algebraic Varieties. Fixing $n>0$, define $Z_{n}(X,t)$ to be the formal power series
$$Z_{n}(X,t)\ \overset{{}_{\text{def}}}{=}\ \sum_{d=0}^{\infty}\big|\text{Chow}_{n,d}(X)(\mathbb{F}_{q})\big|\ \!t^{d}.$$
What is known about this series? Does it have obvious pathological behavior? Are there natural candidates for analogues of the Weil conjectures that one expects for $Z_{n}(X,t)$? Is there an easy way to derive any of these analogues from the Weil conjectures themselves, i.e., from the $n=0$ case?

REPLY [12 votes]: As Jason Starr notes in his comment, the series I ask about cannot be rational in general. There's more to the story though. I recently stumbled across several papers where Elizondo and Kimura study motivic versions of these series:
      [1] Elizondo, E. Javier. The Euler Series of Restricted Chow Varieties.
      [2] Elizondo, E. Javier and Kimura, Shun-Ichi. Irrationality of motivic series of Chow varieties.
      [3] Elizondo, E. Javier and Kimura, Shun-Ichi. Rationality of motivic Chow series modulo $\mathbb{A}^{\!1}$-homotopy.
I share some of their results here.
      Fix a ground field $k$. Let $\mathcal{M}^{\text{rat}}_{k}$ denote the category of Chow motives, and let $K_{0}(\mathcal{M}^{\text{rat}}_{k})$ denote its Grothendieck ring. Let $K(\text{Var}_{k})$ denote the Grothendieck ring of varieties. If $\overline{k}=k$ and $\text{char}_{\ \!}k=0$,then we have a ring homomorphism $K(\text{Var}_{k})\longrightarrow K_{0}(\mathcal{M}^{\text{rat}}_{k})$ taking the class of each smooth projective $k$-variety to the class of its Chow motive. In each of the rings of formal power series $K_{0}(\mathcal{M}^{\text{rat}}_{k})[\![t]\!]$ or $K(\text{Var}_{k})[\![t]\!]$, we can consider a motivic version of the series I ask about, namely
    $$
\sum_{d=0}^{\infty}\text{Chow}_{n,d}(X)\ t^{d},
$$
where the coefficients $\text{Chow}_{n,d}(X)$ denote either the $K_{0}(\mathcal{M}^{\text{rat}}_{k})$- or $K(\text{Var}_{k})$-classes of the corrresponding Chow varieities. In [2, Theorem 4.5], Elizondo and Kimura show that if $n\ge 2$, then the motivic Chow series
    $$
\sum_{d=0}^{\infty}\text{Chow}_{n-1,d}(\mathbb{P}^{n})\ \!t^{d}\ =\ \sum_{d=0}^{\infty}\frac{1-{\mathbb{A}^{\!1}}^{\left(\!\begin{smallmatrix} n+d \\ d\end{smallmatrix}\!\right)}}{1-\mathbb{A}^{1}}\ \!t^{d}
$$
in $K_{0}(\mathcal{M}^{\text{rat}}_{k})[\![t]\!]$ is irrational. But then in [3], they make the following observation: If we imagine giving $\mathbb{A}^{1}$ some family of measures $\mu(\mathbb{A}^{1})$ such that the limit $\mu(\mathbb{A}^{1})\to 1$ makes sense, then in this limit the above series becomes
    $$
\frac{1}{(1-t)^{n+1}}.
$$
In particular, it is rational. This leads to the idea of passing to an "$(\mathbb{A}^{\!1}\to 1)$-limit" , which we can think of as being some kind of motivic version of a deformation retraction of $\mathbb{A}^{1}$ to a point. Concretely, this means passing to one of the quotients $K_{0}(\mathcal{M}^{\text{rat}}_{k})_{\mathbb{A}^{1}}\overset{{}_{\text{def}}}{=}K_{0}(\mathcal{M}^{\text{rat}}_{k})\big/\!\sim$  or $K(\text{Var}_{k})_{\mathbb{A}^{1}}\overset{{}_{\text{def}}}{=}K(\text{Var}_{k})\big/\!\sim$, where "$\sim$" is the relation
    $$
X\times\mathbb{A}^{1}\ \sim\ X.
$$
      Assume $\overline{k}=k$ and $\text{char}_{\ \!}k=0$. If $X$ is a smooth projective $k$-variety, then for each $n\in  \mathbb{Z}_{\ge0}$, let $B_{n}(X)$ denote the commutative monoid of connected components of the scheme $\text{Chow}_{n}(X)\overset{{}_{\text{def}}}{=}\bigsqcup_{d=0}^{\infty}\text{Chow}_{n,d}(X)$. For each $\beta\in B_{n}(X)$, let $\text{Chow}_{n,\beta}(X)$ denote the corresponding component of $\text{Chow}_{n}(X)$. Then it is somewhat more natural to consider the motivic Chow series
    $$
\sum_{\beta\in B_{n}(X)}^{\infty}\!\!\!\!\text{Chow}_{n,\beta}(X)\ t^{\beta}
$$
inside either of the rings
    $$
K(\text{Var}_{k})_{\mathbb{A}^{1}}\big[\!\big[B_{n}(X)\big]\!\big]
\ \ \ \ \ \ \mbox{or}\ \ \ \ \ \
K_{0}(\mathcal{M}^{\text{rat}}_{k})_{\mathbb{A}^{1}}\big[\!\big[B_{n}(X)\big]\!\big]
$$
Elizondo and Kimura prove the following:

If $C_{1}$ and $C_{2}$ are very general curves, then for each $n\ge 0$, the motivic Chow series
$$
\sum_{\beta\in B_{n}(C_{1}\times C_{2})}^{\infty}\!\!\!\!\!\!\!\!\!\text{Chow}_{n,\beta}(C_{1}\times C_{2})\ t^{\beta}
$$
is rational [3, Corollary 3.11].
If $Y_{\Sigma}$ is a smooth projective toric variety, then for each $n\in\mathbb{Z}_{\ge0}$, its motivic Chow series in $K_{0}(\mathcal{M}^{\text{rat}}_{k})_{\mathbb{A}^{1}}\big[\!\big[B_{n}(Y_{\Sigma})\big]\!\big]$ is rational [3, Corollary 4.4], and satisfies the formula
$$
\sum_{\beta\in B_{n}(Y_{\Sigma})}^{\infty}\!\!\!\!\!\!\text{Chow}_{n,\beta}(Y_{\Sigma})\ t^{\beta}
\ \ =
\prod_{\sigma\in\Sigma(n)}\frac{1}{1-t^{\beta(\sigma)}},
$$
where $\beta(\sigma)$ denotes the $B_{n}(Y_{\Sigma})$-class of the torus-orbit corresponding to $\sigma$. Because the Euler characteristic preserves $\mathbb{A}^{\!1}$-homotopy, this provides a motivic argument behind Elizondo's earlier proof that each series
$$
\sum_{d=0}^{\infty}\chi\big(\text{Chow}_{n,d}(Y_{\Sigma})\big)\ \!t^{d}
\ \ \in\ \
\mathbb{Z}[\![t]\!]
$$
is rational [1]. It also generalizes the formula for the motivic L-function, modulo $\mathbb{A}^{\!1}$-homotopy, of a smooth projective toric variety $Y_{\Sigma}$ that comes from the motivic decomposition of $Y_{\Sigma}$.<|endoftext|>
TITLE: When are the braid relations in a quasitriangular Hopf algebra equivalent?
QUESTION [6 upvotes]: Quasitriangular Hopf algebras have to satisfy, amongst other conditions, the following equations:
$$(\Delta \otimes \mathrm{id}) (R) = R_{13} R_{23}$$
$$(\mathrm{id} \otimes \Delta) (R) = R_{13} R_{12}$$
It appears that these two equations are not equivalent, although I don't know a single example for an $R$ that satisfies one of the equations and not the other.
Are there known additional conditions on the Hopf algebra that imply the equivalence of the two braid relations?
Or otherwise:
Is there a Hopf algebra $H$ and an element $R \in H \otimes H$ that satisfies e.g. the first braid relation all further axioms, but not the second braid relation?
Motivation: I think I have discovered a case where they are equivalent, which is the case of a quantum group with half-twist, as described in this article by Noah Snyder and Peter Tingley. It deals with Hopf algebras where $R = \left(t^{-1} \otimes t^{-1}\right) \Delta(t)$ for a suitable element $t$ in the Hopf algebra which is called the half-twist. I'm currently writing this particular instance up, but I'm very curious whether there are other conditions that make the braid relations equivalent, or counterexamples. It's nice if they are equivalent, since it saves you a lot of work when searching for new examples.

REPLY [3 votes]: The condition $R_{21}\,R=I$ (to be triangular) implies the equivalence of the two equations.
For the second question, let $A$ be a finite abelian group, and $H=k^A$ the Hopf algebra of function on $A$. Functions $f:A\times A\to \mathbb{C}^*$ (or equivalently maps from $A$ to $\operatorname{Maps}(A,\mathbb{C}^*$)) are in correspondence with invertible elements in $H^{\otimes 2}$. The first equation is equivalent to $f(a,b+c)=f(a,b)f(a,c)$ and similarly for the second one. Then a map $R: A\to \operatorname{Hom}(A,\mathbb C^*)$ that is not a group homomorphism and $R(0)=0$, works as an example.<|endoftext|>
TITLE: Complex manifolds with spanning sets of holomorphic tensor fields
QUESTION [8 upvotes]: This question is an extension of this one.  Consider a complex manifold $(M^{2n}, J)$.  Fix $1 \leq p \leq n-1$, and suppose that the space of holomorphic sections of $\Lambda^{p,0}$ spans $\Lambda^{p,0}_x$ for all $x \in M$.  (The referenced question above is the case $p=1$.)  How wide a class of manifolds is this?  Certainly complex tori satisfy this, but I imagine there are other examples.  Of particular interest is the case $p=2$, where the question is at least in principle related to the existence of holomorphic Poisson structures.

REPLY [2 votes]: A very incomplete answer: if $c_1<0$ then there are no global sections of these tensor bundles: Kobayashi, First Chern class and holomorphic tensor fields, Nagoya Math. J., vol. 77, 1980, theorem A.
A compact complex manifold admitting a Kaehler metric immerses into a complex torus if and only if its cotangent bundle is spanned by global sections (https://arxiv.org/abs/1702.01701).
On the other hand, Robert Bryant, in his paper Rigidity and quasi-rigidity of Hermitian cycles in Hermitian symmetric spaces, says that 

if a vector bundle $F \to M$ is generated by its global sections and
  $M$ is compact and Kaehler, then, as is well-known, $c_2(F) ≥ 0$. If
  equality holds, then either $F$ is the pullback to $M$ of a holomorphic
  bundle $F' → C$ over a curve $C$ via a holomorphic map $M → C$ or else
  $F = L ⊕ T$ where $L$ is a line bundle and $T$ is trivial. There is a
  similar (though more complicated) characterization when $c_3(F) = 0$.<|endoftext|>
TITLE: When does there exist a convex polyhedron with given edge lengths?
QUESTION [5 upvotes]: Let $n$ be a positive integer, and let $n = \ell_1 + \dots + \ell_k$ be
a partition of $n$. Then there exists a convex polygon with side lengths
$\ell_1, \dots, \ell_k$ if and only if all of the $\ell_i$ are smaller
than $\frac{n}{2}$.
Questions: 

What is the corresponding criterion in $3$ dimensions,
i.e. under which conditions does there exist a convex polyhedron with
edge lengths $\ell_1, \dots, \ell_k$?
How many of the $p(n)$ partitions of $n$ do occur as lists of edge
lengths of some polyhedron -- i.e. which are these values for small $n$,
and what is asymptotically the proportion of such partitions among all
partitions of $n$ when $n$ goes to infinity?

REPLY [7 votes]: Since I expect the answer for all combinatorial types of polytopes to be somewhat obscure, let me give an answer to the more simple question of what happens for the tetrahedron. We denote the six edges of the tetrahedron by $(x,y,z,a,b,c)$ and assume six faces the ones with edges $(a,b,c), (x,y,c), (x,b,z)$ and  $(a,y,z)$. The sextuple $(x,y,z,a,b,c)$ corresponds to such a tetrahedron if and only if 

$(a,b,c), (x,y,c), (x,b,z)$ and  $(a,y,z)$ satisfy the triangle inequality, and 
the Cayley-Menger determinant $$CM(x,y,z,a,b,c):=\left(\begin{array}{rrrrr}
0 & x^{2} & y^{2} & z^{2} & 1 \\
x^{2} & 0 & c^{2} & b^{2} & 1 \\
y^{2} & c^{2} & 0 & a^{2} & 1 \\
z^{2} & b^{2} & a^{2} & 0 & 1 \\
1 & 1 & 1 & 1 & 0
\end{array}\right)$$ is positive. 

This would be a complete answer to your first question if you only asked for the tetrahedron.
So if you consider a partition $(p_1,\dots,p_6)$ of $n$, you have to say which of the $p_i$s correspond to which edge, or just check all permutations of the partition (so perhaps it would be more natural to consider compositions?!). In any case if you do this for the first few $n$, you get a sequence like this (if there is no mistake in my calculations):$$[0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 3, 2, 3, 4, 4, 6, 9, 9, 11, 13, 16, 17, 24, 25, 32, 34, 40, 47, 55, 61, 70, 83, 91, 101, 117, 124, 148, 160, 181, 195, 225, 243, 270, 301, 327, 359, 396, 432, 473, 513, 566, 604, 668, 712, 782, 844, 913, 985, 1056, 1148, 1227, 1313, 1421, 1509, 1640, 1730, 1863, 1985, 2127, 2269, 2410, 2572, 2738, 2905, 3099, 3276, 3490, 3682, 3914, 4128, 4377, 4615, 4892, 5170, 5450, 5753, 6057, 6389, 6725, 7067, 7460, 7807, 8236, 8638, 9077, 9504, 10005, 10461, 10961].$$ 
Compare this to partitions into 6 parts:$$[0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 14, 20, 26, 35, 44, 58, 71, 90, 110, 136, 163, 199, 235, 282, 331, 391, 454, 532, 612, 709, 811, 931, 1057, 1206, 1360, 1540, 1729, 1945, 2172, 2432, 2702, 3009, 3331, 3692, 4070, 4494, 4935, 5427, 5942, 6510, 7104, 7760, 8442, 9192, 9975, 10829, 11720, 12692, 13702, 14800, 15944, 17180, 18467, 19858, 21301, 22856, 24473, 26207, 28009, 29941, 31943, 34085, 36308, 38677, 41134, 43752, 46461, 49342, 52327, 55491, 58767, 62239, 65827, 69624, 73551, 77695, 81979, 86499, 91164, 96079, 101155, 106491, 111999, 117788, 123755, 130019, 136479]$$
If you plot these two, you get:
For a better comparison, I take the fifths root of the values to get this plot (since we now that partitions into 6 parts roughly grow like $n^5$):
From these few values one might conjecture that the blue values also grow roughly like $n^5$, perhaps with a factor $(1/2)^5$ slower than all partitions into 6 parts.<|endoftext|>
TITLE: About positive upper density
QUESTION [11 upvotes]: For $S\subset \mathbb{N}$ define the upper density as $D^{\ast
}(S)=\limsup_{n\rightarrow \infty }\frac{\left\vert S\cap \{1,2,\ldots,n\}   \right\vert }{%
\left\vert n\right\vert }.$
Question: Suppose $D^{\ast
}(S)>0$.  Is there $n\in\mathbb{N}$ such that $(S-n
)\cap S$ has positive upper density?
$S-n=\{s-n| s\in S\}$

REPLY [12 votes]: Suppose that $(S-n)\cap S$ has upper density 0 for all $n\leq N$. Then the density of the set of integers $x$, such that $|[xN, (x+1)N]\cap S|\geq 2$ is also 0, hence the upper density of $S$ is $\leq \frac{1}{N}$. Taking $N\rightarrow\infty$ gives a positive answer to your question.<|endoftext|>
TITLE: Cardinality of connected Hausdorff topologies
QUESTION [5 upvotes]: Let $X$ be an infinite set and let $C(X)$ denote the collection of connected Hausdorff topologies on $X$. Suppose $N\subseteq C(X)$ has the property that whenever $\tau\neq\sigma \in N$ then $(X,\tau)$ and $(X,\sigma)$ are not homeomorphic. In terms of $|X|$, how large can $|N|$ be at most?

REPLY [10 votes]: Let $\kappa:=|X|$.   Then $2^{2^\kappa}$  is an obvious upper bound for the number of topologies on $X$. 
Every ultrafilter on $\kappa$ will give you a Hausdorff topological space on $\kappa+1$; these are $2^{2^\kappa}$ many spaces.  Some of them might be homeomorphic, but there are only $2^\kappa$ many bijections, so you get $2^{2^\kappa}$ many non-homeomorphic spaces.  
Assuming that $\kappa$ is at least continuum, you can add one more point, and copies of the unit interval to each point of your original space; this will now give you a connected (even path-connected) space.  
If $\kappa$ is smaller than continuum, then replace the unit interval by some countable connected Hausdorff space.<|endoftext|>
TITLE: What do globes (used to construct globular sets, $\omega$-categories, etc.) actually look like?
QUESTION [7 upvotes]: Nlab introduces the globular category as a geometrical model to construct certain higher categorical structures (e. g. strict $\omega$-categories), just as quasi-categories, for example, are modelled on simplices. However, I didn't find much information on the geometric intuition behind their definition.
So, what do these globes really look like? More precisely, how is the geometric realization of globular sets obtained?

REPLY [5 votes]: The paper on the fundamental globular groupoid of a filtered space has some pictures, and a definition of the simplicial nerve of a globular $\omega$-groupoid. Relations with other areas are in the paper M. Kapranov, "Membranes and higher groupoids" arxiv 1602.06166 .  My own philosophy is spelled out in this Aveiro presentation. 
See also this discussion.<|endoftext|>
TITLE: Estimate of incomplete binomial integral
QUESTION [5 upvotes]: Let $0\le k \le n$. Prove that
$$
n\binom{n}{k}\int_{0}^{\frac{k}{n+1}}t^k(1-t)^{n-k}\,dt \le 1/2.
$$
As far as I know
1) it is proved for $\frac{k}{n+1}\le 1/2$ and 
2) not proved for $1/2 <\frac{k}{n+1}< 1$.
Is it really so?

REPLY [2 votes]: The following stronger inequality holds for all $k=0,\dots,n$: 
$$
(n+1)\binom{n}{k}\int_{0}^{\frac{k}{n+1}}t^k(1-t)^{n-k}\,dt \le 1/2.
$$
Indeed, the latter inequality means precisely that the median $m$ of the Beta distribution with parameters $a:=k+1\ge1$ and $b:=n-k+1\ge1$ is no less than $\frac{k}{n+1}=\frac{a-1}{a+b-1}$. 
By the well-known "Mode, Median, and Mean Inequalities" for the Beta distribution (see e.g. page 2 in http://arxiv.org/abs/1111.0433v1), 
\begin{equation}
 m\ge\frac{a-1}{a+b-2}\bigwedge\frac{a}{a+b} \ge\frac{a-1}{a+b-1}=\frac{k}{n+1},  
\end{equation}
as desired.<|endoftext|>
TITLE: Finite group action on quasi-projective varieties
QUESTION [5 upvotes]: Let $X$ be a smooth, quasi-projective variety, $G$ be a finite group which acts freely and properly on $X$. Denote by $\alpha:X \to X/G$ the quotient. Is $\alpha$ generically etale? 
Also, as I am new to this topic (of group action on varieties) can someone suggest a good reference for the topic? I am mainly interested in finite group actions.

REPLY [8 votes]: Let $X$ be a normal variety over an algebraically closed field of characteristic 0 with a finite group $G$ acting effectively.  Since $G$ is finite it is reductive and a geometric quotient $X/G$ exists.  The quotient map $X\to X/G$ is étale at a point $x\in X$ if and only if the stabilizer at $x$ is trivial.  Since the equations $gx-x$ are algebraic and define the locus where the action fails to be free, the complement is Zariski open and shows that the quotient map is always generically étale.
See Section 4.3 here.<|endoftext|>
TITLE: Difference between constructive Dedekind and Cauchy reals in computation
QUESTION [9 upvotes]: If the Axiom of Countable Choice (ACC)
$$ \forall n\in \mathbb{N} . \exists x \in X . \varphi [n, x] \implies \exists f: \mathbb{N} \longrightarrow X . \forall n \in \mathbb{N} . \varphi [n, f(n)] $$
is not assumed in constructive mathematics, Dedekind and Cauchy real numbers are not equivalent in general. Even worse, Dedekind real numbers might not even form a set. Besides that, Dedekind cuts are notoriously unpleasant in terms of actual computation which is one of the most attractive parts of constructive mathematics. However, see Efficient Computation with Dedekind Reals by Bauer.
If we think of a Cauchy real as an algorithm that contains a convergence modulus $M_x: \mathbb{Z} \rightarrow \mathbb{N}$ and outputs rational approximations as:
$$ \left( x \in \mathbb{R} \right) \triangleq \left( \forall k \forall n,m \geq M_x(k) . \big| x(n) - x(m) \big| \leq 2^{-k} \right), $$
we then have a straightforward and transparent framework for computations (see, for instance, Schwichtenberg). On the other hand, in presence of the ACC, program extraction from proofs may get more difficult. 
Dedekind reals are often used instead of Cauchy reals when ACC does not hold.
A Dedekind real is a defined as a pair $(L, U)$ of located sets of rational numbers satisfying:

$q \in L \iff \exists r.(q < r \land r \in L)$ and $q \in U \iff \exists r.(r < q \land r \in U)$ 
Both sets are inhabited: $\exists q. q \in L$ and $\exists q. q \in U$ 
Sets are disjoint: $ \neg (q \in U \land q \in L)$
(locatedness): $q < r \implies q \in L \lor r \in U$ 

What is the actual computational content of a Dedekind real in comparison to a (straightforward) Cauchy real? Can an efficient algorithm be derived for every constructive Dedekind real that outputs rational approximations with predictable accuracy (cf. modulus of convergence)? If yes, what is the difference between Cauchy and Dedekind algorithms?

REPLY [6 votes]: As others have mentioned, there are many different variations in the exact notion of Cauchy reals and Dedekind reals which affect the answer.
I will choose variations so that I can offer a counterpoint to Paul Taylor's claim that Dedekind cuts represent a problem that "remains to be solved." If one uses the propositions-as-types correspondence to encode Dedekind cuts as "predicative subsets" $L, U : \mathbb{Q} \to \mathcal{U}$, where, $\mathcal{U}$ is the universe of types, and also reinterprets the rules of Dedekind cuts using propositions-as-types, for instance

inhabitedness: $\sum_{x : \mathbb{Q}} x \in L$ and $\sum_{x : \mathbb{Q}} x \in U$
locatedness: $\prod_{q, r : \mathbb{Q}} q < r \to (q \in L) + (r \in U)$

then Dedekind cuts do represent problems which have already been solved as well. In particular, the inhabitnedess and locatedness proofs above give the computational content. By inhabitedness, one can determine that a real number lies within some finite open interval. Then, by repeatedly using locatedness to cover this interval with several smaller ones, one can narrow down the interval where the real number must be to an arbitrarily small width.
Conversely, we can put the Cauchy definition on the same footing as the Dedekind definition by treating it as a "metric completion" in the sense of Steve Vickers's Localic completion of generalized metric spaces I. In this framework, a Cauchy real is a predicate $B : \mathbb{Q} \times \mathbb{Q}^+ \to \mathcal{U}$ on "formal balls", where $B(q, \varepsilon)$ holds if the real number is (strictly) within $\varepsilon$ from $q$. Then one of the rules which $B$ must satisfy is
$$\prod_{\varepsilon : \mathbb{Q}^+} \sum_{q : \mathbb{Q}} B(q, \varepsilon),$$
which says topologically that for arbitrarily small $\varepsilon$, $\mathbb{R}$ is covered by balls with rational centers and radius $\varepsilon$, and computationally, that we can compute some $q$ within $\varepsilon$ of the real number.
The axioms defining valid Dedekind cuts as well as valid Cauchy predicates on formal balls provide a computational interface for computing with real numbers. The axioms also have a particular "geometric" form, which Vickers explains, and each suffices to define a topological space (or, more accurately, a space within the framework of locale theory or formal topology). Vickers proves (Theorem 26) that these two spaces, the order-theoretic (Dedekind) and metric (Cauchy), are homeomorphic, meaning that it is possible to use one computational interface to implement the other.
In general, the points of spaces formulated in this way may not be sets (predicatively), since one can vary the universe level of the predicative subsets. But $\mathbb{R}$ is in fact small (see Palmgren's Predicativity problems in point-free topology), so in fact the points of $\mathbb{R}$ form a set.
In terms of efficiency, one would probably consider metric API more "efficient", since approximating to a rational with a tolerance of $\varepsilon$ requires 1 "API call", whereas one needs a variable number of uses of the "locatedness" rule for Dededkind cuts, and each additional call provides only one more bit of information.<|endoftext|>
TITLE: Complete the following sequence: point, triangle, octahedron, . . . in a dg-category
QUESTION [25 upvotes]: Let $\mathcal C$ be a pre-triangulated dg-category (or a stable $\infty$-category, if you wish).

An object $X$ in $\mathcal C$ gives a "point":
$$X$$
A morphism $X\xrightarrow f Y$ in $\mathcal C$ gives a "triangle":
$$\begin{matrix}X&&\to&&Y\cr&\nwarrow&&\swarrow\cr&&\operatorname{cone}(f)\end{matrix}$$
A chain of two morphisms $X\xrightarrow fY\xrightarrow gZ$ in $\mathcal C$ gives an "octahedron":
$$\begin{matrix}\operatorname{cone}(g)&&\leftarrow&&Z\cr&\searrow&&\nearrow\cr\downarrow&&Y&&\uparrow\cr&\swarrow&&\nwarrow\cr\operatorname{cone}(f)&&\to&&X\end{matrix}\qquad\qquad\begin{matrix}\operatorname{cone}(g)&&\leftarrow&&Z\cr&\nwarrow&&\swarrow\cr\downarrow&&\operatorname{cone}(gf)&&\uparrow\cr&\nearrow&&\searrow\cr\operatorname{cone}(f)&&\to&&X\end{matrix}$$
(copied from wikipedia, which also has a more three-dimensional rendition of the octahedron).

Given a longer chain $X_0\xrightarrow{f_1}\cdots\xrightarrow{f_p}X_p$ in $\mathcal C$, is there a canonical $(p+1)$-dimensional convex polytope $V_p$ organizing together the natural morphisms between $X_0,\ldots,X_p$ and all possible cones $\operatorname{cone}(f_j\circ f_{j-1}\circ\cdots\circ f_i)$?  Note that it is natural to expect that:

One of the facets of $V_p$ is the $p$-simplex $\Delta^p$ of morphisms $X_0\xrightarrow{f_1}\cdots\xrightarrow{f_p}X_p$.
The polytopes associated to the chains obtained by "forgetting" one of the $X_i$ also occur as facets of $V_p$.

REPLY [11 votes]: I believe these are called 'hypersimplices'. See
1) Gelfand, Manin "Methods of homological algebra", Ex. IV.2 1(c), p. 260.
2) Belinson, Bernstein, Deligne "Faisceaux pervers", Remarque 1.1.14, p. 26.
3) The diagrams for $n$ up to 4: http://students.mimuw.edu.pl/~pa235886/pdf/hypersimplices.pdf

REPLY [10 votes]: As pointed out in Piotr Achinger's answer, what follows is just an attempt to solve Exercise IV.2 1(c) (p. 260) in Gelfand & Manin.
Trying to use the above comments by Dylan Wilson, seems like $V_p$ is the convex hull of edge midpoints of $\Delta^{p+1}$.
Its facets are $p+2$ copies of $V_{p-1}$ (convex hulls of edge midpoints of each of the $p+2$ facets of $\Delta^{p+1}$) and $p+2$ copies of $\Delta^p$ (obtained at each of the $p+2$ vertices of $\Delta^{p+1}$ as a result of truncation (of length up to edge midpoints, which is the harshest possible truncation until truncating hyperplanes begin to intersect inside the polyhedron)).
In addition to the facets described in two bullets at the end of the question, there is one more copy of $V_{p-1}$ corresponding to the composable $(p-1)$-tuple$$\operatorname{cone}(f_1)\to\operatorname{cone}(f_2f_1)\to...\to\operatorname{cone}(f_p\cdots f_1),$$and $p+1$ more copies of $\Delta^p$ corresponding to composable $p$-tuples
\begin{align*}
&\operatorname{cone}(f_1)\to\operatorname{cone}(f_2f_1)\to...\qquad \to\operatorname{cone}(f_{p-1}\cdots f_1)\to\operatorname{cone}(f_p\cdots f_1)\xrightarrow*X_0,\\
&\operatorname{cone}(f_2)\to\operatorname{cone}(f_3f_2)\to...\qquad\qquad\qquad   \to\operatorname{cone}(f_p\cdots f_2)\xrightarrow*X_1\to\operatorname{cone}(f_1),\\
&\operatorname{cone}(f_3)\to\operatorname{cone}(f_4f_3)\to...\qquad\qquad\qquad\ \quad \xrightarrow*X_2\to\operatorname{cone}(f_2f_1)\to\operatorname{cone}(f_2),\\
&\vdots\\
&\operatorname{cone}(f_i)\to...\to\operatorname{cone}(f_p\cdots f_i)\xrightarrow*X_{i-1}\to\operatorname{cone}(f_{i-1}\cdots f_1)\to...\to\operatorname{cone}(f_{i-1}),\\
&\vdots\\
&\operatorname{cone}(f_p)\xrightarrow*X_{p-1}\to...\ \to\operatorname{cone}(f_{p-1}f_{p-2}f_{p-3})\to\operatorname{cone}(f_{p-1}f_{p-2})\to\operatorname{cone}(f_{p-1}),\\
&X_p\to\operatorname{cone}(f_p\cdots f_1)\to...\ \quad\to\operatorname{cone}(f_pf_{p-1}f_{p-2})\to\operatorname{cone}(f_pf_{p-1})\to\operatorname{cone}(f_p), 
\end{align*}
where maps with stars denote maps with degree shift by one; that is, $A\xrightarrow*B$ means a map of the form $A\to\Sigma B$ (or else $A\to B[1]$, or $\Omega A\to B$, or...).
Hope the maps and the patterns are clear from the above. If not, tell me, I'll try to formulate more details.
Here is a projection of $V_3$ to the 3-space ($C_{ij}$ stands for $\operatorname{cone}(f_j\cdots f_i)$; red means degree 1)

One can see it inside out on Wikipedia under the name of rectified 5-cell<|endoftext|>
TITLE: Lower central series quotients in terms of (co)homology
QUESTION [15 upvotes]: Let $G$ be a group. It is well-known that $H_1(G,\mathbb{Z})=G/[G,G]$. Also (at least up to torsion) $[G,G]/[G,[G,G]]=\Lambda^2H^1(G,\mathbb{Z})/H_2(G,\mathbb{Z})$ as explained, for example, in this answer
Denote by $(G_r)$ the lower central series of $G:G_0=G,G_{r+1}=[G,G_r]$. Can further quotients $G_r/G_{r+1}$ be expressed in terms of group (co)homology?
Moreover, if $G=\pi_1(X)$ where $X$ is a compact algebraic variety, quotients $G_r/G_{r+1}$ possess a Hodge structure introduced by Hain. Can these Hodge structures be expressed in terms of the Hodge structure on the cohomology of $X$?

REPLY [3 votes]: Lower central quotients can be extracted from group homology via spectral sequence built up from free simplicial resolution of a group. So, if your complex variety is aspherical, you probably know those Hodge structures because everything is as natural and functorial as it can be.
By a classical result of Magnus, for free groups we have isomorphism between free Lie ring on abelianisation $\mathcal LF_{ab} = \mathcal LH_1(F, \mathbb Z)$ and Magnus Lie ring $LG := \bigoplus L_nF =  \bigoplus \gamma_i(F)/\gamma_{i+1}(F)$.
In general, this morphism is just epi. Now take free simplicial resolution $F_{\bullet} \twoheadrightarrow G$. We can look at exact couple defined by exact sequences  $L_nF \to F/\gamma_{n+1}F \to F/\gamma_{n}F$ and associate with it graded Lie ring spectral sequence converging to $L_n(\pi_0 (F)) = LG$. First sheet is given by $E^1_{p, q} = \pi_q (L_p F)$, and $s$-th differentials have degree $(s, -1)$. Actually, $E^1$ differs from free graded Lie ring on group homology only by torsion (we can check it introducing analogous s. s. for augmentation powers filtration on $\mathbb ZF$ and looking at morphism between them induced by $G \hookrightarrow \mathbb Z G$; rationally it's split injection by PBW, and it's known that $LG \otimes Q \cong \Delta_{\mathbb Q}G$), zeroth row is always free Lie on $G_{ab}$, first column is the shifted by 1 integral homology of $G$, and stripes below $k$ depend only on $H_{\leq k + 1}$.
Also, we instantly prove Stallings' result about maps inducing isomorphism on factors by $\gamma_i$ (for $G \xrightarrow{f} G'$ $G$ is para-$G'$ iff $H_1(f)$ iso and $H_2(f)$ is epi) just by checking differentials degree. Expression for $\gamma_1/\gamma_2$ comes from boundary morphism in this s. s.
(I don't remember reference for that stuff, but it's not hard to convert those speculations to actual proof. See also Cochran&Harvey, http://arxiv.org/pdf/math/0407203.pdf and Ellis, "Magnus-Witt type isomorphism for non-free groups".)<|endoftext|>
TITLE: Are there good bounds on binomial coefficients?
QUESTION [21 upvotes]: Motivated by the central limit theorem, one expects that
  $$\binom{n}{k} \approx \frac{2^n}{\sqrt{\pi n/2}} \exp\left(-\frac{(k-n/2)^2}{n/2}\right).$$
Computations suggest that the ratio of the two sides approaches 1 only for $|k-n/2| < 2\sqrt{n}$, and presumably this will follow from some version of the CLT.
In the literature or standard usage, are there any explicit upper (and lower?) bounds for binomial coefficients with a similar form that are sharp (in the ratio sense) for a wider range of $k$?

REPLY [27 votes]: Let $h(x)=-x\ln x-(1-x)\ln (1-x)$ be the binary entropy function in nats, then for $k\in [1,n-1]\cap \mathbb{Z}$ we have
$$
\sqrt{\frac{n}{8k(n-k)}}\exp\{nh(k/n)\} \leq \binom{n}{k} \leq \sqrt{\frac{n}{2\pi k(n-k)}}\exp\{nh(k/n)\}
$$
where the upper bound approaches equality if $k$ and $n-k$ are both large. This is obtained from Stirling and then some other manipulation, and covers the whole range of $k$.
This result is most certainly not mine, I learned it from Bob Gallager's Information Theory and Reliable Communications.<|endoftext|>
TITLE: $l^1$ versus $l^2$
QUESTION [8 upvotes]: Is there an elementary proof of this Banach space fact?

If the Banach space $V$ is linearly isomorphic to $l^1$, then it does not isometrically contain euclidean spaces of arbitrarily large finite dimension, i.e., a copy of $l^2_n$ for all $n$.

Failing that, good references on the subject?

REPLY [5 votes]: Is it true? $l^1$ is a sum of finite-dimensional $l_n^1$ over $n=1,2,\dots$. In summands you have almost spherical sections of large dimensions by Dvoretzky theorem, this allows to change norm a bit so that unit balls in summands contain large spherical sections.

REPLY [4 votes]: There does not exist an elementary proof. Sadly, there is no proof at all, because the fact is false. Here is a construction of an $\ell^1$-space containing every $\ell^2_n$.
Consider the normalized Gaussian measure $d\mu=\pi^{-1/2}e^{-t^2}dt$. Form its infinite dimensional tensor product $d\mu_\infty$ over ${\mathbb R}^{\mathbb N}$. Now let $E$ denote the vector space of linear functions
$$f_a:x\longmapsto a\cdot x,\qquad x\in{\mathbb R}^{\mathbb N},$$
where $a$ has finite support. $E$ is contained in $L^1(d\mu_\infty)$. One sees easily, from the rotational invariance of $d\mu$, that
$$\|f_a\|_1\left(=\int|f_a(x)|d\mu_\infty(x)\right)=C\|a\|_2$$
where
$$C=\int_{\mathbb R}|t|d\mu(t).$$
Hence $L^1(d\mu_\infty)$ contains every $\ell^2_n$.
Nota. This argument is used to prove that Euclidian spaces satisfy the Hlawka Inequality. The latter is almost trivial in an $\ell^1$ space.<|endoftext|>
TITLE: Can a topological manifold have different tangent bundles?
QUESTION [34 upvotes]: We know that the tangent bundles of the sphere arising from different smooth structures are equivalent as vector bundles. Is it right in general? I want to know the relationship between the set of smooth structures and these tangent bundles.

REPLY [49 votes]: This is answered in [Crowley, Diarmuid J.; Zvengrowski, Peter D, On the non-invariance of span and immersion co-dimension for manifolds, Arch. Math. (Brno) 44 (2008), no. 5, 353–365], see here. 
Specifically, in each dimension $>8$ there is a closed PL manifold admitting two smooth structures whose tangent bundles are non-isomorphic. One tangent bundle is trivial and the other one has nonzero second Pontryagin class. See remark 1.3. 
Such examples do not exist in dimensions $\le 8$ by Corollary 2.6.
In dimensions $\ge 18$ this was known since 1969 and due to Roitberg in
[On the PL noninvariance of the span of a smooth manifold,  Proceedings of the American Mathematical Society
Vol. 20, No. 2 (Feb., 1969), pp. 575-579].<|endoftext|>
TITLE: Is there a smooth manifold which admits only rigid metrics?
QUESTION [24 upvotes]: Does there exist a (finite dimensional) smooth manifold $M$, such that every Riemannian metric on $M$ has no isometries except the identity?
Of course, such a manifold must not admit a diffeomorphism of finite order.
Since a surface $S$ admits a diffeomorphism of order $n$ iff its mapping  class group (MCP) has an element of order $n$ (see here), it follows that if $S$ has the above property, then its MCP has only elements of infinite order.

REPLY [5 votes]: The survey paper "Do manifolds have little symmetry?" by Volker Puppe lists several known results about manifolds with no nontrivial finite group action. The ArXiv link is http://arxiv.org/pdf/math/0606714v1.pdf
In particular

aspherical manifolds with $Z(\pi_1M)=0$ and $Out(\pi_1M)$ torsionfree 

have no finite group action (Borel) 
Examples are

certain mapping tori of nilmanifolds (Conner-Raymond-Weinberger)
certain hypertoral manifolds, i.e., n-manifolds with a degree 1 map to the n-torus (Schultz)
certain 3-manifolds (Edwards)
a certain Bieberbach manifold (Waldmüller)<|endoftext|>
TITLE: Bruhat order of reflection subgroups
QUESTION [9 upvotes]: Let $(W,S)$ be a Coxeter group, $T=\bigcup_{w\in W}wSw^{-1}$ its set of reflections, and $A\subseteq T$. From results of Dyer and Deodhar, we know that the subgroup $W_A$ generated by the elements of $A$ is a Coxeter group, with a canonical system of generators $S_A$.
In the final remarks to his note, Deodhar points out that it would be interesting to compare the Bruhat order on $(W_A,S_A)$ to the restriction of the Bruhat order of $(W,S)$ to $W_A$, and that he plans to take up this project.
My question is: Has this been done? I would be happy with any references, but I am particularly interested in the following special case: Assume that $\beta_1,\ldots, \beta_k$ are positive roots in $W$'s root system. Is it possible to relate elements $w\in W$ satisfying $w\lessdot wr_{\beta_i}$ for all $i=1,\ldots ,k$, where $\lessdot$ denotes covering in the Bruhat order of $(W,S)$ and $r_{\beta_i}$ is the reflection across $\beta_i$ to elements $\widetilde{w}\in W_A$ with the same covering relations, but now in the Bruhat order of $(W_A,S_A)$?
An example: Consider the root system of $G_2$ with $\alpha$ the short, and $\beta$ the long simple root. The only element of $W$ satisfying $w\lessdot wr_\alpha, wr_{3\alpha+\beta}$ is $w=r_\alpha r_\beta r_\alpha r_\beta$, so it is true that $wr_\beta <w$, $wr_{\alpha+\beta} <w$, $wr_{3\alpha+2\beta}<w$, $wr_{2\alpha+\beta}<w$. Now let $\alpha,\beta$ be positive roots forming a sub-root system of type $G_2$ in some large root system. Do the four relations above still hold for the Bruhat order of the large group?

REPLY [5 votes]: The Bruhat graph of a reflection subgroup is an induced subgraph of the Bruhat graph of the larger group. This is proved by Dyer in "On the Bruhat graph of a Coxeter system." Thus if two elements that happen to be in the same reflection subgroup have a covering relation, then (the same) one also covers the other when you restrict to Bruhat order of the reflection subgroup. This actually generalizes the concept of permutation patterns, which was almost the topic of my dissertation (I did something else instead).<|endoftext|>
TITLE: How to estimate a specific infinite matrix sum
QUESTION [6 upvotes]: Let $M$ be an $n$ by $n$ matrix with each diagonal element equal to $k$ and each non-diagonal element equal to $k-1$ where $n$ and $k$ are positive integers. Let $k < n$ and we can assume both $k$ and $n$ are large.
What is
$$S_{M,k} = \sum_{x \in \mathbb{Z}^n} e^{-x^T M x}\;?$$
Is there some way to estimate this sum?
This was also posted to https://math.stackexchange.com/questions/1741157/how-to-estimate-a-specific-infinite-sum a few days ago.
Added examples
If $k=1$ we know $S_{M,1} \approx (\sqrt{\pi} +2\sqrt{n}e^{-\pi^2})^n \approx 1.7726372048^n$.  I don't know a closed form approximation for any other value of $k$.
For $n=12$ and $k = 1, \dots, 12$ using computer code to approximate the sum we get $ 962.58329951, 267.409968069, 196.186732001, 171.404195004, 162.313077353, 158.96911585, 157.738949838, 157.286397212, 157.119912408, 157.058666071, 157.036134803, 157.027846013$.
In general it seems numerically that for every fixed $n$, $S_{M,k}$ converges fairly quickly to some value as $k$ increases towards $n$.

REPLY [4 votes]: Let $s = k-1$, and write $x^T M x = \|x\|^2 + s (e \cdot x)^2$ where $e = (1,\ldots,1)$.  Let $P_j = \{x \in \mathbb Z^n: e \cdot x = j\}$. Each $P_j$ is a translate of $P_0$, which is a subgroup of $\mathbb Z^n$.  Then
$$S_{M,k} = \sum_{j\in \mathbb Z} e^{-s j^2} \sum_{x \in P_j} e^{-\|x\|^2} = \sum_{j \in \mathbb Z} e^{-s j^2} T(j)$$
where $T(j) = \sum_{x \in P_j} e^{-\|x\|^2}$.
Now $P_{j+n} = e + P_j$, and if $x \in P_j$ we have $\|e + x\|^2 = n + 2 j + \|x\|^2$.  Thus
$$T(n+j) = e^{-n-2j} T(j)$$
so that $$T(mn + j) = e^{-m^2 n - 2 mj} T(j)$$
We just have to compute (or approximate) $T(0), \ldots, T(n-1)$, and 
then
$$\eqalign{S_{M,k} &= \sum_{j=0}^{n-1} \sum_{m \in \mathbb Z} e^{-s(j+mn)^2} e^{-m^2n-2mj} T(j)  = \sum_{j=0}^{n-1} e^{-sj^2} T(j) \sum_{m \in \mathbb Z}
e^{-(n^2s+n) m(m+2j/n)}\cr
&= \sum_{j=0}^{n-1} e^{-sj^2} T(j) \theta_3(i(ns+1)j, \exp(-n^2 s))}  $$ 
where $\theta_3$ is a Jacobi theta function.<|endoftext|>
TITLE: Difficulties with descent data as homotopy limit of image of Čech nerve
QUESTION [14 upvotes]: Apologies if this question is inappropriate for MO. It is not a research level question in any of the topics it addresses, I just don't see how a novice can go about answering it alone (I've tried navigating the vast sea of literature).
I'm trying to start learning descent theory, and after seeing how descent along open covers for topological bundles already loosens the sheaf condition and adds to cocycle condition, I'm looking for a conceptual definition with machinery that encodes all the coherence conditions.
This MO question and its answers affirm that descent data is the homotopy limit of the the cosimplicial diagram obtained as the image of the Čech nerve along $\mathcal F$. However, this already leaves me unsure as to what $\mathcal F$ should be in the general setting. For sheaves, it's just a presheaf of sets. For bundles, it seems to be a category fibered in categories (thought of as a presheaf of categories?).
Below are two similar remarks which involve something I don't understand. First, from Zhen Lin's answer to the question linked above:

The category of descent data is indeed the homotopy limit of your cosimplicial diagram. In the case where $F$ actually is fibred in categories (and not higher categories), then you can truncate above degree 2, recovering the classical definition. If $F$ is fibred in sets, then you can even truncate above degree 1, recovering the classical sheaf condition. So, morally, the category of descent data generalises the set of matching elements of a presheaf with respect to a cover.

Second, by David Roberts, in the first comment to this MO question:

If you're interested in $n$-limits, these are more easily treated using homotopy colimits, as we don't understand the $n$-category of $(n−1)$-categories very well, for $n>3$, say. A coequaliser is the homotopy colimit (in a 1-category!) of a simplicial diagram truncated all the way down to a pair of parallel arrows. The "2-" version of this is a slightly less truncated simplicial diagram. Mapping out of these (in an appropriate model structure, in the absence of a good higher category structure) gives the cosimplicial version Street has, and replaces colimits by limits.

The three (main) things I don't understand are:

Why can we "truncate"? The remarks suggest that if $\mathcal F$ is a presheaf taking values in $n$-categories (with $n=0$ being sets), then we "can truncate" the cosimplicial diagram above level $n+1$. I take it this means the (proper notion of a) limit of the entire cosimplicial diagram is the same as that of the truncated one, but I don't understand why this is true.
Why is 'homotopy limit' the correct term here? I have some vague understanding that homotopy limits in the sense of derived functors of functors between homotopical categories somehow present $(\infty,1)$-limits, but this after choosing weak equivalences! Nowhere in the first question are weak equivalences mentioned, so what are they?
The meaning of homotopy limit seems to change depending on context in David Robert's remark: coequalizers, which are just $1$-limits, are said to be homotopy limits of parallel pairs, which suggests the weak equivalences here on $\mathsf{Cat}$ are just taken to be isomorphisms. However, if we're fibered over categories then this no longer makes any sense...

REPLY [13 votes]: To answer your question I'll need to do a fairly long digression on homotopy limits and colimits. Before I delve deep into the topic let me say that there's more than one way to describe this topic, for example some people like model categories, other people might prefer triangulated categories (shudder), I'll simply explain the point of view that has been most helpful for me in the past.
The language of $\infty$-categories
I think that one of the best way to understand homotopy limits and colimits is in the settings of $\infty$-categories. In fact, in the theory $\infty$-categories then homotopy limits and colimits are just like ordinary limits and colimits and I will stop prefixing everything with the word "homotopy" from now on.
My favourite catchphrase is that $\infty$-categories are just categories together with a notion of homotopy beteween morphisms. In fact you need a bit more, basically for every two objects $X$ and $Y$ and every $n\ge0$ you need a set $\textrm{Map}(X,Y)_n$ that morally corresponds to maps $X\times \Delta^n\to Y$ (the so-called higher homotopies) satisfying a set of compatibilities that I will not detail here (a good choice is to ask that $\textrm{Map}(X,Y)_\bullet$ form a Kan complex). To keep this concrete, examples are

Spaces, with the obvious definition of homotopy;
Manifolds and embeddings, with the notion of homotopy given by isotopies;
Chain complexes with the notion of chain homotopy;
Categories, where an homotopy between two functors is a natural isomorphism (and higher homotopies just chains of $n$ composable natural isomorphisms $F_0\to F_1\to\cdots\to F_n$).

(The last example is of course the most relevant for your question).
The important part about $\infty$-categories is that the notion of homotopy should seep through in our definition of commutative diagrams (so-called coherently commutative diagrams or coherent diagrams for short). For example a commutative square
$$\require{AMScd} \begin{CD}
A @>{f}>> B\\
@V{h}VV @VV{g}V\\
C @>>{k}> D
\end{CD}$$
is the datum of four objects, four maps and a homotopy between $gf$ and $kh$.
Once you have set up all of this you can go on and define initial and terminal objects, limits, colimits and so on and so forth exactly like you did for ordinary categories (that is without homotopies). These are often called "homotopy limits" and "homotopy colimits" to distinguish from the limits and colimits computed without paying attention to the homotopies.
Ok, but what's the deal with the Čech complex anyway?
Let's give an example. Suppose that $C,D$ are two categories and $F,G:C\to D$ two functors. If we ignored the homotopies the equalizer would simply be the objects $c\in C$ such that $Fc=Gc$. But we all know that this is a stupid notion. Naturally equivalent functors will give different answers and we do not want to distinguish between naturally equivalent functors. So we use the notion of equalizer in the $\infty$-category of categories.
As you can see immediately from studying the coherent diagrams of the form $*\to C\rightrightarrows D$ the objects of the equalizer in this brave new setting are objects $c\in C$ together with an isomorphism $\alpha:Fc\cong Gc$. Now that's a much better behaved notion!
Similarly, you can see that if $F:\Delta^{op}\to C$ is a functor to an ordinary category we get that
$\lim_{\Delta^{op}} F = \textrm{eq}(F([0])\rightrightarrows F([1]))$
so that $F(X)\to \lim F(\check U_\bullet)$ being an equivalence is just the ordinary sheaf condition. But what happens if we go one categorical level up? Now let $F:\Delta^{op}\to \mathrm{Cat}$ be a functor to categories. Then you can see that an object of $\lim_{\Delta^{op}}F$ is an object $x\in F([0])$ together with an isomorphism $\alpha:d_0^*x\cong d_1^*x$ in $F([1])$ such that $d_0^*\alpha\circ d_2^*\alpha = d_1^*\alpha$. But this is exactly the notion of descent datum.
So the notion of sheaf in this brave new context is just the clasical notion of stack
You might have noticed that we are using just a small portion of the diagram. This is because our categories do not have many "interesting" homotopies. Sets have no homotopies at all and categories have only interesting 1-homotopies, higher homotopies are basically composable sequences of 1-homotopies. In general, the higher the order of the interesting homotopies, the more pieces of $\Delta$ we have to use. (precisely for an $n$-category you just need to go to $[n+1]$. The reason for this is basically Quillen's theorem A, as noted by Dylan Wilson in the comments). For spaces and chain complexes you need to consider the whole of $\Delta$.
Conclusions and references
Ugh that's a long answer. I hope I managed to give at lesat an inking of what's going on without getting bogged by the technical details. If technical details are what you want however the standard references for $\infty$-categories are Luries's Higher Topos Theory and Higher Algebra. They are not an easy read, by anyone's standards.
The model of $\infty$-categories I've used is called fibrant simplicial categories. I think it is an excellent model to develop some intuition, but it is actually quite bad to work with. Most of the people in the industry use quasicategories, which are quite pleasant to work with but I didn't have the time to properly introduce them here.
Appendix: why can we truncate?
Let $E$ be an $n$-category, that is an $\infty$-category such that the mapping spaces $\mathrm{Map}(x,y)$ are $n$-truncated for every $x,y\in E$. Let $j:C\to D$ be a functor such that for every $d\in D$ the geometric realization $|C\times_D D_{d/}|$ is $n$-connected. Then for every functor $F:D\to E$ $\lim_D F$ exists if and only if $\lim_C Fj$ exists and they coincide.
Lemma 1: Let $K$ be a simplicial set such that the geometric realization $|K|$ is $n$-connected. Then for every $e\in E$ the limit of the constant functor $K\to E$ at $e$ exists and coincides with $e$.
Proof: $\mathrm{Map}(e',\lim_K e) \cong \lim_K \mathrm{Map}(e',e) = \mathrm{Map}(K,\mathrm{Map}(e',e))=\mathrm{Map}(e',e)$.
Lemma 2: Let $p:\tilde D\to D$ be a cartesian fibration such that for every $d\in D$ the geometric realization $|\tilde D_d|$ is $n$-connected. Then for every functor $F:D\to E$ the limit $\lim_D F$ exists if and only if the limit $\lim_{\tilde D} Fp$ exists and they coincide.
Proof: By an easy cofinality argument the right Kan extension along a cartesian fibration is obtained by computing the limits fiberwise. Then the thesis follows from Lemma 1.
Proof of the main result: Let $\tilde D\to D$ the cartesian fibration classified by the functor $D^{op}\to \mathrm{Cat}$ given by $d\mapsto C\times_D D_{d/}$. We have a canonical functor $C\to \tilde D$ sending $c$ to $(c,jc=jc)$. A standard cofinality argument implies that the functor $C\to \tilde D$ is coinitial. Then the thesis follows from the previous lemma.<|endoftext|>
TITLE: Operator space structures on CB(H,K) where H and K are Hilbertian operator spaces?
QUESTION [5 upvotes]: (I'd be grateful if anyone thinking of putting MathJax in the question title refrains from doing so.)

By consulting various standard sources (Effros-Ruan's book, Pisier's book, the lexicon of Wittstock's old group) I can find some descriptions of the underlying Banach space of $\mathcal{CB}(H,K)$ where $H$ and $K$ are Hilbert spaces equipped with various operator space structures. However, it's not clear from these accounts if there has been systematic work on studying what $CB(H,K)$ looks like as an operator space in these situations.
Let me raise two particular problems for which I'd be happy to get more information, or pointers to the literature. It could be that as a dilettante in the world of operator spaces I've just missed some basic arguments.
Q1. Let ${\bf B}$ denote $B(\ell_2)$ with its usual OSS and let ${\rm COL}$ be $\ell_2$ with column OSS. Then $CB({\rm COL}, {\rm COL})$ is completely isometrically isomorphic to ${\bf B}$. In some sense this characterises ${\rm COL}$: a result of Blecher (Theorem 3.4 in his 1995 Math. Ann. paper) tells us that if $X$ is a Hilbertian operator space and $CB(X)$ is isomorphic as an algebra and an operator space to ${\bf B}$ then $X$ is isomorphic to column Hilbert space. However, when does the formal identity map give a completely bounded map ${\bf B} \to CB(X)$? What about ${\bf B} \to CB(X,Y)$ where $X$ and $Y$ can be different?
Remark. For the last part of Q1: clearly this is possible if the formal identity map $X\to Y$ factors in a cb way through ${\rm COL}$. I'd be more interested to know if this can happen when either $X$ or $Y$ is incomparable with ${\rm COL}$.
Q2. If $X$ is a homogeneous, Hilbertian operator space then by definition the underlying Banach space of $CB(X)$ is $B(\ell_2)$. This happens, for instance, if $X$ is row, column, the self-dual OSS, MAX, or MIN. Of the 5 possibilities just listed, are any of the resulting o.s. structures on $CB(X)$ comparable (i.e. weaker or stronger as o.s. structures) to ${\bf B}$?

REPLY [4 votes]: Both your questions can be answered by considering rows and columns in $\mathbf B$, and by noticing that in $CB(X,Y)$ colums are completely isometric to $Y$ and rows are completely isometric to $X^*$.
Therefore if $X_1,X_2$ and $Y_1,Y_2$ are operator spaces structures on $\ell_2$ such that $id\colon CB(X_1,Y_1)\to CB(X_2,Y_2)$ is completely bounded, then $id \colon Y_1 \to Y_2$ and $id \colon X_2 \to X_1$ are both completely bounded. The converse is clear, so this is a characterization when $CB(X_1,Y_1)$ and $CB(X_2,Y_2)$ are comparable.
Therefore the answers to your questions are:
Q1 If either $X$ or $Y$ is incomparable with $\mathrm{COL}$, then $\mathbf B$ and $CB(X,Y)$ are incomparable.
Q2: No. If $X$ and $Y$ are homogeneous hilbertian operator spaces, then $CB(X)$ and $CB(Y)$ are comparable if and only if $X$ and $Y$ are completely isomorphic.<|endoftext|>
TITLE: Can a weak fibration category be non saturated?
QUESTION [7 upvotes]: A weak fibration category is a category $\mathcal{C}$ equipped with two subcategories
$$\mathcal{F}, \mathcal{W} \subseteq \mathcal{C}$$
containing all the isomorphisms, such that the following conditions are satisfied:

$\mathcal{C}$ has all finite limits.
$\mathcal{W}$ has the 2-out-of-3 property.
The subcategories $\mathcal{F}$ and $\mathcal{F}\cap\mathcal{W}$ are closed under base change.
Every morphism $f: X \to Y$ can be factored as $X \xrightarrow{g} Z \xrightarrow{h} Y$ where $g \in \mathcal{W}$ and $h \in \mathcal{F}$.

The homotopy category $Ho(\mathcal{C})$ of a weak fibration category  $(\mathcal{C},\mathcal{W},\mathcal{F})$ is obtained from $\mathcal{C}$ by formally inverting the morphisms in $\mathcal{W}$. The weak fibration category is called saturated if every map in $\mathcal{C}$ that maps to an isomorphism in $Ho(\mathcal{C})$, is in $\mathcal{W}$. 
I have two questions:

Does there exist a small weak fibration category $(\mathcal{C},\mathcal{W},\mathcal{F})$ such that $\mathcal{W}$ is closed under retracts, but $\mathcal{C}$ is not saturated?
Does there exist a small weak fibration category $(\mathcal{C},\mathcal{W},\mathcal{F})$ such that $\mathcal{W}$ is closed under retracts, but $\mathcal{W}$ does not satisfy 2-out-of-6?

Of course, if the answer to 2. is yes then the answer to 1. is also yes.

REPLY [10 votes]: It is a result of Cisinski that in a fibration category the three conditions you mention (saturation, 2-out-of-6, weak equivalences closed under retracts) are all equivalent. See Theorem 7.2.7 in this paper.<|endoftext|>
TITLE: Pfaffian equals complex determinant?
QUESTION [11 upvotes]: Let $V$ be a Euclidean vector space and let $V^{\mathbb{C}} = V \oplus V$ be its complexification, with complex structure
$$J = \begin{pmatrix} 0 & -\mathrm{id}\\ \mathrm{id} & 0 \end{pmatrix}.$$
Of course, we can regard $V^{\mathbb{C}}$ also as a real vector space with a canonical orientation (for a basis $v_1, \dots, v_n$ of $V$, the basis $(v_1, 0), (0, v_1), \dots, (v_n, 0), (0, v_n)$ is positively oriented). Now let $A$ be an anti-symmetric endomorphism of $V$ and consider the anti-symmetric endomorphism 
$$\tilde{A} = \begin{pmatrix} A& -\mathrm{id}\\ \mathrm{id} & A\end{pmatrix}$$
of $V^{\mathbb{C}}$. Then $\tilde{A}$ has a well-defined Pfaffian. On the other hand, we have $\tilde{A} = A + i$. Is it true that
$$\mathrm{Pf}(\tilde A) = \det(A+i),$$
at least if $V$ is even-dimensional? What about the odd-dimensional case?

REPLY [16 votes]: The answer is YES in every dimension, up to a sign. Here is the calculation. On the one hand, $\det \tilde A=\det(A^2+I)$ because the blocs commute to each other. Therefore
$${\rm Pf}(\tilde A)^2=\det(I+iA)\det(I-iA).$$
On the other hand
$$\det(I-iA)=\overline{\det(I+iA)}=\det(I+iA)^*=\det(I+iA),$$
yields
$${\rm Pf}(\tilde A)^2=(\det(I+iA))^2.$$
Let us remark that $\det(I+iA)$ is a polynomial in the entries of $A$, with real entries because $I+iA$ is Hermitian. We deduce
$${\rm Pf}(\tilde A)=\epsilon\det(I+iA)$$
for some constant $\epsilon=\pm1$. Taking $A=0_n$ gives $\epsilon=1$.
Remark that ${\rm Pf}(\tilde A)$, as a polynomial in the entries of $A$, is even. In particular, if $n$ is odd, ${\rm Pf}(\tilde A)$ has degree $n-1$ only, whereas if $n$ is even, then ${\rm Pf}(\tilde A)$ has degree $n$. For instance, if $n=3$ and the off-diagonal entries of $A$ are $\pm a,\pm b,\pm c$, then
$${\rm Pf}(\tilde A)=1-a^2-b^2-c^2.$$<|endoftext|>
TITLE: Conditions for existence of dominating $\sigma$-finite measure for all conditional distributions
QUESTION [5 upvotes]: Suppose $X$ and $Y$ are two real-valued random variables with a specified joint probability distribution $P_{X,Y}.$ I wish to determine if there is a $\sigma$-finite measure $\mu$ on the real line such that $P_{Y|X=x} << \mu$ for $P_X$-almost all $x\in\mathbb{R}$. Call this property $Q.$ Property $Q$ does not always hold e.g. if $X$ is a standard normal variable and $Y=X.$
Is there an equivalent description of property $Q$ that is easy to check for a given probability distribution $P_{X,Y}$?

REPLY [4 votes]: In one direction it's clear: If $P_{X,Y}\ll P_X\otimes P_Y$, then there is a jointly measurable density $f$, and (a version of) the conditional distribution $P_{Y|X=x}$ is given by $f(x,y)P_Y(dy)$, so the choice $\mu=P_Y$ suffices.
Conversely, suppose that $Q$ holds. Then there is a  jointly measurable function $g$ such that $P_{Y|X=x}(dy) = g(x,y)\mu(dy)$, for $P_X$-a.e. $x$. Define $h(y)=\int_R g(x,y)\,P_X(dx)$, and observe that if $h(y)=0$ then $g(x,y)=0$ for $P_X$-a.e. $x$. Define $B=\{y: h(y)=0\}$, and $f(x,y) = 1_{B^c}(y)g(x,y)/h(y)$. I claim that $f\cdot P_X\otimes P_Y=P_{X,Y}$. Indeed,
$$
\eqalign{
P_{X,Y}(C)
&=\int\int 1_C(x,y)P_X(dx)P_{Y|X=x}(dy)\cr
&=\int\int 1_C(x,y)P_X(dx)g(x,y)\mu(dy)\cr
&=\int\int 1_C(x,y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy).\cr
}
$$
In particular, taking $C=R\times A$, for $A$ a Borel subset of $R$,
$$
\eqalign{
P_Y(A)=P_{X,Y}(R\times A)
&= \int\int 1_A(y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy)\cr
&= \int 1_A(y)1_{B^c}(y)h(y)\mu(dy)\cr
&= \int 1_A(y)h(y)\mu(dy).\cr
}
$$
That is, $P_Y=h\cdot\mu$. Using this to continue the computation in the last-but-one display (multiply and divide by $h(y)$) we have
$$
\eqalign{
P_{X,Y}(C)
&=\int\int 1_C(x,y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy).\cr
&=\int\int 1_C(x,y)P_X(dx)f(x,y)P_Y(dy),\cr
}
$$
and the asserted absolute continuity.<|endoftext|>
TITLE: Does $a_n=\prod^n_{k=1}(1-e^{k\alpha \pi i})$ converge to zero when $\alpha$ is irrational?
QUESTION [28 upvotes]: I came across a problem concerning about the convergence of products. I wonder if the complex series $a_n=\prod^n_{k=1}(1-e^{k\alpha \pi i})$ converges to zero when $\alpha$ is irrational. Of course, the infinite product doesn't converge to any nonzero number.
It seems that the behavior of $e^{k\alpha \pi i}$ is not quite "predictable". I have no idea how to approach this problem. Thanks in advance!

REPLY [24 votes]: The product does not tend to the limit zero.  For any irrational number $\alpha$ one can show that
$$ 
\limsup_{N\to \infty} \prod_{n=1}^{N} |1- e(n\alpha)| = \infty. \tag{1}
$$
(Here I use the usual notation $e(\alpha) =e^{2\pi i\alpha}$, and the product in the question is stated for $\alpha/2$ rather than $\alpha$).
I'll prove a slightly weaker result; namely I'll show that (1) holds for any irrational $\alpha$ for which there are infinitely many rational approximations $a/q$ with $(a,q)=1$ and
$$ 
\alpha=\frac{a}{q} +\beta, \qquad \text{with}\qquad |\beta| \le \frac{1}{100q^2}. \tag{2}
$$
The $1/100$ is just chosen for ease of exposition, and a more careful argument can make do with $1/\sqrt{5}$ (any constant $<1/2$ is enough), and then it is well known that every irrational number admits infinitely many approximations with $|\alpha -a/q| \le 1/(\sqrt{5}q^2)$. Condition 2 holds for almost all irrational numbers --  it fails only if the continued fraction expansion only uses numbers below $100$.
Suppose then that $q$ is such that (2) holds, and consider the product in question at $N=q-1$.  By the triangle inequality, for $1\le n\le N$,
$$ 
|1-e(n\alpha)| = |(1-e(an/q))+e(an/q)(1-e(\beta n))| \ge |1-e(an/q)| 
\Big(1-\frac{|1-e(n\beta)|}{|1-e(an/q)|}\Big)= 2\Big|\sin\frac{\pi an}{q}\Big| \Big( 1- \frac{|\sin(\pi n\beta)|}{|\sin(\pi an/q)|}\Big).
$$
Now write $\Vert x\Vert= \min_{\ell \in {\Bbb Z}} |x-\ell|$ for the distance from $x$ to the nearest integer.  Note also that for $0\le x\le \pi/2$ one has $2x/\pi \le \sin x \le x$.  Therefore we get for $1\le n\le N$
$$ 
|1-e(n\alpha)| \ge |1-e(an/q)| \Big(1 - \frac{\pi n|\beta|}{2\Vert an/q\Vert}\Big) \ge |1-e(an/q)| \exp\Big( - \frac{\pi q|\beta|}{\Vert an/q\Vert}\Big), \tag{3}
$$
where in the last inequality we used that $\eta =\pi n|\beta|/(2\Vert an/q\Vert) \le \pi q^2|\beta|/2 \le 1/10$, so that $(1-\eta) \ge \exp(-2\eta)$.
Multiplying (3) for $n$ from $1$ to $N=q-1$, we obtain
$$ 
\prod_{n=1}^{N} |1-e(n\alpha)| \ge \prod_{n=1}^{q-1} |1-e(an/q)| 
\exp\Big( - \sum_{n=1}^{N} \frac{\pi q |\beta|}{\Vert an/q\Vert}\Big). \tag{4}
$$
Now
$$ 
\sum_{n=1}^{q-1} \frac{1}{\Vert an/q\Vert} \le 2\sum_{n\le q/2} \frac{q}{n} =2 q \log q +O(q),
$$
and $\prod_{n=1}^{N} |1-e(an/q)| = q$, and so from (4) we conclude that
$$ 
\prod_{n=1}^{q-1} |1-e(n\alpha)| \ge q \exp\Big( - 2\pi q^2 |\beta| \log q +O(q^2|\beta|)\Big) \gg q^{0.9}. 
$$
This proves the claim.  Note that this argument is closely related to the nice attempt of Sangchul Lee.
Added information (October, 2021): From comments on Fedja's recent question Can all partial sums $\sum_{k=1}^n f(ka)$ where $f(x)=\log|2\sin(x/2)|$ be non-negative? I was led to the following paper by D.S. Lubinsky which shows that (see Theorem 1.2 there)
$$
\limsup_{N\to \infty} \frac{\log \prod_{n=1}^{N} |1-e(n\alpha)|}{\log N} \ge 1,
$$
for all irrational numbers $\alpha$, so that the product in the question gets almost as large as $N$ infinitely often.<|endoftext|>
TITLE: What is the Heegaard Floer Homology of a connect sum of $S^2 \times S^1$s?
QUESTION [5 upvotes]: There are many conjecturally-equivalent three-manifold Floer homologies, of which my understanding is the most-computable is Heegaard Floer homology.  

What is the (Heegaard) Floer homology of a connect sum of $k$ copies of $S^2 \times S^1$?  Is it trivial?

I am an outsider, and so please forgive me if my question is not quite well formed.  E.g. I gather that HF depends on a $\mathrm{Spin}^c$ structure, so at best I would understand the answer for all choices; the value of Heegaard Floer homology is really an exact triangle, not a vector space, so I guess I'm asking for that triangle; perhaps there are other technicalities I am unaware of.

REPLY [7 votes]: The explicit calculation of this Heegaard-Floer homology (at least, $HF^-$, and of the unique $\text{Spin}^c$ structure for which $c_1(\mathfrak s) = 0$) was carried out in the paper in which it was introduced, and indeed is an important part of the proof that Heegaard-Floer homology is well-defined. See Lemma 9.1. One also has the isomorphisms $HF^+ \cong \Bbb Z[U,U^{-1}]/U\Bbb Z[U] \otimes H_*(T^g;\Bbb Z)$ and $HF^\infty \cong \Bbb Z[U^{-1},U] \otimes H_*(T^g;\Bbb Z)$. The exact sequence relating the triple ($HF^+, HF^-, HF^\infty)$ is exactly what you think it is. When the $\text{Spin}^c$ structure is nonzero, all the Floer groups are zero. (These four groups are all computable by hand, as done in the original paper.)
In this case, the Seiberg-Witten Floer homology is computable, because $S^2 \times S^1$ has positive scalar curvature, as do all connect-sums with itself by Gromov-Lawson. When $c_1(\mathfrak s)$ is not torsion (in this case, not zero), all the Floer groups are trivial, and in the case that $c_1(\mathfrak s) = 0$ all the Floer groups are explicitly calculable, as is the exact sequence (one of the connecting maps is zero) - for the precise result, see Chapter 36 of Kronheimer-Mrowka, "Monopoles and 3-manifolds".
To compare these results, one should note the relationship (now proved, thanks to Colin-Ghiggini-Honda and Kuthulan-Lee-Taubes) $\overline{HM} \cong HF^\infty$, $\check{HM} \cong HF^-$, $\widehat{HM} \cong HF^+$. $\widehat{HF}$ corresponds to a reduced Seiberg-Witten Floer homology not described in that chapter of Kronheimer-Mrowka, $\widetilde{HM}$.<|endoftext|>
TITLE: What do Hecke eigensheaves actually look like?
QUESTION [20 upvotes]: Let $\mathbb F_q$ be a finite field, $C$ a curve over $\mathbb F_q$ of genus $g\geq 2$, $\rho: \pi_1(C) \to GL_2(\overline{\mathbb Q}_\ell)$ an irreducible local system. The geometric Langlands correspondence constructs a geometrically irreducible Hecke eigensheaf on $Bun_2(C)$ associated to $\rho$, which is a perverse sheaf.
What is this sheaf's generic rank (as a function of $g$, presumably)? What is its singular locus? 
It seems to me that the singular locus should be more-or-less independent of $\rho$ because in the complex-analytic picture it is supposed to depend smoothly on $\rho$, but the coordinates of $\rho$ are $\ell$-adic and the coordinates of the singular locus are in $\overline{\mathbb F}_q$ so there is no natural way for the second to depend naturally on the first other than being locally constant. If the singular locus is locally constant, because the complex-analytic space of local systems is connected, it should be globally constant as well. By similar logic it seems like the generic rank should be independent of the local system.
I could come up with some plausible guess for what the singular locus should be (maybe the unstable locus?), but I have no idea what the generic rank should be (some polynomial in $g$?).

REPLY [8 votes]: The characteristic cycle of Hecke eigensheaves (for irreducible local systems) is the zero fiber of the Hitchin fibration, counting with multiplicity. This is almost obvious for eigen-D-modules constructed from opers in characteristic zero (by Beilinson and Drinfeld), but for $GL_2$ it should not be hard to check this from Drinfeld's formula and verify that the answer still holds, especially if you only care about their smoothness locus and rank. Anyway, the answer for $GL_2$  is as follows:

Generic rank is $2^{\dim H^0(C,\Omega^{\otimes 2})}=2^{3g-3}$
The singularity locus consists of bundles $E$ that admit a nilpotent Higgs field; explicitly, this means that $E$ has a line-subbundle $L\subset E$ such that $H^0(C,L^{\otimes 2}\otimes\det(E)^{-1}\otimes\Omega)\ne 0$, which is equivalent to $H^1(C,\det(E)\otimes L^{\otimes -2})\ne 0$.<|endoftext|>
TITLE: Complexity of solving systems of linear diophantine equations
QUESTION [6 upvotes]: It is "well known" that a matrix system $Ax=b$ where $A\in \Bbb Z^{m\times n}$, $x\in \Bbb Z^n,b\in\Bbb Z^m$ for some $m,n \in \Bbb N$, can be solved in polynomial time, using Smith/Hermite Normal Form computation. By "solved" I mean either "decide, for a given $A,b$, whether a solution exists" or "decide whether a solution exists and output one". My question is: 

Can we claim anything more specific than polynomial, such as
  cubic/quartic in $(m+n+\max{|A_{ij}}|)$? Or at least cubic/quartic if
  we assume that any arithmetic operation can be performed in unit time
  (ignoring possible blowup of the coefficients)?

The closest result I could find is in "Near optimal algorithms for computing Smith normal forms of integer matrices"
 by Storjohann where a nice complexity bound is presented for computing the Smith normal form, but it is not clear from the paper how to recover the unimodular transformation matrices (authors just write "In the future, we will present ... algorithm that compute (them)").
Another close result I found is this paper where a cubic algorithm is presented under the additional assumtion that we can use a unit-time oracle for computing greatest common divisor.
Another reference is this paper which deals with the case of non-singular square matrices. (Is there a simple reduction from a general $m\times n$ matrix to the non-singular square case?)
Another reference is this paper (thanks @Dima Pasechnik) which gives upper bounds on Hermite normal form computation, at least for the non-singular case: in the last section there is an outline how to generalize it for any matrices, but I again don't see how to recover the transformation matrix, as their algorithm does not use only elementary row/column operations.

REPLY [3 votes]: Ok, apparently, professor Storjohann sent me the reference I was looking for: Theorem 19 in A Fast Practical Deterministic Algorithm for Triangularizing Integer Matrices. (Thanks)<|endoftext|>
TITLE: Are all numbers from $1$ to $n!$ the number of perfect matchings of some bipartite graph?
QUESTION [6 upvotes]: Let $f(G)$ give the number of perfect matchings of a graph $G$.
Consider set $\mathcal N_{2n}=\{0,1,2,\dots,n!-1,n!\}$.
Consider collection of all $2n$ vertex balanced bipartite graph to be $\mathcal G_{2n}$.
For every $m\in\mathcal N_{2n}$ is there a $G\in \mathcal G_{2n}$ such that $f(G)=m$?

REPLY [2 votes]: Consider $A$ the order $n$ adjacency matrix which has 1 in the $i,j$th entry if there is an edge in the bipartite graph $G$ from top vertex $i$ to bottom vertex $j$, 0 if not. Then the permanent of $A$ counts the number of perfect matchings in $G$.  One now sees that any such bipartite graph missing $k\lt n$ edges is a multiple of $(n-k)!$, which has some significance when $k$ is small.  In particular when $k=2$, the number of matchings is a multiple of $(n-2)!$. So the feasible values near $n!$ are indeed few in number.
Elsewhere (Number of unique determinants for an NxN (0,1)-matrix) I talk about the determinant spectrum problem for order $n$ $0-1$ matrices; a similar problem for permanents can be and probably has been asked, for which I would like to see references.
Gerhard "Ask Me About Multilinear Functions" Paseman, 2016.05.18.<|endoftext|>
TITLE: $U(n)$-submodules of ${\rm SO}(2n)$-modules
QUESTION [6 upvotes]: Let $\Gamma_{(\lambda_1, \dots, \lambda_{n})}$ denote an irreducible $SO(2n)$-module with highest weight $(\lambda_1, \dots, \lambda_n)$ and let more specifically $X = \Gamma_{(2\lambda, \dots, 0)}$ and $Y = \Gamma_{(2\lambda_1, \dots, 2\lambda_n)}$, where at least one of the $\lambda_j$ with $j>1$ is not equal to $0$. Furthermore, let $W \leq X$ be a (irreducible) $U(n)$-submodule of $X$, and $\tilde{W} \leq Y$ a (irreducible) $U(n)$-submodule of $Y$. My Questions:
1) Is it true that for any $X,Y,W,\tilde{W}$ choosen in such a fashion
\begin{align*} 
W \ncong \tilde{W}?
\end{align*}
2) Is there a branching rule for $U(n)$-submodules of $SO(2n)$-modules? (This obviously relates to Question 1)

REPLY [3 votes]: you can find branching rule ${\rm SO}(2n) \to U(n)$ in the book of Knapp, Lie groups beyond an introduction, and also in some paper of his, and in the book of Zelobenko, compact Lie groups and their rep's, also try the book of Tom Dieck.
best
jorge<|endoftext|>
TITLE: reference for KK theory
QUESTION [5 upvotes]: I wanted to ask you, if you have any good references (book or pdf) to learn about the KK theroy of Kasparov. I think the presentation of Blackadar is too close from the commutative theory. 
I was searching for a C* proof of the Bott periodicity and arrived in this thread. I searched in the references mentionned there but I didn't find anything about KK theory.
Any advice ?
Thanks :)

REPLY [3 votes]: If you want to learn more about KK-theory there is a book by Kjeld Knudsen Jensen and Klaus Thomsen with the title "Elements of KK-theory". You can have a look there.
I would also recommend those surveys:
http://www.personal.psu.edu/ndh2/math/Papers_files/Higson%20-%201990%20-%20A%20primer%20on%20KK-theory.pdf
https://arxiv.org/abs/1304.3802<|endoftext|>
TITLE: Self-intersection of a Cartier divisor
QUESTION [5 upvotes]: Let $X$ be a smooth projective variety, and $D$ a Cartier divisor on $X$ inducing a surjective morphism $f\colon X\rightarrow C$, where $C$ is a curve. 
May we conclude that $D^{2}=0$?

REPLY [5 votes]: The answer is yes if the complete linear system $|D|$ is without fixed components. In such a situation, the fact that $f$ is a morphism onto a curve implies that $|D|$ is composed with a base-point free rational pencil, hence $D^2=0$.
On the other hand, if there are fixed components the answer is in general no. In fact, take as $X$ the Hirzebruch surface $\mathbb{F}_n=\mathbb{P}(\mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(-n))$, where $n \geq 3$, endowed with the fibration $\pi \colon X \to \mathbb{P}^1$. Let $f$ be a fibre of $\pi$ and $C_0$ the unique section such that $(C_0)^2=-n$, and set $D:=C_0+f$. Since $$D C_0 = (C_0+f)C_0=-n+1 <0,$$ it follows that $C_0$ is the fixed component of $|D|$, in other words $|D|=C_0+|f|$. 
Then the map induced by $D$ is the same as the map induced by $f$, which in turn is $\pi \colon X \to \mathbb{P}^1$. On the other hand, we have $$D^2 = (C_0)^2 + 2 C_0f = -n + 2 <0.$$<|endoftext|>
TITLE: A question about (unicity of certain cycles in a Cayley graph of a) symmetric group
QUESTION [6 upvotes]: Let $S=\{(1,2),(1,2,3,\ldots,n),(1,2,3,\ldots,n)^{-1}=(1,n\ldots,2)\}$ be a subset of the symmetric group $S_n$. We know that $(1,2,\ldots,n)(1,2)=(2,3,\ldots,n)$, and thus $$[(1,2,\ldots,n)(1,2)]^{n-1}=(1,2,\ldots,n)\overbrace{(1,2)\cdots(1,2,\ldots,n)}^{2n-2}(1,2)=(1).$$ We want to know whether or not there exists another sequence of elements  $a_1,a_2,\ldots,a_{2n-4}\in S$ such that $$(12\ldots n)a_{2n-4}a_{2n-5}\cdots a_2a_1(12)=(1),$$ where  $a_{i+1}\neq a_i^{-1}$ for $i=0,1,2,\ldots,2n-4$ (putting $a_0=(12)$, $a_{2n-3}=(1,2,\ldots,n)$).
Equivalently, I want to ask if, in the cubic Cayley graph $Cay(S_n,S)$, there is a unique cycle of length $2(n-1)$ passing through $(1)$, $(1,2,\ldots,n)$ and $(1,2)$.

REPLY [8 votes]: The smallest $n$ for which there exist sequences as asked for is $n = 7$:

$(1,2,3,4,5,6,7) \cdot (1,2) \cdot (1,7,6,5,4,3,2) \cdot (1,2)
     \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot$ 
$(1,7,6,5,4,3,2) \cdot (1,2) \cdot (1,2,3,4,5,6,7) \cdot (1,2)
     \cdot (1,7,6,5,4,3,2) \cdot (1,2) = ()$, and
$(1,2,3,4,5,6,7) \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot (1,7,6,5,4,3,2)
     \cdot (1,7,6,5,4,3,2) \cdot$
$(1,2) \cdot (1,2,3,4,5,6,7) \cdot (1,2,3,4,5,6,7) \cdot (1,2) \cdot
     (1,7,6,5,4,3,2) \cdot$
$(1,7,6,5,4,3,2) \cdot (1,2) = ()$.

For $n = 8$ there is no such sequence other than the trivial one
mentioned in the question, for $n = 9$ there is one, for $n = 10$ there
are $18$, for $n = 11$ there are $5$ and for $n = 12$ there are $104$ such
sequences. This has been found with the following GAP function:
SearchXueyiSequences := function ( n )

  local  sequences, search, S;

  search := function ( sequence, a )

    local  b;

    if Length(sequence) = 2*n-3 then
      if sequence[2*n-3] <> (1,2) and Product(sequence)*(1,2) = () then
        Add(sequences,Concatenation(sequence,[(1,2)]));
      fi;
      return;
    fi;
    for b in Difference(S,[a^-1]) do
      search(Concatenation(sequence,[a]),b);
    od;    
  end;

  S := Set(GeneratorsAndInverses(SymmetricGroup(n)));
  sequences := [];
  search([],S[2]);
  sequences := Set(sequences);
  sequences := sequences{[2..Length(sequences)]}; # exclude trivial sequence
  return sequences; 
end;<|endoftext|>
TITLE: Survivor sets for expanding maps of the interval
QUESTION [5 upvotes]: Let $T:[0,1]\to [0,1]$ be a piecewise smooth expanding map, i.e., $|T'(x)|>1$ for all $x$. Let $I_n$ be a sequence of nested intervals (i.e., $I_{n+1}\subset I_n$) such that the length of $I_n$ tends to 0 as $n\to\infty$. 
Define the survivor set for $I_n$ as follows:
$$
\mathcal J(I_n)=\{x \in[0,1] : T^k(x)\notin I_n, \forall k\ge0\}.
$$
Question. Is it true that 
$$
\dim_H \mathcal J(I_n)\to 1, \quad n\to \infty?
$$

REPLY [2 votes]: The answer of your question is yes if $T \in C^2$. It holds since $\dim_H(\mathcal{J}(a,b))$ varies continuously when the end points of the removed intervals varies continuously. Then, by continuity, the Hausdorff dimension of $\mathcal{J}(I_k)$ must tend to $1$ as $k \to \infty$.
This is proved by Urbanski in two old-ish papers: 

Hausdorff Dimension of Invariant Sets for Expanding Maps of a Circle. Ergod. Th. and Dynam. Sys. 6 (1986), 295-309.
Invariant Subsets of Expanding Mappings of the Circle. Ergod. Th. and Dynam. Sys. 7 (1987), 627-645.  

It is worth mentioning that the result also holds if a finite number of intervals are removed.
There is also a relationship between the Hausdorff dimension and the escape rate function. This is in a paper by Keller and Liverani: Rare events, e.scape rates and quasistationarity: some exact formulae. Journal of Statistical Physics 135 (3), 519-534. 
I hope this helps.<|endoftext|>
TITLE: Mahler measure of a totally positive, expanding algebraic integer
QUESTION [5 upvotes]: Consider a degree-$d$ algebraic integer $\alpha$ all of whose conjugates (including itself) are real numbers greater than 1. Its Mahler measure $M(\alpha)$ is simply equal to the norm $N(\alpha)$. Since $N(\alpha-1) \geq 1$, an application of $H\ddot{o}lder's$ inequality gives the lower bound $N(\alpha) \geq 2^d$, with equality achieved only by $\alpha = 2$. 
Question. Does there exist $C > 2$ such that $N(\alpha) \geq C^d$ for all $\alpha \neq 2$?

REPLY [3 votes]: Gypsum's argument is really nice. In the same spirit, we have the following inequality:
$$\ln|x| \geq \frac{2\ln|x-1| + \ln|x-2|}{5} + \ln\sqrt{5}$$
which is achieved at $x = \frac{5\pm \sqrt{5}}{2}$. This way we obtain the optimal constant $C = \sqrt{5}$. Perhaps some further tweaks can yield even larger $C$ with finitely many exceptions, as in the work of Smyth and Flammang. 
I wonder how far this approach can be extended. Smyth showed that for totally positive algebraic integers (whose conjugates are not necessarily greater than 1), their $M(\alpha)^{\frac{1}{d}}$ are dense beyond 1.73, which is very close to the lower bound cited by @BobbyGrizzard. Do we have a similar situation here? As a first step, it would be good to construct infinitely many $\alpha$ that give upper bound to $C$. For example, consider the $n$-th $Chebyshev$ polynomial $T_n(x)$. Then the monic polynomial $(-x)^nT_n(\frac{2}{x}-1)$ has all its roots greater than 1. In this case $N(\alpha) = 2^{2n-1}$, suggesting $C \leq 4$.<|endoftext|>
TITLE: Equivalent descriptions of Hodge conjecture?
QUESTION [33 upvotes]: I would like to know equivalent descriptions of the Hodge conjecture (with references).

Dan Freed's Version:


Consider a topological cycle (boundary less chains that are free to deform) on a projective manifold. The topological cycle is homologous to a rational combination of algebraic cycles, if and only if the topological cycle has rotation number zero.


Deligne's version (Clay's official description):


On a projective non-singular algebraic variety over $\mathbb{C}$ , any Hodge class is a rational combination of classes $\rm{Cl(Z)}$ of algebraic cycles.


nLab ((Pure)Motivic description):


Let $SmProj^{cor}_\mathbf{C}$ denote the category of algebraic correspondences of smooth projective algebraic varieties over the complex numbers.
  Then the canonical functor
$$ SmProj^{cor} \to HS^{pure} $$
to the category of rational pure Hodge structures, given by taking rational Betti cohomology, is full.

Equivalence between statements of Hodge conjecture

REPLY [4 votes]: A possibly less known reformulation of the Hodge conjecture was given by Richard Thomas. He relates to a problem about finding nodal hypersurfaces with sufficient homology.
https://arxiv.org/abs/math/0212216<|endoftext|>
TITLE: If two Hecke characters cut out the same field, are they Galois conjugates?
QUESTION [8 upvotes]: First question on MathOverflow, I hope it is appropriate for this site. There are two related questions.
Let $K$ be a number field, $G_K = Gal(\overline{K}/K)$, $p$ a prime, and
$$\chi_1,\chi_2:G_K\rightarrow\overline{\mathbb{Q}}_p^\times$$
be continuous characters such that $\ker(\chi_1) = \ker(\chi_2)$. 
My first question is, are the two fields generated by the values of $\chi_1$ and $\chi_2$ the same? That is, does 
$$\mathbb{Q}_p(\chi_1(\sigma), \sigma\in G_K) = \mathbb{Q}_p(\chi_2(\sigma),\sigma\in G_K)$$
As far as I can tell, this is true if $\chi_1,\chi_2$ have finite image, but I'm not sure if it's true if they have infinite image.
My second question is, does there exists $\sigma\in G_{\mathbb{Q}_p} = Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ such that for all $\tau\in G_K$, $\chi_1(\tau) = \sigma(\chi_2(\tau))$?
If $E = \mathbb{Q}_p(\chi_1(\sigma), \sigma\in G_K) = \mathbb{Q}_p(\chi_2(\sigma),\sigma\in G_K)$, then one could define $\sigma$ by the rule $\sigma(\chi_2(\tau)) = \chi_1(\tau)$, show $\sigma\in Gal(E/\mathbb{Q}_p)$, and take an extension of $\sigma$ to an element of $Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$. Then the answer to the second question would be yes if the answer to the first question is yes.
I don't know how to show that $\sigma$ defined as above is a well defined element of $Gal(E/\mathbb{Q}_p)$.

REPLY [9 votes]: The answer to both your questions is "no".
Take $K = \mathbf{Q}$. Then there is a unique $\mathbf{Z}_p$-extension of $K$ (contained in $\mathbf{Q}(\zeta_{p^\infty})$) which gives us a surjection $G_K \to \Gamma$ where $\Gamma$ is isomorphic to $\mathbf{Z}_p$.
Now, what are the continuous characters $\mathbf{Z}_p \to \overline{\mathbf{Q}}_p^\times$? It turns out that for every $a \in \overline{\mathbf{Q}}_p^\times$ with $|a - 1| < 1$, there's a unique character sending $1 \in \mathbf{Z}_p$ to $a$. If $a$ is not a root of unity, then this character is an injection.
So pick two values $a, a'$ which aren't roots of unity, don't generate the same extension of $\mathbf{Q}_p$, and aren't Galois-conjugate. Then this gives you two continuous characters $\chi, \chi': G_K \to \overline{\mathbf{Q}}_p^\times$ with the same kernel which are the required counterexamples.
With a little more work you can even find counterexamples with characters that are locally algebraic (and thus arise from algebraic Groessencharacters).<|endoftext|>
TITLE: Languages beyond enumerable
QUESTION [17 upvotes]: A language is a set of finite-length strings from some finite alphabet $\Sigma$.
It is no loss of generality (for my purposes) to take $\Sigma=\{0,1\}$; so a language is a set of bit-strings.
Languages are commonly classified in a hierarchy, with the

enumerable $\equiv$ recursively enumerable $\equiv$ Turing-recognizable

the outermost set, beyond decidability:

     


     
(Figure from Sipser.)

My question is about the terminology extending this hierarchy.
The non-enumerable are clearly outside the enumerable ($\equiv$ Turing-recognizable).
But some languages $L$ are enumerable but their complement $\overline{L}$
is not enumerable (such as the language of all Turing Machines that halt
on a given input),
whereas other languages $L$ are not enumerable, and their complement 
$\overline{L}$ is also not enumerable (such as the pairs of "equivalent" Turing machines).

Q1. Are there standard names for languages that go farther outside of
  the above diagram?

The set of all languages is uncountable, so there is plenty of room for expanding
the diagram. But I haven't seen a clear, "standard" expansion.
I am seeking a progression further and further out into the uncountable to
help students (and me!) see the vastness.
I am seeking the equivalent of this
iconic complexity-theory hierarchy:

     


     
(Image from Seneca ICT.)


Q2. Is there an equivalent diagram in language theory?

REPLY [5 votes]: There are many such classes of languages depending on the approach you want to use.  Let me focus on just one class: the hyperarithmetic (hyperarithmetical?) class, and try to explain why it is a very natural class, by mentioning several equivalent definitions of it:

"Computability in a type 2 functional": a hyperarithmetic set of integers is one that can be computed by a "hyperarithmetical" machine, where a "hyperarithmetical" machine is one that can do everything that a Turing machine can do, but with the additional ability to decide whether a function $f\colon \mathbb{N} \to \mathbb{N}$ takes a nonzero value, where the function $f$ is itself computed by a hyperarithmetical machine.  (In other words, given a hyperarithmetical machine computing $f$, such that $f(n)$ is defined for every $n\in\mathbb{N}$, a hyperarithmetical machine can compute every $f(n)$ at once and decide whether there is one such that $f(n)\neq 0$, in which case of course it can also trivially find the corresponding $n$; if not every $f(n)$ is defined because the machine computing them does not halt, then the overall call will also not halt).  This definition is recursive, of course, and I didn't formalize it completely, but it still makes sense as the smallest which satisfies the conditions.  In the classical computability literature, this description is called "computability [à la Kleene] in the type 2 functional $\mathbf{E}$", but apparently it is never described in terms of "machines" like I tried to sketch.
…In practice, a hyperarithmetical machine is one that can compute not only with integers, but also with exact real numbers (i.e., sequences of integers), not all real numbers but, precisely, those which are hyperarithmetical.  I think this makes the definition fairly natural.
…Yet another way of rephrasing the same definition is that a hyperarithmetical machine is one that can compute infinite conjunctions/disjunctions (logical and/or), provided the terms of the conjunction/disjunction are themselves computed hypearithmetically.  This is just a rephrasing of the above, but it provides a link with certain kinds of infinitary logic.
"Metarecursion": a set of integers is hyperarithmetic when it can be computed by a machine much like a register machine, except that the registers are allowed to hold ordinal values, the ordinals ranging up to the Church-Kleene ordinal (=smallest nonrecursive ordinal) $\omega_1^{\mathrm{CK}}$, and the machine is allowed to "loop" up to that value (I tried to summarize the idea of such ordinal computations here, in which terminology I would be speaking of $(\omega_1^{\mathrm{CK}},\omega_1^{\mathrm{CK}})$-machines).  These machines can compute much more than sets of integers, but those sets of integers which they can compute are precisely the hyperarithmetic ones.
The level $\Delta^1_1$ of the analytical hierarchy is again the class of hyperarithmetic sets: essentially those sets of integers which can be defined using one second-order quantifier, both in an existential and in a universal manner.  This is an analogue (the so-called "lightface" analogue) of one of the definitions of Borel sets in descriptive set theory.
Iterating the Turing jump: a $0$-machine is just a Turing machine; a $0'$-machine is one that has access to an oracle that can tell it whether a $0$-machine halts; a $0''$-machine is one that has access to an oracle that can tell it whether a $0'$-machine halts; a $0^{(n)}$-machine is what you imagine; a $0^{(\omega)}$-machine (or arithmetical machine) is one that has access to an oracle that can tell it, given $n$, whether a $0^{(n)}$-machine halts; it is possible (although not completely trivial) to iterate this over the recursive ordinals, and hyperarithmetic sets are precisely those which are recognized by a $0^{(\alpha)}$-machine for some recursive ordinal $\alpha$.
The sets of integers belonging to the level $L_{\omega_1^{\mathrm{CK}}}$ of Gödel's constructible universe where each level is defined essentially by adding every subset of the previous level that can be defined in it by a first-order formula.  Also, this level $L_{\omega_1^{\mathrm{CK}}}$ is the first one which satisfies a sizable amount of set theory, namely Kripke-Platek.

Since all these definitions conspire to give the same class of hyperarithmetic sets, I think it's fair to say that it's a natural class.  There are plenty of classes both above and below, but I think this one deserves to be better known.
(Also, concerning complexity: there are also plenty of classes between $\mathsf{EXPTIME}$ and $\mathsf{REC}$: there are $\mathsf{ELEMENTARY}$ and $\mathsf{PR}$, but also lots of classes which can be defined between the class $\mathsf{PR}$ of primitive recursive sets/functions and that $\mathsf{REC}$ of recursive sets/functions, and that try to bridge the gap between complexity and computability.  These "subrecursive" hierarchies also deserve to be better known.)<|endoftext|>
TITLE: Computing endomorphism rings of supersingular elliptic curves
QUESTION [15 upvotes]: I would like to know what algorithms there are to compute the linearly independent generators $(1,i,j,k)$ for quaternion algebra containing the endomorphism ring of a supersingular curve. The curve in question is pretty small (around 16 bits, but I have no idea how to perform a brute force search).
One has that the Frobenius endomorphism would give us a generator linearly independent to the integers (that gives us $i^2=-p$), so how can one find the other? I have tried to look at Hilbert polynomials $h_D(X)$ for different $D$'s to see if $j_E$, the $j$-invariant of the elliptic curve, is a root, but there are more than one solution, so could one take the smallest $D$ so that $j^2=-D$?  
Also, once the first part is completed, how does one find the $\mathbb{Z}$-basis that generates the endomorphism ring?
Lastly, how does one check if the endomorphism ring or the quaternion algebra obtained from the algorithms are correct?
(Aside: Does anyone have a reference to show that roots of $D$-Hilbert polynomials are $j$-invariants of elliptic curves with endomorphism ring containing the imaginary quadratic field with discriminant $D$? I've seen it used all the time, but have not seen an explicit reference.)
Edit: I have found a paper by Pizer that contains a proposition (5.1) that describes the quaternion algebra the endomorphism ring lives in. Another proposition (5.2) describes a maximal order, but this still does not help my cause.

REPLY [3 votes]: For the computation of a basis of the endomorphism algebra, you should read David Kohel's thesis, see for instance Theorem 2.
To find the endomorphism ring, you can repeat Kohel's methods to find more endomorphisms, until the ring they generate is a maximal order in the algebra.
In order to check that the endomorphism algebra is correct, you only need to check that the basis elements really are endomorphisms and that they generate an algebra of dimension $4$. If you are only given the isomorphism class (and not actual endomorphisms), you need to check that the algebra is ramified exactly at $\{p,\infty\}$.
Similarly, to check that the endomorphism ring is correct, you only need to check that you are given actual endomorphisms and that they generate a maximal order in a quaternion algebra. If you are only given the isomorphism class, I don't know how to check that it is correct faster than by recomputing it and comparing the results.
For more about quaternion algebras and how to algorithmically perform the tasks I mentionned, see John Voight's book and references therein.<|endoftext|>
TITLE: Law of large numbers for martingales
QUESTION [9 upvotes]: I apologize in advance if this question is too basic, but I've received no response on Math Stack Exchange, so perhaps it is more appropriate here:
Let $X_n$ be a square-integrable martingale with $\mathbb{E}\lbrack X_n^2\rbrack=O(n)$. Is it true that $X_n/n$ tends to $0$ almost surely?
Note that if one demanded instead that $\mathbb{E}\lbrack X_n^2\rbrack =O(n^{1-a})$ for some $a>0$, then the claim would follow for any random variables from Markov's Inequality and the Borel-Cantelli Lemma.

REPLY [10 votes]: The answer is yes, and it is based on an idea by Prokhorov (cf. e.g. Theorem 10 in Section 3 of Ch. IX in [Petrov, V. V., Sums of independent random variables, Springer-Verlag, 1975]). We have $EX_n^2\le Cn$ for some real $C>0$ and all natural $n$. For natural $s$, let 
\begin{equation}
 T_s:=\max_{2^s\le n<2^{s+1}}\frac{|X_n-X_{2^s}|}{2^s}. 
\end{equation}
Then, by [Doob's martingale inequality], for any real $t>0$ 
\begin{equation}
 P(T_s\ge t)\le\frac{E(X_{2^{s+1}}-X_{2^s})^2}{(2^s t)^2}
 \le\frac{EX_{2^{s+1}}^2}{(2^s t)^2}\le\frac{C2^{s+1}}{(2^s t)^2}=\frac{2C}{2^s t^2}, 
\end{equation}
so that $\sum_{s=1}^\infty P(T_s\ge t)<\infty$. So, by the Borel--Cantelli lemma, $T_s\to0$ almost surely (a.s.) as $s\to\infty$. 
Therefore, for any natural $n$ and $r$ such that $2^r\le n<2^{r+1}$ (so that $r=r_n:=\lfloor\log_2 n\rfloor$), one has
\begin{equation}
 \frac{|X_n-X_1|}n\le2 \frac{|X_n-X_1|}{2^{r+1}}
 \le2\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s\to0 \tag{1}
\end{equation}
a.s., since $\sum_{s=0}^r 2^s<2^{r+1}$; cf. e.g. Lemma 9 in Section 3 of Ch. IX in [Petrov, V. V., Sums of independent random variables, Springer-Verlag, 1975]). Thus, $\frac{X_n}n\to0$ a.s., as desired. 

Details on $(1)$: Since $T_s\to0$ a.s., without loss of generality for each real $\epsilon>0$ there is a natural-valued random variable $R_\epsilon$ such that for any natural $s$ the event $s>R_\epsilon$ implies $T_s\le\epsilon$. Therefore,
\begin{equation}
 \frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s
 \le\frac1{2^{r+1}}\,\sum_{s=0}^{R_\epsilon} 2^s T_s+\frac1{2^{r+1}}\,\sum_{s=R_\epsilon+1}^r 2^s \epsilon
 \le\frac1{2^{r+1}}\,\sum_{s=0}^{R_\epsilon} 2^s T_s+\epsilon. 
\end{equation}
So, $\limsup_{r\to\infty}\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s\le\epsilon$, for any $\epsilon>0$, and hence $\lim_{r\to\infty}\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s=0$.<|endoftext|>
TITLE: Factorization system "tilted" by $(L,R)$
QUESTION [6 upvotes]: Suppose you have a pair of orthogonal factorization systems, $(E_0, M_0), (E_1, M_1)$ in a category $\cal C$ such that $M_0\subseteq M_1$; this entails that there is a ternary factorization
$$
X\xrightarrow{e_1} A\xrightarrow{e_0 m_1} B\xrightarrow{m_0} Y
$$
where each arrow is labeled according to the class it belongs to.
Suppose now to have another OFS $(L,R)$ on $\cal C$, and to factor the middle arrow $A\to B$ into $A\xrightarrow{l} S\xrightarrow{r} B$; this gives a factorization $X\to S\to B$.

Does this define a third factorization system $(E_{01}\wr L, M_{01}\wr R)$?

REPLY [8 votes]: Yes.
First note that Galois connections are far more common than field theorists would have you believe: any binary relation gives rise to one and the two fixed sets a closed under any appropriate algebraic structure.
Write

$P\perp Q$ for "every member of $P$ is orthogonal (in the sense of factorisation systems) to every member of $Q$";
$P;Q$ for the set of all composites of members of $P$ followed by members of $Q$;
$L'=L\cap E_0\cap M_1$ and $R'=R\cap E_0\cap M_1$ (to avoid any ambiguity in this argument); and
$E=E_1;L'$ and $M=R';M_0$.

Then

$E_1\perp M_1\supset R'$,
$E_1\perp M_1\supset M_0$,
$L'\subset L\perp R\supset R'$ and
$L'\subset E_0\perp M_0$.

Since orthogonality respects composition, it follows that $E\perp M$, whilst by construction $E;M$ is the entire hom-class of the category.
Therefore $(E,M)$ is a factorisation system.
The things that I have assumed are all proved in my book but I don't have a copy to hand to look up the references.
In the question as stated, it is given that $M_0\subset M_1$, but there is no order-relationship between these and $R$. This is why it was necessary to introduce $R'$.  The construction gives a way of handling expressions in the lattice of factorisation systems on a category, so it would be an interesting exercise in lattice theory to find out whether this is modular or even distributive.<|endoftext|>
TITLE: Conditions for a smooth scheme of finite type with trivial class group to be quasi-affine
QUESTION [7 upvotes]: Let $X$ be a smooth scheme of finite type over an algebraically closed field of characteristic zero and with a trivial class group $Cl(X)=0$. Let $Y$ be a dense open subscheme of $X$ such that: 
1) $Y$ is a quasi-affine scheme s.t. $\Gamma(\mathcal{O}_Y,Y)$ is of finite type; and 
2) $X \setminus Y$ is irreducible of codimension at least two.   
Does these conditions imply that $X$ is itself a quasi-affine scheme? If not, can anyone provide a counter-example please?
Thank you in advance!

REPLY [2 votes]: Here is a counterexample. Unfortunately it is not separated, so I don't know how interesting it is to you.
Consider $X = \mathbf{A}^2 \cup \mathbf{A}^2$, glued along $\mathbf{A}^2 \setminus \{(0,0)\}$, and let $Y$ be one copy of $\mathbf{A}^2$. $X$ is not quasi-affine since it is not separated. $\operatorname{Cl}(X) \cong \operatorname{Cl}(Y) = 0$ since the complement of $Y$ in $X$ is a point, which has codimension 2. Also, $Y$ is affine, and $\Gamma(Y,\mathcal{O}_Y) = k[x,y]$ is of finite type.
Finally, some remarks which I commented earlier:

Schröer has an example of a complete normal variety with $\operatorname{Pic}X = 0$, but I didn't compute its class group since you wanted a smooth example.
Hamm and Lê show that a complex algebraic variety with $H^1 = H^2 = 0$ (the actual condition is weaker) would have trivial class group, so this might be a place where you could find a counterexample that is also a variety.<|endoftext|>
TITLE: If a faithfully flat extension of dg/A_$\infty$-algebra is formal, is the original algebra formal (over positive characteristic)?
QUESTION [8 upvotes]: Proposition 6.2 of Formality of DG algebras (after Kaledin) by Lunts reads (with a few additions to clarify notation): 

Let $k$ be a field of characteristic 0.  Let $A$ be an $A_\infty$ algebra over a commutative $k$-algebra $R$ which is a finite as an $R$-module. Let $R\to Q$ be a homomorphism
  of commutative rings.  Assume that $Q$ is a faithfully flat $R$-module.
  Then $A$ is formal if and only if the $A_\infty$ $Q$-algebra $A\otimes_RQ$ is formal.

Does this theorem fail for a DG-algebra of positive characteristic?  Is there some additional hypothesis I can add to fix it?  Specifically, would it help to know that $A$ was actually the specialization of an algebra $A_{\mathbb{Z}}$, which is formal after base change to a characteristic 0 field?

REPLY [4 votes]: I'm not sure off hand what the situation is for $A_\infty$-algebras, but for $\mathbb{E}_\infty$-algebras it's worth noting that in many cases generic formality does not imply formality at each fiber. For example, consider the cochain algebra $C^*(X,\mathbb{Z})$. The rationalization $C^*(X,\mathbb{Q})$ can often be formal, as is the case for $X$ compact Kähler by the theorem of Deligne-Griffiths-Morgan-Sullivan. However, $C^*(X,\mathbb{F}_p)$ is not formal if it is non-trivial. In fact, it is not even equivalent to a cdga at all! This follows from the the fact that Steenrod operation $Sq^0$ is the identity on cohomology of a space, but acts as zero on $H^i(A)$ for $i\neq 0$ if $A$ is a cdga over $\mathbb{F}_p$. (I learned this originally from Mike Mandell.)
Of course, this does not actually answer your question. The reason that Kaledin's theorem works, as explained by Lunts, is that there is an obstruction to formality in a Hochschild cohomology group that behaves well under base extension. Unfortunately, the definition seems to use the Campbell-Hausdorff formula, so it's not clear to me what happens at all in characteristic $p$.
Yanki Lekili asked me a potentially related question: are the cochain algebras $C^*(G,\mathbb{F}_p)$ formal as dgas not as $\mathbb{E}_\infty$-algebras for compact Lie groups $G$?<|endoftext|>
TITLE: Computing in quantum groups
QUESTION [10 upvotes]: I'd be interested in doing some computations in quantum groups $ U_q(\mathfrak g)$ that are conceptually simple (``is this element 0"?, and $\mathfrak g = sl_5$), but are somewhat lengthy to do by hand. Is there software available that can do this? (I.e. take an element of the quantum group and reduce it to a PBW basis.) I've done some Googling and asked a couple people who I thought might know, but haven't found an answer. (In particular, the QuantumGroups package by Scott Morrison on KnotAtlas doesn't seem to do reductions to a PBW basis.)

REPLY [9 votes]: There is the package QuaGroup by de Graaf for both GAP and Magma: see QuaGroup.  I've used it in both systems and found it to be extremely helpful.  Since there's a GAP package, you also have the option of using it inside Sage.
You will potentially want to be careful about exactly which PBW basis you and the computer are working with, of course.<|endoftext|>
TITLE: Is $\omega^\frac{1}{\omega} > n \forall n \in \mathbb{N}$?
QUESTION [8 upvotes]: I was thinking about $log(\omega)$ which appears to be $\{\mathbb{N}|\omega^{\frac{1}{n}}\}_{n\in\mathbb{N}}\stackrel{?}{=}\omega^\frac{1}{\omega}$. Intuitively, there's the idea that, if the highest power of $\omega$ is greater for one surreal number, then it is greater, but this seems incorrect for two reasons:
$$n<\omega^\frac{1}{\omega}\rightarrow n^\omega <\omega$$
The statement on the right is not true, hence the question. Additionally, from a real analysis perspective,
$$\lim_{n\to\infty}{n^{\frac{1}{n}}}=1$$

REPLY [8 votes]: Every surreal number has a unique representation as a Hahn series of $\omega$ with coefficients being real numbers and exponents being themselves surreal numbers, and they are compared lexicographically as such Hahn series.  (This form is known as a "Conway normal form".  See, e.g., Norman Alling, Foundations of Analysis over Surreal Number Fields (North-Holland Math Studies 141 (1987)), esp., around §6.50, although pretty much the whole book really.)  So $\omega^x$ is greater than all real numbers for any $x>0$, including $1/\omega$.<|endoftext|>
TITLE: Group cohomology of the cyclic group
QUESTION [14 upvotes]: It is well known how to compute cohomology of a finite cyclic group $C_m=\langle \sigma \rangle$, just using the periodic resolution, 
$\require{AMScd}$
\begin{CD}
    \cdots @>N>> \mathbb Z C_m @> \sigma -1>>  \mathbb Z C_m @>N>> \mathbb Z C_m @> \sigma -1>> \mathbb Z C_m  @> >> \mathbb Z 
\end{CD}
Using this resolution, it easy to see that
\begin{align} H^n(C_m; A)= \begin{cases}\{a\in A: Na=0\}/(\sigma-1)A, \qquad &\text{if } n=1, 3, 5, \ldots \\
A^{C_m}/NA, \quad  &\text{ if } n = 2, 4, 6, \ldots,
\end{cases}
\end{align}
where $N= 1+ \sigma + \sigma^2 +\cdots +\sigma ^{m-1}$.
Now, for some applications of group cohomology is important to work with standard cocycles, that is cocycles respect to the standard (also called Bar) resolutions. A construction of quasi-isomorphism from the periodic resolution to the standard resolution can be done as follows: take a section of $\pi$ in the exact sequence 
\begin{CD}
   0 @>>> \mathbb Z @> m >>  \mathbb Z @>\pi>> C_m @>>> 0,  
\end{CD}so we get a $\gamma\in Z^2(C_m,\mathbb{Z})$. For  $Z^1(C_m,A)$ and $Z^2(C_m,A)$ the map can be defined by hand easily. In general we can construct the map  $:Z^1(C_m,A)\to Z^{2n+1}(C_m,A)$ just using the cup product $\alpha\mapsto \gamma^{\cup n}\cup \alpha$ and analogously for $:Z^2(C_m,A)\to Z^{2n}(C_m,A)$. Thus, at the end you find the map from the "periodic" cocycles to the standard cocycles.
My question is: How to define in general the quasi-isomorphisms from the standard cocycles to the "periodic" cocycles?

REPLY [5 votes]: Just stumbled across that old question while researching the subject on the internet, ended up doing the computation myself, and thought I would give an answer in case someone finds themself in the same situation.
A possible quasi-isomorphism $u_n: C^n(C_m,A)\to A$ from the standard complex to the periodic one is given by the following formulas: if $n=2r$, and $\tau$ is a generator of $C_m$,
$$u_{2r}: f\mapsto (-1)^r\sum_{g_1,\dots,g_r\in C_m}\sum_{\varepsilon_1,\dots,\varepsilon_r\in \{0,1\}} (-1)^{\sum \varepsilon_i} f(g_1,\tau^{\varepsilon_1},\dots,g_r,\tau^{\varepsilon_r}),$$
and if $n=2r+1$,
$$u_{2r+1}: f\mapsto (-1)^{r+1}\sum_{g_1,\dots,g_r\in C_m}\sum_{\varepsilon_1,\dots,\varepsilon_{r+1}\in \{0,1\}} (-1)^{\sum \varepsilon_i} f(\tau^{\varepsilon_1},g_1,\tau^{\varepsilon_2},\dots,g_r,\tau^{\varepsilon_{r+1}}).$$
In the example I was interested in, the action of $C_m$ on $A$ was trivial, but I read the computation carefully and I don't think it intervened in this formula.<|endoftext|>
TITLE: representations of $S_k \times S_j$
QUESTION [6 upvotes]: For a group $G$ and a field $K$ let $S(G,K)$ be the sum of the dimensions of the irreducible K representations of $G$. Note that $S(G,\mathbb{C})< |G|$. It's not difficult to prove that if $n \ge 6$ then $S(S_n,\mathbb{C}) < (n-2)!(n-2)-n$. I'm interested in "good" bounds (not necessarily the best but at least significantly better than $|G|$). I need a bound for $S(S_k \times S_{n-k}, \mathbb{Q})$ the best as possible.

REPLY [8 votes]: I guess that $S_n$ denotes the symmetric group. It is well known that all irreps of $S_n$ over $\mathbb C$ are defined over $\mathbb Q$. Therefore $S(S_k\times S_{n-k},\mathbb{C})=S(S_k\times S_{n-k},\mathbb{Q})$. Also the fact that the sum of the squares of the dimensions is the group order and Cauchy-Schwarz immediately imply that $S(G,\mathbb C)\le \sqrt{c(G)|G|}$ where $c(G)$ is the number the conjugacy classes. In your case it means $S(S_n,\mathbb C)\le \sqrt{p(n)n!}$ where $p$ is the partition function.

REPLY [8 votes]: Since, as mentioned by Friedrich Knop, the irreducible $\mathbb{Q}$-representations of $S_{n}$ are absolutely irreducible, ( so, in particular all complex irreducible characters of $S_{n}$ are real valued) it follows that 
$\sum_{\chi \in {\rm Irr}(S_{n})}\chi(1)$ is the number of solutions of $x^{2} = 1$, where $\chi$ runs over the complex irreducible characters of $S_{n}$, which is the same in this case as the sum of the degrees of irreducible rational representations of $S_{n}$. This uses the well-known theory of the Frobenius-Schur indicator.
Hence the sum in question is 
$ 1+ \sum_{k=1}^{\lfloor  \frac{n}{2} \rfloor} \frac{n!}{(n-2k)! 2^{k}k!}$, since the centralizer of a product of $k$ disjoint transpositions is isomorphic to $ S_{n-2k} \times (S_{2} \wr S_{k})$. 
The same reasoning applies to direct products of symmetric groups.
If you wish to approximate sums such as those above, the middle terms tend to dominate  ($k$ close to $\frac{n}{2}$), and just taking $ k = \frac{n}{2}$ if $n$ is even or $k \in \{\frac{n \pm 1}{2}\}$ if $n$ is odd can be used to directly obtain a bound of the form $p(n) \sim e^{c \sqrt{n}}$ for the partition function.

REPLY [5 votes]: If all representations of a group can be realized over $\mathbb{R}$, then $S(G, \mathbb{C})$ equals the number of elements of order 2 in $G$. More generally, the number of solutions of the equation $x^2=g$ in $G$ is $\sum\epsilon_\chi \chi(g)$, where $\epsilon_\chi=1$, if $\chi$ belongs to a real representation, $\epsilon_\chi=-1$, if $\chi$ has non-real values, and $\epsilon_\chi=0$, if all values of $\chi$ are real, but the representation associated to $\chi$ is not real.
Chowla, Herstein and Moore have shown that the number of solutions of $x^2=1$ in $S_n$ is asamptotically equal to
$$
\frac{n^{n/2}e^{\sqrt{n}}}{\sqrt{2}e^{n/2+1/4}},
$$
Müller has shown that there exists an asymptotic expansion in terms of $n^{-1/2}$.<|endoftext|>
TITLE: Error estimate in the spectral theorem of compact operators on a Hilbert space
QUESTION [5 upvotes]: Given a compact self-adjoint operator $K$ mapping $L^2(\mathbb{R}^d) \rightarrow  L^2(\mathbb{R}^d)$ as $f \rightarrow \int K(x,y) f(y) d\mu(y)$,  let us define its eigenvalues $\lambda_i$ and eigen-functions $e^i(x)$. 
Let us define a sequence of "partial sum" operators $K_N(x,y) = \sum_{i=1}^N \lambda_i e^i(x)e^i(y)$. Then this sequence converges to $K(x,y)$ in the product Hilbert space under the product measure.  
More explicitly then,
$$\lim_{N\rightarrow \infty} \int \int |K(x,y) - \sum_{i=1}^N \lambda_i e^i(x)e^i(y)|^2 d\mu(x)d\mu(y) = 0.$$

But is there any example where for some specific $K$ someone has tried estimating this error-integrand $|K(x,y) - \sum_{i=1}^N \lambda_i e^i(x)e^i(y)|^2 $ ? I would be hapy to find any reference along these lines..

REPLY [2 votes]: I'm following Szego's book on orthogonal polynomials.
In chapter III he considers $f \in L^2\left((a,b),d\alpha \right)$, with $-\infty \leq a < b \leq \infty $. There exist a set of orthogonal polynomials $\{p_n \}_{n=0}^{\infty}$ w.r. to the measure $d\alpha$. 
We consider the kernel $K_n(t,x) := \sum\limits_{\nu=0}^{n} \bar{p_{\nu} (t)} p_{\nu} (x) $. Then one can show that $\int\limits_{a}^{b} K_n (t,x) f(t) \, d\alpha (t) = \sum\limits_{\nu = 0}^{n} \hat{f}(n)p_n (x) $, the spectral expension of order $n$. 
So, if we settled that the polynomial expansion of $f$ is an integral transform, there's a detailed discussion in Szego's book, Davis "approximation theory" and many other books about its $L^2$ error.
Edit: If we look in Aronszajn (1950), we can have by Section 9 that the kernels converge in L^2 norm.
Edit 2: I recommend you to read the aforementioned paper, as it deals with the much broader class of reproducing kernels, and give a lot of conditions and results that are relevant to your question.<|endoftext|>
TITLE: A transcendence question involving the exponential function
QUESTION [5 upvotes]: Let $(z_n)$ be a sequence of complex numbers satisfying $|z_n|\to +\infty$ and such that $\{e^{z_n}\mid n \in \mathbb{N}\}$ is infinite.
Is it always true that $\{(z_n,e^{z_n})\mid n \in\mathbb{N}\}$ is Zariski-dense in $\mathbb{C}^2$? In other words, if $p(x,y) \in \mathbb{C}[x,y]$ is a polynomial such that $p(z_n,e^{z_n})=0$ for every $n \in \mathbb{N}$, is it always the case that $p = 0$?
This is clearly true if $z_n$ are taken to be real numbers by an "order of growth" argument. But in general I'm not even completely sure if this should be true. Any ideas?

REPLY [9 votes]: I claim that there are many occasions where a sequence $(z_n,e^{z_n})$ with $z_n\to \infty$ is NOT Zariski dense.
Take for instance the complex function $f(w)=e^{1/w}-1/w$. It has an essential singularity at 0, so by Picard's great theorem we know that for all but one (maybe) values $c$ there exists a sequence $w_n$ converging to 0 with $f(w_n)=c$.
So let's define the polynomial $p(x,y)=y-x-c$ (for a non-stupid choice of $c$) and the sequence $z_n=1/w_n$ (for the corresponding sequence $w_n$).
Note that $z_n\to \infty$, $p(z_n,e^{z_n})=0$ and $e^{z_n}=z_n+c$ is not constant.

REPLY [7 votes]: It is false. Equation $\sin x=1/x$ has infinitely many real solutions $(x_n)$, take $z_n=ix_n$, we get $z_n(e^{2z_n}-1)=-2e^{z_n}$.<|endoftext|>
TITLE: Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?
QUESTION [25 upvotes]: The question is: if I assert in ZF that there exists a Reinhardt cardinal, do I really get a theory of higher consistency strength than when I assert in ZFC that there exists an I0 cardinal (the strongest large cardinal not known to be inconsistent with choice, as I understand)? This is implicit in the ordering of things on Cantor's Attic, for example, but I've been unable to find a proof (granted, I don't necessarily have the best nose for where to look!).
One thing that worries me is that when there is a ZFC analog of a ZF statement, many equivalent formulations of the ZFC statement may become inequivalent in ZFC. So we don't have much assurance that the usual definition of a Reinhardt cardinal is "correct" in the absence of choice.
I think it should be clear that Con(ZF + Reinhardt) implies Con(ZF + I0). But again, it's not clear that ZF+I0 is equiconsistent with ZFC+I0.
It's apparently not possible to formulate Reinhardt cardinals in a first-order way, so I should really talk about NBG + Reinhardt, or maybe ZF($j$) + Reinhardt, where ZF($j$) has separation and replacement for formulas involving the function symbol $j$.
EDIT
Since this question has attracted a bounty from Joseph Van Name, maybe it's appropriate to update it a bit. Now, I'm not actually a set theorist, but it's not even clear to me that Con(ZF + Reinhardt) implies Con(ZFC + an inaccessible). So perhaps the question should really be: what large cardinal strength, if any, can we extract from the theory ZF + Reinhardt?

REPLY [2 votes]: Mohammed Golshani's link doesn't work, so I have reconstructed a sketch of a proof. The key fact is this (You can find a proof in most Set Theory textbooks):
Theorem: If $DC_\omega$ holds and $D$ is an $\omega_1$-complete ultrafilter, then the ultraproduct of $N$ by $D$ is well-founded, for any inner model $N$.
Theorem: If $DC_\lambda$ holds and $\kappa$ is $I0$ (With target $\lambda$) if and only if there is a non-principal $\kappa$-complete $L(V_{\lambda+1})$-ultrafilter on $V_{\lambda+1}$.
Proof. Let $D$ be such an ultrafilter. To verify the existence of a such an ultraproduct, we can code elements of $D$ as function $F: X^{\lt\lambda}\rightarrow X'$ for every set $X'=\{\chi_x|x\in V_{\lambda+1}\}$, there is a function $f: \lambda\rightarrow X$, such that $f(\chi_x)\in\chi_x$, and so some $A\in D$, such that $\{\chi_x|x\in A\}$ admits a Choice function. Then if $M\cong Ult_D(L(V_{\lambda+1}))$ is the ultrapower $M\ni V_{\lambda+1}$. $L(V_{\lambda+1})$ inherits a well-order for each element (A well order not necessarily in $L(V_{\lambda+1})$ from $L(V_{\lambda+1})$. Then by condensation $M= L(V_{\lambda+1})$ and $\kappa$ is the critical point of the ultrapower embedding $k_D: L(V_{\lambda+1})\prec L(V_{\lambda+1})$. For the other direction, define an ultrafilter $D=\{X\subseteq V_{\lambda+1}|j\restriction V_\lambda\in j(X)\}$. This satisfies all the necessary properties.◼
Theorem: If $\kappa$ is Reinhardt, indeed even the critical point of $j: V_{\lambda+2}\prec V_{\lambda+2}$, then $\kappa$ is $I0$.
Proof. Let $\kappa$ Reinhardt as witnessed by $j: V\prec V$, and let $\lambda$ be the least fixed point above $\kappa$. Let $D=\{X\subseteq V_{\lambda+1}|j\restriction V_\lambda\in j(X)\}\cap L(V_{\lambda+1})$. Then $D$ satisfies all the necessary properties. The second case is not much trickier, as it uses the same argument.◼<|endoftext|>
TITLE: Obtaining Lawvere's "State categories and response functors"
QUESTION [7 upvotes]: I'm looking to get my hands on a (.pdf) copy of Lawvere's 1986 preprint State categories and response functors. If someone can post it and answer this question by offering a link, I'd appreciate it.

REPLY [4 votes]: A draft of Lawvere's paper was sent to me by Joachim Kock; it's the best I've found. I put it on my webpage for the convenience of others.<|endoftext|>
TITLE: Eigenvalues of partial Hankel matrices
QUESTION [6 upvotes]: I was wondering if there are closed formulas for the singularvalues of a partial Hankel matrix (by partial I mean $\ell<n$)
\begin{align*}
H=
\begin{bmatrix}
c_1 & c_2 & \ldots & c_\ell \\ c_2 & c_3 & \ldots & c_{\ell+1} \\ 
c_3 & c_4 & \ldots & c_{\ell+2}\\
\vdots & \vdots & \ddots & \vdots\\
c_n & c_{n+1} &\ldots & c_{n+\ell-1}
\end{bmatrix}
\end{align*}
What does it mean for this matrix to be low-rank?

REPLY [2 votes]: Typically, one does "Prony method": considers an infinite (or just long enough) sequence $c=(c_1,c_2,\dots)$ and a system of equations of the form $V(x)Z=c$, with $Z$ a vector of non-0 unknowns, and $V(x)$ the Vandermonde matrix
$$
V(x)=\begin{pmatrix} 
1&1&\dots&1\\
x_1& x_2&\dots &x_k\\
x_1^2& x_2^2&\dots &x_k^2\\
&\dots&\dots&\dots\\
\end{pmatrix} 
$$
By multiplying both sides of $V(x)Z=c$ on the left by the row vectors $a_\ell=(\underbrace{0,\dots,0}_\ell,a_0,a_1,\dots)$, for $\ell\geq 0$, one relates the polynomials $x^\ell a(x)=x^\ell\sum_{i\geq 0} a_i x^i$ with zeros $\{x_1,\dots,x_k\}$ and the kernel of a submatrix of the the infinite Hankel matrix corresponding to the sequence $c$. 
By controlling the degree of $a(x)=k$, you should be able to get something interesting...<|endoftext|>
TITLE: Cotangent complex of certain dg-scheme
QUESTION [6 upvotes]: This is a somewhat embarrassing question, but still I will ask it. Let $V$ be a vector space over $\mathbb C$ of dimension $d$.
Let $X$ be the dg-preimage of $0$ under the natural map $V\to Sym^2(V)$ (i.e. impose equations of the form $f=0$ for all homogeneous polynomials of degree 2 on $V$; we have $d\choose{2}$ equations, so $X$ always has a non-trivial dg-part for $d>1$). 
$\mathbf{Question:}$ What does the cotangent complex of $X$ look like? More precisely, I want to know its restriction to the unique $\mathbb{C}$-point of $X$.
I would be happy to understand the case $d=2$.

REPLY [3 votes]: Well, unless I am mistaken, whenever you have a cdg ring given by dg generators and relations, then the cotangent space at zero is simply the cone of the differential $T^*\text{Spec}(S^*{\text{generators}})\to T^*\text{Spec}(S^*{\text{relations}})$ at the corresponding point. In this case, as $d(\text{anything quadratic}) = 0$ at $0\in V$, this should give a complex with zero differential, $V\oplus S^2(V)[-1]$ at the point (here $S^2(V^*)[1]$ means it's in degree $1$, not a shift).
More concretely in this case, you have a Koszul CDG-free resolution for $X$ as follows: $A = \left(\bigwedge^* S^2(V)\otimes S^*[V], d\right)$ with $S^2(V)$ in degree -1, with differential determined on the degree-minus-one generators by $d(v\otimes w) = vw\in S^*(V)$. It's more intuitive for me to understand the dual problem of computing the tangent rather than the cotangent complex, i.e. derivations $A\to \mathbb{C}$. As a graded space, this is dual to the space of generators, so $V^*\oplus S^2(V^*)[-1]$, where the differential $d:V^*\to S^2(V^*)$ is $d(\partial_\xi) = [d, \partial_\chi] = -\partial_\xi\circ d_A$ acts on $v w\in S^2(V)$ by $d(\partial_\xi)(vw[-1]) = \partial_\xi(vw[0]) = v\partial_xi(w) \pm w \partial_\xi(v)$ (here I'm using the crude notation $[0], [-1]$ to distinguish degree 1 and degree 0 elements), and as both $\partial_\xi(v), \partial_\xi(w)\in \mathbb{C}$ are killed by $V$, this just gives us $d=0,$ in the tangent complex hence the desired result.<|endoftext|>
TITLE: "Diagonalizing" Littlewood-Richardson coefficients
QUESTION [6 upvotes]: Let's consider the Littlewood-Richardson coefficients $c^{\lambda}_{\mu \nu}$ so that
\begin{equation}
V_\mu \otimes V_\nu = \bigoplus_\lambda  V_\lambda^{\oplus c^{\lambda}_{\mu \nu}}
\end{equation}
where $V_\mu$ are representations of $GL_n$.
Usually the basis elements of the (infinite dimensional) vector space of irreducible representations of $GL_n$ are labelled by partitions. I have two questions:

Is there a basis (with basis elements labelled by $i, j, k, \cdots$) that "diagonalizes" the Littlewood-Richardson coefficients? In other words, $c^{i}_{jk} = 0$ unless $i=j=k$?
If so, is there an elegant way of relating such a basis to the basis labelled by partitions?

REPLY [7 votes]: If a diagonal basis existed, tensoring with a fixed representation would kill all but finitely many basis elements. This is not the case because e.g. tensoring with the $1$-dimensional trivial representation doesn't kill anything.

REPLY [5 votes]: The vector space spanned by the irreps of $G=GL_n$ can be identified, by means of the character, with the vector space of $G$-invariant algebraic functions on $G$, for the adjoint action.
If you disregard the distinction between various kinds of functions (algebraic functions, smooth functions, distributions,...), then the Dirac delta functions at the various points of $G/G_{ad}$ can be thought of as a basis of this vector space.<|endoftext|>
TITLE: Characteristically simple locally compact abelian groups
QUESTION [5 upvotes]: Say a topological group $G$ is topologically characteristically simple if there does not exist a closed subgroup $1 < K < G$ such that $K$ is invariant under all automorphisms of $G$ (here `automorphisms' are required to preserve the topology as well as the group structure).
Q1: Has someone written down a classification of locally compact second-countable abelian groups that are topologically characteristically simple?
I tried to derive a classification myself; the case I am having difficulty with is when $G$ is torsion-free and the group $pG$ of $p$-th powers in $G$ is a proper dense subgroup of $G$.  (Another possible division of this case is between those $G$ with a dense divisible subgroup, and those with no divisible subgroup, but I don't know where to go from there.)
I know one such group for each $p$: let $G$ be the group of all functions from $\mathbb{N}$ to $\mathbb{Q}_p$ under pointwise addition such that all but finitely many values are in $\mathbb{Z}_p$, and topologise it so that the group of functions from $\mathbb{N}$ to $\mathbb{Z}_p$ is an open subgroup with the compact-open topology.  This example is topologically characteristically simple with a dense divisible subgroup (the finitely supported functions).
Q2: Are there any other locally compact second-countable abelian groups $G$ such that $G$ is torsion-free, $pG$ is a proper dense subgroup (edit: and $x^{p^n} \rightarrow 1$ for all $x \in G$)?  If so, are any of them topologically characteristically simple?

REPLY [5 votes]: I managed to answer my own question a few years later, here:
https://www.degruyter.com/document/doi/10.1515/jgth-2020-0107/html
C. Reid, A classification of the abelian minimal closed normal subgroups of locally compact second-countable groups. DOI: doi.org/10.1515/jgth-2020-0107
The summary is that the locally compact second-countable abelian groups that are topologically characteristically simple are "the obvious ones"; in particular, the only one locally isomorphic to $\mathbb{Z}^{\mathbb{N}}_p$ is the one I described in the original question.<|endoftext|>
TITLE: How to understand Taubes' moduli space of holomorphic curves?
QUESTION [5 upvotes]: Let $(X, \omega)$ be a closed symplectic 4-manifold. Let $\mathcal{C}=(C_i, mi_i)$ be a holomorphic current in $X$, where $C_i$ is a somewhere injective  $J$-holomorphic curve in $X$ and $m_i$ is positive integer.  Then we can define the ECH index of $\mathcal{C}$ as follow:
$$I(\mathcal{C})= \langle c_1(TX), \mathcal{C}\rangle + \mathcal{C} \cdot \mathcal{C}.$$
This integer comes from  the dimension of the moduli space of solutions of he Seiberg Witten equation. 
Taubes defines his Gromov invariant by counting $I=0$ holomorphic currents. Fix $A \in H_2(X)$, let $\mathcal{M}(X, \omega)=\{\mathcal{C}=(C_i,m_i) : I(\mathcal{C})=0, [\mathcal{C}]=A\}$ to be moduli space of holomorphic currents with $I=0$. Here $I$ only depends on the homology class $A$. If a multiple cover arises, i.e. $m_i>1$ for some $i$, then $I$ doesn't involve any information about the holomorphic map.  In contrast to  the usual moduli space of holomorphic curves (considered as holomorphic maps), this moduli space is quite strange; it only consists of currents. 

Does make sense to talk about the transversality  of  $\mathcal{M}(X, \omega)$, or does    $\mathcal{M}(X, \omega)$   admit a virtual cycle structure?

REPLY [4 votes]: This is not Taubes' moduli space. Taubes has many more constraints on the currents, so that his "moduli space" is really a finite set of points (for generic $J$), and he requires special weightings on the multiply-covered curves (which are necessarily unbranched covers of tori due to his constraints) to get a well-defined count from his set of currents. He can also do this when $I>0$, for which he chops down the set of currents by requiring them to pass through a given set of $I/2$ points in $X$.
Transversality concerns the cokernel of the deformation operators of these currents. When the curves are embedded, you have the ordinary deformation operator for which it has no cokernel when $J$ is generic. For multiply-covered curves, you can pull-back the deformation operator of the underlying embedded curve and analyze the (co)kernels. But you need to be careful about branch points, as this will destroy the equality between virtual dimension and actual dimension of the moduli space.
The point is, a moduli of curves (in general) is usually not a manifold when multiply-covered curves exist, so Taubes abandons that and restricts to embedded curves, but he needs some information about multiple covers (namely the multiplicity of covers of certain holomorphic tori). Now, there is a notion of convergence for currents, and a Gromov compactness statement, and Taubes uses this to argue that his "admissible set of currents" is a finite set (for generic $J$).<|endoftext|>
TITLE: A centralised website for computational attempts in graph theory and metric geometry?
QUESTION [6 upvotes]: The set of questions below stems from this question. 

1) does a website exist that contains (at least links to) code and data files, with the aim to centralise computational results in graph theory (or more generally metric geometry), both definite ones and unsuccessful/partial ones?

(Note that the Open Problem Garden website only lists statements and some recent references that are usually not of a computational flavor. The idea is to avoid wasting time repeating an ultimately unsuccessful approach.).

2) Are there examples of such type of website in other areas of mathematics?

REPLY [11 votes]: There are a few websites with lists and/or databases of graphs, maps and polytopes.

House of graphs has a searchable database of interesting graphs and aims to serve as a repository for lists of graphs and graph generators,
Encyclopedia of graphs is an online encyclopedia of graph collections with some data about different families of graphs,
Database of graphs in combinatorica format,
Encyclopedia of Finite Graphs is a "set of tools and data to compute all known invariants for simple connected graphs".

There are several censuses of symmetric objects (for example Marston Conder, Dimitri Leemans, Primož Potočnik).
I would also like to know about other websites of similar nature out there, since I'm working on one myself. It's still an early stage project, but has some use already. It aims to one day be able to provide a platform for computational attempts for all sorts of discrete objects (you can see some features for the exploratory aspect on the website).<|endoftext|>
TITLE: Applications of space filling curves
QUESTION [14 upvotes]: I am seeking articles where a space filling curve has been used as a theoretical application, such as in the study of general orthogonal polynomials.

REPLY [2 votes]: In a similar manner to the way they are used by Google maps, space filling curves are used for load balancing super computers. Since computational domains are sometimes more refined in certain locations, space filling curves can catch this (by refining the curve as needed). Another advantage of using these curves for load balancing is that communication between computation units is usually better (though this is only a heuristic).<|endoftext|>
TITLE: How many convex shapes can be made with the pieces of the Stomachion?
QUESTION [5 upvotes]: Tangrams are a well-known dissection of the square into seven convex polygons. One fun mathematical question is: how many convex rearrangements of the seven pieces are there?
Answer: there are 12 more. This is a theorem of Fu Traing Wang and Chuan-Chih Hsiung from 1942. 
The Stomachion is a dissection of the square into fourteen pieces, apparently studied by Archimedes. In how many ways can these pieces be reassembled into a convex polygon?
There are at least two versions of this: how many different convex shapes, and then how many rearrangements of the pieces for each shape. (People think that Archimedes studied this second question for the square.)
I'm also interested in the same questions for the Stomach, a closely related 11-piece dissection.

REPLY [8 votes]: The Logelium page of Bernd Karl Rennhak claims that there are 637 convex polygons that can be formed. There are linked pages with pictures of the different types of polygons and the numbers of solutions.
edit: This is not a complete answer to your question. The ones depicted in the pages linked to above are built from the "simplified" tiles of the Stomach puzzle and furthermore satisfy the restriction that the tiles "conform to the base grid". I couldn't find a precise definition but this is described roughly at the bottom of this page. 
It seems that Rennhak found these solutions via his software which is capable of finding solutions to tiling problems that fit on an integer grid.  See these pages for some more details.
In any case it would probably be worth writing him if you are interested in more of the missing details.<|endoftext|>
TITLE: an algebraic variety for a boolean circuit
QUESTION [5 upvotes]: There is a polynomial reduction from a $3-CNF$ $SAT$ problem to some system of polynomial equations over $\mathbb{F}_2$.   
I mean there is  polynomial reduction $F$ such that for every boolean circuit $g(x_1,\ldots, x_n)$ the corresponding system of polynomial equations $F(g) = \begin{cases} f_1(x_1, \ldots, x_n)=0 \\ \ldots \\ f_m(x_1, \ldots, x_n)=0 \end{cases} $ 
satisfies the following property: $ \forall (x_1, \ldots, x_n) \in \mathbb{F}_2^n: g(x_1, \ldots, x_n)=1 \Leftrightarrow (x_1,\ldots, x_n)$ is the solution of $F(g)$. 
So, every boolean circuit corresponds to an algebraic set. To use power of algebraic geometry we wish  this set has some nice properties. Is there a polynomial reduction such that every circuit corresponds to an absolute irreducible algebraic set?
UPD: in fact it is also interesting to find answers on weaker questions:
is there a reduction to an algebraic variety with poly$(n)$ equations, can this reduction be calculated by using poly$(n)$ memory and so on.

REPLY [5 votes]: Are you willing to allow adding additional coordinate variables $y_1,\ldots,y_m$ that are unused in $g$?  (I mean that, for any $x_1,\ldots,x_n,y_1,\ldots,y_m \in \mathbb{F}_2$, the variety will have the $\mathbb{F}_2$-point $(x_1,\ldots,x_n,y_1,\ldots,y_m)$ iff $g(x_1,\ldots,x_n)=1$.  Note that I am asserting the equivalence for an abitrary value of the $y_j$, not just with the existence of such a value: so it is enough if I have an unlimited supply of unused variables to draw from.)
If so, the following can be done: find $f_1,\ldots,f_m \in \mathbb{F}_2[x_1,\ldots,x_n]$ as you described.  We can assume that the $f_j$ have degree at most $1$ in each variable and are linearly independent over $\mathbb{F}_2$.  Define $\widehat{f_j} \in \mathbb{F}_2[x_1,\ldots,x_n,y_1,\ldots,y_m]$ by $\widehat{f_j} := y_j^2 - y_j - f_j$ (the minus signs are purely decorative here, of course, and are merely there to emphasize that this is an Artin-Schreier equation).  Since for $z\in\mathbb{F}_2$ the equation $y^2 - y - z = 0$ in the unknown $y$ over $\mathbb{F}_2$ has a solution iff $z=0$, in which case it has both $y=0$ and $y=1$, the variety $V := \{(\forall j) \widehat{f_j}=0\}$ defined by the $\widehat{f_j}$ has the property I stated above.  It remains to determine whether it is absolutely irreducible.
Now the morphism from $V$ to $\mathbb{A}^n$ consisting of projecting to the first $n$ coordinates is étale.  This is simply by the Jacobian criterion: $\frac{\partial}{\partial y_j}\widehat{f_{j'}} = \delta_{j,j'}$ (with $\delta_{j,j}=1$ and $\delta_{j,j'}=0$ if $j\neq j'$, of course).  See, e.g., Milne's Lectures on Étale Cohomology, prop. 2.1.  In particular, $V$ is smooth (or just use the Jacobian criterion directly for smoothness), so to show that it is geometrically integral, it is equivalent to show that it is geometrically connected (see, e.g., here).
Now this in turn is equivalent to the field extension $L = K(y_1,\ldots,y_m)$ of $K := k(x_1,\ldots,x_n)$ (where $k$ is the algebraic closure of $\mathbb{F}_2$) generated by the equations $y_j^2 - y_j - f_j$ being of the right degree (namely, $2^m$).  But these equations are Artin-Schreier equations, and by Artin-Schreier theory (Lang, Algebra, chapter VI theorem 8.3), it is enough for the $f_j$ to be $\mathbb{F}_2$-linearly independent with a span in direct sum with $\wp K := \{h^2-h : h\in K\}$.  But since we assumed that the $f_j$ were $\mathbb{F}_2$-linearly independent and their span consists entirely of polynomials of degree $<2$ in each variable, which are certainly not in $\wp K$ except for zero, this is indeed the case.<|endoftext|>
TITLE: Is there a $C_c^{\infty}( \mathbb{R}^d)$ function whose Fourier transform we can explicitly write down?
QUESTION [15 upvotes]: I noticed that although $C_c^{\infty}$-functions are dense in some quite large spaces and  well understood (especially their Fourier transform) I have never encountered an explicit example of a Fourier transform $F(f)$ for $f \in C_c^{\infty}$ in my life (there are good reasons for this, as most explicit expression for $C_C^{\infty}$ look rather difficult to integrate). Does anybody know an example, where the Fourier transform can be explicitly calculated or are there just no examples?

REPLY [7 votes]: The simplest infinitely differentiable function with compact support is
$f(x)=\exp\left(-1/x(1-x)\right), \; 0<x<1$. This is the first example given in textbooks.
Its Fourier transform is 
$$F(z)=\int_0^1e^{-izx}f(x)dx.$$
Is this an "explicit" formula? In not, why is the infinite product in Terence Tao's answer "explicit"?
(I think the word "explicit" (and "analytic expression", which some people think is synonimous) are very much abused on this site. These words always require a further explanation of their meaning).
Back to $C^\infty$ functions with compact support. The inverse Fourier transform
of $\mathrm{sinc}$ is the characteristic function of an interval.
Thus the example in Tao's answer is an infinite convolution of these
characteristic functions. (Is this explicit or not?) H/"ormander, Analysis of partial differential operators, vol. I, Chap I, sect 1.3 discusses at great length how to construct $C^\infty$ functions with compact support by taking infinite convolutions of characteristic functions, and surprisingly, obtains a proof of the Denjoy-Carleman quasianalyticity theorem by elementary methods:-)<|endoftext|>
TITLE: Decidability of convex rearrangements of polygons
QUESTION [7 upvotes]: Triggered by the MO question,
"How many convex shapes can be made with the pieces of the Stomachion?," I would like to pose this question:

Q. Given $n$ polygons in a set $S$, say each with integer coordinate vertices,
  is there some algorithm to count/list the number of ways that some subset of $S$
  of these polygons could be fit together, with pairwise disjoint interiors, such that their union is a convex polygon?

The polygons may be rotated and translated in the fitting-together.
One might restrict the polygons in $S$ each to be convex. Specifying integer coordinates is meant to make the input to the question of finite length,
say, $L$ bits.
Examples from the page that j.c. found:

          


          

(Image from Bernd Karl Rennhak.)


The difficulty is that it is not immediately evident how to reduce the problem
to a finite set of possibilities to try, for the gluing-together need
not be whole-edge to whole-edge.
Maybe decidability of 1st-order theories of reals can be applied?

REPLY [10 votes]: $1)$ To answer your finiteness question - this is pretty standard.  There is a finite number of  combinatorial arrangements of tiles.  Each leads to a system of linear equations which can be solved.  Even if you are working over $\Bbb R$, you have only finitely many irrationalities over $\Bbb Q$, so it is still a finite problem. 
More precisely, suppose the polygon side lengths are presented as 
$(p_0/q_0) + (p_1/q_1)\alpha_1 + \ldots  + (p_k/q_k)\alpha_k$ 
with some $p_1,q_i \in \Bbb N$ and rationally independent $\alpha_i$. From the combinatorial arrangement, built a dual network with unknown weights corresponding to intersection lengths.  Write linear equations (side = sum of its parts in the intersections) which can be viewed as a vector equations in $\Bbb Q\langle\alpha_1,\ldots,\alpha_k\rangle$. Solve this system for non-negative variables.  Done.
$2)$ For the complexity part of your question, I am pretty sure the counting is #P-hard.  Suppose for a minute that your tiles are polyominoes.  The only convex regions they can form are rectangles.  Deciding if a set of polyominoes tiles a rectangle is known to be hard for several variations of this problem. Jed Yang proves that rectangular tileability is undecidable when tiles are allowed to be repeated.  Demaine and Demaine prove NP-compleness when the tiles are non repeated, but the rectangle is fixed.  There are many more refs I am missing some mentioned in these papers.  
Back to convex polygons.  Well, subdivide each polyomino into small tiles with irrational edge lengths and side angles so the only way they can fit each other is by forming polyominoes first.  For example, take a point slightly off-center in each polyomino square and take a triangle over an edge which you attach to the triangle on the the other side whenever two squares are adjacent. 
Now, you have a large collections of triangles and quadrilaterals.  Some pairs of small triangles can easily form a convex region; these small tilings are easy to count.  But large convex regions are all rectangles which are hard to count.<|endoftext|>
TITLE: A careful roadtrip from locally symmetric spaces to algebra
QUESTION [13 upvotes]: I'm trying to break the classification of locally riemannian symmetric spaces to little steps to make it more comprehensible (and s.t. the technical details can be verified without drowning completely). 
The first big step which I find difficult to break to concise little pieces is how to get from general riemannian locally symmetric spaces to a product of simply connected indecomposable symmetric spaces. Here's my progress so far:
Let $(M,g,\nabla)$ be a riemannian manifold with a levi-civita connection (of dimension $n$).
For every point $p \in M$ the linear reflection on the tangent space $-Id: T_p M \to T_p M$ extends to define an involution on every normal coordinate  ball $\sigma_p : N_p \to N_p$ which corresponds to reversing the geodesics through $p$.

Definition: A locally symmetric space is a riemannian manifold $(M,g)$ with the property that for every $p \in M$ the geodesic involution $\sigma_p :N_p \to N_p$ is an isometry (or equivalently affine with respect to the connection $\nabla$).

Here's how I understand the steps so far:

$M$ and $\nabla$ are analytic - Any manifold with a connection $\nabla$ for which both torsion and curvature are parallel is analytic and $\nabla$ is analytic. Kobayashi & Nomizu VI 7.7
Assume $M$ is complete - As far as I understand we can't make progress without this. 
Universal cover -  The universal cover $\pi : \tilde M \to M$ inherits a metric structure for which it is an isometry implying $\tilde M$ is again locally symmetric. We assume from now on that $M$ is simply connected.
Involutions extend to global isometries (?) - Kobayashi and Nomizu IV 6.3 - "Let $M \supset U \to N$ be an isometric immersion of connected analytic riemannian manifolds with $M$ complete and $N$ simply connected then $f$ extends to an isometric immersion $f: M \to N$". So the isometry group $I(M)$ acts transitively and $M = I(M)/K$ where $K$ is a compact stabilizer group. Furthermore $K \subset O(T_p M)$ at every point.
Holonomy is subgroup of stabilizer  - Transvections $T^{\gamma}_t := \sigma_{\gamma(t/2)} \circ \sigma_{\gamma(0)}$ along geodesics $\gamma : I \to M$ form one parameter families of isometries (fixing correspondingly the points $\gamma(0)$). Every curve is $C^1$ limit of piecewise geodesic curves. Let $c : I \to M$ be a closed curve which is a limit of closed polygon geodesics. Transvections must fix the edges of the polygons and therefore define in the limit a parallel translation. $K$ Is compact in $I(M)$ in $I(M)$ (being a stabilizer subgroup). Therefore $Hol_p \subset K$.  
De Rahm decomposition theorem (?) - There's a question about this on the site but it hasn't received very useful answers. I'd like to know where to find the statement and the proof of this theorem in the most modern language.  

Adding everything together we have a product decomposition $I(M_0)/H_0 \times \dots \times I(M_k)/H_k$ where $I(M_j)$ acts irreducibly. From here on it's pretty much an algebraic road I think.

Is there something substantial missing/wrong in the above outline?

REPLY [15 votes]: Let me outline a different approach, namely the one É. Cartan himself took, one that doesn't depend on assuming completeness, doesn't rely on ideas about real-analyticity, and side-steps many of the issues that you would have to deal with in fleshing out the outline above.
So suppose that $(M^n,g)$ is a locally symmetric Riemannian manifold, which means that each $p\in M$ has an open neighborhood $U$ that supports a $g$-isometry $\iota_p:(U,g)\to (U,g)$ that fixes $p$ and is such that $(\iota_p)'(p):T_pU\to T_pU$ is minus the identity.
Let $B\to M$ denote the $\mathrm{O}(n)$-bundle of $g$-orthonormal coframes $u:T_pM\to\mathbb{R}^n$.  Let $\omega:TB\to\mathbb{R}^n$ denote the canonical $1$-form and let $\theta:TB\to{\frak{so}}(n)$ be the unique $1$-form with values in skew-symmetric matrices that satisfies the first structure equation
$$
\mathrm{d}\omega = -\theta\wedge\omega.\tag1
$$
This $\theta$ satisfies the second structure equation
$$
\mathrm{d}\theta = -\theta\wedge\theta + R(\omega\wedge\omega)\tag2
$$
where $R:B\to K_n\subset \mathrm{Hom}\bigl({\frak{so}}(n),{\frak{so}}(n)\bigr)$ is the Riemann curvature function, where $K_n \subset \mathrm{Hom}\bigl({\frak{so}}(n),{\frak{so}}(n)\bigr)$ is the subspace of curvature tensors that satisfy the (first) Bianchi identity. Of course, $R$ is $\mathrm{O}(n)$-equivariant and satisfies the third structure equation
$$
\mathrm{d}R = -\theta . R + R'(\omega)\tag3
$$
where $R':B\to \mathrm{Hom}\bigl(\mathbb{R}^n, K_n)$ is the curvature function that represents the first covariant derivative of the Riemann curvature tensor.
Now, the hypothesis that $(M,g)$ is locally symmetric implies that $R'$ vanishes, so let's assume this. (It's not immediately obvious that the converse holds, but it does.)  Since $\mathrm{d}R = -\theta . R$, it follows that $R$ takes values in a single $\mathrm{O}(n)$-orbit in $K_n$ and, in fact, is a submersion of $B$ onto this orbit.  
Let $r\in K_n$ be an element of this orbit, let $H\subset \mathrm{O}(n)$ be its stabilizer, and let $P = R^{-1}(r)\subset B$.  Then $P$ is a principal right $H$-bundle over $M$, and, pulling everything back to $P$, the equation $(1)$ continues to hold, the equation $(2)$ becomes
$$
\mathrm{d}\theta = -\theta\wedge\theta + r(\omega\wedge\omega)\tag{2'}
$$
and $(3)$ becomes 
$$
0 = -\theta.r\tag{3'}.
$$
In other words, $\theta$ (pulled back to $P$) takes values in ${\frak{h}}\subset{\frak{so}}(n)$, the Lie algebra of $H\subset\mathrm{O}(n)$.
Because $r:{\frak{so}}(n)\to {\frak{so}}(n)$ is symmetric (thanks to the first Bianchi identity), its kernel is equal to its image, say $\frak{r}\subset {\frak{so}}(n)$.  Differentiating $(3')$ and using $(2')$ shows that
$$
0 = \bigl(r(\omega\wedge\omega)\bigr).r,
$$
implying that $\frak{r}\subseteq \frak{h}$.  In particular, the right hand side of $(2')$ is a $2$-form that takes values in $\frak{h}$.
At this point, we have a coframing 
$$
\gamma = (\omega,\theta):TP\to \mathbb{R}^n\oplus\frak{h} = \frak{g}
$$
satisfying structure equations $(1)$ and $(2')$ where the $2$-forms on the right hand sides of these equations have constant coefficients in the coframing.  Since $\mathrm{d}^2 =0$ is an identity, it follows that there is a unique Lie algebra structure on $\frak{g}$ such that $(1)$ and $(2')$ are equivalent to 
$$
\mathrm{d}\gamma = -\tfrac12[\gamma,\gamma].
$$
The fact that the equations $(1)$ and $(2')$ would be unaffected by replacing $\omega$ by $-\omega$ implies that the automorphism of $\frak{g}$ that is minus the identity on $\mathbb{R}^n$ and the identity on $\frak{h}$ is an automorphism of the Lie algebra, i.e., the pair $(\frak{g},\frak{h})$ is an orthogonal symmetric pair.
For the converse, you need the existence of a Lie group with a given Lie algebra, etc.
N.B.: The fact that $a\in \frak{h}$ satisfies $a.r = 0$ implies the statement that 
$$
r\bigl([a,y]\bigr) = \bigl[a,r(y)\bigr]
$$
for all $a\in \frak{h}$, so $\frak{r}$ is an ideal in $\frak{h}$.  In fact, one usually (but not always) has $\frak{r}=\frak{h}$, but we don't need this to complete the argument.  
N.B.: Note that $\frak{g}$ is the Lie algebra of the maximal symmetry group of $(M,g)$, while $\frak{u} = \mathbb{R}^n\oplus\frak{r}$ is the Lie algebra of the 'minimal' symmetric representation of $(M,g)$.<|endoftext|>
TITLE: Probability theory without deductive closure
QUESTION [9 upvotes]: Human knowledge is not deductively closed.  Uncertainty can arise from that just as much as from lack of brute facts.  (When a Harvard graduate was reported to have thought that the earth is farther from the sun in the winter than in the summer and that that's why seasons happen, I commented to someone that that was stupid because he ought to have known that when it's winter in the northern hemisphere, it's summer in the southern hemisphere and vice-versa.  I was told that I was making the mistake of expected a person's knowledge to be deductively closed.  I don't altogether agree, since those who make fun of Sarah Palin for (allegedly) not knowing that Africa is a continent were not really expecting all humans to be omniscient.; it's similar to that.)
But the conventional mathematical theory of probability is deductively closed.  As a means of expressing uncertainty, its closure can be useful because in telling us how much uncertainty is justified, it can make us aware of logical connections we might have missed.
However, let's suppose you're wondering if some number with an almost unimaginably large number of digits is prime.  (Say it's $5963$, which, unbeknownst to all humans, can be factored as $67\times89$.)  After many years of work with supercomputers, you've ruled out the possibility of its divisibility by all primes up to and including $17$, which is the largest prime not exceeding its cube root.  If I'm not mistaken, there are ways to estimate the sum of primes bigger than that but not bigger than the square root of the number to be factored (In the present example, $1/19 + 1/23 + 1/29 + 1/31 + 1/37 + 1/41 + 1/43 + 1/47 + 1/53 + 1/59 + 1/61 +  1/67 + 1/71 + 1/73 \approx 0.354$.)  It seems reasonable to consider this a sort of upper bound on the probability that the large number considered is composite.  But in a deductively closed system, that probability is $0$ or $1$, and if it's $1$, then $0.354$ is not an upper bound on it.
Is there any mathematically precise theory of probability without deductive closure?

REPLY [2 votes]: You might like Les Valiant's book "Probably Approximately Correct" which is an approach to explaining human knowledge that's also a busy topic in machine learning.  It goes beyond lack of deductive closure to seeing how learning happens without much deduction at all.  There's a related Wikipedia article: PAC learning.<|endoftext|>
TITLE: Nonexistence of generic objects over $L(\mathbb{R})$
QUESTION [9 upvotes]: A well known result (stated and credited to Todorcevic in "Semiselective Coideals", by Farah, Mathematika, 1997, but with antecedents going back to Mathias) says that, under the appropriate large cardinal hypothesis (enough to get all sets of reals in $L(\mathbb{R})$ to be universally Baire, say), a selective ultrafilter is $L(\mathbb{R})$-generic for $([\omega]^\omega,\subseteq^*)$.
It is also well-known that selective ultrafilters need not exist; Kunen showed that they are destroyed by iterating random forcing over a model of CH. More generally, Miller showed that $Q$-points are destroyed by iterating Laver (or Mathias) forcing, and Shelah produced a model without $P$-points.
Here's my (admittedly broad) question: 

Let $\mathbb{P}$ be a nontrivial (say, infinite, separative) $\sigma$-closed notion of forcing which is in
  $L(\mathbb{R})$, by which I mean the underlying set, its elements, and
  its order are all in $L(\mathbb{R})$. Suppose that under CH
  one can define an ultrafilter $G$ in $\mathbb{P}$ which is generic
  over $L(\mathbb{R})$ (under suitable large cardinal hypothesis). Is
  there a general theorem which tells us that such a $G$ consistently
  does not exist?

REPLY [6 votes]: The book draft linked to below shows that existence of a weakly compact Woodin cardinal implies the existence of $L(\mathbb{R})$-generic filters for the following partial orders (all ordered by containment) : (1) the partial order of countable injections from $\mathbb{R}$ to $\mathbb{R}$; (2) the partial order of countable partial selectors for a countable Borel equivalence relation; (3) the partial order of countable linearly independent sets in a Polish vector space over a countable field of scalars. The first example is Example 6.2.13 and the corresponding fact is Theorem 12.2.3. Examples (2) and (3) are discussed after Definition 6.2.9 (which they satisfy) and the corresponding theorem is Theorem 12.2.5. It should be possible to significantly increase the class of examples. 
Here is the link : https://people.clas.ufl.edu/zapletal/files/balanced8.pdf<|endoftext|>
TITLE: Does anyone recognize this exponential sum?
QUESTION [5 upvotes]: For $a$, $b$ two integers, let $(a,b)$ denotes their gcd. We define the following exponential sum :
$$G_q(n):=\sum_{d|q,~(d,q/d)=1}{e^{2i\pi n\frac{dd'}{q}}}$$ 
for $n$ a non-negative integer and $q$ a positive integer ($d'$ denotes the inverse of $d \pmod{q/d}$).
Does anyone recognize this sum?
Many thanks!

REPLY [10 votes]: Yes, these sums occur as the arithmetic part of the Fourier expansion of
period kernels $\sum_{ad-bc=1}(a\tau+b)^{-k}(c\tau+d)^{-k}$, the analytic part
being J-Bessel functions. The derivation is not difficult. I can send you
the paper where I compute them if you want (from 1980), private e-mail
please.<|endoftext|>
TITLE: Mathematicians with aphantasia (inability to visualize things in one's mind)
QUESTION [44 upvotes]: Are there any mathematicians with aphantasia? If so, could they please elaborate upon what their experience with mathematics is like?
I realize that this question probably falls outside of the scope of Mathoverflow, but it's so shocking that such a fundamental mental difference exists that I think the question is worth asking here anyways. Even if it gets closed, which I suspect it will, if even one mathematician with aphantasia sees this and has the startling revelation that they have aphantasia, I'll be 1000000% glad I posted the question.
*inability to visualize things in one's mind. see this note that went viral recently for a more detailed explanation: https://www.facebook.com/notes/blake-ross/aphantasia-how-it-feels-to-be-blind-in-your-mind/10156834777480504

REPLY [3 votes]: I do not think I have aphantasia, but I want to point out that "mental imagery" is very different from an image that you are directly looking at.  Consider a map of the world.  If you ask me to mentally visualize a map of the world, I believe I can do so, but if you then ask me, "Does China share a border with Kyrgyzstan?" I can't answer that question by just "looking at that part of the map" and getting the answer, which I certainly could do if I were looking at an actual (political) map of the world.  My geographical knowledge of that part of the world is somewhat fuzzy and so my mental image is fuzzy, but it's not fuzzy in the way that an actual blurred visual image is fuzzy.  Furthermore, even when I fancy that I can mentally "see" a border, such as the border between the U.S.A. and Canada, you can easily ask me difficult questions that I can't answer using my "mental image" that I could trivially answer by looking at an actual map.
I have also noticed many people talking about "visualizing" 3-dimensional space, but surely this is another example of a clear difference between mental imagery and actual visual images.  At best, we literally see only two spatial dimensions.  Three-dimensional space is something we conceptually construct, not something we see directly. I'd go further and say that even our mental "images" of two-dimensional projections of three-dimensional objects are quite different from actual two-dimensional images. Take for example the old puzzle where we take a square-based pyramid with equilateral triangles as faces and we glue a regular tetrahedron to one of the faces of the pyramid.  I think I can form a mental image of this object, but suppose you ask me whether it has 7 faces or 5 faces.  With the actual object, if you were unsure, you could just pick it up, look at it from an appropriate angle, and see that the faces are flush.  But with a mental image, if you are unsure, you probably cannot solve the problem just by "looking."  You probably have to think about it and work it out.
All this is to say that I have doubts about how much the ability to conjure up "mental imagery" helps mathematical thinking.  For those with the ability, I am sure it helps to some extent, but I suspect that it helps less than it might seem, and that much of what we attribute to our ability to form mental images is actually a far more complex process than (ahem) meets the eye.<|endoftext|>
TITLE: Nelson's proof of Liouville's theorem
QUESTION [29 upvotes]: The paper "A proof of Liouville's theorem" by E. Nelson, published in 1961 in Proceedings of AMS, contains just one paragraph, giving a (now) standard proof that every bounded harmonic function in $\mathbb{R}^n$ is a constant.
I presume there must be a story behind. First, it is hard to imagine that this proof was unknown before 1961. Second, even if this is the case, it doesn't feel usual, for the author, to submit such a paper and, for the editor, to accept it.
So, can anyone tell that story? Or, to make the question precise:
1) are there any earlier references for this proof?
2) what was/were the standard proof(s) before 1961? 
3) by a very similar reasoning, one obtains $̣||\nabla h||_{\infty,\Omega}\leq C(\Omega,\Omega')||h||_{\infty, \Omega'}$ for $\Omega\subset\Omega'$. Was that argument also unknown until 1961?

REPLY [10 votes]: This proof was new to me when I read it:-) The standard proof, which I teach, and which is given in most books uses Harnack's inequality, which follows from Poisson's formula for the ball, or Poisson's formula directly.
If I were the editor or a referee, I would accept this paper.<|endoftext|>
TITLE: "Identity tensor transpose" as a map $M_n \hat{\otimes} M_n \to M_n \overline{\otimes} M_n$
QUESTION [6 upvotes]: Equipping $M_n$ with its usual operator space structure,
$\newcommand{\ptp}{\widehat{\otimes}}$
 we can form the projective tensor product of operator spaces $M_n\ptp M_n$. In particular this puts a Banach space norm on the algebraic tensor product $M_n\otimes M_n$.
Now consider the transpose map $T:M_n \to M_n$. It is a standard calculation to show that $T$ is not completely bounded,
$\newcommand{\stp}{\overline{\otimes}}$
 and in particular one can show that the map $\iota \otimes T : M_n \stp M_N \to M_n\stp M_n$ has norm $n$, where $\stp$ denotes the spatial tensor product (in this setting the same as the injective tensor product of operator spaces).

Question 1. What is the asymptotic behaviour (as $n\to \infty$) of $\Vert \iota \otimes T :   M_n \ptp M_N \to M_n\stp M_n\Vert$?

Note that I'm asking merely about the norm as a map between two Banach spaces, not about the cb norm. If precise asymptotics are tricky, how about the following sub-question:

Question 2. In particular, does the norm of this map tend to infinity as $n\to\infty$?

This feels like something that should follow by tweaking a standard example or exercise in one of the usual books on Operator Spaces, but I couldn't succeed in converting the usual examples to get something that answers the question above.
Remark: the usual way to get a lower bound on
$\Vert\iota \otimes T : M_n \stp M_N \to M_n\stp M_n \Vert$ is to consider what this map does to the tensor $x = \sum_{i,j=1}^n E_{ij} \otimes E_{ji}$, the point being that $x$ has norm $1$ when viewed as an element of $M_{n^2}$ while $(\iota\otimes T)(x)$ has norm $n$ as an element of $M_{n^2}$. However, since matrix multiplication gives a complete contraction $M_n \ptp M_n \to M_n$, I think it can be shown that $x$ has norm $n$ as an element of $M_n\ptp M_n$.

Update 2016-05-04: I think I've now found a proof that this map is contractive for all $n$,
which moreover works if you replace proj tp with Haagerup tp. Previously I thought that this stronger claim (with the Haagerup tp) was false by adapting the usual argument to show the claim fails for min tp; however, this was based on a stupid miscalculation.
based on an interpolation argument. If the details work then I'll leave them as an answer.

REPLY [3 votes]: Well I guess I should write something quickly, even if it doesn't have all the details, otherwise I'll keep putting it off. And maybe someone will spot a mistake...
Fix Hilbert spaces $V$ and $W$, which we think of as having column OSS. Equip $B(V)$ and $B(W)$ with their usual OS structures. Then the linear map 
$$ \iota\otimes \top : B(V) \otimes B(W) \to B(V \otimes_2 W)$$
extends to a contractive linear map $B(V) \ptp B(W) \to B(V\otimes_2 W)$.
The proof goes in stages.
Step 1. If $x = \sum_i a_i \otimes b_i \in B(V)\otimes B(W)$, then 
$$ \Vert \sum\nolimits_i a_i \otimes b_i^\top \Vert_{B(V\otimes_2 W)}
 = \sup \left\{ \Vert \sum\nolimits_i a_i c b_i \Vert_{HS(W,V)}
\;\colon\; c \in HS(W,V), \Vert c\Vert_{HS(W,V)} \leq 1 \right\} $$
(This is not hard to hack out by hand, but with hindsight can also be found in various places, for instance I think it is in Pisier's book somewhere early on.)
Step 2. Note that there are have natural completely contractive maps$\newcommand{\ptp}{\widehat{\otimes}}$
$$ B(V) \ptp V \to V\quad,\quad W^* \ptp B(W) \to W^* $$
where the first is the usual action $a\otimes v \mapsto av$ and the second is the transposed action $\psi\otimes b \mapsto \psi\circ L_b$, $L_b$ being the action $w\mapsto bw$. Therefore by general stuff on operator space tensor products, we have complete contractions$\newcommand{\itp}{\otimes_{\rm min}}$
$$ B(V) \ptp( V\ptp W^*) \ptp B(W) \to V\ptp W^* $$
$$ B(V) \ptp( V\itp W^*) \ptp B(W) \to V\itp W^* $$
where $\itp$ denotes injective tensor product of operator spaces.
Note that if we identify $V\otimes W^*$ with the space of finite rank operators $W\to V$, then the two maps above just correspond to $a\otimes c \otimes b \mapsto acb$.
Step 3. Under the natural identification of $V\otimes W^*$ with the finite-rank operators $W\to V$, we have isometric isomorphisms $V\ptp W^*\cong S_1(W,V)$ and $V\itp W^* \cong S_\infty(W,V)$, where $S_1$ denotes trace-class operators and $S_\infty$ the compact operators.
Step 4. Let $x$ be as in Step 1 and WLOG assume its norm in $B(V)\ptp B(W)$ is $1$. Let $E_x$ denote the elementary operator on $B(W,V)$ defined by $c\mapsto \sum_i a_icb_i$. We wish to show that the norm of $E_x$ as a map $HS(W,V)\to HS(W,V)$ is $\leq 1$. But now by Steps 2 and 3, we know that $E_x$ is (completely) contractive from $S_1(W,V)$ to $S_1(W,V)$, and (completely) contractive from $S_\infty(W,V)$ to $S_\infty(W,V)$. By classical complex interpolation results, $E_x$ is therefore contractive on all the intermediate Schatten classes, in particular on the Hilbert-Schmidt operators, and we are done.
Remark. Of course we can interpolate in the category of operator spaces. The argument above, if correct, seems to actually show that $E_x$ is completely contractive on $HS(W,V)$ when we equip this space with the "self-dual" OSS. If I denote this operator space by OH temporarily, then we can rephrase this as: $\iota\otimes \top$ is contractive from $B(V)\ptp B(W)$ to $CB(OH)$. It seems plausible that we actually get a complete contraction, but I haven't yet done the book-keeping required to check this.<|endoftext|>
TITLE: Adapting arguments and plagiarism
QUESTION [24 upvotes]: I'm currently working on my PhD thesis. I have several suggested problems to work on, some of them are very similar to some problems that my advisor have worked before and published already, either in his thesis or papers. Basically, the main difference is in the dimension of some singular sets (his works are mainly on the isolated case, but I'm working on a case with a far more hairier, non-isolated singular set), which we were unsure if the argument would hold but it seems that the adaptations I've made were fine.
Not that the nature of the problem matters, but the approach I'm making worries me. It seem to me that if there would be a 'railroad' to prove the results I'm working on, it would be the same path that he followed to write his own results, with different objects. That's the way I've been advised to work, and it's been producing results.
Is this a reproachable approach? Of course, there is the problem of using similar introductions (and in that subject I've read this previous question Does this qualify as "self plagiarism" or something? , only one that got close to my problem) sinde the objects being studied by me and that has been studied by him were so similar.

REPLY [8 votes]: One major aim of a PhD is to study a problem so intently that suddenly, after much hard work and perseverance, the solution becomes obvious - or at least that it becomes obvious what the approaches to the problem would be and why one of these approaches is likely to be superior to the other approaches. Be aware that everyone has good ideas - and groundbreaking approaches tend to build on earlier approaches. My PhD was "simply" the combination of three existing approaches in a novel way. But to get to the point where it was obvious that these three things together were what was necessary or would produce a solution took 2 1/2 years' work.<|endoftext|>
TITLE: moving up a consequence of PFA
QUESTION [8 upvotes]: The Proper Forcing Axiom (PFA) implies that every forcing which adds a subset of $\omega_1$ either adds a real or collapses $\omega_2$. Is it consistent that every forcing which adds a subset of $\omega_2$ either adds a subset of $\omega_1$ or collapses $\omega_3$?

REPLY [7 votes]: Yes, it is consistent. The following theorem is proved by Shelah and myself (our  paper  is not yet complete):

Theorem. Assume $\kappa$ is weakly compact and $\lambda > \kappa$ is 2-Mahlo. Then there is a generic extension in which $\kappa=\aleph_2, 2^{\aleph_1}=\aleph_3=\lambda$ and any forcing notion which adds new subset to $\aleph_2$,  collapses $\aleph_2$ or $\aleph_3.$

Unfortunately the proof is not so easy to state it here.
Note that by a new subset to $\aleph_2$ I mean a subset of $\aleph_2$ which is not in $V$, the ground model, but such that all its initial segments are in $V$.

Remark. It is evident that the above theorem gives the consistency of the requested question. If the forcing does not add subsets to $\aleph_1,$ then it adds new subsets to $\aleph_2$ so the theorem applies.


Added note: The paper is now available at Specializing trees and answer to a question of Williams.<|endoftext|>
TITLE: Fréchet L-Spaces
QUESTION [5 upvotes]: According to the paper The emergence of open sets, closed sets, and limit points
in analysis and topology famous mathematician Maurice Fréchet who introduced the concept of metric spaces has also introduced another similar class of abstract spaces  called Limit spaces based on the primitive idea of the limit of an infinite sequence in 1904, which was defined as follows:

An L-space is a set $X$ together with a function $F : S\to X,$ where $S$ is a set of infinite sequences of members of $X$.

If $(x_n)\in S$, then $F(x_n)$ was said to be the “limit of the sequence $(x_n)$" satisfying following two axioms:

$A_1$: If $(x_n)$ is a constant sequence whose value is $a$, then $F(x_n)=a$.
$A_2$: If $F(x_n)=a$, then for any sub-sequence of $(x_n)$ given by $(x_{n_k})$ we have $F(x_{n_k})=a$.

I would like to know more about mathematics on L-spaces. But I could not find any thing by just Googling.
Where could I find about these spaces?

REPLY [6 votes]: I think that these spaces don't go under the name of $L$ spaces anymore. Actually, I am not sure if there is a consensus on how these structures are called today.
A good place to start is the fairly recent book

Convergence Structures and Applications to Functional
Analysis  by R. Beattie and H.-P. Butzmann, Kluwer, 2002

where these structures are named convergence spaces, more precisely sequential convergence spaces and are studied in chapter 1.7.
Another, slightly older, reference is the article (in French)

Convergence du type $\mathcal{L}$, Jan Kisyński, Colloquium
Mathematicae, 1960

but I haven't read that one.<|endoftext|>
TITLE: Definitions of Hilbert Bundles
QUESTION [8 upvotes]: I have some doubts regarding definitions and conventions on Hilbert Bundles. Some authors like Peter Kuchment (Floquet Theory for Partial Differential Equations) and Serge Lang (Differential and Riemannian Manifolds) use the usual definition of a vector bundles to define Banach Bundles. As such, they usually do not need to worry about measurability issues. Due to presence of the locally trivializing map in their construction, given a connected base space, all the fibers are isomorphic to each other. Therefore, fibers with varying dimensions are not allowed. 
Other authors, like Dautray and Lions, or Birman and Solomjak define measurable Hilbert bundles and do not seem to insist on isomorphic fibers. 
My question is, in the case of isomorphic fibers, why not simply use Bochner spaces like $C(X;L^2(\Omega))$? Reed and Simon (in the fourth volume) insist that the focus is on the fibers rather than on $X$ and promise to cover general Hilbert bundles in Chapter XVI of their series "Methods of Modern Mathematical Physics", but as far as I know, that chapter never appeared.
Some other questions:

Is the "constant rank" condition necessary in order to put a differentiable (or continuous) structure on the bundle? 
In contrast, is it true that the "constant rank" condition is not imposed on measurable Hilbert bundles because measurability is a weaker condition that allows for variation of fibers?
Are "constant rank" fibrations more "natural" in some way in mathematics?

REPLY [3 votes]: "Why not simply use Bochner spaces like $C(X;L^2(\Omega))$?" --- do you mean that this would be the space of continuous sections of the bundle with fiber $L^2(\Omega)$? Yes, that is correct if the bundle is trivial, i.e., a bundle of the form $X \times H$ where $H$ is the fiber Hilbert space. But of course not all bundles have this form. A natural class of examples where the bundle is usually not trivial arises when $X$ is a Riemannian manifold and the fiber at a point is the tangent space there. Each tangent space carries an inner product, so this is a Hilbert bundle.
"Is the constant rank condition necessary in order to put a continuous structure on the bundle?" --- yes. It's hard to see how to reasonably topologize a bundle whose fibers could vary in dimension. One does sometimes consider "dimension drop" conditions where you only consider sections which, say, vanish at some point, or lie in a proper subspace of the fiber at some point.
"Is it true that the constant rank condition is not imposed on measurable Hilbert bundles because measurability is a weaker condition that allows for variation of fibers?" --- again, I basically agree. This setting is very different from the topological setting; measurable sections will be totally insensitive to any global topological features of a bundle. So one can just take a measurable Hilbert bundle over a measure space $X$ to be something of the form $\bigcup (X_n\times H_n)$, with $n$ ranging over $\mathbb{N} \cup \{\infty\}$, $(X_n)$ a measurable partition of $X$, and $H_n$ an $n$-dimensional Hilbert space. (The nonseparable setting introduces some bad pathology, so I prefer to stick to the separable case.)
While I'm on the subject, measurable Hilbert bundles provide a setting for spectral theory that nicely accomodates multiplicity:
$\bullet$ If $A$ is a (bounded or unbounded) self-adjoint operator on a Hilbert space $H$, then there is a measurable Hilbert bundle over the spectrum of $A$, and an isometric isomorphism between $H$ and the $L^2$ sections of this bundle which turns $A$ into multiplication by $x$.
$\bullet$ If $\mathcal{A} \subset B(H)$ is a separable abelian C*-algebra then there is a metrizable locally compact Hausdorff space $X$, a Borel measurable Hilbert bundle over $X$, and an isometric isomorphism between $H$ and the $L^2$ sections of the bundle which turns $\mathcal{A}$ into the set of multiplication operators by functions in $C_0(X)$.
$\bullet$ If $\mathcal{M} \subseteq B(H)$ is an abelian von Neumann algebra, then there is a metrizable locally compact Hausdorff space $X$, a Borel measurable Hilbert bundle over $X$, and an isometric isomorphism between $H$ and the $L^2$ sections of the bundle which turns $\mathcal{M}$ into the set of multiplication operators by functions in $L^\infty(X)$.
I would say that these statements cleanly exhibit the way the abstract C${}^*$- and von Neumann algebras are situated within $B(H)$.
(Details can be found in my book Measure Theory and Functional Analysis.)<|endoftext|>
TITLE: smoothing locally-finite (Borel-Moore chains)
QUESTION [5 upvotes]: Let $M$ be a smooth manifold. As is recorded in (for example) Lee's book, de Rham proved that one can calculated singular homology, $H_*(M)$ using smooth simplices. Does the result extend to Borel-Moore homology $H^{BM}_{*}(M)$? One might guess that the argument in Lee's book even proves this, but there are a lot of details there and even if it goes through, it would be good to have a reference that has already checked every line.

REPLY [3 votes]: I do not know a reference, let me give you some ideas on how to prove it:
1) First let us remark that we have an isomorphism:
$$\rho:C_*^{BM}(M)\rightarrow lim_KC_*(M,M-K)$$
between Borel-Moore chains and the limit of relative chains taken over all $K$ where $K\subset M$ is a compact subset.
2) Similarly you can build an isomorphism  
$$\rho^{sm}:C_*^{BM,sm}(M)\rightarrow lim_KC^{sm}_*(M,M-K)$$
between Borel-Moore smooth chains and the limit of relative smooth chains.
3) For any subset $U\subset M$ you have a canonical inclusion:
$$\phi:C^{sm}_*(U)\rightarrow C_*(U)$$
that induces a canonical morphism
$$\phi:lim_KC^{sm}_*(M,M-K)\rightarrow lim_KC_*(M,M-K).$$
4) For each compact set you have a quasi-isomorphism (by the 5-lemma):
$$C^{sm}_*(M,M-K)\rightarrow C_*(M,M-K)$$
the inverse limit of these morphisms is the map $\phi$. To conclude that the inverse limit is also a quasi-isomorphism you have to use homological properties of the inverse limit functor.<|endoftext|>
TITLE: Embedding property of weakly compact cardinals
QUESTION [6 upvotes]: One of the characterizations of $\kappa$ being a weakly compact cardinal is being inaccessible, and for every $\kappa$-model $M$, there is a [$\kappa$-model] $N$ and an elementary embedding $j\colon M\to N$ with critical point $\kappa$.
I am looking for a reference, or at least a proof, that this is equivalent to another property of weakly compact cardinals (which is not an embedding characterizations), for example the tree property or indescribability, or even the extension property.
However all the papers and books refer to one of two places:

Kanamori,
Cummings' Handbook article (section 16).

A diligent search through Kanamori produced nothing (the second edition, anyway). In Cummings' paper he refers to Kanamori, and only proves an implication from the embedding property to the Hauser embedding property (that we can also assume $j,M\in N$).

Does anyone knows where to find a proof for any equivalence of weak compactness? (Or if it is really folklore, what is the proof?)

It should be easy to prove some characterizations from the embedding property, but I do not quite see how to prove the embedding property from something like compactness, the tree property, indescribability, or end-extensions.
(I should also say that I scoured other standard books like Jech and the Handbook, as well newer books like Schindler's Set Theory. To non-avail.)

REPLY [13 votes]: This is a selection on from my lecture notes text, Lectures on
Forcing and Large Cardinals, which I wrote long ago and which shows
the main equivalences, including the ones you mention.
Theorem. If $\kappa^{<\kappa}=\kappa$, then the following are
equivalent.

(weak compactness property) $\kappa$ is weakly compact. That
is, $\kappa$ is uncountable and every $\kappa$-satisfiable theory
in an $L_{\kappa,\kappa}$ language of size at most $\kappa$ is
satisfiable.
(extension property) For every $A\newcommand\of{\subseteq}\of
V_\kappa$, there is a transitive structure $W$ properly extending
$V_\kappa$ and $A^*\of W$ such that $\langle
V_\kappa,{\in},A\rangle\newcommand\elesub{\prec}\elesub\langle
W,{\in},A^*\rangle$.
(tree property) $\kappa$ is inaccessible and has the tree
property.
(filter property) If $M$ is a set containing at most
$\kappa$-many subsets of $\kappa$, then there is a
$\kappa$-complete nonprincipal filter $F$ measuring every set in
$M$.
(weak embedding property) For every $A\of\kappa$ there is a
transitive set $M$ of size $\kappa$ with
    $\kappa\in M$ and a transitive set $N$ with an embedding $j:M\to N$ with critical point $\kappa$.
(embedding property) For every transitive set $M$ of size
$\kappa$ with $\kappa\in M$ there is a
      transitive set $N$ and an embedding $j:M\to N$ with  critical point $\kappa$.
(normal embedding property) For every $\kappa$-model $M$ there
is a $\kappa$-model $N$ and an embedding
      $j:M\to N$ with critical point $\kappa$, such that $N=\{j(f)(\kappa)\mid f\in M\}$.
(Hauser embedding property) For every $\kappa$-model $M$ there
is a $\kappa$-model $N$ and an embedding $j:M\to N$
      with critical point $\kappa$ such that $j,M\in N$.
(partition property) $\kappa\to(\kappa)^2_2$.

Proof: (weak compactness implies extension property) Assume
that $L_{\kappa,\kappa}$ exhibits the weak compactness property
and suppose $A\of V_\kappa$. First, we argue that $\kappa$ is
inaccessible. The regularity of $\kappa$ follows from our
assumption that
$\newcommand\ltkappa{{{<}\kappa}}\kappa^\ltkappa=\kappa$ (although
even without that assumption, it follows from the weak compactness
property). Suppose now that $2^\beta\geq\kappa$ for some
$\beta<\kappa$. Let $L$ be the language having a constant symbol
$\check\alpha$ for every $\alpha<\kappa$ and a unary predicate
symbol $U$. For any $x\of\beta$, let $\sigma_x$ be the sentence
$(\bigvee_{\alpha\in x}\neg
U(\check\alpha))\vee(\bigvee_{\alpha\in\beta\setminus x}
U(\check\alpha))$. Thus, $\sigma_x$ asserts that $U$ is different
from $x$ on the set $\{\check\alpha\mid \alpha<\beta\}$. Let $S$
be the theory consisting of all $Los\sigma_x$ for $x\of\beta$. This
theory has size $2^\beta$, but the language of $S$ has size only
$\beta$. Since $2^\beta\geq\kappa$, the theory $S$ is
$\kappa$-satisfiable, since any subtheory of size less than
$\kappa$ will omit some $\sigma_x$, and then we can interpret
$\check\alpha$ as $\alpha$ and $U$ as $x$ to build a model. But
clearly $S$ is not satisfiable, since $U$ must pick out some
pattern $x=\{\alpha<\beta\mid U(\check x)\}$ under any
interpretation. So $\kappa$ is inaccessible.
Next, we show that $\kappa$ has the extension property. Let $L$ be
the language with a constant symbol $\check a$ for every $a\in
V_\kappa$, as well as a binary relation symbol $\check\in$, an
additional constant symbol $c$ and a predicate symbol $\dot A$.
Let $R$ be the first order theory $\text{Th}(\langle
V_\kappa,{\in},A,a\rangle_{a\in V_\kappa})\cup\{c\neq\check a\mid
a\in V_\kappa\}$, together with the infinitary assertion
$\sigma=\neg\exists\vec x(\wedge x_{n+1}\in x_n)$, which asserts
that $\check\in$ is well founded. This theory is
$\kappa$-satisfiable, by interpreting $c$ in the structure
$\langle V_\kappa,{\in},A\rangle$ to be one of the $\check a$
missing from the subtheory. By the weak compactness property,
there is a model $\langle W,{\in^*},A^*\rangle$ satisfying $R$.
The relation $\in^*$ on $W$ must be well founded, since the
structure satisfies $\sigma$. By taking the Mostowski collapse, we
may assume that $W$ is a transitive set and $\in^*$ is the $\in$
relation. Furthermore, since the first order theory of $\langle
V_\kappa,\in,A\rangle$ is satisfied, it follows that $V_\kappa\of
W$ and $\langle V_\kappa,{\in},A\rangle\elesub
\langle W,{\in},A^*\rangle$, as desired.
(extension implies tree property) Assume $\kappa$ has the
extension property. We show first that $\kappa$ is inaccessible.
Regularity follows from $\kappa^\ltkappa=\kappa$ (but it also
follows directly from the extension property). If
$2^\beta\geq\kappa$ for some $\beta<\kappa$, then let $\vec
a=\langle a_\alpha\mid \alpha<\kappa\rangle$ be a
$\kappa$-sequence of distinct subsets of $\beta$. By the Extension
property, there is a transitive set $W$ and $\vec a^*$ such that
$\langle V_\kappa,{\in},\vec a\rangle\elesub\langle W,{\in},\vec
a^*\rangle$. Let
$\kappa^*=W\newcommand\intersect{\cap}\intersect\newcommand\ORD{\text{Ord}}\ORD$,
and observe that $\kappa<\kappa^*$. Let $a$ be the $\kappa^{th}$
element of the sequence $\vec a^*$. Thus, $a\of\beta$ and
consequently $a\in V_\kappa$. Since $W$ satisfies that $a$ appears
on the sequence $\vec a^*$, it follows by elementarity that $a$
also appears on $\vec a$. Thus, $a=a_\alpha$ for some
$\alpha<\kappa$, and so $a$ appears twice on $\vec a^*$, at
coordinates $\alpha$ and $\kappa$, contradicting that these were
distinct subsets of $\kappa$. So we have established that $\kappa$
is inaccessible.
Now suppose that $T$ is a $\kappa$-tree. By replacing $T$ with an
isomorphic copy, if necessary, we may assume
$T\of\kappa^\ltkappa\of V_\kappa$. By the Extension property,
there is a transitive set $W$ and a subset $T^*\of W$ with
$\langle V_\kappa,{\in},T\rangle\elesub\langle
W,{\in},T^*\rangle$. It follows that $T^*$ is a tree of height
$\kappa^*=W\intersect\ORD$. Furthermore, for any $\alpha<\kappa$,
the $\alpha^{th}$ level of $T$, which is an element of $V_\kappa$,
is by elementarity also the $\alpha^{th}$ level of $T^*$. Thus,
$T^*$ is an end-extension of $T$. Let $q\in T^*$ be any node on
the $\kappa^{th}$ level of $T^*$, and consider the set of
predecessors $b=\{p\in T^*\mid p<_{T^*} q\}$. The elements of $b$
form a linearly ordered subset of $T^*$ on the levels below
$\kappa$. Thus, $b$ is a $\kappa$-branch through $T$. So $\kappa$
has the tree property.
(tree property implies filter property) Suppose that $\kappa$ is
inaccessible and has the tree property. Suppose that
$\{A_\alpha\mid \alpha<\kappa\}$ is a collection of $\kappa$ many
subsets of $\kappa$. By enlarging the collection if necessary, let
us assume that all singletons $\{\xi\}$, for $\xi<\kappa$, appear
on the list. For each $s\in 2^\beta$, where $\beta<\kappa$, let
$A_s$ be the result of intersecting all $A_\alpha$ or the
complement $\kappa\setminus A_\alpha$, chosen according to the
values of $s(\alpha)$. That is,
$A_s=(\newcommand\Intersect{\bigcap}\intersect\{A_\alpha\mid
s(\alpha)=1\})\intersect(\Intersect\{\kappa\setminus A_\alpha\mid
s(\alpha)=0\})$. Let $T=\{s\in 2^\ltkappa\mid |A_s|=\kappa\}$.
This is clearly a tree. For any $\beta<\kappa$, we may define for
any $\gamma<\kappa$ a sequence $s_\gamma\in 2^\beta$ so that
$s_\gamma(\alpha)=1\iff\gamma\in A_\alpha$ for all $\alpha<\beta$.
In particular, $\gamma\in A_{s_\gamma}$, and so
$\kappa=\newcommand\Union{\bigcup}\Union\{A_s\mid s\in 2^\beta\}$.
Since $2^\beta<\kappa$, it must be that some $A_s$ has size
$\kappa$, and so $T$ has nodes on the $\beta^{th}$ level. Thus,
$T$ is a $\kappa$-tree. By the tree property, there is a
$\kappa$-branch $b\in[T]$. Let $F$ be the filter generated by the
$A_{b\newcommand\restrict{\upharpoonright}\restrict\beta}$ for
$\beta<\kappa$. Thus, $X\in F$ if and only if
$A_{b\restrict\beta}\of X$ for some $\beta<\kappa$. Since $b$
cannot choose to add a singleton $\{\xi\}$, since this does not
have size $\kappa$, it must choose the complement, and so the
filter $F$ is not principal. Since the sequence of
$A_{b\restrict\beta}$ is descending and $\kappa$ is regular, the
filter $F$ is $\kappa$-complete. And since either $A_\alpha\in F$
or $\kappa\setminus A_\alpha\in F$ explicitly at stage $\alpha$,
depending on whether $b(\alpha)=1$ or $0$, the filter $F$ decides
every set in our original family. So $\kappa$ has the filter
property.
(filter property implies weak embedding property) Assume the
filter property and suppose $A\of\kappa$. By the
Löwenheim-Skolem theorem and using $\kappa^\ltkappa=\kappa$,
there is a transitive set $M\elesub H_{\kappa^+}$ with $A\in M$
and $M^\ltkappa\of M$. By the filter property, there is a
$\kappa$-complete nonprincipal filter $F$ deciding every element
of $P(\kappa)^M$. Consider the ultrapower $M^\kappa/F$, where we
use only functions $f:\kappa\to M$ with $f\in M$. The relations
$f=_F g$ and $f\in_F g$ are still equivalence relations, even
though $F$ may not be an ultrafilter, because the corresponding
sets $\{\alpha\mid f(\alpha)=g(\alpha)\}$ and $\{\alpha\mid
f(\alpha)\in g(\alpha)\}$ are in $M$ if $f$ and $g$ are. Since $F$
is $\kappa$-complete and $M$ is closed under $\omega$-sequences,
it follows that $\in_F$ is well founded. Thus, the ultrapower
$N=\newcommand\Ult{\text{Ult}}\Ult(M,F)$ is a transitive set. The
proof that the canonical embedding $j:M\to N$ is elementary
proceeds as in Łos' Theorem, by induction on the complexity of the formula,
appealing to the Axiom of Choice in $M$ in the existential case.
Similarly, using $\kappa$-completeness, one shows for
$\alpha<\kappa$ that if $[f]_F\in[c_\alpha]_F$, then $f=_F
c_\beta$ for some $\beta<\alpha$, and consequently
$j(\alpha)=\alpha$; meanwhile, $\kappa\leq [id]_F<j(\kappa)$, so
the critical point of $j$ is $\kappa$. So $j:M\to N$ is as
desired.
(weak embedding implies embedding property) Assume the weak
embedding property and suppose $M$ is a transitive set of size
$\kappa$. Since $M\in H_{\kappa^+}$, we may code $M$ with a set
$A\of\kappa$, and then find a transitive set $\bar M$ with $M\in
\bar M$ and an embedding $j:\bar M\to
\bar N$ with critical point $\kappa$. The restriction
$j\restrict M:M\to j(M)$ is as desired.
(embedding implies normal embedding property) Assume that $\kappa$
has the embedding property, and suppose that $M$ is a
$\kappa$-model. By the embedding property there is an embedding
$j:M\to N$ with critical point $\kappa$. Let $X=\{j(f)(\kappa)\mid
f\in M\}$ be the seed hull of $\kappa$ via $j$. It follows that
$X\elesub N$, and so if $\pi:X\cong N_0$ is the Mostowski
collapse, we obtain the induced factor embedding, where
$k=\pi^{-1}$ and $j_0=\pi\circ j$.
Note that $N_0=\{\pi(j(f)(\kappa))\mid f\in
M\}=\{j_0(f)(\kappa)\mid f\in M\}$. If $\vec x=\langle
x_\alpha\mid
\alpha<\beta\rangle\in N_0^\ltkappa$, then there are functions
$f_\alpha\in M$ with $x_\alpha=j_0(f_\alpha)(\kappa)$. Since $M$
is a $\kappa$-model, it follows that $\langle f_\alpha\mid
\alpha<\beta\rangle\in M$. Thus, $\langle j_0(f_\alpha)\mid
\alpha<\beta\rangle\in N_0$, and so by evaluation these functions
at $\kappa$, it follows that $\vec x=\langle j_0(f)(\kappa)\mid
\alpha<\beta\rangle\in N_0$, and so $N_0$ is a $\kappa$-model as
desired.
(normal embedding implies Hauser embedding property) Suppose that
$\kappa$ has the normal embedding property and suppose that $M$ is
a $\kappa$-model. Since $M$ has size $\kappa$, there is a relation
$E$ on $\kappa$ such that $\langle M,{\in}\rangle\cong\langle
\kappa,E\rangle$. Thus, $M$ must be the Mostowski collapse of the
relation $E$. By the Löwenheim-Skolem theorem theorem, there
is a $\kappa$-model $\bar M$ with $E\in \bar M$. By the normal
embedding property there is an embedding $j:\bar M\to \bar N$ with
critical point $\kappa$. Since
$j(E)\intersect\kappa\times\kappa=E$, it follows that $E\in\bar
N$, and since the Mostowski collapse of $E$ is unique, that $M$ is
in both $\bar M$ and $\bar N$. By elementarity, if $x\in M$ is
coded by $\alpha$ with respect to $E$, then $j(x)$ is coded by
$j(\alpha)=\alpha$ with respect to $j(E)$. The model $N$ can
therefore reconstruct $j\restrict M$ from $E$ and $j(E)$. Since
$\bar M$ can see that $M^\ltkappa\of M$, it follows that $\bar
N\models j(M)^{< j(\kappa)}\of j(M)$. In particular, $M$ and
$j\restrict M$, which have size $\kappa$ in $N$, must be in
$j(M)$. Thus, $j\restrict M:M\to j(M)$ has the desired Hauser
embedding property.
(embedding properties all equivalent) The Hauser embedding
property immediately implies the weak embedding property, since
any $A\of\kappa$ can be placed into a $\kappa$-model $M$.
(embedding implies weak compactness property) Assume that $\kappa$
has the embedding properties. We show first that $\kappa$ is
inaccessible. Regularity follows from $\kappa^\ltkappa=\kappa$
(although it also follows from the weak embedding property). If
$2^\beta\geq\kappa$ for some $\beta<\kappa$, then there is a
$\kappa$-sequence $\vec a=\langle a_\alpha\mid
\alpha<\kappa\rangle$ of distinct subsets of $\beta$. We may code
$\vec a$ with a subset of $\kappa$, and therefore find a
$\kappa$-model $M$ and an embedding $j:M\to N$ with $\vec a\in M$.
If $a=j(\vec a)(\kappa)$ be the $\kappa^{th}$ element of $j(\vec
a)$, then by $M^\ltkappa\of M$, it follows that $a\in M$. Since
$a\of\beta<\kappa$, it follows that $j(a)=a$. Since $j(a)$ appears
on $j(\vec a)$, it follows by elementarity that $a$ appears on
$\vec a$. So $a=a_\alpha$ for some $\alpha<\kappa$. But since
$j(\vec a)(\alpha)=j(a_\alpha)=a_\alpha$, this means that $a$
appears at least twice on the sequence $j(\vec a)$, at coordinates
$\alpha$ and $\kappa$, contradicting the assumption that the sets
were distinct. So $\kappa$ is inaccessible.
Lastly, we verify the weak compactness property. Suppose that $T$
is a $\kappa$-satisfiable theory in an $L_{\kappa,\kappa}$
language $L$ of size at most $\kappa$. Since $\kappa$ is
inaccessible, it follows that $T$ has size at most $\kappa$. We
may assume that the symbols of $L$ are built from ordinals below
$\kappa$, and so $T$ may be coded with a subset of $\kappa$. Thus,
by the $\kappa$-model embedding property, there is a
$\kappa$-model $M$ with $T,L\in M$ and an embedding $j:M\to N$
with critical point $\kappa$ into a transitive set $N$. Since
$V_\kappa\of M$, the model $M$ has all the subtheories of $T$ of
size less than $\kappa$, and the models that exist for them (for
this conclusion, one needs an infinitary version of the upward
Löwenheim-Skolem theorem, which can be proved by an
infinitary analogue of the classical proof). Thus, $M$ satisfies
that $T$ is $\kappa$-satisfiable. By elementarity, it follows that
$N$ satisfies that $j(T)$ is $j(\kappa)$-satisfiable. But $j$
fixes every element of $T$ individually, so $T=j(T)\intersect
V_\kappa\in N$, and so $T$ is a subtheory of $j(T)$ of size less
than $j(\kappa)$. Thus, $N$ has a model $\cal A$ of $T$.
Furthermore, the satisfaction relation is absolute to transitive
sets, so $\cal A$ really is a model of $T$, as desired.
(embedding implies partition property) Most other accounts derive
the partition property from the tree property, but the embedding
point of view, ever efficacious, seems to simplify things. Assume
that $\kappa$ has the embedding properties and suppose
$F:[\kappa]^2\to 2$. Since $F\in H_{\kappa^+}$, we may find a
$\kappa$-model $M_0$ with $F\in M_0$. By the normal embedding
property there is a $\kappa$-model $N_0$ and an embedding
$j_0:M_0\to N_0$ with critical point $\kappa$. Since the entire
embedding $j_0:M_0\to N_0$ has hereditary size $\kappa$, we may
find a $\kappa$-model $M$ with $M_0$, $N_0$ and $j_0$ all having
size $\kappa$ in $M$. By the Hauser embedding property there is an
embedding $j:M\to N$ such that $j,M\in N$. Applying $j$ to
$j_0:M_0\to N_0$ yields an embedding $h=j(j_0):j(M_0)\to j(N_0)$
and a factor diagram. Let $j^*=h\circ j\restrict M_0$ be the
composition of the diagonal. This diagram commutes because
$j_0(x)=y$ implies $j(j_0)(j(x))=j(y)$. Let
$\mu_0=\{X\of\kappa\mid X\in M_0\And\kappa\in j_0(X)\}$ be the
$M_0$-normal filter induced by $j_0$. Since $\mu_0$ has size
$\kappa$ in $M$, it follows from $P(\kappa)^M\of N$ that $\mu_0\in
N$. Using $j\in N$, it follows also that
$j\newcommand\image{''}\image\mu_0\in N$. Since $\mu_0\of M_0$, it
follows that $j\image\mu_0\of j(M_0)$. Since $M$ knows that
$M_0^\ltkappa\of M_0$, it follows that $N\models j(M_0)^\kappa\of
j(M_0)$, and consequently $j\image\mu_0\in j(M_0)$. This is a
collection of $\kappa$ many elements of $j(\mu_0)$ in $j(M_0)$,
and so by the $j(\kappa)$-completeness of $j(\mu_0)$ over
$j(M_0)$, it follows that $I=\Intersect j\image\mu_0\in j(\mu_0)$.
Fix any $\delta_0\in I$ and observe that $X\in\mu_0\iff\delta_0\in
j(X)$ for $X\in M_0$. Let $\delta_1=j(\kappa)$, and apply $j$ to
the definition of $\mu_0$ to see that $Y\in j(\mu_0)\iff
\delta_1\in h(Y)$ for all $Y\in j(M_0)$.
Suppose $j^*(F)(\delta_0,\delta_1)=i$. We will build an increasing
monochromatic sequence $\langle
\gamma_\alpha\mid \alpha<\kappa\rangle$ such that (i)
$\alpha<\alpha'\implies F(\gamma_\alpha,\gamma_{\alpha'})=i$; and
(ii) $\{\xi\mid F(\gamma_\alpha,\xi)=i\}\in\mu_0$ for every
$\alpha<\kappa$. Suppose that $\langle
\gamma_\alpha\mid \alpha<\beta\rangle$ is defined and we would like to
define $\gamma_\beta$. Consider the set of possible values for
$\gamma_\beta$, namely $X=\{\gamma<\kappa\mid
\forall\alpha<\beta\,\,F(\gamma_\alpha,\gamma)=i\text{ and }\{\xi\mid
F(\gamma,\xi)=i\}\in\mu_0\}$. It suffices to show that $X$ is not
empty. By (ii) for any $\alpha<\beta$, we know that
$j(F)(\gamma_\alpha,\delta_0)=i$, since $\delta_0$ is a seed for
$\mu_0$ via $j$. Similarly, $\{\xi\mid j(F)(\delta_0,\xi)=i\}\in
j(\mu_0)$, since $\delta_1$ is in $h$ of this set, in light of
$j^*(F)(\delta_0,\delta_1)=i$. Thus, $\delta_0\in j(X)$. It
follows that $X$ is unbounded in $\kappa$, and so we may choose
$\gamma_\beta\in X$ and continue the construction. So $F$ admits a
monochromatic set of size $\kappa$.
(partition implies tree property) Assume that
$\kappa\to(\kappa)^2_2$. We first show that $\kappa$ is
inaccessible. Regularity follows from $\kappa^\ltkappa=\kappa$
(but even without this hypothesis, it follows from the partition
property). If $2^\beta\geq\kappa$ for some $\beta<\kappa$, then
let $\langle a_\alpha\mid \alpha<\kappa\rangle$ be a
$\kappa$-sequence of distinct subsets of $\beta$. Let
$F(\alpha,\beta)=0$ if $a_\alpha$ precedes $a_\beta$ in the
lexical ordering of $2^\beta$, and otherwise $1$. It is not
difficult to prove that there is no monotone sequence of length
$\beta^+$ in the lexical order on $2^\beta$. Consequently, there
can be no monochromatic set for $F$ of size $\kappa$. So $\kappa$
is inaccessible.
Now suppose that $T$ is a $\kappa$-tree. We may assume that the
underlying set of $T$ is exactly $\kappa$. For any two nodes
$\alpha$ and $\beta$ in $T$, let us say that $\beta$ is to the
{\df right} of $\alpha$ in $T$, if $\alpha\perp_T\beta$ and
$\alpha'<\beta'$, where $\alpha'\leq_T\alpha$ and
$\beta'\leq_T\beta$ are least in $T$ such that
$\alpha'\perp_T\beta'$. Define $F(\alpha,\beta)=0$ if $\beta$ is
above or to the right of $\alpha$, and otherwise $1$. Suppose that
$H\of\kappa$ is a monochromatic set of size $\kappa$. Suppose
first that the monochromatic value is $0$. Thus, whenever
$\alpha<\beta$ in $H$, then $\beta$ is above or to the right of
$\alpha$ in $T$. By applying the partition property once more, we
may further assume that only one of these answers arises. If any
$\alpha<\beta$ in $H$ has $\alpha<_T\beta$, then the elements of
$H$ are linearly ordered, and $T$ has a $\kappa$-branch.
Otherwise, we assume that whenever $\alpha<\beta$ in $H$, then
$\beta$ is to the right of $\alpha$ in $T$. We may therefore
follow the ``right-most'' path through (the $T$-predecessors of
elements of) $H$. Specifically, using the fact that the levels of
$T$ have size less than $\kappa$ and $\kappa$ is regular, there is
on each level $\xi$ a right-most node $\zeta$ that occurs as a
$T$-predecessor of all sufficiently large elements of $H$. The set
of such nodes is clearly linearly ordered, and therefore forms a
$\kappa$-branch through $T$. The final case, when the
monochromatic value of $F$ on $H$ is $1$, is similar, except that
we follow the left-most branch through $H$ to provide a
$\kappa$-branch through $T$. QED
I believe that it was in these lecture notes that the terms $\kappa$-model and the  "Hauser
property" were introduced, since I remember inventing that
terminology. (I knew Kai Hauser from the time I was an
undergraduate at Caltech, where he was a graduate student.)
Finally, let me mention that Sean Cox, Brent Cody, Thomas
Johnstone and myself have a paper in preparation called, "The
weakly compact embedding property," which is all about the
property when you drop the inaccessibility requirement. There are
some very interesting things to say, and I'l post a link when the
article is available. But meanwhile, I refer you to the slides for
my talk The weakly compact embedding property.<|endoftext|>
TITLE: Lattice n-gons with ordered side lengths 1,2,3,...,n
QUESTION [13 upvotes]: Consider the octagon in the Cartesian plane with vertices at (0,0), (1,0), (1,2), (4,2), (4,6), (7,2), (7,8), and (0,8). 
Are there other (infinitely many) polygons, such as this, lying entirely in the first quadrant of the Cartesian plane with all its vertices at lattice points and whose sides, in anti-clock wise order, are precisely of lengths 1, 2, 3, ... n starting from the side departing at the origin?

REPLY [10 votes]: There are indeed other such polygons.
-- For example there is one for $n = 11$, as follows
(the origin is in the lower left corner):

Also there is one for $n = 15$:

Further there are $21$ such polygons for $n = 16$.
One of them is the following:

These polygons have been found with this GAP function.
There is one such polygon for $n = 8$, $n = 11$ and $n = 15$, each,
there are $21$ such polygons for $n = 16$, and $225$ for $n = 19$.
For all other $n < 20$ there are no such polygons.
The complete list for $n \leq 19$ in GAP-readable form
can be found here.
Added on May 3, 2016: A zip file with *.png images of all
$249$ polygons for $n \leq 19$ can be found here (545KB).
Added on May 4, 2016 (following a request in the comments):
When leaving away the condition that the polygon is entirely in the
first quadrant, we get $3$, $5$, $6$, $584$ and $882$ distinct polygons
for $n = 8$, $n = 11$, $n = 12$, $n = 15$ and $n = 16$, respectively.
A GAP-readable coordinate list can be found here, and
a zip file with *.png images of all these $1480$ polygons
can be found here (2.8MB).<|endoftext|>
TITLE: Generalizing Ramanujan's "1729 story"
QUESTION [6 upvotes]: Whenever I read the anecdote about Hardy, Ramanujan and the taxi number 1729 I'm amazed that it could have occurred to anyone just off the top of their head that 1729 can be written as the sum of two cubes in two different ways -- and that it is the smallest such number.
At all events, there are several ways to look at this in a more general way. For positive integers $n, k$ let us set $$r_n(k) = | \{(x,y): x\leq y \text{ and } x^n + y^n = k\} |.$$
For what, if any, $n,y\geq 2$ is $r_n^{-1}(\{y\})$ is infinite? 
(Also partial answers and/or examples are very welcome.)

REPLY [7 votes]: There are many articles that study the quantity you call $r_n(k)$ using sieve methods. Among them I mention the following, which give highly non-trivial bounds for the number of $k<X$ such that $r_n(k)>1$. If you look at these, and also forward reference them using MathSciNet, you should be able to find the state of the art.

T.D. Browning, Equal Sums of Two kth Powers, Journal of Number Theory, Volume 96, Issue 2, October 2002, Pages 293–318.
C. Hooley. On another sieve method and the numbers that are a sum of two hth powers. Proc. London Math. Soc., 226 (1981), pp. 30–87.
C. Hooley, On another sieve method and the numbers that are a sum of two hth powers: II. Journal für die reine und angewandte Mathematik (1996) Volume: 475, page 55-76<|endoftext|>
TITLE: Is the normal bundle of a torus trivial?
QUESTION [23 upvotes]: Question:
Let $T^k \subseteq \mathbb{R}^n$, $ n > k$, be a smoothly embedded $k$-torus. Is its normal bundle trivial?
What about the normal bundle of $S^k \subseteq \mathbb{R}^n$, $n > k$, the $k$-sphere smoothly embedded in $\mathbb{R}^n$ (I suspect it is not always trivial)?
What I know so far:
For example, I can prove that the normal bundle of $S^1 \subseteq \mathbb{R}^n$, $n > 1$, a smoothly embedded circle in $\mathbb{R}^n$, is trivial. I also know that the normal bundles of tori and spheres are always trivial in the codimension-1 case, since these manifolds are orientable. I'm interested in a more general answer.
Bigger picture:
More generally, I'm interested in techniques that can be used to prove normal bundles are trivial/nontrivial.

REPLY [20 votes]: You should be able to prove that the normal bundles in codimension $2$ are trivial as well. This is a little harder than codimension $1$; you need to know that such bundles are determined by their Euler class. The older literature (see papers cited below) tells you that the normal bundles of spheres are trivial once the codimension is high enough.
The paper "On the normal bundle of a homotopy sphere embedded in Euclidean space" by Hsiang-Levine-Szczarba (Topology Volume 3, Issue 2, April 1965, Pages 173-181) quotes an unpublished result of A. Haefliger that there is an embedding of $S^{11}$ in $R^{17}$ with non-trivial normal bundle. Jerry Levine's paper "A classification of differentiable knots" (Annals 82 (1965) 15-50) determines (see proposition 6.2) the possible normal bundles in some range of codimensions in terms of maps in various exact sequences. It's hard to summarize the results, but he gives a table at the end dealing with relatively low dimensions, where calculations can be made explicit. The case $n=11$ and $k=6$ contains Haefliger's example; in fact there are exactly $5$ possible normal bundles among all embeddings.
The introductory sentence to that last section is a classic Jerry Levine understatement: "By strenuous use of Proposition (6.2), together with results of [1], [9], [10], [13], and [24], computation of many of the geometrically defined groups we have discussed can be carried out for low values of $n$."
It's reasonably likely (but you'd have to check) that some of those non-trivial normal bundles of spheres give rise to non-trivial normal bundles of tori, by taking the connected sum of a standard torus (with a trivial normal bundle) with one of those knotted spheres.<|endoftext|>
TITLE: "Spatial (geometrical)" realization of Elementary topos?
QUESTION [10 upvotes]: It is well known that (Grothendieck) Topos (in fact, Model topos too) has many good geometrical properties. In many senses reflects general forms of generic geometry.
Note: Grothendieck view of Topos is as an "ultimate" generalization of space.
Also, Elementary topos has many good logical properties. I am interested in elementary topos as a formal geometry.

Question: Elementary topos can be seen as a generalized space?

Note: Can "Elementary higher topos" reflects the geometrical nature of objects in mathematics? This could suggest the physical nature of mathematics (this is vague, only a philosophical note).

REPLY [16 votes]: One way in which elementary toposes can be seen as "fully geometric" is by relativizing them.  An elementary topos is a Grothendieck topos if and only if it admits a bounded geometric morphism to the category of sets, and the 2-category of Grothendieck topoi (and geometric morphisms) is equivalent to the slice category of elementary topoi (and geometric morphisms) equipped with a bounded map to Set.  If we replace Set by any other elementary topos $S$, then the 2-category of elementary topoi equipped with a bounded geometric morphism to $S$ behaves very much like the 2-category of Grothendieck topoi, and in particular is very geometric, especially if $S$ has a NNO.<|endoftext|>
TITLE: Comodules of Cosemisimple Hopf Algebras
QUESTION [5 upvotes]: A cosemisimple Hopf algebra is one which is the sum of its cosimple sub-cobalgebras. Is it clear that a comodule of a cosemisimple Hopf algebra always decomposes into irreducible parts? Moreover, will this decomposition obey the property that the type and multiplicity of the irreducible comodules appear be the same in any decomposition.
I am sure that this should be the case but I can't see how to prove it. One thing that confuses me is the prospect of infinite multiplicity in the case of an infinite dimensional comodule.
Note: I previously asked this question on stackexchange, but moved it here after no response.

REPLY [4 votes]: Regarding your first question, I think the following definition and theorem settles the answer to the affirmative: 
Definition: A coalgebra $C$ is called right cosemisimple (or right completely reducible coalgebra) if the Category $M^C$ is a semisimple Category i.e. if every right $C$-comodule is cosemisimple. 
Similarly, left cosemisimple coalgebras are defined by the semisimplicity of the Category of left comodules. 
Theorem: Let $C$ be a coalgebra. The following assertions are equivalent:

$C$ is a right cosemisimple coalgebra
$C$ is a left cosemisimple coalgebra
$C=C_0$
Every left (right) rational $C^*$-module is semisimple

where $C_0=Corad(C)$ i.e. the coradical of $C$, which is the sum of all its simple subcoalgebras. 
For a proof of the above you can see for example Hopf algebras-an introduction, Dascalescu-Nastasescu-Raianu, Ch.3, p.118-119, Theorem 3.1.5. (However this is more or less standard material, you can also find it in Sweedler's book on Hopf algebras, S.Montgomery's book etc). 
Now, regarding your second question, I think you can use the above theorem together with the correspondence between $C$-comodules and rational $C^*$-modules to investigate the situation closer.<|endoftext|>
TITLE: Character Values for Alternating Groups of degree $\geq 7$
QUESTION [7 upvotes]: Is it true that in each row and column of the character table of alternating groups with degree $\geq 7$ there are at most two complex values? Any reference will be highly appreciated.

REPLY [2 votes]: This answer essentially summarizes information from the other answers, hopefully, making the whole picture clear. For each self-conjugate partition $\lambda$ (i.e., $\lambda=\lambda'$) of $n$, the irreducible character $\chi_\lambda$ of $S_n$ is a sum of two irreducible representations of $A_n$: $\chi_\lambda = \chi_\lambda^+ + \chi_\lambda^-$. Also, for each partition with distinct odd parts $\mu=(2m_1+1,\dotsc,2m_d+1)$, where $m_1>\dotsb >m_d$ the conjugacy class $C_\mu$ of elements with cycle type $\mu$ in $S_n$ splits into two $A_n$-conjugacy classes, represented by elements $w_\mu^+$ and $w_\mu^-$.
There is a bijective correspondence $\lambda\mapsto \phi(\lambda)$ from the set of self-conjugate partitions onto the set of partitions with distinct odd parts. The image of $\lambda$ is $\mu=(2m_1+1,\dotsc,2m_d+1)$ if the number of boxes in the Young diagram of $\lambda$ strictly to the right (or strictly below) a box of the form $(i,i)$ is $m_i$.
Theorem. All character values of $A_n$ are real except possibly those of the form $\chi_{\lambda}^\pm(w_\mu^\pm)$, where $\mu=\phi(\lambda)=(2m_1+1,\dotsc,2m_d+1)$.
The character value $\chi_{\lambda^\pm}(w_\mu^\pm)$ is not real if and only if $\sum_{i=1}^d m_i$ is odd.
It follows that, in each row and column of the character table there is at most one pair of irrationalities. The only rows with irrationalities are those corresponding to $\lambda^\pm$ where $\lambda$ is a self-conjugate partitions such that the number of off-diagonal boxes in its Young diagram is odd. The only columns with irrationalities are of the classes of elements with cycle decomposition $(2m_1+1,\dotsc,2m_d+1)$, where $m_1>\dotsb > m_d$, and $\sum_{i=1}^d m_d$ is odd.
The only alternating groups $A_n$ with real character table are for the values $n=2,5,6,10,14$.
James and Kerber discuss the characters of alternating groups, but do not give any interesting examples. Frobenius (Georg Ferdinand Frobenius. 1901. ‘Über die Charaktere der alternirenden Gruppe.’
[On the characters of alternating groups.] S’ber Akad. Wiss. Berlin, 303–15) has computations for $A_8$. My book has many examples and a more detailed explanation in Chapters 4 and 5.<|endoftext|>
TITLE: Table of (integral) cohomology groups of K(Z,n)
QUESTION [19 upvotes]: Can I find somewhere a table of the (first few) cohomology groups of $K(\mathbb{Z},n)$ with integer coefficients?
It seems like a natural counterpart to the table of the homotopy groups of spheres, but I couldn’t find anything. I’m aware of exposé 11, année 7 in the Cartan seminar where the homology of Eilenberg-MacLane spaces is computed, and I guess I could adapt it to the case where the group is $\mathbb{Z}$ and use the universal coefficient theorem to get the cohomology, but it’s not completely trivial, and I would be surprised that nobody thought about doing it before me.

REPLY [5 votes]: There are several computations carried out explicitly in the paper
Samuel Eilenberg and Saunders Mac Lane, MR 65162 On the groups $H(\Pi,n)$. II. Methods of computation, Ann. of Math. (2) 60 (1954), 49--139.
For instance, I once wanted to know the groups $H_{4+i}(K(\mathbb{Z},4);\mathbb{Z})$ for $i\le 3$, and was able to extract the answer from Section 24.<|endoftext|>
TITLE: Constructing the largest finite group with a fixed number of conjugacy classes
QUESTION [10 upvotes]: It is known that there are finitely many finite groups with a given number of conjugacy classes.  How can one construct (or get a character table for) the groups $G$ that realize the maximum possible order among groups with $k$ conjugacy classes?  Help even with $k=4$ or 5 would be very useful.

REPLY [4 votes]: Following @verret comment, here are the list of the largest finite groups $G_k$ with a fixed class number $k\le 14$ (i.e. the number of conjugacy classes of elements, or the number of irreducible complex representations up to equiv.), coming from the papers of Vera-López and Sangroniz MR804489, MR880291, MR880291.
See also OEIS/A002319. We will consider $c_k:=|G_k|^{1/k}$.
$\begin{array}{c|c|c}
k & G_k & |G_k| &  c_k \newline 
\hline                           
1 & C_1 & 1 & 1  \newline 
\hline 
2 & C_2 & 2 & 1.4142\dots \newline 
\hline 
3 & S_3 & 6 & 1.8171\dots \newline 
\hline 
4 & A_4 & 12 & 1.8612\dots \newline 
\hline 
5 & A_5 & 60 & 2.2679\dots \newline 
\hline 
6 & PSL(2,7) & 168 & 2.3490\dots \newline 
\hline 
7 & A_6 & 360 & 2.3183\dots \newline 
\hline 
8 & M_{10} & 720 & 2.2759\dots \newline 
\hline 
9 & A_7 & 2520 & 2.3874\dots \newline 
\hline 
10 & PSL(3,4) & 20160 & 2.6943\dots \newline 
\hline 
11 & Sz(8) & 29120 & 2.5458\dots \newline 
\hline 
12 & M_{22} & 443520  & 2.9551\dots \newline 
\hline 
13 & C_9^2:SL(2,3)  & 1944  & 1.7905\dots \newline 
\hline 
14 & PSU(3,5) & 126000 & 2.3137\dots \newline 
\end{array}$
Monstrous question: Is it true that $G_{194} = M$?
If so, $c_{194} = 1.8960\dots$
Bertram's problem (19.11; B): Is there $\alpha$ such that for any finite group $G$ of class number $k(G)$ then $$|G| \le \alpha^{k(G)}$$ 
Bonus question 1:  Is $c_k \le c_{12}$ for all $k$?  
If so, it would solve Bertram's problem explicitly with $\alpha = c_{12}$.
In particular, we would have $|G|< 3^{k(G)}$ for any $G$ (checked by GAP for $G$ simple with $|G|<10^7$).  
Bonus question 2: Is it true that $c_k \to 1$ when $k \to \infty$?   
If so, it would solve Bertram's problem implicitly.
I guess it is known (from CFSG) that for $G$ simple, $|G|^{1/k(G)} \to 1$ when $|G| \to \infty$. We can reduce to the infinite families, ok for $C_p$ and $A_n$, now if $X(q)$ is a generic finite simple group of Lie type then $\exists n,m \ge 1$ such that $|X(q)| = O(q^n)$ and $k(X(q)) = O(q^m)$, but $q^{n/{q^m}} \to 1$ when $q \to \infty$.
Then, an eventual reduction of Bertram's problem to the simple group case should solve it.<|endoftext|>
TITLE: Triangulation with simplices of same volume
QUESTION [11 upvotes]: Let $M$ be a Riemannian smooth compact manifold.
It is known that $M$ has a triangulation, for any dimension. But do we know if there exists a triangulation such that all simplices have same volume ?
I've been looking for some references, but I can't find any dealing with this problem. I would appreciate if someone had an answer - even partial.

REPLY [16 votes]: I don't know a reference, but I think that we can use a theorem of Moser to get what you want.
Start with any triangulation $\mathcal{T}$ of $M$, whose cardinal is denoted by $k$ and denote by $\omega$ the Riemannian volume form. There is a positive smooth function $f$ such that $\int_M f\omega=\mathrm{vol}(M)$ (i.e. $f$ has average $1$) and such that for all $T\in\mathcal{T}$:
$$\int_T f\omega = \frac{\mathrm{vol}(M)}{k}$$
(simply adjust $f$ in the interior of each simplex).
Now, Moser proved in 1965 (answering a question of a MO user by the way) that, since $f\omega$ and $\omega$ have the same total volume, there is a diffeomorphism $\phi:M\to M$ such that $\phi_*(f\omega)=\omega$. Then by the change of variable formula, 
$$\int_{\phi(T)} 1\,\omega= \int_T f\,\omega = \frac{\mathrm{vol}(M)}{k}$$
so that the collection $\{\phi(T)\}$ is a triangulation of $M$ with all simplices of the same volume.<|endoftext|>
TITLE: K theory for pre $C^*$-algebras
QUESTION [8 upvotes]: In noncommutative geometry when one want to go to the differentiable level, one is forced to work with algebras which are no longer $C^*$. It is nice if we don't loose much information by the replacement of $C^*$-algebras by more general algebra. One possible choice is the so called pre $C^*$-algebra which is by definition a subalgebra $\mathcal{A}$ of a $C^*$-algebra $A$ of which is complete with respect to some locally convex topology finer that the topology of $A$ and is closed under the holomorphic functional calculus. Suppose that we have such $\mathcal{A}$ which is moreover a Fréchet algebra and assume that $\mathcal{A}$ is dense in $A$ (in the norm topology of $A$). The claim is that:  

The inclusion $i:\mathcal{A} \to A$ induces an isomorphism in $K_0$ groups.   

I found the proof of this fact in the book "Elements of Noncommutative Geometry" by Várilly, Gracia-Bondía and Figueroa. There are two moments which are not clear for me:
1. The first is that authors state that it is enough to prove this fact for unital algebras. I believe that it is indeed the case however I'm not sure whether there are no technical problems.
2. The second thing is the end of the proof: let us recall that $K$-theory may be defined in terms of idempotents (not projections, i.e. our idempotents need not to be self-adjoint) in matrix algebras $Q_n(A)=\{e \in M_n(A): e^2=e\}$ (the same for $\mathcal{A}$: one checks that matrix algebras for $\mathcal{A}$ are again Fréchet pre $C^*$-algebras). 
Of this the quotient space is made by equivalence relation where $e \sim f$ iff there is some invertible matrix $g$ (possibly bigger that $e$ and $f$) such that $(e \oplus 0)g=g(f \oplus 0)$. At the end of the proof authors arrived to the conclusion that the inclusions $Q_n(\mathcal{A} \to Q_n(A)$ are all homotopy equivalences and they claim that this finishes the proof provided we are able to prove that two idempotents which are homotopic are equivalent in the above sense. I don't quite see two things:  

Why it is enough to prove this to conclude that the inclusion induces iso on $K_0$ groups?  

And also 

How to prove that being homotopic implies being equivalent (in the context of general idempotents and invertibles)?

REPLY [2 votes]: 1) First question: unital algebras are sufficient.
If $i:K_0(\tilde {\mathcal A}) \rightarrow K_0(\tilde A)$ is an isomorphism, and since $id:K_0(C) \rightarrow K_0(C)$ is an isomorphism, you make a diagram chase by doubling the exact sequence
$$0 \rightarrow K_0(A) \rightarrow K_0(\tilde A) \rightarrow^\rho K_0(C) \rightarrow 0$$
(second case with ${\mathcal A}$) and connect both lines with $i$ to see that $K_0({\mathcal A}) \rightarrow K_0(A)$ is an isomorphism as well. Note that $K_0(A)$ is the kernel of $\rho$ by definition!
3) "How to prove that being homotopic implies being equivalent (in the context of general idempotents and invertibles)?"
Note that ${\mathcal A}$ is a subalgebra of $A$, so you have all the notions from $C^*$-algebras like projections in ${\mathcal A}$ as well. I think the elementary standard proofs from $C^*$-algebras work in ${\mathcal A}$ as well.
I think you must go by yourselves through the proofs.
See Wegge-Olsen Proposition 5.2.10 & 5.2.12 that homotopy-equivalence and unitary equivalence (or invertible equivalence (Proposition 4.2.6)) is the same in matrix algebras.
See Wegge-Olsen 5.B: Every idempotent is homotopic to a projection.
2) "Why it is enough to prove this to conclude that the inclusion induces iso on $K_0$ groups?"
By item 2) we may equivalently switch from invertible equivalence of idempotents to homotopy equivalence of idempotents in $Q$.
If $i$ is a homotopy equivalence on the level $i:Q({\mathcal A}) \rightarrow Q(A)$ then there exists a continuous inverse $k$ such that $i\circ k \sim id$ and $k \circ i \sim id$.
$i$ on the level $K_0$ is injective:
Suppose
$$i(e) - i(f)=0$$
for $e,f \in Q({\mathcal A})$. Then $i(e) \oplus m \sim i(f) \oplus m$ (homotopy is $\sim$; Grothendieck group) for some $m \in Q(A)$. Note $m \sim i \circ k (m)$. Hence
$$i(e) \oplus i(k(m)) \sim i(f) \oplus i(k(m)),$$
or
$$i(e \oplus k(m)) \sim i(f \oplus k(m)).$$
Since $k$ is continuous, it respects homotopy (pathes)!. Thus
$$e \oplus k(m) \sim k(i(e \oplus k(m))) \sim k(i(f \oplus k(m))) \sim f \oplus k(m).$$
Hence
$$e - f = 0.$$
Surjectivity of $i$ is similar.<|endoftext|>
TITLE: Unique factorization of posets
QUESTION [7 upvotes]: Given two finite posets $P$ and $Q$, we can form the direct product poset $P \times Q$ whose elements are pairs $(p,q) \in P \times Q$ with $(p,q) \leq (p',q')$ if $p \leq p'$ and $q \leq q'$. Let us say a finite poset $P$ on $\geq 2$ elements is indecomposable if $P=P_1 \times P_2$ implies that either $P_1$ or $P_2$ is equal to $P$.
Is it true that a finite poset $P$ on $\geq 2$ elements can be written as a product $P = P_1 \times \cdots \times P_n$ of indecomposable posets in a unique way up to permutation of the factors?
Surely this is a classical question/result. I am apparently having trouble coming up with the right terms to google for, however. I would definitely appreciate any pointer to the literature. If the general result fails, I would also be interested in what mild conditions we can put on $P$ (e.g. gradedness) to guarantee a positive result.

REPLY [8 votes]: Basically repeating what I said in the comments:

In full generality, the answer to your question is "No". For a counterexample, see: Tadasi Nakayama and Junji Hashimoto, On a problem of G. Birkhoff, Proc. Amer. Math. Soc. 1 (1950), pp. 141--142.
However, if you assume that $P$ is connected (i.e., the undirected graph obtained by forgetting the directions of the arcs of the Hasse diagram of $P$ is connected), then the answer is "Yes". For a proof, see Junji Hashimoto, On Direct Product Decomposition of Partially Ordered Sets, Annals of Mathematics, Second Series, Vol. 54, No. 2 (Sep., 1951), pp. 315--318.

(Note: I have verified the counterexample, but I haven't looked at the proof.)<|endoftext|>
TITLE: Arbitrarily many primes in a Fibonacci-type sequence
QUESTION [8 upvotes]: It is conjectured that the standard Fibonacci sequence contains infinitely many primes. While this is perhaps too difficult, I am wondering about the following simpler version:
Question. For any $K$, does there necessarily exist positive integers $a, b$ such that the sequence given by $x_1 = a, x_2 = b, x_{n} = x_{n-1} + x_{n-2}~(n \geq 3)$ contains at least $K$ primes? 
I feel this bears some relation to the Green-Tao theorem, but I can't make the connection formal.

REPLY [12 votes]: I think the answer to this question is yes. The theorem of Green, Tao and Ziegler says that a collection of $K$ linear forms over the integers will all take prime values infinitely often provided that they are nondegenerate (which, in the homogeneous case, means not multiples of one another) and there are no local obstructions. Thus you can't make $a, b , a+b$ all prime infinitely often (there's an obstruction mod 2) but you can make $a, b, a+ 2b$ prime infinitely often. 
Let me explain how to find 5 prime elements in such a sequence (the argument clearly generalises).
The nth term of your sequence is $F_{n-2} a + F_{n-1}b$. Since the Fibonacci sequence is periodic modulo any integer $Q$, I can find 5 values $n_1,\dots,n_5$ such that $F_{n_i-2} = 0 \pmod{30}$ and $F_{n_i-1} = 1 \pmod{30}$. The linear forms $F_{n_i-2} a + F_{n_i-1}b$ are a nondegenerate system, and when $b = 1 \pmod{30}$ they all take the value $1 \pmod{30}$. Hence there is no obstruction modulo $2$, $3$ or $5$ to all five of these forms being prime. Since consecutive Fibonacci numbers are coprime, none of the forms $F_{n_i-2} a + F_{n_i-1} b$ vanishes identically $\pmod{7}$. Thus the image of the map $(a,b) \mapsto (F_{n_1-2} a + F_{n_1 - 1}b, \dots, F_{n_5 - 2} a + F_{n_5 - 1} b)$ is a subspace $V$ of $(\mathbb{Z}/7\mathbb{Z})^5$ on which each of the five coordinate maps $\pi_i : (\mathbb{Z}/7\mathbb{Z})^5 \rightarrow (\mathbb{Z}/7\mathbb{Z})$ is nontrivial. By a counting argument, $V$ cannot be the union of the $\ker \pi_i|_V$, so there is some choice of $a,b$ for which the five linear forms are nonzero modulo $7$: that is, there are no local obstructions modulo 7 to all these forms being prime. Same modulo $11$, $13$ etc.<|endoftext|>
TITLE: Injectivity/Surjectivity of $F_A :=\frac{d}{dt} +A(t) $ for a hyperbolic path of matrices $A(t)$ on $H^1 $
QUESTION [5 upvotes]: I am looking for a reference to the following problem:
Given a hyperbolic (no purely imaginary eigenvalues), continuous path of matrices $A(t)$ in $\mathbb{R}$ with hyperolic limits at $\pm \infty $.
Under which assumptions (on A(t)) is the operator $F_A :=\frac{d}{dt} +A(t) :H^1 (\mathbb{R},\mathbb{R}^m )\to L^2 (\mathbb{R},\mathbb{R}^m )$ injective, resp. surjective?

REPLY [6 votes]: It is straightforward to see that if the path is constant then hyperbolicity is the equivalent to the invertibility of the operator for example via the Fourier transform (see Atiyah Patodi Singer (Spectral Asymmetry and Riemann Geometry I) for example).
It is then straightforward to see that if the $A(t)$ is asymptotically hyperbolic then the operator is Fredholm by a parametrix patching argument.
There is then a well-defined index problem.  The index is then equal to the spectral flow of the family $A(t)$, namely the intersection number of the spectrum with the imaginary axis.  This can be easily proved using the homotopy invariance
of the index and reducing to the case where the $A(t)$ are simultaneously diagonalizable with diagonal entries being either constant or $\pm\tanh(t)$.  This can be solved explicitly with kernel or cokernel $\text{sech}(t)$ depending on the sign.
Back to your problem, the sign of the index provides an obstruction to the operator being injective or subject but not a complete one.  For example if $A(t)$
is direct sum of two matrix paths $A_-(t)$ and $A_+(t)$ of spectral flow $-n$ and $n$ respectively, then the corresponding operator has kernel and cokernel of dimension at least $n$ but the total operator has index $0$.  On the other hand, in the one-dimensional case it seems straightforward to see that it is a complete obstruction. Writing $\lambda(t)=a_{11}(t)$ the hyperbolicity assumption means that $a(t)$ converges at $\pm\infty$ to constants $a_\pm$ of the same sign.   An element of the kernel has the form:
$$
f(t)=c+\exp[-\int_0^t \lambda(s)ds]
$$
Regardless of how you choose the constant, $f(t)$ is exponentially growing at 
one end or the other. A similar argument applies to the cokernel.<|endoftext|>
TITLE: A finiteness property for semi-simple algebraic groups
QUESTION [5 upvotes]: Let $G$ be a semi-simple algebraic group over a field $K$, I am considering a question about whether there exists a finite set of semi-simple $K$-subgroups, say $H_1,...,H_r$, such that for any  semi-simple $K$-subgroup $H\subset G$, $H$ is conjugate to one of the $H_i$ by an element in $G(K)$?
I know the answer is "no" for $K=\mathbb{Q}$. In fact, for any quaternion algebra $\mathcal{D}$ over $\mathbb{Q}$, the group $\mathrm{SL}(\mathcal{D})$ can be embedded in $\mathrm{SL}_4(\mathbb{Q})$. As $\mathcal{D}$ varies, we got infinitely many semi-simple subgroups $\mathrm{SL}(\mathcal{D})$ of $\mathrm{SL}_4(\mathbb{Q})$ that are not isomorphic to each other, therefore not conjugate to each other.
Now I would like to ask what happens if $K=\mathbb{R}$, I think the answer is yes, but how to prove it? And what about $K=\mathbb{Q}_p$ or some other fields?
Thanks very much!

REPLY [4 votes]: In case it is useful to the OP, here is a reference: 
Richardson, R. A rigidity theorem for subalgebras of Lie and associative algebras. Illinois J. Math. 11 1967 92–110. 
Proposition 12.1 seems to be the relevant result.<|endoftext|>
TITLE: Which graphs embedded in surfaces have symmetries acting transitively on vertex-edge flags?
QUESTION [8 upvotes]: A vertex-edge flag in a graph is a vertex together with an edge incident to that vertex.  Given a graph $\Gamma$ embedded in a compact oriented surface $S$, when does the group of homeomorphisms of $S$ act transitively on vertex-edge flags of $\Gamma$?
We can start by looking a graphs embedded in the sphere, and start among those by looking at graphs that consist of the vertices and edges of a convex polyhedron.  Among these, the obvious examples on which isometries act transitively on vertex-edge flags are the 5 Platonic solids.  However, there are also two Archimedean solids with this property, namely the cuboctahedron:

and the icosidodecahedron:

I believe these are all the convex polyhedra on which isometries act transitively on vertex-edge flags.  A polyhedron is said to be isotoxal if isometries act transitively on edges.  There are 9
isotoxal convex polyhedra, and among these I see 7 on which isometries act transtively on vertex-edge flags: the 5 Platonic solids and the 2 shown above.
There are also infinitely many other connected graphs embedded in the sphere on which isometries act transitively on vertex-edge flags, namely  the hosohedra, like this:

and the dihedra, like this:

I believe these are all the connected graphs embedded in the sphere on which isometries acts transitively on vertex-edge flags.  If we consider graphs embedded in the sphere that are not connected, we get a host of other examples: for example, a bunch of equal-sized regular $n$-gons embedded in the sphere.  If we further drop the requirement that the homeomorphisms be isometries, we get even more examples: for example, a bunch of copies of the complete graph on 4 vertices embedded in the sphere.  I am happy to restrict attention to connected graphs, to avoid such clutter.  
I am also happy to omit graphs that have self-loops.
So, here is a sub-question that interests me: what are all the connected graphs without self-loops embedded in the sphere on which homeomorphisms of the sphere act transitively on vertex-edge flags?   Have I listed them all, or are there more?
(Whoops, here are some more: the empty graph, the graph with one vertex and no edges, and the graph with two vertices and one edge connecting them.
The last might be considered a degenerate hosohedron.)
When we go to higher genus the situation becomes a lot more complicated and interesting.  For example, in genus 3 we have Klein's quartic curve tiled by 56 triangles meeting 7 at each vertex:

So, I'd be perfectly happy to hear classifications that only handle surfaces of genus less than some fixed value.

REPLY [4 votes]: It sounds like what you're asking for is close to the notion of "regular map":

Jozef Siran, "Regular Maps on a Given Surface: A Survey", Topics in Discrete Mathematics, 2006. (pdf)

A (connected) map $M$ is regular if its automorphism group acts transitively (and hence regularly) on the set of flags.  On the other hand, the definition of "symmetry" you're using in the question is weaker than the standard definition of map automorphism, which in the case of maps embedded on compact oriented surfaces would restrict to orientation-preserving homeomorphisms (that also preserve the graph incidence relationships). This stronger definition rules out the cuboctahedron and icosidodecahedron, and indeed Siran asserts that the only regular maps on the sphere are the five platonic solids, the hosohedra ("$k$-dipoles") and the dihedra ("$k$-cycles") (plus the "semi-stars", if you allow graphs with dangling semi-edges).
Apparently a lot is also known about regular maps in the higher genus case!  But since I'm not at all familiar with this, I'll just refer you to Siran's survey.<|endoftext|>
TITLE: Banach-Mazur distance between the cube and the octahedron
QUESTION [18 upvotes]: The Banach-Mazur distance $d(X, Y)$ between two normed spaces $X, Y$ of the same dimension is defined as $d(X, Y) = \log\inf \|T\| \cdot \|T^{-1}\|$, where the $T:X \to Y$ is a linear and invertible operator. The estimates between classical $\ell_p^n$ spaces are known and in particular we have that $d\left(\ell_1^n, \ell_{\infty}^n\right) \sim \sqrt{n}$ asymptotically. Using Walsh matrices it can be shown that for $n=2^k$ this distance is actually not greater than $\sqrt{n}$.
In the two dimensional case we clearly have $d(\ell_1^2, \ell_{\infty}^2)=1$. I wonder if anybody has managed to provide some more specific results in the three dimensional case. It is equivalent to finding two positively homothetic parallelotopes with one contained in and the second containing the regular octahedron with the homothety ratio as minimal as possible. It seems like pretty basic question in three dimensional discrete geometry, but still I could not find any information and the problem might be highly non-trivial. The only information I found was in the paper of Stromquist ("The maximum distance between two dimensional Banach spaces") where he wrote that this distance is not known. But it was 35 years ago, so I guess that there could be some new results in this direction. If not, maybe at least one could make some good conjecture on what this distance should be like?

REPLY [15 votes]: This can be done as a nonlinear optimization problem: $T$ is a $3 \times 3$ matrix, and if $e_i$  and $v_j$ are the vertices of the unit balls in $\|\cdot\|_1$ and $\|\cdot\|_\infty$ norms respectively, you want to minimize $s$ subject to constraints $\|T e_i\|_\infty \le 1$, $\|T^{-1} v_j\|_1 \le s$ (or equivalently, $\|\text{Adj}(T) v_j\|_1 \le s |\det(T)|$, with $\det(T) \ne 0$).
Maple's Global Optimization Toolbox returns (almost immediately) an approximate solution, which is (up to roundoff error)
$$ T =  \pmatrix{1/3 & 1 & 1\cr -1 & 1 & -1/3\cr 1 & 1/3 & -1\cr},\ s = \dfrac{9}{5}$$
It is easy to check that this does satisfy the constraints.
Thus it appears that the answer is $9/5$.  It should be possible to prove optimality rigorously, if somewhat tediously, using the Karush-Kuhn-Tucker conditions.
EDIT: I also tried the $4$-dimensional case.  The solution Maple obtained was
$$ T = \pmatrix{0 & 1 & -1 & 1\cr -1 & 0 & 1 & 1\cr 1 & -1 & 0 & 1\cr -1 & -1 & -1 & 0\cr}, \ s = 2 $$<|endoftext|>
TITLE: Two notions of bundle of C* algebras
QUESTION [6 upvotes]: One can find in the literature two notions of $C^*$-algebra over a topological space $X$.
The first is as the data of an open surjection $ \pi: B \rightarrow X$ together with the structure of a $C^*$-algebra on each fiber $B_x = \pi^{-1} \{ x \}$ such that:

all the algebraic operations are continuous on the fiber products over $X$.
The norm functions $b \mapsto \Vert b \Vert_{B_x}$ on $B$ is upper semicontinuous.
There is one last condition that roughly says that the topology of $B$ is determined by the topology of $X$ and the norm functions.

One can eventually add additional condition: like that the norm function is continuous, or that there is enough of continuous local sections/global sections of the map $B \rightarrow X$.
One can also find an apparently completely different notion in the litterature:
A bundle of $C^*$-algebras over $X$ is the data of a $C^*$-algebra $A$ together with a continuous map $Prim(A) \rightarrow X$ where $Prim(A)$ is the space of primitive ideal of $A$ with the Jacobson topology.
When $X$ is nice (like locally compact Hausdorff , but completely regular and paracompact should be enough) it is well known that these definitions are equivalent (it follow from the Dauns-hoffmann theorem).
But what happen when $X$ is more general ?
I roughly see that it should works (one should not expect the two notions to be equivalent, but there is way to go from one to the other eventually under some assumptions and this should give a close relationship between the two notions. For example the second notion can only corresponds to bundle where there is enough global sections...) but I haven't look at the details yet.
I wanted to know first if there is some literature about this question ?
(If there is then I don't need to figure it out myself, and if there is none maybe I should figure it out and write something by myself)

REPLY [3 votes]: While I don't know of any reference that answers this question explicitly, I have a few observations that might be helpful.  For an upper-semicontinuous C$^*$-bundle $\pi : \mathcal{A} \to X$ and an open subset $U \subseteq X$, $\Gamma^b (U, \mathcal{A})$  denotes the C$^*$-algebras of continuous sections $s:U \to \mathcal{A}$ that are norm bounded.  $\Gamma_0 (U, \mathcal{A} )$ denotes the C$^*$-subalgebra of $\Gamma^b(U,\mathcal{A})$ consisting of those sections $s$ that `vanish at infinity' on $U$ in the sense that
$$
\{ x \in U : \| s(x) \| \geq \alpha \}
$$
is compact for all $\alpha>0$.

Suppose that $A$ is a C*-algebra and $\phi: \mathrm{Prim}(A) \to X$ is continuous.  Combining Remark 3.7.3 and Theorem 5.6 of the paper "Sheaves of C$^*$-algebras" by Ara and Mathieu, we do get an upper-semicontinuous C$^*$-bundle $\pi: \mathcal{B} \to X$ over $X$ together with a canonical embedding of the multiplier algebra $M(A)$ into the C$^*$-algebra $\Gamma^b (X, \mathcal{B} )$.  If we then regard $A \subseteq \Gamma^b(X , \mathcal{B})$, one could then consider in each fibre $\mathcal{B}_x$ of $\mathcal{B}$ the subalgebra
$$
\mathcal{A}_x:=\{ b \in \mathcal{B}_x : b=a(x)\mbox{ for some }a \in A \}.
$$
It looks plausible that $A$ would then be isomorphic to (a subalgebra of) $\Gamma_0 (X, \mathcal{A} )$ where $\mathcal{A}$ is a sub-bundle of $\mathcal{B}$ with fibres $\mathcal{A}_x$, though I haven't checked the details.
As for the converse, suppose that $A = \Gamma_0 (X , \mathcal{A} )$ for some bundle $\pi : \mathcal{A} \to X$. I think that the existence of a map $\phi : \mathrm{Prim} (A) \to X$ can possibly be deduced from Lemma 2.25 of the paper "C*-algebras over topological spaces : the bootstrap class," by Meyer and Nest (here we need the assumption that $X$ is a sober space): 

Let $\mathcal{O}(X)$ and $\mathcal{I} (A)$ denote the directed sets of open subsets of $X$ and closed-two-sided ideals of $A$ respectively (both ordered with respect to inclusion).  Then there is a bijective correspondance between continuous maps $\mathrm{Prim}(A) \to X$ and maps $\mathcal{O} (X) \to \mathcal{I} (A)$ that commute with arbitrary suprema and finite infima.

Certainly, if $A$ is isomorphic to $\Gamma_0 (X, \mathcal{A} )$ for some u.s.c. C*-bundle $\pi:\mathcal{A} \to X$, then there is a natural map
$\mathcal{O}(X) \to \mathcal{I} (A)$ where $U \subseteq X$ is identified with the ideal 
$$\Gamma_0(U,\mathcal{A})=\{ a \in A: a(x) = 0 \mbox{ for all } x \in X \backslash U \} $$
of $A$, which appears to have the required properties.
It is worth pointing out that when constructing the bundle in point 1. one cannot simply adapt the proof from the case of locally compact Hausdorff $X$ in general (the one used in Appendix C of "Crossed products of C$^*$-algebras" by Dana Williams for example).
Indeed, for $x \in X$ define the ideal
$$
I_x : = \bigcap \{ P \in \mathrm{Prim} (A) : \phi (P)  = x \}
$$
and the quotient C*-algebra $A_x = A/I_x$.  Then there is a natural way to regard each $a \in A$ as a cross section $X \to \coprod_{x \in X} A_x$, letting $a(x)$ be the image of $a$ under the quotient mapping $A \to A/I_x$.  Indeed, for locally compact Hausdorff $X$, this identification gives the required $*$-isomorphism $A \to \Gamma_0(X,\mathcal{A})$.
Returning to the general case, if we take $X= \mathrm{Prim} (A)$ and $\phi$ to be the identity map, this construction yields an upper-semicontinuous C*-bundle with fibres $A_x$ and section algebra $\Gamma_0(X,\mathcal{A})\cong A$ if and only if $\mathrm{Prim} (A)$ is Hausdorff, in which case the bundle is in fact continuous.  Thus it looks like the fibres of the bundle constructed in point 1. will not coincide with $A_x$ in general.<|endoftext|>
TITLE: Can the topological algebra of analytic functions be endowed with a norm that defines the natural topology?
QUESTION [7 upvotes]: Right, so in my research in complex analysis I was puzzled by this question which may have a simple approachable answer that eludes me, but I am truly itching to find out and in need of it so I am requesting help here.
The question I face is:

If we have a domain (like the unit disk) D in the complex plane and we have the topological algebra of analytic functions in one variable on D, is there a suitable norm on this algebra such that it induces the "natural topology" i.e. the topology of convergence on compact sets?

I would truly appreciate a help in the form of a yes/no response with some convincing to settle this for me. I thank all kind helpers.

REPLY [8 votes]: By Montel's theorem, every bounded set (w.r.t. the topology of uniform convergence on compact sets) in the space of holomorphic functions is relatively compact. If the space were normed its closed unit ball would be compact which implies that the space is finite dimensional.<|endoftext|>
TITLE: what is the universal cover of GL(2,R)?
QUESTION [14 upvotes]: In the theory of Bridgeland stability conditions one has an action of the universal cover $G'$ of $G = GL^+(2,\mathbb R)$.

What is G'?

I know there is concrete description in terms of pairs (M,f) with M in G and f a function (see Lemma 2.14 of Huybrechts http://arxiv.org/pdf/1111.1745v2.pdf). But I'd like to know if G' is diffeomorphic to some other manifold.
At some point I convinced myself that G was actually diffeomorphic to $\mathbb C^* \times \mathbb C$ and therefore $G'$ was $\mathbb C^2$. But I've never seen this written down, so I'm getting suspicious.
EDIT: For those who have voted to close: I realize this question was very elementary. However, none of the resources on stability conditions I know of state this fact explicitly, which is quite frustrating. I think it would be very useful for people looking for a quick answer to have a place where this fact appears unambiguously.

REPLY [9 votes]: For the sake of completeness here is an explicit proof. Let $m\in GL_{2}(\mathbb{R})^{+}$, then $m\rightarrow \frac{m}{\det(m)}$ maps it to an element in $SL_{2}(\mathbb{R})$. And we know $SL_{2}(\mathbb{R})\cong \mathbb{R}^2\times \mathbb{S}^1$ via appropriate parametrization. Thus $GL_{2}(\mathbb{R})^{+}\cong \mathbb{R}^{+}\times \mathbb{R}^{2}\times \mathbb{S}^1\cong \mathbb{R}^{3}\times \mathbb{S}^1$ as a manifold. And its universal cover is $\mathbb{R}^4$. 
It might be overkill, but I think others were alluding to Iwasawa decomposition.<|endoftext|>
TITLE: Norm vs A-norm in non-Archimedean Functional Analysis
QUESTION [6 upvotes]: Let $K =(K,| \cdot |)$ be a non-Archimedean valued field.
Let $E$ be a $K$-vector space.  A norm  on $E$ is a map $||\cdot||:E\to[0,\infty)$ such that:

$||x||=0$ if and only if $x=0$,
$||\lambda x||=|\lambda|\,||x||$,
$||x+y||\leq\max\{||x||,||y||\}$,

for all $x,y\in E$.
A function $||\cdot||:E\to[0,\infty)$ satisfying conditions 1, 2 and $||x+y||\leq ||x||+||y||$ is called an A-norm on E.
Clearly every norm is an A-norm, but not conversely.
Why in the literature (van-Rooij, Schikhof, etc,) the study of Banach spaces and the development of non-Archimedean functional Analysis is done with norms instead of A-norms? Does it worth to generalize the results of norms to A-norms?

REPLY [3 votes]: I guess that putting an archimedean norm on a vector space over a nonarchimedean field gives just an uncorrelated product of something archimedean with something nonarchimedean. Number theorists sometimes look at all places of $\mathbb{Q}$ at once: all p-adic valuations and the archimedean valuation. I don't see a reason why one should want to look at just two places. I do not expect it would generate any new theory.
Anyway, the reason why people in nonarchimedean functional analysis work with nonarchimedean norms is that the vector spaces over a nonarchimedean field which turn up naturally, come with a nonarchimedean norm. Like finite dimensional vector spaces or spaces of functions.<|endoftext|>
TITLE: Choosing two-colorable subgraph in a triangulation
QUESTION [7 upvotes]: Consider a planar graph $G$ which is a triangulation. 
Is it possible to find a two-colorable subgraph $H$ of $G$ which has a common edge with every face of $G$?
It is known that it is not always possible to take $H$ to be a spanning tree. See this MO question. 
Note that we do not require $H$ to be connected. 
If it is true, is there an efficient algorithm to find such a subgraph?

REPLY [14 votes]: Yes, such a subgraph always exist.  Let $G$ be a planar triangulation.  By the $4$-colour theorem, $G$ has a $4$-colouring.  We let $H$ be the subgraph consisting of all edges with endpoints coloured $1$ and $2$, or with endpoints coloured $3$ and $4$.  Since every face of $G$ is a triangle, every face must contain a $12$ edge or a $34$ edge, as required.  Also, $H$ is clearly bipartite since $(X,Y)$ is a bipartition of $H$ where $X$ is the set of vertices coloured $1$ or $3$ and $Y$ is the set of vertices coloured $2$ or $4$.  
Regarding the algorithmic question, there is a quadratic algorithm to find such a subgraph.  This follows from this paper of Robertson, Sanders, Seymour, and Thomas, where they present a quadratic algorithm to $4$-colour planar graphs.<|endoftext|>
TITLE: The axiom $I_0$ in the absence of $AC$
QUESTION [16 upvotes]: It is well-known that if $AC$ holds and if $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$ is a non-trivial elementary embedding with $crit(j) < \lambda,$
then $\lambda$ has countable cofinality (and in fact it is the least fixed point of $j$ above $crit(j)$).

Question. Is $AC$ needed to show that $\lambda$ has countable cofinality.

In other words, is it possible to show, just working in $ZF$ that, $cf(\lambda)=\omega.$
Remark. I can show that if we can prove the result without AC, then there are no Reinhardt cardinals in ZF.

REPLY [3 votes]: It is consistent that $AC$ fails and there exists a non-trivial elementary embedding $j: L(V_{\lambda+1}) \to L(V_{\lambda+1})$  with $crit(j) < \lambda,$
and $\lambda$ has uncountable cofinality. See BERKELEY CARDINALS AND THE STRUCTURE OF $L(V_{δ+1})$.<|endoftext|>
TITLE: Representation viewpoint on Chern–Weil (cohomology computations done with rep theory?)
QUESTION [12 upvotes]: $\DeclareMathOperator\Sym{Sym}$Let $G$ be a compact lie group. Chern–Weil theory tells us that there's a homomorphism:
$$H^{*}(BG;\mathbb{R}) \to (\Sym^{\bullet} \mathfrak{g^*})^G$$
which in our case is an isomorphism since $G$ is compact. The procedure I know to prove this is pretty ad-hoc. Starting with a $G$-bundle on a manifold $M$ we pick an arbitrary connection and evaluate invariant polynomials on its curvature form to get characteristic classes.

What is the representation theoretic viepoint on the isomorphism $H^{*}(BG;\mathbb{R}) = (\Sym^{\bullet} \mathfrak{g^*})^G$?
For a finite dimensional lie group $G$ (adding compact here doesn't matter) is there always a canonical way to build $BG$ as a colimit of homogeneous manifolds?

Recently I found that many computations in algebraic topology can be simplified using representation theory. Here are some more complutations I'd like to be able to understand in representation theoretic terms:
1. Cohomology ring of a homogeneous space $H^*(G/H)$.
2. Cohomology ring of a parallel curvature cartan geometry $(P, \omega)$ for the pair $(\mathfrak{g},H)$ with curvature form $K \in \operatorname{Hom}(\bigwedge^2\mathfrak{g}/\mathfrak{h}, \mathfrak{g})$.  (Side question: is $K$ some kind of cocycle here?)
Is there a reference for these kinds of computations?

REPLY [10 votes]: The construction you describe appears in Tamvakis' The connection between representation theory and Schubert calculus (Enseign. Math. 50 (2004), 267-2860). Basically, instead of working with representations of $GL(n)$ he works with representations of Vec, which he then applies to the tautological bundle over the Grassmannian. The main result is that this map corresponds (part of) the basis of irreps with the basis of Schubert classes.<|endoftext|>
TITLE: $G$-invariant holomorphic vs. polynomial functions
QUESTION [6 upvotes]: Let $X\subseteq\Bbb C^n$ be a smooth affine variety and $G$ a complex reductive group acting algebraically on $X$. 
Let $x_0\in X$. If there is a non-constant $G$-invariant holomorphic function $f:X\to\Bbb C$ such that $f(x_0)\neq 0$, must there also exist a non-constant $G$-invariant polynomial $p\in\Bbb C[X]$ such that $p(x_0)\neq 0$?

REPLY [4 votes]: Luna has proved that every holomorphic invariant is of the form $h(p_1,\ldots,p_n)$ where $h$ is a holomorphic function in $n$ variables and $p_1,\ldots,p_n$ are polynomial invariants. By the way, Luna proves the same with "holomorphic" replaced by "continuous". The statement is wrong for real differentiable or real analytic functions, though: Take $G=\{\pm1\}$ and $X=\mathbb C$. Then the ring of invariants is generated by $z^2$ and $f(z):=z\overline z=\|z^2\|$ is a counterexample.
Reference: Luna, Domingo, Fonctions différentiables invariantes sous l'opération d'un groupe réductif. Ann. Inst. Fourier (Grenoble) 26 (1976), no. 1, ix, 33–49<|endoftext|>
TITLE: Homotopy Type of the Based Mapping Space $Map_*^{(k,l)}(\mathbb{C}P^2,BU(2))$
QUESTION [6 upvotes]: Path components of the based mapping space $Map_*(\mathbb{C}P^2,BU(2))$ are indexed by a pair of integers $(k,l)$ determined by the values of the first two Chern classes that a map $f:\mathbb{C}P^2\rightarrow BU(2)$ pulls back, loosely $(f^*c_1,f^*c_2)=(k,l)$. The space $Map_*^{(k,l)}(\mathbb{C}P^2,BU(2))$ is then the classifying space of the based gauge group of the principal $U(2)$-bundle over $\mathbb{C}P^2$ having first and second Chern classes $c_1=k$, $c_2=l$, respectively. 
I am specifically interested in the homotopy type (they share a common type thanks to the coaction of $S^4$ on $\mathbb{C}P^2$) of the components $Map_*^{(1,l)}(\mathbb{C}P^2,BU(2))$. Being even more particular I would like most of all to determine the 2-component of $\pi_3Map_*^{(1,l)}(\mathbb{C}P^2,BU(2))$.
The problem is that I am completely out of ideas as to how to access more homotopic information. The evaluation fibration in which it sits places it between even more mysterious spaces. It fibres over $Map_*^k(S^2,BU(2))\simeq\Omega_0U(2)$ but I have no obvious way to identify the fibre since the pinch map takes $Map_*^l(S^4,BU(2)$ into the wrong component (and i have in fact shown that the fibre must in fact have a different homotopy type).
Any thoughts on how to access some of the homotopy of this space are very welcome.

REPLY [9 votes]: [This version has been updated in response to comments from the OP]
Recall that $B$ gives an endofunctor of the category of abelian topological groups.  We can apply $B$ to the obvious map $\mathbb{Z}\to\mathbb{Z}/2$ to get a homomorphism $S^1=B\mathbb{Z}\to B\mathbb{Z}/2$, then compose with $\det\colon U(2)\to S^1$ to get a homomorphism $U(2)\to B\mathbb{Z}/2$.  The fibre of this is the group $\{(g,u)\in U(2)\times S^1:\det(g)=u^2\}$, which is easily identified with $S^3\times S^1$.  We can now apply $B$ to get a fibration
$$ B\mathbb{Z}/2 \to BS^3\times BS^1 \to BU(2) \to K(\mathbb{Z}/2,2). $$ 
(Note here that $S^1$ is implicitly embedded in $U(2)$ as the centre, rather than the top left copy of $U(1)$.)  We can now apply the functor $\text{Map}_*(\mathbb{C}P^2,-)$ to get a fibration
$$ 0 \to \text{Map}_*(\mathbb{C}P^2,BS^3)\times \mathbb{Z} \to \text{Map}_*(\mathbb{C}P^2,BU(2)) \to \mathbb{Z}/2. $$ 
Now let $P_k$ denote the subspace of $\text{Map}_*(\mathbb{C}P^2,BU(2))$ consisting of maps that have degree $k$ on the bottom cell, and put $M=\text{Map}_*(\mathbb{C}P^2,BS^3)$.  From the above fibration we can see that $P_k\simeq M$ whenever $k$ is even.
Next, we have a cofibre sequence $S^3\xrightarrow{\eta}S^2\to\mathbb{C}P^2$ (where $\eta$ is the Hopf map), giving a fibration 
$$ M \to \Omega^2BS^3 = \Omega S^3 \xrightarrow{\eta^*} \Omega^3BS^3=\Omega^2S^3. $$
It is known that $\Omega^2S^3=S^1\times W$ for a space $W$ whose homotopy and (co)homology groups are all finite.  It is easy to see that $\text{Map}_*(\Omega S^3,S^1)$ is contractible, so $\eta^*$ really lands in $W$.  
The above fibration gives short exact sequences $C_{i+1}\to\pi_iM\to K_i$, where $K_i$ and $C_i$ are the kernel and cokernel of the map $u\mapsto u\circ\Sigma^{i-1}\eta$ from $\pi_i\Omega S^3=\pi_{i+1}S^3$ to $\pi_{i+2}S^3$.  At least the $2$-torsion part of these groups can be read off (in a substantial range) from Toda's book "Composition methods in the homotopy groups of spheres".  If I have got everything straight, the first few nontrivial groups are as follows:

$K_2$ is $\mathbb{Z}$, generated by $2\iota$.
$C_4$ is $\mathbb{Z}/2$, generated by (the image of) $\nu'$.  Here $\nu'\colon S^6=S^3\wedge S^3\to S^3$ can be obtained from the commutator map $S^3\times S^3\to S^3$ by collapsing out the axes; it is conceivable that this description could be illuminating.
$K_5$ is $\mathbb{Z}/2$, generated by $2\nu'$
$K_7$ is $\mathbb{Z}/2$, generated by $\nu'\eta^2$
$C_{11}$  is $\mathbb{Z}/2$, generated by $\epsilon$

This gives $\pi_2M=\mathbb{Z}$ and $\pi_3M=\pi_5M=\pi_7M=\pi_{10}M=\mathbb{Z}/2$ (ignoring any odd torsion).  
You can probably also extract some information about the mod $2$ (co)homology of $M$ using the same fibration.  Recall that $H_*(\Omega S^3;\mathbb{Z})=\mathbb{Z}[x]$ with $|x|=2$, but the multiplication here comes from the loop structure and $\eta^*$ is not a loop map, so it will not give a ring map in homology.  Instead we should use $H^*(\Omega S^3;\mathbb{Z}/2)$, which is generated by classes $u_i$ in degree $2^{i+1}$ (for $i\geq 0$) satisfying $u_i^2=0$.  It is known that $\Omega S^3$ splits stably as a wedge of spheres, and this means that all Steenrod operations are trivial.  On the other hand, $H_*(\Omega^2S^3;\mathbb{Z}/2)$ is polynomial on generators $y_i$ in degrees $2^{i+1}-1$ for $i\geq 0$, and the corresponding cohomology ring has generators $v_{ij}$ in degree $2^j|y_i|$ satisfying $v_{ij}^2=0$.  These support many nontrivial Steenrod operations, which probably forces the map 
$$ (\eta^*)^* \colon H^*(\Omega^2S^3;\mathbb{Z}/2) \to H^*(\Omega S^3;\mathbb{Z}/2)$$
to be mostly zero.  All this story was worked out by Fred Cohen in his section of the book "Homology of iterated loop spaces".
It would also be good to understand $P_k$ when $k$ is odd.  For this, we can use the following framework.  Suppose we have a based space $B$ and a Hurewicz fibration $p\colon E\to B$ with fibre $F=p^{-1}\{*_B\}$.  We can choose any point $x\in F$ and use it as the basepoint for $F$ and also for $E$; this then gives a fibration $\Omega_xF\to\Omega_xE\to\Omega B$ which gives information about the homotopy type of the component of $x$ in $F$.  We can now take $B=\text{Map}_*(S^3,BU(2))$ and $E=\text{Map}_*(S^2,BU(2))$ with $p=\eta^*$; this gives $F=\text{Map}_*(\mathbb{C}P^2,BU(2))$.  If we pick a point $x\in P_1\subset F$, we get a fibration as before, involving $\Omega_xF=\Omega_xP_1$.  Here $E$ can be identified with $\Omega^2BU(2)=\Omega U(2)$, which has a group structure.  This means that $\Omega_xE$ can be identified with the loop space at the usual basepoint, giving $\Omega_xE=\Omega^2U(2)=\Omega^2S^3$ (using $\Omega^2S^1=0$).  Similarly, we have $\Omega B=\Omega^3U(2)=\Omega^3S^3$.  Thus, $\Omega_xP_1$ is the fibre of a certain map $q\colon\Omega^2S^3\to\Omega^3S^3$.  This is probably not exactly the same as $\eta^*$, because of the change of basepoint in $E$.  It should be possible to work out the details, but I have not done so.<|endoftext|>
TITLE: Families of abelian varieties on the line (or more generally simply connected varieties)
QUESTION [7 upvotes]: I'm curious whether the following is true:

Question 1: Let $V/\mathbb{C}$ be a smooth connected variety such that $V^\text{an}$ is simply connected. Then, is every abelian scheme $f:\mathscr{A}\to V$ isotrivial?

Specifically I'm curious about the case when $V=\mathbb{A}^1_\mathbb{C}$.
I think the answer to Question 1 is yes by the following line of reasoning. Deligne shows in Theorie de Hodge II that $\mathscr{A}$ is characterized by $(R^1f^\text{an}_\ast\underline{\mathbb{Z}})^\vee$ as polarizable variation of Hodge structure of type $\{(-1,0),(0,-1)\}$. Now, in our case we know that $(R^1f_\ast^\text{an}\underline{\mathbb{Z}})^\vee$ is a $\mathbb{Z}$-local system and so trivial as such. But, of course, there is no a priori reason to believe that the Hodge filtration is constant.
That said, in this article (see the remarks following Theorem 11) any $\mathbb{Q}$-VHS on a simply connected compactifiable complex manifold is actually constant (as a VHS). In particular, this should imply (since every smooth variety is compactifiable by resolution of singularities), by Deligne's theorem that there is a constant family $\underline{A}$ on $V$ and an isogeny $\varphi:\underline{A}\to\mathscr{A}$. But, $\varphi$ must be $\underline{\mathbb{Z}/n\mathbb{Z}}$ and so $\mathscr{A}$ must be $\underline{A/(\mathbb{Z}/n\mathbb{Z})}$ as desired.
Is this correct?
It then leads to the following natural questions

Question 2: Is every $\mathbb{Z}$-VHS on a simply connected compactifiable (perhaps even algebraic) manifold constant?

and

Question 3: Let $V/\mathbb{C}$ be a smooth connected variety such that $\pi^1_{\acute{e}\text{t}}(V)=0$. Then, is every abelian scheme $f:\mathscr{A}\to V$ isotrivial?

I apologize if these are silly questions. 
Also, of note, there must be something truly algebraic happening here as the 'universal family' $\mathscr{E}/\mathbb{H}$ shows.
Thanks!

REPLY [3 votes]: Here's a bit of a long comment which I hope will help the OP. I also answer some of the questions that arose in the comments.
Firstly, any smooth complex algebraic quasi-projective variety carrying an immersive period map is known to be "hyperbolic" in the sense that 
i) it is Brody hyperbolic,  
ii) all its subvarieties are of log-general type, and
iii) the fundamental group of $X$ is infinite (and more...). 
The hardest of these to show is ii) which follows from a theorem of Kang Zuo; see Theorem 0.1 in https://www.researchgate.net/publication/252994947_On_the_negativity_of_kernels_of_Kodaira-Spencer_maps_on_Hodge_bundles_and_applications
or the  more general
Theorem 0.3 in Brunebarbe's http://sma.epfl.ch/~brunebar/Articles/Symmetric%20differentials%20and%20variations%20of%20Hodge%20structures.pdf
These results are used in Lemma 6.3 of http://arxiv.org/pdf/1505.02249v1.pdf . The statement of Lemma 6.3 is a consequence of the statement in Brunebarbe as any PVHS coming from geometry is unipotent (up to finite etale base-change of the base). The result of Zuo is also used in the proof of Proposition 3.1 in Abromovich--Varilly-Alvarado's http://arxiv.org/pdf/1601.02483v2.pdf . I mention these statements as they provide a more direct link to your questions.
To prove iii) you can use the argument given by Arapura or use Theorem 3.1 in Voisin's second book on Hodge theory (which is the same argument as in Arapura's answer).
The above results of Zuo (and Brunebarbe) tell us that all horizontal (smooth) algebraic  subvarieties of period domains of PVHS's are hyperbolic. This is I think also the best one could hope for: the period domains of PVHS's are not in general "hyperbolic".  
Finally, the argument in Arapura's answer is  applied to the stack $\mathcal {A}_{g,1}^{\textrm{an}}$ or, if stackiness bothers you, to the fine moduli space of $g$-dimensional ppav's with full level $3$-structure. It is, in my opinion, unnatural to consider the coarse moduli space (which is not hyperbolic, as it contains copies of the $j$-line) when studying hyperbolicity of parameter spaces of ppav's (or smooth proper curves of genus at least two, or polarized K3 surfaces, polarized Calabi-Yau manifolds, etc.).  Note that Siegel upper half-space $\mathbb H_g$ is the universal cover of the stack $\mathcal {A}_{g,1}$. The holomorphicity of a lift $\mathbb C\to \mathbb H_g$ for a holomorphic map $\mathbb C\to \mathcal A_{g,1}^{\textrm{an}}$ is therefore immediate. 
I can't give you an   intuitive reason as to why $\mathcal A_{g,1, \overline{\mathbb F_p}}$ is no longer hyperbolic when $g\geq 2$. If $g=2$ the non-hyperbolicity comes from the supersingular locus. Does that help?<|endoftext|>
TITLE: Does $E_8$ know $Spin(7)$?
QUESTION [22 upvotes]: One way to define the compact group $Spin(7)$ is as the stabilizer of a certain 4-form on Euclidean $\mathbb R^8$ (see e.g. this MO question).  This 4-form can be defined in various ways.  For example, it can be interpreted in terms of octonions as $\Omega(w,x,y,z) = \langle w, x(\bar y z) - z(\bar y x)\rangle)$ where $y \mapsto \bar y$ is octonionic conjugation and $\langle,\rangle$ is the standard inner product.
A curious fact about this 4-form is that the indices that appear in each monomial are precisely the (length four) words in the Hamming Code $Ham(8,4)$.  Is this just a coincidence (perhaps because all of these objects relate to the octonions)?  The $E_8$-lattice can be constructed from $Ham(8,4)$ in a routine way.

Is there a machine that, when fed $E_8$ lattice, produces the group $Spin(7)$?

REPLY [3 votes]: After a couple days of thought, I understand how $Ham(8,4)$ knows $Spin(7)$.  I will describe a way here.  Since I am answering my own question, I have marked this answer Community Wiki.  You are explicitly invited to edit the answer however you see fit to improve it.  If you make substantive edits, of course you should probably change this paragraph as well.  --Theo

The real Clifford algebra $Cliff(8)$ is naturally a "twisted group algebra" for $\mathbb F_2^8$ in the following sense: it has an $\mathbb R$-basis $x_\alpha$ indexed by $\alpha \in \mathbb F_2^8$ with $x_\alpha x_\beta = \epsilon(\alpha,\beta) x_{\alpha + \beta}$ where $\epsilon = \pm 1$ is some group cocycle on $\mathbb F_2^8$.  Of course, different people might make different choices for the cocycle $\epsilon$ --- changing it by a coboundary corresponds to changing the basis by a diagonal matrix.  I will fix the following choice: identify $\mathbb F_2^8$ with the power set of $\{1,\dots,8\}$; let $x_1,\dots,x_8$ be the generators of $Cliff(8)$; if $\{i,j,\dots,k\} \subseteq \{1,\dots,8\}$ is ordered (so $i<j<\dots<k$) then $x_{ij\dots k} = x_ix_j\dots x_k$.  (I will not use expressions like "$x_{21}$" or "$x_{11}$".)
The Hamming Code $Ham(8,4) \subseteq \mathbb F_2^8$ is a subgroup isomorphic to $\mathbb F_2^4$.  Let's try to lift this to an inclusion $\mathbb F_2^4 \to Cliff(8)$.
Recall the definition of the Hamming Code: $\emptyset$ and $12345678 \in Ham(8,4)$, and the rest consists of certain 4-element subsets of $\{1,\dots,8\}$; to figure out which ones, run through all 4-element subsets in alphabetical order, and keep all the ones that have at most two elements of overlap with all previously-kept terms.  The end result implies that if $\alpha,\beta \in Ham(8,4)$ and $|\alpha| = |\beta| = 4$, then $|\alpha \cap \beta| = 0$ or $2$.  This implies in turn that if $\alpha,\beta \in Ham(8,4)$ then $x_\alpha^2 = x_\beta^2 = 1$ and $x_\alpha x_\beta = x_\beta x_\alpha$.
There's still the sign problem that in general $x_\alpha x_\beta \neq x_{\alpha + \beta}$.  To fix this, for $\alpha \in Ham(8,4)$, set $y_\alpha = (-1)^{|\alpha|/4}x_\alpha$.  Then
$$ y : \mathbb F_2^4 \cong Ham(8,4) \hookrightarrow Cliff(8) $$
is a multiplicative map.  So a lift exists.
Claim: $Spin(7) \subseteq SO(8)$ is precisely the stabilizer of $\sum_{\alpha \in Ham(8,4)} y_\alpha$ (under the conjugation $SO(8)$-action on $Cliff(8)$).
Since the $SO(8)$ action is inner, the image of $\mathfrak{so}(7)$ under this action is precisely the quadratic part of the centralizer of $\sum_{\alpha \in Ham(8,4)} y_\alpha$.
Proof: In the notation of this MO question, $\sum_{\alpha \in Ham(8,4)} y_\alpha = 1 + x_{12345678} - \Omega_0$.  But $1$ and $x_{12345678}$ are stabilized by all of $SO(8)$.
How many choices were there?  For any choice of four generators of $Ham(8,4)$ (say $1234, 1256, 1278, 1357$) you could set $y'_\alpha = \pm y_\alpha$, but that's it, so there are $2^4$ choices, generated by $4$ of them.  (It must go to the line spanned by $x_\alpha$ and have the same square.)  You can absorb these choices by changing $x_i$ to $-x_i$ for various $i$s.  There are naturally two families of choices distinguished by whether element $12345678 \in Ham(8,4)$ is mapped to $x_{12345678}$ or $-x_{12345678}$.  Any two choices in the same family are related by conjugating by diagonal matrices in $SO(8)$.  (The two families are related by conjugating by diagonal matrices in $O(8)$ but not in $SO(8)$.)
So there's really not much choice, and the Hamming Code does "know" $Spin(7)$.  I think what's going on with the two choices is the following.  Let $\tau : Spin(8) \overset \sim \to Spin(8)$ denote the triality map, and let $\iota : Spin(7) \to Spin(8)$ denote the usual embedding (covering the canonical inclusion $\mathfrak {so}(7) \to \mathfrak {so}(8)$ which is $0$ in the last row and column) and $\pi : Spin(8) \to SO(8)$ the double cover.  Then the composition $\pi \tau \iota : Spin(7) \to SO(8)$ is the embedding in question.  But you could also have used the "other" triality $\tau^{-1}$, which corresponds to choosing the other spin module for $Spin(8)$.<|endoftext|>
TITLE: References for metrics in matrix groups
QUESTION [5 upvotes]: I am studying a very concrete matrix group with a riemaniann (right invariant) metric  for solving a question on Applied Math. I need explicit formulas for the distance between two matrices, geodesics and so on.
The matrix group I am studying is  $UT(n)$, the $n\times n$ upper triangular matrices $\Pi=(\Pi_{i j})$ with positive diagonal elements and $\Pi_{11}=1$; but specially $UT(3)$.
The metric is defined in the book of C.G. Small The Statistical Theory of Shape ( p. 101-106): 
Let $\Pi_x$ and $\Pi_{x+dx}$ in  $UT(n)$ 
and let $\Lambda = \Pi_{x+dx}\cdot \Pi_{x}^{-1}$. The infinitesimal 
distance ds from  $\Pi_x$ to $\Pi_{x+dx}$ is given by the formula
$$ds^2= \frac{\sum (\lambda_j-\bar\lambda)^2}{n}$$ where $\lambda_j$ are the eigenvalues of $\Lambda^T\Lambda$ and $\bar\lambda=\frac {\sum \lambda_j}{n}$.
Looking for Lie groups I found references for metric questions,  but in a very general setting of differential geometry. 
In the other hand, I found introduction references for matrix groups, but they avoid metric considerations. 
And thus my question: Are there references in matrix groups that avoid generalities but have a lot of results in the metric structure?

REPLY [3 votes]: This paper might give you some ideas on how to calculate the geodesics. It is about left invariant metrics on $GL_n(\mathbb{R})$. The geodesics are calculated using their characterization as critical points of the energy functional. 
(This is not exactly your case, but perhaps you can adapt some of the arguments there).
Note however, that even after calculating the geodesics, the actual calculation of the distance can still be intractable.<|endoftext|>
TITLE: List is a monad, but is it a comonad with these natural transformations?
QUESTION [6 upvotes]: List is known to be a monad.  It takes a set and maps it to lists of elements of that set.  The natural transformations are, singleton and flatten, whereby we map a set to a set of singleton lists each with one element, and we take lists of lists and flatten them into lists (somewhat like concatenating).  Suppose we define two more natural transformations, copy and delete.  Copy takes a list and maps it to a list of two lists, they being copies of the original list.  Delete takes a list and maps it to the underlying set of the List functor.  The Delete natural transformation seems a bit dubious.  Does this form a monad that is also a comonad?  Here we have a paper that seems to suggest that non-empty list is a comonad.  Does this apply to the example I am talking about?

REPLY [17 votes]: Todd's comment provides an important limitation on what you can do here, but here's what I think is the most interesting way to answer your question.  Define an endofunctor $L^+$ on the category of sets by
$$
L^+(X) = \sum_{n \geq 1} X^n
$$
for sets $X$, where $\sum$ means coproduct (disjoint union).  Thus, an element of $L^+(X)$ is a nonempty finite list of elements of $X$.
Then $L^+$ can be given the structure of both a monad and a comonad.  
The monad structure is the same as the one you described for possibly-empty lists: the unit sends $x \in X$ to the single-element list $x$, and the multiplication is concatenation ("flattening").  Its algebras are semigroups.
The comonad structure is less well-known (at least, to mathematicians; it seems better known in computer science).  The counit map $L^+(X) \to X$ sends a list $(x_1, \ldots, x_n)$ to $x_1$.  The comultiplication $L^+(X) \to L^+(L^+(X))$ forms the "tails" of a list; for instance,
$$
(x_1, x_2, x_3, x_4)
\mapsto
((x_1, x_2, x_3, x_4), (x_2, x_3, x_4), (x_3, x_4), (x_4))
$$
and the general definition is what you'd guess from this.
(In fact, there's a relationship between the monad structure and the comonad structure on $L^+$, or more exactly a mixed distributive law between them, with $L^+$ itself as its underlying functor.  In the terminology that some people use, this makes $L^+$ into a "bimonad".)
More generally, it sounds like you're looking for examples of monads that are also comonads. There are many of these: e.g. for any monoid $M$, the endofunctor $M \times -$ on the category of sets is both a monad and a comonad in a natural way.<|endoftext|>
TITLE: Schwartz-Zippel lemma for an algebraic variety
QUESTION [5 upvotes]: Let $X $ be a smooth affine subvariety of $(\overline{\mathbb{F}_q})^n$ defined by a prime ideal $I$. Let $f$ $\in \mathbb{F}_q[x_1,\ldots,x_n]$ be a polynomial such that $f \notin I$. 
Let $r_1, \ldots, r_m$ be selected independenty and uniformly among all $\mathbb{F}_{q^k}$-rational points of $X$ for some big $k$.
Is it true that $\Pr[f(r_1)=0, \ldots, f(r_m) =0]$ is small for rather big $m$? 
It would be a generalization of
Schwartz-Zippel lemma:

Lemma (Schwartz, Zippel).
  Let
  $P\in F[x_1,x_2,\ldots,x_n]$
  be a non-zero polynomial of total degree $d \geq 0$ over a field, $F$. Let $S$ be a finite subset of $F$ and let $r_1, r_2, \dots, r_n$ be selected at random independently and uniformly from $S$.
  Then
  $\Pr[P(r_1,r_2,\ldots,r_n)=0]\leq\frac{d}{|S|}.$

UPD: I think I have understood why this must be true: $f \cap V$ is the variety with smaller dimension than $V$ and it has much smaller $\mathbb{F}_{q^k}$-rational points than $V$ (Number of rational points in a non-smooth variety). However I can not get an explicit estimation...

REPLY [6 votes]: Let $d=\dim(X)$ and let $Y=X\cap\{f=0\}$, so $\dim(Y)=d-1$. Then the Lang-Weil estimate says that $\#X(\mathbb F_{q^k})=q^{dk} + O(q^{k(d-1/2)})$, and applying it to each irreducible component of $Y$ gives $\#Y(\mathbb F_{q^k})=O(q^{(d-1)k})$. Your probability is the ratio $\#Y(\mathbb F_{q^k})^m/\#X(\mathbb F_{q^k})^m$, if I understand correctly, which equals $O(C^mq^{-mk})$, which is indeed small when $m$ is large, provided you choose $k$ large enough. If you want to include an error term, which would of course depend on the ideal defining $X$ and the function $f$ defining $Y$, then the big-$O$ constant may be taken to be the number of components $\nu(Y)$ of $Y$, i.e., something like $\text{Prob}\le \nu(Y)^mq^{-mk}+O(q^{-(m+1/2)k})$<|endoftext|>
TITLE: "Small" simplicial complex with torsion trees
QUESTION [5 upvotes]: I am giving an expository talk soon about Duval-Klivans-Martin's paper Simplicial Matrix Tree Theorems, and I've been struggling to find a good example to do at the board. An important aspect of the theory is that some spanning trees have torsion, and I want to highlight this fact. Here is the definition:

If $\Delta$ is a simplicial complex and $\Gamma^d$ is a subcomplex which agrees with $\Gamma$ in positive codimension $\Delta_{(d-1)}=\Gamma_{(d-1)}$, then $\Gamma$ is called a spanning tree if any two (and hence all) of the following conditions hold:

$\tilde H_d(\Gamma,\Bbb Z)=0$
$|\tilde H_{d-1}(\Gamma,\Bbb Z)|<\infty$
$f_d(\Gamma)=f_d(\Delta)-\tilde\beta_d(\Delta)+\tilde\beta_{d-1}(\Delta)$.


Ideally, I'm trying to find a simplicial complex which has both torsion and nontorsion spanning trees, and yet has few enough spanning trees that it could be reasonably computed on the board. Some candidates which don't fulfill these purposes include

A minimal triangulation $\Bbb{RP}^2$ is a torsion tree, which shows existence but is not so interesting.
You can stick a sphere out of $\Bbb{RP}^2$ (like the connect sum but without removing the disk) and this complex has four trees, but alas they are all torsion.
There is a minimal triangulation of $\Bbb{RP}^2$ which spans the $2$-skeleton of the $6$-simplex, but there are far too many spanning trees of this complex to do on the board.

REPLY [7 votes]: Here is an example (which might be be what you are describing in your third bullet point) that I think is both doable and illustrates torsion spanning trees. Consider the minimal triangular of $\mathbb{RP}^2$ shown below.
$\hskip 2in$ 
Let $\Gamma$ denote this simplicial complex which triangulates $\mathbb{RP}^2$.
Let $\Delta = \Gamma \cup \{1,2,3\}$ and let $\Theta = \Delta \setminus \{4,5,6\}$. Then $\Gamma$ is a simplicial spanning tree of $\Delta$ with torsion while $\Theta$ is a simplicial spanning tree of $\Delta$ with out torsion. Further $\Theta' = \Delta \setminus \{i,j,k\}$ for any $\{i,j,k\} \ne \{1,2,3\}$ will be homotopy equivalent to $\Theta$ and thus have the same homology.<|endoftext|>
TITLE: A criterion for orbits of complex reductive group to be closed
QUESTION [5 upvotes]: I am having some trouble understanding Nakajima's proof of the Kempf-Ness theorem in [1]. At the end (proof of Proposition 3.9(6)), his argument is basically the following:

Let $G=K_{\Bbb C}$ be a complex reductive group acting linearly on a complex vector space $V$. Let $v\in V$. If there is $X\in\operatorname{Lie}(K)$ such that the function $t\mapsto\|\exp(itX)\cdot v\|^2$ diverges, then the orbit $G\cdot v$ is closed.

Why is that? I know that if there exists $w\in\overline{G\cdot v}-G\cdot v$ then there is a one-parameter subgroup $\lambda:\Bbb C^*\to G$ such that $\lambda(t)\cdot v$ converges to $w$. But does that imply that $\exp(itX)\cdot v$ converges for all $X\in\operatorname{Lie}(K)$?

More precisely, his proof is:
$$\begin{align}
&\text{The function }p_v:G\to\Bbb R,g\mapsto\|g\cdot v\|^2\text{ has a minimum at }g=e\\
\implies&\exists X\in\operatorname{Lie}(K_v)^\perp\text{ such that }p_v(\exp(itX))\text{ diverges}\\
\implies& G\cdot v\text{ is closed}.
\end{align}$$
The first implication is fully explained and I understand it. But the second implication is mentioned without any justification.

Reference.
[1] Nakajima, H. Lectures on Hilbert schemes of points on surfaces, Vol. 18. Providence, RI: American Mathematical Society, 1999.

REPLY [2 votes]: I like the proof in:
Schwarz, Gerald W., The topology of algebraic quotients. Topological methods in algebraic transformation groups (New Brunswick, NJ, 1988), 135–151, Progr. Math., 80, Birkhäuser Boston, Boston, MA, 1989.
Summary: Suppose $Gv$ is not closed, then there is a 1-parameter subgroup $\lambda:\mathbb{C}^*\to G$ such that $\lim_{z\to 0} \lambda(z)v\notin Gv$.  Then $||\lambda(e^s)v||^2=\sum e^{2n_js}|v_j|^2$ and since $\lim_{s\to -\infty}\lambda(e^s)v$ exists we must have $n_j\geq 0$ for all $j$, with at least one strict inequality since the limit is not in $Gv$. Therefore $\frac{d}{ds}||\lambda(e^s)v||^2\not=0$ at $s=0$, and so $p_v$ is not critical at $g=e$.
Here is one possible way to understand the parts of Nakajima's proof that you highlight:

The divergence of $p_v(\mathrm{exp}\ t\sqrt{-1}\xi)$ as $t\to \infty$ implies the image of $p_v$ is complete (since we have a minimum and the image is path-connected).
If the orbit was not closed, then there is a 1-parameter subgroup $\lambda:\mathbb{C}^*\to G$ such that $\lim_{z\to 0} \lambda(z)v:=w\notin Gv$ (as in the above summary).
Since $p_v$ has a min, $Kv=\{x\in Gv\ |\ ||x||=||v||\}$.  Thus, if there exists $g_0v$ so $||g_0v||=||w||$, since $w$ is a limit point and $Kv$ is closed (since $K$ is compact), we have $w=g_0kv$ for some $k\in K$, a contradiction.
Hence $||w||^2$ is not in the image of $p_v$.  But step 2. showed that to be impossible since the image of $p_v$ is complete.

Remark: As stated in the comments to your question, you are misreading the proof a bit.  The quantifier is "for all" (not "there exists") $\xi\in \mathfrak{k}_v^\perp$ such that $||\xi||=1$.<|endoftext|>
TITLE: Cohomology ring of a fiberwise join
QUESTION [8 upvotes]: I am very interested in the cohomology ring of the following construction. Let $f: Y\to X$ be a map between (connected) topological spaces. Suppose that the image of the map $f^*:H^*(X) \to H^*(Y)$ is isomorphic as a ring to the quotient $H^*(X)/I$. Recall that the fiberwise join $Y\ast_X Y$ is the homotopy pushout of the diagram:
$$
\require{AMScd}
\begin{CD}
Y\times_X Y @>{p_1}>> Y\\
@V{p_2}VV @VVV \\
Y @>>> Y\ast_X Y
\end{CD} $$
Here $Y\times_X Y$ is the homotopy pullback:
$$
\require{AMScd}
\begin{CD}
Y\times_X Y @>{p_1}>> Y\\
@V{p_2}VV @VV{f}V \\
Y @>{f}>> X
\end{CD} $$
Then there exists the natural map $f*f:Y*_XY \to X$. I strongly believe that the image of the map $(f*f)^*:H^*(X)\to H^*(Y*_XY)$ is isomorphic as a ring to $H^*(X)/I^2$. Unfortunately, I have no idea how to prove this.
I am going to use this in the case of the natural map $f: G/T\to BT$. Here  G is a Lie group and T is its maximal torus. In this case we understand a lot (at least in rational cohomology) about the Leray spectral sequence for the map $f$. So maybe, there exists a connection between the Leray spectral sequence for the map $f*f$ and the Leray spectral sequence for the map $f$.

REPLY [11 votes]: What you wish to prove is not true. Namely, if $I=\operatorname{ker} f^*$ then $I^2\subseteq \operatorname{ker}(f\ast f)^*$ but the inclusion may be strict.
The fibred join construction comes up in the study of sectional category. If $f:Y\to X$ is a fibration, the sectional category of $f$, denoted $\operatorname{secat} f$, is the minimal number of open sets needed to cover $X$, such that $f$ admits a local section on each set in the cover. Taking $p:PX\to X$ to be the Serre based path fibration space over $X$, we see that $\operatorname{secat}(p)=\operatorname{cat}(X)$, so that this invariant includes as a special case the Lusternik-Schnirelmann category of topological spaces.
Now, as was shown originally by A. S. Schwarz (who initiated the study of sectional category under the name genus), a fibration $f:Y\to X$ has $\operatorname{secat} f\le k$ if and only if the $k$-fold iterated fibred join of $f$ admits a section. In particular, $\operatorname{secat} f\le 2$ if and only if $f\ast f:Y\ast_X Y\to X$ admits a section.
This means people who are interested in estimating sectional category from below are interested in finding elements in the kernel $\operatorname{ker}(f\ast f)^*$, because these obstruct the existence of a section of $f\ast f$. In fact, elements in this kernel are precisely the elements of sectional category weight at least 2 with respect to the fibration $f$, see 
Michael Farber and Mark Grant, MR 2407101 Robot motion planning, weights of cohomology classes, and cohomology operations, Proc. Amer. Math. Soc. 136 (2008), no. 9, 3339--3349.
This notion generalizes the notion of category weight, due to Fadell and Husseini, and later studied by Rudyak, Strom and others. 
A product of elements in $I$ will have weight at least 2. However, there are  indecomposable classes of weight 2, and these can be produced using stable cohomology operations or Massey triple products. Examples can be found in the above paper, or in the papers of Fadell-Husseini and Rudyak or the thesis of Jeff Strom on category weight. 
To give an explicit example, let $p:PX\to X$ be the based path fibration where $X$ is the complement of the Borromean rings. All cup products of elements in the $\tilde{H}^*(X)=\operatorname{ker} p^*$ are trivial, so $I^2=0$. But there are triple Massey products of category weight 2, and these are in $\operatorname{ker} (p\ast p)^*$, which is non-trivial (hence $\operatorname{cat}(X)\ge 2$).<|endoftext|>
TITLE: From Weyl groups to Weyl groupoids?
QUESTION [11 upvotes]: I'm trying to find a framework where the choices in the classical construction of a root system of a semi-simple lie algebra are not needed.
Let $\mathfrak{g}$ be a semisimple lie algebra. 

Definition (Weyl groupoid): Let $\mathcal{W}$ be the following groupoid:

$Obj(\mathcal{W})=\{\mathfrak{b}\subset \mathfrak{g} | \space \mathfrak{ b} \text{ is a maximal solvable algebra}\}$
$Mor(\mathcal{W}) = \{Ad(e^x):\mathfrak{b}_1 \to \mathfrak{b}_2 \space |\space x \in \mathfrak{g}\}$


Here's what I like about this so far:

Objects in $\mathcal{W}$ are in 1-1 correspondence with choosing: cartan subalgebra & Weyl chamber (root basis). 
Morally there's sort of an exponential map $exp: \mathcal{W} \to
   G/B$ (this is not meant to be a precise statement).

A couple of questions:


Is this the correct categorification? 
How do I get the weyl group out of this? 
I read somewhere that a choice of borel subalgebra is equivalent to a choice of complex structure on the complexified tangent bundle of $G/B$. Can this groupoid be interpreted in terms of a groupoid of complex structures?
Where can I find more about this kind of construction?

REPLY [6 votes]: It's a bit unclear to me precisely what you want, but the usual solution is take the abstract Cartan (discussed a bit here), which is defined to be $\mathbb{T}=B/[B,B]$ for any Borel $B$.  This is canonical, since any two Borels are conjugate, and every Borel is self-normalizing so $\mathbb{T}$ is unique up to the map induced by conjugation by an element of $B$.  The action of the Weyl group is given by looking at all the isomorphisms induced between $\mathbb{T}$ and any fixed torus $T$ by all the different Borels containing $T$ (which are all conjugate under $N(T)$).  Composing one of these isomorphisms with the inverse of another gives all the elements of the Weyl group.<|endoftext|>
TITLE: Intuition for density comonad in relation to lifting problems
QUESTION [9 upvotes]: In Emily Riehl's Categorical Homotopy Theory, there is a section on Garner's Small Object Argument which I'm trying and failing to understand. Originally I followed most of Garner's paper, using the book to try and explicitly understand the stages in the transfinite construction.
I guess I'll try to sketch some details first. Garner's small object argument is a a refinement of Quillen's in that it is universal and converges. The point is in to construct a 'free natural weak factorization system', i.e in reflecting along a semantics functor $\mathcal G:\mathsf{nwfs}(\mathsf{C})\longrightarrow \mathsf{Cat}/\mathsf{C^2}$. This problem is simplified by factoring $\mathcal G$ in a reasonably canonical way, because abstract nonsense gives two of the three reflections needed. The final reflection is then shown to be equivalent to constructing a free monoid, which is then shown to be equivalent to the convergence of a certain transfinite construction called the 'free monoid sequence'.
(Now we come to proposition 4.22 in Garner's paper.)
A convergence criterion in the context of our problem is the condition the functor $R=d_0\circ F:\mathsf{C^2}\longrightarrow \mathsf{C^2}$ preserves $\lambda$-filtered colimits, where $d_0$ is the image of the simplicial $\delta_0$, sending a composite of arrows to the first one. By abstract nonsense again it suffices to prove $L=d_2\circ F$ preserves such colimits. By abstract nonsense, $L$ admits the following explicit description. First, given the object $I:\mathsf{J}\rightarrow \mathsf{C^2}$ we want to reflect, form its left Kan extension along itself $M=\operatorname{Lan}_II:\mathsf{C^2}\longrightarrow \mathsf{C^2}$.
Remark 1. Remark 12.5.2 in the book says that left Kan extending a functor along itself yields (by abstract nonsense) a comonad, called the density comonad.
With the usual coend formula, the component at $f$ of the counit $\epsilon :M\Rightarrow 1_{\mathsf{C^2}}$ is the arrow $$\int^{j\in \mathsf J}\mathsf{C^2}(Ij,f)\circ Ij\longmapsto f$$ which is adjunt to the identity natural transformation on $\mathsf{C^2}(I(-),f)$. This $\epsilon$ can then be factored as the composite $M\overset{\xi}{\Longrightarrow}L\overset{\Phi}{\Longrightarrow}1_{\mathsf{C^2}}$ where the components of $\xi$ are pushouts and the components of $\Phi$ are identities on its domain.
By abstract nonsense, $L$ preserves any colimit $M$ does, so we need to find an ordinal $\lambda$ for $M$. The proof ends by using the smallness condition to interchange some colimits. $\blacksquare$
It looks like section 12.4 of the book explain the process a little, but not enough for me. Suppose we want to factor an arrow $f$. The author says the square (12.2.3) should be thought of as a "generic lifting problem" that tests whether or not $f$ is a fibration, which makes sense. The fill in $\phi_f$ in (12.4.1) is referred to as a lifting function, which also makes sense. In step-one we factor the above square through the pushout of its cospan. This yields an equivalent lifting problem, proving that $f\in \mathcal J^\pitchfork$ iff it has has the right lifting property against $Lf$ in the canonical square (triangle) $Rf\circ Lf=f\circ 1$. We iterate this process, modding out redundancies, to ensure (using smallness) that it converges.
Remark 2. If I understand correctly, Garner says $\epsilon$ is component-wise exactly the transition to the "lifting function" point of view. I am confused about the second sentence regarding the equivalent lifting problem...

I would like to understand what the density comonad is (preferably at a level of generality where I can make out the relevance to this context), why, conceptually, does it encode the passage to an equivalent (but much more canonical) lifting problem, and whether the quotienting described in section 6.5 of Garner's paper, which gives convergence, is also encoded in it.

Finally, let me emphasize I did not learn Quillen's small object argument before, so answers like "this is just a modification of the standard argument" will not help me very much. I am trying to learn this "corrected" version from scratch. I have also never met density comonads or codensity monads before.

REPLY [6 votes]: Maybe it would help to have:

A definition of "$f \in \mathsf{J}^\pitchfork$" when $\mathsf{J}$ is a category equipped with a functor $I : \mathsf{J} \to \mathsf{C}^2$; a definition that does not mention the density comonad of $I$.
An argument showing that $f \in \mathsf{J}^\pitchfork$ is equivalent to having a diagonal filler for the counit $\mathrm{Lan}_II(f) \to f$.

[In the case that $\mathsf{J}$ is discrete the definition in step 1 would be simply that for each $j \in \mathsf{J}$ and each square $j \to f$ there is a diagonal filler. And in step 2 you'd argue that the counit $\mathrm{Lan}_II(f) \to f$ is exactly the square (12.2.3), and that having a diagonal filler for it provides all at once the required diagonals for all the individual squares $j \to f$.]
OK, the definition of $f \in \mathsf{J}^\pitchfork$ is that you can choose for each object $j \in \mathsf{J}$ and each square $\alpha : I(j) \to f$ a diagonal filler $\phi_\alpha$ in such a way that the fillers are compatible with composition in $\mathsf{J}$, namely, given a morphism $g : j \to j'$ and a square $\alpha' : I(j') \to f$ the fillers for $\alpha'$ and $\alpha = \alpha' \circ I(g)$ satisfy $\phi_\alpha = \phi_{\alpha'} \circ I_{\mathsf{cod}}(g)$ (where (1) $I_\mathsf{cod}(g)$ means the bottom arrow of the square $I(g) : I(j) \to I(j')$ --bottom if you draw $I(j)$ going down-- and (2) I apologize for not knowing how to draw diagrams with diagonal arrows on MO).
Now, how do we see that giving these compatible fillers $\phi$ is the same as filling the counit $\mathrm{Lan}_II(f) \to f$? Well, first notice the compatibility with composition of the $\phi$'s can be rephrased as saying that $\phi$ gives a functor from the comma category $I \downarrow f$ to $\mathsf{C}^2$. In fact, the collection of lifting problems $I(j) \to f$ form a cocone with vertex $f$ over the diagram $I \downarrow f \xrightarrow{\pi} \mathsf{J} \xrightarrow{I} \mathsf{C}^2$. And now just recall the colimit formula for the left Kan extension: $\mathrm{Lan}_II(f) = \mathrm{colim}(I \downarrow f \xrightarrow{\pi} \mathsf{J} \xrightarrow{I} \mathsf{C}^2)$. The counit of the density comonad is corresponds to the cocone mentioned above.<|endoftext|>
TITLE: Can we trap light in a polygonal room?
QUESTION [42 upvotes]: Suppose we have a polygonal path $P$ on the plane resulting from removal of an one  of a convex polygon's edges  and a ray of light "coming from infinity" (that is, if we were to trace the path backwards in time, we would end up with an infinite ray) and reflecting from the sides of $P$ according to standard law of reflection (see below for discussion of interaction with vertices). Is it possible that $P$ will stay inside $P$ forever?
More generally, we could ask this for more general polygonal paths and instead of talking about "inside of $P$" we might talk about for example its convex hull. I know that it is possible with more general curves, even with "infinitely-sided" polygons, which can be shown with idea of this answer (illustrated here). The linked question was also a partial motivation to this one.
About corners: the easiest (and preferable for me) way to deal with them is to require that the light should never hit it (the light then disappears or whatever, I leave it up to your interpretation). Another idea I had is to make the light reflect so that it makes equal angles with bisector of the internal angle at the vertex. However, if it helps to construct an example, feel free to suggest another rule.
At first I thought that an example most likely exists, but I couldn't find one (I had to work with pen and paper though, I don't have software with such functionality). I know that the path cannot be eventually periodic, because the process of light traveling is reversible. I see no obvious obstruction which would disallow more complicated infinite paths.
Thanks in advance for all the feedback.
Edit: in the view of Joseph's example, I would like to clarify that I want the endpoints of the polygonal path to also be contained in it and the light is supposed to vanish on them.
Edit2: I don't know how much easier the problem will become, but seeing the paper Joseph refered to in his first answer I thought that the following weakening might have ability to trap a ray of light: instead of polygonal chain, let's just consider a set of closed line segments. The line segments are allowed to intersect, and the light is supposed to never hit an endpoint of an line segment or an intersection point of two segmentss. Let's see if anyone can figure out such a trap...
Edit3: I forgot to add this to the description of the bounty, I plan to award either (these are the weakest possibilities in both directions):

Anything that implies possibility of trapping light with a set of mirrors as described in edit2, or
Anything that implies impossibility of trapping light in a convex polygon as described in the first paragraph.

If neither of the above is achieved by anyone, I am willing to award the bounty to any noteworthy attempt in proving either.

REPLY [9 votes]: I post this in response to Wojowu's request in a comment; I don't really consider this usefully addressing the question raised.
I thought of a simpler construction. Take a square with mirrored sides, and
poke a point-hole in the interior of the top edge. When a ray from outside
with an irrational slope (w.r.t. the box sides) enters through the hole, it is
trapped. For if it escaped through the hole after $k$ reflections, then it
would contradict the irrationality of the slope. It is known that the path of
this "billiard" is aperiodic (a result due to G.A.Galperin, I believe).

          


And of course almost all rays have irrational slope.
But as soon as the point-hole is enlarged to a positive-width hole, the ray
eventually escapes (Poincaré recurrence).<|endoftext|>
TITLE: Polynomial roots in the ring extension
QUESTION [7 upvotes]: Let $R$ be a ring with identity (not necessarily commutative) and $R[x]$ be a ring of polynomials over $R$. We say that a ring $S$ is an extension of $R$ if there is a subring $\tilde{R}$ in $S$ isomorphic to $R$.
Let $S$ be an extension of $R$, and $$\phi: R\to \tilde{R}\subset S$$
be a ring isomorphism.
We say that a polynomial $f(x) = \sum\limits_{j\geq 0}f_jx^j\in R[x]$ has a root $\alpha\in S$ if
$$
\sum\limits_{j\geq 0}\phi(f_j)\alpha^j = 0.
$$
In the case, where $R$ is a commutative ring every monic polynomaial $f(x)\in R[x]$ has a root $[x]_f$ in the extension $S = R[x]/R[x]f(x)$ of $R$. In the case, where $R$ is not commutative the set $R[x]/R[x]f(x)$ is a left $R[x]$-module but not a ring, because an ideal $R[x]f(x)$ is not two-sided ideal, but only one-sided. Also in non-commutative case there are examples such that two-sided ideal, containing $f(x)$ that is an ideal $R[x]f(x)R[x]$ is equal to $R[x]$ and in this case $R[x]/R[x]f(x)R[x]$ isomorphic to zero ring.
I want to prove that for every ring with identity $R$ and every monic polynomaial $f(x)$ over $R$ there exists an extension $S$ of $R$ such that $f(x)$ has a root in $S$.

REPLY [4 votes]: I have found a very simple and demonstrative construction of the ring extension, which came from non-commutative generalization of Hamilton-Caley's Theorem.
Let $R$ be a ring and $f(x) = x^m-\sum\limits_{j=0}^{m-1}f_jx^j\in R[x]$ be a monic polynomial. 
We identify ring $R$ with a subring $\tilde{R} = \{\mathrm{diag}(r,r,\ldots,r): r\in R\}\subset M_m(R)$. 
Then $f(x)$ has a root of the form
$$
\alpha=\left(\begin{array}{llllll}
0& e& 0&\ldots& 0& 0 \\
0& 0& e&\ldots& 0& 0 \\
. & . &  . & . & .& . \\
0& 0&0&\ldots& 0& e \\
f_0& f_1& f_2&\ldots& f_{m-2}& f_{m-1} \\
\end{array}
\right)
$$
That is $\alpha^m-\sum\limits_{j=0}^{m-1}f_j\alpha^j = 0\in M_m(R)$.
But we note that in general:
$f(\alpha^T)\neq 0$.
See article for proof  of non-commutative generalization of Hamilton-Caley's Theorem.<|endoftext|>
TITLE: Is there a subset of $\Sigma_n$ s.t. each pair of elements is once in each pair of positions?
QUESTION [7 upvotes]: Is there a subset $A \subset \Sigma_n$ such that for each pair $(x, y)$ and each pair $(i, j)$, there is exactly one permutation $\sigma \in A$ such that $\sigma(i) = x$ and $\sigma(j) = y$? Remark that it implies that the cardinality of $A$ is $n * (n-1)$.
Is there a simple condition on $n$ such that it is possible?
If n is prime, it is easy, for example with the following method. For each pair $(x, k)$, consider the permutation $\sigma$ such that $\sigma(i) = x + (i - 1) * k$ (modulo $n$). For example, for $n = 5$, you get:

1 2 3 4 5 and its 5 circular permutations,
1 3 5 2 4 and its 5 circular permutations,
1 4 2 5 3 and its 5 circular permutations,
1 5 4 3 2 and its 5 circular permutations.
If n = 4, it is also possible. For example :
1 2 3 4
1 3 4 2
1 4 2 3
2 1 4 3
2 3 1 4
2 4 3 1
3 1 2 4
3 2 4 1
3 4 1 2
4 1 3 2
4 2 1 3
4 3 2 1
For $n = 6$, I suspect that it is not possible, but I am not sure...

REPLY [3 votes]: You are asking for a sharply $2$-transitive set of degree $n$. These objects are closely linked to projective planes of order $n$.
In particular, they exist if $n$ is a prime power, and no example is known with $n$ not a prime power. It also follows from the Bruck–Ryser Theorem that there is no example of degree $6$.
See for example 
http://tinyurl.com/go5el94
and 
http://www.mathematik.uni-wuerzburg.de/~mueller/Papers/sharplyTGPM.pdf<|endoftext|>
TITLE: Definition of discrete spectrum and continuous and basic properties
QUESTION [14 upvotes]: I apologize if this is too basic for MO.
I have an embarrassing admission to make: I don't know the actual definition of the discrete/continuous spectrum of a reductive group $G/\mathbb{Q}$ (in the context of automorphic forms if there is some ambiguity I'm unaware of).In particular, I know what the discrete series representations are for various groups $G$ (e.g. $\text{GL}_2$), and I assume that the discrete spectrum is made up of the discrete series (pluse extra! see the comments below), but I don't know how to define them in a general, 'elegant' way.
I have the, possibly wildly incorrect, impression that the discrete spectrum of $G$ is the largest semisimple $(\mathfrak{g},K)\times G(\mathbb{A}^\infty)$-subrepresentation of $L^2(G(\mathbb{Q})\backslash G(\mathbb{A}))$ and that the continuous spectrum is the orthogonal complement of this which should be some sort of 'direct integral' over continuous parameters. But, to be honest, in the searching that I've done I haven't seen this explicitly stated.
So, with this being said, I have the following three basic questions:

Question 1: What is the rigorous, 'elegant' definition of the discrete and continuous spectrum of a reductive group $G/\mathbb{Q}$?
Question 2: Why are these called 'discrete' and 'continuous'?
Question 3: Can we explicitly describe the discrete and/or continuous spectrum in any reasonable way for general $G$? If not, which $G$ can we describe it for?

NB: I am a relative neophyte, so I would appreciate if any answer could be in as simple language as possible.
Thanks!
EDIT: There's also the following 'bonus question' if anyone feels up to it:

Question 4: How do Eiseinstein representations fit into this? Why do they appear in both the continuous and discrete spectrum?

REPLY [3 votes]: A few comments.
Question 1: When working with reductive groups, it is better to work with the subgroup $G(\mathbb{A})^1\subset G(\mathbb{A})$ or to work with the subspace of $L^2(G(\mathbb{Q})\backslash G(\mathbb{A}))$ transforming under the center according to a fixed character. If you care about the representations occurring in these $L^2$ spaces, then you can define the discrete spectrum as the subspace of $L^2(G(\mathbb{Q})\backslash G(\mathbb{A})^1)$ (say) which decomposes as a direct sum of irreducible sub representations of the action of $G(\mathbb{A})$ by right translation. Then the continuous spectrum is the orthogonal complement of the discrete spectrum as you said, so 
$$
L^2(G(\mathbb{Q})\backslash G(\mathbb{A})^1) = L^2_\mathrm{disc}(G(\mathbb{Q})\backslash G(\mathbb{A})^1)\oplus L^2_\mathrm{cont}(G(\mathbb{Q})\backslash G(\mathbb{A})^1).
$$
The cuspidal spectrum is a subspace of the discrete spectrum, but in general there are discrete spectrum representations which are not cuspidal. The definition of an automorphic representation is more general than these representations "occurring in" $L^2$. For this, see the articles of Borel-Jacquet and Langlands in Corvallis. 
Question 3: It depends on what you would like to know. For example, the Moeglin-Waldspurger classification for $\mathrm{GL}(N)$ describes the discrete spectrum in terms of the cuspidal spectrum of Levi subgroups. Namely, it says that there is a bijection between pairs $(\mu,m)$ and the discrete spectrum of $\mathrm{GL}(N)$, where $m$ divides $N$ and $\mu$ is a cuspidal representation of $\mathrm{GL}(m)$. Recent work of Arthur describes the discrete spectrum of quasi-split orthogonal and symplectic groups in terms of the discrete spectrum for $\mathrm{GL}(N)$. If $G$ is such a classical group, he provides a decomposition of the space $L^2_\mathrm{disc}(G(\mathbb{Q})\backslash G(\mathbb{A}))$ into a direct sum parametrized by representations of $\mathrm{GL}(N)$. This work has been extended by Mok to quasi-split unitary groups. 
As far as the continuous spectrum, one place to start might be Section 7 of Arthur's "An Introduction to the Trace Formula," where he describes Langlands' main theorem on Eisenstein series. He also discusses all the things mentioned above. I won't try to say much else about this. 
Question 4: For classical groups, I do not think that a representation occurring in the continuous spectrum can also occur in the discrete spectrum. I believe this is the problem of "embedded eigenvalues." For this, I would check out Arthur's note "Eigenfamilies, characters and multiplicities" and "The embedded eigenvalue problem for classical groups."<|endoftext|>
TITLE: Compact non-nuclear operators
QUESTION [5 upvotes]: I am not sure if this question makes sense, or if it is trivial,  but does there exists an infinite dimensional Banach space (necessarily without the approximation property) such that no compact, non-nuclear operator is the norm limit of finite rank operators?

REPLY [11 votes]: Pisier constructed a Banach space such that the operator norm is equivalent to the nuclear norm on the finite rank operators.  Consequently, no compact non nuclear operator on his space is the norm limit of finite rank operators.  However, it is open whether there exists a compact non nuclear operator on his space!
Pisier, Gilles Counterexamples to a conjecture of Grothendieck. Acta Math. 151 (1983), no. 3-4, 181–208.<|endoftext|>
TITLE: Average size of $p$-part of the Tate-Shafarevich group for elliptic curves
QUESTION [8 upvotes]: Let $E/\mathbb{Q}$ be an elliptic curve defined over $\mathbb{Q}$. For a given prime $p$, the $p$-Selmer group $\operatorname{Sel}_p(E)$ of $E$ and the $p$-part of the Tate-Shafarevich $Ш_E[p]$ group are related by the following exact sequence:
$$\displaystyle 0 \rightarrow E(\mathbb{Q})/pE(\mathbb{Q}) \rightarrow \operatorname{Sel}_p(E) \rightarrow Ш_E[p] \rightarrow 0,$$
where $E(\mathbb{Q})$ denotes the Mordell-Weil group of the elliptic curve $E$. 
In a series of papers (see below for some references), Bhargava and Shankar showed that the average size of the $p$-Selmer group for elliptic curves over any `large' family of elliptic curves over $\mathbb{Q}$ is equal to $3,4, 6$ respectively for $p = 2,3,5$. Are there any analogous results for the $p$-part of $Ш_E$ for the same primes, perhaps upper or lower bounds?
Both lower bounds and upper bounds for this average have interesting consequences. If one can prove a non-trivial lower bound, for example suppose the average $2$-rank of $Ш_E$ is at least $0.8$, then from the equality
$$\displaystyle r_2(\operatorname{Sel}_2(E)) = r(E) + r_2(E(\mathbb{Q})[2]) + r_2(Ш_E[2])$$
and the fact that the average $2$-Selmer rank is at most $1.5$ and that the average of $r_2(E(\mathbb{Q})[2])$ is 0, we see that the average of the Mordell-Weil rank $r(E)$ is at most $1.5 - 0.8 = 0.7$, which beats the current best bound (also due to Bhargava and Shankar, as a consequence of their proof that the average $5$-Selmer size is 6). If one can prove a non-trivial upper bound for $Ш_E[p]$, then one can use it to produce more curves with positive rank. 
Are such averages known, whether over a `large' family or not? I am particularly interested in the case $p = 2$.
References:
http://annals.math.princeton.edu/2015/181-1/p03
http://annals.math.princeton.edu/2015/181-2/p04
http://arxiv.org/abs/1312.7859

REPLY [7 votes]: What's known is very little, but there is a very compelling conjectural picture of the distribution of Mordell-Weil, Selmer, and Sha in the paper of Bhargava, Kane, Lenstra, Poonen, and Rains:
http://arxiv.org/abs/1304.3971<|endoftext|>
TITLE: Time-Energy Uncertainty Relation in relativistic Quantum Mechanics
QUESTION [6 upvotes]: There is an old  intriguing result in non-relativistic QM, stating (roughly) that there is an Heisenberg Time-Energy Uncertainty Relation. 
Unfortunately, in QM time is not an operator like space, and an old result shows that if there was one, it would imply negative values of the Energy operator. 
However, if I remember well, in the Dirac equation negative energy does  pop up, in the  infamous  Dirac's sea. Moreover, time and space are indeed unified by relativistic constraints.
Thus I wonder: is there some setup of relativistic QM where the above T-E relation pops up in some form or another, as a genuine analogue of the other one X-P between space and momentum?
PS I am well aware that the standard path to relativistic QM is to demote space to the same rank of time, ie as parameters, not to promote time to an operator. But I know that in principle something like this could also be done, thought it would lead to a more complicated picture.  
ADDENDUM The mathematical rationale behind my question is basically this: I would like to think of the E-T pair as some kind of rotation of the X-P operators pair

REPLY [3 votes]: You don't really need quantum mechanics to address this issue, in classical mechanics you would ask whether time and energy can be thought of as canonically conjugate variables (because quantization would then produce an uncertainty relation between these operators). You would then want to treat energy $E$ as a dynamical variable, like position $x$, which means that the Hamiltonian $H(x,E;p,t)=H_0(x,p;t)-E$ must depend explicitly on time $t$, like it does on momentum $p$. The Hamilton equations of motion,
$$\dot{q}=\partial H/\partial p,\;\;\dot{p}=-\partial H/\partial q,$$
$$\dot{E}=\partial H/\partial t,\;\dot{t}=-\partial H/\partial E,$$
tell you that, indeed, $E$ and $t$ are canonically conjugate like $q$ and $p$.
There is a fun discussion along these lines at Physics SE, in response to the question "Is energy actually momentum in the direction of time?"<|endoftext|>
TITLE: What is the group of automorphisms of $l^{\infty}$?
QUESTION [5 upvotes]: What is the group of automorphisms of $l^{\infty}$?
I think it would be the permutations of the integers. Is this right?

REPLY [6 votes]: This is to record what is probably the easiest proof (that the answer is "yes"), following a remark by Nik Weaver.
Noting that $\ell^\infty(\mathbb{Z})\simeq C(\beta\mathbb{Z})$, using Banach-Stone, we see that every auto gives a homeo of $\beta\mathbb{Z}$.
But since $\mathbb{Z}$ is the subset of isolated points, every homeo preseves it, and since it is dense the homeo is determined by its restriction to $\mathbb{Z}$.<|endoftext|>
TITLE: Upper Bound for the Difference of Even Probability and Odd Probability in Hypergeometric Distribution
QUESTION [7 upvotes]: Let $X$ be a random variable following the hypergeometric distribution with parameters $N,K,n$, where 
\begin{equation}
Pr(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}.
\end{equation}
To make things easier, we bring symmetry and avoid the boundry issues by letting $K=\frac{N}{2}$ and $n<K$. For an odd $n$, we know by symmetry
\begin{equation}
Pr(X~even)=Pr(X~odd).
\end{equation}
For an even $n$, my conjecture is that 
\begin{equation}
\vert Pr(X~even)-Pr(X~odd) \vert \leq \frac{\binom{N}{n/2}}{\binom{N}{n}}. 
\end{equation} 
I ran some simulations and the result shows that it should be true, but I have not been able to prove it so far. Could someone help me with that? 
If this question is answered, a subsequent question is the case where $K$ is roughly $\frac{N}{2}$ but not exact. A corresponding condition is $\vert K-\frac{N}{2}\vert \leq c N$, for a constant $0<c<0.5$. We still assume $n<min (K, N-K)$ to avoid boundary issues. The conjecture is still the same in this case.
I try not to be greedy and will be much satisfied if only the first question is answered. Many thanks in advance!

REPLY [7 votes]: In the case $N=2K$ the sum can be evaluated in closed form
using generating functions, and satisfies the desired inequality.
In general, multiply by ${N \choose n} t^k z^n$
and sum over both $k \leq K$ and $n \leq N$ to get 
$$
G_{N,K}(t,z) = (1+tz)^K (1+z)^{N-K}
$$
(by writing $n=k+m$ and factoring into a product of two binomial sums).
The desired alternating sum is then ${N \choose n}^{-1}$ times
the $z^n$ coefficient of $G_{N,K}(-1,z) = (1-z)^K (1+z)^{N-K}$.
When $K=N-K$, this simplifies to $(1-z^2)^K$, so the coefficient is
zero for $n$ odd (as you noted by an antisymmetry argument) and 
$(-1)^n {K \choose n/2}$ for $n$ even.  In particular the absolute
value is less than $N \choose n/2$ as desired (and by a factor of
about $2^{n/2}$).  For general $K,N$, you can bound or estimate
the $z^n$ coefficient using stationary-phase techniques.<|endoftext|>
TITLE: Logarithms of matrices in the disk-algebra
QUESTION [6 upvotes]: It is easy to see that within the disk algebra $A(D)$
  $$\Delta(z):= \begin{pmatrix} 1&0\\z&1 \end{pmatrix}\; \begin{pmatrix} 1&1\\0&1 \end{pmatrix}=
 \begin{pmatrix} 1&1\\z&1+z
 \end{pmatrix} $$
 is a product of two exponential matrices.  Is $\Delta(z)$ itself an exponential matrix of a holomorphic matrix $M(z)$ ?
 I don't think so, but could not come up with a proof. Is there a general method to deal with these questions?

REPLY [2 votes]: $\Delta(z)$ does not have a nonpositive real  eigenvalue for $|z| < 4$, so the principal branch of the logarithm is defined and analytic on a neighbourhood of its spectrum, and thus the holomorphic functional calculus produces the desired analytic logarithm $M(z)$ for $|z| < 4$.<|endoftext|>
TITLE: Link between the hairy ball theorem and the fundamental theorem of algebra
QUESTION [8 upvotes]: I read in the book "Concepts of modern mathematics" by Ian Stewart that it was possible to proof the fundamental theorem of algebra using the hairy ball theorem (complete reference to the page is in the following quora question : https://www.quora.com/How-can-the-hairy-ball-theorem-be-used-to-prove-the-fundamental-theorem-of-algebra). I am not aware of such a proof. It is not obvious to me (compactification does not seem to lead to a satisfactory answer) and I would be delighted if someone could help me with that.

REPLY [5 votes]: Maybe this is an idea. It suffices to prove that any polynomial $f\in \mathbf C[z]$ of even degree $2d\geq2$ has a zero in $\mathbf C$. Assume that it has no zero. Then the rational section
$$
s=f\cdot\left(\tfrac{\partial}{\partial z}\right)^{\otimes d}
$$
of the $d$-th tensor power $T^{\otimes d}$ of the holomorphic tangent bundle on the complex projective line $\mathbf P^1(\mathbf C)$ is a nowhere vanishing global section. This is so since the holomorphic vector field $\partial/\partial z$ does not vanish on $\mathbf C$, has a zero of order $2$ at $\infty$, the polynomial $f$ does not vanish on $\mathbf C$, and has a pole of order $2d$ at $\infty$. Consider the section $s$ as a topological surface in the complex line bundle $T^{\otimes d}$ over $\mathbf P^1(\mathbf C)$. As image of the sphere $S^2$, it is homeomorphic to $S^2$ and does not have any nontrivial topological coverings. Its inverse image under the power-$d$-map
$$
T\rightarrow T^{\otimes d}
$$
from the complex line bundle $T$ into the complex line bundle $T^{\otimes d}$ is, therefore, a disjoint union of $d$ disjoint copies of the sphere $S^2$ in the complement of the zero section of the tangent bundle of $S^2$. Each would define a nowhere vanishing vector field on $S^2$. Contradiction by the hairy ball theorem.<|endoftext|>
TITLE: Rotation invariance of an integral
QUESTION [8 upvotes]: Consider the integral depending on 2 parameters
$$f(\tau,x):=\int_{-\infty}^{+\infty}\frac{dp}{\sqrt{p^2+1}}e^{-\sqrt{p^2+1}\tau+ipx},$$
where $\tau >0,x\in \mathbb{R}$. This integral absolutely converges in this region.
Question. Is it true that $f(\tau,x)$ is $SO(2)$ invariant in the above region, i.e. depends only on $\tau^2+x^2$?
Remark. My motivation to ask this question comes from the well known fact that after the so called Wick rotation $\tau=it$  one gets an integral invariant under the Lorentz transformations in the plane $(t,x)$.

REPLY [13 votes]: Yes. Denote $p=\sinh s$, then $s\in (-\infty,\infty)$, $\sqrt{p^2+1}=\cosh s$, $dp=\cosh s\, ds$. Next, denote $r=\sqrt{\tau^2+x^2}$, $\tau=r\cos \varphi$, $x=r\sin \varphi$, where $\varphi\in (-\pi/2,\pi/2)$. So our integral rewrites as $$\int_{-\infty}^\infty ds \exp(-r\cosh(s-i\varphi)).$$
It does not depend on $\varphi$ as may be seen from the rectangular contour (integrals over vertical sides tend to 0).<|endoftext|>
TITLE: Complex manifold with subvarieties but no submanifolds
QUESTION [36 upvotes]: I previously asked this question on MSE and offered a bounty but received no responses.

There are examples of compact complex manifolds with no positive-dimensional compact complex submanifolds. For example, generic tori of dimension greater than one have no compact complex submanifolds. The proof of this fact, see this answer for example, shows that these tori also have no positive-dimensional analytic subvarieties either (because analytic subvarieties also have a fundamental class). 
My question is whether the non-existence of compact submanifolds always implies the non-existence of subvarieties.

Does there exist a compact complex manifold which has positive-dimensional analytic subvarieties, but no positive-dimensional compact complex submanifolds?

Note, any such example is necessarily non-projective.

REPLY [12 votes]: The theorem that inkspot refers to in his answer is originally from Inoue's paper New Surfaces with No Meromorphic Functions, II which seems like a more complete reference for this question. In particular, Inoue gives an explicit example of a compact complex surface which has an analytic subvariety but no compact complex submanifolds.
If $x$ is a real quadratic irrationality (i.e. a real irrational solution of a real quadratic equation), denote it's conjugate by $x'$. Let $M(x)$ be the free $\mathbb{Z}$-module generated by $1$ and $x$, then set $U(x) = \{\alpha \in \mathbb{Q}(x) \mid \alpha > 0, \alpha\cdot M(x) = M(x)\}$ and $U^+(x) = \{\alpha \in U(x) \mid \alpha\cdot\alpha' > 0\}$. Both $U(x)$ and $U^+(x)$ are infinite cyclic groups and $[U(x) : U^+(x)] = 1$ or $2$.
If $\omega$ is a real quadratic irrationality such that $\omega > 1 > \omega' > 0$, then $\omega$ is a purely periodic modified continued fraction; that is, $\omega = [[\overline{n_0, n_1, \dots, n_{r-1}}]]$ where $n_i \geq 2$ for all $i$, $n_j \geq 3$ for at least one $j$, and $r$ is the smallest period. For every such $\omega$, Inoue constructs a compact complex surface $S_{\omega}$ which is now known as an Inoue-Hirzebruch surface.
There are compact subvarieties $C$ and $D$ of $S_{\omega}$ with irreducible components $C_0, \dots, C_{r-1}$ and $D_0, \dots, D_{s-1}$ respectively; here $s$ is the smallest period of the modified continued fraction expansion of another element $\omega^*$ related to $\omega$ (alternatively, $s$ can be determined from the modified continued fraction expansion of $\frac{1}{\omega}$). When $r \geq 2$, $C$ is a cycle of non-singular rational curves, and when $r = 1$, $C$ is a rational curve with one ordinary double point. Proposition $5.4$ shows that $C_0, \dots, C_{r-1}, D_0, \dots, D_{s-1}$ are the only irreducible curves in $S_{\omega}$.
In the case where $[U(\omega) : U^+(\omega)] = 2$, we have $r = s$. Furthermore, there is an involution $\iota$ such that $\iota(C_i) = D_i$ for $i = 0, \dots, r - 1$. The quotient of $S_{\omega}$ by $\iota$ is denoted $\hat{S}_{\omega}$ and is now known as a half Inoue surface. Note that the images of $C_0, \dots, C_{r-1}$ are the only irreducible curves in $\hat{S}_{\omega}$.
If we can find a real quadratic irrationality $\omega$ such that $\omega > 1 > \omega' > 0$, $r = 1$, and $[U(\omega) : U^+(\omega)] = 2$, then $\hat{S}_{\omega}$ is a compact complex surface containing a unique curve, namely a rational curve with one ordinary double point. In particular, it provides an example of a compact complex manifold with a subvariety but no compact complex submanifolds. One such $\omega$ was given in the paper (end of section 6).

Example. Take $\omega = (3 + \sqrt{5})/2$. Then $[U(\omega) : U^+(\omega)] = 2$ and $\alpha_0 =\ \text{a generator of}\ U(\omega) = (1 + \sqrt{5})/2$, $\alpha = \alpha_0^2 = (3 + \sqrt{5})/2$, $\omega = [[\overline{3}]]$, $r = 1$.
In this case, $b_2(\hat{S}_{\omega}) = 1$ and $\hat{S}_{\omega}$ contains exactly one curve $\hat{C}$. Moreover, $\hat{C}$ is a rational curve with one ordinary double point and $(\hat{C})^2 = -1$.

For those interested in the details, in addition to Inoue's paper, it may also be worth reading the earlier paper Hilbert modular surfaces by Hirzebruch. As mentioned in his paper, Inoue used some methods from Hirzebruch's paper (which gives some indication of why the resulting surfaces are jointly named).<|endoftext|>
TITLE: Poincare-Hopf theorem for polytopes?
QUESTION [6 upvotes]: Is there an analogue of Poincare-Hopf theorem for polytopes?
I want to apply it in the following situation.
I have a polytope in $R^n$ and a smooth explicitly given vector field in $R^n$.
I want to find the number of points where the field is normal to the surface of the polytope (rather, lies in the normal cone). If the polytope was a smooth compact manifold, I'd project the field to the tangent space and apply Poincare-Hopf theorem. Is there a version of Poincare-Hopf theorem which lets me do something similar for polytopes?

REPLY [3 votes]: There are certainly analogous theorems but the most direct analogy you might not find particularly useful.  For example, manifolds with boundary could be considered a step towards a general Poincare-Hopf for polyhedra.  
The most direct version of Poincare-Hopf for manifolds with boundary requires that the vector field be outward-pointing (or inward-pointing) along the boundary -- you do have to be consistent, though, i.e. outward pointing at all boundary facets, or inward pointing on all boundary facets.   The result is the same.    So it is not enough here to state on the boundary facets that your vector field is simply "normal".  This is where the biggest constraint comes in, attempting to generalize to polytopes.
Euler characteristic is "additive", so you could use the above as a guide towards attempting to generalize the theorem. 
Consider for example a polyhedron which is a Y-graph.  The euler characteristic is 1.  But if you demand a vector field to be "normal" on the three "boundary" vertices, the simplest vector field you can imagine has just two zeros, both with the same degree.  
I think the simplest thing to do in your case is to use the fact that your polytope is sitting in Euclidean space.  So take a small regular neighbourhood of it.  Provided your vector field has finitely-many zeros and is outward-pointing on the boundary, the regular Poincare-Hopf theorem applies.  Since your regular neighbourhood has the polytope as a deformation-retract they have the same Euler characteristic, so you have the generalization you desire. 
Ensuring the vector field is outward-pointing on the boundary is the only real technical condition for this theorem.  You can state this as a kind of transversality condition for the vector field on the polytope.  If this is the kind of thing you think would suffice, I could elaborate on this.<|endoftext|>
TITLE: On the eigenvalues' distribution of random unitary
QUESTION [5 upvotes]: Fix an integer $d$, let $\mathbb{U}_d$ be the $d\times d$ unitary group.
For any $U\in \mathbb{U}_d$, define $\Omega(U)$ be the length of the smallest arc containing all the eigenvalues of $U$ on the unit circle.
Now, we are interested in the probability density function of $\Omega(U)$ as $U$ being distributed as Haar measure.
What is the probability that $\Omega(U)$ is smaller than some given constant $\epsilon$?

REPLY [6 votes]: This amounts to the question what is the probability distribution of the largest gap $\Delta=2\pi-\Omega$ between the eigenphases in the circular unitary ensemble. It was addressed in Extreme gaps between eigenvalues of random matrices (2010). Asymptotically for large $d$, the distribution $P(\Delta)$ is peaked around $\Delta_0=d^{-1}\sqrt{32\log d}$, with Poissonian fluctuations around this value.<|endoftext|>
TITLE: Langlands program vs Shimura-Taniyama-Weil conjecture
QUESTION [7 upvotes]: Edward Frenkel said that "we can  see Langlands program as a generalization of Shimura-Taniyama-Weil conjecture in the case of elliptic curves" 
I hope I'm not distorting his phrase, can someone explain what that means. Lets say that I'm little bit familiar with the ingredients used in both conjectures, Galois representations, elliptic curves,...

REPLY [6 votes]: To expand on zeno's answer, a Langlands-type formulation of the modularity conjecture would be:

(Taniyama-Shimura) $L(E,s)=L(f,s)$

Here $L(E,s)$ is the Hasse-Weil L-function of an elliptic curve $E$ over $\mathbb{Q}$, and $L(f,s)$ is the L-function of a modular form $f$ of weight 2 with integral coefficients. In fact you have a complete correspondence: all elliptic curves (over $\mathbb{Q}$) are modular in that sense, and all newforms (of that type) come from elliptic curves.
A "Langlands generalization" of this result is:

(Langlands) $L(V,s)=L(\pi,s)$

Here $L(V,s)$ is the Hasse-Weil L-function of any algebraic variety $V$ over an arbitrary number field, and $L(\pi,s)$ the L-function of some automorphic form $\pi$ over $\mathrm{GL}_n(\mathbb{A})$.
This is a theorem in some cases (other than Taniyama-Shimura), for example abelian varieties over $\mathbb{Q}$, elliptic curves over real quadratic fields and elliptic curves with complex multiplication.
You can further generalize:

(Langlands) $L(M,s)=L(\pi,s)$

With $L(M,s)$ the L-function of a motive. This of course includes the case above, Artin L-functions (non-abelian class field theory) and all the other Galois representations that should be automorphic ($l$-adic, mod p).
In the other direction, the Langlands conjectures restricted to the the case of an elliptic curve over $\mathbb{Q}$ predict precisely the same as the Taniyama-Shimura conjecture.<|endoftext|>
TITLE: Is the equicontinuous weak-star topology locally convex on the dual of an LF-space?
QUESTION [8 upvotes]: The Banach-Dieudonné theorem states that if $X$ is a metrizable locally convex Hausdorff space then the equicontinuous weak-* topology $ew^*$ on $X'$ coincides with the topology of precompact convergence and is therefore a locally convex topology. ($ew^*$ is the final topology on $X'$ coinduced by the inclusions of the equicontinuous sets when equipped with the weak-* topology $w^*$. Note that $ew^*$ is a priori not the locally convex final topology of these inclusions!) If $X$ is complete and thus Fréchet then it also coincides with the topology of compact convergence.
Is this also true in the case that $X$ is a strict inductive limit of a sequence of Banach or Fréchet spaces? In other words, is the equicontinuous weak-* topology on the dual of an LB- or LF-space locally convex or at least linear?
EDIT: I think to have found a counterexample which I try to sketch.
Consider the LF-space $\mathcal{D} := C^\infty_c(\mathbb{R})$ of test functions with the locally convex inductive limit topology and its dual $\mathcal{D}'$ - the space of distributions.

Since $\mathcal{D}$ is Montel, the strong dual $\mathcal{D}'_\beta$ is Montel and thus a sequence in $\mathcal{D}'$ is strongly convergent iff it is weakly*-convergent.
It is not hard to see that if $X$ is the strict inductive limit of separable Fréchet spaces then $ew^*$ is a sequential topology and has the same convergent sequences as $w^*$. Thus, $ew^*$ is the sequential coreflection of $w^*$. (The separability of the Fréchet spaces induce separability of $X$ which in turn is used for the (equicontinuous) polars of a neighborhood base of $0$ in $X$ to be metrizable and thus sequential. I can sketch a more detailed proof.) The space $X = \mathcal{D}$ satisfies these assumptions.
Dudley, "Convergence of Sequences of Distributions" (1971) has shown that $\mathcal{D}'_\beta$ is not sequential and that the topology of all sequentially (strongly) open sets (which by 1. and 2. coincides with $ew^*$) is not a vector topology (addition is merely jointly sequentially continuous).

For my applications it is rather of interest, whether for the LB-space $X = C_c(\mathbb{R})$ the $ew^*$-topology on its dual $X'$ (the space of real Radon measures) is a vector topology. We can't use the above proof since point 1. is not satisfied for $X$, i.e. a weakly* convergent sequence in $X'$ needs not be strongly convergent. (Dudley has stated in his paper that $X'_\beta$ is not sequential, but I can't use this fact to check the linearity of the $ew^*$-topology.)

REPLY [6 votes]: An interesting paper on spaces for which the $ew^*$ topology coincides with the topology of precompact convergence is S-spaces and the open mapping theorem by Taqdir Husain. He calls such spaces S-spaces, a term that doesn't appear to have become standard.
Proposition 2 of that paper shows that a complete locally convex space is an S-space if and only if the $ew^*$ topology is locally convex. Corollary 2.1 that precedes it states that a complete S-space is always B-complete (i.e. fully complete in the sense of Ptak). Section 9 is on inductive limits, and notes that Grothendieck had already given an example of an LF space that is not B-complete. Unfortunately the sufficient condition he gives for strict inductive limits to be S-spaces can never apply to limits of infinite-dimensional Banach spaces.<|endoftext|>
TITLE: Style guide for referring to past work
QUESTION [9 upvotes]: Has anyone written or expressed a coherent position on how to refer to mathematical results (theorems, proofs) by past authors?  Even if there are no hard and fast rules, I find it helpful to have a guide that I can follow in order to speed things along. Here are some of the issues that I regularly find myself dealing with inconsistently, even in the same article:

Do I refer to authors (citet, in natbib, say) or their articles (citep)?  Or in what situations is one preferred?  In the case of citep, is it reasonable to treat the actual reference [ABC+03] as a noun?  Or is that bad style? (Perhaps some journals have specific rules on this.)
Do I refer to the act of proving results in the present tense or past tense?  

Perhaps this relates to how one stands on the question of whether mathematical results are "discovered" or "invented".  But here are five ways to say almost the same thing:

[AB03] proves X.
[AB03] proved X.
Abacus and Bacchus [AB03] prove X.
Abacus and Bacchus [AB03] proved X.
A proof for XYZ appears in [AB03].

(... and perhaps the answers would be different for s/prov/show/.)
In addition to the above, what other inconsistencies have people come across and how do they address them?

REPLY [3 votes]: Well, this is what the American Mathematical Society recommends:

Form I and form II refers to the different ways to present the bibliography, with numbered references (I) or with alphabetically ordered authors (II). So I would say number 4 of your options is recommended.<|endoftext|>
TITLE: Average minimum number of random k-sparse vectors in GF(2) to span the whole space?
QUESTION [6 upvotes]: What is the average minimum required number of independent $k$-sparse (having at most $k$ non-zero elements) random vectors belonging to $\mathbb{F}_2^n$ to span the whole space of $\mathbb{F}_2^n$? Any such vector is uniformly probable to be chosen among the total $\sum_{j=0}^k \binom{n}{j}$ vectors. 
Here are the two extreme cases: 

If $k=n$, this average value is $n+1.6067$ as proved here.
If $k=1$, using coupon collector problem this average value is proved to be larger than $\Theta(n \log n)$.

Can we prove that if $k = \Theta(\log n)$, then this average value is $\Theta(n)$? or something similar? My simulation results show that for a pretty large range of $k$ this average value is $\Theta(n)$.

REPLY [5 votes]: Let $k=\Theta(\log n)$. We will keep adding random $k$-sparse vectors one at a time and stop when we get a spanning set. Assume that currently the vectors span a space $A$ of codimension $j$ for $j\le n/(2k)$. We want to figure out the expected number of steps to reach codimension $j-1$, which is the inverse of the probability that a randomly chosen $k$-sparse vector is outside of $A$.
Since $A$ has codimension $j$, we can find $j$ basis vectors $e_{\ell_1},\ldots,e_{\ell_j}$ such that $A\cap\text{span}(e_{\ell_1},\ldots,e_{\ell_j})=\{0\}$. This means that for any vector $v$ that has $k-1$ 1's, none of them at a position $\ell_i$, we have that at most one of $v,v+e_{\ell_1},v+e_{\ell_2},\ldots,v+e_{\ell_j}$ is in $A$. Thus there are at least $j\binom{n-j}{k-1}$ $k$-sparse vectors outside of $A$, so the probability of a random $k$-sparse vector is outside of $A$ is at least
$$
\frac{j\binom{n-j}{k-1}}{\sum_{i=0}^k \binom{n}{i}}=\Theta\left(\frac{j\binom{n}{k-1}}{\binom{n}{k}}\right)=\Theta\left(\frac{jk}{n}\right)=\Theta\left(\frac{j\log n}{n}\right)
$$
where we used the fact that $j\le n/(2k)$ to get $\binom{n-j}{k-1}=\Theta\left(\binom{n}{k-1}\right)$. This means that going from codimension $j$ to codimension $j-1$ takes an average of $\Theta\left(\frac{n}{j\log n}\right)$ steps, so going from codimension $\lfloor \frac{n}{2k}\rfloor$ to codimension $0$ (spanning set) takes an average of
$$
\sum_{j=1}^{n/(2k)} \frac{n}{j\log n}=\Theta\left(\frac{n\log(n/(2k))}{log n}\right)=\Theta(n)
$$
steps.
If the codimension is $j=\lfloor \frac{n}{2k}\rfloor$, then we derived above that decreasing the codimension by 1 requires an average of $\Theta\left(\frac{n}{j\log n}\right)=\Theta(1)$ steps. Since the number of steps necessary is only smaller when the codimension is greater, it takes us at most $\Theta(n/\log n)$ steps to get to codimension $\lfloor \frac{n}{2k}\rfloor$. Therefore the whole process takes $\Theta(n)$ steps.<|endoftext|>
TITLE: Who proved that $l^1$ and $L^1[0,1]$ are not isomorphic?
QUESTION [22 upvotes]: $l^1$ has the Schur property (every weakly convergent sequence is norm convergent) and $L^1[0,1]$ does not, so the two spaces cannot be isomorphic.
Is this folklore, or is it credited to someone? (Also I wonder whether the original proof of non-isomorphism was this one.)
Edit: to clear up any confusion, I am asking about the nonexistence of an isomorphism, not the nonexistence of an isometric isomorphism (which is basic, as pointed out in the comments).
Edit 2: the comment section seems to have degenerated into attempts to find alternate proofs of this fact (and debate over common misconceptions, such as whether a Banach space which is isomorphic to a dual space must itself be a dual space). That's great, although I still think the Schur property argument is the easiest one. However, this was not my question. Who first proved this?

REPLY [23 votes]: This result was already published in the French (1932) edition of Banach's book "Theory of Linear Operations" (I do not know whether it was published in the Polish (1931) edition). On page 245 in the table you see two reasons why the spaces $\ell_1$ and $L_1(0,1)$ have different linear dimension:

Schur property for $\ell_1$ and no Schur property for $L_1$.
The fact that $\ell_1$ is isomorphically embeddable into each of its infinite dimensional subspaces and the same is not true for $L_1$. 

Apparently for both statements about $L_1$ Banach means that they follow from the Khinchin's inequality (1923), but I did not find the corresponding place in the book.<|endoftext|>
TITLE: Relation between harmonic vector field and harmonic 1-form
QUESTION [5 upvotes]: Definition 1: A unit vector field $X$ side to be harmonic if it is critical point for the following energy function
  $$E(X)=\frac{1}{2}\int_M\|dX\|^2dvol_g=\frac{m}{2}vol(M,g)+\int_M\|\nabla X\|^2dvol_g.$$
Definition 2: A 1-form $\omega$ side to be harmonic if it is in kernel of Laplace operator. i.e. $\Delta\omega=(d\delta+\delta d)\omega=0$.

Question: Is there relation between two above definitions? Please give a simple example.
Update: I find some theorem in this topic:


Theorem 1. If $\omega$ is harmonic and $X$ is the dual vector field, we have that $\mathrm{div}X = 0$.
Theorem 2. If $X$ is a vector field on $(M,g)$ and $\omega(v) = g(X,v)$ is the dual 1-form, then
    $$\mathrm{div}X = −\delta\omega.$$


Thanks.

REPLY [12 votes]: The two notions are related, but they are not the same.  
The condition for a unit vector field $X$ on a Riemannian manifold $(M,g)$ to be harmonic is not the same as the condition that the dual $1$-form $X^\flat$ be harmonic.  The point is that, for unit vector fields, one defines the energy as the energy of the map $X:M\to S(M)$ where $S(M)$ is the unit sphere bundle of $(M,g)$ endowed with the Sasaki metric and one says that a unit vector field is harmonic if it is a critical point of this energy.  This is not the same as the energy of the $1$-form $X^\flat$ in general (though it can be sometimes, for example, if the metric is flat).  
A simple example is to take $(M,g)$ to be $S^3=\mathrm{SU}(2)$ endowed with its natural bi-invariant metric.  Then any unit left-invariant (or right-invariant) tangent vector field $X$ is harmonic in the above sense, but the dual $1$-form $\omega = X^\flat$ is not harmonic as a $1$-form because the only harmonic $1$-form on $S^3$ is the zero $1$-form. (Since $H^1(S^3) = 0$, this follows, for instance, from the Hodge Theorem.)
There are several good sources for study of this notion of harmonicity of unit vector fields.  There is a whole book, Harmonic Vector Fields: Variational Principles and Differential Geometry, by S. Dragomir and Domenico Perrone  (Elsevier, 2012), but there are also articles that you may find useful:  For example, see the survey article Volume, energy and generalized energy of unit vector fields on Berger spheres. Stability of Hopf vector fields by Olga Gil-Medrano and Ana Hurtado (http://www.ugr.es/~ahurtado/PDF/correcciones.pdf) and the references therein.<|endoftext|>
TITLE: Algebraic $K_1$ group for a $C^*$-algebra
QUESTION [9 upvotes]: Let $A$ be a $C^*$-algebra: then one defines topological $K_1$ group as $GL_{\infty}(A^+)/\Big(GL_{\infty}(A^+)\Big)_0$ where $A^+$ denotes $A$ with the unit adjointed (even if $A$ already had a unit: in this case $A^+$ is isomorphic to $A \oplus \mathbb{C}$ but this definition turns out to be equivalent if we insert $A$ instead of $A^+$). The algebraic $K_1(A)$ is defined as $K_1(A)=GL_{\infty}(A)/[GL_{\infty}(A),GL_{\infty}(A)]$ where ${H,H}$ is a commutator subgroup and it is known that it is not the same as topological $K_1$. I read somewhere that $K_1^{top}(A)$ may be defined somehow similarly to the algebraic $K_1$ as $K_1^{top}(A)=GL_{\infty}(A)/ \overline{[GL_{\infty}(A),GL_{\infty}(A)]}$, at least for unital $C^*$-algebras. My question is the following:

How to show that these two aproaches are equivalent?

REPLY [8 votes]: One must show that 
$$
\overline{[GL_\infty(A),GL_\infty(A)]}= (GL_\infty(A))_0.
$$
The proof that the left side is in the right side is the easier one: Let $a$ be in the left side. Then $a=bc$, where $b\in [GL_\infty(A),GL_\infty(A)]$ and $\|c-1\|<1$. Commutators belong to $(GL_\infty(A))_0$, because $K_1(A)$ is abelian (or because of the proof of this fact). So $b\in (GL_\infty(A))_0$. Also, $c=e^h$ for some $h\in M_\infty(A)$, and so it is connected to $1$ by the path $t\mapsto e^{th}$.
Choose now $a\in (GL_\infty(A))_0$. It can be written as finite product of exponentials $e^h$, with $h\in M_\infty(A)$. So it is enough to show that
these exponentials belong to the left side. We first prove this for $h\in M_\infty(A)$ of the form $xy-yx$.
In this case one has
$$
(e^{x/n}e^{y/n}e^{-x/n}e^{-y/n})^{n^2}\to e^{xy-yx}.
$$
The elements on the left side are products of commutators so we get the desired result. Finally, notice that any $h\in M_\infty(A)$ is a limit of commutators. For example, assuming that $h\in A$ for simplicity we can 
express the matrix
$$
\begin{pmatrix}
1 & & & \\
& -\frac 1 n  & &\\
&& -\frac 1 n &\\
&&&\ddots
\end{pmatrix}
$$
as a commutator in $M_{n+1}(\mathbb C)$ then multiply by $h\otimes 1_{n+1}$ to get diag$(h,-h/n,\ldots,-h/n)$ as a commutator.<|endoftext|>
TITLE: natural weak factorization systems
QUESTION [7 upvotes]: I am trying to understand the definition of natural weak factorization systems from this article by Tholen and Grandis, and these notes by Emily Riehl.
In Riehl's notes, the splitting $s,t$ are of $\rho_f,\lambda_g$, and it is explicitly shown (see diagram below) how their existence enables us to solve the lifting problem for $(f,g)$ to prove $f\pitchfork g$. On the other hand, Riehl does not go into much detail about how a natural choice of such splittings gives rise to (co)monads.

Tholen and Grandis's paper, on the other hand, looks at splittings $s,t$ of $\rho_{Lf},\lambda_{Rf}$. They just on to say these splitings are needed for constructing liftings, but I don't understand how. For instance, everything here is part of the data around the arrow $f$ (see diagram below) so how can it solve lifting problems?

So both papers discuss the significance of having a natural choice of splittings, but for different arrows, even though both seem to say their splittings are used to solve lifting problems (show weak orthogonality). Riehl shows this explicitly but does not demonstrate how this data is relevant to the monadic formalism, while Grandis and Tholen do the latter without the former.
Help!!

REPLY [3 votes]: My answer mostly consists of the relevant details of Omar Antolín-Camarena's answer to this MO question.
First of all, all the splittings you mention arise as solutions to canonical lifting problems of arrows against their $(\mathcal L,\mathcal R)$-factorizations. Such solutions are fillers for a commutative square with one edge the identity, and this is what gives the splitting.
A crucial observation is that the existence of such sections for a given arrow is actually equivalent to solving all lifting problems it involves. This equivalence is described in detail in section 12.4 of Emily Riehl's book Categorical Homotopy Theory. She also describes exactly how splittings are necessary for constructing lifts - they solve a generic lifting problem encompassing all the other ones for a given arrow.
Naturality of the splittings is the additional structure which brings in the density comonad. Recall naturality into $\mathsf{C}$ is the same as functoriality into $\mathsf{C^2}$, and it is the filler functor (called $\phi$ in the linked answer) that solves a lifting problem for the counit of the density comonad. So naturality lets us involve density comonads and motivates the monad-heavy definitions that come later. I think such functoriality is reasonable since in the context of cofibrant generation on a set $J$ seen as a discrete subcategory, we mightaswell look at it as the full subcategory of $\mathsf{C^2}$ generated by these arrows, and ask for coherence w.r.t composition in $\mathsf{C^2}$. The formalism then generalizes to general categories over $\mathsf{C^2}$.
Finally, about the specific issue of existence of splittings for $\rho_{Lf},\lambda_{Rf}$ as opposed to $\rho_f,\lambda_g$: these ensure different things. The former splittings must exist because they are equivalent via the generic lifting problem to $\lambda_f\in \mathcal L$ and $\rho_f\in \mathcal R$, which we obviously want to hold. So I think the authors of the paper mean they must exist because they assume $\lambda,\rho$ are indeed $(\mathcal L,\mathcal R)$-factorizations. The existence of the latter splittings is, again, equivalent to generic lifting problems, and decides whether $f\pitchfork g$ or not.<|endoftext|>
TITLE: Is it possible to have a research career while checking the proof of every theorem that you cite?
QUESTION [57 upvotes]: A colleague raised the above question with me; more precisely he said:

Suppose that a mathematician were resolved not to publish any theorems 
  unless they had checked the proof of every theorem that they cite (and 
  recursively the proofs of all the theorems that those rely on etc.). 
  Can they have a career in pure mathematics?

With the obvious proviso:

Of course, there are a few well-known theorems, like the classification 
  of finite simple groups, whose proofs are virtually impossible for any one 
  person to check at all. But one can have a perfectly good mathematical career 
  without ever citing any of those.

It seems to me that complete checking might be possible, though perhaps
only in narrow fields of mathematics, but does anyone know of mathematicians
who actually do it? (or try to)

REPLY [10 votes]: It is worth quoting what Hirsch says about J. H. C. Whitehead (footnote 23 on page 95 of Hirsch's contribution to the Smalefest volume.)
"Whitehead was very good about what he called "doing his homework," that is, reading other people's papers. "I would no more use someone's theorem without reading the proof," he once remarked, "than I would use his wallet without permission." He once published a proof relying on an announcement by Pontryagin, without proof, of the formula $\pi_4(S^2)=0$, which was later shown (also by Pontryagin) to have order 2. Whitehead was quite proud of his footnote stating he had not seen the proof. Smale, on the other hand, told me that if he respected the author, he would take a theorem on trust."<|endoftext|>
TITLE: Picard groups of quartic K3 surfaces
QUESTION [8 upvotes]: Does anyone know where I can find examples of quartic K3 surfaces for which the Picard group is known? I'm really interested in examples where there are explicit constructions of the divisors generating the Picard group (rather than examples where we only know the Picard rank). I appreciate that this is not easy to compute in general, so any useful references you can suggest will be most welcome.

REPLY [2 votes]: There are some simple such examples if you want to know the generators but not explicit equations for the surfaces:

A very general quartic in $\mathbb P^3$ has Picard number $1$ and hence its Picard group is generated by any minimal degree curve.
A quartic, very general among those containing a fixed line has Picard number $2$ and its Picard group is generated by that line and the hyperplane class (or the complementary cubic curve).
A quartic, very general among those containing a fixed conic has Picard number $2$ and its Picard group is generated by that conic and the hyperplane class (or the complementary conic).
A quartic, very general among those containing two fixed skew lines has Picard number $3$ and its Picard group is generated by those lines and the hyperplane class.

If you are only interested in generating $\mathrm{Pic}\otimes\mathbb Q$, then you have that

$\mathrm{Pic}\otimes\mathbb Q$ is generated by the smooth rational and elliptic curves on it for any $K3$ with Picard number at least $5$.
$\mathrm{Pic}\otimes\mathbb Q$ is generated by the smooth rational curves on it for any $K3$ with Picard number at least $12$.<|endoftext|>
TITLE: Is there a smooth polar decomposition for non-invertible matrices?
QUESTION [5 upvotes]: Every $n \times n$ real matrix $A$ has a polar decomposition $A=OP$, where $O \in O_n, P$ is symmetric positive semi-definite.
$P$ is uniquely determined by $A$, by $P(A)=\sqrt{A^TA}$,
and when $A$ is invertible $O$ is also unique, given by $O=A(\sqrt{A^TA})^{-1}$.
When $A$ is not invertible, then $O$ is non-unique.
Question:
Can we choose a smooth polar factor $O(A)$ for all non-zero matrices $A$ with non-negative determinant? 
More precisely, denote by $M_n^+$ the set of matrices with non-negative determinant. 
Is there a smooth function $O:M_n^+ \setminus \{0\} \to O_n $ such that for every $A \in M_n^+ \setminus \{0\}$, $A=O(A)P(A)=O(A)\sqrt{A^TA}$?
Remarks:
1) We know that such a function, if exists, must give to each $A \in GL_n^+$ its unique polar factor. 
2) The reason we exluded the zero matrix is that continuity implies two contradictory evaluations:
$O(0)=\lim_{t \to 0^+} O(\left(\begin{matrix}t & 0 \\ 0 & t\end{matrix}\right))=I$, 
$O(0)=\lim_{t \to 0^+}=O(\left(\begin{matrix}0 & -t \\ t & 0\end{matrix}\right))=\left(\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right)$
3) The reason we had to restrict to $M_n^+ \setminus \{0\}$ (instead of working with all $M_n \setminus \{0\}$) is connectedness issues. 
$M_n\setminus \{0\}$ is connected (for $n >1$). However for $O|_{O_n}=Id_{O_n}$, so if we insisted to take the domain to be all $M_n \setminus \{0\}$, the image would be $O_n$ which is disconnected.
The restriction in fact implies that $O$ is a function into $SO_n$.

REPLY [6 votes]: To take user35593's comment to fruition:

There is a unique continuous extension of the polar decomposition from $GL^+_n$ to the portion of its boundary consisting of matrices with nullity 1. 

Why is the polar decomposition non-unique when $A$ is non-invertible?
Suppose $A = OP$ and $P$ has non-trivial kernel. Then if $\Omega$ is any orthogonal matrix that acts as the identity on $\mathrm{ker}(P)^\perp$, then $O\Omega P$ is another polar decomposition. And it is easy to see that this freedom is the extent of non-uniqueness: if $A = OP = \tilde{O} P$, then $O^{-1} \tilde{O}$ is an orthogonal matrix that acts as the identity on $\mathrm{ker}(P)^\perp$. 
When the kernel of $A$ (equivalently kernel of $P$) is one dimensional, then the only possibility for $\Omega$ is the matrix that is the identity on $\mathrm{ker}(P)^\perp$ and $-1$ on $\mathrm{ker}(P)$. 
Proof of claim
Now, since $\det A = \det O \det P$ and we know that $\det P \geq 0$, when we restrict to $GL_n^+$ we must have that $\det O > 0$. That is, on $M_n^+$ we can require the matrix $O$ in the polar decomposition to be not only orthogonal, but in fact in $SO_n$. And for rank $n-1$ matrices only one of $O$ and $O\Omega$ can fulfill that criterion. And it is easy to check that this is the continuous extension from $GL_n^+$. 
Remark: if you extend from $GL_n^-$ you will pick up the other one. An illustrative example is the rank $1-1 = 0$ matrix $(0)\in M_1$. Approaching from $M_1^+$ the continuous limit of the polar decomposition gives $$ (0) = (1)(0)$$
while approaching from $M_1^-$ the continuous limit of the polar decomposition gives 
$$ (0) = (-1)(0). $$
Differential geometry
To answer this question in the comments: no, when $n > 1$ the set $M_n^+$ is should not be thought of as a (smooth) submanifold with boundary; it is better described as a manifold with corners. This is precisely what you outlined in your post with the different limiting directions approaching the zero matrix. 
But in a neighborhood of (topological) boundary points which have rank $n-1$, the set $M_n^+$ does look like a manifold with boundary: in a neighborhood of a rank $n-1$ matrix, the function $\det: M_n\to \mathbb{R}$ has non-vanishing derivative, and so can be used as a defining function for a hypersurface.<|endoftext|>
TITLE: Swan K-theory of Z/4
QUESTION [15 upvotes]: Given a finite group $G$ and a commutative ring $R$, define the Swan $K$-theory $K_0(G, R)$ to be the Grothendieck group of the category finitely generated projective $R$-modules with $G$-action (with respect to all short exact sequences). When $R = \mathbb{C}$, this recovers the classical representation ring. When $R = \mathbb{F}_p$ and $G$ is a  $p$-group, one gets $\mathbb{Z}$ for the Swan $K$-theory (everything is built up via short exact sequences from the trivial representation). 
Are there computations in the literature when $R$ is non-regular in modular characteristic? I am interested in the example $G  = C_2, R = \mathbb{Z}/4$ for starters. It may be difficult to classify representations of $C_2$ on free $\mathbb{Z}/4$-modules but I am hoping it is a little easier to understand the Swan $K$-theory.

REPLY [11 votes]: There seems to be a classification of representations for the example you mention (and more generally for representations of $C_p$ over $\mathbb{Z}/p^s\mathbb{Z}$) in

V. S. Drobotenko, E. S. Drobotenko, Z. P. Zhilinskaya and E. V. Pogorilyak,
  Representations of cyclic groups of prime order $p$ over rings of residue classes mod $p^s$, Ukrain. Mat. Z. 17 (1965), 28-42. MR0188304.

I've not been able to find a copy of the paper, which is in Russian, but if I understand correctly the review in Math Reviews, it seems as though the indecomposable representations, up to isomorphism, are as follows:

For each monic irreducible polynomial $\varphi(x)$ over $\mathbb{Z}/2\mathbb{Z}$, and each $e\geq1$, a representation sending the generator of $C_2$ to $I+2X$, where $X$ is a companion matrix of $\varphi(x)^e$.
For each monic irreducible polynomial $\varphi(x)$ over $\mathbb{Z}/2\mathbb{Z}$, and each $e\geq1$, a representation sending the generator of $C_2$ to $-I+2X$, where $X$ is a companion matrix of $\varphi(x)^e$.
The regular representation.

These are all non-isomorphic except that the rank one representations given by (1) and (2) are the same.
The representations that are irreducible (in the sense of having no $\mathbb{Z}/4\mathbb{Z}$-free subrepresentations) are those that occur in (1) and (2) for $e=1$. Perhaps $K_0(G,R)$ will be $\mathbb{Z}$-free on the classes of these?
Edit: I don't think the statement of the classification can be quite right, as it seems to me that the representations given in (1) are isomorphic to those given in (2):
Let $X$ be the companion matrix, over $\mathbb{Z}/2\mathbb{Z}$, of a polynomial $p(t)$ and let $Y$ be the companion matrix of $p(t+1)$. Then $I+X$ has minimal and characteristic polynomial $p(t+1)$ and so is similar to $Y$. So $(I+X)A=AY$ for some invertible matrix $A$. 
Lift $A$ to a matrix $\tilde{A}$ over $\mathbb{Z}/4\mathbb{Z}$. Then $\tilde{A}$ is automatically invertible, and $(I+2X)\tilde{A}=\tilde{A}(-I+2Y)$, so $\tilde{A}$ gives an isomorphism between the representation where a generator of $C_2$ acts as $I+2X$ and the one where it acts as $-I+2Y$.<|endoftext|>
TITLE: Does every ring of integers sit inside a monogenic ring of integers?
QUESTION [51 upvotes]: Given a number field $K/\mathbf{Q}$ whose ring of integers $\mathcal{O}_K$ is, in general, not of the form $\mathbf{Z}[\alpha]$ (not monogenic), does there exist an extension $L/K$ which has $\mathcal{O}_L= \mathbf{Z}[\alpha]$ (is monogenic)?
This question is imported from math stackexchange.
As noted by Jake: Kronecker-Weber implies this is true when $K/\mathbf{Q}$ is abelian.

REPLY [2 votes]: [edit: initially I thought there is no local obstruction but now I'm not sure about that so I deleted what I wrote on that]
Although I don't have proof, I suspect that most fields $K$ are counter examples to the OP's question. Empirically, most number fields are not monogenic. If you compute the best generator that polredabs in GP/PARI can find, and find that it does not generate $O_K$, as is almost always the case (except for low degree or cyclotomic, etc), then it is very likely that $O_K$ is not monogenic. If $O_K$ is monogenic but polredabs didn't find a generator for $O_K$, then $O_K$ would have a generator that is much better than the one found by polredabs.  Now polredabs in GP/PARI is based on LLL techniques, and while one can not prove that LLL finds the best element, it is only a bounded factor worse than optimal. This means that the empirical observation that polredabs usually doesn't find a generator for $O_K$ is good empirical evidence that most of these $O_K$ are not monogenetic. It is hard to see how increasing the degree could help globally, so it seems likely that for most fields $K$ (other than very special cases) there will not be an extension $L$ for which $O_L$ is monogenic.<|endoftext|>
TITLE: Complexity of Turing Machine behavior
QUESTION [6 upvotes]: If one looks at the code for a Turing Machine (TM) with
$q$ states and, let's say, $2$ symbols, they all look
pretty much the same:
A list of $5$-tuples:
$$
< state, symbol{-}read, symbol{-}to{-}write, head{-}movement, state > \;.
$$
Some of the $q$-state TMs are, however, rather
more complex than others:
the Busy Beaver (BB) TMs.
These TMs exectute an immense number of steps on an empty
tape before halting.
For example, there is a $6$-state BB that takes
more than $10^{36534}$ steps before halting.

          


          

(Image from 
Jeffrey Shallit's notes (PDF).)


My question is:

Is there some measure that captures the complexity/intricacy of
  a TM's behavior?

One can replace TM by "computer program" here.
I am looking for something beyond the complexity of the 
description of the program, and which instead captures its possibly
complex behavior on certain inputs.
It seems the Kolmogorov complexity would treat a $q$-state
BB as equally complex to a mundane $q$-state TM.
One possibility would be to run the TM on all inputs up to
some length beyond which the TM could not distinguish,
and form a measure from the number of steps before halting.
(Or perhaps: also before looping?)
Have such measures been considered in the literature?

REPLY [7 votes]: Here is one proposal. 
A set of natural numbers is computably enumerable if it can be enumerated by a Turing machine program, that is, if the set is the range of a Turing computable function. Two such sets are said to be Turing equivalent, if each of them can be computed from an oracle relative to the other, and the resulting collection of c.e. Turing degrees is an extremely rich and intensely studied hierarchy. 
Since every Turing machine program can be viewed as enumerating some c.e. set, we can naturally classify the complexity of the programs by the complexity of these c.e. sets that they enumerate, as points in the Turing degrees. 
It was the famous Post's problem that asked whether indeed there were any c.e. degrees other than the decidable sets and the halting problem, but it was found that there is indeed a very rich hierarchy of intermediate c.e. degrees.

REPLY [7 votes]: If you restrict attention to TMs that always halt, then:
One measure of complexity of a Turing machine is its running time, the maximum number of steps taken before it halts on inputs of length $n$, as a function of $n$. If this is $O(n^c)$, i.e., bounded by a polynomial, then we get the notion of polynomial time, and so forth.
That's time complexity. Then there's also space complexity, the maximum number of tape cells inspected during computation.<|endoftext|>
TITLE: Cofibrations in the model structures for non-negative graded (commutative) DG algebras
QUESTION [7 upvotes]: Let $k$ be a field of characteristic 0. Let $\mathtt{DGA}_{k}^{+}$ denote the category of non-negative graded DG algebras and $\mathtt{CDGA}_{k}^{+}$ denote the category of non-negative graded commutative DG algebras. It is well known that there are model structures on them that the weak equivalences are the quasi-isomorphisms and the fibrations are the maps which are surjective in all positive degrees.
We say that a morphism $f: A\rightarrow B$ in $\mathtt{DGA}_{k}^{+}$ (resp., $\mathtt{CDGA}_{k}^{+}$) is an almost free extension if $B\cong A*_{k} T_{k}V$ (resp., $B\cong A\otimes_{k} \Lambda_{k}V$) as graded algebras and the composition of $f$ with this isomorphism is the canonical map $A\rightarrow A*_{k} T_{k}V$ (resp., $A\rightarrow A\otimes_{k} \Lambda_{k}V$).
Is it true that the cofibrations in $\mathtt{DGA}_{k}^{+}$ and $\mathtt{CDGA}_{k}^{+}$ are precisely the retracts of almost free extensions? It seems like that this statement is a little bit different than the statement in nLab that the cofibrations are precisely the retracts of relative Sullivan algebras.
If the statement above is true, then it implies that the cofibrant objects are precisely the retracts of almost free DG algebras in $\mathtt{DGA}_{k}^{+}$ (resp., in $\mathtt{CDGA}_{k}^{+}$). (We say that $R\in \mathtt{DGA}_{k}^{+}$ is almost free if $R\cong T_{k}V$ as graded algebras for some graded vector space $V$, and $R\in \mathtt{CDGA}_{k}^{+}$ is almost free if $R\cong \Lambda_{k}V$ as graded algebras for some graded vector space $V$).
The second question is that if we only know the cofibrant objects are precisely the retracts of almost free DG algebras, can we get that the cofibrations are precisely the retracts of almost free extensions?
Thank you very much.

REPLY [3 votes]: The confusion between your desired statement and the statement from the nlab is arising because of the confusion between chain complexes and cochain complexes. 
The answer to your first questions is "yes." I'm not sure about the second question, but I would guess the answer is "no."

Because your fibrations are surjections in positive degrees, you are working with chain complexes, where the differential reduces degree. On the nlab link, you can see that they specify that they are working with cochain complexes, where the differential goes up in degree, and that their fibrations are surjections in all degrees. The underlying categories, while superficially similar, are quite different, and the model category structures make this more so.
Your model category structure comes from transferring the model category structure on non-negatively graded chain complexes, which is cofibrantly generated. For $i\ge 0$, let $S^i$ denote the chain complex which is $k$ in degree $i$ and let $D^{i+1}$ denote the acyclic chain complex which is $k$ in degree $i$ and $i+1$. Then the cofibrant generators for chain complexes are $J$, the inclusions of $0$ into $D^{i+1}$ (for acyclic cofibrations) and $I$, the inclusions of $S^i$ into $D^{i+1}$ (for cofibrations), along with the extra weird generator $0\to S^0$ (which we could think of as $S^{-1}\to D^0$, I guess). Then applying a free functor to these examples yields the generating cofibrations for the model structure on algebras or commutative algebras. 
Then a $(\mathtt{C})\mathtt{DGA}^+$-map is a cofibration precisely if it is a retract of a relative $F(I)$-cell complex. I'm not so good at messing around with cardinals but it's certainly clear that pushing out along a map in $F(I)$ is an almost free extension. So unless there are problems coming from some large cardinal (I'm pretty sure there are not) you get that relative $F(I)$-cell complexes are (some subset of) almost free extensions. But what subset? 
The point is that if this is going to be a cell complex, you've got to attach your cells in an order, and you can only glue cells onto previously attached cells. This is where the difference between chain complexes and cochain complexes really comes into play. In chain complexes, you can choose to attach your cells inductively, starting with the zero cells from the weird generator and then proceeding up the tower, attaching all $n$-cells in any arbitrary order. This reordering of your cells can't mess anything up because adding $n$-cells can never touch anything above $n$. This gives you an automatic filtration. Your $0$-cells are never attached to anything; your $1$-cells can only be attached to $0$-cells, your $2$-cells can only be attached to things you built out of $0$-cells and $1$-cells, and so on. So in chain complexes, every almost free extension is automatically Sullivan.
This is definitely not true in cochain complexes, where you can find Borromean type examples ($\Lambda(x,y,z)$, all generators in degree $1$, with the differential $dx=yz, dy=zx, dz=xy$). This happens because the differential from degree one can interact with generators from degree $1$.
This is tagged "reference request" but I don't actually know a clean reference; most of the careful expositions are done with unbounded complexes and the few I know that use non-negatively graded complexes seem not to try to characterize the cofibrations. 

For the second question, you would somehow need to be able to see that the inclusions $S^i\to D^{i+1}$ are all cofibrations; I don't see how to do that starting only with the data of the cofibrant objects. All I know how to do is build things with pushouts and sequential colimits and so on. In that situation I don't see any way to obtain cofibrations $X\to Y\cong_{\text{linearly}}X\oplus Z$  such that there is a differential that goes from $Z$ to $X$. I'd love to be wrong about this!<|endoftext|>
TITLE: A question on subsets of $\omega_1$
QUESTION [10 upvotes]: Is there a family $\{A_\alpha:\alpha<2^{\omega_1}\}\subset [\omega_1]^{\omega_1}$ in ZFC such that for each  countable set $I\subset 2^{\omega_1}$ and $\alpha\in 2^{\omega_1}\setminus I$ we have
$$A_\alpha\not\subset\bigcup\{A_\beta:\beta\in I\}.$$
Of course, we can find such a family of size $\omega_2$ instead of $2^{\omega_1}$.

REPLY [6 votes]: Today I could prove  that the existence of such family is not provable from  ZFC. 
Theorem.
If GCH holds in $V$,  and $\mu\ge \omega_3$  then in $V^{Fn(\mu,2)}$ the following holds:
If $\{A_i:i<{\omega}_3\}\subset [{\omega}_1]^{{\omega}_1}$, then there is
 $I\in [\omega_3]^\omega$ and $\alpha\in \omega_3\setminus I $ such that 
 $$A_{\alpha}\subset \bigcup_{i\in I}A_i.$$
Proof.
An ${Fn(I,2)}$-name $\dot B$ of a subset of $\omega_1$ is called 
 nice if for each ${\nu}<\omega_1$ there is an antichain 
$B_{\nu}\subset  {Fn(I,2)}$ such that 
$$
\dot B=\{\langle p,\check{{\nu}}\rangle:{\nu}\in {\omega_1}\land p\in B_{\nu}\}=
\bigcup\{B_{\nu}\times\{\check{{\nu}}\}:{\nu}\in {\omega_1}\}.
$$
We let $supp(\dot B)=\bigcup\{dom(p):p\in\bigcup\limits_{{\nu}<{\omega_1}}B_{\nu}\}$.
It is well-known  that every set of ordinals in  $V[G]$ has a nice name in $V$.
If ${\varphi}$ is a bijection between two sets $I$ and $J$ then
${\varphi}$ lifts to a natural isomorphism between ${Fn(I,2)}$ and
${Fn(I,2)}$, which will be also denoted by ${\varphi}$, as follows: 
for $p\in {Fn(I,2)}$ let $dom({\varphi}(p))={\varphi}''dom(p)$ and
${\varphi}(p)({\varphi}({\xi}))=p({\xi})$. Moreover ${\varphi}$ also
generates a bijection between the nice ${Fn(I,2)}$-names and the nice
$Fn(J,2)$-names: 
if $\dot A$ is a nice $Fn(I,2)$-name then let
${\varphi}(\dot A  )=\{\langle{\varphi}(p),\hat{{\xi}}\>:\langle p,\hat{{\xi}}\rangle \in\dot B\}$.
If $I$ and $J$ are sets of ordinals with the same order type then
  ${\varphi}_{I,J}$ is the natural order-preserving
bijection from $I$ onto $J$.
For each $i<{\omega}_3$ pick a nice $Fn(\mu,2)$-name $\dot A_i$ 
for $A_i$ in $V$. Then  $S_i=supp(A_i)\in [\mu]^{\le \omega_1}$.
Since $2^{\omega_1}=\omega_2$ in $V$ there is a subset 
$K\in [\omega_3]^{\omega_3}$ such that 
(1) $\{S_i:i\in K\}$ forms a $\Delta$-system with kernel $S$, 
(2) every element  of the family $\{S_i:i\in K\}$ has the same
    order type  
(3) $\varphi_{S_i,S_j}$ is the identity on $S$ for all
    $i,j\in K$.  
(4) $\varphi_{S_i,S_J}(\dot A_i)=\dot A_j$ for
    $i,j\in K$.
Let $I\in [K]^\omega$ and $\alpha \in K\setminus I$.
Assume on the contrary that  $p\in Fn(\mu,2)$ and $\zeta<\omega_1$ such that 
$$
p\Vdash \check \zeta\in \dot A_{\alpha}\setminus \bigcup_{i\in I}\dot A_i.  
$$
Pick $\beta \in I$ such that $dom(p)\cap (S_\beta\setminus S)=\emptyset.$
Let 
$$
r=p\cup \varphi_{S_\alpha, S_\beta}(p\restriction S_\alpha).
$$
Since $dom(p)\cap (S_\beta\setminus S)=\emptyset$, $r$ is a condition.
Moreover, $p\Vdash \check \zeta\in \dot A_{\alpha}$,
so $p\restriction S_\alpha \Vdash \check \zeta\in \dot A_{\alpha}$,
and so $\varphi_{S_\alpha, S_\beta}(p\restriction S_\alpha )\Vdash 
\check \zeta\in \dot A_{\beta}$. 
Thus $r\Vdash \zeta\in \bigcup_{i\in I}\dot A_i.$  Contradiction.<|endoftext|>
TITLE: I have a very large sparse matrix, 'A', in Ax = b. What work in advance of getting 'b' can be done to reduce solving time?
QUESTION [9 upvotes]: This question borders between a programming and math question (more math). I have a little matrix knowledge but this is past my ability, so any help is very much appreciated.
Question
I have a very large sparse matrix (say one million rows square). I'd like to solve this for many different sets of constant values (to clarify, by constant values, I mean the 1xn matrix 'b' in Ax = b). Performance is absolutely crucial - so it would make sense to do whatever simplification I can in advance of introducing each set of constants.
Goal
I'd like to get the time complexity for solving each constant set down to O(n)... I may be dreaming however.
Other Info

The matrix really is a million (maybe even 100 million) rows
It's very sparse
It has a bandwidth ranging between 2 and as much as a few hundred
Even for larger bandwidths, each row has max. 7 or 8 non zero entries
It takes an (extremely rough) diagonal form
Each matrix 'b' in Ax = b is actually very sparse itself
Due to the nature of the problem, a solution should always be available 
All nonzero values are real, floating point numbers. They can be negative however.

REPLY [2 votes]: There is a huge amount of research on this problem --- it turns out that solving huge linear systems is pretty ubiquitous in applied maths, because linear algebra is one of the few things that we can do efficiently for large problems.
Sparse LU decompositions have already been mentioned; the other main approach is Krylov subspace methods (such as GMRES) with a vast choice of preconditioners --- there is a whole lot of strategies, and many of them work only for specific problems.
I would suggest contacting someone in person; see who works in numerical analysis at your local institution. $10^6$ to $10^8$ is already quite a large problem, and it is likely that some tricks of the trade are needed to get things working. All this especially if you say that "performance is crucial".
Also, in any case try to use an existing program or library (Matlab, for some quick experimentation if the problem fits in memory for it, or Scipy, PetSC, Trilinos, ...). If you are not an expert, rewriting things is unlikely to match state-of-the-art code in terms of performance.<|endoftext|>
TITLE: Integral representation of adjoint L-factor for GL(2)
QUESTION [8 upvotes]: My question is about a local computation in the paper of Gelbart and Jacquet, "A relation between automorphic representations of GL2 and GL3", from 1978.
Let $\sigma$ be an irreducible smooth complex representation of $GL_2(F)$, where $F$ is a non-archimedean local field (characteristic 0 if it helps), and $\chi$ a smooth character of $F^\times$.
Jacquet has defined a Rankin--Selberg $L$-factor $L(\sigma \otimes \sigma^\vee \otimes \chi, s)$, which is always of the form 1/(polynomial in $q^{-s}$) where $q$ is the size of the residue field. Gelbart and Jacquet define the adjoint L-factor by 
$$ L(Ad(\sigma), \chi, s) := \frac{L(\sigma \otimes \sigma^\vee \otimes \chi, s)}{L(\chi, s)}.$$
In the Gelbart--Jacquet paper, they write down an integral $I(s, f, \Phi, \Psi, W)$ on the metaplectic group $Mp_2(F)$, which depends on $s$ and various choices of auxiliary data ($W$ is a vector in the Whittaker model of $\sigma$, $\Phi$ and $\Psi$ are Schwartz functions on $F$, etc). 
If either:

$\sigma$ is unramified (or a twist of an unramified representation);
or $\sigma$ is ramified, but $\chi$ is much more ramified than $\sigma$ is, so both $L(\sigma \otimes \sigma^\vee \otimes \chi, s)$ and $L(\chi,s )$ are identically 1,

then they show that the auxiliary data can be chosen in such a way that $I(s, \dots) = L(\operatorname{Ad}(\sigma), \chi, s)$.  

Does this hold more generally? Is it true, for arbitrary $\sigma$ and
  $\chi$, that we can choose a finite collection of quadruples $(f_i, \Phi_i, \Psi_i, W_i)$, $i=1 \dots r$, such that $$\sum_{i=1}^r I(s, f_i, \Phi_i, \Psi_i, W_i) = L(\operatorname{Ad} \sigma, \chi, s)?$$

(EDIT: I didn't make it plain originally that I was willing to allow a finite collection of test data, rather than just one. This form of the statement is true, essentially by definition, for the integral representations of the standard L-function on $GL_n$, and the Rankin--Selberg L-function on $GL_m \times GL_n$, so I'm looking for a generalisation of that to the adjoint L-function.)

REPLY [3 votes]: I'm not familiar with Gelbart and Jacquet's integral representation you mention (so this is more of a long comment than an answer). One can see some techniques from the local Rankin--Selberg GLn\times GLm convolutions setting in:

the 2010 dissertation of Kyung-Mi Kim (called Test Vectors of Rankin--Selberg convolutions of general linear groups and available online);
and in my work with Nadir Matringe last year (arXiv.1501.07587v3) which treats the cuspidal case and gives the existence of data such that a single Rankin-Selberg integral realises the L-factor (we give explicit data using Bushnell--Kutzko type theory when the L-factor is non-trivial (hence in particular n=m); that there exists such data when the L-factor is trivial is known from the work of Jacquet--Piatetski-Shapiro--Shalika as we mention in our introduction.)

You can find more details in the introductions to both of the above.  In the Rankin--Selberg case, we expect that an amalgamation of these techniques would answer the general question (giving a finite number of explicit vectors such that the sum of the corresponding integrals realises the L-factor), however determining under which conditions there is a single integral I'm not sure.<|endoftext|>
TITLE: Examples to keep in mind while reading the book 'The Admissible Dual...' by Bushnell and Kutzko and the importance of Interwining of representations
QUESTION [8 upvotes]: I am a beginner in the field of representation theory. I was reading the book 'The Admissible Dual of $GL(N)$ Via Compact Open Subgroups' by  Bushnell and Kutzko. 
Let me first describe the book a little.
The book uses hereditary orders $\mathfrak{A}$ arising from lattice chains and then defines simple  strata using the special integer $k_0(\beta,\mathfrak{A})$ (cf. page $43$). Then in page $95$ the authors construct the subgroups $H^m(\beta,\mathfrak{A})$ and $J^m(\beta,\mathfrak{A})$ and uses them (specifically $J^0$ and $J^1$ ) in the later chapters to extend representations of $J^1$ to $J^0$ and then they define the simple types $(J,\lambda)$ (cf. page $184$). In page $162$ they define the maximal and minimal orders $\mathfrak{B}_M $ and $\mathfrak{B}_m$ and uses them throughout the theory. Chapter $7$ proceeds with the Iwahori decomposition where they define (cf. page $213$) the parabolic subgroups $P$ with Levi decomposition $P=MU$, where $U$ is the unipotent radical of the group $G$ in consideration. The book finishes (chapter $8$) with the classification of smooth representations: the supercuspidals containing a simple type and the 'atypical' representations containing the split types. 
Being a beginner in this field it is necessary for me to have a concrete example to keep in mind while reading the above book because it is too general (construction using lattices and so on). Therefore I would like to have a concrete example of the following:
$(a)$ the hereditary order $\mathfrak{A}$,
$(b)$ the groups $U^n(\mathfrak{A})$ in page $21$,
$(c)$ the integer $k_0(\beta,\mathfrak{A})$,
$(d)$ the groups $H^m(\beta,\mathfrak{A})$ and $J^m(\beta,\mathfrak{A})$, specifically when $m=0,1$,
$(e)$ the utility of the maximal and the minimal hereditary orders $\mathfrak{B}_M$ and $\mathfrak{B}_m$ and there concrete examples,
$(f)$ concrete examples of the groups $P,M,U,U^-$ defined in $(7.1.13)$.
$(g)$ the importance of computing the interwining of representations (cf ($5.1.8$), ($6.1.3)$, e.t.c).
Regarding the Interwining of characters, it is not clear to me why the Interwining is impotant in representation theory. What role does it play? I agree that in chapter $8$ there is a theorem which says that the supercuspidal representations twisted by unramified characters have the same simple types. But I do not understand the impotantance of computing the Interwining of representations.
I know that answering these questions above will consume long typing in latex. I thank you in advance for your efforts.

REPLY [3 votes]: 1) On the groups $U({\mathfrak A}))$ and $U^n ({\mathfrak A})$. Their structure theory are very well explained in:
Bushnell, C. J.; Fröhlich, A. Nonabelian congruence Gauss sums and $p$-adic simple algebras. Proc. London Math. Soc. (3) 50 (1985), no. 2, 207–264. 
But as PL wrote, you should start with the case $GL_p$, $p$ prime. The structure of these groups in that case is explained (maybe with a different notation) in Carayol's paper. 
2) Before tackling Bushnell and Kutzko's theory, except looking at the $GL_p$ case, you should first read introductory papers :
Henniart, Guy Représentations des groupes réductifs $p$-adiques. (French) [Representations of $p$-adic reductive groups] Séminaire Bourbaki, Vol. 1990/91. Astérisque No. 201-203 (1991), Exp. No. 736, 193–219 (1992). 
Bushnell, Colin J.; Kutzko, Philip C. The admissible dual of ${\rm GL}_N$ via restriction to compact open subgroups. Harmonic analysis on reductive groups (Brunswick, ME, 1989), 89–99, Progr. Math., 101,
Kutzko, Philip C. Smooth representations of reductive $p$-adic groups: an introduction to the theory of types. Geometry and representation theory of real and $p$-adic groups (Córdoba, 1995), 175–196, Progr. Math., 158, Birkhäuser Boston, Boston, MA, 1998.<|endoftext|>
TITLE: Is it true that irreducible generic representations of $G_2(F)$ are self-dual?
QUESTION [9 upvotes]: Let $G_2$ be the split exceptional group of type $G_2$ and $F$ be a p-adic field. Is it true that every irreducible smooth representation of $G_2(F)$ is self-contragredient? If the answer is Yes, can anybody give me a reference? If not, is there a way to describe the contragredient, for example, is there a MVW involution like the classical group case?
Edit：
I thank Paul Garrett and Jim Humphreys for their comments and Jeffrey Adams for his nice answer. According to Jeffrey Adams's answer, one expects that each L-packet of $G_2(F)$ is self-dual. On the other hand, according to the general philosophy of Gan–Gross–Prasad, in each L-packet, there should be at most one generic representation (GGP conjectured that in each generic local L-packet, there is at most one representation which has the given Bessel model or Fourier–Jacobi model for classical groups. I do not know if anybody conjectured this for exceptional groups. But I just think that we can expect this once we can define the corresponding model. In particular, one would expect the uniqueness of generic member in each L-packet. For the conjecture of uniqueness of generic element in each L-packet, there is probably early reference, but I learned it from GGP). Thus one would expect that:
Each generic smooth irreducible representation of $G_2(F)$ is self-dual.
Do we expect this or is this also false? If we do expect this, do we know anything related to this?

REPLY [6 votes]: $\DeclareMathOperator\PGSp{PGSp}\DeclareMathOperator\PGL{PGL}$Generic representations of a $p$-adic $G_2$ are indeed self dual. It suffices to prove this for super-cuspidal representations. 
Observe that, for unitary representations, taking dual is the same as taking complex conjugate. 
Now all generic super-cuspidal lift one-to-one to generic representations of $\PGSp_6$ by the exceptional theta correspondence, see Savin–Weissman - Dichotomy for generic supercuspidal representations of $G_2$, Compositio Math (2011) (MSN). Since the exceptional theta correspondence commutes with complex conjugation, the statement follows from self-duality on $\PGSp_6$ side. In fact, since any representation of $G_2$ lifts either to $\PGSp_6$ or to a compact form of $\PGL_3$, one can completely classify representations of $G_2$ that are not self dual: the super-cuspidal representations that correspond to non-trivial representations of the compact form of $\PGL_3$.<|endoftext|>
TITLE: Is the restriction of a graded automorphism linearizable in characteristic zero?
QUESTION [8 upvotes]: This question follows up a previous one which was answered by Todd Leason. I want to impose two new requirements on the setup.
Let $k$ be a characteristic zero field. Let $A=k[x_1,\dots,x_n]$ be the polynomial algebra with the usual grading. Let $g$ be a graded automorphism of $A$ and let $B$ be a graded subring of $A$ such that:

$A$ is integral over $B$.
$B$ is fixed setwise by $g$.
$B$ is itself a polynomial algebra.


Is $g$'s restriction to $B$ linearizable?

By linearizable I mean that there exists a set of algebra generators $f_1,\dots,f_n$ of $B$ such that the $k$-vector space $V = \langle f_1,\dots,f_n\rangle_k$ is invariant under $g$. (So that $B$ can be seen as the symmetric algebra over $V$ and $g|_B$ the automorphism induced on the symmetric algebra by $g|_V$.)
The two new requirements are integrality and characteristic zero.
Todd Leason's response to the previous question shows that without the characteristic zero assumption the answer is no. I think it's probably no in general, since if $B$ is generated in distinct degrees then $g$ must act on its generators diagonally in order to be linearizable, and that seems a lot to ask. But Todd's example used the characteristic $p$-ness in an essential way, so I remain curious.

REPLY [2 votes]: Gregor Kemper answered a related question with a technique that can be used to answer this one affirmatively in the case that $g$ has finite order. If $g$ does not have finite order and we drop the integrality assumption, the answer is negative.
If $g$'s restriction to $B$ has finite order, then its action on $B$ is linearizable.
Proof: Note that $B$ is connected (i.e., its degree 0 component is just $k$) since $A$ is. Let $I$ be $B$'s positively-graded ideal. By the graded Nakayama lemma, any set of homogeneous elements of $I$ that generate the $k$-vector space $I/I^2$ will also generate $I$ as an ideal in $B$. By a standard induction argument, any homogeneous ideal generators for $I\triangleleft B$ are actually algebra generators for $B$. (To express an arbitrary homogeneous element $f$ in $B$, which without loss of generality can be taken to have positive degree since $B$ is connected, as a polynomial in these generators, express it first as a linear combination of them with coefficients in $B$. This is possible since $f$ has positive degree, and is therefore in $I$, and $I$ is an ideal. Since $B$ is graded, this linear relation can be taken to be homogeneous just by discarding all terms of degree different than $f$. But then the coefficients are all homogeneous of lower degree, so apply the induction hypothesis.) Since $B$ is presumed to be a polynomial algebra, $\dim_k I/I^2 = \dim_{\text{Krull}}B$. Thus any homogeneous lifts to $I$ of a $k$-basis for $I/I^2$ will automatically be polynomial-algebra generators for $B$.
Since $g$ acts as a graded automorphism on $B$, it preserves $I$. Therefore, it also preserves $I^2$. Since $k$ is of characteristic zero and $g$ has finite order, we have access to Maschke's theorem, which asserts that $I^2$ has a $g$-invariant complement $V$ in $I$. Because the action of $g$ on $B$ respects the grading, and $I$ and therefore $I^2$ are graded ideals, $V$ is even a graded vector space, i.e., it is the direct sum of its intersections with the graded components of $B$. (Indeed, for any $d\in\mathbb{N}$, we can apply Maschke's theorem in $I\cap B_d$ to find a $g$-invariant complement $V_d$ for $I^2\cap B_d$, and then take $V=\bigoplus V_d$.) Therefore, it has a homogeneous basis $\mathcal{B}$. Because $I=I^2\oplus V$, this basis descends to a basis of $I/I^2$. By the work in the previous paragraph, $\mathcal{B}$ is a set of polynomial algebra generators for $B$, so $V$ is the desired subspace.
Remark: This argument actually also works if $k$ has positive characteristic, as long as the order of $g$'s action on $B$ is not divisible by the characteristic. The argument also makes no use of the assumption that $A$ is integral over $B$, although if we know integrality, then we know that $\dim_{\text{Krull}}B = n$, so we know what to expect for the dimension of $I/I^2$.
If $g$'s restriction to $B$ has infinite order, and we drop the hypothesis that $A$ is integral over $B$, then $g$'s action on $B$ may not be linearizable.
Let $A=\mathbb{C}[x,y]$ and let $B=\mathbb{C}[x,xy]$. Let $g$ act on $A$ by $x\mapsto x$, $y\mapsto x+y$. Then $gB\subset B$ since $xy\mapsto x^2+xy\in B$, and $B\subset gB$ since $xy\in \mathbb{C}[x,x^2+xy]=gB$. Thus $B$ is fixed setwise by $G$. But $B$ is algebra-generated in distinct degrees and $g|_B$ does not act diagonally on the generators, so it is not linearizable.
The previous argument fails in this situation because the conclusion of Maschke's theorem fails: while $I^2$ is still a $g$-invariant subspace of $I$, it does not have a $g$-invariant complement.
But note that in this situation, $A$ is not integral over $B$, as $y$ is not integral over $B$.
I do not know if there is an example of infinite order $g$ and $A/B$ integral in which $g|_B$ is not linearizable.<|endoftext|>
TITLE: Which paths in a graph are orthogonal to all cycles?
QUESTION [22 upvotes]: Start with some standard stuff.  Suppose we have a directed graph $\Gamma$.  I'll write $e : v \to w \,$ when $e$ is an edge going from the vertex $v$ to the vertex $w$.  We get a vector space of 0-chains $C_0(\Gamma,\mathbb{R})$, which are formal linear combinations of vertices, and a vector space of 1-chains $C_1(\Gamma,\mathbb{R})$, which are formal linear  combinations of edges.  We get a boundary operator
$$ \partial : C_1(\Gamma,\mathbb{Z}) \to C_0(\Gamma,\mathbb{Z}) $$
sending each edge $e: v \to w$ to the difference $w - v$.  A 1-cycle is a 1-chain $c$ with $\partial c = 0$.  There is an inner product on 1-chains for which the edges form an orthonormal basis.  
Any path in the graph gives a 1-chain.  When is this 1-chain orthogonal to all 1-cycles?
To make this interesting, I need to rule out some obvious possibilities.  If we have a graph consisting of two triangles connected by an edge, the path consisting of that one edge will be orthogonal to all 1-cycles:  

To rule out this sort of situation, let's suppose $\Gamma$ has no bridges, meaning edges whose removal increases the number of connected components.  The edge with the arrow on it in the picture above is a bridge. 
Question: Suppose $\Gamma$ is a graph with no bridges.  Any path in such an embedded graph gives a 1-chain.  If this 1-chain is orthogonal to all 1-cycles, must it vanish?
To make this precise: I'm defining a path $\gamma$ to be a finite sequence of edges $e : v \to w$ and their formal 'inverses' $e^{-1}: w \to v$, like this:
$$ e_1^{\pm} : v_0 \to v_1, $$
$$ e_2^{\pm} : v_1 \to v_2 , $$
$$ \dots $$
$$ e_n^{\pm} : v_{n-1} \to v_n .$$
The corresponding chain is
$$ c(\gamma) = \pm e_1 \pm e_2 \pm \; \cdots \;\pm e_m. $$
Question: If $\Gamma$ is a graph with no bridges, and $\gamma$ is a path in $\Gamma$ such that the inner product of $c(\gamma)$ with every cycle vanishes, must we have $c(\gamma) = 0$?
I believe someone should have settled this by now, since it sounds easy, and the space of 1-chains orthogonal to all cycles has been studied quite a lot: it's called the cut space of the graph.    
A cut is a partition of the vertices of a graph into two disjoint subsets. Any cut determines a cut-set, the set of edges that have one endpoint in each subset of the partition.  If we take the sum of all those edges, we get a 1-chain orthogonal to all cycles.   It's known that the cut space is spanned by 1-chains coming from cuts in this way.   For example, in the graph I drew, the edge with the arrow on it spans the cut space.  
A proof can be found here:

Norman Biggs, Algebraic Graph Theory.

but I suspect there's a lot more known about this subject!
[Note: I have edited my original question to simplify the hypotheses, and also called elements of $Z_1(\Gamma,\mathbb{R})$ 1-cycles, to distinguish them from cycles in the sense of graph theory.]

REPLY [5 votes]: This answer is an attempt to slightly rephrase Ilya's argument; I would have written it as a comment, but there's not enough room.
Given a path $\gamma$ with endpoints $v_0$ and $v_n$, we have a 1-chain $c(\gamma)$, which is an integer-linear combination of edges of the graph.  Let $E_{c(\gamma)}$ be the set of edges with non-zero coefficients in $c(\gamma)$; this need not include every edge that appears in the path, since some might cancel to zero in the 1-chain.  If we think of these edges as decorated with the integer coefficients they inherit from the 1-chain $c(\gamma)$, we can associate a 1-chain with any subset of $E_{c(\gamma)}$.
It's possible that $E_{c(\gamma)}$ as a subgraph of $\Gamma$ contains several connected components that share no vertices with each other, but since the boundary of $c(\gamma)$ is only non-zero on $v_0$ and $v_n$, and it's impossible for a 1-chain to have a single-point boundary, these components would need to include exactly one whose 1-chain's boundary was non-zero on $v_0$ and $v_n$, and all the rest would have to give 1-cycles.  Since they're all disjoint, any one of the 1-cycles $\chi$ would satisfy $\langle c(\gamma), \chi \rangle = \langle \chi, \chi \rangle \gt 0$.
Assuming now that $E_{c(\gamma)}$ is connected as a subgraph, we can build up a 1-chain $\sigma$ with a positive inner product with $c(\gamma)$ as follows.  Starting at $v_0$, pick an edge $\epsilon_1$ incident on $v_0$, and put $\pm \epsilon_1$ in $\sigma$, with the sign chosen to be the same as the coefficient of $\epsilon_1$ in $c(\gamma)$.  Then advance to the other vertex of $\epsilon_1$, and choose an edge $\epsilon_2$ such that $\epsilon_2 \ne \epsilon_1$, and $\pm \epsilon_2$ with the same sign as the coefficient of $\epsilon_2$ in $c(\gamma)$ gives a boundary for $\pm \epsilon_1 \pm \epsilon_2$ that is zero at the current vertex.  This must be possible (assuming we haven't ended up at an endpoint of the path), since the boundary of $c(\gamma)$ is zero at the current vertex, so the signs of the edge coefficients in $c(\gamma)$ can't all give boundaries of the same sign here.
We continue this process until we reach either $v_n$, the endpoint of the path, or a vertex we've visited before.  If we reach a vertex we've visited before, we can drop any earlier edges from $\sigma$ and obtain a 1-cycle with a positive inner product with $c(\gamma)$.
If we reach $v_n$, there are two possibilities.  If $\sigma = c(\gamma)$ then $\sigma$ describes a simple path $\gamma'$ from $v_0$ to $v_n$ with the same 1-chain as our original path, and we can proceed to use that simple path in place of $\gamma$.  If $\sigma \ne c(\gamma)$, it nonethless has the same boundary.  Since $\sigma$ is supported on a subset of the same edges as $c(\gamma)$, and its coefficients are of the same sign but never greater in magnitude (and must be less on at least one edge), $\langle c(\gamma), \sigma \rangle$ will be strictly less than $\langle c(\gamma), c(\gamma) \rangle$.  So we'll have a non-trivial 1-cycle $\ell = c(\gamma) - \sigma$, and:
$$\langle c(\gamma), \ell \rangle  =  \langle c(\gamma), c(\gamma) \rangle - \langle c(\gamma), \sigma \rangle \gt 0$$
Finally, suppose we have a simple path $\gamma': v_0 \to v_1 \to \dots \to v_n$ with the same 1-chain as $\gamma$.  (The following refinement of Ilya's argument is something that John described to me in correspondence.)  Since the edge $e_n: v_{n-1} \to v_n$ cannot be a bridge, there must be a path joining $v_n$ to $v_0$ that does not include that $e_n$.  If we follow that path only as far as the first of the $v_i$ it reaches, we will have a path $\rho: v_n \to v_i$ which uses no edges of the simple path $\gamma'$.  We can then append the portion of $\gamma'$ that goes from $v_i$ to $v_n$, call it $\gamma'_i$, to obtain a 1-cycle $c(\rho) + c(\gamma'_i)$ that must have a positive inner product with $c(\gamma)$:
$$\langle c(\gamma), c(\rho) + c(\gamma'_i) \rangle = \langle c(\gamma'_i), c(\gamma'_i) \rangle \gt 0$$<|endoftext|>
TITLE: Groebner bases for differential operators with field coefficients (reference request)
QUESTION [5 upvotes]: Let $K$ be a field, $\partial_i$ be commuting derivations on $K$, and consider the ring $R=K[\partial_1\ldots \partial_n]$ (it is implicitly assumed that the derivations do not commute with elements of $K$ and the commutation relation is the obvious $\partial_i k - k\partial_i=\partial_i(k)$).
For example, $R$ can be the ring of differential operators with rational function coefficients, etc.
It appears that much of the theory of Groebner bases for polynomial rings $S=K[x_1\ldots x_n]$ can be trivially restated for $R=K[\partial_1\ldots \partial_n]$, since when we work with a monomial order, we usually track only what happens to the leading monomial, and it is the same for $R$ and $S$. I am pretty sure that this is also true for modules over $R$. (If I am wrong and this is not true, an explanation of this will be a perfect answer to this question:) )
Is there a reference which restates the standard results over $S$ for modules over $R$?
Most of the references I have found deal with the more(?) complicated case of polynomial coefficients. Other than that, I am aware of two references, both by Nobuki Takayama,
"Gröbner basis and the problem of contiguous relations" and a chapter in "Gröbner Bases: Statistics and Software Systems". The first basically introduces a version of Buchberger's algorithm for modules over $R$ and the second briefly goes over the very basic facts for ideals in $R$.

REPLY [2 votes]: Elaborating on Michael's comments and your concerns about Saito, Sturmfels, Takayama, "Grobner Deformations of Hypergeometric Differential Equations", the case this book addresses is the case of the Weyl algebra over a characteristic 0 field $k$. I.e., the noncommutative polynomial ring $$D=k[x_1,\ldots, x_n, \partial_1,\ldots, \partial_n]$$, subject to the commutator relations $[\partial_i, x_j]=\delta_{ij}$. There's also a few points at which they address the larger ring $$k(x_1,\ldots, x_n)\otimes_{k[x_1,\ldots, x_n]} D.$$ Is this what you're looking for?
As for online things, some of the above is also covered in Oaku and Takayama's "Algorithms for D-modules —restriction, tensor product, localization, and local cohomology groups".<|endoftext|>
TITLE: Finding Toeplitz matrix nearest to a given matrix
QUESTION [8 upvotes]: For an arbitrary $N\times N$ Hermitian matrix $A$, I want to derive a Toeplitz matrix from $A$ such that the eigenvectors of both matrices have minimal change. 
Specifically I want find the Toeplitz matrix such that the $L^2$ norm between the eigenvectors of the Toeplitz matrix and eigenvectors of the matrix $A$ is minimal. Is there any alternative method other than searching numerically for the matrix? What is the computational cost of such such search?
I am aware of some work done related to perturbations of Toeplitz matrices, in addition eigenvectors of banded toeplitz matrix is studied, but the matrix I want in my application is not banded. I would appreciate any suggestion.
Edit: Is the problem tractable/solvable/realistic if we are given a sequence of matrices $A^n$ instead of $A$?

REPLY [7 votes]: The set of $n \times n$ symmetric Toeplitz matrices is
$$\left\{ x_1 \mathrm M_1 + x_2 \mathrm M_2 + \cdots + x_n \mathrm M_n  \mid x_1, x_2, \dots, x_n \in \mathbb R \right\}$$
where $\mathrm M_1, \mathrm M_2, \dots, \mathrm M_n$ are $n \times n$ symmetric Toeplitz basis matrices. Let $\mathrm M_1 = \mathrm I_n$ correspond to the main diagonal, whereas the remaining basis matrices correspond to super and sub diagonals.
Let $\mathrm M : \mathbb R^n \to \mbox{Sym}_n (\mathbb R)$ be defined by
$$\mathrm M (\mathrm x) := x_1 \mathrm M_1 + x_2 \mathrm M_2 + \cdots + x_n \mathrm M_n$$

Spectral norm
To complement Suvrit's comment, using the spectral norm, we obtain the following unconstrained optimization problem in $\mathrm x \in \mathbb R^n$
$$\begin{array}{ll} \text{minimize} & \| \mathrm M (\mathrm x) - \mathrm A \|_2\end{array}$$
which can be rewritten as the following semidefinite program (SDP) in $\mathrm x \in \mathbb R^n$ and $t \geq 0$
$$\boxed{\begin{array}{ll} \text{minimize} & t\\ \text{subject to} & - t \,\mathrm I_n \preceq - \mathrm A + x_1 \mathrm M_1 + x_2 \mathrm M_2 + \cdots + x_n \mathrm M_n \preceq t \,\mathrm I_n\end{array}}$$
whose solution can be found numerically.

Frobenius norm
To complement Federico's comment, using the squared Frobenius norm, we obtain the following unconstrained quadratic program (QP) in $\mathrm x \in \mathbb R^n$
$$\begin{array}{ll} \text{minimize} & \| \mathrm M (\mathrm x) - \mathrm A \|_{\text{F}}^2\end{array}$$
where the objective function is
$$\begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix}^\top \begin{bmatrix} \langle \mathrm M_1, \mathrm M_1 \rangle & \langle \mathrm M_1, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_1, \mathrm M_n \rangle\\ \langle \mathrm M_2, \mathrm M_1 \rangle & \langle \mathrm M_2, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_2, \mathrm M_n \rangle\\ \vdots & \vdots & \ddots & \vdots\\ \langle \mathrm M_n, \mathrm M_1 \rangle & \langle \mathrm M_n, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_n, \mathrm M_n \rangle\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix} - 2 \,\begin{bmatrix} \langle \mathrm A, \mathrm M_1 \rangle\\ \langle \mathrm A, \mathrm M_2 \rangle\\ \vdots \\ \langle \mathrm A, \mathrm M_n \rangle\end{bmatrix}^\top \begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix} + \| \mathrm A \|_{\text{F}}^2$$
where $\langle \mathrm M_i, \mathrm M_j \rangle$ denotes the Frobenius inner product of (symmetric) basis matrices $\mathrm M_i$ and $\mathrm M_j$.
Computing the gradient of the objective function and finding where it does vanish, we obtain the following linear system
$$\begin{bmatrix} \langle \mathrm M_1, \mathrm M_1 \rangle & \langle \mathrm M_1, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_1, \mathrm M_n \rangle\\ \langle \mathrm M_2, \mathrm M_1 \rangle & \langle \mathrm M_2, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_2, \mathrm M_n \rangle\\ \vdots & \vdots & \ddots & \vdots\\ \langle \mathrm M_n, \mathrm M_1 \rangle & \langle \mathrm M_n, \mathrm M_2 \rangle & \cdots & \langle \mathrm M_n, \mathrm M_n \rangle\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix} = \,\begin{bmatrix} \langle \mathrm A, \mathrm M_1 \rangle\\ \langle \mathrm A, \mathrm M_2 \rangle\\ \vdots \\ \langle \mathrm A, \mathrm M_n \rangle\end{bmatrix}$$
Fortunately, the basis matrices are orthogonal and, thus, the matrix above is diagonal. Hence,
$$\begin{bmatrix} n & 0 & \cdots & 0\\ 0 & 2(n-1) & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & 2 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots \\ x_n\end{bmatrix} = \begin{bmatrix} \sum_{i=1}^n a_{i,i}\\ 2 \sum_{i=1}^{n-1} a_{i,i+1}\\ \vdots \\ 2 a_{n,n}\end{bmatrix}$$
and, thus, the solutions $x_1, x_2, \dots, x_n$ are the arithmetic means of the $n$ (distinct) diagonals of $\rm A$
$$\begin{array}{rl} x_1 &= \dfrac{\langle \mathrm A, \mathrm M_1 \rangle}{\langle \mathrm M_1, \mathrm M_1 \rangle} = \dfrac 1n \displaystyle\sum_{i=1}^n a_{i,i}\\ x_2 &= \dfrac{\langle \mathrm A, \mathrm M_2 \rangle}{\langle \mathrm M_2, \mathrm M_2 \rangle} = \dfrac 1{n-1} \displaystyle\sum_{i=1}^{n-1} a_{i,i+1}\\ x_3 &= \dfrac{\langle \mathrm A, \mathrm M_3 \rangle}{\langle \mathrm M_3, \mathrm M_3 \rangle} = \dfrac 1{n-2} \displaystyle\sum_{i=1}^{n-2} a_{i,i+2}\\ & \qquad\qquad\quad\vdots\\ x_n &= \dfrac{\langle \mathrm A, \mathrm M_n \rangle}{\langle \mathrm M_n, \mathrm M_n \rangle} = \displaystyle\sum_{i=1}^1 a_{i,i+n-1} = a_{1,n}\end{array}$$
Lastly, the Toeplitz matrix nearest to the given symmetric matrix $\rm A$ is
$$\boxed{\hat{\mathrm X} := \left( \dfrac 1n \displaystyle\sum_{i=1}^n a_{i,i} \right) \mathrm I_n + \left( \dfrac 1{n-1} \displaystyle\sum_{i=1}^{n-1} a_{i,i+1} \right) \mathrm M_2 + \left( \dfrac 1{n-2} \displaystyle\sum_{i=1}^{n-2} a_{i,i+2} \right) \mathrm M_3 + \cdots + a_{1,n} \mathrm M_n}$$<|endoftext|>
TITLE: Cohomology theory "from" Grothendieck's six operations?
QUESTION [10 upvotes]: How, precisely, (as suggested by grothendieck) cohomology theory naturally follows from the Grothendieck's six operations associated to the category derived from the topos?
I would like some references.

REPLY [10 votes]: Assuming you have the usual 6 operations we have the following recipes for co/homologies of $f \colon X \to Y$ with coefficients $\mathcal{F}$ over $X$, $\mathcal{G}$ over $Y$ and $\mathcal{O}$ denoting the monoidal unit.
Cohomology:........................................... $\mathrm{Hom}_X(\mathcal{O}_X,\mathcal{F})$
Homology:.............................................. $\mathrm{Hom}_X(\mathcal{O}_X,f^!\mathcal{G})$
Cohomology with compact support:....... $\mathrm{Hom}_Y(\mathcal{O}_Y,f_!\mathcal{F})$
Borel-Moore homology:......................... $\mathrm{Hom}_Y(\mathcal{O}_Y,f_!f^!\mathcal{G})$
The defining relations between the 6 functors yield the usual dualities and comparison maps between the four theories. Notice that when f is proper $f_! = f_*$ and the latter pair becomes the first.<|endoftext|>
TITLE: Socle of tilting modules in the BGG category $\mathcal{O}$ over a semisimple Lie algebra
QUESTION [6 upvotes]: Suppose that $\mathfrak{g}$ is a finite dimensional, complex, semisimple Lie algebra. Let $\mathcal{O}$ be the BGG category over $\mathfrak{g}$.
Tilting module theory play an important role in the study $\mathcal{O}$. They are sometimes projective covers and hence injective since the duality functor preserve tilting modules. Is it true that every tilting module has simple socle?
Thanks!

REPLY [8 votes]: EDIT: Following a conversation with Ivan Losev, the situation is clearer now. Consider the principal block of $\mathcal{O}$. Recall two facts:
1) the socle of any Verma module is $L_{w_0}$,
2) taking the socle is a left exact functor.
Thus the socle of any object with Verma flag (in particular a tilting module) is isomorphic to a direct sum of copies of $L_{w_0}$.
Now if $T_x$ is an indecomposable tilting module we have
$\dim Hom(L_{w_0}, T_x) = P_{id, xw_0}(1)$
where $P_{y,z}$ is a Kazhdan-Lusztig polynomial. Thus the socle is simple if and only if $P_{id, xw_0} = 1$ which is the case if and only if the Schubert variety $xw_0$ is rationally smooth.
The formula is a consequence of the more general formula (which Peter reminded me of):
$dim Hom(\Delta_x, T_y) = P_{w_0x,w_0y}(1)$
See e.g. "Tilting exercises" or Soergel's papers on tilting modules.
(I deleted the longer version of the answer, because this seems much cleaner than earlier attempts.)<|endoftext|>
TITLE: Union-closed family generated by n 2-sets
QUESTION [6 upvotes]: I asked this question on Stackexchange, but I got no answer, so I ask it here.
Let us define a $2$-set as a set with exactly $2$ elements. For a natural number $n$, let $l(n)$ denote the least possible number of members of a union-closed family of sets generated by $n$ distinct $2$-sets. I'm interested in a useful formula or minoration for $l(n)$.
There are easy majorations, for example $l({r\choose 2}) \leq 2^{r} - 1 - r$ and (for $r \geq 1$) $l({r\choose 2}+1) \leq 2^{r} + 2^{r-1} - 1 - r$, and these upper bounds seem to be exact values for small values of $r$, but I would avoid the task of handling this question if there is literature about it. Do you know ? Thanks in advance.
Note : there is a similar question here :
Kruskal-Katona type question for union-closed families of sets
but not identical.

REPLY [4 votes]: That problem was solved by Uwe Leck, Ian T. Roberts and Jamie Simpson in the paper "Minimizing the weight of the union-closure of families of two-sets.", Australasian Journal of Combinatorics 52 (2012): 67-73.<|endoftext|>
TITLE: $p$-adic Bott periodicity?
QUESTION [18 upvotes]: The Bott periodicity theorem can be formulated as the existence of homotopy equivalences $\Omega^2(KU)\equiv KU$ and $\Omega^8(KO)=KO$. I always wondered whether this theorem could also be transferred to the other completions of $Q$, i. e. the $p$-adic fields. Of course, since the topology of $\mathbb Q^p$ behaves rather incompatibly with the topology of, say, $CW$ complexes, it's not really clear how to make sense of that. (The classifying space of the topological group should have the same homotopy groups as the classifying space constructed from the underlying discrete group, if I'm not mistaken). Therefore a more sophisticated approach (possibly via $p$-adic rigid geometry) might be necessary. So, does there exist anything which could be considered as the $p$-adic analogue of Bott periodicity?
As I understood it, the underlying algebraic reason for the lengths of the period in the real and complex case, respectively) is that the sequence of Clifford algebras $Cl(k,n)$ exhibits the same periodicity behaviour up to Morita equivalence (precisely, $Cl(k+8,\mathbb R )=M(Cl(k,\mathbb R),8)$ and $Cl(k+2,\mathbb R )=M(Cl(k,\mathbb R),2)$. A quick check in the literature shows that the clifford algebras over $\mathbb Q_p$ are also periodic of length $2$ if $p\equiv 3 \mod 4$, $4$ if $p \equiv 1 \mod 4$ and $8$ for $p=2$, so at least at this algebraic level, Bott periodicity is present.

REPLY [18 votes]: The $p$-completed algebraic $K$-theory of the algebraic closure of $\mathbb{Q}_p$, i.e., $K(\bar{\mathbb{Q}}_p; \mathbb{Z}_p)$, is equivalent to its second loop space, up to an issue about path components. This is due to Suslin. The descent to $\mathbb{Q}_p$ is more subtle than the descent from $\mathbb{C}$ to $\mathbb{R}$, because the absolute Galois group of $\mathbb{Q}_p$ is much more complicated than that of $\mathbb{R}$. Still, if you reduce to homotopy with $\mathbb{Z}/p$ coefficients, $K(\mathbb{Q}_p; \mathbb{Z}/p)$ is equivalent to its $(2p-2)$-fold loop space, up to the same issue as before. Boekstedt and Madsen proved this using topological cyclic homology. I did the case $p=2$. Later it followed from the proof of the Lichtenbaum-Quillen conjectures by Voevodsky and others.<|endoftext|>
TITLE: Separating Heavier from the Lighter Balls
QUESTION [5 upvotes]: This Question was originally posted Here, where I'm more interested in the methods for manual solutions yielding $n$ or less moves on average.
I wanted to post it here as well, to see what the people of mathoverflow think about it. 

I think we are familiar with the classic problem where we need to find one heavier ball among the rest identical lighter $n$ amount of balls using a scale and the minimum number of weightings. 
But I'm interested in a variation of this problem.

You have an even number of balls, $2n$ identical balls.   Half of
  them, $n$ amount of balls, are "Heavy Balls" and the other half are
  "Light Balls".
Find a method to separate the balls into the "Heavy" and the "Light"
  box with the least weightings as possible; Using a scale instrument,
  which from you can read exact difference between the total weight of
  the right and the left side of the scale.
What is the minimum number of weightings required if we are given $2n$
  balls?
What is the optimal method we can use for any case of $n$ to separate
  the balls with the least weightings as possible?

For my progress on the specific cases of $n$ so far, check the original question linked Here.

REPLY [7 votes]: The answer is indeed asymptotic to $4n/\log(n)$, but I don't know of an elementary or easy construction for this upper bound. I believe simple probabilistic-method type constructions do not work.
This is also phrased as a "coin-weighing" problem. One point to keep in mind is that we can think of each weighing of $k$ balls as simply learning how many of the $k$ are light versus heavy. Thus you can rephrase your problem as follows: There is a special unknown subset $S$ of $\{1,\dots,2n\}$, where $|S| = n$, and we wish to determine $S$. At each step, we can query any $T \subseteq \{1,\dots,2n\}$ and learn $|S \cap T|$.
I find these references in the tutorial/lecture by Galvin[1], where he attributes the upper bound to (independently) [2,3] and the lower bound to (independently) [4,5]. Note that in his formulation there are only $n$ coins rather than $2n$ (hence the bound is $2n/\log n$), and it is not promised how many are heavy or light (so in particular the  upper bound holds for the case where half are promised to be heavy).
This is also addressed in this MO question[6], but I'm not sure where a factor $2$ has gone missing in that answer.

[1] http://arxiv.org/abs/1406.7872
[2]  D. Cantor and W. Mills, Determination of a subset from certain combinatorial properties, Can. J. Math. 18 (1966), 42–48
[3] B. Lindstr¨om, On a combinatorial problem in number theory, Can. Math. Bull. 8 (1965), 477-490.
[4] P. Erd˝os and A. R´enyi, On two problems of information theory, Publ. Hung. Acad. Sci. 8 (1963), 241–254.
[5]  L. Moser, The second moment method in combinatorial analysis, in “Combinatorial Structures and Their Applications”, 283–384, Gordon and Breach, New York, 1970.
[6] Guessing a subset of {1,...,N}<|endoftext|>
TITLE: What is the motivation behind Ramanujan's conjecture?
QUESTION [10 upvotes]: One motivation I have seen given for Ramanujan's conjecture for the estimate 
$$ |a_p|< C p^{k - \frac{1}{2}} $$
for the Fourier coefficients of a cusp form of weight $2k$ is that it allows one to show that the error term in some formulas for the number of representation of a number by a certain quadratic form (which comes from a cusp form) is dominated by the main term. However, at least in applications in Serre's "a course in arithmetic", it seems that even the Hecke estimate suffices. I know that this is not the case for some more recent applications (say, construction of Ramanujan graphs, `a la Lubotzky-Philip-Sarnak), but these could have hardly been Ramanujan's motivation.
I would like to know what was Ramanujan's original motivation for making this conjecture, and also a hint as to how he may have come up with the $1/2$ power saving term. This question came up in a course on analytic number theory I taught this semester. ANT is not my field of research, so I'd not be surprised if the answer to this question turns out to be well-known or even trivial. I will still appreciate it if an expert could respond or provide a reference.

REPLY [6 votes]: Ramanujan made his conjecture before Hecke gave his bound. As Lucia noted,
the conjecture was based on empirical computation concerning the coefficients $\tau(n)$ of the $\Delta$ modular form, so Ramanujan would have had no reason to propose a conjecture less precise than what was apparently true.
In other words, no one came up with the 1/2 power saving term. Ramanujan
found directly the best possible bound, and the first one who proved
something approaching, Hecke, was off by 1/2.<|endoftext|>
TITLE: First Chern class vanishes on a Lagrangian submanifold
QUESTION [8 upvotes]: Suppose $(M,\omega)$ is a symplectic manifold and $L \subset M$ is a compact Lagrangian submanifold. Is it the case that the first Chern class $c_1(TM) \in H^2(M)$ vanishes when restricted to the Lagrangian submanifold $L$?

REPLY [9 votes]: Yes, if you ignore 2-torsion:  After all, if you give $TM$ a complex structure compatible with $\omega$, say $J$, then you'll have that $TM$ pulls back to $L$ to become isomorphic to $TL\oplus T^*L\simeq TL\oplus J(TL)$, so, as a complex bundle, $TM$ pulls back to be the complexification of $TL$, and hence $$L^*\bigl(c_1(TM)\bigr) = c_1(TL^\mathbb{C}) = -c_1(TL^\mathbb{C}),$$ since $TL^\mathbb{C}$ is isomorphic to its complex conjugate bundle.  Thus
$$
2\,L^*\bigl(c_1(TM)\bigr) =0.
$$
(Note that I use $L^*$ to denote the pullback (aka restriction) map $H^*(M)\to H^*(L)$ in cohomology.)<|endoftext|>
TITLE: Can we cover a set by a particular family of sets?
QUESTION [5 upvotes]: Let $A_1,A_2,\ldots,A_m,B_1,B_2,\ldots, B_m$ be (not  necessarily distinct) subsets of $[n]=\{1,2,\ldots,n\}$. Suppose that each $i\in [n]$ appears in at least  $k$ of these $2m$ sets. 
I want to find a family of $m$ sets $\{C_i\}_{1\le i\le m}$ to cover $[n]$ (that is $C_1\cup\cdots\cup C_m=[n]$), where $C_i=A_i$ or $C_i=B_i$ for each $1\le i\le m$.
How large $k$ (as a function of $n$) is sufficent for the existence of such a family $\{C_i\}_{1\le i\le m}$. I wonder whether $k>\log_2(n)$ is sufficient.
Is it an 'old' problem? Are there any references?
Thanks in advance for any suggestions.

REPLY [14 votes]: $2^k>n$ is sufficuent. Choose $C_i=A_i$ or $C_i=B_i$ at random, with equal probabilities. The probability that we do not cover given element $s$ is at most $2^{-k}$, thus the expectation of the number of not covered elements is at most $n/2^{k}$, strictly less than 1. Therefore sometimes all elements are covered.
But if $n=2^{k}$, $m=k$, it is not always possible. Consider the vertices of a cube $\{0,1\}^k$ and all $2k$ facets ($A_i$, $B_i$ are parallel facets).<|endoftext|>
TITLE: When are growth series rational?
QUESTION [13 upvotes]: For a finitely generated group $G$ with generating set $S$, one can define the word metric. Using this, we are able to define the sequence $\sigma(n)$ by 
 $ \sigma(n) = |\mathcal{S}(e,n)| $, which is the number of elements that are contained in the sphere at a distance $n$ of the identity element (ie. All elements in $G$ that are at a distance precisely $n$ of the identity). Consequently  one can define the growth series $\mathcal{S}(n) = \sum_{n\ge 0} \sigma(n) z^n$. One can then ask if this formal power series is rational (ie. it is the power series associated to a rational function). 
I have two questions concerning this:

Why is it interesting to ask this question? (ie. Why is the fact that the formal power series is rational interesting?)
If one would take two different generating sets $S$ and $S'$, is it possible that the the growth series coming from $S$ is rational and the one coming from $S'$ is not? 

Thank you for your help.

REPLY [6 votes]: I don't know how interesting this is to you, but knowing that $S(z)$ is a rational function is a quick way to show that the exponential growth rate $\lim_{n \rightarrow \infty} \sigma(n)^{1/n}$ of the group with respect to $S$  is an integer algebraic number.  So, for example, the only exponential growth rate you can get out of a hyperbolic group is an algebraic number.  
I'll note that there do exist exponential growth rates that are not algebraic numbers--in fact, there are uncountably many possible exponential growth rates, as shown by Anna Erschler in
A. Erschler. Growth rates of small cancellation groups. In Random walks and geometry, pages 421–430. Walter de Gruyter GmbH & Co. KG, Berlin, 2004.<|endoftext|>
TITLE: Bialgebraic structure of Sklyanin algebra
QUESTION [6 upvotes]: Does Sklyanin algebra (which is an elliptic extension of the quantum group) admit a bialgebra structure or even Hopf algebraic structure? Or is it proved that it is impossible to have such a structure? 
Note that the algebra is defined as four generators $S_0, S_{\alpha = 1,2,3}$ and
\begin{equation}
\begin{aligned}
& \{ S_0, S_{\alpha} \} = 2 J_{\beta \gamma} S_{\beta} S_{\gamma}\\[0.5em]
& \{ S_{\alpha}, S_{\beta} \} = -2 S_0 S_{\gamma}
\end{aligned}
\end{equation}
where $(\alpha, \beta, \gamma)$ is the cyclic permutation of $1,2,3$.
Moreover, if one so far cannot find such bialgebraic structure, is it still possible to define tensor product representation of the algebra?

REPLY [6 votes]: In response to your first question, there is no Hopf structure on any Sklyanin algebra of any dimension. See Corollary 2.8 (i) of the following paper:
https://arxiv.org/pdf/1601.06687v1.pdf<|endoftext|>
TITLE: Why is $\operatorname{nr}_{F[G]}:K_1(F[G])\to Z(F[G])^\times$ a bijection?
QUESTION [6 upvotes]: Let $A$ be a finite dimensional semisimple $F$-algebra and $K_1(A)$ the Whitehead group of $A$.
By splitting $A$ into its Wedderburn components, the reduced norm map $\operatorname{nr}_A:K_1(A)\to Z(A)^\times$ can be defined, where $Z(A)$ is the centre of $A$.
I've read in the literature that if $G$ is a finite group of odd order then the norm map $\operatorname{nr}_{F[G]}$ is bijective.
Is there a simple proof of this?

REPLY [5 votes]: I am assuming that you meant $F$ to be a number field.
Let $A = M_n(D)$ where $D$ is a finite dimensional division algebra over $\mathbb{Q}$ and let $Z = Z(A) = Z(D)$. Then $K_1(A)\cong D^\times/[D^\times,D^\times] = D^\times/SL_1(D)$. By Eichler's theorem, if $A$ is not ramified at any real place of $Z$, then the reduced norm map $\mathrm{nr} : D^\times \to Z^\times$ is surjective. Now if $G$ is a finite group of odd order, then the degree of the Wedderburn components of $F[G]$ are odd (because they always divide the order of $G$), so they cannot be ramified at real places. This proves the result.<|endoftext|>
TITLE: Archimedean fibers "intersecting" curves on arithmetic surfaces
QUESTION [7 upvotes]: Let's fix a number field $K$ with its ring of integers $O_K$. Moreover consider an arithmetic surface $f:S\to \text{Spec } O_K$. For every archimedean place $\sigma$ in $K$, $K_\sigma$ is the completion of $K$ with respect to $\sigma$ (note that $K_\sigma$ is $\mathbb R$ or $\mathbb C$).
We can imagine $\sigma$ as "a point at infinity" of  $\text{Spec } O_K$, and therefore we define the archimedean fiber over $\sigma$ as:
$$S_\sigma:= S\times_{O_K}\text{Spec }K_\sigma$$
I'd like to understand how $S_\sigma$ interacts with $S$ and the (integral) curves on $S$:

Is the canonical morphism $S_\sigma\to S$ a closed embedding? I mean: what can we really recover of $S_\sigma$ on $S$? I would be surprised if $S_\sigma$ could be embedded in $S$.
Let $C\subset S$ be an integral vertical curve on $S$, then my intuition says that "the intersection $S_\sigma\cap C$" should be empty. Is it true? 
On the contrary if $C$ is horizontal, "the intersection  $S_\sigma\cap C$" should be non empty. Is it true?

Edit: Here by $S_\sigma\cap C$ I mean the set of points of $S_\sigma$ mapped in $C$. In other words, they are the points at infinity of $C$.

REPLY [5 votes]: It seems that you are beginning to study Arakelov geometry. In this theory, the archimedean fibers have to be “thought of” fibers, but they can't be in the strict sense of scheme theory. 
In fact, in Arakelov theory, the vertical fiber is never treated as a closed subscheme. Rather, the usual cycles on the arithmetic scheme are enhanced with an archimedean structure (Green currents) which is used to define their intersection as an enhanced cycle. 
In Arakelov's theory for surfaces, these Green currents are canonically chosen; however, the general theory of Gillet-Soulé relaxes this assumption for it is useful to be able to pick Green currents according to one's choice (or to what geometry dictates).<|endoftext|>
TITLE: Elementary reference for the isometry group of $\mathbb{RP}^2$
QUESTION [6 upvotes]: Endow the real projective plane with the distance defined by $d(L,L')$ := "the angle between the lines $L$ and $L'$ ".
It is the case that every isometry from $RP^2$ onto $RP^2$ is induced by an element of the orthogonal group $O(3)$.
I want to use this fact in a research paper. But it turns out that all my considerations are so elementary that, apart from this fact, this paper is readable by a 3rd year student at the University. I would like to have a reference for this result that would be readable by such a student (without resorting to Riemannian structures).
But I have no reference at all for this result actually, so a reference with more fancy technology would already be valuable.
Thank you very much.

REPLY [2 votes]: Robert Bryant's post provides a neat proof: it is a good reference for the result I am interested in. I found another (more conventional: published book vs MO answer) elementary reference for it:  9.7.1 and 19.1.2.2 in Berger's Geometry. This proof is less trickily efficient than that of Robert Bryant, but it is also interesting (geometric insight). I have decided to cite both.<|endoftext|>
TITLE: Perturbations on the pseudoinverse of a matrix
QUESTION [5 upvotes]: Given a matrix $A \in \mathbb{R}^{n\times m}$, and its perturbation 
$$
A_p = A + \Delta
$$
is there a way to represent 
$$
(A_p)^{\star}= (A)^{\star} + f(\Delta)
$$
where $(A_p)^{\star}$ ($(A)^{\star}$) is the pseudo-inverse of $A_p$ ($A$)?
What can be said about the spectral norm of $f(\Delta)$?

REPLY [4 votes]: substituting $A_p=A+\Delta$ into the definition $A_p^\star=\lim_{\epsilon\rightarrow 0}(A_p^\ast A_p+\epsilon I)^{-1}A_p^\ast$, and linearizing in the perturbation $\Delta$, gives $A_p^\star=A^\star+f(\Delta)$ with
$$f(\Delta)= -A^\star\Delta A^\star+
\lim_{\epsilon\rightarrow 0}(A^\ast A+\epsilon I)^{-1}\Delta^*P+{\rm order}(\Delta^2),$$
with $P=I-AA^\star$ the orthogonal projector onto the range of $A$. If $\Delta^\ast P=0$ the simple result
$$A_p^\star=A^\star-A^\star\Delta A^\star+{\rm order}(\Delta^2)$$
is obtained.<|endoftext|>
TITLE: Known results in the Cohomology of finite groups
QUESTION [9 upvotes]: I am learning to compute cohomology of finite groups and came across this survey article http://www.ams.org/notices/199707/adem.pdf "Recent Developments in the cohomology of finite groups" by Alejandro Adem and various other articles by him on calculations of cohomology of finite groups. There is also a book by Adem and Milgram in which they compute cohomology of many finite groups. I would like to know 
To what extent the cohomology of finite groups is known. Is it for example known for every finite simple group (in the classification). Any reference on the list of finite groups with known (unknown) cohomology will be nice. 
I do apologize if the question is too simple to ask for this forum.

REPLY [9 votes]: Two references that naturally come to mind: 

D.J. Benson and S.D. Smith. Classifying spaces of sporadic groups. AMS Mathematical Surveys and Monographs vol. 147  

Describes homotopy colimit decompositions of the $2$-completed classifying spaces of the sporadic groups (based on subgroup complexes from the "$2$-local geometry"). These are relevant for the computation of mod $2$ cohomology. The book also contains remarks on the computations of cup-product structure.

J.F. Carlson, L. Townsley, L. Valeri-Elizondo, M. Zhang. Cohomology rings of finite groups. With an appendix: Calculations of cohomology rings of groups of order dividing 64.  Algebras and Applications, Volume 3, 2003, Springer.

Describes some of the more advanced methods for computing cohomology rings of finite groups. Looking at the appendix also shows that besides the complications that one would naturally expect with the finite simple groups, computing the cohomology rings of $p$-groups can also be tough business.<|endoftext|>
TITLE: Properties of connection Laplacian on vector fields
QUESTION [8 upvotes]: Let $(M,g)$ be a simply-connected compact surface with boundary $\partial M$ and metric $g$. Let $N$ denote the outward unit normal on $\partial M$, $\nabla$ the Levi-Civita connection and $\Delta_g$ the corresponding connection Laplacian (i.e. the trace of the Hessian).
I am interested in the properties of the vector field $X$ that minimises
$\int_M \langle \nabla X, \nabla X\rangle dV$ subject to the boundary condition $\nabla_N X=0$, where $\langle,\rangle$ denotes the extension of g to arbitrary tensor fields. Because of these boundary conditions, $\int_M \langle \nabla X, \nabla X\rangle dV=-\int_M \langle \Delta_g X, X\rangle dV$ and we just need the lowest eigenvalue of the Laplacian with these boundary conditions. 
I am not sure this boundary problem is well posed, but if so, this eigenvalue cannot be zero in general, since there are no parallel vector fields on a curved surface by the Gauss-Bonnet theorem. It is my hope that a general solution to this boundary problem exists, and that the lowest eigenvalue can be expressed in terms of the Gauss curvature of the surface. I would be very grateful for an answer to this question, and ideally a reference that deals with this problem in more detail, since all books I can find deal with the Hodge Laplacian.

REPLY [2 votes]: I don't think such a result is possible without any assumptions on the geometry of the boundary.
Consider the scalar case. If $M$ is a closed manifold with $\operatorname{Ric} \geq (n-1)k g$ for some positive constant $k$, then $\lambda_1(M) \geq nk$. The proof (which is due to Lichnerowicz and is summarized in Obata's paper Certain conditions for a Riemannian manifold to be isometric with a sphere) uses integration by parts, so one has to be careful when $\partial M$ is nontrivial. This was studied by Reilly in Applications of the Hessian operator on a Riemannian manifold. He proved that the same lower bound for the first (Dirichlet) eigenvalue holds as long as $\partial M$ has nonpositive mean curvature.
The proof is just a computation, so it should be straightforward (though tedious) to see if a similar result holds for the connection Laplacian on vector fields.<|endoftext|>
TITLE: Number of polynomials whose Galois group is a subgroup of the alternating group
QUESTION [12 upvotes]: Let $f = x^n + a_{n-1}x^n + \cdots + a_0$ be a monic polynomial of degree $n \geq 2$ with integer coefficients. By $\text{Gal}(f)$ we mean the Galois group over $\mathbb{Q}$ of the Galois closure of $f$. Define $H(f) = \max\{|a_i|\}$ denote the naive or box height of $f$. Hilbert's irreducibility theorem asserts that for most integer $n$-tuples $(a_0, \cdots, a_{n-1})$ with $H(f) \leq B$ say, the corresponding polynomial $f$ is irreducible. In fact it is conjectured that for all but $O(B^{n-1})$ many such polynomials, the Galois group of $f$ is isomorphic to the symmetric group $S_n$. 
The best bounds to date, due to Rankin, assert that all but $O(B^{n-1/2})$ many monic polynomials with box height bounded by $B$ has Galois group isomorphic to $S_n$. David Zywina proved that one can sharpen this bound if one excludes those polynomials whose Galois group is the alternating group. Indeed, he showed that for every $\epsilon > 0$ there exists a number $N$ such that for all $n \geq N$, there are $O(B^{n-1+\epsilon})$ many monic polynomials of height at most $B$ which have Galois group distinct from $S_n$ and $A_n$. 
What about lower bounds? Is it obvious that the alternating group should be the second most common Galois group, at least for all sufficiently large degrees? It is apparently a conjecture of Narkiewicz (from Carl Pomerance's talk https://math.dartmouth.edu/~carlp/PDF/fieldstalk.pdf) that the number of fields of degree $n$ over $\mathbb{Q}$ whose Galois group contains a transposition will constitute a positive proportion of all number fields of degree $n$. The alternating group conspicuously does not have any transpositions and so under this conjecture, $0\%$ of fields have Galois group equal to the alternating group. This to me makes it not so obvious why one would expect more polynomials with alternating Galois group as opposed to some other isomorphism class.
Are there known lower bounds for the number of polynomials with Galois group isomorphic to a subgroup of $A_n$, either with respect to the box height or otherwise?

REPLY [17 votes]: Firstly, the conjecture whereof you speak (with an $\epsilon$ in the exponent) has been proved by yours truly (there is an arXiv.org preprint as of about six months ago). 
Secondly, the most common "exceptional" situation is when the polynomial is reducible. It is clear that at least $O(B^{n-1})$ polynomials are reducible, and this is the truth, asymptotically, for $n>2.$ 
Thirdly, the Galois group is a subgroup of $A_n$ if and only if the discriminant is a perfect square. The obvious heuristic is that the probability that the value of a polynomial of degree $d$ is a perfect square is something like $1/B^{d/2}$ The degree of the discriminant is $2(n-1),$ which would indicate that alternating group is pretty thin on the ground.
ADDED LATER Experimental data (for the probability that a monic irreducible polynomial of degree $n$ and coefficients bounded by $B$ in absolute value has discriminant a perfect square) is consistent with the heuristic above when $n>3$ - the results are not clear for $n=3,$ and the question is vacuous for $n=2$ (a polynomial whose discriminant is a square is reducible).<|endoftext|>
TITLE: Geometric construcion of Proj as a quotient by a $\mathbb{G}_m$ action
QUESTION [9 upvotes]: I'm trying to translate the Proj construction as a kind of quotient by a $\mathbb{G}_m$ action. Here's what I have so far:
Let $X=Spec\,A$ be an affine scheme (after this case is setteled I imagine it would be relatively easy to generalize to relatively affine schemes and get global Proj). Let $\mathbb{G}_m = Spec\,\mathbb Z[x,x^{-1}]$ be the multiplicative group.
The data of an action $\rho :X \times \mathbb{G_m} \to X$ is equivalently a homomorphism $\rho^{\flat}: A \to A[x,x^{-1}]$ which is equivalently a grading on $A$ given by $A=\bigoplus_{d \in \mathbb{Z}} A_d = \bigoplus_{d \in \mathbb{Z}} (\rho^{\flat})^{-1}(Ax^d)$.
This is very similar to a decomposition into irreducible subrepresentations only it doesn't look like a representation but rather a "co-representation".
How should I think of this "co-representation?" Is the theory of comodules over a hopf algebra just "dual" in some sense to the that of modules over a hopf algebra? (meaning everything is practically the same upto flipping arrows).
As far as I understand at this point we'd like to consider the free locus of the action on $X$. This is reasonable (coming from differential geometry where free actions by compact groups always exist) although I wonder if this technique works for any action.

Question 1 : Does the quotient of a scheme by a free action of a group scheme always exist? (as a scheme).

Interpreting the "free locus" of the action in terms of algebra is pretty confusing for me. Having knowledge of the solution these are supposed to be prime submodules of $A$ (prime ideals) which contain the irrelevant ideal $A_+$. The irrelevant ideal corresponds geometrically to the maximal irreducible subset of $X$ on which $\mathbb{G}_m$ acts trivially. So primes containing it correspond to irreducible sets contained in it which we should throw away as well. Is this true/enough?
Having thrown out the problematic points we get a free action on $X\text{ \ }X_p$ which we can now hopefully quotient by the action and get a $\mathbb{G}_m$ torsor. $SpecA \text{ \ } Spec{A_0} \to Proj A$. The fact that orbits of the action are in 1-1 with homogeneous prime ideals (not containing the irrelevant ideal) is a consequence of the fact that the orbit of every point in the free locus is itself a point (irreducible subscheme).
Having a torsor we have an equivalence of categories between $\mathbb{G}_m$-equivariant sheaves on $SpecA \text{ \ } Spec{A_0}$ and sheaves on $Proj A$

Question 2: Is this correct?
Question 3: What's a good reference for equivariant sheaves, quotients by group schemes and quotient stacks in the context of geometric representation theory?

REPLY [8 votes]: First of all, the ideal $I$ corresponding to the fixed point set is generated by all $A_d$ with $d\ne0$. Thus $I=\bigoplus_d I_d$ with $I_d=A_d$ for $d\ne0$ and $I_0=\sum_{d\ne0}A_dA_{-d}\subseteq A_0$.
Secondly, as already stated, the answer to Q1 in this generality is "no" but, under the given circumstances, the answer is actually "yes": If $\mathbb G_m$ (or any torus for that matter) acts locally freely (i.e., finite isotropy subgroups are ok) on a quasiaffine variety $X$ then the quotient exists as a scheme. This follows from the fact that this is true for affine varieties and $X$ can be covered by invariant affine open subsets.
BUT: The quotient will not be separated, in general. Take for example for $X$ the affine plane with $(tx,t^{-1}y)$-action and remove the origin. Then the action is free and the quotient is the affine line with a double point coming from the two components of $\{xy=0\}$. To get a separated quotient one has to throw out something, i.e., pass to an open subset. There are many choices and selecting one is the main point of GIT. There are even in general open subsets not coming from GIT. These yield quotients which are not quasi-projective. There are some papers of Białynicki-Birula and Swiecicka on this topic.<|endoftext|>
TITLE: Geometric or topological results from group theory
QUESTION [8 upvotes]: Do you know interesting examples of purely geometric or topological results which can be proved using group theory? To make precise what I have in mind, let us consider the two following examples:

There does not exist any Riemannian metric on the torus whose sectional curvature is $<0$.

This is a consequence of Milnor's article A note on curvature and fundamental group, where he proves that the fundamental group of a negatively-curved Riemannian manifold has exponential growth. On the other hand, the fundamental group of the torus, namely $\mathbb{Z}^2$, has quadratic growth.

Any compact Riemannian manifold whose sectional curvature is $\equiv 0$ has a torus as a finite cover.

This is a consequence of Bieberbach theorem. 
More recently, showing that quasiconvex subgroups of hyperbolic cubulable groups are separable was the key point in the proof of the virtual Haken's conjecture. However, this is more technical.

REPLY [3 votes]: I'm still a little uncertain about this question, but I'll try to say something about the Virtual Haken conjecture (discussed above) and in the process explain why I think it's a good example.
The Virtual Haken conjecture (now Agol's theorem) can be stated as follows --

Every hyperbolic 3-manifold has a finite-sheeted cover that contains an embedded, incompressible surface.

-- a thoroughly topological statement (though more group-theoretic statements can be given). 
For me, modern (geometric) group theory enters the picture in a truly astonishing way via the following result, Wise's Malnormal Special Quotient Theorem (MSQT).

Let $G$ be a hyperbolic, virtually special group, and $H$ a malnormal, quasiconvex subgroup.  Then, for all sufficiently deep finite-index subgroups $K\lhd H$, the quotient $G/\langle\langle K\rangle\rangle$ is hyperbolic and virtually special.

I won't explain the definitions here, but rather point out that this tells us that we can kill large subgroups and stay in the (very well behaved) world of hyperbolic, virtually special groups.  Note that this is not true of manifolds.  You can't crush an immersed surface in a hyperbolic manifold and get a new manifold.
Although the proofs of the MSQT can (and usually are) phrased in an entirely topological/geometric manner, the key point here is that they concern geometric complexes (more precisely, CAT(0) cube complexes), not manifolds.  The transition from manifolds to more general geometric complexes is surely the hallmark of modern infinite group theory.
Agol's proof, still the only proof we know, makes essential use of the MSQT.  In this sense, the Virtual Haken conjecture is truly a theorem of geometric group theory rather than topology, in the sense that we don't know how to keep the proof purely in the world of topology (ie manifolds) -- you have to pass to the world of CAT(0) cube complexes (ie group theory).<|endoftext|>
TITLE: What Turing degree is this function?
QUESTION [10 upvotes]: Over at http://www.scottaaronson.com/blog/?p=2725#comment-1089004 we had a discussion of intermediate Turing degrees.
The following function came up:

Take Chaitin’s constant, and rearrange its binary digits as follows: for each of the sets {1st digit} {2-3rd digits} {4-7th digits} {2^n-(2^n+1)-1}, order the digits within in ascending order, i.e. zeros then ones.

(This is a number, the function is just {n->nth digit of the number} for natural numbers n)
A later comment says it's non-computable:

because it has unbounded information about omega.
We know that K(Omega_n) >= n + O(1), but knowing how many 0s and 1s are in the second half of n/2 bits would allow you to save about (log n)/2 bits. This gives a contradiction for large enough n of the form n=2^i

It's clearly either of degree 0′ or lower. Which is it? In other words, does an oracle for this function let you solve the halting problem?

REPLY [6 votes]: I think it's strictly below $0'$. Namely let's call your number $\Gamma(\Omega)$ where $\Gamma$ is a Turing functional. Let $\Phi$ be any other Turing functional. Then show that the set
$$
S = \{X: X = \Phi(\Gamma(X))\}
$$
has measure 0 (which is easy since $\Gamma$ erases a lot of information about $X$)
and moreover show that it is a Martin-Löf null set (this requires a bit more care). Then, since $\Omega$ is Martin-Löf random, it follows that $\Omega$ does not belong to $S$. Hence $\Omega$ is not Turing reducible to $\Gamma(\Omega)$.
On the other hand, $\Gamma(\Omega)$ is above another ML-random number in Turing degree, namely
$N(\Omega) := \{n: \Omega$ has at least as many 1s as 0s in the $n$th interval in the definition of $\Gamma(\Omega) \}$.
We should then have
$$
\mathbf 0 < \mathrm{deg}_T(N(\Omega)) <\textrm{deg}_T(\Gamma(\Omega)) <\mathbf 0'
$$
Note however that the Turing degrees $\mathrm{deg}_T(N(\Omega))$ and $\textrm{deg}_T(\Gamma(\Omega))$ presumably depend on the chosen Gödel numbering of the Turing functionals, so part of the answer to the question "what Turing degree does $\Gamma(\Omega)$ have" is "it depends on your Gödel numbering of the Turing functionals".<|endoftext|>
TITLE: Simplest explicit counterexample for $Vect(BG) \ne Rep(G)$ as monoids
QUESTION [9 upvotes]: Let $G$ be a topological group, $Vect(BG)$ the monoid of complex vector bundles over its classifying space (not the stack!) and $Rep(G)$ its monoid of complex representations.
Generally $Vect(BG) \ne Rep(G)$ since by Atiyah Segal completion the grothendiek ring of one is the completion of the other. Unfortunately I don't have any understanding of the Atiyah-Segal theorem so this question is not really about it.

Question: What is the simplest example of a vector bundle over a $BG$ which doesn't come from any continuous representation of $G$? Is there such an example for $\mathbb{CP}^{\infty}=BS^1$?

Edit: I realize now there are many examples for non-compact $G$. Is there a simple example for $G$ compact? (specifically $G=S^1$).

REPLY [3 votes]: Recall that $H^*(BS^3)=\mathbb{Z}[y]$ with $|y|=4$.  For any odd integer $k>0$ there is a map $\psi^k\colon BS^3\to BS^3$ with $(\psi^k)^*(y)=k^2y$.  (I am not actually sure if $k$ needs to be odd.)  There are evident inclusions $S^1\xrightarrow{i}S^3\xrightarrow{j}SU$.  The composite $\psi^k\circ Bi$ just comes from the representation 
$$z\mapsto\left[\begin{array}{cc}z^k&0 \\ 0 & z^{-k}\end{array}\right],$$
and the composite $Bj\circ \psi^k\in[BS^3,BSU]\subset K^0BS^3$ can be described in terms of Adams operations.  However, I do not know a way to construct $\psi^k$ itself without recourse to étale homotopy theory or something like that.  Section 5 of Sullivan's Geometric Topology notes is relevant here.  Anyway, if $V$ is the tautological bundle over $BS^3$ and $k>1$ then $(\psi^k)^*(V)$ is a vector bundle over $BS^3$ which comes from a virtual representation of $S^3$ but not from an honest representation.  More recent work on this sort of thing usually has the phrase "maps between classifying spaces"; there are papers by Jackowski, McClure and Oliver, and other papers by Notbohm.<|endoftext|>
TITLE: $C^{k,\alpha}$ diffeomorphisms and vector fields
QUESTION [9 upvotes]: This feels like something I should know, but I have a hard time finding a definite reference.
Let $M$ be a compact (Riemannian) manifold, $k\ge 1$ be an integer and $\alpha\in(0,1)$. When v is a $C^k$ vector field, plenty of references show that its flow $\Phi^t(x)$ is $C^k$ in $(t,x)$. Now, I would like first to consider the $C^{k,\alpha}$ regularity (I need it not to loose too much when I next solve a simple PDE, where this is the "right" regularity to work with), and second I would like to ensure that the derivative of $\Phi^t$ in $t=0$ is $v$ in the $C^{k,\alpha}$ norm; i.e. I want to ensure that there is a family $\varepsilon_t$ of $C^{k,\alpha}$ vector fields going to $0$ in $C^{k,\alpha}$ norm when $t\to0$, such that
$$\Phi^t(x) = \exp_x(tv+t\varepsilon_t(x)).$$
I don't doubt any of this, but I would like to have a good reference to cite (and read).

REPLY [2 votes]: Edit: unfortunately I did botch the crucial computation. So, surprisingly, the map $f$ defined below need not be locally Lipschitz in the $C^{k,\alpha}$ space.
Let me give an example in $1$-dimension: let $v(x)=\frac23 |x|^{3/2}\in C^{1,\frac12}$, and consider for small $\varepsilon$ the functions $g(x)=\varepsilon +x$ and $h(x)=x$. Then obviously $\lVert g-h\rVert_{C^{1,\frac12}}=\varepsilon$, but we have for $x>0$
$$(v\circ g-v\circ h)'(x)=\sqrt{\varepsilon+x}-\sqrt{x}=\frac{\varepsilon}{\sqrt{\varepsilon+x}+\sqrt{x}} \to \sqrt{\varepsilon} \quad\mbox{when } x\to 0$$
so $\lVert v\circ g-v\circ h\rVert_{C^{1,\frac12}} \ge \sqrt{\varepsilon}$.
Even if this counter-example does not rule out the possibility that what I asked holds (the flow of $v$ above has the wanted property), it still makes me doubt.

Previous version (hopefully still useful in other regularities)
Since references seem elusive, let me propose a simple proof. The idea is simply to use the Picard–Lindelöf theorem to the right object.
First, since $M$ is compact it has positive injectivity radius, and its (global) exponential map $\exp:TM\to M$ induces a diffeomorphism $(x,u)\mapsto (x,\exp_x(u))$ from a neighborhood of the zero section of $TM$, to a neighborhood of the diagonal in $M\times M$ (of course this is already the Picard–Lindelöf theorem, but in the more classical smooth regularity and without the need for the "global derivative" part of the question). Pulling back by $\exp$, we can identify the diffeomorphisms of $M$ which are uniformly close to the identity with vector fields (which are then uniformly close to zero). We use the letter $\Phi$ to denote a diffeomorphism seen as a point the consequent open set $\Omega$ of $\Gamma^{k,\alpha}$ (the space of $C^{k,\alpha}$ vector fields), and the letter $V$ to denote an element of $\Gamma^{k,\alpha}$, seen as a tangent vector to $\Omega$.
Let $v\in \Gamma^{k,\alpha}$ be our given vector field, and define $f:\Omega\to \Gamma^{k,\alpha}$ by
$$f(\Phi) = v\circ \Phi$$
Then $f$ is locally Lipschitz in the $C^{k,\alpha}$ norm, which makes $\Gamma^{k,\alpha}$ a Banach space. This follows from a certainly classical, but somewhat tedious computation I hope I got right (the same needed to show that the $C^{k,\alpha}$ regularity is stable by product and by composition; this last item requires $k\ge1$)
Applying the Picard-Lindelöf theorem to the differential equation $\Phi'(t)=f(\Phi(t))$ in $\Omega\subset \Gamma^{k,\alpha}$ thus yields a unique maximal solution starting at $\Phi(0)=\mathrm{Id}_M$. This solution is obviously the flow of $v$, and now the wanted derivative in $C^{k,\alpha}$ norm follows from the equation itself.<|endoftext|>
TITLE: Factoring Bessel functions into an amplitude and a phase
QUESTION [5 upvotes]: Take some $\nu>0$. Let $J_\nu(x)$ be the Bessel function of the first kind. Let's restrict its domain to $\mathbb R^+$. Is it possible to find a pair of functions $A_\nu(x), \phi_\nu(x):\mathbb R^+\to\mathbb R$ that are real-analytic and completely monotone (i.e. the function itself and all its derivatives are monotone) such that
$$J_\nu(x)=A_\nu(x)\sin(\phi_\nu(x)),\quad A_\nu(x)>0?$$
Is such a pair of functions uniquely determined by $\nu$ (modulo constant term $2\pi n$ in $\phi_\nu$)?

The same question applies to other decaying oscillating functions such as the cosine integral $\operatorname{Ci}(x)$ or the Airy function $\operatorname{Ai}(-x)$.

REPLY [2 votes]: Not an answer but too large for a comment.
In my paper with van de Lune On the exact location of the non-trivial zeros f Riemann's zeta function, it is proved.
Theorem. If $f\colon\mathbf{R}\to\mathbf{C}$ is real analytic, then there are two real analytic functions
$U\colon\mathbf{R}\to\mathbf{R}$ and $\varphi\colon\mathbf{R}\to\mathbf{R}$ such that $f(t)=U(t)e^{i\varphi(t)}$. Given two such representations, $f=U_1e^{i\varphi_1}$ and $f=U_2e^{i\varphi_2}$ we have either $U_1=U_2$ and $\varphi_1-\varphi_2=2k\pi$ or $U_1=-U_2$ and $\varphi_1-\varphi_2=(2k+1)\pi i$ for some integer $k$.
Your question is slightly different. If a  real analytic function can be represented as $f(x)=A(x)\sin(\phi(x))$ with $A$ and $\phi$ real analytic, then $f(x)=\Im (A(x)e^{i\phi(x)})$. The function $F(x) =A(x)e^{i\phi(x)}$ will be a complex real analytic function with $f(x)=\Im(F(x))$. But this $F$ is not unique. Any real and real analytic function $h(x)$ gives us 
$F(x)+h(x)$ with the same property. By the above theorem we will have 
$F(x)+h(x)=B(x)e^{i\phi_h(x)}$ with $B$ and $h$ real real analytic functions. Therefore we obtain 
$f(x)=B(x) \sin(\phi_h(x))$. All representation to your functions are of this type, starting from the one given in the comment by Carlo Beenakker.
The problem whether we may get $\phi_h(x)$ completely monotone appear to be complicated. In simple cases as $\Gamma(s/2)=|\Gamma(s/2)|e^{i\vartheta(t)+i\frac{t}{2}\log\pi}$, we get the representation
$$\Im(\Gamma(\tfrac14+i\tfrac{t}{2}))=|\Gamma(\tfrac14+i\tfrac{t}{2})|\sin(\vartheta(t)+\tfrac{t}{2}\log\pi),\qquad s=\tfrac12+it$$
the phase is not monotonous for $t>0$.  Certainly $\vartheta(t)+\frac{t}{2}\log\pi$ appear to be almost completely monotonous. I will be very surprised if in this case there is a completely monotonous phase.

REPLY [2 votes]: Also too long for a comment. Let us discuss an equivalent representation $J_\nu(x) = A_\nu(x) \cos \phi_\nu(x)$ (with a cosine rather than sine). Theorem 5 in:

K.S. Miller, S.G. Samko, Completely monotonic functions, Integral Transforms and Special Functions 12(4) (2001): 389–402, DOI: 10.1080/10652460108819360

asserts that the function $(A_\nu(x))^2 = (J_\nu(x))^2 + (Y_\nu(x))^2$ is completely monotone. Furthermore, the phase function $\phi_\nu$ satisfies $\phi_\nu'(x) = \tfrac{2}{\pi x} (A_\nu(x))^{-2}$, as it is observed in the reference pointed by Carlo Beenakker in his comment:

M. Goldstein, R.M. Thaler, Bessel Functions for Large Arguments, Mathematical Tables and Other Aids to Computation 12(61) (1958): 18–26, DOI: 10.2307/2002123

Some random remarks:

Complete monotonicity of $(A_\nu(x))^2$ does not automatically imply complete monotonicity of $A_\nu(x)$. This would be the case if $(A_\nu(x))^2$ were a Stieltjes function, but unfortunately it is not.
The function $\phi_\nu(x)$ is a Bernstein function if and only if $\tfrac{1}{x} (A_\nu(x))^{-2}$ is completely monotone. This is only possible when $\nu \leqslant \tfrac{1}{2}$. For $\nu > \tfrac{1}{2}$, $\tfrac{1}{x} (A_\nu(x))^{-2}$ is increasing (possibly a Bernstein function), and therefore $\phi_\nu(x)$ is neither completely monotone nor Bernstein.
A closely related function $\tfrac{1}{x} (A_\nu(\sqrt{x}))^{-2}$ appears in the representation of $\sqrt{x} K_\nu(\sqrt{x}) / K_{\nu-1}(\sqrt{x})$ as a Stieltjes transform, see entry 116 in Section 16.8 in:


R. Schilling, R. Song, Z. Vondraček, Bernstein functions: theory and applications, De Gruyter, 2012.

(In the above I use the term "completely monotone" for $(-1)^n f \geqslant 0$ for $n = 0, 1, \ldots$, and "Bernstein" for $f \geqslant 0$ and $f'$ completely monotone.)<|endoftext|>
TITLE: What is the height (or depth) of $[\mathbb{N}]^\infty$?
QUESTION [7 upvotes]: (This question assumes familiarity with combinatorial cardinal characteristics of the continnum.)
Let $[\mathbb{N}]^\infty$ be the family of infinite subsets of $\mathbb{N}$,
partially ordered by $\subseteq^*$, where $a\subseteq^* b$ means $a\setminus b$ is finite.
Let $\mathfrak{ht}$ (sometimes called $\operatorname{Depth}^+([\mathbb{N}]^\infty)$) be the minimal cardinal number $\kappa$ such that there is no $\subset^*$-decreasing $\kappa$-sequence in $[\mathbb{N}]^\infty$.
Then $\mathfrak{t}<\mathfrak{ht}\le\mathfrak{c}^+$.
A classic result from Kunen's thesis asserts that, when adding $\kappa\ge\aleph_1$ Cohen reals to a model of CH, we obtain $\mathfrak{ht}=\aleph_2$.
Open-ended question. Can the hypothesis $\mathfrak{b}<\mathfrak{ht}$ be expressed using cardinal characteristics other than $\mathfrak{ht}$?
It is easy to see that 
"$\mathfrak{t}=\mathfrak{b}$ or $\mathfrak{b}<\mathfrak{d}$"
implies 
"$\mathfrak{b}<\mathfrak{ht}$". 
Question 1. Is it consistent that 
"$\aleph_1=\mathfrak{t}<\mathfrak{b}=\mathfrak{c}=\aleph_2<\mathfrak{ht}$"?
In the Laver model, we have 
$\aleph_1=\mathfrak{t}<\mathfrak{ht}=\mathfrak{b}=\mathfrak{c}=\aleph_2$.
Question 2. What is the value of $\mathfrak{ht}$ in the Hechler and Mathias models?
Update: Will Brian answers Question 1 in the positive below, in the Hechler model (thus also answering Question 2 for Hechler). 
I thought that $\mathfrak{b}<\mathfrak{ht}$ implies there is a nontrivial set of reals satisfying the selection principle $\operatorname{S}_1(\Gamma,\Gamma)$, by
the the linked paper. But Brian's comments make it clear that I oversimplified the question for this purpose. I opened a new question that fits better the intended application.

REPLY [3 votes]: In the Hechler model, $\aleph_1 = \mathfrak{t} < \mathfrak{b} = \mathfrak{c} = \aleph_2 < \mathfrak{ht}$.
(By "the Hechler model" I mean the result of a length-$\omega_2$ finite support iteration of the forcing to adjoin a dominating real.)
That $\aleph_1 = \mathfrak{t} < \mathfrak{b} = \mathfrak{c} = \aleph_2$ is discussed in Section 11.6 of Blass's handbook article. That $\mathfrak{ht} > \mathfrak{c}$ follows from Theorem 4.1 in


J. Baumgartner and P. Dordal, "Adjoining dominating functions," Journal of Symbolic Logic vol. 50 (1985), pp. 94 - 101, available here.


Interestingly, Baumgartner and Dordal show that, while there are $\subset^*$-decreasing sequences of length $\omega_2$ in this model, none of them are maximal -- all maximal sequences have cofinality $\omega_1$.
This answers question 1 and the first part of question 2. I don't know what $\mathfrak{ht}$ is in the Mathias model, and I also don't know the answer to your open-ended question (though I think it's an interesting one).<|endoftext|>
TITLE: Groups whose finite index subgroups of fixed index are isomorphic
QUESTION [23 upvotes]: I am interested in finitely generated groups $G$ that are residually finite and have the following property: For each $d \geq 1$, $G$ has subgroups of finite index $d$, and all such subgroups are isomorphic.
I know three infinite families of such groups: (1) Free abelian groups $\mathbb Z^n$ (2) Free groups $F_n$ (3) Fundamental groups of closed orientable surfaces $\Gamma_n = \pi_1 (\Sigma_n)$.
The way to see that these are indeed examples is via topology: the classifying spaces of these groups are tori, graphs and surfaces and we understand their covering theory.
[Note: some geometric group theorist claim that these are the easiest torsion-free groups that exist. Their outer automorphism groups are given by $GL_n(\mathbb Z)$, $Out(F_n)$ and $Mod^{\pm}_n$, groups that generated a vast body of research.]
Are there more (than the above mentioned) examples of such groups? 
similar, but in fact a different question

REPLY [4 votes]: A recent preprint of Friedl, Park, Petri, Raimbault, and Ray classifies the compact 3-manifolds with empty or toroidal boundary that have the topological analogue of this property. The authors do not know any examples of infinite residually finite groups with the property beyond the three classes listed in the question.<|endoftext|>
TITLE: Explicit solution to a Rayleigh quotient equation
QUESTION [6 upvotes]: For 5 months! I have been struggling to solve the following equations analytically without numeric method (ie, Newton method):

Main equation:
$$
 \biggl(M^2-\cfrac{\mathbf{x^{\text{T}}}M^2\mathbf{x}}{\mathbf{x^{\text{T}}}\mathbf{x}}E\biggr)\mathbf{x}=\mathbf{1}
$$
Constraint equations:
$$
\begin{cases}
 \mathbf{x^{\text{T}}1}=0 \\
\\
\mathbf{x^{\text{T}}x}=u 
\end{cases} $$
where $\{M,E\}\in\mathbf{R}^{n \times n}$ and $\{\mathbf{1},\mathbf{x}\}\in\mathbf{R}^n$ are defined, then $M$ is an arbitrary symmetric matrix, $E$
  is an identical matrix,
  $\mathbf{1}$ is all one vector, $\mathbf{x}$ is a
  variable vector and $u\in\mathbf{R}$ is a scalar.
  Furthermore, as a knowledge, the below equation form is called Rayleigh
  quotient $R(M^2,\mathbf{x})$:
$$R(M^2,\mathbf{x}):=\cfrac{\mathbf{x^{\text{T}}}M^2\mathbf{x}}{\mathbf{x^{\text{T}}}\mathbf{x}}$$

Now, we attempt to estimate the $\mathbf{x}$. Does the analytic solution or method exist? My ability is shortage but, I guess that this problem has a beautiful solution. Also, main equation is a simultaneous cubic equation. Theoretically, this is solvable. Just, this is my theme question.

REPLY [4 votes]: Since $M^2$ is symmetric, it is diagonalizable with orthogonal  eigenvectors. Also the eigenvalues are positive. Let $(v_i)$ be an orthonormal basis of eigenvectors for $M^2$ with eigenvalues $\lambda_i\ge 0$. Now express $\mathbf 1$ in terms of the eigenvectors as $\sum b_iv_i$ and write $\mathbf x=\sum a_iv_i$. The equations reduce to $a_i(\lambda_i-K)=b_i$, where $K$ is the Rayleigh quotient $(\sum \lambda_i a_i^2)/(\sum a_i^2)$. 
I would approach this by defining for each $t$, $\alpha_i(t)=b_i/(\lambda_i-t)$ and then $F(t)$ to be the Rayleigh quotient for the corresponding family of $(\alpha_i(t))$, that is 
$$
F(t)=\left(\sum \lambda_i \frac {b_i^2}{(\lambda_i-t)^2}\right)\Big /
\left(\sum \frac{b_i^2}{(\lambda_i-t)^2}\right).
$$
If for some $t$, $F(t)=t$, then you have a solution to your equations (namely $a_i=\alpha_i(t)$). Multiplying through by the denominators on both sides, you see that $F(t)$ is a continuous function of $t$, bounded between $\min\lambda_i$ and $\max\lambda_i$. Hence there is always a solution to the equation (by the intermediate value theorem).
But I don't think you should expect to be able to find an analytic solution, as this seems to be essentially the same as solving a typical degree $2n$ polynomial equation.<|endoftext|>
TITLE: General Relativity and Differential Geometry intuitions of Second Bianchi Identity
QUESTION [13 upvotes]: In General Relativity, one uses the Riemann Tensor in its coordinate form $R_{abcd}$, and proves the Second Bianchi Identity-


$R_{abcd;e} + R_{abde;c} + R_{abec;d} = 0$


It is said that this identity amounts, in the Riemmanian geometry sense, to the saying that, loosely, "the boundary of the bounday is null". So my questions are
Question 1: What would be a more rigorous statement of "the boundary of the boundary is null" and how does the above identity is equivalent ot it?
Question 2: Is there an analog to this geometrical intuition in general relativity?
I assume both questions might be answered with a reference, which is even better.

REPLY [5 votes]: The second Bianchi identity can also be viewed as a consequence of the diffeomorphism invariance of the Riemann curvature tensor, i.e. the fact that $\operatorname{Rm}_{\varphi^*g} = \varphi^* \operatorname{Rm}_g$ for any metric $g$ and diffeomorphism $\varphi$. Differentiating with respect to $t$ the equation $\operatorname{Rm}_{\varphi_t^*g} = \varphi_t^* \operatorname{Rm}_g$ , where $\{\varphi_t\}$ is a one-parameter family of diffeomorphisms, yields a proof of the first and second Bianchi identities.
This is discussed in (for instance) Chapter 5.3.1 of The Ricci Flow in Riemannian Geometry: A Complete Proof of the Differentiable 1/4-Pinching Sphere Theorem by Ben Andrews and Christopher Hopper.
In terms of a relativistic interpretation, this is related to the idea of general covariance, which requires the laws of physics to be diffeomorphism invariant.<|endoftext|>
TITLE: Eigenvectors of the Fourier transformation
QUESTION [16 upvotes]: The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$
by
$
\hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx.
$
It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is
$
u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi.
$
The Fourier transformation  can be extended to the tempered distributions
$\mathscr S'(\mathbb R^n)$ with the formula
$$
\langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}
=
\langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}.
$$
Below we note  $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that
$\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.

Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2},
e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case
  $$
T_0=\sum_{k\in \mathbb Z^n}\delta_k,
$$
  where the equality $\mathcal FT_0=T_0$ is the Poisson summation formula. 

I think that, thanks to your answers and a reference, I found the answer: using What are fixed points of the Fourier Transform which deals with the $L^2$ case, one may guess that the fixed points of $\mathcal F$ in $\mathscr S'(\mathbb R^n)$ are
$$
\Bigl\{S+\mathcal F S+\mathcal F^2 S+\mathcal F^3 S
\Bigr\}_{S\in \mathscr S'(\mathbb R^n)}=(Id+\mathcal F+\mathcal F^2+\mathcal F^3)(\mathscr S'(\mathbb R^n)).
$$
Since $\mathcal F^4=Id$ on $\mathscr S'(\mathbb R^n)$, the above distributions are indeed fixed points and if $\mathcal F T=T$, then
$$
T=\frac14\bigl(T+\mathcal F T+\mathcal F^2 T+\mathcal F^3 T\bigr),
$$
conluding the proof.

REPLY [7 votes]: There is a whole chapter in Titchmarsh, Fourier transform, describing all eigenfunctions in great detail. He calls them "self-reciprocal" functions. A more difficult question is about the eigenvectors of discrete Fourier transform, but this one you do not ask.
Edit. Since you are asking now about discrete transform too, it has 4 eigenspaces
corresponding to eigenvalues $1,-1,i,-i$ and the question is how to choose a convenient basis in each. Of course the choice is very non-unique, so many bases were proposed. A survey can be found here:
http://www.cs.princeton.edu/~ken/Eigenvectors82.pdf. Here is a newer paper:
http://www.sciencedirect.com/science/article/pii/S0165168408001783. More can be found by typing "Eigenvectors of the discrete Fourier transform" on Google.<|endoftext|>
TITLE: Reference request: Goodwillie tower of the identity
QUESTION [11 upvotes]: The Taylor (Goodwillie) tower of the identity functor on based spaces has as its $j$-th layer the infinite loop space-valued functor
$$
X\mapsto \Omega^\infty (W_j \wedge_{h\Sigma_j} X^{[j]})
$$
in which $W_j$ is a spectrum with (naive) $\Sigma_j$-action, $X^{[j]}$ is the $j$-fold smash power and ${}_{h\Sigma}$ denotes homotopy orbits. The underlying unequivariant homotopy type of the coefficient spectrum $W_j$ is that of a wedge of $(j-1)!$ spheres of dimension $S^{1-j}$. 
The restriction of the action to the subgroup $\Sigma_{j-1} \subset \Sigma_j$ is known to be given by the permutation action.
I have been told by several folks in the past that the $\Sigma_j$-action actually extends to an action by $\Sigma_{j+1}$.  The way this is somehow done, if I recall, is by considering a certain space of metric trees with $j+1$ leaves. 
Does anyone know of such a construction? Does anyone know a reference?

REPLY [5 votes]: Jean-Louis Loday told me about the extended action by $\Sigma_{j+1}$ in the fall of 1992, after an Oberwolfach talk I gave about the rank filtration of algebraic $K$-theory, where the $\Sigma_j$-representations given by the integral homology of the Goodwillie derivative  spectra $W_j$ played a role. I had shown that these representations were freely generated by $j$-fold Lie brackets, and Loday knew the connection to spaces of trees. Maybe you were there, too?<|endoftext|>
TITLE: A very torsioned closed curve in space
QUESTION [11 upvotes]: Is there a  simple  smooth closed curve $\gamma$ in $\mathbb{R}^{3}$  such that for  all $x,y\in \gamma$  with $x \neq  y $,  $l_{x}$  and  $l_{y}$ are  skew  lines, where $l_{x}  $ and  $l_{y}$ are straight lines tangent to $\gamma$ at $x,y$, respectively? What  about real  analytic  case?

REPLY [21 votes]: If by "skew" we mean nonparallel and nonintersecting, then the answer is NO. Every smooth closed curve in $\mathbf{R^3}$ has uncountably many pairs of intersecting tangent lines. This follows from Poincare-Hopf index theorem; see
M. Ghomi, Tangent bundle embeddings of manifolds in Euclidean space, 
Comment. Math. Helv., 81(2006), 259-270.
The lowest dimensional space where a closed curve with nonparallel and nonintersecting tangent lines may be constructed is $\mathbf{R^4}$. A very natural example is given by
$$
\mathbf{C}\supset\mathbf{S}^1\ni z\longmapsto (z,z^2)\in \mathbf{C}^2\simeq\mathbf{R}^4,
$$
as described in the above paper.
On the other hand, if we require the tangent lines only to be nonparallel, then the answer is YES, such curves do indeed exist in $\mathbf{R}^3$. They were first constructed by Beniamino Segre, to disprove a (short-lived) conjecture of Hugo Steinhaus,  in 1968:
B. Segre, Global differential properties of closed twisted curves, Rendiconti del Seminario Matematico e Fisico di Milano,
December 1968, Volume 38, Issue 1, pp 256-263.
They are not difficult to construct: one starts with a closed spherical curve $T\colon[a,b]\to\mathbf{S}^2$ parametrized by arc length which does not have any self intersections, is disjoint from its antipodal reflection, and contains the origin in the interior of its convex hull. The last property ensures the existence of a (positive) density function $\rho\colon [a,b]\to\mathbf{R^+}$ such that the center of gravity of the weighted curve $\rho T$ is the origin, i.e., $\int_a^b\rho(s)T(s)ds=o$. Then we integrate this curve to obtain:
$$
\gamma(t):=\int_0^t\rho(s)T(s)\,ds.
$$
Choosing $T$ and $\rho$ to be smooth periodic functions will ensure that $\gamma$ is smooth as well.
 The resulting curve $\gamma$ will be closed, due to the center of mass condition, and will have no pairs of parallel tangent lines, because its tangential spherical image is $T$ by construction.
The first explicit example of these curves probably appeared in
M. Ghomi, Shadows and convexity of surfaces, Ann. of Math., 155 (2002) 281-293,
where it was used to solve Henry Wente's shadow problem. The example in the above paper is both simple and analytic (it is given by polynomials). Also it lies on a convex surface. Later, Bruce Solomon and I studied these curves, which we called skew loops, some more in
M. Ghomi and B. Solomon, Skew loops and quadric surfaces, 
Comment. Math. Helv., 77 (2002) 767-782.
where it is shown that the only closed surfaces in $\mathbf{R}^3$ which do not admit any skew loops are ellipsoids. The higher dimensional analogues of these objects are also studied in
M. Ghomi and S. Tabachnikov, Totally skew embeddings of manifolds, 
Math. Z., 258 (2008), 499-512.
Here the term totally skew means nonparallel and nonintersecting.<|endoftext|>
TITLE: Concentration of U-statistics for exchangable distributions (and the unbounded case)
QUESTION [5 upvotes]: Consider the following so-called $U$-statistic of order 2: $$U = \frac1{\binom{m}{2}} \sum_{i < j} h(w_i,w_j)$$ where $w_1,\dots,w_m$ are IID from some distribution and $h$ is symmetric. If $|h(w_1,w_2)| \le B$ a.s., then a classical result gives the following concentration inequality:
\begin{align}
   \mathbb P( | U - \mathbb E U| \ge t) \le 2 \exp(- m t^2 /(8B^2)).
\end{align}
(Maybe there are better constants known here.)
I have the following two questions:

It seems to me that this immediately generalizes to the case where $(w_1,\dots,w_n)$ has an exchangeable distribution, due to de Finetti's theorem saying that any such distribution is a result of a mixture of IID ones. More precisely, there is a random measure $G$, such that conditional on $G$, $w_1,\dots,w_n$ are IID draws from $G$. By conditioning on $G$, using the inequality above for the IID case, and then taking expectations, we get the result for the exchangeable case. Is this argument correct?
What is the state of the art in terms of concentration bounds for such $U$-statistics, esp in the case where $h$ is not bounded? Are there clear generalization known, esp. for the exchangable case? The above argument seems to rely on boundedness of $h$ in a critical way.

EDIT: I should say that for the first point, it might be that we have to assume $h$ to be surely bounded (?)
EDIT: As was pointed out, the argument in point one is flawed. In the hindsight, one needs extra conditions. The case where $w_1=w_2=\dots=w_m$ is an example of an exchangeable distribution for which the concentration (with $m$ in the exponent) need not hold.

REPLY [2 votes]: I do not think the conclusion will hold for a general exchangeable sequence. In a more general case, you have to assume that U-statistics itself are not degenerated ($w_i\neq w_j$ for $i\neq j$) and the concentration bound is still not so good as shown in [Arcones].
However, for the case where exchangeable pairs exists (like independent yet not necessarily identical), [Lester Mackey et.al] Coro 5.2. might be what you want.
I think the following paper Matrix Concentration Inequalities via the Method
of Exchangeable Pairs is what you want to/what you should read. Instead of considering exchangeable random variables, the usual way of thinking independence, which started by C.Stein (Stein's original paper), is to consider independent pairs of random variables. This is also natural from a categorical view since we can only discuss the commutativity of one diagram (commutativity of diagrams is equivalent to exchangeability if we formalized the category appropriately, see Category-theoretic structure for conditional independence.) 
The definition given in [Lester Mackey et.al] is:

Let $Z$ and $Z′$ be random variables taking values in a Polish space
  $\mathcal{Z}$. We say that $(Z, Z′)$ is an exchangeable pair if it has
  the same distribution as $(Z′,Z)$. In particular, $Z$ and $Z′$ must
  share the same distribution.

[Arcones] Arcones, Miguel A. "A Bernstein-type inequality for U-statistics and U-processes." Statistics & probability letters 22.3 (1995): 239-247.
[Lester Mackey et.al] Mackey, Lester, et al. "Matrix concentration inequalities via the method of exchangeable pairs." The Annals of Probability 42.3 (2014): 906-945. https://arxiv.org/pdf/1201.6002.pdf<|endoftext|>
TITLE: BSD and generalisation of Gross-Zagier formula
QUESTION [5 upvotes]: The classical Gross-Zagier formula and the modularity theorem leads to a proof of half-BSD (i.e. an inequality and not equality) for elliptic curves of analytic rank 0.
The Gross-Zagier formula gives something on the BSD conjecture for elliptic curves over $\mathbb{Q}$. In particular the something that interest me is that if an elliptic curve as analytic rank 1, then the elliptic curve as a point of infinite order so as algebraic rank at least 1.
As said in the answers, in the function field case a theorem of Tate and Milne, says that the algebraic rank is smaller than the analytic rank (so BSD conjecture is true for analytic rank 0). So an analogue of the Gross-Zagier formula in the function field case, gives the BSD conjecture for analytic rank 1.
This article of Yun-Zhang pretends :

This identity can be viewed as a function-field analog of the Waldspurger and Gross–Zagier formula for higher derivatives of L-functions.

Does this article as any consequence on the analogue of BSD in function fields ?

REPLY [2 votes]: The BSD conjecture for an abelian variety $A$ over a function field holds if Ш$(A)[\ell^\infty]$ is finite for some prime $\ell$ ($\ell = p$ allowed). This is a theorem by Schneider, Bauer and Kato-Trihan. If $A$ is a constant abelian variety, Ш$(A)$ is finite by Milne's PhD thesis.
Edit: Since the analytic rank $\rho$ is always greater or equal than the algebraic rank, one has BSD if $\rho = 0$ (by the equivalence of weak BSD and the finiteness of an $\ell$-primary component of Sha).  I show this inequality even for Abelian schemes over higher dimensional bases over finite fields in http://kellertimo.name/Height.pdf, Lemma 2.17.<|endoftext|>
TITLE: Duality between topology and bornology
QUESTION [18 upvotes]: I want to understand in what sense topology is dual to bornology at a most basic level. Therefore, I rephrased the definition of a bornology in the following way:
Let $X$ be a set and let $\mathcal{P}(X)$ be the bounded lattice induced by the power set $P(X)$ together with unions, intersections, $X$ and $\emptyset$ as joints, meets, $1$ and $0$. An $\textbf{ideal}$ $\beta \subseteq \mathcal{P}(X)$ is called a $\textbf{bornology}$ if $\bigvee_{B \in \beta} \, B=1.$
This should be equivalent to the usual definition. An obvious dualization of this is a $\textbf{filter}$ $\nu \subseteq \mathcal{P}(X)$, s.t. $\bigwedge_{N \in \nu}=0$. 
This should be seen as the set of all neighbourhoods with respect to some topology on $X$ and the question is now how to extract a topology from it. I know that there is a unique topology for every neighbourhood system $\{N(x)\}_{x \in X}$ and of course I can define a trivial neighbourhood system from a given filter $\nu \subseteq \mathcal{P}(X)$ by setting $N(x)$ to be the subfilter of sets in $\nu$ containing $x$. This leads to the finest possible topology, but what I actually want is a unique coarsest neighbourhood system (s.t. $\nu = \cup_x \, N(x) $)! Unfortunately, I don't really have an idea how to start showing the existence of such a thing.. do you? Of course, any other idea how to define open sets is welcome! 
EDIT 1: it seems that coarsest is not the right property here, since for a given topology, the neighbourhoodfilter is in general not the coarsest neighbourhood system
EDIT 2: I guess I made a really stupid mistake here: the filter $\nu$ cannot distinguish between topologies $\tau,\tilde{\tau}$ with neighbourhood filters $N(x),\tilde{N}(x)$, s.t. $\bigcup_x \, N(x) =\bigcup_x \, \tilde{N}(x)= \nu$..
EDIT 3: For topological vector spaces $(X,\tau),(X,\tilde{\tau})$ it should be true that from $\bigcup_x \, N(x) =\bigcup_x \, \tilde{N}(x)$ follows $N(0)=\tilde{N}(0)$, thus $\tau = \tilde{\tau}$. I used both continuity of multiplication and addition.
Moreover, if $\tau$ is nontrivial, we have $\bigcap_{N \in \bigcup_x N(x)} \, N = \emptyset$. I didn't use that $\tau$ is Hausdorff.

REPLY [2 votes]: Since this query has resurfaced, I would like to suggest that the pro and ind categories of Grothendieck are a suitable framework to discuss the topic.  These are ways to extend a given categories by adding formally projective and inductive limits.
Let me start with the case of vector spaces since the situation here is more transparent and this is the one which has been studied in most detail, at least for bornologies.
If we start with the category of Banach spaces, we obtain respectively that of locally convex spaces and of convex bornological spaces (Buchwalter, Hogbe Nlend).  This suggests that the natural structure of a vector space (of functions, distributions) depends on whether they are natural projective or inductive limits of Banach spaces.  Thus the smooth functions on $I$, the unit interval—a lcs, the distributions thereon—a cbs.  Similarly, the continuous functions on the line—a lcs, the compactly supported measures thereon—a cbs.
These examples suggest continuing by incorporating duality in its functional analytic form.  The basic example is the functor which assigns to each Banach space its dual.  As a contravariant functor, this extends to one from the pro category into the ind category and vice versa, in our case form the category of lcs’s into cbs’s.  The fact that one can regard the dual of a lcs as a cbs (and vice versa) in a natural way has already been mentioned in the above answers and there are many indications external to category theory that is the more natural approach to duality rather than the spandrel of regarding the dual of a lcs as a lcs with various candidate topologies. (Health warning: this is very much a minority opinion).
Turning to the non linear case, the basic category is that of metric spaces with Lipschitz functions as morphisms. (In order to keep things simple, I will the blanket assumption that everything is sight is complete).  The corresponding pro category is that of uniform spaces.  The ind category has not, to my knowledge, been discussed but it can be 
defined directly as a class of bornologies, but enriched by the fact that
each bounded set is provided with a suitable metric.  Everybody has a collection of interesting uniform spaces in his toolbox,  most of the relevant bornological spaces have, in fact, suitable metrics.<|endoftext|>
TITLE: When is there an unbounded tower in $[\mathbb{N}]^\infty$?
QUESTION [5 upvotes]: (Edit: I'm splitting the question, leaving here only what is answered by Ashutosh, and moving the rest to another question.)
This question assumes familiarity with combinatorial cardinal characteristics of the continnum. It is a refined version of an earlier question.
Let $[\mathbb{N}]^\infty$ be the family of infinite subsets of $\mathbb{N}$,
partially ordered by $\subseteq^*$, where $a\subseteq^* b$ means $a\setminus b$ is finite.
Let $\kappa$ be a cardinal number.
A tower of height $\kappa$ is a $\kappa$-sequence 
$\langle\, s_\alpha : \alpha<\kappa\,\rangle$ in $[\mathbb{N}]^\infty$ that
is $\subset^*$-decreasing as the ordinal number $\alpha$ increases.
An element $a\in [\mathbb{N}]^\infty$ is identified with its increasing enumeration. This way, the set $[\mathbb{N}]^\infty$ 
becomes the family of increasing functions in $\mathbb{N}^\mathbb{N}$,
and the standard relation $\le^*$ is defined on 
$[\mathbb{N}]^\infty$.
A set $X\subseteq [\mathbb{N}]^\infty$ is bounded if it is bounded (from above) with respect to $\le^*$.
The general goal is to understand when is there an unbounded tower.
Let us call this axiom BT (and ignore the coincidence).
It is known or easy to see that:

If there is an unbounded tower of any cardinality, then BT holds. (The present proof is dichotomic.)
If $\mathfrak{t}=\mathfrak{b}$ or $\mathfrak{b}<\mathfrak{d}$, then BT holds.

Question. Is BT consistent with "$\aleph_1=\mathfrak{t}<\mathfrak{b}=\mathfrak{c}=\aleph_2$"?
Will Brian's answer for my previous question implies that BT fails in the Hechler model. BT also fails in the Laver model, indirectly by the main result of the linked paper.

REPLY [4 votes]: A model for Question 1: Let $V \models$ MA + $2^{\aleph_0} = \kappa \geq \aleph_2$. Using MA, construct a $\supseteq^{\star}$-chain $\overline{A} = \langle A_i : i < \kappa \rangle$ such that $\mathcal{A} = \{p_{A_i} : i < \kappa\}$ (where $p_X(k)$ is the $k$th member of $X$) is a dominating family in $\omega^{\omega}$. Let $P$ add $\aleph_1$ random reals. Since the random reals added constitute a non null set of size $\aleph_1$, standard arguments show that $V^P \models \mathfrak{p} = \mathfrak{t} = \aleph_1$. Since $P$ satisfies ccc and is $\omega^{\omega}$-bounding, $V^P \models \mathfrak{b} = \kappa = 2^{\aleph_0}$. Finally, $\mathcal{A}$ remains dominating in $V^P$ so the principle BT continues to hold.  So $V^p \models \aleph_1 = \mathfrak{t} < \mathfrak{b} = \kappa = 2^{\aleph_0} $ plus BT.<|endoftext|>
TITLE: Distinctness of products of Fibonacci numbers
QUESTION [21 upvotes]: Suppose that $S$ and $T$ are sets of Fibonacci numbers greater than $1$.  Let $S^*$ be the product of numbers in $S$, and likewise for $T^*$.   If $S^*=T^*$, must $S=T$?

REPLY [43 votes]: (Finite sets, of course, if you don't want to allow $S^* = T^* = \infty$)
Yes.  We may assume wlog $S$ and $T$ are disjoint, and $\max S < \max T$.  By Carmichael's theorem, every Fibonacci number except $1$, $8$ and $144$ has a prime factor that does not divide any earlier Fibonacci number.  So $\max T$ can only be $8$ or $144$, and neither of these work.<|endoftext|>
TITLE: What is the expected value of an N-dim vector of uniform randoms that sum to 1 which have been sorted into descending order?
QUESTION [17 upvotes]: What is the expected value of an N-dimensional vector of uniformly distributed random numbers which sum to 1 and have been sorted in descending order?
Here is the algorithm for drawing a sample from the distribution of this N-dimensional vector:

Create an N-1 dimensional vector of uniformly distributed random numbers between 0 and 1, inclusive. For example, for N=3, [0.564, 0.243].
Append the numbers 0 and 1 to the vector, then sort it in ascending order. For example, [0, 0.243, 0.564, 1].
Compute the 1st-order differences of the vector, yielding an N-dimensional vector of uniformly distributed random numbers that sum to 1. For example, [0.243, 0.321, 0.436].
Sort this vector into descending order. For example, [0.436, 0.321, 0.243].

It is easy to demonstrate by graphical analogy that for N=2, the expected value of the vector is exactly [3/4, 1/4].
For N=3, by sampling I have determined that the expected value is approximately [0.611, 0.277, 0.111].
For N=4, by sampling I have determined that the expected value is approximately [0.520, 0.271, 0.146, 0.063].
What is the analytical form of the expected value for any value of N?

REPLY [15 votes]: Liviu has given an excellent answer with a nice geometric flavor to it. This answer is meant to serve as a complement with a slightly more probabilistic bent. In the process, some of the computations get buried in favor of bringing out the connections to other areas of elementary probability theory.
A well-known alternative mechanism for generating points uniformly on the simplex is as follows: Let $Z_1,\ldots,Z_n$ be iid exponential random variables. Then $(Z_1/Z,\ldots,Z_n/Z)$ is uniformly distributed on the simplex where $Z = Z_1 + \cdots + Z_n$. 
This is a special case of the Dirichlet distribution. In your question, you are effectively interested in computing $\mathbb E \frac{Z_{(k)}}{Z}$ for each $k$ where $Z_{(k)}$ is the $k$th order statistic from your iid exponential sample.
By Rényi representation, we can construct $Z_{(k)}$ directly as follows: Let $Z_{(0)} = 0$, and let $(\widetilde Z_i)_{i \geq 1}$ be a sequence of iid exponentials. Then, form 
$$
Z_{(k)} = Z_{(k-1)} + \frac{\widetilde Z_k}{n-k+1}\>.
$$
Note that $Z = \sum_{k=1}^n Z_k = \sum_{k=1}^n Z_{(k)} = \sum_{k=1}^n \widetilde Z_k$.
Hence
$$
\mathbb E \frac{Z_{(k)}}{Z} = \sum_{i=1}^k \frac{1}{n-i+1}\mathbb E \frac{\widetilde Z_i}{Z} = \frac{1}{n} \sum_{i=1}^k \frac{1}{n-i+1} \>,
$$
since the $\widetilde Z_i/Z$ have the same $\text{Beta}(1,n-1)$ distribution for each $i$.
References

A. Rényi (1953), On the theory of order statistics, Acta Mathematica Hungarica, vol. 4(3–4), pp. 191–231. 
L. Devroye (1986), Non-Uniform Random Variate Generation, Springer-Verlag. (Chapter 5: Uniform and Exponential Spacings).

Postscriptum: Note that the algorithms given in this answer for generating a point on the standard simplex and on its sorted variant are both $O(n)$ compared to the $O(n \log n)$ approach in the question, so if $n$ is large and computing $\log x$ for $0 < x < 1$ is cheap, this may lead to meaningful gains in sampling efficiency.<|endoftext|>
TITLE: Obstruction to rationality of del Pezzo surfaces of degree 4
QUESTION [7 upvotes]: Let $X$ be a del Pezzo surface over a number field $k$. (A del Pezzo surface over $k$ is a smooth, projective, geometrically connected surface whose anti-canonical class $K_X$ is ample.) Let $d := K_X^2$ be the degree of $X$. It is well-known that $d$ satisfies $1 \leq d \leq 9$.
If $d \geq 5$, and $X(k) \neq \emptyset$, then $X$ is birationally equivalent to $\mathbb{P}^2_k$. Moreover, if $d=5$ or $d=7$, the condition on $X(k)$ is automatically satisfied. (This is Theorem 9.4.7 in Bjorn Poonen's Rational Points on Varieties.) In other words, for del Pezzo surfaces of degree $5$ and higher, the only obstruction to $k$-rationality is the possible lack of $k$-rational points.
My question concerns the "first" (i.e. highest-degree) non-trivial case, as far as rationality is concerned, namely the case where $d=4$. In this case, there are examples of $X$ with $X(k)\neq \emptyset$ but where $X$ is non-rational. An obstruction to rationality is given by the (non-trivial part of the) Brauer group $\operatorname{Br}(X)$ of $X$: that is, if $\operatorname{Br}(X)/\operatorname{Br}(k) \neq 0$, then $X$ is not $k$-rational. (To be sure, such $X$ exist, even among those $X$ that have rational points. This is where the $d=4$ case differs from the higher degree cases.) Indeed, this simply follows from the fact that $\operatorname{Br}$ is a birational invariant of smooth, projective, geometrically connected varieties, and from the fact that $\operatorname{Br}(\mathbb{P}^2_k) = \operatorname{Br}(k)$. My question is:

Is the converse true? That is, if $X$ is a del Pezzo surface of degree $4$ over a number field $k$ with $X(k)\neq\emptyset$, and $\operatorname{Br}(X) = \operatorname{Br}(k)$, does it follow that $X$ is a $k$-rational surface?

REPLY [5 votes]: I will supplement Dan's nice answer by claiming that the answer is almost always no.  Specifically, almost every degree 4 del Pezzo surface over the rational numbers with a rational point is non-rational but has trivial Brauer group.
To support this, I wrote a little Magma program that picked two random quadrics in $\mathbb{P}^4$ containing the point $(1:0:0:0:0:0)$ and with integer coefficients in $[-5,5]$.  Let $X$ be the intersection of these two quadrics.  I computed the Fano scheme of lines on $X$.  If the Fano scheme is irreducible over $\mathbb{Q}$, that is, the lines are all Galois-conjugate, then a calculation shows that the Brauer group is trivial and the surface is not rational.  I ran the program 100 times and, apart from 1 time when $X$ was singular, this was always the case.  I'm sure it would be straightforward to show that the Fano scheme is always irreducible outside a thin set of the space parametrising such pairs of quadrics.<|endoftext|>
TITLE: Quotient of Coxeter complex in terms of double cosets?
QUESTION [6 upvotes]: In Victor Reiner's Quotients of Coxeter Complexes and $P$-Partitions, we have the below definition for the quotient complex of a Coxeter complex by a finite subgroup of the Coxeter group. I think there is something wrong with the definition, and would like your help sorting it out.
Given a finite Coxeter system $(W,S)$, with $S$ a set of $n$ Euclidean reflections in some $n$-dimensional real inner product space that generate a finite group $W$, Reiner begins with the following pair of definitions:

$\Sigma(W,S)$ is the triangulation of the unit sphere in $\mathbb{R}^n$ obtained by intersecting it with all the hyperplanes fixed (pointwise) by any reflection in $W$.
Alternatively, $\Sigma(W,S)$ is the poset whose elements are the cosets $wW_J$ of subgroups $W_J$ of $W$ generated by subsets $J$ of $S$, ordered by reverse inclusion.

(This is on p. 9 of AMS Memoirs edition of Reiner's work, but p. 11 of the link above.)
The definitions are equivalent, in that the same abstract simplicial complex has the former definition as its geometric realization and the latter definition as its face poset. Reiner cites Kenneth Brown's book on buildings for this.
So far, so good. I both find the argument in Brown convincing and find that the result matches my hand calculations. Reiner goes on to give a parallel pair of definitions for the quotient of $\Sigma(W,S)$ by the action of a subgroup $G\subset W$.

$\Sigma(W,S)/G$ is the topological quotient of the simplicial complex $\Sigma(W,S)$ by the action of $G$.
$\Sigma(W,S)/G$ is the poset of double cosets $GwW_J$, ordered by reverse inclusion.

(p. 11 in the AMS edition and p. 13 in the link.)
Here, I have a problem. I don't see that the geometric version of the quotient complex (which is still a $\Delta$-complex since $G$ acts by simplicial homeomorphisms, but is no longer a simplicial complex) will have face poset given by the reverse inclusion relation. I agree that the double cosets $GwW_J$ will be in bijection with the simplices of the geometric version, provided we regard cosets with distinct $J$'s as distinct even if they are the same set, but I think that the incidence relations are subtler. My question is, am I right about this, and if so, what's the correct way to define the incidence relation in this combinatorial/group-theoretic definition of the quotient complex?
To make the problem concrete:
Let $W$ be $S_4$, seen as the isometries of a regular tetrahedron in $\mathbb{R}^3$ centered on the origin. Let $S=\{(12),(23),(34)\}$ with appropriate labels on the vertices of the tetrahedron. Then $\Sigma(W,S)$ is the barycentric subdivision of the boundary of the tetrahedron. Let $G = \langle (13),(1234)\rangle = D_4\subset S_4$.
The double cosets $D_4(id.)\langle(12),(23)\rangle$ and $D_4(id.)\langle (23),(34)\rangle$ both contain the entire group $S_4$; so if we order by reverse inclusion, they both sit below everything else, including each other. This doesn't make sense. In the $\Delta$-complex this would correspond to two vertices being incident to both each other (??) and also every edge.
So, to reiterate the question: am I flat wrong? And if not, then what is the partial order on the double cosets $GwW_J$ (regarding $GwW_J \neq Gw'W_{J'}$ unless we have set equality and $J=J'$) that recovers the incidence relation of the faces in the geometric definition of $\Sigma(W,S)/G$?
Addendum 5/12/16: Reiner answers the question below with a reference to a 2004 paper where he gave the right definition. Jim Humphreys has given a link in a comment, but just so the complete question/answer pair can be on this page, here is the answer:

$\Sigma(W,S)/G$ is the poset of ordered pairs $(GwW_J,J)$, ordered by reverse inclusion of both factors.

I.e. $(GwW_J,J) \leq (Gw'W_{J'},J')$ if $J\supset J'$ and $GwW_J\supset Gw'W_{J'}$.

REPLY [7 votes]: I think this is the same issue that I noticed later myself, and addressed in Remark 2.13 on page 228 of the paper that I wrote with Eric Babson: "Coxeter-like complexes" (Discrete Mathematics and Theoretical Computer Science 6, 2004, 223–252).  Let me know, if that is not the issue!<|endoftext|>
TITLE: Hypersurface missing just one point
QUESTION [6 upvotes]: Let $\mathbb F_q$ be a finite field and $n$ an integer.

What is the minimal degree $d = d(q,n)$ of a polynomial $f \in \mathbb F_q[X_1,\dots,X_n]$ such that the set $Z(f)$ of zeros of $f$ in the affine space $\mathbb F_q^n$ has cardinality $q^n-1$, that is misses exactly one point?

It is clear that $d$ is finite, more precisely $d(q,n) \leq n(q-1)$. Indeed, for $f = \prod_{i=1}^n \prod_{\alpha \in \mathbb F_q^\ast} (X_i-\alpha)$, $Z(f)$ is the complement of the origin. 
It is also easy to see that $d(q,n) \geq q$, because if $f$ is a polynomial of degree $d<q$, the cardinality of $Z(f)$ is divisible by the characteristic $p$ of $\mathbb F_q$ (Chevalley-Warning), and $\leq d q^{n-1}$ (Schwartz-Zippel), which gives two independent proofs that it cannot be $q^n-1$.
One has $d(2,2)=2$ (e.g. $f(x,y)=xy$ has three zeros) and $d(2,3)=3$ (the zeros of the degree 3 polynomial $f(x,y,z)=1+x+y+z+xy+xz+yz+xyz$ are everything but the origin, and it is not hard to see that you can't do that with a degree 2 polynomial). 
But what is the value of $d$ in general?

REPLY [4 votes]: An even further generalisation of this result is the Alon-Füredi theorem, which says that if a polynomial $f$ does not vanish completely on the "grid" $S = S_1 \times \dots \times S_n$ where $S_i$'s are finite subsets of an arbitrary field $F$, then $f$ does not vanish on at least $\min \prod y_i$ points of the grid as $y_i$'s ranges over integers satisfying $\sum y_i = \sum |S_i| - \deg f$ and $1 \leq y_i \leq |S_i|$. 
Clearly, when $\deg f < \sum (|S_i| - 1)$, this minimum value is at least $2$, thus proving that a polynomial vanishing on all points of the grid except one must have degree at least $\sum (|S_i| - 1)$ (and you can easily think of examples of this degree satisfying the condition). In general, this theorem gives you an explicit bound (it is easy to describe what the minimum value is) on the number of non-zeros of an arbitrary polynomial that does not vanish on the grid completely. 
Alon-Füredi also works in much more generality, where you can replace $F$ by any commutative ring with identity assuming that for all $i$, for all distinct $a, b$ in $S_i$, the element $a - b$ is not a zero divisor of the ring. In fact, one can generalise further by imposing extra conditions on the degrees of individual variables in the polynomial, see On Zeros of a Polynomial in a Finite Grid. The proof of Alon-Füredi and its generalisation is an easy induction argument using basic properties of single variable polynomials.<|endoftext|>
TITLE: reduced norm from degree 3 division algebra
QUESTION [6 upvotes]: Let $D$ be a degree $3$ division algebra over a field $k$ of char not 2 and 3.
Any such division algebra is cyclic. I am interested in knowing the cases when the reduced norm map $Nrd : D^* \rightarrow k^*$ is surjective. Of course, this happens over $\bar k$ and finite field etc. Here is my explicit question.
I want to relate surjectivity of reduced norm to the finiteness of $k^*/(k^*)^3$. To me it looks like not having enough of degree 3 field extensions is somehow responsible. I would appreciate examples, counterexamples or any reference in this direction.
Thanks a lot.

REPLY [7 votes]: The Merkurjev-Suslin Theorem says that an element $x \in k^*$ is a norm from $D$ if and only if $[D] \cup (x)$ is zero in the Galois cohomology group $H^3(k, \mathbb{Z}/3)$.  Therefore, for the reduced norm to be surjective for every division $k$-algebra of degree 3, it suffices to assume that $H^3(k, \mathbb{Z}/3) = 0$.
Conversely, if $k$ contains a primitive 3rd root of unity, then $H^3(k, \mathbb{Z}/3)$ is isomorphic to $K_3(k)/3$, i.e., degree 3 Milnor $K$-theory mod 3, and in particular is generated as an additive group by symbols of the form $[D] \cup (x)$.  So, in this case, for the reduced norm to be surjective for every division $k$-algebra of degree 3, it is also necessary that $H^3(k, \mathbb{Z}/3) = 0$.<|endoftext|>
TITLE: Do arbitrary $K$-twin primes exist?
QUESTION [8 upvotes]: Polignac's conjecture states that for any positive even integer $K$, there exist infinitely many pairs of primes such that their difference is $K$.
I am interested the status in a much weaker form of the conjecture: 

Is it true that for all even numbers $K$, there exist primes $p,q$
  with $p-q=K$?

REPLY [7 votes]: I think this is open just as much as the original conjecture. In fact, in analytic number theory, we usually prove the existence of certain objects by showing that there are many of them (certainly infinitely many, which can be further defined by density etc.).<|endoftext|>
TITLE: is the maximal tensor product of compact operators an essential ideal?
QUESTION [6 upvotes]: I'm searching for a counterexample for $C^*$-algebras $A$ and $B$ and essential ideals (I assume an ideal to be closed and only two-sided ideals) $I\subseteq A$, $J\subseteq B$ , such that the ideal $I\otimes_{max} J$ is not essential in the (maximal tensor product) -$C^*$-algebra  $A\otimes_{max} B$. I'm not sure if my idea works:
The algebra of compact operators $K(l^2(\mathbb{N}))$ is an essential ideal in the $C^*$-algebra of bounded linear operators on  $l^2(\mathbb{N})$, $B(l^2(\mathbb{N}))$. The reason is that $B(l^2(\mathbb{N}))$ has only the closed ideals $\{0\}$, $K(l^2(\mathbb{N}))$ and $B(l^2(\mathbb{N}))$, thus $K(l^2(\mathbb{N}))\cap M\neq 0$ for all closed nontrivial ideals $M\subseteq B(l^2(\mathbb{N}))$. 
But is $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))$ essential in $B(l^2(\mathbb{N}))\otimes_{max} B(l^2(\mathbb{N}))$? 
I don't think so, but I'm stuck to prove that there must be a nontrivial closed ideal $M$ in $B(l^2(\mathbb{N}))\otimes_{max} B(l^2(\mathbb{N}))$ such that $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))\cap M$ is trivial. For the proof it must be important that one takes the maximal norm-closure of $I\odot J\subseteq A\odot B$ ($\odot$ denotes the tensor product as $*$-algebras), because particularly the spatial tensor product satisfies that if $I$ is essential in $A$, $J$ essential in $B$, then $I\otimes_{min} J$ is essential in $A\otimes_{min} B$ (and here  is $B(l^2(\mathbb{N}))\otimes_{min} B(l^2(\mathbb{N}))\neq B(l^2(\mathbb{N}))\otimes_{max} B(l^2(\mathbb{N}))$. If my ideal don't work, what else can I do?
I appreciate your help.

REPLY [2 votes]: My suggestion (I hope that nothing is wrong):
Let $$F:B(l^2(\mathbb{N}))\otimes_{max} B(l^2(\mathbb{N}))\to B(l^2(\mathbb{N}))\otimes_{min} B(l^2(\mathbb{N}))$$ be the canonical map and define $M:=\ker(F)$. Since $F$ is not injective, $M$ is nontrivial. 
Claim: $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N})) \cap M = 0$.
First note that $K(l^2(\mathbb{N}))$ is nuclear (because it's a inductive limit of nuclear $C^*$-algebras) and simple, therefore $$K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))\cong K(l^2(\mathbb{N}))\otimes_{min} K(l^2(\mathbb{N}))$$ is simple.
It follows 
$K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N})) \cap M=\ker(F_{|K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))})=0$, because $F_{|K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))}$ is nonzero and $\ker(F_{|K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))})$ is an ideal in $K(l^2(\mathbb{N}))\otimes_{max} K(l^2(\mathbb{N}))$.<|endoftext|>
TITLE: Characterizing graphs whose incidence matrix has the all ones vector in its row span
QUESTION [5 upvotes]: Suppose we have a simple connected graph $G=(V,E)$. Then let $A$ be its $|E|\times |V|$ incidence matrix. Here I am considering the unoriented incidence matrix. I want to know when the row span of $A$ contains the all ones vector in $\mathbb{R}^{|V|}$. I believe this happens if and only if $G$ has a spanning regular subgraph. One direction is of course clear. Does anyone know if this is true/ have a  counterexample?

REPLY [5 votes]: No, this is not true.  Let $G$ be the bowtie graph (this is the graph obtained by gluing two triangles at a vertex $u$).  Then, $G$ does not have a spanning regular subgraph, but $\mathbb{1}$ is in the row space of $G$.  Just set $x_e=\frac{1}{4}$ if $e$ is adjacent to $u$ and $x_e=\frac{3}{4}$ for the other two edges.  
The exact characterization of when $\mathbb{1}$ is in the row space of $G$ can be extracted from Chris Godsil's answer.  
Characterization. $\mathbb{1}$ is in the row space of $G$ if and only if each bipartite component of $G$ is balanced (the left and right sides have the same number of vertices).

REPLY [4 votes]: The rank of the incidence matrix is $|V|$ minus the number of bipartite components. I assume the graph is connected. If it is not bipartite, it follows that the row space is $\mathbb{R}^{V}$ and hence it contains the all-ones vector. If the graph is bipartite, the vector that is 1 on the vertices in the first colour class and $-1$ on the other is orthogonal to each row. So the row space is the orthogonal complement to this vector.
[Edit to deal with Tony's point.] If the colour classes are of equal size, this signed vector is orthogonal to the all-ones vector, and so the all-ones vector is in the row space. Otherwise the all-ones vector is not orthogonal to our signed vector, and so it does not lie in the row space.
In summary if a connected graph is not bipartite, or bipartite with colour classes of equal size, the row space contains the all-ones vector. Otherwise it does not.<|endoftext|>
TITLE: Law of sines for tetrahedra
QUESTION [8 upvotes]: Wikipedia gives a generalization of the law of sines to higher dimensions, as defined in F. Eriksson, The law of sines for tetrahedra and n-simplices. However, this generalization misses an important point about the standard law of sines, which relates it to the radius of the circumcircle of the triangle.
Is there a property which generalizes this relation of the 2-dimensional law of sines? In other words: is there a constant relation of this kind that all tetrahedra inscribed in a sphere of the same radius have in common?

REPLY [7 votes]: Yes. See S. Yang's 2004 paper The generalized sine law and some inequalities for simplices.<|endoftext|>
TITLE: Neighbourhood of a word and Levenshtein distance
QUESTION [5 upvotes]: The Levenshtein distance or Edit distance $$ lev(U,V) $$ between two strings $U$ and $V$ over a finite alphabet $\Sigma$ of size $ \left| \Sigma \right| = \sigma ,$ is the minimal number of insertions, deletions  and replacements  to make the strings equal.
For $k \in \mathbb{N} $ and $U \in  \Sigma^*$ we define :
$$ N_k(U) = \{ V \in  \Sigma^* : lev(U,V)  \leq k \}  $$
For example if $\Sigma = \{A,B,L\}$ : 
\begin{eqnarray*}
 N_1(AAA) &=& \{ AAA, AA, AAB, AAL, ABA, ALA, BAA, LAA, AAAA, BAAA,  \\ &&LAAA, ABAA, ALAA, AABA, AALA, AAAB, AAAL \}  \\
N_1(LAB) &=& \{LAB, LA, AB, LB, AAB, BAB, LBB, LLB, LAA, LAL, ALAB, \\
&&BLAB, LLAB, LAAB, LBAB, LALB, LABA, LABB, LABL\} 
\end{eqnarray*}
My goal is to compute for $\Sigma, n $ and $k$ fixed  :
$$ \max_{U \in  \Sigma^* , ~ \left| U \right|  = n}  \left|N_k(U) \right| $$
or an upper bound $f(\Sigma,k,n)$.
I conjecture that the maximum is reached for the words of the form :
$$ c_1c_2 \dots c_{\sigma}c_1c_2 \dots c_{\sigma}\dots $$
where $\Sigma = \{  c_1,c_2, \dots,c_{\sigma}  \}. $
Does anyone have an idea of how to compute $ \max_{U \in  \Sigma^* , ~ \left| U \right|  = n}  \left|N_k(U) \right| $ or to prove this conjecture?
A special case has already been studied in the section 6 of the paper What's Behind Blast by Gene Myers.

REPLY [2 votes]: A recent paper by Hélène Touzet discusses how to compute the size of the neighborhood of a given word via the Universal Levenshtein automaton.
"On the Levenshtein Automaton and the Size of the Neighbourhood of a Word" (unfortunately not freely available online yet).
Maybe it is possible to exploit the structure of the universal DFA or to at least do some practical experiments regarding your conjecture.<|endoftext|>
TITLE: Showing the positivity of a singular integral that came up in circle method
QUESTION [5 upvotes]: Let $F(\mathbf{x}) \in \mathbb{Z}[x_1, ..., x_n]$ be a degree $d$ homogeneous form. Let
$$
I(\alpha) = \int_{[0,1]^n} e^{2 \pi i F(\mathbf{x}) \alpha} dx_1...dx_n.
$$
Then the singular integral is defined as 
$$
\sigma_{\infty} = \int_{\mathbb{R}} I(\alpha) d\alpha.
$$
It says in an article I am reading that $\sigma_{\infty} > 0$
if the equation $F(\mathbf{x}) =0$ has a non-singular real solution in $(0,1)^n$.
I would greatly appreciate if someone could give me a hint or some explanation on how I can prove this statement. Thank you very much.

REPLY [2 votes]: Here's a sketch. First of all, proceeding formally, if we start with the $\alpha$ integral, then we obtain
$$
\sigma_{\infty} = \int \delta(F(x))\, dx ,
$$
and near an $a$ with $F(a)=0$, $\nabla F(a)\not= 0$, this looks like $\delta(v\cdot t)$, so we obtain a positive contribution.
To make a proof out of this, fix a $\varphi\ge 0$ with $\widehat{\varphi}\ge 0$, $\varphi(0)=1$ (Fejer kernel would work), and consider
$$
\int I(\alpha)\varphi(\alpha/n)\, d\alpha = n \int \widehat{\varphi}(nF(x))\, dx .
$$
Now near an $a=(b,c)$ as above, with $b\in\mathbb R$, $c\in\mathbb R^{d-1}$, we find solutions to $F=0$ in a whole neighborhood $U\subseteq \mathbb R^{d-1}$ of $c$ (if $\partial F/\partial x_1(a)\not=0$), by the implicit function theorem. As before, integration over $(a-\delta,a+\delta)\times U$ gives a contribution that stays positive in the limit $n\to\infty$.<|endoftext|>
TITLE: Flat connections on 3-manifold with boundary
QUESTION [8 upvotes]: Suppose $Y$ is a 3-manifold and the boundary $\Sigma:=\partial Y$ is non-empty. Let $G$ be a Lie group with trivial center. Let $\overline {\mathcal A}_{flat}(\Sigma)$ and $\overline {\mathcal A}_{flat}(Y)$ denote the gauge equivalence classes of flat connections on the trivial bundles $\Sigma \times G$ and $Y \times G$ respectively. The space $\overline {\mathcal A}_{flat}(\Sigma)$ is a finite dimensional symplectic manifold with singularitites.  I am trying to understand why the image of the restriction map $\overline {\mathcal A}_{flat}(Y) \to \overline {\mathcal A}_{flat}(\Sigma)$ is a Lagrangian submanifold.
I know why the image is an isotropic subspace, but I am having a problem with the dimensions. In particular, the space
$$\overline {\mathcal A}_{flat}(\Sigma)\simeq \operatorname{Hom}(\pi_1(\Sigma),G)/G$$
has dimension $(2\operatorname{genus}(\Sigma)-2)\dim(G)$, but I can not see why the space $\operatorname{Hom}(\pi_1(Y),G)/G$ has half that dimension.
For completeness, I include a proof of why the image of $\overline {\mathcal A}_{flat}(Y)$ in $\overline {\mathcal A}_{flat}(\Sigma)$ under the restriction map is isotropic.
Let $\mathcal A(Y)$ be the space of connections on $Y \times G$. There is a one-form $\mathcal F$ on $\mathcal A(Y)$ defined as 
$${\mathcal F}_A(\alpha):=\int_Y \langle \alpha \wedge F_A\rangle,$$
where $A \in \mathcal A(Y)$ is a connection, $F_A \in \Omega^2(Y,\mathfrak g)$ is its curvature and $\alpha \in T_A\mathcal A(Y)$. We can see by some calculations that
$$d\mathcal F_A(\alpha,\beta)=\int_\Sigma \langle \alpha \wedge \beta \rangle,$$
where $\alpha, \beta \in T_A\mathcal A(Y)$, which coincides with the symplectic form on $\mathcal A(\Sigma)$ the space of connections on $\Sigma \times G$. When restricted to $\mathcal A_{flat}(Y)$, $\mathcal F$ vansihes, and so does $d\mathcal F$. Therefore the restriction map takes $\mathcal A_{flat}(Y)$ to an isotropic submanifold of $\mathcal A_{flat}(\Sigma)$. By gauge invariance, the restriction map takes $\overline {\mathcal A}_{flat}(Y)$ to an isotropic submanifold of $\overline {\mathcal A}_{flat}(\Sigma)$.

REPLY [8 votes]: Another good reference is Chris Herald's paper. Legendrian cobordism and Chern-Simons theory on 3-manifolds with boundary. Comm. Anal. Geom. 2 (1994), no. 3, 337–413. 
It is an easy exercise with Poincare duality to see that the image of the map in cohomology (so at the level of "Zaraski" tangent spaces)
$$
H^1(Y;ad_\rho) \to H^1(\Sigma;ad_\rho)
$$
is half dimensional.  Indeed this map appears in the long exact sequence 
of the pair $(Y,\Sigma)$ 
$$
H^1(Y;ad_\rho) \to H^1(\Sigma;ad_\rho)\to H^2(Y,
\Sigma;ad_\rho)
$$
and the two arrows are dual by PD so have the same rank. The exactness tells you
the sum of these ranks is $dim(H^1(\Sigma;ad_\rho))$.
Here $\rho$ is a representation of $\pi_1(Y)\to G$ corresponding to the given flat connection and $ad_\rho$ is the local system with fiber $\mathfrak{g}$ corresponding to the adjoint action of $\rho$ on $\mathfrak{g}$.<|endoftext|>
TITLE: On $\eta(6z)\eta(18z)$ and the splitting / modularity of $x^3 - 2$
QUESTION [12 upvotes]: Consider one of the simplest non-abelian examples of modularity. Let $$\eta(6z)\eta(18z) = q\prod_{n=1}^\infty (1 - q^{6n})(1 - q^{18n}) = q - q^7 - q^{13} -q^{19} + q^{25} + 2q^{31} - q^{37} + 2q^{43} - q^{61} - q^{67} - q^{73} - q^{79} + q^{91} - q^{97} - q^{103} \dots = \sum_{n=1}^\infty a_n q^n$$
Then $a_p + 1 =$ the number of solutions of $\{x^3 - 2 = 0\}$ in $\mathbb{Z}/p\mathbb{Z}$, for prime $p > 3$.
1) What might be the best method to prove this? 
2) Is it possible to prove it without knowledge of modular forms? i.e. without using the dimension / basis of the relevant space of moduli forms of some weight and level and nebentypus. For example, can it be explained using Weyl–Kac character formula etc.?
3) Unfortunately LMFDB does not yet have a list of weight $1$ modular forms and their corresponding Artin representations. Where can we find more of such weight $1$ examples?

REPLY [15 votes]: Here's an answer to 2. You can tell me if it's the "best method".
Euler's Pentagonal number theorem gives that
$$ \eta(24z) = \sum_{n \in \mathbb{Z}} (-1)^{n} q^{(6n+1)^{2}}. $$
This yields the formula
$$ \eta(6z) \eta(18z) = \sum_{m, n \in \mathbb{Z}} (-1)^{m+n} q^{\frac{(6n+1)^{2} + 3(6m+1)^{2}}{4}}. $$
If $m=x$ and $n = x+2y$, then $m+n$ is even and we get
$$ \sum_{x, y \in \mathbb{Z}} q^{(6x+3y+1)^2 + 27y^{2}}. $$
On the other hand if $m=x$ and $n=x+2y+1$, then $m+n$ is odd and we get
$$ -\sum_{x,y \in \mathbb{Z}} q^{4(3x+y+1)^{2} + 2(3x+y+1)(2y+1) + 7(2y+1)^{2}}.$$
The upshot (skipping a few small details) is that
$$ \eta(6z) \eta(18z) = \frac{1}{2} \left[ \sum_{x,y \in \mathbb{Z}} q^{x^{2} + 27y^{2}} - q^{4x^{2} + 2xy + 7y^{2}} \right]. $$
Now, Gauss proved that if $p \equiv 1 \pmod{3}$ is prime, then there is some $z$ so that $z^{3} \equiv 2 \pmod{p}$ if and only if $p = x^{2} + 27y^{2}$ for some $x, y \in \mathbb{Z}$. It follows from this that the $p$th coefficient of $\eta(6z) \eta(18z)$ is $2$ if $p \equiv 1 \pmod{3}$ and $2$ is a cube modulo $p$. 
If $p \equiv 1 \pmod{3}$, but $2$ is not a cube modulo $p$, then $p$ is represented by $4x^{2} + 2xy + 7y^{2}$ in two ways, and this means that the $p$th coefficient of $\eta(6z) \eta(18z)$ is $-1$.
If $p \not\equiv 1 \pmod{3}$, the $p$th coefficient of $\eta(6z) \eta(18z)$ is clearly $0$.<|endoftext|>
TITLE: In what sense are operads "better" than PROPs?
QUESTION [9 upvotes]: I'm a newcomer to operads so apologies if this is a naive question.
The standard picture of an operad is of a collection of $n$-ary operations, thought of as objects with $n$ upward-pointing legs (inputs) and one downward-pointing leg (output), which can be composed in a sensible way.  "Algebras over operads" then give rise to all sorts of algebras in the usual sense (associative, $A_\infty$ etc).
My understanding is that a PROP is basically the same thing but with operations allowed to have more than one output.  Algebras over PROPs can also accommodate coproduct-like operations, and hence also include useful things like coalgebras and Hopf algebras.
Historically though, operads came later than PROPs.  So why was it useful to single out the notion of an operad, as opposed to a general PROP?

REPLY [19 votes]: I was friends with Frank Adams and Saunders Mac Lane, who invented PROPs in one of the world's most extensive unpublished collaborations. Saunders once showed me a box full of their correspondence. One reason they never published is that they lacked a way of showing the PROPs they were interested in acted on the things they wanted to have actions.  Operads are of course equivalent to a special kind of PROP, and the specialization made it very much easier to find operad actions.  The connection with monads was intrinsic to the definition of operads (I convinced Mac Lane to switch from "triples'' to "monads" in Categories for the Working Mathematician in large part in order to make the names operad and monad to mesh) and operads allowed tons of explicit computations (in the homology of iterated loop spaces in particular) that would not have not come naturally if at all from PROPs.  Of course, there are interesting examples of PROPs of the more general sort, ones that do not come from an operad, but they are irrelevant to the original work with iterated loop spaces.<|endoftext|>
TITLE: Coefficient-wise powers of matrices. Reference wanted
QUESTION [10 upvotes]: Let $K$ be a commutative field and ${\rm M}_n (K)$ be the ring of $n\times n$ square matrices with coefficients in $K$ ($n\geqslant 1$ is an integer). For $k\geqslant 1$ and $A =(a_{ij})_{1\leqslant i,j\leqslant n}\in {\rm M}_n (K)$, define: $A^{[k]} =(a_{ij}^k )_{1\leqslant i,j\leqslant n}$. 

Is the description of all matrices $A\in {\rm M}_n (K)$ satisfying $A^k =A^{[k]}$,  for all $k\geqslant 1$,  known? If yes do you have a reference ?

REPLY [17 votes]: This is problem 50.4 by Moubinool Omarjee from volume 50 of The Bulletin of the International Linear Algebra Society, with solutions in volume 51 by Eugene Herman and Bojan Kuzma, and further work by Roman Drnovsek in When powers of a matrix coincide with its Hadamard powers.
For real matrices the solution is<|endoftext|>
TITLE: The number of maximal subgroups up to isomorphism
QUESTION [21 upvotes]: Every maximal subgroup of infinite index of a free non-cyclic group $F_k$ is free of countable rank. Thus even though the set of maximal subgroups of $F_k$ is uncountable, there are only countably many isomorphism classes of such subgroups (including subgroups of finite index). The same is true for surface groups of genus at least 2. Many other finitely generated groups have at most countable set of isomorphism classes of maximal subgroups. 
Question: Is there a finitely generated group with uncountably many isomorphism classes of maximal subgroups?

REPLY [26 votes]: Yes. Let $F$ be a nonabelian f.g. free group, then  $F\times F$ has this property. Indeed, for each normal subgroup $N$ of $F$, denote $H_N$ the subgroup $\{(x,y), xN=yN\}$ of $F\times F$ (this is a fibre product $F\times_{F/N} F$). Then $H_N$ contains the normal subgroup $N\times N$, and the quotient $H_N/(N\times N)$ is isomorphic to $F/N$. This normal subgroup can be intrinsically defined in $H_N$, because it is generated by $N\cup\{1\}\cup \{1\}\times N$, which is the set of elements of $H_N$ whose centralizer in $H_N$ is nonabelian. Hence the isomorphism class of $H_N$ determines the isomorphism class of $F/N$. Moreover, if $F/N$ is simple then $H_N$ is maximal in $F\times F$. Since there are uncountably many non-isomorphic 2-generated simple groups, we deduce that $F\times F$ has uncountably many non-isomorphic maximal subgroups.

Edit: here's a reference for the fact that there exists uncountably many non-isomorphic 2-generated simple groups:
Camm, Ruth.
Simple free products. 
J. London Math. Soc. 28, (1953). 66-76.
[MR review behind paywall: (...) we have a continuum of non-isomorphic simple groups $G(\rho,\sigma,\tau)$. Each of these is a free product, with an amalgamated subgroup, of two free groups, and so is, in particular, locally infinite. Moreover, each can be generated by two elements.]<|endoftext|>
TITLE: Is this simple-looking moment inequality true?
QUESTION [7 upvotes]: Let $p \ge 1$ be an integer. Does there exist a constant $C_p$ such that for every random variable $X \ge 0$,
$$
\mathbb{E} \left[ \left(X - \mathbb{E} \left[ X \right] \right)^{2p} \right] \le C_p \mathbb{E} \left[ \left(X^p - \mathbb{E} \left[ X^p \right] \right)^{2} \right] \ \ ?
$$

REPLY [6 votes]: The answer is yes. Let $Y$ be an independent random variable distributed as $X$. We have
$$
\|X - \mathbb{E}[X]\|_{L^{2p}}  = \|\mathbb{E}\left[X - Y \, | \, X\right]\|_{L^{2p}}  \le \|X - Y\|_{L^{2p}}.
$$
Moreover, there exists a constant $C_p$ such that for every $x,y \ge 0$,
$$
|x-y|^p \le C_p |x^p - y^p|.
$$
Indeed, it suffices to verify this for $x = 1$ and $y \in [0,1]$ by homogeneity and symmetry. This is then a simple exercise. As a consequence, we deduce
$$
\|X - \mathbb{E}[X]\|_{L^{2p}} \le C_p \||X^p - Y^p|^{1/p}\|_{L^{2p}} = C_p \|X^p - Y^p\|_{L^{2}}^{1/p}.
$$
We conclude by the triangle inequality.<|endoftext|>
TITLE: What is the meaning of this analogy between lattices and topological spaces?
QUESTION [34 upvotes]: Let me add one more edit to help explain why this is a serious question. Theorem 5 below is a sort of lattice version of Urysohn's lemma, and it has essentially the same proof. Theorem 6, the famous theorem of Raney characterizing completely distributive lattices, is an easy consequence of this result. In other words, a result of basic importance in lattice theory is somehow at heart "the same" as Urysohn's lemma. But "the same" in what sense?

I noticed this weird analogy a long time ago. Is there a good explanation?
A lattice is a partially ordered set in which every pair of elements has a least upper bound and a greatest lower bound. Now let me set up some very nonstandard terminology which will show what I am talking about. Say that a lattice is compact if every subset has a least upper bound (join) and a greatest lower bound (meet). A map between lattices is continuous if it preserves arbitrary (i.e., possibly infinite) joins and meets whenever these exist in the domain. A subset $C$ of a lattice is closed if $$x \in C,\,x \leq y \quad\Rightarrow\quad y \in C$$ and $C$ is stable under the formation of arbitrary meets whenever these exist. A subset $U$ is open if $x \in U,\,x \leq y \,\Rightarrow \,y \in U$ and the complement of $U$ is stable under the formation of arbitrary joins whenever these exist. A subset is clopen if it is both closed and open. A lattice is totally disconnected if for every $x \not\leq y$ there is a clopen subset containing $x$ but not $y$. A lattice is Hausdorff if for every $x \not\leq y$ there exist a closed set $C$ and an open set $U$ whose union is the whole space and such that $x \in U$, $y \not\in C$.
Note that "closed" and "open" do not refer to any topology. For instance, the union of two closed sets need not be closed. I am aware that every poset carries a natural topology, but that does not seem particularly relevant here.
Now here are some theorems.
Theorem 1. Any union of open sets is open, and any intersection of closed sets is closed.
Theorem 2. A map between two spaces is continuous if and only if the inverse image of any closed subset is closed and the inverse image of any open set is open.
Theorem 3. Any continuous image of a compact space is compact.
Theorem 4. A space is totally disconnected if and only if it embeds in a power of the two-element space.
Theorem 5. If a space is compact Hausdorff and $x \not\leq y$ then there is a continuous map into $[0,1]$ taking $x$ to $1$ and $y$ to $0$.
Theorem 6. Every compact Hausdorff space embeds in a power of $[0,1]$.
Theorem 7. Every compact Hausdorff space is the continuous image of a totally disconnected compact Hausdorff space.
(In the usual terminology, a "compact Hausdorff" lattice is a completely distributive complete lattice.)
Each of these statements is true both of topological spaces (replacing $x \not\leq y$ with $x \neq y$ in Theorem 5) and, with my terminology, also of lattices. I am not sure what form a satisfying explanation would take. Is there a broader theory of which topological spaces and lattices are both special cases? Is there some other way of understanding this? Or is it not as remarkable as it seems, and needs no special explanation.
(I'm including a category theory tag because I suspect that may be an arena in which an explanation could be found.)
Edit: I'm getting some feedback (Wallman's generalization of Stone duality, Scott continuity) describing other, as far as I can tell unrelated, connections between lattice theory and topology. Obviously that's not what I'm asking for.
Maybe I should emphasize that Theorem 6 is a serious result about completely distributive lattices, discovered in 1952 by G. N. Raney. From my point of view it is rather easy ... if anyone can show me how to get it out of Wallman's theory, I will retract the preceding comment.

REPLY [10 votes]: EDIT: I mentioned in the comments that Nik's list of definitions and theorems looks to me strikingly similar to Wallman's generalization of Stone duality.
After getting some feedback from Nik and taking some time to think about the problem more, the connection seems less direct than I previously suggested. However, I do still feel that there is a strong connection between Wallman's theory and Nik's, and I have edited my post to explain this as best I can.
This post does not completely answer Nik's question, but I hope it provides a useful partial answer.
Wallman's Construction:
The main difference between Wallman's theory and yours is that you seem to be thinking of lattice elements as points, whereas Wallman thought of them as closed sets.
I can't find a clear and thorough account of Wallman's results online except for his original paper (which is a bit lengthy, and uses some outdated terminology)

H. Wallman, ``Lattices and topological spaces," Annals of Mathematics vol. 39 (1938), pp. 112 - 126, available here.

I'll outline the construction here.
To every "compact, Hausdorff" lattice $L$ we may associate a topological space called $W(L)$. The points of $W(L)$ are not the members of $L$, but rather the ultrafilters on $L$. The idea is to put a topology on $W(L)$ in such a way that $L$ becomes (isomorphic to) the lattice of closed subsets of $W(L)$.
To put a topology on $W(L)$, let each $a \in L$ define a closed set
$$C_a = \{p \in W(L) : a \in p\}.$$
In other words, $C_a$ is the set of all ultrafilters containing $a$. With your definition of a compact lattice, it's not hard to check that these sets constitute the closed subsets of a topology on $W(L)$.
(Sidebar: If you don't require that every subset of $L$ has a greatest lower bound, then this construction will still work. Instead of giving you a topology, the sets $C_a$ will then define a basis for the closed sets of a topology. To see this idea at work, I recommend the first part of section 1 of this paper of Dow and Hart.)
Given a closed set $C$ in $W(L)$, we obtain a "closed" subset of $L$ (in your terminology) by considering $\{a \in L : C \subseteq C_a\}$. Given a "closed" subset $C$ of $L$, we obtain a closed subset of $W(L)$ corresponding to it, namely $\bigcap_{a \in C}C_a$.
As Nik points out in the comments, the converse is not necessarily true: in general, the lattice of closed subsets of a compact Hausdorff space is not necessarily a "compact Hausdorff" lattice.
Why I think Wallman's idea is relevant here:
Using Wallman's construction and known facts from point-set topology, we can derive "near-miss" versions of some of Nik's theorems. For example, we can get a weaker version of Theorem 5, where we drop the continuity requirement but still require our map to be order-preserving:
Theorem: Suppose $L$ is a "compact Hausdorff" lattice and $a \not\leq b$ are in $L$. Then there is an order-preserving map from $L$ into $[0,1]$ taking $a$ to $1$ and $b$ to $0$.
Proof: Let $X = W(L)$ be the compact Hausdorff space whose lattice of closed sets is isomorphic to $L$. In $X$, $a$ and $b$ correspond to the nonempty closed sets $C_a$ and $C_b$, and the relation $a \not\leq b$ translates to $C_a \not\subseteq C_b$. Let $x$ be a point of $X$ that is in $C_a$ but not in $C_b$. Since $X$ is $T_{3\frac{1}{2}}$, there is a continuous function $f: X \rightarrow [0,1]$ such that $f(x) = 1$ and $f(C_b) = 0$. Let $\varphi: L \rightarrow [0,1]$ be defined by
$$\varphi(d) = \sup \{f(z) : z \in C_d\}.$$
Clearly $\varphi(a) = 1$ and $\varphi(b) = 0$, and $\varphi$ is order-preserving. However, it is easy to come up with examples where $\varphi$ is does not preserve arbitrary meets and joins (hence the "near-miss").
Similarly, we can get a near-miss version of Theorem 6:
Theorem: If $L$ is a "compact Hausdorff" lattice, then there is an order-preserving injection from $L$ into a power of $[0,1]$.
Proof: Let $I = \{(a,b) \in L \times L : a \not\leq b\}$. For each $(a,b) = i \in I$, fix an order-preserving function $\varphi_i$ as in the previous theorem. Define $\varphi: L \rightarrow [0,1]^I$ by $\pi_i \circ \varphi(a) = \varphi_i(a)$. This function is clearly order-preserving. For injectivity, consider that for all $a \neq b \in L$, either $a \not\leq b$ or $b \not\leq a$.
We get a near-miss version of Theorem 4 in the same way (first prove a $0$-dimensional analogue of Theorem 5, stating that the function in question can be assumed to be two-valued).<|endoftext|>
TITLE: Is there a complete Riemannian manifold with infinite volume whose the time-one map of the geodesic flow is recurrent?
QUESTION [11 upvotes]: Let M be complete Riemannian manifold M with infinite volume, it is know that the geodesic flow, $\varphi^t:T^1M \rightarrow T^1M$  preserves the Liouville measure $\mu$, that is, $\mu(\varphi^t(A)) = \mu(A)$ for every $t \in \Bbb{R}$ and for all borelian set A. In particular, the time-one map $\varphi^1$ also preserves the measure $\mu$ because $\mu(\varphi^{-1}(A)) = \mu(A)$.
We say that the map $\varphi^1$ is recurrent (or conservative) if all wandering set W for $\varphi^1$( meaning that $W \cap \varphi^{-n}(W) = \varnothing  $ for $n \geq 1$) has necessarily $\mu(W) = 0$.
An another equivalent definition is, 
if A is borelian set with $\mu(A) > 0$  then
$$      A \subset \displaystyle\bigcup_{n \geq 1} \varphi^{-n}(A) \ \  (mod \ \ \mu)                               $$ 
Question: Is there a complete Riemannian manifold M with infinite volume whose the time-one map of the geodesic flow is recurrent ?

REPLY [12 votes]: Take a compact, connected Riemannian manifold $M$ with negative sectional curvature. Then choose any cover $M'$ of $M$ which is connected, Galois, and whose group of deck transformations is $\mathbb{Z}$ or $\mathbb{Z}^2$. 
In dimension $2$ and for a $\mathbb{Z}$ cover, this can be done by taking $\mathbb{Z}$ copies of the manifold $M$, cutting along an essential curve (in dimension $2$). Then we get manifold with boundaries which can be indexed by $\mathbb{Z} \times \{0,1\}$. Glue $(p,0)$ with $(p+1,1)$.
The geodesic flow on $T^1 M'$ behaves roughly like a random walk on $\mathbb{Z}$ (or $\mathbb{Z}^2$) with centered and bounded increments, so it is recurrent.
There are other examples in the same fashion, for instance $\mathbb{C} \setminus \mathbb{Z}$ endowed with a $\mathbb{Z}$-invariant hyperbolic metric.
See for instance:

J. Aaronson, M. Denker, Distributional limits for hyperbolic, infinite volume geodesic flows (Tr. Mat. Inst. Steklova 216 (1997), Din. Sist. i Smezhnye Vopr., 181--192)
J. Aaronson, M. Denker, The Poincar\'e series of $\mathbb{C}\setminus\mathbb{Z}$ (Ergodic Theory Dynam. Systems 19 (1999), no. 1, 1--20.)<|endoftext|>
TITLE: Fourier transform surjective on $L^p(\mathbb{R}^n)$ for $p \in (1,2)$?
QUESTION [5 upvotes]: I know that $F_2:L^2 \rightarrow L^2$ is of course unitary, whereas $F_1:L^1 \rightarrow C_0$ is injective but not surjective. This can be seen by looking at the dual map.
Riesz-Thorin gives us that there is also $F_p: L^p \rightarrow L^q$ for $p \in (1,2).$ Here, the dual map trick does not work, so this transform has a chance of being surjective. Since every $f \in L^q$ is also in $S'$ we can also define a promising candidate $F_{S'}^{-1}(f).$ Unfortunatly, this does not really tell me whether this $F_{S'}^{-1}(f) \in L^p$ again. 
This raises the question whether $F_p$ is actually surjective or not?
Comment on the discussion below: Thanks to everybody participating in the disccusion. Actually this question came to my mind while I was thinking about this problem from PDEs, which would have an easy solution in this case. I have to admit that the fact that $L^p$ is not isomorphic to $L^q$ is indeed something I know, but I have never actually used it, as I am not primarily active in analysis. Probably I should give my questions more thought in the future.Sorry for any inconvenience my question caused.

REPLY [6 votes]: If $1\leq p<2$ then $\mathscr{F}: L^p \to L^{p'}$  is not surjective.  I had this as a homework problem a week back.
The reason is the bounded inverse theorem:  $\mathscr{F}: L^p \to L^{p'}$ is injective, (by fourier inversion on the dense subspace of schwarz functions).  If the map were surjective then there would be an inverse that would be continuous, since $\mathscr{F}$ is an open map under this assumption.  
Thus we just need to prove that there is no bounded inverse:  For $f \in \mathscr{S}$, there is no constant $c$ such that $ ||f||_{p} \leq c ||\hat f||_{p'}$ for $f \in \mathscr{S}$ with the constant only depending on $p$.    
This is easy:  The function $f_\lambda=e^{-\pi i \lambda x^2-\pi x^2}$ satisfies $||f_\lambda||_p=c$ independent of $\lambda$, whereas $||\hat f_\lambda||_{p'} \leq c \lambda^{1/p'-1/2}$.  But there is no constant such that $c \leq \lambda^{1/p'-1/2} $ for all $\lambda >0$.  Therefore the fourier transform is not surjective from $L^p \to L^{p'}$ for $1\leq p<2$<|endoftext|>
TITLE: Homotopy of space of immersions, Smale-Hirsch theorem
QUESTION [10 upvotes]: If $M$ and $N$  are  manifolds with $\dim M< \dim N$, we denote by $Imm\left(M,N\right)$ the space of immersions of $M$ in $N$.
Let $M$ and $M'$ be a simply connected manifolds of dimensions $m>0$. It is true that if $M$ is homotopic to $M'$, then for $k\geq m$, the spaces $Imm\left(M,\mathbb{R}^{m+k}\right)$ and $Imm\left(M',\mathbb{R}^{m+k}\right)$ are homotopic? i.e
if  $k\geq m$, then $M\simeq M'\Rightarrow Imm\left(M,\mathbb{R}^{m+k}\right)\simeq Imm\left(M',\mathbb{R}^{m+k}\right)$?
Thanks
Abdoul

REPLY [2 votes]: Note that when $M$ and $M′$ are simply connected, I have a positive answer in the rational case. I.e  
If $k\geq m$ is an odd integer, then $M\simeq M'\Rightarrow Imm\left(M,\mathbb{R}^{m+k}\right)\simeq_{\mathbb{Q}}Imm\left(M',\mathbb{R}^{m+k}\right).$
Here $"\simeq_{\mathbb{Q}}"$ denotes the rational homotopy equivalence.<|endoftext|>
TITLE: The number of prime factors of a natural number
QUESTION [5 upvotes]: It seems to be true that OEIS sequences A001222 and A257091 are closely related. First one is the number of prime divisors of n counted with multiplicity. The second one is the logarithms to the base 5 of the denominators of the Dirichlet series of zeta(s)^(1/5). Actually first terms of these sequences satisfy A257091(n)-A001222(n)= 1 if n is a multiple of 32, and 0 otherwise. Is it true, obvious, known or somthing else?

REPLY [11 votes]: Robert Israel's guess is almost correct, actually for $n=\prod p_j^{e_j}$ we have
$$
A257091(n)−A001222(n)=\sum_j \nu_5(e_j!)=\sum_j \sum_{m=1}^{\infty} \lfloor e_j/5^m\rfloor
$$
(where $\nu_5(N)$ denotes the largest $k$ for which $5^k$ divides $N$).
Indeed, expanding as binomial each Euler multiple in $\zeta^{1/5}$  we get $$(1-p^{-s})^{-1/5}=\sum_{k=0}^\infty (-1)^k\binom{-1/5}k p^{-ks}.$$
Thus if $n=\prod p_i^{e_i}$, coefficient of $n^{-s}$ in $\zeta(s)^{1/5}$ equals $$\prod_j (-1)^{e_j}\binom{-1/5}{e_j}=\prod_{j}\frac{1(1+5)(1+2\cdot 5)\dots (1+(e_j-1)5)}{5^{e_j} e_j!}$$
This rational number have only 5's in the denominator (because for other prime $p$ we may replace $-1/5$ to some integer like $(p^{4n}-1)/5$ without changing $p$-adic valuation of all multiples $-1/5-t$, $t=0,1,\dots,e_j-1$.) And 5 arises in the power $\sum e_k$ plus exponent of 5 in $\prod e_j!$.

REPLY [5 votes]: The first two exceptions to your rule:  $A257091(243) - A001222(243) = 6 - 5 = 1$ and $A257091(486) - A001222(486) = 7 - 6 = 1$, but $243$ and $486$ are not multiples of $32$. 
EDIT: Of course $243 = 3^5$.  From computing the results up to $n=10000$, it
looks to me like if $n = \prod_{j} p_j^{e_j}$ is the prime factorization of $n$, $A257091(n) - A001222(n) = \sum_j \lfloor e_j/5 \rfloor$.
  Thus $7776 = 2^5 \cdot 3^5$ and 
$A257091(7776) - A257091(7776) = 12 - 10 = 2$.<|endoftext|>
TITLE: Poincaré Duality for non-compact manifolds without Zorn's Lemma
QUESTION [7 upvotes]: Does exists a proof of the Poincaré Duality version for non-compact manifolds without using the Zorn's Lemma? I know that there is a proof using the Whitney embedding theorem, but I don't know this theorem's proof, so I don't know if it uses the Zorn's Lemma too.

REPLY [3 votes]: A slight modification of Hatchers Proof gives you what you want:
Hatcher relies on the fact that if $M$ is a union of opens totally ordered by inclusion, which satisfy PD, then $M$ does, too. It is easy to see that you can weaken that to the opens only being directed. Also, another case tells you that all finite intersections of coordinate balls satisfy PD, so all finite unions do too. Then, at the end instead of using Zorn's Lemma you only have to look at the cover of all finite unions of coordinate balls, which are obviously directed w.r.t. inclusion.<|endoftext|>
TITLE: Verify that a group is hyperbolic via computer algebra
QUESTION [11 upvotes]: I would like to know whether there is some computer algebra software that can be used to verify if a group, given by a finite presentation, is hyperbolic (in the sense that it terminates with "yes" if the group is hyperbolic and otherwise might not terminate). I read that the kbmag software package for GAP could be used for this task. However I did not find any information in the documentation about how this can be accomplished.
I would also be interested to know if there is some software (say a GAP package) to check if a given group presentation satisfies a small cancellation condition (like $C'(\lambda)$).

REPLY [11 votes]: The KBMAG package can be used to verify hyperbolicity of a group defined by a finite presentation. It does it by verifying that geodesic bigons in the Cayley graph are uniformly thin. Then a result of Paposoglu implies that the group is hyperbolic. So it proves hyperbolicity, but it does not provide any useful information aboutn the constant of hyperbolicity.
Unfortunately, whoever designed the GAP interface (well me) did not provide a convenient interface to this functionality. But it is possible to use the GAP Exec command to do this. Here is a simple with the group $\langle x,y \mid x^2=y^3=(xy)^7=1 \rangle$. The KBMAG package needs to be compiled. I am running it on Linux, and I don't know whether it will work properly with other operating systems.
LoadPackage("kbmag");;
F:=FreeGroup(2);; rels:=[F.1^2, F.2^3, (F.1*F.2)^7];; G:=F/rels;;
R:=KBMAGRewritingSystem(G);;
AutomaticStructure(R);;
WriteRWS(R,"237",";");;
progname := Filename(_KBExtDir,"autgroup");;
callstring := Concatenation(progname," 237");;
Exec(callstring);
progname := Filename(_KBExtDir,"gpgeowa");;
callstring := Concatenation(progname," 237");;
Exec(callstring);

If the final command completes successfully then it has succeeded in verifying that geodesic bigons are uniformly thin. The output of the last command should be something like:
#Geodesic word found not accepted by *geowaptr.
#Geodesic word-acceptor with 54 states computed.
#Geodesic pairs machine with 114 states computed.
#Geodesic difference machine with 31 states computed.<|endoftext|>
TITLE: Deformation Quantization
QUESTION [5 upvotes]: I am a beginner and I want to learn about deformation quantization. Please suggest me with which book or notes, I should start?

REPLY [6 votes]: In addition to the references pointed out by Stefan, I would like to add  

Déformation, quantification, et théorie de Lie, by Catteno, Keller, and Torossian (Part I and Part III are actually in English, despite the french title), which you can find e.g. at http://imperium.lenin.ru/~kaledin/math/synthese4.pdf
Deformation quantization: a survey, by Martin Bordemann.<|endoftext|>
TITLE: Finite objects for which isomorphism is NP-hard or harder?
QUESTION [11 upvotes]: Are there finite objects for which deciding isomorphism
is NP-hard or harder?
Graphs and groups are not solutions.
Searching the web didn't return answer for me.

Partial result based on Chow's comment.
From a paper p.16

for IBDDs (Indexed BDDs) ... the equivalence test is coNP-complete even
  if there are only two layers.

Indexed BDD are rooted digraphs, representing boolean function.
"Equivalence" appears very close to isomorphism and means representing the same boolean function.

REPLY [2 votes]: Two candidates for NP-hardness of isomorphism are CIRCUIT ISOMORPHISM
and FORMULA ISOMORPHISM.
For CIRCUIT ISOMORPHISM Joshua Grochow claims.

Circuit Isomorphism is the problem: given two circuits $C_1, C_2$ on $n$ inputs, is there a permutation $\pi \in S_n$ such that $C_1(\vec{x})$ and $C_2(\pi(\vec{x}))$ are equivalent? Circuit Isomorphism has a status similar to that of Graph Isomorphism, but one level higher in $\mathsf{PH}$: it is $\mathsf{coNP}$-hard (by the same reduction above), not known to be in $\mathsf{coNP}$, in $\mathsf{\Sigma_2 P}$, but not $\mathsf{\Sigma_2 P}$-complete unless $\mathsf{PH} = \mathsf{\Sigma_3 P}$

For FORMULA ISOMORPHISM:  Michael Bauland, Edith Hemaspaandra, Isomorphic Implication, Theory Comput Syst (2009) 44: 117–139, doi:10.1007/s00224-007-9038-1, arXiv:cs/0412062, on p.1:

The formula isomorphism problem is in $\Sigma_2^p$ , NP-hard, and unlikely to be $\Sigma_2^p$.-complete

From the abstract:

We study the isomorphic implication problem for Boolean constraints. We show that this is a natural analog of the subgraph isomorphism problem. We prove that, depending on the set of constraints, this problem is in P, NP-complete, or NP-hard, coNP-hard, and in parallel access to NP. We show how to extend the NP-hardness and coNP-hardness to hardness for parallel access to NP for some cases, and conjecture that this can be done in all cases.

These examples appear to contradict Emil's claim that isomorphism of finite object can't be too hard.<|endoftext|>
TITLE: Neron models and ramification
QUESTION [9 upvotes]: I encountered this result while reading a few things, and it was stated without reference. I am having a hard time finding a reference for it (or a simple proof), so maybe you can help me:
let $E$ be an elliptic curve over $\mathbb{Q}$, $\ell$ a prime number, and consider $\rho=\rho_{E,\ell}$ the Galois representation of $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ on the $\ell$-adic Tate module of $E$. Let $p\neq\ell$ be a finite prime number where $\rho$ is ramified. Let $\ell^n$ be the highest power of $\ell$ such that $\rho\mod\ell^n$ is unramified at $p$. 
Then $(\ell)^n=(c_p)$ in $\mathbb{Z}_{\ell}$, where $c_p$ is the Tamagawa factor of $E$ at $p$ (the order of the component group of the special fiber at $p$ of the N\'eron model of $E$ over $\mathbb{Z}$).

REPLY [3 votes]: If $E$ has (split) multiplicative reduction, then you can probably get this by a direct computation on the Tate model. And for the additive reduction cases, one could possibly use explicit models for the various reduction types, a la Tate's algortithm, and do a direct computation. I expect this is fairly easy if $p\ge5$, reasonable if $p=3$, and quite painful (if even possible) for $p=2$. So the SGA reference that Olivier gives is undoubtedly the right approach, but if you want to avoid all the machinery, this might give a way to do it.<|endoftext|>
TITLE: Precipitous ideals and GCH
QUESTION [11 upvotes]: It is well known that ZFC + "There is a measurable cardinal" is equiconsistent with ZFC + "There is a precipitous ideal on $\omega_1$."  Is ZFC + "There is a measurable $\kappa$ such that $2^\kappa > \kappa^+$" equiconsistent with ZFC + "There is a precipitous ideal on $\omega_1$" + CH + $2^{\omega_1}>\omega_2$?

REPLY [6 votes]: Just a measurable cardinal is sufficient.
Assume $\kappa$ is a measurable cardinal and force with $Col(\omega, < \kappa)*\dot{Add}(\kappa, \tau^*)$, where $\tau^* \geq \kappa^{++}$
is suitably defined (see Gitik's paper ``On Generic Elementary Embeddings'').
As it is shown in the above cited paper, there is a nomral precipitous ideal on $\omega_1$ in the resulting generic extension.<|endoftext|>
TITLE: Invariant polynomials under diagonal action of the orthogonal group
QUESTION [5 upvotes]: Consider the diagonal action of the orthogonal group $O(n)$ on $\mathbb{R}^n\times\mathbb{R}^n$ defined as: $U\cdot (x,y) = (Ux,Uy)$ for $U\in O(n)$ and $x,y\in\mathbb{R}^n$. I am looking for a description of the algebra of polynomials in $\mathbb{R}[x,y]$ that are invariant under this action. That is, the subalgebra of all polynomials $p(x,y)$ such that $$p(Ux,Uy) = p(x,y)\, \quad \forall U\in O(n), \forall x, y\in\mathbb{R}^n.$$ 
I'll be interested in the generators of this invariant subalgebra, if possible. I have looked at several sources in Classical Invariant Theory but haven't found what I need yet. Any suggestion or references would be greatly appreciated.

REPLY [10 votes]: This is covered in Chapter 2, Section 9 (starting on page 52) in Weyl's The Classical Groups, Their Invariants and Representations.
As already answered, there you will find:
Theorem (Theorem 2.9.A) Every orthogonal invariant in vectors $x^1,...,x^m$ in $\mathbb{R}^n$ is expressible in terms of the $m^2$ scalar products $(x^i,x^j)$.<|endoftext|>
TITLE: Martin-Solovay Tree of Weakly Homogeneous Tree under $\mathsf{AD}_\mathbb{R}$
QUESTION [5 upvotes]: A tree $T$ on $\omega \times \lambda$ is weakly homogeneous if there is a countable set $\sigma$ of countably complete measures on ${}^{<\omega}\lambda$ so that $x \in p[T]$ if and only if there is a countably complete tower of measures $\bar{\mu} = \langle \mu_i : i \in \omega \rangle$ so that each $T_{x\upharpoonright i} \in \mu_i$ and each $\mu_i \in \sigma$.
Fix such a tree $T$ and a $\sigma$ witnessing weak homogeneity. Fix an enumeration $\langle \mu_i : i \in \omega\rangle$ of $\sigma$ so that projections of $\mu_i$ come before $\mu_i$. 
For an ordinal $\nu$, the Martin-Solovay tree $\mathrm{MS}_\nu(T,\sigma)$ is a tree on $\omega \times \nu$ defined by: $(h,s) \in \mathrm{MS}_\nu(T,\sigma)$ if and only if for all $i < l < |s|$, if $k_i =  \dim(\mu_i)$ and $k_l = \dim(\mu_l)$, $T_{s \upharpoonright k_i} \in \mu_i$, $T_{s \upharpoonright k_l} \in \mu_l$, and $\mu_i$ is a projection of $\mu_l$, then $j_{i,l}(h(i)) > h(l)$, where $j_{i,l}$ is the natural map between the $\mu_i$ ultrapower and the $\mu_l$ ultrapower. 
In Steel's "The Derived Model Theorem" Lemma 1.19, he shows that $p[T] = {}^\omega\omega \setminus p[MS_{\lambda^+}(T,\mu)]$. This is proved by finding a continuous witness to the ill foundedness of all appropriate towers coming from $\sigma$. At least in the Steel's proof provided there, it seems that he is using the axiom of choice. 
So my question is: Under $\mathsf{AD}_\mathbb{R}$, Martin showed that every tree $T$ on $\omega \times \lambda$ where $\lambda < \Theta$ is weakly homogeneous. Fix $T$ and $\sigma$ witnessing weak homogeneity. The Martin-Solovay tree $\mathrm{MS}_{\lambda^+}(T,\sigma)$ can be constructed. Under $\mathsf{AD}_\mathbb{R}$, is $p[T] = {}^\omega\omega \setminus p[MS_{\lambda^+}(T,\sigma)]$ still true?
In Larson's Stationary tower book, he mentioned that it is unknown whether the type of continuous witnesses to illfoundedness of all towers (used in $\mathsf{AC}$ proof) can be proved in $\mathsf{ZF + DC}$. 
In his Games and Scales paper, Steel mentioned that Martin-Solovay trees are studied in the choiceless $\mathsf{AD}$ context. I suspect at least under determinacy the Martin-Solovay tree should project onto the complement of a weakly homogeneous tree. 
Thanks for any information or references to this question.

REPLY [3 votes]: It is a theorem of $\text{ZF+DC}$ that if $T$ is a weakly homogeneous tree on $\omega\times \kappa$ some $\kappa$ with homogeneity system of measures $\vec{\mu}$ and $ms(T,\vec{\mu})$ is the Martin-Solovay tree associated to $T$ then $p[T]$ and $p[ms(T,\vec{\mu})]$ are complements. See Cabal reprints volume 1 Jackson's article theorem 4.10 for a proof. Note that to show weak homogeneity of trees on $\omega\times \kappa$, where $\kappa$ less than the supremum of the Suslin cardinals, one actually only needs $\text{AD}$ (result of Woodin). 
$\text{AD}$ or $\text{AD}_{\mathbb{R}}$ are used to show homogeneity or weak homogeneity of trees. This is done using partition properties in the determinacy context (see Jackson's article mentionned above). These assumptions are not needed to establish that a weakly homogeneous tree $T$ and its Martin-Solovay twin project to complements (if that's what you were looking for).<|endoftext|>
TITLE: Classification of fake (quaternionic, octonionic) projective spaces
QUESTION [23 upvotes]: If $X$ is a closed $n$-manifold, a fake $X$ is another closed manifold homotopy equivalent to $X$. There is some interest in classifying manifolds (up to, say, homeomorphism) homotopy equivalent to a given manifold; the Poincare conjecture is the special case of $X = S^n$ (there are no topologically fake $S^n$s). I'm interested in this in the case of $X = \Bbb{KP}^n$, where $\Bbb K = \Bbb {R,C,H,O}$ (where in the $\Bbb O$ case we fix $n=2$; there are no other objects which deserve to be named "$\Bbb{OP}^n$".) One reason to find this interesting is that fake projective spaces are precisely the manifolds that have a sphere bundle over them with total space a sphere, with as usual the unfortunate exception that this is not true for $\Bbb{OP}^2$. (I would note that this is different than the algebro-geometric usage of "fake projective plane", where to my understanding what was sought was a classification of compact complex manifolds with the same Hodge diamond as complex projective space.)
Both real and complex fake projective spaces have been classified: both computations are carried out in Wall's book on surgery theory (see sections 14C and 14D for the classifications of fake complex and real projective spaces, respectively), and there are nice, briefer descriptions of the real and complex cases on the Manifold Atlas.
I've had some difficulty in finding a classification of fake quaternionic projective spaces or fake octonionic projective planes. Has this been carried out, and if so, what is a reference?

REPLY [4 votes]: I would like to add some more references concerning the (smooth) fake quaternionic projective spaces (FQPS). There is a beautiful paper of Hsiang:
Hsiang, Wu-chung A note on free differentiable actions of S1 and S3 on homotopy spheres. Ann. of Math. (2) 83 1966 266–272
where he proves that there are infinitely many FQPS which can be distinguished by their rational Pontryagin classes. Moreover he shows that every FQPS can be obtained by a smooth and free action of $S^3$ on an exotic sphere.
Finally there is a recent preprint where the authors compute some interesting groups concerning smooth structures on $\mathbb H\mathbb P^ n$ for low $n$.
But in general I believe there is not much known and I think it is interesting subject to think about.<|endoftext|>
TITLE: Can a semigroup with zero be globally isomorphic to a semigroup without zero?
QUESTION [10 upvotes]: This is not a great question for sure and it may even be trivial for all I know, but a couple of years ago, when I still thought I'd be a mathematician, I spent quite a lot of time thinking about it and my advisor wasn't able to help so I'd like to ask it here, even if only for the feeling that my effort wasn't completely wasted if anybody reads it. Hopefully that's OK.
Two semigroups are called globally isomorphic if their power semigroups are isomorphic. The power semigroup $P(S)$ of a semigroup $S$ is the semigroup of all non-empty subsets of $S$ with the natural multiplication.
Mogiljanskaja gave examples of semigroups which are globally isomorphic but not isomorphic, and more examples have been produced since. All of those examples, at least the ones I've seen, are semigroups with zero which differ on their annihilators, say, $\mathrm{Ann}(S) = \{s\in S\,|\,(\forall r\in S)\, rs=0\}$ or ones built from such semigroups. In general, there doesn't seem to be a great variety of examples. Mogiljanskaja asked what other kinds of examples could be found, and the general question seems to be: "what of the semigroup structure does a power isomorphism have to preserve?"
The first question that came to my head was if it was possible to have a semigroup with zero globally isomorphic to a semigroup without zero. Obviously, the power semigroup of a semigroup with zero has a zero as well, so we would have to have a semigroup without zero whose power semigroup has a zero. It's easy to construct such semigroups, and they have been studied. A semigroup whose power semigroup has a zero is called a homogroup and there's a paper by Clifford and Miller from 1948 about them, giving a general construction: Semigroups Having Zeroid Elements, A. H. Clifford and D. D. Miller, American Journal of Mathematics, Vol. 70, No. 1 (Jan., 1948), pp. 117-125. (JSTOR)
They are exactly the semigroups which have a two-sided ideal that's a group. (I know that thanks to this question.) That has to be the least ideal of the semigroup and so its kernel. Any semigroup with zero is a homogroup and $\{0\}$ is that ideal in those.
So the question is whether a semigroup without zero can be globally isomorphic to a homogroup with a bigger kernel. And another question that immediately comes to mind is whether, simply, two globally isomorphic homogroups can have non-isomorphic kernels. 
So these are the questions I'd like to ask. If there are such pairs, I think that would be genuinely interesting. Probably there aren't though and the question is boring, but I wasn't able to prove it.

REPLY [2 votes]: This is a little too long for a comment. 
Here is an idea to prove the minimal ideal can be recovered. Let $G$ be the minimal ideal of $S$ and let $e$ be its identity; it is a central idempotent.  Then $E=\{e\}$ is a central idempotent of $P(S)$ with $EP(S)$ isomorphic to $P(G)$. Since groups are globally determined you can recover $G$ (as the group of units of $P(G)$). So it remains to show the idempotent $E$ can be recovered from the semigroup structure of $P(S)$. My feeling is that it is the unique maximal central idempotent $F$ of $P(S)$ in the usual order on idempotents such that $FP(S)$ is isomorphic to the power set of a group.<|endoftext|>
TITLE: Chern classes and singular hermitian metrics on vector bundles
QUESTION [7 upvotes]: Let $L$ be a holomorphic line bundle on a complex manifold $X$, and assume it is equipped with a singular hermitian metric $h$ with local weight $\varphi$. Then, one can show that the de Rham class of $\frac{i}{\pi}\partial \overline{\partial} \varphi$ coincides with the first Chern class $c_1(L)$ of the line bundle.
Is there a generalization of this result to higher-rank vector bundles? 
For example, let $E$ be a holomorphic vector bundle of rank $r$ on a complex manifold $X$ and let $h$ be a singular hermitian metric on $E$. Can we describe the Chern clases $c_1(E),\ldots,c_r(E)$ in terms of $h$ and its curvature form?

REPLY [5 votes]: As Hassan mentions, in the setting of singular metrics on vector bundles, the notion of curvature appears problematic, which is discussed in the paper of Raufi. 
Still, as is also discussed in that paper, in the case that the singular metric is positively or negatively curved in the sense of Griffiths, one can give a reasonable meaning to $c_1(E,h)$ as a current by $c_1(E,h) := dd^c \log (\det h)$. This gives a way of defining the first Chern class without having defined a curvature matrix, but which coincides with the usual definition when $h$ is smooth.
Me, Raufi, Ruppenthal and Sera do a variant of this also for higher Chern classes, through local regularizations of the metric, under some additional assumption that the singularities of the metric are not too "big". This is so far not completely written up, but a discussion of it can be seen at 
https://www.birs.ca/events/2016/5-day-workshops/16w5080/videos/watch/201605020920-Larkang.html
The precise statement of our main result is discussed at around 37:40, and basically, in the positively curved case, we can define $c_k(E,h)$ if the set $L(\log (\det h^*))$ is contained in a variety of codimension $\geq k$, where $L(\log(\det h^*))$ is the complement of the set where $\log(\det h^*)$ is locally bounded.
By taking the "diagonal" metric $h = e^{-\varphi} Id$ on a trivial vector bundle, where $\varphi$ is a plurisubharmonic function, one sees that one should expect to need some condition on the singularities of $h$, since $(dd^c \varphi)^k$ does typically not have a natural meaning when $\varphi$ is plurisubharmonic and unbounded, and $k$ is too large.<|endoftext|>
TITLE: Combinatorics: set partitions of a poset
QUESTION [5 upvotes]: Let $\pi_n$ be the poset of all set partitions of $\{1,...,n\}$ ordered by refinement, $\sigma = \{B_1,...,B_k\}$ be a set partition with blocks $B_i$, and $\max(B_i)$ be the maximum value in the block $B_i$. I'm trying to prove the following:
$$\Sigma_{\sigma\in\pi_{n-1}}\Pi_{B_i\in\sigma}(-1)^{|B_i|-1}(|B_i|-1)!(n+\max(B_i))=(2n-1)!!$$
I've already coded this in Mathematica to check that it's true until n=11 (after which the numbers become too large):
<<Combinatorica`

f[x_]:=(-1)^(x-1)*(x-1) !

Total[Times @@@ Map[f, Map[Length, SetPartitions[n-1], {2}], {2}]*Times @@@ (Map[Max, SetPartitions[n-1], {2}] + n) ] == Factorial2[2*n-1]

Any thoughts would be greatly appreciated.

REPLY [6 votes]: Let us prove even more general formula that Douglas Zare's one from the comments. Let $a_1>\dots>a_n$ be real numbers. Then
$$
  \sum_{\sigma\in\pi_{n}}
    \prod_{B_i\in\sigma}(-1)^{|B_i|-1}(|B_i|-1)!a_{\min B_i}
  =\prod_{i=1}^n(a_i-i+1).
$$
(The required formula follows by setting $a_i=2(n+1)-i$ and shifting $n$ by $1$.)
To see this, rewrite the right-hand side as
$$
  \prod_{i=1}^n\left(a_i-\sum_{1\leq j<i}u_{i,j}\right),
$$
where $u_{i,j}=1$ for $j<i$, and expand all the brackets. We get the sum of the terms each having the form
$$
  (-1)^{n-|I|}\prod_{i\in I}a_i\prod_{i\notin I}u_{i,j(i)},
$$
where $I$ is a subset of $\{1,\dots,n\}$ containing $1$. To each such term, put into correspondence a partition into $|I|$ parts, where $a_i$ ($i\in I$) are the maximal elements of the parts, and each $u_{i,j}$ is interpreted as `$a_i$ and $a_j$ lie in the same set of partition'. 
This way, each partition $(B_i)$ will correspond to exactly $\prod_i (|B_i|-1)!$ terms, all having the sign we need. Thus, after substituting $u_{ij}=1$ we obtain the required formula.<|endoftext|>
TITLE: Is it usual for a referee to heed updated versions on arxiv?
QUESTION [32 upvotes]: I've put a paper on arxiv one year ago and I've submitted the version 6 to a journal seven months ago.  
During these last seven months, I've given several talks about this work, which led me to update again the paper on arxiv, to fix some typos, polish, add clarification, add a few new applications and questions... nothing very serious, but the last arxiv version (of 51 pages) is surely better than the submitted one (of 47 pages). 
Out of respect for the work of the referee and to not delay the report, I believe (perhaps wrongly) that it is better to not update the submission before the first report; but I was wondering:
Question:  Is it usual for a referee to heed updated versions on arxiv?

REPLY [12 votes]: I am hesitant to try to add anything further to the solid answers here, but here it goes anyway.
In practice, here is what I do:

Once I have a manuscript that I feel is publisher-ready, I send it out to experts that I feel might be interested.  After a week or two I re-read, make updates, and incorporate any feedback I received.
Then I post the manuscript on the arXiv.  I again wait a few weeks to see if any feedback comes.  After that period I make changes as appropriate and give it another read making edits as they arise.
Then I submit to a journal.
Generally speaking I do not update the arXiv between submission and a referee report.  In that time I always find typos and minor edits.  Sometimes new content wants in too.  To address this I make said changes and keep a log, commented out in TeX, at the top of my file, itemizing all changes that I have made.  After the referee report, I address the referee suggestions and if rejected update the arXiv and resubmit.  If accepted, I use the log I created to walk the referee through the changes (to make his/her job easier), and then update the arXiv.

This process has worked for me so far. There have been an occasion or two where I have broke protocol (exceptions always arise), but this is my attempt at a "best practice" from the author's point of view.
As a referee, I only consider the submitted manuscript.  If I were to receive an update in the middle, generally speaking, I would consider the "clock" reset and start over from scratch.  That said, I will sometimes compare the arXiv version to the submitted version.  But I evaluate exclusively on the submitted version.

REPLY [4 votes]: As a referee, I usually stick to the submitted version, which is what I am asked to evaluate.
Often, it happens to encounter an error, an imprecision, or an unclear passage. Sometimes, it is not immediately clear whether it is just a typo, or something the authors really did not consider. In these cases, I usually check for a more recent version on the arXiv, or on the author's webpage, to see whether that part has been somehow improved or corrected.
So, updating papers on the arXiv even after submission could be beneficial to both authors and referees, by saving some time. 
Indeed the possibility to improve the manuscript after the submission is not an excuse to send out poorly revised manuscripts for evaluation.
To comment Friedrich Knop's answer, in my opinion it is a very good (but not widespread) practice to update the arXiv version to fix or improve parts of the manuscript - even after publication. Most of the times, however, the arXiv version is not even updated to reflect the (sometimes substantial) modifications suggested by the referees.<|endoftext|>
TITLE: Combinatorics problem about sum of natural numbers
QUESTION [17 upvotes]: Following combinatorics problem is claimed to be an open problem in "The Princeton Companion to Mathematics" (pp. 6)

Let $a_1,a_2,a_3,...$ be a sequence of positive integers, and suppose that each $a_n$ lies between $n^2$ and $(n+ 1)^2$. Will there always be a positive integer that can be written in a thousand different ways as a sum of two numbers from the sequence?

I am curious to know current status and references for this problem.

REPLY [11 votes]: If the sequence $a_1, a_2,\dots$ has the property that each integer can be written in at most $g$ ways (counting order and allowing repetition), then we call the set $\{a_1,a_2,\ldots\}$ a $B^\ast[g]$ set. I wrote an extensive bibliography about 10 years ago, and published it in the Electronic Journal of Combinatorics section on hot bibliographies. Unfortunately, I've never gotten around to updating it. For some reason, Math Sci-Net never cataloged it, either.
A related and more famous question, asked by Erdos and Turan, is if it is possible for a $B^\ast[g]$ set of nonnegative integers to also be a basis, that is, is it possible for every positive integer to have a representation while no positive integer has more than $g$ representations? The answer is known to be "no" for small $g$; the proofs are essentially a depth first search through $B^\ast[g]$ sets.
To answer the Shahrooz Janbaz's question, there is nothing too special about 1000 except that it removes "just compute all possibilities" as a possible strategy. That is, if "1000" is replaced by 4 or 5 or 6, then we could perhaps just search through all possibilities. Perhaps a clever program (SAT solver?) could handle 10 or 20. But 1000 clearly requires a deeper understanding of addition.<|endoftext|>
TITLE: Abstract Wave Equation and Semigroups
QUESTION [5 upvotes]: If an operator $A$ on a Hilbert space $H$ generates a strongly continuous semigroup, does then the operator $B$ on $H \oplus H$ given by the matrix
$$ B := \begin{pmatrix} 0 & \mathrm{id} \\ A & 0\end{pmatrix}$$
generate a semigroup as well? This would then yield a solution to the wave equation $u^{\prime\prime} = A u$ on $H\oplus H$. 
If this is not generally true, I would be very interested in a counterexample. Also, if this is not generally true, what properties of $A$ are needed in order for this to be true?

REPLY [4 votes]: You need two additional assumptions: the operator $A$ has to be a so-called cosine function generator, and your product space has to be $V\times H$ with a space $V\subset H$. 
Cosine function generator is more than sectorial, it is more or less when the numerical range of $A$ is in a parabola (selfadjoint negative definite is ok, this is discussed in the references Liviu mentioned).
The space $V$ is difficult, but if $A$ is selfadjoint, then it is essentially $D(A^{1/2})$.
See Section VI.3 in

Klaus-Jochen Engel and
  Rainer Nagel, MR 1721989 One-parameter semigroups for linear
  evolution
  equations, ISBN:
  0-387-98463-1. (here a downloadable version)

or Section 7.4 in

Haase, Markus The functional calculus for sectorial operators.
  Operator Theory: Advances and Applications, 169. Birkhäuser Verlag,
  Basel, 2006. xiv+392 pp. ISBN: 978-3-7643-7697-0; 3-7643-7697-X<|endoftext|>
TITLE: A sufficient condition (or not) for positive semidefiniteness of a matrix?
QUESTION [6 upvotes]: Let $A\in M_{n}(\mathbb{C})$ be a Hermitian matrix. If for all $z_1,...,z_n\in\mathbb{C}$, $$\sum_{i,j=1}^{n}A_{ij}z_{i}\overline{z_{j}}\ge 0$$ then A is positive semidefinite. 
I do not think the counterexample to the original question below will answer the following version of the question motivated by the free nonabelian group algebra analogue of of Helton's Theorem originally due to Schmüdgen. I ask the question here in the hope of conclusively recording the state of this question in its most extreme form:

Question: If for all $m\in \mathbb{N}$ and $U_1,...,U_{n}\in M_{m}(\mathbb{C})$ unitary matrices, $$\sum_{i,j=1}^{n}A_{ij}U_{i}U_{j}^{*}$$ is positive semidefinite, then is A positive semidefinite plus a trace zero diagonal matrix?

An affirmative answer to the above question is equivalent to the claim that for every positive semidefinite $n\times n$ matrix $C$ with constant diagonal, there exists $m\in \mathbb{N}$ and $U_{1},...,U_{n}$ unitaries in $M_{m}(\mathbb{C})$ and a matrix $B\in M_{m}(\mathbb{C})$ such that for all $i,j\in\{1,...,n\}$, $$C_{ij}=\frac{1}{m}Tr((U_iB)^{*}(U_{j}B)).$$
This equivalence comes from a result we recently proved (in a paper to appear in Operators and Matrices) that if $\sum_{ij}A_{ij}C_{ij}$ is positive semidefinite for all correlation matrices $C$, then $A$ is positive semidefinite plus trace zero diagonal.
This claim is true for matrices with real entries by our result above together with a nice result of Dykema and Juschenko asserting that every real correlation matrix $C$ can be written in the above way with $B$ the identity matrix. So a counterexample will require $A$ to have complex entries.

The original question asked here, which is trivially false by the counterexample below provided by Noam Elkies, was
Question: If for all $z_1,...,z_{n}\in S^{1}$, $$\sum_{i,j=1}^{n}A_{ij}z_{i}\overline{z_{j}}\ge 0$$ then is A positive semidefinite?
The same counterexample, combined with simultaneous diagonalization answers the following question negatively:
Question: If for all $m\in \mathbb{N}$ and $U_1,...,U_{n}\in M_{m}(\mathbb{C})$ pairwise commuting unitary matrices, $$\sum_{i,j=1}^{n}A_{ij}U_{i}U_{j}^{*}$$ is positive semidefinite, then is A positive semidefinite?

REPLY [6 votes]: No (except of course for $n=1$).  A counterexample for $n=2$ is 
$A = {\rm diag}(2,-1)$.  The corresponding form $2|z_1|^2 - |z_2|^2$ 
is clearly indefinite, but takes the positive value $1$ whenever 
$|z_1|=|z_2|=1$.  This readily generalizes to all $n \geq 2$.<|endoftext|>
TITLE: Expressive power of $\omega$-order logic
QUESTION [5 upvotes]: According to the article Second-order and Higher-order Logic
 from the Stanford Encyclopedia of Philosophy,

there is no need to stop at second-order logic; one can keep going. [...] we can allow quantification over super-predicate symbols. And then we can keep going further.
We reach the level of type theory after ω steps.

I wonder what the expressive power of "$\omega$-order logic" is:

Can you give an example of two structures $\mathcal A$, $\mathcal B$ that satisfy the same $\omega$-order sentences but are not isomorphic?

REPLY [8 votes]: Instead of an example, I give an existence proof:   
Take any finite or countable language, for example the language of equality.  Since all formulas (even in $\omega$-logic) are finite, there are only countably many formulas, hence at most $\mathfrak c:= 2^{\aleph_0}$  many theories.  Find more than continuum many cardinalities (for example $\{\aleph_\alpha:\alpha < \mathfrak c^+\}$), and for each such cardinality $\kappa$ find a structure whose size is $\kappa$. These structures are pairwise non-isomorphic, but there must be two that satisfy the same set of $\omega$-sentences.
(For languages with $\lambda$ many symbols, replace $\mathfrak c^+$ by $(2^\lambda)^+$.)<|endoftext|>
TITLE: A combinatorial identity involving harmonic numbers
QUESTION [15 upvotes]: The harmonic numbers are given by $$H_n=\sum_{k=1}^n\frac{1}{k}.$$
Numerical calculation suggests
$$
\sum_{k=1}^{n}(-1)^k{n\choose k}{n+k\choose k}\sum_{i=1}^{k}\frac{1}{n+i}=(-1)^nH_n.
$$
I can not give a proof of this identity. How to prove it? 
Hints, references or proof are all welcome.

REPLY [12 votes]: First we prove the formula
$$\sum_{k=0}^n (-1)^k\binom{n}{k}\binom {x+k}{k} = (-1)^n\binom xn,\tag{1}$$
which is special case of Vandermonde's theorem:
$$\begin{aligned}
\sum_{k=0}^n (-1)^k\binom{n}{k}\binom {x+k}{k} 
  &= \sum_{k=0}^n \binom n{n-k} \binom{-x-1}{k}\\
  &= \binom{n-x-1}{n} = (-1)^n\binom xn.
\end{aligned}$$
Since $(1)$ is an identity of polynomials in $x$, we may differentiate it with respect to $x$, obtaining 
$$\sum_{k=0}^n (-1)^k\binom{n}{k}\binom {x+k}{k}\sum_{i=1}^k\frac{1}{x+i}
  =(-1)^n\binom xn \sum_{i=0}^{n-1}\frac{1}{x-i}.
$$
The OP's identity follows on setting $x=n$.
The same method can be used to prove many identities involving harmonic numbers.<|endoftext|>
TITLE: Problems and Algorithms Requiring Non-Bipartite Matching
QUESTION [5 upvotes]: While the importance of the non-bipartite matching problem itself from an algorithmic and complexity point of view is well known, applications of non-bipartite matching are hard to find.
I did an online search for hints, but almost always the articles I found lacked problems that demonstrated the need for non-bipartite matching. The few exceptions I found were:  

A scheduling problem [Fujii, Kasami & Ninomiya, 1969] described in Gerard's matching survey 
The oil well drilling problem of [Devine 1973] described in Gerard's matching survey 
The Christofides heuristic for the TSP [Christofides 1976] Wikipedia 
Plotting street maps with minimum pen lifting [Iri&Taguchi 1980] described in Gerard's matching survey 
Kekule structures in chemistry described here 
Nonparametric Tests for Homogeneity [Rosenbaum 2005] described here 
TSP: unraveling of tours with a high number of pairs of crossing edges: 
Generate a graph $\Gamma$ in which the edges correspond to pairs of crossing tour edges and their weight equals the amount by which the tour length decreases when replacing the respective tour edges by a pair of non-crossing ones.
The maximum weight matching in $\Gamma$ identifies the pairs of tour-edges whose exchanging incurs the maximal length reduction.
That matching may have to be applied several times until all crossings have been removed.  
Optimizing Triangulations: take as nodes the triangles and as edges the sides that are shared by two triangles; in that settig a plethora of cost functions can be envisaged for different optimality criteria.
There are two basic operations that can be performed on triangulations, namely 

diagonal-swapping for the purpose of improving a triangulation:   


minimum weight matching with length reduction incurred by swapping diagonals as edge cost when striving for minimum weight triangulations
maximum weight matching with the increment of the geometric angle between edge-adjacent triangles when striving for "smooth" triangulations like e.g. of terrain data  
maximum weight matching with the most acute geometric angle of a triangle as edge cost when striving for triangulations of e.g. terrain data, that are in the spirit of planar Delaunay triangulations.

merging edge-adjacent triangles for the purpose of generating a quadrilateralization:


a minimum weight matching with the circumference of the generated quadrilateral as edge-cost when striving for "fine grained" quadrilateralizations
a maximum weight matching over the most acute geometric angle of a generated quadrilateral when striving for the most orthogonal  quadrilateralization e.g. for use as finite elements or, as patches for spline interpolation of 3D data.
a minimum weight matching with the spatial distance between a quadrilateral's diagonals when striving for quadrilateralizations with planar cells.   



Question
are there further applications for non-bipartite matching, especially newer, examples of different areas of applicability?

REPLY [2 votes]: See "Optimal Nonbipartite Matching and Its Statistical Applications" by Lu et al. in The American Statistician Feb 2011.<|endoftext|>
TITLE: Density of the linear span of products of harmonic polymomials
QUESTION [10 upvotes]: Let $\mathcal{H}$ denote the space of all harmonic polynomials with complex coefficients in $n$ variables $x_1,\ldots, x_n$ in $\mathbb{R}^n$. I'm trying to show that the linear span of the set $\mathcal{M}=\{p\, q: p,q\in\mathcal{H}\}$ is dense in $C(K)$ under supremum norm, where $K$ is a compact set in $\mathbb{R}^n$, for $n\geq 2$. One may assume that $K$ is the closed unit ball in $\mathbb{R}^n$ since once it is proved for the unit ball, the result follows for any compact $K$. This seems to be classical and should be known but I have not found a reference.
When $n$ is even, it is not hard to prove. In fact, if $n=2m$, one may identify $\mathbb{R}^n$ with $\mathbb{C}^{m}$ and use the fact that the linear span of $\mathcal{N}=\{f\,\overline{g}: f, g \text{ are holomorphic polynomials}\}$ is dense in $C(K)$ by Stone-Weierstrass Theorem. Note that for any holomorhpic polynomials $f, g$ in $\mathbb{C}^m$, the polynomials $p=f$ and $q= \overline{g}$ are harmonic in $\mathbb{R}^n$. Therefore, $\mathcal{N}\subset\mathcal{M}$ and hence, $\mathcal{M}$ is dense in $C(K)$.
I have not been able to resolve the case $n$ is odd. In general, $\mathcal{M}$ is not a subalgebra of $C(K)$ even though it is self-adjoint, separates points and vanishes at no point so Stone-Weierstrass Theorem cannot apply directly to $\mathcal{M}$ (even when $n$ is even).
I would highly appreciate it if someone can point out a reference or offer a proof.

REPLY [3 votes]: Induction on $n$. Base $n=2$ is clear, as you said. Let $\nu$ be a non-trivial finite compactly supported (complex) measure in $\mathbb R^n$ orthogonal to any product $uv$. Then we can smear it a bit and get a non-trivial continuous compactly supported function $f$ orthogonal to each product. Now choose any $n-1$ dimensional space and consider the harmonic functions that do not depend on the orthogonal variable. Applying the induction assumption, we see that the integral of $f$ along every line must be $0$ (otherwise we can project $f$ to get something non-trivial in $\mathbb R^{n-1}$). This is enough to conclude that the Fourier transform of $f$ is identically $0$, so $f$ must be $0$.<|endoftext|>
TITLE: Spectrum Problem for Higher-Order Logic
QUESTION [5 upvotes]: Definitions.  Given a sentence $\varphi$ of $n$th-order logic, we define the spectrum of $\varphi$ to be the set of cardinalities of finite structures that satisfy $\varphi$. A set $X\subseteq\mathbb N$ is said to be an $n$th-order spectrum if there is an $n$th-order sentence whose spectrum is $X$. Given any positive integer $n$, let $S_n$ be the set of all $n$th-order spectra. An $\omega$-order spectrum is an element of the set $S_{\omega}:=\bigcup_{n\in\mathbb Z_+}S_n$.
Questions.

Clearly, $S_n\subseteq S_{n+1}$ for each positive integer $n$. One knows that $S_1\subsetneq S_2$ (J. H. Bennett, On spectra, 1962).

Do we have $S_n\subsetneq S_{n+1}$ for each positive integer $n$?


One does not know if the complement of a first-order spectrum is also a first-order spectrum. That is why I wonder:

Is the complement of a second-order spectrum always a second-order spectrum? Is the complement of a third-order spectrum always a third-order spectrum? ...


For each positive integer $n$, every element of $S_n$ is decidable. But we can always construct a decidable set $D_n$ which is not in $S_n$: We simply construct the diagonal set $D_n$ such that $m \in D_n$ if and only if $m$ is not in the $m$th $n$th-order spectrum. (source, page 2)

Is $D_n$ always an element of $S_{n+1}$? Is $D_n$ always an element of some $S_p$, where $p>n$?




Is the complement of an $\omega$-order spectrum always an $\omega$-order spectrum?

REPLY [4 votes]: One can answer questions about higher-order spectra by characterizing $S_d$ as certain complexity classes, based on similar characterizations for classes of structures definable by fragments of higher-order logic.
Let me first recall how things work for the base case (first-order logic, i.e., $S_1$). Let $\phi$ be an FO sentence in a finite relational signature $\sigma=\{R_1,\dots,R_k\}$. If we reconsider each $R_i$ as a second-order variable, we can quantify them away in second-order logic to get the $\Sigma^1_1$ sentence
$$\Phi=\exists R_1\dots\exists R_k\,\phi(R_1,\dots,R_k)$$
in the empty signature with the property
$$\exists M\,(|M|=n\land M\vDash\phi)\iff M_n\vDash\Phi,$$
where $M_n$ denotes the $n$-element model with no extra structure. Conversely, every $\Sigma^1_1$ sentence in the empty language is of the form $\Phi$ for some first-order $\phi$. Thus, FO spectra are exactly sets of the form
$$\{n\in\mathbb N:M_n\vDash\Phi\}$$
for $\Sigma^1_1$ sentences $\Phi$ in the empty signature.
Now, Fagin's theorem states

Theorem: Let $\sigma$ be a finite signature. A class of finite $\sigma$-structures is definable by a $\Sigma^1_1$ sentence iff it is recognizable in NP.

Applying this to empty $\sigma$, we see that FO spectra are sets of the form
$$\{n\in\mathbb N:M_n\in L\},\qquad L\in\mathrm{NP}.$$
Observe that $M_n$ is more-or-less just the unary encoding of $n$ itself. Since conversion from the usual binary representation to unary blows up the input exponentially, it is easy to see that this implies
$$S_1=\mathrm{NE}:=\mathrm{NTime}(2^{O(n)}).$$
We can generalize this to higher-order logic, but first we have to decide what exactly is higher-order logic. There are at least two important choices to be made:

What things can appear in the signature in $d$-th order logic? In FO, we have just predicates $R(x_1,\dots,x_k)$ where $x_i$ are first-order entities. Such predicates are effectively the same as free second-order variables. There are two obvious ways how to generalize this to $d$-th order logic:
a) The signature can contain predicates $R(X_1,\dots,X_k)$, where $X_i$ are variables of any type allowed in the logic; that is, order $d$ or lower. This makes $R$ effectively a free variable of order at most $d+1$.
I will denote spectra for this form of $d$-th order logic $S_d^+$.
b) The signature can only contain predicates $R(x_1,\dots,x_k)$, where $x_i$ are first-order. That is, it is just an FO signature.
I will denote spectra for this form of $n$th order logic $S_d^-$.
Do the higer-order variables have arbitrary arity, or only unary (monadic)? I will consider the former as default ($S_d^+$, $S_d^-$), and denote the spectra for the latter as $MS_d^+$, $MS_d^-$.

Second, we need an analogue of Fagin’s theorem. The following natural generalization is given in Kołodziejczyk [1]. Let me define the iterated exponential function $2_0^n=n$, $2_{d+1}^n=2^{2_d^n}$.

Theorem: Let $\sigma$ be a finite FO signature, $d\ge2$, and $m,k\ge1$. A set of finite $\sigma$-structures is definable by a $\Sigma^d_m$ sentence (that’s $(d+1)$-th order!) using only higher-order variables of arity $\le k$ if and only if it is recognizable in $\Sigma_m\text{-Time}(2_{d-1}^{O(n^k)})$.

Here, $\Sigma_m\text{-Time}(f(n))$ is the class of languages recognizable by an alternating Turing machine in time $f(n)$ with $m-1$ alternations, starting in an existential state.
The case of $\Sigma^1_m$ is more tricky, but as long as we do not restrict the arity, we still have that $\Sigma^1_m$-definable = recognizable in $\Sigma_m\text{-Time}(2_0^{n^{O(1)}})=\Sigma_m^P$.
By the same argument as when going from Fagin’s theorem to characterization of $S_1$, we obtain:

Theorem: Let $d\ge2$.

$S_d^+=\mathrm{NTime}(2_d^{O(n)})$
$S_d^-=\mathrm{AltTime}(O(1),2_{d-1}^{O(n)})$
$\mathrm{NTime}(2_d^{n+O(1)})\subseteq MS_d^+\subseteq S_d^+=\mathrm{NTime}(2_d^{O(n)})$
$MS_d^-=\mathrm{AltTime}(O(1),2_{d-1}^{n+O(1)})$ for $d\ge3$; $\mathrm{NE}\cup\mathrm{AltTime}(O(1),O(2^n))\subseteq MS_2^-\subseteq S_2^-=\mathrm{EH}$
$S_\omega=\mathrm{ELEMENTARY}$, where $S_\omega:=\bigcup_dS_d^+=\bigcup_dS_d^-=\bigcup_dMS_d^+=\bigcup_dMS_d^-$.


Here, $\mathrm{AltTime}(f(n),g(n))$ denotes languages recognizable by an alternating TM with $f(n)$ alternations in time $g(n)$; in particular, $\mathrm{AltTime}(O(1),g(n))=\bigcup_m\Sigma_m\text{-Time}(g(n))$.
An exact characterization of the $MS$ classes gets messy because of the fact that the extra existential quantifiers we get by quantifying away the nonlogical symbols in the sentence do not have bounded arity unlike the rest. However, the above is sufficient for our purposes.
Now, we can answer the questions in turn.
Q1: Using standard variants of the nondeterministic time-hierarchy theorem, we have
$$\begin{gather*}
S_1\subsetneq S_2^+\subsetneq S_3^+\subsetneq S_4^+\subsetneq\cdots\\
S_1\subsetneq MS_2^+\subsetneq MS_3^+\subsetneq MS_4^+\subsetneq\cdots\\
S_1\subseteq S_2^-\subsetneq S_3^-\subsetneq S_4^-\subsetneq\cdots\\
S_1\subseteq MS_2^-\subsetneq MS_3^-\subsetneq MS_4^-\subsetneq\cdots
\end{gather*}$$
The inclusions $S_1\subseteq MS_2^-$ and $S_1\subseteq S_2^-$ are also conjecturally strict. Despite the imprecision in our characterization of $MS_2^-$, the closure properties of $\mathrm{NE}$ ensure
$$S_1=MS_2^-\iff S_1=S_2^-\iff \mathrm{NE=EH}.$$
It is generally assumed that $\mathrm{NE\ne EH}$, but as this implies $\mathrm{NP\ne coNP}$, it remains a difficult open problem.
Q2: The classes $S_d^-$ for $d\ge2$, and $MS_d^-$ for $d\ge 3$ are closed under complement. Under unproven complexity-theoretic assumptions, $S_1$, $S_d^+$, and $MS_d^+$ are not closed under complement, but this is at least as hard to prove as $\mathrm{NP\ne coNP}$. The remaining case is $MS_2^-$; based on the asymmetry in its characterization, I’d think this might also not be closed under complement, but I don’t have a generally accepted complexity assumption to support it.
Q3: This depends wildly on how you define $D_n$, however, diagonalization using a natural and reasonably efficient enumeration of sentences should indeed give $D_n\subseteq S_{n+1}$, with appropriate superscripts and modifiers.
Q4: Yes.
Reference:
[1] Leszek A. Kołodziejczyk, Truth definitions in finite models, Journal of Symbolic Logic 69 (2004), no. 1, pp. 183–200. jstor, preprint